Consider a rocket engine in which the combustion chamber pressure and temperature are 30 atm and 3756 K, respectively. The area

Question

3 years ago

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Consider a rocket engine in which the combustion chamber pressure and temperature are 30 atm and 3756 K, respectively. The area

of the rocket nozzle exit is 15 m2 and is designed so that the exit pressure exactly equals ambient pressure at a standard altitude of 25 km. For the gas mixture, assume that γ = 1.18 and the molecular weight is 20. At a standard altitude of 25 km. Calculate the:
a. specific impulse
b. exit velocity
c. mass flow
d. thrust
e. throat area.

Engineering

1 answer:

Lerok [7] · Answer 1 · 2022-07-29T10:48:17+03:00

Answer:

379.48 s ( specific impulse )
3678.75 m/s ( exit velocity )
263.21 kg/s ( mass flow )
979855.3 N ( thrust )
0.168 $m^{2}$ ( throat area )

Explanation:

A) Specific impulse

standard altitude = 25 km

take free steam pressure ( $p_{0}$ ) = 2527.3 N/ $m^{2}$

Exit pressure = ambient pressure. therefore exit pressure ( $p_{e}$ ) = 2527.3 N/ $m^{2}$

Applying specific impulse equation

$I_{sp}$ = $\frac{1}{g_{0} }$ [ $\frac{2Y T_{0} }{Y - 1} (\frac{R}{M} )[ 1 - (\frac{P_{e} }{p_{0} })^{\frac{y-1}{1} } ]]^{\frac{1}{2} }$ equation 1

given

$g_{0}$ = 9.81 m/ $s^{2}$ , M = 20 kg/kmol, R = 8314.47 j/kmol.k

y = 1.18, $T_{0}$ = 3756 k, $p_{e }$ = 2527.3 N/ $m^{2}$ , $p_{0}$ = 30 atm

substitute the given data into equation 1

$I_{sp}$ ( specific impulse ) = 379.48 s

B) exit velocity

Relating the equation for isentropic process, pressure and temperature at the combustion chamber , the exit temperature (Te) can be calculated using this formula :

$\frac{p_{e} }{p_{0} }$ = $(\frac{T_{e} }{T_{0} })^{\frac{y}{y-1} }$