Chapter 12

Q. 12.13

A Kaplan turbine working under a head of 56 m develops 10 MW. The hub diameter of the runner is 0.35 times the outside diameter. The speed ratio and flow ratio are 2.1 and 0.67 respectively. If the overall efficiency is 85%, find out the diameter of the runner and the speed of the turbine.


Verified Solution

Given: H=6  m ; S P=10 \times 10^6  W ; D_h=0.35 D_o ; \rho=2.1 ; \psi=0.67 ; \eta_o=0.85

Tangential velocity of runner  u=\rho \sqrt{2 g H}=2.1 \times \sqrt{2 \times 9.81 \times 6}=22.785  m / s

Flow velocity  V_f=\psi \sqrt{2 g H}=0.67 \sqrt{2 \times 9.81 \times 6}=7.23  m / s

From the equation for efficiency, \eta_o=\frac{S P}{w Q H}=\frac{10 \times 10^6}{9810 \times Q \times 6}=0.85

∴                Discharge through turbine  Q=\frac{10 \times 10^6}{9810 \times 6 \times 0.85}=199.88  m ^3 / s

Also, discharge is given by the equation:

Q=\frac{\pi\left(D_o^2-D_h^2\right) V_f}{4}=\frac{\pi\left[D_0^2-\left(0.35 D_o\right)^2\right] \times 7.23}{4}

∴        Outer diameter D_o=6.33  m

Hub diameter    D_h=0.35 \times D_o=2.22  m

Mean diameter  D=\left(D_o+D_h\right) / 2=4.275  m

From the equation for tangential velocity, u=\pi D N / 60,  speed of turbine is:

N=\frac{60 u}{\pi D}=\frac{60 \times 22.785}{\pi \times 4.275}=101.8   rpm