2 Basic Fault Calculation&Analysis of Balanced Fault

8/22/2019 2 Basic Fault Calculation&Analysis of Balanced Fault

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GRID

Technical Institute

Basic Fault Calcu lat ionsand Analysis o f Balanced Faults


2/81

> Basic Fault Calculations & Analysis of Balanced Faults2

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


3/81


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


4/81


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


5/81


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


6/81


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


7/81> Basic Fault Calculations & Analysis of Balanced Faults7

Faults A re Mainly Caused By Insulat ion Failure

Underground CablesDiggers

Overloading

Oil Leakage

Ageing

Overhead Lines

Lightning

Kites

Trees

MoistureSalt

Birds

Broken Conductors



Faults A re Mainly Caused By Insulat ion Fai lure

Machines

Mechanical Damage

Unbalanced Load



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



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



Types Of Fault

CHANGING

FAULT INCABLE

a

b c



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



Vectors

Rotating Vectors can be used to represent

Sinusoidal Electrical Quantities

t

+V

= Vsint

-V

V



Vectors

Vector notation can be used to represent phase

relationship between electrical quantities

V

Z

I

V = Vsint = V 0

I = I- = Isin(t-)



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)



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



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



j Operato r

Used to express vectors in terms of real and

imaginary partsZZ :Example

Imaginary

RealZCos

ZSin

Z

jXR

sinZjZcosZZ

Resistance Reactance



j Operato r

0

0VV

)( sin tIII

00VtsinVV:LET

ZVV

ZIMPEDANCE II jX

R

sinjZcosZZZ

planeR/jXonanglesiseanticlockwbydrepresenteVLaggingI



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



a = 1120

Balanced 3 voltages :-

VA

VC = aVA

a2 + a + 1 = 0

VB = a2VA



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


23/81



LINE X

LOADS

LINE Y

3 FAULT

ZLOAD

ZLY

IbF

IcF

IaFZLXZTZG

Ec

Ea

Eb

GENERATOR TRANSFORMER


24/81



Positive Sequence (Single Phase) Circuit :-

Ea

Ec

EaF1

N1

Eb

IbF

Ia1 = IaF

IcF

IaF

ZT1 ZLX1 ZLX2ZG1

ZLOAD


25/81


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


26/81


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


27/81


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


28/81


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


29/81


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


30/81


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


31/81


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


32/81


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



33/81



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


34/81


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


35/81


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


36/81


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


37/81


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


38/81



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


39/81


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



40/81



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



41/81



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


42/81



Transposed Line:-

DAB

DCA

AB

C

DBC

A

B

C

C

C

A

A

B

B



43/81



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


44/81


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


45/81


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


46/81



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.



47/81



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



48/81



Other Base Quantities :-

kAinkV.3

MVACurrentBase

Ohmsin

MVA

)(kVZImpedanceBase

b

bb

b

2b

b



49/81



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.



50/81



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%



51/81


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


52/81


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


53/81


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)



54/81


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



55/81


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



56/81


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



57/81


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


58/81


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


59/81


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


60/81


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)


61/81


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)


62/81


~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)


63/81


Useful when reducing overall system equivalent impedance to

a single value

~

~

~

~

~

ZEQUIV

Star/Delta & Delta/Star Trans fo rm Theorem (2)


64/81


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


65/81


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)


66/81


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)


67/81


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)


68/81


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)


69/81


Fault Between F1 and N1:-

F1

N1

Ignore Load:-

~ E

F1

N1

E~

IF

IF

Pre-Fault Load Condit io ns (1)


70/81


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)


71/81


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


72/81


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


73/81


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


74/81


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


75/81


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 :-



76/81


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



77/81


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



78/81


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 :-



79/81


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



80/81


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


81/81

~

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-

Documents

2 Basic Fault Calculation&Analysis of Balanced Fault