uniten arsepe 08 l5
TRANSCRIPT
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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
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- .- 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
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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
- - -
15
,
- 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
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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
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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
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-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
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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
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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
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protection)
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Directional protection
Directional protection will also
provide similar effect when thecircuit are in parallel
source
c1 c2F
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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
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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
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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
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.
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
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.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
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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
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, , ,
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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THANK YOU
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DISTRIBUTION PROTECTION:
BUSBAR ARC PROTECTION
12 Feb 2008
by Muhamad Subian Sukaimy
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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
BUSBAR PROTECTION USING ARC PROTECTION METHOD
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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BUSBAR PROTECTION USING ARC PROTECTION METHOD
ABB REA 101:
-Can be independent
-Constraint: fibre loop distance
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BUSBAR PROTECTION USING ARC PROTECTION METHOD
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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ARC protection Installation
examples
ARC protection Installation
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ARC protection Installation
examples
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Termination fault isolated by ARCProtection 27 Nov 2007
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1
FUNDAMENTAL OF COORDINATION
12 Feb 2008
by Muhamad Subian Sukaimy
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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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P.U. Calculation (2)
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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P.U. Calculation (3)
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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P.U. Calculation (4)
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.
Thevenin Eqv Network: Fault
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Thevenin Eqv Network: Fault
Simulation (2)
S
R+jXs R+jX
+ seq network
R+ Xs R+ X
FaultPoint
phase to
phase fault:
If
14
- seq network
0 seq network
R+jXs R+jX
-
positive andnegativesequencenetwork only.
Thevenin Eqv Network: Fault
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Thevenin Eqv Network: Fault
Simulation (3)
S
R+jXs R+jX
+ seq network
R+ Xs R+ X
FaultPoint
Single phase toground fault:
- connect allsequence
If
If
15
- 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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Observe:X axis = multiple ofcurrent; increases to
IDMT Curve (3)
10.00
IncreasePlug eg200%
18
the rightY axis = operatingtime; increases
vertical
10800
TX LV
0.10
.
1000.0 10000.0 100000.0
EQUIVALENT FAULT CURRENT AT REFERENCE BASE IN TABLE BELOW
OPERATINGT
IME(S
IDMT Curve (4)
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Observe:X axis = multiple ofcurrent; increases to
IDMT Curve (4)
10.00
IncreaseTMS eg0.2 to 0.4
19
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
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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
Slide No. 2 Copyright 2006 TNB Research
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
Slide No. 3 Copyright 2006 TNB Research
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
Slide No. 4 Copyright 2006 TNB Research
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
Slide No. 5 Copyright 2006 TNB Research
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
Slide No. 6 Copyright 2006 TNB Research
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
Slide No. 7 Copyright 2006 TNB Research
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
Slide No. 8 Copyright 2006 TNB Research
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.
Slide No. 9 Copyright 2006 TNB Research
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
Slide No. 10 Copyright 2006 TNB Research
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
Slide No. 11 Copyright 2006 TNB Research
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
Slide No. 14 Copyright 2006 TNB Research
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.
Slide No. 15 Copyright 2006 TNB Research
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
Slide No. 16 Copyright 2006 TNB Research
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
Slide No. 17 Copyright 2006 TNB Research
TNB Research Sdn. Bhd.
No. 1, Lorong Ayer Itam,
Kawasan Institusi Penyelidikan,
43000 Kajang, Selangor, MALAYSIA
Tel: +603-89268818 Fax: +603-89268828 / 29