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Thread: Calculating Phase Currents for Unbalanced Loads

  1. #101
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    Quote Originally Posted by rattus View Post
    If the phase currents are not separated by 120 degrees, there is no unique solution if we have only the line currents.
    I think the criteria for no unique solution is if the delta load currents do not sum to zero (zero-sequence current <> 0).

  2. #102
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    ............................
    Last edited by Smart $; 08-04-10 at 12:12 AM. Reason: double post
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  3. #103
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    Quote Originally Posted by david luchini View Post
    Well, double check me on this...but if I had a system with Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect an inductive impedance of 15.969<6.9, between B-C, I connect a capacitive impedance of 19.941<-18.56, and between C-A, I connect an inductive impedance of 28.139<13.74. My phase currents should be:

    Iab=13.025<-36.9
    Ibc=10.431<-131.44
    Ica=7.392<76.26

    In rectangular form:

    Iab=10.416-j7.820
    Ibc=-6.904-j7.820
    Ica= 1.756+j7.180

    Iab+Ibc+Ica = 5.268-j8.460 = 9.97<-58.09

    Since Ia=Iab-Ica, Ib=Ibc-Iab & Ic=Ica-Ibc, we get

    Ia=8.660-j15.000
    Ib=-17.320+j0.00
    Ic=8.660+j15.000

    Ia+Ib+Ic= 0+j0 = 0

    This is an unbalanced (and unequal power factor) delta connected loads, where the sum of the line currents equals zero, but the sum of the phase currents does not.
    Well I didn't double check but your math appears good (I'm not getting any bad vibes ). But this makes my [disguised] point... Doesn't the circuit described have circulating current? i.e. there will be energy (which requires current) passed back and forth between the capacitive and inductive elements as long as the load is energized, and that energy is only dissipated upon deenergization of the circuit. That energy comes in on the lines during the first [few?] cycle[s] upon energizing... which we don't discuss, because our modus operandi is after steady operating state has been achieved. I guess this takes us to jghrist last posed and unanswered question...???

    I say if we are starting with a nominally-balanced delta load and trying to determine the initial stage of winding degradation, for example, the centroid solution or the p&n sequences method would suffice. The goal is not one which requires accuracy, but rather a dependable means to determine a moderate degree of unbalance beyond nominal in real-time. But then we have to ask, would not just the line values themselves suffice for such a determination?
    I will have achieved my life's goal if I die with a smile on my face.

  4. #104
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    Quote Originally Posted by jghrist View Post
    Or, a simpler example:

    Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
    Iab+Ibc+Ica=Iab=20.8<-30 not =0

    Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0
    Intuition leads me to believe the load currents sum to zero, but logic says otherwise. Clearly, changing one of the load impedances changes that current only and does not affect the other two. That is, tweaking one load would change the sum of the load currents
    Don't mess with B+!
    (Signal Corps. Motto)

  5. #105
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    Quote Originally Posted by jghrist View Post
    Or, a simpler example:

    Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
    Iab+Ibc+Ica=Iab=20.8<-30 not =0

    Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0
    Intuition leads me to believe the load currents sum to zero, but logic says otherwise. Clearly, changing one of the load impedances changes that current only and does not affect the other two. That is, tweaking one load would change the sum of the load currents
    Don't mess with B+!
    (Signal Corps. Motto)

  6. #106
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    Quote Originally Posted by jghrist View Post
    Or, a simpler example:

    Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
    Iab+Ibc+Ica=Iab=20.8<-30 not =0

    Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0
    And to answer the question about load currents being 120 apart, just add 10 kohm resistors in B-C and C-A. Then these load currents will be 20.8 mA, all power factors the same, all currents 120 apart, but the sum of the currents is still very close to 20.8<-30.

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