joule heating of material with anisotropic electric conducti

[mzenker]

But he has done that already, at least it's in the sif he posted above...

[mahichihi]

Hello everyone

I appreciate all the reply and great feedback. I attempted to compute with a simple shape according to the advice of annier, but when using conductivity with the tensor to calculate, the temperature did not rise(Fig1). However, as long as the conductivity is not expressed in terms of tensor, the temperature rise is calculated (Fig 2). Material and body force section in sif are shown in below.

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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 = Variable Temperature;real;290   926000;1290  834000;end

  Electric Conductivity = 92600      ! S/m   ! isotropic case
!  Electric Conductivity (3,3) = Real \         ! anisotropic case
!                                       926000 0 0  \
!                                       0 926000 0     \
!                                       0  0 92600

   
  !Heat Conductivity = 13.4     ! W/mK
   Heat Conductivity (3,3) = Real \
                                 13.4 0 0  \
                                  0 13.4 0     \
                                  0  0 1.34


 ! Density =8400e-18     ! kg/(um)^3
  Density =8400            ! kg/m^3
  Heat Capacity = 450
End

Body Force 1
  Name = "BodyForce 1"
! Heat Source = Equals Joule Heating
  Joule Heat = True
End

The above result is very strange for me.

best regards
Mahichihi

[raback]

Hi,

Sorry all, didn't properly check the "joule heat" flag. I suggest that you share the full case. The sif file looks ok to me at quick inspection.

-Peter

[mahichihi]

Thanks Peter
I show the complete sif file. Please point out if you notice any problems.

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Header
  CHECK KEYWORDS Warn
  Mesh DB "." "."
  Include Path ""
  Results Directory ""
End

Simulation
  Max Output Level = 5
  Coordinate System = Cartesian
  Coordinate Mapping(3) = 1 2 3
  Simulation Type = Steady state
  Steady State Max Iterations = 100
  Output Intervals = 1
  Timestepping Method = BDF
  BDF Order = 1
  Solver Input File = case.sif
  Post File = case.ep
coordinate scaling = 1e-6
End

Constants
  Gravity(4) = 0 -1 0 9.82
  Stefan Boltzmann = 5.67e-08
  Permittivity of Vacuum = 8.8542e-12
  Boltzmann Constant = 1.3807e-23
  Unit Charge = 1.602e-19
End

Body 1
  Target Bodies(1) = 4
  Name = "Body 1"
  Equation = 1
  Material = 1
  Body Force = 1
  Initial condition = 1
End

Solver 1
  Equation = Static Current Conduction
  Calculate Joule Heating = True
  Calculate Volume Current = True
  Procedure = "StatCurrentSolve" "StatCurrentSolver"
  Variable = Potential
  Exec Solver = Always
  Stabilize = True
  Bubbles = False
  Lumped Mass Matrix = False
  Optimize Bandwidth = True
  Steady State Convergence Tolerance = 1.0e-5
  Nonlinear System Convergence Tolerance = 1.0e-7
  Nonlinear System Max Iterations = 20
  Nonlinear System Newton After Iterations = 3
  Nonlinear System Newton After Tolerance = 1.0e-3
  Nonlinear System Relaxation Factor = 1
  Linear System Solver = Iterative
  Linear System Iterative Method = BiCGStab
  Linear System Max Iterations = 500
  Linear System Convergence Tolerance = 1.0e-10
  BiCGstabl polynomial degree = 2
  Linear System Preconditioning = Diagonal
  Linear System ILUT Tolerance = 1.0e-3
  Linear System Abort Not Converged = False
  Linear System Residual Output = 1
  Linear System Precondition Recompute = 1
End

Solver 2
  Equation = Result Output
  Output Format = Vtu
  Procedure = "ResultOutputSolve" "ResultOutputSolver"
  Output File Name = case
  Scalar Field 3 = potential
  Scalar Field 2 = joule heating
  Scalar Field 1 = temperature
  Vector Field 1 = volume current
  Exec Solver = Always
End

Solver 3
  Equation = Heat Equation
  Variable = Temperature
  Exported Variable 1 = electric conductivity
  Exported Variable 2 = voulume current
  Procedure = "HeatSolve" "HeatSolver"
  Exec Solver = Always
  Stabilize = True
  Bubbles = False
  Lumped Mass Matrix = False
  Optimize Bandwidth = True
  Steady State Convergence Tolerance = 1.0e-5
  Nonlinear System Convergence Tolerance = 1.0e-7
  Nonlinear System Max Iterations = 20
  Nonlinear System Newton After Iterations = 3
  Nonlinear System Newton After Tolerance = 1.0e-3
  Nonlinear System Relaxation Factor = 1
  Linear System Solver = Iterative
  Linear System Iterative Method = BiCGStab
  Linear System Max Iterations = 500
  Linear System Convergence Tolerance = 1.0e-10
  BiCGstabl polynomial degree = 2
  Linear System Preconditioning = Diagonal
  Linear System ILUT Tolerance = 1.0e-3
  Linear System Abort Not Converged = False
  Linear System Residual Output = 1
  Linear System Precondition Recompute = 1
End

Equation 1
  Name = "Equation 1"
  Active Solvers(3) = 1 2 3
End

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 = Variable Temperature;real;290   926000;1290  834000;end

  Electric Conductivity = 92600      ! S/m
!  Electric Conductivity (3,3) = Real \
!                                       926000 0 0  \
!                                       0 926000 0     \
!                                       0  0 92600

   
  !Heat Conductivity = 13.4     ! W/mK
   Heat Conductivity (3,3) = Real \
                                 13.4 0 0  \
                                  0 13.4 0     \
                                  0  0 1.34


 ! Density =8400e-18     ! kg/(um)^3
  Density =8400            ! kg/m^3
  Heat Capacity = 450
  Heat Capacity = 450
End

Body Force 1
  Name = "BodyForce 1"
! Heat Source = Equals Joule Heating
  Joule Heat = True
End

Initial Condition 1
  Name = "InitialCondition 1"
  Potential = 0
  Temperature = 20
End

Boundary Condition 1
  Target Boundaries(1) = 1 
  Name = "BoundaryCondition 1"
  Potential = 0
  Temperature = 20
End

Boundary Condition 2
  Target Boundaries(1) = 2 
  Name = "BoundaryCondition 2"
  Potential = 0.4
  Temperature = 20
End

best regards
Mahichihi

[Franz Pichler]

I also work with these solvers, thats why i am interestet in your problem. Just out of curiosity:
Are your boundaries the complete bottom and top surfaces? because, in the isotropic case, it looks like only the very corners have the (dark blue) 20 degree, and that they get hotter next to the corner.

[mahichihi]

Hi Franz

Thank you for your response.

>Are your boundaries the complete bottom and top surfaces?

I use part of upper part and lower part as boundary condition use the bottom and top surfaces.
I attach the sif file and the mesh file for reference.
If there is a problem with the condition of the attached file, please comment.

sincerely yours
Mahichihi

[annier]

Hi All,
1. I guess mahichihi has mentioned something worthy to note.
I tested with his/her (mahichihi - verify) files and found that the following conditions are not same for the presented solver input file:

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 ! Electric Conductivity = 92600      ! S/m
!or 
!  Electric Conductivity (3,3) = Real \
                                       92600 92600 92600  \
                                       92600 92600 92600     \
                                       92600 92600 92600

The first one i.e. scalar or single value of electric conductivity produces the Effect of Joule heating (i.e. shows an increase in temperature of the medium). The applied voltage of 0.4 produces joule heating .

Simulation condition utilizing scalar value or single denomination for electric conductivity

T-single-electric-conductivity.png (35.84 KiB) Viewed 5384 times

The second condition with the utilization of matrix or tensor for representing electric conductivity does not produce raise in temperature due to joule heating i.e. the result shows only the dirichlet conditons of temperature of the boundaries (20 units) enforced throughout the medium.

Simulation condition utilizing tensor in matrix form for representing electric conductivity

T-electric-conductivity-matrix.png (10.77 KiB) Viewed 5384 times

2. Peter, is there anything else(additional parameters) we need to write in Solver Input File to enforce matrices for material properties?
Yours Sincerely,
Anil Kunwar

[Franz Pichler]

Ok, i think i found the problem,
and its one for the Elmer team.
But correct me if i am wrong, and i have been wrong before about details like that

So the way the Joule heat is implemented here, at some point the differentials.f90 module is active and there is the code part

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    !------------------------------------------------------------------------------
    ! The simplest model just evaluates precomputed elemental heating at integration point
    !------------------------------------------------------------------------------
    IF( JouleMode == 1 ) THEN
      stat = ElementInfo( Element,Nodes,u,v,w,SqrtElementMetric,Basis )
      JouleH = SUM( Basis(1:n) * Jvar % Values( Jvar % Perm( NodeIndexes ) ) )

      ! Make an early exit since we don't need conductivity 
      RETURN
    END IF


    !------------------------------------------------------------------------------
    !  All other models require electric conductivity
    !------------------------------------------------------------------------------
    k = ListGetInteger( CurrentModel % Bodies &
         (Element % BodyId) % Values, 'Material')
    Material => CurrentModel % Materials(k) % Values
    
    ElectricConductivity(1:n) = ListGetReal( Material, &
        'Electric Conductivity',n,NodeIndexes,GotIt )

Here the conductivity is assumed to be of scalar type and not a tensor.

A simple workaround for this would be to use a body force that directly uses the joule heating variable, i.e.

Code: Select all

Heat Source = Equals Joule Heating

After checking everything again i just saw that you had this line in your sif file, but commented.
I just tried your case and it worked for me, with that line.

[raback]

Hi Franz

Yes, there is definately an inconsistency. Thank you for reporting.

The stuff in "Differentials" is an old approach from the era when solvers were not dynamically loadable. I think this has not been touched in ages but definately it assumes isotropic conductivity.

The approach you propose has the shortcoming that for material interfaces a nodal value cannot be discontinuous the way the heating can.

-Peter

[mahichihi]

Hi all

Thank you for supporting this problem.

Please suggestion.
If I want to use the tensor type conductivity, I think that it is necessary to rewrite the code, is that a difficult task?

PS: Franz
When A is used, the temperature rise does not seem to be calculated properly (reise temperature is so high)

best regards
Mahichihi