Induction heating

1. Jul 20, 2011

goedelite

The resistance per unit length of a conducting wire is proportional to the square root of the ratio of permeability to conductivity.

The power generated as heat may be expressed as I-squared x R: also as V-squared / R.

In induction, the EMF induced is determined by the rate of change of the magnetic flux (Faraday). Thus, it would seem to me that the appropriate calculation of the heat produced in the conductor is given by V(induced) - squared / R. As a result, the heat is proportional to the square root of the ratio of conductivity to permeability (since R is in the denominator, above.) Thus, the higher the permeability (as in a ferromagnetic conductor) the lower the heat loss.

This is contrary to the case where the current is constant, determined by an external source of EMF. In that case, the heat generated is proportional to the product of the current squared and the resistance: I - squared x R. In this case, the heat generated is proportional to the square root of the ratio of the permeability to the conductivity.

In discussing why a high permeability is desirable in induction heating, the latter analysis is usually presented. But the induction current is not determined directly by an external source of EMF but instead by the changing magnetic flux from the primary current. It seems to me that the constant quantity is the induced EMF and not the induced current.

2. Jul 20, 2011

Drakkith

Staff Emeritus
I THINK the issue is that the EMF determines the Current in the material which determines the heat generated. But I am not sure.

3. Jul 20, 2011

goedelite

Certainly, you are correct. The EMF does determine the current in the conductor being induction-heated. But, not directly. I have taken up the third-year college physics challenge by treating the problem as if it were two coupled circuits, coupled by their mutual inductance - as if it were a transformer situation. The two circuit equations can be solved for a periodic solution, and the secondary current determined in terms of V, M, L1, L2, mu1, mu2, the conductivities and the frequency, omega. From the secondary current, with reasonable approximations, I expect to see that the constant current is fairly well determined as you wrote by the EMF. However, I am not sure either. Thanks for your reply.

Meanwhile, if anyone has a quick answer that is convincing, please don't be hesitant!

4. Jul 20, 2011

Drakkith

Staff Emeritus
Does the changing magnetic field produce heat by the magnetic domains in the material changing directions?

5. Jul 20, 2011

goedelite

In the application I have in mind, the heating is the ordinary resistance heating. Surely, some of the heat does in fact come from such losses. But my interest is in the skin effect, the confining of the currents to the surface of the heated conductor. That confinement raises the resistance per unit length. In conductors of low permeability the confinement is low.

6. Jul 20, 2011

Drakkith

Staff Emeritus
Ah ok. Well, sorry I couldn't help!

7. Jul 20, 2011

goedelite

Not at all! A little conversation is very motivating. Thanks!

8. Jul 21, 2011

goedelite

The transformer equations are solved and presented on the net for the simple case I had in mind; saves me work. The result, in the form of the secondary current, does not help me in understanding the usual argument regarding induction heating and the need for a high permeability metal on the "burner". No doubt, that is true. Rather than treating the pan or pot like the secondary of a transformer, I should be treating the currents for what they are, eddy currents, rather than currents in a secondary. Therein may be the explanation I am looking for.

9. Jul 21, 2011

goedelite

I have been making the issue more complicated than it is. The usual discussion of induction cooking simply says for a given current induced in the pot, a high permeability pot is desirable. Therefore one calculates the power using Isquare R in comparing rather than Vsquare / R.

I can't discard all my aluminum cookware; so I'll avoid induction cookers. Thanks to all!
Problem solved!