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## Induction Heating, Induction coil

Hi, i'm simulating an induction heating process. For the induction heating i used a mult-turn coil domain of 33 turns and a 7A current. When I get the results and looak at the coil parameters, it says that the resistence of the coil is 3,8 ohms. When I calculate it manually i get a much lower value (0.09 ohms).

How is it posible, is there any factor that i'm missing??

the coil has a diameter of 2mm and the frequency used is 30kHz

10 Replies Last Post 09-Jun-2013 7:45 am EDT
Posted: 4 years ago
Hi,

How you compute the analytical resistance ?
I suppose with R*rho*L/S.
The problem is the choice of S, because the current with 30kHz frequency is not uniform.
So, when you increase the frequency, you decrease S, so you increase R.

Jean-Marc

Posted: 4 years ago
Excuse me, I don't see your different figure.
In this assumption multiturn coil, you neglect the eddy current in the winding.
In the field area, it is the area of one turn coil and not the total one.

Jean-Marc

Posted: 4 years ago
In your case it's not possible to use multiturn coil option due to the skin effect.
This assumption is right when the skin depth is around the diameter of the one turn coil.
It's necessary to use coil group domain and describe all 77 turns.
In this cas you consider all the turn in series.
It's very simple with the menu transform > array of the geometry nodes.

Jean-Marc

Posted: 4 years ago
thanks for the replies, i did what you said and simulate using coil group domain instead of multiturn coil domain.
But i still don't understand why is the coil resistance so high.

For 30kHz the skin depth is 0.37 mm. So the effective area on the conductor will be a ring near the surface of the cable.

The effective area should be:

S=pi*(Rout^2-Rin^2)= pi*(1^2-(1-0.37)^2) =1,89 mm^2

The length of the coil is aprox. 3,23 meters so:

R=rho*L/S = 1.68e-8 * 3,42/1,89e-6 = 0.0304 ohm.

But when i simulate i get a resistance of 5 ohms.

On the other hand, when I use the current for the coil excitation the programs calculate the other coil parameters such as coil voltage or coil power. Is I use that coil voltage for the coil excitation i get completely different results.

The final temperature using current is 270ºC. The temperature using voltage at the same time is 140ºC

Posted: 4 years ago
The resistance you compute with the formula R=rho*L/S is only the copper resistance.
The resitance compute by COMSOL is a global resistance taking into account the susceptor inside the coil.

If you want to compute only the copper resistance, you must integrate resistive heating only in the coils, and with the currents or voltage deduce the copper resistance.

Jean-Marc

Posted: 4 years ago
So if I've unterstood correctly, that resistance is the resistance of the coil considering the effects of the workpiece's current on the coil in the same way the secondary current do in a transformer.

Posted: 4 years ago
Yes totaly exact.
It's the equivalent resistance of your tal circuit and same thing for the inductance.
Your total system is equivalent to a Requivalent and Lequivalent in series.

Jean-Marc

Posted: 4 years ago
You can after compute the efficency of your system.
With the equivalent resistance you can deduce the total active power (power generator) and with the inductance the total reactive power.
JM

Posted: 4 years ago
Hi, thank you for helping me with my induction heating system, I can unterstand it better now. Just another thing. I'm not sure how much do you know about induction heating but here's my question.

I designed the coil based on the AWG wire size table, that relates the conductor size with the maximum current that can flow through it. But when working with high frequencies, the conductor area isn't the hole area but the effective one due to skin effect, so I have to look on the table for the equivalent wire that has the same area, am I right?

The effective area of my coil is A=1.68 mm^2 (1.8 diameter wire; 30kHz), the equivalent coil with that area is the AWG 14, which can resist 5.9A, so the current density would be 5.9/1.68= 3.62 A/mm^2. However I've seen on the results of the simulation that the current density is very high (20 A/mm^2 or even higher) in some points. Can the coil resist so much current??

For the other hand, I've seen that most of the induction coils used for induction heating are cooled with water. Does my coil need to be cooled if the power of my circuit is 100W.

Link to the AWG wire size table: www.powerstream.com/Wire_Size.htm

Posted: 4 years ago
Thank you for helping me with my induction heating system, I can unterstand it better now. Just another thing. I'm not sure how much do you know about induction heating but here's my question.

I designed the coil based on the AWG wire size table, that relates the conductor size with the maximum current that can flow through it. But when working with high frequencies, the conductor area isn't the hole area but the effective one due to skin effect, so I have to look on the table for the equivalent wire that has the same area, am I right?

The effective area of my coil is A=1.68 mm^2 (1.8 diameter wire; 30kHz), the equivalent coil with that area is the AWG 14, which can resist 5.9A, so the current density would be 5.9/1.68= 3.62 A/mm^2. However I've seen on the results of the simulation that the current density is very high (20 A/mm^2 or even higher) in some points. Can the coil resist so much current??

For the other hand, I've seen that most of the induction coils used for induction heating are cooled with water. Does my coil need to be cooled if the power of my circuit is 100W.

Link to the AWG wire size table: www.powerstream.com/Wire_Size.htm

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