# Total Work (Pumping Oil from a Tank).

• amcavoy
Thanks for the catch.In summary, the conversation discusses the calculation of the work required to pump all of the oil from a fuel tank with specific dimensions and weight. The volume and mass of the oil are calculated, and an integral is used to determine the total work required. A discrepancy in the answer is discussed, and the proper integral is explained using the concept of infinitesimally small changes.
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?

alexmcavoy@gmail.com said:
What I did was to first calculate the volume of the oil:

$$V=\pi r^{2}h=150\pi$$
OK.
Next, I calculated the mass of the oil:

$$150\pi*50=7500\pi$$
That's the weight of the oil.

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:

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

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.

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

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

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.

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

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

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

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

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:

## 1. What is total work in pumping oil from a tank?

Total work in pumping oil from a tank refers to the amount of energy required to pump a certain volume of oil from the tank to its destination. It takes into account the distance the oil needs to be pumped, the height it needs to be lifted, and the efficiency of the pumping system.

## 2. How is total work calculated in pumping oil from a tank?

Total work is calculated by multiplying the force required to pump the oil by the distance it needs to be pumped. This force is typically measured in Newtons and the distance in meters. The result is then converted to the appropriate unit of energy, such as joules or kilowatt-hours.

## 3. What factors affect the total work in pumping oil from a tank?

The total work in pumping oil from a tank is affected by the volume of oil being pumped, the distance it needs to be pumped, the height it needs to be lifted, the efficiency of the pumping system, and the density of the oil. Other factors, such as viscosity and friction, may also play a role.

## 4. How can the total work in pumping oil from a tank be reduced?

The total work in pumping oil from a tank can be reduced by improving the efficiency of the pumping system, minimizing friction and other energy losses, and using alternative energy sources such as renewable energy or waste heat from other processes. Additionally, reducing the distance and height the oil needs to be pumped can also decrease the total work.

## 5. How is total work in pumping oil from a tank important in the oil industry?

Total work in pumping oil from a tank is an important concept in the oil industry as it directly affects the cost and efficiency of oil production. By understanding and optimizing the total work, the industry can reduce energy consumption and costs, improve production rates, and minimize environmental impact.

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