Electrical – Harmonic resonance calculation in power-factor correction

Apparently you have a inductive load to a AC power supply, and want to cancel the reactive part of the load by putting a capacitor accross it. You mention values Rs and Rl, but since you provided no definition for these, we can't tell how they are hooked up and therefore their relevance.

However, if I understand the problem correctly, this is mostly about finding the capacitance that cancels the inductance. From basic circuits you know that the impedance magnitude of a capacitance is 1/ωC, and for a inductance is ωL. Setting these two equal and solving for C yields C = 1/ω²L = 1/(2πf)²L = 880 µF.

The only general rule of thumb I can think of is to balance loading whenever possible.

In a three phase system, when harmonics are generated at any point, every mitigating measure is some variation of getting them to dissipate as harmlessly as possibly, through or across some kind of reactance. One way of getting them to dissipate is by simply allowing them to flow through devices that are closer to the non linear load than the service entrance. The load creating a harmonic can only supply it so much power, and by sinking that through another load, it doesn't make it's way back to the mains as well. Some harmonics negate entirely when balanced. Occasionally facilities that get in trouble for harmonic generation or effective power factor can get back within limits simply by balancing their phase loading.

Of course, if this is the strategy, then the linear loads near harmonic generators will have to be rated to handle the extra heat generated.

And it seems like a higher per unit impedance would attenuate harmonics leaving or entering a facility; so generally speaking, I suppose it is safe to say that a higher short circuit capacity would allow higher amplitude harmonics simply by not stopping them.

As far as your gigantic light bulb theory, we can reason it out. If they draw the same amount of instantaneous power, then they draw the same amount of instantaneous current, which suggests they have the same impedance at the fundamental frequency. Of course, that impedance will be higher for higher order harmonics, making the motor less effective at harmonic sinking than the resistive light bulb. In reality, though, that light bulb is enormous, AND you need three of them - and you probably don't need that much light. But, maybe this makes a case for placing lighting loads alongside noisy loads?

I suppose the next question would have to be, can a halogen luminaire bulb handle the harmonics?