Un Symmetrical Faults

8/2/2019 Un Symmetrical Faults

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Un-Symmetrical Faults

Symmetrical components The method of symmetrical components is a powerful

technique for analyzing unbalanced three phase systems.

It is a linear transformation that transforms from phasecomponents to a new set of components called symmetricalcomponents.

The advantage of this transformation for balanced three phasenetworks the equivalent circuit obtained called the sequencenetwork are separated into three uncoupled networks.

For unbalanced three phase systems, the three sequencenetworks are connected only at point of unbalance.

Decoupling a detailed three phase network into three simplersequence networks reveals complicated phenomena.


2/202

Symmetrical components Assume that a set of three phase voltages designated Va, Vb

and Vc is given.

These phase voltages are resolved into the following threesets of sequence components:

a) Zero sequence components: consisting of three phasors withequal magnitudes and zero phase displacement.

b) Positive sequence components: consisting of three phasorswith equal magnitudes and 120 phase displacement andpositive sequence.

c) Negative sequence components: consisting of three phasors

with equal magnitudes and 120 phase displacement andnegative sequence.

Symmetrical components

Zero sequence

components Positive sequencecomponents

Negative sequence

components

Va0Vb0

Vc0

Va1

Vb1

Vc1

Vb2

Vc2

Va2

Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0

Vc = Vc1 + Vc2 + Vc0


3/203


Zero sequence

componentsPositive sequencecomponents

Negative sequencecomponents

Va0

Vb0Vc0

Va1

Vb1

Vc1

Vb2

Vc2

Va2

Va

Vb

Vc


Va = Va1 + Va2 + Va0 Vb = Vb1 + Vb2 + Vb0 Vc = Vc1 + Vc2 + Vc0

Let us define the following operator a as follows:

1201=a 24012

=a

1

2

1 ab VaV = 11 ac aVV=

22 abaVV =

2

2

2 acVaV =

00 abVV = 00 ac VV =

=

2

1

0

2

2

1

1

111

a

a

a

c

b

a

V

V

V

a a

aa

V

V

V


4/204


=

2

2

1

1

111

a a

aa

A

=

aa

a a

A

2

21

1

1

111

3

1

=

c

b

a

a

a

a

V

V

V

aa

a a

V

V

V

2

2

2

1

0

1

1

111

3

1)(

3

10 cbaa VVVV ++=

)(3

1 21 cbaa VaaVVV ++=

)(3

1 22 cbaa aVVaVV ++=


)(3

10 cbaa VVVV ++=

This equation shows that no zero sequence components exists if the sum of

the unbalanced phasors is zero.

Since the sum of the line-line voltage phasors in a three phase system is

always zero (Why?), zero sequence components are never present in the linevoltages regardless of the amount of unbalance.

However, the sum of the three line-neutral voltage phasors is not necessaryzero and hence line-neutral voltages may contain zero sequence components.


5/205


)(3

10 cbaa IIII ++=

)(3

1 21 cbaa IaaIII ++=

)(3

1 22 cbaa aIIaII ++=

The previous set of equations can be written for currents as well as shown below:

021 aaaa IIII ++=

021 bbbb IIII ++=

021 cccc IIII ++=

In a three phase system, the sum of the line currents is equal to the current Inin the return path

ncbaIIII =++ )( 03 an II =

Example 1:

+

=

120277

120277

0277

cn

bn

an

V

V

V

Calculate the sequence components of the following balanced line-neutralvoltages with abc sequence:

Solution:

0)(3

1

0=++=

cbaVVVV

0277)(3

1 21 =++= cba VaaVVV

0)(3

1 22 =++= cba aVVaVV


6/206

Example 2:

+

=

12010

12010

010

c

b

a

I

I

I

A Y connected load has balanced currents with acb sequence as follows, find thesequence currents:

Solution:

0)(3

10 =++= cba IIII

0)(31 2

1 =++= cba IaaIII

010)(3

1 22 =++= cba aIIaII

Example 3:

=

12010

0

010

c

b

a

I

I

I

A three phase line feeding a balanced Y load has on of its phases (phase b)open. The load neutral is grounded and the unbalanced currents are

Calculate the sequence and neutral currents:

Solution:

6033.3)(3

10 =++= cba IIII

067.6)(3

1 21 =++= cba IaaIII

6033.3)(3

1 22 =++= cba aIIaII

6010)( =++=cban

IIII

60103 0 == IIn

OR


7/207

21 aaaoIII ++

Representing the unbalance currents withtheir symmetrical components we get:

cban IIII ++=

21 bbboIII ++ 21 ccco III +++= +nI

coboaoIII ++

4434421

0

111

zer

cbaIII ++

4434421

zero

cbaIII

222+++= +nI

aoI3=nI

Since the positive and negative sequence components are add to zero at theneutral point, therefore there is no positive or negative sequence components

flow from the neutral to the ground.

1. The Sequence circuits for Wye

and Delta connected loads

For the star connected load withgrounded neutral point,

YZaI

bI

cI

YZYZ

abV

bcV

caV

aonI3I ====

anV

nV

n

Under unbalance condition:

aoI3=nI

naonnn ZIZIV 3==

The voltage drop between neutraland ground is:

It is very important to distinguish between voltages to neutral and voltages to ground.

naoannana ZIVVVV 3+=+=

=

c

b

a

V

V

V

cn

bn

an

V

V

V

naoZI3+

+

n

n

n

V

V

V

1

1

1

c

b

a

I

I

I

YZ=

For unbalance threephase system, the phase

voltages are:

YZaI

bI

cI

YZYZ

abV

bcV

caV

aon I3I ====

anV

nV

n


8/208

2

1

a

a

ao

I

I

I

AYZ====

2

1

a

a

ao

V

V

V

A naoZI3++++

1

1

1

Multiplying byA-1

Using the symmetricalcomponents:

2

1

a

a

ao

I

I

I

YZ====

2

1

a

a

ao

V

V

V

naoZI3++++

1

1

1

1A

2

1

a

a

ao

I

I

I

YZ=

2

1

a

a

ao

V

V

V

naoZI3+

0

0

1

aooao

aonYao

IZV

I)Z3Z(V

====

++++====

22

11

aYa

aYa

IZV

IZV

=

=

OREq. 1

Eq. 2

Eq. 3

nYoZZZ 3+=

11 aYa IZV =

aoI

aoV

YZ

nZ3

n

oZ

22 aYaIZV =

1aI

+aV

YZ n

1Z

aonYao IZZV )3( +=

2a

I

aV

YZ n

2Z

Using the three previous

equations (Eqs. 1, 2 and 3)then the Symmetrical circuitsfor a Wye-connected load

with neutral point connectedto ground is:

YZaI

bI

cI

YZYZ

abV

bcV

caV

aon II 3=

anV

nV Positive Sequence Circuit

Negative Sequence Circuit

Zero Sequence Circuit


9/209

1aI

1aV

YZ n

1Z

2aI

2aV

YZ n

2Z

YZ

aI

bI

cI

YZYZ

abV

bcV

caV anV

If the neutral point of a Y-connected load is notgrounded, therefore, no zero sequence current

can flow, and

Symmetrical circuits for Y-connected load

with neutral point is not connected to groundare presented as shown:

====nZ

aoI

ao

V

YZ n

oZ ====nZ

Y-connected load (Isolated Neutral):


Positive Sequence Circuit


The Delta circuit can not provide a path through neutral. Therefore for a Deltaconnected loador its equivalent Y-connectedcan not contain any zero sequence

components.

Delta connected load:

ZZ

Z

aI

bI

cI

abV

bcV

caVabI

bcI

caI

abab IZV ====

bcbc IZV ====

caca IZV ====

0V)VVV(3

10abcabcab ========++++++++

The summation of the line-to-line voltages

or phase currents are always zero

and

Therefore, for a Delta-connected loadswithout sources or mutual coupling there will be no

zero sequence currents at the lines (There are some cases where a circulating currents maycirculate inside a delta load and not seen at the terminals of the zero sequence circuit).

aoI

aoV

nZ 1aI

1aV

n3/Z

2aI

2aV

n3/Z

0I)III(3

10abcabcab ========++++++++

NegativeSequence

Circuit

PositiveSequence

Circuit

ZeroSequence

Circuit


10/2010

The Delta circuit have balanced impedances of21 ohms. Determine the sequence impedances.

Example:

The positive- and negative-sequence circuits have per-phase

impedance

==

73/1 ZZ

==

73/2 ZZ

Solution:

ZZ

Z

aI

bI

cI

abV

bcV

caVabI

bcI

caI

The zero-sequence circuit have per-phaseimpedance of 21 ohms. Thezero sequence current is circulating in Delta circuit.

aoI

aoV

nZaoI

aoV

nZ +aI

+aV

n3/Z

+aI

+aV

n3/Z

aI

aV

n3/ZaI

aV

n3/Z

2. Sequence Circuits ofTransmission Lines

Consider a symmetrical transmission

line where,

Zaa: is the self-impedance and is the same for each phase

Zab: is the mutual-impedance between each two phases

Znn: is the self-impedance of the neutral conductor

Zan: is the mutual-impedance between the neutral and each phase

bI

cI

aI

anZabZ

nI

a

b

c

n

a

b

c

n

aaZ

nnZ

aaZ

aaZ

nnncanbanaan

nanancabbabaaaan

IZIZIZIZ

VIZIZIZIZV

++++++++++++++++==== Using KVL

The voltage drop across the line section is:

nnnancanabbanabaanaanaan I)ZZ(I)ZZ(I)ZZ(I)ZZ(VV ++++++++++++====

Similarly, for phases b and c:

nnnancbanabaanaanaan I)ZZ()II)(ZZ(I)ZZ(VV ++++++++++++==== Eq. 1

ccbbaa ZZZ ========

bcacab ZZZ ========

cnbnan ZZZ ========

212111 IMjILjV ++++====


11/2011

nnnanbaanabcanaanccn I)ZZ()II)(ZZ(I)ZZ(VV ++++++++++++====

nnnancaanabbanaanbbnI)ZZ()II)(ZZ(I)ZZ(VV ++++++++++++====

The neutral current is:)III(I

cban++++++++====

Substituting Eq. 4 intoEqs. 1,2 and 3

Eq. 2

Eq. 3

Eq. 4

)III)(ZZ()II)(ZZ(I)ZZ(VV cbannancbanabaanaanaan ++++++++++++====

cannnabbannnabaannnaanaan I)Z2ZZ(I)Z2ZZ(I)Z2ZZ(VV ++++++++++++++++++++====

cannnabbannnaaaannnabnbbn I)Z2ZZ(I)Z2ZZ(I)Z2ZZ(VV ++++++++++++++++++++====

cannnaabannnabaannnabnccn I)Z2ZZ(I)Z2ZZ(I)Z2ZZ(VV ++++++++++++++++++++====

annnaas Z2ZZZ ++++====

annnabm Z2ZZZ ++++====

Let:

smm

msm

mms

ZZZ

ZZZZZZ

Then, the voltage drops across the lines are

c

b

a

I

II

)ZZ(00

0)ZZ(0

00)ZZ(

ms

ms

ms

====

nccn

nbbn

naan

VV

VVVV

====

cc

bb

aa

V

VV

Using symmetrical components and rearranging the impedance matrix, we get:

=

2

1

0

aa

aa

aa

V

V

V

A

++++

mmm

mmm

mmm

ZZZ

ZZZ

ZZZ

2

1

0

a

a

a

I

I

I

A

++++

111

111

111

Z

100

010

001

)ZZ(A mms1

Multiplying by A-1

2

1

0

a

a

a

I

I

I

A=

2

1

0

aa

aa

aa

V

V

V


12/2012

Or,

=

2'

1'

0

aa

aa

aa

V

V

V

++++

)ZZ(00

0)ZZ(0

00)Z2Z(

ms

ms

ms

2

1

0

c

b

a

I

I

I

msoZZZ 2+=

msZZZ =1

msZZZ =2

annnaas ZZZZ 2++++====

annnabm ZZZZ 2++++====

Where,

Therefore,

Substituting forms ZandZ

)2(2)2( annnabannnaao ZZZZZZZ ++++++++++++====

msoZZZ 2++++====

annnabaao ZZZZZ 632 ++++++++====

aoIa

n

a

n

0anV 0naV oZ

)2()2(1 annnabannnaa ZZZZZZZ ++=

msZZZ =1

abaaZZZ =

1

msZZZZ == 12

abaaZZZ =2

And

++++Z

++++aI

a

n

a

n

++++anV

++++naV

aI

Z

a

n

a

n

anV

naV

The positive and negative sequence impedances are equal and dont include the neutral

conductor impedances . The return path conductors enter into the zero

sequence impedances only.

)ZorZ( annn

Notes:

The ground wires (above overhead TL) combined with the earth works as a neutral

conductor with impedance parameters that effects the zero sequence

components. Having a good grounding (depends on the soil resistively), then the voltages

to the neutral can be considered as the voltages to ground.

)ZandZ( annn


13/2013

3. Sequence Circuits of Synchronous Machines

aI

bI

anE

cnE

cI

nInZ

bnE

aI

bI

anE

cnE

cI

nInZ

bnE

3. Sequence Circuits of Synchronous Machines

Positive Sequence Circuit:

The windings of a synchronous machine are

symmetrical.

Thus the generator voltages are of positive

sequence only.

The positive sequence network consists of an

EMF (equal to no-load terminal voltage) in series

with the positive sequence impedance of the

machine.

++++aI

++++bI

anE

cnE

++++cI

bnE

++++Z

++++Z

++++Z

++++aI

++++bI

anE

cnE

++++cI

bnE

++++Z

++++Z

++++Z

++++aV++++Z

Ean ++++aV++++Z

Ean

The neutral impedance (Zn) does not appear in this

circuit because no positive sequence current will

flow through it.

0111=++

cbaIII


14/2014

aI

aVZ

Negative Sequence Circuit:

The synchronous machine does not generate any negative sequence voltages.

Z

ZZ

aI

bI

cI

Z

ZZ

aI

bI

cI

Zero Sequence Circuit:

No zero sequence voltage is included in a synchronous machine.

goZ

aoI

boI

coI

nZgoZ

goZaoI3

goZ

aoI

boI

coI

nZgoZ

goZaoI3

aoI

aoV

nZ3

goZ

aoI

aoV

nZ3

goZ

aocoboao IIII 3====++++++++

goZ :Zero sequence impedanceper phase.

gono ZZZ ++++==== 3

++++aVEan

++++Z01

aoI

aoV

nZ3

goZ

aI

aVZ

++++aVEan

++++Z01

aoI

aoV

nZ3

goZaoI

aoV

nZ3

goZ

aI

aVZ

aI

aVZ

Zero

Sequence

Negative

Sequence

Positive

Sequence

Symmetrical

Circuits

aI

bI

anE

cnE

cI

nInZ

bnE

aI

bI

anE

cnE

cI

nInZ

bnE

Summary of the threesequence circuits


15/2015

4. Sequence Circuits of Delta and Wye Transformers

A. Wye-Wye Bank, Both

Neutrals GroundedWith both wyes grounded, zeroWith both wyes grounded, zero

sequence current can flow. Thesequence current can flow. The

presence of the current in onepresence of the current in one

winding means that secondarywinding means that secondary

current exists in the other.current exists in the other.

A

B

C

N

a

b

c

n

NZ nZ

210 AAAAIIII ++=

210 BBBB IIII ++=

210 CCCCIIII ++=

++++++++++++==== aa0aa IIII

0AI3 0aI3

A

N

a

n

NZ nZ

++++++++++++==== AA0AA IIII

++++++++++++==== aa0aa IIII

0AI3 0aI3

ANV anV

A

VaV

TheThe flowflow ofof thethe sequencesequence currentscurrents dependdepend onon thethe windingwinding connectionsconnections.. TheThe differentdifferent

installationsinstallations ofofDeltaDelta--WyeWye windingswindings determinedetermine thethe configurationconfiguration ofof thethezerozero sequencesequence

circuitcircuitandand thethephasephase shiftshift inin thethe positivepositive andand negativenegative sequencesequence circuitscircuits..

NANA VVV ++++====

N0AANAN0ANAA0A ZI3)VVV(VVV ++++++++++++====++++++++ ++++++++

The negative- and positive-sequence

voltages to ground are equal to negative-

and positive-sequence voltages to

neutral.

. Eq. 1

. Eq. 2

Two linked windingsTwo linked windings

Similarly, on the low voltage side

n0aanan0anaa0a ZI3)VVV(VVV ++++++++====++++++++ ++++++++

nana VVV ====

The voltages and currents on both sides of the transformer are related by the turns

ration (N1/ N2). Therefore,Eq. 4 can be written as

0A

2

1nAN

1

2AN

1

20AN

1

2aa0a I)

N

N(Z3)V

N

NV

N

NV

N

N(VVV ++++++++====++++++++

++++++++

MultiplyingEq. 5 by (N1/ N2).

0A

2

2

1nANAN0ANaa0a

2

1 I)N

N(Z3VVV)VVV(

N

N++++++++====++++++++

++++++++

Substituting fromEq. 2 for N0AAA0AANAN0AN ZI3VVV)VVV( ++++++++====++++++++ ++++++++

0A

2

2

1n0ANAA0Aaa0a

2

1 I)N

N(Z3IZ3VVV)VVV(

N

N++++++++====++++++++

++++++++

. Eq. 3

. Eq. 4

. Eq. 5

. Eq. 6

Then,

. Eq. 7

By equating voltages of the same sequence, we can write

The negative sign comesfrom the direction of thezero sequence current


16/2016

Eq. 9 represents the relation for the zero sequence. This relation can be represented

as shown in the Fig. When the voltages on both sides of the transformer are

expressed in per unit, the turns ratio becomes unity. The zero sequence

impedance of the circuit, (adding the leakage impedanceZ), is:

21N:N 0a

I

0AV 0aV0AI

n

2

2

1 Z)N

N(3

NZ3

unitperZ3Z3ZZ nN0 ++++++++==== . Eq. 10

Z

0A

2

2

1nN0A0a

2

1 I])N

N(Z3Z3[VV

N

N++++==== . Eq. 9

++++++++==== Aa

2

1 VVN

N

==== Aa2

1 VVN

N

This is similar to a regular transformer, and therefore, the positive

and negative sequence circuits for a transformer are applicable.

. Eq. 8and

ZZ++

Reference Bus

ZZ++

Reference Bus

ZZ--

Reference Bus

ZZ--

Reference Bus

Z

Z

Reference Bus

A. WyeA. Wye--Wye bank with both neutrals groundedWye bank with both neutrals grounded,,

zero sequence current can flow. The presence ofzero sequence current can flow. The presence ofthe current in one winding means that secondarythe current in one winding means that secondary

current exists in the other. The equivalent circuit iscurrent exists in the other. The equivalent circuit is

as shown in the Figure.as shown in the Figure.

B.B. WyeWye--wyewye Bank, One Neutral GroundedBank, One Neutral Grounded

With ungroundedWith ungrounded wyewye, no zero sequence current, no zero sequence current

can flow. No current in one winding means that nocan flow. No current in one winding means that no

current exists in the other.current exists in the other.Reference Bus

C.C. WyeWye--delta Bank, Groundeddelta Bank, Grounded WyeWye

Zero sequence currents will pass through theZero sequence currents will pass through the wyewye

winding to ground. As a result, secondary zerowinding to ground. As a result, secondary zero

sequence currents will circulate through the deltasequence currents will circulate through the delta

winding.winding.No zero sequence current will exist on theNo zero sequence current will exist on thelines of the secondarylines of the secondary..

Reference Bus

ZZ =0

ZZ =0

Z

OC

OC

Z

E. DeltaE. Delta--delta Bankdelta Bank

Since for a delta circuit no return path for zero sequenceSince for a delta circuit no return path for zero sequence

current exists,current exists, no zero sequence current can flow into ano zero sequence current can flow into a

deltadelta--delta bankdelta bank, although it can circulate within the, although it can circulate within the

delta windings.delta windings.

Reference Bus

Z

OCOC


17/2017

Summary of Transformer Sequence Networks

Transformer Zero Sequence Impedance

ExampleExample 44: Draw the sequence circuits for the network.

T1G1

J0.02

T2

220 kV Transmission Line

1

2 34220 kV Transmission Line

G2

J0.03

G1

MVA X+Voltage X - Xo

100 0.2511 kV 0.25 0.05

G2 100 0.2011 kV 0.20 0.05

T1 100 0.0611/220 0.06 0.06

T2 100 0.0711/220 0.07 0.07

Line 100 0.10220kV 0.10 0.30


18/20


19/2019

T1G1

J0.02

T2


1


G2

J0.03

J0.2J0.07J0.25 J0.06

J0.1

J0.1


T1G1

J0.02

T2



G2

J0.03

1

2 3

1J0.05J0.07

J0.09

J0.06

J0.3

J0.3

J0.05

J0.06

4



20/20

Un Symmetrical Faults

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