Planar film heater - QuickField simulation example
Electric current is passing through the contact electrode to the thin film heater.
How to find temperature rise in planar film heaters?
Answer Typical applications Geometry
Given
Thermal conductivity of glass 1 W/m*K.
Task Solution
Joule heat is produced in a very thin film. Film thickness t is much less than that of a glass. We simplify the model for heat transfer analysis and replace the film with the edge with the known heat flux boundary condition:
The convection coefficient value is reciprocal to the surface resistance value:
Result
Glass temperature is calculated using simplified model with the film replaced by the boundary condition.
References:
Engineering question
Set up a plane-parallel QuickField Steady-state Heat Transfer problem for a planar film heater and evaluate temperature rise from computed field results.
planar film heaters, surface heating sheets, thin resistive heaters
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Simulation problem
Problem Type
Plane-parallel problem of DC Conduction and Heat Transfer.
Model thickness Lz = 1 m.
Indium tin oxide film thickness t = 170 nm, sheer resistance Rs = 12.6 Ohm/sq.
Electrical conductivity of copper 56 MS/m.
Electric current I = 5 A.
Inside air temperature +20°C, surface resistance** 0.13 K·m²/W.
Outside air temperature -15°C, surface resistance** 0.04 K·m²/W.
Calculate the heater power and the glass temperature.
First we simulate DC conduction problem and calculate the heat power.
Film resistivity* ρ = Rs * t = 12.6 * 170e-9 = 2.142 μΩ*m. In QuickField DC conduction analysis we specify the electrical conductivity, that is reciprocal to the resistivity:
σ = 1 / ρ = 1/2.124e-6 = 467 kS/m.
Heat power [W/m²] = Volume power [W/m³] * t.
αinside = 1/0.13 W/(K·m²),
αoutside = 1/0.04 W/(K·m²).
Film heater power is 1.848e9 W/m³ * 170nm = 314 W/m².
* Wikipedia, Sheet resistance.
** ISO 10077-2:2012 Thermal performance of windows, doors and shutters.
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