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Main >> Applications >> Sample problems >> Dielectric heating

Dielectric heating

QuickField simulation example

Simulation of the wood drying process* in a high frequency heater. The heater creates an alternating electric field between capacitor plates. Heating occurs due to the dielectric loss in the wood placed inside the capacitor.

Problem Type
Plane-parallel multiphysics problem of AC conduction and transient heat transfer.

Geometry
Lz = 3 m.
High frequency food dryer kiln. Dielectric heating The heater creates an alternating electric field between capacitor plates. 150 mm 100 mm Ø1700 mm Metal sheets 0 0 0 0 U~ U~ U~ Wood Tank Air

Given
Relative permittivity of wood ε = 2, loss tangent tan(δ) = 0.1
Voltage U~ = 1 kV (r.m.s.), frequency f = 6.87 MHz.
Initial wood temperature is +20°C. Maximum heating temperature is +70°C.

Task
Calculate temperature rise in the wood.

Solution
Electrical conductivity of materials should be set in AC conduction problems. We can take into account dielectric losses by specifying the apparent value of conductivity**:
g = 2πf·εε0·tan(δ) = 2·3.142·68700000·2·(8.854·10-12)·0.1 = 76.4 uS/m.
The conductivity of wood may change during the drying process. It may be needed to simulate several problems with different conductivity of the wood. In this example we consider the conductivity to be independent of time.

When temperature hits maximim limit the high frequency source is switched off. After temperature falls to the min limit the high frequency source is switched on.

Results
Distribution of power losses in wood and temperature rise after 2 hours of heating.
High frequency food dryer kiln. Dielectric heating.

References:
*High frequency vacuum wood dryer
**Dielectric loss