Chapter 7.5, Problem 18E

To determine

To Find: the wok that it takes to pump the kerosene to the level of the top of the tank.

Answer to Problem 18E

Less work is required through a hose attached to the bottom of the tank.

Explanation of Solution

Given information:

Height of the cylinder is 30 ft and the diameter is 20 ft.

Weight of the kerosene is $51.2{\text{ lb/ft}}^3$ .

Formula used:

The work to move an object from $x=a$ to $x=b$ is: $W={\displaystyle \underset{a}{\overset{b}{\int }}F(x)dx}$ .

  $F\left(x\right)$ is weight of water above the pump, that is, $F\left(x\right)=\gamma x$ .

Calculation:

The work done is: $\pi {\left(10\right)}^2\left(51.2\right)\left(y\right)△y$

Hence the total work can be calculated by:

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{6}{\int }}4\pi \left(62.4\right)\left(y+15\right)dy}=249.6\pi {\left[\frac{y^2}{2}+15y\right]}_0^6\\ =249.6\pi \left[\frac{{\left(6\right)}^2}{2}+15\left(6\right)-0\right]\\ \approx 84687.3ft-lb\end{array}$

(ii) The work through the rim is

  $\pi {\left(2\right)}^2(62.4)(6+15)△y$

Hence the total work can be calculated by

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{6}{\int }}4\pi (62.4)21dy}=4\pi (62.4)21{\displaystyle \underset{0}{\overset{6}{\int }}dy}\\ =4\pi (62.4)(21){\left(y\right)}_0^6\\ =4\pi (62.4)(21)\left(6-0\right)\\ \approx 98801.8ft-lb\end{array}$

Therefore, the less work is required to use the valve at the bottom.