Question 12.13: A Kaplan turbine working under a head of 56 m develops 10 MW......

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