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.

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.

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

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