ECE 476
POWER SYSTEM ANALYSIS
Lecture 21
Symmetrical Components, Unbalanced Fault Analysis
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements





Be reading Chapters 9 and 10
HW 8 is due now.
HW 9 is 8.4, 8.12, 9.1,9.2 (bus 2), 9.14; due Nov 10 in class
Start working on Design Project. Tentatively due Nov 17 in
class
Second exam is on Nov 15 in class. Same format as first
exam, except you can bring two note sheets (e.g., the one
from the first exam and another)
1
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.
2
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 


3
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
4
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)
5
Example 9.3
6
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
7
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
8
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
9
LL Faults, cont'd
To satisfy

If


I f
&

Vf


Vf
the positive and negative sequence networks must
be connected in parallel
10
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 



11
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 



12
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 )
13
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.
14
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
15
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
16
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
17
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
18
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
19
Generalized System Solution


1.
2.
3.
4.
Assume we know the pre-fault voltages
The general procedure is then
Calculate Zbus for each sequence
For a fault at bus i, the Zii values are the thevenin
equivalent impedances; the pre-fault voltage is the
positive sequence thevenin voltage
Connect and solve the thevenin equivalent sequence
networks to determine the fault current
Sequence voltages throughout the system are
20
Generalized System Solution, cont’d
Sequence voltages throughout the system are given
by
 0 


This is solved


 0 
for each


sequence
V  V prefault  Z   I f 
network!
 0 




 0 
4.
5.
Phase values are determined from the sequence values
21
Unbalanced System Example
For the generators assume Z+ = Z = j0.2; Z0 = j0.05
For the transformers assume Z+ = Z =Z0 = j0.05
For the lines assume Z+ = Z = j0.1; Z0 = j0.3
Assume unloaded pre-fault, with voltages =1.0 p.u.
22
Positive/Negative Sequence Network
24 10 10 
0.1397 0.1103 0.125 


Ybus
 j  10 24 10  Zbus
 j 0.1103 0.1397 0.125 




 10 10 20 
0.1250 0.1250 0.175 
Negative sequence is identical to positive sequence
23
Zero Sequence Network
16.66 3.33 3.33 
0
0
Ybus
 j  3.33 26.66 3.33  Zbus



3.33 6.66 
 3.33
0.0732 0.0148 0.0440 
j 0.0148 0.0435 0.0.292 


0.0440 0.0292 0.1866 
24
For a SLG Fault at Bus 3
The sequence networks are created using the pre-fault
voltage for the positive sequence thevenin voltage,
and the Zbus diagonals for the thevenin impedances
Positive Seq.
Negative Seq.
Zero Seq.
The fault type then determines how the networks are
interconnected
25
Bus 3 SLG Fault, cont’d
I f

If
V
V
1.00

  j1.863
j (0.1750  0.1750  0.1866)


If

0
If
  j1.863
1.00
 0   0.7671
 
 1.00  Zbus
0    0.7671



 

1.00
 j1.863 0.6740 
 0   0.2329 
 
 Zbus
0    0.2329 

 

 j1.863  0.3260 
26
Bus 3 SLG Fault, cont’d
 0   0.0820 
0 
V 0  Zbus
0    0.0544 

 

 j1.863  0.3479 
We can then calculate the phase voltages at any bus
0
 0.3479 


V3  A   0.6740    0.522  j 0.866 




 0.3260 
 0.522  j 0.866 
0.4522
 0.0820 


V1  A   0.7671    0.3491  j 0.866 




 0.2329 
 0.3491  j 0.866 
27
Faults on Lines


The previous analysis has assumed that the fault is
at a bus. Most faults occur on transmission lines,
not at the buses
For analysis these faults are treated by including a
dummy bus at the fault location. How the
impedance of the transmission line is then split
depends upon the fault location
28
Line Fault Example
Assume a SLG fault occurs on the previous system
on the line from bus 1 to bus 3, one third of the way
from bus 1 to bus 3. To solve the system we add a
dummy bus, bus 4, at the fault location
29
Line Fault Example, cont’d
0
30 
 44 10
The Ybus
 10 24 10

0

now has

Ybus
 j
10 25 15 
4 buses
 0
 30

0
15 45

Adding the dummy bus only changes the new
row/column entries associated with the dummy bus

Zbus
0.1397
 0.1103
 j
 0.1250
 0.1348

0.1103 0.1250 0.1348 
0.1397 0.1250 0.1152 

0.1250 0.1750 0.1417 
0.1152 0.1417 0.1593 
30
Power System Protection


Main idea is to remove faults as quickly as possible
while leaving as much of the system intact as
possible
Fault sequence of events
1.
2.
3.
4.
Fault occurs somewhere on the system, changing the
system currents and voltages
Current transformers (CTs) and potential transformers
(PTs) sensors detect the change in currents/voltages
Relays use sensor input to determine whether a fault has
occurred
If fault occurs relays open circuit breakers to isolate fault
31
Power System Protection


1.
2.
Protection systems must be designed with both
primary protection and backup protection in case
primary protection devices fail
In designing power system protection systems
there are two main types of systems that need to be
considered:
Radial: there is a single source of power, so power
always flows in a single direction; this is the
easiest from a protection point of view
Network: power can flow in either direction:
protection is much more involved
32
Radial Power System Protection

Radial systems are primarily used in the lower
voltage distribution systems. Protection actions
usually result in loss of customer load, but the
outages are usually quite local.
The figure shows
potential protection
schemes for a
radial system. The
bottom scheme is
preferred since it
results in less lost load
33
Radial Power System Protection



In radial power systems the amount of fault current is
limited by the fault distance from the power source:
faults further done the feeder have less fault current
since the current is limited by feeder impedance
Radial power system protection systems usually use
inverse-time overcurrent relays.
Coordination of relay current settings is needed to
open the correct breakers
34
Inverse Time Overcurrent Relays


Inverse time overcurrent relays respond instantaneously to a current above their maximum setting
They respond slower to currents below this value
but above the pickup current value
35