Slab induction heating power - QuickField simulation example
A pre-heated steel slab is placed in the inductor. Calculate the Joule heating power.
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
Set up a plane-parallel QuickField AC Magnetics problem for a slab induction heater and evaluate Joule heating power from computed field results.
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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