Natural BC using Cylinder1

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Author Message   Klaus Albers (k_albers) Member Username: k_albers Post Number: 8 Registered: 07-2005 Posted on Monday, March 23, 2009 - 11:49 am:    Hello, when using a point natural BC in a Cylinder1 geometry, I get different (and wrong) results compared to a Cylinder2 geometry. It's just a simple heat radiation problem, the region and BC definition is as follows: REGION 1 { For each material region } START(0) { Walk the domain boundary } point natural (temp)=0 LINE TO (d) point natural (temp)=E_austausch*sigma*((temp+273)^4-(T_umgebung+273)^4)*d/2 Only the multiplication by "d/2" leads to the same results compared to the same geometry in Cylinder2. It's probably s.th. stupid due to the dimensional reduction but I don't get it right away. Does anyone have an idea? Thanks a lot, Klaus   Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1226 Registered: 06-2003 Posted on Monday, March 23, 2009 - 04:00 pm:    It would help if we had the full scripts for both problems. Can you post them?   Klaus Albers (k_albers) Member Username: k_albers Post Number: 9 Registered: 07-2005 Posted on Tuesday, March 24, 2009 - 04:03 am:    Here they are: Cylinder1 waermestrahlungsheizung_kristall.pde (1.3 k) Cylinder2 waermestrahlungsheizung_kristall_2D.pde (1.5 k)   Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1229 Registered: 06-2003 Posted on Tuesday, March 24, 2009 - 08:46 pm:    1. In both 1D and 2D cylindrical geometry, FlexPDE expands the div(grad()) term in cylindrical geometry and multiplies by the cylindrical volume element r*dr*dz*dphi. It must then multiply the specified natural BC by the appropriate surface element, r*dz*dphi. It appears to do this correctly in 2D but not in 1D. If you manually multiply the natural BC by r (ie, your d, not d/2 as you have done), you get the correct answer. 2. You have used "Cylinder2" geometry specification, but this is nowhere documented. It translates internally (for apparently ancient reasons) to "Xcylinder(R,Z)", which means that your Cylinder2 problem is running the heat flow along the axis, not the radius of the cylinder. 3. Your "Threshold" specification is ten times the total final value of temperature, which means you have asked for a very sloppy answer. Your threshold should be the smallest temperature variation that you want to hold to the specified ERRLIM tolerance. "Threshold=1" would be more reasonable. If you multiply the 1D natural by R, and specify "Ycylinder" coordinates in 2D, you get the same answer in both problems. We will try to track down the missing surface area multiplier in 1D cylindrical geometry and include the fix in the 6.05 release. At that time, the manual *R suggested above will again become erroneous and will need to be removed. cyl1d_r.pde (1.5 k) ycylinder.pde (1.5 k)   Klaus Albers (k_albers) Member Username: k_albers Post Number: 10 Registered: 07-2005 Posted on Wednesday, March 25, 2009 - 05:13 am:    Perfect, thank you very much. The programs are pretty sloppy themselves, just a quick test to check my Matlab code :-) Here's the result and that's what I wanted to see: Comparison comparison_matlab_flexpde_analytical.png (48.8 k) It may be useful to add a cylinder1 sample problem to the Help file that uses a non-zero natural BC, just for clarity.   Klaus Albers (k_albers) Member Username: k_albers Post Number: 11 Registered: 07-2005 Posted on Wednesday, March 25, 2009 - 05:18 am:    Ok, I'll try to embed the graphic directly: