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A rotary air compressor working between 1 bar and 2.5 bar has internal and external diameters of impeller as 300mm and 600mm respectively.

The vane angle at inlet and outlet are 30 deg and 40 deg respectively. If air enters impeller at 15m/s. Find the speed of an impeller in r.p.m and work done by the compressor per kg of air.

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Given

$P_1 = 1\hspace{0.05cm}bar,\hspace{0.25cm}P_2 = 2.5\hspace{0.05cm}bar,\hspace{0.25cm}d_1 = 30\hspace{0.05cm}mm = 0.3\hspace{0.05cm}m\\ d_2 = 600\hspace{0.05cm}mm =0.6\hspace{0.05cm}m,\hspace{0.25cm}\beta_1 = 30^\circ,\hspace{0.25cm}\beta_2 = 40^\circ,\hspace{0.25cm}V_{f1} =15\hspace{0.05cm}m/s$

To Find: N = ?, W.D/kg of air = ?

Solutions: $\hspace{5cm}\tan\beta_1 = \frac{V_{f1}}{V_{b1}}\\ \hspace{5.5cm}V_{b1} =\frac{V_{f1}}{\tan\beta_1}=\frac{15}{\tan30} \\ \hspace{5.5cm}V_{b1} = 25.98\hspace{0.05cm}m/s = \frac{\pi d_1N}{60}\\ \hspace{6cm}=\frac{3.14\hspace{0.05cm}\times\hspace{0.05cm}0.3\hspace{0.05cm}\times\hspace{0.05cm}N}{60}\\ \hspace{5.5cm}N = 1654.7\approx1655\hspace{0.05cm}rpm$

Assuming flow velocity is constant throughout the compressor.

$V_{f2} = V_{f1} = 15\hspace{0.05cm}m/s\\ V_{b2} = \frac{\pi d_2N}{60} = \frac{3.14\hspace{0.05cm}\times\hspace{0.05cm}0.6\hspace{0.05cm}\times\hspace{0.05cm}1655}{60} = 51.97\hspace{0.05cm}m/s\\ (V_{b2} -V_{w2}) = \frac{V_{f2}}{\tan\beta_2}=\frac{15}{\tan40} = 17.87\hspace{0.05cm}m/s\\ V_{w2} = V_{b2} - (V_{b2} - V_{w2}) = 51.97 - 17.87 = 34.1\hspace{0.05cm}m/s$

$\textit{W.D/kg of air} = \frac{V_{b2} - V_{w2}}{1000}\\ \hspace{2cm}= \frac{15.97\hspace{0.05cm}\times\hspace{0.05cm}34.1}{1000}\\ \hspace{2cm}= 1.77\hspace{0.05cm}\textit{KJ/kg of air}$