basic question about HID lamps and power issues

I've seen lots of discussion about the harmonics and other issues from HID lamps (or more properly, tehir power supplies).
When I was working at an ISP I noticed some very headachey readings when looking at the supply and neutrals for the racks, but we were well below the current ratings so left things as is.
Anyway... I know just enough about this to realize I don't know enough, so I figured I'd better ask a couple of questions even if they sound foolish.
In my residential building complex we've got about 1,000 apartments, and I doubt there's enough usage of CFLs or HIDs or switching power supplies to be a major issue.
However, we have a standalone muliti-story garage, which is illuminated 24 hrs/day by about 35 kw of high pressure sodium luminaires (with another 5 kw or so at night) getting fed via 3 phase AC. There's perhaps one or two addiotional kw of demand for some office equipment and (fluorescent) lighting. And a short occassional kw when opening/closing the garage doors.
This building is separately metered.
If I ask the utilty if the wave forms are screwing up the meter readings, would they laugh me out of the room? Is there any valid issue here?
Thanks.
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No, they wouldn't but unless they are a rather large utility they probably won't have harmonic analysis equipment.
Do you fear some weird electrical stuff going on with third harmonics?
Some background information. Third harmonics will pass through three pahse transformers and end up on all phases excatly the same. Some say they exist in the neutral but only if you measure using the neutral, as a reference to a phase conductor. This appears as being in the neutral by measurement. To prove this further, if you measure a voltage between phases and do not involved the neutral they usually do not appear as they are common mode. The clue in this is to connect critical loads phase to phase if possible. Power Factor correction capacitors can help absorb some of the harmonics.
In a three phase system there are three waveform starts per cycle. Assume the first at 0 degrees, the second a 120 degrees and the third at 240 degree (equi-distance). Consider each phase to have third harmonics starting at the same time (0,120,240) as their respective cycle crosses zero. The next third harmonic waveform will now cross zero at exactly the same time as the other two phase (fundamental) voltages. This makes them all in phase with each other and looks like it is in the neutral because of measurement techniques as discussed above.

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How does that work? I thought they could only reduce IV phase angles.
Nick
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Capacitive reactance is lower to higher frequencies. Very simple electrical laws.
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Interesting. I wonder how to calculate the power factor improvement for, say, a 4.7 uF cap in parallel with a 150 watt 277 V load with a 10% 3rd harmonic distortion.
The first step might be to calculate the original power factor, with no cap.
This seems different and more promising:
https://www.galco.com/circuit/PFCC_har.htm
Harmonic currents can be significantly reduced in an electrical system by using a harmonic filter.
In its basic form, a filter consists of a capacitor connected in series with a reactor tuned to a specific harmonic frequency. In theory, the impedance of the filter is zero at the tuning frequency; therefore, the harmonic current is absorbed by the filter. This, together with the natural resistance of the circuit, means that only a small level of harmonic current will flow in the network.
A 600 Hz filter might help a power supply that clips 10% of an 60 Hz peak.
Nick
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Notch filters are always good selective units. Our problem is filtering 10MW ++ loads full of harmonics. That is a lot of heat to disapate. hmmmmmm... building heat in the winter and hot water? Why bother when tons of heat is thrown away in the main transformers via cooling fans and pumps. Trouble is the utility buldings run metered in with the big bulk of the consumers and is not apparent to anybody in the utility. **SIGH**
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The R in these series RLC circuts broadens the notch and uses real power. How much would a 4.7 uF cap in parallel with a 150 watt 277 V load with a 10% 3rd harmonic distortion raise the power factor?
With 1/(377C) = 564 ohms at 60 Hz and 277 V and L = 1/((2Pi180)^2x4.7x10^-6) = 0.166 henrys in series to resonate at 180 Hz, limiting 60 Hz ripple current requires Z = sqr(R^2+(wL-1/(wC)^2) = 564, ie R^2+(377x0.166-564)^2 = 564^2, ie R = sqr(564^2-251291) = 258 ohms. With how many real watts?
How much would that raise the power factor?
Nick
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Man, you're pushing my math skills here. I used to gobble this stuff for breakfast nut now I'm not even sure I could do it with a Geritol injection...LOL
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