Electrical Transients in Power System

January 2009 Mehdi Vakilian

Text Books:

1-Transients in Power Systems by: Lou van der Slius, 2001

2- Electrical Transients in Power System

by: Allan Greenwood, 1991

COURSE OUTLINE Fundamental Notions About Electrical Transients

Basic Concepts and Simple Switching Transients

Damping Effect on Switching Transients

Abnormal Switching Transients

Testing of Circuit Breakers

Transient Analysis of 3Ph Power Systems

Course Outline …..continued

Transient Analysis of 3Ph Power Systems Traveling Waves and Other Transients on

Transmission Line Modeling Power Equipments for Transients Numerical Simulation of Elec. Transients Lightning and its Induced Transients Insulation Coordination Protection Against Over Voltages

Evaluation System

Assignments : 10% Mid Term One (items 1 to 4) : 10% Mid Term Two (items 5 to 7) : 10% Final : 60% Class Project : 10%

Chapter One : Fundemental Notions about Electrical Transients Time Scale in Power System Studies: planning, Load Flow, Dynamic Stability Switching, external disturbances Frequency ContentDifferential Equations SolutionDistributed and Lumped ParametersCalculatable,Controllable, PreventableTools for Study

CCT Parameters

In Steady State and Transient

Mathematical Presentation & Physical Interpretation

Simple RC Circuit, Closing Ideal Sw.

Equations of RC Circuit

1dQ dVI Cdt dt

11

dVV RC Vdt

1

1

dV dtV V RC

1V IR IdtC

RC Circuit Response

/1

t RCV V Ae

/1 1[ (0)] t RCV V V V e

.)ln( 1 ConsRCtVV

RC Circuit Discharge

11 0dVRC V

dt

/1 1(0) t RCV V e

Capacitor Voltage of RC CCT

Simple Circuits Characteristic (thumbprint)RC , RL , LC Circuits

Thumbprints:

RC CCT: Time Constant ; RC

RL CCT : Time Constant ; L/R

LC CCT : Period of Oscillation ;

2 LC

Principle of Superposition

If stimulus s1 produces R1 & s2 produces R2 applying s1+ s2 simultaneously responds R1+R2 in Linear System

Linear System: response proportional to : stimulus

S.P. Application in Switching

CCT Detail I1: Pre-opening current I2: Superposed current to simulate current cease

S.P. application in Closing switch V1 : voltage across contacts pre-closing Therefore: -V1 fictitious stimulus superposed simulating the closing action

The LaplaceTransform Method

0

, 0

( ) ( )

lim ( )

st

sta

F t F t e dt

F t e dt

Laplace Transform Continued

1 2 1 2

( ) ( )( ) ( )( ) ( )[ ( ) ( )] ( ) ( )

s jF t f sI t i sV t v sF t F t F t F t

Transform of Simple Functions

00 0

'

'' ' '

20 0

.

( )

stst st

st st

consV

e VV V e dt V e dt Vs s

I t I t

II t I t e dt I te dts

Laplace Transform continued

sin2

j t j te etj

2 2

1 1 1sin ( )2

tj s j s j s

2 2cos sts

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 nF t s F t s F s F F

Laplace Transform Continued 01 1[ ( ) ] ( ) ( )

t

F t dt F t F ds s

01 1[ ( ) ] ( ) ( ) ( )t

I t dt I t I t dt Q ts s

( ) (0)[ ( ) ] ( )t i s QI t dt q s

s s

Solving RC problem with Lap. Trans. In terms of I in the

CCT:

Applying L.P. :

0dI Idt RC

( )( ) (0) 0i ssi s IRC

(0)(0) cV VIR

Continuing RC CCT solution

The L.T. solution:

The time solution:

(0) 1( ) 1cV Vi sR s RC

(0)( ) [ ]t

c RCV VI t eR

RL CCT excited by Battery V

Solving for I in CCT

The L.T. of Eq.:

The response:

dIRI L Vdt

( ) ( ) (0) VRi s Lsi s LIs

1( ) ........ (0) 0[ ]

Vi s IRL s s L

RL Time solution1 1 1 1[ ]

( )s s s s

1 1 1 [1 ]( )

tes s

( ) [1 ]RtLVI t e

R

(0) 0, : (0)tR LI add I e

Example: 377 MVA Gen field windingL=0.638H, Exciter noload:1.2 MW(480V) Energy stored in F.W.:

61.2 10 2500480

I A

2 2 61 1 0.638 2.5 10 1.9942 2

E LI MJ

How must the exciter voltage be changed to reduce the field current to zero in 5 Sec

. .480 0.1922500f WR

0.638 3.3230.192

L sR

53.323(5) 2500 (1 ) 0( )

0.192 exciterVI e V V

617......V Volts

Example on LC CCT Transient

Two energy stored elementsSecond order O.D.E.

cdIL V Vdt

1dIL Idt Vdt C

(0)( )( ) (0) cQi s VLsi s LIsC sC s

(0): (0)cc

Qwhere VC

LC CCT solution Ass. I(0)=0

2 2

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

cV V si s IL s sLC LC

12 020 2 2

0

1(0) 0, ( ) ( )cCV i s VLC L s

12

0( ) ( ) sinCI t V tL

LC CCT cont. solving for Vc

Surge Imp. 12

0 ( )LZC

22 20 02

cc

d V V Vdt

22 2 '0

0( ) ( ) (0) (0)c c cVs v s sV Vs

If I(0)=0 then: V`c(0)=0 and Vc(0)

20

2 2 2 20 0

(0)( )( )

cc

V sVv ss s s

21 0

02 20

1 cos( )

ts s

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

Vc characteristic

Vc Osc. Amp depend on V-Vc(0) Vc starts at Vc(0) as expected Response for : 1-Vc(0)=-V 2-Vc(0)=0 3-Vc(0)=+V/2Voltage and Current Relation

Solution of an RL CCT Stimulatedby an Exp. Drive (Ass. I(0)=0)

( ) tU t V e

( ) ( ) (0) VR i s Lsi s LIs

( )( )( )

Vi sR Ls s

Exp. Stimulated RL CCT, Cont. If /R L

1 1( ) ( )( )Vi s

L s s

( ) ( )( )

t tVI t e eL