Special Topics in Power_1

8/3/2019 Special Topics in Power_1

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Electrical Transients in PowerSystem

January 2009

Mehdi Vakilian


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Text Books:

1-Transients in Power Systems

by: Lou van der Slius, 2001

2- Electrical Transients in Power System

by: Allan Greenwood, 1991


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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


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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


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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%


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Chapter One : FundementalNotions

about ElectricalTransients Time Scale in Power System Studies:

planning, Load Flow, Dynamic Stability

Switching, external disturbances

Frequency Content

Differential Equations Solution

Distributed and Lumped ParametersCalculatable,Controllable, Preventable

Tools for Study


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CCT Parameters

In Steady State and

Transient

Mathematical

Presentation & Physical

Interpretation


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Simple RC Circuit, ClosingIdeal Sw.


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Equations of RC Circuit

1dQ dV I C

dt dt

= =

11

dVV RC V

dt

= +

1

1

dV dt

V V RC=

1V IR Idt

C

= +


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RC Circuit Response

/1

t RCV V Ae=

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

.)ln( 1 ConsRC

tVV +=


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RC Circuit Discharge

11 0

dVRC V

dt

+ =

/1 1(0) t RCV V e=


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Capacitor Voltage of RC CCT


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Simple Circuits Characteristic(thumbprint)

RC , RL , LC Circuits


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Thumbprints:

RC CCT: Time

Constant ; RC

RL CCT : Time

Constant ; L/R LC CCT : Period of

Oscillation ;

2 LC


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Principle of Superposition

If stimulus s1 produces R1

& s2 produces R2

applying s1+ s2 simultaneously

responds R1+R2 in Linear System

Linear System: responseproportional to :

stimulus


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S.P. Application in Switching


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CCT Detail I1: Pre-openingcurrent

I2: Superposedcurrent

to simulatecurrent cease


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S.P. application in Closingswitch V1 : voltage across contacts pre-closing

Therefore:

-V1 fictitious stimulus superposed

simulating the closing action


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The LaplaceTransform Method

0

, 0

( ) ( )

lim ( )

st

st

a

F t F t e dt

F t e dt

=


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Laplace Transform Continued

1 2 1 2

( ) ( )( ) ( )

( ) ( )

[ ( ) ( )] ( ) ( )

s j

F t f sI t i s

V t v s

F t F t F t F t

= +

==

=

+ = +


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Transform of Simple Functions

00 0

'

'

' ' '2

0 0

.

( )

stst st

st st

consV

e VV V e dt V e dt V s s

I t I t

II t I t e dt I te dts

= = = =

=

= = =


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Laplace Transform continued

s in2

j t j t

e etj

=

2 2

1 1 1

sin ( )2t j s j s j s

= = +

2 2coss

t 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 '( ) ( ) ( 0 ) (0 ) . . .n n n nF t s F t s F s F =


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Laplace Transform Continued

0

1 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)[ ( ) ] ( )

ti s Q

I t dt q ss s

= + =


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

CCT:

Applying L.P. :

0dI I

dt RC+=

( )( ) (0) 0

i ssi s I

RC

+ =

(0)(0) c

V VI

R

=


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Continuing RC CCT solution

The L.T. solution:

The time solution:

(0) 1( )

1cV Vi s

R sRC

=

+

(0)

( ) [ ]

tc RC

V V

I t eR

=


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RL CCT excited by Battery V

Solving for I in CCT

The L.T. of Eq.:

The response:

dIRI L V

dt+ =

( ) ( ) (0)V

Ri s Lsi s LIs

+ =

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

[ ]

Vi s I

RL s sL

= =

+


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RL Time solution

1 1 1 1[ ]( )s s s s = + +

1 1 1 [1 ]( )

tes s

= +

( ) [1 ]Rt

LV

I t eR

=

(0) 0, : (0)tRLI add I e


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Example: 377 MVA Gen fieldwinding

L=0.638H, Exciter noload:1.2MW(480V)

Energy stored in F.W.:

61.2 10

2500480I A

= =

2 2 61 10.638 2.5 10 1.9942 2E LI MJ= = =


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How must the exciter voltage bechanged to reduce the field

current to zero in 5 Sec. .

4800.192

2500f WR = =

0.6383.323

0.192

Ls

R = = =

53 . 3 2 3(5 ) 2 5 0 0 (1 ) 0 (

0 . 1 9 2

e x c it e r

VI e V

=

617......V Volts =


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Example on LC CCT Transient


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Two energy stored elementsSecond order O.D.E.

c

dIL V V

dt+ =

1dI

L Idt Vdt C+ =

(0)( )( ) (0) c

Qi s VLsi s LI

sC sC s + + =

(0): (0)c c

Qwhere V

C=


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LC CCT solution Ass. I(0)=0

2 2(0) 1( ) (0)

1 1( ) ()

cV V si s IL s s

LC LC

= +

+ +

12 020 2 2

0

1(0) 0, ( ) ( )c CV i s V LC L s

= = =

+

12

0( ) ( ) sin

C

I t V tL =


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LC CCT cont. solving for Vc

Surge Imp. 12

0 ( )L

ZC

=

22 2

0 02

c

c

d V

V Vdt + =

22 2 '0

0

( ) ( ) (0) (0)c c c

Vs v s sV V

s

+ = + +


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If I(0)=0 then: V`c(0)=0 andVc(0)

2

0

2 2 2 20 0

(0)

( ) ( )

c

c

V sV

v s s s s

= +

+ +

2

1 002 2

0

1 cos( )

ts s

= +

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


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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/2

Voltage and Current Relation


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

( ) tU t V e =

( ) ( ) (0)V

R i s Lsi s LIs

+ =

+

( )( )( )

Vi s

R Ls s =

+ +


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Exp. Stimulated RL CCT, Cont.

If /R L=

1 1( ) ( )( )

Vi s L s s = + +

( ) ( )( )

t tV

I t e eL

=

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