Water is pumped steadily out of a flooded basement at a speed of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$through a uniform hose of radius $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$. The hose passes out through a window $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$ above the waterline. What is the power of the pump?

• Speed of water through the hose is $\mathrm{v}=5.0\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

• Radius of the hose is $\mathrm{r}=0.1\mathrm{cm}=0.01\mathrm{m}$.

• The height of the window above waterline is $\mathrm{h}=3.0\mathrm{m}$.

Use the work-energy theorem, equation of power relating to work and time, and the mass flow rate of water to find the power of the pump. According to the work-energy theorem, the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

Formulae are as follows:

$\mathrm{\Delta W}=\mathrm{\Delta K}+\mathrm{\Delta U}$

$\mathrm{\Delta K}=\frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{\Delta U}=\mathrm{mgh}$

$\mathrm{P}=\frac{\mathrm{\Delta W}}{\mathrm{\Delta t}}$

$\frac{\mathrm{\Delta m}}{\mathrm{\Delta t}}-\mathrm{\rho Av}$

Where the is mass, is velocity, is area, is time, is density, is the kinetic energy, is the potential energy, is the work done, is an acceleration due to gravity, is height, and is the power.

From the work-energy theorem,

$\mathrm{\Delta W}=\mathrm{\Delta K}+\mathrm{\Delta U}$

And, the equation of power is,

$\mathrm{P}=\frac{\mathrm{\Delta W}}{\mathrm{\Delta t}}$

$=\frac{\mathrm{\Delta K}+\mathrm{\Delta U}}{\mathrm{\Delta t}}$

$=\frac{\mathrm{\Delta m}}{\mathrm{\Delta t}}\left(\mathrm{gh}+\frac{1}{2}{\mathrm{v}}^2\right)$

Now, the rate of mass flow of fluid is given by,

$\frac{\mathrm{\Delta m}}{\mathrm{\Delta t}}-\mathrm{\rho Av}$

So, the equation of power will become,

$\mathrm{gh}\left(\mathrm{v}+\frac{1}{2}2\right)$

$=1000\times \mathrm{\pi }\times (0.01\mathrm{m}{)}^2\times (5\mathrm{m}/\mathrm{s})\left(9.8\mathrm{m}/{\mathrm{s}}^2\times 3.0\mathrm{m}+\frac{1}{2}(5.0\mathrm{m}/\mathrm{s}{)}^2\right)$

$=65.8\mathrm{W}$

Hence, power of the pump is, $65.8\mathrm{W}$.