written 7.0 years ago by
teamques10
★ 64k
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modified 7.0 years ago
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Given
Head H = 14 m
Hub diameter $D_b= 0.35\times D_o$
Where $D_o$ is the diameter of runner
Speed N =100 r.p.m
Vane angle at outlet $\phi =20^\circ $
Flow ratio $=\dfrac {Vf_1}{\sqrt{2gH}} =0.6 \\ Vf_1 = 0.6\times \sqrt{2gH} \\ =0.6\times \sqrt{2\times 9.81 \times 14} = 9.94 m/s $
From the outlet velocity triangle $Vw_2=0$
$\tan \phi = \dfrac {Vf_2}{U_2} = \dfrac {Vf_1}{U_2} \space \space \space (\because Vf_2=Vf_1=9.94) \\ \tan 2=^\circ = \dfrac {9.94}{U_2}\\ U_2 = 27.30 m/s $
But for kaplan turbine $U_1=U_2 = 27.30 m/s $
Now using the relation $U_1 = \dfrac {\pi D_0\times N}{60} \\ 27.30 = \dfrac {\pi\times D_0\times 100}{60} \\ D_0= \dfrac {60\times 27.30}{\pi\times 100} = 5.21 m$
$D_b =0.35\times D_o = 0.35\times 5.21=1.82m$
Discharge through turbine is given by equation
$Q= \dfrac \pi4 [D_o^2-D_b^2]\times Vf_1\\ = \dfrac{\pi}4 [5.21^2-1.82^2]\times 9.94 \\ = 186.05 m^3/s$