# Tank in a shape of cone containing fluid, how much work needed to pump it out 3 feet above top

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 $\; 59 \;$ 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. Graph of the tank.

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.

Attempt at solution:

So firstly, I'm slightly confused here. If we slice the tank into slices, and then let's say the upper boundary of oil is at height = 7, so if I move that top slice to the height of 19 (16 + 3), that means I'd need to move it up by 8 ft (19-7). But the next slice, is going to be at height 6, so it will be moved up by 9 ft? How the hell do I figure out a formula for this.

Then when you get to the most-bottom slice, it would have to be moved 18 feet, or maybe even 19 ft if the slices are very thin (Δy).

Am I even correct at understanding how the oil is going to be pumped?

It's just I don't get why one of the suggested solutions to this problem is that you find the upper boundary of oil (let's assume it's 7), and then you take the interval (when you integrate) to be from $0$ to 8 (19-7).

It doesn't make sense, because it assumes that once you move the top slice of oil up, the one in the bottom would magically float up to the same position. I call BS on this.

Solution:

At height h, near the bottom, r will be smaller, so you do the relation: at the top it's: h = 16, r = 8, $\frac{16}{8}$ = $\frac{h}{r}$

r = $\frac{h}{2}$

Work is going to be: $\pi \cdot r^2 \cdot 59lb \cdot (3+16 - h) dx$

Then to find half the volume (your h):

$\frac{1}{3} \pi \cdot r^2 \cdot h = \frac{1}{3} \pi \cdot (\frac{h}{2})^2 \cdot h$

$\frac{59}{4}\!\left(\left(2\cdot 16\cdot 8^{2}\right)^{\frac{1}{3}}\right)^{3}\!\left(\frac{19}{3}-\frac{\left(2\cdot 16\cdot 8^{2}\right)^{\frac{1}{3}}}{4}\right)\pi$

Which I don't quite get.

Your understanding of how the oil is pumped is correct. If we count height $h$ from the bottom of the tank, each bit of oil needs to be lifted to $h=19$ feet. No bit of oil "magically floats up", you need to pump it from where it starts to $19$ feet. The work done on a mass $m$ is $mg\Delta h=mg(19-h_{\text{start}})$, where $g$ is the acceleration of gravity and $h_{\text{start}}$ is the height it starts at. We need to integrate this over the volume of the oil to get the total work done. We integrate because the slices have to be very thin.
The reason you start by finding the height of the upper surface of a half tank is to find out how high the oil reaches at the start, which gives the upper limit of our integration. A slice of oil from $h$ to $h+dh$ will have a radius of $r(h)$, which you should be able to figure out from your diagram and the dimensions you have. The area will be $\pi (r(h))^2$, so the mass is $\rho \pi (r(h))^2dh$ The work done will be $(19-h)\rho \pi (r(h))^2dh$ The total work will be $\int_0^{h_{\text{max}}}(19-h)\rho \pi (r(h))^2dh$
• That is true. That is where the $(19-h)$ term in the integral comes from. $h$ is the starting height and it is pumped up to $19$. $h_{max}$ is the height of the upper surface of the oil before any pumping is done. – Ross Millikan Sep 12 '15 at 22:07
• @Jack: I don't see where the divisions by 3 and 4 are in $h_{max}$. I see they are denominators in the official solution, but without a chain of reasoning I don't know where they come from. – Ross Millikan Sep 28 '15 at 4:13