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 8, Problem 42P

a)

To determine

The entropy change of the refrigerant.

a)

Expert Solution
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Explanation of Solution

Given:

The initial pressure (P1) is 200 kPa.

The initial quality of the steam is 40%.

The final pressure (P2) is 400 kPa.

The volume of the tank (ν) is 0.5m3.

The source temperature (Tsource) is 35°C.

Calculation:

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 200kPa.

  uf=38.26kJ/kgufg=186.25kJ/kg

  sf=0.15449kJ/kgKsfg=0.78339kJ/kgK

  vf=0.0007532m3/kgvg=0.099951m3/kg

Calculate the initial internal energy of the refrigerant (u1).

  u1=uf+x1ufg

  u1=38.26kJ/kg+(0.4)(186.25kJ/kg)=112.76kJ/kg

Calculate the initial entropy of the refrigerant (s1) .

  s1=sf+x1sfg

  s1=0.15449kJ/kgK+(0.4)(0.78339kJ/kgK)=0.4678kJ/kgK

Calculate the initial specific volume of the refrigerant (v1).

  v1=vf+x1(vgvf)

  v1=0.0007532m3/kg+(0.4)(0.099951m3/kg0.0007532m3/kg)=0.04043m3/kg

Specific volume remains constant for a rigid tank (v2=v1).

Refer the Table A-12, “Saturated refrigerant 134a– pressure table”, obtain the following properties at saturated pressure of 400kPa.

  vf=0.0007905m3/kgvg=0.051266m3/kg

  uf=63.61kJ/kgufg=171.49kJ/kg

  sf=0.24757kJ/kgKsfg=0.67954kJ/kgK

Calculate the final quality of the refrigerant using the specific volume.

  v2=vf+x2(vgvf)

  0.04043m3/kg=[0.0007905m3/kg+x2(0.051266m3/kg0.0007905m3/kg)]x2=0.040430.00079050.0512660.0007905=0.7853

Calculate the final internal energy of the refrigerant (u2) .

  u2=uf+x2ufgu2=63.61kJ/kg+(0.7853)(171.49kJ/kg)=198.29kJ/kg

Calculate the final entropy of the refrigerant (s2).

  s2=sf+x2sfgs2=0.24757kJ/kgK+(0.7853)(0.67954kJ/kgK)=0.7813kJ/kgK

Calculate the mass of the refrigerant (m).

  m=νv1m=0.5m30.04043m3/kg=12.37kg

Calculate the expression for the entropy change of the refrigerant (ΔSsystem).

  ΔSsystem=m(s2s1)ΔSsystem=12.37kg(0.7813kJ/kgK0.4678kJ/kgK)=3.878kJ/K

Thus, the entropy change of the refrigerant is 3.878kJ/K.

b)

To determine

The entropy change of the heat source

b)

Expert Solution
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Explanation of Solution

Write the expression for the energy balance equation for closed system.

  EinEout=ΔEsystem        (I)

Here, energy transfer into the control volume is Ein, energy transfer exit from the control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute Ein=Qin, Eout=0 and ΔEsystem=m(u2u1) in Equation (I).

  Qin0=m(u2u1)Qin=m(u2u1)

  Qin=(12.37kg)(198.29kJ/kg112.76kJ/kg)=1058kJ

Heat transfer for the source (Qsource) is equal to heat transfer input (Qin). The heat transfer for the source (Qsource) is equal in magnitude but opposite in direction.

  Qsource=Qin=1058kJ

Calculate the entropy change of the heat source.

  ΔSsource=QsourceTsource        (X)

  ΔSsource=1058kJ35°C=1058kJ(35+273)K=3.435kJ/K

The entropy change of the heat source is 3.435kJ/K.

c)

To determine

The total entropy change during the process.

c)

Expert Solution
Check Mark

Explanation of Solution

Calculate the total entropy change during the process.

  ΔStotal=ΔSsystem+ΔSsourceΔStotal=3.876kJ/K3.434kJ/K=0.442kJ/K

Thus, the total entropy change during the process is 0.442kJ/K.

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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What is entropy? - Jeff Phillips; Author: TED-Ed;https://www.youtube.com/watch?v=YM-uykVfq_E;License: Standard youtube license