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Calculating Power Consumption for 3-Phase Devices with Different Currents: Using U, I & Clamp Meter

Prepond 42203 12
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  • #1 6036232
    Prepond
    Level 19  
    Hello,
    I know that it is probably trivial, but how to count how much power, how many kW, is consumed by a device on 3 phases, with a different current on each phase (I have a clamp meter).

    Because for one phase it is standard P = U * I. How is it for 3?

    Thanks.
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  • #3 6036454
    Prepond
    Level 19  
    How do I have 400V interphase to the formula and I substitute 230? because this is the phase in terms of PE, right?
  • #4 6038247
    Tomisan
    Level 12  
    By writing "device" I suppose that my colleague means one unbalanced 3-phase device connected to the power grid, and the question is how?
    In the star connection arrangement:

    P = Uf * Ip1cos?1 + Uf * Ip2cos?2 + Uf * Ip3cos?3

    Uf = 230V (for whistle calls with neutral connected to
    the neutral conductor, disregarding the conductor impedance
    neutral)
    Ip = line current measured with a clamp meter on the lines
    supply.

    In delta connection arrangement:

    P = Um * If1cos?1 + Um * If2cos?2 + Um * If3cos?3

    Um = 400V
    If = phase current - it should be measured with each load separately,
    however, it is not the current flowing in the power cables (Ip), only
    load current connected between two phases.
  • #5 6040274
    zubel
    Conditionally unlocked  
    You will do it almost without counting. Turn off all appliances in the apartment and record the meter reading. Plug in your contraption for an hour and write off the meter reading. Subtract what's used up and you've got the result
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  • #6 6040767
    Prepond
    Level 19  
    I would subtract but it's a factory :) I found out that the device consumes 133.4kW per hour. The phases were 160,210 and 210 amps respectively.
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  • #7 6041753
    Stary1952
    Level 32  
    Prepond wrote:
    The phases were 160,210 and 210 amps respectively.

    And what a strange device that draws so unevenly electricity from the network. Often, such uneven consumption may indicate a defect of the device. :?:
    Greetings .
  • #8 6042856
    czesiu
    Level 37  
    Prepond wrote:

    Because for one phase it is standard P = U * I. How is it for 3?
    If you simplify this, then for 3 phases you can take: P = U * (I1 + I2 + I3).
    Then you take 230V as U.
  • #9 6042974
    wola
    Level 26  
    Prepond wrote:
    I would subtract but it's a factory :) I found out that the device consumes 133.4kW per hour. The phases were 160,210 and 210 amps respectively.


    Hello A it should not be 133.4kVA; unless the power factor = 1
  • #10 6044994
    Tomisan
    Level 12  
    A colleague did not mention anything about the device itself, what kind of device is it? The colleague measured the current with a clamp meter, probably on the power cables. There was 133.4kW of active power. So maybe a colleague here will present us his calculations?
  • #11 6047399
    zubel
    Conditionally unlocked  
    Write what device it is, if it's not a secret
  • #12 6055557
    Prepond
    Level 19  
    This is a rectifier with an output rating of 22-24V DC and a current of 12,000 A. Consumption measured with an output load of 10,000 A.
  • #13 6056392
    Stary1952
    Level 32  
    Prepond wrote:
    This is a rectifier with an output rating of 22-24V DC and a current of 12,000 A. Consumption measured with an output load of 10,000 A.

    Uneven current draw indicates that one of the rectifier elements in the bridge is most likely damaged. In 6-pulse rectifiers, the current consumption should be almost the same in each phase.
    Greetings .
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Topic summary

The discussion revolves around calculating the power consumption of a 3-phase device with varying currents on each phase. The standard formula for single-phase power (P = U * I) is adapted for three phases, where power can be calculated using either a star or delta connection. In a star connection, the formula is P = Uf * (Ip1cos?1 + Ip2cos?2 + Ip3cos?3), with Uf being the phase voltage (230V). In a delta connection, it is P = Um * (If1cos?1 + If2cos?2 + If3cos?3), where Um is the line voltage (400V). A user reported a device consuming 133.4kW with phase currents of 160A, 210A, and 210A, raising concerns about the uneven current draw, which may indicate a defect in the device. The device in question is identified as a rectifier with an output rating of 22-24V DC and a current of 12,000 A.
Summary generated by the language model.
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