# Total Work (Pumping Oil from a Tank).

1. May 11, 2005

### amcavoy

"A fuel tank is an upright cylinder, buried so that its circular top is 10 feet beneath ground level. The tank has a radius of 5 feet and is 15 feet high, although the current oil level is only 6 feet deep. Calculate the work required to pump all of the oil to the surface. Oil weighs 50 lb/ft^3."

What I did was to first calculate the volume of the oil:

$$V=\pi r^{2}h=150\pi$$

Next, I calculated the mass of the oil:

$$150\pi*50=7500\pi$$

To figure out the total work, I integrated it with bounds from 0 to 25 because that is how far it would take to get it all to the surface:

$$\int_{0}^{25}7500\pi dx=187500\pi\approx 589049$$

The problem I have is that this isn't the answer my book has. It says the answer should be 518363. The only way I get that answer is by making the upper bound 22 rather than 25. Can anyone tell me why it is 22?

2. May 11, 2005

### Staff: Mentor

OK.
That's the weight of the oil.

I don't understand the logic of that integral. But the average height that the oil must be raised is 22 feet; so the work required is mgh = (weight)x(22 ft).

3. May 11, 2005

### Staff: Mentor

If you wanted to do this problem using integration, the integral would look like this:
$$\int_{19}^{25}\rho A x dx$$

where $\rho$ is the weight density (given as 50 lb/ft^3), A is the cross-sectional area of the cylinder ($\pi r^2$), and "x" is the distance the oil needs to be raised. Note that oil at the bottom must be raised 25 ft, oil at the top only 19 ft.

4. May 11, 2005

### amcavoy

So then the total work done would be the average of those integrals when $$x=19$$ and $$x=25$$?

5. May 11, 2005

### Jameson

This integral is more complicated then you might think. I would write it like this:

$$W = \int pAxdy$$

I chose y as the variable of integration to emphasize that you will need to make a substitution for "x" in terms of the height.

Jameson

6. May 11, 2005

### Staff: Mentor

I don't know what you mean by "average of those integrals". The range of the integral that I provided is from x = 19 to x = 25; it represents the range of heights that the oil must be raised.

7. May 12, 2005

### scholzie

Your mistake was calculating the total volume. The whole idea of calculus is dealing with infintessimally small changes, and what happens over those changes. In that same vein, these pumping problems are best thought of as adding the small amounts of work necessary to pump out infintessimally small "layers" of fluid. Think about the general case when the oil level is at any given height. What is the work required to lift THAT layer of liquid out?

Well:
What is the AREA of that extremely thin layer? $A=\pi r^2$. This never changes, so....what does?
What is the VOLUME of that extremely thin layer? $dV=\pi r^2dh$ where $dh$ is the infinitessimally small change in height.
What is the WEIGHT of that extremely thin layer? $\rho=50$ lb/ft^3, times the VOLUME = $\rho dV = \rho \pi r^2 dh$
What is the HEIGHT of that extremely thin layer? Well, that's what is changing, so that's what we need to integrate with respect to.

So $$W(h)=\int_{h_1}^{h_2}F(h)dh$$, and $F=\rho A h$

So $$W(h)=\int_{h_1}^{h_2}\rho A h{ }dh=\rho A\int_{h_1}^{h_2}h{ }dh$$

What are $h_1$ and $h_2$? h1 is the initial height, 15 foot tank - 6 foot level + 10 feet underground = 19. h2 is the final height, 15 full feet (tank is now empty) + 10 feet underground = 25.

Your final Integral is $$\rho \pi r^2 \int_{19}^{25} h{ }dh$$. Plug in your variables and you will get the answer you expect to get.

Last edited: May 12, 2005
8. May 12, 2005

### whozum

Very well written scholzie, the final integral is intended to be

$$\rho\pi r^2 \int_{19}^{25} h dh$$

9. May 12, 2005

### scholzie

oops, didn't catch that. Thanks :)

edit: upon further inspection I made a few other typos... I think I got them all though.

Last edited: May 12, 2005