written 7.5 years ago by
teamques10
★ 67k

•
modified 7.5 years ago

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^2D_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
$N_s = \dfrac {N\sqrt P}{H^{5/4} } = \dfrac {61.06\times \sqrt{7357.5}}{(5.5)^{5/4}}= 621.82$