EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 7.13, Problem 42P

A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 35°C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.

a)

Expert Solution
Check Mark
To determine

The entropy change of the refrigerant.

Answer to Problem 42P

The entropy change of the refrigerant is 3.876kJ/K.

Explanation of Solution

Write the expression to calculate initial specific volume of the refrigerant.

v1=vf+x1(vgvf) (I)

Here, initial specific volume is v1, saturated liquid specific volume is vf, initial vapor quality is x1 and saturated vapor specific volume is vg.

Write the expression to calculate the initial internal energy of the refrigerant.

u1=uf+x1ufg (II)

Here, initial internal energy is u1, saturated liquid internal energy is uf, initial vapor quality is x1  and evaporated internal energy is ufg.

Write the expression to calculate the initial entropy of the refrigerant.

s1=sf+x1sfg (III)

Here, initial entropy is s1, saturated liquid entropy is sf, initial vapor quality is x1 and evaporated entropy is sfg.

Write the expression to calculate the final specific volume of the refrigerant.

v2=vf+x2(vgvf) (IV)

Here, final specific volume is v2, saturated liquid specific volume is vf, final vapor quality is x2 and saturated vapor specific volume is vg.

Write the expression to calculate the final internal energy of the refrigerant.

u2=uf+x2ufg (V)

Here, final internal energy is u2, saturated liquid internal energy is uf, final vapor quality is x2 and evaporated internal energy is ufg.

Write the expression to calculate the final entropy of the refrigerant.

s2=sf+x2sfg (VI)

Here, final entropy is s2, saturated liquid entropy is sf, final vapor quality is x2 and evaporated entropy is sfg.

Write the expression to calculate the mass of the refrigerant.

m=νv1 (VII)

Here, mass of the refrigerant is m, volume of the tank is ν and initial specific volume is v1.

Write the expression to calculate the expression for the entropy change of the refrigerant.

ΔSsystem=m(s2s1) (VIII)

Here, entropy change of the refrigerant is ΔSsystem, mass of the refrigerant is m, initial entropy is s1 and final entropy is s2.

Conclusion:

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

uf=38.26kJ/kgufg=186.25kJ/kg

Substitute 38.26kJ/kg for uf, 0.4 for x1 and 186.25kJ/kg for ufg in Equation (II).

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

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

sf=0.15449kJ/kgKsfg=0.78339kJ/kgK

Substitute 0.15449kJ/kgK for sf, 0.4 for x1 and 0.78339kJ/kgK for sfg in Equation (III).

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

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

vf=0.0007532m3/kgvg=0.099951m3/kg

Substitute 0.0007532m3/kg for vf, 0.4 for x1 and 0.099951m3/kg for vg in Equation (I).

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

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

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

vf=0.0007905m3/kgvg=0.051266m3/kg

Substitute 0.04043m3/kg for v2, 0.0007905m3/kg for vf and 0.051266m3/kg for vg in Equation (IV).

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

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

uf=63.61kJ/kgufg=171.49kJ/kg

Substitute 63.61kJ/kg for uf, 0.7853 for x2 and 171.49kJ/kg for ufg in Equation (V).

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

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

sf=0.24757kJ/kgKsfg=0.67954kJ/kgK

Substitute 0.24757kJ/kgK for sf, 0.7853 for x2 and 0.67954kJ/kgK for sfg in Equation (VI).

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

Substitute 0.5m3 for ν and 0.04043m3/kg for v1 in Equation (VII).

m=0.5m30.04043m3/kg=12.37kg

Substitute 12.37 kg for m, 0.7813kJ/kgK for s2 and 0.4678kJ/kgK for s1 in Equation (VIII).

ΔSsystem=12.37kg(0.7813kJ/kgK0.4678kJ/kgK)=3.876kJ/K

Thus, the entropy change of the refrigerant is 3.876kJ/K.

b)

Expert Solution
Check Mark
To determine

The entropy change of the heat source

Answer to Problem 42P

The entropy change of the heat source is 3.434kJ/K.

Explanation of Solution

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

EinEout=ΔEsystem (IX)

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.

Write the expression to calculate the entropy change of the heat source.

ΔSsource=QsourceTsource (X)

Here, source temperature is Tsource.

Conclusion:

Substitute Qin for Ein, 0 for Eout and m(u2u1) in Equation (IX).

Qin0=m(u2u1)Qin=m(u2u1) (XI)

Here, heat transfer input is Qin, mass of the refrigerant is m, final internal energy is u2, and initial internal energy is u1.

Substitute 12.37 kg for m, 198.29kJ/kg for u2 and 112.76kJ/kg for u1 in Equation (XI).

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

Substitute –1058 kJ for Qsource and 35°C for Tsource in Equation (X).

ΔSsource=1058kJ35°C=1058kJ(35+273)K=3.434kJ/K

The entropy change of the heat source is 3.434kJ/K.

c)

Expert Solution
Check Mark
To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441kJ/K.

Explanation of Solution

Write the expression for the total entropy change during the process.

ΔStotal=ΔSsystem+ΔSsource (XII)

Here, total entropy change during the process is ΔStotal.

Conclusion:

Substitute 3.876kJ/K for ΔSsystem, and 3.434kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876kJ/K3.434kJ/K=0.441kJ/K

Thus, the total entropy change during the process is 0.441kJ/K.

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Chapter 7 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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