Find the work to pump water out of a sphere

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January 22nd 2010, 11:18 AM
DBA

Find the work to pump water out of a sphere

Hi, given is a sphere filled with water. The radius is 3m
The sphere has a spout on top and the hight of the spout is 1m.

What I did

I did draw a circle with r=3 in (0,0)
Then I made a horizontal "cut" somewhere above the x-axis.

The cut gives me a disk with A=pi*r^2

The formula of work we need to use is

Work = weight * displacement
= Lambda * Volume * displacement
= Lambda *Area * thickness * displacement

For the area:

A= pi*x^2 I need to find x for the disk I cut.

The x lies on the circle and the formula of this circel is
x^2+y^2 = 9 because radius of the sphere is 3

then x^2 = 9-y^2

So A = pi * (9-y^2)

Displacement
I am not sure here.
My distance from my cut on the y-axis is Y
The radius of the sphare and therefore of the circle I did draw is 3 also.
Then the distance the water of my disk has to move (3-y) upwards.
But I also have the spout of 1m

Is the displacement then (4-y)?

Thickness

dy, because the thickness of my disk is an "up-down" value or y-value

Integral interval

I am not sure here also, I would say from -3 to 3. Because that is where the water is
Is that right?

Integral I use S for "Integral from -3 to 3"

Work = Lambda * pi S (9-y^2)(4-y) dy

Can someone let me know if I did that right? I do not understand how I get the displacement when I have a spout and what the interval of the integral should be.
Thanks
January 22nd 2010, 11:43 AM
VonNemo19

I think that you are sayinyg that you have imposed your coordinate axes at the center of the sphere? If this is the case, then you should have

$W=62.4\pi\int_{-3}^3(9-y^2)(3-y)dy$
January 22nd 2010, 12:16 PM
DBA

Yes, the center of the sphere is at (0)0).
But what is with the spout (1m height) which is on top of the sphere. Don't I have to add that in the displacement?
January 22nd 2010, 12:20 PM
VonNemo19

Quote:

Originally Posted by DBA

Yes, the center of the sphere is at (0)0).
But what is with the spout (1m height) which is on top of the sphere. Don't I have to add that in the displacement?

Oh, I hadn't noticed that. I apologize. Yes, we would take that into account so that the distance that each $i^{th}$ disc would be moved : $d=4-y$ .

Again, I apologize for the confusion.
January 22nd 2010, 02:40 PM
CaptainBlack

Quote:

Originally Posted by DBA

Hi, given is a sphere filled with water. The radius is 3m
The sphere has a spout on top and the hight of the spout is 1m.

What I did

I did draw a circle with r=3 in (0,0)
Then I made a horizontal "cut" somewhere above the x-axis.

The cut gives me a disk with A=pi*r^2

The formula of work we need to use is

Work = weight * displacement
= Lambda * Volume * displacement
= Lambda *Area * thickness * displacement

For the area:

A= pi*x^2 I need to find x for the disk I cut.

The x lies on the circle and the formula of this circel is
x^2+y^2 = 9 because radius of the sphere is 3

then x^2 = 9-y^2

So A = pi * (9-y^2)

Displacement
I am not sure here.
My distance from my cut on the y-axis is Y
The radius of the sphare and therefore of the circle I did draw is 3 also.
Then the distance the water of my disk has to move (3-y) upwards.
But I also have the spout of 1m

Is the displacement then (4-y)?

Thickness

dy, because the thickness of my disk is an "up-down" value or y-value

Integral interval

I am not sure here also, I would say from -3 to 3. Because that is where the water is
Is that right?

Integral I use S for "Integral from -3 to 3"

Work = Lambda * pi S (9-y^2)(4-y) dy

Can someone let me know if I did that right? I do not understand how I get the displacement when I have a spout and what the interval of the integral should be.
Thanks

Work done = change in potential energy

You are raising the centre of mass of the water a height of $4$ m so change in potential energy is:

$\Delta E=4 \times g \times 1000 \times \frac{4\pi 3^3}{3}$

CB