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Induction heating of a cylinder

  1. Oct 23, 2011 #1
    Hello!

    I am making som calculations on induction heating of a metallic cylinder inside
    a solenoid through which an AC current passes, but the answer I get seems to
    be completely unrealistic (10 kW with a 50 Hz, 5 Amps, 230 V rms), but I can't seem
    to find my mistake.

    What I think is wrong is the expression for the magnetic field inside the solenoid. I
    have used
    [tex] B = \mu n i [/tex]

    But is this really correct for a time varying field?
    As far as I know it is derived using amperes law in the static case when the displacement
    current is zero, but I haven't found any passing comment or remark or anything anywhere
    that this equation is valid only in the static case. So is this formula correct even in the
    time varying (sinusoidal) case? If not, what should it be replaced by?

    Basically, what I did for the rest of the calculation was simply to get the E-field by starting
    from the maxwell equation

    [tex] \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}[/tex]

    and integrating both sides over a circle of radius r. The righthand side was simplified using the expression
    for the B-field above, and the left hand side was simplified using Stoke's Theorem to obtain
    an expression for the E-field at radius r from the cylinder axis.

    Then I calculated the power from

    [tex] P = \iiint_{\text{cylinder}}{\sigma E^2 \text{dV}} [/tex]

    Another thing that worries me is that I find that the E-field is proportional to the distance
    r to the cylinder axis. When I read about induction heating on the internet the skin effect is
    always mentioned, but it didn't pop out my calculations anywhere... But on the other hand,
    these calculations came straight from Maxwell's equations so I find it hard to believe that they are completely wrong...
     
  2. jcsd
  3. Oct 23, 2011 #2

    Low-Q

    User Avatar
    Gold Member

    First, 10kW out of 5A at 230V is impossible.

    An induction heater is working pretty much like an ordinary transformer. You have a primary winding and a secondary winding. However, the secondary is the one you want to heat up instead of transfer electric current through a lamp or a AC motor.


    Vidar
     
  4. Oct 23, 2011 #3
    Okay, I figured it out. It turns out the problem was that the magnetic field inside the solenoid
    was not given by that expression for B, but was actually given by a rather intricate expression
    with bessel functions and such.

    Anyway, here is a link to an article discussing just this, if anyone should be interested :P

    http://jap.aip.org/resource/1/japiau/v17/i3/p195_s1 [Broken]
     
    Last edited by a moderator: May 5, 2017
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