Question

A motor pump is delivering water at a certain rate. In order to increase the rate of delivery by 100%, the power of the motor is to be increased by
A) $300\% $
B) $200\% $
C) $400\% $
D) $800\% $

Answer 1

Hint: The pump is a device used for converting electric energy to mechanical energy to lift the water to a certain height. To solve this problem, this power used by the pump should be expressed in terms of flow rate of the water.

Complete step by step solution:
The power is defined as the rate of work done per unit time. Mathematically, it is the ratio of work done to the time.
$P = \dfrac{E}{t}$
The work done to lift the water is converted to kinetic energy of water. Hence,
$E = \dfrac{1}{2}m{v^2}$
where m = mass of water and v = velocity of water.
The mass of water is equal to the product of its volume V and density, $\rho $.
$m = V\rho $
Also, the velocity of water is equal to distance travelled by the water per time.
$v = \dfrac{s}{t}$
Substituting these values in the equation for kinetic energy, we get –
$E = \dfrac{1}{2}V\rho {\left( {\dfrac{s}{t}} \right)^2}$
$ \Rightarrow E = \dfrac{1}{2}\rho V\left( {\dfrac{s}{t}} \right)\left( {\dfrac{s}{t}} \right)$
$ \Rightarrow E = \dfrac{1}{2}\rho \left( {\dfrac{V}{t}} \right)\left( {\dfrac{{{s^2}}}{t}} \right)$
By grouping, we have the quantity called volume flow rate, represented by Q, which is equal to the volume of fluid discharge per time.
Hence,
$E = \dfrac{1}{2}\rho Q\left( {\dfrac{{{s^2}}}{t}} \right)$
where $Q = \dfrac{V}{t}$
Power,
$P = \dfrac{E}{t} = \dfrac{1}{2}\rho Q\left( {\dfrac{{{s^2}}}{t}} \right) \times \dfrac{1}{t}$
$ \Rightarrow P = \dfrac{1}{2}\rho Q{\left( {\dfrac{s}{t}} \right)^2}$
$ \Rightarrow P = \dfrac{1}{2}\rho Q{v^2}$
Here, we have expressed power of the pump in terms of the volume flow rate.
Given, that the rate of delivery is to be increased by $100\% $ , the new volume flow rate will be –
${Q_1} = 2Q$
Also, the velocity of water will also be twice since the volume flow rate is proportional to the velocity.
Hence, ${v_1} = 2v$
Power required is,
${P_1} = \dfrac{1}{2}\rho {Q_1}v_1^2$
Substituting,
${P_1} = \dfrac{1}{2}\rho {Q_1}v_1^2$
${P_1} = \dfrac{1}{2}\rho \times 2Q \times {\left( {2v} \right)^2}$
$ \Rightarrow {P_1} = 2 \times 4 \times \dfrac{1}{2}\rho Q{v^2}$
$ \Rightarrow {P_1} = 8P$
Thus, power has to be increased by 8 times or 800%.

Hence, the correct option is Option D.

Note: In this problem, the flow rate is automatically considered as volume flow rate since we are using water which is an incompressible fluid. In an incompressible fluid, there is no change in the density during the flow. During compressible fluid, the flow rate should be considered as mass flow rate, which is mass per unit time.