This source provides users with the ability to add Joule Heating contributions to any thermal solver.
The option solves an equation for the electrical potential, V, of the form
\[ \div (\sigma \grad V) = 0 \]
Where \( \sigma \) ( sigma) is the electrical conductivity. The thermal source is then given by
\[ \dot{Q} = (\sigma \grad V ) \dprod \grad V \]
A sample result is shown below
The option is specified using:
heating
{
type jouleHeatingSource;
active true;
jouleHeatingSourceCoeffs
{
anisotropicElectricalConductivity no;
}
}
The electrical conductivity can be specified using either (see usage):
heating
{
type jouleHeatingSource;
active true;
jouleHeatingSourceCoeffs
{
anisotropicElectricalConductivity no;
// Optionally specify sigma as a function of temperature
sigma table
(
(0 127599.8469)
(1000 127599.8469)
);
}
}
Tutorials
Source code
| Would you like to suggest an improvement to this page? | Create an issue |
Copyright © 2018 OpenCFD Ltd.