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.

Solution

1. Calculate Key Velocities

First, we calculate the theoretical velocity from the head, then use the ratios to find the peripheral and flow velocities.

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

2. Calculate Discharge (Q)

First, find the required water power using the overall efficiency, then find the discharge.

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

3. Find the Diameter of the Runner (\(D_o\))

The discharge is related to the annular flow area and the velocity of flow.

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

4. Find the Speed of the Runner (N)

The rotational speed is calculated from the peripheral velocity at the outer diameter.

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