Lecture Notes EEE 360 - Michigan Technological … Circuit Calculations, Unsymmetrical Faults Leonard Bohmann, Michigan State University Elham Makram, Clemson University. 2 ... Lecture

1

SESSION 3

Short Circuit Calculations, Unsymmetrical Faults

Leonard Bohmann, Michigan State University

Elham Makram, Clemson University

2

Short Circuit CalculationsLeonard Bohmann

Michigan State University

3

Line to Ground Short

2003 IEEE T & D CONFERENCE


-3

-2

-1

0

1

2

3

4

5

6

7

0 0.05 0.1 0.15 0.2tA

mp

s(k

A)

Short Circuit Fault Currents

high frequency transient

dc offset decaying sinusoid


Short Circuit Fault Currents (continued)

v(t)

X(t) R

]e)tcos(

)t[cos()t(Z

V)t(i

then

)t(jXR)t(Z

)tcos(V)t(v

if

)t(XtR

0

m

m

ω−θ−α+ω−

θ−α+ω=

+=θ∠α+ω=

-3

-2

-1

0

1

2

3

4

5

0 0.05 0.1 0.15 0.2

6

dc offset

-3

-2

-1

0

1

2

3

4

5

0 0.05 0.1 0.15 0.2

θ−α+ω−θ−α+ω=

ω− XtR

0m e)tcos()tcos(

Z

V)t(i


7

Variable magnitude

-6

-4

-2

0

2

4

6

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

t


)tcos()t(Z

V)t(i m θ−α+ω=

8

Reactance of Synchronous Machine

00.1

0.20.3

0.40.5

0.60.7

0.80.9

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4

t

reac

tan

ce(p

u)

X'd

X"d

Xd


9

• Use sub-transient reactance to calculatefault currents– worst case for when CB is opening

• Do not calculate dc offset directly– use X/R to approximate offset

(asymmetry ratio) at opening time– X/R is the time constant of the decay in

cycles


10

How do youcalculate the

fault current (If)at bus 3?

ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT15

ZL5

5

ZL2

If

ZT23

ZT34


11

Use Thevenin equivalent

VTh

ZTh

3

If

• VTh: open circuitvoltage

– also called the pre-fault voltage, Vf

– often use 1∠ 0°



Use for:

• symmetrical faults

– single phase networks

– three phase faults in three phasenetworks (per phase equivalent)

• concept for solving all faults


• This method calculates the currentgoing into the fault

• Often want the current elsewherein the network (through a circuitbreaker, in bus bars, in other lines,etc.)

14

A current source, with the magnitude ofthe fault current, placed at the fault, hasthe same effect on the rest of the circuitas the short circuit.

Vf

ZTh

3

If

Vf

ZTh

3

If


15

Adding a current source tothe original circuit hasthe same electricaleffect as adding it tothe equivalent circuit


Vf

ZTh

3

If

ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT23ZT15

ZL5

5

ZL2

ZT34

If

16

All the voltages and allthe currents can be foundusing the new circuit.


ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT23ZT15

ZL5

5

ZL2

ZT34

If

Use superposition:– Use the original sources

(or assume currents are 0and voltages are 1∠ 0°)

– Use just the currentsource, gives changesdue to the fault

17

Calculation detailsBus Admittance Matrix: Ybus


I = Ybus V

vector of currentinjections

vector of busvoltages

ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT23ZT15

ZL5

5

ZL2

ZT34

If


ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT23ZT15

ZL5

5

ZL2

ZT34

If

=

555351

4443

35343332

232221

151211

bus

00

000

0

00

00

Y

YYY

YY

YYYY

YYY

YYY

Diagonals:Σ of Y connected at nodeY22 = 1/ZT12 + 1/ZL2 + 1/ZT23

Off diagonals:- of Y connected between nodesY53 = -1/ZT35


Ybus is sparse (lots of zero)

– small size

– easy to build

– quick algorithms to manipulate

20

Bus Impedance Matrix: Zbus


V = Zbus I

vector of currentinjections

vector of busvoltages ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT23ZT15

ZL5

5

ZL2

ZT34

If

Zbus = Ybus-1

21

=

5554535251

4544434241

3534333231

2524232221

1514131211

busZ

ZZZZZ

ZZZZZ

ZZZZZ

ZZZZZ

ZZZZZ


Zbus is a full matrix– slow to invert– slow to manipulate– avoid using if

possible

Diagonal element is the TheveninImpedance at that bus– Z33 is the Thevenin Impedance at bus 3


To find the change in voltages due tothe fault current at bus 3:

⋅

=

∆∆∆∆∆

0

0

I-

0

0

V

V

V

V

V

f

5554535251

4544434241

3534333231

2524232221

1514131211

5

4

3

2

1

ZZZZZ

ZZZZZ

ZZZZZ

ZZZZZ

ZZZZZ

23

To calculate current in line 2-3:


23T

32f23 Z

VVI

∆−∆=∆

1 2

3

4

ZT23

5 If

∆I23f

24

Asymmetry ratio: S

• Method calculates the RMS worst caseof the 60 Hz component– this is needed to set protective relays

• For circuit breaker selection, need toknow the total RMS current (60 Hzand dc offset)– Circuit breakers are rated on symmetrical

fault current, assuming an X/R ratio of 152003 IEEE T & D CONFERENCE


( ) RXt2

ac

RMS

RXt2

acRMS

2RX

t

ac2acRMS

2dc

2acRMS

21isS(t)wheretSI

I

21II

I2II

III

ω−

ω−

ω−

⋅+=

⋅+=

+=

+=

e

e

e

Asymmetry ratio

26

Calculating X/R• Equations presented are

for a single order system• For higher order systems,

– ignore R and findequivalent X,

– ignore X, and findequivalent R

• Actual decay is typicallyquicker than thisapproximation

ZT12

ZT35

1 2

3

4

ZG1

ZG4

ZT23ZT15

ZL5

5

ZL2

ZT34

If


27

Algorithm


• Calculate Thevenin Impedance• Calculate current into fault• Find the needed elements of Zbus (not all

of them)• Find the needed changes in bus voltages• Find the line currents of of interest• Find the dc offset

28

Example• The oneline drawing of a system, along with Zbus

are shown below.

j 0.08 pu

j0.03 pu

1 2

3

4

j0.1pu

j0.1 pu

j0.025 puj0.05 pu

j15pu

5

j10 pu

If

j0.05 pu


puj

=

0799004290064400623005590

0429006580048700450003340

0644004870073100675005020

0623004500067500831005380

0559003340050200538006560

Zbus

.....

.....

.....

.....

.....

i) Calculate the fault current for athree phase solid fault at bus 3.

ii) Calculate the voltages at bus 1,2, and 3 during this fault.

iii) Calculate the current flowingfrom bus 2 to bus 3.

29

Part (i): Calculate the fault current for a threephase solid fault at bus 3.

• Assume all initial busvoltages are 1∠ 0°– therefore all initial

currents are 0

• Use the 3,3 elementof Zbus for theThevenin impedance.

1∠ 0°

j0.731

3

If


puj

j

6813I

07310

1I

f

f

.

.

−=

=

30

Multiply the negative of the fault current by elements inthe 3rd column to get the change in voltages


Part (ii): Calculate the voltages at bus 2, 3, and4 during this fault.

⋅

=

∆∆−

−−

0

0

13.68--

0

0

0799004290064400623005590

0429006580048700450003340

0644004870073100675005020

0623004500067500831005380

0559003340050200538006560

V

V

1

9230

6870

5

4

jj

.....

.....

.....

.....

.....

.

.


Part (ii): Calculate the voltages at bus 2, 3, and4 during this fault. (continued)

pu

pu

pu

0101VVV

07660923001VVV

3130687001VVV

33i3f

22i2f

11i1f

=−°∠=∆−==−°∠=∆−==−°∠=∆−=

..

..

The voltages during the fault are the initial voltagesminus the changes in the voltages


Part (iii): Calculate the current flowing from bus 2to bus 3.

The current is the differencebetween the voltagesdivided by the impedance

pujj

8630250

007660I

Z

VVI

23f

T23

3f2f23f

..

. −=−=

−=

1 2

3

4

5-j13.68 pu

-j3.86 pu

33

Problem

1 2

3

4

j0.1 puj0.1 pu

j0.1 pu

j0.025 puj0.05 pu

j0.05 pu

j 0.08


A small power system is sketched below along with Zbus.

a) What is the fault current for a three-phase fault toground at bus 3?

b) What are the voltages at busses 2, 3, and 4?c) What is the current due to the fault from bus 4 to 3 and

from 2 to 3?

puj 2bus 10

4045106342221742

1063658463432613

4222634353440443

1742261304437834

Z −×

=

....

....

....

....


Unsymmetrical FaultsElham Makram

Clemson University

1

SINGLE LINE TO GROUND FAULT

Fig.1. Single line to ground fault on phase A


Zf

Ibf IcfIaf

a

b

c

n

0f210

affa

210

12

0

21

22

12

22

10212

0

afff

cfbf

3I*ZVVVOr,

I*ZVNow

III

0Ia)a(IThus

II

a)I–(aa)I–(a

0IaaIIaIIaI

equations,currenttheFrom

I*ZV

0II

locationfaultatConditionsBoundary

=++=

===++

==

=++=++

===


SINGLE LINE TO GROUND FAULT (CONTD.)

Fig.2. Sequence connection for A-G fault

I0

To satisfy the voltage and current boundary conditions, the sequence networks should beconnected in series as shown in Fig. 2.

3Zf

ZeroSequenceNetwork

ZeroSequenceNetwork

ZeroSequenceNetwork

I1

I2

V0

+

-

+V1-

V2-

+


LINE TO LINE FAULT

Fig.3. Line to line fault

a

Zf

Ibf IcfIaf

b

c

n

Vaf

+

Vbf Vcf

1f21

22

f12

12

22

10

212

0f212

0

bffcfbf

21

22

12

22

1212

0

cfbf

af

IZVV

a)V-(aZa)I-(aa)V-(a

toleadsThis

VaaVV

)aIIa(IZaVVaV

IZVV

conditionsboundaryVoltage

I-I

a)I(a-a)I(a

0IaaIaIIa

0I

Thus,

I-I

0I

ConditionsboundaryCurrent

+=+=

+++

++=++

+=

=+=+

=+++

=

==


Fig.4. Connection of sequence network for BC fault

To satisfy the voltage and current boundary conditions, the sequence networks shouldbe connected as shown in Fig. 4.

LINE TO LINE FAULT (CONTD.)

Zf

PositiveSequenceNetwork

NegativeSequenceNetwork

I1

I2

+V1-

V2-

+


DOUBLE LINE TO GROUND FAULT

Fig.5. Double line to ground fault

0f10

f22

12

12

0212

0

21

22

10212

0

cfbf

fcfbfbf

210

af

I3ZV-VThus

Z)IaaIaI

Ia2I(aVVaV

Also,

VVThus

VaaVVaVVaV

Thus,

VV

and,

)ZI(IV

conditionsboundaryVoltage

0III

Thus,

0I

location,faultat the

ConditionsboundaryCurrent

=++

++=++

=++=++

=

+=

=++

=

Zf

a

b

c

n

+

Ibf IcfIaf

Vaf Vbf Vcf


DOUBLE LINE TO GROUND FAULT (CONTD.)

Fig.6. Sequence network connection for doubleline to ground fault

Considering the voltage and current boundary conditions, the sequence networks areconnected as shown in Fig. 6.

3Zf

I0

ZeroSequenceNetwork

PositiveSequenceNetwork

NegativeSequenceNetwork

I1

I2

V0

+

-

+V1-

V2-

+


Example 1

Consider the power system shown in Fig. 7.

Ratings

G1 and G2: 100 MVA, 20 kV

X1 = X2 = 0.3 pu

X0 = 0.04 pu

T1 and T2: 100 MVA, 345/20 kV

XT1 = XT2 =0.08 pu

T. L. X1 = X2 = 0.15 pu

X0 = 0.5 pu

Find the voltages at Bus 1 and thecurrents from T1 to P due to

(i) Single line to ground fault at P

(ii) Line to line fault at P

(iii) Double line to ground fault at P

Fig. 7. Example 1

G1 G2

T1 T21 2P T. L.


Positive sequence network

Negative sequence network Zero sequence network

Z0th = j0.08

j0.2213

Pj0.08 j0.5Z1th = j0.2213

P

j0.3 j0.3

j0.08 j0.15 j0.08

1

P

2 j0.53j0.38

P

j0.2213

P

j0.2213j0.15)j0.08(j0.3||j0.8)(0.3Z

01VEEIf

th

21

=+++=°∠===


(i) Single line to ground fault

P--T1fromcomponentsphasecurrentThe

0.15304-(j0.8)j1.913-ZI-V

0.42332-(j0.38)(j1.114)-ZI-V

0.57668j0.38)(-j1.114)(–1.0ZI-EV

:arelocationfaultat thecomponentsvoltagesequenceThe

j1.9134-I

j1.114-I

114.10.38)(0.53

0.53*II

P--T1fromcomponentscurrentsequencethearefollowingThe

j1.9134-j0.08)j0.2213(j0.2213

01

)ZZ(Z

EIII

000

222

1111

0

2

11

2th1th0th021

>===

======

==

−=+

=

>

=++

°∠++

===

j

=

a2

a1

a0

2

2

c

b

a

I

I

I

*

1

1

111

I

I

I

aa

aa


-j0.799c

-j0.799b

-j3.027

c-j5.739= 3 I0

-j2.712

j0.799-j1.913–(-j1.114)aa(-j1.114)I

j0.799-j1.913–a(-j1.114)(-j1.114)aI

j3.027-j1.913-j1.114--j1.114I

2c

2b

a

=+=

=+=

==

j0.866-0.2297

))(-0.42332240(1)(0.57668)120(10.15304-VaaVVV

j0.866--0.2297

))(-0.42332120(1)(0.57668)240(10.15304-aVVaVV

00.42332–0.576680.15304-VVVV

:arelocationfaultat thevoltagesphaseThe

22

10c

212

0b

210a

+=°∠+°∠+=++=

=°∠+°∠+=++=

=+=++=


(ii) Line to line fault

°∠=°∠=

°∠===

==−=

=+=

=+=

==>

−=+

−=>

−=+

°∠=+

=−=

00.5V

0-0.5V

01V

5.0ZI-V

0.5j0.38)(-j1.316)(-1.0ZIEV

:Voltages

2.279(j1.316)a(-j1.316)aI

2.279a(j1.316)(-j1.316)aI

0I

316.1jP)--1T(I

316.1j)0.53(0.38

0.5326.2jP)--1T(I

26.2j0.2213)(j0.2213

01

)Z(Z

EII

:Currents

c

b

a

2a2a2

1a1a1

2c

2b

a

2

1

2th1th21 j


SEQUENCE PHASE SHIFT THROUGH WYE/DELTA TRANSFORMER BANK

•Positive and negative sequence impedance is independent of connection

•These networks ignore phase shift

•When current and voltages are transformed from one side to the other phase shift must

be considered.

•Standard ANSI connection has been shown in figs. 8 (a) and 8 (b).

Fig. 8 (a) Wye/Delta Fig. 8 (b) Delta/Wye

High side Low sideIA

IB

IC

Ia

Ib

Ic

A

B

C

a

b

c

nIA

nIB

nIC

IA

IB

IC

n:1

BC

A

bc

a

High side Low side

Ia

Ib

Ic

IA

IB

IC

a

b

c

A

B

C

Ia/n

Ib/n

Ic/n

n:1

BC

A

bc

a


30VV

unitperInor

30V3na)-(1nV)aV-n(V)V-n(VV

:voltagesequenceNegative

30II

unitperInor

30I3n)a-(1nI)Ia-n(I)I-n(II

:currentsequenceNegative

30VV

unit,perinor

30V3n)a-(1nV)Va-n(VV

:volatgesequencePositive

30II

unit,perIn

30-I3na)-(1nI)aI-n(I)I-n(II

:currentsequencePositive

8(a)fig.Refer to

a2A2

a2a2a2a2b2a2A2

a2A2

A22

A2A22

A2C2A22a

a1A1

a12

aa12

a1A1

a1A1

A1A1A1A1C1A11a

°−∠=

°−∠====

°−∠=

°∠====

°∠=

°∠===

°∠=

°∠====


°−∠=

°−∠=

°−∠==

°∠=

°−∠==

°∠=

°∠===

30VV

voltagesequenceNegative

30II

unitperInor

30In

3)aI-(I

n

1I

:currentsequenceNegative

30VV

unit,perinor,

30Vn

3)aV-(V

n

1V

:voltagesequencePostitive

30II

unit,perinor,

30In

3)Ia-(I

n

1)I-(I

n

1I

:currentsequencePostitive

(b)8fig.Refer to

a2A2

a2A2

a2a2a2A2

a1A1

A1A1A1a1

a1A1

a1a12

a1b1a1A1


°∠=°+∠=

∆∆

30-voltageLowsequenceNegativegeHigh voltasequenceNegative(ii)

30voltageLowsequencePositivegeHigh voltasequencePositive(i)

:sconnectionY-or-Yingeneral,In


Example 2

Find the currents and voltages at G1 of Example 1, due to single line to ground fault at P.

0I

j1.9295)60(1.114)120(1.114

)60-(1.1141201)120-(1.1142401aIIaI

j1.9295-60-j1.114120-j1.114-III

0I

60-j1.11430j1.114-I

120-j1.114-30-j1.114-I

:Currents(i)

cg

ag2ag12

bg

ag21agag

ag0

ag2

1ag

==°∠+°∠=

°∠°∠+°∠°∠=+=

=°∠+°∠=+=

=

°∠=°+∠=

°∠=°∠=


°∠°∠

°−∠=

=

°∠=+=

°∠=

°∠°∠=+=

°−∠=°∠°∠=+=

°∠=

°∠==

==

==

901

12.240577.0

60577.0

V

V

V

*

1

1

111

V

V

V

901.0VaaVV

240.120.577

1500.334-2100.666aVVaV

60577.0300.334-30-0.666VVV

300.334-V

30-0.666V

shift)phaseneglectingNetworksequence(from0.3342-

ZI–V

shift)phaseneglectingNetworksequence(from0.666

j0.3)(-j1.114)(-1.0ZI–EV

:Voltages(ii)

ag2

ag1

ag0

2

2

cg

bg

ag

ag22

ag1cg

ag2ag12

bg

ag2ag1ag

ag2

ag1

g2a2ag2

g1a1ag1

aa

aa

Example 2 ( contd.)


PROBLEM #1

For the system shown below, find the currents in the transformer and generator windings inper unit and amperes for a single-line-to-ground fault at

(i) P

(ii) R

G1

15/115 kVP

G1R

G: 100 MVA, 15 kV

X1=X2= 0.1,

X0 = 0.05

Transformer:

100 MVA, 15/115 kV

X = 0.1

PROBLEMS

19

PROBLEM #2

Solve examples 1 and 2 for a double-line-to-ground fault at P.

PROBLEMS



PROBLEM #3

Solve examples 1 and 2 for a line-line fault at P.

PROBLEMS

Documents

Lecture Notes EEE 360 - Michigan Technological … Circuit Calculations, Unsymmetrical Faults Leonard Bohmann, Michigan State University Elham Makram, Clemson University. 2 ... Lecture