Symmetrical Components and
Unbalanced Faults
By
Dr. E. Matlotse
Symmetrical Components and
Unbalanced Faults
• Different types of unbalanced faults are single
line-to-ground fault, line-to-line fault and
double line-to-ground fault.
• Since the one-line diagram simplifies the
solution of balanced 3-Ф problems, the
method of symmetrical components that
resolves the solution of unbalanced circuit
into a number of balanced circuits is used.
Symmetrical Components and
Unbalanced Faults
• Ultimately, then this approach is
applied to the unbalanced faults,
which permits the treatment of
the problem on a simple per
phase basis.
Fundamental of Symmetrical
Components
• Symmetrical Components allow unbalanced phase
quantities such as currents and voltages to be
replaced by three separate balanced symmetrical
components.
(a)
(b)
(c)
• Fig. 1: Representation of Symmetrical Components
Fundamental of Symmetrical
Components
• Refer to the phasor representation of a 3-Ф
balanced current shown in fig. 1(a).
• By convention, the direction of rotation of the
phasors is taken to be counterclockwise.
• Three phasors are written as


1
1
2 1
I b  I a 240  a I a 
1
1
1 
I c  I a 120  aI a 
I a1  I a1 0  I a1
(1)
Fundamental of Symmetrical
Components
• Where we have defined an operator a that
causes a counterclockwise rotation of 120⁰,
such that
a  1120  0.5  j 0.866 

(2)
a 2  1240  0.5  j 0.866
a 3  1360  1  j 0


• The order of the phasors is abc and this is
designated as the +ve phase sequence
Fundamental of Symmetrical
Components
• When the order is acb as shown in fig. 1(b), it
is called the –ve phase sequence.
• -ve
phase
sequence
quantities
are
represented as


2
2
2 
I b  I a 120  aI a 

I c2  I a2 240  a 2 I a2 
I a2  I a2 0  I a2
(3)
Fundamental of Symmetrical
Components
• When analysing certain types of unbalanced
faults, it will be found that a 3rd set of
balanced phasors must be introduced .
• These are called the zero phase sequence
which are found in phase with each other.
• Zero phase sequence currents, as shown in fig.
1(c) are
0
0
0
(4)
I a  Ib  Ic
Fundamental of Symmetrical
Components
Fig. 2: Resolution of unbalanced phasor into
symmetrical components
Fundamental of Symmetrical
Components
• Supersripts 1, 2 and 0 are being
used to represent +ve, -ve and
zero-sequence
quantities,
respectively.
Fundamental of Symmetrical
Components
• Refer to 3-Ф unbalanced currents, Ia, Ib and Ic
shown in fig. 2.
• We are aiming to find the three symmetrical
components of the current such that
I a  I a0  I a1  I a2 

0
1
2
Ib  Ib  Ib  Ib 
0
1
2
I c  I c  I c  I c 
(5)
Fundamental of Symmetrical
Components
• According to the expression of the
symmetrical components as given by (1), (3)
and (4), we can rewrite (5) all in terms of
phase a components


0
2 1
2
I b  I a  a I a  aI a 
0
1
2 2
I c  I a  aI a  a I a 
I a  I a0  I a1  I a2
(6)
Fundamental of Symmetrical
Components
or
 I a  1 1
 I   1 a 2
 b 
 I c  1 a
1   I a0 
 1

a I a 
a 2   I a2 
(7)
• We rewrite the above equation in matrix
notation as
abc
012
(8)
I
 AI
a
Fundamental of Symmetrical
Components
• Where A is known as symmetrical component
transformation matrix (SCTM) which transforms
phasor currents I abc into component currents I a012 and
is
1 1 1 
(9)
A  1 a 2 a 
1
a
a 2 
• Solving (8) for symmetrical components of current
yields
(10)
I a012  A1 I abc
Fundamental of Symmetrical
Components
• Inverse of A is
A 1
1 1
1
 1 a
3
1 a 2
1
a 2 
a 
(11)
• From (9) and (11), we have
A
1
1 *
 A
3
(12)
Fundamental of Symmetrical
Components
• Substituting for
A1
in (10), we get
 I a0 
1 1
 1 1
 I a   3 1 a
 I a2 
1 a 2
 
1  I a 
a 2   I b 
a   I c 
(13)
• Similar formulae exist for voltage
• Therefore, the unbalanced phase voltage in
terms of the symmetrical components are
Fundamental of Symmetrical
Components


0
2 1
2
Vb  Va  a Va  aVa 

Vc  Va0  aVa1  a 2Va2 
Va  Va0  Va1  Va2
(14)
• In matrix notation
V
abc

012
AVa
(15)
Fundamental of Symmetrical
Components
• Symmetrical components in terms of the
unbalanced voltages
1
 Va  Vb  Vc 
3
1
Va1  Va  aVb  a 2Vc
3
1
Va2  Va  a 2Vb  aVc
3
Va0













• In matrix notation
Va012  A1V abc
(16)
(17)
Sequence Networks of a Loaded
Generator
• Figure below presents a 3-Ф synchronous
generator with neutral grounded through
an impedance Zn.
• Generator (shown on the next slide) is
supplying a 3-Ф balanced load
Sequence Networks of a Loaded
Generator
Fig. 1: 3-Ф balanced source and impedance
Sequence Networks of a Loaded
Generator
• Synchronous machine generates balanced 3-Ф
internal voltages and is represented as a +ve
sequence set of phasors abc  12 
E  a  E a
(1)
 a 
• The machine under consideration is supplying a
3-Ф balanced load.
• Applying KVC to each phase, we get
Va  E a  Z s I a  Z n I n 

Vb  Eb  Z s I b  Z n I n 
Vc  Ec  Z s I c  Z n I n 
(2)
Sequence Networks of a Loaded
Generator
• Substituting for I n  I a  I b  I c and formulating
(2) in matrix form, we have
Va   Ea   Z s  Z n
V    E    Z
n
 b  b 
Vc   Ec   Z n
Zn
Zs  Zn
Zn
Ia 
I 
 b 
Z s  Z n   I c 
Zn
Zn
(3)
• In compact form, we have
V abc  E abc  Z abc I abc
(4)
Sequence Networks of a Loaded
Generator
• Where V abc
is the phase terminal voltage
vector and I abc is the phase current vector.
• Transforming the terminal voltages and
current phasors into their symmetrical
components results in
(5)
AVa012  AE a012  Z abc AI a012
Sequence Networks of a Loaded
Generator
• Multiplying the above by A1 , we have
Va012  Ea012  A1Z abc AI a012
(6)
 Ea012  Z 012 I a012
where
Z 012
1 1
1
 1 a
3
1 a 2
1  Z s  Z n
a 2   Z n
a   Z n
Zn
Zs  Zn
Zn
 1 1
 1 a 2

Z s  Z n  1 a
Zn
Zn
1
a 
a 2 
(7)
Sequence Networks of a Loaded
Generator
• Finally,
Z 012
 Z s  3Z n
  0
 0
0
Zs
0
0  Z 0

0    0
Z s   0
0
Z1
0
0

0
Z 2 
(8)
• Since the generated emf is balanced, there is
only +ve sequence voltage as shown below
0
(9)
E 012   E 
a
 a
 0 
Sequence Networks of a Loaded
Generator
• Substituting for Ea012 and
Va0   0   Z 0
 1   
Va    E a    0
Va2   0   0
    
• In component form
Z 012
0
Z1
0
into (6), we have
0   I a0 
 
0   I a1 
Z 2   I a2 
Va0  0  Z 0 I a0 

1
1 1
Va  Ea  Z I a 
2
2 2 
Va  0  Z I a 
(10)
(11)
Sequence Networks of a Loaded
Generator
• 3 above equations can be represented by
the 3 equivalent sequence networks
shown below
Fig.2: Sequence networks (a) +ve-sequence (b) –vesequence and zero-sequence
Line-to-Ground Fault
• Fig. 1 shows a 3-Ф generator with neutral grounded through
impedance Zn
Fig. 1: Line-to-ground fault on phase a
Line-to-Ground Fault
• Assuming that a line-to-ground fault occurs on
phase a through impedance Zn.
• Also, assuming the generator is initially on noload, the boundary conditions at the fault
point are
(1)
Va  Z f I a
Ib  Ic  0
(2)
Line-to-Ground Fault
• Substituting for Ib  I c  0
, the symmetrical
components of currents are
 I a0 
1 1
 1 1
 I a   3 1 a
 I a2 
1 a 2
 
1  I a 
a 2   0 
a   0 
(3)
• From the above relation, we have
I a0

I a1

I a2
1
 Ia
3
(4)
Line-to-Ground Fault
• Phase a voltage in terms of symmetrical
components
(5)
Va  Va0  Va1  Va2
• Substituting forVa0 , Va1 and Va2 and noting that
I a0  I a1  I a2 , we have
Va  Ea  (Z 1  Z 2  Z 0 ) I a0
(6)
• where Z 0  Z s  3Z n
. Substituting for Va and
noting I a  3I a0 , we have
Line-to-Ground Fault
3Z f I a0  Ea  (Z 1  Z 2  Z 0 ) I a0
 I a0 
Ea
Z  Z  Z  3Z f
1
2
0
(7)
• Fault current is
I a  3I a0
3Ea
 1
Z  Z 2  Z 0  3Z f
(8)
Line-to-Ground Fault
• Resulting network is
Fig. 2: Sequence network connection for line-toground fault
Line-to-Line Fault
• Fig. 1: Line-to-line fault between phases b and c
Line-to-Line Fault
• A 3-Ф generator with a fault through
impedance Zf between phases b and c is
shown in fig. 1.
• Assuming the generator is initially on no-load,
the conditions at the fault point are
Vb  Vc  Z f I b
(1)
(2)
Ib  Ic  0
(3)
Ia  0
Line-to-Line Fault
• Substituting for I a  0
and I c  I b , the
symmetrical components of currents are
 I a0 
1 1
 1 1 
 I a   3 1 a
 I a2 
1 a 2
 
1  0 
a 2   I b 
a   I b 
(4)
• From the above equation,
I a0  0
1
I a1  (a  a 2 ) I b
3
(5)
(6)
Line-to-Line Fault
1
I a2  (a 2  a) I b
3
(7)
I a1   I a2
(8)
Vb  Vc  (a 2  a)(Va1  Va2 )  Z f I b
(9)
• From (6) and (7), we have
• We have
Line-to-Line Fault
• Substituting for Va1 and Va2 and noting
yields
I a2   I a1
,
(a 2  a)[( Ea  Z 1I a1 )  (0  Z 2 I a2 )]  Z f I b
 (a 2  a)[( Ea  Z 1I a1 )  (Z 2 I a1 )]  Z f I b
 (a 2  a)[( Ea  (Z 1  Z 2 ) I a1 ]  Z f I b
(10)
Line-to-Line Fault
• Substituting for I b from (6), yields,
(a  a)[( Ea  ( Z  Z
2
1
 [( Ea  ( Z  Z
1
2
2
) I a1 ] 
) I a1 ]
 3I a1 
 Zf 
2 
(
a

a
)



3I a1
Zf  2
2 
(
a

a
)(
a

a
)

(11)
Line-to-Line Fault
• Since (a  a 2 ) (a 2  a)  3 and solving for
Ea  ( Z  Z
1
2
) I a1
I a1
 3I a1 
 Zf 

3


 Ea  (Z 1  Z 2 ) I a1  Z f I a1
 I a1 
Ea
Z1  Z 2  Z f
(12)
Line-to-Line Fault
• Phase currents are
 I a  1 1
 I   1 a 2
 b 
 I c  1 a
1  0 
a   I a1 
a 2   I a1 
(13)
I b   I c  (a 2  a) I a1
(14)
Line-to-Line Fault
• Resulting circuit is
Fig. 2: Sequence connection for line-to-line fault
Double Line-to-Ground Fault
Fig. 1: Double line-to-ground fault
Double Line-to-Ground Fault
• A 3-Ф generator with a fault on phases b and c
through an impedance Zf to ground is shown
in fig. 1.
• Assuming the generator is initially on no-load,
the conditions at the fault point are
Vb  Vc  Z f ( I b  I c )
(1)
I a  I a0  I a1  I a2  0
(2)
Double Line-to-Ground Fault
• Using the same arguments as cases for the
other faults, the resulting equation for this
kind of a fault is
I a1
Ea

Z1 
Z 2 ( Z 0  3Z f )
Z 2  Z 0  3Z f
(3)
Double Line-to-Ground Fault
• Therefore, resulting network connection is
Fig. 2: Sequence network connection for double
line-to-ground fault
Example
The one-line diagram of a simple power
system in the figure below. The neutral of
each generator is grounded through a currentlimiting reactor of 0.25/3 per unit on a 100
MVA base. The system data expressed in per
unit on a common 100 MVA base is tabulated
below. The generators are running on no-load
at their rated voltage and rated frequency
with their emfs in phase.
Example
• Figure: The one-line diagram
Example
Item
Base MVA
Voltage Rating
X1
X2
X0
G1
100
20 kV
0.15
0.15
0.05
G2
100
20 kV
0.15
0.15
0.05
T1
100
20 /220 kV
0.1
0.1
0.1
T2
100
20 /220 kV
0.1
0.1
0.1
L12
100
220 kV
0.125
0.125
0.3
L13
100
220 kV
0.15
0.15
0.35
L23
100
220 kV
0.25
0.25
0.7125
Example
Determine the fault current for the following
faults:
(a) A single-line-to-ground fault at bus 3
through a fault impedance Zf = j0.10 per unit
(b) A line-to-line fault at bus 3 through a fault
impedance Zf = j0.10 per unit
Solution
• The +ve-sequence network
Solution
• To find the Thevenin’s impedance viewed from
the faulted bus (bus 3), we convert the delta
formed by buses 123 to an equivalent Y
Z1S 
( j 0.125)( j 0.15)
 j 0.0357143 pu
( j 0.125)  ( j 0.15)  ( j 0.25)
Z 2S 
( j 0.125)( j 0.25)
 j 0.0595238 pu
( j 0.125)  ( j 0.15)  ( j 0.25)
Z 3S 
( j 0.15)( j 0.25)
 j 0.0714286 pu
( j 0.125)  ( j 0.15)  ( j 0.25)
Solution
The resulting network is
Solution
The positive and negative-sequence Thevenin
impedances are
1
2
Z 33
 Z33
\ 
( j 0.0357143  j 0.15  j 0.1)( j 0.059524  j 0.1  j 0.15)
 j 0.0714286
( j 0.0357143  j 0.15  j 0.1)  ( j 0.059524  j 0.1  j 0.15)
 j 0.22 pu
Solution
The zero-sequence impedance network
Solution
To establish its Thevenin impedance viewed
from the faulted bus, we convert the delta
formed by buses 123 to an equivalent Y
Z1S 
( j 0.30)( j 0.35)
 j 0.07703595 pu
( j 0.30)  ( j 0.35)  ( j 0.7125)
Z 2S 
( j 0.30)( j 0.7125)
 j 0.156933235 pu
( j 0.30)  ( j 0.35)  ( j 0.7125)
Z 3S
( j 0.35)( j 0.7125)

 j 0.183088774 pu
( j 0.30)  ( j 0.35)  ( j 0.7125)
Solution
The resulting circuit is
Solution
The zero-sequence Thevenin impedance is
0
Z 33

( j 0.077064  j 0.1  j 0.05  j 0.25)( j 0.156881  j 0.1)
 j 0.183088774
( j 0.077064  j 0.1  j 0.05  j 0.25)  ( j 0.156881  j 0.1)
 j 0.35 pu
Solution
Single line-to-ground fault at bus 3 is
3
V
0
1
2
a (0)
I3  I3  I3  1
Z 33  Z 332  Z 330  3Z f
1

  j 0.9174 pu
j 0.22  j 0.22  j 0.35  3( j 0.1)
Solution
The fault current is
 I 3a  1 1
 b 
2
I

1
a
 3 
 I 3c  1 a
 
1   I 30  3I 3a   j 2.7523
 0   

 pu
a I 3    0  
0



2
0


a   I 3   0  
0
Solution
(b) The zero-sequence component of current is
zero
0
I3  0
The +ve and –ve sequence components of the
Fault are
a
V
(0)
I 31   I 32  1 3 2
Z 33  Z 33  Z f

1
  j1.8519 pu
j 0.22  j 0.22  ( j 0.1)
Solution
The fault current is
 I 3a  1 1
 b 
2
I

1
a
 3 
 I 3c  1 a
  
1 
0
  0 
a   j1.8519   3.2075 pu
a 2   j1.8519   3.2075 