Work required to pump - Very tricky

[IntegrateMe]

A right cylindrical tank is filled with sea water. The tank has a radius of 2 feet and a height of 8 feet. If the water level is now 2 feet below the top of the tank, how much work will be required to pump the sea water to the top of the tank? (The weight-density of seawater is 64 Ib/ft³.)

The answer is 24,127 ft-Ibs, can anyone figure out why?

 

[rock.freak667]

How do you think you need to approach the problem?

 

[IntegrateMe]

deltaV = pi(r)^2(thickness) --this is of the cylindrical pieces you cut out of the whole
so r = 2

deltaV = 4pi(deltaY)

F(y) = weight = 64 Ib/ft³ (4pi(deltaY))

..and that's pretty much all i can think of?
It's very confusing, I'm sorry.

 

[rock.freak667]

The water fills the tank to 6ft. You can find the volume of water using the volume of a cylinder $V=\pi r^2 h$. Give volume and density you can get the mass of water. You just need to get the energy needed to raise the water a height of 2 ft.

 

[IntegrateMe]

so using the volume equation i get V = 4pi(6) = 24pi

Then to get the force i use the work equation?

so,

W = fnint ((64 Ib/ft³)(24pi),x,0,2)

Is that it?

 

[HallsofIvy]

The water fills the tank to 6ft. You can find the volume of water using the volume of a cylinder $V=\pi r^2 h$. Give volume and density you can get the mass of water. You just need to get the energy needed to raise the water a height of 2 ft.

No, that's wrong. The water at the top of the tank only has to be raised 2 feet but the water at the bottom of the tank has to be raised 8 feet. That's why IntegrateMe is talking about "cylindrical pieces". Each "piece" has area [/itex]4\pi[/itex] square feet and, if we take the "thickness" of each "piece" to be dx, volume $4\pi dx$ cubic feet and so mass $(64)(4\pi dx)= 256\pi dx$ pounds. If that pieces is at height "x", it has to be raised 8- x feet. The work done in raising that piece is $256\pi (8- x)dx$ and so the total work done is
[tex]256\pi\int_0^6 (8- x)dx[/itex] "foot-pounds".<br /> <br /> Because the density is constant, you <b>could</b> look at the <b>average</b> distance the water must be lifted. The cylinder also has constant cross section so the "average" height of the water is 1/2 way up, at (0+ 6)/2= 3 feet. The water has to be lifted and "average" distance of 8- 3= 5 feet. Find the weight of the entire cylinder of water and multiply by 5 feet, not 2. That will give the same answer as the integral.[/tex]