ELE B7 Power Systems
Engineering
Symmetrical Components
Analysis of Unbalanced Systems



Except for the balanced three-phase fault, faults
result in an unbalanced system.
The most common types of faults are single lineground (SLG) and line-line (LL). Other types are
double line-ground (DLG), open conductor, and
balanced three phase.
The easiest method to analyze unbalanced system
operation due to faults is through the use of
symmetrical components
Slide # 1
Symmetrical Components


The key idea of symmetrical component analysis is
to decompose the unbalanced system into three
sequence of balanced networks. The networks are
then coupled only at the point of the unbalance (i.e.,
the fault)
The three sequence networks are known as the
–
–
–
positive sequence (this is the one we’ve been using)
negative sequence
zero sequence
Slide # 2
Symmetrical Components
Balance
Systems
Unbalance Currents
Unsymmetrical
Fault
IA
IC
IB
zero sequence
Zero
Sequence
Symmetrical
components
Unbalance
System
Sequence Currents
Positive
Sequence
positive sequence
Three
balanced
Systems
Negative
Sequence
negative sequence
Slide # 3
Symmetrical Components
Assuming three unbalance voltage phasors, VA, VB and VC having a positive
sequence (abc). Using symmetrical components it is possible to represent each
phasor voltage as:
VA  VA0  VA  VA

B

B
VB  V  V  V
0
B
VC  VC0  VC  VC
Zero Sequence Component
Positive Sequence Component
Negative Sequence Component
Where the symmetrical components are:
Slide # 4
Symmetrical Components



The Positive Sequence Components (
VA , VB , VC )
Three phasors
Equal in magnitude
Displaced by 120o in phase
Having the same sequence as the original phasors (abc)
The Negative Sequence Components (
VA , VB , VC )
Three phasors
Equal in magnitude
Displaced by 120o in phase
Having the opposite sequence as the original phasors (acb)
0
0
0
The zero Sequence Components ( VA , VB , VC
Three phasors
Equal in magnitude
Having the same phase shift ( in phase)
)
VA
VC
120 o
120 o 120 o
VB
VA
VB
120 o
120 o 120 o
VC
VB0
VA0
VC0
Slide # 5
Example
VA  VA0  VA  VA
VB  VB0  VB  VB
V A0
VB 0
VC 0
Zero
Sequence
VA
VC  VC0  VC  VC
V A0
V A
Positive
Sequence
VC  0
V A
VC 
120 o
V A
120 o
VB 
VA
VC  0
V A
VB
Unbalance
Voltage
Negative
Sequence
120 o
120 o VC 
VB 
VB
Synthesis Unsymmetrical phasors
using symmetrical components
Slide # 6
Sequence Set Representation

Any arbitrary set of three phasors, say Ia, Ib, Ic can
be represented as a sum of the three sequence sets
Ia 
Ib 
Ic 
0
Ia
I b0
I c0




Ia
I b
I c




Ia
I b
I c
where
I a0 , I b0 , I c0 is the zero sequence set
I a , I b , I c is the positive sequence set
  
I a , Ib , I c
is the negative sequence set
Slide # 7
Conversion Sequence to Phase
Only three of the sequence values are unique,
I0a , I a , I a ; the others are determined as follows:
 = 1120
 2  3  0
3  1
I0a  I0b  I0c (since by definition they are all equal)
I b   2 I a
I c   I a
I b   I a
I c   2 I a
0

1
1
1
1
I
 
 a
1 
 Ia 
1










0
+
2
2
 I   I 1  I   I   1 
   Ia 
b
a
a
a





 

 
 2  1   2   I  
 I c 
1
  
 
  a 
Slide # 8
Conversion Sequence to Phase
Define the symmetrical components transformation
matrix
1
1 1


2

A  1 
1   2 


0
0


I
I
a
 Ia 
 
 


Then I  I b  A  I a   A  I   A I s
 
 
 
 I c 
 I a 
 I 
Slide # 9
Conversion Phase to Sequence
By taking the inverse we can convert from the
phase values to the sequence values
1
Is  A I
1
1 1
1 1 
2
with A  1   
3
1  2  


Sequence sets can be used with voltages as well
as with currents
Slide # 10
Example
If the values of the fault currents in a three phase system are:
I A  150 45 I B  250 150
I C  100 300
Find the symmetrical components?
Solution:
VO
Slide # 11
Example
If the values of the sequence voltages in a three phase system are:
Vo  100
V  200 60
V  100 120
Find the three phase voltages
Solution:
V A  200 60  100 120  100
V A  300 60
V B  1 240( 200 60 )  1120( 100 120 )  100
V B  300   60
VC  1120( 200 60 )  1 240( 100 120 )  100
VC  0
Slide # 12
Use of Symmetrical Components
1. The Sequence circuits for Wye and Delta connected loads
Consider the following Y-connected load:
I n  I a  Ib  I c
Vag  I a Z y  I n Z n
Ia
V ab
Vag  ( ZY  Z n ) I a  Z n I b  Z n I c
n
ZY
Ib
Vbg  Z n I a  ( ZY  Z n ) I b  Z n I c
ZY
V ca
Ic
V an
I n  3 I ao
ZY
Zn
V bc
Vcg  Z n I a  Z n I b  ( ZY  Z n ) I c
Vag 
Z y  Zn
 

Vbg    Z n
V 
 Z
n
 cg 

Zn
Z y  Zn
Zn
  Ia 
 
Z n  Ib
 
Z y  Z n   I c 
Zn
Slide # 13
Vn
Use of Symmetrical Components
Vag 
Z y  Zn
Zn
Zn   Ia 
 

 
Z y  Zn
Z n  Ib
Vbg    Z n
 
V 
 Z
  I c 
Z
Z
Z

n
n
y
n
 cg 

V
 Z I , V  A Vs , I  A I s
A Vs  Z A I s

 Z y  3Z n

1
0
A ZA  

0

Vs  A 1 Z A I s
0
Zy
0
0

0
Z y 
Slide # 14
Networks are Now Decoupled
V 0 
 Z y  3Z n 0
 

Zy
0
V   
 

0
0
V
 

Systems are decoupled
V
V 0  ( Z y  3Z n ) I 0
V   Zy I
I ao
Vao
ZY



Zo 


Zero Sequence Circuit
Ia
ZY
n
3Z n
Va 
0
0  I 
 
0  I 
 

Z y  I 
 



Z 


Positive Sequence Circuit
 Zy I
n
Ia
Va 
ZY
n



Z 


Negative Sequence Circuit Slide # 15
Y-connected load (Isolated Neutral):
Ia
I ao
Vca
ZY
Ic
Vao
ZY
Ib



Zo 


Van
Vab
ZY
ZY
If the neutral point of a Y-connected load is not
grounded, therefore, no zero sequence current
can flow, and
Zn  
Symmetrical circuits for Y-connected load
with neutral point is not connected to ground
are presented as shown:
Zn  
Zero Sequence Circuit
Ia
Vbc
n
Va 
ZY
n



Z 


Positive Sequence Circuit
Ia
Va 
ZY
n



Z 


Negative Sequence Circuit Slide # 16
Delta connected load:
The Delta circuit can not provide a path through
neutral. Therefore for a Delta connected load or its
equivalent Y-connected can not contain any zero
sequence components.
Vab  Z  I ab
,
Vbc  Z  I bc
,
Ia
Vab
Vca  Z  I ca
Ib
The summation of the line-to-line voltages
or phase currents are always zero
1
( Vab  Vbc  Vca )  Vab 0  0
3
and
I ab
Vca
Vbc
Ic
n
V ao
Va 
Z
Z
I bc
1
( I ab  I bc  I ca )  I ab 0  0
3
Therefore, for a Delta-connected loads without sources or mutual
will be no zero sequence currents at the lines (There are some
circulating currents may circulate inside a delta load and not seen at
the zero sequence circuit).
Ia
Ia
I ao
Z
Z
/
3

n
Zero
Sequence
Circuit
I ca
Z
Positive
Sequence
Circuit
Va 
coupling there
cases where a
the terminals of
Z / 3
n
Negative
Sequence
Circuit
Slide # 17
Sequence diagrams for lines


Similar to what we did for loads, we can develop
sequence models for other power system devices,
such as lines, transformers and generators
For transmission lines, assume we have the
following, with mutual impedances
Ia
Zaa
a’
Ib
Zaa
b’
a
b
Zab
Ic
c
n
In
Zan
Zaa
Z nn
c’
n’
Slide # 18
Sequence diagrams for lines, cont’d
Assume the phase relationships are
 Va 
 V 
 b
 Vc 
where
 Zs
 Zm

 Z m
Zm
Zs
Zm
Zm   Ia 
Z m   Ib 
 
Z s   I c 
Z s  self impedance of the phase
Zm
 mutual impedance between the phases
Writing in matrix form we have
V
 ZI
Slide # 19
Sequence diagrams for lines, cont’d
Similar to what we did for the loads, we can convert
these relationships to a sequence representation
V
A Vs
 ZI
V
 Z A Is

 Z s  2Z m
A 1 Z A  
0

0

 A Vs
Vs
0
Zs  Zm
0
I  A Is
 A
1
Z A Is
0 
0 

Z s  Z m 
Slide # 20
Sequence diagrams for lines, cont’d
I ao
a
Therefore,
Zo
Zo  Z s  2 Zm
Van0
Z  Zs  Zm
n
Z  Zs  Zm
a
Where,
Van 0
n
Ia
Z s  Z aa  Z nn  2 Z an
Van
Z m  Z ab  Z nn  2 Z an
n
a
The ground wires (above overhead TL) combined with the earth
works as a neutral conductor with impedance parameters
that effects the zero sequence components. Having a good grounding
(depends on the soil resistively), then the voltages to the neutral can
be considered as the voltages to ground.
a
Z
Van 
n
Ia
Van
n
a
a
Z
Van 
n
Slide # 21
Sequence diagrams for generators

Key point: generators only produce positive
sequence voltages; therefore only the positive
sequence has a voltage source
I ao
Ia
Z
Ean
Z go
Va 
Z
Va 
Vao
3Zn
During a fault Z+  Z  Xd”. The zero
sequence impedance is usually substantially
smaller. The value of Zn depends on whether
the generator is grounded
Slide # 22
Sequence diagrams for Transformers

The positive and negative sequence
diagrams for transformers are similar to
those for transmission lines.

The zero sequence network depends upon both how
the transformer is grounded and its type of connection.
The easiest to understand is a double grounded wyewye
Z
Z
+
Reference Bus
Z-
Reference Bus
Z0
Reference Bus
Slide # 23
Transformer Sequence Diagrams
Slide # 24