Slab induction heating power - QuickField simulation example
Joule-heating power distribution in a preheated steel slab placed inside an induction-heating coil under the specified high-frequency excitation.
How to find joule heating power in slab induction heater?
Answer Typical applications Geometry
Given
Task
Solution
Results
We approximate the power distribution in the layers with the rectangular pulses. These pulse magnitudes are listed in the table below and are used for the further thermal analysis.
We are grateful to Selit S.R.L. for providing us with the idea of this model.
Engineering question
Solve the AC-magnetic field with the temperature-dependent steel conductivity and integrate the induced Joule losses to obtain the heating-power distribution across the slab.
slab induction heating systems, longitudinal flux heaters, steel strip heaters
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Simulation problem
Problem Type
Plane-parallel problem of AC magnetics.
Slab width in z-direction is 1200 mm.
Coil current 12.5 kA (r.m.s), frequency 150 Hz,
Relative permeability of steel μ = 1 at the temperature above 760°C.
Electrical resistivity value depends on the temperature.
Calculate the Joule heating power distribution in the steel slab.
In QuickField we specify the electrical conductivity, which is reciprocal to the resistivity.
We do not know the actual temperature distribution, so as an initial approximation we consider the temperature to be 900°C in all parts of the slab.
# r, mm Q, W/cm³
0 55.. 60 34.0
1 50.. 55 27.3
2 40.. 50 19.4
3 30.. 40 11.4
4 20.. 30 5.8
5 10.. 20 2.1
6 0.. 10 0.29
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