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Is this a contradiction between primary assumptions and CFD ? |
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September 12, 2012, 05:03 |
Is this a contradiction between primary assumptions and CFD ?
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#1 |
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mhp
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Hi everyone;
Consider a 2D flow past a circular cylinder ,the geometry is symmetric, all B.C. are symmetric for both upper and lower part, but what happens that the solution is not symmetric for Re>100? How can this problem be explained with Navier Stocks equations? Is there any terms in NS equations that explain this? With CFD programs like fluent it can be solved with both steady and unsteady solutions but the solutions for blunt bodies like cylinder and stream wise bodies like NACA0012 are different.For NACA0012 both solutions are the same but for cylinder aren't. |
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September 12, 2012, 06:09 |
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#2 | ||
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cfdnewbie
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Quote:
again not sure what puzzles you here... |
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September 12, 2012, 06:13 |
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#3 |
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Filippo Maria Denaro
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I don't know to have understood your question but symmetry in the geometry (and BC) does not imply symmetry in the flow solution... the NS equations are non-linear ...
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September 12, 2012, 06:21 |
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#4 |
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Joern Beilke
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The use of symmetrical boundary conditions just violates the reality.
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September 12, 2012, 09:42 |
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#5 | |
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mhp
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Yes you are true, but in the same low Re number (suppose Re=100) you will never see unsteady flow for NACA0012 but it is seen in circular cylinder. more close to our discussion, for a simple cylinder up to Re~40 the flow is steady but with an increase in the Re number flow became unsteady, what is the reason for this ? I read somethings about karman vortex street but what is the main reason for these vortex creation? Last edited by MHP; September 12, 2012 at 09:45. Reason: add a picture |
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September 12, 2012, 10:02 |
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#6 |
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mhp
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So we shall never use symmetric BC for problems and it is wrong to model semi circular cylinder in a flow with symmetry wall in Re>~50, because the result for full cylinder is not symmetric,
Also another question is that what caused the flow pattern became different when the Re number increased while flow is laminar ? Is it true that unsteady flows are unstable flows? |
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September 12, 2012, 11:47 |
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#7 | |
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Filippo Maria Denaro
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in poor words, instability is when a solution, for example steady, when perturbed from its condition moves to a different solution instead of returning to be as the previous. As example, the Poiseuille solution is valid for any Reynolds number but is not stable for all the Re numbers ... |
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September 12, 2012, 15:13 |
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#8 |
Senior Member
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If you want more puzzles, consider also that a general symmetric laminar and steady flow can be unstable and find a more stable state which is asymmetrical (of course, at some level, even if infinitesimal and not deterministically reproducible, there is a perturbation somewhere to have this situation)
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September 13, 2012, 04:30 |
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#9 |
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Joern Beilke
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"Never say never Again" ;-)
Most of the time you are in a loose-loose situation. An one side you know that you can not solve the problem without simplifications. On the other side you know that simplification might lead to uncertain results. Now it's up to you how to go on. Here are some nice animations of flow configurations with symmetrical geometries: http://www.youtube.com/watch?v=5lSvDKv7sTg http://www.youtube.com/watch?v=_TXRXYxUGU8 http://www.youtube.com/watch?v=QXaSmkAkqzc |
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September 13, 2012, 10:24 |
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#10 | |
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mhp
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September 13, 2012, 10:28 |
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#11 | |
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mhp
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September 13, 2012, 21:19 |
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#12 |
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Chris DeGroot
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