Chapter 6.6, Problem 18E

To determine

To Calculate: The work done needed to pump all the water out of a circular swimming pool.

Answer to Problem 18E

The work needed to empty the swimming pool is $339120lb\cdot ft$

Explanation of Solution

Given Information: The dimensions of the tank are

• The diameter of the pool is $24ft$ .
• Height of the pool is $5ft.$
• The depth of water is $4ft$ .
• Density of water is $62.5lb/f{t}^3$ .

Concept Used:The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

The area of a circle of diameter $d$ is $\text{π}{\left(\frac{d}{2}\right)}^2$ .

Calculation: The mass of an infinitesimally thin layer of water of thickness $dh$ and at a depth $h$ from the top of the pool is

  $\begin{array}{l}dm=density\times volume=\rho dv=\rho \times A\times dh=\rho \pi {(}^{\frac{d}{2}}dh\\ =62.5\times \pi \times {(}^{\frac{24}{2}}dh=9000dh\end{array}$

This infinitesimal mass of water moves up a distance equal to its depth $h$ from the top for it to be pumped out. So, the work done in pumping this infinitesimally small mass of water is

  $dW=dm\times h=9000\pi dh\cdot h$

The total work needed to pump the entire water out of the pool is the integral of the above $dW$ from $h=1ft$ (depth of the top most layer of water) to $h=5ft$ (depth of the bottom most layer of water)

Therefore

  $W={{\displaystyle \int }}_1^{5}dW={{\displaystyle \int }}_1^{5}9000\pi h\cdot dh$

  $={9000\pi \frac{h^2}{2}]}_1^5=9000\pi \times (\frac{5^2}{2}-\frac{1^2}{2})$

  $=339120$

Therefore, the total work done is $339120lb\cdot ft$ .