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Inductive Heating 3D/2D Problems: Resistance Losses and Current Flow
Posted Mar 23, 2011, 1:08 p.m. EDT Low-Frequency Electromagnetics, Heat Transfer & Phase Change, Materials, Parameters, Variables, & Functions Version 4.1 2 Replies
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I'm trying to simulate ohmic resistance losses caused by induced currents. So I started with a pretty simple geometry:
3 cylinders (Copper, Air, Aluminium).
The problem I am facing in 3D is that there seem to be no resistant losses in the copper caused by the current. Current and Current density seem to be ok in the copper though (Skin-effect).
I tried the same geometry in 2D axialsym. But here I am not sure how to get the current flow the right way. Current density seems to be wrong, because it's constant in the whole copper.
Btw. how do you figure out the alpha (h in comsol) when you choose convective cooling? At the moment I calculate it depending on the kind of fluid and geometry [Nusseltzahl etc.]. But because I don't know the wall temperature, I have to iterate here.
Can anybody help me to resolve this issues?
Thanks in advance.
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Ja alles is so anders mit deutschen Namen ;)
I havent really used that feature yet, so I'm not a specialist, but
In your 2D-axi model I believe you should turn on the in-plane or "three component" vector (main physics node) to be able to see a magnetic field, then you can get some coupling
Then what annoys me is that you have no "air around your system, so I'm not sure how the field lines might loop around like. but this might just be me not being used to this particular physics set-up
And to make life easier, define your materials then you can simpler refer to them
if you have a current flow along Z then you need to adjust the density w.r.t. "r" yourself in your equation. That is probably why the density is "constant", you defined it that way. This is linked to the implicit x,y,z coordinates in the GUI equations, most variables are fields or functions of spatial and time
--
Good luck
Ivar
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In 2D-axial, I think you could try to add a surface current density according to the example model.
For resistive heating, you could try to find the resistance first and then get the heat by Q=I^2*R.
Hope this helps.
Thanks.