posted
There is an old urban myth (according to Scopes) that say there was a big meeting of Colgate executives arguing furiously over how to increase sales. Nothing seemed to be the correct answer and the meeting lasted well into the night. A cleaning woman happened to be working near by. When she heard the argument she said, "Why don't you make the mouth of the tube wider?"
She is now a multi-milloinare, for with that simple suggestion, the amount of toothpaste used, and wasted, doubled.
This is myth.
But I have seen the cola version of this coming to fruition. Not content to be offering bigger and bigger buckets of soda to their consumers, the soda companies have discovered a new way to increase our consumption.
The straws available at fast food restaurants and at convience stores, have grown extra wide. No advertisement or explanations have been given. No reporters or health officials have been consulted. Yet the straws at these places are super-sized, just so we can gulp down that sugar-loaded, diabetic creating, dangerous and un-healthy super duper biggie gulp all the faster.
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posted
Isn't it so that the wider the straw, the more effort one has to use to suck up the liquid? Something to do with the capillary effect, if I remember my physics correctly... If so, then the new and improved version of the would cause the consumer to spend more energy to down the same amount of sugar-loaded product. To my reckoning this would actually constitute an improvement. If they haven't adjusted the price to make up for the cost of the bigger straw, that is...
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posted
Actually, the reason for the bigger straws is to make it easier to drink the thick milkshakes. It's practically impossible to drink a McDonald's milkshake through a normal-width straw, so McDonald's has had the wider-than-average straws for decades. (Incidentally this made McDonald's straws superior for the "twist-up and snap" trick.) Rather than go to the trouble of providing two sizes of straws and all the manufacturing, sorting, and storing effort that would involve, they just provide the shake-friendly straws.
What you are seeing as a "recent" development is just all the other fast-food places catching up.
For the record, though, I hate the larger straws for drinking sodas, but love them for thick shakes. (Not that I've had many of those recently. )
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quote:Originally posted by Eduardo St. Elmo: Isn't it so that the wider the straw, the more effort one has to use to suck up the liquid? Something to do with the capillary effect, if I remember my physics correctly...
True about the "more effort because of the wider straw", false about it being related to the capillarity effect.
It is harder to suck because there is more quantity to "lift" using the same "breath of air", due to the wider cross section of the straw.
The capillarity effect is almost irrelevant here. If you want to visualize its "action", just observe how much liquid is sticking at the bottom of a straw when you lift it out of the drink, as gently as possible (with the top end uncorked by your finger). I’m talking about the drop that forms like a “cork”, not the individual droplets that stick to the sides of the straw “wetting” it. The wider the straw, the less liquid will stick there (with a wider enough straw, just the droplets would remain) and that is because of the superficial tension in the liquid. The thinner the straw, the easier it is for the tube to “catch” a drop of liquid that would sustain itself because of the superficial tension. The consequence is that with a very thin straw, you would have to put quite a lot of effort in order to drink (i.e. “un-sticking” the liquid once it was “trapped” inside). The wider the straw, the less you have to “worry” about the capillarity effect. Plus, the capillarity only affects the “first drop” that you are drinking, because once “the flow” is started (between the liquid in the cup and the liquid in your moth, via the straw) it is all reduced to liquid density and viscosity.
Does that make sense? It is quite clear in my head
posted
I figured that the bubble teaphenomenon might be having aesthetic sequelae in the rest of the market. Sort of Hello Kitty gone wild.
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I think the bigger straws are easier to et through all the ice, but not easier to drink form. Much easier to get the soda to spew forth onto new or freshly laundered shirts.
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quote:It is harder to suck because there is more quantity to "lift" using the same "breath of air", due to the wider cross section of the straw.
This isn't true. The vacuum that is applied to the surface of the liquid acts over the area of the liquid. Like increasing the bore of a piston at the same pressure increases the overall force available to do work, the increased surface area creates a larger force, given a fixed vacuum applied. Each unit of surface area lifts the same volume of liquid, so the vacuum required to lift a column of liquid a certain distance remains constant, regardless of the diameter of the tube.
Vacuum is measured in "inches of water column." In order to lift water from a cup to your mouth (let's say) 4 inches, all you have to do is to generate 4 inches of water column of vacuum with your mouth in order to overcome gravity.
The limiting factor is the amount of vacuum you can generate with your mouth, not the amount of air you move, unless the volume of the straw is actually larger than the volume of your mouth.
The other limiting factors are viscous forces, which becomes very obvious when you try to drink a shake. These become less significant as the size of the tube increases, since the force applied increases as per the area of the top of the column, while the viscous forces (resistance) only increase as per the circumference of the tube.
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quote:It is harder to suck because there is more quantity to "lift" using the same "breath of air", due to the wider cross section of the straw.
This isn't true. The vacuum that is applied to the surface of the liquid acts over the area of the liquid. Like increasing the bore of a piston at the same pressure increases the overall force available to do work, the increased surface area creates a larger force, given a fixed vacuum applied. Each unit of surface area lifts the same volume of liquid, so the vacuum required to lift a column of liquid a certain distance remains constant, regardless of the diameter of the tube.
Vacuum is measured in "inches of water column." In order to lift water from a cup to your mouth (let's say) 4 inches, all you have to do is to generate 4 inches of water column of vacuum with your mouth in order to overcome gravity.
The limiting factor is the amount of vacuum you can generate with your mouth, not the amount of air you move, unless the volume of the straw is actually larger than the volume of your mouth.
The other limiting factors are viscous forces, which becomes very obvious when you try to drink a shake. These become less significant as the size of the tube increases, since the force applied increases as per the area of the top of the column, while the viscous forces (resistance) only increase as per the circumference of the tube.
Glenn Arnold, do you mind if I ask you how do you define “vacuum”? How do you generate “vacuum” with your mouth?
When you begin drinking a liquid using a straw, you suck the air in the straw and the atmospheric pressure is “pushing the liquid up” inside the straw. So your ability to drink is directly related with the volume of the air that you have to suck in order to have the liquid reach your mouth. The wider the straw, the more air to absorb. That is what I wanted to say.
I agree that the pressures are the same regardless the straw cross section, but your lungs don’t only provide the (same) pressure but also have (the same) limited air capacity. So while the pressure you need is the same, the “harder” part comes with “using more air”. You have to “stretch” your lungs “further” and (for one thing) the muscles doing that don’t have the same force ability at that point. Yet the important “detail” is that you have to stretch farther. So you will notice that “added effort”. If you used a robot to suck liquid in an experiment (no force variation with distance “problem” this way), you’ll actually have the same pressures, but the power consumed (i.e. energy over time) would be greater with a wider straw. But I think that’s quite obvious, and not the issue.
I think we do agree even if we use different terms for this, but remember that my “argument” was presented to refute the idea that “the effort” has “something to do” with the capillarity effect.
Maui Babe, I missed this when you posted your question, but AW has the right link. I'd add, though, that in step two you do the twisting part "hand over hand" until the straw winds up, pressing all the air into the middle. And it's easier if you just hold it like that and get someone else to flick it. If the flicker isn't familiar with the trick they might get really startled (bonus??) since it can be quite loud, especially with McD's straws.
Posts: 6394 | Registered: Dec 1999
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quote:Still interested in the “vacuum” issue though.
True that vacuum is simply negative pressure, but I guess you missed the part about the volume of the straw being smaller than the volume of your mouth. Take a small straw, seal the end, and use your mouth to draw a vacuum on it. All it takes is a small deflection of your cheeks. And notice, the back of your throat is closed, to create a seal on the volume within your mouth.
Now do the same thing with a large diameter straw. Feel any difference? There isn't going to be any, until you have a straw (or hose, perhaps) that's so large that you have to actually move a volume of air out of your mouth to someplace else (actually inhale) in order to achieve the level of vacuum you're looking for.
Now try taking a small but long thin straw (well, a length of tubing), and try to drink something out of a cup that's, oh, say, ten feet below you. Feel the tension in your cheeks as you draw the liquid up the straw. Your ears might pop as well.
You'll get a real good feeling for how much vacuum you need to generate in order to lift a column of water. But the trick is this: you can increase the volume of the tube, and the feeling of tension in your cheeks and ears will be the same at any given column height, even though you are lifting a larger volume of water. Again, the only time this will change is if you have to exhaust the air you've sucked into your mouth in order to draw the liquid up further. The difference in the deflection in your cheeks will become noticeable when you get a really long tube with a wider diameter, but that will only serve to point out how insignificant it is with a tube that's only the length of a straw.
Also: I'd never heard of the bursting straw trick before, but I tried it, and it's pretty cool.
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posted
I am skeptical that a wider diameter straw makes it more difficult to drink quickly. Consider that with the larger straw you bring up more liquid in a given amount of time with constant sucking power (or pressure, if you prefer). This is partly because of the larger cross section and partly because of the smaller resistance from viscosity. It might be a little more awkward to maintain a constant pressure for long because of the increased size, but I'm guessing for most people that is more than made up for by the other factors. (There's also an increase in backwash, I'm guessing, but that's probably pretty easy to minimize with some quick tongue action.)
I also think that any additional effort needed to evacuate the air from the straw is minimal. Sure, it'll be slightly more difficult with the larger straw, but there's something wrong if you're running out of lung capacity to do it. (Hint: don't inhale through your straw; it's just generally a bad idea. )
Along these same lines, when did soda/beer cans start getting wide mouths? It's gotten to the point where cans with the old, smaller-sized openings feel, well, slow.
Posts: 1810 | Registered: Jan 1999
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quote:If the fluid has a high surface tension, such as is the case with water, it is possible to make higher negative pressures than a pure vacuum. What is actually happening here is the water's surface exerts a force on anything contained within the water by sticking to the surface. If the water has gases dissolved some will form bubbles and relieve the force, but if the water has had all the gas removed it is very hard for the water to form a new surface and so a pressure below 0 absolute is possible.
This is how, for example, tall trees can draw water up past the point where atmospheric pressure gives out.
Ooh, a really interesting article on the subject: http://www.aip.org/pt/feb00/maris.htm. And, hot damn, I just realized that the Maris who coauthored this article is the same guy I took mechanics with freshman year. Whoa.
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posted
Glenn Arnold, thanks for the answer. I actually agree with you and acknowledge that is was my poor word choice that lead to "arguing".
Mike, I'd agree that you can drink faster with a wider straw, but when it comes to effort or energy usage, it "costs" more. Doesn’t it?
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All this talk about "vacuum" and liquids (and possibly Mike's second link) made me think of an experiment (mental experiment if you like):
Imagine a tube with one end sealed and a mobile piston, hermetically positioned inside. The tube is provided with a heat source, so it can keep a constant temperature of its walls (piston included). Suppose that that temperature is somewhere above the freezing point of water (0ºC) and below the boiling point of water (100ºC). Now, inside the tube, there is water (and nothing else, air or other substances), at the same temperature as the tube. We take all this in outer space (so no atmosphere, no gravity, only vacuum). Question: What happens when we try to pull out the piston? What amount of force would be necessary?
Of course, this is simply “qualitative”. I’m not interested in detailed calculus concerning the diameter/length of the tube, the temperature or whatever (that would be “quantitative”). I’m curious about what do you think will happen? Is it “easy” or “hard” to pull out the piston? Why?
A.
PS: sorry to derail this thread into a “physics” discussion Posts: 1154 | Registered: Oct 2005
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quote:Now, inside the tube, there is water (and nothing else, air or other substances), at the same temperature as the tube. We take all this in outer space (so no atmosphere, no gravity, only vacuum). Question: What happens when we try to pull out the piston? What amount of force would be necessary?
Am I misunderstanding something? If there's nothing but vacuum outside, and there's liquid water at some temperature inside, then there's at least some vapor pressure inside as well, and the net force would push the piston out. In fact, if there's vacuum on the outside of a piston, the piston will always be pushed out absent friction.
Other than that, when you pull out the piston, the pressure inside would drop at first, but as more water vaporizes, the pressure would rise back up to where it started, as vapor pressure only changes with temperature. This would continue until all the liquid turned into gas, at which point it would obey the ideal gas law like normal.
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posted
That is precisely the situation that is described in the first paragraph of the section "How does a liquid break?" in the article I linked, except for the external pressure (or lack thereof). The lion's share of the force holding the piston in is provided by the surface tension of the water, so I don't expect the situation would be much different.
That is, it is generally very difficult to pull the piston out, but if a tiny bubble forms it will grow spontaneously and ultimately eject the piston. Bubbles will form more readily at higher temperatures and lower pressures (that is, the farther you pull the piston out the more likely it is to break the equillibrium). In fact, I wouldn't be surprised if the whole system is metastable even for temperatures close to freezing.
Edgehopper, there isn't always some vapor pressure on the inside. That would certainly be true if there were a liquid-gas barrier, but in the absense of bubbles, pulling the piston out places the entire volume of water under negative pressure. You might want to read the section of the article on the energy barrier of bubble formation (starting at the third paragraph of the "How does a liquid break?" section).
I must admit, this is really fascinating stuff, and rather non-intuitive at first.
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quote: Mike, I'd agree that you can drink faster with a wider straw, but when it comes to effort or energy usage, it "costs" more. Doesn’t it?
My gut feeling (sorry ) is that it would take less energy to drink a given volume of liquid through a larger straw. It might take more energy to drink for a fixed duration, say, one minute. But I can't really back that up except by intuition.
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posted
For now I'll say that Edgehopper's answer is closer to the complete one.
A hint: look for the "phase diagram of water" and think what would we have to do to set up the experiment (especially having liquid water inside the tube when we start pulling the piston out).
If no one else tries to answer, I'll post the one that I was expecting and probably kill the poor thread at the same time ...
quote:That is, it is generally very difficult to pull the piston out, but if a tiny bubble forms it will grow spontaneously and ultimately eject the piston.
It seems to me that the action of pulling on the piston would allow bubbles to nucleate. Given that the temperature of the walls is held constant, the system is not adiabatic. So pulling on the piston allows bubbles to form, lowering the temperature, heat would then be added, and the bubbles would grow and eject the piston as described above.
Here's another scenario:
Assume you have an insulated flask of liquid nitrogen and you want to dip your finger into the flask for some reason. Coating your finger with vaseline insulates your finger from the temperature of the liquid nitrogen. Should you coat your finger with vaseline? Or dip it into the LN2 with no coating?
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posted
The best straws for the pop! trick are the green Starbucks straws. They take a little more oomph because the plastic is pretty thick, but the result is worth it Posts: 3636 | Registered: Oct 2001
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quote:Originally posted by Glenn Arnold: It seems to me that the action of pulling on the piston would allow bubbles to nucleate. Given that the temperature of the walls is held constant, the system is not adiabatic. So pulling on the piston allows bubbles to form, lowering the temperature, heat would then be added, and the bubbles would grow and eject the piston as described above.
Close, but still not adding more than Edgehopper. Plus, The system is isothermal so moving the piston doesn’t change the temperature at all.
quote:Here's another scenario:
Assume you have an insulated flask of liquid nitrogen and you want to dip your finger into the flask for some reason. Coating your finger with vaseline insulates your finger from the temperature of the liquid nitrogen. Should you coat your finger with vaseline? Or dip it into the LN2 with no coating?
Liquid Nitrogen means a very low temperature, close to zero Kelvin (i.e. zero absolute). Pretty much anything else you put in there will instantly freeze, vaseline and human fingers included. (I assume you are in "normal atmosphere"). Even if you might use vaseline as an “insulator” for the low temperature, it will not manage to protect the finger. It (the vaseline) will freeze and get very cold very fast, so your finger will be quickly damaged (frozen biological cells don’t work too well ). You might get (at least) your finger amputated. Don’t do it!
posted
The wider straw does make a better field expedient oboe. Just bite one end until it flattens into a double blade. Trim the corners of the blade slightely if you like. Then cut six 1/8 inch holes at intervals of aproxiamtely one 1 inch, starting from the bottom. Enjoy. By using fife fingerings you should be able to play a diatonic scale and any tune that requires that pitch set. It's guaranteed to have you "asked to leave" from any sit-down establishment.
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posted
Ok, here it is. I don’t claim to have any authority in the matter, this is my “take” on it. If you have a better answer, prove me wrong!
quote:Originally posted by suminonA: Imagine a tube with one end sealed and a mobile piston, hermetically positioned inside. The tube is provided with a heat source, so it can keep a constant temperature of its walls (piston included). Suppose that that temperature is somewhere above the freezing point of water (0ºC) and below the boiling point of water (100ºC). Now, inside the tube, there is water (and nothing else, air or other substances), at the same temperature as the tube. We take all this in outer space (so no atmosphere, no gravity, only vacuum). Question: What happens when we try to pull out the piston? What amount of force would be necessary?
I’ll start with Egdgehopper’s answer:
quote:Originally posted by Edgehopper: Am I misunderstanding something? If there's nothing but vacuum outside, and there's liquid water at some temperature inside, then there's at least some vapor pressure inside as well, and the net force would push the piston out. In fact, if there's vacuum on the outside of a piston, the piston will always be pushed out absent friction.
Other than that, when you pull out the piston, the pressure inside would drop at first, but as more water vaporizes, the pressure would rise back up to where it started, as vapor pressure only changes with temperature. This would continue until all the liquid turned into gas, at which point it would obey the ideal gas law like normal.
Edgehopper, you were quite close. I didn’t state the experiment as a “trap”, but it is counter-intuitive in some way, because we are used to think in (or close to) "Standard (or Normal) conditions for temperature and pressure". But the experiment was set outside such conditions (namely in vacuum = zero pressure).
So yes, it has to do with “vapor pressure”. But, here’s the whole “story”: Have you noticed that on top of a high mountain (where the atmospheric pressure is less the one atmosphere [about 10k Pa]) the water boils at less than 100ºC? (Try it if you have the chance ). Well, the fact is that the boiling temperature (and the freezing one) for water (as it is for all substances) depends on the pressure.
For each substance there can be drawn a “phase diagram” that represents the stable phases (gas/liquid/solid state) for every couple of values (temperature and pressure). Here is the one for water. Usually the “y axis” (for pressures) is not linear in scale (and the region close to zero is “blown up” a bit). And yes, the “supercritical fluid” state is as counter intuitive as you can get. You can start with liquid and get to gas, changing the temperature and pressure accordingly, without ever passing through “vaporizing/boiling”… )
[Remark: Notice though that the “negative pressures” are not drawn, because there is no such thing as a “natural negative pressure”. If you have on one side of a piston (inside a chamber) zero pressure (vacuum), you cannot “subtract more” by puling the piston “out”, because there is nothing inside to apply that pressure on, so the pressure on the other side will continue to be zero. (Yet the force needed to do so will depend on the “atmospheric” pressure on “your side” of the piston.) If there is at least one molecule of any substance on the other side (inside the hermetically closed chamber) and the temperature is more than absolute zero (-273,15 ºC) then there is a positive pressure inside (because of the molecular motion), which will only help you “pull” the piston out all the more.
One more note: the “negative pressure” in the article linked by Mike is a particular definition of “stress applied to a substance”, but the “exterior pressure” in which the experiments take place is always non-negative.]
What we see is that according to the diagram, at a temperature between 0ºC and 100ºC, and very near to zero pressure (if outer space is not “perfect vacuum”) the stable phase of water is GAS. The experiment stated that when we start, we have liquid water inside. To do that, at said temperature, we had to fill the tube in normal atmosphere and transporting it to space, holding in place the piston (i.e. maintaining the exterior pressure).
Now, if we decide to pull out the piston, it means that first we have to “release the pressure” off the piston, and then see what effort is needed to “complete the task”. We can do it in basically two ways. If we want to control the process, we need to keep pushing. If not, no force is needed. Here’s why:
If we suddenly release the piston (we remove all force), the liquid inside will find itself at zero pressure and at a temperature of more than 0ºC, so will instantly boil (becoming gas), which will happen through a kind of explosion (because of the sudden increase in volume, as gas is less dense than the liquid). [Incidentally, this would make a nice “environment friendly” gun in outer space, if the tube is of finite length. ]
Yet, if we want to avoid that “catastrophe”, we will need to first decrease pressure on the piston (meaning that we continue to push, but less and less strongly, so as to pass through successive equilibrium points) until the pressure equals a certain value, the precise value where the temperature inside is the corresponding “boiling temperature”. [Note that all this time the volume change is virtually zero, there is only liquid inside at constant temperature]. At that point the liquid starts to boil (i.e. transforms into gas, and "the bubbles appear") and of course the volume will begin to increase. As we continue to let the piston get out (i.e. we let the volume increase), we’ll need the same pressure (still pushing!) but the proportion liquid/gas will change, getting more and more gas vs. liquid (we are now at a “special” equilibrium point where liquid and gas exist simultaneously, at the same temperature and pressure). [Note: if, in the middle of this “boiling” we want to “push back” the piston, we can do it without increasing the pressure, but we’ll get more and more liquid and less and less gas inside the chamber.] After all liquid becomes gas (supposing the tube is long enough) we will be “simply” dealing with a “perfect gas” and it will push the piston out by itself, if we don’t keep pushing in. The more the gas volume increases, the less pressure it will apply on the piston (remember, we are always at the same temperature). Therefore, working with a tube of finite length, at one point the piston will reach its end and if we don’t keep pushing in/blocking it, it will “fall out” by itself. If the tube is “infinitely long” (possible, because we are conducting a thought experiment) then the volume will never cease to (try to) increase, the gas pushing (less and less strongly) the piston out. Never reaching infinite volume, so never reaching zero pressure.
That was the long answer. The short one is: we need no force if all we want is to get the piston out, and if we don’t care about the “catastrophic” consequences.
I acknowledge that Edgehopper and Glenn Arnold’s answers are very close, the difference being that the vapor pressure (or bubbles) only gets in the picture when we reach the special pressure of equilibrium (where the liquid boils at the given temperature). The reason for not having vapor before that has to do with the fact that the molecules that would “normally” evaporate (at any given pressure) at our temperature, don’t have “enough space” to do it because of the closed chamber. And this in the case scenario where we try to “control” the piston. Otherwise it is all instantaneous and the vapor “appears” all at once.
If you aren’t bored to death yet (not my intention ), I have one more question for you:
What happens if we try to do the same “here” (as opposed to outer space, so no more “vacuum” )?.
posted
I think that if it's set up as described (That is, the configuration is held stable so that there is no perturbation of the conditions), that you will have a situation similar to the "exploding water" described in this link.
You can't simply assume that there is zero pressure on the water. There is zero pressure on the outside of the piston, but the piston and the walls of the cylinder exert pressure on the water. In fact, the water exerts pressure on the water (which is how the "exploding water" fails to boil even though it's above the boiling termperature). And the water exerts pressure on the piston. So if the water is exerting pressure on the piston, the piston is exerting pressure on the water. It begins as a liquid, so unless something perturbs the system, it will remain liquid. This is the case in the "exploding water" scenario, where the water is above the boiling point in normal conditions, but fails to turn into a gas unless something provides a nucleation site.
In order for the water to begin expanding without changing to a gas, it would have to be in a supercritical state. I've never worked with supercritical water, but supercritical CO2 is pretty common. It has the density and viscosity of a liquid, but it fills it's container as if it's a gas. Thus if the volume of the container increases, the volume of the supercritical liquid increases with it. But water is only supercritical at very high pressure and temperature. It can't happen at low pressures, so it wouldn't happen here.
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posted
It's not just straws that increased in size but many soda cans have larger holes in the top to make drinking a can faster/easier.
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posted
suminonA, I think you are ignoring the fact that, in order for any of the water to boil you need to start out with a bubble, i.e. a nucleation site. Nucleation sites can occur spontaneously, but there is a specific energy threshhold that must be overcome before such a microbubble can become big enough to expand without bound.
Glenn, you might be right that the action of pulling the piston out can cause nucleation sites on the edges of the container, but I wouldn't be surprised if this only occurs with any frequency at relatively high temperatures. I'm assuming here that we're dealing with as close to pure water as possible (no dissolved gasses, etc.) and very smooth container walls.
My intuition that the piston would strongly resist removal is based on this quote from the article I linked earlier:
quote: Marcellin Berthelot claimed in 1850 that he had reached –50 bars in a glass ampoule completely filled with pure water.1 In 1967, Edwin Roedder at the US Geological Survey reached –1000 bars with water inclusions in natural rocks.2 The world’s record now belongs to Austen Angell and his collaborators at Arizona State University, who in 1991 reported achieving –1400 bars with a similar technique but synthetic materials.3
These are the references if anyone cares:
quote:1. M. Berthelot, Ann. Chim. Phys. 30, 232 (1850). 2. E. Roedder, Science 155, 1413 (1967). 3. Q. Zheng, D. J. Durben, G. H. Wolf, C. A. Angell, Science 254, 829 (1991).
(Incidentally, this goes against the common adage that liquids are incompressible. You can change the density of a liquid through external pressure, apparently in both directions.)
So, my question is, how did these people create such astonishingly large negative pressures in liquid water if it is so easy to form nucleation sites? I don't think it's correct to say that "water is only supercritical at very high pressure and temperature", as this is precisely a supercritical condition that is being described. (Note that these pressures are up to 1380 times atmospheric pressure, which is why I think I am justified in neglecting the outside air pressure on the piston or lack thereof.)
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citadel, I've noticed that too.
Posts: 1810 | Registered: Jan 1999
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quote:Originally posted by Glenn Arnold: You can't simply assume that there is zero pressure on the water. There is zero pressure on the outside of the piston, but the piston and the walls of the cylinder exert pressure on the water. In fact, the water exerts pressure on the water (which is how the "exploding water" fails to boil even though it's above the boiling termperature). And the water exerts pressure on the piston. So if the water is exerting pressure on the piston, the piston is exerting pressure on the water.
You might have missed the “detail” in boldface here:
quote:Originally posted by suminonA: […]Now, inside the tube, there is water (and nothing else, air or other substances), at the same temperature as the tube. We take all this in outer space (so no atmosphere, no gravity, only vacuum).[…] [emphasis added]
A liquid in outer space does not exert pressure on itself, other than the one due to superficial tension.
Notes: 1) The superficial tension is due to the different (electrostatic) attraction of the liquid molecules (forming the surface) toward the rest of the liquid vs. molecules “outside” the liquid.
2) At any given positive absolute temperature (i.e. more than 0 Kelvin) the molecules of the liquid are in constant motion (I’ll call that “the Brownian Motion” even though it is not strictly accurate). This allows some molecules to “escape the liquid” even below the boiling temperature, because they pass the “superficial tension barrier (+ the outside pressure)”. [This explains why an uncovered glass of water evaporates in time, even at room temperature. Not the solid glass. The water inside. ]
3) The “boiling temperature” at a given pressure is the temperature at which the Brownian Motion is so important that any molecule can pass the “superficial tension barrier + the outside pressure”. [This is what “liquid/vapour equilibrium” means]
Imagine now that you are on a geostationary space station (i.e. no gravity effects) and that you have a (transparent, so you can see inside) chamber containing a drop of (pure) water, surrounded by air. It will “float” inside and we will try to avoid the situation where the drop sticks to one of the sides of the chamber. Now, using multiple slots in the chamber simultaneously, we begin to “suck the air out” which will decrease the pressure inside the chamber. What will happen with the “floating drop” when the pressure inside the chamber reaches the “boiling pressure” for the present temperature?
According to the notes above, the drop will instantly “explode” (but not violently; we are at an equilibrium pressure) because all the molecules (close enough of the “surface”) will “get free”, as the superficial tension cannot hold the drop together any longer. We have instantly vapours only. I don’t think that there can be a “meta-stable” situation where the drop keeps its liquid form below the boiling pressure, because the Brownian Motion is always “pushing” the molecules out, so they will get out as soon as they get the chance, won’t they? It is perfectly possible to transform the liquid into gas, without the “need” of any single bubble inside (nucleation), because the “drop surface” simply “disappears” and the molecules are free to disperse until they fill out the chamber. This is what “gas” means: the molecules don’t hold together (with each other) at all. [I wonder: Is this true only in “super critical fluid” conditions?]
Now, back to our piston experiment. Having the “drop of water” surrounded not by air but by solid walls (and the piston) changes only the behaviour of the molecules in contact with the solid walls, because they might stick to them (depending on the particular material used). But when there is no more pressure outside the piston, the Brownian Motion will push the molecules apart and set them “free”, because the temperature is sufficiently high so the electrostatic attraction is irrelevant (at that ZERO pressure). [note: at any given time, the pressure has the same value in all the volume of the chamber. That means that when outside the piston there is zero pressure it is like the chamber is OPEN (to vacuum).]
Maybe completely neglecting the “stickiness” of the liquid molecules to the walls is an exaggeration, but we can manage to reduce it using special materials (the same way we don’t need to worry about friction) to the point it becomes irrelevant.
quote:Originally posted by Glenn Arnold: It begins as a liquid, so unless something perturbs the system, it will remain liquid. This is the case in the "exploding water" scenario, where the water is above the boiling point in normal conditions, but fails to turn into a gas unless something provides a nucleation site.
I obviously didn’t make experiments about this, but the only part that “bothers” me is the Brownian Motion factor. Isn’t it a constant “system perturbation” enough?
quote:Originally posted by Glenn Arnold: In order for the water to begin expanding without changing to a gas, it would have to be in a supercritical state. I've never worked with supercritical water, but supercritical CO2 is pretty common. It has the density and viscosity of a liquid, but it fills it's container as if it's a gas. Thus if the volume of the container increases, the volume of the supercritical liquid increases with it. But water is only supercritical at very high pressure and temperature. It can't happen at low pressures, so it wouldn't happen here.
Again, I lack “laboratory practice” with this, so if you speak out of experience, than I can only hope to see an explanation for what (and why it) happens, coming from someone having the knowledge, someone like you.
Please correct me if I’m wrong on any of the points above. Thanks.
--- Sorry to keep posting off topic. Long live the “big straw” conspiracy!
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suminonA, your scenario with the floating droplet is correct. You do not need a nucleation site for the liquid to enter the gaseous phase because there is already a (rather large) surface from which molecules can escape. There is no such surface in the piston experiment. I agree that the dynamics at the edge of the container are different from those in the middle of the liquid, but neglecting the "stickiness" of water to itself is not reasonable, especially on the scales at which bubbles form. (We can certainly posit that the water does not stick to the container walls; this will make it easier for the explosive scenario to occur, but not make it inevitable.)
The phase diagram you linked to earlier represents the stable equilibrium for a given temperature and pressure. There are, however, metastable equilibria: witness, e.g., the water in the microwave which seems to violate the phase diagram.
Did you read the article I linked? You might want to — it addresses your "Brownian Motion" factor and why it is not always enough of a "system perturbation" to do what you expect.
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I think we're on the same page. I'm trying to wrap my mind around the concept of -1500 bars. Negative relative to what? Easier to conceptualize it as "tensile strength of a liquid."
Yeah, given that the experiment is theoretical, I'd agree that the water should be pure and the cylinder is perfectly smooth and perfectly sealed, etc., in which case it'd be harder to create a nucleation site. But as far as temperature is concerned, it seems to me that 100ºC ought to be enough, and the experiment only states that the water is "somewhere above the freezing point of water (0ºC) and below the boiling point of water (100ºC)." To me, it's intuitively logical that if you're close to 100ºC, it shouldn't be too hard to create a bubble.
Your article brought back memories. I used to work with an engineer (Ann Tassin Leger) who's phD thesis involved cavitation. We worked in what I called the "smoke and mirrors lab" where we did imaging and velocimetry of gas flow using a smoke generator and soap bubbles filled with helium. I learned a lot about cavitation damage to impellers, ultrasonic cleaners and sonoluminescence (talk about high energy density!), and other stuff that had nothing to do with what we were supposed to be studying. Fun stuff.
suminonA
quote:We take all this in outer space (so no atmosphere, no gravity, only vacuum).[…] [emphasis added]
Gravity isn't an issue. The only way water can change from a liquid to a gas is that individual molecules have to escape from the liquid. There has to be a place where they can go that allows them to move outside of the liquid regime. The structure of the cylinder itself is a barrier that prevents the water molecules from escaping the liquid.
From the perspective of the water molecules, the vacuum on the other side of the piston doesn't exist. First they have to get the piston moving. Individual molecules may exert pressure on the piston, but only as a liquid, not as a gas, so the water can't expand to follow the piston as it moves. As long as it's in a liquid state, individual molecules don't have enough momentum to overcome the inertia of the piston, because on average, the attractive forces of the water are stronger than the "pressure" of individual molecules. You'd have to get a bunch of molecules working together at the same time (a bubble). The equations in Mike's link describe the energy required.
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quote:Originally posted by Mike: Did you read the article I linked? You might want to - it addresses your "Brownian Motion" factor and why it is not always enough of a "system perturbation" to do what you expect.
Well, I did. Where do you think the idea of the experiment came from?
As for the article, do you mean the part quoted below?
quote:Thermal versus quantum cavitation
How close can experiments come to the spinodal of helium? As the pressure is made more negative, the energy barrier preventing the nucleation of a bubble becomes smaller and smaller. At some distance from the spinodal, the barrier is small enough that there becomes a chance that, with the aid of a thermal fluctuation, a bubble larger than the critical size will be created. Such a bubble will then grow to a macroscopic size. As the temperature is lowered, the thermal fluctuations become weaker, and the probability of bubble nucleation remains small until the pressure is very close to the spinodal. Below a critical temperature Tc, thermally activated passage over the nucleation barrier becomes unimportant compared to quantum tunneling through the barrier. This process is “macroscopic,” in the sense that it requires the cooperative motion of several hundred helium atoms (of course, “macroscopic” is a great exaggeration). The detailed theory of quantum cavitation was worked out by one of us (Maris) in 1995 and by Montserrat Guilleumas and her colleagues in 1996. Both analyses predicted that quantum cavitation should become important in 4He at temperatures below Tc = 0.2 K and at pressures within about 0.3 bars of the spinodal. It is essential in these calculations to allow for the softening of the liquid as the spinodal is approached. This effect was not included in earlier calculations.
[emphasis added]
It does talk about “thermal fluctuation”, but it insists on the conditions (very low temperatures) where the quantum cavitation is more important. So I’d say that in the case of water and greater temperatures (more than 273K, as opposed to close to 0.2 K in the case of helium) the effect of “thermal fluctuations” is what prevails. Or am I misunderstanding it?
quote:The phase diagram you linked to earlier represents the stable equilibrium for a given temperature and pressure. There are, however, metastable equilibria: witness, e.g., the water in the microwave which seems to violate the phase diagram.
Yes, I acknowledge that the meta-stable equilibrium is observable in special conditions, and I think it has to do with the electrostatic attraction of the molecules with each other inside the liquid, attraction that might be a lot greater than what I would anticipate… I also think that the same attraction (stickiness with itself as you call it) is the one responsible for the possibility of reaching “negative pressure” in the experiments in the article.
Thanks for the comments. Confronting my ideas with others is a wonderful opportunity to acquire new knowledge.
quote: It does talk about “thermal fluctuation”, but it insists on the conditions (very low temperatures) where the quantum cavitation is more important. So I’d say that in the case of water and greater temperatures (more than 273K, as opposed to close to 0.2 K in the case of helium) the effect of “thermal fluctuations” is what prevails. Or am I misunderstanding it?
I think you're right. I wouldn't expect any quantum effects in our experiment. Any cavitation we see would be from thermal effects.
Glenn,
quote:I think we're on the same page. I'm trying to wrap my mind around the concept of -1500 bars. Negative relative to what? Easier to conceptualize it as "tensile strength of a liquid."
Good to hear. Yes, thinking of it as tensile strength might make a little more sense, intuitively.
That smoke & mirrors stuff does sound fun...
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Incidentally, how do things change if we are starting out with ice?
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Well, I'm just trying to envision a cylinder of ice made out of ultrapure water.
One of the things we did (or rather tried to do) in the smoke and mirrors lab was to measure gas flow by measuring the concentration of paradichlorobenzene (para for short) in vent gas when flow patterns pass over a "glass flat" plate of para.
It was my job do make the plate of para, but I couldn't manage to cast it without gas bubbles appearing as it solidified. The same is true with ice. When's the last time you saw ice that was perfectly clear and free of bubbles. Doing that work gave me a real appreciation for how soluble gas is in liquids. (I no longer wonder how fish breathe)
As to your question, I think if we start with ice and keep the temperature constant, it will just sit there. The ice sure isn't going to sublime if it has no where to go. I don't think pulling on the piston will do any good either. Somewhere recently I read that ice can be stronger than steel under certain conditions, so the strength of the piston and cylinder materials comes into play.
Any comment about my liquid nitrogen question?
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My guess for the ice variation is that the force required to pull out the piston depends on how strongly the piston/container material sticks to ice. If you were able to generate a gap, say, between the piston and the ice, the vapor pressure of the ice would push out the piston. (Sublime! )
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For the liquid nitrogen question, I have heard that dipping your hand in will instantly boil a thin layer of the nitrogen surrounding your hand. Since nitrogen gas is a poor conductor of heat (and has such a low specific heat capacity), you should remain pretty much unfrozen for a while. Covering your hand in vaseline might prevent the protective layer of nitrogen gas from forming. So the vaseline would freeze rather quickly and probably transport heat away from your hand more efficiently than the nitrogen gas. Leaving you with a popsicle on your arm in short order.
That's just a guess, though.
I remember reading an article on the web a while ago that talked about this, plus dipping your hand (or finger?) into molten lead, droplets of water skittering around in a hot pan, and other similar phenomena.
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When the temperature of an solid object placed in LN2 is more than 50º (F, I think) hotter than the boiling point, it causes "film boiling" where the entire nitrogen contact area flashes into a layer of gas, preventing the solid from coming into contact with the liquid, so heat transfer is limited (it just feels like putting your hand in a freezer). Once the temperature of the solid gets to be closer than 50º, it changes to nucleate boiling, which allows the liquid to be in contact with the solid, and increases heat transfer by an order of magnitude or so.
The vaseline insulates your hand, so the outside surface cools very quickly and allows nucleate boiling to happen, so your hand will freeze fast enough to give you frostbite before you can react.
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quote:Originally posted by Glenn Arnold: The only way water can change from a liquid to a gas is that individual molecules have to escape from the liquid. There has to be a place where they can go that allows them to move outside of the liquid regime. The structure of the cylinder itself is a barrier that prevents the water molecules from escaping the liquid.
From the perspective of the water molecules, the vacuum on the other side of the piston doesn't exist. First they have to get the piston moving. Individual molecules may exert pressure on the piston, but only as a liquid, not as a gas, so the water can't expand to follow the piston as it moves. As long as it's in a liquid state, individual molecules don't have enough momentum to overcome the inertia of the piston, because on average, the attractive forces of the water are stronger than the "pressure" of individual molecules. You'd have to get a bunch of molecules working together at the same time (a bubble). The equations in Mike's link describe the energy required.
I’ve flexed a few more neurons on this. I don’t agree at all with the part: ”From the perspective of the water molecules, the vacuum on the other side of the piston doesn't exist.” I insist on this because the “vacuum” is a key factor in the experiment. This is why we have to use our reasoning before our intuition here.
Here’s my reasoning:
1) The surface of liquid water (at a given temperature below boiling point) is not “a special layer of molecules, locked together as a bag, holding the rest of the molecules inside”. It (the surface) is “made up” of agitated molecules (thermal fluctuations) that “sink back in the liquid” to let others “surface” for a while. A "still" glass o water is "full of motion", even at its surface.
2) Inertia by itself cannot “stop” the moving of an object. Suppose we have a solid object (e.g. the piston) at rest in a given frame of reference. It has a certain inertial mass. Now suppose it is found inside a system where friction and stickiness to other substances was carefully minimized (removed even). Given those conditions, any amount of force applied to it will set it in motion, along the line of the action of the force. Actually it will accelerate it. The force, the mass and the acceleration are related according to a Newtonian law (Second Law of Motion). It doesn’t matter how important the mass is, or how feeble the force is. There is a non-zero acceleration as a result of a non-zero force. [NOTE: here the “zero friction” and “zero stickiness” are ideal conditions that are probably impossible to reach in practice, making them non-negligible at molecular level…]
3) Any liquid, in contact with a solid, is exerting a pressure on it (that is, a force on any given surface), caused by the moving molecules reaching “at the wall”. In a gravitational field, this pressure is negligible compared with the hydrostatic pressure (the deeper the considered point is, the greater the pressure). In a gravitation free environment, the “stickiness” of the liquid to the solid is the main factor in “hiding” the pressure of the liquid. But if the “stickiness” is reduced to zero, only that pressure remains.
All that follows is considering this “ideal situation” where the friction and the stickiness are totally removed.
4) Therefore, having a liquid in a closed chamber, consisting of a tube and a mobile piston, the liquid on the inside of the piston will exert a non-zero pressure on one side, at any given positive temperature. But if on the other side of the piston there is NOTHING (the famous vacuum ) then we find our piston in a non-equilibrium situation. That pressure on the inside of the piston means a force, that force means acceleration, that acceleration means the movement of the piston “being pushed out” by the liquid. Even seen at a molecular lever, the impact of any single molecule with the solid piston is affecting it (conservation of the momentum). The fact that the mass of a single molecule of water is so small compared with the mass of the piston is a non issue. A non zero force produces a non zero acceleration on ANY mass. Plus we have more than one molecules of water and a “continuous bombardment” of the inside wall. If there is no friction, no stickiness and no exterior pressure, there is NOTHING that can (i.e. would) “hold the piston” in place. It will move!
5) If you are going to say that it cannot move, because liquid water is incompressible (and does not stretch, “especially not by itself”), then think of what that movement means: the volume of the closed chamber is increasing, leaving some “void space” inside the chamber. This void is possible because outside we have also void (vacuum) so there is no pressure to minimise the volume of the void inside. This is why I think that the presence of the piston with vacuum on the outside is equivalent to “open tube” (again, if we manage to reduce the liquid/solid stickiness). This means that the water inside is “instantly” at zero pressure (if we don’t keep pushing the piston in).
6) Being at such a low pressure, it finds itself below boiling pressure (where the limit liquid/gas had “disappeared”) and therefore the water molecules will scatter freely (this is the result of “high temperature and low pressure”). Which is perfectly possible as the piston is “already” in motion (giving room) because of the disequilibrium that happened while the liquid was still being a liquid and begun pushing the piston out (the moment we released any force on the piston).
This is also based on the analysis that I’ve made in a previous post, about the “water drop in a transparent, large chamber in zero gravity”. The result that I’m using here is the conclusion that a liquid cannot “hold” an inner pressure by itself, at a given temperature, if there is no exterior “atmospheric” pressure. If we don’t push the molecules in, they scatter (with an explosion if the exterior “pressure fall” is drastic).
Does this work in “ideal conditions” or do you see something wrong with this? I’ll come back to this, considering “non perfect conditions”, where the friction and the stickiness cannot be totally eliminated, but still reduced to a “practical” maximum. Of course, anyone might do that analysis if interested.
BTW, in this “perfect conditions” above, starting with ice instead of liquid water would only change the temperature of our experiment (it would be below 273.15 K). But all the rest of the considerations (the steps of reasoning in this post) remain as valid. Of course, the ice won’t “melt first and evaporate next” because being all instantaneous it is called “sublimation”. So yes, even ice can “explode”.
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I suspect you are right, suminonA, about what happens when we truly have zero stickiness. Which I think I was subconciously hinting at when I proposed the variation with the ice. I think you are also right that zero stickiness is impossible to reach in practice, but I could be wrong about that.
I also think that you use more words than you need to make your points, by an order of magnitude or so. Posts: 1810 | Registered: Jan 1999
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quote:Originally posted by suminonA: PS: this post was intented to be as short as possible
And you prove my point once again.
But seriously (folks), I like seeing the energy and enthusiasm behind long posts. I simply make no guarantee about my ability to read every word with the same enthusiasm.
And now back to your regularly scheduled topic at hand...
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