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Caculation of line current on a 3 phases unbalanced delta

  1. Apr 25, 2015 #1
    Hello,
    I cannot find back from school, the formula to use:
    A 3 phases 220Vac system, no neutral, 3300W of pure resistance load between L2 and L3, the same load between L1 and L3, nothing between L1 and L2.
    What will be I1, I2, I3 the current in the 3 lines of the network?

    Thanks in advance
    --
    Jmp0
     
  2. jcsd
  3. Apr 26, 2015 #2

    Hesch

    User Avatar
    Gold Member

    Assume V1 = 220 / 0, V2 = 220 / 120, V3 = 220 / 240.

    Calculate I23 and I13. (I23 is current from L2 to L3 ).

    I2 = I23, I3 = -I23 - I13 (KCL)

    I1 + I2 + I3 = 0 (KCL)

    Calculate with complex numbers all the way.
     
    Last edited: Apr 26, 2015
  4. Apr 27, 2015 #3
    I agree with Hesch except the voltages are V12=220>0,V23=220>-120,V31=220>-240[or +120] and since the load it is declared as “pure resistance” [cos(fi)=1] the currents will be:

    I12=0;I23=3300/220*[(COS(-120)+SIN(-120)j];I31=3300/220*[(COS(+120)+SIN(+120)j]
    upload_2015-4-28_7-52-7.png

    Considering all currents “entering” position:

    I1=I13=-I31=-15*[(COS(+120)+SIN(+120)j] =7.5-12.99j;I2=I23=-7.5-12.99j;I3=-I1-I2=2*12.99j=25.98j.


    Remark: the imaginary part of the current here it does not mean reactive part. The reactive part will be proportional with the sinus of angle between current and supplied voltage[fi] which remains 0 [in this case].

    One could calculate also-following Hesch logic-using a trigonometric circle and use X and Y coordinates instead of complex number. The circle radius is equal to I=P/V=3300/220=15 [A]. Then:
    upload_2015-4-28_7-52-45.png

    I23X=-15*SIN(30)=-7.5[A]; I23Y=-15*SIN(60)=-12.99 [A]; I31X=-15*SIN(30)=-7.5 [A]; I31Y=+15*SIN(60)=12.99[A].

    I1=I13=-I31 =7.5X-12.99Y;I2=I23=-7.5X-12.99Y;I3=-I1-I2=-(I1+I2)=2*12.99Y=25.98[A].
     
  5. Apr 28, 2015 #4
    Hello,

    Thanks for your answers. This is correct. I found it also (with help) with another method and result are the same:
    I1=15A I2=15A I3=26A
    This method:
    Phase currents = I=P/E, = O amps for L1-L2 (ab) 3300/220 = 15A for L2-L3 (bc) & 15A for L3-L1 (ac)
    Line currents =
    L1 =√(ab²+ac²)+(ab x ac),
    L2 =√(ab²+bc²)+(ab x bc),
    L3 =√(ac²+bc²)+(ac x bc)

    Thanks everybudy
    --
    Jmp0
     
  6. Apr 29, 2015 #5
    It is a very interesting formulas. The complete one it is as following:

    L3 =√(ac²+bc²)-2x(ac x bc)xcos(angle)

    and for "angle" between ac and bc vectors[let's say "phasors"] of 120o turn it in:

    L3 =√(ac²+bc²)+(ac x bc) [as you said]

    As you can see for any other angle an error occurs.
     
  7. Apr 29, 2015 #6

    Hesch

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    Gold Member

    (note my underline).
    Well, in Denmark (Europe) we operate with a phase-system (voltages and currents are as per phase) and a main-system (voltages and currents are phase to phase). So when I read "3 phases 220V", I regard 220V per phase to be meant ( V12 = 220V*√3 )?? In Europa we would write this: A 3×380V system.

    I don't want to introduce a discussion about these different terms in US/EU, this is meaningless. I'm just trying to clarify the reason to some confusion
     
    Last edited: Apr 29, 2015
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