Symmetrical components

Symmetrical Components
 Symmetrical Component Analysis
 Synthesis of Unsymmetrical
Phases from Their Symmetrical
Components
 The Symmetrical Components of
Unsymmetrical Phasors
 Phase Shift of Symmetrical
Components in or
Transformer Banks
 Power in Terms of Symmetrical
Components
Y Y

Symmetrical Components
 Unsymmetrical Series
Impedance
 Sequence Impedance and
Sequence Network
 Sequence Networks of
Unloaded generators
 Sequence Network
 Zero-Sequence Network

Symmetrical Component Analysis
Goal :
Symmetrical component analysis is a very useful tool
for dealing with unbalanced three-phase faults.

Synthesis of Unsymmetrical Phases from
Their Symmetrical Components 1
“An unbalanced system of n related phasors can be resolved into
n systems of balanced phasors called the symmetrical components
of the original phasors. The n phasors of each set of components
are equal in lengths , and the angles between adjacent phasors of
the set are equal.”
by C.L Fortescue , 1918

1. For positive- sequence
components
2. For negative-sequence
components
(1) Positive- sequence components (2) Negative-sequence components
1a 1b
1c
2a
2c
2b
n n

(3) Zero-sequence components
0aV
0bV
0cV
0 For zero-sequence components

021 aaaa VVVV 
021 bbbb VVVV 
021 cccc VVVV 
0aV
2aV
aV
1aV
1bV
2bV
0bV
1cV
bV
0cV cV
2cV

Use a to designate the operator that causes a rotation of in the
counterclockwise direction ,
13601
866.05.02401
866.05.01201
03
02
0



a
ja
ja
0
120
2
a
3
,1 a
2
a a
3
,1 a
a

The Symmetrical Components of
Unsymmetrical Phasors 1
0000
2
2
222
111
2
1
,
,
,
acab
acab
acab
VVVV
VaVVaV
aVVVaV



1a 1b
1c
2a
2c
2b

02
2
1021
021
2
021
021
aaacccc
aaabbbb
aaaa
VVaVaVVVV
VaVVaVVVV
VVVV


































2
1
0
2
2
1
1
111
a
a
a
c
b
a
V
V
V
aa
aa
V
V
V












2
2
1
1
111
aa
aaA











aa
aaA
2
21
1
1
111
3
1































c
b
a
a
a
a
V
V
V
aa
aa
V
V
V
2
2
2
1
0
1
1
111
3
1
* When three phase phasors are balanced , only the
positive-sequence component exists .

1.Sequence component representation of L-L voltage































ca
bc
ab
ab
ab
ab
V
V
V
aa
aa
V
V
V
2
2
2
1
0
1
1
111
3
1
2.Sequence component representation of current































c
b
a
a
a
a
I
I
I
aa
aa
I
I
I
2
2
2
1
0
1
1
111
3
1

)(
3
1
)(
3
1
)(
3
1
0
0
0
cbaa
cabcabab
cbaa
IIII
VVVV
VVVV



No zero-sequence components exist if the sum of the three
phasors is zero.

)(
3
1
0 cbaa VVVV 
00 aV
00 aV 3When is balanced
When 0)(  cba VVV
* If then is unbalanced.
* Unbalanced does not guarantee .
00 aV 3
3 00 aV

)(
3
1
0 cabcabab VVVV 
is always zero whether the three phase system
is balanced or not.
)( cabcab VVV  is always zero (form closed loop)
0abV
a
b
c

aI
bI
cI
aI
bI
cI
aI
bI
cI
nI
0)(
3
1
0  cbaa IIII
( ungrounded Y )
0)(
3
1
0  cbaa IIII
nacba IIIII  03)(
Y with a path to neutral
0)(
3
1
0  cbaa IIII
connected
VECTOR
SOFTWARE

example : 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 assuming that line c is open, find the symmetrical components
of the line currents

Z Z
Z
a
c
b
ampIa
0
010
ampIb
0
18010
0cI
AIa
0
010
AIb
0
18010
AIc 0
The line currents are :

0)018010010(
3
1 00
0 aI
)012018010010(
3
1 000
1 aI
Aj 0
3078.589.25 
)024018010010(
3
1 000
2 aI
Aj 0
3078.589.25 
AIb
0
1 15078.5 
AIb
0
2 15078.5 
00 bI
AIc
0
1 9078.5 
AIc
0
2 9078.5 
00 cI
Since there no neutral current involved ,
should be zero .
0aI

Phase Shift of Symmetrical Components
in or Transformers Banks 1Y Y
The American standard for designating terminal and on or
transformer requires that the positive-sequence voltage drop from to
neutral leads the positive-sequence voltage drop from to neutral by ,
regardless of whether the or winding is on the high tension side .
Similarly, the positive-sequence voltage drop from to neutral leads the
voltage drop from to neutral by and the positive-sequence voltage
drop from to neutral leads the voltage drop from to
neutral by .
1H
1X Y1H
1X 0
30
0
30
3H
2X
3X
2H
0
30
Y
Y

Example :
A
B
C a
c
b1H
3H
2H 2X
1X
3X
A
B
C
a
b
c
1H
3H
2H
1X
2X
3X
1AV 1bV 0
30leads by 1aVleads by1AV 0
30

The American standard for designating terminal and on or
transformer requires that the negative-sequence voltage drop from to
neutral lags the negative-sequence voltage drop from to neutral by ,
regardless of whether the or winding is on the high tension side .
Similarly, the negative-sequence voltage drop from to neutral lags
the voltage drop from to neutral by and the negative-sequence
voltage drop from to neutral lags the voltage drop from to neutral
by .
1H
1X Y1H
1X 0
30
0
30
3H
2X
3X
2H
0
30
Y
Y

A
B
C a
c
b1H
3H
2H 2X
1X
3X
A
B
C
a
b
c
1H
3H
2H
1X
2X
3X
2aV(b) lags by2AV 0
30
2bV(a) lags by2AV 0
30
Example :

A
B
C a
c
b1H
3H
2H 2X
1X
3X
2B
2A
2C
2b
2c
2a
1B
1A
1C
1b
1a
1c
1AVleads by1aV 0
90
2AVlags by2aV 0
90
1bVleads by1AV 0
30
2bVlags by2AV 0
30
Y 

in or Transformers Banks 6
Example 11.7. The resistive Y-connected load bank of Example 11.2 is supplied from the low-voltage
Y-side of a Y- transformer. The voltages at the load are the same as in that example. Find the line
voltages and currents in per unit on the high-voltage side of the transformer.
unitperI a
0)1(
6.439857.0 
unitperI a
0)2(
3.2502346.0 
)(6.439857.0 0)1(
basevoltageneutraltolineunitperV an 
)(3.2502346.0 0)2(
basevoltageneutraltolineunitperV an 
9456.02785.06.739857.0306.439857.0 000)1(
jV A 
1517.01789.03.2202346.0303.2502346.0 000)2(
jV A 
unitperjVVV AAA
0)2()1(
8.828.07939.00994.0 
7138.06798.04.469857.0 0)1(2)1(
jVaV AB 

Y Y

0791.02209.07.192346.0 0)2(2)2(
jVaV AB 
unitperjVVV BBB
0)2()1(
4.4120.17929.09007.0 
2318.09581.06.1939857.0 0)1(2)1(
jVaV AC 
2318.00419.03.1002346.0 0)2(2)2(
jVaV AC 
unitperjVVV CCC
0)2()1(
1800.100.1 
5868.18013.07929.09007.07939.00994.0 jjjVVV BAAB 
)(8.11678.1 0
basevoltagaeneutrallineunitper 
)(8.116
3
78.1 0
basevoltagaelinetolineunitper 
Y Y

7939.09007.10.17939.09007.0 jjVVV CBBC 
)(7.2206.2 0
basevoltagaeneutrallineunitper 
)(7.2219.17.22
3
06.2 00
basevoltagaelinetolineunitper 
97939.00994.17939.00994.00.1 jjVVV ACCA 
)(8.215356.1 0
basevoltagaeneutrallineunitper 
)(8.215783.08.215
3
356.1 00
basevoltagaelinetolineunitper 
unitperIA
0
8.8280.0 
unitperIB
0
4.4120.1 
unitperIC
0
1800.1 
Y Y

A
B
C a
c
b1H
3H
2H 2X
1X
3X
A
B
C
a
b
c
1H
3H
2H
1X
2X
3X
(b) leads by 0
30aV )1(
(a) leads byAV )1( 0
30 AV )1(
aV )1(
Figure 11.23
labeling of lines connected to a three-phase Y- transformer.
Y Y

Power in terms of Symmetrical Components
ccbbaa IVIVIVjQPS ***

   *
012012
*
AIAV
I
I
I
V
V
V
S
T
c
b
a
T
c
b
a






















*
012012
*
012
*
012 3 IVIAAV
TTT

 
*
2
1
0
2103











a
a
a
aaa
I
I
I
VVV
)(3 2
*
21
*
10
*
0 aaaaaa IVIVIV 
IAAT
3, *


Unsymmetrical Series Impedance 1
a
b
c '
c
'
b
'
a
cZ
bZ
aZ
aI
bI
cI
caZ
acZ
abZ































c
b
a
ccbca
bcbba
acaba
cc
bb
aa
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
'
'
'
































2
1
0
2
1
0
'
'
'
a
a
a
ccbca
bcbba
acaba
cc
bb
aa
I
I
I
A
ZZZ
ZZZ
ZZZ
V
V
V
A
































2
1
0
1
2
1
0
'
'
'
a
a
a
ccbca
bcbba
acaba
cc
bb
aa
I
I
I
A
ZZZ
ZZZ
ZZZ
A
V
V
V
Z

ZAAZ 1
012
















)2()2()2(
)2()2()2(
)2()2()2(
001122
220011
112200
MsMsMs
MsMsMs
MsMsMs
ZZZZZZ
ZZZZZZ
ZZZZZZ
Where
)(
3
1
)(
3
1
)(
3
1
2
2
2
1
0
cbas
cbas
cbas
aZZaZZ
ZaaZZZ
ZZZZ



)(
3
1
)(
3
1
)(
3
1
2
2
2
1
0
abcabcM
abcabcM
abcabcM
aZZaZZ
ZaaZZZ
ZZZZ




2111220000
)2()2()2(' aMsaMsaMsaa
IZZIZZIZZV 
2221000111
IZZIZZIZZV 
2001110222
IZZIZZIZZV 

Case 1. If no coupling , 0)(  jiijZ
2112000' asasasaa
IZIZIZV 
)(
3
1
)(
3
1
)(
3
1 2
2
2
10 cbaacbaacbaa ZaaZZIaZZaZIZZZI 
2011022' asasasaa
IZIZIZV 
2210011' asasasaa
IZIZIZV 
)(
3
1
)(
3
1
)(
3
1 2
21
2
0 cbaacbaacbaa ZZaZIZZZIZaaZZI 
)(
3
1
)(
3
1
)(
3
1
2
2
1
2
0 cbaacbaacbaa ZZZIZaaZZIaZZaZI 
then 0210  MMM ZZZ

Case 2 . If
0)(  jiijZ
cba ZZZ 
aaaa
ZIV 11'  aaaa
ZIV 22'  aaaa
ZIV 00' 
Symmetrical components of unbalanced currents flowing in a balanced- load
or in balanced series impedances produce voltage drops of the same sequence ,
provided no coupling exists between phases.
If the impedances are unequal, the voltage drop of any one sequence is dependent on the
current of all three sequences.
If coupling such as mutual inductance exists among the three impedances, then the
formula will become more complicated.
Y
1.
2.
Complete transportation assumed

Assume:
1. No coupling
2.
Positive-sequence currents produce positive-sequence voltage drops.
Negative-sequence currents produce negative sequence voltage drops.
zero-sequence currents produce zero-sequence voltage drops.
cba ZZZ 

Sequence Impedance and Sequence Network 1
The impedance of circuit when positive- sequence
current alone are flowing is called positive-sequence
impedance.
The impedance of circuit when negative-sequence
currents alone are flowing is called negative
sequence impedance.
When only zero-sequence currents are present, the
impedance is called zero sequence impedance.

The single-phase equivalent circuit composed of the impedance to
current of any one sequence only is called the sequence network.
Positive-sequence network contains positive sequence current and
positive sequence impedance only.
Negative-sequence network contains negative sequence current
and negative sequence impedance only.

Sequence network carrying the individual currents ,
and are interconnected to represent various
unbalanced fault condition.
1aI 2aI
0aI
Zero-sequence network contains zero sequence current and
zero sequence impedance only.

Sequence Impedance of Various Devices
Positive Negative Zero
Line same same different
Transformer same same same
Machine different different different**
* Usually they are assumed to be the same

Sequence Networks of Unloaded Generators 1
The generator voltage
are of positive sequence only,
since the generator is designed
to supply balanced three-phase
voltage.
aI
bI
cI
nI
cE
bE
aE
bc
a
++
+
-
--
nZ ),,( cba EEE

2aI
2cI
2bI
a
c
b
2Z
2Z
2Z 2aV
2aI
a
2Z
+
-
Negative-sequence
network
+
-
1aV
aE
1aI
a
1Z
+
-
Positive-sequence
network
Reference
Reference
1aI
1bI
cE
bE
aE
bc
a
+
++
-
--
1Z
1Z1Z
aI
111 ZIEV aaa 
222 ZIV aa 

1cI
0aI
0aI
0aI
a
c
b
0gZ
0gZ 0gZ
0aI
0aV
a
0gZ
+
-
nZ3
0Z
Zero-sequence
network
Reference
only appears in the zero-sequence
network
nZ
000 ZIV aa nZ
03 aI
)3( 00 nga ZZI 
na II 03

Example 11.6. A salient-pole generator without dampers is rated 20 MVA, 13.8kV and has a
direct=axis subtransient reactance of 0.25 per unit. The negative-and-zero-sequence reactance
are, respectively, 0.35 and 0.10 per unit. The neutral of the generator is solidly grounded. With
the generator operating unloaded at rated voltage with , a single line-to-
ground fault occurs at the machine terminals, which then have per-unit voltages to ground,
Determine the subtransient current in the generator and the line-t0-line voltages for subtransient
conditions due to the fault.
unitperEan
0
00.1 
0aV 0
25.102013.1 cV0
25.102013.1 bV
0cI
aI
0bI
na II 
cnE
bnE
anE
bc
a
++
+
-
- -
nZ
n
Figure 11.15

Figure 11.15 shows the line-to-ground fault of phase a of the machine.
unitperjVb 990.0215.0 
unitperjVc 990.0215.0 
unitper
j
j
j
j
j
aa
aa
V
V
V
c
b
a















































0500.0
0643.0
0143.0
990.0215.0
990.0215.0
0
1
1
111
3
1
2
2
)0(
)0(
)0(
unitperj
j
j
Z
V
I
go
a
a 43.1
10.0
)0143.0()0(
)0(




unitperj
j
jj
Z
VE
I
aan
a 43.1
25.0
)0643.0()00.1(
1
)1(
)1(





unitperj
j
j
Z
V
I
a
a 43.1
35.0
)0500.0(
2
)2(
)2(



29.43 )0()2()1()0(
jIIIII aaaaa 
There, the fault current into the ground is
The base current is and so the subtransient current in line a isA837)8.133(000,20 
AjjIa 590,383729.4 

Line-to-line voltage during the fault are
unitperjVVV baab
0
7.7701.1990.0215.0 
unitperjVVV cbbc
0
270980.1980.10 
unitperjVVV acca
0
7.7701.1990.0215.0 
kVVab
00
7.7705.87.77
3
8.13
01.1 
kVVbc
00
27078.15270
3
8.13
980.1 
kVVca
00
3.10205.83.102
3
8.13
01.1 

Before the fault the line voltages were balanced and equal to 13.8kV. For comparison with the line
voltages after the fault occurs, the prefault voltages, with as reference, are given asanan EV 
kVVab
0
308.13  kVVbc
0
2708.13  kVVca
0
1508.13 
Figure 11.6 shows phasor diagrams of prefault and postfault voltages.
Figure 11.6
(a) Prefault (b) Postfault
anV
caVcaV
bcV
bcV
abVabV
aa
n
b
b
c c

The positive-sequence diagram of a generator is
composed of an emf in series with the positive-sequence
impedance of the generator.
The negative and zero-sequence diagrams contain no
emfs but include the negative and zero-sequence
impedances of the generator respectively.

Sequence Networks 1
The matching reactance in positive-sequence network is the subtransient ,transient,
or synchronous reactance, depending of whether subtransient , transient, or
steady- state condition are being studied.
The reference bus for the positive and negative sequence networks is the neutral
of the generator. So for as positive and negative sequence components are
concerned , the neutral of the generator is at ground potential even if these is
connection between neutral and ground.
The reference bus for the zero sequence network is the ground (not necessary
the neutral of the generator).
nZ

Sequence Networks 2
Convert a positive sequence network to a negative sequence
network by changing, if necessary, only the impedance
that represent rotating machine , and by omitting the emf.
The normal one-line impedance diagram plus the induced emf is the
positive sequence network.
Three-phase generators and motors have internal voltage of positive
sequence only.

Example of Positive and Negative-Sequence Network 1
Example: Draw the positive and negative-sequence networks
for the system described as below . Assume that the
negative-sequence reactance of each machine is
equal to its subtransient reactance .Omit resistance.
1T
1M
T
r
p
m nk l
2M

Example of Positive and Negative-Sequence Network 2
++
+
--
-
0857.0j 0915.0j0815.0j
5490.0j
02.0j
2745.0j
2mE1mE
gE
k l m n
p r
0857.0j 0915.0j0815.0j
5490.0j
02.0j 2745.0j
Reference bus
k l m n
p q
(Positive)
(Negative)

Zero sequence Network 1
1 . Zero-sequence network currents will flow only if a return path exists.
2 . The reference bus of the zero-sequence network is the ground.
ZZ
Z
N
Z
Z
Z
N
Z N
R
N
Reference

Z
Z
Z
N
Z
N
R
nN
03 an II 
nZ3
0aI
Z Z
Z
R
Z

Zero-sequence equivalent circuit of three phase transform banks.
Symbols Connection Diagrams
Zero-Sequence
Equivalent Circuit
p
p
p
p p p
Q
Q
Q
Q Q
0Z
0Z
Reference bus
Reference bus

Symbols Connection Diagrams Equivalent Circuit
Zero-Sequence
p
p
p
p
p
p
Q
Q
Q Q
0Z
0Z
Reference bus
Reference bus

Symbols Connection Diagrams Equivalent Circuit
Zero-Sequence
p
p
p
Q Q
0Z
Reference bus

nZ3
0gZ Q
R
S
T
M N
P
(Zero-Sequence)
Example:
nZ
Q S
P
R T
M N

1gZ
Q S
R T
M
NP
2gZ
1gE 1gE
++
--
Q
R T
M
N
P
S
(Positive-Sequence)
(Negative-Sequence)
SEQUENCE
NETWORK
SOFTWARE

Example 11.9. Draw the zero-sequence network for the system described in Example 6.1. Assume
zero-sequence network for the generator and motors of 0.05 per unit. A current-limiting reactor of
is in each of the each of the neutrals of the generator and the large motor. The zero-sequence
reactance of the transmission line is
4.0
km5.1
Generator:
Motor 1:
Motor 2:
unitperX 05.00 
unitperX 0686.0)
8.13
2.13
)(
200
300
(05.0 2
0 
unitperX 1372.0)
8.13
2.13
)(
100
300
(05.0 2
0 
 333.1
300
)20( 2
ZBase
 635.0
300
)8.13( 2
ZBase

unitperZn 900.0)
333.1
4.0
(33 
unitperZn 890.1)
635.0
4.0
(33 
unitperZ 5445.0
3.176
645.1
0 


The zero-sequence network is shown in Fig. 11.28
05.0j
900.0j
0857.0j 5445.0j 0915.0j
l m
k
n
p r
0686.0j
890.1j
1372.0j
reference

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