Circulating currents in a delta winding

In summary, in case 1, the current in the line will sum to zero because the line current and the 3rd harmonic current are the same magnitude. However, in case 2, the current in the neutral wire will be increased because the 3rd harmonic current is higher than the line current.
  • #1
BjornFanden
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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.
 
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  • #3
jim hardy said:
This thread has a simple explanation for third harmonic, see if it helps

https://www.physicsforums.com/threads/odd-harmonics-in-power-system-reduction.882923/#post-5550164

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.
 
  • #4
BjornFanden said:
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.
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
( which i post with trepidation ,,,, o:) shameless self promotion? Sorry about that. )
old jim
 
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  • #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
 
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  • #6
Both of you, thanks.

jim hardy said:
http://www.designnews.com/author.asp?doc_id=273075
( which i post with trepidation ,,,, o:) shameless self promotion? Sorry about that. )
Nah, it was nice reading.EDIT:
Case 1: Let's 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?
 
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  • #7
EDIT 2:

cabraham said:
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.

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?
 
  • #8
jim hardy said:
http://www.designnews.com/author.asp?doc_id=273075
( which i post with trepidation ,,,, o:) shameless self promotion? Sorry about that. )

old jim

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.
 
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  • #9
BjornFanden said:
Both of you, thanks.

Nah, it was nice reading.

EDIT:
Case 1: Let's 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?

Yes. Both primary & secondary line currents will sum to zero for 3 wires.

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?

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
 
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  • #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?
 
  • #11
BjornFanden said:
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?
I knew you'd ask and i have been wrestling with that question.I don't 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
 
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  • #12
@cabraham
i'm still scrambled...
any pointers ?

old jim
 
  • #13
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.
 
  • #14
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
 
  • #15
BjornFanden said:
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?
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?
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.

However if a delta tertiary winding were to be introduced this transformer could supply the load that produces 3rd harmonics, am I correct?
Yes.

:-) Claude
 
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  • #16
cabraham said:
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.

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
 
  • #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?
cabraham said:
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.
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?

cabraham said:
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.

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.
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)
 
  • #18
LagCompensator said:
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)

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 possesses 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
 
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  • #19
Thanks, for your time. No extra elaboration is needed @cabraham.

By the way I found some Spice simulations for the pretty much the same problem, can't believe I have not seen that page before.
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/harmonics-polyphase-power-systems/
 
  • #20
Thanks guys. You answered some nagging questions from my boring forty year old diesel anecdote .old jim
 
  • #21
cabraham said:
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.

Hey, very sorry to rehash this after a couple of years, but I am relatively confused.

- If Y-Y configuration cannot support an unbalanced load, then what happens when there is load unbalance on the secondary? There is always at least some load unbalance I would imagine. It seems impossible that every load on the system would be 100% balanced.

- Were you still speaking of a Y-Y configuration where only one side is grounded or are you speaking of a Y-Y configuration with neither side grounded? In the second case I could understand how there could be no zero-sequence current able to flow during normal operation because it has no where to flow (although this still confuses me because again, because I see no realistic way the load can be exactly the same on each phase). Or if there is unbalance in the system, does the N-G point just shift to balance things out?

-With the secondary grounded at the star point, with no neutral wire, I would expect that triplen harmonics and zero-sequence current could still flow (even though there are only 3 wires) through the ground connection. Therefore, loads could have some unbalance and the únbalance current would have a path to and from ground at the star point.

- On the Y-Y configuration with neither side grounded, during a SLG fault, one of the phases now has a ground connection and zero-sequence current then flows from the line to Earth capacitances through the fault. This is an asymmetrical fault and the three-phase system is now unbalanced, but able to keep serving L-L connected loads… Or is this only when a wye-delta or delta-wye is being used. I think if it was a Y-Y configuration, then a SLG fault on the secondary would be seen on the primary and result in high fault current coming from the primary (if there is generation on the primary side) which would trip ground fault protection. Or is this only when it is a Y-Y with both sides grounded?

Sorry that is a lot of confusion, but maybe you'd be so kind as to clear some of that up for me? :)

-
 

1. How do circulating currents form in a delta winding?

Circulating currents in a delta winding occur when there is an imbalance in the voltages of the three phases, causing a flow of current between the phases.

2. What causes an imbalance in voltages in a delta winding?

There are several factors that can contribute to an imbalance in voltages in a delta winding, such as variations in the impedance of the winding, unbalanced loads, or unequal resistance in the phases.

3. What are the effects of circulating currents in a delta winding?

Circulating currents can cause overheating and damage to the winding and connected equipment, as well as a decrease in efficiency and power quality.

4. How can circulating currents be reduced in a delta winding?

One solution is to use a star-delta transformer, which separates the primary and secondary windings and eliminates the circulating currents. Another option is to use a zigzag grounding transformer to divert the circulating currents to ground.

5. Can circulating currents be completely eliminated in a delta winding?

It is not possible to completely eliminate circulating currents in a delta winding, but measures can be taken to minimize their effects and prevent damage to the equipment.

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