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DISTRIBUTION PROTECTION:

1

TOWARDS OPTIMAL PERFORMANCE

CUSTOMER SATISFACTION

12 Feb 2008

by Muhamad Subian Sukaimy


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Objectives

Overview of Distribution Protection Designing for optimal performance

2

performance

fundamental of OCEF coordination*.

Optimisation project**


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Overview-Type of Protection

Protection system is to protect equipment.

- Reliable/dependable

- Minimum area load disru ted

3

Categorised into:

Unit protectionNon unit protection


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Overview Unit Protection

Unit protection

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- of coverage.

- Also referred to as main protection

- Fast operating time


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Overview-Non unit Protection

Non Unit protection

5

- .- Also known as back-up protection

- Slower operating time


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Main:

- Operating timefixed; sensitivity isadjustable

Overview-Main Vs Backup

Bus-bar

Feeder

HV

6

Transformer

LV Backup:

- Operating timeand sensitivity areto be adjusted


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Overview-Main Vs Backup

HV

Comparison of relay tripping time:-Eqv time X factor

Differential - 50mS 50mS 1

REF - 50mS 50mS 1

7

LV

Solkor Rf - 50mS 50mS 1OCEF (11kV feeder) 0.4 S 400mS 8X

OCEF (33kV feeder) 1.2 S 1200mS 24X

SBEF Stage 1 - 2 S 2000mS 40X


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Bus-bar

Examples are:

- Feeder CurrentDifferential

-

Overview-Main Protection

Feeder

HV

8

LV

differential,Restricted EarthFault

- Busbar High/LowImpedance

Transformer


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Bus-bar

Examples are:

- Over-Current andEarth Fault

-

Overview-Backup Protection

Feeder

HV

9

LV

Fault

- Can be definite

time or InverseDefinite MinimumTime

Transformer


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Bus-bar

OCEF Relay are:

- Either 2 OC & 1 EFor 3 OC & 1 EF

-

Backup - OCEF Protection

Feeder

HV

10

LV

,transformers forvarious backup

purposes

Transformer


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Bus-bar

OC fault will affectwhole system.

EF is normallywithin specific

Fault path

Feeder

11

LV

.

- Thus coordination

of backup isimportant

Transformer

oc

ef


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Bus-bar

Operate for in

zone fault.- Normally operate

more than 1

Main Protection practice

Feeder

HV

12

LV

breaker to isolateequipment

Transformer


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Bus-bar

OCEF Relay are:- Time arranged so that

downstream relayoperates faster

Backup - OCEF practice

Feeder

HV

c

d

13

LV

- coordinated for allvoltage level

- EF required to becoordinated up to thetransformer LV

Transformer

a

b


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Optimal performance

It requires:

Planning criteria:

-

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- few seconds interruption

- normal

- Need to simulate fault condition


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Optimal performance

It requires:

Proper equipment/installation

- - -

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,

- Proper protection scheme chosen-CT,relaying sheme


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Optimal performance

Point to note:

Protection system available are capable ofhandling system design, however systeminstallation and operation may affect

16

system protection


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Zero interruption- parallel

operationFor optimal performance of parallel circuits:

Same source

Equal load sharing between c1 & c2

Max load not exceeding one cable rating

No other load in between

source

c1 c2

17

Proper laying method to eliminate induction Unit protection for cable (or directional

protection if time is not a problem)

Inter-tripping scheme implemented (atsource)

Busbar separation scheme( at load s/s)

Proper back up coordination


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Same source

Paralleling 2

different sourceswill result in thepower cable to act

Diff source

c1 c2

Same source

c1 c2

18

-between the twosources. Maycause power swingand systemtripping if it isabove threshold


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Equal load sharing

Equal load sharing achieved

by having same resistance(i.e. same capacity, samelen th .

Same source

c1 c2

19

This is to prevent uneven

loading which will lead totripping of one cable onover-current


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Cable laying

Improper cable laying

(especially for single corecables) may introducemutual inductance

Same source

c1 c2M

20

This may lead to unbalanced

phase current and mayresult in one cable trippingon earth fault when load

picks up


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Unit protection

Required to prevent double circuit

tripping occurring when a faultoccurs on one cable (since unitrotection is faster than back u

source

c1 c2F

21

protection)


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Directional protection

Directional protection will also

provide similar effect when thecircuit are in parallel

source

c1 c2F

22

However it may mal-operate forfaults happening on other

feeders.


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Inter-tripping scheme

When source substationbackup protection

operated, bus-tie may betripped causing bothcable to be the inter-connector now.

After ope

c1 c2

Before ope

c1 c2

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An inter-tripping schemeneed to be installed todecouple one cable so as

to prevent both cabletripping and/orpreventing the totalsource tripping


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Busbar separation scheme

Busbar separation scheme( at

load s/s) need to be implementedto prevent total load loss forbusbar fault

c1 c2

24


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Proper backup protection

Need to be properly

coordinated tohave trippingse uence that is

After ope

c1 c2

Before ope

c1 c2

25

coordinated fromdownstreamupwards.

Setting should alsocover single cable

outage.


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Few second interruption-

spur and auto transferFor this :

Source can be different

One cable as main, the other as back up.When relay at receiving detect loss ofsupply, it command change over of supply.

source

c1 c2

on on

on off

26

.

Max load not exceeding one cable rating

No other load in between

Unit protection for cable may be an option

Proper back up coordination Idealy 1 set per circuit.


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Minutes interruption- spur

and off-point:

Source can be different One cable as main, the other as

source

c1 c2

on on

on off

27

.either done by manual orsupervisory from control centre

Max load not exceeding one cablerating

Proper back up coordination


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Busbar protection using arc method.@

Optimizing relay maintenance project

Other initiative

28


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Given: 10km 33kV cable feed

into a 33/11kVsubstation

1 transformer in service; 33kV

Tutorial

OCEF 2

OCEF 1

Oc plug 100%

OC tms = 0.475

At 1.2 sec

600/5

600/5

29

, , ,

NER 4 ohmFind: 3 and 1 phase fault

current at PPU 11kV

busbar current and TMS for

overcurrent setting of allrelays

11kV

OCEF 3; t=0.8s

OCEF 4; t=0.4s

1600/5

300/5


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Discuss the problem

with the followingscheme from

33kV

Tutorial11kV

yn

D

S

30

perspective.

11kV

Y

y

Yn

D


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31

THANK YOU


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DISTRIBUTION PROTECTION:

BUSBAR ARC PROTECTION

12 Feb 2008



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BUSBAR PROTECTION USING ARC PROTECTION METHOD

Method Of Sensing:

Detection of light (arcing) + Overcurrent = Trip

Sensing of arc using photo sensors

or bare fibre

Sensing of overcurrent using current input

from OCEF


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typical sensor locations for single bus-bar and double bus-bar switchgears


L or Z typemounting plate

ARC t ti M t Sl


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ARC protection Master-Slave

Vamp 220 + 12CD:

ARC protection Basic MV


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ARC protection Basic MV

application

3

Modular Cable


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ABB REA 101:

-Can be independent

-Constraint: fibre loop distance


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ABB REA 101-103:

- Master-slave

-Both can have diffloop

- 101 trip incomer

ABB REA 101-105:

- Selective

-Both can have diff

loop

-105 trip feeder

-101 trip incomer

ARC protection Installation


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examples



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examples


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Termination fault isolated by ARCProtection 27 Nov 2007


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1

FUNDAMENTAL OF COORDINATION

12 Feb 2008



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OCEF Relay are:

- Star connected

-

Backup OCEF Protection

CT

R

YB

2

Ir+Iy+Ib = 0 OC OC

EFStarpoint


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Earth Fault On B phase

- Highlighted relay circuitwill be energized

-

OCEF Protection-Illustration

CT

R

YB

3

operate OC OC

EFStarpoint


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OCEF Protection Coordination

CT

R

YB

OC element are setbased on 3 phase fault

- EF element are based onearth fault or 1 phase

4

OC OC

EFStarpoint

fault- Coordination are done

using simulator or

coordination software

- Fundamental ??


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OCEF Coordination Fundamental

Network Diagram to be coordinated

Construct Thevenin Eqv diagram Obtain source & circuit impedances

5

per unit system is very useful)

Determine relay operating time

Obtain relay time multiplier

M i P t ti ti


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Bus-bar

Obtain thenetwork to becoordinated

Obtain the short

Main Protection practice

Feeder

HV

S

6

LV

circuit impedancesof eachequipment.

- i.e R+jX+, R+jXoincl length

Transformer

Dyn


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Component Short Circuit

Impedancetypical cable short cct resistance value

unit in ohm per metre

kv type MVA R+ jX+ Ro jXo33 o/h 150 silmalec 18 0.000219 0.000373 0.0008 0.00037

33 u/g 630 Al 1 core 30 0 0.0001296 0 0.0001723 > Fed Pow

11

xlpe 240 sq. mm

3c 0.000129 0.00009 0.00489 0.000031 ]

7

11

xlpe 150 sq. mm

3c 0.000211 0.000096 0.00578 0.000027 ] fujikura

11

xlpe 500 sq. mm

1c 0.0000647 0.000097 0.001298 0.000086 ]

typical transformer

impedance

kv MVA z%

33/11 30 10.06

33/11 15 9.54

11/0.433 0.75 4

Th i E N t k


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Thevenin Eqv Network

S

R+jXs R+jX

+ seq network

R+ Xs R+ X

FaultPoint

8

- seq network

0 seq network

R+jXs R+jX

Z+ = Z-

P U C l l ti (1)


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P.U. Calculation (1)

Declare:

- MVA base (normally 100 MVA)

9

- ase epen on source eve

- Z base (this is calculated; at all voltage

level if required)

- I base (this is calculated; at all voltage

level if required)

P U C l l ti (2)


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Source impedance formulae:

Zs+ pu= j MVA base

10

au

Zs0 pu = j [ {3 x MVA base } - 2Zs+ ]

[ { 1 Fault MVA} ]

P U Calc lation (3)


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Useful formulae:

Z base = (KV base)2 check (V x V)

11

ase x

I base A= (MVA base) check (V x I)

(KV base) (V)

P U Calculation (4)


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Changing of base formulae:

Z pu new = Z pu old x MVA base new

12

ase o

Z pu = Z () I pu = I (A)

Z base I base

Th i E N t k F lt


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Thevenin Eqv Network: Fault

Simulation (1)

S

R+jXs R+jX

+ seq network

R+ Xs R+ X

FaultPoint

3 phase fault:

- short the

If

13

- seq network

0 seq network

R+jXs R+jX

sequencenetwork only.



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Simulation (2)

S

R+jXs R+jX

+ seq network

R+ Xs R+ X

FaultPoint

phase to

phase fault:

If

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- seq network

0 seq network

R+jXs R+jX

-

positive andnegativesequencenetwork only.



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Simulation (3)

S

R+jXs R+jX

+ seq network

R+ Xs R+ X

FaultPoint

Single phase toground fault:

- connect allsequence

If

If

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- seq network

0 seq network

R+jXs R+jX

networ nseries and shortat fault point.

- I fault = 3If

If

IDMT Curve (1)


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Observe:

at tms=1:

At psm=2; t op = 10 sec

IDMT Curve (1)

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At psm=10; t op = 3 sec

For psm > 20 t op = t op at psm=20.

That is the characteristic of anIDMT curve

IDMT Curve (2)


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Observe:X axis = multiple ofcurrent; increases to

IDMT Curve (2)

10.00

17

the rightY axis = operatingtime; increases

vertical

10800

TX LV

0.10

1.00

1000.0 10000.0 100000.0

EQUIVALENT FAULT CURRENT AT REFERENCE BASE IN TABLE BELOW

OPERATINGT

IME(S

IDMT Curve (3)


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IDMT Curve (3)

10.00

IncreasePlug eg200%

18


vertical

10800

TX LV

0.10

.

1000.0 10000.0 100000.0


OPERATINGT

IME(S

IDMT Curve (4)


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IDMT Curve (4)

10.00

IncreaseTMS eg0.2 to 0.4

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vertical

10800

TX LV

0.10

1.00

1000.0 10000.0 100000.0


OPERATINGT

IME(S

IDMT Curve


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Observe from last 3 slides:

by varying these 2 variablesnamely plug (current) setting and

IDMT Curve

20

more relays; the IDMT curve willbe changed!

This is the basis of OCEFCoordination.


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Optimization of Distribution Protection

Relays Maintenance

Involve to Solve

Slide No. 1 Copyright 2006 TNB Research

33.2C

76.9C

40

50

60

70

- ower ng neer ng en re

Project Overview


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Project Overview

Project Background

Maintenance cost of distribution protection relays is quite

sizeable given the large amount of them in the system Large collection of historical test data is available but without

any analysis/trending study

Involve to Solve


With the vast amount of data, there is good prospect tooptimise maintenance scheduling through proper dataanalysis/trending

Project Overview


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Project Overview

Potential Benefit

Currently, relays are required to be maintained once a year

Maintenance Cost: RM140/relay per year

(Kadar Harga Tetap, Panduan Kejuruteraan Bil.A10/2006)

Assume 50,000 relays in the system

Annual Maintenance Cost: RM140x50,000 =

Involve to Solve


RM7million/year

If we can reduce the frequency of maintenance (relay test)by half (ie. maintenance every 2 years);

Saving of RM3.5million/year

Note:CESI Italy recommends that relay maintenance/testing to be done every 3-4years (Internal Working Report IWR 3a-01, 19 Jan. 2006)

Total no of relays as of 1 July 2006: 57,590

Project Overview


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Project Overview

Project Aims & Objectives

To develop an intelligent system in identifying problematicrelays and performance trending

To optimise maintenance schedule of relays

Deliverables

A database to store and manage all relay test data

A system to capture relay test data, with automated/intelligent

Involve to Solve


algorithm for analysis/trending and maintenance optimisation

Intelligent

System

Relay testresults

Relayparameters:

age, make etc

RelayCriticality

Analysis/trending of

relay performance

Relay Maintenance frequency

Proposed Research Methodology


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Proposed Research Methodology

Methodology Flow Diagram

Initial Phase: Lit. Review & Data Identification

Hypothesis Development

Progress Phase: Initial Statistical Data Analysis

START

Data Collection, ie. historical data from relay test sets.

Involve to Solve


Design of Data Structure/Format, System Interfacing

Development of Intelligent System

Final Phase: Final analysis & Report of findings

Adoption strategies

Results validation, modification & improvement

END

Scope of Work


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Scope of Work

Data collection

To study the numerical relay data in Negeri Sembilan

Initial Statistical Data AnalysisEnable identification of patterns/trend from relay test data

Involve to Solve


To investigate factors that contribute towardsoptimisation of relay maintenance eg. Relay parameters,test data, criticality etc.

To develop a model for relay maintenance optimisation(AI algorithm for Data Mining Microsoft SQL Server 2005)

Data Capture


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ata Captu e

How shall the data be obtained?

TNB Distribution (Engineering Services) has given

the consent and commitment to co-operate with datacollection [Tn. Hj. Subian, En. Norazman Atib]

Data to be obtained from distribution maintenance

Involve to Solve


eng neers paper copy

What shall be done to ensure dataintegrity?

Data obtained will be examined and any doubts willbe verified with TNBD (Engineering Services)

Data Requirement eg. Relay Inverse Time Test


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Involve to Solve


Data Collection


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All relay overcurrent and earthfault test datawere collected from the state of Negeri

Sembilan.

The data collected are historical data dated

Involve to Solve

ac rom .

A total of 1316 no of relays data werecollected. These are the historical data usedin the AI Engine.


Data Analysis


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y

How shall the data be analysed?

Depending on the TMS and plug setting, theoperating time T is calculated.

A healthy relay should fall within +-10% of the

Involve to Solve


calculated value.

A collection of historical data for a particular relayshows the deviation (in %) from the referenceoperating time.

The system developed will analyse the data fortrending/pattern

Formula for operating

Time/currentcharacteristics of IDMTrelay


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Where:

T = operating time

I = Fault current

Formula for operating

time of IDMT relay

relay

Involve to Solve


IS = Setting current

TMS = Time multipliersetting

Panduan Kejuruteraan

Bil.A10/2006

Example:

Data Analysis


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Example:

-- Trending for a particular relay using 3 years of test data

Time(s)

Test data year 1

Test data year 2

Test data year 3

Involve to Solve

Slide No. 12

current

10%

Error thickness

(relay failed)

Overview of AI


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The scatter will be thetraining data ArtificialNeural Network

Training data will enablete network to determinethe ri ht wei htin s for

Involve to Solve

Slide No. 13

each perceptron and

control its sensitivity. A trained network would

enable the output toindicate if a relay is proneto failure or otherwise.

The ArtificiaI Intelligence Engine


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Involve to Solve


What can the software reports on?


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Faulty Relay Detector can predict which relay isprone to become faulty for the current year.

Normal listing of relay information, such as:- list all relays based on different makes

-

Involve to Solve

- list all relays in certain areas/SSUs- list all relays that has been replaced and their test

results, and many others.

The reports generated will be updated automaticallyas new relay test data are added in the database.


Conclusions


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This project will save TNB in terms of theirmaintenance cost of protective relays (optimumperiod for routine maintenance)

Reduces unnecessary power outages due to mal-operation of protection relays

Increase work efficienc of ro ec ion en ineers

Involve to Solve


Add value to distribution protection system/protectionrelays test activities by undertaking analysis of testresult

Identification of problematic relay types/model

Enhanced decision-making process related tomaintenance frequency for protection relays

Thank You


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Contacts:-

Involve to Solve


TNB Research Sdn. Bhd.

No. 1, Lorong Ayer Itam,

Kawasan Institusi Penyelidikan,

43000 Kajang, Selangor, MALAYSIA

Tel: +603-89268818 Fax: +603-89268828 / 29

uniten arsepe 08 l5

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