Gas purge of control volume 1

[flapjack1]

Hi there,

I'm working on a gas purging analysis for a rectangular control volume. The control volume is initially full of ambient air; the bottom surface is an inlet for gaseous argon flowing at a constant velocity, and the top surface is an outlet for the mixture of argon and air. The goal is to figure out how long it will take for the concentration of air inside the control volume to drop to 1 ppm. I have results from a CFD simulation and now need to perform a hand calculation to validate those results. However, I'm having trouble finding an equation (or set of equations) that describes this kind of gas mixing/gas-gas two phase flow/dilution purging situation. Does anyone know of a good way to verify my CFD results with hand calculations?

Thanks!

 

[LittleInch]

My initial reaction to your first sentence was "go do a CFD analysis", but then you have.

I think there are far too many variables here to do this by an equation based method.

To get to 1ppm of what? Oxygen/ Nitrogen? "air" is rather vague, is a hard task and the last few hundred ppm could take a long time.

If asked to guess I would have said less than 10 swept volumes of the control volume, but more than 4. But that is based on getting down to <5% of air / oxygen. It will take you 100+ to get to 1ppm

What time / volume are you getting?

The only mixing equations I know of are based on pipes and circular vessels where flow is from one end.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[1503-44]

How close does the hand calculation need to be? Can it be a 2D flow? Are you interested in mixture content at the corners of the box? Do you mean ppm taken as the average over the box's entire volume? It may take some extra time to purge the corners. Assume minimal pressure to avoid compressibility and flow complications?

As a quick comparison, find how long will it take you to inflow a volume equal to 1,000,000 volumes of the box? There's your minimum time approximation to reach 1ppm. Add any time difference it will take you to remove one box volume, if that time is longer.

Try Graham's law. Assume a time step for an introduction of a certain volume of argon equal to the flow rate, calculate the diffusion rate, keep track of mixture content over distance, remove an equal packet of air/mixture from the outlet according to the mix-distance plot, repeat.

Don't forget that air has a tiny little bit of Ar in it.

“What I told you was true ... from a certain point of view.” - Obi-Wan Kenobi, "Return of the Jedi"

 

[jari001]

I've attached a CEP article that talks about inerting vessels with sweep through purging. I don't see a reason this method wouldn't work for a rectangular prism geometry if the mixing/safety factor recommendations for the mixing factor discussed seem relevant to the use of your vessel.

 

[LittleInch]

And purity of the argon in the bottle is only 99.9 something.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[EdStainless]

Decide if you care about N, O, or H2O the most. You need to know what to measure. H2O will be the most difficult because it adheres to surfaces.
The inlet flow must be non turbulent. If it is just a jet then mixing will force you to purge much longer.
We used high flow low pressure regulators and an inlet diffuser. Being nearly quiescent flow will let the density difference assist you.
At some point diffusion back through your vent will limit how low you can get.
We used to purge boxes to <50ppmO2 with about 8-10 swept volumes over about 20min.
If you are working from a liquid Ar source then you will have 99.9995% min purity and at least -70C DP, this will help a lot.

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P.E. Metallurgy, consulting work welcomed

 

[flapjack1]

Hi everyone - thanks for the responses, I'll try to answer these questions.

LittleInch - The goal is to get down to an average of 1 ppm of original air, or up to 999,999 ppm argon. My CFD showed 3.83 hours to purge. The CFD includes an assembly of tubes with venting holes within the control volume and is therefore more complicated, but I'm hoping to do a hand calculation based on the simplified situation described in my original post as a sanity check. My hypothesis is that the simplified purge time should take less time without the flow restrictions in the tube assembly, so if I find that to be true I'll feel good about the CFD results. I don't know the details of the argon we'll use to purge, but I assume it will be highly purified or something... otherwise we'll never get to 999,999 ppm Argon.

ax1e - I think a 2D flow might be an okay approximation since the velocity is constant across the cross-sectional area, but I'm not sure how to use that to my advantage. Is there a 2D formula somewhere? The hand calculation doesn't have to be perfect, so I think localized idiosyncrasies like at the corners can be ignored. I'm also thinking about the comment about finding how long it would take to inflow 1 million volumes of argon. Are you sure that would be the minimum purge time? If the argon were instead water, for instance, you would only need 1 volume of water to displace all the air. Argon is of course a gas and much less dense than water, but intuitively it seems unlikely that it is at least a million times slower at displacing air than water. But I could be wrong, what do you think? I'm also not sure I understand your method in the paragraph starting with Graham's Law. Would it be possible to give an example with sample numbers to illustrate your point?

jari001 - I was excited to see those formulas and found the original source on page 300 of this textbook. My control volume is 139349 cm^3 and my volumetric flow rate is 132937 cm^3/hr. Assuming an initial oxidant concentration of 0.9907 (air is 0.93% Ar) and a final oxidant concentration of 1e-6, my purge time is about 14.5 hours. Per my note to LittleInch, I would have expected purge time for the simplified situation to be less than 3.83 hours, so I think something is still off here. It seems a little odd that formula doesn't at least include densities of the gases involved. I would think that would affect purge time?

EdStainless - someone on our team did a study a while back that the H20 adherence won't be a problem here. It's a good thought; I don't know the details but am working under that assumption for the purposes of this study. The flow velocity is 1.2 m/hr, so I'm hoping I shouldn't have to worry too much about turbulence? I'm also assuming there will be no backflow; constant velocity outward at the outlet.

Thanks!

 

[jari001]

The disparity may be related to the modes of mixing as EdStainless noted but also the physical nature of the control volume doesn't seem to be what I thought. Based on your last response to LittleInch where you mentioned tubes with vent holes within the rectangular control volume, I realize I misunderstood your original post - I was thinking you were working with something like a rectangular atmospheric tank without anything inside of it. Those equations in the CEP article aren't for such "complicated" geometries as I understand the derivation (classic CSTR equation).

 

[flapjack1]

Just to be clear, the hand calculations I'm after are indeed for a simplified scenario that excludes the tube assembly, since no hand calculation could ever hope to capture all of that complexity. I can then hypothesize in what way the exclusion of the tube assembly would cause differences from the full CFD, compare the results, and see if the results match my predictions. In this case, I'm predicting that purging without the tube assembly will take less time, since the gas won't have any hollow tubes to get "stuck" in.

What are the modes of mixing that EdStainless was talking about? I wasn't totally clear on that (sorry, I'm relatively new to most of this).

 

[LittleInch]

Correct me if I'm wrong but it seems you're looking at a box 1.3m^3 and you're flowing at about 1.3m^3/hr?

So one volume per hour.

That sounds rather slow to me and means you're going to get diffusion and mixing of the gases to a significant extent. I suspect most calculators and formulae assume a higher velocity for the gas front.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[flapjack1]

Almost - 0.13 m^3 control volume with a 0.13 m^3/hr flow rate. Either way though yes, 1 volume per hour. If most formulae assume faster flow though, shouldn't I expect my hand calculations to yield an even shorter purge than the CFD, rather than longer? That's why I think something is still off here.

 

[EdStainless]

If I could I would the entire bottom of the box as a diffuse plate (porous ceramic or SS) with a plenum under it for Ar delivery. The more area the better. You should be able to still have laminar flow at 6-8 vol per hour. And a 2 hr purge should get you to terminal.
If your Ar source is 99.995% (industrial high purity liquid) then you are looking at up to 4ppm O2 (1-2ppm typical), with this you will not get below 10ppm. There are grades with higher purity, but they get very expensive.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed