A gas turbine with air as the working fluid has an isentropic effciency of $0.70$ when operating at a pressure ratio of $3$ . Now, the pressure ratio of the turbine is increased to $5,$ while maintaining the same inlet conditions. Assume air as a perfect gas with specific heat ratio $γ = 1.4$ . If the speciic work output remains the same for both the cases, the isentropic efficiency of the turbine at the pressure ratio of $5$ is
(round off to two decimal places)
0.514

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Solution

The correct option is A 0.514
$\frac{P_{3}}{P_{4 s}} = 3$

$\frac{P_{3}^{'}}{P_{4 s}} = 5$
$w_{n e t_{1}} = c_{p} T_{3} {1 - \frac{1}{3^{γ - 1 / γ}}} 0.7$
$w_{n e t_{1}} = c_{p} T_{3} {1 - \frac{1}{5^{γ - 1 / γ}}} η$
$w_{n e t_{1}} = w_{n e t_{2}}$
$\Rightarrow η = \frac{(0.7) {1 - \frac{1}{3^{γ - 1 / γ}}}}{{1 - \frac{1}{5^{γ - 1 / γ}}}} = 0.514$
Consider it is asked only for turbine portion as for full cycle,data is not sufficient

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The correct option is A 0.514P3P4s=3 P′3P4s=5 wnet1=cpT3{1−13γ−1/γ}0.7 wnet1=cpT3{1−15γ−1/γ}η wnet1=wnet2 ⇒η=(0.7){1−13γ−1/γ}{1−15γ−1/γ}=0.514 Consider it is asked only for turbine portion as for full cycle,data is not sufficient