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

Marks: 08

Years: MAY 2016

Head on turbine $H_1 = 20m$

Speed $N_1=200 r.p.m$

Discharge $Q_1= 9m^3/s $

Overall efficiency $N_o = 90\%$ or $0.90$

Performance of the turbine under a head $H_2=20m $ means to find the speed discharge and power developed by the turbine when working under the head of 20 m

Let for the head $H_2 = 20 m$

Speed $=N_2$

discharge $Q_2$

Power $P_2$

Using the relation

$N_o = \dfrac {P}{W.P} = \dfrac {P_1}{\dfrac {p\times g\times Q_1\times H_1}{1000}}\\ P_1 = \dfrac {N_o\times p\times g\times Q_1\times H_1}{1000} = \dfrac {0.90\times 1000\times 9.81\times 9\times 20}{1000}\\ =1589.22Kw \\ \dfrac {N_1}{\sqrt{H_1}} = \dfrac {N_2}{\sqrt{H_2}} \\ N_2 = \dfrac {N_1\sqrt{H_2}}{\sqrt{H_1}} = \dfrac {200\times \sqrt{20}}{\sqrt{20}} = 200r.p.m \\ Also \dfrac {Q_1}{\sqrt{H_1}}= \dfrac {Q_2}{\sqrt{H_2}} \\ Q_2 = Q_1\times \dfrac {\sqrt{H_2}}{\sqrt{H_1}}= 9\times \dfrac {\sqrt{20}}{\sqrt{20}} = 9m^3/s$

And $\dfrac {P_1}{H_1^{3/2}}=\dfrac {P_2}{H_2^{3/2}} \\ P_2 = \dfrac {P_1H_2^{3/2}}{H_1^{3/2}}= P_1 (\dfrac {H_2}{H_1})^{3/2} \\ =1589.22(\dfrac {20}{20})^{3/2} \\ =1589.22 Kw $