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In Induction Heating, what does the heat output depend on?

  1. Feb 9, 2016 #1
    Wasn't sure whether to post this here or in EE forum. I've been learning about induction heating for the past few hours and scoured every video on youtube and every web article I could come across and I still lack an understanding of how to determine the heat output of an induction heater.

    I understand clearly how a resistive heater works - but with induction heating I'm not sure. Does the heating of the conductor depend on the resistance of the conductor itself, or upon the inductor? If the former, could the breaker trip if too large of a piece of metal (inducing a excessively high resistance) was placed in the inductor?

    Basically I am trying to figure out how I can use induction heating to heat a pipe with 1000w of power, but I really do not know which part of the system to vary. For instance, with a resistive heater on 120v, I would just need to make sure the heating element has 14 Ohms of resistance (120v/8.333amps). Yet the calculations for induction heating I have no idea how to configure the system to get my desired output.
  2. jcsd
  3. Feb 10, 2016 #2


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    But don't ask me to do the calculations.
  4. Feb 10, 2016 #3
    I've read the Wiki several times. I have no issue understanding how induction heating works in principle. What I don't understand is where the numbers/calculations come from. Very few guides on the topic go in to any depth about it, they only explain the basic principle of it.
  5. Feb 11, 2016 #4


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    The heating effect depends on the characteristics of the heated material. Mainly its resistance but magnetic materials have some additional losses.
    For a first approximation consider the heated material to be the shorted secondary of a transformer. As you may have noted in the past, a shorted transformer gets HOT. That's largely because of the high current flowing in the resistance of the windings. You have the same effect with inductive heating, you are inducing a voltage into a short circuit. The power dissipated in watts will be E^2 / R. Treat the system as a transformer. If there is good coupling between the coil and the work, the induced voltage is proportional to the turns ratio. Then you need to know the resistance of heated material to get the power.

    Now for the confounding factors.
    1) You won't get "good" coupling between the coil and the work. (80-90% ?)
    2) The resistance of the work will vary with temperature
    3) "Skin Effect" will affect the resistance of both the work and the coil.

    The skin effect is a phenomenon where an alternating current tends to flow near the surface of a conductor. At power line frequency it is on the order of 1/4 to 1/2 inch. That's why you won't find electrical wire much larger than that. The skin depth, how deep into the material current flows, is inversly proportioanl the sqrt(frequency). If you are using RF heating, a thin walled pipe works just as well as a thick walled one because the current is concentrated near the surface. Take this into account when calculating the resistance of the work.
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