Salvador Acevedo

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 V1f, V2f and V3f 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

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Step 1. Pre-Fault Solution

+E1-

zLoad

+E2-

zgt1

z12

z13

zgt2

z23

i12

i13 i23

iG1 iG2

+

V1

-

+

V2

-

+V3-

I1 I2

ygt1

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=0.930

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

V1

V2V3

11 12 13

21 22 23

31 32 33

=

-

-

=

-

-

0 405 5466

0496 3960

0 02 09996

0333 05490

. .

. .

. .

. .

j

j

j

j

=

V1V2

V3

0929 7 10916 10 2

0920 7 2

. .

. .

. .

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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

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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)

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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=I13 2 * Rline13=(0.019) 2(0.05)=0.00002

line 2-3: P=I23 2 * 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.

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Step 2. Fault at Bus 2 (Thvenin Contribution)

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

=-

-

= =+

+

-

0509 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

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Thvenin Contribution (step 2)

DDD

D

D

D

D

DD

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

D

* 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:

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Thvenin Contribution (step 2)

Fault current and changes in voltages are now calculated.The Thvenin 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 75 4 1802 862 754 180

. .

. . ). . )

DDDV = 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

. . )

. .. .. .

- +

= - - = - -

DDD

Vf

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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

=

=

=

= - = - - = - =

D

D

D

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 0 395 830

1497 685

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.