Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
AE1303
AERODYNAMICS – II
3 0 0 100
OBJECTIVE
To understand the behaviour of airflow both internal and external in compressible flow regime with
particular emphasis on supersonic flows
1.
ONE DIMENSIONAL COMPRESSIBLE FLOW
7
Energy, Momentum, continuity and state equations, velocity of sound, Adiabatic steady state flow
equations, Flow through converging, diverging passages, Performance under various back pressures
2.
NORMAL, OBLIQUE SHOCKS AND EXPANSION WAVES
15
Prandtl equation and Rankine – Hugonoit relation, Normal shock equations, Pitot static tube,
corrections for subsonic and supersonic flows, Oblique shocks and corresponding equations,
Hodograph and pressure turning angle, shock polars, flow past wedges and concave corners, strong,
weak and detached shocks, Rayleigh and Fanno Flow. Flow past convex corners, Expansion
hodograph, Reflection and interaction of shocks and expansion, waves, Families of shocks, Methods
of Characteristics, Two dimensional supersonic nozzle contours
3.
DIFFERENTIAL EQUATIONS OF MOTION FOR STEADY COMPRESSIBLE FLOWS
9
Small perturbation potential theory, solutions for supersonic flows, Mach waves and Mach angles,
Prandtl-Glauert affine transformation relations for subsonic flows, Linearised two dimensional
supersonic flow theory, Lift, drag pitching moment and center of pressure of supersonic profiles
4.
AIRFOIL IN HIGH SPEED FLOWS
6
Lower and upper critical Mach numbers, Lift and drag divergence, shock induced separation,
Characteristics of swept wings, Effects of thickness, camber and aspect ratio of wings, Transonic area
rule, Tip effects
5.
HIGH SPEED WIND TUNNELS
8
Blow down, in-draft and induction tunnel layouts and their design features, Transonic, supersonic and
hypersonic tunnels and their peculiarities, Helium and gun tunnels, Shock tubes, Optical methods of
flow visualization
TOTAL: 45
TEXT BOOK
1.
Rathakrishnan, E., “Gas Dynamics”, Prentice Hall of India, 2003.
REFERENCES
1.
Shapiro, A.H., “Dynamics and Thermodynamics of Compressible Fluid Flow”, Ronold Press, 1982.
2.
Zucrow, M.J. and Anderson, J.D., “Elements of gas dynamics”, McGraw-Hill Book Co., New York,
1989.
3.
Mc Cornick. W., “Aerodynamics, Aeronautics and Flight Mechanics”, John Wiley, New York, 1979.
4.
Anderson Jr., D., – “Modern compressible flows”, McGraw-Hill Book Co., New York 1999.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Course Materials
Basic ideas about the fluid flow and historical facts
Fluids moving with high speeds produce resistance to the objects which are not totally related
with the speed of the fluid. This phenomenon can only be explained by the understanding of
the compressibility effect fluids possess.
The theory of resistance was first proposed by Sir Isaac Newton in 1726. According to him,
aerodynamic forces depend upon the density and velocity of the fluid, and the shape and size
of the displacing object.
The difference between the effects a high speed flow and low speed flow do bring to the
objects can only be understood when we start taking fluid as varying density. If we assume
that the density is constant throughout the process, simply high speed can’t explain all the
peculiar behavior of fluid it has at high speed.
This effect of change in density of fluid when it starts encountering the objects while flowing
with high speed is called compressibility effect.
In the compressibility effect, density changes point to point depending upon the local
parameters, but all the changes in density can’t be associated with the compressibility.
Density can be changed by changing temperature of a system too by keeping pressure
constant. So by saying compressibility we mean the change in density due to the change in
pressure.
Fluids flowing with low speed too have got their compressibility effect valid but the effect is
so low that, it loses its significance. So for the purpose of engineering importance, any flow
below M < 0.3 is taken as incompressible. (This value is not a fixed value. It only gives the
rough idea). After this Mach no. compressibility effect becomes appreciable.
The branch of science which studies the fluid dynamics with compressibility effect is called,
Rarified Gas Dynamics or simply Gas Dynamics.
Compressibility is measured by the Bulk Modulus of Elasticity and is defined by
Normally liquids like gases have very high bulk modulus of elasticity, under normal
circumstances, they are taken as incompressible. Water starts showing compressibility only
after coming to a very high pressure (e.g. 1000 atmosphere).
Gases are very much compressible but their flow at low speed too is taken as incompressible
flow. Only high speed flows of gases bring about appreciable changes in density and are
taken as compressible flow.
In a nutshell we can say that compressibility is a phenomenon by the virtue of which the flow
changes its density with the change in speed (dynamic pressure) or simply the pressure.
The quantitative measure of the compressibility is given by the above equation of
compressibility.
For an ideal gas pressure, density and temperature relation is given as
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Or in more general way
Where R is specific while Ru is universal.
Ru = 8.314 KJ/mol-K and
R = 287.26 J/Kg-K for air
Isothermal Compressibility
In any isothermal process, pressure-volume relation is given by the Charles’s Law
Where p is pressure and V is specific volume and the subscripts 1 and 2 indicate the initial
and final conditions.
Now on combining the isothermal relation of pressure-volume change and the bulk modulus
of elasticity, we get that bulk modulus of elasticity of any system at a given pressure is the
pressure itself.
Derivation
Let us use the classical mechanics of considering a small element of medium.
dx
p, V,
Area
p+dp
V+dV
A
Now using pressure – volume change with constant temperature
Adiabatic Compressibility
In the case of adiabatic pressure-volume change, the bulk modulus of elasticity is the γ times
the existing pressure. i.e.
Derivation
This can be also easily derived using the same classical method. For the adiabatic process
pressure volume changes follow another equation.
.
(note: we use the same equation for adiabatic as well as isentropic process.)
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Basic Assumptions
There are some basic assumptions always made in the compressible flows which are follows
•
•
•
•
Conservation of mass
Conservation of energy
Conservation of momentum
Equation of State
(Continuity Equation)
(Energy Equaiton)
(Momentum Equation)
(Ideal Gas Equation)
Velocity of Sound
Sound is a vibration which travels through an elastic medium as wave. Speed of sound
describes how fast this wave travels in the medium. Speed of the sound depends upon the
properties of medium.
In dry air, at a temperature of 20° C (68° F) sound wave travels at a speed of 343 m/s (1,125
ft/s) which is equal to 1,236 KMPH (768 mph). Speed of sound depends upon the elastic
constant (compressibility) and the density of the medium. At the same time it depends upon
the physical state of the medium as well. Speed of sound varies from faster to slower in
Solids, Liquids and Gases.
Speed of sound decreases with density but the vital fact is sound travels faster in water
(almost 4.4 times) than in air. The reason behind this is the compressibility of the water just
makes up the loss in speed caused by its density.
Sound is a pressure pulse created in atmosphere so, sound too travels with the same speed.
Mathematically it is given as
Speed of sound can be found using classical method and calculation. Considering a small
element of the medium in which we want to find the speed of sound
Derivation:
In a constant area duct
Area = A
p, v,
p+ p
v+ v
+
Let us go through the classical mechanism to find the speed of sound in any medium. Let
there be a pressure pulse travelling at a speed of sound while altering all the flow properties.
The conditions across the pressure pulse can be taken as in the figure.
Continuity equation
Using this relation across the pressure pulse
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
………………….(1)
Momentum equation
2
2
2
2
Now on using the outcome of continuity equation
2
The above equation gives the speed at which pressure wave travels. It is also the speed of
sound under similar condition.
Using compressibility equation
OR
So
Now for the isotherm process
For the isentropic process
The pressure volume change while traveling of sound wave in the fluid is isentropic process
so
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Mach No.
It is ratio of velocity and speed of sound.
If has been widely accepted that compressibility can be ignored when the change in density is
less than 5%. i.e.
0.05
This condition applies till M ≤ 0.3. So at the normal temperature and pressure condition, flow
is treated as incompressible if V ≤ 100 m/s.
Subsonic
:
M<1
Sonic
:
M=1
Supersonic
:
M>1
Example
1. For an aircraft flying at 1,250 kmph, find the Mach no. at SL and at tropo-pause level
(11 km).
Temperature at the Sea Level is 15° C and at the Tropo-pause Level it is – 56.5° C.
2. For the same speed Mach no. keeps changing as the altitude changes, why?
3. An airplane flies at an M=0.9 at an altitude where temperature is -17.0° C. Calculate
the VTAS.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Some thermodynamic relations
Some relations:
Where Cp and CV are the specific heat coefficients at constant pressure and constant volume
, so
respectively. Also
(For Air cp = 1005, cv = 718 and R = 287 J/Kg-K)
For the isentropic process
So
Perfect Gas Equation of State
It is the equation which is followed by all the ideal gases. It is basically a relation between
pressure, volume (or specific density) and density. If any of these two is known the third
one can be found out, so ideal gas has only two degrees of freedom in its thermodynamic
behaviour.
Mathematically it is expressed as:
, are the static pressure, total volume of the gas under consideration,
Where, , ,
universal gas constant and absolute temperature and “n” is the number of moles available in
the available volume of the gas. At this point
has the value 8314 J/kmole-K.
Now since normally we deal with the air flow and its properties, we can modify the above
equation with specific value of gas constant also.
Here and both are specific with the unit m3/kg and kJ/kg-K respectively. Now this
specific gas constant for air is equal to 287.26 J/kg-K.
The same equation can be written as:
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Again:
Where is the concentration of the gas (no. of moles per unit volume).
Where, NA and NC are the Avogadro number and number of particles per unit volume and
is the Boltzmann constant.
6.023 x 10
1.38 x 10
/
Enthalpy and Internal Energy
When some system is supplied with heat energy, it either increases the internal energy of the
system (random motion of the molecules) or does work against the external pressure and
increases the volume of the system (work done) or both.
In most of the calculations Pressure and Volume exist together so we have another quantity
defined as Enthalpy.
This enthalpy is also called as thermodynamic potential. This term pV accounts the work
done in pressure volume change.
Also
and
This enthalpy includes all kind of energy and is sufficient to give the entire thermal condition
of any system.
The internal energy, strictly speaking, is the function of two thermodynamic properties, viz.
temperature and pressure. In reality, dependence of the internal energy of the gases upon
pressure is very weak and hence their internal energy is determined by the temperature alone.
Such gases are called Thermally Perfect Gases. So
and the gases with constant sp.
heats are called Calorically Perfect Gases (cp and cv are constants).
The internal energy is the energy contained in the molecules of a gas due to its random
motion. So total energy a gas molecule can contain is determined by the number of modes in
which energy can be stored by the molecules (or atoms) and the amount of the energy that
can be stored in each modes.
This number of modes in which energy can be stored is also known as degree of freedom.
Higher the degree of freedom, larger is the modes in which molecule can do the random
motion and store internal energy.
In normal cases, a gas can store energy in three modes of motion, translation, rotation and
vibration. Out of these three translation is most prominent one as translational mode gets
fully excited at as low as 3 K itself. The rotational mode gets fully excited beyond 600 K and
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
vibration mode gets fully excited beyond 2000 K or so. The bond, which connects two atoms,
is idealized as springs which stores energy by the means of vibration. These springs begin to
vibrate beyond 600 K. Normally vibration mode doesn’t contribute much in the internal
energy as at that high temperature, atoms begin to ionize. These effects do not represent
degrees of freedom so normally we ignore them.
The classical equipartition energy principle states that each degree of freedom, when “fully
excited”, contributes ½ RT to the internal energy per unit mass of the gas. The term “fully
exited” means that no more energy can be stored in these modes.
Now let us discuss about the various gases:
Monatomic gas:
It contains only one atom per molecule so it has only three degrees of freedom and that are all
in translation in all the three directions. So its internal energy will be
3
1
2
3
2
3
,
2
5
3
5
2
1.667
Diatomic gas:
it contains two atoms per molecule so it not only can translate, rather can rotate also about all
the three axes. By considering carefully we can imagine that these two atoms might be
arranged in the shape of a dumb-bell. The rotation of this dumb-bell about two axes will
contribute to the internal energy but rotation about its own axes will not have any effect as
moment of inertia will be very low in that mode. So every di-atomic gas has total five degrees
of freedom (three in translation mode and two in rotation mode). So the internal energy will
be
5
1
2
5
2
5
,
2
7
5
7
2
1.4
Tri­atomic gas:
It will have six degrees of freedom, three in translation mode and three in rotation mode. So
the internal energy will be
6
1
2
6
2
6
,
2
8
6
8
2
1.333
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Adiabatic and Isentropic Processes
An adiabatic process is defined by as an isolated system which doesn’t interact with the
surrounding regarding heat exchange. So there is no energy loss, only conversion of energy
from one form to another is possible. So total heat of the system remains conserved, i.e.
ΔQ=0.
If total heat remains constant then total enthalpy too will remain constant and then stagnation
temperature too will be a constant one.
So
This leads to
………
On the other hand, an isentropic process is defined by a process with constant entropy which
is one step ahead of being adiabatic. In one sentence an isentropic process can be defined as
“a reversible adiabatic process”. So Adiabatic Process is an essential condition for the
Isentropic Process which is obvious from the equation below.
0
0
0
On integrating
ln
ln
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
ln
constant
constat or
……………………
So isentropic process follows the equation (B) and equation (A) merely tells tha the process is
adiabatic, which may and not be isentropic.
Reference States
Normally in the case of high speed aerodynamics, we refer two states very often. 1. Sonic
state (critical state) and 2. Stagnation state
Sonic State
This is the state when local flow velocity becomes equal to speed of sound. This is also called
critical state because across this boundary, nature and behavior of the flow is reversed. A
diffuser for a subsonic flow becomes a nozzle for the supersonic flow and vice versa.
All the parameters related to this state are denoted with a subscript “*”. In the case of
chocked flow this condition exists at the smallest area location (throat).
At this condition, Mach number is always equal to 1 as local velocity is equal to the local
speed of sound, as per definition. The parameter M* is not the Mach number and is equal to
Mach number only at the sonic state.
Sonic state parameters
,
,
,
,
Where A* is a geometric parameter referred to the throat area where sonic state exists in the
chocked condition.
Stagnation State
This state refers to the zero velocity condition. In the pressure manifold where there is no
velocity and high pressure air or gas are at rest, is the only condition where stagnation
condition can be found. Once the flow starts, practically we don’t have any place where we
can have the stagnation condition though theoretically this condition is important to refer
various states.
At the stagnation state, we define that velocity becomes zero, so the whole energy is
converted into the thermal energy and kinetic energy is zero. To achieve this state 100%, we
must diffuse the flow through diffuser of infinite length which is not possible. If the flow is
stopped suddenly by blocking the flow to achieve stagnation, the whole flow stops and there
is no concept of stagnation state in that case.
All stagnation state parameters are denoted by the subscripts “0”.
,
,
,
,
The stagnation state parameters are related to sonic state parameters as below:
1
2
1.2
1
1.2
2
1
1
2
.
1.893
1.2
.
1.577
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Stagnation enthalpy is a free of the process to reach it. As it is directly associated with the
heat content. As long as there is no heat loss, stagnation temperature and hence stagnation
enthalpy remains constant. So
and
is always valid but
are valid only for the isentropic process.
blindly,
So to evaluate the stagnation temperature, we can use the relation
but to evaluate the stagnation pressure or density, we must know the nature of the process, as
is valid only for the isentropic process.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
One Dimensional Compressible Flow
Constant Area Duct
Continuity Equation
In a constant area duct in the absence of any source or sink, the net mass flow rate will
remain constant.
dx
Area
p
Mass flow rate
A
p+dp
= constant
So
is called continuity equation. This is valid only if there is no change
in area of the duct. Otherwise
Momentum Equation
Considering the above elemental fluid in motion under the pressure difference
Acceleration in x-direction is
In the similar fashion we can write the acceleration in y-direction as well as in z-direction.
In the case one dimensional flow, terms containing x and y will just vanish.
So now acceleration in the x-direction becomes
Where term containing time (t) is called local acceleration and term containing spacial
variable (x) is called convective acceleration.
In the case of steady flow, acceleration becomes
Also net force acting is
In the absence of friction and body forces, Newton’s Second Law can be written as
Force = mass * acceleration
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Where dp is
dx so
On integrating with respect to “x”
Above is called, Euler’s equation which is valid for compressible as well as incompressible
flow.
For incompressible flow
= const. So
This equation is called Bernoulli’s equation which is valid only for the incompressible
flows.
For the compressible flow (for pressure volume change to be isentropic), Bernoulli’s
equation takes the shape of
……
(ii) is just another form of the same Bernoulli’s Equation.
Discharge from a reservoir
Energy Equation
Energy equation simply follows the law of conservation of energy which says that total
energy remains conserved i.e. it can neither be created nor be destroyed and only
change of form is possible.
When we say energy, it includes all sort of energies like “kinetic energy, potential energy,
internal energy and thermal energy”.
For low speed subsonic flows, kinetic energy is too low to have any influence on the flow
property as heat content is so large in this case. In this case even if the whole kinetic energy is
converted into heat, it will hardly make any difference in the temperature of the flow.
On the other hand, for high speed flows, kinetic energy content can be so large compared to
its heat content that the variation in temperature due to the variation in speed can be
substantial.
Considering a closed system “heat added to the system minus work done by the system will
be equal to the increase in the internal energy”.
It is also called as “First Law of Thermodynamics”.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
1
2
Capital letters talk about absolute values and small letters denote the mass specific
parameters.
In the above equation, W is the total work done including shaft work. So
1
2
1
2
1
2
Now in the case of an isolated adiabatic system where there is not heat transfer or shaft work
available we have
1
2
Also neglecting the changes in potential energy
0
1
2
Taking specific values
0
1
2
0
Or
1
2
1
2
Or
(HW: This is the most prominent energy equation used in the high speed aerodynamics
problems which can also be derived from the Euler’s equation using isentropic relation of
pressure – volume changes. This is also called Bernoulli’s Equation for the compressible
flows. This relation is valid even for the non-isentropic relations. Though the relations
derived for pressure and density using this relation will be valid only for isentropic flows.)
Subscript “0” indicates the stagnation condition property.
Various forms of energy equation
1
2
and
1
2
1
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
1
2
1
1
1
2
1
and
1
1
2
1
1
1
1
2
1
1
2
Now as
Also as it is possible to convert the whole kinetic energy into thermal energy by stopping the
flow to zero velocity isentropicaly, it is also possible to convert the entire thermal energy into
kinetic energy. At this point we shall get the maximum velocity.
1
2
1
1
2
1
2
1
2
1
2
2
1
Also at the sonic condition, velocity of the flow becomes equal to velocity of sound
1
1
2
.
1
2
1
2
All star “*” subscripts represent the sonic conditions unless mentioned otherwise.
Free Stream Values
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Local Stream values
Local and Free Stream values relations
Critical Values
At the critical point M=1 and all the value relating to critical point are referred as Critical
Values (why?) and they are denoted by subscript (*).
1.2
1.2
.
1.893
1.2
.
1.577
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Examples
1. An A/C flies at 0.8 Mach no. at (1) ISA level, (2) 5000 m altitude. Calculate the
stagnation properties using (1) compressible and then (2) incompressible flow
equation and conclude.
2. The flow velocity at some point on the wing 240 KMPH and the temperature is 275
K. If at some other point, on the same wing, velocity is 400 KMPH, calculate the
temperature and the Mach no. at the second point.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Adiabatic Flow Process
Above equation is also called Adiabatic Flow Process Energy Equation.
In a general process
Also
In the case of adiabatic process δq=0
,
Note: here e, h and v are the specific quantity of internal energy, enthalpy and volume.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Flow through variable area
Area – Velocity relation
Mass rate of flow
0…………………
We know that
From Euler’s equation
On substituting this in (i) we get that
0
1
1
So area and velocity are related to each other through the free stream Mach no. Depending
upon the Mach no. a variable area duct acts as nozzle or diffuser for the flow.
Defining nozzle and diffuser as the duct through which velocity increases and decreases
respectively, we got some peculiar behavior here.
1
……..
1 ……..
For a subsonic flow a convergent section will act as nozzle and divergent section will act as
diffuser. On the other hand for a supersonic flow a convergent section will act diffuser and a
divergent section will act as nozzle.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Flow Direction
Flow Direction
Subsonic → Nozzle
Subsonic → Diffuser
Supersonic → Diffuser
Supersonic → Nozzle
For the supersonic flow diverging section acting as nozzle is a surprising fact, but it is true.
Mass flow rate still remains constant as the increase in velocity is compensated by the
decrease in density.
Pressure – Area relation
Mass conservation tells
0……..
0
From Euler’s equation
0
Also
On substituting these in the equation (A)
0
1
1
1
1
1
1
0
1
0
1
0
1
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
1
So with the increase in area pressure too behaves differently for supersonic and subsonic
flows.
……..
1
……..
1
Conclusion:
1. For subsonic flow with the increase in area pressure increases.
2. For the supersonic flows with the increase in area pressure decreases.
Example: If at a section in stream of flow, the area of the flow increasing by 1.5% when M =
1.6, find the corresponding change in pressure and density.
Pressure – Velocity relation
If there is no any shock then total pressure of the flow remains unchanged, provide there is no
any loss accounted.
From the energy conservation equation in the form of pressure
1
1
2
1
1
1
1
1
2
1
2
2
1
2
1
2
So the above relation shows the variation of pressure with respect to Mach no.
Example: At a point in a flow where M=2, M changes by 20%. Calculate the % change in
pressure at that point.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Area Ratio
Mass conservation tells
……..
(B) is also called Area – Mach no. relation.
This area ratio
can be at the least equal to 1. A* represents the throat area, so for the simple
reason there can’t be any area smaller than this throat. For this ratio equal to unity, we get
only one solution for Mach no. i.e. M=1. For any other value (
1) there well be two
solution for the Mach no. out of which one lies in the subsonic region and other in the
supersonic region.
6,
Example: An air nozzle of the shape shown I the picture has
0.97,
300 ,
corresponding velocities , ,
2 atm. Calculate the Mach nos.
,
7.5 and
,
and the
.
Section 3 is at the free stream condition.
.1
For air with γ
.2
.3
1.4 our area ratio becomes:
So it is clear that for a single exit area to throat area ratio, there can be only one exit Mach
number. So if want to have so many supersonic Mach number in our test – section of wind
tunnel, it is necessary to have separate nozzles for every Mach number of our interest.
However we can have variable area nozzle also but that is highly expensive item. Higher the
exit area, higher the exit Mach number will be.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Sometimes when flow is expanded to get the high speed in the test section, temperature and
density comes down markedly. So it is required to heat up sometimes to the flow at the
stagnation section, so that there will be reasonable temperature at the test – section. The
requirement of maintaining the moderate temperature become important if the temperature in
the test – section goes below -270º C as at such a low temperature metal become brittle and
can’t be used to such a low temperature. Therefore for high Mach number requirement in the
test – section air in the stagnation stage is heated to thousands of degree Celsius before it is
expanded in the wind tunnel.
A sample data for an isentropic flow for the exit Mach number 2 and 4 are given below
2
0.1278
0.5556
4
0.0066
0.2381
If we expand the air taking from stagnation condition at normal sea level condition (say 20º
C) then in the test – section of the tunnel temperature will be -110.2º C and -203.2º C for
Mach number 2 and 4 in the test – section respectively.
In the case of M=4 we must heat the air before expanding as nitrogen in air liquefies at -180º
C and air properties will change in that case.
Chocking of the Nozzle
If we apply a favourable pressure difference across a variable area duct, the flow starts
accelerating. If we keep increasing the pressure difference across the section, flow accelerates
further and mass flow rate increases, till we reach sonic flow at the location of smallest area
in the duct. This minimum area can be anywhere in the duct but not at the entry. This
minimum area is called throat area and the location is called throat.
Once the sonic flow is achieved at the throat, a further increase in the pressure difference
accelerates the flow further but the mass flow rate remains constant. Such a condition of
nozzle is called chocking and it is said that nozzle has got chocked. Further increase in the
mass flow rate is not possible as long as nozzle keeps working at the same stagnation
condition.
Mass flow rate
2
On putting the expression for v and
2
We get that
2
2
2
1
2
1
1
1
1
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
2
………………..….
1
,
Now for
0
Let
then
2
1
0
1
2
0
2
2
1
2
0.528
1
So we get only one non-trivial value for the pressure ratio. At this pressure ratio, the Q and
hence mass flow rate must be maximum (why?).
Also the mass flow rate is maximum where area is low (throat) and at this moment we can
get the maximum Mach no. equal to unity. So the moment flow achieves M=1, flow gets
chocked and further increase in the mass flow rate is not possible.
Mach no. more than unity is not possible to have unless we have a pressure ratio
0.528.
At this pressure ratio flow becomes sonic. It is obvious from the above discussion also that
this pressure p must occur at the least area location (throat). This throat may not be
necessarily be the exit point. If there is further extension of the duct, then it must happen with
the increase in the area (why?). In that extended area, flow can accelerate or decelerate or
both (accelerate and decelerate part wise).
What happens if we get any other area which is lower than throat area in the extended
section?
It is also obvious that this chocked value of the mass flow rate still depends upon the
stagnaton conditions. If the stagnation properties are changed mass flow rate changes.
Now let us find the numerical value of this chocked mass flow rate.
2
1
On putting the value of the pressure ratio
2
2
1
2
1
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
2
2
1
1
1
2
2
2
1
1
2
1
2
1
1
2
2
1
2
1
2
1
2
1
1
0.0404
So when the flow is chocked the above relation is always fulfilled.
.
and mass rate of flow is constant for
It is clear from the above equation that for a given
a given section. If the stagnation conditions change, it changes the mass rate of flow. So
doubling the stagnation pressure doubles the mass rate of flow and doubling the stagnation
temperature would reduce the mass rate of flow by about 29%.
Mass Flow Relation in Terms of Mach number
/
1
2
/
1
1
2
1
/
/
1
2
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
/
Convergent Section
Let there be convergent section attached to a tank at the atmospheric pressure.
Entry
Exit
O
A
B
C
Any nozzle is designated by its exit pressure ratio, throat area, total pressure, critical pressure
and the chocked mass flow rate (
).
, , , ,
Out of these, max. mass flow rate and total pressure are interrelated. Critical pressure and
total pressure too are interrelated.
1. When there is no pressure difference and back pressure is equal to total pressure then
there is not flow and pressure is p0 throughout the duct. This condition is depicted by
“O”.
2. The moment back pressure is reduced a little bit, a finite pressure difference exists
there and flow starts taking place. This condition is depicted by “A”, pe =pb. flow is
not chocked yet and further acceleration and increment in mass flow rate is possible.
0.528.
3. Back pressure is further decreased and flow accelerates further, mass flow rates
increases but exit Mach no. is still not unity. This condition is depicted by “B”, pe =pb.
Pressure ratio
1. Flow is not chocked yet.
0.528,
4. Now back pressure is further decreased, flow accelerates and
1
0.528,
0.0404 condition is attained and flow gets chocked now. pe =pb.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
5. If the back pressure is further decreased, it doesn’t affect the flow through nozzle, and
exit pressure still remains equal to chocked value of , and flow reaches the back
pressure through expansion fans available at the exit.
Convergent – Divergent section
It is obvious that when the flow is accelerated through this convergent nozzle, the maximum
Mach no. which can be attained is unity at the exit. Further acceleration is not possible as the
flow has got chocked now. Then now to get the supersonic flow?
For long people tried to get the supersonic flow in the laboratory by accelerating a subsonic
flow through a subsonic nozzle. In all their tries they were able to get maximum speed which
is sonic (M=1) at the end. So they thought that if they increase the section length a little bit
further then there will be supersonic flow at the end, but in the extended condition, nozzle
attains sonic Mach no. at the exit and inside it, flow is subsonic throughout.
Later on it was discovered that a flow can be accelerated to supersonic region only if we have
a convergent-divergent nozzle.
Figure above shows a typical C-D Nozzle. Entry is exposed the total temperature and
pressure where speed is zero. At the exit there exists a back pressure, which may be ambient
pressure as well. The difference in these two pressures creates flow through the nozzle. This
difference can be created by either by decreasing the back pressure from total pressure to
downward or by simply opening the stop cock and allowing the entry point to take the access
of stagnation pressure from stagnation chamber.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
2
2
The figure above shows how the flow behaves as the back pressure changes, relative to the
chamber pressure.
1. When there is no pressure difference, and ambient pressure (back pressure) is equal to
total pressure, there is not flow and pressure throughout the duct remains unchanged.
2. When there is little pressure drop in the back pressure, a pressure gradient exists and
flow starts taking place with subsonic flow throughout the nozzle. Up to the throat
area, flow accelerates and velocity becomes maximum at the throat area. After this,
flow starts decelerating and pressure starts rising till the exit end. At the exit end
pressure becomes equal to back pressure i.e.
1,
0.528. The exhaust jet is subsonic. This condition is depicted by the curve “a”.
3. Now as the back pressure is further lowered, pressure difference keeps increasing and
flow keeps accelerating in the convergent section and decelerating in the divergent
section. This trend continues till at the throat we get a sonic flow first time and then,
flow still decelerates in the divergent section and pressure at the exit becomes equal to
back pressure i.e.
,
.
. This is called
chocking condition, and we have maximum mass flow rate at this moment, any
higher pressure difference will not change the flow in the convergent section, but
divergent section is still not running at its design conditions. So it is possible to
accelerate the flow in the divergent section, but mass flow rate remains same even in
this section. Flow throughout the nozzle is still subsonic except at the throat where it
has M=1. The exhaust jet is still subsonic. This condition is depicted by the curve “b”.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
4. If back pressure is further lowered, pressure difference across the nozzle increases.
Flow in the convergent section remains unchanged, in the divergent section for some
distance; flow becomes supersonic, and accelerates for some distance. Back pressure
ratio is still above the design exit pressure ratio, so at some point in the divergent
section, pressure will rise suddenly and it is possible only through a shock. So a shock
appears there in the divergent section where flow is supersonic and across which
pressure increases and flow becomes subsonic. After the shock, flow again
decelerates following normal isentropic subsonic flow theory along a divergent
section. Exit pressure is equal to back pressure still and mass flow rate is still the
same as
at chocked condition
,
.
. It is
interesting to see that after chocking, any increment in the pressure difference across
the C-D nozzle doesn’t have any effect on the flow in the divergent section. This still
gives a subsonic exhaust jet. This condition is depicted by the curve “c”.
If the back pressure is further lowered, the shock moves towards the exit, back
pressure starts approaching the design exit pressure, flow in the diverging section
becomes supersonic for larger distance, and this condition persists till shock reaches
the exit point.
5. Once the shock reaches exit end, entire flow throughout the nozzle becomes
isentropic and smooth. Now exit pressure is not equal to back pressure though nozzle
is running at the design condition of pressure difference
. Pressure just after
the exit, rises through shock and velocity comes down so exhaust jet is still subsonic.
,
.
. This condition is
In this condition
depicted by the curve “d”.
Any further decrement in the back pressure will start bending the shock along the jet
and a very complex set of shock and its reflections will exist there. Exhaust jet will
now involve a mixture of subsonic and supersonic flow as shock strength comes
down. Also shock is no longer perpendicular to the flow; it will bend the jet flow
inward. This condition is referred as over-expanded condition of the nozzle. This
condition is depicted by the curve “e”.
6. Further decrement in the back pressure, will further weaken the shock at the exit and
will try to bend it more and more and at one point when back pressure is equal to exit
pressure, shock vanishes, flow is smooth throughout the nozzle and exit jet, and back
pressure is again equal to exit pressure i.e.
,
.
. This condition is depicted by the curve “f”.
This condition is referred to as the design condition of the nozzle.
7. If the back pressure is lowered further, it doesn’t affect the flow throughout the nozzle
but affects the jet of the exit. As the flow must attain the back pressure after exit, there
exists an expansion fan. While passing through this expansion fan, flow gets
expanded and it lowers the pressure and flow reaches the back pressure which lower
than exit pressure now. This condition is called under-expanded condition of the
nozzle. Expansion fans turn the flow outward in a plume and sets up a different type
of complex wave pattern. This condition is depicted by the curve “g”.
Nozzle and its exit conditions regarding flow pattern have been depicted below
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Supersonic flow generation
Example
01. A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75.
The nozzle is supplied from an air reservoir at 68x105 N/m2 (abs). Assuming onedimentional flow, calculate the following:
a) Maximum back pressure to choke the nozzle
b) Range of back pressure over which a normal shock will appear in the nozzle
c) Back pressure for the nozzle to be perfectly expanded to the design Mach number M.
d) Range of back pressure for supersonic flow at the nozzle exit plane.
02. A De Laval nozzle has to be designed for an exit Mach number of 1.5 with an exit
diameter of 200 mm. Find the required ratio of throat area to exit area. The reservoir
conditions are given as p0=2 atm; T0=20°C. Find also the maximum mass flow rate
through the nozzle. What will be the exit pressure and temperature?
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Problems
1. Air is compressed isentropically in a centrifugal compressor from a pressure of 1.0 x105
N/m2 to a pressure of 6.0 x 105 N/m2. The initial temperature is 290 K. Calculate (a) the
change in temperature, (b) the change in internal energy, and (c) the work imparted to the
air, neglecting the velocity change.
[Ans. (a) 193.868 K; (b) 1.39 x 105 Nm/kg; (c) 1.39 x 105 Nm/kg]
2. Air undergoes a change of state isentropically. The initial pressure and temperature are
101 kPa and 298 K, respectively. The final pressure is seven times the initial pressure.
Determine the final temperature. Assume air to be an ideal gas with γ=1.4.
[Ans.519.9 K]
3. If the velocity of sound in an ideal gas with molecular weight of 29 is measured to be 400
m/s at 100° C, determine the cp and cv of the gas at 100° C.
[Ans. cp = 860.1 J/kg-K, cv =573.4 J/kg-K]
4. Air flows isentropically through a nozzle. If the velocity and the temperature at the exit
are 390 m/s and 28°C, respectively, determine the Mach number and stagnation
temperature at the exit. What will be the Mach number just upstream of a station where
the temperature is 92.5°C?
[Ans. 1.12, 103.29°C, 0.387]
5. For the operation of a supersonic test-section, an air flow at the conditions p1, T1 and M1
is led through an inlet section of area A1 to a Laval nozzle which expands the flow to a
pressure of p2 at the test-section. Given p1 = 6.5 atma, T1 = 440 K, M1 = 0.5, A1 = 160
cm2, p2 = 1.0 atma. What will be M2, A*/A2 and ?
[Ans. M2 = 2.0, A*/A2 = 0.6, m
17.54 kg/s]
6. Air at a stagnation temperature and pressure of 200.9 K and 6.895 x 105 N/m2
respectively, flows from a reservoir through a converging nozzle that exhausts into an
atmosphere where the pressure is 1.014 x 105 N/m2. Calculate (a) the pressure in the
nozzle exit plane, (b) the minimum stagnation pressure for which the flow is chocked, and
(c) the exit plane pressure if the stagnation pressure is reduced to 1.724 x 105 N/m2.
[Ans.(a) 3.64 x 105 N/m2, (b) 1.92 x 105 N/m2, (c) 1.01492 x 105 N/m2]
7. An airplane flies at an altitude of 15,000 m with a velocity of 800 km/h. Calculate (a) the
maximum possible temperature at the airplane skin, (b) the maximum possible pressure
intensity on the airplane body, (c) the critical velocity of the air relative to the airplane,
and (d) the maximum possible velocity of the air relative to the airplane.
[Ans. (a) 241.05 K, (b) 1.756 x 105 N/m2, (c) 284.1 m/s, (d) 695.9 m/s]
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Normal, Oblique Shocks and Expansion Waves
Shock Waves and Basic Ideas
Shocks are described as discontinuity in the flow, across which there is a drastic change in
the flow properties. Normally it is associated with the high rise in pressure across it, but it
brings about the changes in almost all the properties of isentropic flow except total
temperature. Thickness of the shock is comparable to the mean free path of the air molecule
and that’s why it is taken as drastic change because in a very small thickness properties
change to a large extent.
Appearance of shock takes place only when there is a supersonic fluid flow which can be
understood with general consideration also.
When flow takes place in any flow field, then it gets information about any
obstacle ahead through the pressure pulses originating from the obstacle.
Pressure pulses tell about the obstacle and flow adjusts itself accordingly.
These pressure pulses are actually pressure wave fronts which travel with the
local speed of sound in the medium. As long as flow is subsonic, they do get
information well in advance to get aware of the obstacle available there in the
path, but the moment flow becomes supersonic, it reaches the obstacle with a
shock and it is taken by the obstacle by surprise. Naturally, flow will never be
able to know about the obstacle and it will have to reduce its speed before it
treats the object lying ahead and this is done when it passes through a shock
wave. This shock wave is essentially a pressure wave which brings the speed
of flow down while passing through it and thus increases the pressure.
The normal shock wave is called so for it stands normal to the flow the remains normal to the
after stream flow as well.
Equations of Motion for Normal Shock Wave
Across shock wave properties of flow change as below:
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
It is clear from the relations that it is a non-isentropic phenomenon across which there is a
loss of momentum but the total thermal energy remains same. Now as it is a non-isentropic
phenomenon, the isentropic flow relations can’t be applied directly to the flow across the
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
shock. Yet the flow before and after the shock can very much be taken as isentropic and they
follow the law of isentropic flows.
Still there are some certain relations which are still valid across the shock which become our
useful tools to solve the shock related problems.
01.
The continuity equation
……..
02.
The momentum equation
……..
Above is also called as Impulse Function. This impulse function “I” remains
constant across the shock. Impulse function is actually given by
and
if it is a constant area flow, we can neglect it.
(Note: during an isentropic flow in a nozzle, the force acting on the nozzle is given by the
difference in the impulse function and the power required is given by the difference in the
total thermal energy “enthalpy”).
03.
The energy equation
……..
04.
Equation of state
……..
05.
Thermal equation
……..
06.
Entropy equation
……..
Apart from these relations, there is one more very important relation relating Mach nos.
across the shock wave. Before we derive that equation, let us know more about the reference
velocities.
Reference Velocities
Absolute velocities
a) Local velocity: it is the local velocity “v” of any flow, which can be used for the
reference.
b) Speed of sound: it is the local velocity of sound “a” at the local condition of
temperature and pressure.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
c) Stagnation speed of sound: it is the speed of sound at stagnation condition.
d) Maximum speed of flow: it is the maximum speed flow can attain. This is attained
when the whole thermal energy is converted into kinetic energy.
2
1
e) Critical speed of sound: this is the velocity of sound under critical condition i.e.
M=1.
2
1
Ratios
a) Mach No.: it is the ratio of local velocity of flow to the speed of sound under local
condition.
b) Mach star (M*): it the ratio of local velocity of flow and speed of sound under
critical condition i.e. at M=1.
(Note: this M* must not be confused with critical Mach No. which is equal to unity.)
The Normal Shock Relations for a Perfect Gas
Prandtl Relation
Considering the above equations which are valid for the shock phenomena
Dividing (B) by (A) we get
2
1
2
1
1
2
1
1
1
2
1
1
1
2
2
1
1
2
2
1
1
2
2
1
2
1
1
1
2
1
2
…………….
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
…………….
OR
…………….
Equation (1) is called the Prandtl Relation which is a very vital relation and is very handy
while solving Normal Shock related problems.
It also infers that the flow through shock must be from supersonic to subsonic or vice versa
but we know that in subsonic flow shocks don’t exist at all. So across a shock velocity of
fluid must be from supersonic to subsonic.
Importance of M Star
This M star and Mach no. are related by the relation:
(HW: Derive the above relation)
When a flow takes place through a nozzle, then theoretically speaking, Mach
no. “M” can range from 0 to ∞. So to help, we do some of the calculation in
the terms of M star rather than in Mach nos. This M star has a finite range.
and
∞
.
Now replacing M star (M*) in the Prandtl Relation using M star (M*) Vs Mach no. relation
(HW: Derive the above relation)
This clearly shows that after stream Mach no. is a solo function of free stream Mach no.
and if M1=1 then M2=1. This is the case of infinitely weak normal shock, which is actually a
Mach wave. As M1 increases, normal shock becomes progressively stronger and M2 becomes
progressively less than 1. In the limiting case when M1 approaches infinity, M2 approaches a
finite minimum value.
1
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
∞
1 /2
0.378
Also
Ratio of the velocities can be written in the form of M* also.
Strength of Shock
Continuity equation
1
1
2
So the density ratio is
From the impulse function
1
1
Using
1
Substituting for v2/v1 we get that
………………….
The above equation gives the relative strength of the shock
1
1
2
.
1
1
1
2
1
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Above relation gives the pressure ratio across a normal shock.
Above equation gives the relation of temperature ratio across the normal shock wave.
(HW: derive the above relation)
Entropy across the shock
We know that entropy change is given by
Also
The above entropy relation can be written as:
ln
ln
ln
ln
1
2
1
ln
1
1
2
2
2
1
1
1
1
ln
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
ln
ln
But since total temperature across a shock remains constant, the change in entropy across a
shock will be given by:
ln
So finally
ln
ln
With slight rearrangement it can be rewritten as:
Above equation gives the entropy change across a normal shock.
Prandtl equation tells that the nature of flow changes across the normal shock and a subsonic
flow becomes supersonic, or a supersonic flow becomes subsonic, but by looking into the
entropy relation, it is obvious that in the appearance of shock, only second condition
(supersonic flow changes to subsonic flow) is possible as other condition will decrease the
entropy.
All the results derived above hold good till M1=5 for air at standard conditions. Beyond Mach
5, the temperature behind the normal shock becomes high enough such that γ no longer
remains a constant.
Limiting conditions
Limiting conditions of parameters as free stream Mach number approaches it boundaries:
Mach number can become zero or infinity. Mach number equal to zero indicates either
velocity is zero (stagnation) or speed of sound is infinity (very high temperature). Practically
it is achieved when stagnation is achieved, because achieving infinite temperature is not
possible. Mach number equal to infinity indicates, either velocity is infinity or speed of sound
is zero (very low temperature).
M1 equal to 1 is a unique case for Mach number. At this condition, process becomes
isentropic and there is no change in the after stream Mach number. This is the case of
weakest shock wave called Mach Wave.
1
lim
2
1
1
lim
lim
0.378
6.0
∞
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
lim
∞
Total Pressure
Total pressure change is given as follows:
/
/
Impulse Function
Impulse function is defined by
.
From the momentum equation (Euler’s equation):
0
0
So impulse function across the shock:
1
1
1
1
1
1
1
1
1
1
1
1
Impulse Function Analysis:
We have three equations with same origin:
1
1.
2
1
2.
1
2
3.
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
First equation is the Bernoulli’s equation for incompressible flow. This is derived using
assumption
, so it is valid only for incompressible flows. Also we can write it
in any form like:
constant as
constant.
Second equation has been derived using isentropic relation of pressure – volume
constant. So it is valid only for the isentropic flows. In this relation we can’t
change
write density with velocity, as it will no longer remain a constant then. i.e.
constant. So, as density is not
constant, but we can for sure write
constant, it is not interchangeable.
Third relation has been derived using constant mass flow rate (i.e.
constant). So we
can’t write
or
. So this is not valid at stagnation condition.
The reason behind this is, it will violate one of our assumption that
constant. If
velocity becomes zero, density must become infinity, which is impractical. So it is expected
to get absurd results if we apply this relation at the stagnation point. Let us check:
1
1
1
Which, is absurd as we know that
1
2
Examples
1. The free stream Mach number, pressure and temperature ahead of a normal shock are
given as 2.0, 0.5 atm, and 300 K, respectively. Determine M2, p2, T2, and v2 behind the
wave.
2. A re-entry vehicle (RV) is at an altitude of 15,000 m and has a velocity of 1,850 m/s.
Bow shock wave envelopes the RV. Neglecting dissociation, determine the static and
stagnation pressure and temperature just behind the shock wave, at the RV centre line
where the shock wave may be treated as normal shock. Assume that the air behaves as
perfect gas, with γ=1.4 and R=287 J/kg-K.
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Pitot Tube
In the gas dynamics, we normally encounter three types of pressures.
Total pressure: It is the pressure a fluid would exert, if it is brought down to rest
isentropically.
Static Pressure: It is the pressure fluid exerts due to its static position. There is not velocity
involved in this. This pressure acts equally in all directions in space.
Dynamic Pressure: It is the pressure a fluid would exert by the virtue of its own dynamic
motion. Speed and mass (density) of the fluid play major role in this type of pressure.
Pitot tube is the tube which brings down the flow to zero isentropically (ideally speaking)
and measures the total pressure.
A pitot-static tube is the device which measures total as well as static pressure and gives the
difference which is equal to dynamic pressure
. This way, it gives the idea about the
velocity of the fluid entering into the tube.
Effect of High speed on dynamic pressure
Low subsonic Flow
For the low subsonic flow, when the flow is slow enough to consider the flow to
incompressible, we use the Bernoulli’s equation to get the velocity of flow.
1
2
And then velocity
Note: the pitot static tube which measures the air speed is calibrated based on the
assumption of incompressible theory. So it is the dynamic pressure which is erred at high
speed flows.
High subsonic Flow
In the case of high subsonic flow, we simply can’t use the Bernoulli’s equation for
incompressible flow. As the flow starts showing appreciable compressibility just at and after
M=0.3 even if the flow is subsonic, we must account for the compressibility while working
out with the high subsonic flows.
By accounting the compressibility of the fluid, total pressure and static pressure are related as
below:
1
1
2
1
1
2
1
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
We know that
Then
2
1
1
1
4
1
2
7/5
Now taking fluid as air
10
7
1
5
1.4
1
40
smaller than that of low
So in the case of high speed subsonic flows, dynamic pressure is
speed flows, where
Now velocity is
Supersonic Flow
In the case of supersonic flow, it is not only the compressibility effect which affects the
dynamic pressure calculation, even total pressure changes due to appearance of a detached
shock which is normal in nature near the opening of the probe. When flow passes through the
normal shock, there is a loss of momentum and total pressure drops.
Instead of
we get
for the flow reaching the pitot tube probe. While passing through
to
. These two total pressures are related as
the shock total pressure drops from
/
/
Also the Mach number approaching the probe will no longer be M∞ rather it will have new
value given by the normal shock relation:
And now q will be given as:
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Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
And velocity is given by:
Where
is given by:
Example:
175 ⁄ ,
1
and
Calculate the dynamic pressure of the flow if
What will be the percentage error if the flow is treated as incompressible?
298 .
Effect of High Speed on Pressure Coefficient
Pressure coefficient is always defined by
.
.
Where p is the static pressure and all other parameters with subscript ∞ gives the free stream
conditions. As it is clear that calculation of dynamic pressure involves the calculation of
dynamic pressure, so coefficient of pressure too will be depending upon the high speed
effects.
Above equation gives the coefficient of pressure during any fluid flow of an incompressible
fluid.
Low subsonic flow
A subsonic flow which is taken as incompressible gives a coefficient of pressure as per the
above equation and it will always be equal to unity “1”.
High subsonic flow
In the case of high subsonic flow when compressibility plays an important role we can’t use
the above formula as such though coefficient of pressure is still defined by the above relation.
So now coefficient of pressure is:
Now recall our free stream relations:
1
1
2
1
1
2
1
page 45
Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
2
2
1
1
2
1
1
2
1
1
So coefficient of pressure at the stagnation point:
2
1
1
2
1
1
4
40
So on accounting compressibility effect, coefficient of pressure at the stagnation point
becomes:
………..
Above equation gives accurate result up to Mach number equal to 2. Beyond that it loses
accuracy. For the Mach numbers below unity, the above equation can be further simplified
without compromising much with the accuracy.
………..
page 46
Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Problems
1. Nitrogen gas passes through a normal shock with upstream conditions of
303 and
923 / . Determine the velocity
and
300
,
upstream of the shock. If the same deceleration from
to
takes place is
entropically, what will be the resultant ?
[Ans. 2.316 MPa, 267.64 m/s, 5.036 MPa]
2. There is a normal shock in the uniform air stream. The properties upstream of the
shock are
412 / ,
92
,
300 . Determine
, , ,
,
and
downstream of the shock. Also calculate the entropy increase across the
shock.
[Ans. 311.98 ms/, 136.66 kPa, 336.51 K, 384.96 K, 218.83 kPa]
3. A normal shock wave forms in an air stream at a static temperature of 22 K. If the
total temperature is 400 K, determine the Mach number and static temperature behind
the shock.
[Ans. 0.3893, 382.8 K]
4. The flow properties upstream of a normal shock are 500 m/s, 100 kPa and 300 K.
Determine the velocity, pressure and temperature of the gas downstream of the shock
and the increase in entropy. Take the gas to be air.
[Ans. 284.25 m/s, 225.25 kPa, 384.21 K, 15.432 J/kg-K]
5. A pitot tube is placed in an air stream of static pressure 0.95 atm. Determine the flow
Mach number if the Pitot tube records (a) 1.1 atm, (b) 2.5 atm, and (c) 10 atm.
[Ans. (a) 0.465, (b) 1.275, (c) 2.79]
6. A C – D nozzle connects two reservoirs at pressures 5 atm and 3.6 atm. If a normal
shock has to stand at the nozzle exit, find the pressure at the nozzle exit, just
downstream of the shock.
[Ans. 2.876 atm]
7. Air at 1 MPa and 300 K enteres the Mach 2 De Laval nozzle of a supersonic wind
tunnel at a low velocity. If a normal shock wave is formed at the nozzle exit plane,
determine the pressure, temperature, Mach number, velocity and the stagnation
pressure of the flow just behind the shock.
[Ans. 0.5751 MPa, 281.3 K, 0.57735, 194.1 m/s, 0.72087 MPa]
8. A blunt nozzle model is placed in a mach 3 supersonic tunnel test – section. If the
settling chamber pressure and temperature of the tunnel are 10 atm and 315 K,
respectively, calculate the pressure, temperature and density at the nose of the model.
Assume the flow to be one dimensional.
[Ans. 332.69 kPa, 315 K, 3.68 kg/m2]
page 47
Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Rankine – Hugoniot Relation
We saw the changes across the shock as functions of only free stream Mach number which
are useful in solving the normal shock related problems. It is also possible to express the
changes across the normal shock using only the thermodynamic variables, without
considering explicit reference to a velocity or Mach number. Following is the procedure for
this:
By continuity equation:
On substituting this in the impulse momentum equation we get that:
and
Now considering the energy equation:
1
2
1
2
1
2
1
2
On substituting for
and
we get:
1
2
1
2
1
2
1
2
1
2
1
1
2
1
1 2
1 2
2
2
1
2
2
1 2
2
2
2
1
1
2
2
1
1
2
So we get the Rankine – Hugoniot Relation which gives equation across normal shock by
involving only the thermodynamic properties:
OR
Where
is density and
is the specific density.
page 48
Notes on Aerodynamics – II, III year, 5th Semester by Mohammad Fazlur Rahman
Oblique Shock
Oblique Shock Wave is a wave which remains inclined with respect to the flow direction. It
is a weaker wave as compared to Normal Shock. It brings about the change in flow
parameters in the similar direction as that of Normal Shock but to a lesser degree.
This wave normally comes into picture when a supersonic flow turns into itself. So flow
changes direction across this shock.
page 49