Calculate blade angle at inlet and work done per kg of air.

Given:

$V_f = 150\hspace{0.05cm}m/s,\hspace{0.25cm}R_d = 50\hspace{0.05cm}\%\\ D_m = 35\hspace{0.05cm}cm,\hspace{0.25cm}N = 15000\hspace{0.05cm}rpm,\hspace{0.25cm}\alpha = 27^\circ$

To Find: $\alpha_2 = \hspace{0.05cm}?,\hspace{0.25cm}\textit{W.D/kg of air} = \hspace{0.05cm}?$

Solution:

$\hspace{5cm}V_b = \frac{\pi D_{mean}.N}{60}\\ \hspace{5.5cm}= \frac{\pi\hspace{0.05cm}\times\hspace{0.05cm}0.35\hspace{0.05cm}\times\hspace{0.05cm}15000}{60}\\ \hspace{5.5cm}= 274.75\hspace{0.05cm}m/s\\ \hspace{5cm}\frac{V_b}{V_f} = \tan\beta_1 + \tan\beta_2\\ \hspace{4.5cm}\frac{274.75}{150} = \tan\beta_1 + \tan(27^\circ)\\ \hspace{5cm}\beta_1 = \alpha_2 = 52.898^\circ$

$\textit{W.D/kg of air} = V_b[V_{w2} - V_{w1}]$

$\tan\alpha_1 = \frac{V_{w1}}{V_{f1}}\\ V_{w1} = V_f.\tan\alpha_1\\ \hspace{0.5cm}= 150\hspace{0.05cm}\times\hspace{0.05cm}\tan(27)\\ \hspace{0.5cm}= 76.43\hspace{0.05cm}m/s$

$\tan\alpha_2 = \frac{V_{w2}}{V_{f2}}\\ V_{w2} = V_f.\tan\alpha_2\\ \hspace{0.5cm}= 150\hspace{0.05cm}\times\hspace{0.05cm}\tan(52.898)\\ \hspace{0.5cm}= 198.32\hspace{0.05cm}m/s$

Therefore,

$\textit{W.D/kg of air} = 274.75[198.32 - 776.43]\\ \hspace{2.2cm}= 33489.28\hspace{0.05cm}\textit{J/kg of air}\\ W.D = 33.489\hspace{0.05cm}\textit{KJ/kg of air}$