A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine ( a ) the work done, ( b ) the heat transfer, ( c ) the exergy destroyed, and ( d ) the second-law efficiency of this process. FIGURE P8–48

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Answer 1

Textbook Question

Chapter 8.8, Problem 47P

A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the work done, (b) the heat transfer, (c) the exergy destroyed, and (d) the second-law efficiency of this process.

FIGURE P8–48

Chapter 8.8, Problem 47P, A pistoncylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20C. Heat is

(a)

Expert Solution

To determine

The work done.

Answer to Problem 47P

The work done is 0.497 kJ.

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

(b)

Expert Solution

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

(c)

Expert Solution

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

(d)

Expert Solution

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

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Answer 2

Textbook Question

Chapter 8.8, Problem 47P

A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the work done, (b) the heat transfer, (c) the exergy destroyed, and (d) the second-law efficiency of this process.

FIGURE P8–48

Chapter 8.8, Problem 47P, A pistoncylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20C. Heat is

(a)

Expert Solution

To determine

The work done.

Answer to Problem 47P

The work done is 0.497 kJ.

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

(b)

Expert Solution

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

(c)

Expert Solution

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

(d)

Expert Solution

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

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Answer 3

Textbook Question

Chapter 8.8, Problem 47P

A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the work done, (b) the heat transfer, (c) the exergy destroyed, and (d) the second-law efficiency of this process.

FIGURE P8–48

Chapter 8.8, Problem 47P, A pistoncylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20C. Heat is

(a)

Expert Solution

To determine

The work done.

Answer to Problem 47P

The work done is 0.497 kJ.

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

(b)

Expert Solution

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

(c)

Expert Solution

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

(d)

Expert Solution

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

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Answer 4

Textbook Question

Chapter 8.8, Problem 47P

A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the work done, (b) the heat transfer, (c) the exergy destroyed, and (d) the second-law efficiency of this process.

FIGURE P8–48

Chapter 8.8, Problem 47P, A pistoncylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20C. Heat is

(a)

Expert Solution

To determine

The work done.

Answer to Problem 47P

The work done is 0.497 kJ.

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

(b)

Expert Solution

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

(c)

Expert Solution

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

(d)

Expert Solution

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The work done.

Answer to Problem 47P

The work done is 0.497 kJ.

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

(b)

Expert Solution

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

(c)

Expert Solution

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

(d)

Expert Solution

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

Answer 8

(a)

Expert Solution

To determine

The work done.

Answer to Problem 47P

The work done is 0.497 kJ.

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The work done.

Answer 13

Answer to Problem 47P

The work done is 0.497 kJ.

Answer 14

Explanation of Solution

Express the boundary work done.

Wb,out=mP2(ν2−ν1) (I)

Here, mass is m, final pressure is P2 and initial and final velocity is ν1 and ν2 respectively.

Conclusion:

Perform the unit conversion of initial pressure and final pressure from kPa to MPa.

P1=100 kPa=100 kPa[MPa1000 kPa]=0.1 MPa

P2=120 kPa=120 kPa[MPa1000 kPa]=0.12 MPa

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to initial pressure of 0.1 MPa and initial temperature (T1) of 20°C.

v1=0.23373 m3/kgu1=248.81 kJ/kgs1=1.0919 kJ/kg⋅K

Here, initial specific volume, internal energy and entropy is v1,u1 and s1 respectively.

Refer Table A-13, “superheated refrigerant-134a”, and write the properties corresponding to final pressure of 0.12 MPa and final temperature (T2) of 80°C using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (II)

Here, the variables denote by x and y is final pressure and final specific volume respectively.

Show the final specific volume at 0.10 MPa and 0.14 MPa as in Table (1).

Final pressure P2 (MPa)	Final specific volume v2 (m3/kg)
0.10 (x1)	0.28465 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	0.20242 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 0.28465 m3/kg for y1 and 0.20242 m3/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(0.20242−0.28465) m3/kg(0.14−0.10) MPa+0.28465 m3/kg=0.23669 m3/kg=v2

Show the final specific internal energy at 0.10 MPa and 0.14 MPa as in Table (2).

Final pressure P2 (MPa)	Final specific internal energy u2 (kJ/kg)
0.10 (x1)	297.10 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	296.77 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 297.10 kJ/kg for y1 and 296.77 kJ/kg for y3 in Equation (II).

y2=(0.12−0.10) MPa(296.77−297.10) kJ/kg(0.14−0.10) MPa+297.10 kJ/kg=296.94 kJ/kg=u2

Show the final specific entropy at 0.10 MPa and 0.14 MPa as in Table (3).

Final pressure P2 (MPa)	Final specific entropy s2 (kJ/kg⋅K)
0.10 (x1)	1.2573 (y1)
0.12 (x2)	(y2=?)
0.14 (x3)	1.2289 (y3)

Substitute 0.10 MPa, 0.12 MPa and 0.14 MPa for x1, x2 and x3 respectively, 1.2573 kJ/kg⋅K for y1 and 1.2289 kJ/kg⋅K for y3 in Equation (II).

y2=(0.12−0.10) MPa(1.2289−1.2573) kJ/kg⋅K(0.14−0.10) MPa+1.2573 kJ/kg⋅K=1.2419 kJ/kg⋅K=s2

Thus, write the values obtained from interpolation method:

v2=0.23669 m3/kgu2=296.94 kJ/kgs2=1.2419 kJ/kg⋅K

Substitute 1.4 kg for m, 120 kPa for P2, 0.23669 m3/kg for v2 and 0.23373 m3/kg for v1 in Equation (I).

Wb,out=(1.4 kg)(120 kPa)(0.23669−0.23373) m3/kg=0.497 kJ

Hence, the work done is 0.497 kJ.

Answer 15

(b)

Expert Solution

To determine

The heat transfer.

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The heat transfer.

Answer 20

Answer to Problem 47P

The heat transfer is 67.9 kJ.

Answer 21

Explanation of Solution

Express heat transfer.

Qin=m(u2−u1)+Wb,out (III)

Conclusion:

Substitute EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 8.8, Problem 47P , additional homework tip 1 for , 0.497 kJ for Wb,out, 296.94 kJ/kg for u2 and 248.81 kJ/kg for u1 in Equation (III).

Qin=(1.4 kg)(296.94 kJ/kg−248.81 kJ/kg)+0.497 kJ=67.9 kJ

Answer 22

(c)

Expert Solution

To determine

The exergy destroyed.

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The exergy destroyed.

Answer 27

Answer to Problem 47P

The exergy destroyed is 14.8 kJ.

Answer 28

Explanation of Solution

Express the exergy destruction.

Xdestroyed=T0Sgen (IV)

Here, entropy generation is Sgen and surrounding temperature is T0.

Express the entropy generation by taking entropy balance on an extended system.

Sin−Sout+Sgen=ΔSsystemQinTsource+Sgen=m(s2−s1)Sgen=m(s2−s1)−QinTsource (V)

Here, net entropy transfer by heat and mass is Sin−Sout, entropy generation is Sgen, change in entropy of system is ΔSsystem and source temperature is Tsource.

Conclusion:

Substitute 1.4 kg for m, 1.2419 kJ/kg⋅K for s2, 1.0919 kJ/kg⋅K for s1, 67.9 kJ for Qin and 150°C for Tsource in Equation (V).

Sgen=(1.4 kg)(1.2419−1.0919) kJ/kg⋅K−67.9 kJ150°C=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ(150+273) K=(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K

Substitute (1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K for Sgen and 25°C for T0 in Equation (IV).

Xdestroyed=(25°C)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(25+273) K[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=(298 K)[(1.4 kg)(0.15 kJ/kg⋅K)−67.9 kJ423 K]=14.8 kJ

Hence, the exergy destroyed is 14.8 kJ.

Answer 29

(d)

Expert Solution

To determine

The second law efficiency of the process.

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The second law efficiency of the process.

Answer 34

Answer to Problem 47P

The second law efficiency of the process is 26.2%.

Answer 35

Explanation of Solution

Express exergy expended.

Xexpended=ηth,revQin=[1−TLTH]Qin (VI)

Here, efficiency for reversible cycle is ηth,rev, temperature of low source is TL and temperature of high source is TH.

Express the second law efficiency.

ηII=1−XdestroyedXexpended×100% (VII)

Conclusion:

Substitute 25°C for TL and 150°C for TH and 67.9 kJ for Qin in Equation (VI).

Xexpended=[1−25°C150°C](67.9 kJ)=[1−(25+273) K(150+273) K](67.9 kJ)=[1−298 K423 K](67.9 kJ)=20.06 kJ

Substitute 20.06 kJ for Xexpended and 14.8 kJ for Xdestroyed in Equation (VII).

ηII=1−14.8 kJ20.06 kJ×100%=0.262×100%=26.2%

Hence, the second law efficiency of the process is 26.2%.

Answer 36

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