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Circulating currents in delta winding

  1. Sep 7, 2016 #1
    Dear PF members.

    I have trouble understanding how zero sequence currents circulate in a delta winding. For example zero sequence currents such as 3rd harmonics, they will circulate in the delta winding and get trapped, why?

    Why can they not go further?

    I have recently started to study transformers on my own, without a strong electrical background, and I feel maybe I lack some fundamental understanding to be able to answer my question?


    Thank you.
     
  2. jcsd
  3. Sep 7, 2016 #2

    jim hardy

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  4. Sep 7, 2016 #3
    Thanks, informative post.

    But I'm not sure why the 3rd harmonic is short circuited in the delta winding, why does it circulate instead of flowing out of the delta winding? I think if i understand this question, i am in a position to understand most of the post you linked.
     
  5. Sep 7, 2016 #4

    jim hardy

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    For third harmonic the windings are in series adding
    so current V3rd is I3rd harmonic X Z = third harmonic current capability of source X impedance of transformer ,
    meaning 3rd harmonic content of your power source is very nearly short circuited to zero by the delta winding.
    If harmonic current capability is 5% and transformer impedance is 10% , 3rd harmonic voltage is reduced to 0.5%

    That's why one must be aware and cautious when he connects a small transformer to a big source.
    If he connects them such that the little transformer tries to short circuit all the third harmonic content of a big source , the poor little transformer will melt.
    If you want to read a real world account of that see this article,,
    http://www.designnews.com/author.asp?doc_id=273075 [Broken]
    ( which i post with trepidation ,,,, o:) shameless self promotion? Sorry about that. )



    old jim
     
    Last edited by a moderator: May 8, 2017
  6. Sep 7, 2016 #5
    In a delta, the 3rd harmonics can only circulate, they cannot exit the delta & cannot appear on the lines. The 3 phases are 120 degrees apart for the fundamental frequencies. The current in a line is the difference between the 2 phase currents summing into that line. For example, Line current "A" consists of delta phase current "CA" minus phase current "AB". ref line current A exiting delta, CA phase current directed towards node A, AB phase current directed away from node A.

    Since these current phasor fundamentals are 120 degrees apart, their difference is a phasor with magnitude sqrt(3) times the phase current magnitude, & phase 30 degrees behind. The 3 line currents must sum to zero, which they do. If the 3 phase currents are 1.0 unit in magnitude, with phases of 0, -120, & +120 deg, the 3 line currents are sqrt(3) in magnitude, with phases of -30, -150, & +90 deg. The 3 line currents have equal magnitude with 120 deg displacement, so that they must sum to 0. It is impossible for a system with only 3 wires to have the 3 currents not sum to 0.

    But the 3rd harmonic is a different beast. The triple frequency waveforms are phase displaced not by 120 deg, but by 3 times 120 deg, or 360 deg. Three sine waveforms displaced by 360 deg are in unison, or in phase if you prefer. Since there are 3 line currents, & all 3 MUST sum to 0, the magnitude must be 0 as well. But inside the delta 3rd harmonics can & do circulate. The 3 line currents are all 0. But each line current is the difference between 2 phase currents. Let's use 0 deg as the phase of all 3 phase currents, with magnitude of 1.0 unit.

    I_CA = I_AB + I_BC = 1.0 at angle 0 deg.
    I_A - I_CA - I_AB = (1.0 + j0) - (1.0 + j0) = 0. Likewise for the other 2.

    The 3rd harmonic currents circulate within the closed loop of the delta, but do not propagate onto the line currents. Did I help?

    Claude
     
  7. Sep 8, 2016 #6
    Both of you, thanks.

    Nah, it was nice reading.


    EDIT:
    Case 1: Lets say we have a delta-delta transformer supplying an unbalanced load, then currents will circulate in the windings in such a way that the sum of the 3 line currents equal zero?

    Case 2: If we now have a wye-delta transformer supplying an unbalanced load, then the currents will circulate in the delta winding, but lead to an increase of current in the neutral wire on the wye side?
     
    Last edited by a moderator: May 8, 2017
  8. Sep 8, 2016 #7
    EDIT 2:

    Did you mean:
    I_CA = I_AB = (+) I_BC = 1.0 at angle 0 deg
    I_A = (-) I_CA - I_AB = (1.0 + j0) - (1.0 + j0) = 0
    (Symbols in parentheses are what you wrote in your post.)

    Or did I missunderstood something?
     
  9. Sep 8, 2016 #8

    Averagesupernova

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    I would have known that was you regardless. The phrases "fry spit", and "a question well stated is half answered" have you written all over them.
     
    Last edited by a moderator: May 8, 2017
  10. Sep 9, 2016 #9
    Yes the currents will circulate in the delta winding. No they will not increase Y side neutral current. If the Y primary has the neutral connected, i.e. a 4-wire Y, the 3rd harmonics see a pair of paths in parallel. The 3rd harmonics can circulate in the delta loop, and in the 4-wire Y, since 3rd harmonics can exist in the 3 primary hot wires, & return on the neutral.

    But when there are 2 parallel paths, the path with lower impedance will have the greater portion of the total current. What impedance does the delta loop present? The copper resistance plus the reactance due to leakage inductance. This is a small value of impedance. What is the impedance on the Y primary path? It is the resistance of the copper from the primary all the way back to the previous transformer secondary, or generator. It could be miles. Likewise for inductive reactance. The cables are specified in ohm/mile & henry/mile, so for many miles the value could be large.

    As a result, the delta loop generally offers a lower impedance path for 3rd harmonics, so most of these currents are in the delta. The Y primary neutral will have a small 3rd harmonic current, but for the most part, the delta provides the path. The greater the delta current, the smaller the Y neutral current.

    Does this help, BR.

    Claude
     
  11. Sep 10, 2016 #10
    Yes, that helps, thank you for your time.

    @cabraham and @jim hardy , one last follow up question:
    If we have a wye-wye transformer where only 1 side is grounded, this means no zero sequence current can flow in the transformer, correct?
    What would happen if this transformer were supposed to supply a load that produces 3rd harmonics i.e. zero sequence currents, would it be impossible for this transformer setup to supply the load?

    However if a delta tertiary winding were to be introduced this transformer could supply the load that produces 3rd harmonics, am I correct?
     
  12. Sep 10, 2016 #11

    jim hardy

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    I knew you'd ask and i have been wrestling with that question.


    I dont yet have a stepwise mental model worked out.
    I apologize, just i'm not fluent in phase sequence components.
    My intuition says it'll distort the voltage wave and give 180hz voltage at the neutral.

    i'm perusing these series trying to improve my own understanding.
    http://www.vias.org/matsch_capmag/matsch_caps_magnetics_chap6_12_08.html
    http://www.hammondpowersolutions.com/files/HPS_article_Harmonics_PhaseShifting.pdf
     
  13. Sep 11, 2016 #12

    jim hardy

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    @cabraham
    i'm still scrambled....
    any pointers ?

    old jim
     
  14. Sep 12, 2016 #13

    CalcNerd

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    Not that I know anything either, but I always thought of the third harmonic (triplen) as a 180 Hz spike where 2/3rds of the wave is flat, because most of the third harmonics (2 out of 3) ARE canceled. It's just every third waveform, gets to add to the other two phases and form a spike. This spike is large and causes a large flow of current. AND because the I^2R factor kicks in, large amounts of heat is generated due to spike currents. So all the conductors are subjected to large spike currents larger than the conductors are typically sized to carry.

    But that is my interpretation.
     
  15. Sep 12, 2016 #14

    jim hardy

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    I appreciate that kind of stepwise thinking.

    I woke up last night thinking about Lundell automobile alternators . They are three phase wye machines connected to full wave rectifiers.
    3rdharmonic_lundell.jpg

    there's the nonlinear load. It flattens the voltage peaks and that's certainly hatmonic-rich.
    What happens at the simplest level is quadrature current cancels field mmf , presumably making stator flux non-sinusoidal
    I've never seen a paper on harmonics in a Lundell but there must be one out there.

    Will look some more.

    I'm not quite up to this one just yet.
    http://portal.unitbv.ro/Portals/0/UserFiles/User334/ICCS_2009_-_Stoia+Cernat+Mailat+Ilea.pdf
    3rdharmonic_lundell2.jpg

    there has to be a painless way to teach it.

    old jim
     
  16. Sep 12, 2016 #15
    Correct.
    Yes, but I need to emphasize the following. "Zero sequence" & "3rd harmonic" currents are NOT one & the same. Zero sequence currents exist when the line and/or load is unbalanced. The triplen harmonic currents are a natural property of the non-linear transformer core. Zero sequence currents are bound by the law of balancing ampere-turns. A zero sequence current on the secondary can exist only if there is an equal amp-turn value on the primary, & vice-versa. The triple harmonics, OTOH, are NOT bound by amp-turn balance. The triple harmonic is a part of the magnetizing current, & magnetizing current does not get balanced.

    Take a single phase unit with no load. The magnetizing current exists in the primary but not in the secondary, thus amp-turns do not balance. The triple harmonic exists as the primary path is 2 wires, so all harmonics have a path. But if the unit is a 3-phase with a Y on both primary & secondary, w/o a delta tertiary, the magnetizing current in the primary will contain non-triplen harmonics, i.e. 1st, 5th, 7th, 11th, etc., with or without a neutral. But the triplens, i.e. 3rd, 9th, 15th, etc., will exist only if a neutral path exists.

    Now getting back to zero sequence current, if the xfmr is Y-Y, no delta tertiary, an unbalanced load on the secondary from line to neutral will result in the 3 phase voltages going out of balance. With 3 phase shell type core, or 3 single phase units ganged together for 3 phase operation, the Y-Y configuration is avoided, since it cannot support an unbalanced load with only 3 wires. A 4th wire, neutral wire, mitigates this problem. But the whole purpose for using 3 phase is to save conductor cost/space. A 3 phase system requires only 75% the aluminum (or copper) of single phase, or any other number of phases. Using a 4th wire for the neutral defeats the savings, so that a 3 phase system consumes 100% the amount of conductor as any other number of phases.

    If the distance is large, a 4th wire adds a lot of cost as well as support structure needed to carry this wire. Also, with a Y-Y, even if the secondary load is balanced, the flux gets distorted by the absence of triplen harmonics. The line to neutral voltages can exhibit peaks that are 40 to 70% higher than a pure sine curve. This is mitigated by a delta tertiary. The delta requires an additional set of windings, light gauge since the current is small. The cost is much less than miles of extra wire. The triplen harmonics circulate in the delta, & are not balanced in the primary.

    For these reasons it has been standard practice in the power industry to avoid a pure Y-Y connection. With 3-phase shell type cores, & 3 individual cores, the Y-Y is very problematic. But if a delta is present, then zero sequence currents due to unbalanced loads circulate in the delta, as well as triplen harmonics. The delta mitigates both problems. Sometimes it is desirable to connect both primary & secondary in Y. A Y can provide a very good ground point, & is robust for high voltage insulation stress relief. But a delta is added via a tertiary to maintain support for unbalanced loading, & harmonics.

    There is, however, another core type in use, the 3-phase core type unit. This is a pair of "E" shaped cores. Each core half has 3 legs of equal width, 2 pieces joined together. This type of core has all 3 phases **magnetically** coupled in addition to electrically coupled. The other 2 have only electrical coupling. This core type behaves as if it had a delta tertiary. The technical reason is involved, I don't have time, but maybe I can write a paper soon. The 3 fluxes are forced to sum to zero. A delta forces the 3 voltages to sum to zero via its closed loop. But voltage is the time derivative of flux, times turns, with a minus sign. Anyway, if the core is a 3-legged "E" type pair, a Y-Y configuration supports unbalanced loading on the secondary, using only 3 primary wires, without a delta.

    Maybe I can explain why later.

    Yes.

    :-) Claude
     
    Last edited: Sep 12, 2016
  17. Sep 12, 2016 #16

    jim hardy

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    Thank you Claude. As i said i'm not expert in phase sequence methods. That was to be my next question.

    I have used the terms interchangeably because so many authors seem to, but last few days have questioned (in my alleged mind) their equivalence .
    Last night it dawned on me if a 60hz 3 phase thought experiment can have positive negative and zero sequence, so can a 180 hz one.

    Perhaps triplen harmonic load currents acquire a sequence when they flow ?

    Every single thing i learn shines a flashlight into the abyss of my ignorance. When this point sinks in it'll be one less dark corner...

    I look forward to that.

    old jim
     
  18. Sep 13, 2016 #17
    Hi, I have also been struggling with this topic, I have some questions.

    If we think of a 3 phase load producing tripling harmonics (current sources in parallel with load, injecting harmonic currents into the system).

    (I've read that harmonics generated due to non-linear loads can be thought of current harmonic sources connected in parallel to the load, so that is how I'm thinking when I'm imagining the following scenario.)

    Scenario 1:
    4 wire wye-delta: Tripling harmonics are traveling from the load to the delta side on the transformer, these currents gets trapped in the delta loop, but how are they amp-turn balanced by the primary side?
    I understand that triple harmonics drawn by the transformer for magnetizing current are not amp-turn balanced, but if we have a load that produces 3rd harmonic currents and injecting them back to delta secondary, how are these currents amp-turn balanced on the primary 4 wire wye side?

    Are you meaning of 3rd harmonics needed to generate a sinusoidal flux? If you're not, and you mean 3rd harmonics produced by the load - I do not see how the 3rd harmonics generated by the load are amp-turn balanced.


    Thanks.

    (I'm amazed by your knowledge cabraham)
     
  19. Sep 16, 2016 #18
    Thanks LagCompensator. You ask a very interesting question. Triplen harmonics, when discussed, are usually taken as magnetizing current, which is not amp-turn balanced. But nobody ever brought up triplen harmonics due to non-linear loading, interesting, & worthy of examination.

    An example very ubiquitous is the classic incandescent lamps. If 3 lamps are connected across a 3-phase line, in Y or delta, what is the result? Let's start with a single lamp with a single phase power bus. An incandescent lamp exhibits an impedance that increases with increasing voltage, assuming constant voltage drive, which is what commercial power is designed to do. So if the lamp voltage is a pure sine wave, the current will be a sine wave with a flattened top & bottom. This will result in a current spectrum that consists of a fundamental, plus odd harmonics. Because the lamp is symmetrical in both directions, the current waveform will possess half wave symmetry, so that no even harmonics will exist.

    Since single phase has 2 wires, all harmonics have a closed path to circulate. But with 3 phase, that is not the case. Taking 3 identical lamps across a 3 phase bus, first we put the lamps in delta connection. In a delta lamp load, each lamp is across the bus line to line, so that the voltage is a pure sine. What currents do we obtain? Well, obviously the 3 line currents cannot contain triplens, i.e. no 3rd, 9th, 15th, etc., harmonics of the current, only fundamental plus 5th, 7th, 11th, etc. But the lamps are in delta so that the triplens must circulate inside this delta loop formed by the lamps. Yet they cannot exist on the 3 phase wires.

    But if the lamps are connected in Y, then there cannot be any triplens at all, since there is no path. In Y, each lamp is not connected directly across the bus, so the lamp voltage will not be a pure sine. With the triplen currents missing, each lamp will have a voltage which contains the fundamental frequency, plus all triplens, i.e. 3rd harmonic, 9th, etc. The lamp voltage is not pure sine.

    Very interesting problem, maybe I can set up a Spice simulation. I would have to brush up on how to define a non-linear load such as a lamp. I hope this helps, let me know if elaboration is needed. Best regards.

    Claude Abraham
     
  20. Sep 17, 2016 #19
  21. Sep 17, 2016 #20

    jim hardy

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    Thanks guys. You answered some nagging questions from my boring forty year old diesel anecdote .


    old jim
     
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