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Fundamental Element Modeling In Laplace
Resistance
+
V(s)
-
I(s)
)s(RI)s(V
1/sC I(s)
s
)0(V)s(I
sC
1)s(V c
c
Capacitance
-
Li(0)
+
I(s)
)0(Li)s(sLI)s(LV
Inductance
+
Vc(0)/s
-
+
Vc(s)
-
Ls
+
VL(s)
-
Patterson Power Engineers, LLC
Transform Pairs
22
cos
s
ssF
ttf
22s
sF
tsintf
Patterson Power Engineers, LLC
22MVs
s
+
L
C
C
CLV
Initial condition is upper cap bank shorted.
C=6.98uF CLV = 650x3 = 1950uF 1314563
2161000VM
Patterson Power Engineers, LLC
22MVs
s
+
L
C
C
CLV
When the switch is opened the current will be assumed to extinguish at current zero. Since the circuit is dominantly capacitive this will occur when the voltage source is at its peak, VM. Note that since the current will be zero there will be no stored energy in the inductor’s magnetic field.
Patterson Power Engineers, LLC
+
C
CLV
Calculate initial voltages across the two capacitors in the circuit prior to opening the switch.
The voltage across C = = 0.996 VM
LV
LVM
CC
CV
The voltage across CLV = = 0.03567 VM
LV
MCC
CV
Patterson Power Engineers, LLC
22MVs
s
+
sL
Laplace Circuit at t=0+
+
-
+
-
sC
1
sC
1
s
0.996VM
s
0.03567VM
LVsC
1
Patterson Power Engineers, LLC
Calculate the current after the top capacitor bank bypass switch is opened.
LV
M
22
M
sC
1
sC
2sL
s
V
377s
sV
sI
LV
2
LC
1
LC
2M Let
22
M
2222
2
M
Ms
LV
)M)(s377(s
LsVsI
M = 3200 rad/s
Patterson Power Engineers, LLC
22
M
2222
2
M
Ms
L
V
)M)(s377(s
LsVsI
Partial fraction expand the first term in I(s)
)M)(s377(s
LsV2222
2
M
jM)(s
*B
jM)(s
B
j377)(s
*A
j377)(s
A
Patterson Power Engineers, LLC
)M377)(s(s
LsV
22
2
M
j
2.85)377(M3772j
LV377
22
M
2
j
A =
s = j377
A =
Patterson Power Engineers, LLC
jM))(s377(s
LsV
22
2
M
j
j734)(2jM)j377(-M
LVM
22
M
2
B =
s = jM
B =
Patterson Power Engineers, LLC
22
M
Ms
MLM
V
jM)(s
*B
jM)(s
B
j377)(s
*A
j377)(s
AI(s)
22s
sF
tsintf
Now but the last term in the I(s) equation into a form that is easily inverse transformerd.
Patterson Power Engineers, LLC
)90cos(377tj85.22ti 0
Inverse Laplace transform back to time domain
in(3200t)1468- s
)90cos(3200tj734-2 0
22 3200s
0)(1468)(320
j3200)(s
j734
j3200)(s
j734-
j377)(s
j85.2-
j377)(s
j85.2I(s)
Patterson Power Engineers, LLC
t)sin()90tcos( 0 Since
t)sin()90tcos( 0
00t)1468sin(32-77t)170.4sin(3-00t)1468sin(32ti
note that damping was neglected, hence no exponential decaying term multiplied against the high-frequency term. Also, two terms cancel
Patterson Power Engineers, LLC
)90cos(377tj85.22ti 0
in(3200t)1468- s
)90cos(3200tj734-2 0
22
M
2222
2
M
Ms
LV
)M)(s377(s
LsVsI
)Ms(s
CLM
MV
)M)(s377(s
CLsVsV
22
M
2222
M
To calculate the voltage waveform across the upper cap bank after the switch is opened - multiply I(s) by 1/sC to get V(s) and inverse transform
Patterson Power Engineers, LLC
Partial fraction expand the first term in V(s)
)M)(s377(s
CLsV2222
M
jM)(s
*B
jM)(s
B
j377)(s
*A
j377)(s
A
* = complex conjugate
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)M377)(s(s
CLsV
22
M
j
33306)M(-3772
CLV
22
M
A =
s = j377
A =
Patterson Power Engineers, LLC
jM))(s377(s
CLsV
22
M
j
32840)(2)j377(-M
CLV
22
M
B =
s = jM
B =
Patterson Power Engineers, LLC
teAtf t cos2)(
js
A
js
AsF
*
expedient Transform Pair
Note that Ө is the angle associated with the factor A. e.g.
0θ|A|
Patterson Power Engineers, LLC
jM)(s
*E
jM)(s
E
s
D
)Ms(s
MCLM
V
22
M
Now expand the second term from V(s)
)M(s
MCLM
V
D22
M
s = 0
65685M
CLV2
M
)jMs(s
CLVM
E
s = jM
328402M
CLV-
)jM(2jM
CLV2
MM
Patterson Power Engineers, LLC
65685 cos(377t)333062tV
Notice that the “B” and “E” terms cancel and then Inverse Laplace transform back to time domain
jM)(s
*E
jM)(s
E
s
D
jM)(s
*B
jM)(s
B
j377)(s
*A
j377)(s
AV(s)
32840B
33306A
32840-E
65685D
6568577t)66612cos(3V(t)
This is the voltage across the top capacitor group from the moment of the bypass switch around it being opened
Patterson Power Engineers, LLC
This voltage oscillates between 0 and 130,370V. In the top capacitor bank the units are rated 15.92kV with 3 in series in each parallel string. Overall string rating would be 3 x 15.92kV = 47.76kVrms
The capacitors will experience rms pu69.1kV76.47
kV8.80
This voltage will last until the DC component discharges across the 8.56 MΩ discharge resistors – on the order of minutes. Following slide suggests they don’t have that long…
kV8.8065685)2
66612(V(t)rms 22
Patterson Power Engineers, LLC
6568577t)66612cos(3V(t)
pu69.1kV76.47
kV8.80The capacitors will experience rms overvoltage
From IEEE Std 1036-2010, “IEEE Guide for Application of Shunt Power Capacitors”
Patterson Power Engineers, LLC
RMS voltage profile of top cap bank over 300 seconds after switch open
80.8kV/47.76kV = 1.69p.u.
At 15 seconds 66.9kV/47.76kV = 1.40p.u.
Same data but only first 15 seconds
Patterson Power Engineers, LLC
0
0.5
1
1.5
2
2.5
0.01 0.1 1 10 100 1000 10000
1036-2010
Simulation (top cap bank RMS)
Patterson Power Engineers, LLC
Comparison of top cap bank RMS voltage profile against IEEE 1036-2010 short-time overvoltage chart
M0.996V6568577t)66612cos(3V(t)
The voltage across the bottom capacitor bank will be similar except it will have the additional term of initial voltage added (see slide 7 for initial conditions).
6524577t)66612cos(3V(t)
This voltage oscillates between 0 and 128,240V. In the bottom capacitor bank the units are rated 19.1kV with 5 in series in each parallel string. Overall string rating would be 5 x 19.1kV = 95.5kVrms
The capacitors will experience
kV5.8065245)2
66612(V(t)rms 22
This equates to 0.84 p.u. of rated.
Patterson Power Engineers, LLC
6568577t)66612cos(3V(t)
Voltage across bottom cap bank = green Voltage across top cap bank = red
Voltages from simulation after bypass switch opens = match closed form solution.
Patterson Power Engineers, LLC
M
LV
22
LV
M
2222
LVMLV V
)Cs(C
C
)Ms(s
LMC
MV
)M)(s377(s
LCsVsV
To calculate the voltage waveform across the LV capacitors after the switch is opened - multiply I(s) by 1/sCLV and add the initial voltage to get VLV(s) and then inverse transform
Patterson Power Engineers, LLC
Partial fraction expand the first term in VLV(s)
)M)(s377(s
LCsV2222
LVM
jM)(s
*B
jM)(s
B
j377)(s
*A
j377)(s
A
* = complex conjugate
Patterson Power Engineers, LLC
)M377)(s(s
LCsV
22
LV
M
j
119)M(-3772
LCV
22
LV
M
A =
s = j377
A =
Patterson Power Engineers, LLC
jM))(s377(s
LCsV
22
LV
M
j
117)(2)j377(-M
LCV
22
LV
M
B =
s = jM
B =
Patterson Power Engineers, LLC
jM)(s
*E
jM)(s
E
s
D
)Ms(s
LCV
22
LV
M
Now expand the second term from V(s)
)M(s
LCV
D22
LV
M
s = 0
235M
LCV2
LVM
)jMs(s
LCV LVM
E
s = jM
1172M
LCV-
)jM(2jM
LCV2
LVMLVM
Patterson Power Engineers, LLC
M
LV
VCC
C234 cos(377t)1192tV
Notice that the “B” and “E” terms cancel and then Inverse Laplace transform back to time domain
M
LV
V)Cs(C
C
jM)(s
*E
jM)(s
E
s
D
jM)(s
*B
jM)(s
B
j377)(s
*A
j377)(s
AV(s)
117B
119A
117-E
234D
234cos(377t)228V(t)
This is the voltage across the LV capacitors from the moment of the bypass switch around the top capacitor group being opened.
Patterson Power Engineers, LLC
Waveform from simulation shows close agreement
Voltage across LV capacitance after top cap bank bypass switch opened.
Patterson Power Engineers, LLC
234cos(377t)228V(t)
Alternate approach to find the voltage across the top capacitor bank
377t)-170.4sin(ti
+
v(t)
-
i(t)
0K1
)( 1
0
vdiC
tv
dt
tdvCti
t
1K1
)(V
diC
t
t
1K377sin4.1701
)(V
dC
t
t
1K)377cos(4.170377
1)(V t
Ct
1K)377cos(64756)(V tt
Patterson Power Engineers, LLC
Find constant of integration from boundary conditions at t=0+
1K)0cos(647560
64756K1
64756-)377cos(64756)(V tt
compares very well (within round off) to the previous solution shown on slide 21
Patterson Power Engineers, LLC
1K)377cos(64756)(V tt