Generator and Power Station Protection_2012.pdf

8/10/2019 Generator and Power Station Protection_2012.pdf

1/62

Generator and

Power Station Protection

Synopsis

Disclaimer

Barrie Moor : 2012 [email protected]

(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION

Barrie Moor, B Eng (Elec)

Slide [email protected]

DisclaimerSeminar Synopsis

[email protected]@powersystemprotection.com.auwww.powersystemprotection.com.au


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 theinformation 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.


2/62

Generator and


Synopsis

Disclaimer


(07) 3298 5260


Seminar Synopsis

Over Current Protection

Differential Protection

High Impedance Differential Protection

Biased Differential Protection

Sequence Components

Motor Protection

Generator Protection

Generator Faults

Generator Events

Power System Events


3/62

Seminar costs will vary depending on attendeenumbers and your individual circumstances.However, by organising your own in-houseseminar:

You can expect savings of between 40%and 65%.

Plus you eliminate travel andaccommodation expenses for all of yourattendees.

We provide:

2 or 3 day seminar presentation All seminar handout material, notes, folders,

CDs, etc Laptop Computer and Data Projector Note that we guarantee that all seminars will be presented personally by

our principal engineer and seminar author, Barrie Moor

Each attendee receives:

Two or three day seminar presentation Hard copy manual with all presentations, plus supporting technical

papers

CD with all printed material, plus considerable extra material and tools,including: pdf of seminar colour slides 2 per page additional technical papers tools for sequence component analysis of single, double and three

phase faults tools for grading of IDMT overcurrent relays tools for distance relay calculations, apparent impedance calculations,

fault resistance, mho and offset mho characteristic load limits

Certificate of attendance

You provide:

Seminar conference room (preferably on-site, within your own facilities) Whiteboard Any catering for lunch and tea breaks

To discuss your requirements, or to obtain a firm price quotation,please contact us at:

[email protected]


4/62

Generator and



Fuses and Contactors


(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION



OVER CURRENTPROTECTION

Overcurrent Relays

Fuses & Contactors

Directional Relays



Over Load Protection

Operation to the thermal capability of plant


Primarily for clearance of faults

Some measure of over load protection may be provided


Discrimination by Time

Setting chosen to ensure CB nearest to the fault opens

first

Often referred to as

Independent Definite Time Delay Relay

Timing intervals selected to ensure upstream relays do

not operate before CBs trip at fault location

Disadvantage

Longest fault clearing time occurs in section closest tothe power source where fault level is the highest


Discrimination by Time

RELAY A RELAY B RELAY C

RELAY C

RELAY B

RELAY A

CURRENT

TIME

0.4 secs

0.4 secs


Discrimination by Current

Apply where fault current varies with fault location due to

intermediate impedance

Set to operate at current values so that only relay

nearest to fault trips its CB

Difficulties

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)



RELAY A RELAY B

Relay A cannot distinguishbetween a fault here, forwhich it needs to operate

And a fault here forwhich it should notoperate


5/62

Generator and





(07) 3298 5260



Significant difference betweencurrents seen for Faults A & B

Set HV OC to 1.3 x maximum

through current for LV Fault

HV OC

FDR OC

FDR OC

FDR OC

FDR OC

A

B


Discrimination by Time & Current

Time and currentcoordination


RELAY C

RELAY B

RELAY A

CURRENT

TIME

ICmax IBmax IAmax

IAmax IBmax ICmax

Instantaneouselement


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


Discrimination w ithInverse Time Over Current Relays

Inverse time andcurrent coordination


RELAY C

RELAY B

RELAY A

CURRENT

TIME

ICmax IBmax IAmax

IAmax IBmax ICmax

Instantaneouselement


Relay Curves to IEC 60255(BS142)

I = Actual relay current

Relay Settings

TMS = Time Multiplier Setting

P = Plug (Current) pickup setting

Usual curve for t ransmission and distribution systems

1P

I

TMS14.0TIME

02.0Inverse_dardtanS

=




Relay Settings



Systems where the fault level decreases significantlybetween relaying points

1P

I

TMS5.13TIME Inverse_Very

=


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Generator and





(07) 3298 5260




Relay Settings



Grading with fuses

1P

I

TMS80TIME

2Inverse_Extremely

=




Relay Settings



Long time thermal protection

Motor & Generator Protection

1P

I

TMS120TIME

Inverse_Time_Long

=


Standard Characteristics to IEC 60255

Long Time (LTI)

Extremely Inverse (EI)

Very Inverse (VI)

Standard Inverse (SI)

Relay Characteristic

1I

TMS14.002.0

1I

TMS5.13

1I

TMS802

1I

TMS120

100 1.103

1.104

0.1

1

10

100

Standard Inverse

Very Inverse

Etremely Inverse

Long Time Inverse

IDMT Relay Grading Curves

Fault Current

Seconds


US Characteristics to IEC 60255

U5 Short Time Inverse

U4 Extremely Inverse *

U3 Very Inverse

U2 Inverse

U1 Moderately Inverse

Relay Characteristic

+1M

0104.00226.0TD

02.0

+1M

95.5180.0TD

2

+1M

88.30963.0TD

2

+1M

64.502434.0TD

2

+1M

00342.000262.0TD

02.0

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


Electro Mechanical Relays

FLUX PRODUCED BY INPUT

TAPPEDCOIL

I

- L

SHADING LOOP

FLUX PRODUCED BY INPUT CURRENT

DISC DISC

FLUX PRODUCED BY SHADING LOOP

kI

L

(1-k) I


7/62


8/62

Generator and





(07) 3298 5260


Instantaneous Element

Reduces tripping time at high fault levels

Allows 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

4IDMT Relay Grading Curves

Fault Current

Seconds

100 1.103

1.104

0.5

1

1.5

2

2.5

3

3.5


Fault Current

Seconds

100 1.103

1.104

0.5

1

1.5

2

2.5

3

3.5


Fault Current

Seconds Set Tx HV inst

element and nowgrade here

OC

OC

GENERATOR and

POWER STATION

PROTECTION




Setting andCoordination

Procedures


Relay Coordination ProcedureCurrent Setting

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 relay


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

relay

Allow for load pickup current


Load Pickup Current

Motor starting current

Auxiliary heaters

Transformer magnetising inrush

Capacitor charging current

Lighting loads - 10s to 100s of msec

Filaments and electrodes heating

Arc lamps starting


9/62

Generator and





(07) 3298 5260


Load Pickup Current

Hot load pickup

Short term loss of supply and subsequent load pickup

currents on return of supply

Cold load pickup

Load pickup, but now with loss of diversity between cyclicloads

Voltage 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


Relay Coordination ProcedureTime Multip lier Setting

Coordinate relays via time multipliers to achieve

appropriate grading margins

Determine, under various system configurations, thevalues of short circuit current that will flow through eachprotective 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)


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)


Relay TMS Grading

Relay TMS Grading

Hence for an E/M relay tr ipping 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 [email protected]

Grading of Parallel Elements

Worst case for grading is with only 1 transformer in service But this will be an unusual operating condition

E/M & Electronic Relays Only a single relay setting is available

Hence, effectively no option but to set for the worst case, namely1 transformer case

And accept slower performance for system normal,namely when both transformers are in service

Microprocessor 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


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Generator and





(07) 3298 5260

Grading of Parallel Elements

Maximum through fault level occurs when both transformers

are in service

But the maximum individual transformer current flows whenthe 2nd transformer is OOS

Need to consider both conditions when grading relays

HV OC

HV OCFdr_1 OC

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

3 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

Imax

Fdr1_Plug10=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:=

Imax

Fdr2_Plug25=

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 800A

and check against maximum fault level of 10kA

SI P TMS, I,( ) 0.14TMS

I

P

0.02

1

:=Relay Characteristic

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, Imax,( ) 0.832= _T 0.411=

100 1 .103

1 .104

1 .105

0

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 1 Tx_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_33kV 20000000

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, Imax,( ) 0.769=

Tx HV Relay_3n

SI Tx_HV_Plug Tx_HV_TMS, I3

n

,

( ):= SI Tx_HV_Plug Tx_HV_TMS, Imax,( ) 1.187= _T 0.418=

100 1.10

3

1.10

4

1.10

50

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 OC

Fdr 1 OC

Tx HV OC

Tx OC Grading (11kV Base Currents)


Sequential Operation of Over CurrentRelays

As CBs trip, fault current magnitudes and flows will change

We need to integrate how far each relay progresses towardstripping 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 thetotal of Relay 2 and the subsequent Relay 3 operations

Relay 3Relay 2

Relay 1


11/62

Generator and





(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION




Directional Relays


Directional Over Current Relays

Extra discrimination may be achieved by making the

response of the relay directional when current can flow in

both directions Achieved via voltage (polarising) connections to the relay

Digital and numeric relay achieve phase displacementsvia software

EM & Static relays require suitable connection of input

quantities to the relay


Directional Over Current RelaysAppl ication to Paral lel 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 2

BA


0.90.9

0.

5

0

.5

0.1 0.1

1.3

1.

7

2.1 2.1

1.

7

1.31.3

Ring Mains Systems

Open the ring at thesupply point

And then gradeclockwise

And then anticlockwise

Source substation relaysdo not HAVE to be

directional Intermediate substation

slower relays do notHAVE to be directional

GENERATOR and

POWER STATIONPROTECTION




Earth Fault Relays


Earth Fault Protection

Implement more sensitive protection responding only to

residual current of the system

Low 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 FLC

Time grade in the same manner as for phase OC relays

Beware 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 relays magnetic circuits


12/62


13/62

Generator and





(07) 3298 5260


Expulsion Fuses

Used where expulsion gases cause no problem such as in overhead

circuits and equipment

Special materials (fiber, melamine, boric acid, liquids such as oil orcarbon tetrachloride ) located in close proximity to fuse element andarc rapidly create gases

These produce a high pressure turbulent medium surrounding the

arc

Expulsion process deionises gases them as well as removing themfrom arc area

In inductive circuits, transient recovery voltage (TRV) will be

maximum at current zero.


Fuses & TRV Performance

0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242

1.5

1

0.5

0

0.5

1

1.5

2

Circuit VoltageFuse Voltage

Current

0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242

1.5

1

0.5

0

0.5

1

1.5

2


Current

0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242

1.5

1

0.5

0

0.5

1

1.5

2


Current

0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242

1.5

1

0.5

0

0.5

1

1.5

2


Current

0 0 .0 02 0 .0 04 0 .0 06 0 .0 08 0 .0 1 0 .0 12 0 .0 14 0 .0 16 0 .0 18 0 .0 2 0 .0 22 0 .0 242

1.5

1

0.5

0

0.5

1

1.5

2


Current

Current lagsVoltage by 90

deg

SystemVoltage

Currentinterrupted at

naturalcurrent zero

TRV acrossblown fuse

element

Fuse Voltage


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 reduced

Fuse element is completely surrounded with f iller 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


Current Limiting Fuses

Tin for M Effect lowoverload fuse performanceSee later




Current Limiting FusesM Effect for low level overloads

M Effect : A.W. Metcalf - 1939


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Generator and





(07) 3298 5260




Grading Relays wi th Fuses

Extremely Inverse curve

follows a similar I2t

characteristic Relay current setting should

be approximately 3 times the

fuse rating

Grading margin of not less

than 0.4 seconds

recommended

Or 15.04.0' + tT

1PI

TMS80TIME

2Inverse_Extremely

=


Grading Relays with Fuses

First relay upstream of the fuse should be set to EI

characteristic

Now 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 characteristic

And subsequent further upstream relays to SI characteristics


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 abovethis point


100 1.103

1.104

1.105

0.01

0.1

1

10

100

Fuse

Relay / ContactorFuse

Fuse Contactors

10kA Contactoroperates for faults

below 7kA

Fuse operates forfaults above 7kA

Fuse operationbelow 2kA is not

permissible


15/62

Generator and

Power Station Protection High Impedance Differential Protection


(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION



HIGH IMPEDANCEDIFFERENTIALPROTECTION

Busbar Protection and

Galvanically Connected

Plant


Synopsis

HZ Differential Protection Principles

Determination of setting voltage

CT requirements

Current operated schemes and stabilising resistors

Limiting secondary system voltages to safe levels

Primary operating current and application of shuntresistors

CT supervision requirements

Application of high impedance differential protection

schemes to other galvanically connected plant


SIMP

LE!!!

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


CT Connections & Polarity

S2

P2

S1

P1

I1

I2

P2

S2

P1

I1 S1

I2


RELAY

Internal Fault


RELAY

External Fault


16/62

Generator and



(07) 3298 5260


3 Phase CT Connections

CT MARSHALLING

DIFFERENTIALRELAY


Current Mismatch

CT Manufacturing Variations

Inequality of CT Burdens

CT Saturation

Highest Fault Current on CT exposed to through fault

Worst possible mismatch is

Total saturation of the CT on the faulted plant

All other CTs transform perfectly


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


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, butrelay operation is assured provided absolutely all CTs

meet the requirement


17/62

Generator and



(07) 3298 5260


CT Selection

All CTs to be the same ratio

All 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 absolutely critical

But highly recommended to specify class PX CTs

0.1 PX 500 R3

Magnetisingcurrent at kneepoint voltage

Magnetisingcurrent at kneepoint voltage

CT knee

point voltage

CT knee

point voltage

CT internalresistance

CT internalresistance


Summary

Ensure Stability under through faults

Ensure Operation for genuine in-zone faults

Beware of short cut methods

Do not simply set

RELAYKNEE V2V

2

VV KNEERELAY=

LEADSCTFAULTRELAY RRIV

Preferably 5 times to optimiserelay performance, but 2 is theabsolute minimum to ensure

reliable relay operation

Preferably 5 times to optimiserelay performance, but 2 is theabsolute minimum to ensure

reliable relay operation


Current Operated Schemes

Voltage operated

Current operated, incl stabilising

resistor

Typical 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


Metrosils

In the case of a heavy internal fault, secondary system voltages maybecome excessive

Implications include damage to equipment and safety of personnel

Empirical Formula

VK = CT RMS knee point voltage

VF = Maximum RMS voltage that would occur if the CT did notsaturate

Install metrosils if this voltage become excessive(eg. >2.8kV peak)

KFKPEAK VVV22V


Metrosil Parameters

Where V & I are PEAK

values

C = Metrosil constant

B = Metrosil constant

(0.2 0.25)

And, because of the

non-linearity

I52.0IRMS

ICV METROSIL NON-LINEAR RESISTORS

V = C . IB

10

100

1000

1104

0.001 0.01 0.1 1 10 100

C = 450 B = 0.25

C = 600 B = 0.25

C = 900 B = 0.25

Metrosil Current - Amps


Metrosil Parameters

Based on the previous equations, the RMS current at the

relay setting voltage Vs(rms) is:-

52.0

I

2

CV RMS

RMSSetting

1

Setting

RMSC

V252.0I RMS


18/62

Generator and



(07) 3298 5260


MAGMETROSILRELAYP INIICTI

RELAYImag

R

shunt

Ishunt

Irelay

Imet r

osil

Imag

N = Number of CTs in parallel

Primary Operating Current


Shunt Resistors

Desensitise scheme

Prevent tripping on open CTs

Primary operating current > maximum plant loading

Effectively becomes a medium impedance scheme


MAGSHUNTMETROSILRELAYP INIIICTI

RELAYImag

R

shunt

Ishunt

Irelay

Imet r

osil

Imag

N = Number of CTs in parallel

Primary Operating Current


CT Supervision

Effect of CT problems when Shunt Resistors are installed

Scheme is part way to a trip condition

Effect of CT problems when BZ Check scheme is installed

Half of the scheme is already tripped

CT Supervision Setting Principles

Set to 50% of minimum load

Operation : initially (eg. < 3 secs)

Nil : allow for correct operation of BZ protection

Operation : short time (eg. > 3 secs)

Alarm

CT Shorting

GENERATOR and

POWER STATION

PROTECTION



HIGH IMPEDANCEDIFFERENTIALPROTECTION

Application to other

Plant


HZ Protn Application to Plant

Requires Galvanic Connection

All CT ratios the same

Can Apply To

Busbars

Transformers

Generators & Motors

Capacitors

Reactors


19/62

Generator and



(07) 3298 5260


Relays

StabilisingResistors

Generators & Motors


DIFF

Auto Transformers

All CT r atio s tobe the same

This CT will carrymaximum currentand hencedictates ALL CTratios

But this CT is internaland may have a singlefixed ratio.Thus, must be specifiedcorrectly at time ofpurchase !!


REF

Restric ted Earth Fault Protection

CT terminalsaway fromprotected objectare connected

CT terminalsnear toprotected objectare connected


DIFFDIFF

A

DIFF

B C

Reactors Earthed Neutral


DIFFDIFF

A

DIFF

B C

Reactors Floating Neutral

Floating neutralbus is also protected


DIFFDIFF

A

DIFF

B C

Capacitors Earthed Neutral


20/62

Generator and



(07) 3298 5260


DIFFDIFF

A

DIFF

B C

Capacitors Floating Neutral

Floating neutralbus is also protected


21/62

Generator andPower Station Protection Sequence Components

Barrie Moor : 2012 [email protected](07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION



SEQUENCECOMPONENTS

A Quick Review


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


Sequence Components

I1 I2 I0

I phase

IC

IB

IA


Sequence Networks

Source

RelayLocation

FaultLocation

PositiveSequenceNetwork

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation

NegativeSequenceNetwork

Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Zero

SequenceNetwork

Z0f

Z0s

I0

V1 = 1 / 0 V2 = 0 V0 = 0


Sequence ComponentsThree phase conditions

Positive sequence only

Three phase load

Three phase fault

No neutral (earth fault) current


In = 0

3 Phase Balanced Current

Balanced currents sum to zero

Positive sequence currents

Negative sequence currents

But zero sequence current will sum to 3.IoIn = 3.Io


22/62




Sequence Networks3 Phase Fault

Source

RelayLocation

FaultLocation


Z1s

Z1f

I1

f1Zs1ZZ

Z

V1II

POS

POS

Fault_Phase_3

=


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 CPhase to Phase fault


Sequence Networks (A phase)Phase Phase fault

A phase

IA1 & IA2 antiphaseSum to zero

B phase

IB1 & IB2 at 60o

C phase IC1 & IC2 at 60o

IB = - IC

Source

RelayLocation

FaultLocation

Positive

SequenceNetwork

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation


Z2f

Z2s

I2

2I1I=


Sequence Networks (A phase)Phase Phase fault

Source

RelayLocation

FaultLocation

Positive

SequenceNetwork

Z1s

Z1f

I1

Source

RelayLocation

FaultLocation


Z2f

Z2s

I2

Since Z1 ~ Z2

|I1| = |I2| = 50%of 3 phase faultlevel

negpos ZZ

V2I1I

=


Sequence ComponentsPhase Phase fault

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

Thus |IB| = |IC| = 86.6% of 3 phase fault level(because of 60o angles)


Sequence ComponentsEarth Fault

A phase positive sequencenegative sequencezero sequence

Equal in magnitudeand phase


23/62




Sequence ComponentsEarth Fault


Sequence NetworksA Phase Earth Fault

Source

RelayLocation

FaultLocation


Z1s

Z1f

I1

Source

RelayLocation

FaultLocation


Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

ZeroSequenceNetwork

Z0f

Z0s

I0


Sequence NetworksA Phase Earth Faul t

Source

RelayLocation

FaultLocation


Z1s

Z1f

I1

Source

RelayLocation

FaultLocation


Z2f

Z2s

I2

Source

RelayLocation

FaultLocation

Ze

roSequenceNetwork

Z0f

Z0s

I0

I = 1.0 I = 1.0 I = 1.0

V1 = 1 / 0 V2 = 0

V0 = -0.15

V0 = -0.50

V2 = -0.10

V2 = -0.25

V0 = 0

V1 = 0.90

V1 = 0.75

= 0.10 = 0.10

= 0.15= 0.15

= 0.15

= 0.35


Sequence NetworksA Phase Earth Fault

PositiveSequence

FaultLocation

NegativeSequence

FaultLocation

ZeroSequence

FaultLocation

RelayLocation

ZS1

Zl

ZS2

Zl

I2I1

RelayLocation

ZS0

Zl

I0

RelayLocation

zeronegpos ZZZ

V0I2I1I

=

0I2I1IIA

0IB=

0IC=

0I3INEUT


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

Io

Io

Io

3Io

GENERATOR and

POWER STATION

PROTECTION



TRANSFORMERSand

SEQUENCECOMPONENTS

DifferentialProtection

Requirements


24/62




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



Star / Star Transformer : LV Earth Fault But, suppose we dont 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 ??



Star / Star Transformer : LV Earth Fault

Retain the delta connected tertiary winding

But, lets reinstate the power system earth

Power system and delta winding zero sequence current flowdistributions will depend on their relative Z0 impedances

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

I0I0


Transformers, Sequence Componentsand Differential Protection

Star/Star transformers, with a delta tertiary winding:

Will have a mismatch between zero sequence current flows onthe HV & the LV windings

It is thus necessary to exclude zero sequence current from thedifferential relay protection algorithms

Star/Star transformers, without a delta tertiary winding:

May still have a mismatch between zero sequence current flowson 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



Delta / Star Transformer : LV Earth Fault Current in corresponding HV winding only

Appears as phase to phase fault from the perspective ofthe HV system

I1 & I2 onlyI1 & I2 only

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

I1, I2 & I0I1, I2 & I0


Positive Sequence Network

LVZS

ZS HVZ1HL

HV

LV

Zfdr

Zfdr


25/62




Negative Sequence Network

ZS

ZS Z1HL

LV

HV

HV

LV

Zfdr

Zfdr


Zero Sequence Network

ZS

ZS Z1HL

LV

HV

HV

LV

Zfdr

Zfdr



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



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


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


Sequence ComponentsTransformer LV ph-ph fault

LV phase to phase fault HV distribution is 1 : 2 : 1 LV distribution is 0 : 1 : 1

So be careful when grading HV & LV IDMT OC Relays |I1| = |I2| = 50% of 3 fault level But note that I1 & I2 are 60o apart on the LV, but are shifted 30o and

are thus, for the 2 phase, in phase on the HV LV current is 86.6% of 3 fault level in both phases HV current in the 2 phase is the same as for LV 3 fault !!

30 deg

LVHV


26/62




Transformers, Sequence Componentsand Differential Protection

Compensate for the transformer phase shift

Exclude 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 deltawindings

Zero sequence current can circulate around delta windings(said to be trapped in the delta)


27/62

Generator and

Power Station Protection Transformer Protection


(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION



TRANSFORMERPROTECTION


Types of Fault

Phase-ground faults - from winding to core or winding to tank

Phase-phase faults - between windings

Interturn faults - between single turns or adjacent layers of the samewinding (Buchholz)

Arcing contacts

Local hotspots caused by shorted laminations

Low level internal partial discharges (moisture ingress or designproblems)

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)


Buchholz Protection

Two floats in the relay:

Upper float

Detects accumulation of gas

Detects loss of oil

Incipient 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 < 100ms

Although 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


Pressure Relief Device (Qualitrol )

Spring assisted pressure relief devices

Relieves pressure impulses due to massive internal fault conditions.

Helps prevent the tank bursting or splitting

Relay 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. Several units may therefore

be required on larger transformers.


Basic Transformer Protection

Fuses

Transformers without CBs

Perhaps to a few MVA

Overcurrent & Earth Fault Protection

Transformers with CBs

Perhaps 5 - 50MVA


Transformers > 10MVA Fast

Can be sensitive

May detect terminal faults also

GENERATOR and

POWER STATION

PROTECTION




Biased Differential

Protection


28/62

Generator and



(07) 3298 5260



DifferentialRelay

P1

S1

P2

S2

SIDE OF CT AWAYFROM PROTECTEDPLANT CONNECTS

TO RELAY

CURRENT FLOWSINTO PLANT

CURRENT FLOWSOUT OF PLANT

CURRENT FLOWSINTO RELAY

CURRENT FLOWSOUT OF RELAY

SIDE OF CT AWAYFROM PROTECTEDPLANT CONNECTS

TO RELAY

TRIPELEMENT

IT IS NOT THE P1/S1 OR P2/S2ORIENTATIONS THAT ARE

IMPORTANT, BUT THEPREFERENCE FOR THE

AWAY SIDES OF THE CTsTO CONNECT TO THE RELAY


Differential Protection of Transformers

11/132kV

2400/1 200/12400A 200A

1A 1A

BIAS orRESTRAINTELEMENT

BIAS orRESTRAINTELEMENT

TRIPPING ELEMENTDETECTS ONLY THE

MIS-MATCH CURRENT

TRIPELEMENT


Transformer Differential Mismatch

Differential CT ratio selection

CT ratios selected to compensate for the transformer turnsratio

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 somesaturation due to DC fault current sets in.

The amount of bias is increased under heavy through faultconditions to compensate for possible CT saturation


Transformer Differential Mismatch

Inrush on energisation (2nd harmonic)

Over excitation (5th harmonic)

Transformer phase shifts

Earth fault (neutral zero sequence) currents


Inrush Current onEnergisation of Transformer

TRIPELEMENT


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 secondharmonics, 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)


29/62

Generator and



(07) 3298 5260


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

Inrush current : with 2nd Harmonic


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 andrun down

Not so necessary for transmission or distribution applications

Frequency will not deviate significantly from nominal


Bias Differential Protection

Allow for Transformer turns ratio

Allow for Transformer phase shifts

Eliminate Zero Sequence currents from the relaying system

OperatingWinding

P1

S1

Bias Windings

P1

S1

GENERATOR and

POWER STATION

PROTECTION



SEQUENCECOMPONENTS

A Quick Review

GENERATOR and

POWER STATION

PROTECTION




Continued


CT Connections and Ratios

Star/Delta and Delta/Star transformers have a 30 degree phase shift

Compensate 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 compensated

Zero sequence currents flowing in the transformer star windings

prevented from entering the relaying system

But how do we get the correct delta connection for our CTs ???


30/62

Generator and



(07) 3298 5260


Determination of CT Connection

Diff Prot

Yd11

D11D11

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


Determination of CT Connection

Dy11

D1D1

Diff Prot

CT Primary is star connectedCT secondary is D1 connectedOverall connection is thus YD1


Star/Delta and Delta/Star TransformersCT Connection Summary

Transformer HV is STAR connected

HV CTs are delta connected (ie. phase shift ing)

HV CTs EQUAL to the transformer phase shift

LV CTs in star

Transformer LV is STAR connected

LV CTs are delta connected (ie. phase shifting) LV CTs OPPOSITE to the transfor mer phase shift

HV CTs in star


CT Connection Summary

Compensates for the phase shift across a star-deltatransformer.

The correct vector group must be chosen for the CTs toensure that through currents balance.

Prevents any zero sequence currents flowing in the starwinding 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


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

S1

C

S2

S1

S2

B

C

B

A

CT YD11 Connections

D11D11






D11

S2

A

S1

S1

C

S2

S1

S2

B

D1

C

B

A

S2B

S2

S1

S1

A

S1C

S2

C

B

A

CT YD1 Connections

D1D1


31/62

Generator and



(07) 3298 5260

A B

Bias Windings

P1 P2

S1 S2

A1 A2

OperatingWindings

C

a2 a1P1P2

S2 S1

Notice that theconnections forthe Delta windingsare the same !!

Away side of CTs connected to relay.Hence, transformer current in or outcorresponds to relay current in or out.

Away side of CTs connected to relay.Hence, transformer current in or outcorresponds to relay current in or out



There must be a path for the current to flow

There must be an Ampere Turns balance

If 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

OperatingWindings

C

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

OperatingWindings

C

a2 a1P1P2

S2 S1

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)

External Phase Phase Fault


Delta CTs and Ratio Selection

CT ratios must allow for the fact

that current flowing into therelay from the delta connected

CTs is 3 times the CTsecondary current

Hence, a standard 1A CT will

result in relay current of 3times the CT secondary current

Thus, 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


Delta CTs and Ratio Select ion

Ia - Ic = 1.732 /-30

CT ratios must allow for the fact

that current flowing into therelay from the delta connected

CTs is 3 times the CTsecondary current

Hence, a standard 1A CT will

result in relay current of 3times the CT secondary current

Thus, CTs with ratios such as

1000/0.577 are, for this reason,quite common.


32/62


33/62

Generator and



(07) 3298 5260


CT Ratio Selection

The CT ratios must be opposite to the transformer ratio

Choose one CT ratio

Base all other CT ratios on this selection and transformerturns ratio

Not on winding MVA !!!

And, as noted before CT ratios must allow for the fact

that current flowing into the relay from the delta connected

CTs is 3 times the CT secondary current


Modern Microprocessor Relays

All CTs connected in Star

Relay has to process phase shifts

Relay has to remove neutral current

P1

P1

P1

S1

S1

S1

S1

S1

S1

P1

P1

P1



P1

P1

P1

S1

S1

S1

S1

S1

S1

P1

P1

P1

ICIAIARELAY

IAIBIBRELAY

IBICICRELAY

IC

IB

IA

110

011

101

3

1

IC

IB

IA

RELAY

RELAY

RELAY

IC

IB

IA

110

011

101

3

1

IC

IB

IA

RELAY

RELAY

RELAY


D1

D11





D11

S2

A

S1

S1C

S2

S1

S2

B

D1

C

B

A

S2B

S2

S1

S1

A

S1C

S2

C

B

A


=

IC

IB

IA

110

011

101

3

1

IC

IB

IA

RELAY

RELAY

RELAY

=

IC

IB

IA

101

110

011

3

1

IC

IB

IA

RELAY

RELAY

RELAY



=

IC

IB

IA

211

121

112

3

1

IC

IB

IA

RELAY

RELAY

RELAY

D12

D1

D11

=

IC

IB

IA

110

011

101

3

1

IC

IB

IA

RELAY

RELAY

RELAY

=

IC

IB

IA

101

110

011

3

1

IC

IB

IA

RELAY

RELAY

RELAY


Modern Microprocessor RelaysD12 Zero sequence current elimination

=

IC

IB

IA

211

121

112

3

1

IC

IB

IA

RELAY

RELAY

RELAY

( )ICIBIAIA33

1=

( )( )ICIBIAIA33

1++=

( )0I3IA33

1=

0IIA=

( )ICIBIA23

1IARELAY =


34/62

Generator and



(07) 3298 5260



All CTs can now be connected in Star

Relay internal processing adjusts for phase angle

Relay internal processing rejects zero sequencecomponents

CT 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


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

Yy0

Software CTx 1.143

Software CTx 1.429

Yd11

00 -300

TAP POSITION

SoftwareCT Ratio

Adju stm ent


Differential Relay Bias Settings

CT ratio selection

Select one ratio to meet load requirements

Base all other ratios on the first (not on load !)

Allow Margin (perhaps 10-15%) allowing

the worst mismatch of transformer ratio

and CT ratios (remember 3 for delta CTs !!!)

To decide worst case - consider the overall scheme

At the top tap position .......... & then

At the bottom tap position.

And both should be the same for a microprocessor based

relay that is correctly mid tap balanced


Differential Relay Bias Settings

Use the relay equations to determined the worstmismatch

Typically

I1 & I2 are the currents into the two sides of thetransformer

2I1IDiff

2

2I1IBias

50%Not Average !!


Tap Changer Position

For any setting of tap changer and through current, andgiven the CT ratios, the values of bias current anddifferential current can easily be calculated.

Base calculation on the relay algorithm not on anarbitrary or simple mismatch calculation Note that the previous equations are not the only options

available to and used by relay manufacturers

Hence use THE ACTUAL RELAY equations !!

Allow the recommended safety margin Probably 10 15%

And check your e lbow !!


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 Setting



Differential

Current

Diff I1 I2+:=

Bias Current BiasI1 I2+

2:=

OPERATEOPERATE

TransformerInternal Fault

Protection Trips

TransformerInternal 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


35/62

Generator and



(07) 3298 5260


CT Requirements

Some CT saturation is permissible for through faults

Such CT saturation occurs mainly due to the DC

component of the fault current

Most manufacturers provide simple equations to

determine CT class

No complex calculations required

GENERATOR and

POWER STATION

PROTECTION




Winding Neutral

End Faults


Protection for neutral end earth faults

Small current Many turns

Large current Few turns


Line current mismatchis too small to trip thedifferential relay

We need to monitorneutral current flow


Transformer Neutral Fault Current & Primary Current

0 10 20 30 40 50 60 70 80 90 1000

2

4

6

8

10

12

14

16

Fault Current in Shorted TurnsPrimary Current

FAULT

CURRENT

multiples

of

rated

current

DISTANCE OF FAULT

FROM NEUTRAL

(Percentage of winding)

Significant current flows in theactual fault near to the neutralend of the transformer winding

But very little linecurrent flows at theHV terminals


REF


CT terminalsaway fromprotected objectare connected

CT terminalsnear toprotected objectare connected


DIFF

Auto TransformerHigh Imp Phase Segregated Protection

All CT r atio s tobe the same

This CT will carrymaximum currentand hencedictates ALL CTratios

But this CT is internaland may only have asingle fixed ratio.Therefore, this must bespecified correctly atthe time of purchase !!


36/62

Generator and



(07) 3298 5260


High Impedance Protection

Especially sensitive in detecting earth faults in the

bottom 25% of the winding

Application to

Auto Transformers

REF earth fault schemes

Tertiary windings not specifically protected

Setting principles same as for bus zone

Select voltage setting to achieve through fault stability

CT Vk > 2.Vset to ensure scheme operation

HighImpedanceRelays

Tertiary

HV CTs LV CTsNeutral CTs One per phase at neutral end But above the star point


Restricted EF ProtRestricted EF Prot

Transformer delta winding and star winding REF protection schemes

LV scheme remainsstable for external EF

HV scheme remainsstable for external EF

HV scheme also remainsstable for all LV EFs

GENERATOR and

POWER STATION

PROTECTION



EARTHINGTRANSFORMERS

Operationand

Protection


Earthing Transformer

LOAD

Earthing Transformer Fault Currents


B

A

C



37/62

Generator and



(07) 3298 5260



LOAD

Earthing Transformer Fault Currents


Earthing Transformers

Provide a good earth reference for a delta winding duringearth faults

Restrict the voltage rise on the healthy phase duringearth faults inoperative during balanced voltageconditions

Carry significant current only during earth faultsie. 3 x I0 only

Earthing transformer and associated power transformerconsidered as a single unit and tripped together


Effectively Earthed System

R0/X1 < 1

X0/X1 < 3

Limit voltage rise on unfaulted phases

80% of rated phase - phase voltage

Otherwise healthy phases can reach 100% of rated phase

- phase voltage

May even exceed this on transients


Protection of Earthing Transformer

Two types of faults we need to consider:

Internal faults - faults inside the earthing transformer, the

result of insulation breakdown.

External faults - faults on the system outside the earthing

transformer. These can cause overheating of the earthing

transformer




Interturn, interwinding or winding-to-core faults

Fed from delta-connected current transformers, so thatearth faults on the system, which generate only zero

sequence current, are not seen

Hence, O/C setting can be very low




Long term low level unbalance may cause thermal

damage

Need to consider and set earth fault below

continuous rating capability

short-time rating capability

Combination of IDMT and definite time functions used


38/62

Generator and



(07) 3298 5260



O/C relay does not operate for externalearth faultsDef Time and IDMT E/F relays operate forexternal earth faults IDMT E/F relay

Def Time E/F relay

O/C relay

LOAD


Overcurrent Setting Must be:

Greater than the magnetising current

Greater than the maximum inrush current. This depends on

Earthing transformers B-H characteristics The point-on-wave of the energisation

The remanence of the core

One common estimate of upper bound is 50x the magnetising

current

Tolerant of Earthing Transformer manufacturing variations

Impedances variations between the individual phases of the

earthing transformer may result in some spill current from the

delta CTs under through EFs



Affected by long term residual voltage, which may cause

thermal damage

remember - no over temperature sensor is provided

Need to consider and set earth fault below

continuous rating capability

short-time rating capability

Combination of IDMT and definite time relays used to do

this

Thermal Protection

10 100 1 103

1 1040.1

1

10

100

1 103

1 104

1 105

EARTHING TRANSF THERMAL PROTECTION

EARTH FAULT CURRENT - AMPS

TIME-SECONDS

30

2300

contrating30A

max E/Fcurrent2300A

adiabatic thermal limit

actual thermal limit

earthing transformer E/F relay - Definite Time

downstream E/F relay

earthing transformerE/F relay - IDMT


Biased Differential Protection

Earthing transformers are included inside the biased differential

zone of their power transformer

Current transformer connections important

LV CTs will be star connected

Thus Io can flow to the differential relay

And earthing transformer now supplies Io

Thus stability for external earth faults becomes an issue

To eliminate Io from the relaying system

YD12 CT connection with microprocessor differential relay

Or utilise additional Earthing Transformer CT connections

0

C

B

A

400/0.577

132 / 33 kV

externalearthfault

1600/1

0

c

b

0a

Earthing Transformer CTsDifferential Trip due to LV I0 Currents


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Generator and



(07) 3298 5260

all 1600/0.333

0

C

B

A

400/0.577

132 / 33 kV

Nexternal

earthfault

1600/1

0

c

b

0a

Earthing Transformer CTsEliminate I0 from the relaying system


Zero Sequence Current Trap

Star connected interposingCTs with neutral connection

Interposing CT secondarywithout neutral connection

CT includes a delta tertiary

winding to trap zerosequence currents

3 wire connection to the relayI1 & I2 only


D1

D11


IC

IB

IA

110

011

101

3

1

IC

IB

IA

RELAY

RELAY

RELAY

IC

IB

IA

101

110

011

3

1

IC

IB

IA

RELAY

RELAY

RELAY

IC

IB

IA

211

121

112

3

1

IC

IB

IA

RELAY

RELAY

RELAY

D12

No phase change

But, removes neutral current

GENERATOR and

POWER STATION

PROTECTION




Delta Windings withEarthed Corner


Delta windings with earthed corner

Typical for transformer tertiary windings

To maintain stability on through faults

Need to include CT in the earthed corner

CT to be the same ratio as the main CT

Connect in parallel with the earthed phase CT

This cannot be overcome with phase shifting and Io

facilities of microprocessor based relays

Earthed Delta Corners inside theDifferential Zone

CT installed on the earthed corner.Same ratio as phase CTsIn parallel with the phase CT onthe earthed phase

This is also required formicroprocessor relay schemes


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Generator and



(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION




Neutral Displacement

Protection


Neutral Displacement Protection

Applicable to delta windings with no earth reference

Connected to open delta of VT secondary

To provide a zero sequence flux path

3 x 1 phase VTs or

5 limb VT

Relay must be immune to 3rd harmonics

Tripping time can be relatively slow

Problems can occur with resonance on energisation



VoltageDisplacement

Relay

VT with open deltasecondary winding



Measures 3.Vo for a solid E/F

Healthy phases rise to full phase-phase potential

Healthy phases are now only 60 deg apart

Thus, for standard 63.5V VT, output voltage is 190.5V

Set relay to 10% say 20V

NeutralDisplacement

Protection

0 V0 V

VB = 3VC = 3

VA = 0

-150+150

63.5 V63.5 V110 V110 V190.5 V190.5 V


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Generator and

Power Station Protection Motor Protection


(07) 3298 5260

GENERATOR and

POWER STATION

PROTECTION



MOTORPROTECTION


Motor Protection

To detect faults within the motor and supply system

Faults on the supply system, cables, etc

Internal faults

Often as a result of previous thermal events

Thermal Withstand

Motor overload

Starting and Stalling currents & times

Unbalanced voltage supply (NPS)

System events

Loss of load, under voltage, etc.


Motor Protection

Thermal Protection

Extended Start Protection

Stalling Protection

Number of Starts Limitation

Negative Sequence Current Detection

Short Circuit Protection

Earth Fault Protection Under Voltage Protection

Loss of Load Protection

Winding RTD measurement and trip

THERMALPROTECTION

GENERATOR and

POWER STATION

PROTECTION



MOTORPROTECTION

Thermal Protection


Thermal Overload Protection

Winding failures are often due to previous heating

Because of the motors thermal mass

Infrequent short term overloads may have no affect

But sustained overloads, of even just a few percent, may

cause aging and premature insulation failure

Insulation life may be halved for operation every 10 deg Cabove rated maximum.


Starting Current

High starting current reducesto full load current as the motor

gains speed

Positive sequence impedance

of the motor at any slip is :-

SYNC

ACTSYNC

N

NNSlip

=

2

1R1S

2

1R1SPOS XX

S

RRZ

0 250 500 750 1000 1250 15000

0.2

0.4

0.6

0.8

1

Z s( )

s

0 250 500 750 1000 1250 15000

1

2

3

4

5

6

I s( )

s


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Generator and



(07) 3298 5260


Negative Sequence Equivalent Circuitof Induction Motor

Positive Sequence Equivalent Circuitof Induction Motor

(Rs2 + Rr2)

(Rs1 + Rr1)

j(Xs2 + Xr2)

(S - 1) Rr2

(2 - S)

j(Xs1 + Xr1)

S

(1 - S) Rr1

Motor Impedances


Positive Sequence Impedance

At any slip

At standstill with slip = 1.0

This is a small impedance, hence high starting current

eg. up to 6 pu

21R1S

2

1R1SPOS XX

SRRZ

2

1R1S

2

1R1SPOS XXRRZ


Negative Sequence Impedance

At any slip

At normal running with slip 0

2

2R2S

2

2R2SNEG XX

S2

RRZ

2

2R2S

22R

2SNEG XX2

RRZ


Motor Impedances

Positive Sequence : Standstill

Negative Sequence : Running

R is small compared with X:

These will be approximately the same !!

2

1R1S

2

1R1SPOS XXRRZ

2

2R2S

2

2R2SNEG XX

2

RRZ

RUNNINGSTANDSTILL NEGPOSZZ =


Motor Impedances Consider an example where ISTART/IFLC = 6

Hence, 1pu positive sequence supply voltage results in:

1 pu current when running at full load

6 pu current at start-up

But the negative sequence impedance at running is thesame as the positive sequence impedance at startup

So, when running, 1 pu NPS supply voltage wouldresult in:

6 pu NPS current

Or, as a typical example, when running, 1% NPSsupply voltage would result in:

6% NPS current


CurrenttartingS

CurrentLoadFull

Z

Z

RUNNING

STARTING

POS

POS=

Motor Impedances

Positive sequence impedances at Start-up & Running

And since:

CurrenttartingS

CurrentLoadullF

Z

Z

RUNNING

RUNNING

POS

NEG=

RUNNINGSTARTING NEGPOSZZ =

eg. 1/6 th

eg. ZNEG = ZPOS / 6

So, what does this

mean for NPS current


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Generator and



(07) 3298 5260


Negative Sequence Current

Negative sequence current will equal:

% of NPS in the supply voltage

Multiplied by the ratio of starting / full load current

Multiplied by full load current

=NPSI

POS

NPS

V

V

FLC

CurrentStarting[ ]

FLCI

Generator and Power Station Protection_2012.pdf

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Transcript of Generator and Power Station Protection_2012.pdf