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stason (Electrical) (OP)
19 Apr 10 10:41
Hello,

I am performing some simulations in PowerFactory where I energize a transformer (13 MVA, 57/10,5 kV, YNd11) on the LV side (delta side) from a 7 MVA synchronous generator (with the star point solidly grounded).

The inrush current causes 2nd and 3rd harmonics to appear in the generator's currents.

If I look at the simulated zero-seqence current of the generator, it equals zero. I figured this could not be true, since the star point of the gen is connected, which would allow the 3rd harmonic currents to flow in the gen's neutral, which they also do. Is this correct? And if it does flow in the neutral, might the ground current relay trip?

Besides, will the 2nd harmonic current in the generator make the differential relay think it could be an internal fault?

Thank you in advance!
Regards,
stason
dpc (Electrical)
19 Apr 10 11:42
Very few 7 MVA generators have solidly grounded neutrals.  

Transformer inrush contains a lot of 2nd harmonics.  

Differential relays used with transformers have harmonic restraint to prevent false operation on inrush.  

I would have low expectations of any simulation software being capable of accurately modeling transformer inrush.  It's a very complex event.   

David Castor
www.cvoes.com

Helpful Member!(2)  jghrist (Electrical)
19 Apr 10 12:09
Differential relays use 2nd harmonic currents to inhibit tripping during inrush.  Third harmonic currents, if they are balanced, are zero-sequence and will not flow because of the delta transformer winding.  During inrush, however, the third harmonic currents are not balanced.  Third harmonic currents are present in the line during inrush even with a delta winding, and they sum to zero.

See the attached graphs which were developed from a Fourier analysis of inrush current in the primary feed to a 100-24.9 kV, 30 MVA delta-wye transformer.  One shows Phase X raw waveform with harmonic RMS components.  One shows the 3rd harmonic RMS in all phases.
 
jghrist (Electrical)
19 Apr 10 12:10
stason (Electrical) (OP)
19 Apr 10 12:52
THank you for your responses.

jghrist, why do the 3rd harmonic currents sum to zero if they are unbalanced?

The question, however, was primarily: will there be a current flow in the generator's neutral due to the 3rd harmonic currents, which would cause the ground current relay at the generator to trip (when the pick-up value is reached)?

And if yes, what is the return path of this current? It can't return to the delta winding of the transformer, right? Or will it return to the HV wye-winding of the transformer, which is solidly grounded?

Concerning the differential relay, I meant the one for the gen's stator differential protection. But I guess I was thinking wrong. The 2nd harmonic currents are negative-sequence ones and they actually appear at the both sides of the stator of same amplitude and they cancel each other at the neutral, since they are 120 degrees apart. Unless they are unbalanced and then have a third harmonic content. So I guess the stator differential protection relay should not react to the negative-sequence currents.

Thanks!
Helpful Member!  electricpete (Electrical)
19 Apr 10 14:15

Quote:

jghrist, why do the 3rd harmonic currents sum to zero if they are unbalanced?
If 3rd harmonic components are balanced they are zero sequence and do not sum to zero. Therefore only unbalanced 3rd harmonic component can sum to 0.

Balanced means there is a time shift between phases which corresponds to 120 degrees of fundamental.  That same time shift corrresponds to 240 degrees of 2nd harmonic and 360 degrees of 3rd harmonic.  So balanced 3rd harmonic is always zero sequence and can't flow where zero sequence can't.  

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jghrist (Electrical)
19 Apr 10 15:39

Quote:

The question, however, was primarily: will there be a current flow in the generator's neutral due to the 3rd harmonic currents, which would cause the ground current relay at the generator to trip (when the pick-up value is reached)?

And if yes, what is the return path of this current? It can't return to the delta winding of the transformer, right? Or will it return to the HV wye-winding of the transformer, which is solidly grounded?
There will not be current flow in the generator neutral. As you say, there is no return path because of the delta winding.

Quote:

Concerning the differential relay, I meant the one for the gen's stator differential protection.
The transformer inrush current is seen by both inputs of the generator differential so it will not cause a trip.

 
Helpful Member!  davidbeach (Electrical)
19 Apr 10 17:58

Quote (jghrist):

There will not be current flow in the generator neutral. As you say, there is no return path because of the delta winding.

But, but, but...

There will be 3rd harmonic in the neutral.  That's the whole basis of the 3rd harmonic 100% stator ground fault detection.  No, it won't flow beyond the delta winding, true, but it is flowing in all the capacitance, both deliberate and parasitic, in the circuit between the neutral and the delta winding.
stason (Electrical) (OP)
19 Apr 10 18:14
Thanks again, guys.

What I don't understand though is that the 3rd harmonic is present in gen's phase currents during the inrush, although the transformer is energized at the delta winding. How can the 3rd harmonic current flow then?

davidbeach, the parasitic capacitance is connected in parallel to the system, which would enable the flow of the current through the neutral, yes. But this current still must return to the transformer, which caused it. Right?

Regards
davidbeach (Electrical)
19 Apr 10 19:00
Generators produce 3rd harmonic.  Generator -> out through capacitances to ground -> up the neutral to the generator.  Complete circuit.
Helpful Member!  slavag (Electrical)
20 Apr 10 3:12
Hi.
Please take in account. Standard generator diff protection, 87G is not include 2-nd harmonic detector/blocking.

What Dvaid saied:
Generators produce 3rd harmonic.  Generator -> out through capacitances to ground -> up the neutral to the generator.  Complete circuit

But isnt influence on the diff protection.

Best Regards.
Slava
stason (Electrical) (OP)
20 Apr 10 3:39
Hmmm, the weird thing is that the 3rd harmonic also appears in another generator, whose star point is not grounded at all.  
AMBMI (Electrical)
20 Apr 10 8:38
Stason, please could you attach your PF .dz file with your project? I hope that your company policy could allow that.
I have some experience with Power Factory and I would like to take a look. Moreover please consider that the PF differential relay models can include 2nd harmonic measurement and differential blocking (obviously if it's present in the "real" relay we are mocking up). It could be very nice to simulate such generator behavior and see if the relay is blocked or not.
jghrist (Electrical)
20 Apr 10 11:43

Quote:

What I don't understand though is that the 3rd harmonic is present in gen's phase currents during the inrush, although the transformer is energized at the delta winding. How can the 3rd harmonic current flow then?
The inrush 3rd harmonic is not balanced, so it is not necessarily zero-sequence.  Because of the delta winding, no zero-sequence current can flow; this necessitates that the 3rd harmonic current not be zero-sequence.
 
stason (Electrical) (OP)
21 Apr 10 3:49
AMBMI, I'd like that too, but unfortunately I am not allowed to post it. One can easily build one though with some transformer that has saturation and maybe even a standard generator out of the library.

jghrist, that makes sense. I think I will go with that. Thanks.
Bronzeado (Electrical)
21 Apr 10 11:04
Hi folks,

jghrist is correct when says that 3rd harmonic is not a zero seguence component.
By definition, zero sequence components in a three-phase system must have the same magnitude, phase and, of course, frequence.
Any harmonic components (in a thre-phase system) that have equal magnitude, phase and frequence are, by definition, zero sequence. The 3rd harmonic (in three-phase system), in general, are, particularly, of sequence zero.
For example, a 3rd harmonic generated in just one phase of a three-phase circuit IS NOT A ZERO SEQUENCE COMPONENT and can flow with or without delta conections.

Now, I wonder if somebody could explain why there is a negative offset appears in the current waveform sent by jghrist.

Regrads,

H. Bronzeado
 
stason (Electrical) (OP)
21 Apr 10 11:19
Bronzeado,

I think the current goes below zero once every cycle as the voltage driving the transformer goes negative.  
Bronzeado (Electrical)
21 Apr 10 11:59
stason,

My guess (only gues!) is that something happens in the magnetic core during inrush, magnetizing it "homopolarly" (sorry for the word I think does not exist in English). It may occur due to the zero sequence magnetic flux, which path leaks from the yokes (say superior yokes), goes into the transformer tank and back to the yokes (in this case, inferior ones).

I wonder if the experts on transformer in this forun could give us a good reason for that phenomena which is not generally seen in simulations.

Regards,

Herivelto Bronzeado    
electricpete (Electrical)
21 Apr 10 13:10
Here's my explanation why there is a dc offset in transformer inrush:

Look at one phase fed by voltage v(t) = cos(w*t + theta0)

simplify it to a linear model and neglect resistance
v = L di/dt  (where L is magnetizing inducatance)
i = 1/L * int(v(t) dtau from tau=0 to t

i = 1/L * int( cos(w*tau + theta0) dtau + i(0)   where i(0) = 0

i = [(1/L) * (cos(w*tau + theta0) ] evaluated at tau =t minus same thing at tau = 0.

i = [(1/L) * (cos(w*tau + theta0) ] evaluated at tau =t minus same thing at tau = 0.

i =  [(1/L) * (cos(w*t + theta0) ] -  [(1/L) * (cos(0*t + theta0) ]

i =  [(1/L) * (cos(w*t + theta0) ] -  [(1/L) * (cos(0*0 + theta0) ]

i =  [(1/L) * (cos(w*t + theta0) ] -  [(1/L) * (cos(theta0) ]

i =  [(1/L) * (cos(w*t + theta0) ] + Idc
where Idc = -  [(1/L) * (cos(theta0) ] is a dc component.

You can see there is a dc component which varies depending on theta0... if theta0=+/- 90 degree  (close at max or min of voltage which is the "natural current zero", then there is no dc offset, otherwise there is a dc offset).

If you added back in the effects of resistance the dc component decays away.

If you add the iron non-linearity, the current is highly non-sinusoidal... the dc component causes much higher saturation current peak on one side of zero than the other.

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electricpete (Electrical)
21 Apr 10 13:13
Small correction in bold

Quote:

v = L di/dt  (where L is magnetizing inducatance)
i = 1/L * int(v(t) dtau from tau=0 to t

i = 1/L * int( cos(w*tau + theta0) dtau + i(0)   where i(0) = 0

i = [(1/L) * (SIN(w*tau + theta0) ] evaluated at tau =t minus same thing at tau = 0.

i = [(1/L) * (SIN(w*tau + theta0) ] evaluated at tau =t minus same thing at tau = 0.

i =  [(1/L) * (SIN(w*t + theta0) ] -  [(1/L) * (SIN(0*t + theta0) ]

i =  [(1/L) * (SIN(w*t + theta0) ] -  [(1/L) * (SIN(0*0 + theta0) ]

i =  [(1/L) * (SIN(w*t + theta0) ] -  [(1/L) * (SIN(theta0) ]

i =  [(1/L) * (SIN(w*t + theta0) ] + Idc
where Idc = -  [(1/L) * (SIN(theta0) ] is a dc component.

You can see there is a dc component which varies depending on theta0... if theta0= 0 degree  (close at max or min of voltage which is the "natural current zero", then there is no dc offset, otherwise there is a dc offset).

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electricpete (Electrical)
21 Apr 10 13:17
For completeness, the current solution should also be divided by a factor of w that I forgot, but doesn't affect the conclusion.

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Bronzeado (Electrical)
21 Apr 10 14:24
Sorry folks! I did not make myself clear.
The offset I talk about is the negative part that appears in the current waveform, which it supose to be zero.

electicpete, could you, please, explain that offset?

Regards,

Herivelto  
electricpete (Electrical)
21 Apr 10 14:41
I was trying to explain why the graph posted 19 Apr 10 12:10 has much larger positive component than negative component.  

The answer to that question has to do with angle of voltage at time of closing as was demonstrated for simple inductive circuit.   

I would be glad to elaborate on that, but I'm not sure I'm answering the question you asked....

Quote:

The offset I talk about is the negative part that appears in the current waveform, which it supose to be zero.
Which current waveform (posted what date)?
I didn't see any where the negative is larger than the positive. (where?)
Why would part of the waveform be zero?

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Bronzeado (Electrical)
21 Apr 10 15:06
electricpete,

Thank you very much for your interest. I think this phenomenon is not so trivial.

I am not talk about "components" but about the Phase X current waveform itself posted on 19 April 10 12:52 by jghrist. In that waveform, a typical transformer inrush current, we can see a negative part which does not appear normally in computer simulations.

Best regards,

Herivelto Bronzeado
davidbeach (Electrical)
21 Apr 10 15:22
Then your computer simulations are wrong.  That's what inrush looks like.
electricpete (Electrical)
21 Apr 10 15:29
Attached is a simulation I did in excel vba using simple R/L circuit... the result in tab "plot sheet" looks very similar to the one posted by jghrist.  

My question to you is why would you expect it to look differently?  It is not a hard matter to construct the curve yourself by mentally integrating (assuming you start with simple model... my preference is to start with simple models... do you think something is missing from simple R/L model that would affect the results?).

My suspicion is that you may be used to viewing transients where the positive part goes much higher so the plot scale gets compressed and the negative portion of the cycle is hard to see.

By the way, you can tweak the parameters in the spreadhseet and re-run the simulation if you'd like.... I think the instructions are self-explanatory.

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electricpete (Electrical)
21 Apr 10 15:34
Another thought – is it possible you had a load on the secondary of transformer during energization?  That would certainly change the waveform dramatically different than the R/L model I suggested.

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electricpete (Electrical)
21 Apr 10 15:49
The last plot closed at an angle midway between zero and peak of the voltage.   

Here is a plot that closes at voltage zero crossing which gives max offset.  I also extended the plot range to steady state.
 

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electricpete (Electrical)
21 Apr 10 15:52
I guess there is one difference between my simulation plot and jghrist's post... mine have a rounded negative portion of the wave and his have a flat portion.  I don't know why that is.  Any ideas?

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DHambley (Electrical)
21 Apr 10 15:58
A simple way to visually see where the 3rd harmonic currents are going is to simply sketch them on a peice of paper with a pencil. Forget the equations for a moment.
1. Draw 3 separate horizontal lines, then 3 waveforms 120deg apart.
2. Now draw in a 3rd harminic for each one, starting a 0 degress for each one.
3. Now compare all of your 3rd harmonics. See how they are all in phase? They can't all flow into the neutral without current flowing in the neutral.  
Bronzeado (Electrical)
21 Apr 10 16:34
Hi davidbeach. I used to say that the only certainty I have about simulations is that they are not the reality! By the way, do you have any ideia about how to explain this phenomenon?

electricpete, your your simulation looks good. However, the jghrist's  current waveform has a flat "zero" offset. I am sending attached another inrush current waveform that shows the same behaviour. The question is why this offset occurs? Please, explain physically, if possible.

Regards,

Herivelto Bronzeado
Bronzeado (Electrical)
21 Apr 10 16:54
electricpete,

As I said, my guess is that this offset may be caused by the an offset magnetizing of the core during the transformer inrush.
In a three-phase transformer with three limbed core, this phenomenon can beseen in the neutral current, indicating that this should be caused by zero sequence magnetic flux (induction), which path goes outside the core.

Regards,

H. Bronzeado
Bronzeado (Electrical)
21 Apr 10 17:18
electricpete,

In your simulation, the transformer seems to be over-excited in the steady-state.

Regards,

H. Bronzeado
electricpete (Electrical)
21 Apr 10 19:48
I didn't give a huge amount of thought to the particular paramater values set for that simulation other than to try to recreate the posted waveform.  I am under the impression that power transformers at no-load would be partially into saturation and some waveform distortion would be evident.

As load increases the current would look more sinusoidal due to:
1 - exciting branch voltage decreases due to voltage drop across leakage reactance. Therefore not as much saturation effects
2 - In addition to magnetizing current we now have load component which is shifted in phase relative to magnetizing component and will have less distortion.

At least that's my thought. It's been 10 years since I've looked at transformer waveforms.  

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stason (Electrical) (OP)
22 Apr 10 3:20
The fact that the inrush current can become negative while the flux stays above zero during the inrush means that each the flux and the current have their own DC offset magnitudes. That I just discovered from looking at some simulations.
stason (Electrical) (OP)
22 Apr 10 3:26
The physical explanation for the DC offset would be the law of the flux linkages, which states that the flux just before the switching event and right after it has to be the same and can't change instantaneously, if that was the question, Bronzeado.
electricpete (Electrical)
22 Apr 10 7:53
Bronzeado – I see now your earlier comments and questions were deeper than the simple aspects I was discussing.  My apologies.  I can imagine there can be zero sequence flux during energization,  but it's beyond my ability to visualize what that would look like in the current waveforms or to connect it with the flat shape on the negative side of the A phase current waveform.    

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