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Question
A water pump lifts water from 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower that the engine should have to do this.
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Solution
Given:
Height through which water is lifted, h = 10 m
\[\text{ Flow rate of water } = \left( \frac{\text{ m }}{\text{ t } } \right)\]
\[ = 30 \text{ kg } /\min = 0 . 5 \text{ kg/s } \]
Power delivered by the engine,
\[\text{ P } = \frac{\text{ mgh } }{\text{ t } }\]
\[ = \left( 0 . 5 \right) \times 9 . 8 \times 10\]
\[ = 49 \text{ W } \]
1 hp = 746 w
So, the minimum horse power (hp) that the engine should possess
\[= \frac{\text{ p } }{746} = \left( \frac{49}{746} \right)\]
\[ = 6 . 6 \times {10}^{- 2} \text{ hp } \]
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