Discussion Forum

Hi

it happens that COMSOL is a general physics programme and that materials are often anisotropic (in all generality) therefore where we "human and physicist explain we have one constant (same diagonal value of an isotropic material) there are in fact a 3x3=9 terms of a 2nd order TENSOR and if you look carefully COMSOL is ALL tensor developed

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Good luck
Ivar

Hi,

so if I want to define a material with the same thermal conductivity in all but one (e.g. z-direction 10 times higher) direction. iI have to define k as follows:

{k11,k21, k31, k12, k22, k32, k13, k23, k33} = {a,a,a,a,a,a,a,a,a*10}

Is that correct?

Michael

Hi

no neither not, if its an isotropic material property with one preferred direction..

The easiest is to select your Heat transfer in Solid node, change the Thermal Conductivity "From material" to "User defined"
Then type 1 in the Isotropic tab entry (this is the default isotropic heat conduction.
If you now change to "Diagonal" you will see the equivalent tensor result based on your initial "1" (all "1" on the diagonal, 0 outside)

Here you could change the Z direction to 10 and you have what I understand as your question for material with strong conduction in one preferred direction. (if you need it along another direction you need to adapt the coordinate term

If you chose fully isotropic and add values outside the diagonal you get coupling therms from one direction into another, I'm not sure what kind of material that is, would have to dig into the theory first ;)

Now this works in 4.3a, hope it's still the case with 4.0 which was an "early version"

You can also define this in the material, in which case you should look after the diagonal terms k11 k22 k33
Now if your material is bending around in space and not aligned with neither x,y, or z then you need th ensure you use a local coordinate system, or a user defined one (Model Definitions Coordinates)

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Good luck
Ivar

Thank you Ivar,

now it makes sense to me.
I will use the diagonal-setting, that is more convenient.

Hower, I could also define k like this:

{k11,k21, k31, k12, k22, k32, k13, k23, k33} = {a,0,0,0,a,0,0,0,a*10}

and would get the same result - correct?

Thanks,
Michael

Hi

if you choose full anisotropic yes,
Check the equations (preferences show equation nodes) to see how this is set up. There has been a change in approach now in v4 latest (>4.3), now you copy all tensor values from your source case to the full 9 components

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Good luck
Ivar