Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.
Heat Transfer: Symmetry & Radiation
Posted Sep 7, 2010, 7:44 a.m. EDT 7 Replies
Please login with a confirmed email address before reporting spam
normally, it is usual to take advantage of symmetry. This works quite well especially for heat transfer, navier stokes ...
BUT
If you enable radiation (Heat Transfer Modul), you can't utilize symmetry, because COMSOL "looks through" the (open) symmetry boundary.
To demonstrate the issue, I have attached a MPH-File. (just click the "solve" button)
Have I done some mistakes? How to set up the model correctly, to take advantage of symmetry even with radiation?
Best regards
Attachments:
Please login with a confirmed email address before reporting spam
I'm afraid you are slightly too quick in your conclusions, there. For me your two cases are different.
First of all you should use a radiation "group number" different for the 1/2 part and for the 2*1/2 parts.
Then if you integrate the total normal heat flux on the three (right) interior boundaries of your 1/2 model and the 2*1/2 model you will find quite a lot of differences.
For me you have no "symmetry" boundary on the left 1/2 model, as you have excluded the opening. There you need a perfect mirror with no absorbsion.
Furthermore if you look at the final temperature with a line cut (i.e. from (-0.2 0.2) to (1.4 0.2) you will see a rather large difference for aour two models
--
Good luck
Ivar
Please login with a confirmed email address before reporting spam
First, you're right. I should use group numbers for radiation. In this case it makes no difference. :)
Second, i know about the differences - thats the problem.
Usually, if you want to model "pure" heat transfer of the right "2*1/2"-model, then you would feed COMSOL with the left 1/2-model, because it HAS a geometrical symmetry boundary . Then (no radiation) the results would be the same - no difference between left and the right model. But if you activate radiation, then you get differences between the left 1/2-model and the right 2*1/2-model, as you told me correctly. Do you know what I mean?
So my problem is, if you have radiation, you can't utilize (geometrical) symmetry boundary. To solve this, i thought about a small black plate on the "open" boundary. The results of the "open-blacked" 1/2-model showed good agreement in comparison with the right 2*1/2-model.
kind regards
Please login with a confirmed email address before reporting spam
you need to define a fully reflective boundary where the radiation is escaping. Your physics is not correct if you do not consider a "closed" system. That is basic Physics, nothing to do really with FEM methodology.
I'm sure you will find a way around (radiation means you must have a closed system, even if one boundary is a fully refolective or "infinite element escaping to the universe average temperature
--
Good luck
Ivar
Please login with a confirmed email address before reporting spam
this is exactly what i have tried: a fully reflective boundary. It works quite well (see temperature_distribution.png), but there are still some differences between the results of the 1/2 model and the fully 2*1/2-model (see temperature_boundary6vs21.png). I adjusted the 1/2 model with a small plate wich covers (almost) the whole hole. The boundaray, which "looks" inside the chamber, has a temperature of 0K and has surface-to-surface-radiation with an emissivity of 0. Physically this should be a fully reflective boundary with no radiation from the plate itself, or?
But how can the difference be explained which are still left? Numerically unprecisions, since the reflective boundary has an radiative flux which is not zero (as you can see in picture radiative_flux_boundary7.png)? I would say, that COMSOL (V3.5a) can't provide an real refelective boundary, even with an emissivity of zero: If radiation energy comes from inside the chamber to the "refelctive" boundary, then the energy is absorbed and immediately emitted. But emittance does not care about the income angle (as it is correct for thermal radiation!), it is distributed along lamberts cosine law: J(alpha) = J0 * cos(alpha). For an real reflective boundary, it should be J_emit = J_absorbed but only in one direction: alpha, respectively alpha'. ???
kind regards
Attachments:
Please login with a confirmed email address before reporting spam
probablky yes it could be numerical issues, as your flux is rather spiky. Have you tried to integrate fully over the area ? and see whathappens when you refine the mesh a few times ?
But it looks far better than without your reflector, no ?
--
Good luck
Ivar
Please login with a confirmed email address before reporting spam
thanks for reply! I always appreciate your help!
I tried serveral things to improve the reflective boundary (Radiation resolution: 4096, abnormaly fine mesh size, perfectly matching reflector size and positioning using insulated pairs, sharpened tolerances, low time stepping) but the results were the same.
So i really think the major problem is, that this is a diffusive reflector and no mirror-like, specular reflector.
best regards
Please login with a confirmed email address before reporting spam
if nothing else works, try to send the issue to support, they are usually good to reply and this is for me a rather important issue how to handle correctly radiative aanlysis.
pls do not forget to report back here for us others ;)
--
Good luck
Ivar
Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.