# A pump installed on ground takes 15 minutes to fill a water tank of 30m³ volume placed at a height of 50 m. If the efficiency of the pump is 30%, the power consumed by the pump will be ?

2
by kunal322

2014-09-05T23:16:20+05:30

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Energy needed to pump the water up to the tank is the change in the potential energy of water.

Actual Energy needed = m g h = 30m³ * 1000kg/m³ * 9.8m/sec² * 50m
E  = 1.47*10^7 Joules

Due to inefficiency of pump, the actual amount consumed = E / efficiency
= 1.47 * 10^7 J * 100/30 = 4.9 * 10^7 J

Power consumed by motor = energy utilized / time duration = 4.9*10^7 J/ 15*60 sec
= 54.4 kW

In case the efficiency is 33 % instead of 30% the Power consumed will be
= 49.45 kW

yeah efficiency must be 33%
No,the efficiency is 30%
then it must be a printing mistake
Never,this question is from nstse and it clearly states that the answer is 49KW
did u find out about the solution and is this method correct ?
2014-09-06T16:21:45+05:30
Power consumed by the pump = P
Power utilized by the pump = 30% of P = (3/10)P
Volume of water raised = v = 30 m³
Density of water = d = 1000 kg/m³
Mass of water raised = m = vd = 30 * 1000 kg
Acceleration due to gravity = g = 9.8 m/s²
Height to which water is raised = h = 50 m
Total work done by the pump = W = mgh = 30 * 1000 * 9.8 * 50 J
Time taken = t = 15 min = 15 * 60 s
Work done per second by the pump = Power utilized by the pump = W/t
= (30 * 1000 * 9.8 * 50)/(15 * 60) Watt = 49000/3 watt
Hence, (3/10) P = 49000/3
=> P = 490000/9 = 54444.44 watt = 54.44 kW
Thanx a lot!!
My pleasure.....
shall i answer ur maths question?