Closed Form Solutions 2-stage cap bank -...

Closed Form Solutions 2-stage cap bank

4-11-2013

Patterson Power Engineers, LLC

[email protected]

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)

-


Transform Pairs

22

cos

s

ssF

ttf

22s

sF

tsintf


22MVs

s

+

L

C

C

CLV

Initial condition is upper cap bank shorted.

C=6.98uF CLV = 650x3 = 1950uF 1314563

2161000VM


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.


+

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


22MVs

s

+

sL

Laplace Circuit at t=0+

+

-

+

-

sC

1

sC

1

s

0.996VM

s

0.03567VM

LVsC

1


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


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


)M377)(s(s

LsV

22

2

M

j

2.85)377(M3772j

LV377

22

M

2

j

A =

s = j377

A =


jM))(s377(s

LsV

22

2

M

j

j734)(2jM)j377(-M

LVM

22

M

2

B =

s = jM

B =


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.


)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)


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


)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


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


)M377)(s(s

CLsV

22

M

j

33306)M(-3772

CLV

22

M

A =

s = j377

A =


jM))(s377(s

CLsV

22

M

j

32840)(2)j377(-M

CLV

22

M

B =

s = jM

B =


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|


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


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


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


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”


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


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)


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.


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.


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


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


)M377)(s(s

LCsV

22

LV

M

j

119)M(-3772

LCV

22

LV

M

A =

s = j377

A =


jM))(s377(s

LCsV

22

LV

M

j

117)(2)j377(-M

LCV

22

LV

M

B =

s = jM

B =


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


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.


Waveform from simulation shows close agreement

Voltage across LV capacitance after top cap bank bypass switch opened.


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


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


1K)377cos(64756)(V tt

Closed Form Solutions 2-stage cap bank -...

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