Domenic's question at Yahoo Answers regarding computing the work to empty a tank

[MarkFL]

Here is the question:

Calculus II question?? Work/pumping fluids? HELP!?


A tank containing oil is in the shape of a downward-pointing cone with its vertical axis perpendicular to ground level. In this exercise we will assume that the height of the tank is h=16 feet , the circular top of the tank has radius r=8 feet, and that the oil inside the tank weighs 55 pounds per cubic foot.
How much work (W) does it take to pump oil from the tank to an outlet that is 3 feet above the top of the tank if, prior to pumping, there is only a half-tank of oil? Note: "half-tank" means half the volume in the tank.
Note: For this problem a pound is to be taken as a unit of weight (not of mass), so it is not necessary to multiply by the gravitational acceleration constant in order to find the work.

PLEASE HELP. Thank you

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello domenic,

I prefer to work problems like this in general terms, and derive a formula we may then plug our data into. Let's let the base radius of the conical tank be $r$, and the height be $h$.

Let's orient a vertical $y$-axis along the axis of symmetry of the tank, where the origin is at the bottom of the tank, and the initial depth of the fluid in the tank is $y_0$. We wish to find the amount of work $W$ is needed to pump all of the fluid to a point some distance $z$ above the top of the tank.

Now, if we imagine slicing the cone of fluid we wish to remove into disks, we may state, using work is force $F$ times distance $d$ for a constant force:

\(\displaystyle dW=Fd=wd\)

The force exerted is equal to the weight $w$ of the slice.

The weight $w$ of the slice is the product of the weight density $\rho$ in \(\displaystyle \frac{\text{lb}}{\text{ft}^3}\), and the volume \(\displaystyle V=\pi r_s^2\,dy\) of the slice, i.e.:

$w=\rho\pi r_s^2\,dy$

The radius of the slice can be found by similarity.

\(\displaystyle \frac{r_s}{y}=\frac{r}{h}\,\therefore\,r_s=\frac{r}{h}y\)

The distance the slice must be vertically moved against gravity is:

$d=h+z-y$

Putting it all together, we have:

\(\displaystyle dW=\frac{\rho\pi r^2}{h^2}\left(hy^2+zy^2-y^3 \right)\,dy\)

Summing up all the work elements through integration, we obtain:

\(\displaystyle W=\frac{\rho\pi r^2}{h^2}\int_0^{y_0} (h+z)y^2-y^3\,dy\)

Applying the anti-derivative form of the FTOC, there results:

\(\displaystyle W=\frac{\rho\pi r^2}{12h^2}\left[4(h+z)y^3-3y^4 \right]_0^{y_0}=\frac{\rho\pi r^2y_0^3}{12h^2}\left(4(h+z)-3y_0 \right)\)

Now, if $V$ is the volume of the tank, let $kV$ be the initial volume of fluid in the tank, where $0\le k\le1$. To find the initial depth $y_0$ of the fluid in terms of $k$, we may use:

\(\displaystyle \pi\left(\frac{r}{h}y_0 \right)^2y_0=k\pi r^2h\)

\(\displaystyle y_0^3=kh^3\)

\(\displaystyle y_0=h\sqrt[3]{k}\)

Hence:

\(\displaystyle W=\frac{\rho\pi hkr^2}{12}\left(4(h+z)-3h\sqrt[3]{k} \right)\)

We now have a formula into which we may plug the given and known data:

\(\displaystyle \rho=55\,\frac{\text{lb}}{\text{ft}^3}\)

\(\displaystyle h=16\text{ ft}\)

\(\displaystyle k=\frac{1}{2}\)

\(\displaystyle r=8\text{ ft}\)

\(\displaystyle z=3\text{ ft}\)

And so we find for this problem:

\(\displaystyle W=\frac{\left(55\,\dfrac{\text{lb}}{\text{ft}^3} \right)\pi \left(16\text{ ft} \right)\left(\dfrac{1}{2} \right)\left(8\text{ ft} \right)^2}{12}\left(4(16+3)-3(16)\sqrt[3]{\frac{1}{2}} \right)\text{ ft}=\frac{28160\pi}{3}\left(19-6\sqrt[3]{4} \right)\text{ ft}\cdot\text{lb}\)

\(\displaystyle W\approx190959.32\text{ ft}\cdot\text{lb}\)