Fundamentals of Power System Protection_2012

Page 1

Course SynopsisFundamental Principles ofPower System Protection

May 2012© Barrie Moor

Slide 1

Fundamental Principlesof

Power System Protection

Slide 1Slide 1

Barrie Moor

[email protected]

bmoor@powersystemprotection.com.auwww.powersystemprotection.com.au

Slide 2

Disclaimer

The material presented in this module is for Educational purposes only.This module contains a summary of information for the protection of various types of electrical equipment. Neither the author, nor anyone acting on his behalf, makes any warranty or representation, express or implied, as to the accuracy or completeness of the information contained herein, nor assumes any responsibility or liability for the use or consequences of the use of any of this information.The practical application of any of the material contained herein must be in accordance with legislative requirements and must give due regard to the individual circumstances.

Page 2

Course SynopsisFundamental Principles ofPower System Protection


Slide 3

Course Synopsis

Fundamental Concepts of Protection DesignFault CalculationsOver Current & Earth Fault ProtectionVTs & CTsFundamentals of Distance ProtectionFundamentals of Protection SignallingFundamentals of High Impedance Differential ProtectionFundamentals of Transformer Biased Differential ProtectionFundamentals of Busbar Biased Differential ProtectionFundamentals of Feeder Differential ProtectionAuto ReclosingCapacitor Bank Protection

Seminar costs will vary depending on attendee numbers and your individual circumstances. However, by organising your own in-house seminar:

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including: pdf of seminar – colour slides 2 per page additional technical papers tools for sequence component analysis of single, double and three

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Page 1

Fault CalculationsSequence Components

Fundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

FAULT CALCULATIONS

An Introduction

Slide 2

Basic Calculations : 3 Phase Fault

ZS ZLVS VR IF

LS

SF ZZ

VI NL

+= −

LFR ZIVNL

•=−

Slide 3

Basic Calculations : Ph - Ph Fault

( )LS

SF ZZ2

VI LL

+•= −

= I3PH * √3 / 2

ZSVS

IF

ZL

Slide 4

Basic Calculations : Earth Fault

GLS

SF ZZZ

VI NL

++= −

ZSVS

IF

ZL

Multiple return paths

Slide 5

System Impedances and Fault Calculations

Transformers– Voltages reflected via turns ratio– Currents reflected inversely to turns ratio– Impedances reflected via (turns ratio)2

Slide 6

Per Unit Values

For those who think this is hard … you always operate in a per unit system !!1 Volt1 Ampere

A current of 1 Ampere flowing through a resistance of 1 Ohm produces a voltage drop of 1 Volt and an energy dissipation of …

SecondJoules

SecondElectrons18

ElectronJoules19 11025.6106.1 ⋅=⋅⋅⋅⋅⋅ −

= 1.6 x 10-19 Joules / Electron= 6.25 x 1018 Electrons / Second

Watt1 ⋅=

Page 2




Slide 7

Per Unit Values

But I manufacture 100MVA 132/66kV Transformers, so these figures of 1 Volt and 1 Ampere don’t reflect how I work

100 MVA = 1 pu– On the 132kV side … 1 pu voltage = 132 kV– On the 132kV side … 1 pu current = 437.4 A– On the 132kV side … 1 pu impedance = 174.24 Ω

– On the 66kV side … 1 pu voltage = 66 kV– On the 66kV side … 1 pu current = 874.8 A– On the 66kV side … 1 pu impedance = 43.56 Ω

Slide 8

Per Unit Values and Transformers

100 MVA132kV 66kV

2Ω12Ω

• Transfer impedances to the HV side (via (turns))2

• Convert impedances to per unit on 100 MVA base

20Ω

132kV 66kV

0.1148 pu

132kV 66kVpu⋅= 1148.0

24.17420

( ) 20212 266

132 =⋅+

Slide 9

Per Unit Values and Transformers

100 MVA132kV 66kV

2Ω12Ω

• Transfer impedances to the LV side (via (turns))2

• Convert impedances to per unit on 100 MVA base

5Ω

132kV 66kV

0.1148 pu

132kV 66kVpu⋅= 1148.0

56.435

( ) 5122 213266 =⋅+

Slide 10

Per Unit Values

So, with the same base MVAAnd base voltage equal to system voltagePer Unit impedances remain the same across the transformer

132kV 66kV11kV

NB !!NB !!

Slide 11

Per Unit Values

So, with the same base MVAAnd base voltage equal to system voltagePer Unit impedances remain the same across the transformer

132kV 66kV11kV

NB !!NB !!

ZTOT = 0.1 puZTOT = 0.1 pu IFAULT = 10.0 puIFAULT = 10.0 puV = 1.0 puV = 1.0 pu

10pu@66kV = 8748A10pu@66kV = 8748A10pu@132kV = 4374A10pu@132kV = 4374A10pu@11kV = 52500A10pu@11kV = 52500A

Slide 12

Per Unit Quantities : Change of Base

Manufacturer base quantities most likely will not match the base values we wish to use.We must use a constant MVA base across the entire model.Base voltages must match system voltages.For example : Consider a generator step up transformer, with the manufacturer nominal values …

– 75 MVA– 11/145 kV– 12.5% impedance on rating

But for fault study simulations, we have chosen– 100MVA base– 132kV system nominal voltage

Page 3




Slide 13


Transformer– 75MVA, 145kV, 12.5% impedance

Calculate base impedance– 1452 / 75 = 280.33 Ω

Convert transformer pu impedance to ohms– 280.33 * 12.5% = 35.04 Ω

System fault study– 100MVA, 132kV

Calculate base impedance– 1322 / 100 = 174.24 Ω

Hence, for our fault study simulation– 35.04 / 174.24 = 20.1%

Slide 14


2

NEW

OLD

OLD

NEWOLDNEW kV

kVMVAMVAZZ ⎟⎟

⎠

⎞⎜⎜⎝

⎛••=

2

NEW 132145

751005.12Z ⎟

⎠⎞

⎜⎝⎛••=

%1.20ZNEW =

Slide 15

Classical Fault Study

Pre-fault voltages set to 1/0°Pre-fault load currents ignoredTransformers on system voltage tap (eg. 132/66kV to match system voltages,even if transformer nominal tap is say 132/69kV)Shunt impedances ignored (Shunt capacitors, etc)Zero ohms fault resistanceGenerator Sub-transient reactance

– Assumes generator contribution to the fault remains at its maximum

This is adequate for setting of protection relays– Relay setting calculations will all be on the basis of the same fault

level data, and hence coordination is achieved– “C” factor of 1.1 usually applied in determining equipment ratings

Slide 16



Slide 16Slide 16

SEQUENCE COMPONENTS

An Introduction

Slide 17

Sequence Components

Positive Sequence– A B C– Equal in magnitude– 120 degrees apart

Negative Sequence– A C B– Equal in magnitude– 120 degrees apart

Zero Sequence– A B C– Equal in magnitude– In phase

V1

V2

V0

I1

I2

I0

Slide 19

Sequence Components

I1 I2 I0

I phase

IC

IB

IA

Page 4




Slide 20

Sequence Components

210 IAIAIAIA ++=

210 IBIBIBIB ++=

01201a ∠=

210 ICICICIC ++=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

2

11

111

IAIAIA

aaaa

ICIBIA

212

0 IAaIAaIA ⋅+⋅+=

22

10 IAaIAaIA ⋅+⋅+=

Slide 21

Sequence Components

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

IAIAIA

AICIBIA

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ICIBIA

aa1aa1111

31

IAIAIA

2

2

2

1

0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2

1

0

2

2

IAIAIA

aa1aa1111

ICIBIA

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ICIBIA

A1

IAIAIA

2

1

0

Slide 22

Sequence Networks

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

V1 = 1 / 0º V2 = 0 V0 = 0

Slide 23

Sequence ComponentsThree phase conditions

Positive sequence only

– Three phase load– Three phase fault– No neutral (earth fault) current

Slide 24

In = 0

3 Phase Balanced Current

Balanced currents “sum to zero”– Positive sequence currents– Negative sequence currents– But zero sequence will sum to 3.Io In = 3.Io

Slide 25

Sequence Networks3 Phase Fault

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

fZsZZ

ZVII

POS

POSFaultPhase

11

1__3

+=

==

Page 5




Slide 26

Sequence ComponentsPhase – Phase fault

Positive and Negative sequence components only– And consider the special case where …

A phase equal in magnitude but opposite in phase

– B to C Phase to Phase fault

Slide 27

Sequence Networks (A phase)Phase – Phase fault

A phase– IA1 & IA2 antiphase

Sum to zeroB phase

– IB1 & IB2 at 60o

C phase– IC1 & IC2 at 60o

IB = - IC

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

21 II =

Slide 28

Sequence Networks (A phase)Phase – Phase fault

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Since Z1 ~ Z2

|I1| = |I2| = 50% of 3 phase fault level

negpos ZZVII+

== 21

Slide 29

Sequence ComponentsPhase – Phase fault

|I1| = |I2| = 50% of 3 phase fault levelThus |IB| = |IC| = 86.6% of 3 phase fault level(because of 60o angles)

Slide 30

Phase to Phase Example

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

V2 = 0V1 = 1 / 0º

= 0.10 = 0.10

= 0.15= 0.15

I = 2.0

V1 = 0.5

V1 = 0.8 V2 = 0.2

V2 = 0.5

Slide 32

Sequence ComponentsEarth Fault

A phase positive sequencenegative sequencezero sequence

Equal in magnitude and phase

Page 6




Slide 33

Sequence ComponentsEarth Fault

Slide 34

Sequence NetworksA Phase Earth Fault

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

Slide 35

Sequence NetworksA Phase Earth Fault

PositiveSequence

FaultLocation

NegativeSequence

FaultLocation

ZeroSequence

FaultLocation

RelayLocation

ZS1

Zl

ZS2

Zl

I2I1RelayLocation

ZS0

Zl

I0RelayLocation

zeronegpos ZZZVIII

++=== 021

021 IIIIA ++=

0=IB

0=IC

03 II NEUT ∗=

Slide 36

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

Phase to Ground Example

= 0.10 = 0.10

= 0.15= 0.15

= 0.15

= 0.35

I = 1.0 I = 1.0 I = 1.0

V1 = 1 / 0º V2 = 0

V0 = -0.15

V0 = -0.50

V2 = -0.10

V2 = -0.25

V0 = 0

V1 = 0.90

V1 = 0.75

Slide 37

Sequence NetworksResistive Earth Fault

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

FaultResistance3 x R_fault3 x R_fault

Earth Fault current is 3 x IoHence, voltage drop across the fault resistance will be 3 x Io x R_faultBut the sequence model only has Io flowingSo, include 3 x R_fault in the model

Slide 38

Sequence ComponentsSummary

Positive Sequence– Balanced three phase load– Balanced three phase fault– No neutral (earth) current

Negative Sequence– Unbalanced load– Phase to phase fault– No neutral (earth) current

Zero Sequence– Earth fault– Neutral current = 3 . Io– Cannot flow into or out of a delta– Can circulate around (within) the delta

Io = 0

Io = 0

Io = 0

IoIoIo

3Io

Page 7




Slide 39

Positive Sequence Network

LVZS

ZS HVZ1HL

HV

LV

Zfdr

Zfdr

Slide 40

Negative Sequence Network

ZS

ZS Z1HL

LV

HV

HV

LV

Zfdr

Zfdr

Slide 41

Zero Sequence Network

ZS

ZS Z1HL

LV

HV

HV

LV

Zfdr

Zfdr

Page 1

Over Current ProtectionFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

Over Current Protection

Overcurrent RelaysDirectional RelaysFuses & Fuse Contactors

Slide 2

Over Current Protection

Over Load Protection– Operation to the thermal capability of plant

Over Current Protection– Primarily for clearance of faults– Some measure of over load protection may be

provided

Slide 3

Discrimination by Time

Setting chosen to ensure CB nearest to the fault opens firstOften referred to as …“Independent Definite Time Delay Relay”Timing intervals selected to ensure upstream relays do not operate before CBs trip at fault locationDisadvantage …Longest fault clearing time occurs in section closest to the power source where fault level is the highest

Slide 4

Discrimination by Time

RELAY ‘A’ RELAY ‘B’ RELAY ‘C’

RELAY ‘C’

RELAY ‘B’

RELAY ‘A’

CURRENT

TIM

E

0.4 secs

0.4 secs

Slide 5

Discrimination by Current

Apply where fault current varies with fault location due to intermediate impedanceSet to operate at current values so that only relay nearest to fault trips its CBDifficulties– Same fault level at the end of one zone and the start

of the next– Fault levels vary with changing source impedance

(eg. As generators come on and go off line)

Slide 6


RELAY ‘A’ RELAY ‘B’

Relay ‘A’ cannot distinguish between a fault here, for which it needs to operate

And a fault here for which it should not operate

Page 2



Slide 7


Significant difference between currents seen for Faults A & BSet HV OC to 1.3 x maximum through current for LV Fault

HV OC

FDR OC

FDR OC

FDR OC

FDR OC

A

B

Slide 8

Discrimination by Time & Current

Time and current coordination


RELAY ‘C’

RELAY ‘B’

RELAY ‘A’

CURRENT

TIM

E

ICmax IBmax IAmax

IAmax IBmax ICmax

Instantaneous element

Slide 10

Inverse Over Current Relays

Time of operation inversely proportional to fault current– Faster operating times at higher fault levels– Faster operating times for faults nearer to the source

Curves generally plotted in log - log or log(current) – linear(time) format

Slide 11

Discrimination withInverse Time Over Current Relays

Inverse time and current coordination


RELAY ‘C’

RELAY ‘B’RELAY ‘A’

CURRENT

TIM

E

ICmax IBmax IAmax

IAmax IBmax ICmax

Instantaneous element

Slide 12

Relay Curves to IEC 60255(BS142)

I = Actual relay currentRelay Settings

– TMS = Time Multiplier Setting– P = Plug (Current) pickup setting

Usual curve for transmission and distribution systems

1PI

TMS14.0TIME 02.0Inverse_dardtanS

−⎥⎦⎤

⎢⎣⎡

•=

Slide 13




Systems where the fault level decreases significantly between relaying points

1PI

TMS5.13TIME Inverse_Very

−⎥⎦⎤

⎢⎣⎡•

=

Page 3



Slide 14




Grading with fuses

1PI

TMS80TIME 2Inverse_Extremely

−⎥⎦⎤

⎢⎣⎡

•=

Slide 15




Long time thermal protection– Motor & Generator Protection

1PI

TMS120TIME Inverse_Time_Long

−⎥⎦⎤

⎢⎣⎡•

=

Slide 16

Standard Characteristics to IEC 60255

Long Time (LTI)

Extremely Inverse (EI)

Very Inverse (VI)

Standard Inverse (SI)

Relay Characteristic

1ITMS14.0

02.0 −•

1ITMS5.13

−•

1ITMS80

2 −•

1ITMS120−•

100 1 .103 1 .1040.1

1

10

100

Standard InverseVery InverseEtremely InverseLong Time Inverse

IDMT Relay Grading Curves

Fault Current

Seco

nds

Slide 18

US Characteristics to IEC 60255

U5 Short Time Inverse

U4 Extremely Inverse *

U3 Very Inverse

U2 Inverse

U1 Moderately Inverse

Relay Characteristic

⎥⎦⎤

⎢⎣⎡

−+•

1M0104.00226.0TD 02.0

⎥⎦⎤

⎢⎣⎡

−+•

1M95.5180.0TD 2

⎥⎦⎤

⎢⎣⎡

−+•

1M88.30963.0TD 2

⎥⎦⎤

⎢⎣⎡

−+•

1M64.502434.0TD 2

⎥⎦⎤

⎢⎣⎡

−+•

1M00342.000262.0TD 02.0

TD = Time dial (TMS)M = Multiple of pick-up current

Slide 19

Electro Mechanical Relays

disc.

FLUX PRODUCED BY INPUT

TAPPEDCOIL Φ I

Φ- L

SHADING LOOP

FLUX PRODUCED BY INPUT CURRENT

DISC DISC

FLUX PRODUCED BY SHADING LOOP

Φk I

Φ L

(1-k)Φ I

Page 4



Slide 20

Electro Mechanical RelaysConstruction

Current (Plug) Settings

Time multiplier (TMS) Settings (continuous adjustable)

Shaded Pole

Trip Contacts andFlag

Trip Contacts

Moving Contact

Slide 21

Electro Mechanical RelaysInverse Definite Minimum Time

Current sensitivity selected by “Plugs” on the magnetic circuit– Higher sensitivity selected via more turns – ie. Same Ampere Turns operating quantity …

more turns = less currentMagnetic circuit saturates at extreme over current

– Limits the minimum operating time– Typically around 20x plug setting– Hence IDMT performance : DM = definite minimum

Time coordination via Time Multiplier setting– Adjusted the starting point of the induction disk wrt the fixed trip

contact– Often called “Lever Setting”

disc.

100 1 .103 1 .1040.1

1

10

100Standard Inverse Relay Grading Curves

Fault Current

Seco

nds

Adjust TMS to achieve time coordination

And since we are usually interested in operating times of 3 seconds or less, we may get a better perception if we use a linear axis for time

100 1 .103 1 .104

0.5

1

1.5

2

2.5

3

3.5

4Standard Inverse Relay Grading Curves

Fault Current

Seco

nds

If we have sufficient

margin here

Then with the same characteristic, we tend to have greater margin at lower currents due to

divergence of the curves

Slide 24

IDMT Curves

Electromechanical relays– Must not pick up at < 1.00 pu current– Must pick up at > 1.30 pu current– May not have well defined characteristics between

1.3 and 2.0 pu currentElectromechanical relays tend to a definite minimum time at high currents, say > 20 x ISET– Due to saturation of their magnetic circuits

Microprocessor based relays will have a genuine definite minimum time.

OC OC

OC

OC

100 1 .103 1 .104

0.5

1

1.5

2

2.5

3

3.5

4IDMT Relay Grading Curves

Fault Current

Seco

nds

0.4 SecondsMargin

0.4 SecondsMargin

Page 5



Slide 26

Instantaneous Element

Reduces tripping time at high fault levelsAllows a the discriminating curves behind the high set element to be lowered

– Grading of upstream relay now occurs at the instantaneous setting and not at maximum fault level

Minimises fault damage in both cases

Beware …Simple E/M instantaneous elements may have a substantial transient overreach on fault currents that include DC offset

OC OC

100 1 .103 1 .104

0.5

1

1.5

2

2.5

3

3.5


Fault Current

Seco

nds

100 1 .103 1 .104

0.5

1

1.5

2

2.5

3

3.5


Fault Current

Seco

nds

100 1 .103 1 .104

0.5

1

1.5

2

2.5

3

3.5


Fault Current

Seco

nds Set Tx HV inst

element and now grade here

OC

OC

Slide 29

Relay Coordination Procedure

Start with selection of relay characteristic– As far as possible, use relays of the same characteristic

Choose current settings– Determine maximum load current limitations– Determine starting current requirements– As far as possible, select operating current of each upstream relay greater

than that of the successive downstream relayCoordinate relays via time multipliers to achieve appropriate grading margins

– Determine, under various system configurations, the values of short circuit current that will flow through each protective device

– Set relays to give minimum operating time at maximum fault currents– Check performance (discrimination) at lower fault levels

Plot and coordinate relay curves on log/log or log/linear format– Plot to a common current base (across transformers)

Earth faults are considered separately and require separate plots

Slide 30

Relay Current Pick-up Setting

Set above maximum load current– Allow for emergency loading conditions– Allow safety margin– Allow for relay reset ratio

Set below the current pickup level of the next “upstream” relayAllow for load pickup current

Slide 31

Load Pickup Current

Motor starting currentAuxiliary heatersTransformer magnetising inrushCapacitor charging currentLighting loads - 10s to 100s of msec– Filaments and electrodes heating– Arc lamps starting

Slide 32

Load Pickup Current

Hot load pickup– Short term loss of supply and subsequent load pickup

currents on return of supplyCold load pickup– Load pickup, but now with loss of diversity between

cyclic loadsVoltage recovery pickup– Pickup currents not as severe as for complete loss of

supply and subsequent hot load pickup– But more motors may still be on-line as under voltage

releases may not have disconnected them

Page 6



Slide 33

Relay TMS Grading

Must provide for– CB tripping time (0.1 sec ??)– Relay timing errors– Relay overshoot– CT errors (10% ??)– Safety margin (10% ??)

A typical figure of 0.3 - 0.4 seconds is usually OK– 0.3 for numerical relays– 0.4 for electromechanical relays

Alternatively calculate a margin– Only necessary for slow tripping times (> 1.0 sec)

Slide 34

Relay TMS Grading

Slide 35

Relay TMS Grading

Hence for an E/M relay tripping in 0.5 seconds– t’ = (7.5 + 7.5 + 10)% x 0.5 + 0.1 + 0.05 + 0.1– t’ = 0.375 seconds

0.30.30.350.4Typical margin (s)

0.030.030.050.1Safety Margin (s)

0.020.020.030.05Overshoot Time (s)

5557.5Timing Error %

NumericalDigitalStaticElecto-Mechanical

Relay Technology

CT Errors

Slide 36

Grading of Parallel Elements

Worst case for grading is with only 1 transformer in serviceBut this will be an unusual operating conditionE/M & Electronic Relays

– Only a single relay setting is available– Hence, effectively no option but to set for the worst case,

namely 1 transformer case– And accept slower performance for system normal,

namely when both transformers are in serviceMicroprocessor based relays

– These relays have multiple setting groups– So, maybe set Group 1 for system normal : 2 transformers– And change to group 2 when one transformer is OOS

Automatically ??Via SCADA & operator intervention ??

OC OC

OC OC

OC

Slide 37

Grading of Parallel Elements

Maximum through fault level occurs when both transformers are in serviceBut the maximum individual transformer current flows when the 2nd transformer is OOSNeed to consider both conditions when grading relays

HV OC

HV OC Fdr_1 OC

3φ Fault Levels• 2 Tx IN : 16000A• 1 Tx IN : 12000A

3φ Fault Levels• 2 Tx IN : 10000A • 1 Tx IN : 7500A

33kV 11kV

20MVA

20MVA

300A FLC

800A FLC

Fdr_2 OCSI 400ATMS 0.2

Fdr1_TMS 0.28=Fdr1_TMS round Fdr1_TMS .003+ 2,( ):=Round Up

Fdr1_TMS 0.276=

Fdr1_TMS 1Fdr1_Tmin

Fdr1_TMS_1⋅:=Hence we can calculate the required TMS to achieve the required tripping time

Fdr1_TMS_1 2.971=This would result in a tripping time of

Fdr1_TMS_1 SI Fdr1_Plug 1.0, Imax,( ):=Assume TMS = 1.0

Fdr1_Tmin 0.821=Fdr1_Tmin Fdr2_Tmin 0.4+:=Required tripping time

ImaxFdr1_Plug

10=Fdr1_Plug 1000:=So select settings for Feeder 1

Fdr2_Tmin 0.421=Fdr2_Tmin SI Fdr2_Plug Fdr2_TMS, Imax,( ):=Tripping time at maximum fault level

Fdr2_TMS 0.2:=

ImaxFdr2_Plug

25=Fdr2_Plug 400:=Given data for Feeder 2

Imax 10000:=Grade Fdr_1 OC over Fdr_2 OC at the maximum through fault level of 10kASet Fdr_1 OC above maximum feeder load of 800Aand check against maximum fault level of 10kA

SI P TMS, I,( )0.14 TMS⋅

IP

⎛⎜⎝

⎞⎠

0.021−

:=Relay Characteristic

Page 7



Feeder 1 Relay_2n SI Fdr2_Plug Fdr2_TMS, I2n,( ):= SI Fdr2_Plug Fdr2_TMS, Imax,( ) 0.421=

Feeder 2 Relay_1n SI Fdr1_Plug Fdr1_TMS, I1n,( ):= SI Fdr1_Plug Fdr1_TMS, Imax,( ) 0.832= ∆_T 0.411=

100 1 .103 1 .104 1 .1050

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

Fdr 2 OCFdr 1 OC

Tx OC Grading (11kV Base Currents)

Tx_HV_TMS 0.36=Tx_HV_TMS round Tx_HV_TMS .003+ 2,( ):=Round up

Tx_HV_TMS 0.355=

Tx_HV_TMS 1Tx_HV_Tmin

Tx_HV_TMS_1⋅:=Hence we can calculate the required TMS to achieve the required tripping time

Tx_HV_TMS_1 3.297=This would result in a tripping time of

Tx_HV_TMS_1 SI Tx_HV_Plug 1.0, Imax,( ):=Assume TMS = 1.0

Tx_HV_Tmin 1.169=Tx_HV_Tmin Fdr1_Tmin 0.4+:=Transformer HV OC

Fdr1_Tmin 0.769=Fdr1_Tmin SI Fdr1_Plug Fdr1_TMS, Imax,( ):=Fdr Tripping time at maximum fault level

Tx_HV_Plug 1500=

Tx_HV_Plug 3 Tx_HV_Plug⋅:=Allow for 33/11kV ratio

Tx_HV_Plug 500:=Set130 %⋅ FLC_33kV⋅ 455=FLC_33kV 350=FLC_33kV20000000

3 33000⋅:=

Imax 12000:=

Grade Transformer HV OC under the maximum current condition, namely with one transformer OOS

Feeder 1 Relay_1n SI Fdr1_Plug Fdr1_TMS, I1n,( ):= SI Fdr1_Plug Fdr1_TMS, Imax,( ) 0.769=

Tx HV Relay_3n SI Tx_HV_Plug Tx_HV_TMS, I3n,( ):= SI Tx_HV_Plug Tx_HV_TMS, Imax,( ) 1.187= ∆_T 0.418=

100 1 .103 1 .104 1 .1050

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

Fdr 2 OCFdr 1 OCTx HV OC

Tx OC Grading (11kV Base Currents)

Slide 42

Resetting of Over Current Relays

Electromechanical relays tend to have a slow reset of their operating mechanism

– eg. For the disk to rotate back to the “stops”Other relays may have

– Disk simulation resetting– Instantaneous resetting

This becomes important in auto-reclosing schemes– E/M relays may not have fully reset before re-application of

a fault via autoreclose– They will thus be partially integrated to the trip and will

require less and less time to reach a trip on each successive reclosure

Slide 43

Sequential Operation of Over Current Relays

As CBs trip, fault current magnitudes and flows will changeWe need to integrate how far each relay progresses towards tripping in each stage

– To determine total tripping times– To ensure relays that should not trip, remain stable

Relay 1 operating time must have a suitable margin above the total of Relay 2 and the subsequent Relay 3 operations

Relay 3Relay 2

Relay 1

Slide 44



Slide 44Slide 44

DirectionalOver Currentand Earth FaultProtection

Page 8



Slide 45

Directional Over Current Relays

Extra discrimination may be achieved by making the response of the relay directional when current can flow in both directionsAchieved via voltage (polarising) connections to the relayDigital and numeric relay achieve phase displacements via softwareEM & Static relays require suitable connection of input quantities to the relay

Slide 46

Directional Over Current RelaysApplication to Parallel Feeders

Apply directional relays at the feeder receiving ends– Typically set to 50% of FLC, TMS = 0.1– Grade below non-directional relays at the source end– Ensure DOC relay thermal rating is OK

OC OC

OCOC

Fdr 1

Fdr 2BA

Slide 49

Earth Fault Protection

Implement more sensitive protection responding only to residual current of the systemLow settings are permissible and beneficial

– Earth faults are the most frequent– Earth faults may be limited by earth fault resistance– Earth faults may be limited by neutral earth impedance

Typical settings 20 - 40% x FLCTime grade in the same manner as for phase OC relaysBeware of the burden that electromechanical relays may place on CTs at low current settings

– Although burden does decrease at very high currents with saturation of the relay’s magnetic circuits

Slide 50


EF Prot

EF Prot

OC

OC

OC

EF Prot

OC

OC

Not suitable for 2:1:1 Current applications(Transformer HV

current in the case of Star/Delta or Delta/Star through phase/phase

faults)

Slide 51

Directional Earth Fault Protection

Voltage Quantity required to polarise relayUse the system residual voltage– This is the vector sum of all three phase voltages– This is thus three times the zero sequence voltage

This voltage will be zero for balanced system voltages– Normal Load conditions– Three phase events– 2 phase events not involving earth

Slide 52

Directional Earth Fault Protection

3 . V0 is obtained from a VT with the secondary connected in a broken deltaPrimary star point of VT must be earthedAnd to provide the path for zero sequence flux …VT must be …– 5 limb core type– 3 x 1 phase units

Va

VbVc

3.Vo

Page 9



Slide 53



Slide 53Slide 53

FusesandFuse Contactors

Slide 54

Fuses

Performance effectively follows I2t law– Pre Arcing time– Arc time

Fuse – Fuse grading requires that the total I2t of the smaller fuse be less than the pre-arcing I2t of the larger fuse

Slide 55

FuseBullrush Curves

Discrimination between fuse links is achieved when the total I2t of the minor fuse link does not exceed the pre-arcing I2t of the major fuse linkBut note that this applies only for high speed operation where there is no heat dissipation …ie. I2t adiabatic performanceAs a starting point, a current rating ratio between fuses of 1.6 - 2 is probably OK (but this depends on the specific fuse design)

100

160

200

250

Fuse Rating

403532 50 63 1 25

80

minimumpre-arcingI2t

maximumtotal I2t

Slide 56

Expulsion Fuses

Used where expulsion gases cause no problem such as in overhead circuits and equipmentSpecial materials (fiber, melamine, boric acid, liquids such as oil or carbon tetrachloride ) located in close proximity to fuse element and arc rapidly create gasesThese produce a high pressure turbulent medium surrounding the arcExpulsion process deionises gases them as well as removing them from ‘arc area‘In inductive circuits, transient recovery voltage (TRV) will be maximum at current zero.

Slide 57

Fuses & TRV Performance

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.0242

1.5

1

0.5

0

0.5

1

1.5

2

Circuit VoltageFuse VoltageCurrent

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.0242

1.5

1

0.5

0

0.5

1

1.5

2


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.0242

1.5

1

0.5

0

0.5

1

1.5

2


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.0242

1.5

1

0.5

0

0.5

1

1.5

2


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0.022 0.0242

1.5

1

0.5

0

0.5

1

1.5

2


Current lags Voltage by 90

deg

System Voltage

Current interrupted at

natural current zero

TRV across blown fuse

element

Fuse Voltage

Slide 58

Current Limiting Fuses (HRC Fuses)

Fuse is designed to insert a large resistance– Hence, prospective level of fault current is reduced– And zero crossing of the current and voltage will be

reasonably in phase – TRV significantly reducedFuse element is completely surrounded with filler material, typically silica sand

– Arc energy melts the sand, thus inserting the required high resistance

But this design may have difficulty interrupting low level overloads.Overcome by …

– M Effect designs– Spring assisted designs

Page 10



Slide 59

Current Limiting Fuses

Tin for “M Effect” low overload fuse performanceSee later

Slide 60


Slide 61

Current Limiting FusesM Effect for low level overloads

M Effect : A.W. Metcalf - 1939

Slide 62


Slide 63

Grading Relays with Fuses

Extremely Inverse curve follows a similar I2t characteristicRelay current setting should be approximately 3 times the fuse ratingGrading margin of not less than 0.4 seconds recommended

Or …

EI P TMS, I,( )80 TMS⋅

IP

⎛⎜⎝

⎞⎠

21−

:=

15.04.0' +⋅≥ tT

Slide 64

Grading Relays with Fuses

First relay upstream of the fuse should be set to EI characteristicNow to coordinate further upstream relays …– Option 1 : Also select EI characteristics– Option 2 : Check also for the possibility of setting

The next relay to a VI characteristicAnd subsequent further upstream relays to SI characteristics

Page 11



Slide 65

Fuse Contactors

High fault level applications eg …– 40kA fault level– Contactor rated to only 10kA– Fuse operates for all faults above say 7 kA– Contactor and associated protection relay operate for

lower fault levels– Warning … the fuse may also have a minimum

breaking capacity and the contactor must be set to operate above this point

Slide 66

100 1 .103 1 .104 1 .1050.01

0.1

1

10

100

FuseRelay / ContactorFuse

Fuse Contactors

10kA Contactor operates for faults

below 7kA

Fuse operates for faults above 7kA

Fuse operation below 2kA is not

permissible

Page 1

Voltage TransformersCurrent Transformers



Slide 1



Slide 1Slide 1

VOLTAGE andCURRENTTRANSFORMERS

Slide 2

DIST

Basic Concepts of Distance Protection

Measure V/IIf this falls below some preset value, a fault is detected and relay operates

faultZ=I

V

Zfault

Slide 3



Slide 3Slide 3

VOLTAGETRANSFORMERS

Specification to AS60044.1

Slide 4

Specification of VTsAS1243 (Superseded by AS60044)

555P

222P

111P

Phase displacement (crad)

Percentage voltage ratio errorProtection Class

1 crad = 34.4 mins1 crad = 34.4 mins

Slide 5

AS60044 : Specification of VTs

Percentage voltagePercentage voltage

2402402404806.06.06.012.06P

1201201202403.03.03.06.03P

FV10052FV10052

Phase displacement(minutes)

Ratio error(percent)

Protection Class

Slide 6


Protection VTs to operate between– V = 0.05 pu– V = Voltage Factor : FV pu

8 h1.9

Phase – earth in isolated neutral system

Continuous1.230 s1.9

Phase – earth in non- effectively earthed neutral system w e/f tripping

Continuous1.230 s1.5

Phase – earth in effectively earthed neutral system

Continuous1.2SystemRated TimeRated Voltage Factor

Page 2




Slide 7


Voltage Error– KN = rated transformation ratio– UP = actual primary voltage– US = actual secondary voltage

Phase displacement– Primary and secondary voltage phase difference– Said to be positive when the secondary voltage leads

the primary

100%__ •⋅−

=P

SNP

UUKUErrorVoltage

Slide 8



Slide 8Slide 8

VOLTAGETRANSFORMERS

Transient Performance

Slide 9

VT Fundamentals

Magnetic VTs– HV systems

Capacitor VTs– EHV systems

Requirements of VT plus Relay– In Zone Faults– Out of Zone Faults– Switching

Slide 10

Magnetic Voltage Transformer Energisation & De-energisation

Minimal problems with magnetic VT transient performance– Transient effects typically short term

Energisation– Flux Doubling depending on POW switching– OK since VT’s are designed to operate at low flux densities

(also minimises errors in normal operation)De-energisation

– Flux cannot immediately decay to zeroPrimary Fault

– Collapse of voltage on fault occurrence– Recovery of voltage on fault clearance

Slide 11

Capacitor Voltage Transformer

Use a voltage divider principle to reduce system HV voltage to a lower levelAnd then use a lower ratio transformer to …– provide final step down ratio to protection relay– provide galvanic isolation

Voltage divider implemented via capacitorsLoading effects eliminated via series tuning choke

Slide 12

V

RL . VRH + RL

RL

RH

N:1

ZL

Resistive losses – heating effectsPerformance varies with load burden

Voltage Divider Principles

Voltage across RH varies with current supplied to VT burden, ZL

Page 3




Slide 13

Thevenin Equivalent Circuit

RL.RHRL+RH

RL.VRL+RH

ZL

Equivalent to resistors in parallel !! Intermediate VT voltage

source reduces as burden current increases => errors !!

Slide 15

V

CH

CL ZL

N:1X

Resistive losses – nilPerformance at 50Hz does not vary with load burden

CH . VCH + CL

Capacitor Voltage Transformers : CVTs

Slide 16

CVT Thevenin Equivalent Circuit

CH+CL

ZLCH.VCH+CL

X

Equivalent to capacitors in parallel !!

• Capacitive divider and series tuning choke with identical impedance at 50Hz.

• Impedances cancel.

Slide 17

CVT Thevenin Equivalent Circuit

CH+CL

ZLCH.VCH+CL

X

• Capacitive divider and series tuning choke with identical impedance at 50Hz.

• Impedances cancel.• CVT loading effects eliminated

CH.VCH+CL

Slide 19

CVT Transient Performance

Resonances– Low frequency transient response between …

Intermediate VT magnetising branchThevenin equivalent of main capacitors

– High frequency transient response between …Tuning chokeCapacitance of intermediate VT

Slide 20

CVT Transient Performance

Resonant effects minimised via simple resistive damping

CH+CL

ZLCH.VCH+CL

N2ZL

High Frequency

Low Frequency

RP XP XS RS

XM RM CM

X

Page 4




Slide 21

CVTs and Distance Relay Performance

Distance Relay Reach Measurement– Accuracy at Zone reach

Distance Relay Directionality– Immunity to tripping on reverse faults

Distance Relay Problems– Spurious operation on transients

Eg. On simple de-energisation of feeder

Modern Relays– May have special facilities to provide for CVTs

Slide 22



Slide 22Slide 22

CURRENTTRANSFORMERS

Specification to AS60044.1andComparison with AS1675

Slide 23

Specification of CTsAS1675 (Superseded by AS60044.1)

Class P CTs– Equivalent to IEC P Class CTs

General purpose protection CT– Not usually used in HV systems– Not usually used in high speed

differential systems– Suitable for slower speed systems

where perhaps a few cycles of distorted output will not seriously affect relay performance (eg IDMT & Def Time relays)

Not generally intended for applications requiring good transient performanceTurns compensation is permissible

10 P 60 F15

Composite error % at accuracy limit current

Composite error % at accuracy limit current

Secondary ref voltage

at ALF

Secondary ref voltage

at ALF

Accuracy limit factorAccuracy limit factor

Slide 24


Class P CTsComposite Error

– RMS value of the errors in the instantaneous values of the actual secondary current expressed as a percentage of the nominal secondary current

– (ie. Basically the CT error, but via instantaneous values to allow for both magnitude and phase errors) .. See next slide !!

Secondary reference voltage– RMS value of the secondary terminal voltage on which the

performance of the CT is basedAccuracy limit factor

– Factor (applied to the rated primary current) for the accuracy limit of the CT

– ie. Factor (applied to the rated primary current) for which the CT will comply with the requirements for composite error

– If not specified … F20 is assumed

Slide 25


Class P CTsComposite Error

KN = Rated transformation ratioIP = RMS value of primary currentiP = instantaneous value of primary currentiS = instantaneous value of secondary currentT = duration of 1 cycle

( )∫ ⋅−⋅⋅⋅=T

PSNP

C dtiiKTI

E0

21100

Square of sum of squares to compute the RMS value of the

difference between the instantaneous values of the nominal and actual currents

Slide 26

Specification of CTsAS1675 comparison with AS60044.1

To convert P Class specification to IEC Specification

– AS1675 Class P CT 100/5 5 P 60 F 20

– IEC Class P CT 100/5 15 VA Cl 5 P 20

Composite Error

Composite Error

Page 5




Slide 27



– AS1675 Class P CT100/5 5 P 60 F 20

– IEC Class P CT100/5 15 VA Cl 5 P 20

Accuracy Limit Factor

Accuracy Limit Factor

Slide 28



– AS1675 Class P CT100/5 5 P 60 F 20

– IEC Class P CT100/5 15 VA Cl 5 P 20

Terminal Voltage Specified

at FAULT Current

Connected Burden Specified

at LOAD Current

Slide 30

CT Assignment : Class P CTs

What is the maximum fault level (Primary Amps) where the CT performance is guaranteed?

A Current transformer, for general purpose use in an application NOT requiring good transient performance has been specified, as per AS60044.1 as …

200/1 5VA Class 5 P20

CTpri x ALF = 200 x 20 = 4000A

Slide 31


At this fault level, what is the maximum secondary burden [ie. leads plus relay(s)] in OHMS that the CT can supply.


200/1 5VA Class 5 P20

Rated Burden = VA x 1 x 1 = 5 x 1 x 1 = 5 OhmsA A 1 1

Slide 32


What will the CT terminal voltage be under this condition


200/1 5VA Class 5 P20

Vct = I x R = (Isec x ALF) x R = (1 x 20) x 5 = 100V

Page 6




Slide 33


What is the equivalent (ie. old) AS1675 specification for this CT


200/1 5VA Class 5 P20

200/1 5P 100 F20

Slide 34


Is this CT adequate to supply the connected burden of 6 ohms at this site specific fault level of 3kA.

At fault levels below ALF, we can confirm operation provided rated terminal voltage is not exceeded

It turns out that at the particular substation where this CT is installed, the maximum fault level is actually only 3kA.The connected burden of leads and relay is however 6 ohms.

200/1 5VA Class 5 P20

Vct = I x R = 3000 x 6 = 15 x 6 = 90V … OK !!200

Slide 35


Comment on the suitability of using this CT at the 3kA fault location, but now with a connected burden of 7 ohms.

At fault levels above ALF, or if rated terminal voltage is exceeded, we also need to know CT internal resistance to confirm suitability.

Now, tests on the CT reveal that it has an internal resistance of 1 ohm.

200/1 5VA Class 5 P20

Vknee = Vterm + Ialf x Rct = 100 + 20 x 1 = 120V

Vreq = I x Rtot = 15 x (7 + 1) = 15 x 8 = 120V … OK

Slide 42

Class PX CTs : AS60044.1

Class PL CTs under AS1675Applications requiring good transient performance

– High accuracy high speed schemes

General– Jointless core wound from

continuous strip– Turns for each section of the

winding to be uniformly distributed

– Turns compensation not permissible

0.1 PX 200 R5

Magnetising current at knee point voltage


CT knee point voltage


CT internal resistance


10% increase in voltage requires a 50% increase in magnetising current

Slide 43

CTs with multiple tappings

Because the Class PX CT has a uniformly wound secondary winding, it may be possible to utilise intermediate ratiosFor CT with n terminals, ½.n.(n-1) ratios may be available

– CT specified as 2400/2000/800/1– May also provide 400/1200/1600/1

Beware of simple interpolation between ratios … but in general

– Resistance α Ratio (number of turns)– Voltage α Ratio (number of turns)– Magnetising current α 1/Ratio (inverse of turns)

Recommended to confirm performance of intermediate ratios with the manufacturerCT continuous rating

– Rating of primary conductor– Rating of secondary : perhaps 2 x In

2400

2000

800

1200/1

Slide 44



Slide 44Slide 44

CURRENTTRANSFORMERS

Transient Performance

Page 7




Basic CT Requirement : No Transients

( )V ICT

R R RKNEEFAULT

RATIOCT LEADS RELAY= ⋅ + +

Ifault

Isec

Rct RleadsVk Rrelay

v t( ) Vm sin ω t⋅( )⋅:= E Ldidt⋅:=

i t( ) Im sin ω t⋅π2

−⎛⎜⎝

⎞⎠

⋅:=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.14

3

2

1

0

1

2

3

4Fault Current : Inductive Power System

Seconds

v t( )

i t( )

t

Current lags voltage by 90ºCurrent cannot changeinstantaneously


Fault_Point 0.025= i t( ) if t Fault_Point< 0, Im sin ω t⋅π2

−⎛⎜⎝

⎞⎠

⋅,⎛⎜⎝

⎞⎠

:=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.14

3

2

1

0

1

2

3


Seconds

v t( )

i t( )

t

OKI = 0 at fault inceptionI lags V by 90 deg



−⎛⎜⎝

⎞⎠

⋅,⎛⎜⎝

⎞⎠

:=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.14

3

2

1

0

1

2

3


Seconds

v t( )

i t( )

t

NOT OKI is not 0 at fault inception


Fault_Point 0.020= i t( ) if t Fault_Point<( ) 0, Im sin ω t⋅π2

−⎛⎜⎝

⎞⎠

⋅ DC_Offset+,⎡⎢⎣

⎤⎥⎦

:=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.14

3

2

1

0

1

2

3


Seconds

v t( )

i t( )

t

DC_Offset 100% Im⋅=

NOW OKI = 0 at fault inceptionI lags V by 90 degBut now there is DC offset



−⎛⎜⎝

⎞⎠

⋅,⎛⎜⎝

⎞⎠

:=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.14

3

2

1

0

1

2

3


Seconds

v t( )

i t( )

t

NOT OKI is not 0 at fault inception

Page 8





Fault_Point 0.030= i t( ) if t Fault_Point<( ) 0, Im sin ω t⋅π2

−⎛⎜⎝

⎞⎠

⋅ DC_Offset+,⎡⎢⎣

⎤⎥⎦

:=

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.14

3

2

1

0

1

2

3


Seconds

v t( )

i t( )

t

DC_Offset 100− % Im⋅=

NOW OKI = 0 at fault inceptionI lags V by 90 degBut now there is DC offset

Slide 56

DC Offset in the fault current waveform

Faults occurring away from voltage peak will result in DC offset in the fault current waveformUp to 100% DC offset is possibleDC offset may be positive or negativeBut the power system is not purely inductive, so the DC offset will not continue but will decay exponentially

Primary Transient Fault Current

System Parameters R 1 L .1 =LR

0.1 sec =.ω L 31.416

=φ 88.177 deg =.ω L

R31.416

θ φ .90 deg =θ 1.823 deg

2

1

0

1

2

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

AC ComponentDC Component

SECONDS



0.03 sec =.ω L 9.425

=φ 83.943 deg =.ω L

R9.425

θ φ .90 deg =θ 6.057 deg

2

1

0

1

2

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2


SECONDS



0.01 sec =.ω L 3.142

=φ 72.343 deg =.ω L

R3.142

θ φ .90 deg =θ 17.657 deg

2

1

0

1

2

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2


SECONDS

Slide 60

DC Offset in the fault current waveform

DC component of the fault current will magnetise the CT coreIf the CT core becomes fully magnetised (ie. above its knee point) it cannot transform the primary current to a proportional secondary quantity.

Page 9




⎥⎥⎦

⎤

⎢⎢⎣

⎡+⋅=+

systempower

systempowerACDCAC R

XpeakPEAK

_

_)( 1φφ

5

0

5

10

15

20

25

30

35

0 0.1 0.2 0.3 0.4 0.5

AC FluxDC FluxTotla Flux

CT Flux (times AC component)

Seconds

Total Flux : AC Component plus DC ComponentTotal Flux : AC Component plus DC Component

AC FluxDC FluxTotal Flux

B

H

V

Imag

CT terminal voltage is low based on E = N dφ/dtie Simply enough to drive current through the connected burdenBut in specifying a high flux capability we, by default, have a high voltage capabilityThe high knee point is required because of the flux, not the voltage requirements

Slide 63

⎥⎦⎤

⎢⎣⎡ +⋅=+ 1

11)( RX

peakPEAK ACDCAC φφ

CT Specification to provide forTotal Flux : AC & DC Components

For purely sinusoidal quantities, the VT voltage and flux requirements are directly related

So, we can effectively specify the AC & DC flux requirements by specifying the proportional AC sinusoidal voltage requirements

NV⋅

=ω

φ maxmax

Slide 64

V V XRKNEE ACpeak peak

= ⋅ +⎡⎣⎢

⎤⎦⎥

1 11

V V XRKNEE ACrms rms

= ⋅ +⎡⎣⎢

⎤⎦⎥

1 11

( )V ICT

R R RKNEEFAULT

RATIOCT LEADS RELAY= ⋅ + +

To provide for the AC Component only

( )V ICT

XR

R R RKNEEFAULT

RATIOCT LEADS RELAY= ⋅ +⎡

⎣⎢⎤⎦⎥⋅ + +1

To provide for the AC & DC Components

⎥⎦⎤

⎢⎣⎡ +⋅=+ 1

11)( RX

peakPEAK ACDCAC φφ

Specify CT Voltage requirement to provide for the AC & DC Flux requirements

5

0

5

10

15

20

25

30

35

0 0.1 0.2 0.3 0.4 0.5

AC FluxDC FluxTotla Flux

CT Flux (times AC component)

Seconds

( )V ICT

XR

R R RKNEEFAULT

RATIOCT LEADS RELAY= ⋅ +⎡

⎣⎢⎤⎦⎥⋅ + +1

AC FluxDC FluxTotal Flux

Slide 66

CT Transient Performance

CT must cope with exponentially decaying DC component of fault currentNormal practice is to allow transient factor of (1 + X/R)– At Relaying Point– Or at Zone 1 Reach Point

Beyond the scope of our discussion– CT saturation, after relay operation, may be acceptable– Modern microprocessor based relay algorithms may

accommodate some CT saturation

Page 10




Slide 67

CT Assignment : Class PX CTs

A distance relay is employed on a simple radial 132kV system, with source and feeder impedances as per sketchAll figures are on 100 MVA base.System operates at nominal voltage – ie. Source voltage = 1 /0ºDistance relay Zone 1 will be set to 85% of the outgoing feeder

ZS = 0.05 /85º Zfdr = 0.10 /75º

132kVDistance Prot

Slide 68


Calculate the relay current (in Amps) for a 3φ fault occurring just in front of it, namely, at the relaying point.Calculate the fault current (in Amps) for a 3φ fault occurring at the proposed 85% zone 1 reach point.

ZS = 0.05 /85º Zfdr = 0.10 /75º

132kVDistance Prot

Slide 69

Ibase100 MVA⋅

132 kV⋅ 3⋅:= Ibase 437.387= Zs rect 0.05 85 deg⋅,( ):=

Zfdr rect 0.10 75 deg⋅,( ):=

Ztot Zs:=

Ifault1

Ztot:= Ifault 1.743 19.924i−= Ifault 20=

arg Ifault( ) 85− deg=

XR tan arg Ifault( )−( ):= XR 11.43=

Ifault Ifault Ibase⋅:= Ifault 8748=

3φ Fault at Relaying Point

Slide 70

Ibase100 MVA⋅

132 kV⋅ 3⋅:= Ibase 437.387= Zs rect 0.05 85 deg⋅,( ):=

Zfdr rect 0.10 75 deg⋅,( ):=

Ztot Zs 85 %⋅ Zfdr⋅+:=

Ifault1

Ztot:= Ifault 1.457 7.29i−= Ifault 7.434=

arg Ifault( ) 78.701− deg=



3φ Fault at Zone 1 85% Reach Point

Slide 71


Lead resistance is 2 ohms (loop total)Connected relay burden is 1 ohmFor the fault at the relaying point, check and comment on the transient performance of the CT.For the fault at the proposed Zone 1 reach point, check and comment on the transient performance of the CT.

ZS = 0.05 /85º Zfdr = 0.10 /75º

132kVDistance Prot

VT = 132000/110 voltsCT = 600/1 0.1 PX 600 R 5VT = 132000/110 voltsCT = 600/1 0.1 PX 600 R 5

3251AX/R=5

8748AX/R=11

Slide 72

CT Performance : 0.1 PX 600 R53φ Fault at Relaying Point

Ztot Zs:=

Ifault1

Ztot:= Ifault 1.743 19.924i−= Ifault 20=

arg Ifault( ) 85− deg=



Half of lead loop resistance for 3φ fault

CT 600V knee point is INADEQUATE

CT6001

:= IrelayIfaultCT

:= Irelay 14.58=

Rct 5:=Leads 2:=Relay 1:=

Vk Irelay 1 XR+( )⋅ Rct 0.5 Leads⋅+ Relay+( )⋅:= Vk 1269=

Page 11




Slide 73

CT Performance : 0.1 PX 600 R53φ Fault at Zone 1 85% Reach Point

Ztot Zs 85 %⋅ Zfdr⋅+:=

Ifault1

Ztot:= Ifault 1.457 7.29i−= Ifault 7.434=

arg Ifault( ) 78.701− deg=



Half of lead loop resistance for 3φ fault

CT 600V knee point is ADEQUATE

CT6001

:= IrelayIfaultCT

:= Irelay 5.419=

Rct 5:=Leads 2:=Relay 1:=

Vk Irelay 1 XR+( )⋅ Rct 0.5 Leads⋅+ Relay+( )⋅:= Vk 228=

Slide 74

Transient Performance Equation Aspects

How to determine X/R of power system?– Current angle from fault study simulation

Transient performance for close-in faults?– High fault level & high X/R

Transient performance for Zone 1 faults?– Lower fault level & lower X/R due to fdr impedance.

Transient performance for 3 phase faults?– No neutral return current in CT secondary wiring

Transient performance for 1 phase faults– Neutral return current in CT secondary wiring

Page 1

Distance ProtectionFundamentals of Performance


October 2010© Barrie Moor

Slide 1



Slide 1Slide 1

DISTANCEPROTECTION

Fundamentals of Performance

VI= +Z Zline loadHealthy Conditions:-

Zs I

V

Z line

Z load

VI= Z faultFault Conditions:-

Zs

V

I

Z fault

Z line

Z load

Slide 4

DIST

TIME

DISTANCE RELAY TIME DISCRIMINATION

LOCATION

ZONE 3

ZONE 2

LOCATION

ZONE 1

DIST

DIST

TIME

ZONE 1

ZONE 2

ZONE 2

ZONE 3

ZONE 1

ZONE 3

Distance Relay Zones

Slide 5

Distance Relay : Basic Scheme

Basic Scheme

ZONE 2MEASURER

ZONE 2TIMEDELAY

ZONE 3MEASURER

ZONE 3TIMEDELAY

ZONE 1MEASURER

>1 TRIP

Slide 6

DIST DIST DIST DISTK L M

Zone 1 Zone 1

Zone 2 Zone 2

Zone 3

Time and Reach Coordinationof Distance Relays

TimeCoordination

TimeCoordination

ReachCoordination

ReachCoordination

Page 2




Slide 7

ZONE 2

NON-SWITCHED DISTANCE RELAY OPERATION

A-B

ZONE 2TIMER

TRIP

ZONE 1C-EA-E B-E A-B B-C C-A A-E B-E C-E B-C

ZONE 3TIMER

A-EB-C C-AZONE 3

B-E C-E A-B C-A

Non Switched Distance Relay

Slide 8

ZONE REACHSWITCHING NETWORK

ZONE 2TIMER

TRIP

B-CA-E B-EZONE 1C-E A-B C-A

ZONE 3

ZONE 3TIMER

A-E C-EB-E C-AB-CA-B

Zone Switched Distance Relay

Fdr

Z1

Z3

Z1/2

Slide 9

ZONE 2TIMER

INCREASEIMPEDANCE

SETTINGTO

ZONE 2 REACH

VOLTAGE&

CURRENTSWITCHINGNETWORK

C

VOLTS

N

TRIP

CURRENT

MEASURER

VOLTS

N

CURRENT

BA

CBA

E

STARTERS

A B C

INCREASEIMPEDANCE

SETTINGTO

ZONE 3 REACH

ZONE 3TIMER

STARTERTIMER

Fully Switched Distance Relay

Slide 15

Primary & Secondary Impedances

Vsecondary = Vprimary / VTratio

Isecondary = Iprimary / CTratio

ratio

ratio

primary

primary

ondarysec

ondarysec

VTCT

IV

IV

•=

ratio

ratioprimaryondarysec VT

CTZZ •=

Slide 16

Simple Distance Relay Comparator

So, firstly, provide a ZSEC replica impedance within the relay to establish the relay’s zone of operation

– A real element of resistive and inductive components in an electromechanical relay

– An algorithm in a microprocessor based relayAnd measure VSEC from the VTAnd measure ISEC from the CTThe actual fault impedance will be given by VSEC / ISEC

If VSEC / ISEC < ZREPLICA, fault is in the zone and relay tripsIf VSEC / ISEC > ZREPLICA, fault is out of zone and relay restrainsBut, how can we easily perform the V / I calculation and comparison with Z

Slide 17

Simple Distance Relay Comparator

Pass CT secondary current (“I”) thru relay replica impedance “Z”– This develops a relay internal, or “replica” voltage “IZ”– So, the R / X diagram has become a IR / IX voltage diagram, with

“I” simply being a constant of proportionality (Note: I at 0º)

R

X ZV

I

II

I

V

Page 3




Slide 18

Operating & Restraining QuantitiesSimple Distance Relay Comparator

Hence, simply compare magnitudes of V and I*Zr– Trip if V < I*Zr– Stable if V > I*Zr

Restraining Quantity : VOperating Quantity : I*Zr

And even though this is a voltage diagram where we compare V and IZ,the IR & IX axes are usually labelled simply as “R” & “X”, since “I” is just a constant of proportionality

VI ZR≤Trip Condition

IR

IZr

X

V

RZIV ×≤Alternatively …

I

I

Slide 19

SimpleBeam BalanceComparator

OperateQuantity

RestrainQuantity

Restrain = VOperate = I * Zr

Simple Amplitude Comparator

Slide 20

Distance Relay Operation

Distance relay does not …– Store Current I– Store Voltage V– Compute ratio V/I– Determine this to be impedance to the fault– Check to see if this impedance is less than some preset value– Trip or not trip accordingly

Distance relay simply …– Trips if the operating quantity lies inside the trip zone– Restrains if the operating quantity lies outside the trip zone– Operates in a simple “GO” / “NO GO” mode

Slide 21

Distance Relay Does :-

Trip if impedance falls inside a specified domain in the impedance plain

– Inside = Trip– Outside = No trip– On the edge = Maybe !!

V

Marginal

TripV

R

V

No Trip

I.Zr

X

Slide 22

Mho Circle Comparator

Consider the quantities– S1 = I.Zr - V– S2 = V

And set the operating criteria to be the angle between these quantities, not their magnitudeSelect 90º as the criteriaRemember that the diameter of a circle always subtends 90º at the circumferenceWe have thus established a circular characteristic, with diameter of : I.Zr

R

X

I.Zr

S2 = V

S1 = IZ - V

Slide 23

Induction Cup Comparator

I2

I1

Induction Cup Angle Comparator

Direction of rotation depends on the phase

angle between S1 & S2, either to open or close

the trip contacts

S1

S2

Page 4




Slide 24

Phase Angle Comparators

Characteristics easily implementedPronounced operation, especially at the relay characteristic anglePolarising is available– Healthy Phase Cross Polarising– Memory Polarising– Positive Sequence Polarising

Slide 25

Cross Polarising S I Zr VS V12= ⋅ −=

A

BC

k.Vbc /+90 deg

Contract for Reverse Fault

X

R

Expand for Forward Fault

S2 = Va + k.Vbc /+90 degS2 = Va + k.Vbc /+90 deg

Slide 26

Close-in Faults : MHO

Close-in faults are difficult to detect– For unbalanced faults, augment

S2 with some healthy phase voltage

– Modern microprocessor based relays will most likely use the positive sequence voltage.(ref slide 39)

– For 3 phase faults, augment S2 with some pre-fault memory voltage

– For 3 phase SOTF events, special SOTF logic is required

X

R

VSVZrIS

=−⋅=

21

Slide 27

R

X

Typical Mho Zones of Protection

Slide 28

R

X

Quadrilateral Characteristic

Slide 29

LOADR

XExport WattsImport Watts

Lagging pf

Leading pfLagging pf

Leading pf

Three Phase Load Limits

Page 5




Slide 30

3 Phase Load Limits

Remember to allow for emergency conditions– eg. A single feeder carrying the load normally shared by 2

feedersExcept for the impedance circle characteristic, load transfer will vary for …

– export and import conditions – and with power factor.

Allow a safety margin– And ensure the system operators are aware that the safety

margin has been includedAnd be aware that some relays have a different characteristic under 3 phase conditions

– This applies especially to old fully switched distance relays

Slide 31

R

QUADRILATERAL

X

Z1

Z2Feeder

Z3

Tall, narrow quadrilateral for long feeders : Good load transfer performance

Quadrilateral Characteristics

Slide 32

QUADRILATERAL

Feeder

X

R

Z2Z1

Z3

Short, wide quadrilateral for short feeders : Good

fault resistance coverage

Quadrilateral Characteristics

Slide 33

Load Encroachment Characteristic

Export Load near to unity pfImport Load near to unity pfCalculate impedance that corresponds to max loadLimit characteristics for export load (3 phase only)Import characteristic is OK in this example

Slide 34



Slide 34Slide 34

DISTANCEPROTECTION

Comparator Connections

Slide 35

Relaying Quantities

What Voltage will we apply to the relayWhat Current will we apply to the relay3 phase faults2 phase faults2 phase to earth faults1 phase to earth faults

Page 6




ZVIL =φ

φ

This could be used for detection of three phase faults.However, [except by default in the Earth Fault Comparator connection (see later)] it is not !

Vb

IcVc

Zl

Ib

IaVa

Zl

ZlVI

Z

ZV

I

ZVI

BC

BL

L

L

= ⋅

=⋅

=

2

2φφ

φ

φφ

φφ

. . . (1)

. . . (2)

First equation is not used, except for starters in older style switched distance relays.The Second equation is usedThis correctly detects three phase faults also.

ZlIcVc

Ia = 0

IbVb

Va

Zl

Zl

Iφφ = Ib – IcBut Ib = – Ic⇒ Iφφ = Ib – Ic = 2 Ib = 2 Iφ

Slide 38

φφ

φφ

IV

Z =BA

BA

IIVV

−−

= A

BC

AB

Phase – Phase Comparatorand 3 Phase Performance

Hence this φ-φ comparator also correctly detects 3 phase faults

A

A

IV

=

°∠⋅⋅°∠⋅⋅

=303303

A

A

IV

Slide 39

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

Phase to Ground Example

= 0.10 = 0.10

= 0.15= 0.15

= 0.15

= 0.35

I = 1.0 I = 1.0 I = 1.0

V1 = 1 / 0º V2 = 0

V0 = -0.15

V0 = -0.50

V2 = -0.10

V2 = -0.25

V0 = 0

V1 = 0.90

V1 = 0.75

Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

At the relay location– Vph = 0.65 pu– Iph = 3 pu– Therefore …

= 0.10 = 0.10

= 0.15= 0.15

= 0.15

= 0.35

I = 1.0 I = 1.0 I = 1.0

V1 = 1 / 0º V2 = 0

V0 = -0.15

V0 = -0.50

V2 = -0.10

V2 = -0.25

V0 = 0

V1 = 0.90

V1 = 0.75

217.0365.0

IV

PH

PH ==

Wrong !?

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅−

⋅⋅⋅+⋅=1

10101 3

3ZZZZIZIV φφ

PositiveSequence

FaultLocation

NegativeSequence

FaultLocation

ZeroSequence

FaultLocation

I1RelayLocation

ZS1

Z1

ZS2

I2RelayLocation

Z2

ZS0

I0RelayLocation

Z0

002211 ZIZIZIV ⋅+⋅+⋅=φ

( ) 00121 ZIZIIV ⋅+⋅+=φ

( ) 10001021 ZIZIZIIIV ⋅−⋅+⋅++=φ

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅−⋅+⋅=33

1

11001 Z

ZZZIZIV φφ

( ) 100 Z3 ⋅⋅⋅+= IKIV φφ

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅−

=1

100 3

where......ZZZK

001 3 IKI

VZ

⋅⋅+=

φ

φ

How will this measure for a 3 phase fault ?

Residual Compensation for Earth Faults

Residually Compensated Phase Current

Phase Voltage

Page 7




Source

RelayLocation

FaultLocation

Posi

tive

Sequ

ence

Net

wor

k

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

Neg

ativ

e Se

quen

ce N

etw

ork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

Seq

uenc

e N

etw

ork

Z0f

Z0s

I0

At the relay location– Vph = 0.65 pu– Iph = 3 pu– Residual Compensation …

– Therefore …

= 0.10 = 0.10

= 0.15= 0.15

= 0.15

= 0.35

I = 1.0 I = 1.0 I = 1.0

V1 = 1 / 0º V2 = 0

V0 = -0.15

V0 = -0.50

V2 = -0.10

V2 = -0.25

V0 = 0

V1 = 0.90

V1 = 0.75

150.014444.033

65.00I0K3I

VPH

PH =••+

=••+

Correct !!

4444.015.03

15.035.01Z31Z0Z0K =

•−

=•−

=

Slide 44

Distance Relay Comparator Connections

ZONE 2


A-B

ZONE 2TIMER

TRIP


ZONE 3TIMER

A-EB-C C-AZONE 3

B-E C-E A-B C-A

001 3 IKI

VZ

⋅⋅+=

φ

φ

φφ

φφ

IV

Z =1

Page 1

Protection SignallingFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

DISTANCE PROTECTION

Protection Signalling

Slide 2

Distance Relay Zone Discrimination

DIST

TIME

DISTANCE RELAY TIME DISCRIMINATION

LOCATION

ZONE 3

ZONE 2

LOCATION

ZONE 1

DIST

DIST

TIME

ZONE 1

ZONE 2

ZONE 2

ZONE 3

ZONE 1

ZONE 3

Slide 3

AEMC Requirements(Australian Energy Market Commission)

National Electricity Rules : NER– Automatic Access Standards– To maintain system stability– To not constrain inter or intra regional power flows

Maximum Fault Clearance Times (milliseconds)

System Voltage kV Faulted End Remote End Breaker Fail

≥400kV 80 100 175

≥250kV to < 400kV 100 120 250

>100kV to < 250kV 120 220 430

≤ 100kV As necessary to prevent plant damage and meet stability requirements

Slide 5

Protection Signalling

AnalogueDigitalCommunications Bearers– Microwave– Fibre Optics (OPGW & ADSS)– Radio– Cable Carrier– Power Line Carrier– External Communications Network

Slide 6

Protection Signalling Equipment

External stand alone equipment– Duplex operation– Single signal– Multiple signals– Maintenance aspects

Built into protection relays– Duplex operation– Multiple signals– Maintenance aspects

Slide 7

Protection Signalling Schemes

Permissive Intertripping– Under Reaching– Over Reaching

Blocking IntertrippingDirect IntertrippingSeries Intertripping

Page 2



Slide 8



Slide 8Slide 8

PROTECTIONSIGNALLING

Permissive Intertripping

Slide 9

Permissive Intertripping

Applies primarily to non-switched distance relaysRemote end relay has a dedicated Zone 2 measurer– Zone 2 measurer has detected the fault but has to wait

for expiration of Zone 2 timer before tripping – Signal sent to ‘permit’ that measurer to trip in fast time

Two schemes in use …– Permissive ‘Underreaching’ ... PIT signal from Zone 1

(will not signal for faults beyond feeder end)– Permissive ‘Overreaching’ … PIT signal from Zone 2

(may signal for faults beyond feeder end)

Slide 10

Non-Switched Distance Relay

ZONE 2


A-B

ZONE 2TIMER

TRIP


ZONE 3TIMER

A-EB-C C-AZONE 3

B-E C-E A-B C-A

At the feeder remote end, one of these Zone 2 elements will have detected the fault, but has to wait for expiration of Zone 2 timer before tripping is allowed

Slide 11

PIR

ZONE 3MEASURER

ZONE 3TIMEDELAY

&

ZONE 2MEASURER

ZONE 1MEASURER

ZONE 2TIMEDELAY >1

PIS

TRIP

Permissive Underreaching

Slide 12

Permissive Underreaching

PUTT : Permissive Underreaching Transfer TripPermissive signal sent via the Zone 1 “underreaching” elementSimply implemented– No concerns since a signal is only sent when the fault

is actually on the protected feeder

Slide 13

ZONE 1MEASURER

ZONE 3MEASURER

ZONE 3TIMEDELAY

ZONE 2MEASURER

PIR &

ZONE 2TIMEDELAY

>1

PIS

TRIP

Permissive Overreaching

Page 3



Slide 14

Permissive Overreaching

POTT : Permissive Overreaching Transfer TripPermissive signal sent via the Zone 2 “overreaching”elementNot simply implemented– Concerns exist since a signal may be sent when the

fault is beyond the protected feeder

Slide 15



Slide 15Slide 15


Blocking Intertripping

Slide 16

Blocking Signalling

Local end relay has Zone 2 set to trip in fast time.Remote end relay sends a signal to inhibit this fast Zone 2 trip.

Signal sent from B to 'Block'the fast Zone 2 of relay A

DISTA

TIME

ZONE 1

Reverse looking Blocking Zone

LOCATION

FAST ZONE 2

B

NORMAL ZONE 2

ZONE 3

Slide 17

>1ZONE 1MEASURER

ZONE 3TIMEDELAY

REVERSELOOKINGZONE 4

ZONE 3MEASURER

ZONE 2TIMEDELAY

ZONE 2SHORTTIMEDELAY

BR

ZONE 2MEASURER

&

BS

TRIP

Distance Relay Blocking Scheme

Slide 18

Blocking Signalling Considerations

Blocking delay timer coordination - Fast Z2 coordination delay setting must allow time for receipt of blocking signalFor security, 2 signals are sent– Different signalling paths

Guard fail scheme provides security in the case of communication system failure

Slide 19

BLK B

SYSTEM B

BLK A

SYSTEM A

Feeder 'Y' Protection

BS B

SYSTEM BSignalling Equipment

BS B

BS A

SYSTEM ASignalling Equipment

BS B

BS APOSBS A

Blocking Send

Page 4



Slide 20

BLK B

SYSTEM B

BLK A

SYSTEM A


BR B

GFB B


BR B

POSGFBGFA

BR A

BR A

GFB A BR B

BR A

Feeder 'Y' ProtectionPOS

DISTPROT

'Y'

OR

AND

≥1

&

≥1

BR_A

BR_B

GF_A

GF_B

InhibitFastZone 2

Blocking Receive & Guard Fail Logic

Slide 21



Slide 21Slide 21


Direct IntertrippingSeries Intertripping

Slide 22

Direct Intertrip

Trips remote CB directlyUsed where security not paramountSingle DIT for back-up protection applications(eg. CB fail protection)Duplicate DIT for primary protection applications(eg. Line end transformer protection)– ‘X’ uses one signalling path– ‘Y’ uses a separate signalling path

Slide 23

CB FailProt DIS DIR

DIT for CB Fail Event

Slide 24

TransfProt Y DIS

TransfProt X

TransfProt Y

TransfProt X

DIS

DIS

DIS

DIR

DIR

DIR

DIR

Duplicate Direct Intertripping

Slide 25

Series Intertrip

Trips remote CB directlySecurity paramount– Increased security over direct intertripping– Reduced reliability compared with direct intertripping

Single SIT for back-up protection applications(eg. CB fail protection)– Two signals over separate paths

Duplicate SIT for primary protection applications(eg. Line end transformer protection)– ‘X’ uses two signals over one signalling path– ‘Y’ uses two signals over a second signalling path

Page 5



Slide 26

CB FailProt SIS SIR

Series Intertripping

Slide 27

SYSTEM B

SIT B

SIT A

SYSTEM A

Intertrip Cubicle

SIS A&BPOS


SIS B

SIS B


SIS A

SIS A

Series Intertrip Send

Slide 28

SYSTEM B

SIT B

SIT A

SYSTEM A


SIR B

SIR B


SIR A

SIR A

SIR A&B

Intertrip CubiclePOS

Series Intertrip Receive

Slide 29

TransfProt X

TransfProt Y

SIS

SIS

SIR

SIR

Duplicate Series Intertripping

Slide 30

SYSTEM B

SIT B1SIT B2

SYSTEM A

SIT A1SIT A2

SIS A2

POS

'Y' Intertrip Cubicle

SIS B

'X' Intertrip Cubicle


SIS B2

SIS B2

SIS B1


SIS A2

SIS B1

SIS APOS

SIS A1

SIS A1

Duplicate Series Intertrip Send

Slide 31

SYSTEM B

SIT B1SIT B2

SYSTEM A

SIT A1SIT A2

SIR B1


SIR B1

SIR B2

SIR B2


SIR A2

SIR A2

POS

SIR B

'Y' Intertrip Cubicle

SIR A

SIR A1

SIR A1'X' Intertrip Cubicle

POS

Duplicate Series Intertrip Receive

Page 6



Slide 32



Slide 32Slide 32


Power Line Carrier

Slide 33

Power Line Carrier

HF signal sent over the transmission lineCoupling Equipment & Line traps– To allow signal injecting– To limit signal distribution

Slide 34

V

CH

CL ZL

N:1X

CH . VCH + CL

CVT’s & PLC Signal Injection

Inject HF Signal

Line Trap

Slide 35

CVT’s & PLC Signal Reception

XN:1

ZLCH + CL

CL CH . V

V

CHLine Trap

Receive HF Signal

Slide 36

Power Line Carrier

Low Power– Cannot be guaranteed to signal through a fault

OK for Blocking SchemesProbably OK for DIT & SIT schemes Some concerns for permissive schemes

– Low Power Permissive schemes do seem to work in any case – Quiescent Scheme

Low power under normal circumstances (guard tone)High Power to guarantee signalling through faults

Page 1

High Impedance Differential ProtectionFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

HIGH IMPEDANCE DIFFERENTIAL PROTECTION

Busbar ProtectionandGalvanically Connected Plant

Slide 2

SIMPLE !!

!Bus Zone Protection Requirements

Dependability– Must trip for all ‘in-zone’ faults

Discrimination– Must not trip for any ‘out-of-zone’ faults

Security– Against all sources of mal-tripping

Speed of operation– As quickly as possible

Dependability & Security

RELAY

Internal FaultSlide 4

CT Connections & Polarity

S2

P2

S1

P1

I1

I2

P2

S2

P1

I1 S1

I2

RELAY

Internal Fault

RELAY

External Fault

Page 2



Slide 8

3 Phase CT Connections

CT MARSHALLING

HZ RELAY

Slide 9

Current Mismatch

CT Manufacturing VariationsInequality of CT BurdensCT Saturation– Highest Fault Current on CT exposed to through fault– Worst possible mismatch is

Total saturation of the CT on the faulted plantAll other CTs transform perfectly

Slide 10

15000A 5000A

RELAY

5000A

External Fault

5000A

External Fault & CT Saturation

RELAY

External Fault

Rlead

Rlead

Rct High Impedance

Relay

CT Saturates :Magnetising branch

impedance becomes zero

( )LEADSCTFAULTRELAY RRIV +⋅=

Slide 12

Setting Voltage and Margins

Fault current comprises …– AC Component– DC Component

Hence, employ a DC Stabilised Relay– No additional margin on the setting is required

And considering 0% / 100% CT saturation case– This in an unrealistically extreme case– 100% safety margin is automatically built in

So, no additional safety margin on setting is required

RELAY

Internal Fault

High Impedance

Relay

RELAYKNEE V2V ⋅≥

CTs will saturate under internal fault conditions.But relay operation is assured provided absolutely all

CTs meet the requirement …

Page 3



Slide 14

CT Selection

All CTs to be the same ratioAll CTs to have Vk ≥ 2.Vsetting

– This is an absolute “MUST”– Preferably Vk ≥ 5.Vsetting

Need to know– Knee Point voltage– CT Resistance

Class Requirements– Not critical– But easiest to specify class “PX” CTs

CTs will almost certainly saturate under in-zone fault conditions– Not suitable for connection to other protection relays

0.1 PX 200 R5







Slide 15

Current Operated Schemes

Voltage operatedCurrent operated, incl stabilising resistorTypical current settings

– as low as possible, but– > 20% of plant rating– < 30% of fault current

20% setting is usually OK– Assuming the CT has

been selected to match plant rating

V = I.R = 0.2 x (200 + 10) = 42 volts

Relay0.2A

10 ohms

200 ohms

Slide 16

Summary

Ensure Stability under through faults

Ensure Operation for genuine ‘in-zone’ faults

Beware of short cut methods

– Do not simply set …

RELAYKNEE V2V ⋅≥

2VV KNEE

RELAY =

( )LEADSCTFAULTRELAY RRIV +⋅=

Preferably 5 times to optimise relay performance, but 2 is the absolute minimum to ensure

reliable relay operation

Preferably 5 times to optimise relay performance, but 2 is the absolute minimum to ensure

reliable relay operation

Slide 17



Slide 17Slide 17

HIGH IMPEDANCE DIFFERENTIAL PROTECTION

Application to other Plant

Slide 18

HZ Prot’n Application to Plant

Requires Galvanic ConnectionAll CT ratios the sameCan Apply To …– Busbars– Transformers– Generators & Motors– Capacitors– Reactors

Slide 19

DIFF

Auto Transformers

All CT ratios to be the same

This CT will carry maximum current and hence dictates ALL CT ratios

But this CT is internal and may have a single fixed ratio.Thus, must be specified correctly at time of purchase !!

Page 4



Slide 20

REF

Restricted Earth Fault Protection

CT terminals “away” from protected object are connected

CT terminals “near” to protected object are connected

Slide 21

DIFFDIFF

A

DIFF

B C

Reactors – Earthed Neutral

Slide 22

DIFFDIFF

A

DIFF

B C

Reactors – Floating Neutral

“Floating” neutral bus is also protected

Page 1

Transformers andSequence Components



Slide 1



Slide 1Slide 1

TRANSFORMERSandSEQUENCE COMPONENTS

DIFFERENTIAL PROTECTIONREQUIREMENTS

Slide 2

Transformer Current Flows

Star / Star Transformer : LV Earth Fault– Current flows in corresponding HV winding– Appears as EF on the HV system also

I1, I2 & I0I1, I2 & I0I1, I2 & I0I1, I2 & I0

Slide 4


Star / Star Transformer : LV Earth Fault– But, suppose we don’t have an upstream power system earth– However, consider the effect of adding a delta connected tertiary winding– HV line current flows in a 2:1:1 ratio– No I0 on the HV system as there is no path for neutral current flow

I1, I2I1, I2

I0I0

I1, I2 & I0I1, I2 & I0

So where did the I0 go ??So where did the I0 go ??

Slide 5


Star / Star Transformer : LV Earth Fault– Retain the delta connected tertiary winding– But, let’s reinstate the generator earth– Power system and delta winding zero sequence current flow

distributions will depend on their relative Z0 impedances

I1, I2 & I0I1, I2 & I0I1, I2, I0I1, I2, I0

I0I0

Slide 6

Transformers, Sequence Componentsand Differential Protection

Star/Star transformers, with a delta tertiary winding:– Will have a mismatch between zero sequence current flows

on the HV & the LV windings– It is thus necessary to exclude zero sequence current from

the differential relay protection algorithmsStar/Star transformers, without a delta tertiary winding:

– May still have a mismatch between zero sequence current flows on the HV & the LV windings

The transformer tank can act as a low quality “tertiary delta winding”

– It is thus still necessary to exclude zero sequence current from the differential relay protection algorithms

Slide 7


Delta / Star Transformer : LV Earth Fault– Current in corresponding HV winding only– Appears as phase to phase fault from the perspective

of the HV system

I1 & I2 onlyI1 & I2 only

So where did the I0 go ??So where did the I0 go ??

I1, I2 & I0I1, I2 & I0

Page 2

Transformers andSequence Components



Slide 8


Delta / Star Transformer : LV phase to phase fault– Current in 2 LV windings – Current in 2 HV windings– Appears as 2:1:1 fault on the HV system

I1 & I2I1 & I2I1 & I2I1 & I2

Slide 9


Star / Delta Transformer : LV phase to phase fault– Current in all 3 LV windings – Current in all 3 HV windings– Appears as 2:1:1 fault on the HV system

I1 & I2I1 & I2I1 & I2I1 & I2

Slide 11

Sequence ComponentsTransformer LV ph-ph fault

30 deg

Consider B-C fault on the LV of either of …– Star Delta transformer– Delta Star transformer

30 deg phase shift– Positive seq components– Negative seq components

… will shift + 30 deg… will shift - 30 deg

Slide 12

Sequence ComponentsTransformer LV ph-ph fault

LV phase to phase fault– A phase I1 & I2 cancel– B & C phase I1 & I2 are 60o apart

HV distribution is 1 : 2 : 1– I1 & I2 are shifted ±30o across the transformer– On two of the HV phases, I1 & I2 are now 120o apart– On one of the HV phases, I1 & I2 are now exactly in phase !!

30 deg

LVHV

Slide 13

Transformers, Sequence Componentsand Differential Protection

Compensate for the transformer phase shiftExclude zero sequence current from the differential relay protection scheme– Zero sequence current can flow into and out of

earthed star windings– Zero sequence current cannot flow into or out of delta

windings– Zero sequence current can circulate around delta

windings (said to be “trapped” in the delta)

Page 1

Transformer ProtectionFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

TRANSFORMER PROTECTION

Slide 3

Types of Fault

Phase-ground faults - from winding to core or winding to tankPhase-phase faults - between windings Interturn faults - between single turns or adjacent layers of the same winding (Buchholz)Arcing contactsLocal hotspots caused by shorted laminations Low level internal partial discharges (moisture ingress or design problems)Bushing faults (internal to the tank)Tapchanger faults (often housed in a separate tank)Terminal faults (external to the tank, but inside the transformer zone)

Slide 4

Buchholz Protection

Two floats in the relay:– Upper float

detects accumulation of gasdetects loss of oilIncipient faults

– Partial discharge– Winding & core overheating– Bad contacts and joints

May alarm only or may be set to trip– Lower float

detects surge in oil < 100msalthough it does take a finite time for pressure waves to initiate Buchholz tripping

To Conservator

Gas Sample

Trip

Alarm

BUCHHOLZ RELAY

Float

To Tank

Float

Slide 5

Pressure Relief Device – (Qualitrol)

Spring assisted pressure relief devicesRelieves pressure impulses due to massive internal fault conditions. Helps prevent the tank bursting or splittingRelay contacts are also connected to trip the transformer.

Since pressure waves travel with a finite velocity, they may rupture the tank locally before the pressure wave has reached the pressure relief device, if it is some distance away. Severalunits may therefore be required on larger transformers.

Slide 6

Basic Transformer Protection

Fuses– Transformers without CBs– Perhaps to a few MVA

Overcurrent & Earth Fault Protection– Transformers with CBs– Perhaps 5 - 50MVA

Differential Protection– Transformers > 10MVA

FastCan be sensitiveMay detect terminal faults also

Slide 7



Slide 7Slide 7


Biased Differential Protection

Page 2



Slide 8

Differential Protection

DifferentialRelay

DifferentialRelay

P1

S1

P1

S1

Slide 9


DifferentialRelay

P1

S1

P2

S2

SIDE OF CT AWAY FROM PROTECTED PLANT CONNECTS

TO RELAY

CURRENT FLOWS INTO PLANT

CURRENT FLOWS OUT OF PLANT

CURRENT FLOWS INTO RELAY

CURRENT FLOWS OUT OF RELAY

SIDE OF CT AWAY FROM PROTECTED PLANT CONNECTS

TO RELAY

TRIPELEMENT

IT IS NOT THE P1/S1 OR P2/S2 ORIENTATIONS THAT ARE

IMPORTANT, BUT THE PREFERENCE FOR THE

“AWAY” SIDES OF THE CTs TO CONNECT TO THE RELAY

Slide 10

Differential Protection of Transformers

132/66kV

100/1 200/1100A 200A

1A 1A

BIAS orRESTRAINTELEMENT

BIAS orRESTRAINTELEMENT

TRIPPING ELEMENT DETECTS ONLY THE

MIS-MATCH CURRENT

TRIPELEMENT

Slide 11

Transformer Differential Mismatch

Transformer turns ratio & tap changingInrush on energisation (2nd harmonic)Over excitation (5th harmonic)CT MismatchSome CT Saturation on through faults Transformer phase shiftsEarth fault (neutral … zero sequence) currents

Slide 12

Inrush Current on Energisation of Transformer

TRIPELEMENT

Slide 13

Second Harmonic on Inrush

Transformer inrush current on energization.– Inrush current produces a current from the energizing

side only, appearing as an internal fault. – Inrush current magnitude can be as great as a through

3 phase fault.– This current is characterized by the appearance of

second harmonics, so additional restraint can be based on this 2nd harmonic “signature”

– Relay setting below the 2nd harmonic level is required(Ratio of 2nd harmonic to fundamental)

Page 3



Slide 14

Transformer Inrush Current

2

0

2

4

6

8

10

12

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Transformer Inrush Current

Current

Seconds

Second Harmonic on inrush

Slide 15

Fifth Harmonic on over excitation

Overfluxing, caused by too high a voltage, or too low a frequency. – Increased magnetising current– This is characterized by third & fifth harmonics. – Fifth harmonic restraint to retrain tripping of the differential

element– Typically no user calculations or settings are required

Sustained overfluxing may damage the transformer– Time delayed V/f tripping function (long time)– Especially applicable to generator transformers

Frequency can be anywhere from zero to nominal during run-up and run down

– Not so necessary for transmission or distribution applicationsFrequency will not deviate significantly from nominal

Slide 16

Unbalance Currents

Mismatched CTs– CTs do not exactly compensate for transformer turns

ratio– Transformer turns ratio changes with tap changing– Implement a “biasing” restraint system

Magnetizing current in the CTs, especially as some saturation due to DC fault current sets in. – The amount of bias is increased under heavy through

fault conditions to compensate for possible CT saturation

Slide 17

Bias Differential Protection

Allow for Transformer turns ratioAllow for Transformer phase shiftsEliminate Zero Sequence currents from the relaying system

OperatingWinding

P1S1

Bias Windings

P1S1

Slide 19



Slide 19Slide 19


CONTINUED

Slide 20

CT Connections and Ratios

Star/Delta and Delta/Star transformers have a 30 degree phase shiftCompensate with CTs connected opposite to the transformer connections. ie:

– Star connected CTs on the delta side of the transformer– Delta connected CTs on the star side of the transformer

Phase shift compensatedZero sequence currents flowing in the transformer star windings prevented from entering the relaying systemBut how do we get the correct delta connection for our CTs ???

Page 4



Slide 21

Determination of CT Connection

Diff Prot

Yd11D11D11

CT Primary is star connectedCT secondary is D11 connectedOverall connection is thus YD11

Slide 22

Determination of CT Connection

Dy11D1D1

Diff ProtCT Primary is star connectedCT secondary is D1 connectedOverall connection is thus YD1

Slide 23

Star/Delta and Delta/Star TransformersCT Connection Summary

Transformer HV is STAR connected– HV CTs are delta connected– HV CTs EQUAL to the transformer phase shift– LV CTs in star

Transformer LV is STAR connected– LV CTs delta connected– LV CTs OPPOSITE to the transformer phase shift– HV CTs in star

Slide 24

CT Connection Summary

Compensates for the phase shift across a star-delta transformer.– The correct vector group must be chosen for the CTs

to ensure that through currents balance.Prevents any zero sequence currents flowing in the star winding from entering the relay– Since they are not present in the line on the delta side.

And for Star / Star transformers ??– It is still necessary to eliminate Io from the relaying

system– Connect CTs delta / delta– Or use the D12 / D12 feature of microprocessor relays

Slide 25

A phase output is at "11 o'clock"

A phase "S1" connects to B phase "S2"B phase "S1" connects to C phase "S2"C phase "S1" connects to A phase "S2"

D11

S2

A

S1

S1C

S2

S1

S2B

C

B

A

CT YD11 Connections

D11D11

Slide 26





D11

S2

A

S1

S1C

S2

S1

S2B

D1

C

B

A

S2B

S2

S1

S1

A

S1CS2

C

B

A

CT YD1 Connections

D1D1

Page 5



A B

Bias Windings

P1 P2

S1 S2

A1 A2

OperatingWindingsC

a2 a1P1P2

S2 S1

Notice that the connections for the Delta windings are the same !!

“Away” side of CTs connected to relay.Hence, transformer current “in” or “out”corresponds to relay current “in” or “out”.

“Away” side of CTs connected to relay.Hence, transformer current “in” or “out”corresponds to relay current “in” or “out”

Slide 28


There must be a path for the current to flowThere must be an Ampere Turns balanceIf there is current flowing in one winding– There must be current in the coupled winding

If there is no current flowing in one winding– There can be no current in the coupled winding

A B

Bias Windings

P1 P2

S1 S2

A1 A2

OperatingWindingsC

a2 a1P1P2

S2 S1

External Phase – Earth Fault

Protection Scheme remains balanced– HV 0:1:1 (HV looks like a phase – phase fault)– LV 0:0:1 (LV is actually a single phase fault)

A B

Bias Windings

P1 P2

S1 S2

A1 A2

OperatingWindingsC

a2 a1P1P2

S2 S1

External Phase – Phase Fault

Protection Scheme remains balanced – HV 1:2:1 (HV has a 2:1:1 current distribution)– LV 0:1:1 (LV is actually a phase – phase fault)

Slide 31

Delta CTs and Ratio Selection

CT ratios must allow for the fact that current flowing into the relay from the delta connected CTs is √3 times the CT secondary currentHence, a standard 1A CT will result in relay current of √3 times the CT secondary currentThus, CTs with ratios such as 1000/0.577 are, for this reason, quite common.

1 / 0

1 / -120

1 / -

240

= 1.732 /-30

1 / 0 – 1/-240

Slide 32

Delta CTs and Ratio Selection

Ia - Ic = 1.732 /-30

CT ratios must allow for the fact that current flowing into the relay from the delta connected CTs is √3 times the CT secondary currentHence, a standard 1A CT will result in relay current of √3 times the CT secondary currentThus, CTs with ratios such as 1000/0.577 are, for this reason, quite common.

1 / 0 – 1/-240

Page 6



Slide 33

Modern Microprocessor Relays

All CTs connected in StarRelay has to “process” phase shiftsRelay has to “remove” neutral current

P1

P1

P1

S1

S1

S1

S1

S1

S1

P1

P1

P1

Slide 34


P1

P1

P1

S1

S1

S1

S1

S1

S1

P1

P1

P1

ICIAIARELAY −=

IAIBIBRELAY −=

IBICICRELAY −=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−•=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ICIBIA

110011101

31

ICIBIA

RELAY

RELAY

RELAY

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡•

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−•=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ICIBIA

110011101

31

ICIBIA

RELAY

RELAY

RELAY

IArelay

IBrelay

ICrelay

⎛⎜⎜⎜⎜⎝

⎞⎟⎟

⎠

1

3

1

1−

0

0

1

1−

1−

0

1

⎛⎜⎜⎝

⎞

⎠⋅

IA

IB

IC

⎛⎜⎜⎝

⎞

⎠:=

relay

D1

D11IArelay

IBrelay

ICrelay

⎛⎜⎜⎜⎜⎝

⎞⎟⎟

⎠

1

3

1

0

1−

1−

1

0

0

1−

1

⎛⎜⎜⎝

⎞

⎠⋅

IA

IB

IC

⎛⎜⎜⎝

⎞

⎠:=

relay





D11

S2

A

S1

S1C

S2

S1

S2B

D1

C

B

A

S2B

S2

S1

S1

A

S1CS2

C

B

A

Modern Microprocessor RelaysSlide 36


All CTs can now be connected in Star– Relay internal processing adjusts for phase angle – Relay internal processing rejects zero sequence

componentsCT ratios mismatches can also now be accommodated– Internal processing within relay then adjusts CT

current to match transformer turns ratio– CTs can be fine tuned to match middle tap position– Allows for more sensitive relay settings

Slide 37

CT Phase and Ratio Adjustment

Dyn120MVA 33/11kV

1500/1400/1

DifferentialElement

350A 1050A

0.7A

-300

0.875A

00

00

1A

00

1A

Transformer Microprocessor Differential Protection Relay

• Magnitudes normalised to transformer FLC• Phase angles compensated• Zero sequence current eliminated

Yy0Software CT

x 1.143Software CT

x 1.429

Yd11

00 -300

TAP POSITION

Software CT Ratio

Adjustment

Transformer Bias Differential Protection

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60

0.5

1

1.5

2

2.5

3

3.5

4

15% Differential Setting25% Differential Setting35% Differential Setting

Differential Current

Diff I1 I2+:=

Bias Current BiasI1 I2+

2:=

OPERATEOPERATE

Transformer Internal Fault

Protection Trips

Transformer Internal Fault

Protection Trips

Through Fault withCT Saturation

Through Fault withCT Saturation

Through FaultMismatch due to CT Ratios &Transformer Tap Changing

Through FaultMismatch due to CT Ratios &Transformer Tap Changing

RESTRAINRESTRAIN

Page 1

Low Impedance Busbar Differential ProtectionFundamental Principles of Power System Protection


Slide 2



Slide 2Slide 2

LOW IMPEDANCEBUSBAR DIFFERENTIAL PROTECTION

Slide 3

Low Impedance Busbar Prot’n

Utilises Bias Restraint– 1 x 3 phase input for every item of plant

Often applied as a retrofitExpensiveAre not simple schemes

Slide 4

Basic Types

Central Unit– GE B30– SEL 487B– REB 670 & RED 521– Areva P746

Bay units with connection to central unit– Areva P740– ABB REB500

Slide 5

Special Features

Mismatched CTs– Applicable where CT ratios vary– Applicable where CT classes vary

Poor quality CTs– Increased operating current pickup– Reduced knee point for increase in bias– Increased bias slope– CT saturation algorithms

Slide 7

LZ Busbar Bias Differential Characteristic

Bias Current

OperatingCurrent

Through Fault

Internal Fault

Reduced knee point for

increase in bias

Increased bias slope

TRIPZONE

Increased operating

current pickup

Slide 8

CT Saturationand Through Fault Performance

Bias Current

OperatingCurrent

Through Fault

Internal Fault

TRIPZONE

CTSaturation

CTSaturation

Monitor the “locus” and use thisto detect CT saturation and restrain the

relay under through fault conditions

Monitor the “locus” and use thisto detect CT saturation and restrain the

relay under through fault conditions

Page 2



Slide 9

Special Features

CT problems accommodated– Mis-matched CTs– Poor quality CTs

Provide for multiple bus zones– One relay covers (say) up to 6 zones– Do not need separate CTs where zones overlap– Do not need separate CTs for Master & Check Zones

Allow for dynamic switching of bus zones– Requires Isolator status (a & b) inputs

Provide CB Fail and CB Fail Bus Trip Facilities

Slide 10

Multiple Bus Zones

Feeder and coupler CTs for BZ1 scheme

Feeder and coupler CTs for BZ2 scheme

Feeder CTs for overall check zone

Slide 12

Dynamic Switching of Bus Zones

Two separate BZ schemes

Checkzone

Slide 13


Diameter closed.Single BZ scheme for entire substation

Checkzone

Slide 14


Bus 2 disconnector now open.BZ schemes reconfigured OK.

Checkzone

Slide 15

Bus Zone CB Fail Protection

CB Fail for a bus zone fault– Fault on bus– CB Failure detected by BZ relay inbuilt CBF feature– BZ relay initiates tripping of remote CB(s)

Remote end CBs for plant connected to busNext bus for coupler or section CB failure

Page 3



Slide 16

CB Fail Protection & CBF Bus Tripping

CB Fail for a plant fault (eg feeder fault)– Plant protection detects fault and initiates tripping of its

CB(s)– Plant protection also initiates BZ relay inbuilt CBF feature

(via opto input)– CB Failure detected by BZ relay inbuilt CBF feature– BZ relay “knows” what bus the plant is connected to– BZ relay “knows” what other plant is connected to that bus– BZ relay initiates CBF Bus Trip of required CBs

Especially important for switched busbars– BZ relay is the only system that “knows” the busbar

topology !!

Page 1

Feeder Differential ProtectionFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

FEEDERDIFFERENTIAL PROTECTION

Pilot Wire Protection

Slide 2

Pilot Wire RelayingLimitations and Requirements

Pilot Length– Pilot Cost– Pilot Wire Resistance

Must not exceed relay design limitationsBut, add separate (padding) resistors to bring the pilots to the relay manufacturer’s design value (say 1000 ohms)

– Pilot Wire CapacitanceMay disable relay operation : circulating current schemeMay unstabilise relay operation : opposed voltage scheme

Relays at each feeder end tripping local CBs for– Strong infeed– Weak infeed– Zero infeed

Two Elements– Operating element - trips relay on mismatch– Bias element - restrains relay on through current

Slide 3

Circulating Current Scheme

Through current results in relay current circulating between line end relays

R R

OO

Slide 4

Circulating Current Scheme

Through current results in relay current circulating between line end relaysFeeder fault current results in current flowing in the relay operating elementsAnd the effect of Pilot capacitance ??

– Desensitises or even disables the relay operating elements

R R

OO

Slide 6

Opposed Voltage Scheme

Through current results in relay current circulating between line end relaysPilot wires crossed to create an opposed voltage schemeAnd reconfigure the operating and restraining elementsAnd the effect of Pilot capacitance ??

– Disables relay restraint : may trip on through faults

R R

OO

limited to relay restraint elements

R

O

R

O

Slide 7

Opposed Voltage Scheme

Feeder fault current results in current flowing in the relay operating elements

R R

OO R

O

R

O

Page 2



Slide 8

Summation Transformer

Allows comparison of composite quantityMust trip for all internal faultsMust be stable for all external faultsVarious sensitivities ..... OK

Slide 9

Summation Transformer

FAULT RELATIVETYPE PICK-UP

A-B 0.8AB-C 1.0AC-A 0.44AA-B-C 0.51AA-N 0.19AB-N 0.25AC-N 0.33A

SUMMATION TRANSFORMER

C

N

3

B

A

1

1.25

This arrangement ensures correct operation under 2:1:1 current distributions

Slide 10

Complete Scheme

PADDINGRESISTOR

SUMMATIONTRANSFORMER

PILOT WIRES

PADDINGRESISTOR

SUMMATIONTRANSFORMER

PILO

T WIR

ER

ELAY

TAPPEDPILOT ISOLATIONTRANSFORMER

TAPPEDPILOT ISOLATIONTRANSFORMER

STABILISINGRESISTOR

STABILISINGRESISTOR

PILO

T W

IRE

RE

LAY

Just a few ohms to improve stability under heavy through fault conditions

Tapped pilot isolation transformers R α Turns2

Padding resistance at each end to bring total pilot resistance to the relay specification requirements

Slide 11



Slide 11Slide 11

FEEDERDIFFERENTIAL PROTECTION

Digital Current Differential Protection

Slide 12

Digital Current Differential Schemes

Digital CommunicationsIndividual measurements per phaseChannel delay automatically compensatedData security checks– CRC & Parity bits

12

12.2 π1

i1

i2i3

i4

i5

i6i7

i8

( )

( )

IN

n t i

IN

i i n t i

S nn

N

CN

nn

N

= ⋅ ⋅ ⋅ ⋅⎡

⎣⎢⎤

⎦⎥

= ⋅ + + ⋅ ⋅ ⋅⎡

⎣⎢⎤

⎦⎥

=

−

=

−

∑

∑

2

22 2

1

1

0

1

1

sin

cos

ω

ω

∆

∆

Page 3



Slide 14

In-Phase & Quadrature Components

Slide 15

Current Differential Data Synchronisation

How can the local and remote end data samples by synchronisedGPSPing Pong Topology– Asynchronous Samples – Synchronised Samples

tA5tA*

tA6

tA4

tA1tB3

tp2

td

Current Vectors

Current VectorstA1

tA3

tB3*

tA2

td

tp1

RELAYA

tA1

tB6

tB5

tB4

tB3

tB*tB2

tB1

RELAYB

( )2

1*21 tdTAtAtptp −−==

Relay ping-pong system determines that it needs a sample at tB3, whereas samples exist at tA3 and tA4 … choose the closest.

Slide 18

Data alignment for non-synchronised schemes

Remote relay

samples

Localrelay

samples

Adjust the non-synchronised samples via

relay computation

algorithm

Adjust the non-synchronised samples via

relay computation

algorithm

Slide 19

Current Differential Data Synchronisation

Ping Pong Topology– Asynchronous Samples

Relay selects nearest available and uses the Ping Pong to adjust the non-synchronised samples via its computation algorithm

– Synchronised SamplesPing Pong system aligns the samples which can then be compared directly

– Both systems require equal send and receive times (paths) GPS Synchronisation

– Truly Synchronised Samples (time tagged)– Send and receive times (paths) do NOT have to match– Ping Pong Back-up

IS1

IS2

k1

k2

TRIPTRIP

( )ZYXBIAS

ZYXDIFF

III5.0I

IIII

++•=

++=

IS1=0.2 puIS1=0.2 pu

IS2=2.0 puIS2=2.0 pu

k1=30%k1=30%

k2=100%k2=100%

NO TRIPNO TRIP

Through Load or

Fault Event

Through Load or

Fault Event

Feeder Fault Event

Feeder Fault Event

BIAS CURRENT

IX IY

IZ

X

Z

Y

DIF

FER

ENTI

AL

CU

RR

ENT

Page 4



Slide 21

Alpha Plane Representation

Some relays now use the ratio of remote to local end feeder currents to define their characteristicThis will ideally be -1 under through load and through fault conditions

Slide 22

Alpha Plane Representation Ziegler : Numerical Differential Protection

RestrainRestrainTripTrip

Increasing Bias SlopeIncreasing Bias Slope “Conventional” Differential vs Restraint characteristics can also be represented on the alpha plane

TripTrip

RestrainRestrain

Slide 23

Alpha PlaneRestrain Zone Requirements

“Restrain Zone” must, as a minimum, be adjusted to provide for magnitude and angular deviation for …– Line charging current– CT saturation

Slide 24

Alpha PlaneBalance : Restrain & Trip Zone Requirements

Relay trip region needs to expand to accommodate angle variation under in-zone fault conditions. Similarly, the restrain zone needs to expand to ensure stability under through load and fault conditions.That is, allow for variations and simultaneously increase both zones to achieve the optimum.

Slide 25

Alpha PlaneSEL311L Recommendation

Trip Zone to ensure correct tripping for all in-zone faultsRestrain Zone to ensure stability for all external load and fault conditions195° recommended

– SEL advises that this allows 35°of margin for other sources of error

Slide 26

Current Differential Signalling

2 ended schemes – single comms3 ended schemes – dual commsDaisy chained schemes

Page 5



Slide 27

2 Ended Scheme – Single Comms

DIFF

DIFF

Slide 28

3 Ended Scheme – Dual Comms

DIFF

DIFF

DIFF

Slide 30

Daisy Chain Scheme : >3 Ended

Schemes covering up to 6 ended tee feeders are available

DIFF

DIFF

DIFF

A

B

C

IA

IA + IB

IC + ID

IB + IC + ID

DIFF

IA + IB + IC

ID

D

Slide 31

Daisy Chain Scheme : with Redundancy

DIFF

DIFF

DIFF

A

B

C

IA

IA + IB

IC + ID

IB + IC + ID

DIFF

IA + IB + IC

ID

Close the Daisy Chain– Redundant path is normally idle– Redundant path becomes active

when any other link is broken

D

Page 1

Auto ReclosingFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

AUTO RECLOSING

EHV SystemsHV SystemsDistribution Systems

Slide 2

Application of Auto Reclosing

Most faults are single phaseMost faults are transientHence, an automatic reclosure is often provided to restore faulted feeders to service– To improve transient and voltage stability of the

system– To restore the system to normal levels of security– To restore supply to customers

Slide 3

Safety Aspects of Auto Reclosing

Auto Reclosing is usually blocked or inhibited for …Cases where the following plant may be affected– Cables– Transformers

Cases where the initial protection operation may have involved backup for …– CB Fail Event– Blind Spot Event– Failure of a remote protection scheme

Slide 4

Auto Reclosing Sequence

Initiate by protection operationCB Auto Recloses after Dead Time

– With checks to ensure the CB was actually previously closed

– With checks to ensure the fault was actually on the feeder and that protection operation is not a back-up function

Scheme resets after Reclaim TimeScheme lockouts

– Transmission schemes typically lockout upon occurrence of a 2nd fault within reclaim time

– Distribution schemes may allow multiple reclosures

Slide 5

Auto Reclosing Initiate Signal

Usually for high speed tripping only– Feeder differential protection– Distance relay Zone 1 protection– Distance relay Fast Zone 2 protection

Usually not for slow speed tripping– eg. Distance relay Slow Zone 2 protection

Usually not for tripping for remote faultsAllow for trip signal resetting– May need to extend the relay trip signal

Slide 6

Blocking of Auto Reclosing

Transformer or Shunt Reactor fault– eg. On receipt of DIT or SIT from remote end

CBF or Blind Spot faultsFollowing manual close of CBAR Check systems– Dead line check to prevent the master end reclose

onto a back-energised system– Live line check to only allow the slave end reclose

onto a healthy feeder– Sync check at the slave end to prevent re-connection

between systems that are now out of synchronism

Page 2



Slide 7

Auto Reclosing Dead Time

Time delay between operation of protection and initiation of CB close - has to provide for …– Fault clearance (local and remote end clearance)– Dissipation of ionised air– Effects of parallel feeders, parallel phases (SPAR) &

couplingInductiveCapacitive

– Resetting of protection relay

Slide 8

Auto Reclosing Dead Time

Beware of AR effects on nearby power stations– AR onto multiphase fault should be avoided until

oscillations of the generator shaft have subsided, perhaps:

5 seconds for double phase to ground faults10 seconds for three phase faults

Slide 9

Auto Reclosing Reclaim Time

To prevent multiple reclosures onto permanent faultsTime delay following autoreclosure during which another fault is considered to be re-occurrence of the original fault– Not to be set too short as reoccurring faults may not

be properly identified– Not to be set too long as totally independent faults

may be incorrectly identified as reoccurrence of the original fault

To ensure the CB capability for Trip - Close - Trip sequences is not be compromised

Slide 10

EHV Auto Reclosing

Weakly interconnected systems– System stability is of prime concern

Strongly interconnected systems– Return of system security is of concern

SPAR and/or TPAR– SPAR for 1 phase faults– TPAR for 2 and 3 phase faults– Perhaps with different dead times

Typically only single shot– Failed reclose attempts will have a serious effect on system

stabilityTrip all 3 poles and lockout if AR is unsuccessful

Slide 11



Slide 11Slide 11

Single Pole Auto Reclosing(SPAR)

Slide 12

Single Pole Auto Reclosing (SPAR)

Advantages– In-service phases maintain system synchronism– In-service phases improve system “robustness”– Less system disturbance on reclose

Requires phase selective circuitry– Tripping, Closing and CB Fail

CB Pole discrepancy– Breaker must accept the single pole open operation

during SPAR

Page 3



Slide 13


Dead Time – May need to be increased over TPAR time

(Due to coupling effects, particularly capacitive coupling)

Weakly interconnected systems : “2 phasing”scenarios– Reasonably short dead time may be required

Strongly interconnected systems– Longer dead times may be acceptable

Slide 14


Minimal affect on power stationsBut beware of possible cross country faults on double circuit feeders– Each feeder experiences a single phase fault– But the system, and nearby power stations, are

subjected to a double phase to ground fault

Slide 15


No requirement for sync check– System synchronism is ensured by the remaining

2 in-service phases Typically, no live line check implemented either– Single phase faults do not pose a significant risk to

system stability or to nearby plant– Common practice is to not implement such check

facilities. – Both ends simply reclose and both will be subject to

the effects of a permanent fault and will then trip (all 3 phases) and lock-out

Slide 16

SPAR & Transformer Ended Feeders

Single pole tripping of transformer ended feeders must not occur– The open phase remains magnetically coupled– 100% voltage is likely on the open phase– Fault is sustained for the complete dead time– Disastrous consequences for the transformer and for system

stabilityFlux from B & C phases continues to energise A phase

A phase fault current continuesWhoops !!

Slide 17



Slide 17Slide 17

Three Pole Auto Reclosing(TPAR)

Slide 18

Three Pole Auto Reclosing (TPAR)

EHV Systems– SPAR often implemented for single phase faults– TPAR implemented for multi-phase faults

Master/Slave system always implemented

HV Systems– SPAR rarely implemented for single phase faults– TPAR usually implemented for all fault types

Master/Slave system usually implemented

Distribution Systems– TPAR implemented for all fault types

System is usually radial

Page 4



Slide 19

Three Pole Auto Reclosing (TPAR)Live Line Check

Simple Voltage check to implement a Master/Slave system– Master end recloses after the dead time– Slave end only recloses if the feeder is healthy– Prevents multiple reclosures onto permanent faults

Master end must have a secure source of supply– Hence, usually the “stronger” end

But near to Power Stations, the remote end may be selected as Master– Minimising the power station effects of reclosure onto

permanent faults

Slide 20

Three Pole Auto Reclosing (TPAR)Synchronism (Sync) Check

Strongly Interconnected Systems– Feeder tripping is unlikely to split the system– Synchronism is maintained– Sync Check is not needed– Dead times can be longer

Weakly interconnected systems– Feeder tripping may split the system– Synchronism may be lost– Sync check is needed– Dead times need to be shorter

Slide 21

Three Pole Auto Reclosing (TPAR)Synchronism (Sync) Check

Sync Check may allow reclosure for:– LLLB : Live Line, Live Bus, in synchronism– LLDB : Live Line, Dead Bus– DLLB : Dead Line, Live Bus– DLDB : Dead Line, Dead Bus

Sync Check monitors both sides of the CB for:– Frequency– Voltage– Phase angle– Rate of change of frequency (df/dt)

(To ensure the systems are not drifting at an unacceptable rate)

Slide 22



Slide 22Slide 22

Distribution SystemAuto Reclosing

Slide 23

Distribution System Auto Reclosing

Longer fault clearance times– Fault damage (eg. arc burning) can cause a transient fault to

become permanentLonger fault clearance times

– Slow clearance can allow a semi-permanent fault to burn clearConnected load

– Industrial CustomersDead times must allow expensive, complex or dangerous processes to become fully disconnected before restoring supply

– Domestic CustomersSimple loss of supply is of prime importanceAutoreclose delay is chosen to optimise protection performance, minimise fault damage, and to automatically return supply to as many customers as possible

Slide 24

Distribution System Auto ReclosingHigh Speed Tripping

High speed tripping results in minimal fault damage and minimises the possibility of transient faults becoming permanentBut protection discrimination is lost

– Downstream faults may result in the rapid and non-selective tripping of upstream circuit breakers

– Auto reclose returns supply to all customersMultiple reclosures are usually implemented

– High speed tripping is inhibited after reclosure– On reclosure, permanent faults will be tripped by time

coordinated schemes, ensuring discrimination – After reclosure, permanent faults then result in tripping of

only the faulted portion of the system

Page 5



Slide 25

Distribution System Auto ReclosingSemi-Permanent Faults

Slower protection operations allow for semi-permanent faults to be burnt away– Perhaps due to contact with a tree

High speed tripping on the initial event– System disturbances minimised– Transient faults cleared and system restored

Slower time graded tripping after reclosure– Slower, coordinated tripping clears just the faulted

portion of the system– Slower tripping allows for semi-permanent faults to be

burnt away– May involve multiple reclosures

Slide 26

Distribution System Auto Reclosing Sectionalisers

Applies to radial distribution systemsFeeder fault is cleared at the source end CB (Recloser) Sectionaliser

– Cannot clear fault current– But counts the number of recloses– And sectionalises the feeder during the open dead time

CB(Recloser)

Sectionaliser Sectionaliser

N = 2N = 3

Slide 27

Distribution System Auto Reclosing Sectionalisers

Fault occurs … N = 1– CB Trips– No action by sectionalisers

Auto RecloseFault re-occurs … N = 2

– CB Trips– N = 2, so sectionaliser opens during the AR dead time

Auto recloseSystem restored, but with faulted section isolated

Sectionaliser Sectionaliser

N = 2N = 3

CB(Recloser)

Slide 28



Slide 28Slide 28

Auto ReclosingandSafety

Slide 29

Auto Reclosing and Safety

In autoreclosing fails, manual closing will be delayed to ensure public safetyAutoreclosing should be turned off in times of power system maintenanceAutoreclosing should not be initiated for faults not likely to be transient

Slide 30

Auto Reclosing and SafetySEF Protection

Distribution system sensitive earth fault protection– Detects very low level earth faults– These pose a significant danger to the public– These are rarely transient events– Faults to trees, fences, even to dry roads– Long time clearance : 10 or more seconds

Auto Reclose Reclaim time must be set longer than SEF protection timesSEF protection trips should inhibit or block any subsequent auto reclosing sequences.

Page 6



Slide 31

Auto Reclosing and Fire Considerations(Victoria Black Saturday)

Of special relevance to distribution and SWER systemsLonger fault clearances increase the likelihood of fire ignition.High speed autoreclosure also increases the likelihood of fire ignition

– The initial event may not cause fire ignition, but it predisposes dry forest fuel to ignition on subsequent events (reclosure)

– Probability of fire ignition is perhaps 3 times greater on reclosure than on the initial event

Slide 32

Auto Reclosing and Fire Considerations(Victoria Black Saturday)

In times of fire risk:– Protection operations need to be high speed– Second shot, slow protection tripping should be avoided– Autoreclosing dead times should be extended (30 secs or

more)– Autoreclosing should be turned off

Periods of extreme fire riskLocations with extreme consequences

Page 1

Capacitor Bank ProtectionFundamental Principles ofPower System Protection


Slide 1



Slide 1Slide 1

CAPACITOR BANK PROTECTION

Slide 2

Capacitor Bank Protection

Overcurrent & Earthfault protection– IDMT– INST

Differential protection– Eg. For HV & EHV applications– HZ or LZ biased schemes

Balance protectionOver Voltage protection

Slide 3

Capacitor Bank Ratings

Maximum continuous operation at 110% Voltage– But the system typically operates 1.05pu voltage or

higher– Thus typically specify capacitor bank to provide for this– Eg. A 20MVar 33kV bank would be specified and

purchased as 24MVar at 36kVMaximum continuous operation at 130% Current– (Extra component on current is to allow for harmonics)

Cfreq21Zc ⋅⋅π⋅

=

Slide 4

In-Rush Current

Add these sinusoidal quantities– Steady state (load) current– Inrush from the system– Inrush from adjacent banks

This is a worst case solutionAs an absolute worst case approach, use this current in determining IDMT O/C relay TMS Settings

( ) ( )2RMS_ADJt2

RMS_SYSt2

LRMS_TOT IeIeII 21 ⋅+⋅+= ⋅α−⋅α−

Slide 5

In-Rush Current - Adjacent Banks

Series Reactor(s) installed to limit inrush current– May be installed at line potential– May be installed at neutral potential, one per phase, above the star

pointApply to (n-1) banks to limit inrush from adjacent banksApply to all banks to limit inrush from the system

– Also limits outrush to system faultsInrush current may also be limited by POW switching

Slide 6

In-Rush Current

Effect of inrush on protection may be eliminated by using stabilised relays (ie. not sensitive to higher frequency components)– IDMT OC Protection

Probably no problems in any caseTypically set to 150%, 0.1 – 0.2 TMS

– High set (Inst) OC ProtectionShould be stabilisedAnd even then, 1 or 2 cycle time delay may be necessary

Page 2



Slide 7


Un-Earthed Capacitor Banks (non-effectively earthed systems)– With no earth connection, phase currents balance and residual

current on inrush is thus small.– Sensitive and fast IDMT EF protection can be applied.

Earthed Capacitor Banks (effectively earthed systems)– EF tripping on in-rush and also on out-rush is likely.– EF protection maybe disabled in such circumstances.– Stabilised EF protection can be simply set above load current.– Setting less than 2% of terminal EF levels not recommended.– Beware of setting electromechanical IDMT relays excessively

sensitive as the large relay burden can cause CT saturation under heavy terminal fault events.

No timing issues, so TMS of 0.1 may be appropriate.

Slide 8


Slide 9


Slide 14

Internally Fused Can

Failed element has negligible effect– Failure mode of a capacitor element is for it to short circuit– Associated fuse blows– Only a small part of capacitor is lost– Adjacent elements

Small current increaseSmall voltage increase

– Requires at least 8 elementsin parallel

Applicable to larger cansDischarge resistor

– Reduce can voltage to 50Vwithin 5 minutes

Slide 15

Externally Fused Can

Failed element shorts the parallel elements– Whole row is shorted and is effectively OOS

Can impedance reducesCan current increases

With further failures, external fuse operatesTypically for lower kVar cans

– Few elements in parallel– Many elements in series

Cap bank made up of series and parallel cans– Parallel cans allow bank to remain in

service with one can out (fuse blown).Discharge resistor

– Reduce to 50V within 5 minutes

Slide 16

Fuseless Can

Failed element shorts the parallel elements– Whole row is shorted and is effectively OOS

Can impedance reducesCan current increases

Typically for lower kVar cans– Few elements in parallel– Many elements in series

HV Capacitor bank– Cans in series, none in parallel

LV Capacitor bank– Cans in parallel, few (or none) in series

Discharge resistor– Reduce to 50V within 5 minutes

Page 3



Slide 17

Balance Protection

Unbalance detected by simple neutral displacement voltage measurementBut this will also be sensitive to system voltage unbalances

– Steady state– During faults

Monitor 3 phase terminal volts also to compensate for any system voltage unbalanceTime delay to allow system faults to clear Voltage

Displacement

Slide 18

CurrentBalance

Balance Protection

Cans positioned at commissioning for minimum neutral unbalance current flowMonitor unbalance neutral currentUnaffected by system unbalances

Slide 19

CurrentBalance

Balance Protection

Note that these two schemes give different results– Unearthed system neutral voltages are locked together, and

will change with unbalance– Earthed system neutral simply remain locked at earth

potential

Slide 20

Bala

nce

A

Bala

nce

Bala

n ce

B C

Balance ProtectionPhase Segregated Scheme

Slide 21

Internally FusedCapacitor Can

1 CapacitorCan

One Phase of a36kV, 24MVAr Capacitor Bank

BalanceProtection

Capacitor Bank Constructionand Failure Mode

Slide 22


1 CapacitorCan


BalanceProtection

Balance ProtectionPrinciples

When initially commissioned, zero current flows via balance protectionOn failure of one element in one can, a small current is now detected

– The parallel elements in that can also now have a small over voltage condition

– Hence, all other things being equal, the most likely subsequent failure is another element in the same row in the same can

– Unbalance current subsequently increases and is detected

Page 4



Slide 23

Balance ProtectionPrinciples

Trip before 10% overvoltage on the parallel cans – small time delay

– Typically about 50% of elements failedAlarm at half this value – small time delay

– Typically about 25% of elements failedUnbalance current is very small … maybe <1A primary

– CT ratio typically 1 / 1A– CT does not need a protection class specification, in fact a

measurement class CT should probably be specified


1 CapacitorCan


BalanceProtection

Slide 24

Over Voltage Protection

To trip the bank if the continuous voltage capability (110%) is exceededTo protect the system from over voltage due to the capacitor banksCoordinate with any nearby generator under excitation protection

– Trip capacitors to reduce system voltage before any generator protections may operate

Staged tripping recommended – To prevent wide spread capacitor tripping and hence

prevent a subsequent under voltage event from occurring– eg. Where more than one bank is installed at a substation– eg. Where banks are installed a nearby substations

Documents

Fundamentals of Power System Protection_2012