Question

A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a pressure ratio of 8, and the temperature of the gases at the turbine inlet is 2200 F. Utilizing the air-standard assumptions, determine (a) the temperature and pressure of the gases at every point of the cycle, (b) the velocity of the gases at the nozzle exit

Answer 1

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, $g_{c}$ = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

$c_{p}$= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}$

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence $v_{a} ^{2} = 0$

$h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} + \frac{v_{1} ^{2} }{2c_{p} }$

$T_{1}$ = 20+460 = 480°R

$T_{a} =480+ \frac{(800)(800}{2(0.240)(25037}$= 533.25°R

Pressure at the inlet of compressor at isentropic condition

$P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}$

$P_{a}$ = $(10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}$= 14.45 psia

$P_{2}= 8P_{a} = 8(14.45) = 115.6 psia$

Answer 2

Answer:

a) The temperature and pressure of the gases at every point of the cycle are

T = 38.23 K

P = 2.91 kpa

Respectively

b) The velocity V of the gasses at the nozzle exit = 3590 m/s

Explanation: Please find the attached files for the solutions