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8/6/2019 Unsymmetrical Fault Analysis 1

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Zero sequence network

Negative sequence network

Positive sequence network

a I

+

a I

0a I

Z0

Z+

Z-

aV

+

aV

0aV

Fig. 1

Distribution systems generally do not satisfythe requirements in (1) and (2) above sincetransposition is not used, that is, distributionsystems are not symmetric.

Transmission systems generally do satisfythe requirements in (1) and (2).2.0 Important concept

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Section 12.2 of your text provides a verygood conceptual description of using

symmetrical components in fault analysis. Icondense this discussion here for you.Please read it, and digest/absorb it!!! SC provides 3 decoupled systems for analysis of unbalanced sources applied to asymmetrical system.

Faulted symmetrical systems (except for 3- phase faults) are not symmetrical systems,so it would appear that SC are not muchgood for SLG, 2LG, and LL faults.

But we can replace the fault with anunbalanced source (substitution theorem),then the network becomes symmetric.

Then get the sequence components of the(fictitious) unbalanced source at the fault point, then you can perform per-phaseanalysis on each sequence circuit.

3.0 Developing sequence networks

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Basic steps in using symmetricalcomponents for assessing faulted conditions

are (all quantities are assumed to be in pu).A. For positive, negative, & zero sequence:

1. Develop the sequence network for the system under analysis.

2. Obtain the Thevenin equivalents lookinginto the network from the fault point.B. Connect the networks to capture the

influence of the particular fault type.C. Compute the fault current from the circuit

resulting from step B.D. From step C, you will also determine the

currents in all three of the networks(positive, negative, and zero sequencecurrents). This enables computation of the phase currents Ia, I b, and Ic from Iabc=AIS.

We discuss each one of these steps in whatfollows.We address loads, lines, transformers, &generators. For each of these, we may derive

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expressions for the 0+- sequenceimpedances via the following steps:

1. Express abc voltages as a function of abccurrents and abc impedances.2. Substitute symmetric components for abc

voltages and currents (e.g., Vabc=AVS andIabc=AIS).

3. Manipulate to obtain the form VS=ZSIS.

If you refer back toSymmetricalComponents2, you will seethat this is the procedure we followed toobtain 0+- sequence impedances for a Y-connected load.

However, in what follows, we will not gothrough the analytical details but will rather just state the results.

3.1 Loads (see section 12.6 of text)

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Recall from SymmetricalComponents2that for a Y-connected, balanced load

grounded through a neutral impedance Znwith impedance ZY per phase, the 0+-sequence circuits are as in Fig. 2.

a I

+

a I

0a I

aV

+

aV

0aV

Fig. 2where we see the sequence impedances are:Z0 =ZY+3Zn (6)Z+=ZY (7)Z- =ZY (8)

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If the neutral is solidly grounded, then Zn=0and eq. (6) above becomes Z0=ZY.

If the neutral is ungrounded, then Zn=, andeq. (6) above becomes Z0=, i.e., the 0-sequence circuit is an open circuit, implyingno 0-sequence current flows in anungrounded Y connection.

For a delta-connected, balanced load, wesimply convert to an equivalent Y usingZY=Z/3 and then apply relations for anungrounded Y connection, resulting inZ0 = (9)Z1 = Z/3 (10)Z- = Z/3 (11)

Example: A delta connected balanced loadwith phase impedance Z is in parallel witha solidly grounded Y-connected load with phase impedance ZY. Draw the sequencenetworks for the entire paralleled load.

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It is possible to develop the sequencenetworks using the 3-step approach given

above (and I have notes that do that).However, this is very painful. Intuition,which suggests that we should just obtainthe sequence networks of the parallelcombination as parallel combinations of theindividual sequence networks, is right.Figure 3 shows the result.

a I

+

a I

0a I

aV

+

aV

0aV

Fig. 3

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3.2 Lines (see section 12.9 of text)

In SymmetricalComponents2, the workwe did to answer the question of What if the load (or line, or load and line) is notsymmetric? led to another question, whichwas: So what are the conditions for the off-diagonal elements of ZS to be 0?We have already reviewed the answer to thisquestion in Section 1.0 above, which was:

Equal diagonal elements (phaseimpedances must be equal), i.e.,

ccbbaaZ Z Z = (1)

Equal offdiagonal elements (offdiagonal phase impedances must be equal), i.e.,bcacab

Z Z Z = (2)In this case, the 0+- sequence impedancesare:

00000 ====== ++++ S S S S S S Z Z Z Z Z Z (3)abaaS Z Z Z 20 += (4)

abaaS S Z Z Z Z ==+ (5)

Equation (3) simply says that all off-diagonal elements on the 0+- sequence

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impedance matrix are zero. Equations (4)and (5) provide the actual expressions that

we need for the 0, positive, and negativesequence impedances. These expressionsapply to transmission lines becausetransposition makes conditions (1, 2) true.

The sequence networks are given in Fig. 4.0

a I

a I

+

a I

aV

+

aV

0aV

Fig. 4

It is interesting to compare eqs. (4) and (5)for a symmetric line, and note that they

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indicate that the zero-sequence quantity islarger than the positive and negative

sequence quantities by 3Zab.A typical overhead line has Zab(2/5)Zaa. Inthis case, it is easy to show that Z0=3Z+,suggesting that finding zero sequenceimpedance 3 times as large as positivesequence impedance, is quite typical (see bottom of page 474 in text).

3.3 Transformers (see Sec. 12.8 of text)

There are five different types of transformerconnections to assess. These are:1. Grounded Y to grounded Y.2. Grounded Y to Y or Y to grounded Y.3. -4. Grounded Y to or to grounded Y5. Y- or -Y.As before, we can perform our 3-step procedure given at the end of Section 3.0

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(pg. 5), where we express abc quantities,substitute in symmetrical components, and

then obtain the decoupled equations. Here,however, we must repeat this for both sidesof the transformer and then relate the twosets of equations.

We will not perform this tedious work butwill instead simply observe general guidesfor drawing appropriate sequence circuits. Informing these guides, we assume:

Exciting current is negligible so shunt pathis infinite impedance and we only have theseries Z (winding resistance and leakagereactance) in our transformer abc model.Transformers in -Y or Y- configurationare always connected so that positivesequence voltages on the high side lead positive sequence voltages on the low side by 30 (per industry convention). See pp.139-140 for more on xfmr 30 phase shift.

General guidelines for transformer 0+-sequence circuits:

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1. Positive and negative sequenceimpedances are equal, i.e.,

Z+

=Z-

=Zserieswhere Zseries is the transformer windingresistance and leakage reactance.

2. For connection types 4 and 5 in the abovelist (pg. 11), the phase shift is includedfrom low side to high side as

+30 for positive sequence-30 for negative sequence

Lets take a brief aside to look at this.a. Why does a -Y or Y- xfmr have a

30 phase shift for positive sequence

quantities? Consider a Y-. Across thewinding, it is Van/VAB, but on the Y-side, the line-line voltage isVab=3Van/_30, so line-line voltageratio is Vab/VAB=3Van/_30/ VAB.

b. Why does negative sequence use -30?Consider the Y-side in our Y-. FromKVL, Vab=Van-V bn. See Fig. 5.

i. Positive sequence: V bn=Van/_-120ii. Negative sequence: V bn=Van/_+120

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1V bn

Fig. 5

3. For zero-sequence network,a. We get a complete open circuit (I0 =0)if there is an ungrounded Y on one or both sides. b. We get isolation of primary fromsecondary if there is a on one or both

sides. This means that connection prevents pass-through of zero-sequencecurrents. However, we may still getzero-sequence current flowing if theother side is grounded Y or .c.We get no isolation if both sides aregrounded-Y.d. Z0 =Zseries+3Znp+3Zns where:

Znp: neutral impedance on primaryZns: neutral impedance on secondary

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Noflow.

Flow,

butno

pass-th

rough..

Flow and

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Two concepts important in understandingthe points under (3) above are:

There must be a connection to ground onthe primary (secondary) side for zero-sequence current to flow between the primary (secondary)-side system and the primary (secondary) side of thetransformer.If zero-sequence currents cannot flowon primary (secondary) side of thetransformer,then because currents on thesecondary (primary) side of thetransformer can only arise throughinduction of currents on the primary(secondary) side of the transformer,zero-sequence currents also cannot flow onthe secondary (primary) side of thetransformer.

So lets draw 0+- sequence circuits for various transformer connections.1. Grounded Y to grounded Y.

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0a I

a I

+

a I

aV

+

aV

0aV

Fig. 6

Here, there is no place for zero-sequencecurrents to flow on the Y side (since there isno neutral and sum of phase currents, whichequals I0, must be 0). Therefore, there can beno zero-sequence currents flowing on theother side either. So I0=0 for this connection.

Y to grounded Y is the same.

3. -

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0a I

a I

+

a I

aV

+

aV

0aV

Fig. 7Here, zero sequence currents cannot enter orleave either winding, so for all practical purposes, the zero-sequence circuit is anopen on both sides.

4. Grounded Y to or to grounded Y

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30:1

30:1

0a I

a I

+

a I

aV

+

aV

0aV

Fig. 8: Grounded Y to

30:1

30:1

0a I

a I

+

a I

aV

+

aV

0aV

0

A I

Fig. 9: to Grounded Y

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In Figs. 8 and 9, we observe that zerosequence currents can flow out of the

grounded Y side, which means they alsomust be able to flow within the (but notout of the ).

We also observe that, in both Figs. 8 and 9:-30 phase shift occurs of low sidequantities relative to high side quantitiesfor positive sequence (which implies highside leads low side by 30 for positivesequence quantities, in conformance withindustry convention)30 phase shift occurs of low sidequantities relative to high side quantitiesfor negative sequence (which implies highside quantities lag low side by 30 for negative sequence, in conformance withindustry convention).

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5. Y- or -Y.

30:1

30:1

0a I

a I

+

a I

aV

+

aV

0aV

Fig. 10: Y to

30:1

30:1

0a I

a I

+

a I

aV

+

aV

0aV

0

A I

Fig. 11: to Y

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Observe that Figs. 10 and 11 are exactly likeFigs. 8 and 9 in the positive and negative

sequence circuits. The only difference is thezero-sequence circuit, where we see that, inFigs. 10 and 11, not only can zero-sequencecurrents not pass through (which is the casein Figs. 8 and 9) but they cannot flow at all.

3.4 Rotating machines (see sec 12.7, text)

Development of sequence impedances forthe synchronous machine requiressignificant effort together with backgroundin two-reactance theory, including the Parkstransformation. We do not have that background. Your text offers some of that background in chapter 7, and in Appendix 5Additional references include [1, chap. 6], [2,chap. 1]. Here we simply provide somecomments.

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Positive sequence reactance: As in thesymmetrical fault analysis, we will just use

Xd, Xd, or Xd, depending on what timeframe of interest we have. This is quitereasonable for smooth rotor machines, butapproximate for salient pole-machines.

Negative sequence reactance: The negativesequence currents set up flux in the air gapthat rotates opposite to the rotor andtherefore sweeps rapidly over the face of therotor, inducing currents in the iron whichcounteract the original flux. This conditionis similar to the rapidly changing fluximmediately upon the occurrence of a shortcircuit at the machine terminals. As a result,the negative sequence reactance is generallyassumed equal to Xd.

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Zero-sequence: Two comments:1. The zero sequence reactance is typically

quite small. The reason for this is that thezero sequence currents in the a, b, and cwindings are in-phase. Their individualfluxes in the air gap sum to zero andtherefore induce no voltage, so the onlyeffect to be modeled is due to leakagereactance. We call this

0

g Z .2. As with loads, if the neutral is grounded

through an impedance Zn, because 3 timesthe zero sequence current flows throughZn, we model 3Zn in the zero sequencenetwork.

Therefore we have, for generators, thatn g Z Z Z 3

00 += (12)Voltage source: Finally, because generators produce balanced positive sequencevoltages, generators produce no negative orzero sequence voltages. Therefore, wemodel a voltage source only in the positivesequence circuit.

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When working in per-unit (as we have beenassuming throughout this discussion), the

positive sequence source voltage is typicallyassumed to be Ean=1.0. Although it actuallymay be something a little different than 1.0,the influence on final short circuit currentscalculated is negligible.

0

a I

+

a I +aV

0aV

Z0

a I aV

Fig. 12Your text provides a table of typicalsynchronous machine reactances, see Table

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12.1 of pp. 468. You can compare this tablewith a similar one below that I obtained

from another reference [3

]. All values are in per-unit on the MVA base of the machine.

SmoothRotor

SalientPole

SynchronousCondensers

Motor

X+Xd 1.1 1.15 1.8 1.2Xd 0.23 0.37 0.4 0.35Xd 0.12 0.24 0.25 0.30

X- 0.13 0.29 0.27 0.35X0 0.05 0.11 0.09 0.16

Homework Assignment: Due MondayIn your text:12.2, 12.3, 12.9, 12.10, 12.11, 12.12, 12.17

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1[[] P. Anderson, Analysis of Faulted Power Systems, Iowa State UnivPress, 1973.2[[] E. Kimbark, Power system stability, Vol. II, Synchronous Machi1995 by IEEE (Originally published in 1955).3[[] J. Glover and M. Sarma, Power system analysis and design, Publishers, 1987