A 137 mm diameter jet of water issuing from a nozzle impinges on the buckets of a Pelton wheel and jet is deflected through an angle of 165oby the buckets. The head available at the nozzle is 400 m. assuming coefficient of velocity as 0.97, speed ratio as 0.46, and reduction in relative velocity while passing through the buckets as 15%. Find:

a) The force exerted by jet on buckets in tangential direction.

b) Power developed.

GIVEN :

Diameter of jet, d = 0.137m

Area of jet = a = $\frac{π×0.137^2}{4}=$ $0.014 m ^2$

Angle of deflection , ($180-β_2$) = 165° = $β_2$=15°

Net head, H = 400m

$C_v$=0.97 , Speed ratio=0.46

Relative velocity at outlet = (1 - 0.15)× Relative velocity at inlet

∴ $Vr_2$ = 0.851$V_1$

$V_1=C_v\sqrt{2gH}=85.93 m/s=V_1$=$Vω _1=85.93 m/s^2$

$U_1=ϕ\sqrt{2gH}=85.93 m/s=U_1=40.75m/s=u2=u$

∴ From inlet velocity triangle, $Vr_1=V1-u$

W.K.T

$Vr_2=0.85 Vr_1$

$Vr_2=38.4m/s$

From outlet velocity triangle,

$Vω_2=Vr_2 cosβ_2-U$

$Vω_2=-3.65m/s$

Since $Vω_2 \lt U$ ,the outlet velocity triangle is modified as follows:

(i) $F_x= δ.a.V_1[Vω_1=Vω _2 ]$

$F_x=104.2 kN$

(ii) Power developed, $P=\frac{F_x × u}{100}$

P = 4247.15 kW