Download - CPT5 - Short Circuit Analysis-6th Batch
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NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER
Certificate in
Power System Modeling and Analysis
Competency Training and Certification Program in Electric Power Distribution System Engineering
U. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTER
Competency Training and Certification Program in Electric Power Distribution System Engineering
T r a i n i n g Co u r s e i n
Sh o r t Ci r c u i t A n a l y s i s
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2
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
Co u r s e Ou t l i n e
1. Sources & Types of Faults
2. Network Reduction Techniques
3. Analysis of Faulted Power System
4. Computer Solution
5. Short Circuit Studies
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3
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
So u r c e s a n d Ty p e o f Fa u l t s
Sources of Short Circuit Currents
Type of Faults
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
So u r c e s o f Sh o r t Ci r c u i t
C u r r e n t s
FaultMV
LV
UtilityG
Fault Current Contributors
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5
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
Ty p e s o f Fa u l t
Shunt Fault: Unintentional Connection betweenphases or between phase and ground.
1. Single Line-to-Ground Fault
2. Line-to-Line Fault
3. Double Line-to-Ground Fault
4. Three Phase Fault
Series Fault: Unintentional Opening of phaseconductors
Simultaneous Fault
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6
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
Shunt Faults
Line-to-Line
Double Line-to-Ground Single Line-to-Ground
Three Phase
Ty p e s o f Fa u l t
7T i i C i Sh Ci i A l i
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7
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
N e t w o r k Re d u c t i o n Te ch n i q u e s
Calculating Short Circuit Currents by
Network Reduction
Reduced Sequence Networks
8T i i C i Sh t Ci it A l i
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
N e t w o r k Re d u c t i o n Te ch n i q u e s
Calculating Short Circuit Currents by Network Reduction
1. Draw the Single Line Diagram.
2. Draw the Impedance Diagram.
3. Convert all parameters to per-unit.4. Reduce the network between the source(s) and the fault
location.
5. Calculate the fault current
equiv
f
f
Z
VI =
9T i i C i Sh t Ci it A l i
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9
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
N e t w o r k Re d u c t i o n Te ch n i q u e s
Combination of Branches in Series
1 2
1 1 2 2
1 2 1 2
( ) ( )
( ) ( )
eqZ Z ZR jX R jX
R R j X X
= += + + +
= + + +
Z1 Z2
Combination of Branches in Parallel
1 2
1 2
1 1 2 2
1 2 1 2
( )( )
( ) ( )
eq
Z ZZ
Z ZR jX R jX
R R j X X
=
++ +
=+ + +
Z1
Z2
10Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
Transforming Wye to Delta
Transforming Delta to Wye
a b b c c aA
a
a b b c c aB
b
a b b c a c C
c
Z Z Z Z Z Z Z
Z
Z Z Z Z Z Z Z Z
Z Z Z Z Z Z Z
Z
+ +=
+ +=
+ +=
B Ca
B C
A Cb
B C
A Bc
A B C
Z ZZ
Z Z Z
Z ZZZ Z Z
Z ZZ
Z Z Z
=+ +
= + +
=+ +
ZC
ZB ZA
Za Z b
Zc
N e t w o r k Re d u c t i o n Te ch n i q u e s
11Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
Example:
Determine the fault current for a three phasebolted fault in each bus for the 4 bus systembelow.
G
1
432
Line1Line
3
Line 5 Line 4
Line
2
4-bus system
LINE FB TB Z(p.u.)
Line1 1 4 j0.2
Line2 1 3 j0.4
Line3 1 2 j0.3
Line4 3 4 j0.5
Line5 2 3 j0.6
The generator is rated 100 MVA, 6.9 kV and has asubtransient reactance of 10%. Base Values: 100 MVA, 6.9 kV
12Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
Solution:
Draw the impedance diagram
E 1.0
0.1
0.20.3
0.6 0.5
0.4
4
3
2
1
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
E 1.0
0.1
0.20.3
0.6 0.5
0.4
43
2
1
If
-
+
= +
= +
=
=
+
=
+
=
= +
= +
=
a1 2 2 3
b 1 3
1 3
c3 4
R edu ce the ne tw o rk
X
0 .3 0 .6
0 .9
X
(0.9)(0.4)0 .9 0 .4
0 .276923
X
0 .276923 0 .5
0 .776923
a
a
b
X X
X X
X X
X X
a) Fault @ Bus 4
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
cd 14
c 14
X XX
X X(0.776923) (0.2)
0.776923 0.2
0.159055
=+
=+
=
d
equiv genX X X
0.1 0.159055
0.259055
= +
= +=
If
-
+
E 1.0
0.25905
f
1.0
I 0.259055
3.860184 p.u.= =
base
f
100 1000I 8367.64 A
3(6.9)I 3.860184 x 8367.64
= 32,300.63 A
x= =
=
15Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Training Course in Short Circuit Analysis
b) Fault @ Bus 3
E 1.0
0.1
0.20.3
0.6 0.5
0.4
432
1
If
-
+
a
23 12X X X0.3 0.6
0.9
= += +
=b
14 34X X X
0.2 0.5
0.7
= +
= +
=
a bequiv 13X (X ||X ) ||X
0.198425
=
=
16Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
g y
gen equivX X X
= 0.1 0.198425
= 0.298425
= ++
If-
+
E 1.0
0.298425f
1.0I 0.298425
= 3.350923 p.u.
=
17Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
g y
c) Fault @ Bus 2
a
14 34X X X
0.2 0.50.7
= +
= +=
E 1.0
0.1
0.20.3
0.6 0.5
0.4
432
1
If
-
+
ab 13
a
13
X XX
X X
(0.7)( 0.4)0.7 0.4
0.254545
=+
=+
=
18Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
g y
= +
= +=
c b
23X X X
0.254545 0.6
0.854545
=+
= +=
cd 12
c
12
X XX
X X
(0.854545)( 0.3)0.854545 0.3
0.222047
= +
=
d
genX X X
0.322047
=
=
f 1.0I 0.322047
3.095525 p.u.
If
-
+
E 1.0
0.322047
19Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
d) Fault @ Bus 1
=
=
genX X
0.1E 1.0
0.1
0.20.3
0.6 0.5
0.4
432
1
If
-
+
=
=
f
1.0I
0.110.0 p.u.
20Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Re d u ce d Se q u e n ce N e t w o r k s
Since we mentioned that various power system
components behave/respond differently to the flowof the currents sequence components, it followsthat the there will be a unique power system modelfor each of the sequence component. These are
called the sequence networks.
Positive-Sequence Network
Negative-Sequence Network Zero-Sequence Network
21Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Re d u ce d Se q u e n ce N e t w o r k s
The Thevenin equivalent of the power system a tt h e f a u l t p o i n t is called the sequence network.
PositiveSequence
NegativeSequence
ZeroSequence
0aVr
F0
N0
Z0
0aIr
+
-
F1
N1
thf VVr
=
Z11aIr
1aVr
+
-
+
-
F2
N2
Z2
2aIr
2aVr
+
-
11ath1a ZIVV
rrr
= 22a2a ZIV
rr
= 00a0a ZIV
rr
=
22Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Eq u i v a l e n t Ci r c u i t o f U t i l i t y
The equivalent sequence networks of the ElectricUtility Grid are:
Utility Thevenin Equivalent Circuits
+
-
gE
r
+
-
R1 +jX1
+
-
PositiveSequence
NegativeSequence
ZeroSequence
+
-
R2 +jX2 R0 +jX0
23Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Eq u i v a l e n t Ci r c u i t o f Ge n e r a t o r s
Positive-Sequence Impedance:
Xd=Direct-Axis Subtransient Reactance
Xd=Direct-Axis Transient Reactance
Xd=Direct-Axis Synchronous Reactance
Negative-Sequence Impedance:
for a salient-pole machine)"X"X(X qd21
2 +=
for a cylindrical-rotor machine"XX d2 =Zero-Sequence Impedance:
"X6.0X"X15.0 d0d
24Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Gr o u n d e d -W y e Ge n e r a t o r
Eq u i v a l e n t Ci r c u i t o f Ge n e r a t o r s
The sequence networks for the grounded-wye
generator are shown below.
jZ0
F0
N0
jZ2
F2
N2
F1
N1
gEr
jZ1+
-
PositiveSequence
NegativeSequence
ZeroSequence
25Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Gr o u n d e d -W y e t h r o u g h a n I m p e d a n ce
Eq u i v a l e n t Ci r c u i t o f Ge n e r a t o r s
If the generator neutral is grounded through animpedance Zg, the zero-sequence impedance is
modified as shown below.
jZ0
F0
N0
3Zg
jZ2
F2
N2
F1
N1
gEr
jZ1+
-
PositiveSequence
NegativeSequence
ZeroSequence
26Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Eq u i v a l e n t Ci r c u i t o f Ge n e r a t o r s
Un g r o u n d e d -W y e Ge n e r a t o r
If the generator is connected ungrounded-wye or
delta, no zero-sequence current can flow. Thesequence networks for the generator are shownbelow.
F1
N1
gEr
jZ1+
-
jZ2
F2
N2
jZ0
F0
N0
Positive
Sequence
Negative
Sequence
Zero
Sequence
27Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Eq u i v a l e n t Ci r c u i t o f T r a n s f o r m e r s
Po s i t i v e & N e g a t i v e Se q u e n ce N e t w o r k s
2Ir
2Z
+
-
+
-PrimarySide SecondarySide
1
Ir
1Z
+
-
+
-PrimarySide SecondarySide
NegativeSequence
Network
PositiveSequence
Network21 ZZ =
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
10 ZZ =
+
-
HVr
+
-
XVr
10 ZZ =
+
-
HVr
+
-
XVr
Ze r o Se q u e n c e N e t w o r k * Transformer Connection Zero-Sequence Network
*Excluding 3-phase unit with a 3-legged core.
Eq u i v a l e n t Ci r c u i t o f T r a n s f o r m e r s
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Transformer Connection Zero-Sequence Network
*Excluding 3-phase unit with a 3-legged core.
10 ZZ =
+
-
HVr
+
-
XVr
10 ZZ =
+
-
HVr
+
-
XVr
Ze r o Se q u e n c e N e t w o r k *
Eq u i v a l e n t Ci r c u i t o f T r a n s f o r m e r s
30Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Transformer Connection Zero-Sequence Network
*Excluding 3-phase unit with a 3-legged core.
10 ZZ =
+
-
HVr +
-
XVr
10 ZZ =+
-
HVr
+
-
XVr
Ze r o Se q u e n c e N e t w o r k *
Eq u i v a l e n t Ci r c u i t o f T r a n s f o r m e r s
31Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Eq u i v a l e n t Ci r c u i t o f L i n e s
If the line is completely transposed,
s0s ZZ = m0m ZZ =0ZZ 2m1m ==0ZZ 2s1s ==
The sequence impedance matrix reduces to
+
=
ms
ms
ms
2
1
0
ZZ00
0ZZ0
00Z2Z
Z
Z
Z
Note: The sequence impedances are completely decoupled.
32Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
For a completely transposed line, the equation inthe sequence domain is
=
2a
1a
0a
2
1
0
2a
1a
0a
I
II
Z00
0Z000Z
V
VV
r
r
r
r
r
r
where
s
m
a21D
DlnksjsrZZ +==
Eq u i v a l e n t Ci r c u i t o f L i n e s
2
ms
3
e
da0DD
Dlnksjsr3srZ ++=
33Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Eq u i v a l e n t Ci r c u i t o f L i n e s
21 II
rr
=
21 ZZ =
+ +
0I
r
0Z
+ +
- - - -
Z e r o - S e q u e n c e
N e t w o r k
Po s i t i v e - & N e g a t i v e -
Se q u e n ce N e t w o r k s
34Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Example: Determine the reduced sequence
networks of the power system shown for a fault aF. Assume Eg = 1.0 p.u.
G
T1 T2Line OpenF
G: X1 = 40% X2 = 40% X0 = 20%
T1, T2: X = 5%
Line: X1= X
2= 15% X
0= 35%
Note: All reactances are in per-unit of a commonMVA base.
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Competency Training & Certi fication Program in
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National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Positive-Sequence Network:
+
-
j0.61aI
r
1.0
N1
1aV
F1j0.05
+
gEr
+
-
rj0.4
-
j0.15Open
j0.05
F1
N1
36Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Competency Training & Certi fication Program in
Electric Power Distribution System EngineeringU. P. National Engineering Center
National Electrification Administration
Negative-Sequence Network:
j0.6
2aIr
N2
F2j0.05 j0.15+
-
2aVr
j0.4
Openj0.05
F2
N2
37Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
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National Electrification Administration
Zero-Sequence Network:
j0.044
0aIr
N0
F0
N0
j0.05 j0.35+
-
0aVr
j0.2
Openj0.05
F0
38Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
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National Electrification Administration
Example: Reduce the sequence networks for thepower system shown if a fault occurs at bus 4.
G1
TL1
L2
G2
L3
2 3
4
T: X=0.08G1: X1=0.40 X2=0.40 X0=0.15
G2: X1=0.50 X2=0.50 X0=0.25
L1: X1=0.40 X2=0.40 X0=0.80L2: X1=0.30 X2=0.30 X0=0.60
L3: X1=0.20 X2=0.20 X0=0.40
39Training Course in Short Circuit Analysis
-
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Competency Training & Certi fication Program in
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National Electrification Administration
Positive-Sequence Network:
F1
1aIr
+
-
N1
j0.4
j0.08
j0.4
j0.3 j0.2
j0.5
1G
Er +
- 2GEr
2 3
4
40Training Course in Short Circuit Analysis
-
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Competency Training & Certi fication Program in
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National Electrification Administration
Negative-Sequence Network:
F2
1aIr
j0.2
j0.08
j0.3
j0.42 3
4
j0.5j0.4
N2
41Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
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National Electrification Administration
Zero-Sequence Network:
F0
0aIr
N0
j0.2
j0.08
j0.8
j0.6 j0.4
j0.252 3
4
42Training Course in Short Circuit Analysis
-
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Competency Training & Certi fication Program in
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National Electrification Administration
A n a l y s i s o f Fa u l t e d Sy s t em
The Fault Point
Three-Phase Fault
Single-Line-to-Ground Fault
Line-to-Line Fault
Double-Line-to-Ground Fault
43Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
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National Electrification Administration
The system is assumed to be balanced, withregards to impedances, except at one point calledthe fault point.
FaultCurrents
a
b
c
aVr
bVr
cVr
Ground
aIr
bIr
cIr
Line-to-ground
voltages
F
Th e Fa u l t Po i n t
44Training Course in Short Circuit Analysis
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Competency Training & Certi fication Program in
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Th r e e Ph a s e Fa u l t
On a balanced three phase system, thesame magnitude of fault currents willflow in each phase of the network if athree phase fault occurs.
Since faults currents are balanced, thefaulted system can, therefore, be
analyzed using the single phaserepresentation.
45Training Course in Short Circuit Analysis
Th r e e - Ph a s e Fa u l t
-
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Competency Training & Certi fication Program in
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Th r e e - Ph a s e Fa u l t
ab
c
aVr
bVr
cVr
Ground
bI
r
cI
r
gZ
aI
r
fZ fZ
gIr
fZ
Note: The system is still balanced. Currents andvoltages are positive sequence only. The groundcurrent is zero.gI
r
46Training Course in Short Circuit Analysis
Sequence Network Interconnection:
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Competency Training & Certi fication Program in
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National Electrification Administration
Sequence Network Interconnection:
F1 F2
N2
Z2
2aIr
2aV
r
+
-
N0
F0
Z0
0aIr
0aV
r
+
-fV
Z1 1aIr
1aV
r
+
+
--
Zf
N1
Sequence currents
f
f
aZZ
VI+
=1
1
r
020 == aa IIrr
47Training Course in Short Circuit Analysis
Three Phase Fault Currents:
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Competency Training & Certi fication Program in
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National Electrification Administration
Three Phase Fault Currents:
f
f
aaaaZZ
VIIII
+=++=
1
210
f
f
aaab
ZZ
VaaIIaII
+
=++=1
2
21
2
0
f
f
aaacZZ
aVIaaIII +=++= 1
2210
48Training Course in Short Circuit Analysis
Example: A three-phase fault occurs at point F
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Competency Training & Certi fication Program in
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National Electrification Administration
Example: A three phase fault occurs at point F.Assuming zero fault impedance, find the faultcurrents at fault point F. Determine the phasecurrents in the line and the generator. Assume Eg= 1.0 p.u.
G
T1 T2Line
OpenF
G: X1 = 40% X2 = 40% X0 = 20%
T1, T2: X = 5%
Line: X1 = X2 = 15% X0 = 35%
Note: All reactances are in per-unit of a common MVA base.
49Training Course in Short Circuit Analysis
Positive-Sequence Network:
-
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49/187
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Competency Training & Certi fication Program in
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National Electrification Administration
Positive Sequence Network:
+
-
j0.61AI
r
1.0
N1
F1
+
-
1AVr
j0.05
j0.4
+
gEr
-
Openj0.05
j0.151AI
r
L1AIr
g1aIr
F1
N1
50Training Course in Short Circuit Analysis
Sequence Network Interconnection:
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Competency Training & Certi fication Program in
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Sequence Network Interconnection:
The sequence fault currents
Zf
=
=
=+=
0
2
11
a
a
f
f
a
I
I
ZZ
V
I
r
r
r
+
-
j0.61AI
r
1.0
N1
F1+
-
1AVr
The phase fault currents
=
==
c
b
a
I
II
51Training Course in Short Circuit Analysis
Si n g l e L i n e - t o - Gr o u n d Fa u l t
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Competency Training & Certi fication Program in
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g
Assuming the fault is in phase a,a
b
c
aVr
bVr
cVr
Ground
aIr bI
r
cIr
fZ
afa IZVrr
=Boundary Conditions: (1)
0== cb IIrr
(2)
52Training Course in Short Circuit Analysis
Transformation: From (2), we get
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Competency Training & Certi fication Program in
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National Electrification Administration
abcIAI
rr
1
012
=
Transformation: From (2), we get
aaaa
2
2
11
111
3
1
2
1
0
a
a
a
II
I
r
r
r
=00
aIr
3
1
=a
a
a
II
I
r
r
r
which meansa
3
12a1a0a IIII
rrrr
===
From (1), we get
)(210210 aaafaaa
IIIZVVVrrrrrr
++=++
or
0210 3 afaaa IZVVVrrrr
=++
53Training Course in Short Circuit Analysis
Sequence Network Interconnection:
-
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Competency Training & Certi fication Program in
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National Electrification Administration
q
F1
N1
fVr
Z1 1aIr
1aV
r
+
-
+
-
F2
N2
Z2
2aIr
2aV
r
+
-
0aV
r
F0
N0
Z0
0aIr+
-
3Zf
The sequence fault currents
f
f
aaaZZZZ
VIII
3210210 +++
===rrr
54Training Course in Short Circuit Analysis
Single-Line-to-Ground Phase Fault Currents:
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Competency Training & Certi fication Program in
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g
f
f
aaaaZZZZ
VIIII
3
3
021
210 +++=++=
0=bI
0=cI
55Training Course in Short Circuit Analysis
Example: A single line-to-ground fault occurs at
-
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Competency Training & Certi fication Program in
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National Electrification Administration
p g gpoint F. Assuming zero fault impedance, find thephase currents in the line and the generator.Assume Eg = 1.0 p.u.
G
T1 T2
Line OpenF
G: X1 = 40% X2 = 40% X0 = 20%
T1, T2: X = 5%
Line: X1 = X2 = 15% X0 = 35%
Note: All reactances are in per-unit of a commonMVA base.
56Training Course in Short Circuit Analysis
Positive-Sequence Network:
-
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Competency Training & Certi fication Program in
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+
-
j0.61aI
r
1.0
N1
1aV
F1j0.05
+
gEr
+
-
rj0.4
-
j0.15
Openj0.05
F1
N1
57Training Course in Short Circuit Analysis
Negative-Sequence Network:
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57/187
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Competency Training & Certi fication Program in
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National Electrification Administration
j0.6
2aIr
N2
F2j0.05 j0.15
+
-
2aVr
j0.4
Openj0.05
F2
N2
58Training Course in Short Circuit Analysis
Zero-Sequence Network:
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58/187
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Competency Training & Certi fication Program in
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National Electrification Administration
j0.044
0aIr
N0
F0
N0
j0.05 j0.35
+
-
0aVr
j0.2
Openj0.05
F0
59Training Course in Short Circuit Analysis
Sequence Network Interconnection:
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59/187
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Competency Training & Certi fication Program in
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N1N2
F1
J0.6 1AIr
+
-
F2
J0.6
2AIr
1.0
N0
F0
J0.044
0AIr
Sequence Fault Currents
)044.06.06.0(j
0.1III
2A1A0A ++===
rrr
p.u.804.0j=
60Training Course in Short Circuit Analysis
Phase Fault Currents
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Competency Training & Certi fication Program in
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p.u.411.2jI3I 0AA ==
rr
0II CB ==rr
Sequence Currents in the Transmission Line
p.u.804.0jII 1AL1A ==rr
p.u.804.0jII 2AL2A ==
rr
p.u.089.0jI4.005.0
05.0I 0AL0A =+
=rr
Phase Currents in the Transmission Line
p.u.696.1jIIII L2AL1AL0AAL =++=rrrr
61Training Course in Short Circuit Analysis
p.u.714.0jIaIaII L2AL1A2
L0ABL =++=rrrr
rrrr
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Competency Training & Certi fication Program in
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p.u.714.0jIaIaII L2A2
L1AL0ACL =++=
rrrr
Sequence Currents in the Generator: Using the 30o
phase shift,
p.u.120804.030II ooL1Ag1a ==rr
p.u.60804.030II
oo
L2Ag2a =+=
rr
p.u.696.0j402.0 =
p.u.696.0j402.0 =0I g0a =
r
Phase Currents in the Generator
p.u.392.1jIIII g2ag1ag0aag =++=rrrr
62Training Course in Short Circuit Analysis
p.u.392.1jIaIaII g2ag1a2
g0abg =++=rrrr
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0IaIaII g2a2
g1ag0acg =++=
rrrr
Three-line Diagram:
H3
H1
H2
A
BC H3
H1
H2
A
BC
X3X
1
X2
a
b
c
T1 T2a
b
j2.411
j1.696
j0.714
j0.714
j1.392
j1.392
0
j0.268
j0.714
j2.1430
63Training Course in Short Circuit Analysis
L i n e - t o - L i n e Fa u l t
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Competency Training & Certi fication Program in
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Assuming the fault is in phases b and c,
0=aIr
a
b
c
aVr
bVr
cVr
Ground
bIr
cIr
fZ
aIr
(1)Boundary Conditions:
cb IIrr
=(2)
fbcb ZIVVrrr
=(3)
64Training Course in Short Circuit Analysis
Transformation: From (1) and (2), we get
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Competency Training & Certi fication Program in
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abcIAI1
012
=
r
aa
aa2
2
1
1
111
3
1
2
1
0
a
a
a
I
I
I
r
r
r
=
b
b
I
Ir
r
0
3
1
=
b
b
IaaIaar
r
)()(
0
2
2
which means00 =aI
r
bbaa IjIaaIIrrrr
3
1231
21 )( ===
65Training Course in Short Circuit Analysis
From (3), we get
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Competency Training & Certi fication Program in
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)( 212
0 aaa VaVaV
rrr
++faaaaaa ZIaIaIVaVaV )()( 21
2
02
2
10
rrrrrr
++=++
Since and , we get
faaa ZIaaVaaVaa 12
2
2
1
2 )()()(rrr
=+
00 =aIr
21 aa IIrr
=
or
faaa ZIVV 121rrr
=
66Training Course in Short Circuit Analysis
Sequence Network Interconnection:
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Competency Training & Certi fication Program in
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F1
N1
fV
Z1 1aIr
1aVr
+
-
+
-
F2
N2
Z2
2aIr
2aVr
+
-
Zf
N0
F0
Z0
0aIr
The sequence fault currents
f
f
aaZZZ
VII
++==
21
21
rr
00 =aIr
67Training Course in Short Circuit Analysis
Line-to-Line Phase Fault Currents:
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Competency Training & Certi fication Program in
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0=aI
21
2
0 aaabaIIaII ++=
( ) 112
11
2 )(0 aaaa jIIaaIaIa ==++=
f
f
bZZZ
VjI++
=21
3
f
f
cZZZ
VjI
+++=
21
3
68Training Course in Short Circuit Analysis
Example: A line-to-line fault occurs at point F.Assuming zero fault impedance find the fault
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Competency Training & Certi fication Program in
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National Electrification Administration
Assuming zero fault impedance, find the fault
currents at fault point F. Assume Eg = 1.0 p.u.
G
T1 T2Line
OpenF
G: X1 = 40% X2 = 40% X0 = 20%
T1, T2: X = 5%
Line: X1 = X2 = 15% X0 = 35%
Note: All reactances are in per-unit of a common MVA base.
69Training Course in Short Circuit Analysis
Sequence Network Interconnection:
F0F1 F2
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Competency Training & Certi fication Program in
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F0F1 F2
N1 N2
J0.6 1AIr
+
-
J0.6
2AIr
J0.044
0AIr
1.0
N0
70Training Course in Short Circuit Analysis
Sequence Fault Currents:
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Competency Training & Certi fication Program in
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71Training Course in Short Circuit Analysis
Phase Fault Currents:
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Competency Training & Certi fication Program in
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National Electrification Administration
72
Training Course in Short Circuit Analysis
D o u b l e - L i n e - t o - Gr o u n d Fa u l t
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7/29/2019 CPT5 - Short Circuit Analysis-6th Batch
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Competency Training & Certi fication Program in
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a
b
c
aVr
bVr
cVr
Ground
bIr
cIr
g
Z
aIr
fZ fZ
cb II
rr
+
Assuming the fault is in phases b and c,
0=aIr
cgbgfb IZIZZVrrr
++= )(
bgcgfc IZIZZVrrr
++= )(
(1)Boundary Conditions:
(2)
(3)
73
Training Course in Short Circuit Analysis
Transformation: From (1), we getrrrr
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Competency Training & Certi fication Program in
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2100 aaaa IIII ++==From
21
2
0 aaab
VaVaVVrrrr
++=
2
2
10 aaac VaVaVVrrrr
++=we get
22
12 )()( aacb VaaVaaVV
rrrr
+=
2120 aaab IaIaII
rrrr
++=
Likewise, from
2
2
10 aaac IaIaIIrrrr
++=
74
Training Course in Short Circuit Analysis
we get
22rrrr
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Competency Training & Certi fication Program in
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Competency Training & Certi fication Program in
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2
2
1
2
)()( aacb IaaIaaII +=From boundary conditions (2) and (3), we get
)( cbfcb IIZVV
rrrr
=Substitution gives
])()[( 22
1
2
aaf IaaIaaZrr
+=2
21
2 )()( aa VaaVaarr
+
Simplifying, we get
2211 afaafa IZVIZVrrrr
=
75
Training Course in Short Circuit Analysis
From boundary conditions (2) and (3), we getrrrr
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))(2( cbgfcb IIZZVV ++=+
2102 aaacb VVVVV
rrrrr
=+2102 aaacb IIIII
We can also show
rrrrr
=+
Substitution gives
)2(2 210210 aaafaaa IIIZVVVrrrrrr
=
)2(2 210 aaag IIIZ
rrr
+
76
Training Course in Short Circuit Analysis
Rearranging terms, we getrrrrr
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11000 422 afaagafa IZVIZIZV =)(2 2122 aagafa IIZIZV
rrrr
++
Earlier, we got
2211 afaafa IZVIZVrrrr
=
021 aaa III
rrr
=+Substitution gives
)(2622 11000 afaagafa IZVIZIZV
rrrrr
=
( )1100 3 afaagfa IZVIZZV
rrrr
=+
77
Training Course in Short Circuit Analysis
Sequence Network Interconnection:
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F1
fV
Z1 1aI
r
1aVr
+
+
-
F2 F0
-
N1 N2
Z2
2aI
r
2aVr
+
-
Z0
0aI
r
0aVr
+
-
ZfZf Zf+3Zg
N0
gfT ZZZZ 300 ++=
fT ZZZ += 11
Let
fT ZZZ += 22
78
Training Course in Short Circuit Analysis
The sequence fault currents
Vr
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TT
TTT
fa
ZZ
ZZZ
VI
20
201
1
++
=r
From current division, we get
120
0
2 aTT
T
aI
ZZ
ZI
rr
+=
From KCL, we get
1
20
20 a
TT
Ta I
ZZ
ZIrr
+=
210 aaa IIIrrr
= or
79
Training Course in Short Circuit Analysis
Double-Line-to-Ground Phase Fault Currents:
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0=aI
21
2
0 aaab aIIaII ++=
( )
TTTTTT
TTf
ZZZZZZ
aZZVj
020121
203
++
=
2
2
10 aaac IaaIII ++=
TTTTTT
TTf
ZZZZZZ
ZaZVj
020121
22
03
++
+=
80
Training Course in Short Circuit Analysis
Example: A double-line-to-ground fault occurs atpoint F. Assuming zero fault impedance, find the
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fault currents at fault point F. Assume Eg = 1.0 p.u.
G
T1 T2Line
OpenF
G: X1 = 40% X2 = 40% X0 = 20%
T1, T2: X = 5%
Line: X1 = X2 = 15% X0 = 35%
Note: All reactances are in per-unit of a common MVA base.
81
Training Course in Short Circuit Analysis
Sequence Network Interconnection:
F0F1 F2
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N1 N2
J0.6 1AIr
+
-
J0.6
2AIr
J0.044
0AIr
1.0
N0
82
Training Course in Short Circuit Analysis
Sequence Fault Currents:
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83
Training Course in Short Circuit Analysis
Phase Fault Currents:
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84
Training Course in Short Circuit Analysis
Com p u t e r So l u t i o n
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Development of the Model
Rake Equivalent Formation of Zbus
Analysis of Shunt Fault
85
Training Course in Short Circuit Analysis
D e v e lo p m e n t o f t h e M o d e l
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Ob s e r v a t i o n s o n M a n u a l N e t w o r k
S o l u t i o n
The procedure is straight forward, yet tedious andcould be prone to hand-calculation error.
Is there a way for a computer to implement thismethodology?
86
Training Course in Short Circuit Analysis
D e v e lo p m e n t o f t h e M o d e l
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Consider the three-bus system shown below. Letus analyze the system for a three-phase fault inany bus.
G1G2L1
L2
1 2
3
G1, G2 : X1=X
2=0.2 X
0=0.1
L1 : X1=X2=0.6 X0=1.2
L2 : X1=X2=0.24 X0=0.5
87
Training Course in Short Circuit Analysis
Positive-Sequence Network:
j0 61 2
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j0.2
1GEr +
-
+
-2GE
r
j0.2
j0.6j0.24
1 2
3
j0.6
3
j0.2j0.2
j0.24
+
-
GEr
1 2Combine the sourcesand re-draw. AssumeE
G
= 1.0 per unit.
88
Training Course in Short Circuit Analysis
For a three-phase fault in bus 1 (or bus 2), we getthe positive-sequence impedance.
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16.0j)]6.02.0//(2.0[jZ1 =+=
25.6jZ
1
Z
E
I11
G
F ===
r
For a three-phase fault in bus 3, we get
4.0j)]6.02.0//(2.024.0[jZ1 =++=
5.2jZ
1
Z
EI
11
GF ===
r
89
Training Course in Short Circuit Analysis
Let us connect a fault switch to each bus. In orderto simulate a three-phase fault in any bus, close
th f lt it h i th t b
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Next, use loop currents to
describe the circuit with allfault switches closed.Since there are four loops,
we need to define fourloop currents.
the fault switch in that bus.
j0.6
3
j0.2j0.2
j0.24
+
-
GEr
1 2
1I
r
3I
r
2I
r
4Ir
4
90
Training Course in Short Circuit Analysis
The loop equations are
)(2.00.1 431 IIIjrrr
+=loop 1:rr
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loop 2:
loop 3:
)(2.06.0)(2.00 314442 IIIjIjIIjrrrrrr
+++=
)(2.00.1 42 IIjrr
+=
3431 24.0)(2.00.1 IjIIIjrrrr
++=
loop 4:
or
0.12.02.02.0
2.044.002.0
2.002.00
2.02.002.0
0
0.1
0.1
0.1
4
3
2
1
I
I
I
I
r
r
r
r
= j
91
Training Course in Short Circuit Analysis
Current I4 is not a fault current. It can be eliminatedusing Krons reduction. We get
IZV )1(rr
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IZV )1(bus=where
3
1
421
)1(
bus ZZZZZ
=and
44.002.0
02.002.002.0
Z1 = j
2.0
2.02.0
Z2 = j
Z3 = j[-0.2 0.2 -0.2 ] Z4 = j[1.0]
92
Training Course in Short Circuit Analysis
Substitution gives
16004016001 1I
r
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40.004.016.0
04.016.004.016.004.016.0
= j
0.1
0.10.1
3
2
1
I
II
r
r
IZV )1(busrr
=
Note:(1) The equation can be used to analyze a three-
phase fault in any bus (one fault at a time).
)1(busZ(2) is called the positive-sequence bus-
impedance matrix, a complex symmetric matrix.
93
Training Course in Short Circuit Analysis
Ra k e Eq u i v a l e n t
Consider the matrix voltage equation
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Consider the matrix voltage equation
232212
131211
ZZZ
ZZZ
One possible equivalentcircuit is shown. This circuitis called a rake-equivalent.
+
-
Z11 Z22 Z33
Z12 Z23
Z13
1.0
1Ir
2Ir
3Ir
332313 ZZZ=0.1
0.1
0.13I
2
1
I
I
r
r
r
Suppose we are asked tofind a circuit that satisfiesthe matrix equation.
94Training Course in Short Circuit Analysis
Consider again the three-bus system. The circuit isdescribed by the matrix equation
r
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40.004.016.0
04.016.004.0
16.004.016.0
= j
0.1
0.1
0.1
3
2
1
I
I
I
r
r
The rake equivalent is
shown. The diagonalelements of the matrix areself impedances while the
off-diagonal elements aremutual impedances.
+
-
j0.16
1.0
1Ir
2Ir
3Ir
j0.16 j0.4
j0.04j0.04
j0.16
95Training Course in Short Circuit Analysis
For the three-bus system, assume a fault in bus 3.The equation for bus 3 is
I40jI040jI160j01rrr
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321 I4.0jI04.0jI16.0j0.1 ++=
Since only bus 3 is faulted,
I1=I2=0. We get
3I4.0j0.1r
=or
5.2j4.0j
1I3 ==r
+
-
j0.16
1.0
3Ir
j0.16 j0.4
j0.04j0.04
j0.16
2Vr
+
-1Vr
+
-
From KVL, we get the voltage in bus 1.
6.0Z
Z0.1IZ0.1V
33
133131 ===rr
96Training Course in Short Circuit Analysis
Similarly from KVL, we get the voltage in bus 2.
90Z
Z
01IZ01V23
3232
rr
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9.0Z0.1IZ0.1V 333232 ===
Note: Once the voltages in all the buses are
known, the current in any line can be calculated.In general, for a three-phase fault in bus k of asystem with n buses, the fault current is
kk
kZ1I =
r
k=1,2,n
The voltage in any bus j is given by
kk
jk
jZ
Z0.1V =
r
j=1,2,n
97Training Course in Short Circuit Analysis
The current in any line, which is connected frombus m to bus n, can be found using
VVrr
r
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mn
nmmn
z
VVI =
where zmn is the actualimpedance of the line.j0.2
1GE
r+
-
+
-2GE
r
j0.2
j0.6
j0.24
1 2
3
-j2.0
-j2.5
-j0.5
5.0j6.0j
6.09.0z
VVI21
1221 ===
rr
r
For example, the
current in the linebetween buses 2 and1 is
98Training Course in Short Circuit Analysis
Fo r m a t i o n o f Zb u s
Zbus can be built one step at a time by adding one
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Zbus can be built, one step at a time, by adding onebranch at a time until the entire network is formed.
The first branch to be added must be a generatorimpedance. This is necessary in order to establishthe reference bus.
Subsequent additions, which may be done in any
order, fall under one of the following categories:
(1) Add a generator to a new bus;
(2) Add a generator to an old bus;
(3) Add a branch from an old bus to a new bus;
(4) Add a branch from an old bus to an old bus.
99Training Course in Short Circuit Analysis
Assume thatat the current
stage the n22221
n11211
ZZZ
ZZZ
01
0.1
2
1
I
Ir
r
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+
-
Z11 Z22 Zkk
Z12 Z2k
1.0
1Ir
2Ir
kIr
Znn
Zkn
nIr1 2 k n
stage, thedimension ofZbus is n.
nn2n1n
n22221
ZZZ
ZZZ
0.1
0.1
n
2
I
I
r
=
oldbusZ
Let us examineeach category inthe addition of a
new branch.
100Training Course in Short Circuit Analysis
Type 1: Add a generator to a new bus
- 1 0 Let Zg be the impedance of
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+
Z11 Z22 Zkk
Z12 Z2k
1.0
1Ir
2Ir
kIr
Znn
Zkn
nIr1 2 k n
g pthe generator to be added.
Zg
1nI +r n+1
g
nn2n1n
n22221
n11211
Z000
0ZZZ
0ZZZ
0ZZZ
0.1
0.1
0.1
0.1
1n
n
2
1
I
I
I
I
+
r
r
r
r
=
The dimensionis (n+1).
101Training Course in Short Circuit Analysis
Type 2: Add a generatorto an old bus k
Let Zg be the +
-
1.0
w
Ir
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Let Zg be theimpedance of thegenerator to beadded.
Z11 Z22 Zkk
Z12 Z2k
1Ir
2Ir
kIr
Znn
nIr1 2 k n
The new current in impedance Zkk is (Ik+Iw). Thenew equations for buses 1 to n are
nn1wkk1212111 IZ...)II(Z...IZIZ0.1rrrrr
++++++=
nn2wkk2222121 IZ...)II(Z...IZIZ0.1rrrrr
++++++=
nnnwknk22n11n IZ...)II(Z...IZIZ0.1rrrrr
++++++=
Zg
102Training Course in Short Circuit Analysis
For the added generator loop, we get
wgnknwkkk22k11k IZIZ...)II(Z...IZIZ0
rrrrrr
+++++++=
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IZIZ...)II(Z...IZIZ0 +++++++=In matrix form, we get
wknkk2k1k
nknnnk2n1n
k2n2k22221
k1n1k11211
ZZZZZ
ZZZZZ
ZZZZZZZZZZ
0
0.1
0.10.1
=
w
n
2
1
I
I
II
r
r
r
r
where Zw=Zkk+Zg. The last row is eliminated usingKrons reduction. The dimension remains as n.
103Training Course in Short Circuit Analysis
Type 3: Add a branchfrom an old bus k to
a new bus +
-
Z Z
1.0
Z
-
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Z11 Z22 Zkk
Z12 Z2k
1Ir
2Ir
kIr
Znn
Zkn
nIr
1 2 k n
The new current in impedance Zkk is (Ik+In+1). Thenew equations for buses 1 to n are
nn11nkk1212111 IZ...)II(Z...IZIZ0.1rrrrr
++++++= +
nn21nkk2222121 IZ...)II(Z...IZIZ0.1rrrrr
++++++= +
nnn1nknk22n11n IZ...)II(Z...IZIZ0.1rrrrr
++++++= +
Zb
1nI +r
n+1
104Training Course in Short Circuit Analysis
For the new bus, we get
...)II(Z...IZIZ0.1 1nkkk22k11k +++++= +rrrr
IZIZ ++rr
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In matrix form, we get
wknkk2k1k
nknnnk2n1n
k2n2k22221
k1n1k11211
ZZZZZ
ZZZZZ
ZZZZZZZZZZ
0.1
0.1
0.10.1
=
1n
n
2
1
I
I
II
+
r
r
r
r
1nbnkn IZIZ +++
where Zw=Zkk+Zb. Krons reduction is not required.The dimension increases to (n+1).
105Training Course in Short Circuit Analysis
Type 4: Add abranch from an oldbus j to an old bus k
+
-
1.0 wIr
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Z11
Z22
Zjj
Z12 Z2j
1Ir
2Ir
kIr
Zkk
Zkn
nIr1 2
jk
Znn
jIr n
The new current in impedance Zjj is (Ij+Iw). Thenew current in impedance Zkk is (Ik-Iw). The newequations for buses 1 to n are
)II(Z...IZIZ0.1 wjj1212111rrrr
++++=
nn1wkk1 IZ...)II(Zrrr
+++
w
Zb
106Training Course in Short Circuit Analysis
)II(Z...IZIZ0.1 wjj2222121rrrr
++++=
nn2wkk2 IZ...)II(Z
rrr
+++
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)II(Z...IZIZ0.1 wjnj22n11nrrrr
++++=
)( +++
nnnwknk IZ...)II(Zrrr