Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Design a 1D compressed air rocket engine. Assume standard atm conditions and ise

ID: 1863157 • Letter: D

Question

Design a 1D compressed air rocket engine. Assume standard atm conditions and isentropic flow. k=1.4, R=0.287


Determine:

1. Tank pressure required producing a Machof 1,3,5, 7, and 9.


2. A throat area of 0.5m^2 and a tank stagnation temp of 1000K, determine mass flow rates for each of the design Mach numbers.


3.The final nozzle area for each Mach number


4. The exit temperatures for all Mach numbers.


5. The exit velocity for all Mach nmbers


6 The amount of mass required for each mach number to achieve a burn time of 30 secs, 1 minute, 5 min, 10 min.


7 Size of tank which can hold the required mass of air for each burn time (Volume).


8 The total force generated by the rocket engine for each speed.

Explanation / Answer

Standard atmospheric pressure = 101.325 kPa


1.

Ma = 1

P0 / P = (1 + 0.2*Ma^2)^3.5

P0 / 101.325 = (1 + 0.2*1^2)^3.5

P0 = 191.8 kPa


Ma = 3

P0 / P = (1 + 0.2*Ma^2)^3.5

P0 / 101.325 = (1 + 0.2*3^2)^3.5

P0 = 3721.9 kPa


Ma = 5

P0 / P = (1 + 0.2*Ma^2)^3.5

P0 / 101.325 = (1 + 0.2*5^2)^3.5

P0 = 53610 kPa


Ma = 7

P0 / P = (1 + 0.2*Ma^2)^3.5

P0 / 101.325 = (1 + 0.2*7^2)^3.5

P0 = 419469 kPa


Ma = 9

P0 / P = (1 + 0.2*Ma^2)^3.5

P0 / 101.325 = (1 + 0.2*9^2)^3.5

P0 = 2138288 kPa


2.

Mass flow m = 0.6847*P0*A_throat / (R*T0)^0.5


For Ma = 1,

m = 0.6847*191.8*10^3*0.5 / (287*1000)^0.5

m = 122.568 kg/s


For Ma = 3,

m = 0.6847*3721.9*10^3*0.5 / (287*1000)^0.5

m = 2378.45 kg/s


For Ma = 5,

m = 0.6847*53610*10^3*0.5 / (287*1000)^0.5

m = 34259 kg/s


For Ma = 7,

m = 0.6847*419469*10^3*0.5 / (287*1000)^0.5

m = 268058.3 kg/s


For Ma = 9,

m = 0.6847*2138288*10^3*0.5 / (287*1000)^0.5

m = 1366455.8 kg/s


3.

A / A_throat = (1 + 0.2*Ma^2)^3 / (1.728*Ma)


For Ma = 1,

A / 0.5 = (1 + 0.2*1^2)^3 / (1.728*1)

A = 0.5 m^2


For Ma = 3,

A / 0.5 = (1 + 0.2*3^2)^3 / (1.728*3)

A = 2.117 m^2


For Ma = 5,

A / 0.5 = (1 + 0.2*5^2)^3 / (1.728*5)

A = 12.5 m^2


For Ma = 7,

A / 0.5 = (1 + 0.2*7^2)^3 / (1.728*7)

A = 52.07 m^2


For Ma = 9,

A / 0.5 = (1 + 0.2*9^2)^3 / (1.728*9)

A = 163.59 m^2


4.

T = T0 / (1 + 0.2*Ma^2)


For Ma = 1,

T = 1000 / (1 + 0.2*1^2)

T = 833.3 K


For Ma = 3,

T = 1000 / (1 + 0.2*3^2)

T = 357.1 K


For Ma = 5,

T = 1000 / (1 + 0.2*5^2)

T = 166.6 K


For Ma = 7,

T = 1000 / (1 + 0.2*7^2)

T = 92.6 K


For Ma = 9,

T = 1000 / (1 + 0.2*9^2)

T = 58.1 K


5.

V = Ma*(kRT)^0.5


For Ma = 1,

V = 1*(1.4*287*833.3)^0.5

V = 578.6 m/s


For Ma = 3,

V = 3*(1.4*287*357.1)^0.5

V = 1136.4 m/s


For Ma = 5,

V = 5*(1.4*287*166.6)^0.5

V = 1293.6 m/s


For Ma = 7,

V = 7*(1.4*287*92.6)^0.5

V = 1350.2 m/s


For Ma = 9,

V = 9*(1.4*287*58.1)^0.5

V = 1375.1 m/s


6.

Mass requred = mass flow rate* time


For 30 sec,


For Ma = 1:

Mass required = 122.568*30 = 3677 kg


For Ma = 3:

Mass required = 2378.45*30 = 71353.5 kg


For Ma = 5:

Mass required = 34259*30 = 1027770 kg


For Ma = 7:

Mass required = 268058.3*30 = 8041749 kg


For Ma = 9:

Mass required = 1366455.8*30 = 40993674 kg


7.

Volume V = m*R*T0 / P0


For Ma = 1,

Volume = 3677*287*1000 / (191.8*10^3)

V = 5502 m^3


For Ma = 3,

Volume = 71353.5*287*1000 / (3721.9*10^3)

V = 5502 m^3


For Ma = 5,

Volume = 1027770*287*1000 / (53610*10^3)

V = 5502 m^3


For Ma = 7,

Volume = 8041749*287*1000 / (419469*10^3)

V = 5502 m^3


For Ma = 9,

Volume = 40993674*287*1000 / (2138288*10^3)

V = 5502 m^3



8.

Force = mass flow rate * velocity


For Ma = 1,

F = 122.568*578.6 = 70917.8 N


For Ma = 3,

F = 1136.4*2378.45 = 2702870 N


For Ma = 5,

F = 1293.6*34259 = 44317442 N


For Ma = 7,

F = 1350.2*268058.3 = 361932316 N


For Ma = 9,

F = 1375.1*1366455.8 = 1879013370 N