Download Special Topics in Power

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



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Assignments
Mid Term One (items 1 to 4)
Mid Term Two (items 5 to 7)
Final
Class Project
: 10%
: 10%
: 10%
: 60%
: 10%
Chapter One : Fundemental Notions
about Electrical Transients
Time Scale in Power System Studies:
planning, Load Flow, Dynamic Stability
Switching, external disturbances
Frequency Content
Differential Equations Solution
Distributed and Lumped Parameters
Calculatable,Controllable, Preventable
Tools for Study

CCT Parameters

In Steady State and
Transient

Mathematical
Presentation & Physical
Interpretation
Simple RC Circuit, Closing Ideal Sw.
Equations of RC Circuit
1
dQ
dV 1
C
V  IR   Idt I 
dt
dt
C
dV 1
V  RC
V1
dt
dV 1
dt

V  V 1 RC
RC Circuit Response
t
ln( V  V1 )  
 Cons.
RC
V 1  V  Ae
 t / RC
V 1  V  [V  V 1(0)]e
 t / RC
RC Circuit Discharge
dV 1
RC
V1  0
dt
V 1  V 1(0)e
 t / RC
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

 F (t )   F (t )  e  dt
 st
0
lim ,a0  F (t )  e
 st
 dt
Laplace Transform Continued
s    j
 F (t )  f ( s )
 I (t )  i ( s )
 V (t )  v( s)
 [ F 1(t )  F 2(t )]   F 1(t )   F 2(t )
Transform of Simple Functions
cons.V


 st 
e
 V   V  e dt  V  e dt  V
s
0
0
 st
 st
I (t )  I t
'


'
I
 I 't   I 't  e  st dt  I '  te  st dt  2
s
0
0
0
V

s
Laplace Transform continued
e jt  e jt
sin t 
2j
1
1
1

 sin t  (

) 2 2
2 j s  j s  j s  
s
 cos t  2
2
s 
Laplace Transform Application
 F (t )  s F (t )  F (0)
'
 F (t )  s  F (t )  sF (0)  F (0)
''
2
'
 F (t )  s  F (t )  s F (0)  s F (0)  ...  F (0)
( n)
n
n 1
n2
'
n 1
Laplace Transform Continued
t
0
1
1
 [  F (t )dt ]   F (t )   F ( )d
s
s 

t
0
1
1
 [  I (t )dt ]   I (t )   I (t )dt   Q(t )
s
s 

t
i ( s) Q(0)
 [  I (t )dt ] 

 q( s)
s
s

Solving RC problem with Lap. Trans.

In terms of I in the
CCT:
dI
I

 0
dt
RC
Applying L.P. :
i( s)
si( s)  I (0) 
0
RC
V  Vc (0)
I (0) 
R
Continuing RC CCT solution

The L.T. solution:
V  Vc (0)
1
i(s) 

R
s 1

RC
The time solution:
V  Vc (0)  t RC
I (t )  [
]e
R
RL CCT excited by Battery V

Solving for I in CCT

The L.T. of Eq.:
dI
RI  L  V
dt
V
Ri ( s)  Lsi( s)  LI (0) 
s

The response:
V
1
i ( s) 
........I (0)  0
L s[s  R ]
L
RL Time solution
1
1 1
1
 [ 
]
s( s   )  s s  

1
1
1
 [1  e t ]
s( s   ) 
V
 Rt
I (t )  [1  e L ]
R
I (0)  0, add : I (0)e
R t
L
Example: 377 MVA Gen field winding
L=0.638H, Exciter noload:1.2 MW(480V)

Energy stored in F.W.:
1.2 106
I
 2500 A
480
1 2 1
E  LI   0.638  2.52  106  1.994MJ
2
2
How must the exciter voltage be changed
to reduce the field current to zero in 5 Sec
R f .W .
480

 0.192
2500
L 0.638
 
 3.323s
R 0.192
V
5
I (5)  2500 
(1  e 3.323 )  0(Vexciter  V )
0.192
V  617......Volts
Example on LC CCT Transient
Two energy stored elements
Second order O.D.E.
dI
L
 Vc  V
dt
dI
1
L

dt C
 Idt  V
i ( s) Qc (0) V
Lsi ( s)  LI (0) 


sC
sC
s
Qc (0)
where :
 Vc (0)
C
LC CCT solution Ass. I(0)=0
V  Vc (0)
1
i ( s) 
2
L
s (1
Vc (0)  0, 1
LC
s
 I (0) 2
)
s ( 1
LC )
C 12 0
   i( s)  V ( )
LC
L
s 2  02
2
0
C 12
I (t )  V ( ) sin 0t
L
LC CCT cont. solving for Vc

Surge Imp.
L 12
Z0  ( )
C
2
d Vc
2
2
 0 Vc  0 V
2
dt
( s   )vc ( s) 
2
2
0
V
2
0
s
 sVc (0)  V (0)
'
c
If I(0)=0 then: V`c(0)=0 and Vc(0)

1
02
s( s   )
2
2
0
 1  cos 0t
V
sVc (0)
vc ( s) 
 2
2
2
2
s( s  0 ) s  0
2
0
Vc (t )  V (1  cos 0t )  Vc (0)cos 0t  V  [V  Vc (0)]cos 0t
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

Solution of an RL CCT Stimulated
by an Exp. Drive (Ass. I(0)=0)
U (t )  V  e t
V
R  i ( s)  Lsi ( s)  LI (0) 
s 
V
i( s) 
( R  Ls)( s   )
Exp. Stimulated RL CCT, Cont.

If
  R/ L
V
1
1
i( s) 
(

)
L(   ) s   s  
V
 t
 t
I (t ) 
(e  e )
L(   )