ECE4762007_Lect19

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Transcript ECE4762007_Lect19

ECE 476
POWER SYSTEM ANALYSIS
Lecture 19
Fault Analysis
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements

Homework 9 is 7.1, 7.17, 7.20, 7.24, 7.27
–


you do not need to turn it in, but should do it before the
exam
Second exam is Tuesday Nov 13 in class
Design Project 2 from the book (page 345 to 348)
was due on Nov 15, but I have given you an
extension to Nov 29. The Nov 29 date is firm!
1
Fault Analysis


The cause of electric power system faults is
insulation breakdown
This breakdown can be due to a variety of different
factors
–
–
–
–
lightning
wires blowing together in the wind
animals or plants coming in contact with the wires
salt spray or pollution on insulators
2
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
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 lineto-ground (DLG) faults, and balanced three phase
faults
3
Lightning Strike Event Sequence
1.
Lighting hits line, setting up an ionized path to
ground



2.
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.
Conduction path is maintained by ionized air after
lightning stroke energy has dissipated, resulting in
high fault currents (often > 25,000 amps!)
4
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

4.
Circuit breakers open to de-energize line in an
additional one to two cycles


5.
nearby locations see decreased voltages
breaking tens of thousands of amps of fault current is no
small feat!
with line removed voltages usually return to near normal
Circuit breakers may reclose after several seconds,
trying to restore faulted line to service
5
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 insure devices
can survive the fault
need to determine the maximum current the circuit
breakers (CBs) need to interrupt to correctly size the CBs
6
RL Circuit Analysis

To understand fault analysis we need to review the
behavior of an RL circuit
v(t ) 
2 V cos( t   )
Before the switch is closed obviously 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
7
RL Circuit Analysis, cont’d
1. Steady-state current component (from standard
phasor analysis)
iac (t ) 
where Z 
I ac
2V cos(t   )
Z
R 2  ( L)2  R 2  X 2
V

Z
8
RL Circuit Analysis, cont’d
2. Exponentially decaying dc current component
i dc (t )  C1e
t
T
where T is the time constant, T  L R
The value of C1 is determined from the initial
conditions:
t
2V
i (0)  0  i ac (t )  i dc (t ) 
cos(t     Z )  C1e T
Z
2V
C1  
cos(   Z ) which depends on 
Z
9
Time varying current
10
RL Circuit Analysis, cont’d
Hence i(t) is a sinusoidal superimposed on a decaying
dc current. The magnitude of i dc (0) depends on when
the switch is closed. For fault analysis we're just
2V
concerned with the worst case: C1 
Z
i (t )  i ac (t )  i dc (t )
i (t )


2V
2V t T
cos(t ) 
e
Z
Z
t
2V
(cos(t )  e T )
Z
11
RMS for Fault Current
t
2V
The function i(t) 
(cos(t )  e T ) is not periodic,
Z
so we can't formally define an RMS value. However,
as an approximation define
I RMS (t ) 

2
2
iac
(t )  idc
(t )
2
I ac
2t
2  T
 2 I ac e
This function has a maximum value of 3 I ac
Therefore the dc component is included simply by
multiplying the ac fault currents by 3
12
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
'
Ea
13
Generator Modeling, cont’d
The time varying reactance is typically approximated
using three different values, each valid for a different
time period:
X"d
 direct-axis subtransient reactance
X 'd
 direct-axis transient reactance
Xd
 direct-axis synchronous reactance
14
Generator Modeling, cont’d
For a balanced three-phase fault on the generator
terminal the ac fault current is (see page 245)
iac (t ) 
 1  1
1  
 ' 

e
X
 Xd Xd 
'  d
2Ea 
 t "
1 
 1
Td

e
 X " X ' 
d 
 d
t
Td'



 sin( t   )



where
Td"  direct-axis subtransient time constant (  0.035sec)
'
Td
 direct-axis transient time constant (  1sec)
15
Generator Modeling, cont'd
The phasor current is then
 1  1
1  
 ' 

e
X
 Xd Xd 
'  d
I ac  Ea 
 t "
1 
 1
Td
 X "  X '  e
d 
 d
t
Td'







The maximum DC offset is
2 Ea' 
I DC (t ) 
e
"
Xd
t
TA
where TA is the armature time constant (  0.2 seconds)
16
Generator Short Circuit Currents
17
Generator Short Circuit Currents
18
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
X d"  0.15, X d'  0.24, X d  1.1 (all per unit)
Td"  0.035 seconds, Td'
 2.0 seconds
TA  0.2 seconds
19
Generator S.C. Example, cont'd
Substituting in the values
1   t 2.0 
1  1


e



1.1  0.24 1.1 

I ac (t )  1.05 

 1  1  e  t 0.035

 0.15 0.24 

I ac (0)  1.05
 7 p.u.
0.15
I base

500  106
 14,433 A I ac (0)  101,000 A
3
3 20  10
I DC (0)  101 kA  2 e
t
0.2
 143 k A I RMS (0)  175 kA
20
Generator S.C. Example, cont'd
Evaluating at t = 0.05 seconds for breaker opening
1   0.05 2.0 
1  1


e



1.1  0.24 1.1 

I ac (0.05)  1.05 

 1  1  e 0.05 0.035

 0.15 0.24 

I ac (0.05)  70.8 kA
I DC (0.05)  143  e
0.05
0.2
kA  111 k A
I RMS (0.05)  70.82  1112  132 kA
21
Network Fault Analysis Simplifications

To simplify analysis of fault currents in networks
we'll make several simplifications:
1.
2.
3.
4.
5.
Transmission lines are represented by their series
reactance
Transformers are represented by their leakage
reactances
Synchronous machines are modeled as a constant
voltage behind direct-axis subtransient reactance
Induction motors are ignored or treated as synchronous
machines
Other (nonspinning) loads are ignored
22
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
generator has 1.05
terminal voltage &
supplies 100 MVA
with 0.95 lag pf
Convert to per unit: X line
19.5

 0.1 per unit
2
138
100
23
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 VT  1.05, SG  1.018.2
*
I Gen
1.018.2 


  0.952  18.2
 1.05 
E 'a  1.1037.1
24
Network Fault Example, cont'd
The motor's terminal voltage is then
1.050 - (0.9044 - j 0.2973)  j 0.3  1.00  15.8
The motor's internal voltage is
1.00  15.8  (0.9044 - j 0.2973)  j 0.2
 1.008  26.6
We can then solve as a linear circuit:
1.1037.1 1.008  26.6
If 

j 0.15
j 0.5
 7.353  82.9  2.016  116.6  j9.09
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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.
2.
Direct method: Use prefault conditions to solve for the
internal machine voltages; then apply fault and solve
directly
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
26
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
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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!
28
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.
I g  I (1)  I g(2)
g
Im  I m(1)  I m(2)
(2)
(2)
I f  I (1)

I

0

I
f
f
f
29
Two Bus Superposition Solution
Before the fault we had E f  1.050,
I (1)  0.952  18.2 and I m(1)  0.952  18.2
g
Solving for the (2) network we get
Ef
1.050
(2)
Ig


  j7
j0.15
j0.15
E f 1.050
(2)
Im


  j 2.1
j0.5
j0.5
I (2)
f
  j 7  j 2.1   j 9.1
I g  0.952  18.2  j 7  7.35  82.9
This matches
what we
calculated
earlier
30
Extension to Larger Systems
The superposition approach can be easily extended
to larger systems. Using the Ybus we have
Ybus V  I
For the second (2) system there is only one voltage
source so I is all zeros except at the fault location


 0 


I   I f 


 0 


However to use this
approach we need to
first determine If
31
Determination of Fault Current
Define the bus impedance matrix Z bus as
Z bus
 Z11
Then 

 Z n1
1
Ybus
V  Z busI
(2) 

V


1
 (2) 


Z1n  0
V2 



  I   

 f  

Z nn   0  V (2) 
n 1

  (2) 
Vn 
For a fault a bus i we get -If Zii  V f  Vi(1)
32
Determination of Fault Current
Hence
Vi(1)
If 
Zii
Where
Zii
driving point impedance
Zij (i  j )
transfer point imepdance
Voltages during the fault are also found by superposition
Vi  Vi(1)  Vi(2)
Vi(1) are prefault values
33