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Copper inductor with a steel tube

A 2-turns copper indutor is used to heat the cylindrical steel part.

Problem Type:
Axisymmetric problem of the magneto-thermal coupling.


solenoid model

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
Induction heating simulation: Current density distribution

Temperature distribution in the steel part
Induction heating simulation: Temperature distribution

See the Inductor_magn.pbm and inductor_heat.pbm problems for the magnetic and thermal analysis respectively.

View movie Video:
1.Magnetic problem setup.
2.Coupled thermal problem setup.
See more videos in QuickField Analysis for Induction heating webinar.

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Download icon Download simulation files (files may be viewed using any QuickField Edition).

There are no restrictions applied to the QuickField Student Edition postprocessors.
You can view field maps, make plots, calculate integrals and print pictures in the same way that the Professional Edition users do.