0
13kviews
Penstock supplies water from a resevoir to the pelton wheel with a gross head of 500 m. One third of the gross head is lost in friction in the penstock.

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

3
1.8kviews

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

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