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A water pump driven by petrol raises water at a rate of 0.5 m' min from a depth of 30 m. If the pump is 70% efficient, what power is developed by the engine? |
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Answer» Solution :Here mass/MIN. `(dm)/(dt)=500 kg min^(-1)` =`(500)/(60)kg s^(-1)` Also `(70)/(100)xxP=(mgh)/(t)` or`P=(100)/(70).(mgh)/(t)` =`(100)/(70)xx(500)/(6)xx9.8xx30` =3500 W. |
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