Chapter 6, Problem 29RE

(a)

To determine

To calculate: the work required to pump the water out of the tank.

Answer to Problem 29RE

The required work to pump the water out of the tank is $\underset{\_}{8378\text{ft-lb}}$.

Explanation of Solution

Given information:

The shape of the tank is paraboloid.

Height of the tank is $4\text{ft}$.

Radius of the tank is $4\text{ft}$.

Weight density of water is $\delta =62.5{\text{lb/ft}}^{\text{3}}$

Calculation:

Sketch the parabolic shape tank for dimension as shown below:

Show the equation of parabola as shown below:

$y=\frac{1}{4}{x}^2$ (1)

The radius of the cross section at y as follows:

$\begin{array}{c}y=\frac{x^2}{4}\\ x=\sqrt{4y}\end{array}$

Show the area of cross section as shown below:

$A=4\pi y$

Expression of volume at an equipotential in the gravitational field:

$\begin{array}{c}dV=Ady\\ =4\pi ydy\end{array}$

Express to find the mass as shown below:

$\text{Mass=density×volume}$ (2)

Substitute $62.5{\text{lb/ft}}^{\text{3}}$ for density, and $4\pi ydy$ for volume in equation (2).

$\begin{array}{c}\text{Mass=62}\text{.5}\times \text{4}\pi \text{y}\text{dy}\\ \text{=250}\pi \text{y}\text{dy}\end{array}$

The distance of the layer each particle approximately $\left(4-y\right)$

The work done $W_i$ to raise this layer, the product of the force F, and the distance is (4-y).

Expression to find the element of work:

$W_i=F\times d$ (3)

Here, F is the force, and d is the distance.

Substitute $250\pi y$ for F, $\left(4-y\right)$ for d in Equation (3).

$\begin{array}{c}{W}_i=F\times d\\ =250\pi y\left(4-y\right)dy\end{array}$

Express to find the total work as shown below:

$W={\displaystyle \underset{a}{\overset{b}{\int }}\left(W_i\right)dy}$ (4)

Substitute 0 for a, 4 for b, and $250\pi y\left(4-y\right)dy$ for $W$ in Equation (4).

$\begin{array}{c}W={\displaystyle \underset{0}{\overset{4}{\int }}250\pi y\left(4-y\right)dy}\\ =250\pi {\left[2y^2-\frac{1}{3}{y}^3\right]}_0^4\\ =250\pi \left[\left(2{\left(4\right)}^2-\frac{1}{3}{\left(4\right)}^3\right)-\left(2{\left(0\right)}^2-\frac{1}{3}{\left(0\right)}^3\right)\right]\\ =250\pi \left[\left(32-\frac{64}{3}\right)-0\right]\end{array}$

$\begin{array}{c}=250\pi \left(\frac{32}{3}\right)\\ =2513.27\\ =8377.58\\ \text{=8378}\text{ft-lb}\end{array}$

Therefore, required work to pump the water out of the tank is $\underset{\_}{8378\text{ft-lb}}$.

(b)

To determine

To find: the remaining depth of water in the tank after $4000\text{ft-lb}$ of work done.

Answer to Problem 29RE

The remaining depth of water in the tank after $4000\text{ft-lb}$ of work done is $\underset{\_}{2.1\text{feet}}$

Explanation of Solution

Given information:

The work done is $4000\text{ft-lb}$.

Calculation:

Apply the integral equation to find the remaining depth of water after 4000 ft-lb of work done.

$\begin{array}{c}4000={\displaystyle \underset{d}{\overset{4}{\int }}250\pi y\left(4-y\right)dy}\\ 4000=250\pi {\left[2y^2-\frac{1}{3}{y}^3\right]}_d^4\\ \frac{16}{\pi }=\left(32-\frac{64}{3}\right)-\left(2d^2-\frac{1}{3}{d}^3\right)\end{array}$ (5)

Both sides divided by $250\pi$.

$\begin{array}{c}\frac{16}{\pi }-\frac{32}{3}=\frac{1}{3}{d}^3-2d^2\\ \frac{1}{3}{d}^3-2d^2+5.573=0\\ d^3-6d^2+16.72=0\end{array}$

Solving the Equation to get d.

$d=\left(2.059,5.434,-1.5\right)$

The depth value of -1.5 ft is negative.

The depth value of 5.434 ft greater than the given depth value.

Therefore, the remaining depth of water in the tank after $4000\text{ft-lb}$ of work done is $\underset{\_}{2.1\text{feet}}$