Short Circuit Analysis

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

Short Circuit Analysis

Training Course in

2

Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration



Training Course in Short Circuit Analysis

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

3






� Sources of Short Circuit Currents

� Types of Fault

� The Fault Point

� Three-Phase Fault

� Single-Line-to-Ground Fault

� Line-to-Line Fault

� Double-Line-to-Ground Fault

Analysis of Faulted Power System by Symmetrical Components

4






FaultMV

LV

UtilityG

Fault Current Contributors

Sources of Short Circuit Currents

5






Shunt Fault: Unintentional Connection between phases or between phase and ground.

1. Single Line-to-Ground Fault

2. Line-to-Line Fault3. Double Line-to-Ground Fault 4. Three Phase Fault

Series Fault: Unintentional Opening of phase conductors

Simultaneous Fault

Types of Fault

6






Shunt Faults

Line-to-Line

Double Line-to-Ground Single Line-to-Ground

Three Phase

Types of Fault

7






The system is assumed to be balanced, with regards to impedances, except at one point called the fault point.

Fault Currents

ab

c

aVr

bVr

cVr

Ground

aIr

bIr

cIrLine-to-

ground voltages

F

The Fault Point

8






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.

• Positive-Sequence Network• Negative-Sequence Network• Zero-Sequence Network

The Fault Point

9






The Thevenin equivalent of the power system at the fault point is called the sequence network.

F1

N1

thf VVr

=

Z11aIr

1aVr

+

-

+

-

F2

N2

Z2

2aIr

2aVr

+

-

0aVr

F0

N0

Z0

0aIr

+

-

11ath1a ZIVVrrr

−= 22a2a ZIVrr

−= 00a0a ZIVrr

−=

Positive Sequence

Negative Sequence

Zero Sequence

The Fault Point

10






� 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.

Three Phase Fault

11






Three-Phase Fault

Note: The system is still balanced. Currents and voltages are positive sequence only. The ground current is zero.gI

r

ab

c

aVr

bVr

cVr

Ground

bIr

cIr

gZ

aIr

fZ fZ

gIr

fZ

12






Sequence Network Interconnection:

F2

N2

Z2

2aIr

2aVr+

-

F1

fV

Z1 1aIr

1aVr+

-

+

-

N1 N0

F0

Z0

0aIr

0aVr+

-

Sequence currents

f

fa ZZ

VI

+=

11

r020 == aa II

rr

Zf

13






Three Phase Fault Currents:

f

faaaa ZZ

VIIII

+=++=

1210

f

faaab ZZ

VaaIIaII

+=++=

1

2

212

0

f

faaac ZZ

aVIaaIII

+=++=

12

210

14






Example:

Determine the fault current for a three phase bolted fault in each bus for the 4 bus system below.

G

1

432

Line 1Line 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 a subtransient reactance of 10%. Base Values: 100 MVA, 6.9 kV

15






Solution:

Draw the impedance diagram

E 1.0

0.1

0.20.3

0.6 0.5

0.4

43

2

1

16






E 1.0

0.1

0.20.3

0.6 0.5

0.4

43

2

1If

-

+

a) Fault @ Bus 4= +

= +

=

=+

=+

=

= +

= +

=

a1 2 23

b 13

13

c34

R edu ce th e n e tw o rk

X

0 .3 0 .6 0 .9

X

(0 .9 )(0 .4 )

0 .9 0 .4 0 .2 76 9 2 3

X

0 .2 7 6 92 3 0 .5 0 .7 7 6 92 3

a

a

b

X X

X XX X

X X

17






cd 14

c14

X XX

X X

(0.776923) (0.2)

0.776923 0.2 0.159055

=+

=+

=

dequiv genX X X

0.1 0.159055 0.259055

= +

= +

=

If

-

+E 1.0

0.25905

f

1.0I

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= =

=

18






b) Fault @ Bus 3

E 1.0

0.1

0.20.3

0.6 0.5

0.4

432

1If

-

+

a23 12X X X

0.3 0.6 0.9

= +

= +

=

b14 34X X X

0.2 0.5 0.7

= +

= +

=

a bequiv 13X (X ||X ) ||X

0.198425

=

=

19






gen equivX X X

= 0.1 0.198425= 0.298425

= +

+

f

1.0I

0.298425= 3.350923 p.u.

=

If

-

+E 1.0

0.298425

20






c) Fault @ Bus 2

E 1.0

0.1

0.20.3

0.6 0.5

0.4

432

1If

-

+

a14 34X X X

0.2 0.50.7

= +

= +

=

ab 13

a13

X XX

X X

(0.7)( 0.4)

0.7 0.40.254545

=+

=+

=

21






= +

= +

=

c b23X X X

0.254545 0.6 0.854545

=+

=+

=

cd 12

c12

X XX

X X

(0.854545)( 0.3)

0.854545 0.3 0.222047

= +

=

dgenX X X

0.322047

=

=

f

1.0I

0.322047 3.095525 p.u.

If

-

+E 1.0

0.322047

22






d) Fault @ Bus 1

E 1.0

0.1

0.20.3

0.6 0.5

0.4

432

1

If

-

+

=

=

genX X

0.1

=

=

f

1.0I

0.1 10.0 p.u.

23






Example: A three-phase 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.

G: X1 = 40% X2 = 40% X0 = 20%T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%

G

T1 T2Line Open

F

Note: All reactances are in per-unit of a common MVA base.

24






Positive-Sequence Network:

+

-

j0.6 1AIr

1.0

N1

F1+

-

1AVr

j0.05

j0.4+

gEr

-

Openj0.05

j0.15 1AIr

L1AIr

g1aIr

F1

N1

25







Zf

The sequence fault currents

=

=

=+

=

0

2

11

a

a

f

fa

I

I

ZZ

VI

r

r

r

+

-

j0.6 1AIr

1.0

N1

F1+

-

1AVr

The phase fault currents

=

=

=

c

b

a

I

I

I

26






Single Line-to-Ground FaultAssuming the fault is in phase a,

Boundary Conditions: (1) afa IZVrr

=

ab

c

aVr

bVr

cVr

Ground

aIr bI

rcIr

fZ

(2) 0== cb IIrr

27






abcIAIrr

1012

−=

Transformation: From (2), we get

aa

aa2

2

1

1

111

3

1

2

1

0

a

a

a

I

I

I

r

r

r

=

0

0aIr

31

=

a

a

a

I

I

I

r

r

r

which means a31

2a1a0a IIIIrrrr

===

From (1), we get

)( 210210 aaafaaa IIIZVVVrrrrrr

++=++

or0210 3 afaaa IZVVV

rrrr=++

28







F1

N1

fVr

Z1 1aIr

1aVr+

-

+

-

F2

N2

Z2

2aIr

2aVr+

-

0aVr

F0

N0

Z0

0aIr+

-

3Zf


f

faaa ZZZZ

VIII

3210210

+++===

rrr

29






Single-Line-to-Ground Phase Fault Currents:

f

faaaa ZZZZ

VIIII

3

3

021210

+++=++=

0=bI

0=cI

30






Example: A single line-to-ground 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.

G: X1 = 40% X2 = 40% X0 = 20%T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%

G

T1 T2Line Open

F


31







+

-

j0.6 1aIr

1.0

N1

F1+

-

1aVr

j0.05

j0.4+

gEr

-

Openj0.05

j0.15

F1

N1

32






Negative-Sequence Network:

j0.62aI

r

N2

F2+

-

2aVr

j0.05

j0.4

Openj0.05

j0.15

F2

N2

33






Zero-Sequence Network:

j0.0440aI

r

N0

F0+

-

0aVr

N0

j0.05

j0.2

Openj0.05

j0.35

F0

34







N1 N2

F1

J0.6 1AIr

+

-

F2

J0.62AI

r

1.0

N0

F0

J0.044

0AIr

Sequence Fault Currents

)044.06.06.0(j0.1

III 2A1A0A++

===rrr

p.u.804.0j−=

35






Phase Fault Currents

p.u. 411.2jI3I 0AA −==rr

0II CB ==rr

36






Line-to-Line FaultAssuming the fault is in phases b and c,

Boundary Conditions: (1) 0=aIr

(2) cb IIrr

−=

ab

c

aVr

bVr

cVr

Ground

bIr

cIr

fZ

aIr

(3) fbcb ZIVVrrr

=−

37






abcIAI 1012

−=r

Transformation: From (1) and (2), we get

aa

aa2

2

1

1

111

31

2

1

0

a

a

a

I

I

I

r

r

r

=

b

b

I

Ir

r

−

0

31

=

b

b

Iaa

Iaar

r

)(

)(

0

2

2

−

−

which means00 =aI

r

bbaa IjIaaIIrrrr

312

31

21 )( =−=−=

38






From (3), we get

)( 212

0 aaa VaVaVrrr

++

Since and , we get

faaa ZIaaVaaVaa 12

22

12 )()()(

rrr−=−+−

faaaaaa ZIaIaIVaVaV )()( 212

022

10

rrrrrr++=++−

00 =aIr

21 aa IIrr

−=

or

faaa ZIVV 121

rrr=−

39







F1

N1

fV

Z1 1aIr

1aVr+

-

+

-

F2

N2

Z2

2aIr

2aVr+

-

N0

F0

Z0

0aIr

Zf


f

faa ZZZ

VII

++=−=

2121

rr00 =aI

r

40






Line-to-Line Phase Fault Currents:

212

0 aaab aIIaII ++=

0=aI

( ) 112

112 )(0 aaaa jIIaaIaIa −=−=−++=

f

fb ZZZ

VjI

++−=

21

3

f

fc ZZZ

VjI

+++=

21

3

41






Example: A line-to-line fault occurs at point F. Assuming zero fault impedance, find the fault currents at fault point F. Assume Eg = 1.0 p.u.

G: X1 = 40% X2 = 40% X0 = 20%T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%

G

T1 T2Line Open

F


42







N1 N2

F1

J0.6 1AIr

+

-

F2

J0.62AI

r

1.0

N0

F0

J0.044

0AIr

43






Sequence Fault Currents:

44






Phase Fault Currents:

45






ab

c

aVr

bVr

cVr

Ground

bIr

cIr

gZ

aIr

fZ fZ

cb IIrr

+

Double-Line-to-Ground FaultAssuming the fault is in phases b and c,

Boundary Conditions: 0=aIr

cgbgfb IZIZZVrrr

++= )(

bgcgfc IZIZZVrrr

++= )(

(1)

(2)

(3)

46






Transformation: From (1), we get

2100 aaaa IIIIrrrr

++==

From

212

0 aaab VaVaVVrrrr

++=

212

0 aaab IaIaIIrrrr

++=

we get2

210 aaac VaVaVV

rrrr++=

22

12 )()( aacb VaaVaaVV

rrrr−+−=−

Likewise, from

22

10 aaac IaIaIIrrrr

++=

47






22

12 )()( aacb IaaIaaII

rrrr−+−=−

we get

From boundary conditions (2) and (3), we get

)( cbfcb IIZVVrrrr

−=−

Substitution gives

])()[( 22

12

aaf IaaIaaZrr

−+−=2

21

2 )()( aa VaaVaarr

−+−

Simplifying, we get

2211 afaafa IZVIZVrrrr

−=−

48






From boundary conditions (2) and (3), we get

))(2( cbgfcb IIZZVVrrrr

++=+

Substitution gives

2102 aaacb VVVVVrrrrr

−−=+

We can also show

2102 aaacb IIIIIrrrrr

−−=+

)2(2 210210 aaafaaa IIIZVVVrrrrrr

−−=−−

)2(2 210 aaag IIIZrrr

−−+

49






Rearranging terms, we get

11000 422 afaagafa IZVIZIZVrrrrr

−=−−

)(2 2122 aagafa IIZIZVrrrr

+−−+

Earlier, we got

2211 afaafa IZVIZVrrrr

−=−

021 aaa IIIrrr

−=+

Substitution gives

)(2622 11000 afaagafa IZVIZIZVrrrrr

−=−−

( ) 1100 3 afaagfa IZVIZZVrrrr

−=+−

50







F1

fV

Z1 1aIr

1aVr+

-

+

-

N1

F2

N2

Z2

2aIr

2aVr+

-

N0

F0

Z0

0aIr

0aVr+

-

ZfZf Zf+3Zg

Let gfT ZZZZ 300 ++=

fT ZZZ += 11

fT ZZZ += 22

51







TT

TTT

fa

ZZ

ZZZ

VI

20

201

1

++

=r

210 aaa IIIrrr

−−=

From current division, we get

From KCL, we get

120

02 a

TT

Ta I

ZZ

ZI

rr

+−=

120

20 a

TT

Ta I

ZZ

ZI

rr

+−=or

52






Double-Line-to-Ground Phase Fault Currents:

212

0 aaab aIIaII ++=

0=aI

( )

TTTTTT

TTf

ZZZZZZ

aZZVj

020121

203++

−−=

22

10 aaac IaaIII ++=

( )TTTTTT

TTf

ZZZZZZ

ZaZVj

020121

22

03++

−+=

53






Example: A double-line-to-ground fault occurs at point F. Assuming zero fault impedance, find the fault currents at fault point F. Assume Eg = 1.0 p.u.

G: X1 = 40% X2 = 40% X0 = 20%T1, T2: X = 5% Line: X1 = X2 = 15% X0 = 35%

G

T1 T2Line Open

F


54







N1 N2

F1

J0.6 1AIr

+

-

F2

J0.62AI

r

1.0

N0

F0

J0.044

0AIr

55






Sequence Fault Currents:

56






Phase Fault Currents:

57






� Development of the Model

� Rake Equivalent

� Formation of Zbus

� Analysis of Shunt Fault

Bus Impedance Matrix Method

58






Observations on Manual Network Solution

The procedure is straight forward, yet tedious and could be prone to hand-calculation error.

Is there a way for a computer to implement this methodology?

Development of the Model

59






Development of the ModelConsider the three-bus system shown below. Let us analyze the system for a three-phase fault in any bus.

G1, G2 : X1=X2=0.2 X0=0.1

G1G2

L1

L2

1 2

3

L1 : X1=X2=0.6 X0=1.2L2 : X1=X2=0.24 X0=0.5

60







j0.2

1GEr +

-

+

-2GE

rj0.2

j0.6j0.24

1 2

3

Combine the sources and re-draw. Assume EG = 1.0 per unit.

j0.6

3

j0.2j0.2

j0.24

+

-

GEr

1 2

61






For a three-phase fault in bus 1 (or bus 2), we get the positive-sequence impedance.

16.0j)]6.02.0//(2.0[jZ1 =+=

25.6jZ1

ZE

I11

GF −===r

For a three-phase fault in bus 3, we get

4.0j)]6.02.0//(2.024.0[jZ1 =++=

5.2jZ1

ZE

I11

GF −===r

62






Next, use loop currents to describe the circuit with all fault switches closed. Since there are four loops, we need to define four loop currents.

Let us connect a fault switch to each bus. In order to simulate a three-phase fault in any bus, close the fault switch in that bus.

j0.6

3

j0.2j0.2

j0.24

+

-

GEr

1 2

1Ir

3Ir

2Ir

4Ir

4

63






The loop equations are)(2.00.1 431 IIIj

rrr−+=loop 1:

loop 2:

loop 4:

loop 3:

)(2.06.0)(2.00 314442 IIIjIjIIjrrrrrr

−−+++=

or

0.12.02.02.02.044.002.0

2.002.002.02.002.0

−−

−

−

00.10.10.1

4

3

2

1

IIII

r

r

r

r

= j

)(2.00.1 42 IIjrr

+=

3431 24.0)(2.00.1 IjIIIjrrrr

+−+=

64






Current I4 is not a fault current. It can be eliminated using Kron’s reduction. We get

31

421)1(

bus ZZZZZ −−=

IZV )1(bus

rr=

where

44.002.002.002.002.0

Z1 = j

2.02.02.0

−

−

Z2 = j

Z3 = j[-0.2 0.2 -0.2 ] Z4 = j[1.0]

and

65






Substitution gives

40.004.016.004.016.004.016.004.016.0

= j

0.10.10.1

3

2

1

III

r

r

r

IZV )1(bus

rr=

)1(busZ

(1) The equation can be used to analyze a three-phase fault in any bus (one fault at a time).

(2) is called the positive-sequence bus-impedance matrix, a complex symmetric matrix.

Note:

66






One possible equivalent circuit is shown. This circuit is called a rake-equivalent.

+

-

Z11 Z22 Z33

Z12 Z23

Z13

1.0

1Ir

2Ir

3Ir

Consider the matrix voltage equation

332313

232212

131211

ZZZZZZZZZ

=

0.10.10.1

3

2

1

III

r

r

r

Suppose we are asked to find a circuit that satisfies the matrix equation.

Rake Equivalent

67






Consider again the three-bus system. The circuit is described by the matrix equation

40.004.016.004.016.004.016.004.016.0

= j

0.10.10.1

3

2

1

III

r

r

r

The rake equivalent is shown. The diagonal elements of the matrix are self impedances while the off-diagonal elements are mutual impedances.

+

-

j0.16

1.0

1Ir

2Ir

3Ir

j0.16 j0.4j0.04j0.04

j0.16

68






For the three-bus system, assume a fault in bus 3. The equation for bus 3 is

321 I4.0jI04.0jI16.0j0.1rrr

++=

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 j0.4j0.04j0.04

j0.16

2Vr+

-1Vr+

-

From KVL, we get the voltage in bus 1.

6.0ZZ

0.1IZ0.1V33

133131 =−=−=rr

69






In general, for a three-phase fault in bus k of a system with n buses, the fault current is

kkk Z

1I =r

k=1,2,…n

Similarly from KVL, we get the voltage in bus 2.

9.0ZZ

0.1IZ0.1V33

233232 =−=−=rr

Note: Once the voltages in all the buses are known, the current in any line can be calculated.

The voltage in any bus j is given by

kk

jkj Z

Z0.1V −=

rj=1,2,…n

70






The current in any line, which is connected from bus m to bus n, can be found using

mn

nmmn z

VVI

rrr −

=

where zmn is the actual impedance of the line.

j0.2

1GEr +

-

+

-2GE

rj0.2

j0.6j0.24

1 2

3-j2.0

-j2.5

-j0.5

5.0j6.0j

6.09.0z

VVI

21

1221 −=

−=

−=

rrr

For example, the current in the line between buses 2 and 1 is

71






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.

Formation of Zbus

72






+

-

Z11 Z22 Zkk

Z12 Z2k

1.0

1Ir

2Ir

kIr

Znn

Zkn

nIr1 2 k n

Let us examine each category in the addition of a new branch.

Assume that at the current stage, the dimension of Zbus is n.

nn2n1n

n22221

n11211

ZZZ

ZZZZZZ

0.1

0.10.1

n

2

1

I

II

r

r

r

=

……

…

… … …

oldbusZ

73






+

-

Z11 Z22 Zkk

Z12 Z2k

1.0

1Ir

2Ir

kIr

Znn

Zkn

nIr1 2 k n

Type 1: Add a generator to a new bus

Let Zg be the impedance of the generator to be added.

Zg

1nI +

r n+1

g

nn2n1n

n22221

n11211

Z0000ZZZ0ZZZ0ZZZ

0.10.10.10.1

1n

n

2

1

IIII

+

r

r

r

r

=

…

…

…The dimension is (n+1).

74






Type 2: Add a generator to an old bus k

Let Zg be the impedance of the generator to be added.

+

-

Z11 Z22 Zkk

Z12 Z2k

1.0

1Ir

2Ir

kIr

Znn

nIr1 2 k n

The new current in impedance Zkk is (Ik+Iw). The new 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

wIr

75






For the added generator loop, we get

wgnknwkkk22k11k IZIZ...)II(Z...IZIZ0rrrrrr

+++++++=

In matrix form, we get

wknkk2k1k

nknnnk2n1n

k2n2k22221

k1n1k11211

ZZZZZZZZZZ

ZZZZZZZZZZ

00.1

0.10.1

=

w

n

2

1

II

II

r

r

r

r… …… …

… …… …

…… … …

where Zw=Zkk+Zg. The last row is eliminated using Kron’s reduction. The dimension remains as n.

76






Type 3: Add a branch from an old bus k to a new bus +

-

Z11 Z22 Zkk

Z12 Z2k

1.0

1Ir

2Ir

kIr

Znn

Zkn

nIr1 2 k n

The new current in impedance Zkk is (Ik+In+1). The new 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

77






where Zw=Zkk+Zb. Kron’s reduction is not required. The dimension increases to (n+1).


… …

wknkk2k1k

nknnnk2n1n

k2n2k22221

k1n1k11211

ZZZZZZZZZZ

ZZZZZZZZZZ

0.10.1

0.10.1

=

1n

n

2

1

II

II

+

r

r

r

r

… …

… …… …

…… … …

For the new bus, we get...)II(Z...IZIZ0.1 1nkkk22k11k +++++= +

rrrr

1nbnkn IZIZ +++rr

78






Type 4: Add a branch from an old bus j to an old bus k

+

-

Z11 Z22 Zjj

Z12 Z2j

1.0

1Ir

2Ir

kIr

Zkk

Zkn

nIr1 2

jk

Znn

jIr n

The new current in impedance Zjj is (Ij+Iw). The new current in impedance Zkk is (Ik-Iw). The new equations for buses 1 to n are

)II(Z...IZIZ0.1 wjj1212111

rrrr++++=

nn1wkk1 IZ...)II(Zrrr

++−+

wIr

Zb

79






)II(Z...IZIZ0.1 wjj2222121

rrrr++++=

)II(Z...IZIZ0.1 wjnj22n11n

rrrr++++=

nn2wkk2 IZ...)II(Zrrr

++−+

nnnwknk IZ...)II(Zrrr

++−+

For the added loop, we get

)II(Z)II(Z...IZIZ0 wkjkwjjj22j11j

rrrrrr−+++++=

...IZIZ[IZIZ... 22k11kwbnjn ++−+++rrrr

]IZ...)II(Z)II(Z nknwkkkwjkj

rrrrr++−+++

80







vknjn2k2j1k1j

nknjnn2n1n

k2j2n22221

k1j1n11211

ZZZZZZZZZZZZ

ZZZZZZZZZZ

−−−

−

−

−

00.1

0.10.1

w

n

2

1

II

II

r

r

r

r

=

…

…

…

…

…… …

where Zv=Zjj+Zkk-2Zjk+Zb. The last row is eliminated using Kron’s reduction. The dimension remains as n.

81






Example: For the network shown, use the step-by-step building algorithm to form the bus impedance matrix.

Step 1. Add generator G1 to bus 1.

1Xbus = [0.2]1


Xbus =2.00

02.0

1 2

1

2

j0.2+

-

+

-

j0.2

j0.6j0.24

1 2

3 1.01.0

82






Step 3. Add the line from bus 1 to bus 2.

0.12.02.02.02.00

2.002.0

−

−Xnew =

1 2

1

2

*

*

Apply Kron’s reduction to eliminate the last row and column. We get

=−3

142 XXX

2.02.0

−[0.2 -0.2]

83






04.004.004.004.0

−

−=−

31

42 XXX

We get

31

421bus XXXXX −−= =

1 2

1

2 16.004.004.016.0

Step 4. Finally, add the line from bus 1 to bus 3.

4.004.016.004.016.004.016.004.016.0

Xbus =

1 2

1

2

3

3

No Kronreduction is required.

84






Example: Determine the positive-sequence bus-impedance matrix for the four-bus test system shown.

G1T

L1

L2

G2

L3

2 3

4

G1: X1=0.40 X2=0.40 X0=0.15G2: X1=0.50 X2=0.50 X0=0.25L1: X1=0.40 X2=0.40 X0=0.80

T: X=0.08

L2: X1=0.30 X2=0.30 X0=0.60L3: X1=0.20 X2=0.20 X0=0.40

1

85






+

- N1

j0.4

j0.08

j0.4

j0.3 j0.2

j0.5+

-1.0 1.0

21 3

4Positive-sequence network

1. Add G1 to bus 1.

1Xbus = [0.4]1

2. Add the transformer from bus 1 to bus 2.

Xbus =48.04.04.04.0

1 2

1

2

88.048.04.048.048.04.04.04.04.0

Xbus =

1 2

1

2

3

3

3. Add the line from bus 2 to bus 3.

86







38.188.048.04.088.088.048.04.048.048.048.04.04.04.04.04.0

Xnew =

1 2

1

2

3

3

*

*

Apply Kron’s reduction.

38.11

31

42 XXX =−

88.048.04.0

[0.4 0.48 0.88]

87






3188.01739.01449.01739.03130.02609.01449.02609.02841.0

Xbus =

1 2

1

2

3

3

=−3

142 XXX

5612.03061.02551.03061.01670.01391.02551.01391.01159.0

We get

The new bus impedance matrix is

31

421bus XXXXX −−=

88







6130.01739.03130.02609.01739.03188.01739.01449.03130.01739.03130.02609.02609.01449.02609.02841.0

Xbus =

1 2

1

2

3

3

4

4


784.04391.01449.01391.01159.04391.06130.01739.03130.02609.01449.01739.03188.01739.01449.0

1391.03130.01739.03130.02609.01159.02609.01449.02609.02841.0

−

−

1 2

1

2

3

3

4

4

*

*

Xnew=

89






Apply Kron’s reduction. We get

Note: This is the positive-sequence bus-impedance matrix for the four-bus test system.

3671.02551.02351.01959.02551.02920.01996.01664.02351.01996.02884.02403.01959.01664.02403.02669.0

1 2

1

2

3

3

4

4

=)1(busX

90






Negative- and Zero-Sequence Zbus

The same step-by-step algorithm can be applied to build the negative-sequence and zero-sequence 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 negative-sequence and zero-sequence bus-impedance matrices can also be described by a rake equivalent circuit.

91






Example: Find the zero-sequence bus-impedance matrix for the four-bus test system.

Zero-sequence network

N0

j0.15

j0.08

j0.8

j0.6 j0.4

j0.252

1

3

4

1. Add G1 to bus 1.

1Xbus = [0.15]1

2. Add the transformer from bus 1 to bus 2.

Note: The impedance is actually connected from bus 2 to the reference bus.

Xbus =08.00015.0

1 2

1

2

92






88.008.0008.008.000015.0

Xbus =

1 2

1

2

3

3

3. Add the line from bus 2 to bus 3.


13.188.008.0088.088.008.0008.008.008.0000015.0

Xnew =

1 2

1

2

3

3

*

*

93







=−3

142 XXX

6853.00623.000623.00057.00000

1947.00177.000177.00743.000015.0

Xbus =

1 2

1

2

3

3

The new bus impedance matrix is

94







6743.00177.00743.000177.01947.00177.000743.00177.00743.0000015.0

Xbus =

1 2

1

2

3

3

4

4


2336.16566.0177.00566.006566.06743.00177.00743.00

177.00177.01946.00177.000566.00743.00177.00743.00000015.0

−

−

1 2

1

2

3

3

4

4

*

*

Xnew=

96






Positive-Sequence Zbus

The positive-sequence bus-impedance matrix describes the positive-sequence network.

)1(nn

)1(2n

)1(1n

)1(n2

)1(22

)1(21

)1(n1

)1(12

)1(11

ZZZ

ZZZZZZ …

…

……=)1(

busZ

Rake Equivalent

+

-1.0

1 2 k n

N1

)1(11Z )1(

22Z )1(kkZ )1(

nnZ

)1(12Z )1(

k2Z )1(knZ

95







Note: This is the zero-sequence bus-impedance matrix for the four-bus test system.

3248.01119.00442.001119.01693.00258.000442.00258.00717.0000015.0

1 2

1

2

3

3

4

4

=)0(busX

97






Negative-Sequence Zbus

)2(nn

)2(2n

)2(1n

)2(n2

)2(22

)2(21

)2(n1

)2(12

)2(11

ZZZ

ZZZZZZ …

…

……=)2(

busZ

The negative-sequence bus-impedance matrix describes the negative-sequence network.

Rake Equivalent

N2

1 2 k n

)2(11Z )2(

22Z )2(kkZ )2(

nnZ

)2(12Z )2(

k2Z )2(knZ

98






Zero-Sequence Zbus

)0(nn

)0(2n

)0(1n

)0(n2

)0(22

)0(21

)0(n1

)0(12

)0(11

ZZZ

ZZZZZZ …

…

…

…=)0(busZ

The zero-sequence bus-impedance matrix describes the zero-sequence network.

Rake Equivalent

1 2 k n

)0(11Z )0(

22Z )0(kkZ )0(

nnZ

)0(12Z )0(

k2Z )0(knZ

N0

99






Analysis of Shunt Faults

The bus-impedance matrices can be used for the analysis of the following shunt faults:

1. Three-Phase Fault2. Line-to-Line Fault

3. Single Line-to-Ground Fault4. Double Line-to-Ground Fault

Since the bus-impedance matrix is a representation of the power system as seen from the buses, only bus faults can be investigated.

100






Three-phase Fault at Bus k

)1(11Z )1(

22Z )1(kkZ )1(

nnZ

N1

1 2 k n

)1(kk

k Z1

I =r

The fault current is

kk

jkj Z

Z0.1V −=

r

The voltage at any bus is

The current in any line is mn

nmmn z

VVI

rrr −

=

101






Example: Consider a three-phase fault at bus 4 of the four-bus test system. Find all line currents.

The positive-sequence bus-impedance matrix is

3671.02551.02351.01959.02551.02920.01996.01664.02351.01996.02884.02403.01959.01664.02403.02669.0

1 2

1

2

3

3

4

4

=)1(busX

The fault current is

7241.2j3671.0j1

Z1

I )1(44

F −===r

102






4663.03671.01959.0

1V1 =−=r

3595.03671.02351.0

1V2 =−=r

3051.03671.02551.0

1V3 =−=r

0V4 =r

The bus voltages are

kk

jkj Z

Z0.1V −=

rj=1,2,…n

103






The line currents are given bymn

nmmn z

VVI

rrr −

=

3344.1j4.0j4663.01

I 1G −=−

=r

3897.1j5.0j3051.01

I 2G −=−

=r

3342.1j08.0j

3595.04663.0I12 −=

−=

r

1360.0j4.0j

3051.03595.0I23 −=

−=

r

104






1984.1j3.0j

03595.0I24 −=

−=

r

5257.1j2.0j

03051.0I34 −=

−=

r

+

- N1

j0.4

j0.08 j0.4j0.3 j0.2

1.0

2

1 3

4

+

-

j0.5

1.0

1GIr

2GIr

12Ir

23Ir

24Ir 34I

rFIr

105






Line-to-Line Fault at Bus k

)1(11Z )1(

22Z )1(kkZ )1(

nnZ

N1

1 2 k n

)2(11Z )2(

22Z )2(kkZ )2(

nnZ1 2 k n

N2


0I 0a =r

)2(kk

)1(kk ZZ

1+

=2a1a IIrr

−=

Sequence Voltages at bus j

0V 0a =r

)1(jk1a1a ZI1V

rr−=

)2(jk2a2a ZIV

rr−=

1aIr

2aIr

106






Example: Consider a line-to-line fault at bus 4 of the four-bus test system. Find the phase currents in lines L2 and L3.

3671.02551.02351.01959.02551.02920.01996.01664.02351.01996.02884.02403.01959.01664.02403.02669.0

1 2

1

2

3

3

4

4

=)1(busX

The positive-sequence bus-impedance matrices is

For this power system, )2(bus

)1(bus XX =

107






The sequence fault currents are

362.1j)3671.0(2j

1ZZ

1II )2(

44)1(

442a1a −==

+=−=

rr0I 0a =

r

The sequence voltages in bus 4 are

0V 40a =−

r

)1(441a41a ZI1V

rr−=−

5.0ZIV )2(442a42a =−=−

rr5.0)3671.0j)(362.1j(1 =−−=

108







0V 20a =−

r

6798.0ZI1V )1(241a21a =−=−

rr

3202.0ZIV )2(242a22a =−=−

rr


0V 30a =−

r

6526.0ZI1V )1(341a31a =−=−

rr

3474.0ZIV )2(342a32a =−=−

rr

109






The sequence currents in line L3 are

0I 3L0a =−

r

7628.0j2.0j

5.0653.0I 3L1a −=

−=−

r

7628.0j2.0j

5.0347.0I 3L2a =

−=−

r

The phase currents in line L3 are

0IIII 3L2a3L1a3L0a3La =++= −−−−

rrrr

3213.1IaIaII 3L2a3L1a2

3L0a3Lb −=++= −−−−

rrrr

3213.1IaIaII 3L2a2

3L1a3L0a3Lc =++= −−−−

rrrr

110







0I 2L0a =−

r

5992.0j3.0j

5.068.0I 2L1a −=

−=−

r

5992.0j3.0j

5.032.0I 2L2a =

−=−

r


0IIII 2L2a2L1a2L0a2La =++= −−−−

rrrr

0378.1IaIaII 2L2a2L1a2

2L0a2Lb −=++= −−−−

rrrr

0378.1IaIaII 2L2a2

2L1a2L0a2Lc =++= −−−−

rrrr

111






SLG Fault at Bus kSequence Fault Currents

2a1a0a IIIrrr

==

)2(kk

)1(kk

)0(kk ZZZ

1++

=

Sequence Voltages at bus j

)0(jk0a0a ZIV

rr−=

)1(jk1a1a ZI1V

rr−=

)2(jk2a2a ZIV

rr−=

1 2 k n

)0(11Z )0(

22Z )0(kkZ )0(

nnZ

N0

)2(11Z )2(

22Z )2(kkZ )2(

nnZ1 2 k n

N2

)1(11Z )1(

22Z )1(kkZ )1(

nnZ1 2 k n

N1

1aIr

2aIr

0aIr

112






Example: Consider a single line-to-ground fault at bus 4 of the four-bus test system. Find the phase currents in lines L2 and L3.


3067.0ZIV )0(440a40a −=−=−

rr

6534.0ZI1V )1(441a41a =−=−

rr

3466.0ZIV )2(442a42a −=−=−

rr

The sequence fault currents are

2a1a0a IIIrrr

==

9443.0jZZZ

1)2(

44)1(

44)0(

44

−=++

=

113







0417.0ZIV )0(240a20a −=−=−

rr

778.0ZI1V )1(241a21a =−=−

rr

222.0ZIV )2(242a22a −=−=−

rr


1057.0ZIV )0(340a30a −=−=−

rr

7591.0ZI1V )1(341a31a =−=−

rr

2409.0ZIV )2(342a32a −=−=−

rr

114







4417.0j6.0j

3067.00417.0I 2L0a −=

+−=−

r

4154.0j3.0j

6534.0778.0I 2L1a −=

−=−

r

4154.0j3.0j

3466.0222.0I 2L2a −=

+−=−

r


2725.1jIIII 2L2a2L1a2L0a2La −=++= −−−−

rrrr

0262.0jIaIaII 2L2a2L1a2

2L0a2Lb −=++= −−−−

rrrr

0262.0jIaIaII 2L2a2

2L1a2L0a2Lc −=++= −−−−

rrrr

115







5026.0j4.0j

3067.01057.0I 3L0a −=

+−=−

r

5289.0j2.0j

6534.07591.0I 3L1a −=

−=−

r

5289.0j2.0j

3466.02409.0I 3L2a −=

+−=−

r


5603.1jIIII 3L2a3L1a3L0a3La −=++= −−−−

rrrr

0262.0jIaIaII 3L2a3L1a2

3L0a3Lb =++= −−−−

rrrr

0262.0jIaIaII 3L2a2

3L1a3L0a3Lc =++= −−−−

rrrr

116






Double Line-to-Ground Fault at Bus k


)Z//Z(Z1

I )0(kk

)2(kk

)1(kk

1a+

=r

)1(11Z )1(

kkZ )1(nnZ

N1

1 k n

)0(11Z )0(

kkZ )0(nnZ

N0

1 k n

)2(11Z )2(

kkZ )2(nnZ

N2

1 k n1aI

r2aI

r0aI

r

117






Sequence Voltages at bus j)0(

jk0a0a ZIVrr

−=

)1(jk1a1a ZI1V

rr−=

)2(jk2a2a ZIV

rr−=

1a)2(kk

)0(kk

)0(kk

2a IZZ

ZI

rr

+−=

1a)2(kk

)0(kk

)2(kk

0a IZZ

ZI

rr

+−=

118






Example: Consider a double line-to-ground fault at bus 4 of the four-bus test system. Find the phase currents in lines L2 and L3.

8538.1j)Z//Z(Z

1I )0(

kk)2(

kk)1(

kk1a −=

+=

rSequence Fault Currents

8703.0jIZZ

ZI 1a)2(

kk)0(

kk

)0(kk

2a =+

−=rr

9835.0jIII 2a1a0a =−−=rrr


3195.0ZIVVV )0(440a42a41a40a =−=== −−−

rrrr

119







0435.0ZIV )0(240a20a =−=−

rr

5641.0ZI1V )1(241a21a =−=−

rr

2046.0ZIV )2(242a22a =−=−

rr


1101.0ZIV )0(340a30a =−=−

rr

5271.0ZI1V )1(341a31a =−=−

rr

222.0ZIV )2(342a32a =−=−

rr

120







46.0j6.0j

3195.00435.0I 2L0a =

−=−

r

8155.0j3.0j

3195.05641.0I 2L1a −=

−=−

r

3828.0j3.0j

3195.02046.0I 2L2a =

−=−

r


0273.0jIIII 2L2a2L1a2L0a2La =++= −−−−

rrrr

6764.0j0378.1I 2Lb +−=−

r

6764.0j0378.1I 2Lc +=−

r

121







5235.0j4.0j

3195.01101.0I 3L0a =

−=−

r

0383.1j2.0j

3195.05271.0I 3L1a −=

−=−

r

4874.0j2.0j

3195.0222.0I 3L2a =

−=−

r


0273.0jIIII 3L2a3L1a3L0a3La −=++= −−−−

rrrr

799.0j3213.1I 3Lb +−=−

r

799.0j3213.1I 3Lc +=−

r

122






� Thevenin Equivalent Circuit

� Three-Phase Line Segment Model

� Transformer Generalized Matrices

� Analysis of Faulted Unbalanced Feeder

Short Circuit Analysis of Unbalanced Distribution System

123






Equivalent system:

Source ABCsysZ

ABCI

ABCLNE ABC

LNV

abcI

abcLNV

Thevenin equivalent circuit @ secondary bus:

thE thZ

abcI

abcLNV

Thevenin Equivalent Circuit

124






Thevenin Equivalent CircuitPrimary transformer equivalent line-to-neutral voltages

abctABCsys

ABCLNABC

ABCsys

ABCLN

ABCLN IdZEIZEV ⋅⋅⋅⋅⋅⋅⋅⋅−−−−====⋅⋅⋅⋅−−−−====

Secondary line-to-neutral voltage:

abctABCLNt

abcLN IBVAV ⋅⋅⋅⋅−−−−⋅⋅⋅⋅====

Substituting,

{{{{ }}}} abctabctABCsys

ABCLNt

abcLN IBIdZE AV ⋅⋅⋅⋅−−−−⋅⋅⋅⋅⋅⋅⋅⋅−−−−⋅⋅⋅⋅====

Thevenin equivalent voltages & impedances:

ttABCsyst

abcth

ABCLNt

abcth

BdZAZ

EAE

++++⋅⋅⋅⋅⋅⋅⋅⋅====

⋅⋅⋅⋅====

125






Three-Phase Line Segment Model

zaa

zbb

zcc

zab

zbc

zca

½ Yabc½ Yabc

clineI

blineI

alineI

cmI

bmI

amI

cnI

bnI

anI

mcgV

mbgV

magV

ncgV

nbgV

nagV

abcmc,Iabc

nc,I

abc

abcabc

abcm

abcmLG

abcnLG

Zb

YZUa

IbVaV

=

⋅⋅+=

⋅+⋅=

21

,,

abcabc

abcabcabcabc

abcm

abcmLG

abcn

YZUd

YZYYc

IdVcI

⋅⋅+=

⋅⋅⋅+=

⋅+⋅=

21

41

,

Voltages & currents at node n in terms of the voltages & currents at node m:

Node n Node m

126






Three-Phase Line Segment Model

abcn

abcnLG

abcm

abcn

abcnLG

abcmLG

IdVcI

IbVaV

⋅+⋅−=

⋅−⋅=

,

,,

Voltages & currents at node m in terms of the voltages & currents at node n:

Voltages at node m as a function of voltages at node n and currents entering node m:

baB

aA

IBVAV

⋅=

=

⋅−⋅=

−

−

1

1

,,abcm

abcmLG

abcmLG

127






Transformer Generalized Matrices

abctabcLNtABC

abctabcLNt

ABCLN

IdVcI

IbVaV

⋅⋅⋅⋅++++⋅⋅⋅⋅====

⋅⋅⋅⋅++++⋅⋅⋅⋅====

abctABCLNt

abcLN IBVAV ⋅⋅⋅⋅−−−−⋅⋅⋅⋅====

Generalized three-phase transformer bank:

H1

H2

H3

H0

X1

X2

X3

X0

aI

bI

cI

nI

AI

BI

CI

NI

ANV

BNV

CNV

anV

bnV

cnV

H1

128







Delta – Grounded Wye Step-Down Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−

−−−−

−−−−

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⋅⋅⋅⋅−−−−

====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅

−−−−====

====

101

110

0111

d

000

000

000

c

side voltage-low the to referred are ,,

02

20

20

3b

012

201

120

3a

sidelow rated ,

side high rated ,

ttt

ct

bt

at

bt

at

ct

at

ct

bt

tt

tt

LN

LLt

n

ZZZ

ZZ

ZZ

ZZnn

V

Vn

129







Delta – Grounded Wye Step-Down Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−

−−−−

−−−−

⋅⋅⋅⋅====ct

bt

at

tt

t

Z

Z

Z

n00

00

00

B

110

011

1011

A

130







Ungrounded Wye – Delta Step-Down Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−

−−−−

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−

−−−−

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−

−−−−

−−−−

⋅⋅⋅⋅====

====

012

021

011

31

d

000

000

000

c


02

02

0

3b

101

110

011

a

sidelow rated ,

side high rated ,

ttt

cat

bct

abt

cat

cat

bct

bct

abt

abt

tttt

LL

LNt

n

ZZZ

ZZ

ZZ

ZZn

n

V

Vn

131







Ungrounded Wye – Delta Step-Down Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−−−−−

−−−−−−−−

−−−−++++

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

024

0422

0222

91

B

201

120

012

31

A

cat

abt

cat

abt

cat

bct

cat

bct

abt

bct

bct

abt

t

tt

ZZZZ

ZZZZ

ZZZZ

n

132







Grounded Wye – Gounded Wye Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

====

100

010

0011

d

000

000

000

c


00

00

00

b

100

010

001

a

sidelow rated ,

side high rated ,

ttt

ct

bt

at

ct

bt

at

tttt

LN

LNt

n

ZZZ

Z

Z

Z

nn

V

Vn

133







Grounded Wye – Grounded Wye Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

ct

bt

at

tt

t

Z

Z

Z

n00

00

00

B

100

010

0011

A

134







Delta – Delta Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−

−−−−−−−−

−−−−−−−−

⋅⋅⋅⋅====

====

cat

bct

abt

abcttV

tt

LL

LLt

Z

Z

Z

n

n

V

Vn

00

00

00

Z

100

010

001

A

201

120

012

31

W

211

121

112

3a

sidelow rated ,

side high rated ,

135







Delta – Delta Connection

1

1

1

GZWB

221

121

112

31

A

100

010

0011

d

GZAWb

0

0

01

G

⋅⋅⋅⋅⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−

−−−−−−−−

−−−−−−−−

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡⋅⋅⋅⋅====

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−−−−−

++++

−−−−

⋅⋅⋅⋅++++++++

====

abctt

tt

tt

abctVt

bct

bct

abt

cat

abt

cat

bct

cat

cat

bct

abt

nn

ZZZ

ZZZ

ZZ

ZZZ

136







Open Wye – Open Delta Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡−−−−⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡−−−−

−−−−

⋅⋅⋅⋅====

====

000

100

0011

d

000

000

000

c

side voltage-low the to referred are ,

000

00

00

b

000

110

011

a

sidelow rated ,

side high rated ,

ttt

bct

abt

bct

abt

tttt

LL

LNt

n

ZZ

Z

Z

nn

V

Vn

137







Open Wye – Open Delta Connection

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−

−−−−−−−−

−−−−

⋅⋅⋅⋅====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−

−−−−⋅⋅⋅⋅====bct

abt

bct

abt

bct

abt

tt

t

ZZ

ZZ

ZZ

n20

0

02

31

B

021

011

012

31

A

138






Short-Circuit Analysis of Unbalanced Feeders

ABCsysZ ABC

subZ ABCeqSZ abc

xfmZ abceqLZ

SystemVoltageSource

EquivalentSystem

Impedance

SubstationTransformer

TotalPrimary

LineSegment

Impedance

In-lineFeeder

Transformer

TotalSecondary

LineSegment

Impedance

1 2 3 4 5

System: (((( )))) (((( ))))

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

++++−−−−−−−−

−−−−++++−−−−

−−−−−−−−++++

⋅⋅⋅⋅====

ΩΩΩΩ−−−−====ΩΩΩΩ====φφφφφφφφ

011010

100110

101001

)(

11

2

03

2

1

2

2

2

31

Z

23

Z

ZZZZZZ

ZZZZZZ

ZZZZZZ

ZMVAkV

MVAkV

Z

ABCapproxsys

LLLL

139






� Thevenin equivalent voltages at points 2 and 3: computed by multiplying the system voltages by the generalizedtransformer matrix At of the substation transformer.

� Thevenin equivalent voltages at points 4 and 5: the voltage at node 3 multiplied by the generalized transformermatrix At of the in-line transformer.

� Thevenin equivalent phase impedance matrices:sum of the phase impedance matrices of each device between thesystem voltage source and the point of fault.


140






EcEa Eb

Thevenin equivalent circuit:

ZTOT

Zf

Zf

Zf

afI

bfI

cfI

axV

bxV

cxV

xgV

a

b

c

x

g

Ea, Eb, Ec = Thevenin equiv. line-to-ground voltages @ the faulted nodeZTOT = Thevenin equiv. phase impedance matrix @ the faulted nodeZf = fault impedance


141






⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡++++

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡++++

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡++++

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

xg

xg

xg

cx

bx

ax

cf

bf

af

f

f

f

cf

bf

af

cccbca

bcbbba

acabaa

c

b

a

V

V

V

V

V

V

I

I

I

Z

Z

Z

I

I

I

ZZZ

ZZZ

ZZZ

E

E

E

00

00

00

xgabcxabcfF

abcfTOTabc VVIZIZE ++++++++⋅⋅⋅⋅++++⋅⋅⋅⋅====

In compressed form,

Combining terms,

{{{{ }}}} xgabcxabcfEQxgabcx

abcfFTOTabc VVIZVVIZZE ++++++++⋅⋅⋅⋅====++++++++⋅⋅⋅⋅++++====

Solving for the fault currents,

1ZY

VYVYEYI−−−−====

⋅⋅⋅⋅−−−−⋅⋅⋅⋅−−−−⋅⋅⋅⋅====

EQ

xgabcxabcabcf


142






Define abcabcP EYI ⋅⋅⋅⋅====

Substituting & rearranging,

xgabcxabcf

abcP VYVYII ⋅⋅⋅⋅++++⋅⋅⋅⋅++++====

Expanding,

(((( ))))

(((( ))))

(((( )))) xga

Scxaabxcbaxcacf

cP

xga

Scxbcbxbbaxbabf

bP

xga

Scxacbxabaxaaaf

aP

xg

xg

xg

cccbca

bcbbba

acabaa

cx

bx

ax

cccbca

bcbbba

acabaa

cf

bf

af

cP

bP

aP

VYVYVYVYII

VYVYVYVYII

VYVYVYVYII

V

V

V

YYY

YYY

YYY

V

V

V

YYY

YYY

YYY

I

I

I

I

I

I

++++++++++++++++====

++++++++++++++++====

++++++++++++++++====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡++++

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡++++

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡


143






where

acabaaa

S

bcbbbab

S

acabaaa

S

YYYY

YYYY

YYYY

++++++++====

++++++++====

++++++++====

3 equations, 7 unknowns - xgcxbxaxcf

bf

af VVVVIII ,,,,,,

cP

bP

aP III ,, are functions of the total impedance &

the Thevenin voltages and are known

Needed: 4 additional equations


144






Three-Phase Faults:

0

0

====++++++++

============

cba

cxbxax

III

VVV

Three-Phase-to-Ground Faults:

0

0

====++++++++

================

cba

xgcxbxax

III

VVVV

Line-to-Line Faults (assume i-j fault with phase k unfaulted):

0

0

0

====++++

====

========

jf

if

kf

jxix

II

I

VV


145






Line-to-Ground Faults (assume phase k fault withphases i and j unfaulted):

0

0

========

========

jf

if

xgkx

II

VV


146






7 equations in matrix form:

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

−−−−−−−−−−−−−−−−−−−−−−−−−−−−

−−−−−−−−−−−−−−−−−−−−−−−−−−−−

−−−−−−−−−−−−−−−−−−−−−−−−−−−−

−−−−−−−−−−−−−−−−−−−−−−−−−−−−====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

xg

cx

bx

ax

cf

bf

af

S

S

S

cP

bP

aP

V

V

V

V

I

I

I

YYYY

YYYY

YYYY

I

I

I

3333231

2232221

1131211

100

010

001

0

0

0

0

In condensed form:

XCI ⋅⋅⋅⋅====sP


147






Solving,sPICX 1 ⋅⋅⋅⋅==== −−−−

Example: 3-phase fault

1

1 1 1

737271

665544

============

============

CCC

CCC

All of the other elements in the last 4 rows of C will be set to zero.


148






ExampleInfinite

bus

1 2 3 4

ABCI abcI

ABCeqSZ abc

eqLZ

Compute the short-circuit currents for a bolted line-to-line fault between phases a and b at node 4.

SLNtth EAE 4, ⋅⋅⋅⋅====

Line-to-neutral Thevenin voltage at node 4:

Thevenin equiv. impedance at secondary terminals (node 3):

abctt

ABCeqStthZ ZdZA3, ++++⋅⋅⋅⋅⋅⋅⋅⋅====

149






Total Thevenin impedance at node 4:

abceqLthTOTth ZZZZ 3.4, ++++========

Equivalent admittance matrix at node 4:1

4, ZY −−−−==== TOTeq

Equivalent injected currents at point of fault:

4,4, EYI theqP ⋅⋅⋅⋅====

For the a-b fault at node 4,

0

0

0

========

====

====++++

bxax

cf

bf

af

VV

I

II

150






Unknowns are computed as

SPICX 1 ⋅⋅⋅⋅==== −−−−

Suppose that the phase impedance matrices for the 2 line segments are:

Ω

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

Ω

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

4970.01921.01931.00614.01751.00598.0

1931.00614.04885.01939.02302.00607.0

1751.00598.02302.00607.05035.01907.0

5353.01414.02955.00361.02752.00361.0

2953.00361.05353.01414.03225.00361.0

2752.00361.03225.00361.05353.01414.0

jjj

jjj

jjj

jjj

jjj

jjj

abceqL

ABCeqS

Z

Z

151






The transformer bank consists of three single-phase transformers eachrated: 2000 kVA, 12.47-2.4 kV, Z = 1.0 + j6.0 %

Source line segment:

[ ]0

5353.01414.02955.00361.02752.00361.0

2953.00361.05353.01414.03225.00361.0

2752.00361.03225.00361.05353.01414.0

100

010

001

1

1

11

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡===

c

Zb

Uda

jjj

jjj

jjj

ABCeqS

152






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

⋅=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

−

−

5353.01414.02955.00361.02752.00361.0

2953.00361.05353.01414.03225.00361.0

2752.00361.03225.00361.05353.01414.0

100

010

001

11

11

111

jjj

jjj

jjj

baB

aA

153






[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

4970.01921.01931.00614.01751.00598.0

1931.00614.04885.01939.02302.00607.0

1751.00598.02302.00607.05035.01907.0

100

010

001

0

4970.01921.01931.00614.01751.00598.0

1931.00614.04885.01939.02302.00607.0

1751.00598.02302.00607.05035.01907.0

100

010

001

2

2

2

2

22

jjj

jjj

jjj

jjj

jjj

jjj

B

A

c

b

da

Load line segment:

154






Transformer:Transformer impedance in ohms referenced to the low-voltage winding

Ω+=⋅+=

Ω=⋅

=

1728.00288.088.2)06.001.0(Z

88.22000

10004.2 2

jj

Z

lowt

base

Transformer phase impedance matrix

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+

+

+

=

1728.00288.000

01728.00288.00

001728.00288.0

j

j

jabctZ

Turns ratio:

1958.54.247.12

==tn

Transformer ratio:

9998.24.23

47.12=

⋅=ta

155






Generalized matrices are:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

−−−−

−−−−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⋅

−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−

−−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡⋅

−=

000

000

000

02993.00499.05986.00998.0

5986.00998.002993.00499.0

2993.00499.05986.00998.00

02

20

20

3

07319.14639.3

4639.307319.1

7319.14639.30

012

201

120

3

t

tt

tt

ttt

t

tt

jj

jj

jj

ZZ

ZZ

ZZn

n

c

b

a

156






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+

+

+

==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

⋅=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

⋅=

1728.00288.000

01728.00288.00

001728.00288.0

1925.01925.00

01925.01925.0

1925.001925.0

110

011

1011

1925.001925.0

1925.01925.00

01925.01925.0

101

110

0111

j

j

j

n

n

abctt

tt

tt

ZB

A

d

157






The infinite bus balanced line-to-line voltages are 12.47 kV, which leads tobalanced line-to-neutral voltages at 7.2 kV:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

−∠

∠

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

−∠

∠

=o

o

o

sLNo

o

o

sLL1206.7199

1206.7199

06.7199

V

150470,12

90470,12

30470,12

,, EE

The line-to-neutral Thevenin circuit voltages at node 4 are:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

−∠

−∠

=⋅=o

o

o

sLNtth902400

1502400

302400

,4, EAE

158






The Thevenin equivalent impedance at the secondary terminals (node 3)of the transformer consists of the primary line impedances referredacross the transformer, plus the transformer impedances:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−−−

−−+−−

−−−−+

=

+⋅⋅=+⋅⋅=

1906.00366.00071.00039.00106.00039.0

0071.00039.01886.00366.00086.00039.0

0106.00039.00086.00039.01921.00366.0

3,

jjj

jjj

jjj

abctt

ABCeqSttt

ABCeqStth ZdZABdZAZ

Total Thevenin impedance at node 4:

Ω

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+++

=

+==

6876.02287.01860.00575.01645.00559.0

1860.00575.06771.02305.02216.00568.0

1645.00559.02216.00568.06955.02273.0

3,4,

jjj

jjj

jjj

abceqLthTOTth ZZZZ

159






Equivalent admittance matrix at node 4:

S

4532.14843.03133.01145.02510.00688.0

3133.01148.05280.15501.03907.01763.0

2510.00688.03907.01763.04771.15031.0

14,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+−

+−−+−

+−+−−

=

= −

jjj

jjj

jjj

TOTeq ZY

The equivalent injected currents at the fault point:

A

4.169.4440

0.1389.4878

4.968.4466

4,4,⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

∠

−∠

=⋅=o

o

o

theqp EYI

Sums of each row of the equivalent admittance matrix:

S

8889.03007.0

8240.02590.0

8353.02580.03

1,

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

== ∑= j

j

j

Yk

ikeqsY

160






For the a-b fault at node 4,

0 0

0 0

==

==+

bxax

cf

bf

af

VV

III

The coefficient matrix

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−+−−−

−+−−+−

−+−+−−

=

0001000

0010000

0000100

0000011

889.0301.0452.1484.0313.0115.0251.0069.0100

824.0259.0314.0115.0528.1550.0390.0176.0010

835.0258.0252.0069.0390.0176.0477.1501.0001

jjjj

jjjj

jjjj

C

161






The injected current matrix:

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∠

∠

−∠

=

0

0

0

0

4.169.4440

0.1389.4878

4.968.4466

o

o

o

spI

The unknowns are computed by:

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

∠

−∠

∠

−∠

=⋅= −

o

o

o

o

sp

1.899.2587

0

0

6.904.7740

0

6.1717.8901

4.87.8901

1 ICX

The interpretation of the results are:

oxg

cxcf

bxob

f

oax

oaf

XV

XVXI

XVXI

XVXI

1.899.2587

0 0

0 6.1717.8901

6.904.7740 4.87.8901

7

63

52

41

∠==

====

==∠==

−∠==−∠==

163






Fault Current at Different Times

Clearing Time of High Voltage

Breakers

Contact Opening Time

of High Voltage Breakers

Clearing Time of

Fuse

Clearing Time of Molded Breakers

Fault Current that upstream overcurrent

devices must withstand while downstream devices

isolate the fault

164







� 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

165






� 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


166






Application of Short Circuit Analysis

� Comparison of Closed-and-Latch (Momentary or Making) and Interrupting (Breaking) Duties of Interrupting Devices

� Comparison of Short-time 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

167






Characteristic of Short Circuit Currents

( )sindi

Ri L E tdt

ω φ+ = +

( )2 2 2 2

sin sin( ) R tXE t E

i eR X R X

ωω θ φ θ φ −+ − −= +

+ +

R L

E sin (ωωωωt+φφφφ)

168






Characteristic of Short Circuit Currents

( ) ωω θ φ θ φ −+ − −= +

+ +2 2 2 2

sin sin( ) R tXE t E

i eR X R X

FactoralAsymmetricII RMS lsymmetricaRMStotal •=,

169






ANSI/IEEE and IEC Standards

� ANSI/IEEE: American National Standards Institute/ Institute of Electrical and Electronics Engineers

� IEC: International ElectrotechnicalCommission

Prescribes Test Procedures and Calculation Methods

171






1.5-4 Cycle Network: the network used to calculate interrupting short-circuit current and protective device duties 1.5-4 cycles after the fault.

N/ARelay

N/ASwitchgear and MCC

N/AFuse

N/ALow Voltage CB

Interrupting CapabilityHigh Voltage CB

DutyType of Device

ANSI/IEEE Calculation Method

170






½ Cycle Network: also known as the subtransientnetwork because all rotating machines are represented by their subtransient reactances


Type of Machine XscUtility X”Turbo generator Xd”Hydro-generator with amortisseur windings Xd”Hydro-generator without amortisseur windings 0.75 Xd’Condenser Xd”Synchronous motor Xd”Induction Machine > 1000 hp @ 1800 rpm or less Xd” > 250 hp @ 3600 rpm Xd” All other ≥ 50 hp 1.2 Xd” < 50 hp 1.67 Xd”

Xd” of induction motor = 1/(per-unit locked-rotor current at rated voltage)

173






30 Cycle Network: also known as the steady-state network

InfinityInduction Machine

InfinitySynchronous Motor

InfinityCondenser

Xd’Hydro-generator w/o Amortisseur Winding

Xd’Hydro-generator w/ Amortisseur Winding

Xd’Turbo Generator

X’’Utility

XscType of Machine


174






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 short-circuit current is the portion of the short-circuit 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.


175






Momentary (1/2 Cycle) Short-Circuit Current

Peak Momentary Short-Circuit Current

π−

= ⋅

⎛ ⎞= +⎜ ⎟⎜ ⎟

⎝ ⎠

, , ,

2 1

mom peak p mom rms symm

X Rp

I MF I

MF e


176






Momentary (1/2 Cycle) Short-Circuit Current

Asymmetrical RMS value of Momentary Short-Circuit Current

π

−

−

=

= ⋅

= +

, ,

, , , ,

2

3

1 2

pre faultmom rms symm

eq

mom rms asymm m mom rms symm

X Rm

VI

Z

I MF I

MF e


177






High Voltage Circuit Breaker Interrupting Duty (1.5-4 Cycle)

Adjusted RMS value of Interrupting Short-Circuit Current (for total current basis CBs)

−=int, ,

3pre fault

rms symm

eq

VI

Z


symmrmsiadjrms IAMFI ,int,,int, ⋅=

( )lrli MFMFNACDMFAMF −+=where

178






No AC Decay (NACD) Ratio

The NACD ratio is defined as the remote contributions to the total contributions for the short-circuit current at a given location


total

remote

IINACD =

• Total short circuit current Itotal = Iremote + Ilocal• NACD = 0 if all contributions are local• NACD = 1 if all contributions are remote

179







π−

= +

4

1 2t

X RrMF e


Circuit BreakerRating in Cycles

Contact PartingTime ( t ) in Cycles

8 45 33 22 1.5

180






High Voltage Circuit Breaker Interrupting Duty (1.5-4 Cycle) Calculation


Multiplying factors (total current basis CBs) MFr for 3-phase & line-to-ground faults.

181








Multiplying factors (total current basis CBs) MFl for 3-phase faults.

182








Adjusted RMS value of Interrupting Short-Circuit Current (for symmetrically rated CBs)

SIAMF

I symmrmsiadjrms

,int,,int,

⋅=

Circuit Breaker ContactParting Time (Cycles)

S Factor

4 1.03 1.12 1.2

1.5 1.3

183






Low Voltage Circuit Breaker Interrupting Duty (1/2 Cycle) Calculation

Adjusted asymmetrical RMS value of Interrupting Short-Circuit Current

−=int, ,

3pre fault

rms symm

eq

VI

Z


symmrmsadjrms IMFI ,int,,int, ⋅=

184







π

π

π

π

−

−

−

−

+=

+

+=

+

( )

( )

2(1 )

2(1 )

1 2

1 2

test

test

X R

X R

X R

X R

eMF

e

eMF

e

Unfused power breakers

Fused power breakers &

Molded Case


Note: If calculated MF < 1.0, set MF = 1.0

185








Circuit Breaker Type (X/R)testPower Breaker (Unfused) 6.59Power Breaker (Fused) 4.90Molded Case (> 20 kA) 4.90Molded Case (10.001 – 20 kA) 3.18

Molded Case (10 kA) 1.73

186






Fuse Interrupting Short-Circuit Current Calculation

- same procedure as Circuit Breaker Interrupting

Duty calculation.


187






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 zero-sequence network.

Calculations consider the electrical distance from the fault location to synchronous generators.

188






Initial Symmetrical Short-Circuit Current (I’’k) RMS value of the AC symmetrical component of an available short-circuit current applicable at the instant of short-circuit if the impedance remains at zero time value.

Peak Short-Circuit Current (ip)Maximum possible instantaneous value of the available short-circuit current.

Symmetrical Short-Circuit Breaking Current (Ib)RMS value of an integral cycle of the symmetrical AC component of of the available short-circuit current at the instant of contact separation of the first pole of a switching device


189






Steady-state Short Circuit Current (Ik)RMS value of the short-circuit current which remains after the decay of the transient phenomena.

Subtransient Voltage (E’’) of a Synchronous MachineRMS value of the symmetrical internal voltage of a synchronous machine which is active behind the subtransientreactance Xd’’ at the moment of short circuit.

Far-from-Generator Short-Circuit Short-circuit condition to which the magnitude of the symmetrical ac component of the available short-circuit current remains essentially constant


190






Near-to-Generator Short-Circuit Short-circuit condition to which at least one synchronous machine contributes a prospective initial short-circuit current which is more than twice the generator’s rated current or a short-circuit condition to which synchronous and asynchronous motors contribute more than 5% of the initial symmetrical short-circuit current (I”k) without motors.

Subtransient Reactance (Xd’’) of a Synchronous MachineEffective reactance at the moment of short-circuit. 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.

( )

ϕ

= +

=+

''

max''1 sin

K G d

nG

r d r

Z K R jX

ckVK

kV x

kVn = nominal voltage of the terminal buskVr = motor rated voltageXd” = subtransient reactanceϕr = machine rated power factor


191






Minimum Time Delay (Tmin) of a Circuit Breaker

Shortest time between the beginning of the short-circuit 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 sourcefor the minimum and maximum current calculations

1.001.10> 35 KV to 230 KV

1.001.10> 1 kV to 35 kV

1.001.05Other LV up to 1 KV

0.951.00230/400 V

Min SC CalculationMax SC Calculation

Voltage FactorVoltage Factor


192







k

nk Z

UcI3

=′′ Zk = equiv. Impedance at fault point

kp Iki ′′= 2 k = function of system R/X at fault location

kb II ′′= for far-from-generator fault

kb II ′′μ=for synch. machines, for near-to-generator

faults

kb IqI ′′μ= for ind. machines, for near-to-generator

faults

193







HV CB asymmetrical breaking & dc current rating

⎟⎠⎞

⎜⎝⎛ π

−=

⎟⎠⎞

⎜⎝⎛ π−+=

RXtfII

RXtfII

symmbdc

symmbasymmb

/ 2exp2

/ 4exp21

min,

min,,

f = system frequencytmin = minimum delay timeIb,symm = AC breaking current

X/R = calculated based on testing PF of 7% at 50 Hz

194







LV CB asymmetrical breaking current rating

⎟⎠⎞

⎜⎝⎛ π−+=

RXtfII symmbasymmb / 4exp21 min

,,

f = system frequencytmin = minimum delay timeIb,symm = AC breaking currentX/R = calculated based on testing PF given by IEC

195







Fuse asymmetrical breaking current rating

⎟⎠⎞

⎜⎝⎛ π−+=

RXtfII symmbasymmb / 4exp21 min

,,

f = system frequencytmin = assumed to be a half cycleIb,symm = AC breaking currentX/R = calculated based on testing PF of 15%

196






Selection of Device Duties

40.277428013.8 KV13.8 – 1000

30.45813.56013.8 KV13.8 – 750

19.637214013.8 KV13.8 – 500

46.97848.6804.16 KV4.16 – 350

33.25835604.16 KV4.16 – 250

10.11910.5204.16 KV4.16 – 75

Short-Circuit Capability

(Symmetrical RMS Current at 3-Cycle Parting

Time

Closing and Latching

Capability (Total First Cycle RM

Current)

Interrupting Rating (Total

RMS Current at 4-cycle Contact-

Parting Time

Momentary Rating (Total 1st-Cycle RMS

Current

Example Maximum System

Operating Voltage

Circuit Breaker Nominal Size Identification

8-Cycle Total-Rated Circuit Breakers (KA)

5-Cycle Symmetrical-Rated Circuit Breakers (KA)

198






Short Circuit Analysis

Documents

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