EE 0308 POWER SYSTEM ANALYSIS
CHAPTER 4
SEQUENCE NETWORKS AND
UNSYMMETRICAL FAULTS ANALYSIS
1
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
Ia
I
Ic
=
I (0)
b
I (0)
c
+
Ib
I (2)
a
I (1)
a
I (1)
c
(0)
a
+
I
(1)
b
I (2)
b
I (2)
c
Fig. 1
4
Defining operator ‘ a ‘ as
a = 1 120 0
(1)
it is to be noted that
a 2  1240 0 ;
a 3  1360 0  1
Also a = - 0.5 + j 0.866 ;
Hence 1  a  a 2  0
a 2   0.5  j 0.866
(2)
(3)
(4)
5
Ia
I
I (0)
a
Ic
=
I (0)
b
I (0)
c
+
Ib
I (2)
a
I (1)
a
(1)
c
+
I
(1)
b
I (2)
b
I (2)
c
Further referring Fig. 1
2 (1)
I (1)
b  a Ia
(1)
I (1)
c  a Ia
(5)
(2)
I (2)
b  a Ia
2 (2)
I (2)
c  a Ia
Therefore
(1)
(2)
I a  I (0)
a  Ia  Ia
(1)
(0)
2 (1)
(2)
I b  I (0)
 I (2)
b  Ib
b  Ia  a Ia  a Ia
(1)
(0)
(1)
2 (2)
I c  I (0)
 I (2)
c  Ic
c  Ia  a Ia  a Ia
Thus
I a  I (0)
a 
I (1)
a 
I (2)
a
2
(1)
(2)
I b  I (0)
a  a Ia  a Ia
(1)
2 (2)
I c  I (0)
a  a Ia  a Ia
(6)
6
1 1
I a 
i.e. I b  = 1 a 2
1 a
 I c 
I (0)

a
 (1) 
I a 
I (2)

 a 
1
a 
a 2 
i.e. I a , b , c  A I 0 , 1 , 2
(7)
i.e. I 0 , 1 , 2  A 1 I a , b , c
(8)
The inverse form of the above is
I (0)

a
1
 (1) 
I
=
 a 
3
I (2)

 a 
1 1
1 a

1 a 2
1
a 2 
a 
I a 
I 
 b
 I c 
Similarly, corresponding to voltage phasors
Va , b , c  A V0 , 1 , 2
and
(9)
V0 , 1 , 2  A 1 Va , b , c
(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 ba z bb z bc 
 z ca z cb z cc 
Then
v a,b,c  z a,b,c i a,b,c
(11)
i.e. A v 0,1,2  z a,b,c A i 0,1,2
v 0,1,2  A 1 z a,b,c A i 0,1,2
v 0.1,2  z 0,1,2 i 0,1,2
where
z 0,1,2 
A 1 z a,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 = A 1 z a,b,c A
(12)
8
For power system components, sequence impedance z 0,1,2 will be decoupled as
z 0,1,2
z (0)

=  0
 0

0
z (1)
0
0 

0 
z (2) 
(13)
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 Z n as shown in Fig. 2. Assume that the generator is designed to
generate balanced voltage.
9
a
+
Ea n
Zn
In
Ia
Ec n
+
+
Eb n
c
Ib
b
Ic
Fig. 2
Let E an be its generated voltage in phase a . Then
E a 
1
E  = a 2  E
an
 b
 
 E c 
 a 
Ec
Ea
This gives
Eb
10
E (0)

1 1
a
1 
 (1) 
=
E
 a 
1 a
3
2
E (2)


1
a
a



1
a 2 
a 
1
 0 
a 2  E = E 
  an
 an 
 a 
 0 
(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 a( 1 )
a
I a( 1 )
Z1
Zn
+
I n  0 Ec n
Ea n
Z1
+
Z1
+
Eb n
Z1
I
(1)
b
b
I (c1 )
c
Note that In = 0
Ea n
Va( 1 )
+
__
Reference bus ( Neutral )
Positive sequence network
11
a
Zn
I a( 2 )
I a( 2 )
Z2
Z2
In  0
Z2
Z2
c
Va( 2 )
I b( 2 )
b
Reference bus ( Neutral )
I c( 2 )
Negative sequence network
Note that In = 0
a
Zg0
Zn
In  3I
c
I a( 0 )
(0)
a
I a( 0 )
Zg0
n
Zg0
I b( 0 )
Zg0
Z0
Va( 0 )
3 Zn
b
I c( 0 )
Note that In = 3 Ia(0)
Fig. 3
Reference bus ( Ground )
Zero sequence network
12
Z 1 and Z 2 are the positive sequence and negative sequence impedance of the
generator. Z g 0 is the zero sequence impedance of the generator. Total zero sequence
impedance Z 0 = Z g 0 + 3 Z n .
Sequence components of the terminal voltage are
Va( 0 ) 
 Z 0 I (a0 )
Va( 1 )  E a n  Z 1 I (a1 )
Va( 2 ) 
(15)
 Z 2 I (a2 )
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
Reference
Fig. 5
14
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 Z 0 accounts for
the leakage impedance Z and the neutral impedances 3 Z N and
3 Z n where applicable.
15
Connection diagrams
Zero sequence equivalent circuit
P
Q
Z0
P
Q
ZN
Zn
Reference
P
Q
Z0
P
Q
ZN
Reference
16
P
P
Z0
Q
Q
Reference
P
P
Z0
Q
Q
ZN
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.
T2
T1
M1
G
M2
Fig. 7
18
Per unit reactances are:
Generator
X g0  0.05 ; X n  0.32 ; X1  0.2 ; X 2  0.25
Transformer T1
X 0  X1  X 2  0.08
Transformer T2
X 0  X1  X 2  0.09
Transmission line
X 0  0.52; X1  X 2  0.18
Motor 1
X m o  0.06; X n  0.22; X1  X 2  0.27
Motor 2
X m o  0.12; X1  X 2  0.55
19
T2
T1
M1
G
M2
Positive sequence network
j0.08
j0.18
j0.09
j0.2
j0.55
j0.27
Eg
+
Em1
+
+
Em 2
Reference
20
T2
T1
M1
G
M2
Negative sequence network
j0.08
j0.18
j0.09
j0.25
j0.27
j0.55
Reference
21
T2
T1
M1
G
M2
Zero sequence network
j0.08
j0.52
j0.09
j0.05
j0.06
j0.12
j0.66
j0.96
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
Va(1)  E a n  I (1)
a Z1
(16)
Va(2)   I (2)
a Z2
(17)
Va(0)   I (0)
a Z0
(18)
23
Va(1)  E a n  I (1)
a Z1
(16)
Va(2)   I (2)
a Z2
(17)
Va(0)   I (0)
a Z0
(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
(1)
(2)
and sequence voltages
I (0)
Va(0) , Va(1) , Va(2) . Sequence components
a , Ia , Ia
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.
Ia
a
Zf
+
Zn
_E a n
Ec n
+
+
Eb n
c
Ib
b
Fig. 9
Ic
25
The fault conditions are
Ib  0
Ic  0
Va  Z f I a
(19)
(20)
(21)
I (0)
a  1/3 ( I a  I b  I c )  I a /3
2
I (1)
a  1/3 (I a  a I b  a I c )  I a /3
2
I (2)
a  1/3 ( I a  a I b  a I c )  I a /3
(1)
(2)
Thus I (0)
a  Ia  Ia
(22)
Further from eqn. (21)
(1)
(2)
(1)
Va(0)  Va(1)  Va(2)  Z f ( I (0)
a  I a  I a )  3 Zf I a
(23)
Using eqns. (16) to (18) in the above
(1)
(2)
(1)
 I (0)
i.e.
a Z 0  Ea n  I a Z1  I a Z 2  3 Z f I a
(1)
(1)
(1)
 I (1)
i.e.
a Z 0  Ea n  I a Z1  I a Z 2  3 Z f I a
I (1)
a 
Ea n
Z1  Z 2  Z 0  3 Z f
(24)
26
I
(1)
a

Ea n
(24)
Z1  Z 2  Z 0  3 Z f
Then the sequence networks are to be connected as shown in Fig. 10.
Z1
+
Va(1)
_
I (1)
a
+
Ea n
_
Z2
I
(2)
a
I
(0)
a
+
Va(2)
_
3 Zf
Z0
+
Va(0)
_
Fig. 10
27
Line to line fault
The circuit diagram is shown in Fig. 11
Ia
a
+
Zn
Ea n
_
Eb n
c
+
+
Ib
Ec n
b
Ic
Zf
Fig. 11
The fault conditions are
Ia  0
Ib  Ic  0
Vb  Z f I b  Vc
(25)
(26)
28
(27)
Ia  0
Ib  Ic  0
Vb  Z f I b  Vc
(25)
(26)
(27)
Then I (0)
a  1/3 ( I a  I b  I c )  0
(28)
2
2
I (1)
a  1/3 (I a  a I b  a I c )  I b /3 ( a  a )
2
2
I (2)
a  1/3 ( I a  a I b  a I c )  I b /3 ( a  a )
I (0)
Va(0) = - Z0 I (0)
a = 0 ,
a = 0
Since
(1)
Further
I (2)
a   Ia
From eqn. (27)
(29)
(30)
(2)
(0)
Va(0)  a 2 Va(1)  a Va(2)  Z f ( a 2 I (1)
 a Va(1)  a 2 Va(2)
a  a I a )  Va
2
(2)
( a 2  a ) Va(1)  Z f ( a 2  a ) I (1)
a  ( a  a) Va
(2)
Thus Va(1)  Z f I (1)
a  Va
(31)
(1)
(2)
(1)
From the above eqn. Ea n  Z1 I (1)
a  Zf I a   Z 2 I a  Z 2 I a
Ea n  ( Z1  Z 2  Z f ) I (1)
a
i.e.
Therefore
I (1)
a 
Ea n
Z1  Z 2  Z f
(32) 29
Therefore
I (1)
a 
Ea n
(32)
Z1  Z 2  Z f
(1)
I (2)
a = - Ia
and I (0)
a = 0;
Va(0) = 0
Sequence networks are to be connected as shown in Fig. 12.
Zf
Z1
+
Ea n
_
Z2
I (1)
a
Z0
I (2)
a
Va(1)
Va(2)
I (0)
a
Va
(0)
=0
Fig. 12
30
Double line to ground fault
The circuit diagram is shown in Fig. 13.
a
Ia
+
E
_ an
Eb n
Zn
c
+
+
Ib
Ec n
b
Ic
Zf
Fig. 13
The fault conditions are
Ia  0 ;
Vb  Z f ( I b  I c ) and
Vc  Z f ( I b  I c )
(33)
31
The fault conditions are
Ia  0 ;
Vb  Z f ( I b  I c ) and
Vc  Z f ( I b  I c )
(33)
(0)
Because of I (0)
,

1/3
(
I

I

I
)
I

I

3
I
a
a
b
c
b
c
a
Therefore
Vb  3 Z f I (0)
a
(34)
Vc  3 Z f I (0)
a
(35)
Va(1)  1/3 ( Va  a Vb  a 2 Vc )  1/3 [ Va  ( a  a 2 ) Vb ]
Va(2)  1/3 ( Va  a 2 Vb  a Vc )  1/3 [ Va  ( a 2  a ) Vb ]
Therefore
Va(1)  Va(2)
(36)
32
Further Va(0)  1/3 ( Va  Vb  Vc ) i.e.
(0)
i.e. 2 Va(0)  2 Va(1)  6 Z f I (0)
3 Va(0)  Va(0)  Va(1)  Va(2)  3 Z f I (0)
a  3 Zf I a
a
(0)
(0)
i.e. Va(1)  Va(0)  3 Z f I (0)
  Z 0 I (0)
a
a  3 Zf I a   ( Z0  3 Zf ) I a
i.e.
Va(1)   ( Z 0  3 Z f ) I (0)
a
(37)
(1)
(2)
From eqn. (33) I (0)
i.e.
a  Ia  Ia  0
Va(1)
Va(2)
(1)

 Ia 
 0
Z 0  3Z f
Z2
i.e.
Va(1)
Va(1)
(1)

 Ia 
 0
Z 0  3Z f
Z2
Z 2  Z 0  3Z f
1
1

)  Va(1)
Z 0  3Z f Z 2
Z 2 ( Z 0  3Z f )
Z 2 ( Z 0  3 Z f ) (1)

Ia
Z 2  Z0  3 Zf
Z 2 ( Z 0  3 Z f ) (1)
 I (1)
Z

Ia
a
1
Z 2  Z0  3 Zf
Ea n

Z ( Z0  3 Zf )
Z1  2
Z 2  Z0  3 Zf
Therefore I (1)
 Va(1) (
a
i.e.
Va(1)
i.e.
Ea n
Thus
I (1)
a
(38)
(39)
33
Thus
I
(1)
a
Ea n

Z ( Z0  3 Zf )
Z1  2
Z 2  Z0  3 Zf
From eqn. (38)

Va(1)  Va(2)   Z 2 I (2)
a
(39)
Z 2 ( Z 0  3 Z f ) (1)
Ia
Z 2  Z0  3 Zf
Z0  3 Zf
(40)
Z2  Z0  3 Zf
Again substituting eqn. (37) in eqn. (38)
Z 2 ( Z 0  3 Z f ) (1)
Z2
(0)
(1)

I
I


I
Thus
(41)
 ( Z 0  3 Z f ) I (0)
a
a
a
a
Z 2  Z0  3 Zf
Z 2  Z0  3 Zf
For this fault, the sequence networks are to be connected as shown in Fig. 14.
(1)
I (2)
a   Ia
Therefore
Z2
Z1
Ea n
+
_
I (1)
a
Va(1)
Z0
I (2)
a
(2)
a
V
I (0)
a
Va(0)
3 Zf
Fig. 14
34
3 SUMMARY OF UNSYMMETRICAL FAULTS AT THE GENERATOR
TERMINALS
For any unsymmetrical fault
Va(1) = E a - Z 1 I (1)
a
I a = I (0)
a +
+
I (1)
a
I (2)
a
Va(2) = - Z 2 I (2)
a
2 (1)
+ a I (2)
I b = I (0)
a + a Ia
a
Va(0) = - Z 0 I (0)
a
+ a I (1)
+ a 2 I (2)
I c = I (0)
a
a
a
Single line to ground fault
Z1
Ia
I (1)
a
+
Va(1)
_
Ea n
Ib
Zf
Z2
Ic
Fault conditions are:
Ib = 0
Ic = 0
Va = Z f I a
I
(2)
a
Z0
I (0)
a
+
Va(2)
_
3 Zf
+
Va(0)
_
35
I
(1)
a

Ea n
Z1  Z 2  Z0  3 Zf
;
Corresponding phase components are
(1)
I (2)

I
a
a ;
(1)
I (0)

I
a
a
I a , I b and I c
(1)
I

I

3
I
Fault current f
a
a
Va(1) = E a - Z 1 I (1)
a
Va(2) = - Z 2 I (2)
a
Va(0) = - Z 0 I (0)
a
Corresponding phase components are
Va , Vb and Vc
36
Line to line fault
Fault conditions are:
Ia
Ia = 0
Ib
Ic = - Ib
Zf
Vb - Z f I b = Vc
Ic
Zf
Z1
Ea n
Z2
I
(1)
a
Z0
I (2)
a
Va(1)
I (0)
a
Va(2)
37
I
(1)
a

Ea n
Z1  Z2  Zf
;
(1)
I (2)


I
a
a ;
I (o)
a  0
Corresponding phase components are I a , I b and I c
Fault current
(0)
(2)
(1)
(1)
2 (1)
2
I f  I b  I a  a I a  a I a  (a  a) I a   j 3 I a
Va(1) = E a - Z 1 I (1)
a
Va(2) = - Z 2 I (2)
a
Va(0) = - Z 0 I (0)
a
Corresponding phase components are
Va , Vb and Vc
38
Double line to ground fault
Fault conditions are:
Ia
Ia  0
Ib
Vb  ( I b  I c ) Z f
Zf
Ic
Z1
Ea n
+
-
Vc  ( I b  I c ) Z f
Z2
I (1)
a
Va(1)
Z0
I (2)
a
(2)
a
V
I (0)
a
Va(0)
3 Zf
39
I
(1)
a
I
(0)
a
Ea n

Z 2 ( Z0  3 Zf )
Z1 
Z 2  Z0  3 Zf
I
(1)
a
Z2
Z 2  Z0  3 Zf
(1)
I (2)


I
a
a
Z 0  3 Zf
Z 2  Z0  3 Zf
(0)
Fault current I f  I b  I c  3 I a
Corresponding phase components are I a , I b and I c
Va(1) = E a - Z 1 I (1)
a
Va(2) = - Z 2 I (2)
a
Va(0) = - Z 0 I (0)
a
Corresponding phase components are
Va , Vb and Vc
40
Example 2
The reactances of an alternator rated 10 MVA, 6.9 kV are
X 1 = X 2 = 15 % and X g0 = 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
= 0.15 p.u.
10
= 0.0798 p.u.
X n = 0.38 x
6.9 2
X1
=
X2
X 0 = X g0 +3 X n
= 0.05 + 0.2394 = 0.2894 p.u.
(2)
I (1)
a = Ia
= 1.0 / j ( 0.2894 + 0.15 + 0.15 ) = - j 1.6966 p.u.
=
I (0)
a
Corresponding phase components are
I a = -j 5.0898 p.u.
Ib
=
Ic
=0
41
10 x 1000
Base current =
3 x 6.9
Line currents are
Fault current,
Ia
=
If
= 836.7 A
= - j 4258.8 A ;
Ia
Ib
=
Ic
=0
= - j 4258.8 A
= 1.0 – ( j 0.15 ) (- j 1.6966 ) = 1.0 – 0.2545 = 0.7455 p.u.
Va(2) = - ( j 0.15 ) (- j 1.6966 ) = - 0.2545 p.u.
Va(0) = - ( j 0.2894 ) (- j 1.6966 ) = - 0.491 p.u.
Va(1)
Corresponding phase components are
0
0
Va = 0 ; Vb = 1.1386   130.38 p.u. ; Vc = 1.1386 130.38 p.u.
Multiplying by
Va
= 0;
Vb
6.9
3
= 4.5359   130.38 0 kV ;
Vc
= 4.5359 130.38 0 kV
42
Example 3
The reactances of an alternator rated 10 MVA, 6.9 kV are
X 1 =15 %; X 2 = 20 % and X g0 = 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
X1
= 0.15 p.u. ;
I (1)
a =
I (2)
a
X2
= 0.2 p.u. ;
XF
= 0.15 p.u.
X0
=?
1.0 / j ( 0.15 + 0.2 + 0.15 ) = - j 2 p.u.
=-
I (1)
a
= j 2 p.u.
and
I (0)
a
=0
Corresponding phase components are
Ia = 0 ;
Ib
= - 3.4641 p.u. ;
Ic
= 3.4641 p.u.
43
Base current = 836.7 A
Line currents are
Ia
=0;
Fault current
Ib
= - 2898.4 A
If
=
Ib
= - 2898.4 A ;
Ic
= 2898.4 A
= 1.0 – ( j 0.15 ) (- j 2 ) = 0.7 p.u.
- ( j 0.3 ) ( j 2 )
= 0.4 p.u.
Va(2) =
Va(0) = 0
Corresponding phase components are
Va(1)
Va
= 1.1 ;
Vb
= 0.6083
Multiplying by
6.9
3
,

 154.72 0 p.u. ; Vc = 0.6083 154.72 0 p.u.
Va
= 4.3821 kV
= 2.4233   154.72 0 kV
154.72 0 kV
Vc = 2.4233
Vb
44
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
11 2
Base impedance =
= 12.1 Ω ;
10
Z1
= j 0.09917 p.u. ;
Z2 Z0
Z2  Z0
Z2
= j 0.07438 p.u. ;
Z1
+
I (1)
a
= 1.0/ j 0.10234 = -j 9.7714 p.u.
Z0
= j 0.00331 p.u.
= j 0.10234 p.u.
I
(2)
a
0.00331
= j 9.7714
= j 0.4163 p.u.
0.07769
I
(0)
a
0.07438
= j 9.7714
= j 9.3551 p.u.
0.07769
45
I (1)
a
= 1.0/ j 0.10234 = -j 9.7714 p.u.
I
(2)
a
0.00331
= j 9.7714
= j 0.4163 p.u.
0.07769
I
(0)
a
0.07438
= j 9.7714
= j 9.3551 p.u.
0.07769
Corresponding phase components are
Ia
=0;
Ib
= 16.5758 122.16 0 p.u. ;
Base current =
10 x 1000
3 x 11
=
Va(2)
=
Va(0)
= 16.5758 57.84 0 p.u.
= 542.86 A
Current in faulted phases are
Va(1)
Ic
Ib
= 8998.3 122.16 0 A
Ic
= 8998.3 57.84 0 A
= - ( j 0.07438 ) ( j 0.4163 ) = 0.03096 p.u.
Voltage of the healthy phase
Va
= 0.09288
x
11
3
= 0.5899 kV
46
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 Between
Off nominal
Line Impedance
HLCA
No.
buses
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
1
1
YBus =
2
3
 0.5583  j 2.575
 0.5583  j 2.582


0
2
3
 0.5583  j 2.582
0

0.5583  j 10.0938 j 8.2716
j 8.2716
 j 9.09 
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
1
2
3


 

 2  j 6 2.7  j 8  0.7  j 2
2


3   1  j 3  0.7  j 2 1.7  j 5 
1
The slack bus voltage is 1.04 0 0 . 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.
2.
V1 = 1.04 0 0 ;
V2 = 1.0 0 0 ;
PI2 = 0.7 – 0.2 = 0.5;
V2(1) =
V3 = 1.0 0 0
QI2 = 0.1 – 0.3 = - 0.2; PI2 + j QI2 = 0.5 – j 0.2
1 PI2  QI 2
[
 Y2 1 V1  Y2 3 V3 ]
(0) *
Y22
V2
=
1
[ (0.5  j 0.2)  (2  j 6) (1.04)  (0.7  j 2)
2.7  j 8
=
(0.5  j 0.2)  (2.08  6.24)  (0.7  j 2)
3.28  j 8.04
=
2.7  j 8
2.7  j 8
= 1.0265 + j 0.0636 = 1.0284 3.54 0
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 E a n
Positive sequence pre-fault voltage is Vf
Positive sequence impedance is Z 1
Thevenin’s
equivalent
impedance
between the fault point and the
reference bus in the positive sequence
network is Z 1
Negative sequence impedance is Z 2
Thevenin’s
equivalent
impedance
between the fault point and the
reference
bus
in
the
negative
sequence network is Z 2
Zero sequence impedance is Z 0
Thevenin’s
equivalent
impedance
between the fault point and the
reference bus in the zero sequence
network is Z 0
51
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.
52
5 UNSYMMETRICAL FAULT ANALYSIS USING Z bus 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.
q
p
z
a
p
V
a,b,c
pq
b
p
V
c
q
c
p
V
V
b
q
V
Vqa
Fig. 15
It can be described as
v a,p qb,c  z a,p qb,c i a,p qb,c i.e.
 v ap q 
 b 
v p q  =
 v cp q 


z ap aq
 ba
z p q
z cpaq

z ap bq
z bp bq
z cpbq
z ap cq 

z bp cq 
z cpcq 
(42)
i ap q 
b 
i p q 
i cp q 
 
(43)
53
Voltages at bus p and q can be denoted as
 Vpa 
 Vqa 
 
 
(44)
Vpa,b,c  Vpb  ;
Vqa,b,c  Vqb 
 Vpc 
 Vqc 
 
 
Considering the impedance of each three phase element as z a,p qb,c , using building
algorithm, the bus impedance matrix of transmission-generator can be obtained
as
1
2
N
Z
Z

Z
Z
Z a,busb,c 
 

 a,b,c
Z a,N2b,c
N Z N1
The bus impedance matrix
 Z
Since
element
1
2
impedance
a,b,c
11
a,b,c
21
of
a,b,c
12
a,b,c
22
any


 Z 
where Z a,i jb,c  i
 
a,b,c 
 Z NN 
Z a,busb,c will be normally full
a,b,c
1N
a,b,c
2N
in
sequence
j
 Z ai ja
 ba
Z i j
 Z ci ja

Z ai jb
Z bi jb
Z ci jb
Z ai jc 

Z bi jc 
Z ci jc 
with non-zero entries.
frame, z p0,1,2
q
is
decoupled,
0,1,2
computationally it is advantage to use the matrix Z bus
instead of Z a,busb,c .
54
For two bus system
1
(0)
1 Z 11
 (0)
2 Z 21
0
Z bus

2
(0)

Z 12
 ;
Z (0)
22 
Z 1bus
1
(1)
1 Z 11

 (1)
2 Z 21
2
(1)

1
Z 12
2
;
Z


bus
Z (1)
2
22 
1
2
(2)
(2)
Z 11

Z 12
 (2)
(2) 
Z 21 Z 22 
Then
1
1
0,1,2
Z bus

0
(0)
0 Z 11
1
1
2
2
(1)
Z 11
2
2
(1)
Z 12
0 Z (0)
21
1
1
(2)
Z 11
2
2
0
(0)
Z 12
(2)
Z 12
(45)
Z (0)
22
Z (1)
21
Z (1)
22
Z (2)
21
0
2
, Z1bus and Z bus
Normally Z bus
Z (2)
22
are constructed and stored independently. It is
0,1,2
evident that as compared to Z a,busb,c , construction of Z bus
requires less computer
time and less core storage. For a 100 bus system, Z a,busb,c will be a 300 x 300 full
0,1,2
matrix; whereas for Z bus
, we need 3 numbers of 100 x 100 matrices. Thus only
0,1,2
1/3 rd of the core storage is required for Z bus
as compared to Z a,busb,c . Hence for
0,1,2
unsymmetrical fault analysis, use of Z bus
is more advantages than Z a,busb,c .
55
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.
0,1,2
In order to obtain Z bus
, 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.
56
Example 5
Consider the system described in Example 1. Obtain the matrices
0
2
.
Z bus
, Z1bus and Z bus
Solution
Required bus impedance matrices can be constructed using bus impedance
building algorithm.
First consider the zero sequence network shown in Fig. 8(a).
1
j0.08
2
j0.52
j0.05
3
j0.09
4
j0.06
j0.12
j0.66
j0.96
0
Reference
57
Element 0-1 is added:
Element 0 – 2 is added:
Element 0 – 3 is added:
0
Z bus
 j 1 1.01
0
Z bus
 j
0
Z bus
 j
1
2
1
2
3
1
2
0 
1.01
 0
0.08

1
2
3
0
0 
1.01
 0

0.08
0


 0
0
0.09
Element 2 – 3 is added: With  th bus
1
Z
0
bus
 j
2
3


1
2
3
0
0
0 
1.01
 0

0.08
0
0.08


 0
0
0.09  0.09


0.08  0.09 0.69 
 0
58
0
Eliminating the  th bus, Z bus
 j
1
2
3
1
2
3
0
0 
1.01
 0

0.07
0.01


 0
0.01 0.08
0
Element 0 – 4 is added: The final Z bus
is obtained as
1
Z
0
bus
 j
2
3
4
1
2
3
4
0
0
0 
1.01
 0

0.07
0.01
0


 0
0.01 0.08
0 


0
0
0
0.72


Consider the positive sequence network shown in Fig. 8(b)
59
j0.08
1
j0.2
Eg
j0.09
j0.18
2
3
4
+
j0.55
j0.27
Em1
+ +
Em 2
Reference
1 0.2
1
Element 0 – 1 is added: Z bus
 j
1
Element 1 – 2 is added: Z bus
 j
1
Element 2 – 3 is added: Z bus
 j
1
2
1
2
3
1
2
0.2 0.2 
0.2 0.28


1
2
3
0.2 0.2 0.2 
0.2 0.28 0.28


0.2 0.28 0.46
60
1
Element 3 – 4 is added: Z
1
bus
 j
2
3
4
Element 0 – 4 is added:
1
1
Z bus

j
2
3
4

1
2
3
4
0.2 0.2 0.2 0.2 
0.2 0.28 0.28 0.28


0.2 0.28 0.46 0.46


0.2
0.28
0.46
0.55


It has an impedance of j0.18. With the  th bus
1
2
3
4

0.2
0.2
0.2
 0.2 
 0.2
 0.2

0.28
0.28
0.28

0.28


 0.2
0.28
0.46
0.46  0.46


0.2
0.28
0.46
0.55

0.55


 0.2  0.28  0.46  0.55 0.73 
Eliminating the  th bus, final
1
2
1 0.1452 0.1233
2 0.1233 0.1726
1

Z bus = j
3 0.0740 0.1036

4 0.0493 0.0690
1
is obtained as
Z bus
3
4
0.0740 0.0493
0.1036 0.0690
0.1701 0.1134

0.1134 0.1356
61
Similarly, considering the negative sequence network shown in Fig. 8(c)
1
j0.08
2
j0.18
3
j0.09
4
j0.25
j0.27
j0.55
Reference
0
2
Its bus impedance Z bus
can be obtained as
1
2
3
4
1 0.1699 0.1442 0.0866 0.0577 
2 0.1442 0.1904 0.1143 0.0761
2


Z bus = j
3 0.0866 0.1143 0.1765 0.1177 


4 0.0577 0.0761 0.1177 0.1384
62
0,1,2
6 UNSYMMETRICAL FAULT ANALYSIS USING Z bus
MATRIX
0,1,2
1
For unsymmetrical fault analysis using Z bus
the first step is to construct Z bus
,
2
0
and Z bus
by considering the positive sequence, negative sequence and zero
Z bus
sequence network of the power system. They are
1
2
1
Z bus

p
N
1
(1)
 Z 11
 (1)
 Z 21
 
 (1)
 Z p1
 
 (1)
Z N1
2
p
(1)
(1)
Z 12
 Z 1p


Z (1)
 Z (1)
22
2p


Z (1)
 Z (1)

p2
pp


(1)
Z (1)

Z

N2
Np
N
(1)

Z 1N

Z (1)
2N 
 
(1) 
Z pN 
 

(1)
Z N N 
(46)
63
2
Z bus
0
Z bus
1
(2)
1  Z 11
 (2)
2  Z 21
 


p  Z (2)
p1
 
 (2)
N Z N1
1
(0)
1  Z 11
 (0)
2  Z 21
 


p  Z (0)
p1
 
 (0)
N Z N1
2
p
(2)
(2)
Z 12
 Z 1p


Z (2)
 Z (2)
22
2p


Z (2)
 Z (2)

p2
pp


Z (2)
 Z (2)

N2
Np
2
p
(0)
(0)
Z 12
 Z 1p


Z (0)
 Z (0)
22
2p


Z (0)
 Z (0)

p2
pp


Z (0)
 Z (0)

N2
Np
N
(2)

Z 1N

Z (2)
2N 
 
and
(2) 
Z pN 
 

(2)
Z N N 
(47)
N
(0)

Z 1N

Z (0)
2N 
 

Z (0)
pN 
 

(0)
Z N N 
(48)
64
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
Vf .
The currents flowing out of the original balanced system from phases a, b and c
at the fault point are designated as I f a , I f b and I f c . We can visualize these
currents by referring to Fig. 16 which shows the three lines a, b and c of the
three phase system where the fault occurs.
P
a
If a
b
If b
c
Fig. 16
If c
65
The
currents
flowing
out
in
hypothetical
stub
are
I f a , I f b and I f c .
The
(1)
(2)
corresponding sequence currents are I (0)
f a , I f a and I f a . These sequence currents
(2)
(0)
I (1)
f a , I f a and I f a are flowing out as shown in Fig. 17. The line to ground voltages
at any bus j of the system during the fault are Vj a , Vj b and Vj c . Corresponding
(1)
(2)
sequence components of voltages are Vj(0)
,
and
V
V
a
ja
ja .
1
- Vf +
- Vf +

- Vf +
Positive
sequence
network
having
bus
impedance
matrix
2
p
I (1)
fa
N
Z (1)
bus
66
1
1
Negative
sequence
network
having
bus
impedance
matrix
Zero
sequence
network
having
bus
impedance
matrix
2
p
N
I (2)
fa
Z (2)
bus
Z (0)
bus
2
p
N
I (0)
fa
Fig. 17
67
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
2
(1)
Vbus

p
N
 V1a(1) 
 (1) 
 V2a 
  
 (1) 
 Vp a 
  
 (1) 
VN a 
(49)
can be obtained using Superposition Theorem.
68
It is to be noted that
1
2
I (1)
bus 
p
N
 0 
 0 


  


0


(1)
 I f a 


0


  


 0 
and
pre-fault voltage =
1
1
 

 
1 V
1 f
 
1

 
1
(50)
Using these we get
(1)
 V1a(1) 
 Z 11
 (1) 
 (1)
V
 2a 
 Z 21
  
 
 (1)  =  (1)
 Vp a 
 Z p1
  
 
 (1) 
 (1)
VN a 
Z N1
(1)
Z 12
Z (1)
22

Z (1)
p2

Z (1)
N2
(1)
 Z 1p
 Z (1)
2p

 Z (1)
pp

 Z (1)
Np
(1)

 Z 1N

 Z (1)
2N 
 

 Z (1)
pN 
 

(1)
 Z N N 
 Vf
 0 
1

 0 
1
 Vf


 

  

 (1)  +   Vf = 
 Vf
 I fa 
1

  




 
 0 
1
Vf
(1) (1)
 Z 1p
If a 
(1) 
 Z (1)
I
2p f a 


(1) 
 Z (1)
I
pp f a 



(1)
(1)
 Z N p I f a 
69
(51)
Consider the Negative Sequence and Zero Sequence Networks:
In a much similar manner, the negative sequence and the zero sequence bus
(2)
(0)
voltages in the faulted system, namely Vbus
and Vbus
, 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) (2)
 V1a(2) 
  Z 1p
If a 
 (2) 

(2) (2) 
V

Z
2p I f a 
 2a 

  



 (2)  = 
(2) (2) 
V

Z
 pa 

pp If a 
  



 (2) 


(2)
(2)
VN a 
 Z N p I f a 
and
 V1a(0) 
 (0) 
 V2a 
  
 (0)  =
 Vp a 
  
 (0) 
VN a 
(0) (0)
  Z 1p
If a 

(0) (0) 

Z
2p I f a 





(0) (0) 

Z

pp If a 





(0)
(0)
 Z N p I f a 
(52)
70
When the fault occurs at bus p , it is to be noted that only the p th column of
1
2
0
, Z bus
and Z bus
are involved in the calculations. If the symmetrical
Z bus
(2)
(0)
I (1)
f a , I f a and I f a , are known, than the
sequence voltages at any bus j can be computed from
components of the fault currents , namely
(0)
Vj(0)
  Z (0)
a
jp I f a
(53)
(1)
Vj(1)
 Vf  Z (1)
a
jp I f a
(54)
(2)
Vj(2)
  Z (2)
a
jp I f a
(55)
(2)
(0)
It is important to remember that the I (1)
f a , I f a and I f a 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.
71
General procedure for unsymmetrical
fault analysis
when fault occurs at a point in a
power system
72
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
 

2
 I f b  = 1 a
I f c 
1 a
 
1
a 
a 2 
 I f a (0) 
 (1) 
 Ifa 
I (2) 
 fa 
73
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
 Vpa 
1 1


1 a 2
V
=
p
b



V p c 
1 a


1
a 
a 2 
 Vp a (0) 

(1) 
V
 pa 
 V (2) 
 pa 
74
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
 V ja 
1 1


1 a 2
=
V
 jb 

 V jc 
1 a


1
a 
a 2 
 V j a (0) 
 (1) 
 Vj a 
 V (2) 
 ja 
75
Single line to ground fault
I (f 1)a
+
Vf
Z (1)
pp
Vp(1)a
_
I
Z
(2)
(0)
I (1)
f a = If a = If a
(2)
fa
3 Zf
(2)
pp
Vp(2)a
I (f 0)a
Z (0)
pp
Vp(0)a
76
Line to line fault
Zf
Z (2)
pp
Z (1)
pp
+
Vf
_
I (1)
fa
I (2)
fa
Vp(1)a
I (0)
fa
Vp(2)a
77
Double line to ground fault
Z (1)
pp
+
Vf
Z (2)
pp
I (1)
fa
Vp(1)a
Z (0)
pp
I (2)
fa
Vp(2)a
I (0)
fa
Vp(0)a
_
3 Zf
78
SINGLE LINE TO GROUND FAULT
For a single line to ground fault through impedance Z f , the hypothetical stubs
on the three lines will be as shown in Fig. 18 The fault conditions are
P
a
Zf
If a
b
If b
c
If c
Fig. 18
If b  0
If c  0
(56)
Vp a  Z f I f a
79
Using the above conditions ( similar to conditions for the LG fault at generator
terminals through impedance ) and also knowing that
(1)
Vp(1)a  Vf  Z (1)
pp If a
(57)
(2)
Vp(2)a   Z (2)
pp If a
(58)
(0)
Vp(0)a   Z (0)
pp If a
(59)
from eqns. (51) and (52), similar to eqns. (22) and (23), we can get the relations
I (1)
= I (2)
= I (0)
and
fa
fa
fa
(60)
Vp(1)a  Vp(2)a  Vp(0)a  3 Z f I (1)
fa = 0
(61)
Therefore
I (1)
fa 
Z
(1)
pp
Z
(2)
pp
Vf
 Z (0)
p p  3 Zf
(62)
The above relationships are satisfied by connecting the sequence networks as
shown in Fig. 19
80
I (f 1)a
+
Vf
Z (1)
pp
Vp(1)a
_
I
Z
(2)
(0)
I (1)
f a = If a = If a
(2)
fa
3 Zf
(2)
pp
Vp(2)a
I (f 2)a
Z (0)
pp
Vp(0)a
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.
81
(2)
(0)
Once the currents I (1)
,
and
I
I
fa
fa
f a are known, the sequence components of
voltage at the faulted bus are calculated as
(1)
Vp(1)a  Vf  Z (1)
pp If a
(2)
Vp(2)a   Z (2)
pp If a
(63)
(0)
Vp(0)a   Z (0)
pp If a
Thereafter the sequence components of voltage at any bus j can be calculated as
(1)
(1)
Vj(1)
a  Vf  Z j p I f a
(2)
(2)
Vj(2)
a   Z jp I f a
j  1,2,....... N
(0)
(0)
Vj(0)
a   Z jp I f a
jp
(64)
Phase components of voltage and current can be calculated from the relations
Va , b , c  A V0 , 1 , 2
I a , b , c  A I 0 , 1, 2
82
Example 6
The positive sequence, negative sequence and zero sequence bus impedance
matrices of a power system are shown below.
(2)
j
Z (1)
bus  Z bus 
1
2
3
4
Z (0)
j
bus 
1
2
3
4
1
0.1437
0.1211

0.0789

0.0563
1
0.1553
0.1407

0.0493

0.0347
2
3
4
0.1211 0.0789 0.0563
0.1696 0.1104 0.0789
0.1104 0.1696 0.1211

0.0789 0.1211 0.1437 
2
3
4
0.1407 0.0493 0.0347
0.1999 0.0701 0.0493
0.0701 0.1999 0.1407

0.0493 0.1407 0.1553
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.
83
Solution
(2)
(0)
I (1)
I
I
=
=
fa
fa
fa =
Z (1)
33
Vf
(0)
 Z (2)
33  Z 33
Let Vf  1.0 0 0
Then I
(1)
fa
= I
(2)
fa
= I
(0)
fa
1.00 0
=
=  j1.8549
j (0.1696  0.1696  0.1999 )
(1)
(2)
(1)
The fault current I f a  I (0)
f a  I f a  I f a  3 I f a = - j 5.5648; I f b = I f c = 0
The sequence components of voltage at bus 3 are calculated as
(0)
V3(0)a   Z (0)
I
33
f a   ( j 0.1999 ) (  j1.8549 )   0.3708
(1)
V3(1)a  Vf  Z (1)
33 I f a  1.0  ( j 0.1696 ) (  j1.8549 )  0.6854
(2)
V3(2)a   Z (2)
33 I f a   ( j0.1696 ) (  j1.8549 )   0.3146
84
Phase components of line to ground voltage of bus 3 are computed as
 V3 a 
1 1


1 a 2
=
V
3
b



 V3 c 
1 a


1
a 
a 2 
 0.3708
 0.6854  =


 0.3146
0
1.0292   122.710
1.0187 122.710
The sequence components of voltage at bus 4 are calculated as
(0)
V4(0)a   Z (0)
I
43
f a   ( j 0.1407 ) (  j1.8549 )   0.2610
(1)
V4(1)a  Vf  Z (1)
43 I f a  1.0  ( j 0.1211 ) (  j1.8549 )  0.7754
(2)
V4(2)a   Z (2)
43 I f a   ( j0.1211 ) (  j1.8549 )   0.2246
Phase components of line to ground voltage of bus 4 are computed as
 V4 a 
1 1


1 a 2
V
=
4
b



 V4 c 
1 a


1
a 
a 2 
 0.2610
 0.7754  =


 0.2246
0.2898 00
1.0187   121.8 0
1.0187 121.8 0
85
LINE TO LINE FAULT
To represent a line to line fault through impedance Z f the hypothetical stubs on
the three lines at the fault are connected as shown in Fig. 20.
P
a
If a
b
If b
Zf
c
(
If c
Fig. 20
Fault conditions are
I fa  0
If b  If c  0
Vp b  Z f I f b  Vp c
(65)
86
Using the above conditions along with the relations
(1)
Vp(1)a  Vf  Z (1)
pp If a
(2)
Vp(2)a   Z (2)
pp If a
(66)
(0)
Vp(0)a   Z (0)
pp If a
we can get the following relations.
I (0)
fa  0
(67)
(2)
I (1)


I
fa
fa
(68)
(2)
Vp(1)a  Z f I (1)
f a  Vp a
(69)
and hence
I (1)
fa 
Z
(1)
pp
Vf
 Z (2)
pp  Zf
(70)
87
To satisfy the above relations the sequence networks are to be connected as
shown in Fig. 21.
Zf
Z
+
Vf
_
Z (0)
pp
Z (2)
pp
(1)
pp
I (1)
fa
Vp(1)a
I (2)
fa
I (0)
fa
Vp(2)a
Fig. 21
(2)
(0)
Once I (1)
are calculated,
f a , I f a and I f a
Vp(1)a , Vp(2)a and Vp(0)a can be computed from
(2)
(0)
eqn. (63). Thereafter Vj(1)
a , Vj a and Vj a for j  1,2,...., N ; j  p can be calculated
using eqn. (63). The corresponding phase components are then calculated using
the symmetrical component transformation matrix.
88
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 I (0)
fa  0
(2)
I (1)


I
fa
fa 
Vf
1.0

  j 2.9481
(1)
(2)
j 0.1696  j 0.1696
Z33  Z33
The phase components of the currents in the fault are
(1)
(2)
I f a  I (0)
f a  I fa  I f a  0
(2)
(1)
I f b  a 2 I (1)
f a  a I f a   j 3 I f a  - 5.1061
I f c   I f b  5.1061
Sequence components of voltage at bus 3 are
V3(0)a  0 ;
(2)
V3(1)a  V3(2)a   Z (2)
3 3 I f a   ( j 0.1696 ) (j 2.9481 )  0.5
89
Phase components of voltage at bus 3 are
V3 a  V3(0)a  V3(1)a  V3(2)a  1.0
V3 b  V3 c  a 2 V3(1)a  a V3(2)a   V3(1)a  -0.5
Sequence components of voltage at bus 4 are
(0)
V4(0)a   Z (0)
I
43
fa  0
(1)
(1)
V4a
 Vf  Z (1)
I
43
f a  1  ( j 0.1211 ) (  j 2.9481 )  0.643
(2)
V4(2)a   Z (2)
I
43
f a  ( j 0.1211 ) ( j 2.9481 )  0.357
Phase components of voltage at bus 4 are
V4a  V4(0)a  V4(1)a  V4(2)a  1.0
V4 b  a 2 V4(1)a  a V4(2)a  - 0.5
V4 c  a V4(1)a  a 2 V4(2)a  - 0.5
90
DOUBLE LINE TO GROUND FAULT
For a double line to ground fault, the hypothetical stubs are connected as shown
in Fig. 22.
P
a
If a
b
If b
(
c
If c
Zf
Fig. 22
The relations at the fault bus are
I fa  0
Vp b  Z f ( I f b  I f c )
Vp c  Z f ( I f b  I f c )
(71)
91
Further the relations are also applicable.
(1)
Vp(1)a  Vf  Z (1)
pp If a
(2)
Vp(2)a   Z (2)
pp If a
(72)
(0)
Vp(0)a   Z (0)
pp If a
Using eqns. (71) and (62) the following relations can be obtained.
Vp(1)a  Vp(2)a
Vp(1)a  Vp(0)a  3 Z f I (0)
fa
(1)
(2)
I (0)

I

I
fa
fa
fa  0
On further simplification we get
I (1)
fa 
Z (1)
pp 
Vf
(0)
Z (2)
pp ( Zpp  3 Zf )
(73)
(0)
Z (2)

Z
pp
pp  3 Zf
92
To represent the above relations, the sequence networks must be interconnected
as shown in Fig. 23.
Z (1)
pp
I (1)
fa
+
Vf
Z (2)
pp
Vp(1)a
Z (0)
pp
I (2)
fa
Vp(2)a
I (0)
fa
Vp(0)a
_
3 Zf
(0)
The sequence currents I (2)
f a and I f a
I
(2)
fa
I
(1)
fa
I
(0)
fa
I
(1)
fa
Z (0)
p p  3 Zf
Z
(2)
pp
Z
(0)
pp
 3 Zf
(74)
 3 Zf
(75)
Z (2)
pp
Z
(2)
pp
Z
(0)
pp
Fig. 23
can be obtained from
93
(2)
(0)
Knowing I (1)
,
and
sequence components of voltage at fault point are
I
I
fa
fa
fa
calculated from
(1)
Vp(1)a  Vf  Z (1)
I
pp
fa
(2)
Vp(2)a   Z (2)
pp If a
(76)
(0)
Vp(0)a   Z (0)
pp If a
Thereafter, sequence components of voltage at any other bus can be obtained
from
(1)
(1)
Vj(1)
a  Vf  Z j p I f a
(2)
(2)
Vj(2)
a   Z jp I f a
j  1,2,....... N
(0)
(0)
Vj(0)
a   Z jp I f a
jp
(77)
Knowing the sequence components, corresponding phase conponents are obtained
as
I a,b, c  A I 0,1, 2 or
Va,b, c  A V0,1, 2
(78)
94
Example 8
The positive sequence, negative sequence and zero sequence bus impedance
matrices of a power system are shown below.
(2)
j
Z (1)
bus  Z bus 
1
2
3
4
1
j
Z (0)
bus 
2
3
4
1
0.1437
0.1211

0.0789

0.0563
2
3
4
0.1211 0.0789 0.0563
0.1696 0.1104 0.0789
0.1104 0.1696 0.1211

0.0789 0.1211 0.1437 
1
2
3
4
0
0
0 
0.19
 0

0.08
0.08
0


 0
0.08 0.58
0 


0
0
0
0.19


95
A double line to ground fault with Z f  0 occurs at bus 4. Find the fault
current and voltages at the fault bus.
Solution
Sequence components of fault current are
I (1)
fa 
Z (1)
44
I
(2)
fa
I
(0)
fa
I
Vf
1.0

  j 4.4342
(2)
(0)
(
j
0.1437
)
(
j
0.19
)
Z Z
j 0.1437 
 (2)44 44(0)
j 0.1437  j 0.19
Z 44  Z 44
(1)
fa
I
(1)
fa
Z (0)
j0.19
44


(

j4.4342
)
 j2.5247
(2)
(0)
j0.1437

j0.19
Z 44  Z 44
Z (2)
j0.1437
44


(

j4.4342
)
 j1.9095
(0)
j0.1437

j0.19
Z (2)

Z
44
44
96
Phase components of current at the fault bus are
(1)
(2)
I f a  I (0)
f a  If a  If a  0
2
(1)
(2)
0
0
I f b  I (0)
f a  a I f a  a I f a  j1.9095  4.4342150  2.5247210
= - 6.0266 + j 2.8642
(1)
2
(2)
0
0
I f c  I (0)
f a  a I f a  a I f a  j1.9095  4.434230  2.5247  30
= 6.0266 + j 2.8642
Fault current I f  I f b  I f c  j 5.7285
Sequence components of voltage at the faulted bus are calculated as follows.
Noting that Z f  0
(1)
V4(1)a  V4(2)a  V4(0)a  Vf  Z (1)
44 I f a  1.0  ( j0.1437 ) (  j4.4342 )  0.3628
Phase components of faulted bus voltage are:
V4 a  V4(0)a  V4(1)a  V4(2)a  1.0884
V4 b  0
V4 c  0
97
PROBLEMS – UNSYMMETRICAL FAULTS
1.
In an unbalanced circuit the three line currents are measured as
I a  7.031159.85 0
Ib  4.3733203.840
Ic  2.6810160.750
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
21 6 6 
z
 j  6 21 6 
a,b,c
 6 6 21
Calculate its sequence impedances.
98
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.
99
9.
Two
synchronous
machines
are
connected
through
three-phase
transformers to the transmission line as shown.
T1
1
T2
2
3
1
4
2
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, Zbus1 and Zbus2.
100
10.
The one-line diagram of a power system is shown below.
T1
1
T3
3
2
5
6
3
1
T2
4
2
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, Zbus1 and Zbus2.
101
11.
The single line diagram of a small power system is shown below.
T1
1
2
P
T2
3
S
4
1
Switch open
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.
102
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. Ia0  Ib0  Ic0  21200 ; Ia1  3.5300 ; Ib1  3.52700 ; Ic1  3.51500
Ia2  3600 ;
Ib2  31800 ;
Ic2  33000
2. Ia  7.031159.850 ; Ib  4.3733203.840 ; Ic  2.681160.750
3. Z 0  j 33; Z 1  Z 2  j15
4. Ia = -j 3586.1 A
Va = 0
Ib = 0
Ic = 0
If = -j 3586.1 A
Vb = 8.0694   102.220 kV
Vc = 8.0694 102.220 kV
Vab = 8.0694  77.780 kV; Vbc= 15.7724   900 kV; Vca= 8.0694 102.220 kV
5. Ia = 0
Ib = -2415.5 A
Va = 9.2948 kV
Vab = 13.9422 kV
Ic = 2415.5 A
Vb = -4.6474 kV
Vbc = 0
If = -2415.5 A
Vc = -4.6474 kV
Vca = 13.9422 1800 kV
103
6. Ia = 0
Ib = 4020.95 132.220 A
Va = 5.6720 kV
Vb = 0
Vab = 5.6720 kV
If = 5956.08 900 A
Vc = 0
Vbc = 0
7. Ia = 3347   900 A
Ic = 4020 47.780 A
Vca = 5.6720 1800 kV
Ib = 3347 1500 A
Ic = 3347 300 A
If = 3347   900 A
Va = Vb = Vc = 0
Vab = Vbc = Vca = 0
8.
0.9655 Ω
9. Zero sequence network:
1
j0.08 2
j0.5
3
j0.04
j0.08
4
j0.04
j0.15
j0.15
104
Reference
Positive sequence network:
1
j0.08
2
j0.15
3
j0.08
4
j0.2
j0.2
+
+
-
-
Reference
Negative sequence network:
1
j0.08
2
j0.15
3
j0.2
j0.08
4
j0.2
Reference
105
1 0.19
2  0

= j
3  0

4  0
Zbus0
0 
0 
0.08 0.58
0 

0
0
0.19
0
0
0.08 0.08
1
1
Zbus1 = Zbus2 = j
2
3
4
10.
2
0.1437
 0.1211

0.0789

0.0563
3
4
0.1211 0.0789 0.0563
0.1696 0.1104 0.0789
0.1104 0.1696 0.1211

0.0789 0.1211 0.1437
Zero sequence network:
1
j0.12 2
j0.5671
1
j0.4764
3
j0.075
5
j0.1276
6
j0.24
4
j0.072
j0.15
j0.225
j0.45
106
Reference
2
Zbus0 =
j
3
4
5
0
0
0
0
0.3

 0 0.104468 0.031065 0 0.031065


 0 0.031065 0.177871 0 0.177871


0
0
0
0.6
0


 0 0.031065 0.177871 0 0.654271
Negative sequence network:
1
j0.12 2
j0.18
j0.12
3
5
j0.1276
6
j0.24
4
j0.25
j0.15
j0.2
Reference
107
Zbus1 = Zbus2 =
1
1
2
j
3
4
5
0.167640
0.128108
0.068808

0.031276
 0.048041
0.025959

2
3
4
5
6
0.128108 0.068808 0.031276 0.048041 0.025959
0.189500 0.101836 0.046289 0.071101 0.038419

0.101836 0.151377 0.068808 0.105690 0.057109

0.046289 0.068808 0.140367 0.048041 0.025959
0.071101 0.105690 0.048041 0.157574 0.085145
0.038419 0.057109 0.025959 0.085145 0.114956
6
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.
108