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Planar heater 3D

QuickField simulation example

The planar heater is placed between two layers of glass of irregular shape. The heater consists of a serpentine stripe carrying electric current. QuickField is used to calculate the temperature distribution in the glass.

Problem Type
A 3D problem of Heat Transfer.

Geometry
There are 10 mm paddings between the glass edges and the heater.
Planar heater The planar heater is placed between two layers of glass of irregular shape. The heater consists of a serpentine stripe carrying electric current. QuickField is used to calculate the temperature distribution in the glass. Glass Indoor Outdoor Current 511 mm 4 mm 6 mm 995 mm

Given
Heater power 600 W/m².
Indoor air temperature is +20°C, convection coefficient is 8 W/m²*K.
Outdoor air temperature is -15°C, convection coefficient is 23 W/m²*K.

Task
Calculate temperature distribution in the glass.

Solution
Film thickness (~0.01 mm) is much less than the glass thickness. So we do not model film as a body, but as a surface with a surface power heat specified instead. For simplicity we assume that the heating power is evenly distributed in the serpentine stripe.

Result
Glass inner surface average temperature is +5.3°C, outdoor surface average temperature is +2.4°C.

Average heat flux generated by the planar heater is (171.76 - 51.02) / 0.42977 = 281 W/m².
Planar heater