In a simple Brayton cycle, the pressure ratio is $8$ and temperature at the entrance of the compressor and turbine are $300 K$ and $1400 K$ respectively. Both compressor and gas turbine have isentropic efficiencies equal to $80$ %. For the gas, assume a constant value of $c_{p}$ (specific heat at constant pressure) equal to $1 k J / k g K$ and ratio of specific heats as $14$ . Neglect changes in kinetic and potential energies

The power required by the compressor in $k W / k g$ of gas flow are is

194.7

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243.4

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304.3

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378.5

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Solution

The correct option is C $304.3$
$Given data:$

$Pressure ratio,$

$r_{p} = 8$

$T_{1} = 300 K; T_{3} = 1400 K$

$η_{c} = η_{T} = 0.8$

$C_{p} = 1 k J / k g K$

$γ = 1.4$

$For adiabatic isentropic process 1 - 2 s$

$\frac{T_{2 s}}{T_{1}} = {(\frac{p_{2}}{p_{1}})}^{\frac{γ - 1}{γ}}$

$\frac{T_{2 s}}{300} = (r_{p})^{\frac{γ - 1}{γ}} = (8)^{\frac{1.4 - 1}{1.4}} = (8)^{0.2857}$

$\frac{T_{2 s}}{300} = 1.811$

$T_{2 s} = 300 \times 1.811 = 543.3 K$

$η_{C} = \frac{Isentropic rise in temperature}{Actual rise in temperature}$

$= \frac{T_{2 s} - T_{1}}{T_{2} - T_{1}}$

$∴, 0.8 = \frac{543.3 - 300}{T_{2} - 300}$

$T_{2} - 300 = \frac{243.3}{0.8} = 304.12$

$T_{2} = 604.12 K$

$Power required by compressor,$

$W_{C} = c_{p} (T_{2} - T_{1})$

$= 1 (604.12 - 300)$

$= 304.12 k J / k g$

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