CPT5 - Short Circuit Analysis-6th Batch

7/29/2019 CPT5 - Short Circuit Analysis-6th Batch

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NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER

Certificate in

Power System Modeling and Analysis

Competency Training and Certification Program in Electric Power Distribution System Engineering

U. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTER

Competency Training and Certification Program in Electric Power Distribution System Engineering

T r a i n i n g Co u r s e i n

Sh o r t Ci r c u i t A n a l y s i s


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2

Competency Training & Certi fication Program in

Electric Power Distribution System EngineeringU. P. National Engineering Center

National Electrification Administration




Training Course in Short Circuit Analysis

Co u r s e Ou t l i n e

1. Sources & Types of Faults

2. Network Reduction Techniques

3. Analysis of Faulted Power System

4. Computer Solution

5. Short Circuit Studies


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3








So u r c e s a n d Ty p e o f Fa u l t s

Sources of Short Circuit Currents

Type of Faults


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4








So u r c e s o f Sh o r t Ci r c u i t

C u r r e n t s

FaultMV

LV

UtilityG

Fault Current Contributors


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5








Ty p e s o f Fa u l t

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

1. Single Line-to-Ground Fault

2. Line-to-Line Fault

3. Double Line-to-Ground Fault

4. Three Phase Fault

Series Fault: Unintentional Opening of phaseconductors

Simultaneous Fault


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Shunt Faults

Line-to-Line

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

Three Phase

Ty p e s o f Fa u l t

7T i i C i Sh Ci i A l i


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7








N e t w o r k Re d u c t i o n Te ch n i q u e s

Calculating Short Circuit Currents by

Network Reduction

Reduced Sequence Networks

8T i i C i Sh t Ci it A l i


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8









Calculating Short Circuit Currents by Network Reduction

1. Draw the Single Line Diagram.

2. Draw the Impedance Diagram.

3. Convert all parameters to per-unit.4. Reduce the network between the source(s) and the fault

location.

5. Calculate the fault current

equiv

f

f

Z

VI =

9T i i C i Sh t Ci it A l i


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9









Combination of Branches in Series

1 2

1 1 2 2

1 2 1 2

( ) ( )

( ) ( )

eqZ Z ZR jX R jX

R R j X X

= += + + +

= + + +

Z1 Z2

Combination of Branches in Parallel

1 2

1 2

1 1 2 2

1 2 1 2

( )( )

( ) ( )

eq

Z ZZ

Z ZR jX R jX

R R j X X

=

++ +

=+ + +

Z1

Z2

10Training Course in Short Circuit Analysis


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10








Transforming Wye to Delta

Transforming Delta to Wye

a b b c c aA

a

a b b c c aB

b

a b b c a c C

c

Z Z Z Z Z Z Z

Z

Z Z Z Z Z Z Z Z

Z Z Z Z Z Z Z

Z

+ +=

+ +=

+ +=

B Ca

B C

A Cb

B C

A Bc

A B C

Z ZZ

Z Z Z

Z ZZZ Z Z

Z ZZ

Z Z Z

=+ +

= + +

=+ +

ZC

ZB ZA

Za Z b

Zc




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11








Example:

Determine the fault current for a three phasebolted fault in each bus for the 4 bus systembelow.

G

1

432

Line1Line

3

Line 5 Line 4

Line

2

4-bus system

LINE FB TB Z(p.u.)

Line1 1 4 j0.2

Line2 1 3 j0.4

Line3 1 2 j0.3

Line4 3 4 j0.5

Line5 2 3 j0.6

The generator is rated 100 MVA, 6.9 kV and has asubtransient reactance of 10%. Base Values: 100 MVA, 6.9 kV



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12








Solution:

Draw the impedance diagram

E 1.0

0.1

0.20.3

0.6 0.5

0.4

4

3

2

1



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E 1.0

0.1

0.20.3

0.6 0.5

0.4

43

2

1

If

-

+

= +

= +

=

=

+

=

+

=

= +

= +

=

a1 2 2 3

b 1 3

1 3

c3 4

R edu ce the ne tw o rk

X

0 .3 0 .6

0 .9

X

(0.9)(0.4)0 .9 0 .4

0 .276923

X

0 .276923 0 .5

0 .776923

a

a

b

X X

X X

X X

X X

a) Fault @ Bus 4



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cd 14

c 14

X XX

X X(0.776923) (0.2)

0.776923 0.2

0.159055

=+

=+

=

d

equiv genX X X

0.1 0.159055

0.259055

= +

= +=

If

-

+

E 1.0

0.25905

f

1.0

I 0.259055

3.860184 p.u.= =

base

f

100 1000I 8367.64 A

3(6.9)I 3.860184 x 8367.64

= 32,300.63 A

x= =

=



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b) Fault @ Bus 3

E 1.0

0.1

0.20.3

0.6 0.5

0.4

432

1

If

-

+

a

23 12X X X0.3 0.6

0.9

= += +

=b

14 34X X X

0.2 0.5

0.7

= +

= +

=

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

0.198425

=

=



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g y

gen equivX X X

= 0.1 0.198425

= 0.298425

= ++

If-

+

E 1.0

0.298425f

1.0I 0.298425

= 3.350923 p.u.

=



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g y

c) Fault @ Bus 2

a

14 34X X X

0.2 0.50.7

= +

= +=

E 1.0

0.1

0.20.3

0.6 0.5

0.4

432

1

If

-

+

ab 13

a

13

X XX

X X

(0.7)( 0.4)0.7 0.4

0.254545

=+

=+

=



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g y

= +

= +=

c b

23X X X

0.254545 0.6

0.854545

=+

= +=

cd 12

c

12

X XX

X X

(0.854545)( 0.3)0.854545 0.3

0.222047

= +

=

d

genX X X

0.322047

=

=

f 1.0I 0.322047

3.095525 p.u.

If

-

+

E 1.0

0.322047



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d) Fault @ Bus 1

=

=

genX X

0.1E 1.0

0.1

0.20.3

0.6 0.5

0.4

432

1

If

-

+

=

=

f

1.0I

0.110.0 p.u.



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Re d u ce d Se q u e n ce N e t w o r k s

Since we mentioned that various power system

components behave/respond differently to the flowof the currents sequence components, it followsthat the there will be a unique power system modelfor each of the sequence component. These are

called the sequence networks.

Positive-Sequence Network

Negative-Sequence Network Zero-Sequence Network



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Re d u ce d Se q u e n ce N e t w o r k s

The Thevenin equivalent of the power system a tt h e f a u l t p o i n t is called the sequence network.

PositiveSequence

NegativeSequence

ZeroSequence

0aVr

F0

N0

Z0

0aIr

+

-

F1

N1

thf VVr

=

Z11aIr

1aVr

+

-

+

-

F2

N2

Z2

2aIr

2aVr

+

-

11ath1a ZIVV

rrr

= 22a2a ZIV

rr

= 00a0a ZIV

rr

=



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Eq u i v a l e n t Ci r c u i t o f U t i l i t y

The equivalent sequence networks of the ElectricUtility Grid are:

Utility Thevenin Equivalent Circuits

+

-

gE

r

+

-

R1 +jX1

+

-

PositiveSequence

NegativeSequence

ZeroSequence

+

-

R2 +jX2 R0 +jX0



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Eq u i v a l e n t Ci r c u i t o f Ge n e r a t o r s

Positive-Sequence Impedance:

Xd=Direct-Axis Subtransient Reactance

Xd=Direct-Axis Transient Reactance

Xd=Direct-Axis Synchronous Reactance

Negative-Sequence Impedance:

for a salient-pole machine)"X"X(X qd21

2 +=

for a cylindrical-rotor machine"XX d2 =Zero-Sequence Impedance:

"X6.0X"X15.0 d0d



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Gr o u n d e d -W y e Ge n e r a t o r


The sequence networks for the grounded-wye

generator are shown below.

jZ0

F0

N0

jZ2

F2

N2

F1

N1

gEr

jZ1+

-

PositiveSequence

NegativeSequence

ZeroSequence



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Gr o u n d e d -W y e t h r o u g h a n I m p e d a n ce


If the generator neutral is grounded through animpedance Zg, the zero-sequence impedance is

modified as shown below.

jZ0

F0

N0

3Zg

jZ2

F2

N2

F1

N1

gEr

jZ1+

-

PositiveSequence

NegativeSequence

ZeroSequence



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Un g r o u n d e d -W y e Ge n e r a t o r

If the generator is connected ungrounded-wye or

delta, no zero-sequence current can flow. Thesequence networks for the generator are shownbelow.

F1

N1

gEr

jZ1+

-

jZ2

F2

N2

jZ0

F0

N0

Positive

Sequence

Negative

Sequence

Zero

Sequence



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Eq u i v a l e n t Ci r c u i t o f T r a n s f o r m e r s

Po s i t i v e & N e g a t i v e Se q u e n ce N e t w o r k s

2Ir

2Z

+

-

+

-PrimarySide SecondarySide

1

Ir

1Z

+

-

+

-PrimarySide SecondarySide

NegativeSequence

Network

PositiveSequence

Network21 ZZ =



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10 ZZ =

+

-

HVr

+

-

XVr

10 ZZ =

+

-

HVr

+

-

XVr

Ze r o Se q u e n c e N e t w o r k * Transformer Connection Zero-Sequence Network

*Excluding 3-phase unit with a 3-legged core.




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Transformer Connection Zero-Sequence Network


10 ZZ =

+

-

HVr

+

-

XVr

10 ZZ =

+

-

HVr

+

-

XVr

Ze r o Se q u e n c e N e t w o r k *




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Transformer Connection Zero-Sequence Network


10 ZZ =

+

-

HVr +

-

XVr

10 ZZ =+

-

HVr

+

-

XVr

Ze r o Se q u e n c e N e t w o r k *




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Eq u i v a l e n t Ci r c u i t o f L i n e s

If the line is completely transposed,

s0s ZZ = m0m ZZ =0ZZ 2m1m ==0ZZ 2s1s ==

The sequence impedance matrix reduces to

+

=

ms

ms

ms

2

1

0

ZZ00

0ZZ0

00Z2Z

Z

Z

Z

Note: The sequence impedances are completely decoupled.



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For a completely transposed line, the equation inthe sequence domain is

=

2a

1a

0a

2

1

0

2a

1a

0a

I

II

Z00

0Z000Z

V

VV

r

r

r

r

r

r

where

s

m

a21D

DlnksjsrZZ +==


2

ms

3

e

da0DD

Dlnksjsr3srZ ++=



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21 II

rr

=

21 ZZ =

+ +

0I

r

0Z

+ +

- - - -

Z e r o - S e q u e n c e

N e t w o r k

Po s i t i v e - & N e g a t i v e -

Se q u e n ce N e t w o r k s



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Example: Determine the reduced sequence

networks of the power system shown for a fault aF. Assume Eg = 1.0 p.u.

G

T1 T2Line OpenF

G: X1 = 40% X2 = 40% X0 = 20%

T1, T2: X = 5%

Line: X1= X

2= 15% X

0= 35%

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



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Positive-Sequence Network:

+

-

j0.61aI

r

1.0

N1

1aV

F1j0.05

+

gEr

+

-

rj0.4

-

j0.15Open

j0.05

F1

N1



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Negative-Sequence Network:

j0.6

2aIr

N2

F2j0.05 j0.15+

-

2aVr

j0.4

Openj0.05

F2

N2



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Zero-Sequence Network:

j0.044

0aIr

N0

F0

N0

j0.05 j0.35+

-

0aVr

j0.2

Openj0.05

F0



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Example: Reduce the sequence networks for thepower system shown if a fault occurs at bus 4.

G1

TL1

L2

G2

L3

2 3

4

T: X=0.08G1: X1=0.40 X2=0.40 X0=0.15

G2: X1=0.50 X2=0.50 X0=0.25

L1: X1=0.40 X2=0.40 X0=0.80L2: X1=0.30 X2=0.30 X0=0.60

L3: X1=0.20 X2=0.20 X0=0.40



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F1

1aIr

+

-

N1

j0.4

j0.08

j0.4

j0.3 j0.2

j0.5

1G

Er +

- 2GEr

2 3

4



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F2

1aIr

j0.2

j0.08

j0.3

j0.42 3

4

j0.5j0.4

N2



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F0

0aIr

N0

j0.2

j0.08

j0.8

j0.6 j0.4

j0.252 3

4



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A n a l y s i s o f Fa u l t e d Sy s t em

The Fault Point

Three-Phase Fault

Single-Line-to-Ground Fault

Line-to-Line Fault

Double-Line-to-Ground Fault



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The system is assumed to be balanced, withregards to impedances, except at one point calledthe fault point.

FaultCurrents

a

b

c

aVr

bVr

cVr

Ground

aIr

bIr

cIr

Line-to-ground

voltages

F

Th e Fa u l t Po i n t



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Th r e e Ph a s e Fa u l t

On a balanced three phase system, thesame magnitude of fault currents willflow in each phase of the network if athree phase fault occurs.

Since faults currents are balanced, thefaulted system can, therefore, be

analyzed using the single phaserepresentation.


Th r e e - Ph a s e Fa u l t


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Th r e e - Ph a s e Fa u l t

ab

c

aVr

bVr

cVr

Ground

bI

r

cI

r

gZ

aI

r

fZ fZ

gIr

fZ

Note: The system is still balanced. Currents andvoltages are positive sequence only. The groundcurrent is zero.gI

r


Sequence Network Interconnection:


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F1 F2

N2

Z2

2aIr

2aV

r

+

-

N0

F0

Z0

0aIr

0aV

r

+

-fV

Z1 1aIr

1aV

r

+

+

--

Zf

N1

Sequence currents

f

f

aZZ

VI+

=1

1

r

020 == aa IIrr


Three Phase Fault Currents:


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Three Phase Fault Currents:

f

f

aaaaZZ

VIIII

+=++=

1

210

f

f

aaab

ZZ

VaaIIaII

+

=++=1

2

21

2

0

f

f

aaacZZ

aVIaaIII +=++= 1

2210


Example: A three-phase fault occurs at point F


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Example: A three phase fault occurs at point F.Assuming zero fault impedance, find the faultcurrents at fault point F. Determine the phasecurrents in the line and the generator. Assume Eg= 1.0 p.u.

G

T1 T2Line

OpenF

G: X1 = 40% X2 = 40% X0 = 20%

T1, T2: X = 5%

Line: X1 = X2 = 15% X0 = 35%

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




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Positive Sequence Network:

+

-

j0.61AI

r

1.0

N1

F1

+

-

1AVr

j0.05

j0.4

+

gEr

-

Openj0.05

j0.151AI

r

L1AIr

g1aIr

F1

N1




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The sequence fault currents

Zf

=

=

=+=

0

2

11

a

a

f

f

a

I

I

ZZ

V

I

r

r

r

+

-

j0.61AI

r

1.0

N1

F1+

-

1AVr

The phase fault currents

=

==

c

b

a

I

II


Si n g l e L i n e - t o - Gr o u n d Fa u l t


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g

Assuming the fault is in phase a,a

b

c

aVr

bVr

cVr

Ground

aIr bI

r

cIr

fZ

afa IZVrr

=Boundary Conditions: (1)

0== cb IIrr

(2)


Transformation: From (2), we get


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abcIAI

rr

1

012

=

Transformation: From (2), we get

aaaa

2

2

11

111

3

1

2

1

0

a

a

a

II

I

r

r

r

=00

aIr

3

1

=a

a

a

II

I

r

r

r

which meansa

3

12a1a0a IIII

rrrr

===

From (1), we get

)(210210 aaafaaa

IIIZVVVrrrrrr

++=++

or

0210 3 afaaa IZVVVrrrr

=++




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q

F1

N1

fVr

Z1 1aIr

1aV

r

+

-

+

-

F2

N2

Z2

2aIr

2aV

r

+

-

0aV

r

F0

N0

Z0

0aIr+

-

3Zf


f

f

aaaZZZZ

VIII

3210210 +++

===rrr


Single-Line-to-Ground Phase Fault Currents:


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g

f

f

aaaaZZZZ

VIIII

3

3

021

210 +++=++=

0=bI

0=cI


Example: A single line-to-ground fault occurs at


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p g gpoint F. Assuming zero fault impedance, find thephase currents in the line and the generator.Assume Eg = 1.0 p.u.

G

T1 T2

Line OpenF

G: X1 = 40% X2 = 40% X0 = 20%

T1, T2: X = 5%

Line: X1 = X2 = 15% X0 = 35%

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




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+

-

j0.61aI

r

1.0

N1

1aV

F1j0.05

+

gEr

+

-

rj0.4

-

j0.15

Openj0.05

F1

N1




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j0.6

2aIr

N2

F2j0.05 j0.15

+

-

2aVr

j0.4

Openj0.05

F2

N2




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j0.044

0aIr

N0

F0

N0

j0.05 j0.35

+

-

0aVr

j0.2

Openj0.05

F0




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N1N2

F1

J0.6 1AIr

+

-

F2

J0.6

2AIr

1.0

N0

F0

J0.044

0AIr

Sequence Fault Currents

)044.06.06.0(j

0.1III

2A1A0A ++===

rrr

p.u.804.0j=


Phase Fault Currents


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p.u.411.2jI3I 0AA ==

rr

0II CB ==rr

Sequence Currents in the Transmission Line

p.u.804.0jII 1AL1A ==rr

p.u.804.0jII 2AL2A ==

rr

p.u.089.0jI4.005.0

05.0I 0AL0A =+

=rr

Phase Currents in the Transmission Line

p.u.696.1jIIII L2AL1AL0AAL =++=rrrr


p.u.714.0jIaIaII L2AL1A2

L0ABL =++=rrrr

rrrr


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p.u.714.0jIaIaII L2A2

L1AL0ACL =++=

rrrr

Sequence Currents in the Generator: Using the 30o

phase shift,

p.u.120804.030II ooL1Ag1a ==rr

p.u.60804.030II

oo

L2Ag2a =+=

rr

p.u.696.0j402.0 =

p.u.696.0j402.0 =0I g0a =

r

Phase Currents in the Generator

p.u.392.1jIIII g2ag1ag0aag =++=rrrr


p.u.392.1jIaIaII g2ag1a2

g0abg =++=rrrr


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0IaIaII g2a2

g1ag0acg =++=

rrrr

Three-line Diagram:

H3

H1

H2

A

BC H3

H1

H2

A

BC

X3X

1

X2

a

b

c

T1 T2a

b

j2.411

j1.696

j0.714

j0.714

j1.392

j1.392

0

j0.268

j0.714

j2.1430


L i n e - t o - L i n e Fa u l t


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Assuming the fault is in phases b and c,

0=aIr

a

b

c

aVr

bVr

cVr

Ground

bIr

cIr

fZ

aIr

(1)Boundary Conditions:

cb IIrr

=(2)

fbcb ZIVVrrr

=(3)


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


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abcIAI1

012

=

r

aa

aa2

2

1

1

111

3

1

2

1

0

a

a

a

I

I

I

r

r

r

=

b

b

I

Ir

r

0

3

1

=

b

b

IaaIaar

r

)()(

0

2

2

which means00 =aI

r

bbaa IjIaaIIrrrr

3

1231

21 )( ===


From (3), we get


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)( 212

0 aaa VaVaV

rrr

++faaaaaa ZIaIaIVaVaV )()( 21

2

02

2

10

rrrrrr

++=++

Since and , we get

faaa ZIaaVaaVaa 12

2

2

1

2 )()()(rrr

=+

00 =aIr

21 aa IIrr

=

or

faaa ZIVV 121rrr

=




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F1

N1

fV

Z1 1aIr

1aVr

+

-

+

-

F2

N2

Z2

2aIr

2aVr

+

-

Zf

N0

F0

Z0

0aIr


f

f

aaZZZ

VII

++==

21

21

rr

00 =aIr


Line-to-Line Phase Fault Currents:


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0=aI

21

2

0 aaabaIIaII ++=

( ) 112

11

2 )(0 aaaa jIIaaIaIa ==++=

f

f

bZZZ

VjI++

=21

3

f

f

cZZZ

VjI

+++=

21

3


Example: A line-to-line fault occurs at point F.Assuming zero fault impedance find the fault


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Assuming zero fault impedance, find the fault

currents at fault point F. Assume Eg = 1.0 p.u.

G

T1 T2Line

OpenF

G: X1 = 40% X2 = 40% X0 = 20%

T1, T2: X = 5%

Line: X1 = X2 = 15% X0 = 35%




F0F1 F2


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F0F1 F2

N1 N2

J0.6 1AIr

+

-

J0.6

2AIr

J0.044

0AIr

1.0

N0


Sequence Fault Currents:


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Phase Fault Currents:


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72


D o u b l e - L i n e - t o - Gr o u n d Fa u l t


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a

b

c

aVr

bVr

cVr

Ground

bIr

cIr

g

Z

aIr

fZ fZ

cb II

rr

+

Assuming the fault is in phases b and c,

0=aIr

cgbgfb IZIZZVrrr

++= )(

bgcgfc IZIZZVrrr

++= )(

(1)Boundary Conditions:

(2)

(3)

73


Transformation: From (1), we getrrrr


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2100 aaaa IIII ++==From

21

2

0 aaab

VaVaVVrrrr

++=

2

2

10 aaac VaVaVVrrrr

++=we get

22

12 )()( aacb VaaVaaVV

rrrr

+=

2120 aaab IaIaII

rrrr

++=

Likewise, from

2

2

10 aaac IaIaIIrrrr

++=

74


we get

22rrrr


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2

2

1

2

)()( aacb IaaIaaII +=From boundary conditions (2) and (3), we get

)( cbfcb IIZVV

rrrr

=Substitution gives

])()[( 22

1

2

aaf IaaIaaZrr

+=2

21

2 )()( aa VaaVaarr

+

Simplifying, we get

2211 afaafa IZVIZVrrrr

=

75


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


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))(2( cbgfcb IIZZVV ++=+

2102 aaacb VVVVV

rrrrr

=+2102 aaacb IIIII

We can also show

rrrrr

=+

Substitution gives

)2(2 210210 aaafaaa IIIZVVVrrrrrr

=

)2(2 210 aaag IIIZ

rrr

+

76


Rearranging terms, we getrrrrr


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11000 422 afaagafa IZVIZIZV =)(2 2122 aagafa IIZIZV

rrrr

++

Earlier, we got

2211 afaafa IZVIZVrrrr

=

021 aaa III

rrr

=+Substitution gives

)(2622 11000 afaagafa IZVIZIZV

rrrrr

=

( )1100 3 afaagfa IZVIZZV

rrrr

=+

77




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F1

fV

Z1 1aI

r

1aVr

+

+

-

F2 F0

-

N1 N2

Z2

2aI

r

2aVr

+

-

Z0

0aI

r

0aVr

+

-

ZfZf Zf+3Zg

N0

gfT ZZZZ 300 ++=

fT ZZZ += 11

Let

fT ZZZ += 22

78



Vr


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TT

TTT

fa

ZZ

ZZZ

VI

20

201

1

++

=r

From current division, we get

120

0

2 aTT

T

aI

ZZ

ZI

rr

+=

From KCL, we get

1

20

20 a

TT

Ta I

ZZ

ZIrr

+=

210 aaa IIIrrr

= or

79


Double-Line-to-Ground Phase Fault Currents:


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0=aI

21

2

0 aaab aIIaII ++=

( )

TTTTTT

TTf

ZZZZZZ

aZZVj

020121

203

++

=

2

2

10 aaac IaaIII ++=

TTTTTT

TTf

ZZZZZZ

ZaZVj

020121

22

03

++

+=

80


Example: A double-line-to-ground fault occurs atpoint F. Assuming zero fault impedance, find the


80/187







fault currents at fault point F. Assume Eg = 1.0 p.u.

G

T1 T2Line

OpenF

G: X1 = 40% X2 = 40% X0 = 20%

T1, T2: X = 5%

Line: X1 = X2 = 15% X0 = 35%


81



F0F1 F2


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N1 N2

J0.6 1AIr

+

-

J0.6

2AIr

J0.044

0AIr

1.0

N0

82


Sequence Fault Currents:


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83


Phase Fault Currents:


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84


Com p u t e r So l u t i o n


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Development of the Model

Rake Equivalent Formation of Zbus

Analysis of Shunt Fault

85


D e v e lo p m e n t o f t h e M o d e l


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Ob s e r v a t i o n s o n M a n u a l N e t w o r k

S o l u t i o n

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

Is there a way for a computer to implement thismethodology?

86


D e v e lo p m e n t o f t h e M o d e l


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Consider the three-bus system shown below. Letus analyze the system for a three-phase fault inany bus.

G1G2L1

L2

1 2

3

G1, G2 : X1=X

2=0.2 X

0=0.1

L1 : X1=X2=0.6 X0=1.2

L2 : X1=X2=0.24 X0=0.5

87



j0 61 2


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j0.2

1GEr +

-

+

-2GE

r

j0.2

j0.6j0.24

1 2

3

j0.6

3

j0.2j0.2

j0.24

+

-

GEr

1 2Combine the sourcesand re-draw. AssumeE

G

= 1.0 per unit.

88


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


88/187







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

25.6jZ

1

Z

E

I11

G

F ===

r

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

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

5.2jZ

1

Z

EI

11

GF ===

r

89


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

th f lt it h i th t b


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Next, use loop currents to

describe the circuit with allfault switches closed.Since there are four loops,

we need to define fourloop currents.

the fault switch in that bus.

j0.6

3

j0.2j0.2

j0.24

+

-

GEr

1 2

1I

r

3I

r

2I

r

4Ir

4

90


The loop equations are

)(2.00.1 431 IIIjrrr

+=loop 1:rr


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loop 2:

loop 3:

)(2.06.0)(2.00 314442 IIIjIjIIjrrrrrr

+++=

)(2.00.1 42 IIjrr

+=

3431 24.0)(2.00.1 IjIIIjrrrr

++=

loop 4:

or

0.12.02.02.0

2.044.002.0

2.002.00

2.02.002.0

0

0.1

0.1

0.1

4

3

2

1

I

I

I

I

r

r

r

r

= j

91


Current I4 is not a fault current. It can be eliminatedusing Krons reduction. We get

IZV )1(rr


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IZV )1(bus=where

3

1

421

)1(

bus ZZZZZ

=and

44.002.0

02.002.002.0

Z1 = j

2.0

2.02.0

Z2 = j

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

92


Substitution gives

16004016001 1I

r


92/187







40.004.016.0

04.016.004.016.004.016.0

= j

0.1

0.10.1

3

2

1

I

II

r

r

IZV )1(busrr

=

Note:(1) The equation can be used to analyze a three-

phase fault in any bus (one fault at a time).

)1(busZ(2) is called the positive-sequence bus-

impedance matrix, a complex symmetric matrix.

93


Ra k e Eq u i v a l e n t

Consider the matrix voltage equation


93/187







Consider the matrix voltage equation

232212

131211

ZZZ

ZZZ

One possible equivalentcircuit is shown. This circuitis called a rake-equivalent.

+

-

Z11 Z22 Z33

Z12 Z23

Z13

1.0

1Ir

2Ir

3Ir

332313 ZZZ=0.1

0.1

0.13I

2

1

I

I

r

r

r

Suppose we are asked tofind a circuit that satisfiesthe matrix equation.


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

r


94/187







40.004.016.0

04.016.004.0

16.004.016.0

= j

0.1

0.1

0.1

3

2

1

I

I

I

r

r

The rake equivalent is

shown. The diagonalelements of the matrix areself impedances while the

off-diagonal elements aremutual impedances.

+

-

j0.16

1.0

1Ir

2Ir

3Ir

j0.16 j0.4

j0.04j0.04

j0.16


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

I40jI040jI160j01rrr


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321 I4.0jI04.0jI16.0j0.1 ++=

Since only bus 3 is faulted,

I1=I2=0. We get

3I4.0j0.1r

=or

5.2j4.0j

1I3 ==r

+

-

j0.16

1.0

3Ir

j0.16 j0.4

j0.04j0.04

j0.16

2Vr

+

-1Vr

+

-

From KVL, we get the voltage in bus 1.

6.0Z

Z0.1IZ0.1V

33

133131 ===rr


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

90Z

Z

01IZ01V23

3232

rr


96/187







9.0Z0.1IZ0.1V 333232 ===

Note: Once the voltages in all the buses are

known, the current in any line can be calculated.In general, for a three-phase fault in bus k of asystem with n buses, the fault current is

kk

kZ1I =

r

k=1,2,n

The voltage in any bus j is given by

kk

jk

jZ

Z0.1V =

r

j=1,2,n


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

VVrr

r


97/187







mn

nmmn

z

VVI =

where zmn is the actualimpedance of the line.j0.2

1GE

r+

-

+

-2GE

r

j0.2

j0.6

j0.24

1 2

3

-j2.0

-j2.5

-j0.5

5.0j6.0j

6.09.0z

VVI21

1221 ===

rr

r

For example, the

current in the linebetween buses 2 and1 is


Fo r m a t i o n o f Zb u s

Zbus can be built one step at a time by adding one


98/187







Zbus can be built, one step at a time, by adding onebranch at a time until the entire network is formed.

The first branch to be added must be a generatorimpedance. This is necessary in order to establishthe reference bus.

Subsequent additions, which may be done in any

order, fall under one of the following categories:

(1) Add a generator to a new bus;

(2) Add a generator to an old bus;

(3) Add a branch from an old bus to a new bus;

(4) Add a branch from an old bus to an old bus.


Assume thatat the current

stage the n22221

n11211

ZZZ

ZZZ

01

0.1

2

1

I

Ir

r


99/187







+

-

Z11 Z22 Zkk

Z12 Z2k

1.0

1Ir

2Ir

kIr

Znn

Zkn

nIr1 2 k n

stage, thedimension ofZbus is n.

nn2n1n

n22221

ZZZ

ZZZ

0.1

0.1

n

2

I

I

r

=

oldbusZ

Let us examineeach category inthe addition of a

new branch.


Type 1: Add a generator to a new bus

- 1 0 Let Zg be the impedance of


100/187







+

Z11 Z22 Zkk

Z12 Z2k

1.0

1Ir

2Ir

kIr

Znn

Zkn

nIr1 2 k n

g pthe generator to be added.

Zg

1nI +r n+1

g

nn2n1n

n22221

n11211

Z000

0ZZZ

0ZZZ

0ZZZ

0.1

0.1

0.1

0.1

1n

n

2

1

I

I

I

I

+

r

r

r

r

=

The dimensionis (n+1).


Type 2: Add a generatorto an old bus k

Let Zg be the +

-

1.0

w

Ir


101/187







Let Zg be theimpedance of thegenerator to beadded.

Z11 Z22 Zkk

Z12 Z2k

1Ir

2Ir

kIr

Znn

nIr1 2 k n

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

nn1wkk1212111 IZ...)II(Z...IZIZ0.1rrrrr

++++++=

nn2wkk2222121 IZ...)II(Z...IZIZ0.1rrrrr

++++++=

nnnwknk22n11n IZ...)II(Z...IZIZ0.1rrrrr

++++++=

Zg


For the added generator loop, we get

wgnknwkkk22k11k IZIZ...)II(Z...IZIZ0

rrrrrr

+++++++=


102/187







IZIZ...)II(Z...IZIZ0 +++++++=In matrix form, we get

wknkk2k1k

nknnnk2n1n

k2n2k22221

k1n1k11211

ZZZZZ

ZZZZZ

ZZZZZZZZZZ

0

0.1

0.10.1

=

w

n

2

1

I

I

II

r

r

r

r

where Zw=Zkk+Zg. The last row is eliminated usingKrons reduction. The dimension remains as n.


Type 3: Add a branchfrom an old bus k to

a new bus +

-

Z Z

1.0

Z


103/187







Z11 Z22 Zkk

Z12 Z2k

1Ir

2Ir

kIr

Znn

Zkn

nIr

1 2 k n

The new current in impedance Zkk is (Ik+In+1). Thenew equations for buses 1 to n are

nn11nkk1212111 IZ...)II(Z...IZIZ0.1rrrrr

++++++= +

nn21nkk2222121 IZ...)II(Z...IZIZ0.1rrrrr

++++++= +

nnn1nknk22n11n IZ...)II(Z...IZIZ0.1rrrrr

++++++= +

Zb

1nI +r

n+1


For the new bus, we get

...)II(Z...IZIZ0.1 1nkkk22k11k +++++= +rrrr

IZIZ ++rr


104/187







In matrix form, we get

wknkk2k1k

nknnnk2n1n

k2n2k22221

k1n1k11211

ZZZZZ

ZZZZZ

ZZZZZZZZZZ

0.1

0.1

0.10.1

=

1n

n

2

1

I

I

II

+

r

r

r

r

1nbnkn IZIZ +++

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


Type 4: Add abranch from an oldbus j to an old bus k

+

-

1.0 wIr


105/187







Z11

Z22

Zjj

Z12 Z2j

1Ir

2Ir

kIr

Zkk

Zkn

nIr1 2

jk

Znn

jIr n

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

)II(Z...IZIZ0.1 wjj1212111rrrr

++++=

nn1wkk1 IZ...)II(Zrrr

+++

w

Zb


)II(Z...IZIZ0.1 wjj2222121rrrr

++++=

nn2wkk2 IZ...)II(Z

rrr

+++


106/187







)II(Z...IZIZ0.1 wjnj22n11nrrrr

++++=

)( +++

nnnwknk IZ...)II(Zrrr

Documents

CPT5 - Short Circuit Analysis-6th Batch