Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
8th Edition
ISBN: 9780077624811
Author: CENGEL
Publisher: MCG
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Chapter 9.12, Problem 147P
To determine

The exergy destruction for each of the processes of the cycle

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Answer to Problem 147P

The exergy destruction associated with process 1-2 is 19.2kJ/kg.

The exergy destruction associated with process 5-3 is 38kJ/kg.

The exergy destruction associated with process 3-4 is 16.1kJ/kg.

The exergy destruction associated with process 6-1 is 85.5kJ/kg.

The exergy destruction associated at regenerator is 0.41kJ/kg.

Explanation of Solution

Draw the Ts diagram of the regenerative Brayton cycle as shown in Figure (1).

Thermodynamics: An Engineering Approach, Chapter 9.12, Problem 147P

Write the expression for the temperature and pressure relation for the isentropic process 1-2.

T2s=T1rp(k1)/k (I)

Here, the pressure ratio is rp, and the specific heat ratio is k.

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

ηC=cp(T2sT1)cp(T2T1) (II)

Here, the specific heat at constant pressure is cp.

Write the expression for the temperature and pressure relation ratio for the expansion process 4-5s.

T5s=T4(1rp)(k1)/k (III)

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

ηT=cp(T4T5)cp(T4T5s) (IV)

Apply first law to the heat exchanger.

mcp(T3T2)=mcp(T5T6)(T3T2)=(T5T6)T6=T5+T2T3 (V)

Write the expression for the net work done per kg for the Brayton cycle with regeneration (wnet).

wnet=cp(T4T5)cp(T2T1)                                   (VI)

Here, the specific heat at constant pressure is cp.

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

qin=cp(T4T3)                                       (VII)

Write the expression for rate of heat rejection in the regenerative Brayton cycle (qout).

qout=cp(T6T1) (VIII).

Write the expression for the exergy destruction during the process of as steam from an inlet to exit state.

xdestroyed=T0sgenxdestroyed=T0(sesiqinTsource+qinTsink)

Write the expression of exergy destruction for process 1-2 (xdestroyed,1-2).

xdestroyed,1-2=T0(cplnT2T1RlnP2P1) (IX)

Here, pressure at state 2 is P2, the pressure at state 1 is P1 and surrounding temperature is T0.

Write the expression of exergy destruction for process 5-3 (xdestroyed,5-3).

xdestroyed,5-3=T0(cplnT3T5RlnP3P5qinTsource) (X)

Here, pressure at state 5 is P5 and the pressure at state 6 is P6.

Write the expression of exergy destruction for process 3-4 (xdestroyed,3-4).

xdestroyed,3-4=T0(cplnT4T5RlnP4P3) (XI)

Here, pressure at state 7 is P7

Write the expression of exergy destruction for process 6-1 (xdestroyed,6-1).

xdestroyed,6-1=T0(cplnT1T6RlnP1P10qoutTsink) (XII)

Here, pressure at state 10 is P10 and temperature at state 10 is T10.

Write the expression of exergy destruction for regenerator (xdestroyed,regen).

xdestroyed,regen=T0(Δs25+Δs46)xdestroyed,regen=T0(cplnT5T2+cplnT6T4) (XIII)

Conclusion:

Substitute 20°C for T1, 8 for rp and 1.4 for k in Equation (I).

T2s=(20°C)(8)1.41/1.4=(20+273)K(8)1.41/1.4=530.8K

Substitute 20°C for T1, 530.8K for T2s and 0.87 for ηC in Equation (II).

T2=20°C+585K20°C0.87=(20+273)K+585K(20+273)K0.87=566.3K

Substitute 800°C for T4, 8 for rp and 1.4 for k in Equation (III).

T5s=(800°C)(18)1.41/1.4=(800+273)K×(18)1.41/1.4=592.3K

Substitute 800°C for T4, 592.3K for T5s and 0.93 for ηT in Equation (IV).

T5=800°C(0.93)(800°C592.3K)=(800+273)K(0.93)[(800+273)K592.3K]=625.9K

The cold airstream (T3) leaves the regenerator 10°C cooler than the hot airstream (T5) at the inlet of the regenerator.

T3=T510 (XIV).

Substitute 625.9K for T5 in Equation (XIV).

T3=(625.9K273°C)10°CT3= 352.9°C10°C

T3= 342.9°C=(342.9+273)K= 615.9K

Substitute 615.9K for T3, 625.9K for T5 and 566.3K for T2 in Equation (V).

T6=625.9K+566.3K615.9K=576.3K

Substitute 1.005kJ/kgK for cp, 800°C for T4, 625.9K for T5, 566.3K for T2 and 20°C for T1 in Equation (VI).

wnet=(1.005kJ/kgK)[(800°C625.9K)][(566.3K20°C)]=(1.005kJ/kgK)[((800+273)K625.9K)][(566.3K(20+273)K)]=174.7kJ/kg

Substitute 1.005kJ/kgK for cp, 800°C for T4 and 615.9K for T3 in Equation (VII).

qin=(1.005kJ/kgK)(800°C615.9K)=(1.005kJ/kgK)[(800+273)K615.9K]=459.4kJ/kg

Substitute 1.005kJ/kgK for cp, 576.3K for T6 and 20°C for T1 in Equation (VIII).

qout=(1.005kJ/kgK)(576.3K20°C)=(1.005kJ/kgK)[576.3K(20+273)K]=284.7kJ/kg

Substitute 20°C for T0, 1.005kJ/kg for cp , 566.3K for T2 , 20°C for T1 , 0.287kJ/kgK for R and 8 for (P2P1) in Equation (IX).

xdestroyed,1-2=20°C((1.005kJ/kg)ln566.3K20°C(0.287kJ/kgK)ln(8))=(20+273)K((1.005kJ/kg)ln566.3K(20+273)K(0.287kJ/kgK)ln(8))=19.2kJ/kg

Thus, the exergy destruction associated with process 1-2 is 19.2kJ/kg.

Here, (P3=P5)

Substitute 293K for T0, 1.005kJ/kgK for cp , 1073K for T3 , 615.9K for T5 ,0.287kJ/kgK for R , 459.4kJ/kg for qin and 1073K for Tsource in Equation (X).

xdestroyed,5-3=293K((1.005kJ/kgK)ln1073K615.9K(0.287kJ/kgK)ln(P5P5)459.4kJ/kg1073K)=293K((1.005kJ/kgK)ln1073K615.9K0459.4kJ/kg1073K)=38kJ/kg

Thus, the exergy destruction associated with process 5-3 is 38kJ/kg.

Substitute 625.9K for T4, 1.005kJ/kgK for cp , 1073K for T5 , 293K for T0 , 0.287kJ/kgK for R and 18 for (P4P3) in Equation (XI).

xdestroyed,3-4=293K((1.005kJ/kgK)ln625.9K1073K(0.287kJ/kgK)ln(18))=16.1kJ/kg

Thus, the exergy destruction associated with process 3-4 is 16.1kJ/kg.

Here,(P1=P6)

Substitute 520R for T0, 1.005kJ/kgK for cp , 576.3K for T6 , 520R for T1 , 0.287kJ/kgK for R , 284.7kJ/kg for qout and 293K for Tsink in Equation (XII).

xdestroyed,6-1=293K((1.005kJ/kgK)ln293K576.3K(0.287kJ/kgK)lnP6P6qout293K)=293K((1.005kJ/kgK)ln293K576.3K0284.7kJ/kg293K)=85.5kJ/kg

Thus, the exergy destruction associated with process 6-1 is 85.5kJ/kg.

Substitute 615.9K for T5, 1.005kJ/kgK for cp , 625.9K for T4 , 576.3K for T6 , 566.3K for T2, and 293K for T0 in Equation (XIII).

xdestroyed,regen=(293K)×1.005kJ/kgK(ln615.9K566.3K+ln576.3K625.9K)=0.41kJ/kg

Thus, the exergy destruction associated at regenerator is 0.41kJ/kg.

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Chapter 9 Solutions

Thermodynamics: An Engineering Approach

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