An aircraft flies with a Mach number Ma1 = 0.9 at an altitude of 7000 m where the pressure is 41.1 kPa and the temperature is 242.7 K. The diffuser at the engine inlet has an exit Mach number of Ma2 = 0.3. For a mass flow rate of 38 kg/s, determine the static pressure rise across the diffuser and the exit area.
The static pressure rise across the diffuser and the exit area.
Answer to Problem 128RP
The pressure rise across the diffuser in the aircraft is 24.2 kPa.
The exit area of the diffuser is 0.463 m2.
Explanation of Solution
Write the formula for velocity of sound at the inlet conditions.
c1=√kRT1 (I)
Here, speed of sound at the inlet condition is c1, gas constant of air is R, actual temperature of air at the inlet is T1 and ratio of specific heats is k.
Write formula for the velocity of air at inlet.
V1=Ma1c1 (II)
Here, velocity of air after the normal shock is V2.
Write the formula stagnation temperature of air at the inlet.
T01=T1+V212cp (III)
Here, actual (static) temperature of air at the inlet is T1, specific heat at constant pressure is cp and stagnation temperature of air at inlet is T01.
Consider the flow is isentropic.
Write the formula for stagnation pressure of air at inlet (isentropic condition).
P01=P1(T01T1)k/(k−1) (IV)
Here, the static pressure of air at inlet is P1, and stagnation pressure of air at inlet is P01.
Consider the diffuser is adiabatic and the inlet and exit enthalpies are equal.
h01=h02cpT01=cpT02T01=T02
When the stagnation temperatures are equal, the stagnation pressure are also equal.
P01=P02
Here, the static pressure of air at inlet is P1, and stagnation pressure of air at exit is P02.
Write the formula for velocity of sound at the exit conditions.
c2=√kRT2 (V)
Here, speed of sound at the exit condition is c2, gas constant of air is R.
Write formula for the velocity of air at exit.
V2=Ma2c2=Ma2√kRT2 (VI)
Here, the exit Mach number is Ma2.
Write the formula stagnation temperature of air at the exit.
T02=T2+V222cp (VII)
Here, actual (static) temperature of air at the inlet is T1, specific heat at constant pressure is cp and stagnation temperature of air at inlet is T01.
Write the formula for static pressure of air at exit (isentropic condition).
P2=P02(T2T02)k/(k−1) (VIII)
Here, the static pressure of air at inlet is P1, and stagnation pressure of air at exit is P02.
Write the formula for static pressure difference of air.
ΔP=P2−P1 (IX)
Write the formula for mass flow rate of air at exit condition.
˙m=A2V2v2=A2V2P2RT2 (X)
Here, the exit cross sectional area is A2 and exit specific volume is v2.
Rearrange the Equation (X) to obtain A2.
A2=˙mRT2V2P2 (XI)
Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.
The gas constant (R) of air is 0.287 kJ/kg⋅K.
Refer Table A-2, “Ideal-gas specific heats of various common gases”.
The specific heat at constant pressure (cp) of air is 1.005 kJ/kg⋅K.
The specific heat ratio (k) of air is 1.4.
Conclusion:
Substitute 0.287 kJ/kg⋅K for R, 1.4 for k, and 242.7 K for T1 in Equation (I).
c1=√1.4×0.287 kJ/kg⋅K×242.7 K=√97.51686 kJ/kg×1000 m2/s21 kJ/kg=312.2769 m/s
Substitute 312.2769 m/s for c1, and 0.9 for Ma1 in Equation (II).
V1=0.9×312.2769 m/s=281.0492 m/s≃281 m/s
Substitute 242.7 K for T1, 281 m/s for V1 , and 1.005 kJ/kg⋅K for cp in Equation (III).
T01=242.7 K+(281 m/s)22×1.005 kJ/kg⋅K =242.7 K+78961 m2/s2×1 kJ/kg1000 m2/s22.01 kJ/kg⋅K =242.7 K+39.28 K=282 K
Substitute 41.1 kPa for P1, 282 K for T01, 1.4 for k and 242.7 K for T1 in Equation (IV).
P01=41.1 kPa(282 K242.7 K)1.4/(1.4−1)=41.1 kPa(1.1619)3.5=41.1 kPa(1.6908)=69.5 kPa
Here,
T01=T02=282 KP01=P02=69.5 kPa
Substitute 0.3 for Ma2, 0.287 kJ/kg for R, and 1.4 for k in Equation (VI).
V2=0.3×√1.4×0.287 kJ/kg⋅K×T2=0.3×√0.4018 kJ/kg⋅K×1000 m2/s21 kJ/kg×T2=0.3×20.04√T2=6.01√T2 m/s (XII)
Substitute 282 K for T02, 6.01√T2 m/s for V2, and 1.005 kJ/kg⋅K for cp in
Equation (VII).
282 K=T2+(6.01√T2 m/s)22×1.005 kJ/kg⋅K 282 K=T2+36.12T2 m2/s2×1 kJ/kg1000 m2/s22.01 kJ/kg⋅K 282 K=T2+0.01797 T2 282=1.01797 T2
T2=2821.01797=277.0219 K≃277 K
Substitute 1.4 for k, 282 K for T02, 277 K for T2 and 69.5 kPa for P02 in
Equation (VIII).
P2=69.5 kPa(277 K282 K)1.4/(1.4−1)=69.5 kPa(0.9823)3.5=69.5 kPa(0.9393)=65.28 kPa
Substitute 65.28 kPa for P2, and 41.1 kPa for P1 in Equation (IX).
ΔP=65.28 kPa−41.1 kPa=24.2 kPa
Thus, the pressure rise across the diffuser in the aircraft is 24.2 kPa.
Substitute 277 K for T2 in Equation (XII).
V2=6.01√277 K m/s=100.0263 m/s≃100 m/s
Substitute 38 kg/s for ˙m, 65.28 kPa for P2, 100 m/s for V2, 0.287 kJ/kg⋅K for R, and 277 K for T2 in Equation (XI).
A2=(38 kg/s)(0.287 kJ/kg⋅K)(277 K)(100 m/s)(65.28 kPa)=(38 kg/s)(0.287 kJ/kg⋅K×1 kPa⋅m31 kJ)(277 K)6528 kPa⋅ m/s=3020.962 kPa⋅m3/s6528 kPa⋅ m/s=0.4628 m2
≃0.463 m2
Thus, the exit area of the diffuser is 0.463 m2.
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Thermodynamics: An Engineering Approach
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