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Q & A: unstable balloons

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Most recent answer: 01/29/2016
Q:
Why does this happen? There are two inflated ballons attach to a tube (one on each side) the tube ha a faucet and one of the balloons is less inflated than the other. when i open the faucet the air from the less inflated ballooons passes to the balloon that is more inflated.
- Estherika
Israel
A:

 That’s a really interesting question. Here’s what I think the answer is.

The air will move around to minimize the total free energy of the air and the balloons, the same way a fluid will flow downhill as far as it can, minimizing gravitational energy. Let’s make some approximations to see whether this weird behavior follows from simple ingredients or requires some special properties of the balloons.

Let’s say that the total volume of air is pretty much fixed. Then the free energy changes come from the stretching of the rubber in the balloons. Let’s say that, like most springy things, the excess free energy is proportional to the square of how far they’re stretched. Now the volume goes as radius cubed, but the free energy goes as radius squared. So the excess free energy is proportional to the balloon volume to the 2/3 power. The rate of change of free energy with respect to volume increase (the slope or derivative) then is inversely proportional to volume to the 1/3 power. So the more-inflated balloon increases free energy less for a given volume of added air. So free energy goes down when air flows from the less inflated to the more inflated balloon. The process will stop when the less inflated balloon isn’t stretched any more.

Mike W.


(published on 12/08/2007)

Follow-Up #1: unstable pair of balloons

Q:
Why air goes from a small ballon into a big one?Thank u
- sarah (age 37)
iran
A:

This old answer was hard to find, but I think it's what you need.

Mike W.


(published on 07/26/2015)

Follow-Up #2: wrong ideas about balloons

Q:
That is a Balooney of an answer. She is simply asking why does the flow go from a low pressure to a higher pressure balloon? The answer is that there is nothing strange about this! In fact the counter example is right there in the experiment! The air flows from the stronger and smaller mouth-balloon to the weaker larger rubber-balloon. Otherwise, if you doubt that the larger balloon has higher pressure, then here is the proof: At one point the two (identically-constructed) balloons were equal in size and hoop tension. Then one balloon received more air, grew, stretched more, and got more hoop tension force, i.e., more pressure per square inch at the surface, and (according to Pascal's law) this (air fluid) pressure at the surface transfers equally to everywhere in the air (inside the balloon) including at the nozzle. So now, forget about the balloon size, etc., just focus on the nozzle pressures. The question simplifies to: Why flow takes place from the lower pressure nozzle to the higher pressure nozzle? I suspect that the answer is that the experiment is a magic show setup, i.e., tricky. I suspect the larger balloon is of a thinner, easier stretchable material than the smaller one, hence, the smaller one is more forceful -- just as in the mouth example. In fact, when you blow air out of your mouth (such as when you cough) the air flows from a smaller mouth-balloon to a much larger atmosphere-balloon.
- Anonymous
A:

Seriously? You think air spontaneously flows from low to high pressure? The fundamental laws of thermodynamics are all wrong? 

I suggest you try it with ordinary balloons. Maybe you'll change your mind.

Mike W.


(published on 01/26/2016)

Follow-Up #3: unstable balloons revisited

Q:
No! That is a complete reversal of what I said -- may be my phrasing was not clear. She is THINKING (i.e., it seems to her) that "the flow goes from a low pressure to a higher pressure balloon" but in reality it does not. It is a trick of observation, i.e., actually the smaller balloon has higher pressure. Let me try it this way. the flow (as observed) is from the small-balloon side of the nozzle to the large-balloon side. Agreed? OK. This means there is less pressure on the large-balloon side. Agreed? OK. This means there is less pressure everywhere in the large balloon (as per Pascal's law of pressure in a static fluid). Agreed? OK. This means there is less pressure on the surface of the large balloon. Agreed? OK. This means there is less hoop tension in the large balloon skin. Agreed? OK. Assumption 1: Two balloons are identically constructed. This means the large balloon's skin is stretched less than the small balloon. Agreed? OK. This means large balloon is smaller than the small balloon. Contradiction. Hence, Assumption 1 is incorrect, i.e., two balloons are NOT identically constructed, i.e., they are magic balloons, i.e., smaller balloon's skin is stiffer than the large balloon. I then gave the example of mouth as a small balloon that inflates a large rubber balloon because mouth's skin is stiffer. Also whereas the mouth exhales into the atmosphere, hence, the mouth is a stronger balloon than the atmosphere. May be my wording was confusing. But (I think) the logic is correct. Now, my argument is based on a number of tacit assumptions. You may want to debate these assumptions: Assumption 2) Low pressure at the nozzle is the same low pressure everywhere in the large balloon -- otherwise a current would obtain which will eventually settle once the pressure equalizes everywhere. Assumption 3) Hoop tension is directly related to the internal pressure, i.e., hoop tension increases (or decreases) as does the air pressure inside. Assumption 4) As the hoop (rubber skin) tension decreases so does the surface of the balloon. This assumption is based on the elastic behavior of material -- such as the rubber skin of the balloon. Assumption 5) As the surface area decreases so does the diameter, i.e., the size of the balloon. This is just a geometry fact. So please, refute these assumption and / or the argument. Using free energy (or conservation laws) and other principles which are not directly and graphically related to the experiment leaves the questioner wanting. If you can prove your point in one way (using energy argument) you ought to be able to prove it the other way, i.e., using air pressure, hoop tension, elasticity, and so forth. Thanks.
- Anonymous
A:

Your clarification does an excellent job of clearing up the logic of the problem, including where you took a wrong turn. We're exactly with you up to "This means there is less hoop tension in the large balloon skin." It was that very point which we addressed indirectly in our original answer. As you say, our argument may have been too technical sounding, but you've provided a nice structure to discuss it further.

What's the relation between the pressure difference across the balloon skin and the radius of the balloon? As you say, as the radius gets bigger, the tension in the skin goes up. How does that tension get translated into a pressure difference? In equilibrium the force on any little patch is zero. So the force from the tension must cancel the force from the pressure difference. And here's the key point you missed. If the skin were simply flat, no amount of tension in it would create any force at right angles to the skin! The only reason there's a force in that direction is that the surface curves, so that neighboring patches pull in a little on the patch we're considering. The force depends on the product of the tension and the curvature. Yes, as the balloon gets bigger the tension goes up, but the curvature goes down. In terms of the radius, R, the curvature goes as 1/R and the tension goes about as R. So it turns out that this line of argument is not getting us a unique answer.

Here's our simpler argument:
1. In equilibrium in the room, there's some volume V of air inside the balloons, at room temperature.
2. So let's assume that the air has that fixed volume and figure out how to distribute that V to minimize the free energy.
3. At fixed V and T, the distribution of V has no effect on the air free energy.
4. So the distribution is determined by minimizing the elastic free energy of the balloons, under the fixed-total-V condition.
So that's what we did.

Mike W.


(published on 01/27/2016)

Follow-Up #4: figuring out how to teach

Q:
The problem that many students or in general people who are studious (such as those asking questions on this forum -- not those who goof around) is that a logical line of explanation gets polluted by cult-like statements such as "minimization of energy principle" (as in formation of soap bubbles) "light takes the shortest path." How, do soap bubbles or light beam know calculus? That is when the studious students lose interest, because they cannot follow the line of reasoning. I wish, forums such as Physics Van and others, state assumptions and principles that are intuitive, at least as far as classical physics is concerned -- quantum confusion is another matter. So I could follow "sensibly" your statement up to your point about larger balloon has flatter surface and hence requires larger hoop tension for the same pressure. But then, that voodoo stuff about air "minimize[ing] the free energy" showed up its ugly head. That is when students lose their train of logic. If teachers (or expounders like yourself) could carry on the common-sensical explanation (at least in classical physics questions) then we mortals won't get confused. Thank you.
- Anonymous
A:

Thanks for your constructive thoughts on how to convey these ideas. It ain't always easy! If I figure out a way to take your advice and rephrase our argument in terms of forces without getting something wrong, I'll do it.

You do raise an interesting question, that applies to all of physics, not just some special little category. "How, do soap bubbles or light beam know calculus?" The amazing thing is that nature seems to know calculus and much fancier branches of math. In fact, it doesn't seem to know anything else.

Meanwhile, I think that for this not-too-fancy experiment, the advice to try it yourself is the key!

Mike W.


(published on 01/28/2016)

Follow-Up #5: axioms and theorems

Q:
Thanks for your explanation, I am now clear about the experiment. However, it is hard to swallow the notion that nature knows calculus (or fancier math.) Nature works with the immediate (differential) action / reaction forces. The global principles such as "minimization of energy" are integral of these differentials. In fact the full statement of the above principle is "minimization of the integral of energy differentials." And this requires calculus of variation -- the fancy math you alluded to. Such fancy-math based statements, are the upshot of all the millions of tedious little interactions and their primitive maths. The best, we can say is that: 'Such global (integral) statements as minimization of energy are the equivalent of pages of computation at the local (differential) level'. But now, having explicitly and formally made this (in single quotes) statement, one has to prove it. So, (at last, after this tortured thread of discussion) here is my question: Has there been any methodical program of proving each of such grandiose global statements (as the minimization of energy), for a general case, starting from the first principles (as forces, Newton's laws, etc.), in other words, changing them to theorems rather than the (present) principles? Remember a sound axiomatic system (which physics and other sciences try to be) minimizes its axioms by replacing them with theorems.
- Anonymous
A:

Well, speaking of axioms, your fundamental one is false. You write "Nature works with the immediate (differential) action / reaction forces. " That's a specific type of local realist theory. The violations of the Bell Inequalities (see e.g. https://van.physics.illinois.edu/qa/listing.php?id=30737&t=bell-inequalities) prove that no such theory can describe our world.

But ok, let's confine our attention to the subset of phenomena in which a local-realist classical approximation works fairly well. All of standard intermediate mechanics (Hamiltonian and Lagrangian) is based on formal proofs of the equivalence of the global principles to the local description  you mention.

In our balloon case, however, none of that is particularly relevant. We're asking what condition things settle down to, not what the detailed motions are on the way. The little forces you describe obey laws that are the same if you reverse the direction of time. Have you ever observed any phenomenon that would be the same backward in time? I doubt it, since that would mean all the light rays would be going back into the sun or the light bulb, etc.  So if you want to ask the question "what do things settle down to look like" you absolutely need the time-irreversible entropy maximization principle, called the Second Law of Thermodynamics.  It basically says that given a chance nature wanders with equal probability through all the available quantum states. It turns out that the vast majority of these states look similar- e.g. with balls at the bottom of cups and not rolling around the rim. So our predictions of what we'll see as things settle down are based on state-counting, not dynamical details. Dynamical details cannot tell you whether balls spontaneously jump up. State-counting does tells you that big ones don't, but molecule-size ones do.

Incidentally, even the tension in rubber-like polymers is primarily driven by configurational entropy maximization and not by energy minimization. (And energy minimization is itself just another manifestation of more general entropy maximization.) The curled up polymer has more choices of ways to curl up than the stretched-out one has of being stretched. That's why in equilibrium you find them curled up. So even the most mechanical-seeming aspect of this problem is in fact statistical.

Mike W.


(published on 01/29/2016)

Follow-up on this answer.