osram Posted August 28, 2004 Share Posted August 28, 2004 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 Link to comment Share on other sites More sharing options...
ydoaPs Posted August 28, 2004 Share Posted August 28, 2004 really? if u hold a 2-liter upside-down, it will leak Link to comment Share on other sites More sharing options...
osram Posted August 28, 2004 Author Share Posted August 28, 2004 not if the hole is small enough. think 5 mm is enough Link to comment Share on other sites More sharing options...
ydoaPs Posted August 28, 2004 Share Posted August 28, 2004 if u punch a hole in the lid of the 2-liter cap, then turn it upside-down it will leak Link to comment Share on other sites More sharing options...
osram Posted August 28, 2004 Author Share Posted August 28, 2004 test for yourself, it won't Link to comment Share on other sites More sharing options...
ydoaPs Posted August 28, 2004 Share Posted August 28, 2004 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? Link to comment Share on other sites More sharing options...
osram Posted August 28, 2004 Author Share Posted August 28, 2004 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. Link to comment Share on other sites More sharing options...
ydoaPs Posted August 28, 2004 Share Posted August 28, 2004 o, like if you suck liquid into a straw and cover one end with your thumb, it wont leak.(unless it is carbonated) Link to comment Share on other sites More sharing options...
osram Posted August 28, 2004 Author Share Posted August 28, 2004 exactly, maybe my description of the problem was bad... Link to comment Share on other sites More sharing options...
Primarygun Posted August 29, 2004 Share Posted August 29, 2004 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. Link to comment Share on other sites More sharing options...
Thales Posted August 29, 2004 Share Posted August 29, 2004 edit: Pulkit explains it better below... Link to comment Share on other sites More sharing options...
pulkit Posted August 29, 2004 Share Posted August 29, 2004 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 Link to comment Share on other sites More sharing options...
Primarygun Posted August 29, 2004 Share Posted August 29, 2004 The pressure acting towards the hole is 1 atm + the height of water ( mm water) no matter with or without air, right? Link to comment Share on other sites More sharing options...
osram Posted August 29, 2004 Author Share Posted August 29, 2004 Someone that knows how to calculate the surface tension? Link to comment Share on other sites More sharing options...
pulkit Posted August 29, 2004 Share Posted August 29, 2004 [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) Link to comment Share on other sites More sharing options...
toolman Posted August 29, 2004 Share Posted August 29, 2004 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? Link to comment Share on other sites More sharing options...
swansont Posted August 29, 2004 Share Posted August 29, 2004 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. Link to comment Share on other sites More sharing options...
osram Posted August 29, 2004 Author Share Posted August 29, 2004 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? Link to comment Share on other sites More sharing options...
Primarygun Posted August 30, 2004 Share Posted August 30, 2004 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. Link to comment Share on other sites More sharing options...
swansont Posted August 30, 2004 Share Posted August 30, 2004 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. Link to comment Share on other sites More sharing options...
osram Posted August 30, 2004 Author Share Posted August 30, 2004 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? =) Link to comment Share on other sites More sharing options...
swansont Posted August 30, 2004 Share Posted August 30, 2004 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. Link to comment Share on other sites More sharing options...
osram Posted August 30, 2004 Author Share Posted August 30, 2004 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. Link to comment Share on other sites More sharing options...
Primarygun Posted August 31, 2004 Share Posted August 31, 2004 Where do the water flow if there is a vacuum machine extracting air outside? | connect to machine | | vauum | |_______| | water | Link to comment Share on other sites More sharing options...
pulkit Posted August 31, 2004 Share Posted August 31, 2004 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. Link to comment Share on other sites More sharing options...
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