Symmetrical Components I 2004

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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?

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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



What Type of Fault?

-10000

0

10000

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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



What Type of Fault?

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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



What Type of Fault?

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5000

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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



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



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



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



Basic Course Topics

Terminology

Phasors

EquationsFault Analysis Examples

Calculations



Unbalanced Fault

Ia

Ib

Ic

Ia

Ib

Ic

Ia

Ib

Ic



Symmetrical Component

PhasorsThe unbalanced three phase system

can be transformed into threebalanced phasors.

–Positive Sequence

–Negative Sequence

–Zero Sequence



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



Positive Phase Sequence

The positivesequencequantities have a-

b-c, counter clock-wise, phaserotation. V

a1

Vb1

Vc1



Negative Phase Sequence

Each have thesame magnitude.

Each negative

sequence voltageor current quantityis displaced 120°from one another.

Va2

Vc2

Vb2



Negative Phase Sequence

The negativesequencequantities have a-

c-b, counter clock-wise, phaserotation. V

a2

Vc2

Vb2



Zero Phase Sequence

Each zerosequence quantityhas the same

magnitude. All three phasors

with no angulardisplacement

between them, allin phase.

Va0

Vb0

Vc0



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.



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



Rotation of the a Operator

120° counter clock-wise rotation.

A vector multiplied by 1 /120° results inthe same magnitude rotated 120°.

Va

aVa



Rotation of the a2 Operator

240° counter clock-wise rotation.

A vector multiplied by 1 /240° results inthe same magnitude rotated 240°.

Va

a2Va



B-Phase Zero Sequence

We replace theVb sequenceterms by Va

sequence termsshifted by the a operator.

Vb0 = Va0Va0

Vb0

Vc0



B-Phase Negative Sequence

We replace the Vbsequence terms byVa sequence terms

shifted by the a operator

Vb2 = aVa2 Va2

Vc2

Vb2



C-Phase Zero Sequence

We replace theVc sequenceterms by Va

sequence termsshifted by the a operator.

Vc0 = Va0Va0

Vb0

Vc0



C-Phase Negative Sequence

We replace theVc sequenceterms by Va

sequence termsshifted by the a operator

Vc2 = a2Va2

Va2

Vc2

Vb2




Synthesis Equations

Va = Va0 + Va1 + Va2

Vb = Va0 + a2Va1 + aVa2

Vc = Va0 + aVa1 + a2Va2




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




Analysis Equations - 1/3 ??Va1= 1/3 (Va + aVb + a2Vc)

Adding the phases

Va




Analysis Equations - 1/3 ??Va1= 1/3 (Va + aVb +

a2Vc)

Adding the phases yields

Va

aVb

Vc

Va

Vb



Example Vectors

An Unbalanced Voltage

Va

Vc

Vb

Va = 13.4 /0°Vb = 59.6 /-

104°

Vc = 59.6 /104°



S t i l C t




Present During Shunt Faults

Three phase fault

–Positive

Phase to phasefault

–Positive

–Negative

Phase toground fault

–Positive

–Negative

–Zero



Th Ph F lt Ri ht?



Three Phase Fault, Right?

-25025

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2500

-2500

0

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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



A t G d F lt Ok ?



A to Ground Fault, Okay?

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0

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0

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0

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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

A Symmetrical Component View



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



A t B F lt E ?



A to B Fault, Easy?

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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



A Phase Symmetrical Component

View of an A to B Phase Fault


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



C Phase Symmetrical Component

View of an A to B Phase Fault


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



B t C t G d



B to C to Ground

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I A

I B

I C

I R

C y c l e s

IA IB IC IR



Again What Type of Fault?



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



One Phase Open (Series)



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?



What About This One?

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I A

I B

I C

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C y c l e s

IA IB IC IR



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!



Finally, The Last One!

-250

0

250

-250

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-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



Use of Sequence Quantities in



Use of Sequence Quantities in

RelaysZero Sequence filters

–Current

–Voltage

Relay operating quantity

Relay polarizing quantity



Zero Sequence Current

Ia

IbIc

Direction of the

protected line

Ia+Ib+Ic3I0



Zero Sequence Voltage

VaVaVc

Vb

Va

Vb

3Vo

Va

Vb



Sequence Operating Quantities

Zero and negative sequence currentsare not present during balancedconditions.

Good indicators of unbalanced faults



Sequence Polarizing Quantities

Polarizing quantities are used todetermine direction.

The quantities used must provide a

consistent phase relationship.





Polarizing 3Vo is out of phase with Va

-3Vo is used to polarize for ground faults

Va

Vb

3Vo



Learning Check

Given three current sources

How can zero sequence be producedto test a relay?

How can negative sequenceproduced?

How can zero sequence be



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



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



Advanced Course Topics

Sequence Networks

Connection of Networks for Faults

Per Unit System

Power System Element Models



References

Symmetrical Components for PowerSystems Engineering, J LewisBlackburn

Protective Relaying, J Lewis Blackburn

Power System Analysis, Stevenson

Analysis of Faulted Power System,Paul Anderson

Symmetrical Components I 2004

Documents

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