Question:

Consider a Pelton wheel of 1 m diameter. The magnitude of relative velocity of water at the bucket inlet is the same as the magnitude of relative velocity of water at the bucket exit. The absolute speed of water at the bucket inlet is 125.66 m/s\(^{-1}\). For maximum power output from the Pelton wheel, the rpm of the Pelton wheel is (rounded off to 1 decimal place).

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For maximum power output from a Pelton wheel, the tangential velocity of the wheel is half of the velocity of the water entering the bucket. Use this relationship to find the rotational speed (rpm).
Updated On: Apr 15, 2025
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Solution and Explanation

To calculate the rpm of the Pelton wheel for maximum power output, we need to consider the relationship between the velocity of water, the diameter of the Pelton wheel, and the rpm. Given: \[ D = 1 \ {m}, \quad V_1 = 125.66 \ {m/s} \] For maximum power output in a Pelton wheel, the bucket speed \( u \) is: \[ u = \frac{V_1}{2} = \frac{125.66}{2} = 62.83 \ {m/s} \] The linear speed \( u \) is related to rotational speed \( N \) by: \[ u = \frac{\pi D N}{60} \Rightarrow N = \frac{60u}{\pi D} \] Substituting the known values: \[ N = \frac{60 \cdot 62.83}{\pi \cdot 1} = \frac{3769.8}{\pi} \approx 1200.0 \ {rpm} \] 
Correct Answer: 1200.0 rpm

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