01f symmetrical components

DMAM of 19 Symmetrical Components

Symmetrical Components The method of symmetrical components first discussed by C. L. Fortescue in 1918 at a meeting of the American Institute of Electrical Engineers. The method of symmetrical components is a powerful tools or techniques for analyzing or dealing with unbalanced poly-phase (such as three-phase) systems. Synthesis of Unsymmetrical Phasors from Their Symmetrical Components [Ref. 1, p. 275] Fortescue’s work proves that “An unbalanced system of n related phasors can be resolved into n systems of balanced phasors called the symmetrical components”. The n phasors of each set of components are equal in length (or magnitude), and the angle between adjacent phasors of the set are equal. Simply Fortescue defined a linear transformation from phase components to a new set of components called symmetrical components. According to Fortescue’s theorem, three unbalanced phasors of a three-phase system can be resolved into three balanced system of phasors. The balance sets of components are:

1. Positive-sequence components (as shown in Fig. 4.1 a) consisting of three phasors equal in magnitude, displaced from each other by 120o in phase, and having the same phase sequence as the original phasors. In positive sequence phase b lagging phase a by 120°, and phase c lagging phase b by 120°.

2. Negative-sequence components (as shown in Fig. 4.1 b)consisting of three phasors equal in magnitude, displaced from each other by 120o in phase, and having the same phase sequence opposite to that of the original phasors. In negative sequence phase b leading phase a by 120°, and phase c leading phase b by 120°.

3. Zero-sequence components (as shown in Fig. 4.1 c) consisting of three phasors equal in magnitude and with zero phase displacement from each other. Zero sequence sets have neutral current.

(a) Positive sequence components

(b) Negative sequence components

(c) Zero sequence components

(d) Phase a (e) Phase b (f) Phase c Fig. 4.1 [Ref. 3. p. 399] Resolving phase voltages into three sets of sequence components.

Similar diagram can be obtained for three-phase unbalanced current.


When solving a problem by symmetrical components, to designate the three-phases of the system as a, b, and c in such a manner that the phase sequence of the voltages and currents in the system is abc. Thus the phase-sequence of the positive-sequence components of the unbalanced phasors is abc, the phase-sequence of the negative-sequence components of the unbalanced phasors is acb. If the original phasors are voltages, they may be designated by Va, Vb, and Vc. The three sets of symmetrical-components are designated by the additional subscript 1 for the positive-sequence components, 2 for the negative-sequence components, and 0 for the zero-sequence components. The positive-sequence components of Va, Vb, and Vc are Va1, Vb1, and Vc1. Similarly, the negative-sequence components are Va2, Vb2, and Vc2, and the zero-sequence components are Va0, Vb0, and Vc0.Fig. 4.1 shows three such sets of symmetrical components. Phasors representing current will be designated by I with subscripts as for voltages. The synthesis of a set of three unbalanced phasors from the three sets of symmetrical components of Fig. 4.1 is shown in Fig. 11.2.

Fig. 4.2 [Ref. 1, p. 277] Graphical addition of the components shown in Fig. 4.1 to obtain three unbalanced phasors. Similar diagram can be obtained for three-phase unbalanced current.

Since each of the original unbalanced phasors is the sum of its components, the original phasors expressed in terms of their components are:

)1.4(021 aaaa VVVV ++= )2.4(021 bbbb VVVV ++= )3.4(021 cccc VVVV ++=

The symmetrical component method is basically a modeling technique that permits systematic analysis and design of three-phase systems. Decoupling a detailed three-phase three-phase network into three simpler sequence network results can be superposed to obtain three-phase network results. Advantage of this transformation For balance three-phase networks the equivalent circuits obtained for the symmetrical components, called sequence networks, are separated into three uncoupled networks. For unbalanced three-phase systems, the three sequence networks are connected only at points of unbalance. Sequence networks for many cases of unbalances three-phase systems are relatively easy to analyze and leads to accurate prediction of system behavior. The values of current and voltage at various points in the system under unsymmetrical fault can be easily found since this method consists in finding the symmetrical components of the current at the unsymmetrical fault. Important Observations


The sequence components do not exist as physical quantities in the network. A balance system has no negative and zero sequence components therefore actual balance system equals to positive sequence system. The generated emf is balanced, and therefore positive phase sequence only. In a three-phase three-wire system, there are no zero sequence components because there is no neutral connection. Operators [Ref. 1, p. 277] In order to express the sequence components algebraically, the letter a is commonly used to designate the operator that causes a rotation of 120o in the clockwise direction. Such an operator is a complex number of unit magnitude with an angle of 120o and is defined by:

866.05.023

21)3/2sin()3/2cos()3/2(11201 jjjjea +−=+−=+==°∠= πππ

866.05.023

21)3/4sin()3/4cos()3/4(124012 jjjjea −−=−−=+==°∠= πππ

101)0sin()0cos()3/6(10136013 =+=+==°∠=°∠= jjjea π Table 4.1 [Ref. 3, p. 400]

°∠== 12014 aa 01 2 =++ aa

°∠=− 3031 a °−∠=− 3031 2a °∠=− 27032 aa

°∠= 2101ja

°∠=−=+ 60121 aa °−∠=−=+ 6011 2 aa

°∠=−=+ 180112 aa

Fig. 4.3 [Ref. 1, p. 278] Phasor diagram of the various powers of the operator a.

The Symmetrical Components of Unsymmetrical Phasors [Ref. 1, p. 278] According to Fig. 4.1, we obtain:

00

22

12

1

aVbVaaVbVaVabV

=

=

=

00

22

2

11

aVcVaVacV

aaVcV

=

=

=

(4.4)

Repeating Eq, (4.1) and substituting Eqs. (4.4) in (4.2) and (4.3) yeld )5.4(021 aaaa VVVV ++=

)6.4(201 abab VaVVaV ++=

)7.4(022

1 aaac VVaaVV ++= In matrix form

)8.4(11

111

2

1

0

2

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

a

a

a

c

b

a

VVV

aa

aa

VVV


Let, ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

c

b

a

pabc

VVV

VV ; ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2

2

11

111

aa

aaA ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

2

1

0

012

a

a

a

sa

VVV

VV

)1.8.4(012 sapabc AVAVVV ===

pabc VV = is the column vector of phase voltages, sa VV =012 is the column vector of sequence voltages, and A is 3×3 transformation matrix.

)11.4(11

111

31 2

22

1

0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

a

a

a

VVV

aa

aa

VVV

)1.11.4(11012 pabcsa VAVAVV −− ===

Where, ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=−

aa

aa 2

2

1

11

111

31A

( )

( )

( ) )14.4(31

)13.4(31

)12.4(31

22

21

0

cbaa

cbaa

cbaa

aVVaVV

VaaVVV

VVVV

++=

++=

++=

If required, the components Vb0, Vb1, Vb2, Vc0, Vc1, and Vc2, can be found by Eqs. (4.4) Equation (4.12) shows that no zero-sequence components exist if the sum of the unbalanced phasors is zero. Since the sum of the line-to-line voltage phasors in a three phase system is always zero, zero-sequence components are never present in the line voltages, regardless of the amount of unbalance. The sum of the three line-to-neutral voltage phasors is not necessarily zero, and voltages to neutral may contain zero-sequence components. Sequence Components of Current Because some of the preceding equations are so fundamental, they are summarized for currents:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

2

11

111

a

a

a

c

b

a

III

aa

aa

III

; where, ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

c

b

a

pabc

III

II ; ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

2

1

0

012

a

a

a

sa

III

II

sapabc AIAIII === 012

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

a

a

a

III

aa

aa

III

2

22

1

0 11

111

31 where, pabcsa IAIAII 11

012−− ===

)15.4(021 aaaa IIII ++=

)16.4(201 abab IaIIaI ++=

)17.4(022

1 aaac IIaaII ++=


( )

( )

( ) )20.4(31

)19.4(31

)18.4(31

22

21

0

cbaa

cbaa

cbaa

aIIaII

IaaIII

IIII

++=

++=

++=

)21.4(ncba IIII =++ )22.4(3 0an II =

00

22

12

1

aIbIaaIbI

aIabI

=

=

=

00

22

2

11

aIcIaIacI

aaIcI

=

=

=

In the absence of a path through the neutral of a three-phase system, In is zero, thus the line currents contain no zero-sequence components where neutral connection is absent. A ∆-connected load provides no path to neutral, and the line currents flowing to a ∆-connected load can contain no zero-sequence components. Example 1: Calculate the sequence components of the following balanced line-to-neutral voltages with abc sequence. Va = 277∠0o, Vb = 277∠−120o, and Vc = 277∠120o.

( ) ( ) 0120277120277027731

31

0 =°∠+°−∠+°∠=++= cbaa VVVV

( ) ( )

( ) a

cbaa

V

VaaVVV

=°∠=°∠+°∠+°∠=

°∠×°+∠+°−∠×°+∠+°∠=++=

027702770277027731

12027724011202771201027731

31 2

1

( ) ( )

( ) 0240277120277027731

12027712011202772401027731

31 2

2

=°∠+°∠+°∠=

°∠×°+∠+°−∠×°∠+°∠=++= cbaa aVVaVV

Similarly, 00 =bV ; bb VV =1 ; 02 =bV ; 00 =cV ; bc VV =1 ; 02 =cV Thus a balance three-phase system with abc sequence (or positive sequence) have no zero-sequence or negative-sequence components. Example 2: Calculate the sequence components of the following balanced line-to-neutral voltages with acb sequence. Va = 277∠0o, Vb = 277∠120o, and Vc = 277∠−120o.

( ) ( ) 0120277120277027731

31

0 =°−∠+°∠+°∠=++= cbaa VVVV

( ) ( )

( ) 0120277240277027731

12027724011202771201027731

31 2

1

=°∠+°∠+°∠=

°−∠×°+∠+°∠×°+∠+°∠=++= cbaa VaaVVV


( ) ( )

( ) a

cbaa

V

aVVaVV

=°∠=°∠+°∠+°∠=

°−∠×°+∠+°∠×°∠+°∠=++=

027702770277027731

12027712011202772401027731

31 2

2

Similarly, 00 =bV ; 01 =bV ; bb VV =2 ; 00 =cV ; 01 =cV ; bc VV =2 Thus a balance three-phase system with acc sequence (or negative sequence) have no zero-sequence or positive-sequence components. Example 4.1 [Example 11.1, Ref. 1, p. 280] One conductor of a three-phase line is open. The current flowing to the ∆-connected load through line a is 10 A. With the current in line a as reference and assumeing that line c is open, find the symmetrical components of the line currents. Solution: Fig. 4.4 is a diagram of te circuit. The line currents are

A010 °∠=aI A18010 °∠=bI A0=cI ( ) ( ) 0018010010

31

31

0 =+°∠+°∠=++= cbaa IIII

( ) [ ] °−∠=−=+°+°∠+°∠=++= 3078.589.250)120180(10010

31

31 2

1 jIaaIII cbaa

( ) [ ] °∠=+=+°+°∠+°∠=++= 3078.589.250)240180(10010

31

31 2

2 jaIIaII cbaa

00

15078.52

15078.51

=

°∠=

°−∠=

bIbIbI

00

9078.52

9078.51

=

°−∠=

°∠=

cIcIcI

Fig. 4.4 Circuit for Example 11.1.

Components Ic1 and Ic2 have definite values although line c is open and can carry no net current. As is expected, therefore, the sum of the components in line c is zero. Of course, the sum the components in line a is 10∠0o A, and the sum of the components in line b is 10∠180o A. Example 3: In an unbalanced three-phase system: Va = 4.0∠0o, Vb = 3.0∠−90o, and Vc = 8.0∠143.1o. Find all the voltage components of the corresponding positive, negative and zero-sequence components, and draw the phasors.

( ) ( ) [ ]°∠=+−=

+−+−=°∠+°−∠+°∠=++=

1.1430.1)6.08.0

)6.08.0(0.80.30.4311.1430.8900.300.4

31

31

0

j

jjVVVV cbaa


( ) ( )

( ) °∠=+=°∠+°∠+=

°∠×°+∠+°−∠×°+∠+°∠=++=

4.189.455.165.41.238303431

1.1430.82401900.3120100.431

31 2

1

j

VaaVVV cbaa

( ) ( )

( ) °−∠=−=°∠+°−∠+=

°∠×°+∠+°−∠×°∠+°∠=++=

2.8615.215.2153.01.26382103431

1.1438120190324010431

31 2

2

j

aVVaVV cbaa

°∠=°+°−∠==

°−∠=°+°∠==

°∠==

80.3315.2)1202.86(15.222

6.1019.4)4.18240(9.412

1

1.1430.100

aaVbVaVabV

aVbV

°−∠=°+°−∠==

°∠=°+°∠==

°∠==

2.20615.2)2402.86(15.222

2

4.1389.4)4.18120(9.411

1.1430.100

aVacV

aaVcVaVcV

°∠=+=+−+−++=++= 00.4099.3)6.08.0()15.2143.0()55.165.4(210 jjjjVVVV aaaa

Power in Terms of Symmetrical Components [Ref. 1, p. 288] The total complex power into a three-phase circuit through three lines a, b, and c is

***ccbbaa IVIVIVjQPS ++=+=

[ ]

*

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=+=

c

b

a

cba

III

VVVjQPS ; let,

*

2

1

0

2

1

0

;⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

a

a

a

a

a

a

III

VVV

IV

[ ] *****][ IAAVIAVAAIAV TTTTTjQPS ===+=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

100

010

001

311

11111

111

2

22

2*

aa

aa

aa

aaT AA

Noting that AT = A and that a and a2 are conjugate, we obtain


[ ] [ ]*22

*11

*00

*

2

1

0

210 3100

010

001

3 aaaaaa

a

a

a

aaa IVIVIVIII

VVVS ++=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

So complex power is [ ]*

22*11

*00

*** 3 IVIVIVIVIVIVjQPS ccbbaa ++=++=+= Unsymmetrical Series Impedances [Ref. 1, p. 289] Fig. 11.12 shows the unsymmetrical part of a system with three unequal series impedances Za, Zb, and Zc. The voltage drops across the part of the system shown are given by the matrix equation

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

c

bc

ca

bc

b

ab

ca

ab

a

cc

bb

aa

III

ZZZ

ZZZ

ZZZ

VVV

'

'

'

And in terms of symmetrical components of voltage and current

Fig. 11.12 Portion of a three-phase system showing three unequal series impedance

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2'

1'

0'

a

a

a

c

bc

ca

bc

b

ab

ca

ab

a

aa

aa

aa

III

ZZZ

ZZZ

ZZZ

VVV

AA

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

2

1

01

2'

1'

0'

a

a

a

c

bc

ca

bc

b

ab

ca

ab

a

aa

aa

aa

III

ZZZ

ZZZ

ZZZ

VVV

AA

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

12

02

21

1

01

20

10

0

2'

1'

0'

a

a

a

aa

aa

aa

III

ZZZ

ZZZ

ZZZ

VVV

221210202'

212110101'

202101000'

aaaaa

aaaaa

aaaaa

IZIZIZVIZIZIZVIZIZIZV

++=++=++=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

2

22

2

1

2

12

02

21

1

01

20

10

0 11

11111

111

31

aa

aa

ZZZ

ZZZ

ZZZ

aa

aa

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

ZZZ

c

bc

ca

bc

b

ab

ca

ab

a

c

bc

ca

bc

b

ab

ca

ab

a

AA

Diagonal Sequence impedances

[ ] )22.2.8(22231

0 cabcabcba ZZZZZZZ +++++=

[ ] )23.2.8(31

21 acbcabcba ZZZZZZZZ −−−++==

Off-diagonal sequence impedances

[ ] )24.2.8(31 22

2001 cabcabcba ZaZaZaZZaZZZ −−−++==

[ ] )25.2.8(31 22

1002 cabcabcba aZZZaZaaZZZZ −−−++==


[ ] )26.2.8(22231 22

12 cabcabcba ZaZaZaZZaZZ +++++=

[ ] )27.2.8(22231 22

21 cabcabcba aZZZaZaaZZZ +++++=

Unsymmetrical Series Impedances without Mutual Inductances If we assume no mutual inductances (no coupling) (i.e. Zab = Zbc = Zca = 0) between the three impedances, then the above equation becomes

[ ] )1.22.2.8(31

210 cba ZZZZZZ ++===

[ ] )1.24.2.8(31 2

2001 cba aZZaZZZ ++==

[ ] )1.25.2.8(31 2

1002 cba ZaaZZZZ ++==

[ ] )1.26.2.8(31 2

12 cba aZZaZZ ++=

[ ] )1.27.2.8(31 2

21 cba ZaaZZZ ++=

Symmetrical Series Impedances [Ref. 1, p. 289] A symmetrical load is defined as a load whose sequence impedance matrix is diagonal; that is, all mutual impedances are zero. Equation these mutual impedances to zero and solving, the following conditions for a symmetrical load are determined. When both

loadlsymmetricaaforConditions⎭⎬⎫

====

cabcab

cba

ZZZZZZ


aba ZZZ 20 +=

aba ZZZZ −== 21 Off-diagonal sequence impedances

0211210022001 ====== ZZZZZZ

Symmetrical Series Impedances without Mutual Inductances A symmetrical load is defined as a load whose sequence impedance matrix is diagonal; that is, all mutual impedances are zero. Equation these mutual impedances to zero and solving, the following conditions for a symmetrical load are determined. When both

loadlsymmetricaaforConditions⎭⎬⎫

====

cabcab

cba

ZZZZZZ


aa ZZZZ === 20 Off-diagonal sequence impedances

0211210022001 ====== ZZZZZZ Sequence Network of Impedance Loads [Ref. 3, ch. 8, p. 404]


Fig. 4.5 shows a balanced Y impedance load. The impedance of each phase is designated ZY, and a neutral impedance Zn is connected between the load neutral and ground. The line-to-ground voltage for phase a can be written as

nInZaIYZagV +=

)( cIbIaInZaIYZagV +++=

cInZbInZaInZYZagV +++= )( (8.2.1)

Similarly,

cInZbInZYZaInZbgV +++= )( (8.2.2)

cInZYZbInZaInZcgV )( +++= (8.2.3)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

nY

n

n

n

nY

n

n

n

nY

cg

bg

ag

III

ZZZZ

ZZZ

Z

ZZ

ZZ

VVV

abcpppabcgp IZIZVV === (8.25)

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

==

cg

bg

ag

abcgp

VVV

VV ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+=

nY

n

n

n

nY

n

n

n

nY

p

ZZZZ

ZZZ

Z

ZZ

ZZZ ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

c

b

a

abcp

III

II Fig. 4.5 Balanced Y-impedance load

sps AIZVA =

Where, Zp is called phase impedance matrix. ( ) sp

-s IAZAV 1=

sss IZV = Weher, AZAZ p

-s

1= is called sequence impedance matrix.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ +=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

Y

Y

nY

nY

n

n

n

nY

n

n

n

nY

s

ZZ

ZZ

aa

aa

ZZZZ

ZZZ

Z

ZZ

ZZ

aa

aa 0

0

0

0

00

311

11111

111

31

2

22

2

Z ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ +=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

1

0

00

0

0

00

3

III

ZZ

ZZ

VVV

Y

Y

nY

)13.2.8()3( 0000 IZIZZV nY =+=

)14.2.8(1111 IZIZV Y ==

)15.2.8(2222 IZIZV Y ==


Where, nY ZZZ 30 += is called zero-sequence impedance; YZZ =1 is called positive-sequence impedance; YZZ =2 is called negative-sequence impedance;

Fig. 4.6 [Ref. 3, F. 8.4, p. 407] Sequence network of a balanced Y-connected load.

Eqs. (8.2.13) – (8.2.15) can be represented by the three networks shown in Fig. 4.6. These networks are called the zero-sequence, positive-sequence, and negative-sequence networks. As shown, each network is separate, uncoupled from the other two. This separation underlies the advantages of symmetrical components. When the neutral of the Y-connected load is solidly grounded with a zero-ohm conductor, the neutral impedance Zn is zero but I0 is not zero and the term 3Zn in the zero-sequence network becomes short-circuit. If there has no neutral path, then the neutral impedance Zn is infinite and the term 3Zn is the zero-sequence network becomes an open circuit. Under this condition of an open neutral, no zero-sequence current exists (I0 = 0). The zero-sequence network is open at the ∆-connected circuit. Zero-sequence current may circulate inside the ∆-circuit since the ∆ is a closed series circuit fir circulating single-phase currents. Such current would have to be produced in the ∆, however, by induction from an outside source or by zero-sequence generated voltages. A ∆ circuit and its zero-sequence network are shown in Fig. 4.7.

Fig. 4.7 [Ref.1, F. 1.17, p. 298] ∆-connected load and its

zero-sequence network.

Fig. 4.8 [Ref. 3, F. 8.5, p. 408] Sequence network for an equivalent Y representation of a balanced ∆-

connected load.


Even when zero-sequence voltages are generated in the phases of the ∆, no zero-sequence voltages exists between the ∆ terminals, for the rise in voltage in each phase of the generator is matched by the voltage drop in the zero-sequence impedance of each phase. The sequence network of the equivalent Y load corresponding to a balanced-∆ load are shown in Fig. 4.8. Example 8.4 [Ref. 3, p. 408] A balanced Y load is in parallel with a balanced ∆-connected capacitor bank. The Y load has an impedance ZY = (3+j4) Ω per phase, and its neutral is grounded through an inductiove reactnce Xn = 2 Ω. The capacitor bank has a reactance Xc = 30 Ω per phase. Draw sequence network for this load and calculate the load sequence impedances. Solution:

Z0= ZY + 3Zn = 3 + j4 + 3(j2) = 3+j10 Ω Z1=ZY//(Z∆/3)=(3+j4)(-j30/3)/[3+j4–(j30/3)]= 7.454∠26.57o Ω Z2= Z1= 7.454∠26.57o Ω

Sequence Network of Unloaded Generators [Ref. 1, 291] A synchronous generator, grounded through a reactor, is shown in Fig. 11.13. When a fault (not indicated in the figure) occurs at the terminals of the generator, current Ia, Ib, and Ic flow in the lines. If the fault involves ground, the current flowing into the neutral of the generator is designated In and the line currents can be resolved into their symmetrical components regardless of how unbalanced they may be.

0=−−+− agVaIgZaEnInZ

aEnInZaIgZagV +−−=;

Let cIbIaInI ++=

aEcInZbInZaInZgZaEcIbIaInZaIgZagV

+−−+−=

+++−−=

)(

)(

bEcInZbInZgZaInZbEcIbIaInZbIgZbgV

+−+−−=

+++−−=

)(

)(

cEcInZgZbInZaInZcEcIbIaInZcIgZcgV

++−−−=

+++−−=

)(

)(

Fig. 11.13 [Ref. 1, p. 292] Circuit diagram of an

unloaded generator.


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

c

b

a

ng

n

n

n

ng

n

n

n

ng

cg

bg

ag

EEE

III

ZZZZ

ZZZ

Z

ZZ

ZZ

VVV

pabcppppabcgp EIZEIZVV +−=+−==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

==

cg

bg

ag

abcgp

VVV

VV ; ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

c

b

a

p

EEE

E ; ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+=

ng

n

n

n

ng

n

n

n

ng

p

ZZZZ

ZZZ

Z

ZZ

ZZZ ;

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

c

b

a

abcp

III

II

psps AEAIZVA +−=

( ) p-

sp-

s EAIAZAV 11 +−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

a

a

a

a

a

a

ng

n

n

n

ng

n

n

n

ng

a

a

a

aEEa

E

aa

aa

III

aa

aa

ZZZZ

ZZZ

Z

ZZ

ZZ

aa

aa

VVV

22

22

1

0

2

22

22

1

0 11

111

31

11

11111

111

31

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ +−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

0

000

0

0

00

3

2

1

0

2

1

0

a

a

a

a

g

g

ng

a

a

a

EIII

ZZ

ZZ

VVV

Let, gggngg ZZZZZZ ==+= 210 ;3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

1

0

2

1

0

00

0

0

00

0

0

a

a

a

a

a

a

a

III

ZZ

ZE

VVV

222111000 ;; aaaaaaa IZVIZEVIZV −=−=−= The equations developed to this point are based on a simple machine model which assumes the existence of only fundamental components of currents; on this basis the positive- and negative sequence impedance are found to equal to one another but quite difference from the zero-sequence impedance. In fact, however, the impedances of rotating machines to currents of the three sequences will generally be different fro each sequence. The mmf produced by negative-sequence armature current rotates in the direction opposite to that of the rotor which has the dc field winding. Unlike the flux produced by positive-sequence current, which is stationary with respect to the rotor, the flux produced by the negative-sequence current is sweeping rapidly over the face of the rotor. The current induced in the field and damper windings counteract the rotating mmf of the armature and thereby reduce the flux penetrating of the rotor. This condition is similar to the rapidly changing flux immediately upon the occurrence of a short circuit at the terminals of a machine. The flux paths are the same as those encountered in evaluating subtransient reactance. So, in a cylindrical-rotor machine subtransient and negative-sequence reactances are equal. The reactances in both the positive- and negative-sequence circuits are often taken to be equal to the subtransient or transient reactance, depending on whether subtransient or transient conditions are being studied. When only zero-sequence current flows in the armature winding of a three-phase machine, the current and mmf of one phase are a maximum at the same time as the current and mmf of each of the other phases. The windings are so distributed around the circumference of the armature that


the point of maximum mmf produced by on e phase is displaced 120 electrical degrees in space from the point of maximum mmf of each of the other phases.

(a) Zero-sequence current paths (b) Zero-sequence network

(c) Positive-sequence current paths (d) Positive-sequence network

(e) Negative-sequence current paths (f) Negative-sequence network Fig. 11.14 Paths for current of each sequence in a generator and the corresponding sequence networks. If the mmf produced by the current of each phase had a perfectly sinusoidal distribution in space, a plot of mmf around the armature would result in three sinusoidal curves whose sum would be zero at every point. No flux would be produced across the air gap, and the only reactance of any phase winding would be that due to leakage and end turns. In an actual machine the winding is not distributed to produce perfectly sinusoidal mmf. The flux resulting from the, sum of the mmfs is very small, which makes the zero-sequence reactance the smallest of the machine's reactances -just somewhat higher than zero of the ideal case where there is no air-gap flux due to zero-sequence current. Sequence Model of Three-phase Transformer [Ref. 3, p. 422] No current flows in the primary of a transformer unless current flows in the secondary, if we neglect the relatively small magnetizing current. The primary current is determined by the secondary current and the turns ratio of the windings, again with magnetizing current neglected. These principles guide us in the analysis of individual cases.

The positive- and negative-sequence of three-phase transformer impedances of each type of connection are identical.


A phase shift is included in positive- and negative-sequence networks. For the American Standard, the positive-sequence voltage and currents on the high-voltage side of Y-∆ transformers lead the corresponding quantities on the low voltage side by 30o. For negative sequence, the high-voltage quantities lag by 30o. Zero-sequence current in the Y winding if there is a neutral connection, and corresponding zero-sequence currents flow within the ∆ winding. However, no zero-sequence current enter or leaves the ∆ winding.

Symbols Connection Diagrams Zero-sequence Circuits

Fig. 11.18 Zero-sequence equivalent circuits of three-phase transformer banks, together with diagrams of connections and the symbols for one-line diagram.


Five possible connections of two-winding transformers will be discussed. These connections are summarized, along with their zero-sequence circuits, in Fig. 11.18. The arrows on the connection diagrams of the figures to follow show the possible paths for the flow of zero-sequence current. Absence of an arrow indicates that the transformer connection is such that zero-sequence current cannot flow. The zero-sequence equivalent circuits are approximate as shown since resistance and the magnetizing-current path are omitted from each circuit. The letters P and Q identify corresponding points on the connection diagram and equivalent circuit. The reasoning to justify the equivalent circuit for each connection follows. Case 1: Y-Y bank, One Neutral Grounded If either one of the neutral of Y-Y bank is ungrounded, zero-sequence current cannot flow in either winding. The absence of path through one winding prevents current in the other. An open circuit exists for zero-sequence current between the two parts of the system connected by the transformer. Case 2: Y-Y bank, Both Neutral Grounded In this connection a path through the transformer exists for zero-sequence current in both-windings. The zero sequence current can flow in both windings of transformer. Case 3: Y-∆ bank, Grounded Y If the neutral of a Y-∆ is grounded, zero-sequence currents have a path to ground through the Y because corresponding induced current can circulate in the ∆. The equivalent circuit must be provided for a path from the line on the Y side through the equivalent impedance to the reference bus. An open circuit must exist between the line and the reference bus on the ∆ side. Case 4: Y-∆ bank, Ungrounded Y An ungrounded Y zero-sequence current cannot flow in the transformer winding. Case 5: ∆-∆ bank Since a ∆ circuit provides no return path for zero-sequence current, no zero-sequence current can flow into a ∆-∆ bank, although it can circulate within the ∆ windings. Zero-sequence equivalent circuits determined for various parts of the system separately are readily combined to form the complete zero-sequence network as shown in Figs. 11.19 and 11.20.

Fig. 11.19 One-line diagram of a small power system and the corresponding zero-sequence network.


Fig. 11.20 One-line diagram of a small power system and the corresponding zero-sequence network.

Example 11.4 [1, p.301] Draw (i) the positive-sequence network, (ii) negative-sequence network , and (iii) zero-sequence network for the system shown in the figure: G1: 300 MVA, 20 kV, X′′ = 20%, zero-sequence reactance, X0= 5%, current limiting reactors Zn= 0.4 Ω Transmission line: 64 km (4- mi), series reactance of the transmission line is 0.5 Ω/km, zero-sequence reactance = 1.5 Ω/km. T1: 350 MVA, 230/20 kV, X′′ = 10% T2: 300 MVA, 220/13.2 kV, X′′ = 10% M1: 200 MVA, 13.2 kV, X′′ = 20%, zero-sequence reactance, X0=5%, current limiting reactors Zn= 0.4 Ω M2: 100 MVA, 13.2 kV, X′′ = 20%, zero-sequence reactance, X0= 5% Select the generator rating as base in the generator circuit.


Solution: Base MVA, Sbase, 3φ = 300 MVA Base kV, VbaseLL = 20 kV

G1: Ω=== 333.1300

2)20(2)(

MVAbaseLLkVbase

baseZ

9.0333.1

4.033 ==nZ pu

X′′ = 0.1 pu; X0 = 0.05 pu

T1: 0857.03503002

2302301.0'' =⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=X pu

Transmission Line: Given, series reactance of the transmission line is 0.5 Ω/km. Reactance of transmission line, Ω×= 640.5X

Base impedance of transmission line, Ω=== 3.176300

2)230(2)(

MVAbaseLLkVbase

baseZ

Per-unit reactance of transmission line, 1815.03.176645.0

=×

==baseZXX pu

5445.03.176645.1

0 =×

=X pu

T2: 0915.03003002

13.813.21.0'' =⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=X pu

Motor M1: Ω=== 635.0300

2)8.13(2)(

MVAbaseLLkVbase

baseZ

2745.02003002

13.813.22.0'' =⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=X pu

0686.02003002

13.813.205.00 =⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=X pu

89.1635.04.033 ==nZ pu


Motor M2: 5490.01003002

13.813.22.0'' =⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=X

1372.01003002

13.813.205.00 =⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=X pu

Fig. 11.21 Sequence networks of Example 11.4

References [1] Willaim D. Stevenson, Elements of Power System Analysis, Fouth Edition, McGraw-Hill International Editions, Civil Engineering Series, McGraw-Hill Inc. [2] John J. Grainger, William D. Steevnson, Jr., Power System Analysis, McGraw-Hill Series in Electrical and Conputer Engineering, McGraw-Hill Inc. [3] J. Duncan Glover, Mulukutla S. Sharma, Thomas J. Overbye, Power System Analysis and Design, Fouth Edition (India Edition), Course Technology Cengage Learning [4] Hadi Saadat, Power System Analysis, Tata McGraw-Hill Publishing Company Limited [5] I J Nagrath, D P Lothari, Modern Power System Analysis, Second Edition, Tata McGraw-Hill Publishing Company Liited [6] V. K. Mehta, Rohit Mehta, Principles of Power System, Multicolor Illustrative Edition, S. Chand and Company Limited

01f symmetrical components

Documents