Symmetrical Components I An Introduction to Power System Fault Analysis Using Symmetrical Components 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
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What Type of Fault?
-250
25
-250
25
-250
25
-2500
0
2500
-2500
02500
-2500
0
2500
1 2 3 4 5 6 7 8 9 10 11
V
A
V
B
V
C
I A
I B
I C
C y c l e s
V A V B V C IA IB IC
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What Type of Fault?
-10000
0
10000
-10000
0
10000
-10000
0
10000
-10000
0
10000
1 2 3 4 5 6 7 8 9 1 0 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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What Type of Fault?
-10000
0
10000
-10000
0
10000
-10000
0
10000
1 2 3 4 5 6 7 8 9 1 0 11
I A
I B
I C
C y c l e s
IA IB IC
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What Type of Fault?
-5000
0
5000
-5000
0
5000
-5000
0
5000
-2500
0
2500
1 2 3 4 5 6 7 8 9 10 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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What Type of Fault?
-100
0
100
-100
0
100
-100
0
100
-200
-0
200
1 2 3 4 5 6 7 8 9 10 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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What Type of Fault?
-200
0
200
-200
0
200
-200
0
200
-500
0
500
1 2 3 4 5 6 7 8 9 1 0 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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What Type of Fault?
-250
0
250
-250
0
250
-250
0
250
-100
0
100
1 2 3 4 5 6 7 8 9 10 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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Basic Course Topics
Terminology
Phasors
EquationsFault Analysis Examples
Calculations
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Unbalanced Fault
Ia
Ib
Ic
Ia
Ib
Ic
Ia
Ib
Ic
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Symmetrical Component
PhasorsThe unbalanced three phase system
can be transformed into threebalanced phasors.
–Positive Sequence
–Negative Sequence
–Zero Sequence
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Positive Phase Sequence (ABC)
-1.0
-0.5
0.0
0.5
1.0
0.000 0.017 0.033 0.050
Time
M a g n i t u d e
Va Vb Vc
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Positive Phase Sequence
The positivesequencequantities have a-
b-c, counter clock-wise, phaserotation. V
a1
Vb1
Vc1
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Negative Phase Sequence
Each have thesame magnitude.
Each negative
sequence voltageor current quantityis displaced 120°from one another.
Va2
Vc2
Vb2
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Negative Phase Sequence
The negativesequencequantities have a-
c-b, counter clock-wise, phaserotation. V
a2
Vc2
Vb2
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Zero Phase Sequence
Each zerosequence quantityhas the same
magnitude. All three phasors
with no angulardisplacement
between them, allin phase.
Va0
Vb0
Vc0
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Symmetrical Components
EquationsEach phase quantity is equal to the
sum of its symmetrical phasors.
Va = Va0 + Va1 +Va2
Vb = Vb0 + Vb1 +Vb2
Vc = Vc0 + Vc1 +Vc2
The common form of the equationsare written in a-phase terms.
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The a Operator
Used to shift the a-phase terms tocoincide with the b and c-phase
Shorthand to indicate 120° rotation.
Similar to the j operator of 90°.
Va
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Rotation of the a Operator
120° counter clock-wise rotation.
A vector multiplied by 1 /120° results inthe same magnitude rotated 120°.
Va
aVa
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Rotation of the a2 Operator
240° counter clock-wise rotation.
A vector multiplied by 1 /240° results inthe same magnitude rotated 240°.
Va
a2Va
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B-Phase Zero Sequence
We replace theVb sequenceterms by Va
sequence termsshifted by the a operator.
Vb0 = Va0Va0
Vb0
Vc0
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B-Phase Negative Sequence
We replace the Vbsequence terms byVa sequence terms
shifted by the a operator
Vb2 = aVa2 Va2
Vc2
Vb2
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C-Phase Zero Sequence
We replace theVc sequenceterms by Va
sequence termsshifted by the a operator.
Vc0 = Va0Va0
Vb0
Vc0
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C-Phase Negative Sequence
We replace theVc sequenceterms by Va
sequence termsshifted by the a operator
Vc2 = a2Va2
Va2
Vc2
Vb2
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Symmetrical Components
Synthesis Equations
Va = Va0 + Va1 + Va2
Vb = Va0 + a2Va1 + aVa2
Vc = Va0 + aVa1 + a2Va2
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Symmetrical Components
Analysis Equations - 1/3 ??
Where does the 1/3 come from?
Va1= 1/3 (Va + aVb + a2Vc)
Va = Va0 + Va1 + Va2
When balanced
0 0
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Symmetrical Components
Analysis Equations - 1/3 ??Va1= 1/3 (Va + aVb + a2Vc)
Adding the phases
Va
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Symmetrical Components
Analysis Equations - 1/3 ??Va1= 1/3 (Va + aVb +
a2Vc)
Adding the phases yields
Va
aVb
Vc
Va
Vb
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Example Vectors
An Unbalanced Voltage
Va
Vc
Vb
Va = 13.4 /0°Vb = 59.6 /-
104°
Vc = 59.6 /104°
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S t i l C t
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Symmetrical Components
Present During Shunt Faults
Three phase fault
–Positive
Phase to phasefault
–Positive
–Negative
Phase toground fault
–Positive
–Negative
–Zero
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Th Ph F lt Ri ht?
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Three Phase Fault, Right?
-25025
-250
25
-25025
-2500
0
2500
-2500
0
2500
-2500
0
2500
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V
A
V
B
V
C
I A
I B
I C
C y c l e s
V A V B V C IA IB IC
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A t G d F lt Ok ?
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A to Ground Fault, Okay?
-10000
0
10000
-10000
0
10000
-10000
0
10000
-10000
0
10000
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I A
I B
I C
I R
C y c l e s
IA IB IC IR
A Symmetrical Component View
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A Symmetrical Component View
of an A-Phase to Ground Fault
Component Magnitude Angle
Ia0 7340 -79
Ia1 6447 -79
Ia2 6539 -79
Va0 46 204
Va1 123 0
Va2 79 178
0
45
90
13 5
18 0
22 5
27 0
31 5
I0I1I2
V0 V1V2
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A t B F lt E ?
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A to B Fault, Easy?
-10000
0
10000
-10000
0
10000
-10000
0
10000
1 2 3 4 5 6 7 8 9 1 0 11
I A
I B
I C
C y c l e s
IA IB IC
A Phase S mmetrical Component
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A Phase Symmetrical Component
View of an A to B Phase Fault
Component Magnitude Angle
Ia0 3 -102
Ia1 5993 -81Ia2 5961 -16
Va0 1 45
Va1 99 0
Va2 95 -117
0
45
90
13 5
18 0
22 5
27 0
31 5
I1
I2
V1
V2
C Phase S mmetrical Component
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C Phase Symmetrical Component
View of an A to B Phase Fault
Component Magnitude Angle
Ic0 3 138
Ic1 5993 279Ic2 5961 104
Vc0 1 -75
Vc1 99 0
Vc2 95 2.5
0
45
90
13 5
18 0
22 5
27 0
31 5
I1
I2
V1V2
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B t C t G d
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B to C to Ground
-5000
0
5000
-5000
0
5000
-5000
0
5000
-2500
0
2500
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I A
I B
I C
I R
C y c l e s
IA IB IC IR
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Again What Type of Fault?
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Again, What Type of Fault?
-100
0
100
-1000
100
-100
0
100
-200
-0
200
1 2 3 4 5 6 7 8 9 10 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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One Phase Open (Series)
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One Phase Open (Series)
Faults Voltage
– No zero sequence voltage
– Negative 90 out of phase with positivesequence
Current
– Negative and zero sequence 180 out of phase with positive sequence
What About This One?
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What About This One?
-200
0
200
-200
0
200
-200
0
200
-500
0
500
1 2 3 4 5 6 7 8 9 1 0 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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Ground Fault with Reverse Load
Ic0 164 -22
Ic1 89 -113Ic2 41 -6
Vc0 4 -123
Vc1 38 0
Vc2 6 -130
0
45
90
13 5
18 0
22 5
27 0
31 5
I0
I1
I2
V0
V1
V2
Finally The Last One!
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Finally, The Last One!
-250
0
250
-250
0
250
-250
0
250
-100
0
100
1 2 3 4 5 6 7 8 9 10 11
I A
I B
I C
I R
C y c l e s
IA IB IC IR
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Use of Sequence Quantities in
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Use of Sequence Quantities in
RelaysZero Sequence filters
–Current
–Voltage
Relay operating quantity
Relay polarizing quantity
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Zero Sequence Current
Ia
IbIc
Direction of the
protected line
Ia+Ib+Ic3I0
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Zero Sequence Voltage
VaVaVc
Vb
Va
Vb
3Vo
Va
Vb
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Sequence Operating Quantities
Zero and negative sequence currentsare not present during balancedconditions.
Good indicators of unbalanced faults
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Sequence Polarizing Quantities
Polarizing quantities are used todetermine direction.
The quantities used must provide a
consistent phase relationship.
Zero Sequence Voltage
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Zero Sequence Voltage
Polarizing 3Vo is out of phase with Va
-3Vo is used to polarize for ground faults
Va
Vb
3Vo
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Learning Check
Given three current sources
How can zero sequence be producedto test a relay?
How can negative sequenceproduced?
How can zero sequence be
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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 paralleland set their amplitude and the
phase angle equal to one another
–The sequence quantities will be equal toeach source output
How can negative sequence
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How can negative sequence
produced? A single source provides positive, negative
and zero sequence
– Each sequence quantity will be 1/3 of the totalcurrent
Set the three source’s amplitude equal toone another and the phase angles toproduce a reverse phase sequence (Ia at
/0o
, Ib at /120o
and Ic at /-120o
)
– Only negative sequence will be produced
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Advanced Course Topics
Sequence Networks
Connection of Networks for Faults
Per Unit System
Power System Element Models
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References
Symmetrical Components for PowerSystems Engineering, J LewisBlackburn
Protective Relaying, J Lewis Blackburn
Power System Analysis, Stevenson
Analysis of Faulted Power System,Paul Anderson