Simple Tank Underground Problem Make No Sense

I have trouble making since of this but say you to compute the work to take some liquid from a tank underground to ground level laying horizontally

Well work = force * distance(displacment)

I have seen this basic formula done twice for and cylinder and ellopsoid
to find force you need the mass * g then you partition the tank into slices
which seems to be rectangular slices

lets say in this example i have a cylinder lying on its side radius 4 ft, length 12 ft, 10 ft underground

the formula for the rectangular slice is 2x * length * thickness

but would your length be x which is 12 and the thickness is just some change in y, your left with width which seems to be would be some y value

but they took the equation of a circle

x^2 + y^2 = r^2

where in this case y was equal to the distance/displacment which was (14-y)

i dont see how you can just plug these random things in, this post is a mess

 

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Hmm, not sure if I understand what you want to do, but if you have a tank full of water and you want all the water lifted (pumped) to ground level that's equivalent to raising the entire volume of water so the "bottom" is at ground level.
Work is also equal to the work a pump has to do to empty the tank.

But you can skip all that and just use the weight of a given volume (in this case the tank volume) which is the 'force component', and the distance is the depth of the tank.

 

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I think he's trying to find the total mass of an object using a triple integral utilizing a density function. Then using that total mass to determine the work done, but there's no way to tell because it's too convoluted.

 

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The water at the bottom of the tank has to be lifted more than the water at the top of the tank.

 

Pete: if you have a tank - say a swimming pool - full of water so the surface is level with the ground, that's a volume of water at ground level = 0.

Say your pool happens to have a bottom which is a large plate, supported by hydraulic rams that lift the bottom of the tank to ground level.
The work the plate + hydraulic system does lifting the entire volume is equivalent to work done pumping it out say, a bucketful at a time, or by using a bucket + rope, or just letting the water evaporate.

The work done on the water is the same "path", and is independent of the "means" used to lift it. Different methods will be more or less efficient as work will be done in each case, that won't move any water (for instance if you use a bucket, the work you do manipulating it when it's empty is "wasted").

Maybe the question is about calculating a pumping rate, given a pipe diameter etc, but the OP is a little vague...

 

Of course. But your previous post is misleading.
"Raising the entire volume of water so the "bottom" is at ground level" implies that the entire pool is lifted while maintaining its shape, not just poured out to ground level.

And your suggested solution...

...works for that interpretation (entire pool lifted), but not for the problem at hand (poured out on the ground), because the force (the weight being lifted) is not constant.

 

Ok... "so the bottom of the liquid is at ground level, so everything above the tank bottom, i.e. the water is moved".
I mistakenly assumed that my first sentence was intransitive.

P.S. The "lifting rule" of entropy here is related to something else, that some other math-head might know more about than me. But work is a symmetric function.

 

The horizontal "slice" of the water's volume is a rectangle, with length equal to the length of the cylinder's axis, and width equal to the horizontal chord of the circle describing the cylinder's cross section perpendicular to its axis. It varies by depth.

 

You want the Laplacian for the volume form on C.

 

OK, a cylindrical tank, on it's side. The top of the tank is 10ft under.

So the cylindrical tank is partitioned into thin horizontal slices.
The volume \(\Delta V\) of a slice is length x width x thickness.

• The length of each slice is a constant 12 ft (the length of the cylinder).
• The width of each slice will be different for each slice, varying from zero at the bottom of the tank to 8ft in the middle and back to zero at the top. You can express this as 2x, where x is half the width of the tank at that particular y value.
• The thickness of each slice is the distance from the start of one slice to the start of the next; an arbitrarily small constant \(\Delta y\). y is the height of the slice from some vertical reference (like the middle of the tank).

\(\Delta V=L \times 2x \times \Delta y\)

But to make this work, you have to make y the only variable. You can't directly integrate a function involving both x and \(\Delta y\).
That means that 2x has to be expressed as a formula involving y.

Right, that's what you need to change 2x to a function of y.
\(x = \sqrt {r^2-y^2}\)
Now we can get the volume in terms of y:
\(\Delta V=2L \sqrt {(r^2-y^2)} \Delta y\)

y is the distance from the middle of the tank to the slice. The displacement involved in lifting that slice to ground level is (14-y) if the middle of the tank is 14ft down.

So... if my quick and very nasty manipulations are right...
For one slice, we begin with:
work = force x distance
distance = 14-y
\(F = mg = \rho \Delta V g\)

(\(\rho\) = Density of water)

Put them together...
\(W = \rho \Delta V g (14-y)\)

And remembering that...
\(\Delta V = 2L \sqrt {(r^2-y^2)} \Delta y\)

We get...
\(W = 2 \rho Lg \sqrt {(r^2-y^2)} (14-y) \Delta y\)

And for the entire cylinder, we sum all the slices:
\(W = \int_{-4}^4 2 \rho Lg \sqrt {(r^2-y^2)} (14-y) dy\)


I haven't given this a reality check, so treat with caution.

 

http://en.wikipedia.org/wiki/Volume_form

In this case you have all the boundaries and M is a cylindrical volume of fluid.
(the leftright-corner thingy that Tex is having trouble with is I think called a connection form)

http://en.wikipedia.org/wiki/Lie_derivative

Note: Wikipedia is a good reference when decent mathematicians contribute understandable rhetoric.

 

That's also assuming the tank is symmetrical on a horizontal plane. Which of course it is, in this case.