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A kaplan turbine runner is to be designed to develop 7357.5 Kw shaft power. The net available head is 5.5 m. Assume that the speed ratio is 2.09 and flow ratio is 0.68.

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

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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

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

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