CONNECT FOR THERMODYNAMICS: AN ENGINEERI
CONNECT FOR THERMODYNAMICS: AN ENGINEERI
9th Edition
ISBN: 9781260048636
Author: CENGEL
Publisher: MCG
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Chapter 11.10, Problem 117RP
To determine

The process that causes the greatest amount of exergy destruction.

Expert Solution & Answer
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Answer to Problem 117RP

The process that causes the greatest amount of exergy destruction is process41(6.44kJ/kg).

Explanation of Solution

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

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 11.10, Problem 117RP

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

h3h4 (I)

Here, specific enthalpy at state 3 and 4 is h3andh4 respectively.

Express specific enthalpy at state 2 using compressor efficiency.

h2=h1+h2sh1ηC (II)

Here, specific enthalpy at state 2, 1 and 2s is h2,h1andh2s respectively and Carnot efficiency is ηC.

Express temperature at state 3.

T3=Tsat@700kPaTsubcooling (III)

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

Express quality at state 4.

x4=h4hf@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=h1h4 (VI)

Here, specific enthalpy at state 4 is h4.

Express heat added in the condenser.

qH=h2h3 (VII)

Express the exergy destruction during process 1-2.

xdestroyed,1-2=T0(s2s1) (VIII)

Here, surrounding temperature is T0, specific entropy at state 1 and 2 is s1ands2 respectively.

Express the exergy destruction during process 2-3.

xdestroyed,2-3=T0[s3s2+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(s4s3) (X)

Here, specific entropy at state 4 is s4.

Express the exergy destruction during process 4-1.

xdestroyed,4-1=T0[s1s4qLTL] (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=244.55kJ/kgs1=sg=0.9378kJ/kgK

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

Perform unit conversion of pressure at state 2 from kPatoMPa.

P2=700kPa[MPa1000kPa]=0.7MPa

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

Write the formula of interpolation method of two variables.

y2=(x2x1)(y3y1)(x3x1)+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/kgK)

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 0.9314kJ/kgK,0.9378kJ/kgKand0.9642kJ/kgK for x1,x2andx3 respectively, 268.47kJ/kg for y1 and 278.59kJ/kg for y3 in Equation (XII).

y2=[(0.93780.9314)kJ/kgK][(278.59268.47)kJ/kg](0.96420.9314)kJ/kgK+268.47kJ/kg=270.45kJ/kg=h2s

Thus, the specific enthalpy at state 2s is,

h2s=270.45kJ/kg

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

Tsat@900kPa=26.7°C

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

T3=26.7°C2.7°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=84.98kJ/kgs3=sf=0.3195kJ/kgK

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

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

h4=84.98kJ/kg

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

h2=244.55kJ/kg+270.45kJ/kg244.55kJ/kg0.85=275.02kJ/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=38.53kJ/kghfg@10°C=206.02kJ/kgsf@-10°C=0.1549kJ/kgKsfg@-10°C=0.7828kJ/kgK

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

x4=84.98kJ/kg38.53kJ/kg206.02kJ/kg=0.225

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

s4=(0.1549kJ/kgK)+(0.225)(0.7828kJ/kgK)=0.3314kJ/kgK

Refer Table A-13, “superheated refrigerant 134a”, and write the specific entropy at state 2 corresponding to pressure at state 2 of 0.7MPa and specific enthalpy at state 2 of 275.02kJ/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/kgK)

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=0.9526kJ/kgK

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

qL=244.55kJ/kg84.98kJ/kg=159.57kJ/kg

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

qH=275.02kJ/kg84.98kJ/kg=190.04kJ/kg

Perform unit conversion of temperature from °CtoK

T0=TH=22°C=(22+273)K=295K

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

xdestroyed,1-2=295K[(0.95260.9378)kJ/kgK]=4.369kJ/kg

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

xdestroyed,2-3=(295K)[(0.31950.9526)kJ/kgK+190.04kJ/kg295K]=3.293kJ/kg

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

xdestroyed,3-4=295K[(0.33140.3195)kJ/kgK]=3.5105kJ/kg

Take the freezing temperature of water as,

TL=0°C=(0+273)K=273K

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

xdestroyed,4-1=(295K)[(0.93780.3314)kJ/kgK159.57kJ/kg273K]=6.44kJ/kg

Hence, the process that causes the greatest amount of exergy destruction is process41(6.44kJ/kg).

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Chapter 11 Solutions

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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