How caluclate the time for which a motor pump of 10Hp and efficiency 80% be switched on so as to pump 20m3 ?

Question

. [density of water=1000 kgm-3,g=a0ms-2,1Hp=750W]

Answer 1

Mass m = volume * density = 20*1000 = 20,000 kg

Height h = 20 m

Work is done against gravity.

Work done = mgh = 20,000 * 10 * 20 = 4 * 10^6 Joule

Motor pump has power of 10HP and efficiency of 80%.

So, effective power = 80% if 10 HP = 10 * 80/100 = 8 HP

1 hp = 750 W

Therefore, effective power = 8 * 750 W = 6,000 W

Time = work done/effective power

= 4*10^6/6000 sec = 4*10^3/6 sec = 4000/6 sec = 2000/3 sec = 666.67 sec