Calculating Work Required to Pump Water from a Hemi-spherical Tank

[cybercrypt13]

Homework Statement

A hemi-spherical tank with a 5 foot radius is full of water. Given than water weighs 62.5 pounds per cubic foot, find the work required to pump the water out of the tank

Homework Equations

Radius = 5 so range will be 0 to 5. Formula should be pi * r^2 for each slice of water, but need help figuring the integration formula I should use...

The Attempt at a Solution

I know how to solve if I can get past the setup...

Thanks,

cybercrypt

 

[Dick]

Hopefully you are think of the integration range 0 to 5 representing the depth (h) of a circular cross-section of water of thickness dh. What is r in terms of h? What is the volume of that slice? How much work does it take to move that slice to the top of the hemisphere?

 

[cybercrypt13]

Well, I think we are going to take the weight of 62.5 * (25pi) * delta x. This gives me the area of a circular region, its thickness multiplied by the thickness of a region, between 0 and 5.

So my integral would be from 0 to 5 for int( 62.5*25pi*x, dx)

Does this sound right?

Thanks,

cybercrypt

 

[Dick]

Well, I think we are going to take the weight of 62.5 * (25pi) * delta x. This gives me the area of a circular region, its thickness multiplied by the thickness of a region, between 0 and 5.

So my integral would be from 0 to 5 for int( 62.5*25pi*x, dx)

Does this sound right?

Thanks,

cybercrypt

Nope. Doesn't sound right. I liked your pi*r^2 for the area of the disc (so the volume is pi*r^2*dx. If we call the depth x like you did, then if x=0 then I think r=5, if x=5 then r=0. Are we thinking of the same picture? So the radius is a function of x. What function?

 

[cybercrypt13]

Well, I'm sort of stuck at the moment because I've never had a question like this in our exercises and can't find any references on the internet. I think I'm wrong in the integration part because you can't integrate the entire circular region so I guess i have to break things up to only working to the right of the y axis... Just not sure how to proceed.

Thanks,

glenn

 

[Dick]

You need to find the area of the disk as a function of its depth. From the center of the hemisphere go down vertically a distance x. Now go horizontally some distance r until you meet the hemisphere. These are the two legs of a right triangle with hypotenuse being the radius of the hemisphere. Using Pythagoras can you find r as a function of x?

 

[cybercrypt13]

I'm sorry, but that just doesn't make sense to me. You are describing coming down the Y axis (center) and across horizontal (which would be the radius). You say the piece joining the two lines would be the hyp and it would be the radius but that isn't the case.

So obviously I'm not picturing what you're describing. Can you explain to me a little more?

Thanks,

cybercrypt

 

[Dick]

There are two different radii, the radius of the hemisphere (5!) and radius of your disk (which is variable). Apparently I'm quite inept at explaining geometry in words.

 

[HallsofIvy]

One thing YOU haven't told us is the complete geometry of the situation. The tank is hemispherical but is the flat side on top or beneath? You also haven't told us what your variables mean. You talk about int( 62.5*25pi*x, dx) but that makes no sense when we don't know what x represents on the sphere. I am going to assume that this hemisphere has its flat side on the bottom and that x represents a height above that base.

Imagine a thin layer (thickness dx) of water at height x above the base. All of that layer must be lifted a specific distant, h, so the work done on that layer will be (62.5)(area*dx)(h) where "area" is the area of that disk and the total work done will be
[62.5\int_0^5 area(x)h(x)dx[/itex]
Now, what is h, the heigth the water must be lifted- as a function of x? What is the area of that disk?

 

[cybercrypt13]

well, its full of water so I'm assuming the flat part is at the top otherwise it would have said it had a hole in it. I gave you the entire problem exactly as I have it in my original post. I have nothing else...

So first I'd flip your image over as yours makes no sense to me unless we're going to start cutting holes. I'm thinking like a bit round pot with legs on it. So, My original numbers were only my thoughts and they came from the original equation so i didn't think I needed to list them out, but here goes.

I know that Pi * r^2 will give me the area of a circle and I know that each slice I take of the water will be a circle. So that is the equation I'll use for the slices. I also know that the radius is 5ft, so I'm going to have a range from 0-5. I know the water weighs 62.5lb/ft^3 so I'm thinking my equation of integrations will be as follows:

int( 62.5 * Pi * x * ( 5 - x )^2, x = 0..5 ); Maple format...

Would this seem correct? x is obviously my position along the curve as I integrate.

Thanks,

cybercrypt

 

[Dick]

Almost ok. You've put r=(5-x). That varies LINEARLY from r=5 at x=0 to r=0 at x=5. Does that seem right? Sounds more like a cone to me.