EE 369 POWER SYSTEM ANALYSIS Lecture 18 Fault Analysis Tom Overbye and Ross Baldick 1.
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Transcript of EE 369 POWER SYSTEM ANALYSIS Lecture 18 Fault Analysis Tom Overbye and Ross Baldick 1.
EE 369POWER SYSTEM ANALYSIS
Lecture 18 Fault Analysis
Tom Overbye and Ross Baldick
1
AnnouncementsRead Chapter 7. Homework 12 is 6.43, 6.48, 6.59, 6.61,
12.19, 12.22, 12.20, 12.24, 12.26, 12.28, 12.29; due Tuesday Nov. 25.
Homework 13 is 12.21, 12.25, 12.27, 7.1, 7.3, 7.4, 7.5, 7.6, 7.9, 7.12, 7.16; due Thursday, December 4.
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 TypesThere 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 Sequence1. 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’d3. 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 AnalysisFault 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 AnalysisTo 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 willhave two components: 1) a steady-state value2) 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
VI
Z
Z R L R X
L R
ac
stantaneous current is:
2 cos( )( ) ZV t
i tZ
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
tT
dc
tT
Z
i t C e
LT T RC
Vi i t i t t C e
Z
VC
Z
cos( ) which depends on Z 10
Time varying currenti(t)
time
Superposition of steady-state component andexponentially 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 12
urrent occurs for: = ,
( ) ( ) ( )
2 2( ) cos( )
2( cos( ) )
ac dc
tT
tT
VC
Zi t i t i t
V Vi t t e
Z Z
Vt e
Z
12
RMS for Fault CurrentThe interrupting capability of a circuit breaker is
specified in terms of the RMS current it can interrupt.
2The function ( ) ( cos( ) ) is
not periodic, so we can't formally define an RMS value
tTV
i t t eZ
.
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 Current2 2
RMS
22 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 tT T
ac dc ac
tT
ac ac
ac
I I
V VI I e I e
Z Z
I I e
I
ltiplying the ac fault currents by 3.14
Generator Modeling During FaultsDuring 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:
'aE
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
tT
d dda t
T
d d
d
eX XX
i t E t
eX X
T
'
btransient time constant ( 0.035sec)
direct-axis transient time constant ( 1sec)dT
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
tT
d dda t
T
d d
tTa
d
A
eX XX
I E
eX X
EI t e
X
T
18
Generator Short Circuit Currents
19
Generator Short Circuit Currents
20
Generator Short Circuit ExampleA 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
ac0.035
ac
6
base ac3
0.2DC
Substituting in the values
1 1 11.1 0.24 1.1
( ) 1.051 1
0.15 0.24
1.05(0) 7 p.u.0.15
500 1014,433 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
RMSA (0) 175 kAI 22
Generator S.C. Example, cont'd
0.052.0
ac 0.050.035
ac
0.050.2
DC
RMS
Evaluating at t = 0.05 seconds for breaker opening
1 1 11.1 0.24 1.1
(0.05) 1.051 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
reactance2. Transformers are represented by their leakage reactances3. Synchronous machines are modeled as a constant voltage
behind direct-axis subtransient reactance4. Induction motors are ignored or treated as synchronous
machines5. Other (nonspinning) loads are ignored
24
Network Fault ExampleFor the following network assume a fault on the terminal of the generator; all data is per unitexcept for the transmission line reactance
2
19.5Convert to per unit: 0.1 per unit
138100
lineX
generator has 1.05terminal voltage &supplies 100 MVAwith 0.95 lag pf
25
Network Fault Example, cont'dFaulted 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 18.20.952 18.2 1.103 7.1
1.05
T G
Gen a
V S
I E
26
Network Fault Example, cont'dThe 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:
1f
j j
j j
I
.103 7.1 1.008 26.60.15 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 ApproachFaulted Condition
Exact Equivalent to Faulted ConditionFault is representedby two equal andopposite voltage sources, each witha magnitude equalto the pre-fault voltage
29
Superposition Approach, cont’dSince this is now a linear network, the faulted voltagesand currents are just the sum of the pre-fault conditions[the (1) component] and the conditions with just a singlevoltage 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 sourceat the fault location, with a magnitude equal to thenegative of the pre-fault voltage at the fault location.
(1) (2) (1) (2)
(1) (2) (2)0
gg g m m m
f f f f
I I I I I I
I I I I
31
Two Bus Superposition Solutionf
(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 07
j0.15 j0.15
E 1.05 02.1
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 82.9j
This matcheswhat wecalculatedearlier
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
fI
I
However to use thisapproach we need tofirst determine If
33
Determination of Fault Currentbus
1bus 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
if
ii
ii
ij
i i i i
VI
Z
Z
Z i j
V V V V
35