The pressure rise due to compression in the compressor is limited to 1.6 bar. Take the mechanical efficiency of compressor as 80%.

Given:

$V_s = 6\hspace{0.05cm}m^3/min,\hspace{0.25cm}P_1 = 2.2\hspace{0.05cm}bar,\hspace{0.25cm}P_2 = 2.2\hspace{0.05cm}bar\\ P_i = 1.6\hspace{0.05cm}bar,\hspace{0.25cm}\eta_{mech} = 0.8$

To Find: $\textit{Power required} = \eta = ?$

Solution:

Power required = (Area A + Area B)

$\textit{Area A} = \frac{\Upsilon}{\Upsilon - 1}P_1V_5[(\frac{p_i}{P_1})^\frac{\Upsilon - 1}{\Upsilon} - 1]\\ \hspace{1cm} = \frac{1.4}{0.4}\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}\times\hspace{0.05cm}\frac{6}{60}\hspace{0.05cm}\times\hspace{0.05cm}[(\frac{1.6}{1})^\frac{0.4}{1.4} - 1]\\ \hspace{1cm}= 5030.22\hspace{0.05cm}W\approx 5.03\hspace{0.05cm}KW\\ \textit{Area B} = V_b = V_s(\frac{P_1}{p_i})^\frac{1}{r} = \frac{6}{60}\hspace{0.05cm}\times\hspace{0.05cm}(\frac{1}{1.6})^\frac{1}{1.4} = 0.07148\hspace{0.05cm}m^3/s\\ \hspace{1cm}= (P_2 - P_i)V_b\\ \hspace{1cm}=(2.2\hspace{0.05cm}\times\hspace{0.05cm}1.6)\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}0.07148 = 4288.8\hspace{0.05cm}W = 4.288\hspace{0.05cm}KW$

Power required = 5.03 + 4.288 = 9.318 KW

$P_{act} = \frac{9.318}{0.8} = 11.647\hspace{0.05cm}KW$

$P_{ideal} = \frac{\Upsilon}{\Upsilon - 1}P_1V_1[(\frac{P_2}{P_1})^\frac{\Upsilon}{\Upsilon - 1} - 1]\\ \hspace{1cm} =\frac{1.4}{0.4}\hspace{0.05cm}\times\hspace{0.05cm}1\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}\times\hspace{0.05cm}\frac{6}{60}[(\frac{2.2}{1})^\frac{0.4}{1.4} - 1]\\ \hspace{1cm} = 8.843\hspace{0.05cm}KW$

Compressor efficiency

$\eta_{comp} = \frac{P_{ideal}}{P_{actual}} = \frac{8.843}{11.647}\\ \eta_{comp} = 75.92\hspace{0.05cm}\%$