# BCs and correctBoundaryConditions() for an electric field and current solver

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January 21, 2014, 06:56
BCs and correctBoundaryConditions() for an electric field and current solver
#1
Senior Member

Join Date: Oct 2013
Posts: 397
Rep Power: 17
Hello,

I'm working on a solver for calculating electric field and current in an inhomogenous conductivity field.

I'm currently evaluating the validity of the solution and I'm wondering which boundary conditions make most sense, and which fields need to use field.correctBoundaryConditions(). To my understanding correctBoundaryConditions() adjusts the values of the boundary so they match their gradient conditions after the values of the cells were (explicitly) modified.

Here's a simplified excerpt of the solver code:

Code:
```forAll(sigma, iter)
{
sigma[iter] = someValues;
}
sigma.correctBoundaryConditions();
int i = 0;
while(i < 10)
{
i++;

//quasistatic electric potential
fvScalarMatrix phiEqn
(
fvm::laplacian(sigma, phiE)
);
Foam::solverPerformance phiSolve = phiEqn.solve();

//electric field
//EField.correctBoundaryConditions();

//current density
j = sigma * EField;
//j.correctBoundaryConditions();

Info << "phiE: " << min(phiE) << "/" << max(phiE) << endl;

if(phiSolve.nIterations() == 0 && phiSolve.converged())
{
Info << "phi equation converged in " << i << " iterations" << endl;
break;
}
}```
Now I am examining the current through electrodes, which are modelled as boundary conditions. I can either specify zeroGradient BC, i.e. same conductivity as the neighbouring cell, or specify the conductivity of the metal used in the electrode (which is usually much much higher). The latter method produces no realistic results at all.

Electric field and current both use zeroGradient BC, although I would prefer to use a BC that has zeroGradient in normal direction and fixedValue 0 in tangential direction. Is there such a BC?

For testing, I've used zeroGradient for the conductivity and I'm having some numerical problems. Below you can see the output of the calculated current using different methods, with j being the current density:

Quote:
 without correctBoundaryConditions(only sigma), bc for sigma at electrodes: zeroGradient Sf() & j over electrode1: 0.0555727 Sf() & j over electrode2: -3.00958e-08 Sf() & j over electrode3: -0.0555726 Difference between electrodes:-1.62838e-09(-2.93018e-06%) Integral(j) over electrode1: (-3.69829e-19 -0.000252525 -0.0555727) Integral(j) over electrode2: (-5.99056e-23 -5.249e-09 -2.61724e-08) Integral(j) over electrode3: (5.1536e-17 -0.0555726 -0.00882784) Difference between electrodes5.11661e-17 -0.0558252 -0.0644005) Magnitude(Integral(j)) over electrode1: 0.0555732 Magnitude(Integral(j)) over electrode2: 2.66935e-08 Magnitude(Integral(j)) over electrode3: 0.0562694 Difference between electrodes: -0.000696218(-1.25279%) with correctBoundaryConditions(sigma, EField, j), bc for sigma at electrodes: zeroGradient Sf() & j over electrode1: 0.0555727 Sf() & j over electrode2: -2.98116e-08 Sf() & j over electrode3: -0.053637 Difference between electrodes:0.00193558(3.48298%) Integral(j) over electrode1: (-3.69829e-19 -0.000252525 -0.0555727) Integral(j) over electrode2: (-5.99056e-23 -5.13672e-09 -2.60005e-08) Integral(j) over electrode3: (5.1536e-17 -0.053637 -0.00882784) Difference between electrodes5.11661e-17 -0.0538896 -0.0644005) Magnitude(Integral(j)) over electrode1: 0.0555732 Magnitude(Integral(j)) over electrode2: 2.65031e-08 Magnitude(Integral(j)) over electrode3: 0.0543587 Difference between electrodes: 0.00121455(2.18549%)
I expected to see the same results for "Sf() & j" (inner product with the surface area vector) and Magnitude(Integral(j)), because the current should be orthogonal to the surface, since the electric potential is constant on the surface.
I also expect that the difference between the electrodes should be (close to) zero, meaning that the current is conserved. However, there is a significant error.

I am using a simple wedge mesh generated with blockMesh. The electrodes have different angles towards each other, one at the top, one at the bottom, and one at the side of the cylindrical mesh, separated by isolating material with zero conductivity BC. Right now the cells are completely orthogonal.

I believe this could be caused by the somewhat unfitting zeroGradient BC for electric field and current, but I don't know if there is such a condition as described above.

I would be very grateful if someone could give me some hints.
It is important to me to get current conservation (atleast to a certain accuracy).

By the way, can anyone explain to me the meaning of phiSolve.nIterations() and phiSolve.converged() ?
I suspect this is related to the way the solution matrix is approximated, but if it is, why does the loop in the solver above has to run multiple times until the solution converges? I would expect the solver to iterate until conversion is reached. Is there some artificial limit that stops it before convergence?

Last edited by chriss85; January 22, 2014 at 03:55.

 May 2, 2022, 21:25 #2 New Member   Mostafa Join Date: Oct 2021 Posts: 20 Rep Power: 3 hello chriss85, I know it has been a long time, but I'm also solving electric field after computing electric potential in multiregion solver. i could not get continous electric field across interface. do you have any suggestion how to solve the problem thanks in advance

 May 3, 2022, 10:32 #3 Senior Member   Join Date: Oct 2013 Posts: 397 Rep Power: 17 I don't remember from my head, I think it was about prescribing either current continuity or potential. It might be easier if you have a current source and know what current needs to be pass through the interface in total. In that case you could solve one region first to calculate the potential and then apply that potential to the other region. This will only work if you can assume a vastly different conductivity so that you have an equipotential surface at the interface. Otherwise I think you will need some iterative solution. Another possibility would be using a third mesh that encompasses both meshes, solve the equation on that mesh and then project the solution on the other two regions.

May 3, 2022, 12:29
#5
New Member

Mostafa
Join Date: Oct 2021
Posts: 20
Rep Power: 3
Quote:
 Originally Posted by chriss85 Another possibility would be using a third mesh that encompasses both meshes, solve the equation on that mesh and then project the solution on the other two regions.
i tried something like that

i calculated the electric potential over fluid and solid and coupled them using compressibleBaffle class then i mapped the phi on the solid mesh and in the fluid mesh over on mesh contianing both meshes to have one elecctric potential field using MapField , after thaty i took the gradient successfully and got the right Electric field

the problem is mapField took long time about 3 hrs to map field (phi) from solid domain to its projection on the full mesh and other 3 hrs to map field (phi) from fluid domain to it's projection on the full mesh , so i cannot do this all time