Chapter 8.5, Problem 24E

To determine

The time to fill the tank.

Answer to Problem 24E

The time to fill the tank is $31$ hours.

Explanation of Solution

Given information :

Diameter of pipe $4in$

Diameter of the base of tank $20ft$

Height of the tank $25ft$

The base of the tank is $60ft$ above the ground.

Depth of the well is $300ft$ ,the pump is $3HP$ pump.

Formula used :

  $\Delta V=\pi r^2\times \text{ thikness}$

  $\text{Force}=\text{ weight }\times \Delta V$

  $W_{total}=W_{pipe}+W_{\tan k}$

Calculation :

Call $y$ as the height of the pipe being filled with water. The height of the pipe is $300\text{ ft+60 ft =360 ft}\text{.}$ Then, the height of the pipe that is not being filled with water is $360-y$ .

Here, the force required to fill the tank is in lb unit. To calculate this, multiply the water’s density by the volume of the pipe that is not filled with water. The volume of a cylinder is $V=\pi r^{2}h$ . Plug in the number to have

  $V\left(y\right)=\pi {\left(2\text{ in}\times \frac{1\text{ ft}}{12\text{ in}}\right)}^2\left(360-y\right)=\frac{1}{36}\pi \left(360-y\right){\text{ft}}^3$

Then the force required to fill the tank is

  $F\left(y\right)=62.4\frac{\text{lb}}{{\text{ft}}^3}\times \frac{1}{36}\pi \left(360-y\right){\text{ft}}^3=\frac{26}{15}\pi \left(360-y\right)\text{lb}$

To calculate the work, integrate $F\left(y\right)$ indefinitely with the range as the height of the pipe: $0\le y\le 360$

  $\begin{array}{c}{W}_{pipe}={\displaystyle \underset{0}{\overset{360}{\int }}\frac{26}{15}\pi \left(360-y\right)dy}\\ =\frac{26}{15}\pi {\left(360y-y^2\right)|}_0^{360}\\ =\frac{26}{15}\pi \left(360\left(360\right)-\frac{{360}^2}{2}\right)\\ \approx 352864\text{ lb}\text{.ft}\end{array}$

Call $y$ as the height of the tank being filled with water. The height of the tank from the pump is $300\text{ ft }+\text{ 60 ft}\text{}=358\text{ ft}$ . Then, the height of the tank that is not being filled with water is $385-y$ .

Here, the force required to fill the tank is in lb unit. To calculate this, multiply the water’s density by the volume of the tank that is not filled with water. The volume of a cylinder is $V=\pi r^{2}h$ . Plug in the number to have $V\left(y\right)=\pi {\left(10\right)}^2\left(385-y\right)=100\pi \left(385-y\right){\text{ ft}}^3$

Then the force required to fill the tank is

  $F\left(y\right)=62.4\frac{\text{lb}}{{\text{ft}}^3}\times 100\pi \left(385-y\right){\text{ft}}^3=6240\pi \left(385-y\right)\text{ lb}$

To calculate work, integrate $F\left(y\right)$ indefinitely with the range as the height of the tank $0\le y\le 25$

  $\begin{array}{c}{W}_{\tan k}={\displaystyle \underset{0}{\overset{25}{\int }}6240\pi \left(360-y\right)}dy\\ =6240\pi {\left(385y-\frac{y^2}{2}\right)|}_0^{25}\\ =6240\pi \left(385\left(25\right)-\frac{{25}^2}{2}\right)\\ \approx 182557949\text{ lb}\text{.ft}\end{array}$

Here, the force required to fill the tank is in lb unit. To calculate this, multiply the water’s density by the volume of the tank that is not filled with water. The volume of a cylinder is $V=\pi r^{2}h$ . Plug in the number to have

  $V=\pi {\left(10\right)}^2\left(385-y\right)=100\pi \left(385-\pi \right){\text{ ft}}^3$

Then the force required to fill the tank is :

  $F\left(y\right)=62.4\frac{lb}{f{t}^3}\times 100\pi \left(385-y\right)f{t}^3=6240\pi \left(385-y\right)lb$

To calculate work, integrate $F\left(y\right)$ indefinitely with the as the height of the tank $0\le y\le 25$

  $\begin{array}{c}{W}_{\tan k}={\displaystyle \underset{0}{\overset{25}{\int }}6240\pi \left(360-y\right)}dy\\ =6240\pi {\left(385y-\frac{y^2}{2}\right)|}_0^{25}\\ =6240\pi \left(385\left(25\right)-\frac{{25}^2}{2}\right)\\ \approx 182557949\text{ lb}\text{.ft}\end{array}$

Add the works together

  $\begin{array}{c}{W}_{total}=W_{pipe}+W_{\tan k}\\ =352864\text{ lb}+182557949\text{ lb}\text{.ft}\\ \text{=182910813 lb}\text{.ft}\end{array}$

To calculate the time to fill the tank, divide the work by the and convert the time to hour:

  $\text{182910813 lb}\text{.ft}\times \frac{\sec }{1650\text{ lb}\text{.ft}}\times \frac{1\text{ hr}}{3600\text{ sec}}\approx 31\text{ hrs}$