Question

A water pump driven by petrol rises water at a rate of 0.5 m3 min–1 from a depth of 30 m. If the pump is 70% efficient, what power is developed by the engine?

Answer 1

Given in the question :-
Mass, pump Flow rate = 30 kg / min. = 30 kg / 60 s. = 0.5 kg/s.
Height h = 10 m.

we know, pump water has negligible velocity , then it's K.E is zero.
Pump work to increase the level of water above 10 m.

Now weight of the water = mg
= 0.5 × 9.8 
= 4.9 N.
Now power = mgh
P= 4.9 × 10 m
P= 49 W.

Now we know 1 hp = 746 W.
So minimum power = 49 / 746 
=0.066
= 6.6 × 10⁻² hp.
Hence the minimum horsepower is 6.6 x 10⁻² hp.


Hope it helps