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May 1, 2016, 14:49 |
Lid Driven Cavity - Finite Differences BCs?
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#1 |
New Member
Join Date: Dec 2015
Posts: 4
Rep Power: 11 |
Hello all,
I thought I'd set up the steady-state LDC problem correctly using finite differences, but upon further review...I'm pretty sure there's something wrong here. I'm using finite differences with a staggered grid to discretize the [0,1] x [0,1] square. I've got dirichlet BC's all the way around except on the lid where u = 1, v = 0. I've prescribed "fictitious" boundary nodes around the boundary in cases where u or v doesn't fall on the boundary. I've attached a figure which shows the values I have given to the fictitious boundary nodes. The idea is that the average of the inside node and outside node would satisfy the BCs. The problem is that the vortex seems to grow wildly with the Reynold's number. The figures I'm seeing elsewhere don't do this. I've attached to figures that show what I'm getting. One with Re = 500 and the other with Re = 1000. Any ideas? I'd really appreciate some input. I've been beating my head against a wall for too long with this already. Thanks! |
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May 1, 2016, 19:39 |
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#2 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,896
Rep Power: 73 |
I need some further infos..
1) you are using the u,v,p formulation, the node for pressure is the green circle, u is the red and v the blue, right? 2) you are using linear extrapolation for the velocity values in the ghost cells, right? therefore the values for u on the top row are uex=-uint+2*uwall 3) the BC.s enters only by the diffusion of u, what if you try the case Re=1? 4) what about the threshold for the steady state convergence? |
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May 1, 2016, 21:18 |
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#3 | |
New Member
Join Date: Dec 2015
Posts: 4
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Quote:
2) yes, I believe so. For the top fictitious boundary nodes I have, uex = 2 - uint so that when I average (uex + uint)/2 = 1 = u for lid (top wall). For all other walls I have uex = - uint or vex = -vint as appropriate. 3) I have attached a figure showing the case Re=1. The problem only occurs as I increase the Reynolds number. But, it shouldn't take over the whole bottom portion of my figure, right? 4) The tolerance for my linearization (Picard Iteration) convergence is set to 1e-7. Thanks so much for your input! I've really got to figure this one out, pronto. |
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May 1, 2016, 23:50 |
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#4 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,896
Rep Power: 73 |
Re=1 seems correct, i suggest:
- check the Div v=0 constraint in each cell, for staggered grid and second order discretization it must be zero at machine accuracy - check the solution for refined grids, at higher Re, the non linear terms require a good grid - check again the tolerance for the steady state, it Could be necessary to decrease the value |
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Tags |
boundary conditions, finite differences, lid driven cavity, navier stokes |
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