The process that causes the greatest amount of exergy destruction.

Question

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Answer 1

Question

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

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Answer 2

Question

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

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Answer 3

Question

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

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Answer 4

Question

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

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Answer 5

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Answer 6

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Answer 7

Chapter 11.10, Problem 117RP

To determine

The process that causes the greatest amount of exergy destruction.

Answer 8

Expert Solution & Answer

Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Show the T-s diagram for refrigeration system as in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 11.10, Problem 117RP

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

h3≅h4 (I)

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

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2s−h1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1 and h2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700 kPa−Tsubcooling (III)

Here, saturation temperature at pressure of 700 kPa is Tsat@700 kPa and temperature of sub-cooling is Tsubcooling.

Express quality at state 4.

x4=h4−hf@−10°Chfg@−10°C (IV)

Here, specific enthalpy at saturation liquid and temperature of −10°C is hf@−10°C and specific enthalpy at evaporation and temperature of −10°C is hfg@−10°C.

Express specific entropy at state 4.

s4=sf@-10°C+x4sfg@-10°C (V)

Here, specific entropy at saturation liquid and temperature of −10°C is sf@−10°C and specific entropy at evaporation and temperature of −10°C is sfg@−10°C.

Express heat rejected in the evaporator.

qL=h1−h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2−h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2−s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1 and s2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3−s2+qHTH] (IX)

Here, specific entropy at state 3 is s3 and temperature of ambient air is TH.

Express the exergy destruction during process 3-4.

xdestroyed,3-4=T0(s4−s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1−s4−qLTL] (XI)

Here, freezing temperature of water is TL.

Conclusion:

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to initial temperature (T1) 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.

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

P2=700 kPa[MPa1000 kPa]=0.7 MPa

Refer Table A-13, “superheated refrigerant 134a”, and write the specific enthalpy at state 2s corresponding to pressure at state 2 of 0.7 MPa and specific entropy at state 2 (s2=s1) of 0.9378 kJ/kg⋅K using interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XII)

Here, the variables denote by x and y is specific entropy at state 2 and specific enthalpy at state 2s respectively.

Show the specific enthalpy at state 2s corresponding to specific entropy as in Table (1).

Specific entropy at state 2 s2(kJ/kg⋅K)	Specific enthalpy at state 2s h2s(kJ/kg)
0.9314 (x1)	268.47 (y1)
0.9378 (x2)	(y2=?)
0.9642 (x3)	278.59 (y3)

Substitute 0.9314 kJ/kg⋅K, 0.9378 kJ/kg⋅K and 0.9642 kJ/kg⋅K for x1, x2 and x3 respectively, 268.47 kJ/kg for y1 and 278.59 kJ/kg for y3 in Equation (XII).

y2=[(0.9378−0.9314) kJ/kg⋅K][(278.59−268.47) kJ/kg](0.9642−0.9314) kJ/kg⋅K+268.47 kJ/kg=270.45 kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45 kJ/kg

Refer Table A-12, “saturated refrigerant-134a-pressure table”, and write the saturated temperature corresponding to pressure of 700 kPa.

Tsat@900 kPa=26.7°C

Substitute 26.7°C for Tsat@700 kPa and 2.7°C for Tsubcooling in Equation (III).

T3=26.7°C−2.7°C=24°C

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the specific enthalpy and entropy at state 3 corresponding to temperature at state 3 (T3) of 24°C.

h3=hf=84.98 kJ/kgs3=sf=0.3195 kJ/kg⋅K

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

Substitute 84.98 kJ/kg for h3 in Equation (I).

h4=84.98 kJ/kg

Substitute 244.55 kJ/kg for h1, 270.45 kJ/kg for h2s, and 0.85 for ηC in Equation (II).

h2=244.55 kJ/kg+270.45 kJ/kg−244.55 kJ/kg0.85=275.02 kJ/kg

Refer Table A-11, “saturated refrigerant-134a-temperature table”, and write the properties corresponding to final temperature (T4) of −10°C.

hf@−10°C=38.53 kJ/kghfg@−10°C=206.02 kJ/kgsf@-10°C=0.1549 kJ/kg⋅Ksfg@-10°C=0.7828 kJ/kg⋅K

Substitute 84.98 kJ/kg for h4, 38.53 kJ/kg for hf@−10°C and 206.02 kJ/kg for hfg@−10°C in Equation (IV).

x4=84.98 kJ/kg−38.53 kJ/kg206.02 kJ/kg=0.225

Substitute 0.1549 kJ/kg⋅K for sf@-10°C, 0.7828 kJ/kg⋅K for sfg@-10°C and 0.225 for x4 in Equation (V).

s4=(0.1549 kJ/kg⋅K)+(0.225)(0.7828 kJ/kg⋅K)=0.3314 kJ/kg⋅K

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7 MPa and specific enthalpy at state 2 of 275.02 kJ/kg using an interpolation method.

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

Specific enthalpy at state 1 h2(kJ/kg)	Specific entropy at state 2 s2(kJ/kg⋅K)
268.47 (x1)	0.9314 (y1)
275.02 (x2)	(y2=?)
278.59 (x3)	0.9642 (y3)

Use excels and tabulates the values from Table (2) in Equation (XII) to get,

s2=0.9526 kJ/kg⋅K

Substitute 244.55 kJ/kg for h1 and 84.98 kJ/kg for h4 in Equation (VI).

qL=244.55 kJ/kg−84.98 kJ/kg=159.57 kJ/kg

Substitute 275.02 kJ/kg for h2 and 84.98 kJ/kg for h3 in Equation (VII).

qH=275.02 kJ/kg−84.98 kJ/kg=190.04 kJ/kg

Perform unit conversion of temperature from °C to K

T0=TH=22°C=(22+273) K=295 K

Substitute 295 K for T0, 0.9526 kJ/kg⋅K for s2 and 0.9378 kJ/kg⋅K for s1 in Equation (VIII).

xdestroyed,1-2=295 K[(0.9526−0.9378) kJ/kg⋅K]=4.369 kJ/kg

Substitute 295 K for T0, 0.3195 kJ/kg⋅K for s3, 0.9526 kJ/kg⋅K for s2, 190.04 kJ/kg for qH and 295 K for TH in Equation (IX).

xdestroyed,2-3=(295 K)[(0.3195−0.9526) kJ/kg⋅K+190.04 kJ/kg295 K]=3.293 kJ/kg

Substitute 295 K for T0, 0.3314 kJ/kg⋅K for s4 and 0.3195 kJ/kg⋅K for s3 in Equation (X).

xdestroyed,3-4=295 K[(0.3314−0.3195) kJ/kg⋅K]=3.5105 kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273) K=273 K

Substitute 295 K for T0, 0.9378 kJ/kg⋅K for s1, 0.3314 kJ/kg⋅K for s4, 159.57 kJ/kg for qL and 273 K for TL in Equation (XI).

xdestroyed,4-1=(295 K)[(0.9378−0.3314) kJ/kg⋅K−159.57 kJ/kg273 K]=6.44 kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process 4−1(6.44 kJ/kg).

Answer 12

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The process that causes the greatest amount of exergy destruction.

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The process that causes the greatest amount of exergy destruction.

Answer to Problem 117RP

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