The process that causes the greatest amount of exergy destruction.

Question

Want to see more full solutions like this?

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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

6. A part of the structure for a factory automation system is a beam that spans 30.0 in as shown in Figure P5-6. Loads are applied at two points, each 8.0 in from a support. The left load F₁ = 1800 lb remains constantly applied, while the right load F₂ = 1800 lb is applied and removed fre- quently as the machine cycles. Evaluate the beam at both B and C. A 8 in F₁ = 1800 lb 14 in F2 = 1800 lb 8 in D RA B C 4X2X1/4 Steel tube Beam cross section RD

30. Repeat Problem 28, except using a shaft that is rotating and transmitting a torque of 150 N⚫m from the left bear- ing to the middle of the shaft. Also, there is a profile key- seat at the middle under the load.

28. The shaft shown in Figure P5-28 is supported by bear- ings at each end, which have bores of 20.0 mm. Design the shaft to carry the given load if it is steady and the shaft is stationary. Make the dimension a as large as pos- sible while keeping the stress safe. Determine the required d = 20mm D = ? R = ?| 5.4 kN d=20mm Length not to scale -a = ?- +а= a = ? + -125 mm- -250 mm- FIGURE P5-28 (Problems 28, 29, and 30)

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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684, 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

Ch. 11.10 - Why do we study the reversed Carnot cycle even...Ch. 11.10 - Why is the reversed Carnot cycle executed within...Ch. 11.10 - A steady-flow Carnot refrigeration cycle uses...Ch. 11.10 - Refrigerant-134a enters the condenser of a...Ch. 11.10 - Does the ideal vapor-compression refrigeration...Ch. 11.10 - Why is the throttling valve not replaced by an...Ch. 11.10 - In a refrigeration system, would you recommend...Ch. 11.10 - Does the area enclosed by the cycle on a T-s...Ch. 11.10 - Consider two vapor-compression refrigeration...Ch. 11.10 - It is proposed to use water instead of...

The process that causes the greatest amount of exergy destruction.

Concept explainers

Videos

The process that causes the greatest amount of exergy destruction.

Answer to Problem 117RP

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions