A Kaplan turbine develops 9000 kW under a net head of 7.5 m. Mechanical efficiency of the wheel is 86%. The speed ratio based on the outer diameter is 2.2 and the flow ratio is 0.66. Diameter of the boss is 0.35 times the external diameter of the wheel. Determine the diameter of the runner and the specific speed of the runner.

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Kaplan Turbine Design Calculation

Problem Statement

A Kaplan turbine develops 9000 kW under a net head of 7.5 m. Mechanical efficiency of the wheel is 86%. The speed ratio based on the outer diameter is 2.2 and the flow ratio is 0.66. Diameter of the boss is 0.35 times the external diameter of the wheel. Determine the diameter of the runner and the specific speed of the runner.

Given Data & Constants

  • Shaft Power, \(P_s = 9000 \, \text{kW}\)
  • Net Head, \(H = 7.5 \, \text{m}\)
  • Speed ratio, \(K_u = 2.2\)
  • Flow ratio, \(K_f = 0.66\)
  • Overall efficiency, \(\eta_o = 86\% = 0.86\) (*Assumed as per hint*)
  • Boss diameter ratio, \(D_b = 0.35 D_o\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

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.

$$ \text{Theoretical Velocity, } V_{th} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 7.5} \approx 12.13 \, \text{m/s} $$ $$ \text{Peripheral Velocity, } u = K_u \times V_{th} = 2.2 \times 12.13 \approx 26.69 \, \text{m/s} $$ $$ \text{Velocity of Flow, } V_f = K_f \times V_{th} = 0.66 \times 12.13 \approx 8.00 \, \text{m/s} $$

2. Calculate Discharge (Q)

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

$$ \text{Water Power, } P_w = \frac{\text{Shaft Power}}{\eta_o} = \frac{9000000 \, \text{W}}{0.86} \approx 10465116 \, \text{W} $$ $$ Q = \frac{P_w}{\rho g H} = \frac{10465116}{1000 \times 9.81 \times 7.5} \approx 142.23 \, \text{m}^3/\text{s} $$

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

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

$$ Q = \text{Area of flow} \times V_f = \frac{\pi}{4}(D_o^2 - D_b^2) \times V_f $$ $$ \text{Substitute } D_b = 0.35 D_o \implies D_b^2 = 0.1225 D_o^2 $$ $$ Q = \frac{\pi}{4}(D_o^2 - 0.1225 D_o^2) \times V_f = \frac{\pi}{4}(0.8775 D_o^2) \times V_f $$ $$ 142.23 = \frac{\pi}{4}(0.8775 D_o^2) \times 8.00 $$ $$ 142.23 \approx 5.51 D_o^2 $$ $$ D_o^2 = \frac{142.23}{5.51} \approx 25.81 \implies D_o \approx 5.08 \, \text{m} $$

4. Find the Speed of the Runner (N)

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

$$ u = \frac{\pi D_o N}{60} \implies N = \frac{60 u}{\pi D_o} $$ $$ N = \frac{60 \times 26.69}{\pi \times 5.08} \approx 100.4 \, \text{r.p.m.} $$

5. Find the Specific Speed of the Runner (\(N_s\))

The specific speed is a key parameter for classifying turbines.

$$ N_s = \frac{N \sqrt{P_s}}{H^{5/4}} \quad (\text{where P is in kW}) $$ $$ N_s = \frac{100.4 \sqrt{9000}}{(7.5)^{5/4}} = \frac{100.4 \times 94.87}{12.57} $$ $$ N_s \approx 758.4 $$
Final Design Parameters:

Diameter of the runner: \( D_o \approx 5.08 \, \text{m} \)

Specific speed of the runner: \( N_s \approx 758.4 \)

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