Redo the previous problem for a large stationary Brayton cycle where the low T heat rejection is used in a process application and thus has a non-zero exergy.

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The thermal efficiency (first law) is Eq.12.1

[latex]\eta_{ TH }=\frac{\dot{ W }_{ ext {net }}}{\dot{ Q }_{ H }}=\frac{ w _{ ext {net }}}{ q _{ H }}=1- r _{ p }^{-( k -1) / k }[/latex]

The corresponding [latex]2^{ ext {nd }} ext { law }[/latex] efficiency is

[latex]\eta_{ II }=\frac{ w _{ ext {net }}}{\Phi_{ H }-\Phi_{ L }}=\frac{ h _3- h _2-\left( h _4- h _1 ight)}{ h _3- h _2- T _{ o }\left( s _3- s _2 ight)}[/latex]

where we used

[latex]\begin{aligned}& \Phi_{ H }= ext { increase in flow exergy }=\Psi_3-\Psi_2= h _3- h _2- T _{ o }\left( s _3- s _2 ight) \& \Phi_{ L }= ext { rejection of flow exergy }=\Psi_4-\Psi_1= h _4- h _1- T _{ o }\left( s _4- s _1 ight)\end{aligned}[/latex]

Now divide all terms with the difference [latex]h _3- h _2[/latex] to get

[latex]\begin{aligned}& w _{ ext {net }} /\left( h _3- h _2 ight)= \eta _{ TH } ; \& \Phi_{ H } /\left( h _3- h _2 ight)=1- T _{ o }\left( s _3- s _2 ight)\left( h _3- h _2 ight)=1- T _{ o } \ln \left( T _3 / T _2 ight)\left( T _3- T _2 ight)^{-1} \& \Phi_{ L } /\left( h _3- h _2 ight)=\frac{ h _4- h _1}{ h _3- h _2}\left[1- T _{ o } \frac{ s _4- s _1}{ h _4- h _1} ight]=\left(1- \eta _{ TH } ight)\left[1- T _{ o } \frac{\ln T _4 / T _1}{ T _4- T _1} ight]\end{aligned}[/latex]

Substitute all terms to get

[latex]\eta_{ II }=\frac{\eta_{ TH }}{1- T _{ o } \frac{\ln \left( T _3 / T _2 ight)}{ T _3- T _2}-\left(1-\eta_{ TH } ight)\left[1- T _{ o } \frac{\ln \left( T _4 / T _1 ight)}{ T _4- T _1} ight]}[/latex]

Comment: Due to the temperature sensitivity of [latex]\Phi_{ H } ext { and } \Phi_{ L }[/latex] the temperatures do not reduce out from the expression.

Question 12.CSGP.123: Redo the previous problem for a large stationary Brayton cyc......

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