Question:

The ratio of speed of a Pelton wheel and the speed of the jet for maximum head efficiency is

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For maximum power output and efficiency in an impulse turbine like a Pelton wheel, the blade speed (or wheel speed) should be half the jet speed. This optimal ratio ensures the maximum transfer of kinetic energy from the jet to the wheel.
Updated On: May 22, 2025
  • \( 1.0 \)
  • \( 0.25 \)
  • \( 0.333 \)
  • \( 0.5 \)
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The Correct Option is D

Solution and Explanation

Step 1: Understand the parameters for a Pelton wheel.
A Pelton wheel is an impulse turbine. In a Pelton wheel, a high-velocity jet of water impinges on a series of buckets mounted on the periphery of a runner.
Let:
\( u \) = tangential velocity of the wheel (bucket speed)
\( v \) = velocity of the jet
Step 2: Recall the condition for maximum hydraulic efficiency of a Pelton wheel.
For maximum hydraulic efficiency (also known as head efficiency) in a Pelton wheel, the tangential velocity of the wheel should be half the velocity of the jet. This condition ensures that the relative velocity of the jet with respect to the bucket is maximized for energy transfer, and the water leaves the bucket with minimum absolute velocity. The condition for maximum efficiency is: \[ u = \frac{v}{2} \]
Step 3: Calculate the ratio of wheel speed to jet speed.
The question asks for the ratio of the speed of a Pelton wheel to the speed of the jet for maximum head efficiency, which is \( \frac{u}{v} \). From the condition for maximum efficiency: \[ \frac{u}{v} = \frac{1}{2} \] \[ \frac{u}{v} = 0.5 \] The final answer is $\boxed{\text{4}}$.
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