Question

A pump is required to lift $ 600{\text{ kg}} $ of water per minute from a well $ 25{\text{ m}} $ deep and to eject it with a speed of $ 50{\text{ m/s}} $ . Calculate the power required to perform the above task.

Answer 1

Hint: Power is the amount of energy transferred or converted per unit time. It is also defined as the work done per unit time. The S.I unit of power is watt which is equal to one joule per second. The dimensions of power are $ M{L^2}{T^{ - 3}} $ .

Complete step by step answer:
From the definition of power,
 $ P = \dfrac{W}{{\Delta t}} $
Work done is the measure of the transfer of energy that occurs when a body is moved over a distance by external force.
Here two kinds of work are done.
i) In lifting the water at a height of $ 25{\text{ m}} $
ii) in ejecting the water with a speed of $ 50{\text{ m/s}} $
Work done in case (i)
 $ W = Mgh $ where $ M $ is the mass of the water, $ g $ is the acceleration due to gravity, $ h $ is the depth of the well.
 $ $ $ \Rightarrow {W_1} = 600 \times 10 \times 25 = 150000{\text{ J}} $
Work done in (ii)
 $ W = \dfrac{1}{2}M{v^2} $ $ W = \dfrac{1}{2}M{v^2} $ where $ v $ is the speed with which the water is to be ejected.
 $ \Rightarrow {W_2} = \dfrac{1}{2} \times 600 \times 50 \times 50 = 750000{\text{ J}} $
Total work done,
 $ W = 150000 + 750000 = 900000{\text{ J}} $
Therefore,
 $ P = \dfrac{W}{{\Delta t}} = \dfrac{{900000}}{{60}} = 15000{\text{ W}} $
The power required to pump the $ 600{\text{ kg}} $ of water per minute from a well $ 25{\text{ m}} $ deep and to eject it with a speed of $ 50{\text{ m/s}} $ is $ 15000{\text{ W}} $ .

Note:
Instead of calculating the total work done we can also calculate the total energy by adding the potential energy of water at the mouth of the well at a height of $ 25{\text{ m}} $ and the kinetic energy of the water ejecting with a speed of $ 50{\text{ m/s}} $ .