Question

An area of land is an average of $2m$ below sea level. To prevent flooding, pumps are used to lift rainwater up to sea level. What is the minimum pump output power required to deal with $1.3 \times {10^9}kg$ rain per day?
A) $15KW$
B) $30KW$
C) $100KW$
D) $300KW$

Answer 1

A water pump is a device used to operate water and to lift it up. Power is the rate at which the energy flows in a given time. Power of a pump is defined as the minimum electric power required to operate a pump, so that it can lift the water up.

Complete step by step solution:
Step I: Energy is to be supplied to the water pump so that it can work. Therefore, work is said to be done.
Given that the
amount of water is $m = 1.3 \times {10^9}kg$
Since the water is lifted up, there will be a force of gravity also. So, $g = 10m/{s^2}$
$h = 2m$
$t = 1day$
$1day = 24hours = 60 minutes = 60 seconds$
$t = 1 \times 24 \times 60 \times 60$

Step II: The formula of power is written as ,$Power = \dfrac{{Energy}}{{time}}$
Since gravitational potential energy is the energy stored in an object due to its height, so the above formula becomes
$P = \dfrac{{mgh}}{t}$
Where m is the mass
$g$ is acceleration due to gravity
$h$ is the height and
$t$ is the time
Substitute the values in the above equation and solving for the value of P
$P = \dfrac{{1.3 \times {{10}^9} \times 10 \times 2}}{{1 \times 24 \times 60 \times 60}}$
$P = 300 \times {10^3} = 300kW$

Step III: The minimum output power required will be $300kW$

Hence, The correct option is (D).

Note: It is important to note that though energy and power are related to each other, they are different terms. Energy is the ability to cause any change, but power is how much time it takes to supply or provide energy. There are different forms of energy. The energy can not be created or destroyed but it can be transferred into other forms of energy like kinetic energy, potential energy and thermal energy etc.