Air (ideal gas) enters a perfectly insulated compressor at a temperature of $310 K$ . The pressure ratio of the compressor of is $6$ . Specifci heat at constant pressure for air is $1005 J / k g . K$ and ratio of specific heats at constant pressure and constant volume is $1.4$ . Assume that specific heats of air are constant. If the isentropic efficiency of the compressor is $85$ percent, the difference in enthalpies of air between the exit and the inlet of the compressor is
\(kJ/kg) (round off of nearest integer).
245

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Solution

The correct option is A 245

$T_{1} = 310 K$
$r_{p} = \frac{P_{2}}{P_{1}} = 6 η_{s} = 0.85$
$c_{p} = 1.005 k J / k g K$
$γ = 1.4$
$∴ \frac{T_{2}}{T_{1}} = {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$\Rightarrow \frac{T_{2}}{310} = (6)^{(1.4 - 1) / 1.4}$
$\Rightarrow T_{2} = 517.225 K$
Now, $η_{i s e n} = \frac{W_{i s e n}}{W_{a c t u a l}} = \frac{h_{2} - h_{1}}{h_{2}^{'} - h_{1}}$
$= 0.85 = \frac{c_{p} (T_{2} - T_{1})}{h_{2}^{'} - h_{1}}$
$\Rightarrow h_{2}^{'} - h_{1} = \frac{1.005 (517.22 - 310)}{0.85}$
$= 245 k J / k g$

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