The thermal efficiency of an air-standard Brayton cycle in terms of pressure ratio $r_{p}$ and $γ (= c_{p} / c_{v})$ is given by

1 - \frac{1}{r_{p}^{γ - 1}}

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1 - \frac{1}{r_{p}^{1}}

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1 - \frac{1}{r_{p}^{1 / γ}}

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1 - \frac{1}{r_{p}^{(γ - 1) / γ}}

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Solution

The correct option is D $1 - \frac{1}{r_{p}^{(γ - 1) / γ}}$

$\frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{4}} \Rightarrow \frac{T_{4}}{T_{1}} = \frac{T_{3}}{T_{2}}$
Thermal efficiency
$=Work\,doneheat\,supplied=1−Heat\,rejectedHeat\,supplied$
$= 1 - \frac{c_{p} (T_{4} - T_{1})}{c_{p} (T_{3} - T_{2})} = \frac{1 - (\frac{T_{4}}{T_{1}} - 1)}{T_{2} - (\frac{T_{2}}{T_{2}} - 1)}$
$η_{t h} = 1 - \frac{T_{1}}{T_{2}} = 1 - \frac{1}{(r_{p})^{\frac{γ - 1}{γ}}}$
$⎡ ⎢ ⎢ ⎢ ⎣ \frac{T_{1}}{T_{2}} = {(\frac{p_{1}}{p_{2}})}^{\frac{γ - 1}{γ}} ⎤ ⎥ ⎥ ⎥ ⎦$

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\frac{C_{p}}{C_{v}} = γ

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