joule heating of material with anisotropic electric conducti
Re: joule heating of material with anisotropic electric conducti
But he has done that already, at least it's in the sif he posted above...
Re: joule heating of material with anisotropic electric conducti
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.
The above result is very strange for me.
best regards
Mahichihi
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.
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 = 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
best regards
Mahichihi
- Attachments
-
- Fig. 2 (use of isotropic electrical conductivity)
- Fig. 2.png (13.96 KiB) Viewed 5400 times
-
- Fig. 1 (use of anisotropic electrical conductivity)
- Fig. 1.png (7 KiB) Viewed 5400 times
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Re: joule heating of material with anisotropic electric conducti
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
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
Re: joule heating of material with anisotropic electric conducti
Thanks Peter
I show the complete sif file. Please point out if you notice any problems.
best regards
Mahichihi
I show the complete sif file. Please point out if you notice any problems.
Code: Select all
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
Mahichihi
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Re: joule heating of material with anisotropic electric conducti
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.
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.
Re: joule heating of material with anisotropic electric conducti
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
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
- Attachments
-
- aniso elmer_1.zip
- (1.49 MiB) Downloaded 318 times
Re: joule heating of material with anisotropic electric conducti
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:
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 .
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.
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
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:
Code: Select all
! Electric Conductivity = 92600 ! S/m
!or
! Electric Conductivity (3,3) = Real \
92600 92600 92600 \
92600 92600 92600 \
92600 92600 92600
Yours Sincerely,
Anil Kunwar
Last edited by annier on 28 Feb 2017, 16:34, edited 1 time in total.
Anil Kunwar
Faculty of Mechanical Engineering, Silesian University of Technology, Gliwice
Faculty of Mechanical Engineering, Silesian University of Technology, Gliwice
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Re: joule heating of material with anisotropic electric conducti
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
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.
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.
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
Code: Select all
!------------------------------------------------------------------------------
! 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 )
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
I just tried your case and it worked for me, with that line.
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Re: joule heating of material with anisotropic electric conducti
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
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
Re: joule heating of material with anisotropic electric conducti
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
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