120V from both legs

I was wondering if there is a simple, correct, and safe way to draw 120V
equally from both legs of US standard two leg household service. What I
want to do is reduce my electricity bill by eliminating uneven draw from
each leg. US meters run as if the maximum draw from either leg is being
pulled from both legs. When your refrigerator compressor runs, you pay for
the same amount of current being pulled from the other leg as well.
Most electricians will say that by putting an even number of 120V breakers
on each leg will balance your draw on both legs, but I think that is a
crock. Draw on either leg will almost never be balanced with this method.
What do I do, wait for the refrigerator compressor to turn on one leg so I
can run my shop vac on the other leg?
I thought that if I could change the phase of one leg 180 degrees, I could
connect it in parallel with the other leg.
How would I go about this?
Help please
Reply to
John Doe
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WRONG! You have been fed a pack of lies and have believed them. You can draw all of the current from one leg, half from each, or any other combination and the meter will measure the correct usage. Anyone with any metering knowledge at all knows this and anyone with the simplest meter test equipment can prove it.
The way to lower your bill is to use less electricity.
Charles Perry P.E. who is never amazed at the outrageous utility meter stories he hears, but is always amazed that people believe them
Reply to
Charles Perry
OK. I guess I'll do some experimenting. I know that I sould take everything with a grain of salt, but the source that's telling me it's bunk (usenet) is the same place that I read this (mis)information in the first place.
The truth is out there, but where?
Reply to
John Doe
"Handbook for Electricity Metering", Edison Electric Institute, 1992 (although there is a newer version now). It is the bible for electric metering.
Also, for about $1000 a day, I will be more than happy to test your theory. I always warn customers ahead of time when it is obvious that the testing will not get the answer they want, but if they insist, I take the money and do the testing.
Charles Perry P.E.
Reply to
Charles Perry
Sounds like this may be an urban legend created as a perversion of the following reasoning:
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thread thl3419501793d
Reply to
John Doe
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google thread thl3419501793d
Could be (I didn't actually read Dan Lanciani's spiel in your link - though if it makes you more comfortable you can consider that Dan Lanciani acknowledges having little knowledge while Mr. Perry and others here are very much knowledgeable). I'll second the debunking. The meter knows how much energy you are using and you get billed accordingly. Certainly as far as unbalanced load is concerned anyways. A simple way to verify it is to just call the utility. I'm sure they'll happily tell you on the phone that it works that way, and they may be able provide some document, or website, to that effect. I am assuming you would be content that the utility would not outright lie to you on the matter.
j
Reply to
operator jay
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> google thread thl3419501793d
By the way, unless you have some reason to believe otherwise, your home probably has reasonably balanced load. Utilities like for loads to be reasonably well balanced so that their lines, transformers, generators, etc. can carry maximal loads and provide balanced voltages. If your loads were not all that well balanced you probably wouldn't have much to worry about.
j
Reply to
operator jay
to draw 120V
service.
sure you can, thats how most residential services are set up. Commercial service will often be three phase with one line (the stinger) higher than 120v to neutral..be careful with those.
uneven draw from
leg is being
runs, you pay for
as well.
120V breakers
that is a
with this method.
on one leg so I
Thats about it.
degrees, I could
Thats tricky... its never done...but it might work. its probably not legal though..and screw ups would tend to blow the utilitity company transformer off the pole...they just HATE when that happens.
You probably shouldnt..but yer mind is sharp...maybe you can come up with another approach a way to isolate yer strategy so if it gets crossed up it wont take out the pole transformer.
What you could do is fit a 240 v X 120 volt transformer across the incoming line.. then the output, 120 volts would be all from the common balanced source. See a wiring diagram on the bonding and neutral issues... those are important.
You would loose about 3% in the transformer... there should be a net savings.
Phil Scott
Reply to
Phil Scott
to draw 120V
service. What I
uneven draw from
leg is being
runs, you pay for
as well.
them. You can
other
Anyone with any
simplest meter test
There are harmonics issues with unbalanced loads that can screw up the power factor and run the tab up on yer bill dramatically... not something I was overly familiar with but recent review has turned up some good books on the subject.. primarily applicable to large commcl or industrial installations.. not dramatically to the typical home.
Phil Scott
he hears, but is
Reply to
Phil Scott
perversion of the
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> google thread thl3419501793d
your link - though
Lanciani
others here are
meter knows how
Certainly as far
verify it is to
the phone that
document, or website,
utility would
I tend to agree...however Ive had them lie to me regarding demand charge and power factor adjustments at industrial facilities nation wide, repeatedly...many dont even have a clue what those factors are... the companies in some cases have even changed the terms on the billing statement to occlude demand charges.
You can save an industrial firm a lot of money in many cases by getting rid of a bad harmonics situation (current wasted to the neutral and between lines due to wave distortions etc...back emf from motors and transformers...and such as large scale transormer stage lighting). I dont do a lot of this. when I do I spend $500 and rent a set up that clamps to all 3 lines for amperage, reads voltage and the neutral... and computes the power factor issues hour to hour over the duration of the test.. I've seen some horrendous losses due to these factors...some back fed distortions from other nearby facilities. Desk top computer power supplies can backfeed serious distortion into a buildings power grid. Not single home though..but a problem in a large building.
I had been entirely unware of the problem until a client raised the issue and I studied up... it can get complex. But there are books on the situation and instrument rental outfits that specialize in the test eq (25k for a real good set up)...rental is much less.
Phil Scott
Reply to
Phil Scott
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> > google thread thl3419501793d
P.P.S. "Top-posting" is generally frowned upon, FYI.
Reply to
operator jay
The presence of unbalanced loads does not create, or indicate, the presence of harmonics. Also, residential customers are only billed for kWh usage, so power factor is not an issue. Severe cases of harmonic distortion can increase I^2 R losses, but our testing and research shows that these losses are small.
Charles Perry P.E.
Reply to
Charles Perry
Back emf from motors causing harmonics?
You do realize that harmonics have a negligable effect on kWh, and no affect on displacement power factor? I really don't know of an industrial metering setup that penalizes for harmonics. Perhaps you could point one out.
Charles Perry P.E.
Reply to
Charles Perry
The effects of one house with an unbalanced load on a utility system stop at the first distribution transformer that serves that house, it is not passed along to the rest of the system.
The utility itself may have an unbalance problem with the load or number of houses on each of the 3 three legs of a three phase feeder but that is the problem of the utility (not customer caused).
BTW, 20 or 30 years ago I recall hardware stores selling "light-bulb savers". These were nothing more than small diodes that you would insert in the light socket (incandescent, versus compact fluourescent being the standard at the time), and then screw the bulb in over the saver. The claim was that the bulb would last a lifetime and energy would be saved.
Well, it was true, the bulbs, did last longer, but the lights only burned at half brightness because they received half-wave rectification of the current.
This discussion of accurate meters with no neutral got me to thinking. Would a standard meter accurately measure the power consumed by one of these things (because of the pulsating DC component)?
Reply to
Beachcomber
There is no reason to look at the neutral. If x current is flowing in phase A and y current is flowing in phase B they simply add x+y to get the load. Think of it as a 120v equivilent. If x is significantly different than y the difference is coming back on the neutral so you only get charged once. The part that is coming back on y represents a 240v load and you get charged twice for that. It may be an oversimplification but the model works.
Reply to
Greg
Harmonics caused by non-linear loads do *not* affect the kwh meter used energy metering. Having a screwed up power factor does not 'run the tab up on yer bill dramatically.' Only in very rare situations, when a very poor pf requires the utility to take extraordinary measures do they add *additional* metering and bill for it. PF correction is usually an issue for large installations so they can reduce the size of the service equipment. But a residential kwh meter will only register the true power component, a poor pf does not affect your bill (it can affect other appliances if it is harmonics and you have a relatively high source impedance, but that's another story)
daestrom
Reply to
daestrom
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google thread thl3419501793d
Having read through 'Dan Lanciani's discussion, I can say that what he's describing is *not* a metering problem. He discusses if the voltage drop from the meter to the load is significant, how a balanced or unbalanced 120/120 system would behave. But the difference in the two situations he discusses (50A load on each leg vs. 100A load on one leg and neutral) is not a metering issue. It is a *line loss* issue. And the line that has the losses is not the utility's, it is in the homeowner's with a voltage drop from the meter to the load that is 'significant'.
In the one case of a balanced 50A load on each leg, he assumes a voltage drop of 0.5 volt in each leg (resistance of 0.01 ohm). So the power delivered to a resistive load is 239V * 50A = 11.95 kW. But then he incorrectly calculates the load as seen by the meter as 239V * 50A. This is wrong because he assumed the meter was *upstream* of that 0.5 volt drop so the meter registers 240V * (50A +50A)/2 = 12.0 kW. The difference is simply the line losses (I^2*R) 50*50*.02 = 50 watts (or put another way, E^2/R 1.0*1.0/.02 = 50 watts)
Then, he calculates the condition of 100A on one leg and 0A on the other leg. In this situation, the voltage drop on one leg is 1V and in the neutral is 1V. So the power delivered to a resistive load is 118V * 100A = 11.8kW. But again, he incorrectly calculates the load as seen by the meter as 239V*50A. This is wrong for the same reason as before. It should be calculated as 240V * (0A + 100A)/2 = 12.0 kW. Same metered load as before (just 100A on one leg instead of 50A on two legs). The line losses this time are 100A*100A* .02ohms = 200watts.
Dan Lanciani notes the drop in delivered power (150 watts less power delivered in the second case), and wonders "where did it go??" The simple answer is that the second scenario quadruples the power losses in the customer's (not the utility's) long line from the meter to the load. It went from 50watts to 200 watts (difference of 150watts). This is reasonable since the line resistance is the same, yet the current is doubled, and since I^2*R losses are proportional to current squared, double current means quadruple the losses.
But the losses that are increased are in the customer's system, not the utility's. The power delivered *
at the point of trade* is the same in both cases, 12.0kW. The later case is just the customer wasting more of their purchase in 'long lines from the meter to the load'.
True, the losses in the utility equipment are different also, but they are upstream of the meter and we assume the voltage at the meter is constant. If we don't assume a constant voltage at the meter the numbers change only slightly, but the results are similar. The difference between what is registered at the meter and what is delivered to the load is the line losses in the customer's long line.
And if you have an average distance between the meter and the load, the resistance becomes much smaller and so does the voltage drop and the losses. If you have a particular service in mind, calculate the resistance in the two legs between the meter and the load distribution point. Then you can calculate for yourself the difference in line losses between a perfectly balanced and unbalanced system.
daestrom
Reply to
daestrom
Short answer?? YES. The modern kwh meter (either electromechanical or electronic) is just so d___ good at doing what it is designed for you'd be amazed. High harmonics (non-linear loads), half-wave, displacement power factor, unbalanced legs, voltage variations, you name it, it still comes out right.
daestrom
Reply to
daestrom
message
regarding
cases
wasted to
Oh yes... like a barber shop quartet. Searching googles web tab I got 45,000 hits on the following key word string. 'Electric, harmonics, motors, line' You could trim the list down by adding in EMF, phase distortion, and power factor.
This one was typical..it includes a section deeper in the body of text that addresses these issues.
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clamps to
and
due to
nearby
single
But
outfits
kWh, and no affect
No I don't. I have actually an entirely oposite view. But no use in hassling each other, there is plenty written on these issues..its not something new at all, with industrial applications at least... it might be new to the residential sector though.
I will decline to comment further in the interests of peace and harmony and recommend that anyone interested in these issues has available to them limitless information on the internet by searching Googles 'web' tab with the key word string mentioned above.
formatting link
hit the 'web' tab on the top left of the box... then type in yer key words.
one out.
There is not such an animal to my knowledge, nor should there be... but the power factor is calculated and the bill based on that. Power factor deviations stem from harmonic disturbances (and a few other things). Utility companies have a concern with EMF impedance, and the repercussions beyond a facility getting to the larger grid, and will address that with owners in number of ways... power factor correction capacitors included...and those costs are passed on to the customer.
To address one of your earlier remarks.. I dont know if this is suitable but it is quite common to install power factor correction capacitors on industrial motors, boosting the net motor efficiency and reducing utility costs, not just the cost of running the motor but also providing a better pf to the rest of the equipment in the facility.
These issues are not commonly known but the information is available as mentioned above.
Phil Scott
Reply to
Phil Scott
Well, now, there are utilities that systematically meter and charge penalties for power factor (for non-residential), such as my local utility. In these locations, pf correction / filtering can be installed with paybacks as low as 1 year. The utility may do this in part because they get a lot of power via dc link. They have a certain amount of capacitance in the filters at their inverter, but don't have the luxury of increasing a field current to compensate for more inductive loading.
But a residential kwh meter will only register the true power
For interest: I heard harmonics may hurt (or help) to a very minor extent. The +ve sequence components of the harmonic currents could cause a small additive torque to the spinning disk, and the negative sequence components could cause a small parasitic torque to slow the disk. I can't confirm this. I believe the utility here had some students look into it for a thesis. They were, as I recall, wondering about revenue loss due to 5th harmonic current in the increasingly electronic load of residences.
j
Reply to
operator jay

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