Chapter 6.6, Problem 22E

To determine

To Calculate: The work done needed to pump all the water out of a tank from a spout located at the top.

Answer to Problem 22E

The work needed to empty the tank is $45000lb\cdot ft$

Explanation of Solution

Given Information: The dimensions of the tank are

• Height of the tank is $6ft$ .
• Length of the tank at the top is $12ft$ and at the bottom is $0ft$ .
• Width of the tank is $10ft$ .
• 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

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

Calculation: At a depth $h$ from the top of the tank (not spout), its height with respect to the bottom is $6-h$ .

The length $l$ of the tank at any level is proportional to the height at that level.

So, to find the width $l$ at any depth $h$ , equate the ratio of widths and heights to write

  $\frac{l_r}{l_{top}}=\frac{heigh{t}_r}{heigh{t}_{top}}$

  $\Rightarrow \frac{l}{12}=\frac{6-h}{6}$

  $\Rightarrow l=12-2h$

Now, the mass of an infinitesimally thin layer of water of thickness $dh$ and at a depth $h$ from the top of the tank is

  $\begin{array}{l}dm=density\times volume=\rho dv=\rho \times A\times dh\\ =\rho \cdot (w\times l)\cdot dh=\rho w(12-2h)dh\\ =625\times (12-2h)dh\end{array}$

This infinitesimal mass of water located at the depth $h$ moves up a distance equal to its depth $h$ .

So, the work done in pumping this infinitesimally small mass of water out is

  $dW=dm\times h=625\times (12-2h)dh\cdot h$

So, the total work needed to pump the entire water out of the tank is the integral of the above $dW$ from $h=0$ (depth of the top most layer of water) to $h=8$ (depth of the bottom most layer of water).

Therefore

  $W={{\displaystyle \int }}_0^{6}dW=625{{\displaystyle \int }}_0^6(12h-2h^2)dh$

  $\begin{array}{l}=625{\left[12\cdot \frac{h^2}{2}-\frac{2h^3}{3}\right]}_0^6\\ =625\left(6\times 6^2-\frac{2\times 6^3}{3}\right)\\ =45000\end{array}$

Therefore, the total work done is $45000lb\cdot ft$ .