# Work required to pump - Very tricky

• IntegrateMe
In summary, 24,127 foot-pounds of work is required to pump the sea water to the top of the cylindrical tank.
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?

How do you think you need to approach the problem?

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.

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.

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?

rock.freak667 said:
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".

Because the density is constant, you could look at the average 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.

## What is the definition of work required to pump?

The work required to pump is the amount of energy or force needed to move a fluid from one location to another.

## How is work required to pump calculated?

The work required to pump can be calculated by multiplying the force or pressure applied to the fluid by the distance the fluid is pumped.

## What factors affect the work required to pump?

The factors that affect the work required to pump include the type and viscosity of the fluid being pumped, the distance the fluid needs to be pumped, and the efficiency of the pumping system.

## What is the unit of measurement for work required to pump?

The unit of measurement for work required to pump is joules (J) or foot-pounds (ft-lb).

## How can the work required to pump be reduced?

The work required to pump can be reduced by optimizing the pumping system, using a more efficient pump, and choosing a fluid with lower viscosity. Additionally, reducing the distance the fluid needs to be pumped can also decrease the work required.

Replies
2
Views
1K
Replies
1
Views
2K
Replies
10
Views
1K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
50
Views
5K
Replies
4
Views
992
Replies
5
Views
1K
Replies
2
Views
2K
Replies
5
Views
1K