Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 17.7, Problem 38P

(a)

To determine

The critical temperature of air.

The critical pressure of air.

The critical density of air.

(a)

Expert Solution
Check Mark

Answer to Problem 38P

The critical temperature of air is 337 K.

The critical pressure of air is 140kPa.

The critical density of air is 1.45kg/m3.

Explanation of Solution

Write the formula to calculate the stagnation temperature of ideal gas.

T0=T+V22cp (I)

Here, the static temperature of ideal gas is T, the specific heat of pressure for ideal gas is cp, and the velocity of the ideal gas flow is V.

Write the formula to calculate the stagnation pressure of ideal gas.

P0=P(T0T)k/(k1) (II)

Here, the static pressure of ideal gas is P and the specific heat ratio of ideal gas is k.

Write the formula to calculate the density of the ideal gas.

ρ0=P0RT0 (III)

Here, the pressure of the ideal gas is P.

Write the formula to calculate the critical temperature at the throat of nozzle.

T=T0(2k+1) (IV)

Here, the stagnation temperature of ideal gas is T0.

Write the formula to calculate the critical pressure at the throat of nozzle.

P=P0(2k+1)k/(k1) (V)

Here, the stagnation pressure of ideal gas is P0.

Write the formula to calculate the critical density at the throat of nozzle.

ρ=ρ0(2k+1)1/(k1) (VI)

Here, the stagnation density of ideal gas is ρ0.

Conclusion:

Refer Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of air at 300K temperature as 0.287kJ/kgK, 1.005kJ/kgK and 1.4.

Substitute 100°C for T, 250m/s for V, and 1.005kJ/kgK for cp in Equation (I).

T0=(100°C)+(250m/s)22×(1.005kJ/kgK)=(100°C)+(250m/s)22×(1.005kJ/kg°C)=(100°C)+(62500m2/s2)×(1kJ/kg1000m2/s2)(2.01kJ/kg°C)=131.1°C

    =131.1°C+273.2=404.3K

Substitute 200 kPa for P, 100°C for T, 404.3K for T0, and 1.4 for k in Equation (II).

P0=(200kPa)×(404.3K100°C)1.4/(1.41)=(200kPa)×(404.3K100°C+273.2)1.4/(1.41)=(200kPa)×(1.32333)=264.7kPa

Substitute 264.7kPa for P0, 0.287kJ/kgK for R, and 404.3 K for T0 in Equation (III).

ρ0=264.7kPa(0.287kJ/kgK)(404.3K)=264.7kPa(0.287kJ/kgK)×(1kPam31kJ)(404.3K)=264.7kPa(116.034kPam3/kg)=2.281kg/m3

Substitute 404.3K for T0 and 1.4 for k in Equation (IV).

T=(404.3K)×(21.4+1)=(404.3K)×(0.8333)=337K

Thus, the critical temperature of air is 337 K.

Substitute 264.7kPa for P0 and 1.4 for k in Equation (V).

P=(264.7kPa)×(21.4+1)1.4/(1.41)=(264.7kPa)×(0.528282)=140kPa

Thus, the critical pressure of air is 140kPa.

Substitute 2.281kg/m3 for ρ0 and 1.4 for k in Equation (VI).

ρ=(2.281kg/m3)(21.4+1)1/(1.41)=(2.281kg/m3)(0.633938)=1.45kg/m3

Thus, the critical density of air is 1.45kg/m3.

(b)

To determine

The critical temperature of helium.

The critical pressure of helium.

The critical density of helium.

(b)

Expert Solution
Check Mark

Answer to Problem 38P

The critical temperature of helium is 241K.

The critical pressure of helium is 104kPa.

The critical density of helium is 0.208kg/m3.

Explanation of Solution

Conclusion:

From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of helium at 300K temperature as 2.0769kJ/kgK, 5.1926kJ/kgK and 1.667.

Substitute 40°C for T, 300m/s for V, and 5.1926kJ/kgK for cp in Equation (I).

T0=(40°C)+(300m/s)22×(5.1926kJ/kgK)=(40°C)+(300m/s)22×(5.1926kJ/kg°C)=(40°C)+(90000m2/s2)×(1kJ/kg1000m2/s2)(10.3852kJ/kg°C)=48.66°C

    =48.66°C+273.2=321.866K321.9K

Substitute 200 kPa for P, 40°C for T, 321.9K for T0, and 1.667 for k in Equation (II).

P0=(200kPa)×(321.9K40°C)1.667/(1.6671)=(200kPa)×(321.9K313.2K)1.667/(1.6671)=(200kPa)×(1.071)=214.2kPa

Substitute 214.2 kPa for P0, 2.0769kJ/kgK for R, and 321.9 K for T0 in Equation (III).

ρ0=214.2kPa(2.0769kJ/kgK)(321.9K)=214.2kPa(2.0769kJ/kgK)×(1kPam31kJ)(321.9K)=214.2kPa(710.0921kPam3/kg)=0.320kg/m3

Substitute 321.9K for T0 and 1.667 for k in Equation (IV).

T=(321.9K)×(21.667+1)=(321.9K)×(0.749906)=241.39K241K

Thus, the critical temperature of helium is 241K.

Substitute 214.2kPa for P0 and 1.667 for k in Equation (V).

P=(214.2kPa)×(21.667+1)1.667/(1.6671)=(214.2kPa)×(0.487092)=104.34kPa104kPa

Thus, the critical pressure of helium is 104kPa.

Substitute 0.320kg/m3 for ρ0, and 1.667 for k in Equation (VI).

ρ=(0.320kg/m3)(21.667+1)1/(1.6671)=(0.320kg/m3)(0.649537)=0.208kg/m3

Thus, the critical density of helium is 0.208kg/m3.

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Chapter 17 Solutions

Thermodynamics: An Engineering Approach

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