Cp1 B9 Lecture No. 4 - Short Circuit Analysis.pdf

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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



Three-Phase Fault



Single-Line-to-Ground Fault



Line-to-Line Fault



Double-Line-to-Ground Fault U. P. National Engineering Center National Electrification Administration

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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 Line-to-Ground Fault 2. Line-to-Line Fault 3. Double Line-to-Ground 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

Line-to-Line

Double Line-to-Ground

Single Line-to-Ground

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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

Line-to- 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. • Positive-Sequence Network • Negative-Sequence Network • Zero-Sequence 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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Three-Phase 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

Note: The system is still balanced. Currents and voltagesr are positive sequence only. The ground current I g is zero. U. P. National Engineering Center National Electrification Administration

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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 4-bus 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 = 0.259055∂ = 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 = (Xa||Xb ) ||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 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. 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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Positive-Sequence 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 Line-to-Ground 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 −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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Single-Line-to-Ground 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 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. 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 per-unit of a common MVA base. U. P. National Engineering Center National Electrification Administration

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Positive-Sequence 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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Negative-Sequence Network: F2 j0.05

j0.15 j0.05

Open

r Ia2

F2 +

j0.6

j0.4

r Va 2 -

N2

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Zero-Sequence 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)

= − j0.804 p.u. U. P. National Engineering Center National Electrification Administration

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Phase Fault Currents

r r IA = 3IA0 = − j2.411 p.u. r r IB = IC = 0

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Line-to-Line 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 = − I c r r r (3) Vb − 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 = − I a 2 =

Vf Z1 + Z 2 + Z f

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Line-to-Line 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 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.

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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Double-Line-to-Ground 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

or

r I a0 = −

r Z 2T I a1 Z 0T + Z 2T

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Double-Line-to-Ground 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 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.

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 hand-calculation error. Is there a way for a computer to implement this methodology?

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Development of the Model Consider the three-bus system shown below. Let us analyze the system for a three-phase 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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Positive-Sequence 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 re-draw. Assume EG = 1.0 per unit.

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1

j0.24

2

3

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

Z1 = j[0.2 //(0.2 + 0.6)] = j0.16

r EG 1 IF = = = − j6.25 Z1 Z1 For a three-phase fault in bus 3, we get

Z1 = j[0.24 + 0.2 //(0.2 + 0.6)] = j0.4

r EG 1 IF = = = − j2.5 Z1 Z1 U. P. National Engineering Center National Electrification Administration

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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. -

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 positive-sequence 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 rake-equivalent. U. P. National Engineering Center National Electrification Administration

-

Z12 Z22

r I2

Z23 Z33 Z13 r

I3

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Consider again the three-bus 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 off-diagonal 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 three-bus 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 three-phase 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 (Ik-Iw). 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 − 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+Zkk-2Zjk+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 step-bystep 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 positive-sequence busimpedance matrix for the four-bus 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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Positive-sequence 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 Kron’s reduction.

X2X 4−1X3 =

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 − 0.1449

4

0.2609 0.3130

0.1739

0.6130

* 0.1159 0.1391 − 0.1449 0.4391 U. P. National Engineering Center National Electrification Administration

0.4391 0.784

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Apply Kron’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 positive-sequence bus-impedance matrix for the four-bus test system.

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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 busimpedance matrices can also be described by a rake equivalent circuit.

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Example: Find the zero-sequence bus-impedance matrix for the four-bus test system. 4

Zero-sequence 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 Kron’s reduction. We get

0

0

0

X2X 4−1X3 = 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 − 0.177

4

0

0.0743 0.0177

0.6743 0.6566

*

0

0.0566 − 0.177 0.6566 1.2336

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Positive-Sequence Zbus 1) 1) … Z(21 Z(22 Z(21n)

=

(1) bus

Z

…

The positive-sequence bus-impedance 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 Kron’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 zero-sequence bus-impedance matrix for the four-bus test system.

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Negative-Sequence Zbus (2) (2) … Z11 Z12 Z1(2n)

The negative-sequence bus-impedance 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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Zero-Sequence Zbus Z(210) Z(220) … Z(20n)

…

The zero-sequence bus-impedance 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 bus-impedance matrices can be used for the analysis of the following shunt faults: 1. Three-Phase Fault 2. Line-to-Line Fault 3. Single Line-to-Ground Fault 4. 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.

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Three-phase 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 − Zkk

r The current in any line is Imn

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r r Vm − Vn = zmn

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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 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) = = − 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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r IG2

j0.5 + 1.0 -

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

The positive-sequence bus-impedance 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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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 line-to-ground fault at bus 4 of the four-bus 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 Line-to-Ground 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 line-to-ground fault at bus 4 of the four-bus 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



Three-Phase Line Segment Model



Transformer Generalized Matrices



Analysis of Faulted Unbalanced Feeder

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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 line-to-neutral voltages ABC ABC ABC ABC ABC VLN = ELN − Z sys ⋅ IABC = ELN − Z sys ⋅ dt ⋅ Iabc

Secondary line-to-neutral 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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Three-Phase 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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Three-Phase 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 three-phase 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 Step-Down 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 Step-Down 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 Step-Down 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 Step-Down 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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Short-Circuit 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

In-line 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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Z 0 − Z1 2Z 1 + Z 0 Z 0 − Z1

Z 0 − Z1 ⎤ Z 0 − Z1 ⎥ ⎥ 2Z1 + Z 0 ⎥⎦

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Short-Circuit 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 in-line 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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Short-Circuit 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. line-to-ground 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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Short-Circuit 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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Short-Circuit 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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Short-Circuit 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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Short-Circuit Analysis of Unbalanced Feeders

Three-Phase Faults: Vax = Vbx = Vcx = 0

I a + I b + Ic = 0 Three-Phase-to-Ground Faults: Vax = Vbx = Vcx = V xg = 0 I a + I b + Ic = 0 Line-to-Line Faults (assume i-j 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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Short-Circuit Analysis of Unbalanced Feeders

Line-to-Ground Faults (assume phase k fault with phases i and j unfaulted): Vkx = V xg = 0 I fi = I fj = 0

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Short-Circuit 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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Short-Circuit Analysis of Unbalanced Feeders Solving, X = C−1 ⋅ IPs Example: 3-phase 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 short-circuit currents for a bolted line-to-line fault between phases a and b at node 4. Line-to-neutral 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: −1 Yeq ,4 = ZTOT

Equivalent injected currents at point of fault:

IP = Yeq ,4 ⋅ Eth ,4 For the a-b 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 single-phase transformers each rated: 2000 kVA, 12.47-2.4 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 low-voltage 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 line-to-line voltages are 12.47 kV, which leads to balanced line-to-neutral 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 line-to-neutral 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 a-b 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 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

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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 1.5-4 Cycle Network: the network used to calculate interrupting short-circuit current and protective device duties 1.5-4 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 Hydro-generator with amortisseur windings Hydro-generator 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/(per-unit locked-rotor 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’

Hydro-generator w/ Amortisseur Winding

Xd’

Hydro-generator 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 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. U. P. National Engineering Center National Electrification Administration

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ANSI/IEEE Calculation Method Momentary (1/2 Cycle) Short-Circuit Current

Peak Momentary Short-Circuit 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) Short-Circuit 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 (1.5-4 Cycle)

Adjusted RMS value of Interrupting Short-Circuit 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 short-circuit 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 (1.5-4 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 (1.5-4 Cycle) Calculation

Multiplying factors (total current basis CBs) MFr for 3-phase & line-to-ground faults. U. P. National Engineering Center National Electrification Administration

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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.5-4 Cycle)

Multiplying factors (total current basis CBs) MFl for 3-phase faults. U. P. National Engineering Center National Electrification Administration

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ANSI/IEEE Calculation Method High Voltage Circuit Breaker Interrupting Duty (1.5-4 Cycle)

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

I int,rms ,adj

AMFi ⋅ 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 Short-Circuit 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 – 20 kA)

3.18

Molded Case (10 kA)

1.73

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ANSI/IEEE Calculation Method Fuse Interrupting Short-Circuit 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 zero-sequence network. Calculations consider the electrical distance from the fault location to synchronous generators.

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IEC Calculation Method 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 U. P. National Engineering Center National Electrification Administration

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IEC Calculation Method 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 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.

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

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IEC Calculation Method 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 Machine

Reactance

(Xd’’)

of

a

Synchronous

Effective 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.

(

ZK = KG R + jX d''

KG =

)

cmax kVn kVr 1 + xd'' sin ϕr

U. P. National Engineering Center National Electrification Administration

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IEC Calculation Method 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 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 far-from-generator fault

I b = μI k′′

for synch. machines, for near-to-generator faults

I b = μqI k′′

for ind. machines, for near-to-generator 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 8-Cycle Total-Rated Circuit Breakers (KA)

5-Cycle Symmetrical-Rated Circuit Breakers (KA)

Momentary Rating (Total 1st-Cycle RMS Current

Interrupting Rating (Total RMS Current at 4-cycle ContactParting Time

Closing and Latching Capability (Total First Cycle RM Current)

Short-Circuit Capability (Symmetrical RMS Current at 3-Cycle 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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