Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 4, Problem 106RQ

(a)

To determine

The pressure of the steam using the ideal gas equation.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The temperture of the steam is (T) is 400°C.

The specific volume of the steam (v) is 0.02m3/kg.

Calculation:

Refer to Table A-1 to find the gas constant (R), the critical pressure (Pcr), and the critical temperature (Tcr) of water as 0.4615kPam3/kgK, 22.06MPa, and 647.1K.

Determine the pressure of the steam using the ideal gas equation.

  P=RTv

  P=(0.4615kPam3/kgK)(400°C)0.02m3/kg=(0.4615kPam3/kgK)(400°C+273)0.02m3/kg=15,529kPa

Thus, the pressure of the steam using the ideal gas equation is 15,529kPa_.

(b)

To determine

The pressure of the steam using the compressibility chart.

(b)

Expert Solution
Check Mark

Explanation of Solution

Determine the reduced temperature of the steam.

  TR=TTcrTR=400°C647.1K=400°C+273647.1K=1.040

Determine the reduced specific volume of the steam.

  vR=vactualRTcr/PcrvR=(0.02m3/kg)(0.4615kPam3/kgK)(647.1K)/(22,060kPa)=(0.02m3/kg)(0.013537m3/kg)=1.4771.48

Determine the pressure of the steam using the compressibility chart.

  P=PR×PcrP=0.57×(22,060kPa)=12,574kPa

Thus, the pressure of the steam using the compressibility chart is 12,574kPa_.

(c)

To determine

The pressure of the steam using the steam table.

(c)

Expert Solution
Check Mark

Explanation of Solution

Refer to Table A-6, “Superheated water”, obtain the below properties at the specific volume 0.02m3/kg using interpolation method of two variables.

Show the pressure at 12.5MPa and 15MPa as in Table (1).

S. No

 specific volume m3/kg

(x)

Pressure, MPa

(y)

10.02003012.5
20.02y2=?
30.01567115.0

Calculate pressure at the specific volume 0.02m3/kg for liquid phase using interpolation method.

Calculate the pressure of the steam using interpolation method of two variables.

  y2=(x2x1)(y3y1)(x3x1)+y1y2=(0.015671m3/kg0.020030m3/kg)(15.0MPa12.5MPa)(0.015671m3/kg0.020030m3/kg)+12.5MPa=12.517MPa

Here, the variables denote by x and y are temperature and final specific volume.

From above calculation the pressure of 0.02m3/kg at specific volume is 12.517MPa.

Thus, the pressure of the steam using the steam table is 12.517MPa_.

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Chapter 4 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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