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 3.8, Problem 82P

Determine the specific volume of superheated water vapor at 15 MPa and 350°C using (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the steam tables. Also determine the error involved in the first two cases.

(a)

Expert Solution
Check Mark
To determine

The specific volume of superheated water vapour based on the ideal gas equation.

The error involved.

Answer to Problem 82P

The specific volume of superheated water vapour based on the ideal gas equation is 0.01917m3/kg_.

The error involved is 67.0%_.

Explanation of Solution

Write the equation of specific volume of superheated water using ideal gas equation of state.

v=RTP (I)

Here, gas constant is R, pressure and temperature of R-134a are P and T respectively.

Calculate the percentage of error involved.

Error=vcalculatedv@15MPa,350°Cv@15MPa,350°C×100% (II)

Here, specific volume at pressure and temperature of 15 MPa and 350°C is v@15MPa,350°C.

Conclusion:

Refer to Table A-1, obtain the gas constant, R of water as 0.4615kPam3/kgK.

Substitute 0.4615kPam3/kgK for R, 350°C for T, and 15 MPa for P in Equation (I).

v=(0.4615kPam3/kgK)(350°C)15MPa=(0.4615kPam3/kgK)(350+273)K15MPa×103kPaMPa=0.01917m3/kg

Thus, the specific volume of superheated water vapour based on the ideal gas equation is 0.01917m3/kg_.

Refer to Table A-6, obtain the value of v@15MPa,350°C at pressure and temperature of 15 MPa and 350°C as 0.011481m3/kg.

Substitute 0.01917m3/kg for vcalculated and 0.011481m3/kg for v@15MPa,350°C in Equation (II).

Error=0.01917m3/kg0.011481m3/kg0.011481m3/kg×100%=67.0%

Thus, the error involved is 67.0%_.

(b)

Expert Solution
Check Mark
To determine

The specific volume of superheated water vapour based on the generalized compressibility chart.

The error involved.

Answer to Problem 82P

The specific volume of superheated water vapour based on the generalized compressibility chart is 0.01246m3/kg_.

The error involved is 8.5%_.

Explanation of Solution

Calculate the reduced pressure.

PR=PPcr (III)

Here, pressure of superheated water vapour is P and critical pressure is Pcr

Calculate the reduced temperature.

TR=TTcr (IV)

Here, temperature of superheated water vapor is T and critical temperature is Tcr

Calculate the specific volume of superheated water vapour based on the generalized compressibility chart.

v=Zvideal (V)

Here, the specific volume at ideal condition is videal.

Conclusion:

Refer to Table A-1, obtain the critical pressure and critical temperature of water.

Pcr=22.06MPaTcr=647.1K

Substitute 22.06 MPa for Pcr and 15 MPa for P in Equation (III).

PR=15MPa22.06MPa=0.679

Substitute 647.1 K for Tcr and 350°C for T in Equation (IV).

TR=350°C647.1K=(350+273)K647.1K=0.962

Refer to figure A-15, “The compressibility chart”, obtain the compressibility factor, Z by reading the calculated reduced pressure and reduced temperature as 0.65.

Substitute 0.65 for Z and 0.01917m3/kg for videal in Equation (V).

v=(0.65)(0.01917m3/kg)=0.01246m3/kg

Thus, the specific volume of superheated water vapour based on the generalized compressibility chart is 0.01246m3/kg_.

Substitute 0.01246m3/kg for vcalculated and 0.011481m3/kg for v@15MPa,350°C in Equation (II).

Error=0.01246m3/kg0.011481m3/kg0.011481m3/kg×100%=8.5%

Thus, the error involved is 8.5%_.

(c)

Expert Solution
Check Mark
To determine

The specific volume of superheated water vapour based on the data from tables.

Answer to Problem 82P

The specific volume of superheated water vapour based on the data from table is 0.01148m3/kg_.

Explanation of Solution

Refer to Table A-6, obtain the specific volume at pressure and temperature of 15 MPa and 350°C as 0.01148m3/kg.

Thus, the specific volume of superheated water vapour based on data from steam tables is 0.01148m3/kg_.

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Chapter 3 Solutions

Thermodynamics: An Engineering Approach

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