Slab induction heating power
QuickField simulation example
A pre-heated steel slab is placed in the inductor. Calculate the Joule heating power.
We are grateful to Selit S.R.L. for providing us with the idea of this model.
Problem Type
Plane-parallel problem of AC magnetics.
Geometry
Slab width in z-direction is 1200 mm.
Given
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.
Task
Calculate the Joule heating power distribution in the steel slab.
Solution
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.
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.
| # | 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 |
- Video: Slab induction heating. Watch on YouTube
- Download simulation files (files may be viewed using any QuickField Edition).