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April 3, 2010, 17:12 |
Can we solve directly Stokes equations?
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#1 |
New Member
the
Join Date: Jan 2010
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Hi everyone.
Since the steady (2D) Stokes equations are linear, I have formed a matrix for both U, V and P and solved directly. The staggered grid was used. To avoid the case of singular matrix, pressure is fixed at a given point by add one more row(with only one non-zero element) to the matrix(so the matrix is non-square but still can solve) I used this method to solve for a simple cavity problem with Utop = Uo. When Uo = constant, I can get a good result. But if Uo is changed along the top edge (Uo = Uo(x)) the result looks terrible! Could you please give me any comments on the method I have used? If my method is acceptable, please give me some suggestions to deal with the case of Uo = Uo(x). Thank you very much. Best, The Last edited by [thelight; April 3, 2010 at 17:33. |
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April 4, 2010, 15:06 |
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#2 |
New Member
the
Join Date: Jan 2010
Posts: 2
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Hey friends, where are you? Could you let me know your comments, anything is also good to me.
Thanks alot! |
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