Copper inductor with a steel tube
QuickField simulation example
A 2-turns copper inductor is used to heat the cylindrical steel part.
Problem Type
Axisymmetric multiphysics problem of AC magnetics coupled to Transient heat transfer.
Geometry
Given
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.
Task
Calculate the temperature distribution in the steel part after 10 seconds of heating.
Solution
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.
Results
- Video: Copper inductor with a steel tube. Watch on YouTube
- Download simulation files (files may be viewed using any QuickField Edition).