A 2-turns copper inductor is used to heat the cylindrical steel part.
Inductor current I = 1000 A;
Inductor frequency f = 22 kHz;
Relative permeability of air and copper μ = 1;
Relative permeability of steel μ = 400;
Electric conductivity of steel σ = 5000000 S/m;
Thermal conductivity of steel λ = 20 W/K*m;
Volume mass density of steel ρ = 7800 kg/m³;
Specific heat of steel C = 200 J/K*kg;
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.
Current density distribution in the steel part
Temperature distribution in the steel part
See the Inductor_magn.pbm and inductor_heat.pbm problems for the magnetic and thermal analysis respectively.
Coupled thermal problem