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Boundary conditions for pressure-correction method |
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May 22, 2018, 03:19 |
Boundary conditions for pressure-correction method
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#1 |
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Moshe De Leon
Join Date: Nov 2017
Location: Portugal
Posts: 31
Rep Power: 8 |
i want to learn about boundary conditions for a compact Poisson stencil. after searching through this forum it seems that people are using ghost points when handling pressure BCs for a collocated arrangement (cf. pressure boundary conditions for collocated grid for Navier-Stokes) is there a way that avoids having a layer of ghost points? this ghost point approach seems to have difficulties. After some google searching I found some slides that talks about pressure boundary conditions for a compact stencil, but it seems to be incorrect (slide 5 of https://www3.nd.edu/~gtryggva/CFD-Co...re-15-2017.pdf) The professor explicit sets the bc rather than modifying the PPE at the boundary.
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May 22, 2018, 04:25 |
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#2 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
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There is no need to prescribe ghost nodes, you can find similar post in this forum
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May 22, 2018, 13:47 |
pression eqn
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#3 |
Senior Member
Selig
Join Date: Jul 2016
Posts: 213
Rep Power: 11 |
Attached is a complete pressure bc based on Dr. Denaros approach. I have not tested this as I have switched to a staggered based solver since I was unfortunately never able to get my collocated solver to work properly.
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May 22, 2018, 13:56 |
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#4 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,849
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On staggered grid this issue is also more simple, a good explanation is given in the JCP paper of Kim and Moin on the fractional step method.
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May 22, 2018, 14:02 |
pressure bc
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#5 |
Member
Moshe De Leon
Join Date: Nov 2017
Location: Portugal
Posts: 31
Rep Power: 8 |
i think this formulation is wrong as you only have 1 diagonal entry in the iteration routine, no?
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May 25, 2018, 13:47 |
thread bump
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#6 |
Member
Moshe De Leon
Join Date: Nov 2017
Location: Portugal
Posts: 31
Rep Power: 8 |
don't mean to bump this, but when we write the laplacian operator at i = 2, we iterate at i = 1, which means we only have 1 diagonal term since the other diagonal term dpdy|j=2 is moved to the right-hand side.
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