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Buoyancy/convection problem - advice sought

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Old   April 23, 2009, 12:53
Default Buoyancy/convection problem - advice sought
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njc
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Hi all.
Firstly: I'm a total CFD novice.

I have a problem I suspect just might be appropriate for a CFD approach, since it's non-trivial to estimate on the back of an envelope, but I have no CFD expertise and am not sure if CFD will give me the answers I want, and whether or not it will take me many weeks of effort to achieve it...

The problem, in a nutshell: a vacuum vessel will be let up to atmospheric pressure with nitrogen gas, after which some flanges will be removed and replaced ("quickly", meaning minutes of open time, during which the holes may/should be blanked off somehow). Some of these flanges are 600 mm diameter, and almost all of them are in a vertical plane. The vessel will then be pumped back down again to fairly high vacuum (1e-8 mbar). We wish to minimise the amount of water vapour entering the vessel during the open time.

Back-of-envelope calculations indicate that diffusion of damp air into the vessel is negligible.
My very half-baked attempts to guesstimate the convective flows (based on nitrogen being about 3% lighter than damp air) show that for a 600 mm port, it might be expected that multiple litres per second of nitrogen (outwards) and damp air (inwards) will occur. This is enough to bother us. The question then arises: if we overpressurise the vessel to cause a net outward flow of nitrogen, how many litres per second of nitrogen outflow are required to overcome the inward flow of damp air? Wild guesses are all I have for the answer to this question.

My first idea would be to model this (in 3D) with a simple wall: on one side is nitrogen, on the other side is air, and there is a hole in the wall... then model the evolution with time of the flow. I guess this is possible?

Bearing in mind that I have no CFD expertise, advice would be welcome

Thanks in advance!
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Old   April 23, 2009, 14:24
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I would think that as soon as the pressure inside the container is in equilibrium with the surrounding pressure you would have no outflow - and any forced out flow after that would be a control-volume mass flow balance - mass,in = mass,out unless you are choking your exit - if I am understanding your problem correctly
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Old   April 24, 2009, 04:31
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Quote:
Originally Posted by mettler View Post
I would think that as soon as the pressure inside the container is in equilibrium with the surrounding pressure you would have no outflow - and any forced out flow after that would be a control-volume mass flow balance - mass,in = mass,out unless you are choking your exit - if I am understanding your problem correctly
OK, had to scratch my head for a minute to figure out quite where the problem lay here.

You're right that the pressure would equalise - but only at one height.
I imagine that the height where the pressures would balance would be approximately half-way up the face of the big open port.

Above that height, the density and thus pressure of the nitrogen falls, but more slowly (because it's lighter) than the density of the air (it's a little heavier). So at any height above the balancing point, the nitrogen density will be higher and the nitrogen will want to push outwards. Similarly, below: the air will be denser and will want to push inwards.

I guess what I'm saying is that this (buoyancy) effect is all about the (very tiny) density & pressure gradient induced by gravity. (Unless I'm talking complete bananas here, in which case somebody please slap me )

By the way, the basis of my half-baked calculation for out/in-flow was this:
Take a 600mm diameter port, and imagine that the in and out flows will have a characteristic dimension (ignoring eddies which I can't do on the back of an envelope ) of rather less than half the dimension of the hole... Say about 100mm or so, to attempt to err on the low side for the flow rate.
Then work out the terminal velocity for a packet of nitrogen based on the buoyancy force, assuming it's about a 1 litre volume (using the 100 mm dimension and cubing it), and roughly a spherical shape.
This gives a velocity of a little over 0.2 m/s. Thus the packet covers its own length more than twice per second, thus I conjure up a flow rate of multiple litres per second. In reality I have to imagine I'm making a significant underestimate of course, but that's just my guess.
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Old   April 24, 2009, 11:37
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This (light gas/fluid below a more dense fluid) is known as Rayleigh-Taylor instability. Is it possible to fill the vacuum chamber with a denser gas (carbon dioxide perhaps, maybe just a sufficiently cold gas)?
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Old   April 24, 2009, 11:50
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Quote:
Originally Posted by jed View Post
This (light gas/fluid below a more dense fluid) is known as Rayleigh-Taylor instability. Is it possible to fill the vacuum chamber with a denser gas (carbon dioxide perhaps, maybe just a sufficiently cold gas)?
Ah, sorry, I guess I maybe fumbled the description somewhat (pictures tell a thousand words and all that ).
The holes in the wall are mainly in a vertical plane. So while we could potentially fill the chamber with a heavy gas, that would only help with the ports on the top plate of the chamber.
The big ports are on what we could refer to as the side walls (i.e. vertical walls) of the chamber, so unfortunately the problem can't be circumvented by this means.
(But thanks for your suggestion.)
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