The exergy destruction for each of the processes of the cycle

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Answer 1

Question

Chapter 9.12, Problem 147P

To determine

The exergy destruction for each of the processes of the cycle

Expert Solution & Answer

Answer to Problem 147P

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

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

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

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

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

Explanation of Solution

Draw the T−s diagram of the regenerative Brayton cycle as shown in Figure (1).

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 9.12, Problem 147P

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

T2s=T1rp(k−1)/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(T2s−T1)cp(T2−T1) (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)(k−1)/k (III)

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

ηT=cp(T4−T5)cp(T4−T5s) (IV)

Apply first law to the heat exchanger.

mcp(T3−T2)=mcp(T5−T6)(T3−T2)=(T5−T6)T6=T5+T2−T3 (V)

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

wnet=cp(T4−T5)−cp(T2−T1) (VI)

Here, the specific heat at constant pressure is cp.

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

qin=cp(T4−T3) (VII)

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

qout=cp(T6−T1) (VIII).

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

xdestroyed=T0sgenxdestroyed=T0(se−si−qinTsource+qinTsink)

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

xdestroyed,1-2=T0(cplnT2T1−RlnP2P1) (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(cplnT3T5−RlnP3P5−qinTsource) (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(cplnT4T5−RlnP4P3) (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(cplnT1T6−RlnP1P10−qoutTsink) (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(Δs2−5+Δs4−6)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)1.4−1/1.4=(20+273)K (8)1.4−1/1.4=530.8 K

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

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

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

T5s=(800°C)(18)1.4−1/1.4=(800+273)K×(18)1.4−1/1.4=592.3 K

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

T5=800°C−(0.93)(800°C−592.3 K)=(800+273)K−(0.93)[(800+273)K−592.3 K]=625.9 K

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

T3=T5−10 (XIV).

Substitute 625.9 K for T5 in Equation (XIV).

T3=(625.9 K−273 °C)−10°CT3= 352.9°C−10°C

T3= 342.9 °C=(342.9 +273) K= 615.9 K

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

T6=625.9 K+566.3 K−615.9 K=576.3 K

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

wnet=(1.005 kJ/kg⋅K)[(800°C−625.9 K) ]−[(566.3 K−20°C)]=(1.005 kJ/kg⋅K)[((800+273)K−625.9 K) ]−[(566.3 K−(20+273)K)]=174.7 kJ/kg

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

qin=(1.005 kJ/kg⋅K)(800°C−615.9 K)=(1.005 kJ/kg⋅K)[(800+273)K−615.9 K]=459.4 kJ/kg

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

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

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

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

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

Here, (P3=P5)

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

xdestroyed,5-3=293 K((1.005 kJ/kg⋅K)ln1073 K615.9 K−(0.287 kJ/kg⋅K)ln(P5P5)−459.4 kJ/kg1073 K)=293 K((1.005 kJ/kg⋅K)ln1073 K615.9 K−0−459.4 kJ/kg1073 K)=38 kJ/kg

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

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

xdestroyed,3-4=293 K((1.005 kJ/kg⋅K)ln625.9 K1073 K−(0.287 kJ/kg⋅K)ln(18))=16.1 kJ/kg

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

Here,(P1=P6)

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

xdestroyed,6-1=293 K((1.005 kJ/kg⋅K)ln293 K576.3 K−(0.287 kJ/kg⋅K)lnP6P6−qout293 K)=293 K((1.005 kJ/kg⋅K)ln293 K576.3 K−0−284.7 kJ/kg293 K)=85.5 kJ/kg

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

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

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

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

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