Componentes simétricas

Symmetrical Components

Symmetrical Component AnalysisSynthesis of Unsymmetrical Phasors from

Their Symmetrical ComponentsThe Symmetrical Components of Unsym

metrical PhasorsPhase Shift of Symmetrical Components i

n or Transformer BanksPower in Terms of Symmetrical Compone

nts Y Y

Symmetrical ComponentsUnsymmetrical Series ImpedanceSequence Impedance and Sequence

NetworkSequence Networks of Unloaded gen

eratorsSequence NetworkZero-Sequence Network

Symmetrical Component Analysis

Goal :

Symmetrical component analysis is a very useful tool

for dealing with unbalanced three-phase faults.

Synthesis of Unsymmetrical Phases from Their Symmetrical Components 1

“An unbalanced system of n related phasors can be resolved into

n systems of balanced phasors called the symmetrical components

of the original phasors. The n phasors of each set of components

are equal in lengths , and the angles between adjacent phasors of

the set are equal.”

by C.L Fortescue , 1918


1. For positive- sequence

components

2. For negative-sequence

components

(1) Positive- sequence components (2) Negative-sequence components

1a 1b

1c

2a2c

2b

n n


(3) Zero-sequence components

0aV0bV

0cV

0 For zero-sequence components


021 aaaa VVVV

021 bbbb VVVV

021 cccc VVVV

0aV

2aV

aV

1aV

1bV

2bV0bV

1cV

bV

0cV cV

2cV


Use a to designate the operator that causes a rotation of in the

counterclockwise direction ,

13601

866.05.02401

866.05.01201

03

02

0

a

ja

ja

0120

2a

3,1 a

2a a

3,1 a

a

The Symmetrical Components of Unsymmetrical Phasors 1

0000

22

222

1112

1

,

,

,

acab

acab

acab

VVVV

VaVVaV

aVVVaV

1a 1b

1c

2a

2c

2b


022

1021

0212

021

021

aaacccc

aaabbbb

aaaa

VVaVaVVVV

VaVVaVVVV

VVVV

2

1

0

2

2

1

1

111

a

a

a

c

b

a

V

V

V

aa

aa

V

V

V


2

2

1

1

111

aa

aaA

aa

aaA2

21

1

1

111

3

1

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

* When three phase phasors are balanced , only the

positive-sequence component exists .


1.Sequence component representation of L-L voltage

ca

bc

ab

ab

ab

ab

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

2.Sequence component representation of current

c

b

a

a

a

a

I

I

I

aa

aa

I

I

I

2

2

2

1

0

1

1

111

3

1


)(3

1

)(3

1

)(3

1

0

0

0

cbaa

cabcabab

cbaa

IIII

VVVV

VVVV

No zero-sequence components exist if the sum of the three

phasors is zero.


)(3

10 cbaa VVVV

00 aV

00 aV3When is balanced

When 0)( cba VVV

* If then is unbalanced.

* Unbalanced does not guarantee .

00 aV 3

3 00 aV


)(3

10 cabcabab VVVV

is always zero whether the three phase system

is balanced or not.

)( cabcab VVV is always zero (form closed loop)

0abV

ab

c


aI

bI

cI

aI

bI

cI

aI

bI

cInI

0)(3

10 cbaa IIII

( ungrounded Y )

0)(3

10 cbaa IIII

nacba IIIII 03)(

Y with a path to neutral

0)(3

10 cbaa IIII

connected


example : One conductor of a three-phase line is open. The current flowing to the

-connected load through line a is 10 A. With the current in line a as

reference and assuming that line c is open, find the symmetrical components

of the line currents

Z Z

Z

a

c

b

ampIa0010

ampIb018010

0cI

AIa0010

AIb018010

AIc 0

The line currents are :


0)018010010(3

1 000 aI

)012018010010(3

1 0001 aI

Aj 03078.589.25

)024018010010(3

1 0002 aI

Aj 03078.589.25

AIb0

1 15078.5

AIb0

2 15078.5

00 bI

AIc0

1 9078.5

AIc0

2 9078.5

00 cI

Since there no neutral current involved ,

should be zero .0aI

Phase Shift of Symmetrical Components in or Transformers Banks 1Y Y

The American standard for designating terminal and on or

transformer requires that the positive-sequence voltage drop from to

neutral leads the positive-sequence voltage drop from to neutral by ,

regardless of whether the or winding is on the high tension side .

Similarly, the positive-sequence voltage drop from to neutral leads the

voltage drop from to neutral by and the positive-sequence voltage

drop from to neutral leads the voltage drop from to

neutral by .

1H

1X Y1H

1X030

030

3H

2X

3X

2H

030

Y

Y


Example :

A

B

C a

c

b1H

3H2H

2X1X

3X

A

B

C

a

b

c

1H

3H2H

1X

2X

3X

1AV 1bV 030leads by 1aVleads by1AV 030


The American standard for designating terminal and on or

transformer requires that the negative-sequence voltage drop from to

neutral lags the negative-sequence voltage drop from to neutral by ,

regardless of whether the or winding is on the high tension side .

Similarly, the negative-sequence voltage drop from to neutral lags

the voltage drop from to neutral by and the negative-sequence

voltage drop from to neutral lags the voltage drop from to neutral

by .

1H

1X Y1H

1X030

030

3H

2X

3X

2H

030

Y

Y


A

B

C a

c

b1H

3H2H

2X1X

3X

A

B

C

a

b

c

1H

3H2H

1X

2X

3X

2aV(b) lags by2AV030

2bV(a) lags by2AV 030

Example :


A

B

C a

c

b1H

3H2H

2X1X

3X

2B

2A

2C

2b

2c

2a

1B

1A

1C

1b

1a

1c

1AVleads by1aV 090

2AVlags by2aV090

1bVleads by1AV 030

2bVlags by2AV 030

Y

Phase Shift of Symmetrical Components in or Transformers Banks 6

Example 11.7. The resistive Y-connected load bank of Example 11.2 is supplied from the low-voltage

Y-side of a Y- transformer. The voltages at the load are the same as in that example. Find the line

voltages and currents in per unit on the high-voltage side of the transformer.

unitperI a0)1( 6.439857.0

unitperI a0)2( 3.2502346.0

)(6.439857.0 0)1( basevoltageneutraltolineunitperV an

)(3.2502346.0 0)2( basevoltageneutraltolineunitperV an

9456.02785.06.739857.0306.439857.0 000)1( jV A

1517.01789.03.2202346.0303.2502346.0 000)2( jV A

unitperjVVV AAA0)2()1( 8.828.07939.00994.0

7138.06798.04.469857.0 0)1(2)1( jVaV AB

Y Y


0791.02209.07.192346.0 0)2(2)2( jVaV AB

unitperjVVV BBB0)2()1( 4.4120.17929.09007.0

2318.09581.06.1939857.0 0)1(2)1( jVaV AC

2318.00419.03.1002346.0 0)2(2)2( jVaV AC

unitperjVVV CCC0)2()1( 1800.100.1

5868.18013.07929.09007.07939.00994.0 jjjVVV BAAB

)(8.11678.1 0 basevoltagaeneutrallineunitper

)(8.1163

78.1 0 basevoltagaelinetolineunitper

Y Y


7939.09007.10.17939.09007.0 jjVVV CBBC


)(7.2219.17.223


97939.00994.17939.00994.00.1 jjVVV ACCA


)(8.215783.08.2153


unitperI A08.8280.0

unitperIB04.4120.1

unitperIC01800.1

Y Y


(b) leads by 030aV )1((a) leads byAV )1( 030 AV )1(aV )1(

Figure 11.23

labeling of lines connected to a three-phase Y- transformer.

Y Y

A

B

C a

c

b1H

3H2H

2X1X

3X

A

B

C

a

b

c

1H

3H2H

1X

2X

3X

Example :

Power in terms of Symmetrical Components

ccbbaa IVIVIVjQPS ***

*012012

*

AIAV

I

I

I

V

V

V

S T

c

b

a

T

c

b

a

*012012

*012

*012 3 IVIAAV TTT

*

2

1

0

2103

a

a

a

aaa

I

I

I

VVV

)(3 2*

21*

10*

0 aaaaaa IVIVIV

IAAT 3, *

Unsymmetrical Series Impedance 1

a

b

c 'c

'b

'a

cZ

bZ

aZ

aI

bI

cI

caZacZ

abZ

c

b

a

ccbca

bcbba

acaba

cc

bb

aa

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

'

'

'


2

1

0

2

1

0

'

'

'

a

a

a

ccbca

bcbba

acaba

cc

bb

aa

I

I

I

A

ZZZ

ZZZ

ZZZ

V

V

V

A

2

1

01

2

1

0

'

'

'

a

a

a

ccbca

bcbba

acaba

cc

bb

aa

I

I

I

A

ZZZ

ZZZ

ZZZ

A

V

V

V

Z


ZAAZ 1012

)2()2()2(

)2()2()2(

)2()2()2(

001122

220011

112200

MsMsMs

MsMsMs

MsMsMs

ZZZZZZ

ZZZZZZ

ZZZZZZ

Where

)(3

1

)(3

1

)(3

1

22

21

0

cbas

cbas

cbas

aZZaZZ

ZaaZZZ

ZZZZ

)(3

1

)(3

1

)(3

1

22

21

0

abcabcM

abcabcM

abcabcM

aZZaZZ

ZaaZZZ

ZZZZ


2111220000)2()2()2(' aMsaMsaMsaaIZZIZZIZZV




Case 1. If no coupling , 0)( jiijZ

2112000' asasasaaIZIZIZV

)(3

1)(

3

1)(

3

1 22

210 cbaacbaacbaa ZaaZZIaZZaZIZZZI



)(3

1)(

3

1)(

3

1 221

20 cbaacbaacbaa ZZaZIZZZIZaaZZI

)(3

1)(

3

1)(

3

12

21

20 cbaacbaacbaa ZZZIZaaZZIaZZaZI

then 0210 MMM ZZZ


Case 2 . If

0)( jiijZcba ZZZ

aaaaZIV 11'

aaaa

ZIV 22' aaaaZIV 00'

Symmetrical components of unbalanced currents flowing in a balanced- load

or in balanced series impedances produce voltage drops of the same sequence ,

provided no coupling exists between phases.

If the impedances are unequal, the voltage drop of any one sequence is dependent on the

current of all three sequences.

If coupling such as mutual inductance exists among the three impedances, then the

formula will become more complicated.

Y

1.

2.

Complete transportation assumed


Assume:

1. No coupling

2.

Positive-sequence currents produce positive-sequence voltage drops.

Negative-sequence currents produce negative sequence voltage drops.

zero-sequence currents produce zero-sequence voltage drops.

cba ZZZ

Sequence Impedance and Sequence Network 1

The impedance of circuit when positive- sequence

current alone are flowing is called positive-sequence

impedance.

The impedance of circuit when negative-sequence

currents alone are flowing is called negative

sequence impedance.

When only zero-sequence currents are present, the

impedance is called zero sequence impedance.


The single-phase equivalent circuit composed of the impedance to

current of any one sequence only is called the sequence network.

Positive-sequence network contains positive sequence current and

positive sequence impedance only.

Negative-sequence network contains negative sequence current

and negative sequence impedance only.


Sequence network carrying the individual currents ,

and are interconnected to represent various

unbalanced fault condition.

1aI 2aI

0aI

Zero-sequence network contains zero sequence current and

zero sequence impedance only.


Sequence Impedance of Various Devices

Positive Negative Zero

Line same same different

Transformer same same same

Machine different different different**

* Usually they are assumed to be the same

Sequence Networks of Unloaded Generators 1

The generator voltage

are of positive sequence only,

since the generator is designed

to supply balanced three-phase

voltage.

aI

bI

cI

nI

cEbE

aE

bc

a

++

+

-

--

nZ),,( cba EEE


2aI

2cI

2bI

a

c b

2Z

2Z2Z 2aV

2aIa

2Z

+

-

Negative-sequence

network

+

-

1aVaE

1aIa

1Z

+

-

Positive-sequence

network

Reference

Reference

1aI

1bIcE

bE

aE

bc

a

+

++

-

--

1Z

1Z1Z

aI

111 ZIEV aaa

222 ZIV aa


1cI

0aI

0aI

0aI

a

c b

0gZ

0gZ 0gZ

0aI

0aV

a0gZ

+

-nZ3

0Z

Zero-sequence

network

Reference

only appears in the zero-sequence network

nZ

000 ZIV aa nZ

03 aI)3( 00 nga ZZI

na II 03


Example 11.6. A salient-pole generator without dampers is rated 20 MVA, 13.8kV and has a

direct=axis subtransient reactance of 0.25 per unit. The negative-and-zero-sequence reactance

are, respectively, 0.35 and 0.10 per unit. The neutral of the generator is solidly grounded. With

the generator operating unloaded at rated voltage with , a single line-to-

ground fault occurs at the machine terminals, which then have per-unit voltages to ground,

Determine the subtransient current in the generator and the line-t0-line voltages for subtransient

conditions due to the fault.

unitperEan000.1

0aV025.102013.1 cV

025.102013.1 bV

0cI

aI

0bI

na II

cnEbnE

anE

bc

a

++

+

-- -

nZ

n

Figure 11.15


Figure 11.15 shows the line-to-ground fault of phase a of the machine.

unitperjVb 990.0215.0

unitperjVc 990.0215.0

unitperjj

j

Z

VI

go

aa 43.1

10.0

)0143.0()0()0(


unitperjj

j

Z

VI

aa 43.1

35.0

)0500.0(

2

)2()2(

29.43 )0()2()1()0( jIIIII aaaaa

There, the fault current into the ground is

The base current is and so the subtransient current in line a isA837)8.133(000,20

AjjIa 590,383729.4


Line-to-line voltage during the fault are

unitperjVVV baab07.7701.1990.0215.0

unitperjVVV cbbc0270980.1980.10

unitperjVVV acca07.7701.1990.0215.0

kVVab00 7.7705.87.77

3

8.1301.1

kVVbc00 27078.15270

3

8.13980.1

kVVca00 3.10205.83.102

3

8.1301.1


Before the fault the line voltages were balanced and equal to 13.8kV. For comparison with the line

voltages after the fault occurs, the prefault voltages, with as reference, are given asanan EV

kVVab0308.13 kVVbc

02708.13 kVVca01508.13

Figure 11.6 shows phasor diagrams of prefault and postfault voltages.

Figure 11.6

(a) Prefault (b) Postfault

anV

caVcaV

bcVbcV

abVabV

aan

b b

c c


The positive-sequence diagram of a generator is composed of an emf in series with the positive-sequence

impedance of the generator.

The negative and zero-sequence diagrams contain no

emfs but include the negative and zero-sequence

impedances of the generator respectively.

Sequence Networks 1

The matching reactance in positive-sequence network is the subtransient ,transient,

or synchronous reactance, depending of whether subtransient , transient, or

steady- state condition are being studied.

The reference bus for the positive and negative sequence networks is the neutral

of the generator. So for as positive and negative sequence components are

concerned , the neutral of the generator is at ground potential even if these is

connection between neutral and ground.

The reference bus for the zero sequence network is the ground (not necessary

the neutral of the generator).

nZ

Sequence Networks 2

Convert a positive sequence network to a negative sequence

network by changing, if necessary, only the impedance

that represent rotating machine , and by omitting the emf.

The normal one-line impedance diagram plus the induced emf is the

positive sequence network.

Three-phase generators and motors have internal voltage of positive

sequence only.

Example of Positive and Negative-Sequence Network 1

Example: Draw the positive and negative-sequence networks for the system described as below . Assume that the

negative-sequence reactance of each machine is equal to its subtransient reactance .Omit resistance.

1T1M

T

r

pm nk l

2M

Example of Positive and Negative-Sequence Network 2

++

+

---

0857.0j 0915.0j0815.0j

5490.0j

02.0j

2745.0j

2mE1mEgE

k l m np r

0857.0j 0915.0j0815.0j

5490.0j02.0j 2745.0j

Reference bus

k l m n

p q

(Positive)

(Negative)

Zero sequence Network 1

1 . Zero-sequence network currents will flow only if a return path exists.

2 . The reference bus of the zero-sequence network is the ground.

ZZ

Z

N

Z

Z

Z

N

Z N

R

N

Reference


Z

Z

Z

NZ

N

R

nN

03 an II

nZ3

0aI

Z Z

Z

R

Z


Zero-sequence equivalent circuit of three phase transform banks.

Symbols Connection DiagramsZero-Sequence

Equivalent Circuit

p

pp

p p p

QQ Q

Q Q

0Z

0Z

Reference bus

Reference bus


Symbols Connection Diagrams Equivalent CircuitZero-Sequence

pp

p

p

pp

QQ

Q Q

0Z

0Z

Reference bus

Reference bus


Symbols Connection Diagrams Equivalent CircuitZero-Sequence

pp

pQ Q

0Z

Reference bus


nZ3

0gZQ

R

S

T

M NP

(Zero-Sequence)

Example:

nZ

Q S

P

R T

M N


1gZ

Q S

R T

MNP

2gZ

1gE 1gE++

--

Q

R TM

NP

S

(Positive-Sequence)

(Negative-Sequence)


Example 11.9. Draw the zero-sequence network for the system described in Example 6.1. Assume

zero-sequence impedance for the generator and motors of 0.05 per unit. A current-limiting reactor of

0.4 Ω is in each of the each of the neutrals of the generator and the large motor. The zero-sequence

reactance of the transmission line is 1.5 Ω/km.

Generator:

Motor 1:

Motor 2:

unitperX 05.00

unitperX 0686.0)8.13

2.13)(

200

300(05.0 2

0

unitperX 1372.0)8.13

2.13)(

100

300(05.0 2

0

333.1300

)20( 2

ZBase

635.0300

)8.13( 2

ZBase

Generator

Motor


unitperZn 900.0)333.1

4.0(33

unitperZn 890.1)635.0

4.0(33

unitperZ 5445.03.176

645.10

The zero-sequence network is shown in Fig. 11.28

05.0j

900.0j

0857.0j 5445.0j 0915.0j

l m

kn

p r

0686.0j

890.1j

1372.0j

reference

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