Copper inductor with a steel tube - QuickField simulation example
A 2-turns copper inductor is used to heat the cylindrical steel part.
How to calculate the temperature distribution in the steel part after 10 seconds of heating?
Answer Typical applications Geometry
Given
Task
Solution
Results
Engineering question
Solve the AC-magnetic problem for eddy-current losses in the steel, transfer the losses to a transient heat-transfer model, and calculate the temperature after 10 seconds.
induction heating coils, steel billets, induction heating systems
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Simulation problem
Problem Type
Axisymmetric multiphysics problem of AC magnetics coupled to Transient heat transfer.
Inductor alternating current I = 500 A, frequency f = 22 kHz.
Relative permeability of steel μ = 400, electric conductivity is 5 MS/m.
Thermal conductivity of steel λ = 20 W/K*m, density ρ = 7800 kg/m³, specific heat is 420 J/K*kg.
Initial temperature is +25°C. Emissivity coeffitient is 0.5, convection coeffitient is 5 W/K*m.
Calculate the temperature distribution in the steel part after 10 seconds of heating.
Penetration depth in the steel part is less than 0.1 mm at this frequency. To get accurate results it is required to build very dense mesh on the surface of the steel part.
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