CPT5 - Short Circuit Analysis- July 25, 2005

Competency Training and Certification Program in Electric Power Distribution System Engineering

Certificate in

Power System Modeling and Analysis Training Course in

Short Circuit Analysis

U. P. NATIONAL ENGINEERING CENTER NATIONAL ELECTRIFICATION ADMINISTRATION

Training Course in Short Circuit Analysis

2

Course Outline 1. Short Circuit Currents 2. Power System Models for Short Circuit Analysis 3. Short Circuit Calculation by Network Reduction 4. Analysis of Faulted Power System 5. Computer Solution 6. Protective Device Duties U. P. National Engineering Center National Electrification Administration

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Short Circuit Currents ‰ Types ‰ The

of Fault

Shunt Fault Point

‰ Sources

of Short Circuit Currents

‰ Characteristics ‰ Short

of Short Circuit Currents

Circuit Studies

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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 [Not included in this course] Simultaneous Fault [Not included in this course] 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

Shunt Faults U. P. National Engineering Center National Electrification Administration

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The Shunt Fault Point

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

r Ia

Line-to- r r r ground Va Vb Vc voltages

r Ib

a b

r Ic

c Fault Currents

Ground

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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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Characteristic of Short Circuit Currents R

L

E sin (ωt+φ)

Ri + L

i =

E sin (ωt + θ − φ ) 2

R +X

2

+

di = E sin (ωt + φ ) dt

E sin(θ − φ ) 2

R +X

2

e

−R

X

ωt

I Asym = Isym AssymetryFactor U. P. National Engineering Center National Electrification Administration

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Training Course in Short Circuit Analysis

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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

Source: Cooper Power Systems U. P. National Engineering Center National Electrification Administration

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Characteristic of Short Circuit Currents Positive Sequence Impedance of Generator The AC RMS component of the current following a three-phase short circuit at no-load condition with constant exciter voltage and neglecting the armature resistance is given by

⎛ −t ⎞ E ⎛ E E ⎞ ⎟⎟ ⎟⎟ exp⎜⎜ + ⎜⎜ − I( t ) = X ds ⎝ X d ' X ds ⎠ ⎝τ d' ⎠ ⎛ E ⎛ −t ⎞ E ⎞ ⎟⎟ exp⎜⎜ ⎟⎟ + ⎜⎜ − ⎝ X d" X d' ⎠ ⎝τ d" ⎠

where E = AC RMS voltage before the short circuit. U. P. National Engineering Center National Electrification Administration

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Training Course in Short Circuit Analysis

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Characteristic of Short Circuit Currents

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Short Circuit Studies ‰

Comparison of Momentary and Interrupting 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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Training Course in Short Circuit Analysis

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Power System Models for Short Circuit Analysis ‰ Sequence

Networks of Power System

‰ Equivalent

Circuit of Utility

‰ Equivalent

Circuit of Generators

‰ Equivalent

Circuit of Transformers

‰ Equivalent

Circuit of Lines

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Sequence Networks of Power System 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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Sequence Networks of Power System F

F

F

+

+

+

Ia1

Ia2

Z1

Va1

Va2

Ia0

Z2

Va0

Z0

+ Vf -

-

V =V – I Z a1

f

a1

1

Positive Sequence

-

V =-I Z a2

a2

2

Negative Sequence

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Vao = - I ao Z o

Zero Sequence

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Sequence Components of Electric Currents From Symmetrical Components: Any Unbalance System of Phasors can be resolved into three balanced system of phasors a) POSITIVE-SEQUENCE PHASOR b) NEGATIVE-SEQUENCE PHASOR c) ZERO-SEQUENCE PHASOR

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Sequence Components of Short Circuit Currents If the fault is balanced (Three Phase Fault) only the Positive Sequence Current exists. If the fault is unbalanced (e.g, SLGF, LLF & DLGF) Positive Sequence Current, Negative Sequence Current exists. Zero Sequence Current exists depending on the connection of generators and transformers.

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Sequence Components of Short Circuit Currents ZERO-SEQUENCE CURRENTS: I c0 c

Ia0

a b 3Io

a

Ia0

Ib0 b

Ic0

c

Ib0 The neutral return carries the in-phase zero-sequence currents. U. P. National Engineering Center National Electrification Administration

Zero-sequence currents circulates in the delta-connected transformers. There is “balancing ampere turns” for the zero-sequence currents. Competency Training & Certification Program in Electric Power Distribution System Engineering

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Equivalent Circuit of Utility Positive & Negative Sequence Impedance From Three-Phase Fault Analysis

I TPF =

Z1 =

V

[V ]

2

f

S TPF

Z1

= V f I TPF =

f

Z1

Where, Z1 and Z2 are the equivalent positive2 sequence and kV = Z 2 negative-sequence Fault MVA 3φ impedances of the utility

[ ]

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Equivalent Circuit of Utility Zero Sequence Impedance From Single Line-to-Ground Fault Analysis

I SLGF =

3V f Z1 + Z2 + Z0

2Z1 + Z0 =

[ ]

3 Vf

S SLGF = V f I SLGF =

[ ]

3Vf

2

2Z 1 + Z 0

Z1 = Z2

2

SSLGF

Resolve to real and imaginary components then solve for Zo

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Equivalent Circuit of Utility The equivalent sequence networks of the Electric Utility Grid are: +

+

R2 +jX2

R0 +jX0

+

r + Eg -

R1 +jX1

-

-

Positive Sequence

Negative Sequence

-

Zero Sequence

Utility Thevenin Equivalent Circuits U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Generators Positive-Sequence Impedance: Xd”=Direct-Axis Subtransient Reactance Xd’=Direct-Axis Transient Reactance Xd=Direct-Axis Synchronous Reactance Negative-Sequence Impedance:

X2 = 12 (X d "+ X q " ) for a salient-pole machine for a cylindrical-rotor machine X2 = X d " Zero-Sequence Impedance:

0.15X d " ≤ X0 ≤ 0.6X d " U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Generators Grounded-Wye Generator The sequence networks for the grounded-wye generator are shown below.

r + Eg -

jZ1

jZ0

jZ2 N2

N1

Positive Sequence

F0

F2

F1

Negative Sequence

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N0

Zero Sequence

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Equivalent Circuit of Generators Grounded-Wye through an Impedance If the generator neutral is grounded through an impedance Zg, the zero-sequence impedance is modified as shown below.

r + Eg -

jZ1

F0

F2

F1

jZ0

jZ2

3Zg N2

N1

Positive Sequence

Negative Sequence

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N0

Zero Sequence

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Equivalent Circuit of Generators Ungrounded-Wye Generator If the generator is connected ungrounded-wye or delta, no zero-sequence current can flow. The sequence networks for the generator are shown below.

r + Eg -

jZ1

jZ0

jZ2 N2

N1

Positive Sequence

F0

F2

F1

Negative Sequence

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N0

Zero Sequence

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Equivalent Circuit of Transformers Positive– & Negative Sequence Networks Z2

Z1 + Primary Side

-

r I1

+

+

Secondary Side

-

Positive Sequence Network

Primary Side

-

Z1 = Z2

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r I2

+ Secondary Side

-

Negative Sequence Network

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Equivalent Circuit of Transformers Zero Sequence Network* Transformer Connection

Zero-Sequence Network

Z0 = Z1

+ r

+ r

VH

VX

-

-

Z0 = Z1 + r

+ r

VH

VX

-

-

*Excluding 3-phase unit with a 3-legged core. U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Transformers Zero Sequence Network * Transformer Connection

Zero-Sequence Network

Z0 = Z1

+ r

+ r

VH

VX

-

-

Z0 = Z1 + r

+ r

VH

VX

-

-

*Excluding 3-phase unit with a 3-legged core. U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Transformers Zero Sequence Network * Transformer Connection

Zero-Sequence Network

Z0 = Z1

+ r

+ r

VH

VX

-

-

Z0 = Z1 + r

+ r

VH

VX

-

-

*Excluding 3-phase unit with a 3-legged core. U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Lines Phase to Sequence Impedances Consider a transmission line that is described by the following voltage equation:

or

r ⎡Va ⎤ ⎡ Z aa ⎢r ⎥ ⎢Z V = ⎢ rb ⎥ ⎢ ab ⎢Vc ⎥ ⎢⎣ Z ac ⎣ ⎦

Z ab Z bb Z bc

r Z ac ⎤ ⎡ I a ⎢r ⎥ Z bc ⎥ ⎢ I b r Z cc ⎥⎦ ⎢⎣ I c

⎤ ⎥ ⎥ ⎥ ⎦

volts

r r Vabc = Z abc I abc

From symmetrical components, we have

r r Vabc = AV012

and

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r r I abc = A I 012

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Equivalent Circuit of Lines Substitution gives or

r r AV012 = Z abc AI 012 r r −1 V 012 = A Z abc A I 012

which implies that

Z 012 = A −1 Z abc A Performing the multiplication, we get

⎡ Z 0 ⎤ ⎡ Z s 0 + 2 Z m0 ⎢Z ⎥ = ⎢ Z − Z m1 ⎢ 1 ⎥ ⎢ s1 ⎢⎣ Z 2 ⎥⎦ ⎢⎣ Z s 2 − Z m 2

Z s2 − Z m2 Z s 0 − Z m0 Z s 1 + 2 Z m1

Z s 1 − Z m1 ⎤ Z s 2 + 2 Z m 2 ⎥⎥ Z s 0 − Z m0 ⎥⎦

Note: Z012 is not symmetric. U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Lines It can be shown that

Z s 0 = 31 ( Z aa + Z bb + Z cc )

Z s 1 = 31 ( Z aa + aZ bb + a 2 Z cc ) Z s 2 = ( Z aa + a Z bb + aZ cc ) 2

1 3

Z m 0 = 31 ( Z ab + Z bc + Z ca ) Z m 1 = ( a Z ab + Z bc + aZ ca ) 1 3

2

Z m 2 = 31 ( aZ ab + Z bc + a 2 Z ca ) U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Lines If the line is completely transposed,

Z s0 = Z s

Z m0 = Z m

Z s1 = Z s 2 = 0

Z m1 = Z m 2 = 0

The sequence impedance matrix reduces to

⎡Z 0 ⎤ ⎡Z s + 2Z m ⎢Z ⎥ = ⎢ 0 1 ⎢ ⎥ ⎢ ⎢⎣ Z 2 ⎥⎦ ⎢⎣ 0

0 Zs − Zm 0

⎤ ⎥ ⎥ Z s − Z m ⎥⎦ 0 0

Note: The sequence impedances are completely decoupled. U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Lines For a completely transposed line, the equation in the sequence domain is r r

V a0 ⎡Z 0 r V a 1 = ⎢⎢ 0 r ⎢⎣ 0 Va2

where

0

Z1 0

0 ⎤ ⎡ I a0 ⎢r ⎥ 0 ⎥ ⎢ I a1 r Z 2 ⎥⎦ ⎢⎣ I a 2

Dm Z 1 = Z 2 = ra s + jωks ln Ds Z 0 = ra s + 3rd s + jωks ln

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De

⎤ ⎥ ⎥ ⎥ ⎦

Ω 3

D s Dm

2

Ω

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Equivalent Circuit of Lines Sequence Capacitance Using matrix notation, we have

r r r r I abc = jωCabcVabc I abc = YabcVabc r r r r From Vabc = AV012 and I abc = YabcVabc, we get r r A I 012 = jω C abc A V012 r r or −1 I 012 = jωA Cabc AV012 Thus, we have

C012 = A −1Cabc A U. P. National Engineering Center National Electrification Administration

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Equivalent Circuit of Lines For a completely transposed line,

Cs0 = Caa = Cbb = Ccc C m 0 = C ab = C bc = C ac

Substitution gives C012 or

0 0 ⎤ ⎡( Cs0 − 2Cm0 ) ⎥ ⎢ = 0 ( Cs0 + Cm0 ) 0 ⎥ ⎢ ⎢⎣ 0 0 ( Cs0 + Cm0 )⎥⎦ C0 = Cs0 − 2Cm0

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Short Circuit Calculation by Network Reduction ‰ Thevenin

Equivalent

‰ Network

Reduction Techniques

‰ Reduced

Sequence Networks

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Thevenin Equivalent Thevenin’s Theorem states that, with respect to a given pair of terminals, any electric circuit can be represented by a single voltage source in series with a single impedance.

r Vth = voltage from terminal a r to b at open-circuit Vth r Zth = ratio of Vth to the short circuit current from terminal a to b

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Zth

+

a

-

b

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Thevenin Equivalent Recall: Fault MVA The short circuit current at any point in the power system is generally expressed in terms of a fault MVA. By definition, where

MVAF = 3 (kVL ) (kA)

kVL = nominal line-to-line voltage in kV kA = short circuit current in kA Note that

Vth =

kVL x 1000 3

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Volts

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Thevenin Equivalent From Zth we get

Vth = Isc

Zth =

kVL x 1000 3(kA) x 1000

=

kVL 3(kA)

From the fault MVA, we get

kA = Substitution gives

Zth

MVA F

3(kVL )

(kVL )2 = MVA F

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Thevenin Equivalent Example: The three-phase fault MVA for a 138-kV bus is given to be 5,975. Find the positivesequence representation of the system at the bus.

Vth = Zth

kVL x 1000 3

=

1 3

(138,000) = 79,674 V

(kVL )2 1382 = = = 3.19 Ω MVA F 5,975

r Vth

Zth

+

a

-

b U. P. National Engineering Center National Electrification Administration

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Network Reduction Combination of Branches in Series

Zeq = Z1 + Z2

Z1

Z2

= (R1 + jX1 ) + (R2 + jX2 )

= (R1 + R2 ) + j( X1 + X2 ) Combination of Branches in Parallel Z1Z2 Zeq = Z1 Z1 + Z2

Z2

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Network Reduction Transforming Wye to Delta Z A = Za Zb + Zb Zc + Zc Za Za

Za Zb + Zb Zc + Zc Za ZB = Zb ZC =

Za Zb + Zb Zc + Za Zc Zc

Za = Zb =

Transforming Delta to Wye U. P. National Engineering Center National Electrification Administration

Zc =

ZA

ZBZC + ZB + ZC

Z

A

Z AZC + ZB + ZC

ZA

Z AZB + ZB + ZC

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Network Reduction 1. Draw the Single Line Diagram. 2. Draw the Impedance Diagram. 3. Convert all parameters to per-unit. 4. Reduce the network between the source(s) and the fault location. 5. Calculate the fault current

Vf If = Z equiv U. P. National Engineering Center National Electrification Administration

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Example: Three Phase Fault Determine the fault current for a three phase bolted fault in each bus for the 4 bus system below. G

LINE

1 3

Line 5 2

Line 2

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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Reduce the network

a) Fault @ Bus 4

X a = X12 + X 23 = 0.3 + 0.6 = 0.9

E

1.0

+ 0.1

Xb

If

(0.9)(0.4) = 0.9 + 0.4 = 0.276923

1 0.3

0.2 0.4

2

0.6

0.5 3

X a X 13 = X a + X 13

4

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X c = X b + X 34 = 0.276923 + 0.5 = 0.776923 Competency Training & Certification Program in Electric Power Distribution System Engineering

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Xd = = =

Xequiv =

Xc X14 Xc + X14 (0.776923) (0.2) 0.776923 + 0.2 0.159055

X gen + Xd

48

E 1.0 +

If

0.25905

= =

0.1 + 0.159055 0.259055 100 x1000 = 8367.64 A Ibase = 1.0 3(6.9) If = 0.259055∂ If = 3.860184 x 8367.64 = 3.860184 p.u. = 32,300.63 A U. P. National Engineering Center National Electrification Administration

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b) Fault @ Bus 3

Xa = X23 + X12 = 0.3 + 0.6 = 0.9

E

1.0

+ 0.1

Xb = X14 + X34

If

= 0.2 + 0.5

1 0.3

= 0.7

0.2 0.4

2

0.6

3

0.5

4

Xequiv = (Xa||Xb ) ||X13 = 0.198425

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X=

Xgen +

50

Xequiv

= 0.1 + 0.198425 = 0.298425

1.0 If = 0.298425 = 3.350923 p.u.

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E 1.0 +

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 X = c X + X12 d

(0.854545)( 0.3) = 0.854545 + 0.3 = 0.222047

X = Xgen + Xd = 0.322047 U. P. National Engineering Center National Electrification Administration

0.322047

1.0 If = 0.322047 = 3.095525 p.u. Competency Training & Certification Program in Electric Power Distribution System Engineering

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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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Network Reduction Reduced Sequence Networks Applying what we have learned so far: 9 There are three distinct network models 9 Every complex network can be reduced into an equivalent circuit. Therefore, we can come up with the following: • Reduced Positive-Sequence Network • Reduced Negative-Sequence Network • Reduced Zero-Sequence Network

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Network Reduction Where in the entire network do we apply this? It was earlier mentioned that we assume the network to be balanced except at the fault point. Therefore we reduce the networks as seen at the fault point

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Network Reduction The Thevenin equivalent of the power system at the fault point is called the sequence network. Positive Sequence F1

r Ia1 Z + r 1 + r Va1 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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Analysis of Faulted Power System ‰ Methodology ‰ Single

Line-to-Ground Fault

‰ Line-to-Line ‰ Double ‰ Three

Fault

Line-to-Ground Fault

Phase Fault

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Methodology Objective: to analyze each of these types of faults 1. Setup the boundary conditions 2. Analyze using symmetrical components 3. Derive expressions relating the symmetrical networks 4. As appropriate, derive an equivalent circuit for the symmetrical networks interconnection 5. Solve the symmetrical components of the fault currents 6. Solve the phase fault currents U. P. National Engineering Center National Electrification Administration

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Methodology Shunt Faults

1. Single Line-to-Ground Fault 2. Line-to-Line Fault 3. Double Line-to-Ground Fault 4. Three-Phase Fault

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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 Ia r r (2) Ib = Ic = 0 U. P. National Engineering Center National Electrification Administration

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Transformation: From (2), we get

r r −1 I012 = A Iabc

r Ira0 1 1 1 Ira1 = 1 a 3 1 a2 Ia2

1 a2 a

r r Ira Ia 1 0 = 3 Ira Ia 0

r r r which means Ia0 = Ia1 = Ia2 =

r 1 I 3 a

From (1), we get

r r r r r r Va0 + Va1 + Va2 = Z f ( Ia0 + Ia1 + Ia2 )

or

r r r r Va0 + Va1 + Va2 = 3Z f Ia0 U. P. National Engineering Center National Electrification Administration

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Sequence Network Interconnection: F1

r r Z1 Ia1 Va1 + r Vth +

-

-

F0

F2

r Va2

r Va0

Z2

Z0

3Zf

-

-

N1

r Ia0

+

r Ia2

+

N0

N2

The sequence fault currents

r r r Ia0 = Ia1 = Ia2

r Vth = Z0 + Z1 + Z2 + 3Zf

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Example 2: 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 j0.4

r Eg

+ -

j0.15

r IA1L

r IA1

F1

j0.05

Open j0.6 +

r Ia1g

1.0 -

r + IA1 r VA1 -

N1 N1

r r r r Note: IA1L = IA1 but IA1L ≠ Ia1g U. P. National Engineering Center National Electrification Administration

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Negative-Sequence Network: F2 j0.05

r Ia2g

j0.4

j0.15

r IA2L

r IA2 j0.05

Open

r IA2

F2 +

j0.6

r VA2 -

N2

N2

r r r r Note: IA2L = IA2 but IA2L ≠ Ia2g U. P. National Engineering Center National Electrification Administration

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Zero-Sequence Network: F0 j0.05

r Ia0g j0.2

j0.35

r IA 0L

r IA 0 j0.05

Open

r IA 0

F0 +

j0.044

r VA 0 -

N0

N0

r r r Note: IA0L ≠ IA0 and Ia0g = 0 U. P. National Engineering Center National Electrification Administration

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Sequence Network Interconnection: F1

r J0.6 IA1 + -

F0

F2

1.0

N1

r IA2

r IA 0

J0.6

J0.044

N0

N2

Sequence Fault Currents

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

Sequence Currents in the Transmission Line

r r IA1L = IA1 = − j0.804 p.u. r r IA2L = IA2 = − j0.804 p.u. r r 0.05 IA 0L = IA 0 = − j0.089 p.u. 0.05 + 0.4

Phase Currents in the Transmission Line

r r r r IAL = IA 0L + IA1L + IA2L = − j1.696 p.u. U. P. National Engineering Center National Electrification Administration

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r r r r 2 rIBL = rIA 0L + arIA1L + a2 rIA2L = j0.714 p.u. ICL = IA 0L + a IA1L + a IA2L = j0.714 p.u.

Sequence Currents in the Generator: Using the 30o phase shift,

r r Ia1g = IA1L ∠ − 30o = 0.804∠ − 120o p.u. = −0.402 − j0.696 p.u. r r Ia2g = IA2L ∠ + 30o = 0.804∠ − 60o p.u. = 0.402 − j0.696 p.u. r Ia0g = 0

Phase Currents in the Generator

r r r r Iag = Ia0g + Ia1g + Ia2g = − j1.392 p.u.

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r r r r 2 Ibg = Ia0g + a Ia1g + a Ia2g = j1.392 p.u. r r r r 2 Icg = Ia0g + a Ia1g + a Ia2g = 0

Three-line Diagram: a

j1.392

a

X1

X3 c

0 b

j1.392 0

T1

C

X2

j1.696

A

H1

H3

j0.714

H1

j2.411 H2 B j0.714

C

A

T2

H3

H2 B

b

j0.268

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j0.714

j2.143

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Example 3: The following data apply to the power system shown: T1: 150/150/45 MVA 138-69-13.8 kV Ynyn0d1 %Z: H-X @ 150 MVA=14.8 H-Y @ 45 MVA=21.0 X-Y @ 45 MVA=36.9

L1 138 kV

13.8 kV T1

T3: 20 MVA 67-13.8 kV Dyn1 X=8% F T4: 6.5 MVA 13.8-2.4 kV Ynd1 X=5.85%

Utility Bus

69 kV

T4

L2 T3

G1 T2

2.4 kV

13.8 kV U. P. National Engineering Center National Electrification Administration

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T2: 20/15/6 MVA 67-13.8-2.4 kV Ynyn0d1 %Z: H-X @ 15 MVA=6.65 H-Y @ 6 MVA=4.69 X-Y @ 6 MVA=1.51 G1: 5 MVA 2.4 kV X1=X2=32% X0=10% L1: X1=X2=8.26 Ω X0=31.46 Ω L2: X1=X2=2.6 Ω X0=7.8 Ω The Fault MVA at the 138-kV utility bus is 875.3 for a three-phase fault and 617.2 MVA for a single line-to-ground fault. Assume X1=X2 for the utility bus. Analyze the system for a single line-to ground fault at point F. Neglect the fault impedance. Use 100 MVA and 138 kV as bases at the utility bus. U. P. National Engineering Center National Electrification Administration

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Representation of the Utility’s System: From the three-phase short-circuit MVA,

(138)2 X1 = X2 = = 21.76Ω 875.3 From the single line-to-ground short-circuit MVA, we get the SLG fault current

IF = Since

617.2

3(138)

= 2.58 kA

3Vth 3(138) IF = = X1 + X2 + X0 2(21.76) + X0

we get X0 = 49.05 Ω. U. P. National Engineering Center National Electrification Administration

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Base kV = 138 at the utility bus and line L1 = 69 in transmission line L2 = 13.8 at the location of T4 = 14.2 at the distribution feeders = 2.47 at the location of G1 Base Current = = = = = =

Base MVA(1,000) 3(Base kV) 418.4 Amps at 138 kV 836.7 Amps at 69 kV 4,062 Amps at 14.2 kV 4,184 Amps at 13.8 kV 23,359 Amps at 2.47 kV

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(Base kV)2 Base Impedance = Base MVA = 190.44 Ω at 138 kV = 47.61 Ω at 69 kV For transmission lines L1 and L2:

Zpu

Actual Impedance = Base Impedance

Conversion of the Impedances to the new Bases: 2

Z pu ( new )

⎛ kVold ⎞ ⎛ MVAnew ⎞ ⎟⎟ ⎟⎟ x ⎜⎜ = Z pu ( old ) ⎜⎜ ⎝ kVnew ⎠ ⎝ MVAold ⎠

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Autotransformer T1: At 100 MVA 138-69-13.8 kV, ZHX = 14.8(100/150) = 9.87% ZHY = 21.0(100/45) = 46.67% ZXY = 36.9(100/45) = 82.0% ZH = 0.5(9.87+46.67-82.0) = -12.75% ZX = 0.5(9.87-46.67+82.0) = 22.60% ZY = 0.5(-9.87+46.67+82.0) = 59.40% Transformer T2: At 100 MVA 69-14.2-2.47 kV, ZHX = 6.65(67/69)2(100/15) = 41.8% ZHY = 4.69(67/69)2(100/6) = 73.7% ZXY = 1.51(67/69)2(100/6) = 23.73% U. P. National Engineering Center National Electrification Administration

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ZH = 0.5(41.8+73.7-23.73) = 45.9% ZX = 0.5(41.8-73.7+23.73) = -4.09% ZY = 0.5(-41.8+73.7+23.73) = 27.82% Transformer T3: At 100 MVA 69 kV, Z = 8.0(67/69)2(100/20) = 37.72% Transformer T4: At 100 MVA 13.8 kV, Z = 5.85(100/6.5) = 90% Generator G1: At 100 MVA 2.47 kV, X1 = X2 = 32(2.4/2.47)2(100/5) = 603.4% X0 = 10(2.4/2.47)2(100/5) = 188.6% U. P. National Engineering Center National Electrification Administration

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Transmission Line L1: X1 = X2 = 8.26/190.44 = 4.34% X0 = 31.46/190.44 = 16.52% Transmission Line L2: X1 = X2 = 2.6/47.61 = 5.46% X0 = 7.8/47.61 = 16.38% Utility’s System: X1 = X2 = 21.76/190.44 = 11.43% X0 = 49.05/190.44 = 25.76%

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Positive Sequence Network

j0.1143

The equivalent impedance ZX = j0.256 j0.0434 ZY = j6.771

r Ia1 Z + 1 r + Va1

1.0

-

-

r Ia1

1.0 138

-j0.127

Z1=j0.055+ZX//ZY) = j0.302 F1

+

j0.226

F1

j0.055

j0.594

13.8

N1

69

j0.459 j0.278 j6.03 j0.377 14.2

N1 U. P. National Engineering Center National Electrification Administration

-j0.041

2.47

+

1.0

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Negative Sequence Network The negative-sequence impedances are the same as the positivesequence impedances.

j0.0434 138 -j0.127

Z2=j0.302

r Ia2

F2 +

Z2

r Va2 -

N2

r Ia2

j0.1143

j0.226

F2

j0.055

j0.594

13.8

N2

69

j0.459 j0.278 j6.03 j0.377 14.2

U. P. National Engineering Center National Electrification Administration

-j0.041

2.47

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Zero Sequence Network

j0.2576

XX = 0.2958//0.594=0.197 XY = 0.459+0.278=0.737 j0.1652 XW = (XX+0.226)//XY=0.269 -j0.127

Z0=j(0.164+XW)=j0.433

r Ia0

F0 +

Z0

r Va0 -

N0

j0.226

r F0 Ia0

138

j0.164

13.8

j0.594

j0.9

N0

69

j0.459 j0.278

j1.89

j0.377 14.2

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-j0.041

2.47

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Sequence Network Interconnection: F1

r j0.302 IA1 + N1

F0

F2

Vth

r IA2

r IA 0

j0.302

j0.433

N2

N0

r Sequence Fault Currents: Let Vth = 1.0∠90o p.u. r r r 1.0∠90o Ia0 = Ia1 = Ia2 = j2(0.302) + j0.433 = 0.965 p.u. U. P. National Engineering Center National Electrification Administration

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83

r r IaF = 3Ia0 = 2.896 p.u.

r r IbF = IcF = 0

Fault Currents at 69-kV (from the Utility)

r r IA1 = IA2 =

r 6.771 Ia1 = 0.93 p.u. 6.771 + 0.256 r r 0.737 IA 0 = Ia0 = 0.613 p.u. 0.737 + 0.423 r r r r IA = IA 0 + IA1 + IA2 = 2.474 p.u. r r r r 2 IB = IA 0 + a IA1 + a IA2 = −0.317 p.u. r r r r 2 IC = IA 0 + a IA1 + a IA2 = −0.317 p.u.

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Fault Currents at 69-kV (from generator G1)

r r IA1 = IA2 = 0.965 − 0.93 = 0.035 p.u. r IA 0 = 0.965 − 0.613 = 0.352 p.u. r r r r IA = IA 0 + IA1 + IA2 = 0.423 p.u. r r r r 2 IB = IA 0 + a IA1 + a IA2 = 0.317 p.u. r r r r 2 IC = IA 0 + a IA1 + a IA2 = 0.317 p.u.

Fault Currents at the 2.4-kV side of generator G1

r Ia1 = 0.035∠ − 30o p.u. r Ia2 = 0.035∠ + 30o p.u.

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r Ia0 = 0

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r r r r Iag = Ia0g + Ia1g + Ia2g = 0.061 p.u. r r r r Ibg = Ia0g + a2 Ia1g + a Ia2g = −0.061 p.u. r r r r 2 Icg = Ia0g + a Ia1g + a Ia2g = 0 Phase Currents at the 138-kV side

r r IA1 = IA2 = 0.93 p.u.

r IA 0 =

0.594 (0.613) = 0.410 p.u. 0.594 + 0.295 r r r r IA = IA 0 + IA1 + IA2 = 2.27 p.u. r r r r r 2 IB = IA 0 + a IA1 + a IA2 = −0.521 p.u. = IC U. P. National Engineering Center National Electrification Administration

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Three-line Diagram for transformer T2: A

B

2.474 2,070

0

C

0.317 265

0.317 265

0

2.896 2,423

0.423 354

0.317 265 H2

0.061 1,425

0.317 265 Y1

Y2 0

Fault

Note: The X side is not shown.

b

H3

H1

884 A

U. P. National Engineering Center National Electrification Administration

c

Y3 0.061 1,425

a

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Balancing Ampere Turns:

r r NH IH = NY IY or

r NH r IH = IY NY r 67 / 3 Phase a: (354) = Iab = 5,700 A 2.4 r r 67 / 3 (265) = Ibc = Ica = 4,275 A Phases b,c: 2.4 Check KCL: Nodes a and b: 1,425 + 4,275 = 5,700 Amps (ok) Node c: 1,425 – 1,425 = 0 (ok)

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Three-line Diagram for transformer T1:

B’

a

0 Y2

0.521 218

Y1 H2

C’

U. P. National Engineering Center National Electrification Administration

0 Y3

47.4 X2 2.27 47.4 1,120 950 H3 X1 X3 A’ H1 514 1,025 0.521 218

0

b

c B

0.317 265 C

0.317 265 A

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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

Ground

r Ic Zf

r Boundary Conditions: (1) Ira = 0 r (2) Ib = − Ic r r r (3) Vb − Vc = IbZf U. P. National Engineering Center National Electrification Administration

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

r I012 = A − 1I abc

r Ira0 1 1 1 Ira1 = 1 a 3 1 a2 Ia2 which means

1 a2 a

0 0 r r 1 2 (a − a )Ib Ib = r 3 r 2 (a − a)Ib − Ib

r Ia0 = 0

r r r 2 Ia1 = − Ia2 = 13 (a − a )Ib = j

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r Ib

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From (3), we get

r r r ( Va0 + a2 Va1 + a Va2 ) r r r r r r 2 2 − (Va0 + aVa1 + a Va2 ) = ( Ia0 + a Ia1 + a Ia2 )Z f

r r r Since Ia0 = 0 and Ia1 = − Ia2 , we get r r r 2 2 2 (a − a)Va1 + (a − a )Va2 = (a − a)Ia1Z f or

r r r Va1 − Va2 = Ia1Z f

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Sequence Network Interconnection: F1

r r Z1 Ia1 Va1 + r Vth +

-

-

Zf

F2

F0

r IA 0

r Ia2

+

r Va2

Z2

Z0

-

N1

N2

N0

The sequence fault currents

r Ia0 = 0 r r Ia1 = − Ia2 =

r Vth Z1 + Z2 + Zf

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Example: The data for this power system was given in the previous example. Utility Assuming zero fault impedance, Bus analyze the system for a line-toline fault at point F. L1 138 kV

Note: The sequence networks for this power system were determined in the previous 69 kV example. L2 F T3

13.8 kV T1 T4 G1 T2

2.4 kV

13.8 kV U. P. National Engineering Center National Electrification Administration

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Sequence Network Interconnection: F1

Zf

F2

r + r Ia1 j0.302 r +r Va2 Va1 Vth

r Ia2

+

-

-

N1

F0

j0.302

r IA 0 j0.433

-

N2

N0 r Sequence Fault Currents: Let Vth = 1.0∠0o p.u. r Ia0 = 0 r r 1.0∠0o Ia1 = − Ia2 = = − j1.658 p.u. j2(0.302) U. P. National Engineering Center National Electrification Administration

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Phase Fault Currents

r IaF = 0

r r r IbF = − IcF = − j 3 Ia1 = −2.872 p.u.

Fault Currents at 69-kV (from the Utility)

r r r 6.771 Ia1 = − j1.597 p.u. IA1 = − IA2 = 6.771 + 0.256 r IA 0 = 0 r r r r IA = IA 0 + IA1 + IA2 = 0 r r r r IB = IA 0 + a2 IA1 + a IA2 = −2.767 p.u. r r r r IC = IA 0 + a IA1 + a2 IA2 = 2.767 p.u. U. P. National Engineering Center National Electrification Administration

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Fault Currents at 69-kV (from generator G1)

r r IA1 = − IA2 = − j1.658 − (− j1.597) = − j0.06 p.u. r IA 0 = 0 r r r r IA = IA 0 + IA1 + IA2 = 0 r r r r 2 IB = IA 0 + a IA1 + a IA2 = −0.105 p.u. r r r r 2 IC = IA 0 + a IA1 + a IA2 = 0.105 p.u.

Fault Currents at the 2.4-kV side of generator G1

r Ia1 = − j0.06∠ − 30o p.u. r Ia2 = j0.06∠ + 30o p.u.

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r Ia0 = 0

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r r r r Iag = Ia0g + Ia1g + Ia2g = −0.06 p.u. r r r r Ibg = Ia0g + a2 Ia1g + a Ia2g = −0.06 p.u. r r r r 2 Icg = Ia0g + a Ia1g + a Ia2g = 0.121 p.u. Phase Currents at the 138-kV side

r IA1 r IA 0 r IA r IB r IC

r = − IA2 = − j1.597 p.u.

=0 r r r = IA 0 + IA1 + IA2 = 0 r r r 2 = IA 0 + a IA1 + a IA2 = −2.766 p.u. r r r 2 = IA 0 + a IA1 + a IA2 = 2.766

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Three-line Diagram for transformer T2: A

2.872 2,403

B

C

2.767 2,315

0

2.767 2,315

0 2.872 2,403

0.105 87.6 H2

0

Fault

Note: The X side is not shown.

H1

U. P. National Engineering Center National Electrification Administration

H3

0.060 1,412

0.105 87.6 Y1

Y2 0.121 2,825

b

c

Y3 0.060 1,412

a

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Balancing Ampere Turns:

r r NH IH = NY IY or

r NH r IH = IY NY r r 67 / 3 Phases b,c: (87.6) = IcB = Ica = 1,412 A 2.4 Check KCL: Nodes a and b: 1,412 –1,412 = 0 (ok) Node c: 1,412 + 1,412 = 2,824 (ok)

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Three-line Diagram for transformer T1:

B’

a

0 Y2

2.766 1,157

Y1 H2 1,158

A’

H1

X1

2.766 1,157 U. P. National Engineering Center National Electrification Administration

0 Y3

X2

0 C’

0

X3

1,158 H3

b

c B

2.767 2,315 C

2.767 2,315 A

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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

Ground

r Zf Zf Ic r r Zg Ib + Ic

r Boundary Conditions: (1) Ira = 0 r r (2) Vb = (Zf + Zg )Ib + Zg Ic r r r (3) Vc = (Zf + Zg )Ic + Zg Ib U. P. National Engineering Center National Electrification Administration

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Transformation: From (1), we get

r r r r Ia = 0 = Ia0 + Ia1 + Ia2

From

we get

r r r r 2 Vb = Va0 + a Va1 + a Va2 r r r r 2 Vc = Va0 + aVa1 + a Va2

r r r r 2 2 Vb − Vc = (a − a)Va1 + (a − a )Va2

Likewise, from

r r r r 2 Ib = Ia0 + a Ia1 + a Ia2 r r r r Ic = Ia0 + a Ia1 + a2 Ia2

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we get

r r r r 2 2 Ib − Ic = (a − a)Ia1 + (a − a )Ia2

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

r r r r Vb − Vc = Z f ( Ib − Ic )

Substitution gives

r r 2 (a − a)Va1 + (a − a )Va2 r r = Z f [(a2 − a)Ia1 + (a − a2 )Ia2 ] 2

Simplifying, we get

r r r r Va1 − Z f Ia1 = Va2 − Z f Ia2

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

r r r r Vb + Vc = (Z f + 2Z g )( Ib + Ic )

We can also show

r r r r r Vb + Vc = 2Va0 − Va1 − Va2 r r r r r Ib + Ic = 2 Ia0 − Ia1 − Ia2

Substitution gives

r r r r r r 2Va0 − Va1 − Va2 = Z f (2 Ia0 − Ia1 − Ia2 ) r r r + 2Z g (2 Ia0 − Ia1 − Ia2 )

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Rearranging terms, we get

r r r r r 2Va0 − 2Z f Ia0 − 4Z g Ia0 = Va1 − Z f Ia1 r r r r + Va2 − Z f Ia2 − 2Z g ( Ia1 + Ia2 )

Earlier, we got

r r r r Va1 − Z f Ia1 = Va2 − Z f Ia2 r r r Ia1 + Ia2 = − Ia0

Substitution gives

r r r r r 2Va0 − 2Z f Ia0 − 6Z g Ia0 = 2(Va1 − Z f Ia1 ) r r r r Va0 − ( Z f + 3Z g )I a0 = Va1 − Z f I a1

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Sequence Network Interconnection: Zf + F1

r r Z1 Ia1 Va1 + r Vth -

-

Zf F2 +

r Va2

F0

r Ia2

r IA 0

+

r Va0

Z2

Z0

-

-

N1

Let

Zf+3Zg

N2

N0

Z0 T = Z0 + Zf + 3Zg Z1T = Z1 + Zf U. P. National Engineering Center National Electrification Administration

Z2 T = Z2 + Zf

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The sequence fault currents

r Ia1 =

Z1T

r Vth Z0 T Z2 T + Z0 T + Z2 T

From current division, we get

r Ia2 = −

r Z0 T Ia1 Z0 T + Z2 T

From KCL, we get

r r r Ia0 = − Ia1 − Ia2

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Example: The data for this power system was given in a previous example. Utility Assuming zero fault impedance, Bus analyze the system for a double line-to-ground fault at point F. L1 138 kV

Note: The sequence networks for this power system were determined in a previous 69 kV example. L2 F T3

13.8 kV T1 T4 G1 T2

2.4 kV

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Sequence Network Interconnection: F1

r j0.302 IA1 + -

F0

F2

Vth

r IA2

r IA 0

j0.302

j0.433

N0 r Sequence Fault Currents: Let Vth = 1.0∠90o p.u. r 1.0∠90o Ia1 = 0.302(0.433) j0.302 + j 0.302 + 0.433 = 2.087 p.u. N1

N2

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r Ia2 = − r Ia0

r 0.433 Ia1 = −1.23 p.u. 0.433 + 0.302 r r = − Ia1 − Ia2 = −0.857 p.u.

Phase Fault Currents

r IaF = 0 r r r r 2 IbF = Ia0 + a Ia1 + a Ia2

= −0.857 + 2.087∠ − 120o − 1.23∠120o

r IcF

= −1.285 − j2.872 = 3.146∠ − 114.1o p.u. r r r 2 = Ia0 + a Ia1 + a Ia2 = 3.146∠114.1o p.u.

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Fault Currents at 69-kV (from the Utility)

r IA1 =

r 6.771 Ia1 = 2.01 p.u. 6.771 + 0.256 r r 6.771 Ia2 = −1.185 p.u. IA2 = 6.771 + 0.256 r r 0.737 IA 0 = Ia0 = −0.544 p.u. 0.737 + 0.423 r r r r IA = IA 0 + IA1 + IA 2 = 0.281 p.u. r IB = −0.957 − j2.767 = 2.928 ∠ − 109 .1o p.u. r IC = 2.928∠109.1o p.u. U. P. National Engineering Center National Electrification Administration

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Fault Currents at 69-kV (from generator G1)

r IA 0 = −0.857 + 0.544 = −0.313 p.u. r IA1 = 2.087 − 2.01 = 0.076 p.u. r IA2 = −1.23 + 1.185 = −0.045 p.u.

r IA = −0.281 p.u. r IB = −0.328 − j0.105 = 0.345∠ − 162.3o p.u. r IC = 0.345∠162.3o p.u.

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Fault Currents at the 2.4-kV side of generator G1

r Ia0 = 0 r Ia1 = 0.076∠ − 30o p.u. r Ia2 = −0.045∠ + 30o p.u.

r Iag = 0.027 − j0.061 = 0.066∠ − 65.9o p.u.

r Ibg = −0.027 − j 0.061 = 0.066∠ − 114.1o p.u. r Icg = j0.076 + j0.045 = j0.121 p.u.

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Phase Currents at the 138-kV side

r IA1 = 2.01 p.u. r IA2 = −1.185 p.u. r IA 0 =

0.594 (−0.544) = −0.364 p.u. 0.594 + 0.295

r IA = 0.462 p.u. r IB = −0.776 − j2.767 = 2.874∠ − 105.7o p.u. r IC = −0.776 + j2.767 = 2.874∠105.7o p.u. U. P. National Engineering Center National Electrification Administration

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Three-line Diagram for transformer T2: A

235

801+ j2315

B

801j2315

C

All currents in Amperes

1076-j2403 1076+j2403 0

235

275 +j88 H2

632+ j1413

275 -j88 Y1

Y2 j2825

Fault

Note: The X side is not shown.

b

H1

U. P. National Engineering Center National Electrification Administration

H3

c

Y3 632j1413

a

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Example: For the power system shown, a double line-to-ground fault occurs at bus 4. Assuming zero fault impedance and neglecting the loads, find the phase currents in transmission lines L2 and L3. T 2

L1

G1

X=0.08 X1=0.40 X1=0.50 X1=0.40

G2 L3

L2

T: G1: G2: L1:

3

4

X2=0.40 X2=0.50 X2=0.40

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X0=0.15 X0=0.25 X0=0.80

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L2: L3:

X1=0.30 X1=0.20

117

X2=0.30 X2=0.20

X0=0.60 X0=0.40

Positive-Sequence Network: The network cannot be simplified using seriesparallel combination. We have to use delta-wye transformation.

j0.4 r +

F1

r Ia1

4

j0.2

j0.3 j0.08 2

j0.4

EG1

-

3

j0.5 +r -

EG2

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118

0.2

Xb 0.4

Xa Xc

0.2(0.3) Xa = = 0.067 p.u. 0.2 + 0.3 + 0.4 0.3(0.4) Xb = = 0.133 p.u. 0.2 + 0.3 + 0.4 0.2(0.4) Xc = = 0.088 p.u. 0.2 + 0.3 + 0.4

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Equivalent Circuit: XL = 0.4+0.08+Xb = 0.6133

F1

XR = 0.5+Xc = 0.5888

4

Xa

X1 = Xa+XL//XR

j0.08

= 0.3671 F1

jX1

+ -

j0.4

r IA1

r Ia1

r + EG1

r Ix 2

Xc

Xb r

Iy

-

Vth

N1 U. P. National Engineering Center National Electrification Administration

3

j0.5

+r -

EG2

N1

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Zero-Sequence Network:

0.6

0.4

Xb 0.8

Xa = 0.133 Xb = 0.267 Xc = 0.178

F0

Xa Xc

r Ia0

4

j0.4

j0.6 j0.08

j0.2

2

j0.8

3

j0.25

N0

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Equivalent Circuit: XL = 0.08+Xb = 0.3467

F0

XR = 0.25+Xc = 0.4278 X1 = Xa+XL//XR = 0.3248

r Ia1

4

Xa j0.08

r Ix 2

Xc

Xb r

Iy

3

j0.25

N0

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Sequence Network Interconnection: F1

r j0.3671 IA1 + -

F0

F2

Vth

r IA2

r IA 0

j0.3671

j0.3248

N0 r Sequence Fault Currents: Let Vth = 1.0∠0o p.u. r 1.0∠0o Ia1 = 0.3671(0.3248) j0.3671 + j 0.3671 + 0.3248 = − j1.8538 p.u. N1

N2

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r Ia2 = − r Ia0

r 0.3248 Ia1 = j0.8703 p.u. 0.3248 + 0.3671 r r = − Ia1 − Ia2 = j0.9835 p.u.

Phase Fault Currents

r IaF = 0 r r r r 2 IbF = Ia0 + a Ia1 + a Ia2

r IcF

= −2.3591 + j1.4753 p.u. r r r 2 = Ia0 + a Ia1 + a Ia2

= 2.3591 + j1.4753 p.u. U. P. National Engineering Center National Electrification Administration

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Sequence Voltages at the Fault (bus 4)

r ( 4) r ( 4) r ( 4) r Va0 = Va1 = Va2 = − Ia0Z0

= − j09835( j0.3248) = 0.3195 p.u. Positive Sequence Voltages at buses 2 and 3

r Ix = −

r 0.5889 Ia1 = − j0.9081 p.u. 0.5889 + 0.6133

r (2) Va1 = 1.0 + j0.9081( j0.48) = 0.5641 p.u.

r Iy = − j1.8538 + j0.9081 = − j0.9457 p.u.

r (3) Va1 = 1.0 + j0.9457( j0.50) = 0.5271 p.u. U. P. National Engineering Center National Electrification Administration

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Negative Sequence Voltages at buses 2 and 3

r Ix = −

r 0.5889 Ia2 = j0.4263 p.u. 0.5889 + 0.6133

r (2) Va2 = − j0.4263( j0.48) = 0.2046 p.u.

r Iy = j0.8703 − j0.4263 = j0.4440 p.u.

r (3) Va2 = − j0.4440( j0.50) = 0.2220 p.u.

Zero Sequence Voltages at buses 2 and 3. Can show that r (2)

Va0 = 0.0435 p.u. r (3) Va0 = 0.1101 p.u.

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Sequence Currents in Transmission Line L2 2

X1=X2=0.3

r (2) Va0 = 0.0435 p.u. r (2) Va1 = 0.5641 p.u. r (2) Va2 = 0.2046 p.u.

X0=0.6

4

r (4) r (4) r (4) Va0 = Va1 = Va2 = 0.3195 p.u.

r 0.0435 − 0.3195 Ia0 = = j0.46 p.u. j0.60 r 0.5641 − 0.3195 Ia1 = = − j0.8156 p.u. j0.30 U. P. National Engineering Center National Electrification Administration

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r 0.2046 − 0.3195 Ia2 = = j0.3828 p.u. j0.30 Phase Currents in Transmission Line L2

r r r r Ia = Ia0 + Ia1 + Ia2 = j0.0273 p.u. r r r r 2 Ib = Ia0 + a Ia1 + a Ia2

= −1.0378 + j0.6764 p.u. r r r r 2 Ic = Ia0 + a Ia1 + a Ia2 = 1.0378 + j0.6764 p.u.

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Sequence Currents in Transmission Line L3 3

X1=X2=0.2

r (3) Va0 = 0.1101 p.u. r (3) Va1 = 0.5271 p.u. r (3) Va2 = 0.2220 p.u.

X0=0.4

4

r (4) r (4) r (4) Va0 = Va1 = Va2 = 0.3195 p.u.

r 0.1101 − 0.3195 Ia0 = = j0.5235 p.u. j0.40 r 0.5271 − 0.3195 Ia1 = = − j1.0383 p.u. j0.20 U. P. National Engineering Center National Electrification Administration

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r 0.2220 − 0.3195 Ia2 = = j0.4874 p.u. j0.20 Phase Currents in Transmission Line L3

r r r r Ia = Ia0 + Ia1 + Ia2 = − j0.0273 p.u. r r r r 2 Ib = Ia0 + a Ia1 + a Ia2

= −1.3213 + j0.7990 p.u. r r r r 2 Ic = Ia0 + a Ia1 + a Ia2 = 1.3213 + j0.7990 p.u.

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Example: An unloaded generator has the following impedances: X1=0.40

X2=0.40

X0=0.20

a) Assume that the generator is connected in wye and the neutral is solidly grounded. For a bolted fault at the generator’s terminals, which shunt fault will yield the largest fault current?. b) Assume that the generator is connected in wye but the neutral is ungrounded. For a bolted single line-to-ground fault at the generator’s terminals, find the voltages from phases b and c to ground. Neglect all sequence capacitances.

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Sequence Networks F1

+

r Ia1 + r Va1

-

-

j0.4 1.0

N1

r Ia2

F2 +

j0.4

r Va2

r Ia0

F0 +

j0.2

r Va0 -

-

N0

N2

a) Single line-to-ground fault

r r r Ia0 = Ia1 = Ia2 =

1.0 = − j1.0 p.u. j[2(0.4) + 0.2]

IF = Ia = 3Ia0 = 3.0 p.u. U. P. National Engineering Center National Electrification Administration

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Line-to-line fault

r Ia0 = 0 r r Ia1 = −Ia2 =

1.0 = − j1.25 p.u. j2(0.4)

IF = Ib = 3Ia1 = 2.165 p.u. Double line-to-ground fault

r Ia1 =

r Ia2

1.0 = − j1.875 p.u. j[0.4 + 0.4 // 0.2] 0.2 =− (− j1.875) = j0.625 p.u. 0.4 + 0.2

r r r Ia0 = −Ia1 − Ia2 = j1.25 p.u. U. P. National Engineering Center National Electrification Administration

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r r r r 2 Ib = Ia0 + a Ia1 + a Ia2 = 2.864∠139.1o p.u.

IF = Ib = 2.864 p.u. Three-phase fault

r r Ia0 = Ia2 = 0 r 1.0 Ia1 = = − j2.5 p.u. j0.4

IF = Ia1 = 2.5 p.u. Conclusion: The single line-to-ground fault yields the largest fault current. U. P. National Engineering Center National Electrification Administration

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b) Sequence Network Interconnection F1

+

r Ia1 + r Va1

-

-

j0.4 1.0

N1

r Ia2

F2 +

j0.4

r Va2

r Ia0

F0 +

j0.2

-

r Va0 -

N0

N2

Because of the open-circuit in the zero-sequence r r network, r

Ia0 = Ia1 = Ia2 = 0 r r Va1 = 1.0 Va2 = 0

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r Va0 = −1.0

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Phase Voltages at the Fault

r r r r Va = Va0 + Va1 + Va2 = 0 r r r r 2 Vb = Va0 + a Va1 + aVa2 = 1.732∠ − 150o p.u. r r r r 2 Vc = Va0 + aVa1 + a Va2 = 1.1124∠142.05o p.u.

Comments: 1. Despite the fault in phase a, no current flows. 2. The magnitude of the line-to-neutral voltages in phases b and c is line-to-line. 3. There will be possible damage to equipment that are connected from line to ground. U. P. National Engineering Center National Electrification Administration

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Three Phase Fault a b c +

+ + Va Vb Vc -

Ia

Ib

Ic Ig

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Three Phase Fault ¾ 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. ¾ 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 Ia Zf Ib Zf Ic Zf r Zg Ig

Ground

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

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Sequence Network Interconnection: F1

F2

+

r r Z1 Ia1 Va1 + r Vth -

+

Zf

r Va2

F0

r Ia2 Z2

N1

N2

Sequence currents

r Ia1 r Ia0

+

r Va0

Z0

-

-

-

r Ia0

N0

r Vth = Z1 + Zf r = Ia2 = 0

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Example: The data for this power system was given in a previous example. Utility Assuming zero fault impedance, Bus analyze the system for a threephase fault at point F. L1 138 kV

Note: The positive-sequence network for this power system was determined in a previous 69 kV example. L2 F T3

13.8 kV T1 T4 G1 T2

2.4 kV

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Sequence Network Interconnection: F1

j0.302

+ -

F0

F2

r IA1 Vth

N1

r IA2

r IA 0

j0.302

j0.433

N2

N0

r Fault Currents: Let Vth = 1.0∠90o p.u. r r 1.0∠90o IaF = Ia1 = = 3.316 p.u. j0.302 r r Ia0 = Ia2 = 0 U. P. National Engineering Center National Electrification Administration

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r IAF = 3.316(836.7) = 2,774 A Fault Contribution from the Utility (138 kV side)

r 6.771 IA1 = (3.316) = 3.195 p.u. 6.771 + 0.256 r IAF = 3.195(418.4) = 1,337 A Fault Contribution from G1 (2.4 kV side)

r Ia1 = 3.316 − 3.195 = 0.121 p.u. r IaF = 0.121(23,359) = 2,825 A

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Computer Solution ‰ Development ‰ Rake

of the Model

Equivalent

‰ Formation ‰ Analysis

of Zbus

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

j0.6

2

j0.2

j0.2

r + EG1

-

3

+r

-

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

U. P. National Engineering Center National Electrification Administration

-

EG2

+

r EG

j0.2

j0.2 j0.6

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. -

Next, use loop currents to describe the circuit with all fault switches closed. j0.2 Since there are four loops, we need to define four loop currents. 1

r I1

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+

r EG

j0.6 3

r I3

j0.2

r I4 2

r I2

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The loop equations are r r r loop 1: 1.0 = j0.2(I1 + I3 − I4 )

r r loop 2: 1.0 = j 0.2(I2 + I4 ) r r r r loop 3: 1.0 = j 0.2(I1 + I3 − I4 ) + j 0.24I3 r r r loop 4: 0 = j0.2(I2 +rI4 ) +r j0.r6 I4 + j0.2(I4 − I1 − I3 )

or

0.2 0 0.2 − 0.2 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

1.0 1.0

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r I1 r I2 r I r3 I4

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

r r 1) V = Z(bus I

1) Z(bus = Z1 − Z2Z4−1Z3

and

− 0.2

0.2 0 Z1 = j 0 0.2

0.2 0

0.2

0.44

0

Z2 = j 0.2

Z3 = j[-0.2 0.2 -0.2 ] U. P. National Engineering Center National Electrification Administration

− 0.2 Z4 = j[1.0]

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Substitution gives

0.16 0.04 0.16 1.0 = j 0.04 0.16 0.04 1.0 0.16 0.04 0.40 r r (1) V = Zbus I 1.0

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) bus

(2) Z 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

r I1 r I2 r I3

Z11 Z12 Z13 1.0 = Z12 Z22 Z23 Z13 Z23 Z33 1.0

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

1.0 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

+

j0.04 j0.16

r I2

1.0 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 + j0.04I2 + j0.4I3

Since only bus 3 is faulted, I1=I2=0. We get or

r 1.0 = j0.4I3

r I3 =

1 = − j2.5 j0.4

+

j0.16 + r

V1 -

j0.04 j0.16 + r

1.0 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 U. P. National Engineering Center National Electrification Administration

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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

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

1

j0.6 -j0.5

-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 1.0

Z11 Z12 Z21 Z22 Zn1

1.0

Zn2

+

Z12 Z22

Z11

r I1

1

r I2

Z2k 2

Znn

r In

…

old Zbus

1.0 Zkn

Zkk

r Ik

Z1n Z2n

…

…

=

… …

r I r1 I2

…

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

158

k

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Znn

r In

n

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

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Add a generator 1.0 to a new bus: 1.0 Let Z be the g

1.0

impedance of the generator to be added.

159

r I1

1

r I2

2

0 0

2n

Zn2 … Znn 0 0 0 Zg

0

r I r1 I2 r I r n In + 1

The dimension is (n+1).

+

Z2k

22

= Z n1

-

Z11

Z11 Z21

1.0

Z12 Z22

Z12 … Z1n Z …Z

1.0 Zkn

Zkk

r Ik

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k

Znn

r In

Zg n

r In +1

n+1

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-

Add a generator to an old bus k: Let Zg be the impedance of Z11 the generator r to be added. I1

+

Z12 Z22 1

r I2

Z2k 2

r Iw

1.0

Zg

Zkk

r Ik

k

Znn

r In

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 = Zn1 I1 + 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 Zn2… Znk … Znn Znk Zk1 Zk 2… Zkk … Zkn Zw

…

…

0

…

… 1.0

=

r I1 r I2 r I rn 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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Add a branch from an old bus k to a new bus:

+

Z12 Z22

Z11

r I1

-

1

r I2

Z2k 2

1.0

Zkk

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 = Zn1 I1 + 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 I rn 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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Add a branch from an old bus j to an old bus k:

+

Z12 Z22

Z11

r I1

-

1

r I2

r Iw

1.0

Zkn

Z2j Zjj 2

r Ij

Zb j

Zkk

r Ik

k

Znn

r In

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 + Z I + ... + Z ( I + I ) 12 2 1 j j w 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 = Zn1 I1 + 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 − [Zk1 I1 + 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 1.0

…

Z1n Z2n

…

=

Z12 Z22

…

0

Zn1 Zn2 … Znn Znj − Znk Zj1 − Zk1 Zj2 − Zk2 … Zjn − Zkn Zv

r In r Iw

Z11 Z21

…

1.0

Z1j − Z1k Z2j − Z2k

r I1 r I2

…

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 G1 to bus 1. j0.2 j0.2 1

Xbus =

1

[0.2]

+

1.0

3

+

1.0

-

-

Step 2. Add generator G2 to bus 2.

Xbus =

1

2

1

0 .2

0

2

0

0 .2

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

2

2

0.2 0

0 0.2

*

0.2 − 0.2

1

Xnew =

*

0.2 − 0.2 1.0

Apply Kron’s reduction to eliminate the last row and column. We get −1 4

X2 X X3 =

0.2 [0.2 -0.2] − 0.2

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X2 X X3 =

0 .04 − 0 .04

169

− 0 .04 0 .04

We get −1 4

Xbus = X1 − X2X X3 =

1 2

1

2

0 .16 0 .04

0 .04 0 .16

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

3

2

0.16 0.04 0.16 0.04 0.16 0.04

3

0.16 0.04

1

Xbus =

2

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0.4

No Kron reduction is required.

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Example: For the same network, use a different sequence of addition in forming the bus impedance matrix. Step 1. Add generator G2 to bus 2. 2

Xbus =

2

[0.2]

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

Xbus =

1

2

0 .2 0 .2

1

0 .2 0 .8

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Step 3. Add generator G1 to bus 1. 2

*

1

0.2 0.2 0.2 0.2 0.8 0.8

*

0.2 0.8 1.0

2

Xnew =

1

2

Xbus =

1

2

0 .16

0 .04

1

0 .04 0 .16

Step 4. Finally, add the line from bus 1 to bus 3. 2

3

1

0.16 0.04 0.04 0.04 0.16 0.16

3

0.04 0.16

2

Xbus =

1

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0.4

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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

j0.08

[0.4]

j0.4 2. Add the transformer + 1.0 from bus 1 to bus 2. 1 2

Xbus =

1 2

0 .4 0 .4 0 .4 0 .48

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

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j0.2

j0.3

1

Xbus =

4

2

1

j0.4 N1

3

j0.5 + 1.0 -

1

2

3

1

0.4

0.4

0.4

2

0.4 0.48 0.48 0.4 0.48 0.88

3

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Step 4. Add generator G2 to bus 3. 1 1

Xnew =

2 3

*

2

3

*

0.4 0.4 0.4 0.4 0.4 0.48 0.48 0.48 0.4 0.48 0.88 0.88 0.4 0.48 0.88 1.38

Apply Kron’s reduction.

0.4

X2X 4−1X3 =

1 1.38

0.48 [0.4 0.48 0.88] 0.88

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We get

175

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

3

2

0.2841 0.2609 0.1449 0.2609 0.3130 0.1739

3

0.1449 0.1739 0.3188

1

Xbus =

2

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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 .3130

0 .1739

0 .3130

3

0 .1449

0 .1739

0 .3188

0 .1739

4

0 .2609

0 .3130

0 .1739

0 .6130

Step 6. Add the line from bus 3 to bus 4. 1

Xnew=

2

3

4

*

1

0.2841 0.2609

0.1449

0.2609

0.1159

2

0.2609 0.3130

0.1739

0.3130

0.1391

3

0.1449 0.1739

0.3188

0.1739 − 0.1449

4

0.2609 0.3130 0.1739 0.6130 0.1159 0.1391 − 0.1449 0.4391

*

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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

j0.08 2

[0.15] j0.15

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

Xbus =

j0.4

j0.6 j0.8

3

j0.25

N0

1

2

1

0 .15

0

2

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 =

180

1 1 2

3

0.15 0 0 0 0.08 0.08 0

3

Step 4. Add generator G2 to bus 3. 1

2

2

0.08 0.88 3

*

0.15 0 0 0 2 0 0.08 0.08 0.08 Xnew = 3 0 0.08 0.88 0.88 0 0.08 0.88 1.13 * 1

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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 1 1

Xbus =

2 3

2

3

0.15 0 0 0 0.0743 0.0177 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 .0743

0 .0177

0 .0743

3

0

0 .0177

0 .1947

0 .0177

4

0

0 .0743

0 .0177

0 .6743

Step 6. Add the line from bus 3 to bus 4.

Xnew=

1

2

3

4

1

0.15

0

0

0

2

0

0.0743 0.0177

0.0743 0.0566

3

0

0.0177 0.1946

0.0177 − 0.177

4

0 0

0.0743 0.0177 0.6743 0.6566 0.0566 − 0.177 0.6566 1.2336

*

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* 0

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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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Positive-Sequence Zbus (1) (1) … Z11 Z12 Z1(1n) 1) 1) … Z(21 Z(22 Z(21n)

1) Z(bus =

…

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

1) Z(n11) Z(n12) … Z(nn

- N1

1.0

+ (1) 11

Z

1

(1) Z12 1) Z(22

2

(1) 2k

Z

(1) kk

Z

k

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Z(kn1) 1) Z(nn

Rake Equivalent

n

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Negative-Sequence Zbus (2) … Z12 Z1(2n) Z(222) … Z(22n)

…

(2) Z11 The negative-sequence bus-impedance matrix Z(212) (2) describes the negative- Zbus = sequence network.

Z(n21) Z(n22) … Z(nn2)

N2

(2) 11

Z

1

(2) 12

Z

(2) 22

Z

2

(2) 2k

Z

(2) kk

Z

k

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(2) kn

Z

(2) nn

Z

Rake Equivalent

n

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Zero-Sequence Zbus (0 ) … Z12 Z1(0n) Z(220) … Z(20n)

…

(0 ) Z11 The zero-sequence bus-impedance matrix (0) Z(210) Zbus = describes the zerosequence network.

Z(n01) Z(n02) … Z(nn0)

N0

(0 ) 11

Z

1

(0 ) 12

Z

(0 ) 22

Z

2

(0 ) 2k

Z

(0 ) kk

Z

k

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(0 ) kn

Z

(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) 11

(1) 22

Z

Z

1

2

r Z jk Vj = 1.0 − Zkk

r The current in any line is Imn

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(1) kk

Z

(1) nn

Z

k

n

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 1

X

(1) bus

=

2 3 4

2

3

4

0.2669 0.2403 0.1664 0.1959 0.2403 0.2884 0.1996 0.2351 0.1664 0.1996 0.2920 0.2551 0.1959 0.2351 0.2551 0.3671

The fault current is

r 1 1 IF = (1) = = − j2.7241 j0.3671 Z44

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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

r 1 − 0.4663 = − j1.3344 IG1 = 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 I24 = = − j1.1984 j0.3 r 0.3051 − 0 I34 = = − j1.5257 j0.2 r 4 r IF I34 r I24 j0.2 j0.3 1 j0.08

j0.4 + 1.0 -

r r IG1 I12

j0.4

r I23

2

N1

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3

r IG2

j0.5 + 1.0 -

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Line-to-Line Fault at Bus k N2

N1

(1) Z11

1) Z(22

Z(kk1)

1

2

k

Sequence Fault Currents r

Ia0 = 0 r r Ia1 = − Ia2 =

Z(kk1)

1) Z(nn

(2) 11

Z

r Ia 1 n

1

(2) 22

Z

2

(2) kk

Z

k

(2) nn

Z

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 (1)

(2)

For this power system, Xbus = Xbus

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The sequence fault currents are

r Ia1

r Ia0 = 0 r 1 1 = − Ia2 = (1) = = − j1.362 (2) j2(0.3671) Z44 + Z44

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 Ic −L 2 = Ia0 −L 2 + a Ia1−L 2 + a2 Ia2 −L 2 = 1.0378

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SLG Fault at Bus k

N1

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 Ib −L 2 = Ia0 −L 2 + a2 Ia1−L 2 + a Ia2 −L 2 = − j0.0262 r r r r Ic −L 2 = Ia0 −L 2 + a Ia1−L 2 + a2 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

r n Ia 1

(2) Z(kk2) Z11

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 =

Z(kk1)

1 + (Z(kk2) // Z(kk0) )

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r Ia2 = −

r Z(kk0) I (0 ) (2) a1 Zkk + Zkk

r Ia0 = −

r Z(kk2) I (0 ) (2) a1 Zkk + Zkk

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 rcurrentsrin line L3 r are

Ia−L 3 = Ia0 −L 3 + Ia1−L 3 + Ia2 −L 3 = − j0.0273 r Ib −L 3 = −1.3213 + j0.799 r Ic −L 3 = 1.3213 + j0.799

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Protective Device Duties ‰ Fault

Current at Different Times

‰ ANSI/IEEE

and IEC Standards

‰ ANSI/IEEE

Calculation Method

‰ IEC

Calculation Method

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Fault Current at Different Times

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Fault Current at Different Times ‰ First

Cycle Fault Current

ƒ Short circuit ratings of low voltage equipment ƒ Ratings of Medium Voltage (MV) to High Voltage (HV) switch and fuse ƒ Close & Latch (Making) capacity or ratings of HV Circuit Breakers ƒ Maximum Fault for coordination of instantaneous trip of relays ƒ Momentary Short Circuit Current (ANSI) ƒ Initial Symmetrical Short Circuit Current (IEC) U. P. National Engineering Center National Electrification Administration

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Fault Current at Different Times ‰ 1.5

to 4 Cycles Fault Current

ƒ Interrupting (breaking) duties of HV circuit breakers ƒ Interrupting magnitude and time of breakers for coordination ƒ Interrupting Short-Circuit Current (ANSI) ƒ Symmetrical Short-Circuit Breaking Current (IEC)

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Fault Current at Different Times ‰ 30

Cycles Fault Current

ƒ For time delay coordination ƒ Steady State Short-Circuit Current (ANSI) ƒ Steady State Short-Circuit Current (IEC)

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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 ½ Cycle Network: the network used to calculate momentary short-circuit current and protective device duties at the ½ cycle after the fault. Type of Device

Duty

High Voltage CB

Closing and Latching Capability

Low Voltage CB

Interrupting Capability

Fuse

Interrupting Capability

Switchgear and MCC

Bus Bracing

Relay

Instantaneous settings

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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

Xsc

Utility

X’’

Turbo Generator

Xd’’

Hydro-generator w/ Amortisseur Winding

Xd’’

Hydro-generator w/o Amortisseur Winding

0.75 Xd’

Condenser

Xd’’

Synchronous Motor

Xd’’

Induction Machine

(1.2 – 1.67) Xd’’

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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 1.5-4 Cycle Network: also known as the transient network because all rotating machines are represented by their subtransient reactances Type of Machine

Xsc

Utility

X’’

Turbo Generator

Xd’’

Hydro-generator w/ Amortisseur Winding

Xd’’

Hydro-generator w/o Amortisseur Winding

0.75 Xd’

Condenser

Xd’’

Synchronous Motor

1.5 Xd’’

Induction Machine

(1.5 – 3.0) Xd’’

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ANSI/IEEE Calculation Method 30 Cycle Network: the network used to calculate the steady-state short-circuit current and protective device duties 30 cycles after the fault. Type of Device

Duty

High Voltage CB

N/A

Low Voltage CB

N/A

Fuse

N/A

Switchgear and MCC

N/A

Relay

Overcurrent settings

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ANSI/IEEE Calculation Method 30 Cycle Network: also known as the steadystate network because all rotating machines are represented by their subtransient reactances 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 than1.5 times the generator subtransient reactance. Otherwise the contribution is defined as remote. 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

Iremote NACD = Ilocal Itotal = Iremote + Ilocal

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

Assymetrical 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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ANSI/IEEE Calculation Method Momentary (1/2 Cycle Short-Circuit Current Calculation (Buses and HVCB)

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

Interrupting Short-Circuit Current

Iint,rms, symm =

Vpre − fault 3Zeq

MF =

2(1 + e 2(1 + e

MF =

1 + 2e 1 + 2e

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−

−

−

π X R

)

Unfused power breakers

π ( X R )test

−

)

π X R

π ( X R )test

Fused power breakers & Molded Case

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

AMFi = MFl + NACD ( MFr − MFl ) Iint,rms, adj = AMFi ⋅ Iint,rms, symm

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

Symmetrical RMS value of Interrupting ShortCircuit Current

Iint,rms, symm =

Vpre − fault

3Zeq

MFr = 1 + 2e

−

4π t X R

AMFi = MFl + NACD ( MFr − MFl ) Iint,rms, adj = AMFi ⋅ Iint,rms, symm U. P. National Engineering Center National Electrification Administration

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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

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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

Reactance (Xd’’) Machine

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''

)

cmax kVn KG = 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

>400 V to 1 KV

1.05

1.00

1 kV to 35 kV

1.10

1.0

35 KV to 230 KV

1.10

1.00

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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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