The final mass ( m 2 ) in the tank is 3 .90 kg _ .

Question

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Answer 1

Question

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K

=0.0556 kJ/K

Substitute 25°C for T0 and 0.0556 kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556 kJ/K)=[(25+273) K](0.0556 kJ/K)=(298 K)(0.0556 kJ/K)=16.6 kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6 kJ_.

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Answer 2

Question

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K

=0.0556 kJ/K

Substitute 25°C for T0 and 0.0556 kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556 kJ/K)=[(25+273) K](0.0556 kJ/K)=(298 K)(0.0556 kJ/K)=16.6 kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6 kJ_.

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Answer 3

Question

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K

=0.0556 kJ/K

Substitute 25°C for T0 and 0.0556 kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556 kJ/K)=[(25+273) K](0.0556 kJ/K)=(298 K)(0.0556 kJ/K)=16.6 kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6 kJ_.

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Answer 4

Question

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K

=0.0556 kJ/K

Substitute 25°C for T0 and 0.0556 kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556 kJ/K)=[(25+273) K](0.0556 kJ/K)=(298 K)(0.0556 kJ/K)=16.6 kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6 kJ_.

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Answer 5

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K

=0.0556 kJ/K

Substitute 25°C for T0 and 0.0556 kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556 kJ/K)=[(25+273) K](0.0556 kJ/K)=(298 K)(0.0556 kJ/K)=16.6 kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Answer 6

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

Answer 7

Chapter 8.8, Problem 75P

a)

To determine

The final mass (m2) in the tank is 3.90 kg_.

Answer 8

a)

Expert Solution

Answer to Problem 75P

The final temperature of the helium is 321.7 K_.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the final mass (m2) stored in the tank.

m2=Vv2 (I)

Here, volume of the rigid tank is V.

Conclusion:

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

vf=0.0008457 m3/kg and vg=0.025645 m3/kguf=94.80 kJ/kg and ug=246.82 kJ/kgsf=0.35408 kJ/kg⋅K and sg=0.91853 kJ/kg⋅K

Here, internal energy of saturated liquid and saturated vapor is uf and ug respectively, specific volume of saturated liquid and saturated vapor is vf and vg respectively, and specific entropy of saturated liquid and saturated vapor is sf and sg respectively.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

he,hf@800 kPa=95.48 kJ/kgse,sf@800 kPa=0.35408 kJ/kg⋅K

Here, enthalpy of saturated liquid is hf.

Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the following properties at the pressure of 800 kPa.

v2,vg@800 kPa=0.025645 m3/kgu2,ug@800 kPa=246.82 kJ/kgs2,sg@800 kPa=0.91853 kJ/kg⋅K

Substitute 0.1 m3 for V and 0.025645 m3/kg for v2 in equation (1).

m2=0.1 m30.025645 m3/kg=3.899 kg≅3.90 kg

Thus, the final mass (m2) in the tank is 3.90 kg_.

Answer 13

b)

To determine

The reversible work (Wrev,out) associated with the process

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K

=0.0556 kJ/K

Substitute 25°C for T0 and 0.0556 kJ/K for Sgen in Equation (VII).

Wrev,out=(25°C)(0.0556 kJ/K)=[(25+273) K](0.0556 kJ/K)=(298 K)(0.0556 kJ/K)=16.6 kJ

Thus, the reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Answer 14

b)

To determine

The reversible work (Wrev,out) associated with the process

Answer 15

b)

Expert Solution

Answer to Problem 75P

The reversible work (Wrev,out) associated with the process is 16.6 kJ_.

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the mass balance for the control volume system.

min−mout=Δmsystemme=m1−m2 (II)

Here, mass of the refrigerant at the inlet is min, mass of the refrigerant at the exit is mout, change in mass within the system is Δmsystem, initial and final mass of the refrigerant is m1 and m2 respectively, and mass at the exit is me.

Write the expression for the energy balance for the system.

Ein−Eout=ΔEsystemQin=mehe+m2u2−m1u1 (III)

Here, net energy transfer in to the control volume is Ein, net energy transfer exit from the control volume is Eout, change in internal energy of system is ΔEsystem, amount of heat transfer to the tank from the source is Qin, mass of refrigerant is m, initial internal energy is u1 and final internal energy is u2.

Write the expression for the initial mass (m1) stored in tank.

m1=mf+mg=0.3Vfvf+(1−0.3)Vgvg=0.3Vfvf+0.7Vgvg (IV)

Here, mass of the liquid refrigerant is mf, mass of the vapor refrigerant is mg, volume of the liquid refrigerant is Vf and volume of the vapor refrigerant is Vg.

Write the expression for the initial internal energy (U1) stored in tank.

U1=m1u1=mfuf+mgug=0.3Vfvf(uf)+0.7Vgvg(ug) (V)

Write the expression for the initial entropy (S1) stored in tank.

S1=m1s1=mfsf+mgsg=0.3Vfvf(sf)+0.7Vgvg(sg) (VI)

Write the expression for the entropy balance for refrigerant.

Sin−Sout+Sgen=ΔSsystemQinTsource−mese+Sgen=m2s2−m1s1Sgen=m2s2−m1s1+mese−QinTsource (VI)

Here, entropy generation is Sgen, change of entropy is ΔSsystem, entropy at inlet condition is Sin, entropy at outlet condition is Sout, and source temperature is Tsource.

Write the expression for the reversible work (Wrev,out) associated with the process.

Wrev,out=Xdestroyed+Wact,out

Here, exergy destroyed is Xdestroyed and actual work associated with the process is Wact,out.

Since, the process does not involve any actual work, substitute 0 for Wact,out.

Wrev,out=Xdestroyed+0Wrev,out=Xdestroyed=T0Sgen (VII)

Here, dead state temperature is T0.

Conclusion:

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg and 0.1 m3 for Vg in Equation (IV).

m1=0.3(0.1 m3)0.0008457 m3/kg+0.7(0.1 m3)0.025645 m3/kg=0.03 m30.0008457 m3/kg+0.07 m30.025645 m3/kg=38.202 kg

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 94.80 kJ/kg for uf, 246.82 kJ/kg for ug and 0.1 m3 for Vg in Equation (V)

U1=m1u1=0.3(0.1 m3)0.0008457 m3/kg(94.80 kJ/kg)+0.7(0.1 m3)0.025645 m3/kg(246.82 kJ/kg)=3,362.73 kJ+673.81 kJ=4,036.4 kJ

Substitute 0.1 m3 for Vf, 0.0008457 m3/kg for vf, 0.025645 m3/kg for vg, 0.35408 kJ/kg⋅K for sf, 0.91853 kJ/kg⋅K for sg, 0.1 m3 for Vg in Equation (VI)

S1=m1s1=0.3(0.1 m3)0.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.7(0.1 m3)0.025645 m3/kg(0.91853 kJ/kg⋅K)=0.03 m30.0008457 m3/kg(0.35408 kJ/kg⋅K)+0.07 m30.025645 m3/kg(0.91853 kJ/kg⋅K)=12.55 kJ/K+2.507 kJ/K

=15.067 kJ/K

Substitute 38.202 kg for m1 and 3.90 kg for m2 in Equation (II).

me=m1−m2=38.202 kg−3.90 kg=34.30 kg

Substitute 34.30 kg for me, 95.48 kJ/kg for he, 38.202 kg for m1, 3.90 kg for m2, 246.82 kJ/kg for u2 and 4036.4 kJ for m1u1 in Equation (III) .

Qin=mehe+m2u2−m1u1=[(34.30 kg)(95.48 kJ/kg)]+[(3.90 kg)(246.82 kJ/kg)]−4,036.4 kJ=3,274.96 kJ+962.598 kJ−4,036.4 kJ=201 kJ

Substitute 3.899 kg for m2, 15.067 kJ/K for m1s1, 34.30 kg for me, 0.35408 kJ/kg⋅K for se, 0.91853 kJ/kg⋅K for s2, 201 kJ for Qin and 60°C for Tsource in Equation (VI).

Sgen={[(3.899 kg)(0.91853 kJ/kg⋅K)]−15.067 kJ/K+[(34.30 kg)(0.35408 kJ/kg⋅K)]−201 kJ60°C}=3.5813 kJ/K−15.067 kJ/K+12.1449 kJ/K−201 kJ(60+273) K=0.6592 kJ/K−201 kJ333 K=0.6592 kJ/K−0.6036 kJ/K