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Electrical Transients in Power System

January 2009Mehdi Vakilian

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Text Books:

1-T ransients in Power Systemsby: Lou van der Slius, 200 1

2- Electrical Transients in Power Systemby: Allan Greenwood, 199 1

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COUR SE OU TLINE

F undamental Notions About E lectrical Transients

Basic Concepts and Simple Switching Transients

Damping Eff ect on Switching Transients

Abnormal Switching Transients

Testing o f Circuit Breakers

Transient Analysis o f 3Ph Power Systems

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C ourse O utline ..continued

Transient Analysis o f 3Ph Power SystemsTraveling Waves and Other Transients onT

ransmission LineModeling Power Eq uipments f or TransientsNumerical Simulation o f E lec. TransientsLightning and its Induced TransientsInsulation CoordinationProtection Against Over Voltages

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Evaluation System

Assignments :10%Mid Term One (items 1 to 4) : 10%

Mid Term Two (items 5 to 7) : 10%F inal : 60%Class Project : 10%

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Ch apter O ne : Fundemental Notions

about Electrical TransientsTime Scale in Power System Studies:planning, Load F low, Dynamic Stability

Switching, external disturbancesF re quency Content

Diff erential Eq uations Solution

Distributed and Lumped ParametersCalculatable,Controllable, PreventableTools f or Study

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CC T Parameters

In Steady State andTransient

MathematicalPresentation & PhysicalInterpretation

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Simple RC C ircuit, C losing Ideal Sw.

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RC C ircuit R esponse

/1

t RC V V Ae!

/1 1[ (0)] t RC V V V V e!

.)ln( 1 Cons RC t

V V !

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RC C ircuit Disc h arge

11 0

dV RC V

d t !

/1 1(0) t RC V V e!

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Simple C ircuits Ch aracteristic

(th

umbprint)RC , R L , LC C ircuits

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Th umbprints:

R C CC T: TimeConstant ; R CR L CC T : Time

Constant ; L/ RLC CC T : Period o f Oscillation ;

2 L C T

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S.P. Application in Switc h ing

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CC T Detail I1: Pre-opening current

I2: Superposed currentto simulate current cease

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S.P. application in C losing switc h

V1 : voltage across contacts pre -closingThere f ore:

-V1 f ictitious stimulus superposedsimulating the closing action

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Th e LaplaceTransform Met h od

0

, 0

( ) ( )

lim ( )

st

st a

F t F t e d t

F t e d t X

^ g

p g p

!

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Laplace Transform C ontinued

1 2 1 2

( ) ( )( ) ( )

( ) ( )

[ ( ) ( )] ( ) ( )

s j

t f s I t i s

V t v s

t t t t

W [

^ ^

^

^ ^ ^

!

!

!

!

!

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Transform of Simple Functions

00 0

'

'' ' '

2

0 0

.

( )

st

st st

st st

consV

e V V V e dt V e dt V s s

I t I t

I I t I t e dt I t e dt s

^

^

! ! ! !

!

! ! !

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Laplace Transform continued

sin2

j t j t e et

j

[ [

[ !

2 2

1 1 1sin ( )2t j s j s j s

[

^ [ [ [ [ ! !

2 2cos st s^ [

[ !

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Laplace Transform Application' ( ) ( ) (0) F t s F t F ^ ^ !

'' 2 '( ) ( ) (0) (0) F t s F t sF F ^ ^ !

( ) 1 2 ' 1( ) ( ) (0) (0) ... (0)n n n n n F t s F t s F s F F ^ ^ !

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Laplace Transform C ontinued01 1

[ ( ) ] ( ) ( )t

t dt t d s s

^ ^ X X

!

0

1 1[ ( ) ] ( ) ( ) ( )t

I t d t I t I t d t t s s^ ^ ^ g g! !

( ) (0)[ ( ) ] ( )t

i s I t d t q s s s^ g! !

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Solving RC problem wit h Lap. Trans.

In terms o f I in theCC T:

Applying L.P. :

0d I I d t R C

!

( )( ) (0) 0

i s si s I

RC

!

(0)(0) c

V V I

R!

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C ontinuing RC CC T solution

The L. T. solution:

The time solution:

(0 ) 1( )

1c

V V i s

R s R C !

(0)( ) [ ]

t

c RC V V

I t e R!

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R L CC T excited by Battery V

Solving f or I in CC T

The L. T. o f Eq .:

The response:

d I R I L V d t

!

( ) ( ) (0)V

Ri s L si s L I s

!

1( ) ........ (0) 0

[ ]

V i s I

R L s s L

! !

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R L Time solution

1 1 1 1[ ]( ) s s s sE E E

!

1 1 1 [1 ]( )

t e

s s

E^ E E

!

( ) [1 ] R t

LV

I t e R

!

(0) 0, : (0)t R L I add I e{

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Example: 377 MVA Gen field winding

L=0.638H, Exciter noload:1.2 MW(480V)Energy stored in F .W.:

6

1.2 10 25 00480

I Av! !

2 2 61 10.63 8 2 .5 10 1 .9942 2 E L I ! ! v v v !

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How must t h e exciter voltage be c h angedto reduce t h e field current to zero in 5 Sec

. .

4 8 00 .1 9 2

2 5 0 0 f W R ! ! ;

0.63 83 .3230.192

L s RX ! ! !

53.323(5) 25 00 (1 ) 0( )

0.192excit er

V I e V V ! ! !

6 17 ......V Volts( !

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Example on L C CC T Transient

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Two energy stored elementsSecond order O .D.E.

cd I

L V V d t

!

1dI L Id t V d t C

!

(0)( )( ) (0) c

Qi s V L si s L I

sC sC s !

(0 ): (0 )c c

Qw h ere V

C !

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LC CC T solution Ass. I(0)=0

2 2

(0) 1( ) (0)1 1( ) ( )

cV V si s I L s s LC L C

!

12 020 2 2

0

1(0) 0, ( ) ( )c C V i s V L C L s

[ [

[

! ! p !

12

0( ) ( ) sinC

I t V t L [ !

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LC CC T cont. solving for Vc

Surge Imp. 12

0 ( ) L

Z C

!

22 2

0 02 c c

d V V V dt

[ [ !

22 2 '0

0( ) ( ) (0) (0)

c c c

V s v s sV V

s

[ [ !

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If I(0)=0 t h en: V`c(0)=0 and Vc(0)

2

02 2 2 2

0 0

(0)( ) ( )

cc

V sV v s s s s

[

[ [ !

2

1 002 2

0

1 cos( )

t s s

[ ^ [ [

!

0 0 0( ) (1 cos ) (0) cos [ (0)] cosc c cV t V t V t V V V t[ [ [ ! !

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Vc ch aracteristic

Vc Osc. Amp depend on V -Vc(0)Vc starts at Vc(0) as expectedR

esponse f or :1- Vc(0)= -V2-Vc(0)=03-Vc(0)= +V/2

Voltage and Current R elation

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Exp. Stimulated R L CC T, C ont.

If / R LP !

1 1( ) ( )( )

V i s L s sE E

!

( ) ( )( )t t V

I t e e

LP E

E P

!