EXTROVERT
Space Propulsion 02
Thrust, Rocket Equation, Specific
Impulse, Mass Ratio
1
EXTROVERT
Space Propulsion 02
Thrust
T  m f Ue  ( Pe  Pa ) Ae
Thrust comes from:
a) Increase in momentum of the propellant fluid
(momentum thrust)
b) Pressure at the exit plane being higher than the
outside pressure (pressure thrust).
Where does the thrust act?
In the rocket engine, the force is felt on the nozzle and the combustor walls,
and is transmitted through the engine mountings to the rest of the vehicle.
Effective Exhaust Velocity
ce  U e 
Ae
( pe  pa )
m
2
EXTROVERT
Space Propulsion 02
Delta V and Mass Ratio
Consider a rocket with effective exhaust velocity ce. As propellant is blasted
out the exhaust nozzle, the mass of the vehicle decreases. This is
substantial in the case of the rocket as compared to air-breathing engines,
because all the propellant comes from inside the vehicle. From Newton's
Second Law,
dM
dV  ce
M
DV  ce
M2

M1
dM
M
 M1 
Dv  ceLoge 

M
 2
where M1 is the initial mass, which includes the propellant, and M2 is the
mass after the propellant has been used up to achieve the velocity
increment DV.
3
EXTROVERT
Space Propulsion 02
Specific Impulse and Mass Ratio
Specific Impulse of the system is
ce
Isp 
g
where g is the standard value of acceleration due to gravity at sea-level (9.8m/s2). Note that
the unit of Specific Impulse is seconds. Using this definition,
M1 
DV  gIsp log  

M 2 
Mass Ratio of a rocket is

M1
e
M2
DV
gIsp
Note: Some
organizations
express Specific
Impulse without
dividing by g
4
EXTROVERT
Space Propulsion 02
Example:
For missions from Earth's surface to escape from earth's gravitational field, Mass Ratio is
large.
For specific impulse of 390 s, g = 9.8 m/s2, and DV = 11,186 m/s (36700 fps), the mass ratio
is 18.67.
This means that the rocket at launch time must be at least 18.67 times as big as the
spacecraft which is left after all the fuel is burned. To get a high specific impulse like 390 s,
we have to use a costly system like liquid hydrogen - liquid oxygen.
For earth orbit, the velocity increment DV needed is 25,000 fps, while 36,700fps will enable
escape from Earth's gravitational field.
5
EXTROVERT
Space Propulsion 02
Newton's Law of Gravitation
To find the velocity increment required for various missions, we must
calculate trajectories and orbits. This is done using Newton's Law of
Gravitation:
Here the lhs is the "radial force" of attraction due to gravitation, between two
bodies; the big one of mass M, and the little one of mass m.
The universal gravitational constant G is 6.670 * 10-11 Nm2/kg2.
6
EXTROVERT
Space Propulsion 02
Rocket Equation Including Drag and Gravity:
Ref: Hill & Peterson, Chapter 10.
The rate of acceleration of the vehicle is
.
dM D
du  ce
 dt  gCos dt
M M
Neglecting the air drag and gravity terms, we get the
Ideal Rocket Equation
dM
du  ce
M
7
EXTROVERT
Space Propulsion 02
Drag Term in the Rocket Equation
D  0.5 U 2 A f CD
where atmospheric density above Earth varies roughly as
 (h)  aExp(bh1.15 )
With density in kg/m3 and h in meters, a = 1.2 and b = 2.9 x 10-5
Roughly, density at 30,000 meters is about 1% of its sea-level value.
Drag Coefficients (typical)
Note: drag coefficient peak is reached at around Mach 1.2.
Inclination, deg. to flt. direction,
CD low-speed
CD peak @1.2
CD @ Mach 2
0
0.06
0.15
0.13
4
0.185
0.16
8
0.23
0.2
8
EXTROVERT
Space Propulsion 02
Gravity Term
 Re 
g  ge 

 Re  h 
2
At 100 miles above the surface the change from the surface is still only about 5%.
9
EXTROVERT
Space Propulsion 02
Example
Specific impulse of 390 s,
g0 = 9.8 m/s2,
and DV = 11186 m/s (36700 fps),
Mass ratio is 18.67.
This means that the rocket at launch time must be at least 18.67
times as big as the spacecraft which is left after all the fuel is burned.
To get a high specific impulse like 390 s, we have to use a costly
system like liquid hydrogen - liquid oxygen.
Velocity increment DV for Low Earth Orbit: ~ 25,000 fps,
Escape from Earth's gravitational field
~ 36,700fps
10
EXTROVERT
Space Propulsion 02
Single Stage Sounding Rocket
Altitude at burnout, assuming it
goes straight up:
tb
hb   Udt
0
Neglecting drag
M
U  ce ln
 g et
M0
If
me is constant,
t
M t   M 0  M 0  M b 
tb
(Sounding rocket: not quite single-stage)
NASA Goddard Space Flight center
http://www.gsfc.nasa.gov
11
EXTROVERT
Space Propulsion 02
Single-Stage Sounding Rocket Going Straight Up (cont’d)
Define Mass Ratio
M0
R
Mb
Substituting,
  1 t 
U  ce ln 1  1     get
  R  tb 
ln R
1
2
hb  cetb
 cetb  getb
R  1`
2
12
EXTROVERT
Space Propulsion 02
Expression for Maximum Altitude Reached
Note that at burnout, the sounding rocket is still moving fast upward.
Equating kinetic energy at burnout with change in potential energy of the final mass
Ub2
Mb
 M b g e hmax  hb 
2
hmax
Ub2
 hb 
2 ge
hmax
ce 2 ln R 
 R


 cetb 
ln R  1
2 ge
 R 1

2
13
Mb
EXTROVERT
Example
Space Propulsion 02
ce 2 ln R 
 R


 cetb 
ln R  1
2 ge
 R 1

2
hmax
Values given:
Ce = 3000 m/s
R = 10
tb = 30s
Hmax = ??
14
EXTROVERT
Space Propulsion 02
Chemical Rockets
Burn time of existing rockets is ~ 30 to 200 seconds.
M0  M L  M p  M s
ML
Payload Mass
Mb  M L  M s
Ms
Structure (incl. engine) Mass
Mp
M
M0
R 0 
Mb ML  Ms
Propellant Mass
Definitions
 Payload ratio
Structure coefficient
Thus,

ML
ML

M0  M L M p  Ms

Ms
M  ML
 b
M p  Ms M0  M L
R
1 
 
15
EXTROVERT
Space Propulsion 02
Multistaging -1
M 0i
M bi
Total initial mass of i-th stage prior to firing, include its effective
payload.
Total mass of i-th stage after burnout, include its effective payload.
ML
Payload of last stage.
M si
Structural mass of i-th stage; include engine controls, instruments.
16
EXTROVERT
Space Propulsion 02
Payload of
i
th
Multistaging - 2
stage is mass of all subsequent stages.
Structural coefficient of
i 
If
i thstage:
M si
M 0i  M 0i 1
i th stage contains no propellant at burnout:
i 
M bi  M 0i 1
M 0i  M 0i 1
17
EXTROVERT
Space Propulsion 02
Multistaging - 3
Mass ratio of i th stage
Ri 
M 0i
M bi
Ri 
i.e.,
.
1  i
 i  i
DU i  Cei ln Ri
n
U n   Cei ln Ri
i 1
Similar stages: same

and

 1  
U n  nCe ln 

   
18
EXTROVERT
Space Propulsion 02
Multistaging - 4
M 01 1  1

M 02
1
M 02 1  2

M 03
2
M 01 n  1  i 


  
M L i 1 i 
M 0n 1  n

ML
n
If i are equal,
M 01  1   


ML   
n


1


n
 M 01 






Mi 
Un

 n ln 

1
Ce


 

n

M


  01   1  1
 
  M i 
 
 

19
EXTROVERT
Space Propulsion 02
Multistaging - 5
 DU

M engine  M tan k 
ML
 1  e Ce
1
 1 


M0
M0
M
p







Structural coefficient


M tan k  M engine
M p  M tan k  M engine
 DU
M tan k 
Ce  M engine
1 e


M p 
M0

 DU

 M tan k  M engine
C
e

1  e
1 


M p 
M0


20
EXTROVERT
Space Propulsion 02
Apollo engines (Source: Hill Peterson, page 479)
Engine
J2
H1
Thrust, kN
1023
1023
Fuel
Hydrogen
Hydrocarbon
Engine Mass, Millions of grams
1.622
0.921
Engine Mass Fraction
0.024
0.014
Tank Mass/ Propellant Mass
0.046
0.016
Eq. Exhaust Vel. m/s
4175
2891
Specific Impulse, seconds
426
295
21
EXTROVERT
Space Propulsion 02
Saturn V Apollo 11 Flight Configuration
Stage
1
2
3
Engine
F-1
J-2
J-2
Fuel
RP1 Hydrocarbon
LH2
LH2
Number of engines
5
5
1
Total thrust, Newton
33 Million
4.45 Million
0.89 Million
Total Initial Mass,
Kg
2.780 Million
0.677 Million
0.215 Million
Propellant mass, kg
1.997 Million
0.429 Million
0.109 Million
Structure & engine,
kg
0.106 Million
0.0326 Million
0.0257 Million
Structure mass
fraction
0.05
0.071
0.191
Payload fraction
0.321
0.466
0.603
22
EXTROVERT
 F 
CF  

P0 At 
Space Propulsion 02
The idea of “Thrust Coefficient”
where F is the thrust, At is the nozzle throat area and P0 is the combustion chamber stagnation
pressure. Basically a higher thrust coefficient means a better usage of the available stagnation
pressure in converting to thrust.
The thrust coefficient has values ranging from 0.8 to 1.9.
Note also that a plot of thrust coefficient vs. altitude for a given nozzle will give the variation of
thrust with altitude for a given chamber pressure and nozzle throat area.
The thrust coefficient is also used to compare different nozzle designs for given constraints.
In the following we will use gas dynamics to derive expressions for the thrust coefficient in terms
of gas properties.
23
EXTROVERT
Space Propulsion 02
 F 
CF  

P0 At 
Thrust Coefficient - 1
where At is nozzle throat area and p0 is chamber pressure (N/m2)
Thus,
ut   RTt
For sonic conditions at the throat,
and
1/( 1)
 2 
t   0 



1


1/ 2
 2T T  2 1/  1   p  1 /   
e


F  At p0 R  t 0 
1




  p0 
 
   1    1 
 ( pe  p0 ) Ae
24
EXTROVERT
Space Propulsion 02
Thrust Coefficient - 2
Using isentropic flow relations,
2
 1
1  2

2


1
t 


  1 
  1 
 p0   2
TtT0  



R


1


 t 
1
0 2

 2  2
F  At p0 
   1


1




 1 1/ 2


 
 1
 1 
 pe
1

  
  p0  



 ( pe  pa ) Ae
and Thrust Coefficient
1/ 2
 1
 1
 2




1


 pe 
 2  2 

CF  
1

   1    p  


1

   0  




( pe  pa ) Ae
p0 At
Depends entirely on nozzle characteristics. The thrust coefficient is used to evaluate
nozzle performance.
25
EXTROVERT
Space Propulsion 02
Characteristic Exhaust Velocity c*
Used to characterize the performance of propellants and combustion chambers
independent of the nozzle characteristics.
 1


2


1
p0 At   2    
p0 At
m

 


 RT0     1 
 RT0



where
So
is the quantity in brackets. Note:
p0 At
m

a0
Characteristic exhaust velocity
a0   RT0
p0 At
c* 
m
Assuming steady, quasi-1-dimensional, perfect gas. The condition for maximum thrust is
ideal expansion: nozzle exit static pressure being equal to the outside pressure. In other
words,
pe  pa
26
EXTROVERT
Space Propulsion 02
End of Section 2
We’ll end Lecture 2 here, and go on to discuss orbits before getting back to compressible
Flow and chemistry considerations. The purposes are:
1.
2.
To enable mission calculations.
To give everyone a chance to look at the content so far and see how much they need
review of compressible flow material. Please browse the web links in the first
lecture.
27