Transcript Traveling Wave - Petra Christian University

Traveling Wave
Transient Overvoltages
1.Introduction
• Transient Phenomenon :
– Aperiodic function of time
– Short duration
• Example :Voltage & Current Surge :
(The current surge are made up of charging or discharging capacitive
currents that introduced by the change in voltages across the shunt
capacitances of the transmission system)
– Lightning Surge
– Switching Surge
Impulse Voltage Waveform
2.Traveling Wave
•
•
•
•
Disturbance represented by
closing or opening the switch S.
If Switch S closed, the line
suddenly connected to the
source.
The whole line is not energized
instantaneously.
Processed :
–
–
–
•
When Switch S closed
The first capacitor becomes
charged immediately
Because of the first series
inductor (acts as open circuit),
the second capacitor is delayed
This gradual buildup of voltage
over the line conductor can be
regarded as a voltage wave is
traveling from one end to the
other end
Voltage & Current Function
•
•
•
•
•
•
vf=v1(x-t)
vb=v2(x+t)
 = 1/(LC)
v(x,t)=vf + vb
vf=Zcif
vb=Zcib
•
•
•
•
•
Zc=(L/C)½
If=vf/Zc
Ib=vb/Zc
I(x,t)=If + Ib
I(x,t)=(C/L) ½
[v1(x-t) -v2(x+t)]
2.1 Velocity of Surge Propagation
• In the air = 300 000 km/s
•  = 1/(LC) m/s
• Inductance single conductor Overhead Line (assuming
zero ground resistivity) :
L=2 x 10-7 ln (2h/r)
H/m
C=1/[18 x 109 ln(2h/r)] F/m
• v
1/ 2
7

 2  10 ln 2h / r   
1
 
 
9
LC  18  10 ln 2h / r   
1
• In the cable :  = 1/(LC) = 3 x 108 K
K=dielectric constant (2.5 to 4.0)
m/s
2.2 Surge Power Input & Energy Storage
•
•
•
•
P=vi Watt
Ws= ½ Cv2 ; Wm= ½ Li2
W=Ws+Wm = 2 Ws = 2 Wm = Cv2 = Li2
P=W  = Li2 /(LC) = i2 Zc = v2 / Zc
2.3 Superposition of Forward and
Backward-Traveling Wave
3. Effects of Line Termination
• Assuming vf, if,vb and ib are the instantaneous
voltage and current.
Hence the instantaneous voltage and current at
the point discontinuity are :
•
•
•
•
v(x,t)=vf + vb and
I=vf/Zc - vb/Zc and
v + iZc= 2vf so
vf = ½ (v+iZc) and
I(x,t)=If + Ib
iZc=vf – vb
v=2vf=iZc
vb = ½ (v+iZc) or
vb= vf-iZc
3.1 Line Termination in Resistance
v  iR
2
i
vf
R  Zc
R  Zc
2R
R  Zc
vb 
vf
R  Zc
vf 
Pf 
v 2f
Zc
vb2
Pb 
Zc
2

v

 vb 
v
f
R
R
Pf  Pb  PR
PR 
2
3.2 Line Termination in Impedance (Z)
2Z
v
vf
Z  Zc
Z  Zc
vf 
v
2R
Z  Zc
vb 
vf
Z  Zc
v  v f
vb  v f
2Z

Z  Zc
Z  Zc

Z  Zc
2
i
if
Z  Zc
• Line is terminated with its characteristic impedance :
– Z=Zc
–  =0, no reflection (infinitely long)
• Z>Zc
– vb is positive
– Ib is negative
– Reflected surges increased voltage and reduced current
• Z<Zc
– vb is negative
– Ib is positive
– Reflected surges reduced voltage and increased current
• Zs and ZR are defined as the sending-end and receiving
end.
•
Z s  Zc
Z R  Zc
s 
; R 
Z s  Zc
Z R  Zc
3.3 Open-Circuit Line Termination
•
•
•
•
Boundary condition for current i=0
Therefore if=-ib
Vb=Zcib=Zif=vf
Thus total voltage at the receiving end
v=vf+vb=2vf
• Voltage at the open end is twice the
forward voltage wave
3.4 Short Circuit Line Termination
•
•
•
•
Boundary condition for current v=0
Therefore vf=-vb
If=vf/Zc=-(vb/Zc)=ib
Thus total voltage at the receiving end
v=if+ib=2if
• Current at the open end is twice the
forward current wave
3.5 Termination Through Capacitor
2Z

Z  Zc
2(1 / Cs )

Z c  1 / Cs
v  v f
2(1 / Cs ) v f 2v f
1
v( s ) 

Z c  1 / Cs s
s Z cCs  1
2v f
1 / Z cC
1
1

 2v f 
s s  1 / Z cC
s s  1 / Z cC
So :
v(t )  2v f (1  e t / Z cC )
i (t ) 
2v f
Zc
e t / Z c C
vb (t )  v f (1  2e t / Z cC )
3.6 Termination Through Inductor
v (t )  2 v f e  ( Z c / L ) t
i (t ) 
2v f
Zc
(1  e
( Z c / L )t
)
vb (t )  v (t )  v f (t )
vb (t )  v f ( 2e
( Z c / L )t
 1)
4. Junction of Two Line
if 
vf
v f  vb  v
Z c1
i f  ib  i
vb
ib 
Z c1
vb
v


Z c1 Z c1 Z c 2
v
i
Zc2
 Z c1 
v
2v f  1 
 Zc2 
vf
2Z c 2
v
vf
Z c1  Z c 2
2 Z c1
i
if
Z c1  Z c 2
Z c 2  Z c1
vb 
vf
Z c1  Z c 2
Z c1  Z c 2
ib 
if
Z c1  Z c 2
Pf 
v
2
f
Z c1
2
v
P
Zc2
2
b
v
Pb 
Z c1
5. Junction of Several Line
Example:
2v f
Zc2
v
Z c1  Z c 2 / 2 2
2Z c1
v
if
Z c1  Z c 2 / 2
if 
2v f
Z c1  Z c 2 / 2
6. Bewley Lattice Diagram