Download - EE493 543 Lecture 1 Sym Fault
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
1
Overview of Fault
Fault means current flowing through a path it isnot intended to flow through.
Temporary faults are caused by lighting, birds,animals.
Permanent faults are caused by insulationfailure, birds, animals.
If a temporary fault occurs, the goal of theprotection scheme is to restore the service afterthe fault clears by itself.
If a permanent fault occurs, the goal of theprotection scheme is to isolate the fault fromthe system and keep the outage area to aminimum.
Sustained fault can result in equipment damageand instability.
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Overview of Fault
Open Circuit Faults: Broken conductor
Short Circuit Faults: Balanced faults:
Three-phase fault (5%, least frequent)
Unbalanced faults:
Single line to ground faults (SLG-70~80%- most frequent)
Line to line faults (LL)
Double line to ground faults (LLG) We shall study short circuit faults in
this course
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Overview of Fault
Depending on the fault location inthe system, either a three-phasefault or a single phase fault canbe the most severe.
There-phase fault is the easiesttype of fault to be analyzed,because it is a balanced fault.
To analyze unbalanced faults, we
need to use symmetricalcomponents
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Transients in R-L Circuit
The first part of the solution is a sinusoid, the second part is
an exponentially decaying DC component. The amount of DC component depends on the switchinginstant, and the rate of decay depends the ratio R/L.
Next three slides show the solution plotted for VRMS= 230 kV,XL=160 (L = 0.4244 H), R = 32 .
For this data, |Vmax|/|Z| = 2.
)sin(max tV~
R
Li
dt
diLRitV )sin(max
)1()]sin()[sin(||
max
T
t
etZ
Vi
andLRZ 22 )(|| sec),/(tan 1
R
LTRL
Solution
Equation
Where,
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Transients in R-L Circuit
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-4
-3
-2
-1
0
1
2
Time (sec)
current(kA)
Current profile for switching RL circuit at different instants
alpha=theta
alpha=theta+pi/4
alpha=theta+pi/2
Observe:When =,dc offset iszero (Noasymmetry)
When-=/2,asymmetryis maximum.
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Amount of dc Offset
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-4
-3
-2
-1
0
1
2
Time (sec)
current(kA)
Current break-up for switching RL circuit at alpha=theta+pi/2
DC Part
Resultant
AC Part
Observe:When -=/2, at t=0, themagnitude of dc offset equalsthe peak of symmetrical accurrent, making the totalinstantaneous current equal tozero - This is dictated by thephysics of the circuit.
However, the current rises fastand the peak goes to morethan 150% of the peak ofsymmetrical current in the
case shown here.
This peak will depend on the rate of decay of dc current. Slower decaymeans the peak would be higher and vice versa (see next slide).
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Transients in R-L Circuit
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-4
-3
-2
-1
0
1
2
Time (sec)
current(kA)
Effect of R on the decay of DC component
AC Com ponen t
R=52 Ohm
R=42 Ohm
R=32 Ohm
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Math shows what figure shows Recall:
If -=/2, then
])2/[sin(||
max LRt
etZ
Vi
At t=0,
Observe: the dc offset is worst - it equals the peak of thesymmetrical current. But the total current at t=0 is zero.
0||||
maxmax Z
V
Z
Vi
If -=0, then
There is no dc offset. Calculate the current (dc, ac, and total) when -=/2, at
t=8 ms.
)][sin(||
max tZ
Vi
)1()]sin()[sin(||
max
T
t
etZ
Vi
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Analyzing Asymmetrical Waveform
At any given moment,
Where Iac is the RMS value of the symmetrical part of the current. We saw that in the worst case scenario, maximum value of Idc
occurs at t=0, and it equals the peak of the symmetrical part of accurrent, i.e., 2Iac.
Also, Idc decays according to a time constant T = L/R (see equation-1, slide#4), or T = X/2fR
Therefore, for the worst case scenario, idc(t)= 2Iace-t/T. This
means:
currentassymetrictheofvalueRMStheis)(where
)()(
)()()(
22
tI
tiItI
OR
tititi
RMS
dcacRMS
dcac
cyclesintimeftwhereeItI
eIeIItI
RX
acRMS
T
t
acT
t
acacRMS
,21)(
212)(
)/(
4
22
2
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Analyzing Asymmetrical Waveform
ac
RX
acRMS IKeItI )(21)( )/(
4
From the previous page,
K() is called the asymmetry factor. The maximum value of K() is at t=0, when the value is 3
Look at Ex# 7.1, page# 360-361 (4th ed.),page#324,325 (3rd ed.)
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Modeling Armature Reaction in Synchronous Generator
2sin
11111)(
'''
''''max
t
Xe
XXe
XXEti
d
T
t
dd
T
t
dd
acdd
The phenomena is mathematically modeled by the followingequation:
Where Td< Td< Td. Let us see how this equation models thephenomena.
At the time of the short circuit, t=0, hence
''''max ")0(,,
2sin)0(
d
RMSRMS
d
acX
EIIort
X
Ei
After a certain time t, when t>>Td, but t
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Modeling a Fault
This circuit models the pre-fault and fault conditions by
two sources of value VPF, the pre-fault voltage at P. When the switch is open, the circuit models pre-fault orload condition, and when close, it models the faultcondition.
XdG
G TL M
XdMXTL
IL
P Suppose a 3-ph fault occurs at P inthis system.
+
-
+
-
P
Neutral
jXdG jXdM
jXTL
EGEM
VPF
VPF
+
-
-
+
IF
IL IL
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Modeling a Fault
fault+
- -
+
P
Neutral
jXdG jXdM
jXTL
EGEM
IMIG
IF
+
-
+
-
P
Neutral
jXdG jXdM
jXTL
EGEM
VPF
VPF
+
-
-
+
IF
Pre-fault
Switchopen
Switchclose
Thus, fault issimulated by thesummation of thesetwo circuits.For small systems,the pre-fault circuitcan be solved withcircuit analysistechniques, but forlarge power systems,it is solved usingload-flow.Fault circuit can besolved usingThevenins theoremfor small systems andusing ZBUS for largesystems.
+
-
+
-
+
-
P
Neutral
VPFjXdG jXdM
EM
ILIL jXTL
EG
IFG=IG+ ILIFM=IM- IL
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PFTH
dmdgTL
TH VEjXjXjX
Z
,11
1
dMdGTL
dMFG
jXjXjX
jXII
)(
Analyzing the fault conditions
+
- -
+
P
Neutral
jXdG jXdM
jXTL
EGEM
IMIG
IF
P+
-
ETH=VPF
ZTHjXTL+ jXdG
jXdM
IF
IG IM
FTH
PFF
TH
PFF
ZZ
VI
Z
VI
then,ZisimpedancefaulttheIf F
dMdGTL
dGTLFM
jXjXjX
jXjXII
)(
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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dMdGTL
dMFG
jXjXjX
jXII
)(
The actual fault currents comingfrom generator and motor are the
sum of currents before fault and dueto fault.
Finding the Total Fault Current
+
- -
+
P
Neutral
jXdG jXdM
jXTL
EGEM
IMIG
IFFTH
PFF
ZZ
VI
dMdGTL
dGTLFM
jXjXjX
jXjXII
)(
L
dMdGTL
dMFLGFG I
jXjXjXjXIIII
)(
L
dMdGTL
dGTLFLMFM I
jXjXjX
jXjXIIII
)(
fault
Pre-fault
+
+
-
+
-
+
-
P
Neutral
VPFjXdG jXdM
EM
ILIL jXTL
EG
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Example
XdG=j0.2
G TL M
XdM=j0.2X=j0.1
IL
P
Assume EG=1.02/_40 PU, EM=1/_0
0 PU.
3-ph fault occurs at P. Find the totalfault current, fault contribution fromgenerator and motor, and post-faultvoltage at the generator bus.
+
-
+
-
+
-
P
Neutral
VPFJ0.2 J0.2
1/_00
ILIL J0.1
1.02/_40
Solve the pre-fault Circuit
PUj
IL0
00
13.830.14665.0
01402.1
PU
jVPF0
00
1.621.0074
)2.0()13.830.1466(01
VGPF
PU
jVGPF0
00
2.421.0114
)2.0()13.830.1466(402.1
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Example continued
+
- -
+
P
Neutral
J0.2 J0.2
J0.1
EGEM
IMIG
IF
P+
-ETH
=1.0074/_1.620
ZTHj0.3
J0.2
IF
IG IM
PUVE
PUjjj
Z
PFTH
TH
0
1
62.10074.1
12.02.0
1
3.0
1
PUZ
VI
TH
PFF
038.883951.8
PUj
jII
PUj
jII
FM
FG
0
0
38.885.0375.0
3.0
38.883.3585.0
2.0
FFMFG
LMFM
LGFG
IIICHECK
PUjIII
PUIII
:
50
864.3 0
Solve the fault Circuit
Add currents from both circuitsa l g e b r a i c a l l y
Observe: There is very little differencebetween IFG&IG and IFM&IM. Therefore,pre-fault currents are often neglected.
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Example continued
Solve the fault Circuit with Actual currents to get post-fault voltage at the generator bus
+
- -
+
P
Neutral
J0.2 J0.2
J0.1
EGEM
IFMIFG
IF
VFG
PUPU
VVV
PUjjIV
FGGPFG
FGFG
0
00
00
1.620.6716434.042.21.0114
434.0)1.0(864.30)1.0(0
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Some Observations
As seen in previous slides, load current is negligiblecompared to fault current in most cases. Therefore,the pre-fault voltage VPF (=ETH) is taken as 1/_0
0 PUand IL is neglected. Thus, load flow results are notused in fault calculations.
Sometimes, fault path has arc-resistance. In thatcase, it is included in the Thevenin model.
It is difficult to find Thevenin equivalent impedance ofa large power system in conventional way. As we shallsee, at bus i, the element ZBUS(i,i,) is equal to theThevenin equivalent impedance. Since ZBUS of anysystem is known, it is quite simple to form theThevenin equivalent circuit.
We solved the example using steady state reactances(Xd) for machines. We can find subtransient ortransient currents in the same way, using thecorresponding machine reactances (Xd or Xd) .
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Using ZBUS for Fault Analysis
As mentioned before, the fault is simulated by two sources of ValuesVPF and VPF in series. The figure above also shows fault impedanceZF.
The fault can be seen as a superposition of the pre-fault and faultconditions. Pre-fault condition has three sources actingsimultaneously: EG, EM and VPF. VPF is short-circuited.
-VPF is the only source working during fault condition, and EG, EM and
VPF are short-circuited. We find bus-voltages in both cases and add the results algebraically toget the final solution. Use these voltages to find line currents.
This is shown on the next slide.
+
-
+
-
jXdG jXdM
jX12
EGEM
VPF
VPF
+
-
-
+
IF
IL
IL1
23
GZF
jX23
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Viewing Fault as Superposition of Sources
Circuit with Fault
Fault
+
Pre-fault voltages are V1, V2, V3.
Due to VPF acting alone, thevoltage changes are V1, V2, V3.
Final bus-voltages will be V1+ V1, V2+ V2 and V3+ V3.
jX12IL
+
-
+
-
jXdG jXdM
jX12
EG EM
VPF
+
-
IL
IL
1 2 3
G
V1
V2 V3
Pre-fault
ZF
jX23
jX23
G
jXdG jXdMVPF +
--IF
1 2 31 2 3V1
V2
V3
jX12
IF1-2 IF3-2
ZF
jX23
+
-
+
-
jXdG jXdM
EGEM
VPF
VPF
+
-
-
+
IF
IL
12
3
GZF
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Analyzing the Fault Circuit
In this circuit, the only current injection into
a bus is -IF at bus-2. Writing a loop equation at bus-2,
0
0
333231
232221
131211
3
2
1
3
2
1
333231
232221
131211
3
2
1
FI
ZZZ
ZZZ
ZZZ
V
V
V
I
I
I
ZZZ
ZZZ
ZZZ
V
V
V
)2(,, 323222121 FFF IZVIZVIZV
Now using the ZBUS model of the system,
G
jXdG jXdMVPF +
--IF
1 2 31 2 3V1
V2 V3jX12
IF1-2 IF3-2
ZF
jX23
G
jXdG jXdMVPF +
--IF
1 2 31 2 3V1
V2 V3jX12
IF1-2 IF3-2
ZF
jX23
)1(0)( 2
2
VVZI
IZ
VV
PFFF
F
F
PF
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Analyzing the Fault Circuit (Cont.)
)3(
0)(
22
22
F
PFF
FPFFF
ZZ
VI
IZVZI
Thus, to solve a three-phase fault, use (3) to find IF and use (2) to find
voltage changes at all buses.
In general, in a n-bus system, for a fault at bus k with fault impedance ZF,
.....1,, niIZVandZZ
VI Fiki
Fkk
PFkF
G
jXdG jXdMVPF +
--IF
1 2 31 2 3V1
V2 V3jX12
IF1-2 IF3-2
ZF
jX23
G
jXdG jXdMVPF +
--IF
1 2 31 2 3V1
V2 V3jX12
IF1-2 IF3-2
ZF
jX23 Using equations (1) and (2) from the previousslide:
FTH
PFF
ZZ
VI Compare (3) with this equation we derived on slide#18:
We can see here that Thevenin impedance at a node is the same asthe diagonal element of ZBUS corresponding to that node.
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Example Continued
Bus-2 has a 3-Ph fault. Fault impedance ZF=0
. This systemhas a ZBUS of order-2. We will get YBUS and invert it to get ZBUS.
PUjjj
YBUS0095115
1.0
1
2.0
1)1,1(
+
-
+
-
J0.2 J0.2
1/_00
J0.1
1.02/_40
1 2
G
+
-
+
-
J0.2 J0.2
1/_00
J0.1
1.02/_40
1 2
G
PUjjj
YBUS0095115
1.0
1
2.0
1)2,2(
PUjj
YY BUSBUS0090110
1.0
1)1,2()2,1(
PUjYBUS
1510
1015PUjYZ BUSBUS
12.008.0
08.012.01
PUjZZ
VI
F
PFF
00
22
22 38.88395.8
012.0
62.10074.1
0
0
02
1
38.1780074.1
38.1786716.0
38.88395.8
0
12.008.0
08.012.0j
V
V
I12IM
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Example Continued
Now add the pre-fault and fault voltages to get final bus voltagesusing (3). Use (4) to calculate line currents.
PUV
V
V
V
V
V
PF
PF
0
0
0
0
0
0
2
1
2
1
2
1
00
434.0
38.1780074.1
38.1786716.0
62.10074.1
42.20114.1
PUjz
VVII FG
00
12
2121 864.3
1.0
0434.0
PUjz
VVI MFM
00
12
2 9052.0
001
+
-
+
-
J0.2 J0.2
1/_00
J0.1
1.02/_40
1 2
G
+
-
+
-
J0.2 J0.2
1/_00
J0.1
1.02/_40
1 2
G
I12IM
Compare these currents with answers on slide#21 and voltages withanswers on slide#22. They tally exactly. Observe that in this method,we use the final voltages to find line currents. Therefore, we do notneed to add the pre-fault load currents separately.
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EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Fuses (Continued)
Special Types of Fuses
Fuses for capacitor banks with high frequency inrush current Fuses with very high interruption ratings
Electronic fuses
Characteristics of Fuses
Minimum Melting Characteristics - An average melting time where arcingdoes not occur (link melts). See next slide
Total Clearing Characteristics Time taken to clear the arc - Used incoordinating against minimum melting characteristic of a larger fuse,located toward the current source. See next slide
E-rated fuses are very commonly used in distribution systems.
Fuses are available with rated continuous current ratings of: 0.5, 1, 2, 3,5, 7, 10, 15, 20, 30, 40, 50, 65, 80, 100, 125, 150, 200, 300, and 400,amperes
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Fuse Characteristics
For the denominator:
Use 300 seconds for links rated upto 100 amperes
Use 600 seconds for links rated 140and 200 amperes
K and T Types of Links:
Speed Ratio of K links are in therange 6.0 to 8.1
Speed Ratio of T links are in therange 10.0 to 13.0
In the given characteristics, thespeed ratio is around 210/31=6.8.
seconds600or300atCurrentMelting
seconds0.1atCurrentMeltingRatioSpeed
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Different MM Characteristics
EE 493/543 -Power Systems III, Symmetrical Faults,Dr. Sukumar Brahma, NMSU
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Expulsion Power Fuse 7.5kV. to 69kV. 100 to400 Amp