D. P. Mishra
3.3 THRUST EQUATION OF ROCKET ENGINES
An expression for the thrust developed by a rocket engine under static condition can be obtained by applying the momentum equation. For this, let us consider a control volume (CV), as shown in Figure 3.1. The pro- pulsive thrust “F” acts in a direction opposite to Ve. The reaction to the thrust “F” on the CV is opposite to it. The momentum equation for such CV is given by [4]
d
dt V dV V V n dA F
cv x
cs
x x x
ò
r +ò (
r ×)
= å (3.1)CV Ae
Ve
Me
F y
x
FIGURE 3.1 Control volume of rocket engine.
As the flow is steady in nature, we can neglect the unsteady term in Equation 3.1:
d
dt V dV
cv
ò
r x =0Let us now evaluate the momentum flux term and sum of forces acting on CV of Equation 3.1, as given in the following:
cs
x x x e
V V n dA V dm mV
ò (
r ×)
=ò
= (3.2)Fx = +F P A P Aa e- e e
å
(3.3)where
F is the thrust
Vx is the velocity component in x-direction Ve is the velocity component at exit of nozzle m is the mass flow rate of propellant
Pe is the pressure at exit plane of nozzle Pa is the ambient pressure
By combining Equations 3.3, 3.2 and 3.1 for steady flow, we can have, F mV= e+
(
P P Ae- a)
e (3.4) where the first and second terms represent the momentum contribution and pressure components of thrust, respectively.3.3.1 Effective Exhaust Velocity
We know that velocity profile at the exit of nozzle need not be one- dimensional in nature in the practical situation. It is quite cumbersome to determine the velocity profile at the exit of a rocket nozzle. In order to tackle this problem, we can define effective exhaust velocity by using Equation 3.4 as given below:
V F
m V P P m A
eq e e a
= = +
(
-)
e(3.5)
where Veq is the effective exhaust velocity that would be produced equiva- lent thrust which could have produced due to both momentum and pres- sure components of thrust. Note that when the exit pressure is same as that of the ambient pressure, the effective exhaust velocity Veq becomes equal to
nozzle exit velocity Ve. Otherwise, the effective exhaust velocity Veq is less than the nozzle exit velocity Ve. Note that the second term in Equation 3.5 is smaller than the nozzle exit velocity Ve. Hence, the effective exhaust velocity Veq is quite close to the nozzle exit velocity Ve. Note that the effec- tive exhaust velocity Veq is often used to compare the effectiveness of dif- ferent propellants in producing thrust in rocket engine. It can be applied to all mass expulsion systems that undergo expansion through a nozzle. It can be evaluated experimentally by measuring the thrust and total propel- lant flow rate. Keep in mind that it can be easily related to specific impulse and characteristic velocity, which will be discussed in a subsequent section.
In recent times, some groups are advocating the use of effective exhaust velocity Veq in place of specific impulse for the purpose of comparison.
3.3.2 Maximum Thrust
We need to evaluate the condition under which maximum thrust can be achieved for a given chamber pressure and mass flow rate with a fixed throat area of nozzle. We can obtain this condition by differentiating the thrust expression, Equation 3.4, given in the following:
dF mdV= e+
(
P P dA A dPe- a)
e+ e e (3.6) Note that mass flow rate remains constant (m). By invoking momentum equation for one-dimensional steady inviscid flow, we can havemdV e = -A dPe e (3.7)
By clubbing Equations 3.6 and 3.7, we can get
dF=
(
P P dAe- a)
e (3.8) The condition Pe = Pa is called optimum expansion because it corresponds to maximum thrust for the given chamber conditions. The maximum thrust can be obtained only when the nozzle exhaust pressure is equal to the ambient pressure (Pe = Pa), as given in the following:dF
dA P P P P
e =
(
e- a)
=0; when e = a (3.9) Then, thrust expression for maximum thrust can be expressed asF mV= e (3.10)
This condition provides a maximum thrust for a given chamber pressure and propellant mass flow rate. The rocket nozzle in which this condition is achieved is known as optimum expansion ratio nozzle.
3.3.3 Variation of Thrust with Altitude
We can also note from Equation 3.4 that the thrust of a rocket engine is independent of the flight velocity, unlike the gas turbine engine. However, the thrust varies with altitude particularly while operating within the earth’s atmosphere, as there will be changes of ambient atmospheric pressure with altitude. When the rocket propels, the thrust increases with increase in alti- tude because the atmospheric pressure decreases with increase in altitude.
It can be noted that around 10%–30% of overall thrust changes may occur due to change in altitude. Let us see the variation of the thrust and the specific impulse Isp with altitude, shown in Figure 3.2, for a typical rocket engine. It can be observed that the thrust and specific impulse increase with altitude to respective asymptotic value at high altitude, indicating that thrust remains invariant beyond the earth’s atmosphere envelope.
3.3.4 Effect of Divergence Angle on Thrust
Recall that we have derived the thrust expression (Equation 3.4) assuming the exhaust velocity at nozzle exit to be parallel to its axis. It would not be true particularly for conical nozzle as divergence angle in the range of 10°–20° [3,4] is used in rocket nozzle. In order to determine the effects of divergence angle on the ideal thrust, we can consider flow in diver- gent portion of nozzle as shown in Figure 3.3, in which all streamlines
Thrust
Specific impulse T=105 kN
T= 128 kN Isp= 310 s
Isp= 275 s
Altitude (km)
FIGURE 3.2 Variation of thrust of a typical rocket engine with altitude.
are considered to be straight that intersect at the point O, in the upstream region of nozzle exit. Note that mass flow crosses only through the spheri- cal exit segment of nozzle, as shown in Figure 3.3. The cross-sectional area segment at nozzle exit can be evaluated as
dA=2pRsinq qRd (3.11)
The velocity at nozzle exit in the axial direction would be Ve cos θ. By using Equations 3.1 and 3.10, we can have expression for thrust under steady- state condition:
F=
ò
mVe R Rd +(
P P Ae- a)
e 02
a
q p q q
cos sin (3.12)
By integrating this equation, we get
F=
(
1+)
mVe+(
P P Ae- a)
e = dmVe+(
P P Ae- a)
e2 cosa
l (3.13)
where λd is the divergence correction factor whose values are closer to unity. For example, for nozzle divergence angle of 15°, λd happens to be around 0.983. Hence, for preliminary calculation of the rocket engine, the divergence correction factor can be taken to be unity.
Example 3.1
A booster rocket engine with nozzle exit diameter of 225 mm is designed to propel a satellite to altitude of 20 km. The chamber pressure at 12 MPa is expanded to exit pressure and temperature of
Ve
α
D/2
O θ
∆θ Vx= Ve cos θ
FIGURE 3.3 Schematic of divergent flow in a conical nozzle.
105 kPa and 1400 K, respectively. If the mass flow rate happens to be 15 kg/s, determine the exit jet velocity, effective jet velocity, and thrust at Al = 20 km. MW = 25 kg/kmol.
Solution
At 20 km altitude, Pa = 5.53 kPa.
Assuming one-dimensional flow at the exit of nozzle, and assum- ing ideal gas law, we can evaluate exhaust velocity from mass flow rate at exit as
V m
A
R T m MWP D
e e e
e e
e e
= = = ´ ´ ´
´ ´ ´
r p
4 4 8 314 1400 15
25 105 3 14 0 225
2 .
.
((
.)
2 =1673 65. m/sBy using Equation 3.5, we can evaluate effective velocity as
V V P P m A
eq e e a
= +
(
-)
e= +
(
-)
´( )
=
1673 65 105 5 53 10 15
3 14 4 0 225 1
3 2
. . . .
6673 65 263 5 1937 15. + . = . m/s
Note that the equivalent velocity is almost the same as the exit veloc- ity, indicating that thrust due to momentum will be predominant. We can evaluate total thrust as
F mV= eq =29057 25 29 06. = . kN.