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Question 7.10: Determine for the data in example 7.9 the power transmitted ......

Determine for the data in example 7.9 the power transmitted for u = 4.5 m/s and u = 3 m/s.

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(i) u = 4.5 m/s, h_{f} = (0.014 × 3600 × 4.5²)/(2 × 9.81 × 0.25) = 208.07 m

Power = (π × 0.25²/4) × 4.5 × 1000 × 9.81 (450 – 208.07) = 524246 W = 524.25 kW

(ii) u = 3 m/s, h_{f} = (0.014 × 3600 × 3²)/(2 × 9.81 × 0.25) = 92.48 m

Power = (π × 0.25²/4) × 3 × 1000 × 9.81 (450 – 92.48) = 516493 W = 516.49 kW

This brings out clearly that the maximum power for a given diameter and head is when the frictional drop equals one third of available head.

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