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June 2, 2015, 14:36 |
Scarborough criteria of boundedness
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#1 |
Senior Member
Join Date: Jun 2010
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Hi,
In the equation for Scarborough criteria , which can be found on wiki: http://en.wikipedia.org/wiki/Scarborough_criterion what do the straight lines around a'_p and a_nb mean? I thought they meant absolute but in every example I have seen, the coefficients are used as they are. For instance where a'p = ap = aW+aE (which are the coefficients obtained after discretisation), they are simply summed and NOT |ap| = |aW|+|aE|. (example in Versteeg book - An intro to computational fluid dynamics, page 112) Please help! This is annoying me alot! Thanks! |
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June 3, 2015, 01:20 |
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#2 |
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It looks like absolute values to me. Though it doesn't matter after all, because the coefficients in cfd should be positive anyway. I think some implicit higher order schemes can produce negative values which is part of why I don't like them.
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June 3, 2015, 05:08 |
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#3 |
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say I was modelling a 1D convection - diffusion problem, my coefficients when using central differencing would be:
aW = (1/2) rho*v+D/L aP = 2D/L aE = -(1/2)*rho*v+D/L where D is diffusivity coefficient, L is control volume length, and rho is density which is constant. So for a unidirectional flow with positive velocity, aW+aE = ap but |aW|+|aE| does not necessarily equal |ap|, does it? I most test books this Criteria is said to be satisfied in this case, but then the other condition of boundedness which is for all coefficients to be positive is not, from which is it derived that the Peclect number must be less than 2. I would appreciate some quick help with this. Thank you for your time. |
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June 3, 2015, 05:25 |
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#4 |
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This shows just the problem with negative coefficients. The coefficients don't match the criterion if Peclect number is bigger than 2. In this case the coefficient matrix doesn't match the criterion, the solver could converge but probably won't. Even if it does, your solution will be most likely crap. The solution in these cases are some kind of wild oszillation, which are actually a valid solution for the discretized equation. I think in Patankars book, there is an example where he shows how a completely wrong solution actually is an exact solution for the central differencing. What I like to recommend is to use first order upwind with some kind of deffered correction for the higher order term.
What I forgot: You're right |aW|+|aE| does not necessarily equal |ap|. On boundaries and for transient calculations ap is usually larger, if aW and aE are not too bad. |
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June 3, 2015, 06:27 |
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#5 |
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Thanks beer, I understand the consequences of not having bounded solutions, and when this criteria is not satisfied. What I am not getting is why in books like the Versteeg book or even in wiki, it is said that coefficients satisfy the criteria.
With first order upwind, for the 1D conv-diff equation, one has aW = max(rho*v,0)+(D/L) ap = max(rho*v,0)-min(rho*v,0)+(2*D/L) aE = -min(rho*v,0)+(D/L) where for v<0 and a unidirectional flow, aE+aW = ap but |aE|+|aW| does not necessarily equal |ap|, but first order upwind is said to be unconditionally bounded. I would appreciate a precise response, my question is only about the Scorborough Criteria and if an absolute sign is there? Thanks. |
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June 3, 2015, 06:34 |
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#6 |
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I have to disagree. If you just calculate the coefficients for v = -1 and v=1 the coefficients |aE|+|aW| are exactly |ap|.
And AFAIK the criterion has to be used with the absolute sign. I'm not a mathematician, so I can't exactly tell you why and how. It's just what I check for when using some formulation. Regards |
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Tags |
discretization, discretization schemes, finite volume method |
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