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Shunt Reactor Switching: Theory and Practice

David Peelo

DF Peelo & Associates Ltd

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Shunt Reactor Switching

• Shunt reactor circuit representation and oscillation calculation

• Generalized circuit breaker TRV calculation

• Type testing

• Practical cases

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Reactor Circuit Representation

• Reactors are represented by series RLC oscillatory circuit with a pre-charged capacitor

• The circuit oscillation is underdamped with a high amplitude factor of 1.9 pu due to the reactors being low loss devices

• Frequency of the oscillation is a few kHz for applications at 72.5 kV and above and tens of kHz at 52 kV and below

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Reactor Oscillation Calculation

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Series RLC with C precharged to Vo

Solve for the capacitor voltage v:

01

2

2

vLCdt

dv

L

R

dt

vd

LCc

L

Rba

1 1

LCL

R 1

2

R

L

V0C

i

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6

Underdamped case: >

2

1

tktkev t sincos 21

Initial conditions:

• at t 0, v Vo k1 Vo

• at t 0, 0 k2 Vodt

dv

tteVv t

o sincos

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Now to generalize the equations: damping coefficient and

generic time

1. Damping coefficient ds: define as /

Cs RRd

where RC is the resistance value which gives critical

damping

C

LR

LCL

RC

C 21

2

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2. Generic time tg: define as where tp is the angular

frequency period without damping

LCtp

pg ttt

gtLCL

Rt

2

g

C

tR

R

gstd

gtLC

tt

ptt

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9

gs

s

sgs

td

pu tdd

dtdev gs 1sin

1 1cos 2

2

2

Underdamped case: R < RC ds < 1

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Generic damping curves: vpu vs tg

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

Generic time tg (Radians)

Oscilla

tio

n a

mp

litu

de (

pu

) R/Rc = 10

R/Rc = 5

R/Rc = 2

R/Rc = 1.5

R/Rc = 1

R/Rc = 0.75

R/Rc = 0.5

R/Rc = 0.2

R/Rc = 0.1

R/Rc = 0

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Generalized Circuit Breaker TRV Calculation

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Calculation of application requirements and overvoltages:

• Given the reactor rating and type plus the grounding

arrangement, what are the requirements to be

specified for the circuit breaker?

• What are the magnitudes of the overvoltages

generated by the switching event?

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Suppression peak overvoltages

• Circuit breakers have no difficulty interrupting the current, typically a few hundred amperes at high voltages

• Current is actually forced to zero prior to the natural zero crossing and phenomenon known as current chopping

• This leaves a trapped current and energy in the reactor resulting in the suppression peak overvoltage

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To calculate the magnitude of the suppression peak overvoltage, consider the

general case for shunt reactor switching

Given the above three-phase representation for shunt reactor, we need now to derive the

first-pole-to-clear representation. CN is considered to have no influence

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Reactor switching general case: first pole to clear considerations

chopping current ic significant

K > 0

arc voltage a significant

calculate the value of the suppression peak overvoltage Vma (ka in pu)

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First pole to clear equivalent circuit:

Looking back from load circuit arc voltage is additive since it is in phase

with current and of opposite polarity to Vo

CL L(1+K)

CB

VO(1+K)

iC

va

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kin = initial voltage at current

interruption (pu)

kin =

=

a = Vo(kin – 1)

o

ao

V

V

o

a

V1

VO

Arc voltage

va

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Source voltage Vs at instant of current chopping is given by:

Vs = Vo (1 + K) + Vo (kin – 1)

= Vo (kin + K)

CL L(1+K)

CB

VO(1+K)

iC

va

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Maximum overvoltage Vm occurs when all of the available

energy is stored in CL:

2c

2

inoL2mL i K1L

2

1KkVC

2

1VC

2

1

o

mb

V

Vk In pu:

2

c2oL

2

in iVC

)K1LKk

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Suppression peak overvoltage to ground ka:

Kkk ab

o

maa

V

Vk w here

Kkk ba

KiVC

)K1LKk 2

c2oL

2

in

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Three cases:

1. Reactor neutral grounded through a neutral reactor

(LN) 0 < K < 0.5:

General equation for ka applies

L

LN

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2. Reactor neutral directly grounded K 0:

L

2

o

c2ina

C

L

V

ikk

L

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3. Reactor neutral ungrounded K 0.5:

5.0V

i

C

L5.15.0kk

2

o

c

L

2

ina

L

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Now to consider:

1. What is the transient recovery voltage across

the circuit breaker?

2. In the event of a reignition, what are the

associated overvoltages?

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After current interruption the load side circuit will ring down and the breaker will be stressed by the voltage difference between

the source and the load sides

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Load side oscillation circuit

Cp1 Lp1 Rp1

CB

RL

Lp2

CS

LS

CL L

Rp2

VS

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Load side oscillation general case

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

Generic time tg

Oscilla

tio

n a

mp

litu

de (

pu

) R/Rcd = 10

R/Rcd = 5

R/Rcd = 2

R/Rcd = 1.5

R/Rcd = 1

R/Rcd = 0.75

R/Rcd = 0.5

R/Rcd = 0.2

R/Rcd = 0.1

R/Rcd = 0

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

Generic time tg

Oscilla

tio

n a

mp

litu

de (

pu

) R/Rcd = 10

R/Rcd = 5

R/Rcd = 2

R/Rcd = 1.5

R/Rcd = 1

R/Rcd = 0.75

R/Rcd = 0.5

R/Rcd = 0.2

R/Rcd = 0.1

R/Rcd = 0

+1 pu = ka

0 pu = - Κ

-1 pu = - (ka +Κ)

+1 pu = ka + Κ

0 pu

-1 pu = - (ka +Κ)

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Circuit breaker TRV

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General equation for TRV

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TRV equation

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Equation shows the interactive nature of the duty:

• V peak voltage at the reactor at current interruption

• K relates to the earthing of the shunt reactor neutral

• ds relates to the damping in the circuit (can be taken as

0.033 for amplitude factor of 1.9 pu)

• tg relates to the reactor sizing and type (oil-filled or dry type)

• ka relates to the shunt reactor application details and

characteristics of the circuit breaker

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Reignitions: sequence and magnitudes

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Reignition: first parallel oscillation circuit

Cp1 Lp1 Rp1

CB

RL

Lp2

CS

LS

CL L

Rp2

VS

Reignition first phase: grading or local capacitance discharges

through the interrupter; frequency is in the order 1 to 10 MHz and the

current will not be interrupted by the breaker

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Reignition: second parallel oscillation circuit

Cp1 Lp1 Rp1

CB

RL

Lp2

CS

LS

CL L

Rp2

VS

Reignition second phase: involves the local source and load side

capacitance circuits; frequency is the range 50 kHz to 1 MHz and

some breakers may interrupt the current possibly leading to

voltage escalation

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Reignition: main circuit oscillation circuit

Reignition final phase: this phase leads directly to another loop of

power frequency current; the oscillation is complex with a

frequency in the range of 5 to 20 kHz but actual occurrence is

dependent on the relative values of CS and CL

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Second and main circuit oscillations

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Reignition overvoltages

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Chopping number approach: applies for all circuit breaker

types except vacuum and is based on the following

equation:

where the chopping number in AF-0.5

N number of interrupters in series on the

breaker

CT total capacitance in parallel with the

breaker

Tc NCi

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LS

LsPT

CC

CCCC

LS

LSP

22c

CC

CCCNi

CB

CPCS CL

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V.A)in (Q IV3Q L

L

V5.1

L2

V

2

3V3

2ooo

Q

5.1

V

L or

L

V5.12o

2o

LI

VL

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We can now rewrite the general equation for ka:

KCC

C

C

C

Q

N K15.1Kkk

LS

S

L

P

22

ina

givingcase) onerous (most CC for 1CC

CSL

LS

S

K1C

C

Q

N K15.1Kkk

L

P2

2

ina

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As before above equation applies for the neutral reactor

grounding case

Reactor neutral directly ground K 0:

1

C

C

Q

N5.1kk

L

P2

2ina

Reactor neutral ungrounded K 0.5:

0.512.25

0.5L

P

22

ina

C

C

Q

Nkk

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Suppression peak overvoltage equation for SF6 circuit breaker

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K1C

C

Q

N K15.1Kkk

L

P2

2

ina

• kin takes circuit breaker arc voltage into account

• K is the neutral shift (zero if directly grounded, 0.5 pu if

ungrounded and usually around 0.3 pu if grounded through

neutral reactor)

• N is the number of interrupters in series on the circuit breaker

• λ is the chopping number of the circuit breaker

• Q is the rating of the shunt reactor

• ω is the angular frequency

• CP is the capacitance in parallel with the circuit breaker

• CL is the load side capacitance including that of the shunt

reactor.

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Chopping numbers vary by circuit breaker type and typical

values for single interrupters are:

The chopping number is not necessarily a constant

but can be dependent on the arcing time

Circuit Breaker TypeChopping Number

(AF-0.5)

Minimum oil 6 104 to 10 104

Air blast 15 104 to 20 104

SF6 puffer 4 104 to 20 104

SF6 self-blast 3 104 to 10 104

SF6 rotating arc 0.4 104 to 1 104

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0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40

Chopping number (AF0.5

)

Su

pp

res

sio

n p

ea

k k

a (

pu

)

5 Mvar

10 Mvar

20 Mvar

50 Mvar

100 Mvar

200 Mvar

500 Mvar

X 104x 104

Chopping number AF-0.5

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Ungrounded case considerations:

• Applies mainly at 52 kV and below but is common in

some countries up to 170 kV

• In some cases, the second term under the root sign

may be negligible and ka = kin i.e. only the arc voltage

is of significance

• Remember that the chopping number approach does

not apply to vacuum breakers and the basic equations

only should be used for these breakers

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Reignitions are detrimental to the circuit breakers. Why and what is the solution?

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SF6 circuit breaker cross-section

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Fault current interruption

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Reignitions during shunt reactor switching

• With fault current interruption, reignitions will occur between the arcing contacts due to the remnant arc channel between the contacts

• Shunt reactor currents are in the order a few hundred amperes at high voltages and no remnant arc channel exists between the arc contacts after current interruption with the result that reignitions can occur anywhere such as between a fixed arcing contact and a moving main contact

• Stray reignitions cause damage to nozzles and may lead to failure to interrupt the current

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Field trace: switching out a 525 kV shunt reactor using controlled switching

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Trace courtesy of BC Hydro

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Circuit breaker damage due to reignitions

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Upper photographs: nozzles

after 1500 operations no

control switching

Lower photographs: same

nozzle type after 650

operations with controlled

switching

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Type Testing

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TRV type test requirements for circuit breakers rated 100kV and above from IEC 62271-110

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For breaker selection at HV and EHV, the following information is needed:

• Laboratory reactor switching test report at a current equal to or less than the application reactor current. The test shall be done using a reactor and not a reactor loaded transformer

• The chopping number characteristic of the breaker as derived in the laboratory test. This is based on a statistical analysis of the suppression peak overvoltages and is usually expressed as an equation relating the chopping number to the arcing time!

• Proof of mechanical opening time consistency over the operating temperature range if controlled switching is to be considered

Note: Some means of overvoltage limitation should be provided.

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Shunt reactor switching tests

CL L

CB I

CSV

Test 1: I = 337 A

V = 141 kV rms

L = 1.125 H

CS = 20 nF

CL = 2 nF

Test 2: I = 84 A

V = 141 kV rms

L = 4.5 H

CS = 10 nF

CL = 1 nF

Courtesy of ABB Sweden

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Tests at 337 A and 84 A: chopped current levels

0

0.5

1

1.5

2

2.5

3

3.5

4

0 5 10 15

Arcing time (ms)

Ch

op

ped

cu

rren

t (A

)

337 A

84 A

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Chopping numbers tests combined

y = 7755.5x - 19859

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 2 4 6 8 10 12 14 16

Arcing time (ms)

Ch

op

pin

g n

um

be

r (A

F0.5

)

337 and 84 A

Linear (337 and 84 A)

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Transient recovery voltage imposed on the circuit breaker

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Transient recovery voltages (TRVs) have a power frequency and a transient component:

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Fault currents: system delivers both the

power frequency and transient

components and the TRV is the multiple

of the two components

Shunt reactor switching: system delivers

the power frequency component and the

shunt reactor circuit the transient

component and TRV is the difference of

the two components

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Practical Cases

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Case 1: 30 kV 100 Mvar 50 Hz reactor with vacuum circuit breaker and grounded neutral point

• Arc voltage negligible means kin = 1

• K = 0

• Calculate current, inductance and AC voltage peak value

• Calculate value for 1 pu tg with CL = 500 pF

• Write the equation!

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Case 2: 30 kV 100 Mvar 50 Hz reactor with vacuum circuit breaker and grounded neutral point ungrounded

• Arc voltage negligible means kin = 1

• K = 0.5

• Calculate AC voltage peak value

• Calculate value for 1 pu tg with CL = 500 pF

• Write the equation!

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TRV: 30 kV, 100 Mvar, 1925 A shunt reactor with earthed neutral and unearthed neutral with no remedial action

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Case 3: 30 kV 100 Mvar 50 Hz reactor with vacuum circuit breaker, RC damper and grounded neutral point

• Arc voltage negligible means kin = 1

• K = 0

• Calculate AC voltage peak value

• Calculate ds for R = 20 Ω and C = 0.15 µF

• Calculate value for 1 pu tg

• Write the equation!

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TRV: 30 kV, 100 Mvar, 1925 A shunt reactor with earthed neutral with remedial action

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Case 4: 525 kV 135 Mvar 60 Hz reactor with grounded neutral point

• Take CL = 1750 pF, CP/CL = 0.5, N = 2, λ = 61844 A/√F, kin = 1.08

• Calculate value for ka (K = 0)

• Calculate current, inductance and AC voltage peak value

• Calculate 1 pu tg value

• Write the equation!

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Case 5: 525 kV 135 Mvar 60 Hz reactor with neutral point grounded through a 1600 Ω reactor

• Take CL = 1750 pF, CP/CL = 0.5, N = 2, λ = 61844 A/√F, kin = 1.08

• Calculate value for K and ka

• Calculate AC voltage peak value

• Calculate 1 pu tg value

• Write the equation!

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TRV: 525 kV 135 Mvar shunt reactor with earthed neutral and grounded through a 1600 Ω reactor

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Read more …

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1. IEC 62271-306 High-voltage switchgear and controlgear – Part 306: Guide to

IEC 62271-100, IEC 62271-1 and other IEC standards related to alternating

current circuit–breakers.

2. IEC TR 62271-110 High-voltage switchgear and controlgear – Part 110:

Inductive load switching.

3. IEEE C37.015 Application Guide for Shunt Reactor Switching.

4. D.F. Peelo, R.P.P. Smeets and B.S. Sunga, “Shunt Reactor Stresses due to

Switching”, Paper No. A2-304, Cigre 2008 Session.

5. David F Peelo, “Current Interruption Transients Calculation”, (Book) John

Wiley & Sons Ltd., April 2014. (Second edition in March 2020).

6. Rene Smeets, Lou van der Sluis, Mirsad Kapetanovic, David Peelo and Anton

Janssen, “Switching in Electrical Transmission and Distribution Systems”,

(Book), John Wiley & Sons Ltd., August 2014.

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Thanks for listening!

Questions?

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