# Thread: Find the work to pump water out of a sphere

1. ## 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

2. 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

$\displaystyle W=62.4\pi\int_{-3}^3(9-y^2)(3-y)dy$

3. 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?

4. 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 $\displaystyle i^{th}$ disc would be moved : $\displaystyle d=4-y$.

Again, I apologize for the confusion.

5. 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 $\displaystyle 4$m so change in potential energy is:

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

CB

,

,

,

,

,

,

,

,

,

,

,

### one hemisphere radius 3 cm full of water ane pump who drop water of hemisphere 15g/is.how much time in hemisphere in water out.

Click on a term to search for related topics.