chapter 4 sequence networks and unsymmetrical faults …
TRANSCRIPT
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CHAPTER 4
SEQUENCE NETWORKS AND
UNSYMMETRICAL FAULTS ANALYSIS
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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
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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.
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Sequence voltages and currents
According to symmetrical components method, a three phase unbalanced system of
voltages or currents may be represented by three separate system of balanced
voltages or currents known as zero sequence, positive sequence and negative
sequence as shown in Fig. 1
I a
I a(2)
(0) (1) I (1)
Ia I c a
I (0)
I b(2)
= b + + I c (0)
Ic
I (1) I c(2)
b I b
Fig. 1
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Defining operator ‘ a ‘ as
a = 1 1200
(1)
it is to be noted that
a2 1240
0 ; a
3 1360
0 1 (2)
Also a = - 0.5 + j 0.866 ; a2 0.5 j0.866 (3)
Hence 1 a a2
0 (4)
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I a (0) (1) I (1) I a(2)
I c a
I
a
I (0)
I b(2)
I c =
(0) b + +
Ic
I (1) I c(2)
b I b
Further referring Fig. 1
I(1) a2 I (1)
b a
I(1) a I(1) c a
(5) I(2) a I(2)
b a
I(2) a2 I (2)
c a
Therefore I
a I(0) I(1) I(2)
a a a
I b
I(0) I(1) I(2) I(0) a2 I(1) a I(2)
b b b a a a
I c
I(0) I(1) I(2) I (0) a I(1) a2 I(2)
c c c a a a
Thus
I a I(0) I(1) I(2)
a a a
I b
I(0) a2
I(1) a I(2) a a a
I c
I(0) a I (1) a2 I(2) a a a
(6)
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I a 1 1 1 I a(0 )
i.e.
=
a 2
a (1 )
I b
1
I a
I 1 a a 2 I (2 )
c a
The inverse form of the above is
I a(0 )
1
1 1 1 I a (1 ) 2 I
a =
1
a a
I
b 3
I (2 ) 1 a 2
a I
a c
i.e.
Ia , b , c A I
0 , 1, 2
i.e. I0 , 1, 2 A1
Ia , b , c
(7)
(8)
Similarly, corresponding to voltage phasors
Va , b , c
A V0 , 1, 2
and V0 , 1, 2 A1
Va , b , c
(9)
(10)
Matrix A is known as symmetrical component transformation matrix. Similar expressions can be written for line to line voltages and phase currents also.
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Sequence impedances and sequence networks
The impedance of any three phase element is of the form
z aa
z ab
z ac
z a,b,c
= z bb
z
ba z
bc (11) z
ca
z cb
z cc
Then
va,b,c
za,b,c
ia,b,c
i.e. A v
0,1,2
za,b,c
A i0,1,2
v0,1,2 A1
za,b,c A i0,1,2 v
0.1,2
z0,1,2
i0,1,2
where z 0,1,2 A1
za,b,c A
Thus for any three phase element having the impedance z a,b,c the corresponding
sequence impedance z 0,1,2 can be obtained from
z 0,1,2 = A1
za,b,c A (12)
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For power system components, sequence impedance z 0,1,2 will be decoupled as
z (0) 0 0 z
0,1,2
= 0 z (1) 0
(13)
0 0 z (2)
For static loads and transformers z(0) z(1) z
(2) .
For transmission lines z(1) z(2) and z (0) > z (1) .
For rotating machines z(0) , z(1) and z (2) will have different values.
The single phase equivalent circuit composed of sequence voltages, sequence
currents and impedance to current of any one sequence is called the sequence
network for that particular sequence. The sequence network includes any
generated emf of like sequence.
Consider a star connected generator with its neutral grounded through an
impedance Zn as shown in Fig. 2. Assume that the generator is designed to generate
balanced voltage.
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a
I a
Zn +
I n
Ec n
Ea n
+
+
Eb n I b
c
b
I c
Fig. 2
Let Ean
Ea Eb
= E
c
be its generated voltage in phase a . Then Ec
1 Ea
2 Ean This gives
Eb a
a
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E(0) 1 1 1 1 0
(1)a
=
1 2 2
Ean =
E
a
1 a a
a
E
a n 3
E (2) 1 a2 a a 0
a
(14)
This shows that there is no zero sequence and negative sequence generated voltages.
The sequence networks of the generator are shown in Fig. 3.
I ( 1 )
a a
Ia( 1 )
Zn
Z1
+
E
Z1 ( 1 )
In 0
a n
Ec n
Va
+
+
Z1
Z1
Eb n
( 1 )
E a n
+
Ib
__
b
c
Ic( 1 )
Reference bus ( Neutral )
Note that In = 0 Positive sequence network
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a
Ia( 2 )
Zn Z 2
In 0
Z 2
c Z 2
Ib( 2 )
b
Ic( 2 )
Note that In = 0
a Ia( 0 )
Zn
Zg 0
In 3 Ia
(0 )
Zg 0
c
Zg 0
Ib( 0 )
b
Ic( 0 )
Note that In = 3 Ia (0)
Fig. 3
Ia( 2 )
Z 2 Va( 2 )
Reference bus ( Neutral )
Negative sequence network
Ia( 0 )
Zg 0
V ( 0 )
n Z0 a
3 Zn
Reference bus ( Ground )
Zero sequence network 12
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Z1 and Z 2 are the positive sequence and negative sequence impedance of the
generator. Zg 0 is the zero sequence impedance of the generator. Total zero sequence
impedance Z0 = Zg 0 + 3 Zn .
Sequence components of the terminal voltage are
V ( 0 )
a
V ( 1 )
a V
( 2 )
a
Z0 I(a0 )
Ea n Z1 Ia(1 )
(15)
Z 2
I( 2 ) a
As far as zero sequence currents are concerned, the three phase system behaves as
a single phase system. This is because of the fact that at any point the zero
sequence currents are same in magnitude and phase. Therefore, zero sequence
currents will flow only if a return path exists.
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The connection diagram and the zero sequence equivalent circuit for star
connected load is shown in Fig. 4.
Z Z Z
3 Zn Z
Zn
Reference
Fig. 4
The connection diagram and the zero sequence circuit for delta connected load
is shown in Fig. 5.
Z
Z Z
Z
14 Reference
Fig. 5
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Special attention is required while obtaining the zero sequence network
of three phase transformers. The zero sequence network will be
different for various combination of connecting the windings and also
by the manner in which the neutral is connected.
The zero sequence networks are drawn remembering that no current
flows in the primary of a transformer unless current flows in the
secondary
( neglecting the small magnetizing current ).
Five different cases are considered and the corresponding zero
sequence network are shown in Fig. 6. The arrows in the connection
diagram show the possible path for the flow of zero sequence current.
Absence of arrow indicates that the zero sequence current can not flow
there. Impedance Z0 accounts for the leakage impedance Z and the
neutral impedances 3 ZN and 3 Zn where applicable.
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Connection diagrams Zero sequence equivalent circuit
P
Q
Z0
P
Z N
Q
Z n
Reference
P Q
Z0
P Q
Z N
Reference
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P
Q
Z 0
P Q
Reference
P Q
Z0
P Q
Z N
Reference
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P Q
Z0
P Q
Reference
Fig. 6
Example 1
For the power system shown in Fig. 7, with the data given, draw the zero sequence,
positive sequence and negative sequence networks.
T1 T2
M1
G
M 2
Fig. 7
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Per unit reactances are:
Generator
Transformer T1
Transformer T2
Transmission line
Motor 1
Motor 2
Xg0 0.05; Xn 0.32 ; X1 0.2; X2 0.25 X0
X1 X2 0.08
X0 X1 X2 0.09
X0 0.52; X1 X2 0.18
Xm o 0.06; Xn 0.22; X1 X2 0.27
Xm o 0.12; X1 X2 0.55
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T1 T2
M1
G
M 2
Positive sequence network
j0.08 j0.18 j0.09
j0.2
j0.55
j0.27
+
Em 1
+ + Em 2
Eg
Reference
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T1
T2
M1
G
M 2
Negative sequence network
j0.08 j0.18 j0.09
j0.25
j0.27 j0.55
Reference
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T1
T2
M1
G
M 2
Zero sequence network
j0.08 j0.52 j0.09
j0.05 j0.06 j0.12
j0.96 j0.66
Reference
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2 UNSYMMETRICAL FAULTS AT GENERATOR TERMINALS
Single line to ground fault ( LG fault ), Line to line fault ( LL fault ) and Double line to
ground ( LLG fault ) are unsymmetrical faults that may occur at any point in a power
system. To understand the unsymmetrical fault analysis, let us first consider these faults
at the terminals of unloaded generator. This treatment can be extended to
unsymmetrical fault analysis when the fault occurs at any point in a power system.
Consider a three phase unloaded generator generating balanced three phase voltage.
The sequence components of the terminal voltages are
V(1) E a n
I (1) Z 1
(16) a a
V(2) I(2)
Z 2
(17) a a
V(0) I(0)
Z 0
(18) a a
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V(1) E a n
I (1) Z 1
(16) a a
V(2) I(2)
Z 2
(17) a a
V(0) I(0) Z 0
(18) a a
The above three equations apply regardless of the type of fault occurring at the terminals of the generator.
For each type of fault there will be three relations in terms of phase components of
currents and voltages. Using these, three relations in terms of sequence components of
currents and voltages can be obtained. These three relations and the eqns. (16), (17) and
(18) are used to solve for the sequence currents
I(0)a , I(1)a , I(2)a
and
sequence
voltages
Va(0) , Va(1) , Va(2) .
Sequence
component
s
relationship will enable to interconnect the sequence networks to represent the particular fault.
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Single line to ground fault ( LG fault )
The circuit diagram is shown in Fig. 9.
I a
a
Zf
Zn
+
_E
a n
Ec n +
+ E
b n I b c
b
Fig. 9 I c
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The fault conditions are
Ib 0
Ic 0
Va Zf Ia
Ia(0) 1/3 ( Ia Ib Ic ) Ia /3
Ia(1) 1/3 (Ia a Ib a
2 Ic ) Ia /3
Ia(2)
1/3 ( Ia a2 Ib a Ic ) Ia /3
Thus Ia(0)
Ia(1)
Ia(2)
Further from eqn. (21)
Va(0)
Va(1)
Va(2)
Zf ( I(0)
a I(1)
a I(2)
a ) 3 Zf I(1)
a
Using eqns. (16) to (18) in the above
Ia(0)
Z0 Ea n Ia(1)
Z1 Ia(2)
Z2 3 Zf Ia(1)
i.e.
Ia(1)
Z0 Ea n Ia(1)
Z1 Ia(1)
Z2 3 Zf Ia(1)
i.e.
Ia(1)
Ea n
Z1 Z2 Z0 3 Zf
(19)
(20)
(21)
(22)
(23)
(24) 26
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E
I(1) a n (24)
a
Z1 Z2 Z0 3 Zf
Then the sequence networks are to be connected as shown in Fig. 10.
Z1
+ I a(1)
Ea n
_
Z 2
I (2)
a
Z0
I (0)
a
+
V (1) a
_
+
3 Zf (2)
Va
_
+
V (0) a
_
Fig. 10 27
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Line to line fault
The circuit diagram is shown in Fig. 11
I a
a
+
Zn
Ea n
_
E
b n +
I b
c
E
c n
+
b
I c
Zf
Fig. 11
The fault conditions are I
a 0 (25)
Ib Ic 0 (26) Vb Zf Ib Vc (27)
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Ia 0
Ib Ic 0 V
b Zf Ib Vc
Then Ia(0) 1/3 ( Ia Ib Ic ) 0
Ia(1)
1/3 (Ia a Ib a2
Ic ) Ib /3 ( a a2 )
I(2) 1/3 ( I a
a2
I b
a I c
) I b
/3 ( a2 a )
a
Since I (0)
= 0 , V (0)
= - Z0 I (0)
= 0 a a a
Further Ia(2)
Ia(1)
From eqn. (27)
V(0) a2 V(1) a V(2)
Z f
( a2
I(1) a I(2)
) V(0)
a V(1) a
2 V(2)
a a a a a a a a
( a2 a ) V
(1) Z
f
( a2 a ) I
(1) ( a
2 a) V
(2)
a a a
Thus V(1)
Z f
I (1) V(2)
a a a
From the above eqn. Ea n Z1 Ia(1)
Zf Ia(1)
Z2 Ia(2)
Z2 Ia(1)
i.e.
Ea n ( Z1 Z2 Zf ) Ia
(1)
Therefore
Ia(1)
Ea n
Z1
Z2 Zf
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32) 29
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Therefore
Ia(1)
Ea n
(32) Z1 Z 2 Zf
I (2) = - I (1) a a
and I(0)
= 0; V(0)
= 0 a a
Sequence networks are to be connected as shown in Fig. 12.
Z
f
+
Z1
Z 2
Z
0
I a(1)
I a(2)
I a(0)
E V (1)
Va(2) Va (0) = 0
a n_ a
Fig. 12
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Double line to ground fault
The circuit diagram is shown in Fig. 13.
a
I a
+
Zn E
a n
_
Eb n
+
I b
E
c n
c +
b I c
Zf
Fig. 13
The fault conditions are
Ia 0 ; Vb Zf ( Ib Ic ) and Vc Zf ( Ib Ic ) (33)
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The fault conditions are
Ia 0 ; Vb Zf ( Ib Ic ) and Vc Zf ( Ib Ic )
Because of I(0)
1/3 ( I a
I b
I c
) , I b
I c
3 I(0)
a a
Therefore
V 3 Z f
I(0) b a
V 3 Z f
I(0) c a
V(1) 1/3 ( V a V a2
V ) 1/3 [ V ( a a2 ) V ]
a a b c a b
V(2) 1/3 ( V a2 V a V ) 1/3 [ V ( a
2 a ) V ]
a a b c a b
Therefore V(1) V(2) a a
(33)
(34)
(35)
(36)
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Further V(0) 1/3 ( V V V ) i.e. a a b c
3 V(0) V(0) V(1) V(2) 3 Z f
I (0) 3 Z f
I(0) i.e. 2 V(0) 2 V(1) 6 Z
f I(0)
a a a a a a a a a
i.e. V(1)
V(0) 3 Z f
I(0)
Z 0 I
(0) 3 Z
f I(0) ( Z
0 3 Z
f ) I(0)
a a a a a a
i.e. V(1)
( Z 0 3 Z
f ) I(0) (37)
a a
From eqn. (33) I(0) I(1) I(2) 0 i.e. a a a
V(1) V(2) V(1) V(1)
a I(1) a 0 i.e. a I(1) a 0
Z0 3Zf
a
Z2
a
Z2
Z0 3Zf
Therefore I(1) V(1) ( 1 1 ) V(1) Z2 Z0 3Zf
a a Z0
3Zf
Z2
a Z2 ( Z0 3Zf )
i.e. V(1)
Z2 ( Z0 3 Zf ) I(1)
a Z2 Z0
3 Zf
a
i.e. E I(1) Z Z2 ( Z0 3 Zf ) I (1)
a n 1
a
a
Z2 Z
3 Zf
0
Thus Ia(1)
Ea n
Z
Z2 ( Z0 3 Zf )
1
Z0 3 Zf
Z2
(38)
(39)
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Thus Ia(1)
Ea n
(39)
Z1 Z2 ( Z0 3 Zf )
Z2 Z0 3 Zf
From eqn. (38) V(1) V(2) Z I(2) Z2 ( Z0 3 Zf ) I(1)
2
a a a Z2 Z0 3 Zf
a
Therefore I(2) I(1)
Z0 3 Zf (40)
a a Z2 Z0 3 Zf
Again substituting eqn. (37) in eqn. (38)
( Z0 3 Zf ) Ia(0)
Z2 ( Z0 3 Zf ) Ia
(1) Thus Ia
(0) Ia
(1) Z2 (41)
Z2 Z0 3 Zf Z2 Z0 3 Zf
For this fault, the sequence networks are to be connected as shown in Fig. 14.
Z1
Z 2 Z0
+
I a
(1) I (2)
I a
(0)
Ea n
Va(1) a
Va(2)
Va(0)
_
3 Zf
34 Fig. 14
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General procedure for unsymmetrical fault analysis
when fault occurs at a point in a power system
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PRELIMINARY CALCULATIONS
1. Draw the positive sequence, negative sequence and zero sequence
networks.
2. Using bus impedance building algorithm, construct ZBus (1)
, ZBus(2)
and
ZBus(0)
.
DATA REQUIRED
Type of fault, fault location (Bus p) and fault impedance (Zf)
TO COMPUTE FAULT CURRENTS I f a , I f b and I f c
1. Extract the columns of ZBus (1)
, ZBus(2)
and ZBus(0)
corresponding to the
faulted bus.
2. Depending on the type of fault interconnect the sequence networks.
3. Calculate I f a(1)
, I f a(2)
and I f a(0)
4. Compute the corresponding phase components I f a. I f b and I f c using I f a
1
1
=
a 2
If b
1
I 1 a
f c
1 I f a
(0)
a (1)
If a
a
2 (2)
I
f a 36
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TO COMPUTE FAULTED BUS VOLTAGES V p a, V p b and V p c
1. Compute the sequence components V p a
(1) (2) (0)
from , V p a and V p a (!) (1) (1)
V p a = Vf – Z p p I f a (2) (2) (2)
V p a = – Z p p I f a (0) (0) (0)
V p a = – Z p p I f a
2. Calculate the corresponding phase components V p a, V p b and V p c from
Vp a 1 1
=
a 2
V
p b 1
a V
p c
1
1 V p a
(0)
a (1)
Vp a
a
2 (2)
V
p a
37
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TO COMPUTE BUS VOLTAGES AT BUS j i.e V j a, V j b and V j c 1.
Compute the sequence components V j a(!)
, V j a(2)
and V j a(0)
from
(!) (1) (1)
V j a = Vf – Z j p I f a
V j a(2)
= – Z j p (2) I f a (2)
V j a(0)
= – Z j p (0) I f a (0)
2. Calculate the corresponding phase components V j a, V j b and V j c from
V 1 1 1 V (0) ja
ja
= 2
a (1)
V
jb 1 a V
ja
V 1 a a2 V
ja
(2)
jc
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Single line to ground fault
I (f
1)a
+ Zp(1)
p Vp
(1)a
V
f _
I f(1)
a = I f(2)
a = If(0)
a
(2)
I f a
3 Zf
Zp(2)
p
V (2)
p a
I (f
0)a
Z(0)
p p Vp(0)a
39
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Line to line fault
Zf
Zp(1)
p
Zp(2)
p
+ I f(1)
a I f(2)
a I f(0)
a
Vf _ Vp(1)
a Vp(2)
a
40
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Double line to ground fault
Zp(1)
p Zp(2)
p Zp(0)
p
+
I f(1)
a
I f(2)
a
I f(0)
a
Vp(0)
a Vf
Vp(1)
a
Vp(2)
a
_
3 Zf
41
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