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Eddy currents induced in let's say a metal increase with the increase of rate of change of magnetic field. They increase with a square of both peak magnetic field and frequency.

Now imagine a system with one coil and a metal being put next to it so that the eddy currents are induced inside of it that are heating the metal. If I have the coil connected to a power grid with a 50 Hz frequency, the eddy currents will be smaller than if I had it on 100 Hz. But what puzzles me with this is that if I just increase the frequency of the power grid, the eddy current losses will skyrocket, while power expended calculated by P=U*I will stay practically the same.

If I increase the frequency by a lot I would by this logic break the 100 % efficiency. So what actually happens here? Where does the additional power come from when I greatelly increase the frequency?

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  • \$\begingroup\$ The inductive reactance of the metal also increases with frequency so it's not as clear cut as you think. \$\endgroup\$
    – Andy aka
    Commented Mar 9, 2017 at 11:35
  • \$\begingroup\$ I know this. Yet the eddy currents Will keep increasing with the frequency. To my knowledge the amplitude of input current does not increase. So HOW does the grid provide more energy if not by the current amplitude? \$\endgroup\$
    – MaDrung
    Commented Mar 10, 2017 at 6:50

2 Answers 2

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Eddy losses increase input resistance (reducing input power efficiency) and increase output resistance (reducing output power efficiency).

If you hold the input voltage constant, the input current will decrease, and the output power will decrease.

If you hold the input power constant, the output power will decrease.

If you hold the input current constant, the input voltage will increase, and the output power will decrease.

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  • \$\begingroup\$ Ok, so I imagine that resistance felt because of the load is not Real resistance, but is caused by the counter emf which is dependent on the input frequency? If so, then why even bother to increase the frequency if that will not change the heating performance? As I see it, in induction cooking it does. But there they use a resonator to store that energy in the system. \$\endgroup\$
    – MaDrung
    Commented Mar 31, 2017 at 5:53
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There is no 100% efficiency. There is a input power, output power and loss. A loss is further split into a copper loss and iron loss. The copper loss is caused by the resistance of the wires/windings, while the iron loss is caused by magnetizing the iron core. Eddy currents are one another type of loss in the iron core. The core made of steel laminations reduces eddy currents. The answer of your question is: There is a decrease of the output power or there is an increase of input power, to maintain the energy balance: \$P_{in}=P_{out}+P_{loss}\$

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  • \$\begingroup\$ I know all of that. I asked what actually happens, when frequency of the input power is increased. It should increase eddy currents, but input power doesn't care about frequency since it's calculated by I*U at every point of time. \$\endgroup\$
    – MaDrung
    Commented Mar 9, 2017 at 9:47
  • \$\begingroup\$ I don't understand what you mean by input power doesn't care ...U*I...Also the output power and power loss doesn't care, but the power balance is always achieved, no matter what goes in,out. \$\endgroup\$
    – Marko Buršič
    Commented Mar 9, 2017 at 10:10
  • \$\begingroup\$ I know that. I'm asking how. :D Because if I increase the frequency of the input, the eddy currents Will increase. But the input current stay bassicaly the same?? only the frequency increases. And if you calculate the power draw it is the same, yet the output power caused by eddy currents isn't. :) There is something foundamental I don't understand here. I want to know what. Your comments about input=output don't help with that. I know that. \$\endgroup\$
    – MaDrung
    Commented Mar 9, 2017 at 14:59

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