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March 29, 2006, 04:20 |
VOF - cell overfilling
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#1 |
Guest
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Hi,
I have 2 questinons for you all. 1. Is it possible to overfill cells in the VOF method, and if so, will this be convected as an overfilled interphase? 2. Is the VOF method prone to numerical diffusion? tnx / Sara |
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March 29, 2006, 05:17 |
Re: VOF - cell overfilling
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#2 |
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Not entirely sure what you mean by overfilled. If you mean phi > 1, then you would hope not. The solution must surely be bound by 0 < phi < 1. You'd hope so anyway. If this isn't what you meant, then my apologies.
The VOF method is prone to diffusion of the interface, which is purely the result of using big mechanical cells to model a small molecular interface. In my experience, the interface is smeared across about 3 or 4 cells. Using different descritisation methods could improve things. |
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March 29, 2006, 05:35 |
Re: VOF - cell overfilling
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#3 |
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Hi Jim,
What I have is a reaction in the cell where one phase transfers mass to the other phase. The two phases are of different density and so the volume of the system (cell) will increase. Since I don't want to tamper with the numerical domain the cell volume stays the same. As a result the cell will be overfilled (but the mass remains the same). I was hoping this "extra" would be transported out of the cell at the next sweep. |
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March 29, 2006, 06:11 |
Re: VOF - cell overfilling
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#4 |
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OK, I see your problem. Never had this problem my self, so I can't tell you how the VOF is going to react to having sources of volume in the cell.
Is this an industrial piece of work or an academic one? If it's industrial, and the difference in density isn't significant, I'd be tempted to keep the densities the same. Of course, it might not be necessary, it might work just fine with no problems, but until you try it (or find someone who knows the more about this than I do), you're not going to know. My best advice is come up with a simple 2D test case that you can run quickly, and give it a go. |
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March 29, 2006, 06:33 |
Re: VOF - cell overfilling
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#5 |
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It's a combined academic/industrial project and regrettably the density difference is significant. I use a commercial solver and I have seen how it reacts to overfilled cells, however I am not sure if this is correct, hoped for a general answer here
Anyways, thank you. |
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March 29, 2006, 09:13 |
Re: VOF - cell overfilling
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#6 |
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Surely, you should be able to look at the equations and see that they make it impossible for phi to go below zero or above one - if not, the equations themselves are wrong. If you have specified the law for mass exchange between the two phases, check what happens with your source/sink terms at the physical bounds of phi - they should go to zero for both phi = 0 and phi = 1.
Onc that bit is right, it only remains to implement it correctly - are you doing that yourself as well or just using "standard" user coding practices? Hrv |
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March 29, 2006, 11:42 |
Re: VOF - cell overfilling
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#7 |
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Sorry, but I can't see where is the problem regarding transporting phi greater than 1 or even lesser than 0.
VOF and level set methods can be thought as very similar methods employed to transport any scalar field. In level set method this scalar field is, in general, a distance field ranging from -phi to +phi from the interface - assuming the interface as the level set 0 and phi as the signed distance function from the interface. Furthermore, even the way of computing fluid properties are similar. In VOF you readily use the value picked up from your computed field to evaluate the properties through the following equations: rho(phi) = rho_1 * phi + rho_2 * ( 1 - phi ) mu(phi) = mu_1 * phi + mu_2 * ( 1 - phi ) In LS you'd use the same equations but employing a Heaviside function to "filter" you advected field, so: rho(phi) = rho_1 * H(phi) + rho_2 * ( 1 - H(phi) ) mu(phi) = mu_1 * H(phi) + mu_2 * ( 1 - H(phi) ) where H is the Heaviside function, or: H(phi) = 0 if phi < 0 H(phi) = 0.5 if phi = 0 (interface) H(phi) = 1 if phi > 0 Regarding "overfilled" cells, I think it would mean that this "extra" quantity would be "poured" to the neighbor cells, and maybe, the density differences would be reflected in some buoyancy effect. Regarding the numerical diffusion, the VOF method is supposed to be more affected by numerical diffusion since it transports immiscible fluids embedded in a sharp gradient function that only admit values ranging from 0 to 1. This problem is reduced in LS since the scalar field is a smooth field with constante gradient. In fact, most of the widely used discrete methods will suffer, more or less, with numerical diffusion (Galerking discontinuous methods, I guess, can overcome numerical diffusion in some sence...) It was my view point of your problem considering two-phase immiscible flows... Hope it'll help you in something... Cheers Renato. |
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March 29, 2006, 14:43 |
Re: VOF - cell overfilling
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#8 |
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Sorry to interrupt, but it has something to do the VOF you are talking about with the V2F model? We are talking about the same or they are different things?
Thanks and excuse me the interruption. |
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March 29, 2006, 14:56 |
Re: VOF - cell overfilling
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#9 |
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I'm not sure about V2F. Isn't it a turbulence model?
VOF and Level Set are methods to solve free surface flows... Regards Renato. |
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March 29, 2006, 14:59 |
Re: VOF - cell overfilling
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#10 |
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I'm not sure about V2F. Isn't it a turbulence model?
VOF and Level Set are methods to solve free surface through multi-phase flows... Regards Renato. |
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March 29, 2006, 15:25 |
Re: VOF - cell overfilling
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#11 |
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1) I am talking about VOF, not level set
2) The phi I am talking about is the volume fraction, i.e. a fraction of the cell filled with fluid A. 3) If phi < 0 or phi > 1 (in a two-phase mixture, for simplicity), the volume fraction of one of the two phases would be negative. 4) You cannot have -4 kg of water. For an exercise, consider this: phi = 1.05, rho1 = 1 kg/m^3, rho2 = 1000 kg/m^3 rho = phi*rho1 + (1 - phi)*rho2 = 1.05*1 + (1 - 1.05)*1000 = 1.05 - 0.05*1000 = 1.05 - 50 = -48.95 kg/m^3 In other words, an overshoot or undershoot of phi may cause negative density. Booooooom! Hope this is now clear, Hrv |
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March 29, 2006, 15:59 |
Re: VOF - cell overfilling
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#12 |
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For sure, I agree with you that mass or volume fraction can't give < 0 or > 1 and It was not my intention to say nothing against it or you. Sorry, but I was only saying that you can transport any value you like, no problem. I only used the LS example to show that you can transport whatever you want. It will depend how you're interpreting the values returned by the scalar field transported ("and" in the level set case, these values make sense since the method does not transport fractions). As I said, the "overfilled cell" idea does not make sense for me. I guess that this mass generation could be interpreted as a mass source (an inlet of one phase).
Regards Renato. |
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March 29, 2006, 16:39 |
Re: VOF - cell overfilling
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#13 |
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Thank you for your interesting input. I have an idea on how I will transport the excess before the next computational sweep. By doing this detached from the main algorithm it will never know that there has been overfilled cells in the first place.
Best Regards Sara |
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March 30, 2006, 05:02 |
Re: VOF - cell overfilling
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#14 |
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I reckon Renato has got a good point there. Make sure that the volume is conserved, and then add back whatever loss of mass you get as a source term. So long as your time step isn't to big, it should be able to cope with that.
The formula for converting between mass and volume fractions is (for a suspension) phi1=(w1*rho2)/(rho1+w1(rho2-rho1) So I reckon you: Do your reaction calc Look at the resultant phi1+phi2 Fudge it so that phi1+phi2=1 Then take what's left and through it in as a source of mass. That way the code should be able to cope with the overfilling. Best of luck |
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April 1, 2006, 15:33 |
Re: VOF - cell overfilling
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#15 |
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What commercial solver are you using, Sara, and ar you not able to inquire of the vendor for assistance?
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