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 × 10³ N-m.

B.      Work done to provide KE to water = W V² /2g

                                                                      = (6000 × 4²)/2

                                                                        = 48 × 10³  N- m.

Whole amount of energy spent = PE + KE

                                                     = (2352 + 48) × 10³  N-m.

                                                            = 2400 × 10³  N-m.

H. P. utilized in (S. I. units) = (2400 × 10³) / 746

                                            = 3217 units

If pump efficiency = 0.8 or 80%

H. P. required for pumps

= 3217 /0.8  = 4021 units.