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May 22, 2012, 03:41 

#21 
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Filippo Maria Denaro
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May 22, 2012, 14:44 

#22 
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Yeah the nuemann condition part is can be tricky. I think a lot of people set one of the boundary values to zero instead of using a nuemann condition. You mentioned that you solved this problem in rectangular coordinates. Did you need any special treatment for the neumann condition then?


May 23, 2012, 11:39 

#23  
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Peter
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Quote:
Quote:
In cylindrical coordinates, I have made the Laplacian matrix, except in contrast to the cartesian version it isn't symmetric but is still near singular. I have done the same thing as before and replaced pressure outside the boundary with the first interior point for all the boundary finite differences. Then I try solving the system Ax = b where A is the near singular discrete laplacian matrix, and b is the divergence of the intermediate velocity field. I use the same GMRES method that I used in the cartesian case but I still produce a field that has a divergence on the order of 1E2 to 1E3 when it should be much lower. 

May 23, 2012, 12:01 

#24 
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Well first off if you are solving a poisson type problem with nueman conditions it is 100% singular not just nearly. Try solving a 1D poisson problem with nuemann conditions using gaussian elimination and you'll see what i mean.
Suggestions 1. Examine the residual r = Axb. See if it is maximum where your divergence errors are max. 2. Look at the pressure field itself to see if you really are achieving a nuemann condition on the pressure. 3. Post your MATLAB code, meshplots of pressure, divergence etc. 

May 23, 2012, 13:34 

#25  
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Peter
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2. I don't ever calculate the pressure points on the boundary or outside because I am computing on a staggered grid. However based on my formulation of the finite difference matrix I believe that this is enforced. 3. My code can be found in the .zip folder in the link below. If you run the script cartesian_sliding_cavity.m, it will compute the cartesian equivalent to my problem. If you run axisymmetric_sliding_cylinder.m, it will run the code I am currently trying to debug. If anything is unclear, please ask since it will force me to justify my actions and perhaps help me better understand what I am doing wrong. Thanks for all the help https://www.dropbox.com/s/1tiuq4aobu...ric%20Code.zip 

May 23, 2012, 13:52 

#26 
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Filippo Maria Denaro
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do you modify congruently near the boundaries the source term in the Poisson equation?


May 23, 2012, 14:20 

#27 
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Program is surprisingly short, I guess all the built in MATLAB functions help but its also makes it very tough to read/understand.
Anyways % pressure correction rhs = reshape(diff([uW;U;uE])/hx+diff([vS' V vN']')'/hy+avg([vS' V vN']')'./rvp,[],1); [p,flag(k)] = gmres(Lp,rhs,[],[],length(Lp)); . . . div = div_cyl(nx,ny,hx,hy,Ue,Ve,X); maxdiv(k) = max(max(abs(div(2:end1,2:end1)))); It looks like you are using two different methods of calculating the divergence. One way is using diff and that is fed into the poisson solver then you use div_cyl to check the divergence. If you check the diveregence after with the diff method it apperas to be ok. 

May 23, 2012, 16:13 

#28  
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Peter
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You are right. I didn't consider that. I guess what confuses me right now is that I expect that the area under the velocity curve should be constant in time. However as time progresses the area under my velocity curve increases. I thought that it was a divergence problem, but could it be due to something else? I'm sorry, but I'm not exactly sure what you mean. 

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