PROF. K. RAMAMURTHI
PROF. S. VARUNKUMAR
Aerospace Engineering
Indian Institute of Technology Madras
INDEX
S. No
Topic
Page No.
Week 1
1
2
3
Introduction
Motion in Space
Rotational Frame of Reference and Orbital Velocities
01
25
47
Week 2
4
5
6
Velocity Requirements
Theory of Rocket Propulsion
Rocket Equation and Staging of Rockets
74
99
126
Week 3
7
8
9
Review of Rocket Principles: Propulsion Efficiency
Examples Illustrating Theory of Rocket Propulsion
and Introduction to Nozzles
Theory of Nozzles
158
190
209
Week 4
10
11
12
Nozzle Shape
Area Ratio of Nozzles: Under Expansion and Over
Expansion
Characteristic Velocity and Thrust Coefficient
232
258
286
Week 5
13
14
15
Divergence Loss in Conical Nozzles and the Bell Nozzles
Unconventional Nozzles and Problems in Nozzles
Criterion for Choice of Chemical Propellants
16
17
18
Choice of Fuel-Rich Propellants
Performance Prediction and Analysis
Factors Influencing Choice of Chemical Propellan
Week 6
312
335
363
388
412
435
Week 7
19
20
21
Low energy liquid propellants and hybrid propellants
Introduction to Solid Propellant Rockets
Burn Rate of Solid Propellants and Equilibrium Pressure
in Solid Propellant Rockets
462
488
511
Week 8
22
23
24
Design Aspects of Solid Propellant Rockets
Burning Surface Area of Solid Propellant Grains
Ignition of Solid Propellant Rockets
Week 9
537
563
587
25
26
27
Review of Solid Propellant Rockets
Feed Systems for Liquid Propellant Rockets
Feed System Cycles for Pump Fed Liquid Propellant
Rockets
618
648
684
Week 10
28
29
30
Analysis of Gas Generator and Staged Combustion Cycles
and introduction to injectors
Injectors, Cooling of Chambers and Mixture Ratio
Distribution
Efficiencies due to Mixture Ratio Distribution and
Incomplete Vaporization
708
745
790
Week 11
31
32
33
Pumps and Turbines: Propellant Feed System at Zero “g”
Conditions
Review of Liquid Bi-propellant Rockets and Introduction
to Mono-propellant Rockets and Hybrid Propellant
Rockets
Combustion Instability in Rockets
819
846
884
Week 12
34
35
36
Principles of Electrostatic and Electromagnetic Rockets
Electrical Thrusters
Electrical and Nuclear Rockets; Advanced Propulsion
909
931
966
Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 01 .
Introduction
Good morning. This will be our first class on rocket propulsion and in this class we will
look at what is this subject on rocket propulsion; how it differs from other propulsion
subjects. We will go through the course contents and then see the books that we must be
referring to. May be the introductory part will take some something like ten fifteen
minutes and then we will get started with the course.
(Refer Slide Time: 00:51)
Let us first take a look on what this subject on Rocket propulsion is about. The word
propulsion comes from the word pro pellerie. The word pellerie in Greek means push,
therefore we are talking of something pushing. Pro, as you know, means something like
forward or before, therefore the word propulsion means push forward and therefore,
whenever we talk of any subject on propulsion, what we mean pushing forward.
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(Refer Slide Time: 01:54)
Let us take the simple example of say, a car which has to climb up a hill, up an inclined
plane. The engine of the car pushes it forward; and by pushing you are changing the
velocity of the car or you are imparting a momentum to the car. What is momentum?
We say the car has a velocity v, it has the mass, if you say mass into velocity is what is
momentum and if you want to change the momentum, you have to change the velocity of
the car since the mass of the car is about a constant.
You give some velocity and you have what we call as change of momentum. Now you
know in deep space, we do not have atmosphere and therefore, the act of imparting
momentum to the object in space is what we deal with rocket propulsion. I will again
repeat: propulsion means pushing objects and when we say rocket propulsion, we are
dealing with pushing objects in space.
Space means anything beyond us, may be up there. Therefore, may be first we must get
an idea on what space is about. But before getting more into defining space, let us
quickly get some idea on what this whole course is on and how we are going to organize
ourselves in the next thirty six classes.
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(Refer Slide Time: 04:14)
May be in the first class starting today, we will look at what we meet by the word
space, what constitutes space, in what way motion in space is going to be
different than motion on the ground. That means, we talk about motion in space
and once we know how motion takes place in space; may be, we will able to find
out what is the exact requirements of a rocket.
To be able to make a rocket -- a rocket could be something very small or it could
be huge -- I must know what is the requirement of motion of a rocket in space?
So the first chapter will deal with motion in space and how we go about
converting this motion in space to the requirement of a rocket. Once the
requirement of a rocket is clear to us, the second chapter will deal with let us say,
the theory of rockets.
You know, I would like to ask you a question: why should a theory of rocket be
different from a theory of a car or let us say, theory of a gun? I fire a gun - the
bullet leaves the gun. In what way is rocket different from a bullet? What is your
thinking on it?
You say it is a non air breathing system. So what, we will get into some details of
air breathing, non-air breathing later. Let us say, I have a gun, I fire a bullet from
a gun and the bullet leaves the gun at high velocity.
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Let us assume a rocket is like this. There is some mass which is available in the
rocket. The mass which is available in a rocket, must be able to push it forward.
That means, it propels the object and what is the object, you have something like
a space capsule which it pushes forward. The mass in the rocket is what propels
and we call it as a propellant.
By propellant in a rocket, you mean the substance used for pushing up the rocket
or propelling the rocket. Now in the case of a rocket, the propellant is
continuingly getting exhausted, it leaves the rocket and therefore, the weight of
the rocket keeps coming down, and therefore compared to a car in which I carry
something like ten litres of petrol or something near it, I carry tons and tons of
propellant which is ejected out and therefore, the theory of a rocket is different
from a car or a bullet.
And therefore, we have to look at the theory of rockets which will be the second
chapter. After finishing the second chapter, since we need to give change of
momentum and therefore,
(Refer Slide Time: 07:37)
we will go into nozzles which produce high velocity and help us to achieve a large
change or impart a large change of momentum to the object in space. You all have
studied nozzles in your gas dynamics course. If necessary, we will start from the basics,
go through the basics of nozzles, advances in nozzles but it is an involved chapter.
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If you have a rocket something like this: may be the internal configuration. I have
something like a nozzle here. The flow must run full otherwise if some portion does not
run full, I could get something like a side force in addition getting a force in this
direction. Therefore, the theory of nozzles is quite involved and the third chapter we
should be doing is on nozzles.
In the fourth chapter, we will get back into the propellants or what is used for propelling.
It could either be a solid propellant, could be a liquid propellant, could be a gaseous
propellant, could be a hybrid a combination of these things, could be electricity itself,
could be nuclear, could be anything. And therefore, in the fourth chapter we will study
about the different propellants.
What are the characteristics required to make a good rocket and towards this we will
study about propellant solid, liquid, gas, hybrid, electric may be nuclear propellants.
Once we are clear about the propellants, we can go into the details of the rockets
(Refer Slide Time: 09:24)
The fifth chapter would be solid propellant rockets. And this type of rockets has been
used very extensively in India both for GSLV, PSLV and we will have to look at it, look
at the design considerations of a solid propellant rocket.
The sixth chapter, would be on liquid propellant rockets. They are more versatile and in
this chapter, we will basically look at what are the cycles of operation, in what way it
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differs from a gas turbine combustor, may be an IC engine and what are the modelling
features. The liquid propellant rocket is continuously evolving and when we talk of
cryogenic propellants, it is again a form of liquid propellant.
Having studied solid and liquid propellant rockets, in the seventh chapter, we study about
hybrid rockets and some rockets which use a single propellant, what we call as mono
propellant rocket. This would finish the different type of rockets, how to make them,
what are their features and what are the problem areas in rockets. And once we are clear
about it, we go to an advanced subject which is combustion instability.
This chapter on unstable combustion on instability is particularly important for PG and
research students. Since, we are going to look at what causes unstable or oscillatory
thrust or movement instead of having a steady and uniform operation and burning
steadily, we will address if it will explode under some conditions or lead to failure.
The ninth chapter, will deal with electrical rockets. What do you mean by electrical
rockets? We told ourselves for any rocket, we need to push an object in space. We will
try to see how we can generate electrical forces. Different forms of generating electrical
forces using electrostatics and electromagnetics.
The tenth chapter would on be nuclear rockets and other advanced rockets. What do we
mean by nuclear; I could use nuclear energy to generate a force. I could also use space
time curvature like relativity to generate a force. And some of these things will consider
as the last segment of this course.
Let me now briefly talk about one or two books which I will be following: A good book
for this particular course.
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(Refer Slide Time: 13:04)
is by Sutton; it is on Rocket Propulsion Elements. I think the publisher is Wiley, year of
publication is 2001. This book gives a very good description of rockets but the
mathematics of rockets is somewhat missing and I have published a book, the name of
which is Rocket Propulsion. It was published by Macmillan, in 2010. This was based on
my teaching of the course over the last 6 to 7 years.
The third book which gives the good description about the different rockets is by H S
Mukunda; the name of the book is Understanding Aerospace Propulsion and is published
by Interline, Bangalore in 2004. One important book which I should have said at the
beginning is a book by Hill and Patterson, the name of the book is Thermodynamics of
Propulsion, the publisher being Reading. It is an old book; it was first published, I think
in 1970 or so but the second edition is published in 1992. We have copies of these books
in the library. This book deals with the thermodynamics of propulsion, including many
of the propulsion elements. I will introduce more books on specific subjects as we go
along.
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(Refer Slide Time: 16:13)
Let us now get started. What do we mean by motion in space? Anybody would like to
guess in what way it will be different than motion on ground.
We need to first define what is space: Is anything above us: space? Say, we go on up
and up or we go sideways go to infinity. Is it space? We talk of space capsules, we talk
of planets, we talk of galaxies. How would you define space? You are telling me that
anything outside the atmosphere is space.
What really is space about and what do you mean by atmosphere? We are here, in
Chennai. Chennai is normally a hot place and let us try to plot the temperature in the air
or in the atmosphere above Chennai as a function or let us say altitude z. We know since
the sun is heating the earth, earth tends to get hot.
May be the temperature at the surface of the earth is around 35 to 40 degree Centigrade.
As we go higher and higher up, that means as we increase the altitude, the temperature
decreases until at an altitude of around, let us say around 10 to 11 kilometres, the
temperature is around minus 50 degree Centigrade. These are notional numbers.
Thereafter the temperature begins to increase again. That means, temperature drops to
minus 50 degree Centigrade at an altitude of around 10 kilometres and then begins to
increase again.
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Why does the temperature decrease? The Earth receives radiation from the Sun and gets
heated and the surface of the Earth is relatively warmer and as you proceed away from
the surface, the temperature drops and this zone where in the temperature drops is known
as troposphere.
A jet aircraft flies at an altitude of around between 8 to 10 kilometres. Let us say, this is
where the jet aircraft flies and it flies where the ambient temperature is between minus
40 to minus 50 degrees Centigrade. You would have heard this announcement while
flying in an aeroplane that your aircraft is cruising at an altitude of around 10 kilometres
where in the ambient temperature is of the order of minus 45 degree centigrade or minus
50 degrees Centigrade.
If you go up still further after the temperature drop, the temperature increases, the
increase in temperature is because in this area you have lot of ozone available. The ozone
sort of gets heated by the solar radiation, it absorbs the solar radiation and decomposes;
the temperature increases and this increase manifests for another 40 to 50 kilometres.
Let us say upto 50 kilometres we have the region of the increasing temperature; this
region is what we call as the stratosphere but when we go to still higher altitude let us
say higher than 50 kilometres, the pressure in air is so small or the molecules of air are so
small that they are unable to absorb any significant radiation from the Sun.
Therefore, the temperature again drops with further increase of altitude from around 100
kilometres. But if you go to still higher altitudes you have the molecular oxygen reacting
with other molecules and you have the temperature going up.
The region wherein the temperature drops again is what we call as a mesosphere and the
region of temperature increase because the individual atoms and molecules are getting
heated by reacting with each other to increase its temperature is what we call as the
ionosphere. This continues for something like another 100 to 200 kilometres.
If I were to plot the pressure of air, may be as a function of an altitude z in let us say in
kilometres. At the surface of the Earth, the pressure is around 100 kiloPascal and the
pressure monotonically drops, keeps on falling until maybe at the ionosphere, you hardly
have any air left. This is where we said that the temperature increases.
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The concept of temperature fails in this region of ionosphere because there is no
continuum. And it is the individual molecules of some of these gases which tend to get
heated to high value. Now the question is what do we define as space? Is it anything
above the surface of the Earth going through the troposphere, stratosphere, mesosphere,
and ionosphere and beyond?
That means, space is sort of endless. It keeps on going till infinity. Not being able to
define space precisely in terms of extent, let us examine what is there in space?
(Refer Slide Time: 22:39)
We cannot define something which is endless. We have to look on what constitutes it?
And if you go back and see what is there in space; you see that there are something like
1011 galaxies not in total space but in the space which we can observe. In the observable
space, we say we have something 1011 galaxies. That means, space is still beyond but
what I can see is only this.
What are galaxies? Gravitationally bounded system of stars and each galaxy has a system
of stars and lot of may be some dark matter, something like gas, dust, etc., in it.
Let us consider one galaxy to which we belong, which is the Milky Way galaxy.
Therefore, what we have said is, space is endless, in the near observable space, we have
something like 1011 galaxies and our attention has now come to the galaxy which we
belong to or in which we live, which is called as the Milky Way galaxy.
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Why the Milky Way? It comes from the some Roman or Greek mythology, which says
that the colour is something like milk and there are some stories around it. We will not
get into those details but just say when we talk of number like 1011, we are talking of a
very large number. At the beach you have lot of sand particles and I go to all the beaches
around the world and collect all the sand particles, then we are nearing a value of 1011 .
One particle among all of them corresponds to our Galaxy out of all the galaxies.
Therefore, we have shrunk ourselves and you can see, you how small we are in relation
to space.
If now you really see what constitutes the Milky Way galaxy: well you have a large
number of stars in it. And what are stars? Massive objects in which nuclear reactions are
taking place. They emit light and heat. Then you could also have something like dark
matter between the stars, you could have some gas and stellar dust or dust: could have
gas, you could also have lot of different things in it.
You would have heard of black holes; what are they? You could have it in in our Milky
Way galaxy. You know what happens is that sometimes the stars shrink to very small
size after their life time is over. Therefore, you have infinite mass concentrated in a very
small volume and when we have large mass concentrated in a small volume, it is capable
of attracting, that is the gravitational pull will be large.
We will get into the gravitational pull towards the end of this class and therefore, what
we say is we have black holes, we could also have other objects like quasars. What do
you mean by quasar: Quasi stellar radio sources. These quasars are objects which are
again travelling at near about the speed of light itself.
Therefore, you have lot of things in our Milky Way galaxy and out of all the stars in it,
let us figure out one star which we call as the Sun. And therefore, we focus ourselves
maybe from a large number of stars to one star. We note that we have come to one
galaxy out of all the galaxies and now we talk of the Sun which is a single star and it is
about this which we will be basically interested in for the present discussion.
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(Refer Slide Time: 27:38)
Let us put this together with some dimensions. If you look at the Milky Way galaxy, let
us see the extent of this; it somewhat cylindrical in shape. The diameter is around 10,000
light years. Why do I say light year and not say kilometres because it so huge. And what
is the magnitude of a light year? The distance travelled by light in one year.
The speed of light as you know is 3 into 10 to the power of 8 meters per second. In 1
year have 365 days multiplied by 24 hours multiplied by 60 per minute, 60 per second.
And therefore, one light year will therefore correspond to so many meters which should
be around 9.5x1011 kilometres or so.
Therefore, we are talking of a large diametrical content of the cylindrical Milky Way
galaxy and its height is something like 2500 light years. And in this you have number of
stars. We are getting focussed around one particular star called the Sun or the solar
system.
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(Refer Slide Time: 29:16)
We are concerned with the solar system. From the large number of galaxies in the
observable part of endless space and we come down to the solar system.
What does the solar system consists of? It consists of the Sun, a single star out of the
large number of stars in our galaxy and you have planets going round the Sun. May be
starting from Mercury, then you have Venus little bigger. Then we have the Earth, what
is next Mars. What would be the next one, Jupiter. Next one Saturn, two more Uranus
and Neptune. Therefore, you have 3 plus 5, 8 planets which are going around. You know
previously we had included another planet known as Pluto giving nine planets or
“navgraha” but Pluto has been decommissioned as a planet because it is not fully formed.
It is something like a loose mass which is still going along with a belt here which known
as a Kuiper belt.
I will come back to this belt because it gives inputs regarding some asteroids coming
and hitting earth. All what we said is in the solar system, we have something like eight
planets going around the sun and this is the part of the space with which we are
immediately interested.
This solar system also consists of something like 31 moons. How do we define a moon?
There are certain objects which go around the planets and they go around like satellites
around it. There are 31 moons and Earth has a particular moon which is going around the
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Earth and this is the moon of the Earth. So also, we will have moon for Saturn, we have
moons for Jupiter and there are something like 30 other moons in the Solar system.
When we are dealing with all these eight planets going around the Sun, it is necessary to
have some idea of what constitutes these planets; may be Earth, let us put down the mass
of the Earth, let us put down the diameter of the Earth. The mass of the Earth is 5.974
into 1024 kg, the diameter of the earth is 12,756 kilometres. These numbers are important
and I will circulate a table to you giving the mass of the different planets and their
diameters. But just to get an idea: mercury is about the smallest planet around one-third
the mass and diameter of the earth.
The largest would be something like Jupiter. These planets go around the sun and it is
the motion of planets which provided or which prompted Newton to formulate the
universal law for gravitation. I think I should repeat this point in a slightly different way
after consolidating what is said so far:
(Refer Slide Time: 33:49)
Let us therefore quickly revise through what we have done so far: we said propulsion or
rocket propulsion deals with pushing in space. For pushing in space, one of the forces
which we could consider is the gravitational force.
Let us start with the gravitational force. It becomes necessary for us to go back into the
solar system to understand it. Look at the revolution of the different planets around the
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sun. Now let us just take one particular case: I take the Earth as shown in this figure and
it is going around the sun.
Earth is going around the sun. You know people have been watching the motion of
planets around the Sun for years together and around the year between 1570 and 1610,
we had a famous person by name, Johanas Keplar. He introduced three laws which
govern the motion of planets like Earth around the Sun.
(Refer Slide Time: 35:26)
The three laws were: 1. All planets move in elliptical path, i.e., have elliptical orbits. By
orbit, I mean the path of the planet around the Sun. But if you read the newspapers
around the month back, there was news that the orbits are not really elliptical but they are
wavy orbits something like a wavy elliptic orbit.
As per Johanas Keplar: we have the orbits in an elliptical path which is the first law of
orbital motion of planets. The second law was on equal areas. Suppose we join the centre
of the sun with the centre of the Earth with a straight line. We find out what is the area
swept by this line during the elliptical orbit. We find that the equal areas are swept out in
equal times.
Let us put it together: this is the Sun which is at the focus and you have something like
an elliptical path. Let say this is the major axis and then you have an elliptical path this is
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the second foci. This is the Earth going around the Sun. First law says it is an elliptical
path, the second law says equal areas are swept out for equal times.
If this is the time taken for this area is swept out by the imaginary line joining the centre
of the earth with the centre of the sun. And similarly in an equal area will be swept out
during the motion from here to her for the same timee. All the second law tells is, if I
come to this particular path in which the minor axis, it’ll travel a longer path here
compared to a shorter path along the major axis.
The third law is one which deals about symmetry of orbits. All what it tells is you have
the Earth, you have Mercury, you have Neptune far away. It tells that the time for 1
revolution divided by the radius, is such that the orbital time t square divided by the
distance cube is a constant (t2/R3 is a constant).
That is the distance from the Sun to the different planets divided by the time of orbit of
the particular planet is the particular constant. These are the three laws of the orbital
motion as formulated by Johanas Keplar.
Why are we getting into orbital motion? We want to understand something about
gravitation. And Newton comes out with the Universal Law of Gravitation based on
these laws of orbital motion of planets. What does Newton find?
(Refer Slide Time: 38:43)
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Well. The story goes like this. He watches an apple fall on the ground from a tree. He
immediately connects the apple falling from a tree to the elliptical orbits or orbital
motion of the different planets around the sun. What is the commonality? How can we
identify what is the common factor between these two. Let us take another look at the
planetary motion that we were dealing with it. This is the Earth as it is going around the
sun in an elliptic orbit.
May be the Earth as it is travelling some distance like this, it falls through some distance
because the trajectory is elliptical. Again the Earth let us say, it would had a horizontal
velocity it will go like this but in the process of going horizontally, it falls through some
particular distance. Again it goes through some it comes over here, again it comes like
this. In another words, if I had given a horizontal velocity to the earth, it keeps on falling
towards the Sun at each instant of time as it progresses at constant horizontal velocity. If
the Earth were to go horizontally at a given velocity, it falls by a certain distance as it
travels.
That means, we have a constant linear velocity and it keeps falling on to the Sun. It
looks as if the Earth is freely and continuously falling. It is no different from an apple
which falls on to the ground from the apple tree. Actually, if we look at ourselves today
all of us are freely falling towards the Sun. Just in the same way, as a fruit or a stone is
falling. Therefore, Newton is able to relate the commonality between an apple falling to
the ground and the planets falling towards the Sun and it becomes the Universal Law.
Newton did not do the experiments himself. He did nothing to really say that, I derive the
gravitational law like this, or that. He based it on observations of Johanas Keplar and
others who preceded him like Galileo Galilee who as you know dropped a piece of a
feather and an iron ball and found that in vacuum both of them will take the same time to
come to the ground. Just based on the observations, he was able to formulate the
Universal Law for Gravitation. But before I get into the gravitational law, its necessary
for me to go into some more details like how do you measure forces, distances and
velocities and their units. We need to be clear on what are the parameters and the units.
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(Refer Slide Time: 41:50)
We wanted to describe motion in space. I think we must be very clear because a time has
come when I need to put some numbers, like to apply or derive equations. We can say
we are all engineers, we know about mass, length and time. I can use these three
fundamental quantities and describe motion in space. How will you describe mass?
Quantity of matter, unit is kilogram. But what is a kilogram. It is some reference kept in
a lab near in Severs near Paris since let say, 1819 or so. Some standard is kept which we
call as kilogram. It is kept very carefully, you know in a desiccator under very controlled
condition such that it will not get worn out nor will form scales on it.
It is a platinum rhodium alloy which has a mass of 1 kg and it has been duplicated at
different places and used as a standard. Well. We say this is 1 kg. So, also when I say
length is in metres. What is a meter, again a reference kept at the same lab since may be
last 150 years or so. And this is the particular length scale which is given; the length of
this standard. It’s again an exotic alloy of platinum rhodium but then there are some
problems; though it is stored in the best of conditions and you have some duplicates in
some other countries also, it keeps eroding or scales are formed on it. Some changes do
take place over a long period of time. Therefore, it is not a good standard and we need a
better standard. How will you have a better standard? In the year 1982, i.e., quite
recently, scientists suggested to express the length standard through physical constants.
What is the physical constant to be used? Whatever happens the length would be the
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same, it cannot change may be after a million years the meter will still be the same since
a physical constant is used.
(Refer Slide Time: 44:35)
C
The physical constant used for defining the length is velocity of light in vacuum. All of
us know that, light is propagated as an electromagnetic wave and the speed at which the
light is propagated, we say is C meters per second. The question is can we use this
constant to define length? It is a constant because the electromagnetic waves propagate
through vacuum at a constant speed. Let us say C meters per second and rather than
define the length in terms of a standard like what we considered viz., a standard of
length, which is kept near Paris, we would like to define it with respect to this constant
velocity of light in vacuum. How do we do it? We say the distance travelled by light is
C meters in one second. In 1/C seconds, the distance travelled will be one meter. The
duration, we are considering, is one over C (1/C) so many seconds. Second and second
get cancelled and we get one meter. Therefore, the more recent definition of length scale
is with reference to the velocity of light in vacuum which is C meters per second. The
precise value of C is:
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(Refer Slide Time: 46:24)
299,792,458 meters per second i.e. about 3x108 m/s. Therefore, the definition of length
of one meter is the distance travelled by light in vacuum over the duration of
1/299,792,458 seconds. This is how we define the length scale of a meter.
Therefore, what is it we have done so far? We have considered something like the
definition of mass as a standard kilogram, may be length has a standard meter but now
we are telling ourselves meter corresponds to the distance travelled by light in vacuum
for a duration of one over let us say 299,792,458 seconds. But then, we have still not
defined time.
How do we define time? We must be very clear. Time is something related the duration:
let us say we have a pendulum; the pendulum goes up and down. The duration of one
cycle of the pendulum; pendulum starts here, goes here, comes back here is what we say
is a duration of one cycle of this pendulum which we could say as one second. But then,
it is difficult to have a period like a standard pendulum being used to describe the time
scale and its easier for us to define the time scale based on the duration of an event such
as a solar day.
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(Refer Slide Time: 48:15)
This is the most simplest to describe. We have the Earth, may be rotating on its axis as it
revolves around the sun. One revolution of the earth around its axis is what we call as
one day. May be we call mid-day (middle of the day) as when sun is vertically above us.
When the sun is vertically above a particular part of the Earth, we call it as mid-day at
that part of the Earth. And we go to the next mid-day, next time the sun is vertically
above at that point and this corresponds to 1 period of rotation and we call it as one solar
day . That means, in 1 solar day, a day consists of 24 hours, again each hour consists of
60 minutes and each minute consists 60 seconds.
And therefore one solar day consists of something like let us say 86400 seconds and we
can therefore define the time. There is one problem in this definition though. We said
earlier that the Earth is revolving in an elliptical orbit around the sun and it is also
rotating on its axis as it is revolving around the sun. The period of rotation is therefore
not exactly 1 day but is slightly shorter because as it is rotating on its axis as it is also
revolving. And therefore, the exact period of one rotation on its axis is not exactly 24
hours but it is something like 23 hours 56 minutes and 45 seconds and this duration is
what is known as a sidereal day. There is a difference between solar day and the sidereal
day.
Therefore, we have complicated the day, instead of having one solar day. We need a side
real day which is smaller than 86400 seconds. Now there are perturbations in the rotation
21
and also in the path of the Earth around the Sun. it. And therefore, it is very difficult to
really define time absolutely very accurately in terms of either sidereal day or in terms
of solar day.
(Refer Slide Time: 51:12)
It becomes necessary to have some other standard for defining time. This is based on the
period of a wave. Cesium133 material is an isotope element and it emits radiation. It
emits radiation in different bands. What we mean when it emits radiation is that it gives
out packets of photons as energy. These radiations are emitted at discrete frequencies i.e.,
each one having a specific period. So, many periods of radiation are getting emitted.
Therefore, we look at one specific band namely at the ground state of cesium. And at
this ground state of cesium, you take one hyper fine level and you say, many number of
periods which are emitted. That means, you say specific number of periods of radiation
which is emitted at the ground state in this particular hyper fine level is what we will call
as one second. Therefore, how can we represent this hyper fine level?
22
(Refer Slide Time: 52:59)
We just said that at this ground state Cesium is emitting radiation. Each wavelength
corresponds to a certain time. I count a large number of these wavelengths and the
number of periods or the number of wavelengths, amounting to something like
9,192,631,770 periods or wavelengths equals one second. Corresponding to this ground
state we have 1 period of radiation or 1 wavelength is 1/ 9,192,631,770 seconds.
And this is how we define time: namely in this ground state 9,192,631,770 periods of
radiation emitted is what constitutes 1 second. And this is an accurate way of defining
time.
To recap: we have defined mass in kg as a standard, we have defined length as meter as a
standard, the standard length being on the basis of the velocity of light. Then we defined
time seconds as a standard.
23
(Refer Slide Time: 55:03)
And now we can derive a set of units, the length or distance divided by time has units
meter per second and this is what we call as velocity. When I say distance: it is a vector,
and therefore, velocity is a vector. And if, I say momentum, we said mass into velocity
or rather the units is equal to kilogram into meter per second is becomes momentum. We
say change of momentum is impulse and therefore, impulse will have units to be same as
momentum namely kilogram meter per second. We also said rate of change of
momentum is what constitutes force or rather the impulse divided by the time is force.
And therefore, force could be defined as rate of change of momentum, that is 1 over
second into we have momentum change as kilogram meter per second or rather the units
of force becomes kilogram meter per second square (kg m/s2) which is what we call as
Newton. Therefore, we have defined through these three basic definition of mass, length
and time, the velocity in meter per second, momentum in kilogram meter per second,
impulse again kilogram meter per second and force which is kilogram meter per second
square.
Having defined these quantities may be its time to go forward and examine how we can
describe using these units the motion in space of the different bodies and this is what we
will do in the next class.
24
Rocket Propulsion
Prof. K Ramamurthi
Department of Mechanical Engineering
Indian institute of Technology, Madras
Lecture No. 02
Motion in Space
(Refer Slide Time: 00:16)
Good morning. In the last class we considered motion in space. For pushing forward we
have to give or provide a change of momentum. Change of momentum is what we call as
impulse and it has units of momentum namely kilogram meter per second. We also
considered what are the parameters used to quantify motion. The length in meters is
based on a standard, but ever since 1982 the standard is based on physical constant
which is the velocity of light in vacuum. Mass in kilograms is, however, still based on
the reference standard.
Currently, research effects are on to define a physical constant rather than a reference
object by which we could define mass. It is not yet done. When we say time, we mean a
period and that period is second. How do we define the direction of time? The direction
of time comes from the second law thermodynamics which says that time progresses in
the direction in which the entropy increases. May be we will try to take a look at it in the
25
subsequent classes because thermodynamics forms the basis of the entire rocket
propulsion.
Having defined these three quantities mass, length and time we talked of change of
momentum. Velocity is defined as meter per second. Since distance is a vector, velocity
is also a vector. Momentum p equals to mass into velocity and is therefore also a vector.
To repeat, momentum p is also a vector being product of mass and velocity - kilogram
meter per second. A question immediately arises why use the term momentum when I
can use velocity. Momentum is a more fundamental quantity compared to the velocity. It
can be seen from the following example. If I have an iron ball which travels at one meter
per second and hits me and let us say I have a feather which travels at the same velocity
and hits me; the feather does not produce any major sensation while, iron ball leads to
significant effect. Therefore, in subjects dealing with motion of molecules or classical
mechanics, we deal with the quantitative momentum.
The other quantity as we saw is acceleration, which is meter per second square (m/s2).
To summarize, in order to be able to push and change the momentum we provide an
impulse. While talking of a change of momentum and let us introduce the term Force.
Let us consider the following example.(Refer Slide Time: 03:43)
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When people fancied in terms of going to the moon may be some 2000 years ago, a
method suggested was if the sea became very rough and a boat tossing in the rough sea
could be caught up in very rough and strong waves. The boat could be catapulted up by
the waves and if the force of the push from the very rough sea was very intense; the boat
could reach the moon. This was the first science fiction article on going from the Earth
to the moon. The idea of being pushed up by storm in the sea namely when you have
huge waves in the sea, that is huge tidal waves such as when a storm occurs over sea and
a boat being pushed upwards towards the moon was proposed by Lucian. He was a
Greek philosopher and satirist who live in the period around 40 BC. Therefore, we are
talking of something being pushed with a large force. Can we quantify push in terms of
impulse that is change of momentum? Or rather describe the push in terms of change of
momentum itself.
We therefore look at the change of momentum which is a vector, may be as a function of
time. Assume a body is traveling with constant momentum and after a short duration of
time, we change the momentum to a slightly a larger value. How do we change it? May
be in the example that we consider, we change it gradually from the steady value by
increasing its velocity as shown. The final momentum again reaches a steady constant
value. Initially the momentum is a constant value and it changes over a time period delta
t; that means, in the plot the momentum is changing from p at time t to p + Δp at time t +
Δt. (Refer Slide Time: 06:03)
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Therefore what is the impulse associated with the change? The impulse is equal to the
value of momentum p +Δ p at time t +Δ t minus p at time t so much kilogram into meter
per second.
Therefore, the definition of impulse is just a change of momentum that you give to a
body. If we were to ask what is the rate of change of momentum? I plot the rate of the
momentum change for the momentum change shown earlier. How will the rate look like?
We find that the momentum remains at p over time t right from zero onwards to this
particular time at t. In other words, dp/dt is zero up to this particular point in time t.
Thereafter it increases reaches a maximum and then goes back to zero at time t + Δt. In
other words over a period of Δt it increases, reaches maximum (at the inflexion point on
the momentum curve) and comes back to zero. The curve is the type of signature that we
would get for dp/dt over the duration delta t.
If now we state that the rate of change of momentum is force, the force due to the change
of momentum is not something, which is a constant over the duration Δt, but keeps
varying. However, whenever I change the momentum and I ask myself what is a force?
It is difficult for me to specify the force since this force is going to vary with
time. Therefore, what we say is that the force is continually varying during the duration
of the change of momentum and it would be better for me to take an average value and
say this is my average value of force for the momentum change over the time duration.
Force is vector like momentum and is an averaged out value. When we talk of force, we
imply this average value. Force is a derived unit. It is not something fundamental like
momentum and we must keep in mind that the force during a particular change of
momentum continually varies.
28
(Refer Slide Time: 08:49)
But I take the average value and what will this average value of be? Let us put it down as
F and we say it is a vector. It is equal to the average value of dp/dt over the duration Δt.
The momentary value of dp/dt, averaged over the duration Δt by integrating it over the
time Δt and dividing it by Δt This must be the averaged value of the rate change of
momentum, i.e., averaged force over a period of delta t.
Therefore, you see that force is something which is not that that good a unit compared to
momentum change or impulse as some averaging are involved. If we have to write an
expression for Force in terms of impulse; how could we do it? We can write Force into
Δt as change in impulse over the duration. Denoting the change in impulse by I we find it
is equal to Force into Δt. Because we find force is equal to this impulse divided by Δt or
equivalently Force into Δt is the impulse or the change of momentum. This is the
connection between impulse, force and change of momentum. If we are very clear about
the above relationships, it is about time to get into the universal law of gravitation. What
did we discuss about planets orbiting about the Sun in the last class?
29
(Refer Slide Time: 11:14)
We told that all planets as they go around the sun are all freely falling bodies. A
planet such as the Earth is continually falling on to the sun. So, also an apple is
falling down and this is the observation which Newton had and he says a heavy body
like the Earth attracts a body which is up in space in the same way that the much
heavier Sun is attracting the Earth. And if we say this is the mass of the Sun and
this is mass of the Earth; he said the force with which it attracts is given by a
constant multiplied by the mass of the attracting body, the mass of the body which is
attracted divided by the square of the distance between them. In other words it need not
be only the Sun and the Earth. It could be any heavenly body or otherwise. Let us say
mass m1 attracts a small body of mass m2. Let us say that r is the distance between
them. Therefore, the force is equal to the product of m1 and m2 divided by r2 into a
constant.
There is therefore a field of attraction around a body of mass m and it is the field
from the heavier mass that attracts the lighter mass. And this relation becomes the
universal law for gravitation. The constant G becomes the universal gravitational
constant. You know this law is important as we shall apply it in moment or two.
Let us also remember that force is a vector.
30
(Refer Slide Time: 12:50)
Therefore, I should have really written this as the force is equal to may be a body one
being attracted towards body two or equivalently let us say the body two attracted
towards one. Therefore, I have the body two which is a light body being attracted by a
heavy body m goes as a gravitational constant into r2. Therefore, I put it as r is a vector to
the power 3 mod of this vector into r bar; that means, m1 into m2 by r2 into G and it is
being attracted. Therefore, we have a negative sign for the attraction.
This is the universal law for gravitational. G is the gravitational constant and therefore,
the unit for G should be what? It should have the units of force divided by kilogram
square multiplied by radius square or meter square. What is the unit for force then? We
should be clear about it. We told that force is the rate of change of momentum; force is
equal to dp/dt. Therefore, force is equal to 1 over time multiplied by kilogram meter per
second which is equal to kilogram meter per second square (kgm/s2). And this particular
unit kilogram meter per second square is what we call as Newton.
Therefore, the force has units of Newton, which is kilogram meter per second square (kg
m/s2). It comes from rate of change of momentum and therefore, we have the units of G
as Newton meter square by kilogram square and the value is something like 6.671 into
ten to the power minus eleven (6.671x10-11) Nm2/kg2. This is the constant in the
universal law for gravitation.
31
In our solar system we had the eight planets going around the Sun. You also have some
loose objects like asteroids, which are also going around, but they do not have a welldefined path as the planets have.
One such asteroid is likely to hit the planet Earth may be in the year 2036 and if if does
not hit the Earth it is likely to come back and again hit it in 2039. What is it we are
talking of? May be some of these asteroids may come and collide with the atmosphere
above the Earth. The question is how to prevent an asteroid from hitting the Earth? What
is the type of propulsion system that we could design such that we prevent the collision?
Can we think of it from the universal law for gravitational forces? People talk of
difference strategies how to how to prevent some of these things happening and let us
take some time off and let us try to solve this problem if it helps us in applying the
universal law for gravitation.
(Refer Slide Time: 16:49)
Let us say one of the thinking is that we launch a rocket on to space and we keep on
accumulating satellite over here; I make a heavy mass over here and when the asteroid
comes over here; this mass being heavy compared to the mass of the asteroid, will attract
it towards the heavy mass and instead of the asteroid going in a given particular direction
deflect it. In this way we change the direction and it will miss the earth. This is known as
a gravity tractor. That means, we put a mass in space and make sure that this mass is
32
large and its distance from the asteroid small, it gets attracted towards it and the asteroid
instead of coming like this can get deflected away.
You know these are all possible. That means, you know the law is not only doing
problems in mechanics; but can be applied for changing the trajectories and changing the
trajectories is as good as giving some propulsion element to it. The gravitational force
what we are talking of or rather the gravitational field is a weak force but it persists over
a very long distance. Like let us say I have the Earth here. May be a mass near the
surface of the Earth is attracted with a higher force than something which is very far
away because the field decreases as the distance from the Earth increases. The attractive
field from the Earth decreases as we move away from the earth.
Let us do one problem to be able to assess the gravitational field and I take a model
problem again of an asteroid and let us calculate what is the force exerted by this asteroid
on something which is moving near it. This would help us understand the magnitude of
the gravitational field for some space related problems.
(Refer Slide Time: 19:14)
Let us do the problem of a space probe by name Rosetta that was launched to study the
asteroids in the space between let us say Mars and Jupiter. This particular space capsule
Rosetta was used to study a particular asteroid by name Steinz.
33
Now, Steinz, let us assume a mass around 1.208x1011 kg. You know the asteroids are
somewhat composed of loose material and they do not have a particular fixed path in
space. The space capsule Rosetta cannot be very heavy and let us assume the mass to be
500 kg. To get any meaningful attraction, it must be brought as near to Steinz as possible
and the nearest distance it came near to this asteroid Steinz was something like 800
kilometers. We would like to know when this space capsule Rosetta is 800 kilometers
from the asteroid Steinz, what is the attractive force exerted by this asteroid on this space
capsule Rosetta? The nearest distance between the space capsule and the asteroid is 800
kilometers.
Therefore, what is the force which the asteroid exerts or pulls the space capsule? We say
force is equal to G multiplied by the mass of the asteroid into mass of this space capsule
Rosetta divided by r square. G is 6.67×10−11. Mass of the space capsule is 500 kg and
this is divided by the distance square. The distance between the two is let us say nearest
position is 800 kilometers and the diameter of the asteroid can be assumed to be 1200
kilometers.
We can neglect the diameter of the space capsule. The total distance from the center of
the space capsule to the center of the asteroid to the center of the space capsule is 600
plus 800 kilometers into 103 squared in meter square and therefore, we are getting a force
of the order of 6.67× 10−11 × 1.208 × 1011 × 500 divided by (1400 × 1000)2. Let us take
look at the units. It is Newton meter square by kilogram square into kilogram into again
kilogram i.e., kilogram square. Denominator is meter square and therefore, we have so
much Newton as attractive force.
When we look at it, the type of force that we get is of the order of 10 to the power minus
6 of a Newton which is something like a micro Newton. That means, the attractive force
exerted by this asteroid on this space capsule is something like a micro Newton; which is
negligibly small. However, in space even small forces are of interest and therefore we
find that weak gravitational forces attract the space capsule. However, if we had a very
massive satellite like the gravity tractor then we could get more gravitational pull.
What is the gravitational field and how is it expressed? You say acceleration due to
gravity. What is acceleration due to gravity? How would you define the field? What do
you mean by gravitational field and what is acceleration due to gravity? How to define
34
it? We say the force experienced by a mass m in a gravitational field g is equal to m g.
What is the unit of g?
(Refer Slide Time: 21:19)
Meter per second square we call it as acceleration, but, how can Earth give something
like an acceleration? Acceleration is rate of change of velocity. You know we go back to
the universal law for gravitational and then we write the force F12 is equal to the mass of
a body attracted by the mass of Earth of ME. Let the mass of this body be m. Therefore,
the force by which this body is pulled towards the Earth as per the universal law for
gravitational forces is − G × m × ME ÷ r2 where r is now the radius of the earth RE plus
the height of the body above the surface of the Earth h. The force is so much Newton.
Now I want to simplify this expression. Therefore, I write this is as equal to minus G ME
by RE square and then I write within the bracket (1+h/RE). I expand out the term
(1+h/RE)2 for height h above the earth to be is very much smaller than RE.
35
(Refer Slide Time: 27:18)
And therefore, I can write this expression as force is equal to − G × ME ÷ RE2 into mass
of the object × (1 − 2 h/RE) and the subsequent higher orders of h/RE. Anyway h is
smaller than RE and the higher order terms could be neglected.
And therefore, I get the value of force is equal to minus G ×ME ÷ RE2 into the mass. We
have the mass of the earth as 5.974 × 1024 kg and the value of G was equal to 6.671× into
10−11 and the value of the radius of the earth was equal to its diameter 12756 kilometers
divided by 2. This is multiplied by the mass m of the body.
Now, we simplify this. I find out that this will come out to be − 9.81 × m which is equal
to force F. This value of 9.81 is a constant since the mass of the earth is a constant, G is a
constant, radius of the earth is a constant and this is why we denote it by g and we say
that F = − mg. Therefore, we are really not telling acceleration due to gravity we just tell
ourselves as per the universal law of gravitation whenever there is a heavy mass it
attracts a lighter mass and it is a field.
36
(Refer Slide Time: 30:00)
What is the unit of g? We put the expression down. The unit of G was Newton meter
square by kilogram square. This is multiplied by mass of Earth in kilogram and divided
by the square of the radius of Earth in m. Meter square, meter square gets cancelled.
Kilogram comes here, you have Newton per kilogram. Newton is equal to kilogram
meter per second square divided by one over kilogram this is equal to meter per second
square. Therefore, the unit of the gravitational field comes out to have units of
acceleration and therefore, many people refer it as acceleration due to gravity whereas, it
is just a field attracting or interacting with a particular mass.
We had said that Newton formulated the law based on observation and since it is based
on observation the law should not be extended beyond the realm of observations. Since it
is based on phenomena it is something like a phenomenological law. When we talk of
bodies in space such as Quasars that travel at a speed near to velocity of light, the law
breaks down. The universal gravitational law is no longer valid. Therefore, whenever we
base anything on observations; the conditions of observations must be related to the
particular law and it has to be applied with caution for conditions outside the realm of
observations.
Though we have seen the law for gravitational field, we have not really pondered over
the question of why a heavy mass should attract a lighter mass. How can you justify it?
Can we do an experiment to show why?
37
(Refer Slide Time: 32:47)
We have scientists, some very famous scientist like Stephen Hawking. He recently
published a wonderful book known as the Grand Design. He talks in terms of a unified
model for explaining the laws of nature. In fact, in this particular book The Grand
Design, he talks about the phenomenological theories of Newton. Then he goes ahead to
Einstein’s theory and the pioneering work of Feynman. It will be nice to read through.
But, all what I want it to say is Stephen Hawking among other great scientists has looked
at the forces in nature but the reason why such forces exist is still not firmly clear.
Einstein gave an explanation for us to understand why such attractive forces should exist
in the region of a body and the attraction of a lighter body by a heavier body.
38
(Refer Slide Time: 34:19)
The reasoning was like this. Supposing I hold something like a towel or rubber sheet
tight like is shown. In the center of the rubber sheet I put an iron ball. What would
happen? The sheet would immediately sag as the heavy body pulls down the sheet. Now,
if we were to place a small ball adjacent to the heavy ball. It will roll towards the heavy
ball; that means, this heavy ball creates a field, which attracts the lighter ball. This helps
in visualizing why a gravitational field should exist.
But, precise reason for gravitational field is still not understood. We can only understand
it through an example like given above. May be when I have a heavy mass I have
something like a gradient and that gradient attracts a smaller mass.
39
(Refer Slide Time: 35:33)
There is one problem, which we have not yet thought of regarding velocities. The
problem is when we see an object traveling at a particular velocity v, how do we define
its absolute velocity? We are on the Earth, the Earth is rotating and when Earth is
rotating I am also rotating along with the Earth. The Earth is also revolving around the
Sun. That means, we are also moving as we measure the velocity of the object in space.
Our velocity is something like 0.46 kilometers per second. This is the speed with which
we are moving because the Earth rotates once in 24 hours.
Now, if we are traveling at 0.46 kilometer per second and as I am rotating this body is
moving. I am only able to relate the velocity of this body with respect to me? It is going
to be difficult to even determine its absolute velocity. And therefore, I need to have
something like a frame of reference in mind so that somebody else does not contradict
my findings. How do I define a velocity? Is it relative velocity or absolute velocity?
40
(Refer Slide Time: 37:10)
You will immediately tell me well, if I have if I am absolutely stationary like for instance
I am standing here. This is my coordinate system. May be I stand over here I am
absolutely stationary and watch a body that is moving; then I can say that distance
travelled by the body divided by time gives me the velocity.
(Refer Slide Time: 37:30)
But if I am on Earth like all of us are, what is my speed? Mind you 0.46 is kilometer per
second will translate into something like 1600 kilometers per hour which is going to be
faster than the fastest car that I can imagine. I am also moving on the surface of the
41
Earth as an object is moving in space and my appreciation of the distance travelled by
the body depends on my movement.
(Refer Slide Time: 37:59)
Therefore, we find everything is relative and we need a frame of reference to be able to
describe the motion of bodies in space.
(Refer Slide Time: 38:15)
Therefore, in order to be able to do I must either be totally stationary which is difficult or
else I must also tell myself if I am not stationary, if I am moving at a particular constant
velocity, I can record changes in velocity of bodies clearly.
42
Supposing, I were to move at some constant (linear) velocity and I am observing a body
at moving at a different velocity; the change in velocity of this with respect to my
constant velocity will always give me the same change. Therefore, my frame of
reference, whether I consider to be stationary or to move at constant velocity will provide
me the change in the velocity of this body. Otherwise, if I am here on this Earth and it is
rotating and all that my translation velocity is continually changing, I cannot really
monitor the change in momentum or change of velocity of the body. Let us put this
down.
(Refer Slide Time: 39:26)
A frame of reference which is either stationary or is moving at constant linear velocity is
known as the inertial frame of reference. This would provide the velocity changes and
momentum changes which would be independent of the frame of reference.
Many of you are studying the subject of combustion. I will ask you a question?
What is the difference between flame speed and burning velocity? Are they the same?
No Answers? Now, frame of reference is extremely important in any engineering
problem. That could be a clue.
43
(Refer Slide Time: 40:39)
Let us say I have tube filled with a combustible gases and that is how we measure the
flame speed in this tube after igniting the gases. When is it that I measure the flame
speed or burning velocity? What gives me the flame velocity?
Suppose I am standing here i.e., I am stationary and the apparatus is also not moving. I
measure the flame speed. I observe it from my frame of reference. I call this as flame
velocity. But, if I were to sit over on the flame, I will find the gases coming towards me
with a velocity. In the frame of reference of the flame, I determine the velocity with
which the gases are coming towards the flame. This is what is called as burning velocity.
What is the difference? What is it that I have done? Should it still be the same or should
it be different? Now, I pose this question to you; I have the tube, I have the burnt gases
and the flame which are pushing the unburned gases ahead of it. The flame now moves
in a region in which the gases also move. When I look at the flame from outside I am
looking at the flame as it were moving in the tube. That means, I am looking at the speed
of this flame as it moves along with the unburned gas. This will be higher than what I
would measure sitting on the flame. Therefore, we have to be clear about the frame of
reference with respect to which measurements are done.
44
(Refer Slide Time: 43:16)
The inertial frame of reference is either a stationary frame of reference or a frame of
reference which moves at constant linear velocity. If we are in a rotational frame of
reference, we need to be more careful to describe the motion correctly. We will look at
this in the next class.
To summaries then; what is it we have done so far?
(Refer Slide Time: 43:29)
We looked at the parameters used for describing motion in space. We looked at the
constituents of space. We talked in terms of planets moving round the Sun in the Solar
45
system. We found that the planets are in a state of continuous fall on to the Sun and the
planets are all freely falling towards the Sun. And Newton saw the commonalty between
a freely falling planet and a body like an apple falling to the ground and formulated the
universal law for gravitation. We looked at the value of g which we said is the
gravitational field in m/s2. Towards the end we looked at the frames of reference for
describing motion in space and we said that to be able to measure momentum or
momentum change; it is necessary to have an inertial frame of reference.
I continue with this in the next class. We will get into a rotating frame of reference, what
are the corrections required to describe the motion correctly in the rotating frame of
reference and that will help us to define orbital velocities and also the requirements of
rockets.
46
Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. #03
Rotational Frame of Reference and Orbital Velocities
(Refer Slide Time: 00:25)
Good morning. In today’s class we will address the orbital requirements of satellites and
therefore, determine what rocket should be doing. In last two classes we learnt that there
are something like eight planets revolving round the sun. We said these planets consisted
of Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. We also said that
there was one planet called Pluto which is no longer considered to be a planet because it
is not dense. Having said that you know there is lot of interest in exploring the different
planets and I will take two examples
47
(Refer Slide Time: 00:55)
However, before doing so let me repeat of what I just now told. The Sun which is shown
in red here; you have about it may be mercury as after that Venus then Earth and the
different planets. Seen in between you also see some asteroids which we talked about in
the last class I will come back to this point a little later,
(Refer Slide Time: 01:18)
You have a rocket launch vehicle known as New Horizon; it is a rocket which goes and
finds out what is happening in the Kuiper belt beyond Pluto. If you take the time taken to
go to Pluto - this was launched in January 2006 and it is supposed to reach there in
48
another 4 or 5 years; something like it takes almost ten years to reach Pluto and what
does it consists of? It consists of series of rockets one about the other and the purpose of
this course is to be able to size the rockets required. We must be able to have rockets
such that we can achieve a specific mission and that is why I showed this rocket.
(Refer Slide Time: 02:00)
And then next slide shows the space craft which this particular rocket launches and it
goes round and round the planet in which we are interested. This spacecraft should orbit
the Kuiper belt that is beyond Pluto
(Refer Slide Time: 02:16)
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The next slide that I show is a launch of a rocket which took place some on August 5,
2011; it was an Atlas Centaure rocket and this is again from U S. It is supposed to be for
a Juno mission i.e., go to Jupiter. As per Greek mythology Jupiter is supposed to be a
God and his wife is known as Juno. Therefore, they have named it as Juno mission.
(Refer Slide Time: 02:52)
We also looked at the different planets. We looked at the mass of the different planets the
diameter and the distance from the Sun and so on. One thing we must keep in mind is the
Earth’s mass is around 5.974×1024 kilograms and its diameter is 12,756 km. The smallest
planet is Mercury whose diameter is about 1/3 of Earth. The largest we said was Jupiter.
We also talked of moons and the moon of Earth is smaller than Mercury. There are 31
moons which are available in the solar system
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(Refer Slide Time: 03:37)
We also looked at the atmosphere of the Earth and the temperature variations in it. When
I wanted clarifications on what is the difference between a rocket and an aeroplane and
other modes of propulsion, you told me that a rocket flies in vacuum; it is really not a
must that it travel in vacuum. Anything which goes in space is what we called as a
rocket and we talked in terms of the temperature variations above the surface of the
Earth. The temperature in the plot is given along the X axis and altitude along the Y axis.
We said because the Earth gets heated by the Sun, the layer of air above the Earth gets
heated and therefore, the temperature drops from a high value of around 40 degree
Centigrade at the surface of the earth to a low value around minus 60 degree Centigrade
at a height of around 10 kilometres
It is at this height that a jet aircraft flies (at altitude between 8 and 10 kilometre). Above
this height the temperature increases. In the troposphere just above the surface of the
earth the temperature drops and after some particular height between 10 and 15
kilometres the temperature increases again. The increase is because the ozone gas which
is available there absorbs the heat energy radiating from the Sun and dissocites giving
heat.
But then you keep on going to yet higher and higher altitudes, say around 50 kilometres;
beyond that you know the amount of air available is negligibly small and therefore there
is nothing really to absorb the heat from the Sun. The temperature begins to rise again.
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Whereas, when you go to extremely high values around 80 to 85 kilometres and so on,
the individual atoms like oxygen atom combines with each other to generate heat and
again the temperature increases. But then the pressure monotonically falls and in
thermosphere there is hardly any pressure therefore, the concept of temperature no longer
holds good.
It is the individual molecules which are at high temperature. There is no continuum; we
cannot even talk of temperature. The rockets goes through the troposphere, stratosphere,
mesosphere into the thermosphere and further.
I think this must be clear to each one of us. Why does the temperature decreases above
the surface of the Earth, the height at which a jet aircraft flies is around 8 to 10
kilometres where the temperature is around minus 45 to minus 50 degree centigrade.
Then you have the stratosphere, mesosphere where the temperature of air again increases
and decrease.
What else did we do in last class?
(Refer Slide Time: 06:43)
We said that planets move in elliptical paths around the Sun and they trace out the equal
areas in equal times.
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(Refer Slide Time: 06:51)
But there are reports in the science magazines over the last two three months about
undulations which are there around the elliptical orbit. The green colours shows the real
orbit.
(Refer Slide Time: 07:11)
Even within the undulations there are further undulations like turbulence.
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(Refer Slide Time: 07:30)
And we had Johannes Kepler who proposed three laws describing the motion of the
palnets — laws for planetary motion. The first law states that all planets move in
elliptical orbit around the Sun; the second law said that equal areas are traced in equal
amount of time in this figure the sun is at one of the focus and therefore, you see the
focus on the left hand side around which those white patches show equal areas in equal
time. The third law which is law of harmonies related the time period t divided by R
namely, t2/R3 is a constant. Therefore you have very well synchronized motion of the
planets around the sun and therefore, this is useful. Newton formulated the universal law
for gravitation based on the planetary motion.
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(Refer Slide Time: 08:26)
How did he go about doing it? Before Newton we said that Johannes Kepler was in the
period 1570 to 1630 and then comes Galileo Galilee who was also interested in looking
at the planets; You see him in this slide on the left hand side gazing at the planets and he
is supposed to have done an experiments in which a feather and a steel ball fall together
under evacuated or vacuum conditions. Newton did not do any experiment but relied on
these data.
(Refer Slide Time: 09:02)
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He relied on these observations and it is stated that he sees an apple falling from a tree he
uses the information which Johannes Kepler gathered (mainly you have elliptical motion
you have equal areas you have t2/R3 is a constant) and then there was this person Robert
Hooke. You have all read about Hooke’s law in Mechanics (stress – strain) and Hooke
also was addressing the free fall of planets and their motion. Newton put everything
together and he said well an apple gets attracted by Earth because Earth is a large mass
just like the Sun which attracts the different planets and he formulates the Universal Law
for Gravitational Forces. Planets fall freely onto the Sun just like a stone falls on the
Earth.
(Refer Slide Time: 10:02)
To summarize again, we said that the Law states that if I have a heavy mass m1 as shown
over here and then I have a light mass m2 which is at a distance r away then I have the
attraction force with which the heavy body attracts the lighter body as F as equal to G
into m1m2 by R square. This is the universal law for gravitational forces. We said G has
units of Newton meter square by kilogram square; the unit is 6.670×10−11. But then we
were not clear and why this gravity force must exist. We talked of very powerful minds
like Stephen Hawking looking into the reasons or theoretical derivation for the Law. As a
lay person, we said that if I stretch a sheet and if I put a heavy mass over the stretched
sheet, the heavy mass deflects the sheet. If we now put a light mass, it rolls towards the
heavy mass. This we said could be due to the field from the heavy body. And we went
further; we expanded the value of the universal law F = Gm1m2/R2 in terms of the heavy
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mass being the Earth and R being the radius of the earth plus the altitude h and we found
that F= − mg.
But you know the gravitational field g is not constant all along the surface of the Earth
because Earth is little bit chubby in shape. It is not a pure sphere. Therefore you find that
with the angle of inclination or the latitude and the height, the gravity or the gravitational
field g keeps changing. The Universal Law F = G m1m2/R2 remains the same and it is the
gravitational field of a particular planet that varies. The units of g was derived as meter
per second square (m/s2) which why we call this as acceleration due to gravity.
(Refer Slide Time: 12:25)
The derivation of F = − mg from the universal law for gravitational forces is shown
above. It came out to be 9.8 m/s2. We had also derived this in the last class.
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(Refer Slide Time: 12:47)
We defined an inertial frame of reference since all of us are moving, the Earth and all the
galaxies are moving, the Sun is also moving. Therefore we said either the Frame of
Reference must be stationary or must move at constant velocity along a straight line for
the change in the in motion of the body to be determined.
The inertial frame of reference is a frame of reference which is either stationary or it
moves at constant linear velocity. Based on this inertial frame of reference we talk in
terms of Newton’s laws of motion which is precisely what we have been talking all along
namely inertia i.e., a body continues to remain in a state of rest or of uniform motion
unless it is constrained otherwise by an external force and this makes sense.
The second law of Newton we say is rate of momentum is proportional to force or force
is equal to rate of change of momentum. Force F is equal to d/dt (mv). You take constant
mass outside you have d/dt of v which is the acceleration. Therefore, second law of
Newton can be expressed either as a relation between force and change of momentum or
between mass and acceleration. We can state that acceleration goes inversely as the mass
of the body and directly as the force on a body.
We will be using these laws. The third law states that action and reaction are equal and
opposite; but mind you the important qualification is inertial frame of reference. These
laws are valid only if the frame of references is inertial. As an example, I want to find the
temperature distribution in a rotor in a machine; therefore, I am looking at the blade as it
58
is rotating; to be able to measure the temperature of a blade, I will be sitting on the blade
and monitoring the temperature. The rotating blade need not be in an inertial frame of
references. Therefore the momentum equation for a rotating body need not follow the
Newton’s law.
(Refer Slide Time: 15:17)
Now let me just get into this last slide corresponding to what we did in the last class. We
said to be able to put anything in space you need to give a force. Lucien, around 40 BC
fancied a giant storm pushing a ship to the moon. But when we looked at the Science
fiction author Jules Verne; he wrote about a capsule which is contained in the barrel of a
canon and forced through to space.
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(Refer Slide Time: 16:00)
We also talked in terms of asteroids which are loose bodies in space; one such asteroid
missed collision with the space station around two months back. The space station is a
satellite which is in low earth orbit and it is used for scientific experiments. Whenever
some asteroid is likely to hit it, the position of the space station slightly shifted so that
while in orbit the asteroid does not hit it. You could have these misses and collisions
with bodies in space
(Refer Slide Time: 16:36)
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But something which is disturbing and about which we talked in the last class is an
asteroid hitting the Earth. I took this slide from the Newspaper “The Hindu”. The
incident was in the news in February this year. It says that on April 13, 2036 one asteroid
by name Apophis is likely to hit the earth. And if this asteroid were to enter the
atmosphere of Earth and hit it, the asteroid would get rapidly heated due to friction; it
would explode and form a blast waves and the resulting severe wind and extremely high
temperatures might cause the end of civilization itself. And in fact some 6 million years
ago the extinction of dinosaurs was because of an impact of an asteroid with the Earth.
(Refer Slide Time: 16:59)
Therefore, the question is how to deflect away the asteroid away from the Earth. One of
the things which are talked of is to put up something like a like a heavy satellite near to
the asteroid and now the asteroid has a certain mass and because of the gravitational
force exerted by the heavy satellite, the asteroid slightly deflects away.
This summarises what we have discussed so far.
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(Refer Slide Time: 17:55)
If we were to consider an orbiting spacecraft about the Earth, i.e., revolving round the
Earth as it were; the bottom picture shows the INSAT (Indian National Satellite) satellite
going around the earth. If we throw a stone with a particular velocity the stone would
keep on rotating round and round the earth. It is like the case of the planets going round
the Sun. Now, the pull of gravity attracts it and causes it to fall freely while the forward
velocity component shifts it away from falling onto the earth and the body gets into a
circular orbit for a given value of the velocity.
(Refer Slide Time: 18:37)
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Now to describe the motion of the body as it is going around the Earth, the frame of
reference could be inertial like when I am standing on the ground and am watching the
body revolve around the Earth. I could also sit on the body and describe the motion in
which case the frame of reference for describing the motion is not along a line and is not
inertial.
Therefore, I would like to get back and ask a question; can I get out of this inertial frame
of reference and describe the motion with the frame of reference being the body?
(Refer Slide Time: 19:28)
Let me put the argument through with an example. Many of us go to watch circus and
you know one of the things which is sometimes shown in a circus is a motor cyclist or a
few of the motor cyclists going round and round in a spherical cage. How does he
support himself when he is on top of the cage in an inverted position? He does not fall
down. If I were to consider myself sitting along with the cyclist and I want to describe
my motion and that is how a motion of a satellite could be described in the frame of
reference of the rotating satellite. I would like to know what is the type of forces which
are acting on me and with this I should be able to correctly describe the motion of bodies
as they are rotating when I am on the body itself.
We say that in the frame of reference of the rotating body; it is not something which is
moving at constant velocity or which is stationary i.e. we have a non-inertial frame of
reference with which I would like to describe the motion of the body. In the inertial
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frame of reference, (with the observer on the ground), the body has a velocity dx/dt in
the x direction and dy/dt in the y direction.
(Refer Slide Time: 20:42)
We will address this problem of describing the motion in the rotating frame of reference
along with that we will be able to find out what we mean by orbital velocity.
(Refer Slide Time: 21:00)
Therefore, I have something like an orbiting mass something is going round and round
about the centre over here; alternatively I have the centre over here let us say a ball is
going round and round. It is in this rotating frame of reference rotating at a speed omega,
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that I am interested in finding out the equation of motion. Is there something different
that I have to do when I consider the rotating frame of reference? As long as I considered
myself to be stationary on the ground, the frame of reference was inertial and the
Newton’s laws of motions could be applied. Now I am rotating with the rotational
velocity omega; I am no longer looking at the body from the inertial frame of reference
since I am on the body; I want to look at the body whose cordinate I call as x’ which is a
rotational frame of reference. In other words I also sit on this body and I want to describe
my motion properly.
Let me say that instead of me standing here and watching this mass go round I tell
myself well I am on this body X’ and I want to describe my motion properly. In other
words you know my motion about the body is zero since I am on the body. With respect
to the body; my displacement or velocity of motion is dx’/ dt is 0. There is no change;
but how do I describe this motion?
Let us say that I move with an angular velocity ω and let us say at time t my position is
X’. At this particular point ωt is equal to the particular angular movement θ. I am here
at X’ at a time duration t. I have reached this particular point and therefore, I can write
the x coordinate and the y coordinate. Now x prime is equal to the radius of the circle R.
I can write x as equal to R cosωt and y is equal to R sinωt.
The x component of radius is x cosθ and the y component of the radius is y = R sinθ and
theta is equal to ωt. I differentiate it to get dx/dt is equal to − Rω sin ωt and I get dy/dt is
equal to Rω cosωt. Instead of taking the velocity along x and along y, if I want to find
the acceleration along x and y. I get d2x/dt2 = − ω2Rcosωt and the second equation I get
is d2y/dt2 = − ω2Rsinωt.
In other words when a body is rotating I find I can describe the acceleration along x is
pointed towards the centre; it is minus. It is in this direction equal to – ω2R cosωt. I also
have acceleration in the y direction is equal to − ω2R sinωt and the net acceleration is:
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(Refer Slide Time: 25:52)
If we were to write the net acceleration (a), we get it equal to the under root of these two
terms squared added together, namely ω2R. This is in the inertial frame of reference. The
sum of sin2ωt + cos2ωt = 1 and therefore this is equal to ω2R. This is the acceleration.
That means when the body is rotating, I have a net acceleration which is along the radius
pointing towards the centre O.
Now from second law we know if a rotating body has an acceleration in some direction,
it should have a force. The force is in the direction of the acceleration and equal – mω2R.
The force is towards the centre O.
What is the name of this force? Centripetal force.
What we find is we are considering a body which is rotating and we find that there is a
net force which act towards the center and this force is equal to mω2R. We see this in the
inertial frame of reference.
However, when I am sitting on this body, i.e., in the frame of reference of the rotating
body X’ itself, I am not moving with respect to the body. Since my movement with
respect to the body is zero, my velocity dx’/dt and my acceleration d2x’/dt2 is zero. This
means that the acceleration of the body in the frame of reference of the body is zero. But
then I find in the inertial frame of reference that I am talking of a centripetal force which
is acting on me. Therefore, if I have to describe my motion correctly, i.e., correctly sort
66
of predict my motion when I am on the body, what should I do? It is necessary for me to
put a force on the body equal and opposite to – mω2R. This force is not real but is one
which is required to correctly predict my motion in the rotating frame of reference. It is
something which is virtual or which we call as a pseudo force. And why do I have to put
this force? In order to be able to correctly describe the motion of the body in the context
of the body itself so that there is no net force because with respect the body; the body is
not moving. I am sitting with the body. If I say that I am not moving with respect to the
body, it is only possible when I put a force like this such that the centripetal force and
this pseudo force are same and opposite to each other. This pseudo force is what we call
as centrifugal force.
Let us repeat the requirement of the pseudo force in the non-inertial rotating frame of
reference. In an inertial frame of reference Newton’s laws are valid. Now I consider a
frame of reference which is not an inertial frame of reference, but it is a rotating frame of
reference. Now I am looking at the motion from the perspective of the body which
means is I am sitting on the body. If I am sitting on the body and I am rotating then
(Refer Slide Time: 30:30)
to be able to correctly define my coordinate, i.e., a pseudo force has to be put on me to be
able to correctly describe my motion, because I find that d2x’/dt2 is equal to 0.
In other words I have a centripetal force minus omega square R and I have to put a force
equal to plus mω2R to be able to correctly determine my motion. We call this as a pseudo
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force since it is not real, but it helps me to define my motion corectly in the in the noninertial frame of reference.
With this background, let us find out velocities in orbit.
(Refer Slide Time: 32:06)
Let us consider the Earth to be here let us consider a satellite or some other body of mass
m going round earth in a spherical orbit of radius R. Now, I want to find out the motion
of this particular body. We assume that the body rotates with an angular velocity ω. With
respect to the body, i.e., in the rotating frame of reference we need to introduce a pseudo
force which we call as centrifugal force to correctly define its motion in its own frame of
reference . This pseudo force is mω2R which is acting along the radius in the outward
direction.
But what could balance this pseudo force the application of which correctly defined its
motion?
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(Refer Slide Time: 33:39)
Gravitational force of attraction by the Earth of the body of mass m. The gravitational
forces balances the pseudo force. In other words the pseudo acts in this direction equal to
mω2R. And what is the force in the opposite direction? The gravitational force which is
given by universal law for gravitational forces and is equal to the mass of the body into
mass of the Earth divided by R2 into the gravitational constant G.
We take the force balance. I have the pseudo force or centrifugal force of mω2R equal to
G into m (mass of the body) multiplied by mass of Earth divided by the square of
distance from the centre of earth to this particular body. And now I find out the angular
velocity omega. m and m cancels you find that irrespective of mass of body, the value of
the angular velocity is the same and we get it equal to √GME/R3. This is what gives us
the angular velocity of rotation of a body around the Earth. If I were to consider instead
of the earth that I go around the Sun, let us write the equation for the angular velocity of
orbit.
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(Refer Slide Time: 35:04)
We have the sun of mass Ms over here I have a body of mass m going round the Sun at a
distance R from the centre of the Sun. Then I have in the frame of reference of the mass
m, the gravitational force that is F = GmMs/R2 and I have the pseudo force which is
equal to mω2R. Therefore, I get the angular velocity ω2 as equal to G into mass of Sun
divided by the cube of the distance from the canter of the Sun and the body. (G
Ms/R3)1/2.
Once I know the angular velocity I can readily find out velocity of orbit.
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(Refer Slide Time: 35:59)
The velocity of the orbit is the velocity with which it is rotating viz., V0. This equals Rω.
The orbital velocity becomes equal to √GME/R. The unit is metres per second. This is
how we calculate the orbital velocity of anybody like the moon is going around the Earth
we know the mass of the moon we know the mass of the Earth we know the distance
between centre of moon and centre of Earth. We can find out the orbital velocity i.e., the
velocity at which the moon is orbiting or travelling provided that the orbit is circular.
(Refer Slide Time: 37:12)
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Let us do a simple example to illustrate this. Let us find out the velocity of a spacecraft
orbiting the Earth in a circular orbit at a height let us say of 100 km above the Earth. I
want to find out velocity of the orbit V0. I make use of this equation V0 =√GME/R =
√6.670x10−11,the unit Newton meter square by kilogram square × the mass of the Earth
viz., 5.974×1024 kg divided by the radius R we said was hundred kilometre above the
surface of the Earth; therefore, R is equal to the radius of the Earth which is 6380
kilometre plus 100 kilometres which is equal to 6480 kilometres into 10 to the power 3
meters. I am left with kg multiplied by kg. Newton is equal to kilogram meter per
second square. Therefore, this becomes under root meter square by second square and
the unit is meter per second.
And if you calculate the value, this will come out to be about 7.76×103 meter per second
or 7.76 kilometres per second. This is typically the velocity of a body orbiting the earth
at a distance of 100 km above the surface of Earth. This is how we calculate the orbital
velocity.
Can we find anything elese from the expression for orbital velocities?
(Refer Slide Time: 39:47)
We find that as the height above the Earth increases the value of orbital velocity will
keep coming down. When I reach infinite height the orbital velocity is zero.
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In other words whenever we are considering any planet and a body is orbiting around a
planet, we can calculate the orbital velocity V0 in meters per second i.e., the orbital
velocity of the rotating body. We talked in terms of INSAT spacecraft which is going
round the earth and therefore we say well, it is at this height and therefore it is rotating at
this particular speed.
In the next class we will go into some details of the total velocity required to orbit a
spacecraft at a given height. You can see from this figure that the orbital velocity keeps
falling as you go higher up but then you expect more velocity should be required to go to
higher orbits. We will also go into more details of different orbits and then fix the
requirements for a rocket to be able to put a spacecraft into different orbits.
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Rocket Propulsion
Prof.K.Ramamurthi
Department of Mechanical engineering
Indian Institute of Technology, Madras
Lecture No. # 04
Velocity Requirements
(Refer Slide Time: 00:18)
Good morning. We will be looking at the orbital requirements and also will examine the
different orbits in the class today. To be able to do this, let us just briefly recap where we
were in the last class. We found that a body goes around the Earth, an object going
around the Earth for which we would like to have a frame of reference on this body itself.
That means, I am sitting on this body as it were and looking at my rotating body, it is not
that I am in an inertial frame where in, I sit here on the ground and watch this. But I have
what we said is the rotating frame of reference and what did we find? If I have a rotating
frame of reference it is necessary for me to put a fictitious force and this fictitious force
we called as a pseudo or virtual force: we call this pseudo force as centrifugal force. We
found that this centrifugal force is equal to mω2R, where ω is the rotational angular
velocity of the body as I am sitting on it and R is the radius from the centre over here. So
far so good.
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The mass of this object is m. You know this force was required to correctly predict the
motion of the body in the frame of reference of the body itself. How did this force arise?
Well, in the perspective of me sitting on the body the body is not moving and therefore, I
had to put a force to correctly define the motion of the body. Namely, we said X’ is the
coordinate of the body there is no change in X’ and therefore, we had d2x’/dt2 was equal
to 0. Using this pseudo force, we wrote an equation in which this is balanced, we said, by
the earth attracting this body due to the universal law of gravitational forces; we wrote it
as G into mass of the body into ME divided by R squared. ME is the mass of the earth. We
consider the product of mass of body and mass of Earth divided by R2 into the
gravitational constant G is equal to the psuedo force which is acting namely mω2R.
(Refer Slide Time: 03:18)
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And therefore we were able to get the angular velocity of rotation, ω = √GME/R3. So
many radians per second and this is what we did. We also went one step further wanted
to find out what is the velocity V0 of this object as it is rotating around. We told that V0
is equal to Rω and therefore, V0 = √GME/R. We had √R2 in the denominator
cancelling with the R in the numerator. If I express G in Newton metre square by
kilogram square, mass of the Earth in kilograms and R in metre, the unit we got was
metre per second. This is the orbital velocity.
Now, we go one step further. Supposing this body is rotating, what is the period of
rotation? What is meant by a period; time required to complete one rotation. I find that it
travels through 360 degrees or 2π radians and therefore, I say the period of rotation must
be 2πR is the total distance it travels divided by V0. Therefore, the distance travelled in
one orbit is equal to πd or 2πR divided by V0. And therefore, the period of rotation will
come out to be equal to 2πR divided by √GME/R. And what does that give you? The
period of rotation which let us call tau τ is equal to so many seconds. So many seconds is
the period of one rotation. Therefore, what is it you find from this? Let us do one or two
small examples. Let us find the period of rotation for two distances:
(Refer Slide Time: 06:00)
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Let us take the problem which we did the other day. When the height above the earth is
100 kilometres we found out the value of the orbital velocity is 7.4 kilometres per second.
I want to find out what is the period and therefore the period in seconds τ = 2πR divided
by the velocity V0. 6380 km is the radius of the earth. Let me write the radius RE as
radius of the earth is 6380 plus, I have 100 km as height above the Earth, so R is 6480
kilometres into 103 in meters divided by G the value of G the gravitational constant is
6.670×10−11 and the mass of the earth is equal to 5.974×1024 kg. This is the time taken to
complete one orbit at a distance of 100 kilometres above us. When I calculate the value I
find this comes out to be something like 5194 seconds or something like equal to 1.44
hours.
Instead of a circular orbit of 100 kilometres height I go to a higher orbit, instead of being
100 kilometres above the earth, I go to a distance let us say 50,000 kilometres above the
earth. And now, I ask what is the value of the time period? If I do the same set of
calculations, τ in seconds at a distance of h equal to 50,000 kilometres will give me a
much larger value. Again we put 2π into 6380 radius of Earth plus 50,000 in kilometres
into 103 divided by √ 6.670×10−11 into the mass of the earth which is equal to 5.974×1024
÷ the radius. And what does this come out to be? This comes out to be something like
35.7 hours. All what we are doing is, we wanted to find the time for going through
one orbit at different heights over the Earth.
At a height of 100 kilometres we find the period is something like an hour and odd and if
have the orbit which is at a height of 50000 km height, it is something like 35 hours. And
if I want to plot this what is it that we get? I keep this figure and erase this part.
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(Refer Slide Time: 09:23)
Now I plot the radius or height above the earth as a function of time period. On
the surface of earth which is something like 6300 km, it will be little bit lower than
at a height of 100 kilometres where we got a value equal to 1.4 hours or so. At a
height of 50,000 kilometres I got it as equal to something like 35.7 or 36 hours.
And the graph shows that the time for one orbit is R3/2. Therefore, the graph shows this
somewhere in between 1.4 hours and something like 35.7 hours. We have the
value corresponding to a single rotation of the earth and that is 24 hours. That means, I
have 24 hours is the period of rotation of the earth. How do you define the period of
rotation of the Earth? We define it as the time between when the sun appears vertically
overhead at a given time (midday) today and sun is vertical at the same time (midday)
tomorrow (the next day); that is the period and that we say is one day or 24 hours. This is
known as solar day.
But actually what is happening? You have the Sun at the centre of the solar system, you
have the Earth rotating in an elliptical fashion and the earth is also revolving on its axis as
it is rotating around the Sun. Therefore, the solar day is going to be different from the
actual rotation time. The period of one rotation is going to be slightly less than 24 hours
and therefore, because it is revolving and as it is moving it is rotating and therefore, the
period cannot be 24 hours.
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(Refer Slide Time: 11:38)
And therefore we define a sidereal day instead of a solar day; that means, 24 hours
is the period corresponding to solar day while the real actual time taken for a
rotation which we call as sidereal day is slightly less; it is something like 23 hours 56
minutes and 4.1 seconds. Therefore, whenever we say one rotation; what is
happening is the Earth is rotating on its axis and as it is rotating it is also revolving
and therefore, one rotation corresponds to not one solar day. Therefore one works
with what we call as a sidereal day which is, 23 hours 56 minutes and this works
out to be something like equal to 86164.1 second.
I do not think we are going to get into that depth of trying to find out the
difference between the solar day and the sidereal day and we will assume that the
Earth rotates once in 24 hours. And having said that you know somewhere in between
36 hours and 1.4 hours we will find that you have the time of one orbit as 24 hours.
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(Refer Slide Time: 13:01)
And at this point when the orbital time is 24 hours what is going to happen? You know
we showed the Earth over here; the Earth rotates on its axis and you have the body.
When the body or the satellite moves with respect to the earth and the rotation of the
body with respect to the earth is the same, in other words the time taken for one
revolution let us say of the body is on this plane the body is moving, let us say from east
to west and the time taken for one orbit as it goes around is one day which is the same
rate at which the Earth is rotating on its axis. In other words the period of rotation of
this body is synchronous with the rotation of the earth and we call this orbit as
geosynchronous orbit.
In other words the Earth rotates from east to west and if the body rotates in a plane
which also rotates once per day we say that the rotation of this orbiting object or body
or satellite is synchronous with respect to the Earth, we call it as geosynchronous orbit.
If now the body is rotating on the east west axis namely on the equatorial plane of the
Earth and the period is 24 hours just the same way as the Earth rotates in 24 hours, then
any point on the surface of the Earth since the Earth is also rotating once in 24 hours
and this is rotating in the equatorial plane at the same rate as once in the 24 hours, the
satellite will always appear stationary at all points on the surface of the Earth. And such
an orbit is known as geostationary orbit. We call is as geostationary because for all
points on the surface of the Earth, the satellite appears stationary. But if by chance the
orbit is not along the equatorial plane, but it is in some other latitude or in some other
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plane, we call it simply as geosynchronous, but not geostationary. Therefore, the
distinction between geostationary and geosynchronous is that the geostationary is in the
equatorial plane having a rotational period same as the rotational period of the Earth
whereas geosynchronous could be at any other plane with a period of 24 hours.
Having understood this, it is not necessary that, we should have Geosynchronous and
geostationary orbits only for the Earth. It is possible to have the stationary and
synchronous orbits for any other planet so long as the planet is rotating at a reasonable
speed. Any other heavenly body if it is rotating and the satellite or object is moving at
the same rate as the rotation of the given planet, we could have a stationary or
synchronous orbit.
(Refer Slide Time: 16:30)
For instance, with planet Mars we could have a synchronous and a stationary orbit.. You
know, all what I am I am trying to say is the value of the height and I call this height
plus the radius of Earth as radius of geosynchronous orbit which is equal to the radius of
the Earth plus the height of the orbit at which, the period is going to be 24 hours. I also
qualify by saying 24 hours is the solar day and I must distinguish it from the sidereal
day which is slightly smaller, which is what you must be actually using. May be for a
person who works on the mission he must take into consideration the sidereal day of 23
hours and may be 56 minutes and 4.1 second.
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(Refer Slide Time: 18:00)
Now, I want to find out the height of the geosynchronous orbit or geostationary orbit?
The difference between the two is in the plane of the equator (equatorial plane) for
geostationary whereas, geosynchronous could be in any plane. We do the same thing as
before viz., rotational period of the Earth is through 360 degrees that is 2π radians in 24
hours that is, 60 into 60 seconds is the angular velocity and what is the angular velocity at
a height which will be RG; that means, I am considering the Earth over here I am
considering orbits going round over here may be on the equatorial plane as it were. I am
considering the height as hg and the radius if I consider the radius say RG. RG is equal to
the radius of the earth plus the value of h corresponding to the geostationary orbit. And
therefore, we have the expression for ω corresponding to 24 hours for one orbit.
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(Refer Slide Time: 19:30)
I get √G ME /RG3 cube is equal to 2π divided by 24 into 60 into 60. And you know the
value of the gravitational constant G is 6.670×10−11. We have the mass of the earth as
5.974×1024 kg and therefore, if now I find the value of RG, it works out to be something
like 42164 kilometres. To summarise, we had 2πR as the distance, we had mω2R which
is equal to the psuedo force and we said it is balanced by the by the gravitational force of
attraction due to the Earth and that is how we got this expression. And therefore, we got
the angular rotation of the orbit is this much and the angular rotation of the Earth which
was equal to it goes to 360 degrees or 2 pi radians in the same time and therefore, we get
the distance at which the period of rotation of the object and the period of rotation of the
Earth are same as equal to 42164 km.
Now the height of the geosynchronous orbit is therefore, equal to RG minus RE (the radius
of the Earth). RE we said is 6378 kilometres. Rather the height at which a space craft
appears stationary is equal to 42164 minus 6378 which is equal to 35786 kilometres.
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(Refer Slide Time: 22:10)
Let us draw the Earth and look at the rotation of the Earth on its axis and the rotation of
the spacecraft. The earth is rotating once in 24 hours the angular velocity of rotation is 2π
divided by 24 into 60 into 60 radians per second. And may be the period at which if we
are at a height above the Earth which is 35800 km, the period of rotation of the spacecraft
and the period of rotation of the Earth are the same. If further, the orbit is in the
equatorial plane, the spacecraft will appear to be stationary.
This concept was not developed by rocket engineers or people working in space missions,
but was told by the famous science fiction author Arthur Clark. Arthur Clark has been
writing a lot of science fiction books. In fact, he settled down in Srilanka. He passed
away several years ago, He was a very prolific writer. And he proposed this orbit in the
year around 1945 and therefore, the geostationary orbit is also referred to as Clark orbit
after him. Therefore, let me just repeat this yet again, once again. All what we are telling
is that we have the earth rotating east to west once in 24 hours.
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(Refer Slide Time: 23:43)
And now if I have a plane which is on the equator, equatorial plane and in this plane we
have an orbit - a circular orbit- in which a spacecraft rotates once in 24 hours, the height
that we are talking is something like 35786 kilometres. And at this point the rotating
object will appears stationary. In fact, this was recognised and it has been the effort to
have such orbits for communication satellites. The first geostationary satellite; that was
developed was known as Sincom 2. It was launched by US in 1963; I think July 26, 1963.
Wee had this geosynchronous satellite there at that height may be looking at the Earth
always there and that was the year when Tokyo Olympics were held and it was the first
time we had live TV communication from Tokyo and people could watch it live in the
US and may be in some other countries also. Therefore, the geostationary satellite is one
for which the period of a single rotation is same as the period of the rotation of the earth
on its axis and the orbit is in the equatorial plane. Therefore, what we have done so far is
we started with the velocity of the orbit. Let us just again put down the equations clearly
such that we are very clear.
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(Refer Slide Time: 25:24)
The velocity of the orbit V0 is equal to √GME/R. The angular rotation of a satellite in
orbit is equal to √GME/R3. When we are looking at the period of rotation, when the
period of rotation of the Earth on its axis and this value is the same, we get the value of
RG as equal to 42000 or the height above the earth as equal to something like 35800 km.
And this was postulated by the science fiction author Arthur Clark; it is also known as the
Clark orbit. Now, many countries have their geostationary satellites. In India we have
INSAT satellites. The first one was from US viz., Syncom in the year 1963. Having seen
the geostationary orbits let us go to some other orbits
(Refer Slide Time: 26:30)
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Let us see if we can have a satellite which goes from north to south; that means, this is
the Earth, with the north pole, the south pole, the satellite to go round from north to
south. And there is one advantage, if a satellite can go round like this in a circular orbit
between the poles. We are still talking of circular orbits; the Earth, this is the east and
west, earth is rotating like this. If we have the satellite which goes round and round
between the poles, as the earth rotates this particular satellite can see all parts of the earth.
This orbit going from pole to pole that is between the north pole to south pole, is known
as polar orbit. The equations are exactly identical. Supposing, I want to put it at a
distance R from the centre of the earth, I calculate the value of ω and I find out the period
or I calculate the value of the velocity of the orbit.
Therefore, we have polar orbit. But then, the polar orbit is really not 90o to the equatorial
plane because the Earth is little chubby not really a sphere. Supposing you want the
sunrays to come; that means, we are talking of the imaginary line between the centre of
the earth to the centre of the sun, this is the Earth to Sun line or axis and we have the
orbital plane over here. If the angle between the orbital plane and the imaginary centre
line of between the Sun and Earth is kept constant we call the orbit it as sun synchronous
polar orbit. Why do we need this? You know if the line joining the centre of the Earth
and Sun here and the polar orbit makes the same angle, the intensity of the sunlight which
the satellite sees on the Earth will be the same. And therefore, I can compare the reading
of what I take today and may be some other day and this is known as sun synchronous
polar orbit. And we have in our country, Indian Remote Sensing satellites (IRS) which
keep sensing or keep looking at the Earth.
Now let us divert our attention a little bit more and ask ourselves, why all these different
orbits? You know if I say geosynchronous orbit that means, I have the Earth as it were
traveling from east west and the object or satellite at a height of 35800 km. You know the
satellite is always appearing stationary with respect to the Earth if it is in the equatorial
plane. If on a clear day, we go out at night and we can see the satellite: we can see
INSAT geostationary satellite. And when I look at it, I will see the stars and all that you
know we said the stars are in a state of continuous motion. We will see the stars and other
heavenly objects drifting across, but this satellite will be dead stationary. That is because
both are rotating at the same speed. And because it is stationary may be I have the INSAT
satellite pointing may be towards the centre of India may be at Nagpur and it is able to
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cover the entire Indian sub-continent and it is able to provide communication, telephony
may be TV programs and other services.
When I talk in terms of polar orbits and talk in terms of sensing the earth why do we need
this? You know, supposing, we say some crop is grown in some part of Andhra Pradesh
and I want to find out if the crop is healthy or not. I can think in terms of crop is not
healthy then, it withers. The frequency what I see, or the colour what I see is going to be
different. I can find out from this particular satellite the nature of the crop and I can warn
the people, look here, your crop is not doing well or may be if somebody wants to catch
fish; fish always prefer to be in the ocean when the temperatures are little higher may be
I monitor the temperature of the ocean and give a message to the to the fisherman to go to
the warm waters and catch fish. And that is how you use the remote sensing polar orbit
and which is again if it is sun synchronous I will get the same illumination, I will be able
to compare the data obtained on different days and we talk therefore, in terms of may be
polar orbit and a geostationary orbit – these are the two major orbits.
(Refer Slide Time: 31:10)
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Let us go one step further. We could have a low earth orbit around the Earth we could
have something like medium Earth orbit. In other words, I have the Earth may be I go to
low altitudes. The low earth orbits are normally used for scientific studies. You all would
have heard of scientific studies being conducted and what are the scientific studies about?
May be we talk in terms of troposphere, we talk in terms of stratosphere. In
this stratosphere we have the wind, you would like to find out the wind velocity. We
were also told there are some charged particles which are available in the stratosphere. I
would like to measure some of these things and that is the way lower orbit is used. But
there is a limit to the height of the lower orbit. Typically, it has to be greater than
200 to 300 kilometres otherwise, the air or atmosphere which is there will cause drag on
the satellite.
Therefore, can I assume that we are fairly clear at this point in time relating to
circular orbits above the earth or for that matter why should it be Earth alone? If I go and
have a stationary satellite about Jupiter, it should be on the same lines. I take the mass of
Jupiter, I take the radius of Jupiter and I can find out at what height I must have.
Therefore, I think at this point let us ask ourselves, are there any other orbits other
than circular orbits? You will recall when we looked at the orbital velocities of the
planets around the Sun, we said all orbits are elliptical. What is the difference between
a circular orbit and an elliptical orbit? In other words, instead of the earth being here, let
us say smaller, if I say its circular, if I say it is elliptical, may be I am talking in
terms of elliptical, something like this. In other words, how do we define an ellipse?
We define something like a foci 1 and 2 and you have major axis which is equal to 2a
and a minor axis which is equal to 2b. Therefore, we have, may be Sun at a focal point.
We told that the Sun is at the foci and Earth is rotating around it. Similarly, if I have the
Earth here and the satellite is in elliptical orbit, I have Earth as the foci and the satellite
going around this elliptical orbit.
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(Refer Slide Time: 34:01)
Therefore, you define two terminologies; eccentricity of elliptic orbit, this is the distance
between the 2 foci Lc. That means, this is the Lc divided by the major axis 2a. And you
find for a circular orbit, Lc is 0 because I have a centre here and the eccentricity becomes
equal to 0 for a circular orbit.
There is another point which we must keep in mind and that is the orbit need not always
be east to west, or need not always be polar; we could have in between. May be an orbit
could be like this; may be at an inclination this is the orbital plane, this is the east west, I
say this is the inclination θ; that means, we say the angle between the orbital plane and
the equatorial plane is what we call the inclination of the orbit.
Let me take you through an example:
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(Refer Slide Time: 35:39)
We take a country like Russia. And this particular country is in the northern hemisphere
and if I were to have something like an equatorial orbit; they are not able to see the
satellite distinctly over a sufficient time. And supposing if I have something like an
elliptical orbit at an angle of something like may be like 63.4 degrees. I will come back to
this particular inclination a little later. And then I have an orbit which has a longer travel
distance here at the northern end and a much smaller distance here at the southern end
that is, the orbit is something like elliptical orbit. I find that the space craft or the object
which is rotating, spends much more time in the northern latitude and correspondingly
smaller time in the southern latitude. And what was the second law of Joahanes Kepler?
Equal area swept in the same time. And therefore, it spends much more time in the
northern hemisphere; the satellite can spend something like 23 hours out of 24 hours.
And this particular type of highly elliptical orbits is what the Russians call as Molniya
orbit.
It has an inclination of something like 63.4o. The distance to the top most point from the
centre is 46000 kilometres and the distance from the centre to the nearest point on the
orbit is something like only 6800. That is, you see the distance between the centre of the
Earth and the furthest point from the Earth which is over here, and this is the furthest
point away from the Earth over here. The distance between the centre of Earth and the
nearest point is what we call as perigee; that means, this is the point which is nearest to
the Earth. The point furthest away from the Earth is known as apogee.
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Therefore, in an elliptical orbit we also define something known as perigee and apogee.
Perigee is something which is the orbital distance nearest to the centre of the Earth and
apogee is the farthest distance from the Earth. These are for the case of orbits which are
elliptical.
To summarize; we have to put a psuedo force to balance the attractive or universal law
for attraction of the object to the centre of the planet and then figure out the radius of the
orbit. But something which we have really not done is about orbital velocity. What is this
orbital velocity?
(Refer Slide Time: 38:52)
We found that if we plotted orbital velocity V0 (mind you we did it in the last class) and
we had the orbital velocity V0 in metres per second equal to √GME/R. We find, if the
radius is infinite the orbital velocity is 0. If the distance from the centre of the Earth
increases, the orbital velocity keeps falling and reaches 0 as the distance goes to infinity.
Why should the velocity of the orbit goes to 0 at infinity? How would you explain? Is
there any suggestion ? Why should it be 0? Any thinking on it.?
We did tell that, supposing the spacecraft were to leave the Earth i.e., escape from the
Earth; well it has to go to infinity. Therefore, when I talk of infinity you have no
attractive force due to the Earth and therefore, it is not in orbit anymore and therefore,
you find that the orbital velocity keeps decreasing and ultimately becomes zero.
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Therefore, the orbital velocity V0 keeps decreasing as I increase the height because as I
increase the height above the Earth the attraction by the Earth keeps decreasing. Now the
next question is, from the surface of the earth you have to go to this height which we
have not yet considered. That means, I require a certain velocity or potential energy to be
given to reach this point or I have to do some work in taking a mass from the surface of
the earth to go to the particular orbit at a distance let us say R or I have to increase the
height from the surface of the earth by h. How do I include this? See in other words, so
far I have talked only of the orbital velocity V0: we have not considered how much
velocity is required to start from the surface of the Earth and go to this height and then
provide the necessary orbital velocity.
Therefore, let us now find out what is the total velocity required for orbiting viz., taking
the space craft from the surface of the Earth or some other planet to the particular radius
of the orbit and then injecting it with the given orbital velocity. How do I do it? If we
have understood what is discussed so far, we must be able to do it very readily.
(Refer Slide Time: 41:27)
Let us try to do it. We want the total velocity which includes the orbital velocity plus the
velocity required to take the object from the surface of the Earth let us say at RE to the
particular orbit at R. How do I determine it? I have the Earth here; I want to take the
spacecraft above the earth to a height h. And I have to insert into orbit for which I have
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give an orbital velocity and then it will continuously rotate. Therefore, what is the total
velocity which I must give? How do I determine it using the same set of equations?
(Refer Slide Time: 42:29)
We talk in terms of the universal law of gravitational forces. We ask ourselves what is the
work required to be done to increase the height of the spacecraft or to take the spacecraft
or the object from RE to R. Let us consider the Earth and what I am asking is I have the
earth here, I have RE here, I want to take the spacecraft of mass m from RE to RE + height
h. Let us say this final distance is R and this height is h how do I do this? What is the
work required to be done. What is the potential energy required to take this spacecraft
from the surface of the earth to a height h. How do I do it? Any suggestions on how will
you solve this problem? I want the work which is required and work is equal to force
into distance. What is the force? Yes. The force to be overcome is the gravitational force
of attraction due to the Earth.
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(Refer Slide Time: 44:00)
Therefore, let us put it together. We find that the force is equal to G × mass of the earth
ME × mass of the body m divided by at any radius R2. What is the value of the work
done, when it moves through a small distance let us say dr. We assume that it goes from a
height R to a height R plus dr. What is the small work which is required to be done? We
call this small work as dW and is equal to the force into the small displacement dr and
therefore, dW must be equal to GmME/R2 × dr. What is the work required as I go from
may be the surface of the earth having radius RE to a height R? I just have to integrate out
or the total work required as equal to integral from the surface of the earth to the radius
R. But mind you; let me qualify again that we are illustrating with respect to the earth. I
could have all these things for any planet or body. Supposing somebody wants to go from
the surface of the moon to some height. I just have to take the mass of the moon the
radius of the moon plus the particular height is what is to be considered. Therefore, I have
GmME/R2 ×dr and integrate it from the value RE to R as I go through the height h. And
this work must be equal to the potential energy available at the height h because you have
increasing the height by h, Therefore, I have higher value of potential energy. But then,
we also know that we need to provide orbital velocity V0; that means, I have to give some
kinetic energy and what is the total energy of an orbiting satellite at a height h? I have,
this potential energy plus I have the kinetic energy.
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(Refer Slide Time: 46:06)
And therefore, I can say that total energy of a rotating spacecraft in orbit must be equal to
integral RE to R of GmME/R2 ×d r plus the kinetic energy of a rotating body equal to ½
mV02 , where V0 is the orbital velocity. And what is V0 square? We have derived it as
√GME/R. And therefore, we can say that the total energy in orbit is therefore equal to: let
us integrate this out, mass of the object is constant, mass of the earth is constant
therefore, we have G gravitational constant m ME into we have integral RE to the value of
R of dr by R2 plus I have ½ m into V02 viz., GME by the particular orbital radius R. Is it
all right?
And if I get started by giving a kinetic energy that is I supply some kinetic energy to this
satellite and I call it as the total velocity VT: this must be equal to ½ m into VT square,
that is the total velocity I give to the spacecraft. This total kinetic energy must equal
potential energy required to increase the height plus the kinetic energy due to orbit. And
therefore, the total velocity what I give to the spacecraft can be derived. Let us do it. m
cancels out over here.
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(Refer Slide Time: 48:04)
And therefore, we get VT square is equal to lets us get this here; half VT square by 2 is
equal to GME into 1 by RE minus 1 by R plus half of GME by R. Therefore, this become
with R equal to RE + h and add the two terms to give one over RE minus 1 over two R.
We substitute R as equal to the radius of the earth RE + height h. Now, I have taking 2RE
outside, the term within brackets as 1 minus RE by 2R. With R equal to RE plus h, we get
RE plus h in the denominator and RE plus 2 h in the numerator. Therefore the total
velocity VT square divided by 2, 2 and 2 gets cancelled and therefore, the total velocity
VT2 = GME/RE × [(RE +2h)/(RE +h)).
(Refer Slide Time: 50:28)
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When the spacecraft is orbiting at a height h above, the total velocity required to be
provided is given by VT equal to GME/RE × (RE +2 h) ÷ (RE + h). Now, what is it we
find? While the orbital velocity keeps decreasing as the height above the earth increases,
we find the total velocity is added by 2h in the numerator and added by h in the
denominator.
The numerator being multiplied by 2h, implies that the value in the numerator increases
faster as compared to the denominator as h increases and the total velocity therefore
increases as h increases. That means, as I go higher and higher I need to give higher
values of total velocities to the spacecraft which is going to be much higher than the
orbital velocity and this difference is what constitutes the potential energy required. Is it
alright?
In the next class we will continue with some problems involving the potential energy and
kinetic energies and solve for different orbits.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 05
Theory of Rocket Propulsion
(Refer Slide Time: 00:20)
In the class today, we will look at Orbit velocities and illustrate it with some examples.
Since you mention something about Escape Velocities, we will see that it is the velocity
required to escape from let say the Earth or from some other planet. Let us discuss
escape velocities first. I have the Earth here and I want to escape from this Earth. That
means, I want the force with which the Earth is attracting me to vanish; that means I
escape. In other words, the gravitational force is given by GMME/R2 in which ME is the
mass of Earth, m is the mass of the body and is divided by R square; This means that R
must become very very large or rather infinity for which this force would vanish.
In other words, I am looking for a body to escape from the surface of Earth i.e., radius
RE; I want to go to infinity to escape from the Earth and I have to determine the
corresponding velocities; and that velocity required to be provided and becomes the
Escape velocity. Therefore, escape velocity is the velocity required such that I escape
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from the attractive force or gravitational field of the Earth. Therefore, what must be the
value of R to escape? I want the force to be zero. I want to escape from the attraction;
therefore, R must be infinity in order to have the Escape velocity. How do I do this
problem?
(Refer Slide Time: 02:11)
I am looking at the force, which is equal to the mass of the body m, mass of the earth ME
divided by R square into G, and what is the work which I must do to take it from to
infinity. The work, which I must do when I travel a small distance dR for a radius R is
equal to GM ME/R2 × dR, is the small amount of work when the distance traveled is dR.
And now I want to escape from let us say from the surface of the earth having radius RE;
I want to go to infinity and therefore, this must be the work that is done. And how do I
do this work? I give the kinetic energy to the body therefore, I give ½ m ×V escape
square; and it is this velocity which is the escape velocity.
And let us find out what this is? We again the find that mass of the body cancels out; and
when we say GME/ R2 square, which is again equal to GME and here I write the value of
distance going from RE to infinity. You see this small increment in the radius dR.
Integrating, we get minus of one over RE and this become equal to GME by RE. And what
is the value I get for (V escape)2 divided by 2?
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(Refer Slide Time: 03:56)
We get the Escape velocity VE2 is equal to 2 G ME by the value of the radius of the Earth
RE. In fact it is very interesting to note that when the Earth was born, we had lot of
hydrogen, which was available around the earth. Hydrogen is the very light gas and
when the gas is light we shall see later on that a lighter gas provides much higher
velocities for a given value of energy when we get into theory of Rocket propulsion a
little later. The velocity of hydrogen is greater when it is hot and the Earth was hot when
it was formed. Since the hot hydrogen moves at high velocities, which is greater than the
escape velocity, we lost the hydrogen. Anything which travels at a velocity greater than
the escape velocity escapes from the surface of the earth or more correctly from the
gravitational field of the Earth.
Let us calculate the value of this Escape Velocity on the surface of Earth. Escape
velocity from the surface of earth is equal to √2 ×gravitational constant 6.670 ×10−11
Newton meter square by kilogram square × mass of the earth 5.974 ×1024 kilogram ÷ the
radius of the earth 6380 into 103 meters. And this is the Escape velocity from the surface
of the earth. You calculate it to be something like 11.17 kilometers per second.
Supposing you want to go to the moon you have to escape from the Earth’s gravitational
field; that means, you need to have the Escape velocity to get out of the Earth’s
gravitational field, then we get into the gravitational field of the moon. And supposing I
want to come back, I need to be provided with the Escape velocity to leave the
gravitational field of the moon and enter the gravitational field of Earth.
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And this is how we work out the total orbital requirements or the total velocity
requirements to put a spacecraft or any body to go to different planets.
(Refer Slide Time: 06:57)
Let me take one or two small examples. We will do these one or two small problems
such that we are very clear. However, before doing that, I also want to tell something
about freely falling bodies. What do we mean by freely falling bodies? What did we say
earlier? We have the eight planets which are going around the Sun in elliptical orbits. We
told that all these planets are just falling freely on to the surface of the Sun, falling
towards the Sun, but why is it not just crashing; because by the time its falls it goes
through some distance corresponding to the orbital velocity and again it goes through
some distance as it falls, it is always falling toward the center of the Sun but never
reaches the sun. Therefore, all planets and all of us are freely falling on to the Sun as it
is.
And how does a spacecraft orbit the Earth? I say this is the Earth; let us for ease consider
something like a circular orbit. As the spacecraft orbits, it is going at a constant linear
speed; however, it tends to fall towards the center of the Earth as it goes horizontally
because of the orbital velocity. It therefore goes horizontally as shown but it falls in the
process. Therefore, all the bodies which are in orbit are freely falling bodies. It is as good
as I drop a stone and it falls freely. So, also all the bodies, which are in orbit are freely
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falling objects, and what is the function or what is the thing, which we understood by
freely falling bodies.
(Refer Slide Time: 08:32)
Let us take two examples: in one a stone falls freely to the surface of Earth; it falls freely,
but let us assume that there is no resistance due to air, because we are talking of space.
Therefore, it just keeps falling freely. I also take an example of an elevator or lift, and let
us say the elevator falls freely. Let us take an example, I am in the elevator which is
falling freely, I am going down, and I am holding in my hand a cup of tea; we all would
have noted this. Now, what happens to the teacup which I am holding? What sensation
do I have?
This stone is freely falling. The frame of reference of the stone is not in an inertial frame
of reference, because it is picking up acceleration. Therefore, it is something like a
linearly accelerating frame of reference; It is different from the rotational frame of
reference which we considered while deriving the orbital velocities.
Now if I am sitting on the stone or if I am standing the lift with a tea cup in my hand
what is the force which I will experience or which the tea cup will experience? Can
somebody tell me? I have the Earth attracting me and therefore I have to move. But I am
on the stone or I am in the elevator and I cannot move with respect to the stone or with
respect to the elevator. And therefore, I have to correct my motion if I were to refer my
motion with respect to the stone or with reference to the elevator, because it is not in an
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inertial frame of reference. I have to put some corrective force. May be the stone is freely
falling; I have to put some force here because I am not moving at all with respect to the
stone. Therefore, I have to put a pseudo force opposite to the attractive force of the Earth
to describe my state of motion correctly with respect to the stone. Attractive force of the
earth is GMEm by R2, and I have to put this pseudo force here, which is equal to the
above force. With this correction I do not move in the given frame of reference.
And the moment I put a pseudo force over here my motion is taken care of I am able to
describe my motion correctively because I am not moving with respect to the stone. I am
not moving with respect to this lift which is correct; but when I put a force equal and
opposite to the force with which I am getting attracted, the net force on me become zero.
And when the force on me becomes zero, I am weightless, or I am in a state of
weightlessness. Why is it? It is not that I have lost my mass. I have my mass, but to be
able to correctively define my motion with respect to the stone or elevator , because I am
dropping along with them, I am sitting on the stone; the stone is coming down, but I have
to correct my motion because I am not moving with respect to the frame of reference of
this stone. Therefore, I need to put the pseudo force, vertical and opposite to the
gravitational force of the Earth so that I am not moving with respect to the stone on
which I am sitting.
Therefore, the moment I put the pseudo force, I do not have any force or weight, I am in
a state of weightlessness when I consider my reference to be the stone or the elevator.
Some people call it as zero “g”; actually it is not zero g; g is the gravitational field; the
gravitational field is always there, but a body in orbit which is also a freely falling body,
is in a state of weightlessness.
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(Refer Slide Time: 12:16)
That is a freely falling body is in a state of weightlessness. It does not seem to have any
weight. And so also if in the case of the lift or elevator coming down I hold a tea cup and
come down, you will feel it is not heavy at all, it is as if it is very light in the
confinement of the elevator as it is descending. I have to correct my motion using a
pseudo force, and that is why whenever you see the picture of astronauts in space, you
see that they are all floating around with respect to the space capsule they are in. It is
because you have to correct their motion with respect to the space capsule by a pseudo
force which makes them appear weightless. And you know to be able to drink a cup of
water while I am orbiting up in space, supposing I were to go there and I am very thirsty,
is going to be difficult. The water does not settle to the bottom of the cup or tumbler.
Therefore, it will be freely floating; that means, I have water, but it will just be floating,
therefore what has to be done is to arrest it somewhere, then may be put a straw and suck
it through then only even drinking little bit of water in space while orbiting is possible.
This is what we call as a state of weightlessness or some people call it as zero “g”.
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(Refer Slide Time: 13:57
But let us not confuse: g is always there; in fact since, many of you are from mechanical
engineering and working in combustion, if you look at a candle flame; the flame rises
because of convection and it is like this. There are some experiments done in a space
capsule while in orbit. What does the flame look in an environment of weightlessness?
Any guesses on what should be the shape of the flame?
The rising flame is because the gas becomes lighter on being heated and the lighter gas
rises and you have the candle flame like this on the ground. If I were to look at it in a
spacecraft in which the objects are in a state of weightlessness, there is nothing to rise up
and the candle flame must be a pure sphere. It does not have any light weight or strong
weights rising up or coming down; it just is a perfect sphere and that is what my
equations, in the absence of the gravity term, give. I can solve for it in a spherical frame
of reference.
I can match my solution for the diffusion and chemical reaction equations and for the
energy release equation in the state of weightlessness by neglecting the gravitational
field. I am able to find out what is the mechanism of diffusion flame which is different
from an experiment on the ground where gravity influences the shape and properties.
And, may be later on, I will show some pictures to show how may be an astronaut drink
water in space, how does a plant and how does a flower look like when grown in an
orbiting spacecraft up in space or how it will be different from that on Earth.
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Therefore, what is it we have done so far? Let us quickly summarize and do one or two
small problems, which will make sure that we have understood this subject.
(Refer Slide Time: 15:44)
We talked in terms of orbits of the different planet in our solar system. We went ahead,
we formulated the universal law for gravitation as determine by Newton. We told that all
planets are freely falling just as an apple is falling from the tree to the ground. A heavier
body attracts a lighter body and you have the universal gravitational law. We used the
gravitational law and we found out the orbit velocity V0 is equal to √the square root of
GME/ R, where R is the orbit radius. We also found out that the total velocity required to
orbit for a circular orbit is equal to √GME/RE ×(RE + 2 h) ÷ (RE + h).
We also talked of geo-synchronous orbit, polar orbit which is used for remote sensing, so
that I can see entire Earth as the spacecraft rotates; for communication and weather
prediction may be geo-synchronous is better suited. I can also have low Earth orbit
around the Earth and we found out the total velocity requirements for orbits at different
heights. We also talked in terms of the escape velocity, we said it is equal to √2GME/RE
from the surface of the Earth. If we have a spacecraft which is orbiting at its distance R
from the center of the earth and I want to push it to infinity then in this case R has to be
substituted in place of RE; and this is all what we have done so far.
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(Refer Slide Time: 17:33)
Let us first take a look at this power point presentation. Here, we see the low Earth orbit
i.e., a body going around the Earth as it circles around it.
(Refer Slide Time: 17:45)
This shows the geostationary orbit, you have the orbit in the equatorial plane of the Earth
i.e. along the East to West along the equator and you find that the spacecraft is going
around at a radius of 42,164 kilometers; this is the Earth and this dotted line is the orbit.
You subtract the radius of the earth from the radius of the orbit and that is the height of
the stationary orbit.
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(Refer Slide Time: 18:12)
Geostationary, this again shows the satellite in equatorial plane going around east to
west.
(Refer Slide Time: 18:23)
And supposing we have an orbit let us say instead of going from east to west, I go from
west to east. I am going against the rotation of the earth. The orbit is no longer
synchronous and such orbits are known as retrograde orbits. It is not useful at all because
why should I go against the rotation and not get any benefit at all.
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(Refer Slide Time: 18:52)
Let us go to the next one; this show Syncom 2, the first geostationary satellite, which
was launched by US. Syncom 1 was not successful and the second one was successful.
The launch was on 26 July 1963, and Syncom 2 was used to relay the Tokyo Olympics.
(Refer Slide Time: 19:09)
This shows the polar orbit; the orbit is in north – south direction. The inclination to the
equatorial plane is not exactly 90o , but little more than 90o.
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(Refer Slide Time: 19:24)
This shows the highly elliptical orbits; this is the perigee, this is the apogee. This
distance we set for the apogee is something like 42000 km; the perigee is of the order of
6000 km, and this is an elliptical orbit.
(Refer Slide Time: 19:39)
And now that you know about orbits and we want to launch satellites in a given orbit,
say into geostationary orbit, we first take off from the ground, we put the spacecraft in an
elliptical orbit; we put the apogee equal to something like 40,164 km highly elliptical
orbit and then it comes to the apogee; we make sure we fire a rocket and circularize it
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and make sure it goes along the geostationary orbit. We call the initial elliptical orbit as a
transfer orbit.
(Refer Slide Time: 20:08)
Suppose, we are interested in a mission to the moon; something like Chandrayaan-1 of
ISRO which orbited around the moon. From the Earth, we keep going through a series of
elliptical orbits, we escape from the Earth i.e., escape the gravitational field of the Earth,
and get inserted on to the moon’s gravitational field and thereafter orbit around the
moon. If we want to come back to Earth from the moon, we again come out of the
moon’s gravity, we escape from the moon and reenter Earth’s gravitational field.
In the previous slide, we had the transfer orbit; that means we do not take the satellite
directly to the geostationary orbit. But to be able to go to this, we first put it in a transfer
orbit, and then when the apogee is the radius of the geosynchronous orbit, we circularize
it. Well these are all about the different orbits and it is about time for us to do a problem
or two.
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(Refer Slide Time: 20:48)
And I take a problem, which is something related to a recent rocket. An innovator by
name Sir Richard Branson wants to ferry tourists to space.
(Refer Slide Time: 21:12)
He take off from the surface of the Earth into deep space so that the tourists could go
above the Earth and see the Earth as it where from space; how it looks and it seems to be
very fascinating. Therefore, what he did is that he starts with an aircraft from the ground;
this aircraft is known as White Knight. It carries a rocket and a space capsule. The
aircraft takes off from the surface of the earth goes to a height to a something like 15
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kilometers, and it returns. At 15 kilometers, you fire a rocket, separate it from the aircraft
and it takes you to a distance of something like a 100 kilometers.
Therefore, the problem that I pose to you is for the rocket to traverse from 15 kilometer
above the surface of the earth to 100 kilometers, what is the value of velocity to be
provided by this rocket? Because, the velocity required from the surface of the earth to
15 km is given by the aircraft. The rocket goes from 15 kilometers to a distance of 100
kilometers; that means, I am looking at this up to 15 kilometers the aircraft White Knight
2 is used. And from here to a height of may be 100 kilometers the rocket is used, I want
to know what is the velocity, which must be provided by the rocket.
(Refer Slide Time: 23:14)
Let us calculate it. We know, that ½ m into V square is the kinetic energy what we are
giving. This must be equal to the work done or the energy required to start from 15
kilometers above the surface of the earth; that means, RE + 15 kilometers, I go a distance
of RE + 100 kilometers. And what I do? I give this kinetic energy, which will give me the
work done in traversing from 15 km to 100 km; and what is the work done? It is
GMEm/R2 × dR for a small distance dR. The total work is the integral from RE +15 km to
RE + 100 km.
And now how to solve this; I have to integrate this equation and find out. What is the
value? How do I do it? I find that m and m gets cancelled, GME/R2 gives me minus 1/R.
Therefore, we get V2/2 is therefore equal to GME × 1 over radius of the earth RE plus 15
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kilometers is 6383 minus 1 over RE plus 100 km; that is 6468. We substitute the value of
G = 6.670 ×10−11, the mass of the Earth and we get the velocity. And this velocity will
come out to be something like 1.278 kilometers per second. And this is the velocity,
which is required.
Now the question is if we had started from the surface of the earth, which is we say
6368 km. I find that the difference velocity is going to be very small. We readily do not
see any advantage in launching from an airplane and going up. But there is something
which we seem to forget. When an aircraft flies, it also gives a horizontal component
namely an orbital velocity component; and that what make it advantageous. You know
we have some rockets that are air launched and one of the rockets is known as Pegasus
rocket. What is done in this rocket is you take the rocket and the space craft in an aircraft
to height of something like 10 to 15 kilometers till atmosphere is available and then
launch the rocket and that way the rocket need not be very powerful, but it can much
smaller to do job.
We call such rockets launched from the air as air-launched rockets. It is not mandatory
that rockets are launched from the ground. It might as well be launched from under the
sea; we have sea launch. We have a missile known as the Polaris Missile, which is
launched from the submarine from under the water: it comes up to the surface of water
and it propels through air. So, wherever we want a launch, we need the value of the
velocity that is to be provided such as the orbital velocity, total velocity, etc.
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(Refer Slide Time: 27:07)
May be I will take a next example which is again very illustrative. We talked in terms of
geostationary orbits. And you know nowadays, we find many countries wanting to
launch spacecraft into geostationary orbit, because this is very useful for communication
purposes. As far as India is concerned, the satellite in geostationary orbit at the given
altitude points towards Nagpur, which is the centre of the India; India gets covered by
the spacecraft and TV programs among others are relayed by the spacecraft. The
program is beamed to the satellite from a given place and it beams it back throughout the
country.
Let us consider the Indian National Satellite INSAT. Many INSAT satellites have been
launched. Can we keep on going continuously using a satellite or is there a life for a
satellite? And if there is a life to a satellite, why should it have a life? This is because
electronics can continue to function for 100’s of years therefore why should there is be a
life? What is your opinion? People say the satellite has a life time of 15 years, some say
it has 20 years, some say it is only 5 years.
What decides the life time of the satellite? Because I keep telling everything is freely
falling; everything is vacuum, everything is going round and round in perfect orbits.
Why should there be something like a life of a geostationary satellite?
You are telling may be the satellite may deviating from its path or orbit? Why should it
deviate? You are partially correct, but then why it should it deviate?
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There are many other forces like for instance, you have the Sun’s gravitational field, you
have solar flares, the gravitational field of the Sun is changing, may be we have the
moon somewhere near at a geosynchronous radius of something like 43000 kilometers;
there is the moon’s attraction that is also a variable. Therefore, it is quite possible that
there are perturbations or changes in the forces on the spacecraft as it orbits. How do you
take care of these perturbations? We have the satellite in the form of a box and in it we
have will have something like 16 rockets placed at the corners or at some other locations.
And whenever you find something is changing have to fire these rockets and generate a
force or momentum which can overcome the disturbance. That means I have to do
something like what we call as station keeping to keep the satellite in its orbit. And make
sure it as always pointed as required, If there is a drift, I have to correct it. This means
that attitude, position, and orbit need corrections. I need energy for the corrections that
are required.
Therefore, for all these things, we fire rockets and therefore, I have to keep in the
satellite and use it as and when required. And once my fuel is over, the life of the
satellite is over; and that is the reason for the life of a spacecraft. We keep talking in
terms of the exotic propulsion like a electrical propulsion, which may not have so much
requirement of a fuel, which can be there for much longer time and therefore we will
cover this things as we go along.
(Refer Slide Time: 30:48)
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Let us consider a satellite in geosynchronous orbit at a radius of about 43,000 km. And
let us say that the lifetime of the geostationary spacecraft is over. If it is going to be left
to orbit it may be pose a danger to other geosynchronous satellites as it may collide with
it. If we have old non-functional satellites it may be hazard for the others.me. This means
that there is the problem even in space even though the space is so large.
Therefore, it is necessary that once the life time of a satellite is over, to push it out with
escape velocity such that it goes into deep space and I have no such problems. How I do
it? In other words, before the INSAT satellite life is over, I should make sure that with
the remaining fuel, I must push it out of the geostationary orbit.
Therefore, let us do this problem. All what I am saying is I have a geostationary
satellite; and now I want to push it out of the orbit; that means, I have to push it to
infinity.
So what is the velocity required to push a satellite out of the geostationary orbit into deep
space? I have to escape from the orbit; let as forget about the pull of the moon, other
planets and all that and let as assume only Earth is attracting. Therefore, I want to find
out what is the escape velocity? Escape velocity is √2GME/R. What is the R now? I want
to escape from this orbit, and what is that R? R is equal to what we said was something
like 42,000 kilometer. I put the value of R, substitute the value of G and Mass of Earth
ME .
And we find that we still require to push it out with a velocity of something like 4.347
kilometer per second. Let us puts the numbers: 2×6.670 ×10−11 × mass of the earth
5.974×1024 divided R which equal to 42178×103 meters. And this comes out to the 4.347
km/s. That means, I must keep some fuel reserve such that with this fuel, I will be able to
push it out; and to keep this amount of fuel reserve is mandatory.
Well, this is all about orbits. I think be have covered it to some extent. Why do we need
rockets? We now go back and ask how to push in space?
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(Refer Slide Time: 34:11)
Joules Verne in his Science Fiction book “From Earth to Moon” suggested we have a
canon; in the cannon, you put a spacecraft you push it out with extremely high velocities
such that it gets into the orbit. What is the type of typical orbital velocities for
geostationary orbit? It is around 13 kilometers per second or let us say that we need an
orbital velocity of around 10 kilometers per second that is 10000 meters per second.
Supposing the mass of the body, which I want to orbit is around we say 1000 kilogram,
because 1000 kilogram is the least required, wherein I can account for some equipment,
may be one or two people can be there; we have something for life support and all that
and 1000 kilogram seems reasonable.
Therefore, what is the energy I must have? The kinetic energy is ½ × 1000 kilogram ×
100002 so much Joules. This is the energy, what I have to give to the body as kinetic
energy. And what is this value? Supposing, I have to launch it instantly using a canon in
something like in 0.1 milli second or let us say 1 milli second, because it has to get out of
the canon fast. Therefore, the power required is equal to ½ × 1000 × 10,0002 ÷ 10−3. And
what is the number we are now talking of? We talking of 500 × 108 divided by 10−3. We
are talking of the huge numbers something like 1013 watts. If you take the entire
electricity which is generated in a super thermal power plant, it is very much lower than
this; therefore, we cannot use such a canon for launching the space capsule.
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(Refer Slide Time: 36:40)
And let us take a look at what others were suggested in very early times. Imagine a ship
is sailing on the sea; there is the giant storm and you get a high velocity waves over the
sea. The ship gets launched by the high velocity wave motion into the atmosphere. Even
if by chance, I get a high velocity, when the body with the high velocity is traveling
through atmosphere, it will get burnt out, because you have frictional resistance of the
air; therefore, we need some other type of launching. When we imply rocket propulsion
all what we mean is, you have continuous ejection of mass from the rocket at high
velocities. What it does is that it provides momentum or rather some change of
momentum. And what do we mean by change of momentum? We call it as impulse (I).
What is the unit of a momentum: momentum is kilogram meter per second. Therefore,
impulse has the same unit kilogram meter per second, but Newton is equal to kilogram
meter per second square; therefore, impulse also has units of Newton second. Please be
careful about units. Therefore, when we launch a body by a rocket, we give some
impulse continuously to the body. During the process the mass of the rocket keeps
decreasing as it ejects mass out and the velocity of the body increases rapidly.
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(Refer Slide Time: 38:07)
This brings us to the theory of rocket propulsion. All what are telling is that we must
give some momentum to the body; and how do I give it a change of momentum? I throw
some mass out of the body at sufficient velocity and cause a momentum change in the
body. Therefore, let us take an example; we will start with this example; it is a very
fascinating example. I borrow this example from my teacher who taught mechanics to
the first level students. Supposing we have something like a rigid sled; what is the sled?
Sled is something on which you slide down the slope of a mountain. And this sled, let us
say, is on level ground and we presume that there is no gravitational field. There are no
external forces present on the sled.
On this sled we have two boys; the sled is stationary and these two boys want to move
the sled; the sled is on a level ground let say the ground is so slippery that there is no
friction. We idealize this situation. This sled is stationary, these two boys find that there
is all ice all around - there is no friction between the ground and the sled; they do not
want to get out; but they want to move this sled. Therefore, they say let us provide some
impulse or let us provide some change of momentum to the sled.
Therefore, in the example, we consider each of the boys carry a stone of mass m. Let the
mass of the stones, the two boys and the sled be M kg; lets the mass of each stone be m
kg. Now, the boys want to move, how do they move? They say let both of us throw the
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stone out simultaneously at a velocity v0, so many meters per second. Therefore, both the
boys simultaneously through this mass with a velocity v0 meters per second.
Now, I want to find out whether this sled will move or not. How do I solve? I go back to
my inertial frame of reference. What I do is stand outside the sled; and I am in the
inertial frame of reference, because I am moving at constant velocity and therefore, I
describe the motion of the body. I am watching these things happen. Now, I want to
know the velocity at which the sled moves. The two boys throw the stone in a given
direction with the velocity v0; let me assume that the sled also moves in this same
direction at a velocity V meters per second. I want to determine the value of V. I am
looking at it from the inertial frame of reference and therefore, what will be the equation
for change of momentum?
(Refer Slide Time: 41:19)
The initial momentum of the sled, the boys and the stone put together, they are all at rest
and the velocity V is equal to 0 initially. And therefore, the value of the initial
momentum is 0. What is the final momentum? The momentum is conserved in the
inertial frame of reference. Therefore, what is the final momentum? Let us calculate it.
Now the final mass of the sled is M − 2m since the two stones have left. The sled has the
velocity V, the stones are hurled with a velocity v0. What is the velocity of the stone as I
am watching it from the inertial frame of reference? The sled is moving with the velocity
V, stones are moving with the velocity v0 and therefore, the velocity as seen by the
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observer in the inertial plane of reference is V+ v0 . The momentum is therefore 2 m × (V
+ v0) in the inertial frame of reference. Momentum is conserved; initially it 0 and final
must be equal to this, because I am talking of the inertial frame of reference. And
therefore the initial momentum is 0 and is equal to the final value of (M − 2m) × (V +
v0). 2mV gets cancelled. We want to find out the final velocity of this sled, which is
capital V? What is the value that we get? V is equal to minus 2m/M × the velocity v0
with which the stones are thrown. What does the negative sign in this equation tell us? If
the stones are thrown out in a given direction, the velocity will be in the opposite
direction.
Therefore, just through the action of these two boys throwing the stone, they are able to
move this sled at this velocity. Now, I ask the second question. What is the relative
velocity of the stone? V + V0. Therefore, I take v0 outside into (1 - 2m/M) × v0.
Refer Slide Time: 44:38)
The next question we ask is if these two boys, given the same set of stones, and throwing
the stones at the same velocity can they move the sled even faster. The first boy throws
the stone with velocity V0 and this is followed after some time by the second boy now
throwing the second stone with the same velocity V0. In fact, they do not throw the
stones simultaneously that is 2m mass of stone thrown together, but rather one stone after
the other. If the stones are thrown one after the other, what will be the final velocity of
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the sled? Let us call it as V’’. What should be the value of V’’? Let us write the equation
of motion again.
Again I am in the inertial frame of reference. I stand over here; watch the fun in the
inertial frame of reference. I find that the first boy throws the stone and let the value of
V’ be the velocity of the sled after the first stone is thrown. And what is the relative
velocity? The relative velocity is equal to V’ + V0. Following from the last example the
value of V’ is – m/M× v0. The relative velocity of the stone which was thrown is
therefore V’ + v0 which is equal to v0 (1- m/M). At this point in time the second boy
throws the stone and therefore, what is the final velocity of the sled? When the first stone
is thrown you have V’ as the velocity of the sled, when the second stone is thrown, you
get V’’. Let us balance the momentum in the inertial frame of reference. We get:
(Refer Slide Time: 46:37)
Initially the momentum is zero. The final momentum is (M−2m) × V’’. The first stone
goes with a velocity of relative velocity in the inertial frame of reference, which is equal
v0 into 1− m/M. And the second stone goes with a momentum of m into the velocity
equal to V’’+ v0 relative velocity; This becomes my equation for momentum balance. If I
solve this, what do I get? I want to find out what is the final velocity V’’? Minus
2m×V’’ + m ×V’’ gives −m× V’’. The next term is simplified as m into v0 × M − m by
M and the third term is left with m × v0..
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What do we get for the value of V’’? V’’ = m/(M−m) × v0 + (m/M)× v0 or rather we get
it as equal to v0 × [ m/M + m/(M−m)]. Now, we find that when one stone is thrown after
the other, I get this velocity whereas when both the stones are thrown together, we had
the velocity as (2m/M)× v0; in both cases we should have had the negative sign since the
direction of velocity of the sled is opposite to the direction in which the stone is thrown
And therefore, what is the comparison? We have m/M plus m/M when both the stones
are thrown together. We have m/M + m/(M−m) when one stone follows the other. M
minus m is smaller than M. Therefore, in the second case the velocity of the sled will be
greater; therefore, the throwing one stone after the other gives a higher velocity than
when both the stones are thrown together. Now we can generalize, instead of having two
stones, I keep on throwing one stone after the other, what is going to happen? I will get
the velocity, which is much better than the spontaneous throwing of all these stones
together. And this is the basis of the rocket propulsion.
What we do in a rocket is keep on ejecting mass till we achieve the required velocity. I
will continue with this; in the next class. We will derive Tsiolkovsky’s equation, which
is known as the rocket equation following this analogy. But this is basically the principle
using which we must be able to design new forms of rockets.
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Rocket Propulsion
Prof. K Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. 06
Rocket Equation and Staging of Rockets
In today’s class we continue with the theory of rocket propulsion. In order to focus
ourselves and just make sure we are on the right track, I will go through a few slides
which will illustrate what we did in the earlier class. We will also find out that not only
rockets eject momentum to propel; but in nature itself we have some creatures which
make use of the same principle.
Let us briefly go through some of the slides. Rockets are used to launch satellites or may
be objects in space. We talked in terms of circular orbits, we talked in terms of
geosynchronous orbits, polar orbits, sun synchronous orbits and retrograde orbits;
different orbits.
(Refer Slide Time: 01:18)
We also talked in terms of elliptical orbits and we said elliptical orbit is one wherein you
have two foci or focal points. This is shown here; the Earth is at the focal point and this
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spacecraft going round in an elliptic orbit. We defined something known as eccentricity
which was the distance between the two foci and the major axis which is 2a over here.
(Refer Slide Time: 01:38)
We also said for an elliptical orbit you could have an inclination, even a circular orbit
could have an inclination and the inclination is between the equatorial plane and the
plane of the orbit. I want to make it clear because we defined one orbit known as a
Molniya orbit and we told that in a country like Russia in the northern hemisphere
wherein the satellite has to stay in the orbit above the northern hemisphere for a longer
time; we have highly elliptical orbits wherein the apogee is at a distance of something
like almost like 60000 to 70000 kilometers and the perigee is quite small of the order of
6000 kilometers.
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(Refer Slide Time: 02:23)
This Molniya orbit is illustrated with the northern hemisphere in the upper portion, the
satellite is in the northern hemisphere for a much longer time. Something like 11 hours it
is over this northern part and only one hour in the southern region. The countries in the
upper north can view the satellite for a longer time. This Molniya orbit is particularly
important and was developed by Russia. The apogee is of the order of 70,000 kilometers.
The inclination of this orbit was something like 63.4 degrees.
We also talked in terms of the rocket principle and in the example of the sled we
reviewed the velocities achieved when stones were thrown out simultaneously and one
after the other.
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(Refer Slide Time: 03:11)
We got the net velocity of the sled was something like the mass of the stones thrown
divided by the total mass into the velocity with which the stones were thrown. This was
when the stones were thrown simultaneously. The important thing here is when we are
looking at the sled which is moving, we are talking with the inertial frame in mind. I look
at the sled. I am standing outside as shown by the observer.
(Refer Slide Time: 03:33)
You know, I am observing it from the inertial frame of reference and writing the
momentum conservation equation.
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(Refer Slide Time: 03:44)
When the two stones are thrown not together but one after the other; we got a higher
value of velocity because instead of 2m/M when the two stones were thrown
simultaneously, we got m/M + m/(M−m) when the two stones were thrown one after the
other. Since M−m is less than small M, we got an increased velocity for the sled. And so
if I have a series of stones thrown one after the other; well I can get a higher velocity
than when the stones thrown together. The question is why is this so?
Because when the net mass is decreasing I accelerate a lesser mass and I get a higher
value of velocity. Therefore, we find that when I throw one stone after the other, I do not
have something like stones being thrown once at the initial time in which case I have the
net velocity is equal to u + at where a is the acceleration; but, since I am accelerating
gradually in view of the reduced mass, I get much higher velocity.
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(Refer Slide Time: 04:52)
Let us look at some examples. I chose the above example from the national geographical
magazine. What is shown here is something known as a squid. Squid is like a large fish
something like 10 meters long and is found off the coast of Japan and off New Zealand.
It periodically visits these coastal areas and it is an endangered specious and it propels
using the rocket principle. Let us look at the parts of this particular squid. You have
something like a mantle or a funnel here through which it sucks in water.
(Refer Slide Time: 05:39)
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The above slide shows a squid which was caught off the coast of new Zealand and you
see the length compared to this man; it is something like 10 meters long.
(Refer Slide Time: 05:50)
Let us go to the principle of is motiont. You know there is a funnel here at the rear of the
squid in this slide. To move itself it opens its funnel or mantle and gulps in water. As it
gulps in water it gulps in some sand may be eggs may be small fish or whatever it is
available around it in the water. Then it closes the funnel i.e., it closes its gate to the
funnel or mantle or the mouth as it were and then it through it contracts the muscles such
that it builds some pressure of water which it has gulped.
And then when it wants to move, it just opens the gate again and spouts out the water in
this direction and when it spouts out the water in this direction it moves in the opposite
direction. It keeps on squirting the water and it propels itself. Therefore, when it squirts
out the water, it moves. Whenever it wants to move; it expels water, waste eggs and all
that and that is how it moves. The principle is very similar to a rocket. It collects water,
pressurizes the water, releases the water gradually and it moves.
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(Refer Slide Time: 07:05)
And the type of velocity what it gets is quite substantial. It pressurizes the water to
around 0.4 times the atmosphere and once when it pushes it out, it is able to leap
something like a 50 meters and gets a velocity of something like 2.5 kilometers per hour.
This is quite phenomenal when you consider that this is in water that is a viscous liquid.
It is able to go at that speed because of the gradual release of pressurized water from its
mouth i.e., funnel or mantle
Now I show the funnel again. It opens its mouth, water enters, it contracts itself and then
it opens it, closes the gate, compresses it, releases the pressure, then the water squirts out
and whatever is available along with the water is squirted out and it moves.
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(Refer Slide Time: 08:01)
And this is the principle of what we call as the GIANT SQUID. In fact in US we have
had rocket projects named as Project Squid. In what way do you think is the motion of a
Squid different from a fish?
(Refer Slide Time: 08:24)
A fish has fins by which it displaces the water. That means it slowly displaces the water
like I go on a boat let us say. I am sitting in the boat I have an oar. With the oar, I
displace water. I displace some water i.e., provide velocity to the water but, the mass of
water displaced is large though the velocity during the displacement is small.
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In the case of a squid: It takes in a small amount of water, it pushes it out at high velocity
and its able to do a much better job. This same principle is used in a water rocket. I show
a water rocket; if you want to make one, all what you do is take one of these bottles,
partially fill in with water and then invert it and pressurize the water. I remove the cap
and when I do so water squirts out as a jet and the water rocket moves up.
The principle of any rocket is quite identical. Only major difference is that you need
higher velocities. Therefore, you put more enthalpy into the working medium of the
rocket, which is then expanded out. The medium, which is expanded would have a
higher value of velocity if its enthalpy is higher. Therefore, we get higher velocity with a
gas heated in chemical or nuclear rockets.
(Refer Slide Time: 10:02)
Having observed about giant squid, I discuss another example. There is a creature by
name bombardier beetle. You know what a beetle is: in Tamil we call as “vandu” and in
Malayalam it is known as “nandu”. It is a harmless creature. It cannot fly well as it is
bulky and it goes round either by flight or by walk. Whenever it comes into our house,
all what we do is push it out by placing it on a piece of cardboard or thick paper and
throwing it out.
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(Refer Slide Time: 10:39)
A particular form of beetle known as bombardier beetle has a different means of
locomotion. Beetle we had said is a harmless creature; ants sting it and nature has given
its some means to protect itself.
(Refer Slide Time: 10:56)
It has something like two stomachs, and after the two stomach there is another receptacle
which is like a third stomach. In one of these stomachs, it secretes hydrogen peroxide
which is an oxidizer. In the second stomach it secretes hydroquinone which is a fuel.
Hydrogen peroxide being an oxidizer and hydroquinone being a fuel can react to form
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hot gases. When the bombardier beetle is chased by the ants or when it is bitten by the
ants it immediately squirts the hydrogen peroxide and hydroquinone into its third
stomach which is coated with enzymes. The fuel and oxidizer react to generate hot gases
in the presence of the enzymes and these hot gases are forced out in the form a jet which
kills the ants or else they are chased out as the bombardier beetle moves forward.
(Refer Slide Time: 11:49)
Let us examine the processes again. In the above slide, the two stomachs are shown
which form hydrogen peroxide and hydroquinone. They are secreted here. In the third
stomach below is a lining of mucus (an enzyme) which acts as a catalyst. The catalyst
promotes the reaction and whenever it is attacked it just squirts out the hot gases. This is
similar to the processes of combustion and expansion in a liquid propellant rocket.
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(Refer Slide Time: 12:07)
What does a liquid propellant rocket consist of? We are yet to study it. You have a fuel
tank, you have an oxidizer tank, you pump the fuel and oxidizer into it, you ignite it and
you push the gases out. So, this small insect which is available in nature works on the
principle of a liquid propellant rocket or rather the rocket works on the principle of the
bombardier beetle.
(Refer Slide Time: 12:33)
I think we should look at nature to understand many of the things what we are studying.
Everything is there in nature and we have to be more observant. Having said that let us
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get into some more details. Let us illustrate the rockets used in satellites because we told
that whenever a satellite is there in space in geostationary orbit and its life is nearing
completion, we have to push it out. A satellite has something like 16 rockets which are
there at the edges of this satellite and these are used for correcting the attitude. May be
for station keeping of the satellite and whenever the life time of the rocket is near to
being over, we fire some of these rockets such that we remove it from the geostationary
orbit and push it into deep space. That means we make it escape to deep space.
(Refer Slide Time: 13:20)
Based on the above background how to develop the theory of rocket propulsion which
leads to the rocket equation? The rocket equation was developed not very early, only in
the year 1903 and that also by a Russian school teacher by name Tsialkowski. Now, let
us see how this is done and what is this rocket equation also referred to as Tsialkowski
equation.
Let us derive it first. We will follow the same procedure what we adopted while finding
out the velocity gain by the sled, these two boys standing on it, throwing one stone after
the other. We will make some simplifying assumptions.
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(Refer Slide Time: 14:31)
Let us assume I have a rocket as shown in this figure. We will assume rocket has a shape
something like this. It need not really be the only shape and may be you will be come out
with better configurations of rockets. Let say at time t it is moving with a velocity let us
say V. Let its mass at time t be M. After a small time Δτ after time t i.e., time t +Δτ let it
gains a small velocity ΔV.
Now how does it gain this velocity? It is moving forward and then during this small time,
a small mass Δm is being ejected out. The rocket is ejecting matter and in a small time
Δτ a small mass Δm is ejected out with a velocity VJ. I do not know the direction in
which the mass is ejected out. We will presume it to be in the direction of motion of the
rocket.
And therefore, the final mass of this rocket at time t +Δτ is going to be M−Δm. Let us
presume that the velocity of the rocket at time t +Δτ is V +ΔV. Since the rocket has lost a
mass delta m in this small time, the final mass of the rocket is M −Δm.
Now we talk in terms of inertial frame of reference and therefore we watch the rocket
from the ground. I watch rocket go up with a velocity V at time t and then after a time Δτ
I am looking at it going with a velocity V +ΔV. Now, I do the momentum balance in the
inertial frame of reference, and what is it I get?
The initial momentum of the rocket is equal to MV. It is mass into velocity at time t.
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(Refer Slide Time: 17:50)
Now what is the momentum of this rocket at time t +Δτ as a I am seeing from the inertial
frame of reference? I have (M−m) × (V+ΔV) plus you also find that Δm has been pushed
out of the rocket with velocity VJ.. The velocity VJ is with respect to the rocket and from
the inertial frame of reference when the rocket is moving with velocity V+ΔV, the mass
Δm will appear to leave with a velocity V+V+VJ. This is what is shown here. That is the
velocity of this parcel of mass Δm and is the relative velocity VJ + V +Δ V. This is how
we wrote the equation for the sled. We ignore the gravitational field and the resistance to
motion of the rocket by the air and we get the momentum to be conserved exactly in the
same way as in the sled problem. In the case of the sled which was initially stationary,
the initial momentum was 0. In this case, the initial momentum is MV. This equals the
momentum after t +Δτ seconds.
If we simplify, what is it that get. MV is equal to MV−ΔmV. Then we get M ΔV − ΔmV.
And Δm × (VJ +V +ΔV). I should have written here small δV instead of ΔV because I
want to reserve the capital delta for something else.
And now I find that this Δm δV and Δm δV cancels. So also MV and MV cancels.
Therefore, I am left with the term M into δV plus delta mVJ is equal to zero. This gives
δV= − Δm VJ / M.
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(Refer Slide Time: 23:12)
Therefore, we get δV =− (Δm/M) VJ.
We need to solve the above equation. What is the value of Δm? Let us assume that the
mass which gets exhausted from the nozzle is something you are constantly pushing out.
Mass at the rate let us say mdot; m°. Therefore, the value of Δm should be equal to m°
into the small time Δτ. You know the rate at which mass is leaving the nozzle is m° over
a small time Δτ; it is equal to m°Δτ and what should be the value of M? Capital M must
be equal to the initial mass of the rocket at time t which is equal to its mass at time zero
minus m° into t. This is the mass of the rocket at time t.
In other words at time t is equal to 0; the rocket had a mass equal to the initial mass; it
continues to eject mass at the constant rate m° and therefore, at time t its value M =
initial mass Mi − m° t.
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(Refer Slide Time: 25:13)
And therefore, now I can erase this part and we can write the value of delta V as equal to
minus VJ into Δm which is equal to m°×Δτ ÷ (Mi − m° t).
I want to integrate this equation and if I have to integrate this equation from initial time
of 0 to a final time tf at which I get the total velocity increment of ΔV. I know the value
of m°. The mass at time t is the initial mass Mi − m°t. The value of ΔM equals −m°×Δτ.
With the two negatives the minus sign will not be there in the expression for delta V. I
show the derivation of ΔM as being − mdot × Δτ in the following.(Refer Slide Time:
27:16)
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The minus sign shows that the mass has left the system and therefore, when I substitute
the value of ΔM, I should have substituted − m° ×Δτ; that means this should have been a
negative sign and this negative sign and this negative sign would have given me a
positive sign. The same is seen from the expression for ΔM.
To summarize: We solved the momentum equation in the inertial frame of reference and
balanced initial momentum MV with final momentum (M−ΔM) ×(V+ δV) + ΔM×(VJ +
V +δV) and then we got an expression which gave us δV as equal to − VJ ×ΔM ÷M. We
find δM is equal to − m°×δτ and therefore we got it as VJ m°×Δτ ÷ Mi − m°t.
We now integrate it to determine the net velocity gained by the rocket.
(Refer Slide Time: 29:28)
And what is the expression we get now? Let us assume that VJ which is the velocity at
which the gases leave the rocket is a constant. The limits of integration are from t = 0 to
t= tf. Integral of m° divided by (Mi − m°t) dt is natural log of Mi − m° t. We also get a
negative sign from the − m° t in the denominator and this expression is between the
limits t = 0 and t= tf.
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(Refer Slide Time: 30:29)
The value of the expression Mi − m°t at t= 0 is Mi viz the initial mass of the rocket while
Mi − m° t at t= tf is its final mass Mf. And therefore, the value ΔV is equal to minus VJ
into natural logarithm of Mf by Mi. The negative sign can be removed by inversing the
term within the logarithm. The velocity change or increment provided by the rocket is
therefore VJ into logarithm of the initial mass to the final mass of the rocket. This
velocity change is spoken of as incremental velocity and the equation is known as the
rocket equation.
All what the rocket equation tells is when I burn a quantity of fuel between the initial
value and the final value and I am exhausting it out at a constant velocity VJ, the final
velocity of the rocket is equal to the velocity with which I am ejecting matter out into
natural logarithm of the initial mass to the final mass. This is what we call as the rocket
equation. The Russian school teacher Tsialkowski derived it and postulated that a high
value of jet velocity VJ is required and a large value of mass ratio (Mi to Mf) is desirable
if we have to go into interplanetary missions because all what we want is we want a jet
velocity and the mass should keep getting depleted and this is what is the theory of
rocket propulsion states.
Can I repeat it again because this forms the basis and you should know the limitations?
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(Refer Slide Time: 32:45)
The jet velocity or the velocity which with mass is being removed into natural logarithm
of the initial mass divided by the final mass decides the velocity provided by a rocket
(ΔV=VJ ln(Mi/Mf). The value of final mass of a rocket to the initial mass of the rocket is
also called as the mass ratio of a rocket. We did not consider the gravitational forces nor
the drag forces in the derivation of ΔV.
Therefore, ΔV is also spoken of as ideal velocity increment. Now, in the earlier classes
we saw that the rocket has to supply the necessary orbital velocity and the total velocity
for which expressions were derived.(Refer Slide Time: 34:21)
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If I can write it, VT =√ (G ME /RE)× (RE +2 h)÷ (RE + h); so many meters per second is
the velocity what was a required.
Now, if I have a rocket and on top of it I put whatever I want to launch; I give it a
velocity I can put it in orbit. But we found that we need a velocity of the order of
something like 10 to 12 kilometers per second. Therefore, what it is required in order to
achieve high velocities? See, you have to have high value of mass ratios which means
that the ratio of initial mass to the final mass must be a large. That means that difference
between the initial and the final must be large and also the VJ must be a large value. That
means the jet velocity is a controlling parameter and higher the jet velocity you can have
higher delta V..
Therefore, the figure of merit of a rocket if somebody were to ask us we may say well
one is the jet velocity, the other is something related to the masses. Therefore, let us go
back and look at this term because this is fairly clear to us. Supposing, I were to exhaust
at some jet velocity and I can get as high a value as possible. Apparently you cannot get
to the speed of light, you cannot get more than some amount. But, I have some
limitations. I will come back to these limitations. In addition I am talking in terms of Mi
by Mf should be a large number.
Let us just examine this number before we can design a rocket because as of today you
cannot get more than something like 3000 to 5000 meters per second as jet velocity. We
will find out where and what the limitation are there but, let us first examine this.
(Refer Slide Time: 37:02)
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Let us therefore write the ideal velocity increment as equal to jet velocity into natural
logarithm (ln) of the initial mass to the final mass of the rocket. What does the initial
mass of the rocket consist of? It will anyway have the useful part of a rocket? What is the
useful part of a rocket? The object which is going round and round that is the useful part
of the rocket which is we call as payload. Then we have the structure of the rocket that
means it must have some metal and other structural materials including inert materials to
contain it and protect it from heat such as insulation. Thus we would have a structural
mass.
Plus it should also have some fuel and we said fuel is used for propelling the rocket and
is known as propellant. Therefore, we have a mass corresponding to mass of propellant
Mp. That means that the initial mass of a rocket comprises of the payload mass, the mass
of the structure plus the mass of the propellant and we say Mi is equal to Mu plus the
structural mass plus the fuel or the propellant. This is the initial mass. When the rocket
has done its job, what must be the final mass of the rocket? It has done its job that means
all the propellant has burned out. Therefore, the final mass will be the mass of the useful
component viz., the payload plus the mass of the structure. But, it is also possible that the
structure could be removed the payload after the rocket functions and thrown out. But,
otherwise the structure will remain as part of the final mass of the rocket. This is
generally the case. Therefore, the initial mass Mi = Mu + Ms + Mp while the final mass
is Mu plus Ms.
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(Refer Slide Time: 39:33)
Now, let us analyze the masses. The incremental velocity ΔV = VJ ln [(Mu + Ms + Mp)
÷ (Mu + Ms). Let us express these mass as non-dimensional terms or as mass fractions.
Let us call the useful mass of a rocket Mu divided by the total initial mass namely Mu ÷
Mi + Ms + Mp as equal to alpha (α). The value of α is proportional to the initial mass of
the rocket and is the non-dimensional payload mass. Similarly, we denote structural mass
divided by the initial mass of the rocket as the structural mass fraction and call it as β.
The propellant mass fraction is the ratio of the propellant mass divided by the initial
mass of the rocket. This is denoted by γ.
(Refer Slide Time: 40:48)
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Dividing the mass terms within the logarithmic term by initial mass of the rocket, we get
ΔV = iVJ ln [(α+β+γ)/(α+β)]. However, the sum of the masses is the total mass of the
rocket and therefore the sum α+β+γ = 1. And therefore, we get this equation for ΔV =
VJ× ln (1/(α+β).
Now, what is it that we want to do in a rocket? We want to have as much payload as
possible. May be we would like the useful mass to be high. Let us say what is the
fraction of the useful mass α.
(Refer Slide Time: 42:01)
I get therefore, from the above expression one over alpha plus beta is equal to
exponential of ΔV by VJ. This is done by taking exponential on both the sides and the
exponential of ln becomes unity. Taking the inverse on both sides we get α+β is equal to
the exponential –ΔV/VJ. The payload mass fraction α is therefore equal to exponential of
the negative of delta V by VJ minus the structural mass fraction β
Now, let us examine under what conditions will we get the value of the useful payload
mass fraction α to be high. Let us plot it out for different values of VJ at given values of
velocity increments. The velocity increment required could be between 8 to 12 km/s as
seen earlier and depends on the mission. There might be some variations in the jet
velocity between 3000 and 5000 meters per second. What are the values of payload
fraction that we get? Let us just plot it out and see.
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(Refer Slide Time: 43:36)
This is shown in the above wherein alpha is plotted as a function of ΔV/VJ. As the
incremental velocity increases for a given value of VJ, the value of alpha decreases for a
given specified value of β. As the structural mass fraction β increases, the useful payload
fraction alpha decreases. If the efflux velocity VJ is higher for a given incremental
velocity delta V, the payload fraction α increases. And generally this value of the
payload fraction might be around 0.04 and keeps falling. In the operable regions, the
fraction could be around 0.1 or even lower depending on the structural mass fraction β.
Further, as β increases α decreases. We have still not put any numbers here because we
do not know what it is the range of β. But, generally beta should be around let us say 0.1.
As mass of structure increase β increases to around 0.12 to 0.15. That means the payload
fraction will keep decreasing as the structural mass increases. If we can have a high
value of VJ then we can get a higher value of the payload fraction α. Or if we have a
rocket or object, which requires more ideal velocity then I get a lower value α of
payload. You know we are just looking at the rocket equation and trying to draw some
conclusions from it. The conclusions that we draw are if we want to put payload of
higher mass then I need a structure which much be very light. I must have a large value
of jet velocity VJ or else if I can have rocket which does not have to go very far away it
and the orbit is nearby then I can carry a higher mass. Well this is all about the rocket
equation and the conclusions from it.
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(Refer Slide Time: 45:32)
But then the problem is that we need an incremental velocity of about 10 to 12
kilometers per second. We called it as VT. This is what we said as the total velocity to
climb up and orbit. But, then to get a reasonable value for the payload mass fraction for
this incremental velocity with the existing jet velocities VJ and structural mass fraction β
is very difficult, if not impossible. If I have a single rocket, top of which I have a
payload, I may not be able to get a useful mass fraction for the payload, because I have a
definite mass of the structure. I have a limitation on VJ and therefore, to be able to launch
a payload into orbit using a single rocket is difficult. I use the word difficult since it
appears to be impossible at this time. So far it has been impossible but the quest for the
rocket engineer is to make a single rocket do the job.
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(Refer Slide Time: 46:50)
But the job can be done by using multiple rockets. Let us examine this point. Let us say
instead of having a single rocket, I make two rockets. I put a rocket on the top of another
one and so on. This is my first rocket, this is my second rocket. Now, the first rocket
gives a value of the ideal velocity ΔV1. The second rocket already has a ΔV1 when its
starts functioning. It gives me a value of velocity ΔV2 and the total velocity of this
composite two stage rocket, gives me a ΔV = ΔV1 + ΔV2. Therefore, by putting one
rocket on top of the other, we are able to achieve higher incremental velocities. We call
these rockets as multi stage rockets and most of the rockets used today are multi stage
rockets. That means you want a velocity increment of something like ten kilometers per
second; may be the first one could give you 1 km/s, the second one could give you 3km/s
and the third one could be still higher at 6 km/s and therefore, you keep on adding stages
of a rocket and this is what we say as multi stage rockets.
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(Refer Slide Time: 48:21)
For a multistage rockets, we have ΔV = VJ of the first rocket into logarithm of the initial
mass to the burn out mass of the first rocket + VJ of the second stage multiplied by the
logarithm of the initial mass to final mass of the second rocket and so on. I get the final
ideal velocity. Maybe we should we should try to analyze this in some detail. This is
about staging of rockets. We have something as the base or core stage, then on the top of
it we have the first stage, then the second stage and so on.
(Refer Slide Time: 49:32)
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Let us taken an example: Let us examine the construction of India’s GSLV rocket,
which is very much in the news. We know that this rocket consist of a core rocket, it
consist of four rockets attached to the core stage, then it consists of the second stage and
the third stage on the top of the third stage sits the satellite.
Therefore, you have the first stage, second stage, third stage. This gives you ΔV1, ΔV2,
ΔV3, the sum of which gives you the velocity to put it into orbit. Therefore, we talk in
terms of staging. Staging means one after the other but, what about these four rockets
attached to the core? Why should they be required?
We have put one stage on the other to get higher incremental velocity; however, in the
process we have increased the mass of the total rocket. When you have increased the
mass and you want this to be lifted, the core stage should develop sufficient force.
However, the single core stage may not be able to generate that level of forces.
Therefore you need additional rockets so that the force or the thrust is able to take off
from the ground and that is why we put rockets together and this is known as clustering.
Why do we need clustering of rockets? To provide sufficient force for propelling. Even
the upper stages may need clustering.
Let us just put things together and summarize what we have learnt so far.
(Refer Slide Time: 51:16)
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We found that a rocket gives you incremental velocity ΔV. How does it give you ΔV?
Because of change of momentum, produced by the efflux of the jet. We wrote the
momentum balance equation from an inertial frame of reference and found out the value
of ΔV. What is the change of momentum known as? It is known as impulse. What is the
unit of impulse? Same as momentum viz., kilogram meter per second. But, kilogram
meter per second can also be written as kilogram meter per second square into second
which is same as Newton second. Therefore, I can write the impulse as equal to Newton
second.
Impulse in Newton second is what gives ΔV; therefore, what is the force with which the
rocket is pushed up? Rate of change of momentum means d/dt of mv or d/dt of Impulse
I. That is, we get so much force, which is equal to d by dt of momentum; this is equal to
mass flow rate m° of the exhaust which is going out with velocity VJ. This equals m°
into VJ and therefore, I can also write the force is equal to m°VJ (Newton) or compared
to momentum which is equal to mV, I write force is equal to m°V and this is the force
pushing the rocket. There is a limit to the mass m° that can be released.
(Refer Slide Time: 53:31)
And therefore, we allow more mass to be going through by clustering and thus achieve
the desired force. That means that we need a larger force to push and that is why we
require clustering. Sometimes, we have a booster rocket to whose sides we attach two
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rockets. These are like straps. I strap something on to it. The side rockets are also known
as strap-on.
Let me take one or two examples. May be I will show it through slides when we meet in
the next class. But, to be able to just conclude at this point of time, all what I would like
to say is we derive the rocket equation from the change of momentum. We looked at the
inertial frame of reference, watched the rocket go up and we found out what is the ideal
velocity increment.
We also discussed about some creatures in universe which make use of the rocket
principle. Then we found out that the structural mass of the rocket plays an important
role just as much as the jet velocity plays a role. Then to get a high value of ΔV, we
found the need to operate in stages and to be able to take off with the larger mass of the
stages we needed some additional side rockets, which are known as clustering and strapon.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No: 07
Review of Rocket Principles: Propulsion Efficiency
(Refer Slide Time: 00:17)
Good morning! In today’s class, we will look at the following. We will look at the theory
of rocket propulsion again. We derived the rocket equation in the last class. We also
looked at staging; we looked at clustering of rockets and also the strap on rockets and
what function they do. But, we did not really calculate what is the type of acceleration
what we can get from a rocket at takeoff; We will illustrate this in a better way; why we
need additional straps or additional clustering for a rocket.
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(Refer Slide Time: 01:01)
Let us consider this example. Let us say I have a rocket, whose initial mass is Mi. And,
let us say it allows mass efflux at the rate m° kg/s. Let us also assume that the rate at
which the efflux leaves the rocket (nozzle) is at velocity VJ m/s. Let us put the units
together; m°, so much kilogram per second; VJ, so much meter per second. I want to be
able to calculate, what is the initial acceleration of this rocket. How do we do it?
In the last class, we told that the force or the thrust with which a rocket is pushed up is
equal to rate of change of momentum or d/dt of mv. And, here it is the momentum mVJ.
And, therefore the force or thrust is equal to m°VJ.
Now, can you tell me what will be the initial acceleration of this rocket? Therefore, the
initial acceleration should be equal to the initial mass of the rocket into the acceleration
a. That means, force is equal to mass into acceleration or rather acceleration is equal to
m°VJ divided by the value of the initial mass. And, what did we tell in the last class? As I
keep on adding more and more mass to the rocket, the initial mass increases and it
becomes impossible for the rocket to accelerate.
We will work out an example, a numerical example to be able to figure out how we
calculate the acceleration and how we decide what must be the level of acceleration; this
is because acceleration is also important. Suppose some human beings are sitting in a
rocket; it cannot take off at a very high acceleration. The human beings will be adversely
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influenced by the large acceleration. Therefore, there has to be some control on
acceleration.
The ΔV or incremental velocity is equal to summation of; we now use simplified
nomenclature. VJ depends on the particular stage and we have natural logarithm ln of the
ratio of the initial to the final mass of the corresponding stage. That is i=1,2,3…as i goes
from first stage to the second stage and so on. And, this is how we calculate.
(Refer Slide Time: 04:08)
We can try to simplify and get the terms together. And to simplify, we write the
summation as: ΔV is equal to VJ1 1n of (1 over the mass ratio of stage one). Does it make
sense? We said mass ratio of a rocket Rm is equal to the final mass divided by the initial
mass. Therefore it is Rm1. For the second stage we get ΔV2 is VJ2 ln of the second stage
into 1 over the mass ratio of the second plus and so on. Supposing we have rockets in
which the jet velocities are the same for all the stages, VJ1 is equal to VJ2 is equal to VJ3
and so on.
And, further we could have the mass ratios of each stage to be also the same for all the
rocket stages. Supposing, if each stage has a same mass ratio and we say Rm1 is equal to
Rm2 is equal to Rm3 and so on. Then, what do we get for ΔV. And this expression would
be true only when the jet velocity and mass ratio are all the same. Then, we get ΔV is
equal to n times the value of VJ ln of 1 over the mass ratio viz., ΔV = n VJ ln (1/Rm).
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That means, by increasing the number of stages we are able to increase the delta V
corresponding to the number of stages used. But, this is not possible in practice because
the mass ratios of the individual stages may not be the same and the jet velocity of all
these stages may also not be the same. You could relax these things by considering VJ to
be different, you could consider Rm to be different, and we can keep on getting different
ideal velocities. May be we will do a homework problem a little later and try to find out
how to calculate the jet velocity taking into consideration a number of stages of rocket
together. Well, this is all about the rocket equation and the number of stages, clustering
of rockets, and adding a strap-on in a rocket.
(Refer Slide Time: 06:28)
We will look at some examples. But, before looking at examples, I thought can we
extend the rocket principle to say a toy rocket used as fire crackers during Deepavali
festival. How do we launch this toy rocket? We have a bottle which is used as for
launching rockets; we put that stabilizer which is the wooden stick to the fire cracker
which is filled with some black powder. We will look at its composition later on. And,
you have a small squib over here and you light it and zoom it goes up in a particular
direction, if not vertically up. Supposing I want to write the equation for this fire cracker
rocket. It should basically be same as the rocket equation?
Therefore, here also we note that the initial mass of the rocket will include the stick, the
paper which binds the black powder or gun powder together, and the mass of the gun
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powder or black powder used. This is initial mass. And, when the rocket is all consumed,
we are left with this wood and left with paper which has still not got burnt. This will be
the final mass. And therefore, what is the value of VJ? A small hole is provided here
through which the gases are escaping and the velocity of gases is the value of VJ..
But the mass of powder used as a propellant is very small. And rather, if I were to go
back and write the equation what I wrote in the last class namely ΔV =
VJ ln
[(α+β+γ)/(α+β)] where α was the payload mass fraction, β was structural mass fraction
and γ the propellant mass fraction. The amount of gun powder or the amount of black
powder which I keep is very small compared to the weight of the stick and the cracker
assembly. And therefore, gamma tends to be negligibly small. And, in fact it is not only
in the Deepavali rocket that it is small but also in a booster or first stage of a multistage
rocket.
(Refer Slide Time: 08:57)
But, if I have to look at another example say that of the multi-stage rocket. We have one
rocket stage after the other. I say let us have a four stage rocket. The initial mass of this
rocket is going to be Mi summed over i = 1 to 4. This would be mass of first stage MI +
the mass of the second stage MII + the mass of the third stage which is MIII + the mass of
the fourth stage which is MIV. Now, we have some propellant in the first stage and
propellant in the successive stages. Whatever be the propellant we have for the first
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stage, since the total mass of the rocket is large, the mass ratio of the propellant of the
first stage divided by the initial mass of the rocket Mi would be small.
So, also in these cases of multi stage rockets, the first stage, which we sometimes call as
“Booster”; what is the “Booster”? It boosts the rocket; it allows the rocket to takeoff
from the ground. It boosts the acceleration of the rocket. Therefore, it is known as
“Booster stage”. And in a Booster stage, the mass of the propellant divided by the initial
mass is the small number. If it is a small number can we find out whether I can simplify
the rocket equation.
(Refer Slide Time: 11:11)
Let us go back and see what happens? Let us write that equation again. ΔV =VJ ln [1 +
γ/(α+β)]. This was the general rocket equation, which holds good for each of the stages.
Now, I come back to the Booster stage or to the case of a Deepavali rocket. For that, I
say the propellant mass fraction γ divided by α+β should be small. It is small because
there is so much of mass above it. The mass of the propellant is going to be small over
here. Therefore, let us say this γ/(α+β) is equal to x, which is small. Therefore, now I get
the equation; ΔV = VJ ln (1 + x), where x is a small number. And, what is ln of (1 + x)? x
– x2/2 + x3/3 + …. .But, x is a small number. Therefore, I can as well forget about
square, cube etc., of x and write ΔV for a rocket, like a booster rocket or the bottle rocket
which we used for Deepavali time as VJ × x or VJ ×γ / (α+β).
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(Refer Slide Time: 12:54)
This is something like ΔV = VJ × mass of the propellant divided by mass of the useful
part plus mass of the structure. I am just writing these terms here. And, what is the mass
of the propellant? Mass of the propellant is equal to the volume of the propellant into
density of the propellant. And therefore, this becomes equal to density of the propellant
into jet velocity into the volume of the first stage rocket divided by Mu plus Ms. In other
words, we find that in comparison with the rocket equation wherein ΔV was VJ ×ln
(Mi/Mf), we now get it as density of propellant into VJ into volume divided by the
masses. Rather, instead of VJ, density of propellant into VJ becomes influential.
Therefore, for a Deepavali cracker or for a booster stage rocket, instead of VJ being a
figure of merit, it works out that the density of propellant into VJ becomes a figure of
merit and that is the difference. And, of course this becomes the volume proportional to
mass of the propellant and Mu by Ms. This must be kept in mind when we design the
boosters. I will come back to this point when I show some slides. But, this is something
which is important, which comes very simply by looking at the expansion of this
particular expression containing the terms within logarithm.
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(Refer Slide Time: 15:20)
When we want to make a rocket, there are two important parameters that we must
consider as we have discussed. One is VJ. It must be very high. The ratio of Mi by Mf
must also be large. But, when we make a booster rocket or when we are making a fire
cracker rocket, then in that case what is going to be different? Instead of VJ, I would like
the density of the propellant into the value of VJ to be large. And in fact, as we go along
we will see, since the density of hydrogen is small, the use of hydrogen and oxygen as in
cryogenic rockets is not that advantageous for boosters. Whereas if we use solid
propellant, which is a dense material, may be it is better for the booster stages. Let us
keep this in our minds.
Well, this is all about the theory of rocket propulsion, rocket equations, staging, etc. Let
us go back and refresh ourselves on what we have learnt through some slides. And then,
we will come back and see what we mean by propulsion efficiency and then we will
solve one or two problems.
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(Refer Slide Time: 16:52)
We go through a few slides. You know, here we see a two stage rocket. May be the first
stage gives you a velocity ΔV1, the second stage gives you a velocity ΔV2. The total
velocity of this combination of the first and second stage is ΔV = ΔV1 + ΔV2. This is a
two stage rocket.
(Refer Slide Time: 17:10)
Let us go to the next one. This is the first rocket which was designed in ISRO at
Trivandrum. This was in the period, may be nineteen seventy to seventy eight. And, the
first successful launch was in 1980. It was a simple rocket, you know. It consists of four
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stages. And, the all the stages used solid propellants. I will give you a problem involving
this rocket, and we will do it today. This is the second stage, this is the third stage, this is
the fourth stage. The total velocity, which the rocket gives is ΔV1 for the first one; ΔV2
for the second one; ΔV3 for the third one; ΔV4 for this fourth stage.
We had four stages here. You know it is very deceptive to think that you can make a
good rocket to give the desired incremental velocity by increasing the number of its
stages. You know sometimes we feel we can keep on increasing number of stages, looks
very straight-forward. The more stages we have, the more commands we have to give to
the rocket. We have to ignite this second stage and the subsequent stages, we have to
separate it out; it becomes more complicated. The reliability of the rocket comes down.
And therefore, the trend today is to go only for two stage to orbit (TSTO) or three stage
to orbit. People are still working on a single stage to orbit.
(Refer Slide Time: 18:31)
We go on to the next slide. The SLV 3 rocket, what I showed you here, it can only take a
payload of something like 40 kilogram. It is very small. Therefore, it was necessary to go
to higher payloads. With higher payload, you cannot have the vehicle to accelerate
adequately and provide the incremental velocity required; you need to increase the
propellant weight. It was necessary to put two straps. Therefore, we have a strap on
either side. The straps are same as the first stage in this vehicle( you have two strap on)
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and then the second stage, the third stage, fourth stage. This is known as Augmented
SLV or “ASLV”.
(Refer Slide Time: 19:11)
And, now I show some more examples. These are the current rockets which fly;
“ARIANE V”, by which we launched several INSAT satellites. We have two straps; first
stage and the second stage
(Refer Slide Time: 19:30)
This is the “SPACE SHUTTLE”. You know, it has been work a workhorse for US Space
program though it has been decommissioned now. The last flight of “SPACE
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SHUTTLE” is over. In this we have the space plane, which comes back. It has something
like three liquid engines, all clustered together so that, you get the high thrust. And, you
also have two straps. One strap here, the other strap over here; that means two straps for
giving the initial acceleration. It starts off with the three liquid engines and straps
burning and it pushes itself up. The brown thing what you see is the hydrogen tank,
which stores hydrogen which is required for the liquid propellant rockets in the space
plane.
(Refer Slide Time: 20:18)
This shows the Space Shuttle taking off. These are the two boosters. Solid rocket straps
as I said and you have three engines which generate the thrust.
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(Refer Slide Time: 20:29)
Another work horse of US is the DELTA vehicle. Again you have a number of straps
over here or cluster of engines.
(Refer Slide Time: 20:37)
This slide shows “SOYUZ”, a Russian rocket. Here you have straps here or a cluster of
engines here and the main engine is firing. The exhaust from the several rockets is
interacting to give the shape of something like a ball. And, “SOYUZ” was used for
launching our first experimental satellite namely “Aryabhatta”. This was in nineteen
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seventy five to seventy six time period. The slide below shows a powerful Russian rocket
by name “Proton”. The multiple stages and cluster of stages are seen clearly.
(Refer Slide Time: 21:15)
All what I want to communicate is that, most of the vehicles have a number of rockets
which are clustered along with a number of stages. This is why I showed these examples.
(Refer Slide Time: 21:29)
We told had said that a rocket can be launched from the sea viz., from a submarine it
takes off and you see the water droplets splashing over here.
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(Refer Slide Time: 21:43)
The slide above shows Saturn rocket for which I will give a problem which we will try to
solve. This was one of the very powerful rockets; which was used to take men to the
moon. It was known as Saturn V. And, here again and you have a number of stages. I
think it is almost like a five stage vehicle; the ground having straps, then one after the
other. Then, you have the spaceship module on top which carry the astronauts and which
comes back. Maybe we will take a re-look when we are solving the problem.
(Refer Slide Time: 22:15)
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This is our PSLV; Polar Satellite Launch Vehicle of India again. And here, again you
find straps around the first stage - you have straps. Six straps or four straps could be put.
Then, you have the second stage, third stage.
(Refer Slide Time: 22:32)
And, of course this is the GSLV. The movie shows GSLV going up. Let us watch it
closely. First see the configuration. The configuration consists of, as you see it has four
straps or cluster of four engines; one, two, three, four. Then, there is a central engine.
First, the four straps fire. It generates thrust, the vehicle takes off. Then, the core also
fires and then you have a huge thrust that keeps accelerating it. And, once the four stages
clustered to the first stage have finished their operation, they are separated and falls
down to the ground. So also the first booster stage. Then, the second stage fires. This is
the inter stage. Then, it keeps going further. Then the third stage fires and the rocket
keeps accelerating. After all stages have fired the payload which is a spacecraft gets
separated and proceeds forward. And, once it reaches the particular orbit, it sort of
deploys. This is how we get the rocket goes up. Therefore, staging and clustering are
very important in the configuring a rocket.
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(Refer Slide Time: 24:01)
And, this is the GSLV. As we just now saw, these were the four straps. This is the first
stage, second stage and the third stage. First, these four stages are ignited. Four straps are
ignited, gives you the thrust to take off. Immediately after takeoff, the core is also
burning. Therefore, you have the huge thrust which pushes it. Then the second stage
fires, then the third stage fires. And, this is how a rocket functions.
(Refer Slide Time: 24:30)
Well, we also talked in terms of a water rocket; wherein we could have water and I could
pressurize it and launch it. Maybe we will solve this problem in class a little later.
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(Refer Slide Time: 24:47)
Well, I think by now we should be very clear about what is the principle of rocket, what
is the rocket equation as it were, what will be the rocket equation modified for a booster,
in which case rho VJ becomes more important than VJ itself. And then, we talked in
terms of staging, clustering or and also straps.
If the above is clear, let us go to the next part. We will address efficiencies.
(Refer Slide Time: 25:40)
We would like to know what is the efficiency of a rocket. What do you understand by
efficiency? We are looking at the rocket flying up; therefore I want to find out how
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effectively or efficiently it is being propelled or pushed. Therefore, we talk in terms of
propulsive efficiency. How do I define it? It has to do something about the forces, how it
is moving up or the power which you are giving, the power which is being used.
Therefore, I say, well, propulsion efficiency is something like what part of the power
generated by a rocket is converted to useful work done by the rocket per unit time. What
would it be? What is your guess? Useful work that a rocket does while it goes up. And,
what should it be? The actual work done by rocket. Work done by rocket per unit time
plus the work wasted by rocket is the power generated. That is the total work which is
done by the rocket per unit time. How to put these aspects together as an efficiency?
(Refer Slide Time: 27:32)
The work done by the rocket is equal to force into distance. The force of the rocket, we
found is equal to m° × VJ × distance, let say L. Therefore, I say work done by the rocket
per unit time is equal to m° × VJ × the velocity of the rocket. And, therefore we say so
much joules per second or so much watts is the useful work done by the rocket. Is it all
right? Useful work; or the useful power.
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(Refer Slide Time: 28:34)
Let us try to sketch the factors involved. We have a rocket going up. It goes with the
velocity V. It is pushed up by a force. Therefore, the useful work which is done by the
rocket is the force into the velocity per unit time. In the process is anything getting
wasted? What is the waste? How do we get the waste energy or work?
As the rocket is getting pushed, the plume from the rocket is going down. Again, we
picture the rocket going up. I am in the inertial frame of reference. I am standing here,
watching the fun of the rocket going up. What do I see? I see that this plume is now
going down with the velocity VJ with respect to the rocket or rather if V is the velocity of
the rocket is going up, the plume is going up with a velocity V minus VJ as I see it from
the inertial frame of reference.
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(Refer Slide Time: 30:11)
I will give you an example to illustrate. See, on some days, maybe early morning say 5:
30 or 6: 00AM you watch a jet aircraft up in skies. May be some of these jets skip
Chennai and you will see the trail behind it as the aircraft is moving. You see the aircraft
is going and the trail follows it.
Let us try to picture it out. Let us picture the trail. Aircraft is moving. You see the aircraft
going, and then you will find the whitish trail behind. The trail is also following at a
slower speed. Why does it have to happen? Maybe because this fellow is leaving the
aircraft with the velocity VJ; the aircraft is moving with the velocity and therefore you
see this particular jet or plume as it were following it a velocity V − VJ.
Therefore, what is being wasted in the rocket? The energy content of this is getting
wasted because it is getting lost. And, what do I see from the inertial frame of reference?
In the inertial frame of reference, we look at the work done by the rocket per unit time.
But, we also see that this work of the plume is getting wasted. And, what is my waste?
That kinetic energy is getting wasted or ½ mass of this into (V − VJ)2 squared or what is
the rate at which I am seeing is: ½ m° × (V − VJ )2. That is the waste. That means the
rocket is going up; this plume is still following it up like this. And therefore, this is
waste. It need not have got wasted.
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(Refer Slide Time: 31:37)
And therefore, how will I put the expression for propulsive efficiency together? I will
now write the efficiency as ηp (propulsive efficiency) is equal to the useful work done
equal to m°×VJ×V. This is divided by useful work m°×VJ×V + the kinetic energy per
unit time viz., ½ m° × (V − VJ )2 .
Please let us be very clear. You know, we will have to define the efficiency of the scram
jet. We will have to define the propulsive efficiency of an airplane. We will find that
there are some optimum values of efficiencies. And, I find some research work going on.
I will refer you to a paper in today’s class itself. The way people tend to think; can we
improve the rocket by looking the propulsive efficiency?
Let us first simplify this equation. This is equal to m° VJ V divided by the term. We
bring 2 on top. In the denominator, we get 2 m° VJ V + m°V2 + m° VJ2 − 2 m° V VJ. Is it
all right? V2 − 2 V VJ + VJ2. You find that this 2 m°VVJ gets cancelled; m dot gets
cancelled in the numerator and denominator. And, what is the propulsive efficiency
therefore equal to?
Propulsive efficiency is therefore equal to 2V VJ / (V2 + VJ2 ). Is it alright? V2 + VJ2 in
the denominator. Let us simplify it. Let us divide the numerator and denominator by VJ2
square and we get ηp = 2 V/VJ ÷ 1 + (V/VJ)2 .
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(Refer Slide Time: 33:54)
Now I want to ask you, when will the propulsive efficiency be a maximum. Just look at
this expression. Just be unbiased and tell me whether I can identify a condition for the
propulsive efficiency to be a maximum. We will anyway solve for the maximum. We
will find out the maxima and get the condition. But, by looking at this expression can
you tell me when should the efficiency become maxima?
Let us substitute V/VJ by x. All what we are saying is ηp = 2 x/ ( 1 + x2 ). What should
be the value of x which for which eta p is the maximum? Efficiency cannot be greater
than 1. It has to be 1. And, we find that the moment x is 1 or V by VJ is 1. It becomes 2
by 1 plus 1; 2. Therefore, by inspection itself I can say when V by VJ is equal to 1, then
the propulsive efficiency will be a maximum of 1. And, how do we do it? Normally, we
find the maxima.
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(Refer Slide Time: 36:02)
Let us differentiate it. Let us determine the condition for dηp by d(V/VJ) must be equal to
0 to give the maxima. And for that, you would say: denominator into differential of
numerator minus numerator into differential of denominator divided by denominator
square must be zero. Therefore 1 + (V/VJ)2 × differential of numerator which is 2 −
numerator which is 2 V/VJ × differential of denominator which is 2 V/VJ. And, this must
be equal to zero. Therefore, I am not really bothered about the denominator of the
differential and I need write 1 + (V/VJ)2 over here. And therefore, what does it give me?
It gives me 2 × (1 +(V/VJ)2) − 4 (V/VJ)2 = 0. What does this give you? 1 + 2 V/VJ +
(V/VJ)2 − 2 (V/VJ)2 or rather it gives 1 – (V/VJ)2 = 0. This means V/VJ must be equal to
1 to get the maximum.
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(Refer Slide Time: 37:57)
Therefore, if we were to plot the propulsive efficiency of a rocket ηp as a function of its
flying velocity divided by the jet velocity, the efficiency becomes a maximum when the
velocity ratio is 1. It increases initially till the ratio is one and thereafter it begins to
decrease. Therefore, this gives us some suggestion viz., that if I can have the exhaust
velocity equal to the velocity of the rocket which is high then the rocket flight will have
maximum efficiency when the rocket speed is also at this high value. Or rather, as the
velocity of the rocket changes, if I can somehow keep on changing my exhaustive
velocity, I operate the rocket at its maximum efficiency.
But then, you know that this is just not possible; a rocket is rapidly accelerating and the
condition is very difficult to meet this. But, there is some very interesting work on this
topic. And, one paper which deals with this and which is very exciting to read is this one.
I will give you the reference. May be you should take a look at it. It is by “King Jr MK”.
The title of the paper is “Rocket Propulsion Strategy based On Kinetic Energy
Management”. It appeared in “Journal of Propulsion and Power”. I just write it down
here; “Journal of propulsion and power” of AIAA. The volume number is 14. It is in the
year 1998 and the page number is 272-273. I would request each of you to take a look at
it.
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(Refer Slide Time: 40:39)
See, we know it is not possible to meet the condition of V = VJ.. But can you somehow
get something to do with VJ and try to see whether you can get a better propulsive
efficiency. Such ideas are useful. You know, because later on we will get into electrical
propulsion and nuclear propulsion. We will try to see, whether we can somehow make a
rocket more efficient by tailoring the jet velocity to be near to its speed of the rocket;
because as of today even to go to Jupiter, we saw it takes something like five years. If we
have to go to the Kuiper belt it takes something like ten years. Whether for galactic
missions - going to different galaxies- we can progressively change the exhaust
velocities. Therefore, this article by King is something which is useful. It says it is a blue
eyed tutorial.
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(Refer Slide Time: 41:44)
But having said that, I want to ask you one question; Why is it when V = VJ the
propulsive efficiency is a maximum? We had said that the rocket is going up and we see
the trail following it. What is the condition of the trail or the plume when V by VJ is 1?
That means the kinetic energy of the plume is what? Zero. What happens to the trail? I
see a rocket going up; the plume is also going up, what will happen to that plume when
propulsive efficiency is one or maximum? That means V minus VJ is 0; that means it will
be static; that means it has no energy at all.
In other words we are saying is, the plume as seen from the inertial frame of reference
does not follow the rocket. It just stays put at the given location. And in other words, we
have made use of the total kinetic energy for pushing the rocket up. And this is the
reason for the propulsive efficiency to be a maximum.
I shall now go through two or three case studies. I will start with the SLV 3 rocket on
how we calculate the masses, payloads, etc., in the first part. And then we shall do a
simple problem of the water rocket. But, since this water rocket is here on this particular
slide itself, maybe I get started with this water rocket.
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(Refer Slide Time: 43:18)
Let us try to solve this problem of trying to figure out the size of the neck or vent through
which water should leave the rocket in order to achieve a given value of acceleration.
We have the bottle which contains water; we have high pressure gases above it. I want to
push out the water out through the neck (vent) using compressed air. The compressed air
pressure is told to be something like 0.35 Mega Pascal. That is, 3.5 atmospheres. I want
to find out the size of the neck such that the rocket leaves the ground at a given
acceleration. The volume of the bottle is given, but I need to know the size of the vent or
hole such that the rocket can leave with the given acceleration. Let us do this problem.
(Refer Slide Time: 44:44)
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Therefore, let me write this problem on the board. You have in a bottle, 0.5 kilogram of
water. The air is relatively massless which is above the water and the pressure of air is
3.5 ×100 kilo Pascal. That is the 0.35 mega Pascal. You know, it is also told that the
mass of the bottle that is the structure containing the water is also 0.5 kilogram. And, all
what we are interested to know is, we would like this rocket to leave with a given value
of acceleration.
The level of acceleration is to be 0.5 g where g is the gravitational field due to the Earth.
The g value is 9.81 meter per second square. With respect to ‘g’, it is half. That is the
value of acceleration with which it must get pushed up. Now the question asked is, what
must be the size of the diameter of the vent or the hole by which the water should escape
from the bottle?
You have the mass of water 0.5 kilogram. Let us say the mass of water is mass of
propellant which is used for pushing up 0.5kg of the bottle. The mass of the structure is
equal to 0.5 kilogram. There is no payload in this problem. Just the bottle is moving up.
Therefore, the useful payload Mu is equal to 0 kilogram. Now, you can find out what
must be the rate at which water is getting pushed out and if you determine the velocity at
which water is getting pushed out, we can find out the diameter of this hole.
Therefore, how should I do this problem? What are the things that I should do? Let us
say I first want to find out what is the velocity with which the water will leave this
particular hole. In other words, I am interested in finding out the jet velocity VJ. How do
I get VJ? The gas pressure is 350 kPa; the ambient pressure is 100 kilo Pascal. That is,
the air pressure pa is equal to 100 kilo Pascal. Water is incompressible. Let us neglect the
height of water. May be from Bernoulli’s equation, we can say p for the compressed air
by rho plus what?
The pressure differential driving the water is p-pa which is 350 minus 100 which equals
250 kPa. From Bernoulli equation velocity square by the 2 is equal to delta p divided by
rho. And, that would have given us 3.5 minus 1 i.e., 2.5 into 10 to the power 5 divided by
the density of water; 1000 kilogram per meter cube. Then, what is the value coming out
to be? If you calculate, you get it to be equal to 22.36 meters per second.
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We have neglected the height of water in the bottle because the height is small and
because you have the pressure which is so high, then the height will not really matter.
But in a real problem, yes, I would like the height of water to be considered.
Now, we would like to make use of this jet velocity and find out what must be the
diameter of the hole. But, what is given to me? Something important is given to me. It is
told that the rocket should leave with an acceleration of 0.5 g or rather with an
acceleration of 4.905 m/s2. How do I get this? That means I must be able to calculate the
force. And, that force I have to convert it to acceleration and make sure I get this
acceleration. Let us revise what we have just now done and do this problem.
(Refer Slide Time: 52:50)
I would like to find out what is the force, which is generated by the particular rocket
divided by the initial mass of the rocket, which is the acceleration of the rocket. And, the
rocket is going up. Therefore, what is the acceleration with which it is going up? It is
equal to F minus the gravitational force of the mass divided by mi (weight of water); is
the acceleration with which it is going up. See, this differentiation is important. I say
force; force is what it is pushing it up. As it is pushing up; the gravitational field is also
exerting a force is equal to mi × g on the mass of the body. Therefore, F minus mi g
divided by mi must be acceleration and not Force/mi shown here. This later one is what
would be the acceleration when we have no gravitational field.
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Therefore, let us get back to the problem. The acceleration a is given to be as 0.5 g. Let
me erase this portion and write VJ is equal to 22.36 meter per second.
(Refer Slide Time: 54:15)
What is the value of F? Let us assume, let water flow at the rate of m° kilogram per
second; because if I know the water flow rate, I can calculate this diameter. Therefore, F
is equal to m° kilogram per second into what? How do you calculate the force? We have
done it. Change of momentum, m VJ is momentum; force is equal to d/d t of m VJ which
is equal to m°VJ. And, VJ we have already calculated. It is equal 22.36 into m° is equal to
the force.
Now, what is the force? I go back to the equation what I wrote here. I get (F − mi×g)
divided by mi is equal to 0.5 g. Here g is 9.81. What is the value of initial mass of the
rocket? 0.5 kg plus the structural mass is 0.5 kg, which is one kg. Therefore, F is equal to
0.5 g plus mi into g. mi is 0.5 plus 0.5 is 1. It becomes 1.5 g; that is 9.81 and mi was 1.
Therefore, force is equal to 1.5 into 9.81. I put it over here.
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(Refer Slide Time: 57:06)
And, what I get is the value of m dot. And, if I know m dot, how do I calculate the hole
diameter or diameter of the vent? m dot is equal to; we get the value, the density of water
into the velocity of water into the area. Area is equal to pi by 4 into the hole diameter
square. You have already calculated the mass flow rate; density of water is 1000
kilogram per meter cube. You have calculated the VJ. This is the velocity with which the
liquid is leaving; VJ square pi by 4. Therefore, the only unknown is the diameter of the
hole or vent.
I think I leave it as carryover homework for you to complete. All what I want to tell you
is that it is possible to calculate the thrust. You need the value of VJ; VJ we find through
simple calculation involving Bernoulli’s equation. And, once you know this, I can
always relate it to the acceleration.
I would like to do a problem on the masses; structural mass, may be the propellant mass
and the acceleration for a multi stage rocket vehicle. May be in the next class, I will just
go through one or two small examples on it and then we go to the next topic, which is on
nozzles.
.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 08
Examples Illustrating Theory of Rocket Propulsion
and Introduction to Nozzles
Good morning: First let us recap what we have done so far in a couple of minutes. And
then, we will solve one or two small problems, such that we are fully aware of the theory
of rocket propulsion and then move over to a new chapter on nozzles.
(Refer Slide Time: 00:31)
We discussed the Rocket Equation first formulated by the Russian school teacher
Konstantin Tsialkowsky. We derived the rocket equation and found that the ideal
velocity increment of a rocket is given by the efflux velocity or the jet velocity
multiplied by the natural logarithm of the ratio of the initial mass to the final mass of the
rocket. We also found that for a rocket with a smaller amount of mass of propellant
compared to the total mass of it, we found that this equation gets slightly simplified and
density of propellant times the jet velocity becomes important parameter and this
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multiplied by the volume of the propellant divided by the initial mass of the rocket gives
the velocity increment by the rocket. Therefore, we will try to do one or two small
problems, such that we illustrate how this ideal velocity works. We have seen what the
payload mass fraction is, we saw what is the propellant mass fraction and we saw the
structural mass fraction. We related these fractions in the earlier classes. Let us go ahead
and solve one or two small problems.
(Refer Slide Time: 02:04)
I take a problem of the 4 stage rocket to begin with similar to SLV 3 about which we
discussed in the last class. Let us say, I have a booster stage which is the first one, which
first takes off from the ground and then on top of it I have the second stage, then I have
the third stage and then I have the fourth stage and on top of it fix my useful mass which
we call as payload. Let me take a typical example, in which the top mass that is the
useful mass, which we called as useful mass, is equal to 40 kg.
The propellant mass of the first stage, is equal to 9000 kg and the mass of the structure;
that means, the casing, insulation and inert of the first stage is equal to 1500 kg. The jet
velocity of the first stage is equal to 2200 meters per second. Now, on top of this stage is
the second stage.
The second stage has a propellant mass equal to 3500 kg and the mass of the structure of
this stage is equal 550 kg. On top of the second stage, you have the third stage, for which
the mass of the propellant is equal to 1700 kg, and its structural mass is 250 kg. Then we
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said a fourth stage whose mass of propellant is 260 kg and the mass of the hardware
including the structure is equal to 40 kg.
The jet velocity of the first stage is 2200 m/s, the jet velocity of the second stage is
slightly higher at 2400 meter per second. For the third stage, it is 2500 meter per second
and the fourth stage the jet velocity is 2750 meters per second. Therefore, we have this 4
stage rocket, on top of which you have a useful mass or payload mass of 40 kg. It takes
off from the ground. We would like to determine the value of the delta V of this rocket or
the total value of the ideal velocity, which the rocket gives.
(Refer Slide Time: 05:59)
The first slide at 05.59 gives the velocity increment (ideal velocity) of ΔVI, that is
available to the rocket when the second stage starts and so on and therefore the net ideal
velocity of the rocket is equal to what is provided by the first stage plus what is provided
by the second stage plus what is provided by the third stage plus what is provided by the
fourth stage. Therefore, our main effort has to find out what is the delta V of the first
stage, second, third, fourth stages. I just arithmetically add it all up and I will know what
will be the ideal velocity given to the payload by this fourth stage vehicle.
Let us do the calculations. Let us do for the first stage and similarly we can do for the
second, third and the fourth stage. The first stage, you find that the exit or the jet velocity
is equal to VJ is equal to 2200 m/s. Therefore, we have 2200 meters per second into
logarithm of the initial mass of the rocket to the final mass of the rocket. What is the
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initial mass of the rocket at this point in time? It includes all the mass of propellants, all
the mass of structures plus the payload.
Therefore, if we were to write the expression for the initial mass, which as we saw in this
expression for ΔV to be the numerator, it is equal to for the first stage, the propellant
mass is 9000, the structural mass is 1500, now we add the second stage which is equal to
3500 plus the structural mass is 550, we keep adding up for the third stage, the propellant
mass is 1700 plus we have 250 for the mass of the structure plus the last stage we have
260 plus 40 plus and the useful payload which is plus 40. Now, what is the final mass of
this rocket after the first stage is performed? What must, I remove or what must be done?
First stage propellant gets burnt out and therefore, the total mass minus the mass of
propellant in the first stage which is knocked off. Alternatively, I have 1500 kg as the
mass of the first stage structure plus we add the mass of the other stages viz., 3500 plus
550 for the second stage plus we have for the third stage which is equal to 1700 plus 250
plus we have 260 plus 40 for the fourth stage plus we have the value of 40kg for the
payload. In this way we can get the value of ΔVI. Let us substitute it. We therefore get,
ΔVI is equal to 2200 ln of the masses in the numerator 9000 + 3500 + 550 + 1700 + 250
+ 260 + 40 + 40 and this gives me the total initial mass. We get this value to be 16840
kg. If we were to subtract from it the mass of propellant in the first stage, we get the final
mass of the rocket at the end of the first stage of operation as subtract the value of 9000,
that is 16840 minus 9000 which is 7840 kg. Hence ΔVI is therefore 2200 ln(16840/7840)
and this gives the velocity to be 1.682 kilometers per second or 1682 meters per second.
Let us put the value here. We get the velocity of the first stage ΔVI as equal to 1.682
kilometers per second.
Similarly, we do for the second stage. I know what the initial mass of the rocket at the
start of the second stage is. Similarly, we get the final mass when its propellant is burnt
out. We start with the initial mass of the rocket which should be the second stage, plus
third stage, plus fourth stage, plus the payload weight. And what is the mass of the
second stage? 3500 plus 550 that is 4050. Now, the third stage 7700 plus 250 that is
1950, the stage fourth stage 260 plus 40 which is 300 kg plus useful 40 kg, And what is
the mass at when the second stage has stop functioning? It would be less by the second
stage propellant mass of 3500 kg. Hence, ΔVII of the second stage, is equal to the jet
velocity which is 2400 meters per second multiplied by the logarithm of the ratio of these
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mass before and after. This will give me a value of 1.927 kilometers per second.
Therefore, the second stage gives a ΔVII of 1.927 km/s. Now, we repeat the calculations
for the third stage and how do we get the calculations for the third stage? The jet velocity
is 2500 into logarithm of the mass of this plus the fourth stage plus the payload which is
1700 + 250 + 260 + 40 + 40 which is the initial mass. At the end of the third stage I have
depleted this amount of propellant for the third stage. I am left it 250 + 260 + 40 + 40.
This gives me the velocity increment of the fourth stage which gives velocity of let us
put this over here. We have a slightly higher velocity of something like 3.39 kilometers
per second.
(Refer Slide Time: 13:36)
And let us do for the last stage i.e., the fourth stage gives a velocity increment of ΔVIV.
And we find that the jet velocity of the fourth stage is 2750 meters per second. That is
2.750 kilometers per second into logarithm of the initial mass is 260 is the mass of the
propellant + 40 is the structure + plus 40 and what is left after the fourth stage burns? 40
plus 40 is what is left. And this gives me the velocity as equal to something like 3.979
kilometers per second. The fourth stage gives me a velocity of 3.979 km/s.
Now, let us try to draw some inferences from the above. We find that the first stage gave
a velocity increment of 1.682 kilometers per second; the second stage gives me a slightly
higher value 1.927; third stage gives a significantly higher value 3.39, while the fourth
stage gives even higher value of 3.979 km/s. That means, the upper stages contribute
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more ΔV than the lower stages: the reasons being, one is the jet velocity of the lower
stages is generally smaller than the upper stages though larger quantities of propellant are
burnt. And th second point is we have the benefit of a lighter rocket towards the end
which can give you a higher ΔV. And therefore, what is that total delta V of the rocket
we just add the velocities.
That means, the total delta V which is the increment provided by the rocket, that is the
addition of 1.628 km/s provided by the first stage + the second stage which gave us
1.927 + the third stage which gave us 3.39 + the last stage which gave us something like
3.98 kilometers per second. And the total velocity what we get is therefore equal to
10.978 kilometers per second. We must remember, that we have neglected the velocity
gains when the discarded stages are removed. And this is the velocity which is available
to you for orbiting plus the potential energy or the velocity to increase the height of the
rocket from the ground to the particular orbit. And how do you match it with the orbital
velocity and the total velocity?
(Refer Slide Time: 16:33)
You already derived the expression, that the total velocity provided by the rocket can be
written as √ GME/RE × (RE + 2 h) ÷ (RE + h) and this will tell you at what height this
particular rocket can launch the particular satellite. And this is how we do any velocity
increment for any multistage rocket.
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To be able to illustrate about initial acceleration and the need of a strap-on or cluster of
rockets, I give you some more data and let us do a small problem as an extension of the
above problem. Suppose, we want to find the acceleration of the first stage rocket, when
it takes off from the ground.
(Refer Slide Time: 17:22)
The first stage operates for let us say, 50 seconds. And the question posed now is; if the
first stage operates for 50 seconds determined the initial acceleration of the vehicle or
this particular rocket. Now, how could we do this? That means, each stage operates may
be, the first stage operates for 50 seconds, the second stage for 35 seconds and so on.
May be the final stage operates for let say another 80 seconds. But the data which is
given to you is that the first stage which has a propellant mass of 9000 kg, operates for a
time of 50 seconds. The flow rate should be given; let us presume that the flow rate of
the mass leaving the rocket is a constant.
Therefore, we can say the rate at which propellant gets depleted that is dMp/dt is
constant and is Mp divided by 50 seconds which is equal to 9000 kg is the mass of the
propellant and it is getting depleted over 50 seconds. Therefore, the rate at which the
propellant is leaving the nozzle is equal to something like 180 kilograms per second.
You find the mass flow rate from a nozzle is quite high. We are talking of several
hundred kilograms per second and when the rocket is huge, like what we said is the
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moon rocket it will be very much higher. Therefore, the mass propellant mass flow rate
or rather dMp by dt, which I can also write as m° is equal to this value.
Now, we know the value of VJ to be 2200 m/s. I want to find out the initial acceleration
of the rocket. To be able to find the acceleration, I need the force produced or the thrust
of the rocket and what is the force? The force F which the rocket gives is equal to d by d
t of change of momentum, and the change of momentum is equal to Mp into VJ. The
thrust is equal to m° into VJ, which equals 180 kg/s × the value of VJ of 2200 m/s. what
is the unit? Kg per second into meter per second; kg meter per second2; this is in
Newton.
(Refer Slide Time: 20:43)
We find that the force which the rocket develops or the thrust is equal to 180 into 2200
Newtons. Therefore, what is the acceleration at take off? We find acceleration of this
particular vehicle at take off is equal to the force minus the initial mass which is
subjected to the initial gravitational field of the earth, divided by Mi. Mass into
acceleration is equal to the net upward force. And therefore, the acceleration at take off
for the rocket is equal to 180×2200 − the initial mass of the rocket 16840 kg × the
gravitational field of 9.81 m/s2 divided by 16840. This gives me the value of acceleration
as equal to 13.705 meter per second square. This is Newton divided by kilogram and the
unit is meter per second square.
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Therefore, you find that the rocket is leaving the ground with an acceleration which
is something like may be 1.4 to 1.5 g. This is the level with in g with which it is leaving.
Is there anything else we could do? Supposing, I want the acceleration when the fourth
stage just starts operation i.e., it fires. Let us say, I know the time over which the
fourth stage operates and once I know the value of m°, I can calculate the force. I
correct for the attraction of this initial mass by the Earth, divided by the total mass
that will give me the acceleration. I can therefore, find out the acceleration at any
particular time and this is how we determine the acceleration. The last thing,
which I would also like to know is what is the payload mass fraction of this rocket?
(Refer Slide Time: 23:03)
Let us say, what is the value of α which we said is the payload mass fraction? Can you
tell me what it should be? Useful payload, 40 kg divided by total vehicle mass and that
we said is 16840 kg. It comes out to be a very small number. Let us put down this
number. It is 2.37 into 10−3. Or rather the net useful fraction that comes out of this rocket
is something like 0.2 percent. You find that in rockets lot of energy gets expended.
And the aim of a rocket designer is to improve this fraction to something like 2 percent,
3 percent, which is what we will be considering in the subsequent classes. We talked of
the payload mass fraction. Supposing we ask in terms of the structural mass fraction let
say β. But then, we have four stages, first stage, second stage, third stage, fourth stage.
Let me say, I am interested in the structural mass fraction of the first stage. What will be
the value? Yes, the structural mass is 1500 kg divided by what: here we should be a little
careful, I am asking for the first stage.
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The first stage consist of 1500 kg as structural mass and the balance 9000 kg as the
propellant mass. Therefore, the total mass of the first stage is 1500 plus 9000 =
10,500 kg and the structural fraction mass of the first stage is 1500 divided by 10,500.
If I consider the second stage well it is going to be something like 550 divided by 3500
plus 550. We can calculate for this third stage and for fourth stage. And this is how we
calculate the payload fraction, structural fraction, may be a propellant fraction. We also
calculated that the total velocity and the acceleration.
Having done a problem for a rocket let us do one for the satellite using the same theory.
We had said that a satellite carries some amount of fuel or propellant and the movement
propellant is consumed, the useful life of a satellite is over. Therefore, let us take an
example.
(Refer Slide Time: 26:21)
Let me take an example of let say INSAT. The INSAT satellite consists of something
like a cube or box like structure: this is the basic structure, inside this is housed the
electronics and propellant tank. And we use the propellant for correction of the orbit of
the satellite. We will get into details of this when we do the liquid propellant rockets.
We attach a series of rockets at the edges over here, a lot of them something like 16 of
them, such that, we can make the small corrections in velocity whenever required in
space.
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Supposing, we need to determine the quantity of propellant to be carries in the
INSAT spacecraft. The dry mass of the satellite is given as let us say 800 kg. What is
this dry mass? Mass of the structure, mass of the tank, maybe I could have
something like an antenna, solar panels, sensors, etc.
To be able to maintain this satellite let say for 10 years or 15 years, we need to be able to
make the necessary corrections for its attitude, orbit and also to push it out of the
geosynchronous orbit once its life time is over. I need to be able to configure the rockets
such eventualities and the total ΔV required for the corrections and eventualities is given.
Let us assume that this velocity is 950 meters per second. This is equivalent to the
incremental velocity provided by the rockets.
I should qualify the correction velocity that is required to be provided. Whenever you
have the satellite not pointing correctly, I must give a small impulse to the satellite; that
means, I have to give some change of momentum to the satellite to deflect it. I know the
mass of the satellite therefore, I can find out what is the corresponding delta V which I
must give to the satellite. I keep on adding the different ΔV for attitude control, for
station keeping, etc., including the final push away out of the geosynchronous orbit. I
find that during the total life period of this satellite (over several years) I need to give a
delta V of 950 meters per second. Now, as a designer we should know how much fuel or
how much propellant do we carry in this space craft? This is what I want to determine.
In other words, all what we are saying is, the ΔV to be provided by the rockets is equal to
950 meters per second. And then, we need to know, what is the jet velocity of the
different rockets and then we say of the initial mass divided by the final mass of the
satellite. Initially we have some mass while after propellants are exhausted we have a
final mass. What is given is that dry mass? Dry mass is the final mass when there is no
propellant. Therefore Mf is given as equal to 800 kg. The rockets are designed to give a
certain VJ. Let us presume that the value of VJ is equal to 2500 meters per second.
In the next class we will find out how we calculate this value; Therefore, now my
question is what must be the mass of propellant which I carry in this particular rocket?
The initial mass of the satellite will also contain the propellants in it.
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(Refer Slide Time: 30:24)
Therefore, the initial mass is equal to mi kg, dry mass of 800 kg is the final mass. The
initial mass is therefore 800 plus the mass of the propellant. And we find that ΔV what is
required is 950 meters per second should be equal to the jet velocity is 2500 into
logarithm of 800 + the mass of the propellant ÷ 800. And we know, what is the mass of
propellant to be taken in the satellite. Let solve this equation, we get 800+Mp (mass of
propellant in kg) / 800 kg is equal to e950/2500. We take exponential on both sides to get
the above. 950/2500 = 1.462. Or rather, I get the value of Mp that is the propellant mass
as 370 kg.
If the satellite has to be operational for 20 years or more, we may require more of
fuel because more corrections are required. If we want it for one year I can have much
lower mass. Whenever we read in the news that the initial orbit has a significant
error and therefore the life time of this satellite comes down. Why does it come
down? Initially, itself I take some of this fuel and use it for the corrections and
therefore, the propellant available for the station keeping of this satellite and the attitude
orbital corrections are decreased. Are there any questions?
Your question is, how did I get 950 meters per second over a given period? See you
know that, the satellite is slightly getting drifted, because I have the gravity of the moon,
may be gravity of some other planet which is there or else some aerodynamics itself.
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Therefore, I know the mass of the satellite, I know what is the force which is required to
give the orientation. Therefore, I can calculate what is the value of impulse for particular
correction, once I know the correction at that movement of time, I know what must be
the initial mass. Therefore, I know what is the value of correction required; may be for
the first correction like that you know daily or once in a few days we require corrections.
Therefore, I keep on adding all these corrections and the sum of all these things is what
gives me a value of 950 meters per second. We do not really do a force balance; we just
specify the ideal velocity required for the corrections
Let summarize once again. In order to appreciate the points made so far, I show a scale
model of the GSLV rocket of the Indian Space Research Organization. Here we see that,
we have a core stage at the bottom followed by the second stage over here, followed by
the third stage over here. The core stage is surrounded by 1 2 3 and 4 straps; that means,
you have a core stage followed by 1 2 3 4 rockets, these 4 rockets are a cluster and are
known as straps.
At the beginning of the mission, the core and the straps either fire together or the straps
fire before and immediately the core fires so that, you get a huge thrust, which can carry
the entire mass of the launch vehicle. Once, the firing of the straps are over, they are
discarded and thereafter the core fires for a small additional time and then this is also
removed. The second stage then operates and once the second stage operation is over, it
is also removed and it is thrown out and then the third stage takes over. The third stage
fires and takes this space capsule which is sitting on the top of the third stage and puts it
into orbit; but before putting into orbit, the third stage is also removed.
Therefore, we see in this particular rocket you have something like a core rocket, you
have four straps, you have the second stage rocket, you have the third stage rocket on
which you have the space capsule i.e., the satellite and it is this satellite which is put into
orbit. The satellite, which is put in orbit, is the INSAT satellite. I show a scale model of
it; the satellite consists of a box like structure, which is shown in brown over here. You
have the solar panel here, which takes the energy from the sun converts it to electricity.
You have a balance for the solar panel mass - a solar boom over here and you have the
antennas.
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But what I really wanted to show was, you have in red 1 2 3 4 similarly, you have 1 2 3
4 something like 16 to 18 thrusters which are small rockets which are mounted at the
edges of the satellite, which will correct for the position of the satellite, may be the
orbit and also the position of the satellite. Therefore, even a satellite in orbit has
rockets attached to it which give it some impulse. (Refer Slide Time: 37:43)
VJ is a very important quantity that is the efflux velocity, unit is meter per second. We
talked in terms of Tsialkowsky’s equation or Rocket equation, which told ΔV is equal to
VJ into the natural logarithm of initial mass by final mass. We had the term impulse of a
rocket, what is impulse? The change of momentum. This equals mass of the propellant
into VJ and is the impulse, what is the unit here? We are talking of kilogram meter per
second. This could also be written as kilogram meter per second square into second,
which is same as Newton second.
Let us be very clear about units: We have kilogram meter per second which is
impulse. We can also write it as kilogram meter per second square into second which
is nothing but Newton second. Therefore, impulse could expressed in kilogram meter per
second or Newton second. This is the total impulse which is given by the Mp × VJ,
because this is what is moving out. Now, I ask myself a question. Can we say impulse
per unit mass of propellant, which could call as specific, instead of calling as impulse
call it as specific
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impulse that is impulse per unit mass and that gives me the value as VJ. And what is the
unit I get? Newton second by kg and therefore, you find that the jet velocity and specific
impulse are the same.
We will make some corrections for the mechanism of flow taking place in a rocket. We
will find that it may not be identical, but this is the way to go about it. You know, in
many text books, they express the value of specific impulse in seconds which is really
not correct. We are talking of force divided by mass flow rate; because I could have also
written Isp as equal to force divided by m°. Why do I write it? Force into time is impulse
and therefore, this is same thing as impulse over total propellant mass.
Therefore, if we write instead of mass the weight or rather the weight rate of flow of the
propellant or we alternatively express force in kg instead of in Newton, I am left with a
unit of Isp in seconds which is really not right because the unit of force should have been
Newton and the units should be Newton divided by kilogram per second, which gives me
Newton second by kilogram. In fact, units are important. Therefore, please be careful
when you read a book. If the unit is specified as seconds for specific impulse may be the
unit being considered for force is kilograms. Therefore, it becomes necessary for us to
multiply it by 9.81 and then have this unit of Isp as N s per kg.
May I will take an example as I go long. You know, see there is always a problem.
People talk in terms of mass flow rate, they talk in terms of weight flow rate as you are
mentioning, but can we say weight flow rate? It is actually mass flow rate. You cannot
have weight flow rate since you cannot have force which is flowing out. We must
distinguish between mass and weight correctly. Whenever I measure a mass by a spring
balance, it is the attraction and therefore, we measure a force that is a weight, where as
when we consider quantity of matter it is mass or quantity of matter.
Let us keep our definitions clear for as masses always imply quantity of
matter which is Kg; impulse is Newton second, impulse per unit mass of propellant is
Newton second by kilogram or force by mass flow rate over here. I think these
definitions are important.
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(Refer Slide Time: 42:42)
Let us now come back to the next part, namely we ask how can we obtain a high value of
VJ? We said that it essential to get a high jet velocity. We had mentioned that when
Tsiakowsky derived the rocket equation, since the cathode tubes were coming up in
those days, his idea was to use the high jet velocity of electrons. In fact around that time
Robert Goddard in US was also thinking in terms of high velocity electrons to be used in
a rocket. But then electrons have a very low mass and therefore force that we get is
small.
In fact, you also had that the third rocket pioneer by name Herman Oberth at about the
same time. He was from Austria. He wrote a beautiful book on rocket propulsion and
you will be surprised many of the things we do in rocket today remains exactly similar to
what he suggested at that point in time. And in fact, what he said was you put one stage
after other and I can get a high velocity like what we did in staging of rockets. And he is
also a great rocket pioneer.
I would like to now address on how to get the high jet velocity VJ.
Consider a chamber in which I have a high-pressure gas filled with pressure Pc. Now I
make a hole in the chamber. I know the gas will escape out. I want to find out the jet
velocity or the efflux velocity of the gas. Let me take you through a small example, I
have a balloon, because this tends to be somewhat easy to illustrate. Therefore, what I
am trying to consider is I have a balloon filled with air, at a pressure Pc. I make a small
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opening here, and I want to find out the velocity with which the gases are going out. And
this is exactly what we do in rocket propulsion. In other words, we allow high pressure to
be built up in a chamber and allow the high pressure gas to escape through a vent. The
gases escape at some velocity and the balloon goes up. It is pushed up similar to a rocket
being pushed up. If I could have a controlled opening, I could get continuous thrust.
And therefore, let us again fill the balloon with air by closing this vent or hole. I now
open this vent, and I find the air going out at a certain velocity. I want to calculate the jet
velocity of this particular air, which is leaving the balloon. Let us do this problem. We
may increase the pressure; we increase the flow rate and the velocity. Our aim is through
this small opening what we have here, what is the value of Vj that we get in meter per
second. We have done such problems in the thermodynamics course in the first year
engineering, but let us repeat it.
(Refer Slide Time: 46:52)
I draw a huge reservoir, which I call as a chamber. The pressure in it is Pc; I provide a
vent over here. I allow to this particular area of opening for the vent. I want to calculate
the value of velocity of the jet Vj, when the pressure at the exit of the vent is pe. The
vent or hole is spoken of as a nozzle.
What do I have to solve for: I know the pressure here, I know the pressure at the exit, I
want to know the jet velocity. I am interested in this particular vent or nozzle, which has
a given shape. I do not know the shape, what I just showed you something like this. Air
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enters at the pressure Pc; air leaves at a pressure Pe I am interested to calculate the value
of Vj.
This is a control volume, why do I say a control volume; I have a fixed volume in space.
And what is happening in the control volume is that air is entering at a high pressure and
leaving at a low pressure. And therefore, I solve for a control volume. And to be able to
solve this problem, I have to make some assumptions. What are the assumptions I could
probably make?
Let us say this is the vent. Let us assume the flow to be adiabatic, that means the vent is
such that there is no heating of the air in the vent or no heat comes from outside into the
nozzle. In other words, I say Q which is entering this particular boundary of the control
volume is zero. Let us for the present also assume that this vent is rigid; if it is rigid, it
cannot expand; it cannot do any work. Therefore, the work done by this vent that is Wx
is equal to zero.
See, I could have something like a flexible vent, which could move and it can do some
work. But I assume that it is rigid, when it is rigid I have the work done is zero since
there is no displacement. Therefore the assumptions are that the vent is such that it is
adiabatic, that heat transfer is zero and the work done by the vent is zero. Across the
surface what we get the work done is zero. The flow can also be assumed to be steady.
Now I have to write the steady flow energy equation for the control volume.
(Refer Slide Time: 50:01)
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The flow has been assumed as being steady. What do you mean by that the flow is
steady? The mass, which is entering by vent and the mass leaving the vent is the same. If
we have Q the rate at which heat is entering the vent and Wx rate at which work is done
by the air, what would happen? Let us say that the mass of gas, which is entering the
vent and leaving the vent is m°. You have enthalpy, which is entering and which is
leaving and because you have some heat which is coming in and the work is done at the
vent surface. The energy entering is the enthalpy hi plus kinetic energy plus potential
energy at the entry to the vent and what is leaving is the enthalpy he at the exit plus I
have a kinetic energy at the exit plus I have a potential energy at the exit.
Why do we write enthalpy here to consider the heat energy, because you have internal
energy and it also has a specific volume that has some flow work? Enthalpy is equal to
internal energy plus p into specific volume. Therefore, we have for the energy balance: Q
minus Wx is equal to what is the enhancement in its energy; it has got in the enthalpy,
kinetic energy and potential energy during its motion through the control volume. Q and
Wx being zero are dropped and this left hand side is zero; since the flow rate at the entry
and exit are the same, m° also cancels out. .
So if we have he plus the kinetic energy per unit mass, it is equal to VJ2/2 where Vj the
exit velocity. I have potential energy, potential energy is equal to g into ze or the height
above the datum at the exit. At the entry, we have hi plus kinetic energy at entry plus
potential energy at the entry. What is the velocity with which the gases leave the
chamber? See, the chamber is huge, this is small; therefore the velocity what we get at
entry into the vent is almost zero, that means I can I can neglect the velocity at the entry.
The potential energy per unit mass is g zi where zi is the height above the datum at the
entry.
If the nozzle or the vent is small, the change in height will be very small and I can cancel
out the g zi terms containing potential energy. And we get VJ2 divided by two is on the
other side and we get it equal to hi enthalpy at the entry minus enthalpy at the exit he.
We will continue with this in the next classes and try to see under what conditions can I
get a high jet velocity. To summarize, we looked at two problems illustrative of the
principle of rocket propulsion. And then we went ahead and try to find out what is the jet
velocity with which a pressurized gas will squirt out of a vent.
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Rocket propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 09
Theory of Nozzles
(Refer Slide Time: 00:14)
Good morning. We will develop the equation for VJ. We have a high pressure container.
It contains a gas at chamber pressure Pc. We have something like a hole or a vent
through which the gas squirts out at velocity VJ.
We wish to find out the efflux velocity VJ, which is called as jet velocity or we call it as
efflux velocity VJ. However, before I do this since there was a question of specific
impulse, and unit of specific impulse let us just spend some 2 or 3 minutes on this issue.
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(Refer Slide Time: 00:57)
What was specific impulse? We told that specific impulse is equal to impulse I divided
by the mass of the propellant, that is the impulse generated from unit Mp mass of
propellant. Impulse is change of momentum and therefore is equal to Mp into the
velocity VJ divided by Mp which is equal to VJ . As per this logic, the unit of specific
impulse should be meter per second, which is same as the efflux velocity VJ. But how
did we define specific impulse? It was defined it as impulse per unit mass of propellant
or equivalently thrust per unit mass flow rate of propellant. In other words impulse has
unit of momentum, change momentum viz., kilogram × meter per second ÷ kilogram and
when we say specific impulse we get back m/s. From force considerations, we get force
into a given time; that means, we can write the change of momentum as equal to
kilogram, meter per second square into second divided by kilogram and this is Newton
second by kilogram. Therefore I also see that unit of specific impulse can be expressed in
Newton second by kilogram which gets reduced to m/s.
So far so good. Let us derive the units of specific impulse by expressing specific
impulse as equal to impulse per unit time (Force) divided by mass flow rate of propellant
per unit time. And impulse per unit time is force, force into time is impulse, in other
words I have force divided by m°. In other words force as unit of Newton, mass flow rate
of unit kilogram per second, therefore the unit of impulse specific impulse comes out to
be Newton second by kilogram which again reduces to m/s.
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Therefore, whether I express specific impulse as impulse per propellant mass or specific
impulse as force per unit mass flow rate we get the same unit as Newton second by
kilogram, therefore the unit for specific impulse is Newton second by kilogram. The unit
for impulse should have been Newton × second, that is what gives the value of impulse
kilogram meter per second as Newton second. Therefore let us keep ourselves very clear;
impulse has unit of force into second: Newton second, specific impulse has unit of
Newton second by kilogram, but well there are many text books which specify the
specific impulse not in meter per second, not in Newton second by kilogram, but as
second.
(Refer Slide Time: 04:31)
See when we say the specific impulse is so many meters per second, we could have
multiplied both numerator and denominator by kilogram and kilogram, and what is it we
get; we get the unit of VJ as equal to kilogram meter per second, divided by kilogram.
Now we again multiply the numerator and denominator by second to gives us kilogram
meter per second square × second per unit kilogram. This gives us Newton second by
kilogram. Therefore meter per second is actually identical to Newton second by
kilogram. Either of the units is ok,
Why is it some people use the unit as seconds. They specify the force in kilogram or
pounds, therefore pounds and pounds gets cancelled and second is left. Therefore
whenever somebody gives the units in second it is our duty on Earth to multiply by the
211
gravitational field gc, and then use it in Newton second by kilogram. May be you all
should go through it, but in this class we will always address specific impulse in Newton
second by kilogram or in terms of meter per second both of which you see have the same
identical units.
Units are very important in engineering. With wrong units we will be talking of
something but getting of some other numbers. Let us get back to our problem on VJ. We
want to find the VJ. We derived the equation for control volume we said Q minus the
work done by the particular vent Wx divided by let say m° is equal to the enthalpy which
is leaving (let say he) + the kinetic energy per unit mass which is leaving – (hi + the
kinetic energy per unit mass that is entering the vent. We told that the change potential
energy between the exit and the entry is zero and thereforewe can neglect it.
(Refer Slide Time: 07:09)
And we told that we are to looking at hi per unit mass, this is enthalpy hi for unit mass
and let us put the units clearly; the unit of he is therefore joule per kilogram, the kinetic
energy is equal to V2 divided by 2 and that is equal to meter squared per second squared.
We multiply numerator and dominator as usual by kilogram, meter square by second
square, this is equal to kilogram meter square per second square. This gives Newton
meter per kilogram, which is equal to joule per kilogram. Therefore the kinetic energy
V2/2 has units of joule per kilogram; that means we are taking per unit mass of the gas
which is moving and determinig its kinetic energy.
212
Therefore let us puts things together. We told that the process of expansion is adiabatic
in the vent, it is not something which can dilate the nozzle and work can be done by it. If
you were to apply the same problem to our heart valve, see heart also pumps our blood,
but the valve is also something like flexible when we write the same equation for the
control volume for the flow by blood through one of the arteries and valves in our body
for which Wx is not 0. May be that is what is makes modelling of the heart more difficult
compared to a vent over here. Therefore now let us put things together: we have VJ2
which is equal to the exit velocity squared and that is equal to 2 of the enthalpy which is
entering minus enthalpy with exiting the nozzle i.e., VJ2 = 2 (hi –he).
Now, we want to solve this equation and we want make sure that we solve it in terms of
the properties of a particular gas. What are the gas properties? They could be the
temperature, could be the molecular mass, could be some other property which we need
to consider. Now to be able to solve this we have to make further assumptions. Let us
assume that the gas is ideal and what do you mean by the gas is ideal? A gas is ideal
when the enthalpy per unit mass or the specific enthalpy and specific internal energy are
only functions of temperature. I think this definition is important. Let me just briefly go
through the definition of the enthalpy is only a function of temperature and the internal
energy is only a function of temperature for which we say that the gas is ideal.
(Refer Slide Time: 10:10)
213
And what is the consequence of this, we have the definition that h minus u are rather we
define enthalpy as equal to u + p × specific volume v, and therefore we find for an ideal
gas if h is function of temperature, internal energy is the function of temperature plus p
v, are rather we get p into the specific volume is only a function of temperature, and
therefore we write it as R T. Therefore an ideal gas for which enthalpy and internal
energy are only the function of temperature has an equation of state which specifies p v
is equal to R T, and this R is what we call as specific gas constant. Let us again just
repeat what we said.
(Refer Slide Time: 11:04)
We told that for an ideal gas may be h is only a function of temperature, similarly u is
function of temperature and now if we take the slope anywhere we get the value of
specific heat Cp is equal to dh by dT and we get Cv is equal to du by dT. How did this
come for any system? We have δQ minus δW is equal to du and for a constant volume
system work done is 0, therefore, heat required per unit temperature change per the unit
mass is therefore the value Cv is equal to du by dT. That means for a constant volume
system work done is zero and I get this value. For a constant pressure system what did
we do we defined dh as instead of du?
I write now du as dh − d (pv) and the last term gives p dv and v dp. I have p dv over here
it cancels with p dv of du and I get Cp for a constant pressure process as equal to dh by
dT. We are still considering an ideal gas with specific gas constant R and what is a unit
214
for R, let us put it down pressure, Newton/ meter2 × volume, meter cube by kilogram by
its specific volume ÷ Kelvin giving Newton meter viz., joule per kilogram per Kelvin;
that means we have joule per kilogram Kelvin, which for air is about 287. It is a value
specific to the gas, air has value 287 and may be CO2 will have a lower value, may be
helium may have a higher value, this is R depends on the type of gas which we use.
Therefore we say well I have the equation of state of this gas given p v is equal to R T, or
if I consider the volume of a gas V which has a certain mass m, we can write the same
equation pV = m R T, and this is what we have been studying in thermodynamics.
Let us now go forward. We find h is the function of temperature, u as a function of
temperature, the curve is varying, Cp is also a function of temperature and Cv keeps
varying with temperature. If I have to write from this equation, the value of dh, we get as
equal to Cp dT or rather the enthalpy of the mass of gas m is given by H = m Cp T. But
Cp changes with temperature now my problem is going to become more complicated
because I have to consider the Cp variation with temperature.
Therefore, I put another idealisation and in this idealisation I say that the gas is even
better than ideal; I call it as a perfect gas. A perfect gas is one for which Cp and CV are
constant in addition to the value of h and u being as a function of temperature. This
implies one more assumption and we find that the functional dependence of enthalpy on
temperature is not like this but the functional dependence is straight line. Cp and Cv are
constants for a perfect gas.
Therefore, we will solve this equation assuming let say a perfect gas. If the gas pressure
is pC and it temperature is Tc and if the exit temperature at this plane be say Te, my job
is clear and I need to derive the value of VJ.
215
(Refer Slide Time: 15:59)
We now write VJ2 = 2 ( hi – he ); with Cp is a constant VJ2 = 2 Cp × the change in
temperature. The value hi − he, hi corresponds to the temperature is Tc, while he
corresponds to temperature Te. Now we want find out what is the value of Cp and we
want to know the properties of Cp of gas in terms let us say the gas constants may be in
terms of the a molecular mass of the gas. Therefore we again go through the relation Cp
by Cv is equal to gamma viz., the specific heat ratio. Also from equation h − u is equal to
p v is equal to R T, we get the expression Cp minus Cv is equal to R. How did this
come? h − u = p v = R T. I take differential dh by dT, du by dT is equal to R and
therefore Cp –Cv = specific gas constant R. Now we find R, the specific gas constant,
keeps changing with type of gas and supposing we keep changing the gas here and I have
to change the value R. Why not I express the R in terms of the universal gas constant,
which is same for all gases namely R0.
216
(Refer Slide Time: 17:57)
And now how do I define R0 with respect to R? Let us again talk in terms of little more
basics over here. What did we talk about it when we started this course. We told that the
quantity of matter could be expressed in terms of kg, I have a given amount of matter
and how do I say kg - well somebody maybe in 1827 or 1830 he kept some mass in some
lab in France which said the mass to be 1 kg. The quantity of matter is what I express in
kg; but why it should be kg? It could also be another unit. Instead of saying that the mass
of the duster is let us say 500 grams why not express it in some other unit, let us say the
molecular mass of wood. If the molecular mass of wood is something like 500 grams per
mole for this particular duster, I can as well say that this duster contain one mole of the
wood. Instead of defining the mass of this duster as 500 grams, I can say the duster
consists of 1 mole of the substance wood. Therefore I can also define the quantity of this
duster in terms of a mole just as we define it in terms of mass of 500 grams.
Getting back to some more details we must remember that this mole is different from the
number of molecules. The number of molecules in 1 mole is what we call as Avogadro’s
number.
217
(Refer Slide Time: 19:51)
And the number of molecules in 1 mole is 6.023 into 1023. In other words 1 mole of any
substance has something like 6.023 into 1023 molecules. Now we must be a little more
clear and let me take one more example. Supposing I consider let us say a box; an empty
box into which I introduce a mass of 1 kg of oxygen. Now instead of saying 1 kg of
oxygen, I did rather describe this quantity oxygen in terms of moles of oxygen. We all
know that the oxygen O2 has a molecular mass of 32 grams per mole, therefore instead of
saying 1 kg of oxygen, I can as well say I have 1000 divided by 32 so much moles of
oxygen. Therefore to state the amount of matter which is there or amount of matter
which is available, we can express it either in mass or in terms of moles.
218
(Refer Slide Time: 21:40)
We saw the equation p × the specific volume v; p is a pressure with units Newton/meter2,
specific volume has units of meter3/kilogram, as equal to R T, where R is the specific gas
constant. R has units of joule per kilogram Kelvin since temperature has a unit of Kelvin.
Now instead of expressing this specific volume in meter3/kilogram, we can also write it
as meter3/mole. With this unit my right hand side becomes R0 T where R0 for all gases is
the same. This is known as the universal gas constant. In other words we write pV =
nR0T instead of pV = mRT. If we consider m in kilograms I have RT if I consider this
term in mole I get it as R0T with the same value of R0 for all gases which now become
universal. We say R0 is universal gas constant which should have the units as joule per
mole Kelvin and the value is 8.314 joule per mole Kelvin. Therefore we just refreshed
ourselves with a little bit of thermodynamics.
We say we are talking in terms of a perfect gas for which Cv, Cp are constants we also
learn to distinguish between R and R0 which is very primary but which is very essential.
We can now go back to this particular equation = 2 Cp (Tc – Te). Note that we said
Cp−Cv = R and Cp/Cv = γ.
219
(Refer Slide Time: 24:07)
And therefore we could write that same equation as Cp into 1 minus Cv by Cp, Cp by Cv
is gamma and therefore Cp( 1 – 1/γ) is equal to specific gas constant R. Let us keeps
track of the units. R has the unit of joule per kilogram Kelvin, Cp has units of heat
required per unit mass per unit Kelvin, and hence Cp and R have same units.
We can write the value of Cp as equal to γ R / (γ−1). Supposing we want to write it in
terms of moles insatead of mass, all that we do is instead of writing R in joule per
kilogram we have to write R in terms of joule per mole Kelvin. Therefore we say R by
comparing p V is equal to m R T, p V is equal to n R0 T, we get R is equal to R0 by the
the molecular mass M, where M is the molecular mass in kilogram per mole. Rather if
we now simplify the equation, we get the value of Cp in joules per kilogram Kelvin as
equal to γR0/M(γ−1).
220
(Refer Slide Time: 25:44)
I just swallowed 1 or 2 small steps. What are the steps that I did not show? We could
write p × V volume in meter cube is equal to m R T are rather this could also be = nR0T,
where R0 is in joule per mole Kelvin. R in joule per kilo gram Kelvin to convert to R0,
we hav eto multiply R by kg/mole M. That means R0 divided by molecular mass is equal
to R, and this R0 has the unit of joule per mole Kelvin.
(Refer Slide Time: 26:23)
And if I have to convert joule per kilogram Kelvin into joule per mole Kelvin, multiply it
by the molecular mass in kg per mole.
221
Therefore we have R is equal to R0 divided by M, the the molecular mass of the
particular gas.
(Refer Slide Time: 27:28)
Therefore, let us now substitute this in the expression here for VJ2/2 = γR0/[(γ-1)M] (Tc –
Te). Taking Tc outside, we have VJ2 = [2γR0/(γ−1)M]Tc(1−Te/Tc). This is the value of
the efflux velocity squaredor jet velocity squared.
So far we have nade no assumptions of the nature of the expansion process. Since we do
not know what is the exit temperature but we know the exit pressure, we would like to
convert Te by Tc in terms of pressures. Let us assume the next assumption as flow
through the vent or flow through the hole is adiabatic. We have already assumed it
earlier and said there is no heat transfer. Let us make one more small assumption.
Let us assume that the flow through the particular vent is quite slow. What do I mean by
slow? I mean its slow enough such that the flow is reversible. What do you mean the
flow is reversible; in other words the flow is not so fast that it cannot retrace back. It
goes through the series of equilibrium states and therefore I say that the flow through the
vent is adiabatic and reversible or rather the expansion process is isentropic.
222
(Refer Slide Time: 30:00)
And if we have the isentropic flow in the vent, the equation for the process involving this
isentropic process is going to be pvγ = constant. With v being the specific volume i.e.,
1/density ρ, the equation becomes p/ργ =constant But we know that the gas is ideal gas or
rather perfect. This is already assumed. Based on this, we can write the equation of state
as p/ (rho ρ × T ) is a constant. We take to the power of gamma of this equation viz.,
pγ/(ργTγ) = constant and compare this equation along with the isentropic equation. I
divide one by the other and get p1−1/γ/Tγ is a constant.
(Refer Slide Time: 31:08)
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And therefore we can write the equation as the value of Te/Tc is equal to (pe/pc)γ−1/γ .
Therefore the expression for jet velocity VJ = √2γR0Tc/(γ−1)M{1-(pe/pc)(γ−1)/γ , so many
meters per second and this is the expression for the efflux jet velocity. That means we are
able to derive an expression based on the assumption for perfect gas, the flow through he
vent is adiabatic and it is reversible i.e., it is a slow flow. It gives us the jet velocity as so
many meters per second. If this expression is clear, may be we can draw some
conclusions based on this equation and that is what we will be doing in the next couple
of minutes. Is it clear how we get the jet velocity. The only assumption which we made
is flow through the vent is adiabatic and also reversible because we set it as a slow
process.
We will keep these assumptions in mind. Let us discuss the equation. When will we get a
high value of VJ? When the temperature Tc is very high. And what does this tell me? It
tells me if I use something like a cold gas such something like
(Refer Slide Time: 33:29)
let us say I take the same balloon and use it as a rocket, my balloon over here, I have this
particular vent over here, I have the balloon which was initially filled with air at 35
degree centigrade; that means if my temperature is small, I do not get a very high value
of VJ. If I can increase this temperature in some way I can get a much higher VJ, and that
is why a hot gas is better than a cold gas. I could have a cold gas rocket, I could have a
hot gas rocket but hot gas is definitely better. Therefore, first thing we say is Tc should
224
have the highest possible. But is there some limitation? The material must withstand the
high temperature and therefore there is a limits to this temperature. We normally use
temperatures of 3000 to 4000 Kelvin. And how do we generate it? We burn fuel and
oxygen which we call as propellants; that means we use chemical reactions to generate
high temperature or rather get a high value of Tc.
Therefore, a rocket could be a cold gas rocket, could be a hot gas rocket. In a chemical
rocket you generate high temperature with chemical reactions, or we could introduce
resistance wire to electrically heat the gases to a high temperature Tc. Cold gas rockets
are also used wherein we need small jet velocities. We allow gas in a chamber and allow
the gas to expand in a vent to give thrust. You have chemical rockets. You could also
have nuclear reactions, and in nuclear reactions we can get a even higher temperature
and can get a higher jet velocity. We have classification of rocket as cold gas rocket, hot
gas rocket, chemical reaction rocket, electrical rocket, nuclear rocket and we say that
temperature is the one of the major parameters that contribute to jet velocity.
Let us examine the other parameters. Consider the exit pressure. If exit pressure is very
small, well the fraction pe by pc becomes small. Therefore, we would like pe must be
small, or if my exit pressure cannot be small I can set pc must be large. That means if we
can store gas at very high pressure such that we can get a higher value of the pressure
ratio; in other words the ratio pe by pc must be small are rather the chamber pressure
must be high and the exit pressure must be small. This is the second conclusion from this
equation for the jet velocity VJ.
Let us take a look at the molecular mass of the gases which are exhausted out through the
vent. We find VJ is equal to 1/√M i.e., one over under root of molecular mass; that
means, if I can have a gas which is very light i.e., small molecular mass like hydrogen or
let say or helium well my VJ will be higher, but how do I get a light gas. Let us take an
example.
225
(Refer Slide Time: 37:08)
Supposing I have a chemical rocket in which I take carbon, I burn it with oxygen and I
get product combustion of carbon dioxide, C + O2 = CO2. We assume complete
combustion. I have another rocket, I have Hydrogen H2 + ½ O2 = H2O; or 2 H2 + O2 = 2
H2O. What is the difference in these two rockets, as far as VJ is concerened? The first
rocket gives a temperature Tc around 3200 Kelvin while the second one gives around
3300 Kelvin at the same pressure conditions. The temperatures are not very much
different. We look at the molecular mass of carbon dioxide; the molecular mass is equal
to 12 plus 32 = 44 gram per mole, while for water the molecular mass is 16 plus 2 = 18
g/mole. In other words if we burn hydrogen and oxygen we get a very low molecular
mass and the VJ which is directly proportional to the under root of molecular mass will
therefore be large and that is why we find hydrogen is a prefered fuel. Even if we have a
solid propellant; we would like the solid propellant to contain as much hydrogen as
possible. Since to cary hydrogen gas inbulk is very difficult, we liquefy hydrogen at low
temperatures and carry it as liquid hydrogen. This is what we mean by crygenic liquid
propellant. Cryogenic propellant rockets have better performance. For the cryogenic
propellant rockets, the specific impulse is about 4600 Newton second by kilogram
whereas for an ordinary fuel it is of the order of 3500 Newton second by kilogram. You
get immense benefit out of the smaller value of molecular mass.
226
Therefore we infer th eparameters 1. Temperature, 2. Pe/pc and 3. Molecular mass which
must be as low as possible. Can we say what must gamma be like? Should gamma be
small or large and if so why?
(Refer Slide Time: 39:58)
We find VJ goes as √γ/(γ−1). Therefore would we like a small value of gamma or large
value of γ? Let us divide numerator and denominator by γ to give 1/(1 −1/γ) within the
underroot sign. We find if we have the smaller value of gamma, we subtract a larger
number in the denominator and this gives a higher value for VJ. A small value for γ is
preferable. Therefore gamma must be small. Because if gamma is small, we subtract a
larger quantity and my denominator comes down and for the same numerator VJ is
larger.
Therefore we also tell that γ must be small. What is the sensitivity of gamma? It is not
inversely proportional as molecular mass because it is γ/(γ−1); γ is not very much
influential. Let us take an example; how do I make gamma small, how do I make gamma
large and how does gamma depend on the gas?
227
(Refer Slide Time: 41:26)
Let us consider helium; this is a mono atomic gas has a value of γ is equal to 1.67 or
rather is equal to 5 by 3. If we take air γ = 1.4. If I take nitrogen or oxygen γ = 1.4. If we
consider CO2, γ ≈ 1.35. If the gas molecule is more complicated like we have Freon gas,
used as refrigerant, γ ≈ 1.1; that means as the molecular mass of the gas increases the
gamma becomes smaller. If the molecule of the gas is more complex, gamma is smaller,
and therefore the influence on VJ will be better.
But when we looked at the molecular mass; if the exhaust gases have lower molecular
mass, we obtained higher VJ. This is contradictory from viewpoint of γ. The role of
molecular mass and gamma is just the opposite, but it so happens the effect of gamma is
much smaller than the molecular mass and therefore we would still like to have a lighter
gas to be exhausted out.
I will just repeat the four salient conclusion which I draw from this particular jet velocity
equation.
The temperature of the gases must be as high as possible; the ratio pe/pc must be small
are rather the ratio of the chamber pressure to the exit pressure as must be large as
possible; molecular mass of the gas which is coming out must be as small as possible and
lastly we would like gamma to be small, but this gamma being small contradicts the
requirements of the small molecular mass, and therefore we do not really pay much
attention to gamma since it is not as influential as molecular mass.
228
(Refer Slide Time: 44:07)
Let we quickly go through the conclusions on the above slide, because I plotted
the equations for different values of temperature. In the first slide, I have plotted the
value of VJ is meters per second as the function of temperature, the temperature varies
from 300 to 3800 Kelvin. The range of Tc in chemical rockets is of the order of may be
something like 3000 to 4000 Kelvin, much lower than 4000 Kelvin, and therefore I
restrict myself to 3800 here. I also plot for different values of pe by pc. You find as the
value of pe by pc decreases I get the higher value of jet velocity, and this the conclusion
that we drew looking at the equation.
If I have say a gas at a low temperature such as used in a cold gas rocket and if I decrease
the value of pe by pc from 0.1 to 0.001 I do not really get much benefit, where as if I
have a high temperature rocket; Tc is high and I get much larger benefit. Therefore if we
were to design cold gas rocket I can make a small rocket and I do not need to really
expand it much. I can as well have a lower chamber pressure.
Therefore a conclusion that we could draw is may be for a cold gas rocket, and let us put
it down as an important conclusion: for a cold gas rocket for which Tc is quite small we
do not really require a small value of pe by pc.
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(Refer Slide Time: 45:38)
Because you find when the temperature is small, the gain what we got in expanding the
gas by a large amount is very small while if I have really a high temperature chemical
rocket this gives me higher temperature and much better gain in Vj.
In the next slide, the value of the jet velocity in meter per second as a function of the
molecular mas is given. I find when the molecular mass is small, we get the value of VJ
which is higher, and of course the same trend continues. This is at a mean temperature of
3000 Kelvin. This tell me very clearly that as the molecular mass is smaller, we get a
higher value of VJ.
In the last slide, I show the influece of γ; we show the value of VJ as a function of
gamma what we find is at a low value of pe/pc of 0.1, gamma really does not influence
VJ. We find that the curve is quite flat and it is independent of gamma as it were.
However, when the value of pe/pc is quite small of the order of 0.001, I find as gamma
increases I get a smaller value of VJ. The conclusions which we drew that as gamma
decreases, the value of VJ increases is seen to be more effective at smaller value
pe/pc.Whereas when I have a higher value of pe/pc such as 0.1, the effect of gamma
changes does not influence VJ.
If we have a rocket for which the expansion ratio is not very high like in a cold gas
rocket, we can even use helium for which gamma is 1.67. Whether I use helium or we
use air with gamma of 1.4, it really does not make things worse, and therefore a cold gas
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rocket normally uses a light gas like helium, and helium has low molecular mass around
4 gram per mole, and therefore we get the benefit of the molecular mass and we do not
lose any effect due to gamma.
Therefore what is it we have done thus far. We derived this particular equation for VJ, we
looked at the effect of temperature, molecular mass, expansion ratio and also γ on VJ. We
found for a cold gas rocket pc need not be very high and the effect of gamma is not
dominant. But if we were to look at the effect of temperature in a chemical rocket for
which temperature is high, we can operate at a much lower value of pe/pc and gamma
effects will also becomes significant. Therefore by now we must be very clear that if
where to have a choice of propellants for my rocket, we must have propellants which
will generates a high temperature and a low molecular mass gas. We do not have much
control over γ; γ is not very controlling provided the pressure ratio is not too low.
I think this is all about this jet velocity and we have to now relate it to choice of
propellants viz., solid propellant rockets, liquid propellant rockets and other forms of
rockets. In the next class what we do is address the shape of this vent which we give us a
high value of velocity. In other words we move into the chapter of nozzle, shaping of
nozzle and what are the problems with nozzles.
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Rocket Propulsion
Prof. K. Ramamurthi
Department Of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 10
Nozzle Shape
(Refer Slide Time: 00:16)
We will continue with the subject of nozzle today. If we recall what we did in the last
class, we had a chamber and this chamber had a gas at a pressure pc, temperature Tc, the
molecular mass of this gas was M gram per mole. And we expanded this gas out through
a small opening, which we called as a vent; we were able to calculate the exhaust jet
velocity in meters per second. We found out the condition for which VJ is quite large; we
found that the temperature of the gas must be large, the value of the chamber pressure
must be large, and the molecular mass of the gas must be small. We also examined the
variation with respect to gamma, found that gamma is not very influential, especially
when pc is small or the ratio of the exit pressure pe to the value of pc is somewhat high
or equivalently the pc value is small compared to pe, then gamma is not influential; this
was seen from the expressions which we derived.
Now, for a rocket, impulse(I) is equal to mass which is ejected out multiplied by the jet
velocity. We also said that the force is equal to d/dt of I, which is equal to m° into VJ.
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Therefore, we are interested in a high value of this velocity with which the mass efflux
leaves the chamber.
(Refer Slide Time: 02:02)
Therefore, the question is whether there anything like a configuration or shape of vent,
which can give a high value of jet velocity. And this is where we get started with and
afterwards we will try to find out the conditions for this high jet velocity to be realized.
We will have to look at matching of the exit pressure with respect to the chamber
pressure and that is what I propose to do in today’s class. We will also physically try to
understand, if there is anything like some information transfer between the outside and
inside of a chamber. Let’s get started with this background.
We have a nozzle and we are interested in the shape of this nozzle; therefore let us have
shape like this. The shape is such that at a distance, let us say this is x and is equal to 0 at
the beginning, at a distance x from the initial origin the area of the vent is A. Let the
density of flow through this be ρ, let the velocity of flow be V, at this particular section x
at a distance x which is the reference plane. Let us consider the variation in the properties
at a distance x plus dx.
The area is different from A, and I want to find out the configuration of this particular
vent which gives me the maximum VJ; therefore let the area at x + dx be A + dA, let the
density at this section be ρ + dρ and the velocity at this section V + dV. My main aim is
to find out what must be the shape, such that I get a high value of VJ. Therefore I just
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look at these two sections; let say section one at x, at which the area is A, density is ρ,
and velocity is V. At section two, dx away from this section the area is A + dA, small
change in area, I think according to this figure dA should be negative, but I just have a
general notation A plus dA, let the density be ρ + dρ, and the velocity V + dV. Now I say
that the flow is steady or constant, in other words whatever comes in at x here flows out
through this particular opening or vent. The mass flow rate m° in kg per second is equal
to ρ A V
(Refer Slide Time: 04:48)
through the section one and the same thing flows through the section two, which is equal
to ρ + dρ, that is density at the section two × A + dA × V + dV; and this is mass balance
equation. Since the flow is steady, the same mass flow rate flows through this section
and this section. Let us solve this equation and what do we get? Mass flow rate is a
constant or rather d (m°) must be 0, because mass flow m is constant. Therefore, I get the
expression d(ρ A V) is equal to 0. And therefore, now if we expand this expression we
get ρA dV + AV dρ + ρV dA = 0.
Or rather I divide this entire equation by ρAV, which is a constant and we get dV/V +
dρ/ρ + dA/A = 0 and this becomes my mass balance equation in the differential form. I
could have derived this expression by saying ρAV is a constant, therefore logarithm of
ρAV is constant and differentiating this would have given me dV/V + dρ/ρ + dA/A is
equal to 0; which we call as the continuity or the mass balance equation. But, why did we
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have to derive this. We wanted to it find out what is the change in area, which will give
us high value of velocity and therefore, our aim was to relate dV by V with dA by A.
Unfortunately, we are left with dρ/ρ and we would like to get rid of this term dρ/ρ.
Because the flow, we assume is compressible, the density is changing; and therefore, I
am left with this one term here dρ/ρ, which we should express every in terms of dV/V or
dA/A to be able to find the dependence of velocity on the change in area. Therefore to do
that I again ask, can I write one more equation; let say the momentum equation.
What does the momentum equation tell? The momentum equation specifies that the rate
of change of momentum must be equal to the impressed pressure or rather impressed
force.
(Refer Slide Time: 07:45)
Therefore, let us write the momentum equation; the mass flow rate is m°, the change in
velocity is from V to V plus dV, that means V plus dV minus V. The rate of change of
momentum is therefore m° × dV.
And this is balanced by the force and what is the force that we get? we have the change
in pressure across is dp and the area is A, we find dp is higher therefore, the force acts in
the direction of change of momentum and therefore, we have change of momentum is
equal to the force. And what is m°? m° = ρAV. A and A and also V and V get canceled.
ρV dV is equal to minus A of dp, and therefore we get ρVdV plus dp is equal to 0, which
becomes the momentum equation or pressure balance equation.
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We must be able to generally derive the momentum equation by just applying the
Newton’s second law. Seeing that, I have a mass which flows across, the change in
velocity is dV, the rate of change of momentum is m° into dV and that you have the
pressure force, which acts on the mean area into dp is the change in the force and
therefore, this is the force balance equation. Therefore, now we find that though we
wanted to get rid of dρ/ρ, we have got an expression in terms of dp. So, how do I still get
rid of this term in some way or the other. To able to do so, we look at the sound speed.
What is this sound speed or velocity? You know it is very central to gas dynamics and
compressible fluid mechanics. I talk to you and when I talk to you the sound waves
travel, we say at the speed of sound; and this we say is the speed of sound denoted by a
m/s.
(Refer Slide Time: 10:12)
I have a sound wave, which is travelling at a speed of a; and when you hear the sound,
you get a little bit of pressure on the ear drum which you perceive depending on the
response of the ear drum. How you get some velocity from the pressure? May be, I get
some velocity change from the pressure change. I could even get a very small
temperature change from the pressure change. If the sound intensity is large, may be the
temperature change could be high; therefore, let us take look at what this sound wave
means and how do we introduce sound wave into that equation. And towards the end of
this lecture, we will try to see the importance of sound when we talk of the flow.
236
Let us imagine that we have a pipe and a planar sound wave propagates in this pipe. I
take the same medium of a gas, let the pressure of this medium be p, let the density of
this medium be ρ and the temperature of this medium be T. The medium, let it be
stagnant like the room in which we are seated. This means that the velocity V is equal to
0. In other words, I am just looking at the sound wave propagating in a stationary
medium (at rest-the velocity is zero), of pressure p, density ρ and temperature T. And
what happens when the sound wave propagates through the medium? Let say this sound
wave, it increases the pressure little bit by p plus dp. Initially, velocity is 0, the sound
induces a small velocity dv. we have the density now to be ρ + dρ and temperature could
also change.
Now, I want to write an equation for this particular change. What is happening? I am
standing over here, this is my frame of reference. I am watching the sound wave go by,
the sound wave processes this medium, which is initially stationary, increases the
pressure by dp, increases the velocity from 0 to dv of the particles, may be the density
changes from ρ to ρ + dρ and that is what I am watching. And for me, to write the
equations standing here to see the wave traveling at the velocity of sound a meters per
second is difficult because the sound wave is also moving as also the particles in the flow
are moving. And therefore, we transform the frame of reference; instead of me standing
here and watching the wave go by, I position myself on the wave.
And if I stand on the wave and I am moving along with the wave, I see the gas coming
towards me with a speed a meters per second, because I have told that the sound wave
moves with a velocity a. And now, the conditions ahead of me here are p, rho and
perhaps T, and velocity is a. And we had when the velocity was 0, we had dv therefore,
the velocity here is a plus dv, the pressure is still the same p plus dp, the density is ρ +
dρ. I think such transformations are important in the sense that I initially watched the
sound wave go by but if I stand on the wave, this is that I see? Since I am moving, I put
myself on the wave therefore, the medium is coming over here and I have the changes
happening behind. Now, I write the equations for this frame of reference. What will be
the equations that we will get?
237
(Refer Slide Time: 13:59)
I have over here; on the right hand side the undisturbed medium, this is the disturbed
medium and I stand on the wave between the two. The equation that I get for mass
balance is rho into a into cross sectional area. I take the same cross sectional area, and we
get ρ × a = (ρ + dρ) × (a + dV). This is the mass balance or continuity equation. You see
the similarity of this equation to the earlier equation, wherein you have ρ A V being
conserved with area change, I just wanted find out how the sound wave travels and
therefore wrote the equation for constant cross sectional area. Let us simplify it. I have
ρa is equal to ρ + dρ into a + dv. But mind you we are talking of sound waves; sound
wave is travelling at a speed a meters per second.
We also know that the pressure change from the sound wave dp would be small, and
similarly dV is small, the density change dρ is small, therefore the product of these small
quantities for all practical purposes can be neglected. And therefore, what is it that we
get from this equation? We find that ρ a and ρ a get cancelled and we get the value of
ρ dv = −a dρ, this is the continuity or mass balance equation across a sound wave. This
equation is derived as I sit on this sound wave and I see the medium being processed by
it.
I now want to write the momentum equation. What should it be? Well, I am standing
here on the wave, I see the change of momentum: that means, I see ρa is mass flux,
which is coming over here and what is happening? I have the velocity changing from a to
238
a plus dv giving ρ a dv as the value of the rate of change of momentum; and that is
balanced by the change in pressure.
I do not need to repeat this again viz., as rate of change of momentum per unit area is the
change of pressure and this becomes the momentum equation for the sound wave. (ρa dV
+ dp = 0)
Now, we look at this equation, which we call as equation three, because we have already
derived the continuity equation which was Eq. 1 and we have the momentum equation
that we call it as Eq. 2 for the flow through the section. Looking at the mass and
momentum equation for sound wave, which is Eq. 3 for mass balance and Eq. 4 is the
momentum equation. Therefore, if I have solve these two equations (3 and 4) for the
sound wave together, I have the following:
(Refer Slide Time: 17:11)
ρdv = − dp/a; therefore, if we substitute ρ dv over here from the mass balance equation 3
as −a dρ what is it I get? Minus dp/a is equal to minus a dρ. It gives us the dp by dρ is
equal to a2. That means, the velocity of sound square is equal to the ratio of the
differential of pressure to the differential of density. That is when I am talking to you; the
perturbations in pressure across the sound wave to the perturbations in density across it
equals to the sound velocity squared. I call this relation as Eq. 5.
239
Let us come back to what we set out to do i.e., get rid of term dρ by ρ. We have dρ is
equal to dp / a2.
Getting back to the equation dv/V + dρ/ρ + dA/A = 0 and substituting for dρ by ρ, we get
for dρ is equal to dp / a2, we get dp/ρa2 + dV/V + plus dA by A = 0.
But then from the momentum equation 2, we have dp = − ρV dV and therefore, if we
substitute the value of dp as − ρ V dV what is it we get? We get the equation as dV/V – ρ
V dV and ρ and ρ gets cancelled therefore we get V and on top that is dp is equal to V by
a2 into the value of dV plus the last term that is equal to dA /A = 0.
What is it that we have done? We substituted the value of dρ by ρ in terms of dp by a2
into 1 / ρ, because we found the dp by dρ is equal to a2 . And then we wanted to write get
rid of dp in this expression and therefore, we use the momentum equation, which we
wrote as dp = − ρv dv. So, dρ/ρ became v into dv by a2 with a minus sign.
(Refer Slide Time: 20:43)
Let us simplify and write it down over here. We get dV/V × (1+ plus an expression
involving V and a. Since we took V outside, in the denominator we get V2 and it is
divided by a2. We still have the term plus dA/A, the sum of which with the previous term
is equal to 0.
240
Let us call as Mach number M the ratio of velocity of the medium divided by the sound
velocity. I will come back to the physical significance of this later. Therefore, I can write
dV/V = − dA /A × 1/{1−M2}.
Therefore this is the final expression that we get. Now, whenever we derive an
expression, we must analyze what the expression means or signifies. What do we find? If
dA is negative like what we have drawn earlier and what did we draw? We say area is
decreasing as x is progressing. If dA is negative, the term becomes positive. And if we
say M is less than one, when dA is negative, then dV is positive. What does this mean? If
we want the velocity to increase as the flow progresses, if the Mach number M is less
than one, we get this to be a positive number. Unless we have area which is decreasing as
x proceeds, I cannot have an increasing value of V.
(Refer Slide Time: 23:23)
In other words, we tell that flow will accelerate or flow will increase in velocity, if the
Mach number is less than one; dV will be positive, if we have dA to be negative. In other
words, all what we are saying is the cross sectional area, if we have a subsonic flow with
Mach number less than one, then we should have something like this converging section
for velocity here at section two to be greater than at section one. That means, here it
enters at lower velocity V, thereafter dV gets enhanced and we have higher velocity
over here. That means, for the case of a subsonic flow or a flow for which Mach number
is less than one, flow will accelerate in converging passage.
241
On the contrary, using the same set of arguments, if we have a case wherein the flow
takes place in diverging configuration, and if we have Mach number greater than one,
then what happens? Mach number is greater than one; this becomes negative, negative
and negative gets cancelled, therefore, dV by V goes as positive of dA by A. In other
words, the flow accelerates, if Mach number is greater than one, dV is positive since the
change in velocity dV is positive. See through the simple argument of looking at the
mass balance and the momentum equation, we are able to come out with a conjecture
that in a converging section velocity will increase only if Mach number is less than one.
On the other hand, if we have a diverging section and if Mach number is greater than
one, then only the flow velocity will increase.
(Refer Slide Time: 25:25)
What will happen if the Mach number is one? If Mach number is one, the equations sort
of break down. In fact, we find that if Mach number is one, then we have one over 0 and
unless I have dA by A equal to 0, this equation cannot predict anything at all. Therefore
if Mach number is one, maybe we must have a constant area; that means, A is constant or
rather dA must be equal to 0 for flow to take place.
242
(Refer Slide Time: 26:06)
So far we looked at the continuity and momentum equations, and found that in a vent or
in a small opening, we should have initially the flow should come like this, it should
come to a value of M is equal to one. And then if you pass through the divergent, such
that dA is positive here, we have dV is positive and therefore, I can have acceleration of
flow. And therefore, along this particular length if I plot the velocity V will keep
increasing, and at dA = 0 the velocity must be equal to the sound velocity. Therefore, in
order to get a high value of jet velocity, what we require is we must have a minimum
area and this minimum area like a constriction; we call it as throat. Therefore, we start
with a large area, we converge it, we increase the velocity to a value is equal to the sound
speed at this section (the Mach number is equal to one) and thereafter when the Mach
number is greater than one and the flow velocity increases and therefore, I can have high
value of jet velocity.
Therefore the configuration of the vent should be a convergent followed by a constant
section which we call as throat followed by divergent, if we have to get a high value of
jet velocity. Supposing by chance, the mass flow rate is such that (like when we
considered the balloon) which had a pressure of Pc, a temperature Tc and a molecular
mass of gas with density ρc; if the Mach number at the smallest section throat is less
than one, then what is going to happen?
243
(Refer Slide Time: 27:56)
Let us say that we have the same configuration of the vent or nozzle; in other words we
have the convergent followed by the throat, followed by a divergent. If, the velocity of
flow is less at the throat giving the Mach number to be less than one at the throat section,
then what is going to happen? Well, we now plot velocity as a function of distance, the
velocity keeps increasing up to the throat, but I find that V is still less than the sound
speed i.e. Mach number is less than one. And therefore, the velocity begins to drop as
flow progresses further into the divergent. That means, in the convergent the velocity
increases in the divergent velocity decreases; and this portion divergent is what we call
as diffuser.
A contraption, which decreases the velocity and enhances the pressure, is what we call as
a diffuser and a contraption, which increases the velocity is what we call as a nozzle.
And this total is what we call as a nozzle. If I can have a Mach number one at the throat
and then I have a convergent followed by divergent, I can get a high value of jet velocity
since the velocity increases in both the convergent and the divergent; that means, Mach
number at the exit will be a large value much greater than one.
244
(Refer Slide Time: 29:36)
Therefore, from the continuity and momentum equations, we find it necessary to have a
convergent section, followed by a throat section, followed by a divergent section and this
is what gives me a high velocity and this is what I call as a nozzle. For getting the high
velocity, the Mach number at the throat should be unity. This is the convergent divergent
nozzle, which in some text books is also referred to as de Laval Nozzle.
But let us be very clear, if by chance I do not get the Mach number at the throat as equal
to one, well the buildup of velocity in the convergent is a lost in the divergent and the
velocity drops. And this is the configuration of a nozzle - a convergent divergent nozzle.
Let us go through an exercise involving variation of parameters, because the convergent
divergent nozzle is central to having a high jet velocity. Let us find out the variations of
parameters across a convergent divergent nozzle.
245
(Refer Slide Time: 30:39)
Let me first sketch a nozzle, converging till the throat, followed by a divergent and let us
assume that the flow rate through the nozzle is such that the Mach number is equal to one
at the throat. Now, let us first find out what is a change in dV by V along the length. We
just now saw that dV by V increases up to the throat since the Mach number is less than
one and the variation in dA is negative as x increases.
And then it further increases in the divergent, because the Mach number is one at the
throat and characteristic of the equation changes as 1 − M2 becomes negative and
therefore, dV by V increases. If dV by V is given by this trend, well V the value of
velocity would also keep changing and what is it that we get? I get the velocity to
continually increase.
If the velocity increases, what is going to the happen to the pressure and the density. Let
us now plot a few more parameters, instead of dV by V, we are going plot the variation
of pressure that is let us say dp by p. How should it look like? To determine this let us
first determine the variation of dρ/ρ, because we already have an expression, which we
derived for it.
246
(Refer Slide Time: 32:14)
We had the expression dV/V +dρ/ρ + dA/A = 0 or rather we have dρ by ρ is equal to
minus of dV/V + dA/A. I know the value of dV/V increases based on the earlier
discussions and therefore, dρ by ρ will be negative. Though dA by A in the convergent is
negative, dV by V has a higher value since it is divided by a fraction given by 1−M2. In
other words, if we say this is my throat region and we have Mach number equal to one at
the throat here, this is my length x along the nozzle which I am considering, we find
therefore, the density will keep falling. And what happens at the throat region? At the
throat region, we had one minus Mach number squared and therefore, I have something
like a step gradient in velocity; we have very step region dρ by ρ also. This is important.
In other words, the density keeps decreasing more rapidly at the throat region where the
Mach number is around unity. The region at the throat is a region of decreasing density
as the flow progresses. In fact, we will find and when we get into this problem of
combustion instability, we will find that when we have a rocket nozzle, let say have a
rocket with a convergent divergent section like this. The rapid decrease in density or the
change in density over here acts as a sort of reflecting surface with the disturbances in
the chamber being reflected back into the chamber. We will come back to this point later
on. That means, at the throat portion, you have region of decreasing densities and
therefore, if we now plot the density as a function of x here, we have the throat here well
the density keeps decreasing. Therefore, for a convergent divergent nozzle, for which the
247
Mach number is equal to one at the throat velocity change increases subsequently while
the density change decreases.
Now, there are two other parameters which are left. What is going to happen to the
temperature? What is happen going to happen to the pressure? We have made an
assumption that the flow through the nozzle or vent is adiabatic and is reversible.
(Refer Slide Time: 34:44)
Therefore, p/ργ is a constant. I know density changes and therefore, the pressure if I have
the Mach number at the throat equal to one, well the pressure should also decrease this
will be my variation of pressure with respect to x.
And we also have derived an expression in which we found T1/T2 = (p1/p2)(γ-1)/γ .
Therefore, the temperature, if I have the throat here should also decrease and this will be
by temperature variations with respect to the distance x.
The net result is that the pressure decreases in a nozzle monotonically, the density
decreases in a nozzle with rapid changes in density taking place at the throat. Well the
velocity increases; this is all what we have deduced thus far.
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(Refer Slide Time: 36:08)
Now, let us include some other parameters. If we want to plot the sound velocity
variations what will it look like? In other words, we have the expression for sound
velocity a. We want to plot how the sound velocity will vary along the length and at the
throat the Mach number is one. We again go through the expression what we derived
today. We had dp/dρ is equal to a2. Now we told that the nozzle flow is isentropic that is
adiabatic and reversible therefore, we have p/ργ is a constant. Therefore, what is a
square?
(Refer Slide Time: 36:52)
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Let us write the term p = constant c × ργ. Therefore, dp/dρ = γ ργ−1 × by constant C. And
therefore, now if I want write the value of the constant C, it is p/ργ. Hence, the expression
for dp/dρ becomes γ p ρ−γ ργ-1 and this is equal to γp/ρ. Therefore, we find that the sound
speed a2 = γp/ρ and for perfect gas or rather for an ideal gas p is equal ρRT. The sound
speed is therefore equal to √γRT.
(Refer Slide Time: 37:50)
Therefore how would the sound speed vary occur along the length of the nozzle? Well
the sound speed starts with a high value of sound speed corresponding to the high
chamber temperature; it keeps on coming down and it like this along the length of the
nozzle as the temperature drops.
One last parameter, I can still think of is the Mach number variations along the length of
the nozzle. I know the velocity variations which we had previously determined along the
length x over here in this particular form and the value increases. We also found that the
sound speed keeps coming down along x; and therefore, if I find out the Mach number
variations as a function of x, what is it that we get? The numerator is increasing,
denominator is decreasing as the Mach number is equal to V by a, and therefore, there is
a much greater variation of Mach number along the length of the nozzle as is shown like
this.
The above parameters are central to nozzle flows. Let us repeat it again. The change dV
by V progressively increases along the nozzle length when the Mach number is one at
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the throat. The velocity increases while the density decreases and pressure also
decreases. The temperature decreases and the sound speed decreases along the length of
the nozzle. The Mach number increases along the length. And we are interested in a high
value of Mach number or jet velocity at the exit.
And if by chance the Mach number at the throat is not one, that means, we have
insufficient pressure over here to give at the throat high flow velocity equal to the sound
speed; in other words, I now get the Mach number to be less than one, what is going to
happen? dV by V is going to increase over here in the convergent, it is still less than the
sound speed that is V is less than a and therefore, it falls down in the divergent portion;
Velocity increases up to the throat, but it is still less than the sound speed and therefore,
it begins to droop in the divergent portion. What happens to the density? Density falls up
to the throat and thereafter increases; therefore the divergent acts as a diffuser.
When the Mach number is less than one at the throat, the pressure recovers in the
divergent part here i.e., instead of falling it increases. Similarly, the temperature recovers
in the divergent. The sound speed falls up to the throat and recovers in the divergent. In
essence, we have the Mach number going up in the convergent and coming down in the
divergent, if the value of Mach number at the throat is less than one.
I would request you to go back and study these figures again, because this is relevant in
the study of nozzles. What we find is we must have a convergent section followed by a
divergent section and in between I must have a constant area section, which we call as a
throat. And the Mach number at the throat must be one for continued expansion to get
high jet velocities at the exit of the nozzle.
251
(Refer Slide Time: 41:19)
If we want to have a high value of jet velocity, it is essential for us to have a convergent
followed by a throat section, followed by a divergent section, with the condition that the
Mach number must be one viz., velocity must be equal to a sound speed at the throat.
And with this condition the velocity keeps increasing. We can get a high jet velocity at
the exit and this is what the convergent divergent nozzle does.
If we find that we have inadequate mass flow rate like for instance, I have a cold gas I
have inadequate pressure and I cannot have a Mach number equal to one at the throat or
velocity is equal to sound speed at the throat, then its better my nozzle is like this –
consisting of convergent portion alone, such that my jet velocity is still less than the
sound speed. That means, if I have inadequate pressure or inadequate conditions here,
such that I cannot effectively use the divergent get then I must do away with the
divergent section. Because if I have this divergent section over here, I am really loosing
the velocity and I am really not gaining anything. Therefore, whenever rocket nozzle has
to be designed, we have to ensure that the flow velocity at the throat must be equal to the
sound speed at the throat.
Now, we ask one last question: What is the significance of sound speed? Is there
something very significant about it and the velocity of flow being equal to it at the
throat?
252
(Refer Slide Time: 42:58)
What we did while understanding the convergent divergent shape was that we introduced
the sound speed through the equation for conservation of mass in order to eliminate dρ/ρ
in that expression. We substituted dρ by dp by noting that dp/dρ = a2 i.e., sound speed
square and then got al the parametric variations in the nozzle. But does the sound speed
have some implication. To be able to answer the question, let’s do a simple experiment
or let us say a thought experiment and what is this thought experiment? Let us say we
have a tank or a chamber something like this; it is at ambient pressure - one atmosphere
pressure, I am really not bothered about temperature at this point in time. Maybe we
attach a convergent divergent nozzle to it this is my thought experiment; let us draw it
properly.
Now, what we do with in this experiment? We attach a vacuum pump here at the exit of
the divergent and we suck the air out of the chamber through this particular nozzle. And
what is the construction of this nozzle? It has this convergent, divergent portion attached
to the tank, which initially is at atmospheric pressure. And then we start pumping out or
start sucking the air out of this tank through the nozzle.
Now, what is going to happen? Now the pressure in the tank is initially one atmosphere,
and this is the same as the pressure outside Pe or we say that the chamber pressure
denoted by say Pc is one atmosphere, let us say Pe is also one atmosphere to begin with.
When the pressure Pc and pressure here Pe are same, obviously, there is no mass flow
253
rate. When we attach a pump downstream of the nozzle and start sucking out the air from
this chamber, this may provide us with some clues on what really the flow velocity equal
to the sound speed means.
(Refer Slide Time: 45:20)
In this experiment, we plot the mass flow rate through the nozzle, that means, the rate at
which mass is getting sucked out as a function of let us say the value of Pe that is the exit
pressure to the chamber pressure as it is progressively decreased by the running of the
vacuum pump.
Initially the pressures Pc and Pe are the same and therefore the mass flow rate is zero.
We thereafter start sucking of the air by decrease the pressure Pe; if we start decreasing
the value of Pe, the mass flow rate will increase. That means, as we decrease the value of
Pe, the mass flow rate will increase and therefore, do you think that as I keep on
increasing the vacuum level i.e., decrease the value of Pe, the flow rate should keep on
increasing or what should happen?
Let start our thinking process. Initially when the pressure outside is one atmosphere,
pressure inside is one atmosphere; the velocity at throat is equal to 0. Then we start
sucking air out, as we keep decreasing the pressure Pe. The flow velocity at the throat of
the nozzle Vt will keep on increasing till it reaches a value Vt is equal to the sound speed
at the throat at . In other words, let us just plot the mass flow rate versus the pressure till
that time. We keep on decreasing the value of Pe, that means we keep moving on this
254
sloping curve here till the time a stage is reached at which given the value of velocity at
the throat Vt is equal to the sound speed at. Therefore, at that point in time what is
happening?
(Refer Slide Time: 47:16)
Let us plot the value of the velocity at the throat I call it as VT. What happens to it? Now
in the same figure on the X axis, we show the value of Pe by Pc, as a function of VT.
What we find is initially the velocity is zero for Pe/Pc equal to one. As Pe by Pc
decreases, the velocity VT keeps increasing till the time for a given value of Pe reaches a
critical value of Pe*. At this ratio of Pe* by Pc, the value flow velocity at the throat is
equal to the sound speed at the throat.
What is the implication of this? Now, what I find is the flow velocity at this particular
point for this condition, when I have reduce the pressure to Pe*, the velocity at the throat
is equal to the sound speed. Now, let us ask ourselves some foolish questions. What is
that drives the mass flow rate from this chamber to the outside? Because I suck
something, when I suck something what happens? I reduce the pressure here and this
information of a reduced pressure is transmitted through the divergent portion, the throat
of the nozzle and the convergent part to the chamber, which causes the mass flow. That
means, I suck the air out, something the information is supposed to reach the chamber
and that is why some mass is flowing out. As I keep on decreasing the pressure, this tank
senses viz., it gets some information of the lowering of this downstream pressure being
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lower and therefore, the mass flows out. This is how information is fed to the tank for
drawing an increased mass from the tank.
Now, as Pe is reduced progressively, a stage comes, when the flow velocity at the throat
VT is equal to the sound velocity at the throat. The information or disturbance travels at
the speed of sound. Since the flow at this critical state Pe* by Pc is same as the speed of
sound, any information here is travelling at the speed of sound. But if the gas is flowing
at the speed of sound then no information on need for flow can reach it from the nozzle.
But if the gas is not flowing at a velocity VT equal to sound speed at the throat and is
much lower, it can access the communication that gas is required to satisfy the pressure
conditions. The information that additional flow rate is required can be reached against
the flow till the sound speed is less than the sound speed at the throat. When the flow
speed and sound speed are the same speed, no information can be passed on and
therefore the chamber becomes isolated. The chamber over here cannot get any
information of the reduce pressure here, when the flow velocity at the throat is same as
the sound speed at, because information or disturbances travel at the speed of sound.
Therefore when the Mach number is equal to one at the throat, what is going to happen?
You know I keep on decreasing the pressure Pe, but the flow velocity here is same as the
sound speed and it effectively isolates and therefore, the tank is unable to know that a
low pressure exists downstream calling for additional flow. And therefore, what is going
to happen? The mass flow rate cannot increase any further as the reduced pressure is not
able to communicate with the chamber. Therefore I have something like this. If the
downstream pressure is such that I get the flow velocity at the throat to be less than the
sound speed, the mass flow rate increase. Thereafter the mass flow rate is a constant
since for any further reduction in Pe as a value of flow velocity reaches the sound peed. .
And that means even though I am trying to suck or pull the gas, the information is not
given to my tank. It supplies at the same constant rate. We call this condition as choking
and say that the throat is choked and this is known as choked flow. This reasoning of
choked flow comes from the flow velocity being same as sound speed. Therefore, when
we have a convergent divergent nozzle and at the throat the value of the Mach number is
equal to one, it corresponds to a choked flow through the nozzle.
256
(Refer Slide Time: 51:00)
In other words, any downstream disturbances cannot really affect the flow for the choked
condition; in other words, the concept of the flow through the nozzle throat at Mach
number of unity viz., choked flow is central to the rocket nozzles. I will continue with
this in the next class. In the next class, what we do is we will find out the value of the
density corresponding to the choked flow, pressure corresponding to the choked flow and
there after we will relate it to the area ratio of the nozzle.
257
Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 11
Area Ratio of Nozzles: Under Expansion and Over Expansion
(Refer Slide Time: 00:16)
Good morning. We will continue with the portion on nozzles. We will find out what is
the effect of area ratio and define area ratio of a nozzle. We will also find out how the
nozzle operates at different attitudes and look at some typical results. But, before getting
into it let us quickly recap where we so that we can connect it with what we are going do
today.
We said in the last class that if we need to have a high jet velocity; we need to have a
convergent, it should have a throat for which the Mach number is equal to one and then
we should have a divergent. We were very clear about the throat and we said it is the
place where the velocity is sonic; that means the gas flow velocity at the throat is the
sonic velocity: Vt = at.
We also found that any disturbance generated downstream of the throat; suppose I stand
on nozzle here and make a loud noise or we make some disturbance, this disturbance
cannot enter the convergent and therefore, the chamber is isolated. The reason is that the
velocity at the throat is equal to the velocity of sound.
258
Disturbances generated downstream cannot travel upstream. This is because the
disturbances travel at the sound speed. Any disturbance downstream of the throat
cannot enter the chamber.
And therefore, the sonic throat essentially decouples the convergent and the chamber
from the downstream portion. Second point was that the throat is choked. In other
words, we told that for the given mass flow rate we can have a maximum velocity,
which corresponds to sound speed at the throat and if we want to have higher
pressure in the chamber or if the gases are sucked it at lower pressure we cannot
exceed this condition of sonic velocity. This means that the throat always will have
Mach number equal to 1 or the velocity here should be the sound speed. I think these
findings are important.
We also derived an expression for the jet velocity VJ at the exit, which we found VJ
as equal to √(2γ/γ−1))RTc[1−(pe/pc)(γ−1)/γ]. We did not consider the convergent
divergent shape while deriving this equation; we just said that if the chamber
pressure is Pc and if the exit pressure is Pe then you have the pressure ratio alone
which is important. The temperature in the chamber was Tc and this is how we
derived the expression for VJ; so many meters per second. Is it alright?
And what we started was with a vent, we derived the velocity and then looked at the
shape of the vent; it was necessarily for us to have a convergent followed by the
divergent such that if we were to plot it we have a convergent divergent nozzle.
259
(Refer Slide Time: 03:59)
The velocity of the gases as a function of the length of the nozzle mind you increases
along the length of the nozzle. Initially we have a converging shape then I have a
throat then we have the diverging shape. I get the sonic velocity at the throat and
therefore, the velocity will keep increasing along the nozzle in the divergent.
The moment we have at the throat the velocity less than sonic velocity, the velocity
thereafter the drops. Therefore the necessity to have Mach number one at the throat
was essential. I think this we must remember. Having said that, in today’s class we
will try to see instead of mentioning that the exit pressure is Pe can I put it in terms
of the area ratio and area at the exit Ae. What must be the area Ae such that the
nozzle will give me the required velocity and that is we want to do today.
Let me repeat again; see when we realize a hardware we do not know the value of
Pe; all what we know is that we must have a configuration like this. We must have a
diameter over here of a given size, I must also have a given the diameter at the exit
or rather the exit area ratio. Therefore it becomes essential for me to define
something like area ratio of a nozzle to be able to give me the value Pe such that I
can get the jet velocity or rather I want to know the configuration of a nozzle which
will give me the required velocity.
260
(Refer Slide Time: 05:58)
Let us put it this way: I have convergent, I have the throat, I have the divergent. I
want know what will be my exit area here such that I can get the VJ what we want.
We want to find out the expression for the jet velocity in terms of a diameter or an
area ratio rather than put it in terms of the value of the pressure at the exit, which we
called as Pe. To be able to do that we are looking at the exit area, I would like to
define the area at the throat, because I know that at the throat the Mach number is
always equal to 1, therefore I can define it as a critical or an unique particular area
for reference and I call the nozzle area ratio as equal to the exit area divided by the
area at the throat.
Well. If I have to define something like an area ratio maybe I should think in terms
of the area of the chamber over here and relate it to the area at the exit. But whatever
be the area here the reference is the throat because that is where the velocity is
always equal to the sound speed or Mach number is one. And it is related to the area
of the chamber. The gas accelerates from Mach one at the throat in the divergent to
high velocities. Therefore we define the nozzle area ratio as the exit area divided by
the throat area and it is denoted by Epsilon (ε). Ε = Ae/At. Is it okay? Having said
that area ratio of a nozzle is Ae /At, we want to derive an expression how the area
ratio will affect the jet velocity or how should the performance of a nozzle be linked
to the exit area ratio.
261
Let me repeat the problem such that it becomes further clear. Supposing, I have a
small rocket, I have the throat Mach number as one. I could also have a small area
ratio or a large value of area ratio. I could also have a same rocket in which now I
again draw this nozzle over here, I could have a very large area ratio and how
do I find out and compare the performance of a nozzle with a nozzle of exit area
Ae1 with a nozzle of exit area Ae2 and if the throat area is the same in the two
cases; I have area ratio in one case which is equal to Ae1 by at At. In the second
case, I have area ratio is equal to Ae2 by At. I want to compare which one gives
me higher velocity, I want to compare these two nozzles and therefore, we define
area ratio as exit area divided by the throat area.
I could also have had a larger chamber something like this with much
larger chamber, but still even for a larger chamber the throat area would describe the
same flow condition namely Vt is equal to the sound velocity at. The area ratio
would be Ae/At. Area ratio is always defined with respect to the throat that is exit
area divided by the throat area. (Refer Slide Time: 09:36)
We want to determine VJ and therefore the ratio of Pe/Pt as a function of the value of
Ae/At for the nozzle. We just look at the continuity of flow; we look at the mass
which is entering the nozzle, mass which is passing through the throat, mass which
ispassing through the exit and I write m° is equal to the mass which is passing
through throat area × density at the throat ρt × the velocity at the throat is equal to
the area at the exit × ρ at the exit × V at the exit.
262
Therefore, we find that we need the condition at the throat namely the ρt at the
throat. I need to be able to find out ρe. I do not know Ve; but Ve is the velocity with
which the gas is exiting the nozzle. It will be equal to the VJ, which I have already
derived. Therefore, I need to find the conditions at the throat. Therefore, let us first
spend a couple of minutes on deriving the expression for the conditions at the throat,
which are critical to a nozzle.
The condition at the throat will specify the mass flow rate, because it is choked here
and I will clarify this later on. Let us first find out what are the density, pressure and
temperature at the throat.
(Refer Slide Time: 11:09)
We define the density at the throat as ρt, velocity by Vt, pressure at the throat as pt,
and temperature at the throat is Tt. This means that subscript t denotes the throat
condition. And how did we define the chamber conditions? The density is ρC,
velocity in the chamber VC, which we said was equal to 0 pressure in the chamber pc
and temperature in the chamber Tc. At the exit ρe, exit velocity Ve , which is equal
to the jet velocity VJ at the exit, pe is the pressure at the exit; well these are all the
variables what we have. We want us to find out the value of ρt as a function of ρc,
may be pt as a function of pc and Tt as a function of Tc. Therefore, we again just
look at the flow conditions. We are interested in the condition at the throat, the
conditions are given by subscript t over here for ρ and V. We treat this convergent as
a control volume or we
263
are considering our attention only in this small region, which I show hatched over
here. Gas enters at a pressure pc at velocity 0 at a temperature Tc at it leaves at the
throat with a condition of ρt at a velocity equal to the sound velocity. The pressure is
pt and the temperature is Tt.
Let us write the expression for this control volume. Let us again assume adiabatic
condition and therefore we can write the enthalpy entering is hc plus kinetic energy 0
is equal to h at the throat + we have Vt2/2 which is kinetic energy per unit mass. We
must be able to write this steady flow energy equation; same mass flow is here,
enthalpy in the chamber corresponding to this initial kinetic energy of 0 while at the
throat the enthalpy is ht and Vt2/2 is the kinetic energy.
(Refer Slide Time: 13:55)
Now let us simplify this equation. We get from this equation Vt2/2 is equal to hc
minus ht. And what is the difference in this enthalpy? It is equal to: Cp ✕ (Tc – Tt).
Therefore what is Tc minus Tt; tt is equal to Vt2/2 Cp. What is the value of C p in
terms of gamma: γR/(γ−1). How did this come? We had derived in the earlier class:
Cp minus Cv is R, Cp by Cv is gamma and therefore, Cp is equal to γR/(γ−1).
Therefore, we can write this expression as equal to (γ−1)/2 ×Vt2/ γR.
264
And therefore, we can now write the value of Tc: We take Tt on the other side to
give Tt × {1 + (γ – 1)/2 ×Vt2 / γ R Tt} . What did we do? We have taken Tt at the
denominator, and therefore, I have gamma γ−1 into Vt2/ γRTt. We know that γRT is
the sound speed square or γR × Tt is a sound speed at the throat square. Vt is also
equal to the sound speed at the throat and therefore this is the Mach number of one
square and we get 1 + (γ−1)/2 × Mach number square. Mach number is one and
therefore I get the value of Tt × (1 + (γ−1)/2). This gives us the value of the
temperature at the throat as a function of the chamber temperature Tc. Tc/Tt =
(γ+1)/2 or rather Tt/Tc = 2/(γ+1).
(Refer Slide Time: 16:28)
Therefore, for gamma of 1.4, we find that the temperature at the throat is
approximately 1 over 1.2 times that in the chamber. If the chamber temperature is
2000 K then it will be something near to 1650 degrees; in other words, if gamma is
equal to 1.4 the value of Tt by Tc is equal to 1 over 1.2. Therefore, the temperature
falls at the throat and it is less than the value in the chamber.
265
(Refer Slide Time: 17:36)
Now what will be the value of pt by pc? We have derived the expression in the last
but one class. It is equal to (Tt/Tc)γ−1/γ . Please look back in your notes. Let us see
how we got this value. We had p/ργ is a constant for an isentropic flow and p by ρ×T
is a constant for an ideal gas from the equation of state. Solving for this we got this
particular expressions. In fact, you will remember in the expression for VJ, we had
the expression 1 − (Pe/Pc)γ−1/γ and how did it come, this was essentially Te by Tc
and we expressed it in terms of the pressure ratio. We therefore have pt /pc =
[2/(γ+1)]γ/(γ−1).
What is the value of ρt/ρC? This equals [2/(γ+1)]1/(γ−1). This comes from the ientropic
relation p/ργ is a constant.
We have derived the conditions at the throat namely the value of the temperature at
the throat, the value of pressure at the throat and the value of density at the throat as
a function of the conditions in the chamber pressure and which is known to us. The
chamber conditions are given to us.
I want us to go back and apply these three relations since we know the density at the
throat may be we have to find out the density at the exit. And then find out the value
of the exit area ratio as a function of the exit pressure or alternatively the exit
pressure as a function of the area ratio that we are interested in.
266
(Refer Slide Time: 19:53)
Therefore, I hope that by now we know how to evaluate the conditions at the throat
of a nozzle. We have the throat conditions from a chamber pressure pc and
temperature Tc and density ρc. We know how to find out the conditions of pt, Tt and
ρt .
Now, let us go back and solve the continuity equation; we said area at the throat×
velocity at the throat × the density at the throat is equal to area at the exit ×velocity
at the exit × density at the exit or rather from this I get the area ratio ε = Ae/At = Vt ×
ρt ÷ (Ve × ρe).
I want to substitute the values. We know the value ρt in terms of ρc; it is equal to
[2/(γ+1)]1/(γ−1) . Now we have the value of ρt/ ρc into Vt. We know that Ve is equal to
VJ. What is the value of VJ? VJ = √2γRTC/(γ−1)×{1−(pe/pC)(γ-1)/γ}.
Now, we would like to somehow get rid of ρe and also Vt. We can write Vt as the
sound speed and this is equal to at and therefore we can write it as√γRTt viz., equal
to under root gamma into specific gas constant R into temperature at the throat.
Please be careful since these are all simple algebraic expressions and we are
substituting one into the other and in the process we are also learning how the
properties are varying.
267
(Refer Slide Time: 22:32)
Let solve the equation for the area ratio ε: It is equal to ρt/ρe × Vt/Ve and equals
[2/(γ+1)]1/(γ−1) ×ρc ; ρc can be written as pc/ RTc from the ideal gas equation p = ρRT.
And now, we have γR and let us strike of some of the numbers in numerator and
denominator; √γ and √R go. We will take Pc outside; √2/γ−1×[1− (pe/pc)γ−1/γ . We
have √Tt /√Tc. This is divided by ρe.
We now have an expression for area ratio as given by the above expression viz., ε =
[2/(γ+1)]1/(γ−1) × pc/(RTc) ×√Tt/TC ×1/ρe ÷ √2/(γ+1)[1−(pe/pC)(γ−1)/γ . We would like to
simplify the expression by expressing it as ratio of pressures so that we can write it
as a function of the pc/pe alone. For this purpose, let us take a look at ρe. We can
write the value of ρe in terms of the pressure at the exit. The pressure at the exit
divided by density is equal to specific gas constant into the temperature at the exit. I
simplify this expression to give me ρe = pe/(RTe). Substituting this value of ρe, we get
an expression in terms of pc/pe. We can also make some changes for the value of
Tt/Tc that is the temperature at the throat and chamber.
268
(Refer Slide Time: 26:04)
And as we had seen earlier that the temperature at the chamber divided by the
temperature at the throat is equal to 1 + (γ−1)/2 into Mach number squared and the
Mach at the throat is one. The value of Tc/Tt = (γ+1)/2.
Now, these two equations namely the value of temperature at the throat in terms of
the chamber temperature and the exit gas density in terms of Pe by RTe are
substituted in this particular expression for ε. We therefore get the area ratio ε =
2/(γ+1)1/(γ−1) ×pc/RTc × √Tt /Tc. We can also write the value √Tt/Tc, as [2/(γ+1)]1/2 .
We had got from ρe which was equal to RTe/pe and this is divided by the same value
√2/ (γ−1) ×(1 – (pe/pc) γ−1/γ). Now, let us simplify this: R and R gets cancelled and
we get pc/pe and Te/Tc. If we were to put it in terms of Te by Tc in terms of pc by pe
269
(Refer Slide Time: 28:59)
we get an expression for epsilon ε = {2/(γ+1)}1/(γ−1) × (pc/pe) and R got cancelled
out. Therefore, now we are left with Te, that is the exit temperature and there is
nothing else left. Let us write the value of Te over here and we have taken Pe inside
and here we have Tc. And this × 2/((γ+1)1/γ −1 and this [2/(γ+1)]1/2 ÷ under root of the
denominator. This comes out as √2/(γ−1)[1−(pe/pc)(γ−1)/γ] .
Now immediately we see that 2/(γ+1), 2 gets cancelled and γ+1 comes on top in the
denominator and therefore now I can write the denominator as equal to √(γ+1)
÷(γ−1) × [1 – (pe/pc)(γ−1)/γ}.
Let us now simplify the numerator; we have [2/γ+1]1/(γ−1) .
270
(Refer Slide Time: 31:21)
And now we can express these terms: express Te/Tc in terms of pc/pe. Please
observe that we have been doing this by setting Te by Tc using the isentropic
expansion process as (pe/pC)γ-1/γ. You will recall we have done this several times and
therefore, if now we say pC/pe and (pe/pC)(γ−1)/γ ( multiply this together), we will get
(pC/pe)1−(γ−1)/γ, which is equal to (pC/pe)1/γ because 1 – (γ −1)/γ gives 1/γ.
(Refer Slide Time: 32:26)
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And therefore, we write the area ratio ε = (pC/pe)1/γ × (2/(γ+1)1/(γ−1) ÷√(γ+1)/(γ−1)[1(pe/pc)(γ−1)/γ] .
What does this expression tell us? This expression tells us that the area ratio of a
nozzle increases as the chamber pressure increases or rather as the ratio of pC by pe
increases. The increase in Pc/Pe can come about either by increasing the chamber
pressure or by decreasing the exit pressure. If we have a very low value of exit
pressure my pressure ratio is larger and we require a larger area ratio nozzle. Of
course, gamma also plays a role, but only a secondary role. The main aspect of area
ratio comes from the change from the variations in the value of the chamber pressure
to the exit pressure.
(Refer Slide Time: 33:50)
Let us take a look at some of the values which I have plotted in the slides which will
be now presented. First, we see that area ratio is defined by the value of Ae by At.
The expression we had got we had the jet velocity VJ or Ve given by
√2γ/(γ−1)×RTC[1−(pe/pC)(γ-1)/γ.
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(Refer Slide Time: 34:11)
And thereafter we wrote the area ratio in terms of these parameters and got this
particular value, which worked out to be {2/(γ+1)}1/(γ−1) and a whole series of
gamma terms.
(Refer Slide Time: 34:17)
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We also had the expression in the denominator as √(γ+1)/(γ−1)×[1− (pe/pC)γ−1/γ] . When we plot
the expression for ε, we get: (Refer Slide Time: 34:41)
(Refer Slide Time: 34:53)
the area ratio as a function of pC by pe. As pc/pe increases, the area ratio increases.
Further, as the value of γ decreases from gamma of 1.4 to 1.1, we find a larger area
ratio is required to give the same value of pC by pe.
274
(Refer Slide Time: 35:14)
This slide gives area ratios for larger value of pressure ratio. Again, the value pC/pe
is expressed on the X axis while the area ratio ε is shown on the Y axis. You find
that as γ decreases we need a larger value of the area ratio for the same pressure
ratio. In other words what it tells me is if the gases have a smaller value of γ then I
need a larger area ratio to give me the same value of pressure ratio. This is all about
area ratio.
(Refer Slide Time: 35:49)
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What is it we have done? We found out the value of the area ratio and related it to
the value of the chamber pressure divided by the exit pressure. In general area ratio
of most of the nozzles are between 15 to 400. If we were to look at the expression
for area ratio you find that when pe becomes zero, we need area ratio, which is
something like infinity. We cannot have infinity, because we cannot construct a
rocket, which gives me a very large value of area ratio going to infinity. We cannot
keep on extending because the mass of my rocket will keep on increasing;
therefore, the general practice is to have area ratios between 15 and 400, 15 for
those rockets, which operate within the atmosphere or which operate near to the
Earth and 400 or values around this for rockets, which operate in the vacuum
regions. Therefore, the question which now crops up is if I have a rocket whose
nozzle whose area ratio is either too small or large, how does it perform?
(Refer Slide Time: 37:08)
Suppose, I have a rocket nozzle, which let us say has a small value of Ae by At: this
is the value ε1 which is equal to Ae1 divided by At1. For the same condition of the
throat, I also have another rocket, which has a larger area ratio, let us say Ae2 for the
same value of let us say At1. The latter is ε2 = Ae2/ At1. Now suppose the chamber
pressure is the same in both the cases. What we going to get is a smaller value of pe2
as compared to pe1 since the area ratio ε2 is more. Stated in the reverse, pe1 > pe2. The
smaller nozzle expands to a higher value of pressure; if area ratio increases as in the
case of nozzle with area ratio ε2, the value of pe2 is less than pe1. We also know that
the ambient pressure decreases as the altitude above the surface of the Earth
increases. At sea level, the ambient pressure is one atmosphere i.e., 105
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Pascal. As the altitude increases, the pressure decreases till at an altitude
corresponding to one at which at the edge of the atmosphere let say around 50 or 60
kilometers altitude, the pressure will go down to a very small value and when we go
to geosynchronous altitudes its almost perfect vacuum. The ambient pressure with
respect to the exit pressure of the nozzle is expected to play a role. Let us assume
that in the specific case pe1 for the nozzle with area ratio ε1, the ambient pressure is
Pa.
(Refer Slide Time: 39:25)
Let us show the values or pressures in the figure. In the figure here, we show the
pressure variation along the nozzle. In the first case with ε1 area ratio, the pressure in
the nozzle continually decreases. It starts with the value of pc comes down to a value
of pe1. In the second case, for the same chamber pressure pc, it starts from pc, but
continues further till we get a much lower value of pe2 at the exit.
Let us consider a situation where in the ambient pressure is equal to Pa. I show Pa to
be somewhat less than the value pe1 this figure.
The pressure is varying in the nozzle along its length. The ambient value of pressure
Pa in the first case of the small rocket nozzle is less than pe1. In the second case, the
rocket nozzle is bigger and therefore the gases expand further with the exit pressure
being less than the ambient pressure Pa.
277
(Refer Slide Time: 40:21)
Therefore in the first case the pressure at the nozzle exit is greater than the ambient
pressure? In the second case, the pressure at the exit pe2 is less than the ambient
pressure Pa. In the first case the expansion is not completed; therefore, we call this
nozzle as being an under-expanded nozzle. In the second case, we expand it over and
above the ambient pressure; therefore, we call this particular nozzle as an over
expanded nozzle.
Are there any problems with these two nozzles? What we have done when the nozzle
area ratio is small, that is, the area ratio is to a lower pressure than what is possible,
the expansion is lower than what could have been possible. Therefore we are not
able to get a high jet velocity because the exit pressure has still not been able to
match the ambient value. The expansion is incomplete. We could have got much
more jet velocity had we really expanded it a little bit more come till the ambient
pressure. We are losing some velocity.
What is the problem with over expanded nozzle? Well the pressure here itself within
the nozzle is equal to the ambient pressure. At the exit, the pressure is going to be
even lower. However, at the exit, the ambient pressure is higher. Mind you, the flow
in the divergent is supersonic and does not know the conditions existing ahead of it.
All of a
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sudden the supersonic flow finds a higher pressure because it has already been
expanded to a lower pressure and this is clearly not possible. Therefore, it is
necessary that something like a shock stands over here within the nozzle; that means,
I have supersonic flow it is not able to see anything before it, but all of sudden when
the flow reaches it sees a higher pressure and therefore something like a shock is
required to match the exit pressure.
The situation is like the following: I have a nozzle here and now the pressure has
come to the ambient over here itself and therefore, if I were to plot the pressure, the
pressure is going to decrease further. I need to have a shockwave and the flow downs
stream of the shockwave is subsonic. The divergent nozzle considering the subsonic
flow will act as a diffuser instead of a nozzle and the pressure will increase till it
reaches the ambient value at the exit. That means there is going to be a shock and the
adverse pressure because of the shock would cause flow to separate at the walls of
the nozzle. Since we have a higher pressure at the nozzle wall, the performance of
this nozzle may be even better than had the flow not separated. But, normally this
flow separation does never happen symmetrically, and it’s leads to something like
side forces, and therefore over expansion is never preferred at all. I will get back to
this point a little later; this point may not be clear at this point in time.
(Refer Slide Time: 44:34)
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What we are discussing is that if the nozzle area ratio ε is not properly tailored and
we get the pressure at the exit of the nozzle pe to be greater than the ambient pressure
Pa, we have under expanded nozzle. In this case the nozzle performance is rather
poor; we do not get the high value of jet velocity which is possible by further
expansion to the ambient pressure. But, in case the nozzle exit pressure pe is less
than the ambient pressure Pa, we will have something like a shock. The increase in
pressure at the shock and in the subsonic flow subsequently will lead to flow
separation at the walls of the nozzle. And the flow gets separated from the walls due
to the adverse pressure gradient. Flow separation does not take place symmetrically
along the circumference, with result that in some regions we have higher pressure,
where flow separation takes place. In regions where the flow is not separating, the
pressure is lower. The low and high pressure distributions along the circumference
which give rise to side forces and this is not desirable.
(Refer Slide Time: 45:50)
We will come back to this point after seeing a few pictures on flow separation in
nozzles. What do we really mean? Let us go back and make a plot of pressure
distribution with flow separation or let us say an under expanded nozzle. How does
the flow behave? Let us consider this diagram. A chamber, a nozzle and the center
line of the nozzle. We said that for the under expanded nozzle pe is greater than Pa.
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Therefore the flow comes here, it meets lower pressure; therefore I have something
like flow is going to expand out; as if it bellows out. And how does the flow expand
out; we have rarefaction fans or expansion waves which are generated at the nozzle
exit. And similarly over here that means the plume comes here, this is the lowpressure region; this is higher value of pressure and when the flow expands
out as a series of expansion fans. So we have the expansion waves being shed at
the nozzle exit while the plume spills out. I have here the back-pressure, which is
the ambient pressure surrounding the expanded plume. The same expansion fans
are shown for a two dimensional geometry..
After the expansion waves, pressure in the plume matches with the lower pressure
ambient. At the center the flow velocity is still higher. Therefore, at the center line,
we have expansion, therefore here the pressure is going to be less than the value of
Pa; that is the nozzle with under-expansion forms expansion fans that meets the
expanded boundary of the plume. The expansion waves are reflected back from the
plume surface as compression waves, the compression waves converge to form
shock waves as shown. The pressure behind the shock waves increases more than the
ambient pressure and the shock waves intersect as shown. We have compression
region after the shocks. The compression waves subsequently hit the plume as
shown. They are reflected back as expansion waves. And therefore now, the pressure
decreases from the expansion waves. In this way a series of zones of pressures more
than the ambient pressure and less than the ambient pressure are formed in the plume
from the nozzle. The formation of these zones is due to the interaction of the
rarefaction fans and shocks with the boundary of the plume. In regions wherein
pressure is high the temperature is also high. If the temperature is higher, the plume
becomes luminous and you can see the pattern with alternate bright and dark
zones.
We find that because of under expansion, there is further sudden expansion that
means there is an expansion fans and this expansion fans impinges on the plume
surfaces. And when the expansion fans impinge on the plume surface, the expansion
waves are reflected as a compression that means as weak oblique shocks. These
oblique shocks further compress the medium. The interaction of the compression
waves with the plume surface forms expansion waves and the process of
compression and expansion continues.
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If we were to have an over-expanded nozzle, we will have different flow pattern in
the plume. A shock is formed within the nozzle and this compresses the initially
expanded flow. This is because pe is less than Pa. The flow being supersonic, we
need a shock which will match the higher value of pressure. Therefore, what is going
to happen is the plume boundary will come down like this since the ambient
pressure is higher. The shock waves interact with the plume boundary and are
reflected back as expansion waves. The plumes expand following the expansion
or rarefaction fans. The expansion fans are reflected from the plume as
compression waves and the pressure in the plume thereafter increases. And so the
processes of compression and expansion continue along the plume.
In other words, in the case of overexpansion, we get a higher-pressure region little
bit away from the nozzle exist. In the case of the under-expanded nozzle, we get a
high-pressure region just at the nozzle exist, that means over here, I get a highpressure region following the oblique shock waves in the case of under-expanded
nozzle.
(Refer Slide Time: 52:11)
To be able to appreciate this point, I show some slides of the nozzle plume and
this will become clear to you now.
282
(Refer Slide Time: 52:15)
This shows a particular second stage rocket of PSLV, and here you see this is the
divergence portion of the nozzle. This is the combustion chamber. And if we take the
inside configuration of the nozzle, it will have a throat and will come back with a
convergent shape like this to the chamber. Therefore, we are looking at the outer
portion of the nozzle here.
When the rocket fires for some time, the nozzle runs hot and become red hot. We are
looking at this hot diverging nozzle. It becomes at red hot as time progresses,
because it is heated by the hot gases. And then hot gases are converging towards the
center after leaving the nozzle like in an over expanded nozzle. You see the plume
becoming luminous after the shock wave. The downstream is not clear, because
water is sprayed to cool the plume.
283
(Refer Slide Time: 53:02)
Let me go to the next one, I show the same nozzle again. It is red hot. The white part
is the luminous zone after the oblique shock waves. The oblique shocks are seen and
they interact along the base. (Refer Slide Time: 53:23)
Let us go to some other experimental firing; this shows the engine test wherein at the
exit of the nozzle the flow is probably over expanded. And therefore, you
have something like shock waves, which increase the pressure and temperature in
certain regions of the plume. These regions become luminous. And what happens is
the high pressure region over here, gives you a higher pressure and higher
temperature. If I have oblique shocks like this, which give us
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a high temperature region. It looks like a shock diamond you know, I have a high
pressure region which is luminous. Afterwards, the oblique shocks comes here, I
have rare fractions fans coming another oblique shocks coming; I have another white
patch over here. Again the process, I have something like a series of shock diamonds
from the shocked high temperature gases.. (Refer Slide Time: 54:15)
In continuation, this slide shows a space plane SR-71 We see the shock diamonds in
the plume in this particular case. And we continue with this, this is a test of
an engine. And here you find, there is a shock here and something like this.
This is because the exit condition is over expanded. (Refer Slide Time: 54:24)
We will continue with nozzles in the next class. We will review overexpansion and underexpansion and then proceed further.
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Rocket Propulsion
Prof. K. Ramamurthy
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 12
Characteristic Velocity and Thrust Coefficient
(Refer Slide Time: 00:26)
Good morning. We continue with what we were doing with nozzle area ratios. The
plume from an under-expanded and over-expanded nozzle is shown here. When under
expanded, the exit pressure was higher than the ambient pressure. So, the flow continues
to expand; how does the flow expand when its exits a nozzle? You have series of rare
fraction fans and when these rare fraction fans hits the boundary of the hot plume, they
are reflected as compression waves which merge to form oblique shock waves. These
oblique shock waves when they when they interact with boundary of the hot plume from
the nozzle, re reflected back as rarefaction fans which again forms oblique shock waves
and this process continues.
Therefore, when I look at an under expanded jet i.e., jet formed from the under-expanded
nozzle as it were, I start a high value of Pe compared to Pa. I get first a set of expansion
fans, which are followed by a oblique shock waves. Behind the oblique shock waves, we
have compression. We therefore get a higher temperature, and this higher temperature
286
region shows as a luminous zone. Following this, we have rarefaction fan, again
compression waves and oblique shock waves. Therefore another luminous zone is
obtained here.
Similarly, if we have an over expanded nozzle, the value of pressure at the exit is than
the ambient pressure. Therefore, I form something like a shock wave which matches the
pressure and therefore, to be able to get the value Pe equal to Pa. I have something like
an oblique shock wave. The oblique shock waves continue to interact and we have a
system of interacting oblique shocks. However, along the center line, after the interacting
oblique shocks, I need to have the velocity at the center which is still going axially
straight and the compression process causes the pressure to exceed the ambient pressure.
This is followed by an expansion following the high pressure region. Therefore, in this
particular zone, I have something like a high pressure and a high temperature region. The
high pressure and high temperature region promotes chemical reaction and emits light
and shows up as a luminous zone.
Therefore, what is happening in an under expanded nozzle? We have a set of initial rare
fraction fans, followed by oblique shock wave. In the case of over expanded nozzle, I
have oblique shock waves, which thereafter result in the expansion fan. Therefore, there
is a distinct change in the distribution of the luminous zones.
If the absence of expansion i.e., in the zones of compression, we should get some
brighter spots light these bright zones are due to the shock wave heating and these are
known as shock diamonds. We get this both for the under expanded flow as well as the
over expanded flow. The only difference is that there is a phase difference in the location
of the shock diamond. If we have an over expanded flow, the shock diamond comes
much earlier as shown because we do not initially get the expansion fan. We start with
the oblique shock wave. If we have under expanded nozzle, the luminous zone is after
the first set of rarefaction fans where the pressure is higher than the ambient pressure.
Therefore, a trained eye can determine if a nozzle is under-expanded or over-expanded
by looking at the plume.
287
(Refer Slide Time: 04:02)
See in this case, we have oblique shock wave at the nozzle exit and the plume is seen to
collapse inward. The shock diamond is formed a little later as is seen here. But in this
particular test water is used for cooling the plume so that subsequent shock diamonds
were not visible.
In the plume from the exhaust of the nozzle, we saw one shock diamond followed by
many shock diamonds. With an inviscid flow, you could have infinite number of these
diamonds. But with viscosity, there is some dissipation and that is where I showed this
particular slide which shows the space plane SR 71 in which case may be you do see a
shock diamond followed by second one, third one, fourth one and keep on going.
Therefore, the shock diamonds are the reflection of under expansion and over expansion
in a nozzle. It is very fascinating to see some of these pictures and try to conjuncture
what is really happening. But in practice there is another problem. With oblique shocks,
when they interact at high incidence, Mach stem shocks are formed.
And we have instead of having a regular reflection like what we have discussed so far,
we have Mach reflection and therefore another shock is formed. And this shock wave is
known as a Mach shock wave. That means, we have an incident shock, we have a
reflected shock and a Mach stem shock in between. I have the incident shock over here,
reflective shock and the vertical Mach stem shock. Thus the shock diamond pattern is
different with a Mach stem shock being formed.
288
And that is, how you see the diamonds in particular pictures? The diamond pattern is
actually something like this wedged shape pattern. Normally, I would have expected may
be incident waves like this, what we said was the diamond pattern is something like the
interacting shock pattern but most often we have a Mach stem shock. We do not observe
this regular reflection pattern.
(Refer Slide Time: 06:47)
And when we get a diamond pattern it is of interest to see that whether it is due to underexpansion or over-expansion in the nozzle. The optimum is when the exit pressure of the
nozzle is equal to ambient pressure in which case it goes straight without the formation
of the shock diamonds.
Are there are some problems we have with a under expanded and over expanded
nozzles?
289
(Refer Slide Time: 07:30)
Supposing, we have a rocket nozzle with say an area ratio of let us say ten or so, which
gives an exit pressure equal to the ambient pressure on the ground. Thereafter, this
particular nozzle operates at altitude where the ambient pressure is lower than on ground.
i.e., Pa is lower at a higher altitude than the on the ground. The exit pressure of the
nozzle remains the same at the altitude as on the ground. However, the ambient pressure
has reduced. The nozzle becomes under expanded at the altitude. And therefore, we will
not be able to get a high value of VJ which would have been possible, had we used a
higher value of expansion ratio or equivalently a higher area ratio nozzle.
Whereas, if on the other hand, we have a nozzle of area ratio of let us say forty which
gives me an exit pressure which is much lower than the ambient pressure on the ground,
the nozzle is over expanded. We will have flow separation which is again not desirable. I
have shocks and I have as asymmetric flow separation and in essence I have side forces
on the nozzle.
Therefore, we find that in general, we cannot always have optimum expanded nozzles as
we use fixed area ratio nozzles. We have to live with under expansion, and because of
this we loose out on jet velocity. But at the same time I cannot have over expansion
because I cannot deal with flow separation as it introduces side forces. These are some
problems in rocket nozzles.
290
(Refer Slide Time: 09:26)
And therefore, supposing I have say a booster stage of a rocket and we define booster
stage as a ground stage which has to fly let us say between zero kilometer to something
say like up to forty kilometers or so. Then I cannot have a nozzle, which will be perfect
or optimum for all the entire range of altitude. In other words, at zero kilometers I have
ambient pressure of hundred kilo Pascal whereas at forty to fifty kilometer, maybe the
ambient pressure may be something like four or five kilopascals or might be even lower.
Therefore, my ambient pressure is decreasing and therefore, it is not possible for me to
make a nozzle for ground condition with a single value of epsilon (ε) compatible over the
range of the altitudes. We will lose too much on the jet velocity VJ.
Therefore, I make the nozzle optimum for an intermediate altitude; let us say between
these two extremes - maybe the rocket has to operate from zero to forty kilometers. I
design my nozzle for ten kilometer altitude in which case, the pressure may be somewhat
different from hundred kilo Pascal and may be its something like nearer to let us say 30
kilo Pascal; that means, the exit pressure of the nozzle for the particular area ratio is such
that the exit pressure of the nozzle Pe has a value of 30 kilo Pascal.
291
(Refer Slide Time: 11:19)
Now, this particular nozzle when the rocket takes off, what will happen? Is it over
expanded or under expanded? Now let us show it the figure over here. The Y axis shows
the altitude and we say this design for an altitude of ten kilometers. The rocket has to
operate upto an altitude let us say of forty kilometers. This is where viz., at 10 km
altitude I have the optimum nozzle therefore, initially the nozzle performs in an over
expanded mode and thereafter, it performs in under expanded mode.
We are not getting the high value of jet velocity which we could have got instead of
choosing a value of area ratio ε which gives me the exit pressure corresponding to ten
kilometer beyond 10 kilometers. Had I chosen a value of epsilon which was
corresponding to forty or fifty kilometers, I would get a much higher value of VJ. But
then we cannot also afford to have an over expand the nozzle for the lower altitude of
operation and get into side thrust problems. It is always that a given nozzle operates
either in an under expanded mode or an over expanded mode, but we try to decrease the
extent of these modes. You have seen in the earlier figures where we have shown a
booster stage has a small area ratio nozzle. Whereas, if the rocket is going to operate at
higher altitude, we will design a nozzle with larger value of area ratio such that it is more
in the area of near to optimum throughout its flight altitudes. The optimum shifts to
higher altitude for the upper stages and this something which we need to keep in mind.
292
(Refer Slide Time: 13:14)
We had said that the nozzle, if over expanded, implies that Pe is less than the value of Pa.
But in general, gases flow at high velocity in a nozzle divergent and therefore, we have
the inertial force available as well. And in practice, when we have the exit pressure less
than something like 0.4 times the ambient pressure, then only we have this problem of
flow separation and shock formation even though, ideally we said that Pe just less than
Pa (the ambient pressure) causes flow separation. Experiments have shown because of
the inertial forces the rocket nozzle can operate without flow separation at a much lower
value of exit pressure Pe than the ambient pressure. The value for flow separation is Pe <
0.4 × Pa. This condition was devised by Summerfield and it is known as Summerfield
criterion. What is the implication of this?
293
(Refer Slide Time: 14:49)
Supposing I have to make a nozzle perform on the ground as well as at altitude. The
choice of the area ratio is such that the exit pressure would be about 40 kPa instead of
100 kPa. It is not necessary to have the nozzle designed for an area ratio ε for an exit
pressure of hundred kilo Pascal. But it is okay for me to have a much larger area ratio
such that the exit pressure is equal to much lower value of forty kilo Pascal. The inertial
forces help in delaying the onset of separation. Having said this, it is also necessary to
note that, this is a very accurate condition but only a sort of a thumb rule. In fact the local
Mach number at the zone of flow separation affects the flow separation criterion. And
the value is given by the value of Pe by Pa at the zone of separation is equal to Pe/Pa =
(1.88M − 1)−0.64. In fact based on experiments, it is not only the inertial force which
delays the flow separation as a constant value of 0.4 times the ambient pressure, but it
depends on the local Mach number at which the flow separation takes place. This
criterion based on Mach number at the zone of separation is again based on experiments.
Therefore, I hope by now we get a feel for nozzles and the problems involved with it. We
find that the nozzle area ratio ε depends on the ambient pressure. Ambient pressure keeps
varying in the flight and therefore, we need variable area ratio nozzles. The moment we
talk of fixed area ratio nozzles, it is also necessary for us to consider the problem of
under expansion and over expansion. We lose performance by under expansion. We get
side loads from over-expansion, which is harmful. Having seen these aspects, let us go to
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the next phase wherein, we also would like to know a little bit more about, what is the
thrust? Or what is the thrust generated by the expansion in the nozzle?
(Refer Slide Time: 17:22)
So, far we have considered the jet velocity as a function of the chamber properties; we
said a convergent divergent nozzle is required; we look at the area ratio Ae/At of the
nozzle. We talked in terms of the under expanded, over expanded and optimum nozzles.
We are clear about this, but now the question is, what is the thrust generated by nozzle?
How is the thrust generated? Thrust from change of momentum or momentum thrust
m°×VJ. But we have been telling that depending on the nozzle area ratio ε, the exit
pressure Pe could be different from ambient pressure Pa.
Thrust could come from pressure also, because I am not able to fully expand the gas.
Therefore, I have some pressure thrust just as momentum thrust. Let us write an
expression for the force developed by a rocket, which we have said so far is equal to rate
of change of momentum.
295
(Refer Slide Time: 18:29)
Let us put the diagram in the form of a control volume. We again start from basics. Let
us imagine, we have a rocket as shown here; let the pressure at the nozzle exit be at a
pressure Pe. Now what I do is, I want to find out what is the force is generated?
Therefore, we clamp the rocket on the ground and we attach it to a fixture over here. We
hold it on firmly with a force F such that the rocket is stationary while it is clamped but
in operation. Now, we want to find out this force F which rocket develops? The exit
pressure is not balanced by the ambient pressure all along the rocket and we make a
control volume about the rocket as shown by the straight dotted lines.
Now we want to find out the force F. We find that everywhere on the on the walls of the
control diagram, that is this imaginary dotted lines, where the pressure is the ambient
pressure? Let us call these lines as AB , B C, CD and DA. The nozzle exit is EF along
line BC. We find all along the lines the pressure is the same and the only place where we
have a difference in pressure is in this region EF. That means, over this particular area
corresponding to nozzle exit area EF, we have ambient pressure Pa is acting upstream of
it and Pe acting downstream. Along all other lines, the pressure Pa is balanced acting in
the opposite directions such as on AB and CD and similarly on left and right sides of the
lines outside the boundary of the rocket.
296
Now let us write the force equation for this control volume. The momentum thrust is
balanced by the pressure forces and the restraining force F the momentum component
which is issuing out of the nozzle = F + force from the pressure unbalance; that is: m°VJ
= F + (Pa –Pe) ×exit area of the nozzle Ae. Rather the thrust F = m°VJ – (Pa − Pe)Ae.
The momentum thrust is partially offset by the unbalanced value of pressure force at line
EF giving a value of force (Pa – Pe) × exit area of the nozzle in the opposite direction.
Therefore, the force or rather the net thrust is equal to m°×VJ + (Pe−Pa)Ae. We have now
modified the thrust equation in which we originally considered only the momentum
thrust and we now incorporate the pressure thrust in it.
(Refer Slide Time: 22:07)
To repeat; we had got an expression as F is equal to m°×VJ. Now we add the pressure
term (Pe –Pa) × the value of the exit area. Well, if we want to get the thrust developed by
the rocket; we need the expression for m°VJ. We have already derived the value of VJ.
This was √2γRTc/(γ−1)[1-(pe/pc)(γ-1)/γ]. Therefore, if we can derive an expression for
m°, we can find out the momentum thrust to which we can add the pressure thrust. Let us
now calculate the value of m° that is required.
297
(Refer Slide Time: 22:55)
m° = ρt Vt At and is better that we use the reference as At, because at the throat the flow
Mach number is unity. Therefore, we substitute Vt = at viz., sound speed at the throat.
Now can we write it in a form which is easy to determine. Let us do it. We have ρt/ρC =
[2/(γ+1)]1/(γ−1) . We derived this expression for ρt/ρC in the last class from Pt/Pc and Tt
/Tc .
What is ρC? Pc/RTc from the ideal gas equation. The sound speed at =√γRTt and we need
to consider the local condition here at the throat. At is known.
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(Refer Slide Time: 24:51)
Therefore, what is the expression for the mass flow rate m°? It is equal to
√(2/γ+1)1/(γ−1)×pc /(RTC ) ×At ×√γRTt. Tt can be written Tt/TC×TC and Tt/TC = 2/(γ+1).
√RTc in the numerator with RTc in the denominator gives √ RTC in the denominator. We
therefore get the value of m° = √γ × [2/(γ+1)]γ/(γ−1)×Pc×At×1/√RTc × √Tt/Tc .
We have taken √RTc outside and are therefore left with √Tt/Tc.
Now instead of √Tt/Tc, we can write it as √2/(γ+1) and we now we get the final
expression. It equals √γ × [2/(γ+1)](γ+1)/2(γ−1) × 1/√RTc ×Pc×At. The terms containing γ
are function of γ alone. Let us therefore define √γ × [2/(γ+1)](γ+1)/2(γ−1) = Γ.
299
(Refer Slide Time: 28:20)
In which case the value of m° = Γ/√RTc ×Pc×At. This gives us the net flow rate m° and
is the mass flow rate through the nozzle.
You find that for a given mass flow rate in kilograms per second, if we were to use this
expression, the term capital gamma divided by under root √RTc transfers the mass flow
rate into pressure Pc. We have something like a transfer function. What do you mean by
this transfer function? If I were to rearrange this equation, we get Pc as equal to m°÷At,
that is mass flux through the throat, × (1/Γ)/ √RTc or rather √RTc/Γ. We can interpret
this by the following: for a given mass flux through the nozzle this represents something
like as a transfer function which will give me the chamber pressure; that means, √RTc/Γ
is a function which converts the mass flux into pressure.
Let us examine the unit of √RTc. R has units of Joule per kilogram Kelvin, Tc is in K.
the unit is therefore Joule which is Newton meter. Newton meter is equal to kilogram
meter per second squared into meter and therefore the unit is meter per second. The unit
is of velocity. Γ does not have any units. Therefore this transfer function given by
√RTc/Γ by capital gamma has unit sof velocity and is called as characteristic velocity
C*.
300
(Refer Slide Time: 30:58)
Therefore we can now write, the value of m° = Pc × At ÷ C*, where C star (C*) is
defined as a characteristic velocity which is equal to √RTc/Γ.
We have just defined the transfer function term C* and because we find that the unit is
velocity and we call it as characteristic velocity. And therefore, given a rocket; assume
have a rocket chamber with the throat of area At m2, if we know that the mass flow
through the nozzle is so many kilograms per second, we can go back using this transfer
function, find out what is the value of chamber pressure by using the transfer function
which we call as characteristic velocity C*. It is an extremely important parameter, to
characterize the mass generation rate of a rocket. We will come back to this later on.
Therefore, what is it that we ended up doing? We wanted to find out the value of m° and
the value of m° was √γ ×[2/(γ+1)](γ+1)/2(γ−1) × 1/√RTc ×Pc ×At. This multiplied by VJ +
(Pe –Pa) ×Ae is the force or thrust generated by the rocket.
Slide Time: 33:35)
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Let us simplify this expression and get the value for the thrust generated by a rocket. We
find that F = √γ ×[2/(γ+1)](γ+1)/2(γ−1) × 1/√RTc ×Pc ×At ×VJ given by √2γRTc/(γ−1)[1(Pe/Pc)(γ-1)/γ] + (Pe –Pa) × Ae .
Therefore, now if we take PcAt outside and we simplify this whole expression again,
can we remove some terms? We get RTc over here and this was equal to R into chamber
temperature Tc. Therefore, RTc gets cancelled and therefore the expression for the
thrust is F = Pc×At×{√γ ×[2/(γ+1)](γ+1)/2(γ−1) ×√2γ/(γ−1)[1-(Pe/Pc)(γ-1)/γ] + (Pe/Pc –Pa/Pc)
× ε }, where ε =Ae/At .
Let us understand each term properly. This whole term within curly bracket ×{√γ
×[2/(γ+1)](γ+1)/2(γ−1) ×√2γ/(γ−1)[1-(Pe/Pc)(γ-1)/γ] + (Pe/Pc –Pa/Pc) × ε }if denoted by a
coefficient CF; we can say the force generated by a rocket or thrust generated is equal to
CF into chamber pressure into At, where CF =×{√γ ×[2/(γ+1)](γ+1)/2(γ−1) ×√2γ/(γ−1)[1(Pe/Pc)(γ-1)/γ] + (Pe/Pc –Pa/Pc) × ε }.
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(Refer Slide Time: 37:25)
Let us first find out the value if we have the value of Pe by Pc as an optimum i.e., when
the expansion in the nozzle is to the ambient pressure or Pe = Pa. The value of CF when
Pe is equal to Pa i.e., for an optimally expanded nozzle. There in this case this pressure
term becomes zero. We denote CF under optimum expansion as CF0.
This pressure term is out, but then you are increasing the value of Pa. The value of CF0
corresponding to the value of Pe is equal to Pa will be a maximum and that is denoted by
CF0 which corresponds to the optimum nozzle. This can be shown by differentiating with
respect to Pe/Pa and equating to zero to find the maximum CF. We get CF0 = {√γ
×[2/(γ+1)](γ+1)/2(γ−1) ×√2γ/(γ−1)[1-(Pe/Pc)(γ-1)/γ]}.
If we have the exit pressure not being equal to the ambient pressure i.e., we have let us
say an over expanded nozzle. We get a negative pressure thrust term. If we have an
optimum expansion in which Pe is equal to Pa, this becomes zero. If we have an under
expanded nozzle, we have more thrust compared to cases when Pe is equal to Pa. Hence
we get the maximum thrust F when Pe is same as Pa.
This is because of the contribution of momentum thrust and the pressure thrust. The
condition CF is equal to CF0 implies the maximum thrust when the nozzle is optimally
expanded. This means that an adopted nozzle always gives maximum thrust. Otherwise
because of under expansion, we lose thrust. We are not able to get sufficient momentum
thrust and the pressure thrust is not adequate to give a thrust greater than the deficit.
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Let us take a look at some other results from this expression. You know, all we did was
to derive this value of CF0, which we got for the optimum value.
(Refer Slide Time: 40:38)
That is CF0, the thrust coefficient also represented as CF0 for the optimum thrust
conditions. What is it we find? As we increase the value of Pe/Pc, that means in this slide
it is the inverse of Pc/Pe, we have higher expansion ratios, higher values of Pc or lower
values of Pe, we get a higher and higher value of CF0. The values, which are realizable
for values of γ = 1.4 are shown. It is seen that at the lower value of γ, we have a higher
values of CF0 especially at smaller values of Pc/Pe. The value of the thrust coefficient is
sensitive to the value of gamma, in addition to being a function of Pc/Pe.
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(Refer Slide Time: 41:28)
The same results are plotted on this particular slide, wherein the optimum thrust
coefficient is shown as a function of γ. We find that at large values of Pe by Pc, γ does
not influence the thrust coefficient CF0 coefficient at all. If we have a small values Pe/Pc,
variations in γ cause a considerable change in the thrust coefficient. That means, a
decreased value of gamma will give us a higher value of CF0..
(Refer Slide Time: 42:13)
Therefore, we have looked at the thrust coefficient for optimum thrust viz., when you
have a case of Pe = Pa, for which the thrust F = CF0 × Pc × At, we find that the value of
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CF is also, function of gamma in addition to being a function of Pe by Pc. A small value
of Pe/Pc or a larger value of chamber pressure gives us a higher thrust coefficient. The
value of gamma also influences the CF. If the value of Pc/Pe is not large, for instance we
have a low chamber pressure, then gamma is not very influential. A small value of
gamma will give a higher value of CF0. This is all about thrust generated in a nozzle.
(Refer Slide Time: 43:13)
We told that m° viz., the mass flow through a nozzle is m° = (1/C*) × chamber pressure
Pc ×throat area At, where C* is equal to √RTc/Γ, the unit being meter per second. And
we said, force in the nozzle is equal to CF (whether under expanded, optimum or over
expanded conditions) into Pc into At. Therefore, whenever we make a rocket, we
evaluate the it for CF and C*. But C* does not come from the nozzle, since we are talking
of the transfer function between mass flow rate and the chamber pressure. That means
C* comes from the chamber. Whereas, the thrust coefficient CF tells me what a nozzle is
doing; it takes the chamber pressure multiplied by a coefficient, it gives me the force.
Can we put the performance of a nozzle together? I think we need to go a little deeper
into these two particular expressions. Therefore, let us write out these two expressions in
a slightly different form and then infer. So far we assumed the flow in a nozzle is
adiabatic and reversible i.e., isentropic flow and also one dimensional flow.
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(Refer Slide Time: 44:52)
And for one-dimensional flow, we drive all the expression for VJ, we derive the
expression for under expanded, over expanded, may be flow in the plume outside the
nozzle and all that. And we find that we can write m° is equal to 1 over C* into Pc into
At. We also derived the expression for the thrust and we find it is equal to the thrust
coefficient CF into Pc into At. Now if we play with these two equations, I find over here
Pc into At is common for both these equations. Therefore, can we somehow put this
together? Can we substitute Pc into At from mass flow equation into the force equation?
If we were to do it, we get Pc ×At equal to m° into C* and substituting in the expression
for thrust F, we have F = CF × m° ×C* .
And what is F divided by m°? What is the thrust per unit mass flow rate? That is we are
talking of F d t impulse divided by mass of the propellant m°dt, or impulse per unit mass
is a specific impulse Isp. We therefore have Isp = CF × C*. This is an expression for
specific impulse.
307
(Refer Slide Time: 46:31)
The expression for Isp was VJ. What was VJ? VJ was, when the exit pressure was
matching the ambient pressure that is when we derived the expression for the jet
velocity. Now you have the contribution coming from the exit pressure also.
Therefore, what is it the net inference? The specific impulse Isp depends on the
performance of the chamber, and what does a chamber do? Chamber is generating or
producing high pressure gases. It gives C*. And what does a nozzle do? It expands the
gases and provides a value of CF? CF is representative of the nozzle converting this high
pressure into high velocity. Something like the nozzle effectiveness or the nozzles role
and therefore CF is the figure merit of the nozzle. It is an index of how effectively, it
converts the high pressure gas into velocity
And C* is a figure of merit of the chamber, in that it tells us how high pressure is made
available in nozzle? Therefore, you have a composite index of C* and CF. Therefore the
specific impulse or Isp is a product of pressure generating capacity in the rocket and the
effectiveness in the gases being expanded.
And how is pressure generated?
It is through C*. Whereas, once the pressure is
generated, the nozzle helps in generating the high jet velocity. Therefore, you have both
the chamber and the nozzle contributing to the jet velocity or specific impulse. There is a
propellant to generate hot gas and CF which means the effectiveness of the nozzle.
Therefore, Isp =CF ×C*.
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But so far, we considered only ideal one dimensional isentropic flow and it is necessary
for me to go back and apply corrections since everything cannot be ideal. Therefore,
there has to be some something like an efficiency.
(Refer Slide Time: 48:49)
Therefore, can we talk in terms of efficiencies of mass generation and efficiencies for
thrust production? The mass generation, we decided in terms of C*; therefore, if we have
to have an efficiency for mass generation, it is equal to C star which is experimentally
observed divided by the ideal value which we derived in this class to be ideal value and
equal to √RTc/ Γ.
Therefore, all what we do is an experiment; we have a rocket and find out the rate mass
at which the mass is delivered out. Find out the ideal using the calculation, and you say
this is eta C* (ηC*). We will find that the values are quite high of the order of 98 to 99
percent. We will do some problems on this later on.
How do you get the thrust efficiency? We call it as thrust correction factor ζ. And you
denote it by ζF, a correction factor which is equal to CF actually measured in a rocket
chamber divided by CF which we calculated under ideal conditions. And we use the ideal
value, to find out the efficiency or the correction coefficient. And again these are quite
large for nozzles of the order of again about 97 to 98 percent. Therefore, what we have
done in today class is, we looked at the high temperature and high pressure gases
generated by the propellant and being expanded in a nozzle.
309
(Refer Slide Time: 50:45)
We expressed the mass generation rate in terms of C*; that means, C star tells us what is
the rate at which hot gases get generated from the propellant? And therefore, we wrote it
as m° = (1/C*)×Pc×At and we called C* as a transfer function. And the expression for
this is extremely simple, is equal to √RTc/Γ.
We also talked in terms of the thrust coefficient CF, which described the thrust developed
by a rocket; F = CF × Pc ×At. We determined the expression for CF, and also how CF
varies? And we talked in terms of a thrust correction factor ζF, which is equal to CF
actual divided by CF which is calculated based on ideal, one dimensional isentropic flow
in a nozzle.
310
(Refer Slide Time: 51:43)
Sometimes an effective jet velocity is defined in the literature, to determine the
composite of VJ or the velocity thrust or the momentum thrust and the pressure thrust.
Let us examine what this effective jet velocity is? Let say that the total thrust is given by
the momentum thrust which is equal to m° into VJ and the pressure thrust is pressure (Pe
–Pa) × Ae. This becomes the pressure thrust. Now if we have to put both the momentum
and pressure thrust in terms of an effective jet velocity, we could say F is equal to m° ×
V effective and this is equal to m° × VJ + (Pe – Pa) × Ae. And therefore, the effective jet
velocity is equal to the jet velocity plus, you have (Pe –Pa) ÷ m° × Ae. This is defined as
effective jet velocity and when a nozzle is not adapted to the ambient pressure.
This means that when the exit pressure is different from its ambient pressure at that
particular altitude, the effective jet velocity is different from the jet velocity at the exit of
the nozzle; this is to take care of the contribution of pressure thrust in addition to the
momentum thrust. We will continue with nozzles in the next class, but in the next class
we shall try to take a look on how to shape a nozzle? And about the deviations from the
ideal cases that we have studied and the approximations.
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Rocket Propulsion
Prof. K . Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. #13
Divergence Loss in Conical Nozzles and the Bell Nozzles
Good morning. In today’s class, we continue with nozzles. First, I will review what we
have done so far and then, we will see whether there are any gaps, anything which has
happened in nozzle development, which we have not covered. We said that nozzle is
something like a vent, and we first learnt how to calculate the jet velocity. How did we
calculate? We said that we have a chamber, which is at a pressure pc. Suppose, at the
exit I have pressure pe; I can calculate the jet velocity VJ. We wanted VJ to be as high as
possible and to be able to get a high value we need this vent or opening to be in the form
of a convergent divergent shape and we also put a condition that the minimum area
which we called as a throat, we should have a Mach number equal to 1.
(Refer Slide Time: 00:28)
We called this nozzle as a de Laval nozzle or a convergent divergent nozzle. In the
convergent part, the Mach number was less than 1; in the divergent the Mach number
was greater than 1. After having done this, we said at the exit I have exit pressure which
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is pe; the ambient pressure is pa and if the exit pressure does not match to the ambient
pressure i.e., is not equal to the ambient pressure, we could have some shortcomings in
the nozzle performance either due to under expansion or due to over expansion.
After this, we took a look at what is the flow through the nozzle, we got the equation for
m° = pc ×At/C* as the mass flow rate. We got an expression say m°/At which is mass
flux as equal to chamber pressure pc × 1 over C*. C* had units of meter per second and
we were able to correlate it with the pressure built in the chamber.
The pressure built in the chamber is related to the mass generated or rather to the mass
flow rate through the nozzle. We saw C* as a transfer function to develop pressure in a
rocket chamber. Let us take one example just to clarify things. Suppose, I have a rocket
in which the mass of propellant is let us say Mp kg.
(Refer Slide Time: 02:17)
Let the rocket fire steadily for a period of t seconds. The mass flow rate through the
nozzle is therefore equal to Mp/t kg/s since Mp is in kilograms and the flow rate is in
kilogram per second. If the pressure built in the rocket chamber is pc and if the nozzle
throat area is At, we can directly write that Mp/t = m° = (1/C*) × chamber pressure pc ×
At. We can therefore determine this value of C*. If we do an experiment and determine
the value of C star based on the measured value of mass of propellant, time and pressure
and then we compare the value of C* which I actually measure to the C* which we
derived ideally, both would not be equal. We calculated C* is equal to √RTc/Γ; we
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would find that the ideal value may be a little more than the actual experimental value
because, some flow losses are taking place. The measured value to the ideal theoretical
value was called it as C star efficiency of a rocket or rocket chamber.
We also said that the nozzle has a divergent portion and we can also write the thrust of a
nozzle as equal to the chamber pressure × At × a coefficient called thrust coefficent. We
derived the equation for this coefficient. We called this as CF0 ideal when pe =pa.
Further, we found that since the thrust also depends on the exit pressure and the ambient
pressure, and I show exit pressure pe and the ambient pressure pa here, pe may not be
equal to pa, but the thrust is a maximum in pe is equal to pa; we called this ideal thrust
coefficient as CF0 when the exit pressure was equal to pa.
(Refer Slide Time: 03:56)
In a similar manner, when we have a rocket, which is firing or operating, we measure the
pressure using transducers. We measure what is the thrust that is generated by rocket and
determine the measured value of CF. We also calculate the value of CF, which we did
using the ideal theory. The ratio of the experimental to the theoretical value was called as
the thrust correction factor ζF. This is a summary of what we did. We also did something
which was important. We said that instead of specifying the exit pressure and the
chamber pressure, we can also specify a rocket nozzle in terms of a nozzle exit area Ae
divided by the throat area At which we said was ε.
314
(Refer Slide Time: 05:54)
Are there any questions on what we have done so far? Mind you all that we have done is
for an ideal case, an adiabatic nozzle. And one-dimensional flow you always said that
flow is going straight like this to the right. The mass flow rate is contributing to the
thrust and expansion. Are there any questions so far?
Your question is why we need to couple the C* with CF to determine the value of
specific impulse when the specific impulse can be readily evaluated based on nozzle
flow.
Let us first clarify that C* is something which tells how much chamber pressure is
developed when we provide a certain mass flow rate through the nozzle. The nozzle is
identified by the throat area. In other words, the transfer function between mass per unit
flow rate through the nozzle or mass flux through the nozzle at the throat to the chamber
pressure gives me the value of the C*. We could get the chamber pressure for a given
mass flux at the throat and this is the transfer function. What does it tell us? When we
looked at the expression for C*, it was √RTc/Γ where Γ is a function of γ viz., √γ ×
[2(γ+1)]2(γ+1)/(γ−1). What does it really tell?
315
It tells, supposing I have a mass flow rate through the nozzle - mass flux through the
nozzle throat,
(Refer Slide Time: 06:54)
what is the value of chamber pressure that we get? To get a high value of VJ, I need a
high value of chamber pressure. Therefore, this C* tells you the capacity of whatever
propellant you have in the chamber to generate a high pressure. Therefore, C star is not a
function of nozzle performance, but more like what how a chamber can build up high
pressure. All what it tells you is let us take one or two small examples.
Let us take an example of a rocket, which burns a liquid fuel. We will study about
propellants in the next series of classes.
Suppose, I have a tank containing let us say liquid kerosene. I have another tank
containing oxygen. I introduce them or rather force or push them in a chamber and allow
it to burn to generate a high value of chamber pressure pc. The value of C* for the
kerosene and oxygen, introduced in the chamber, will tell the capacity of propellants
kerosene and oxygen to generate a pressure pc in the chamber.
316
(Refer Slide Time: 09:07)
A higher value of chamber pressure pc will be obtained with a higher value of C star. But
then it is for the chamber. However, we did specify the mass flow rate and the throat
diameter. We did not look at the divergent part of the nozzle even though we did solve
for the nozzle flow. We had written that the mass flow rate m° is equal to ρt × At × Vt.
Therefore, we did not really look at the divergent part of the nozzle; we looked only up
to the throat. Therefore, C* is representative of the chamber to be able to generate high
pressure gases when some mass is flowing. If instead of having kerosene and oxygen
suppose, I have say liquid hydrogen and liquid oxygen may be the C* could be higher.
Therefore, we would prefer this propellant combination to kerosene oxygen. C*
therefore becomes a capacity of the chamber for a given propellant to generate high
pressure gases.
Normally, the value of C star is around 2000 to 3500 m/s; with a lower performing
rocket or propellant that is not that good, will give a lower value of C*. Propellants,
which are extremely energetic, will give higher values of C*.
Now, what is CF? We defined CF as equal to thrust divided by pc × At. In other words, if
we had terminated the rocket at the nozzle throat itself we would perhaps have got a
lower thrust than in a convergent divergent nozzle. Let us qualify this further. If I have a
rocket nozzle and I terminate it at the throat, where the Mach number is equal to 1, what
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would be the thrust? We have chamber pressure pc and now for all practical purposes pc
is acting on all this area. The pressure is also acting normally over the head end over here
and over the convergent part of the nozzle. The force generated from the pressure gets
cancelled as shown, except over the throat area.
The thrust from the unbalanced pressure is over the throat area and is equal to pc × At.
This gives the order of magnitude since we did not consider the variations in pressure
along the length of the nozzle. But we have the divergent part like this because of which
the thrust would have increased. The increase comes from the pressure acting on the
walls of the divergent.
(Refer Slide Time: 11:38)
We can write F is equal to CF × pc × At. The value of pc × At was the thrust if the nozzle
was truncated at the throat. Therefore, CF is something like thrust magnification due to
the divergent part of the nozzle and therefore CF is a quality factor for nozzle. In fact the
values of CF for most nozzles are between 1.2 to something like 3 or 4. We will work
through some examples in the later part of this class. Therefore, we conclude that CF is a
quality factor for a nozzle while C* is a quality factor on the capacity to generate the
pressure in the combustion chamber of a rocket. The product of CF and C* is the net
specific impulse Isp. What is Isp? It is the total thrust divided by the mass flow rate m°.
318
(Refer Slide Time: 13:34)
Let us put the expressions down again. Force or thrust = thrust coefficient CF × pc × At.
Well, pc can be written as in terms of mass flow rate m° = (1/C*) × pc × At. Therefore
pc can be written as equal to m° × C* ÷ At. Substituting this value of pressure in the
expression for thrust F, we observe that At and At get cancelled and we have thrust or
force divided by m° is equal to CF × C*. The value of mass of propellant is m° into time
while force is impulse per unit time. Either way the specific impulse is impulse per unit
mass of propellant or thrust per unit mass flow rate and works out to be the product of CF
× C*.
Therefore, the specific impulse of a rocket has two attributes or properties in it; 1. the
capacity of chamber to generate high pressure and high temperature gases and 2. how
you expand the gases to get high velocity. Therefore, let us keep this terminology very
clear. I have nozzle factor and a chamber factor, which gives us the net Isp. I will dwell
on this further after a couple of minutes. But does this answer your specific question,
why C*? Why CF? And what is the relation? How I got the specific impulse to depend on
C* and CF?
Let us take one example: let me take the example of a particular nozzle which operates
let us say in vacuum, and let me take another nozzle which operates on the ground, let us
say at Chennai which is at sea level. I have a chamber, which generates high pressure
and high temperature gases. The exit pressure is equal to say pe. At sea level the ambient
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pressure is equal to pa which is equal to 100 kPa or 0.1 MPa. Now, I want you to tell me
what is the relation between let us say Isp when the rocket operates at sea level and in
vacuum. I want Isp at sea level condition and at very high altitude conditions where the
pressure is almost zero.
We can the thrust developed as equal to m dot VJ plus I have pe minus pa into what? Ae.
The thrust comes from the momentum thrust plus the exit pressure minus pa into the exit
area where Ae is the exit area of the nozzle. Mind you we derived this, and we said we
had control volume and therefore, we found a pressure thrust in addition to momentum
thrust.
(Refer Slide Time: 15:33)
Now, we would likw to write the equation for thrust of the rocket using the same nozzle,
when it operates in vacuum instead of operating on the ground at sea level conditions.
Let us say the nozzle now operates in the vacuum, the same nozzle, the same area ratio
this is the value of the exit pressure is the same and it is pe. The chamber pressure
remains the same at pc. The thrust now becomes m dot into VJ plus the pressure thrust pe
into Ae.
Now, I want to find out what is the specific impulse at sea level. Specific impulse at sea
level is therefore equal to VJ + [(pe – pa) /m°] Ae .
320
(Refer Slide Time: 17:24)
Now, what would be the value of specific impulse for the same nozzle functions in
vacuum; I call it as vacuum specific impulse. Vacuum is equal to what would be the
value of pa; it is 0. Therefore, we have VJ + (pe/m°) × Ae. In other words, the same
nozzle when it is fired in vacuum gives me a higher thrust because, minus pa is missing
in the above expression. Can we relate these two?
Let us say Isp at vacuum with the Isp at sea level. I find therefore, the specific impulse
of a given rocket operating in vacuum is greater than when it operates on the ground. In
other words, the specific impulse corresponding to operation in vacuum is greater than
when the same rocket operates at sea level conditions. Now, we want to derive a slightly
modified relationship relating the two.
Therefore, we write Isp at vacuum Isp,vac = VJ + pe / m° × Ae and what is the value of
m°? m° = 1/C* × pc × At. If in terms of the C star, the Isp becomes the following: I have
the value of Ae; Ae by At is the nozzle area ratio ε. The vacuum specific impulse
becomes VJ + C* × pe/pc × ε . This is the nozzle area ratio ε in this expression. In other
words, compared to a nozzle which gave me VJ plus this value of (pe – pa)/pc×ε at sea
level, we get a much higher value. If this particular nozzle at sea level was such that I
have optimum expansion namely pe was equal to pa, the Isp at sea level would have
been just VJ alone.
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All what I am telling is, if the nozzle was such that that the exit pressure was same as the
ambient pressure for which we told ourselves the CF is a maximum, we would have got
the value Isp = VJ. But the same nozzle when operating in vacuum, we get an additional
contribution C*×pe/pc×ε. Therefore, now the question comes how do I specify the
specific impulse? If we tell that the rocket is operating in vacuum, we get a higher value
of specific impulse. If it is tested on ground, we get a different value. Therefore, we must
be clear in our terminology and therefore, two types of specific impulses are given; one
is Isp corresponding to sea level operation and the second is Isp corresponding to
vacuum. Therefore, whenever the performance of a rocket is specified we must be
careful to know whether sea level or vacuum operation is being specified.
(Refer Slide Time: 20:00)
Therefore, there are two ways of specifying the specific impulse whether vacuum
specific impulse or sea level specific impulse; but then there is another problem. If we
have a higher value of chamber pressure, we get a higher value of expansion ratio and I
can get a higher value of the specific impulse. Therefore, we also need some terminology
which says a standard chamber pressure and the standard chosen is we specify specific
impulse for pc equal to 7 MPa or 70 bar pressure; that means, specific impulse is
normally specified when the chamber pressure is equal to 70 bar. The choice of 70 bar
comes as it is about 1000 psi in the FPS system of units.
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(Refer Slide Time: 22:40)
When the ambient pressure is equal to 1 bar or 100 kPa pressure, we imply sea level
conditions whereas, when we talk in terms of vacuum, we imply very rarified
atmosphere. However, for each of these two conditions, we specify chamber pressure as
7 MPa. A rocket can fire at different pressures, but if we are to compare something, we
need some standards and the standard is a chamber pressure of 70 bar and an ambient
pressure of 1 bar for sea level Isp and 0 bar for vacuum specific impulse. The vacuum
specific impulse is higher than the sea level specific impulse, which is VJ when the exit
pressure is equal to ambient pressure. Are there any other questions on what we have
done?
See so far, we have been talking of only one - dimensional flow in the nozzle. We
discussed the divergent part and sketched it as a diverging cone.
We have the convergent part, but I am really not bothered about convergent because,
anyway at throat I had a Mach number of one and the flow lines stream out almost
axially. We had a chamber of pc. Now, in the divergent part flow is diverging out, and if
we look at the flow which is taking place, the gas which is flowing near to the wall will
have a direction along the wall while for the gas along the center line the flow would be
along the axis as per symmetry. Therefore, we may not be justified in assuming one dimensional flow at the exit of the nozzle. It is really not correct and we have to make
some corrections for may be the radial flow or for the divergence in the flow.
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There is a simple way of doing this. The thrust is not going to be in the axial direction
according to the figure. A component of thrust is going in this direction normal to the
center line; it gets balanced out and only the effective thrust is in the axial direction.
How do I get that value?
Well, there is an actual flow taking place along the nozzle; let us assume that the half
divergent angle of the nozzle divergent is alpha (α). Now, the flow near the wall will be
alpha. On an average, the mean flow direction could be alpha by 2 because here, it is
alpha along the wall and zero along the center line of symmetry. On an average the mean
flow we can assume makes an angle of alpha by 2. I can also define a small element and
do the problem by integrating it out, but the above approximation is sufficient for me to
give an answer. In other words, on an average the flow leaves at at an angle equal to α/2.
Is it ok?
We would like to determine the thrust. Let us assume that the nozzle is adapted; that
means, ambient pressure is equal to pe here; F is equal to m dot into VJ over here. What
is the mass which flows along the axis? It is equal to m° × cos (α/2), that is the actual
mass flow rate along the axis because on an average some mass flows at alpha, some
flows at 0 with the mean direction being α/2.
(Refer Slide Time: 24:00)
The average mass flowing along the axis is m°cosα/2; VJ is again corresponding to this
α/2 as the mean direction of the velocity giving the axial component of velocity as VJ cos
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α/2. Therefore, the thrust due to the divergence being at an angle α will therefore be F =
m°cos α/2 × VJ cos α/2 i.e., m°VJ cos2α/2. Is it all right? All what we told was flow is not
all along the axis. Flow along the wall is at an angle α; on an average the flow is at α/2
and therefore, the mass component along the axis is equal to m°cosα/2. The axial
velocity on an average is equal to VJ cosα/2. Therefore, the product of m°cosα/2 into VJ
cos α/2 make this cos2α/2.
We would like to simplify this expression. We use the trigonometric expression Cos2θ =
2 cos2θ – 1. This gives cos2θ = (cos2θ+1)/2. And therefore, we can write cos2α/2 =
(1+cosα)/2. This trigonometric manipulation is done because we can express it in terms
of the half divergence angle of the nozzle. Mind you the total divergence is 2α; we said
that α is equal to half divergence angle of the divergent.
(Refer Slide Time: 27:35)
Therefore we get the thrust along the axis of a rocket F is equal to m°VJ (1 + cos α) / 2.
The term (1+cosα)/ 2, shown within the circle, is the loss factor due to the divergence. It
is denoted by the Greek symbol lambda λ; and we say λ corresponds to divergence loss
or F = m° × VJ × λ. Well, λ is something we say loss due to divergence, but I am not
really looking at a loss; see actually, we are just multiplying it by a factor and it is the
diverging loss factor. Therefore, if we have to have a loss, the loss should be something
different.
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We say λ denotes the availability of the axial thrust out of the total due flow vectoring.
What is the non-available part? The thrust not available is equal to Δ =1 −λ. In other
words, λ tells us the fraction of the available thrust, which we call as divergence loss
factor. The loss of thrust expressed explicitly is equal to Δ =1 − λ. Therefore, we have
defined two terms and what are the two terms for the actual divergence effects? We
defined λ as the divergence loss factor or rather we have to multiply the value of m° VJ
by λ to determine be the thrust in the convergent divergent nozzle.
(Refer Slide Time: 30:40)
What is not available? The thrust, if the entire flow was axial everything would have
been m°VJ therefore, 1−λ is not available. Therefore, we define capital delta as 1 −λ for
losses. Now, why are we doing all this? We would like to have a nozzle in which we do
not have too much of loss due to the divergent. Therefore, let us put some numbers for
the losses.
If the half divergence angle alpha of a nozzle is 0 and the loss would be zero. This is not
possible because, I have a parallel walls for the nozzle. Alpha could be 5 degrees; it
could be 10; it could be 15; it could be 20; it could be 25 or let us say 30. In other words,
we are looking at different nozzles for which let us say half divergence angle varies from
0 degree in which case I have no divergence to other angles.
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When α = 0, cos 0 is 1; therefore, the value of 1 plus cos α/2 is 1; that means, the entire
thrust is available and the loss coefficient is 0 or in terms of percentage loss it is 0
percent.
When I have the value of alpha equal to 5o, the value of lambda is 1 plus cos 5o / 2; Cos
5o is around 0.99 and the value of λ comes out to be 0.9988, and if I have Δ = 1 minus
lambda, it is equal to 0.12% ; that means, 0.0012. If the angle α is 10 degrees, λ is equal
to 0.9924 and the loss Δ comes out to be 0.76 percent. Let us put few more values: for 15
degrees, the value λ is 0.9830 and the loss is 1.7 percent. If α is 20 degrees λ is 0.9699; Δ
= 1 − λ gives me value of around 3 percent. If it is 25 degrees, λ is 0.9537; I will qualify
these numbers and the loss is 4.63 percent or 0.0463. Well, the last value of α of 30
degrees for which lambda is 0.933 and the loss 1 −λ comes out to be 6.7 percent. If we
were to put one more angle let us say 35 degrees, the value of λ is 0.9066 and the value
of loss Δ is 9.04 percent. What is it that we are tabulating here? We are considering the
divergent angle of the nozzles to vary from 5 degrees to 35 degrees and for each of the
values, we get the divergence coefficient and also the loss in thrust in percentage. You
find when I go from a semi divergent angle of 5o, I am losing just 0.12 percent thrust; for
10o I am losing 0.76 percent thrust; when I come to 15, I have lost already 1.7 percent
thrust; when I go to 20o the loss is quite high. We have lost 3 percent of the thrust; when
α goes to 25o it becomes 4, 6 and so on.
In other words, it does not appear meaningful to have any semi-divergence angle greater
than 20 since the losses become substantial. In fact, we were to compare 10 and 15o
nozzle divergents, the loss for the 15o nozzle is 1.75 times more.
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(Refer Slide Time: 32:16)
If the half divergent value was 20o, the loss is quite heavy. Therefore, the general
practice therefore is to adopt some value around 15 degrees such that the loss is
somewhat small. What loss? The divergence loss, but that is not the only reason. Let us
try to put one more reason on to it. Let us consider the divergent part that I show here.
(Refer Slide Time: 37:27)
We have seen that if the divergence angle is very small say of 5 degrees or even 1
degree, the loss factor is almost going to be negligibly small. Therefore, why not have
such a small angle. There is another implication as shown in the figure. We show the
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center line alone along the axis of the nozzle; we have the throat; the throat radius is rt;
the exit value of radius is re and what is α? tanα is a value of this angle: (re − rt) ÷ the
length of the divergent Ld. The divergent length Ld is equal to (re − rt)/tanα. Is it all
right? All what we are saying is, if I have a small angle for alpha, a nozzle for the same
exit diameter and throat diameter will be much longer. If the angle is 0o, the length will
be infinity. Therefore, let us put the length here into this Table. What do we find? Well,
we just put the length of the divergent for a particular case of throat and exit diameter of
the nozzle i.e., Ld divided by re at exit − radius rt at the throat. This for the zero degree
nozzle 0 is infinity.
If we have a semi divergent angle of 5o, the value becomes 11.43. If it 10o, it is 5.67. If it
is 15o, it is 3.73. If it is 20o, it is 2.75; 25o it is 2.14; 30o it is 1.73 and if it is 35o the value
is 1.43. What does this mean? The length of the nozzle is very large if the angle is very
small and the nozzle length reduces as α increases., If we compare the ratio for a 15o and
20o degree nozzle, we have values of 3.73 with 2.75; the change is not as rapid as it is
between 11.53 and 5.67 for smaller values of α. As the nozzle length becomes longer and
longer, the mass of the nozzle becomes larger. We had also found earlier that ΔV, the
ideal velocity provided by a rocket, is equal to you Isp or VJ into natural logarithm of
initial mass to final mass of the rocket. The mass of the rocket will go up as the length of
the nozzle increases. The mass of a nozzle and therefore of a rocket of small angle of
nozzle divergence will be more. And as the inert mass of the rocket increases, the ideal
velocity provided by the rocket will decrease. The general practice is therefore to choose
a divergence angle around 15o for a conical divergent. Mind you it is just based on the
premise that we do not lose any further. We do not lose too much of thrust because of
enhanced angle, but at the same time, we do not enhance too much the mass of the
nozzle. We have lost only 1.7 percent of thrust and and the nozzle weight does not go up
drastically as it is if we go for smaller angles. Therefore, based on this divergence
analysis, we can summarize that a conical nozzle will normally have a semi divergence
angle of about 15o. We will not go for smaller angles because in that case, the nozzle
becomes long and mass would go up; we will not go for larger values of angle because,
if we go for larger angles, we will lose more of the thrust. Therefore, the optimum for a
conical nozzle is generally kept at 15o semi divergent angle.
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(Refer Slide Time: 41:20)
Does it make sense? Now, if this part is clear I just have a few more things to tell in a
nozzle. The question is, why did we address this divergence problem in such a major
way? We said that the divergent part of nozzle is so chosen for α such that do not loose
thrust and therefore require we smaller value of alpha. What prevents us from having a
nozzle in which we can bring it back to give a very small divergence angle α at the exit?
We can initially expand it out with larger angles and reduce the angle at the exit. We
have the throat here; I have a conical nozzle. If we could have a small value of
divergence angle at the exit, we would not loose out by the divergence loss.
(Refer Slide Time: 42:28)
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What is being said is that in the initial stages of the divergent portion, we provide a
larger divergence angle and reduce it later on; that means, we have initially a higher rate
of expansion followed by a smaller rate of expansion. The shape of the divergent then
looks something like a bell: the shape of the nozzle looks like a bell here. This is the
center line. Initially, we have larger expansion angle. We decrease the divergence angle
such that the flow goes out more axially.
In other words, we have a contour for the shape of the nozzle and such nozzles are
known as contour nozzles or simply as bell nozzles. I think we could discuss further for a
couple of minutes on the contour nozzles. We initially expand out the gases using larger
values of divergence angles. Let us plot the pressure distribution along the length of the
nozzle. We know how to do it. The pressure at the throat is equal to [2/(γ+1)]γ/(γ−1). Let
us first consider a conical nozzle divergent with a divergence angle of 15 degrees. We
are only following up with the one - dimensional analysis. We plot the value of pressure
in the nozzle as a function of distance over here. This is from the chamber; this is at the
throat “t” followed by the divergent portion.
Now, the pressure keeps falling as we progress towards the nozzle exit. Let us say this is
the chamber pressure value at the throat; we know how to calculate it. The pressure
keeps falling further and this is the exit value of the pressure. This is for a conical nozzle.
At the entrance to the divergent, wherein the pressure is still quite high, we have a more
rapid expansion in the case of a contour nozzle. Since the pressure is high the flow
cannot readily separate from the walls of the divergent. The divergent walls of the nozzle
will continues to guide the flow. Afterwards, the angle is reduced and the flow is guided
to be more axial.
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(Refer Slide Time: 44:34)
In this case it is seen possible for us to even reduce the length of the nozzle further and
therefore, these bell nozzles are specified in terms of let say 80 percent bell or they say
70 percent bell. What is meant is the following. A bell nozzle whose divergent length is
80 percent or 70 percent of the divergent length of a conical nozzle is all what is
required. We can terminate the length here itself because we are able to get the exit
pressure or equivalently the exit area ratio. We can more effectively employ a bell
nozzle than a conical nozzle.
(Refer Slide Time: 46:51)
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The exit divergence angle of a bell nozzle could be between 2 degrees to something like
5 degrees, thus giving a very low divergence loss. Initially, we expand it rapidly; here we
provide a large value may be 20 degrees to 50 degrees. We can make the nozzle a little
more stubby or shorter with than the length of the bell nozzle being a fraction of the
conical nozzle. To repeat, 80 percent bell nozzle means, the length of the bell is 0.8 times
that of an equivalent conical nozzle. What do we do in a bell nozzle? We immediately
expand downstream of the throat where the pressure is higher and allow the divergence
to be smaller in the region of the exit such that we have less divergence loss. In fact, one
paper, which we could read on this subject is by G V R Rao. He worked on it at
Rocketdyne a long time ago. The paper is on exhaust nozzle contour for optimum thrust
and is refereed to as Rao nozzle.
(Refer Slide Time: 48:16)
This was published in the journal ‘Jet Propulsion’ which preceded the AIAA journal.
The volume number is 38 and the year of publication is 1958. The page number is 377 to
381.
In a bell nozzle we initially expand the gas rapidly; we have something like rapid
expansion and then we have something slow expansion which does not lead to adverse
pressure gradients near the walls. We can afford to have a shorter nozzles compared to a
conical nozzles. Most rockets make use of the bell nozzles. If we have a bell nozzle,
none of the earlier criterion like flow separation are relevant because we have higher
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pressure gradient along the wall. Therefore, whenever we talk of Sumerfield criterion
saying exit pressure is 0.4 times the ambient; it is more applicable for conical nozzle and
not exactly for a bell nozzle. I think this is all about nozzles; conical nozzle and contour
nozzle. I want to spend another few minutes on different types of nozzles.
(Refer Slide Time: 50:13)
The ambient pressure decreases as the rocket moves up to higher altitudes. We had said
that for maximum thrust coefficient, the pressure at the nozzle exit should be the same as
the ambient pressure. Is it possible to have a different type of nozzles which can adapt to
the altitude of operation. Can we make a nozzle to adapt to different altitudes starting
from 0 kilometers and keep going up to 10 kilometers or 100 kilometers height? We
shall deal with this in the next class and work out one or two small problems.
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Rocket Propulsion
Prof.K.Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 14
Unconventional Nozzles and Problems in Nozzles
Good morning. In the last class, we discussed about contour nozzles. The shape of the
contour nozzle is in the form of a bell. Initially, you expand the flow by a large angle and
you compress the flow later on, such that you get a very small value of divergence angle
of let say 2 to 5 degrees at the nozzle divergent. Initially, you expand the flow at a larger
angle say between 20 to 50 degrees and this shape of this contour is something like a
parabola. You can fit with a second order parabolic equation for the shape or contour of
the bell nozzle.
(Refer Slide Time: 00:15)
This is how a contour nozzle looks like. In today’s class, let us look at some
unconventional nozzles and examine whether there are nozzles other than conical nozzle
and contour nozzle.
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(Refer Slide Time: 01:38)
Also, we would like to note that the changes from conventional conical and contour
nozzles must be such that they must give better performance. We keep this in our mind.
Well, a nozzle operates at its optimum when the value of pressure at its exit is equal to
the value of the ambient pressure. In other words, if we want to make a nozzle, we
design it for operation, let say at 15 kilometres. But the rocket operates between 0 km
and 30 km. We start operating the rocket on ground i.e., from 0 kilometres and the rocket
goes up to 30 kilometres. During the initial stages of its operation, viz., between 0 km
and 15 km, it is not optimum as the exit nozzle pressure would be less than the ambient
value. At 15km, it is optimum and beyond 15 km again the nozzle exit pressure is greater
than the ambient. It is under expanded and not optimum. We start off with the nozzle in
an over-expanded mode which after the design point operates in an under-expanded
mode. Therefore, can we have say a nozzle in which we can have an extendable nozzle
with varying area ratios in which the exit pressure is matched to the ambient pressure as
the rocket moves up?
Let me give you an example. Suppose, we have a nozzle and this nozzle is operating at
lower altitudes or higher at values of ambient pressures. Now, we want to make it
optimum and therefore we have to shorten it with a lower value of the exit pressure at the
nozzle exit. May be we have to expand it out to larger values of area ratios when the
rocket operates at higher altitudes. Therefore, we initially have a nozzle something like
this. we extend the last part over here and then lock it. The area ratio has now increased
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and the nozzle has become longer. This initial lower area nozzle operates at low altitude
while the larger area ratio nozzle operates at higher altitude.
(Refer Slide Time: 03:00)
When the rocket reaches higher altitude, we shift the second half of the nozzle divergent
through some mechanism. We shift this over here and thus increase the area ratio of the
nozzle. That means, this becomes our initial low area ratio nozzle and correspondingly
we have a higher area ration nozzle. That means, we extend the nozzle to provide higher
area ratios during the flight. At the lower altitude, we use the smaller area ratio. During
this time we keep the larger area ratio segment on top of the lower area ratio portion.
When the rocket reaches a particular altitude, we push it out. The nozzle length increases
and the area ratio increases and this is what we call as an extendible nozzle. This was
tried in a flight. It is not used in practice even though it has been tried. We call it as
extendible nozzle. We could have several segments coalescing one on top of the other at
lower altitudes and being pushed up as required by a mechanism in an extendible nozzle.
If, instead of having an extendable nozzle, are there other alternatives? One such
alternative is a double bell nozzle or something like a dual bell nozzle.
In this we have a nozzle like this and I want to increase the area ratio still further. What
we do is that we put something like a step at the exit of the lower area ratio and then we
continue the nozzle profile like this. Now, the centre line of the nozzle is the same. Both
the portions are permanently in place. Now, what is going to happen? At lower altitudes,
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the flow expands to the ambient pressure and flow separates at the step and flows over.
At higher altitude, because the pressure is very much higher than the ambient pressure,
the flow reattaches at the junction or step and flows into the second part of the bell.
Therefore, we can get area ratio ε1 and area ratio ε2 corresponding to the lower and
higher altitude of operation respectively. This is known as a dual bell nozzle. In fact, this
month’s issue of AIAA journal has a paper on this dual bell nozzle looking at the
optimum conditions. That means, work is still pursued with the dual bell nozzles and it
may have some promise.
(Refer Slide Time: 03:50)
Therefore, this is the second unconventional nozzle. First is the extendable nozzle and
second is a dual bell nozzle. The third is something like a radial flow nozzle. By the
radial flow nozzle; what do we mean? Something an expansion deflection nozzle. Let us
sketch a throat followed by a divergent. Let us say, this the convergent and we have a
divergent following the throat. This is the centre line. At the throat, put a block centrally
making the flow in the throat to be annular; something like a plug in the throat. We allow
the flow to take place in the annular space between the blockage and the throat and what
happens? The flow is guided by the contour wall in the divergent part. At the centre, it is
not guided. Therefore, we have something like an expansion wave. By this centre
expansion, which is avalable, the nozzle is able to adapt to different altitudes.
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Therefore, by putting this centre blockage, I can make this particular nozzle operate at
different altitudes. In other words, I have the outer wall, which guides the flow. Inner
part is free. It can adapt different altitudes and therefore, this is known as an Expanded
Deflection (ED) nozzle or expanded deflection nozzle. What we have to do is that we
want the nozzle to operate at different altitudes.
(Refer Slide Time: 05:03)
Therefore, what we have done is, we have introduced a plug in the throat region and we
allow the contour to change the pressure of the outer flow. But, the inner region of the
flow is keept free to expand. This is the principle of the expansion deflection nozzle.
As an extension to this type of the plug nozzle, we could also have a different type of
plug nozzles. Instead of having plug here and allowing the outer divergent contour to
guide the flow in a nozzle, we could have a nozzle and I can have this plug in the form of
a contour over here. Centre contour along the plug as shown. What is it that we do? We
guide the flow over here, from the chamber. In other words, I have a annular chamber
which leads to the annular throat over here and we allow the shaped plug to guide the
flow. We allow the flow to come along the plug. We allow the inner contour instead of
the outer contour from the divergent to balance the flow. Keep the outer open such that
the flow is free to expand. This is again a case of an adapted nozzle.
The shaped plug at the annular throat can adapt to any ambient pressure condition. I
could have the contour of the plug in the form of a spike, in which case, we call it as a
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spike nozzle or simply we call it as a plug nozzle. I could still have some more
variations.
This particular plug which is at the centre: we could terminate it a little earlier instead of
ending up at a point. We do not allow the total spike. In other words, we have a primary
flow along this and I have the shock waves here and the secondary flow here. I have
recirculation and a base pressure here and this becomes what we call as the aero-spike
nozzle. What we are saying is that the combustion is happening over here; in an annular
combustion chamber. We push the flow along the spike or contour and we allow the
spike to expand the flow along the contour. The free expansion in the outer portion
adapts the flow to the ambient pressure.
Instead of having an outer boundary which regulates the pressure, we have an inner
boundary which corresponds to a spike, which we call as a plug nozzle or a spike nozzle.
Of course, this is a plug and the same plug which we used in the case of a contour
nozzle; Here, the outer is free such that the nozzle can adapt to different ambient
pressures of operation. This is therefore the case of an adapted nozzle.
Aero-spike again: we would have additional thrust coming from the base of the plug if
truncated as shown and which we call as a Aero-spike. But, why should we always think
in terms of a cylinder or a bell or something like that. Why not open out the bell. Make it
something like linear or planar. If we do not have a cylinder, but have an opened out
cylinder we call it as a linear nozzle. What do we mean by a linear nozzle? Well, we
open out the nozzle something like a two dimensaional sheet and we have the surface
such as a ramp in the shape of a contour. Now, we allow the flow over this contour
surface, this ramp as it were, and we use the ramp to expand the flow to the ambient
pressure.
In other words, I have a two dimensional surface as shown. Let say, over here, it could
be a shape of something like an aircraft wing. Flow comes along this, guides along the
surface and comes out. The shaped surface is not confined and it could as well be a part
of an aeroplane, like let say a wing or a fuselage.
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(Refer Slide Time: 09:18)
We allow the gas to come along it and flow along the surface. It is known as linear
nozzle. This has been used for space plane. These follow the same principle what we
have discussed on nozzles so far.
(Refer Slide Time: 10:38)
We now summarize what we have learnt in nozzles through a series of slides. In the first
slide, we look at the divergence losses, may be, alpha and how we got the divergence
losses coefficient.
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This was the value of Δ, which we decided as percentage loss in thrust versus the angle
α.
(Refer Slide Time: 10:47)
We derived this and we had said that the nozzle half angle is around 15 degrees or so for
a conical nozzle. We talked in terms of a contour nozzle instead of having a conical
nozzle wherein we initially expand the flow to αi before bringing it back to αe.
(Refer Slide Time: 11:05)
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The value of αi is between 20 to 50 degrees and then, bring it back. We have a low angle
over here at the exit of something like 2 to 5 degrees or so. Smaller the angle at the exit,
smaller is the loss.
This is the extendable segment.
(Refer Slide Time: 11:30)
Initially, we have a small area ratio corresponding to exit area Ae1.We store another
segment on top of this a segment between area ratio Ae2 and Ae1 and at higher altitude
deploy or push itto get the larger area ratio nozzle.
(Refer Slide Time: 11:42)
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This is a dual bell nozzle. As you see along the divergent contour we have a step at
which for higher ambient pressure the flow separates but at lower ambient pressures
follows the contour of the second part of the nozzle. As I told you, there was a research
paper in the AIAA Journal this month in which dealt with the dual bell nozzle.
(Refer Slide Time: 12:10)
Those who are interested should go through it. This is a plug nozzle. I put a plug in the
throat to make an annular throat. We have an outer surface, which guides the flow. The
inner surface is free therefore, it can adapt to the ambient pressure. What you have is an
annular throat, instead of having a cylindrical throat.
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(Refer Slide Time: 12:13)
This is what we called as a plug nozzle or a spike nozzle. We have a spike following the
annular throat. The flow comes from the annular combustion chamber. The hot gas is
generated in an annular chamber instead of a cylindrical chamber. We push the flow onto
this inner contour surface and this surface guides the flow. Outer surface is free;
therefore, the expansion can adapt to the altitude. It is not used in practice. This
summarizes what we learnt about nozzles.
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(Refer Slide Time: 12:37)
A nozzle runs hot as the hot gases in it are at high temperatures. Therefore, to protect the
nozzle, we provide insulation on the inner surface. This is the conical nozzle. I give
something like a carbon phenolic composite material as an insulation, which can
withstand a high temperature The composite materials such as carbon phenolic are
known as ablative materials. I will come back to it, when we deal with cooling of
rockets. I will get back to this slide a little later in the course. But, this is how the
construction of a nozzle looks like. This is the outer surface and this is the inner wall of
the nozzle.
(Refer Slide Time: 13:09)
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I repeat the case of a hot nozzle in this slide. A nozzle is firing for a certain amount of
time. This is the conical nozzle. You see, that the nozzle runs red hot.
(Refer Slide Time: 13:26)
Well, with that we close the portion on nozzles. But, it will be useful to do a couple of
problems on nozzles. One of the problems which we do is related to this particular rocket
shown in this slide. This rocket is known as Saturn 5. Saturn 5 rocket was used to put
first men on the moon. We call it as Saturn 5 launch vehicle, which puts the Apollo
capsule carrying three men on the moon. This is, by far the biggest rocket ever made in
the history of rockets. It is the most powerful rocket and what does it consist of?
The first stage of the rocket, in the lower portion, consists of five rockets clustered
together. Each one of these rockets is known as F1 rocket. It consists of five F 1 rockets
clustered together. These rockets use liquid kerosene and liquid oxygen as propellants.
Kerosene as fuel and liquid oxygen as oxidizer. The second stage consists of five rockets
again. It is known as J 2 rocket. we will get back into the details of this later on while
studying liquid propellant rockets. 5 J 2 rockets clustered together for the second stage.
They use liquid hydrogen and liquid oxygen. The third stage consists of one single J 2
rocket. Therefore, what is it we are talking of? The Saturn 5 rocket consists of the first
stage, which consists of five rockets and these are 5 F 1 rockets clustered together. The
second stage similarly, consists of a cluster of 5 J 2 rockets. J 2 rocket uses liquid oxygen
and liquid hydrogen as fuel. This first stage uses kerosene and liquid oxygen On the third
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stage, you have a single J 2 rocket and on top of this sit the particular capsule, which is
the Apollo capsule, where the three astronauts who travel to the moon are housed.
(Refer Slide Time: 13:55)
We would like to do an sample problem. Let us take an example of F 1 rocket. However,
before that, let us put some numbers. J 2 rocket in the third stage has a thrust of
something about 1100 kilo Newton. Each of the J 2 rockets here, mind you, the same J 2
rockets when used for the second stage, has a thrust of 1000 kilo Newton. Each of the F
1 rockets has a thrust of something like 3800 kilo Newton. Let me make sure about the
numbers. This has something like a thrust of something like 110 ton thrust. Because kilo
Newton, therefore, we are talking of 10 Newton is equal to 1 kilogram. Therefore, we are
talking of a huge force here. Therefore, why is it that the same engine when used in
second stage produces less thrust than when used for the third stage? Altitude. That
means, higher the altitude, I get more specific impulse and therefore we get more thrust.
Let us work out a problem concerning the F 1 rocket. Out of all these five, let us do the
nozzle problem related to one F 1 rocket. The thrust of this rocket is equal to 3800 kilo
Newton; is that what was said? No my numbers are not correct. The thrust is very much
higher. 6800 kilo Newton, I am sorry for the wrong numbers.
Each F 1 rocket has a thrust of 6800 kilo Newton. The mass flow rate through the nozzle
is equal to 2600 kilogram per second. These are typical numbers. We should keep in
mind that we are not talking of 1 kilogram per second. We are talking of something like,
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almost 3 tons of propellant gases going through the nozzle per second. The area of the
nozzle is equal to 0.65 meter square. That means, if we consider the diameters, a man
can easily stand at the throat or walk through through the throat of this nozzle.
The molecular mass of gases which are passing through the nozzle is equal to 22 grams
per mole. The temperature of the combustion products in the chamber is equal to 3300
Kelvin and the chamber pressure is equal to 6.65 Mega Pascals. That means, something
like 66 bar. This is little below the standard pressure of 7 MPa, which we are talking of
as a standard value of pressure for specific impulse.
Mind you, this rocket was developed in the period of 1960s and we had the moon
mission by 1969. Therefore, we are talking of an old rocket. But, mind you, it is still the
most powerful rocket ever developed in the history of rockets and that is where I
thought, maybe we should do a problem on this rocket.
(Refer Slide Time: 17:50)
Now, I want to find out for this F 1 rocket, the value of ηC*, the value of Isp and the
value of thrust correction factor ζF. Let us do it. We should be able to do it since the area
is available and you know how much propellans are burnt per second.
To be able to get the value of ηC*, I must get the value of C*, which is actual and I must
also get the value of C star, which is calculated under ideal conditions. The ratio of these
two (measured/calculated ideal value) is the c star efficiency ηC*. How do I get the ideal
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value? Well, we already know it is equal to √RTc/ Γ. The value of capital gamma Γcan
be determined. We get the value of the specific gas constant R as equal to the universal
gas constant divided by the molecular mass of the gas and multiply this with the value of
Tc to find RTc. Now capital gamma Γ is equal to √γ × (2/γ+1)(γ+1)/2(γ−1). The value of
gamma for the gases is equal to 1.22. Therefore, we substitute the value of gamma is
equal to 1.22 and the value of capital gamma works out to be equal to 0.652; √1.22 ×
2/(1.22 + 1)2.22/(2×0.2).
To get C* ideal. For C* let us substitute the value, R0 the universal gas constant 8.314
joule per mole Kelvin, and molecular mass as 22 g/mole..Please write the units whenever
we do a problem. The value of Tc for this particular propellant combination is given as
3300 K. The value of the molecular mass is equal to 22 grams per mole but, I am talking
in terms of joule which is related to kilogram. Therefore, we take 0.022 kilogram per
mole.
(Refer Slide Time: 20:10)
This is important. Many of us, mainly I find students just putting 22 here, which is not
right because, when I say mole, I want the value of the soecific gas constant R in joule
per kilogram Kelvin. Therefore, it must be kilogram per mole. The value of capital
gamma, we already said is equal to 0.652 and this must be the numerator for this one.
The value of C* ideal becomes 1713 m/s.
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This is how we calculate the ideal C*. That indicates the capacity of the propellants in
the chamber to generate hot gases at high pressure in the chamber. Let us repeat this. The
capacity of kerosene and liquid oxygen to generate chamber pressure is given by C*
ideal and this capacity is 1713 meters per second. Now, we want to get the measured
value of C*. We have to calculate the actual expermiental value. How would I do it? I go
back to look at the problem. The mass flow rate is given to us. The mass flow rate is
given as 2600 kg/s equal to 1/C* × pressure × throat area. Pressure is given as 6.65×106
Pascal. The throat area At is given as 0.65 square meters. Therefore, the value of actual
C* can be calculate from these values. This is 2600 kilogram per second. C* will come
out to be equal to 6.65 × 106 × 0.65/ 2600. This is equal to 1663 meters per second. The
value of ηC* is therefore 1663 ÷ 1713 = 0.97.
In fact, you find that the c star efficiency is quite high even for a rocket made in the
1960s. The present rockets like the space shuttle main engine, has a C* efficiency of the
order of 0.99. This is the way they are and are very efficient. There is hardly any room
for improving the combustion any further. We have to understand that, when we do
liquid propellant rockets, we will try to understand how come we get such values and
what are the factors which govern it.
(Refer Slide Time: 24:13)
May be, we should use some of these tactics in the other propulsive devices also.
Therefore, we have done one part of it, namely, what is the c star efficiency of this
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particular F 1 engine. The next one, I would like to find out what is the value of Isp. How
do I do it? What is the specific impulse of this engine? Yes, I know the thrust is 6800
kilo Newton and I know the mass flow rate is 2600 kg per second. Well, it is simple; is it
not? Specific impulse is equal to Impulse I over mass of propellant Mp which is equal to
I over t divided by Mp dot. The specific impulse Isp is therefore equal to force or thrust
divided by mass flow rat eof propellants. This is already avalable. Th evalu eo fspecific
impulse is therefore 6800 into 1000 Newton divided by 2600. The value of specific
impulse comes out to be equal to what 2710 Newton secod per kilogram. This is the
value of Isp.
(Refer Slide Time: 26:26)
Now, we want to get the value of the thrust correction coefficient ζF. For this, we need to
do little more calculation. Let me erase this part of the board. We get ζF is equal to the
ratio of actual thrust and ideal thrust. How do I get the ideal thrust?
This particular rocket develops a thrust of 6600 kilo Newton, when the exit pressure is
also sea level or rather, when the value of pe of this rocket is equal to pa which is equal
to 0.1 MPa because, it is tested under sea level conditions. The test has been done at sea
level, for which the exit pressure is equal to pa. We are assuming here that the nozzle
exit pressure is equal to the ambient pressure.
Therefore, for this condition, the value of F ideal is equal to m° × VJ because, there is no
pressure thrust coming; pe minus pa is 0 because pe is equal to pa and how do we get
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the value of VJ? We have derived the expression VJ2 = square is equal to 2 of the
enthalpy difference which came out to be equal to √2γR0Tc/(γ−1)M{1−(pe/pc)(γ−1/γ)}.
Here pe is equal to the ambient sea level pressure. Put in the numbers, R0 is 8.314 joule
per mole kelvin, Molecular mass is equal to 0.022 kg/mole and temperature is given
3300 and gamma is given 1.22.
(Refer Slide Time: 28:05)
The value of pe is equal to 0.1 MPa, pc is equal to the value what is given here that is
6.65 MPa. You substitute it and you get the VJ is equal to 2710 meter per second.
Therefore, what is the value of ideal thrust? Thrust is equal to m° × 2600 kilogram per
second multiplied by this value of VJ. Therefore, the value of F ideal is equal to VJ
multiply it m° which gives you the 2600 into 2710, which is equal to 7046×103 Newton.
What is the actual value of thrust? 6800 kilo Newton. But, how do I get the value of zeta
F? Therefore, our immediate reaction or anybody’s immediate reaction would be to take
the value of ζF, i.e., the thrust correction factor is equal to F actual divided by F ideal. F
actual is equal to 6800 kilo Newtons.
The ideal value is some what larger, that is 7406 kilo Newton and therefore, you will tell
me, that this value is equal to 0.965. This is what one expects. But actually, you know,
we have is an actual rocket, in which we must also consider the effect of C* efficiency.
ζF is actually the thrust correction factor. In other words, Isp at the actual thrust goes as
ηC* × the thrust correction factor ζF into the value of Isp.
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Therefore, if I were to correct for theefficiency of C*, I should have ζF = 0.965/ ηC*,
which we got as 0.97. Rather, this works out to be 0.995, because this is only for the
nozzle. We looked at that total problem and the total problem gave us this value and we
have to isolate the correction to apply for the nozzle. Therefore, the correction factor for
the nozzle is 0.995. Whereas, the contribution from the combustion or from the value of
pressurization or c star is 0.97.
(Refer Slide Time: 31:13)
I think this is how we get the efficiencies. Well, I would be happy even if we put a
number 0.965. But, let us keep in mind that 0.965 also includes the value of ηC*. That is
why, I had to remove it and that is where I got this particular number 0.995.
Let us take one more problem that is problem of rocket being propelled at different
altitudes. Let me pose this problem to you first. Yes, let us say a booster rocket operates
between sea level (0 kilometres) to 30 kilometres altitude and the chamber pressure of
this rocket pc is given to be 7 MPa.
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(Refer Slide Time: 33:38)
The specific heat ratio of the combustion products is given as 1.2 and the throat area At
is equal to 0.1 meter square. Since, this rocket performs between 0 to 30 kilometre
altitude, the designers felt that the nozzle could be designed for mean altitude of
operation, say 15 kilometre altitude. Now, I want to determine the following: First, the
nozzle expansion ratio, ε, and the value of the exit area Ae. Second, the value of the
thrust coefficient CF at 30 kilometre altitude and what is the optimum value of CF at 30
kilometre altitude and third, we also want to know, till what height or till what altitude
will flow separation occur. In other words, we assume that we have a conical nozzle
which operates between 0 and 30 kilometres. The nozzle is designed for an altitude of 15
kilometres. We want to know till what height flow separation takes place in this conical
nozzle. We also want to find out the area ratio, the area at the exit and the thrust
coefficient at 30 kilometre, optimum value of thrust coefficient at 30 kilometres and to
what altitude will flow separation persist. Let us do this problem.
We need the data and the data on ambient pressure which are normally avalable as ICAO
tables. In these Tables, the height in altitude versus the ambient pressure is given. ICAO
stands for international civil aviation organisation. This gives some standards and they
will list the altitude versus the pressure, ambient pressure in Newton per meter square,
temperature , density. If at sea level, the pressure in Newton per meter square is 101325
Newton per meter square. If the altitude is 4 meters height, the value is 61660 Newton
per meter square. If the altitude is 8 kilometres, the value is 36651. You see, the value
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keeps decreasing. Let us put two or three more values of ambient pressure at the different
altitudes as shown in the following slide:
(Refer Slide Time: 36:20)
12 kilometres, the value is 19399 and if the altitude is 16 kilometres, it is 10353 N/m2. If
it is 20 kilometres, it is 10353 N/m2. If the alttude is 20 km, the ambient pressure is 5529
N/m2 and if it is 30 kilometres, the value of pressure is equal to 1186 N/m2. Since, I do
not give the value of ambient pressure at 15 km height, let us assume that the nozzle is
designed for 16 kilometre altitude instead of 15 km. So that, the ambient pressure table is
available to us.
We would like to first calculate the area ratio of the nozzle and the exit area. What do
we tell? We say, well, the nozzle is designed for 16 kilometre altitude and therefore, for
16 kilometre altitude, we have pe is equal to pa. Therefore, what should be the value of
pe? For the nozzle? 10353 N/m2 or Pa, because the nozzle is designed for this particular
altitude.
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(Refer Slide Time: 38:42)
That means, we have pe as 10353 Pa. Other data viz., the value of pc is equal to 7 MPa.
We say 7 ×106 Newton per meter square. The value of γ is given as 1.2. Therefore, we
immediately write out the expression for the expansion ratio ε. Let us go back to your
notes.
Epsilon is equal to [2/(γ−1)]1/(γ−1) ×(pc/pe)1/γ ÷√[(γ+1)/(γ−1)]{1−(pe/pc)(γ−1)/γ}. You can
easily derive it out. It is not difficult. I do not want us to memorize anything. You
substitute the values and you get the value as equal to 52.5. Area ratio of the nozzle is
therefore 52.5. The value of the exit area Ae/At is equal to 52.5. Rather since At is given
to you as 0.1 meter square, the value of Ae is 5.25 meter square. Is it alright? It is simple.
You know, the calculations for rockets tend to be extremely simple.
In fact, rockets are very simple. In India, we still have not made good diesel engine or
internal combustion engine or gas turbine engine. We have been taking time to do it and
we have still to do it on our own. Whereas, rockets being easier to do, we see spectacular
progress in making of rockets.
Therefore, you have Ae is equal to 5.25 meter square.
Let us go to the next part of the problem. CF at 30 kilometres. How do we evaluate it?
What will be involved in this? We had derived the expression for thrust coefficient CF.
Let us go back and take a look at it. CF0 = √2γ2/(γ−1)[2/(γ+1)(γ+1)/(γ−1){1−(pe/pc)(γ−1)/γ .
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(Refer Slide Time: 41:30)
You will recall, we did this earlier. pa by pc and ε. Please check as to what are the
values. I am interested at thrust coefficient at 30 kilometres. What are the values that we
substitute? Well, γ = 1.2. What is the value of pe and what is the value of pc and what is
the value of pa? Epsilon ε? We have already determined it as equal to 52.5. What is the
value of pe? Which value to take? Yes, the nozzle has been designed for 16 kilometre
altitude and that is what the exit pressure should be. Because, it is now opearting at a
higher altitude but, the nozzle exit pressure will not change. Therefore, pe is equal to
10353 Pa. Your answer is correct. pc, we know is 7 into 10 to the power 6 Newton per
meter square or Pa. pa at the current altitude of 30 kilometres, 11806 Pa. We substitute
these values in the expression for CF and we get the value of CF as equal to, I use the
other side of the board, 1.828 for CF0 plus 0.0687 for the pressure contribution
{(pe/pc)−(pa/pc)} × ε, the total being equal to 1.896.
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(Refer Slide Time: 44:07)
Please check these numbers. Now, what is the optimum value at 30 kilometres? Let us go
back to the value of pe at optimum. From this expression itself you can tell me that
gamma is still the same and what will be the optimum value at 30 kilometre altitude.
(Refer Slide Time: 45:22)
Well, pe must equal to pa. Therefore, for optimum, this term will get knocked out and
what will be the value of pe, if it is optimum. Take a look at this table. Optimum at 30
kilometres, that means, I will have a nozzle which gives me this value, 1186 Pa.
Therefore, I now change the value of pe as equal to 1186 and the pressure term is no
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longer there. The value of CF which now represented as becomes CF0, will now become
something like 2.246. You see, the thrust coefficient is typically around 2 to 3.
What is the percentage reduction from optimum? You had a nozzle. I think I will erase
this out now. You had a nozzle which was designed for 16 kilometres. You are operating
it at 30 kilometres. If it was designed for 30 kilometres, it would have been optimum at
30 kilometres. We would have the optimum CF0 as 2.246. But, the nozzle is designed for
a lower value of area ratio and correspondingly a lower altitude, we get the lower value
of the pressure thrust as 0.0687 and we do not get the net momentum thrust possible. We
get the value of CF as equal to 1.896.
(Refer Slide Time: 45:50)
Therefore, we can now determine the percentage reduction from optimum is equal to
2.246 minus 1.896 divided by 2.246. In other words, I have something like 0.156 or
something like 15.6 percent reduction from the optimum. Therefore, you see the
importance. You know that we are not able to get the nozzle to expand to the ambient
pressure at the given altitude and in fact we are having an under expanded nozzle. That is
why, I am losing 15.6 percent thrust. Had the nozzle been designed for 30 kilometre
altitude, we would have got a higher thrust. But then, we would have got a problem of
over expansion and flow separation at the lower altitudes.
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We now go to the next part of the problem and determine the altitude till which the
nozzle is over expanded or the altitude till which the flow separation takes place. In other
words, we want to determine the altitude till which the nozzle is over expanded.
For this, we apply Summerfield criterion, which states that, when the exit pressure of the
nozzle is less than or equal to 0.4 times the ambient pressure, then the flow is over
expanded. We have looked at a other criterion namely, involving Mach number also.
(Refer Slide Time: 47:48)
Let us use the Summerfield criterion. When the ambient pressure pa is greater than or
equal to pe divided by 0.4, then we can say that flow separates in the conical divergent.
Therefore, let us determine the altitude below which flow sepration is possible. Let us
examine the changes in the ambient pressure with respect to altitude. Let me address the
Table again to determine this specific altitude of interest. Let us make a table of altitude
in kilometre and the ambient pressure pa in Pascal. Let us plot it for something like two
or three altitudes for which we are interested.
At the altitude of 8 kilometres, the ambient pressure is 36651 Pascal. At the altitude of
12 kilometres, the value of the ambient pressure is now 19399 Pa. It has reduced,
because the altitude has gone up. At 16 kilometres, for which this particular nozzle is
designed, the ambient pressure is 10353 Pascal. The question is, the nozzle is designed
for 16 kilometres and therefore, the exit pressure of the nozzle is 10353 Pascal, we want
to find out the altitude at which the flow begins to separate or the nozzle gets to be over
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expanded. Therefore, we have to state here that flow separation, we have just written, pa
must be greater than or equal to pe divided by 0.4 and pe for the nozzle is defined or
defined as the ambient pressure at 16 km altitude viz., a pressure of 10353 Pascal.
Therefore, pa must be greater than or equal to 10353 divided by 0.4 and this is equal to
0.259 into 10 to the power of 5 Pascal. Now, the question is what is the altitude when the
ambient pressure is equal to or just greater than this value?.
Now, when pa is equal to 25900 Pa, it is somewhere between 8 and 12 kilometres.
Therefore, we say at 8 kilometres, the ambient pressure is 36651 minus the value at 16
kilometres is 10353. But, we are interested in the altitude at 0.259 into the 10 to the
power 5. Therefore, we have the value of 25900 minus the value at 16 kilometres which
is 10353 Pa and the change in kilometre is from 16 to 8 corresponding it to a value of
something like 8 kilometres. Therefore, we have 8 kilometres plus their change is 25 to
10353 and therefore, the value is 8 plus this so much kilometres.
(Refer Slide Time: 49:15)
This works out to be something like 8 plus 2.16 which is equal to 10.16 kilometres.
Therefore, this is the altitude at which flow separation ceases or thereafter the nozzle is
either runs full or is under expanded. This is all about nozzles.
In the next class, we will start with chemical propellants. We will again keep it very very
simple in the sense, we will look at what are the requirements of chemical propellants
and then see, what are the propellants that, we must use.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Module No. # 01
Lecture No. # 15
Criterion for Choice of Chemical Propellants
Good morning, today we will start a new topic on chemical propellants.
(Refer Slide Time: 00:19)
Let us first be clear what we mean by a propellant. We said that any substance used for
propulsion of a rocket is known as a propellant. It could be anything, it could be
something like a gas, the gas could be cold at high pressure, the gas could be hot. We
even considered these boys throwing stones and stones were the propellant. It could be
anything; it could be a charged particle, it could be a plasma. But what we consider in the
next three classes is chemical propellants.
This brings us to chemical rockets, which use chemical propellants. The chemical
rockets are solid propellant rockets, liquid propellant rockets, hybrid rockets and so on.
Therefore, let us see what are the requirements of propellants in these rockets since
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before we study the propellants; we must know what are the requirements. What is the
requirement of a propellant? I think that is the basic with which we should get started.
We studied that nozzles were an integral part of a rocket; let us say that this is a
chamber and you have a nozzle which is connected to it and the aim is to get as high as a
jet velocity as possible. And to get a high jet velocity we found that we required a high
value of the chamber pressure and a high value of the chamber temperature.
We also found that we require a small molecular mass of the gasses, which are being
expanded out. In this case I will get a higher jet velocity and what was the term that we
used? We used the term C* which is a transfer function between what is sent into the
combustion chamber and the high pressure, high temperature, low molecular mass
combustion gases which are generated. And what was the expression for C*? We had the
expression, C* = √RTc/ Γ. Here R is the specific gas constant which in terms of the
universal gas constant R0, we could write it as R0/M where M is the molecular mass of
the gases which are sent out through the nozzle. Tc is the temperature of the hot gases.
And Γ is a function of γ. Therefore, this expression for C* should tell us what we really
require of a propellant in a rocket.
What we require is that this transfer function which tells the capacity of this particular
chamber to generate high pressure gases must be high or rather this characteristic
velocity C* must be a large number. The transfer function C star of the propellant of the
chemical propellant must be large.
If C star has to be large obviously Tc must be large. The molecular mass of the gas
escaping through the nozzle must be small. Anything else? We have the combination of γ
in Γ: we found that gamma must be small because Γ = √γ(2/γ+1)(γ+1)/2(γ−1). The
requirement was that gamma should be small.
Γ, however, is not as sensitive and influential as Tc and molecular mass of the gas.
Therefore, basically we are looking at the following parameters: Tc to be large and the
molecular mass of the gases to be small; may be the specific heat ratio also to be a small
number. If I can have propellants which could generate gases such that the temperature
of the hot gases and the pressure of the gases could be large. The molecular mass of the
gases should be small and the specific heat ratio of the gases must be small. This is what
a propellant be capable of.
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(Refer Slide Time: 02:24)
Let us first take a look at temperature Tc: What will give me a high value of Tc? If we
have a chemical propellant and now we must say we are talking of chemicals, there are a
large number of chemicals available. And why do we use chemical substances? May be I
want the chemicals to react with each other and generate hot gases. Why hot gases? If I
have hot gases we have a temperature Tc, which could be a high. Therefore, basically we
are looking at chemicals, which could react and generate hot gases and these hot gases
could be at a high temperature. To generate a high temperature gases the heat released in
the chemical reactions should be large.
We consider unit mass of propellant and denote heat generated by unit mas of propellant
in the chemical reaction. The heat release per unit mass is required to be large. If the heat
release is divided by the mean value of the specific heat of the gases, say at constant
pressure, we get the temperature increase from the chemical reactions of the propellants.
If the specific heat at constant pressure is small, we can have a larger value of
temperature Tc for the same heat release.
We have now introduced one more term viz., specific heat of the gases as a requirement.
What we are saying is if we have chemicals these chemicals react generate to hot gases
at high temperature. When do we get high temperature? If the heat release per unit mass
of the chemical propellant is a large number and if the specific heat Cp of the burnt gas is
small.
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Therefore, we now state that the specific heat of the gas generated by the chemical
reactions must be also small.
These are the requirements that we are looking for are that the chemicals should have a
large value of heat release, a small value of specific heat and a small molecular mass of
the product gases that are generated at high temperature and also perhaps a small value
of γ.
(Refer Slide Time: 07:19)
If we can address all the above parameters together, maybe we can narrow down a few
chemicals out of the millions chemicals that are available which can be used as chemical
rocket propellants or as rocket propellants. Which of the chemicals are suited to be
rocket propellant? And this is what we are going to address in this class. Therefore, let
us ask how we will get high values of heat release, small value of molecular mass of the
gas and low specific heat of the reacted gas? Let us start with something simple; under
what conditions will we get low molecular mass of the gases that are generated in the
chemical reaction?
Let us say we have a chemical substance and the reaction of the substance generates hot
gas. Therefore, basically we must take a look at the atomic mass or the mass of the
elements in the chemical substance. If the elements are of small atomic mass, the product
gases formed by chemical reaction would also be small.
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(Refer Slide Time: 08:06)
If the atomic mass of the elements in the propellant is small the product gases will also
have low atomic mass. First let us take a look at the periodic table. What is there in a
periodic table? All elements are arranged in terms of their atomic number and this is
what we will first take a look at it. Let us take a look at the periodic table starting with
elements having an atomic number of 1. The atomic mass of hydrogen is unity and its
atomic number is 1. The next is helium the atomic number is 2 and its atomic mass is 4.
Next is lithium. The atomic number is 3 and the mass of lithium atom is 6.9. Beryllium 4
atomic mass is 9. Boron 5 the atomic mass is 10 and so on carbon 6, atomic mass 12 and
so on.
Next then to oxygen, nitrogen, fluorine we have neon an inert gas, then we have sodium,
magnesium, aluminum, silicon, phosphorous, sulphur and chlorine. We stop at atomic
number of 17 because already the atomic number has increased to 17 and as you see the
atomic mass has increased from 1 to almost 36. The values of the atomic mass are given
are with respect to hydrogen. Beyond this the atomic mass of elements becomes large
such that that any product which is formed will have larger values of the molecular mass
i.e., become very heavy. What does this Table tell us? Let us take a particular case may
be hydrogen with the lowest atomic mass. Even if I take a hydrogen molecule H2, the
molecular mass is 2.
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The molecular mass is lower than other molecules or substances. Therefore, it is a very
viable substance that could be used as a propellant from the molecular mass point of
view. We next go to helium. Helium is inert. It can be used as an inert gas but cannot be
used as a chemical propellant. Next, lithium is used in solid propellants because of its
low atomic mass. We will have to take a look at it when we study solid propellants.
Beryllium and boron can be used; the masses are still small with atomic mass of 9 and
10. Carbon is a part of any hydrocarbon; carbon and hydrogen together. Well we cannot
escape from hydrocarbon and is suited as the atomic mass is not too large. Fortunately
for us carbon has an atomic mass of 12, which is still not very bad. Next in the periodic
table is nitrogen; it is inert, but most of the substances in nature are associated with
nitrogen. The atomic mass of nitrogen is 14.
Oxygen is a powerful oxidizer. It has an atomic number of 8 and an atomic mass as 16.
The molecular mass of oxygen O2 is 32. Fluorine is a very reactive oxidizer, much more
reactive then oxygen and its molecular mass is very near to oxygen itself; its atomic mass
is 19. Neon is inert I cannot consider it for chemical reactions. Sodium is a very reactive
metal. If we drop sodium in water it just explodes and to use it would be impossible.
Magnesium: you would have seen magnesium ribbon being used as a diwali cracker, you
could light it and it burns as a reactive metal. It is quite reactive therefore, it may be
difficult to use magnesium as it is.
Aluminum is a light metal of atomic mass 27. If we have to use a metal in a propellant, it
appears better to suggest aluminium. Other metals would be very much heavier. Only
aluminum is used. Iron is seldom used. Silicon is a light material, but it is not reactive.
We will not consider phosphorous as it is very reactive. We cannot consider sulphur and
chlorine since these also have high atomic mass of 32 and 35.5 respectively.
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(Refer Slide Time: 08:47)
Beyond this you know you go to argon and other elements and they becomes
progressively more and more heavier and they cannot be used. Therefore, all what we
say is out of all the chemicals that we can think of using, we can now isolate those
containing elements upto chlorine which has an atomic number of 17.
From the molecular mass point of view, therefore, we prefer the lighter elements. I show
these elements again in the next slide where in we see the relatively lighter elements
hydrogen, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, aluminum,
sulphur and chlorine.
Well these elements, if contained in the chemicals, would be better since low value of
molecular mass in the reacted products is more desirable. Therefore, we address the first
point of choice of the substances from the molecular mass point of view. When will the
molecular mass of my products be small? When the atomic mass of the elements that
constitute it are small.
Fluorine is very reactive; it is more reactive than oxygen. In fact it was tried for rockets
but it was very reactive and would corrode even the propellant tanks. It was used in one
of the Delta rockets; however, it has not been subsequently used. We will keep it in mind
and see under what conditions fluorine can be used.
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(Refer Slide Time: 12:44)
Oxygen, fluorine, chlorine are oxidizers. Hydrogen, carbon, lithium, beryllium, boron,
and aluminum are fuels. Aluminum is a metal, which can burn and release heat. All of
you would have observed a sparkler during Diwali festival. The sparkler consists of a
composition which is known as a black powder composition and it is coated on a on a
rod may be a steel rod or something like it.
In some instances metal powder or filings are added to the composition in the sparkler.
When the sparkler with the metal filings or metal powder burns, it burns much more
violently. This is because the burning or combustion of the metal releases much more
energy. Therefore, metals such as aluminum, boron, beryllium can also be used very
effectively as fuel. Therefore, we now conclude by stating that low atomic mass elements
are more desirable in a propellant from molecular mass point of view of the products.
Let us now take a look from the temperature point of view on whether we should have
some of these elements or what should be the composition of the chemicals in the
propellant such that we get a high value of temperature.
May be if we go through this aspect, about half of our work in choosing a propellant will
be over. For high temperature, we needed high heat release and low values of specific
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heats. Let me come to the second part on specific heats before looking at heat release
from chemical reactions in a propellant.
We had said that the products of burning or of chemical reaction in the combustion
chamber of a rocket are released through the nozzle. The temperature of the hot gases Tc
goes as heat released divided by the specific heat and therefore we would like to have Cp
as small as possible. Single atom like oxygen atom, hydrogen atom i.e., mono atomic
species have a specific heat of the order of something like 20 joule per mole Kelvin. Unit
of specific heat is per mole per Kelvin. If we have di atomic molecule like O2, hydrogen
H2 or OH that is two atoms of the elements, the specific heat increases to something like
35 joule per mole Kelvin. If we still have more complicated molecule like CO2, 3 of
atoms together, the specific heat increases to almost like 62 or 63.
Let us say that the specific heat of the triatomic molecule is 65 joule per mole Kelvin.
Why should specific heat increase as the molecule changes from mono atomic to di
atomic to tri atomic? What will be your reaction? Why should it increase? Mind you the
unit is per mole Kelvin.
(Refer Slide Time: 15:50)
Why should the specific heat go up? If we take a simple atom, let us say like oxygen
atom it is just O, if I take a molecule O2: 2 well it has two O atoms. We are not bothered
about double bond single bond and how the two O atoms are bonded together. When we
heat say the O atom, it has smaller degrees of freedom and it therefore absorbs less
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amount of energy. This will absorb more energy if it could have more degrees of
freedom. A relatively more complex molecule like CO or CO2 has more bonds and more
degrees of freedom. It can absorb more energy. Therefore, the energy absorbed per mole
of a mono atomic substance is less, di atomic is more, tri atomic is still higher and so on
it increases.
Therefore, let us see the variation of the specific heats. A mono atomic substance has the
lowest, di atomic has higher value, tri atomic has still has a higher value and as we go on
the value of the specific heat keeps increasing. Therefore, from this point of view we
should say if the product gases are all mono atomic, we are better off. A di atomic gas is
to be preferred to a mono atomic gas. What this implies is that the product of combustion
or reaction of the propellants must be simple not complex, in which case I can have a
smaller value of the specific heat in Joules per mole per Kelvin.
I show some of the values of Cp in the next slide. Helium is mono atomic. At the two
temperatures of 2000 Kelvin and 3000 Kelvin there is hardly any change in specific heat.
Therefore, irrespective of temperature for the mono atomic may be helium, hydrogen
atom, oxygen atom the value is around 20 Joule per mole Kelvin. If we go to di atomic
gases; hydrogen H2, hydroxyl OH or may be HCl or may be N2 or CO, all have a ball
park number of around 35 Joules per mole kelvin. A change in temperature does not
markedly change the value of specific heat.
When we have tri atomic gases, the values of specific heats are even higher; well of the
order of 60 to 65 Joules per mole Kelvin - something like 60. Water is around 51 to 58
Joule per mole Kelvin, CO2 is between 60 and 63 may be around 63 to 64. We find that
mono atomic substances have specific heats around 20 Joule per mole Kelvin, diatomic
around 35 to 36 while triatomic substances have specific heats around 60 to 65 Joule per
mole Kelvin. Therefore, this is the range and based on this it is preferable for the product
gases to be more dissociated if we want high temperatures.
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(Refer Slide Time: 19:19)
If we keep on increasing the value of specific heats, we will not be able to get a high
temperature and therefore, we said that the specific heat of the product gases must be
small. We have looked at two criteria namely we looked at the criterion of molecular
mass and find that the atomic mass must be small or atomic number must be low. Second
we found that from the specific heat point of view of the product gases (which result
from the chemical reactions of the chemicals) must be somewhat simple or disssociated.
Let us go to the next criterion before we come back to the temperature and heat release.
Let us take a look at γ, the ratio of specific heats. We had said that γ should also be small
in order to get a high value of jet velocity VJ.
What does this small value of γ imply? If we go through thermodynamics and the kinetic
theory of gases, we find that γ is defined as Cp/Cv, which in terms of the degree of
freedom (n) of the molecule can be written as 1 + 2/(n+3). The value of n shows the
degrees of freedom of the gas over and above the translational modes. Let us illustrate
this. If we have an oxygen atom; it can either move in the three directions, along X, Y
and Z. This means it has 3 degrees of freedom along the three translational axes. If we
have an oxygen molecule well it has it has in addition to translation in the 3 directions
rotational motion also. The oxygen atoms in it could also vibrate.
Therefore, it has additional degree of freedom of 2, compared to the atom, which only
translates. I could also have atoms in which we have may be C3H8, which becomes more
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complicated it could have many more degrees of freedom compared to simple
translation. And for instance if I take CCl4, carbon tetrachloride, the degrees of freedom
are almost something like 13 or 14. As the molecule becomes more complex, it has more
degrees of freedom.
Therefore, the value of gamma value for a mono atomic gas, for which the number of
degrees of freedom is 3 is determined by the following: the value n gives the degrees of
freedom. We say for mono atomic gases like let us say γ for helium is equal to 1+2/3,
which is equal to 1.67.
For di atomic gases like oxygen, hydrogen, nitrogen molecules we have two additional
degrees of freedom from rotation and vibration. The value of γ is 1+2/5, which is 1.4.
May be as it becomes very complex like γ for CCl4, carbon tetrachloride, is going to be
something like 1+2/(13+3) or so giving γ about 1.13. That means that more complex the
molecule, it has a lower value of gamma. If therefore γ is required to be small, then the
gases, which are passing through the nozzle should have should be more complex.
(Refer Slide Time: 20:21)
The requirement for γ to be small calls for the gases to be more complex, but for Cp to
be small the gases must be simple. These two criteria call for just the opposite
requirement. Similarly, if you want the atomic mass to be small; we are also thinking of
simple product gases therefore, we find that gamma requirement is somewhat contrary to
the requirements of Cp and also molecular mass in the product gases. Therefore there is a
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problem and we also know based on what we have studied so far from the nozzle theory
that VJ is not very sensitive to gamma. Therefore, we will not give very much
importance to gamma as for specific heat and molecular mass.
To summarize, the Cp of the product gases should be small which necessitates that the
product gases must be in a dissociated form as simple molecules. The molecular mass of
the gases would then also be small. However, then a small value of gamma cannot be
obtained as the value of gamma decreases as the complexity of the gas increases. We
give less weight age to gamma because Cp directly impacts into temperature and C star
is inversely proportional to the molecular mass.
We are left with heat release to be able to determine the value of temperature Tc. Let us
take a look at heat release in a chemical reaction.
Let us have a chemical substance and we call it as C1. This chemical gets converted to
gas like a product let us say P1. We are looking in this case of this chemical by itself
reacting to give P1 or else we have substances C1 plus C2 reacts to give us P1, product 1,
P2, product 2… The question is how do we determine the heat released in these
reactions? These are all chemical reactions of substances giving the products or chemical
reaction between two chemicals, which give us products. And we are interested in
determining the energy from the chemical reaction.
We have studied about this both in the combustion course and in the course on explosion
physics. We had said that any substance would have its own internal energy? We called
it as chemical internal energy of the substance. Now, if it gets converted to products and
it will also have some energy like chemical internal energy of the products. In a chemical
reaction, the reactants get converted to products. If you have more energy of the reacting
chemicals compared to the products, the deficit of energy is what is obtained as heat of a
reaction.
Let us elaborate. Suppose we have some chemicals; these chemicals have some energy.
As an example, this duster has some energy. Where does the energy come? I have all
these bonds together in the wood; it has some energy. Now I burn it and I get carbon di
oxide, carbon mono oxide or whatsoever it may be. If the energy which is available
before the burning (i.e., the reaction) is more than the energy of the final products, which
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are formed in the reaction, then since the energy cannot get destroyed, the deficit or
reduction manifests itself as heat and that is the heat of a reaction.
(Refer Slide Time: 20:25)
Therefore, basically we need to find out what this internal chemical energy or what is the
chemical energy available in the chemical, which we call as internal chemical energy.
You know the word given to describe this internal chemical energy is heat of formation.
And how to define the heat of formation? The heat required to form a substance - any
substance - at standard state from its elements again at standard state.
We need to be able to describe what is the energy available in a given chemical and to be
able to do so we need certain standard conditions.
Well. At the standard condition what is the energy available in the chemical substance.
This substance could be let us say hydrogen, carbon, nitrogen or some arbitrary
substance like HaCbNc. What are the elements, which constitute this arbitrary substance?
They are again hydrogen, carbon as a solid, nitrogen combined in a certain way and mind
you these come from elements or elemental substances. Hydrogen is a gas under
standard conditions, carbon is a solid under standard conditions, nitrogen is a gas under
normal conditions.
And the standard condition is taken as 25oC and one atmosphere pressure. In other words
if we want to form a substance at the standard condition from its elements again at the
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same standard condition, the energy required to form the substance is known as heat of
formation. The notation for the heat of formation is as follows: we have to give some
heat, which is enthalpy to form Hf. We have to give enthalpy, some increment Δ in
enthalpy to form the substance from its elements. We form it at the standard condition
from its elements at the same standard condition. Superscript ‘0’ shows the standard
condition. Therefore, ΔHf0 denotes the heat at standard condition of the substance and
the elements from which it is formed and subscript of “f” is the notation for heat
formation of a substance.
(Refer Slide Time: 29:11)
If we know the heat of formation of a substance, then we have its internal chemical
energy. If a substance C1, which is a chemical substance has a certain heat of formation
and if the product formed has a given heat of formation, the decrease or the reduction in
the heat of formation from reactant to product gives the energy released in the reaction.
The decrease, in the heat of formation, shows that heat is generated in the reaction. The
reaction, we say, is exothermic; we get some heat from the chemical reaction. This is
how we go about finding out of the heat generated in a reaction.
Let us assume that I have a fuel F, I have an oxidizer O and their combination causes a
chemical reaction, which produces the products. Fuel could be any fuel may be I take
wood as a fuel or oxygen as oxidizer or use the air as an oxidizer and I get some products
it like CO2 or may be CO.
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Therefore, now we say that the chemical reaction between fuel plus oxygen gives us say
CO2 plus CO. If the fuel contains hydrogen also may react with oxygen to form H2O as
products. Let us say the reaction takes place between a1 moles of the fuel with a2 moles
of the oxidizer to give b1 moles of the first product (say CO2), b2 moles of second
product (say CO) and b3 of the third product (say H2O). Therefore, what is the energy
released in this particular reaction? We have to consider the moles of the substances and
suitably define the heat of formation in terms of joules per mole. If the net heat of
formation of the products is less than the net heat of formation of the reactants, heat is
generated in the reaction. Let us therefore look at heat of formation of a substance.
The heat of formation of a substance ΔHf0 is the heat required to form 1 mole of the
substance at standard condition from its elements again at the same standard conditions.
Heat of formation has units of joule per mole. And this is the mole; a1, a2, b1, b2, b3. The
heat of formation of the products is b1 moles into heat of formation for CO2, plus b2
moles into heat of formation of CO plus b3 moles into heat of formation of H2O. This is
the value of the total heat of formation of the products over here. What is the heat of
formation of the reactants? It is equal to a1 moles into heat of formation of fuel F plus a2
moles of oxidizer O into heat of formation of the oxidizer.
We have standard sign “0” to indicate the standard conditions. Since heat energy is
released when there is a net decrease in the heat of formation, we denote the heat energy
released in the reaction with a minus of the decrease. The heat energy is equal to minus
of heat of formation of the products minus the heat of formation of the reactants.
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(Refer Slide Time: 31:41)
We would like to know which are the chemicals, which can release lot of heat energy
thereby making a good choice of the chemicals to be used for the propellants. Could they
be characterized in terms of their heats of formation at the standard conditions? Let us
therefore take a few examples. Let us form carbon dioxide from carbon. We do an
experiment. We take carbon at the standard state. And carbon at the standard state of 1
atmosphere pressure at temperature of 25oC is a solid.
We react it with oxygen, which is a gas and the oxygen is again at 25oC and one
atmosphere pressure and I from carbon dioxide. Again we form it at 25oC and at one
atmosphere pressure. The energy required to form carbon dioxide gas from its elements
C in solid viz., C(s) and O2 as gas will give us the heat of formation. But all of us know,
if we take carbon and burn, it gives out some amount of energy, which is quite
significant amount of energy. And the energy which we get from burning 1 kilogram of
carbon is something like is 32,800 kilo joules; kilo joules per kilogram of carbon burnt.
That means when we burn element carbon with element oxygen, then per kilogram of
carbon burnt, we get something like 32,800 kilo joules of energy. Now, what is going to
happen? This 32,800 kilo joules of energy is not going to form carbon di oxide at 25
degrees; rather the temp of carbon di oxide will go up. If the product carbon dioxide has
to be at 25 degrees centigrade, what is it we have to do? We have to remove this heat
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from this reaction. This means I have to remove i.e., minus 32,800 kilo joule per kg of
carbon burnt so, that the product CO2 can be at 25oC.
Carbon and Oxygen are elements at the standard state. Since the heats of formation are
defined with reference to the elements, the standard heats of formation of the elements at
the standard condition would be zero. Therefore, the standard heat of formation of CO2
should be – 32,800 kilo joules per kg of carbon burnt. But there is something wrong with
the units here. We defined standard heat of formation as the heat required to form one
mole of the substance from its elements when both the substance and the elements are at
the standard conditions. We cannot state it in terms of kilogram of carbon but rather must
be expressed in terms of one mole of carbon di oxide. Let us try to remedy the situation.
For expressing in terms of one mole of carbon di oxide, we note that for every mole of
carbon burnt, one mole of carbon di oxide is formed. This implies that for every 12 g of
(0.012 kg) carbon, 1 mole of CO2 is formed. The energy, which is released is equal to 1
mole of carbon di oxide is therefore equal to 32800×0.012 = 397 kilo joules per mole.
What is it that we had to do while forming carbon di oxide at the standard conditions?
We had to remove the heat and that the heat of formation at standard state of carbon di
oxide is therefore equal to −397 kilo joules / mole of carbon dioxide.
(Refer Slide Time: 35:15)
By now you would have guessed that for any reaction which is exothermic, the net heat
of formation of the products will be less than the net heat of formation of the reactants.
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Large negative values of the standard heats of formation of the products will be
favorable for more heat release in a reaction. Let us take one or two small examples
because this is something, which is basic. Let us take the reaction of 1 mole of carbon as
a solid at element level reacting with half mole of oxygen both at standard state forming
1 mole of carbon mono oxide. Now, I have carbon mono oxide here instead of carbon di
oxide. It is not fully oxidized. The heat, which is generated in the reaction, if we do an
experiment is something like 9208 kilo joules per kilogram of carbon burnt.
Therefore, we quickly convert it for per mole of CO formed. In order to form one mole
CO, we need to burn 0.012 kilogram of carbon to get 9208×0.012 - so much kilo joules
per mole of CO. This is because one mole of carbon monoxide is formed from one mole
of carbon. And this is equal to 110.5 kilo joule/mole. Since heat is getting generated as
CO is formed, we have to bring it back to the same standard condition of 25oC degrees of
oxygen element and 25oC of carbon element. The heat of formation of CO is therefore
−110.5 kilo joule/mole.
Hydrogen and oxygen react to form water. The reaction is given by 1 mole of hydrogen
reacts with half mole of oxygen giving 1 mole of water (H2O). Why do we take this
example? Hydrogen is an element gas at 25oC , oxygen is an element, a gas at 25oC, but
water should be a liquid at 25oC. We can say that the heat of formation water as a liquid
at standard condition should be equal to the negative of the heat release in this reaction,
which is 286 kilo joule/mole of water.
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(Refer Slide Time: 39:39)
Now we are expressing heat of formation in terms of per mole; heat of formation of H2O
as a liquid (water) is equal to −286 kilo joule/mole. 1 mole of hydrogen forms 1 mole of
water and this is how we determine the heat of formation. Please let us not forget that the
heat of formation of a substance is defined with respect to the elements that constitute it.
The heats of formation of the elements themselves are therefore 0 at the standard state.
One last substance I should consider with a positive heat of formation. Let us take the
formation of hydrogen as an atom. We could consider 1 mole of H2 dissociating to give 2
moles of H atom.
What we do in this case? We need to supply heat to be able to form hydrogen atom from
the hydrogen molecule, which occurs naturally. The amount of heat required to
dissociate 1 mole of hydrogen is about − 435 kilo joules per mole of hydrogen. And the
reaction is endothermic. Therefore, heat of formation of H is equal to + 435 for 2 moles
of hydrogen. For each mole of hydrogen atom, the heat of formation is therefore + 217.5
kilo joules/mole. And this is plus because the formation of hydrogen atom from the
elements is endothermic I have to supply heat to form hydrogen atom from the naturally
occurring element hydrogen and it is + 217.5 kilo joules/mole.
Therefore this is how the standard heats of formation of different substances are
determined. However, we need not do an experiment to determine the heat of formation
of a substance. If we have, let us say a hydrocarbon. The bonds between carbon and
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hydrogen and between the carbon atoms are known. We know the energy of each of the
bonds then we know the energy of the bonds of the basic elements. We subtract the bond
energy of the product or the substance from that of the element and we get the value of
the heat of formation. But there are certain problems, which come while estimating heat
of formation from bond energies. A substance does not only have energy of the bonds, it
could have energy in some resonance modes.
It is necessary to have bond energy plus resonance energy of the substance minus the
bond energy of the elements, which will give us this is the way of theoretically
calculating the heats of formation. Details of estimation of heats of formation from bond
energies are given in the textbook on “Chemical Problems in Jet Propulsion” by Penner.
(Refer Slide Time: 42:45)
He gives a good treatment of heat of formation in the book. With the understanding that
we have developed for heats of formation, let us see if we can make some
recommendations for the choice of propellants.
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(Refer Slide Time: 45:54)
I will quickly go through heat of formation of some of the substances in this slide. We
have fuels and let us consider hydrocarbons. A hydrocarbon could be saturated, it could
be unsaturated, it could be aliphatic, it could be aromatic. What do we mean by all this,
that you would have studied in your high school chemistry? All what we mean is if the
carbon atom in the hydrocarbon are fully saturated that means the C C and CH all are
single bonds. We say the hydrocarbon is an aliphatic substance. If you say aromatic well
you have something like a change a benzene ring of 6 carbon atoms with alternate double
bonds.
(Refer Slide Time: 46:13)
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Substances, which have this benzene ring structure are known as aromatic substances.
Getting back to saturated hydrocarbons. The simplest hydrocarbon is methane, the next
one is ethane that is CH4, C2H6. The next is propane C3H8 and then butane C4H10 and so
on. The chain keeps increasing and we come to kerosene - it is little longer chain Dodecane C12H26 and further up we have lubricating oils and so on.
If we determine the heat of formation of let us say methane; it is −74.9 kJ/mole, ethane is
−84.7 kJ/mole. Propane, which is C3H8, is −103.9 kJ/mole while butane is −124.7
kJ/mole. Kerosene has a value of −293 kJ/mole. This means a fuel as it becomes more
and more complicated in structure or more and more longer in chain has a higher
negative value of heat of formation. If we go to a polymer and what is a polymer? It
consists of chains of carbon and hydrogen and perhaps oxygen and nitrogen and its heat
of formation is something like −60 kJ/mole. But the polymer does not come in this
particular family of the saturated hydrocarbons. It consists of unsaturated double bonds
and we will be dealing with it when we study solid propellants.
Let us summarize the trends in the values of heat of formation of fuels. For simple
substances with minimum saturated bonds, the heat of formation has a small but negative
value. The value of heat of formation becomes more negative as the substance becomes
more and more complex with a large number of bonds. That means that we can write it
as large negative values; this is just based on methane ethane propane and all that up to
kerosene.
If you take a substance like hydrogen, which is an element, the standard heat of
formation is 0. There are certain substances, which are known as explosives. Explosives
are substances that have in built oxygen in them. This means that explosives contain both
fuel and oxygen within it as compared to a fuel that reacts with an extraneous oxidizer to
form products of combustion.
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(Refer Slide Time: 49:00)
Hydrogen peroxide H2O2 can be considered as an explosive since it dissociates to form
water and hydrogen. It has inbuilt fuel hydrogen and oxygen. Similarly, you have
substances containing nitrogen such as hydrazine N2H4 and this is an explosive because
by itself it could react to form products.
The explosive hydrogen peroxide has a heat of formation of −187.8 kJ/mole while
hydrazine N2H4 has a small but positive value of heat of formation. It is + 50 kJ/mole.
And what are the values of heats of formation if we were to consider some other
explosive like nitroglycerin. What is nitro glycerine? Nitroglycerine is glycerine known
as propane triol and the propane triol has chemical formula C3H5(OH)3. We replace OH
by NO2 to get nitroglycerine C3H5(NO2)3. Its heat of formation is − 370 kJ/mole.
(Refer Slide Time: 50:52)
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And based on the standard heats of formation of the different substances, we would like
to find out which chemical when it reacts gives maximum heat. We will continue with
this in the next class and we will try to find out what are the chemical substances which
are most viable as propellants for rockets. We will try to zero down the number of
chemicals, which can be used for rockets, to something like seven or eight.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Module No. # 01
Lecture No. # 16
Choice of Fuel-Rich Propellants
We continue with chemical propellants in this class, namely the criterion for choice of
propellants. What did we learn so far? We found that the propellants, which are
chemicals must have low atomic mass, such that we have low molecular mass of
products formed from combustion or chemical reaction. This was point 1. Point 2 we
said was that the products could be dissociated. What do we mean when we say that the
products of combustion must be dissociated? Instead of having water if I could have
something, like H atom or O atom or OH atom, well the specific heat will be smaller and
the molecular mass will be smaller.
Third, we told ourselves from point of view of γ, it may be better to have more complex
products of combustion. This complex product is against what we decided in point 2.
Towards the end of last class, we also defined standard heats of formation and heat
release from combustion. What did we say heat of formation is? We defined the standard
heat of formation of a substance as the heat required to form one mole of the substance at
the standard condition, the standard condition being one atmosphere pressure and say
25oC. The heat was required to form the substance at these standard conditions from the
elements, which constitute the substance again at the standard condition. This is how we
defined the heat of formation. This was the way we defined it for products, for chemicals
or any substance.
The heat, which is released in a chemical reaction, we determined in the following way.
If we have products being formed in a chemical reaction, the sum of the heat of
formation of the products minus the sum of the heat of formation of the reactants with a
minus sign gave the heat released. If the products consists of n1 moles of substance 1, n2
moles of the second substance and so on and each of these substances have the standard
heat of formation, which is given by ΔHf0 corresponding to the particular substance, we
have the net value of heat of formation of the products as the summation of the moles
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and the corresponding heats of formation. Well, this defined the heat of formation of the
products.
Now, we subtract from it the summation of heat of formation of the reactants. If there is
a decrease in the heat of formation, we said that energy is released in the combustion.
Let us be clear about this notation. If, we have for the reactants, n1 mole of chemical 1,
may be n2 moles of chemical 2, etc., forming, let us say n1 moles of product 1 plus n2
moles of product 2 and so on. All what we say is for any specie i going from 1 to n for
the reactants and ‘i’ going from 1 to n for the products and multiplying each for the
corresponding mole with the corresponding standard heat of formation, we get the heat
of combustion as the negative value of the difference.
(Refer Slide Time: 00:20)
What is it that we want? We want this heat or which we also called as q to be as large as
possible and for this we looked at heat of formation of different substances. We looked at
heat of formation of let us say methane, ethane, propane, butane and all that up to
kerosene, which we called as do-decane C12H26. We found that the heat of formation
keeps increasing in the negative direction for this series of hydrocarbons. Therefore, we
told that if the substance or the chemical is little more complex, may be the heat of
formation is higher but negative. We also observed this trend in the example of CO2 and
CO. We found that the heat of formation of CO2 was something like − 386 or −387
kJ/mole whereas for CO we found it was − 105.5 kJ/mole.
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In fact, we used the reaction to find out the heat of formation of CO, heat of formation of
CO2 and as the product gets to be more complex in its molecular structure, the heat of
formation was higher. Mind you it was negative or − 387 and − 105.5 kJ/mole.
Therefore, we would like to know the conditions for the chemicals to produce maximum
heat release.
What should be the choice of the substance with either large or small values of heats of
formation, positive or negative? This is what we are trying to get it. Once we do this, we
could be a little more wiser in the choice of propellants to be used for as rockets.
Let us get back to the slides. What I have shown here is the standard heat of formation
for fuels such as methane −75 kJ/mole, ethane −85 kJ/mole, propane −104 kJ/mole,
butane −125 kJ/mole. See it keeps on increasing and till we come to kerosene, it has
increased and it is a much larger negative quantity equal to −293 kJ/mole. That means
increasingly negative quantities as the molecule becomes longer.
(Refer Slide Time: 05:54)
If we have a polymer; what is a polymer? Polymer is a slightly different animal, in the
sense we are looking at something like a chain of C, H, O and perhaps N. The chain gets
replicated a number of times. We find that the heat of formation of some of these
polymers and we will look at polymers in some detail like poly butadiene later on when
we deal with solid propellants. The polymer has heat of formation as shown in this slide
of about − 80 kJ/mole.
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When we talk of other fuels like hydrogen; hydrogen is an element and at the standard
condition the heat of formation is zero. We conclude by saying that for different fuels,
the negative values of heat of formation keeps increasing as the complexity of the
substance increases. For a polymer, it is around −80 kJ/mole. For hydrogen, which is an
element and again at the standard condition, it is 0 kJ/mole.
Now, we would like to include some more substances. In the last class, we said that there
are some substances, which are known as explosives. We keep on reading about
explosions. What is the difference between explosive and a fuel? When we have fuel and
oxidizer already mixed together; mixed very well or if not premixed with the oxidizer
and fuel are in the molecule itself. That means fuel and oxidizer are an integral part,
either extremely well mixed or else it is a part of the substance itself.
Let us take one or two such explosives. In the last class, we dealt with nitroglycerine.
When we say nitroglycerine is basically glycerin derived from propane. We make
propane triol by hydrolyzing propane C3H8 and form C3H5(OH)3. That is we take 3 of the
8 H atoms in propane and substitute it by OH. This is known as propane triol or
glycerine. That means it is just an alcohol of propane. Now we substitute OH by a nitro
radical ONO2, we form the explosive. You get C3H5(ONO2)3 and this becomes nitro
based on glycerine which is propane triol and this is known as nitroglycerine.
This has a heat of formation of 370 kJ/mole. How do you get it? You do an experiment
by combining it or forming it with substances whose heats of formation are known. Why
did I take this particular example of nitroglycerin. I want to know whether nitroglycerin
can act as an explosive. It has oxygen, it has fuel, it can burn together to give me CO2
plus H2O plus may be CO. Nitrogen, which is inert, is also present. However, if we look
at the elements or atoms, which are there in nitroglycerin, we find it has 3 of carbon C, 5
of hydrogen H, 3 of nitrogen N, and 9 of oxygen O. That means it has 9 atoms of
oxygen, 5 of hydrogen, 3 of carbon.
Now, if we want to oxidize the 3 atoms of carbon, we need something like 6 atoms of
oxygen to form CO2. We require two and half atoms of oxygen to form H2O. Since we
have 3 atoms of carbon, 5 atoms of hydrogen and 9 atoms of oxygen in one molecule of
nitroglycerin, if we want all the carbon atoms to form carbon-dioxide, we need
something like 6 O atoms. If I want to oxidize all the 5 atoms of hydrogen to form water,
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we need two and half atoms of oxygen. Therefore, we require for complete oxidization
8½ O atoms; but I have 9 O atoms. Therefore, nitroglycerin could still act as an oxidizer
even though it is an explosive. It has some oxygen left in it, which can still be used for
oxidizing a fuel. Nitroglycerin we say is an oxidizing agent.
(Refer Slide Time: 07:00)
I will repeat this because it is something central to the choice of an oxidizer and a fuel in
a propellant. Let us take a substance like nitric acid HNO3. If we take nitric acid, we
have one atom of hydrogen which requires half atom of oxygen for its oxidation.
Therefore, we are still left with two and half atoms of oxygen. Therefore, nitric acid can
be used as an oxidizer. That means it is an oxidizer even though it has fuel atom
hydrogen in it.
If we have a substance like ammonium perchlorate; all of you would have heard of it. It
is a very widely used oxidizer for solid propellant rockets. The formula for ammonium
perchlorate is NH4ClO4. We have 4 atoms of oxygen, but I have 4 atoms of hydrogen
requiring only 2 atoms of oxygen for oxidization. I have chlorine which is again
oxidizer. Therefore, it has excess oxidizers in it and is an oxidizer. Similarly, if we have
nitroglycerin, nitroglycerin can react by itself but can also provide oxygen. We can use it
as a propellant directly, but I can also use it as an oxidizer in combination with some
other fuel. In other words, we can use it as an oxidizer or else we can also use it in
isolation as nitroglycerin itself.
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(Refer Slide Time: 12:16)
Similarly, if I take H2O2, which is hydrogen peroxide; we do not need both the O atoms
in it to form water. We are left with one O after the fuel H is consumed and therefore
H2O2 act as an oxidizer. Mind you H2O2 is an explosive just as nitroglycerin is an
explosive. All these are all substances, even though they contain fuel is it, functions as an
oxidizer.
Let us take one example of a fuel that is used with nitroglycerin. The example of this fuel
also be an explosive, an explosive which can act as a fuel. The simplest one is may be
like this wood, which is a cellulous material. The molecular formula for cellulose is
C6H10O5 and it consists of several such molecules to give its molecular formula as
[C6H10O5]n, where n is a large number.
Now, we can also write the above formula for cellulose as [C6H5(OH)5]n. Therefore, we
say that it consists of n number of these molecules together. This is the equation to
cellulous or formula for cellulose such as paper. Suppose, we nitrate it. That means I
want to make nitrocellulose. We take some of the OH out and substitute it by ONO2 and
what we get is some part of the OH is left; but some are substituted by ONO2. If of the 5
OH, x are removed and replaced by ONO2, the chemical formula for nitrocellulose
becomes [C6H5(OH)5-x(ONO2)x]n.
We now have the formula for nitrocellulose which is now C6H5, x of ONO2 nitrate and
5-x of OH. Now, if we look at this, you know that the 6 carbon atoms will require 12 of
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oxygen atoms to form CO2. The 5 - x + 5 H atoms will require 5-x/2 atoms of oxygen to
form water H2O. The total requirement of O atoms is therefore 17 –x/2 of oxygen atoms.
But the oxygen atoms, which is available, is only 3x plus 5 minus x, which is 5 plus 2x.
The maximum value of x can only be 5. While the total requirement is 14½ , only 10 O
atoms are available. Therefore, the availability of oxygen in nitrocellulose is much lower
than the amount required for the oxidation of carbon and hydrogen present in nitrocellulose.
(Refer Slide Time: 14:30)
Therefore, in a sense the oxygen available within the molecule is much less than that
required to oxidize the fuel component of carbon and hydrogen. Therefore, nitrocellulose
is fuel - rich. It can dissociate by itself using the small amount of oxygen, but it cannot
form completely oxidized species since carbon and hydrogen are more than the oxygen
available in it. Therefore, it is also used as a fuel. We use it as a fuel because the
component of fuel in the nitrocellulose is much greater than the amount of oxidizer in it
and this nitrocellulose if you were to go back and look at what is its heat of formation, it
has a large negative value, which is −670 kJ/mole.
We talked of hydrazine N2H4 in the previous lecture. Hydrazine which is again an
explosive has the standard heat of formation of which is + 53 kilo joules per mole. We
therefore observe, that explosives and other substances could act either as a fuel or an
oxidizer depending on the relative amounts of oxidizer and fuel components in it. We
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also see the heat of formation varies from something like a positive number of + 50 to a
large negative value of − 670 kJ/mole.
We next take a look at the heat of formation of the oxidizers. An oxidizer could be
oxygen. Oxygen is an element at standard condition. The heat of formation is zero. If we
consider nitric acid, we just saw it is an oxidizer. Its heat of formation is −171 kJ/mole. If
we remove the fuel component H from it and if we make into di nitrogen tetra oxide
N2O4, which is a volatile liquid. The heat of formation is + 90.63 kilojoules per mole.
We talked in terms of other oxidizers solid ammonium perchlorate whose chemical
formula we said was NH4ClO4. NH4ClO4 has a heat of formation of −295 kJ/mole. If we
instead of the perchlorate radical, we use the nitrate radical and get the oxidizer
ammonium nitrate NH4NO3, the heat of formation is −365 kJ/mole.
(Refer Slide Time: 17:40)
Therefore, you see the heat of formation widely varies for oxidizers also. In the case of
nitroglycerin, it is −370 kJ/mole, for hydrogen peroxide it is −187 kJ/mole. Nitro
nitroglycerin as a large negative value while N2O4 as a slight positive value. This is the
variation in the heats of formation for oxidizers.
(Refer Slide Time: 19:54)
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Similarly, for other chemical species such as the products from combustion or from a
reaction; what are variations in the heats of formation? Well, carbon gets oxidized to
CO2 or CO. Hydrogen gets oxidized to H2O. Therefore, the products are essentially CO2,
may be CO, may be H2O and so on. If we have aluminum in the metal, we could form
aluminum oxide and these are some of the products, with which we are interested. And if
you look at the heat of formation of some of these products which we worked out in the
last class, it was something like −387 kJ/mole for CO2, −110 kJ/mole for CO and −296
kJ/mole for water.
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Why do we say water and not steam or vapor? Because we are looking at the standard
condition of 25oC while the pressure is not that important for a liquid. Therefore, under
standard condition, it is water. Therefore, the water has the standard heat of formation of
−296 kJ/mole. H atoms dissociated was found to have a value of +217 kJ/mole. The
heat of formation of OH was again high at +395 kJ/mole, but if we take aluminum oxide,
it has an extremely large negative value of −1670 kJ/mole. Well, these are some values
of heat of formation of fuels, oxidizers and products.
Now, we are interested in rocket propellants or chemical propellants, which will give as
much heat as possible and therefore give a high value of temperatures. The negative of
the difference between the net heats of formation of the products minus net heat of
formation of the reactants is what gives us the value of the heat released (q). Therefore,
we see that if the products could have individually negative values and if these negative
values are large, we could have high value of heat release q.
Therefore, one of the requirements of chemicals, which can be used as propellants is that
they must form products which should have large negative values of heat of formation.
Mind you when we say products, by products we are not talking of chemicals. We are
talking of may be p1, p2, p3 or rather we are looking at CO2, H2O, CO., etc. The products
must have large negative values of heat of formation. Is it ok?
(Refer Slide Time: 21:00)
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Similarly, if we talk in terms of the reactants; which are essentially the unreacted
propellants, what should their heats of formation be? We have heat release is – {heat of
formation of products – heat of reaction of reactants}. Therefore heat release goes as
product of minus and minus which is positive. If the reactants would have positive value
of heat of formation, it is better for us because we have a more positive number and
greater heat release. Therefore, based on this logic, all what we say is if we have a
propellant as a single chemical or a single substance or a combination of a fuel and
oxidizer they must have small negative values of heats of formation or better to have
large positive value of the heat of formation.
Why did we write small negative values? It is because if the heat of formation is positive,
the substance is basically unstable. Why is it unstable? Because you are supplying heat to
form the substance from its elements at the naturally occurring state and that it cannot
remain so in the standard condition. Therefore, the general requirement is a small
negative value, if possible instead of a large positive value, which is not possible.
(Refer Slide Time: 23:00)
In general, some of the substances like N204 have small positive values of heat of
formation; some of the explosives have positive values. We had considered these
explosives earlier, but in general, most of the substances like kerosene have negative
values. However, the desirable feature for a propellant is for large positive values of
heats of formation though this is not possible in practices in view of such substances
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being unstable. We sort of compromise with a small positive value or a small negative
value. This tells us what is the choice of propellants from the heat of formation point of
view.
If this part is clear, may be subsequent things are quite simple. Therefore, we tell that the
choice of propellants for rockets should be such that they have positive value of heat of
formation or small negative values and the products that they form should have large
negative values of heat of formation.
We will do one or two problems towards the end of this class. It will become further
clear to all of us. Let us consider a fuel like butane; unfortunately, butane is a gas and it
is difficult to use, but let us take this example. We react it with oxygen as the oxidizer.
Butane has the formula C4H10 plus oxygen O2. The propellants we consider are therefore
butane and oxygen. Let us say the propellants completely burn into carbon dioxide and
water. Therefore we get 4 CO2 plus 5 H2O. Now, we want to balance this reaction. We
require 8 plus 5 oxygen atoms giving 13 O atoms. Therefore, I get 13/2 of oxygen O2.
So, we can write this reaction as 2 moles of C4 H10 plus 13 moles of O2 give 8 moles of
CO2 plus 10 moles of H2O. What is this reaction? In this reaction we form completely
oxidized products of combustion. We cannot oxidize water any further. We cannot
oxidize carbon dioxide further than this. Therefore, these are all completely oxidized
products, completely oxidized or finished as it were. When we form a reaction in which
the products are completely oxidized, we call the reaction to be stoichiometric.
What do you mean by stoichiometric reaction? The word stoichio means element and
metric means proportion in Greek. Therefore, we are talking proportion of the fuel and
oxidizer such that we form completely oxidized products of combustion. This is what we
mean by a stoichiometric reaction. But the question is if we have for propellant butane as
a fuel and oxygen as an oxidizer, is it possible that either more or less of oxygen than a
stoichiometric reaction will give better value of C* through better values of temperature
or smaller values of molecular mass of products. What will happen if instead of using 13
moles of oxygen, we were to have 15 moles of oxygen for every 2 moles of butane? In
other words, we would like to consider proportion of fuel and oxidizer which is best
suited for the propellant combination.
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(Refer Slide Time: 24:44)
When we studied the subject of combustion, we talked in terms of equivalence ratio,
which was defined as the fuel air ratio divided by fuel air ratio under stoichiometric
conditions. In rocket propulsion, we use the word mixture ratio and mixture ratio is
defined as of mass of oxidizer divided by mass of fuel in the propellant combination.
Let us illustrate it. If we want to find out what is the mixture ratio for this stoichiometric
reaction between butane and oxygen. What is the mixture ratio? Mixture ratio for
stoichiometric combustion of butane with oxygen is equal to mass of oxygen = 13×32.
The amount of fuel is 2 × (12×4 which is 48 plus 10 which gives 58). The mass of
oxidizer is 13×32. The mixture ratio is 13×32 ÷ 2×58. That is the mixture ratio for this
reaction = 3.6. Therefore, if we use a fuel in the proportion of oxygen to fuel of 3.6:1, we
get completely oxidized products of combustion.
How do you calculate the heat release in the reaction? All what we do is the heat release
for this reaction is equal to – {(8 × −387, the value of the standard heat of formation of
CO2 + 10 × −286, he heat of formation of water) – (2 × −124.7, the heat of formation of
butane + 13 × 0 the heat of formation of oxygen being zero since it is an element). The
energy liberated in the reaction is therefore –{ (−8×387 −10×286) – (−2×124.7)}. This is
the heat liberated in the reaction.
There is decrease in heat of formation as we go from reactants to products. we have to
look at this from the net value of heat of formation of the products and the net value of
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the heat of formation of the reactants. We find that there is a decrease and we have a
minus sign. Therefore, we get 8×397 + 10×286 − 2 ×124.7 - so many kilojoules of
energy, which is liberated. This is how we calculate the heat liberated in this
stoichiometric reaction.
(Refer Slide Time: 28:16)
Instead of having stoichiometric composition, let us introduce extra oxygen into the
reaction. We take the number of moles of oxygen to be 15 instead of 13 for the 2 moles
of C4H10. Therefore 2 moles of C4H10 + 15 moles O2 are the reactants. What is this
reaction going to give as products? We have excess oxygen, therefore we still get 8 CO2
+ we get 10 H2O + we are left with 2 of oxygen O2. This is because we have more
oxygen than is required? The oxygen oxidizes the carbon and the hydrogen to form
carbon dioxide and water and the balance O2 is left in the products.
What is the mixture ratio for this reaction? It is equal to 15×32 mass of oxidizer mass ÷
2×58 for mass of fuel. We just said the molecular mass of fuel is 48 plus 10 giving 58.
The mixture ratio is 15×32/2×58 = 4.14. That means the mixture ratio has gone up from
the stoichiometric value of 3.6 to a value of 4.14. Is the heat release in this reaction going
to be different from the stoichiometric value. It will be same because oxygen here has
zero heat of formation. Therefore, the heat of reaction is still at the same value.
Let us now consider the third case in which we have less of oxygen available than
stoichiometric reaction. We take the same reaction of 2 moles of C4H10 plus instead of
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giving 13 moles for stoichiometric reaction we have 11 moles of oxygen. That means we
are starved for oxygen. If we are starved for oxygen, what is going to happen? You know
I cannot get all CO2, I cannot get 8 moles CO2, I cannot get 10 moles of H2O. The reason
being we need 16 plus 10 atoms of O, i.e., 26 whereas we have only 22 of O. Therefore,
it is not possible to get completely burnt products and balance the atoms on the left and
right side of the reaction.
One of the ways we could do is to be able to find out what are the products that we will
get? If we cannot get all carbon dioxide and all water, would we get CO, OH and other
substances because there is inadequate oxygen to form carbon dioxide and water.
Now, how do we determine this? We cannot just like that determine the products of the
reaction. We would have to do an analysis for the equilibrium composition of the
products at a given pressure and temperature which means we have to use chemical
thermodynamics to be able to determine this composition. However, this is involved and
instead of analyzing the equilibrium of the products, there is an approximate or slightly
easier method of doing this problem.
We say hydrogen is very reactive and therefore, all the 20 atoms of hydrogen, they
search for oxygen and get converted into something like 10 H2O, that is 20 of H pick up
the 10 oxygen from the original 22 oxygen that we have. Since we have removed 10
from the 22 atoms of O, we are left with 12 O atoms. That means I have 20 atoms of
hydrogen requiring 10 atoms of oxygen for forming 10 H2O because hydrogen is very
reactive. But we find that we have 8 atoms of carbon and we cannot form 8 moles of CO2
because this will require 16 atoms of oxygen. We have only 12 O atoms.
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(Refer Slide Time: 31:32)
Since we have 8 atoms of carbon, let us first use the 8 of 12 atoms of O to form 8 CO. If,
after doing this we are still left with O atoms, part of the CO will get oxidized to CO2.
We are left with 4 atoms of oxygen and what we do is use these 4 of oxygen to oxidize
four of the eight CO to CO2. Of the 8 CO, we remove 4 to form 4 CO2 and therefore, the
reaction will be: 2C4H10 +11 O2 = 10 H2O + 4 CO2 + 4 CO.
Let me repeat it. Some of you have done this method of calculating the products of fuel
rich explosives in the explosion course. Since there is insufficient oxygen to form
completely oxidized products of combustion, first the hydrogen attacks the oxygen
because hydrogen is very reactive or rather the hydrogen removes part of the oxygen to
form water. The balance of oxygen oxidizes the carbon to form carbon monoxide and if
some oxygen is still left, the balance or the part of the carbon monoxide is converted to
CO by the left over oxygen.
What is the heat release in this reaction? What is the mixture ratio of this particular
reaction? Mixture ratio of this reaction is equal to 11× 32 ÷ 2 × 58 and this equals 3.03.
What is the heat liberated in this reaction? The heat liberated in this reaction is – {(10 ×
the heat of formation of H2O + 4 × heat of formation of CO2 + 4 × the heat of formation
of CO) – (2 ×heat of formation of butane)}. Is it going to be higher or lower compared to
stoichiometric reaction? It will be lower because CO has the heat of formation, which is
−110 kilo joule per mole while CO2 has a higher negative value of − 397.
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Therefore, you find that when a reaction is fuel rich or equivalently oxygen - lean, it is
short of oxygen and the value of heat release comes down. If it is oxygen rich, then the
heat release from the reaction is same as stoichiometric and this value is the maximum
heat which is possible in a chemical reaction.
(Refer Slide Time: 38:21)
Let us ask one last question. Oxygen rich means the mixture ratio is greater than mixture
ratio corresponding to the stoichiometric composition. Fuel rich means mixture ratio less
than mixture ratio stoichiometric. If we plot the heat release from the chemical reaction,
how will it look like? So, let us plot it. Our aim is to get a high value of temperature or
we are still debating what must be the choice of the proportion of fuel and oxidizer to be
used as rocket propellant.
On the Y axis, we show the heat released in the reaction. On the X axis we show the
mixture ratio. ? Suppose this is stoichiometric mixture ratio. I plot the heat release in the
reaction as a function of mixture ratio. Based on our discussions, we find anything more
than the stoichiometric mixture gives us the maximum value of heat release whereas,
below this I keep on dropping because unoxidized or not completely oxidized products
of combustion are being formed.
Now, we want to convert the value of heat release into temperature. How will I convert it
to temperature? I tell myself well, this is the fuel rich part, this is the oxidizer rich part
and we want to convert it to temperature. We calculated in fact the heat of combustion or
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the heat, which is liberated in the chemical reaction by looking at the products and their
heats of formation. We said it must be less than the heat of formation of the reactants and
the deficit is the heat, which is generated. Therefore, if we were to divide it by the
summation of the mole and the corresponding specific heats, this will give me something
like the temperature increase. It will give us the combustion temperature. That means
specific heat × the number of moles of the products × the temperature increase is the heat
release.
(Refer Slide Time: 39:29)
Therefore, the temperature of the combustion products in the combustion chamber Tc is
equal to q / (Cp × the corresponding moles in the products). Again, we take the mixture
ratio at stoichiometric condition and this is mixture ratio scale. Now, this is the
temperature scale. Here, we have moles, which are coming in addition to the value of Cp.
We need to be able to convert it because we have different number of moles. A direct
comparison from this to the temperature may be a little difficult at the beginning.
Therefore, what we could probably do is convert the heat release into heat release per
mole.
Let us do this exercise. If you were to calculate it and we will go to the left side and plot
the heat release q per unit mole of the products and this we note is similar q per unit
mass. We plot this as a function of mixture ratio. Again this is the value of mixture ratio
at stoichiometry. How will the curve for heat release translate into heat release per
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mole? Let us do this exercise. Let us again go back into these equations and see for
stoichiometric conditions. You have C4H10 plus 13 O2. When it was oxygen rich, the
number of moles increased and since the heat release q is same as it become more and
more oxidizer rich. We plotted the heat release q so much kilojoules as a function of
mixture ratio and the mixture ratio is at this point corresponds to mixture ratio
stoichiometric. This corresponds to the oxidizer rich because mixture ratio is defined as
mass of oxidizer divided by mass of fuel. This is oxidizer rich zone and this is the fuel
rich zone.
We found when the oxygen content was more than what is required for stoichiometric
mixture ratio, the heat content does not change. In fact, it remains same whereas, in the
fuel rich side, since we are not able to burn all the carbon and hydrogen atoms, the heat
release keeps coming down.
Instead of plotting the heat release q on the Y axis, supposing we want to plot q divided
by the number of moles of products which are formed. What is the type of trend, which
we could expect? Ultimately, we are interested in finding out the temperature. Therefore,
we want to find out for per unit mass or per unit mole, if I can divided this by specific
heat, we get the temperature and therefore, let us first find out what is the value of heat
release per unit mole in the product.
The X axis is mixture ratio and this point is the mixture ratios corresponding to
stoichiometry. Now, what is happening as the oxidizer quantity increases; I am left with
more of the products that is the number of moles of the product increases. Therefore, the
value of heat release per mole begins to drop because the number of moles is increasing
in the product after the point of the stoichiometric mixture ratio.
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(Refer Slide Time: 42:06)
How about in the fuel – rich case? If the mixture is fuel rich, we are not able to form that
much of moles now. Therefore, the number of moles of the product could decrease and
therefore, this curve would become a little less drooping than what it was earlier. The
peak value of heat release per unit mole is still at stoichiometric and the curve drops on
either side in the fuel rich side and in the oxidizer rich side.
In other words, the amount of heat release per unit mole gives the maximum value at the
stoichiometric mixture ratio and falls on either side of it. Instead of expressing heat
release per unit mole, I can also have a similar figure for heat release per unit mass of
products. That means q so much kilojoules per kilo gram of product plotted as a function
of mixture ratio. Well, it will be exactly similar. The peak heat release corresponds to the
stoichiometric mixture ratio.
Going one step further, I divide this q per unit kgs something like kilojoules per kilogram
by the specific heat in kilo joules per kg Kelvin and therefore, now I can get the value of
temperature verses mixture ratio. This is what I show in the next figure namely, I get a
plot wherein the temperature varies with mixture ratio as shown. We must be able to
differentiate between the total heat release and the heat release per unit mass and this
heat release per unit mass when divided by the mean value of specific heat will give me
the value of the temperature which has a behavior something like this curve with the
peak value at stoichiometric mixture ratio.
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There is a subtle difference when we look at specific heats of substances, which are
formed in the fuel rich conditions. Under fuel rich conditions, we are forming substances
which are less oxidized like CO instead of CO2. We had noted earlier that diatomic
species have higher value than monatomic species. Triatomic species have still higher
values of specific heat per mole. That means, as the diatomic species becomes triatomic
at stoichiometric condition, the specific heat increases. Therefore, we find that specific
heat is slightly lower in this fuel rich region compared to the stoichiometric. Therefore, if
we plot mixture ratio stoichiometric here, this is the value of mixture ratio stoichiometric
at which we obtained maximum heat release per mole. In the case of temperature Tc,
considering the lowering of specific het in the fuel rich region and the flatness of the heat
release curve at the stoichiometric mixture ratio here the maximum temperature Tc will
get slightly shifted to the fuel rich condition and we will a some shape as shown.
In essence, when the propellants are fuel rich, we form more of the smaller elements CO
instead of CO2 and since CO has less specific heat compared to CO2, we get a lower
value of mean specific heat. Since we divide the heat release near the peak by the value
of specific heat, even though the heat release remains about the same in the
neighborhood of the stoichiometric mixture ratio, the value of peak temperature now
shifts to the fuel-rich region. This may not very noticeable, but still we must remember
the trends. The peak temperature occurs not at stoichiometric, but at slightly fuel rich
conditions.
That means peak temperature occurs over here to the left of the stoichiometric mixture
ratio. What is going to happen to the mean value of the molecular mass of the reactants
and molecular mass of the products? If mixture ratio is equal to stoichiometric, what is
the value of the mean molecular mass of the products that we got? We get 8 CO2 + 10
H2O. Therefore, the molecular mass is equal to (8 × 44 + 10 × 18) ÷ (8 + 10).
If the mixture ratio was more than stoichiometric, we had the mean molecular mass of
the products as (8 × 44 + 10 ×18 + 2 × 32) ÷ (8 + 10 + 2).
We are looking at the mean molecular mass of the products. If we had a mixture ratio
which was less than mixture ratio stoichiometric, what is the value of the molecular
mass? We get ( 10 ×18 + 4 × 44 + 4 × 28) ÷ (10 + 4 + 4).
(Refer Slide Time: 49:45)
408
What are the values? Let us put down the values. It is 26 g per mole for stoichiometric
mixture ratio. Let us make an assessment rather than have the numbers. What we find is
for a stoichiometric we have this value of 26 g/mole. When the mixture ratio is greater
than stoichiometric, we are adding substances of higher molecular mass. Therefore, the
molecular mass is higher. If we have a mixture ratio less than stoichiometric, we are
adding moles of substances, which have lower value of molecular mass at the expense of
higher molecular mass and therefore the molecular mass of the products decrease. We
can plot the molecular mass of the products as a function of the mixture ratio. We find
that as mixture ratio increases the molecular mass also increase. The fuel rich mixture
ratios give lower molecular mass for the combustion products as compared to
stoichiometric mixture ratio.
What is it we were ultimately interested in? We were interested in the value of C*
=√RTc ÷ Γ. The specific gas constant R is R0 by molecular mass. If we were to consider
Tc by molecular mass M verses mixture ratio, we find that the temperature peaks in the
slightly fuel rich region. The molecular mass increases as the mixture ratio becomes
increasingly fuel rich; that is as the mixture ratio keeps decreasing. That means we will
have a higher value of C* in the fuel rich region compared to stoichiometric and oxygen
rich mixture ratios.
Refer Slide Time: 52:25)
409
Let us re-plot this figure of C* versus Mixture ratios. The peak value of C* occurs for
mixture ratios less than stoichiometric in the fuel rich side.
Why it is higher in the fuel rich side? Because the molecular mass of the products is
smaller in the fuel rich side. We also find that the maximum temperature also occurs
little bit on the fuel rich side and therefore, the net effect is we have higher performance
in the fuel rich. Therefore, one of the criterion for choice of propellants is that the
propellant must be fuel rich. Generally all propellants used in rockets are fuel rich
propellants.
In other words, if we have stoichiometric reaction of fuel H2 plus oxidizer O2 giving me
H2O, what is the stoichiometric mixture ratio? The stoichiometric reaction is 1 mole of
H2 reacting with half mole of O2. The mixture ratio in this case is ½ ×32 ÷ 2 = 8. That
means we are talking of mixture ratio of 8, which is stoichiometric. In practice what we
use is mixture ratio between 5 and 6. The reason being we get advantages of the lower
molecular mass of the products and also to some extent, higher temperatures at the
mixture ratios less than stoichiometric, the dominant factor however, being the molecular
mass.
410
(Refer Slide Time: 55:20)
Therefore, we find that it is better to have fuel rich propellants. That means mixture ratio
less than mixture ratio stoichiometric.
(Refer Slide Time: 55:45)
In the next class, we will take a small example and also analyze the performance of
rockets.
411
Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Module 01
Lecture 17
Performance Prediction and Analysis
Good morning. In the class today, we will extend what we did in the last class and look
at the reaction of hydrogen and oxygen to form water. When we use hydrogen as a
propellant, it is not used a gas at the standard conditions but as a very low temperature
liquid. Similarly, oxygen is used as a liquid at low temperature.
(Refer Slide Time: 00:23)
We use hydrogen at a temperature about 20 K, oxygen at let us say 80 K; our interest is
to determine what will be the temperature of water which is formed. Mind you, as per
these reaction: H2 + ½ O2 = H2O. The question is to estimate the temperature when the
initial condition of hydrogen and oxygen is not standard, but drastically different from
the standard conditions.
However, before we do this, let us again be very clear of we did in the last class. We had
said that the value of C* is a maximum when the mixture ratio is less than mixture ratio
corresponding to stoichiometry. That means in the region of fuel rich conditions, we
have higher value of c star compared in the oxidizer rich and stoichiometric condition.
412
It is very rare that rockets are operated under oxidizer rich condition because you have
higher molecular mass and you do not get the advantage of temperature either.
Therefore, we normally choose compositions, which are fuel rich.
Since we studied what are the conditions required in the choice of propellants, let us just
put down some five or six propellants which we can say are viable propellants. One is let
us say hydrogen and oxygen to give me H2O as products; but we want to operate it under
fuel rich conditions. Therefore, I will get something like H2O plus H plus OH and so on
because we do not have sufficient oxidizer to form water and further at the high
temperatures water formed could dissociate.
Therefore, we say the hydrogen with oxygen could be one of the propellants and let us
do a small problem under stoichiometric conditions and then extend it to fuel rich
condition. The other propellant we said could be hydrazine N2H4. We said oxidizer
could be nitric acid HN03 or better still compared to nitric acid, we said N2O4, which has
a small positive value of the heat of formation. Therefore, hydrazine and N2O4 is a good
combination and this would again give some products. Why not we think in terms of
kerosene? Kerosene had a large negative value of the heat of formation, but not too very
large. It has a heat of formation of about −200 kJ/mole and you could react it with
oxygen to form products. Again I choose fuel rich condition and this could be a good
propellant.
You find that substances like nitric acid may be N2O4 are widely used as oxidizers.
Hydrazine is also used and we will see the advantages of using hydrazine with nitric acid
and N2O4. Instead of hydrazine N2H4, we could remove one of the hydrogen atoms
giving N2H3 and substitute it with the methyl radical and it becomes one or mono
methyl hydrazine. It is known as mono methyl hydrazine (MMH), The combination of
mono methyl hydrazine and N2O4 is a very popular propellant combination. Mind you, it
is also used as fuel rich and not oxidizer rich or stoichiometric. We can keep on adding
different fuels and oxidizers to form propellant combinations.
When the fuel and oxidizer are in gaseous state at ambient temperatures like hydrogen
and oxygen, we could liquefy them and use them as a liquid propellants.
(Refer Slide Time: 01:23)
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The above fuels and oxidizers were in liquid phase; we could also have solid fuels and
solid oxidizers including solid explosives. What could be the solid fuel and a solid
oxidizer? We had ammonium per chlorate, which was an oxidizer plus a polymeric fuel
or an aluminum metal. We also mentioned in the last class, about nitro cellulous could be
used as a fuel or else polymer could be used as a fuel. We normally do not combine the
solid oxidizer ammonium per chlorate with aluminum and nitrocellulous but rather we
combine nitrocellulose and nitroglycerin to give a solid propellant. We also combine
ammonium per chlorate plus aluminum plus polymer to form a good solid propellant.
Well, the selection of the number of chemicals or chemical substances in liquid and solid
forms to give liquid and solid propellants is somewhat limited. We cannot have an
infinite number of propellants because we need a high value of C*. In the case of solid
propellants, we could have used either nitrocellulose or nitroglycerine singly by itself in
which case we say they are single base propellants. When used in combination i.e.,
nitrocellulose and nitroglycerine together, they become a double base propellant. There
are two bases. Each one could be a propellant, but a combination could also be a
propellant. We could also a have composite mixture of ammonium per chlorate
aluminum and polymer to give me something like a solid propellant.
(Refer Slide Time: 04:20)
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We have talked in terms of liquid and solid propellants. We could also think having a
solid fuel with a liquid oxidizer and vice versa in which case we are considering hybrid
propellants. We take the polymer and cast it over as a solid. We allow the acid (liquid
oxidizer) to fall on it. The combination of liquid and solid and this gives us, what we call
as a hybrid propellant. We make sure that the rate of reaction in the fuel rich region is
such that we get a high value of performance parameter C*. This is all about the different
types of propellants but I think we have to get into some more depth of how the initial
conditions of temperature and pressure influence the C star; how do you get the
temperature of the combustion products?
We have a chamber in which chemical reactions are taking place. Supposing we have
high pressures; very high pressure in the combustion chamber. If we have high pressure,
the amount of dissociation that takes place would be less. What do we mean by
dissociation. The CO2 formed in the products becoming CO plus O or rather let us say
into CO + ½ O2 or C + O2. You know if we have a high pressure environment, we
cannot increase the number of molecules that easily and to make gases dissociate or
breakup at high pressures is more difficult than at low pressures. This is because pressure
tries to reduce the volume and does not favor an increase in the number of molecules.
We have to increase the volume to accommodate the increasing number of molecules for
which we essentially need low pressures. (Refer Slide Time: 06:10)
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But we have still not considered the effect of pressure. It must come through dissociation
of the products. Let us see some results on what will be the effect of dissociation. All
what we say is a dissociation reaction is one in which we are breaking up the more
complex molecules into simpler ones. What is breaking up? May be water breaks up into
O plus OH. Why does it break up? The temperature is so high and it has so much of
energy in it that it tries to break up.
Similarly, I have CO2. It is trying to break up into C + CO + O. Let us say you know
when this breakup will be possible. At extremely high temperatures, at the temperatures
encountered. However, if we have a high chamber pressure, the pressure will try to
prevent the increase in the number of molecules. The amount of dissociation will be
much less at high pressure than at low pressure. At low pressures the resistance to an
increase is less and the products can dissociate much more easily.
At high pressure, this is not possible and therefore, I find that high pressure, the
dissociation is small. If the dissociation is small at high pressures, well energy is not lost.
In this case, since dissociation is less, the temperature Tc will be higher. But since
dissociation is small, the molecular mass will also be large. However, the effect of
temperature Tc will much be more dominant than the increase in the molecular mass and
therefore the increase of pressure could still have a favorable effect on C*. (Refer Slide
Time: 07:22)
416
Let us take a look at typical results in a rocket using hydrogen and oxygen. We are still
have to do the theory part of it; nevertheless let us examine some results. The
temperature of the products of combustion as a function of mixture ratio for a fuel rich
hydrogen oxygen mixture at a pressure of one 1 MPa, 10 MPa, 100 MPa and 1000 MPa
is plotted in the following slide as a function of mixture ratio.
(Refer Slide Time: 08:56)
You find that at a given mixture ratio, as the pressure increases, the temperature increases
because at high pressure, the amount of dissociation is less. More completely oxidized
products of combustion having more negative heats of formation are formed.
(Refer Slide Time: 09:45)
417
You also find that the molecular mass of gases increases with mixture ratio as
determined by us earlier. But we also find that as the pressure increases, the molecular
mass increases. Why is that? At higher pressures we have less dissociated gases and
dissociated gases have a low molecular mass. Is it clear?
Let us go to the next one. If now I say the value of gamma, we said gamma is higher for
the mono atomic species, 1.67 for helium, for very complicated CCl4 it is about 1.13.
Therefore, you find that as the chamber pressure increases, the dissociation is less and
therefore, the gamma value is smaller. With increase of mixture ratio we have already
seen the variation of gamma with mixture ratio. If it is more oxidizer rich, we have
heavier products, which are formed and therefore, the gamma is decreasing, but more
importantly now we are considering the effect of pressure. As pressure increases, the
value of gamma decreases.
Why does this happen? Because I have more complex molecules at higher pressures,
dissociation is less. If we put all the factors together, we find the C* or equivalently the
Isp increases. This is because the Isp is equal to C* into the nozzle performance. We also
find that as pressure increases, the optimum value of the mixture ratio for which I get the
maximum value of C* keeps changing.
(Refer Slide Time: 11:00)
418
Therefore, there it is not only just the mixture ratio alone, which decides the maximum
value of specific impulse, but the pressure is important because pressure influences the
amount of dissociation. This is for hydrogen oxygen propellant combination. Therefore,
we have to take a look at dissociation and these are results for which we are not yet
equipped to do the analysis, but we will be doing it. Therefore with this background into
what happens when we have a chemical reaction at different values of pressures for the
reaction of hydrogen and oxygen, let us find out the temperature of the combustion
products.
How do we do this problem? We would first like to understand what is the heat
generated in this reaction. Hydrogen and oxygen are not at the standard state of 298
Kelvin, but hydrogen is a liquid. we use it as a liquid at 20 Kelvin; we use oxygen as a
liquid at 80 Kelvin instead of a gas at 298 Kelvin. Therefore, we have to first convert
these two substances, which are at a reduced temperature to 298 Kelvin. In other words, I
would like to take hydrogen, which is a liquid at 20 Kelvin, convert it ultimately to
hydrogen as a gas at 298 Kelvin before we can use the standard heat of formation.
(Refer Slide Time: 11:54)
419
Similarly, I have to heat liquid oxygen, which at 80 Kelvin, convert it into oxygen gas at
298 Kelvin. Once I have converted this, then the standard heat of formations can be used.
Thereafter, we can take this liquid oxygen, which is at 298 Kelvin and gaseous hydrogen,
which is at 298 Kelvin. And combine them to form water H2O, which is a liquid at 298
Kelvin.
Heat is generated in the reaction between hydrogen and oxygen. This heat converts the
water to H2O vapor at high temperature Tf or Tc and it is this temperature that we require
to determine.
(Refer Slide Time: 13:11)
420
No heat is lost to the ambient nor is heat gained from it. Therefore, the process is
adiabatic. We call the temperature, if the process is adiabatic; as adiabatic flame
temperature. Therefore, we would like to find out the adiabatic flame temperature of the
products of combustion when the reactants are not at the standard state, but in a state of
liquid at low temperatures.
Therefore, what is it we have to do? I have to first take this hydrogen which is a liquid,
take it to its boiling point at which it is still a liquid and boiling point of hydrogen is
about 22 K. That means this is sensible heat wherein we still have the liquid phase and
then once it has reached the boiling point at 22 K, we supply the latent heat or heat of the
vaporization and convert it to H2 vapor again at the boiling temperature itself. This is a
constant temperature process taking place at the boiling temperature Tb, which we said
we said is 22K at the stated pressure. Once it has become vapor, again I provide sensible
heat and increase the temperature from 222 to 298 K.
Therefore, heating of the liquid, conversion of the liquid to vapor and heating of the
vapor to the standard condition is what is required here. Similarly, for oxygen: oxygen is
a liquid till the temperature of boiling, temperature of oxygen is around 90 Kelvin and
then e convert it using latent heat into oxygen vapor. Thereafter we increase the vapor
temperature to the standard at 298 K.
What is the process with respect to the products? We have H2O as a liquid which is
formed at 298 K. Mind you, 298 K is 25oC. We have to heat it as water again till the
boiling temperature of water. Let us assume the boiling temperature to be 100oC and at
100oC what do have to do? We have vaporization, that is heat has to be supplied for the
latent heat to form water vapor. Now, this conversion is again at 100oC and then we
convert the 100oC vapor to the final adiabatic temperature Tf.
Therefore, the heat, which is generated by the reaction at the standard state, should be
able to supply heat for the conversion of liquid hydrogen and liquid oxygen to a vapor at
the standard state plus taking the water to this particular temperature. Therefore, let us
write the equations. I think this energy equation is important because we do not ever use
propellants always at 25oC. Sometimes in a cold condition, like for instance, whenever
we use N2O4, we chill the N2O4 and use it. Otherwise, it tends to vaporize.
So, can we start with the energy balance relation? Is it clear? All what we did while
estimating the heat release is that the propellants were originally at the standard state and
the products were also formed at the standard state and we calculated the heat which is
421
generated at the standard state. We had for the heat release from the standard heat of
formation of water as ΔHf0 = − 286 kJ/mole. The standard heat of formation of the
reactants hydrogen and oxygen was 0 giving the decrease in the heat of formation of
products as – (−286 +0) = 286 kJ/mole of water formed. Therefore, the heat generated is
equal to 286 kJ in this reaction.
We never bothered about the heat of this reaction, which goes into conversion of the
reactants to standard state and the products to the final temperature. Let us put this data
clearly. We have hydrogen as a liquid at 20 K, the boiling temperature is 22 K and we
need heat for this change. The specific heat of the liquid hydrogen is equal to 20 joule
per mole Kelvin. The latent heat of conversion of liquid hydrogen to hydrogen vapor at
the constant temperature of 22 Kelvin is equal to 890 joule per mole. We prefer to use
mole as the unit of the quantity of a substance because it tends to be simpler because
chemical rate equations are written in terms of moles.
Similarly, for oxygen the specific heat of liquid oxygen is equal to 29 joule per mole.
The latent heat that is equal to 6800 joule per mole compared to 890 J for hydrogen. We
need two more things. We need the specific heat of oxygen as it increases in temperature
as a vapor. That means we are looking at specific heat of hydrogen as a gas and
similarly, specific heat of oxygen as a gas.
(Refer Slide Time: 14:06)
The specific heat of hydrogen as a gas is equal to 30 Joule per mole Kelvin and the
specific heat of oxygen is equal to 35 J/(mole K). We should have had the word Kelvin
here. Whenever we have specific heat, we are talking per unit temperature change and
similarly, when we talk in terms of water, we have specific heat of water and steam.
422
The specific heat of water is equal to 90 J/(mole K), the latent heat hfg for water is equal
to 35 kJ/mole. Again we check these numbers. Yes, 35000 Joule per mole and specific
heat of vapor is equal to 58 J/(mole K).
We have the heat determined under standard conditions; We were able to write if H2 is at
298 Kelvin and it reacts with half of oxygen, again at 298 Kelvin, we form water one
mole of water at 298 Kelvin. In this case, the heat of formation of water is equal to minus
286 kilo joules per mole and we got the energy release as 286000 joules of energy per
mole of water formed.
This heat raises the state of liquid hydrogen and liquid oxygen to the standard state and
raises the temperature of water to the final temperature. Let us put this in a better form.
The heat given to one mole of liquid hydrogen starts with 20 K, goes to boiling point at
22 Kelvin. This corresponds to Cp of liquid hydrogen into 2. Plus we have hfg
corresponding to the hydrogen going from liquid to vapor plus we have the temperature
rising between 22 and 298 K. The value of Cp corresponding to gaseous hydrogen in
joules per mole K is used here. We repeat the same for oxygen. I have from 80 Kelvin to
the value at 90 which is the boiling temperature is equal to Cp corresponding to liquid
oxygen plus hfg corresponding to oxygen at 80 Kelvin plus the sensible heat in going
from 80 to 298 K with the corresponding to Cp of oxygen gas.
Whatever we have written here through arrows is expressed in the above. We have
assumed Cp to be constant in this particular region and whenever we say Cp is a
constant, we assume it something like a perfect substance.
Let me not do this exercise by putting the actual numbers. I will just call it as h1. What is
h1; so many joules, corresponding to the sensible and latent heats. Now, I want to know
the value for water. What happens to water? Again, one mole of water is formed. I think
we forgot something. See this is per mole, per mole, per mole is when I have one mole of
oxygen. It is half mole here. We have half mole because hfg was per mole. Therefore, h1
should contain half mole of oxygen and one mole of hydrogen over here. (Refer Slide
Time: 21:42)
423
Let us remember that we defined Cp as J/(mole K). We define the latent heat of
vaporization as J/mole. If you are going to use the mass units and specific heats as joule
per gram Kelvin or joule per kilo gram Kelvin, better to convert the mole into gram or
mole to mass and then use this relation. However, working with moles is very much
simpler.
Now, for water formed as combustion products; we again take one mole of water is
formed at 298 K. Therefore, you have to heat water from 25oC to the boiling temperature
of water assumed as100oC. If the pressure is higher, the boiling temperature will go up.
Cp for water as a liquid which we know × (100 −25) plus hfg for water which is given in
J/mole + we have the sensible heat from 100oC to the value of final temperature Tf,
which the adiabatic flame temperature. Here Cp is the specific heat of the vapor. Let us
call this as h2.
The heat generated in the chemical reaction provides for h1 and h2 under adiabatic
conditions. We therefore equate 28600 to h1 + h2. The only unknown is the value of Tf
which was in the expression Cp for H2O vapor into Tf minus the boiling temperature of
water because all other quantities are known. We can thus get the value of Tf in oC and
convert it to K. This is how we calculate the adiabatic flame temperature.
What is the molecular mass? Well, molecular mass is equal to 18 g per mole and
therefore, you can find out the C* and you can find out the performance of the
propellant.
424
Therefore, for any propellant at any temperature, you have to convert it to the standard
and then, evaluate based on the difference at the standard condition between products
and the reactant, the heat generated in the reaction and convert the products into
something like a final temperature viz., the adiabatic flame temperature. This is how you
calculate the flame temperature or the adiabatic flame temperature.
You have a question. I took half mole of oxygen from the initial temperature of 80 K to
90 K. It is the liquid Cp, multiplied by temperature increase of 10 into half mole, convert
it to vapor. It is again at 90 Kelvin. This is because we said that the boiling temperature
is 90 Kelvin. That means, this should have been at latent heat of vaporization at the
boiling temperature of 90 Kelvin. Thereafter the vapor is heated from 90 K to 298 K.
Your question is whether a chemical reaction will take place at 25 degrees centigrade.
No, let us be very clear. The reaction need not take place at 25oC. See we are not talking
of a chemical reaction at all. We are doing some equilibrium analysis. We say that when
a substance gets converted to products, that means the substance which is reacting is at
the standard state. It has certain heat of formation. When have products, which are again
at the same standard state which is same as the standard state of the reactants. It has some
heat of formation. When we go from this structure of reactants to products at the standard
state, there is some change in internal chemical energy, which is its heat of formation.
Under the standard conditions, there is a decrease in the chemical energy. That energy is
the heat of the reaction at the standard condition.
We are not talking of a chemical reactions taking place at 25o or 100o or 200oC. This is
just the case, wherein we are equating the change in the chemical energy as the heat: heat
= deficit in a chemical energy between reactants and products. This is all what we
consider. What is the heat? Deficit in chemical energy, which is the decrease in the heat
of formation? It ultimately comes as the heat of combustion because this deficit goes to
heat. The products are formed at high temperature from the heat release in the reaction.
We are not telling about rate of reaction that comes through totally different source such
as chemical kinetics and concentration.
Let me take one example. If I have a tank, let say this is a tank. This tank has a volume of
let us say 1 m3. Into this tank I add hydrogen. The amount of hydrogen I add is let us say
two-thirds of the volume i.e., I put 2/3 m3 of hydrogen at atmospheric pressure. I also add
1/3 m3 of oxygen at atmospheric pressure. I mix the two. Nothing is going to happen at
25oC.
425
(Refer Slide Time: 31:00)
The gas is at one atmospheric pressure and 25oC. We are not going to get water at all.
But if I were to overcome by initial barrier say increasing the temperature by supply of
energy. What is the initial barrier I am talking of? Both hydrogen and oxygen have some
energy levels near a datum and if we excite it or activate it. That means I provide
ignition energy to get the reaction going. After the reaction is over we form the products.
That means we have to supply some ignition energy to it. Therefore, I put a spark and I
create a high temperature environment and then what is happening because of this high
temperature; we make hydrogen and oxygen react. May be we have to go into chemical
kinetics and may be let us see if we should do it later. Therefore, this reacts and forms
H2O, which is in the vapor state because high temperature gases are formed.
Then, what is it that we do? We cool the products of combustion into something like we
form it as water at the standard state and now we find that the change in the energy for
this is corresponding to – (the heat of formation of the product minus the heat of
formation of the reactants). I am not talking of reaction taking place at any temperature.
All what we say is heat of formation is at a standard condition and at this standard
condition so much heat is formed. What is the mechanism of temperature increase? This
heat goes to increase the products to the higher temperature and if the reactants are at
different temperature, again it supplies or removes the heat such that the reactant is at the
same standard condition.
426
A word about pressure: since we are considering ideal gases, the enthalpy is independent
of pressure and it therefore does not enter the calculations for a given substance. The
amount of species formed depends on pressure as we discussed under dissociation.
In most of the rockets, say liquid propellant rockets we will see when we supply fuel and
oxygen into the combustion chamber. We could have mono methyl hydrazine as a fuel.
N2O4 could be the oxidizer. You mix the two and burn them to generate hot gases. If you
calculate the temperature, the temperature may be something like 3200 K and this hot
gas is expanded through the nozzle.
What are the products you are getting? You have carbon and hydrogen in the reactants.
You could get CO2, you get CO, you get may be some dissociated species, you get
water, you have OH. These are the products you get. Now, you know the chamber runs
hot. Therefore, you rather take the fuel and instead of injecting it into the chamber
directly you use it for cooling the chamber. The fuel then gets pre heated. Instead of the
fuel being injected at room temperature we inject it at a higher temperature of something
like 80oC.
I use the liquid fuel to cool the chamber and then inject it. Now, it is at a higher
temperature. Therefore, it has higher energy and this energy also contributes to an
increase of the temperature of the products. Therefore, in this case we get h1 to be
negative because this is bringing in more heat into the combustion and adds to the heat of
formation.
Similarly, N2O4 may be at a temperature of something like 5oC instead of being at 25oC.
Therefore, heat has to be supplied to make it 25 degree centigrade. Heat has to be added
to make it to 25 degree centigrade and therefore, I have less heat here which goes to
decrease the effective energy release for the high temeparture. (Refer Slide Time: 33:55)
427
In fact, by preheating using the hot chamber, we are using a regeneration process. We are
reusing the heat from the hot combustion chamber and such type of cooling is known as
regenerative cooling. We will address this later.
Therefore, we know how to calculate the properties of the products for a set of
propellants, which have their initial conditions different from the standard conditions.
What is the heat liberated? We did a set of calculations for hydrogen oxygen at
stoichiometric mixture ratio, which gave H2O. The mixture ratio was mass of oxygen
divided by mass of fuel, which was equal to 16 by 2 which is equal to 8.
What is the molecular mass of the products; 18 gram per mole. Supposing we have to
consider a fuel rich reaction H2 plus we cannot have ½ O2. May be I have ¼ O2. In this
case, what is going to happen? Let me just have guess over here. I cannot form H2O
because I have only quarter mole of oxygen. Therefore, I form ½ mole of H2O. We get
the balance hydrogen. What is there is 1 – ½ mole hydrogen which is left. Please check.
What is the mixture ratio of this reaction? It is equal to (¼)×32 ÷ 2, which is equal to 4.
Is it not?
What is the molecular mass of the products? We have half into H2O: (½)×18 + ½ ×2 = p
10. What is the energy liberated in this reaction? If it was stoichiometric, what is the
energy liberated in this reaction? Let us put that number down. Also, what is the heat of
formation of water? −286 kJ/mole. Therefore, the heat of reaction in this case is equal to
delta heat of combustion is equal to 286 kilo joules in the case of stoichiometric reaction.
(Refer Slide Time: 36:10)
428
What is the heat in this reaction having mixture ratio of 4? At standard state it is equal to
143 kilo joule and this came as half of 286 since only half a mole of water was formed.
If we consider the reaction of H2 plus O2; this would give us H2O plus half O2. In fact
we are assuming this since the product could get dissociated. Mind you we have to do an
equilibrium chemical analysis and find out what are the constituents, but this may not be
very bad to begin with.
Therefore, what is the mixture ratio? 32 by 2 which is 16. What is the value of the
molecular mass of the products? Yes, we are talking in terms of 18+16 ÷1.5. This is
equal to 2 by 3 into 44, which is equal to 29.3. This is for oxidizer rich condition. We
have seen fuel rich and stoichiometric and let us now compare the results.
(Refer Slide Time: 39:10)
What happens to the molecular mass? As we go from fuel rich to stoichiometric to
oxidizer rich, we find that for oxidizer rich is 23 , for fuel rich it was around 10 and for
stoichiometric, it is 18. Therefore, molecular mass is the least for the fuel rich condition.
We find that the heat release comes down when we go from the stoichiometric to fuel
rich condition. The heat release for oxidizer rich is same as stoichiometry because
oxygen does not matter. It is again 286 kilo joule for this reaction, but if I were to put it
in terms of per unit mass or per unit mole, I have half mole increasing over here and
therefore, heat release per mole comes down.
(Refer Slide Time: 42:21)
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Let us plot the general results. What is it we get for the hydrogen oxygen reaction? We
get the molecular mass as this is stoichiometric. This is fuel rich condition. Now, for fuel
rich it is less and increases for oxidizer rich condition. The q value for stoichiometric is
mixture ratio is here; It is constant 286 here. It drops if we were to put it per mole, moles
keeps increasing in this direction because I add more and more oxygen and therefore, if
we plot heat release q per mole, the value decreases.
Therefore, it is better to operate under fuel rich condition. We learnt how to calculate the
temperature. Let us now go to the next part namely, how do I calculate all the products
and the heat release in a better way. We need to consider chemical equilibrium. What do
you mean by chemical equilibrium?
At a given value of chamber pressure, at a given value of temperature, products of
combustion are formed. Can they exist in equilibrium? Can the products exist in
equilibrium? What is the composition of the products? Supposing, I react hydrogen plus
oxygen at stoichiometric conditions. Is it that only water will be in equilibrium or is it
some hydrogen will also be in equilibrium with it? If so, how many moles of water?
How many moles of hydrogen? How many moles can stay together at this specified
value of pressure and temperature? This is a much better way rather than assuming
hydrogen is more reactive and therefore, first it reacts and then, carbon reacts. Why not
we do the real analysis and this is what we say as chemical equilibrium analysis.
(Refer Slide Time: 42:43)
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To be able to do the chemical equilibrium analysis, you must first be clear about
equilibrium. I think we talked about it in one of the classes. What do you mean by
equilibrium of a substance or a system being in equilibrium at a given temperature and
pressure? How do you look at it? Let say first take a general example.
Supposing, we consider this room as a system and let us say this room is beautifully
insulated. Nothing can come in, no power can come in, no mass can come in, nothing
will come in. What is the system we are considering? The system is may be all these
lights, may be all of us, may be these chairs, which are the attributes of the system.
Therefore, I consider this room as a system and then, we say it is totally isolated. That
means I consider an isolated system. I allow the attributes of this isolated system to be
there for infinite time, long time. What is going to happen? No change is possible.
Everything is finished. All of us have spent such a lot of time with nothing. All of us are
in the dead state.
This final state of a system wherein no further changes are possible is what we call as
equilibrium. No more changes are possible. Therefore, the concept of equilibrium is very
profound in thermodynamics. We tell ourselves, no further change in the state of a
system is possible. Why do we say that? Because we have prevented any further changes
taking place. May be for sometime it evolved, but afterwards there is no change possible.
Therefore, this final equilibrium dead state is when no further changes are possible. We
call it as equilibrium, but even though I say it takes infinite time for a system to reach
equilibrium, it takes very few nano seconds or a few milli seconds for a gas to reache the
equilibrium state.
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(Refer Slide Time: 45:00)
Therefore, it is not necessary that I allow infinite time, but to be able to appreciate the
problem well, an isolated system left to itself for long time, reaches us state of
equilibrium. Having said that, how do you define equilibrium? No further changes.
In the context of the performance of a rocket, we consider the rocket combustion
chamber where the propellants burn to form products of combustion. The velocities in
this region are small so that there is adequate time for the combustion to take place. We
say that that the residence time or stay time is significant for equilibrium of the process
or chemical equilibrium to take place. We calculate the products based on equilibrium.
However, when the products of combustion are expanded in the nozzle, the flow
accelerates. The flow velocities, we have seen, are subsonic before the throat in the
convergent portion and supersonic in the divergent portion.
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In the convergent portion where the velocities are less than the sound speed, the
residence time is still significant and we can presume that chemical equilibrium will
prevail. But in the divergent portion of the nozzle, the flow is at Mach number greater
than 1 and the flow velocity is greater than the local sound velocity. The flow has also
expanded; i.e., the temperature of the gases have reduced. Considering the small
residence times and the lower temperatures, equilibrium or complete combustion is not
possible. The products remain the same as before or the composition is frozen.
The flow in the nozzle is such that the composition keeps shifting with the temperature
change in the convergent i.e., we have shifting equilibrium in the convergent whereas it
tends to be frozen in the divergent.
When we calculate the nozzle performance i.e., the exit jet velocity VJ or equivalently
the specific impulse, we presumed the specific heat ratio γ and the molecular mass of the
products, etc., are constants. This implies that the composition of the gas remains fixed
or frozen. We found that Isp is a maximum in the fuel rich region.
If the composition shifts throughout the nozzle, i.e., we have equilibrium, which results
in shifting composition, further reactions take place and heat gets released from the
recombination of the dissociated gases. Hence the performance viz., Isp predicted by the
shifting equilibrium will be higher. The specific impulse obtained assuming shifting
equilibrium flow in the nozzle will be higher than if frozen flow is assumed in the
nozzle.
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But then we also observe that we could have frozen flow in the divergent and hence
the actual specific impulse would be in between the two. This is shown in between the
two by the dotted line. This is observed in practice.
We could reasonably assume that in nozzle flows, we have shifting equilibrium in the
convergent and frozen composition in the divergent. But then this is not true for all
propellant compositions and depends on the product gases generated.
This is all what would like to discuss in analysis of performance and prediction of
performance.
Thank you.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Module No. # 01
Lecture No. # 18
Factors Influencing Choice of Chemical Propellants
(Refer Slide Time: 00:15)
Good morning. In today’s class, we will discuss the choice of chemical propellants. By
choice I mean, the criterion what we need to follow. How we say this propellant is
chosen as suitable because there are so many chemicals available. What are the
chemicals, which we can use? We will see with respect to solid propellants, liquid
propellants and any other form of propellants like hybrid. We will make a judicious
choice and then go into details of the propellant used in a solid propellant rocket. We will
then look at liquid propellant rockets and the other types of rockets.
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(Refer Slide Time: 01:06)
Before I get into the topic, I would like to clarify something from the last class. When we
consider ramjets and scramjets, does dissociation play a role like for instance in rockets?
Let us make a diagram of the performance of a scramjet or a ramjet on a T S diagram.
What do you do? In the intake we, assuming an ideal condition, isentropically compress
the air from a temperature one to two. Then, what is it we do? We add heat at let us say
constant pressure and then expand the gasses again. We expand the gasses again in an
ideal isentropic process.
Now, if we were to enhance compressing the gasses more and more, we would have the
temperature after the compression, which is higher. If we increase the compression still
further we end up in still higher temperature after the compression process. In other
words, as we go to higher and higher temperature, the air will get dissociated as the
temperature will be high. If the temperature is so high that we cannot add any more heat
in the combustion process, we cannot have any useful engine. But the question is if we
still increase the amount of heat addition, even though the temperature here is high, can
we still have an engine, which develops thrust?
If we increase the heat addition to a very high value, we get intense dissociation. The
chemical energy is not fully available, as completely burnt gases are not formed.
Therefore, you do get some bounds of possible temperatures. Dissociation plays a role in
the choice of the upper bound of temperatures in which some of these engines can
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operate. Let us keep it in mind. Dissociation is not very peculiar to rockets alone, but to
all other engines. As per the ideal Carnot engine, we would like to have the high
temperature to be as large as possible. That is when we get maximum efficiency.
If the highest temperature is increased, we get a higher efficiency because the efficiency
of a Carnot engine is equal to 1 minus low temperature divided by the high temperature.
If we can make the high temperature to be near infinity, we get a very higher value of
efficiency. But to be able to achieve these temperatures would be difficult as the hot gas
begins to dissociate. That means I will be using my chemicals or reactants very poorly.
Let us keep in mind with respect to the restriction of high value of temperature.
(Refer Slide Time: 04:00)
Another point which I also thought we must note is it is not only the species which are
getting dissociated when we go to high temperatures; but we could form ions. We can
get H+ ions. We can get the plasma and we can write the equation for these plasmas
using the equilibrium constant. We can also find out how much of the atoms get into
different forms of excitation.
Having said the above, let us keep the focus of this class very clear. We want to examine
the different choice of chemicals which can be used for solid propellants, liquid
propellants and hybrid propellants. What are the points we have done so far which will
help us?
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(Refer Slide Time: 04:49)
We would like to get a high value of heat release. We would like to get a high value of
temperature. We would like to get a low value of molecular mass. Well, γ really did not
matter very much.
Well, we addressed the above points earlier; but a chemical must be capable of being
handled. That means chemical must be capable of being stored. It must not readily
absorb moisture from the air; sort of being hygroscopic. It must be stable. It must not
happen before I start using the chemical, it begins to spontaneously react. Therefore,
there are other factors in addition to the performance which also plays a role and which
we will be addressing as we study the different propellants.
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(Refer Slide Time: 05:39)
Let us get started with some details of solid propellants. We have already seen the
different chemicals in solid phase. We said that we could use nitrocellulose, which we
said was an explosive. What was the formula for nitrocellulose? We had cellulose as
C6H10O5. Cellulose is like paper. This could be written as C6H5(OH)5 and this comes as
a chain with m of them. For nitro cellulose, we substitute part of OH by the nitrate
radical. That means out of 5 OH replace x by nitrate i.e., (ONO2)X and balance 5-x of
(OH) to give [C6H5 (ONO2)x(OH)5−x]m.
We had talked of this earlier in the class. We had said that this contains both oxygen and
carbon, is terribly fuel rich, but can still dissociate to release heat. May be CO, may be
other species could come out as products and this could be used as a propellant. In other
words, we could call it as a single base propellant because it is a single substance which
dissociates and does the job.
We could have nitroglycerine and we had said that nitroglycerin is little better than
nitrocellulose because nitrocellulose had a very large and negative value of heat of
formation. Nitroglycerin was better that way.
What was nitroglycerine? It was derived from propane C3H8. We take propane, and
instead of 3 of hydrogen substitute it by OH to form an alcohol propane triol C3H5(OH)3.
This is also known as glycerine. We now substitute OH by nitro radial. That means we
have C3H5(ONO2)3 and this is what is nitroglycerine.
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Again in the earlier class, we found this is slightly oxidizer rich, but it contains O, C and
H. It can be used by itself. Well, this could also a single base propellant. We call it as a
single explosive propellant or a single base propellant just like nitro cellulose, which was
fuel rich and could be used. In this case, it is slightly oxidizer rich. However, both
nitrocellulose and nitroglycerine are seldom by themselves.
(Refer Slide Time: 08:54)
What is done is we mix nitrocellulose that is fuel rich and nitroglycerine that is oxidizer
rich. That means you have two constituents or bases and this double base is what is
known as a double base propellant. Therefore, the first solid propellant what we study,
viz., a double base propellant.
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(Refer Slide Time: 19:16)
It consists of nitrocellulose and nitroglycerine. Nitrocellulose is fuel rich and
nitroglycerin is slightly oxidizer rich and you get much better performance than if these
were individually used as single base propellants.
How do we make it? Well, we take nitrocellulose. We put nitroglycerine in it, make it as
a colloidal solution and add some more additives to it. What are additives? You know
whenever we make a propellant; it is not just mixing and forming some hard solid
material because we want it as a solid. What do we add? We add some plasticizer, so that
it becomes more plastic. We add additives, which will make it flow so that we would like
the two to mix together. We make it into a colloid, we cure it, make it as a solid block
and this solid block what we call as a propellant grain or propellant block.
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(Refer Slide Time: 13:51)
Let us list a couple of names of the plasticizers, which are added. Plasticizers used are
triacetin, diethyl phthalate, etc.
What do you mean by a plasticizer? You know something which makes it a little more
plastic. That means it makes it little less sensitive to impact. A rigid solid, when hit,
forms a spark. The plastic sort of absorbs the impact, but at the same time, it gives some
more consistency to it and therefore, the first propellant we say consists of nitrocellulose
and nitroglycerin with a small amount of plasticizer, such that we can make a solid
block. In this the nitrocellulose and nitroglycerin are mixed so perfectly that it is
homogenous or molecularly mixed and the double base propellants are also called
homogenous propellants.
The mixture is essentially something like a colloid. Colloid means suspension in liquid
and therefore, this is also known as colloidal propellants. The trade name for a double
base propellant is Cordite. We have a factory at Aravankadu in Nilgiris in South India,
which specifically manufactures Cordite propellant for use in defense. Therefore, we say
that the first solid propellant that we consider is the double base propellant of
nitroglycerin and nitrocellulose.
Individually, these are single base propellants and together they form the double base
propellant because you have two bases and we also add some amount of trace quantities
of plasticizer and some more additives to make sure that the final product which is
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workable with suitable mechanical properties. I would like to machine the solid block to
have fixed dimensions. I would like to have a rocket with this solid block or grain.
Let us go to the next type of solid propellant. There are only four types out of which the
second one about which we will discuss is very widely used. Unlike a homogenous
propellant like the double base propellant, that means everywhere it is exactly same and
we cannot find out the differences in composition or otherwise, the second propellant is
what we call as heterogeneous propellant. This is also known as composite as distinct
from the double base propellants, which are uniform, which are something like a single
molecule or uniform throughout. The composite propellants are heterogeneous. We add
oxidizer in a solid form, we add aluminum again as a fuel and we glue it together using
some particular fuel.
What do we understand by this? You know all of us have talked in terms of this black
powder in the course on explosions. What was black powder? We had KNO3, which was
ground, we had carbon, which was ground and we had sulfur, which was also ground.
We mixed the three, put gum and make it as a paste and use it when dry as an explosive
for fireworks.
So also a composite propellant consists of a solid oxidizer, may be a fuel and may be the
material that bonds these together. Material that bonds is called as binder and could is a
fuel. Therefore, a composite propellant will consist of an oxidizer, which is something
like a solid crystal say solid ground crystals, a fuel which could be aluminum, and a fuel
which binds the solid crystals and aluminum together. The binder fuel could be
something like a hydrocarbon.
Therefore, as distinct from homogenous propellants, a composite propellant consists of
an oxidizer or fuel and this fuel glues the particles of the oxidizer together. Let us take
some examples and the composite propellant and this will become further clear to all of
us. Let us first see the types of oxidizers we could use in a composite propellant.
(Refer Slide Time: 16:13)
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One of the oxidizer, which is very widely used is Ammonium Perchlorate. We call it as
AP, and the chemical formula is NH4ClO4. Mind you, it consists of hydrogen, but
amount of oxygen that it contains is higher so that it is still in oxidizer. It also contains
chlorine as an oxidizer. May be this is why it is very widely used. It dissociates easily
and it is not hygroscopic. It can be easily ground and can be made as small particles
without any problem and that is why ammonium perchlorate is the most preferred choice
of oxidizer for solid heterogeneous propellants.
We have still more energetic oxidizers like hydrazinium perchlorate. Let me get the
chemical formula of it correctly. It should contain some hydrazine. That means it should
be N2H5ClO4. Another one is hydrazinium nitroformate and the formula for this will
again be hydrazinium N2H5 into CNO2 three times. You know these two solid oxidizers
are being investigated upon, but the only oxidizer, which is widely used today is still AP.
We have coated AP instead of having the raw AP, such that it is more easily processable.
We will come back to this point a little later.
Therefore, we say, we could have choice of different types of oxidizers, but the oxidizer,
which is generally used is AP. We could use Ammonium nitrate NH4NO3. It also
contains oxygen, but it is hygroscopic. Not much energy as with AP. We could in fact
use Potassium nitrate KNO3, but again the potassium K has a larger value of atomic mass
and therefore, it is not a preferred oxidizer. Therefore, whatever be the oxidizer for a
composite propellant, ultimately what is being used today is only ammonium
perchlorate. These are about oxidizers for composite propellant.
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Let us take a look at the fuels, those that could be possible contenders and then, we will
put things together and see what is involved in a composite propellant.
(Refer Slide Time: 18:58)
When we say fuel, it could be a hydrocarbon fuel. Let us see what are the different types
of hydrocarbons and how we could use hydrocarbon as fuel.
Let us start with the classification of the hydrocarbons. We know hydrocarbons consist
of aliphatic compounds and a little bit of review of organic chemistry is always useful.
Aromatic hydrocarbons are those having a strong smell. Why does the smell come?
Because you have the benzene molecule in it whereas aliphatic compounds are either
straight chain or cyclic chain compounds. This could be either saturated or unsaturated.
The saturated chain hydrocarbons are known as alkanes.
What do you mean by saturated or straight chain compounds? Carbon chains are all
single bonds. The second we said is alkenes having one double bond and the remaining
as single bonds. The third one is alkadiene, which have two double bonds. May be
something like as shown in the slide. The two double bonds and a single bond here for
the C atoms.
In summary, for single chain hydrocarbon structure, we could have single bonds, which
are alkanes, one double bond along with the remaining single bonds that are alkenes and
we could have two double bonds, which are alkadienes.
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We could also have one triple bond like acetylene C2H2. These are known as alkynes.
You know the triple bond is highly unstable and acetylene is an explosive. Therefore, we
cannot use the alkynes. We are trying to find out how we choose the fuel in a propellant
and that is why I am going through this. If you have alkanes, well the example is
methane CH4, C2H6 ethane, C3H8 propane. Mind you, all these are gases. I cannot use the
gases anywhere, but there is a tendency today to liquefy the gases like liquid methane,
liquid propane and use it for liquid propellants. We will come back to this at the end of
this class.
Therefore, we say the alkanes are gases are not so useful for solid propellants. How
about butane? Butane is also a gas C4H8. Let us take the butane form of alkanes, alkenes
and alkadiene viz., butane, butane and butadiene.
(Refer Slide Time: 21:55)
Butane has the formula is C4H10. This is the structure of butane. Butane is also a gas.
Let us go to the next form viz., alkenes. Well, I have double bond in butene C4H8.
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(Refer Slide Time: 22:45)
If I now have butadiene with two double bonds, well I have two double bonds and the
formula will be C4H6 because I proportionally have decreased the hydrogen.
Now amongst these choices in aliphatic compounds; we are talking of the differences
between alkanes, alkenes and alkadienes? You know this structure of butadiene is
something, which you do not lose so much of heat of formation. See another thing which
I forgot to tell you was we found that the heat of formation of methane was small
negative value. The heat of formation keeps getting increasingly negative and therefore,
very complex substances having large number of carbon atoms are not desirable.
But butane and butene are gases. With butadiene we are still with a simple hydrocarbon.
With a poly butadiene molecule we make a closed chain and this is very widely used as a
fuel. But poly butadiene is just carbon with hydrogen and to be able to form a good fuel
out of poly butadiene, we alter the end chains. We now come to the actual fuel that we
use as the binding agent. I want to make it into a good polymer, which has a long chain
which can embed the oxidizer particles in it.
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(Refer Slide Time: 24:29)
We take these poly butadiene whose structure we have just seen. Butadiene is a linear
chain. It can stretch freely. I need to make it a little strong, so as that it can hold things
together. Therefore, I add something like acrylic acid and acrylonitrile at the ends of the
chain.
What is acrylic acid? It is CHCH2COOH giving carbo oxalic acid. The component Ch
CH2 is may be x times as shown. This is what acrylic acid is. Similarly, acrylonitrile is
CHCH2CN. May be y molecules of acrylonitrile. We do have this chain of poly butane:
C C C C is a polybutane as shown with m molecules of it of it; at the end of the chain we
add acrylic acid CHCH2COOH may be a few of them, then we add CHCH2CN viz.,
acrylonitrile. What does this do? Though something like a chain, the COOH and CN
cross link it and give it rigidity in the lateral direction. We thus have poly butadiene
acrylic acid acrylonitrile.
That means, we add the acrylonitrile and acrylic acid at the ends of the poly butadiene
chain. I call this as PBAN - poly butadiene acrylic acid acrylonitrile and this is one of the
fuels which are used for solid propellants. You would have read about it. PBAN is used
in the space shuttle and in the huge solid rockets. Why it is used? It is possible to make it
as a binder and it is possible to have it with more hardness and strength. This is one way
that we use the poly butadiene.
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(Refer Slide Time: 26:54)
Therefore, the first fuel for solid propellant is poly butadiene acrylic acid acrylonitrile
which we call as PBAN. PBAN is hard due to the cross-linking. We also see that it has
lot of inert nitrogen atoms in it. If is it possible to somehow get rid of nitrogen, we could
have a better fuel. Just substitute the end groups by the carboxylic acid namely, the acid
part of it. That means, I terminate the poly butadiene with a carboxylic acid. This type of
chain termination of poly butadiene is known as carboxy terminated poly butadiene
(CTPB). That is the second hydrocarbon fuel.
There are not many variants: just some two or three of them. You know all what we have
done is we removed the acrylonitrile part of it, keep the carboxyl acid here and stop the
chain here. May be it gives little better properties and has more energy than PBAN
propellant; but the choice for major booster today is still PBAN. Let us see whether some
improvements in CTPB are possible.
The question is why not have a poly butadiene in which we do not terminate the chain
with COOH, but we just put OH at the ends. We put OH as shown in the slide. In other
words, we have hydroxyl radicals which are used at the ends of the chain. We have poly
butadiene C C C C again. May be n molecules of it in the chain. At the end of the chain,
we put hydroxyl radical and we call it as hydroxy terminated poly butadiene (HTPB).
The advantage of using OH is that more hydrogen is available. It is much lighter,
stronger and is more energetic. Therefore, the most popular propellant today or rather the
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most popular fuel for propellant is HTPB. The reason being is more energetic, but for
boosters, especially for very large boosters, we still find PBAN continuing to be used.
The most popular fuels are PBAN and HTPB and we must be aware of the reasons. Yes.
HTPB is more energetic because it contains H compared to C and N. But basically we
have the poly butadiene chain, which is terminating in these type of radicals. These are
the type of fuels what we use.
In fact, HTBPB is very widely used in the tyre industry. There are one or two exotic
fuels, which are being talked presently. Fuels, that we should be aware of.
(Refer Slide Time: 30:21)
One is GAP, known as Glycidyl Azide Polymer and this has a formula something like
CHNO. In different proportions, it has been tried something like C5H9NO. But we must
keep in mind that GAP contains little more hydrogen than even HTPB. The current line
of research is to increase the energy content of the fuels. How do you increase the energy
content? Have more hydrogen on the lines of hydroxyl terminated poly butadiene.
The oxidizers for composite propellant could be AP. The fuels could be something like
PBAN, CTPB, HTPB or GAP. We could also use aluminum powder as fuel. What is the
advantage of aluminum powder? It could form Al2O3 in the products. We said Al2O3 has
a very large negative value of heat of formation and we could increase the energy release
in the combustion of solid propellants.
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(Refer Slide Time: 31:57)
Therefore, a composite propellant basically consists of these three constituents.
Aluminum powder, which we said is a fuel, a binder, such as HTPB as a fuel and AP for
the oxidizer. The fuel binder is also used for binding aluminum and AP crystals. This is
what constitutes a composite propellant.
In the absence of aluminum, the solid propellant could still consist of HTPB or CTPB or
PBAN plus ammonium perchlorate. These are what are called as composite propellant
because ammonium perchlorate comes as crystals, small spheres, or particles and this
you bind together using the binder fuel. This is a non-aluminized composite propellant. If
we add aluminum powder to it, we get this particular aluminized composite propellant.
These are the composite propellants.
So, whenever we talk of composite propellants, we have in mind something like a
polymer, ammonium per chlorate and may be some metal powder. It may not always
have a metal. If you want more energy, we add a metal. With metal in the propellant, we
have high temperature exhaust and the plume becomes very noticeable. Radar can
observe it and therefore, for strategic purposes either a double base or some other form
of propellants, which does not have such thermal signatures are used. We will take a look
at some of the requirements.
In the past, instead of PBAN or CTPB or HTPB, polyvinyl chloride PVC has been used
as a fuel. What is PVC? It consists of CHCH2 with Cl as a large chain, but these are not
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energetic. Nowadays, it is not used. People tried poly sulphide, people tried foam, which
is polyurethane. These have been tried, but these are all very low energy fuels and
therefore, it is not used significantly. What is used is essentially HTPB, PBAN and to a
certain extent CTBP.
Therefore, we summarize the second form of propellant as a composite propellant and it
consists of a composite of oxidizer particles bound by fuel. May be metal particles like
aluminum and a binder or a polymer, which keeps the composite together.
(Refer Slide Time: 34:41)
Therefore, we talk in terms of double base propellants and composite propellants. With
composite propellants, there is something about particle sizes that we should know.
(Refer Slide Time: 35:02)
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We have oxidizer particles in the propellant. Let us presume that we have spherical
oxidizer particles. We have AP as shown in the figure. If we have a single size of AP, the
amount of AP that can put together in a given volume will be small whereas, if I can
have two sizes of particles one large and the other small, the smaller one can occupy the
gap between the large one and the loading of AP can be increased. Let us say we have
AP size of let say 300 microns. Then, if we try to fill with AP and we have binder, which
binds all these particles together, it is going to bind it like this. Here I have polymer as
shown. The amount of AP, which we can put in the propellant is restricted. We call the
amount of AP in a solid composite propellant as solid loading in a propellant.
Now, instead of having say 300 micron size of AP, we make AP of two sizes mainly, we
make 300 micron size and another one as 30 micron size; then, the small particles can be
put over here in the gaps. That means, we can increase loading of AP in the propellant.
Therefore, in almost all the composite propellants that we use today, we just do not use a
single size of AP, but we use bi-model size typically between 300 and 30 microns. Why
it is used? I want to increase the amount of AP as much as possible and when we use
aluminum; aluminium is even smaller size, typically around 5 micron size of aluminum.
Why aluminum? Aluminium is again a solid. I can improve the loading of solid and
therefore, I can have a dense propellant as it were and this is all about composite
propellants.
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The other two types of solid propellants in are derived from these two. We say composite
modified double base propellant. We have already seen double base propellant is
nitrocellulose and nitroglycerin.
(Refer Slide Time: 36:50)
Supposing, we want to increase its energy. We add oxidizer ammonium perchlorate used
in composite to double base propellants and make it more oxidizer rich. We could also
add explosive to enhance the energy and the explosive we add is something like an
explosive known as HMX. What is an explosive? It consists of fuel and oxidizer
together. This HMX consists of something like cyclo tetra methylene tetra nitramine.
HMX stands for Her Majesty’s Explosive because it came from UK. It consists of a box
like structure with chain C CH3 . You have N here and NO2 . The methylene radical CH3
with C and you have NNO2. This is just an explosive and I do not think we should spend
much time on it. All what you do is you add a HMX to a double base propellant. It
becomes composite of a double base and this particular HMX explosive and this is also
known as composite modified double base propellant (CMDB).
It is used for strategic purposes in the defense because they would like to have their
missiles, which are more energetic than double base propellant. They would like to have
properties of the propellant, which can be modified as per their requirement.
(Refer Slide Time: 39:34)
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Therefore, the third propellant we say is a mixture of a double base and AP or an
explosive.
The last type of solid propellant is what we call as nitramine propellant. You know all
these propellants have distinct characteristics. You know why? They have a burn rate
that changes with pressure and this varies with the type of propellant. We will consider
this when we study the burn rate of solid propellants. We are just trying to take a look at
what are the constituents of the different solid propellants and what are these different
solid propellants.
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(Refer Slide Time: 40:16)
Nitramine propellant has a binder like HTPB which is a polymer. To this is added HMX
which is an explosive or a slightly less energetic explosive like RDX which is cyclo tri
methylene tri nitramine. What is the difference in structure? It is tri therefore, you have
three of CCH3, three of NNO2.
Therefore, you add an explosive into the HTPB binder and what you get is a nitramine
propellant. What is the advantage? See HMX is fuel rich, HTPB is fuel rich, therefore
what happens is you do not get high temperature, but it is able to generate gas at low
temperature and still propel the rocket. Therefore, an enemy cannot see that a rocket is
going up because he uses radar to see the temperature of the plume. Therefore, each of
the propellant has its own advantages plus the burning rate will change. We will look at
it when we deal with burn rate of solid propellants.
In addition to adding HMX, RDX known as Research and Development explosive which
is cyclo tri methylene tri nitramine is also used. You add almost 80 to 85 percent of these
explosives to the fuel binder. We find that AP is also being added to give AP plus HMX
or RDX and HTPB to give nitramine propellants.
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(Refer Slide Time: 42:29)
The research in propellants is on fuel GAP viz., glycidyl azide polymer since it has more
hydrogen and is a more energetic fuel. You add HMX or RDX and it could not therefore
be only HTPB, but glycidyl azide polymer (GAP), which could also be used instead.
We will not get into too much of the details because our main concentration would be the
composites, may be the double base propellants. How they burn and how they are
applied in rockets. Therefore, we say solid propellants essentially consists of double
base, composite, composite modified double base and nitramine propellants. We will
look at their characteristics when we study the solid propellant rockets.
Let us now move to liquid propellants. They are extremely simple, much simpler than
what we talked in terms of solid fuels and oxidizers and they are classified in a slightly
different way. We do not classify them into four such categories such as we did with
solid propellants.
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(Refer Slide Time: 43:43)
Liquid propellants are categorized into three categories. These are low energy
propellants, medium energy propellants and high energy propellants.
May be by now you can tell me which will be high energy and low energy. Energy is a
misnomer in rocketry because it is not only energy, which is important, but a low
molecular mass is also important. We had observed that hydrogen and oxygen produces
products with a low molecular mass, even though the energy may not be high yields high
values of Isp. When we say high energy propellants, we always mean high Isp
propellants. Similarly medium energy and low energy propellants imply medium Isp
propellants and low Isp propellants respectively. Propellants, which have sea level Isp
less than 3000 Newton second / kilogram are known as low energy propellants. The
propellants having Isp between 3000 and 4000 Newton second / kilogram are known as
medium energy propellants and those in excess of 4000 Newton second by kilogram are
known as high energy propellants.
Well! hydrogen oxygen combination is by far the best from Isp point of view and
therefore, the high energy propellants are hydrogen and oxygen. Perhaps hydrogen and
chlorine would also be good. Yes, it is even more energetic, but chlorine is highly
reactive and therefore, it was tried in one rocket mission and dropped thereafter. But
hydrogen and oxygen cannot be used as room temperature as they are gases. It has to be
liquified by reducing the temperature to low values. Hydrogen liquefies at 20 Kelvin
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while oxygen liquefies at 80 Kelvin. Being liquids at these low temperatures, wherein
they are used, we call these high energy propellants as cryogenic propellants. This is
because, they have to be kept as liquids under refrigerated conditions at cryogenic
temperatures (less than about 123 K) and we call them as cryogenic propellants.
Therefore, if we want to make a missile, I cannot use a cryogenic propellant because a
missile must be ready for launch any time. We cannot wait during a war. Enemy can
strike at any time. Therefore, these cryogenic propellants are normally used for launch
vehicles in which you can fill the liquid propellant slowly and with care as you want.
You have all the time in the world, but if you want to have propellants that can be readily
stored and used, we need to look at other fuels and oxidizers. Let us take a look at some
of the low energy propellants first and then address the medium energy propellants.
(Refer Slide Time: 46:25)
What are low energy propellants? Low energy we said those having specific impulse less
than 3000 Newton second per kilogram. Let us put the names of a few propellants - few
of them. All of us had have heard of V 2 rocket which was the first rocket ever made and
it was by the Germans around 1945 during the second world war. It used liquid oxygen
and alcohol. The alcohol is ethyl alcohol and ethane is C2H6. May be you remove some
of the H and put OH in its place. You have ethyl alcohol. But it is very poor performing,
very low in specific impulse when used with liquid oxygen. The liquid oxygen is a
cryogenic liquid while alcohol is liquid at room temperature.
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Therefore, you have a cryogenic oxidizer with a room temperature fuel and therefore,
this combination will be a semi cryogenic propellant. It is not totally cryogenic. Why
does it have poor properties? Alcohol contains oxygen also because you have OH. That
means it is not a good fuel. Why not use a better fuel like hydrocarbon? Why not use
kerosene? Kerosene has molecular formulas something like dodecane C12H26 as a linear
chain. It is a much better fuel and therefore, the present trend is used semi cryogenics
consisting of liquid oxygen and kerosene together. Kerosene forms soot and it cokes
when heated. It can evaporate quite fast. It is volatile. We would like to make it suitable
for rocket and the kerosene, which is modified by adding additives known as a rocket
propellant.
Kerosene has been extensively used with liquid oxygen in rockets. It is also known as
Rocket Propellant (RP). It is little different from the pure kerosene in that it is somewhat
modified by additives. What are the changes? Flash point of kerosene is around 38oC but
you would like to increase it to something like 55oC. Therefore, you put some more
additives in it and you increase the flash point. You make sure it does not gel or it does
not form gum or does it coke when heated. We make it more conducting to prevent
accumulation of electrical charge in it. It is used with liquid oxygen and the values of sea
level specific impulse are not very high of the order of 3000 Newton second per
kilogram. Therefore, it is one of the low energy propellants.
Alcohol is no longer used, but we use liquid oxygen with kerosene. It is a rocket
propellant, but unfortunately there are some problems with kerosene. See kerosene is not
a pure chemical. Depending on the crude, it may contain more of paraffin or of
aromatics. And these influence its combustion.
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(Refer Slide Time: 49:29)
Therefore, there has been an attempt to get kerosene to be synthesized for use in rockets.
The Russians have done it and called it as SINTIN. But SINTIN is costly. It is made
from carbon and hydrogen at high pressure in the lab. It is not very widely used though is
used by the Russians. We should have some control over the kerosene used and we say
rocket propellant is something like kerosene, almost kerosene, but with some additives
added to it.
If we can strain a hydrocarbon molecule, i.e., if the bond can be strained, it locks more
energy and the kerosene becomes more energetic fuel. We change the bond
characteristics. We can make kerosene to be more energetic. But the fuel may not be
very stable.
Maybe there are a few more low energy propellants. I will go through it in the next class,
but what we did today is that we looked at solid propellants, we classified them into four
categories and we started with liquid propellants. We saw the high performance liquid
propellants to be liquid oxygen and liquid hydrogen. Then, we started with low energy
propellants and we will continue with this in the next class.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 19
Low energy liquid propellants and hybrid propellants
Good morning. In the last class we were dealing with the liquid propellants and we
continue on it. But before continue on it, I would like to clarify that the classification of
liquid propellants was distinctly different from the way we classified solid propellants.
While we classified solid propellants into double base, composite, mixture of composite
and double base i.e., composite modified double base and nitramine propellants, we had
somewhat different classification for liquid propellants.
(Refer Slide Time: 00:47)
The liquid prpellants were classified as low energy propellants, medium energy
propellants and high energy propellants. In today’s class, we continue with this
classification and also we will try to see whether any other classification is also possible,
so as to cover the entire spectrum of the different liquid fuels and liquid oxidizers. These
liquid fuels and liquid oxidizers comprise the liquid propellants. For high energy liquid
propellants, we found that hydrogen and oxygen, which generate products with low
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molecular mass, give high Isp not because it has a high energy, but due to the low value
of the molecular mass of the products.
Hydrogen with oxidizer fluorine will also have high energy but fluorine being very
reactive cannot be used as a propellant. This was about high energy propellants and we
said that high energy propellants are those which have specific impulse greater than 4000
Newton second by kilogram. We then came to the low energy propellants. We said these
are propellants which have specific impulse is less than about 3000 Newton second by
kilogram. And for this we considered liquid oxygen LOX with alcohol which was used
in the V 2 rocket in 1945. It was developed by the Germans. I always take this is an
example but alcohol contains oxygen also in addition to hydrocarbon and it is not that
energetic.
From LOX alcohol, we went to LOX kerosene. When we look at kerosene we find well,
kerosene is something like an aliphatic compound; we said something like C12H26 and
since it comes from a petroleum base, it also contains some other constituents and is not
a pure chemical. And therefore, while using it we need to be careful. Raw kerosene has
a flash point around 38oC. And we wanted it to increase it for which we added additives.
Additives are also added to modify other properties of kerosene. The resulting
keroseneused with liquid oxygen was called as rocket propellant (RP).
Flash point temperature is the temperature of the liquid fuel at which the vapor forms a
flame when mixed with air but the flame cannot persist since the vapor gets consumed.
Fire point is the temperature of the liquid fuel wherein so much of copious vapour is
produced such that flame persists even when you remove the ignition source. The flash
point of kerosene is around 38 degrees Celsius and we increase by adding additives.
Further, when you heat liquid kerosene in the absence of oxygen it should not form solid
residues like coke and coking temperature temperature is the temperature at which such
residues are formed. By adding additives we make sure the no such coking takes place
till a temperature of around 550 Kelvin. This is because during the passage of kerosene
in cooling channels or injection ports, it should not block the flow due to the coking if it
is heated.
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Though, basically when we say LOX kerosene propellant, we refer to it as rocket
propellant RP. It is not correct to say a combination because we could use it at different
mixture ratios.
Let us see what are the other combinations of liquid oxidizers and liquid fuels which
could be the low energy propellants.
(Refer Slide Time: 04:56)
We considered LOX as an oxidizer. Why should LOX alone be an oxidizer? Any
substance which has excess oxygen could be an oxidizer. We could have HNO3, nitric
acid. And if you see HNO3, it consists of one H and one N. N any way is inert. We have
three of oxygen what you need is half O to form H2O. That means we still have two of
oxygen. Therefore, nitric acid can be used as an oxidizer and it is used as an oxidizer.
We could also consider the other substance maybe N2O4 which have more oxygen in it
than HNO3. In the case of HNO3, we saw while we looking at the heat of formation it has
something like minus 170 kilo joules per mole for the heat of formation. Whereas, N2O4
had slightly positive heat of formation. We also noted that for propellant, if the fuel and
oxidizer have slightly positive or small negative values of heat of formation, it is to be
preferred as the energy released in the reaction increases. Therefore, N2O4 tends to a
better oxidizer compared to nitric acid.
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In HNO3 you are losing some oxygen to the hydrogen whereas, in the case of N2O4 all
the oxygen is available for tne burning. N2O4 is also used as an oxidizer. Let us go little
deeper into HNO3 as it was used extensively and is still used in strategic rockets.
HNO3 is used as such; but if we add NO2 to it, we add more oxygen to it and it becomes
more energetic. Therefore, HNO3 is sort of energised by addition of something like
fifteen percent NO2 and what happens? You know you have fumes of NO2 coming out
the moment you open the lid of the HNO3 tank and therefore HNO3 with NO2 added to it
is known as red fuming nitric acid or rather RFNA. These fumes of NO2 coming are
orangish or reddish in colour and hence the name RFNA. We will see some photo graphs
of some test firings and the fumes. You will see some orangish fumes in the plume. If
you add a smaller amount of NO2 say 0.5 percent, the HNO3 still fumes but not as red
fumes. It is known as white fuming nitric acid WFNA.
Both RFNA and WFNA are used in practice as oxidizers; but then this acid is very
corrosive. Both the red fuming nitric acid and white fuming nitric acid are extremely
corrosive it tends to corrode the vessel in which it is stored. Therefore, what is done is
that we add something like 0.5 percent of hydro fluoric acid to say RFNA and this
inhibits the corrosive nature of the oxidizer. This inhibited RFNA is known as inhabited
red fuming nitric acid IRFNA.
Therefore, you have nitric acid as red fuming nitric acid, white fuming nitric acid and
with additive hydro fluoric acid HF added to it such that the corrosion of the tank is
suppressed it becomes IRFNA or IWNA. Therefore, while considering nitric acid let us
keep in mind RFNA, red fuming nitric acid; WFNA, white fuming nitric acid and may be
modified or inhabited red fuming nitric acid. You could also have inhabited white
fuming nitric acid. These are the different forms in which nitric acid is used. It was used
extensively in defense and in other establishments but now the preferred oxidizer is
N2O4. This is more energetic considering the slightly positive heat of formation of N2O4.
Well these are about the only oxidizers which are used in addition to liquid oxygen.
Hydrogen proxide H2O2 is also an oxidizer because we find there is still excess oxygen
over hydrogen H2O. We are still left with one O; but it is a very mild oxidizer with small
oxygen content. It has a heat of formation which is again terribly negative. Something
like −180 kJ/mole and therefore, it is not an effective oxidizer. In fact H2O2 can directly
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also dissociate into H2O + 1/2 O2 and this reaction is exothermic. It is used more as a
single propellant i.e., mono propellant; as a single in built explosive consisting of oxygen
and fuel rather than as an oxidizer.
Therefore, the two oxidizers other than LOX which we can consider are HNO3 and
N2O4. Now, let us list the fuels which they are used with. We note that these oxidizers
will not be as strong as liquid oxygen for the simple reason you are having some little bit
of nullifying effect of hydrogen and nitrogen. When we use HNO3 with kerosene or
N2O4 with kerosene the performance is going to be even poorer than LOX with kerosene
and therefore, combination of these oxidizers with the fuels what we considered like
kerosene will be still be low energy propellants. Let us take look at fuels, other than
kerosene.
(Refer Slide Time: 11:10)
We can talk in terms of what we talked earlier hydrazine, N2H4. It is a very popular fuel.
You know this also had a slightly positive heat of formation and therefore is a good fuel.
You have lot of hydrogen in it and therefore, you could have molecular mass of the
products which could still be small.
We can take one of the H from hydrazine and substitute it with a methyl radical. In other
words from N2H4, we take one of the H out, we are left with 3 H, and we substitute the H
with CH3. In other words we substitute one methyl radical in hydrazine and we get what
is known as mono methyl hydrazine N2H3CH3 (MMH). While talking of different
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chemicals you will recall, we said hydrazine is an explosive in the sense it can directly
dissociate whereas, mono methyl hydrazine is also about the same. It has about the same
heat of formation around – 50 kJ/mole but it has a higher specific heat and it is not as
reactive as hydrazine. Therefore, the preference is to use mono methyl hydrazine rather
than hydrazine. But hydrazine is also used. Being more reactive requires some more care
as it is more prone to combustion instability.
Hydrazine tends to be a little unstable compared to mono methyl hydrazine. We can keep
on evolving around hydrazine. Instead of taking one hydrogen atom in hydrazine and
substituting it by a methyl radical, on one side ot itself i.e., unsymmetrically we put one
more methyl radical here – two of them, and this is known as unsymmetrical dimethyl
methyl hydrazine UDMH (N2H2(CH3)2]. It is not as strong as hydrazine because now
you are adding more carbon molecules to it but it is a very powerful fuel. Most of the
liquid propellant boosters that we are using in our country make use of unsymmetrical
dimethyl hydrazine for fuel.
Therefore, the fuels which we have considered so far are UDMH, MMH and hydrazine.
These three fuels can be used with RFNA or with white fuming nitric acid which is quite
rare or with N2O4.
(Refer Slide Time: 14:18)
We use these oxidizers with hydrazine or mono methyl hydrazine or UDMH. Mono
methyl hydrazine has a high specific heat and more expensive compared to UDMH.
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Therefore, wherever smaller quantities of hydrazine fuel are required we use mono
methyl hydrazine like in upper rocket stages where we do not want to use so much of
fuel. Hydrazine is also used especially in spacecrafts, but as I told you it is little more
reactive whereas UDMH is more widely used for wherever large propellant requirements
are there.
All these three fuels can cause cancer and there is a trend in today’s world to get back
and substitute it by kerosene or some other fuel. We call these fuels as carcinogenic. But
in rocketry since only small quantities are required we take adequate precautions, we
continue to use these fuels.
(Refer Slide Time: 15:40)
Therefore, we summarize the low energy fuels: In addition to LOX kerosene, we could
have N2O4 with mono methyl hydrazine or with unsymmetrical dimethyl hydrazine i.e.,
either MMH with N2O4 or UDMH with N2O4. There is also one last fuel which we call
as a mixture of fifty percent UDMH plus fifty percent hydrazine. This is known as AZ
fifty and is used with N2O4. It is used in some of the rockets developed in US. Therefore,
the low energy propellants could be LOX alcohol, well this is outdated and alcohol is
replaced by kerosene. HNO3 in the form of red fuming nitric acid and N2O4 which more
energetic, and fuels hydrazine, MMH and UDMH. These are the low energy liquid
propellants.
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It may be useful to go back in history and address a few other fuels used in rockets for
missiles. We still see some other fuels being used earlier especially at DRDL when they
were making missiles.
(Refer Slide Time: 17:28)
One is known as aniline. See, while all the fuels what we considered where somewhat
based on aliphatic compounds like kerosene we say C C all in the straight chain, aniline
is derived from an aromatic compound. Aromatic compound corresponds to the benzene
chain. You have six carbon atoms with alternate double bonds. What do you do in aniline
is to substitute one of the hydrogen atoms in the benzene molecule by the amine radical
NH2. This becomes aniline. Let us keep this in mind because we will come back to this
NH2 amine radical. We have NH2 in hydrazine and NH2 in MMH and UDMH. This
amine radical is very reactive and gives an important property to the propellant.
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(Refer Slide Time: 18:56)
That the moment the fuel containing NH2 comes in contact with let us say HNO3 or with
N2O4, it ignites and reacts. Such of the propellants which contain this amine radical
readily react with HNO3 and N2O4. And such propellant combinations of N2O4 or HNO3
with a fuel like aniline or with hydrazine or UDMH or MMH will not therefore, require
an ignition source to start the combustion process.
All what we do is to introduce the fuel, one of these fuels with the oxidizer into the
chamber. Immediately by itself it will ignite and these are known as hypergolic
propellants. Why is it hypergolic? The amine radical in these fuels readily reacts with the
acid or the N2O4 to form hot combustion products. Therefore, the design of the rocket
become much simpler. All what we need to do is to push this fuel having the amine
radical in the chamber, push the oxidizer into the chamber and when the two mix it burns
and releases hot gas. We do not even need to ignite it extraneously. We do not need an
auxiliary igniter whereas, if we consider even liquid hydrogen or kerosene with liquid
oxygen and N2O4, we need to ignite it because we have to overcome the energy barrier
before a chemical reaction can take place.
propellants.
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These are all hypergolic low energy
(Refer Slide Time: 20:48)
And we call those propellants which require an ignition source for initiating reactions or
combustion as being non-hypergolic. These are liquid oxygen with liquid hydrogen,
liquid oxygen with kerosene.
The only exception is liquid fluorine and liquid hydrogen. Liquid fluorine is so reactive
that it will react with anything including the container in which it is stored. Therefore, it
is also comes in the category of hypergolic propellant. But mind you, it does not contain
an amine radical. It comes from the reactivity of fluorine itself.
Aniline was extensively used in the seventies and eighties. As a slight improvement over
aniline one very last fuel of which we talk is known Xyladiene.
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(Refer Slide Time: 21:40)
And in xyladiene we have the same aromatic structure of benzene, you have alternate
double bonds, you have NH2 here as in aniline. We take two more of the hydrogen atoms
and replace it by methyl radicals as shown. This substance is known as xyladiene. Again
it consists of an amine radical and therefore, it is hypergolic with N2O4, RFNA and
WFNA. Aniline and xyladiene are used as liquid fuels for some the missiles. They are
low performing with low energies.
(Refer Slide Time: 22:49)
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We can say that IRFNA with xyladiene or aniline or better still with N2O4 and N2O4 with
hydrazine and UDMH and with MMH and LOX with kerosene are the low energy
propellants with progressively increasing Isp values. However, the maximum Isp is less
than about 3000 Newton second by kilogram.
These are the low energy propellants and many of these low energy propellants
especially those using the nitric acid and N2O4 are hypergolic which means they are
much easier to use. This is why we see these fuels being repeatedly used in spite of
being carcinogenic and having low performance. The rocket construction gets simpler.
You do not need an auxiliary igniter. We can whenever we want to you just fire it.
Supposing we have a space craft; let us say that in the space craft you have a fuel tank
and an oxidizer tank. The oxidizer is N2O4 and fuel is MMH.
(Refer Slide Time: 24:31)
A spacecraft such as INSAT has a number of small rockets for attitude and orbit
corrections. We have some 12 such rockets and these are connected to the fuel and
oxidizer tanks through flow control valves and regulators. Whenever you want to have
some correction to be done in the spacecraft for which a particular rocket has to operate,
we feed the propellants into it. We just have to open the valve for a given duration. It
will give an impulse which is used for the correction. We do not need any other ignition
device to be able to initiate the combustion in the rocket and that is the advantage of
many of the low energy propellants.
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For large boosters for which fuel and oxidizer are required in abundance LOX and
kerosene are desired. Kerosene is readily available it is a very cheap fuel again and that
is why the majority of the boosters of the launch vehicles, which require large quantities
of fuel make use of liquid oxygen and kerosene. We are still to start with these activities
in our country. The only problem is that it becomes a semi cryogenic propellant.
At this point in time I will slightly divert the discussions and return to the medium
energy propellants a little later.
(Refer Slide Time: 26:04)
We covered hypergolic, non-hypergolic propellants and this could as well be a
classification of the propellants. The classification of propellants can also be done be
under how you can store the propellants. Why do we say storing is so important?
Suppose we have a missile to take off; see a missile must be ready for launch at any
point in time. That means the tanks of the missile must be always filled with the fuel and
the oxidizer. If a propellant could be stored on Earth we say the propellant is Earth
storable. If we want to use it in a space craft and the space craft goes round for twenty
years or twenty five years, the propellant must be there in the spacecraft during the
period of the orbit. That means the propellant must be storable in space. The distinction
between Space storable and Earth storable is in the conditions experienced in space and
on the Earth.
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If we have LOX-kerosene we cannot keep it on ground all the time because LOX has to
be kept under refrigerated conditions or under insulated conditions for a short time.
Otherwise it will evaporate. Therefore, this is not even Earth storable leave alone space
storable. You know similarly, LOX and liquid hydrogen are cryogenic fuels requiring
special storage. Semi cryogenic and cryogenic fuels are not Earth storable and only the
low energy propellants, such as red fuming nitric acid or N2O4 as oxidizer with
hydrazine, UDMH and MMH are Earth storable. Space storable poses harsher criterion;
it becomes more difficult to store. In space we can get much lower temperatures and
therefore, not all Earth storable propellants can be space storable. For instance let us take
an example N2O4 as a very good oxidizer.
Let us consider the freezing point of N2O4; the freezing point of N2O4 is around −9oC
and whenever we have a spacecraft, when it is not looking at the sun the temperature
comes down. Therefore, we need heaters to keep this warm; but if we were to add
something like three per cent of nitric oxide that is NO to it, that means three per cent
nitric oxide to N2O4 we can decrease the freezing point. The freezing point instead of
being −9oC can be decreased to −15oC. Therefore, if we have to make the oxidizer N2O4
perform in space one of the things to be done is may be we should add some additives
like nitric oxide. If we add three per cent, we have a little lower freezing point. The
addition of NO to N2O4 makes it into mixed oxides of nitrogen and the mixture is known
as Mixed Oxides of Nitrogen MON. When we add three per cent NO, we call it as MON
3.
When we use N2O4 as an oxidizer and if the same N2O4 has to be used in space for a
prolonged period then we have to decrease its freezing point. We do that by adding NO
to it. If we add three per cent it is known as MON 3. If I add something like 25 percent of
N O to N2O4 we can decrease the freezing point to – 55oC, which is even better.
Therefore, in space we can use the fuel as MMH, but we substitute N2O4 partly by NO to
get MON. This is the difference between Earth storable and space storable propellants;
cryogenic and semi cryogenic propellants are those which are neither earth storable or
space storable. They can be used in launch vehicle and conditioned may be a day or a
few days before the launch.
The cryogenic and semi-cryogenic propellants cannot be used for missiles nor for
satellite propulsion and control. These are the different classification of propellants.
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(Refer Slide Time: 30:46)
Medium energy propellants would have Isp in the range between 3000 Newton second
by kilogram and 4000 Newton second by kilogram. We talk in terms of sea level specific
impulse because vacuum specific impulse is much higher than what we evaluate on the
ground. Therefore, whenever sea level specific impulse is less than three thousand, we
say it is low energy, when it is greater than 4000, we say high energy propellant. But
there are hardly any good medium energy propellants.
The only exception which we can think of is instead of having LOX kerosene, if we use
instead of LOX, liquid fluorine, it gives an specific impulse greater than 3000; something
like 3200 and it could be a good candidate. But liquid fluorine cannot be used as it is
extermely reactive.
Oxidizers like RFNA, N2O4 are inferior to LOX because they contain some other
elements in addition to oxygen. Therefore, if we can use LOX instead of N2O4 with one
of the fuels say UDMH or let us say combination of UDMH and hydrazine, Aerozine 50
or let us say with MMH, then we get higher performance in excess of 3000 Newton
second per kilogram. And these are the medium energy propellants. But in practice this
propellant combination LOX and UDMH has been used only in Russia.
Therefore, with liquid propellants, we find only a few propellants in the three categories
of low energy, medium energy and high energy are used. Let us now summarize it.
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(Refer Slide Time: 33:06)
High energy propellants are neither storable on Earth or in Space. It could be LOX
(liquid oxygen) and liquid hydrogen. The low energy propellants are generally storable
on Earth and many of them are storable in space with little modifications such as MON 3
or MON 25 instead of N2O4. These are essentially hypergolic whenever the fuel
contained the amine radical in it and the oxidizer was N2O4 or HNO3. The low energy
propellants liquid oxygen and kerosene is not Earth storable neither is it space storable.
You have the medium energy propellants which is again semi cryogenic liquid oxygen
and UDMH. This is the only propellant in the medium energy category which has been
used in a rocket so far. Well, these are the classifications according to energetics of the
propellant. We could classify the propellants into hypergolic or non-hypergolic.
Hypergolic are ones which in which the oxidizer and the fuel when they come in contact
it automatically burns. We will consider this aspect when we deal with liquid propellant
rockets.
Non-hypergolic propellants need an igniter to get the combustion started in the rockets.
The last classification of propellants was based on being Earth storable liquid and space
storable. The semi cryogenic and cryogenic are neither earth storable nor space storable.
This is all about the different liquid propellants that we considered.
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(Refer Slide Time: 35:14)
We have considered solid propellants and liquid propellants; let us take a look at what
are the hybrid propellants. Thereafter, we can solve one or two small problems. What do
we mean by hybrid propellants? Hybrid propellants are those propellants which are in a
mixed phase: may be the oxidizer could be a solid, fuel could be a liquid whereas, the
other one could be a gas or a liquid. Usually the fuels of hybrid propellants use HTPB.
What was HTPB? Well! it is poly butadiene. You have C C C and C two double bonds
and a single bond between C atoms and between C and H. This is what was poly
butadiene and it was a chain as it is, may be m times over. The chain is terminated by
hydroxyl radiacal and we get hydroxy terminated poly butadiene. That means we have
one OH here, I have one OH here. This could be a fuel. The advantage we had with
hydroxy terminated poly butadiene was lighter hydrogen in it compared to PBAN which
was poly butadiene acrylic acid acrylo nitrile.
Therefore, a typical fuel solid fuel for a hybrid rocket is something like HTPB grain or
solid and what is done is you form a solid as shown with a hole in it and then put it in a
cylindrical case and attach it to a nozzle. The HTPB is the solid fuel. Now we require an
oxidizer. We could inject the liquid oxidizer on it. The oxidizer could be anyone of the
oxidizers we have considered.
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(Refer Slide Time: 37:15)
The oxidizers could be liquid oxygen, could be N2O4, could be red fuming nitric acid or
better still inhibited red fuming nitric acid. Well use one of these oxidizers and spray
them on the solid fuel. The fuel and oxidizer begin to react and hot gases are generated.
The hybrid combination is generally a solid fuel such as HTPB and one of the three
liquid oxidizers. We do not use a gaseous oxidizer, because we need a large volume to
store it. There was a time in 80’s when there was lot of interest in hybrid rockets but
again now a days we see the private companies in US developing hybrid rockets for
space tourism. These are manned vehicles which will take tourists from ground to space
using hybrid rockets. When we talk of liquid oxidizers, liquid oxygen is powerful
enough.
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(Refer Slide Time: 38:50)
Can we make the value of the specific impulse of let us say liquid oxygen with a solid
fuel to be higher in some way than by modifying the property of the liquid oxygen itself?
The specific impulse can be increased by either increasing the combustion temperature
or reducing the molecular mass of the combustion products. Either you increase the
temperature of the combustion products or you reduce the molecular mass of the
combustion products. Let us examine the following: Supposing we add liquid fluorine to
liquid oxygen; then in that case, when the oxidizer is liquid oxygen and the solid fuel is
hydrocarbon, the product of combustion is water from the hydrogen in hydrocarbon.
Now, with some liquid fluorine added to liquid oxygen and mind you it is quite easy to
add liquid fluorine to liquid oxygen, because the boiling point of liquid fluorine and
liquid oxygen are about the same in addition to getting water since I add fluorine, I also
get hydrogen fluoride. Therefore, the products of combustion now with liquid fluorine
added to liquid oxygen are in addition to H2O,
hydrogen fluoride HF also. The
molecular mass of water namely H2O is 18 g per mole: 2 +16 =18 g/mole. The molecular
mass of hydrogen fluoride is 10 g/mole. Therefore, you find that the act of adding liquid
fluorine to liquid oxygen results in getting more of hydrogen fluoride as we increase the
quantity liquid fluorine to oxygen and therefore the net value of the molecular mass of
products that we get decreases. Therefore, since this molecular mass decreases, we get
higher value of specific impulse.
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(Refer Slide Time: 40:59)
Therefore, what is it that we do? May be we add lliquid flourine to liquid oxygen: say
10% of liquid fluorine to a mixture of liquid oxygen and liquid fluorine. That 10% liquid
fluorine in this particular mixture; it is known as FLOX 10. FLOX stands for fluorine
and liquid oxygen. If we add something like seventy percent of liquid fluorine to the
liquid oxygen mixture then we call it as FLOX 70. Therefore, by using FLOX instead of
liquid oxygen LOX, we get much higher specific impulses and therefore, these are
considered to be higher energetic oxidizers. The main aim therefore is by using FLOX
instead of liquid oxygen LOX, we get a higher value of specific impulse. The higher
value of specific impulse does not come from the energetics of propellant alone. But by a
reduction in the molecular mass of the products which gives us the higher value of
specific impulse. Let us keep this in mind; we can improve the performance of
propellants by reducing the molecular mass of the products.
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(Refer Slide Time: 42:24)
We said that propellants could either be solids, could be liquids or could be a hybrid
combination of fuel and oxidizer. We classified solids into four categories and what were
the four categories? We said it could be a double base or a single base, it could be a
composite in which case you have crystals of solid oxidizer dispersed in fuel
heterogeneously, it could be a combination of the above two: composite modified double
base. It could also be a nitramine propellant. For liquid propellants liquids we said could
be space storable, earth storable, hypergolic, non-hypergolic. We also said: low energy,
medium energy and high energy and of course, we looked at hybrid propellants.
These are about the only propellants which are in use today. And when we consider solid
rockets, solid propellant rockets, liquid propellants rockets and hybrid propellant rockets
we will consider the details of how to incorporate the propellant to make these rockets.
There are other factors we will be looking at; both the chemistry and the ballistics. This
what we will be doing from the next class onwards.
I would like to give you one reference which is extremely fascinating to read about
propellants. It is a book called Ignition. It is in our library the exact name of the book is
Ignition - an informal history of liquid rocket propellants. It is by John D Clarke. Quite
an old book though written up in 1962. But it provides a very comprehensive coverage. It
not only deals with the performance, but storability, ability to handle, dangerousness,
ignitability, corrosiveness, etc.
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Let us quickly do one or two small problems such that we revise this particular area of
propellants. The first one I choose is extremely simple but illustrative.
(Refer Slide Time: 45:20)
Supposing we use a MMH and N2O4 as propellants and we use it at a mixture ratio of
1.5 that means, the moment I choose a propellant, I also choose a mixture ratio. Why?
Because choice of mixture ratio governs Isp or C* value; Isp and C* will be maximum in
the fuel rich region. We always choose fuel rich mixture ratio. Normally even though I
say 1.5, may be here we choose this because for simplicity in doing a problem.
Sometimes we will use mixture ratio in which the volume quantities of the fuel and
oxidizer are the same. The volume of N2O4 is same as MMH so that we just need to
develop one tank and a similar tank can be used for both the MMH and N2O4.
Let us start with this problem. We need to make a rocket using MMH and N2O4 at a
mixture ratio of 1.5; We would also be given the value of thruct. Well, the thrust is 6.7
kilo Newton. Now we ask what must be the mass flow rate of MMH into the chamber?
What must be the mass flow rate of N2O4 in the chamber to give us a thrust of 6.7 kilo
Newton? Therefore, how to estimate these quatities ?
The propellant combination is known.
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(Refer Slide Time: 47:08)
I know the formula for MMH we said it is CH3N2H3. It reacts with N2O4 at the particular
mixture ratio. We can write out the equation for the chemical reaction and find out the
temperature of the products. We can find out the composition of the products using the
methods discussed or if you want to simpler or approximate analysis one assume some
products and do the problem. Assume the hydrogen to be much more reactive than
carbon and therefore hydrogen first consumes the oxygen and only the balance of oxygen
is left for carbon to react. The nitrogen in the substance is obtained in the products as N2.
Therefore, the procedure is first we use the hydrogen in the fuel which searches for the
oxygen, consumes it, the balance oxygen is now available for carbon to react either fully
to carbon dioxide or partially to carbon monoxide. If oxygen is in short supply it may
not oxidize carbon and the nitrogen in the fuel or oxidizer is available in the products as
N2.
This approximate procedure is quite useful in determining the combustion products for
fuel rich mixtures. However, we could do the detail problem using chemical equilibrium.
We will find that the temperature of the flame or temperature of the combustion products
is 3028 Kelvin; the molecular mass of the products is equal to 20.39 g per mole. The
value of γ is 1.235. Once we know this, we can get the value of C*.
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(Refer Slide Time: 49:07)
C* = √(R0Tc/M) / Γ, where Γ = √γ×(2/(γ+1))(γ+1)/2(γ−1) . We substitute the values and the
C*= 1737 m/s. Most of the propellants have C* around 1800 to 2000 m/s. If we assume a
C star efficiency of 0.96 for the present since we have not yet studied how to estimate it,
and also take the thrust coefficient of the nozzle to be 1.95, something which is
realizable, we can calculate the specific impulse.
(Refer Slide Time: 50:33)
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To calculate the value of specific impulse: Isp = C* × CF which is 1.95. C star is
efficiency is 0.96. The theoretical value of C* = 1737 m/s. Hence Isp = 0.96×1737×1.95.
We want a thrust = 6.7 kilo Newtons. The the value of thrust divided by the mass flow
rate of the fuel and oxidizer (total propellant) is equal to Isp. From this we get the value
of mass flow rate of the propellant m° = 2 kg/s.
(Refer Slide Time: 51:36)
What is the flow rate of oxidizer and flow rate of fuel? You know the mixture ratio is
given as 1.5. m°ox /m°F = 1.5. Therefore, we can get the mass flow rate of oxideizer and
fuel which are 1.2 and 0.8 kg/s. This is how we solve for the propellants required in a
rocket.one solves.
Let us consider another problem involving a solid propellant. Let us say I have a solid
propellant which consists of ammonium per chlorate as oxidizer and HTPB as the fuel.
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(Refer Slide Time: 52:49)
We specify that the solid loading in this propellant is 75 percent. We know the molecular
mass of the HTPB, the molecular mass of AP namely NH4ClO4 and the mass of AP in
the mass of propellant viz., the solid loading is given. We have to find out whether this
propellant is fuel rich or oxidizer rich.
Since the molecular mass of HTPB and AP is available to us, we can find out the number
of moles of HTPB and AP in 100 kg of the propellant. We divide 25 by the molecular
mass of HTPB and 75 by the molecular mass of AP. From this we get the value of the
number of moles of HTPB required for a single mole of AP. Then from this we write the
equation for the chemical reaction and find out whether it is oxidizer rich or fuel rich.
You will find even with this 75 percent solid loading the propellant is still fuel rich. All
propellants tend to be fuel rich.
Please complete this problem as a homework problem?
The solid propellants are made as a solid block and in the next class we will see how to
make a solid motor out of this block. Similarly, for the different liquid fuels what are the
cycles which we can use such that we can get them to burn at high pressures and provide
us with high values of jet velocity? And similarly, for hybrids. We will start with solid
propellant rockets in the next class.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No # 20
Introduction to Solid Propellant Rockets
(Refer Slide Time: 00:21)
Today we will start with solid propellant rockets and this will be an extremely simple
topic. We will take a look at what are the elements of a solid propellant rocket, but
primarily, we start from the propellant point of view, because that is what we have
covered so far. We told that solid propellants could be double base, may be composite,
may be composite modified double base, and also nitramine propellants. Whenever these
propellants are made, you cure the substances and make it into something like a solid
form, and this solid is what constitutes let us say a solid propellant. We would like to use
it to generate thrust.
Therefore, what is it we do? We take, let us say a case, which could be cylindrical. We
put the solid propellant inside the case. How do you put it inside the cylinder or a case?
We keep the case vertically, we put all the ingredients into it; we make the propellant
mixture into a sort of a liquid or a slurry, then we cure it and we get something like a
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solid which is formed within the cylindrical case. We then attach a nozzle to it and this
becomes a solid propellant rocket.
Therefore, in a solid propellant rocket we have a structural member which is a case and
which supports the propellant. Either we pour the slurry of the mixture may be a
composite slurry consisting of the binder, the ammonium per chlorate and aluminum in
into in this. Then we take it into a furnace and cure it at a temperature of between 70 and
200oC depending on the composition. We form a block may be a composite solid
propellant block which we call as a propellant grain. And when we cure the slurry or the
mixture in it and make it as a solid inside a case the rocket or solid rocket is known as a
case bonded rocket.
(Refer Slide Time: 02:50)
It is also possible to make the propellant in the form of a paste and extrude it out just like
we squish out the tooth paste from its container. We could thereafter cure this squished
out paste using in a furnace and enclose it in within a covering. Thereafter we could
introduce it viz., charge it into the case. In this case, it is known as a free standing
propellant grain. And what is free standing? The grain propellant inside it is freely
standing it is known as a free standing propellant grain. It is known as a case bonded
propellant grain when the propellant is cast in the case. Therefore, without any
qualification, we have just introduced the word propellant grain. What does grain mean?
The solid propellant block of a particular configuration in a case is what we call as a
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grain. This grain or propellant grain could either be directly cured in the case itself in
which case it is known as case bonded.
And when it is cured outside and we machine it and then push it into the case the
propellant grain is known as free-standing. We have both free standing and case bonded
grains being used, but when we talk of large solid propellant rockets invariably we cast it
in the case itself. But there are other things required. As the propellant burns, the
hardware or the metal case or whatever be the case will get heated and therefore it is
necessary for us to protect the case with a liner. This liner could be an insulator on the
inner walls of the case and then we put the propellant into it and then we attach a nozzle
onto the case.
Therefore, we have a case, may be an insulator or something which will which will make
sure that when something burns the hot gases generated by the solid propellant do not
come and melt the motor case and then we have the nozzle and this what constitutes the
solid propellant rocket. But what is it we want a rocket to do? We want the rocket
essentially to generate some thrust. The thrust of the rocket is equal to m° ×VJ or m° ×
specific impulse Isp. We learnt how to calculate the specific impulse and supposing we
want to generate a large force, we need to generate more mass flow from the burning
solid propellant.
Therefore, in today’s class we will first examine how you design the solid propellant
grain, which could either be case bonded or free standing. If we ignite it and start
burning it, how do we ensure that we have sufficient amount of mass getting generated
from the propellant. We have just studied the chemistry of the propellants, but we have
not studied anything about how it will burn. We want to get some feel for how a
propellant surface will burn. How do we calculate the mass generation rate and what
should be the surface area such that we get the concerned thrust and this is what we will
be discussing today.
Let us again look at the problem in a slightly different scenario. However, before that to
repeat if we have a composite propellant like we have a polybutadiene which is a resin
and the resin, which is in liquid form, is added with the solid crystals of AP and
aluminum powder. The slurry so formed is a mixture of all these three constituents and
we heat it to a high temperature to form a solid block. This solid block is covered with a
490
liner of insulation and kept inside the case. And what is this resin? It is something, which
when heated sets, and this resin is known as thermosetting resin. This is because when
heated it becomes hard. Resin is a polymer polybutadiene; like glue.
I will show you some examples. But if you have plastic which is also a resin and if we
heat it, then it softens therefore we cannot use such resins for composite propellants and
the same thing is true when I use nitrocellulose and nitroglycerine or the insulation. We
make it as a liquid, cure it into a solid and make it as a block and this block is known as a
propellant grain. Therefore, we are interested in finding the burn rate; the rate at which it
will burn or the rate at which mass gets generated from a grain. Is it clear? We must
differentiate between a propellant block that is the solid propellant which is obtained
after curing.
A propellant grain gives us the necessary m dot of gases generated for providing the
thrust. And this propellant grain could either be directly bonded in a case or the hardware
of a rocket or it could be prepared elsewhere in which case we just introduce the grain
inside the case. The later is known as free standing while the former is known as case
bonded grain.
(Refer Slide Time: 08:27)
This is all about the terminologies. Let us get back and ask ourselves if we have a case
over here in which we have the propellant block inside it and let us say we have no
nozzle. We want to generate hot gases from propellant burning. May be this is the
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exposed surface area of the propellant. It is this surface, which burns and we call it as
burning surface area and denote it as Sb. Supposing this surface is ignited. We would like
to know the rate at which we are generating the hot gases from it. And if the surface
burns we would like to have something like r so many meters per second is the rate at
which a propellant burns. What is this rate?
Supposing, I have a surface, which is like this, let us say a convex surface and this is the
propellant. We can take a flat surface of a propellant, which is the propellant surface
here. Or the other extreme is we take a concave surface something like this and this is the
propellant here. We could have different configurations of the surface. Or we could have
hole right through. Let me consider these cases of flat, concave, convex surfaces of
propellant burning.. At the next instant after burning commences, the flame advances by
a given distance as shown.
In other words the burning progresses into the propellant. The rate at which the surface
regresses is what we call as the burning rate. Rate at which surface regresses, but it must
be normal to the surface and that is why we made these different configurations over
here. The surface regresses in this direction over here; the surface regresses in this
direction normal to the surface in this direction. In all cases the surface evolves normal to
it and this constitutes the burning. Therefore, we call burning rate as normal to the
surface.
(Refer Slide Time: 10:48)
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When we say a propellant burns at the rate r meters per second, we mean that the burning
is normal to the surface. Let us spend some more time on this. Supposing we have a
propellant let us say a block something like this at a corner. The shaded part is the
propellant. We start the burning over the exposed surface. How will the burning take
place? The burning will take place normal to this vertical surface along X axis. The
burning will take place normal to this horizontal surface along the Y axis. And what
happens to this contact point? Well! In other words over here since it is normal to this
point it will go as a circle (quadrant of a circle) over here. This is how normal to the
surfaces that the regression proceeds. May be we will have to look at this in some detail
when we come to different configurations of the propellant grain.
Therefore, we define a burning rate in terms of meter per second, which is normal to the
surface. The gas generated viz., rate of mass of gas which is generated by the burning in
kg/s is equal to r, the burn rate so, many meters per second × the burning surface area so,
many meter square; this gives us meter per second into meter square which is meter cube
per second that is the volume rate at which the burning of the solid propellant. Knowing
the volume rate at which the regression takes place we multiply it by the density of the
propellant, which is so much kilogram per meter cube. We get meter cube and meter
cube gets cancelled and we get kg/s. Hence m° = r × burning surface area × ρρp (so,
much kilograms per second).
The rate at which mass gets generated from a surface can be written in terms of the linear
regression rate known as burn rate. In terms of a linear dimension into the burning
surface area in this case the volume of burning.
The burning surface area is this surface initially and at the next instant of time may be it
is over here this is the burning surface area and you have burning surface area into the
burn rate into the propellant density as the rate at which mass of gas is generated. And
burn rate is the rate of regression of the surface. Therefore, we say mass generated by
burning is equal to the burn rate in meters per second ×surface area in meter square ×
propellant density in kilogram per meter cube. Let us illustrate it through an example,
because we also you will recall we have been talking of black powder in the course on
explosion also. We said that black powder consists of an oxidizer like KNO3 and it also
consisted of something like charcoal you say carbon and some sulfur.
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(Refer Slide Time: 14:14)
We had said that it consists of something like 75 percent potassium nitrate maybe 15
percent of carbon and sulfur is the remaining 10 percent. And how do you form these
crackers or sparklers, which use this black powder. You have these solids this is an
oxidizer and fuel you grind them together and make a good mixture of it may be add
some glue so that becomes wet. You sort of cure it make it as a solid and that is what
constitutes a composition. Well this is also a sort of composite propellant.
Therefore, let us take this example just to understand about burn rate. I brought a
sparkler with me. We will see the burn rate of this composition. I will just ignite it I
brought two sparklers, because I thought the effect of aluminum must also be clear to us.
Let us just ignite it and see what happens? Well! This is ignition right. You light a
candle. Well, this is a sparkler. It consists of as I said the composition, and the
composition is again carbon, you have sulfur and you have KNO3 and this it will burn by
itself without the support of air. It is a propellant and instead of using polybutadiene it
has resin that is the glue which is used.
Well, you ignite it at the surface tip. How does burning take place? This surface, which
gets ignited and this ignited surface burns the next surface next adjacent to it and so on.
We find that this surface is regressing at a particular rate. Well, in this case the whole
surface is exposed, but because there is air over here the flame expanded slightly here.
The burning surface area is perpendicular to this plane and the burning progresses
494
axially. This is the case of this composition without metal in it. Let us see what is the
effect when we have a metal in this sparkler, the other composition being about the same.
You see the distinct effect of the metal. The intensity is very much higher. See the metal
having so much energy you find the heat, which is much higher. Metal has a higher heat
of combustion and that is why metals are included in solid propellant. Again we are
talking of the same surface regressing as it is going on and we find that the propellant
regresses; i.e., it burns. And what is happening is gas is getting generated the rate at
which the surface is regressing is the rate at which the composition is getting consumed.
The energetics is such that you know the sparks fall on me I feel little warm and is
harmless but with metals in it I get burnt. Therefore, you find that metals have a definite
role to play and I think this is all what I wanted to illustrate. We are talking of a burning
rate, which is linear regression rate of the propellant. We will come back to our sparklers
when we talk of combustion instability. We will see how even this can make a cavity
unstable. But, when burnt a propellant regresses at a given rate and as it regresses mass
gets generated. This is all about burning rate.
What will be the typical burning rate of the sparkler? We had a length something like
almost twenty centimeters, it burnt over a period of one minute. Therefore, the mean
burn rate r is equal to 200 millimeters divided by 60 seconds, which is about 3 mm per
second or 3.3 mm per second. And this is how burning rate is calculated; you have a
solid propellant strand; you ignite it and measure the burning rate. Therefore, with this
background I think it is time to go ahead and examine the burn rate of let us say the
double base propellants and may be the composite propellants. We will try to get into
some details and we will try to make some sense of how the composition influences the
burn rate.
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(Refer Slide Time: 19:34)
Let us say that we have a double based propellant here, and just the same way we ignited
this sparkler we place a fire over here or a heat source over here start the burning. What
was the composition of double base propellant? We told that it consist of nitroglycerine
which is C3H5(ONO2)3 and nitrocellulose C6H5(OH) and ONO2. We had something like
five of this OH in cellulose we put x of ONO2 we removed x of OH and put ONO2 there
we had 5 minus x over here giving C6H5(OH)5-x(ONO2)x.
Now, we ignite the double base propellant and what happens. The region very near to the
surface gets heated, because all these propellants are insulators. Heat does not conduct
deeply into it. And therefore, at the other end, if we consider this as the cordinate x, the
temperature will be almost the ambient value. Only a small thickness of the propellant
near the surface gets heated. And therefore, a plot of the temperature in the propellant
with the temperature shown on the y axis and this is my distance along will be something
like this. Let us show it in red; the depth of the propellant is really not heated. We have
the temperature at the surface as Ts and this is the original ambient temperature.
Let us say that the initial temperature of the propellant Ti, all what we have done is just
like we ignited the sparkler we ignite this particular surface of the propellant. And
therefore, the surface temperature increases heat gets conducted over a certain thickness
while at the end the temperature has the original value. These are insulators. We could
have held the sparkler along the charge because of its poor thermal conductivity.
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Now, what happens when we heat this particular zone to a higher temperature say Ts.
This particular zone begins to foam as it were begins and begins to evolve some hot
gases. And the gases, which are evolved are essentially aldehydes which are COH; could
be formaldehyde could be acetaldehyde CH3COH. We could get some CO, we could also
get some NO. These are the gases, which are evolved. Because something is getting
heated these gases tend to come out, and when the gases come out they react further; that
means, I have a zone adjacent to the surface which is now at temperature Ts which forms
hot gases and the temperature continues to increase in the gas region, in this particular
zone.
Adjacent to this particular zone of propellant, which is getting heated which foams and
forms hot gases; this zone is known as foam zone and is the solid zone, which is heated.
And while it is getting heated it decomposes and generates these gases like aldehydes
may be NO2 may be NO may be CO. These gases are coming out and they react and
increase the temperature from the surface temperature Ts to a higher value of
temperature which we call as T1, the temperature T1. These gases do not completely burn
and when the pressure over here is less than, let us say something like 10 MPa that is
about 100 bar.
What happens is the gases are not at very high pressure and the process of intermediate
products being formed continues without final products being formed. The gases relax
for some time they begin to mix you know it takes some time for the gases to burn and
therefore the temperatures remain at T1 level for quite some time. That means, hot gases
are generated, it reacts increases the temperature from the surface temperature Ts to T1.
The gases are still reacting and what are the reactions which could take place? The
reactions, which could take place are CO which is formed reacting with aldehydes; even
NH2 is formed reacting with CO plus NH2, but these are not very exothermic reactions.
And therefore, the temperature remains fairly constant till the time final recombination
reactions take place. This constant temperature zone is a precursor to the final reaction.
After the reaction takes place, the final flame temperature Tf is reached.
We have seen in the last few classes how to calculate the flame temperature. We can
calculate this flame temperature Tf. But this flame temperature when the ambient
pressure is less than about ten MPa goes in steps initially it goes to T1, then it reacts, but
the reactions are not very exothermic and thereafter it increases this in this zone.
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Actually, CO2, H2O which are complete combustion products are formed and you get the
maximum temperature Tf. Of course, you have CO, because these are all fuel rich
substances and therefore, you have something like a two step reaction process in the gas
medium or gas phase.
This was the within the solid region; this was outside. We should extend the gas phase
right up to the boundary over here, this is the gas phase and this is the solid. I will make
it clear in the next diagram.
(Refer Slide Time: 25:55)
What is it we are talking? We consider a propellant solid block something like this;
Initially the surface gets heated. And there after some chemical reactions are occurring in
this zone and the surface temperature increases to Ts. In the gas phase, the temperature
increases from the surface temperature Ts to T1 over here. Then I have another zone over
here in which no increase in temperature we said T1 remains constant over a certain zone
and there after the final chemical reactions take place it leads to temperature Tf. Let us
plot the temperature distribution again over here as a function of x.
We said at the surface this is the temperature starts from the initial value in this zone it
increases, then it increases still further then, we have a zone in which the temperature is
more or less constant and then the temperature shoots to the Tf over here. This is the way
a reaction takes place and temperature evolves. This is all the gas part of the combustion.
This is the solid part of it. This is the solid reaction zone; we already called it as foam
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zone. The zone in which the initial temperature increases in the gas phase is known as
the fizz zone, because you are having some reactions zone it is known as fizz zone. This
is a luminescent zone, because chemical reactions are happening just like we saw in this
sparkler, some luminosity. This luminous zone is known as the hot zone or second
luminous zone.
The fizz zone is followed by a dark zone. Why it is dark? There is no further increase in
temperature. It looks dark at the low temperature. There after you have this zone which is
the second luminous zone, because again high temperature are achieved giving a second
luminous zone. This is how a double base propellant burns to give you the high
temperature Tf. If we have very high pressures like exceeding 100 atmospheres that is 10
MPa, the chemical reaction continues in the dark zone or heat release continues here.
And I could have the temperature going straight away to Tf at high values of pressures.
That means, the dark zone will be absent when you have high pressure. We will have this
luminous zone viz., the first luminous zone followed by the second luminous zone going
to the temperature final temperature, which is Tf. That means, the temperature increases
like this goes to the surface and then you have this zone and the temperature increases
over here. This is at very high pressures whereas at moderate pressures up to 100
atmospheres well you have this dark zone and after which the temperature increases to
the final value of Tf.
Well, this picture should be sufficient for us to be able to calculate the burn rate. I have a
high temperature T1 over here. The surface temperature is lower at Ts and heat gets
conducted from the surface. It is this heat, which further heats this propellant surface and
causes it to gasify and foam. Therefore, I am interested in events in the fizz zone wherein
hot gases or heat from this zone gets conducted over to the surface.
What is the model or what is my mental picture with which I can solve for the burn rate?
It is necessary to first consider the physical problem. Like we say in the case of the
sparkler. We looked at how regression took place.
The heat is generated here. The temperature in the fizz zone is T1 and the heat gets
conducted from T1 to the surface, which is at temperature Ts; and if we can calculate this
heat, which is conducted then, we can write the equation at the surface and then get the
burn rate. This is my mental picture of the problem. The mental picture might be wrong.
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If my mental picture is wrong, the burn rate what I get would be wrong, But the picture
seems to be reasonable and has been experimentally verified. But this thickness is what
we are talking of. Even though I show it like this, the total thickness of the fizz zone
followed by dark zone followed by the second luminous zone is something like fraction
of a mm. That means, it is all over here just like what happened in the sparkler.
(Refer Slide Time: 30:33)
You had a rod. You coated the propellant over it. And what was this propellant? KNO3
plus carbon plus sulfur all bonded together. We found the reaction distances are all
within a fraction of a mm. We are talking of distances which are fraction of a millimeter
thick.
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(Refer Slide Time: 31:08)
Therefore, let us write an equation for it. It is not difficult at all. We are going to be
interested in the fizz zone, which is the zone wherein temperature increases form Ts to
T1. Why are we not considering the dark zone? See the temperature is constant here. If
temperature is constant here nothing is going to come over here into the fizz zone from
the hot zone. If we were say that the temperature at the edge of the fizz zone is T1 and at
the surface the temperature is Ts, we have heat conduction from T1 to Ts coming here
and still getting conducted in the solid.
We know that heat is getting conducted along in the fizz zone to the surface of the
propellant; but in the dark zone since the temperature is constant no heat is getting
conducted. Therefore, my condition here is T1 over here, and the heat gets conducted
from T1 as the temperature falls to Ts at the surface. We would like to write the equation
for the heat conduction.
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(Refer Slide Time: 32:33)
This is the mental picture or the model that we have constructed. Well, we are interested
only in the fizz zone; let us forget about the dark zone and the second luminous zone. We
now say that let the length of the fizz zone be capital L. We recognize that we are talking
of dimensions which are fraction of a mm thick. Let us say we have the surface at x is
equal to 0; this is the surface over here. We are interested in let us say a small element
over here within the fizz zone at a distance x from the surface and of width dx.
We would like to write an equation for this element and then integrate it to find out how
the burning progresses. And this is how we do any heat transfer problem. Therefore, we
are interested in finding out for the surface area over here equal to unity - 1 meter square
- such that I do not need to carry the value of surface area as I am doing the problem. Let
us say there is some heat conduction q° here. What is the value of heat conduction? Rate
of heat conduction at x; q° = − thermal conductivity of the gas (kg) × dT/dx by
theFourier law. At x plus dx, the rate q° = the small increment over dx which is d/dx of
q° + dx plus q°. The net heat which is leaving this is equal to minus kg× d2T/dx2 × dx
and is the net difference between the heat which is entering and leaving. Heat which is
entering is equal to −kg dT/dx; heat which is leaving is – [kg dT/dx + a small increment,
which is d/dx into kg × dT/dx × dx. Therefore, the net accumulation of heat in a control
volume of width dx is equal to minus kg d2T/dx2 × dx.
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We should learn to write these equations at any time for any problem. We will again be
coming to such formulations when wish to calculate how much ignition energy is
required to ignite a propellant. When burning takes place at the surface, heat from the
burning of gases supplies energy to the propellant. We will erase this part in the fizz zone
and assume the rate of chemical reaction heat release in the reactions to be equal to
q°chem. So much let us say joules per volume. This is joules per meter cube and is the rate
at which heat is getting generated. And therefore, the amount of heat which is getting
generated in this fizz volume is going to be q°chem × the volume of this element. And the
volume of this element is unit area into dx giving heat generated as q°chem × dx. This is
the rate at which heat is getting generated in this volume of thickness dx.
Is there anything else I have to consider in the heat balance relation? When the propellant
burns, we considered the surface to be the frame of reference. With respect to this
surface propellant, which is regressing or burning at a value r, hot gases are getting
generated. The gases generated at the surface are moving out at a velocity ug and this ug
will be different from r. This is because r is for a solid while ug is for a gas. Let the mean
density of the gases be ρg. Therefore, the mass flow rate at which the gases are entering
into this control volume is equal to m°g per unit area. This is equal to ρg into ug that
means, I have so much kilogram per meter cube into meter per second therefore, I have
so much kilogram per meter square second. This is the mass flux.
(Refer Slide Time: 36:54)
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Since we are considering unit area, the mass rate at which gas is moving or the mass flux
is equal to ρg into ug. And what is the enthalpy that is gained as it moves through the
elemental volume.
The gas is moving at this particular x where temperature is T the temperature and at the
next x+dx is equal to T + d T, because there is a small increase in temperature. And
therefore, the increase in enthalpy of the gases is going to be ρg × ug ×into specific heat
let us say at constant pressure into the temperature at exit which is T +dT minus
temperature T at x. Because for this element at the left hand side the temperature is T it
has increased in temperature over a small distance dx by dT to T plus dT. Therefore, the
increase in energy or increase in enthalpy of the gases is equal to ρg × ug × Cp × d T.
Where does this energy come from? The energy comes from the chemical reactions
which generate so much of heat so much joules per meter cube and and also some heat
which is getting conducted.
(Refer Slide Time: 38:49)
The enthalpy gained plus the heat transferred from the element must be the net heat that
is generated. This is expressed by the the equation as −kg × d2T/dx2 × d x, which is the
heat transfer + heat which is getting generated q°chem × d x, which is the net heat
accumulation. And this heat supplies the enthalpy = ρg × ug × Cp × d T. This becomes
the energy balance equation and we must be able to write such energy balance for any
problem. We can write this equation as kg × d2T/dx2, we take it on the right hand side +
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ρg × ug × Cp × dT/dx. dx = q°chem, where q°chem is the energy release per unit volume in
the fizz zone so much joules per meter cube.
Solving this equation, we get the temperature distribution, in the fizz zone. If we know
the value of ρg into ug and q°chem, then only I can solve for temperature distribution. Once
temperature distribution is known, I can find out the temperature at the surface Ts. We
can also get the value of dT/dx at x = 0 viz., the surface. And then find out the rate of
heat transfer. Therefore solution will give us the temperature distribution, but this
temperature distribution is going to be a function of the value of the thermal conductivity
of the gas in the fizz zone, the density in the fizz zone, ug in the fizz zone and the value
of q°chem.
We examine these quantities. The thermal conductivity of the gas kg should be a
function of pressure of the gas. As the pressure increases the conductivity of the gas will
go up, because we have more number of molecules in it. The density of the gas ρg is
again a function of pressure. ug has to be related it to the burn rate at the surface and if
we know it we will get the value of the burn rate r. For a given ug we can solve this
problem for a given q°chem. But what is this rate of heat release? The rate of heat release
q°che in the gas depends on the pressure pm × e−E/(R0×T) where E is the activation energy,
R0 is the universal gas constant and T is the temperature. We have a steric factor
depending on the molecules here.
We therefore find that q°chem is a strong function of pressure., ρg is a function of pressure.
The thermal conductivity is also a function of pressure. Therefore, we can say that ug,
that we wanted to find out should also be a function of pressure.
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(Refer Slide Time: 42:06)
In addition to pressure, we find that ug should be a function of activation energy and this
activation energy should be a function of the composition. May be the density could be
other parameters which are there. But what we get is that the temperature distribution
and the velocity of the gases ug would be a function of pressure.
What is the value of ug? The gas velocity ug in the fizz zone of density rho g equals the
mass, which is moving out and must be equal to the density of propellant into the linera
burning rate of the propellant.
We have a solid, which regresses at a rate r and which the propellant burn rate r. The
mass which is getting generated is equal to ρp × r, because we have unit surface area and
the rate at which gas is getting generated is equal to ρg × ug per unit surface area.
Therefore, the value of the burn rate r = (ρg /ρp) ×ug.
But ug is seen to be a strong function of pressure and therefore the burn rate is going to
be again a strong function of pressure and the effect of the other constituents like the
composition, activation energy etc.. And therefore, it is possible for us to write the
burning rate r in let us say meters per second as equal to a constant × pressure to the
power n, where this constant will take care of activation energy, the composition and
other features.. It is possible for us to write the burn rate for the double base propellant r
= constant × pressuren.
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(Refer Slide Time: 44:21)
And this form of burn rate is what is known as Vielle’s law. It is also known as Saint
Robert’s law and is r = a pn.
Therefore, what is it we have done so far? We said that the surface regression rate of a
double base propellant could be written in terms of the ambient pressure p at which it
burns in the form r is equal to a pn a will have the compositional features may be may
be the initial temperature may be the final temperature. Maybe we should have written
here ug is going to be a function of the initial temperature, may be the temperature T1 and
so on.
Therefore, we put all the effects other than pressure into this a factor ‘a’, that is the pre
exponent and I have the burn rate law during r = a pn. It becomes much easier to use a
law like this rather than keep trying to solve this equation which is a little bit more
difficult. And in practice we use this law. How do we determine the value of n? We do
an experiment at pressure p1 and measure the burn rate r1 which can be written as a p1n .
We then measure the burn rate r2 at a pressure which is little bit away at p2. We get r2 as
equal to a p2n. We use these two equations to find out the value of n. We get n = (ln r2 −
ln r1) / (ln p2 − ln p1).
Once we get the value of n, we determine the value of a and this is how we determine the
burn rate law as r = a pn. Just like the sparkler experiment, may be at an ambient
pressure, we put a strand of propellant in a vessel and do the experiment. We measure
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the time taken to regress a particular distance, measure the burn rate, then determine the
value of n and a.
(Refer Slide Time: 46:47)
If we have the ambient pressure in excess of let us say 10 MPa, then we had said that
there is no such thing as a intermediate T1 zone. The temperature directly goes to
something like Tf over here and more heat is conducted. And therefore, if we have to plot
the burn rate r let us say meters per second, as a function of pressure at low pressures we
have n which is small like this: when pressure goes up it also increases like this
exponentially as shown in this plot.
This is how the burn rate varies with pressure when the pressure corresponds to within
where the dark zone prevails. However, when the dark zone is absent at higher values of
pressure, something greater than about 10 MPa, the heat transfer increase rapidly and the
burn rate increases much faster. A plot of burn rate as a function of pressure would
steepen as shown. But when we add additives to the propellant, then the dependence
changes.
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(Refer Slide Time: 47:54)
Let me again repeat this point. We found that the burn rate of a double base propellant
can be written as r is equal to a pn, where a is a factor which considers all factors other
than pressure like temperature of the gases T1, the surface temperature Ts, the activation
energy etc.. And pressure is directly taken to a power n. Then we said that at low
pressure the value of n is smaller while at high pressures for which there is no dark zone
and you have higher temperatures driving the heat conduction the value of n becomes
greater. n is larger when p is greater than about 10 MPa.
But we could also add some salts, like lead salts to the double base propellant. Then what
happens is instead of getting a curve like the one earlier, we get a different pattern. As a
function of pressure, the burn rate initially picks up and there after it does not change
with pressure that means, you have a plateau over here. In other words, we have burn
rate increasing up to a pressure and it becomes a plateau. Such type of propellants which
exhibit this feature of n = 0 over a certain range of pressures are known as plateau
burning propellants.
Well, some propellants have even a slightly different behavior and some may have their
burn rates going down. We have r versus p. In fact, we should have written ln r versus ln
p over here and that is when I would have a straight line. So, this should have been ln r
and ln p. Sometimes the burn rate begins to fall at higher pressures. It all depends on the
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chemical constituent in the propellant and such propellants which exhibit a drop in
burning rate with increase in pressure are known as mesa burning propellants.
Therefore, what is it we did in this class today? We looked at the rate at which mass gets
generated from a surface. We found out that the surface area is important - Sb is
important. Then we found for the double base propellant we have two or three zones
during their burning: like a foam zone followed by a fizz zone in which temperature
increases to some intermediate T1, then a dark zone in which the temperature is more or
less constant followed by a zone in which temperature further increases.
We looked at the events happening at the at the fizz zone and not in the foam zone. We
wrote an equation and we were able to say that the velocity of the gases in this zone are
dependent on pressure and the initial temperature may be the T1, may be the surface
temperature Ts, activation energy E. And since the velocity of gases is directly
proportional to the burn rate r of the propellant, we said that the law is r = a pn which is
known as Saint Robert’s law or Vielle law.
Thereafter we found that since we have the reactions happening directly at high
pressures; that means, a dark zone is not present at high pressures and therefore at high
pressure the temperature increases from surface temperature directly to the flame
temperature and more heat gets conducted. Therefore, n is higher at the higher pressures.
Therefore you have small n followed by a large n for the double base propellants. If we
add some salts and alter the mode of burning we could have something like a plateau
burning propellant or a mesa burning propellant.
In the next class, we will take a look at composite propellants, and then put the whole
thing of burn rate together and then address the design of a solid propellant rocket.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 21
Burn Rate of Solid Propellants and Equilibrium Pressure in Solid Propellant
Rockets
Good afternoon. We will quickly recap what we were doing earlier, and then go to the
burn rate of composite propellants, and also address whether there are some particular
values of n, the exponent in burn rate law r is equal to a pn ,which is necessary, when we
design the solid propellant rockets.
(Refer Slide Time: 00:35)
What we did in the last class was that we considered the propellant grain which is inside
the case and the nozzle, as shown above, which constitutes the solid propellant rocket.
We ignite the surface of the propellant, the surface burns or regresses at a particular rate
which we call as the burn rate.
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(Refer Slide Time: 00:51)
And what did we do for the case of double base propellants? We told that we have
nitrocellulose and nitroglycerin in the double base propellant, we had NO2, aldehydes
and also some other constituents, which are formed during the combustion. The
temperature increased from the surface temperature to the temperature at the edge of the
fizz zone. This was the foam zone, which is preheated solid surface of the propellant
from which the gases got generated. And then you had the fizz zone where the
temperature went to T1; thereafter it remained constant at this value in this dark zone
where essentially the reaction between NO and NH2, NO and CO do not generate much
heat. But thereafter the reactions generate heat and you get CO2, CO, H2O and in this
zone the temperature went to the final value Tf. This zone, this particular zone of dark
zone is there only for pressures less than around 10 MPa or 100 atmospheres. And at
higher values of pressures, what we got was that the temperature directly goes to the
value of Tf.
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(Refer Slide Time: 01:55)
Having said that we wanted to translate this information into the burn rate and therefore
we again looked at the preheated zone, a foam zone, a fizz zone, a dark zone and a
luminous zone as shown in the slide above.
(Refer Slide Time: 02:09)
We wrote the equation in this particular zone, the fizz zone, and what we wrote was for a
small element in the fizz zone. This was at x from the surface of the propellant though
the, fizz zone extended from x = 0 upto x = L. The length of the fizz zone is a fraction of
a mm. We just show it exaggerated to be able to visualize the phenomenon. We had this
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small element here with unit surface area and we said heat enters and heat leaves it. The
heat leaving is in excess by some amount. The enthalpy, per unit volume is equal to ρg
into ug into specific heat into temperature. The gases leave the element at a temperature
T plus dT and therefore, we equated the excess heat transfer and the heat generated in the
fizz zone and we were able to write the equation for heat balance.
(Refer Slide Time: 02:55)
And based on this equation, we said we could get the temperature profiles for given
values of ug, pg and ρg. We related the burn rate to the pressure.
(Refer Slide Time: 03:06)
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We also found that in the region of pressure for which we have the dark zone, the
variation of the logarithm of the burn rate versus logarithm of pressure was slightly less,
because the temperature T1 which supplies heat to the surfaces is lower than the final
temperature Tf supplying heat in the fizz zone. And therefore, we have low exponent
here at low pressures a higher exponent at higher pressures as shown.
(Refer Slide Time: 03:38)
We would like to do a similar analysis for a composite propellant. But a composite
propellant is distinctly different from a double base propellant. What is the distinction
between the two?
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(Refer Slide Time: 03:51)
Let us put it on the board. We have in composite propelalnts, AP in the form of crystals.
And this AP is contained in the polybutadiene. We could also have aluminum just like
we saw in this sparkler when there is metal in it and the temperatures are much higher. It
is much hotter and that is why metals like aluminum are introduced in it. Therefore, you
have something like solid crystals of AP and in between may be it is bounded by the
polybutadiene, let say PBAN. Instead of PBAN we could have HTPB or some other
polybutadiene. It could be polybutadiene acrylic acid acrylonitrile or it could be HTPB, it
could be CTPB and these are the fuels or the binders.
And when we supply heat and start making the propellant to burn, AP is NH4ClO4. It
contains oxygen, it contains hydrogen, therefore, we could get a flame something like a
monopropellant flame; this AP itself will burn. But the hydrocarbon which is over
adjacent to it may be the polybutadiene cannot burn because it is just hydrogen and
carbon, it does not contain oxygen. Therefore, the volatiles of hydrogen and carbon from
the polybutadiene are generated above it.
Therefore, let me use a slightly different color chalk over here to show this; may be this
is the fuel vapor which is coming out; this is a monopropellant flame and this flame is
typically at a temperature of around 1600 Kelvin. Maybe whenever we are talking of the
combustion taking place we presume that combustion takes place at a pressure of the
order of greater than around 10 atmospheres. Because rockets do not generally operate at
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chamber pressure less than this value. Therefore, we are talking of something like 1 MPa
pressure. And at this pressure may be we have the flame of AP which is coming over
here, we have the vapor of fuel coming over here and this is the picture that we can
visualize in our mind of what takes place.
Mind you, in practice when I look at the propellant, I will have AP all over the place. I
have small particles of AP and I could have aluminum in between. I have all these
particles and we are just magnifying this zone and expanding this distance and putting
this scheme to be able to formulate a model. We are interested in having a model, which
we can express as an equation and solve the equation for burn rate. That is we have fuel
vapor coming over here. We have oxidizer rich vapor burning and when the fuel vapor
the burning oxidizer rich gases meet a little later on, we could have something like a
zone of burning again. We have a zone over here wherein again burning will take place
between the oxidizer rich flame which is coming from AP. and here mixing will be
taking place between the oxidizer rich gases and the fuel vapor or a mixing dominated
combustion will take place. Whereas in the AP flame above the AP crystal, it is just
premixed.
Whatever AP decomposes, its vapor comes out as decomposed products of combustion.
i.e.,as premixed combustion; there is no mixing involved. At the edges of the AP crystal,
the fuel vapor is formed from the binder and the oxidizer rich vapor are transported from
AP flame. They meet, mix and burn and therefore we call this mixing dominated
combustion as diffusion flame. Essentially the flame is controlled by diffusion process.
Therefore, above the vapor we have premixed combustion, at the edges of AP wherein
the fuel vapor comes and meets these hot gases we have mixing dominated combustion.
And the temperatures here are typically at these pressures - greater than about 10
atmospheres - around something like 3200 Kelvin. Well, the gases are still reacting,
these reacting gases meet the oxidizer rich gases again and products of combustion are
again formed. Products from this lower diffusion flame come over here again meet the
oxidizer rich gases and therefore we have another mixing combustion here on top. This is
the final diffusion flame. And the temperature here is typically around 3500 Kelvin.
Therefore what is the picture that we are trying to create? We are trying to draw a picture
wherein we picture in our mind viz., , the mind model. We have AP giving the oxidizer
rich products, which are essentially decomposition products and are premixed. They
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form a diffusion flame at the edges. The first diffusion flame is formed at the edges and
we again get a diffusion flame over the AP flame, a final flame. And typically, the
temperature of the first diffusion flame is about 3200 Kelvin, while for the AP flame the
temperature is around 1000 Kelvin. The highest temperature is reached in the final
diffusion flame. This could be my mental picture or the model of combustion or burning
taking place in a composite propellant.
Well, this is distinctly different from what we had for the double base propellant,
wherein we had aldehydes, NH2, NO reacting to give a fizz zone, you had a dark zone,
you had the second luminous zone. In the case of composite propellants, we have a
premixed zone, a first diffusion flame and a second diffusion flame. Now, how do we
solve this? It becomes a little bit tricky.
(Refer Slide Time: 09:45)
For the composite propellant, it is becoming a little complicated. Because we have a
series of oxidizer particles, we have something like premixed combustion taking place
just above. It is associated with low temperature. We have something like a diffusion
flame over at the several edges above AP crystals over here, may be something like this
as shown. And then we have a final diffusion flame on top.
Now, what is it that we are discussing? We have AP here, the polybutadiene over here
and this is what I expect. If we have AP of smaller particle sizes, the zones will be
something like this near to the surface and the final diffusion flame and heat release zone
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will be nearer to the surface. In other wards it tells us that the height of the final diffusion
flame will be proportional to the size of AP. If AP size is less, then the final diffusion
term will be nearer. We still have these diffusion flames over here, the premixed flame
over here. Therefore, AP size will decide the distance at which the final diffusion flame
or heat release will take place.
Now, let us go through the assumptions. Let us assume that the final temperature of the
combustion products is Tf. And now I want make an assumption involving the three
flames and how do we do it? We cannot have all these premixed and diffusion flames
and solve the equations. Why not make an assumption? I say this my propellant surface,
it is a composite propellant; I know the size of the AP crystals are around 300 microns,
the fine AP is around 30 microns. We cannot see that closely at the micro level. Now,
we tell that the final diffusion flame formed is at a temperature Tf and it is formed at
some distance away something like let say X*, a standoff distance. It forms after some
distance from the propellant surface. Why does it form after some distance? Why does it
not form at the surface?
Well, we first have the A P flame first which forms over AP crystals, then the mixing
takes place then the secondary mixing takes place and then the final diffusion flames
forms at some distance away. But this distance again is a fraction of an mm or so in
practice, just like in the sparkler. It looks as if the burning is at the surface, but if we take
a magnifying glass and see well there is a distance above the surface at which the
burning takes place.
Therefore, now we are in a position to write an equation for this simplified scheme. The
final temperature is Tf. This is the propellant surface. We are interested in writing an
equation for the burn rate so many meters per second.
We can readily do this. Let us let us look at the slide again. This is where we have three
AP crystals over here. We have the shaded portion which is the monopropellant flame or
the premixed flame. And then I have the vapor coming from this hydrocarbon, this is
pure vapor when it mixes with the oxidizer rich gases, I have a zone of diffusion flame
over here. And the products from this diffusion flame and the AP monopropellant give
me a final diffusion flame here. And the distance between this final diffusion flame and
the surface is what we call as the standoff distance or the height X*.
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(Refer Slide Time: 13:19)
If this part is clear, let us go to the next slide. We say that this my final zone of
combustion at a standoff distance X star and this is my surface and the surface
temperature is Ts. Mind you the propellant get heated here from initial temperature to the
surface temperature and in this gas zone, I say that the temperature varies between Ts to
the final value Tf. This is a mental picture that we use or model for the combustion or
burning of a composite propellant. At this point I thought I should illustrate what
happens, many of you are working in combustion. These are the experiments which we
were do in the lab; I have something like butane gas coming over here.
And when we ignite butane gas issuing from an orifice, at the orifice itself we seem to
have a flame like this. We increase the velocity of butane gas flows and combustion
takes off after a certain distance. I have mixing taking place in this zone and thereafter I
have combustion. At still at higher velocity, we have this zone of standoff wherein
mixing taking place is at high velocity. The heat is insufficient to propagate the flame
into the stand off distance. We have something like a standoff distance here. Well, the
standoff distance is something on these lines, but not precisely because of velocity in the
case of the composite propellants. But it takes a certain distance for the final diffusion
flame to form.
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(Refer Slide Time: 14:45)
Therefore, I would like to write an equation for this configuration. We have the
propellant surface here. We have the final diffusion flame over here at a distance X*
away. The temperature of the final diffusion flame is Tf and the surface temperature is
Ts. If we were to plot the temperature distribution along this stand off distance; this is the
propellant surface at temperature Ts; this as the distance over here; in the depth of the
propellant the temperature is the initial value that is the initial temperature. And then
what happens near the surface, the temperature increases to Ts at the surface and then in
the flame zone, it further increases to something like Tf. Mind you this is my increasing
direction of T and this is the distance from the in-depth of the propellant wherein at the
depth of the propellant is still at the initial value of the temperature. We have an
increasing temperature to Ts and then goes to the gas value Tf.
Therefore if we want to know what is the heat, which is coming on the surface, we need
the value of thermal conductivity of the gas above the surface kg × dT/dX at the surface
of the propellant. And the gradient in temperature can be approximated by the
temperature increase to Tf from the surface value Ts over a distance of X star to be
linear. We use this X* as the flame standoff distance and we shall see what it represents
subsequently. The heat transfer to the surface per unit area is equal to kg ×(Tf – Ts)/X*.
This is a very simple way of estimation where we assume the gradient to be linear. There
are various flame models which are used for describing the combustion of composite
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propellants. But this particular one was formulated by Professor Hermance at the
University of Waterloo in Canada. It is very simple and is very illustrative.
There are various models; we have the granular diffusion model and other models, but
this simple model gives a reasonable picture of the combustion behavior. This is the rate
of heat which is coming to the surface equal to q°s. We consider unit surface area and
therefore it is so much Joules per second per meter square. This is the unit of heat flux.
Mind you, we consider unit surface area. Where does this heat go? The heat goes to
increase the surface temperature from the initial value Ti to the Ts. If we say that the rate
at which the propellant regresses or burns is equal to m° then what happens? The heat
corresponding to temperature increase into the specific heat into it has to be accounted
for.
Suppose, at the surface I have some endothermic reactions taking place because I have
the binder, which we said is polybutadiene. It has to get heated. It also requires heat to
vaporize it. We need to supply this rate of heat q°chem to it. So, some endothermic heat of
reaction q chemical is supplies to the surface.
(Refer Slide Time: 18:13)
Therefore, we can write the heat balance equation as thermal conductivity of the gas kg ×
the flame temperature − the surface temperature of the propellant ÷ the flame standoff
distance X* = mass which is getting released at the surface × the specific heat of the
propellant × the surface temperature changes to Ts from the initial value Ti plus the heat
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which gets released at the surface might be an endothermic reaction at the surface q°chem.
And therefore, this gives us the energy balance equation. Namely, this is the heat, which
gets transmitted from the flame to the surface. And this helps to increase the surface
temperature to Ts from the initial value and also supplies the energy required to vaporize
this surface or to convert the surface from solid to vapor through a set of endothermic
reactions.
Therefore, now, using this particular equation, if we have write the value m°: , m° =
kg(Tf−Ts)/X* ÷ {Cps(Ts−Ti) + qchem} .. Actually, in this we need not even put q° here
because it is just the magnitude of heat release and therefore, maybe we should do away
with the dot here. And just write kg( Tf− Ts)/X* ÷ the sensible heat required plus the
endothermic reactions at the surface.
Now, we would like to discuss this particular value of mass release rate. The mass
release rate at the surface m° per unit area comes from regression of the propellant and is
equal to ρp into the burn rate r. Or rather the burn rate r can now be written as r = [kg(Tf
−Ts)/X*]/ {ρp [Cs(Ts−Ti) + qchem]}. The value of r is in meters per second, ρp in
kilograms per meter cube.
Now, let us examine this equation under different conditions. We find that there is a
flame, which is standing off at a distance X* from the propellant surface. Let us again
sketch it. We have seen this sketch several times during this class. We have the flame at
a temperature Tf standing at a distance X* away from the propellant surface. If the size
of the ammonium perchlorate particles are small, then the mixing will take place
immediately near to the surface and the flame will be near. X star would then be small.
This is point one. If the chamber pressure, on the other hand, or if the pressure at which
the burning rate is evaluated is high, we have more number of molecules which can react
and therefore, the X* will come down.
Therefore, what is the inference which can we can draw? Well, if we have fine AP
particles, fine ammonium perchlorate particles in the composite propellant, then it is
quite possible that X* will be smaller and the burn rate will be higher. If the value of
pressure at which the burning takes place is somewhat higher or large let us say, then
chemical reactions get finished in a very short time and therefore, X* will be small. And
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if X* is small well, the burn rate will be higher. Therefore, a higher value of chamber
pressure also leads to a higher value of the burn rate r.
If the pressure is higher, the thermal conductivity of the gas will be higher and therefore,
again burn rate will be higher. ρp is the density of propellant and does not change. The
surface temperature effects will come in, may be the endothermic reaction activation
energy will come in, but we find that pressure is a major factor because pressure decides
the value of X*.
When we did double base propellants, we found that the burn rate law could be
expressed as in terms of Saint Robert’s law as r is equal to a constant ‘a’ × pressuren. We
find here also as pressure increases the flame comes nearer and therefore, a similar law
viz., r = apn can be used to determine the burn rate of composite propellants.
Now, depending on the value of n the effects of pressure are modeled, but what is the
value of ‘a’? a will depend on the size of ammonium perchlorate particle size. The
activation energy of the endothermic reactions and may be some of the compositional
aspects like initial temperature of the propellant. As the initial temperature increases, we
find that the denominator decreases and r increases. Therefore, the effect of the initial
temperature and the activation energy would also influence the burn rate. The
ammonium perchlorate particle size and the other parameters are embedded in the value
of ‘a’. Therefore, we can conclude that the same law r = apn which describe the burn rate
of double base propellants could also used can be used for composite propellants.
The burn rate of composite propellants can therefore be expressed in this particular form
r = apn. Very simple;but a very illustrative derivation is used. We must be able to write
such expressions for any system such as wood smoldering, may be carbon or may be
charcoal burning.
Let us let us summarize it again. We said we had a flame zone at a distance X* away
and can be assumed though we had two zones like premixed zone and the diffusion
zones. And we got the expression for r and let us now go back and examine the value of
r, how should it change with pressure? Could I rub this off? Is it is it clear?
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(Refer Slide Time: 25:36)
Now, we would like to use the equation and plot the value of the burn rate, logarithm of
burn rate versus logarithm of pressure. Because we said that for composite propellant
also the burn rate can be described by Vielle or the Saint Robert’s law which is r = apn.
Well logarithm of r as a function of logarithm of pressure should be a straight line as per
this equation. But if we burn the propellant at low pressures, i.e., operate in the low
pressure region of pressures less than some threshold value, p*. We are yet to define this
value of pressure; however, we say at low pressures. What happens to the burn rate at
low values of pressure? May be AP will decompose and the hydrocarbon vapors are
being generated.
And what happens to the rate of decomposition at low pressures? It is limited by the
kinetics; why kinetics? We write m°chem = Apm exp(−E/R0T) and therefore, at lower
pressure, we have lesser amount the decomposition and less of energy getting liberated.
Or this becomes the controlling parameter because the decomposition generates vapor,
but the vapors are in short supply. Therefore at low pressures, we presume that chemical
kinetics and AP decomposition controls the rate of burning. Sufficient molecules are not
available for the reaction.
When we talk of higher pressures; the diffusion process is independent of pressure and at
higher pressures we get copious amount of fuel and oxidizer vapor coming out because
the reactivity has increased. But the mixing of gases is independent of pressure and
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therefore, mixing now is an impediment and controls the burning. And therefore, in this
zone of high pressures, we have something like mixing controlling the final diffusion
flame.
All what we now state is we may not get the same value of n over the entire region of
pressure as was done earlier. At low pressures, we get premixing or a premixed flame
dominating and this is the limiting factor, which controls the burn rate. Whereas, at high
pressures, diffusion is limited and therefore it becomes the controlling factor limiting the
burn rate. The diffusion process does not allow the availability of fuel and oxidizer
vapors to mix and burn. Therefore we say that the burn rate at high pressures is diffusion
or mixing controlled.
In other words, if we were to again re-plot burn rat edependence on pressure, we have a
region in which we have premixed combustion controlling. And at higher pressures, we
have something like diffusion mechanism controlling. And this pressure is of the order of
15 atmospheres or so. And what happens is whenever it is premixed control, we have a
higher value of n because it goes the rate of a reaction goes as p to the power of an
exponent And if it is diffusion controlled we have something like n which is very much
lower.
(Refer Slide Time: 29:04)
Therefore, for composite propellants, what happens? It is different from double base
propellants in that if I were to plot the value of logarithm of burn rate versus logarithm of
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pressure, I have something like high value of ‘n’ at the lower pressures and smaller value
of ‘n’ at the higher pressures. Or rather the burn rate goes something like convex
upwards. What did we have in the case of double base propellant? It was just the
opposite; we had at higher values of ‘n’ at higher pressures due to the absence of the dark
zone. In the case composite propellants, it is the reverse and we have at the lower
pressures a higher value of n while at higher pressures, we have a lower value of n.
This distinction must be clear. But normally most of the rockets are operated at pressures
excess of this threshold value of pressure. And therefore, we will say that n typically is
around 0.25 to 0.35 for the composite propellants. In the lower pressure region, wherein
n is nearer to something like 0.4 to 0.5, we have premixed combustion dominating. I
show this in this particular slide wherein at a pressure less than a threshold value, we
have premixed combustion dominating with a higher value of n followed by diffusion
combustion dominating with a lower value of n.
With this background, let us determine the choice of ‘n’ necessary in the choice of a
propellant.
(Refer Slide Time: 31:00)
What must be the value of n which gives stable burning? Based on our understanding of
the burn rates, can we recommend certain values on n? Before answering this question,
what should be the burn rate law for a composite modified double base? I know for a
double base propellant, the burn rate law. I know the law for a composite propellant. We
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can say composite modified double base should incorporate both these feature and give
something like average of this two. In other words, composite modified double base will
give a value of n which is slightly lower than for double base, for double base you know
it is all premixed combustion.
The value of ‘n’ is high for double base; for composite at the regions of interest ‘n’ is
less. For the composite modified double base propelalnt “n’ will be smaller than for
double base propellant. If we were to consider something like nitramine propellants well,
in nitramine propellants we had the explosive HMX, we had binder which were mixed
together both of them are premixed and therefore the value of ‘n’ should be high. That
means, in the burn rate law, r = apn, n is higher for nitramine propellants, for CMDB it is
less than for double base propellants, but higher than for composite. In this way one
could address the values of the burning rate index ‘n’.
We should work out some numerical problems for burn rate and put things together. But
what we have done is that for double base propellants, we looked at the evolving layers
like fizz zone, dark zone and also the second luminal zone. For composite propellants,
we used the simple Hermance model wherein we talk of a flame standing away from the
surface by a standoff distance X*. We also know what this standoff means. When a gas
burns, it take some time for the constituents to chemically react and form a flame.
If the burn rate r is equal to a p to the power n, can we now assess whether n should be
small n or should be large? Should we have n equal to infinity or should we have n is
equal to 0, what is the value, which will give proper burning in a rocket?
Let us consider a solid propellant rocket. Well! this is the propellant we have in a rocket
as shown. This is the burning surface area Sb. In other words I have if we take a section
over here, this is my surface, this is my burning surface area Sb meter square. And this is
where the burning takes place and gas is getting released from the surface. And the gas
gets pushed out through the particular nozzle.
Let us put some numbers down. We will call rate at which mass is getting generated
from this surface area as equal to m° generated m°g. The rate at which gases are leaving
the nozzle, we will call it as m°n.
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Can we write an equation for the dependence on m°g and m°n? Let us consider the case
wherein we have a certain volume. Let say this particular volume of the port be V. Let
the pressure in this volume be p and the temperature of the gases be T. How do we write
an equation for mass balance? Mass gets generated over here m°g it leaves the control
volume or this volume at a rate m°n. What is the equation that we get?
(Refer Slide Time: 35:47)
You immediately tell me that m°g −m°n = the rate of accumulation of the mass of gas in
the particular volume or the chamber. What is the value of mass from gas equation? We
have p V = m R T. The pressure is p, the volume is V and the temperature is T.
Therefore, m = pV/RT. Therefore, we can now write d/dt [pV/ R T] = rate of
accumulation of mass in the volume.. Is it alright?
Let us simplify this term before I come to the other terms? How will I simplify this
particular term? Temperature is a constant viz., the flame temperature; pressure is a
variable. As the propellant regresses the volume varies. Therefore, we have two variables
pressure and volume while the temperature is constant.
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(Refer Slide Time: 37:36)
Therefore, for temperature being a constant or a specific gas constant R, we can write
this equation d/dt(pV/RT) = 1/RT × {V ×dp/dt + p ×dV/dt}. Again we simplify this and
write it as V/RT ×dp /dt + p/RT×dV/dt. What is p/RT equal to? Let us take a look at the
gas equation or equation of state pV = mRT, giving p by R T as equal to mass/volume
which is the gas density. We can write this as equal to V/RT × dp/dt + density of the gas
ρg ×dV/dt.
What is dV by dt? Rate at which the volume is increasing and what is the rate at which
the volume is increasing? The burning surface area is Sb, regression rate is r and
therefore, dV/dt = burning surface area × the propellant burn rate so many meter2 ×
meter/second, i.e., meter3 by second. This is dV/dt. In one second, the distance would
have been moved by r. Therefore, the increase in volume dV/dt = r ×Sb. Therefore, we
write the left side of the equation as V/RT { dp/dt plus the term ρg ×Sb × r}.
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(Refer Slide Time: 40:08)
This is equal to what m°generated minus m°nozzle which is leaving.
The value m°g can be expressed in terms of pressure or in terms of burn rate? The
burning surface area Sb × r × ρp because Sb × r tells the volumetric rate at which the
propellant is getting consumed and if it is multiplied by the propellant density ρp, it
represents the mass at which is getting generated. This is equal to Sb×ρp × a pn . Here the
burn rate law r = a pn is used.
What is the rate at which mass is leaving the nozzle? We have been doing it all along.
We had defined C* as solid propellant property that is m°n = pC At / C*. . C* star is the
transfer function between chamber pressure and mass flow rate per unit throat area. And
therefore, let us just substitute it and get the equation as:
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(Refer Slide Time: 42:41)
We substitute this again to get V/RT × [dp/dt] = Sb×ρp × a pn − p ×At/C*. Here we
wrote r = apn from the burn rate law. What is it we get? We get dp/dt = RT/V{Sb×a pn
×(ρp − ρg) − p At /C*}. All what we have done is that we brought ρg on the right side,
because it is with the minus sign now.
Now, we want to know what happens under steady conditions. By steady conditions, we
mean the pressure with respective time is a constant, pressure does not change and
therefore, dp/dt = 0. We therefore get:
(Refer Slide Time: 44:20)
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The value of Sb×an × (ρp − ρg) = p At/C*. If we were to call this steady value of chamber
pressure which, is actually the steady equilibrium value in the chamber, the value of
equilibrium pressure p equilibrium is determined as {Sb×a × ( ρp − ρg) ×C star /At}1/(1−n).
We take here 1 minus n and this becomes 1 over 1 minus n. Please check whether it is
right. We take p on this left side, it becomes p into 1 minus n and I still retain Sb a ρp −
ρg and I take C star upstairs over here and At downstairs and this is equilibrium pressure.
When we burn a propellant in a rocket chamber, whatever be the configuration of the
grain, the equilibrium pressure or steady state pressure is given in terms of the burning
surface area, the pre-exponent a, the difference between propellant density and the gas
density and the C star of the propellant divided by At raised to the power 1/(1-n). This
gives the value of the equilibrium pressure.
Normally the gas density will be very much lower than propellant density. In fact, had
we neglected the gas density, we would have got this expression by equating the mass
generation with the mass leaving through the nozzle. We would have got this to be equal
to (Sb×a × ρp × C* /At)1/(1−n). This should have been the value of the equilibrium
pressure. The gas density would be around one thousandth of the solid propellant
density. Therefore, what are the conclusions that could be drawn?
(Refer Slide Time: 46:46)
If n has a value near 1, what will happen to this equation? If n has a value of 1 well, we
have infinite pressure it is just not possible. If n is around 0.995 very near to 1, then we
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have a really large exponent and small perturbations or changes in this will give me a
high value. Therefore, a value n near 1 is not acceptable. That means, n must be as small
as possible. And any changes in burning surface area while the propellant is burning,
some changes in the gas density while it is burning, some changes in C* or throat area
should not lead to an explosion by giving an abnormally high value of pressure.
Therefore, we tell ourselves the one of the quality is required in the burning rate of a
propellant r = a pn, is thatn must be a small number. A number around 1 is just not
acceptable. And since for composite propellants the value of ‘n’ is around 0.25 to 0.4
while for double base it is around near to 0.4 to 0.5, therefore composites are better off
than double base propellants.
(Refer Slide Time: 48:21)
Let us also plot the rate of mass generation and mass leaving the nozzle. We have the
mass generation rate of propellant as a function of pressure. m°g = apn × Sb × ρp.
Therefore, it will go up progressively as shown. If we have n which is greater than 1,
this will be my shape. If I have n less than 1, what will be the shape of the curve?
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(Refer Slide Time: 49:25)
We must be able to do these things that is the basic aim of the course. The curve begins
to droop as the pressure increases to higher values.
Now, what is the value of m°n? m°n is linear function of p viz., p ×At /C*. This is a
straight line in the plot versus p. Now, we put both m°g and m°n on the same plot and let
the two intersect at a point. Let us take a look at this point of intersection. What is this
point wherein both are equal; steady state i.e., p equilibrium. And we know the
expression for the p equilibrium, we have just derived it as saying Sb×a×C*/At to the
power 1/(1−n).
Now, let us examine the characteristics around this particular point for n > 1. Suppose, if
by chance as the motor is functioning, there is a small dispersion in pressure and the
pressure slightly increases. If pressure slightly increases the mass generation rate
increases. The pressure therefore further increases, it goes like this leading to very high
pressures and the rocket explodes. If pressure slightly falls, the mass flow through the
nozzle is higher compared to mass generation rate and therefore the pressure comes
down. If the pressure decreases, the mass generation rate further decreases and the
process continues till the rocket ceases to function. Therefore, when we have n > 1, we
cannot get equilibrium, the motor either quenches or it explodes. Therefore values of n
greater than one cannot be recommended.
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Let us get back to the curves for n < 1 and examine it again. If there is a slight increase in
pressure, what is going to happen? The pressure has increased the nozzle flow rate has
increased compared to the mass generation rate and therefore, the pressure again comes
to the point. If the pressure falls slightly, the mass generation rate is higher, it gets back
to this point. Therefore, the point of intersection becomes a stable operating condition.
Therefore, now, we are very clear that ‘n’ must be very much less than 1 and the reason
for it. We learnt how to do this using these two plots. We also learnt how to do this using
the equations wherein we had the expression raised to the power 1/(1 – n). And this is
what how we decide the choice of n.
I will continue with this in the next class. But to summarize, we derived an expression
for the burn rate of composite propellants using the standoff distance X* . We found that
the burn rate can still be modeled using the Saint Robert’s law namely r = a pn. We
related it to a rocket and we said well n must be very much less than 1. In the next class,
we will look at the temperature sensitivity of r, we will also look at some other
parameters, which are important for the burn rate r, and then go to designing a solid
propellant rocket. And what do we do in a design? We have to have some particular
burning surface area; it becomes a simple geometric problem, and this what we will do in
the next class.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 22
Design Aspects of Solid Propellant Rockets
We will continue with solid propellant rockets. What is it we have done so far? Let us
take a quick review; we know that the propellant could be composite, it could be double
base or it could be nitramine or it could be composite modified double base propellant.
(Refer Slide Time: 00:18)
We can write the burn rate as a linear regression rate, r = a pn. We also said that if we
have a rocket in which the burning surface area is Sb, the equilibrium value of pressure
can be derived as the burning surface area into this particular constant a in the burn rate
law into ρp into C* divided At to the power 1 /(1 – n). How did this come?
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(Refer Slide Time: 01:25)
We said if we had a rocket and we considered a simple scheme where in we had
propellant, which was enclosed in a case, we have something like a nozzle attached to it,
we said the rate at which the mass leaves through the nozzle m°n can be written as p
At/C*, where p is the pressure.
And we said the nozzle is always choked at the throat. Therefore, this is the mass being
generated and this is the mass, which is leaving through the nozzle. And the rate at which
the mass is getting generated from the burning of propellants, we got from the regression
rate is r and that was equal to Sb × the propellant density × the burn rate r, which is equal
to a pn so much kilograms per second. We equated the two viz., mass generation and
mass leaving rates and got the value of equilibrium pressure p1−n = Sb × a × ρp × C* / At.
What does this tell us? Let us take a relook at this equation. We looked at it from the
point of view of n and found that n cannot be anywhere near 1, because then what
happens for any small change in the parameters, there is a large magnification here. And
therefore, n should very much less than 1, because if n is near 1, I get a very large
exponent and a small change can magnify into a large value o pressure. Therefore, we
say from stable considerations, n must be very much less than 1, but we also learnt to
look at it graphically.
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(Refer Slide Time: 03:16)
When we have n in the burn rate equation r = a pn, if n > 1, how does the burn rate law
change as pressure changes. Let us make a plot: as burn rate increases, if you have a
given burning surface area and a given density of propellant we can plot the rate at which
mass is generated due to burning. In other words, the rate at which mass flow gets
generated depends on the burning surface area into the density of the propellant into the
burn rate law; we want to plot it as a function of pressure.
Then if n > 1; well it keeps increasing higher than a linear value i.e., concave
upwards.and the mass generation rate rapidly increases rather exponentially. If however,
n < 1, then the mass generation rate will have something like a drooping characteristic
i.e., convex upwards. This is for n less than 1 with variation of pressure; this is for n
greater than 1. How does the mass flow rate, which leaves the nozzle change with change
in pressure? We have been writing this on and off as m° nozzle is equal to (1/C*) p×At.
The mass flow rate through the nozzle with respect to pressure will increase as a straight
line.
Therefore, let us now plot the mass which leaves a nozzle and the mass generation rate
for ‘n’ is greater than 1 and ‘n’ is less than 1 and see whether we can conclude on the
type of exponent ‘n’ which we require.
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(Refer Slide Time: 05:13)
Let me now plot all the three together in a single figure. Y axis which is mass generation
or let us say mass which is leaving the nozzle while pressure on the X axis. Let me put
the value first for n less than 1 we get a curve like this drooping one. If we have n greater
than 1, the curve exponentially increases. And what is the rate at which mass leaving the
nozzle? I show it by a white line - a straight line like this with respect to pressure. Now
we find the point at which the mass rate of generation for n > 1, and the mass, which is
leaving through the nozzle are the same at this point of intersection. Therefore this will
correspond to let us say equilibrium pressure for the case of n > 1. The red line is for n >
1 therefore p equilibrium corresponding to let us say case 1.
Let me also say tell that may be for n < 1 this is the p equilibrium value. We would like
to examine whether are these two equilibrium pressures are possible? Well theoretically,
this is the rate at which mass is leaving the nozzle, the rate at which mass is getting
produced in the chamber and therefore this is equilibrium pressure. Similarly, for n < 1
this is the pressure for mass balance. Let us try to get some idea whether these two points
are possible and if so are there some problems with the equilibrium pressures?
When n > 1, let us say we have a small perturbation in pressure. A small perturbation can
always come and let say that the pressure reaches this higher value. That means the
pressure is slightly higher than the equilibrium value. Since the pressure is slightly
higher, what we find at this point the mass generation rate is higher than the mass which
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is leaving the nozzle. Therefore, the pressure will increase further; when pressure further
increases, the mass generation made is further increased. Therefore, this point cannot be
a stable equilibrium pressure as any small perturbation will make the pressure increase
further and further till the rocket explodes.
If we considered points to the left of the equilibrium point by saying that by some chance
there is a small pressure perturbation and the pressure in the motor falls to a value less
than the equilibrium pressure pe1, then what happens? The mass generation rate is lower
than mass, which is flowing out through the nozzle. In other words mass flowing out
through the nozzle is more than what is generated. Therefore, the pressure further falls,
the pressure continues to fall till the rocket is extinguished.
Therefore, we tell that in case n > 1, we cannot really get an equilibrium pressure since
with small changes in it due to perturbations, the chamber will either explode or pressure
will become zero. Therefore, we tell that n greater than 1 is not desirable or not possible.
Let us examine the case, when n in the burn rate law, what was the burn rate law? r = a
pn . We would like to know if the equilibrium pressure obtained when n < 1 is possible.
Let us have the same set of arguments again. If the pressure falls slightly less than the
equilibrium value; we now find that the mass generation rate is higher than the mass
which is leaving the nozzle. Therefore the pressure will go back to the equilibrium
value.
If by chance the pressure exceeds the equilibrium value i.e., moves to the right of the
equilibrium point. Here again we find the mass leaving the nozzle is higher than the mass
generation rate and pushes the back to the equilibrium value. Therefore this point
becomes a stable point. The value of n < 1 is therefore possible and it gives to rise to
what we say is a stable situation for equilibrium pressure. Therefore, in the burn rate law
r = a pn ; ‘n’ must be less than 1. This is what we have found.
(Refer Slide Time: 10:00)
541
We need to put everything together and design a solid propellant rocket. What is meant
by design of a rocket? We must be able to generate a given thrust from the rocket, and
what is the thrust? We said it is equal to CF × p × At. We drop the subscript c in pC. We
can express in terms of the nozzle effectiveness into chamber pressure and At and this is
the thrust which is developed. Therefore, if we want a rocket to develop a particular
thrust, we know that p equilibrium goes as Sb × the constant a × the propellant density ×
C* / throat area to the power 1/(1 – n). All what we need to do is to configure the
burning surface area Sb such that we obtain the desired value of thrust. But it is not that
easy as we shall see in the in the subsequent class. We can write the thrust as CF × the
value of p equilibrium from here × throat area; so we get the thrust as as Sb1/(1−n) .
We take p equilibrium as Sb1/(1−n) . Then we write the other terms together namely and
get At. Now At is in the denominator here to the power 1 /(1 – n). And then we solve the
other parameters namely a value of ρp into C* to the power 1 / (1 – n). Therefore, we
find for a given constant throat area, if we know the evolution of burning surface area as
the surface regresses, we can find out the value of thrust varying with time. This is how a
solid rocket is designed. It is a simple geometric problem of evolution of the burning
surface area Sb. And how Sb, the burning the surface area in meter squared, evolves with
time would be dealt with in the class today.
(Refer Slide Time: 12:30)
542
But before we do that, let us recall that the burn rate law r = a constant ‘a’ into p to the
power ‘n’. We considered explicitly the effect of pressure alone. But we said ‘a’ includes
the effect of the initial temperature of the propellant. Let us take an example. We can
consider a temperature of the propellant to be ambient that is a rocket motor is tested
today. The temperature is quite hot today may be 32o C. Well, you all would read about a
solid propellant rocket, which misbehaved in one of the space shuttle flight. It was a
challenger rocket, which was launched on a very cold day, when the temperature was
around 0oC. And we will look at the failure after completing the portion on solid
propellant rockets. Well, we could have a missile, which is operated from mountains in
the Himalayan ranges where the temperature could be as low as minus 50oC.
What is the effect of burn rate on temperature that is the initial temperature of the
propellant itself? We are not looking at the flame temperature all. What we say is that we
have a propellant block, the initial temperature of this block before it burns or just begins
to burn is what we call as initial temperature of the propellant. We would like to know
the effect of initial temperature of the propellant on the regression rate of the particular
propellant.
(Refer Slide time: 14:00)
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We define a term known as temperature sensitivity factor for the propellant. Let us see,
what it is. We would like to know how sensitive the burn rate is to the propellant
temperature changes. Therefore, we were interested in finding out the change in burn rate
with temperature dr/dT. And we are considering the effect of temperature alone. This
implies that the pressure is fixed; we are considering at constant ambient pressure or
otherwise. But then instead of just saying burn rate variations with temperature or the
variations in burn rate due to unit change of temperature, we now say fractional variation
in burn rate. This is known as temperature sensitivity factor for burn factor r. In other
words, we define a term πr as equal to this dr/r i.e., d ln r divided by dT at constant
pressure. This is defined as the temperature sensitivity factor for a solid propellant.
And just like how we determined ‘n’ by conducting two experiments on burn rates at
pressures p1 and p2 and measured the burn rate r1 and r2 and n = (ln r1 − ln r2) / (ln p1 −
ln p2), in the same way the temperature sensitivity of burn rate is measured at different
temperatures at the specified pressure. The factor πr, which defines the sensitivity to
temperature is determined. The value is around 3 × 10−3. What should be units?
Well ln r has no units; dr by r; the units cancelled and it is only dT. Therefore oC inverse
are the units. Typically for most composite problems the value is around 3×10−3 oC−1 and
about 5 ×10−3 oC−1 for double base propellants. And for HMX based propellants, it is
even lower; it is 2 ×10−3 oC−1 . This is one of the reasons for the choice of HMX
propellants for missiles.
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We can integrate this equation for temperature sensitivity and find out explicitly, how the
burn rate changes with temperature. Let us do it. We take the expression for πr. We write
d ln r = πr × dT at constant pressure. We have taken the change in the logarithmic burn
rate is equal to πr × dT. Let us solve this equation. If we have at temperature T1 the burn
rate as r1 and at temperature T2 the burn rate is r2 and we are interested in finding out the
burn rate at a temperature T2; and therefore, we just integrate out this to get ln r2 − ln r1 =
πr (T2 − T1).
(Refer Slide Time: 15:53)
Therefore ln (r2 / r1 ) = πr (T2 − T1) or rather we get r2 / r1 = exponential of πr (T2 − T1).
Therefore, if I know the burn rate at temperature T1, using the value of the temperature
sensitivity factor we can find out the burn rate at a temperature T2 and this is how the
effect of the variations of temperature are taken into account.
We are now in a position to design solid rockets. This means essentially we find out how
much burning surface area is required and how it should evolve with time. However, let
us ensure whether there any concerns about the burn rate equation which is needed for
the design.
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(Refer Slide Time: 19:00)
The burn rate is expressed in millimeters per second or meters per second and is given by
a×pn. What is the unit for r? We say meters per second, millimeter per second or
centimeters per second. The burn rate is the rate of regression of the propellant surface.
What is the unit of pressure? Could be Pascal. It may be mega Pascal, could be
atmosphere also. Then what is the unit for a? The constant ‘a’; it may be noted, depends
on temperature and on composition of the propellant including the AP particle size.
The unit of a is clumsy; it becomes meter per second divided by let us say Pascal raised
to the power n. This is not a correct way of expressing a constant. We have a constant,
which is a function of the parameters and the units of pressure. We cannot say that the
constant is so much meter per second to the power of pressure to the exponent ‘n’. How
do we get over this problem? The form of equation is, however, correct. If we can find
the burn rate let us say at pressure, which we call as reference pressure, and we evaluate
the burn rate r with reference to this particular value p reference to the power n, the
problem can be overcome.
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(Refer Slide Time: 20:44)
But we are interested in burn rate r at any pressure p as a into pressure the power n. We
can write the value of r as equal r reference × p by p reference to the power n and this is
one way we get over the units of pressure in the constant ‘a’. In other words, the constant
‘a’ is the burn rate at the given reference pressure provided the reference pressure is used
for non dimensionalising the value of pressure. This is how the burn rate is expressed
through a non-dimensional pressure, which is based on the reference pressure. The
reference pressure is normally taken as 70 atmospheres, which is about 1000 psi.
You may recall that when we studied nozzle we said under sea level conditions and
vacuum conditions for evaluation of the specific impulse; we took the chamber pressure
as 70 atmosphere. This is about the pressure at which a solid rocket or a good performing
solid rocket works. This is equal to 7 MPa. And therefore, the burn rate law can now be
written as r at a reference pressure 70 atmospheres or 7 MPa into pressure divided by 70
or 7 provided p is in atmosphere or MPa to a power n. This is at 7 Mega Pascal pressure.
Some books write the value of ‘a’ as r at 7 MPa or 70 atmospheres into p divided by 7 or
70 to the power n.
The constant ‘a’ is therefore denoted by a7 or a70, which is the burn rate at the reference
pressure of 7 MPa or 70 atmospheres. This is all about burn rates, effect of temperature
on burn rates, etc., but there are many more problems which would be considered later
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like for instance in a rocket chamber; there could be velocity at the propellant surface,
there could be thermal radiation in the chamber, there could be external heating.
We now return to the design of a solid propellant rocket for which, we wanted to find out
the burning surface area and the evolution of the burning surface area. Let us do a simple
problem, and then go to the evolution of the burning surface area to give a certain thrust
and thrust profile. Let us consider a propellant block, which is contained in the rocket
case and we put insulation and connect a nozzle here. Well I have a solid propellant
rocket.
If this propellant block is ignited over here towards the nozzle, and it burns from the
exposed end, that means the end of the propellant is ignited. We call it as end burning,
because it burns from one end to the other; it does not burn from the sides here, because
it is prevented from burning from the sides. The burning or the flame can go normally in
this particular direction. Now supposing the throat area of the nozzle is At. What is the
value of pressure in the cavity? We have already done it as equilibrium pressure. It is
equal to let us write it down: (ρp × a × C* × Sb / At)1/(1−n). The burning surface area is
known, and we determine the pressure in the cavity. The value of burn rate r is equal to a
pn and knowing the pressure, we can find the burn rate. And therefore, if the propellant
grain has a length L, the time taken for the propellant to be consumed can be determined
as L / burn rate. The burning surface area Sb is a constant and therefore the pressure p
remains the same.
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(Refer Slide Time: 24:11)
And therefore, the burning time tb is equal to the value of L, the length of the grain
divided by the burn rate r and is equal to L divided by a pn. The thrust developed by this
particular end burning is CF into pressure into At. We know the value of pressure, the
throat area and the burning surface area and the thrust developed can be determined.
We said that solid rockets are generally used when we want large thrust as in booster
stage. Suppose we want a thrust of several 1000 tones. Then in that case, the burning
surface area and hence the diameter of the solid propellant rocket is going to be
extremely large.
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(Refer Slide Time: 27:22)
Therefore to be able to get some meaningful values of large burning surface area, we
need some innovations. We have the propellant block; the same propellant block as in
the end burning solid propellant rocket. But instead of burning it from the end, we burn it
radially. We make a cylindrical hole along the axis in the propellant and again put it in a
motor case with the nozzle. The propellant block now looks as shown with a central hole
and we coat the propellant block at the nozzle edge with some material, which is an
insulator. This will prevent it from burning on this side. We call this insulator, which
prevents or inhibits the burning viz., an inhibitor. We coat it with the inhibitor, which
will prevent the propellant from burning on the nozzle face of the propellant. We ignite
this inner cylindrical surface, which now becomes the burning surface area.
If we take a cross section, what is it we get? We get this outer surface, we have the case
over here, and then we have the inner diameter over here, and this is my propellant in
between. We ignite this internal surface of the propellant, and the propellant burns
normal to the surface; it therefore burns radially outward, and this type of burning is
known as radial burning. The propellant block is in the form of a cylindrical annulus
between the outer and inner diameters, and now we ignite the inner surface area of this
particular annulus of propellant. And then what happens is the burning will progress, let
us say normal the inner cylindrical surface towards the outer cylindrical surface.
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(Refer Slide Time: 30:25)
And therefore, now we have the initial burning surface area. If we were to write an
expression for this burning surface area; this is the length L of the grain; the annulus is
between the inner and the outer diameters. We take inner diameter as Di, outer diameter
is Do, and the length of the propellant grain is L. So, how does the burning take place? It
takes place radial, normal in other words to the surface as burning proceeds the burning
surface area changes. We need to determine the pressure.
The initial burning surface area is equal to the perimeter into length. It is equal to πDi
which is the perimeter × L. And what is the final burning surface area? It is equal to π Do
the perimeter × length L. And therefore the equilibrium pressure to begin with
corresponds to (πDi L ×a × C* / At)1 /(1 − n).
The throat area of the nozzle is equal to At. What is the change we are made as compared
to the end burning grain? We now have the entire length of the inner perimeter πDi as the
burning surface area to begin with. Therefore, we could have a longer grain to give a
much greater burning surface area. And in radial burning grain, which burns in the radial
direction, we can get Sb to be much larger than in an end burning grain of the same
diameter. And therefore, we can get a large value of thrust. This is the modification that
could be done; but you know there is some limit to the burning surface area. We still
need to explore if for a particular diameter and length it is possible to increase this
551
surface area even further. In other words all what we are asking is whether for a radial
burning grain is it possible to increase this surface area by some means?
How can we do it? If we can wrinkle the surface; we can wrinkle it in some form, and
how do I wrinkle it? I show in this small model, this was the original circular perimeter
over here. We wrinkle this surface i.e., we make stars or some other shapes in the inner
surface. In other words instead of having something like a cylinder over here, I make the
inner surface in the form of a few star. Now I find that this surface has something like 5
vertices; 5 vertex star. And therefore, now I find my surface area has increased
enormously, and therefore now I must be able to evaluate how the burning surface
comprising the wrinkled surface will evolve as it continues to burn.
In other words this my outer diameter; the outer diameter will come over here, and the
burning surface will evolve along these surfaces, and this is the problem which we must
do. However, before doing this problem, let us do the simple problem of may be a radial
burning grain burning from inside to outside in a cylindrical configuration. What is the
value of pressure and what is the time required for burning this cylindrical grain. In the
case of end burning grain, we got the burn time tb = length ÷ a pn; we knew how to
calculate the pressure and hence the duration of burning. We wish to follow a similar
procedure for radial burning.
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(Refer Slide Time: 34:34)
Let us also find out how the pressure will change since the burning surface area is
changing. We have the initial value of pressure is equal to whatever we have written
here, let us rewrite it : ( πDi which is the initial perimeter × L and this gives the initial
burning surface area × ‘a’ × ρp × C* / At ) to the power 1/(1 – n). Now, what is the final
value of pressure when the burning has progressed? In other words the burning progress
from the inner cylinder and reaches the outer cylinder. What is the value the final value
of burning surface area: πDo is the perimeter into L is the final surface area. This
multiplied by a × ρp × C* / At to the power of 1/ (1 – n) gives the final value of pressure.
Now, is the pressure constant like as in the end burning rocket grain? It is a variable. Di
has increased to Do and so the pressure also increases. If we were to plot the pressure,
the pressure initially corresponds to diameter Di, while when the rocket burns out, the
diameter is Do. When the diameter is Di, the pressure is pi and when the diameter is Do
the pressure is po. The value of po > pi.
We know that Di < Do; therefore, initially the pressure is less, let us say when it reaches
the final value the pressure is higher, therefore the pressure increases. In the end burning
grain we had the same pressure throughout.
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If instead of having the grain start burning at the inner diameter and progressively burn
to the outer diameter, we somehow put the case over here and ignite the outer surface,
and then the flame propagates inward.
Then what is going to happen? We will get just the opposite evolution of pressure. We
start with a value of po of pressure; that means the initial value is now higher and the
pressure drops to the value pi. In other words while the end burning grain gave us a
constant pressure; and therefore a constant value of thrust, a cylindrical grain burning
from inside to outside gave a progressive increase in pressure or a progressive increase in
thrust. A cylindrical grain burning from outside to inside gives a progressive decrease of
pressure and thrust.
This means that a cylindrical propellant grain burning from outside to inside gave us a
falling pressure; and therefore, a decreasing thrust. We could have three types of thrust
evolution in such rockets. When the pressure is constant, we call as neutral burning.
What should be the name for this progressive increase of thrust? This is progressive
burning. If the pressure and thrust keeps decreasing; that is regressive burning.
If we have a propellant, which burns from end to end viz., end burning, the type of
burning is neutral and we get constant pressure constant thrust. If the burning is radial
from inside to the outside the pressure keeps increasing and the burning is progressive.
While if the radial burning is from outside to inside, the burning could be regressive.
Therefore, we could think in terms of three types of burning; neutral burning rockets,
progressive burning rockets and regressive burning rockets. If this is clear, the question
is what type of burning is required for solid propellant rockets? We cannot think of a
rocket, which is progressive burning as when the rocket takes off the thrust must be high
and as it goes up the thrust can come down.
Therefore, may be something like this with higher initial thrust may be better than
progressive or regressive, because when the rocket goes up it is mass is higher and it flies
in the atmosphere where there is drag. We cannot expect it to go up with a higher
acceleration in the beginning itself. Therefore, we have to somehow get a thrust pattern,
which is desirable. We also require higher thrust to begin with and for which we just
took the inner surface and wrinkled it. The wrinkled shape was in the form of a star. We
could have wrinkled into some other shape instead of giving the shape of a star. We
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could have given a different shape something like this. We could have given any shape
and we want to calculate, how does the burning rate evolve around this surface. We find
out how the burning surface Sb changes with time. And once we know how the Sb
changes, we know how the pressure changes with time. We can find out how the force or
the thrust which the rocket will develop with time, and that is all what we need to do in
the design of propellant grains in solid propellant rockets. Therefore to be able to pursue
on this, let us start with a simple example.
(Refer Slide Time: 41:00)
Let us find out the time taken for a cylindrical grain to burn. What do we mean by a
cylindrical grain; the first example; we have an end as shown. We make a hole here I
have radial burning from inside to outside. It burns through from inner surface to outer
surface. The inner diameter is let us say Di, the outer diameter is Do, the length is L. We
are repeating this figure. The question is can we predict how the pressure will change
with time and the time for burning. And how do we do it? It is a simple problem.
Let us consider a small part of the propellant between Di and Do gets burnt; and let the
small part of thickness δ over a time; let us say small time dt. At the beginning of dt let
the pressure be p corresponding to the initial burning surface area which is πDL. At the
end of dt, we will calculate the pressure again. The value of diameter at the burn out of
this element will be the original diameter plus twice delta. In this way we can
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progressively calculate the pressure for each time step and the new value of the burning
surface.
We know the pressure to begin with. It is this value at the burning surface given by the
surface at Di. What is the pressure when the diameter is Di plus 2δ. We know the new
value of perimeter and the length and hence the new value of the burning surface area.
We can determine the value of p at Di+2δ. What is the value of p delta. The value is Di
plus 2 delta into these terms viz., ‘a’ burn rate constant, rho p, C star divided by throat
area to the power 1 by 1 minus n. We know the value of pressure this point. We want to
know the time taken for consuming delta of the propellant.
(Refer Slide Time: 43:53)
Therefore, the mean pressure between this and the earlier value pi which is (pi + p
delta)/2. And therefore, the mean burn rate between the initial at Di and Di plus 2 delta is
determined equal to r bar, the mean value. And what is the mean value of burn rate equal
to? a pin + a pdeltan / 2.
When the burning has progressed by a distance delta from the initial diameter Di, we
find that the pressure has increased. The burn rate has also gone up. The time taken to
burn the small quantity between Di and Di plus 2 delta is t delta and is equal to delta
divided by r bar. We continue with the same process further. We now take this as initial
condition and go to the next step of delta and find the value the value of pressure and the
time taken to consume the element. In this way we march ahead till we reach the outer
556
diameter. We can also determine by summation the total time required for the evolution
of pressure starting from pi to pf.
In an inner burning rocket we can follow this procedure, but this is numerical way of
doing the problem. There is no other way of doing when we have complex configuration
with wrinkled inner surface. We can find out how the surface should evolve with time.
And this is the method to calculate the variation of pressure with time. Once we know
the variation in pressure with time, we can readily go ahead and determine the thrust
variations with time. This is how a solid propellant rocket propellant grain is analyzed.
(Refer Slide Time: 46:00)
Let us do one small problem. Suppose we are asked to find the time taken to burn a
propellant that burns radially outward. Now we are using radial grain between diameter
Di and Do, the length of the grain being L. What we consider is the initial diameter of
the cylindrical grain is Di; the final diameter or the outer diameter is Do and the length is
L. We want to find the time of burn time when a nozzle of throat diameter At is
connected to it. We can follow the procedure outlined earlier. But in this case a simple
analytical solution is possible.
Let us consider any diameter D between Di and Do. Let us find out the time taken for the
this diameter D in between Di and Do to increase from the value D to a value D plus a
small change over here say dD, and this we say the diameter has increased to D plus dD.
557
If we can find the time taken to burn this part we can integrate out between the initial Di
and the outer Do and find the time taken. And that is what we are going to do.
Let the time required to burn the grain between diameter D and diameter D plus dD be
dt. Since the distance dD is very small, the variation in pressure while burning between
D and D plus dD will be very small. . Therefore, the pressure could be assumed to be the
value at D; and therefore, the pressure is equal to. Sb is equal to pi×D, the perimeter ×
length L is the burning surface area × ρp × ‘a’ × C* / At to the power 1/(1 – n).
(Refer Slide Time: 48:00)
Therefore, what is the burning rate r given by the expression? It is to a pn. We substitute
the value of p as a function of D in the burn rate law to give ‘a’ into the factor pi into D
into L into rho p into a into C star by At to the power n by 1 minus n. The time taken to
burn a small distance dD by 2 at the diameter D i.e., time taken for diameter to burn
from D to D plus dD is therefore this small thickness is equal to dD /2 divided by r which
is the time taken dt. And what does is this come out to be.
We find π is a constant, length L is a constant, D is variable, ρp is a constant, C* and At
are also constants and we can write this as equal to (dD/2) /constant A × D to the power
n /(1−n). We have this a in the denominator. And then we write it as D to the power n by
1 minus n. We first check, if what we are expressing has units of length say meters
divided by meter per second and this is the time taken. Here you have dD by 2 is the
distance propagated divided by a into all the factors equal to π L C* / A to the power
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n/(1 – n) that is p to the power n. Let us say the time taken to burn is dt. The total time to
burn from diameter Di to Do is denoted by tb
(Refer Slide Time: 50:44)
We get dt as it goes from start to the end, which is the burn time from zero to tb which
must be now equal to the diameter going from initial diameter Di to the outer diameter
Do. This equals dD/2 divided by D n/(n−1) and we can take A since these are constants.
The limits of integration are that the diameter varies from Di to Do. What do we get on
integration, tb minus 0; therefore, tb is equal to 1 over 2 (A × a the exponent of the
burning rate law). Now we integrate D−n/(1−n). This becomes 1 – n/(1−n). We have
D1−n/(1−n) and divided by 1 − n / (1 – n).
What is the final value? Therefore, this is equal to 1 / (2 A a ) ×(1 – n)/ (1 − 2 n) × Do (1 −
2 n)/ (1 – n)
− Di(1 − 2 n ) /(1 – n). This is the time taken tb for the burning.
559
(Refer Slide Time: 53:42)
Therefore, we are able to find out the burn time of the cylindrical grain. If we have for
the exponent ‘n’ in the burn rate law n = 0.5, what is the time for burning? We said that
as long as n is less than 1, it is usable. If n equals 0.5 what would be the value of burning
time? Lets do it 1 minus 2n is 0, so there is a problem. What is wrong? The expression is
derived correctly.
But it is not working at n = 0.5. Can somebody come out with an answer and sort out the
problem. I think we should be able to analyze it. Let us write it as 1 / Aa × integral of Di
to Do of dD / 2 D to the power 0.5 divided by 0.5; and therefore this equal to dD/D. We
take 2 outside, and get for n is equal to 0.5, tb is equal to 2aA ln of Do by Di. That means
what happens for 0.5 is that we need to use a logarithmic form for n is equal to 0.5. The
problem is not physical like for n equal to 1 but only mathematical.
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(Refer Slide Time: 55:37)
We are able to find the time of burning and this is how a cylindrical grain will be
analyzed and designed. What is it we have done in today’s class? We looked at the effect
of temperature on burn rate, we also defined the value of the constant in burn rate law a
as a70 at a reference pressure of 70 bar. Then we learnt how to develop an equation for
thrust and pressure for an end burning grain. We also determined the pressures and burn
times for radial burning. We found for radial burning grain, the pressure evolution had to
be done by in increments as the burning progresses.
(Refer Slide Time: 56:18)
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We could solve analytically for the cylindrical radial burning grain. We shall continue
with this in the next class and look at the evolution of burning surface area from
something like a star grain, which was wrinkled to give a large burning surface area. We
will also look at the different forms of grain shapes, which are used in practice, and the
reasons for it. After that we will summarize the solid propellant rockets by incorporating
the igniter in it along with the other aspects.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 23
Burning Surface Area of Solid Propellant Grains
(Refer Slide Time: 00:15)
In today’s class, we continue with our discussions on solid propellant rockets. By now,
we know what the burning rate is; it is the linear regression law given by r = a pn; this is
applicable for both composite and double base propellants and also the composite
modified double base and also nitramine propellants. What we were discussing is how do
we assemble the propellant grain and what should be the configuration of the grain. We
saw in the last class was if we have something like a radial burning grain, we would like
to wrinkle this surface into something like a star or some other configuration, such that
we get increased burning surface area and therefore, increased pressure and therefore
increased thrust. The aim is to get a large thrust.
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(Refer Slide Time: 01:19)
Therefore, let us quickly recap where we were. We talked in terms of let say neutral
burning grain; that means, the burning surface area is constant as it keeps burning. We
called it as neutral burning, because both the pressure and the thrust were constant at all
times. We also discussed about radial burning grain, burning from inside to outside and
what did we find? We found that the pressure progressively increased and the thrust
progressively increased. In other words, burning started at the inner surface and
progressed to the outer surface; we called this as progressive burning. And, if we could
somehow get the burning to start from the outer and progress inward, the pressure will
keep falling with time, we called it as regressive. Therefore, we talked in terms of neutral
burning, progressive burning and regressive burning.
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(Refer Slide Time: 02:15)
And, thereafter, we also evaluated how we could determine the pressures at different
instance of time for the radial inward burning grain and also the time taken to consume
the propellant from the inner diameter Di to the outer diameter Do.
(Refer Slide Time: 02:32)
Having said that it is not possible for us to determine the pressure variation along the
grain analytically for non-circular shapes, let us look at a few definitions. The distance
between the inner surface and the outer surface is what we call as the thickness of the
grain. The minimum thickness is known as a web thickness. And, if we say this is the
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inner diameter this is the outer diameter, the pressure keeps increasing and therefore, as
distance along the grain increases, the time increases. Therefore, we were able to plot
pressure as a function of time and we found that the thrust of this rocket, thrust going as
CF into p into At keeps increasing, with time as is shown here. This is the progressive
burning grain.
(Refer Slide Time: 03:22)
We had also addressed wrinkling of propellant surfaces. I show a picture of a wrinkled
propellant grain burning radially outwards. Instead of having the grain, which was
circular on the inside, we sort of wrinkle the surface such that we have something like a
star shape and this star shape is throughout the grain surface. And, now what is going to
happen as the burning progresses? The inner burning surface area Sb is going to be much
larger, in other words Sb is going to be the perimeter of this particular star shape
multiplied by the length along the grain that is going to be the initial burning surface
area. And, therefore, the pressure will be much higher, the thrust will be much higher
compared to what it would have been had it been for a circular geometry of diameter D.
These are wrinkled surfaces. And, in today’s class we will calculate how the burning
surface will evolve for this particular star grain.
But a wrinkled radial burning propellant grain need not be a star alone.
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(Refer Slide Time: 04:12)
We could have grains of different shapes; maybe we could have lobe like this instead of
having a star over here. The burning surface area, this is the inner surface of the grain,
this is the length of the grain in this direction, the evolution of surface area would keep
on evolving like this until it touches the outer surface of the grain.
The minimum thickness from this root to the case is what is known as the web thickness.
Mind you, there are several thicknesses. It could be from the vertex to the outer diameter
or it could be from the root or bottom part to the outer diameter. The minimum thickness
is what we call as the web thickness. The same grain I show over here just to make sure
we understand. We find that the inner surface has something like a projection here, a
valley here, again a projection here; this projection is all along the surface and the grain
burns radially. That means, it burns into the grain in this particular fashion and all along
the surface, the surface keeps regressing like this. And, we will consider a few examples
such that we are very clear how to calculate the burning surface areas.
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(Refer Slide Time: 05:22)
The grain surface need not always be a star or a shape, like what I just now showed
which was a lobe like a wagon wheel. But it could be any shape other shape; could be a
dendrite, dendrite is a crystal shape which is like this. And, burning surface starts or the
inner burning surface to begin with; this is the perimeter into the length of the grain over
here. And burning proceeds in this direction; this is known as a dendrite grain. You have
a wagon-wheel in the shape of the wheel of a wagon, see you have the inner surface and
the length of this is along this particular direction. Therefore, the perimeter along this is
not quite large compared to what would have been for a circular diameter over here,
multiplied by the length is what gives me the burning surface area and this is known as a
wagon-wheel.
We could have the shape of an anchor, instead of having a circle like this, I sort of
extended make it shape like an anchor and what is an anchor? You drop an anchor when
a ship is not sailing and this is the shape of an anchor. And if this burns, well the burning
shape will keep on evolving along the anchor surfaces and I can find out how the burning
shape evolves with time and calculate how the thrust as keeps burning. I could have
dendrite, I could have wagon-wheel shape, I could also have the shape of a bone, the
dog-bone where this is the shape of the bone which is shown. This is the inner diameter.
Now, how is this grain going to evolve? It is going to burn perpendicular to the face of
the bone shape. We should be able to calculate it progressively till this the surface
touches here and thereafter the burning surface area decreases.
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(Refer Slide Time: 07:10)
We could also have something like a cylindrical shape and at the end of which we have a
cone; that means, I have a cone within a cylinder and this is known as conocyl; that
means, I have a cone within a cylinder. How is this going to evolve? The surface would
keep on evolving parallel to the initial surface. And it is the evolution of burning surface
area what we are basically interested. In addition to having sort of a cone in a cylinder, I
could also have a cone like this and this cone has this cylindrical portion; in the
cylindrical portion I make ribs like what I show here.
(Refer Slide Time: 07:58)
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In the section shown, we make ribs like this; something like fins. Therefore, in addition
to having a cone on which this is situated, we have something like a cone over here,
coming to a cylindrical portion on which we have ribs; therefore, here also burning could
evolve along the cylinder, cone and fins and we would like to maximize the perimeter
and the burning surface area. When we have a fin in a cone with cylinder arrangement,
the grain is known as Finocyl; that is we have fin in the cylindrical portion. Therefore,
there are various grain shapes, but what is maximally used amongst these is a star grain.
We will see the reason in this class. We note that different grain shapes are used.
Therefore, we could have any shape of the inner surface? We want to increase the
burning surface area to the maximum possible extent, initial burning surface area and the
progression of the burning surface with respect to time. In other words if we are
interested as time or as burning proceeds, the burning surface area should initially be
large such that I get a thrust; you could either evolve progressively increasing like this,
or it could decrease or it could be a constant. In other words the shape could provide
progressive, regressive or neutral burning. The burning surface area directly translates
into pressure and pressure directly translates into thrust F which is equal to CF × chamber
pressure × At. And, what was chamber pressure p? We got an expression in terms of
Sb1 /(1 - ‘n’) and therefore, it was directly connected.
Let us now go to the star grain, which is of primary interest. I forgot to mention one
grain, which is known as a slotted grain. See, so far when we considered these different
grain shapes; we essentially considered let say a cylindrical grain; wherein I have
something like a cylinder, this is the length, may be this is my outer case over here and
here I put my nozzle over here, we told ourselves this end is insulated it does not burn
over here. Burning takes place along this cylindrical surface. Therefore, the burning
surface area progresses radially and it is outward from the centre. Burning takes place
from the inside surface to the outside. If we ignite it on the outer surface may be we have
to allow a gap between the case which is insulated and we allow it to burn inward;
therefore, we say radial inward. Radial outward was progressive and radial inward was
regressive.
We also had axial burning; burning takes place along the axis of the grain. We had
something like a case in which I put the grain over here; we allow the burning to take
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place normal to this surface axially; here also the burning is normal to this particular
surface over here. But it is in the radial direction while this is in the axial direction. And,
for the neutral burning, that is end burning i.e., regressing from from end to end, it is sort
of end burning which is neutral. It is also possible to have some configuration like
combining radial and axial. Let slightly modify this configuration.
If we make some slots on a cylindical grain. How is burning going to proceed? Burning
goes radially here, burning progresses axially here; that means, at the next instant of time
the surface is going to be something like this, coming over here radially, surface comes
here axially. It goes both axially and radially and these are known as radial cum axial
burning or three dimensional burning surfaces. In other words, it is both axial and
axisymmetric and therefore, these are known as three dimensional burning surfaces. And
we show a three dimensional surface over here which is a slot. It continuous to burn in
this direction along the slot and cylinder, but the problem is a simple geometric problem.
We want to find out how this initial burning surface area which is the perimeter into the
length keeps increasing as the regression of the surface continues? This is the slot. If we
put a number of these cylindriacal grains together and allow them to burn both on the
inner and outer surfaces we are no longer having a particular burning. The burning is
unrestricted because it is burning from radially inward and radially outward and we get a
large burning surface area. Control of burning is difficult with unrestricted burning but is
used in practice for small duaration of large thrust requirements such as rocket assisted
takeoff for planes.
Therefore, to summarize we talk in terms of neutral burning, progressive burning,
regressive burning, unrestricted burning and everything decides on how the surface area
keeps changing. Having said that let us come to this particular problem of a star grain.
See, here I show in the end view a star grain.
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(Refer Slide Time: 14:30)
We have a grain like this; this is the length of the grain, propellant grain over here and
the nozzle is integrated to give me the thrust. We were able to calculate the pressure and
thrust when we have a circular section corresponding to the outer diameter and a circular
hole at the centre. And now the initial burning surface area is in the form of a star, this is
the length of the grain L, this was the mean inner diameter and this was the outer
diameter Do.
Now, if we sort of wrinkle this surface and we showed how wrinkling is done, instead of
having this surface we have a multi star configuration. That means, I have 1 2 3 4 5 6 7 8
or 9, 9 star, 9 vertices of the star. We remove the inner diameter and introduce the 9
vertices star. And we find that we have a much larger burning surface area and how is it
going to evolve at the next instant of time as the propellant burns? The flat surface
evolves as such with the vertex evolving as a circle. And, therefore, we find it regresses
in a slightly different shape and we are interested in finding how the perimeter and
burning surface area evolves with time. Question is whether we will get progressive
burning, regressive burning or neutral burning. It is possible to configure the star grain to
give both neutral as well as progressive or regressive burning. Let us do it.
And, how do we predict the thrust developed by a star grain? If we take a look at this
figure again, what is it that we see? This is the star grain. In practice to get a point is
difficult and therefore it is slightly curved at the vertex. And, therefore, the next line over
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here shows, after sometime when the burning has progressed it goes like this. And,
therefore, I need to calculate this perimeter and if I calculate this perimeter and multiply
by the particular length, I get the burning surface area Sb after some time t. At the next
instant of time, well the surface is again evolving; we can calculate this perimeter and so
on. Towards the end, this is the shape; see, initially I have these surfaces, but as it
progresses it becomes something like a circle. And, why does it become a circle, because
a point when it evolves, it progresses into a circular fashion. A normal to the circle is
along all directions. And, therefore, to be able to understand the evolution of the star
surface, we first deal with a simpler case of a square opening in a cylindrical grain.
(Refer Slide Time: 17:39)
We will calculate how the surface evolves with time and based on the experience that we
gain in this particular case, we will take a look at how a burning in a star propellant grain
takes place.
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(Refer Slide Time: 18:12)
We would like to find out, how the burning surface area progresses for a star grain; what
is a star gain? It is one in which the internal surface is sort of wrinkled such that we have
larger number of surfaces. If we can estimate the how to find out the evolution of the
burning surface for a star grain and if we can do one or two cases, we can find out how
the thrust of a solid propellant rocket can be varied.
Let us do this simple configuration of a square in a cylindrical grain. We show a cut view
of the grain Instead of having a circle at the centre, which we have already done; radial
burning and radial burning is normal to the circle, we now consider the case wherein at
the centre we put a square hole instead of a circular hole. We have a square hole of
dimension b. The grain is of length L. We have a nozzle of throat area At connected to
the case. We would like to know, how this square surface evolves? How these four
straight lines which constitute the perimeter evolve?
We said that the length of the sides are b. Therefore, the surface area of each side is L ×
b. The total surface area at the start of burning will be 4 b which is the perimeter × the
length L. What is the surface area which we are igniting; the inner surface area
corresponds to b into L, b into L, b into L, b into L or rather perimeter is 4 b × length;
that is the initial surface area which begins to burn.
How does burning proceed? Burning takes place linearly at the surface. I want to find
out, what will be the burning surface area after a certain time; let say when the surface
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moves through a distance let us say Δ. What would be the burning surface area? Initial
value is 4 b into L. Is this clear, are there any questions?
(Refer Slide Time: 21:35)
Let me draw a slightly bigger figure. Now, this is the centre, all sides are b. Let us
assume that the grain moves through a distance or regresses through a distance Δ.
Therefore, this straight line shifts by Δ, this straight line moves normally by Δ; all four
lines of the perimeter get shifted by Δ.
How does this corner point evolve? We said burning is always normal, therefore, how
should it evolve? Let say, we have this vertex here, burning surface is normal here,
burning surface is normal here; this is the surface which burns. How will this point
evolve? Normal? Point will evolve like what? How would you look at this problem?
Two things; this surface goes straight normal to the surface. How will a point burn as it
proceeds? When we have a point then it should be normal to the point; that means,
burning should take place here, what happens over here? That means, a point will go as a
curve of particular radius; that means, when this flat surface moves through a particular
distance Δ, it will form at the corner a quadrant because for a point we say burning is
always normal to it. Normal to the point could be in any direction, therefore, normal to
the point will go as a circle, but if we look at the flat surface, it evolves parallel to itself.
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Similarly this point will evolve as a circle over here; that means, only a quadrant till it
meets lines. What is that radius of this quadrant? The distance moved by the flat surface.
Therefore, radius is delta again. That means, if we have a point and the burning rate is r,
in the first second it comes to r, second second it comes to 2 r, third second it comes to 3
r and so on. Therefore, this has now regressed by delta in all the directions delta, delta
over here. Therefore, when the surface has sort of gone from the initial point to the final
point which is delta away, what is the value of the burning surface area?
(Refer Slide Time: 25:03)
We said that at the beginning the perimeter was just 4b and the burning surface area was
4Lb. Each b remains as it is and what is the additional value perimeter that we now get?
We get four quadrants, four into each quadrant with radius Δ; that means radius is Δ;
therefore, 2πr, 2πΔ divided by 4 is the perimeter of each quadrant. We get the burning
surface area, when the grain has regressed by delta as equal to (4 b + 2 πΔ) × the length
L, so many meter square. If we have something like a square with a vertex over here, the
vertex evolves as circle when it regresses. But the flat surface evolves as a plane because
burning is normal to it.
Therefore, the vertex as it regresses meets it - meets these two adjacent surfaces.
Therefore, we have a quadrant, the perimeter of this quadrant is equal to 2 π × radius or
2πΔ/4 is the length of the quadrant. let say from 1 to 2, from 3 to 4, the length is 2 π Δ/4,
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for 5 to 6 and for 7 to 8; whereas, surfaces 2 -3 is equal to the initial perimeter, giving the
total sutface arae as (4 b + 2 πΔ)× L meter2.
Therefore, we find that the burning surface area now keeps increasing for each additional
value of delta and how long will it keep increasing? We find that the surface for some
distance keeps evolving in the same way; we are very clear about it. If the burning
process let say by some amount Δ over here, now my new Δ is Δ1, we have another arc
of a circle, another arc of a circle, straight line, arc of a circle straight line, straight line
over here, arc of a circle. And, therefore, the perimeter corresponding to burning surface
area in this case is 1 2 3 4 which is same as b + the four arcs of the quadrants.
And, we find Δ keeps increasing as the burning proceeds. Therefore, Sb now keeps
increasing with time; rather instead of having the grain wherein diameter, if I had
something like a circular hole, you know the diameter directly increases. In this case, we
get smaller effect. The burning surface area would keep on increasing still further till it
reaches the limit when this particular arc or vertex comes and hits the outer diameter.
Then we have a circle of diameter over here and the plane surface comes over here.
Similarly, the circumference of this quadrant comes and touches the outer diameter over
here, for I have a circle like this; plane surface continues. And, this is the limit till the
burning surface area keeps increasing as the perimeter 4b plus 2π × value of Δ or portion
of the propellant which gets burnt. Once, this happens, the perimeter and the associated
surface area begins to decrease. This is because the propellant surface has already
reached the case and there is no propellant left in the quadrant to further evolve. The rate
of increase first decreases and therafter the value decreases. Therefore, once the vertex
corresponding to the middle of the quadrant comes and hits the case, thereafter the
burning surface area would decrease. After this the perimeter corresponding the the
evolving quadrant decreases. And, therefore, the burning surface area may be decreases
like this and ultimately becomes 0. This is the way they burning surface area evolves or
changes.
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(Refer Slide Time: 30:17)
And therefore, the pressure in this particular motor which has a square cavity in it will
evolve with time something like shown in the figure. But the minimum distance between
the surface or a point on this surface and the case we defined as the web thickness. Let us
again take a look at what I mean by this web thickness, I will show it in this figure again.
If we have something like a outer diameter over here and the inner diameter over here Di
and the outer diameter is Do, the thickness of the grain is equal to (Do − Di)/2. This is
the thickness of the grain; that means, the grain starts burning here, when the thickness is
so much it gets totally consumed.
Now, if instead of a circle of diameter Di, supposing we were to put a square over here at
the centre. In other words, now the grain is of this shape. Now, what is the web
thicknesses that I can talk of? This thickness is a little larger, this along the vertex if I
join a line from the center to the vertex and then to the case, the distance between the
vertex and the case along this line will be the minimum distance. As the grain evolves, it
goes as a circle and first touches the outer daimeter of the grain. The thickness of the
propellant is more along the other lines from the center. And therefore, this thickness
between the corner point or vertex to the casing is the minimum thickness and this
minimum thickness is what is known as a web thickness. Why do we call it as web
thickness, because till the web thickness is consumed, the burning is progressive or the
burning surface area keeps increasing and after the surface comes over here the burning
surface area decreases.
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Thereafter, what is going to happen this particular perimeter? It keeps decreasing once
the point reaches the outside diameter; whereas, the straight line portion is still constant.
Therefore, the burning surface area will begin to decrease and therefore the progressive
burning is up till the web gets consumed. The region of incresing pressure is known as
web burning and this progressive part corresponds to the web being consumed. However,
some propellant is still left and is known as sliver or left over sliver burning. Therefore,
in such square hole grains, you find that sliver burning occupies quite some time and
leads to low pressure and low thrust. The low pressure comes from the smaller burning
surface area. This is the reason why such square cavity grains are not used in practice.
If now we go to a star propellant grain with which we are interested, we had something
like a circle, we wrinkle the circle such that we get a star shape. We have a number of
star points over the circle. The web thickness will be from the vertex to the outer
diameter of the grain. This is the minimum thickness and is the web thickness. At the
other places, the thickness is much larger. Therefore, I am interested in making sure that
the web burning distance is large with respect to the propellant size so that the sliver,
sliver is the length left over after the burning, is quite small. Inadvertently we introduced
some words like web burning and left over propellant after web burning as sliver. With
this background, I think we are in a position to be able to calculate how the burning
should proceed for a let say a star grain.
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(Refer Slide Time: 35:02)
We wrinkle the surface such that we get something like a star surface; let us have 1 2 3 4
5 that is of 5 vertex star. I could choose any number of vertex in the star, it could be 7, it
could be 8. How do I characterize the evolution of the burning surface from the star
surface as the burning progresses towards the outer diameter of the propellant grain or
the case diameter? We are interested in finding out how the burning surface area keeps
evolving with time. Let us say that the sides of the star are of length s and how does the
grain look in the three dimensional plane? Well, this is my outer surface; all these are
points here. And, therefore, if we were to make a plan view of this, we get a cylinder, we
get this as the center line, corresponding to one vertex we get a line over here it should
be dotted, we get another line over here corresponding to the other vertex. We are
interested how the surface keeps evolving? Now, you could tell me, the initial value
burning surface area.
580
(Refer Slide Time: 36:27)
Let us have a star grain with n vertices. We have something like n of these vertices. The
vertex A is between the two straight lines of length s. Therefore, what is the initial
burning surface area? It is an n pointed star. The length of the grain is L.
The initial burning perimeter is 2sn and the initial surface is 2nsL. You are correct.
We want to know what will be the shape of the perimeter or what will be the value of
this length s when the regression progresses through a distance Δ. Let say the burning
surface area now comes over here. It moves through a distance Δ. We want to calculate
what will be the configuration of this particular point? Can we choose some axis of
symmetry and do the problem? Let us examine this figure again; may be it will become a
little more clear. We say this is the point A, the vertex; it will evolve as circle. Here on
the flat surface the regression will be normal to the surface. It means that this perimeter
should become like this. But then the length will change as we cannot have interference.
But this is going to be my access of symmetry here; and it is sufficient if we consider the
evolution of a single line of initial length s.
Therefore, we have an axis of symmetry here and now with this axis of symmetry the
center is over here. All sides of the star are symmetrical. Therefore, if we can determine
the changes for this single line when the burning progresses by a distance delta, the
length of the line can be calculated. The burning surface area will equal the new value of
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s into L into now we have 2 n of these surfaces. How do we calculate the length of s as it
regresses?
The change is Δ normal to the surface and the line moves by distance Δ. Therefore, if AB
is the original length, when it moves by delta normal to itself, it comes from AB to CD
as shown in the figure. What happens to this particular point viz., the vertex? Again, we
find this is going a circular arc. The new length of this line multiplied by 2 n times,
multiplied by the grain length is the new surface area.
How does this point A evolve for the regression by delta? It evolves as a circle of radius
delta. It evolves as a circle. This becomes delta here; the arc is EC. Therefore, I find that
this particular line on the inner surface now becomes partly the same straight line parallel
to AB and partly this circular arc.
How do I calculate this particular total length E C to D? Is the problem clear? We have n
vertices. We choose a symmetry line so that we need to examine a single line and
multiply it by 2n to get the perimeter. And, whatever happens to this line will happen to
all the other 2n minus 1 lines. We would like to find out the length of ECD.
It is a geometric problem; there is no burning. How do we determine the value of E C? I
know that the radius is delta. If we specify this angle as β radians we can now say EC is
equal to delta into beta. And, we should also calculate the value of CD. To get the value
of EC, we need the value of angle β?
For the particular star let the total angle between the sides of the star be θ, in other words
we have this star; let this angle between its sides be θ; therefore, this half angle which is
bisected by the line of symmetry is theta/2. In other words this small angle is equal to
theta/2.
What is the angle at the center, which is included by this particular side AB of lengths?
The n vertices have an angle equal to 360 degrees at the center, that is 2π radians. We
have n vertices and now each of the n vertices consist of 2 lines; therefore, the angle
included is by line AB is 2π/2 n or rather π/n is the included angle for this side at the
center. What is the value of this angle adjacent to θ/2? This angle we said θ/2 giving the
angle as 180 or π − θ/2. If this angle is π − θ/2 and this angle is π/n, what is the value of
this angle χ in the triangle?
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The sum of the angles in a triangle is 180 degrees or pi radians. The angle χ = π – (π –
θ/2 –π/n) = θ/2 –π/n.
(Refer Slide Time: 45:10)
This is right angle adjacent to χ; therefore β = 2 π – π/2 −χ. Since χ = was equal to pi by
n minus theta by 2, beta is equal to π/2 – π/n + θ/2. Therefore, we have found out this
angle β and therefore, the length of the line EC is equal to Δ×( π/2 – π/n + θ/2).
Let us check the angles. The angle made at the center is π/n, this angle becomes π – θ/2;
therefore, the sum of these angles become π − θ2 +π/n. See, this value of χ = π − θ/2 –
π/n. And, this is θ/2 − π/n. The angle beta therefore is equal to π − (pi/2 + π/n − θ/2).
What is the value of the straight part CD? How do I get this value? We find that this is
also right angle and we can have a right angle for which the perpendicular distance is
delta. The base of this triangle, is equal to cot θ/2 × Δ to give Δ cot θ/2. The value of AB
minus Δ cot θ/2 would be the length of the straight line CD. Let us make this clear. The
perpendicular distance is delta, this angle is θ2; therefore, the base is equal to Δ cot θ/2.
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(Refer Slide Time: 49:01)
The length of single stretch of line ECD equals Δ × π/2 + π/n – θ/2 + s − Δ cot θ/2.
Therefore, what is the new surface area Sb when the grain has receded by delta? Sb is
equal to 2 n of the s perimeters × the length and this particular value is what gives me
the new burning surface area.
If we want to form a grain a particular star grain, which should have neutral burning, the
evolution of the line should not change their length. It must always be s; in other words
for neutral burning it is necessary for us to have Δ × π/2 + π/n – θ/2 – Δ cot θ/2 is equal
to zero. For this condition, we will have neutral burning. And, what is it we get? We
solve this equation, we find Δ and Δ gets canceled and rather we get cot θ/2 – π/2 + π/n –
θ/2 =0.
And, since it involves a cot term we cannot do it directly. We do it numerically. May be
we use the method of steepest descent or a Newton Raphson scheme. We find that theta
is going to be a function of the number of vertices in the star.
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(Refer Slide Time: 50:51)
If number of vertices are 6, the value of θ = 67 degrees. If the number of vertices is 8, the
value of θ = 67 degrees. In other words depending on the angle θ, we could have neutral
burning. And if the angle < θ, what happens? If the angle is less than θ, the surface area
keeps decreasing because we are subtracting a larger quantity corresponding to cot θ/2.
Therefore, if theta is less than 67 degrees for something like a 6 vertex star, we get
regressive burning.
Whereas, when theta is greater than 67 degrees; that means for n is equal to 6 we get
progressive burning. Therefore, a star grain depending on the value of θ and the number
of vertex, we could have either neutral burning or progressive burning or regressive
burning and that is why star grains are quite useful. By suitable wrinkling, we can make
it burn the way we want.
Let us just conclude. For a grain with a central square cavity, when the corner of the
square meets the outer grain diameter then the balance is what is known as a sliver and
you have web burning up to this particular time. We thus differentiated between web
burning and sliver burning. And, then we analyzed the star grain, we determined the
evolution of the burning surface. We found that this is equal to the length of each side of
star (s − Δcot θ/2 + Δ × π/2 + π/n – θ/2) × 2n × length L of the star grain. Using the
burning surface area, the equilibrium pressure and the thrust of the rocket are calculated.
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And, then we got the value of θ for neutral burning as equal to 67 degrees for neutral
burning when the number of vertex n the star was 6. For larger values of theta we get
progressive burning while for smaller θ we get regressive burning. Therefore, a star grain
could be designed for whatever be the type of burning we desire. This is all about
evolving burning surface area in different propellant grains. In the next class, we will
look at the other elements of solid propellant rocket.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 24
Ignition of Solid Propellant Rockets
(Refer Slide Time: 00:19)
We have seen how the burning surface area can be calculated as a function of time and
therefore how the thrust of a solid propellant rocket will change with time. This is
because once you know the burning surface area, you can calculate the value of
equilibrium pressure in the rocket, and equilibrium pressure × thrust coefficient × the
throat area is equal to the thrust of the rocket. And how did we calculate the equilibrium
pressure; based on the burning surface area? We considered it the last but one class. We
have considered propellant grains of different shapes.
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(Refer Slide Time: 01:08)
And a nozzle is connected to the case containing the propellant grain. We said the grain
could be radial burning or the end burning. Now the question is how do we ignite the
propellant grain. How do we start the burning in the grain? We must use a heat source,
something like an igniter, which will start the burning process. But then we also know
that the grain surface is quite large; may be it could be something like several meters,
like for instance if we take an example of the world’s largest solid propellant rocket. We
call it as solid rocket booster for the space shuttle. It is something like 40 meters long.
Therefore, the question is how do we make sure that the grain surface ignites, and that is
what we will be dealing with in the first half of the class.
588
(Refer Slide Time: 02:10)
What should be the attributes of an igniter? Namely how do you make an igniter? Well,
we can immediately say igniter must be capable of catching fire easily. And therefore,
maybe we will use something like a black powder, which is used for making fire
crackers, and this consist of potassium nitrate, some amount of fuel carbon and some
amount of fuel sulphur. Typically around 15 percent carbon, 10 percent sulphur and the
balance viz., 75 percent of KNO3 is used. And this compostion is easily ignitable. You
may recall that we light fire crackers with match sticks and it begins to flare up. We have
something like a flower pot type of cracker in which have the black powder; we light it
with a match stick and we get sparkles coming out. Therefore, may be this could be one
of the contenders for igniters. If we were to use it, how could it be adapted for use it in
solid propellant rockets? Maybe I could have a small bag or container and in the bag we
have this particular composition of black powder.
But we want to ignite it. How do we ignite it? May be we take a resistance wire. May be
an electrical resistance wire like a thin nichrome wire. And why nichrome wire? A thin
wire of nichrome has high electrical resistance. If we pass a current through it is gets red
hot very soon.
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(Refer Slide Time: 03:44)
And therefore, we take this particular resistance wire, coat it or we coat on its surface
some easily ignitable composition may be black powder or something like KCl which
immediately catches fire. And then may be surrounding it we put more of this
composition, black powder or equivalent.
When we pass a current through the wire, it gets heated and KCl or black powder begins
to burn. It generates heat, a flame, and this flame could be used for ignition of solid
propellants. This particular arrangement of a resistance wire heated by electricity or
electrical energy and using some easily burning composition such as black powder is
what we call as squib. But whenever we use electrical current for heating, it is also
possible that such electric current could be accidentally generated when we have some
electrostatic or electromagnetic disturbances. We could then have a current and even
when we do not want to ignite, we could have a small current which could heat the wire.
Therefore, it is necessary to ensure we have current greater than some threshold value
only for which the composition like KCl or black powder will ignite. And this threshold
value of current is known as all fire current. And if the current is less than the threshold
value formed accidently by stray electrostatic discharge, it will not catch fire and the
rocket cannot be ignited.
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(Refer Slide Time: 05:52)
We want to ignite or burn the surface of a solid propellant rocket. And to be able to
ignite the solid propellant surface, we need to add some heat or some energy to the
propellant surface. And therefore, all what we do is to release the energy in the
propellant cavity. We cannot go and put a fire inside it. Therefore we have a composition
which is easily ignitable with a nichrome wire; start the ignition process by passing
current through the wire, this generates heat. But then a squib has only a small quantity
of charge like KCl or black powder. We add some more charge surrounding the initial
charge and when we pass a current greater than some threshold value of current. It starts
a chemical reaction of the black powder which is easily ignitable. The black powder or
some composition around the squib ignites, and a fire is formed and this fire impinges on
the propellant surface and makes it catch fire. That means, we have a squib surrounded
by some of these easily ignitable powders. We call the easily ignitable composition as
pyrotechnic composition or pyrotechnic powder.
Such igniters, which make use of a squib with pyrotechnic composition around it to
generate sufficient energy and ignite the solid propellant rockets grains, are known as
pyrotechnic igniters. The igniter seems to be a simple device.
591
(Refer Slide Time: 07:54)
Since we need electrical current, we have a battery or some other source of electric
current. We put something an igniter over here in the cavity of the propellant grain. And
then we pass a current through the squib and ignite the squib. The squib ignites the
powder, which is around it, and it sprays the flame that is a plume which is formed
ignites the propellant surface. The volume of the cavity gets pressurized and the flames
spreads over the surface and in this way the propellant surface ignites.
We had discussed the internal surface of the grain, the outer surface of the grain; and just
for the sake of simplicity taking a radial grain, this is the nozzle; we put an igniter over
here, as shown. We pass a current and a flame or plume originates from the igniter, it
impinges over the propellant surface. It ignites the surface over which it impinges and
when this surface ignites the next or the adjacent surface gets ignited and so on till the
entire surface ignites.
In other words, we have first something like local ignition, where in the sparklers or the
plume or incendiary impinges on it. Then we have a flame, which is spreading over the
surface, and once the flame spreads the pressure may still not be the equilibrium value
corresponding to the burning surface area. Thereafter, as the last part chamber fills up
and gets pressurized to equilibrium value.
These are the 3 events, which would happen: local ignition, flame spread and chamber
filling to equilibrium. If we have a bag igniter like this and we have a small rocket like
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an end burning charge or an end burning grain let us look at the sequence of events
during the ignition process.
(Refer Slide Time: 09:51)
We have a solid propellant in an end burning configuration; we have the nozzle over here
downstream of the grain. We want to ignite it. From the nozzle side we introduce a bag
of pyrotechnic charge, inside the bag we have a squib; we ignite the pyrotechnic charge
over here, a flame is formed and it impinges over the surface of the propellant and the
surface catch fire.
Whereas if we something like a radial burning grain or a grain like a star grain which
burns from inside to the outside. We put the igniter in the cavity volume; maybe we
would like a part of the exposed propellant surface to catch fire. And then this fire
spreads over here till the entire propellant surface catches fire. This is how we ignite the
total propellant surface. If we have a bag of pyrotechnic powder, we cannot have
controlled burning and therefore very often the pyrotechnic composition is compressed
and made in the form of pellets. What do you mean by pellets? You know we take this
Anacin or Aspro for mild headache and the tablet is in the form of some pellets. We say
Aspirin tablets. Instead of having a powder charge; you form pellets like this small
Aspirin pellets like this. And what is the advantage of having solid pellets like this. The
burning surface area that burns can be controlled; it is not like a powder, which
immediately burns. It takes some time for a pellet to burn and therefore, it can give better
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ignition. That means, we can sustain the ignition source for some time. Instead of having
a bag containing powder, we could have something like a firm bag or a let us say a
cylindrical tube containing the pellets as shown.
We put a lot of pellets in it of the pyrotechnic powder. We put the squib and we ignite
the pellets. We make some holes here, through which the flame or the plume goes out
and impinges on the propellant surface and ignites it. This is the local ignition and is
followed by flame spread. This is how the igniter functions. And what is the requirement
of an igniter? To transfer heat or energy to the propellant surface for which the black
powder could be used
(Refer Slide Time: 12:38)
And black powder consists of KNO3, carbon and sulphur. You know the products of
burning are essentially gaseous except for little bit of carbon in solid phase. But all of us
know that if we can put some metal into it like let us say I put iron powder or I put
something like copper filings into it; when we heat iron or copper to high temperatures,
we get molten iron or be molten copper. And what is the advantage of using this metal
powder or filing in the igniter composition? A gas cools down when it expands whereas
a solid retains heat for some time. And also if I have a surface and on the surface may be
a molten iron falls it will transfer the heat of the molten iron into the surface more
effectively then a gas will transfer it.
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And a molten iron or a molten hot substance is in better contact with the surface, and it is
able to able to conduct the heat to the surface much more effectively than a gas.
Therefore most of the igniter compositions also have metal powder. And why do we add
metal powder? The reason is a hot liquid metal conducts heat much better onto a surface
than a gas.
The simple experiment, which we saw when we lit a sparkler; the composition that did
not have metal it was not that violent, but when we had metal filings in it, and when the
sparkles fell on my hand it got burnt. The reason is that the hot metal is able to conduct
heat much better. Therefore, the pyrotechnic composition will consist of may be some
metal powders and metal powders which are used include aluminum and boron. These
are also used in the composition of pellets of the pyrotechnic powder.
(Refer Slide Time: 14:45)
We therefore also include some metal powder in the igniter charge. However, there are
some basic issues and let us let us try to resolve some of these issues. You know instead
of having metals in the pyrotechnic charge, there are certain substances known as
thermites. Thermite reactions are those in which we have metals reacting, metals reacting
with oxides. If we take rust Fe2O3 and react it with aluminum Al what we get is 2 Fe +
Al2O3. This reaction is very exothermic and what we form is molten iron and aluminum
oxide. If we can use this as an igniter well it has a metal constituent in it in molten form
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that will touch the propellant surface. It will ignite much easier. Such reactions are
known as thermite reaction and these substances are known as thermites.
You know this is the time we should be looking at thermites as we find research work
going on in the area of nano thermites, which are more effective in producing heat.
(Refer Slide Time: 16:44)
If, instead of a solid propellant rocket, we want to ignite let us say a liquid hydrogen
liquid oxygen rocket. I have hydrogen oxygen which must be ignited. And one of the
contenders for this you just use the thermite mixture as an igniter. We have an igniter
over here. We spray Fe2O3 plus molten aluminium and ignite the mixture of liquid
hydrogen and liquid oxygen. We get molten substances and molten substances retain
heat for a long time and it will ensure that the mixture gets ignited. Therefore, such of the
igniters using thermite mixtures are called as a thermite igniters.
Therefore what is it we have considered so far? We said an igniter could consist of a
composition which generates something like a plume or a hot gas jet or if we were to put
metal in it will also create some metal or some hot molten metal which will transfer heat
to the surface better and give rise to ignition of a solid propellant surface.
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(Refer Slide Time: 17:43)
When we say ignition of a solid propellant, what is the mechanism by which a solid
propellant ignites? When heat is transferred to a solid propellant, some vapors of the
hydrocarbon are generated from the fuel and AP dissociates into a mono propellant
flame. The products so formed could mix together and form a flame. And therefore, the
process of ignition could happen in the gas phase, wherein vapors ignite. It could happen
at this at the solid propellant surfaces, wherein the products of dissociation formed in the
gas phase could readily react at the surface, giving rise to surface reactions. Reactions
could also occur in the solid phase. All these three reactions viz., gas, surface and solid
phase are possible, but it is difficult to say which one dominates and under what
conditions. And we will assume that all the three reactions take place viz., gas phase,
heterogeneous at the surface and in solid phase, which lead to ignition of a solid
propellant.
That means a solid propellant, if we have a slab of solid propellant, this is the surface and
we consider transfer some energy to it, in the solid part of it some reactions take place, in
the gas phase above the surface some reactions takes place, some surface reactions also
take place. I could model it using any of these three theories or combination of two or
three theories and I could find out what is the critical condition for ignition. I will not get
into details other than say that we should supply some ignition energy greater than some
threshold limit so that a propellant ignites. This is a subject by itself. But we know that
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if the energy is sufficient, the plume ignites the surface of propellant over which it is
incident and ignition is achieved.
However, if the pressure in the cavity is small, the ignition energy must be large; this is
because if pressure is higher the flame surface will be nearer the propellant surface. We
therefore should ensure that a minimum pressure is formed in the cavity by the igniter.
(Refer Slide Time: 20:16)
This brings us to the point on the role of the igniter. It supplies the necessary ignition
energy to the propellant surface and also pressurizes the chamber cavity to some
threshold value of pressure. The pressure is about 15 atmospheres. What happens is
when the pressure is low? We noted that the flame standoff is higher at lower pressures
while when pressure is higher the standoff is lower. I make sure that the standoff
distance is small such that the propellant surface ignition is sustained much better and
this we will see when we study the combustion instability of solid propellant rockets. If
the pressure is greater than some threshold value, the solid propellant combustion is
more stable and therefore the role of an igniter is to ensure that it pressurizes the
chamber to a value greater than some threshold limit. It also supplies some energy
greater than some threshold value for ignition to occur. This is all what is required from
an igniter.
598
And what does the igniter therefore do. We have a propellant grain and we have an
igniter here; it generates a plume may be from the pellets within it burn and the plume
impinges over part of the propellant surface.
(Refer Slide Time: 21:20)
If the energy transferred to this surface is greater than the energy required for ignition
and if the pressure in this cavity is greater than some limit, the propellant surface ignites.
We have something like local ignition of a small part of the propellant surface over
which the igniter plume impinges. And when this part ignites, heat which is generated in
this zone plus the heat which is generated by the igniter helps to supply the necessary
energy to ignite the adjacent surface and therefore, the flame keeps on spreading. And
we call this spreading of the flame as flame spread. That means flame spreads from the
local ignition area over the entire surface of the propellant. But the pressure is still not
equal to the equilibrium value and therefore thereafter the pressure increases to the
equilibrium value. And we calculated the equilibrium value as equal to we said = (Sb × a
× ρp × C*/ At )1/(1−n).
599
(Refer Slide Time: 23:03)
Therefore let us put this down on a figure such that we clearly understand it. What is it
we were discussing in the earlier classes? We said that the pressure with respect to time
would be the equilibrium value if we consider a radial burning grain? This is the pressure
corresponds to t0 when the surface is ignited and at burn out the pressure corresponds to
time tf. The pressure increases monotonically between t0 and tf because of the
progressive burning.
Now when we ignite the motor, we start with ambient value of pressure and the pressure
increases to the value at t0.
We have the igniter composition increasing the pressure to some threshold value. Let us
say from 1 when it increases the value to a threshold value and heat transfer takes place,
we move from 1 to 2, which is local ignition of a small part of the surface. That means,
only a part of the propellant surface gets ignited initially, because the plume is impinging
on it, the hot metals impinge on it and ignites it. The energy is further released from here
and flow takes place and ignites the balance surface of the propellant. Therefore, we
have something like a flame spread from 2 to 3 and flame spreads over the entire surface
over the propellant. We call this zone as flame spread and when once flame spreads over
the surface, we have reached this pressure, which is still less than the equilibrium
pressure. And then the pressure in the chamber increases or there is something like
pressurization in the cavity.
600
These are the 3 processes namely local ignition followed by flame spread followed by
cavity pressurization to the equilibrium value. And this is the equilibrium value to get
started with and thereafter progressive burning of the propellant grain takes place. We
would like to write equations for these 3 phases so that we can find out the time required
for ignition and the transient. How do we do it? How do we determine this rate of
pressure evolution and this portion wherein ignition takes place? Rather we would like to
determine the transient during ignition? Let us try to write an equation for some of these
processes. It is quite simple if we really get into the details. Nothing complicated and we
have done much more difficult problems trying to determine how the pressure should
evolve in a complicated grain shape.
(Refer Slide Time: 25:41)
We would like to solve the mass balance equation. We say the rate at which mass is
added by the igniter; during the process of flame spread, mass is not only added by the
igniter and also the ignited burning surface area is also adding mass as the flame spreads.
We are adding more and more mass as the more of surface of the propellant is getting
ignited.
Therefore, the rate at which mass is added to the propellant dm/dt = the rate at which the
igniter adds mass + the rate at which the burning propellant adds mass; let us call it as
m°ig + m°p and this minus the mass flow rate which nozzle leaves through the nozzle
mn°. When we talk of local ignition the only the m° is from igniter adding mass, because
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we are still in the first phase wherein the propellant surface is not yet ignited. It is only
dm/dt the mass added by the igniter is equal to m°ig. When we talk of the second phase,
we have m°ig + m°p, which is changing with time as the flame is spreading. We should
have m°, which is leaving through the nozzle as − m°n which is leaving. And during the
pressurization to equilibrium time, the igniter function is over we just have m°p when the
whole surface of the propellant grain is burning − m°n, which is leaving which is equal
to dm/dt. If the igniter mass is present during this last phase, it would be very much
smaller than the mass generated by the surface of the propellant.
During the local ignition phase followed by the flame spread phase, the igniter is still
supplying energy while part of the propellant surface which is burning is also supplying
energy or supplying mass to the hot gases. And some hot gases are leaving through the
nozzle. We have in the final phase, when the entire propellant surface catches fire, the
pressurization of the cavity. These are the equations that we could solve to determine the
variation of pressure with time during the process of ignition.
(Refer Slide Time: 27:56)
From ideal gas equation, we get m = PV/RT, where V is the volume of the cavity or the
cavity volume and therefore we can write the equation for dm/dt. The volume V during
ignition is about a constant since there is hardly any significant regression of the surface.
We can take volume V as a constant and also temperature of the products T as a
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constant. We therefore write dp/dt ×V/RT = m°ig + m°p −m°n. But m°n = 1/C* × p × At
where the chamber pressure is denoted by p.
And now if we are given the rate at which the igniter is supplying mass, we know m°p is
equal to the mass from the burning surface area, we can find the value of dp/dt. The burn
rate is apn. We can solve for dp/dt and we can determine the pressure and also the thrust.
Now, let us take a look at this figure once again, and try to draw some inferences and see
if some rational approximations may be introduced.
(Refer Slide Time: 29:14)
Supposing as a first case we consider a small solid propellant rocket; and let us consider
for ease an end burning grain. This is the propellant grain, the cavity volume is quite
small. In the second case, we consider is a huge solid propellant rocket, which is very
much larger. And when we say the initial volume is large it means that we have
something like with a radially burning propellant grain. We want a large burning surface
area. Correspondingly, we have a larger area for the cavity or port volume V. We know
the volume case 2 is very much larger than the volume for case 1.
We put an igniter; a controlled igniter with pellets in the case of the end burning small
rocket. We make sure that the pellets squirts fire and firebrands on the end - burning
surface of the propellant grain. Now what is going to be the change in the transient for
pressure for the small solid propellant rocket compared to a large solid propellant rocket?
Apparently in the small end burning grain, the surface will locally ignite. The entire
603
surface of the propellant grain is directly ignited by the igniter in the case of the small
end - burning rocket. This is what we would expect.
(Refer Slide Time: 30:50)
The sequence of events may be seen in the pressure versus time figure. At the start of
ignition, the pressure immediately goes up and reaches the equilibrium value for neutral
burning. The local ignition mainly constitutes the process of ignition. In other words, the
plumes from the igniter directly impinge and ignite the propellant surface. There is
hardly any flame spread over the propellant surface. We can say that 0 to 1 will be now
the local ignition, 1 to 2 is the flame spread which is very small region compared to 0 to
1, then we have the equilibration to final pressure taking place which is again very small.
This last part is negligible because the volume V is small.
In other words for a small solid propellant rocket the ignition process is governed by
local ignition. If we have a large cavity volume and a large rocket, well local ignition
first takes place. Since we have a large volume, it takes time to pressurize it therefore,
the pressure build up starts very slowly. We need a minimum pressure and minimum
ignition energy to initiate the burning. Therefore, from 0 to 1 which now we call as local
ignition slowly takes place. Then the flame spreads over the surface. The surface area
being large, the flame spread continues to spread at low pressure till flame spreads over
the total surface of the propellant and we reach point 2. After this we have to the pressure
604
from pressure at 2 changing to the equilibrium value and we have something like
pressurization of the cavity between 2 and 3. At 3 equilibrium pressure is reached.
Therefore the sequence is consisting of local ignition followed by a significant portion of
flame spread compared to a small rocket. And this portion of chamber filling to the
equilibrium value also tends to be a large fraction. We would like to predict what must
be the shape of this curve viz., how the pressure increases. Let us address this particular
curve. We write a simple mass balance equation.
(Refer Slide Time: 33:31)
What happens during the chamber pressurization in the case of the large rocket? Well
igniter has finished its job; it has ignited the surface and flame spread has happened.
Therefore, we can write dm/dt as corresponding to the entire surface of the propellant
catching fire. Therefore, the burning surface area × apn × propellant density is the rate of
mass generation. Even if the igniter continues to supply mass, it will be negligibly small
compared to this large value of mass flow rate from the propellant surface. The mass
which is leaving is – 1/C* × p × At where p is the chamber pressure. We can call the
chamber pressure as pc or p. Now dm/dt has already been written as equal to dp/dt ×
V/RT. Therefore, we have RT/V × Sb × a pn × ρp − 1/C* × p × At. This is what the mass
balance equation gives. Let us solve this equation. But since we have so many variables
can we take the variables out and put it in terms of some non-dimensional numbers?
605
(Refer Slide Time: 34:38)
We already know that (C* × ρp × a ×Sb / At)1/(1−n) = equilibrium pressure peq. We also
know that the value of C* = √RT/Γ where Γ = √γ × (2/(γ+1))(γ−1)/2(γ+1). And we can write
the non dimensional pressure in the chamber at any time as the ratio of p by p
equilibrium instead of p.
We write the value as dp/dt is equal to: instead of RT we write Γ2 ×C*2 and simplifying
by taking C* outside Γ2C* we Sb a C* ρp ; we also take At outside. Therefore we are left
with the pressure p left over. We find C star and C star gets cancelled. We have volume
of the cavity divided by the throat area. This gives us the length called as L*. We define
it as equal to initial volume of cavity or port volume divided by throat area (L*) or the
initial value of L*. And therefore, we can write this equation as equal to dp/dt = Γ2C*/L*
and C*/L* represents a velocity divided by length which corresponds to one over a
reference time. We have Sb × a × C* × ρp / A t is equal to equilibrium pressure peq1−n.
This is because the above term to the power 1 over 1 minus n is p equilibrium.
The burn rate of the propellant is given by a pn. Therefore we have pn over here. We have
S b a ρp C*/At as peq1−n. Simplifying, we have: dp/dt = Γ2C*/L*{pn(peq)1−n p} as shown
in the slide.
606
(Refer Slide Time: 38:47)
We non dimensionalize p by the equilibrium pressure peq and call it as p-. And now we
have time t in dp/dt. We non dimensionalize t with respect to a characteristic time. We
have L*, Γ is anyway a constant and is divided by the characteristic velocity C*. This
represents a time which is again a characteristic time. Therefore, now we say t by t
characteristic is equal to a non-dimensional time t-. The characteristic time is equal to
L*/ C*Γ2. Length divided by velocity gives unit of time and we call this as characteristic
time. We will develop on this further when we study combustion instability in rockets.
And now if were to introduce these two non-dimensional terms in the equation, what is
the final equation that we get?
607
(Refer Slide Time: 40:02)
We get d (p/peq) on the left. And since we have brought 1 over peq on the left therefore,
it should also divide the terms on the right. Dividing t by t characteristic gives dt- in the
denominator on the left side. Since we divided the right side by peq, this is equal to pn /
peqn and the next term is p by p equilibrium. Or rather this equation is now telling us
that d of non-dimensional pressure divided by d of non-dimensional time = p-n – p-. This
is a final equation that get. And now we can solve this equation by writing as dp bar by
dt bar as equal to p bar to power n minus p bar. Let us integrate this equation from a
value of p bar at time of flame spread to the equilibrium value at time when the
equilibrium pressure is reached.
608
(Refer Slide Time: 41:16)
Let us put the value pressure and time. At time t2 the non-dimensional pressure is p2 bar
while at time t3 it is 1. As a function of the non dimensional time starting from a value t2
bar, it goes the equilibrium value. Equilibrium value is p bar is defined as a pressure by
equilibrium value which is 1. We have t2 as the time at which the flame spread is
completed. And between 2 to 3 wherein we get the equilibrium condition we are
interested in finding out the time taken namely from the value of t2 to the value of t3.
The pressure value is to be integrated from p2 bar to one. Let us find out what is the time
required for the pressure to go from the end of p2 bar to one. That is the entire propellant
grain has ignited to the condition when there is there is some pressure p bar in the
system. Now that means, I am interested in this particular time t bar to reach the
equilibrium value from the initial time t2 bar.
609
(Refer Slide Time: 42:38)
We integrate the equation between p2 bar and a value of p bar corresponding to time
between t2 bar and t bar. The time changes from t 2 over here to a value t and if the time
is also non dimensional t 2 bar to t bar. Therefore if we want to integrate we have on the
left side dp bar by p bar to the power n minus p bar. And this is a standard integral which
can be integrated by parts. And we get the value as natural logarithm of 1 divided by 1
minus p bar to the power 1 minus n. And this changes from p2 bar to a value of p bar.
And the net value will therefore, be equal to ln of 1 minus p 2 bar to the power 1 minus
n, divided by 1 minus p bar to the power 1 minus n. When we look at the right hand side
it are quite simple I get the value between t bar minus t 2 bar over here. The above
expression is shown in the slide
610
(Refer Slide Time: 44:39)
Therefore, the final expression what we get is tbar minus t2bar (non dimensional times)
is equal to natural logarithm of 1 minus p2bar to the power 1 minus n divided by 1 minus
pbar to the power 1 minus n. If we are interested in finding out the time to reach
equilibrium pressure, well I substitute the value as 3 over here; I substitute the value as
equilibrium or p3 here which in non-dimensional form is 1. Therefore, the denominator
becomes 0 and therefore, the time taken to reach the equilibrium state is infinite.
Therefore the trend of variation of pressure is such that initially there is progressive
increase thereafter it droops and it takes infinite time or a long time to reach the
equilibrium pressure.
611
(Refer Slide Time: 45:44)
The ignition transient is such that the pressure starts off and increases but it droops later
and approaches the equilibrium value. Using this trend we will try to define some
characteristic times for burning and ignition subsequently. The ignition transient is seen
to have a characteristic pattern, which starts increasing initially but then droops and
reaches the equilibrium value. And thereafter if it is neutral burning it goes like this, if it
a progressive burning it increase thereafter while if we have regressive burning, the
pressure drops further. And this is how we predict the ignition transient.
Therefore we talked in terms of pyrotechnic charge; we talked in terms of metal
powders, we talked in terms of thermite igniter. Initially for a large rocket, local ignition
takes place followed by flame spread and then transition to equilibrium like the rise
followed by the droop. If we have a very large solid propellant rocket and most of the
solid propellant rockets whether it is for missiles or whether it is a launch vehicle, the
initial port volume is quite large.
The amount of charge also increases with the volume. Would we be able to calculate the
mass of charge required in the igniter?
612
(Refer Slide Time: 47:23)
The mass of charge should be capable of increasing the pressure in the cavity to a given
stable value of burning pressure and also ensure that sufficient energy is imparted for
local ignition of a part of the grain surface. If we have larger cavity, we must have more
charge and as the volume keeps on increasing, the mass of charge also increases. And
you cannot have large amount of pyrotechnic charge as its burning is not well controlled.
If we need mass of igniter as a large mass, we can as well use a small rocket itself for the
rocket igniter.
That means, that we use the plume from small solid propellant rocket to ignite the large
solid propellant rockets. We get the plume coming over the propellant surface, it ignites
the surface of the large solid propellant rockets. The solid propellant rockets, which are
used as igniters for large rockets, are known as pyrogen igniters. And almost all the solid
propellant rockets, which are developed make use of pyrogen igniters. This is true for
both the booster rockets and rockets for upper stages.
Pyrogen igniter is a small solid propellant rocket, which is used for igniting a large
rocket, but this rocket should not contain aluminum, because the nozzle will tend to get
clogged. And therefore, pyrogen igniters use non-aluminized propellants. It could be
HTPB it could be PBAN or CTPB based. Therefore, this is all about igniters.
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(Refer Slide Time: 49:48)
When we ignite a solid propellant rocket, we know how to calculate the equilibrium
pressure and the transient during ignition. When the rocket ceases to burn, how does the
pressure fall? And we use the same theory again.
We say dm/dt after burn out of the rocket motor, is equal to well igniter is not there. All
the propellant has ignited it is also not there. And only thing which is there is m°n which
is leaving the nozzle and therefore, the dm/dt = − m°n. let us quickly integrate it and
examine the result.
(Refer Slide Time: 50:36)
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We have dm/dt = V/RT ×dp/dt. What has happened the entire propellant has burnt out
but the chamber pressure is still quite high. The mass leaving through the nozzle is minus
chamber pressure into At by C star. The volume in the chamber does not change and V is
a constant because all the propellant has burnt. We have the temperature is still the
higher value over here and it could be assumed as a constant. Therefore, we can write it
as V/RT × dp/dt = − m°n = − p × At / C*.
(Refer Slide Time: 51:43)
Let’s let us solve this equation. We have dp/dt = − RT/V × p × At /C*. The product of
RT can be written in terms of C* since C* = √RT/Γ. Therefore, we substitute Γ2C*2 for
RT. This is divided by V and multiplied by At and divided by C* and multiplied by p.
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(Refer Slide Time: 52:36)
We substitute V by At as equal to L* and retain the dimensional form of the equation, we
can write it as dp/dt = − Γ2C*/L* ×p. We therefore get dp/p = − Γ2C*/L* ×dt. And on
integration, we get the value of natural logarithm of p divided by the equilibrium value
which at this particular point of burnout is equal to p4 viz., ln(p/p4) = − Γ2C*/L* × (t−t4),
where t4 corresponds to the time of equilibrium pressure at which the grain got burnt out.
(Refer Slide Time: 53:27)
And therefore, what is it we see? The pressure keeps coming down logarithmically or
rather the pressure should come down and to reach ambient pressure it is going to take
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infinite time. Rather the slope would decrease with time. This is all about the changing
pressure in the rocket chamber once the propellant has burnt out.
We are still left with one or two small things regarding the solid propellant rockets. The
characteristic times involved in a rocket and some examples of some big solid propellant
rockets. We shall address them in the next class. But in the class today, we looked at
ignition and igniters and we also covered briefly about how long it takes to reach the
ambient pressure once the propellant grain has burnt out.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 25
Review of Solid Propellant Rockets
We will finish our discussions on the solid propellant rockets today. Let us start with
something amusing.
(Refer Slide Time: 00:22)
Solid propellants rocket, abbreviated as SPR are known as solid propellant rocket
motors. Whereas, when we talk of liquid propellant rockets, liquid propellant rockets are
known as liquid propellant rocket engines. What do you think is the reason calling one as
motor and the other as engine? You know in many textbooks you find this solid propellant
rocket referred to as a motor whereas, liquid propellant rocket referred to as an engine.
What do you think would be the reason?
Let us go back and look at the construction of a solid propellant rocket. We have a
case in which we put some insulation. We will revise it again towards the end of the
class. Then I have a nozzle. The propellant grain, which could be a radial burning grain
could be a star or something else is contained within the case.
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And what else we did. An igniter is placed in the port volume of the grain. It
generates a hot plume and ignites the propellant grain. If we talk in terms of a liquid
propellant rocket engine, we should have tanks, which carry the liquid, we have the
propellant lines to the chamber. And to be able to pump the liquid propellants, let us say
the liquid fuel we need a pump, we similarly need a pump here for the oxidizer.
Therefore, we have moving parts such as pumps, which moves whereas to drive the pump
we need again a turbine. we have moving parts in a liquid propellant rocket. Whereas, a
solid propellant rocket has no moving parts; it is just simple case enclosing the propellant
and igniter. Therefore, for some reason or the other a solid propellant rocket, because it
has no moving part is referred to as a motor. In fact, the case is referred to as a motor
case. A liquid propellant rocket considering that it has moving parts is referred to as
engine.
(Refer Slide Time: 02:54)
Now with this introduction let us see where we were in the last class. We
discussed the igniters. And let me just briefly go through what we covered in igniters
again. See before discussing igniters we were very clear how to be able to define the
burning surface area; the configuration of the grain according to the amount of thrust
which is required. We could design the grain. The igniter jet, impinges on the propellant
surface, pressurizes the cavity and then ignition takes place. We had different types of
igniters: one was a pyrotechnic igniter in which we have a charge which is easily ignitable
whereas we also talked in terms of a pyrogen igniter wherein we put a small rocket motor
itself as the igniter in a larger rocket.
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What is the principle? Let us just say whenever we make a fire - like for instance I
want to light a candle let us say. I use a matchstick and light this candle. I cannot use this
matchstick if I were to light a sparkler. I do not generally use a matchstick, because it
requires more sustained flame to ignite it. And therefore, I use a candle for lighting this
sparkler. Now again I say I have something like a Bengal pot, it is something like a mud
pot in which I put some pyrotechnic composition. I cover it over here it is something like
this, I light it over here and we produce a plume of sparkles. This is known as a Bengal
pot, because this type of fire cracker originated in India in Bengal and therefore, to light
it, I use a sparkler. I show the sparkler here and light this. What is it we see? A small fire
is required to make a little bigger fire. And a bigger fire is required to initiate an yet
bigger fire? And with this bigger fire we can make a still bigger fire.
That means, in practice to be able to ignite anything a small fire is required to
make a bigger fire and so on. A bigger fire is required to ignite a still bigger fire; and that
is how things are if we have a furnace we do not put an electrostatic spark to ignite the
fuel air mixture in it. We create a pilot flame with the pilot flame we ignite it and so on.
And so also in solid propellant rockets what we do is we use a small rocket motor over
here. And that small rocket motor has a nozzle; let us say this is the case over here, we put
a small rocket over here. That means, we have something like a nozzle over here, we have
another igniter for it, and this will contain a squib for its igniter. It is similar to a small fire
makes a bigger fire, maybe makes a still bigger fire and so on. The pyrogen igniter and
makes a bigger fire which ignites the rocket.
(Refer Slide Time: 06:10)
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Therefore a pyrogen is a small rocket, which ignites the main rocket. And what
did we note? If we have a large rocket and if we look at the pressure time trace, we have
something like it initially ignites over a small surface locally. What the igniter does is it
pressurizes the chamber to some small value. And also transfers heat over here and
therefore, we have local ignition from let us say 0 to 1. And then the flame spreads over
the surface to 2 and after the flame has spread it reaches the equilibrium pressure which
we say is p equilibrium.
We derived expressions and we found that the transients could be easily be
predicted. How did we predict the pressures? We had dm/dt that is the rate of mass
accumulated or developed in this cavity. The rate of change of mass is equal to the rate at
which the igniter supplies the mass at that particular time plus the contribution, which
comes from the burning of the propellant and the spread of the propellant minus the rate
at which the flow takes place through the nozzle. And we were able to say m is equal to
PV/RT. And we took the simple case where in dm/dt corresponds to the condition when
the surface entire surface of the propellant has just got ignited that is local ignition of a
small surface followed by the entire surface getting ignited and we were able to get the
equation to this curve and what was the equation?
(Refer Slide Time: 07:47)
We wrote an expression for dm/dt. And you see how simple it is; m is equal to
PV/RT that is dp/dt × volume (volume is constant) V/RT = the entire surface is burning
Sb × r viz., ‘a’ × pn × ρp i.e., the rate at which mass is getting generated – 1/C* × p × At.
Therefore when we solve this equation, we had this within the bracket, Sb, ‘a’, p to the
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power n density of the propellant minus 1/C* × p × At. And what else did we do? We said
C* was equal to √RT/Γ, where Γ = √γ (2/(γ+1)(γ−1)?2(γ+1).
We replaced RT by Γ2/C*2 in terms of capital gamma squared into C star squared and we
were equation we were able to get the equation in a non-dimensional form. How did we
get p bar? We said p bar is equal to pressure at any point in time p divided by equilibrium
pressure and this was the final steady state value at burnout. And we defined a
characteristic length which came from V/At = L*. L*/C* we said has a unit of time; we
called it as characteristic time. And we said we will take a look at it when we study
combustion instability. We also said that the non-dimensional time t bar is equal to t by t
characteristic. And therefore, we were able to get say t characteristic here by t bar. We got
dp bar by dt bar is equal to p bar to power n – p bar.
(Refer Slide Time: 10:02)
We integrated this expression and got it in the ln form; that means pressure at any
point or the time after event 2 between let us say between 2 to 3 was derived as logarithm of 1
minus p at 2 non dimensionlized to the power 1 minus n, to the power 1 minus p at anytime to
the power 1 minus n is the expression for the time. Or rather we found it droops after some
particular time. We followed the same logic to be able to find out what will be the variation of
pressure after the propellant burns out in a rocket. Let us say p over here, t over here on the X
axis; and we said the motor ignites, keeps on burning till all the propellant gets burnt. What
will be the signature for the pressure transient after the propellant gets burnt? What will be
the equation to describe this event? Does the pressure go like this or does it go exponentially
like this; we were interested in the shape. And the equation we got for this was quite similar
for the ignition events. All the propellant is getting consumed over here. Therefore, the
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equation for that particular case of depletion that we derived in the last class was dp by dt was
again equal to RT by V into the rate of mass depletion by nozzle which is minus p × At / C*
i.e., m°n.
(Refer Slide Time: 11:28)
And we followed the same non-dimensional procedure. RT is equal to capital
gamma squared × C*2. And therefore we got gamma squared into C star squared divided
by V, we took At outside, and the value of C* and p and this negative sign of mass leving
the rocket. And with V/At = L* i.e., volume by throat area, which is the characteristic
length at burn out of the propellant grain. And we had C* over whose unit is velocity and
with V/At = L*, we could also write this particular equation in the form as 1 over L* ÷ C*
which is equal to the characteristic time with a negative sign; i.e., 1 over characteristic
time, because length over velocity has a unit of time. And therefore for dp/dt = p divided
by the characteristic time.
(Refer Slide Time: 12:49)
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We can write this equation as dp bar by dividing both the sides by equilibrium
pressure. By bringing the characteristic time tch to the left side, we get dt bar that is non
dimensional time. This is equal to p bar. Or rather this equation expresses the differential
of non-dimensional pressure with respect to non-dimensional time. Mind you there is a
minus sign.
(Refer Slide Time: 13:11)
And this tells me that dt is equal to − dp/p. Rather the time taken after the burning
of the propellant is completed, tb is when burning gets completed; we have t − tb.
Therefore, we get t − tb is equal to natural logarithm of the pressure by pressure at burn
out. The decay is of pressure would be exponential and to reach zero value, it is going to
take a very longtime. And we did it dimensionally the other day. We would have got Γ 2 C*
/L*. We must be able to do this in different ways. We could have got dp/dt = −Γ 2 C*/L*.
And if we were to integrate, we get ln p = − Γ 2C*/L*(t-tb). That means, the pressure
continuously decays with time.
(Refer Slide Time: 16:05)
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Why are we repeating this? See there must be some reason. The reason is
whenever a rocket motor ignites, the pressure changes with time, starts slowly
building up as flame spread progresses, then reaches after chamber filling interval the
neutral, progressive, or regressive equilibrium value. This particular zone is
equilibrium pressure; that means a steady state pressure. At the end of this what
happens is well all the propellant gets consumed and the chamber pressure and thrust
decays out. We have the transient for ignition; the ignition transient followed by a
period of steady or equilibrium burning which will be much longer, and then the time
when it burns out or you have we called it as tail off. How do we use this signature of
the transients and equilibrium burning to define the effective burning times? It
becomes a little complicated or subjective and there are standard procedures to do it.
Let us consider the case of neutral burning. We plot a tangent to this particular
pressure time trace in the zone of neutral equilibrium burning. So also we plot a
tangent to the curve in the ignition transient zone. We get a particular point of
intersection of the lines here. Similarly, we plot a tangent here in the tail off curve
and the point of intersection of the tangent of the tail off curve and the equilibrium
burning curve is obtained. We call these points as ‘a’ and ‘b’ respectively. And we
now see, burning is taking place during the end of the ignition transient and the initial
phase of the tail off and the precise points of the start and stop could not be found.
After the intersection of the tangents, we get the points a and b as the start of burning
and end of burning and the time from a to b is called the burn time. And it is denoted
by the symbol tb. And this is how we characterize a solid propellant rocket motor for
the burn time of the propellant. This means that the burn time of this motor is so
many seconds or so many minutes or so.
But we also realize during the period before start of the burn time, the rocket
motor is still giving us impulse or some momentum. It is contributing to impulse
even earlier and after the burn time. Therefore, when we want to define a mission and
for a mission a certain impulse is required. We therefore take the maximum pressure
value at the point of the intersection of the tangent line to the ignition transient and
the equilibrium pressure curves. This pressure is denoted by pa. We divide it by 1 by
10 that is 10 percent of the value. Similarly, I get the value of maximum pressure at
burnout pb. We take 1 by 10th of this value. That means, the pressure corresponding
to this decaying pressure point is equal to pressure corresponding to the b divided by
10. And now here also we get the certain impulse here. The impulse may be small.
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The particular time between 10% of the maximum pressure at the start and at the end
is called as the time of action of the motor or action time. We denote it by ta. In other
words, when we have to plan a mission, we get thrust over the action time. However,
to characterize the motor in a test or otherwise, we are interested in the burn time.
And we see that the action time is greater than the burn time.
(Refer Slide Time: 20:27)
Now let us examine one or two small problems we can have in solid propellant
rockets. Why we are considering this is whenever we make a rocket; let us say we
have a rocket as sketched here. Let us say it has a radial burning grain. We add an
igniter to it. And we told ourselves the other day most of the igniters are pyrogen
igniters, because normally the rocket motors are quite large.
We were quite clear how to go about making an igniter. The igniter must
pressurize this cavity to some value, not very high value, such that a flame can be
near the surface. The igniter must also give some energy to the propellant surface and
we said propellant requires some minimum energy for ignition. We have plumes
from the igniter and it ignites a particular surface. These were all the requirements
and thereafter the flame spread and pressurization take over. We had this particular
transient curve for local ignition flame spread and the pressurization of the cavity;
But sometimes when we do a test or an experiment we had the pressure going up
instead of following the ignition transient curve shown earlier.
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(Refer Slide Time: 21:53)
Very often the pressure goes like this in the form of a spike and comes down after
spiking to the equilibrium pressure. In other words we get a peak of pressure much
greater than the equilibrium value. That means, we get something like an ignition
peak in the process of ignition transient. The pressure peak can burst the case or
provide an unplanned thrust and acceleration and this is detrimental. Why should
such an event take place?
(Refer Slide Time: 22:25)
Let's take a look and address the parameters. How do we make a propellant grain?
We take a case motor case. Inside it we put a mandrel; if we want to make a
cylindrical grain we put a cylindrical rod, pour the propellant slurry between it and
the case. If we want to make a star grain well the shape of this mandrel would be star
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shaped. The slurry is then cured and we remove this mandrel. We have this particular
shape of the grain. And some times to remove the grain is difficult and therefore we
use some agents which a like silicon oil which are essentially insulators to be able to
easily remove the mandrel from the grain.
Now if the surface of the grain so formed is such that it is not easily ignitable.
What happens is that we are transferring energy the grain; it gets heated and as it
continues to get heated its temperature increases. When it begins to burn it starts
burning at a higher temperature, and since it starts burning at a higher temperature
the value of the burn rate is now influenced by the temperature sensitivity factor. And
therefore the burn rate is higher since burning takes place at a high temperature. It
produces much higher rate of mass or it burns with a higher speed. And therefore,
you have a higher amount of mass and energy, which is getting released and
therefore, the pressure could go up. This is one of the reasons for the spike in
pressure. The second reason could be, we have higher velocities especially towards
the nozzle end of the grain surface which if high enough can enhance the burn rates.
(Refer Slide Time: 24:19)
But, to be able to prevent the pressure spike what this normally done is we take
something like an emery paper and remove the surface defects and ensure that the
surface of the propellant is easily ignitable. Let us put down the points to prevent
ignition spike ignition in a solid propellant rocket. What we do is we emery a
surface. Take an emery paper may be make the surface make sure oxidizer and fuel
are readily available, and it catches fire easily. You have to make sure, that the
surface is such that some other reason like burning due to increased velocities, which
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we call as erosive burning are not possible. We will consider shortly about erosive
burning.
We can always tolerate a small value of the spike; but must guard against it.
Something that we missed out was that we often require to ignite a motor under low
ambient pressures or in vacuum of space.
(Refer Slide Time: 25:26)
Since we need to build pressure in the port volume, we put something like a
closure at the nozzle. When pressure builds up and the motor is ignited, the closure is
thrown out. We make sure that after adequate pressure is generated around 5 bar to 6
bar, the closure is dislodged. This closure is known as a nozzle closure.
This is all about solid propellant rockets. We have considered the propellant burn
rates, we have considered how to go about making grains of different configurations
to get thrust. And then we looked at igniter; we looked at the action time and the burn
time. And therefore, maybe we should put things together at this point in time before
we close our discussions on the solid propellant rockets.
Let us start with propellant burn rate r. How did we define the burn rate or how
did we determine the burn rate. We said that we make a propellant strand may be
something like a cm in diameter. We could put it in a chamber and pressurize the
chamber to whatever pressure we are interested in. Then we ignite the surface and
measure the burn rate when the burn propagates through a particular distance. We
control the pressure in this chamber. This particular chamber in which such strands
are burnt is known as Crawford bomb. It is something like a bomb type of a
calorimeter in which we burn the propellant, but all what we do is we put a series of
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fuse wires at known distance apart along the line of burning. An electronic timer is
used to find the time taken to burn between the individual segments of the strand.
The distance L divided by the time is the burn rate r at the particular chamber
pressure.
(Refer Slide Time: 26:21)
Can we use this burn rate in a rocket motor? Let us say the same end burning
configuration of propellant grain is used in the rocket motor. What we have is the
diameter of the grain is D and the throat diameter is d t. We want to find out the burn
rate of the grain the rocket chamber. We measure over here it gives me let us say 4
millimeters per second at a pressure of let us say 5 or let us say standard pressure 7
MPa. The question is will we get the same value of r as in the Crawford bomb or
should we get a different value. What is your take on this? Should it be the same as
measured in a strand at the same pressure? How would you look at this problem? We
again go back and write the simple equation, that we derived for burn rate.
(Refer Slide Time: 28:42)
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n
We got the equation r is equal to ap . How did we get this equation? We have the
surface and the flame standing off at a distance X* from the surface. And how did we
get the burn rate r. It is equal to the heat which is given over here, thermal
conductivity of the gas above the surface, into the temperature of the flame minus
temperature at the surface, divided by X* is the heat which is conducted divided by
ρp into specific heat into surface temperature minus the initial temperature plus the
exothermic heat release at the surface; we derived this based on the simple model.
Now can we look at this, for the experiment in the rocket and the experiment in
the strand burner? And would it be different in the two cases or should it be the
same? This is a perennial problem we have with solid propellant rockets. Now what
is happening in the Crawford bomb is that the ambient is all cold gas even though it
is at the same pressure. Here the ambient is hot gas therefore, we will have heat
radiation coming on the surface. In other words when we test a motor we will have
something like q radiation coming on the propellant surface. We could also have in a
radial burning grain q due to convection coming on the surface in addition to the
radiation.
And therefore, the burn rate in a motor should be higher than in when it is tested
in a strand burner. And therefore, to determine the burn rate what is done is you have
to test it in a small solid propellant rocket and these propellant grains are known as
control blocks or control rounds, because I cannot really use this standard Crawford
apparatus to determine the burn rate. I have to use a rocket configuration, because it
is more representative of the actual. May be when I am developing different
propellant formulations, we can screen them in a strand burner. But the final burn
rate is always derived with a small solid propellant rocket itself. Which is known as a
control round; in India we call control round as a Agni round, but let us not confuse it
with Agni missile.
The small rockets used for burn rate measurements may be of diameter around
200 mm and length around 400 mm with cylindrical burning. Burn rate is equal to
web thickness divided by the web burn time. And that is how we determine the burn
rates in practice. The problem now gets more confusing for the following reasons.
Let us consider two cases of solid rocket motors having the same composition of the
propellant grain. The solid propellant rocket used in space shuttle is a very large with
diameter is around 3.8 meters diameter the length is around 40 meters.
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(Refer Slide Time: 31:41)
It has a star shape grain, but we assume a simple radial burning. Let us say that I
have another motor of smaller dimensions using the same PBAN propellant as the
solid rocket booster. Let us say that this small rocket has a length of 10 meters long
and diameter of 1 meter. If the pressures in both are the same, will the burn rate
determined at the same pressure be the same or different. Again we look at the
radiation heat transfer; it depends on the mean beam length of radiation. Therefore, I
expect the burn rate in a larger motor to be different, but it is not necessarily true;
there are other factors like mechanical properties of the propellant. Why I say
mechanical properties mechanical properties could be hardness could be tensile
strength, could be the ductility of the propellant. And during burning I could have
some deformation taking place all those things are going to influence the web
thickness.
And therefore, scaling of burning rate with size of the motor or size of the rocket
is always of interest and a problem of interest. We observe as we go from a medium
size motor to larger size, the increase in burn rate increases by something like 4 to 6
percent generally. And after particular size, it is not significantly influenced by the
size. But we have to verify it through models. And what are the models we use; we
go back to our basics, write the equation find out what is the role of convection and
radiation and solve the problem.
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(Refer Slide Time: 33:52)
This brings me to the last point, namely if we were to have something like a long
propellant grain. Like let us say we have a internal burning grain, let us say radial and the
initial cavity or port diameter which is also defined as port of a rocket grain is of small
diameter. And as it burns at the surface let us say this is the propellant grain, gas is
coming out let us plot the value of velocity of the gases which is moving as a function of
length starting from the head end towards the nozzle end.
(Refer Slide Time: 36:00)
At the head end there is hardly any velocity, but as more and more gas is
generated, the velocity is increasing towards the nozzle end. Velocity is a maximum at the
nozzle end. Now what does velocity does to a surface which is burning. Well it can erode
the surface like in a river. Let us say a river is flowing and what does current of velocity
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do? It drags the sand from the river. So, also I could have something like erosion. Let us
write it down. The velocity could erode the propellant surface; in other words I could
have something like an erosive burning. Mind you propellant was heterogeneous;
composite, it is sort of eroding the surface or erosive burning, but more than erosive
burning we find velocity here is higher therefore, the Nusselt number or the Reynolds
number will be greater. We have Reynolds number as a function of length and it increases
with velocity. If Reynolds number increases, well the Nusselt number or heat transfer
coefficient is bound to go up.
Therefore we are also going to get increased convective heat transfer. In other
words we can talk in terms of erosive burning arising from convective heat transfer, and
when we do such a modeling and calculate the new value of heat transfer coefficient by
convection, it is equal to function of Reynolds number into Prandtl number. And this we
write in terms of Nusselt number as hd by k. And therefore, we can always find out the
Nusselt number and once we know heat transfer coefficient, we can find out what is that q
convection and find out the increase in heat transfer rate. When we do this we find that
the burning rate can be expressed in terms of a constant into something like a Mach
number into the pressure to the power c (r ≈ Mc_.
(Refer Slide Time: 36:40)
The value of the exponent c is typically between 0.7 and 0.8, which is something
like in the standard correlation for Nusselt number. It is equal to 0.023 into Reynolds
number to the power 0.8 into Prandtl number to the power 1 by 3 for turbulent flow. This
suggests that convective heat transfer does play a role. In addition to pressure we have
Mach number effects and this is what gives us the erosive burning.
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(Refer Slide Time: 37:11)
And this erosion effect, because it increases the burn rate can also lead to the
ignition spike since at this time the port volume is small and velocities would be large.
The larger mass burning rates could provide the spike.
(Refer Slide Time: 37:52)
Supposing let us say we are launching a particular solid propellant rocket. And to
be able to stabilize it sometimes is spun i.e., it rotates on its axis. We have let us say an
end burning grain, something like this it is burning over here, it is being launched like
this. We also have a spinning of a radial burning grain. And why do you spin to make it
stable like just like we have a top which when it spins is stable.
Now what is happening is the burning surface area. In the frame of reference of
the burning surface, we have aluminum, which is burning over here; it gets pushed
towards the surface. Therefore, I get the effect of local acceleration and the effect of
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acceleration is to be able to push it towards the surface. And therefore, we can say that
acceleration will affect the X* viz., the flame standoff and therefore, it will also affect the
burn rate. We can thus find the effect of acceleration due to the spinning. All we have to
find out is how the acceleration influences the stand of distance of the flame. If we can
find it out through a simple model, we can find out the influence on burn rate. We know
the centrifugal force from the acceleration. We know the mass of aluminum particles,
which are burning and therefore we can do this problem. Let us quickly revise through
and then address one or two of the very major issues which was faced in solid propellant
rockets namely the control of thrust. Let us quickly revise in two or three slides what we
have been talking of.
(Refer Slide Time: 39:58)
This is an igniter may be a pyrotechnic igniter. It produces these plumes which
impinges on the propellant surface, ignites the surface also pressurizes the cavity. These
are the individual plumes, which are igniting the surface. The flame spreads and then the
gases move out through the nozzle and this is how ignition takes place.
(Refer Slide Time: 40:26)
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Let us go to the next one. This is a pyrogen igniter, a small solid propellant rocket.
This has a pyrotechnic igniter here, squib over here burns here, ignites this surface and
flame moves forward. That means a pyrogen igniter is a small solid propellant rocket.
(Refer Slide Time: 40:54)
See this is how pyrogen propellant grain looks. It is like a solid propellant rocket
only you do not need to generate thrust. We provide a multi point star shown in red. When
the propellant here is ignited, it generates hot gases that ignite the main motor.
(Refer Slide Time: 41:13)
We also said that the hot gases must be contained within this port volume or cavity
to ensure ignition. We use a nozzle closure and the moment pressure builds up well this is
ejected out. And therefore, flow through the nozzle gets started after we make sure that
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the ignition takes place in a chamber. The port volume is sort of enclosed with this nozzle
closure. It is made of some ablative material and is bonded over here by glue. The
moment pressure is developed in the port volume, it is pushed out. We use such nozzle
closures in liquid propellant rocket engines also.
(Refer Slide Time: 41:56)
We talked in terms of burn time. Two tangents intersecting; this is shown for
progressive case: A to B is the burn time. The time of one- tenth of the pressure at ignition
to one-tenth of this pressure at burn out over here, is what is the action time. Well these
are all about the solid propellant rockets.
(Refer Slide Time: 42:15)
Now having done all this, let us put all components of a solid propellant rocket
together in a single figure. What are the components of a solid propellant rocket? Well
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propellant is basic and it is contained within an insulation or a liner. After this insulation
we have another liner, make sure that it is compatible with this insulation such that heat
does not get conducted and weaken the motor case or cause the case to burn off.
Therefore, we have a propellant, we have a case, we have insulation and liner is also a
form of insulation. And then we have a nozzle; the nozzle could be sunk into the
propellant and it could be made to flex. We have seen when we talked of nozzles. We
have a nozzle closure. We have an igniter, which could be a pyrogen igniter or a
pyrotechnic igniter. Well this is all what a solid propellant rocket consists of. And we said,
the solid propellant rocket is called a motor because there are no moving parts in a solid
propellant rocket. Having done all this I thought let us review two practical problems,
which have been encountered, during the history of development of different rockets. And
I just choose two of them, because all of us would have heard of these problems and let
try us clarify what really happened.
(Refer Slide Time: 43:37)
One is we talk in terms of solid rocket booster for space shuttle. Our interest in
this was because it is the world’s largest solid propellant rocket. It uses PBAN,
polybutadiene acrylic acid acrylonitrile as a propellant. This is the fuel binder. Of course,
it contains AP and large amount of aluminum as in all solid propellants. And what was the
problem? In one of the shuttle launches in 1986 in the flight Challenger the motor
misbehaved. And the entire crew of 7 died. It was the first time that they took a civilian
into space, they took a school teacher along. What was the problem?
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(Refer Slide Time: 44:54)
Let us try to understand what really went wrong. Well this figure shows the space
shuttle. What does the space shuttle consists of? It consists of a central engines which are
hydrogen oxygen cryogenic engines, there are 3 of them here clustered together and they
burn simultaneously. And behind the space plane we have a huge liquid hydrogen tank
and at the bottom of it you have the liquid oxygen tank, this is the huge liquid hydrogen
tank you need a huge tank, because liquid hydrogen is not very dense. You have two solid
rocket boosters. First what is done is these 3 liquid engines fire, it is ensured that adequate
thrust is developed, because you can always switch on and switch off a liquid propellant
rocket. And once it has developed a particular thrust, the two solid rocket boosters are
fired. Mind you this what we said is around 3.8 diameters and around 40 meters in height.
They begin to fire and in this particular launch it happened on a cold day.
(Refer Slide Time: 45:57)
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The temperature of the ambient was around minus 1 degree; it was around in the
morning around 8 clock or so that the launch was to take place. The previous night the
temperatures went down as low as minus 15 degree Celsius. And this shows the perfect
launch at take off. It takes off beautifully, but then after sometime around 0.6 seconds
after ignition of the solid rocket booster in this region in the right side engine little bit of
gas was found to escape.
(Refer Slide Time: 46:45)
Whenever we have such huge boosters, the construction of the propellant block as
a single grain is very difficult. The propellant is cast into different blocks and then
assembled together. Each block is known as a segment. And now therefore you make
small segments of the same diameter and the solid propellant segments in the case of
space shuttle consists of something like 6 segments. And what is done at the factory
where in these segments are made? Few of the segments are assembled in the factory
such that it is still transportable. The final segments are assembled together at the launch
site. How do you assemble the segments? We have the case over here, you need to make
sure that these two are put together or joined together such that the no leakage of high
pressure high temperature gas is possible once the huge motor ignites.
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(Refer Slide Time: 48:03)
Between one segment and the other may be have to put something some
insulation, then we have to put the other segment over here, this has to be joined together
maybe I should be able join it in some form at the interface. And how is it joined? We use
‘O’ rings. Let us try to make a sketch of how the O rings function. See how do you
assemble an ‘O’ ring in a groove? We insert the O ring in the grove. When we assemble
the cylindrical face of the other segment the O ring being flexible, it flows and makes this
junction to be air tight or leak tight. And therefore, 2 O rings are used and these O rings
are of rubber. The rubber O rings which were meant for this had not been tested for
temperatures less than 15 degree Celsius. The previous night was cold this particular
launch was on hold for some time. And therefore, what happened was that the O ring
which is resilient at ambient temperature becomes rigid and hard and when it becomes
hard it does not seal the joint properly and allows the gas to flow by.
(Refer Slide Time: 48:55)
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And therefore, at the segment joint some little bit of the high temperature gas from
the chamber begins to leak. And this was observed within 0.6 seconds after ignition. This
was observed and some minute amount of gas was beginning to escape. But you know
that the propellant used is highly aluminized. Therefore, what does aluminum oxide do, it
goes and blocks the leak path and prevents any further leak. The motor is still safe it
keeps on firing further and further.
(Refer Slide Time: 49:23)
From 0 to 0.6 seconds at which the “O” rings have failed up to something like 60
seconds or 62 seconds, the flight or operation of the solid rocket booster was perfect. The
aluminum oxide has blocked the leak path and the chamber remains normal.
(Refer Slide Time: 49:46)
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It then goes through the atmosphere at around may be something like 13 kilo
meters height Whenever we fire a missile or a rocket and it goes up through the
atmosphere, we have wind in a particular direction. But in some locations we have bottom
layer of wind in the opposite direction to the next upper layer. We called is as a wind
shear. Some layer of wind moves in this direction the adjacent layer of wind moves in the
opposite direction. When the space shuttle is moving up let us again put the events
together. Now the vehicle is moving up, and what is happening? It goes though wind
shear, it get’s shaken and therefore, at that point in time the accumulated aluminium oxide
which hold the leak gets breached.
(Refer Slide Time: 50:20)
At the bottom on the right hand side the leakage path opens out some flame comes
out. And when this flame comes out it hits against one of the attachments, which secures
the solid rocket booster to the main core rocket. And that gives way and the solid rocket
booster comes out and it gives a thrust in some other direction. In addition, the flame hits
against what we said is the hydrogen tank, ruptures it and spills the hydrogen. This
happened at a height of around 14 kilometers. And well the hydrogen mixed with air and
there is a huge fire ball and the entire mission is a failure.
Therefore, we see the corrective action of aluminum oxide in sealing the hole or
the leak path which is disrupted by the wind shear. In fact one of the recommendation is
whenever you use a pyrogen igniter we should not use considerable aluminium for the
propellant as the nozzle will get clogged.
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(Refer Slide Time: 51:11)
But in the case of solid rocket boosters it helped. But the failure was because of the O
rings which were not doing the job and the wind shear which dislodged the sealing by
aluminium oxide..
(Refer Slide Time: 51:17)
We will now consider one last example, which is interesting. We will perhaps
cover the details of when we look at instability in rockets.
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(Refer Slide Time: 51:35)
You know this example relates to the second largest solid rocket motor. We just
picked on these two; while the solid rocket boosters of space shuttle have something like
500 tones of propellant, this second biggest rocket has about 280 tones of propellant. This
is used in the launch vehicle Arianne. Arianne is a French rocket. And the solid propellant
rocket uses a HTPB based propellant, hydroxyl terminated polybutadiene. And in this
particular case, what happened is again being a large rocket with a number of segments.
And how do you assemble the segments? Well in between the segments, we put glue or
some inhibitor, join it together so that it is perfectly leak tight.
(Refer Slide Time: 52:17)
And let us now consider a segment joint. We have one segment over here; this is
the inner diameter; we have the next segment coming over her. We have the inhibitor or
joining glue over here in between. Now when the propellant burns the propellant burns
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fast where as this fellow viz., the inhibitor does not burn as fast. Therefore, after sometime
we have the inhibitors standing like this over the propellant surface. The flow of the gas in
the port is obstructed and eddies are found. These eddies in the flow have a characteristic
frequency and these disturbances get amplified and the thrust instead of being steady starts
oscillating. And the oscillation is because of eddies that are formed, because of the
projection of the inert in the propellant. Well we have this protrusion here, which causes
the pressure to oscillate. We will study the mechanism of oscillations in the chapter on
combustion instability. Well, this is all about solid propellant rockets; maybe we should try
different problems.
(Refer Slide Time: 53:47)
To sum up, if we are given the thrust of a solid propellant rocket, which is to be
made, and given the specific impulse of the rocket, we can find out what is a mass flow
rate required. We can use the C* of the propellant to find out what is the value of the
nozzle throat area At. We choose the pressure of operation of the solid propellant rocket p.
n
And the burn rate of the propellant is known; r = ap . We can find out the burn rate and
then solve for the burn surface area. And in the assignments, I have given you something
like 10 problems, which you should do. In the next class we will start with liquid
propellant rockets.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 26
Feed Systems for Liquid Propellant Rockets
We will start this new chapter on liquid propellant rockets.
(Refer Slide Time: 00:22)
When we discussed solid propellant rockets, we had a solid propellant grain which burns
at the surface, you had a flame near to the surface. And once a solid propellant rocket
begins to burn, it is almost impossible to extinguish it. Once you ignite the grain, there is
no way you can quench it, it continues to burn. Of course, there is some work done to see
under what conditions you can stop the burning. And by rapid depressurization you can
quench it, but this has not reached a stage of being applied for a rocket. Therefore, we
say a solid propellant rocket, once it gets started, you cannot control it, it burns and
burns, that is about it. Whereas, when we talk of a liquid propellant rocket, we have
considered the different liquid propellants and we inject the liquid propellant into the
chamber. We can always control it and therefore, there are certain advantages that liquid
propellant rockets have.
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But to get started with this liquid propellant rockets, I thought let us look at nature as we
did examine it when we considered the theory of rocket propulsion.
(Refer Slide Time: 01:45)
We talked of a small insect known as the bombardier beetle, you will recall. What did we
tell at that time? This particular insect beetle is something smallish may be an inch to 2
inches in size. And it is somewhat heavy and we find it all the time at all places, it sort of
unwieldy and it cannot fly that rapidly. And it is invariably attacked by insects. But it is
harmless, it does not bite, it cannot fly much. We just pick it up on a piece of paper and
throw it out whenever it flies in. One particular form of beetle, known as bombardier
beetle, has been investigated for the last 10 to 15 years. And an such article came in the
proceedings of the U S National Academy of Sciences in 1997. It is in volume 94. It runs
over 5 pages from pages 1692 to 1697. It makes very interesting reading.
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(Refer Slide Time: 02:46)
As I was telling you, this particular bombardier beetle is attacked by ants, and what does
ant do? It stings. Why do feel a sting? It injects some formic acid into our system.
Formic acid is something like carboxylic acid and that is why you feel a sting and that is
the same as bee also stings. We feel pain because something is injected into us.
(Refer Slide Time: 03:31)
In fact, the ant in Greek is known as Formica because when it stings us, it injects the
formic acid into us. The ant also pesters this beetle by stinging it. And what mechanism
does this beetle have to escape from it or to get rid of the ant? What it has is, it has one
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chamber in its stomach wherein it produces hydrogen peroxide. In the other chamber, it
produces a fuel like hydroquinone, which is hydrocarbon. And whenever something
attacks it flexes its muscles and pushes the hydrogen peroxide and hydroquinone into a
third chamber of its stomach. This third chamber is coated with some enzyme, which is
catalyst.
Enzyme is something you know like even when we prepare the batter for making idly
you know we put some yeast into it and it foams. We have this enzyme coated third
chamber, wherein the hydrogen peroxide and hydroquinone are mixed together. And in
the presence of the catalyst, the hydrogen peroxide and the hydroquinone they react and
form hot gases and the bombardier beetle squirts it, squirts it on the ant and therefore, the
ant gets driven away. And this is nature’s way of protecting the bombardier beetle. It is a
very unique evolution of species. All the species if you look at Darwin’s theory has come
through some hierarchy, but it seems this particular insect seems to be different from the
other species, but it is very illustrative of something, which I will now tell you.
(Refer Slide Time: 04:46)
Let us go ahead and see what really happens. In one stomach H2O2, hydrogen peroxide.
Is formed. In the other chamber, hydroquinone, which is the fuel is formed. When the
bombardier beetle wants to squirt hot gases out of it, it pushes this hydroquinone and
hydrogen peroxide into the third chamber, which is coated with enzyme, and it
chemically reacts, to form hot gases, which is squirted out.
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(Refer Slide Time: 05:16)
Well, the experiments which I told you of at the National Academy of Science are one of
a number of experiments done on a bombardier beetle. They take something like a fork
here, and they try to place it mimicking the ant and immediately it squirts out the hot
gases. They put this particular intrusive fork over here, mimicking the ant and it pushes
the gas. Therefore, it is able to squirt the hot gases in the different directions. You put
this fork here and it squirts out the hot gases in this direction and so on.
This feature reminds us of Leonardo Di Vinci. You know he was the painter, who lived
in the late sixteen century. And you know he made different sorts of things. He sort of
looked at the birds flying, he tried to construct an airplane, he made bridges of course, he
was a famous painter, you will remember the Mona Lisa painting. And all throughout his
life and we should remember this, he opened his eyes to the natural world and he never
stopped thinking about why things happened the way they did. And so also may be if one
had seen this insect with interest long back, maybe we would have seen something like
an oxidizer and fuel injected into a chamber and a liquid propellant rocket long long ago.
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(Refer Slide Time: 06:37)
And if you get to see what really happens in the case of the bombardier beetle? This is
again about squirting hot gases. And squirting is in different directions.
(Refer Slide Time: 06:49)
And if we were to apply the principle of what the bombardier beetle does to a device; we
have a chamber containing oxidizer, a chamber containing fuel, a third chamber in which
it reacts and forms hot gases and squirts it out. Well, a liquid propellant rocket consists
of a fuel in a tank, an oxidizer in a tank; you pump the fuel and the oxidizer into a
chamber. And how do you pump it? You pump it at high velocities through an injector
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may be you break it up into fine droplets; you vaporize it which is all happening in the
third chamber of the bombardier beetle. And then you have combustion taking place and
you expand through the nozzle. We have an injector and a combustion chamber. And the
nozzle is what creates a thrust. The combination of injector, chamber and nozzle is also
known as thrust chamber. And therefore, a liquid propellant rocket consisting of all these
gadgets is nothing very much different from the bombardier beetle, which is there in
nature.
(Refer Slide Time: 07:47)
And therefore, I think we should learn to look at nature and understand more about
nature. And say now we come to a large rocket, what does the large rocket consists of?
This consists of N2O4 as the oxidizer and UDMH as a fuel. And what is done is you
squirt it out through from the tank by a pump; you force the N2O4 into the chamber, you
force the UDMH into the chamber make it burn over here, expand the burnt products
through the nozzle and this is what a liquid propellant rocket consists of. We see the
oxidizer N2O4 tank, the UDMH tank at the bottom, you have a series of pumps, which
push the propellant into the chamber and you get the thrust. Well, this is the liquid
propellant rocket, which is much akin to the mechanism by which the bombardier beetle
generates hot gases.
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(Refer Slide Time: 08:51)
Let us take a look at few liquid propellant rockets. Well, this is the thrust chamber of the
liquid propellant rocket engine, which was seen earlier. It is taller than this man over
here, this is the cylindrical combustion chamber, this is the pump above it, this is the
nozzle and this slide shows the rocket firing.
(Refer Slide Time: 09:07)
Well, the same engine assembled together as a rocket stage is shown with large tanks to
carry fuel and oxidizer. We have the oxidizer tank over here, I have the fuel tank, I have
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the pump over here below the tanks that push the fuel and oxidizer it into the chamber
and it generates thrust using expansion in the nozzle.
(Refer Slide Time: 09:23)
Let us take one more example. If instead of using N2O4, we use liquid oxygen and in
place of UDMH we use liquid hydrogen.
(Refer Slide Time: 09:33)
The oxidizer in the earlier case was N2O4 and the fuel was UDMH. We also talked in
terms of fuel could be liquid hydrogen; the oxidizer could be liquid oxygen. And when
we carry the propellants like liquid oxygen and liquid hydrogen well, liquid hydrogen is
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not very dense and we need a larger tank. We have this liquid hydrogen here, we have
liquid oxygen here, a pump over here which pushes them into the combustion chamber
where it burns. This is what a cryogenic liquid propellant rocket is like.
(Refer Slide Time: 10:08)
Well, this slide shows a cryogenic propellant rocket known as an RL 10. This was the
first cryogenic propellant rocket to be developed and this was in United States. This was
in the period 1960s and here again you have the pump pushing the liquid hydrogen and
liquid oxygen into the combustion chamber, which is here. And you see the nozzle is a
relatively large portion of the liquid propellant rocket, I think this gives us some feel for
the sizes of the different constituents in a liquid propellant rocket.
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(Refer Slide Time: 10:43)
A UDMH / N2O4 rocket is shown in this slide. This shows the chamber, the nozzle and
these are the pumps supported above the chamber.
(Refer Slide Time: 11:02)
Therefore, what is it that we could infer from the few examples, which we saw? We need
a fuel tank, an oxidizer tank, a pump to increase the pressure from low value at the tank
to a significant value at the supply point to the chamber. At this pressure, we must be
able to supply the required quantities and break up the liquid into droplets or particulates.
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That means, we need something like an injector, which allows the required quantity of
propellants to come into the chamber and to break it into droplets. And we have the
chamber wherein reaction takes place and then I have this huge nozzle wherein we
expand the gases. Well, this becomes the liquid propellant rocket. We have to put some
flow control valves here to start the supply and stop of the propellants whenever we
want. And maybe we have let say the liquid fuel over here, the oxidizer over here and
here you have the thrust chamber wherein high pressure gases are generated by
combustion and thrust is developed.
Now, you may immediately ask a question. Why do we need to have pumps in the first
place? After all we have supply of the fuel and why not we supply it directly like when
we do an experiment in our lab using diesel combustion and an internal combustion
engine. We could have something like an overhead tank just as we have a diesel tank for
an experiment and we directly connect it to the IC engine. Why not we directly connect
the fuel and oxidizer and may be to push it through using a gas bottle here containing
high-pressure gases?
And if we put a pressure regulator in which case we can reduce the pressure of the gases
pressurizing the liquid to any level that we require. We have a reduction mechanism for
pressure through the pressure regulator. We have a high pressure source of gas so that we
could do with a smaller gas bottle. We pressurize the propellants to the required high
pressure and send it into the chamber. Therefore, we need not always have a pump, but
to be able to understand whether pumps are really required, we need to do a little more
exercise. Let us again re-examine the issue.
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(Refer Slide Time: 13:35)
We have something like a gas bottle, which stores gas at high pressure. We have a
pressure regulator downstream of it and what does the pressure regulator do? Maybe
from high pressure of the stored gas at maybe 30 MPa, maybe 300 bars to 400 bars, since
we do not want such high pressures and we want much lower pressure maybe 10 MPa or
1 MPa. Therefore, we put a pressure regulator, which will reduce the pressure by
regulating the opening through a spring, it is a feedback control to give constant supply
pressures. And then we connect this gas to the fuel tank and to the oxidizer tank, which
contain propellants and then we supply them into the combustion chamber at the required
pressures.
We are talking of having a high-pressure gas bottle and regulating the pressure to a lower
value to push the fuel and oxidizer into the combustion chamber. Well, this becomes
something like we use a cold gas, a high pressure gas and this what we call as a regulated
gas pressure fed system.
We started with a pump. Let us see under what conditions we would really require a
pump. Under what conditions can we use a regulated pressure system? Well, it may not
even be necessary to have a gas bottle and the regulator and we can still think in terms of
a simpler scheme or configuration.
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(Refer Slide Time: 15:20)
We could take a tank as such may be a fuel tank, may be an oxidizer tank and do not
have to fill the propellants fully in the tank. Let say this is the fuel tank with liquid fuel
filled in to a certain height. We have the oxidizer tank with the liquid oxidizer over here.
We fill high pressure gas in the free volume over the liquid and have a valve downstream
of the tank. Once the valve is opened, immediately propellant flows. When we want to
stop the flow of propellants, we close this valve. And therefore in this system, the gas is
contained in the tank itself and it blows the propellant into the combustion chamber. This
scheme is known as a blow-down system.
(Refer Slide Time: 16:33)
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In the case of the liquid propellant rockets, we need to supply the propellants to the
combustion chamber. We started looking at the beetle and told that a pump may be
required. But we also told that may be a gas, a cold gas at high pressure can possibly be
used. This gas can be directly filled, and what will happen when it blows down? The
pressure keeps decreasing and therefore, the thrust keeps decreasing. Whereas, if we give
something like a regulated gas we maintain constant pressure over here and therefore we
get a constant thrust.
Therefore the gas fed, high pressure gas fed system could either be in a blow-down mode
or in a regulated mode. For instance, if you have a satellite which is orbiting in space and
we were to use liquid propellant rockets for its control, which one would you prefer?
(Refer Slide Time: 17:40)
The satellite has a box-like structure. It has solar sails protruding over the box structure.
And when the light falls on the solar sails, you know it creates pressure so we have
booms to balance this pressure. We will study this method of propulsion using sails and
pressure during the end of this course. You have to balance the pressure from the
sunlight falling on this; you put something like a balance over here such that it does not
get tilted over here. Then to control the attitude and orbit, we wanted small rockets
placed over the different faces of the box. And how do we supply propellant to these
small rockets? Inside this structure we have a propellant tank and oxidizer tank; at the
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bottom of this oxidizer tank we have the fuel tank may be in the common bulk head
configuration.
That means we have a single connection from the tank for the fuel and oxidizer. We
could introduce some gas pressure over the liquid column in the tank. We could have the
liquid lines to the different rockets with a flow control valve on each of the rocket.
Whenever we want a particular rocket to operate, we open the particular valve and it
fires. And this is the model of this particular blow-down mode. And if you are going to
see what is inside the box, we have among the electronic packages, the fuel tank and
oxidizer tank. And these red knobs correspond to the thrusters or liquid propellant
rockets.
And what is done is, whenever you want the rocket to be fired you just open the valve of
the particular rocket. The tank is already pressurized. The liquid fuel and liquid oxidizer
flows into these thrusters and they generate thrust and that is how you control the
spacecraft. This is how the liquid propellant rockets are used in satellites. But there is a
problem. Whenever we have fuel and oxidizer and a spacecraft is revolving in a given
orbit at a constant velocity, we found that in the frame of reference of the satellite it has a
centrifugal force, which is balanced by the gravitational field. Therefore, it is in a state of
weightlessness in its own frame of reference and it creates new problems.
Therefore, we have to see under weightlessness how do we supply the propellant? We
will take it up after some time. That means if I put a propellant in a tank that the
propellant may sit on top, it may not really come at the bottom. And when I pressurize it
with the gas, the gas may come out while the liquid may not come out. That means
supply of a propellant when we are in a state of weightlessness is again going to be a
problem. May be we will take a look at it later on. But right now, let us take a look at
what constitutes, let say a gas fed system, which can either be in a blow-down mode, a
regulated mode and then let us come back to the pump.
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(Refer Slide Time: 21:03)
To be able to do this, let us first try to estimate how much gas, how much of high
pressure gas is required. The first thing we have to determine is the quantity or rate of
supply the propellants to the chamber for a given value of thrust. We must know what
must be the mass flow rate of propellants and how do we calculate it? Let us refresh
ourselves.
(Refer Slide Time: 21:39)
We suppose the thrust of the rocket as 6 kilo Newton. Let say the specific impulse of the
rocket is equal to 3000 Newton second by kg. Then we know the mass flow rate of total
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propellant fuel and oxidizer is equal to 6000 Newton divided by 3000, which is equal to
2 kg per second. If the duration of operation of the rocket is 200 seconds let say, we take
a small firing rocket for a small time of firing let say 200 seconds, then I need to supply a
total mass of 400 kg during the 200 seconds. We talk of 6 kilo Newton, 6000 Newton
thrust rocket with a specific impulse of 3000 Newton second per kg requiring a flow rate
of 2 kg per second with the total mass of propellant being 400 kg. We want to know how
much gas is required, how much mass of gas do we require to expel 400 kg of propellant.
If the gas requirement is large, well the rocket cannot takeoff.
How do we estimate it? We again follow the same figure, but now let us simplify it. Let
us simplify this and say well I have gas pressure over the propellant and we also told that
gas can be stored at fairly high pressures.
(Refer Slide Time: 23:09)
We have a gas bottle. It contains let say m0 kilograms, m0 kilograms of air or gas. This is
what we are required to find out in order to expel the propellant from the tank into the
chamber. Let say it is at a high pressure P0 and at the ambient temperature T0 because we
do not know at this point in time whether a hot gas or a cold gas is to be preferred. May
be when we derive the expression, it will be also clear to us what type of gas is required
and what must be the temperature of the gas.
What is it that we do from this high-pressure gas source? We put a pressure regulator
here and reduce the pressure from the value of P0 and it must communicate the two
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tanks. We put the two tanks together as one. Let say the total volume of propellant to be
expelled is VP and this is the volume that must be supplied to the thrust chamber.
Let us say the pressure at which we would like to supply is a pressure pp, pp let say in
Newton per meter square or Pascal. And we have to reduce the pressure from P0 to pp in
this type of regulator. Now we want to find the value of m0 kg. It is a straightforward
thermodynamic problem, but before doing this problem it is also required for us to put
the parameters required to solve the problem.
Let us do it as a general derivation. The quantity of propellant liquid, which is to be
supplied at constant pressure pp is VP meter3. After the volume is expelled at constant
pressure pp, some quantity of gas at pressure pr would be left behind in the gas bottle.
(Refer Slide Time: 33:22)
At the end of operation, let assume the mr kg of gas is left in the gas bottle. Let the
temperature of this residual gas be Tr and the pressure would be pr. At the end of the
depletion process, the volume occupied by the propellant in the tank will be filled with
the gas at the pressure pp. We have the same tank, Vp m3 of the propellants are expelled
and the volume it is now filled with gas. And since we have the regulator, we have
reduced the pressure to pp. When the propellant has just got out of it, it will be filled with
gas at a pressure pp and the temperature Tp, which is the temperature of the propellant.
Since, it is a slow depletion, the gas reaches the temperature of the propellant Tp.
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Let us make the figure complete. We have exhausted or pushed out, a volume of
propellant VP of propellant and we are left with the gas occupying this volume in
addition to this volume of gas in the gas bottle. This is the final condition of the gas. We
want to write an expression and determine the value of m0 kg.
How do we do this problem? Let us write try to write an equation for the expansion
process of the gas and determine the value of m0.
Let us say that the system does not have any heat transfer taking place i.e., the tanks are
fairly well insulated, gas bottle is insulated there is no heat transfer taking place. And
therefore, we ask what is the work, which is done by the gas in pushing out a volume VP
of the propellants.
(Refer Slide Time: 27:02)
Let us calculate VP for this particular problem, Vp, if I take the density of the propellant
to be same as water, density is equal to 1000 kg per meter cube that is the density of
water 1 gram per cc. And therefore, the volume VP over here is going to be 400 divided
by 1000 meter cube which is equal to 0.4 meter cube. In other words, for this rocket we
have to get rid of or expel 0.4 meter cube at a given constant value of pressure.
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(Refer Slide Time: 27:46)
We consider the system of the high pressure gas. The initial state of the gas over and the
final state are known. The gas has moved from the bottle and occupied the propellant
tank. We note there is no heat transfer; Q is zero and the expansion process of gas is
adiabatic. During the expansion, the system does some work at the boundaries by
expelling the propellants at constant pressure. We can write Q minus W is equal to the
change in internal energy for a system. Nothing is getting in, some work is being done
and that change is what is the internal energy. We need to find out the value of W and
the value of change in internal energy.
(Refer Slide Time: 28:48)
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Let us write the expressions for these two. The work done by the gas is in expelling the
liquids. Well, we have a constant pressure, which is applied over here during the
displacement VP, the work done is equal to pp, the pressure into volume of the liquid Vp.
Is it ok? What is the change in internal energy? It is final minus initial value of internal
energy. Well, finally, I am left with mr of the gas over here; we know that du/dT where u
is the specific internal energy and T is the temperature is CV. And dh/dT = Cp. Since we
are talking of internal energy therefore, u is equal to CV into T. And therefore the change
in internal energy is equal to the final gas mr which is available in the gas bottle into CV
into Tr. This corresponds to the final mass which is left in the gas bottle; we say it is at a
reduced pressure and the reduced temperature because it has expanded. Some gas comes
over in the volume originally occupied by the propellants, which is the mp into CV into
TP. The final value of internal energy is the sum of these two internal energies. We
presume that the volumes of the lines over here, plumb lines, are very much smaller and
can be neglected. And what is the value of the initial internal energy minus the final
value? What will be the initial value of the internal energy of the gas in the gas bottle? It
is m0 into CV into initial temperature T0. The work done − W is equal to delta U i.e., the
change of the internal energy. W represents the work done by the system.
(Refer Slide Time: 31:00)
We use this part of the board. We therefore have mr × CV × Tr + mp × CV × Tp − m0 × CV
× T0 = − W and W = pp × Vp. Is it all right? Q minus W is equal to the change in internal
energy. We want to solve this equation, to be able to solve this let us see what CV is, we
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know Cp − Cv = the specific gas constant R. And therefore, Cv (γ – 1) = R into gamma
or Cv = R/(γ – 1). The specific gas constant has units of Joule per kilogram Kelvin and
Cv also has units of Joule per kilogram Kelvin.
Now, what is the value of mr? From the gas equation, we have pr ×V0 = mr × R × Tr. Or
rather I can write the value of the final mass after expansion left over in the gas bottle is
equal to pr × V0 / (R Tr)/ Similarly, we can write an expression for mp = pp × Vp / (R ×
Tp).
We need to find the value of m0 therefore, let us not touch this. Let us substitute these
two values of mr and mp and the value of Cv in this expression and try to get the value of
m0. Let us say mr is equal to pr ×V0 / R Tr and Cv = R/ (γ – 1) and similarly we get pp ×
Vp / (RTp). We get the value of m0 into R/ ( γ – 1) × T0 and equal to minus pp into Vp
over here.
Now we find that R Tr, R Tr cancels; R Tp and R Tp cancels. And now if we were to
simplify further we take gamma minus 1 on the right hand side because all the three have
(γ – 1) and therefore we get pr V0 + pp Vp − m0 × R T0 = − (γ −1) × pp Vp. . pp Vp cancels
on both sides. With V0 = m0 RT0/p0, and mRT0 – pp V0 simplifies as m0 {RT0(1− pr/p0)}
(Refer Slide Time: 36:17)
We are left with γ pp Vp on the left and hence mass of gas m0 = γppVp/{RT0[1−pr/p0]}.
This was done by collecting the similar terms. We get R T0 which will come in the
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denominator. We have in the numerator the value γ on the right hand side into Pp Vp. In
the term R T0 we have (1 − pr / P0). Please check how we come to this expression.
Now, we wish to examine under what conditions it holds good. We say had the process
being isothermal, what would be gas that was required? m0 will be equal to Pp ×Vp/ RT0
because it is an isothermal process.
(Refer Slide Time: 37:52)
Therefore, what happens when gas is expanding and that also irreversibly? Instead of a
mass of gas under isothermal expansion of ppVp/RT0, we have an amplification taking
place by {γ/(1 − pr /P0)}. We write pp Vp by R T0 because m is equal to pV by R T0 and
temperature is a constant for an isothermal process. The term in brackets is the
multiplication factor for the actual mass of gas required.
Now, we want to find out may be for this engine, which we said is 6 kilo Newton the
mass of the gas which is required. Let us go through this example and then we can
extend it for different pressures and different thrust. Let say in this case the volume is
equal to 0.4 meter cube. How much gas do we require? Well, this expression also tells
us, what is the type of gas, which we require? Do we require a light gas or a heavy gas?
Can we all infer something from the equation? Should we pressurize the initial volume
with a high molecular gas, low molecular mass gas or at a high temperature or under
what conditions should I do? All what I can tell you is the value of P0 is limited because I
need a gas bottle which can hold certain pressure and that pressure cannot be greater than
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about 300 to 400 bar. Because beyond that to make something hold such high pressure
gas is difficult and the mass of the bottle goes up. Normally the value of P0 is around 30
to 40 MPa.
How can we choose the other parameters like gamma, R and temperature, which conveys
the type of gas which is required? We do one more exercise let say R is equal to
universal gas constant divided by the molecular mass; 8.314/ molecular mass. If the
molecular mass of the gas is smaller, we get a very large value of R. If we get a large
value of R, we can have a smaller mass of the gas. Therefore, this immediately tells me
the lighter the gas the better it is. Hydrogen is the lightest with a molecular mass of 2
gram per mole; whereas helium has a molecular mass of 4 gram per mole. But hydrogen
is very flammable. Therefore, invariably we would like to choose helium, which is a
light gas for the pressurization. You may observe that the expression is able to tell us
what we are looking for.
But helium has large volume of gamma 1.67 compared to air or preferably of nitrogen of
1.4. The value of 1.67 to 1.4 is in the numerator whereas R could be something like 4
and 28 seven times more in the denominator. Therefore, we choose a light gas for
pressurization. But this expression also tells us a higher temperature is suited because
higher temperature means we have a smaller value of mass of the gas. Rather than
choose a cold gas, a higher temperature gas is advantageous.
(Refer Slide Time: 41:04)
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That means we initially said that a cold gas is used for pressurization. Why not use a hot
gas? Why not have some chemically reacting gases taking in the combustion chamber to
be used for the pressurization? Well. This is also an option and in addition to a cold gas
for the blow-down mode and regulated mode, it is also possible for us to have hot gas.
And the hot gas is more efficient. But in practice it has been difficult to use it. Several
countries have tried, but it has not yet found a commercial application. But we should
keep the advantage in mind and in future it may be a strong contender for a gas pressure
system.
Having said that let us do a small numerical problem on the amount of gas? You know
we should have some feel whether we need a few kilogram of gas or whether we need a
few grams.
(Refer Slide Time: 41:56)
For the particular example, we found Vp to be 0.4 meter cube. Let us assume the
chamber pressure to be small; chamber pressure let say is 1 mega Pascal. If the chamber
pressure is 1 MPa, we have to force propellant into it let say Pp which we assume as
equal to 1.2 MPa. The final pressure, the lowest pr is equal to this supply pressure of 1.2
MPa. The volume Vp is known. The value of the specific gas constant R for gas helium is
equal to 8.314/0.004 so much Joules per kilogram Kelvin. The initial temperature of the
gas is 300 K. We substitute the values and get the value of mass. And this value of mass
for this particular thrust when the supply pressure is 1.2 MPa works out to be 1.34 kg.
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If we have a similar rocket, which is developing a thrust of 6 kilo Newton, what is the
mass of gaseous helium required for pressurization? It is twice the value.
(Refer Slide Time: 43:07)
But if the chamber pressure instead of being 10 bar or 1 Mega Pascal is now changed to
pressure is equal to 10 MPa and the supply pressure therefore, becomes let say 12 MPa,
the mas of the helium gas becomes directly 10 times that is something like 13.4 kg. Now,
if we talk in terms of thrust which is much larger, may be 600 kilo Newton, the quantity
of gas increases so also for the chamber pressure. Let me summarize it through slides; a
regulated pressure system and a blow-down system.
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(Refer Slide Time: 43:44)
.
(Refer Slide Time: 43:52)
We wanted to find the mass of gas required in the gas bottle for a regulated cold gas
pressure fed rocket. The cold gas pressurizes the oxidizer and fuel to supply them to the
combustion chamber.
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(Refer Slide Time: 43:58)
And we simplified it. The initial condition of the gas is in the bottle while the final
condition is it is in the bottle and propellant tanks.
(Refer Slide Time: 44:04)
And we found that mo mass of the gas required is equal to pp Vp by R T0 into gamma by
1 − pr / P0 and then we found that hot gas pressurization is a better option.
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(Refer Slide Time: 44:17)
And then now we calculate the value mass of the pressuring gas when we have a low
pressure engine of thrust of 6 kilo Newton, the chamber pressure is 1 MPa, the supply
pressure is 1.2 MPa, the mass of gas required is 1.34 k g. Then the same 6 kilo Newton is
operated at a high pressure of 10 MPa, the supply pressure is 12 MPa, the mass of gas
becomes instead of 1.34 kg becomes 21.4 kg.
(Refer Slide Time: 45:00)
You see when the chamber pressure increase by 10, the quantity of gas is increased by
almost twenty times. Whereas, if I have a high thrust engine of 600 kilo Newton at the
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same pressure value, the quantity of gas required is something like 2000 kg. And
therefore, it is not humanly possible to have such large amounts of gas being carried in a
rocket and that is where it becomes necessary for us to have a pump, which can supply
the propellants to it. This explains the necessity of a pump for larger thrust engines.
(Refer Slide Time: 45:29)
If we go back and compare the different systems that we discussed; a blow-down, a
stored gas regulated, and hot gas, we find when the thrust is small and the burning time is
small, we can manage to have blow-down system like what was used in the spacecraft.
We need very small thrust and the rockets are operated at low chamber pressure. When
we want to operate at slightly larger time, slightly larger thrust, may be something like a
regulated gas system is required. But if we want still better performance, may be a hot
gas system would be advantageous.
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(Refer Slide Time: 45:31)
But beyond some thrust and beyond some burning time, we need to operate in this region
which can only be done using turbo pump or a pump system. Therefore, the feed system
of a liquid propellant rocket could consist of different options depending on the thrust,
duration of operation and chamber pressure.
(Refer Slide Time: 46:36)
We could have gas pressurization that means, we have a gas bottle containing a cold gas
or a hot gas. The cold gas could either be in a blow-down mode or in a regulated mode.
But if that duration of the rocket operation is for a longer time and if the thrust is large or
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the chamber pressure is large, we essentially need a pump to take the fuel from the tank
into the chamber and this is what we must remember.
Let us now calculate what is the power of the pump, which we require in case we need a
pump for the pressurization of the propellants. Let us let us take this example.
(Refer Slide Time: 47:26)
The thrust of the engine is equal to 60 kilo Newton or better still 600 kilo Newton. We
had seen this engine and it used UDMH and N2O4 at a thrust of around 600 kilo Newton.
And the specific impulse of this engine is 300 Newton second by kilogram. Therefore,
the mass of mass flow rate of propellant is equal to something like 200 kg per second.
Therefore, if we have to supply this quantity to the combustion chamber, let us again
take the density of the propellants same as the density of water. The rate at which the
volume has to be supplied is something like 200 divided by 1000 meter cube per second
which is equal to 0.2 meter cube per second. What is the power required for a pump to
supply 0.2 meter cube per second at a given pressure? Let us take the same value of the
supply pressure of 12 mega Pascal.
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(Refer Slide Time: 48:52)
What is the work to be done by the pump, which is to supply flow rate of 0.2 meter cube
per second at a pressure of 12 MPa. The rate of work done is equal to pp into Vp and this
comes out to be 12 into 10 to the power 6 into Vp is 0.2 meter cube that is equal to 2.4
mega Joule per second which is equal to 2.4 Megawatts. Can we have some feel for this
number? The power produced in the power plant at Ennore, which supplies entire
electricity to Chennai is something like 450 Megawatt. If we take a super thermal power
plant like Ramagundam, the power generation is like 2400 or 2600 Megawatt. That
means the power we are talking is enormous. We cannot have a battery or an electrical
power for running such pumps.
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(Refer Slide Time: 50:20)
And therefore, what we normally do is, we need something like a gas generator, which
can supply work to a turbine and using the turbine we drive the pump. It becomes a
pump and a turbine and is known as a turbo pump. I require another gas source, but to
have a dedicated gas source is going to be a problem.
(Refer Slide Time: 50:39)
And therefore, what we do is that we take the fuel, we take the oxidizer and increase the
pressure. We remove some fuel from the fuel line, we remove some oxidizer from the
oxidizer line, we burn these gases separately in an auxiliary chamber and generate hot
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gases. These hot gases are used to run a turbine, generate work, which drives the fuel
pump and oxidizer pump and supplies the fuel and oxidizer to the chamber. And the
exhaust from the turbine is left out through an auxiliary nozzle. This means that part of
the fuel is used in generating high temperature gases, not very high because the turbine
cannot take very high temperatures and thereafter we leave it to the ambient. And this
becomes what is known as a gas generator cycle. We have a turbine running the pumps
and we use hot gas from a gas generator for running the turbine. This is the gas generator
cycle for feeding propellants into the chamber.
(Refer Slide Time: 51:32)
To sum up, all what we did in today’s class is we looked at the bombardier beetle, and
said well a liquid propellant rocket is quite similar to it. We talked in terms of cold gas
pressurization consisting of blow-down, regulated, and also hot gas pressurization, and
then found that when the duration of the rocket operation is large and the thrust is large
or the chamber pressure is large, it is just not humanly possible to carry such huge
masses of gas which are required. We need a pump, but then we found a pump demands
huge amount of energy to run it. We needed a gas generator to run a turbine, which in
turn ran the pump and this is what we called as a gas generator cycle. In the next class,
we will see what are the other cycles for the feed system and calculate what is the
performance of a rocket corresponding to the cycles. And then we will go into the
components of a rocket.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 27
Feed System Cycles for Pump Fed Liquid Propellant Rockets
Good morning. Let us continue with liquid propellant rockets.
(Refer Slide Time: 00:14)
Let us quickly recap where we were in the last class. We addressed the feeding of liquid
propellants in a liquid propellant rocket and saw the different ways. One was using a
cold gas. The cold gas could be used either in the blow-down mode or in a regulated
mode. The difference is that in a regulated mode, we draw gas at a high pressure,
maintain a constant lower pressure on the propellant side while pushing the propellant
into the chamber. In the blow-down mode, in the tank itself, we have some volume of
pressurized gas above the liquid, known as ullage volume, which is pressurized initially
and you allow the gas pressure to gradually push the propellant. In this case the pressure
at which propellant is supplied into the thrust chamber will keep decreasing with time.
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While deriving an equation for the mass of gas required in the gas bottle, we found if the
temperature of the gas is higher it is better therefore, we also talked in terms of hot gas
pressurization. But we told, even though it is advantageous, it has not been implemented
so far. It may be worthwhile to do so. In fact there was one of the French engine, which
tried this; but they did not follow it up. But I think it is a strong contender.
We also found out that when the time of operation of a rocket is small or when the thrust
is small or when the chamber pressure is small we can go for something like cold gas
pressurization in either the blow-down mode or a regulated mode or even a hot gas
pressurization mode. But the moment the pressure of the chamber is large or the thrust is
large or the time is large, we necessarily have to go for a pump fed system.
We also found that the amount of power required for the pump is enormous like for a
600 kilo Newton engine we found that power required is about 3 Megawatt. And we said
even a large power plant like the Ennore power plant at Chenai generates about 450
megawatts of power therefore, we are talking of huge power. And therefore, we cannot
drive the pump using a battery, or an electrical supply.
(Refer Slide Time: 02:56)
Therefore, in order to drive the pumps, we looked at this figure in the last class also. We
have a fuel tank, we have an oxidizer tank and both are liquids. You sort of push the
propellants into a pump maybe we need a gas bottle which pressurizes it initially to a
small value and pushes it into the pump. We need power or work to rotate these two
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pumps for which we require a turbine. The turbine generates power and drives the two
pumps. We needed a gas generator or a hot gas source, which could power the turbine
and the spent hot gases are exhausted out. The pressure built in these two pumps pushes
the liquid oxidizer, pushes the liquid fuel into what we said was the thrust chamber.
Now, you know to have an additional device for a generating gas and we need another
set of lines for the fuel and oxidizer for the gas generator.
(Refer Slide Time: 03:56)
And the normal practice is to make use of the existing fuel and oxidizer itself. How do
we do it? We also looked at it in the last class. After pressurizing the liquid fuel, we take
a little bit of the liquid fuel into something like a pre burner or a burner, which we call as
a gas generator. We also take the oxidizer also mix the two together in the burner to
generate hot gases, push the hot gases into the turbine and exhaust the output from the
turbine into the ambient. Therefore, you have something like a gas generator, which is
driving the turbine, which in turn drives the fuel pump and the oxidizer pump and
supplies the fuel and oxidizer into the combustion chamber.
Such a feed system is essentially configured around the gas generator and how does the
generator get the fuel and oxidizer? Part of the fuel and oxidizer is drawn from the high
pressure lines and is used to generate hot gases. There is a problem with this
arrangement. The turbines consist of blades and it could be impulse turbine, reaction
turbine maybe we have to look at it, but whatever said and done, whenever we admit
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very hot gases into the turbine from the stoichiometric combustion of the fuel and
oxidizer, the gases are at high temperatures around 3000 to 3600 K.
And therefore we cannot supply these hot gases for running the turbine. The output of
the gases from the gas generator must essentially be at a much small temperature maybe
around an upper limit of about 900 K. Therefore, how do we get these low temperatures
in the gas generator or burner? We use very fuel rich mixtures or very oxidizer rich
mixtures to the gas generator and supply the moderately hot gases to the turbine. The
work done by the turbine is equal to the work done by the two pumps. This cycle is what
we called as the gas generator fed or gas generator cycle for feeding propellants into the
combustion chamber.
(Refer Slide Time: 05:54)
Unfortunately, in thermodynamics we say cycle is something wherein during a process
the medium comes back to the initial state. But in the present case, we are feeding
propellants and the word cycle is adopted here. It is not related to the thermodynamics
cycle such as Otto cycle or Joule cycle or Brayton cycle. It is just a gas generator fed
system, but conventionally it is known as a gas generator cycle for feeding propellants
and we use the same terminology.
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(Refer Slide Time: 06:51)
The temperature of the gases as a function of mixture ratio is such that the maximum
temperature occurs just below the stoichiometric. We had a slight shift; instead of
maximum temperature occurring at stoichiometric it occurs at slightly fuel-rich
conditions. But when we need a much lower temperature maybe we have to operate the
gas generator in a very fuel-rich region such that we get only around 800 to 900 K i.e., a
very oxidizer-rich region wherein I get a low temperature.
Oxidizer-rich region is seldom used, but some of the Russian engines do use oxidizerrich mixture. What is the reason? Why oxidizer-rich is not conventionally used this? If
you have something rich in oxygen, it can always oxidize a metal; whereas, if the
mixture ratio is fuel-rich, it cannot oxidize a metal. Therefore, the trend is to use a fuelrich mixture. This was in the context of the choice of the mixture ratio for the gas
generator. We reiterate that a fuel-rich mixture is used for the gas generator and the
maximum outlet temperature from the gas turbine or the inlet to the turbine is around 900
K.
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(Refer Slide Time: 08:05)
The above figure shows a schematic of a gas generator cycle. We have a small gas bottle,
which pressurizes the fuel and the oxidizer. The fuel and oxidizer enter the pumps. We
also bleed a little bit of fuel, a little bit of oxidizer after the pump and burn it in a gas
generator. We generate hot gases and these hot gases drive the turbine. And when they
drive the turbine in this particular schematic you have a gear train for adjusting the speed
of the pumps. These two pumps supply the propellant to the engine and the same pumps
also bleed or supply propellants for the gas generator.
You may ask how do we start the engine in the first place. After all we need a higher
pressure to supply the propellants to the gas generator and the engine. A bottle of high
pressure gas could be used to initially run the turbine or a slug of the hot gas can be used
to initially use rotate the turbine and once it rotates, it starts pumping and then we shift to
the present arrangement after the initial transient. Well this is all about the gas generator
cycle. We will get back into it in some detail to find out what is it is performance after
we have discussed the other feed system cycles.
Though the term cycle is a misnomer while discussing feed systems as it usually refers to
a thermodynamics cycle, there is however a thermodynamic cycle known as Topping
cycle used in power plants and followed in feed systems of liquid propellant rockets.
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(Refer Slide Time: 09:23)
The Rankine cycle, which is used in power plants, has a boiler, which generates steam.
The boiler runs a turbine that generates high pressure steam that runs a turbine and that is
what generates the power. And then the outlet from the turbine is fed into the condenser,
wherein water is condensed and then you have a pump, which pushes back the water into
the boiler. The water is heated to form steam in the boiler. Very often what is done to
improve the efficiency is, we know Carnot efficiency is maximum when the upper
temperature is the highest; we increase the temperature of the cold water by supplying
heat from the steam. If we were to bleed part of the steam after a part of the expansion,
we bleed some of the steam and take it into a feed water heater. And use this for heating
the feed water in a heater. In this way we supply heated water to the boiler. Well the
steam temperature will go up and the efficiency of the Rankine cycle will increase.
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(Refer Slide Time: 10:42)
And this is what we call as the regenerative Rankine cycle. Just like you have
regenerative process; you heat the water before the boiler using the heat, that is normally
wasted, in the process itself; and how do you heat it? We bleed some of the partially
expanded gases in the first turbine, allow it to heat the feed water and then you pump the
feed water. Therefore, you are essentially able to get a higher efficiency and these cycles
are known as regenerative cycles. We also learnt in terms of cogeneration cycles. And
what do you mean by cogeneration? We bleed some of the hot gases from of steam
coming over here and use it to heat houses or use it for some other purposes than power
generation. We use the heat effectively instead of wasting it. This is what is known as
cogeneration.
And how do we dissipate heat in a condenser. We have something like a tower, in which
we allow the water to drip, we allow the hot steam to go through the tower; the hot steam
condenses and it heats the environment of water. Instead of doing this and loose the
energy, we could allow the steam to be usefully used for heating houses, for heating
some places maybe some industry for heating and that is known as cogeneration. The
regenerative cycle is also known as the Topping cycle. Why do say topping? Because we
use some of the process heat for “topping up” the temperature of steam in the
regenerative Rankine cycle. The cogeneration cycle is known as “bottoming” cycle.
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We talk of “Topping” cycle; we also talk in terms of “Bottoming” cycle. If we have a
furnace, which is generating very high temperature gases and we use this furnace to heat
some product such as for tempering steel or maybe steel making. And instead of
allowing the exhaust gases after the useful work to be dissipated into the environment,
we use it to generate steam; then it is a bottoming process. But we use the waste heat
from the furnace to generate steam and therefore to generate power or to generate
electricity. This is known as the Bottoming cycle.
Can we use some element of Topping cycle in the gas generator cycle, which we just
discussed. Let us be very clear about it because the feed system cycles are supposed to be
very efficient.
(Refer Slide Time: 13:42)
We have a gas bottle, two tanks one for fuel and the other oxidizer. We take the fuel into
the pump, the oxidizer into the oxidizer pump and then we push it out at higher
pressures. Why is this gas bottle required? We need some minimum pressure to push the
liquid into the pump. Therefore, now the requirement of gas is much smaller because my
outlet pressure is going to be smaller. And then we take the pressurized fuel and oxidizer,
part of it, into something like a gas generator or a pre-burner, or something which
generates hot gases at a mixture ratio which produces not a very high temperature. The
balance of the propellants is supplied into the main combustion chamber.
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Therefore in the process, we have a given mass of fuel m°f, which is supplied by the tank
and mass of oxidizer m°o, which is also supplied by the tank. Part of it is bled into the
gas generator. The part of the fuel bled into the gas generator is m°f gg and the part of
the oxidizer m°o gg is taken to the gas generator. Now, the balance what comes over to
the chamber here is m°f − m°f gg, that has been supplied to the gas generator. Similarly,
m°o that is the main oxidizer supply − m°o gg into the gas generator is what gets into
the main combustion chamber. We have the overall mixture ratio R equal to m°o/m°f
viz., the mass of oxidizer divided by mass of fuel.
What is the mixture ratio of the gas generator? R for gas generator is equal to m°o gg
which gets into the gas generator ÷ m°f gg . This mixture ratio must be terribly on the
fuel rich side so that the temperature is low. We had plotted the temperature versus
mixture ratio. If we were to denote the mixture ratio in the gas generator by Rgg; if R gg
is stoichiometry we get about the maximum temperature. When we want a reduced
temperature, we operate in a fuel-rich region or in the oxidizer-rich region. The value of
mixture ratio in the main chamber which generates the thrust RMC is equal to (m°o −
m°o gg) ÷ by (m°f − m°f gg).
Therefore, we have three mixture ratios to contend with: R overall, Rgg and RMC. We
would like to choose the mixture ratios such that we get maximum specific impulse. But,
if we look at the overall system, we would like to choose the mixture ratio R such that
we get a good performance. We have to calculate this a little more carefully.
With this background, let us finish the Topping cycle and then determine the value of
mixture ratio in a gas generator cycle engine. What did this gas generator do? It draws in
part of the propellants, generates low temperature gases for running a turbine. In the
turbine we have high pressure gases and these are expanded to low value and work is
done. Then we exhaust it out through a nozzle. The nozzle is an auxiliary nozzle.
The temperature at the inlet to the turbine or outlet of the gas generator we said is
typically less than about 900 K. Therefore, the temperature at the outlet of your turbine
will be even less maybe around 400 K. And therefore, the type of specific impulse,
which this expansion in the auxiliary nozzle can give is going to be a small number. Let
us call it as specific impulse from the gas generator, Isp,gg which comes from the
exhaust of the turbine which is expanded through a nozzle. Whereas, the specific impulse
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corresponding to the main chamber, we call it as specific impulse of the main chamber is
denoted by Isp,MC. Therefore, what will be the total impulse of the engine? It will be the
fraction of propellant at this specific impulse Isp,gg plus fraction of the propellant in the
main chamber at this impulseIsp,MC. And therefore the net specific impulse Is,net is
calculated.
(Refer Slide Time: 19:41)
The net value Is,net is going to be fraction of the gases f of the total that means m°o +
m°f which is the total supply in the denominator and m°o corresponding to the gas
generator plus m°fuel corresponding to the gas generator in the numerator. This is how
we define the fraction f. And therefore, this fraction × Isp of expansion in the gas
generator exit from the turbine + (1−f) that is the part of the propellant which flows
through the main engine and has a specific impulse Isp,ME × Isp,ME. We noted that the Isp
from the gas generator side would be small because the supply temperature is small and
the expansion ratio of the gases is small. And therefore, the net specific impulse of this
gas generator fed engine is going to be less than the specific impulse what we would
have otherwise got had we not had this gas generator, turbine and auxiliary nozzle
arrangement.
Therefore, the question is whether this heat, which is being wasted in a poor way, can
somehow come into the chamber, to do useful work? This could be very similar to the
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regenerative process in the Rankine cycle. It could be a Topping cycle instead of the gas
generator cycle.
(Refer Slide Time: 21:26)
We again have a gas bottle. It supplies the two tanks: a fuel tank may be an oxidizer
tank. And then we take the propellants out to over fuel pump increases the pressure and
an oxidizer pump, which again increases the pressure. The lines now are at high pressure
following the two pumps. And then what is it we do? We take the some of the fuel at
high pressure over here take it into the gas generator. We also take a little of the oxidizer
and allow it to come into the gas generator. And then we use the hot gases generated to
drive a turbine by the expansion of the hot gases to low pressure.
Now, the temperature of the gas generator is typically around 900 K, (80 to 900 K). 900
K is the upper limit; let say 700 to 900 K. And what do you do with the turbine exhaust?
Let us see how to regenerate it?
The balance of the fuel and oxidizer come into your main chamber. Now, what do you
do with this turbine exhaust? We bring the turbine exhaust and admit it into the main
chamber. That means the heat which is left out in the turbine is again re generatively
used here and this is what constitutes the Topping cycle used for a feed system.
In other words the exhaust is not wasted with low impulse as in a gas generators cycle,
but is fed back into the main chamber for secondary combustion. First combustion takes
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place here at low temperature then turbine expansion is then mixed with the balance
propellants in the main chamber and burnt again. And therefore the combustion now
takes place in two stages first in the gas generator and then in the main chamber and
therefore the Topping cycle is also referred to as combustion taking place in stages, or a
staged combustion cycle.
Let us find out what is the value of the over all mixture ratio and values of mixture ratio
of the gas generator and the value of the mixture ratio in the main engine? Just like we
did for gas generator, let us do it for the stage combustion cycle.
Therefore in stage combustion cycle, we have m° oxidizer, m° of the fuel, we supply
m°o gg to the gas generator and m°f gg into the gas generator. And then for the balance
propellant in the feed line to the main chamber, we have m°f − m°f gg and m°o − m°o
gg. However, this m dot o, gg and m dot f, gg is coming back into the main chamber and
therefore the net value what enters in the main chamber is still m°o and m°f. Or rather
the overall mixture ratio is going to be m°o/m°f. The mixture ratio in the main engine or
main chamber is again the same value as the overall mixture ratio. And what is the
mixture ratio in the gas generator? It is equal to m°o gg/m°f gg.
This is the main distinction with a gas generator cycle and we call it as a stage
combustion cycle or topping cycle. And the gas generator in the stage combustion cycle
is very often it referred to as a pre-burner because it first burns the propellants and again
we have a second burn in the main chamber.
Well! we need not even have a stage combustion cycle when we talk of volatile fuels; we
mean fuels like liquid hydrogen or let say liquid methane or liquid propane. It is possible
for us to have another cycle. Let us quickly investigate it.
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(Refer Slide Time: 26:30)
We again start with a gas bottle. We have volatile fuel, let say liquid hydrogen or liquid
methane and we have oxidizer like liquid oxygen. In this case we have a pump for the
fuel and another pump for the oxygen. Now, what we do is this particular liquid
hydrogen, which we have to supply to the combustion chamber, wherein it burns with
oxygen and the combustion chamber runs hot, we use it for cooling the chamber. And
therefore the hydrogen flows and cools the chamber and gets heated. And this heated
hydrogen is in the form of a gas. We use it to run the turbine. The hot gases such as the
hot hydrogen vapor, generated during cooling drives the turbine.
And after driving the turbine, we take the exhaust gases and put it into the chamber. We
also introduce the oxidizer into the chamber. And in other words what is it we have
done? We have turbine, which is run by the heated fuel or heated oxidizer, whatever be
it, as long as it is possible to generate a vapor. We generate power in a turbine and drive
the pump of the fuel and the pump of the oxidizer. And in other words just by using the
hot chamber, we generate hot gases for expansion in a turbine and we run the pumps.
Such a cycle is a derivative of stage combustion cycle, but without a pre-burner and is
known as an expander cycle.
Therefore, the pump fed systems are basically classified as belonging to gas generator
cycle in which case we allow the exhaust from the turbine to be expanded into the
ambient. The next is stage combustion cycle wherein we have a pre-burner and take the
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combustion to occur in two stages. If not, we use the hot vapor generated on the outside
of the chamber to run the turbine and introduce the exhaust into the main chamber. May
in the next class we will do the regenerative cooling and it will become little more clear
at that time. We use the vapor formed during heating of the volatile fuel while cooling of
the chamber to run a turbine. This turbine then runs these two pumps. This is an
expander cycle.
We could have combinations or variations of some of these cycles. We could allow the
exhaust from the gas turbine to be introduced in the nozzle divergent and thus generate
more thrust instead of being expanded in an auxiliary nozzle. In this case the cycle
known as gas generator with bleed; what is bleed? We allow some of the outlet gases
from turbine to come and generate little more thrust by injecting it into the nozzle here.
You could keep on devising cycles or maybe we could have something like a combustion
taking place in the chamber. We allow the gases to come from the chamber and drive the
turbine and this known as the combustion tap off cycle, but it has not been used in
practice. What has been used in practice are expander; you will recall the cryogenic
engine RL 10 and this uses the expander cycle. We said this was the first engine, first
cryogenic engine developed in US.
Many liquid propellant engines use the gas generator cycle but the stage combustion
cycles are more powerful and have very high performance. Let us take a look on the
merits and performance and see what is normally preferred.
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(Refer Slide Time: 30:46)
But, to be able to surmise the limits and performance, let us summarize whatever we
have understood so far. We said a liquid propellant rocket could either be pump fed or be
gas fed. For gas fed we know how to calculate the amount of gas either in the blow-down
mode, regulated mode or in the hot gas mode. When we talk of pump fed we found it
could operate as gas generator cycle, stage combustion cycle or topping cycle and the
expander cycle. You would like to know under what conditions and when we could use
these cycles. We could have derivatives gas generator with bleed and variations of the
staged and expander cycles.
If we allow the products of combustion from the main chamber to drive the turbine and
expand the turbine gases, we have a combustion tap off cycle. We use instead of having
a separate pre-burner, some of the products from the combustion. We need to have
reproducible temperatures with the same consistent mixture ratio, which is difficult. We
have several other cycles, which use part expander and part of different cycles. We must
keep our minds open and try to see how best we can improve these cycles. If this part is
clear, we can go back and analyze what cycle should we use and when and the merits?
Let us do this exercise.
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(Refer Slide Time: 32:20)
We find that in the gas generator cycle, a fraction f of the propellant in the gas generator
is not very efficiently used. Can we calculate the value of f?
(Refer Slide Time: 32:39)
But, before we do this let us realize that whatever we are doing is related to the feed
system and this is important in the topping cycle as well.
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(Refer Slide Time: 32:45)
In the expander cycle you have fuel and oxidizer. The oxidizer is pumped into the
chamber. The fuel is the volatile fuel. Therefore, while it cools the chamber, the hot
vapor so generated runs the turbine and the exhaust from the turbine is fed back after the
expansion process in the turbine and that is why we call it as the expander cycle. The
vapor is generated during cooling of the particular thrust chamber. Well this is gas
generator with bleed, the exhaust instead of being expanded through an auxiliary nozzle
comes back into the nozzle in the divergent portion to enhance the Isp.
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(Refer Slide Time: 33:36)
Let us now put things together. This figure looks clumsy but it is exactly what we have
been writing on the board. The mass of fuel, mass of oxidizer in different parts is shown
in this particular figure. This we show for the gas generator cycle on this left side; you
have mass of oxidizer coming into the turbine, the part is taken to the gas generator. We
consider this specific case of liquid oxygen, liquid hydrogen as the propellants. Mass of
hydrogen is coming into the pump over here. This is taken into the gas generator burns
here and drives the turbine. The balance what comes in here is only fuel that is m°
hydrogen − m° which goes into your gas generator. Similarly, you have oxygen, which
comes in here which is equal to m°oxygen − m° which goes into the gas generator.
In the GG cycle, to repeat again, you have three mixture ratios; an over all mixture ratio,
mixture ratio for the gas generator and another mixture ratio for the main chamber. In the
stage combustion cycle, the overall mixture ratio is same as the mixture ratio for the
main chamber with the mixture ratio R for the gas generator being different. Is this part
clear?
If this part is clear, let us quickly do this exercise of finding out what will be the value of
f. The turbine generates power, is rotating; the turbine rotates the oxidizer pump the
turbine rotates this fuel pump. The power is mechanically transmitted; the power
generated in the turbine is running these two pumps.
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Therefore, let us find out how what is the fraction of propellant which is required to drive
the turbine. We need to be able to find out how much pump power is required. Therefore,
let us put it down.
(Refer Slide Time: 35:41)
What is the power developed by the turbine and where does the power developed by the
turbine go? It runs these two pumps. The total power required of the two pumps equals
the power required for fuel pump and the oxidizer pump. What is the power developed
by the turbine? We told said that the outlet temperature from the GG is let us say eight
hundred to nine hundred Kelvin. We call the outlet temperature as TGG and this is the
temperature of gases, which enters into turbine. Let us take the outlet temperature from
the turbine after the expansion as Tout. We again repeat TGG is the temperature at which
the hot gases from the gas generator enter the turbine, some power is generated in the
turbine during the expansion of the gases. The turbine may be an impulse turbine. We are
expanding the gases and at the outlet of the turbine, the temperature falls to a value Tout.
Therefore, what is the work done by the turbine? Rate of work is equal to m°o gg the
mass flow rate of oxidizer + mass dot of fuel m°f gg into the gas genrator that is the total
propellant flow rate into the GG. The total propellant flow equals the gas flow rate
generated m° gg and this into the value of CP into the temperature is the enthalpy and
therefore, this is going to be so much Watts. And m° × CP × temperature difference
between the inlet of the turbine and the outlet is the rate of work done in the turbine. And
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since, we are talking of mass flow rates, we talk in terms of the rate of work done i.e.,
power. The turbine has some efficiency. Let us assume that the efficiency of the turbine
is ηt and therefore (m°o gg + m°f gg) × Cp × temperature drop in the turbine (TGG−Tout)
×ηt is the useful power produced by the particular turbine. Let us make sure that
whatever we are writing is the total mass flow rate into CP into delta T across the turbine,
which is the enthalpy change.
The rate of enthalpy drop in the turbine, which is the rate of work done by the turbine
and we can also write as W°t. Can you tell me what is a work done by two pumps? We
have been doing it in the last class also. Rate at which work is done by the two pumps is
equal to the power of your two pumps is watts, let say the unit is watts viz., joule per
second. How do we write it? Let us take a look at this pump; it takes fuel increases the
pressure from this value to this value, let us say that the increase in pressure is ΔP, so
many Newton per meter square. This is same as Pascal. And what is the rate of work
done by the two pumps?
(Refer Slide Time: 39:28)
Now, the rate of work done by a pump can be written as ΔP into the volume flow rate,
where ΔP is the pressure rise across the pump. The volume flow rate through the pump
corresponds to the flow of the oxidizer and the flow of fuel namely, m°o as so many
kilograms per second divided by the density of the oxidizer + m°f divided by density of
the fuel. This is the volume flow rate of the fuel, this is the volume flow rate of the
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oxidizer and this therefore, multiplied by ΔP across the pump is equal to the work
required to run the pumps per unit time. If we look at the units; ΔP has the units of
Newton per meter square and m° have units of kilogram per second, and density has
units of kilogram per meter cube.
That is equal to Newton meter per second or joules per second or it is equal to Watts; this
is the rate of work required for the pumps. For the present analysis, we have assumed
that the pressure increase in the two pumps is the same; otherwise we have to have a
separate expression for the oxidizer and a separate expression for the fuel pump. We are
just trying to illustrate the method. We have assumed that the pressure increase across
the two pumps is the same, which need not be true.
(Refer Slide Time: 41:31)
If the efficiency of the pumps is ηP and the total work required would be more so we
have to divide it by the efficiency. And therefore, now we equate the rate of power
required for the pump with what is the work which is actually done. And therefore, now
we write ΔP × [m°o/ρo + m°f/ρf]×1/ηP = rate of work required by the pumps and is equal
to whatever we got for the turbine. Let us simplify it and write it is equal to (m°o gg +
m°f gg) × CP × ηT × TGG( 1 − Tout/ TGG.
And this becomes my expression. How do we get the fuel fraction from this? Fuel
fraction f was equal to the fraction which is going into your gas generator divided by the
total. Let simplify this little expression further. Let us rewrite it in the following way:
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m°f [1/ρf + (m°o/m°f)/ρo] /ηP = m°f [1/ρf + (R /ρo] / ηP where the overall mixture ratio
R = m°o/m°f is your overall mixture ratio. We follow the same procedure for the rat eof
work in the turbine and get: m°f going through the GG × (1 + RGG) × CP × ηT ×
temperature difference (TGG − Tout).
If we assume isentropic expansion in the turbine Tout/TGG and since for T1/T2 =
(P1/P2)(γ−1)/γ the temperature difference term becomes (1 − the pressure ratio to the
power γ−1/γ. How did we do this? We know, that PV by T is a constant from the
equation of state for an ideal gas. For isentropic process we have PVγ is a constant. From
the gas equation we get Pγ Vγ / Tγ is a constant. And if we divide one by the other and
eliminate V, we get Pγ−1/Tγ is a constant or rather P1(γ−1)/γ/T1 = P2(γ−1)/γ/T2 .
Therefore, if now we have to simplify this expression we get the value of m°f/m°f gg as
equal to the expression in the following slide. We take η to be the product of the turbine
efficiency and your pump efficiency, i.e., efficiency of the turbopump.
(Refer Slide Time: 45:25)
That means mf/m°f GG is equal to η and η is equal to pump efficiency into the turbine
efficiency × (1+ RGG viz.,mixture ratio in the gas generator) × CP × TGG × (1 − 1÷the
pressure ratio in your particular pump to the power gamma minus 1 by gamma) and this
is divided by (1/ρf + R the overall mixture ratio/ ρo) × ΔP.
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We are able to find out the value of m°f/m°f gg through the gas generator. But what is it
that we want? We want to find out the fraction of the fuel and oxidizer, which is going
through the gas generator. We need m° oxidizer which is going through the gas generator
+ m° fuel which is going through the gas generator divided by m° o + m° fuel. That
means, the fraction propellant that is going through the gas generator divided by the
total. Now, this we again simplify as [m°fGG × (1 + RGG)] / [(m°f × (1 + the overall
mixture ratio R)]. Please let us be very clear about it and this is equal to the fraction f.
But what is it we have got here? We have got m°fGG/m°f. Hence we substitute this value
of m°f gg/m°f and get the value of fraction f = by m dot o, which can be written as (1/ρf
+ R/ρo) ×P ÷ η (1 + RGG) × CP ×TGG × (1 – 1/rP(γ−1)/γ) and is multiplied by
(1+RGG)/(1+R). We find that 1 + RGG and 1 + RGG gets cancelled and therefore, f is equal
to (1/ρf + mixture ratio R/ρo) × the pressure increase in the pump ΔP ÷ the net efficiency
of the pump and turbine into CP TGG × (1 − 1 by pressure ratio in the turbine to the
exponent.
Now, we need to discuss these results. We will do it in the next class. What is it we did
in this class? We looked at the fraction of the propellant, which flows through the gas
generator and we have got an expression. We also addressed the different feed system
cycles.
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Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian institute of Technology, Madras
Lecture No. # 28
Analysis of Gas Generator and Staged Combustion Cycles and introduction to injectors
In the last class, we derived an expression for the fraction of the total propellant flow into
the gas generator as a function of R, the overall mixture ratio; the overall mixture ratio
means from the tank whatever be the oxidizer, which is being supplied to the fuel, which
is supplied from the tank. We also had the other parameters viz., the temperature in the
gas generator which depends in the mixture ratio in the gas generator, the density of the
fuel, which we called ρf, density of the oxidizer ρo, the pressure increase across the
pump Δp so many Newton per meter2, the value of Cp and the expansion ratio in the
turbine. (Please note that TGG is missing in the following slide in the denominator).
(Refer Slide Time: 00:58)
What does this expression tell us? Immediately we see f increases as Δp goes up, where
Δp is the pressure increase across the pump. That means, for a high-pressure engine and
a high pressure engine will demand a higher value of pressure at the inlet to the engine.
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Therefore, the increase in the pressure ΔP must be high. Therefore, the fraction of the
propellant f, which goes into the gas generator, must be high. If we want to plot, the
value of f as a function of delta p across the pump, the trend of the change of delta p
should be similar, to the trend of the change of chamber pressure.
We can write here pc as the chamber pressure instead of delta p on the x axis. The
fraction of propellant, which is required to flow through the gas generator, should
increase as the chamber pressure increases; this is first observation; is this alright?
How will the fraction f change with the overall mixture ratio? If overall mixture ratio is
higher that means, the value of ‘f’ will be smaller, because we have R in the
denominator. This R in the numerator is modulated by the density and multiplied by
some number and added to a quantity; therefore, the R in the denominator tends to be
stronger or rather the value of f will decrease as R increases.
(Refer Slide Time: 02:37)
And therefore, we can represent the influence of the overall mixture ratio if we plot Δp
over here or which is same as we said as pc; may be we will get a series of lines for
different values of R and as R increases the value of f decreases. Let us try to interpret
these two graphs, which I have just drawn. We find that as the pump pressure increases
or equivalently the chamber pressure increases, we need more of the fraction of the
propellant to be introduced through the gas generator and what is the implication.
709
(Refer Slide Time: 03:17)
The total impulse now I call it is total specific impulse is equal to f through the gas
generator × specific impulse of the turbine exhaust IGG + (1 – f) of the specific impulse
of the main chamber.
(Refer Slide Time: 03:37)
And we found in the case of a gas generator cycle, if we increase this f the Isp,T will
decrease. In the gas generator cycle, if we plot the total specific impulse of the total
engine system as a function of let us say the value of pc, in the case of the gas generator
cycle net Isp will fall with pressure if the specific impulse of an engine will not increase
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with chamber pressure.
The net effect of the increased fraction ‘f’ is such that the net specific impulse decreases.
Why should the fraction of the propellant which flows through the gas generator
decrease with increase of mixture ratio?
(Refer Slide Time: 04:15)
We find for the specific case of let us say liquid hydrogen as fuel, liquid oxygen as
oxidizer, the density of liquid hydrogen is very much smaller than liquid oxygen. And
therefore, if the mixture ratio increases, we have more of oxygen and therefore, oxygen
is easier to pump compared to light very light density liquid hydrogen, which calls for a
large volume. And therefore, more pump power and that is why this dependence.
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(Refer Slide Time: 04:55)
Therefore, let us summarize these two observations, which I show through these slides
here. We had derived the expression that f is equal to this expression, which I had written
on the board earlier.
(Refer Slide Time: 05:01)
And we said as ‘f’, the fraction of the propellant, which flows through the gas generator,
as a function of p increases. We are considering the Δp as 0.1 MPa, 1 MPa, 10 MPa, 100
MPa and f increases. You know it is a linear with respect to Δp, but since we use a
logarithmic scale the higher values of pressures get compressed as they increase on the X
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axis and hence the curve. As R increases the value of ‘f’ decreases.
(Refer Slide Time: 05:34)
If the chamber pressure is chosen as a parameter and we plot fraction f as a function of
the overall mixture ratio, for a high chamber pressure we have a large fraction ‘f’. The
mixture ratio of the gas generator is assumed to be 0.6. And for η for the turbine pump
system as 0.6; the temperatures of gas generator as 900 Kelvin, we find that for a high
value of chamber pressure we require large flow rates through the gas generator whereas,
if the chamber pressure is small we need a small flow rate.
What is the implication of this? I think this is something which you all can readily work
out and see? The implication is if my chamber pressure is small, then what is it we find?
The value of ‘f’ is negligibly small.
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(Refer Slide Time: 06:48)
If ‘f’ is small you know I do not really spend so much of propellant in the gas generator
and therefore, the Isp is not much adversely affected. However, if the value of ‘f’ is
going to be large, we are pumping so much fuel and oxidizer into the gas generator, that
the net Isp comes down. Rather if I have to make a plot now of the value of the net Isp as
a function of let us say the chamber pressure, I find that Isp monotonically decreases
with increase of pressure if the influence of pressure on specific impulse is not
accounted. If Isp is plotted as a function of the overall mixture ratio, we find that at low
chamber pressures the Isp decreases with mixture ratios. At higher pressure, we get an
increase followed by a drop. This is because, though the Isp is more due to the enhanced
value of pressure, the large fraction of propellants used in the gas generator at the rich
mixture ratio causes the Isp curve to droop.
Whereas if we allow the net propellant into the main chamber like we have in a staged
combustion cycle, may be in that case, we will get a small increase for the stage
combustion cycle. This is value is at a chamber pressure of 1 MPa i.e.,10 bar. But if we
have to operate the engine at a value of let us say 10 MPa, which is slightly higher
pressure, the GG cycle will give a performance over here, slightly higher performance,
but my stage combustion cycle is going to give me a performance, which is going to be
very much higher, because this increase of Isp came from pressure. In fact, for the GG
cycle, the performance drops off rapidly with overall mixture ratio since the mixture
ratio in the main chamber becomes very much oxidizer rich. The stage combustion cycle
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will give a high performance even at the higher mixture ratios. If we go to still higher
pressures, this we are talking of 10 MPa, and if we go to something like 20 MPa or 200
bar may be the G G cycle will come down like this, because I am losing lot of lot of my
impulse in the auxiliary nozzle, whereas, my stage combustion cycle will be much better.
In other words at low pressures by operating a gas generator cycle we do not lose much;
whereas at high pressure we keep losing more and more to the extent that the G G cycle
is not competitive any more.
(Refer Slide Time: 09:32)
Therefore, we can say that a gas generator cycle is more suited for low pressure engines
whereas, the stage combustion cycle or an expander cycle which uses all the propellant
in the main chamber is more adapted for high pressure engines. Of course, if we talk in
terms of this stage combustion cycle, we need a high pressure pump and maybe we will
examine it when we talk in terms of pumps and turbines. To repeat again: let us go
through this in the slides, because this tends to be important in practice.
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(Refer Slide Time: 10:17)
If we operate an engine using the gas generator cycle at a small value of chamber
pressure, this slide shows the net value of specific impulse that we get and if we operate
the same engine on a stage combustion cycle at the same low pressure, we get a slightly
better performance. This is because we have not lost very much, because f is small. I
have lost something from stage combustion cycle to gas generator cycle; therefore I still
find gas generator cycle is lower than stage combustion cycle, but the loss in
performance or the decrease in the value of specific impulse of the gas generator cycle is
small.
The loss is small, because f might be something like 0.01 or something of this order. If
we go to higher pressure what is it we find? At higher pressure the gas generator,
because the pressure is high, we get a slightly higher value of specific impulse, but at the
same value the stage combustion gives me a much higher value of specific impulse. That
means, by operating at something like 100 bar, I lose if I were to operate the rocket in a
gas generator cycle and I will have a lower value of specific impulse whereas, if I
operate it in a stage combustion cycle, I get a higher value. Please remember that the x
axis in this graph represents, the overall mixture ratio R, and as R increases, the quantity
of the oxidizer increases, since we are in an oxidizer rich region there is a fall in pressure
as the mixture ratio increases.
If I go to something like 20 MPa say 200 bar, because of the very high value of ‘f’ the G
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G cycle has this low performance whereas, we do not lose anything in stage combustion
cycle and the performance is very high in terms of specific impulse. We loose a lot by
operating a liquid propellant engine in a gas generator cycle at high pressures.
If we have a low pressure engine, may be a G G cycle is adequate while if we have a
high pressure engine it is necessary to go for stage combustion cycle.
(Refer Slide Time: 12:20)
And generally for cryo engines a chamber pressure upto about 10 MPa or 100 bar seems
to be the limit for a gas generator cycle, above this to operate a gas generator cycle you
will lose a lot. And this follows from a cycle analysis.
To be able to complete the cycle analysis, we must also find out why the droop or fall in
Isp with the overall mixture ratio happens especially for the gas generator cycle. It fell so
rapidly at higher pressures. The reason is that the overall mixture ratio in the main
combustion chamber becomes very oxidizer rich since the gas generator demands fuel
rich mixtures.
717
(Refer Slide Time: 13:01)
Like for instance we had a gas bottle; from the gas bottle, we had the tanks. What did we
do? We took little bit of the oxidizer, little bit of the fuel into the gas generator, and this
is mind you very fuel rich. And therefore, we are bleeding more and more of it. What
happens when I bleed more and more of the fuel rich mixture, the mixture ratio of the
main engine keeps increasing, because I am drawing lot of fuel into gas generator and
starving the main engine of the fuel. Therefore, this becomes oxidizer rich and then
again what is the dependence C star or Isp with respect to mixture ratio. It comes down
after an optimum mixture ratio. And that is why the droop in the curve.
(Refer Slide Time: 13:56)
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Therefore, let us quickly derive an expression what will be the value of the mixture ratio
in the main thrust chamber for a gas generator cycle a function of R and ‘f’.
The value of mixture ratio in the main chamber is required. Let us picture this gas
generator cycle in our minds. We say Rmc, the mixture ratio in the main chamber, is
equal to m°o − m°o gg in the gas generator which is not available in the main chamber ÷
m°f − m°f gg through the gas generator. We consider the specific case of hydrogen
oxygen as propellants. And this we can now write as equal to m°o×(1 × 1 −m°o gg/m°o).
Similarly, we express hydrogen as m°h×(1 − m°h gg through GG/ m°h). And how do I
get this value of m°o gg which is going through the gas generator or m°h gg which is
going through the gas generator to the total oxygen and hydrogen flow. We have already
done something very similar, in the last class.
(Refer Slide Time: 15:38)
Let us take a look at m°h gg through gas generator + m°o gg through gas generator = the
total propellant flow in the gas generator. This is equal to m°h through the gas generator
× (1 + RGG). And similarly, we can we can write an expression for m°o + m°h, which is
the total mass of propellants as equal to m°h × (1 + R). And now we know what is the
fraction f. Fraction f = m°h gg×(1 + RGG)/ [m°h×(1 + R)].
719
(Refer Slide Time: 17:04)
Or rather from this we get m°h gg / m°h = f × (1 + R)/ (1 + RGG). And now we can also
write, if this is ok, m°o GG / m°o. How do we convert m°h,GG to this m°o,GG. Multiply it
by RGG. Therefore m°o GG becomes equal to m°h GG × RGG. And m°o/m°h = R therefore,
m°o = R × m°h. Therefore, m°o gg/m°o = RGG/R into the same value viz., f
×(1+R)/(1+RGG) . And now we substitute these values of m°h, GG by m°h from the first
expression, and we take m°o GG/m°o from the second expression in the expression for the
mixture ratio in the main chamber.
(Refer Slide Time: 18:18)
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And I get the value of R for the main chamber Rmc = (m°o − m°o GG)/ (m°h − m°h,GG).
This is equal to R from mass of oxidizer by fuel × (1 − I take from the second
expression, RGG/R × f × (1+R)/(1+RGG) ÷ (1−f) × (1 + R)/(1+RGG). I think we must learn
to do such derivations for an analysis.
We have obtained the mixture ratio in the main engine as a function of the mixture ratio
in the gas generator and the value of ‘f’. And for combination of parameters like RGG is
being about 0.6 and for different values of R, we can plot RMC as a function of f. We find
as ‘f’ increases the value of the mixture ratio in the main chamber keeps increasing and
as the value of R increases.
If the value when f is equal to 0, the mixture ratio in the main chamber is same as the
overall mixture ratio. This is the condition for the staged combustion cycle, and the
expander cycle engine, wherein there is no loss in the gas generator, because gas
generator supplies the propellant back into the main chamber. That means, when f is
equal to 0, we regain the solution. As ‘f’ increases the mixture ratio in the main chamber
keeps increasing. And if it increases to a very large value you come to a situation where
in you cannot operate the engine.
(Refer Slide Time: 20:40)
In other words what we just now told was R versus the specific impulse or C* goes like
this. We start operating at these low values and that is why the specific impulse or
equivalently C* or the total performance keeps falling. This is how we compare the
721
different feed system cycles such as the gas generator cycle, the stage combustion cycle,
the expander cycle, etc. We will quickly sum up by telling the following: for pump feed
systems, we could operate the liquid propellant rocket as a gas generator feed system, as
a stage combustion cycle or as an expander cycle or some other cycle
(Refer Slide Time: 21:17)
We found that the stage combustion cycle is something like a topping cycle; we find the
G G cycle suffers at high chamber pressures, because the value of the fraction of the
propellant is used to drive the turbine is not properly utilized. And what is driven out is
at a low value of expansion through an auxiliary nozzle; whereas, in the topping cycle
the mixture ratio of the main chamber is same as the overall mixture ratio. In the case of
the GG cycle, the mixture ratio in the main engine and the overall mixture ratio are
related through f and RGG.
For high pressure of operation, the staged combustion cycle is particularly useful because
we gain the advantages of high pressure, but we will have to take a look at the design of
pumps after two or three classes. I think this is all about the gas generator and the stage
combustion cycle and the expander cycle. We cannot operate the expander cycle at high
values of pressure, because we use only a vapor that is generated by heating with the
chamber, but this also has some powerful implications. Maybe we will take a look at
some examples on why we cannot operate the expander cycle at high pressure.
In an expander cycle, we are using a chamber which runs hot. We have limited amount
722
of heat transfer possible in a chamber. And therefore, we cannot have very high power
and since I cannot have very high power. The expander cycle operates at low chamber
pressures.
But, its performance will be very much higher than the gas generator cycle, because we
do not use any amount of propellant in the gas generator, which is not effectively
expanded. I think this is all about the feed system cycles.
Let us go to the next element of our discussion namely the thrust chamber.
(Refer Slide Time: 23:37)
We need a gas bottle and we could find out how much mass of gas is required. We said
well propellants are stored in tanks, then we said we need something like a pump for
fuel, we need a pump for oxidizer, which is driven by the turbine and we still have to
cover this part on pumps. We come to the chamber wherein fuel and oxidizer are injected
into the chamber. That means, we have a fuel to be injected into the chamber, similarly
the oxidizer. We need to know how combustion takes place in a chamber and of course,
we have considered the nozzle expansion earlier.
We would like to concentrate on this thrust chamber in this class and may be first half of
the next class. What does the thrust chamber consist of? It consists of a device to inject
the liquid into it into it. May be the liquid must evaporate get mixed together and burn
and the products of combustion must get expanded. Therefore, let us consider the first
723
part namely the injection device. How do you inject the high pressure fuel into the
chamber?
(Refer Slide Time: 24:56)
The injector admits the requisite quantities of liquid fuel and liquid oxidizer at the given
mixture ratio into the chamber. That means, it must have some way of control of the
mixture ratio. It must the required quantity of propellants into the chamber. Not only
does it does it admit the liquid fluid at the given mixture ratio, it must also sort of
increase the surface area of the liquid or it must atomize the liquids. What do we mean
by atomize. It must disintegrate the liquid fuel into something like fine droplets or let us
say droplets, which can easily evaporate. Not only must the injector assist in
vaporization, but the third point is it must help the evaporated vapor of fuel and oxidizer
to mix together. It must ensure that it will push the fuel and oxidizer in some way such
that the vapor of both will mix together. Once mixed, the fuel and oxidizer can
chemically react and burn. In some cases you need an igniter to start the burning, but
once started the hot environment can always promote the chemical reactions and the
burning. Therefore, the requirement of an injector is it must admit suitable quantities to
give the correct mixture ratio and mix the vapors.
724
(Refer Slide Time: 26:37)
And we are interested in a given mixture ratio. Let us not forget this graph of C* or Isp
as a function of mixture ratio. It is in the fuel rich region that we get a much higher
specific impulse. Therefore, we are interested in this value of mixture ratio. The injector
must also admit the required amount of fuel oxidizer and fuel such that we get the thrust
as desired, get this mixture ratio and it must also break up the liquid into fine droplets.
And mix them together this is what an injector should do. Therefore, let us start with the
simplest form of injectors, which we are familiar and let us let us build up on it.
(Refer Slide Time: 27:14)
725
In the figure, I show a shower head. We use it daily for bathing..
(Refer Slide Time: 27:23)
Let us take a look at the streams of water generated. The streams get broken into drops
later on.
(Refer Slide Time: 27:31)
We have something like a like a head and a number of holes or orifices. I brought a
shower head and take a look at it.
You know this is something like what we use in our shower you know you have the
726
water coming from the water line, water collects in this region known as manifold and
then you have a series of holes or orifices. And this is what we called as a manifold in
which water collects. The pressure in the manifold is higher than the ambient pressure
and water is forced through these orifices. Let me just sketch this shower head on the
board. It tends to be very illustrative of the different types of injectors, which we use.
(Refer Slide Time: 28:20)
I have a surface with a number of fine holes. This is what was said to be a head, which
has lot of orifices. You have something like a place through which the liquid is admitted.
And we have spacing between these two holes or streams. And this is where we admit
the water and this is what we called as a manifold. What does the manifold do? It admits
and maintains the pressure over here such that water squirts out through these holes here.
And you know that very often you find if your shower head is not properly designed and
you are taking a bath let us say, you do not get the streams of water hitting you. If your
shower head is very well designed you find the jets of water come like this as laminar
streams.
If it is not very well designed, you find some drop drops of water coming like this, may
be at the same value of velocity. We would like to have an injector, which is something
like a shower head, but which is able to produce droplets and this is one type of injector.
And this type of injector is known as a shower head injector. Let us again go through a
shower head injector. We have manifold in which the water collects and forces through
727
the orifices.
(Refer Slide Time: 30:06)
Therefore, we are looking for something like flow through an orifice or a hole. The
shower head consist of lot of these holes through which may be water is being pushed
through when we are taking a bath. And a similar scheme can be used in case of liquid
propellants. You have the manifold here. We have the set of orifices here. We could
divide it the manifold into partitions and admit the fuel in some region and admit the
oxidizer in the other portions. We allow them to mix in this region and evaporate and
burn. This becomes the shower head injector. Now, we would like to find out about the
flow through the orifices?
We show one such hole or orifice, this is the manifold here and we have lot of such
holes. This is the manifold; since its volume is large, the liquid as such collects in this
manifold here. That is in the region in the chamber preceding the orifices therefore, the
pressure in the chamber is the supply pressure to the hole or orifice. The velocity is
almost zero here considering the larger volume. And the water squirts out through these
particular holes.
Therefore, we have the manifold; in the manifold be the pressure is p and the velocity is
almost zero. And in the case of rockets, the liquid get supplied into the chamber wherein
the pressure corresponds to the chamber pressure from these orifices. We have the
chamber pressure downstream of the orifice and the supply pressure ahead of it in the
728
manifold. We are interested in finding out the flow through the orifices.
(Refer Slide Time: 32:35)
If we have let us say nf orifices for fuel and I have no orifices for oxidizer, we want to
find the flow rate through the orifices and the mixture ratio. We would like to find out
the flow per orifice and multiply by the number of orifices to determine the net flow
rates.
How do we find the flow through the orifices. We look at this scheme again. We find
that there is lot of science even in a small orifice flow, which we need to understand. We
have an orifice whose edges are sharp. We call it as a sharp edged orifice. How do we
make an orifice for the particular shower head? Each of the holes is drilled on a plate.
If we drill a hole and remove the burs at the edges, we get the sharp-edged orifice. When
the fluid enters the sharp edge from the manifold wherein the velocity is almost zero, it
accelerates and sort of contracts; The liquid separates from the walls of the orifice,
contracts to a minimum and then reattaches later to the walls. The minimum contracted
area is called as veena contracta. In other words this is the where the liquid is flowing. If
I have my shower head which has a very small dimension that means a very thin plate
instead of having a given length of the orifice. The separated flow leaves the orifice.
The flow is coming from the manifold contracts and goes straight out. That means, the
flow does not reattach back to the wall of the orifice. In fact, the flow is separated and
729
the effective area of flow is going to be much lower than the area of the orifice. If we
denote the area of orifice by A0, the area of flow is going to be much lower. How do we
write out the expression for the mass flow through the orifice. Let us try to derive a
simple expression.
(Refer Slide Time: 35:32)
You would have done this in your fluid mechanics class; but let us just do it again. We
have a manifold and we are considering the case of a single orifice. The pressure in the
manifold is higher than the chamber pressure. The difference in pressure is equal to Δp.
We want to write an equation for the velocity here at the exit of the orifice when the
entrance velocity at the manifold is zero. We denote the exit velocity by V. We use the
Bernoulli equation; the flow is liquid at small values of velocities therefore it is
incompressible and its density ρ is a constant. Therefore, we have p/ρ + V12/2 + gz1,
where z1 is height above datum at the entry to the orifice. V1 = 0. The value of pressure
at the exit is pC and the expression becomes pC/ρ + V2/2 + gz2 where z2 is the height
above datum over here. Since the orifice is small in length, we can take gz1 is equal to
gz2, because there is not much change between z1 and z2. And therefore, we immediately
get the square of the velocity of the stream leaving the orifice equal to √2 Δp/ρ. That is V
is √2(p−pC)/ρ.
The mass flow rate is equal to the above velocity multiplied by the area of the orifice into
into density. If the orifice cross sectional area is A0, and we get the mass flow rate as A0
730
×√2 Δp × ρ. This is the mass flow rate for a simple sharp edge orifice.
But we just saw that the orifice sometimes flows full like it is attached over here, there is
some friction over here and sometimes it flows separated from the walls of the orifice.
(Refer Slide Time: 37:44)
Therefore, there are different regimes of flow. Depending on the type of flow through the
orifice, whether attached in which case a it runs full or whether it is detached in which
case it does not run full, the cross sectional area of the flow will change. We find that
based on the orifice area, which is A0, we can define a coefficient namely a discharge
coefficient as equal to flow based on the ideal flow or ideal mass flow what we could
have and the actual flow. We call it as a discharge coefficient and what is the ideal flow?
When the entire area of the orifice that is the flow is running full when there is no
friction at the wall, I could have the total flow corresponding to Δp. In practice we have
friction at the wall, sometimes the flow is separated and the flow may not be totally
axial. The actual flow will be less than ideal flow and therefore, we have a discharge
coefficient Cd which will always be less than one.
To be able to arrive at this value of discharge coefficient we write in terms of mass flow
rate m° in the ideal case. What is the value of the ideal mass flow rate? We have flow
runs full through the area A0, the velocity of flow depends on the pressure drop namely
√2 Δp/ρ. And therefore, we get the ideal flow rate as equal to A0 × √2 Δp × rho. In
practice since you do get separated flow and we have frictional effects you do not know
731
what this area of this separated flow, you base your total flow on the total area A0. The
actual flow would therefore be equal to m° = Cd × A0 × √2 Δp×ρ. The flow is
incompressible and therefore the density is constant. This is the value of the flow, which
takes place, is given by this particular expression. Let us again recall that A0 is the area
of the orifice through which flow is taking place, Cd is the discharge coefficient, ρ is the
density of the liquid, and Δp is the pressure drop across the particular orifice or hole.
(Refer Slide Time: 40:51)
It must be remembered that the value of the discharge coefficient Cd depends on the
regimes of flow through the orifice. What do we mean by regimes of flow? The flow
sometimes runs full such as it happens when the orifice is large, sometimes with
cavitation it gets separated or when the orifice is very very thin, the flow cannot reattach.
And for the different conditions we would like to examine the value of Cd.
732
(Refer Slide Time: 41:28)
We should not spend too much time on it, because we are shifting the topic from liquid
propellant rockets to once particular element of it. This shows your injector head or
shower head wherein you have lot of these small holes through which flow is taking
place.
(Refer Slide Time: 41:32)
733
(Refer Slide Time: 41:45)
If we do experiments and allow flow at different values of Reynolds number, we have as
flow rate increases, an increase of the Reynolds number. We see in the figure the jet
issuing from an orifice at Reynolds number of about 16,000, 32,000 and 33,000. You
find that at smaller values of Reynolds number the texture of the jet is quite smooth; like
you stand under a shower you can see silvery water coming down. At some Reynolds
number it tends to become a little turbulent and rough. At yet higher values of Reynolds
number it becomes violently rough. Let us examine the flow further.
(Refer Slide Time: 42:15)
734
When we have a long orifice the veena contracta is followed by the flow subsequently
attaching to the orifice walls. Sometimes even for the same length the flow goes straight
through; it does not attach. Whereas for a small length the separation is understandable,
because if we were to cut the orifice much before it attached, well there is no way of
reattachment. What could be the reason for flow not attaching for the longer orifices?
(Refer Slide Time: 42:41)
When flow is taking place at high velocities, and we further enhance the velocity by
increase of the pressure drop, the static pressure of the liquid in the orifice decreases. If
the pressure decreases to a value equal to or less than the vapor pressure of the liquid
itself then cavities begin to form in the liquid. Once cavities begin to form in the liquid, a
reattachment like this is not possible and the flow separates out even for the longer
orifices. Such type of flow is known as cavitated flow, and some books call it as super
cavitation. Though the flow should have reattached, the pressure here has gone to a low
level wherein the pressure of the liquid is equal to or less than the vapor pressure of the
liquid and vapor gets generated and the flow separates.
Therefore, essentially we talk in terms of three types of flows. Reattached flow when we
have long value of length to diameter orifices, a flow which is separated when we have
high velocities or cavitation taking place and separated flow for small length to diameter
orifices. Therefore, we could get different values of discharge coefficients accordingly.
735
(Refer Slide Time: 44:01)
And if we do an experiment starting at low value of Reynolds number wherein we get an
attached flow, we get a high value of discharge coefficient since the flow is the attached
and the whole flow area is contributing to the flow. Thereafter the discharge coefficient
decreases as the Reynolds number increases because of the frictional losses and losses
due to turbulence. At some value of Reynolds number cavitation starts and the flow
separates to give a smaller discharge coefficient. The small discharge coefficient persists
at still larger Reynolds numbers.
And now if we start reducing the pressure or reducing the velocity from the high
Reynolds numbers, the discharge coefficient does not increase in the same way it
decreased with increased Reynolds numbers. The flow retains the memory of the
separated flow and continues to be separated and abruptly jumps back to give high
values of discharge coefficients. That means, even a sharp edged orifice which makes the
shower head injector can give a multitude of discharge coefficients.
736
(Refer Slide Time: 44:49)
When we start the experiment and measure the Cd as the function of Reynolds number
we initially get attached flow for which Cd is near to 1. This is because the flow comes
and gets attached and the fully attached flow gives discharge coefficients with losses due
to friction drop and turbulence. But when cavitation starts, the flow separates. The flow
separates giving low Cd when in the forward direction of increasing Reynolds number.
When we reduce the pressure the discharge coefficient does not trace back the original
values and we have a zone for which we get two values of discharge coefficients. This
zone is known as the hysteresis zone.
If we have the length of the orifice to the diameter of some value wherein it is just near
the attachment, the jet issuing from it would have certain characteristics. At the threshold
value the flow attaches and reattaches with the result that there is a flip in the jet and
change in the discharge coefficient. The flip is between attached region and detached
regions of flow.
737
(Refer Slide Time: 46:29)
Therefore, even to choose a shower head we need to understand the mechanics of flow.
Normally the shower heads are such that the length of your orifice is greater than the
diameter so as to get the attached flow. The orifice length L to the diameter of the orifice
D is known as aspect ratio of the orifice and is about 2.
For control purposes whenever you want a controlled flow experiment we use a very thin
orifice with razor type of blade in which the length is a very small number compared to
the diameter. And the flow in this case is always detached. Therefore, you must choose
whether you want detached flow or attached flow and accordingly choose the
dimensions. And therefore, we say that flow through orifices depends on the length to
diameter ratio, because the Cd depends on it.
738
(Refer Slide Time: 47:16)
We normally choose sharp edged orifices. If you go to the market and you want to buy a
shower head for bathing, why not choose a shaped orifice which provided streamline
flow? This could provide attached flow without any flow separation compared to the
sharp edged orifices.
(Refer Slide Time: 47:35)
Let say an orifice at the exit of the manifold could be shaped for a smooth entry. Why
not make such orifice which provides smooth streamline along the flow. This will give
full flow. But to fabricate such orifices is difficult especially in large numbers. We could
739
have an orifice which is like this; however, the next one would be different and to obtain
reproducibility in something like a shaped orifice is more difficult. And we are going to
have 40 holes in a rocket we may have 80 holes or 100 holes or 200 holes. To get so
many shaped holes drilled with shaped orifices is difficult and therefore, we normally
use sharp edged orifices.
We have a manifold in which we fuel and oxidizer. We have a series of orifices through
which oxidizer flows and fuel flows, and that is how we make a shower head injector.
We would like to calculate the mixture ratio formed by the injector.
(Refer Slide Time: 48:56)
The mixture ratio R is equal to the mass flow rate of oxidizer to the mass flow rate of
fuel. Therefore the value of R is equal to Cd for the orifice × the number of oxidizer
orifices × the area of each oxidizer orifice × √2×Δp×ρox ÷ Cd of the fuel orifice ××
number of fuel orifices × area of the each fuel orifice × √2×Δp×ρf. If we have a shower
head which has common Δp and Cd for both the fuel and oxidizer orifices, the terms
cancel out. This is how we obtain the mixture ratio.
This is all about the simple way of injecting fuel in a rocket chamber using what we call
as a shower head. This shower head teaches us one more lesson. We had said that this is
the manifold and this is the orifice. We have a particular manifold volume. What should
be this volume in a liquid propellant rocket. Should it be large or small? From fluid
mechanical considerations if the volume of manifold is large then we will have the same
740
pressure for all the holes over here. If we have a very small manifold volume, the holes
which are at the center near the tube inlet will get the high pressure the others will get a
low pressure.
Therefore, from fluid mechanical considerations we should have a volume of manifold
which is let us say large in order to get all orifices achieve same inlet pressure and form
similar jets. Let us take an example.
(Refer Slide Time: 51:22)
Supposing we have a multi-story building and we want to supply water from the top
which is on the 10th floor. And we want to supply water uniformly to all the floors i.e. to
all the apartments. If we put a common manifold tube or a tube for supplying water, a
person on the 1st floor will get water at high pressure while a person on the top floor will
hardly get water. How do we ensure uniform supply and this is the same problem for the
different orifices in the injector. How to configure the manifold? The flow resistance for
the lower floors has to be increased by reducing the diameter of the pipe conveying water
to the lower floors.
Or else we could introduce resistance by placing filters or gauges. We place a filter with
larger holes on the top floors. At the bottom floors we place finer filter such that we
introduce some pressure drop such that the supply pressure is same. And so also in
rocket injectors whenever we have a manifold we cannot have a large manifold for the
simple reason that lot of propellant collects before it can be injected.
741
(Refer Slide Time: 52:28)
And when we stop the flow of the propellant into the chamber, the huge quantity of
propellant will continue to dribble. What is going to happen? We have the thrust or we
have the chamber pressure and when the propellant flow is terminated the dribbling of
propellants will continue to burn for a long time. And you would have seen this when
you close the valve of the shower and the water continues to dribble. This is because of
the large dribble volume or the volume of your manifold is high.
Whereas, if we have a very small manifold then immediately the thrust terminates
because there is nothing left to burn. Even though we would like to have a large volume
of the manifold such that we can supply to all the orifices at constant pressure, from
considerations of dribble volume it becomes essential to keep the manifold volume to be
small. But if we have to keep the manifold volume small, then how do we ensure
uniform flow in all the orifices?
742
(Refer Slide Time: 53:43)
We place something like a sieve over the holes near the inlet and decrease the pressure at
the inlet to the orifices and in regions away on the periphery we communicate the
pressure without any obstruction. These are some common methods used in the design
of shower head injector.
(Refer Slide Time: 54:08)
But shower head injector is something like a weak injector, because as you know the jets
formed are all parallel. It takes some definite time for it to atomize and form droplets, but
you want droplets as early as possible. Mixing of the fuel and oxidizer is also difficult.
743
And though some of the earlier designs in rockets liquid propellant rockets use shower
head injectors, at the present point in time we never use the shower head injector.
What is it that we do to improve it? Instead of jets being parallel we make them interact
with each other. That means we have impinging jets. And once you impinge jets we form
something like a fan. We will look at the different injection devices in the next class and
also look at some of the problems which we face in the combustion chamber.
In today’s class, we started with the gas generator cycle. We looked at the deficiencies of
a gas generator cycle. Namely some propellant gets wasted which is not fully utilized. A
staged combustion cycle and expanded cycle are preferred especially at high pressures.
Then we just started with the injectors we looked at the shower head injectors. We will
build up on this and look at the other injectors, which are used in liquid propellant
rockets in the next class.
744
Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 29
Injectors, Cooling of Chambers and Mixture Ratio Distribution
(Refer Slide Time: 00:18)
We will continue with the discussion on liquid propellant rockets. In the last class, we
were looking at injectors. And what does an injector do? It admits the required liquid
fuel and liquid oxidizer in the combustion chamber, breaks it up into particles; why do
you have to break it up? So that the surface area increases, it will evaporate. It mixes the
fuel and oxidizer vapor, and after mixing the propellant vapors burn. Therefore the
purpose of injectors what we said in the last class is maybe it fragments the liquid, then
once it is fragmented; that means, the liquid is made into fine drops such that the surface
area increases, then there is something like an evaporation taking place. And then it
mixes the vapors of the liquid fuel and the liquid oxidizer, the vapor of the fuel and
oxidizer are mixed; and then once they mix they chemically react and they burn in the
combustion chamber.
745
In first half of today’s class maybe I will discuss further on the injectors; and what did
we do in injector in the last class? We told we could have something like a shower head
injector which we normally use for bathing. It consists of a series of orifices or holes,
and then you have the water, which comes from your tap into the shower head, and in the
manifold where water collects, and then the jets are generated as parallel streams and this
is what was a shower head. But, we had said that the flow of the water in the shower
head is such that if we consider a single hole, it is not uniform flow throughout its length
in the orifice. We have a vena contracta wherein the flow contracts and thereafter it
diverges and gets attached to the walls of the orifice. If the aspect ratio of the orifice is
small we have a detached flow after the vena contracta. And we could have different
flow patterns, and if the flow cavitates it detaches from the walls. Therefore, the flow
does not happen fully through the orifice.
How did we calculate the flow of let us say the oxidizer and the flow of fuel through this
particular set of shower head orifices? We spent time on the shower head, because it is
central to the others and something which we can readily visualize. Therefore, let us say
the mass of oxidizer that we want to calculate could be had from the following.
(Refer Slide Time: 02:59)
We need to know the velocity of the flow, and we derived an expression using Bernoulli
equation that velocity is equal to √2 ×the pressure drop /divided by density. The pressure
746
drop was Pressure at in the manifold minus the pressure in the combustion chamber is
denoted by Δp.
Therefore, V = √2 Δp/ρ , and to be able to get the quantity of flow which is taking place;
we had all these holes nox for the oxidizer, and the area of each hole was Aox. So the flow
rate of oxidizer was delta p is in Newton per meter2 Aox m2, the number nox and flow in
meter3/second. But mind you, we are still missing something what is that missing?
(Refer Slide Time: 04:10)
You know when flow is taking place through the orifice, very often flow gets detached
here and sometimes the flow gets detached attached. That means, this is the flow taking
as detached and the entire orifice is not fully flowing, and but still I have to base it on the
actual area of the orifice. The real area of flow is only this. The apparent area is the area
of the orifice, which for oxidizer is Aox.
And therefore, the volume rate that flows through the orifice is much lower than what we
have calculated. Not only for this detached flow but even when the flow gets attached;
let us say we have the orifice like this; we have a vena contracta and it gets attached.
There is some frictional drop here the flow may not be axially progressing. It may go at
an angle and therefore, I still have loses and therefore, the losses apparently lower value
of the flow rate. But we are basing on the orifice area. And therefore, we said that the Q
actual which is flowing is multiplied by a coefficient which I call as discharge coefficient
Cd. The volume flow rate is therefore area of the orifice A into the value n of the number
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of holes into √2 Δp/ρ × the discharge coefficient Cd or rather this is the rate of flow in
meter cube per second. We convert it to mass flow rate and we get the mass flow rate
mutilying by density to give m° = Cd×A×n×√ 2Δp×ρ so many kilograms per second.
We did so far in the last class. We can calculate the total propellant flow and the mass
flow rate of fuel and oxidizer. And therefore, we have a number of holes for the oxidizer
and a number of holes for the fuel and we can get the required flow. The total oxidizer
flow will be equal to Cd for the oxidizer orifices into area of the oxidizer orifices into the
number of oxidizer orifices into √ 2 delta p across the oxidizer orifices into the density of
the oxidizer.
And similarly, for fuel we have its mass flow rate as Cd of the fuel orifice into area of
the fuel orifice into the number of fuel orifices and the other terms like pressure drop
across fuel orifice and density of fuel. What was the mixture ratio? Mixture ratio is this
quantity of oxidizer divided by the mass flow rate of the fuel.
(Refer Slide Time: 07:49)
You have the number of holes in the shower head. We compartmentalize it for the
oxidizer and fuel. We have oxidizer orifices and fuel orifices.
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(Refer Slide Time: 08:05)
And then, what we have is something like let us say a shower you have jets of liquid
going like this and you find that these are individual jets which are going, and they break
into droplets. But, there is one problem with shower head; we have something like jets
coming like this from the holes in the orifices and these become droplets.
(Refer Slide Time: 08:27)
The direction of flow is axial like this. If I have one fuel jet and one oxidizer jet, I am not
promoting the mixing, I am not giving it an angle such that the vapors can come and mix.
Therefore, mixing is poor in a shower head injector, and we need good mixing for
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combustion to take place? We need to have some strategies such that we can have the
fuel vapor formed by evaporation of the fuel droplets to mix with the oxidizer vapor
formed from oxidizer droplets.
If the jets were to impinge on each other such that we could get something like a mixing
zone. If we can get a mixing zone then we will be better off. And therefore, I
schematically show this scheme of impinging jets. Let us say for the present that we
have this as the jet which is coming from a shower head, then we reorient the second jet
and make the two jets impinge on each other at this particular point.
(Refer Slide Time: 08:18)
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(Refer Slide Time: 09:45)
And when we make the jets impinge at a point; we have momentum change, rate of
change of momentum taking place at the point of impingement. We have something like
a jet coming over at the point of impingement; one liquid jet coming from one orifice. I
say that the mass of flow is m° here, let the velocity of the flow is V. Therefore, the
momentum or the rate of of momentum along this is equal to m° V. We take another jet
with mass flow rate m°2 and velocity V2 which impinges on the previous jet. And at the
point where these two jets are impinging, let us say they impinge, we have something
like a pressure which is built up at this contact point due to rate of change of momentum.
At this particular impingement point we have a higher pressure and therefore, the liquid
jets spread out; I get something like a sheet, which is known as a fan.
Therefore, if we look at it the two jets impinging on each other symmetrically, we get a
fan like this in a plane normal to the plane of the jets. In the plane of the jets we cannot
see the fan. Let us try to draw it such that it is clear. I have two jets just like I told you; I
have a jet coming over here; the two jets are impinging, and therefore, I get a fan like
this. If I were to draw it in the plan view I am looking from the top I see only one jet,
because the other jet is behind and I get something like a fan over here, and this is the fan
which is a liquid sheet which is formed.
Therefore, if instead of using single jets in a shower head we have lot of impinging jets
like this and form many thin liquid fans at different orientations which then breaks into
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droplets. In this way instead of having something like a shower head we have impinging
jets which form fine droplets from the thin fans and also mix the fuel and oxidizer.
We can make this final jet or fan in any direction we want. How do we do it? Let us
consider a small example. Let us say the angle of this first jet with respect to the
horizontal is α1, the angle of the second one with respect to the horizontal is α2. Let me
draw it distinctly on the other side of the board.
(Refer Slide Time: 12:42)
We have one jet, the angle with the horizontal is α1; let the mass flow rate of this jet be
m°1 let its velocity be V1. I have another jet which impinges it over here, let the angle be
α2; let the mass flow rate be m°2 and the velocity be V2. Rather than the mass we
consider mass flow rate m°1 and m°2.
Now, depending on the rate of change of momentum we get a force here, and we can
make this final or resultant jet at any angle we want. How do we do that? That means,
the resultant jet or the resultant fan or a sheet is formed.. It could be at any angle what we
want and this angle β will depend on the momentum and the angles of these two incident
jets. We want to determine this angle β. The rate of change of momentum in the axial
direction is equal to m°2V2 into cosα2 for jet 2 and for jet 1 the change is in the same
direction. Therefore, we have + m°1V1 cosα1. What is the vertical component? The
vertical component is m°2V2sinα2 − in the downward direcm m°1 V1 sinα1. The tangent
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of the angle β would be the resulting vertical component divided by the horizontal
component.
(Refer Slide Time: 14:36)
Therefore, depending on the angle what we choose and the momentum of the two jets we
can get any angle beta that we desire. And therefore, if we have something like these
holes; I have another set of hole like this. By changing the angle we can make one spray
fan like this; the other spray fan like this and I can mix them better. And therefore, this is
the principle of impinging jets or impinging jet injectors. And therefore, let me
summarize what we said. We have one jet impinging with the other to form a fan, and
this angle we can choose and therefore the mixing in such type of injectors will be much
better.
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(Refer Slide Time: 15:15)
Let us take one or two experiments, which we did on this impinging injectors. Well, we
have one jet here; the other jet here, equivalent to saying one jet comes here; the other jet
comes over here. We have the fan which is formed along this plane perpendicular to the
plane of the two jets. How does the fan look? When we take these two jets which are
meeting at a point; that means, the two jets are impinging. We have an impingement
point and therefore, I get something like a fan over here. The fan is typically of this
particular shape over here, and when we look at this particular fan over here the two, it is
in the plane of the paper and therefore, I get droplets like this. When I look at the side
view, we get a fan like this; and I get a series of droplets. This is the principle of
impinging jets which are used for injection.
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(Refer Slide Time: 16:00)
I show the impinging jets and the formation of the fan with the incident jets at different
Reynolds numbers. At low Reynolds number when the jets are clean, when the jets are
smooth I get neat fan like this; it forms a fan over here, and disintegrates into droplets. I
go into the very turbulent regime at higher Reynolds number; the incident jets are that
smooth, but still I get fine droplets. But, I told you about cavitation flow wherein the
disturbances in the incident jets are quite profound. In this case we do not form a clear
fan and fine droplets. Therefore, we have to choose a proper Reynolds number or a
proper Weber number to make sure that we get fine droplets. This is the principle of
impinging jet injectors.
Now we can go a little faster and we can tell that injectors can be classified into some
simple schemes, and what are the simple ones that we can talk of?
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(Refer Slide Time: 16:49)
We can say injector could either be the simple shower head injector, but we said shower
head is not good because mixing is poor and therefore, it is not very widely used. Instead
of using a single series of straight jets, we can have something like impinging jets and in
both these cases what is it we use for forming the jets? We use the pressure of the liquid
in forming the jets and breaking up the liquid into droplets and these are also known as
pressure induced atomization. What we use is impinging jets. Impinging jet could be two
jets impinging on each other. When we say two jets we call it as doublet, because you
have two of them; that means, I have one jet; I have another jet; I have a fan and this fan
breaks into droplets over here.
We call it as doublet;. could we do the function with a single jet? If we use a single jet
and make it impinge on a plate i.e., make it splash on this plate and what we get is
droplets here and this is known as a splash plate injector. Splash plate injector was used
originally by the Germans, but now they are used in some missiles; it is not very
efficient, because compared to doublet it does not have flexibility. I think the doublet is
much more important and doublets have been used in lox kerosene rockets. You
remember I showed you the F 1 engine which is we said is a very high thrust engine; it
uses doublet injectors. What is done is you choose the dimensions of the sharp edged
orifice in between a fraction of an mm to several mm depending on the size of the
engine, may be upto 4 mm 5 mm. You put a lot of these orifices together and may be
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impinge them on one on each other and you call them as a doublet injector. Well, we will
just see one more example.
(Refer Slide Time: 19:11)
This is how we mount the doublets. We put one orifice by the side of the other; create a
number of fans, make the fans interact with the other.
(Refer Slide Time: 19:34)
Well, this is about the doublet injector. If we vary the mixture ratio what would happen
to the fans? We have designed the fan based on the mass flow rate and velocity of the
incident jets. If we vary the mixture ratio, we would be varying the quantity of fuel with
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respect to the oxidizer and therefore we will have a different type of a fan coming.
Therefore, doublet is not good when the mixture ratio varies. For a fixed mixture ratio
and for a fixed flow doublet is good enough. But then, we must also remember doublet is
very efficient, because I can get whatever droplet size that we want. But at the injector
head, if atomization is taking place and combustion proceeds then the injector head of
the rocket tends to get heated. Let us say this is the combustion chamber; this is the
injector. We have fine droplets here; combustion takes place here; injector gets heated.
Therefore, the manifold must have a high velocity such that the injector remains cool.
(Refer Slide Time: 20:33)
Therefore, we observe that impinging jet injectors could be doublet. In the doublet we
could impinge fuel jet on fuel jet or we could also take an oxidizer jet and impinge it on
an oxidizer jet; these are known as a like doublets. When a fuel jet impinges on an
oxidizer jet, it is known as an unlike doublet. This is because we take unlike substances
fuel and oxidizer and impinge them and what happens in a unlike doublet? When we
have a jet of fuel with a jet of oxidizer and because of chemical reactions which can
occur we could have something like a “blowing apart” of the two jets and this is
problematic. The formation of the fan and fine droplets get drastically affected.
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(Refer Slide Time: 21:27)
Instead of using two jets, which do not function adequately under different mixture ratio
conditions, we can use three impinging jets. What I do is have a central jet may be of
oxidizer; I have fuel jets coming over here on either side of the oxidizer jet; and now I
have a fan in this particular direction along the central oxidizer jet. The use of three jets
impinging at a point is known as triplet element.
(Refer Slide Time: 21:42)
This is again very widely used in the industry, because we have a stable fan which is
formed. Let us take the configuration of a triplet.
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(Refer Slide Time: 21:58)
I have in the above figure, a fuel jet impinging on two oxidizer jets. We form a fan. I
look at it from the top, I see a spray fan. The impingement of three jets gives the triplet
impinging injector. The advantage of a triplet is when I change mixture ratio, I do not
really change the direction of the fan which as before is in the plane bisecting the outer
jets along the central jet. Therefore, the spray fan does not change its angle and that is the
advantage with triplets.
(Refer Slide Time: 22:26)
760
And I show an example of a triplet. This is where you have may be a fuel jet coming in
blue, the oxidizer jet coming in red; here you form a fan and it breaks into droplets in the
combustion chamber.
(Refer Slide Time: 22:41)
This is a particular engine which was used as an upper stage. This is the injector, when I
look at the injector here have a series of small holes three holes together which form the
triplet elements. We will disregard the other two injectors and address them later.
(Refer Slide Time: 23:06)
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This is about triplet injectors. We can also think instead of having triplet why not use
four of them i.e., four jets. It has also been tried; it has been used; it is known as
quadruplet. Its performance is not as good as the triplet, but it has been used and also I
could use five jets, in which case it is known as pentad. I have not seen injectors use
more than four jets - a quadruplet, but quadruplet as we say is not as efficient as the
triplet, because you do not really gain much. How does the quadruplet look like?
(Refer Slide Time: 23:43)
We show it in the above figure. I have something like a hole at the four corners of a
square on a flat head through which four jets issue. The four jets come and impinge on
each other and form a fan. In the case of five jets, we have one additional jet issuing at
the center of the square and we have these five jets impinging as shown in the figure on
the right. All five jets impinge to generate high pressure and form a fan. Actually, what I
should have shown are the two jets in another plane. We will have one central and then
these two on a different plane. Let us sketch it
762
(Refer Slide Time: 24:29)
I show on the left a quadruplet element. With a hole at the center we get the pentad
element. We have a hole here at the center from which one jet; surrounding it are four
holes from each of which issues a jet. All five impinge together to form a pentad
injection element.
(Refer Slide Time: 25:10)
Therefore the injectors what we have studied so far use the liquid pressure and we had
impinging jets and splash plate injectors. In the impinging jets we had the doublet, like
doublet; unlike doublet, triplet, quadruplet and pentad with 2 may be 3, 4, 5 jets
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impinging respectively. All these come in the category of pressure driven injectors, since
the jets were formed under pressure right from a shower head onwards. In the case of a
shower head we do not have impingement of the jets.
Now, the question that comes up is whether we really need liquid jets? In most cases we
have been forming fans or thin liquid sheets that break into droplets. We were forming
first a liquid jet; something like what I was trying to show is from the shower head; you
have something like a jet which issues out. Either this is impinged on a plate or you
impinge against each other and form a fan.
(Refer Slide Time: 26:16)
It is also possible for us to have another construction, wherein you allow the liquid to
come from the orifice, you put a shaped body here and when you put a shaped body, the
liquid jet gets diverted to form something like a water bell. The water bell is something
like a bulbous liquid sheet and this sheet breaks into droplets; that means, you convert
the liquid jet into a sheet. Rather than a jet, we transform it into a sheet and form
droplets.
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(Refer Slide Time: 26:42)
Maybe we have something like a shaped body at the exit of the orifice and form a liquid
sheet here. We have an annulus surrounding the central orifice from which we get a
cylindrical sheet which diverges out. These two sheets break up into droplets. We could
also allow the two sheets to impinge on each other and we get an impinging sheet
injector. This is shown on the right side. We have the coaxial liquid sheets which breaks
into droplets, and these are known as coaxial injectors because the two sheets are coaxial
around the central axis.
(Refer Slide Time: 27:12)
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We could therefore think of having injectors, wherein instead of jets we form sheets, I
have something like a coaxial configuration, wherein I have a conical sheet; maybe I
have another conical sheet which impinges on it and breaks into droplets, or I have the
central sheet like this inner cone; outer sheet and so on. These are known as coaxial
injectors, because these are coaxial with each other. Maybe we will take view of an
injector which uses the coaxial elements. A number of coaxial elements are arranged in
an injector head to give the required flow rates. We have central elements surrounded by
coaxial elements each issuing sheets and breaking them into droplets. In the picture of
injectors of the upper stage engine, coaxial elements injector was seen below the picture
of the impinging jet injector.
(Refer Slide Time: 27:55)
Let us take a look at the coaxial injectors. We have a sheet which comes out like this, a
diverging cone and this sheet breaks into droplets over here. Similarly, surrounding this I
could have another sheet which comes parallel to this and breaks into droplets. The
droplets evaporate, mix and burn. This injection scheme is what we call as coaxial. But,
if the pressure is not sufficient and we do not get a particular diverging geometry of the
sheet, and instead we get something like a bulb or a tulip shape sheet it does not give
good atomization. Therefore, in the case of coaxial sheet injectors, it is necessary that we
have something like a divergent sheet.
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(Refer Slide Time: 28:31)
Having said this we could, instead of having the liquid come in axially, rotate the liquid.
How do we rotate the liquid? We admit it tangentially. Therefore, when we admit it
tangentially the liquid rotates and when a liquid rotates because of the centrifugal force
in the frame of its own reference, the jet is thrown off.
(Refer Slide Time: 28:51)
And therefore, how do I rotate it? Either we admit the liquid tangentially into the orifice
as was shown earlier. Or else, we place something like a spiral or shaped vanes in the
path of the liquid flow. The spiral or swirler or vane induces rotational motion in the
767
liquid. We could also place a ribbon or something protruding spirally on the surface of
the orifice make the liquid rotate.
(Refer Slide Time: 29:08)
Here again I show shaped vanes being placed within the central outer casing which
causes the flow of liquid through it to have rotational motion. We place the casing shown
in the center over the vane shown on the left and achieve the rotational motion of the
liquid. In the picture shown on the right, tangential holes are made in the walls to provide
the rotation.
(Refer Slide Time: 29:19)
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And then, instead of getting something like a jet which atomizes we get a very diverging
jet due to the rotation of the liquid.
(Refer Slide Time: 29:25)
This finer atomization and the increases divergence of the jet when rotated is seen in the
above figure. When we rotate a liquid, we say that the liquid is swirled.
(Refer Slide Time: 29:35)
And therefore, such injectors that incorporate rotation of the liquid propellant are known
as coaxial swirl injectors as compared to a simple straight flow coaxial injector. All these
cofigurations are used in practice. People who come from the gas turbine discipline into
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rocket design prefer the use of swirl coaxial injectors. A coaxial injector also does the
job well.
(Refer Slide Time: 30:17)
I think this is all about injectors. But I have still not completed the portion, because I can
always think in terms of different ways of atomization and different designs. I now show
an injector that we experimented with in our lab here at IIT. You know what we do is
instead of admitting the liquid, we put some gas in the liquid and again make it
disintegrate into droplets and these are known as effervescent injectors, we are still
working towards its improvement.
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(Refer Slide Time: 30:43)
Therefore, we can keep on adding different types of injectors and all have a requirement
of generating fine droplets which can evaporate and the motion given to the droplets
brings about good mixing and burning. The injector, let us not forget, also meters the
required quantity of fuel and oxidizer into the chamber.
(Refer Slide Time: 30:59)
Now, we formal classify the different types of injectors. They could be pressure or
pressure atomizing injectors, these could be shower head; could be doublet. The doublet
could be like or unlike. We could have triplet, quadruplet, pentad; all these use liquid
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jets. We could use the same pressure to create a sheet. We could have coaxial injector,
wherein we have the liquid oxidizer and liquid sheets impinge on each other or they may
not impinge on each other; but just breaking into droplets, or we could also have swirl
which is used co-axially. These are all about injectors which we use in liquid rockets.
But, you would ask me, in cryogenic propellant rockets hydrogen is used. Hydrogen is a
liquid only at very low temperatures and by the time hydrogen comes into the
combustion chamber it could be a gas.
(Refer Slide Time: 32:13)
Also we talked of staged combustion cycle engine wherein products from the exhaust of
the turbine are admitted into the combustion chamber as a hot gas. Can I use the hot
turbine exhaust gas or the hydrogen gas to atomize the liquid oxidizer? In other words,
we have a liquid jet, issuing from an orifice; what we do is on the outside of this jet just
surrounding it just after the metal portion, by the hydrogen jet or the gas from the turbine
to co-flow at high velocities. This means we force the hot gas over the liquid surface at
high velocities. And when we force gas over liquid surface, it picks up the liquid because
of the shear and we get the droplets. This way of forming droplets is known as gas assist
atomization and such injectors are known as gas assist injectors. We use a gas for
atomization and hence gas assist. This gas assist could again be simple coaxial. Or, we
could rotate the liquid in which case I have swirl coaxial; we could rotate the gas and we
could have different configurations of gas assist injectors..
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(Refer Slide Time: 33:39)
Let us quickly go through a few examples of injectors. This picture shows the multiple
elements in a coaxial injector. You have liquid coming through these holes.
(Refer Slide Time: 33:47)
When we talk of gas assist, well I have gaseous hydrogen which is coming through the
outside shown in blue. I have the liquid oxygen which is coming through the central
orifice shown in red. We have liquid oxygen jet over which hydrogen flows at high
velocities causing it to atomize..
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(Refer Slide Time: 34:05)
Well, when we have liquid flowing and we have gas coming on the outside, we want the
gas to have high velocities compared to the liquid so that it can shear the liquid. We
allow the liquid to be within the gas flow orifices such that the gas velocity is felt by the
liquid. We say the gas must not relax such that we can use the high velocity of the gas.
The distance between the outlet of the liquid orifices and the gas orifice is what is known
an as a recess length. Recess essentially makes sure that the gas velocity is available for
atomization of the liquid in the gas assist injector.
(Refer Slide Time: 34:32)
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And therefore, we can talk in terms of shear coaxial, swirl coaxial; maybe we could have
a recessed configuration; we could give rotation to the gaseous hydrogen; we could give
rotation to the liquid oxygen. We could also make the make the passages of the orifices
conveying the liquid oxygen to flare or make a step in it to reduce the injection velocities
of the liquid. In this way the differential velocity between the gas and the liquid increases
and atomization improves. You can keep on innovating and a study of injector by itself
becomes a major portion of the liquid propellant rocket.
(Refer Slide Time: 34:57)
I show the flare and the step in the liquid orifice in this figure. In addition to providing
better atomization in view of the higher velocity differential, the reduction in velocity
provides better flame stabilization and prevents quenching of the flame. To sustain a
flame at high velocities is difficult.
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(Refer Slide Time: 35:18)
Therefore, injector design I would say is the heart of liquid propellant rockets, because
this is what produces the droplets, makes it evaporate, mix and burn. I think I will stop
with the injectors here; I will get back to injectors when we talk of combustion instability
and to a certain extent when we talk about efficiency of combustion.
(Refer Slide Time: 35:45)
Let us look at cooling of liquid propellant rockets which is again closely related to
injectors that we have studied.
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(Refer Slide Time: 35:52)
We have considered a combustion chamber in a cylindrical configuration. We had an
injector on top of the chamber over here. We considered different ways of injection. We
also found how to calculate the mixture ratio and the mass flow rates. Now hot gases are
produced in the chamber and it expands out in the nozzle. We told that the mixture ratios
are such that the temperatures in the combustion chamber are quite high of the order of
3000 to 3600 Kelvin and learnt how to calculate it. If this be so, the material used to
form the combustion chamber will burn out of within seconds, because the gases are
extremely hot.
Therefore, the question is how do we cool the combustion chamber? One of the ways is
may be if we can admit some fuel near the wall of the chamber; We admit a quantity of
the fuel near the wall, give it some velocity, and rotate it such that we make sure it sticks
to the wall.. We have lot of liquid in the wall region near the injector and the length of
the combustion chamber is small. Therefore, turbulence cannot be fully developed. This
liquid, in the wall region, is entrained by the gas and accelerates. I have a certain
thickness, the liquid thickness, and this liquid film keeps the combustion chamber cool
then it vaporizes. When it vaporizes is still much cooler than the combustion gases and
then the vapor can still further continue to cool the combustion chamber. The chamber
wall thus remains cool. In other words, I have a film of liquid, which is injected along
the walls of the combustion chamber and this method of cooling is what we call as film
cooling.
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(Refer Slide Time: 37:27)
Let us take an example. We have a liquid coolant being injected along the cylindrical
walls of the chamber; hot gases are evaporating it; vapor is formed and the vapor further
cools the combustion chamber and nozzle and this is known as film cooling.
Instead of film cooling, we could also make sure that in the combustion chamber we
could have more of the fuel vapor in the wall region. We make the mixture ratio to be a
little bit fuel rich. If it is fuel rich we can determine the temperature of the gases from my
diagram of temperature versus mixture ratio. I know at near stoichiometry I get
maximum flame temperature; actually slightly less than stoichiometric. If we can keep
the gases in the wall region very fuel rich, then we have low temperature over here we
can keep the chamber walls cool. That means, we form a barrier of fuel rich mixture over
the walls and this method of cooling is what we call as barrier cooling.
To a certain extent the film cooling incorporates barrier cooling, because what is it we
are doing? We have a barrier of the fuel vapor over here; maybe it also reacts to some
extent, but it is still much cooler than the core gases and therefore, we keep the wall cool.
778
(Refer Slide Time: 39:07)
How do you implement it in practice? Well, I show a figure here. I have something like
co-axial injectors which are participating in combustion. I give a series of holes over here
in which I admit the fuel; I direct it on the wall of the chamber and that is what is film
cooling. This can be easily predicted; what we do is we calculate the heat transfer
coefficients.
(Refer Slide Time: 39:26)
We can write the equation for the evolution of the thickness from the initial thickness of
the film. We know the heat transfer coefficient between the hot gases and the film. We
779
can calculate the rate at which it evaporates. I know the heat transfer coefficient between
the film and the wall. I do a composite calculation. I can calculate the thickness of the
film. I can do the same type of calculation for a gas film.
Film cooling is used in rockets, and why it is very effective? The combustion chamber
length is small, you do not have full turbulence being developed within the short length
and we can very effectively use it. The disadvantage is - you know immediately that we
loose some amount of fuel in the film cooling process.
(Refer Slide Time: 40:10)
Therefore, to some extent the specific impulse gets affected. Therefore, I have a
performance loss; that means, the specific impulse is little poorer, and second point is
when we have a nozzle and we have the coolant coming over here, the velocity in the
wall region at the exit may not be very large, and also these are may be fuel rich
hydrocarbon gases in it. Since the velocity is not large, it is not thrown off along the
plume; it comes back to the surfaces and when we use these film cooling for the space
craft, it sometime contaminates the glass surfaces of the sensors in the spacecraft. A
spacecraft consist of sensors; what are sensors? You have optical quality glass encasing
the sensor which measures radiation. The Earth sensor, as an example, monitors radiation
from the Earth and helps to point the spacecraft towards the Earth. It should not get
contaminated because of film cooling.
780
(Refer Slide Time: 41:07)
We will keep this in mind. Well, this all about film cooling; other cooling methods you
all are very well aware of. But, let us come back to this figure on loss of performance
due to film cooling after we finish discussions on efficiencies; but just to highlight when
we use something like 30 percent film of fuel for cooling, we get the efficiencies as a
function of mixture ratio as shown by the dark line. If we increase the cooling from 30
percent to 37 percent, the peak value of efficiency shifts to somewhat in the fuel rich
region, because we make the core less fuel rich. The core is still not oxidizer rich. If we
provide 50 percent of the fuel for film cooling the efficiency falls rapidly as mixture ratio
increases the core becomes oxidizer rich. This is an experiment that we conducted. We
will revisit this graph a little later.
781
(Refer Slide Time: 41:52)
The next type of cooling is what we call as regenerative cooling. I will not use the
blackboard I will show it with the slide. In regenerative cooling, like we have the
regenerative Rankine cycle which is used for the feed water heating, we admit the fuel at
the end of the nozzle. The liquid fuel gets heated while cooling the chamber. The heated
fuel being much warmer has more enthalpy as sensible heat and it is used re-generatively
in the chamber. Therefore, the next type of cooling which we can write is regenerative
cooling; it is a regenerative process.
(Refer Slide Time: 42:32)
782
Like feed water heating in boilers we use heating of propellants by regenerative cooling
before being admitted into the chamber. Invariably, regenerative cooling is always used
with film cooling. The reason is that in regenerative cooling we form channels for the
cooling or have tubes mounted over the chamber. If in some region the cooling is not
effective, that portion becomes red hot and burns off. However, if film cooling is also
provided, it can smear out the temperature increases in the local hot zones and protect the
chamber.
Therefore, this is regenerative cooling. There are problems related to very high
conductive materials and the thermal stresses. With repeated operation over a large
number of cycles, the stresses induced cause cracking. This is known as ratcheting and
we will not go into these details.
(Refer Slide Time: 43:17)
Let us get into the other forms of cooling. Just like the body sweats and keeps as our
temperature cool so also we can have the nozzle wherein we admit the liquid coolant in
this particular direction through small orifices in the high heat flux region. The liquid
coolant, which could be fuel, evaporates and keeps the temperature within limits. This
mode of cooling has been used extremely well in the case of the cryogenic rockets for
the injector cooling.
783
(Refer Slide Time: 43:47)
Let us again make a schematic of the injector. This is the injector block; I have the
combustion chamber here; I have let us say a series of coaxial injectors over here, and
now this surface runs hot; maybe I make this of a material which can ooze out may be
the liquid hydrogen over here, and when it oozes out over here, it is like our body
perspiring or transpiration taking place through these pores. The hydrogen absorbs the
heat and keeps the injector cool. And the particular configuration of the porous material
is known as a Rigimesh. It is a commercial name, but all what it consists of is a series of
meshes which allow the liquid to get sweated out through this part and as the liquid
evaporates it absorbs heat and cools the injector plate. This is what we call as
transpiration cooling or sweat cooling.
784
(Refer Slide Time: 44:43)
Well, we have a few more cooling strategies. We could use an ablative material for
cooling. You remember we talked of ablative materials when we discussed the nozzles.
What was it? We have something like resin, which is a some form of glue which is cured
as a solid. If we heat it, its temperature increase and it softens and evaporates at higher
range of temperatures. We increase the strength of it by introducing some fibers in it. We
therefore have something like a composite material. A composite material when exposed
to high temperature evaporates at the ablation temperature; at a particular temperature
where it forms vapors is known as ablation temperature. This gives rise to evaporative
cooling and is known as ablative cooling. The material is sacrificed but protects the
metal.
Instead of cooling we could allow the walls to run hot provided we have a strong
material of construction. In this case we allow radiation itself to dissipate the heat and we
can talk in terms of radiative cooling.
Or if we have want a rocket only for a ground application or test, we can make the rocket
so heavy so much of material such that we operate the rocket only during the transient
heating time of the chamber and nozzle. Let us say the we provide large thicknesses for
the chamber and nozzle walls of the rocket. Maybe I operate it for a second. In a second
the mass of the rocket or the thermal mass absorbs the heat and is known as a heat sink
rocket.
785
(Refer Slide Time: 46:53)
Let us quickly review some of the cooling strategies with a view to application. You
have an ablative material which evaporates when it when exposed to the hot gas flow and
maintains the integrity and the temperature of the liner.
(Refer Slide Time: 47:06)
And this is what we said when we studied about nozzles, we could have the ablative
materials which will evaporate. We use both carbon phenolic and silica phenolic
ablatives, with carbon as a fiber and silica is a fiber. Carbon is more prone to oxidation.
Therefore, silica is better for the cooling, because silica is an inert material.
786
(Refer Slide Time: 47:23)
In this figure we see a metal nozzle cooled by film cooling running red hot and it
radiates away the heat.
(Refer Slide Time: 47:31)
Therefore, to summarize we can use something like an insulation material like ablatives
for cooling. We use a hot structure that means we use high temperature materials like
carbon-carbon composites; may be high temperature resisting materials and radiatively
cool the hardware. We could use regenerative cooling along with film cooling or barrier
cooling and also transpiration cooling. These are the different types of cooling strategies.
787
(Refer Slide Time: 48:10)
Having talked of cooling and injection, what is the problem we have now created? We
said well cooling is important to protect the hardware of the rocket. Let us quickly go
through the cooling procedures again before we put them together. We said the cooling
could be film; could be barrier; could be regenerative; could be ablative; could be
radiation or could be heat sink. Well, heat sink is only for ground test for short duration
transient heating. Radiation cooling is also possible but we need high temperature
resistant materials. Ablative cooling may be difficult to use as the rocket becomes heavy.
Well, regenerative, barrier and film cooling have more positive features. Therefore, in
practice we make use of film cooling, barrier cooling, regenerative cooling and also
radiation cooling. Having said that now what is the problem of reduced performance that
we have got into?
788
(Refer Slide Time: 50:16)
We are injecting fuel into the chamber m°f; we are injecting oxidizer m°o into
the chamber. We determine the mixture ratio in the chamber R or rather MR is
equal to m°o/m°f. But, what is the problem we have now created? We now say at the
wall the mixture ratio is fuel rich if I have barrier cooling or film cooling.
Depending on the injector we distribute the fuel and the oxidizer non-uniformly
along the radius of the chamber; that means, if we take a section near the wall, we have
fuel rich zone. We may have a oxidizer rich zone in the core region. We have zones
which are different in the mixture ratios. The mixture ratio is distributed. We do not
have uniform mixture ratio distribution in the chamber.
In other words, if we were to now define a parameter like let us say a mixture
ratio distribution which we call as a DR, then this DR will represent the difference
between the local mixture ratio Ri at any point i, and the overall mixture ratio R0 and that
will affect the combustion efficiency, and this is what we will do in the next class. And
we will take a look what are the penalties we pay because of the mixture ratio
distribution. If we have a finite length of the chamber is it possible to get combustion
completed or will we have another efficiency influencing combustion?.
To sum up, in today’s class, we revisited injectors, we defined different types
of injectors, then we went into cooling strategies. And then we have now posed a
problem which we must solve in the next class namely, what is the impact of mixture
ratio distribution in a chamber and how to look at the effective specific impulse that we
get.
789
Rocket Propulsion
Prof. K. Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No. # 30
Efficiencies due to Mixture Ratio Distribution and Incomplete Vaporization
(Refer Slide Time: 00:16)
In the last class, you will recall I told you in a thrust chamber of a liquid propellant
rocket, the mixture ratio is not distributed uniformly. Why was this? We said maybe fuel
is used for cooling or you could have barrier cooling. We said in some injectors like
shower head, the droplets travel straight into the chamber from the injector; whereas, in
some cases the droplets are given an orientation. Therefore, it is really difficult to
determine the performance of a chamber since the mixture ratio is varying in it. In the
earlier classes, we found the value of C* as a function of mixture ratio and we wanted to
operate it in the fuel rich region. Let us say this is stoichiometric mixture ratio. We have
maximum value of C* in the fuel rich region. We would choose an injection mixture
ratio to be in the slightly fuel rich region.
That means, we inject some mass flow rate of fuel m°f in kilograms per second and some
mass flow rate of oxidizer m°o into the combustion chamber, and the mixture ratio that
790
we get at injection, let us call it as nominal or injection mixture ratio ‘o’. It is Ro =
m°o/m°f. We would like the C star to be a maximum. Therefore, maybe we adopt this as
the mixture ratio and we call this as mixture ratio corresponding to injection. But what is
it we found? We found that the mixture ratio at the walls would be different; mixture
ratio in the core would be different; and it would be different at the different places. Now
how do you compute the C*? We want to calculate the C* for this chamber and since C*
× the thrust coefficient of the nozzle = Isp, well the Isp and C* are going to be different
from the value corresponding to the mixture ratio value at injection.
(Refer Slide Time: 03:22)
Therefore, we must find out what is the effect of variation of mixture ratio in the
chamber? I think that is the one thing which we have to do without it we will not be
calculating the parameters like Isp and C* correctly. Therefore, first I would like to say is
that we need something like a distribution parameter distribution of mixture ratio in the
chamber. Some injectors generate such fine droplets and the fans are so well distributed,
whereas, in some cases it may not be that well distributed. When we have film cooling,
we have fuel rich zone near the wall and maybe the core might be different, therefore, we
need to have some method of characterizing the distribution of mixture ratio. How can
we do it?
Why not we compartmentalize the mixture ratios. Maybe we look at a zone here, maybe
I look at a zone surrounding this, maybe we look at a zone still surrounding the outer
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zone and so on and we find the mixture ratio at the different zones and then have
something like a distribution parameter.
Let me explain it through another figure. This is the chamber and we have a nozzle. We
are looking at the distribution of mixture ratio; maybe in this zone of the core we have a
particular mixture ratio; and let us say the surrounding core has another mixture ratio and
then it evolves further in the same way. The same evolution extends into the nozzle since
the length of the chamber is small. So on we have different regions of mixture ratios.
Maybe this is the third region of mixture ratios. We could have a central core at a given
mixture ratio; I have an annulus of a different mixture ratio; I have different annuli of
different mixture ratios. What would happen to C star with a distribution in Mixture ratio
like the above?
(Refer Slide Time: 04:34)
When we inject the fuel and oxidizer into the chamber, the droplets of fuel and oxidizer
are initially travelling in all direction therefore we have a very strong three dimensional
zone of varying parameters. And what do you mean by a three dimensional zone?
Droplets are travelling both radially and axially and also in other directions and
vaporization is taking place, and therefore, to quantify this zone is difficult. Therefore we
say immediately downstream of injection because we inject in different directions we
have let us a three dimensional zone and the process of evaporation and mixing would be
three dimensional. But we also know when we come to the throat of the nozzle, the flow
792
is sort of axial. If the flow is going to be axial at the throat somewhere upstream the flow
should have become axial, in other words if the flow is along the axis at the throat maybe
the flow should have come like this as streamlines.
In other words even though in the zone downstream of injection we have a highly
turbulent three dimensional zone, somewhere along we should have started getting flow
along the axis because at the throat it is anyway axial. Therefore we say initially we have
a three dimensional zone, but subsequently we can think in terms of tubes or stream
tubes which are progressing axially. Now I slightly extend this figure I say this is my
injector over here, I have initially a zone wherein the three dimensional flow takes place,
and thereafter since the flow at the throat is one dimensional I have something like tubes
or something like stream tubes in which combustion takes place.
Now a rocket chamber is not that long after all and we do not have sufficient length for
one stream tube to get effected by the neighboring stream tube, because it takes some
time for fully developed turbulent to manifest itself. Therefore, we can say in each of
these small stream tubes I can construct as many stream tubes as I like, in each of the
stream tubes I could have something like a laminar combustion taking place and the
stream tubes do not mix with each other. Why they do not mix? The chamber is short
and we do not have sufficient time for turbulence to develop and intermixing between
stream tubes to take place. Therefore we make these assumptions and these are quite
valid because of the short length of the combustion chamber.
793
(Refer Slide Time: 07:11)
And now we can think in terms of the total combustion in the chamber proceeding in a
series of stream tubes. We say this is the centerline of the chamber. We have initially one
central core and let us say it has mixture ratio R1, surrounding it we have another stream
tube this is now an annular stream tube with mixture ratio R2, surrounding it we have
another stream tube with mixture ratio R3 and so on. And when it comes to the nozzle the
stream tubes converge and diverge out; maybe you have a series of stream tubes which
burn, therefore, this is how it burns and gets exhausted through the nozzle. Now the core
having a mixture ratio R1, the annular stream tube surrounding it at a mixture ratio R2,
the next with a mixture ratio R3, or rather the ith stream tube that we consider has a
mixture ratio Ri which is the local mixture ratio in the ith stream tube.
Now, we want the net composite performance of a series of stream tubes. With reference
to the figure, the core has a certain mixture ratio R1, maybe the outer most near the wall
has a mixture ratio let us say R5. In between, we have the third stream tube with mixture
ratio R3. Therefore, now we want to put this in terms of a single parameter say a
distribution parameter. The mixture ratio at injection it is MR0, this is the value which we
are injecting which is the ratio of mass of oxidizer to mass of fuel injected.
794
(Refer Slide Time: 09:23)
Therefore now, we can say that the mean mixture ratio at injection is equal to MR0
whereas each of the stream tubes has a value Ri, with ‘i’ depending on the number of the
stream tube. The total propellant, which is injected is equal to mp° so many kilograms
per second is equal to m° of fuel + m° of oxidizer. Now the total quantity of propellant,
which is injected is shared between the different stream tubes. Let us say the stream tube
1 has a mass of propellant flow rate mp°1, the stream tube two has a propellant flow rate
mp°2, mp°3 is the propellant flow rate in the third stream tube and so on. We can divide
the flow rates in the individual stream tubes by the total propellant flow rate mp° and
call these values of the fraction; that means, x1 is mp°1/mp°, x2 is mass flow rate of
propellant mp°2/mp° and so on. We therefore have fractions x1, x2, x3 that is the fraction
of propellant flowing in stream tube 1, 2, 3 and so on.
795
(Refer Slide Time: 10:31)
What would the summation of x be? Summation of the mass fraction of mass flow rate
going from let us say stream tube 1 to the nth stream tube when we have ‘n’ streamtubes
is one. This is my first equation.
And my second equation concerns the mixture ratio and C*. We have mixture ratio over
here corresponding to a lower mixture ratio or corresponding to a different values of
mixture ratio in each stream tube. Let us say denote the value of C* at a mixture ratio Ri
as Ci*. It is not going to be the optimum value, but it is going to be a different value for
each of the stream tubes. The mass weighted C* for the chamber would therefore be and
the summation of the fractional mass flow rate in the individual stream tubes multiplied
by the relevant C* (C* = ΣxiCi*). The fractional mass xi in the ‘i’th stream tube into the
value of C* at this value of mixture ratio Ri for this stream tube which is Ci star, summed
over all i’s as ‘i’ goes from 1 to n gives the net mass weighted value of C*.
Now, this is going the value of C* instead to of the value of C*, corresponding to MR0
that is the mean mixture ratio at injection. Let us say the value at MR0 is C0*. Then the
value of efficiency what we get is equal to xi into Ci star as ‘i’ goes from 1 to n divided
by the value of C0*. This is due to the mal distribution or distribution of mixture ratio.
What is it we do? We calculate the performance of each stream tube as if it were an
individual rocket; we calculate the fraction of the propellant in the stream tube and the
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summation gives us the mass weighted C*. The mass weighted or averaged out value of
C* divided by the conditions at injection is what gives us the value of C* efficiency.
I think this concept is useful and we tried it around 10 to 12 years back and found it to be
useful. We can derive an idea on how to distribute mass of fuel and oxidizer such that we
can get the optimum value of C*. This is because we have to distribute the propellants in
the chamber and when you distribute it you get variable mixture ratio, and the
assumption is individual stream tubes do not mix with each other, and why they do not
mix? Because the length of the combustor is short, the turbulence still does not develop
and therefore, mixing between stream tubes is not important. And now how we distribute
the propellant mass and the mixture ratio? Can we characterize a distribution parameter?
(Refer Slide Time: 13:46)
What should the distribution parameter consist of? Let us erase this out and put the
distribution parameter.
797
(Refer Slide Time: 15:00)
In the stream tube i, the value of the mixture ratio is let us say Ri, the overall mixture
ratio is let us say MR0 and the change from the mean is equal to Ri minus MR0. This is
the change in mixture ratio in the ‘i’th tube from the mean value at injection. It could
either be positive or negative value depending on whether the local mixture ratio in the
‘i’th tube is less than or greater than MR0. The change from the mean is the modulus of
this value is the value. The fraction of the propellant in the stream tube is xi. We can
represent a distribution parameter in mixture ratio as DR as equal to the fraction of the
propellant in the particular stream tube xi into the modulus of the deviation in this stream
tube summed over all the stream tube going from 1 to n, and we divide it by the mean
value of the mixture ratio MR0. In other words, we get the distribution parameter DR as
equal to the dispersion multiplied by the fraction divided by the mean mixture ratio.
If we have a perfectly homogeneous mixture ratio distribution everywhere, we get the
same value, well my value comes out to be 0. Therefore, I have to make some changes in
the definition. That means, I am looking at a distribution of mixture ratios. Therefore, DR
should be equal to one minus this value. If everywhere we have the same mixture ratio
well my distribution parameter is unity. If we have at different places different mixture
ratios, the deviation from unity will tell me what is the distribution index, which could be
termed as mal distribution or distribution parameter.
798
(Refer Slide Time: 16:11)
The distribution parameter DR; if it is 0, it tells us that we have uniform mixture ratio,
whereas, the deviation is > 0, and it is always going to be less than 1, tells that it is not
uniformly distributed. The amount by which it is less than 1 tells us how much it is
deviating from a uniform distribution. And therefore, we say DR is a non uniformity
parameter for mixture ratios.
Now, we would like to again revisit the problem of C star efficiency, We found it to be
given by ηC* is equal to sum of the xi’s into Ci*’s divided by C0*. We expect it to be less
than one. There is no way it can be greater unless something wrongly specified and you
have a lower value for the mean. How does ηC* depend on DR? Well, let us plot it out.
799
(Refer Slide Time: 17:26)
We have the value of C star as a function of DR. When the distribution parameter is unity
we get the value corresponding to the maximum. As the distribution parameter decreases
the C* value falls down, but when we looked at this particular figure which I keep
erasing out each time; what was the shape like? It is like this. This is the nominal value
of let us say MR0 which I choose. I find that the slope of this in the fuel rich region is
somewhat steeper than in the oxidizer rich region. This suggest that as MR0 increases,
the change in C* from the maximum will be smaller and the line will be little less steep.
Therefore, if we increase the value of the nominal mixture ratio at injection MR0, we will
get curves do not change very much from the nominal value as the non uniformity of
distribution increases. The shift is a function of the mixture ratio.
All what I am saying is if we were to have the nominal here in the oxidizer rich region,
then the change in C* due to mal-distribution of mixture ratio is little bit less, whereas, if
we choose a value here in the fuel rich region for which the gradient in C* is higher, the
value of C* is more adversely affected by the mal-distribution of mixture ratios. The
mal-distribution causes a penalty in C*.
800
(Refer Slide Time: 18:51)
The mal distribution or distribution of mixture ratio in the combustion chamber or thrust
chamber is of cause of concern and leads to a penalty. How do you qualify this penalty?
In terms of efficiency which is the real value of C* due to distribution divided by the
value of C* at the nominal injection conditions. And how do you get this? Let us denote
the actual value as C*net or C* equivalent that you calculate from the C* corresponding
to the individual stream tubes and the fraction of mass of propellant flowing through the
stream tubes. I hope this is clear. You may not find this in text books, but it is something
which we must understand.
Can we restate your question; you want to know how to determine the distribution of
mixture ratio? Well you know the injector type that you use, you know how it is forming
droplets, you know the distribution of droplets at the head, you know what type of
droplets of oxidizer and fuel are coming out of the injector. But how do we estimate the
value of let us say xi for each of the stream tubes and the value of Ci*. You know how C
star varies with mixture ratio. The only thing you need is what is the value of xi? You
know you can think in terms of a simple experiment, let us think of it, and such
experiments are known as patternation; that means, you would like to study the pattern
from the injector.
What you do is you have an injector set up above a table. It could be any form of injector
with either coaxial elements, impinging jet or whatever be the configuration. We
801
simulate the oxidizer by a particular fluid, simulate the fuel by another fluid and now we
do a cold test by passing the simulated fluids through this injector may be at ambient
conditions or hold the chamber under pressurized condition. We put a series of test tubes
here under the injector at different axial locations. We put a series of test tubes all along
a little bit away in the three dimensional zone, and collect the mass of the fuel and mass
of the oxidizer at the different places. We can study the pattern formed by the injector,
and this is what is used to calculate the xi. Once you know xi at the different locations we
can find out the non uniformity parameter and the efficiency due to the non uniformity.
(Refer Slide Time: 22:00)
Are there any other questions?
The next problem that we consider is somewhat a little more challenging. What is it? We
have the thrust chamber and we inject let us say a mass of fuel at a particular flow rate
like kg per second, mass of oxidizer at a certain rate into it, and these are both liquid
propellants that we are injecting. But what really reacts is not liquid with liquid; we form
droplets the droplets have to evaporate, the vapor has to mix and burn. Therefore, what
will react is mass of the vapor formed from the liquid fuel let us say m°vf and m° vapor
of the oxidizer.
Let us make this even clearer. Instead of using this symbol we say out of liquid fuel
which is sent to the chamber, not everything evaporates, and the unevaporated part of the
liquid drains out. The entire liquid must evaporate if it is used for generating thrust. Of
802
m°f the liquid fuel which is injected, only a certain portion evaporates and this mixes
with the oxidizer and releases energy. Similarly, of the liquid oxidizer injected, only a
certain fraction evaporates, because of the limited length of the chamber. We cannot
keep on increasing the chamber to infinity to make sure everything evaporates and
everything burns.
Therefore, it is quite possible that not all the fuel which is injected vaporizes and burns
and not all the oxidizer which is injec