IEEE PES Boston Chapter Technical Meeting … · 2017-09-29 · Symmetrical Components IEEE PES Boston Chapter Technical Meeting September, ... IEEE-PES-Boston-2017-Fall Symmetrical

Symmetrical Components

IEEE PES Boston Chapter

Technical Meeting

September, 19 2017

National Grid, Waltham, MA

[email protected]

mailto:[email protected]

IEEE-PES-Boston-2017-Fall Symmetrical Components 2

• History and Description

• The General Method of Symmetrical Components – N-Phase Systems

– 3-Phase Systems

• Circuit Element Sequence Representations

• Fault Analysis Using Symmetrical Components

Symmetrical Components

Discussion Topics

_______________________________________ 1J. Lewis Blackburn and Thomas J. Domin, Protective Relaying Principles and Applications, 3rd Ed., CRC Press, 2007

. 2John, A Horak, Derivation of Symmetrical Component Theory and Symmetrical Component Networks, Georgia Tech

protective Relaying Conference, Atlanta, GA, April 2005, http://www.basler.com/downloads/Symmcomp.pdf

http://www.basler.com/downloads/Symmcomp.pdf


Symmetrical Components – History and Description

• The method of symmetrical components provides a tool to study systems with unbalanced phasors.

• Developed by Charles Fortescue in 1913, who presented a paper entitled’ “Method of Symmetrical Co-ordinates Applied to the Solution of Polyphase Networks.”3

• In mathematics terms, a linear transformation4 is a mapping of quantities between two vector spaces, in this case a physical domain (ABC) and a sequence domain (012).

• The method simplifies circuit analysis of a three-phase mutually coupled circuit by transforming it into 3 single phase circuits with no mutual coupling.

_______________________________________

3Fortescue’s paper is available from the University of Waterloo from

http://thunderbox.uwaterloo.ca/~ccanizar/papers/classical/Fortescue.pdf.

4Rowland, Todd and Weisstein, Eric W. "Linear Transformation." From MathWorld—A Wolfram Web

Resource http://mathworld.wolfram.com/LinearTransformation.html

http://thunderbox.uwaterloo.ca/~ccanizar/papers/classical/Fortescue.pdf

http://mathworld.wolfram.com/topics/RowlandTodd.html

http://mathworld.wolfram.com/about/author.html

http://mathworld.wolfram.com/

http://mathworld.wolfram.com/LinearTransformation.html


A Tool for Simplifying Fault Analysis

• A balanced system is easily analyzed because only one phase needs to be considered.

• Unbalanced systems require a full circuit analysis of all three phases, neutral and ground elements.

• Therefore transforming an unbalanced system into balanced systems promises to simplify our analysis.


The General Method of Symmetrical Components

• Method for 3-phase power system is a subset of the more general transformation method. The general method resolves N unbalanced phasors which share the same reference plane into N sets of balanced phasors, with each set having N members.

• Within each set, each of the phasors has the same magnitude and successive phases have the same phase angle separation between them.

• Let α be the angle shift between successive phases in an N-phase system. Then let’s define a useful operator dubbed the “a” operator. The “a” operator is a unit phasor (magnitude = 1) with an angle equal to α. Multiplying a phasor by a simply rotates that phasor by α degrees in the counterclockwise direction.

• Within each sequence network, the angular displacement of successive phasors is -α∙n where n is the phase sequence network number and where n = 0, 1, 2, …N-1.

N

360 1a Rotation operator


Note: All system rotation,

even “negative-sequence”

phasors, is counterclockwise!

3-Phase System Example Seq #0 System (Zero Seq)Physical System

Seq #1 System (Pos Seq)

Seq #2 System (Neg Seq)

a0,

b0,

c0

a1

b1

c1

a

b

c

a2

c2

b2

N = 3

n = 0, 1, 2, …N-1

α = 120 deg

α


4-Phase System Example Seq #0 SystemPhysical System

Seq #1 System

Seq #2 System

Seq #3 System

a

b

c

d

a1

b1

c1

d1

a2,

c2

b2,

d2

a3

d3

c3

b3

a0,

b0,

c0,

d0

N = 4

n = 0, 1, 2, …N-1

α = 90 deg

α


5-Phase System Example Seq #0 System

a0,

b0,

c0,

d0,

e0

Physical System

Seq #1 System

Seq #2 System

Seq #3 System

Seq #4 System

a

b

c

de

a1

b1

c1

d1

e1

a4

e4

d4

c4

b4

a2

d2

b2

e2

c2

a3

c3

e3

b3

d3

N = 5

n = 0, 1, 2, …N-1

α = 72 deg

α


6-Phase System Example

A more detailed development of 6-phase sequence network

applications can be found in the following paper:

5Bhatt, Navin B., Six-Phase (Multi-Phase) Power Transmission

Systems: Fault Analysis, IEEE Transactions on Power

Apparatus and Systems, Vol. PAS-96, No. 3, May/June 1977,

http://www.libsou.com/pdf/01601991.pdf

Trivia Factoid:

When N is prime, each set where n>0 will form a regular N-

sided polygon. Consider n=0 as being a 1-sided polygon.

For N=6, prime factors are 1, 2, 3 and 6. Can you find a 1-

sided, 2-sided, 3-sided and 6-sided polygon among these

sets? Try this for any value of N.

Seq #0 System

a5

f5e5

d5

c5 b5

a2,

d2

b2,

e2

c2,

f2

a1

b1c1

d1

e1 f1

a4,

d4

b4,

e4

c4,

f4

a3,

c3,

e3

b3,

d3,

f3

a0,

b0,

c0,

d0,

e0,

f0

a

bc

d

e f

Physical System

Seq #1 System

Seq #2 System

Seq #3 System

Seq #4 System

Seq #5 System

N = 6

n = 0, 1, 2, …N-1

α = 60 deg


Back to the 3-Phase System Example Seq #0 System (Zero Seq)Physical System

Seq #1 System (Pos Seq)

Seq #2 System (Neg Seq)

a0,

b0,

c0

a1

b1

c1

a

b

c

a2

c2

b2

N = 3

n = 0, 1, 2, …N-1

α = 120 deg

α

210

210

210

cccc

bbbb

aaaa

VVVV

VVVV

VVVV

22

11

00

a

a

a

VV

VV

VV

We can relate physical domain quantities to

sequence domain quantities by superposition.

This transforms 3 quantities into a total of 9 quantities.

2

2

10

21

2

0

210

VaaVVV

aVVaVV

VVVV

c

b

a

Let’s define quantities V0, V1 and V2 in the sequence domain using

a-phase (or specifically Va) as a reference and along with the “a”

operator substitute the new “sequence” quantities into b and c-

phases.

Simplified?


3-Phase System Example Using Matrix Notation

2

2

10

21

2

0

210

VaaVVV

aVVaVV

VVVV

c

b

a

2

1

0

2

2

1

1

111

V

V

V

aa

aa

V

V

V

c

b

a

012VAVabc

Expressing the transformation in matrix notation makes the transformation format easier

to remember.

Matrix notation also lets us easily derive an inverse transform.

abcVAV 1

012

c

b

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1


3-Phase System Example Using Matrix Notation

2

2

10

21

2

0

210

IaaIII

aIIaII

IIII

c

b

a

2

1

0

2

2

1

1

111

I

I

I

aa

aa

I

I

I

c

b

a

012IAIabc

Naturally, other phasor quantities such as current or flux have the same transformation

form as voltage phasors.

abcIAI 1

012

c

b

a

I

I

I

aa

aa

I

I

I

2

2

2

1

0

1

1

111

3

1


Symmetrical Component Transformation

Worked Example

6This example is from Washington State University’s March 2011 Hands-On Relay School



Worked Example (Continued)










• 7Steven Blair at the University of Strathclyde has a good interactive

tool for visualizing symmetrical components:

http://personal.strath.ac.uk/steven.m.blair/seq/


Symmetrical Component Transformation – Visualization

http://personal.strath.ac.uk/steven.m.blair/seq/


Circuit Element Sequence Representations



Sequence Networks

• Elements of a power system are represented by impedances in each of sequence networks. For a 3-phase power system:

– Zero sequence represents impedances of the system to equal (in-phase) currents in all three phases.

– Positive sequence represents impedances of the system to normal (balanced) currents in all three phases.

– Negative sequence represents impedances of the system to currents with reversed phase sequence.



Sequence Network Independence

• Each of the sequence networks is independent of the others.

• For a balanced network:

– Sequence currents produce only like sequence network voltage drops.

• Thus the sequence networks are not connected to each other.

• Unbalanced sources resolve to zero, positive and negative sequence sources.

• For unbalanced network:

– Sequence currents can produce voltage drops in any or all three sequence networks.

• Thus, we model unbalances by setting up independent sequence networks and interconnecting them at the point of the unbalance (i.e. the fault location).

• For these studies we assume the rest of the system and all the sources are balanced (i.e. no sources in the zero and negative sequence networks).



3-Phase System with Sending and Receiving End Voltages

RA

RB

RC

jXAA

jXBB

jXCC

RG

VCSVBSVAS VCRVBRVARjXAG jXCG

jXBG

jXAB

jXBC

jXCA

IC

IA

IB


• The voltage drop equations in matrix form for our system are

• Starting with KVL around the loop, the self impedance of loop A (i.e. the voltage drop

in loop A resulting from the current in loop A) is

• ZBB and ZCC in the impedance matrix are similarly defined.

C

B

A

CCCBCA

BCBBBA

ACABAA

CR

BR

AR

CS

BS

AS

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

AGAAGA

A

IARAS

AA

AAGAAAGAIARAS

XXjRRI

VVZ

IXXjIRRVV

A

A


Self Impedances


• Again, our voltage drop equations in matrix form are

• The mutual impedance from Loop A to Loop B (i.e. the voltage drop in loop A resulting

from the current in Loop B) is

• The XAG term might look like a typo but recall the following for flux linkage in Phase A:

C

B

A

CCCBCA

BCBBBA

ACABAA

CR

BR

AR

CS

BS

AS

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

AGABG

B

IARAS

AB XXjRI

VVZ B

)(

)()()()(

)()()()(

)()()()()(

tiLLtiLLtiLLt

titititiwhere

tiLtiLtiLtiLt

CAGACBAGABAAGAAA

CBAG

GAGCACBABAAAA


Mutual Impedances


• Start with our voltage drop equations in matrix form

• Multiply both sides of the expression by [A-1] which preserves the equality and multiply

[ZABC] by [A-1]·[A] which is the same as multiplying it by the identity matrix [I].

• This leaves us with the voltage drop expressions in the sequence domain.

2

1

0

222120

121110

020100

2

1

0

2

1

0

1111

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

I

I

I

AA

ZZZ

ZZZ

ZZZ

A

V

V

V

A

V

V

V

A

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

R

R

R

S

S

S

C

B

A

CCCBCA

BCBBBA

ACABAA

CR

BR

AR

CS

BS

AS

C

B

A

CCCBCA

BCBBBA

ACABAA

CR

BR

AR

CS

BS

AS


ABC-to-012 Conversion Using Voltage Drop Expressions


2

22

22

1

22

21

0

22

20

2

11

12

1

11

11

0

11

10

2

00

02

1

00

01

0

00

00

2

1

0

222120

121110

020100

2

1

0

2

1

0

210

210

210

)()()(

)()()(

)()()(

I

VVZ

I

VVZ

I

VVZ

I

VVZ

I

VVZ

I

VVZ

I

VVZ

I

VVZ

I

VVZ

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

V

V

V

IRSIRSIRS

IRSIRSIRS

IRSIRSIRS

R

R

R

S

S

S

• Let’s examine each of the impedance terms of Z012.

• See how the diagonal terms indicate impedances with no coupling between

the sequence networks whereas the off-diagonal terms indicate impedances

with coupling between the sequence networks.


Sequence Independence and Sequence Coupling


Cases of Impedance Matrices with High Symmetry

Case 1 – Symmetrical Passive Elements

)(

)(

GMG

ACCBBACABCABM

GSGP

CCBBAAS

CGBGAGG

ACCBBACABCABM

CCBBAAS

CBAP

XXjR

ZZZZZZZ

XXjRR

ZZZZ

XXXX

XXXXXXX

XXXX

RRRR

• In the case of transmission lines, common assumptions include symmetrically spaced

phase conductors with regular conductor transposition. This yields the following

equalities:

• This in turn enables the introduction of new

variables representing self and mutual

impedances.



Case 1 – Symmetrical Passive Elements

• Now we substitute ZS for the self impedance and ZM for the mutual impedances in the

ZABC impedance matrix.

• And convert to Z012.

• Notice that Z012 contains only diagonal elements (i.e. no coupling between sequence

networks) and that Z11 = Z22. The more common form of expression is that Z1 = Z2.

MS

MS

MS

SMM

MSM

MMS

SMM

MSM

MMS

CCCBCA

BCBBBA

ACABAA

ABC

ZZ

ZZ

ZZ

A

ZZZ

ZZZ

ZZZ

AZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

Z

00

00

0021

012




Case 2 – Rotating Machines

)()(

)(

GMG

ACCBBAM

GMG

CABCABM

GSGP

CCBBAAS

CGBGAGG

ACCBBAM

CABCABM

CCBBAAS

CBAP

XXjR

ZZZZ

XXjR

ZZZZ

XXjRR

ZZZZ

XXXX

XXXX

XXXX

XXXX

RRRR

• In the case of rotating machines (motors and generators), mutual coupling between

phases includes rotation as well as physical geometry. This differs from non-rotating

circuit elements because now ZAB ≠ ZBA, ZBC ≠ ZCB and ZCA ≠ ZAC.

• This in turn enables the introduction of new variables

representing self and mutual impedances.



Case 2 – Rotating Machines

• Now we substitute ZS for the self impedance and ZM for the mutual impedances in the

ZABC impedance matrix.

• And convert to Z012.

• Notice that Z012 still contains only diagonal elements (i.e. no coupling between

sequence networks) but this time Z11 ≠ Z22 or more commonly Z1 ≠ Z2.

MMS

MMS

MMS

SMM

MSM

MMS

SMM

MSM

MMS

CCCBCA

BCBBBA

ACABAA

ABC

ZaaZZ

aZZaZ

ZZZ

A

ZZZ

ZZZ

ZZZ

AZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

Z

2

21

012

00

00

00



• You can often get Thevenin impedances (Z0 and Z1) directly from short circuit programs like CAPE or Aspen. An equivalent source is nonrotating so Z2 = Z1.

• Otherwise, you can calculate Z0 and Z1 from fault duties, expressed in MVA or kVA based on the system base voltage and the available fault current. You need both 3-phase and SLG fault duties to calculate Z0 (see next slide). With this method assume impedances to be all reactive.

• For hand calculations you can often assume an ideal voltage source, a.k.a. an infinite bus, (i.e. Z0 = Z1 = Z2 = 0).


System Equivalent Sources


Following is a calculation method for Z0, Z1 and Z2 from given 3ph (S3ph) and SLG (S1ph) fault duties in MVA. Resulting impedances are in per unit.

See Blackburn Appendix 4.1 for a detailed derivation.


System Equivalent Sources

ph

Base

S

SZZ

3

21

ph

Base

ph

Base

g

ph

Baseg

S

S

S

S

ZZZZ

ZZZS

SZ

31

210

210

1

23

3



Synchronous Generators

For a sustained fault on the terminals of an unloaded generator, the

armature current has a decrement as shown below. A piece-wise

linear approximation of this nonlinear response is achieved using

separate positive sequence reactances: Xd”, Xd’ and Xd.


• Reactance increases with time after a short circuit because of the demagnetizing effect of the fault current on the air-gap flux. For round rotor machines, typical positive sequence reactances are as follows:

0.95 Xd 1.45

0.12 Xd’ 0.28 pu on generator base

0.07 Xd’’ 0.17

• Negative-sequence reactance is often approximated by equating it to the subtransient reactance.

X2 = Xd’’


Synchronous Generators


• Induction machines (motors or generators) are often ignored as fault current sources for relaying purposes.

• The fault current contribution from an induction machine typically decays in a few cycles.

• Induction machine contribution to fault current (subtransient reactance) may be considered when performing maximum instantaneous fault current studies for bus and switchgear rating purposes.


Generators – Induction Machines


Z0

3Zn Z1

E

Z2

REF REF REF

Pos. Seq. Neg. Seq. Zero Seq.

(Wye-Connected)

Zero-sequence impedance depends on the manner in which the

generator is grounded. Any impedance in the neutral circuit (Zn) is

represented as three times this value (3Zn) in the zero-sequence

model since 3I0 flows through the neutral.

External neutral

impedance.

For solidly

grounded neutral

3Zn= 0.

For ungrounded

neutral 3Zn = ∞


Balanced 3-Phase Sources

Z0

REF

Zero Seq.

(Delta-Connected)


• Positive and negative sequence impedances of lumped loads are generally equal. These are shown on a reference phase basis in sequence networks.

• For synchronous motors, particularly those that are designed with salient poles, the negative sequence impedance generally lies between Xd’ and Xd’’.

• Zero sequence impedance of loads depends on the manner in which they are connected and grounded as shown on the following page.


Load Impedances

MS

MS

MS

SMM

MSM

MMS

ZZ

ZZ

ZZ

A

ZZZ

ZZZ

ZZZ

AZ

00

00

0021

012

MMS

MMS

MMS

SMM

MSM

MMS

ZaaZZ

aZZaZ

ZZZ

A

ZZZ

ZZZ

ZZZ

AZ2

21

012

00

00

00


Z

Z

Z N

Z

Z

Z N

Z

Z

Z N

Z

Z

Z

3I0

Z

REF

Z

REF

Z

3Zn

REF

Z0

REF

3I0

Connection

Arrangement

Zero Sequence

Equivalent Circuits


Zero Sequence Models for Different Load Connections

Ia0 = I0

Ib0 = I0

Ic0 = I0

Ia0 = I0

Ib0 = I0

Ic0 = I0

Zn

I0

I0


• In sequence network calculations, the per-phase values of line resistance, reactance and shunt capacitive susceptance are used.

– These values can be found from conductor tables or calculated by hand using Carson’s equations.

– There are also several software apps available for deriving the per-phase line constants from physical dimensions of the line or cable.

• Recall from earlier expressions that for lines and cables, ZL1 = ZL2 while ZL0 > ZL1.


Transmission Lines and Cables

MS

MS

MS

SMM

MSM

MMS

ZZ

ZZ

ZZ

A

ZZZ

ZZZ

ZZZ

AZ

00

00

0021

012


• The zero sequence impedance, ZL0, of an overhead line depends on

several factors which can result in wide variation.

– The use of overhead shield wires and the type of tower grounding and

counterpoise.

– Ground resistivity.

– Zero-sequence impedance is usually 2 to 3.5 times the

positive-sequence impedance.


ZL0 for Overhead Transmission Lines

Check out the Z0,

Z1 and Z2 values in

the sample EMTP

line model data.

Impedance matrix, in units of [ohms/mile ] for the system of equivalent phase conductors. Rows and columns proceed in the same order as the sorted input. ZAA 1 2.780137E-01 1.099286E+00 ZBA ZBB 2 2.305740E-01 2.910487E-01 5.180598E-01 1.085324E+00 ZCA ZCB ZCC 3 2.230820E-01 2.305750E-01 2.780138E-01 4.412611E-01 5.181495E-01 1.099286E+00 Impedance matrix, in units of [ohms/mile ] for symmetrical components of the equivalent phase conductor Rows proceed in the sequence (0, 1, 2), (0, 1, 2), etc.; Columns proceed in the sequence (0, 2, 1), (0, 2, 1), etc. Z00 0 7.385127E-01 2.079613E+00 Z10 Z12 1 -2.156088E-02 -4.804473E-02 -4.577000E-03 2.854904E-02 Z20 Z22 Z21 2 1.471788E-02 5.428174E-02 4.869572E-02 -1.642828E-02 6.021421E-01 2.742369E-02


MS

MS

MS

ZZ

ZZ

ZZ

Z

00

00

002

012

AGAAGAAA XXjRRZ


Sample ATP-EMTP Output for OH Transmission Lines

AGABGAB XXjRZ

Calculated impedances are for a typical flat profile 345kV transmission line (left-most

tower configuration) with two bundled conductors per phase and no transposition.

Notice that ZAA = ZCC ≠ ZBB

and ZBA = ZCB ≠ ZCA.


• For cables, the type of sheath or pipe used in the construction is a

major factor in the zero-sequence impedance.

• In addition, the placement of the phase conductors relative to each

other affects the amount of current flow in the cable sheaths or pipe

and thus has a significant impact on the zero-sequence impedance.


ZL0 for Transmission Line Cables


R0

/R1

Total Pipe Current - Amperes

0

5

10

15

20

25

30

35

40

10 100 1000 10000 100000


Pipe-type Cables – R0 and X0

X0

/X1

Total Pipe Current - Amperes

1

2

3

4

5

6

7

8

10 100 1000 10000 100000

• R0 and X0 in pipe type cables vary with current loading.

• Most fault studies will choose a current level at which to study the

cable, usually a level corresponding with the available fault current.


Core-type

(3-legged) Shell-type


Transformers – Core Type vs Shell Type

Single Phase Units

3-Phase Units


Regardless of whether the transformer is a bank of three (3) single-phase units or a 3-phase transformer having either a shell-form or a core-form core:

• Z1 and Z2 are equal and are the leakage impedance of the transformer.

• Positive-sequence voltages and currents shift by 30 when passing through a Delta - Wye transformer bank. The sign of the phase shift angle is determined by the transformer connections and is equal to that of the physical voltage and current phase shift angle.

• The sign of the negative sequence voltage and current phase shift angle is opposite that of the positive sequence phase shift angle.

• No phase shift occurs in the zero sequence network.


Transformers – Positive and Negative Sequence

Core-type

(3-legged)


• Where a 3-phase transformer bank is arranged without interlinking magnetic flux (three-phase shell type, or three single-phase units) and provided there is a path for zero sequence currents, then Z0 = Z1.

• With 3-phase core type units, the zero sequence fluxes produced by zero sequence currents can find a high reluctance path through the tank, the effect being to reduce Z0 to about 85-90% of Z1. With a grounded wye-wye configuration, this appears as a phantom or “tank” tertiary winding in the zero sequence network.

• With hand calculations, it’s not uncommon to ignore this variation and consider the positive and zero sequence impedances to be equal.

• When using short circuit software, it’s common to assume a zero-sequence impedance as indicated above, if manufacturer provided zero sequence values are not available.


Transformers – Zero Sequence

Shell-type

Source: 8Areva Network Protection and Automation Guide, Chapter 5, page 57.


Circuit Element Sequence Representations 2-Winding Transformers – Zero Sequence

This represents a solidly grounded

neutral (i.e. 3Zn = 0).

Replace with 3Zn for an impedance

grounded neutral.

Replace with an open (i.e. 3Zn = ∞) for

an ungrounded neutral.

The connection from internal

impedance to the reference

and the open connection to the

external terminal represents a

delta connection.



3-Winding and Autotransformers

A 3-winding transformer can be represented with a T-model.

Recall that impedances derived from measurements (i.e test reports) are

those between pairs of windings with the third winding being an open circuit.

Thus we can relate the values ZPS, ZPT and ZST to effective individual winding

impedances ZP, ZS and ZT as follows:

With some algebraic manipulation we can arrive at the following:


Circuit Element Sequence Representations 3-Winding and Autotransformers – Positive and Negative Sequence

The positive and negative sequence representations are the same and simply

include solid connections from the internal impedances ZP, ZS and ZT to the

external terminals P, S and T as shown below right.

Ref.


Circuit Element Sequence Representations 3-Winding and Autotransformers – Zero Sequence

Note for N-Winding Transformers:

Representation of transformers with

more than 3 windings (i.e. N-winding

transformers) follow the same process

and connection rules as 3-winding

transformers.

Zero sequence representations use the same internal connection rules for

each winding configuration as seen in the 2-winding example.

• Textbook diagrams often present transformer zero sequence

diagrams in a misleading way.


Now For a Word About Textbook Lookup Tables …

• They often put neutral connections

where delta connections go.

• They often depict that which is

physically outside the transformer

as being inside the transformer.

• This can make representing

shared neutrals (or neutral buses)

seem harder than it really is.

You can also refer to Figures A4.2-1 and A4.2-3 in

Blackburn (Reference #1).



A Common but Incorrect Mapping of I0 to IN

Current across shunt elements in the zero sequence representation corresponds with

neutral current.

If the zero sequence is open, as for a delta-delta transformer or ungrounded wye

winding, clearly there is no neutral current. When there is only a series path there is

zero sequence current flow but not through the neutral.

3I0

3I0 ??!

This is just

wrong.

This is just

wrong.


The connection to reference, making Zt0 a shunt element, models the delta connection.

I0 flowing through the shunt connection models I0 inside the delta while I0 leaving Zt0

models I0 in the wye winding.

Consider a similar mapping for wye-wye windings. Notice that when there is only a

series path there is zero sequence current flow through the windings but the net zero

sequence current through the neutrals is zero.

I0

I0

I0


The Correct Mapping of I0 to IN

• If remember these simple connection rules for wye grounding and

delta arrangements, you may never need a lookup table again.

• Always draw what’s physically outside the transformer as being

outside the transformer in your sequence representations.



Shared Neutral Bus Example

H XT

H0 X0T0

H XT

H0 X0T0

jXH XH jXX

jXX

T

jXH XjXX

jXX

T

3RG

RG

See the problem?


T2-A

Station A

115kV Bus T1-A

Station A

23kV Bus 1

Station A

23kV Bus 2 T2-B

T1-B

Station B

23kV Bus 1

Station B

23kV Bus 2

Line 1

Line 2

Station B

115kV Bus

RGA

RG1B

RG2B


Shared Neutral Bus Example

T1-AjXGA

T2-A

Station A

23kV Bus 1

Station A

23kV Bus 2

Station A

115kV Bus

3RGA

Zero Sequence Reference

Line 1

Line 2

Station B

23kV Bus 1

Station B

23kV Bus 23RG2B

T1-B

T2-B

Station A

Neutral Bus3RG1B

Station B

115kV Bus

jXGBI0I0


T1-AjXGA

T2-A

Station A

23kV Bus 1

Station A

23kV Bus 2

Station A

115kV Bus

3RGA

Zero Sequence

Reference

Line 1

Line 2

Station B

23kV Bus 1

Station B

23kV Bus 2 3RG2B

T1-B

T2-B

Station A

Neutral Bus3RG1B

Station B

115kV Bus

jXGB

T1-A

T2-A

Station A

115kV Bus

jXGA

Positive Sequence

Reference

Station A

23kV Bus 1

Station A

23kV Bus 2

Line 1

Line 2

Station B

23kV Bus 1

Station B

23kV Bus 2

T1-B

T2-B

Station B

115kV Bus

jXGB

T1-A

T2-A

Station A

115kV Bus

jXGA

Negative Sequence

Reference

Station A

23kV Bus 1

Station A

23kV Bus 2

Line 1

Line 2

Station B

23kV Bus 1

Station B

23kV Bus 2

T1-B

T2-B

Station B

115kV Bus

jXGB

I0I1

I2


Fault Analysis Using Symmetrical Components



Example System (Fault at Bus C)



3-Phase (3PH) Fault

At the fault point Va = Vb = Vc = 0 and Ia = Ib = Ic

Since Va = Vb = Vc = 0

From the definitions: V0 = V1 = V2 = 0

Since Ia = Ib = Ic

From the definitions: I0 = 0 and I2 = 0

As expected, these relationships

suggest only the positive sequence

network to be connected at the

fault point.

REFERENCE EQUATIONS

cba

cba

cba

aVVaVV

VaaVVV

VVVV

2

2

2

1

0

3

1

3

1

3

1

cba

cba

cba

aIIaII

IaaIII

IIII

2

2

2

1

0

3

1

3

1

3

1

2

2

10

21

2

0

210

VaaVVV

aVVaVV

VVVV

c

b

a

2

2

10

21

2

0

210

IaaIII

aIIaII

IIII

c

b

a

a

b

c



3-Phase (3PH) Fault

1

1

0

Z

VI

III

VVV

a

cba

cba

a

b

c



3-Phase (3PH) Fault

Z1

V1 V

I1

Z2

V2

I2

Z0

V0

I0

POS

NEG

ZERO

0

0

0

0

20

20

1

II

VV

V

VVV

III

cba

cba

The above fault record is from Washington State University’s March 2011 Hands-On Relay School

Notice we observe an extra

component of V1 and I1

because of load current and

source impedance.



Single-Line-to-Ground (SLG) Fault

a

b

c

REFERENCE EQUATIONS

cba

cba

cba

aVVaVV

VaaVVV

VVVV

2

2

2

1

0

3

1

3

1

3

1

cba

cba

cba

aIIaII

IaaIII

IIII

2

2

2

1

0

3

1

3

1

3

1

2

2

10

21

2

0

210

VaaVVV

aVVaVV

VVVV

c

b

a

2

2

10

21

2

0

210

IaaIII

aIIaII

IIII

c

b

a

At the fault point Va = 0 and Ib = Ic = 0

Since Va = 0 V0 + V1 + V2 = 0

Since Ib = Ic = 0 I0 + a2I1 + aI2 = 0 and

I0 + aI1 + a2I2 = 0

Which simplifies to I0 = I1 = I2

These relationships suggest that

all three sequence networks are

connected in series at the fault

point.




021

0021

021

021

33

0

0

ZZZ

VIIIII

ZZZ

VIII

II

V

aaaaa

aaa

cb

a




210

210 0

0

0

III

VVV

II

V

cb

a

Notice we observe an

extra component of V1

and I1 because of load

current and source

impedance.

I1 Z1

V1 V

Z2

V2

I2

Z0

V0

I0

POS

NEG

ZERO




Line-to-Line (LL) Fault

At the fault point Ia = 0 and Ib = -Ic and Vb = Vc

Since Vb = Vc V0 + a2V1 + aV2 = V0 + aV1 + a2V2

Which simplifies to V1 = V2

Since Ib = -Ic I0 + a2I1 + aI2 = -(I0 + aI1 + a2I2)

Which simplifies to I1 = -I2

Since Ia = 0 I0 + I1 + I2 = 0 where I1 = -I2

Which simplifies to I0 = 0

REFERENCE EQUATIONS

cba

cba

cba

aVVaVV

VaaVVV

VVVV

2

2

2

1

0

3

1

3

1

3

1

cba

cba

cba

aIIaII

IaaIII

IIII

2

2

2

1

0

3

1

3

1

3

1

2

2

10

21

2

0

210

VaaVVV

aVVaVV

VVVV

c

b

a

2

2

10

21

2

0

210

IaaIII

aIIaII

IIII

c

b

a

a

b

c

These relationships suggest the

positive and negative sequence

networks are connected in

parallel at the fault point.




21

21

21

0

aa

aa

cb

cb

a

VV

ZZ

VII

VV

II

I




21

0

21

0

0

II

I

VV

VV

II

I

cb

cb

a

Z1

V1 V

I1

Z2

V2

I2

Z0

V0

I0

POS

NEG

ZERO





current and source

impedance.



Double Line-to-Ground (DLG) Fault

At the fault point Ia = 0 and Vb = Vc = 0

Since Vb = Vc = 0 V0 + a2V1 + aV2 = 0 and

V0 + aV1 + a2V2 = 0

Which simplifies to V0 = V1 = V2

Since Ia = 0 I0 + I1 + I2 = 0

Or rewritten as I1 = -(I0 + I2) These relationships suggest the

zero, positive and negative

sequence networks are all

connected in parallel at the fault

point.

REFERENCE EQUATIONS

cba

cba

cba

aVVaVV

VaaVVV

VVVV

2

2

2

1

0

3

1

3

1

3

1

cba

cba

cba

aIIaII

IaaIII

IIII

2

2

2

1

0

3

1

3

1

3

1

2

2

10

21

2

0

210

VaaVVV

aVVaVV

VVVV

c

b

a

2

2

10

21

2

0

210

IaaIII

aIIaII

IIII

c

b

a

a

b

c




21

21

21

0

aa

aa

cb

cb

a

VV

ZZ

VII

VV

II

I

02

021

1

021

0

0

ZZ

ZZZ

VI

VVV

VV

I

a

aaa

cb

a




Z1

V1 V

I1

Z2

V2

I2

Z0

V0

I0

POS

NEG

ZERO

201

210

0

0

III

VVV

VV

I

cb

a





current and source

impedance.



Phase Shifting Across Delta-Wye Transformers




For this example HV leads LV by 30°. Let the system

voltage ratio (N) equal 1. Consequently, the turns

ratio (n) must be 1/3. We know that

Because the sequence networks are independent, we

can apply them individually adding the results by

superposition.

Starting with positive sequence values we get the

following for voltage and current.

Pictures and equations are from Appendix 4.3 in Blackburn.




Next applying negative sequence values we get the

following for voltage and current.

Recall from before that

Pictures and equations are from Appendix 4.3 in Blackburn.

2 2




• Blackburn Appendix 4.3 provides a second transformer,

Example (b), which has a high side delta, low side wye with

the HV side similarly leading the low side by 30°.

• The process for determining the positive and negative

sequence shift angles is the same as Example (a) as are the

actual shift angles.



Example System (Fault at Bus C)

1. Interconnect the sequence networks

to model the particular fault type and

solve the network to determine the

total fault current components I012

and convert to IABC.

2. Next, determine other internal values

for particular elements – say the

current through Generator GB and

the voltage behind its reactance – by

simply determining the current

through GB in each network (IGB0,

IGB1 and IGB2) as well as the 3

voltages across GB (VGB0, VGB1 and

VGB2).

3. Now convert IGB_012 and VGB_012 to

IGB_ABC and VGB_ABC.


A B

No

Ground

Sources

G L

• X0C is generally assumed equal

to X1C and X2C.

• X0C, X1C and X2C are very large

compared to the series

reactances of the system so X1C

and X2C are shorted by the

XFMR and GEN X1 and X2.

• Note that the location of the fault

has little effect on the magnitude

of the capacitive fault current

since the capacitance is

distributed throughout the

system and the values of the

series reactances are small.

Sequence Networks for a UG System With an SLG Fault



This is simply not suitable on 3PH-4W (grounded) systems

Common Question From Projects Regarding

Open Delta VTs versus Wye-Wye VTs

Why do I have to reject proposals to use two (2) VTs in an open delta configuration instead

of three (3) VTs in a wye-wye configuration?

9Refer to IEEE Std. C37.230-2007, Guide for Protective Relay Applications to Distribution Lines, Figure 6-5.

IEEE-PES-Boston-2017-Fall Symmetrical Components 81 81

Questions?

Documents

IEEE PES Boston Chapter Technical Meeting … · 2017-09-29 · Symmetrical Components IEEE PES Boston Chapter Technical Meeting September, ... IEEE-PES-Boston-2017-Fall Symmetrical