Joule Heating of Tungsten Wire


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.

Load the finite element package.

Multiphysics Model

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.

Set up material parameters for the tungsten wire.

Electrostatics Model

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 Guass's law the stationary potential field satisfies Poisson's equation:

Set up a 3D electrostatics model.

Heat Transfer Model

The heat equation (1) is used to solve for the temperature field in a heat transfer model:

For a steady-state heat transfer model the transient term in (2) are set to zero. Since a solid is modeled, the internal velocity also vanishes and the heat equation simplifies to:


is the density ,
is the heat capacity ,
is the thermal conductivity ,
is the electrical conductivity .

Set up a 3D stationary heat transfer model.


The radius of the tungsten is . The simulation domain and the geometric shape of the wire are described in the following plot.


Specify the parameters of the geometry.

Next we construct the 3D domain by specifying the path of the tungsten wire.

Specify and visualize the centerline of the wire.
Construct the 3D tungsten wire.

Now the boundary mesh region and the full element mesh can be generated. In order to get a good result, a finer than the default grid is used for the mesh generation.

Discretize the simulation domain.

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.

Set up the functions leftEndQ and rightEndQ to represent the wire's end point locations.

A tolerance of is applied at both ends to account for numerical error in the discretized domain.

Solve the PDE Model

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.

Electrostatics Model

There are two types of the boundary conditions involved in the electrostatics model. At both ends of the wire an electric potential difference is applied.

Set the electric potential boundary conditions at both ends of the wire.

On the remaining boundary a default electrical insulation boundary condition: is implicitly used.

Specify the electrostatics PDE with the model parameters.
Solve the electrostatics PDE model with NDSolve.

Next we visualize the electric potential field within the tungsten wire.

Set up a legend bar and ContourPlot options for the electric potential field plot.
Visualize the electric potential field.

Since the electric conductivity is homogeneous throughout the wire, the electric potential decreases linearly from left to right .

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.

Heat Transfer Model

The electrical heat generation is coupled to the stationary heat equation (3) as the source term on the right hand side, which is known as the electromagnetic heat source.

Joule's first law (4) states that the heat generated within an object is equivalent to the product of its conductivity and the square of the potential gradient.

Set up the electromagnetic heat source .

Before solving the heat transfer PDE model, it is intuitive to inspect the distribution of the electromagnetic heat source .

Set a legend bar and ContourPlot options for the electromagnetic heat source .
Inspect the electrical heat generation on the cross section of the wire ().

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.

Calculate the total electrical heat generation .

The total heat generation is shown to be . Next, the heat transfer model is constructed to solve for the steady-state temperature distribution.

There are three types of the boundary conditions involved in the heat transfer model. At both ends of the wire the temperature is fixed at the ambient temperature .

Set up an ambient temperature.
Set up a temperature surface boundary condition on both ends of the wire.

On the side surfaces the tungsten wire is losing heat through both thermal convection and radiation. The Stefan-Boltzmann constant , the emissivity and the heat transfer coefficient are given by:

Specify the heat transfer PDE with the model parameters.
Solve the heat transfer PDE and monitor the time/memory usage.

Post-processing and Visualization

To inspect the effect of the Joule heating, we visualize the temperature distribution within the tungsten wire.

Set up a legend bar and ContourPlot options for the temperature field plot.
Visualize the temperature field.

In the middle portion of the wire the temperature reaches its maximum value at . At both ends, however, the temperature is held at the ambient temperature .

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

Construct the boundary element mesh for the boundary integration.
Calculate the heat loss through thermal conduction at the left end.
Calculate the heat loss through thermal conduction at the right end.

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.

Calculate the heat loss through thermal radiation on the side surfaces of the wire.
Calculate the heat loss through thermal convection on the side surfaces of the wire.
Calculate the total heat loss of the wire.

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 Loss37.13100

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 .

Inspect the total electrical heat generation previously computed.

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.