Industrial

Educational

Scientific

Engineering questions

Sample problems

Examples gallery

Step-by-step tutorials

Verification examples

Programming examples

Distributive examples

Success stories

Customers

Main >> Applications >> Sample problems

Inductively Heated Ceramic - QuickField simulation example

This example is prepared by Didier Werke AG, InduCer Group, Abraham Lincoln Str. 1,65 189 Wiesbaden, Germany

Ceramic nozzle of a Tundish, mounted in the mould wall of a horizontal continuous steel caster, heated by an air-cooled induction coil. Conventional heating (e.g. gas burners) cannot keep steel from freezing in a Tundish nozzle during casting, so the nozzle itself is made electrically conducting and inductively heated - a patented technology (InduCer) developed by Didier Werke AG.

Engineering question

How to calculate impedance and heat-source distribution in inductively heated ceramic?

Answer
Solve the axisymmetric AC-magnetic problem to obtain electrical impedance and spatial electromagnetic losses, then transfer the losses to transient heat transfer for the ceramic temperature field.

More questions →

Typical applications
induction-heated ceramics, dielectric ceramic heaters, ceramic processing heaters

Inductively Heated Ceramic

Download

Simulation problem

Problem Type
Axisymmetric multiphysics problem of AC magnetics coupled to Transient heat transfer.

Geometry
Inductively heated ceramic nozzle of a Tundish Vertical Tundish vessel with didurit and rubinit insulation lining, also under the liquid steel, and a sectioned nozzle with its 7-turn induction coil steel shell didurit rubinit liquid steel induction coil nozzle 1 2 3 4 5 6 7

Given
Induction current I = 1880 A per turn, frequency f = 10 kHz.
Liquid steel temperature T = 1850°C.
Convection coefficient in the coil cooling channels α = 1200 W/m²K.

Task
Calculate the coil impedance and spatial heat-source distribution in the inductively heated ceramic nozzle, and use it to find the resulting temperature field.

Solution
The problem is solved in two coupled steps: an axisymmetric AC magnetics problem gives the eddy-current losses and the coil impedance. These losses are then applied as a heat source in a transient heat transfer problem.
The copper coil is air-cooled and split into two parallel cooling lines: turns 1-4 and turns 5-7. The air heats up along each line. The convection boundary condition on every turn used its own local air temperature.
More detailed description (in PDF format).

Results
The magnetic field strength |H| reaches about 115 kA/m near the coil turns, and the induced losses (hence the heating) concentrate in the mould and steel parts closest to the winding.
The two coil cooling circuits (turns 1-4 and 5-7) heat the cooling air to about 460 K and 545 K respectively.
inductively heated ceramic

Related examples