The greatest amount of exergy destroyed among the processes of system.

Question

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Answer 1

Question

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94 kW.

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Answer 2

Question

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94 kW.

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Answer 3

Question

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94 kW.

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Answer 4

Question

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94 kW.

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Answer 5

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Answer 6

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Answer 7

Chapter 11.10, Problem 117RP

To determine

The greatest amount of exergy destroyed among the processes of system.

Answer 8

Expert Solution & Answer

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94 kW.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 1

Sketch the T−s diagram of the two-stage compression refrigeration as in Figure (2).

EBK THERMODYNAMICS: AN ENGINEERING APPR, Chapter 11.10, Problem 117RP , additional homework tip 2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8−h5)=m˙2(h1−h4)

m˙6=m˙2h1−h4h8−h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=h−hfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7−h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2−h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(se−si−qinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60 kJ/kgs1=sg=0.92711 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400 kPa and specific entropy (s2) of 0.92711 kJ/kg⋅K as 281.56 kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25 kJ/kgs3=sf=0.45325 kJ/kg⋅K

Substitute 127.25 kJ/kg for h3 in Equation (I).

h4=127.25 kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85 kJ/kghfg=191.69 kJ/kgsf=0.24751 kJ/kg⋅Ksfg=0.6796 kJ/kg⋅K

Substitute 63.85 kJ/kg for hf, 191.69 kJ/kg for hfg, and 127.25 kJ/kg for h4 in Equation (III).

x4=127.25 kJ/kg−63.85 kJ/kg191.69 kJ/kgx4=0.3307

Substitute 0.24751 kJ/kg⋅K for sf, 0.6796 kJ/kg⋅K for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751 kJ/kg⋅K+0.3307(0.6796 kJ/kg⋅K)s4=0.47225 kJ/kg⋅K

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400 kPah5=hf=63.91 kJ/kgs5=sf=0.24752 kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91 kJ/kg for h5 in Equation (I).

h6=63.91 kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of −32°C as follows:

hf=10.09 kJ/kg hfg=220.83 kJ/kg sf=0.04249kJ/kg⋅K sfg=0.9157 kJ/kg⋅K

Substitute 63.91 kJ/kg for h6, 10.09 kJ/kg for hf, and 220.83 kJ/kg for hfg in Equation (III).

x6=63.91−10.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249 kJ/kg⋅K for sf, and 0.95819 kJ/kg⋅K for sg in Equation (IV).

s6=0.04249 kJ/kg⋅K+(0.2437)0.9157 kJ/kg⋅K=0.26559 kJ/kg⋅K

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of −32°C as follows:

h7=hg=230.93 kJ/kgs7=sg=0.95819 kJ/kg⋅K

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400 kPa and specific entropy (s8) of 0.95819 kJ/kg⋅K as 264.51 kJ/kg.

Substitute 2 kg/s for m˙2, 255.60 kJ/kg for h1, 127.25 kJ/kg for h4, 264.51 kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2 kg/s×255.60 kJ/kg−127.25 kJ/kg264.51 kJ/kg−63.91 kJ/kgm˙6=1.280 kg/s

Substitute 230.93 kJ/kg for h7, and 63.91 kJ/kg for h6 in Equation (V).

qL=230.93 kJ/kg−63.91 kJ/kgqL=167.02 kJ/kg

Substitute 127.25 kJ/kg for h3 and 281.56 kJ/kg for h2 in Equation (VI).

qH=281.56 kJ/kg−127.25 kJ/kgqH=154.31 kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3−s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, 0.92711 kJ/kg⋅K for s2, 154.31 kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2 kg/s×20°C×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg20°C)=2 kg/s×(20+273)K×(0.45325 kJ/kg⋅K−0.92711 kJ/kg⋅K+154.31 kJ/kg(20+273)K)=30.94 kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4−s3) (IX)

Substitute 2 kg/s for m˙2, 20°C for T0, 0.45325 kJ/kg⋅K for s3, and 0.47225 kJ/kg⋅K for s4 in Equation (IX).

X˙dest,3-4=2 kg/s×20°C×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=2 kg/s×(20+273)K×(0.47225 kJ/kg⋅K−0.45325 kJ/kg⋅K)=11.03 kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6−s5) (X)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.24752 kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280 kg/s×20°C×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=1.280 kg/s×(20+273)K×(0.26559 kJ/kg⋅K−0.24752 kJ/kgK)=6.793 kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7−s6−qLTL) (XI)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.26559 kJ/kg⋅K for s6, and 0.95819 kJ/kgK for s7, 167.02 kJ/kg for qL, and −22°C for TL in Equation (XI).

X˙dest,6-7=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22°C))=1.280 kg/s×20°C×(0.95819 kJ/kgK−0.26559 kJ/kg⋅K−167.02 kJ/kg(−22+273)K)=10.18 kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5−s8)−m˙2(s1−s4)] (XII)

Substitute 1.280 kg/s for m˙6, 20°C for T0, 0.24752 kJ/kgK for s5, 0.95819 kJ/kg for s8, 2 kg/s for m˙2, 0.47225 kJ/kg⋅K for s4, and 0.92711 kJ/kg⋅K for s1 in Equation (XII).

X˙dest,separator=20°C[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=(20+273)K[1.280 kg/s(0.24752 kJ/kgK−0.95819 kJ/kg)−2 kg/s(0.92711 kJ/kg⋅K−0.47225 kJ/kg⋅K)]=0.197 kW