Unbalanced Faults

8/10/2019 Unbalanced Faults

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GRIDTechnical Institute

Analysis of Unbalanced Faults

> Analysis of UnbalancedFults2

Fault Types

Line - Ground (65 - 70%)

Line - Line - Ground (10 - 20%)

Line - Line (10 - 15%)

Line - Line - Line (5%)

Statist ics publi shed in 1967 CEGB Report , but are similar todayall over the world .


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Fault Incidence

85% of faults are overhead line faults.

50% of these due to ligh tning strikes.


Unbalanced Faults (1)

In three phase fault calculations, a single phaserepresentation is adopted

3 phase faults are rare

Majori ty of faults are unbalanced faults

UNBALANCED FAULTS may be classified intoSHUNT FAULTS and SERIES FAULTS

SHUNT FAULTSLine to GroundLine to LineLine to Line to Ground

SERIES FAULTSSingle Phase Open CircuitDouble Phase Open Circuit


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LINE TO GROUNDLINE TO LINELINE TO LINE TO GROUND

Causes :

1) Insu lat ion Breakdown

2) Lightning Discharges and other Overvoltages3) Mechanical Damage



OPEN CIRCUIT OR SERIES FAULTS

Causes :

1) Broken Conductor 2) Operation of Fuses3) Maloperation of Single Phase Circuit Breakers

DURING UNBALANCED FAULTS, SYMMETRY OF SYSTEMIS LOST

SINGLE PHASE REPRESENTATION IS NO LONGER VALID


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Analysed using :-

Symmetrical Components

Equivalent Sequence Networks of Power System

Connection of Sequence Networks appropriate to Typeof Fault



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 unidi rectional phasors

V A = V A1 + V A2 + V A3 + V A4 - - - - - V A(n-1) + V AnVB = VB1 + VB2 + VB3 + VB4 - - - - - VB(n-1) + VBnVC = VC1 + VC2 + VC3 + VC4 - - - - - VC(n-1) + VCnVD = VD1 + VD2 + VD3 + VD4 - - - - - VD(n-1) + VDn- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Vn = Vn1 + Vn2 + Vn3 + Vn4 - - - - - Vn(n-1) + Vnn

(n-1) x Balanced 1 x ZeroSequence


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Unbalanced 5-Phase System of Voltages (1)

This can be resolved into :-

First Set of Balanced Second Set of Phasors Balanced Phasors

V A1

2VE1

VD1 VC1

VB1

V A2

VC2

VE2 VB2

VD2



Third Set of Balanced Fourth Set of Phasors Balanced Phasors

4

3

V A4

VB4

VC4 VD4

VE4

V A3

VD3

VB3 VE3

VC3


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Fifth Set of ZeroSequence Phasors V A5

VB5

VD5VE5

VC5



This can be resolved into : -

First Set of Balanced Second Set of Phasors Balanced Phasors

V A1

VC1

120

VB1

V A2

VB2 VC2

240


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Third Set of ZeroSequence Phasors

V A3VB3

VC3


Unbalanced 3-Phase System (1)

V A = V A1 + V A2 + V A0VB = VB1 + VB2 + VB0

VC = VC1 + VC2 + VC0

Posit ive Sequence Negative Sequence

V A1

VC1

120

VB1

V A2

VB2 VC2

240


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Unbalanced 3-Phase System (2)

Zero Sequence

V A0VB0

VC0



Phase Positive + Negative + Zero

V A V A1 + V A2 + V A0VB VB1 + VB2 + VB0

VC VC1 + VC2 + VC0

V A

VB

VC

+ +

VB1

VC1

V A1

VB2

VC2

VC0VB0V A0V A2

VB1 = a 2 V A1 VB2 = a V A2 VB0 = V A0VC1 = a V A1 VC2 = a 2 V A2 VC0 = V A0

=

==


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Converting from Sequence Componentsto Phase Values

V A = V A1 + V A2 + V A0

VB = VB1 + VB2 + VB0 = a 2V A1 + a V A2 + V A0

VC

= VC1

+ VC2

+ VC0

= a V A1

+ a 2V A2

+ V A0

V A0

VC1

VC

V A2

V A1

V A

VC0

VC2

VB2

VB0VB1

VB


Converting from Phase Valuesto Sequence Components

V A1 = 1/3 {V A + a V B + a 2VC}

V A2 = 1/3 {V A + a 2VB + a V C}

V A0 = 1/3 {V A + VB + VC}

VC3V A0

VB

V A0

V A


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Add V A , VB, VC vectorially

V A = V A1 + V A2 + V A0VB = a 2V A1 + a V A2 + V A0VC = a V A1 + a 2V A2 + V A0

V A + VB + VC = 0 + 0 + 3V A0

V A0 = 1/3 (V A + VB + VC )

Add V A , aV B and a 2VC vectorially

V A = V A1 + V A2 + V A0a V B = V A1 + a 2V A2 + a V A0a 2VC = V A1 + a V A2 + a 2V A0

V A + aV B + a 2VC = 3V A1 + 0 + 0

V A1 = 1/3 (V A + a V B + a 2 VC )


Add V A, a2VB and aV C vectorially

V A

= V A1

+ V A2

+ V A0

a 2VB = a V A1 + V A2 + a2V A0

a V C = a2V A1 + V A2 + a V A0

V A + a2VB + VC = 0 + 3V A2 + 0

V A2 = 1/3 (V A + a2VB + aV C )


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V A = V A1 + V A2 + V A0

VB = a2 V A1 + a V A2 + V A0

VC = a V A1 + a2 V A2 + V A0

I A = I A1 + I A2 + I A0

IB = a2 I A1 + a I A2 + I A0

IC = a I A1 + a2 I A2 + I A0

V A1 = 1/3 {V A + a V B + a2 VC }

V A2 = 1/3 {V A + a2 VB + a V C }

V A0 = 1/3 {V A + VB + VC }

I A1 = 1/3 { I A + a I B + a 2 IC }I A2 = 1/3 { I A + a

2 IB + a I C }

I A0 + 1/3 { I A + IB + IC }


Residual Current

Used to detect earth f aults

* I RESIDUAL is zero for :- * I RESIDUAL is present for :-

Balanced Load /E Faults3 Faults //E Faults/ Faults

I A

IRESIDUAL = I A + IB +IC

= 3 I0

IB

IC

E/F


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Residual Voltage

Used to detect earth f aultsResidual voltage is measured from Open 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 supp ly

side of VT with earthed source)

VRESIDUAL

= V A + VB +VC= 3V 0


Example

Evaluate the posi tive, negative and zero sequence componentsfor the unbalanced phase vectors :

V A = 10

VB = 1.5 -90

VC = 0.5 120

VC

V A

VB


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Solution

V A1 = 1/3 (V A + aV B + a 2VC)

= 1/3 [ 1 + (1 120) (1.5 -90)

+ (1 240) (0.5 120) ]

= 0.965 15

V A2 = 1/3 (V A + a 2VB + aV C)

= 1/3 [ 1 + (1 240) (1.5 -90)

+ (1 120) (0.5 120) ]

= 0.211 150

V A0 = 1/3 (V A + VB + VC)

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

= 0.434 -55


Positive Sequence Voltages

VC1 = aV A1

15

V A1 =

0.965 15

VB1 =a 2V A1


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Zero SequenceVoltages

Negative Sequence

Voltages

V A0 = 0.434 -55VB0 = -VC0 = -

-55 VC2 = a2 V A2

150

VB2 = a V A2

V A2 =0.211 150



VC1

V A1

VB1

VC2

V A2

VB2

VC0

V A0

VB0

V A2VC2

VB2

VC

V A

V0

VB


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Example (1)

Evaluate the phase quantities I A , I B and I C from thesequence components

I A1 = 0.6 0

I A2 = -0.4 0

I A0 = -0.2 0

Solution

I A = I A1 + I A2 + I A0 = 0

IB = a2I A1 + a I A1 + I A0

= 0.6240 - 0.4 120 - 0.2 0 = 0.91 -109

IC = a I A1 + a 2I A2 + I A0= 0.6120 - 0.4 240 - 0.2 0 = 0.91 +109


Example (2)

IC

IB

109

109


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POSITIVE

I A1

NEGATIVE

IB1IC1

I A2

IB2

IC2

IC1

IB1

I A1

I A2

IB2

IC2


PHASE

ZERO

I A0

IB0

IC0

IB

IC

IC

IB

I A0IB0IC0


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Sequence Components (1)

Any 3 phase system o f vector s may be represent ed as t he sum o f3 sets of symmetrical vectors :-

3 PHASEVECTORS

EQUIVALENT SYMMETRICAL COMPONENTS

POSITIVE PHASESEQUENCE (PPS) ZERO PHASESEQUENCE

I2

BALANCED LOAD

OR 3-PHASE FAULT/I A/ = / IB/ = / IC/ = IF

I0 = 0I2 = 0I1 = IF

NEGATIVE PHASESEQUENCE (NPS)

I1

I A

IC IB

I0

I A1

IC1 IB1



PHASE-PHASEFAULT / I A/=/IB/ = IF

I A

IB

I0 = 0

I A1

IC1

IB1

I A2

IC2

IB2I1 = IF

3

I2 = IF3


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PHASE-EARTHFAULTS/I A/ OR / IC/ = IF

I A

IC

IB2

IC2 I A2

I A0 IC0IB0I A1

IC1 IB1

I A2

IB2 IC2

I A1

IC1 IB1

I1 = IF3

I2 = IF3

I0 = IF3

I A0 IC0IB0


Sequence Networks

It can be shown that provid ing the system impedances arebalanced from the points o f generation r ight up to the fault, each

sequence 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 nointerconnection between the sequence networks r ight up to thepoint of fault.


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Unbalanced Voltages and Currents acting onBalanced Impedances

ZSI AV A

ZMZMZSIBVB

ZMZSICVC

V A = I AZS + I BZM + I CZM

VB = I AZM + I BZS + I CZM

VC = I AZM + I BZM + I CZS

In matrix form

V A ZS ZM ZM I A

VB = ZM ZS ZM I B

VC ZM ZM ZS I C


Resolve V & I phasors into symmetrical components

1 1 1 V 0 ZS ZM ZM 1 1 1 I 01 a 2 a V 1 = ZM ZS ZM 1 a 2 a I 11 a a 2 V2 ZM ZM ZS 1 a a 2 I 2

Multiply by [A] -1

V0 1 1 1 Z S ZM ZM 1 1 1 I 0V1 = 1 a 2 a Z M ZS ZM 1 a 2 a I 1V2 1 a a 2 ZM ZM ZS 1 a a 2 I 2

V0 1 1 1 Z S ZM ZM 1 1 1 I 0V1 = 1/3 1 a a 2 ZM ZS ZM 1 a 2 a I 1V2 1 a 2 a Z M ZM ZS 1 a a 2 I 2

V0 ZS + 2Z M ZS + 2Z M ZS + 2Z MV1 = 1/3 Z S - ZM ZM + a Z S + a 2 ZM ZM + aZ M + a 2 ZS

V2 ZS - ZM ZM + a 2 ZS + a Z M ZM + a 2 ZM + a Z S

1 1 1 I 01 a 2 a I 11 a a 2 I2

-1


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V0 ZS + 2Z M 0 0 I 0V1 = 0 Z S - ZM 0 I 1V2 0 0 Z S - ZM I 2

V0 Z0 0 0 I 0V1 = 0 Z 1 0 I 1V2 0 0 Z 2 I 2

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 o runbalanced loading is introduced.


Phase Sequence Equivalent Circuits (1)

Positive Sequence Impedance

E

QPa 2E

aE

Z1

= E/IQ1P 1

I

a 2 I

aI


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Negative Sequence Impedance

For static non-rotating plant :- Z 2 = Z 1

E

QP

a 2E

aE

Z2 = E/IQ 2P 2

I

a 2 I

aI



Zero Sequence Impedance

I

I

I3IE QP

Z0 = E/IP 0 Q0


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Example

V1 = Positive sequence /N voltage at fault point

I 1 = Positive sequence phase current flowing into F 1V1 = E 1 - I 1 (ZG1 + Z T1 + Z L1)

E1F1I1ZL1

ZG1 ZT1N1

(N1)

V1

LineF

N

E

R

Generator Transformer


Negative Sequence Diagram

1. Star t wi th neu tr al point N 2- All generator and load neutrals are connected to N 2

2. No EMFs included

- No negative sequence voltage is generated!

3. Impedance network

- Negative sequence impedance per phase

4. Diagram f inishes at fault point F 2

Z2N2 F2


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Example

V2 = Negative sequence /N voltage at fault point

I 2 = Negative sequence phase current flowing into F 2

V2 = -I2 (ZG2 + Z T2 + Z L2)

F2I2ZL2ZG2 ZT2N2

(N2)

V2

Line FN

E

R


System Single LineDiagram

Negative Sequence Diagram


Zero Sequence Diagram (1)

For In Phase (Zero Phase Sequence) currents to flow in eachphase of the system, there must be a four th connection (this istypically the neutral or earth connection).

I A0 + IB0 + IC0 = 3 I A0

I A0N

IB0

IC0


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Zero Sequence Diagram (2)

Resistance Earthed System :-

3 A0

N

E

R

Zero s equence vol tage between N & E given by

V0 = -3 A0 . RZero sequence impedance of neutral to earth path

Z0 = V0 = 3R

- A0


Transformer Zero Sequence Impedance

QP

ZT0aa

QP

bb

N0


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General Zero Sequence Equivalent Circuitfor Two Winding Transformer

SecondaryTerminal'a' 'a'

PrimaryTerminal

'b' 'b'

N0

ZT0

On appropriate side of transformer :

Earthed Star Winding - Close link a

Open link b

Delta Winding - Open link aClose link b

Unearthed Star Winding - Both links open


Zero Sequence EquivalentDy Tx (1)

3I0

No zero sequencein line connectionon side

I 0

I 0

I0


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Zero Sequence EquivalentDy Transformer (2)

sideterminal

sideterminal ZT0

N0(E0)

I0

Thus, equivalent s ingle phase zero sequence diagram :-


Zero Sequence Equivalent Circui ts (1)

S 0ZT0

N0

P 0

P S

aa

b b


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S 0ZT0

N0

P 0

P S

aa

b b



S 0ZT0

N0

P 0

P S

aa

b b


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S 0ZT0

N0

P 0

P S

aa

b b


Zero Sequence Diagram

V0

= Zero sequence PH-E voltage at fault point

I 0 = Zero sequence current flowing into F 0V0 = -I 0 (ZT0 + Z L0 + 3R T)

F0I 0ZL0ZG0 ZT0N0

N0

V0

Line FN

E

R


System Single Line

Diagram

Zero Sequence Network

3R

E0

3R T

RT


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Zig-Zag Earthing Transformers (1)

Positive (and negative) sequence impedance is very high.

Equivalent Circu it :

P

N

E

R

P 1

N1


Zig-Zag Earthing Transformers (2)

Zero sequence impedance is very low.

Equivalent Circu it P0

ZT0

3R

N0(E0)


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Summary of Sequence Diagrams (1)

System Single Line Diagram

R

E

N

GENERATOR TRANSFORMERLINE F



Positive Sequence

(N1)

V1

F1

ZL1ZT1ZG1E1N1

I1


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Negative Sequence

(N2)

V2

F2ZL2ZT2ZG2N2 I2



Zero Sequence

E0(N0)

V0

F0ZL0ZT0ZG0

3R

N0 I0


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Interconnection of Sequence Networks

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

F1

POSITIVESEQUENCENETWORK

N1F2

NEGATIVESEQUENCENETWORK

N2F0

ZEROSEQUENCENETWORK

N0

I2

V2

I0

V0

I1

V1


Interconnection of Sequence Networks

For any given fault t here are 6 quantities to be cons idered at the fault point

i.e. V A VB VC I A IB IC

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

V1, V2, V0 and I1, I 2 , I 0

This is possible if :-

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

o r 2) 2 are known and 2 o ther s are known to have a spec if icrelationship.

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

The connection of the sequence networks prov ides a single phaserepresentation (in sequence terms) of the fault.


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To derive the system cons traints at the fault terminals :-

I A

V A

IB IC

VB VC

F

Terminals are connected to represent the fault.

A

B

C


I A

V A

IB IC

VB VC

Phase to Earth Fault on Phase A

At fault poin t :-

V A = 0VB = ?VC = ?

I A = ?I B = 0I C = 0

A

B

C


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Phase to Earth Fault on Phase A

At fau lt po in t

V A = 0 ; I B = 0 ; I C = 0

but V A = V1 + V2 + V0

V1 + V2 + V0 = 0 ------------------------- (1)

I 0 = 1/3 ( I A + I B + I C ) = 1/3 I A

I 1 = 1/3 ( I A + a I B + a 2I C) = 1/3 I A

I 2 = 1/3 ( I A + a 2I B + a I C) = 1/3 I A

I 1 = I 2 = I 0 = 1/3 I A ------------------------- (2)

To comply with (1) & (2) the sequence networks must be connected in series :-

I1 F1

N1

V1

+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0

V0ZeroSeqN/W


Example : Phase to Earth Fault

SOURCE LINE F

132 kV2000 MVAZS1 = 8.7

ZS0 = 8.7

A - GFAULTZL1 = 10

ZL0 = 35

Total impedance = 81.1 I1 = I2 = I0 = 132000 = 940 Amp s

3 x 81.1IF = I A = I1 + I 2 + I0 = 3 I0 = 2820 Amps

IF

8.7 10 I1 F1

N1

8.7 10 I2 F2

N2

8.7 35 I0 F0

N0


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I A

V A

IB IC

VB VC

Earth Fault with Fault Resistance

At fault point : -

V A = I AZF

VB = ?

VC = ?

I A = ?

I B = 0

I C = 0

ZF


Earth Fault with Fault Resistance

At fault poin t

V A = I AZF ; I B = 0 ; IC = 0

I0

= 1/3 (I A

+ IB

+ IC)

=1/3 I

A

I 1 = 1/3 ( I A + a IB + a 2IC) = 1/3 I A

I 2 = 1/3 ( I A + a 2IB + a IC) = 1/3 I A

I 1 = I2 = I 0 ------------------------- (1)

Since V A = I AZF :-

V1 + V2 + V0 = (I 1 + I 2 + I 0) ZF

= 3 I 0ZF

V1 + V2 + V0 = 3 I 0ZF ------------ (2)

F1POSITIVESEQUENCENETWORK

N1F2

NEGATIVESEQUENCENETWORK

N2F0

ZEROSEQUENCE

NETWORKN0

I2

V2

I0

V0

I1

V1

3Z F


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I A

V A

IB IC

VB VC

Phase to Phase Fault :- B-C Phase

At faul t po in t : -

V A = ?

VB = VC

I A = 0

I B + I C = 0


Phase to Phase Fault :- B-C Phase

At fau lt po int

VB = VC ; I A = 0; I B + I C = 0

I 0 = 1/3 ( I A + I B + I C ) = 0

I 0 = 0 ---- (1)

I 1 = 1/3 ( I A + a I B + a 2I C ) = 1/3 (a - a 2 )I B

I 2 = 1/3 ( I A + a2

I B + a I C) = -1/3 (a - a2

)I B I 1 + I 2 = 0 ------- (2)

V1 = 1/3 (V A + aV B + a 2VC ) = 1/3 (V A - VB)

V2 = 1/3 (V A + a 2VB + aV C ) = 1/3 (V A - VB)

V1 = V2 ------------------------ (3)

From equations (1), (2) and (3) the positi ve and negative sequence networks areconnected in parallel and the zero sequence network is unconnected.

I1F1

N1

V1+ve

SeqN/W

I2F2

N2

V2

-ve

SeqN/W

I0F0

N0

V0

ZeroSeqN/W


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40/44


I A

V A

IB IC

VB VC

Phase to Phase to Earth Fault :- B-C-E

At fault poin t : -

V A = ?VB = 0VC = 0

I A = 0

I B = ?

I C = ?


Phase to Phase to Earth Fault :- B-C-E

At fau lt poin t

VB = 0 ; V C = 0 ; I A = 0

V1 = 1/3 (V A + aV B + a 2VC ) = 1/3 V A

V2 = 1/3 (V A + a 2VB + aV C ) = 1/3 V A

V0 = 1/3 (V A + VB + VC ) = 1/3 V A

V1 = V2 = V0 = 1/3 V A -------------------- (1)

I A = I 1 + I 2 + I 0 = 0 ------------------------------- (2)

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

I1F1

N1

V1+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0

V0

ZeroSeqN/W


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I A

V A

IB IC

VB VC

Phase to Phase to Earth FaultB-C-E with Fault Resistance

At fault poin t : -

V A = ?VB = (I B + I C) ZFVC = (I B + I C) ZF

I A = 0I

B= ?

I C = ?IB + I C

ZF


Phase to Phase to Earth FaultB-C-E with Resistance

At fau lt poin t

I A = 0 ; V B = VC = ( I B + IC)ZF

I A = I 1 + I 2 + I0 = 0 --------------------- (1)

I0 = 1/3 ( I A + IB + I C) = 1/3 ( IB + I C)

I B + I C = 3 I0

V1 = 1/3 (V A + aV B + a 2VC ) = 1/3 [V A + (a 2 + a)V B] = 1/3 (V A - VB)

V2 = 1/3 (V A + a 2VB + aV C ) = 1/3 [V A + (a 2 + a)V B] = 1/3 (V A - VB)

V1 = V2 --------------------- (2)

V0 = 1/3 (V A + VB + VC ) = 1/3 (V A + 2V B)

V0 - V1 = 1/3 (V A + 2V B ) - 1/3 (V A - VB)

= VB = (IB + IC)ZF = 3I 0ZF

V1 = V0 - I0 (3Z F) --------------------- (3)

I1F1

N1

V1+veSeqN/W

I2F2

N2

V2-veSeqN/W

I0F0

N0

V0ZeroSeqN/W

3ZF


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4-wire Representation of/E Fault (1)

Consider an earthed source.

Total sequence impedances to fault

= Z1, Z2, Z 0 i.e. Source + line

E A

EB

ECC

B

AZ1, Z2, Z0


4-wire Representation of /E Fault (2)

Sequence Networks

I1Z1

I2

I0

Z2

Z0

E

3 ZZ

Z

E

3 ZZ

Z

3 Z2Z

3 ZZZ

E

3

)ZZ(Z E

101

A

101A

01A

021A

1021A

0211

+=

+=

+=

++=

=++=++=

4 Wire Equivalent Circu it

E A

EB

EC

Z1

Z1

Z1

ZN = Z 0 - Z13

10

N

N1

AF

3Z-Z

Z

where ZZ

E I I

=

+

==


43/44


3 Versus 1 Fault Level (1)

XgXTE

Xg XT

E Z1 IF

3

1TgF Z

E XX

E +=



Z0

IF

1 Xg1 XT1

E

Z2 = Z 1

Z1

Xg2 XT2

Xg0 XT0

I = +3E

F 2Z Z1 0


44/44



LEVEL FAULT 3 LEVEL FAULT 1

Z Z IF

Z2Z3E

LEVEL FAULT 1

Z2Z3E

3Z3E

ZE

LEVEL FAULT 3

10

01

1111

>

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Unbalanced Faults