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Time to run out water (Posted on 2006-07-13) Difficulty: 3 of 5
Consider a solid sphere (capable of withstanding full vacuum) of 3m diameter filled completely with water resting at sea level. It has a 10cm hole at the bottom with a cork on it. If you open the cork, what is the time taken for water to completely drain out.

What happens for higher diameter spheres?

No Solution Yet Submitted by Salil    
Rating: 1.2500 (4 votes)

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re: solution (maybe) | Comment 5 of 11 |
(In reply to solution (maybe) by Daniel)

The problem with your calculation becomes that there's an opposite force to the drainage in the vaccuum that is being created as the water flows out.  In other words, freefall equations can't be used without taking into consideration that the larger the vaccuum, the slower the water will flow.

Think about why a glass ketchup bottle sometimes won't let ketchup out if you hold it completely upside down... there's a force, a vaccuum at the bottom of the bottle, which holds the ketchup and keeps it from flowing.

This is also why if you empty a water bottle by holding it upside down, you won't be able to keep bubbles from entering the bottle due to this vaccuum.

Hope that helps because I'm completely clueless about how to solve this.  All I know is that the vaccuum's absolute pressure multiplied by the surface area of the inside water must be LESS than the weight of the water left at every point, or else it just won't flow.

(1)Pressure = Force/Area

Force = Pressure * Area

If at equillibrium...

(2)Force = mass of water * g

Force = volume of water at time t * density * g

So in order to have something flowing (this is discounting that air could go in as water drains...

(3) Vaccuum Pressure * Surface Area < Volume of water * density * g

Now what really startles me is that technically, since the way I'm saying this happens is that no air comes in as the water comes out, this would also mean that the volume approaches zero, the surface area does too, and the pressure does too, then it would also mean that at one point, the pressure inside might be a perfect vaccuum when there's a film of water left on the hole, and that's impossible since the differential of pressure on both sides of the film is enough to burst the film and therefore the sphere would never empty.

Which would make me assume that the sphere emptying DOES reach equillibrium before it even empties and so it doesn't empty, and then I can't assume anything about larger spheres, except that they don't empty either... damn.

But if we were to analyze it in another fashion, we could also say that:

(4)P2 = P1 + density*g*h at equillibrium

Where P2 is atmospheric pressure, and P1 is the pressure inside the sphere on the surface of the water.  So in order for the draining to occur:


And so, if the sphere empties, that is, reaches a complete vaccuum pressure (0, zilch, nothing):


And if we take random numbers, density = 1000 kg/m3, g = 9.81m/s2 and P2 = 1atm, h would have to be...

h > (101,325 N/m2)/(1000kg/m3*9.81)

h > 10.33 meters.........

So............. the sphere doesn't drain!?  .......Looks like that!

It seems that the only way the sphere will drain is if its diameter is greater than 10.33 meters, but then again it would only drain up till that point........... Something is definitely wrong in this statement.  And I'm too lazy to figure it out.

  Posted by Alexis on 2006-07-14 03:59:25
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