A Kaplan turbine working under a head of 29 m develops 1287.5 kW S.P. If the speed ratio is equal to 2.1, flow ratio = 0.62, diameter of boss = 0.34 times the diameter of the runner and overall efficiency of the turbine = 89%, find the diameter of the runner and the speed of turbine.

$$ \text{Theoretical Velocity, } V_{th} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 29} \approx 23.85 \, \text{m/s} $$ $$ \text{Peripheral Velocity, } u = K_u \times V_{th} = 2.1 \times 23.85 \approx 50.085 \, \text{m/s} $$ $$ \text{Velocity of Flow, } V_f = K_f \times V_{th} = 0.62 \times 23.85 \approx 14.787 \, \text{m/s} $$

$$ \text{Water Power, } P_w = \frac{\text{Shaft Power}}{\eta_o} = \frac{1287500 \, \text{W}}{0.89} \approx 1446629 \, \text{W} $$ $$ Q = \frac{P_w}{\rho g H} = \frac{1446629}{1000 \times 9.81 \times 29} \approx 5.087 \, \text{m}^3/\text{s} $$

$$ Q = \text{Area of flow} \times V_f = \frac{\pi}{4}(D_o^2 - D_b^2) \times V_f $$ $$ \text{Substitute } D_b = 0.34 D_o \implies D_b^2 = 0.1156 D_o^2 $$ $$ Q = \frac{\pi}{4}(D_o^2 - 0.1156 D_o^2) \times V_f = \frac{\pi}{4}(0.8844 D_o^2) \times V_f $$ $$ 5.087 = \frac{\pi}{4}(0.8844 D_o^2) \times 14.787 $$ $$ 5.087 \approx 10.27 D_o^2 $$ $$ D_o^2 = \frac{5.087}{10.27} \approx 0.4953 \implies D_o \approx 0.704 \, \text{m} $$

$$ u = \frac{\pi D_o N}{60} \implies N = \frac{60 u}{\pi D_o} $$ $$ N = \frac{60 \times 50.085}{\pi \times 0.704} \approx 1358 \, \text{r.p.m.} $$