2 Unbalanced Faults

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GRID

Technical Institute

Analysis o f Unbalanced Faults


2/87

> Analysis of Unbalanced Fults2

Fault Types

Line - Ground (65 - 70%)

Line - Line - Ground (10 - 20%)

Line - Line (10 - 15%)

Line - Line - Line (5%)

Statistics published in 1967 CEGB Report, but are similar todayall over the world.


3/87


Fault Inc idence

85% of faults are overhead line faults.

50% of these due to lightning strikes.


4/87


Unbalanced Faults (1)

In three phase fault calculations, a single phase

representation is adopted

3 phase faults are rare

Majority of faults are unbalanced faults

UNBALANCED FAULTS may be classified into

SHUNT FAULTS and SERIES FAULTS SHUNT FAULTS

Line to Ground Line to Line Line to Line to Ground

SERIES FAULTS

Single Phase Open Circuit Double Phase Open Circuit


5/87



LINE TO GROUND

LINE TO LINE

LINE TO LINE TO GROUND

Causes :

1) Insulation Breakdown

2) Lightning Discharges and other Overvoltages

3) Mechanical Damage


6/87



OPEN CIRCUIT OR SERIES FAULTS

Causes :

1) Broken Conductor2) Operation of Fuses

3) Maloperation of Single Phase Circuit Breakers

DURING UNBALANCED FAULTS, SYMMETRY OF SYSTEMIS LOST

SINGLE PHASE REPRESENTATION IS NO LONGER VALID


7/87> Analysis of Unbalanced Fults7


Analysed using :-

Symmetrical Components

Equivalent Sequence Networks of Power System

Connection of Sequence Networks appropriate to Typeof Fault



Symmetr ica l Components

Fortescue discovered a property of unbalanced phasors

n phasors may be resolved into :-

(n-1) sets of balanced n-phase systems of phasors, each sethaving a different phase sequence

plus

1 set of zero phase sequence or unidirectional phasors

VA = VA1 + VA2 + VA3 + VA4 - - - - - VA(n-1) + VAn

VB = VB1 + VB2 + VB3 + VB4 - - - - - VB(n-1) + VBn

VC = VC1 + VC2 + VC3 + VC4 - - - - - VC(n-1) + VCn

VD = VD1 + VD2 + VD3 + VD4 - - - - - VD(n-1) + VDn- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Vn = Vn1 + Vn2 + Vn3 + Vn4 - - - - - Vn(n-1) + Vnn

(n-1) x Balanced 1 x ZeroSequence



Unbalanced 5-Phase Sys tem of Vo ltages (1)

This can be resolved into :-

First Set of Balanced Second Set ofPhasors Balanced Phasors

VA1

2VE1

VD1 VC1

VB1

VA2

VC2

VE2 VB2

VD2




Third Set of Balanced Fourth Set of

Phasors Balanced Phasors

4

3

VA4

VB4

VC4 VD4

VE4

VA3

VD

3

VB3 VE3

VC3




Fifth Set of Zero

Sequence Phasors VA5

VB5

VD5

VE5

VC5




This can be resolved into :-

First Set of Balanced Second Set of

Phasors Balanced Phasors

VA1

VC1

120

VB1

VA2

VB2 VC2

240




Third Set of Zero

Sequence Phasors

VA3

VB3

VC3



Unbalanced 3-Phase System (1)

VA = VA1 + VA2 + VA0

VB = VB1 + VB2 + VB0

VC = VC1 + VC2 + VC0

Positive Sequence Negative Sequence

VA1

VC1

120

VB1

VA2

VB2 VC2

240



Unbalanced 3-Phase System (2)

Zero Sequence

VA0

VB0

VC0




Phase Positive + Negative + Zero

VA VA1 + VA2 + VA0

VB VB1 + VB2 + VB0

VC VC1 + VC2 + VC0

VA

VB

VC

++

VB1VC1

VA1

VB2

VC2

VC0VB0VA0VA2

VB1 = a2 VA1 VB2 = a VA2 VB0 = VA0

VC1 = a VA1 VC2 = a2

VA2 VC0 = VA0

=

=

=



Conver ting from Sequence Componentsto Phase Values


VB = VB1 + VB2 + VB0 = a2VA1 + a VA2 + VA0

VC = VC1 + VC2 + VC0 = a VA1 + a2VA2 + VA0

VA0

VC1

VC

VA2

VA1

VA

VC0

VC2

VB2

VB0VB1

VB



Convert ing from Phase Valuesto Sequence Components

VA1 = 1/3 {VA + a VB + a2

VC}VA2 = 1/3 {VA + a

2VB + a VC}

VA0 = 1/3 {VA + VB + VC}

VC3VA0

VB

VA0

VA



Add VA, VB, VC vectorially


VB

= a2VA1

+ a VA2

+ VA0

VC = a VA1 + a2VA2 + VA0

VA + VB + VC = 0 + 0 + 3VA0

VA0 = 1/3 (VA + VB + VC )Add VA, aVB and a

2VC vectorially


a VB = VA1 + a2VA2 + a VA0

a2VC = VA1 + a VA2 + a2VA0

VA + aVB + a2VC = 3VA1 + 0 + 0

VA1 = 1/3 (VA + a VB + a2 VC )



Add VA, a2VB and aVC vectorially


a2VB = a VA1 + VA2 + a2VA0

a VC = a2VA1 + VA2 + a VA0

VA + a2VB + VC = 0 + 3VA2 + 0

VA2 = 1/3 (VA + a2VB + aVC )




VB = a2 VA1 + a VA2 + VA0

VC = a VA1 + a2 VA2 + VA0

IA = IA1 + IA2 + IA0

IB = a2 IA1 + a IA2 + IA0

IC = a IA1 + a2 IA2 + IA0

VA1 = 1/3 {VA + a VB + a2 VC }

VA2 = 1/3 {VA + a2 VB + a VC }

VA0 = 1/3 {VA + VB + VC }

IA1 = 1/3 { IA + a IB + a2 IC }

IA2 = 1/3 { IA + a2 IB + a IC }

IA0 + 1/3 { IA + IB + IC }



Residu al Current

Used to detect earth faults

*IRESIDUAL is zero for :- *IRESIDUAL is present for :-

Balanced Load /E Faults

3 Faults //E Faults

/ Faults

IA

IRESIDUAL = IA + IB +

IC

= 3I0

IB

IC

E/F


23/87


Residual Voltage

Used to detect earth faults

Residual voltage is measured fromOpen Delta or Broken Delta VT

secondary windings.

VRESIDUAL is zero for:-

Healthy unfaulted systems

3 Faults

/ Faults

VRESIDUAL is present for:-

/E Faults//E Faults

Open Circuits (on supply

side of VT with earthed source)

VRESIDUAL= VA+ VB+VC

= 3V0


24/87


Example

Evaluate the positive, negative and zero sequence components

for the unbalanced phase vectors :

VA = 10VB = 1.5 -90

VC = 0.5 120

VC

VA

VB


25/87


Solut ion

VA1 = 1/3 (VA + aVB + a2VC)

= 1/3 1 + (1 120) (1.5 -90)+ (1 240) (0.5 120)

= 0.965 15

VA2 = 1/3 (VA + a2

VB + aVC)= 1/3 1 + (1 240) (1.5 -90)

+ (1 120) (0.5 120) = 0.211 150

VA0 = 1/3 (VA + VB + VC)

= 1/3 (1 + 1.5 -90 + 0.5 120)= 0.434 -55

S


26/87


Pos it ive Sequence Voltages

VC1 = aVA1

15

VA1 =

0.96515

VB1 =

a2VA1


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Zero SequenceVoltagesNegative Sequence

Voltages

VA0 = 0.434-

55VB0 = -

VC0 = -

-55VC2 = a2 VA2150

VB2 = a VA2

VA2 =0.211150

S t i l C t


28/87



VC1

VA1

VB1

VC2

VA2

VB2

VC0

VA0

VB0

VA2VC2

VB2

VC

VA

V0

VB

E l (1)


29/87


Examp le (1)

Evaluate the phase quantities IA, IB and IC from the

sequence componentsIA1 = 0.6 0IA2 = -0.4 0IA0 = -0.2 0

Solution

IA = IA1 + IA2 + IA0 = 0

IB = a2IA1 + aIA1 + IA0

= 0.6240 - 0.4120 - 0.20 = 0.91-109IC = aIA1 + a

2IA2 + IA0

= 0.6120 - 0.4240 - 0.20 = 0.91+109

E l (2)


30/87


Examp le (2)

IC

IB

109

109


31/87


POSITIVE

IA1

NEGATIVE

IB1

IC1

IA2

IB2

IC2

IC1

IB1

IA1

IA2

IB2

IC2


32/87


PHASE

ZERO

IA0

IB0IC0

IB

IC

IC

IB

IA0IB0IC0

Sequence Components (1)


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Any 3 phase system of vectors may be represented as the sum of3 sets of symmetrical vectors :-

3 PHASE

VECTORS

EQUIVALENT SYMMETRICAL COMPONENTS

POSITIVE PHASE

SEQUENCE (PPS)

ZERO PHASE

SEQUENCE

I2

BALANCED LOADOR 3-PHASE FAULT

/IA/ = /IB/ = /IC/ = IFI0 = 0I2 = 0I1 = IF

NEGATIVE PHASE

SEQUENCE (NPS)

I1

IA

IC IB

I0

IA1

IC1 IB1



34/87



PHASE-PHASE

FAULT /IA/=/IB/ = IF

IA

IB

I0 = 0

IA1

IC1

IB1

IA2

IC2

IB2

I1 = IF

3

I2 = IF

3



35/87



PHASE-EARTH

FAULTS

/IA/ OR /IC/ = IF

IA

IC

I

B2

IC2 IA2

IA0 IC0IB0IA1

IC1 IB1

IA2

IB2 IC2

IA1

IC1 IB1

I1 = IF

3

I2 = IF

3

I0 = IF

3

IA0 IC0IB0

Sequence Networks


36/87


Sequence Networks

It can be shown that providing the system impedances arebalanced from the points of generation right up to the fault, eachsequence current causes voltage drop of its own sequence only.

Regard each sequence current flowing within its own networkthro impedances of its own sequence only, with no

interconnection between the sequence networks right up to thepoint of fault.

Unbalanced Voltages and Currents act ing on


37/87


Unbalanced Voltages and Currents act ing onBalanced Impedances

ZSIAVA

ZMZMZSIBVB

ZMZSICVC

VA = IAZS + IBZM + ICZM

VB = IAZM + IBZS + ICZM

VC = IAZM + IBZM + ICZS

In matrix form

VA ZS ZM ZM IA

VB = ZM ZS ZM IB

VC ZM ZM ZS IC


38/87


Resolve V & I phasors into symmetrical components

1 1 1 V0 ZS ZM ZM 1 1 1 I0

1 a2 a V1 = ZM ZS ZM 1 a2 a I1

1 a a2 V2 ZM ZM ZS 1 a a2 I2

Multiply by [A]-1

V0 1 1 1 ZS ZM ZM 1 1 1 I0

V1 = 1 a2 a ZM ZS ZM 1 a

2 a I1

V2 1 a a2 ZM ZM ZS 1 a a

2 I2

V0 1 1 1 ZS ZM ZM 1 1 1 I0

V1 = 1/3 1 a a2 ZM ZS ZM 1 a

2 a I1

V2 1 a2 a ZM ZM ZS 1 a a

2 I2

V0 ZS + 2ZM ZS + 2ZM ZS + 2ZM

V1 = 1/3 ZS - ZM ZM + a ZS + a2 ZM ZM + aZM + a

2 ZS

V2 ZS - ZM ZM + a2 ZS + a ZM ZM + a

2 ZM + a ZS

1 1 1 I0

1 a2 a I1

1 a a2

I2

-1


39/87


V0 ZS + 2ZM 0 0 I0

V1 = 0 ZS - ZM 0 I1

V2 0 0 ZS - ZM I2

V0 Z0 0 0 I0

V1 = 0 Z1 0 I1

V2 0 0 Z2 I2

The symmetrical component impedance matrix is adiagonal matrix if system is symmetrical.

The sequence networks are independent of each other.

The three isolated sequence networks are

interconnected when an unbalance such as a fault orunbalanced loading is introduced.

Phase Sequence Equ ivalent Circuits (1)


40/87



Positive Sequence Impedance

E

QPa2E

aE

Z1 = E/I

Q1P1

I

a2 I

aI



41/87



Negative Sequence Impedance

For static non-rotating plant :- Z2 = Z1

E

QP

a2E

aE

Z2 = E/I

Q2P2

I

a2 I

aI



42/87



Zero Sequence Impedance

I

I

I3I

E QP

Z0 = E/IP0 Q0

Sequence Networks


43/87


Sequence Networks

+ve, -ve and zero sequence networks are drawn for areference phase. This is usually taken as the A phase.

Faults are selected to be balanced relative to the referenceA phase.

e.g. For /E faults consider an A-E fault

For / faults consider a B-C fault

Sequence network interconnection is the simplest for thereference phase.

Pos it ive Sequence Diagram


44/87


Pos it ive Sequence Diagram

1. Start with neutral point N1

- All generator and load neutrals are connected to N1

2. Include all source EMFs

- Phase-neutral voltage

3. Impedance network

- Positive sequence impedance per phase

4. Diagram finishes at fault point F1

Z1N1E1

F1

Example


45/87


Example

V1 = Positive sequence /N voltage at fault point

I1 = Positive sequence phase current flowing into F1

V1 = E1 - I1 (ZG1 + ZT1 + ZL1)

E1F1I1

ZL1ZG1 ZT1N1

(N1)

V1

Line

FN

E

R

Generator Transformer

Negative Sequence Diagram


46/87



1. Start with neutral point N2

- All generator and load neutrals are connected to N2

2. No EMFs included

- No negative sequence voltage is generated!

3. Impedance network

- Negative sequence impedance per phase

4. Diagram finishes at fault point F2

Z2N2 F

2

Example


47/87


Example

V2 = Negative sequence /N voltage at fault point

I2 = Negative sequence phase current flowing into F2

V2 = -I2 (ZG2 + ZT2 + ZL2)

F2I2ZL2ZG2 ZT2N2

(N2)

V2

Line F

N

E

R


System Single LineDiagram


Zero Sequence Diagram (1)


48/87



For In Phase (Zero Phase Sequence) currents to flow in each

phase of the system, there must be a fourth connection (this istypically the neutral or earth connection).

IA0 + IB0 + IC0 = 3IA0

IA0N

IB0

I

C0



49/87



Resistance Earthed System :-

3IA0

N

E

R

Zero sequence voltage between N & E given by

V0 = -3IA0. RZero sequence impedance of neutral to earth path

Z0 = V0 = 3R

-IA0

Trans form er Zero Sequence Impedance


50/87


Trans form er Zero Sequence Impedance

QP

ZT0

aaQP

bb

N0

General Zero Sequence Equ ivalent Circu it


51/87


q qfor Two Wind ing Transfo rmer

Secondary

Terminal'a' 'a'

Primary

Terminal

'b' 'b'

N0

ZT0

On appropriate side of transformer :

Earthed Star Winding - Close link a

Open link b

Delta Winding - Open link a

Close link b

Unearthed Star Winding - Both links open

Zero Sequence Equ ivalent


52/87


q qDy Tx (1)

3I0

No zero sequence

in line connection

on side

I0

I0

I0

Zero Sequence Equ ivalent


53/87


q qDy Transfo rmer (2)

side

terminal

side

terminal

ZT0

N0

(E0)

I

0

Thus, equivalent single phase zero sequence diagram :-

Zero Sequence Equ ivalent Circu its (1)


54/87


q q ( )

S0ZT0

N0

P0

P S

aa

b b



55/87


q q ( )

S0ZT0

N0

P0

P S

aa

b b



56/87


q q ( )

S0ZT0

N0

P0

P S

aa

b b



57/87


q q ( )

S0ZT0

N0

P0

P S

aa

b b

Zero Sequence Diagram


58/87


V0 = Zero sequence PH-E voltage at fault point

I0 = Zero sequence current flowing into F0

V0 = -I0 (ZT0 + ZL0 + 3RT)

F0I0ZL0ZG0 ZT0N0

N0

V0

LineF

N

E

R


System Single LineDiagram

Zero Sequence Network

3R

E0

3RT

RT

Zig-Zag Earthing Trans formers (1)


59/87


Positive (and negative) sequence impedance is very high.

Equivalent Circuit :

P

N

E

R

P1

N1

Zig-Zag Earthing Trans formers (2)


60/87


Zero sequence impedance is very low.

Equivalent Circuit P0

ZT0

3R

N0

(E0)

Summary o f Sequence Diagrams (1)


61/87


System Single Line Diagram

R

E

N

GENERATOR TRANSFORMERLINE F



62/87


Positive Sequence

(N1)

V1

F1ZL1ZT1ZG1

E1N1 I1



63/87


Negative Sequence

(N2)

V2

F2ZL2ZT2ZG2N2 I2



64/87


Zero Sequence

E0(N0)

V0

F0ZL0ZT0ZG0

3R

N0 I0

Interconnect ion o f Sequence Netwo rks


65/87


Consider sequence networks as blocks with fault terminals F & N forexternal connections.

F1

POSITIVE

SEQUENCE

NETWORK

N1

F2

NEGATIVE

SEQUENCE

NETWORK

N2

F0

ZERO

SEQUENCE

NETWORK

N0

I2

V2

I0

V0

I1

V1

Interconnect ion o f Sequence Netwo rks


66/87


For any given fault there are 6 quantities to be considered at the fault point

i.e. VA VB VC IA IB IC

Relationships between these for any type of fault can be converted into anequivalent relationship between sequence components

V1, V2, V0 and I1,I2 ,I0

This is possible if :-

1) Any 3 phase quantities are known (provided they are not all

voltages or all currents)

or 2) 2 are known and 2 others are known to have a specific

relationship.

From the relationship between sequence Vs and Is, the manner inwhich the isolation sequence networks are connected can be determined.

The connection of the sequence networks provides a single phase

representation (in sequence terms) of the fault.


67/87


To derive the system constraints at the fault terminals :-

IA

VA

IB IC

VB VC

F

Terminals are connected to represent the fault.

A

B

C

Phase to Earth Fault on Phase A


68/87


IA

VA

IB IC

VB VC

At fault point :-

VA = 0

VB = ?

VC = ?

IA = ?

IB = 0

IC = 0

A

B

C

Phase to Earth Fault on Phase A


69/87


At fault point

VA = 0; IB = 0 ; IC = 0

but VA = V1 + V2 + V0 V1 + V2 + V0 = 0 ------------------------- (1)I0 = 1/3 (IA + IB + IC ) = 1/3 IA

I1 = 1/3 (IA + aIB + a2IC) = 1/3 IA

I2 = 1/3 (IA + a2IB + aIC) = 1/3 IA

I1 = I2 = I0 = 1/3 IA ------------------------- (2)To comply with (1) & (2) the sequence networks must be connected in series :-

I1 F1

N1

V1+veSeqN/W

I

2

F2

N2

V2-veSeqN/W

I0F0

N0

V0ZeroSeqN/W

Example : Phase to Earth Fault


70/87


SOURCE LINE F

132 kV

2000 MVA

ZS1 = 8.7

ZS0 = 8.7

A - GFAULTZL1 = 10

ZL0 = 35

Total impedance = 81.1I1 = I2 = I0 = 132000 = 940 Amps3 x 81.1IF = IA = I1 + I2 + I0 = 3I0 = 2820 Amps

IF

8.7 10 I1 F1

N1

8.7 10 I2 F2

N2

8.7 35 I0 F0

N0

Earth Fault with Fault Resistance


71/87


IA

VA

IB IC

VB VC

At fault point :-

VA = IAZF

VB = ?

VC = ?

IA = ?

IB = 0

IC = 0

ZF

Earth Fault with Fault Resistance


72/87


At fault point

VA = IAZF ; IB = 0 ; IC = 0

I0 = 1/3 (IA + IB + IC) = 1/3 IA

I1 = 1/3 (IA + aIB + a2IC) = 1/3 IA

I2 = 1/3 (IA + a2IB + aIC) = 1/3 IA

I1 = I2 = I0 ------------------------- (1)Since VA = IAZF :-

V1 + V2 + V0 = (I1 + I2 + I0) ZF

= 3I0ZF

V1 + V2 + V0 = 3I0ZF ------------ (2)

F1

POSITIVE

SEQUENCE

NETWORK

N1

F2

NEGATIVE

SEQUENCE

NETWORK

N2

F0

ZERO

SEQUENCE

NETWORK

N0

I2

V2

I0

V0

I1

V1

3ZF

Phase to Phase Fault :- B -C Phase


73/87


IA

VA

IB IC

VB VC

At fault point :-

VA = ?

VB = VC

IA = 0

IB + IC = 0

Phase to Phase Fault :- B -C Phase


74/87


At fault point

VB = VC ; IA = 0; IB + IC = 0

I0 = 1/3 (IA + IB + IC ) = 0

I0 = 0 ---- (1)I1 = 1/3 (IA + aIB + a

2IC ) = 1/3 (a - a2)IB

I2 = 1/3 (IA + a2IB + aIC) = -1/3 (a - a

2)IB

I1 + I2 = 0 ------- (2)V1 = 1/3 (VA + aVB + a

2

VC ) = 1/3 (VA - VB)V2 = 1/3 (VA + a

2VB + aVC ) = 1/3 (VA - VB)

V1 = V2 ------------------------ (3)From equations (1), (2) and (3) the positive and negative sequence networks areconnected in parallel and the zero sequence network is unconnected.

I1F1

N1V1

+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0V0

ZeroSeqN/W

Phase to Phase Fault Current Level


75/87


For a / Fault ,assuming Z1 = Z2 :-

I1 = -I2 = E

2Z1IA = I1 + I2 + I0 = 0

IB = a2I1 + aI2 + I0 = (a

2 - a)I1 = -j3 . E

2 Z1

IC = aI1 + a2I2 + I0 = (a - a

2)I1 = j3 . E2 Z1 IF(/) = 0.866 IF(3)

E

I1

I2

Z1

Z2

A

B

C

I

BIC

EC

EB

EA

Example : Phase to Phase Fault


76/87


SOURCE LINE F

132 kV

2000 MVA

ZS1 = ZS2 = 8.7

B - CFAULTZL1 = ZL2 = 10

8.7 10

8.7 10

I1

I2

F1

N1

F2

N2

132000

3

Total impedance = 37.4 IB = a2I1 + aI2I1 = 132000 = 2037 Amps = a

2I1 - aI13 x 37.4 = (a2- a) I1I2 = -2037 Amps = (-j) . 3 x 2037

= 3529 Amps

Phase to Phase Fault with Resistance


77/87


IA

VA

IB

IC

VB VC

At fault point :-

VA = ?

VB - VC = IBZF

IA = 0

IB + IC = 0

ZF

Phase to Phase Fault w ith Resistance


78/87


At fault point

IA = 0; IB + IC = 0 ; VB - VC = IBZF

I0 = 1/3 (IA + IB + IC ) = 0

I1 = 1/3 (IA + aIB + a2IC) = 1/3 (a - a

2) IB

I2 = 1/3 (IA + a2IB + aIC) = - 1/3 (a - a

2) IB

I0 = 0 ----------- (1)I1 + I2 = 0

IB = a2I1 + aI2 + I0 = (a

2 - a) I1

VB - VC = (a2V1 + aV2 + V0) - (aV1 + a

2V2 + V0)

= (a2- a) V1 - (a2- a) V2

but VB - VC = IBZF = (a2- a) I1ZF

(a2- a) I1ZF = (a2- a) V1 - (a2- a) V2 V1 - V2 = I1ZF ------------ (2)

I1F1

N1V1

+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0

V0ZeroSeqN/W

ZF

Phase to Phase to Earth Fau lt :- B -C-E


79/87


IA

VA

IB IC

VB VC

At fault point :-

VA = ?

VB = 0

VC = 0

IA = 0

IB = ?

IC = ?

Phase to Phase to Earth Fau lt :- B -C-E


80/87


At fault point

VB = 0; VC = 0 ; IA = 0

V1 = 1/3 (VA + aVB + a2VC ) = 1/3 VA

V2 = 1/3 (VA + a2VB + aVC ) = 1/3 VA

V0 = 1/3 (VA + VB + VC ) = 1/3 VA

V1 = V2 = V0 = 1/3 VA -------------------- (1)IA = I1 + I2 + I0 = 0 ------------------------------- (2)

From equations (1) & (2) the sequence networks are connected in parallel.

I1F1

N1

V1+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0V0

ZeroSeqN/W

Phase to Phase to Earth FaultB C E with Fault Resistance


81/87


IA

VA

IB IC

VB VC

B-C-E with Fault Resistance

At fault point :-

VA = ?

VB = (IB +IC) ZF

VC = (IB +IC) ZF

IA = 0

IB = ?IC = ?IB+ IC

ZF

Phase to Phase to Earth FaultB -C-E with Resis tance


82/87


B-C-E with Resis tanceAt fault point

IA = 0 ; VB = VC = (IB + IC)ZF

IA = I1 + I2 + I0 = 0 --------------------- (1)

I0 = 1/3 (IA + IB + IC) = 1/3 (IB + IC)

IB + IC = 3I0V1 = 1/3 (VA + aVB + a

2VC ) = 1/3 [VA + (a2 + a)VB] = 1/3 (VA - VB)

V2 = 1/3 (VA + a2VB + aVC ) = 1/3 [VA + (a

2 + a)VB] = 1/3 (VA - VB)

V1 = V2 --------------------- (2)V0 = 1/3 (VA + VB + VC ) = 1/3 (VA + 2VB)

V0 - V1 = 1/3 (VA + 2VB ) - 1/3 (VA - VB)

= VB = (IB + IC)ZF = 3I0ZF

V1 = V0 - I0 (3ZF) --------------------- (3)I1

F1

N1

V1+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0

V0ZeroSeqN/W

3ZF

4-w ire Representat ion of/E Fau lt (1)


83/87


/E Fau lt (1)

Consider an earthed source.

Total sequence impedances to fault

= Z1, Z2, Z0 i.e. Source + line

EA

EB

EC

C

B

A

Z1, Z2, Z0

4-w ire Rep resentation o f /E Faul t (2)


84/87


Sequence Networks

I1Z1

I2

I0

Z2

Z0

E

3

ZZZ

E

3ZZZ

3

Z2Z

3

ZZZE

3

)ZZ(ZE

101

A

101A

01A

021A

1021A

0211

I

I

I

I

IIIII

I

4 Wire Equivalent Circuit

EA

EB

EC

Z1

Z1

Z1

ZN = Z0 - Z13

10

N

N1

AF

3

Z-ZZ

where

ZZ

EII

3 Versus 1 Fau lt Level (1)


85/87


Xg

XTE

Xg XT

EZ1

IF

3

1TgF

Z

E

XX

E



86/87


Z0

IF

1 Xg1 XT1

E

Z2 = Z1

Z1

Xg2

XT2

Xg0 XT0

I

3EF

2Z Z1 0



87/87

LEVELFAULT3LEVELFAULT1

ZZIF

Z2Z3ELEVELFAULT1

Z2Z

3E

3Z

3E

Z

ELEVELFAULT3

10

01

1111

2 Unbalanced Faults

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