Question

A water pump of 1 horse power is required to flow water through a cylindrical pipe line at \[1{\text{ }}m/s\]. If flow rate of water is to be doubled then the power of new water pump required is:
$\left( A \right)1HP$
$\left( B \right)2HP$
$\left( C \right)4HP$
$\left( D \right)8HP$

Answer 1

Hint: In the above question the power of the first water pump is given. Power is the work by time ratio. The flow rate of the first pump and the second pump is given in the above data. Apply the relation between the power and flow rate between the two pumps to obtain the power of the new pump. Here the work done is in the form of potential energy.

Formula used:
$\dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)^3}$
In this formula ${P_2}$ and ${P_1}$ are the power of the second and first water pump respectively. ${V_2}$ and ${V_1}$ are the flow rate of the second and first water pump.

Complete step by step solution:
It is the work/time ratio. The SI unit of power is the Watt. Watt is equivalent to a Joule/second.
Energy that is stored in an object is called the potential energy. The stored energy in a system is based on its position, arrangement or state of the object.
Minimum power required to move water is called horsepower. The flow rate of the water and the pressure required to produce that flow is determined by horsepower.
Power of first and second water pump and the flow rate of water is given
$\dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)^3}$
In this formula ${P_2}$ and${P_1}$ are the power of the second and first water pump respectively.
Also, ${V_2}$ and ${V_1}$ are the flow rate of the second and first water pump.
${P_2} = {\left( {\dfrac{2}{1}} \right)^3} \times 1 = 8HP$
${V_2}$ is equal to two because in the question it is given that ${V_2}$ is double to that of ${V_1}$.

Hence option $\left( D \right)$ is the correct option.

Note: Rate of change of work with respect to time is called Power. Power input is always greater than the Power output, as energy is lost during conversion in various forms, such as heat and friction. None of the pumps can convert all of its mechanical power into water power. Some of the power is lost in the pumping process due to friction losses and other physical losses.