Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780073398174
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Chapter 9.12, Problem 144P
To determine

The exergy destruction associated with each process of the Brayton cycle and the exergy of the exhaust gases at the exit of the regenerator.

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Answer to Problem 144P

The exergy destruction associated with process 1-2 of the given Brayton cycle is 38.91kJ/kg.

The exergy destruction associated with process 3-4 of the given Brayton cycle is 35.83kJ/kg.

The exergy destruction associated with regeneration process of the given Brayton cycle is 4.37kJ/kg.

The exergy destruction associated with process 5-3 of the given Brayton cycle is 58.09kJ/kg.

The exergy destruction associated with process 6-1 of the given Brayton cycle is 142.6kJ/kg.

The exergy of the exhaust gases at the exit of the regenerator is 143.2kJ/kg.

Explanation of Solution

Show the regenerative Brayton cycle with air as the working fluid, on Ts diagram as in Figure (1).

Thermodynamics: An Engineering Approach, Chapter 9.12, Problem 144P

For the given Brayton cycle with regeneration and air as the working fluid Ti, Pi, hi and Pri are the temperature at ith state, pressure at ith state, specific enthalpy at ith state, and relative pressure at ith state respectively.

Write the pressure and relative pressure relation for the process 1-2.

Pr2=P2P1Pr1 (I)

Write the pressure and relative pressure relation for the process 3-4.

Pr4=P4P3Pr3 (II)

Write the expression of efficiency of the compressor (ηC).

ηC=h2sh1h2h1 (III)

Write the expression of efficiency of the turbine (ηT).

ηT=h3h4h3h4s (IV)

Write the expression of net work output by the gas turbine (wnet).

wnet=wT,outwC,in

wnet=(h3h4)(h2h1) (V)

Here, work done by the turbine is wT,out, and the work input to the compressor is wC,in.

Write the expression of effectiveness of the regenerator (ε).

ε=h5h2h4h2 (VI)

Write the expression of heat input to the regenerative Brayton cycle (qin).

qin=h3h5 (VII)

Write the expression of heat rejected by the regenerative Brayton cycle (qout).

qout=qinwnet (VIII)

Write the expression of thermal efficiency of the given turbine (ηth).

ηth=wnetqin (IX)

Write the energy balance equation on the heat exchanger.

h5h2=h4h6 (X)

Write the expression of exergy destruction associated with the process 1-2 of the given Brayton cycle (xdestroyed,12).

xdestroyed,12=T0(s2s1)

xdestroyed,12=T0(s2s1RlnP2P1) (XI)

Here, the temperature of the surroundings is T0, the gas constant of air is R, entropy of air at state 2 as a function of temperature alone is s2, and entropy of air at state 1 as a function of temperature alone is s1.

Write the expression of exergy destruction associated with the process 3-4 of the given Brayton cycle (xdestroyed,34).

xdestroyed,34=T0(s4s3)

xdestroyed,34=T0(s4s3RlnP4P3) (XII)

Here, entropy of air at state 3 as a function of temperature is s3, and entropy of air at state 4 as a function of temperature is s4.

Write the expression of exergy destruction associated with the regeneration process of the given Brayton cycle (xdestroyed,regen).

xdestroyed,regen=T0[(s5s2)+(s6s4)]

xdestroyed,regen=T0[(s5s2)+(s6s4)] (XIII)

Here, entropy of air at state 5 as a function of temperature alone is s5, and entropy of air at state 6 as a function of temperature alone is s6.

Write the expression of exergy destruction associated with the process 5-3 of the given Brayton cycle (xdestroyed,53).

xdestroyed,53=T0(s3s5+qinTH)

xdestroyed,53=T0(s3s5RlnP3P5qinTH) (XIV)

Here, the temperature of the heat source is TH.

Write the expression of exergy destruction associated with the process 6-1 of the given Brayton cycle (xdestroyed,61).

xdestroyed,61=T0(s1s6+qoutTL)

xdestroyed,61=T0(s1s6RlnP1P6+qoutTL) (XV)

Here, the temperature of the sink is TL.

Write the expression of stream exergy at the exit of the regenerator (state 6) (ϕ6).

ϕ6=(h6h0)T0(s6s0) (XVI)

Here, the specific enthalpy of the surroundings is h0.

Write the expression of change entropy for the exit of the regenerator (s6s0).

s6s0=s6s0RlnP6P1 (XVII)

Here, entropy of air at the surroundings as a function of temperature alone is s0, and the pressure of air at the surroundings is P0.

Conclusion:

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of 310 K (T1).

h1=310.24kJ/kgPr1=1.5546s10=1.73498kJ/kgK

Substitute 7 for P2P1, and 1.5546 for Pr1 in Equation (I).

Pr2=(7)(1.5546)=10.88

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a relative pressure of 10.88 (Pr2).

h2s=541.26kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of 1150 K (T3).

h3=1219.25kJ/kgPr3=200.15s30=3.129kJ/kgK

Substitute 17 for P4P3, and 200.15 for Pr3 in Equation (II).

Pr4=(17)(200.15)=28.59

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a relative pressure of 28.59 (Pr4).

h4s=711.8kJ/kg

Rearrange Equation (III), and substitute 310.24kJ/kg for h1, 541.26kJ/kg for h2s, and 0.75 for ηC.

h2=h1+h2sh1ηC=310.24kJ/kg+(541.26310.24)kJ/kg0.75=618.26kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a enthalpy of 618.26kJ/kg (h2).

s20=2.42763kJ/kgK

Rearrange Equation (IV), and substitute 1219.25kJ/kg for h3, 711.8kJ/kg for h4s, and 0.82 for ηT.

h4=h3ηT(h3h4s)=1219.25kJ/kg(0.82)(1219.25711.8)kJ/kg=803.14kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a enthalpy of 803.14kJ/kg (h4).

s40=2.69407kJ/kgK

Substitute 1219.25kJ/kg for h3, 618.26kJ/kg for h2, 803.14kJ/kg for h4, and 310.24kJ/kg for h1 in Equation (V).

wnet=(1219.25803.14)kJ/kg(618.26310.24)kJ/kg=108.09kJ/kg

Substitute 0.65 for ε, 803.14kJ/kg for h4, and 618.26kJ/kg for h2 in Equation (VI).

0.65=(h5618.26kJ/kg)(803.14618.26)kJ/kgh5=738.43kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a enthalpy of 738.43kJ/kg (h5).

s50=2.60815kJ/kgK

Substitute 1219.25kJ/kg for h3, and 738.43kJ/kg for h5 in Equation (VII).

qin=(1219.25738.43)kJ/kg=480.82kJ/kg

Substitute 108.09kJ/kg for wnet, and 480.82kJ/kg for qin in Equation (VIII).

qout=(480.82108.09)kJ/kg=372.73kJ/kg

Substitute 108.09kJ/kg for wnet, and 480.82kJ/kg for qin in Equation (IX).

ηth=108.09kJ/kg480.82kJ/kg=0.225=22.5%

Substitute 738.43kJ/kg for h5, 803.14kJ/kg for h4, and 618.26kJ/kg for h2 in Equation (X).

738.43kJ/kg618.26kJ/kg=803.14kJ/kgh6h6=682.97kJ/kg

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a enthalpy of 682.97kJ/kg (h6).

s60=2.52861kJ/kgK

Substitute 290 K for T0, 1.73498kJ/kgK for s1, 2.42763kJ/kgK for s2, 0.287kJ/kgK for R, and 7 for P2P1 in Equation (XI).

xdestroyed,12=(290K)[(2.427631.73498)kJ/kgK(0.287kJ/kgK)ln(7)]=38.91kJ/kg

Thus, the exergy destruction associated with process 1-2 of the given Brayton cycle is 38.91kJ/kg.

Substitute 290 K for T0, 3.129kJ/kgK for s3, 2.69407kJ/kgK for s4, 0.287kJ/kgK for R, and 17 for P4P3 in Equation (XII).

xdestroyed,34=(290K)[(2.694073.129)kJ/kgK(0.287kJ/kgK)ln(17)]=35.83kJ/kg

Thus, the exergy destruction associated with process 3-4 of the given Brayton cycle is 35.83kJ/kg.

Substitute 290 K for T0, 2.60815kJ/kgK for s5, 2.69407kJ/kgK for s4, 2.42763kJ/kgK for s2, and 2.52861kJ/kgK for s6 in Equation (XIII).

xdestroyed,regen=(290K)[(2.608152.42763)kJ/kgK+(2.528612.69407)kJ/kgK]=4.37kJ/kg

Thus, the exergy destruction associated with regeneration process of the given Brayton cycle is 4.37kJ/kg.

Substitute 290 K for T0, 3.129kJ/kgK for s3, 2.60815kJ/kgK for s5, 0.287kJ/kgK for R, 1500 K for TH, and 480.82kJ/kg for qin in Equation (XIV). For the process 5-3, pressure remains constant, hence substitute 0 for lnP3P5.

xdestroyed,53=(290K)[(3.1292.60815)kJ/kgK(0.287kJ/kgK)(0)480.82kJ/kg1500K]=58.09kJ/kg

Thus, the exergy destruction associated with process 5-3 of the given Brayton cycle is 58.09kJ/kg.

Substitute 290 K for T0, 1.73498kJ/kgK for s1, 2.52861kJ/kgK for s6, 0.287kJ/kgK for R, 290 K for TL, and 372.73kJ/kg for qout in Equation (XV). For the process 6-1, pressure remains constant, hence substitute 0 for lnP1P6.

xdestroyed,61=(290K)[(1.734982.52861)kJ/kgK(0.287kJ/kgK)(0)+372.73kJ/kg290K]=142.6kJ/kg

Thus, the exergy destruction associated with process 6-1 of the given Brayton cycle is 142.6kJ/kg.

Refer Table A-17, “Ideal gas properties of air”, obtain the properties of air at a temperature of 290 K (T0).

h0=290.16kJ/kgs00=1.66802kJ/kgK

Substitute 2.52861kJ/kgK for s6, and 1.66802kJ/kgK for s0 in Equation (XVII). Also, at the exit of the regenerator, pressure remains constant, hence substitute 0 for lnP6P1.

s6s0=(2.528611.66802)kJ/kgK=0.86059kJ/kgK

Substitute 682.97kJ/kg for h6, 290.16kJ/kg for h0, 290 K for T0, and 0.86059kJ/kgK for (s6s0) in Equation (XVI).

ϕ6=(682.97290.16)kJ/kg(290K)(0.86059kJ/kgK)=143.2kJ/kg

Thus, the exergy of the exhaust gases at the exit of the regenerator is 143.2kJ/kg.

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Chapter 9 Solutions

Thermodynamics: An Engineering Approach

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