EE 0308 POWER SYSTEM ANALYSIS - SRM University Special attention is required while obtaining the...

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EE 0308 POWER SYSTEM ANALYSIS

CHAPTER 4

SEQUENCE NETWORKS AND UNSYMMETRICAL FAULTS ANALYSIS

2

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

3

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.

4

bI

aI

cI

(1)

bI

(1)

aI

(2)

cI

(2)

bI (0)

cI

(0)

bI

(0)

aI

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

(1)

cI

= + +

Fig. 1

(2)

aI

5

Defining operator ‘ a ‘ as

a = 01201 (1)

it is to be noted that

13601a;2401a0302 (2)

Also a = - 0.5 + j 0.866 ; 0.866j0.5a2 (3)

Hence 0aa12 (4)

6

bI

cI

(1)

bI

(1)

aI

(2)

cI

(2)

bI (0)

cI

(0)

bI

(0)

aI

aI

(1)

cI

= + +

Further referring Fig. 1

(1)

a

2(1)

b IaI

(1)

a

(1)

c IaI

(2)

a

(2)

b IaI

(2)

a

2(2)

c IaI

Therefore (2)

a

(1)

a

(0)

aa IIII

(2)

a

(1)

a

2(0)

a

(2)

b

(1)

b

(0)

bb IaIaIIIII

(2)

a

2(1)

a

(0)

a

(2)

c

(1)

c

(0)

cc IaIaIIIII

Thus (2)

a

(1)

a

(0)

aa IIII (2)

a

(1)

a

2(0)

ab IaIaII (6) (2)

a

2(1)

a

(0)

ac IaIaII

(5)

(2)

aI

7

i.e.

c

b

a

I

I

I

=

2

2

aa1

aa1

111

(2)

a

(1)

a

(0)

a

I

I

I

i.e. 2,1,0c,b,a IAI (7)

The inverse form of the above is

(2)

a

(1)

a

(0)

a

I

I

I

= 3

1

aa1

aa1

111

2

2

c

b

a

I

I

I

i.e. c,b,a

1

2,1,0 IAI (8)

Similarly, corresponding to voltage phasors

2,1,0c,b,a VAV (9)

and c,b,a

1

2,1,0 VAV (10)

Matrix A is known as symmetrical component transformation matrix. Similar

expressions can be written for line to line voltages and phase currents also.

8

Sequence impedances and sequence networks

The impedance of any three phase element is of the form

cb,a,z =

cccbca

bcbbba

acabaa

zzz

zzz

zzz

(11)

Then cb,a,cb,a,cb,a, izv

i.e. 0,1,2cb,a,0,1,2 iAzvA

0,1,2cb,a,

1

0,1,2 iAzAv

0,1,20,1,20.1,2 izv

where z0,1,2 AzA cb,a,

1

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

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

0,1,2z = AzA cb,a,

1 (12)

9

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

0,1,2z =

(2)

(1)

(0)

z00

0z0

00z

(13)

For static loads and transformers (2)(1)(0)zzz .

For transmission lines (2)(1)zz and (0)

z > (1)z .

For rotating machines (1)(0)z,z and (2)

z 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 nZ as shown in Fig. 2. Assume that the generator is designed to

generate balanced voltage.

10

+

+

+

b

nZ

c

a

ncE

nbE

cI

bI

aI

nI

bE

aE

cE

Fig. 2

Let anE be its generated voltage in phase a . Then

c

b

a

E

E

E

=

a

a

12

anE This gives

naE

11

1Z 1Z

1Z

)1(

aI

1Z

Reference bus ( Neutral )

+

+

+

b

nZ

c

a

ncE

nbE

) 1 (

cI

0In

naE

)1(

aI

)1(

bI

(2)

a

(1)

a

(0)

a

E

E

E

= 3

1

aa1

aa1

111

2

2

a

a

12 anE =

0

E

0

an (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.

__

+

)1(

aV

Positive sequence network

naE

Note that In = 0

12

n

nZ3

0gZ nZ

)0(

an I3I

0gZ

0gZ

0gZ

2Z

2Z

2Z

)2(

aI

2Z

Reference bus ( Neutral )

b

nZ

c

a

0In

)2(

aI

)2(

bI

)2(

cI

)0(

aI

0Z

Reference bus ( Ground )

b c

a )0(

aI

)0(

bI

)0(

cI

Fig. 3

)2(

aV

Negative sequence network

)0(

aV

Zero sequence network

Note that In = 0

Note that In = 3 Ia(0)

13

1Z and 2Z are the positive sequence and negative sequence impedance of the

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

impedance 0Z = 0gZ + nZ3 .

Sequence components of the terminal voltage are

)2(

a2

)2(

a

)1(

a1na

)1(

a

)0(

a0

)0(

a

IZV

IZEV

IZV

(15)

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.

14

nZ

Z

The connection diagram and the zero sequence equivalent circuit for star

connected load is shown in Fig. 4.

Z Z

Z

Fig. 4

The connection diagram and the zero sequence circuit for delta connected

load is shown in Fig. 5.

Z Z

Fig. 5

Reference

Z

nZ3

Reference

Z

15

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 0Z accounts for

the leakage impedance Z and the neutral impedances NZ3 and

nZ3 where applicable.

16

NZ nZ

NZ

Connection diagrams Zero sequence equivalent circuit

P Q

P Q

P Q

P Q

Reference

0Z

Reference

0Z

17

P Q

P Q

P Q

P Q

NZ

Reference

0Z

Reference

0Z

18

P Q

P Q

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.

1T 2T

Fig. 7

Reference

G

1M

2M

0Z

19

Per unit reactances are:

Generator 0.32X;0.05X ng0 ; 0.25X;0.2X 21

Transformer 1T 0.08XXX 210

Transformer 2T 0.09XXX 210

Transmission line 0.18XX0.52;X 210

Motor 1 0.27XX0.22;X0.06;X 21nom

Motor 2 0.55XX0.12;X 21om

20

+ + +

gE

1T 2T

G

1M

2M

2mE 1mE

j0.18 j0.09 j0.08

j0.55 j0.27 j0.2

Reference

Positive sequence network

21

j0.09 j0.18 j0.08

j0.55 j0.27

j0.25

1T 2T

G

1M

2M

Reference

Negative sequence network

22

j0.52 j0.09 j0.08

j0.12 j0.06

j0.66

j0.05

j0.96

1T 2T

G

1M

2M

Reference

Zero sequence network

23

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

1

(1)

ana

(1)

a ZIEV (16)

2

(2)

a

(2)

a ZIV (17)

0

(0)

a

(0)

a ZIV (18)

24

1

(1)

ana

(1)

a ZIEV (16)

2

(2)

a

(2)

a ZIV (17)

0

(0)

a

(0)

a ZIV (18)

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 (2)

a

(1)

a

(0)

a I,I,I and sequence voltages (2)

a

(1)

a

(0)

a V,V,V . Sequence components

relationship will enable to interconnect the sequence networks to represent the

particular fault.

25

+ nbE

aI

b c

a

cI

bI

Single line to ground fault ( LG fault )

The circuit diagram is shown in Fig. 9.

nZ

_

Fig. 9

+

+

ncE naE

fZ

26

The fault conditions are

0Ib (19)

0Ic (20)

afa IZV (21)

/3I)III(1/3I acba

(0)

a

/3I)IaIa(I1/3I ac

2

ba

(1)

a

/3I)IaIaI(1/3I acb

2

a

(2)

a

Thus (2)

a

(1)

a

(0)

a III (22)

Further from eqn. (21)

(1)

af

(2)

a

(1)

a

(0)

af

(2)

a

(1)

a

(0)

a IZ3)III(ZVVV (23)

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

(1)

af2

(2)

a1

(1)

ana0

(0)

a IZ3ZIZIEZI i.e.

(1)

af2

(1)

a1

(1)

ana0

(1)

a IZ3ZIZIEZI i.e.

f021

na(1)

aZ3ZZZ

EI

(24)

27

1Z

(2)

aI

(1)

aI

0Z

2Z

naE

(0)

aI

f021

na(1)

aZ3ZZZ

EI

(24)

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

+

_

Fig. 10

+ (0)

aV

_

+ (2)

aV

_

+ (1)

aV

_

fZ3

28

aI

nZ

c

a

nbE

b

bI

cI

Line to line fault

The circuit diagram is shown in Fig. 11

_

Fig. 11


0Ia (25)

0II cb (26)

cbfb VIZV (27)

+ ncE

naE

fZ

+

+

29

0Ia (25)

0II cb (26)

cbfb VIZV (27)

Then 0)III(1/3I cba

(0)

a (28)

)aa(/3I)IaIa(I1/3I2

bc

2

ba

(1)

a

)aa(/3I)IaIaI(1/3I2

bcb

2

a

(2)

a

Since (0)

aI = 0 , (0)

aV = - Z0 (0)

aI = 0 (29)

Further (1)

a

(2)

a II (30)

From eqn. (27)

(2)

a

2(1)

a

(0)

a

(2)

a

(1)

a

2

f

(2)

a

(1)

a

2(0)

a VaVaV)IaIa(ZVaVaV

(2)

a

2(1)

a

2

f

(1)

a

2Va)a(I)aa(ZV)aa(

Thus (2)

a

(1)

af

(1)

a VIZV (31)

From the above eqn. (1)

a2

(2)

a2

(1)

af

(1)

a1na IZIZIZIZE

i.e. (1)

af21na I)ZZZ(E

Therefore f21

na(1)

aZZZ

EI

(32)

30

(1)

aI (2)

aI

1Z 2Z

naE

Therefore f21

na(1)

aZZZ

EI

(32)

(2)

aI = - (1)

aI

and (0)

aI = 0; (0)

aV = 0

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

+

_

Fig. 12

Va(0)

= 0 (2)

aV (1)

aV

fZ

0Z

(0)

aI

31

aI

nZ

c

a

nbE

b cI

bI

Double line to ground fault

The circuit diagram is shown in Fig. 13.

_

Fig. 13


0Ia ; )II(ZV cbfb and )II(ZV cbfc (33)

+

+

+

ncE

naE

fZ

32


0Ia ; )II(ZV cbfb and )II(ZV cbfc (33)

Because of )III(1/3I cba

(0)

a , (0)

acb I3II

Therefore (0)

afb IZ3V (34)

(0)

afc IZ3V (35)

]V)aa(V[1/3)VaVaV(1/3V b

2

ac

2

ba

(1)

a

]V)aa(V[1/3)VaVaV(1/3V b

2

acb

2

a

(2)

a

Therefore (2)

a

(1)

a VV (36)

33

Further )VVV(1/3V cba

(0)

a i.e.

(0)

af

(0)

af

(2)

a

(1)

a

(0)

a

(0)

a IZ3IZ3VVVV3 i.e. (0)

af

(1)

a

(0)

a IZ6V2V2

i.e. (0)

af

(0)

a

(1)

a IZ3VV (0)

af

(0)

a0 IZ3IZ (0)

af0 I)Z3Z(

i.e. (1)

aV (0)

af0 I)Z3Z( (37)

From eqn. (33) 0III(2)

a

(1)

a

(0)

a i.e.

0Z

VI

3ZZ

Vi.e.0

Z

VI

3ZZ

V

2

(1)

a(1)

a

f0

(1)

a

2

(2)

a(1)

a

f0

(1)

a

Therefore )3ZZ(Z

3ZZZV)

Z

1

3ZZ

1(VI

f02

f02(1)

a

2f0

(1)

a

(1)

a

i.e. (1)

a

f02

f02(1)

a IZ3ZZ

)Z3Z(ZV

(38)

i.e. 1

(1)

ana ZIE (1)

a

f02

f02 IZ3ZZ

)Z3Z(Z

Thus

f02

f02

1

na(1)

a

Z3ZZ

)Z3Z(ZZ

EI

(39)

34

0Z

(0)

aI (2)

aI (1)

aI

1Z 2Z

naE

Thus

f02

f02

1

na(1)

a

Z3ZZ

)Z3Z(ZZ

EI

(39)

From eqn. (38) (2)

a2

(2)

a

(1)

a IZVV (1)

a

f02

f02 IZ3ZZ

)Z3Z(Z

Therefore (1)

a

(2)

a II f02

f0

Z3ZZ

Z3Z

(40)

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

(0)

af0 I)Z3Z( (1)

a

f02

f02 IZ3ZZ

)Z3Z(Z

Thus

f02

2(1)

a

(0)

aZ3ZZ

ZII

(41)

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

_

Fig. 14

+

fZ3

(0)

aV (1)

aV (2)

aV

35

1Z

(2)

aI

(1)

aI

0Z

2Z

naE

(0)

aI

3 SUMMARY OF UNSYMMETRICAL FAULTS AT THE GENERATOR

TERMINALS

For any unsymmetrical fault

(1)

aV = aE - 1Z (1)

aI aI = (0)

aI + (1)

aI + (2)

aI

(2)

aV = - 2Z (2)

aI bI = (0)

aI + 2a

(1)

aI + a (2)

aI

(0)

aV = - 0Z (0)

aI cI = (0)

aI + a (1)

aI + 2a

(2)

aI

Single line to ground fault

Fault conditions are:

bI = 0

cI = 0

aV = fZ aI

cI

bI

aI

fZ

+ (2)

aV

_

fZ3

+ (0)

aV

_

+ (1)

aV

_

36

f021

na(1)

aZ3ZZZ

EI

; (1)

a

(0)

a

(1)

a

(2)

a II;II

Corresponding phase components are cba IandI,I

Fault current (1)

aaf I3II

(1)

aV = aE - 1Z (1)

aI

(2)

aV = - 2Z (2)

aI

(0)

aV = - 0Z (0)

aI

Corresponding phase components are cb,a VandVV

37

0Z

naE

(0)

aI (2)

aI (1)

aI

1Z 2Z

Line to line fault


aI = 0

cI = - bI

bV - fZ bI = cV

cI

bI

aI

fZ

fZ

(1)

aV (2)

aV

38

0I;II;ZZZ

EI

(o)

a

(1)

a

(2)

a

f21

na(1)

a


Fault current

(1)aI3j

(1)aIa)

2a

(2)aIa

(1)aI

2a

(0)aIbI

fI (

(1)

aV = aE - 1Z (1)

aI

(2)

aV = - 2Z (2)

aI

(0)

aV = - 0Z (0)

aI


39

naE

0Z

(0)

aI (2)

aI (1)

aI

1Z 2Z



fZ)II(V

Z)II(V

0I

cbc

fcbb

a

-

cI

bI

aI

fZ

+

fZ3

(0)

aV (1)

aV (2)

aV

40

f02

f021

na(1)

a

Z3ZZ

)Z3Z(ZZ

EI

f02

f0(1)

a

(2)

aZ3ZZ

Z3ZII

f02

2(1)

a

(0)

aZ3ZZ

ZII

Fault current

(0)

acbf I3III


(1)

aV = aE - 1Z (1)

aI

(2)

aV = - 2Z (2)

aI

(0)

aV = - 0Z (0)

aI


41

Example 2

The reactances of an alternator rated 10 MVA, 6.9 kV are

1X = 2X = 15 % and g0X = 5 %. The neutral of the alternator is

grounded through a reactance of 0.38 . Single line to ground

fault occurs at the terminals of the alternator. Determine the line

currents, fault current and the terminal voltages.

Solution

1X = 2X = 0.15 p.u.

nX = 0.38 x 26.9

10 = 0.0798 p.u.

0X = g0X +3 nX = 0.05 + 0.2394 = 0.2894 p.u.

(1)

aI = (2)

aI = (0)

aI = 1.0 / j ( 0.2894 + 0.15 + 0.15 ) = - j 1.6966 p.u.

Corresponding phase components are

aI = -j 5.0898 p.u. bI = cI = 0

42

Base current = 6.9x3

1000x10 = 836.7 A

Line currents are aI = - j 4258.8 A ; bI = cI = 0

Fault current, fI = aI = - j 4258.8 A

(1)

aV = 1.0 – ( j 0.15 ) (- j 1.6966 ) = 1.0 – 0.2545 = 0.7455 p.u. (2)

aV = - ( j 0.15 ) (- j 1.6966 ) = - 0.2545 p.u. (0)

aV = - ( j 0.2894 ) (- j 1.6966 ) = - 0.491 p.u.


aV = 0 ; bV = 1.1386 0

130.38 p.u. ; cV = 1.1386 0130.38 p.u.

Multiplying by 3

6.9

aV = 0; bV = 4.5359 0130.38 kV ; cV = 4.5359

0130.38 kV

43

Example 3

The reactances of an alternator rated 10 MVA, 6.9 kV are

1X =15 %; 2X = 20 % and g0X = 5 %. The neutral of the

alternator is grounded through a reactance of 0.38 . Line to line

fault, with fault impedance j 0.15 p.u. occurs at the terminals of

the alternator. Determine the line currents, fault current and

the terminal voltages.

Solution

1X = 0.15 p.u. ; 2X = 0.2 p.u. ; FX = 0.15 p.u. 0X = ?

(1)

aI = 1.0 / j ( 0.15 + 0.2 + 0.15 ) = - j 2 p.u.

(2)

aI = - (1)

aI = j 2 p.u. and (0)

aI = 0


aI = 0 ; bI = - 3.4641 p.u. ; cI = 3.4641 p.u.

44

Base current = 836.7 A

Line currents are aI = 0 ; bI = - 2898.4 A ; cI = 2898.4 A

Fault current fI = bI = - 2898.4 A

(1)

aV = 1.0 – ( j 0.15 ) (- j 2 ) = 0.7 p.u. (2)

aV = - ( j 0.3 ) ( j 2 ) = 0.4 p.u. (0)

aV = 0


aV = 1.1 ; bV = 0.6083 0

154.72 p.u. ; cV = 0.6083 0154.72 p.u.

Multiplying by 3

6.9, aV = 4.3821 kV

bV = 2.4233 0

154.72 kV

cV = 2.4233 0

154.72 kV

45

Example 4

An unloaded, solidly grounded 10 MVA, 11 kV generator has

positive, negative and zero sequence impedances as j 1.2 Ω,

j 0.9 Ω and j 0.04 Ω respectively. A double line to ground fault

occurs at the terminals of the generator. Calculate the currents in

the faulted phases and voltage of the healthy phase.

Solution

Base impedance = 10

112

= 12.1 Ω ;

1Z = j 0.09917 p.u. ; 2Z = j 0.07438 p.u. ; 0Z = j 0.00331 p.u.

1Z + 02

02

ZZ

ZZ

= j 0.10234 p.u.

(1)

aI = 1.0/ j 0.10234 = -j 9.7714 p.u.

(2)

aI = j 9.7714 0.07769

0.00331 = j 0.4163 p.u.

(0)

aI = j 9.7714 0.07769

0.07438 = j 9.3551 p.u.

46

(1)

aI = 1.0/ j 0.10234 = -j 9.7714 p.u.

(2)

aI = j 9.7714 0.07769

0.00331 = j 0.4163 p.u.

(0)

aI = j 9.7714 0.07769

0.07438 = j 9.3551 p.u.


aI = 0 ; bI = 16.5758 0122.16 p.u. ; cI = 16.5758 0

57.84 p.u.

Base current = 11x3

1000x10 = 542.86 A

Current in faulted phases are bI = 8998.3 0122.16 A

cI = 8998.3 0

57.84 A

(1)

aV = (2)

aV = (0)

aV = - ( j 0.07438 ) ( j 0.4163 ) = 0.03096 p.u.

Voltage of the healthy phase aV = 0.09288 x 3

11 = 0.5899 kV

EE 0308 POWER SYSTEM ANALYSIS

SURPRISE TEST 1 March 2010

1 Obtain the bus admittance matrix of the transmission system with the

following data.

Line data

Line

No.

Between

buses Line Impedance HLCA

Off nominal

turns ratio

1 1 – 2 0.08 + j 0.37 j 0.007 ---

2 3 – 2 j 0.133 0 0.909

Shunt capacitor data

Bus No. 3 Admittance j 0.0096

Line No. Between Line admittance HLCA Yp q / a Yp q / a2

1 1 - 2 0.5583 – j 2.582 j 0.007 ----- ----

2 3 – 2 - j 7.5188 ---- - j 8.2715 - j 9.0996

Y11 = 0.5583 – j 2.582 + j 0.007 = 0.5583 – j 2.575

Y22 = 0.5583 – j 2.582 - j 7.5188 + j 0.007 = 0.5583 – j 10.0938

Y33 = - j 9.0996 + j 0.0096 = - j 9.09

9.09j8.2716j0

8.2716j10.0938j0.55832.582j0.5583

02.582j0.55832.575j0.5583

1

1

2

2 3

3

YBus =

2. In a three bus power system, bus 1 is slack bus and buses 2 and 3 are P-Q

buses. Its bus admittance matrix is

5j1.72j0.73j1

2j0.78j2.76j2

The slack bus voltage is 1.04 00 . At bus 2, real power generation is 0.7,

real power load is 0.2, reactive power generation is 0.1 and reactive power

load is 0.3. Taking flat start and using Gauss Seidel method, find the bus

voltage V2 after first iteration.

3

1

1

2

2 3

2. V1 = 1.04 00 ; V2 = 1.0 00 ; V3 = 1.0 00

PI2 = 0.7 – 0.2 = 0.5; QI2 = 0.1 – 0.3 = - 0.2; PI2 + j QI2 = 0.5 – j 0.2

V2(1) = ]VYVY

V

QIPI[

Y

1332112*(0)

2

22

22

= 2)j0.7((1.04)6)j2(0.2)j(0.5[8j2.7

1

= 8j2.7

2)j0.7(6.24)2.08(0.2)j(0.5

=

8j2.7

8.04j3.28

= 1.0265 + j 0.0636 = 1.0284 03.54

51

4 UNSYMMETRICAL FAULTS ON POWER SYSTEMS

In a general power system fault can occur at any bus p. In such case, the fault

analysis discussed in previous section can be extended following one-to-one

correspondence shown below.

Fault at the terminals of the

generator

Fault occurs at bus p in the power

system

Positive sequence pre-fault voltage is naE Positive sequence pre-fault voltage is fV

Positive sequence impedance is 1Z

Thevenin’s equivalent impedance

between the fault point and the

reference bus in the positive sequence

network is 1Z

Negative sequence impedance is 2Z



reference bus in the negative

sequence network is 2Z

Zero sequence impedance is 0Z



reference bus in the zero sequence

network is 0Z

52

Note that the Thevenin’s equivalent circuit of different sequence networks will

be similar to the sequence networks of the generator. Thevenin’s equivalent

circuit of the sequence networks are interconnected, much similar to the case

of fault occurring at the generator terminals, to represent different types of

faults.

This method is not suitable for large scale power systems as it involves

network reduction in positive, negative and zero sequence networks.

53

5 UNSYMMETRICAL FAULT ANALYSIS USING busZ matrix

When an unsymmetrical fault occurs in a power system, three phase network has

to be considered. Any three phase element can be represented as shown in Fig.

15.

Fig. 15

It can be described as

cb,a,

qp

cb,a,

qp

cb,a,

qp izv i.e. (42)

c

qp

b

qp

a

qp

v

v

v

=

cc

qp

bc

qp

ac

qp

cb

qp

bb

qp

ab

qp

ca

qp

ba

qp

aa

qp

zzz

zzz

zzz

c

qp

b

qp

a

qp

i

i

i

(43)

c

qV

cb,a,

qpz

c

pV

q p

b

pV

a

pV b

qV a

qV

54

Voltages at bus p and q can be denoted as

c

p

b

p

a

p

cb,a,

p

V

V

V

V ;

c

q

b

q

a

q

cb,a,

q

V

V

V

V (44)

Considering the impedance of each three phase element as cb,a,

qpz , using building

algorithm, the bus impedance matrix of transmission-generator can be obtained

as

1 2 N

cb,a,

busZ

cb,a,

NN

cb,a,

N2

cb,a,

N1

cb,a,

2N

cb,a,

22

cb,a,

21

cb,a,

1N

cb,a,

12

cb,a,

11

ZZZ

ZZZ

ZZZ

where cb,a,

jiZ i

cc

ji

bc

ji

ac

ji

cb

ji

bb

ji

ab

ji

ca

ji

ba

ji

aa

ji

ZZZ

ZZZ

ZZZ

The bus impedance matrix cb,a,

busZ will be normally full with non-zero entries.

Since impedance of any element in sequence frame, 0,1,2

qpz is decoupled,

computationally it is advantage to use the matrix 0,1,2

busZ instead of cb,a,

busZ .

N

1

2

j

55

2

1

2

1

2

1

For two bus system

1 2 1 2 1 2

0

busZ

(0)

22

(0)

21

(0)

12

(0)

11

ZZ

ZZ ; 1

busZ

(1)

22

(1)

21

(1)

12

(1)

11

ZZ

ZZ ; 2

busZ

(2)

22

(2)

21

(2)

12

(2)

11

ZZ

ZZ

Then

0,1,2

busZ

(2)

22

(2)

21

(1)

22

(1)

21

(0)

22

(0)

21

(2)

12

(2)

11

(1)

12

(1)

11

(0)

12

(0)

11

ZZ2

ZZ1

ZZ0

ZZ2

ZZ1

ZZ0

210210

(45)

Normally 2

bus

1

bus

0

bus ZandZ,Z are constructed and stored independently. It is

evident that as compared to cb,a,

busZ , construction of 0,1,2

busZ requires less computer

time and less core storage. For a 100 bus system, cb,a,

busZ will be a 300 x 300 full

matrix; whereas for 0,1,2

busZ , we need 3 numbers of 100 x 100 matrices. Thus only

1/3 rd of the core storage is required for 0,1,2

busZ as compared to cb,a,

busZ . Hence for

unsymmetrical fault analysis, use of 0,1,2

busZ is more advantages than cb,a,

busZ .

1

1

2

2

56

Further, when unsymmetrical faults occur, the currents

and voltages are unbalanced and using symmetrical

components transformation, we can handle them

conveniently. Therefore, symmetrical components are

invariably used in the study of unsymmetrical fault analysis.

In order to obtain 0,1,2

busZ , first the three sequence networks

are be drawn as discussed earlier. Considering the zero

sequence, positive sequence and negative sequence networks

separately, using bus impedance building algorithm,

are to be constructed independently . Of course special

attention is necessary while drawing the zero sequence network.

57

j0.52 j0.09 j0.08

j0.12 j0.06

j0.66

j0.05

j0.96

Example 5

Consider the system described in Example 1. Obtain the matrices 2

bus

1

bus

0

bus ZandZ,Z .

Solution

Required bus impedance matrices can be constructed using bus impedance

building algorithm.

First consider the zero sequence network shown in Fig. 8(a).

Reference

1 2 3 4

0

58

3

1

2

2

1

Element 0-1 is added: 0

busZ j 1 1.01

1 2

Element 0 – 2 is added: 0

busZ j

0.080

01.01

1 2 3


busZ j

0.0900

00.080

001.01

Element 2 – 3 is added: With th bus

1 2 3

0

busZ j

0.690.090.080

0.090.0900

0.0800.080

0001.01

1

2

3

59

3

1

2

1 2 3

Eliminating the th bus, 0

busZ j

0.080.010

0.010.070

001.01

Element 0 – 4 is added: The final 0

busZ is obtained as

1 2 3 4

0

busZ j

0.72000

00.080.010

00.010.070

0001.01

Consider the positive sequence network shown in Fig. 8(b)

1

2

3

4

60

j0.18 j0.09 j0.08

j0.55 j0.27 j0.2

+ + +

1 2 3 4

gE


busZ j 1 0.2

1 2


busZ j

0.280.2

0.20.2

1 2 3


busZ j

0.460.280.2

0.280.280.2

0.20.20.2

2

1

1

3

2

2mE 1mE

Reference

61

1 2 3 4


busZ j

0.550.460.280.2

0.460.460.280.2

0.280.280.280.2

0.20.20.20.2

Element 0 – 4 is added: It has an impedance of j0.18. With the th bus

1 2 3 4

1

busZ j

0.730.550.460.280.2

0.550.550.460.280.2

0.460.460.460.280.2

0.280.280.280.280.2

0.20.20.20.20.2

Eliminating the th bus, final

1

busZ is obtained as

1 2 3 4

1

busZ = j

0.13560.11340.06900.0493

0.11340.17010.10360.0740

0.06900.10360.17260.1233

0.04930.07400.12330.1452

1

2

3

4

1

2

3

4

1

2

3

4

62

j0.55 j0.27

j0.25

Similarly, considering the negative sequence network shown in Fig. 8(c)

j0.08 j0.18 j0.09

Its bus impedance 2

busZ can be obtained as

1 2 3 4

2

busZ = j

0.13840.11770.07610.0577

0.11770.17650.11430.0866

0.07610.11430.19040.1442

0.05770.08660.14420.1699

1

2

3

4

Reference

1 2 3 4

0

63

N

6 UNSYMMETRICAL FAULT ANALYSIS USING 0,1,2

busZ MATRIX

For unsymmetrical fault analysis using 0,1,2

busZ the first step is to construct 1

busZ ,

2

busZ and 0

busZ by considering the positive sequence, negative sequence and zero

sequence network of the power system. They are

1 2 p N

1

busZ

(1)

NN

(1)

Np

(1)

N2

(1)

N1

(1)

pN

(1)

pp

(1)

p2

(1)

p1

(1)

2N

(1)

2p

(1)

22

(1)

21

(1)

1N

(1)

1p

(1)

12

(1)

11

ZZZZ

ZZZZ

ZZZZ

ZZZZ

(46)

1

2

p

64

1 2 p N

2

busZ

(2)

NN

(2)

Np

(2)

N2

(2)

N1

(2)

pN

(2)

pp

(2)

p2

(2)

p1

(2)

2N

(2)

2p

(2)

22

(2)

21

(2)

1N

(2)

1p

(2)

12

(2)

11

ZZZZ

ZZZZ

ZZZZ

ZZZZ

and (47)

1 2 p N

0

busZ

(0)

NN

(0)

Np

(0)

N2

(0)

N1

(0)

pN

(0)

pp

(0)

p2

(0)

p1

(0)

2N

(0)

2p

(0)

22

(0)

21

(0)

1N

(0)

1p

(0)

12

(0)

11

ZZZZ

ZZZZ

ZZZZ

ZZZZ

(48)

1

2

p

N

1

2

p

N

65

Suitable assumptions are made so that prior to the occurrence of the fault, there

will not be any current flow in the positive, negative and zero sequence networks

and the voltages at all the buses in the positive sequence network are equal to

fV .

The currents flowing out of the original balanced system from phases candba,

at the fault point are designated as cfbfaf IandI,I . We can visualize these

currents by referring to Fig. 16 which shows the three lines candba, of the

three phase system where the fault occurs.

Fig. 16

P

bfI

afI

cfI

c

a

b

66

2

The currents flowing out in hypothetical stub are cfbfaf IandI,I . The

corresponding sequence currents are (0)

afI , (1)

afI and (2)

afI . These sequence currents

(1)

afI , (2)

afI and (0)

afI are flowing out as shown in Fig. 17. The line to ground voltages

at any bus j of the system during the fault are ajV , bjV and cjV . Corresponding

sequence components of voltages are (0)

ajV , (1)

ajV and (2)

ajV .

(1)

afI

1

- V f +

- V f +

- V f +

Positive

sequence

network

having

bus

impedance

matrix

(1)

busZ

N

p

67

2 2

Fig. 17

(2)

afI

1

Negative

sequence

network

having

bus

impedance

matrix

(2)

busZ

N

p

(0)

afI

1

Zero

sequence

network

having

bus

impedance

matrix

(0)

busZ

N

p

68

Consider the Positive Sequence Network:

In the faulted system, there are two types of sources.

1 Current injection at the faulted bus.

2 Pre-fault voltage sources.

The bus voltages in the faulted system namely

(1)

busV

(1)

aN

(1)

ap

(1)

2a

(1)

1a

V

V

V

V

(49)

can be obtained using Superposition Theorem.

N

1

2

p

69

It is to be noted that

(1)

busI

0

0

I

0

0

0

(1)

af

and pre-fault voltage =

1

1

1

1

1

1

fV (50)

Using these we get

(1)

aN

(1)

ap

(1)

2a

(1)

1a

V

V

V

V

=

(1)

NN

(1)

Np

(1)

N2

(1)

N1

(1)

pN

(1)

pp

(1)

p2

(1)

p1

(1)

2N

(1)

2p

(1)

22

(1)

21

(1)

1N

(1)

1p

(1)

12

(1)

11

ZZZZ

ZZZZ

ZZZZ

ZZZZ

0

I

0

0

(1)

fa

+

1

1

1

1

fV =

(1)

af

(1)

pNf

(1)

af

(1)

ppf

(1)

af

(1)

2pf

(1)

af

(1)

1pf

IZV

IZV

IZV

IZV

(51)

p

1

2

N

70

Consider the Negative Sequence and Zero Sequence Networks:

In a much similar manner, the negative sequence and the zero sequence bus

voltages in the faulted system, namely (2)

busV and (0)

busV , can be obtained considering

the negative sequence and the zero sequence networks. Knowing the pre-fault

voltages are zero in the negative and zero sequence networks we get

(2)

aN

(2)

ap

(2)

2a

(2)

1a

V

V

V

V

=

(2)

af

(2)

pN

(2)

af

(2)

pp

(2)

af

(2)

2p

(2)

af

(2)

1p

IZ

IZ

IZ

IZ

and

(0)

aN

(0)

ap

(0)

2a

(0)

1a

V

V

V

V

=

(0)

af

(0)

pN

(0)

af

(0)

pp

(0)

af

(0)

2p

(0)

af

(0)

1p

IZ

IZ

IZ

IZ

(52)

71

When the fault occurs at bus p , it is to be noted that only the thp column of

1

busZ , 2

busZ and 0

busZ are involved in the calculations. If the symmetrical

components of the fault currents , namely (1)

afI , (2)

afI and (0)

afI , are known, than the

sequence voltages at any bus j can be computed from

(0)

af

(0)

pj

(0)

aj IZV (53)

(1)

af

(1)

pjf

(1)

aj IZVV (54)

(2)

af

(2)

pj

(2)

aj IZV (55)

It is important to remember that the (1)

afI , (2)

afI and (0)

afI are the symmetrical

component currents in the stubs hypothetically attached to the system at the

fault point. These currents take on values determined by the particular type of

fault being studied, and once they are calculated, they can be regarded as

negative injection into the corresponding sequence networks.

72

General procedure for unsymmetrical fault analysis

when fault occurs at a point in a

power system

73

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

cf

bf

af

I

I

I

=

2

2

aa1

aa1

111

(2)

af

(1)

af

(0)

af

I

I

I

74

TO COMPUTE FAULTED BUS VOLTAGES V p a, V p b and V p c 1. Compute the sequence components V p a

(1), V p a(2) and V p a

(0) from V p a

(!) = Vf – Z p p(1) I f a

(1)

V p a(2) = – Z p p

(2) I f a(2)

V p a(0) = – Z p p

(0) I f a(0)

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

cp

bp

ap

V

V

V

=

2

2

aa1

aa1

111

(2)

ap

(1)

ap

(0)

ap

V

V

V

75

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

V j a

(!) = Vf – Z j p(1) I f a

(1)

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

cj

bj

aj

V

V

V

=

2

2

aa1

aa1

111

(2)

aj

(1)

aj

(0)

aj

V

V

V

76

fV

(2)

afI

1)(

afI

0)(

afI

Single line to ground fault

(1)

afI = (2)

afI = (0)

afI

(2)

ppZ

(1)

ppZ +

_

fZ3

(0)

ppZ

(1)

apV

(2)

apV

(0)

apV

77

(0)

afI (2)

afI (1)

afI

(1)

ppZ (2)

ppZ

Line to line fault

(2)

apV (1)

apV _

+

fV

fZ

78

(0)

ppZ

(0)

afI (2)

afI (1)

afI

(1)

ppZ (2)

ppZ


fV _

+

fZ3

(0)

apV (1)

apV (2)

apV

79

SINGLE LINE TO GROUND FAULT

For a single line to ground fault through impedance fZ , the hypothetical stubs

on the three lines will be as shown in Fig. 18 The fault conditions are

Fig. 18

0I bf

0I cf (56)

affap IZV

a

afI fZ

P

c

b bfI

cfI

80

Using the above conditions ( similar to conditions for the LG fault at generator

terminals through impedance ) and also knowing that

(1)

af

(1)

ppf

(1)

ap IZVV (57)

(2)

af

(2)

pp

(2)

ap IZV (58)

(0)

af

(0)

pp

(0)

ap IZV (59)

from eqns. (51) and (52), similar to eqns. (22) and (23), we can get the relations

(1)

afI = (2)

afI = (0)

afI and (60)

(1)

aff

(0)

ap

(2)

ap

(1)

ap IZ3VVV = 0 (61)

Therefore

f

(0)

pp

(2)

pp

(1)

pp

f(1)

afZ3ZZZ

VI

(62)

The above relationships are satisfied by connecting the sequence networks as

shown in Fig. 19

81

fV

(2)

afI

1)(

afI

2)(

afI

(1)

afI = (2)

afI = (0)

afI

Fig. 19

The series connection of Thevenin equivalents of the sequence networks, as

shown in the above Fig. 18, is a convenient means of remembering the equations

for the solution of single line to ground fault.

(2)

ppZ

(1)

ppZ +

_

fZ3

(0)

ppZ

(1)

apV

(2)

apV

(0)

apV

82

Once the currents (1)

afI , (2)

afI and (0)

afI are known, the sequence components of

voltage at the faulted bus are calculated as

(1)

af

(1)

ppf

(1)

ap IZVV

(2)

af

(2)

pp

(2)

ap IZV (63)

(0)

af

(0)

pp

(0)

ap IZV

Thereafter the sequence components of voltage at any bus j can be calculated as

(1)

af

(1)

pjf

(1)

aj IZVV

(2)

af

(2)

pj

(2)

aj IZV (64)

(0)

af

(0)

pj

(0)

aj IZV

Phase components of voltage and current can be calculated from the relations

2,1,0c,b,a VAV

2,1,0c,b,a IAI

pj

N.1,2,......j

83

Example 6

The positive sequence, negative sequence and zero sequence bus impedance

matrices of a power system are shown below.

1 2 3 4

(2)

bus

(1)

bus ZZ j

0.14370.12110.07890.0563

0.12110.16960.11040.0789

0.07890.11040.16960.1211

0.05630.07890.12110.1437

1 2 3 4

(0)

busZ j

0.15530.14070.04930.0347

0.14070.19990.07010.0493

0.04930.07010.19990.1407

0.03470.04930.14070.1553

A bolted single line to ground fault occurs on phase ‘a’ at bus 3. Determine the

fault current and the voltage at buses 3 and 4.

1

2

4

3

1

2

3

4

84

Solution

(1)

afI = (2)

afI = (0)

afI = (0)

33

(2)

33

(1)

33

f

ZZZ

V

Let 0

f 01.0V

Then (1)

afI = (2)

afI = (0)

afI = )0.19990.1696(0.1696j

01.00

= 1.8549j

The fault current (1)

af

(2)

af

(1)

af

(0)

afaf I3IIII = - j 5.5648; bfI = cfI = 0

The sequence components of voltage at bus 3 are calculated as

0.3708)1.8549j()0.1999j(IZV(0)

af

(0)

33

(0)

a3

0.6854)1.8549j()0.1696j(1.0IZVV(1)

af

(1)

33f

(1)

a3

0.3146)j1.8549()j0.1696(IZV(2)

af

(2)

33

(2)

a3

85

Phase components of line to ground voltage of bus 3 are computed as

c3

b3

a3

V

V

V

=

2

2

aa1

aa1

111

0.3146

0.6854

0.3708

=

The sequence components of voltage at bus 4 are calculated as

0.2610)1.8549j()0.1407j(IZV(0)

af

(0)

43

(0)

a4

0.7754)1.8549j()0.1211j(1.0IZVV(1)

af

(1)

43f

(1)

a4

0.2246)j1.8549()j0.1211(IZV(2)

af

(2)

43

(2)

a4

Phase components of line to ground voltage of bus 4 are computed as

c4

b4

a4

V

V

V

=

2

2

aa1

aa1

111

0.2246

0.7754

0.2610

=

0.289800

1.01870

121.8

1.01870

121.8

0

1.0292 0122.71

1.0187 0122.71

86

LINE TO LINE FAULT

To represent a line to line fault through impedance fZ the hypothetical stubs on

the three lines at the fault are connected as shown in Fig. 20.

Fig. 20

Fault conditions are

0I af

0II cfbf (65)

cpbffbp VIZV

( c

b

a

bfI

afI

fZ

P

cfI

87

Using the above conditions along with the relations

(1)

af

(1)

ppf

(1)

ap IZVV

(2)

af

(2)

pp

(2)

ap IZV (66)

(0)

af

(0)

pp

(0)

ap IZV

we can get the following relations.

(2)

ap

(1)

aff

(1)

ap

(2)

af

(1)

af

(0)

af

VIZV

II

0I

(68)

and hence f

(2)

pp

(1)

pp

f(1)

afZZZ

VI

(70)

(67)

(69)

88

(0)

afI (2)

afI (1)

afI

(1)

ppZ (2)

ppZ

To satisfy the above relations the sequence networks are to be connected as

shown in Fig. 21.

(0)

ppZ

Fig. 21

Once (1)

afI , (2)

afI and (0)

afI are calculated, (2)

ap

(1)

ap V,V and (0)

apV can be computed from

eqn. (63). Thereafter (2)

aj

(1)

aj V,V and (0)

ajV for pj;N1,2,....,j can be calculated

using eqn. (63). The corresponding phase components are then calculated using

the symmetrical component transformation matrix.

(2)

apV (1)

apV _

+

fV

fZ

89

Example 7

Consider the power system described in example 6. A bolted line to line fault

occurs at bus 3. Determine the currents in the fault, voltages at the fault bus

and the voltages at bus 4.

Solution

For line to line fault 0I(0)

af

2.9481j0.1696j0.1696j

1.0

ZZ

VII

(2)

33

(1)

33

f(2)

af

(1)

af

The phase components of the currents in the fault are

(2)

af

(1)

af

(0)

afaf IIII 0

(1)

af

(2)

af

(1)

af

2

bf I3jIaIaI - 5.1061

bfcf II 5.1061

Sequence components of voltage at bus 3 are

0V(0)

a3 ; 0.5)2.9481(j)0.1696j(IZVV(2)

af

(2)

33

(2)

a3

(1)

a3

90

Phase components of voltage at bus 3 are

a3V (0)

a3V (2)

a3

(1)

a3 VV 1.0

(1)

a3

(2)

a3

(1)

a3

2

c3b3 VVaVaVV -0.5

Sequence components of voltage at bus 4 are

0IZV(0)

af

(0)

43

(0)

a4

0.643)2.9481j()0.1211j(1IZVV(1)

af

(1)

43f

(1)

4a

0.357)2.9481j()0.1211j(IZV(2)

af

(2)

43

(2)

a4

Phase components of voltage at bus 4 are

(2)

a4

(1)

a4

(0)

a44a VVVV 1.0

(2)

a4

(1)

a4

2

b4 VaVaV - 0.5

(2)

a4

2(1)

a4c4 VaVaV - 0.5

91

fZ

DOUBLE LINE TO GROUND FAULT

For a double line to ground fault, the hypothetical stubs are connected as shown

in Fig. 22.

Fig. 22

The relations at the fault bus are

)II(ZV

)II(ZV

0I

cfbffcp

cfbffbp

af

(71)

cfI

( c

b

a

bfI

afI

P

92

Further the relations are also applicable.

(1)

af

(1)

ppf

(1)

ap IZVV

(2)

af

(2)

pp

(2)

ap IZV (72)

(0)

af

(0)

pp

(0)

ap IZV

Using eqns. (71) and (62) the following relations can be obtained.

(2)

ap

(1)

ap VV

(0)

aff

(0)

ap

(1)

ap IZ3VV

0III(2)

af

(1)

af

(0)

af

On further simplification we get

f

(0)

pp

(2)

pp

f

(0)

pp

(2)

pp(1)

pp

f(1)

af

Z3ZZ

)Z3Z(ZZ

VI

(73)

93

(0)

ppZ

(0)

afI (2)

afI (1)

afI

(1)

ppZ (2)

ppZ

To represent the above relations, the sequence networks must be interconnected

as shown in Fig. 23.

Fig. 23

The sequence currents (2)

afI and (0)

afI can be obtained from

f

(0)

pp

(2)

pp

f

(0)

pp(1)

af

(2)

afZ3ZZ

Z3ZII

(74)

f

(0)

pp

(2)

pp

(2)

pp(1)

af

(0)

afZ3ZZ

ZII

(75)

fV _

+

fZ3

(0)

apV (1)

apV (2)

apV

94

Knowing (1)

afI , (2)

afI and (0)

afI sequence components of voltage at fault point are

calculated from

(1)

af

(1)

ppf

(1)

ap IZVV

(2)

af

(2)

pp

(2)

ap IZV (76)

(0)

af

(0)

pp

(0)

ap IZV

Thereafter, sequence components of voltage at any other bus can be obtained

from

(1)

af

(1)

pjf

(1)

aj IZVV

(2)

af

(2)

pj

(2)

aj IZV (77)

(0)

af

(0)

pj

(0)

aj IZV

Knowing the sequence components, corresponding phase conponents are obtained

as

21,0,cb,a, IAI or 21,0,cb,a, VAV (78)

pj

N.1,2,......j

95

Example 8

The positive sequence, negative sequence and zero sequence bus impedance

matrices of a power system are shown below.

1 2 3 4

(2)

bus

(1)

bus ZZ j

0.14370.12110.07890.0563

0.12110.16960.11040.0789

0.07890.11040.16960.1211

0.05630.07890.12110.1437

1 2 3 4

(0)

busZ j

0.19000

00.580.080

00.080.080

0000.19

1

2

4

3

1

2

3

4

96

A double line to ground fault with 0Zf occurs at bus 4. Find the fault

current and voltages at the fault bus.

Solution

Sequence components of fault current are

4.4342j

0.19j0.1437j

)0.19j()0.1437j(0.1437j

1.0

ZZ

ZZZ

VI

(0)

44

(2)

44

(0)

44

(2)

44(1)

44

f(1)

af

j2.5247j0.19j0.1437

j0.19)j4.4342(

ZZ

ZII

(0)

44

(2)

44

(0)

44(1)

af

(2)

af

j1.9095j0.19j0.1437

j0.1437)j4.4342(

ZZ

ZII

(0)

44

(2)

44

(2)

44(1)

af

(0)

af

97

Phase components of current at the fault bus are

(2)

af

(1)

af

(0)

afaf IIII 0

00(2)

af

(1)

af

2(0)

afbf 2102.52471504.4342j1.9095IaIaII

= - 6.0266 + j 2.8642 0

302.5247304.4342j1.9095IaIaII0(2)

af

2(1)

af

(0)

afcf

= 6.0266 + j 2.8642

Fault current cfbff III j 5.7285

Sequence components of voltage at the faulted bus are calculated as follows.

Noting that 0Zf

0.3628)j4.4342()j0.1437(1.0IZVVVV(1)

af

(1)

44f

(0)

a4

(2)

a4

(1)

a4

Phase components of faulted bus voltage are:

(2)

a4

(1)

a4

(0)

a4a4 VVVV 1.0884

b4V 0

c4V 0

98

PROBLEMS – UNSYMMETRICAL FAULTS

1. In an unbalanced circuit the three line currents are measured as

0

c

0

b

0

a

160.752.6810I

203.844.3733I

59.857.0311I

Obtain the corresponding sequence components of currents and draw

them to scale.

2. For the sequence components calculated in Problem 1, find the

corresponding phase components of line currents and verify the results

graphically.

3. A three phase transmission line has the phase impedance of

2166

6216

6621

jcb,a,

z

Calculate its sequence impedances.

99

4. A 20 MVA, 13.8 kV alternator has the following reactances:

X1 = 0.25 p.u. X2 = 0.35 p.u. Xg0 = 0.04 p.u. Xn = 0.02 p.u.

A single line to ground fault occurs at its terminals. Draw the

interconnections of the sequence networks and calculate

i) the current in each line

ii) the fault current

iii) the line to neutral voltages

iv) the line to line voltages

Denoting the neutral point as n and the ground as o , draw the phasor

diagram of line to neutral voltages.

5. Repeat Problem 4 for line to line fault.

6. Repeat Problem 4 for double line to ground fault.

7. Repeat Problem 4 for symmetrical three phase fault.

8. Consider the alternator described in Problem 4. It is required to limit the

fault current to 2500 A for single line to ground fault. Find the additional

reactance necessary to be introduced in the neutral.

100

9. Two synchronous machines are connected through three-phase

transformers to the transmission line as shown.

The ratings and reactances of the machines and transformers are:

Machines 1 and 2: 100 MVA, 20 kV, X1 = X2 = 20 %, Xm0 = 4 %, Xn = 5 %

Transformers T1 and T2: 100 MVA, 20 Δ / 345 Y kV, X = 8 %

On a chosen base of 100 MVA, 345 kV in the transmission line circuit, the

line reactances are X1 = X2 = 15 % and X0 = 50 %. Draw each of the three

sequence networks and find Zbus0, Zbus

1 and Zbus2.

2 1

1

T1

4 3

2

T2

101

10. The one-line diagram of a power system is shown below.

The following are the p.u. reactances of different elements on a common

base.

Generator 1: Xg0 = 0.075; Xn = 0.075; X1 = X2 = 0.25

Generator 2: Xg0 = 0.15; Xn = 0.15; X1 = X2 = 0.2

Generator 3: Xg0 = 0.072; X1 = X2 = 0.15

Transformer 1: X0 = X1 = X2 = 0.12

Transformer 2: X0 = X1 = X2 = 0.24

Transformer 3: X0 = X1 = X2 = 0.1276

Transmission line 2 – 3 X0 = 0.5671; X1 = X2 = 0.18

Transmission line 3 – 5 X0 = 0.4764; X1 = X2 = 0.12

Draw the three sequence networks and determine Zbus0, Zbus

1 and Zbus2.

6 5

4

3 2 1

1 3

2

T1

T2

T3

102

11. The single line diagram of a small power system is shown below.

Generator: 100 MVA, 20 kV, X1 = X2 = 20 %, Xg0 = 4 %, Xn = 5 %

Transformers T1 and T2: 100 MVA, 20 Δ / 345 Y kV, X = 10 %

On a chosen base of 100 MVA, 345 kV in the transmission line circuit, the

line reactances are:

From T1 to P: X1 = X2 = 20 %; X0 = 50%

From T2 to P: X1 = X2 = 10 %; X0 = 30%

A bolted single lone to ground occurs at P. Determine

i) fault current IfA, IfB and IfC.

ii) currents flowing towards P from T1.

iii) currents flowing towards P from T2.

iv) current supplied by the generator.

Note that the positive sequence current in Δ winding of transformer lags

that in Y winding by 300; the negative sequence current in Δ winding leads

that in Y winding by 300.

2 1

1

T1

4 3

T2

P

S

Switch open

103

12. In the power system described in Problem 10, a single line to ground fault

occurs at bus 2 with a fault impedance of j0.1. Determine the bus currents

at the faulted bus and the voltages at buses 1 and 2.

ANSWERS

1. 0

c0b0a0 1202III ; 0

a1 303.5I ; 0

b1 2703.5I ; 0

c1 1503.5I

0

a2 603I ; 0

b2 1803I ; 0

c2 3003I

2. 0

a 59.857.0311I ; 0

b 203.844.3733I ; 0

c 160.752.681I

3. 15jZZ;33jZ 210

4. Ia = -j 3586.1 A Ib = 0 Ic = 0 If = -j 3586.1 A

Va = 0 Vb = 8.0694 0102.22 kV Vc = 8.0694 0102.22 kV

Vab = 8.0694 077.78 kV; Vbc= 15.7724 090 kV; Vca= 8.0694 0102.22 kV

5. Ia = 0 Ib = -2415.5 A Ic = 2415.5 A If = -2415.5 A

Va = 9.2948 kV Vb = -4.6474 kV Vc = -4.6474 kV

Vab = 13.9422 kV Vbc = 0 Vca = 13.9422 0180 kV

104

j0.5 j0.08 j0.08

j0.04

j0.15

j0.04

j0.15

4 3 2 1

6. Ia = 0 Ib = 4020.95 0132.22 A Ic = 4020 047.78 A If = 5956.08 090 A

Va = 5.6720 kV Vb = 0 Vc = 0

Vab = 5.6720 kV Vbc = 0 Vca = 5.6720 0180 kV

7. Ia = 3347 090 A Ib = 3347 0150 A Ic = 3347 030 A

If = 3347 090 A

Va = Vb = Vc = 0

Vab = Vbc = Vca = 0

8. 0.9655 Ω

9. Zero sequence network:

Reference

105

1 2 j0.15 j0.08 j0.08

j0.2 j0.2

3 4

1

+

2

-

j0.15 j0.08 j0.08

j0.2 j0.2

3 4

-

+

Positive sequence network:

Negative sequence network:

Reference

Reference

106

4

3

2

1

j

4

3

2

1

j

1 2 3 4

j0.5671

1

j0.4764 j0.12

j0.075

j0.225

4

3 2 1

j0.15

5 6 j0.1276

j0.24

j0.45

j0.072

Zbus0 =

0.19000

00.580.080

00.080.080

0000.19

Zbus1 = Zbus

2 =

0.14370.12110.07890.0563

0.12110.16960.11040.0789

0.07890.11040.16960.1211

0.05630.07890.12110.1437

10. Zero sequence network:

Reference

107

2

3

4

5

j0.18 j0.12 j0.12

j0.25 4

3 2 1

j0.2

5 6 j0.1276

j0.24

j0.15

Zbus0 = j

0.65427100.1778710.0310650

00.6000

0.17787100.1778710.0310650

0.03106500.0310650.1044680

00000.3

Negative sequence network:

Reference

108

1 2 3 4 5 6

1

2

3

4

5

6

Zbus1 = Zbus

2 =

j

0.1149560.0851450.0259590.0571090.0384190.025959

0.0851450.1575740.0480410.1056900.0711010.048041

0.0259590.0480410.1403670.0688080.0462890.031276

0.0571090.1056900.0688080.1513770.1018360.068808

0.0384190.0711010.0462890.1018360.1895000.128108

0.0259590.0480410.0312760.0688080.1281080.167640

11. IfA = - j 2.4195 p.u.; IfB = 0; IfC = 0

IA = - j 1.9356 p.u.; IB = j 0.4839 p.u.; IC = j 0.4839 p.u.

IA = - j 0.4839 p.u.; IB = - j 0.4839 p.u.; IC = - j 0.4839 p.u.

Ia = - j 1.3969 p.u.; IB = j 1.4969 p.u.; IC = 0

12. I2a = j 3.828162 p.u.; I2b = 0; I2c = 0

V2a = 0.382815 p.u.; V2b = 0.950352 245.680 p.u.

V2c = 0.950352 114.320 p.u.

V1a = 0.673054 p.u.; V1b = 0.929112 248.760 p.u.

V1c = 0.929112 111.2360 p.u.

EE 0308 POWER SYSTEM ANALYSIS - SRM University Special attention is required while obtaining the...

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Transcript of EE 0308 POWER SYSTEM ANALYSIS - SRM University Special attention is required while obtaining the...