# Thread: Work Required To Pump Water

1. ## Work Required To Pump Water

A reservoir has the shape of a vertical cylinder with height 3m and radius 4m and it is filled with water.

i) Let x be the height in meters measured from the bottom of the reservoir. The weight in Newtons of a thin layer of water between x and x+delta x is approximately P(x)(delta x). What is P(x)?

ii) The work in Joules required to pump the thin water layer to 3 meters above the reservoir is approximately w(x)(delta x). What is w(x)?

iii) Using previous results, find, in Joules, the amount of work required to pump all the water 3 meters above the reservoir.

I am unable to solve part (i), so if someone could shed a little light on this question, I may be able to solve the rest, thanks!

2. ## Re: Work Required To Pump Water

Since when is weight measured in Newtons? Oh wait, force due to mass and gravity, yep a force, ok Newtons works

Do you have a formula to find weight from a volume?

3. ## Re: Work Required To Pump Water

Originally Posted by Prove It
Since when is weight measured in Newtons? Oh wait, force due to mass and gravity, yep a force, ok Newtons works

Do you have a formula to find weight from a volume?
The only things we are given are density = 1000 kg/m^3 and g = m/s^2. What I did was find the volume of the cylinder, and then multiply that by 1000 and 9.8 in an attempt to find the weight, but I kept getting the wrong formula. :s

4. ## Re: Work Required To Pump Water

I don't know calculus but here is my attempt
mass of water in tank (kg)*9.8* integral from 0 to 3 of (6-x) dx = W in joules

5. ## Re: Work Required To Pump Water

This is easier than you think.

1. The weight of water in a layer dx thick is its mass times g, and the mass is the volume times density. So: $dw = \rho A g dx$, where A is the area of the cylinder. Here I'm using lower case w for weight.
2. The work to raise water of weight dw to by height h is dW = hdw. Here I'm using upper case W for work. So for a slice of water at height x, the distance it must be raised is 6-x, and the work required is dW=(6-x)dw.
3. The total amount of work to pump all the water to a height of 6 meters is $W = \int _0 ^3 (6-x)dw = \int _0 ^3 (6-x)\rho A g dx$

6. ## Re: Work Required To Pump Water

Originally Posted by ebaines
This is easier than you think.

1. The weight of water in a layer dx thick is its mass times g, and the mass is the volume times density. So: $dw = \rho A g dx$, where A is the area of the cylinder. Here I'm using lower case w for weight.
2. The work to raise water of weight dw to by height h is dW = hdw. Here I'm using upper case W for work. So for a slice of water at height x, the distance it must be raised is 6-x, and the work required is dW=(6-x)dw.
3. The total amount of work to pump all the water to a height of 6 meters is $W = \int _0 ^3 (6-x)dw = \int _0 ^3 (6-x)\rho A g dx$
If you don't mind me asking, why are you including the area in the formula then, if I need to use volume x density?

I have tried inputting the following:
P(x) = pg(pi 4^2 x)
P(x) = pg[2pi(4)(x)+2pi(4^2)]
obviously substituting the density and gravity for their respective values... And we are not supposed to include the delta x at the same.

I am ripping my hair out because of this question haha.

7. ## Re: Work Required To Pump Water

I have stumbled upon the solution on my own finally.. It turns out, I was not expected to include the height variable in my equation as it was already supplied outside of the answer box. The answer for part 1 was in fact 156800pi; volume of cylinder (without the x variable for height) x density of water x gravitational force.

### weight of the water in newtons in cylinder

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