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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
=