CHAPTER 4

SEQUENCE NETWORKS AND

UNSYMMETRICAL FAULTS ANALYSIS

SEQUENCE NETWORKS AND UNSYMMETRICAL FAULTS ANALYSIS

1 SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS

2 UNSYMMETRICAL FAULTS AT THE GENERATOR TERMINALS

3 UNSYMMETRICAL FAULTS ON POWER SYSTEMS

4 CONSTRUCTION OF BUS IMPEDANCE MATRICES OF

SEQUENCE NETWORK

5 UNSYMMETRICAL FAULTS ANALYSIS

2

1 SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS

When a symmetrical three phase fault occurs in a three phase system, the power

system remains in the balanced condition. Hence single phase representation can be

used to solve symmetrical three phase fault analysis. But various types of

unsymmetrical faults can occur on power systems. In such cases, unbalanced

currents flow in the system and this in turn makes the bus voltages unbalanced.

Now the power system is in unbalanced condition and single phase representation

can not be used.

Three phase unbalanced currents and voltages can be conveniently handled by

Symmetrical Components. Therefore unsymmetrical faults are analyzed using

symmetrical components. Some of the important aspects of symmetrical

components are presented in brief.

3

Sequence voltages and currents

According to symmetrical components method, a three phase unbalanced system of

voltages or currents may be represented by three separate system of balanced

voltages or currents known as zero sequence, positive sequence and negative

sequence as shown in Fig. 1

I a

I a(2)

(0) (1) I (1)

Ia I c a

I (0)

I b(2)

= b + + I c (0)

Ic

I (1) I c(2)

b I b

Fig. 1

4

Defining operator ‘ a ‘ as

a = 1 1200

(1)

it is to be noted that

a2 1240

0 ; a

3 1360

0 1 (2)

Also a = - 0.5 + j 0.866 ; a2 0.5 j0.866 (3)

Hence 1 a a2

0 (4)

5

I a (0) (1) I (1) I a(2)

I c a

I

a

I (0)

I b(2)

I c =

(0) b + +

Ic

I (1) I c(2)

b I b

Further referring Fig. 1

I(1) a2 I (1)

b a

I(1) a I(1) c a

(5) I(2) a I(2)

b a

I(2) a2 I (2)

c a

Therefore I

a I(0) I(1) I(2)

a a a

I b

I(0) I(1) I(2) I(0) a2 I(1) a I(2)

b b b a a a

I c

I(0) I(1) I(2) I (0) a I(1) a2 I(2)

c c c a a a

Thus

I a I(0) I(1) I(2)

a a a

I b

I(0) a2

I(1) a I(2) a a a

I c

I(0) a I (1) a2 I(2) a a a

(6)

6

I a 1 1 1 I a(0 )

i.e.

=

a 2

a (1 )

I b

1

I a

I 1 a a 2 I (2 )

c a

The inverse form of the above is

I a(0 )

1

1 1 1 I a (1 ) 2 I

a =

1

a a

I

b 3

I (2 ) 1 a 2

a I

a c

i.e.

Ia , b , c A I

0 , 1, 2

i.e. I0 , 1, 2 A1

Ia , b , c

(7)

(8)

Similarly, corresponding to voltage phasors

Va , b , c

A V0 , 1, 2

and V0 , 1, 2 A1

Va , b , c

(9)

(10)

Matrix A is known as symmetrical component transformation matrix. Similar expressions can be written for line to line voltages and phase currents also.

7

Sequence impedances and sequence networks

The impedance of any three phase element is of the form

z aa

z ab

z ac

z a,b,c

= z bb

z

ba z

bc (11) z

ca

z cb

z cc

Then

va,b,c

za,b,c

ia,b,c

i.e. A v

0,1,2

za,b,c

A i0,1,2

v0,1,2 A1

za,b,c A i0,1,2 v

0.1,2

z0,1,2

i0,1,2

where z 0,1,2 A1

za,b,c A

Thus for any three phase element having the impedance z a,b,c the corresponding

sequence impedance z 0,1,2 can be obtained from

z 0,1,2 = A1

za,b,c A (12)

8

For power system components, sequence impedance z 0,1,2 will be decoupled as

z (0) 0 0 z

0,1,2

= 0 z (1) 0

(13)

0 0 z (2)

For static loads and transformers z(0) z(1) z

(2) .

For transmission lines z(1) z(2) and z (0) > z (1) .

For rotating machines z(0) , z(1) and z (2) will have different values.

The single phase equivalent circuit composed of sequence voltages, sequence

currents and impedance to current of any one sequence is called the sequence

network for that particular sequence. The sequence network includes any

generated emf of like sequence.

Consider a star connected generator with its neutral grounded through an

impedance Zn as shown in Fig. 2. Assume that the generator is designed to generate

balanced voltage.

9

a

I a

Zn +

I n

Ec n

Ea n

+

+

Eb n I b

c

b

I c

Fig. 2

Let Ean

Ea Eb

= E

c

be its generated voltage in phase a . Then Ec

1 Ea

2 Ean This gives

Eb a

a

10

E(0) 1 1 1 1 0

(1)a

=

1 2 2

Ean =

E

a

1 a a

a

E

a n 3

E (2) 1 a2 a a 0

a

(14)

This shows that there is no zero sequence and negative sequence generated voltages.

The sequence networks of the generator are shown in Fig. 3.

I ( 1 )

a a

Ia( 1 )

Zn

Z1

+

E

Z1 ( 1 )

In 0

a n

Ec n

Va

+

+

Z1

Z1

Eb n

( 1 )

E a n

+

Ib

__

b

c

Ic( 1 )

Reference bus ( Neutral )

Note that In = 0 Positive sequence network

11

a

Ia( 2 )

Zn Z 2

In 0

Z 2

c Z 2

Ib( 2 )

b

Ic( 2 )

Note that In = 0

a Ia( 0 )

Zn

Zg 0

In 3 Ia

(0 )

Zg 0

c

Zg 0

Ib( 0 )

b

Ic( 0 )

Note that In = 3 Ia (0)

Fig. 3

Ia( 2 )

Z 2 Va( 2 )

Reference bus ( Neutral )

Negative sequence network

Ia( 0 )

Zg 0

V ( 0 )

n Z0 a

3 Zn

Reference bus ( Ground )

Zero sequence network 12

Z1 and Z 2 are the positive sequence and negative sequence impedance of the

generator. Zg 0 is the zero sequence impedance of the generator. Total zero sequence

impedance Z0 = Zg 0 + 3 Zn .

Sequence components of the terminal voltage are

V ( 0 )

a

V ( 1 )

a V

( 2 )

a

Z0 I(a0 )

Ea n Z1 Ia(1 )

(15)

Z 2

I( 2 ) a

As far as zero sequence currents are concerned, the three phase system behaves as

a single phase system. This is because of the fact that at any point the zero

sequence currents are same in magnitude and phase. Therefore, zero sequence

currents will flow only if a return path exists.

13

The connection diagram and the zero sequence equivalent circuit for star

connected load is shown in Fig. 4.

Z Z Z

3 Zn Z

Zn

Reference

Fig. 4

The connection diagram and the zero sequence circuit for delta connected load

is shown in Fig. 5.

Z

Z Z

Z

14 Reference

Fig. 5

Special attention is required while obtaining the zero sequence network

of three phase transformers. The zero sequence network will be

different for various combination of connecting the windings and also

by the manner in which the neutral is connected.

The zero sequence networks are drawn remembering that no current

flows in the primary of a transformer unless current flows in the

secondary

( neglecting the small magnetizing current ).

Five different cases are considered and the corresponding zero

sequence network are shown in Fig. 6. The arrows in the connection

diagram show the possible path for the flow of zero sequence current.

Absence of arrow indicates that the zero sequence current can not flow

there. Impedance Z0 accounts for the leakage impedance Z and the

neutral impedances 3 ZN and 3 Zn where applicable.

15

Connection diagrams Zero sequence equivalent circuit

P

Q

Z0

P

Z N

Q

Z n

Reference

P Q

Z0

P Q

Z N

Reference

16

P

Q

Z 0

P Q

Reference

P Q

Z0

P Q

Z N

Reference

17

P Q

Z0

P Q

Reference

Fig. 6

Example 1

For the power system shown in Fig. 7, with the data given, draw the zero sequence,

positive sequence and negative sequence networks.

T1 T2

M1

G

M 2

Fig. 7

18

Per unit reactances are:

Generator

Transformer T1

Transformer T2

Transmission line

Motor 1

Motor 2

Xg0 0.05; Xn 0.32 ; X1 0.2; X2 0.25 X0

X1 X2 0.08

X0 X1 X2 0.09

X0 0.52; X1 X2 0.18

Xm o 0.06; Xn 0.22; X1 X2 0.27

Xm o 0.12; X1 X2 0.55

19

T1 T2

M1

G

M 2

Positive sequence network

j0.08 j0.18 j0.09

j0.2

j0.55

j0.27

+

Em 1

+ + Em 2

Eg

Reference

20

T1

T2

M1

G

M 2

Negative sequence network

j0.08 j0.18 j0.09

j0.25

j0.27 j0.55

Reference

21

T1

T2

M1

G

M 2

Zero sequence network

j0.08 j0.52 j0.09

j0.05 j0.06 j0.12

j0.96 j0.66

Reference

22

2 UNSYMMETRICAL FAULTS AT GENERATOR TERMINALS

Single line to ground fault ( LG fault ), Line to line fault ( LL fault ) and Double line to

ground ( LLG fault ) are unsymmetrical faults that may occur at any point in a power

system. To understand the unsymmetrical fault analysis, let us first consider these faults

at the terminals of unloaded generator. This treatment can be extended to

unsymmetrical fault analysis when the fault occurs at any point in a power system.

Consider a three phase unloaded generator generating balanced three phase voltage.

The sequence components of the terminal voltages are

V(1) E a n

I (1) Z 1

(16) a a

V(2) I(2)

Z 2

(17) a a

V(0) I(0)

Z 0

(18) a a

23

V(1) E a n

I (1) Z 1

(16) a a

V(2) I(2)

Z 2

(17) a a

V(0) I(0) Z 0

(18) a a

The above three equations apply regardless of the type of fault occurring at the terminals of the generator.

For each type of fault there will be three relations in terms of phase components of

currents and voltages. Using these, three relations in terms of sequence components of

currents and voltages can be obtained. These three relations and the eqns. (16), (17) and

(18) are used to solve for the sequence currents

I(0)a , I(1)a , I(2)a

and

sequence

voltages

Va(0) , Va(1) , Va(2) .

Sequence

component

s

relationship will enable to interconnect the sequence networks to represent the particular fault.

24

Single line to ground fault ( LG fault )

The circuit diagram is shown in Fig. 9.

I a

a

Zf

Zn

+

_E

a n

Ec n +

+ E

b n I b c

b

Fig. 9 I c

25

The fault conditions are

Ib 0

Ic 0

Va Zf Ia

Ia(0) 1/3 ( Ia Ib Ic ) Ia /3

Ia(1) 1/3 (Ia a Ib a

2 Ic ) Ia /3

Ia(2)

1/3 ( Ia a2 Ib a Ic ) Ia /3

Thus Ia(0)

Ia(1)

Ia(2)

Further from eqn. (21)

Va(0)

Va(1)

Va(2)

Zf ( I(0)

a I(1)

a I(2)

a ) 3 Zf I(1)

a

Using eqns. (16) to (18) in the above

Ia(0)

Z0 Ea n Ia(1)

Z1 Ia(2)

Z2 3 Zf Ia(1)

i.e.

Ia(1)

Z0 Ea n Ia(1)

Z1 Ia(1)

Z2 3 Zf Ia(1)

i.e.

Ia(1)

Ea n

Z1 Z2 Z0 3 Zf

(19)

(20)

(21)

(22)

(23)

(24) 26

E

I(1) a n (24)

a

Z1 Z2 Z0 3 Zf

Then the sequence networks are to be connected as shown in Fig. 10.

Z1

+ I a(1)

Ea n

_

Z 2

I (2)

a

Z0

I (0)

a

+

V (1) a

_

+

3 Zf (2)

Va

_

+

V (0) a

_

Fig. 10 27

Line to line fault

The circuit diagram is shown in Fig. 11

I a

a

+

Zn

Ea n

_

E

b n +

I b

c

E

c n

+

b

I c

Zf

Fig. 11

The fault conditions are I

a 0 (25)

Ib Ic 0 (26) Vb Zf Ib Vc (27)

28

Ia 0

Ib Ic 0 V

b Zf Ib Vc

Then Ia(0) 1/3 ( Ia Ib Ic ) 0

Ia(1)

1/3 (Ia a Ib a2

Ic ) Ib /3 ( a a2 )

I(2) 1/3 ( I a

a2

I b

a I c

) I b

/3 ( a2 a )

a

Since I (0)

= 0 , V (0)

= - Z0 I (0)

= 0 a a a

Further Ia(2)

Ia(1)

From eqn. (27)

V(0) a2 V(1) a V(2)

Z f

( a2

I(1) a I(2)

) V(0)

a V(1) a

2 V(2)

a a a a a a a a

( a2 a ) V

(1) Z

f

( a2 a ) I

(1) ( a

2 a) V

(2)

a a a

Thus V(1)

Z f

I (1) V(2)

a a a

From the above eqn. Ea n Z1 Ia(1)

Zf Ia(1)

Z2 Ia(2)

Z2 Ia(1)

i.e.

Ea n ( Z1 Z2 Zf ) Ia

(1)

Therefore

Ia(1)

Ea n

Z1

Z2 Zf

(25)

(26)

(27)

(28)

(29)

(30)

(31)

(32) 29

Therefore

Ia(1)

Ea n

(32) Z1 Z 2 Zf

I (2) = - I (1) a a

and I(0)

= 0; V(0)

= 0 a a

Sequence networks are to be connected as shown in Fig. 12.

Z

f

+

Z1

Z 2

Z

0

I a(1)

I a(2)

I a(0)

E V (1)

Va(2) Va (0) = 0

a n_ a

Fig. 12

30

Double line to ground fault

The circuit diagram is shown in Fig. 13.

a

I a

+

Zn E

a n

_

Eb n

+

I b

E

c n

c +

b I c

Zf

Fig. 13

The fault conditions are

Ia 0 ; Vb Zf ( Ib Ic ) and Vc Zf ( Ib Ic ) (33)

31

The fault conditions are

Ia 0 ; Vb Zf ( Ib Ic ) and Vc Zf ( Ib Ic )

Because of I(0)

1/3 ( I a

I b

I c

) , I b

I c

3 I(0)

a a

Therefore

V 3 Z f

I(0) b a

V 3 Z f

I(0) c a

V(1) 1/3 ( V a V a2

V ) 1/3 [ V ( a a2 ) V ]

a a b c a b

V(2) 1/3 ( V a2 V a V ) 1/3 [ V ( a

2 a ) V ]

a a b c a b

Therefore V(1) V(2) a a

(33)

(34)

(35)

(36)

32

Further V(0) 1/3 ( V V V ) i.e. a a b c

3 V(0) V(0) V(1) V(2) 3 Z f

I (0) 3 Z f

I(0) i.e. 2 V(0) 2 V(1) 6 Z

f I(0)

a a a a a a a a a

i.e. V(1)

V(0) 3 Z f

I(0)

Z 0 I

(0) 3 Z

f I(0) ( Z

0 3 Z

f ) I(0)

a a a a a a

i.e. V(1)

( Z 0 3 Z

f ) I(0) (37)

a a

From eqn. (33) I(0) I(1) I(2) 0 i.e. a a a

V(1) V(2) V(1) V(1)

a I(1) a 0 i.e. a I(1) a 0

Z0 3Zf

a

Z2

a

Z2

Z0 3Zf

Therefore I(1) V(1) ( 1 1 ) V(1) Z2 Z0 3Zf

a a Z0

3Zf

Z2

a Z2 ( Z0 3Zf )

i.e. V(1)

Z2 ( Z0 3 Zf ) I(1)

a Z2 Z0

3 Zf

a

i.e. E I(1) Z Z2 ( Z0 3 Zf ) I (1)

a n 1

a

a

Z2 Z

3 Zf

0

Thus Ia(1)

Ea n

Z

Z2 ( Z0 3 Zf )

1

Z0 3 Zf

Z2

(38)

(39)

33

Thus Ia(1)

Ea n

(39)

Z1 Z2 ( Z0 3 Zf )

Z2 Z0 3 Zf

From eqn. (38) V(1) V(2) Z I(2) Z2 ( Z0 3 Zf ) I(1)

2

a a a Z2 Z0 3 Zf

a

Therefore I(2) I(1)

Z0 3 Zf (40)

a a Z2 Z0 3 Zf

Again substituting eqn. (37) in eqn. (38)

( Z0 3 Zf ) Ia(0)

Z2 ( Z0 3 Zf ) Ia

(1) Thus Ia

(0) Ia

(1) Z2 (41)

Z2 Z0 3 Zf Z2 Z0 3 Zf

For this fault, the sequence networks are to be connected as shown in Fig. 14.

Z1

Z 2 Z0

+

I a

(1) I (2)

I a

(0)

Ea n

Va(1) a

Va(2)

Va(0)

_

3 Zf

34 Fig. 14

General procedure for unsymmetrical fault analysis

when fault occurs at a point in a power system

35

PRELIMINARY CALCULATIONS

1. Draw the positive sequence, negative sequence and zero sequence

networks.

2. Using bus impedance building algorithm, construct ZBus (1)

, ZBus(2)

and

ZBus(0)

.

DATA REQUIRED

Type of fault, fault location (Bus p) and fault impedance (Zf)

TO COMPUTE FAULT CURRENTS I f a , I f b and I f c

1. Extract the columns of ZBus (1)

, ZBus(2)

and ZBus(0)

corresponding to the

faulted bus.

2. Depending on the type of fault interconnect the sequence networks.

3. Calculate I f a(1)

, I f a(2)

and I f a(0)

4. Compute the corresponding phase components I f a. I f b and I f c using I f a

1

1

=

a 2

If b

1

I 1 a

f c

1 I f a

(0)

a (1)

If a

a

2 (2)

I

f a 36

TO COMPUTE FAULTED BUS VOLTAGES V p a, V p b and V p c

1. Compute the sequence components V p a

(1) (2) (0)

from , V p a and V p a (!) (1) (1)

V p a = Vf – Z p p I f a (2) (2) (2)

V p a = – Z p p I f a (0) (0) (0)

V p a = – Z p p I f a

2. Calculate the corresponding phase components V p a, V p b and V p c from

Vp a 1 1

=

a 2

V

p b 1

a V

p c

1

1 V p a

(0)

a (1)

Vp a

a

2 (2)

V

p a

37

TO COMPUTE BUS VOLTAGES AT BUS j i.e V j a, V j b and V j c 1.

Compute the sequence components V j a(!)

, V j a(2)

and V j a(0)

from

(!) (1) (1)

V j a = Vf – Z j p I f a

V j a(2)

= – Z j p (2) I f a (2)

V j a(0)

= – Z j p (0) I f a (0)

2. Calculate the corresponding phase components V j a, V j b and V j c from

V 1 1 1 V (0) ja

ja

= 2

a (1)

V

jb 1 a V

ja

V 1 a a2 V

ja

(2)

jc

38

Single line to ground fault

I (f

1)a

+ Zp(1)

p Vp

(1)a

V

f _

I f(1)

a = I f(2)

a = If(0)

a

(2)

I f a

3 Zf

Zp(2)

p

V (2)

p a

I (f

0)a

Z(0)

p p Vp(0)a

39

Line to line fault

Zf

Zp(1)

p

Zp(2)

p

+ I f(1)

a I f(2)

a I f(0)

a

Vf _ Vp(1)

a Vp(2)

a

40

Double line to ground fault

Zp(1)

p Zp(2)

p Zp(0)

p

+

I f(1)

a

I f(2)

a

I f(0)

a

Vp(0)

a Vf

Vp(1)

a

Vp(2)

a

_

3 Zf

41