Lecture 18 Fault Analysis Professor Tom Overbye Department of Electrical and Computer Engineering...
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Transcript of Lecture 18 Fault Analysis Professor Tom Overbye Department of Electrical and Computer Engineering...
Lecture 18 Fault Analysis
Professor Tom OverbyeDepartment of Electrical and
Computer Engineering
ECE 476
POWER SYSTEM ANALYSIS
2
Announcements
Homework 8 is 7.1, 7.17, 7.20, 7.24, 7.27 Should be done before second exam; not turned in
Be reading Chapter 7 Design Project is assigned today (see website for
details). Due date is Nov 20. Exam 2 is Thursday Nov 13 in class. You can
bring one new note sheet as well as your first exam note sheet.
3
4
Limiting Carbon Dioxide Emissions
• There is growing concern about the need to limit carbon dioxide emissions.
• The two main approaches are 1) a carbon tax, or 2) a cap-and-trade system (emissions trading)• The tax approach is straightforward – pay a fixed rate
based upon how the amount of CO2 is emitted. But there is a need to differentiate between carbon and CO2 (related by 12/44).
• A cap-and-trade system limits emissions by requiring permits (allowances) to emit CO2. The government sets the number of allowances, allocates them initially, and then private markets set their prices and allow trade.
5
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
6
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 line-to-ground (DLG) faults, and balanced three phase faults
7
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!)
8
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
9
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
10
RL Circuit Analysis
To understand fault analysis we need to review the behavior of an RL circuit
( )
2 cos( )
v t
V t
Before the switch is closed obviously i(t) = 0.When the switch is closed at t=0 the current willhave two components: 1) a steady-state value2) a transient value
11
RL Circuit Analysis, cont’d
ac
2 2 2 2
1. Steady-state current component (from standard
phasor analysis)
2 cos( )i ( )
where ( )
ac
V tt
Z
Z R L R X
VI
Z
12
RL Circuit Analysis, cont’d
dc 1
1
ac dc 1
1
2. Exponentially decaying dc current component
i ( )
where T is the time constant,
The value of is determined from the initial
conditions:
2(0) 0 i ( ) i ( ) cos( )
2
tT
tT
Z
t C e
LT RC
Vi t t t C e
Z
VC
Z
cos( ) which depends on Z
13
Time varying current
14
RL Circuit Analysis, cont’d
dc
1
Hence i(t) is a sinusoidal superimposed on a decaying
dc current. The magnitude of i (0) depends on when
the switch is closed. For fault analysis we're just
2concerned with the worst case:
( )
VC
Zi t
ac dci ( ) i ( )
2 2( ) cos( )
2(cos( ) )
tT
tT
t t
V Vi t t e
Z Z
Vt e
Z
15
RMS for Fault Current
2 2RMS
22 2
2The function i(t) (cos( ) ) is not periodic,
so we can't formally define an RMS value. However,
as an approximation define
I ( ) ( ) ( )
2
This function has a maximum va
tT
ac dc
tT
ac ac
Vt e
Z
t i t i t
I I e
lue of 3
Therefore the dc component is included simply by
multiplying the ac fault currents by 3
acI
16
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
'aE
17
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
18
Generator Modeling, cont’d
'
"
''
ac
" '
"d
For a balanced three-phase fault on the generator
terminal the ac fault current is (see page 245)
1 1 1
i ( ) 2 sin( )1 1
where
T direct-axis su
d
d
tT
d dda t
T
d d
eX XX
t E t
eX X
'd
btransient time constant ( 0.035sec)
T direct-axis transient time constant ( 1sec)
19
Generator Modeling, cont'd
'
"
''
ac
" '
'
DC "
A
The phasor current is then
1 1 1
1 1
The maximum DC offset is
2I ( )
where T is the armature time constant ( 0.2 seconds)
d
d
A
tT
d dda t
T
d d
tTa
d
eX XX
I E
eX X
Et e
X
20
Generator Short Circuit Currents
21
Generator Short Circuit Currents
22
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
" '
" '
A
0.15, 0.24, 1.1 (all per unit)
0.035 seconds, 2.0 seconds
T 0.2 seconds
d d d
d d
X X X
T T
23
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 10I 14,433 A (0) 101,000 A
3 20 10
I (0) 101 kA 2 143 k
t
t
t
e
I t
e
I
I
e
RMSA I (0) 175 kA
24
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
I (0.05) 143 kA 111 k A
I (0.05
e
I
e
I
e
2 2) 70.8 111 132 kA
25
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
26
Network Fault Example
For 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
27
Network Fault Example, cont'd
Faulted network per unit diagram
*'a
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 E 1.103 7.1
1.05
T G
Gen
V S
I
28
Network Fault Example, cont'd
f
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:
1I
j j
j j
.103 7.1 1.008 26.60.15 0.5
7.353 82.9 2.016 116.6 9.09
j j
j
29
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
30
Superposition Approach
Faulted Condition
Exact Equivalent to Faulted ConditionFault is representedby two equal andopposite voltage sources, each witha magnitude equalto the pre-fault voltage
31
Superposition Approach, cont’d
Since 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!
32
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)
I
0
gg g m m m
f f f f
I I I I I
I I I I
33
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 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
34
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
35
Determination of Fault Current
bus
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
36
Determination of Fault Current
(1)
f
ii
ij
(1) (2) (1)i
Hence
I
Where
Z driving point impedance
Z ( ) transfer point imepdance
Voltages during the fault are also found by superposition
V are prefault values
i
ii
i i i
VZ
i j
V V V