Joule Heating of Tungsten Wire
|Introduction||Solve the PDE Model|
|Multiphysics Model||Post-processing and Visualization|
The Joule heating effect (or resistive heating effect) denotes a phenomenon where electric energy is converted into thermal energy as an electric current flows through an object. This effect is commonly found in devices such as electric heaters, incandescent light bulbs and fuses.
In the following model a voltage of is applied to a tungsten wire. Heat is then generated within the wire due to the Joule heating effect. Part of the generated heat dissipates from the side surfaces by both thermal convection and radiation to the surrounding air. The remaining heat propagates towards electrode pads at both ends and leaves the wire by thermal conduction. The electrode pads on both ends act like heat sinks and cool the wire down to the ambient temperature .
The electric potential field and the total heat generation is simulated by an electrostatics model. Based on these result the temperature distribution will then be modeled by a heat transfer model. Finally, the amount of heat loss through each surface will calculated and compared.
The symbols and corresponding units used here are summarized in the Nomenclature section.
Please refer to the information provided in "Heat Transfer" for a more general theoretical background for heat transfer analysis.
There are two physical domains involved in this application: the electrostatics model and the heat transfer model. The coupling between two models is one-way only since the temperature field depends on the electric potential but the electric potential field does not depend on the temperature. This type of problem is considered as a sequential multiphysics model, and will be solved in two separate steps.
First the electrostatics model model is built to simulate the potential field in the wire. The heat transfer model is then constructed and uses the potential field previously computed to show the Joule heating effect of the wire.
In this model the electric potential field is assumed to be independent of the temperature . That is, the electrical conductivity is kept at a constant value at all time. By Gauss’s law the stationary potential field satisfies Poisson's equation:
The heat equation (1) is used to solve for the temperature field in a heat transfer model:
In the following sections we will set up the boundary conditions and calculate the heat flux at both ends and the side surfaces. To do so the functions leftEndQ and rightEndQ are defined to determine whether a point locates on the corresponding surface.
In the following section the electrostatics model will be solved first to simulate the potential field of the wire. The heat transfer model is then constructed to show the heating effect of the electric current.
With the potential field in hand we can calculate the corresponding heat generated by the electric current. The steady-state temperature field is then solved for by the heat transfer model in the following section.
Note that most heat is generated at the inner corners of the wire due to the higher current density in these regions. The total electrical heat generation can then be calculated by the volume integration.
Next, by computing the boundary integral we can calculate and compare the heat loss on each surface of the wire. As shown in the heat transfer tutorial the heat flux by thermal conduction, convection and radiation are given by
Since the temperature distribution is symmetric along the wire, the amount of the heat conduction should be the same at both ends. Here the difference of is the numerical error that can be reduced by using a finer mesh.
The following table summarizes the heat loss through thermal conduction, convection and radiation. It is seen that the tungsten wire loses the majority of the heat it generated through heat conduction at both ends of the wire.
|Power [W]||Ratio [%]|
|Total Heat Loss||37.13||100|
The total heat loss of the tungsten wire is found as . This value can be verified by utilizing the law of energy conservation. That is, in a steady state the total heat loss must equal to the internal heat generation .
Note that the net heat generation matches with the total heat loss . A small deviation of is a numerical difference introduced by the discrete nature of the finite element mesh used in both cases to compute the heat generation. Using a finer mesh will reduce the numerical error.