0
1.1kviews
Axial flow compressor has a constant axial velocity of 150 m/s and 50% reaction.The mean diameter of the blade ring is 35 cm and speed is 15,000 rpm.The exit angle of the blade is 27deg.

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

0
41views

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,

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