Total power in unbalanced wye

Discussion in 'General Electronics Chat' started by jfrost, Dec 22, 2010.

  1. jfrost

    Thread Starter New Member

    Dec 22, 2010
    6
    1
    Hi,

    I am trying to determine the cost of having lots of 277 vac flourescent lights on when not necessary. They are all fed from the 3 phase to neutral legs of of a transformer's wye connected secondary (3 ph. 480/277 delta/wye transformer).

    With all the lights on, I measured the following phase-neutral voltages and phase currents with a true rms multimeter and clamp-on style ammeter:
    Code ( (Unknown Language)):
    1.  
    2. Measured Ph. Voltage (vac)    Measured Ph. Current (amps)
    3. Phase A - N = 282             Phase A = 91        
    4. Phase B - N = 281             Phase B = 88
    5. Phase C - N = 279             Phase C = 93
    All of the lights have 0.9 power factor ballasts

    I'm not sure how to calculate the total kW. I assume that the product of my measured values of voltage and current E x I would give the volt-amps for each phase. And multiplying that times the power factor and dividing by 1000 would give the kilowatts for each phase, correct? But, then how do I combine the three phase kW numbers to get the total kW - just add them? - use phasors???

    If I call what I measured with my meters, E-meas and I-meas, does the power company's meter (what we're getting billed for) measure the lower valued kWh [(E-meas x I-meas x power factor)/1000 x hrs], or the higher valued kVAh [(E-meas x I-meas)/1000 x hrs]?

    I'm just not sure about how to properly total those unbalanced 3-phase loads, and what type of power (real or apparent) the power companies base their charges on.

    Thanks a lot for any help with this!
     
  2. #12

    Expert

    Nov 30, 2010
    16,295
    6,807
    Ask the power company how their meter works. Some meters do, and some meters don't. We can't guess from here any better than you can guess.
     
  3. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
    Thats is actually pretty good advice.

    Your bill should have a formula on it for how your power is billed.

    I would check your power companies website to see if they have more in-depth info.

    Mine does. Check under areas for "customer generators" and the such to find out what their "buy-back" formula is as well as their other billing formulas and fees.
     
  4. jfrost

    Thread Starter New Member

    Dec 22, 2010
    6
    1
    Thanks. I'll try to check for more info from the bills. I don't have access right now to the accounting dept - I'm in electrical maintenance at a large factory :)
     
  5. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    You can't tell by a reading there, you need to know how much each fixture draws then multiply that out.

    You've also got the power factor part of the equation backwards as most people do, the lower it is the more you're penalized if you fall under a certain percent.

    Most power companies will gladly pay a free visit and explain it all to you.
     
  6. jfrost

    Thread Starter New Member

    Dec 22, 2010
    6
    1
    I know better than how much each fixture draws. I know how much they all draw for each phase-neutral circuit they are fed from. Nothing else is on the wye secondary of the transformer. Every light is either on phase A, B, or C with their common neutral N being the grounded common tie point of the wye.

    What's wrong with watts = volts x amps x pf? That's just real power in watts is equal to the apparent power in volt-amps times the power factor (which equals Cos phi).

    [From Wikipedia article - AC Power]
    [​IMG]
    My measured RMS voltages are what is actually supplying the entire group of lights that are on that particular phase-neutral circuit. The measured amps are what that group of lights are actually drawing. I take the product of those to get the volt-amps (i.e. apparent power) for the phase - the vector S in the diagram. The real power in watts (P in the vector diagram) should be S times the power factor which is the same as the ratio of real to apparent power and also equal to Cos phi in the vector diagram. Every light is identical and has a ballast power factor rating of 0.9.

    Also, I'm under the impression that most power company electro-mechanical meters have a disk whose instantaneous rate of spin is linearly proportional to the instantaneous wattage (i.e. real power). It's also connected to a time accumulating "odometer". Since power is energy per unit of time, the "odometer" merely has the effect of integrating instantaneousness real power over time to yield a running tally of total energy consumed in kWh. At least, that was how I thought it all worked. Maybe I'm wrong?
     
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