Moving slab heating in the inductor - QuickField simulation example
Temperature distribution in a steel slab moving at 6 m/min through an inline induction heater, showing progressive heating along the travel direction.
How to find temperature distribution in moving steel slab induction heater?
Answer Typical applications Geometry
Given
Task
Solution
The heating time is t = Inductor length / slab speed = 0.6/0.1 = 6 seconds. We specify time-dependent power sources in transient thermal problem: Q * impulse(t,0,6). The solution time is set to be 10 seconds, so that we can simulate the heating and the subsequent cooling in the same problem.
Results
We are grateful to Selit S.R.L. for providing us with the idea of this model.
Engineering question
Follow one slab cross-section through the inductor, apply the layer heat sources for the transit time in a transient thermal model, and calculate its temperature distribution during heating and cooling.
moving slab induction heaters, strip processing induction lines, steel slab heaters
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Simulation problem
Problem Type
Plane-parallel problem of Transient heat transfer.
Inductor length is 0.6 m, slab speed is 6 m/min (0.1 m/s).
Power density in the slab layers was calculated in the related AC magnetic example: Q0 = 34.0 MW/m³, Q1 = 27.3 MW/m³, Q2 = 19.4 MW/m³ Q3 = 11.4 MW/m³ Q4 = 5.8 MW/m³ Q5 = 2.1 MW/m³ Q6 = 0.29 MW/m³.
Slab initial temperature is +900°C. Slab surface emissivity coefficient ε = 0.8, ambient temperature is +20°C.
Density of steel ρ = 7880 kg/m³.
Calculate the temperature distribution in the steel slab.
We analyze the temperature distribution in one slab cross-section which moves inside the inductor.
We neglect the heat redistribution along the slab and consider the heating of a cross-section due to the internal heat sources and cooling by the radiation from the surface only.
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