Sequence Networks

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mityeltu

mityeltu

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I am a recent grad and am about ton start working in design engineering. I am told I will be doing a number of calulations with ETAP. No problem there, I have used it in the past. However, I have been in controls for a while and not using what I learned in school from my power courses, so I'm brushing up a little.

I am revisiting the sequence networks used for unbalanced fault analysis by rereading what I am lead to believe is a suitable text: Blackburn's Symmetrical Components for Power Systems Engineering.

I am looking at the networks for a line to ground fault. I understand that the networks should be connected in series but I don't understand why. Cn anyone shed some light on this subject for me?

Thanks.
 
I had to get out my copy of Stevenson?s ?Elements of Power Systems Analysis,? a textbook from a course now 20+ years in my past. :roll: What I (re)learned is that the connection of sequence networks is merely a convenient tool for remembering the equations. You start the analysis by writing down the symmetrical components of current, then you go through a series of matrix manipulations I no longer recall how to do. You end up with an expression for the fault current in terms of the voltage source and the sequence impedances. In the case of a single line to ground fault, the solution looks exactly like the solution you can deduce, without any matrix math, by drawing three sequence networks in series. In other words, we put them in series for this type of fault because someone first obtained the solution the hard way, and realized that the same solution can be obtained an easier way.
 
mityeltu said:
I am looking at the networks for a line to ground fault. I understand that the networks should be connected in series but I don't understand why. Cn anyone shed some light on this subject for me?

Thanks.
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Broadly speaking the faults can be classfied as:

1- Shunt faults ( short Circuits )
2- Series faults (open conductor )

Shunt type of faults involve power conductor or conductors to ground or short circuit between conductors. When circuits are controlled by fuses or any device which does not open all three phases, one or two phases of the circuit may be opened while the other phases or phase is closed. These are called series type of faults. These faults may also occur with one or two broken conductors. Shunt faults are characterised by increase in current and fall in voltage and frequency whereas series faults are characterised by increase in voltage and frequency and fall in current in the faulted phases.


Shunt type of faults are classfied as

1- line to ground fault
2- line to line fault
3- double line to ground fault
4- 3-phase fault.

The first three are the unsymmetrical faults as the symmetery is distubed in one or two phases. The method of symmetrical components will be utilized to analyse the unablancing in the system. The 3-phase fault is a balanced fault which could also be analysed using symmetrical components.


The series faults are classified as
1- one open conductor
2- two open conductors

These faults slso disturb the symmetry in one or two phases and are, therefore unbalanced faults. The method of symmetrical components can be used for analysing such situations in the system.


Somebody stop me!
 
Hameedulla-Ekhlas said:
Somebody stop me!
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You will have to do that part yourself. But may I suggest you take another look at the question. The question was related to the series connection of the sequence networks. A person has to have studied rather deeply into the field of power systems analysis, before even encountering the phrase "sequence network." I would conclude that the originator of this thread already understands the basics of what a fault is, and knows the difference between series and shunt faults, and knows about SLG, LLG, and other types of faults. Otherwise, he would not have been able to ask about sequence networks.

Now, where in your answer did you address the question that was being asked?:confused:
 
True, I understand the nature of how the faults are unbalanced and why the use of the symmetrical component networks makes the fault calculation easier. I have bee re-reading all this for a couple of weeks now and working through some basic problems (basic because any real sohpisticated problems will b ehandled by the computer - I just want the understanding of what the computer is doing)


So, charlie, if I understand you correctly, if I were to solve this problem using simultaneous equations (in symbolic rather than numeric form) My final equations shoud look like the networks were all in series for L-N fault and parallel for L-L faults correct?

I'll have to investigate that. Sounds reasonable enough.

Thank you.
 
mityeltu said:
So, charlie, if I understand you correctly, if I were to solve this problem using simultaneous equations (in symbolic rather than numeric form) My final equations shoud look like the networks were all in series for L-N fault and parallel for L-L faults correct?
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Correct, with one minor nitpicky comment. The technique of symmetrical components uses matrix algebra, rather than seeking solutions to "N simultaneous equations in N unknowns."
 
Right, I agree with that, though in my mind they are one and the same. once i have the simultaneous equations, I convert to matrices for solutions.
 
Yes, you are right. We are on the same page. I like the fact that I don't have to do any of that sort of thing anymore (not since passing the related classes so many years ago). ETAP (we use SKM) does all that for us. But I heartily agree with you that unless the user of such programs has an understanding of what they do, and how they go about doing it, they run the risk of getting a wrong answer (because of a wrong input), and not being able to discern the error.
 
charlie b said:
[SIZE
Now, where in your answer did you address the question that was being asked?:confused:
Click to expand...


Ok charli: let me try again

sequence network analysis is only for fault analysis

Originally Posted by mityeltu
I am looking at the networks for a line to ground fault. I understand that the networks should be connected in series but I don't understand why. Cn anyone shed some light on this subject for me?

Thanks.
Click to expand...

he has clearly mentioned regarding to line to ground fault.

I am not going in detail
the condition for fault from line to eath are represented by the equations Ib = 0, Ic = 0 and Va = 0. the sequence networks will be connected in series. The current and voltage conditions are the same when considering an open-circuit fault in phases b and c.

So, it means we consider open circuit in other phase b and c

I have already mentioned what are these.


The series faults are classified as
1- one open conductor
2- two open conductors
Click to expand...
 
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Mityeltu - I'll try to be brief, simple and correct and see if I can remember what Lew Blackburn taught me years ago. (My copies of his books are autographed. If I could become 10% of the engineer that he was, I would feel a lot a better about my designs and my contribution to the art.)

A three phase fault is symmetrical so there are no negative or zero sequence currents, so there are no networks to connect.

A single line to ground fault on phase A, neglecting load current should have Ia= If= fault current, and Ib=Ic=0. Look at equations 4.8 -4.10 in Lew's book.

The zero sequence current is Io= 1/3(Ia+Ib+Ic). Plug the current values into the equation and we get Io= 1/3(Ia +0+0) = 1/3 Ia. Similarly, solve for I1 and I2, the negative and positive sequence currents. They are = 1/3Ia also. To get the currents to be equal in a circuit, the networks have to be in series. The resulting fault current is = 3Io = Ia. Draw the networks and reduce them to the positive, zero, and negative sequence network equivalent impedances, add the numbers in series and divide into the voltage to get Io. Multiply by three to get Ia, the fault current.

This network example is good for explaining why the single phase to ground fault on a delta wye secondary transformer is greater than the three phase fault. The three phase fault current is equal to voltage divided by the positive sequence impedance. If = V/Z1

For most systems the positive and negative impedances are equal and the zero sequence impedance is in the same range or larger. But a delta-wye transformer breaks the zero sequence impedance circuit. The zero sequence impedance of all upstream equipment has no effect on the fault level, because the delta winding disconnects the zero sequence.

Assume Z1=Z2 and there value includes the transformer impedance and the upstream utility impedance. Zo will be less than that because it does not have the upstream impedance and the transformer Zo may be lower than Z1. The line-ground fault will be If =3 Io = 3 x [V/(Z1+ Z2 + Zo)]. If Z1=Z2=Zo, the result would be = 3V/3Z = V/Z, same as for 3-phase fault. But since Zo is much less, the single phase to ground fault is greater than a 3-phase fault. (Get a little ways away from the transformer and the zero sequence impedance grows back intothe same range or larger than the Z1 & Z2 , reducing the ground fault.)

For Line- Line Faults, start with a B-C fault. Then, Ia = 0, & Ib = Ic = If, and Vbf = Vcf. Plug those values into the equations and you see that we need to connect the positive and negative sequences in parallel to satisfy the voltage equations.

Welcome to the profession. Don't get discourages, we all struggle with these ideas. Just imagine how hard it would be to analyze three phase unbalanced circuits without this math concept!
 
Thank you. That was an excellent and clear example. I try not to get discouraged but at times a feel dumber AFTER I ask for help than before. Happily, this is not one of those occasions.
 
Since you have brought it up in your examples, let me ask. In the phase-phase fault on b-c, the current in phase a does not equal zero. The fault contribution from phase a equals zero. Is this right, I hope?
 
mityeltu said:
Since you have brought it up in your examples, let me ask. In the phase-phase fault on b-c, the current in phase a does not equal zero. The fault contribution from phase A equals zero. Is this right, I hope?
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Correct. All the equations we were looking at are based on the fault point and ignore load. Phase A is not involved in the fault so the current is zero.

For those that are lost on this subject: Symmetrical Components is a mathematical way of analyzing unbalanced three-phase systems by breaking it into three balanced systems with positive phase sequence A-B-C, negative (reverse) sequence A-C-B, and zero or no phase sequence (all voltages are in phase).

With balanced systems, I don't have to worry about the voltage drop in B phase caused by the current in A phase and the mutual coupling between the two wires.

When we get the answers for the positive, negative and zero sequence currents or voltages, we add them back together vectorially to get the "real" voltages and currents at the selected point in our system.

It is something like trying to describe a line on a graph (think Etch-A-Sketch). You could list for each point the direction (angle) and distance to the next point:

(Go five units at 53 degrees from horizontal, then 1.41 at 45.).

Or you could resolve it into the horizontal and vertical motion:
(Go 3 clicks right and 4 click up, then one click right and one up.)

Both get you to the same point but it is a lot easier to add the horizontal and vertical steps to find out where you are. (Four right and five up)

Symmetrical components does the same thing for our unbalanced three phases: breaks them down into something we can easily work with and then adds them back to the right answer.

It helps us go from rocks to hammers and then bigger hammers.

Without this math trick, engineers would never have been able to design, analyze, and build the extensive three phase electrical power system we have today.
 
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