From: dseeman@novell.com (Daniel Seeman)
Newsgroups: sci.physics
Subject: Re: bubble in container
Date: 29 Dec 92 01:12:15 GMT
Organization: Novell Inc., San Jose, Califonia

In article <1992Dec29.002632.22407@sfu.ca> Leigh Palmer <palmer@sfu.ca> writes:
>In article <1992Dec28.214917.27561@CSD-NewsHost.Stanford.EDU> Vaughan R. 
>Pratt, pratt@Sunburn.Stanford.EDU asks several questions about this 
>fascinating problem, for which I thank him:
>
>>How do you justify hydrostatic reasoning in a hydrodynamics problem?
>>If the bubble is on its way up the system is not in equilibrium.
>
>My hydrostatic solution applies to the initial and final states of the 
>system, both of which are static. The original question asked if the 
>pressure changed as the bubble rose. I take it that if the initial and 
>final pressures differ, then the pressure must have changed. I will note 
>that the solution applies to any intermediate height of the bubble as 
>well.
>
>>[interesting argument on another, slightly related, hydrodynamic problem 
>>omitted here, since I'm not solving a hydrodynamic problem]
>>
>>...The above argument shows that your formula is not valid (except at the
>>bottom) for sufficiently low surface tension, viscosity etc.  Can you
>>give limiting conditions under which your formula *is* valid?  E.g. an
>>arbitrarily small air bubble, arbitrarily narrow container, arbitrarily
>>high viscosity,...
>
>These are not limits to the validity of my solution in the ideal case. I 
>invented the inverted test tube variant for a practical barostat (note
>the 
>"stat" suffix, as in "thermostat") because when containers are not 
>absolutely rigid and fluids are slightly compressible, it helps to have a 
>larger bubble.
>
>>Viscosity:  The problem with increasing the viscosity would seem to be
>>that it retards transmission of the pressure defect to the side of the
>>container.  It is not at all clear to me whether more or less viscosity
>>is better for your formula.
>
>Viscosity plays no role in a static problem, but hydrostatic equilibrium 
>obtains during the bubble's rise in the limit of zero viscosity.
>
>>Width:  If the water is much wider than it is deep then the time of
>>transit of the bubble would be short compared to the time for its
>>pressure differential to move out to the side, and conversely for a
>>narrow container.  Thus a narrow container would seem to help your
>>formula.
>
>I guess that is correct for the hydrodynamic case.
>
>>Surface Tension:  Enough surface tension will hold the bubble together
>>and prevent the "tube of air," presumably helping your formula.
>
>The surface tension does not matter at all because, in the case of either
>a 
>bubble or a test tube, the geometry of the surface is the same anywhere
>it 
>is brought to rest. Any surface tension contribution to the pressure will 
>be constant, which I will subsume into my "P".
>
>>The combination of high surface tension, low to medium viscosity, and a
>>narrow container would seem like the optimal combination for which your
>>formula would be a good approximation.  There should be a single
>>formula combining these factors to give a measure of goodness of your
>>approximation.  Finding a reasonable such formula seems like an
>>extremely hard hydrodynamics problem.
>
>I guess that's why I solved an easier problem. :-))
>
>I infer that Sushil had already discovered the result I've derived for
>you 
>before he asked his question. He just wanted someone to confirm it for
>him. 
>I felt the same way the first time I figured it out while contemplating a 
>Cartesian diver.
>
>Am I correct, Sushil?
>
>Leigh

Hi,

Let us (for fun of course) add on this extra twist.  Make a glass bubble whose
inner cavity is filled with air (at say 1 atm).  Then, drop that "bubble" in a
container of liquid (water would be fine) and close the top.  Assume there is an
air gap at the top of the container that separates the water surface from the
top lid of the container. Now, increase the pressure in the container until the
"bubble" sinks.  As soon as the "bubble" hits the bottom, it breaks (as-
sume there are a bunch of VERY SHARP pins on the bottom and they rupture the 
"bubble" on contact).  What then happens to the air that was originally contain-
ed in the glass "bubble?"  What happens to the overall pressure in the container?

Just thought I would add a bit to the puzzle...

dks.
Newsgroups: sci.physics