Engineering, 06.05.2020 04:59 tilly40oooo

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.

Answers: 1

Show answers

Answers

Answer from: Neon29

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 :

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

SUBSTITUTE the value of the above parameters

make (Te) subject of the equation Te = 1273.11 k

next calculate for constant specific pressure ( Cp )

Cp = $\frac{y}{y-1}( \frac{R}{M})$

substitute the value of the above parameters into the equation

Cp = 2725.3 J/kg.k

Finally with Te and Cp known calculate the exit velocity

Ve = $\sqrt{2Cp(To - Te)}$

Cp = 2725.3 J/kg.k

To = 3756 k

Te = 1273.11 k

substitute the given values into the equation

Ve = 3678.75 m/s ( exit velocity )

C ) Mass flow

firstly calculate the density of the gas at exit ( De )

De = $\frac{Pe}{(\frac{R}{M}) } Te$

substituting the values of the parameters into the equation

De ( density of gas at exit ) = 0.00477 kg/m $^{3}$

finally calculate the Mass flow at the exit

$m_{e}$ = De*Ve*Ae

De = 0.00477 kg/ $m^{3}$

Ve = 3678.75 m/s

Ae = 15 $m^{2}$

therefore $m_{e}$ = 263.21 kg/s

D ) thrust

calculate the weight flow rate rate ( w ) firstly

w = mg ( mass flow * speed of gravity )

= 263.21 * 9.81 = 2582.1 N/s

finally calculate thrust by applying this equation for thrust

T ( thrust ) = specific impulse * weight flow rate

= 379.48 * 2582. 1 = 979855.3 N

E ) throat area

calculate the throat area using this equation

m = $\frac{PoA}{\sqrt{To} } \sqrt{\frac{y}{(\frac{R}{M} }) } (\frac{2}{y+1} )^{\frac{(Y+1)}{(y-1)} }$

substituting the values of : m , M , R, To, y, and Po into the equation and making A subject of the equation

A = $\frac{263.21 kg/s}{1568.1\sqrt{N.kg /m^{2} \sqrt{m} } }$ = 0.168 $m^{2}$ ( throat area )

Answer from: Quest

i mean i gues

explanation:

Answer from: Quest

answer:

the army does care about its soilders sir i think yout getting off on the wrong foot sir you need to respect our country were keeping the peole that want to d harm towards you out. so dont go saying we don't care cause we do

Another question on Engineering

Engineering, 04.07.2019 18:10

Burgers vector is generally parallel to the dislocation line. a)-true b)-false

Answers: 2

Answer

Engineering, 04.07.2019 18:10

Slip occurs via two partial dislocations because of (a) the shorter path of the partial dislocation lines; (b) the lower energy state through partial dislocations; (c) the charge balance.

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Manometers are good examples of measuring instruments, nowadays they are not as common as before. a)-capacitive probe gauges b)-gravitational gauges deformation ) gauges d)-digital gauges

Answers: 1

Answer

Engineering, 04.07.2019 18:20

Air flows over a heated plate àt a velocity of 50m/s. the local skin factor coefficient at a point on a plate is 0.004. estimate the local heat transfer coefficient at this point.the following property data for air are given: density = 0.88kg/m3 , viscosity 2.286 x 10 ^-5 kgm/s , k = 0.035w/mk ,cp = 1.001kj/kgk. use colburn reynolds analogy.

Answers: 1

Answer

You know the right answer?

Consider a rocket engine in which the combustion chamber pressure and temperature are 30 atm and 375...

Questions