TRANSMISSION &
DISTRIBUTION
A Division of Global Power
POWER SYSTEM STABILITY CALCULATION TRAINING
D5
FltAli
D
ay
5
-
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A
na
l
ys
i
s
July10,2013
Prepared by: Peter Anderson
eBook for You
OUTLINE
2
OUTLINE
• Symmetrical Components
• Sequence Impedances
• Analysis of Fault Conditions
RttifFlt
•


given

set

of

Unbalanced

Three
phase

Vectors

can be represented by the sum of three sets of
Balanced or Symmetrical vectors
AA ACB
BC
B
C
PositiveSe
q
uence Ne
g
ativeSe
q
uence
Z
eroSe
q

I
+
I
=
I
I+I+I=I
A
B
C
NegativeSequence
2
0C2C1CC
0
B
2
B
1
B
B
I+I+I=I
I
I
I
I
B
C
2A
2
2C2A2B2A
Ia=I,aI=I,I

=
I
,
I
02
2
1C
021B
I+Ia+aI=I
I
+
aI
+
I
a
=
I
0
A
0
C
0
A
0
B
0
A
I
I
,

PositiveSequenceCurrentsaredeterminedsolelybythePositive
SequenceDrivingVoltagesproducedbythePowerSources,thePositive
SequencevoltageatthePointofFaultandtheSystemPositiveSequence
Impedance
NegativeSequenceCurrentsaredeterminedsolelybytheNegative
SequencevoltageatthePointofFaultandtheSystemNegativeSequence
Impedance
Impedance
ZeroSequenceCurrentsaredeterminedsolelybytheZeroSequence
voltage at the Point of Fault and the System Zero Sequence Impedance
voltage

at

the

Point

of

Fault

and

the

System

Zero


equence
N
e
t
wor
k
con
t
a
i
ns
th
e
Z
ero
S
equence
I
mpe
d
ances
t
o
thePointofFaultandanyconnectionstoEarth(NoDrivingVoltages)
eBook for You
POSITIVE SEQUENCE IMPEDANCE
9
POSITIVE

SEQUENCE

SEQUENCE

IMPEDANCE
Transformers
I
I
E
I
I
I
I
I
I
Z0=E/I
Ampere‐turnsareequalineachHVandLVWinding
eBook for You
ZERO SEQUENCE IMPEDANCE
12
ZERO

SEQUENCE

IMPEDANCE
Transformers
I
I
L
I
E
I

14
CONNECTIONS
H
L
H
L
H
L
H
L
GroundedWye/Delta Delta/GroundedWye
H L
H L
T
Grounded
Wye
/Grounded
Wye
T
Grounded
Wye
/Grounded
Wye
/Delta
Grounded

Wye
/Grounded

Wye

Rg
3Rg
X0
T
‐
N
H L
3Rg X0
H‐N
N
X0
L‐N
T
X0
T‐N
CORRECTMODEL
eBook for You
ANALYSIS OF FAULT CONDITIONS
ANALYSIS

OF

FAULT

CONDITIONS
16
eBook for You
ANALYSIS OF SHORT
CIRCUIT CONDITIONS
17

0
=
I
3
+
)
a
+
a
+
1
(
I
+
)
a
+
a
+
1
(
I
0=)I+Ia+aI(+)I+aI+Ia(+)I+I+I(
0
=
I
+
I
+
I

B
A
0=I,0=)a+a+1(Since
0
=
I
3
+
)
a
+
a
+
1
(
I
+
)
a
+
a
+
1
(
I
0
2
0
2
1

CIRCUIT CONDITIONS
18
ANALYSIS

OF

SHORT
-
CIRCUIT

CONDITIONS
Three Phase Fault
Three

Phase

Fault
0=I,0=I,
Z
E
=I
02
1
1
E
F1
Z1
F2
Z2 F0
Z0

CB
A
PhasetoPhaseFault
E
I
I
PhasetoPhasetoGroundFault
0
2
E
)
Z+Z
(
0=I
Z+Z
=
I
‐=
I
0
21
21
0
100221
0
2
1
EZ‐
=
I

OF

SHORT
-
CIRCUIT

CONDITIONS
Fault Positive Negative Zer o
3Ф
3Ф
F1
F2
F0
Ф‐E
F0F2F1
Ф‐Ф
F0F2F1
Ф‐Ф‐E
F0F2F1
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