0
\$\begingroup\$

I have been taught in my machine's class that eddy current losses are proportional to the square of frequency*flux density, this seemed so fair that I never questioned this. But my machine design book mentions that

Induced emf = 4.44*freq*Max Flux Density * Area of flux path * No of turns, hence as long as supply voltage is maintained constant, the product of flux density and frequency remains constant even if frequency is changed, and hence there is no change in eddy current losses with frequency.

Now this is also in accordance with emf equation of transformer, at least it seems so, but how is that possible? If that's the case why would we go for high frequencies in eddy current heaters?

I mean, suppose I have source 1 of 230V, 50Hz supplying a coil wound on iron core, now if I replace source 1 with 230V, 100kHz, will the eddy current heating remain same? How that could be the case? Where is my (or book author's) interpretation of equations wrong?

\$\endgroup\$
  • 2
    \$\begingroup\$ Induced EMF =! Eddy current losses. \$\endgroup\$ – winny Mar 17 '18 at 15:49
  • \$\begingroup\$ @winny yes but if induced emf is constant, which shall be case if supply voltage is constant, will then the product of Bm and f remain constant? In that case how would eddy losses change? Because they're directly proportional to square of the product? \$\endgroup\$ – Deep Mar 17 '18 at 15:57
  • \$\begingroup\$ At higher frequencies, the induced current uses a thinner region of the metal (right at the surface ---- check into Skin Effect) and that region becomes hotter with less power needed. \$\endgroup\$ – analogsystemsrf Mar 17 '18 at 16:02
  • \$\begingroup\$ @analogsystemsrf but is the current same as in previous case? or is the induced emf same? or is it due to change in impedance of eddy current paths? Sry but can you please explain little bit in detail ^^". \$\endgroup\$ – Deep Mar 17 '18 at 16:26
2
\$\begingroup\$

Eddy losses is a component of core losses. It essentially occurs when there's a change in the magnetic field of said core. We engineers would perform eddy-current testing to see and measure these "cracks" within the core to determine how much eddy current there is.

When we detect some flaws, heat will be generated. This means that there will be kinetic energy that will convert into thermal energy. This is known as eddy losses.

The textbook equation for eddy losses is \$ P_e=K_eB_m^2t^2f^2V\$ watts.

Eddy losses depend on frequency, thus you will get a different value when you have 50Hz and 100kHz.

Someone in the comments mentioned the Skin Effect. This is a phenomenon where at high frequencies, the magnetic field will not be able to penetrate the inside of the core. The equation I mentioned a few lines higher is only true if you disregard the Skin Effect. HOWEVER, if you keep the same magnetic field with increasing frequency, it will in fact increase the eddy current generated.

\$\endgroup\$
  • \$\begingroup\$ I wish some author of my books had mentioned that the formula is not general one and applies only to certain range of frequencies till the skin effect is negligible, the fact that at my level books only mentions the formula directly without any derivations mad it hard for me to realize the fact it's not the general one \$\endgroup\$ – Deep Mar 17 '18 at 16:57
  • \$\begingroup\$ Well in textbooks, and I'm assuming that your textbook is power systems and distribution, it wouldn't mention the Skin Effect because power generation around the world only relies on 50Hz or 60Hz depending on where you are and it would be irrelevant. But once you talk about wireless transmissions like radio waves and microwave technology, the Skin Effect is more prevalent. \$\endgroup\$ – KingDuken Mar 17 '18 at 17:04
  • \$\begingroup\$ Ah, you're right. \$\endgroup\$ – Deep Mar 17 '18 at 17:17
2
\$\begingroup\$

Induced EMF is certainly proportional to frequency and flux density. That's bound up in standard transformer theory.

Faraday's law: Induced voltage = N\$\frac{d\phi}{dt}\$

The higher the frequency the higher \$\frac{d\phi}{dt}\$ is.

Now if that induced emf connects to a resistor, the power dissipated is proportional to \$V^2\$. That should be also fairly obvious.

So, if power is proportional to voltage squared then it is also proportional to both frequency and flux density squared.

There are complications in that skin effect does play a part at higher frequencies and, a shorted turn of low resistance of course has inductance and that inductance will resist the flow of current more and more at higher frequencies.

\$\endgroup\$
  • \$\begingroup\$ So the guy behind increased heating with higher frequencies is actually the skin effect, right? And if frequency is increased but not to such extent so as to cause any appreciable skin effect, the eddy current losses would remain essentially constant, right? \$\endgroup\$ – Deep Mar 17 '18 at 16:54
0
\$\begingroup\$

Voltage (Back-EMF) and Power (Losses due to eddy current) are not linearly related.

P=V^2/R, so power losses are proportional to the square of back-EMF.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.