Question #129099

Water is supplied to an axial flow turbine under a Total head of 35 m. The mean diameter of the runner is 2 m and is rotates at 145 rev/metre. Water leaves the guide vanes at 30 degrees to the direction of runner rotation at it mean radius and angle of the runner blade at outlet is 28 degrees. If 7% of the total head is lost in the casing and guide vanes and the relative velocity is reduced by 8% due to friction in the runner, determine the blade angle at inlet (at mean radius) and the hydraulic efficiency of the turbine.

Expert's answer

Given;

H = 35 m , D = 2 m N = 145 rev/ min, alpha = 30^{0 }, beta = 28^{0}_{ , }h_{g }=7% of H

Head available H = 0.93 * 35 = 32.6 m

inlet velocity V_{1} = 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

V^{2}_{r1} = 25.3^{2} + 15.2^{2}- (2*15.2*25.3 cos 30) = 14.3 m/s

again

15.2^{2} = 14.3^{2} + 25.3^{2} - ( 2*14.3*25.3 cos (3.14-alpha)) = 148^{0}

Therefore blade angle is 148^{0}

Hydraulic efficiency

C_{1} = 0.92 sqrt 2gH = 0.92 sqrt 2*9.81*32.6 = 23.26m/s

C_{w1} = C_{1} cos alpha = 23.26 cos 30 = 20.15 m/s

nh = C_{w1}U/gH = (20.15*15.2)/9.81*32.6 = 0.9577 or 95.8%

Therefore efficiency = 95.8%

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