Components of symmetric fault analysis

7/17/2019 Components of symmetric fault analysis

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Topic 2

Symmetrical Components

0




Outline

• Concept and definition of symmetrical components

• Sequence networks of impedance loads

• Sequence networks of series impedances

•

Sequence networks of transmission and distribution lines• Sequence networks of rotating machines

• Sequence networks of transformers

• Power in sequence networks

• Main reference: Ch 8 Textbook


1



Outline




•

Sequence networks of transmission and distribution lines• Sequence networks of rotating machines



• Main reference: Ch 8 Textbook


2



Symmetric Components

Charles FortescueDeveloped symmetrical component analysis


3




Introduced to simplify calculation of unbalancednetworks, in particular networks with unbalanced short

circuits and other unbalanced fault conditions.


4



The Problem

Analysis of balanced three phase networks is simple

Analysis of unbalanced three phase networks is difficult

Solution

Adapt balanced analysis to use for unbalanced systems


5



The Principle

Any unbalanced set of three phase phasor quantities (V or I)

can be resolved into three balanced systems of phasor

quantities

Sequence components are the most commonly used set of

symmetrical components


6




• The key idea of symmetrical component analysis is todecompose the system into three sequence networks.

The networks are then coupled only at the point of the

unbalance (i.e., the fault)

•

The three sequence networks are known as the – positive sequence (this is the one we’ve been

using)

– negative sequence

– zero sequence


7



Positive Sequence Sets

• The positive sequence sets have three phase

currents/voltages with equal magnitude, with phase b

lagging phase a by 120°, and phase c lagging phase b by

120°.

• We’ve been studying positive sequence sets


8

Positive sequence

sets have zero

neutral current



Negative Sequence Sets

•

The negative sequence sets have three phasecurrents/voltages with equal magnitude, with phase b

leading phase a by 120°, and phase c leading phase b by

120°.

•

Negative sequence sets are similar to positive sequence,except the phase order is reversed


9

Negative sequencesets have zero

neutral current



Zero Sequence Sets

• Zero sequence sets have three values with equal magnitude

and angle.

• Zero sequence sets have neutral current


10



Symmetrical Components Combination


11

Symmetrical Components combined to create a set of unbalanced

three phase phasor quantities

Ia1 Ia2

Ia0



Sequence Set Representation

• Any arbitrary set of three phasors, say I a , I

b , I

c, can be

represented as a sum of the three sequence sets


12

0 1 2

0 1 2

0 1 2

0 0 0

1 1 1

2 2 2

where

, , is the zero sequence set, , is the positive sequence set

, , is the negative sequence set

a a a a

b b b b

c c c c

a b c

a b c

a b c

I I I I

I I I I

I I I I

I I I I I I

I I I

S i l C



Mathematical Operators


13

2

3

2

3

2

1 90

1 120 0.5 0.866

1 240 1 1201 0

1 0

j

j

e j

2

1 = 3

S t i l C t



Conversion from Sequence to Phase


14

0 1 1

0 0 02 2

1 1 1 1 2 2 1 2

Only three of the sequence values are unique,

, , ; the others are determined as follows:

(since by definition they are all equal)

a a a

a b c

b a c a b a c a

a

b

c

I I I

I I I

I I I I I I I I

I

I

I

0

2 2a0 a1 2 1

2 22

1 1 1 11 1

I 1 I 1

1 1

a

a a

a

I

I I

I

S t i l C t





15

2

2

0 0

1 1

2 2

Define the symmetrical components transformation

matrix

1 1 1

1

1

Thenaa

b a s

c a

I I I I I I

I I I

A

I A A A I

S t i l C t





16

0 0

1 1

2 2

0 1 2

20 1

(removing subscribe )

Sequence vector consisting of sequence currents

Thus

a

s a

a

a

b

I I

I I a

I I

I I I I

I I I

I

2

20 1 2c

I

I I I I




Conversion from Phase to Sequence


17

1

1 2

2

By taking the inverse we can convert from the

phase values to the sequence values

1 1 11

with

Sequence sets can be used with voltages as well

as with curr

1

en

1

ts

3

s

I A I

A




Conversion from Phase to Sequence


18

1 1 2

2

0

21

22

1 1 11

with 1 implies3

1

( ) / 3

( ) / 3

( ) / 3

s

a b c

a b c

a b c

I I I I

I I I I

I I I I

I A I A




Neutral Currents in 3ph Systems


19

0

0

0

In a 4 wire Y-connected system, 3 phases are connectedto the neutral and the neutral current is

3

since ( ) / 3

If the system is balanced, then 0 and

0

This is true for

n a b c

a b c

a b c

I I I I I

I I I I

I I I

I

any 3-ph system with no neutral path,

eg 3ph -connnected or 3 wire Y-connected systems.




Symmetrical Component Example 1


20

1 2s

2

10 0

Consider 10 Then

10

1 1 1 10 01

1 10 10 03

10 01

There is only positive sequence component for since

a

b

c

a b

I

I

I

I I

I

I A I

I

and are balanced positive sequencec I






21

s

10 0 0

If 10 then 0

10 10 0There is only negative sequence component for since

and are balanced negative sequencea b c I I I

I I

I






22

1 2s

2

0

Let unsymmetrical 3ph voltages

Then

1 1 1 0 01

13

6.121There are 3 sequence comp

a

b

c

V

V

V

V

V A V

onents for since

, and are unsymmetrical 3ph voltagesa b cV V V

V






23

0

1

2

2

2

10 0Let 10 with 3 sequence components

Then1 1 1 10 0

1 10

1

contains unbalanced 3ph curren

s

s

I I

I

I

I AI

ts






24





y p

25





y p

26




Outline




• Sequence networks of transmission and distribution lines

• Sequence networks of rotating machines



•Main reference: Ch 8 Textbook

y p

27




Sequence Networks of Impedance Loads

• A note about symbols

y p

28




g − ground V ng = I n Z n

N


• Consider the following Y-connected load:

29

( )

Similarly

( )

( )

n a b c

ag a y n n

ag Y n a n b n c

bg n a Y n b n c

cg n a n b Y n c

I I I I

V I Z I Z

V Z Z I Z I Z I

V Z I Z Z I Z I

V Z I Z I Z Z I





30

In matrix form

Write and in sequence component form

ag y n n n a

bg n y n n b

ccg n n y n

s

s

V Z Z Z Z I

V Z Z Z Z I

I V Z Z Z Z

IV Z

V Z I

V I

V A V

I A I





31

1

1

Then can be written as

3 0 0

0 0

0 0

sequence component impedance matrix

s s

s s s s

y n

s y

y

s

Z Z

Z

Z

V ZI

AV ZAI

V A ZAI Z I

Z A ZA

Z





32

1 2 2

2 2

2 2

2 2

2

1 1 1 1 1 111 1

31 1

31 1 111 3

31 3

3 0 00 0 , using 1+

0 0

y n n n

n y n n

n n y n

y n y y

y n y y

y n y y

y n

y

y

Z Z Z Z Z Z Z Z

Z Z Z Z

Z Z Z Z

Z Z Z Z

Z Z Z Z

Z Z Z

Z

A ZA

30, 1





33

0 0

1 1

2 2

0 0

1 1

2 2

3 0 00 0

0 0

Systems are decoupled( 3 )

y n

y

y

y n

y

y

Z Z V I V Z I

V I Z

V Z Z I

V Z I

V Z I

• With balanced impedances,

the three phase Y-connected

load can be represented by

three completely decoupled

zero, positive and negativesequence networks





34

I 1

I 2

I 0

Z y

Z y

Z y 3Z n

Z 1 = Z y Z 2 = Z y Z 0 = Z y + 3 Z n

• If neutral is not grounded

via Z n then the zero

sequence is open circuit





35

Balanced -connected impedance load and its Y-equivalent

Z /3 Z

/3

Z /3

Z Z

Z

•

Balanced

-connected impedance loads





36

Sequence networks for the Y-equivalent of a balanced load

• Since the -connected load

does not have a neutralconnection, the equivalent

Y-connected load has an

open circuited neutral

I 1

I 2

I 0

Z /3

Z /3

Z /3




Sequence Networks Example

37

Sequence networks: balanced Y and balanced loads

A balanced Y-load is in parallel with a balanced -connected

capacitor bank. The Y-load has an impedance Z y = (3 + j4)

per phase, and its neutral is grounded through an inductive

reactance X n= 2 . The capacitor bank has a reactance X c =

30 per phase. Draw the sequence networks for this loadand calculate the load sequence impedance.

Solution





38

Equivalent circuit and sequence networks

//

//

//





39





40




Outline









41




Sequence Networks of Series Impedance

Elements

42

Z a, Z b and Z c – self-impedances for each phase

Z ab, Z bc and Z ac – mutual-impedances between phases

n





Elements

43

The voltage drops across the series-phase impedances [betweenthe buses (a,b,c) and (a′,b′,c′ )] are given by

Transforming to symmetrical components gives

Or equivalently

'

'

'

an a n a ab ac a

bn b n ab b bc b

cn c n ac bc c c

V V Z Z Z I

V V Z Z Z I

V V Z Z Z I

0 0' 0

1 1' 1

2 2' 2

a ab ac

ab b bc

ac bc c

V V Z Z Z I

A V A V Z Z Z A I V V Z Z Z I





Elements

44

'

0 0' 01

1 1' 1

2 2 ' 2

'

1

2

2

1 1 11

13

Sequence Imped

1

ance:

s s s s

a ab ac

ab b bc

ac bc c

s s s s

a ab ac

s ab b bc

ac bc c

V V Z Z Z I V V A Z Z Z A I

V V Z Z Z I

Z Z Z

A Z Z Z A

Z Z Z

V V IZ

V V Z I

Z

2

2

1 1 1

1

1

a ab ac

ab b bc

ac bc c

Z Z Z

Z Z Z

Z Z Z





Elements

45

2 2

2 2

0

1

2

1 1 1 1 1 11

1 13

1 1

0 0 2 0 0

0 0 0 0

0 0 0 0

a ab ac

s ab b bc

ac bc c

a ab

a ab

a ab

Z Z Z

Z Z Z

Z Z Z

Z Z Z

Z Z Z

Z Z Z

Z

For symmetrical series impedances

,a b c ab ac bc Z Z Z Z Z Z

Sequence Impedance Matrix is thus given by





Elements

46

0 0' 0 0 0

1 1' 1 1 1

2 2 ' 2 2 2

( 2 )

( )

( )

a ab

a ab

a ab

V V Z I Z Z I

V V Z I Z Z I

V V Z I Z Z I

• The three sequence

circuits are decoupled




Outline









47




Ground Current Paths

48

• The zero sequence currents of the phases of overhead transmission lines

cause a field that links the earth and the associated earth currents

• A “remote Equipotential Ground” conductor is derived that has an

equivalent effect to the ground current

Eddy currents reduceground penetration

H field due to line current

IL




Symmetrical Component of Distribution Lines

49

• Distribution lines

usually do not have

earth wire/s

• The ground currents

only flow in theground

• Zero sequence

causes earth

currents




Symmetrical Component of Distribution Lines

50

• A current Ig occurs in the ground due to the zero

sequence currents.

• The magnetic flux associated with the ground current

interacts with the magnetic flux associated with the phase

currents.




Symmetrical Component of Transmission Lines

51

• Transmission lines

usually have overhead

earth wires.

• These earth wires

shield the line fromlightening strikes.

• They act as a parallel

path to the earth for

“Ground-currents”





52





53

• Ground-current flows in the ground and earth wires• The ground-current defines the impedance of the ground

path

Z g = R g + j X g

• Z g modifies the self and mutual impedances of the series

element such that

'

'

a a a g

ab ab ab g

Z Z Z Z

Z Z Z Z





54

Therefore the sequence impedance matrix

0

1

2

0 0 2 0 0

0 0 0 0

0 0 0 0

a ab

s a ab

a ab

Z Z Z

Z Z Z

Z Z Z

Z

becomes

' ' '0

' ' ' '

1

' ' '2

0 0 2 0 0

0 0 0 00 0 0 0

a ab

s a ab

a ab

Z Z Z

Z Z Z Z Z Z

Z





55

Practical implication' '

0

' '1

2 2 2a ab a ab g

a ab a ab g

Z Z Z Z Z Z

Z Z Z Z Z Z

Generally Z 0 > Z 1

Examples

66kV line – no earth wires 100km

Z 1= 0.85 + j1.1, Z 0 = 1.25 + j4.3

500kV line 2 earth wires 150km

Z 1= 0.001 + j0.016, Z 0 = 0.008 + j0.038




Outline








•

Main reference: Ch 8 Textbook

56




Sequence Networks of Generators

57

Y-connected synchronous

generatorSequence networks of Y-connected

synchronous generator




Sequence Networks of Generators

58

Positive Sequence

• V1 sequence rotates in the same direction as the

generator rotor

• Generates the rotating magnetic field

• Only sequence with a source Eg1

Negative Sequence• V2 sequence rotates in the opposite direction to the

generator rotor

• Induces negative sequence rotor current with twice

frequency (eg 250Hz = 100Hz) that reduces flux penetration of rotor, resulting in reduced reactance, ie

• Zg2 < Zg1




Sequence Networks of Synchronous Motors

59

• Currents flow into the

networks – absorbing power




Sequence Networks of Induction Motors

60

• Positive network has

no voltage source sinceinduction motor does

not have back emf




Sequence Networks of Rotating Machines Example 1

61

Example 8.5

A balanced, positive sequence, Y-connected generator withinternal voltage E ab = 4800° V is applied to a balanced -

load with Z =3040° . The line impedance between the

source and the load is Z L=185° for each phase and the

transmission line mutual coupling is ignored. Assume that thegenerator is grounded via an impedance Z n= j10 and that the

generator sequence impedances are Z g 0= j1 , Z g 1= j15 and

Z g 2= j3 .

• Draw the sequence networks

• Calculate the sequence component of the line current





62

0 1 2

480

480 0 303

1 , 1 85 , 30 40

1 , 15 , 3

ab a b a

n L

g g g

E E E E

Z j Z Z

Z j Z j Z j

Solution





63

0

21

22

1

2 0

Sequence component voltages of the generator

1 1 1

11

31

The source is positive-sequence and balanced with phase voltage

48030

3

0

g a

g b

cn g

a

g a

g g

E E

E E

E E

E

E E

E E





64

From





65

and

we get

Z ∆ Z ∆

Z ∆





66

I 1

I 2

I 0

Z /3

Z /3

Z /3





67

0 2

1

1

1 1

0

/ 3

11.68 100.94

g

g L

I I

E I

Z Z Z

1

48030

3 g E





68

Example 8.6

If a solidly grounded Y-connected voltage source with the

unbalanced terminal voltages given below is applied to the

balanced load in the previous example, calculate the phase

currents.

0ag

bg

cg

V

V

V





69

Solution The sequence components of the source voltages are

02

1

22

1 1 11

13

1

1 1 1 01

1 120 1203

1 120 120

62.11

9.218

ag

bg

cg

V V

V V

V V





70

0

11

1

22

1

0 02

1 1

2

2 2

0 an open circuit

/ 3

25.82 45.55

/ 30.86 172.82

1 1 1

1

146.76

L

L

a

b

c

I

V I

Z Z

V I

Z Z

I I I

I I I

I I I

A

2=

9.218∠− 143.41°

0=

15.2∠62.11°




Outline









71

S N t k f T f




Sequence Networks of Transformers

72

Single-line

diagram

Symbols






• Transformers are a series element

• Positive and negative sequences are the phase impedancesrepresenting flux leakage effects and the resistance of the

windings

73

• The excitation branch

of the transformermodel is neglected

when considering

sequence models for

the purposes of faultcalculations

Zero Sequence Networks of Transformers





•Zero sequence equivalent circuit is determined by thewinding arrangement, ie Y or and earthing of Y

connected windings

•Zero sequence currents, I a0 , I b0 and I c0 are in phaseThus, a delta winding acts as a short circuit for zero

sequence currents

74






•There can only be a current flowing in a particular phaseof a primary winding, if it is possible for the equivalent

current to flow in the corresponding winding of the

transformer secondary

•

Zero sequences currents cannot flow in the secondary because no neutral and hence no primary current

• Zero sequence open circuit on secondary

75






• Zero sequences currents can flow in the secondary becauseof earthed neutral

• Induces primary current which flows because of earthed

neutral

• Zero sequence series circuit primary to secondary

(Must include neutral grounding Z if neutral grounded

through impedance)

76






• Zero sequences currents can flow through primary and

within the secondary because of delta

• But no flows in secondary connection lines because current

confined to delta, ie open circuit on secondary side

• If neutral grounded through impedance Z n, then a 3 Z n is in

series connection with the Z 0 of the transformer

77






• Zero sequences currents cannot flow in the primary since

unearthed neutral

No primary current and no secondary current

Open circuit on secondary side

78






79

Dual to Y-∆ cases : circuits flipped






• Zero sequences currents flow in the secondary because ofdelta

• Induces currents in primary delta

• But open circuit on primary and secondary connection

lines because current in connection lines is confined todelta

80

Transformer Sequence DiagramsSymmetrical Components



Transformer Sequence Diagrams

81

Transformer Phase Shifts





•

Previous models ignore the phase shift in Y-

and

-Y• Include phase shift in Y- and -Y based on the

convention:

o The positive sequence voltages and currents for the high

voltage side of the Y-

transformer lead thecorresponding quantities on the low voltage side of the

transformer by 30°

o Hence the negative sequence voltages and currents for

the high voltage side of the Y- transformer must lag

the corresponding quantities on the low voltage side by

30°

82






• Note that although in practice the zero sequence

impedance of a transformer may differ slightly to the

positive and negative sequence impedances, often the same

numeric value is used for all three impedances

83


Sequence Networks of Transformers Example



Example 8.7

84


02

1

22

1 1 1 62.1111

39.2181

ag

bg

cg

V V V V

V V

From Example 8.6 the sequence voltages are





85


Fig on p70.





86






87






88






89


O tli




Outline

• Concept and definition of symmetrical components• Sequence networks of impedance loads






•

Main reference: Ch 8 Textbook

90


Power in Sequence Networks



91


Total power delivered to the 3-phase load





92


*

*

* * * * *

*

* * *

Total power is given by

Using and gives

( ) ( )

T p

p

p

a

T

p ag a bg b cg c ag bg cg b p p

V c

I

p s s

T T

p s s s s

S

I

S V I V I V I V V V I V I

I

V V I I

S V I V I

A A

A A AA





93


*

* 2 2

2 2

2 2

2 2

*

0

* * * * *

0 1 2 1 0 0 1 1 2 2

*

2

1 1 1 1 1 1

1 1

1 1

1 1 1 1 1 1 1 0 0

1 1 3 0 1 0

1 1 0 0 1

3 3 3( ) 3T

p s s s

I

S V I V V V I V I V I V I S

I

AA


Power in Sequence Networks Example



94


Example 8.9






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Components of symmetric fault analysis