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Lost
Picture of kkina
posted
Two child's party balloons, initially identical. One, we'll call it Balloon A, is inflated around 75% (i.e. more air will go in without danger of bursting). The other, Balloon B, is filled with air to about half-capacity. They are connected to each other via a 2-way valved tube.


Question:
What happens when the valve is opened?

Choices:
Nothing happens.
Air will flow from Balloon A to Balloon B until both are about the same size.
Air will flow from A to B until most of the total air is in B.
Air will flow from B to A until most of the total air is in A. (i.e. A gets even bigger, and B gets even smaller.)
Other

 
 
Posts: 10410 | Location: SF Bay Area | Registered: December 11, 2003Reply With QuoteReport This Post
Ammoholic
Picture of Skins2881
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Since I don't know the answer, I guess that all things like to be in a state of equilibrium and it they will both end up 5/8 full.



Jesse

A couple SIGs and a few others
 
Posts: 10024 | Location: Loudoun County, Virginia | Registered: December 27, 2014Reply With QuoteReport This Post
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Picture of Ken226
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If the balloons are of identical construction, and assuming they are of typical rubber type material, the materials stretch/force will be governed by a modulus of elasticity. IE the more it's stretched, the more force is required to further stretch it.

The material of balloon A is stretched more, exerting more force on the contained gas and as such it is under more pressure than the gas in balloon B.

When the valve is opened, gas will flow from the balloon of higher pressure to the balloon of lower pressure untill there balloons pressures are nearly equalized.


Machine Shop
07/02
 
Posts: 1164 | Location: Top Left Corner | Registered: December 23, 2000Reply With QuoteReport This Post
Lost
Picture of kkina
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quote:
Originally posted by Ken226:
If the balloons are of identical construction, and assuming they are of typical rubber type material, the materials stretch/force will be governed by a modulus of elasticity. IE the more it's stretched, the more force is required to further stretch it.

The material of balloon A is stretched more, exerting more force on the contained gas and as such it is under more pressure than the gas in balloon B.

When the valve is opened, gas will flow from the balloon of higher pressure to the balloon of lower pressure untill there balloons pressures are nearly equalized.

So which answer did you choose?
 
Posts: 10410 | Location: SF Bay Area | Registered: December 11, 2003Reply With QuoteReport This Post
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Picture of Ken226
posted Hide Post
quote:
Originally posted by kkina:
quote:
Originally posted by Ken226:
If the balloons are of identical construction, and assuming they are of typical rubber type material, the materials stretch/force will be governed by a modulus of elasticity. IE the more it's stretched, the more force is required to further stretch it.

The material of balloon A is stretched more, exerting more force on the contained gas and as such it is under more pressure than the gas in balloon B.

When the valve is opened, gas will flow from the balloon of higher pressure to the balloon of lower pressure untill there balloons pressures are nearly equalized.

So which answer did you choose?


The second option

Air will flow from A to B untill they are almost equal.

They will not become equal in size because at some point the pressure differential, decreasing asymptotically, will be insufficient to overcome the energy requirements of moving more air through the tubes/valve. But, they will become very close in size.

This is entirely dependant on the nature of the balloon material though. If at some point as the material stretches beyond it's yield, it's the air flow will reverse direction and balloon A will likely burst.


Machine Shop
07/02
 
Posts: 1164 | Location: Top Left Corner | Registered: December 23, 2000Reply With QuoteReport This Post
Lost
Picture of kkina
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^^Thank you. For purpose of this quiz, neither of the balloons is ever stretched close to bursting.
 
Posts: 10410 | Location: SF Bay Area | Registered: December 11, 2003Reply With QuoteReport This Post
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Picture of 2012BOSS302
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#2 - higher pressure always flows to the lower pressure region (A to B), pressure will equalize they will be roughly the same size.


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Posts: 1810 | Location: Idaho | Registered: January 26, 2014Reply With QuoteReport This Post
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Picture of maladat
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quote:
Originally posted by Ken226:
This is entirely dependant on the nature of the balloon material though. If at some point as the material stretches beyond it's yield, it's the air flow will reverse direction and balloon A will likely burst.


It isn't exactly because of yield strength, it is because as the balloon gets more inflated, it gets thinner, and getting thinner means it gets easier to stretch (an extreme example of engineering stress/strain vs true stress/strain).

There's a point at which inflating the balloon more actually decreases the pressure inside the balloon.

https://en.m.wikipedia.org/wik...o-balloon_experiment

"For many starting conditions, the smaller balloon then gets smaller and the balloon with the larger diameter inflates even more. This result is surprising, since most people assume that the two balloons will have equal sizes after exchanging air."

Here's a short video where you can see it happen: https://youtu.be/Mcf7uSgB5B4
 
Posts: 3824 | Location: TX | Registered: January 24, 2011Reply With QuoteReport This Post
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Picture of Sock Eating Golden
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Answered, nothing happens. Then thought about blowing up a balloon and quickly realised that I answered wrong. And no-one has answered correctly yet. Verified my corrected answer is correct.

Good luck all.


Nick




David Russell Gorham 12/25/2009 - 08/26/2012
"The LORD is my shepherd..." Psalm 23
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Posts: 5546 | Location: NE Ohio | Registered: November 17, 2003Reply With QuoteReport This Post
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Nothing happens. Both balloons are already in equilibrium with atmospheric pressure, assuming they’re both in the same environment. At sea level atmospheric pressure = 14.7 lbs/square inch or 1 standard atmosphere.
 
Posts: 1393 | Location: Mid-Atlantic | Registered: December 27, 2002Reply With QuoteReport This Post
Don't Panic
Picture of joel9507
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Thought experiment.

Take two identical balloons, pump one up to twice the size. Which one has more internal pressure? Another way to think about it....when you blow up a balloon, does it get easier or harder as you go?

Hint. The force on the balloon skin is partly due to the atmosphere, and partly due to the elastic compression of the balloon material trying to get back to normal. The more it stretches, the more pressure it takes to push it further out.

The air inside the larger balloon is under higher pressure as it's equilibrated against both atmospheric pressure and the balloon itself. When it gets connected to the wimpier balloon, the pressure will force air into the smaller balloon until the pressure is equal. Small balloon expands, large balloon compresses.
 
Posts: 11605 | Location: North Carolina | Registered: October 15, 2007Reply With QuoteReport This Post
Seeker of Clarity
Picture of r0gue
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quote:
Originally posted by Dakor:
Nothing happens. Both balloons are already in equilibrium with atmospheric pressure, assuming they’re both in the same environment. At sea level atmospheric pressure = 14.7 lbs/square inch or 1 standard atmosphere.


No, I don't think. The balloons are initially identical. So to make one bigger, a higher pressure had to have been delivered. When the valve is opened, they must equalize.



Staring out the windows is for love songs and house flies. -- Jay Electronica
 
Posts: 8284 | Location: PA | Registered: August 02, 2004Reply With QuoteReport This Post
A Grateful American
Picture of sigmonkey
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I'm gonna guess 6.25% will transfer from the 75% to the 50% and equilibrium will result.

That's the "knee jerk" reaction that popped into my head.




"the meaning of life, is to give life meaning" I could explain it to you, but I can't understand it for you.
 
Posts: 36275 | Location: My Happy Little Tire Swing | Registered: December 20, 2008Reply With QuoteReport This Post
Staring back
from the abyss
Picture of Gustofer
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Answer B. Or...42.


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"How dreadful are the curses which Mohammedanism lays on its votaries! Besides the fanatical frenzy, which is as dangerous in a man as hydrophobia in a dog, there is this fearful fatalistic apathy." Winston Churchill
 
Posts: 13382 | Location: Montana | Registered: November 01, 2010Reply With QuoteReport This Post
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Picture of Ken226
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quote:
Originally posted by maladat:
quote:
Originally posted by Ken226:
This is entirely dependant on the nature of the balloon material though. If at some point as the material stretches beyond it's yield, it's the air flow will reverse direction and balloon A will likely burst.


It isn't exactly because of yield strength, it is because as the balloon gets more inflated, it gets thinner, and getting thinner means it gets easier to stretch (an extreme example of engineering stress/strain vs true stress/strain).

There's a point at which inflating the balloon more actually decreases the pressure inside the balloon.

https://en.m.wikipedia.org/wik...o-balloon_experiment

"For many starting conditions, the smaller balloon then gets smaller and the balloon with the larger diameter inflates even more. This result is surprising, since most people assume that the two balloons will have equal sizes after exchanging air."

Here's a short video where you can see it happen: https://youtu.be/Mcf7uSgB5B4


I would submit that your being correct about the thinner part, is why im correct. To go over simplistic, if P=f/a, (I'm using P because I don't know how to type sigma on a Droid) then as the material stretches and gets thinner, the 'a' portion of it's cross section becomes a smaller number causing the 'P' in P=f/a to become a larger number. The larger value for P increases the tensile stress in the material. If the value for P exceeds the materials maximum tensile strength (I think yeild is the right term, but it's been awhile), the things get wrapped around the axle.

Obviously this is a drastic oversimplification

I wouldn't know where to find the tensile strength of bsloon rubber in order to calculate it's actual hoop stress.


Machine Shop
07/02
 
Posts: 1164 | Location: Top Left Corner | Registered: December 23, 2000Reply With QuoteReport This Post
His Royal Hiney
Picture of Rey HRH
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Assuming everything else being equal except for the amount of air blown inside each balloon, when the valve is open, the pressure will equalize.

There's higher pressure in balloon A given that there's more air inside balloon A. Therefore, air will move from balloon A to balloon B until pressure is the same throughout the two balloons joined by the tube.



"It did not really matter what we expected from life, but rather what life expected from us. We needed to stop asking about the meaning of life, and instead to think of ourselves as those who were being questioned by life – daily and hourly. Our answer must consist not in talk and meditation, but in right action and in right conduct. Life ultimately means taking the responsibility to find the right answer to its problems and to fulfill the tasks which it constantly sets for each individual." Viktor Frankl, Man's Search for Meaning, 1946.
 
Posts: 13334 | Location: Bay Area, CA | Registered: March 24, 2011Reply With QuoteReport This Post
Lost
Picture of kkina
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And now the shocking answer...the correct answer is the 4th one: air will flow from the small balloon to the big one until most of the total volume is in the big one (balloon A in my diagram). Don't believe me? Check out this video, and watch what happens when he opens that second valve:



Props to Maladat for getting the correct answer as well as posting his own supporting links. It does have to do with the elastic properties of the balloon material (rubber), within the region of "ideal" behavior (which is why I specified non-extreme limits). You can check out the full Wikipedia for some theory and equations.

Only a couple people got the correct answer (and I was not one of them). More than 80% thought it would be the second choice, that the two balloons would wind up the same size. What does this say about the assumptions we make in the day-to-day world, even though we like to think of ourselves as a technologically and scientifically advanced species?
 
Posts: 10410 | Location: SF Bay Area | Registered: December 11, 2003Reply With QuoteReport This Post
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Picture of konata88
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Is there an inflection point in the larger balloon’s inflation beyond which this is true but below which equilibrium would be correct?




"Those who would give up essential Liberty, to purchase a little temporary Safety, deserve neither Liberty nor Safety." - B.Franklin
"Wrong does not cease to be wrong because the majority share in it." L.Tolstoy
 
Posts: 6714 | Location: In the gilded cage | Registered: December 09, 2007Reply With QuoteReport This Post
Lost
Picture of kkina
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quote:
Originally posted by konata88:
Is there an inflection point in the larger balloon’s inflation beyond which this is true but below which equilibrium would be correct?

If I understand your question, the answer is yes, there is a point of extreme when the pressure inside the larger balloon does indeed go up again. (You can check out the reasons why in the Wikipedia article under "Non-ideal Balloons"). Again, this is why I was careful to specify non-extreme parameters, e.g. saying 75% inflation, well inside the curve.
 
Posts: 10410 | Location: SF Bay Area | Registered: December 11, 2003Reply With QuoteReport This Post
Staring back
from the abyss
Picture of Gustofer
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Yeah, should have thought about it more.

Wiki has a confusing explanation of compliance/elastance.

BTW, I like the crazy Russian videos. His ones on beekeeping crack me up.


________________________________________________________

"How dreadful are the curses which Mohammedanism lays on its votaries! Besides the fanatical frenzy, which is as dangerous in a man as hydrophobia in a dog, there is this fearful fatalistic apathy." Winston Churchill
 
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