BusBar

In this example, we will model a busbar simply modeled as rectangular parallelepiped.A differential electrical potential is applied to the entry/exit of the busbar. We note respectively \(V_0\) the electrical potential on the entry and and \(V_1\) on the exit.

1. Running the case

The command line to run this case in 2D is

mpirun -np 4 feelpp_toolbox_electric --case "github:{path:toolboxes/electric/busbar}" --case.config-file 2d.cfg

The command line to run this case in 3D is

mpirun -np 4 feelpp_toolbox_thermoelectric --case "github:{path:toolboxes/electric/busbar}" --case.config-file 3d.cfg

2. Geometry

The busbar conductor consists in a rectangular cross section extruded along the x axis.+ In 2D, the geometry is as follow: In 3D, this is the same geometry, but extruded along the z axis.

3. Input parameters

Name Description Value Unit

\(Lx\)

internal radius

1

\(m\)

\(Ly\)

external radius

2

\(m\)

\(Lz\)

angle

\(\pi/2\)

\(rad\)

\(V_D\)

electrical potential

9

\(V\)

3.1. Model & Toolbox

  • This problem is fully described by the Electric model, namely a poisson equation for the electrical potential \(V\) with Dirichlet boundary conditions on entry /exit.

  • toolbox: electric

3.2. Materials

Name Description Marker Value Unit

\(\sigma\)

electric conductivity

omega

\(4.8e7\)

\(S.m^{-1}\)

3.3. Boundary conditions

The boundary conditions for the electrical probleme are introduced as simple Dirichlet boundary conditions for the electric potential on the entry/exit of the conductor. For the remaining faces, as no current is flowing througth these faces, we add Homogeneous Neumann conditions.

Marker Type Value

V0

Dirichlet

0

V1

Dirichlet

\(V_D\)

Lside, Rside, top*, bottom*

Neumann

0

*: only in 3D

4. Outputs

The main fields of concern are the electric potential \(V\), the current density \(\mathbf{j}\) and the electric field \(\mathbf{E}\). // presented in the following figure.

5. Verification Benchmark

The analytical solution is given by:

\[V(x) = V_0 + (\frac{x}{L}-1) (V_1-V_0)\]

Note that we may get an expression for the resistance \(R\) of the busbar from this equation. We recall that \(R\) is defined as \(R = V_D/I\) where \(I\) stands for the total current flowing in the busbar (\(V_D\) corresponds to the differential applied voltage).

By definition:

\[I = \int_{V0} \mathbf{j} \cdot \mathbf{n} \,d\Gamma\]

From Gauss law we have: \(\mathbf{j} = \sigma\,\mathbf{E} = -\sigma \nabla V\). It follows:

\[R = \frac{1}{\sigma} \frac{Lx}{S}\]

with \(S=Ly*Lz\).

We will check if the approximations converge at the appropriate rate:

  • k+1 for the \(L_2\) norm for \(V\)

  • k for the \(H_1\) norm for \(V\)

  • k for the \(L_2\) norm for \(\mathbf{E}\) and \(\mathbf{j}\)

  • k-1 for the \(H_1\) norm for \(\mathbf{E}\) and \(\mathbf{j}\)