Question

A pump is used to pump a liquid of density $\rho$ continuously through a pipe of cross section area A. If liquid is flowing with a speed V, then power of pump is:
A) $\dfrac{1}{3}\rho A{V^2}$
B) $\dfrac{1}{2}\rho A{V^2}$
C) $2\rho A{V^2}$
D) $\dfrac{1}{2}\rho A{V^3}$

Answer 1

Hint: Power is defined as the rate of energy spent or the work done per unit time. Even though by definition, it is the ratio of work to time, there are various forms of equations to calculate the power. One of them is given here, which is required for the problem.
Power,
$P = pAV$
where p = pressure, A = area of flow and V = velocity of flow.

Complete step by step answer:
The power is defined as the amount of energy transferred per unit time.
$P = \dfrac{W}{t}$
where W = work and t = time.
The unit of power is watt (W).
Let us derive the equation of power in terms of a fluid flow.
The work done by a fluid when it flows through a distance s when applied a force of F, is given by –
$W = Fs$
Since this force is applied uniformly on the fluid, we can consider the force being equal to pressure of the fluid and the area A.
$F = p \times A$
Substituting the force in the equation for work, we have –
$\Rightarrow$ $W = pAs$
The power is calculated as –
$\Rightarrow$ $P = \dfrac{W}{t} = \dfrac{{pAs}}{t} = pA\dfrac{s}{t}$
Here, the rate of change of the displacement of the fluid per time is equal to the velocity of the fluid. Hence,
The power of fluid flow,
$\Rightarrow$ $P = pAV$
The Bernoulli’s equation gives the following relation between the pressures, during the flow of a fluid –
$\Rightarrow$ $P + \dfrac{1}{2}\rho {V^2} + \rho gh = C$
where $\rho $= density of the fluid and g = acceleration due to gravity.
where,
i) P – Static pressure: The mechanical pressure applied on the fluid.
ii) $\dfrac{1}{2}\rho {V^2}$ – Kinetic pressure: The pressure which arises due to the flow of the fluid.
iii) $\rho gh$ – Hydraulic pressure : The pressure which arises due to the height level of the fluid.
In this case of the pump, the power consumed only depends on the kinetic pressure which is related to the flow of the fluid.
Hence, we have to substitute the value of p in the equation of power for –
$\Rightarrow$ $p = \dfrac{1}{2}\rho {V^2}$
Substituting we get –
$\Rightarrow$ $P = pAV$
$ \Rightarrow P = \dfrac{1}{2}\rho {V^2} \times AV$
$\therefore P = \dfrac{1}{2}\rho A{V^3}$
Hence, the power required by the pump is equal to, $P = \dfrac{1}{2}\rho A{V^3}$.

Hence, the correct option is Option D.

Note: In order to be sure of the answer that we have obtained, let us perform dimensional analysis and validate our answer.
$P = \dfrac{1}{2}\rho A{V^3}$
Substituting the units, we get –
$P = \dfrac{{kg}}{{{m^3}}} \times {m^2} \times {\left( {\dfrac{m}{s}} \right)^3} = \dfrac{{kg}}{{{m^3}}} \times {m^2} \times \dfrac{{{m^3}}}{{{s^3}}}$
Rearranging,
$P = \dfrac{{kg \times m}}{{{s^2}}} \times m \times \dfrac{1}{s}$
$ \Rightarrow P = N \times m \times \dfrac{1}{s}$ , where N – newton
$ \Rightarrow P = J \times \dfrac{1}{s}$ , where J – joule
$\therefore P = 1W$
So, we obtained the units of power as watt by this relation. Hence, we can validate that our answer obtained is correct.