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 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;

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
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.

• Video:
  
  
  Magnetic problem
  
  Coupled thermal problem
  
  See more videos in QuickField Analysis for Induction heating webinar.
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