joule heating of material with anisotropic electric conducti

[mahichihi]

I am new user of Elmer. I want to calculate joule heating for an electric anisotropic material. However I cannot obtain the result if I use the anisotropic parameter. Can I calculate the joule heating of this situation by using Elmer.We will attach the sif file used for reference so please advise.

[annier]

Hi mahichihi,
Yes Electric conductivity can be represented as anisotropic tensor, as it is writtten in the snippet of source code of StatCurrentSolve.F90.

Code: Select all

         !------------------------------------------------------------------------------
         !      Read conductivity values (might be a tensor)
         !------------------------------------------------------------------------------

         CALL ListGetRealArray( Model % Materials(k) % Values, &
             'Electric Conductivity', Cwrk, n, NodeIndexes )

         Conductivity = 0.0d0
         IF ( SIZE(Cwrk,1) == 1 ) THEN
           DO i=1,3
             Conductivity( i,i,1:n ) = Cwrk( 1,1,1:n )
           END DO
         ELSE IF ( SIZE(Cwrk,2) == 1 ) THEN
           DO i=1,MIN(3,SIZE(Cwrk,1))
             Conductivity(i,i,1:n) = Cwrk(i,1,1:n)
           END DO
         ELSE
           DO i=1,MIN(3,SIZE(Cwrk,1))
             DO j=1,MIN(3,SIZE(Cwrk,2))
               Conductivity( i,j,1:n ) = Cwrk(i,j,1:n)
             END DO
           END DO
         END IF

Now, in the Material n section of your solver input file you need to write a 3*3 matrix in three rows separated by \ as for Elmer sif convention as:

Code: Select all

Material 1
  Name = "Material 1"
  Reference Temperature = 290
  !Electric Conductivity = Variable Temperature;REAL;290 9.26e11;1290 8.34e11 ;END
  Electric Conductivity (3,3) = Real \ 
                                  8.34e11 0 0   \
                                  0 8.34e11 0   \ 
                                  0 0 8.34e9
  !Heat Conductivity = 13.4e6
  Heat Conductivity (3,3) = Real \
                                   13.4e6 0 0  \
                                     0 13.4e6 0     \
                                     0  0 13.4e4
  Density = 8400e-18
End

Since you are expressing electric conductivity and heat conductivity as orthotropic tensors, what would happen if you simply write them as:

Code: Select all

Material 1
...
Electric Conductivity  (3) = Real 8.34e11 8.34e11 8.34e11
...
Heat Conductivity  (3) = Real 13.4e6 13.4e6 13.4e6
...
End

Yours Sincerely,
Anil Kunwar

[mahichihi]

Hi annier. Thank you for your reply.
I retried calculation at the advice of you. As a result, I could obtain the joule heating, volume current and potential, but th temperature remained the initial condition.I will attach the result.
Please point out if there are still points to improve

Note that, I wrote 3*3 matrix following formula;

Electric Conductivity(3,3)=REAL 8.34e11 0 0 \
0 8.34e11 0 \
0 0 8.34e9

Heat Conductivity(3,3)=REAL 3.4e6 0 0 \
0 13.4e6 0 \
0 0 13.4e4

since if I use your suggestion the Elmer solver outputs an error message.
ERROR:: SectionContents: Problem reading real keyword: electric conductivity: \ 8.34e11 0 0.

Yours Sincerely,
Mahichihi

[annier]

Hi Mahichihi,
Please put one \ in the first line after Real when representing the anisotropic tensor matrix as in:

Code: Select all

Material 1
    Electric Conductivity (3,3) = Real \ 
                                  8.34e11 0 0   \
                                  0 8.34e11 0   \ 
                                  0 0 8.34e9
...
  Heat Conductivity (3,3) = Real \
                                   13.4e6 0 0  \
                                     0 13.4e6 0     \
                                     0  0 13.4e4
...
End

I have updated it in the second post as well . The numbers of the matrix follow the next line of the sif file.

Yours Sincerely,
Anil Kunwar

[mahichihi]

Hi annier.

I modified the conductivity tensor according to your advice as follows

Code: Select all

  
Material 1
  Name = "Material 1"
  Reference Temperature = 290
  !Electric Conductivity =  Variable Temperature; Real MATC "(-9E+07)tx+1E+12"
  !Electric Conductivity = Variable Temperature;real;290   9.26e11;1290  8.34e11;end
  !Heat Conductivity = 13.4e6
  
  Electric Conductivity (3,3) = Real \ 
                                8.34e11 0 0   \
                                0 8.34e11 0   \ 
                                0 0 8.34e9

  Heat Conductivity (3,3) = Real \
                            13.4e6 0 0  \
                            0 13.4e6 0   \
                            0  0 13.4e4

   Density = 8400e-18
End

however the results were same as before.

I still can not understand why this kind of thing happens.
Do i miss up anything?

Sincerely yours
Mahichihi

[annier]

Hi mahichihi,
The two Electrical conductivity elements of the matrix E_11, E_22,are equal that is 8.34E+11. Make them different to visualize anisotropic effect.
The same reasoning with heat conductivity. In numerical simulation, there are many things we need to adjust.


Yours Sincerely,
Anil Kunwar

[Franz Pichler]

Hi there,

what is the dimension (meters/millimeters) of your geometry? Often you don't see a change in some quantity when there is a scaling mistake.
For example when you want the geometry to be in millimeters, but you got meters then the heat source is simply not big enough to cause a significant change in temperature.
Then you would need to add a Coordinate Scaling = 1.e-3 in the first section of the sif file.

This is just a wild guess tho.

best regards
Franz

[mahichihi]

Hi annier.

My question is, rather than not be able to see the anisotropy, is that the temperature can not be calculated at why calculated Joule heat. For example, if the electric conductivity is isotropic and incorporates anisotropy in the thermal conductivity, the temperature can be calculated.

The following is the sif file used.

Code: Select all

Material 1
  Name = "Material 1"
  Reference Temperature = 290
  !Electric Conductivity =  Variable Temperature; Real MATC "(-9E+07)tx+1E+12"
  !Electric Conductivity = Variable Temperature;real;290   9.26e11;1290  8.34e11;end
  !Electric Conductivity (3,3) = Real \ 
  !                              8.34e11 0 0   \
  !                             0 8.34e11 0   \ 
  !                              0 0 8.34e9
  Electric Conductivity = 8.34e11

 !Heat Conductivity = 13.4e6
  Heat Conductivity (3,3) = Real \
                                 13.4e6 0 0  \
                                  0 13.4e6 0     \
                                  0  0 13.4e4
  Density = 8400e-18
End

Sincerely yours
Mahichihi

[mahichihi]

Hi Franz.
Thank you for your comment.

My geometry has dimensions um. So I rewrie the physical constants.
For examaple, on the Heat Conductivity, I employ 13.4e6 pW/ um K instead of 13.4 W/m K.

Sincerely yours
Mahichihi

[mzenker]

Hi,

did you also adjust the vacuum permittivity (it is in As/Vm by default)?
I assume that you know what you are doing when working in um whereas Elmer assumes m by default. You will have to check all parameters, material properties etc. very carefully...

Alternatively, say

Coordinate Scaling = 1e-6

in the Simulation section, as Franz suggested.

HTH,

Matthias