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120V from both legs

Discussion in 'Electrical Engineering' started by John Doe, Oct 4, 2004.

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  1. John Doe

    John Doe Guest

    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
     
  2. 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
     
  3. John Doe

    John Doe Guest

    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?
     
  4. "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.
     
  5. John Doe

    John Doe Guest

  6. operator jay

    operator jay Guest

    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
     
  7. operator jay

    operator jay Guest

    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
     
  8. operator jay

    operator jay Guest

    P.P.S. "Top-posting" is generally frowned upon, FYI.
     
  9. 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.
     
  10. 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.
     
  11. Beachcomber

    Beachcomber Guest

    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)?
     
  12. daestrom

    daestrom Guest

    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
     
  13. daestrom

    daestrom Guest

    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
     
  14. daestrom

    daestrom Guest

    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
     
  15. operator jay

    operator jay Guest

    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
     
  16. That link does not support your claims. Motors create very low levels of
    harmonics. Filtering these harmonics is NOT generally worth the cost in
    energy savings. Correcting power factor (displacement power factor) with
    capacitors is a good idea since it reduces the magnitude of the current
    flowing to the motor, reducing losses in the entire system.
    Harmonics have no effect on displacement power factor. Period. Quite simple
    actually. Displacement power factor is defined as the phase angle between
    the voltage and current at the fundamental frequency. By definition,
    harmonics are not even included.

    Displacement power factor excludes harmonics. Poor displacement power
    factor is caused by inductive, or capacitive, loads. Yes, I have seen
    facilities with leading power factors. It is just as troubling as lagging,
    sometimes even worse.
    True, utilities are concerned with the power factor of large customers, the
    displacement power factor. This is corrected using capacitors. Harmonic
    filtering is sometimes required in such installations, but not for the
    reasons you state. It is done to avoid resonances that can damage the
    capacitor bank or other equipment.
    They are quite commonly known in the power quality and power metering
    fields.

    BTW, you might be surprised to find the source of much of the information in
    the link you provided ;-)

    Charles Perry P.E.
     
  17. Dan Lanciani

    Dan Lanciani Guest

    |
    | | > Sounds like this may be an urban legend created as a perversion of the
    | > following reasoning:
    | >
    | http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&selm=
    | > 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).

    Umm, minor nit. I said that I do not have full knowledge of the operation
    of the meter's clever magnetic circuit. I said nothing about having "little
    knowledge." :) I suggest that you read the spiel before dismissing it.
    The example requires only a very basic level of electrical knowledge to
    follow. (Which is not to claim that the conclusion is correct. As I said,
    maybe they do something clever to compensate to a first order. But in the
    5 or so years since I gave the example I've yet to have anybody explain *how*
    such compensation could work.)

    | I'll second the debunking. The meter knows how
    | much energy you are using and you get billed accordingly.

    But how does it know? To compute power you need both the current and the
    potential (voltage). The product can then be accumulated/integrated over
    time to give energy. Every split phase meter I've seen is a 4-terminal
    device with no connection to the neutral. It can measure the current in
    each leg and it can measure the potential between the legs, but it has no way
    to measure the potential between the neutral and either leg. The neutral's
    potential cannot be assumed to sit half way between those of the legs. An
    unbalanced load anywhere on the same transformer can drive it closer to either
    leg.

    For the simple case where the load being served by the meter in question
    is the only load on the transformer it appears to me that you pay for the
    energy you use on your side of the meter plus 150% of the energy wasted as
    heat in the neutral on the utility's side of the meter. For more complicated
    cases where somebody else is pulling the neutral off center you might be
    charged for less energy than you actually use. Note that we are talking
    about rather small numbers here. It is not, in any case, something to get
    excited about.

    Dan Lanciani
    [email protected]*com
     
  18. Dan Lanciani

    Dan Lanciani Guest

    |
    | | > Sounds like this may be an urban legend created as a perversion of the
    | > following reasoning:
    | > http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&selm=
    | > 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,

    No, I clearly stated that the long run is from the meter to the transformer.
    The voltage drop that is significant is from the transformer to the meter,
    not from the meter to the load. You can't change the example circuit and
    then complain that the math is no longer correct. :)

    [analysis and "corrections" deleted since they do not apply to the example
    as posed]

    | 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.

    But that assumption is false. Not only are the voltages not constant but
    the two line to neutral voltages likely aren't even equal.

    | If we don't assume a constant voltage at the meter the numbers change only
    | slightly, but the results are similar.

    Except that the customer pays for some of the utility's line loss, absent
    some clever correction that has yet to be described.

    Dan Lanciani
    [email protected]*com
     
  19. Dan Lanciani

    Dan Lanciani Guest

    |Dan the meter doesn't need to measure neutral current.

    I never said it does. What it needs to know is the potential of the neutral
    relative to a leg. To compute power you need both the current *and* the
    voltage.

    |Mr Kirchoff says any
    |current going in will come out somewhere. If it comes out the opposite phase
    |that is a 240v load, any unbalance is a 120v load. You have the current and the
    |voltage.

    No, you don't have the voltage. You are assuming that the potentials
    between each leg and neutral are equal (much as the meter must). But if
    there is any current in the neutral, those potentials are not equal. That's
    where the error comes from. You cannot just take the leg-to-leg potential
    and divide by 2 to get the leg-to-neutral potential.

    Dan Lanciani
    [email protected]*com
     
  20. I think you may need to get a copy of IEEE std 100. I will give you a
    little clue, there is more than one kind of power factor.

    Charles Perry P.E.
     
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