Power developed by a Pelton wheel is maximum if ___________.

Explanation:

Hydraulic or blading efficiency of Pelton turbine

\({\eta _h} = \frac{{2u\left( {{V_1} - u} \right)\left( {1 + k\cos{\beta _2}} \right)}}{{{V_1}^2}} = 2\left( {ρ - {ρ ^2}} \right)\left( {1 + k\cos{\beta _2}} \right)\)

where u = bucket speed, V1 = jet speed

For maximum efficiency: 

\(\frac{d}{{dρ }}({\eta _h}) = 0\; \to \;ρ = \frac{u}{{{V_1}}} = \frac{1}{2}\)

where ρ is bucket to jet-speed ratio.

So for maximum power:

\(ρ = \frac{u}{V_1} = \frac{1}{2} = 0.5\)

∴ Power developed by a Pelton wheel is maximum if jet velocity is equal to twice of bucket velocity.