CONNECT FOR THERMODYNAMICS: AN ENGINEERI
CONNECT FOR THERMODYNAMICS: AN ENGINEERI
9th Edition
ISBN: 9781260048636
Author: CENGEL
Publisher: MCG
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Chapter 17.7, Problem 128RP

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.

Expert Solution & Answer
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To determine

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 24.2kPa.

The exit area of the diffuser is 0.463m2.

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+V122cp (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/(k1) (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=Ma2kRT2 (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/(k1) (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=P2P1 (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=m˙RT2V2P2 (XI)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant (R) of air is 0.287kJ/kgK.

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) of air is 1.005kJ/kgK.

The specific heat ratio (k) of air is 1.4.

Conclusion:

Substitute 0.287kJ/kgK for R, 1.4 for k, and 242.7K for T1 in Equation (I).

c1=1.4×0.287kJ/kgK×242.7K=97.51686kJ/kg×1000m2/s21kJ/kg=312.2769m/s

Substitute 312.2769m/s for c1, and 0.9 for Ma1 in Equation (II).

V1=0.9×312.2769m/s=281.0492m/s281m/s

Substitute 242.7K for T1, 281m/s for V1 , and 1.005kJ/kgK for cp in Equation (III).

T01=242.7K+(281m/s)22×1.005kJ/kgK=242.7K+78961m2/s2×1kJ/kg1000m2/s22.01kJ/kgK=242.7K+39.28K=282K

Substitute 41.1kPa for P1, 282K for T01, 1.4 for k and 242.7K for T1 in Equation (IV).

P01=41.1kPa(282K242.7K)1.4/(1.41)=41.1kPa(1.1619)3.5=41.1kPa(1.6908)=69.5kPa

Here,

T01=T02=282KP01=P02=69.5kPa

Substitute 0.3 for Ma2, 0.287kJ/kg for R, and 1.4 for k in Equation (VI).

V2=0.3×1.4×0.287kJ/kgK×T2=0.3×0.4018kJ/kgK×1000m2/s21kJ/kg×T2=0.3×20.04T2=6.01T2m/s (XII)

Substitute 282 K for T02, 6.01T2m/s for V2, and 1.005kJ/kgK for cp in

Equation (VII).

282K=T2+(6.01T2m/s)22×1.005kJ/kgK282K=T2+36.12T2m2/s2×1kJ/kg1000m2/s22.01kJ/kgK282K=T2+0.01797T2282=1.01797T2

T2=2821.01797=277.0219K277K

Substitute 1.4 for k, 282K for T02, 277K for T2 and 69.5kPa for P02 in

Equation (VIII).

P2=69.5kPa(277K282K)1.4/(1.41)=69.5kPa(0.9823)3.5=69.5kPa(0.9393)=65.28kPa

Substitute 65.28kPa for P2, and 41.1kPa for P1 in Equation (IX).

ΔP=65.28kPa41.1kPa=24.2kPa

Thus, the pressure rise across the diffuser in the aircraft is 24.2kPa.

Substitute 277K for T2 in Equation (XII).

V2=6.01277Km/s=100.0263m/s100m/s

Substitute 38kg/s for m˙, 65.28kPa for P2, 100m/s for V2, 0.287kJ/kgK for R, and 277K for T2 in Equation (XI).

A2=(38kg/s)(0.287kJ/kgK)(277K)(100m/s)(65.28kPa)=(38kg/s)(0.287kJ/kgK×1kPam31kJ)(277K)6528kPam/s=3020.962kPam3/s6528kPam/s=0.4628m2

0.463m2

Thus, the exit area of the diffuser is 0.463m2.

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Chapter 17 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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