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  1. Chris1234
    Online

    Chris1234 Regular EF Member

    Location:
    Manchester, UK
    I’ve been asked to calculate power usage of a machine to works out how much it costs to run.

    Current draw is:

    L1: 9.71
    L2: 9.61
    L3: 11.14

    N: 7.78

    Obviously when designing a circuit you would base it in the highest phase ampage.
    But is power usage worked out the same way? Or are three phases added together because of the unbalanced load?

    KVA= 11.14 x 400 x 1.732 x / 1000 x unit price

    Or is it:

    KVA= 3x 10.15 x 400 x 1.732 / 1000 x unit price

    Never had to do this before!
    Thanks.
     
  2. Pete999
    Online

    Pete999 Trusted Advisor

    Location:
    Northampton
    Business Name:
    None
    https://www.dataforth.com/catalog/pdf/an110.pdf
    https://www.dataforth.com/catalog/pdf/an110.pdf
     
  3. Lucien Nunes
    Offline

    Lucien Nunes Mercury Arc Rectifier Trusted Advisor

    Location:
    London
    The power consumption is the total of the three phases - it doesn't matter whether it's balanced or not. So:
    P=(9.71+9.61+11.14) x 400/1.73=7.0kVA
    This is the apparent power in kVA; you would need to know (or guess) the power factor in order to obtain the real power in kW, from which to calculate the running cost.

    Worst case, at unity pf, the machine uses 7.0kW and therefore 7.0 x unit price per hour. If the load is mainly heating this will be close. If it is mainly induction motor load, the pf might be more like 0.8, in which case the meter will register only 0.8 x 7.0=5.6kW. Not knowing the machine, it looks to me from the fact that two lines have similar current, that the main 3-phase load is around 9.7A, to which 1.5A of single-phase control system is added on one phase. In which case, go with 0.8 or 1.0 according to whether that main load is motors or heaters, or hedge your bets and call it 0.9 giving a real power of 6.3kW.
     
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    Last edited: Feb 13, 2018
  4. static zap
    Offline

    static zap Regular EF Member

    Location:
    west midlands
    Do you know how you are billed ,is there any mention of Peak Demand ?
    Adding new equipment could influence this in an industrial setting.
     
  5. davesparks
    Offline

    davesparks Trusted Advisor

    Location:
    guildford
    What is the nature of the load? That neutral current seems a bit odd for a ‘normal’ load.
     
    • Like Like x 1
  6. Lucien Nunes
    Offline

    Lucien Nunes Mercury Arc Rectifier Trusted Advisor

    Location:
    London
    Good point, I think I read that as 1.78A by mistake. Looks like there's significant harmonic distortion as it's high relative to the difference between the lines.
     
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