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This document is the exclusive property of Alstom Grid and shall not betransmitted by any means, copied, reproduced or modified without the priorwritten consent of Alstom Grid Technical Institute. All rights reserved.
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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> Analysis of UnbalancedFults3
Fault Incidence
85% of faults are overhead line faults.
50% of these due to ligh tning strikes.
> Analysis of UnbalancedFults4
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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> Analysis of UnbalancedFults5
Unbalanced Faults (2)
LINE TO GROUNDLINE TO LINELINE TO LINE TO GROUND
Causes :
1) Insu lat ion Breakdown
2) Lightning Discharges and other Overvoltages3) Mechanical Damage
> Analysis of UnbalancedFults6
Unbalanced Faults (3)
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SINGLE PHASE REPRESENTATION IS NO LONGER VALID
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> Analysis of UnbalancedFults7
Unbalanced Faults (4)
Analysed using :-
Symmetrical Components
Equivalent Sequence Networks of Power System
Connection of Sequence Networks appropriate to Typeof Fault
> Analysis of UnbalancedFults8
Symmetrical 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 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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> Analysis of UnbalancedFults9
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
> Analysis of UnbalancedFults10
Unbalanced 5-Phase System of Voltages (2)
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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> Analysis of UnbalancedFults11
Unbalanced 5-Phase System of Voltages (3)
Fifth Set of ZeroSequence Phasors V A5
VB5
VD5VE5
VC5
> Analysis of UnbalancedFults12
Unbalanced 3-Phase System of Voltages (1)
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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> Analysis of UnbalancedFults13
Unbalanced 3-Phase System of Voltages (2)
Third Set of ZeroSequence Phasors
V A3VB3
VC3
> Analysis of UnbalancedFults14
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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> Analysis of UnbalancedFults15
Unbalanced 3-Phase System (2)
Zero Sequence
V A0VB0
VC0
> Analysis of UnbalancedFults16
Symmetrical Components
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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> Analysis of UnbalancedFults17
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
> Analysis of UnbalancedFults18
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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> Analysis of UnbalancedFults19
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 )
> Analysis of UnbalancedFults20
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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> Analysis of UnbalancedFults21
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 }
> Analysis of UnbalancedFults22
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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> Analysis of UnbalancedFults23
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
> Analysis of UnbalancedFults24
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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> Analysis of UnbalancedFults25
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
> Analysis of UnbalancedFults26
Positive Sequence Voltages
VC1 = aV A1
15
V A1 =
0.965 15
VB1 =a 2V A1
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> Analysis of UnbalancedFults27
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
> Analysis of UnbalancedFults28
Symmetrical Components
VC1
V A1
VB1
VC2
V A2
VB2
VC0
V A0
VB0
V A2VC2
VB2
VC
V A
V0
VB
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> Analysis of UnbalancedFults29
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
> Analysis of UnbalancedFults30
Example (2)
IC
IB
109
109
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> Analysis of UnbalancedFults31
POSITIVE
I A1
NEGATIVE
IB1IC1
I A2
IB2
IC2
IC1
IB1
I A1
I A2
IB2
IC2
> Analysis of UnbalancedFults32
PHASE
ZERO
I A0
IB0
IC0
IB
IC
IC
IB
I A0IB0IC0
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> Analysis of UnbalancedFults33
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
> Analysis of UnbalancedFults34
Sequence Components (2)
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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> Analysis of UnbalancedFults35
Sequence Components (3)
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
> Analysis of UnbalancedFults36
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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> Analysis of UnbalancedFults37
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
> Analysis of UnbalancedFults38
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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> Analysis of UnbalancedFults39
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.
> Analysis of UnbalancedFults40
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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> Analysis of UnbalancedFults41
Phase Sequence Equivalent Circuits (2)
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
> Analysis of UnbalancedFults42
Phase Sequence Equivalent Circuits (3)
Zero Sequence Impedance
I
I
I3IE QP
Z0 = E/IP 0 Q0
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> Analysis of UnbalancedFults45
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
> Analysis of UnbalancedFults46
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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> Analysis of UnbalancedFults47
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
Generator Transformer
System Single LineDiagram
Negative Sequence Diagram
> Analysis of UnbalancedFults48
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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> Analysis of UnbalancedFults49
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
> Analysis of UnbalancedFults50
Transformer Zero Sequence Impedance
QP
ZT0aa
QP
bb
N0
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> Analysis of UnbalancedFults51
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
> Analysis of UnbalancedFults52
Zero Sequence EquivalentDy Tx (1)
3I0
No zero sequencein line connectionon side
I 0
I 0
I0
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> Analysis of UnbalancedFults53
Zero Sequence EquivalentDy Transformer (2)
sideterminal
sideterminal ZT0
N0(E0)
I0
Thus, equivalent s ingle phase zero sequence diagram :-
> Analysis of UnbalancedFults54
Zero Sequence Equivalent Circui ts (1)
S 0ZT0
N0
P 0
P S
aa
b b
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> Analysis of UnbalancedFults55
Zero Sequence Equivalent Circui ts (2)
S 0ZT0
N0
P 0
P S
aa
b b
> Analysis of UnbalancedFults56
Zero Sequence Equivalent Circui ts (3)
S 0ZT0
N0
P 0
P S
aa
b b
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> Analysis of UnbalancedFults57
Zero Sequence Equivalent Circui ts (4)
S 0ZT0
N0
P 0
P S
aa
b b
> Analysis of UnbalancedFults58
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
Generator Transformer
System Single Line
Diagram
Zero Sequence Network
3R
E0
3R T
RT
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> Analysis of UnbalancedFults59
Zig-Zag Earthing Transformers (1)
Positive (and negative) sequence impedance is very high.
Equivalent Circu it :
P
N
E
R
P 1
N1
> Analysis of UnbalancedFults60
Zig-Zag Earthing Transformers (2)
Zero sequence impedance is very low.
Equivalent Circu it P0
ZT0
3R
N0(E0)
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> Analysis of UnbalancedFults61
Summary of Sequence Diagrams (1)
System Single Line Diagram
R
E
N
GENERATOR TRANSFORMERLINE F
> Analysis of UnbalancedFults62
Summary of Sequence Diagrams (2)
Positive Sequence
(N1)
V1
F1
ZL1ZT1ZG1E1N1
I1
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> Analysis of UnbalancedFults63
Summary of Sequence Diagrams (3)
Negative Sequence
(N2)
V2
F2ZL2ZT2ZG2N2 I2
> Analysis of UnbalancedFults64
Summary of Sequence Diagrams (4)
Zero Sequence
E0(N0)
V0
F0ZL0ZT0ZG0
3R
N0 I0
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> Analysis of UnbalancedFults65
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
> Analysis of UnbalancedFults66
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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> Analysis of UnbalancedFults67
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
> Analysis of UnbalancedFults68
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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> Analysis of UnbalancedFults69
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
> Analysis of UnbalancedFults70
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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> Analysis of UnbalancedFults71
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
> Analysis of UnbalancedFults72
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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> Analysis of UnbalancedFults73
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
> Analysis of UnbalancedFults74
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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> Analysis of UnbalancedFults79
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 = ?
> Analysis of UnbalancedFults80
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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> Analysis of UnbalancedFults81
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
> Analysis of UnbalancedFults82
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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> Analysis of UnbalancedFults83
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
> Analysis of UnbalancedFults84
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
=
+
==
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> Analysis of UnbalancedFults85
3 Versus 1 Fault Level (1)
XgXTE
Xg XT
E Z1 IF
3
1TgF Z
E XX
E +=
> Analysis of UnbalancedFults86
3 Versus 1 Fault Level (2)
Z0
IF
1 Xg1 XT1
E
Z2 = Z 1
Z1
Xg2 XT2
Xg0 XT0
I = +3E
F 2Z Z1 0
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> Analysis of UnbalancedFults87
3 Versus 1 Fault Level (3)
LEVEL FAULT 3 LEVEL FAULT 1
Z Z IF
Z2Z3E
LEVEL FAULT 1
Z2Z3E
3Z3E
ZE
LEVEL FAULT 3
10
01
1111
>