Symmetrical Components I
An Introduction to Power System Fault
Analysis Using Symmetrical Components
Dave Angell
Idaho Power
21st Annual
Hands-On Relay School
What Type of Fault?
V A
25
0
-25
V B
25
0
-25
V C
VA
25
0
-25
VB
VC
IA
IB
IC
IA
2500
0
IB
-2500
2500
0
IC
-2500
2500
0
-2500
1
2
3
4
5
6
Cy c les
7
8
9
10
11
What Type of Fault?
IA
IB
IC
IR
IA
10000
0
-10000
IB
10000
0
-10000
IC
10000
0
-10000
IR
10000
0
-10000
1
2
3
4
5
6
Cy c les
7
8
9
10
11
What Type of Fault?
IA
IB
IC
IA
10000
0
-10000
IB
10000
0
-10000
IC
10000
0
-10000
1
2
3
4
5
6
Cy c les
7
8
9
10
11
What Type of Fault?
IA
IA
5000
IB
IC
IR
0
IB
-5000
5000
0
IC
-5000
5000
0
IR
-5000
2500
0
-2500
1
2
3
4
5
6
Cy c les
7
8
9
10
11
What Type of Fault?
IA
IB
IC
IR
IA
100
0
-100
IB
100
0
-100
IC
100
0
-100
IR
200
-0
-200
1
2
3
4
5
6
Cy c les
7
8
9
10
11
What Type of Fault?
IA
IB
IC
IR
IA
200
0
IB
-200
200
0
IC
-200
200
0
IR
-200
500
0
-500
1
2
3
4
5
6
Cy c les
7
8
9
10
11
What Type of Fault?
IA
IB
IC
IR
IA
250
0
IB
-250
250
0
IC
-250
250
0
-250
IR
100
0
-100
1
2
3
4
5
6
Cy c les
7
8
9
10
11
Basic Course Topics
 Terminology
 Phasors
 Equations
 Fault
Analysis Examples
 Calculations
Unbalanced Fault
Ia
Ia
Ib
Ib
Ic
Ic
Symmetrical Component
Phasors
 The
unbalanced three phase system
can be transformed into three
balanced phasors.
– Positive Sequence
– Negative Sequence
– Zero Sequence
Positive Phase Sequence (ABC)
Va
Vb
Vc
1.0
Magnitude
0.5
0.0
0.000
0.017
0.033
-0.5
-1.0
Time
0.050
Positive Phase Sequence
Each have the
same magnitude.
 Each positive
sequence voltage
or current quantity
is displaced 120°
from one another.

Vc1
Va1
Vb1
Positive Phase Sequence

The positive
sequence
quantities have ab-c, counter clockwise, phase
rotation.
Vc1
Va1
Vb1
Reverse Phase Sequence (ACB)
Va
Vb
Vc
1.0
Magnitude
0.5
0.0
0.000
0.017
0.033
-0.5
-1.0
Time
0.050
Negative Phase Sequence
Each have the
same magnitude.
 Each negative
sequence voltage
or current quantity
is displaced 120°
from one another.

Vb2
Va2
Vc2
Negative Phase Sequence

The negative
sequence
quantities have ac-b, counter clockwise, phase
rotation.
Vb2
Va2
Vc2
Zero Phase Sequence
Va
Vb
Vc
1.0
Magnitude
0.5
0.0
0.000
0.017
0.033
-0.5
-1.0
Time
0.050
Zero Phase Sequence
Each zero
sequence quantity
has the same
magnitude.
 All three phasors
with no angular
displacement
between them, all
in phase.

Vc0
Vb0
Va0
Symmetrical Components
Equations
 Each
phase quantity is equal to the
sum of its symmetrical phasors.
 The
 Va
= Va0 + Va1 +Va2
 Vb
= Vb0 + Vb1 +Vb2
 Vc
= Vc0 + Vc1 +Vc2
common form of the equations
are written in a-phase terms.
The a Operator
 Used
to shift the a-phase terms to
coincide with the b and c-phase
 Shorthand to indicate 120° rotation.
 Similar to the j operator of 90°.
Va
Rotation of the a Operator
120° counter clock-wise rotation.
 A vector multiplied by 1 /120° results in
the same magnitude rotated 120°.

aVa
Va
Rotation of the a2 Operator
240° counter clock-wise rotation.
 A vector multiplied by 1 /240° results in
the same magnitude rotated 240°.

Va
a2Va
B-Phase Zero Sequence
 We
replace the
Vb sequence
terms by Va
sequence terms
shifted by the a
operator.
 Vb0
= Va0
Vc0
Vb0
Va0
B-Phase Positive Sequence
We replace the Vb
sequence terms by
Va sequence terms
shifted by the a
operator
 Vb1 = a2Va1

Vc1
Va1
Vb1
B-Phase Negative Sequence
We replace the Vb
sequence terms by
Va sequence terms
shifted by the a
operator
 Vb2 = aVa2

Vb2
Va2
Vc2
C-Phase Zero Sequence
 We
replace the
Vc sequence
terms by Va
sequence terms
shifted by the a
operator.
 Vc0
= Va0
Vc0
Vb0
Va0
C-Phase Positive Sequence
We replace the Vc
sequence terms by
Va sequence terms
shifted by the a
operator
 Vc1 = aVa1

Vc1
Va1
Vb1
C-Phase Negative Sequence
 We
replace the
Vc sequence
terms by Va
sequence terms
shifted by the a
operator
 Vc2 = a2Va2
Vb2
Va2
Vc2
What have we produced?
 Va
= Va0 + Va1 + Va2
 Vb
= Va0 + a2Va1 + aVa2
 Vc
= Va0 + aVa1 + a2Va2
Symmetrical Components
Equations
 Analysis
– To find out of the amount of the
components
 Synthesis
– The combining of the component
elements into a single, unified entity
Symmetrical Components
Synthesis Equations
 Va
= Va0 + Va1 + Va2
 Vb
= Va0 + a2Va1 + aVa2
 Vc
= Va0 + aVa1 + a2Va2
Symmetrical Components
Analysis Equations
 Va0
= 1/3 ( Va + Vb + Vc)
 Va1= 1/3
(Va + aVb + a2Vc)
 Va2= 1/3
(Va + a2Vb + aVc)
Symmetrical Components
Analysis Equations - 1/3 ??
 Where
does the 1/3 come from?
 Va1= 1/3
(Va + aVb + a2Vc)
0
0
 Va = Va0 + Va1 + Va2
 When
balanced
Symmetrical Components
Analysis Equations - 1/3 ??
 Va1= 1/3
(Va + aVb + a2Vc)
 Adding the phases
Va
Symmetrical Components
Analysis Equations - 1/3 ??
 Va1= 1/3
a2Vc)
 Adding
(Va + aVb +
Vc
Va
the phases yields
Vb
Va
aVb
Symmetrical Components
Analysis Equations - 1/3 ??
 Va1= 1/3
a2Vc)
 Adding
(Va + aVb +
Vc
Va
the phases yields
3 Va.
 Divide by the 3 and now
Va = Va1
Va
aVb
Vb
a2Vc
Example Vectors
An Unbalanced Voltage
Vc
 Va
Va
Vb
= 13.4 /0°
 Vb = 59.6 /104°
 Vc = 59.6 /104°
Analysis Results in These
Sequence Quantities
Vc1
Va0
Vb0
Vc0
Vc2
Va2
Va1
Vb2
Vb1

Va0 = -5.4
s
Va1 = 42.9
s
Va2 = -24.1
Synthesize by Summing the
Positive, Negative and …
Vc2
Vc1
Va2
Va1
Vb1
Vb2
Zero Sequences
Vc0
Vc2
Vc1
Va0 Va2
Va1
Vb1
Vb0
Vb2
The Synthesis Equation Results
in the Original Unbalanced
Voltage
Vc0
Vc
Vc2
Vc1
Va0 Va2
Va1
Va
Vb1
Vb
Vb2
Vb0
Symmetrical Components
Present During Shunt Faults
 Three
phase fault
– Positive
 Phase
fault
to phase
– Positive
– Negative
 Phase
to
ground fault
– Positive
– Negative
– Zero
Symmetrical Component
Review of Faults Types
 Let’s
return to the example fault
reports and view the sequence
quantities present
Three Phase Fault, Right?
V A
25
0
-25
V B
25
0
-25
V C
VA
25
0
-25
VB
VC
IA
IB
IC
IA
2500
0
IB
-2500
2500
0
IC
-2500
2500
0
-2500
1
2
3
4
5
6
Cy c les
7
8
9
10
11
A Symmetrical Component View
of an Three-Phase Fault
90
135
Component Magnitude
45
Angle
Ia0
7.6
175
Ia1
2790
-64
Ia2
110
75.8
Va0
0
0
Va1
18.8
0
Va2
0.7
337
V1
180
225
I1
270
0
315
A to Ground Fault, Okay?
IA
IB
IC
IR
IA
10000
0
-10000
IB
10000
0
-10000
IC
10000
0
-10000
IR
10000
0
-10000
1
2
3
4
5
6
Cy c les
7
8
9
10
11
A Symmetrical Component View
of an A-Phase to Ground Fault
90
135
Component Magnitude
45
Angle
Ia0
7340
-79
Ia1
6447
-79
Ia2
6539
-79
Va0
46
204
Va1
123
0
Va2
79
178
V2
180
V1
V0
225
315
I1 I0
I2
270
0
Single Line to Ground Fault
 Voltage
– Negative and zero sequence 180 out of
phase with positive sequence
 Current
– All sequence are in phase
A to B Fault, Easy?
IA
IB
IC
IA
10000
0
-10000
IB
10000
0
-10000
IC
10000
0
-10000
1
2
3
4
5
6
Cy c les
7
8
9
10
11
A Phase Symmetrical Component
View of an A to B Phase Fault
90
135
Component Magnitude
45
Angle
Ia0
3
-102
Ia1
5993
-81
Ia2
5961
-16
Va0
1
45
Va1
99
0
Va2
95
-117
V1
180
0
I2
V2
225
315
I1
270
C Phase Symmetrical Component
View of an A to B Phase Fault
90
I2
135
Component Magnitude
Ic0
3
Ic1
5993
Ic2
5961
Vc0
1
Vc1
99
Vc2
95
Angle
138
279
104
-75
0
2.5
45
V2 V1
180
225
0
315
I1
270
Line to Line Fault
 Voltage
– Negative in phase with positive
sequence
 Current
– Negative sequence 180 out of phase
with positive sequence
B to C to Ground
IA
IA
5000
IB
IC
IR
0
IB
-5000
5000
0
IC
-5000
5000
0
IR
-5000
2500
0
-2500
1
2
3
4
5
6
Cy c les
7
8
9
10
11
A Symmetrical Component View
of a B to C to Ground Fault
90
135
Component Magnitude
I2
Angle
Ia0
748
97
Ia1
2925
-75
Ia2
1754
101
Va0
8
351
Va1
101
0
Va2
18
348
45
I0
V2 V0
180
225
I1
270
V1
315
0
Line to Line to Ground Fault
 Voltage
– Negative and zero in phase with positive
sequence
 Current
– Negative and zero sequence 180 out of
phase with positive sequence
Again, What Type of Fault?
IA
IB
IC
IR
IA
100
0
-100
IB
100
0
-100
IC
100
0
-100
IR
200
-0
-200
1
2
3
4
5
6
Cy c les
7
8
9
10
11
C Symmetrical Component View
of a C-Phase Open Fault
90
135
Component Magnitude
Ic0
69
Ic1
101
Ic2
32
Vc0
0
Vc1
79
Vc2
5
Angle
184
4
183
162
0
90
45
V2
180
I0
I1
I2
V1
225
315
270
0
One Phase Open (Series)
Faults

Voltage
– No zero sequence voltage
– Negative 90 out of phase with positive
sequence

Current
– Negative and zero sequence 180 out of
phase with positive sequence
What About This One?
IA
IB
IC
IR
IA
200
0
IB
-200
200
0
IC
-200
200
0
IR
-200
500
0
-500
1
2
3
4
5
6
Cy c les
7
8
9
10
11
Ground Fault with Reverse Load
90
135
Ic0
164
-22
Ic1
89
-113
Ic2
41
-6
Vc0
4
-123
Vc1
38
0
Vc2
6
-130
45
I2
180
V1
V2 V0
I0
I1
225
315
270
0
Finally, The Last One!
IA
IB
IC
IR
IA
250
0
IB
-250
250
0
IC
-250
250
0
-250
IR
100
0
-100
1
2
3
4
5
6
Cy c les
7
8
9
10
11
Fault on Distribution System
with Delta – Wye Transformer
90
135
Component Magnitude
45
Angle
Ic0
45
40
Ic1
153
-4
Ic2
132
180
Vc0
0.5
331
Vc1
40
0
Vc2
0.5
93
V2
180
I0
I2
V0
225
I1
V1
315
270
0
Use of Sequence Quantities in
Relays
 Zero
Sequence filters
– Current
– Voltage
 Relay
operating quantity
 Relay polarizing quantity
Zero Sequence Current
Ia
Ib
Ic
Direction of the
protected line
Ia+Ib+Ic
3I0
Zero Sequence Voltage
(Broken Delta)
Va
Vb
Vc
3V0
Zero Sequence Voltage
3Vo
Vc
Va
Vb
Sequence Operating Quantities
 Zero
and negative sequence currents
are not present during balanced
conditions.
 Good indicators of unbalanced faults
Sequence Polarizing Quantities
 Polarizing
quantities are used to
determine direction.
 The quantities used must provide a
consistent phase relationship.
Zero Sequence Voltage
Polarizing
3Vo is out of phase with Va
 -3Vo is used to polarize for ground faults

3Vo
Va
Vb
Learning Check
 Given
three current sources
 How can zero sequence be produced
to test a relay?
 How
can negative sequence
produced?
How can zero sequence be
produced to test a relay?
A
single source provides positive,
negative and zero sequence
– Note that each sequence quantity will
be 1/3 of the total current
 Connect
the three sources in parallel
and set their amplitude and the
phase angle equal to one another
– The sequence quantities will be equal to
each source output
How can negative sequence
produced?

A single source provides positive, negative
and zero sequence
– Each sequence quantity will be 1/3 of the total
current

Set the three source’s amplitude equal to
one another and the phase angles to
produce a reverse phase sequence (Ia at
/0o, Ib at /120o and Ic at /-120o)
– Only negative sequence will be produced
Advanced Course Topics
 Sequence
Networks
 Connection of Networks for Faults
 Per Unit System
 Power System Element Models
References
 Symmetrical
Components for Power
Systems Engineering, J Lewis
Blackburn
 Protective Relaying, J Lewis Blackburn
 Power System Analysis, Stevenson
 Analysis of Faulted Power System, Paul
Anderson
Conclusion
 Symmetrical
components provide:
– balanced analysis of an unbalanced
system.
– a measure of system unbalance
– methods to detect faults
– an ability to distinguish fault direction