Find out the amount of energy spent:
6 cubic metre of water is to be lifted to a height of 40 metres and delivered having velocity of 4 metres/sec. If this operation takes 12 mintues, find out
1. the amount of energy spent
2. H. P. utilized
3. H. P. of water-pump whose efficiency is 80%
Solution
Work is needed to be done on water to
1. raise its potential energy, and
2. Develop KE of water.
A. Potential Energy = W × H
Here, W = Weight of water lifted in Newtons
One litre of water = 1 kg = 9.8 Newtons
∴ 6 cu. m. of water = 6000 litres.
∴ W = 6000 g N
∴ Enhance in PE of water = 6000 g × 40
= 2352 × 10^{3} N-m.
B. Work done to provide KE to water = W V^{2} /2g
= (6000 × 4^{2})/2
= 48 × 10^{3} N- m.
Whole amount of energy spent = PE + KE
= (2352 + 48) × 10^{3} N-m.
= 2400 × 10^{3} N-m.
H. P. utilized in (S. I. units) = (2400 × 10^{3}) / 746
= 3217 units
If pump efficiency = 0.8 or 80%
H. P. required for pumps
= 3217 /0.8 = 4021 units.