Symmetrical Components I: Dave Angell Idaho Power 21st Annual Hands-on Relay School
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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 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
I1
225
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
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 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
I1
225
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
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