# VARS and metering

As we all know, the average power factor on domestic electrical loads is not unity so the power company has to deliver on average some
lagging VARS to consumers. The I^2.R transmission losses from this, as well as the larger installed generating capacity, is charged out to all consumers either as higher KWH cost or delivery surcharges, etc. There's no free lunch!
I happen to have a few AC capacitors to hand, not many... about 100 MFD in total, which I calculate to be about 1/2 KVAR (leading) on a 120 VAC, 60 Hz North American supply (not quite twice this in Europe on 240 VAC and 50 Hz.) Here are some questions:
Q1. Should I connect them across my supply permanently as leading VARS? (about 4 1/2 amps AC all the time at a PF of zero, i.e. all leading VARs) Q2. What would that do to my house metering? At a given KW demand: A) would my electric meter register more slowly with the leading VARS connected, or B) read even higher due to the reactive current being there all the time, even when I am not consuming any real power, or C) read just the same as it reads only KW regardless of PF? (My knowledge of current metering technology is somewhat deficient!) Q3. I figure that, at \$0.06 per KWH, and counting only, say, 5% of the leading VARS for the I^2R losses (fuel costs), my leading 1/2 KVAR would be worth about \$0.04 per day to the power company at the margin since I am "soaking up" (providing?) 1/2 KVAR of my neighbour's lagging VARS that the power company no longer would have to supply to heat the wires, i.e. install capacity and burn fuel to generate. What's the chance they would knock \$1.20 per month off my bill? OK, silly question! Q4. Is the above a correct analysis... if not, what have I missed?
Cheers, Roger
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In article

Since you are asking these questions, you obviously do not fully understand what you are doing.
Before getting into such things tell us what your capacitors are. in particular, what kind of service are the supposedly designed for? What voltage? What current?
Gill
--
As the years go by, dying just before having to fill out a tax return has merit.

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On Sun, 8 Nov 2009 12:50:58 -0800 (PST), Engineer

No.
Nothing. Residential meters don't register reactive power, only real. Nothing you do with the PF is going to make any measurable difference in your meter.

Clearly not.

Clearly not.

Bingo. NA power meters are just that; power meters. They don't measure VARs at all.

Indeed it is.

That your power meter doesn't measure reactive power at all. That you're increasing the odds of burning down your house or electrocuting someone.
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Typically,reactive *energy* meters (kVarh) are installed in large consumers, factories etc. and *always* in medium-voltage consumers. There's no benefit for you, even if you used a commercially-available capacitor bank (capacitors are switched off when there's no reactive load, because they consume *real*power, being no ideal capacitors).
--
Tzortzakakis Dimitrios
major in electrical engineering
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Engineer wrote:

Ah, but US service entrance is really 240 VAC. Or would you only connect to the center-tap/neutral for 120? Why? What is the voltage rating of your 'AC capacitors'? This is a key bit of information.

To what end? It will not affect your metering (see Q2). The risk to your home would depend on how well your work is done and protective devices you use. (you were planning on fuses or ckt breakers, weren't you??)

C is correct. Residential metering reads only kWhr, regardless of PF.

A bit faulty analysis here. Your kVAR is still creating current flow between your house and the neighbors, so that heating is still there. The only 'savings' is the difference in cable between the neighbor and the transformer versus the neighbor and you. This could be zero if the transformer is between you and the neighbor (or even negative since your supply cable now carries current 24/7). On the other side of the transformer, your 4.5 amps of current would be proportionately smaller and the losses pretty much nil (if transformer steps from 240V/4.1kV (17.3:1), then the I^2R savings are 1/300 of what they are on the low side (for the same R). When stepped up to transmission voltages, the 4.5 amps is 1/480th and the I^2R losses are 1/230400th (if the R were the same).

None. Besides, you haven't really figured out that you're saving them anything.

The losses drop in the hi-voltage supply by turns-ratio squared. So the savings upstream of the local distribution transformer are very, very tiny for your 4.5 amp kVAR 'generation'.
The actual part of the circuit that has a reduced current is only a sub-section of the line between you and the neighbor and the distribution transformer. So the I^2R losses you are saving is only in that subsection and is pretty small (no where near the 5% of 0.5kVAR you estimated).
Worst case, you'll actually increased the energy losses on the system by causing current flow in your service feeder 24/7, even when the neighbor's kVAR load is nil.
The rate schedule does not provide for any such setup, you would never see a dime.
(of course there are professionally designed systems that the utility uses for such 'voltage support' and under the right operation / design, they *do* provide savings)
daestrom
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Therefore, the best way to save money with these capacitors: Sell them to a fool who thinks they can use them to lower their residential electric bill! (or to an industrial customer who does have a meter that measures VARS)
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In article

No. A static capacitive load is not going to help. Current in a house varies wildly throughout the day. All you'll be doing is causing current to flow even when all your electrical devices are turned off, which wastes energy.

No. Electric meters are built to measure and bill you for real kWh. You don't pay for the reactive element because the meter doesn't see it.