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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