EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 17.7, Problem 85P

Air flowing steadily in a nozzle experiences a normal shock at a Mach number of Ma = 2.5. If the pressure and temperature of air are 10.0 psia and 440.5 R, respectively, upstream of the shock, calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions.

Expert Solution & Answer
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To determine

The pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock and for helium undergoing a normal shock under the same conditions.

Answer to Problem 85P

The actual temperature of air after the normal shock is 942R.

The actual pressure of air after the normal shock is 71.25psia.

The stagnation pressure of air after the normal shock is 85.3psia.

The Mach number value of air after the normal shock is 0.513.

The velocity of air after the normal shock is 772ft/s.

The Mach number of helium gas after the normal shock is 0.553.

The actual temperature of helium after the normal shock is 1233R.

The actual pressure of helium after the normal shock is 75.6psia.

The stagnation pressure of air after the normal shock is 115.46psia.

The velocity of air after the normal shock is 2794ft/s.

Explanation of Solution

Refer Table A-32, “One-dimensional isentropic compressible-flow functions for an ideal

gas with k=1.4”, write the following expressions of temperature ratio, pressure ratio, stagnation pressure ratio, and Mach number after the shock for a Mach number of 2.5.

T2T1=2.1375 (I)

P2P1=7.125 (II)

P02P01=8.5262 (III)

Ma2=0.513

Here, actual temperature after the shock is T2, actual temperature before the shock is T1, actual pressure after the shock is P2, actual pressure before the shock is P1, stagnation pressure  after the shock is P02, stagnation pressure before the shock is P01, and Mach number after the shock is Ma2.

Write the expression to calculate the velocity of sound after the normal shock.

c2=kRT2 (IV)

Here, velocity of sound after the shock is c2, and gas constant of air is R.

Write the expression to calculate the velocity of air after the normal shock.

V2=Ma2c2 (V)

The value from the table is not considered for Mach number (Ma2), because the value for ratio of specific heats k is not 1.4 for helium.

Write the expression to calculate the Mach number for helium after the normal shock.

Ma2=(Ma12+2/(k1)2Ma12k/(k1)1)1/2 (VI)

Here, Mach number of helium before the normal shock is Ma1, Mach number of helium before the normal shock is Ma2, and ratio of specific heats for helium is k.

Write the expression to calculate the actual pressure of helium gas after the normal shock.

P2P1=1+kMa121+kMa22 (VII)

Here, actual pressure of helium after the shock is P2, and actual pressure of helium before the shock is P1.

Write the expression to calculate the actual temperature of helium gas after the normal shock.

T2T1=1+Ma12(k1)/21+Ma22(k1)/2 (VIII)

Here, actual temperature of helium after the shock is T2, and actual temperature of helium before the shock is T1.

Write the expression to calculate the actual pressure of helium gas after the normal shock.

P02P01=1+kMa121+kMa22(1+(k1)Ma222)k/(k1) (IX)

Here, stagnation pressure of helium after the shock is P02, and stagnation pressure of helium before the shock is P01.

Write the expression to calculate the velocity of sound after the normal shock for helium.

c2,helium=kRT2 (X)

Here, velocity of sound after the shock for helium is c2,helium, and gas constant of helium is R.

Write the expression to calculate the velocity of helium after the normal shock V2,helium.

V2,helium=Ma2c2,helium (XI)

Conclusion:

For air:

Refer Table A-2E, “Ideal-gas specific heats of various common gases”, obtain the following properties for air at room temperature.

R=0.06855Btu/lbmRk=1.4

Substitute 440.5R for T1 in Equation (I).

T2440.5R=2.1375

T2=440.5R×2.1375=941.56R=942R

Thus, the actual temperature of air after the normal shock is 942R.

Substitute 10psia for P1 in Equation (II).

P210psia=7.125P2=10psia×7.125P2=71.25psia

Thus, the actual pressure of air after the normal shock is 71.25psia.

The actual pressure before the normal shock (P1) is the equal as the stagnation pressure before the normal shock (P01), since the flow through the nozzle is isentropic,

Substitute 10psia for P01 in Equation (III).

P0210psia=8.5262

P2=10psia×8.5262=85.262psia=85.3psia

Thus, the stagnation pressure of air after the normal shock is 85.3psia.

From Table A-32, “One-dimensional isentropic compressible-flow functions for an ideal

gas with k=1.4”, write the following expressions of Mach number after the shock for a Mach number of 2.5.

Ma2=0.513

Thus, the Mach number value of air after the normal shock is 0.513.

Substitute 1.4 for k, 0.06855Btu/lbmR for R, and 942R for T2 in Equation (IV).

c2=1.4×0.06855Btu/lbmR×942R=1.4×0.06855Btu/lbmR×942R(25,037ft2/s21Btu/lbm)=1504.47ft/s

Substitute 0.513 for Ma2, and 1504.47ft/s for c2 in Equation (V).

V2=0.513×1504.47ft/s=771.7ft/s=772ft/s

Thus, the velocity of air after the normal shock is 772ft/s.

For helium:

Refer Table A-E, “Ideal-gas specific heats of various common gases”, obtain the following properties for helium.

R=0.4961Btu/lbmRk=1.667

Substitute 2.5 for Ma1, and 1.667 for k in Equation (VI).

Ma2=(2.52+2/(1.6671)2×2.52×1.667/(1.6671)1)1/2=(9.24830.24)1/2=0.553

Thus, the Mach number of helium gas after the normal shock is 0.553.

Substitute 1.667 for k, 2.5 for Ma1, and 0.553 for Ma2. in Equation (VII).

P2P1=1+1.667×(2.5)21+1.667×(0.553)2

P2P1=7.5631 (XII)

Substitute 1.667 for k, 2.5 for Ma1, and 0.553 for Ma2. in Equation (VIII).

T2T1=1+(2.5)2(1.6671)/21+(0.553)2(1.6671)/2

T2T1=2.7989 (XIII)

Substitute 1.667 for k, 2.5 for Ma1, and 0.553 for Ma2. in Equation (IX).

P02P01=1+1.667×(2.5)21+1.667×(0.553)2(1+(1.6671)(0.553)22)1.667/(1.6671)

P02P01=11.546 (XIV)

Substitute 10psia for P1 in Equation (XII).

P210psia=7.5632P2=10psia×7.5632P2=75.6psia

Thus, the actual pressure of helium after the normal shock is 75.6psia.

Substitute 440.5R for T1 in Equation (XIII).

T2440.5R=2.7989

T2=440.5R×2.7989=1232.91R=1233R

Thus, the actual temperature of helium after the normal shock is 1233R.

Since the flow through the nozzle is isentropic (P01=P1).

Substitute 10psia for P01 in Equation (XIV).

P0210psia=11.546P02=10psia×11.546P02=115.46psia

Thus, the stagnation pressure of air after the normal shock is 115.46psia.

Substitute 1.667 for k, 0.4961Btu/lbmR for R, and 1233R for T2 in Equation (X).

c2,helium=1.667×0.4961Btu/lbmR×1233R=1.667×0.4961Btu/lbmR×1233R(25,037ft2/s21Btu/lbm)=5052.72ft/s

Substitute 0.553 for Ma2, and 5052.72ft/s for c2 in Equation (XI).

V2,helium=0.553×5052.72ft/s=2794.15ft/s=2794ft/s

Thus, the velocity of air after the normal shock is 2794ft/s.

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Chapter 17 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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