Question 12.CSGP.122: The conversion efficiency of the Brayton cycle in Eq.12.1 wa......
The conversion efficiency of the Brayton cycle in Eq.12.1 was done with cold air properties. Find a similar formula for the second law efficiency assuming the low T heat rejection is assigned zero exergy value.
The thermal efficiency (first law) is Eq.12.1
\eta_{ TH }=\frac{\dot{ W }_{\text {net }}}{\dot{ Q }_{ H }}=\frac{ w _{\text {net }}}{ q _{ H }}=1- r _{ p }^{-( k -1) / k }
The corresponding 2^{\text {nd }} \text { law } efficiency is
\eta_{\text {II }}=\frac{w_{\text {net }}}{\Phi_{ H }}=\frac{ h _3- h _2-\left( h _4- h _1\right)}{ h _3- h _2- T _{ O }\left( s _3- s _2\right)}
where we used Φ_H = increase in flow exergy =\Psi_3-\Psi_2= h _3- h _2- T _{ o }\left( s _3- s _2\right) .
Now divide the difference h _3- h _2 out to get
Comment: Due to the temperature sensitivity of Φ_H the temperatures do not reduce out from the expression.