As I guess everyone knows, when a full bottle with just a small opening is turned upside down it won't leak.

Not sure exactly why it's like that but I can guess.

 

Now to the question:

If the hole is to large, the water will leak out. What's the maximum diameter of the hole to keep the water in the bottle? Is that dependent of the volume of water or height of waterpillow?

 

Think you understand, just to make sure:

_____
|     |
|-----| <- height of water...
|     |
|     |
|     |
\    /
 \  /
 |-| <- not leaking

really? if u hold a 2-liter upside-down, it will leak

not if the hole is small enough.

think 5 mm is enough

if u punch a hole in the lid of the 2-liter cap, then turn it upside-down it will leak

test for yourself, it won't

i'll do it tommarrow

 

u sure if i get a 2-liter of Mt. Dew and put a nail through the cap, it won't leak?

It depends very much on how big the hole is, and as I said I'm not sure if the volume will affect it. But I guess it won't leak (think the hole must be round though).

However, if you make a hole in the bottom of the bottle too (the side that is up when the bottle is upside down =) ) it will leak, cause the pressure from above will push it through.

o, like if you suck liquid into a straw and cover one end with your thumb, it wont leak.(unless it is carbonated)

exactly, maybe my description of the problem was bad...

Yes, I think it won't leak when the hole is very small.

I think that uses the surface tension of the water molecule, however, the height(not the weight ,right?Since the pressure is P=F/a) of water and the hole must not be too large.

The pressure outside I think this is not relevant.

edit: Pulkit explains it better below...

The bottle does not leak because the surface tension of the water bubble formed at the aperture is greater than the weight of the liquid column above.

 

If you were to take a constant hole size and keep increasing your bottle content, it will eventually leak.

 

To get the critical hole size, merely equate the forces at the hole interface

The pressure acting towards the hole is 1 atm + the height of water ( mm water) no matter with or without air, right?

Someone that knows how to calculate the surface tension?

[MATH]h \rho g + \frac{2 \sigma}{r} =p_{o} [/MATH]

h is height of water column

[MATH]\rho[/MATH] is water density

[MATH]\sigma[/MATH] is surface tension of water

r is radius of aperture

[MATH]p_{o}[/MATH] is outside pressure (atmospheric pressure)

g is acceleration due to gravity

 

Note that critical part is height of column and not total weight of water.

 

Assumpttions :

a) Closed bottle

b) No air present in bottle (If any then it has negligible pressure)

o, like if you suck liquid into a straw and cover one end with your thumb, it wont leak.(unless it is carbonated)

 

What makes the carbonated liquid leak? I guess its the gas right?

What makes the carbonated liquid leak? I guess its the gas right?

 

Right. The gas comes out of solution and displaces the liquid.

 

One thing to remember is that for a drop to come out, you have to create that volume inside the bottle. Carbonation does that without any gas coming in from the outside.

Thanks for the answer, I'm trying to test what you say.

 

[MATH]\sigma = 728[/MATH] for [MATH]20^{o} C[/MATH] water

http://www.engineeringtoolbox.com/24_597.html

Not sure with the decimals. Not sure of anything actually, don't like my results.

 

[MATH]\rho = 1 kg/dm^{3}[/MATH]

[MATH]\sigma = 728 dynes/cm[/MATH]

[MATH]r = 0,001 m[/MATH]

[MATH]p_{o} = 101300 Pa[/MATH]

[MATH]g = 9,82 m/s^{2}[/MATH]

 

Changed your formula like this:

 

[MATH]h \rho g + \frac{2 \sigma}{r} =p_{o} [/MATH]->[MATH]h = \frac{p_{o}}{\rho g}-\frac{2 \sigma}{r \rho g}[/MATH]correct?

 

With the above values h resulted -274,949 m, doesn't seem realistic =/.

 

What am I doing wrong?

b) No air present in bottle (If any then it has negligible pressure)

Doesn't no matter where there is air or not, the water still have 1 atm?

When there is no air, but the oringinal pressure in the top of the water is 1 atm,right?

I think I may be wrong.

Thanks for the answer' date=' I'm trying to test what you say.

 

[MATH']\sigma = 728[/MATH] for [MATH]20^{o} C[/MATH] water

http://www.engineeringtoolbox.com/24_597.html

Not sure with the decimals. Not sure of anything actually, don't like my results.

 

[MATH]\rho = 1 kg/dm^{3}[/MATH]

[MATH]\sigma = 728 dynes/cm[/MATH]

[MATH]r = 0,001 m[/MATH]

[MATH]p_{o} = 101300 Pa[/MATH]

[MATH]g = 9,82 m/s^{2}[/MATH]

 

Changed your formula like this:

 

[MATH]h \rho g + \frac{2 \sigma}{r} =p_{o} [/MATH]->[MATH]h = \frac{p_{o}}{\rho g}-\frac{2 \sigma}{r \rho g}[/MATH]correct?

 

With the above values h resulted -274,949 m, doesn't seem realistic =/.

 

What am I doing wrong?

 

Well, for starters, you need to put all of your values in the same unit convention. You have MKS, CGS and "other" all mixed together. (who uses dm for anything?) I get something close to 10m.

Well, for starters, you need to put all of your values in the same unit convention. You have MKS, CGS and "other" all mixed together. (who uses dm for anything?) I get something close to 10m.

Well, I confused the units for surface tension cause I used multiple pages to look up the answer. According to the page however I think it should be 728 N/m? That doesn't change the answer. So what am I doing wrong?

You must have changed something to the better? =)

Well' date=' I confused the units for surface tension cause I used multiple pages to look up the answer. According to the page however I think it should be 728 N/m? That doesn't change the answer. So what am I doing wrong?

You must have changed something to the better? =)[/quote']

 

728 dynes/cm is 0.728 N/m - the two values are off by three orders of magnitude. That does change the answer.

728 dynes/cm is 0.728 N/m - the two values are off by three orders of magnitude. That does change the answer.

Ok, thank you.

 

I actually thought the SI-unit for density was [MATH]kg/dm^{3}[/MATH] but I looked that up and found it isn't, it's [MATH]kg/m^{3}[/MATH]. That changed the answer to the ~10 m you are talking about.

 

But I get ~10 m with every radius I'm testing with >0,0005, that can't be correct.

Where do the water flow if there is a vacuum machine extracting air outside?

| connect to machine |

| vauum |

|_______|

| water |

Where do the water flow if there is a vacuum machine extracting air outside?

| connect to machine |

| vauum |

|_______|

| water |

 

Could you rephrase ?

 

I can't understand where the water is and where you are creating vaccuum.