Discussion Forum

Posted: 1 decade ago 18 févr. 2011, 02:32 UTC−5

Hi I'm interested to understand and hear in which version, and how you import the mesh and how you then get COMSOL to use it: as geometry or as entities or as mesh ? -- Good luck Ivar

Posted: 1 decade ago 18 févr. 2011, 02:52 UTC−5

Hi Ivar, I import the mesh as an .mphtxt file - which is generated by Simpleware and I use the imported mesh to solve the model . I am using V3.5a However, the same question about calculating the average element size applies to any simple geometry drawn and meshed in Comsol. I know I can set the element size when generating the mesh in Comsol. But how do I check that the actual element size in the mesh I generate in Comsol matches what I specified? Thanks

Posted: 1 decade ago 18 févr. 2011, 04:17 UTC−5

Hi "h" is the mesh element size but it's local to be understand as "h_i" where _i represent every "i" mesh element. So basically I would say it's taking the average of "h" over the domain, something like intop1(h)/intop1(1), where intop1() is your Integration Coupling variable (ported back to 3.5, that's getting oldish for me ;) in V4 it's simply aveop1(h) you have also something like a "dvol" if I remember right in addition to "qual" in 3.5 -- Good luck Ivar

Posted: 1 decade ago 2 juin 2011, 22:05 UTC−4

Hi, I am using free Triangular mesh, and need to determine the area of each triangle for calculating ds. I have tried to use average h_i(mesh element size), but it looks like it is not exact solution. Can you please let me know how I can calculate the area of each triangle within Triangular mesh? Thanx, Shahed

Posted: 1 decade ago 12 juil. 2011, 03:49 UTC−4

i'm interested how can you handle the obtained mesh from simpleware in Comsol??

Posted: 1 decade ago 27 avr. 2012, 14:40 UTC−4

Can I have a question that what does it mean by "mesh size" when describing a Free Tetrahedral mesh? Which dimension is it talking about since it has a unit of "m"? I prefer to know the volume of each mesh in my case for calculating the total enthalpy in the final product. Thanks for answering Ivar.

Posted: 1 decade ago 27 avr. 2012, 16:40 UTC−4

Hi there is something I do not understand, why do you all want to know the volume or area of the mesh elements ? In Comsol one define the physics on the geoemtrical Entities,so the results are also obtained by integrating over the entities (doamins and boundaries). A mesh is finally only the discretisation used, and it might change, the entities, far less. It's like when you digitalise a analogue signal, for example some music. Finally one want the melody as results, not the size of each sample used to discretized the signal. If the quality is not good enough, OK one decides to use a faster and more detailed discretisation, but the end esire is the quality of the final sound. or have I missed something ? -- Good luck Ivar

Posted: 1 decade ago 30 avr. 2012, 18:02 UTC−4

Thanks for your reply. In my case, I would like to know the total enthaply inside my food material in joule heating module. After building up a heat transfer module and solved, I can see the temperature distribution, and also, I can export the enthalpy data. However, the unit of "enthalpy" and "total enthalpy" are both with a unit of J/kg. So I am wondering shall I know the volume of each mesh, and then calculate the mass of each mesh, in order to obtain the enthalpy within each mesh and add them up? or just simply add all them up and them multiply by the total mass of product? Thanks, let me know if it is not clear enough. Yang [QUOTE] Hi there is something I do not understand, why do you all want to know the volume or area of the mesh elements ? In Comsol one define the physics on the geoemtrical Entities,so the results are also obtained by integrating over the entities (doamins and boundaries). A mesh is finally only the discretisation used, and it might change, the entities, far less. It's like when you digitalise a analogue signal, for example some music. Finally one want the melody as results, not the size of each sample used to discretized the signal. If the quality is not good enough, OK one decides to use a faster and more detailed discretisation, but the end esire is the quality of the final sound. or have I missed something ? -- Good luck Ivar [/QUOTE]

Posted: 1 decade ago 1 mai 2012, 00:52 UTC−4

Hi Well forget the "mesh" it's "only" the discretization. COMSOL looks after the integration. You should work on the "entities" where you have defined your physics, in particular the "domains" (volume in 3D, surfaces in 2D ...) and the boundaries (surfaces in 3D, edges in 2D ...) If you solve the HT and want the total entalpy, you integrate over the domain(s) of interest "Cp*rho*T" You shoul note that COMSOl calles these dependent variables "variables" but they are in fact "fields" its Cp(x,y,z,t) and rho(x,y,z,t) and T(x,y,z,t) (or whatever corresponds to your coordinate system and if used time response). And these are in dimension of the mesh elements (hence often you see densities per kg or per m^3, so most "variables" must be integrated over the domains (volumes) to get macroscopic values out. For COMSOl you define the physics and BC on the Entities, and you postprocess the entities, the mesh is the discretization or sampling of your "signal/model" it must meet the Nyquist of mesh quality criteria, that is all). You can concentrate on the true physics and not on the FEM math with COMSOL, that is what is so nice with it ;) -- Good luck Ivar

Posted: 1 decade ago 1 mai 2012, 13:26 UTC−4

Hi Ivar, Thanks. As per my understanding, the mesh we made for discretilization is not important at all, can you give me an example of how to calculate the total energy absorbed in my material? The following data is exported as "total enthalpy" with a unit of J/kg. I took two time point for reference. As far as I know the enthalpy distribution is fairly different at every different location in the material, but how to integrate them in comsol or any other postprocessing software? Can you show me simply how to do it? Thanks, % x 0.071 0.071 0.069781614 0.066606098 0.070207608 0.068316062 y -0.04 -0.04 -0.044783645 -0.041805078 -0.043901277 -0.046817275 z 0.1116 0.107269484 0.1116 0.1116 0.106316936 0.107818497 jh.H0 (J/kg) @ t=0 520322.6718 520327.721 520337.4573 520306.691 520351.3471 520336.7744 jh.H0 (J/kg) @ t=900 858315.0763 862184.7099 860672.0054 861906.5826 865079.9856 864478.9603

Posted: 1 decade ago 1 mai 2012, 15:22 UTC−4

Hi I do normally everything within the GUI, check the example, if you need the time evolution, or difference between two times you need to fidle with the at() operator in the postprocessing -- Good luck Ivar

Attachments:

• HTenthalpy.mph

Posted: 1 decade ago 2 mai 2012, 13:41 UTC−4

I am sorry Ivar, I am not open the file you attached. I tried several different machine and it still does not work. I am using Comsol 4.2, Can you check your file and upload it again? Thanks!!!

Posted: 1 decade ago 2 mai 2012, 17:26 UTC−4

Hi sorry I have only 4.2a installed, I agree the chase of versions is tricky and makes life difficult for exchanges, computer time ... -- Good luck Ivar

Posted: 1 decade ago 2 mai 2012, 19:29 UTC−4

Thanks anyway, hope I can find some other people open it for me.. Can you try post a picture for the procedures doing that, I mean, calculate the enthalpy in GUI? Thanks so much, Ivar!

Posted: 1 decade ago 23 févr. 2013, 01:12 UTC−5

Hi Ivar, I am using artificial diffusion in comsol . as the effect of the articial diffusion is correlated with the mesh size, therefore I am really interested in how can I know the mesh size. I have tired the method you described here but no luck. Is there any embedded way that allows us to obtain the average mesh size?

Posted: 1 decade ago 23 févr. 2013, 07:41 UTC−5

Hi I believe you have only the average "h" mesh size (field its h(x,y,z) ;) at disposal, as meshes can be rather anisotropic, sometimes I do too miss some sort of hx,hy,hz, the three components of "h" COMSOL itself have different artificial diffusion methods, depending on the physics, check carefully the solver section of the doc -- Good luck Ivar