The rate of flow of water through the nozzle fitted at the end of the penstock is $2.0m^3/s $ The angle of deflection of jet is $165^\circ $. Determine the power given by the water to the runner and also hydraulic efficiency of the pelton wheel. Take speed ratio =0.45 and $C_v=1.0$

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

Marks: 05

Years: DEC 2015

Given

Gross head $Hg=500m$

Head lost in friction $hf=\dfrac {Hg}3 =\dfrac {500}3 =166.7m $

$\therefore $ Net head $H=Hg-hf=500-166.7=333.30 m$

Discharge $Q=2.0 m^3/s $

Angle of Deflection $=165^\circ $

$\therefore $ Angle $\phi=180^circ -165^\circ =15^\circ$

Speed ratio = $0.45$

Coefficient of velocity $C_v=1.0$

Velocity of jet $V_1= C_v \sqrt{2gH} \\ =1.0\sqrt{2\times 9.81\times 333.3}\\ =80.86 m/s $

Velocity of wheel $u= \text {speed ratio }\times \sqrt{2gH}\\ u=u_1=u_2 \\ =0.45\times \sqrt{2\times 9.81 \times 333.3}\\ =36.387 m/s\\ Vr_1 = V_1- U_1 = 80.86-36.387\\ =44.473 m/s $

Also $Vw_1=V_1 =80.86 m/s $

From outlet velocity triangle we have

$Vr_2=Vr_1 = 44.473 \\ Vr_2 \cos \phi = U_2 + Vw_2\\ 44.473 \cos 15^\circ = 36.3877 + Vw_2 \\ Vw_2 =44.473 \cos 15^\circ -36.387 \\ = 6.57 m/s $

Work done by the jet on the runner per second is given by equation

$fav_1 [Vw_1+Vw_2]\times u = PQ [Vw_1+Vw_2]\times U \because (av_1=Q) \\ =1000\times 2.0 [8.86+6.57]\times 36.387 \\ =6362630 Nm/s$

$\therefore $ Power given by the water to ther runner in Kw

$=\dfrac {\text{work done per second }}{1000} = \dfrac {6362630}{1000} \\ =6362.63 Kw$

Hydraulic efficiency of the turbine is given by equation

$N_n= \dfrac {2[Vw_1 +Vw_2]\times u}{V_1^2} = \dfrac {2[80.86+6.57]\times 36.387}{80.86\times 80.86} \\ 0.9731 \space or\space 97.31\%$