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Residence Power Meter And Reactive Loads Question ?

Discussion in 'Electrical Engineering' started by Bob, Dec 24, 2009.

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  1. Bob

    Bob Guest


    Think I have a very basic understanding of VA, Power Factor, etc. but
    was wondering about this:

    Even though most of the load in a typical residence is probably all
    resistive, how does the power meter outside "decide" what's reactive and
    what's resistive ?

    It measures the total current going into the house, doesn't it ?

    But you should, I think, be charged Only for the current
    consumed/required by all the resistive loads. Is this true ?

    Does it just measure total VA, and assume a Power Factor ?

    If someone a lot better at this stuff could explain this for me, would
    be most appreciative.

    Thanks, and Happy Holidays,
  2. What makes you think that it needs to?
    No, it measures power.
    No. You get charged for your power consumption. What makes you think
    that your reactive consuming loads should be able to operate for free?
    Old power meters had issues with different loads. Newer meters are
    quite capable of determining just how much you use, regardless of what
    type of load is presented to the load side of the meter.
  3. Dean Hoffman

    Dean Hoffman Guest

    There was a long thread here "Balancing the Breaker Box"
    recently. Read that. It should pretty much answer your questions.
  4. James Sweet

    James Sweet Guest

    The meter measures true power, so the power factor of the load has no
    effect on the charges.

  5. There was a time when the number of capacitors hanging on the line
    determined how well aluminum disc power meters would "read" inductive

  6. Note that I am referring to the utility provider side, up on the local
    HV sub for the cap banks.
  7. No, the basic mechanism APPLIED a torque to an Alunimnum plate, and it
    had error when the current and the voltage were not in phase.

    The torque applied is less when the two are out of phase, and the
    subsequent turns per watt/hour will be less as well.

    The power company puts the cap banks on the line to offset their losses
    on those meters.

    They do not need to any more as the meters no longer present the
    It only needed to be 2 wire, and even if all use was on one side, it
    STILL reads resistive accurately, and inductive less so if the phase
    correction is not just right.
  8. krw

    krw Guest

    You surely are an idiot, DinBulb. AlwaysWrong too.
  9. Liar.
  10. No, the meter reads the number of turns the disc gets put to it. When
    reading resistive loads, it reads Watt/hours (you were wrong again, as
    usual), when it is responding to an inductive load, it reads LESS
    Watt/hours than it should. That is one reason why that type of meter has
    been phased out of service, idiot.

    If a load turns it less than usual, that means that the METER fails to
    make a valid claim of what was actually consumed.

    I know it is hard for you to understand, K-Tard, but less is less, and
    the number of functioning brain cells between your ears is no exception.
  11. krw

    krw Guest

    Like I said, folks, AlwaysWrong is *ALWAYS* wrong. He didn't get the
    name because he knows anything.
  12. krw

    krw Guest

    You really don't have a clue. The torque (thus power) is proportional
    to the watts. The integral of that (the accumulator gears) is energy
    Wrong again, AlwaysWrong. The meter reads *exactly* the power
    consumed. It doesn't care about inductance, capacitance, or anything
    else, but WATTs.
    You've been proven wrong again, AlwaysWrong, though you will never
  13. You really do not get it. The torque is LESS when the two coils that
    drive it are out of phase. That means that the same POWER will yield
    LESS metered reporting of it. It does not get much more simple than
    that, K-Tard.

    See if you can wrap your single neuron around that, dumbfuck.
  14. krw

    krw Guest

    No, AlwaysWrong, it is *you* who doesn't get it (or anything). The
    torque is LESS because the POWER is lower when the current in the
    coils is out of phase.
    Now, DimBulb, the POWER is NOT the same when the current in the coils
    is out of phase as it is when they are. The meter reads the POWER, no
    matter what the phase relationship.
    Evidently it does, because *you* are too simple to understand simple
    electricity, AlwaysWrong.
    Once again, AlwaysWrong proves how wrong he always is.

    Keep it up DimBulb, you're making a real fool out of yourself. Still.

  15. You really do not understand the relationship between torque and total
    number of turns over time.

    Less torque will mean slower rotation. It really is THAT simple, you
    moronic sub-human chump.
  16. krw

    krw Guest

    You're a dumbass, DimBulb. Do you have any clue what in integral is?
    No, I suppose not.
    Integrated over time is fewer turns, so? The TORQUE is a function of
    POWER, moron.

    Don't try anything as technical as brushing your teeth, AlwaysWrong.
    You're not up to it. Oh, that's right. You don't.

    Have fun in mommy's hamper, DimBulb.
  17. Guest

    Guest Guest

    a) The induction disk meter "reads" the real power only.
    b)whatever is on the line "upstream" of the meter has no bearing on the
    Capacitors "downstream" will reduce the inductive load (which is being
    ignored by the meter except for a decrease in losses and possible harmonic
  18. Guest

    Guest Guest

    Errr, Not true. If it were, the meters would be useless as the load pf
    is not constant. The meter inherently reads the <real power>, ignoring the
    reactive component.
    Are you thinking of the "lag coil"? This is not a power factor correction

    Nor is the torque "applied". A rotating magnetic field is applied and the
    reaction of induced eddy currents in the disc with this field produces (or
    generates) torque.

    When the two are out of phase, the torque at a given current and voltage
    will be lower, as you indicate, because the torque is proportional to the
    component of current in phase with the voltage- ie -the real power which
    will be lower than the VI product. Voltage*current is, contrary to what you
    appear to believe, not power (watts) for AC.
    The turns per watt-hour will not change.
    .. The speed of rotation will decrease because the real power and hence the
    torque will decrease.
    No. They put capacitors on the line to provide vars locally to compensate
    for inductive loads, reducing line losses "upstream" of the capacitors, or
    to improve the voltage. Note that these capacitors will have absolutely no
    effect on the meter and meter losses will not be corrected by such

  19. The retarded **** has not ever once said a goddamned thing about it,
    you stupid ****.

    Now, IF he comes back with some brains, instead of his stupid, peanut
    gallery dumbfuck crap, MAYBE your remarks would have some merit.

    Since all he does is parrot the same old, tired, retarded horseshit,
    both he, and his retarded butt buddy (you) can shove it up each others
    asses for all I care.

    So, his CRAP is meaningless without stating what his refutation is.

    And since you do not even know what fucking torque is, I doubt
    seriously that you have any real depth of knowledge as it relates to ANY
    mechanical device in existence. This meter is a mechanical device,

  20. Dumbass. The field IS APPLYing the torque force.

    Yes, it IS applied. Whenever a shaft is rotated by a force, torque has
    been applied. And whenever a shaft is turned, torque IS the force that
    was applied to the shaft, by whatever means, to turn it. Go back to
    basic physics classes, dimwit.

    So, if the shaft turns at one speed under an in phase resistive load,
    and it does... and IF the shaft gets slightly less torque applied to it
    by the eddy currents when "looking at" an inductive load, then ANY IDIOT
    can declare that the number of turns will be less, despite the fact that
    the same power was consumed, if the torque is less due to the two coils
    being out of phase, and as it happens, that IS the case with inductive

    So, same power consumed in both cases, and the meter will read the
    inductive load as having been less. Period. The difference was not
    enough for them to be concerned about, and it was made up for ON the FEED
    side (utility) by putting cap banks up on the poles to shift the phase so
    that an inductive load would make the metering a bit more accurate on
    such loads.

    It is not just for reducing line losses, it is also for reducing the
    error in the meters on inductive loading.

    You all need to go back to school... very basic school.
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