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ECE 476
POWER SYSTEM ANALYSIS
Lecture 20
Symmetrical Components, Unbalanced Fault Analysis
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements

Homework 8 is 7.1, 7.17, 7.20, 7.24, 7.27



Design Project has firm due date of Dec 4.
Exam 2 is Thursday Nov 13 in class.



Should be done before second exam; not turned in
Closed book, closed notes, except you can bring one
new note sheet as well as your first exam note sheet.
One short answer problem is based on a case study
article from the pertinent chapters (6, 7, 11).
After exam be reading Chapters 8 and 9.
1
Analysis of Unsymmetric 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.
System is only unbalanced at point of fault!
The easiest method to analyze unbalanced system
operation due to faults is through the use of
symmetrical components
2
Symmetric Components


The key idea of symmetrical component analysis is
to decompose the system into three sequence
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
3
Positive Sequence Sets


The positive sequence sets have three phase
currents/voltages with equal magnitude, with phase
b lagging phase a by 120°, and phase c lagging
phase b by 120°.
We’ve been studying positive sequence sets
Positive sequence
sets have zero
neutral current
4
Negative Sequence Sets


The negative sequence sets have three phase
currents/voltages with equal magnitude, with phase
b leading phase a by 120°, and phase c leading
phase b by 120°.
Negative sequence sets are similar to positive
sequence, except the phase order is reversed
Negative sequence
sets have zero
neutral current
5
Zero Sequence Sets


Zero sequence sets have three values with equal
magnitude and angle.
Zero sequence sets have neutral current
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
I a  I a0  I a  I a
I b  I b0  I b  I b
I c  I c0  I c  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 , I b , I c is the negative sequence set
7
Conversion from Sequence to Phase
Only three of the sequence values are unique,
I0a , I a , I a ; the others are determined as follows:
 1120
0
0
0
Ia  I b  Ic
I b   2 I a
 2  3  0
3  1
(since by definition they are all equal)
I c   I a
I b   I a
I c   2 I a
0

1
1
1
1
I
 
 a
1 
 Ia 
1
 I   I0 1  I +  2   I      1  2    I  


b
a
a
a
a






 



2
2







I
1

 c 
 
  1     I a 
 
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 
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
10
Symmetrical Component Example 1
 I a   100 
Let I   I b   10 
Then
  

 I c   10 
1   100    
1 1
1
1
2
I s  A I  1    10    100 

 

3
1  2    10   0 


 100 
 0 
If I  10   
Is   0 




10  
100
11
Symmetrical Component Example 2
Va   0 
Let V  Vb     
  

Vc     
Then
1   0   0 
1 1
1
1
2
Vs  A V  1         

 

3
1  2        6.12 


12
Symmetrical Component Example 3
 I 0   100 
  
Let I s   I   10


     

 I  
Then
1   100    
1 1


2
I  AI s  1 
  10    

 

1   2       


13
Use of Symmetrical Components
Consider the following wye-connected load:
I n  I a  Ib  I c
Vag  I a Z y  I n Z n
Vag  ( ZY  Z n ) I a  Z n I b  Z n I c
Vbg  Z n I a  ( ZY  Z n ) I b  Z n I c
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
14
Use of Symmetrical Components
Vag 
Z y  Zn
Zn
 

Z y  Zn
Vbg    Z n
V 
 Z
Zn
n
 cg 

V
 Z I V  A Vs
A Vs  Z A I s

 Z y  3Z n

1
A ZA  
0

0

  Ia 
 
Z n  Ib
 
Z y  Z n   I c 
I  A Is
Zn
Vs  A 1 Z A I s
0
Zy
0
0

0
Z y 
15
Networks are Now Decoupled
V 0 
 Z y  3Z n 0
 

0
Zy
V   
 

0
0
V 

Systems are decoupled
V 0  ( Z y  3Z n ) I 0

V  Zy I
0

I
0
 
0  I 
 

Z y  I 
 
V
 Zy I

16
Sequence diagrams for generators

Key point: generators only produce positive
sequence voltages; therefore only the positive
sequence has a voltage source
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
17
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 wye-wye
18
Transformer Sequence Diagrams
19
Unbalanced Fault Analysis

The first step in the analysis of unbalanced faults is
to assemble the three sequence networks. For
example, for the earlier single generator, single
motor example let’s develop the sequence networks
20
Sequence Diagrams for Example
Positive Sequence Network
Negative Sequence Network
21
Sequence Diagrams for Example
Zero Sequence Network
22
Create Thevenin Equivalents

To do further analysis we first need to calculate the
thevenin equivalents as seen from the fault location.
In this example the fault is at the terminal of the
right machine so the thevenin equivalents are:
Zth  j 0.2 in parallel with j0.455
Zth  j 0.21 in parallel with j0.475
23
Single Line-to-Ground (SLG) Faults


Unbalanced faults unbalance the network, but only
at the fault location. This causes a coupling of the
sequence networks. How the sequence networks
are coupled depends upon the fault type. We’ll
derive these relationships for several common
faults.
With a SLG fault only one phase has non-zero fault
current -- we’ll assume it is phase A.
24
SLG Faults, cont’d
 I af
 f
 Ib
 f
 I c
 ? 
  
  0 
 0 
  
Then since
 I 0f 
1  ? 
1 1
 
1
1 f

2 
0


 I f   1    0  I f  I f  I f  I a
 
3
3
 
2
1 
 0 


 I f 


25
SLG Faults, cont’d
f
Va
Vaf
 f
Vb
 f
Vc

f
Z f Ia

1 1


2

1




1 


0

V
1 f
 
  V f 
2 
  V f 
 
This means Vaf  V f0  V f  V f
The only way these two constraints can be satisified
is by coupling the sequence networks in series
26
SLG Faults, cont’d
With the
sequence
networks in
series we can
solve for the
fault currents
(assume Zf=0)
I f
1.050

  j1.964  I f  I 0f
j (0.1389  0.1456  0.25  3Z f )
I  A I s  I af   j 5.8 (of course, Ibf  I cf  0)
27
Line-to-Line (LL) Faults

The second most common fault is line-to-line,
which occurs when two of the conductors come in
contact with each other. With out loss of generality
we'll assume phases b and c.
Current Relationships: I af  0,
Voltage Relationships:
I bf   I cf ,
I 0f  0
Vbg  Vcg
28
LL Faults, cont'd
Using the current relationships we get
 I 0f 
1  0
1 1
 
1

2 f
 I f   1     I b
3
 
1  2     I f

 I f 

 b





I 0f  0
I f

1 f
 Ib    2
3

I f

1 f 2
 Ib   
3

Hence I f   I f
29
LL Faults, con'td
Using the voltage relationships we get
f 
V f0 

1  Vag
1 1
 
 


1
V f   1   2  Vbgf  
3
 
 
2


f
1


V f 

V

  cg 
Hence




1 f
2
f 

Vag     Vbg


3
1 f

Vf 
Vag   2   Vbgf 

3

Vf

V f  V f
30
LL Faults, cont'd
To satisfy

If


I f
&

Vf


Vf
the positive and negative sequence networks must
be connected in parallel
31
LL Faults, cont'd
Solving the network for the currents we get
I f
1.050

 3.691  90
j 0.1389  j 0.1456
 I af
 f
 Ib
 f
 I c

1 
1 1
0
  0 



2
   6.39 

1


3.691


90




 


1   2   3.69190   6.39 



32
LL Faults, cont'd
Solving the network for the voltages we get

Vf
 1.050  j 0.1389  3.691  90  0.5370
V f   j 0.1452  3.69190  0.5370
Vaf
 f
Vb
 f
Vc

1   0   1.074 
1 1



2
  0.537    0.537 
  1 

 


1   2  0.537   0.537 



33
Double Line-to-Ground Faults

With a double line-to-ground (DLG) fault two line
conductors come in contact both with each other
and ground. We'll assume these are phases b and c.
I af  0
Vbgf  Vcgf  Z f ( Ibf  I cf )
34
DLG Faults, cont'd
From the current relationships we get
 I af
 f
 Ib
 f
 I c

1 1


2

1




1 


Since I af  0

0

I
1 f
 
  I f 
2 
  I f 
 
I 0f  I f  I f  0
Note, because of the path to ground the zero
sequence current is no longer zero.
35
DLG Faults, cont'd
From the voltage relationships we get
f 
V f0 

V
1  ag
1 1
 
 


1
V f   1   2  Vbgf  
3
 
 
2


  Vbgf 
V f 
1 
 
V f  V f
Since Vbgf  Vcgf 
Then Vbgf  V f0  ( 2   )V f
But since 1      0 
2
    1
2
Vbgf  V f0  V f
36
DLG Faults, cont'd
 V f0  V f
Vbgf
 Z f ( I bf  I cf )
Also, since
f
Ib

0
If
2 
 I f

I f
I cf  I 0f   I f   2 I f
Adding these together (with    2  -1)
Vbgf
 Z f (2 I 0f  I f  I f )
with I 0f   I f  I f
V f0  V f  3Z f I 0f
37
DLG Faults, cont'd

The three sequence networks are joined as follows
Assuming Zf=0, then

V
1.050

If  


0
Z  Z ( Z  3Z f ) j 0.1389  j 0.092
 4.547  0
38
DLG Faults, cont'd
V f  1.05  4.547  90  j 0.1389  0.4184
I f  0.4184 / j 0.1456  j 2.874
I 0f   I f  I f  j 4.547  j 2.874  j1.673
Converting to phase: I bf  1.04  j 6.82
I cf  1.04  j 6.82
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Unbalanced Fault Summary



SLG: Sequence networks are connected in series,
parallel to three times the fault impedance
LL: Positive and negative sequence networks are
connected in parallel; zero sequence network is not
included since there is no path to ground
DLG: Positive, negative and zero sequence
networks are connected in parallel, with the zero
sequence network including three times the fault
impedance
40