Chapter 7.5, Problem 24E

To determine

To Find: The time to be taken to fill the tank first time.

Answer to Problem 24E

The required time will be $32\text{hr}$ .

Explanation of Solution

Given information:

The figure is:

  

Height of the well 300 ft.

Length of the pipe 4 in.

Diameter of the cylindrical base, 20 ft.

Height of the same is 25 ft.

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, $F\left(x\right)=\gamma x$ .

  $\text{Power}=\frac{\text{Work Done}}{\text{Time}}$

Calculation:

Since there is two cross-sectional area, the work done will be:

  ${{\displaystyle \int }}_0^{360}\gamma Axdx+{{\displaystyle \int }}_{360}^{385}\gamma A^{\prime }xdx={{\displaystyle \int }}_0^{360}\gamma \pi {\left(2\text{in}\right)}^{2}xdx+{{\displaystyle \int }}_{360}^{385}\gamma \pi {\left(10\text{ft}\right)}^{2}xdx$

  $\begin{array}{c}=\gamma \pi \left({{\displaystyle \int }}_0^{360}{\left(\frac{1}{6}\right)}^{2}xdx+{{\displaystyle \int }}_{360}^{385}{\left(10\right)}^{2}xdx\right){\text{ft}}^4\\ =\gamma \pi \left(\frac{1}{36}{{\displaystyle \int }}_0^{360}xdx+100{{\displaystyle \int }}_{360}^{385}xdx\right){\text{ft}}^4\end{array}$

  $=\gamma \pi \left(\frac{1}{36}{\left[\frac{x^2}{2}\right]}_0^{360}+100{\left[\frac{x^2}{2}\right]}_{360}^{385}\right){\text{ft}}^4$

  $=\gamma \pi \left(\frac{1}{36}\left[\frac{{\left(360\right)}^2}{2}-0\right]+100\left[\frac{{\left(385\right)}^2-{\left(360\right)}^2}{2}\right]\right){\text{ft}}^4$

  $=\left(62.4{\text{lb/ft}}^3\right)\pi \left(\frac{1}{36}\left[\frac{{\left(360\right)}^2}{2}-0\right]+100\left[\frac{{\left(385\right)}^2-{\left(360\right)}^2}{2}\right]\right){\text{ft}}^4$

  $=\left(62.4\right)\pi \left(\frac{1}{36}\left[\frac{{\left(360\right)}^2}{2}\right]+100\left[\frac{\left(385-360\right)\left(360+385\right)}{2}\right]\right)\text{lb-ft}$

  $\text{=1}\text{.82911}\times {\text{10}}^8\text{lb-ft}$

Time for filling the tank and pipe:

  $\begin{array}{c}\frac{\text{Work Done}}{1650}=\frac{1.8911\times {10}^8}{1650}\\ =114612.121\text{sec}\end{array}$

For nearest hour $\frac{114612.121}{3600}\text{hr}=31.8367\sim 32\text{hr}$ .