In a gas turbine plant working on Brayton cycle, hot combustion gases from combustion chamber enters into gas turbine at 1450 K and 20 bar and exits at 1 bar. The isentropic efficiency of the gas turbine is 0.90. The work developed by turbine per kg of gas flow is ______(kJ/kg)

(Correct upto two decimal places)

Takes $C_{p} = 1.00 k J / g$ and $C_{v} = 0.769 k J / k g k$
651.3084

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Solution

The correct option is A 651.3084
$Given :$

$T_{3} = 1450 K, C_{p} = 1.00 k J / k g K,$

$C_{v} = 0.769 k J / k g K$

$(Adiabatic index)= \frac{C_{p}}{C_{v}} = \frac{1}{0.769}$

$γ = 1.3$

$\frac{T_{3}}{T_{4 s}} = {(\frac{P_{3}}{P_{4}})}^{\frac{γ - 1}{γ}}$

$\Rightarrow \frac{1450}{T_{4 s}} = {(\frac{20}{1})}^{\frac{1.3 - 1}{1.5}}$

$\Rightarrow T_{4 s} = 726.324 K$

$Isentropic efficiency, η_{T} = \frac{(T_{s} - T_{4})}{(T_{s} - T_{4 s})}$

$0.90 = \frac{(1450 - T_{4})}{(1450 - 726.324)}$

$T_{4} = 798.6916 K$

$Work developed by turbine, W = C_{p} (T_{s} - T_{4})$

$W = 1 (1450 - 798.6916)$

$W = 651.3084 k J / k g$

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