Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 4, Problem 89P
To determine

The final temperature using the ideal gas equation, the compressibility chart, and steam table.

Expert Solution & Answer
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Explanation of Solution

Given:

The temperature of the saturated water vapor (T) is 350°C.

The constant pressure process.

The final specific volume (v2)  is equal to double the initial specific volume (v1) .

v2=2v1.

Calculation:

Refer to Table A-1, “Molar mass, gas constant, and critical-point properties”, to obtain the gas constant (R), the critical temperature (Tcr), and the critical pressure of water (Pcr) as 0.4615 kPam3/kgK, 647.1 K, and 22.06 MPa.

Write the expression to obtain the final temperature using the ideal gas equation.

  T2=(T1)×(v2v1)T2=(350°C+273)K×(2v1v1)=1246 K

Thus, the final temperature using the ideal gas equation is 1246 K.

Refer Table A-4, “Saturated water: Temperature table”, to obtain the pressure at 350°C temperature and quality of 1 as 16,529 kPa.

Write the expression to obtain the initial reduced pressure (PR1).

  PR1=P1PcrPR1=16529kPa(1 MPa103kPa)22.06MPa=0.749

Write the expression to obtain the initial reduced temperature (TR1).

  TR1=T1TcrTR1=623K647.1K=0.963

Refer Table A-15, “Nelson-Obert generalized compressibility chart”, obtain the value of the initial compressibility factor (Z1) and initial pseudo-reduced specific volume (vR1) corresponding to the initial reduced pressure (PR1) of 0.749 and the initial reduced temperature (TR1) of 0.963 as 0.593 and 0.75 respectively.

Write the expression to obtain the final pseudo-reduced specific volume (vR2).

  vR2=(1.8)vR1vR2=(2)(0.75)=1.5

Obtain the final reduced pressure (PR2).

  PR2=PR1=0.749

Refer Table A-15, “Nelson-Obert generalized compressibility chart”, obtain the value of the final compressibility factor corresponding to the final reduced pressure (PR2) of 0.749 and the final pseudo-reduced specific volume (vR2) of 1.5 as 0.88.

Write the expression to obtain the final temperature (T2).

  T2=P2v2Z2R=P2Z2vR2TcrPcr

  T2=16529 kPa×1.5×647.1 K0.88×22060 kPa=826 K

Thus, the final temperature using the compressibility chart is 826 K.

Refer Table A-4, “Saturated water: Temperature table”, to obtain the pressure and specific volume at 350°C temperature and quality of 1 as 16,529 kPa and 0.008806 m3/kg.

Obtain the final specific volume (v2).

  v2=2v1=2(0.008806 m3/kg)=0.017612 m3/kg

Refer to Table A-6, “Superheated water”, obtain the final temperature (T2) as 477°C corresponding to pressure at 16,529 kPa and specific volume at 0.017612 m3/kg.

Convert the temperature from °C to K.

  T2=(477°C+273) K=750 K

Thus, the final temperature using the superheated steam table is 750 K.

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Chapter 4 Solutions

Fundamentals of Thermal-Fluid Sciences

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