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Currents on a three phase system

  1. Nov 5, 2014 #1
    Hi everyone,

    is it possible to sum at any given moment in time L1 + L2 + L3, algebrically speaking? If so, what would this value represent?

    Thanks
     
  2. jcsd
  3. Nov 5, 2014 #2
    If L1,L2,L3 are currents than their sum will represent neutral current (if there is a neutral conductor in 3- phase system, that current flows through it).
    Generally, one third of the phasors sum of all 3 phasors represent zero sequence phasor (in treatment by symmetrical system components)
     
  4. Nov 5, 2014 #3
    Hi, thanks for answering. The neutral current, however, is a vector sum, am I correct? What if I sum L1 + L2 + L3, just the values?
     
  5. Nov 5, 2014 #4
    Values must be taken at same instant of time, then the above holds.
     
  6. Nov 5, 2014 #5
    The problem I'm having is this, the documentation of the device I'm monitoring is giving me "L1-N", "L2-N", "L3-N", and all three are positive values. Maybe I'm missing what the "-N" means here. That's why I wanted to know what would I be getting by adding them up. Additionally, the vector sum is also provided as a different value, which is clearly different.
     
  7. Nov 5, 2014 #6
    I'm reading that in order to calculate how much power the system is using you actually use all three phases:

    current L1 * voltage L1 +. ... + currentL3 * voltageL3 (the formula is more complex, but I'm interested only on the phases)

    so the currents in the single phases are relevant. Logically speaking, if I can use all three phases to calculate how much power the system is using, shouldn't I be able to add them up and say how much current is going into the system and thus used by the system?

    Thanks
     
  8. Nov 5, 2014 #7
    Ok
    This sound very fishy/wrong to me. Show me calculation you have on mind on certain example.
     
  9. Nov 5, 2014 #8
    P1 = V1 x I1
    P2 = V2 x I2
    P3 = V3 x I3

    Total power = P1 + P2 + P3

    is this formula/approach correct?
     
  10. Nov 5, 2014 #9
    P1 = V1 x I1 x cos φ1
    P2 = V2 x I2 x cos φ2
    P3 = V3 x I3 x cos φ3

    Total power = P1 + P2 + P3

    Now, it is correct
     
  11. Nov 5, 2014 #10
    Perfect, thanks.
     
  12. Nov 5, 2014 #11
    One more thing, regarding this point. What I meant was, what happens if I sum only the current values, without takng into account the phase shift, only the magnitudes.

    I mean, even if th system is perfectly balances and the neutral is zero, there's still current going in, right?
     
  13. Nov 5, 2014 #12
    Why would you do that? For a balanced 3-phase system you'll get 3x higher line current and do what with it?

    The phases have currents of equal magnitude. Phase shift between any two is 1200 el.
     
  14. Nov 5, 2014 #13
    The way the currents ( and voltages) are related needs to be looked at as vectors ( a magnitude and an angle) - if you ONLY sum the magnitudes - you can (with a lot of assumptions) estimate the total power used, but this will really tell you nothing about the neutral current, because you can have a "balanced" load with different phase currents - and no neutral current - or this sum could be All Phase 1 - and all of that current is flowing in the neutral. There is just too many things to consider - to be able to use 3 Magnitudes ( scalars ) to tell the whole story.
     
  15. Nov 6, 2014 #14
    I agree this is a complex matter. The point is, I have a professional power distribution meter that's giving me currents on L1, L2 and L3 (all positive) and kWh consumption on L1, L2 and L3. These current values are positive and not identical, so it pushed me to believe that somehow the currents on the single phases could be treated separately.

    I mean, if in the end you have kWh for each phase, then you must also have currents on each phase, this is common sense. By looking at the kWh consumption for each phase, L1 is increasing by 3 kWh at each reading, L2 by 1 and L3 by 2. Different kWh, different currents. From my ignorant perspective, this means I can actually monitor the different currents and voltages that are causing this.
     
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