The exergy destruction associated with each of the processes of the Brayton cycle.
Answer to Problem 142P
The exergy destruction associated with process 1-2 of the given Brayton cycle is 28.08 kJ/kg.
The exergy destruction associated with process 2-3 of the given Brayton cycle is 100.3 kJ/kg.
The exergy destruction associated with process 3-4 of the given Brayton cycle is 32.86 kJ/kg.
The exergy destruction associated with process 4-1 of the given Brayton cycle is 197.93 kJ/kg.
Explanation of Solution
Show the simple Brayton cycle, and with air as the working fluid on T−s diagram.
For the given simple Brayton cycle with air as the working fluid, let Ti, Pi, hi and Pri be the temperature, pressure, specific enthalpy and relative pressure at ith state respectively.
Write the expression of pressure ratio relation for the process 1-2.
Pr2=P2P1Pr1 (I)
Write the relation of pressure ratio and pressure for the process 3-4.
Pr4=P4P3Pr3 (II)
Write the expression of efficiency of the compressor (ηC).
ηC=h2s−h1h2−h1 (III)
Write the expression of efficiency of the turbine (ηT).
ηT=h3−h4h3−h4s (IV)
Write the expression of heat input to the Brayton cycle (qin).
qin=(h3−h2) (V)
Write the expression of heat rejected by the Brayton cycle (qout).
qout=(h4−h1) (VI)
Write the expression of exergy destruction associated with the process 1-2 of the given Brayton cycle (xdestroyed,12).
xdestroyed,12=T0(s2−s1)
xdestroyed,12=T0(s∘2−s∘1−RlnP2P1) (VII)
Here, temperature of the surroundings is T0, gas constant of air is R, entropy of air at state 2 as a function of temperature alone is s∘2, and entropy of air at state 1 as a function of temperature alone is s∘1.
Write the expression of exergy destruction associated with the process 2-3 of the given Brayton cycle (xdestroyed,23).
xdestroyed,23=T0(s3−s2+−qinTH)
xdestroyed,23=T0(s∘3−s∘2−RlnP3P2+−qinTH) (VIII)
Here, temperature of the heat source is TH, entropy of air at state 2 as a function of temperature alone is s∘2, and entropy of air at state 3 as a function of temperature alone is s∘3.
Write the expression of exergy destruction associated with the process 3-4 of the given Brayton cycle (xdestroyed,34).
xdestroyed,34=T0(s4−s3)
xdestroyed,34=T0(s∘4−s∘3−RlnP4P3) (IX)
Here, entropy of air at state 3 as a function of temperature is s∘3, and entropy of air at state 4 as a function of temperature is s∘4.
Write the expression of exergy destruction associated with the process 4-1 of the given Brayton cycle (xdestroyed,41).
xdestroyed,41=T0(s1−s4+qoutTL)
xdestroyed,41=T0(s∘1−s∘4−RlnP1P4+qoutTL) (X)
Here, temperature of the sink is TL, entropy of air at state 1 as a function of temperature alone is s∘1, and entropy of air at state 4 as a function of temperature alone is s∘4.
Conclusion:
Refer Table A-17E given in the textbook to find the properties of air at 295 K (T1), which gives 295.17 kJ/kg for h1, 1.3068 for Pr1 and 1.68515 kJ/kg⋅K for s∘1.
Substitute 10 for P2P1, and 1.3068 for Pr1 in Equation (I).
Pr2=(10)(1.3068)=13.07
Refer Table A-17E given in the textbook to find the properties of air at 13.07 (Pr1), which gives 570.26 kJ/kg for h2s, and 564.9 K for T2s.
Refer Table A-17E given in the textbook to find the properties of air at 1,240 K (T3), which gives 1,324.93 kJ/kg for h3, 272.3 for Pr3 and 3.21751 kJ/kg⋅K for s∘3.
Substitute 110 for P4P3, and 272.3 for Pr3 in Equation (II).
Pr4=(110)(272.3)=27.23
Refer Table A-17 given in the textbook to find the properties of air at 27.23 (Pr4), which gives 702.07kJ/kg for h4s, and 689.6 K for T4s.
Rearrange Equation (III) and substitute 295.17 kJ/kg for h1, 570.26 kJ/kg for h2s, and 0.83 for ηC.
h2=h1+h2s−h1ηC=295.17 kJ/kg+(570.26−295.17) kJ/kg0.83=626.60 kJ/kg
Refer Table A-17E given in the textbook to find the properties of air at 626.60 kJ/kg (h2), which gives 2.44117 kJ/kg⋅K for s∘2.
Rearrange Equation (IV) and substitute 1,324.93 kJ/kg for h3, 702.07 kJ/kg for h4s, and 0.87 for ηT.
h4=h3−ηT(h3−h4s)=1,324.93 kJ/kg+(0.87)(1,324.93−702.07) kJ/kg=783.04 kJ/kg
Refer Table A-17 given in the textbook, to find the properties of air at 783.04 kJ/kg (h4), which gives 764.4 K for T4 and 2.66807 kJ/kg⋅K for s∘4.
Substitute 1,324.93 kJ/kg for h3, and 626.60 kJ/kg for h2 in Equation (V).
qin=(1,324.93−626.60) kJ/kg=698.3 kJ/kg
Substitute 783.04 kJ/kg for h4, and 295.17 kJ/kg for h1 in Equation (VI).
qout=(783.04−295.17) kJ/kg=487.9 kJ/kg
Substitute 295 K for T0, 1.68515 kJ/kg⋅K for s∘1, 2.44117 kJ/kg⋅K for s∘2, 0.287 kJ/kg⋅K for R, and 10 for P2P1 in Equation (VII).
xdestroyed,12=(295 K)[(2.44117−1.68515 )kJ/kg⋅K−(0.287kJ/kg⋅K)ln(10)]=28.08 kJ/kg
Thus, the exergy destruction associated with process 1-2 of the given Brayton cycle is 28.08 kJ/kg.
Substitute 295 K for T0, 3.21751 kJ/kg⋅K for s∘3, 2.44117 kJ/kg⋅K for s∘2, 0.287 kJ/kg⋅K for R, 1,600 K for TH, and 698.3 kJ/kg for qin in Equation (VIII). For the process 2-3, pressure remains constant, hence substitute 0 for lnP3P2.
xdestroyed,23=(295 K)[(3.21751−2.44117 )kJ/kg⋅K−(0.287kJ/kg⋅K)(0)−698.3 kJ/kg1,600 K]=100.3 kJ/kg
Thus, the exergy destruction associated with process 2-3 of the given Brayton cycle is 100.3 kJ/kg.
Substitute 295 K for T0, 3.21751 kJ/kg⋅K for s∘3, 2.66807 kJ/kg⋅K for s∘4, 0.287 kJ/kg⋅K for R, and 110 for P4P3 in Equation (IX).
xdestroyed,34=(295 K)[(2.66807−3.21751 )kJ/kg⋅K−(0.287kJ/kg⋅K)ln(110)]=32.86 kJ/kg
Thus, the exergy destruction associated with process 3-4 of the given Brayton cycle is 32.86 kJ/kg.
Substitute 295 K for T0, 1.68515 kJ/kg⋅K for s∘1, 2.66807 kJ/kg⋅K for s∘4, 0.287 kJ/kg⋅K for R, 295 K for TL, and 487.9 kJ/kg for qout in Equation (X). For the process 4-1, pressure remains constant, hence substitute 0 for lnP1P4.
xdestroyed,41=(295 K)[(1.68515−2.66807)kJ/kg⋅K−(0.287kJ/kg⋅K)(0)+487.9 kJ/kg295 K]= 197.93 kJ/kg
Thus, the exergy destruction associated with process 4-1 of the given Brayton cycle is 197.93 kJ/kg.
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