I recently had an question from a customer about how to set a boundary with a specified thermal flux. Although there is no explicit command for boundary flux in TDiff and HeatWave, there is a simple procedure using the volumetric source capabilities of the programs that is quite versatile. Using it, you can set normal flux over surfaces of any shape.
The TDiff example shown in the first figure illustrates the method and gives an opportunity to check the code accuracy. Consider heating a 5.0 cm radius aluminum sphere. In the mesh, the sphere is divided into two regions: 1) the bulk of the volume and 2) an outer layer of thickness Δ = 1.0 mm. The natural boundary condition at the outer radius is a perfect insulator (i.e., zero heat flux). To begin, we need to do a little homework. Clicking on the Thermal properties command in the Help menu of TDiff, we find the following physical properties for aluminum:
k (thermal conductivity): 210 W/m2-degK
ρ (mass density): 2700 kg/m3
Cp (specific heat) 900 J/kg-degK
The 5.0 cm sphere has the following properties:
V (volume): 5.236E-4 m3
S (surface area): 3.1416E-2 m2
M (mass) = 1.4137 kg
Q (total energy added)
In the first solution, there is a uniform volumetric heat source q (W/m3) in both regions. In this case, the exact solution for the time-dependent temperature is:
M*Cp*(dT/dt) = Q,
where Q is the total input power, Q = q*V. For a temperature change of 100 °C in 10 s, we find that Q = 1.2723E4 W and q = 2.430E7 W/m3. The blue line of the second figure shows the resulting temperature at the center of the sphere, a linear rise to 100 °C.
In the second solution, we want to introduce the same total power Q as a uniform thermal flux on the surface of the sphere. We define a volumetric source q = Q/(SΔ) = 4.0498E8 W/m3 over the surface region. There is no volumetric source in the inner region. Because the outer surface of the solution volume is an insulator, all power is transferred to the aluminum body. In this case, we expect to see a temperature variation in space because of the thermal conduction time. The first figure shows the temperature distribution at t = 10.0 s. The black line of the second figure shows the variation of temperature with time at the center of the sphere. After an initial lag, the temperature rises with the same slope as the uniform-source solution.
In summary, here's how to define a thermal flux incident on a region:
- Create a new region, a thin layer on the surface of the original region.
- Assign the same thermal properties to the new region and introduce a volumetric source q = Q/(SΔ), where Q is the total power added,Δ is the layer thickness and S is the layer surface area.
If you want to try to the TDiff solution, here are links to the input files.
thermalflux.min
thermalflux.tin
Note that the MIN file includes the DCorrect command to ensure that the calculation is stable.
Thermal flux example, temperature at the center of the sphere.
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