Solved with COMSOL Multiphysics 5.

I n d uc t i v e He a ti ng of a Cop p er
Cylinder
Introduction
The induced currents in a copper cylinder produce heat, and when the temperature
rises, the electric conductivity of the copper changes. Solving the heat transfer
simultaneously with the field propagation is therefore crucial for an accurate
description of this process.

The heating caused by the induced currents is called inductive heating. Generally
heating due to currents is also called resistive heating or ohmic heating.

A challenge in induction heating is that the high current in the induction coils requires
active cooling. This can be obtained by making the coil conductors hollow and
circulating water inside. Even for rather modest flow rates, the coolant flow becomes
highly turbulent which makes the heat transfer between conductor and fluid very
efficient. This example illustrates a simplified way of modeling water cooling based on
the assumption of turbulent flow and instantaneous mixing.

For mechanical support and electrical insulation, the cylinder and coil are embedded
in FR4 composite material.

Model Definition
The system to be solved is given by

–1
jωσ ( T )A + ∇ × ( μ ∇ × A ) = 0
∂T
ρC p – ∇ ⋅ k ∇T = Q ( T , A )
∂t

where ρ is the density, Cp is the specific heat capacity, k is the thermal conductivity, and
Q is the inductive heating.

The electric conductivity of copper, σ, is given by the expression

1
σ = ---------------------------------------------------
[ ρ 0 ( 1 + α ( T – T0 ) ) ]

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 Solved with COMSOL Multiphysics 5.1

where ρ0 is the resistivity at the reference temperature T0 = 293 K, α is the

temperature coefficient of the resistivity, and T is the actual temperature in the domain.

The time average of the inductive heating over one period, is given by

1 2
Q = --- σ E
2

The coil conductor is cooled by a turbulent water flow in an internal cooling channel.
This is emulated by a combination of a high effective thermal conductivity and a
homogenized out-of-plane convective loss term:

dM
C (T – T)
d t p in
Q c = ----------------------------------------
2πrA
dM
where is the water mass flow, Tin is the water inlet temperature, r is the radial
dt
coordinate and A is the cross-section area of the cooling channel.

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Results and Discussion

The temperature after 10 h is shown in Figure 1. The average temperature of the
copper cylinder has increased from 293 K to 346 K during this time. The current in
the coil has an amplitude of 2 kA.

Figure 1: Temperature distribution after 10 h.

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 Solved with COMSOL Multiphysics 5.1

Figure 2: The plot shows the temperature evolution in the center of the copper cylinder and
in the cooling channel.

Application Library path: ACDC_Module/Electromagnetic_Heating/

inductive_heating

Modeling Instructions
From the File menu, choose New.

NEW
1 In the New window, click Model Wizard.

MODEL WIZARD
1 In the Model Wizard window, click 2D Axisymmetric.
2 In the Select physics tree, select Heat Transfer>Electromagnetic Heating>Induction
Heating.

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3 Click Add.
4 Click Study.
5 In the Select study tree, select Preset Studies for Selected Physics
Interfaces>Frequency-Transient.
6 Click Done.

GLOBAL DEFINITIONS

Parameters
1 On the Home toolbar, click Parameters.
2 In the Settings window for Parameters, locate the Parameters section.
3 In the table, enter the following settings:

Name Expression Value Description

I0 2e3[A] 2000 A Current
T0 293[K] 293 K Reference temperature
r0 1.754e-8[ohm*m] 1.754E-8 Ω·m Resistivity at T=T0
al 0.0039[1/K] 0.0039 1/K Temperature coefficient
Rc 5[mm] 0.005 m Cooling channel radius
Ac pi*Rc^2 7.854E-5 m² Cooling channel x-section
Mt 1[kg/min] 0.01667 kg/s Cooling water mass flow rate
Tin 10[degC] 283.2 K Cooling water inlet
temperature

GEOMETRY 1

Rectangle 1 (r1)
1 On the Geometry toolbar, click Primitives and choose Rectangle.
2 In the Settings window for Rectangle, locate the Size section.
3 In the Width text field, type 0.2.
4 In the Height text field, type 0.5.
5 Locate the Position section. In the z text field, type -0.25.
6 Right-click Component 1 (comp1)>Geometry 1>Rectangle 1 (r1) and choose Build
Selected.

Rectangle 2 (r2)
1 On the Geometry toolbar, click Primitives and choose Rectangle.

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2 In the Settings window for Rectangle, locate the Size section.

3 In the Width text field, type 0.03.
4 In the Height text field, type 0.1.
5 Locate the Position section. In the z text field, type -0.05.
6 Right-click Component 1 (comp1)>Geometry 1>Rectangle 2 (r2) and choose Build
Selected.

Circle 1 (c1)
1 On the Geometry toolbar, click Primitives and choose Circle.
2 In the Settings window for Circle, locate the Size and Shape section.
3 In the Radius text field, type 0.01.
4 Locate the Position section. In the r text field, type 0.05.
5 Right-click Component 1 (comp1)>Geometry 1>Circle 1 (c1) and choose Build
Selected.

Circle 2 (c2)
1 Right-click Component 1 (comp1)>Geometry 1>Circle 1 (c1) and choose Duplicate.
2 In the Settings window for Circle, locate the Size and Shape section.
3 In the Radius text field, type Rc.

Form Union (fin)

1 Right-click Component 1 (comp1)>Geometry 1>Circle 2 (c2) and choose Build
Selected.
2 In the Model Builder window, under Component 1 (comp1)>Geometry 1 right-click
Form Union (fin) and choose Build Selected.

ADD MATERIAL
1 On the Home toolbar, click Add Material to open the Add Material window.
2 Go to the Add Material window.
3 In the tree, select Built-In>FR4 (Circuit Board).
4 Click Add to Component in the window toolbar.

ADD MATERIAL
1 Go to the Add Material window.
2 In the tree, select AC/DC>Copper.
3 Click Add to Component in the window toolbar.

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MATERIALS

Copper (mat2)
1 In the Model Builder window, under Component 1 (comp1)>Materials click Copper
(mat2).
2 Select Domains 2 and 3 only.
3 In the Model Builder window, expand the Copper (mat2) node, then click Linearized
resistivity (ltr).
4 In the Settings window for Property Group, locate the Output Properties and Model
Inputs section.
5 Find the Output properties subsection. In the table, enter the following settings:

Property Variable Expression Unit Size

Reference resistivity rho0 r0 Ω·m 1x1
Resistivity temperature alpha al 1/K 1x1
coefficient
Reference temperature Tref T0 K 1x1

ADD MATERIAL
1 Go to the Add Material window.
2 In the tree, select Built-In>Water, liquid.
3 Click Add to Component in the window toolbar.

MATERIALS

Water, liquid (mat3)

1 In the Model Builder window, under Component 1 (comp1)>Materials click Water,
liquid (mat3).
2 Select Domain 4 only.
The built-in water material does not provide the electric permittivity and the magnetic
permeability. Add those values.

3 In the Settings window for Material, locate the Material Contents section.

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4 In the table, enter the following settings:

Property Name Value Unit Property group

Relative permittivity epsilo 80 1 Basic
nr
Relative permeability mur 1 1 Basic

Increase the thermal conductivity of the water to model the efficient heat transport in
turbulent flow.

5 In the table, enter the following settings:

Property Name Value Unit Property group

Thermal conductivity k 1e3 W/ Basic
(m·K)

6 On the Home toolbar, click Add Material to close the Add Material window.

MAGNETIC FIELDS (MF)

Add a separate Ampère's Law feature in the copper regions to specify a
temperature-dependent resistivity.

Ampère's Law 2
1 On the Physics toolbar, click Domains and choose Ampère's Law.
2 Select Domains 2 and 3 only.
3 In the Settings window for Ampère's Law, locate the Conduction Current section.
4 From the σ list, choose Linearized resistivity.

Single-Turn Coil 1
1 On the Physics toolbar, click Domains and choose Single-Turn Coil.
2 Select Domain 3 only.
3 In the Settings window for Single-Turn Coil, locate the Single-Turn Coil section.
4 In the Icoil text field, type I0.

H E A T TR A N S F E R I N S O L I D S ( H T )
Set up the Heat Transfer boundary conditions.

Temperature 1
1 On the Physics toolbar, click Boundaries and choose Temperature.
2 Select Boundaries 2, 7, and 9 only.

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3 In the Settings window for Temperature, locate the Temperature section.

4 In the T0 text field, type T0.

Heat Source 1
1 On the Physics toolbar, click Domains and choose Heat Source.
2 Select Domain 4 only.
3 In the Settings window for Heat Source, locate the Heat Source section.
4 In the Q0 text field, type Mt*ht.Cp*(Tin-T)/(2*pi*r*Ac).

MESH 1
In the Model Builder window, under Component 1 (comp1) right-click Mesh 1 and choose
Build All.

STUDY 1

Step 1: Frequency-Transient
1 In the Model Builder window, expand the Study 1 node, then click Step 1:
Frequency-Transient.
2 In the Settings window for Frequency-Transient, locate the Study Settings section.
3 Click Range.
4 In the Range dialog box, type 600 in the Step text field.
5 In the Stop text field, type 36000.
6 Click Replace.
7 In the Settings window for Frequency-Transient, locate the Study Settings section.
8 In the Frequency text field, type 500[Hz].
9 On the Home toolbar, click Compute.

RESULTS

Temperature, 3D (ht)
The revolution plot shows the temperature distribution after 36000 seconds; compare
with Figure 1.

Create point data sets for plotting the temperature evolution in the copper cylinder and
in the cooling channel.

Data Sets
1 On the Results toolbar, click Cut Point 2D.

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2 In the Settings window for Cut Point 2D, locate the Point Data section.
3 In the r text field, type 0.
4 In the z text field, type 0.
5 Right-click Results>Data Sets>Cut Point 2D 1 and choose Duplicate.
6 In the Settings window for Cut Point 2D, locate the Point Data section.
7 In the r text field, type 0.05.

1D Plot Group 5
1 On the Results toolbar, click 1D Plot Group.
2 On the 1D Plot Group 5 toolbar, click Point Graph.
3 In the Settings window for Point Graph, locate the Data section.
4 From the Data set list, choose Cut Point 2D 1.
5 Click Replace Expression in the upper-right corner of the y-axis data section. From
the menu, choose Component 1>Heat Transfer in Solids>Temperature>T -
Temperature.
6 Right-click Results>1D Plot Group 5>Point Graph 1 and choose Duplicate.
7 In the Settings window for Point Graph, locate the Data section.
8 From the Data set list, choose Cut Point 2D 2.
9 On the 1D Plot Group 5 toolbar, click Plot.
The plot shows the temperature evolution in the center of the copper cylinder and
in the cooling channel; compare with Figure 2.
Finish the modeling session by saving a representative model thumbnail.

Temperature, 3D (ht)
Click the Zoom Extents button on the Graphics toolbar.

ROOT
1 In the Model Builder window, click the Root node.
2 In the Settings window for Root, locate the Presentation section.
3 Find the Thumbnail subsection. Click Set from Graphics Window.

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