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Vlykarye

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## Homework Statement

I am trying to understand the logic behind the water pump that requires a user to take into consideration the volume of water at sub-intervals of depth of a tank. There are two types of problems, one where the volume changes at different depths, and one where it doesn't. In both cases, the user also has to consider the height the volume of water is moving. This height also seems to change at different depths of the tank. The end result is the calculation of work required to move all water out of the tank.

Ex: There is cylindrical tank with a height of 10 meters and radius of 5 meters. The tank is standing on its base, so the volume of water does not change at different depths. The volume is described by the equation: [itex]A(y) = \pi r^2 \delta h[/itex] where the height is approaching 0 and is dealt with through integration.

I do not need help solving this word problem. Rather, I need help understanding the logic behind it. We are using a water pump to move the water inside the cylindrical tank. Now, the most logical way one could visualize this problem is to imagine a pump that has a hose reaching to the bottom of the tank. In this case, the height the water must travel is constant (the length of the hose). However, the example uses a different method. So far, I have started a theory that the pump is moving along the height of the tank, in order to constantly be touching the uppermost sub-interval of water in the tank.

I don't understand the logic behind this method, because such an application seems impractical in the real world. I have come to a conclusion that either these examples are using this impractical form of pumping water, or that I'm confused on how the pump is actually being used.

## Homework Equations

[itex]A(y) = \pi r^2 \delta h[/itex]

[itex]W = \int_a^b ρ g A(y) D(y) dy [/itex]

## The Attempt at a Solution

Here is our tank: http://grockit.com/blog/wp-content/uploads/2010/03/11.jpg

Ex, Method 1:

[itex]A(y) = 25 \pi[/itex] | [itex]D(y) = y[/itex] | [itex]ρ = 1000 kg / m^3[/itex] | [itex]g = 9.8 m / s^2[/itex]

Therefore, [itex]W = 25 \pi (1000) (9.8) \int y dy[/itex] from 0 to 10

[itex]W = 12,250,000\pi J ≈ 38,484,510 J[/itex] for the book's method.

Ex, Method 2:

Now, let's consider the method we come up with using logic. Using this method, our pump is pumping water through a hose or tube that reaches the bottom of the tank. The height the water is being lifted stays constant, the height of the tube/tank.

[itex]A(y) = 25 \pi[/itex] | [itex]D(y) = 10[/itex] | [itex]ρ = 1000 kg / m^3[/itex] | [itex]g = 9.8 m / s^2[/itex]

Therefore, [itex]W = 25 \pi (1000) (9.8) \int 10 dy[/itex] from 0 to 10

[itex]W = 24,500,000\pi J ≈ 76,969,020 J[/itex] for this method.

From a quick glance, it's obvious that our second method is using an average height based formula. My question is not whether one way is better than the other (that's easy) nor whether one way is right or wrong ([strike]both ways must be right[/strike]), but rather: why would the first method ever be used to pump water from a tank, and how? Wouldn't the first method require that we are constantly moving the pump or a hose to adjust to the fact that the water level is decreasing? And if so, how do we calculate the extra work being done to move the pump or hose? And lastly, why is this fact never mentioned in any of the examples that I have found?

Side note: in the second method, I do not consider that the change in volume might affect the work required/done. In fact, I cannot consider this due to my lack of physics knowledge. If someone could clarify this, I'd appreciate it.

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