joule heating of material with anisotropic electric conducti

[mahichihi]

Hi Matthias
Thank you for your opinion.

I added "coordinate scaling = 1.0e^-6" to the sif file, but I cannot calculate the temperature.

Code: Select all

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

Simulation
  Max Output Level = 10
  Coordinate System = Cartesian
  Coordinate Mapping(3) = 1 2 3
  Coordinate Scaling = 1e-6
  Simulation Type = Scanning
  Steady State Max Iterations = 100
  Output Intervals = 1
  Timestepping Method = BDF
  BDF Order = 2
  Timestep intervals = 5
  Timestep Sizes = 1
  Solver Input File = case.sif
  Post File = case.ep
End

Constants
  Gravity(4) = 0 -1 0 9.82
  Stefan Boltzmann = 5.67e-08                        ! (pW)/(um)^2 /K^4=W/m^2 /K^4
  Permittivity of Vacuum = 8.8542e-12              ! F/m
  !Permittivity of Vacuum = 8.8542e-6	   ! pF/um
  Boltzmann Constant = 1.3807e-23
  Unit Charge = 1.602e-19
End

Body 1
  Target Bodies(1) = 6
  Name = "Body Property 1"
  Equation = 1
  Material = 1
  Body Force = 1
End

Solver 2
  Equation = SaveScalars
  Variable 2 = volume current
  Filename = save scalars.dat
  Save Flux = True
  Procedure = "SaveData" "SaveScalars"
  Operator 2 = area
  Operator 1 = max
  Variable 1 = potential
  Exec Solver = After Timestep
End

Solver 1
  Equation = Static Current Conduction
  Calculate Joule Heating = True
  Calculate Volume Current = True
  Procedure = "StatCurrentSolve" "StatCurrentSolver"
  Variable = Potential
  !Exec Solver = After Timestep
  Exec Solver = Always
  Stabilize = True
  Bubbles = True
  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 3
  Equation = Result Output
  Output Format = Vtu
  Procedure = "ResultOutputSolve" "ResultOutputSolver"
  Output File Name = case
  Scalar Field 3 = joule heating
  Scalar Field 2 = temperature
  Scalar Field 1 = potential
  Vector Field 1 = volume current
  Exec Solver = After Timestep
End

Solver 4
  Equation = Heat Equation
  Variable = Temperature
  Procedure = "HeatSolve" "HeatSolver"
  !Exec Solver = After Timestep
  Exec Solver = Always
  Stabilize = True
  Bubbles = True
  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(4) = 2 1 3 4
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 = 926000      ! S/m
   Electric Conductivity (3,3) = Real \
                                       926000 0 0  \
                                       0 926000 0     \
                                       0  0 926000

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


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

Body Force 1
  Name = "BodyForce 1"
  Joule Heat = True
End

Initial Condition 1
  Name = "InitialCondition 1"
  Temperature = 290
End

Boundary Condition 1
  Target Boundaries(1) = 1 
  Name = "BoundaryCondition 1"
  !Current Density = variable time;real;1 0;2 1e8;3 2e8;4 3e8;5 4e8;end
  Potential=variable time;real;1 0.1;2 0.2;3 0.3;4 0.35;5 0.4;end
  Temperature = 290
  Save Scalars = True
End

Boundary Condition 2
  Target Boundaries(1) = 2 
  Name = "BoundaryCondition 2"
  Potential = 0
  Temperature = 290
  Save Scalars = True
End

Also checked total heating power and effective resistance in the solver log were 4.5230602241880828E-003 and 35.374280259273135.
The result was consistent with following hand calculation.
The predicted resistance from the length of 900 [um], cross-section area of 25 [um^2] and electric conductivity of 926000 [S/m] is 38.87689 [ohm]. Also the heating power V^2/R is 0.0041156 [J].

So I believe that there is no mistake, such as the parameters used.
What is the problem ?

Sincerely yours
Mahichihi

[mzenker]

Hi,

upon closer reading of your sif file, I think I have found a possible cause of your problem:
You need to set the Simulation Type to transient instead of scanning in the Simulation section.

HTH,

Matthias

[mahichihi]

Hi Matthias

I tried on transition, but I got the same result as scanning. Also the steady state was similar.
I will consider further. Also, please let me know your opinion if time permits.

Best regards.

[mzenker]

Hi,

so you have joule heating but no temperature change?
Does the joule heating correspond to your expectation?
Could you estimate the expected temperature change?
Are there any warnings or error messages in the solver output?

Matthias

[Franz Pichler]

Correct me if i am wriong , but I did not see a "Heat Capacity" in your material section. You only have density and heat conductivity. I think this is essential for the heat solver.

[mzenker]

Ooops, I overlooked that one when browsing the sif file. Franz, you are absolutely right!! I wonder if the Heat Solver throws an error or at least a warning in such cases.
mahichihi, could you examine your Elmersolver log or show it to us?
And yes, you absolutely need the heat capacity.

Matthias

[mahichihi]

Hi Franz and Matthias

Thank you as always.
I confirmed your comments. I have a question. Is there heat capacity required if using "steady state" on the simulation line ?
I think the heat capacity is not necessary if I calculate stady state. Is this interpretation correct ?

I also introduced a specific heat in steady state calculation to make sure, but I did not succeed in calculating the temperature.
(Total heating power and effective resistance are reasonable value compared with the result of manual calculation)

The sif is shown below. And I attach the log file.

Code: Select all

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

Simulation
  Max Output Level = 10
  Coordinate System = Cartesian
  Coordinate Mapping(3) = 1 2 3
  !Simulation Type = scannig
  Simulation Type = stady state
  Steady State Max Iterations = 100
  !Output Intervals = 1
  !Timestepping Method = BDF
  !BDF Order = 2
  !Timestep intervals = 5
  !Timestep Sizes = 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
!Permittivity of Vacuum = 8.8542e-6	   ! pF/um
End

Body 1
  Target Bodies(1) = 6
  Name = "Body Property 1"
  Equation = 1
  Material = 1
  Body Force = 1
  Initial condition = 1
End

Solver 2
  Equation = SaveScalars
  Variable 2 = volume current
  Filename = save scalars.dat
  Save Flux = True
  Procedure = "SaveData" "SaveScalars"
  Operator 2 = area
  Operator 1 = max
  Variable 1 = potential
  Exec Solver = After Timestep
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 = True
  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 3
  Equation = Result Output
  Output Format = Vtu
  Procedure = "ResultOutputSolve" "ResultOutputSolver"
  Output File Name = case
  Scalar Field 3 = joule heating
  Scalar Field 2 = temperature
  Scalar Field 1 = potential
  Vector Field 1 = volume current
  Exec Solver = After Timestep
End

Solver 4
  Equation = Heat Equation
  Variable = Temperature
  Procedure = "HeatSolve" "HeatSolver"
  Exec Solver = Always
  Stabilize = True
  Bubbles = True
  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(4) = 2 1 3 4
End

Material 1
  Name = "Material 1"

Heat Capacity =450 J/ks

!Heat Conductivity = 13.4     ! W/mK

  Heat Conductivity (3,3) = Real \
                                 13.4 0 0  \
                                  0 13.4 0     \
                                  0  0 13.4
!Electric Conductivity =  Variable Temperature; Real MATC "(-9E+07)tx+1E+12"

  !Electric Conductivity = ! Variable Temperature;real;290   926000;1290  834000;end;      ! S/m

  Electric Conductivity (3,3) = Real \
                                       926000 0 0  \
                                       0 926000 0     \
                                       0  0 926000

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

Body Force 1
  Name = "BodyForce 1"
  Joule Heat = True
End

Initial Condition 1
  Name = "InitialCondition 1"
  Temperature = 290
End

Boundary Condition 1
  Target Boundaries(1) = 1 
  Name = "BoundaryCondition 1"
  !Current Density = variable time;real;1 0;2 1e8;3 2e8;4 3e8;5 4e8;end
  !Potential = variable time;real;1 0.1;2 0.2;3 0.3;4 0.35;5 0.4;end
  Potential = 0.4
  Temperature = 290
  Save Scalars = True
End

Boundary Condition 2
  Target Boundaries(1) = 2 
  Name = "BoundaryCondition 2"
  Potential = 0
  Temperature = 290
  Save Scalars = True
End

[mzenker]

Hi,

I agree that physically the heat capacity does not have any influence on the result for the steady state case. But I would put it nevertheless (it doesn't hurt ), I am not sure what the heat solver does technically if it doesn't find the heat capacity.
Now whan you say you cannot calculate the temperature, do you mean that the variable exists, but its value doesn't change?
Can you try to estimate the expected temperature change? Do you still get a temperature change when the conductivity is isotropic?

Matthias

[annier]

Hi Mahichihi,
1. As said by Matthias, it is better if you put a value for heat capacity in the solver input file.

2. In Elmer Models Manaual, Model 1, the heat Eqn. is given as :

Heat Equation of Elmer Models Manual

heatEqn.png (35.78 KiB) Viewed 5341 times

So, when you model steady state conduction with no fluid based convection heat transfer, not adding the heat capacity will cause no error.

3. The geometry and mathematical formulation for your FEA seems quite complicated. Though i do not understand your purpose, i assume the series of pipes bent and such configuration possess a challenge to the numerical calculation.


4. Please create a model in simplified geometry first.


Yours Sincerely,
Anil Kunwar

[raback]

Hi

You need to tell HeatSolver to use the Joule heating as a source term. This should be achieved with keyword in Body Force section

Code: Select all

Joule Heat = True

This activates an internal logic that takes the "Potential" and "Electric Conductivity" and in each Gaussian integration point evaluates the joule heating.

-Peter