Appa-module 6-Fault Current Analysis

8/10/2019 Appa-module 6-Fault Current Analysis

1/65

Module 6.0

Fault Current Calculation

By: Dr. Hamid Jaffari

Power system Review


2/65

Fault Currents

Symmetrical Fault

Asymmetrical fault

Power System Review


3/65

Faul t Analys is

Analysis Type

Power Flow: normal operating conditions Faults: abnormal operating conditions

Fault Types

Balanced or Symmetr ical Fault

Three Phase Short Circuit

Unbalanced or Unsymmetr ical Faul ts

Single line-to-ground

Double line-to-ground

Line-to-line What are the results used for?

o Determining the circuit breaker rating

o Protective Relaying settings


4/65

Various Types o f Fau lts

FaultlSymmetrica)a

FaultcalUnsymmetri)b

Faultline-to-line Faultground-to-linedouble Faultground-to-line

fault1

F

ZZ

V)(3I

l-faultSymmetrica

fault021

Ffault

3Z)3Z(ZZ

3Vground)-to-(LineI

nZ

fault21

Ffault

ZZZ

V3line)-to-(lineI

j

a

b

c

a

b

c

a

b

c

a

b

c

a

b

c


5/65

Asymmetrical

Fault Calculation

Power System Review


6/65

R-L Circu i t Trans ien tsR

+

-0

@

t

ClosedSW

L

)sin(2)( wtVte

0)sin(2)()(: ttVtRidt

tdiLEquation

])sin()[sin(2)()()(: Tt

etZ

VtititiSolution dcac

amptZ

V

tiac )sin(2)(

T

t

eZ

Vtidc

)sin(2)(

2222 )( lRXRZ

R

wltg

R

Xtg 11

fR

X

R

X

R

LT

2

:)(/ forcedCurrentFaultStateSteadyFaultlSymmetrica :)(transientCurrentOffsetdc

Solutionforced Solutionnatural


7/65

Asymmetr ical faul t

])sin()[sin(2)()()( Tt

etZ

Vtititi dcac

Dc offset Magnitude depends on angle :

acIoffsetdc 20 )2(

Z

VcurrentfaultacrmsIwhere ac )(:

])2

[sin(2)()()(

)2

(:

T

t

etItititi

Set

acdcac

In order to get the largest fault current:


8/65

Asymmetr ical faul t

Note: i(t) is not completely periodic. So, how do weget the rms value of i(t) ?

Assume :

Now calculate the RMS Asymmetrical Fault Current:

)constant(Ce Tt

AmpeIeIIIIti Tt

acT

t

dcacdcacrms

2

222221]2[][)()()(

cyclesintimeiswhere

f

t

fR

X

R

X

R

X

R

LTNote

;&

2

:

AmpeIeIeIti RXacfR

Xf

acT

t

acrms)/(

4

2

2

2

212121)(

UnitPer21factoralasymmetric)(:)()(

)/(

4

RX

acrms ekwhereIkI


9/65

Asymmetr ical Faul t Calculat ion

Example:In the following Circuit, V=2.4kV, L=8mH,R=0.4, and =260 rad/sec. Determine (a) the rms

symmetrical fault current; (b) the rms asymmetrical fault

current; (c) the rms asymmetrical fault current for .1 cycle

& 3 cycle after the switch closes, assuming the maximumdc offset.

+

-0@ tClosedSW

mHL 20

)sin(24002)( wtte

4R


10/65

Asymmetr ical Fau lt Calcu lat ion

Solution:

4.82042.3

4.82042.3016.34.0)108)(602(4.0)() 3

ZZ

jxjLjRjXRZa

A95.788042.3

2400

volts

Z

VIac

A46.13662195.788)0()0(;0@) kIItb acrms

00.110739.6121)3(

641.1693.1121)1.0(

54.74.0

016.3)()

354.7

)3(4

54.7

)1.0(4

xecyclek

ecyclek

RatioR

Xc

A95.788)3()3(

A69.294,1641.1)1.0()1.0(

cyclekIcycleI

xcyclekIcycleI

acrms

acrms

+

- 0@

t

ClosedSW

mHL 20

)sin(400,22)( wtte

4R


11/65

Asymmetr ical Fau lt-Unloaded

Synch ronous Mach ine

Three Stages: Subtransient, Transient, and Steady State

constanttimearmatureT

/I

/I

/I

:

offsetdcMaximum22)(

)2

sin(]1

)11

()11

[(2)(

CurrentousInstantane)()()(

A

ReactanceStateSteadyReactance/sSynchronouaxixdirect

''ReactanceTransientaxixdirect

'

ReactancentSubtransieaxixdirect"

/"/

"

''"

""

'"

dgd

dgd

dgd

TtTt

d

g

dc

d

T

t

dd

T

t

dd

gac

XEX

XEX

XEX

Where

eIeX

E

ti

tX

eXX

eXX

Eti

tititi

AA

dd

dcac

Add

ddd XXX

T,T,TConstantsTime

&ReactancesMachine

:provideeresManufactur:Note

'"

,'

,"

Stator

Uniform air-gap

Stator winding

Rotor

Rotor winding

N

S

d-axis

q-axis

axisquadratureaxisqaxisdirectaxisd


12/65

Synch ronous Mach ine

Asymmetr ical Faul t Envelopes

Asymmetry Sources: (1) Open Phase and (2) SLG Fault

d

g

X

EI

"

"

d

g

X

EI

'

'

d

g

X

EI

CurrentfaultntSubtransie

CurrentfaultTransient

CurrentfaultS.S

)(tiac

t

envelopescurrentAC

AA T

t

T

t

d

geIe

XE

"

"MAX-dc 22(t)i

"2I

I2

'2I


13/65

ntSubtransie

Transient

StateSteady

offsetdc

FaultalAsymmetric

GeneratornearFaultalAsymmetricofStages

d

g

X

EI

"

"

"2I

d

g

X

EI'

'

'

2I

d

g

X

EI

"2I

'2I


14/65

Fault Current

Calculation

Power System Review


15/65

Fault Current Analys is

Power System Review

Four methods to calculate the fault current:

1.Ohmic Method (not preferred)

2.Infinite Bus Method (Convenient & Easy)3.Per Unit Method (Most Common)

4.MVA Method (Quick & Easy)

Note: This co urse wi l l focus on PU & MVA Methods


16/65

Fault Curren t Analys is

Power System Review

Ohmic Method


17/65

Ohm ic Method

Power System Review

This Method Requires:Transferring all impedances to high/low

voltage side of transformer using square

of XFMR turn ratio

Using your AC circuit theory knowledge

Voltage & Current dividers

Thevenin & Norton equivalentsKramers Rule, etc

2

1

22

2

1

N

NOR

N

N


18/65

Infinite Bus method

Power System Review



19/65

In f in i te Bus Calcu lat ion

SCBsae

pu

puSC

rtransformeutilityputotal

IxI

Z

ZZ

actualI:Step4kVx3

3KVAICalculate:Step3

0.1ICalculate:Step2

ZCalculate:Step1

SC

LL

Base

)(

Infinite Bus calculation is a convenient way to

estimate the maximum 3fault current flow on the secside of the transformer

The following steps are necessary to calculate the ISC

100

%&

;

KnownisCircuitShortUtilityIf:Note1)(

ZZ

MVA

MVAZ

where

ZZZ

rtransforme

SC

baseutility

rtransformeutilityputotal

0&100

%

;

UnknownisCircuitShortUtilityIf:Note2

utilityrtransforme

rtransformetotal

ZpuZ

Z

where

ZZ

totalZ


20/65

7.5%Z

kVkV/4.1613.8

KVA5000

VS

In f in i te Bus Calcu lat ion

Unknown Uti l ity SC Data

A4.925295.693333.13actualI:Step4

A95.693

16.43

5000

kVx3


333.13075.

0.10.1ICalculate:Step2

075.0100

5.7

100

Z%ZCalculate:Step1

SC

LL

Base

pu

xIxIkVx

Z

pu

SCBsae

pu

puSC

Example1:Calculate the maximum 3fault current on 5000 KVATransformers secondary bus.

DataSourceNo


21/65

7.5%Z

kVkV/4.1613.8

KVA5000

VS

Inf in i te Bus Calcu lat ion

w i th Known Uti li ty SC Data

A642695.69326.9actualI:Step4

A95.69316.43

5000

kVx3


26.9108.0

0.10.1ICalculate:Step2

108.0075.033.0ZCalculate:Step1

SC

LL

Base

)(

total

xIxI

kVx

Z

puZZ

SCBsae

putotal

puSC

rtransformeutility

Example2:Calculate the maximum 3fault current on 5000 KVA

Transformers secondary bus.

150MVASC

puZ

Z

puxS

S

kV

kVZZ

pu

rtransforme

Oldbase

Newbase

new

oldpuUtility

SC

baseutility

OldNew

075.0100

5.7

100

%

033.150

5

16.4

16.41

1

150

150

MVA

MBAZ

ZZZCalculate

22

rtransformeutilitytoal

pu108.0330.00.075Ztotal utilityZ

:StepsnCalculatio


22/65


Power System Review

Per-Unit Method

F lt C t A l i


23/65

Power System Review

Fault Current Analys is:

Per-Unit Method

PU analysis is used for both symmetrical &unsymmetrical fault calculations.

All components are defined in PU system.

Analysis is performed using equivalent per phase

circuit modeling.

Requires knowledge of symmetrical components

Requires selecting two system bases for

calculating all base & PU quantities:kVBase & MVAbase

F lt C t A l i


24/65

Power System Review

Fau lt Curren t Analys is:

Per-Unit Method

This Method requires:Knowledge of symmetrical components

Positive sequence (+ SEQ)

Negative sequence(-SEQ)Zero sequence (0 SEQ)

Interconnecting positive, negative, and

zero networks for calculating the variousunsymmetrical faults(LG, LL/LLG, and 3)


25/65

Symmetr ical Components

Power System Review

Steps involved:

1. Draw a single-line diagram of the desired

power system(equivalent per phase)

2. Define zones using transformation point asa point of demarcation

3. Select a common MVAbase for all zones

4. Select a kVBasefor one zone & Calculatea. kVBase for other zones

b. Zbase, and Ibasefor all zones


26/65

Symmetr ical Components..con t

Power System Review

6. Replace each component with itsequivalent reactance in per-unit

7. Draw sequence networks(+, -, 0)

8. Use (+)SEQ network for SymmetricalFault analysis

9. Combine appropriate networks for

calculating various UnsymmetricalFault analysis


27/65

Symmetrical

Fault Calculation

Power System Review


28/65

3Symmetr ical Faul t Analys is(PU Method)

Symmetrical Fault refers to a balanced 3fault, in a balanced 3system operating in

steady state, which is either :

Bolted fault: LLLG fault with Zfault=0Non-Bolted fault: LLLG fault with Zfault0

Only the (+)SEQ network exists. (0)SEQ & (-)SEQ currents are equal to Zero.

Power System Review


29/65

Symmetr ical Fau lt Model ing

for a Bo lted Fau lt (PU Method)

Z0eq

Note:VF=Pre Fault Voltage

+

_Vo=0

Z2 eq

VF

Z1eq

+

_

+

_V1=0

I0=0

I1 Ia

Ib

Ic

VcVb

Va+

+

+

_ _ _

Ib = -Ia = Ic = ISCVbg = Vag = Vcg =0

Phase

g

+

_V2=0

I2=0

)(1

)()(1PUeq

PUf

ZVI PUfault

02I

00I

SEQ

SEQ)(

SEQ)(SEQ)0(


30/65

Prac t ice Example (PU Method):

In the following power system Calculate(a)3Symmetrical

fault current @ Bus3 and select an appropriate BreakerSize @ Bus 3

G1G2

PU.150X

kV13.8

MVA500

"

.15PU0T1

115kV/13.8kV

MVA500

"X

PU.200X

kV13.2

MVA750

"

.18PU0T2

kV8.13/115kV

MVA750

"X

6XT1

2X 13T

17.63Zbase

115kVKvbase

MVA750

Sbase

1Bus 2Bus

3Bus

4X 23T

.254Zbase

13.8kVKvbase

MVA750

Sbase

.254Zbase

13.8kVKvbase

MVA750

Sbase

MVA750SBase


31/65

Breaker Selec t ion

Modern Circuit Breaker standards are designed based on

ISymmetrical. The following steps are required to determine anappropriate breaker size:

1. Use E/X method to calculate the minimum ISymmetrical.

2. Calculate X/R ratio:

1. If X/R 15 It means the dc offset has not decayed

to an acceptable level. Thus, calculate IAsymmetrical.

3. Calculate IAsymmetricalat calculated fault location.

4. Breaker Interrupting Capability should be 20%greater

than the calculated fault current.


32/65

Breaker Selec t ion Cri ter ion

Generator/ Synchronous Motor/Large Induction motors

Breakers: Use subtransient Reactance Xdto calculate ISymmetrical.

Use 2 cycle Breaker

TransmissionBreakers:

Use 3 cycle Breakers if X/R>15

Use 5 cycle Breaker if X/R


33/65

A2.614,16

8.0

13,291.2I CapabilityngInterruptiBreaker

G1 G2

PU.150X

kV13.8

MVA500

"

.15PU0T1

115kV/13.8kV

MVA500

"

X

PU.200X

kV13.2

MVA750

"

.18PU0T2

kV8.13/115kV

MVA750

"

X

6XT1

2X 13T

17.63Zbase

115kVKvbase

MVA750

Sbase

1Bus 2Bus

3Bus

4X 23T

.254Zbase

13.8kVKvbase

MVA750

Sbase

.254Zbase

13.8kVKvbase

MVA750

Sbase

MVA750SBase

kV115:ClassVoltageBreaker:SelectionBreaker

cycle3:CycleBreaker

Prac t ice Example (PU Method):

A2.291,13I lSymmetrica

S t i l F l t C t


34/65

Symmetr ical Fault Current

AnalysisMVA-Method

MVA Method

Power System Review

F lt C t C l l t i


35/65

Fau lt Curren t Calcu lat ion-MVA Method This method follows a four steps process:

1. Calculate the Admittance of every component in its own

infinite bus.

2. Multiply the calculated admittances in step(1) by the

MVA rating of each component to get MVASC.

3. Combine short-circuit MVAs &follow the Admittance

series & parallel rules:

4. Convert MVAs to Symmetrical fault current

Power System Review

%

100)Admittance(

ZY

)Admittance(YxMVAMVAsc

ntotal MVAMVAMVAMVA ........

:MVAsParallela)

21 ntotal MVAMVAMVAMVA

1........111

:MVAsSeriesb)

21

llkVx

TotalMVAscalIsymmetric

3

)(


36/65

MVA Equ ivalent Network

ntotal MVAMVAMVAMVA ........

:MVAsParallel

21

ntotal MVAMVAMVAMVA

1........

111

:MVAsSeries

21

1MVA 2MVA 3MVA

1MVA 2MVA 3MVA

321 MVAMVAMVAMVAtotal

321

1111

MVAMVAMVAMVAtotal

TotalMVA

TotalMVA


37/65

Why Use the MVA Method?

This method is internationally used and accepted by most

protection engineers.

The network set up is easier than Ohmic or PU method.

You can calculate Ifaultin a shorter time period. This method makes it easier to see the fault contributions

@ every point in the system.

Calculation accuracy is within 3% to 5% compared to PU &

Ohmic method.

Power System Review


38/65

MVA Method Assumpt ions

Power System Review

10.1 R

X

Two Conditions must be satisfied:

OperationStateSteady.2


39/65

Symmetr ical Fau lt Current

Analysis.. .MVA-Method

Power System Review

)(3: KAIxkVxMVAscMVAUtility scllfault

)(:

2

Z

kV

MVACable

ll

fault

%

100:/

"Gend

fault

XxMVAMVAMotorusSycnhroonoGenerator

%

100:

xfmr

fault

ZxMVAMVArTransforme

Formulas:

Note:Impedances (Z) are steady state values


40/65

Where: Xd=direct-axis Subtransient Reactance

Xd

= IFull-load amp

/ILocked Rotor amp

Power System Review

Symmetr ical Fau lt Current


%

100:

"Gend

motorfault

XxMVAMVAMotor

amploadfull

rotorlockedmotorfault

IIxMVAMVAMotorInduction

:

:Motor

Symmetr ical Fau l t Curren t


41/65

Summary:

Power System Review

Symmetr ical Fau lt Curren t


LLkVxMVAI

totalKAfault

3)(

)]/1()/1()/1[(

1

21 nMVAMVAMVAtotalseriesMVA

nMVAMVAMVAtotalparallelMVA 21

E l 1 F lt C l l t i (MVA th d )


42/65

Example1:Fault Calcu lat ion (MVA method )

Generator

M

Utility Source

13.8kV, 15KA fault current

Motor

2MVA Y

4.16kV

Xd=0.25pu

3-500McM cables, 2000 ft

Z=0.2

Transformer

7MVA

13.8kV/4.16kV

Z=9%

1.5MVA Y

4.16kV

Xd=0.15pu

Bus 1 13.8kV

Bus 2 4.16kV

In the following Power System, Calculate the fault current @ Bus2 & fault current

contributions from both Gen & Motor?

S


43/65

Step1:Network Model ing(MVA Method)

Generator

M

Utility Source


Motor2MVA Y

4.16kV

Xd=0.25


Z=0.2

Transformer

7MVA

13.8kV/4.16kV

Z=9%

1.5MVA Y

4.16kV

Xd=0.15

MVAxxMVAsource 5.358)kA15()kv8.13(3

MVAx

ZxMVAMVA

xfmr

rtransforme 77.779

1007

%

100

MVAx

X

xMVAMVAd

Generator 10

15.0

15.1

1"

MVAxXxMVAMVA dMotor 825.0

12

1"

52.358

77.77

10

8

MVAZ

kVMVA

line

Line 53.862.0

)16.4(22

53.86

Bus1 13.8kV

Bus2 4.16kV

St 2 N t k R d t i


44/65

Step 2: Network Reduc t ion(MVA Method)

52.358

77.77

10

8

53.86 10

8

76.36

53.86

1

77.77

1

52.358

11

:MVAsSeries

totalMVA

76.36

53.86

1

77.77

1

52.358

1

1

MVAtotal

321

:

MVAMVAMVAMVA

MVAsParallel

total

76.54876.3610 totalMVA

76.54

MVAFault

St 3 F lt MVA C i t I


45/65

Step 3:Fau l t MVA Conversion to Ifault

76.54faultMVA

6003.7)16.4(3

76.54

)16.4(3

)3()(

LLLL

faultfault

kVxkVx

MVAkAI

AmplSymmetricaIfault 3.600,7)(

kVkVll 16.4

:2QuantitiesBus

Bus2 Fault Current:

Example1:Faul t Analysis (PU Method)


46/65

Generator

M

Utility Source


Motor2MVA Y

4.16kV

Xd=0.25


Z=0.2

Transformer

7MVA

13.8kV/4.16kV

Z=9%

1.5MVA

Y

4.16kV

Xd=0.15

Bus1 13.8kV

Bus2 4.16kV

puVf 0.1

2)( BusforNetworkSEQ

utilityZ

XfmrZ

LineZ

GenZmotorZ

In the following Power System, Calculate the fault current @ Bus2 & fault current

contributions from both Gen & Motor using PU Method?

Example1:Faul t Analysis(PU Method)

Examp le 1: Symmetr ical Fault Current


47/65

Examp le 1: Symmetr ical Fault Current

Calculat ion Comparison between

PU & MVA Methods

AmpI Busfault 3.600,72@

A7.605,7A879,13548.0)(2@ xxIpuII basefaultBusfault

:ncalculatiomethodMVA

:ncalculatioMethodUnitPer

E 1 M t /G F lt C t ib t i


48/65

76.36

8

10

AkVx

MVAI

onContributiGenerator

Genfault 9.387,116.43

10

:

GenMVA

MotorMVA

)( LineXfmrUtilityMVA

AkVx

MVAI

onContributiMotor

motorfault 3.110,116.43

8

:

Ex1: Moto r/Gen Fau lt Con tr ibu t ion

(MVA Method)

AkVx

MVAI

onContributiUtility

fault 102,5205.7

76.36

16.43

76.36

:

A2.600,73.110,19.387,1102,5

:

Genfutilityfmotorffau lt IIII

CurrentFaultTotal

Ex1:Symmetr ical Fau l t Current Analys is


49/65

Ex1:Symmetr ical Fau lt Current Analys is

PU & MVA Methods Comparison

AmpI motorf 3.110,1

A110,1f-motorI

:ncalculatiomethodMVA

:ncalculatioMethodUnitPer

Symmetr ical Fau l t Curren t Calculat ion


50/65

Symmetr ical Fau lt Curren t Calculat ion

MVA MethodExample2: Calculate the Symmetrical fault current @ Bus2 using the MVA Method

Generator

M

M

Generator

Utility Source


Transformer

20MVA Delta-Yn

22.86/4.16kVZ=9% 5MVA

4.16kV

Z=12%

Transformer

3.5MVA Delta-Yn

4.16kV/480V

Z=7%

Motor

2MVA Y

4.16kVZ=15%

Motor

1.5MVA Y

480V

Z=16%

Y

Y


Z=.18

Bolted Fault

Generator

2MVA

480 V

Z=14%

BUS 1

BUS 2

903.5931586.223 kAxkVxMVA LLfault

22.903,218.0

)86.22( 22

kV

Z

kVMVA

line

fault

50

07.0

5.3

100%

222.22209.0

20

100

%

Z

MVAMVA

Z

MVAMVA

Xfmrfau lt

Xfmrfau lt

MVAZ

MVAGMVA

MVAZ

MVAGMVA

fault

fault

286.1414.0

2

100

%)2(

667.4112.0

5

100

%)1(

MVAZ

MVAGMVA

MVAZ

MVAMMVA

fau lt

fau lt

375.916.0

5.1

100

%)2(

333.1315.0

2

100

%)1(


51/65

22.86 kV Utility Source:

Line:

Transformers:

Power System Review

903.5931586.223 kAxkVxMVA LLfault

22.903,218.0

)86.22(22

kV

Z

kVMVA

line

fault

5007.0

5.3

100

%

222.22209.0

20

100

%

Z

MVAMVA

Z

MVAMVA

Xfmrfault

Xfmrfault

Solut ion to Example2 (MVA method):


52/65

Solut ion to Example2 (MVA method):

Generators:

Motors:

Power System Review

MVA

Z

MVAGMVA

MVAZMVAGMVA

fault

fault

286.14

14.0

2

100%

)2(

667.4112.05

100

%)1(

MVAZ

MVAGMVA

MVAZ

MVAMMVA

fault

fault

375.916.0

5.1

100

%)2(

333.1315.0

2

100

%)1(

Example 2:Symmetr ical Fau l t Current


53/65

Example 2:Symmetr ical Fau lt Current

Calculat ion (MVA-method)

Power System Review

593.903MVA

2903.220MVA

222.222MVA

41.667

MVA

50 MVA 13.333MVA

9.375MVA

14.286MVA

BUS 1

BUS 2

ModelingNetwork:Step1

Symmetr ical Faul t Curren t


54/65

Symmetr ical Fault Curren t

AnalysisMVA-Method

Series MVAs:

Parallel MVAs:

Power System Review

)]/1()/1()/1[(

1

21 nMVAMVAMVAtotalseriesMVA

nMVAMVAMVAtotalparallelMVA 21

ReductionMVANetwork:Step2

E l 2 S t i l F l t C t


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Example2: Symmetr ical Fau lt Current

AnalysisMVA-Method

MVAseries:

MVA=1/[(1/593.903)+(1/2,903.220)+(1/222.222)]

MVA=1/[(.0017)+(.0003)+(.0045)]=153.846

Bus1(parallel)=153.846+41.667+13.333=208.846

MVA series@Bus2:

MVA=1/[(1/208.846)+(1/50)]MVA=1/[(.0048)+(.0200)]=40.323

Power System Review

ReductionMVANetwork:Step2

50 MVA

9.375MVA

14.286MVA

BUS 2

208.846MVA

Ex2: Short Circu i t MVA Calcu lat ion


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@ Bus 2(MVA method)

153.846MVA

41.667MVA

50MVA

13.333MVA

9.375MVA

14.286MVA

BUS 1

BUS 2

846.153)]22.222/1()22.903,2/1()903.593/1[(

1

totalseriesMVA

50MVA

9.375MVA

14.286MVA

BUS 2

MVA846.208333.13667.41846.153 parallelMVA

208.846MVA

nCalculatioMVAFault:Step3



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40.323MVA

9.375MVA

14.286MVA

BUS 2

MVA984.63375.9286.14323.402@ BusMVA

323.40)]50/1()846.208/1[(

1

seriesMVA

MVA984.632@ BusMVA fault


@ Bus 2(MVA method)

Example2: Symmetr ical Fau l t


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Example2: Symmetr ical Fau lt

Current AnalysisMVA-Method

Bus2(total) = 40.323+14.286+9.375=63.984 MVA

Now, Calculate the Short Circuit MVA @Bus1?

Power System Review

Available Fault Current @Bus 2:

Ifault=63.984 MVA/[ x 0.48kV]=76,963 A3

Ex2:Calcu late Short Circu it MVA@ Bus1


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Ex2:Calcu late Short Circu it MVA@ Bus1

(MVA method)

153.864MVA

41.667

MVA

50 MVA

9.375MVA

14.286MVA

13.333MVA

195.531MVA

13.333MVA

50MVA

9.375+14.286=23.661MVA

208.864+16.051=224.915MVA

208.864= 195.531+13.333MVA

1/[(1/50)+(1/23.661)]=1/.0623=16.051MVA

Power System Review

BUS 1 BUS 1

BUS 1

BUS 1

BUS 2

MVA531.195667.41846.153 parallelMVA

MVA915.2241@ BusfaultMVA

Ex2: Calcu late Sho rt Circu it MVA


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S.C or Fault MVA @ Bus1:

S.C or Fault MVA= 224.915

Ifault @Bus1= 224.915 MVA/( x4.16kV)

Power System Review

@Bus 1 (MVA method)

3

Available Fault Current at Bus 1:

I fault @Bus1=31,216 A

E l 3 S t i l F l t A l i


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Example 3: Symmetr ical Faul t Analys is

Power System Review

Source M

1500 MVA

Fault69 kV

X=2.8

10 MVA

X=8.5%69kV/-n 13.8kV

13.2 kV

X=0.2

Calculate the symmetrical fault current at the secondary terminals of a 10 MVA XFMR

using both the PU-Method & the MVA Method. Use 15 MVA & 69 kV base values for

the transmission line.

5 MVA -n

Zone 1 Zone 2

kVV 691Base-lL

AkVx

SIBase 57.6273

1Base

Base2 4.31715

69

S

2

Base

Base2

Base1

1

1

kV

Z

MVAS 15Base MVAS 15Base

7.1215

8.132

Base2Z

kVV 8.132Base-lL

Example3: Symmetr ical Faul t


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Example3: Symmetr ical Fault

Analys is(MVA-method)

Power System Review

1700.36MVA

1500MVA

117.65MVA

27.32MVA

5 MVA_____

(13.2/13.8)x0.2=27.32

102.52MVA

27.32MVA

MVA Fault= 102.52+27.32

= 129.84

Ifault=129.84/(1.732x13.8)

= 5,432.3 Amps

65.117

1

36.1700

1

1500

11

MVA

rTransforme

Line

Source

Motor

Example 3:


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Symmetr ical Fau lt Calcu lat ion

Compar ison Between PU & MVA

Methods

I fault=5,410.3 Amp

I fault = 5,432.3 Amp

:methodPU

:methodMVA

Example 3:

Power System Review

R f


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References

1. J.D. Golver, M.S. Sarma, Power System Analysis and design,

4thed., (Thomson Crop, 2008).2. M.S. Sarma, Electric Machines, 2nded., (West Publishing Company,

1985).

3. A.E. Fitzgerald, C. Kingsley, and S. Umans, Electric

Machinery, 4th

ed. (New York: McGraw-Hill, 1983).4. P.M. Anderson, Analysis of Faulted Power systems(Ames, IA: Iowa

Satate university Press, 1973).

5.W.D. Stevenson, Jr., Elements of Power System Analysis, 4th

ed. (New York: McGraw-Hill, 1982).

Solut ion

Break Time !!!!!


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Solut ion

Answer: 37.5 KVA

Break Time !!!!!

Appa-module 6-Fault Current Analysis

Documents

Transcript of Appa-module 6-Fault Current Analysis