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Thread: Kilowatt hour reading for 3 phase lighting circuits

  1. #1
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    Kilowatt hour reading for 3 phase lighting circuits

    I have a Levington meter series 2,000 hooked up on 6 circuits in a 208y/120volt 3 phase 4 wire system. total connected load for the circuits is 25.8amps @116volts. In 21 days the meter read 998kwhrs. By my calculations it should have read 1,452 kwhrs. if I cube it 1.73 I get what the meter reads but I dont think that that I applies to this case? If it does then my other meter is off. Any suggestions or is the meter malfunctioning or installed incorrectly. Thankyou in advance!

  2. #2
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    How did you come up with a single current on six circuits?

    Did you add all six together?

    Are any of them line to line loads?

    Are they all on the same phase?

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    the phaseing is circuits #2,8a phase #4,10b phase #6,12 c phase under 3 doghnuts. I took amp readings on each circuit and added them together
    25.8 then averaged the voltage reading on the 3 phase to 116volts(garage flourescent lights are the loads)

    25.8*116=2,992.8watts/1,000=2.9kw

    2.9kw*504 hours=1,508 kw hrs

  4. #4
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    actual readings from amp probe is


    • #2 3.2amps
    • #4 5.7amps
    • #6 3.6amps
    • #8 4.7amps
    • #10 3.5amps
    • #12 5.1amps

  5. #5
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    Quote Originally Posted by highpowered View Post
    the phaseing is circuits #2,8a phase #4,10b phase #6,12 c phase under 3 doghnuts. I took amp readings on each circuit and added them together
    25.8 then averaged the voltage reading on the 3 phase to 116volts(garage flourescent lights are the loads)

    25.8*116=2,992.8watts/1,000=2.9kw

    2.9kw*504 hours=1,508 kw hrs
    There are a couple of issues to consider with your methodology.

    The first is your averaging of the voltage readings. You should use the voltage reading for each phase with its associated load current to be more precise. If the voltages were 112, 116 and 120 (averaging 116) with the associated load currents, you would get a different result than if the voltages were 114, 116 and 118 (also averaging 116.)

    The second issue is that 25.8A * 116V = 2.99kVA, not kW. To calculate the kWH accurately, you would need to consider what the power factor is for each load.

    The third issue is whether or not the load currents are constant/unchanging for the total 504 hours of measurement.

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    Quote Originally Posted by david luchini View Post
    There are a couple of issues to consider with your methodology.

    The first is your averaging of the voltage readings. You should use the voltage reading for each phase with its associated load current to be more precise. If the voltages were 112, 116 and 120 (averaging 116) with the associated load currents, you would get a different result than if the voltages were 114, 116 and 118 (also averaging 116.)

    The second issue is that 25.8A * 116V = 2.99kVA, not kW. To calculate the kWH accurately, you would need to consider what the power factor is for each load.

    The third issue is whether or not the load currents are constant/unchanging for the total 504 hours of measurement.
    Thank you for taking time to respond I appreciate it
    The parking garage lights are always on 24hrs a day. I'm not sure of how to obtain the power factor for each load please explain? I agree about the voltage my goal is to get my calculations close to the actual reading on the meter. I dont need to be dead on. Is'nt watts I*E? thanks again
    Last edited by highpowered; 07-26-12 at 04:35 PM.

  7. #7
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    120726-1602 EDT

    highpowered:

    Your method is reasonable for a ball park figure. But you need an estimate of the power factor. From the values you provided it looks to be about 0.69 .

    We can assume the lights have a lagging power factor, and 0.69 would not be unlikely. Not enough information on the lamps.

    In an AC circuit where the load is not a pure resistance, then you can expect the energy consumed to be less than predicted by V*I * time.

    If you take a good quality capacitor and measure current, voltage, and power you might find the real power to be 1% of the volt-amperes measured.

    .

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    Quote Originally Posted by gar View Post
    120726-1602 EDT

    highpowered:

    Your method is reasonable for a ball park figure. But you need an estimate of the power factor. From the values you provided it looks to be about 0.69 .

    We can assume the lights have a lagging power factor, and 0.69 would not be unlikely. Not enough information on the lamps.

    In an AC circuit where the load is not a pure resistance, then you can expect the energy consumed to be less than predicted by V*I * time.

    If you take a good quality capacitor and measure current, voltage, and power you might find the real power to be 1% of the volt-amperes measured.

    .
    Ok what would the equation be using the power factor estimate?

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    Quote Originally Posted by highpowered View Post
    Ok what would the equation be using the power factor estimate?
    25.8A * 116V * 0.69pf = 2,065 W/1,000 = 2.065kW

    2.065kW * 504 hours = 1,041 kWH

  10. #10
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    Quote Originally Posted by david luchini View Post
    25.8A * 116V * 0.69pf = 2,065 W/1,000 = 2.065kW

    2.065kW * 504 hours = 1,041 kWH
    thankyou so much! my last question because obviously Im undereducated about power factor in the future how do I find this for my equations?

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