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Certificate in
Power System Modeling and Analysis Training Course in
Short Circuit Analysis
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Course Outline 1. Analysis of Faulted Power System by Symmetrical Components 2. Bus Impedance Matrix Method 3. Short Circuit Analysis of Unbalanced Distribution Feeders 4. Short Circuit Studies
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Analysis of Faulted Power System by Symmetrical Components
Sources of Short Circuit Currents
Types of Fault
The Fault Point
ThreePhase Fault
SingleLinetoGround Fault
LinetoLine Fault
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Sources of Short Circuit Currents G
Utility
MV
Fault LV
Fault Current Contributors U. P. National Engineering Center National Electrification Administration
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Types of Fault Shunt Fault: Unintentional Connection between phases or between phase and ground. 1. Single LinetoGround Fault 2. LinetoLine Fault 3. Double LinetoGround Fault 4. Three Phase Fault Series Fault: Unintentional Opening of phase conductors Simultaneous Fault U. P. National Engineering Center National Electrification Administration
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Types of Fault
Three Phase
LinetoLine
Double LinetoGround
Single LinetoGround
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The Fault Point The system is assumed to be balanced, with regards to impedances, except at one point called the fault point. F a b
r Ia
Lineto r r r ground Va Vb Vc voltages
r Ib
r Ic
c Fault Currents
Ground
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The Fault Point Since we mentioned that various power system components behave/respond differently to the flow of the currents’ sequence components, it follows that the there will be a unique power system model for each of the sequence component. These are called the sequence networks. • PositiveSequence Network • NegativeSequence Network • ZeroSequence Network
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The Fault Point The Thevenin equivalent of the power system at the fault point is called the sequence network. Positive Sequence F1
r Ia1 Z + r r+ 1 V a1 V f = Vth 
N1
Negative Sequence
r Ia2
F2 +
Z2

r r r Va1 = Vth − Ia1Z1
r Va2
Zero Sequence
r Ia0
F0 +
Z0 

N2
r r Va2 = − Ia2Z2
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r Va0
N0
r r Va0 = − Ia0Z0
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Three Phase Fault On a balanced three phase system, the same magnitude of fault currents will flow in each phase of the network if a three phase fault occurs. Since faults currents are balanced, the faulted system can, therefore, be analyzed using the single phase representation.
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ThreePhase Fault a b c
r r r Va Vb Vc
r r r I a Z f Ib Z f Ic Z f r Zg Ig
Ground
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Sequence Network Interconnection: F1 +
r Z1 Va1 
+ 
F2
r I a1
Vf
+
Zf
r Va 2
r Ia2 Z2

N1
F0 +
r Va 0
r Ia0
Z0
N2
N0
Sequence currents
r I a1 =
Vf Z1 + Z f
r r I a0 = I a2 = 0
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Three Phase Fault Currents:
I a = I a 0 + I a1 + I a 2 =
Vf Z1 + Z f
I b = I a 0 + a I a1 + aI a 2 = 2
I c = I a 0 + aI a1 + a I a 2 = 2
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Example: Determine the fault current for a three phase bolted fault in each bus for the 4 bus system below. G
LINE 1
Line 5 2
Line 2
3 e Lin
Lin e
1
Line 4
3 4bus system
4
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 a subtransient reactance of 10%. Base Values: 100 MVA, 6.9 kV U. P. National Engineering Center National Electrification Administration
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Solution: Draw the impedance diagram
E
1.0 0.1 1
0.3
0.2 0.4
2
0.6
0.5
4
3
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R e d u c e t h e n e tw o r k
a) Fault @ Bus 4
X a = X 12 + X 23 = 0 .3 + 0 .6 = 0 .9
E
1.0
+ 0.1
Xb
If
1 0.3
( 0 .9 ) ( 0 .4 ) = 0 .9 + 0 .4 = 0 .2 7 6 9 2 3
0.2 0.4
2
0.6
0.5 3
X a X 13 = X a + X 13
4
Xc = X
b
+ X 34
= 0 .2 7 6 9 2 3 + 0 .5 = 0 .7 7 6 9 2 3
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Xd = = = Xequiv = =
Xc X14 Xc + X14 (0.776923) (0.2) 0.776923 + 0.2 0.159055 X gen + Xd 0.1
+
17
E 1.0 +
If
0.25905
0.159055
=
0.259055 100 x1000 Ibase = = 8367.64 A 1.0 3(6.9) If = 0.259055âˆ‚ = 3.860184 x 8367.64 If = 3.860184 p.u. = 32,300.63 A U. P. National Engineering Center National Electrification Administration
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b) Fault @ Bus 3
X a = X 23 + X 12 = 0.3 + 0.6 = 0.9
E
1.0
+ 0.1
Xb = X14 + X34
If
= 0.2 + 0.5 = 0.7
1 0.3
0.2 0.4
2
0.6
3
0.5
4
Xequiv = (XaXb ) X13 = 0.198425
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X=
Xgen +
= 0.1
+
19
Xequiv 0.198425
= 0.298425
E 1.0
1.0 If = 0.298425 = 3.350923 p.u.
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If
0.298425
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c) Fault @ Bus 2
Xa = X14 + X34 E
= 0.2 + 0.5 = 0.7
1.0
+ 0.1
If
1 0.3
a X X13 b X = a X + X13
0.2 0.4
2
0.6
3
0.5
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(0.7)( 0.4) = 0.7 + 0.4 = 0.254545 Competency Training & Certification Program in Electric Power Distribution System Engineering
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Xc = Xb + X23 = 0.254545 + 0.6 = 0.854545
E 1.0 If +
c X X12 d X = c X + X12
0.322047
(0.854545)( 0.3) = 0.854545 + 0.3 = 0.222047 X = Xgen + Xd = 0.322047 U. P. National Engineering Center National Electrification Administration
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d) Fault @ Bus 1
X = Xgen E
= 0.1
1.0
+ 0.1
If
1.0 If = 0.1 = 10.0 p.u.
1 0.3
0.2 0.4
2
0.6
3
0.5
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Example: A threephase fault occurs at point F. Assuming zero fault impedance, find the fault currents at fault point F. Determine the phase currents in the line and the generator. Assume Eg = 1.0 p.u. T1 G
Line
F T2 Open
G:
X1 = 40% X2 = 40% X0 = 20% T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%
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PositiveSequence Network: F1 j0.05 j0.4
r + Eg 
j0.15
r IA1L
r IA1
F1
j0.05
Open j0.6 +
r Ia1g
r + IA1 r VA1
1.0 

N1 N1
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Sequence Network Interconnection: F1 j0.6 +
r + IA1 r VA1
The sequence fault currents Zf
1.0 

N1
r I a1 = r I a2 = r I a0 =
Vf Z1 + Z f
=
The phase fault currents
Ia = Ib = Ic = U. P. National Engineering Center National Electrification Administration
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Single LinetoGround Fault Assuming the fault is in phase a, a b c
r r r Va Vb Vc
r Ia
Zf
r Ib
r Ic
Ground
r r Boundary Conditions: (1) V a = Z f I a r r (2) I = I = 0 b c U. P. National Engineering Center National Electrification Administration
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Transformation: From (2), we get
r r âˆ’1 I 012 = A I abc
r I a0 r 1 I a1 = 3 r Ia2
1 1 1 a 1 a2
1 a2 a
r Ia 1 r 0 = 3 Ira Ia 0
r Ia
r r r which means Ia0 = Ia1 = Ia2 =
r 1 I 3 a
From (1), we get
r r r r r r V a 0 + V a1 + V a 2 = Z f ( I a 0 + I a 1 + I a 2 )
or
r r r r V a 0 + V a1 + V a 2 = 3 Z f I a 0 U. P. National Engineering Center National Electrification Administration
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Sequence Network Interconnection: F1
r r Z1 I a1 + r Va1 Vf +


N1
F0
F2
r Ia2
+
r Va 2
r I a0
+
r Va0
Z2
Z0
3Zf


N0
N2
The sequence fault currents
r r r I a 0 = I a1 = I a 2 =
Vf Z 0 + Z1 + Z 2 + 3Z f
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SingleLinetoGround Phase Fault Currents:
I a = I a 0 + I a1 + I a 2 =
3V f
Z 1 + Z 2 + Z 0 + 3Z f
Ib = 0 Ic = 0
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Example: A single linetoground fault occurs at point F. Assuming zero fault impedance, find the phase currents in the line and the generator. Assume Eg = 1.0 p.u. T1 G
Line
F T2 Open
G: X1 = 40% X2 = 40% X0 = 20% T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35% Note: All reactances are in perunit of a common MVA base. U. P. National Engineering Center National Electrification Administration
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PositiveSequence Network: F1 j0.05
F1
j0.15 j0.05
j0.4
r Eg
Open j0.6 +
+
r + I a1 r Va1
1.0 


N1 N1
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NegativeSequence Network: F2 j0.05
j0.15 j0.05
Open
r Ia2
F2 +
j0.6
j0.4
r Va 2 
N2
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ZeroSequence Network: F0 j0.05
j0.35 j0.05
Open
r I a0
F0 +
j0.044
j0.2
r Va 0 
N0
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Sequence Network Interconnection: F1
F0
F2
r J0.6 IA1 +
1.0
r IA 2
r IA 0
J0.6
J0.044
N1
N0
N2
Sequence Fault Currents
r r r IA0 = IA1 = IA2 =
1.0 j(0.6 + 0.6 + 0.044)
= âˆ’ j0.804 p.u. U. P. National Engineering Center National Electrification Administration
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Phase Fault Currents
r r IA = 3IA0 = âˆ’ j2.411 p.u. r r IB = IC = 0
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LinetoLine Fault Assuming the fault is in phases b and c, a b c
r r r Va Vb Vc
r Ia
r Ib
r Ic Zf
Ground
r Boundary Conditions: (1) Ira = 0 r (2) I b = âˆ’ I c r r r (3) Vb âˆ’ Vc = I b Z f U. P. National Engineering Center National Electrification Administration
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Transformation: From (1) and (2), we get
r I 012 = A −1 I abc
r I a0 r 1 I a1 = 3 r Ia2 which means
1 1
1 a
1 a2
1 a2
a
0 r 1 Ib = 3 r − Ib
0 r 2 (a − a ) I b r 2 (a − a) I b
r Ia0 = 0
r r r 2 I a1 = − I a 2 = 13 (a − a ) I b = j
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From (3), we get
r r r 2 (Va 0 + a V a1 + a Va 2 ) r r r r r r 2 2 − (Va 0 + aVa1 + a Va 2 ) = ( I a 0 + a I a1 + aI a 2 ) Z f
r r r Since I a 0 = 0 and I a1 = − I a 2 , we get r r r 2 2 2 ( a − a )Va1 + ( a − a )Va 2 = ( a − a ) I a1 Z f or
r r r V a1 − V a 2 = I a1 Z f
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Sequence Network Interconnection: F1
r I a1
+
r Z1 Va1 
+ 
F2
Zf
Vf
r Ia0
r Ia2
+
r Va 2
N1
F0
Z2
Z0
N2
N0
The sequence fault currents
r I a0 = 0 r r I a1 = âˆ’ I a 2 =
Vf Z1 + Z 2 + Z f
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LinetoLine Phase Fault Currents:
Ia = 0 I b = I a 0 + a 2 I a1 + aI a 2 = 0 + a 2 I a1 + a (− I a1 ) = (a 2 − a ) I a1 = − jI a1
Ib = − j 3 Ic = + j 3
Vf Z1 + Z 2 + Z f Vf Z1 + Z 2 + Z f
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Example: A linetoline fault occurs at point F. Assuming zero fault impedance, find the fault currents at fault point F. Assume Eg = 1.0 p.u.
T1 G
Line
F T2 Open
G:
X1 = 40% X2 = 40% X0 = 20% T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%
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Sequence Network Interconnection: F1
F0
F2
r J0.6 IA1 +
1.0
r IA 2
r IA 0
J0.6
J0.044
N1
N2
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Sequence Fault Currents:
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Phase Fault Currents:
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DoubleLinetoGround Fault Assuming the fault is in phases b and c, a b c
r r r Va Vb Vc
r Ia
r Ib
Zf Zg
Ground
r Z f Ic r r Ib + Ic
r Boundary Conditions: (1) Ira = 0 r r (2) Vb = ( Z f + Z g ) I b + Z g I c r r r (3) Vc = ( Z f + Z g ) I c + Z g I b U. P. National Engineering Center National Electrification Administration
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Transformation: From (1), we get
r r r r I a = 0 = I a 0 + I a1 + I a 2
From
we get
r r r r 2 Vb = V a 0 + a V a 1 + a V a 2 r r r r 2 Vc = Va 0 + aVa1 + a Va 2 r r r r 2 2 Vb − Vc = ( a − a )Va1 + ( a − a )Va 2
Likewise, from
r r r r 2 I b = I a 0 + a I a1 + a I a 2 r r r r 2 I c = I a 0 + a I a1 + a I a 2
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we get
47
r r r r 2 2 I b − I c = ( a − a ) I a1 + ( a − a ) I a 2
From boundary conditions (2) and (3), we get
r r r r Vb − Vc = Z f ( I b − I c )
Substitution gives
r r 2 ( a − a )Va1 + ( a − a )Va 2 r r 2 2 = Z f [( a − a ) I a1 + ( a − a ) I a 2 ] 2
Simplifying, we get
r r r r Va 1 − Z f I a 1 = V a 2 − Z f I a 2
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From boundary conditions (2) and (3), we get
r r r r Vb + Vc = ( Z f + 2 Z g )( I b + I c )
We can also show
r r r r r Vb + Vc = 2Va 0 − Va1 − Va 2 r r r r r I b + I c = 2 I a 0 − I a1 − I a 2
Substitution gives
r r r r r r 2Va 0 − Va1 − Va 2 = Z f ( 2 I a 0 − I a1 − I a 2 ) r r r + 2 Z g ( 2 I a 0 − I a1 − I a 2 )
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Rearranging terms, we get
r r r r r 2Va 0 − 2 Z f I a 0 − 4 Z g I a 0 = Va1 − Z f I a1 r r r r + V a 2 − Z f I a 2 − 2 Z g ( I a1 + I a 2 )
Earlier, we got
r r r r Va1 − Z f I a1 = Va 2 − Z f I a 2 r r r I a1 + I a 2 = − I a 0
Substitution gives
r r r r r 2Va 0 − 2 Z f I a 0 − 6 Z g I a 0 = 2(Va1 − Z f I a1 ) r r r r Va 0 − (Z f + 3Z g )I a 0 = Va1 − Z f I a1
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Sequence Network Interconnection: Zf
r I a1
+ F1
r Z1 Va1 
+ 
Vf
Zf F2 +
r Va 2
Z2
r Ia0
F0 +
r Va 0
Z0


N1
Let
r Ia2
Zf+3Zg
N2
N0
Z 0 T = Z 0 + Z f + 3Z g Z1T = Z1 + Z f Z 2T = Z 2 + Z f U. P. National Engineering Center National Electrification Administration
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The sequence fault currents
r I a1 =
Vf Z 0T Z 2T Z1T + Z 0T + Z 2T
From current division, we get
r I a2 = −
r Z 0T I a1 Z 0T + Z 2T
From KCL, we get
r r r I a 0 = − I a1 − I a 2 U. P. National Engineering Center National Electrification Administration
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r I a0 = −
r Z 2T I a1 Z 0T + Z 2T
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DoubleLinetoGround Phase Fault Currents:
Ia = 0 I b = I a 0 + a 2 I a1 + aI a 2 =−j 3
V f (Z 0T − aZ 2T ) Z1T Z 2T + Z1T Z 0T + Z 2T Z 0T
I c = I a 0 + aI a1 + a 2 I a 2 =+j 3
(
V f Z 0T − a 2 Z 2T
)
Z1T Z 2T + Z1T Z 0T + Z 2T Z 0T
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Example: A doublelinetoground fault occurs at point F. Assuming zero fault impedance, find the fault currents at fault point F. Assume Eg = 1.0 p.u.
T1 G
Line
F T2 Open
G:
X1 = 40% X2 = 40% X0 = 20% T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%
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Sequence Network Interconnection: F1
F0
F2
r J0.6 IA1 +
1.0
r IA 2
r IA 0
J0.6
J0.044
N1
N2
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Sequence Fault Currents:
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Phase Fault Currents:
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Bus Impedance Matrix Method
Development of the Model
Rake Equivalent
Formation of Zbus
Analysis of Shunt Fault
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Development of the Model Observations on Manual Network Solution The procedure is straight forward, yet tedious and could be prone to handcalculation error. Is there a way for a computer to implement this methodology?
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Development of the Model Consider the threebus system shown below. Let us analyze the system for a threephase fault in any bus. 1
2
L1
G2
G1 L2 3
G1, G2 : L1 : L2 :
X1=X2=0.2 X1=X2=0.6 X1=X2=0.24
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X0=0.1 X0=1.2 X0=0.5
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PositiveSequence Network: 1
j0.2
r + EG1
j0.6
2
j0 .2 4

j0.2 3
+r

+
r EG
EG2

j0.2
j0.2 j0.6
Combine the sources and redraw. Assume EG = 1.0 per unit.
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1
j0.24
2
3
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For a threephase fault in bus 1 (or bus 2), we get the positivesequence impedance.
Z1 = j[0.2 //(0.2 + 0.6)] = j0.16
r EG 1 IF = = = âˆ’ j6.25 Z1 Z1 For a threephase fault in bus 3, we get
Z1 = j[0.24 + 0.2 //(0.2 + 0.6)] = j0.4
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Let us connect a fault switch to each bus. In order to simulate a threephase fault in any bus, close the fault switch in that bus. 
r EG
Next, use loop currents to + describe the circuit with all fault switches closed. j0.2 j0.2 4 Since there are four loops, j0.6 r we need to define four j0. I4 24 loop currents. 1
r I1
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3
r I3
2
r I2
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The loop equations are r r r loop 1: 1.0 = j 0.2( I1 + I 3 − I 4 )
r r loop 2: 1.0 = j 0.2( I 2 + I 4 ) r r r r loop 3: 1.0 = j 0.2( I1 + I 3 − I 4 ) + j 0.24 I 3 r r r r r r loop 4: 0 = j 0.2( I 2 + I 4 ) + j 0.6 I 4 + j 0.2( I 4 − I1 − I 3 )
or
1.0 0 .2 0 0.2 − 0.2 1.0 0 0.2 0 0 .2 =j 1.0 0 .2 0 0.44 − 0.2 0 − 0.2 0.2 − 0.2 1.0 U. P. National Engineering Center National Electrification Administration
r I1 r I2 r I3 r I4
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Current I4 is not a fault current. It can be eliminated using Kron’s reduction. We get
r r (1) V = Zbus I
where
1) Z(bus = Z1 − Z2Z4−1Z3
and
Z1 = j
0.2
0
0.2
0
0.2
0
0.2
0
0.44
Z3 = j[0.2 0.2 0.2 ] U. P. National Engineering Center National Electrification Administration
− 0.2 Z2 = j 0.2 − 0.2 Z4 = j[1.0]
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Substitution gives
0.16 0.04 0.16 1.0 1.0 = j 0.04 0.16 0.04 1.0 0.16 0.04 0.40 r r 1) V = Z(bus I
r I1 r I2 r I3
Note: (1) The equation can be used to analyze a threephase fault in any bus (one fault at a time). (1)
(2) Zbus is called the positivesequence busimpedance matrix, a complex symmetric matrix. U. P. National Engineering Center National Electrification Administration
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Rake Equivalent Consider the matrix voltage equation
1.0 1.0 = 1.0
r I1 r I2 r I3
Z11 Z12 Z13 Z12 Z22 Z23 Z13 Z23 Z33
1.0
Suppose we are asked to find a circuit that satisfies the matrix equation.
+
Z11
One possible equivalent r circuit is shown. This circuit I 1 is called a rakeequivalent. U. P. National Engineering Center National Electrification Administration

Z12 Z22
r I2
Z23 Z33 Z13 r
I3
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Consider again the threebus system. The circuit is described by the matrix equation
0.16 0.04 0.16 1.0 1.0 = j 0.04 0.16 0.04 1.0 0.16 0.04 0.40 The rake equivalent is shown. The diagonal elements of the matrix are j0.16 self impedances while the r offdiagonal elements are I1 mutual impedances.
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r I1 r I2 r I3

1.0 +
j0.04 j0.16
r I2
j0.04 j0.4 j0.16 r
I3
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For the threebus system, assume a fault in bus 3. The equation for bus 3 is
r r r 1.0 = j0.16 I1 + j0.04 I2 + j0.4I3

1.0
Since only bus 3 is faulted, I1=I2=0. We get
r 1.0 = j0.4I3
or
r I3 =
1 = − j2.5 j0.4
+
j0.16 + r
V1 
j0.04 j0.16 + r
j0.04 j0.4 j0.16
V2 
r I3
From KVL, we get the voltage in bus 1.
r r Z13 V1 = 1.0 − Z13 I3 = 1.0 − = 0.6 Z33
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Similarly from KVL, we get the voltage in bus 2.
r r Z23 V2 = 1.0 − Z23 I3 = 1.0 − = 0.9 Z33
Note: Once the voltages in all the buses are known, the current in any line can be calculated. In general, for a threephase fault in bus k of a system with n buses, the fault current is
r 1 Ik = Zkk
k=1,2,…n
The voltage in any bus j is given by
r Z jk Vj = 1.0 − Zkk
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The current in any line, which is connected from bus m to bus n, r canrbe found using
r Vm − Vn Imn = zmn
1
where zmn is the actual impedance of the line. j0.2 r + For example, the EG1 current in the line between buses 2 and 1 is r r
j0.6 j0 .2 j0.5 4
j2.0
3
j2.5
2
j0.2 +r
EG2

r V2 − V1 0.9 − 0.6 I21 = = = − j0.5 z21 j0.6
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Formation of Zbus Zbus can be built, one step at a time, by adding one branch at a time until the entire network is formed. The first branch to be added must be a generator impedance. This is necessary in order to establish the 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. U. P. National Engineering Center National Electrification Administration
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1.0
Z11
1.0
Z21 Z22
… …
Z2n Znn
r In
Z1n
Zn1
1.0
…
=
Z12
r I r1 I2
…
…
Assume that at the current stage, the dimension of Zbus is n.
72
Zn2

…
old Zbus
1.0 +
Z12 Z22
Z11
r I1
1
r I2
Zkn
Z2k
Znn
Zkk
2
r Ik
k
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r In
n
Let us examine each category in the addition of a new branch.
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Type 1: Add a generator to a new bus 
Let Zg be the impedance of the generator to be added.
1.0 +
Z12 Z22
Z11
r I1
1
1.0
r I2
Zkn
Z2k
Znn
Zkk
2
Z11
r Ik
k
Z12 … Z1n Z …Z
r In 0
Zg
n
r I1 r I2 r In
Z21 0 22 2n 1.0 = Zn1 Zn2 … Znn 0 r 0 0 0 Zg In + 1 1.0 1.0
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r In +1
n+1
The dimension is (n+1).
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Type 2: Add a generator to an old bus k Let Zg be the impedance of the generator to be Z11 added.
r I1

r Iw
1.0 +
Z12 Z22 1
r I2
Z2k
Zg
Zkk
2
r Ik
k
r In
Znn n
The new current in impedance Zkk is (Ik+Iw). The new equations for buses 1 to n are
r r r r r 1.0 = Z11 I1 + Z12 I2 + ... + Z1k (Ik + Iw ) + ... + Z1n In r r r r r 1.0 = Z21 I1 + Z22 I2 + ... + Z2k (Ik + Iw ) + ... + Z2n In r r r r r 1.0 = Zn1I1 + Zn2 I2 + ... + Znk (Ik + Iw ) + ... + Znn In U. P. National Engineering Center National Electrification Administration
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For the added generator loop, we get
r r r r r r 0 = Zk1 I1 + Zk 2 I2 + ... + Zkk (Ik + Iw ) + ... + Zkn In + Zg Iw In matrix form, we get
Z11 Z12… Z1k … Z1n Z1k Z21 Z22… Z2k … Z2n Z2k
1.0 1.0
…
Zn1 Zk1
Zn2… Znk … Znn Znk Z … Z …Z Z k2
kk
kn
w
…
…
… 1.0 0
=
r I1 r I2 r In r Iw
where Zw=Zkk+Zg. The last row is eliminated using Kron’s reduction. The dimension remains as n. U. P. National Engineering Center National Electrification Administration
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Type 3: Add a branch from an old bus k to a new bus
1.0 +
Z12 Z22
Z11
r I1

1
r I2
Z2k Zkk 2
r Ik
Zkn Zb k
r In
Znn n
r In +1
n+1
The new current in impedance Zkk is (Ik+In+1). The new equations for buses 1 to n are
r r r r r 1.0 = Z11 I1 + Z12 I2 + ... + Z1k (Ik + In +1 ) + ... + Z1n In r r r r r 1.0 = Z21 I1 + Z22 I2 + ... + Z2k (Ik + In +1 ) + ... + Z2n In r r r r r 1.0 = Zn1I1 + Zn2 I2 + ... + Znk (Ik + In +1 ) + ... + Znn In U. P. National Engineering Center National Electrification Administration
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For the new rbus, wer get
r r 1.0 = Zk1I1 + Zk 2 I2 + ... + Zkk (Ik + In+1 ) + ... r r + Zkn In + Zb In +1
In matrix form, we get
Z11 Z12… Z1k … Z1n Z1k Z21 Z22… Z2k … Z2n Z2k
1.0 1.0
Zn1 Zn2… Znk … Znn Znk Zk1 Zk 2… Zkk … Zkn Zw
…
…
…
… 1.0 1.0
=
r I1 r I2 r In r In+1
where Zw=Zkk+Zb. Kron’s reduction is not required. The dimension increases to (n+1). U. P. National Engineering Center National Electrification Administration
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Type 4: Add a branch from an old bus j to an old bus k

r Iw
1.0 +
Z12 Z22
Z11
r I1
1
r I2
Zkn
Z2j Zjj 2
r Ij
Zb j
r Ik
Zkk k
r In
Znn n
The new current in impedance Zjj is (Ij+Iw). The new current in impedance Zkk is (IkIw). The new equations for buses 1 to n are
r r r r 1.0 = Z11 I1 + Z12 I2 + ... + Z1 j(Ij + Iw ) r r r + Z1k (Ik âˆ’ Iw ) + ... + Z1n In U. P. National Engineering Center National Electrification Administration
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r r r r 1.0 = Z21 I1 + Z22 I2 + ... + Z2 j(Ij + Iw ) r r r + Z2k (Ik − Iw ) + ... + Z2n In r r r r 1.0 = Zn1I1 + Zn2 I2 + ... + Znj(Ij + Iw ) r r r + Znk (Ik − Iw ) + ... + Znn In For the added loop, we get
r r r r r r 0 = Z j1 I1 + Z j2 I2 + ... + Z jj(Ij + Iw ) + Z jk (Ik − Iw ) r r r r + ... + Z jn In + Zb Iw − [Zk1I1 + Zk 2 I2 + ... r r r r r + Zkj(Ij + Iw ) + Zkk (Ik − Iw ) + ... + Zkn In ] U. P. National Engineering Center National Electrification Administration
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In matrix form, we get
1.0
Z11
Z12
…
1.0
Z21
Z22
…
Z2n
Z2j − Z2k
…
0
Z1j − Z1k
…
…
1.0
=
Z1n
Zn1
r I1 r I2
Zn2
…
Znn
Zj1 − Zk1 Zj2 − Zk2 … Zjn − Zkn
Znj − Znk Zv
r In r Iw
where Zv=Zjj+Zkk2Zjk+Zb. The last row is eliminated using Kron’s reduction. The dimension remains as n.
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Example: For the network shown, use the stepbystep building algorithm to form the bus impedance matrix. 2 1 j0.6 Step 1. Add generator j0 .2 G1 to bus 1. j0.2 4 j0.2 1
Xbus =
1
[0.2]
+
3
1.0
+
1.0 

Step 2. Add generator G2 to bus 2. 1 1
Xbus =
2
2
0 .2 0 0 0 .2
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Step 3. Add the line from bus 1 to bus 2.
Xnew =
1
2
*
1
0.2
0
0.2
2
0
0.2
− 0.2
*
0.2 − 0.2
1.0
Apply Kron’s reduction to eliminate the last row and column. We get −1 4
X 2 X X3 =
0.2 [0.2 0.2] − 0.2
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X 2 X X3 =
0.04 − 0 .04
83
− 0 .04 0.04
We get
1 −1 4
Xbus = X1 − X2X X3 =
1 2
2
0 .16 0 .04 0 .04 0 .16
Step 4. Finally, add the line from bus 1 to bus 3. 1
Xbus =
2
3
1
0.16 0.04 0.16
2
0.04 0.16 0.04
3
0.16 0.04
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0.4
No Kron reduction is required.
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Example: Determine the positivesequence busimpedance matrix for the fourbus test system shown. 1 T 2 3 L1
G1
L3
L2
T: G1: G2: L1: L2: L3:
X=0.08 X1=0.40 X1=0.50 X1=0.40 X1=0.30 X1=0.20
G2
4
X2=0.40 X2=0.50 X2=0.40 X2=0.30 X2=0.20
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X0=0.15 X0=0.25 X0=0.80 X0=0.60 X0=0.40
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Positivesequence network 1. Add G1 to bus 1. 1
[0.4]
j0.4 2. Add the transformer + 1.0 from bus 1 to bus 2. 1 2 1
Xbus =
2
3. Add the line from X bus = bus 2 to bus 3.
2
1
j0.4
N1
0 .4 0 .4 0 .4 0 .48
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j0.2
j0.3 j0.08
1
Xbus =
4
3
j0.5 + 1.0 
1
2
3
1
0.4
0.4
0.4
2
0.4 0.48 0.48
3
0.4 0.48 0.88
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Step 4. Add generator G2 to bus 3.
Xnew =
1
2
3
*
1
0.4
0.4
0.4
0.4
2
0.4 0.48 0.48 0.48
3
0.4 0.48 0.88 0.88 0.4 0.48 0.88 1.38
*
Apply Kronâ€™s reduction.
X2X 4âˆ’1X3 =
1 1.38
0.4 0.48 [0.4 0.48 0.88] 0.88
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We get
0.1159 0.1391 0.2551 X2X 4−1X3 = 0.1391 0.1670 0.3061 0.2551 0.3061 0.5612 The new bus impedance matrix is
Xbus = X1 − X2X 4−1X3 1
Xbus =
2
3
1
0.2841 0.2609 0.1449
2
0.2609 0.3130 0.1739
3
0.1449 0.1739 0.3188
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Step 5. Add the line from bus 2 to bus 4.
Xbus =
1
2
3
4
1
0 .2841
0 .2609
0 .1449
0 .2609
2
0 .2609 0 .1449 0 .2609
0 .3130 0 .1739 0 .3130
0 .1739 0 .3188 0 .1739
0 .3130 0 .1739 0 .6130
3 4
Step 6. Add the line from bus 3 to bus 4. 1
Xnew=
2
3
4
* 0.1159
1
0.2841 0.2609
0.1449
0.2609
2 3
0.2609 0.3130 0.1449 0.1739
0.1739 0.3188
0.3130 0.1391 0.1739 âˆ’ 0.1449
4
0.2609 0.3130
0.1739
0.6130
* 0.1159 0.1391 âˆ’ 0.1449 0.4391 U. P. National Engineering Center National Electrification Administration
0.4391 0.784
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Apply Kronâ€™s reduction. We get 1
X
(1) bus
=
2
3
4
1
0.2669 0.2403 0.1664 0.1959
2 3
0.2403 0.2884 0.1996 0.2351 0.1664 0.1996 0.2920 0.2551
4
0.1959 0.2351 0.2551 0.3671
Note: This is the positivesequence busimpedance matrix for the fourbus test system.
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Negative and ZeroSequence Zbus The same stepbystep algorithm can be applied to build the negativesequence and zerosequence bus impedance matrices. The first branch to be added must be a generator impedance. This is necessary in order to establish the reference bus. The negativesequence and zerosequence busimpedance matrices can also be described by a rake equivalent circuit.
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Example: Find the zerosequence busimpedance matrix for the fourbus test system. 4
Zerosequence network
1. Add G1 to bus 1.
1
1
Xbus =
1
2
[0.15] j0.15
1
Xbus =
2
j0.8
3
j0.25
2. Add the transformer from bus 1 to bus 2. 1
j0.4
j0.6 j0.08
N0 2
0 .15 0 0 0 .08
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Note: The impedance is actually connected from bus 2 to the reference bus.
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3. Add the line from bus 2 to bus 3.
Xbus =
1
2
3
1
0.15
0
0
2
0
0.08 0.08
3
0
0.08 0.88
Step 4. Add generator G2 to bus 3. 1
Xnew =
2
3
*
0
0
0
1
0.15
2
0
0.08 0.08 0.08
3
0 0
0.08 0.88 0.88 0.08 0.88 1.13
*
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Apply Kronâ€™s reduction. We get
0
0
0
X2X 4âˆ’1X3 = 0 0.0057 0.0623 0 0.0623 0.6853 The new bus impedance matrix is
Xbus =
1
2
3
1
0.15
0
0
2
0
0.0743 0.0177
3
0
0.0177 0.1947
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Step 5. Add the line from bus 2 to bus 4.
Xbus =
1
2
3
4
1
0 .15
0
0
0
2
0 0 0
0 .0743 0 .0177 0 .0743
0 .0177 0 .1947 0 .0177
0 .0743 0 .0177 0 .6743
3 4
Step 6. Add the line from bus 3 to bus 4.
Xnew=
1
2
3
4
*
1
0.15
0
0
0
0
2 3
0 0
0.0743 0.0177 0.0177 0.1946
0.0743 0.0566 0.0177 âˆ’ 0.177
4
0
0.0743 0.0177
0.6743 0.6566
*
0
0.0566 âˆ’ 0.177 0.6566 1.2336
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PositiveSequence Zbus 1) 1) … Z(21 Z(22 Z(21n)
=
(1) bus
Z
…
The positivesequence busimpedance matrix describes the positivesequence network.
(1) (1) … Z11 Z12 Z1(1n)
1) Z(n11) Z(n12) … Z(nn
 N1
1.0 + (1) 12
Z (1) 11
Z
1
(1) 2k
Z
Z (1) 22
Z
2
(1) kn
(1) kk
Z
k
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(1) nn
Z
Rake Equivalent
n
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Apply Kronâ€™s reduction. We get 1
X
(0) bus
=
2
0
3
0
4
1
0.15
0
2 3
0 0
0.0717 0.0258 0.0442 0.0258 0.1693 0.1119
4
0
0.0442 0.1119 0.3248
Note: This is the zerosequence busimpedance matrix for the fourbus test system.
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NegativeSequence Zbus (2) (2) … Z11 Z12 Z1(2n)
The negativesequence busimpedance matrix (2) Z describes the negativebus = sequence network.
Z(212) Z(222) … Z(22n)
… Z(n21) Z(n22) … Z(nn2)
N2
(2) 11
Z
1
(2) Z12 Z(222)
2
Z(22k) Z(kk2) k
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Z(kn2) Z(nn2)
Rake Equivalent
n
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ZeroSequence Zbus Z(210) Z(220) … Z(20n)
…
The zerosequence busimpedance matrix (0) Zbus = describes the zerosequence network.
(0 ) (0 ) … Z11 Z12 Z1(0n)
Z(n01) Z(n02) … Z(nn0)
N0 ( 0) 12
Z ( 0) 11
Z
1
(0) 2k
Z (0) 22
Z
2
(0) kn
Z (0) kk
Z
k
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(0) nn
Z
Rake Equivalent
n
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Analysis of Shunt Faults The busimpedance matrices can be used for the analysis of the following shunt faults: 1. ThreePhase Fault 2. LinetoLine Fault 3. Single LinetoGround Fault 4. Double LinetoGround Fault Since the busimpedance matrix is a representation of the power system as seen from the buses, only bus faults can be investigated.
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Threephase Fault at Bus k The fault current is
r 1 Ik = (1) Zkk The voltage at any bus is
N1
(1) Z11
1) Z(22
Z(kk1)
Z(nn1)
1
2
k
n
r Z jk Vj = 1.0 âˆ’ Zkk
r The current in any line is Imn
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r r Vm âˆ’ Vn = zmn
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Example: Consider a threephase fault at bus 4 of the fourbus test system. Find all line currents.
The positivesequence busimpedance matrix is 1
X
(1) bus
=
2
3
4
1
0.2669 0.2403 0.1664 0.1959
2 3
0.2403 0.2884 0.1996 0.2351 0.1664 0.1996 0.2920 0.2551
4
0.1959 0.2351 0.2551 0.3671
The fault current is
r 1 1 IF = (1) = = âˆ’ j2.7241 Z44 j0.3671
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The bus voltages are
r Z jk Vj = 1.0 − Zkk
j=1,2,…n
r 0.1959 V1 = 1 − = 0.4663 0.3671 r 0.2351 V2 = 1 − = 0.3595 0.3671 r 0.2551 V3 = 1 − = 0.3051 0.3671 r V4 = 0 U. P. National Engineering Center National Electrification Administration
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r The line currents are given by Imn
r r Vm − Vn = zmn
103
r 1 − 0.4663 IG1 = = − j1.3344 j0.4 r 1 − 0.3051 IG2 = = − j1.3897 j0.5
r 0.4663 − 0.3595 I12 = = − j1.3342 j0.08 r 0.3595 − 0.3051 I23 = = − j0.1360 j0.4 U. P. National Engineering Center National Electrification Administration
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r 0.3595 − 0 = − j1.1984 I24 = j0.3 r 0.3051 − 0 I34 = = − j1.5257 j0.2 r 4 r IF I34 r I24 j0.2 j0.3 1 j0.08
j0.4 + 1.0 
r r IG1 I12
j0.4
r I23
2
N1
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3
r IG2
j0.5 + 1.0 
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LinetoLine Fault at Bus k N2
N1
(1) 11
Z
1
(1) 22
Z
(1) kk
Z
2
k
Sequence Fault Currents r
Ia0 = 0 r r Ia1 = − Ia2 =
Z(kk1)
(1) nn
Z
r Ia1 n
(2) Z11
Z(222)
Z(kk2)
1
2
k
Z(nn2)
r n Ia 2
Sequence Voltages at bus j
1 + Z(kk2)
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r Va0 = 0 r r (1) Va1 = 1 − Ia1Z jk r r (2) Va2 = − Ia2Z jk
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Example: Consider a linetoline fault at bus 4 of the fourbus test system. Find the phase currents in lines L2 and L3.
The positivesequence busimpedance matrices is 1
X
(1) bus
=
2
3
4
1
0.2669 0.2403 0.1664 0.1959
2 3
0.2403 0.2884 0.1996 0.2351 0.1664 0.1996 0.2920 0.2551
4
0.1959 0.2351 0.2551 0.3671
For this power system, Xbus = Xbus (1)
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(2)
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The sequence fault currents are
r Ia1
r Ia0 = 0 r 1 1 = − Ia2 = (1) = = − j1.362 (2) Z44 + Z44 j2(0.3671)
The sequence voltages in bus 4 are
r Va0 − 4 = 0 r r (1) Va1− 4 = 1 − Ia1Z44
r Va2 − 4
= 1 − (− j1.362)( j0.3671) = 0.5 r (2) = − Ia2Z44 = 0.5
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The sequence voltages in bus 2 are
r Va0 −2 = 0 r r (1) Va1−2 = 1 − Ia1Z24 = 0.6798 r r (2) Va2 −2 = − Ia2Z24 = 0.3202
The sequence voltages in bus 3 are
r Va0 −3 = 0 r r (1) Va1−3 = 1 − Ia1Z34 = 0.6526 r r (2) Va2 −3 = − Ia2Z34 = 0.3474
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The sequence currents in line L3 are r
Ia0 −L 3 = 0 r 0.653 − 0.5 Ia1−L 3 = = − j0.7628 j0.2 r 0.347 − 0.5 Ia2 −L 3 = = j0.7628 j0.2
The phase currents in line L3 are
r r r r Ia −L 3 = Ia0 −L 3 + Ia1−L 3 + Ia2 −L 3 = 0 r r r r 2 Ib −L 3 = Ia0 −L 3 + a Ia1−L 3 + a Ia2 −L 3 = −1.3213 r r r r Ic −L 3 = Ia0 −L 3 + a Ia1−L 3 + a2 Ia2 −L 3 = 1.3213
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The sequence currents in line L2 are r
Ia0 −L 2 = 0 r 0.68 − 0.5 Ia1−L 2 = = − j0.5992 j0.3 r 0.32 − 0.5 Ia2 −L 2 = = j0.5992 j0.3
The phase currents in line L2 are
r r r r Ia −L 2 = Ia0 −L 2 + Ia1−L 2 + Ia2 −L 2 = 0 r r r r 2 Ib −L 2 = Ia0 −L 2 + a Ia1−L 2 + a Ia2 −L 2 = −1.0378 r r r r 2 Ic −L 2 = Ia0 −L 2 + a Ia1−L 2 + a Ia2 −L 2 = 1.0378
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N1
SLG Fault at Bus k Sequence Fault Currents
r r r (1) Ia0 = Ia1 = Ia2 Z11 1 1 = ( 0) Zkk + Z(kk1) + Z(kk2) Sequence Voltages (2) Z at bus j 11
r r (0) Va0 = − Ia0Z jk r r (1) Va1 = 1 − Ia1Z jk r r (2) Va2 = − Ia2Z jk
1
1) Z(22
Z(kk1)
2
k
N2
Z(222)
Z(kk2)
2
k
N0 (0 ) Z11
Z(220)
Z(kk0)
1
2
k
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1) Z(nn
r Ia 1
n
Z(nn2)
r Ia 2
n
Z(nn0)
r Ia 0
n
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Example: Consider a single linetoground fault at bus 4 of the fourbus test system. Find the phase currents in lines L2 and L3.
The sequence r r fault r currents are
Ia0 = Ia1 = Ia2 1 = ( 0) = − j0.9443 (1) (2) Z44 + Z44 + Z44
The sequence voltages in bus 4 are
r r (0) Va0 − 4 = − Ia0Z44 = −0.3067 r r (1) Va1− 4 = 1 − Ia1Z44 = 0.6534 r r (2) Va2 − 4 = − Ia2Z44 = −0.3466
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The sequence voltages in bus 2 are
r r (0) Va0 −2 = − Ia0Z24 = −0.0417 r r (1) Va1−2 = 1 − Ia1Z24 = 0.778 r r (2) Va2 −2 = − Ia2Z24 = −0.222
The sequence voltages in bus 3 are
r r (0) Va0 −3 = − Ia0Z34 = −0.1057 r r (1) Va1−3 = 1 − Ia1Z34 = 0.7591 r r (2) Va2 −3 = − Ia2Z34 = −0.2409
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The sequence currents in line L2 are
r − 0.0417 + 0.3067 Ia0 −L 2 = = − j0.4417 j0.6 r 0.778 − 0.6534 Ia1−L 2 = = − j0.4154 j0.3 r − 0.222 + 0.3466 Ia2 −L 2 = = − j0.4154 j0.3
Therphase currents r rin line L2 r are
Ia −L 2 = Ia0 −L 2 + Ia1−L 2 + Ia2 −L 2 = − j1.2725 r r r r 2 Ib −L 2 = Ia0 −L 2 + a Ia1−L 2 + a Ia2 −L 2 = − j0.0262 r r r r 2 Ic −L 2 = Ia0 −L 2 + a Ia1−L 2 + a Ia2 −L 2 = − j0.0262
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The sequence currents in line L3 are
r − 0.1057 + 0.3067 Ia0 −L 3 = = − j0.5026 j0.4 r 0.7591 − 0.6534 Ia1−L 3 = = − j0.5289 j0.2 r − 0.2409 + 0.3466 Ia2 −L 3 = = − j0.5289 j0.2
Therphase currents r rin line L3 r are
Ia −L 3 = Ia0 −L 3 + Ia1−L 3 + Ia2 −L 3 = − j1.5603 r r r r Ib −L 3 = Ia0 −L 3 + a2 Ia1−L 3 + a Ia2 −L 3 = j0.0262 r r r r Ic −L 3 = Ia0 −L 3 + a Ia1−L 3 + a2 Ia2 −L 3 = j0.0262
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Double LinetoGround Fault at Bus k N2
N1
(1) Z(kk1) Z11
1
k
1) Z(nn
(2 ) Z(kk2) Z11
r n Ia 1
1
k
N0
Z(nn2)
r n Ia 2
(0 ) Z(kk0) Z11
1
k
Z(nn0)
r n Ia 0
Sequence Fault Currents
r Ia1 =
(1) kk
Z
1 (2) (0) + (Zkk // Zkk )
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r Ia2 = −
r Z(kk0) I (0) (2) a1 Zkk + Zkk
r Ia0 = −
r Z I ( 0) (2) a1 Zkk + Zkk (2) kk
Sequence Voltages at bus j
r r (0) Va0 = − Ia0Z jk r r (1) Va1 = 1 − Ia1Z jk r r (2) Va2 = − Ia2Z jk
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Example: Consider a double linetoground fault at bus 4 of the fourbus test system. Find the phase currents in lines L2 and L3.
Sequence Fault Currents
r Ia1 =
r Ia2 r Ia0
1 = − j1.8538 (1) (2) ( 0) Zkk + (Zkk // Zkk ) r Z(kk0) = − (0) I = j0.8703 (2) a1 Zkk + Zkk r r = − Ia1 − Ia2 = j0.9835
The sequence voltages in bus 4 are
r r r r (0) Va0 − 4 = Va1− 4 = Va2 − 4 = − Ia0Z44 = 0.3195 U. P. National Engineering Center National Electrification Administration
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The sequence voltages in bus 2 are
r r (0) Va0 −2 = − Ia0Z24 = 0.0435 r r (1) Va1−2 = 1 − Ia1Z24 = 0.5641 r r (2) Va2 −2 = − Ia2Z24 = 0.2046
The sequence voltages in bus 3 are
r r (0) Va0 −3 = − Ia0Z34 = 0.1101 r r (1) Va1−3 = 1 − Ia1Z34 = 0.5271 r r (2) Va2 −3 = − Ia2Z34 = 0.222
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The sequence currents in line L2 are
r 0.0435 − 0.3195 Ia0 −L 2 = = j0.46 j0.6 r 0.5641 − 0.3195 Ia1−L 2 = = − j0.8155 j0.3 r 0.2046 − 0.3195 Ia2 −L 2 = = j0.3828 j0.3
The phase currents inr line L2rare r r
Ia −L 2 = Ia0 −L 2 + Ia1−L 2 + Ia2 −L 2 = j0.0273 r Ib −L 2 = −1.0378 + j0.6764 r Ic −L 2 = 1.0378 + j0.6764
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The sequence currents in line L3 are
r 0.1101 − 0.3195 Ia0 −L 3 = = j0.5235 j0.4 r 0.5271 − 0.3195 Ia1−L 3 = = − j1.0383 j0.2 r 0.222 − 0.3195 Ia2 −L 3 = = j0.4874 j0.2
Therphase rcurrentsr in line L3 r are
Ia −L 3 = Ia0 −L 3 + Ia1−L 3 + Ia2 −L 3 = − j0.0273 r I = −1.3213 + j0.799 rb −L 3 Ic −L 3 = 1.3213 + j0.799
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Short Circuit Analysis of Unbalanced Distribution System
Thevenin Equivalent Circuit
ThreePhase Line Segment Model
Transformer Generalized Matrices
Analysis of Faulted Unbalanced Feeder
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123
Thevenin Equivalent Circuit Equivalent system: I ABC
Iabc
ABC Z sys
Source ABC ELN
abc VLN
ABC VLN
Thevenin equivalent circuit @ secondary bus: Iabc Eth
Z th abc VLN
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Thevenin Equivalent Circuit Primary transformer equivalent linetoneutral voltages ABC ABC ABC ABC ABC VLN = ELN − Z sys ⋅ IABC = ELN − Z sys ⋅ dt ⋅ Iabc
Secondary linetoneutral voltage: abc ABC VLN = A t ⋅ VLN − Bt ⋅ Iabc
Substituting,
{
}
abc ABC ABC VLN = At ⋅ ELN − Z sys ⋅ dt ⋅ Iabc − Bt ⋅ Iabc
Thevenin equivalent voltages & impedances: ABC abc Eth = At ⋅ ELN abc ABC Zth = At ⋅ Z sys ⋅ dt + Bt U. P. National Engineering Center National Electrification Administration
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ThreePhase Line Segment Model Node n n Vag n Vbg n Vcg
I na I nb I nc
Iabc c,n
a I line b I line c I line
a Im
zaa zbb
zab
zcc
zbc
½ Yabc
b Im
zca
c Im
Iabc c,m
Node m m Vag m Vbg m Vcg
½ Yabc
Voltages & currents at node n in terms of the voltages & currents at node m: abc abc abc VLG = a ⋅ V + b ⋅ I m ,n LG,m
abc abc Iabc d I = c ⋅ VLG + ⋅ n m ,m
a = U + 21 ⋅ Z abc ⋅ Yabc
c = Yabc + 41 ⋅ Yabc ⋅ Z abc ⋅ Yabc
b = Z abc
d = U + 21 ⋅ Z abc ⋅ Yabc
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ThreePhase Line Segment Model Voltages & currents at node m in terms of the voltages & currents at node n: abc abc abc VLG = a ⋅ V − b ⋅ I n LG,n ,m abc abc Iabc = − c ⋅ V + d ⋅ I m n LG,n
Voltages at node m as a function of voltages at node n and currents entering node m: abc abc abc VLG = A ⋅ V − B ⋅ I m LG,m ,m
A = a −1 B = a −1 ⋅ b
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Transformer Generalized Matrices Generalized threephase transformer bank: Ia IA V AN Van H1 X1 IB Ib H2 VBN Vbn X2 IC Ic VCN H3 X3 Vcn IN In H0
X0
ABC abc VLN = at ⋅ VLN + bt ⋅ Iabc abc I ABC = c t ⋅ VLN + dt ⋅ Iabc abc ABC VLN = A t ⋅ VLN − Bt ⋅ Iabc
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Transformer Generalized Matrices Delta – Grounded Wye StepDown Connection VLL, rated high side nt = VLN , rated low side ⎡0 2 1⎤ − nt ⎢ ⋅ 1 0 2⎥ at = ⎥ 3 ⎢ ⎢⎣2 1 0⎥⎦
⎡ 0 − nt ⎢ a ⋅ ⎢ Zt bt = 3 ⎢2Z ta ⎣
2Z tb 0 Z tb
Z tc ⎤ ⎥ 2Ztc ⎥ 0 ⎥⎦
Z ta , Ztb , Z tc are referred to the low  voltage side ⎡0 0 0 ⎤ c t = ⎢0 0 0 ⎥ ⎢ ⎥ ⎣⎢0 0 0⎥⎦ U. P. National Engineering Center National Electrification Administration
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Transformer Generalized Matrices Delta – Grounded Wye StepDown Connection ⎡ 1 0 − 1⎤ 1 ⎢ At = ⋅ − 1 1 0 ⎥ ⎥ nt ⎢ ⎢⎣ 0 − 1 1 ⎥⎦
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⎡Zta ⎢ Bt = ⎢ 0 ⎢ ⎣0
0 Z tb 0
0⎤ ⎥ 0⎥ Z tc ⎥⎦
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Transformer Generalized Matrices Ungrounded Wye – Delta StepDown Connection VLN , rated high side nt = VLL, rated low side ⎡ 1 −1 0 ⎤ a t = nt ⋅ ⎢ 0 1 − 1⎥ ⎢ ⎥ 1 ⎥⎦ ⎢⎣ − 1 0
⎡ Z tab n ⎢ bt = t ⋅ ⎢ Ztbc 3 ⎢ − 2Ztca ⎣
− Ztab 2Z tbc − Z tca
0⎤ ⎥ 0⎥ 0⎥⎦
Z tab , Z tbc , Ztca are referred to the low  voltage side ⎡0 0 0 ⎤ c t = ⎢0 0 0 ⎥ ⎢ ⎥ ⎢⎣0 0 0⎥⎦ U. P. National Engineering Center National Electrification Administration
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Transformer Generalized Matrices Ungrounded Wye – Delta StepDown Connection ⎡2 1 0 ⎤ 1 ⎢ ⋅ 0 2 1⎥ At = ⎥ 3nt ⎢ ⎢⎣ 1 0 2⎥⎦ ⎡ 2Ztab + Ztbc 1 ⎢ bc Bt = ⋅ ⎢2Zt − 2Ztca 9 ⎢ Ztab − 4Z tca ⎣
2Z tbc − 2Z tab 4Z tbc − Ztca − Z tab − 2Z tca
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0⎤ ⎥ 0⎥ 0⎥⎦
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Transformer Generalized Matrices Grounded Wye – Gounded Wye Connection VLN , rated high side nt = VLN , rated low side ⎡ 1 0 0⎤ a t = nt ⋅ ⎢0 1 0⎥ ⎢ ⎥ ⎢⎣0 0 1⎥⎦
⎡Z ta ⎢ bt = nt ⋅ ⎢ 0 ⎢0 ⎣
0 Z tb 0
0⎤ ⎥ 0⎥ Ztc ⎥⎦
Z ta , Ztb , Z tc are referred to the low  voltage side ⎡0 0 0 ⎤ c t = ⎢0 0 0 ⎥ ⎢ ⎥ ⎢⎣0 0 0⎥⎦ U. P. National Engineering Center National Electrification Administration
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Transformer Generalized Matrices Grounded Wye – Grounded Wye Connection ⎡ 1 0 0⎤ 1 ⎢ A t = ⋅ 0 1 0⎥ ⎥ nt ⎢ ⎢⎣0 0 1⎥⎦
⎡Z ta ⎢ Bt = ⎢ 0 ⎢ ⎣0
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0 Z tb 0
0⎤ ⎥ 0⎥ Ztc ⎥⎦
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Transformer Generalized Matrices Delta – Delta Connection VLL, rated high side nt = VLL, rated low side ⎡ 2 − 1 − 1⎤ nt ⎢ a t = ⋅ − 1 2 − 1⎥ ⎥ 3 ⎢ ⎢⎣ − 1 − 1 2 ⎥⎦ ⎡ 1 0 0⎤ AV = nt ⋅ ⎢0 1 0⎥ ⎢ ⎥ ⎢⎣0 0 1⎥⎦
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⎡2 1 0 ⎤ 1 ⎢ W = ⋅ 0 2 1⎥ ⎥ 3 ⎢ ⎢⎣ 1 0 2⎥⎦ ⎡Z tab 0 ⎢ abc Z t = ⎢ 0 Z tbc ⎢ 0 0 ⎣
0 ⎤ ⎥ 0 ⎥ Ztca ⎥⎦
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Transformer Generalized Matrices Delta – Delta Connection G1 =
Z tab
1 + Z tbc + Z tca
⎡ Z tca ⎢ ⋅⎢ Z tca ⎢ − Ztab − Z tbc ⎣
− Ztbc Z tab + Ztca − Ztbc
0⎤ ⎥ 0⎥ 0⎥⎦
bt = W ⋅ AV ⋅ Ztabc ⋅ G1 ⎡ 1 0 0⎤ 1 dt = ⋅ ⎢0 1 0⎥ ⎥ nt ⎢ ⎢⎣0 0 1⎥⎦
⎡ 2 − 1 − 1⎤ 1 ⎢ ⋅ − 1 2 − 1⎥ At = ⎥ 3nt ⎢ 2 ⎥⎦ ⎢⎣ − 1 2
Bt = W ⋅ Ztabc ⋅ G1
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Transformer Generalized Matrices Open Wye – Open Delta Connection nt =
VLN , rated high
side
VLL, rated low
side
⎡1 − 1 0 ⎤ a t = nt ⋅ ⎢0 1 − 1⎥ ⎢ ⎥ 0 ⎥⎦ ⎢⎣0 0
⎡Ztab ⎢ bt = nt ⋅ ⎢ 0 ⎢ 0 ⎣
0 0 ⎤ ⎥ 0 − Z tbc ⎥ 0 0 ⎥⎦
Z tab , Z tbc are referred to the low  voltage side ⎡0 0 0 ⎤ c t = ⎢0 0 0 ⎥ ⎢ ⎥ ⎢⎣0 0 0⎦⎥ U. P. National Engineering Center National Electrification Administration
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Transformer Generalized Matrices Open Wye – Open Delta Connection 1 0⎤ ⎡2 1 ⎢ ⋅ − 1 1 0⎥ At = ⎥ 3nt ⎢ ⎢⎣ − 1 − 2 0⎥⎦
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⎡ 2Z tab 1 ⎢ Bt = ⋅ ⎢ − Z tab 3 ⎢ − Z tab ⎣
0 − Z tbc ⎤ ⎥ 0 − Z tbc ⎥ 0 2Z tbc ⎥⎦
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ShortCircuit Analysis of Unbalanced Feeders ABC Z sys
System Voltage Source
Equivalent System Impedance
System:
1
ABC Z sub
2
Substation Transformer
ABC ZeqS
Total Primary Line Segment Impedance
2 ( kVLL ) Ω Z1 =
MVA 3φ
ABC Z sys ( approx )
4
Z abc xfm
Inline Feeder Transformer
abc ZeqL
5
Total Secondary Line Segment Impedance
3(kVLL ) − 2Z 1 Ω Z0 = MVA 1φ 2
⎡ 2Z 1 + Z 0 1 = ⋅ ⎢ Z 0 − Z1 3 ⎢ ⎣⎢ Z 0 − Z1
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3
Z 0 − Z1 2Z 1 + Z 0 Z 0 − Z1
Z 0 − Z1 ⎤ Z 0 − Z1 ⎥ ⎥ 2Z1 + Z 0 ⎥⎦
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ShortCircuit Analysis of Unbalanced Feeders
Thevenin equivalent voltages at points 2 and 3: computed by multiplying the system voltages by the generalized transformer matrix At of the substation transformer. Thevenin equivalent voltages at points 4 and 5: the voltage at node 3 multiplied by the generalized transformer matrix At of the inline transformer. Thevenin equivalent phase impedance matrices: sum of the phase impedance matrices of each device between the system voltage source and the point of fault.
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ShortCircuit Analysis of Unbalanced Feeders Thevenin equivalent circuit:
ZTOT
Ea
Eb
Zf
I fa
Zf
I fb
Zf
I fc c
a
Vax b
Vbx
x
Vcx
V xg
Ec g
Ea, Eb, Ec = Thevenin equiv. linetoground voltages @ the faulted node ZTOT = Thevenin equiv. phase impedance matrix @ the faulted node Zf = fault impedance U. P. National Engineering Center National Electrification Administration
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ShortCircuit Analysis of Unbalanced Feeders ⎡E a ⎤ ⎡ Z aa ⎢E ⎥ = ⎢ Z ⎢ b ⎥ ⎢ ba ⎢⎣Ec ⎥⎦ ⎢⎣ Zca
Z ab Z bb Z cb
Z ac ⎤ ⎡ I fa ⎤ ⎡Z f ⎢ b⎥ ⎢ ⎥ Z bc ⎢I f ⎥ + 0 ⎢ ⎥ c Z cc ⎥⎦ ⎣⎢ I f ⎥⎦ ⎢⎣ 0
0 Zf 0
0 ⎤ ⎡ I fa ⎤ ⎡Vax ⎤ ⎡V xg ⎤ ⎢ b⎥ ⎢ ⎥ ⎢ ⎥ 0 ⎢If ⎥ + Vbx ⎥ + ⎢V xg ⎥ ⎥ c ⎢ ⎥ Z f ⎥⎦ ⎢⎣ If ⎦⎥ ⎢⎣Vcx ⎥⎦ ⎢⎣V xg ⎥⎦
In compressed form, Eabc = ZTOT ⋅ Ifabc + Z F ⋅ Ifabc + Vabcx + Vxg
Combining terms, Eabc = {ZTOT + Z F } ⋅ Ifabc + Vabcx + Vxg = Z EQ ⋅ Ifabc + Vabcx + Vxg
Solving for the fault currents, Ifabc = Y ⋅ Eabc − Y ⋅ Vabcx − Y ⋅ Vxg −1 Y = Z EQ U. P. National Engineering Center National Electrification Administration
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ShortCircuit Analysis of Unbalanced Feeders abc
Define IP
= Y ⋅ E abc
Substituting & rearranging, IPabc = Ifabc + Y ⋅ Vabcx + Y ⋅ Vxg Expanding, ⎡I Pa ⎤ ⎡ I fa ⎤ ⎡Yaa Yab Yac ⎤ ⎡Vax ⎤ ⎡Yaa Yab Yac ⎤ ⎡V xg ⎤ ⎢ b⎥ ⎢ b⎥ ⎢ ⎥ ⎢V ⎥ ⎥ ⎢V ⎥ + ⎢Y I I Y Y Y Y Y = + bb bc ⎥ ⎢ bx ⎥ ba bb bc ⎥ ⎢ xg ⎥ ⎢ P ⎥ ⎢ f ⎥ ⎢ ba ⎢ ⎢I Pc ⎥ ⎢ I fc ⎥ ⎢⎣Yca Ycb Ycc ⎥⎦ ⎢⎣Vcx ⎥⎦ ⎢⎣Yca Ycb Ycc ⎥⎦ ⎢⎣V xg ⎥⎦ ⎣ ⎦ ⎣ ⎦
I Pa = I fa + (YaaVax + YabV bx+YacVcx ) + YSaV xg
I Pb = I fb + (YbaVax + YbbV bx+YbcVcx ) + YSaV xg I Pc = I fc + (YcaVax + YcbV bx+YaaVcx ) + YSaV xg U. P. National Engineering Center National Electrification Administration
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ShortCircuit Analysis of Unbalanced Feeders a
where YS = Yaa + Yab + Yac
YSb = Yba + Ybb + Ybc YSa = Yaa + Yab + Yac
3 equations, 7 unknowns  I fa , I fb , I fc ,Vax ,Vbx ,Vcx ,V xg I Pa , I Pb , I Pc are functions of the total impedance & the Thevenin voltages and are known Needed: 4 additional equations
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ShortCircuit Analysis of Unbalanced Feeders
ThreePhase Faults: Vax = Vbx = Vcx = 0
I a + I b + Ic = 0 ThreePhasetoGround Faults: Vax = Vbx = Vcx = V xg = 0 I a + I b + Ic = 0 LinetoLine Faults (assume ij fault with phase k unfaulted): Vix = V jx = 0 I fk = 0 I fi + I fj = 0 U. P. National Engineering Center National Electrification Administration
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ShortCircuit Analysis of Unbalanced Feeders
LinetoGround Faults (assume phase k fault with phases i and j unfaulted): Vkx = V xg = 0 I fi = I fj = 0
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ShortCircuit Analysis of Unbalanced Feeders
7 equations in matrix form: ⎡I Pa ⎤ ⎡ 1 0 0 Y11 Y12 Y13 YS1 ⎤ ⎡ I fa ⎤ ⎢ b⎥ ⎢ 2 ⎥⎢ b ⎥ ⎢I P ⎥ ⎢0 1 0 Y21 Y22 Y23 YS ⎥ ⎢ I f ⎥ ⎢I Pc ⎥ ⎢0 0 1 Y31 Y32 Y33 YS3 ⎥ ⎢ I fc ⎥ ⎥ ⎢ ⎥ ⎢ ⎥⎢ − − ⎥ ⎢Vax ⎥ − ⎢ 0 ⎥ = ⎢− − − − ⎢ 0 ⎥ ⎢− − − − − − ⎥ ⎢Vbx ⎥ − ⎥ ⎢ ⎥ ⎢ ⎥⎢ − − ⎥ ⎢Vcx ⎥ − ⎢ 0 ⎥ ⎢− − − − ⎢⎣ 0 ⎥⎦ ⎢⎣ − − − − − − ⎥⎦ ⎢⎣V xg ⎥⎦ − In condensed form:
IPs = C ⋅ X U. P. National Engineering Center National Electrification Administration
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ShortCircuit Analysis of Unbalanced Feeders Solving, X = Câˆ’1 â‹… IPs Example: 3phase fault C44 = 1 C55 = 1
C66 = 1
C71 = C72 = C73 = 1 All of the other elements in the last 4 rows of C will be set to zero.
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Example Infinite bus
1
I ABC
Iabc
ABC Z eqS
Z abc eqL
2
3
4
Compute the shortcircuit currents for a bolted linetoline fault between phases a and b at node 4. Linetoneutral Thevenin voltage at node 4:
Eth ,4 = At ⋅ ESLN Thevenin equiv. impedance at secondary terminals (node 3): ABC Z th ,3 = A t ⋅ Z eqS ⋅ dt + Ztabc
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Total Thevenin impedance at node 4: abc Zth ,4 = ZTOT = Z th.3 + Z eqL
Equivalent admittance matrix at node 4: âˆ’1 Yeq ,4 = ZTOT
Equivalent injected currents at point of fault:
IP = Yeq ,4 â‹… Eth ,4 For the ab fault at node 4,
I fa + I fb = 0 I fc = 0 Vax = Vbx = 0 U. P. National Engineering Center National Electrification Administration
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Unknowns are computed as
X = C−1 ⋅ ISP Suppose that the phase impedance matrices for the 2 line segments are:
ABC Z eqS
Z abc eqL
⎡0.1414 + j 0.5353 0.0361 + j 0.3225 0.0361 + j 0.2752 ⎤ = ⎢ 0.0361 + j 0.3225 0.1414 + j 0.5353 0.0361 + j 0.2953 ⎥ Ω ⎢ ⎥ ⎢⎣ 0.0361 + j 0.2752 0.0361 + j 0.2955 0.1414 + j 0.5353 ⎥⎦ ⎡0.1907 + j 0.5035 0.0607 + j 0.2302 0.0598 + j 0.1751⎤ = ⎢0.0607 + j 0.2302 0.1939 + j 0.4885 0.0614 + j 0.1931⎥ Ω ⎢ ⎥ ⎢⎣ 0.0598 + j 0.1751 0.0614 + j 0.1931 0.1921 + j 0.4970 ⎥⎦
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The transformer bank consists of three singlephase transformers each rated: 2000 kVA, 12.472.4 kV, Z = 1.0 + j6.0 % Source line segment:
⎡1 0 0⎤ a1 = d1 = U = ⎢0 1 0 ⎥ ⎢ ⎥ ⎢⎣0 0 1⎥⎦ ABC b1 = Z eqS
⎡0.1414 + j 0.5353 0.0361 + j 0.3225 0.0361 + j 0.2752 ⎤ = ⎢ 0.0361 + j 0.3225 0.1414 + j 0.5353 0.0361 + j 0.2953 ⎥ ⎢ ⎥ ⎢⎣ 0.0361 + j 0.2752 0.0361 + j 0.2955 0.1414 + j 0.5353 ⎥⎦ c1 = [0] U. P. National Engineering Center National Electrification Administration
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⎡ 1 0 0⎤ A1 = a1−1 = ⎢0 1 0⎥ ⎢ ⎥ ⎢⎣0 0 1⎥⎦ B1 = a1−1 ⋅ b1 ⎡0.1414 + j 0.5353 0.0361 + j 0.3225 0.0361 + j 0.2752 ⎤ = ⎢ 0.0361 + j 0.3225 0.1414 + j 0.5353 0.0361 + j 0.2953 ⎥ ⎢ ⎥ ⎢⎣ 0.0361 + j 0.2752 0.0361 + j 0.2955 0.1414 + j 0.5353 ⎥⎦
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Load line segment: ⎡1 0
a2
b2 c2
0⎤ = d2 = ⎢0 1 0⎥ ⎢ ⎥ ⎢⎣0 0 1⎥⎦ ⎡0.1907 + j 0.5035 0.0607 + j 0.2302 0.0598 + j 0.1751⎤ = ⎢0.0607 + j 0.2302 0.1939 + j 0.4885 0.0614 + j 0.1931⎥ ⎢ ⎥ ⎢⎣ 0.0598 + j 0.1751 0.0614 + j 0.1931 0.1921 + j 0.4970 ⎥⎦ = [0]
⎡ 1 0 0⎤ A 2 = ⎢0 1 0⎥ ⎢ ⎥ ⎢⎣0 0 1⎥⎦ ⎡0.1907 + j 0.5035 0.0607 + j 0.2302 0.0598 + j 0.1751⎤ B 2 = ⎢0.0607 + j 0.2302 0.1939 + j 0.4885 0.0614 + j 0.1931⎥ ⎢ ⎥ ⎢⎣ 0.0598 + j 0.1751 0.0614 + j 0.1931 0.1921 + j 0.4970 ⎥⎦ U. P. National Engineering Center National Electrification Administration
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Transformer: Transformer impedance in ohms referenced to the lowvoltage winding
Z base
2.4 2 ⋅ 1000 = = 2.88 Ω 2000
Z tlow = (0.01 + j 0.06 ) ⋅ 2.88 = 0.0288 + j 0.1728 Ω Transformer phase impedance matrix
0 0 ⎡0.0288 + j 0.1728 ⎤ ⎥ ⎢ Z abc = 0 0 . 0288 + j 0 . 1728 0 t ⎥ ⎢ 0 0 0.0288 + j 0.1728 ⎥⎦ ⎣⎢ Turns ratio:
Transformer ratio:
12.47 nt = = 5.1958 2.4 U. P. National Engineering Center National Electrification Administration
at =
12.47 = 2.9998 3 ⋅ 2.4
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Generalized matrices are:
− 3.4639 − 1.7319 ⎤ 0 ⎡ − nt ⎢ − 1.7319 − 3.4639 ⎥ 0 at = ⎢ ⎥ 3 0 ⎢⎣− 3.4639 − 1.7319 ⎥⎦ Zt ⎤ − nt bt = 0 2Zt ⎥ ⎥ 3 Zt 0 ⎥⎦ − 0.0998 − j 0.5986 − 0.0499 − j 0.2993 ⎤ ⎡ = ⎢− 0.0499 − j 0.2993 0 − 0.0998 − j 0.5986 ⎥ ⎢ ⎥ 0 ⎥⎦ ⎢⎣− 0.0998 − j 0.5986 − 0.0499 − j 0.2993 ⎡0 0 0 ⎤ c t = ⎢0 0 0 ⎥ ⎢ ⎥ ⎢⎣0 0 0 ⎥⎦ ⎡0 2 ⋅ ⎢1 0 ⎢ ⎢⎣2 1 ⎡ 0 ⋅ ⎢ Zt ⎢ ⎢⎣2Zt 0
1⎤ 2⎥ = ⎥ 0⎥⎦ 2Zt
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⎡1 1 ⎢ ⋅ 0 dt = nt ⎢ ⎢⎣− 1 ⎡1 1 ⎢ ⋅ −1 At = nt ⎢ ⎢⎣ 0 Bt = Z abc t
156
− 1 0 ⎤ ⎡ 0.1925 − 0.1925 0 ⎤ 1 − 1⎥ = ⎢ 0 0.1925 − 0.1925 ⎥ ⎥ ⎥ ⎢ 0 1 ⎥⎦ ⎢⎣− 0.1925 0 0.1925 ⎥⎦ − 0.1925 ⎤ 0 − 1⎤ ⎡ 0.1925 0 ⎥ 0 1 0 ⎥ = ⎢− 0.1925 0.1925 ⎥ ⎢ ⎥ − 1 1 ⎥⎦ ⎢⎣ 0 − 0.1925 0.1925 ⎥⎦
0 0 ⎡0.0288 + j 0.1728 ⎤ ⎥ =⎢ 0 0.0288 + j 0.1728 0 ⎢ ⎥ 0 0 0.0288 + j 0.1728 ⎥⎦ ⎢⎣
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The infinite bus balanced linetoline voltages are 12.47 kV, which leads to balanced linetoneutral voltages at 7.2 kV:
⎡ 12,470 ∠30 o ⎤ ⎢ ⎥ ELL,s = ⎢12,470 ∠ − 90 o ⎥ V ⎢ 12,470∠150 o ⎥ ⎣ ⎦
⎡ 7199 .6∠0 o ⎤ ⎢ ⎥ ELN ,s = ⎢7199 .6∠ − 120 o ⎥ ⎢ 7199 .6∠120 o ⎥ ⎣ ⎦
The linetoneutral Thevenin circuit voltages at node 4 are:
⎡ 2400 ∠ − 30 o ⎤ ⎢ o⎥ Eth,4 = A t ⋅ ELN ,s = ⎢2400 ∠ − 150 ⎥ ⎢ 2400 ∠90 o ⎥ ⎣ ⎦
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The Thevenin equivalent impedance at the secondary terminals (node 3) of the transformer consists of the primary line impedances referred across the transformer, plus the transformer impedances: ABC ⋅ dt + B t = Z th,3 = A t ⋅ Z eqS ⎡ 0.0366 + j 0.1921 = ⎢− 0.0039 − j 0.0086 ⎢ ⎢⎣− 0.0039 − j 0.0106
ABC ⋅ dt + Z abc A t ⋅ Z eqS t − 0.0039 − j 0.0086 − 0.0039 − j 0.0106 ⎤ 0.0366 + j 0.1886 − 0.0039 − j 0.0071⎥ ⎥ − 0.0039 − j 0.0071 0.0366 + j 0.1906 ⎥⎦
Total Thevenin impedance at node 4:
Z th,4 = ZTOT = Z th,3 + Z abc eqL ⎡0.2273 + j 0.6955 0.0568 + j 0.2216 0.0559 + j 0.1645 ⎤ = ⎢0.0568 + j 0.2216 0.2305 + j 0.6771 0.0575 + j 0.1860 ⎥ Ω ⎢ ⎥ ⎢⎣0.0559 + j 0.1645 0.0575 + j 0.1860 0.2287 + j 0.6876 ⎥⎦ U. P. National Engineering Center National Electrification Administration
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Equivalent admittance matrix at node 4: −1 Yeq,4 = ZTOT
⎡ 0.5031 − j1.4771 − 0.1763 + j 0.3907 − 0.0688 + j 0.2510 ⎤ = ⎢− 0.1763 + j 0.3907 0.5501 − j1.5280 − 0.1148 + j 0.3133 ⎥ S ⎢ ⎥ ⎣⎢− 0.0688 + j 0.2510 − 0.1145 + j 0.3133 0.4843 − j1.4532 ⎥⎦ The equivalent injected currents at the fault point:
⎡4466 .8∠ − 96.4 o ⎤ ⎢ o ⎥ I p = Yeq,4 ⋅ Eth,4 = ⎢ 4878 .9∠138.0 ⎥ A ⎢ 4440 .9∠16.4 o ⎥ ⎣ ⎦
Sums of each row of the equivalent admittance matrix:
⎡0.2580 − j 0.8353 ⎤ Ys = ∑ Yeq,ik = ⎢0.2590 − j 0.8240 ⎥ S ⎢ ⎥ k =1 ⎣⎢0.3007 − j 0.8889 ⎥⎦ 3
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For the ab fault at node 4,
I fa + Ifb = 0 Vax = 0
I fc = 0 Vbx = 0
The coefficient matrix ⎡1 ⎢0 ⎢ ⎢0 C = ⎢1 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣0
0 0 1 0 1 0 0 0
0 1 0 1 0 0
0.501 − j 1.477
− 0.176 + j 0.390 − 0.069 + j 0.252 0.258 − j 0.835 ⎤ − 0.176 + j 0.390 0.550 − j1.528 − 0.115 + j 0.314 0.259 − j 0.824 ⎥ ⎥ − 0.069 − j 0.251 − 0.115 + j 0.313 0.484 − j 1.452 0.301 − j 0.889 ⎥ ⎥ 0 0 0 0 ⎥ ⎥ 0 0 0 0 ⎥ 0 1 0 0 ⎥ ⎥⎦ 1 0 0 0
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The injected current matrix:
The unknowns are computed by:
⎡4466 .8∠ − 96.4 o ⎤ ⎢ o ⎥ ⎢ 4878 .9∠138 .0 ⎥ ⎢ 4440 .9∠16.4 o ⎥ ⎥ ⎢ s Ip = ⎢ 0 ⎥ ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎣ ⎦
⎡ 8901 .7∠ − 8.4 o ⎤ ⎢ o ⎥ ⎢ 8901 .7∠171 .6 ⎥ ⎥ ⎢ 0 ⎥ ⎢ X = C −1 ⋅ Isp = ⎢7740 .4∠ − 90.6 o ⎥ ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢ o ⎥ ⎢ ⎣ 2587 .9∠89.1 ⎦
The interpretation of the results are:
I fa = X1 = 8901 .7∠ − 8.4 o
Vax = X 4 = 7740 .4∠ − 90.6 o
I fb = X 2 = 8901 .7∠171.6 o
Vbx = X 5 = 0
I fc = X 3 = 0
Vcx = X 6 = 0 Vxg = X 7 = 2587 .9∠89.1o
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Fault Current at Different Times Clearing Time of Molded Breakers
Clearing Time of Fuse
Clearing Time of High Voltage Breakers
Contact Opening Time of High Voltage Breakers
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Fault Current that upstream overcurrent devices must withstand while downstream devices isolate the fault
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Fault Current at Different Times
First (1/2) Cycle Fault Current Short circuit ratings of low voltage equipment Ratings of High Voltage (HV) switch and fuse Close & Latch (Making) capacity or ratings of HV Circuit Breakers Maximum Fault for coordination of instantaneous trip of relays
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Fault Current at Different Times
1.5 to 4 Cycles Fault Current Interrupting (breaking) duties of HV circuit breakers Interrupting magnitude and time of HV breakers for coordination
30 Cycles Fault Current For time delay coordination
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Application of Short Circuit Analysis
Comparison of ClosedandLatch (Momentary or Making) and Interrupting (Breaking) Duties of Interrupting Devices
Comparison of Shorttime or withstand rating of system components
Selection of rating or setting of short circuit protective devices
Evaluation of current flow and voltage levels in the system during fault
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Characteristic of Short Circuit Currents R
L
Ri + L
E sin (ωt+φ)
i =
E sin (ωt + θ − φ ) R +X 2
2
+
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di = E sin (ωt + φ ) dt
E sin(θ − φ ) R +X 2
2
e
−R
X
ωt
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Characteristic of Short Circuit Currents i =
E sin (ωt + θ − φ ) R2 + X 2
+
E sin(θ − φ ) R2 + X 2
e
−R
X
ωt
I total , RMS = I symmetrica l RMS • Asymmetric al Factor
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ANSI/IEEE and IEC Standards
ANSI/IEEE: American National Standards Institute/ Institute of Electrical and Electronics Engineers
IEC: International Electrotechnical Commission
Prescribes Test Procedures and Calculation Methods U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method 1.54 Cycle Network: the network used to calculate interrupting shortcircuit current and protective device duties 1.54 cycles after the fault. Type of Device
Duty
High Voltage CB
Interrupting Capability
Low Voltage CB
N/A
Fuse
N/A
Switchgear and MCC
N/A
Relay
N/A
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ANSI/IEEE Calculation Method ½ Cycle Network: also known as the subtransient network because all rotating machines are represented by their subtransient reactances Type of Machine Utility Turbo generator Hydrogenerator with amortisseur windings Hydrogenerator without amortisseur windings Condenser Synchronous motor Induction Machine > 1000 hp @ 1800 rpm or less > 250 hp @ 3600 rpm All other ≥ 50 hp < 50 hp
Xsc X” Xd” Xd” 0.75 Xd’ Xd” Xd” Xd” Xd” 1.2 Xd” 1.67 Xd”
Xd” of induction motor = 1/(perunit lockedrotor current at rated voltage) U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method 30 Cycle Network: also known as the steadystate network Type of Machine
Xsc
Utility
X’’
Turbo Generator
Xd’
Hydrogenerator w/ Amortisseur Winding
Xd’
Hydrogenerator w/o Amortisseur Winding
Xd’
Condenser
Infinity
Synchronous Motor
Infinity
Induction Machine
Infinity
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ANSI/IEEE Calculation Method ANSI Multiplying Factor: determined by the equivalent X/R ratio at a particular fault location. The X and the R are calculated separately.
Local and Remote Contributions A local contribution to a shortcircuit current is the portion of the shortcircuit current fed predominantly from generators through no more than one transformation, or with external reactance in series which is less than 1.5 times the generator subtransient reactance. Otherwise the contribution is defined as a remote contribution. U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method Momentary (1/2 Cycle) ShortCircuit Current
Peak Momentary ShortCircuit Current
Imom, peak = MFp ⋅ Imom,rms, symm π − ⎛ ⎞ X R MFp = 2 ⎜ 1 + e ⎟⎟ ⎜ ⎝ ⎠
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ANSI/IEEE Calculation Method Momentary (1/2 Cycle) ShortCircuit Current
Asymmetrical RMS value of Momentary ShortCircuit Current
Imom, rms, symm =
Vpre − fault 3Zeq
Imom, rms, asymm = MFm ⋅ Imom,rms, symm MFm = 1 + 2e
−
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2π X R
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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.54 Cycle)
Adjusted RMS value of Interrupting ShortCircuit Current (for total current basis CBs)
Iint, rms, symm =
Vpre − fault 3Zeq
I int,rms ,adj = AMFi ⋅ I int,rms ,symm where AMFi = MFl + NACD (MFr − MFl ) U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method No AC Decay (NACD) Ratio
The NACD ratio is defined as the remote contributions to the total contributions for the shortcircuit current at a given location
I remote NACD = I total • Total short circuit current Itotal = Iremote + Ilocal • NACD = 0 if all contributions are local • NACD = 1 if all contributions are remote U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.54 Cycle)
MFr = 1 + 2e
−
4π t X R
Circuit Breaker Rating in Cycles 8 5 3 2 U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.54 Cycle) Calculation
Multiplying factors (total current basis CBs) MFr for 3phase & linetoground faults. U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.54 Cycle)
Multiplying factors (total current basis CBs) MFl for 3phase faults. U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.54 Cycle)
Adjusted RMS value of Interrupting ShortCircuit Current (for symmetrically rated CBs)
I int,rms ,adj
AMFi â‹… I int,rms ,symm = S
Circuit Breaker Contact Parting Time (Cycles) 4 3 2 1.5 U. P. National Engineering Center National Electrification Administration
S Factor 1.0 1.1 1.2 1.3
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ANSI/IEEE Calculation Method Low Voltage Circuit Breaker Interrupting Duty (1/2 Cycle) Calculation
Adjusted asymmetrical RMS value of Interrupting ShortCircuit Current
Iint, rms, symm =
Vpre − fault 3Zeq
I int,rms ,adj = MF ⋅ I int,rms ,symm
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ANSI/IEEE Calculation Method Low Voltage Circuit Breaker Interrupting Duty (1/2 Cycle) Calculation
MF =
2(1 + e 2(1 + e
MF =
−
1 + 2e 1 + 2e
−
−
π X R
)
Unfused power breakers
π ( X R )test
−
)
π X R Fused power breakers & Molded Case
π ( X R)test
Note: If calculated MF < 1.0, set MF = 1.0 U. P. National Engineering Center National Electrification Administration
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ANSI/IEEE Calculation Method Low Voltage Circuit Breaker Interrupting Duty (1/2 Cycle) Calculation Circuit Breaker Type Power Breaker (Unfused) Power Breaker (Fused) Molded Case (> 20 kA)
(X/R)test 6.59 4.90 4.90
Molded Case (10.001 â€“ 20 kA)
3.18
Molded Case (10 kA)
1.73
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ANSI/IEEE Calculation Method Fuse Interrupting ShortCircuit Current Calculation
 same procedure as Circuit Breaker Interrupting Duty calculation.
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IEC Calculation Method An equivalent voltage source at the fault location replaces all voltage sources. A voltage factor c is applied to adjust the value of the equivalent voltage source for minimum and maximum current calculations. All machines are represented by internal impedances Line capacitances and static loads are neglected, except for the zerosequence network. Calculations consider the electrical distance from the fault location to synchronous generators.
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IEC Calculation Method Initial Symmetrical ShortCircuit Current (Iâ€™â€™k) RMS value of the AC symmetrical component of an available shortcircuit current applicable at the instant of shortcircuit if the impedance remains at zero time value.
Peak ShortCircuit Current (ip) Maximum possible instantaneous value of the available shortcircuit current.
Symmetrical ShortCircuit Breaking Current (Ib) RMS value of an integral cycle of the symmetrical AC component of of the available shortcircuit current at the instant of contact separation of the first pole of a switching device U. P. National Engineering Center National Electrification Administration
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IEC Calculation Method Steadystate Short Circuit Current (Ik) RMS value of the shortcircuit current which remains after the decay of the transient phenomena.
Subtransient Voltage (E’’) of a Synchronous Machine RMS value of the symmetrical internal voltage of a synchronous machine which is active behind the subtransient reactance Xd’’ at the moment of short circuit.
FarfromGenerator ShortCircuit Shortcircuit condition to which the magnitude of the symmetrical ac component of the available shortcircuit current remains essentially constant
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IEC Calculation Method NeartoGenerator
ShortCircuit
Shortcircuit condition to which at least one synchronous machine contributes a prospective initial shortcircuit current which is more than twice the generator’s rated current or a shortcircuit condition to which synchronous and asynchronous motors contribute more than 5% of the initial symmetrical shortcircuit current (I”k) without motors.
Subtransient Machine
Reactance
(Xd’’)
of
a
Synchronous
Effective reactance at the moment of shortcircuit. MS value of the symmetrical internal voltage of a synchronous machine which is active behind the subtransient reactance Xd’’ at the moment of short circuit.
(
ZK = KG R + jX d''
KG =
)
cmax kVn kVr 1 + xd'' sin ϕr
U. P. National Engineering Center National Electrification Administration
kVn = nominal voltage of the terminal bus kVr = motor rated voltage Xd” = subtransient reactance ϕr = machine rated power factor Competency Training & Certification Program in Electric Power Distribution System Engineering
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IEC Calculation Method Minimum Time Delay (Tmin) of a Circuit Breaker Shortest time between the beginning of the shortcircuit current and the first contact separation of one pole of the switching device
Voltage Factor (c)
Factor used to adjust the value of the equivalent voltage source for the minimum and maximum current calculations
Voltage Factor
Voltage Factor
Max SC Calculation
Min SC Calculation
230/400 V
1.00
0.95
Other LV up to 1 KV
1.05
1.00
> 1 kV to 35 kV
1.10
1.00
> 35 KV to 230 KV
1.10
1.00
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IEC Calculation Method c Un I k′′ = 3Z k
i p = 2 k I k′′
Zk = equiv. Impedance at fault point
k = function of system R/X at fault location
I b = I k′′
for farfromgenerator fault
I b = μI k′′
for synch. machines, for neartogenerator faults
I b = μqI k′′
for ind. machines, for neartogenerator faults
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IEC Calculation Method HV CB asymmetrical breaking & dc current rating
⎛ 4 πf t min ⎞ I b ,asymm = I b ,symm 1 + 2 exp⎜ − ⎟ X /R ⎠ ⎝ ⎛ 2πf t min ⎞ I dc = I b ,symm 2 exp⎜ − ⎟ ⎝ X /R ⎠ f = system frequency tmin = minimum delay time Ib,symm = AC breaking current X/R = calculated based on testing PF of 7% at 50 Hz U. P. National Engineering Center National Electrification Administration
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IEC Calculation Method LV CB asymmetrical breaking current rating
I b ,asymm = I b ,symm
⎛ 4 πf t min ⎞ 1 + 2 exp⎜ − ⎟ X /R ⎠ ⎝
f = system frequency tmin = minimum delay time Ib,symm = AC breaking current X/R = calculated based on testing PF given by IEC
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IEC Calculation Method Fuse asymmetrical breaking current rating
I b ,asymm = I b ,symm
⎛ 4 πf t min ⎞ 1 + 2 exp⎜ − ⎟ X /R ⎠ ⎝
f = system frequency tmin = assumed to be a half cycle Ib,symm = AC breaking current X/R = calculated based on testing PF of 15%
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Selection of Device Duties 8Cycle TotalRated Circuit Breakers (KA)
5Cycle SymmetricalRated Circuit Breakers (KA)
Momentary Rating (Total 1stCycle RMS Current
Interrupting Rating (Total RMS Current at 4cycle ContactParting Time
Closing and Latching Capability (Total First Cycle RM Current)
ShortCircuit Capability (Symmetrical RMS Current at 3Cycle Parting Time
4.16 KV
20
10.5
19
10.1
4.16 – 250
4.16 KV
60
35
58
33.2
4.16 – 350
4.16 KV
80
48.6
78
46.9
13.8 – 500
13.8 KV
40
21
37
19.6
13.8 – 750
13.8 KV
60
13.5
58
30.4
13.8 – 1000
13.8 KV
80
42
77
40.2
Circuit Breaker Nominal Size Identification
Example Maximum System Operating Voltage
4.16 – 75
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