Power developed by a Pelton wheel is maximum if ___________.

  1. jet velocity is equal to four times of bucket velocity. 
  2. jet velocity is equal to twice of bucket velocity.
  3. jet velocity is equal to half of bucket velocity.
  4. jet velocity is equal to bucket velocity.

Answer (Detailed Solution Below)

Option 2 : jet velocity is equal to twice of bucket velocity.

Detailed Solution

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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.

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