GATE 2017 ME (ME1) – Question 44
Air passes through a gas turbine with pressure ratio 10, inlet temperature 1200 K, $c_p=1040$ J/(kg K), $\gamma=1.4$ and isentropic efficiency 0.9. Find actual exit temperature in K.
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Correct answer: 675 to 684
Explanation
The isentropic exit is $T_{2s}=T_1(p_2/p_1)^{(\gamma-1)/\gamma}=1200(1/10)^{0.4/1.4}=621.537$ K. Turbine efficiency is $\eta_t=(T_1-T_2)/(T_1-T_{2s})$. Hence $T_2=1200-0.9(1200-621.537)=679.383$ K. It is higher than the reversible exit because less work is extracted.
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