A kaplan turbine runner is to be designed to develop 9100 kw. The net available head is 5.6m, If the speed ratio = 2.09, flow ratio = 0.68, overall efficiency = 86% and the diameter of the boss is 1/3 the diameter of the runner, Find the diameter of the runner, its speed and the specific speed of the turbine.

Given:

Power, P = 9100 kw

Net Head, H = 5.6 m

Speed ratio = 2.09

Flow ratio = 0.68

Overall efficiency, $n_o = 0.86%$

Diameter of boss = $\frac{1}{3}$ of diameter of runner.

$B_b = \frac{1}{3} D_o$

Now, speed ratio = $\frac{u_1}{\sqrt{2gH}}$

$\therefore$ $u_1 = 2.09 \times \sqrt{2 \times 9.81 \times 5.6}$

= 21.95m/s

Step No (1)

Flow ratio = $\frac{v_{f1}}{\sqrt{2gH}}$

$v_{f1} = 0.68 \times \sqrt{2 \times 9.81 \times 5.6}$

= 7.12 m/s

Step No (2)

overall efficiency, $n_o = \frac{P}{(\frac{p \times 9 \times Q \times H}{1000})}$

$Q = \frac{p \times 1000}{p \times g \times H \times n_o}$

$= \frac{9100 \times 1000}{1000 \times 9.81 \times 5.6 \times 0.86}$

$= 192.5 m^3/s$

Step No (3) By kaplan turbine.

$q = \frac{\pi }{ 4} (Do^2 - Db^2) \times xf_1$

$192.5 = \frac{\pi }{4} [ Do^2 - (\frac{Do}{3})^2] \times 7.12$

$ = \frac{\pi }{4} [ 1 - \frac{1}{9} ] Do^2 \times 7.12$

$D_o = \sqrt{\frac{4 \times 192.5 \times 9}{\pi \times 8 \times 7.12}}$

$D_o = 6.21m$

Step No (4) The speed of turbine,

$u_1 = \frac{\pi DN}{60}$

$N = \frac{60 \times u_1}{\pi \times D}$

$N = \frac{60 \times 21.95}{\pi \times 6.21} = 67.5 r.p.m.$

The specific speed is given by,

$N_s = \frac{N\sqrt{P}}{H^5/4}$

$= \frac{67.5 \times \sqrt{9100}}{5.6^5/4}$

$N_s = 746$