A 0.5-m 3 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.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

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

To determine

The entropy change of the refrigerant.

Answer to Problem 42P

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

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

b)

Expert Solution

To determine

The entropy change of the heat source

Answer to Problem 42P

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

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

c)

Expert Solution

To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441 kJ/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 3.876 kJ/K for ΔSsystem, and −3.434 kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Solve for the support reactions at A and B. C 3 kN/m B -1.5 m- -1.5 m 1.5 m- 1.5 m-

4. Solve for the support reactions at A and B. W1 600 lb/ft W2 150 lb/ft A B

In cold isostatic pressing, the mold is most typically made of which one of the following: thermosetting polymer tool steel sheet metal textile rubber

Answer 2

Textbook Question

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

To determine

The entropy change of the refrigerant.

Answer to Problem 42P

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

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

b)

Expert Solution

To determine

The entropy change of the heat source

Answer to Problem 42P

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

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

c)

Expert Solution

To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441 kJ/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 3.876 kJ/K for ΔSsystem, and −3.434 kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

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

To determine

The entropy change of the refrigerant.

Answer to Problem 42P

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

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

b)

Expert Solution

To determine

The entropy change of the heat source

Answer to Problem 42P

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

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

c)

Expert Solution

To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441 kJ/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 3.876 kJ/K for ΔSsystem, and −3.434 kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

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

To determine

The entropy change of the refrigerant.

Answer to Problem 42P

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

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

b)

Expert Solution

To determine

The entropy change of the heat source

Answer to Problem 42P

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

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

c)

Expert Solution

To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441 kJ/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 3.876 kJ/K for ΔSsystem, and −3.434 kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a)

Expert Solution

To determine

The entropy change of the refrigerant.

Answer to Problem 42P

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

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

b)

Expert Solution

To determine

The entropy change of the heat source

Answer to Problem 42P

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

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

c)

Expert Solution

To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441 kJ/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 3.876 kJ/K for ΔSsystem, and −3.434 kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Answer 8

a)

Expert Solution

To determine

The entropy change of the refrigerant.

Answer to Problem 42P

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

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The entropy change of the refrigerant.

Answer 13

Answer to Problem 42P

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

Answer 14

Explanation of Solution

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

v1=vf+x1(vg−vf) (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(vg−vf) (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(s2−s1) (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 200 kPa.

uf=38.26 kJ/kgufg=186.25 kJ/kg

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

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

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

sf=0.15449 kJ/kg⋅Ksfg=0.78339 kJ/kg⋅K

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

s1=0.15449 kJ/kg⋅K+(0.4)(0.78339 kJ/kg⋅K)=0.4678 kJ/kg⋅K

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

vf=0.0007532m3/kgvg=0.099951 m3/kg

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

v1=0.0007532m3/kg+(0.4)(0.099951 m3/kg−0.0007532m3/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 400 kPa.

vf=0.0007905 m3/kgvg=0.051266 m3/kg

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

0.04043m3/kg=0.0007905 m3/kg+x2(0.051266 m3/kg−0.0007905 m3/kg)x2=0.04043−0.00079050.051266−0.0007905 =0.7853

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

uf=63.61 kJ/kgufg=171.49kJ/kg

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

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

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

sf=0.24757 kJ/kg⋅Ksfg=0.67954 kJ/kg⋅K

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

s2=0.24757 kJ/kg⋅K+(0.7853)(0.67954 kJ/kg⋅K)=0.7813 kJ/kg⋅K

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

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

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

ΔSsystem=12.37 kg(0.7813 kJ/kg⋅K−0.4678 kJ/kg⋅K)=3.876 kJ/K

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

Answer 15

b)

Expert Solution

To determine

The entropy change of the heat source

Answer to Problem 42P

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

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The entropy change of the heat source

Answer 20

Answer to Problem 42P

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

Answer 21

Explanation of Solution

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

Ein−Eout=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(u2−u1) in Equation (IX).

Qin−0=m(u2−u1)Qin=m(u2−u1) (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, 198.29 kJ/kg for u2 and 112.76 kJ/kg for u1 in Equation (XI).

Qin=(12.37 kg)(198.29 kJ/kg−112.76 kJ/kg)=1058 kJ

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=−1058 kJ

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

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

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

Answer 22

c)

Expert Solution

To determine

The total entropy change during the process.

Answer to Problem 42P

The total entropy change during the process is 0.441 kJ/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 3.876 kJ/K for ΔSsystem, and −3.434 kJ/K for ΔSsource in Equation (XII).

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Answer 23

c)

Expert Solution

Answer 24

c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The total entropy change during the process.

Answer 27

Answer to Problem 42P

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

Answer 28

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

ΔStotal=3.876 kJ/K−3.434 kJ/K=0.441 kJ/K

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

Answer 29

Ch. 7.13 - Prob. 1P Ch. 7.13 - Does the cyclic integral of heat have to be zero...Ch. 7.13 - Is a quantity whose cyclic integral is zero...Ch. 7.13 - Prob. 4P Ch. 7.13 - Prob. 5P Ch. 7.13 - How do the values of the integral 12Q/T compare...Ch. 7.13 - The entropy of a hot baked potato decreases as it...Ch. 7.13 - When a system is adiabatic, what can be said about...Ch. 7.13 - Prob. 9P Ch. 7.13 - A pistoncylinder device contains helium gas....

Concept explainers

Videos

Answer to Problem 42P

Explanation of Solution

Answer to Problem 42P

Explanation of Solution

Answer to Problem 42P

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions