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power factor - round 2

Discussion in 'Home Power and Microgeneration' started by philkryder at gmail, Mar 21, 2013.

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  1. A couple years back some folks in this group helped me to use a KillaWatt device to deterimine powerfactor on some "lighly loaded" ac motors.


    Time has passed.

    I installed corrective capacitance based on my spreadsheed and formulas provided by others in the group. Thanks for all the help.

    More Time passed.

    What we learned recently is that though the AVERAGE powerfactor is "low" - ..5 or less sometimes,
    there are SPECIAL PEAK LOAD EVENTS that may last for a few seconds (less than 5) every minute or two.


    Coincident with those peak load events, control circuits in the same "neighborhood" - i.e. cicuit need to fire control solenoids for hydraulic controls. The motor is a half-horse 120v 60 cycle that drives a hydraulic pump. Onaverage - 95% of the time - there is no load - but, when there is a a solenoid opens a valve to a hydralic motor and then, other solenoids open valves to hydraulic cylinders...

    You can guess where this is going.


    When the load is applied voltage drops from 120 nominal to 106 or so. Solenoids fail to fire. Power factor changes dramatically.

    So, given that my KillaWatt only does averages, we rented a Fluke meter that records mins and maxes within each one-minute interval.

    I'm inclined to re-tune the capacitors based on highest power Factor seen over an entire 10 hour shift, rather than the average seen by the Killawattwhile I happen to be looking.

    Any other suggested actions?

    Also, "what happens" with a corrected circuit at a 98% power factor when the load increases and the motor's power factor rises from .5 to .85 causing the circuit to be "over corrected"

    thanks in advance.
    Phil
     
  2. g

    g Guest

    lots of inductive loads, 0.5 _is_ rather lousy
    what is a SPECIAL peak load event???
    control solenoids for hydraulic controls. The motor is a half-horse
    120v 60 cycle that drives a hydraulic pump.
    On average - 95% of the time - there is no load - but, when there is a a
    solenoid opens a valve to a hydralic motor
    and then, other solenoids open valves to hydraulic cylinders...
    Provide more information re how long the voltage stays low.

    If the voltage drops that much to a constant low 106 volts, then the
    supply to the motor is dangerously under-dimensioned. Update the supply
    to a larger cable and fuse. If you are talking about a short time
    interval (less than 1 seconds) for the voltage drop then that is normal
    due to the momentary high start current of the motor. The voltage should
    come up to close to nominal after the motor runs at rated speed. But
    since you are stating that the solenoids fail to fire, it indicates that
    the voltage is too low over the time the motor runs...

    //g
     
  3. mike

    mike Guest

    Your objective is unclear.
    When you get the power bill, are you billed for volt-amps or watts?
    If it ain't broke, don't fix it.

    The KillAwatt is limited to 2kw or so. Anything that can be measured
    with that
    is IRRELEVANT.

    When there's no load, the power factor is IRRELEVANT.

    If you are billed for volt-amps or you just want to be a "good citizen",
    You want to provide power factor correction at EACH point of load under
    normal conditions for that load.

    Judging from your statements, your problem is RESISTANCE in the wiring
    or motors severely overloaded...or both.

    Don't think it makes a lot of sense to try to power correct a transient
    load at turn-on unless you switch it in ONLY during turn-on.

    A 120V solenoid shouldn't fail to energize at 106V. I expect your
    transient goes much lower than that.

    You need a storage oscilloscope and a current/voltage probe to determine the
    transient conditions that you experience.

    Think the end result will be bigger wiring.
     
  4. Jim Wilkins

    Jim Wilkins Guest

  5. Guest


    more background data.

    we are off grid.

    We're not trying to save money.
    We're trying to run more reliably.

    Our initial problem was popped breakers due to the high current flow due tothe imaginary current.

    thus the power factor correction.

    step 1 was to remove the caps. - done.

    I still have the question "what happens when the caps are in place and the motor's power factor improves to .85 from the average .5?
     
  6. j

    j Guest

    That power factor was inductive before. What you don't know for sure is
    whether with the caps and the loaded motor whether it had become
    capacitive. I think it may have, particularly if you had compensated to
    a PF of 1.

    You have a complex system,if you were on grid the grid would absorb the
    resulting voltage spike. But your inverter has to deal with that and
    what it does is unknown. Have you considered a DC pump?
     
  7. Jim Wilkins

    Jim Wilkins Guest

    Or an accumulator?
    http://www.engineersedge.com/hydraulic/accumulator_equations.htm
     
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