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highly conductive layer

Posted 19 oct. 2012, 13:30 UTC−4 Version 4.2a, Version 4.3a 5 Replies

Susant Patra
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Hi,

I tried to do a simple transient heat transfer model as follows:
one model with a thin solid domain on top of a thick domain and
another model with highly conductive layer on thick domain.

The thickness in the highly conductive layer in the 2nd model is same as that of thin solid domain of 1st model.
I used anisotropic thermal conductivity property for the highly conductive layer and for the tin domain: K=[1700W/m*K,1700W/m*K,8W/m*K]

I noticed that the result in thin domain do not match. The highly conductive layer provides a peak temperature significantly lower.

Question: Does the highly conductive layer model only works for the isotropic material only?

Regards,
Susant
5 Replies Last Post 20 oct. 2012, 06:31 UTC−4
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 19 oct. 2012, 15:57 UTC−4
Hi

there should not be any large difference, but some probably yes.

The highly conductive layer conducts normally as an isotropic, diagonal or anisotropic layer, assuming both horizontal and through the thickness (normal) conductivities.

While a highly resisitve layer, can only be applied to internal boundaries and assumes no in line (tangeantial) conduction, only heat loss across the boundary normal and this BC node accepts only a scalar normal heat conduction ks. A highly resistive layer has an up(T) and down(T) temperature as well as a T temperature being the average of the up(T) and down(T), so the layer temperature loss is truely up(T)-down(T). A thin resitive layer also is ignored for all dynamic effects

an example (but 4.3a)

--
Good luck
Ivar
Hi there should not be any large difference, but some probably yes. The highly conductive layer conducts normally as an isotropic, diagonal or anisotropic layer, assuming both horizontal and through the thickness (normal) conductivities. While a highly resisitve layer, can only be applied to internal boundaries and assumes no in line (tangeantial) conduction, only heat loss across the boundary normal and this BC node accepts only a scalar normal heat conduction ks. A highly resistive layer has an up(T) and down(T) temperature as well as a T temperature being the average of the up(T) and down(T), so the layer temperature loss is truely up(T)-down(T). A thin resitive layer also is ignored for all dynamic effects an example (but 4.3a) -- Good luck Ivar

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Susant Patra
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Posted: 1 decade ago 19 oct. 2012, 16:25 UTC−4
Hi Ivar,

Unfortunately, I can not open 4.3 file. However, I have attached 4.2a model that you can open. The temperature difference is significant.

Regards,
Susant
Hi Ivar, Unfortunately, I can not open 4.3 file. However, I have attached 4.2a model that you can open. The temperature difference is significant. Regards, Susant

Attachments:

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 19 oct. 2012, 17:14 UTC−4
Hi

indeed you see differences with your model, but

1) if you use the same material for the thin layer (and forget your "Copper")
2) if you use the same thickness !! you have a factor 1000 difference because of the units, geometry in [mm], highly condutive layer in [m] !
3) you use a same mesh density (copy the mesh from one domain to the other)

You will get far better results, almost identical

And anyhow if you use it in this direction (tangeant) then the anisotropy is not really relevant

--
Good luck
Ivar
Hi indeed you see differences with your model, but 1) if you use the same material for the thin layer (and forget your "Copper") 2) if you use the same thickness !! you have a factor 1000 difference because of the units, geometry in [mm], highly condutive layer in [m] ! 3) you use a same mesh density (copy the mesh from one domain to the other) You will get far better results, almost identical And anyhow if you use it in this direction (tangeant) then the anisotropy is not really relevant -- Good luck Ivar
Susant Patra
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Posted: 1 decade ago 19 oct. 2012, 17:30 UTC−4
Hi Ivar,

Thanks! The layer dimension was at error. The results are quite close.

I noticed that you can plot both case studies on the same axis. How did you do that? I would appreciate if you can point me to the relevant document.

Regards,
Susant
Hi Ivar, Thanks! The layer dimension was at error. The results are quite close. I noticed that you can plot both case studies on the same axis. How did you do that? I would appreciate if you can point me to the relevant document. Regards, Susant
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 20 oct. 2012, 06:31 UTC−4
Hi

that is easy, when you select a line plot, and you select edges, by default COMSOL assumes these are joined as a long (poly-) line, if not, that means these are dsjoint, then the default "arc-length" for the horizontal axis cannot be defined.

You simply go the the horizontal axis tab of the line plot sub node and set in a user expression such as x, or r (in 2D-axi) or y,z ... even you can plot flux versus T temperature if that makes some sens (typically with table plots). Then your edgs might be fully disjoint.

Something similar happens too in 2D, i.e. the cut-surface data set ndes, you need sometimes to arange how to show the surface x,y in absolute or elative surface length values

--
Good luck
Ivar
Hi that is easy, when you select a line plot, and you select edges, by default COMSOL assumes these are joined as a long (poly-) line, if not, that means these are dsjoint, then the default "arc-length" for the horizontal axis cannot be defined. You simply go the the horizontal axis tab of the line plot sub node and set in a user expression such as x, or r (in 2D-axi) or y,z ... even you can plot flux versus T temperature if that makes some sens (typically with table plots). Then your edgs might be fully disjoint. Something similar happens too in 2D, i.e. the cut-surface data set ndes, you need sometimes to arange how to show the surface x,y in absolute or elative surface length values -- Good luck Ivar

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