The cooling load and the COP.

Question

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Answer 1

Question

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).

X˙evap=(135.96−244.55) kJ/kg−(298 K)(0.5252−0.9378) kJ/kg⋅K=−108.59 kJ/kg−(298 K)(−0.4126) kJ/kg⋅K=14.37 kJ/kg

Substitute 14.37 kJ/kg for X˙evap and 6.56 kJ/kg for Exdest,evap in Equation (XIX).

ηII,evap=1−6.56 kJ/kg14.37 kJ/kg×100%=54.4%

Hence, the second-law efficiency of the evaporator is 54.4%.

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Answer 2

Question

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).

X˙evap=(135.96−244.55) kJ/kg−(298 K)(0.5252−0.9378) kJ/kg⋅K=−108.59 kJ/kg−(298 K)(−0.4126) kJ/kg⋅K=14.37 kJ/kg

Substitute 14.37 kJ/kg for X˙evap and 6.56 kJ/kg for Exdest,evap in Equation (XIX).

ηII,evap=1−6.56 kJ/kg14.37 kJ/kg×100%=54.4%

Hence, the second-law efficiency of the evaporator is 54.4%.

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Answer 3

Question

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).

X˙evap=(135.96−244.55) kJ/kg−(298 K)(0.5252−0.9378) kJ/kg⋅K=−108.59 kJ/kg−(298 K)(−0.4126) kJ/kg⋅K=14.37 kJ/kg

Substitute 14.37 kJ/kg for X˙evap and 6.56 kJ/kg for Exdest,evap in Equation (XIX).

ηII,evap=1−6.56 kJ/kg14.37 kJ/kg×100%=54.4%

Hence, the second-law efficiency of the evaporator is 54.4%.

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Answer 4

Question

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).

X˙evap=(135.96−244.55) kJ/kg−(298 K)(0.5252−0.9378) kJ/kg⋅K=−108.59 kJ/kg−(298 K)(−0.4126) kJ/kg⋅K=14.37 kJ/kg

Substitute 14.37 kJ/kg for X˙evap and 6.56 kJ/kg for Exdest,evap in Equation (XIX).

ηII,evap=1−6.56 kJ/kg14.37 kJ/kg×100%=54.4%

Hence, the second-law efficiency of the evaporator is 54.4%.

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Answer 5

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).

X˙evap=(135.96−244.55) kJ/kg−(298 K)(0.5252−0.9378) kJ/kg⋅K=−108.59 kJ/kg−(298 K)(−0.4126) kJ/kg⋅K=14.37 kJ/kg

Substitute 14.37 kJ/kg for X˙evap and 6.56 kJ/kg for Exdest,evap in Equation (XIX).

ηII,evap=1−6.56 kJ/kg14.37 kJ/kg×100%=54.4%

Hence, the second-law efficiency of the evaporator is 54.4%.

Answer 6

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

Answer 7

Chapter 11.10, Problem 32P

(a)

To determine

The cooling load and the COP.

Answer 8

(a)

Expert Solution

Answer to Problem 32P

The cooling load and the COP is 108.6 kJ/kg and 2.506 respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Show the T-s diagram for ideal vapor-compression refrigeration cycle as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 32P

From Figure (1), write the specific enthalpy at state 3 is equal to state 4 due to throttling process.

h3≅h4

Here, specific enthalpy at state 3 is h3.

Express the heat removed from the cooled space.

qL=h1−h4 (I)

Here, specific enthalpy at state 1, 3 and 4 is h1,h3 and h4 respectively.

Express heat supplied to the cooled space.

qH=h2−h3 (II)

Here, specific enthalpy at state 2 is h2.

Express the work input.

win=h2−h1 (III)

Express the COP of the cycle.

COP=qLwin (IV)

Express pressure at state 2 and state 3.

P2=P3=Psat@57.9°C (V)

Here, pressure at state 2 and 3 is P2 and P3 respectively and saturated pressure at temperature of 57.9°C is Psat@57.9°C.

Express quality at state 4.

h4=hf@-10°C+x4hfg@-10°C (VI)

Here, specific enthalpy at saturated liquid and evaporation and −10°C is hf@−10°C and hfg@−10°C respectively.

Express specific entropy at state 4.

s4=sf@−10°C+x4sfg@−10°C (VII)

Here, specific entropy at saturated liquid and evaporation and −10°C is sf@−10°C and sfg@−10°C respectively.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature of −10°C.

h1=hg=244.55 kJ/kgs1=sg=0.9378 kJ/kg⋅K

Here, specific entropy at state 1 is s1, specific enthalpy and entropy at saturated vapor is hg and sg respectively.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the pressure state 2 and 3 corresponding to temperature of 57.9°C using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (VIII)

Here, the variables denote by x and y is temperature and saturated pressure respectively.

Show the saturated pressure corresponding to temperature as in Table (1).

Temperature (°C)	Saturated pressure P2=P3(kPa)
56 (x1)	1529.1 (y1)
57.9 (x2)	(y2=?)
60 (x3)	1682.8 (y3)

Substitute 56 °C, 57.9 °C and 60 °C for x1, x2 and x3 respectively, 1529.1 kPa for y1 and 1682.8 kPa for y3 in Equation (VIII).

y2=(57.9°C−56°C)(1682.8 kPa−1529.1 kPa)(60°C−56°C)+1529.1 kPa=1600 kPa=Psat@57.9°C

Substitute 1600 kPa for Psat@57.9°C in Equation (V).

P2=P3=1600 kPa

Perform unit conversion of pressure at state 2 from kPa to MPa.

P2=1600 kPa[MPa1000 kPa]=1.6 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2 corresponding to pressure at state 2 of 1.6 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Show the specific enthalpy at state 2 corresponding to specific entropy as in Table (2).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2 h2(kJ/kg)
0.9164 (x1)	280.71 (y1)
0.9378 (x2)	(y2=?)
0.9536 (x3)	293.27 (y3)

Use excels and substitutes the value from Table (2) in Equation (VIII) to obtain the specific enthalpy at state 2.

h2=287.89 kJ/kg

Refer Table A-12, “saturated refrigerant 134a-pressure table”, and write the properties corresponding to pressure at state 3 of 1600 kPa.

h3=h4=hf=135.96 kJ/kg

s3=sf=0.4792 kJ/kg⋅K

Here, specific enthalpy and entropy at saturated liquid is hf and sf respectively and specific entropy at state 3 is s3.

Refer Table A-11, “saturated refrigerant-134a-tempertaure table”, and write the properties corresponding to temperature of −10°C.

hf@-10°C=38.53 kJ/kghfg@-10°C=206.02 kJ/kgsf@−10°C=0.15496 kJ/kg⋅Ksfg@−10°C=0.7828 kJ/kg⋅K

Substitute 135.96 kJ/kg for h4, 38.53 kJ/kg for hf@-10°C and 206.02 kJ/kg for hfg@-10°C in Equation (VI).

135.96 kJ/kg=38.53 kJ/kg+x4(206.02 kJ/kg)x4=0.4729

Substitute 0.15496 kJ/kg⋅K and 0.7828 kJ/kg⋅K for sf@−10°C and sfg@−10°C respectively and 0.4729 for x4 in Equation (VII).

s4=0.15496 kJ/kg⋅K+(0.4729)(0.7828 kJ/kg⋅K)=0.5252 kJ/kg⋅K

Here, specific entropy at state 4 is s4.

Substitute 224.55 kJ/kg for h1 and 135.96 kJ/kg for h4 in Equation (I).

qL=244.55 kJ/kg−135.96 kJ/kg=108.6 kJ/kg

Hence, the cooling load is 108.6 kJ/kg.

Substitute 287.89 kJ/kg for h2 and 135.96 kJ/kg for h3 in Equation (II).

qH=287.89 kJ/kg−135.96 kJ/kg=151.9 kJ/kg

Substitute 287.89 kJ/kg for h2 and 244.55 kJ/kg for h1 in Equation (III).

win=287.89 kJ/kg−244.55 kJ/kg=43.34 kJ/kg

Substitute 108.6 kJ/kg for qL and 43.34 kJ/kg for win in Equation (IV).

COP=108.6 kJ/kg43.34 kJ/kg=2.506

Hence, the COP of the cycle is 2.506.

Answer 13

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

Answer 14

(b)

To determine

The exergy destruction in each component of the cycle and the total exergy destruction in the cycle.

Answer 15

(b)

Expert Solution

Answer to Problem 32P

The exergy destruction in compressor is 0, condenser is 15.27 kJ/kg, expansion valve is 13.69 kJ/kg, evaporator is 6.56 kJ/kg and the total exergy destruction in the cycle is 35.52 kJ/kg.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

For compressor:

Express the exergy destruction in compressor.

Exdest,comp=T0sgen,1−2=T0(s2−s1) (IX)

Here, surrounding temperature is T0, entropy generation during process 1-2 is sgen,1−2.

For condenser:

Express the exergy destruction in condenser.

Exdest,cond=T0sgen,2−3=T0[(s3−s2)+qHTH] (X)

Here, entropy generation during process 2-3 is sgen,2−3 and high temperature medium is TH.

For expansion valve:

Exdest,exp val=T0sgen,3−4=T0(s4−s3) (XI)

For evaporator:

Express the exergy destruction in evaporator.

Exdest,evap=T0sgen,4−1=T0[(s1−s4)−qLTL] (XII)

Here, entropy generation during process 4-1 is sgen,4−1 and low temperature medium is TL.

Express the total exergy destruction in the cycle.

E˙xdest,total=Exdest,comp+Exdest,cond+Exdest,exp val+Exdest,evap (XIII)

Conclusion:

Perform unit conversion of surrounding temperature from °C to K.

T0=25°C=(25+273) K=298 K

Perform unit conversion of high temperature medium from °C to K.

TH=25°C=(25+273) K=298 K

Perform unit conversion of low temperature medium from °C to K.

TL=5°C=(5+273) K=278 K

Substitute 298 K for T0 and 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (IX).

Exdest,comp=(298 K)(0.9378−0.9378) kJ/kg⋅K=0

Hence, the exergy destruction in compressor is 0.

Substitute 298 K for T0, 0.4792 kJ/kg⋅K and 0.9378 kJ/kg⋅K for s3 and s2 respectively, 151.9 kJ/kg for qH and 298 K for TH in Equation (X).

Exdest,cond=(298 K)[(0.4792−0.9378) kJ/kg⋅K+151.9 kJ/kg298 K]=(298 K)(0.05125 kJ/kg⋅K)=15.27 kJ/kg

Hence, the exergy destruction in condenser is 15.27 kJ/kg.

Substitute 298 K for T0 and 0.5252 kJ/kg⋅K and 0.4792 kJ/kg⋅K for s4 and s3 respectively in Equation (XI).

Exdest,exp val=(298 K)(0.5252−0.4792) kJ/kg⋅K=13.69 kJ/kg

Hence, the exergy destruction in expansion valve is 13.69 kJ/kg.

Substitute 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 respectively, 108.6 kJ/kg for qH and 278 K for TL in Equation (XII).

Exdest,evap=(298 K)[(0.9378−0.5252) kJ/kg⋅K+108.6 kJ/kg278 K]=(298 K)(0.02202 kJ/kg⋅K)=6.56 kJ/kg

Hence, the exergy destruction in evaporator is 6.56 kJ/kg.

Substitute 0 for Exdest,comp, 15.27 kJ/kg for Exdest,cond, 13.69 kJ/kg for Exdest,exp val and 6.56 kJ/kg for Exdest,evap in Equation (XIII).

E˙xdest,total=0+15.27 kJ/kg+13.69 kJ/kg+6.56 kJ/kg=35.52 kJ/kg

Hence, the total exergy destruction in the cycle is 35.52 kJ/kg.

Answer 20

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).

X˙evap=(135.96−244.55) kJ/kg−(298 K)(0.5252−0.9378) kJ/kg⋅K=−108.59 kJ/kg−(298 K)(−0.4126) kJ/kg⋅K=14.37 kJ/kg

Substitute 14.37 kJ/kg for X˙evap and 6.56 kJ/kg for Exdest,evap in Equation (XIX).

ηII,evap=1−6.56 kJ/kg14.37 kJ/kg×100%=54.4%

Hence, the second-law efficiency of the evaporator is 54.4%.

Answer 21

(c)

To determine

The second-law efficiency of the compressor, the evaporator, and the cycle.

Answer 22

(c)

Expert Solution

Answer to Problem 32P

The second-law efficiency of the compressor is 100%, the evaporator is 54.4%, and the cycle is 18%.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Express the exergy of the heat transferred from the low temperature medium.

ExqL=−qL[1−T0TL] (XIV)

Determine the second law efficiency of the cycle.

ηII=ExqLwin×100% (XV)

Express the total exergy destruction in the cycle.

Exdest,total=win−ExqL (XVI)

Express the second law efficiency of the compressor.

ηII,comp=W˙revW˙act,in=m˙[h2−h1−T0(s2−s1)]m˙(h2−h1)×100% (XVII)

Here, rate of work done on reversible process is W˙rev and rate of actual work input is W˙act,in.

Express the exergy difference in evaporator.

X˙evap=X˙4−X˙1=h4−h1−T0(s4−s1) (XVIII)

Here, rate of exergy difference during process 1-4 is X˙4−X˙1.

Express the second law efficiency of the evaporator.

ηII,evap=1−Exdest,evapX˙evap×100% (XIX)

Conclusion:

Substitute 108.6 kJ/kg for qL, 298 K and 278 K for T0 and TL respectively in Equation (XIV).

ExqL=−(108.6 kJ/kg)[1−298 K278 K]=7.813 kJ/kg

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XV).

ηII=7.813 kJ/kg43.34 kJ/kg×100%=18%

Hence, the second-law efficiency of the cycle is 18%.

Substitute 7.813 kJ/kg for ExqL and 43.34 kJ/kg for win in Equation (XVI).

Exdest,total=43.34 kJ/kg−7.813 kJ/kg=35.53 kJ/kg

Substitute 0.9378 kJ/kg⋅K for s2 and s1 respectively in Equation (XVII).

ηII,comp=m˙[h2−h1−T0(0.9378−0.9378) kJ/kg⋅K]m˙(h2−h1)×100%=m˙[h2−h1]m˙(h2−h1)×100%=1×100%=100%

Hence, the second-law efficiency of the compressor is 100%.

Substitute 135.96 kJ/kg and 244.55 kJ/kg for h4 and h1 respectively, 298 K for T0, 0.9378 kJ/kg⋅K and 0.5252 kJ/kg⋅K for s1 and s4 in Equation (XVIII).