the overall efficiency is 60%. The diameter of the Boss is $\dfrac13$ of the diameter of the runner. Find the diameter of the runner, its speed, and its specific speed

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 08

Years: DEC 2014

Given :

Power P=7.57.5 Kw

Net head H = 5.5 m

Speed ratio = 2.09

Flow ratio = 0.68

Overall efficiency $N_0=60\% = 0.6$

Diameter of boss $=\dfrac 13$ of diameter of runner

$D_b= \dfrac 13 D_o$

Now speed ratio $= \dfrac {U_1}{\sqrt{2gH}} \\ \therefore U_1 = 2.09 \times \sqrt{2\times 9.81\times 5.5} =21.71 m/s $

Flow ratio $= \dfrac {Vf_1}{\sqrt{2gH}} \\ \therefore Vf_1 = 0.68\times \sqrt{2\times 9.81\times 5.5} = 7.06m/s $

The overall efficiency is given by $N_0 = \dfrac {P}{\dfrac {p\times g.Q.H}{1000}}\\ or \space Q = \dfrac {P\times 1000}{p\times g\times H\times N_0} = \dfrac {7357.5\times 1000}{1000\times 9.81\times 5.5\times 0.6} \\ =227.27m^3/s $

The discharge through a kaplan turbine is given by

$Q= \dfrac \pi4 [ D_o^2-D_b^2 ]\times Vf_1 \\ 227.27 = \dfrac \pi 4 [D_o^2 - (D_o/3)^2]\times 7.06\\ = \dfrac \pi4 [1-\dfrac 19]\times D_o^2 \times 7.06\\ D_o =\sqrt{\dfrac {4\times 227.27\times 9}{\pi\times 8\times 7.06}} = 6.79 m $

Speed pf turbine is given by $U_1 =\dfrac {\pi DN}{60} \\ N = \dfrac {60\times U_1}{\pi\times D} = \dfrac {60\times 21.71}{\pi\times 6.79}=61.06 r.p.m $

The specific speed is given by

$N_s = \dfrac {N\sqrt P}{H^{5/4} } = \dfrac {61.06\times \sqrt{7357.5}}{(5.5)^{5/4}}= 621.82$