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Skin effect, frequency and heating

  1. Oct 8, 2012 #1
    Hi.

    I'm trying to understand why smaller skin depths are better for induction heating.

    Skin effect means that the highest the frequency is, the thinest the skin depth is (or the depth is simply smaller because of the material). That means the section of the workpiece with electric current in it is smaller, and according to R=ρ*L/S (ρ - electrical resistivity, L - length, S - section area) the resistance is higher, thus heating more.

    But if the resistance is higher, the current decreases so that the power remains the same, right? So what makes the difference when the skin depth is changed?
     
  2. jcsd
  3. Oct 8, 2012 #2

    Think of it like this:

    The resistance only gets higher because the temperature rises, due to increased current. Yes the current will decrease with increased temperature and resistance, but as soon as this current decreases the temperature decreases and the resistance will follow and the current will once again increase. The temperature will now increase again.

    This isn't exactly what is going on but it gives you an idea.
     
  4. Oct 8, 2012 #3
    Ok but what difference makes a change in skin depth so that lower skin depths are better for heating?
     
  5. Oct 9, 2012 #4
    For a start, higher frequencies with shorter wave lengths will propagate less, in terms of distance than a lower frequency wave. Think of it like this if because higher frequency waves are oscillating more rapidly their energy dissipates sooner, thus they travel less than lower frequency with longer wave lengths.

    Knowing this if we consider a high frequency EM wave impinging on the surface of a block of metal, you will see eddy currents form, a number of skin depths deep into the metal. The number of skin depths will depend on the resistivity of the metal block and the frequency of the impinging wave. High resistivity i.e. poorer conductivity means that the current will get attenuated more and as a result the temperature will increase. Yes the current will reduce, but voltages will "build up" because of the "electron pile up" caused by the high resistivitiy due to the temperature increase. As a result the power will remain constant (I*V), and it is the power dissapation that causes the heating effect.

    In addition these swirling eddy currents create their own magnetic field that consequently opposes the magnetic field that created it. Thus it effectively acts as an insulator ensuring that the impinging magnetic field does not penetrate deeper and further into the metal and cause eddy currents at greater thicknesses.

    With all this current confined to the surface of the metal the current density is much greater at the surface and as a result heating will take place as you have so much current passing through a relatively thin metal conductor.

    Hope this helps.


    J
     
    Last edited: Oct 9, 2012
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