EE 369
POWER SYSTEM ANALYSIS
Lecture 18
Fault Analysis
Tom Overbye and Ross Baldick
1

Announcements
Read Chapter 7.
Homework 12 is 6.59, 6.61, 12.19, 12.22,
12.24, 12.26, 12.28, 7.1, 7.3, 7.4, 7.5, 7.6,
7.9, 7.12, 7.16; due Thursday, 12/3.
2

Transmission Fault Analysis
The cause of electric power system faults is
insulation breakdown/compromise.
This breakdown can be due to a variety of
different factors:
– Lightning ionizing air,
– Wires blowing together in the wind,
– Animals or plants coming in contact with the
wires,
– Salt spray or pollution on insulators.
3

Transmission Fault Types
There are two main types of faults:
– symmetric faults: system remains balanced;
these faults are relatively rare, but are the
easiest to analyze so we’ll consider them first.
– unsymmetric faults: system is no longer
balanced; very common, but more difficult to
analyze (considered in EE 368L).
The most common type of fault on a three
phase system by far is the single line-to-
ground (SLG), followed by the line-to-line
faults (LL), double line-to-ground (DLG)
faults, and balanced three phase faults. 4

Lightning Strike Event Sequence
1. Lighting hits line, setting up an ionized path to
ground
 30 million lightning strikes per year in US!
 a single typical stroke might have 25,000 amps,
with a rise time of 10 s, dissipated in 200 s.
 multiple strokes can occur in a single flash, causing
the lightning to appear to flicker, with the total
event lasting up to a second.
2. Conduction path is maintained by ionized air
after lightning stroke energy has dissipated,
resulting in high fault currents (often > 25,000
amps!) 5

Lightning Strike Sequence, cont’d
3. Within one to two cycles (16 ms) relays at both
ends of line detect high currents, signaling
circuit breakers to open the line:
 nearby locations see decreased voltages
4. Circuit breakers open to de-energize line in an
additional one to two cycles:
 breaking tens of thousands of amps of fault current
is no small feat!
 with line removed voltages usually return to near
normal.
5. Circuit breakers may reclose after several
seconds, trying to restore faulted line to service.
6

Fault Analysis
Fault currents cause equipment damage due
to both thermal and mechanical processes.
Goal of fault analysis is to determine the
magnitudes of the currents present during
the fault:
– need to determine the maximum current to
ensure devices can survive the fault,
– need to determine the maximum current the
circuit breakers (CBs) need to interrupt to
correctly size the CBs.
7

RL Circuit Analysis
To understand fault analysis we need to
review the behavior of an RL circuit
( )
2 cos( )
v t
V t
 


(Note text uses sinusoidal voltage instead of cos!)
Before the switch is closed, i(t) = 0.
When the switch is closed at t=0 the current will
have two components: 1) a steady-state value
2) a transient value.
R L
8

RL Circuit Analysis, cont’d
2 2 2 2
1
1. Steady-state current component (from standard
phasor analysis)
Steady-state phasor current magnitude is ,
where ( )
and current phasor angle is , tan ( / )
Corresponding in
ac
Z Z
V
I
Z
Z R L R X
L R

  


   
 
ac
stantaneous current is:
2 cos( )
( ) Z
V t
i t
Z
  
 

9

RL Circuit Analysis, cont’d
1
1
ac dc 1
1
2. Exponentially decaying dc current component
( )
where is the time constant,
The value of is determined from the initial
conditions:
2
(0) 0 ( ) ( ) cos( )
2
t
T
dc
t
T
Z
i t C e
L
T T
R
C
V
i i t i t t C e
Z
V
C
Z
  




      
  cos( ) which depends on
Z
  

10

Time varying current
i(t)
time
Superposition of steady-state component and
exponentially decaying dc offset.
11

RL Circuit Analysis, cont’d
dc
Hence ( ) is a sinusoidal superimposed on a decaying
dc current. The magnitude of (0) depends on when
the switch is closed. For fault analysis we're just
concerned with the worst case.
Highest DC c
i t
i
Z 1
2
urrent occurs for: = ,
( ) ( ) ( )
2 2
( ) cos( )
2
( cos( ) )
ac dc
t
T
t
T
V
C
Z
i t i t i t
V V
i t t e
Z Z
V
t e
Z
  




 
 
  
   12

RMS for Fault Current
The interrupting capability of a circuit breaker is
specified in terms of the RMS current it can interrupt.
2
The function ( ) ( cos( ) ) is
not periodic, so we can't formally define an RMS value
t
T
V
i t t e
Z


  
.
However, if then we can approximate the current
as a sinusoid plus a time-invarying dc offset.
The RMS value of such a current is equal to the
square root of the sum of the squares of the
indivi
T t
dual RMS values of the two current components.
13

RMS for Fault Current
2 2
RMS
2
2 2
I ,
2
where , 2 ,
2
This function has a maximum value of 3 .
Therefore the worst case effect of the dc
component is included simply by
mu
ac dc
t t
T T
ac dc ac
t
T
ac ac
ac
I I
V V
I I e I e
Z Z
I I e
I
 

 
  
 
ltiplying the ac fault currents by 3.
14

Generator Modeling During Faults
During a fault the only devices that can
contribute fault current are those with energy
storage.
Thus the models of generators (and other
rotating machines) are very important since they
contribute the bulk of the fault current.
Generators can be approximated as a constant
voltage behind a time-varying reactance:
'
a
E
15

Generator Modeling, cont’d
"
d
'
d
d
The time varying reactance is typically approximated
using three different values, each valid for a different
time period:
X direct-axis subtransient reactance
X direct-axis transient reactance
X dire


 ct-axis synchronous reactance
Can then estimate currents using circuit theory:
For example, could calculate steady-state current
that would occur after a three-phase short-circuit
if no circuit breakers interrupt current. 16

Generator Modeling, cont’d
'
"
'
'
ac
" '
"
For a balanced three-phase fault on the generator
terminal the ac fault current is (see page 362)
1 1 1
( ) 2 sin( )
1 1
where
direct-axis su
d
d
t
T
d d
d
a t
T
d d
d
e
X X
X
i t E t
e
X X
T
 


 
 
  
 
 
 
 
 
 
 
 

 
 
 
 

'
btransient time constant ( 0.035sec)
direct-axis transient time constant ( 1sec)
d
T

  17

Generator Modeling, cont'd
'
"
'
'
ac
" '
'
DC "
The phasor current is then
1 1 1
1 1
The maximum DC offset is
2
( )
where is the armature time constant ( 0.2 seconds)
d
d
A
t
T
d d
d
a t
T
d d
t
T
a
d
A
e
X X
X
I E
e
X X
E
I t e
X
T



 
 
  
 
 
 
 
  
 
 

 
 
 
 


18

Generator Short Circuit Currents
19

Generator Short Circuit Currents
20

Generator Short Circuit Example
A 500 MVA, 20 kV, 3 is operated with an
internal voltage of 1.05 pu. Assume a solid
3 fault occurs on the generator's terminal
and that the circuit breaker operates after
three cycles. Determine the fault current.
Assume
" '
" '
0.15, 0.24, 1.1 (all per unit)
0.035 seconds, 2.0 seconds
0.2 seconds
d d d
d d
A
X X X
T T
T
  
 

21

Generator S.C. Example, cont'd
2.0
ac
0.035
ac
6
base ac
3
0.2
DC
Substituting in the values
1 1 1
1.1 0.24 1.1
( ) 1.05
1 1
0.15 0.24
1.05
(0) 7 p.u.
0.15
500 10
14,433 A (0) 101,000 A
3 20 10
(0) 101 kA 2 143 k
t
t
t
e
I t
e
I
I I
I e


 
 
  
 
 
 
  
 
 

 
 
 
 
 

  

   RMS
A (0) 175 kA
I 
22

Generator S.C. Example, cont'd
0.05
2.0
ac 0.05
0.035
ac
0.05
0.2
DC
RMS
Evaluating at t = 0.05 seconds for breaker opening
1 1 1
1.1 0.24 1.1
(0.05) 1.05
1 1
0.15 0.24
(0.05) 70.8 kA
(0.05) 143 kA 111 k A
(0.05
e
I
e
I
I e
I



 
 
  
 
 
 
  
 
 

 
 
 
 

  
2 2
) 70.8 111 132 kA
  
23

Network Fault Analysis Simplifications
 To simplify analysis of fault currents in
networks we'll make several simplifications:
1. Transmission lines are represented by their series
reactance
2. Transformers are represented by their leakage
reactances
3. Synchronous machines are modeled as a
constant voltage behind direct-axis subtransient
reactance
4. Induction motors are ignored or treated as
synchronous machines
5. Other (nonspinning) loads are ignored 24

Network Fault Example
For the following network assume a fault on the
terminal of the generator; all data is per unit
except for the transmission line reactance
2
19.5
Convert to per unit: 0.1 per unit
138
100
line
X  
generator has 1.05
terminal voltage &
supplies 100 MVA
with 0.95 lag pf
25

Network Fault Example, cont'd
Faulted network per unit diagram
*
'
To determine the fault current we need to first estimate
the internal voltages for the generator and motor
For the generator 1.05, 1.0 18.2
1.0 18.2
0.952 18.2 1.103 7.1
1.05
T G
Gen a
V S
I E
   

 
       
 
  26

Network Fault Example, cont'd
The motor's terminal voltage is then
1.05 0-(0.9044 - 0.2973) 0.3 1.00 15.8
The motor's internal voltage is
1.00 15.8 (0.9044 - 0.2973) 0.2
1.008 26.6
We can then solve as a linear circuit:
1
f
j j
j j
I
     
    
   

.103 7.1 1.008 26.6
0.15 0.5
7.353 82.9 2.016 116.6 9.09
j j
j
    

        
27

Fault Analysis Solution Techniques
 Circuit models used during the fault allow the
network to be represented as a linear circuit
 There are two main methods for solving for fault
currents:
1. Direct method: Use prefault conditions to solve for
the internal machine voltages; then apply fault and
solve directly.
2. Superposition: Fault is represented by two opposing
voltage sources; solve system by superposition:
– first voltage just represents the prefault operating point
– second system only has a single voltage source. 28

Superposition Approach
Faulted Condition
Exact Equivalent to Faulted Condition
Fault is represented
by two equal and
opposite voltage
sources, each with
a magnitude equal
to the pre-fault voltage
29

Superposition Approach, cont’d
Since this is now a linear network, the faulted voltages
and currents are just the sum of the pre-fault conditions
[the (1) component] and the conditions with just a single
voltage source at the fault location [the (2) component]
Pre-fault (1) component equal to the pre-fault
power flow solution
Obvious the
pre-fault
“fault current”
is zero!
30

Superposition Approach, cont’d
Fault (1) component due to a single voltage source
at the fault location, with a magnitude equal to the
negative of the pre-fault voltage at the fault location.
(1) (2) (1) (2)
(1) (2) (2)
0
g
g g m m m
f f f f
I I I I I I
I I I I
   
   
31

Two Bus Superposition Solution
f
(1) (1)
(2) f
(2) f
(2)
Before the fault we had E 1.05 0 ,
0.952 18.2 and 0.952 18.2
Solving for the (2) network we get
E 1.05 0
7
j0.15 j0.15
E 1.05 0
2.1
j0.5 j0.5
7 2.1 9.1
0.952
g m
g
m
f
g
I I
I j
I j
I j j j
I
  
        
 
   
 
   
    
  18.2 7 7.35 82.9
j
      
This matches
what we
calculated
earlier
32

Extension to Larger Systems
bus
bus
The superposition approach can be easily extended
to larger systems. Using the we have
For the second (2) system there is only one voltage
source so is all zeros except at the fault loca

Y
Y V I
I tion
0
0
f
I
 
 
 

 

 
 
 
 
I
However to use this
approach we need to
first determine If
33

Determination of Fault Current
bus
1
bus bus
(2)
1
(2)
11 1 2
(2)
1 1
(2)
(1)
f
Define the bus impedance matrix as
0
Then
0
For a fault a bus i we get -I
bus
n
f
n nn n
n
ii f i
V
Z Z V
I
Z Z V
V
Z V V



 
 
 
 
   
 
   

  
   
 
   
   
 
 
   
 
   
Z
Z Y V Z I
34

Determination of Fault Current
(1)
(1) (2) (1)
Hence
Where
driving point impedance
( ) transfer point imepdance
Voltages during the fault are also found by superposition
are prefault values
i
f
ii
ii
ij
i i i i
V
I
Z
Z
Z i j
V V V V


 
35