- #1

unggio

- 23

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**need help, VERY CHALLENGING**

a hemispherical tank of radius 6 meters, is posiitoned so that its base is circular. how much work is required to empty the tank?

liquid density is 100 kg/ m^3

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- Thread starter unggio
- Start date

- #1

unggio

- 23

- 0

a hemispherical tank of radius 6 meters, is posiitoned so that its base is circular. how much work is required to empty the tank?

liquid density is 100 kg/ m^3

Last edited:

- #2

unggio

- 23

- 0

i use equation of a circle to calculate the radius

circle is

[tex]x^2 + (y-6)^2=6^2=36[/tex]

therefore

[tex]r=\sqrt {36 - (y-6)^2} [/tex]

[tex]volume= \pi r^2 [/tex]

F=mg

F=vol * density * gravity

[tex]9.8 * 100 * \pi \int_0^6 (36 - (y-6)^2)y dy[/tex]

ansewr i get is: 16625308 joules

circle is

[tex]x^2 + (y-6)^2=6^2=36[/tex]

therefore

[tex]r=\sqrt {36 - (y-6)^2} [/tex]

[tex]volume= \pi r^2 [/tex]

F=mg

F=vol * density * gravity

[tex]9.8 * 100 * \pi \int_0^6 (36 - (y-6)^2)y dy[/tex]

ansewr i get is: 16625308 joules

Last edited:

- #3

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

I presume you mean "the radius of the circle formed by the surface of the water at different heights".

And, by the way, the problem is not "well formed". If you just punch a hole in the bottom, the tank will empty without**any** work being done. I presume that the problem says to empty the tank by pumping it out the top.

Here's how I would think about it:

Of course, the work done lifting a mass m a height h is mgh. Imagine a "layer of water" of thickness dz at height z. That layer will be a circle of radius, say, r. It will have volume [tex]\pi r^2 dz[/tex] and so weight [tex]98\pi r^2dz[/tex]. That water has to be lifted a distance 6-z to the top of the tank: the work done on that one "layer" of water is [tex]\pi (6-z)r^2 dz[/tex].

Now we need to determine r^{2}: x^{2}+ y^{2}+ z^{2}= 36. Since r is measured parallel to the xy-plane, r^{2}= x^{2}+ y^{2}= 36- z^{2}. Putting that into the formula above:

[tex]\pi (6-z)(36-z^2)dz[/tex].

We need to integrate that from z= 0 to z= 6 to cover the entire tank:

[tex]\int_0^6\pi(6-z)(36-z^2)dz= \pi\int_0^6(216- 36z-6z^2+ z^3)dz[/tex].

That is**not** what you have. Did you forget the height lifted?

And, by the way, the problem is not "well formed". If you just punch a hole in the bottom, the tank will empty without

Here's how I would think about it:

Of course, the work done lifting a mass m a height h is mgh. Imagine a "layer of water" of thickness dz at height z. That layer will be a circle of radius, say, r. It will have volume [tex]\pi r^2 dz[/tex] and so weight [tex]98\pi r^2dz[/tex]. That water has to be lifted a distance 6-z to the top of the tank: the work done on that one "layer" of water is [tex]\pi (6-z)r^2 dz[/tex].

Now we need to determine r

[tex]\pi (6-z)(36-z^2)dz[/tex].

We need to integrate that from z= 0 to z= 6 to cover the entire tank:

[tex]\int_0^6\pi(6-z)(36-z^2)dz= \pi\int_0^6(216- 36z-6z^2+ z^3)dz[/tex].

That is

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- #4

unggio

- 23

- 0

i don't think you did it correctly. i inverted my coordinates upside down so that the +y direction is towards the base of the hemisphere, also there's no need to use spherical coordinates, using 2-d is simpler. W=F*D, my D (dist) is y. i start out pumping the top layer, then i move down to the bottom layer. as i pump out each layer the water travels a distance of y from the origin. origin is the top of the tank.

i'm surprised that i only received one response, i thought everybody in the math forum was good at this stuff.

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