Given;
H = 35 m , D = 2 m N = 145 rev/ min, alpha = 300 , beta = 280 , hg =7% of H
Head available H = 0.93 * 35 = 32.6 m
inlet velocity V1 = square root of 2gH = sqrt 2*9.8*32.6 = 25.3 m/s
u = (3.14*2*145)/60 = 15.2 m/s
From inlet triangle of velocity
V2r1 = 25.32 + 15.22- (2*15.2*25.3 cos 30) = 14.3 m/s
again
15.22 = 14.32 + 25.32 - ( 2*14.3*25.3 cos (3.14-alpha)) = 1480
Therefore blade angle is 1480
Hydraulic efficiency
C1 = 0.92 sqrt 2gH = 0.92 sqrt 2*9.81*32.6 = 23.26m/s
Cw1 = C1 cos alpha = 23.26 cos 30 = 20.15 m/s
nh = Cw1U/gH = (20.15*15.2)/9.81*32.6 = 0.9577 or 95.8%
Therefore efficiency = 95.8%
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