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
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