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Basic Fault Calcu lat ionsand Analysis o f Balanced Faults
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Power System Faul t Analys is
All Protection Engineers should have an understanding
To:-
Calculate Power System Currents and Voltages duringFault Conditions
Check that Breaking Capacity of Switchgear is Not
ExceededDetermine the Quantities which can be used by Relays to
Distinguish Between Healthy (i.e. Loaded) and FaultConditions
Appreciate the Effect of the Method of Earthing on theDetection of Earth Faults
Select the Best Relay Characteristics for Fault Detection
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Power System Faul t Analys is
Ensure that Load and Short Ratings of Plant are Not
Exceeded
Select Relay Settings for Fault Detection andDescrimination
Understand Principles of Relay Operation
Conduct Post Fault Analysis
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Power System Faul t Analysis A lso Used To:-
Consider Stability Conditions
Required Fault Clearance Times
Need for 1 Phase or 3 Phase Auto-Reclose
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Compu ter Fault Calcu lat ion Prog rammes
Widely available, particularly in large power utilities
Powerful for large power systems
Sometimes over-complex for simple circuits
Not always user friendly
Sometimes operated by other departments and notdirectly available to protection engineers
Programme calculation methods:- understanding isimportant
Need for by hand spot checks of calculations
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Pocket Calcu lator Methods
Adequate for the majority of simple applications
Useful when no access is available to computers andprogrammes e.g. on site
Useful for spot checks on computer results
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Faults A re Mainly Caused By Insulat ion Failure
Underground CablesDiggers
Overloading
Oil Leakage
Ageing
Overhead Lines
Lightning
Kites
Trees
MoistureSalt
Birds
Broken Conductors
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Faults A re Mainly Caused By Insulat ion Fai lure
Machines
Mechanical Damage
Unbalanced Load
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Types Of Fault
/ E
/ / E
/
3
3/E
a
bc
e
a
bc
e
a
bc
e
e
a
b
c
e
a
b
c
e
a
b
c
a
b
c
a
b
c
a
b
c
a
b
ce
a
b
c
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Types Of Fault
OPEN
CIRCUITS
CROSS
COUNTRY
FAULT
OPENCIRCUIT
+
/ E
a
bc
FAULTBETWEEN
ADJACENT
PARALLEL
LINES
a
bc
e
a'
b'c'
e
a
b
c
e
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Types Of Fault
CHANGING
FAULT INCABLE
a
b c
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Voltage Convent ion
In power system analysis it is important to maintain a
consistent approach when relating currents and voltages
This is achieved by maintaining a VOLTAGECONVENTION:
VAB= Voltage of A above B = + IZ
~
I
EAB
A
Z VAB
+
-B
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Vectors
Rotating Vectors can be used to represent
Sinusoidal Electrical Quantities
t
+V
= Vsint
-V
V
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Vectors
Vector notation can be used to represent phase
relationship between electrical quantities
V
Z
I
V = Vsint = V 0
I = I- = Isin(t-)
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Vector Mul t ip l icat ion And Divis ion
Vector Multiplication
A
B
A.B
A + B
B
A
A.B = A A . B B
= A.B (A + B)
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Vector Mul t ip l icat ion And Divis ion
AB
Example:
ZV
0V
VZ 0
0
III
Vector Division
= A A / B A
= (A - B)
B
A
B
A
B
A
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j Operato r
Rotates vectors by 90 anti-clockwise :
Used to express vectors in terms of real andimaginary parts.
1
9090
9090
j = 1 90
j2 = 1 180= -1
j3 = 1 270
= -j
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j Operato r
Used to express vectors in terms of real and
imaginary partsZZ :Example
Imaginary
RealZCos
ZSin
Z
jXR
sinZjZcosZZ
Resistance Reactance
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j Operato r
0
0VV
)( sin tIII
00VtsinVV:LET
ZVV
ZIMPEDANCE II jX
R
sinjZcosZZZ
planeR/jXonanglesiseanticlockwbydrepresenteVLaggingI
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a = 1120
Rotates vectors by 120 anticlockwise
Used extensively in Symmetrical Component Analysis
120
120 1
120
2
3j
2
1-1201a
2
3j
2
12401a 2
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a = 1120
Balanced 3 voltages :-
VA
VC = aVA
a2 + a + 1 = 0
VB = a2VA
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Balanced (3) Faul ts
RARE :- Majority of Faults are Unbalanced
CAUSES :-
1. System Energisation with Maintenance EarthingClamps still connected.
2. 1 Faults developing into 3 Faults
3 FAULTS MAY BE REPRESENTED BY 1 CIRCUITValid because system is maintained in a BALANCED stateduring the fault
Voltages equal and 120 apart
Currents equal and 120 apart
Power System Plant Symmetrical
Phase Impedances Equal
Mutual Impedances Equal
Shunt Admittances Equal
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Balanced (3) Faul ts
LINE X
LOADS
LINE Y
3 FAULT
ZLOAD
ZLY
IbF
IcF
IaFZLXZTZG
Ec
Ea
Eb
GENERATOR TRANSFORMER
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Balanced (3) Faul ts
Positive Sequence (Single Phase) Circuit :-
Ea
Ec
EaF1
N1
Eb
IbF
Ia1 = IaF
IcF
IaF
ZT1 ZLX1 ZLX2ZG1
ZLOAD
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Maximum Fault Level
Three Phase Fault Level:
IF(max) = VA Fault level / (3 x Line voltage)
= (MVA x 1000) / (3 x kV)
Typical Maximum Fault Levels:
System Voltage kV MVA Fault Level Max. Fault Current(kA)
400 35 000 50
132 5 000 22
33 1 000 17.5
11 250 13.1
400V 25 55Single Phase Fault Level:
Can be higher than 3 fault level on solidly-earthed systems;
Check that switchgear breaking capacity > maximum fault level forall types
G S C C
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Generator Short Circu i t Current
The AC Symmetrical component of the short circuit current varies with timedue to effect of armature reaction.
Magnitude (RMS) of current at any time t after instant of short circuit :
where :
I" = Initial Symmetrical S/C Current or Subtransient Current= E/Xd"
I' = Symmetrical Current a Few Cycles Later orTransient Current = E/Xd'
I = Symmetrical Steady State Current = E/Xd
)e-'()e'-"( t/Td'-t/Td"-ac
i
TIME
Si l G t M d l
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Simp le Generator Models
Generator model to calculate the initial symmetrical S/Ccurrent or subtransient current
jXd
Generator model to obtain the S/C current a few cycles later,
IE; the transient current
Generator model to obtain the steady state current
jXd
E
E
E
jXd
V i t i Of G t F lt C t A ft F lt
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Variat ion Of Generator Fault Current A fter FaultInstant No Voltage Regu lat ion
Iac = (IdId)e-t/T + (Id- Id)e-t/T + Id
CURRENT
T
TIdId
Id
0.368 [I
d I
d]
0.368 [Id Id]
TIME
Eff t f A t ti V l t R l t i O Sh t
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Effect of Automatic Voltage Regu lat ion On Sho rtCircu it Current Of Generators
With AVR
Without AVR
MultiplesOf RatedCurrent
TIME IN SECS1 2 3 4
2
4
6
8
P ll l G t
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Parallel Generator s
If both generator EMFs are equal they can be thoughtof as resulting from the same ideal source - thus thecircuit can be simplified.
11kV
20MVA
XG=0.2pu
11kV 11kV
j0.05 j0.1
XG=0.2pu
20MVA
P U Di
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P.U. Diagram
IF
j0.05 j0.1
j0.2
1.01.0
j0.2
IF
j0.05 j0.1
j0.2j0.2
1.0
P it i S I d f T f
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Posit ive Sequence Impedances o f Transfo rmers
2. Winding Transformers
ZP = Primary Leakage Reactance
ZS = Secondary LeakageReactance
ZM = Magnetising impedance= Large compared with ZP
and ZS
ZM Infinity Represented byan Open Circuit
ZT1 = ZP + ZS = PositiveSequence Impedance
ZPand ZSboth expressed
on same voltage
base.
S1P1
P S
P1 S1ZP ZS
ZM
N1
N1
ZT1 = ZP + ZS
Posit ive Sequence Impedances o f Transfo rmers
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Posit ive Sequence Impedances o f Transfo rmers
3. Winding Transformers
ZP, ZS, ZT = Leakage reactances of Primary, Secondary and Tertiary Windings
ZM = Magnetising Impedance = Large Ignored
P S
T
P
T
SZP
ZM
N1
ZT
ZS
P
T
SZP
N1
ZT
ZSZP-S = ZP + ZS = Impedance
between Primary (P) and
Secondary (S) where ZP & ZS are
both expressed on same voltagebase
Similarly ZP-T = ZP + ZT and ZS-T =
ZS + ZT
T
Auto Transformers
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Auto-Transformers
Equivalent circuit is similar to that of a 3 winding transformer
H L
H
T
LZH1
N1
ZT1
ZL1
ZM = Magnetising Impedance = Large Ignored
T
ZHL1 = ZH1 + ZL1 (both referred tosame voltage base)
ZHT1 = ZH1 + ZT1 (both referred to
same voltage base)
ZLT1 = ZL1 + ZT1 (both referred to
same voltage base)
ZM1
H
T
LZH1
N1
ZT1
ZL1
Motors
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Motors
Fault current contribution decays with time
Decay rate of the current depends on the system. From tests,typical decay rate is 100 - 150ms.
Typically modelled as a voltage behind an impedance
Xd"
M 1.0
Indu ct ion Motors IEEE Recommendat ions
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Indu ct ion Motors IEEE Recommendat ions
Small Motors
Motor load 35kW SCM = 4 x sum of FLCM
Large Motors
SCM motor full load ampsXd"
Approximation : SCM = locked rotor amps
SCM = 5 x FLCM assumes motorimpedance 20%
Synchronous Motors IEEE Recommendat ions
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Synchronous Motors IEEE Recommendat ions
Large Synchronous Motors
SCM 6.7 x FLCM for Assumes X"d = 15%1200 rpm
5 x FLCM for Assumes X"d = 20%514 - 900 rpm
3.6 x FLCM for Assumes X"d = 28%450 rpm or less
Overhead Line Impedances
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Overhead Line Impedances
R = AC RESISTANCE
L = SERIES INDUCTABLE
C = SHUNT CAPACITANCE
G = SHUNT CONDUCTANCE (NEGLIGABLE)
Series elements are of greatest interest to protectionengineers
R L
C G
SERIES
SHUNT
Symmetr ica l Circu i t
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Symmetr ica l Circu i tA
D
C
D
B
D
Carsons Equations
mile
mile
/
/
D
Delogj0.00466f0.00159f
ZImpedanceMutual
Delog0.00466fj0.00159R
ZImpedanceSelf
10
M
G10
P
)/mileG
10Mp21
Dlog0.00466fjRZ-ZZZ
Where: -
R = Conductor a.c. resistance
G = Geometric mean radius of a single conductorD = Spacing between parallel conductors
De = Equivalent spacing of the earth return path (ft)
= 2160 e/f
E = Earth resistivity (ohm metres)
/mile.D
Delog0.01398f.j0.00477fR2ZmZZ
2G
310po
Overhead Line Impedances
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Overhead Line Impedances
Positive Sequence
Ea
Eb
Ec
Ia1
Ib1
Ic1
Zp
Zp
Zp
Zm
Zm
Zm
Ea = Ia1 Zp + Ib1 Zm + Ic1 Zm
= Ia1 {Zp Zm}
Z1 = = Zp ZmEa
Ia1
Overhead Line Impedances
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Overhead Line Impedances
Zero Sequence
E = I Zp + 2 I Zm
= I { Zp + 2 Zm }
ZO = Zp + 2 Zm + ZOM
Single
Circuit
Double
Circuit Line (Single Circuit)
E
I
I
I
Zp
Zp
Zp
Zm
ZmZm
Z0 = = Zp + 2ZmEaI
Non Symmetr ica l Circui ts
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Non Symmetr ica l Circui ts
Transposed Line:-
DAB
DCA
AB
C
DBC
A
B
C
C
C
A
A
B
B
Non Symmetr ica l Circui ts
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Non Symmetr ica l Circui ts
ADAB.DBC.DC
conductorsbetweendistancemeanGeometricD
:w here
.D
Delog0.00869f.j0.00296fR
De0.01398f.j0.00477fRZ
Dlog0.0029f.jR
/mile)(D
log0.00466f.jRZZ
3
2G
10
G
O
G10
G1021
)/(
)/(
.
log
)/(
3
2310
Km
mile
D
Km
Typ ical GMD Values For 3Overhead L ines
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Typ ical GMD Values For 3 Overhead L ines
100 200 300 400 500
2
4
6
8
10
12
14
16
GMD
(Metres)
Sys tem Operat ing Voltage (kV -)
Posit ive Sequence Induc t ion Of Overhead
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Posit ive Sequence Induc t ion Of OverheadTransm iss ion L ines Typic al Values
System Voltage kV
km
mH
100 200 300 400
1
1.5
D Increasing
G ~ Constant
0
G Increasing Faster Than D
G
Dlog10x
Base Quanti t ies and Per Unit Values
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Base Quanti t ies and Per Unit Values
Particularly useful when analysing large systems withseveral voltage levels
All system parameters referred to common base quantities
Base quantities fixed in one part of system
Base quantities at other parts at different voltage levelsdepend on ratio of intervening transformers
11 kV20 MVA
O/H LINE
11/132 kV50 MVA
ZT = 10%ZT = 10%
132/33 kV50 MVA
FEEDER
ZL = 8ZL = 40ZG = 0.3 p.u.
Base Quanti t ies and Per Unit Values
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Base Quanti t ies and Per Unit Values
Base Quantities Normally Used :-
BASE MVA = MVAb = 3 MVA
Constant at all voltage levels
Value ~ MVA rating of largest itemof plant or 100MVA
BASE VOLTAGE = KVb = / voltage in kV
Fixed in one part of system
This value is referred through
transformers to obtain base
voltages on other parts of system
Base voltages on each side of
transformer are in same ratio as
voltage ratio
Base Quanti t ies and Per Unit Values
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Base Quanti t ies and Per Unit Values
Other Base Quantities :-
kAinkV.3
MVACurrentBase
Ohmsin
MVA
)(kVZImpedanceBase
b
bb
b
2b
b
Base Quanti t ies and Per Unit Values
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Base Quanti t ies and Per Unit Values
Per Unit Values = Actual Value
Base Value
CurrentUnitPer
)(kV
MVA.Z
Z
ZZImpedanceUnitPer
KVKVkVVoltageUnitPer
MVA
MVAMVAMVAUnitPer
b
ap.u.
2
b
ba
b
ap.u.
b
ap.u.
b
ap.u.
Base Quanti t ies and Per Unit Values
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Base Quanti t ies and Per Unit Values
Per Unit Values = Actual Value/ Base Value
The base or reference value is generally related to theequipment rating
E.g. Current rating = 600A600A = 1.0 pu
Sometimes specified in percentage, where 1 pu = 100%
Base Quanti t ies and Per Unit Values
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Q
For Impedances
Per unit value = Ohmic value / Base value
Base Value = (kV2)/ MVA
If a 11kV / 440V transformer is rated at 500kVA, with an
impedance 0.1 pu (10%) :
Ohmic value = Per unit value x Base value
= 0.10 x (0.4402 / 0.500)
= 0.10 x 0.387= 0.039 Viewed from the LV side
= 24.2 Viewed from the HV side
Trans form er Unit Impedance
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p
Z Per Unit = Z p.u.
Z Percentage = 100 x Z p.u.
Z p.u. = Za/Zb = Zactual / Zbase
Za =
Zb = = =
Zp.u. = =
kV / (TEST)
I (RATED)ZP ZS
ZM
kV/ (TEST)
3. I(RATED)
kV/ (TEST)2
kV/ (TEST)2
kV/ (TEST)
MVA(RATED) 3. kV/ (RATED) . I(RATED) 3. I(RATED)
Za kV/ (TEST)
Zb kV/ (RATED)
Trans form er Percentage Impedance
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g p
Increase Vpuntil I (RATED) flows in secondary
Z (Percentage) =
Zp.u. =
Vp
I (RATED)ZP ZS
ZMS/C
Secondary
Vp x 100%
V (RATED)
Vp
V (RATED)
Trans form er Percentage Impedance
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g p
If ZT = 5%
with Secondary S/C5% V (RATED) produces I(RATED) in Secondary.
V (RATED) produces 100 x I(RATED)5
= 20 xI(RATED)
If Source Impedance ZS = 0
Fault current = 20 x I(RATED)
Fault Power = 20 x kVA (RATED)
ZT is based on I(RATED) & V (RATED)
i.e. Based on MVA (RATED) & kV (RATED)
is same value viewed from either side of transformer
Trans form er Percentage Impedance
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g p
Example (1)
Per unit impedance of transformer is same on each side ofthe transformer.
Consider transformer of ratio kV1 / kV2
Actual impedance of transformer viewed from side 1 = Za1
Actual impedance of transformer viewed from side 2 = Za2
MVA
1 2
kVb = kV1 kVb = kV2
Trans form er Percentage Impedance
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g p
Zp.u.1 = Za1 = Za1 x MVA
Zb1 kV12
Zp.u.2 = Za2 = Za2 x MVAZb2 kV2
2
BUT Za2 = Za1 x kV22
kV12
Zp.u.2 = Za1 x kV22 x MVA
kV12 kV2
2
= Za1 x MVA
kV12
= Zp.u.1
Trans form er Percentage Impedance
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g p
Example (2)
Base voltage on each side of a transformer must be in the
same ratio as voltage ratio of transformer
Incorrect selectionof kVb 11.8kV 132kV 11kV
Correct selection 132x11.8 132kV 11kVof kVb 141
= 11.05kV
Alternative correct 11.8kV 141kV 141x11 = 11.75kVselection of kVb 132
11.8kV 11.8/141kV 132/11kVOHL Distribution
System
Convers ion of Per Unit Values from One Set ofQ ti t i t A th
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Quant i t ies to Another
Zp.u.2Zp.u.1
Zb1 Zb2
MVAb1 MVAb2
kVb1 kVb2
Actual Z = Za
2
b2
2b1
b1
b2p.u.1
2b2
b2
b1
2b1
p.u.1
b2
b1p.u.1
b2
ap.u.2
b1
ap.u.1
)(kV
)(kVx
MVA
MVAxZ
)(kV
MVAx
MVA
)(kVxZ
ZZxZ
ZZZ
Z
ZZ
Convers ion of Per Unit Values from One Set ofQ ti t i t A th
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Quant i t ies to Another
kVb
MVAb
132
50
349
219 A
33
50
21.8
874 A
Z
V
11
50
2625 A
2.42
11 kV20 MVA
132/33 kV50 MVA
10%40
11/132 kV50 MVA
10% 8
3FAULT
0.3p.u.
Zb= 2
MVAb
kVb
Ib= MVAb3kVb
p.u. 0.3 x 5020
= 0.75p.u.
0.1p.u.40
349= 0.115p.u. 0.1p.u.
8
21.8= 0.367p.u.
1p.u.
1.432p.u.
IF = 1 = 0.698p.u.
1.432
I11 kV = 0.698 x Ib =
0.698 x 2625 = 1833A
I132 kV = 0.698 x 219 = 153A
I33 kV = 0.698 x 874 = 610A
Circui t Laws
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Three laws from which all circuit theorems have been derived: -
Ohms Laws Z
V
I
V = IZ
Kirchoffs Junction Law
I2
I1
I3
I = 0I
1+ I
2+ I
3= 0
Kirchoffs Mesh Law
E1
Z1 Z2
Z31 2E2
i1 i2
Round Any Mesh E = IZ
e.g. E1 = i1Z1 + i1Z3 i2Z3 etc.
Thevenins Theorem (1)
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Useful for replacing part of a system which is not of particular interestby a single equivalent voltage and series impedance
For Example: -
Replacement o f power system s upp ly ing an indu str ial network
~E1 ~
Z1 Z2
Z3 E2
E1 ~
Z1
Z3 E1= Open Circuit Voltage
= Z3 . E1
Z1 + Z3
Thevenins Theorem (2)
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~E1 ~
Z Z2
E2
Z1
Z3Z = Impedance with E1 S/C
= Z1 Z3
Z1 + Z3
Thevenin equivalent circuit :-
Star/Delta & Delta/Star Trans fo rm Theorem (1)
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Useful when reducing overall system equivalent impedance to
a single value
~
~
~
~
~
ZEQUIV
Star/Delta & Delta/Star Trans fo rm Theorem (2)
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Z10
Z30 Z20
1
23
312312
311210
ZZZ
Z.ZZ
312312
231220
ZZZ
Z.ZZ
312312
312330
ZZZ
Z.ZZ
30
2010201012
Z
Z.ZZZZ
10
3020302023
Z
Z.ZZZZ
20
1030103031
Z
Z.ZZZZ
Z31 Z12
Z23
1
23
Superposi t ion Theorem
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Z3
~E1 ~ E2
Z1 Z2
E1 ~
Z1 Z2
~ E2
Z1 Z2
I3
Z3
I31
I32
+
Z3
I3 = I31 + I32
Reducing Sys tem to a Single Source (Ex = EY)
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Healthy Loaded System
~EX~
F1
N1
Faulted System (3 Fault):-
EX~
F1
N1
EY
EY~IF
Reducing Sys tem to a Single Source (Ex = EY)
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If EX = EY = E
E ~
F1
N1
Ignore Load:-
E
~
F1
N1
I
F
IF
Reducing Sys tem to a Sing le Sourc e (ExEY)
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EX ~
F1
N1
Healthy Loaded System:-
F1
N1
VPF = E ~ EY
Thevenise between F1 and N1:-
~ E
Reducing Sys tem to a Sing le Sourc e (ExEY)
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Fault Between F1 and N1:-
F1
N1
Ignore Load:-
~ E
F1
N1
E~
IF
IF
Pre-Fault Load Condit io ns (1)
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VPF~
I3LIIL
Z2
I2L
Z1 Z2 Z4 Z6
P1
N1
I4L
Z5
I5(L)
I6L
~Ex EY
Pre-Fault Load Conditions: -
For a fault at P
Pre-fault voltage = VPF
Pre-fault Load Currents = I. (L)
Pre-Fault Load Condit io ns (2)
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P1
Z3 Z4 Z6
N1
I4(F)I3(F)
Z5
I
5(F)
I6(F)
~ EY
Three Phase Fault At P: -Using exact values of EX and EY
Total current in each branch = I(F)
~
II(F)
I2(F)
Z1
EXZ2
IF
Three Phase Fau lt A t P
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P1
I1(T)
I2(T)
Z1 Z3 Z4 Z6
N1
I4(T)I3(T)
Z5
I5(T)
I6(T)
~ VPF
Thevenised Circuit Used To Calculate FaultCurrent IFCurrent in each branch of system is not necessarily the totalcurrent in the branch
Current in branches = I-(T)
I-(T)I-(F)
Z2
IT = IF
Pre Fault Load Circu i t
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This can be represented by:-
Z1II(L)
EYEX
I2(L)
Z3 Z4 Z6I4(L)
Z5
I5(L)
I6(L)
Z2~
I3(L)
~
N1
VPF
P1
Z1
II(L)
EYEX
I2(L)
Z3 Z4 Z6I4(L)
Z5
I5(L)
I6(L)
Z2~
I3(L)
~
N1
VPF
P1
Theoretical generator VPF does not affect the current distribution
Calculate load current followed by pre-fault voltage using superposition
~
Example Calcu lat ion
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LOAD
Power System
A B
~
3 FAULT
~
O
N1
VPF
EE
1.6
A B2.5
0.75 0.45
0.45
18.85~ ~
O
Examp le Calcu lat ion
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Convert impedances A0, B0 + N10 to equivalent impedancesZAN1 = ZA0 + ZN10 +
= 0.75 + 18.85 + = 51
B0
N10A0
Z
.ZZ
0.45
18.85x0.75
N1
VPF
EE
1.6
A B2.5
0.75 0.45
0.45
18.85~ ~
O
This gives :-
Examp le Calcu lat ion
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ZAB = 1.22
ZAN1 = 51
ZBN1 = 30.6
N1
VPF
EE
1.6
A B2.5
0.4
~ ~
Circuit with equivalent impedances:
By Thevenins Theorem the network can be represented by:-
1.22
51 30.6
N1
VPF
1.6
A B2.5
0.4
Examp le Calcu lat ion
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Reducing the network to a single voltage and impedance
A
1.55VPF1.215~
N1
VPF
A
1.55
0.82
0.345~
N1
VPF
0.682~
A
N1
Examp le Calcu lat ion
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Assuming IF = 1
0.682
N1
VPF ~
A
A
IF=1.0
1.2151.55VPF ~
N1
0.56
2.765
1.2150.44
1.0
N1VPF
A
1.55
0.82
0.345~
0.560.44
1.0
This gives :-
Examp le Calcu lat ion
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553.0
31
6.30x56.0
0.56-0.376=0.184
0.44-0.427=0.013
427.0
6.52
51x44.0
1.22
51 30.6VPF
1.6
A B2.5
0.4
0.56-0.553=0.007
376.072.3
5.2x56.0
1.0
N1
VPF
A
1.55
0.82
0.345~
0.560.44
1.0
Now convert impedances AN1, BN1, and ABback to equivalent impedances:
N1
Examp le Calcu lat ion
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0.553 - 0.184=0.369
0.553
0.184
0.389 0.369=0.02
1.0
1.0 (0.427+0.184)= 1.0 0.611= 0.389
Positive Sequence
Distribution Factors
0.427
N1
VPF
EE
1.6
A B2.5
0.75 0.45
0.4
18.85~ ~
O
D.C. Trans ients & Offsets
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~
R + jL
VM sin (t + )iF
ZFault Applied
iF
ia.c.
id.c.
iF = Id.c. + Ia.c.
)-(wtsinZ
Ve)-(sin
Z
V- MRt/L-M
D.C. Transient Symmetrical A.C.Component Component
Max. value when ( ) = or
If = 90 this occurs when = 0 or 180 i r when the fault occurs at
=
2
3
2-