ECE 524:
Transients in Power Systems
Session 18; Page 1/8
Spring 2014
ECE 524: Lecture 18
Define units:
MW  1000kW
MVA  MW
f  60Hz
ω  2 π f
6
μsec  10
s
MVAr  MVA
rad
ω  376.99 
s
t  0 1.5μsec 
1
sec
60
Transient recovery voltage reduction with resistive switching
VLL  34.5kV
L1  3.157mH
If we include
source resistance
with X/R=12
Cpara  12.7nF
ωn 
fn 
2
V
3 LL
Vm 
1
ω  L1
12
 0.0992 ohm
5 rad
ωn  1.58  10 
L1 Cpara
ωn
s
fn  25.14 kHz
2 π
With no TRV reduction (and neglecting circuit resistance)
vcNoR ( t) 
Worst case:
ωn
2
2
ωn  ω
2
Vm  cos ( ω t)  cos  ωn t 
Vcpk 
ωn
2
2
ωn  ω
2
 Vm  2
RRRV  2 Vm 
fn
0.5
Vcpk  56.34 kV
RRRV  2.83
kV
μsec
ECE 524:
Transients in Power Systems
Session 18; Page 2/8
Spring 2014
Select a TRV reduction resistor to reduce the TRV by 40%
We know (if the resonant frequency is high and the response is underdamped which
would be for this amount of reduction):


vc ( t) = Vm  1  e
 α t
1
2 R C
So we want to see:
where:
α=




MAX Vm 1  e
 α t


 cos  ωd t 
ωd =


  cos  ωd t 
2
α

sin  ωd t 
ωd

ωn  α
2
α

sin  ωd t  = ( 1  0.4) 2pu = 1.2
ωd

The maximum occurs when the derivative with respect to t is equal to zero:


α exp ( α t)  cos  ωd  t 
α

 sin  ωd t   exp ( α t)  sin  ωd t ωd  α cos  ωd t 
ωd

This simplifies to:
e
 α t
 α2

  sin  ωd  t  ωd  sin  ωd t  = 0
 ωd

This is true whenever
which is when:
ωd  t
0 = sin  ωd t
is an integer multiple of . Use the first one.
So we want:
1e
for
 α t


 cos  ωd  t 
α

 sin  ωd t  = 1.2
ωd

ωd  t = π
Then we have:
1e
 α t
( 1  0) = 1.2
which simplifies to:
e
 α t
= 0.2
ECE 524:
Transients in Power Systems
Session 18; Page 3/8
Spring 2014
We now have three equations and three unknowns
Initial conditions
Rdamp  100ohm
ωd  260000
rad
s
tpk  10μsec
Given
2
1
 =ω

ωn  
d

 2 Rdamp Cpara 
2
 tpk


 2 Rdamp Cpara 
e
 = 0.2
ωd tpk = π
 Rd 


t
 peak   Find  Rdamp tpk ωd
 omega 
d

Rdamp  Rd
Rdamp  546.75 Ω
tpk  tpeak
tpk  22.35 μsec
5 rad
ωd  omegad
ωd  1.41  10 
fd 
α 
1


 α t
2 π
fd  22.37 kHz
α  72007.53
2  Rdamp Cpara
vc ( t)  Vm  1  e
ωd
s


 cos  ωd  t 
1
s
α

sin  ωd t 
ωd

ECE 524:
Transients in Power Systems
Session 18; Page 4/8
Spring 2014
Plots of MathCAD Results:
Tn 
4
610
1
fn
 Tn 
  56.34 kV
2
4
vcNoR ( t) 410
vcNoR 
4
210
0
0
5
5
410
810
t
4
410
Td 
1
fd
4
310
vc ( t)
 Td 
  33.8 kV
2
vc 
4
210
4
1.2Vm  33.8 kV
110
0
0
5
410
RRRVdamp  1.2 Vm 
5
810
t
RRRVdamp  1.51
ATP Results (note that source resistance included):
UI
0.0992 ohm
3.157 mH VC
546.75ohm
UI
FAULT
VS
28.169kV pk
12.7nF
RBREAK
2
Td
kV
μsec
ECE 524:
Transients in Power Systems
Session 18; Page 5/8
Spring 2014
Switch voltage without resistor in the circuit (with source resistance included)
60
*10 3
40
20
0
-20
-40
-60
0.00
0.02
(f ile TRV.pl4; x-v ar t) v :VC
0.04
-FAULT
0.06
0.08
*10
0.10
v :VS
Zoom in on the peak voltage (if no source resistance-slight decrease Vmax=56.16kV))
60
Vpk  56.340kV
*10 3
50
40
30
Vs  28.168kV
20
10
0
4.16
4.17
4.18
(f ile TRV.pl4; x-v ar t) v :VC
-FAULT
4.19
4.20
4.21
4.23 *10 -3 4.24
4.22
v :VS
Resonant frequency (estimated from time between peaks)
T1  3.9659 10
3
sec
T2  4.0057 10
3
sec
f0 
1
T2  T1
f0  25.13 kHz
ECE 524:
Transients in Power Systems
Session 18; Page 6/8
Spring 2014
Repeat with a DC offset in the current when breaker clears:
50
*10 3
40
30
20
10
0
-10
-20
0.00
0.02
(file Example.pl4; x-var t) c:VC
0.04
0.06
0.08
0.10
-FAULT
60
*10 3
40
20
0
-20
-40
-60
35
45
55
(file Example.pl4; x-var t) v:RBREAK-FAULT
Vpk  53.773kV
65
75
*10 -3
85
ECE 524:
Transients in Power Systems
Session 18; Page 7/8
Spring 2014
Now add the TRV resistor
35
Vmax  33.691kV
*10 3
Vmax
32
Vm
29
26
23
20
3.6
3.7
(f ile TRV.pl4; x-v ar t) v :VC
3.8
-FAULT
3.9
4.0
4.1
4.3 *10 -3 4.4
4.2
v :VS
 EMTDC Results
546.75 [ohm]
0.0127 [uF]
R=0
IF
VC1
BRK2
0.003157 [H]
IR
VS 0.0992 [ohm]
BRK2
Timed
Breaker
Logic
Closed@t0
BRK1
BRK1
Timed
Breaker
Logic
Closed@t0
 1.2
ECE 524:
Transients in Power Systems
Session 18; Page 8/8
Spring 2014
Main : Graphs
60
VC1
VS
50
40
30
y (kV)
20
10
0
-10
-20
-30
-40
0.1020
0.1025
0.1030
0.1035
0.1040
0.1045
0.1050
0.1055
0.1060
Main : Graphs
40
VC1
VS
30
y (kV)
20
10
0
-10
0.10390
0.10392
0.10394
0.10396
0.10398
0.10400
0.10402
0.10404