à During Short Circuit Studies, power systems faults at ... · Short Circuit Studies àDuring Short Circuit Studies, power systems are solved to obtain current magnitudes during

@ Salvador Acevedo, 2000

Short Circuit Studies

à During Short Circuit Studies, power systems are solved to obtain current magnitudes during faults at different points in the network.MFault: “Failure in a circuit which interferes with

the normal flow of current”

à Purposes of a Short Circuit StudyTo design a PROTECTION scheme to prevent damage to the electric equipment in case of the occurrence of a fault.ð Location of breakersð Selection of breakers

o Ratings of breakers

ð Proper adjustment of breakersð Coordination of the Protection

o Interruption of the currento Isolation of the faulto Sequence of operationo Protection backup

à When is a Short Circuit Study performed?ð When designing the electrical installationð When changing operating conditions ð When installing or removing equipmentð When planning expansion


Types of Faults

à Symmetrical Faultsð Faults involving the three-phases

o about only 5% of the cases

ð Easiest to evaluateð Required in a Short Circuit Study because they

are commonly the worst case

à Unsymmetrical Faultsð Faults involving some unbalance

o Line to ground faults (one phase to ground)à about 70% of the faults

o Line-to-line faults (between two phases)à about 25% of the faults are line-to-line faults

ð To solve for these faults, we require the use of symmetrical components and sequence networks


Faults in a Three-Phase Line

abc

abc

Solid three-phase fault Three-phase to ground fault

Fault impedance

abc

Line-to-line fault Line-to-line to ground fault

abc

Fault impedance

Line to ground fault Line to ground fault through impedance

abc

Fault impedance

abc


Transients in RL Circuits

+

Vs

-

R

Li(t)

( )

( )

The so lu t i o n f o r t h e cur ren t c o n t a i n s a fo r ced r e sponse ( s t e ady s t a t e ) ,and a t r an s ien t resp onse (na tu ral) :i(t) = i i

T h e s t e a d y - s ta te or f o rced re sp o n s e c a n b e o b t a i n e d u s i n g p h aso r s :

I =V sZ

i

T h e t r a n s ie n t r e s p o n se i s the na tu ra l r e s p o n s e o f t he c i r cu it , w h ich i s t h e

so lu t ion t o t h e h o m o g e n e o u s d if fe ren t ia l e q u a t i o n :

s t eady -state t r ans i en t

s t e a d y -s ta te

s t e a d y -s ta te

+

=∠

∠=

∠ − = ∠

= + −

= + =

+ =

=

−

−

VZ

VZ

I

I w t

w h e r e Z R w L a n dw LR

R i Ld id t

i K et r a n s i e n t

R

m a x m a xm a x

m a x sin

: tan

αθ

α θ θ

α θ

θ2 2 2 1

0

( )

( )( ) ( )

( ) ( )

Lt

RL

t

RL

t

K e I w t

I K e

K I I

I e w t

T h e r e f o r e , the to ta l r e s p o n s e i s :

i(t) =

if i (0) = 0 , then 0 =

a n d

i(t) =

−

−

+ + −

− +

= − − = −

− + + −

m a x

m a x

m a x m a x

m a x

sin

sin

sin sin

sin sin

α θ

α θ

α θ θ α

θ α α θ

0

vs Ri Ldidt

i

V wt Ri Ldidt

= + =

+ = +

( )

sin( )max

0 0

α


RL circuit

( ) ( )

Plot for the current

i(t) = I e I wtR

Lt

max maxsin sin−

− + + −θ α α θ

+

=

total current

natural

0 0.05 0.10

20

40

60

forced

0 0.05 0.1-100

-50

0

50

100

0 0.1-100

-50

0

50

100


Short Circuit Current in a Synchronous Generator

-600

-400

-200

0

200

400

600

transient steady-statesubtransient

Subtransient generator equivalent

+Ed”

-

jXd” ra +Vt-

Ia

Ed” : Internal subtransient voltageVt : Terminal voltageXd ”: Subtransient reactance

Transient generator equivalent

jXd’ ra+Ed’

-

+Vt-

Ia

Ed’ : Internal transient voltageVt : Terminal voltageXd’ : Transient reactance

Steady-state generator equivalent

jXd ra+Ed-

+Vt-

Ia

Ed : Internal steady-state voltageVt : Terminal voltageXd: Steady-state reactance

Line current during a three-phase short circuit

Xd > Xd’ > Xd”


Short-Circuit Example: Unloaded Generator

Three-phaseshort circuit

GENERATOR50 MVA, 13.8 KV

xd"=15%xd'=25%xd=80%

TRANSFORMER50 MVA, 13.8-69 KV

x=10%

Find the subtransient, transient, and steady-state generator current when a three-phase short circuit occurs at the high-voltage transformer terminals. Before the fault, there is no load connected and the open circuit voltage at the line terminals is 69KV. Neglect all resistances.


Solution

The equivalent circuits for subtransient, transient, and steady-state periods are shown below. Solving each one will give the magnitude of the fault current during its corresponding stage (subtransient, transient, and steady-state).

+1 p.u.-

+1 p.u.-

+1 p.u.-

j0.15 j0.10

j0.25 j0.10

j0.80 j0.10

If ”

If ’

If

Subtransient solution:

Transient solution:

Steady-state solution:


Short-Circuit Currents for the Example

Since the generator is not supplying any current, we assume Ed=Ed’=Ed”=100%

( )

For each circuit, we find the short - circuit current as:

Subtransient: If ' '=1

j0.25 p.u.

Transient: If '=1

j0.35 p.u.

Steady - state: If =1

j0.90 p.u.

Generator currents are found using the base current:

Ibase1=50000

2091.85 amperes

Generator current magnitudes are:Subtransient: If ' '= 4.000 x Ibase1 = 8367 amperesTransient: If '= 2.857 x Ibase1 = 5976 amperesSteady - state: If = 1.111 x Ibase1 = 2324 amperes

= −

= −

= −

=

j

j

j

4

2 857

1111

3 138

.

.

.


Loaded Machine under Fault Conditions

Three-phaseshort-circuit

GENERATOR50 MVA, 13.8 KV

xd"=15%xd'=25%xd=80%ra=2%

TRANSFORMER50 MVA, 13.8-69 KV

x=10%r=1%

P

Load50MVA69KV

pf=0.9 (-)

Assume a three-phase short-circuit occurs at point ‘P’. To evaluate a fault during the subtransient or transient period, we need to know the pre-fault current value IL.

+Ed-

ra jXd rt jXt

Z

P

IL Switch'S'

Switch ‘S’ is normally open. Close it to simulate a fault at point ‘P’.

Impedance diagram for the circuit before fault:


Pre-Fault Conditions (subtransient)

The load current (pre-fault current) will help us determine the internal voltage for the subtransient, before the fault:

+Ed"

-

ra jXd" rt jXt

Z

P

IL +Vf-

+Vt-

E d V t jXd I V f j X t X d IL L' ' ' ' ( ' ' )≅ + = + +

+Ed"

-

jXd" jXt

Z

+Vt-

P

IL If "+Vf-

Ed Vt ra jXd I V f ra rt j X t Xd IL L' ' ( ' ' ) [( ) ( ' ' )]= + + = + + + +

Neglecting transformer and generator resistances:

Before the fault, If ”=0.


Pre-Fault Conditions (transient)

E d V t j X d I V f j X t X d IL L' ' ( ' )≅ + = + +

+Ed'-

jXd' jXt

Z

P

IL If '+Vf-

+Vt-

The corresponding internal voltage Ed’ for the transient is:

Fault Current

Before the fault, If ’=0.

To simulate the fault, switch ‘S’ is now closed.

Subtransient short - circuit current:

Transient short - circuit current:

Steady - state short - circuit current:

IfEd

j Xd Xt

IfEd

j Xd Xt

IfEd

j Xd Xt

""

( " )

''

( ' )

( )

=+

=+

=+


Multi-Machine System

Three-phaseshort-circuit

GENERATOR SynchronousMOTOR

P

TRANSFORMER

+Eg-

jXg jXt P

IL IL+

Em-

jXm

+Vf-

Equivalent circuit before the fault:

+Eg'-

jXg' jXt P

IL IL+

Em'-

jXm'

+Vf-

+Eg"

-

jXg" jXt P

IL IL+

Em"-

jXm"

+Vf-

Steady-state

Transient

Subtransient


Subtransient Short-Circuit Solution

1. Evaluate subtransient internal voltages for generator and motor under the operating conditions (switch ‘S’ open).

Eg Vf j Xt Xg ILEm Vf jXm IL

" ( " )" "= + += −

2. Close switch ‘S’, and find the current contribution from generator and motor to the fault.

+Eg"

-

jXg" jXt P

Ig" Im" +Em"

-

jXm"

If "

+Eg"

-

jXg" jXt P

IL IL+

Em"-

jXm"

+Vf-S

If "=0

IgEg

j Xt Xg"

EmjXm

If Ig

""

( " )Im

""

" " Im"

=+

=

= +


Thévenin Equivalent Method

Same multi-machine example (2 machines).Combining steps 1 and 2.

IgEg

j Xt XgVf j Xt Xg IL

j Xt Xg

IgVf

j Xt XgIL

"EmjXm

Vf jXm ILjXm

VfjXm

IL

If IgVf

j Xt XgIL

VfjXm

IL

IfVf

j Xt XgVf

jXm

""

( " )( " )

( " )

"( " )

Im""

""

Im""

" " Im"( " ) "

"( " ) "

=+

=+ +

+

=+

+

= =−

= −

= + =+

+ + −

=+

+

Igf ’’ (lets name this term: Igf ”)

{{

Imf ’’ (lets name this term: Imf ”)

Igf ’’

{ {

Imf ’’


Thévenin Equivalent (continued)

IfVf

j Xt XgVf

jXm"

( " ) "=

++

Igf ’’ Imf ’’

{ {

This expression can be represented in the following circuit:

IfVf

j Xt XgVf

jXmVf

j Xt Xg jXm

If VfjXtg jXm

IfVfZth

Zth jXtg XmXtg Xm

"( " ) " ( ' ' ) ' '

' '"

"( )( ' ' )

' '

=+

+ =+

+

= +

= =+

1 1

1 1where: X tg = Xt + Xg"

where:

Vf is the pre-fault voltage at the fault point (Thévenin voltage)Zth is the Thévenin Impedance seen from the fault point.If ’’ is the subtransient fault current.

jXg" jXt P

Igf " Imf "

jXm"

If "

-Vf+

+Vf-

If "

j(Xg"+Xt)

jXm"

Igf "

Imf "


Thévenin Equivalent (continued)

Remember that the total subtransient generator current is

Ig ” = Igf ”+IL

and the subtransient motor current is

Im” = Imf ” - IL

The same problem can be solved applying superposition. The Pre-fault Solution plus the Thévenin Equivalent Solution

+Eg"

-

jXg" jXt P

Ig" Im" +Em"

-

jXm"

If "+Vf-+

-Vf-

Vf is the pre-fault voltage at ‘P’ and the short-circuit is represented by two opposed Vf sources connected in series.


Superposition and Thévenin

2. The contribution to the fault is obtained with -Vf only.This will make

Ig”2=Igf ”, Im”2=Imf ”, If ”=Igf ”+Imf ”.(Here is where we use the Thévenin equivalent)

+Eg"

-

jXg" jXt

Ig"1=IL Im"1=-IL +Em"

-

jXm"

If "=0

+Vf-

IL

To obtain the total solution, we apply superposition: 1. The pre-fault solution is obtained with Eg”, Em” and Vf.

This will makeIg”1=IL, Im”1=-IL, If ”=0.

3. Add steps 1 and 2. This will give the total fault currentsIg”=Igf ”+IL, Im”=Imf ”-IL, If”=Igf ”+Imf ”.

+Eg"=0

-

jXg" jXt

Ig"2=Igf" Im"2=Imf " +Em"=0

-

jXm"

If "=Igf "+Imf "

+-Vf-

-Vf+


Summary of Fault Analysis Using Thévenin Method

à Locate the fault point ‘P’.à Represent the system in admittance form.

ð Convert synchronous machines to their Norton equivalents.

ð Build admittance matrix for nodal analysis [YBUS].

G1

G2

G3

Gi

Generators,Transformers,

Loads,Transmission

Lines, etc.

...

Load 1Load 2Load 3..Load j

FAULTED BUS ‘P’

If+Vf-

If switch open If = 0If switch closed Vf = 0


Step 1

à Find the pre-fault operating conditions.ð Use steady-state values.ð Name Vf the pre-fault voltage at point ‘P’.ð With nodal analysis or Load Flow analysis obtain

prefault voltages V1o, V2

o, V3o….Vf

ð Calculate currents Ig1, Ig2,… Iline1, Iline2, …...

+Vf = Vpre-fault-

→

→

→

→

System matrix[Ybus]

According to the method used todetermine the

pre-fault operatingconditions

...

Load 1Load 2Load 3..Load m

‘P’

If = 0

J1

J2

J3

Ji

If = Fault current


Step 1 (continued)

+Vf = Vpre-fault voltage-

→

→

→

→

System matrix[Ybus]...


‘P’

If = 0

J1

J2

J3

Ji

[ ]Y V J V Y J Z J

V

V

V f

V

Y Y Y YY Y Y Y

Y Y Y Y

Y Y Y Y

JJ

Jp

Jn

B U S B U S B U S

o

o

n

p n

p n

p p p p p n

n n n p n n

= = =

=

=

−

−

1

1

2

0

1 1 1 2 1 1

2 1 2 2 2 2

1 2

1 2

12

0.

.

. . . . . .

. . . . . ..

. . . . . ..

. . . . . .

.

.

1

=

=

=

V

V

V f

V

Z Z Z ZZ Z Z Z

Z Z Z Z

Z Z Z Z

JJ

Jp

Jn

o

o

n

p n

p n

p p p p p n

n n n p n n

1

2

0

1 1 1 2 1 1

2 1 2 2 2 2

1 2

1 2

12

0.

.

. . . . . .

. . . . . ..

. . . . . ..

. . . . . .

.

.

This term is zerobecause before the faultthere is no fault current(switch is open)


Step 2

à Find the Thévenin contribution to the fault.ð Set all sources to zero (including synchronous motors

internal sources).ð Use the subtransient, transient or steady-state

impedances depending on the solution desired.ð Apply a voltage source ‘-Vf ’ at point ‘P’ and solve

the network with this source only. This source injects current ‘-If ’ into faulted node.o This will give the fault current If and all changes

in voltages and currents needed.àName voltage changes ∆V1, ∆V2, ∆V3…àName current changes ∆Ig1, ∆Ig2,….,

∆Iline1, ∆Iline2…

Machine Impedances(for Subtransient, Transient,

or Steady-state),Transformers,

and TransmissionLines

...


If ’’If ’If

-Vf+

‘P’

J1=0

J2=0

J3=0

Ji=0


Step 2 (continued)

Machine Impedances (for Subtransient, Transient,

or Steady-state),Transformers,

and TransmissionLines

...


If ’’If ’If

-Vf+

‘P’

J1 = 0

J2 = 0

J3 = 0

Ji = 0

∆∆

∆

∆

∆∆

∆

VV

Vp

Vn

VV

Vf

Vn

Z Z Z ZZ Z Z Z

Z Z Z Z

Z Z Z Z

p n

p n

p p pp pn

n n np nn

12

12

00

11 12 1 1

21 22 2 2

1 2

1 2

.

.

.

.

... ...

... ....

... ....

... ...

.

=−

=

− If.0

This matrix ZBUS is formed using the appropriate impedances (subtransient, transient or steady-state) to form YBUSbefore inverting.

This term equals -Vf because -Vf is the voltagethat we need to add tothe prefault voltage Vfto have a zero voltageat point ‘P’


Step 2 (continued)

[ ]

The last equations can be sim plified to:

or

from w h ich

where is the fault current

is the pre - fault voltage at poi

∆∆

∆

∆

∆

∆

∆

VV

Vf

Vn

ZZ

Z

Z

If

V Z If

V Z If

Vp Vf Z If

Vn Z If

V Z I IV

Z

I

V

p

p

pp

np

p

p

pp

np

f pp f ff

pp

f

f

12

1

2

1

2

1

2

.

.

.

.

,

,

. . . . .,

. . . .

−

=

−

= −

= −

= − = −

= −

= =

nt ' P '

is the Thévenin Im pedanceZ pp


Step 3

à Add solutions for steps 1 and 2.ð This is equivalent to solving the network with

the switch closed.

à Voltages during the fault areV1

f=V10+∆V1, V2

f=V2+∆V2, V3f=V3+∆V3, …

à at the fault Vp=Vf +(-Vf)=0à Currents during the fault are

Ig1+∆Ig1, Ig2+∆Ig2…Iline1+∆Iline1, Iline2+∆Iline2….

G1

G2

G3

Gi

System matrix[Ybus]...


If+Vf = 0-

‘P’


Study Case: General Solution for Symmetrical Faults

The power system shown operates under steady-state

conditions with Eg1=1 ∠0° p.u. and Eg2=0.9 ∠30° p.u. when a solid three-phase fault occurs at node 2. Obtain the transient short-circuit currents in lines, generators and transformers. Evaluate the transient node voltages V1

f, V2f and V3

f during the fault (transient period).

GENERATOR-1Xd=85%Xd'=25%Xd"=10%

ra=1%

TRANSFORMER-1Y-Y

Zt=0.01+j0.15 p.u.

TRANSFORMER-2Y-Y

Zt=0.01+j0.20 p.u.

GENERATOR-2Xd=120%Xd'=40%Xd"=20%

ra=2%

Transmission line 1-2Z=0.03+j0.4p.u.

Transmission line 1-3Z=0.05+j0.5 p.u.

LOADR=10 p.u.

Transmission line 2-3Z=0.05+j0.5 p.u.

1 2

3

Fault


Step 1. Pre-Fault Solution

+E1-

zLoad

+E2-

zgt1

z12

z13

zgt2

z23

i12

i13 i23

iG1 iG2

+

V1

-

+

V2

-

+V3-

I1 I2ygt1

y12

y13

ygt2

y23

zgt1=zg1+zt1=0.02+j1z12=0.03+j0.4z13=z23=0.05+j0.5zgt2=zg2+zt2=0.03+j1.4zLoad=10E1=1

E2=0.9∠30°

ygt1=0.02-j0.9996y12=0.187-j2.486y13=y23=0.198-j1.980ygt2=0.0153-j0.714yload=0.1I1=0.02-j0.9996I2=0.333-j0.549

Impedance diagram (all values in p.u.)

Admittance diagram (all values in p.u.)

To solve it, we use nodal analysis.

Y Y YY Y YY Y Y

V1V2V3

J1J2J3

-0.187 + j2.486 -0.198 + j1.980-0.187 + j2.486 0.399 - j5.180 -0.198 + j1.980

- 0.198 + j1.980 - 0.198 + j1.980

V1V2V3

11 12 13

21 22 23

31 32 33

=

−

−

=−−

0 405 5466

0 496 3960

0 02 0 99960 333 0549

0

. .

. .

. .

. .j

j

jj

=∠ °∠ °∠ °

V1V2V3

0 929 710 916 10 20 920 7 2

. .

. .

. .


Pre-Fault Currents

Line Currents:Line 1-2I12=y12 (V1-V2)

I12 =(0.187-j2.486)(0.929∠ 7.1°- 0.916∠ 10.2°)

I12 = 0.129∠ −151.9° p.u.

Line 1-3I13=y13 (V1-V3)

I13 =(0.198-j1.980)(0.929∠ 7.1°- 0.920∠ 7.2°)

I13 = 0.019∠ −87.4° p.u.

Line 2-3I23=y23 (V2-V3)

I23 =(0.198-j1.980)(0.916∠ 10.2°- 0.920∠ 7.2°)

I23 = 0.096∠ −18.9° p.u.

Generator Currents:Generator 1

IG1= I12 + I13 = 0.129∠ −151.9° + 0.019∠ −87.4°

IG1= 0.138 ∠ −144.6°

Generator 2

IG2= -I12 + I23 = -0.129∠ −151.9° + 0.096∠ −18.9°

IG2= 0.224 ∠ −24.2°


Power Balance (Pre-Fault)

To verify the solution, a real Power Balance is now calculated (as an exercise):

Generated PowerGenerator 1 + Transformer 1:

SG1=V1 IG1*= (0.929∠ 7.1° )(0.138 ∠ 144.6° )SG1= -0.113 + j 0.061PG1= - 0.113 (where the minus sign means this generator

absorbs P=0.113 p.u. and therefore is acting as a motor)

QG1=0.061

Generator 2 + Transformer 2:

SG2=V2 IG2*= (0.916∠ 10.2°)(0.224 ∠ 24.2° )SG2=0.1986 - 0.0495iPG2= 0.1986QG2= - 0.0495 (this generator absorbs Q = 0.0495 p.u. and

still generates P = 0.1987 p.u., therefore this machine acts as a generator)


Power Balance (continued)

Absorbed PowerLoad:PLoad=(V3)2/Rload=(0.920) 2*0.1=0.0846 p.u.

Real Power dissipated in lines:line 1-2: P=I12

2 * Rline12=(0.129) 2(0.03)=0.0005

line 1-3: P=I132 * Rline13=(0.019) 2(0.05)=0.00002

line 2-3: P=I232 * Rline23=(0.096) 2(0.05)=0.00046

The power balance is:

Pgenerated=Pabsorbed

PG2=PLoad+Pline12+Pline13+Pline23+Pabsorbed-G1

0.1986=0.0846+0.0005+0.00002+0.0004+0.113=0.1986

Note: Nodal analysis has been used to find the operating conditions of the system before the fault. In practice, a Load-flow solution would have been used instead.


Step 2. Fault at Bus 2 (Thévenin Contribution)

To simulate a Fault at Bus 2, we will add the pre-fault response to the Thévenin Contribution.

We use a source equal to the pre-fault voltage at point 2 and set all the original sources to zero. To solve this network for the transient period, we require the use of transient values for the machines impedances.

The machine impedances for the transient period are:zg1’=0.01 + j 0.25, zg2’=0.03 + j0.4

Including transformers:zgt1’=(0.01+0.01)+j(0.25+0.15)=0.02 + j 0.40zgt2’=(0.02+0.01)+j(0.40+0.20)=0.03 + j 0.60

The matrix [YBUS] and its inverse [ZBUS] become:

[ ]

Yj

j

Z Yj

j

BUS

BUS BUS

=−

−

= =+

+

−

0 509 6 960

0 496 3 960

0 020 0 275

0 061 0 487

1

. .

. .

. .

. .

- 0.187 + j2.486 - 0.198 + j1.980- 0.187 + j2.486 0.468 - j6.129 - 0.198 + j1.980

- 0.198 + j1.980 - 0.198 + j1.980

0.014 + j0.186 0.023 + j0.2290.014 + j0.186 0.024 + j0.319 0.025 + j0.2510.023 + j0.229 0.025 + j0.251


Thévenin Contribution (step 2)

∆∆∆

∆

∆

∆

∆

∆∆

VVV

VVfV

Z If

VVfV

Z Z ZZ Z ZZ Z Z

If

V Z Z If Z Z IfV Vf Z

BUS

123

1

3

0

0

1

3

0

0

1 0 02

11 12 13

21 22 23

31 32 33

11 12 13 12

= −

= −

−

=

−

= ⋅ + − + ⋅ = −= − =

from where:( )

12 22 23 22

31 32 33 32

22 22

22

0 03 0 0

⋅ + − + ⋅ = −= ⋅ + − + ⋅ = −

=−−

= =

=

( )( )

Z If Z Z IfV Z Z If Z Z If

IfVfZ

VfZ

VfZ

Z Z

Thev

Thev

∆

* Note that only elements from column ' P' are needed.

From the second equation:

where is the Thevenin impedance for a fault at node 2.

Fault current and changes in voltages are now obtained inin the following way:


Thévenin Contribution (step 2)

Fault current and changes in voltages are now calculated.The Thévenin Impedance for a fault at bus 2 is:

Z = Z = 0.024 + j0.319

Using the pre - fault voltage at node 2:V = V = 0.916 10.2

we find the fault current:

I'VZ

0.916 10.2 0.024 + j0.319

0.916 10.2 0.320 85.6

The voltage changes at the other nodes are found from:V = Z (-I' ) = (0.014 + j0.186)(V = Z (-I' ) = (0.024 + j0.319)(

22 Thev

f 2

ff

22

1 12 f

2 22 f

∠ °

= =∠ °

=∠ °∠ °

= ∠ − °

∠ − °+ °∠ − °+ °

2 862 754

2 862 754 1802 862 754 180

. .

. . ). . )

∆∆∆V = Z (-I' ) = (0.025+ j0.251)(

V = Z (-I' )V = Z (-I' ) =V = Z (-I' ) =

3 32 f

1 12 f

2 22 f

3 32 f

2 862 754 180

0533 169 70 916 169 80 723 1712

. . )

. .. .. .

∠ − °+ °

= ∠ − °∠ − °= −∠ − °

∆∆∆

Vf


Step 3. Fault Conditions

Adding results from steps 1 and 2, we obtain the faulted voltagesat each node:

V V + V = 0.929 7.1 +0.533 -169.7 = 0.398 2.7

V V + V = 0.916 10.2 +0.916 -169.8 = 0

V V + V = 0.920 7.2 +0.723 -171.2 = 0.199 1.3

The current contributions from the lines during the fault are:I' V V 0.398 2.7I' V

f 01

f 02

3f 0

3

12f 1f 2f

32f

1

2

3

1

2

12

32

0187 2 486 0 0 992 82 9

= ∠ ° ∠ ° ∠ °

= ∠ ° ∠ °

= ∠ ° ∠ ° ∠ °

= − = − ∠ °− = ∠ − °=

∆

∆

∆

y jy

( ) ( . . )( ) . .( 3f 2f

g2f f 12f 32f

V 0.398 2.7

The generator contribution is found by Kirchhoff Currents Lawat the faulted node:I' I' I' I'

− = − ∠ °− = ∠ − °

= − − = ∠ − °

) ( . . )( ) . .

. .

0198 1980 0 0 395 83 0

1497 68 5

j

1

2

3

I12f '

I32f 'Ig2f '

If '=2.862∠−75.4°

All quantities have been calculated in per unit.Results are for phase ‘a’.

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