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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.

Geometry:

solenoid model

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;

Problem:
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
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.

Download Download simulation files