A Kaplan turbine runner is to be designed to develop 7357.5 kW S.P. The net available head is 10 m. Assume that the speed ratio is 1.8 and flow ratio is 0.6. If the overall efficiency is 70% and diameter of the boss is 0.4 times the diameter of the runner, find the diameter of the runner, its speed and specific speed.

Leave a Comment / Fluid Mechanics, Numerical (Water) / By
Kaplan Turbine Design Calculation

Problem Statement

A Kaplan turbine runner is to be designed to develop 7357.5 kW S.P. The net available head is 10 m. Assume that the speed ratio is 1.8 and flow ratio is 0.6. If the overall efficiency is 70% and diameter of the boss is 0.4 times the diameter of the runner, find the diameter of the runner, its speed and specific speed.

Given Data & Constants

  • Shaft Power, \(P_s = 7357.5 \, \text{kW}\)
  • Net Head, \(H = 10 \, \text{m}\)
  • Speed ratio, \(K_u = 1.8\)
  • Flow ratio, \(K_f = 0.6\)
  • Overall efficiency, \(\eta_o = 70\% = 0.70\)
  • Boss diameter ratio, \(D_b = 0.4 D_o\)
  • 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.

$$ \text{Theoretical Velocity, } V_{th} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 10} \approx 14.007 \, \text{m/s} $$ $$ \text{Peripheral Velocity, } u = K_u \times V_{th} = 1.8 \times 14.007 \approx 25.21 \, \text{m/s} $$ $$ \text{Velocity of Flow, } V_f = K_f \times V_{th} = 0.6 \times 14.007 \approx 8.40 \, \text{m/s} $$

2. Calculate Discharge (Q)

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

$$ \text{Water Power, } P_w = \frac{\text{Shaft Power}}{\eta_o} = \frac{7357500 \, \text{W}}{0.70} \approx 10510714 \, \text{W} $$ $$ Q = \frac{P_w}{\rho g H} = \frac{10510714}{1000 \times 9.81 \times 10} \approx 107.14 \, \text{m}^3/\text{s} $$

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

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

$$ Q = \text{Area of flow} \times V_f = \frac{\pi}{4}(D_o^2 - D_b^2) \times V_f $$ $$ \text{Substitute } D_b = 0.4 D_o \implies D_b^2 = 0.16 D_o^2 $$ $$ Q = \frac{\pi}{4}(D_o^2 - 0.16 D_o^2) \times V_f = \frac{\pi}{4}(0.84 D_o^2) \times V_f $$ $$ 107.14 = \frac{\pi}{4}(0.84 D_o^2) \times 8.40 $$ $$ 107.14 \approx 5.54 D_o^2 $$ $$ D_o^2 = \frac{107.14}{5.54} \approx 19.34 \implies D_o \approx 4.4 \, \text{m} $$

4. Find the Speed of the Runner (N)

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

$$ u = \frac{\pi D_o N}{60} \implies N = \frac{60 u}{\pi D_o} $$ $$ N = \frac{60 \times 25.21}{\pi \times 4.4} \approx 109.6 \, \text{r.p.m.} $$

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

The specific speed is a key parameter for classifying turbines.

$$ N_s = \frac{N \sqrt{P_s}}{H^{5/4}} \quad (\text{where P is in kW}) $$ $$ N_s = \frac{109.6 \sqrt{7357.5}}{(10)^{5/4}} = \frac{109.6 \times 85.77}{17.78} $$ $$ N_s \approx 528.3 $$
Final Design Parameters:

Diameter of the runner: \( D_o \approx 4.4 \, \text{m} \)

Speed of the runner: \( N \approx 109.6 \, \text{r.p.m.} \)

Specific speed: \( N_s \approx 528.3 \)

Related Posts

Water is flowing over a sharp-crested rectangular weir of width 50cm into a tank with cross-sectional area 0.6m2. After a period of 30s the depth of water in the tank is 1.4m. Assuming a discharge coefficient of 0.9, determine the height of the water above the weir. If the rectangular weir is replaced by a 900 notch weir with the same head and a discharge coefficient of 0.8, calculate the depth increase of the water in the tank after 30s.

Leave a Comment / Flow measurement, Fluid Mechanics, Numerical (Water) / By /

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top