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Joule heating

This source provides users with the ability to add Joule Heating contributions to any thermal solver.

Properties​

The option solves an equation for the electrical potential, V, of the form

∇ ∙ (σ∇V)=0 \div (\sigma \grad V) = 0

Where σ\sigma (sigma) is the electrical conductivity. The thermal source is then given by

Q˙=(σ∇V) ∙ ∇V \dot{Q} = (\sigma \grad V ) \dprod \grad V

A sample result is shown below Joule heating example

Usage​

The option is specified using:

heating
{
    type            jouleHeatingSource;
    active          true;

    jouleHeatingSourceCoeffs
    {
        anisotropicElectricalConductivity no;
    }
}

The electrical conductivity can be specified using either (see usage):

  • If not present the sigma field will be read from file (standard field sigma)
  • If the sigma entry is present the electrical conductivity is specified as a (potentially uniform) function of temperature using a Function1 type, e.g.
heating
{
    type            jouleHeatingSource;
    active          true;

    jouleHeatingSourceCoeffs
    {
        anisotropicElectricalConductivity no;

        // Optionally specify sigma as a function of temperature
        sigma           table
        (
            (0      127599.8469)
            (1000   127599.8469)
        );
    }
}

  • If the anisotropicElectricalConductivity flag is set to 'true', sigma should be specified as a vector quantity.

Further information​

Tutorials

Source code