Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 17.7, Problem 90P
To determine

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

Expert Solution & Answer
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Answer to Problem 90P

The Mach number value of air after the normal shock through the nozzle is 0.5039.

The actual temperature of air after the normal shock through the nozzle

is 604.34K.

The actual pressure of air after the normal shock through the nozzle is 447.76kPa.

The stagnation pressure of air after the normal shock though the nozzle

is 532.5kPa.

The velocity of air after the normal shock through the nozzle is 248.31m/s.

Thus, the Mach number of helium gas after the normal shock through the nozzle

 is 0.5455.

Thus, the actual temperature of helium after the normal shock through the nozzle is 799.362K.

Thus, the actual pressure of helium after the normal shock through the nozzle

 is 475.7kPa.

Thus, the stagnation pressure of helium after the normal shock though the nozzle

is 602.5kPa.

Thus, the velocity of helium after the normal shock through the nozzle is 907.5m/s.

Comparison between the obtained results of air and helium is shown in below Table:

Parameters/ ConditionsAirHelium
Mach number value0.50390.5455
Actual temperature of air604.34K799.362K
Actual pressure of air447.76kPa475.7kPa
Stagnation pressure532.5kPa602.5kPa
Velocity248.31m/s907.5m/s

Explanation of Solution

Write the expression for the velocity of sound after the normal shock.

c2=kRT2 (I)

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

Write the expression for the velocity of airafter the normal shock.

V2=Ma2c2 (II)

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

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

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 for the actual pressure of helium gas after the normal shock.

P2P1=1+kMa121+kMa22 (IV)

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

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

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

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

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

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

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

Write the expression for the velocity of sound after the normal shock.

c2=kRT2 (VII)

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

Write the expression for the velocity of helium after the normal shock.

V2=Ma2c2 (VIII)

Conclusion:

Refer to Table A-33, “One-dimensional normal-shock functions for an ideal gas with k 5 1.4”, obtain the expressions of temperature ratio, pressure ratio, stagnation pressure ratio, and Mach number after the shock for a Mach number of 2.6 before the shock.

Ma2=0.5039

Thus, the Mach number value of air after the normal shock through the nozzle is 0.5039.

T2T1=2.2383 (IX)

P2P1=7.7200 (X)

P02P01=9.1813 (XI)

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

Substitute 270K for T1 in Equation (IX).

T2270K=2.2383T2=270K×2.2383T2=604.34K

Thus, the actual temperature of air after the normal shock through the nozzle

is 604.34K.

Substitute 58kPa for P1 in Equation (X).

P258kPa=7.7200P2=58kPa×7.7200P2=44.76kPa

Thus, the actual pressure of air after the normal shock through the nozzle is 447.76kPa.

The actual pressure before the normal shock is the same as the stagnation pressure before the normal shock (P1=P01) , as the flow through the nozzle is isentropic.

Substitute 58kPa for P01 in Equation (XI).

P0258kPa=9.1813P2=58kPa×9.1813P2=532.5kPa

Thus, the stagnation pressure of air after the normal shock though the nozzle

is 532.5kPa.

Refer to thermodynamics properties table and interpret the value of k, and R for a temperature of 270K, and a pressure of 58kPa for air as 1.4, and 0.287kJ/kgK.

Substitute 1.4 for k, 0.287kJ/kgK for R, and 604.34K for T2 Equation (I).

c2=1.4×0.287kJ/kgK×604.34K=492.27m/s

Substitute 0.5039for Ma2, and 492.27m/s for c2 in Equation (II).

V2=0.5039×492.27m/s=248.31m/s

Thus, the velocity of air after the normal shock through the nozzle is 248.31m/s.

Substitute 2.6for Ma1, and 1.667 for k in Equation (III).

Ma2=(2.62+2/(1.6671)2×2.62×1.667/(1.6671)1)1/2=0.5455

Thus, the Mach number of helium gas after the normal shock through the nozzle

 is 0.5455.

Substitute 1.667 for k, 2.6for Ma1, and 0.5455 for Ma2. in Equation (IV).

P2P1=1+1.667×(2.6)21+1.667×(0.5455)2

P2P1=8.2009 (XII)

Substitute 1.667 for k, 2.6for Ma1, and 0.5455 for Ma2. in Equation (V).

T2T1=1+2.62(1.6671)/21+0.54552(1.6671)/2

T2T1=2.9606 (XIII)

Substitute 1.667 for k, 2.6for Ma1, and 0.5455 for Ma2. in Equation (VI).

P02P01=1+1.667×2.621+1.667×0.54552(1+(1.6671)0.545522)1.667/(1.6671)

P02P01=10.389 (XIV)

Substitute 270K for T1 in Equation (XIII).

T2270K=2.9606T2=270K×2.9606T2=799.362K

Thus, the actual temperature of helium after the normal shock through the nozzle is 799.362K.

Substitute 58kPa for P1 in Equation (XII).

P258kPa=8.2009P2=58kPa×8.2009P2=475.7kPa

Thus, the actual pressure of helium after the normal shock through the nozzle

 is 475.7kPa.

Since, (P1=P01)

Substitute 58kPa for P01 in Equation (XIV).

P0258kPa=10.389P2=58kPa×10.389P2=602.5kPa

Thus, the stagnation pressure of air after the normal shock though the nozzle

is 602.5kPa.

Refer Table A–1, “Molar mass, gas constant, and critical2point properties”, obtain

the value of k, and R for a temperature of 270K, and a pressure of 58kPa for helium as 1.667, and 2.0769kJ/kgK respectively.

Substitute 1.667 for k, 2.0769kJ/kgK for R, and 799.362K for T2 Equation (VII).

c2=1.667×2.0769kJ/kgK×799.362K=1663.59m/s

Substitute 0.5455 for Ma2, and 1663.59m/s for c2 in Equation (VIII).

V2=0.5455×1663.59m/s=907.5m/s

Thus, the velocity of air after the normal shock through the nozzle is 907.5m/s.

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Chapter 17 Solutions

Thermodynamics: An Engineering Approach

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