Question

A pump motor is used to deliver water at a certain rate from a given pipe. To obtain n times water from the same pipe at the same time, by what amount should the power of the motor be increased?
$\begin{aligned} & A.\text{ }{{\text{n}}^{2}}\text{ times} \\\ & \text{B}\text{. }{{\text{n}}^{3}}\text{ times} \\\ & C.\text{ n times} \\\ & \text{D}\text{. }{{\text{n}}^{3/2}}\text{ times} \\\ \end{aligned}$

Answer 1

In this question, we have given initial and final condition, calculate mass flow rate for initial condition and final condition. Write an equation for getting power in n time’s water in the same n times. Expression for power gives relation between force and velocity. So calculate power for initial and final condition, take ration and hence we will get an amount of power.

Complete step by step answer:
We have been provided a pump motor which is used to deliver water at a certain rate. Now we have to calculate the amount of power to which the power of the motor will be increased, to obtain n times water from the same pipe at the same time. Let initial mass flow rate is given by $\dfrac{dm}{dt}.......\left( 1 \right)$
Let, $\rho$ is the density of water, A is the area of cross section and V is the velocity at which water is coming out, we know that,
Density $\left( \rho \right)$ = $\dfrac{mass\text{ (m)}}{volume\text{ (v)}}$
Therefore, equation (1) can be written as,
Mass flow rate = $\dfrac{dm}{dt}=\dfrac{\rho dv}{dt}........\left( 2 \right)$
Now we know that volume can be written as $v=Adx$
Therefore, equation (2) implies
Mass flow rate = $\rho A\dfrac{dx}{dt}$
Since δ and A is constant and $\dfrac{dx}{dt}=v$ (i.e. velocity)
Mass flow rate = $\rho$ A v
Now if we want n times water in the same n times then, new mass flow rate is given by,
${{\left[ \dfrac{dm}{dt} \right]}^{'}}=n\left[ \dfrac{dm}{dt} \right]......\left( 3 \right)$
Let $\delta '$ be the new density of water, $A'$ be the new area of cross section and $v'$ be the new velocity of water then,
${{\left( \dfrac{dm}{dt} \right)}^{'}}=\rho 'A'v'$
So, equation (3) implies,
$\rho 'A'v'=n\text{ }\rho A\text{ }v......\left( 4 \right)$
Now density of water and area of cross section of pipe is be same i.e. $\rho '=\rho$ and $A'=A$ then term of density and area of cross section will cancel out in equation (4), we get
$v'=nv........\left( 5 \right)$
Let, P be the power and $P'$ be the new power then P and $P'$ are given as
$P=Fv$, $P'=F'v'$
Taking ratio, we get
$\dfrac{P'}{P}=\dfrac{F'v'}{Fv}.........\left( 6 \right)$
Now we know that, expression of the force is given by,
$\begin{aligned} & F=v\dfrac{dm}{dt} \\\ & F'=v'{{\left( \dfrac{dm}{dt} \right)}^{'}} \\\ \end{aligned}$
Put in equation (6), we get
$\begin{aligned} & \dfrac{P'}{P}=\dfrac{v'{{\left( \dfrac{dm}{dt} \right)}^{'}}v'}{v{{\left( \dfrac{dm}{dt} \right)}^{}}v}=\dfrac{nv\text{ }n\left( \dfrac{dm}{dt} \right)nv}{v\left( \dfrac{dm}{dt} \right)v} \\\ & \text{(from }\left( 3 \right)\text{ and }\left( 5 \right)) \\\ & \dfrac{P'}{P}={{n}^{3}} \\\ & P'={{n}^{3}}P \\\ \end{aligned}$

So, the correct answer is “Option B”.

Note: Amount of the power of the motor depends on mass flow rate, force acting by water on pipe and velocity of water, mass flow rate is nothing but the rate of movement of liquid pass through a unit area. And it depends on density of water, area of cross section of pipe as well as velocity of water through a pipe.