Chapter 7.5, Problem 17E

$\left(a\right)$

To determine

To find: The work done to empty the tank by pumping the water back to ground level once the tank is full.

Answer to Problem 17E

  $1,497,600\text{ft-lb}\text{.}$

Explanation of Solution

Given information: The rectangle tank, with its top at ground level, is used to catch runoff water. Assume the water weighs $62\cdot 4{\text{lb/ft}}^{\text{3}}$ .

How much work does it take to empty the tank by pumping the water back to ground level once the tank is full.

The length and width are $10\text{ft}$ and $12\text{ft}$ .

The typical slab between the planes at $y$ and $y+△y$ has a volume of:

  $\Delta V=\left(10\right)\left(12\right)\Delta y=120\Delta y$

Since the Force $F\left(y\right)$ required to lift the slab is equal to the weight, it follows that:

  $\begin{array}{l}F=62\cdot 4\Delta V\\ =62\cdot 4\times 120\Delta y\text{lb}\end{array}$

Assume there are $n$ slabs with the partition of $\left[0,20\right]$ and $y=y_k$ denotes the plane associated with $k^{th}$ slab of thickness $△y_k$ , now we can approximate the work done in lifting all the Riemann sum as follows:

  $\begin{array}{l}W=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\left(62\cdot 4\right)\left(120\right)}\left(y_k\right)\Delta y_k\\ ={\displaystyle \underset{0}{\overset{20}{\int }}\left(62\cdot 4\right)\left(120\right)ydy}\\ =\left(62\cdot 4\right)\left(120\right){\left[\frac{y^2}{2}\right]}_0^{20}\\ =\left(62\cdot 4\right)\left(120\right)\left[\frac{{20}^2}{2}-0\right]\\ =\left(62\cdot 4\right)\left(120\right)\left[\frac{400}{2}\right]\\ =1,497,600\end{array}$

Hence, the required work done to pump the water from the tank is $1,497,600\text{ft-lb}\text{.}$

$\left(b\right)$

To determine

To find: The time taken to empty the full tank if water is pumped to ground level.

Answer to Problem 17E

  $\approx 100\min$

Explanation of Solution

Given information: If the water is pumped to ground level with a $\left(5/11\right)-$ horsepower motor (work output $250\text{ft}\cdot \text{lb/sec}$ ). How long it will take to empty the full tank (to the nearest minute).

The time taken to empty the full tank with $\left(5/11\right)-$ horsepower motor is:

  $\begin{array}{l}T=\frac{W}{250}\\ =\frac{1497600}{250}\\ =5990\cdot 4\sec \\ \approx 100\min \end{array}$

$\left(c\right)$

To determine

To show: The pump in part $\left(b\right)$ will lower the water level $\text{10}\text{ft}$ (halfway) during the first $25\text{min}$ of pumping.

Explanation of Solution

Given information: the pump in part $\left(b\right)$ will lower the water level $\text{10}\text{ft}$ (halfway) during the first $25\text{min}$ of pumping.

Following the step of part $\left(a\right)$ , work done to lower the water level $\text{10}\text{ft}$ (halfway) during the first $25\text{min}$ of pumping.

  $\begin{array}{l}W={\displaystyle \underset{0}{\overset{10}{\int }}\left(62\cdot 4\right)\left(120\right)ydy}\\ =\left(62\cdot 4\right)\left(120\right){\left[\frac{y^2}{2}\right]}_0^{10}\\ =\left(62\cdot 4\right)\left(120\right)\left[\frac{{10}^2}{2}-0\right]\\ =\left(62\cdot 4\right)\left(120\right)\left[\frac{100}{2}\right]\\ =374,400\text{ft-lb}\end{array}$

Time taken to empty full tank will be:

  $\begin{array}{l}t=\frac{W}{250}\\ =\frac{374400}{250}\\ =1497\cdot 6\min \\ \approx 25\min \end{array}$

$\left(d\right)$

To determine

To find: The answers to part $\left(a\right)$ and $\left(b\right)$ , in a location where water weighs $62\cdot 26{\text{lb/ft}}^{\text{3}}$ and $62\cdot 5{\text{lb/ft}}^{\text{3}}$ .

Answer to Problem 17E

The work done is $1494240\text{ft-lb}$ and time taken is $99\min$ approximately where water weighs $62\cdot 26{\text{lb/ft}}^{\text{3}}$ and the work done is $1500000\text{ft-lb}$ and time taken is $100\min$ approximately where water weighs $62\cdot 26{\text{lb/ft}}^{\text{3}}$ .

Explanation of Solution

Given information: Because of differences in the strength of Earth’s gravitational field, the weight of a cubic foot of water at sea level can vary from as little as $62\cdot 26\text{lb}$ at the equator to as much as $62\cdot 59\text{lb}$ near the poles, a variation of about $0\cdot 5\%$ . A cubic foot of water that weighs $62\cdot 4\text{lb}$ in Melbourne or New York City will weighs $62\cdot 5\text{lb}$ in Juneau or Stockholm. What are the answers to parts $\left(a\right)$ and $\left(b\right)$ in a location where water weighs $62\cdot 26{\text{lb/ft}}^3?$

  $62\cdot 5{\text{lb/ft}}^3?$

The answers to part $\left(a\right)$ and $\left(b\right)$ , in a location where water weighs $62\cdot 26{\text{lb/ft}}^{\text{3}}$ is calculated as follows:

  $\begin{array}{l}W={\displaystyle \underset{0}{\overset{20}{\int }}\left(62\cdot 26\right)\left(120\right)ydy}\\ =\left(62\cdot 26\right)\left(120\right){\left[\frac{y^2}{2}\right]}_0^{20}\\ =\left(62\cdot 26\right)\left(120\right)\left[\frac{{20}^2}{2}-0\right]\\ =\left(62\cdot 26\right)\left(120\right)\left[\frac{400}{2}\right]\\ =1494240\text{ft-lb}\end{array}$

The time taken will be:

  $\begin{array}{l}\text{Time}=\frac{1494240}{250}\\ =5976\cdot 96\sec \\ \approx 99\min \end{array}$

The answers to part $\left(a\right)$ and $\left(b\right)$ in a location where water weighs $62\cdot 5{\text{lb/ft}}^{\text{3}}$ is calculated as follows:

  $\begin{array}{l}W={\displaystyle \underset{0}{\overset{20}{\int }}\left(62\cdot 5\right)\left(120\right)ydy}\\ =\left(62\cdot 5\right)\left(120\right){\left[\frac{y^2}{2}\right]}_0^{20}\\ =\left(62\cdot 5\right)\left(120\right)\left[\frac{{20}^2}{2}-0\right]\\ =\left(62\cdot 5\right)\left(120\right)\left[\frac{400}{2}\right]\\ =1500000\text{ft-lb}\end{array}$

The time taken will be:

  $\begin{array}{l}\text{Time}=\frac{1500000}{250}\\ =6000\sec \\ \approx 100\min \end{array}$