Question

A pump on the ground floor of a building can pump up water to fill a tank of volume $30{m^3}$ in $15\,\min .$If the tank is $40m$ above the ground and the efficiency of the pump is $30\% ,$ how much electric power is consumed by the pump?
(Take$g = 10m{s^{ - 2}}$)
A. $36.5\,KW$
B. $44.4\,KW$
C. $52.5\,KW$
D. $60.5\,KW$

Answer 1

Hint: In this, first we will find the work done by pump to lift the water. Then we obtain the power of the pump by dividing the work done by time taken. Now multiply by 100/30 to get electric power consumed.
Complete step by step answer:
Volume of water lifted $V = 30{m^3}$
volume of water lifted $V = 30{m^3}$
Let us convert it to mass.
Mass of water lifted $M = volume \times density$
Now, density of water is $P = {10^3}Kg/{m^3}$
So mass of water lifted $ = 30 \times {10^3} = 3 \times {10^4}Kg...........(1)$
Height through which water is lifted $(h) = 40m$
Time in which water is lifted $ = t = 15\min $
Let us convert to SI unit, seconds
$t = 15 \times 60 = 900s.$
So, work done in lifting the water,
$W = M\,\,g\,\,h$
$ = 3 \times {10^4} \times 10 \times 40$ {Given$g = 10m{s^{ - 2}}$}
$W = 12 \times {10^6}J$
Power $ = \dfrac{w}{t} = \dfrac{{12 \times {{10}^6}}}{{900}} = \dfrac{4}{3} \times {10^4}w = \dfrac{{40}}{3}KW$
So, $\dfrac{{40}}{3}KW$ is power required to lift the water, but the pump has only $30\% $ efficiency. This means that if a pump consumes $100kw$ then it does work equal to $30kw$ only.
So, in this case power consumed by pump $ = \dfrac{{40}}{3} \times \dfrac{{100}}{{30}} = 44.44KW$
So, the correct option is (B), $44.44KW$

So, the correct answer is “Option B”.

Note:
1. Sometimes, question is given in terms of hp (horse power), than we have to convert to SI unit, Watt by using expression, $1\,\,horse\,\,power\, = 746W.$
2. Another way to understand efficiency: pump with 30 % efficiency means that pump takes 100 W power, wastes 70 W power while converting and does work equal to only 30 W.