Delta phase current trouble

Discussion in 'Homework Help' started by elecidiot, Mar 20, 2013.

  1. elecidiot

    Thread Starter New Member

    Jun 29, 2010
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    Suppose we know only the line currents of a delta connected source. How to calculate the phase currents for that source from the measured line currents? Is there a unique solution? What are the other conditions needed for a unique solution?
    Thanks in advance
     
    Last edited: Mar 20, 2013
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I believe the only case for which a solution can be found is for the ("trivial" ) case of balanced line currents with equal magnitudes. This would give a balanced 3 phase set of currents flowing in the delta. In that situation we know (at least) that the phase current magnitude is the line current magnitude divided by √3.

    A requirement for the determination of currents in any but a "trivial" case would be on the basis of the relationships ...

    I_phase_1 = I_phase_3 + I_line_1
    I_phase_2 = I_phase_1 + I_line_2

    ..... etc

    Finding a particular phase current requires knowledge of one other phase current - hence (logically) for any case other than the "trivial" one, it would seem that a unique solution is not possible.
     
    Last edited: Mar 21, 2013
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  3. elecidiot

    Thread Starter New Member

    Jun 29, 2010
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    Thanks t_n_k for ur info. But I am afraid that even the balanced line currents(of equal magnitude and phase shift of 120deg from each other) do not guarantee us a unique solution. ("trivial" as u noted). Let me justify with an example.
    Say Ia=173.2 angle -30° ; Ib=173.2 angle -150° ; Ic=173.2 angle -270°

    According to you the unique solution for the above case is
    Iab=100 angle 0° ; Ibc=100 angle -120° ; Ica=100 angle -240°.

    But the following is also one of the infinte solutions for the above ("trivial")case
    Iab=105 angle 0° ; Ibc=97.596 angle -117.457° ; Ica=97.596 angle 117.457°.

    Kindly clarify where I am being misleaded.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I suspect one must turn to the question of the minimum instantaneous energy state which will satisfy the indicated steady-state conditions.

    Edit:

    Perhaps I should attempt to elaborate rather than make cryptic statements - it's a bad habit of mine.

    For the sake of some plausible reasoning method let me assume that the Delta configured source elements each have equal resistive components as part of their internal source impedance. One would also have to make other assumptions such as equal voltage magnitude and symmetrical phase displacements. So we have symmetrical delta configuration. Hopefully this is a reasonable assumption for us to proceed upon.

    My expectation would be that any physical system, such as the delta source by way of example, would "adjust" itself such that its energy state is at a minimum level whilst supplying the energy demand required of it.

    So you propose a delta source supplying given line currents. The question is whether there is an infinity of possible solutions for the delta currents or one unique solution.

    I proposed a "trivial" situation of balanced line currents and (probably like many electrical engineers) blithely assumed that the delta currents would also be balanced - "that's how it always works" I think to myself. Interesting how notions become so embedded in one's thinking that we often fail to remember that a justification might be helpful.

    I digress - so I return to my justification of minimum energy state.

    As a basis I'll use your example of a solution ...

    Iab=105 angle 0° ; Ibc=97.596 angle -117.457° ; Ica=97.596 angle 117.457°.

    The symmetrical delta internal source losses with for any currents Iab, Ibc & Ica would be proportional to a factor K, where

    K={\| I_{ab} \|}^2+{\| I_{bc} \|}^2+{\| I_{ca} \|}^2

    For your proposed particular solution (one of an infinite no. of possibilities) we have a value for K of

    K=105^2+97.596^2+97.596^2=30,074.958

    If one now takes the "trivial" option of balanced delta currents [100A in each delta branch] we obtain a value for K of exactly 30,000. This suggests that a lower energy (loss) state would exist in the delta for balanced currents than would be the case for your particular solution.

    I imagine one would discount any other solution on this basis.

    Whether you accept the premise that the minimum energy state argument is a credible one is a matter for your consideration.
     
    Last edited: Mar 21, 2013
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  5. elecidiot

    Thread Starter New Member

    Jun 29, 2010
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    I am amazed by ur view of the prob in other dimension. But when we add resistances of equal magnitudes in the source, we automatically get a unique solution(there is no possibility of infinite solutions) since we get another (4th) equation by applying KVL around the delta. But the concept of minimal energy state remains true in all scenarios. Thanks for ur clarification.
     
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