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In simple language,the book providesa modernintroductionto power
system operation,controland analysis.
Key Features of the Third
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New chapters added on
) .PowerSystemSecurity
) State Estimation
) Powersystem compensationincludingsvs and FACTS
) Load Forecasting
) VoltageStability
New appendices on :
Ff
> MATLABand SIMULINKdemonstrating
their use in problemsolving.
) Real time computercontrol of power systems.
From
the
Third
Editior;
llf
L' r
J( ^r*I-
^..!
r\(J lltd,fl
lJ Nagrath
Reviewen.,
The book is very comprehensive,well organised,up-to-dateand (above
all) lucid and easyto follow for self-study.lt is ampiy illustratedw1h solved
examplesfor everyconceptand technique.
lMr
ll--=i
iltililttJffiillililil
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern PorroerSYstem
Third Edition
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
About the Authors
D P Kothari is vice chancellor, vIT University, vellore.
Earlier,he was
Professor, Centre for Energy Studies, and Depufy Director
(Administration)
Indian Instituteof Technology,Delhi. He has uiro t."n
the Head of the centre
for Energy Studies(1995-97)and Principal(l gg7-g8),Visvesvaraya
Regional
Engineeringcollege, Nagpur. Earlier
lflaz-s: and 19g9),he was a visiting
fellow at RMIT, Melbourne, Australia. He obtained
his BE, ME and phD
degreesfrom BITS, Pilani. A fellow of the Institution
Engineers(India), prof.
Kothari has published/presented450 papers in national
and international
journals/conferences.He has authored/co-authored
more than 15 books,
including Power system Engineering, Electric Machines,
2/e, power system
Transients, Theory and problems of Electric Machines,
2/e., and. Basic
Electrical Engineering. His researchinterestsinclude power
system control,
optimisation,reliability and energyconservation.
I J Nagrath is Adjunct Professor,BITS Pilani and retired
as professor of
Electrical Engineeringand Deputy Director of Birla Institute
of Technology
and Science, Pilani. He obtained his BE in Electrical
Engineering from the
university of Rajasthanin 1951 and MS from the Unive.rity
of Wi"sconsinin
1956' He has co-authoredseveral successfulbooks
which include Electric
Machines 2/e, Power system Engineering, signals and
systems and.systems:
Modelling and Analyns. He has also puulistred ,"rr.ui
researchpapers in
prestigiousnationaland internationaljournats.
Modern Power System
Analysis
Third Edition
D P Kothari
Vice Chancellor
VIT University
Vellore
Former Director-Incharge, IIT Delhi
Former Principal, VRCE, Nagpur
I J Nagrath
Adjunct Professor, and Former Deputy Director,
Birla Ins1i1y7"of Technologt and Science
Pilani
Tata McGraw Hill Education private Limited
NEWDELHI
McGraw-Hill Offices
New Delhi Newyork St Louis San Francisco Auckland Bogot6
Caracas
KualaLumpur Lisbon London Madrid Mexicocity Milan Montreal
San Juan Santiago Singapore Sydney Tokyo Toronto
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Preface to the Third Edition
Information contained in this work has been obtained by Tata McGraw-Hill, from
sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors
guaranteethe accuracy or completenessof any information published hereiir, and neittier
Tata McGraw-Hill nor its authors shall be responsiblefor any errors, omissions, or
damages arising out of use of this information. This work is publi'shed-with the
understandingthat Tata McGraw-Hill and its authorsare supplying information but are
not attempting to render enginecring or other professionalservices. If such seryicesare
required, the assistanceof an appropriate professional should be sought
TataMcGraw-Hill
O 2003,1989,1980,TataMcGrtrwI{ill EducationPrivateI-imited
Sixteenthreprint2009
RCXCRRBFRARBQ
No part of this publicationcan be reproducedin any form or by any -"un,
withoutthe prior writtenpermissionof the publishers
This edition can be exportedfrom India only by the publishers,
TataMcGraw Hill EducationPrivateLimited
ISBN-13: 978-0-07-049489-3
ISBN- 10: 0-07-049489-4
Publishedby TataMcGrawHill EducationPrivateLimited,
7 WestPatelNagat New Delhi I l0 008, typesetin TimesRomanby Script Makers,
Vihar,New Delhi ll0 063 andprintedat
Al-8, ShopNo. 19, DDA Markct,Paschim
Delhi ll0 053
GopaljeeEnterprises,
Coverprinter:SDR Printcrs
www.muslimengineer.info
Sincethe appearanceof the secondedition in 1989,the overall energy situation
has changed considerably and this has generatedgreat interest in nonconventionaland renewableenergy sources,energyconservationand management, power reforms and restructuringand distributedarrddispersedgeneration.
Chapter t has been therefore,enlargedand completely rewritten. In addition,
the influences of environmentalconstraintsare also discussed.
The present edition, like the earlier two, is designed for a two-semester
course at the undergraduatelevel or for first-semesterpost-graduatestudy.
Modern power systemshave grown larger and spreadover larger geographical areawith many interconnections
betweenneighbouringsystems.Optimal
planning,operationand control of such large-scalesystemsrequire advanced
computer-basedtechniquesmany of which are explainedin the student-oriented
and reader-friendlymannerby meansof numericalexamplesthroughout this
book. Electric utility engineerswill also be benefittedby the book as it will
preparethem more adequatelyto face the new challenges.The style of writing
is amenableto self-study.'Ihe wide rangeof topicsfacilitatesversarileselection
of chaptersand sectionsfbr completion in the semestertime frame.
Highlights of this edition are the five new chapters.Chapter 13 deals with
power system security. Contingency analysis and sensitivity factors are
described.An analytical framework is developedto control bulk power systems
in sucha way that securityis enhanced.Everythingseemsto have a propensity
to fail. Power systemsare no exception.Power systemsecuritypracticestry to
control and operatepower systemsin a defensivepostureso that the effects of
theseinevitable failures are minimized.
Chapter 14 is an introduction to the use of stateestimationin electric power
systems.We have selectedLeast SquaresEstimationto give basic solution.
External system equivalencing and treatment of bad data are also discussed.
The economics of power transmissionhas always lured the planners to
transmit as much power as possible through existing transmission lines.
Difficulty of acquiring the right of way for new lines (the corridor crisis) has
always motivated the power engineersto develop compensatorysystems.
Therefore, Chapter 15 addressescompensationin power systems.Both series
and shunt compensationof linqs have been thoroughly discussed.Concepts of
SVS, STATCOM and FACTS havc-been briefly introduced.
Chapter 16 covers the important topic of load forecasting technique.
Knowing load is absolutelyessentialfor solving any power systemproblem.
Chapter 17 dealswith the important problem of voltagestability.Mathematical formulation, analysis, state-of-art, future trends and challenges are
discussed.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Wl
Preracero rne lhlrd Edrtion
for powersystemanalysisare
MATLAB andSIMULINK, idealprograms
illustrating
includedin thisbookasan appendixalongwith 18solvedexamples
tem problems. The help rendered
tive
theiruse in solvin
by Shri Sunil Bhat of VNIT, Nagpur in writing this appendix is thankfully
acknowledged.
Tata McGraw-Hill and the authors would like to thank the following
reviewers of this edition: Prof. J.D. Sharma,IIT Roorkee; Prof. S.N. Tiwari,
MNNIT Allahabad; Dr. M.R. Mohan, Anna University, Chennai; Prof. M.K.
Deshmukh,BITS, Pilani; Dr. H.R. Seedhar,PEC, Chandigarh;Prof.P.R. Bijwe
and Dr. SanjayRoy, IIT Delhi.
While revising the text, we have had the benefit of valuable advice and
suggestionsfrom many professors,studentsand practising engineerswho used
the earlier editions of this book. All these individuals have influenced this
edition.We expressour thanks and appreciationto them. We hope this support/
responsewould continue in the future also.
D P Kors[m
I J Nlcn+rn
www.muslimengineer.info
Preface to the First
Mathematicalmodellingand solutionon digital computersis the only practical
approach to systems analysis and planning studies for a modern day power
system with its large size, complex and integrated nature. The stage has,
therefore,been reachedwhere an undergraduatemust be trained in the latest
techniquesof analysisof large-scalepower systems.A similar needalso exists
in the industry wherea practisingpower systemengineeris constantlyfacedwith
the challengeof the rapidly advancingfield. This book has bedndesignedto fulfil
this need by integratingthe basic principlesof power systemanalysisillustrated
through the simplestsystemstructurewith analysistechniquesfor practical size
systems.In this book large-scalesystemanalysisfollows as a naturalextension
of the basicprinciples.The form and level of someof the well-known techniques
are presented in such a manner that undergraduatescan easily grasp and
appreciatethem.
The book is designedfor a two-semestercourse at the undergraduatelevel.
With a judicious choice of advancedtopics, some institutionsmay also frnd it
useful for a first course for postgraduates.
The readeris expectedto have a prior grounding in circuit theory and electrical
machines. He should also have been exposed to Laplace transform, linear
differential equations, optimisation techniquesand a first course in control
theory. Matrix analysisis applied throughoutthe book. However, a knowledge
of simple matrix operations would suffice and these are summarisedin an
appendixfbr quick reference.
The digital computerbeing an indispensabletool for power systemanalysis,
computationalalgorithms for various systemstudiessuch as load flow, fault level
analysis,stability, etc. have been included at appropriateplacesin the book. It
is suggestedthat where computerfacilities exist, studentsshouldbe encouraged
to build computer programs for these studies using the algorithms provided.
Further, the students can be asked to pool the various programs for more
advancedand sophisticatedstudies,e.g. optimal scheduling.An importantnovel
featureof the book is the inclusion of the latestand practicallyuseful topics like
unit commitment, generation reliability, optimal thermal scheduling,optimal
hydro-thermalschedulingand decoupledload flow in a text which is primarily
meantfor undergraduates.
The introductory chapter contains a discussion on various methods of
electricalenergygenerationand their techno-economic
comparison.A glimpse is
given into the future of electricalenergy.The readeris alsoexposedto the Indian
power scenariowith facts and figures.
Chapters2 and3 give the transmissionline parametersand theseare included
for the sakeof completnessof the text. Chapter4 on the representation
of power
systemcomponentsgives the steadystatemodelsof the synchronous
machineand
the circuit modelsof compositepower systemsalong with the per unit method.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Contents
preface ro rhe Frrst Edition
W
Chapter5 deals with the performanceof transmissionlines. The load flow
problem is introducedright at this stagethroughthe simple two-bus systemand
basicconceptsof watt and var control are illustrated.A brief treatmentof circle
concept of load flow and line compensation. ABCD constants are generally well
covered in the circuit theory course and are, therefore, relegated to an appendix.
Chapter 6 gives power network modelling and load flow analysis, while
Chapter 7 gives optimal system operation with both approximate and rigorous
treatment.
Chapter 8 deals with load frequency control wherein both conventional and
modern control approaches have been adopted for analysis and design. Voltage
control is briefly discussed.
Chapters 9-l l discuss fault studies (abnormal system operation). The
synchronous machine model for transient studies is heuristically introduced to
the reader.
Chapter l2 emphasisesthe concepts of various types <lf stability in a power
system. In particular the concepts of transient stability is well illustrated through
the equal area criterion. The classical numerical solution technique of the swing
equation as well as the algorithm for large system stability are advanced.
Every concept and technique presented is well supported through examples
employing mainly a two-bus structure while sometimes three- and four-bus
illustrations wherever necessary have also been used. A large number of
unsolved problems with their answers are included at the end of each chapter.
These have been so selected that apart from providing a drill they help the
reader develop a deeper insight and illustrate some points beyond what is directly
covered by the text.
The internal organisation of various chapters is flexible and permits the
teacher to adapt them to the particular needs of the class and curriculum. If
desired, some of the advanced level topics could be bypassed without loss of
continuity. The style of writing is specially adapted to self-study. Exploiting this
fact a teacher will have enough time at his disposal to extend the coverage of
this book to suit his particular syllabus and to include tutorial work on the
numerous examples suggestedin the text.
The authors are indebted to their colleagues at the Birla Institute of
Technology and Science, Pilani and the Indian Institute of Technology, Delhi
for the encouragement and various useful suggestionsthey received from them
while writing this book. They are grateful to the authorities of the Birla lnstitute
of Technology and Science, Pilani and the Indian Institute of Technology, Delhi
for providing facilities necessary for writing the book. The authors welcome
any constructive criticism of the book and will be grateful for any appraisal by
the readers.
I J NlcRArH
D P KorHlnr
www.muslimengineer.info
Preface to First Edition
vn
1 . Introduction
I
1 . 1 A Perspective I
1 . 2 Structureof Power Systems I0
1 . 3 Conventional Sourcesof Electric Energy I3
r . 4 RenewableEnergy Sources 25
1 . 5 Energy Storage 28
1 . 6 Growth of Power Systemsin India
1 . 7 Energy Conservbtion 3I
29
r . 8 Deregulation 33
1 . 9 Distributed and DispersedGeneration 34
1 . 1 0Environmental Aspects of Electric Energy Generation 35
1 . 1 1Power SystemEngineersand Power SystemStudies 39
T . I 2Use of Computers and Microprocessors 39
1 . 1 3ProblemsFacing Indian Power Industry and its Choices 40
References 43
2. Inductance and Resistance of Transmission Lines
45
2 . 1 Introduction 45
2 . 2 Definition of Inductance 45
2 . 3 Flux Linkages of an Isolated
Current-CtrryingConductor 46
2.4 Inductanceof a Single-PhaseTwo-Wire Line 50
2 . 5 ConductorTypes 5I
2 . 6 Flux Linkages of one Conductorin a Group 53
2 . 7 Inductanceof CompositeConductorLines 54
2 . 8 Inductanceof Three-PhaseLines 59
2 . 9 Double-CircuitThree-PhaseLines66
2 . 1 0Bundled Conductors 68
2 . l I Resistance 70
2 . r 2 Skin Effect and Proximity Effect 7I
Problems 72
References 75
3. Capacitance of Transmission Lines
3.1
3.2
Introduction 76
Electric Field of a Long Straight Conductor 76
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
76
fW
. contents
3.3
3.4
3.5
3.7
o
Potential Diff'erencebetweentwo Conductors
of a Group of Parallel Conductors 77
Capacitanceof a Two-Wire Line 78
Capacitanceof a Three-phaseLine
with Equilateral Spacing B0
6.4
6.5
6.6
6.7
UnsymmetricalSpacing BI
Effect of Earth on TransmissionLine capacitance g3
6.9 Control of Voltage Profile 230
Problems 236
l t - t l - - l
a
r . ^
/ r .
J.o
"
rvleln(Jo or \rlvll-, (vloollled)
3.9
BundledConductors 92
Problems 93
References 94
4.3
4.4
4.5
4.6
4.7
IIYJEIETLLCJ
95
6.1
6.2
lntrotluction 184
NetworkModel Formulation I85
www.muslimengineer.info
242
7.I
1.2
7.3
7.4
1.5
7.6
7.7
Introduction 242
Optimal Operation of Generatorson a Bus Bar 243
Optimal Unit Commitment (UC) 250
ReliabilityConsiderations 253
Optimum GenerationScheduling 259
Optimal Load Flow Solution 270
OptimalSchedulingof HydrothermalSystem 276
Problems 284
References 286
291'l
8. Automatic Generation and Voltage Control
128
5.1 Introduction 128
5.2 Short TransmissionLine 129
5.3 Medium TransmissionLine i37
5.4 The Long Transmission
Line-Rigorous Solution I 39
5.5 Interpretationof the Long Line Equations 143
5.6 FerrantiEffect 150
5.1 TunedPowerLines 151
5.8 The Equivalent Circuit of a Long Line 152
5.9 Power Flow througha TransmissionLine I58
5 .1 0 M e th o d so l ' Vo l ra g eC o n t rol 173
Problems 180
References 183
6. Load Flow Studies
)20
LJ7
7. Optimal System Operation
Introduction g5
Single-phaseSolutionof Balanced
Three-phaseNetworks 95
One-LineDiagram and Impedanceor
ReactanceDiagram 98
Per Unit (PU) System 99
ComplexPower 105
SynchronousMachine 108
Representation
of Loads I2I
Problems 125
References 127
5. Characteristics and Performance of power
Transmission Lines
Load Flow Problem 196
Gauss-SeidelMethod 204
Newton-Raphson(NR) Method 213
DecoupledLoad Flow Methods 222
D^t--^---.
yl
4. Representation,of Power System Components
4.1
4.2
.
8.1 Introduction 290
8.2 Load FrequencyControl (Single Area Case) 291
8.3 Load FrequencyControl and
Economic DespatchControl 305
8.4 Two-Area Load FreqlrencyControl 307
8 . 5 Optimal (Two-Area) Load FrequencyControl 3I0
8 . 6 Automatic Voltage Control 318
8 . 7 Load Frequency Control with Generation
Rate Constraints(GRCs) 320
8 . 8 SpeedGovernor Dead-Bandand Its Effect on AGC
8 . 9 Digital LF Controllers 322
8 . 1 0DecentralizedControl 323
Prohlents 324
References 325
321
9. Symmetrical Fault Analysis
t84
9.1 Introduction 327
9.2 Transienton a TransmissionLine 328
Machine
9.3 ShortCircuit of a Synchronous
(On No Load) 330
9.4 Short Circuit of a LoadedSynchronousMachine 339
9.5 Selectionof Circuit Breakers 344
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
327
confenfs
rffi#q
I
'
9.6 Algorithm for ShortCircuit Studies 349
9.7 Zsus Formulation 355
Problems 363
References 368
12.10MultimachineStabilitv 487
Symmetrical Com
10.1 Introduction 369
10.2 SymmetricalComponentTransformation370
10.3 Phase Shift in Star-DeltaTransformers 377
10.4 SequenceImpedancesof TransmissionLines 379
10.5 SequenceImpedancesand SequenceNetwork
of Power Systern 381
10.6 SequenceImpedancesand Networks of
SynchronousMachine 381
10.7 SequenceImpedancesof TransmissionLines 385
10.8 SequenceImpedancesand Networks
of Transformers 386
10.9 Constructionof SequenceNetworks of
a Power System 389
Problems 393
References 396
ll.
Unsymmetrical Fault Analysis
Problems 506
References 508
13. Power System Security
13.1 Introduction 510
13.2 SystemStateClassification 512
13.3 SecurityAnalysis 512
13.4 ContingencyAnalysis 516
13.5 SensitivityFactors 520
13.6 Power System Voltage Stability 524
References 529
14. An Introduction to state Estimation of Power systems
397
11.1 Introduction 397
11.2 SymmetricalComponentAnalysis of
UnsymmetricalFaults 398
11.3
Single
Line-To-Ground(LG) Fault 3gg
,
11.4 Line-To-Line (LL) Fault 402
11.5 Double Line-To-Ground(LLG) Fault 404
11.6 Open ConductorFaults 414
11.1 Bus ImpedanceMatrix Method For Analysis
of UnsymmetricalShuntFaults 416
Problems 427
References 432
12. Power System Stability
12.1 Introduction 433
12.2 Dynamics of a SynchronousMachine 435
12.3 Power Angle Equation 440
12.4 Node Elimination Technique 444
I2.5 SimpleSystems 451
12.6 Steady State Stability 454
12.7 Transient Stability 459
I2.8 Fq'-ralArea Criterion 461
www.muslimengineer.info
15. Compensation in Power Systems
433
531
l4.l Introduction 531
I4.2 Least SquaresEstimation: The Basic
Solution 532
14.3 Static StateEstimation of Power
Systems 538
I4.4 Tracking State Estimation of Power Systems 544
14.5 SomeComputationalConsiderations 544
14.6 External System Equivalencing 545
I4.7 Treatmentof Bad Dara 546
14.8 Network observability and Pseudo-Measurementss49
14.9 Application of Power SystemStateEstimation 550
Problems 552
References 5.13
15.1 Introduction 556
15.2 Loading Capability 557
15.3 Load Compensation 557
15.4 Line Compensation 558
15.5 SeriesCompensation 559
15.6 ShuntCornpensators 562
I5.7 ComparisonbetweenSTATCOM and SVC 565
15.8 Flexible AC TransmissionSystems(FACTS) 566
15.9 Principle and Operationof Converrers 567
15.10FactsControllers 569
References 574
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
550
16. Load Forecasting Technique
16.1 Introduction 575
16.2 ForecastingMethodology 577
timation of Averageand Trend Terms 577
1 6 . 4 Estimation of Periodic Components 581
1 6 . 5 Estimationof y., (ft): Time SeriesApproach 582
1 6 . 6 Estimationof StochasticComponent:
Kalman Filtering Approach 583
16.7 Long-Term Load PredictionsUsing
EconometricModels 587
r 6 . 8 ReactiveLoad Forecast 587
References 589
17. Voltage Stability
591
1 1 .1 In tro d u c ti o n 5 9 1
17.2 Comparisonof Angle and Voltage Stability 592
17.3 ReactivePower Flow and Voltage Collapse 593
11.4 MathematicalFormulationof
Voltage Stability Problem 593
11.5 Voltage Stability Analysis 597
17.6 Preventionof Voltage Collapse 600
ll.1 State-of-the-Art,Future Trends and Challenses 601
References 603
I.T
Appendix A: Introduction to Vector and Matrix Algebra
605
Appendix B: Generalized Circuit Constants
617
Appendix C: Triangular Factorization and Optimal Ordering
623
Appendix D: Elements of Power System Jacobian Matrix
629
Appendix E: Kuhn-Tucker Theorem
632
Appendix F: Real-time Computer Control of power Systems
634
Appendix G: Introduction to MATLAB and SIMULINK
640
Answers to Problems
Index
www.muslimengineer.info
A PERSPECTIVE
Electric energy is an essentialingredient for the industrial and all-round
developmentof any country.It is a covetedform of energy,becauseit can be
generatedcentrallyin bulk and transmittedeconomicallyover long distances.
Further, it can be adaptedeasily and efficiently to domestic and industrial
applications, particularly for lighting purposes and rnechanical work*, e.g.
drives. The per capita consumptionof electricalenergyis a reliable indicator
of a country's stateof development-figures for 2006 are615 kwh for India
and 5600 kWh for UK and 15000 kwh for USA.
Conventionally,electricenergy is obtainedby conversionfiom fossil fuels
(coal,oil, naturalgas),and nuclearand hydro sources.Heatenergyreleasedby
burning fossil fuels or by fission of nuclearmaterialis convertedto electricity
by first convertingheatenergyto the mechanicalform througha thermocycle
and then converting mechanicalenergy through generatorsto the electrical
form. Thermocycleis basicallya low efficiency process-highestefficiencies
for modern large size plantsrange up to 40o/o,while smallerplants may have
considerably lower efficiencies. The earth has fixed non-replenishableresourcesof fossil fuels and nuclear materials,with certain countries overendowedby natureand othersdeficient. Hydro energy,thoughreplenishable,is
also limited in terms of power. The world's increasingpower requirementscan
only be partially met by hydro sources.Furthermore,ecologicaland biological
factorsplace a stringentlimit on the use of hydro sourcesfor power production.
(The USA has already developed around 50Vo of its hydro potential and
hardly any further expansionis plannedbecauseof ecologicalconsiderations.)
x Electricity is a very inefficient agent for heating purposes,
becauseit is generatedby
the low efficiency thermocycle from heat energy. Electricity is used for heating
purposesfor only very special applications, say an electric furnace.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Introduction
with the ever increasing
_ _ _ __ _ _ - - ^ D per capita venergy
^,vt6J
vconsumption
urrJLurlp[rult
g
xpongnlla
iand
ltlu
exponentially
rising population, technologists already r* the end
of the earth,s
s n
nc
onfuel reso.urces*.
oil crisis of the 1970s has dramatically
-The
llfenislable
drawn attentionto this fact. In fact,we can no lon
tor generation of electricity. In terms of bulk electric energy
generation, a
distinct shift is taking place acrossthe world in favour of coalLJin
particular
Cufiailment
of enerry
consumption
The energyconsumptionof most developcdcorrntries
has alreaclyreachecla
level, which this planet cannot afford. There is, in fact,
a need to find ways and
meansof reducingthis level. The developingcountries,on
the other hand,have
to intensify their efforts to raisetheir level of energyproduction
to provide basic
amenities to their teeming millions. of course,-in doing
,o th"y need to
constantly draw upon the experiencesof the developed countries
and guard
againstobsoletetechnology.
rntensification
of effofts to develop alternative
sources of
enerw including
unconventional
sources like solan tidal
energy, etc.
Distant hopesare pitched on fusion energy but the scientific
and technological
advanceshave a long way to go in this regard. Fusion when
harnessedcould
provide an inexhaustiblesourceof energy. A break-through
in the conversion
from solar to electric energy could pr*io" another answer
to the world,s
steeply rising energy needs.
Recyclingr
of nuclear
wastes
Fast breederreactortechnologyis expectedto provide the answer
for extending
nuclear energy resourcesto last much longer.
D e velopm ent an d applicati
intense pollution in their programmes of energy development.Bulk power
generating stations are more easily amenable to control of pollution since
centralized one-point measurescan be adopted.
consumptionon a worldwidebasis.This figure is expectedto rise as oil supply
for industrial usesbecomesmore stringent. Transportationcan be expected to
go electric in a big way in the long run, when non-conventionalenergy
resourcesare we[ developedor a breakthroughin fusion is achieved.
To understand some of the problems that the power industry faces let us
briefly review some of the characteristic features of generation and transmission. Electricity, unlike water and gas, cannot be storedeconomically (except
in very small quantities-in batteries),and the electric utility can exerciselittle
control over the load (power demand) at any time. The power system must,
therefore, be capable of matching the output from generatorsto the demand at
any time at a specifiedvoltageand frequency.The difficulty encounteredin this
task can be imagined from the fact that load variations over a day comprises
three components-a steady component known as base load; a varying
componentwhose daily patterndependsupon the time of day; weather, season,
a popular festival, etc.; and a purely randomly varying componentof relatively
small amplitude. Figure 1.1 shows a typical daily load curve. The characteristics of a daily load curve on a gross basis are indicatedby peak load and the
time of its occurrence and load factor defined as
averageload
= less than unity
maximum (peak)load
B
100
E
B o
c
o
tr
6
:o
6 0
E
l + o
x
o
on of an ttpollu tion techn ologries
In this regard, the developing countries already have the
example of the
developed countries whereby they can avoid going through
the phases of
*varying estimatcshave
bccn put forth for rescrvcsol'oil, gas and coal ancllissionable
rnaterials'At the projectedconsumptionrates,oil and gases
are not expectedto last
much beyond 50 years; severalcountries will face serious
shortagesof coal after 2200
A'D' while fissionable materials may carry us well beyond
the middle of the next
century. These estimates,however, cannot be regarded
as highly dependable.
www.muslimengineer.info
Fig. 1.1 Typical daily load curve
The average load determines the energy consumption over the day, while
the peak load along with considerations of standby capacity determines plant
capacity for meeting the load.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
(xawrrg
. A h i g h l o a d f a c tuvrl,;
o r h eurl p
s i n d rmore
a w nenergy
g m o rlrom
e e naegiven
r g y *installation.
' n u , ' u " n , n , . uI * I ' , i
Asindividuar
road
centris
have
rJir o*n
;;;ilnffi
generalhave a time dJveniry,which when"r,#u","'.i.,ii,
;61il ;&;';;"irili,jij
|
I
mterconnection,rgreajly
aidsin jackinguF tn,a factorsat er in&viJJp
i
excessPowerof a plant audng tight"toaaperiods
is evacuated
throughlong
distancehigh voltagetransmissionlo"r,*hl"
h"""itr;;;pj;;,:;";#
'"*-rwervcs
Prur
power.
"
Diversity
i
|
Factor
Tt:"9:"lT:f:,_Tj_.:llT:11,:::."1:,1.j:ij:::.,j:j:*T:j
andthewages
andthereforeon th€tuel charges
o:rynd..o_ltheunitsproduced
of the.station
.staff.
maybe suchasto influencethe loadcurveandto improve
Tariff structures
s"j:9.fi",.:''
]
Tariff should consider the pf (power factor) of the load of the consumer.
If it is low, it takes more current for the same kWs and hence Z and D
i
i;;;;.i."
and distribution) losses are conespondingly increased. The
rhisisdenned
asthesum
ofindividual
maximum
demands
ontheconsumers, i
::m:r,:_".*'Ji:[1t
z%""l,i""irT3H::Tgl""rii:'J],f;fi:_'J""1tr
divided
by,r,"**i-u--i;#"#UH:ffiTTfi,1"ff":"il:ffi:
I
:#:*'::i,:"ff"Hf,.j,f.',.d1:$:"*Yil'.:fff:fii*,:.;iff;
,."'.::"'?;
;3:":ffii"T,"-jT,ffiH:
i:s,?"
--
::::*1":i""1".:1t:i1;ft;,"
transmission
prant.
rf authedemands ;;"
i:"::19'c,.r"d
i.e.unitv
divenitv
"r-"
ru"to',
tr,"'iotJ
;;';";;;il.;;;;;;ilTffi?
more. Luckily, rhefactoris much highertrranunity,
loads.
"il*t,
i
ili"tH *
;xl#1rJ,ffi?,1?Til'jl[,"ff"tJ'trJ;
tr;'3f:l
-:,':T":.-"::i,"":"^::::":--":J
_ ;;ff;
;;"";;'--
I
,
(iiD the consumermay be asked to use shunt capacitorsfor improving the
i
,;"'11'f,;:
i
f-;;#;
A high diversity factor could be obtained bv;
1' Giving incentivesto farmers and/or some industries
to use electricity in
the
u r u rnight
u E i r r or
lighr
ur U
B I | L rload
o a o periods.
pefloos.
2 using daylight saving as in many other counfies.
3' staggering the offrce timings
m
lil
tharee,lfe
::"T:l:_^,:Tl it1:Y_T^:*
tlil a pf penaltyclausemay be imposedon the consumer.
po*er factor of his installations.
I
Llg4"
1'1
L------
A factory to be set up is to have a fixed load of 760 kw gt 0.8 pt. The
electricrty board offeri to supplJ, energy at the following alb;ate rates:
(a) Lv supply at Rs 32ftvA max demand/annum + 10 paise/tWh
(b) HV supply at Rs 30/kvA max demand/annum + l0 paise/kwh.
rne lrv switchgear costs Rs 60/kvA and swirchgear losses at full load
i
amount to 5qa- Intercst depreciation charges ibr the snitchgear arc l29o of the
I
capital cost. If the factory is to work for 48 hours/week, determine the more
e.conomical tariff.
Actual energyproduced
7@
m a x i m u m p o s s i b l e e ' m -s o f u t i o n M a x i m u m d e m a n d = 0 3 = 9 5 0 k v A
@asedon instarelptant capaciiyy
Loss in switchgear= 5%
4' Having different time zones in the country
like USA, Australia, etc.
5' Having two-part tariff in which consumer
has to pay an amount
dependent on the maximum demand he makes, plus
u
fo.
unit of energyconsumed.sometimes consumer
"h.g;o? tt" u"si,
"u"t
ii charged
of kVA demand instead of kW to penalize to"O.
of to'* lo*",
tin"tor.
other factors used frequently
enuy 3re:
are:
plant capacity
foctor
L
_ Average demand
Installedcapacity
..
Plant usef(tctor
Actual energyproduced
_ _ --_
(kWh)
-. . -,:- -plant capacity(kw) x Time (in
hours)th" plunrh^
I
'
b;i"
il;;ti""
InPut dematrd=
j-950 =
1000 kvA
of switchgear= 60 x 1000 = Rs 60,000
"ost
Annual chargeson degeciation= 0.12 x 60,000= Rs 7,200
Annual fixed chargesdue to maximum demand correspondingto tariff (b)
Tariffs
= 30 x 1.000= Rs 30,000
The cost of electric power is normally given by the
expression(a + D x kW
+ c x kWh) per annum,where 4 is a rixea clarge
f_ ,f," oiifif,'ina"p".a"*
of the power output; b dependson the maximum
demandon tir" syrie- ano
www.muslimengineer.info
i
Annual running chargesdue to kwh consumed
= 1000x 0.8 x 48 x 52 x 0.10
Rs 1. 99. 680
هاجتإلا يمالسإلا- تالاصتإلاو
ةنجل روبلا
=
Introduction
WI
I
t
Total charges/annum
= Rs 2,36,gg0
Max. demand correspondingto tariff(a) j
.'.600 +0.03
or
950 kVA
l"'
4E-too-o.r3dE= o
dP
dP
dE=3m dP
dE=dPxt
From triangles ADF and ABC,
Annual running chargesfor kWh consumed
=950x0.8x48x52x0.10
= Rs t,89,696
Total= Rs 2,20,096
Therefore, tariff (a) is economical.
5,00,000-P _ 3000
8760
5,00,000
P = 328,say 330 MW
Capacityof thermalplant= 170MW
Energy generatedby thermal plant =
A region has a maximum demand of 500 MW
at a road factor of 50vo. The
Ioad duration curve can be assumedto be
a triangle. The utility has to meet
this load by setting up a generating system,
which is partly hydro and partry
thermal. The costs are as under:
Hydro plant: Rs 600 per kw per annum
and operating expensesat 3p
per kWh.
Thermal plant: Rs 300 per kw per annum
and operating expensesat r3p
Determinethe
170x3000x1000
= 255 x106 kwh
Energy generatedby hydro plant = 1935 x i06 kwh
Total annual cost = Rs 340.20 x 106/year
cost=
overallgeneration
x 100
###P
= 15.53paise/kWh
hydroprT!, rheenergygenerated
annuallyby
each, and overall :ffily:f
generationcost per kWh.
Solution
Total energy generatedper year = 500 x 1000
x 0.5 x g760
- 219 x 10' kwh
Figure 1.2 shows the load duration
curve. Since operatingcostof hydro plant
is low, the base load would be supplied
from the hydro plant and peak load from 500,000- P
the thermal plant.
Ler the hydro capacity be p kW and
the energy generaredby hydro plant E
kWh/year.
Thermal capacity = (5,00,000_ p) kW
Thermal energy = (2lg x107 _ E) kwh
B
Hours afoo
0
Annual cost of hydro plant
=600P+0.03E
Fig. 1.2 Load durationcurve
Annual cost of thermarplant = 300(5,00,000- p)
+ 0.r3(zrg x r07_ n)
Total cost C = 600p + 0.038 + 300(5,00,000_ p)
I
+ 0.t3(219 x 107_ E)
For minimum cost, 4Q- = 0
dP
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A generatingstation has a maximum demandof 25 MW, a load factor of 6OVo,
a plant capacity factor of 5OVo,and a plant use factor of 72Vo.Find (a) the
'the reserve capacity of the plant, and (c) the
daily energy produced, (b)
maximum energy that could be produceddaily if the plant, while running as
per schedule,were fully loaded.
Solution
averagedemand
Load factor =
maximumdemand
averagedemand
25
Average demand= 15 MW
0.60 =
Plant capacity factor =
0.50=
averagedemand
;#;;..0".,,,
l5
installedcapacity
Installedcapacity= += 30 MW
0.5
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Introduction
Reserve capacity of the plant = instalredcapacity - maximum demand
=30-25=5MW
Daily energyproduced= flver&g€demandx 24 = 15 x 24
= 360 MWh
Energycorresponding
to installedcapacityper day
=24x30_720MWh
axlmum energy t
be produced
_ actualenergyproducedin a day
plant use factor
= :9 = 5ooMWh/day
0.72
From a load duration curve, the folrowing data are obtained:
Maximum demandon the sysremis 20 Mw. The load suppliedby the two
units is 14 MW and 10 MW. Unit No. 1 (baseunit) works for l00Vo of the
time, and Unit No. 2 (peak load unit) only for 45vo of the time. The energy
g e n e r a t e d b y u n iIt i s 1 x 1 0 8 u n i t s , a n d t h a t b y u n izt i s 7 . 5 x 1 0 6 u n i t s . F i n d
the load factor, plant capacityfactor and plant use factor of each unit, and the
load factor of the total plant.
Solution
Annual load factor for Unit 1 =
1 x 1 0 8 x 1 0 0:81.54V o
14,000
x 8760
The maximumdemandon Unit 2 is 6 MW.
Annual load factor for Unit 2 = 7 . 5 x 1 0x61 0 0 = 14.27Vo
6000x 8760
Load factor of Unit 2 for the time it takesthe load
7.5x106x100
6000x0.45x8760
I
1 . 0 7 5 x 1 0 E x 1=0 6135%o
0
The annual load factor of the total plant =
20,000x 8760
CommentsThe various plant factors, the capacity of base and peak load
units can thus be found out from the load duration curve. The load factor of
than that of the base load unit, and thus the
peak load unit is much higher than that
the
from
generation
cosf of power
from the baseload unit.
i;;;";-l
'- '-.--*-"
i
There are three consumersof electricity having different load requirementsat
different times. Consumer t has a maximum demand of 5 kW at 6 p.m. and
a demand of 3 kW at 7 p.m. and a daily load factor of 20Vo.Consumer 2 has
a maximum demandof 5 kW at 11 a.m.' a load of 2 kW at 7 p'm' and an
averageload of 1200 w. consumer 3 has an averageload of I kw and his
(b)
maximum demandis 3 kW at 7 p.m. Determine:(a) the diversity factor,
the load factor and averageload of each consumer, and (c) the average load
and load factor of the combined load.
Solution
LF
3kw
MD5KW
(a) ConsumerI
ZOVo
a
t
T
p
m
at6pm
Average load
2
k
w
MD5KW
Consumer2
1\2 kW
a
t
T
p
m
at 11 am
load
Average
MD3KW
Consumer3
1kw
atTpm
Maximum demand of the system is 8 kW at 7 p'm'
=
sum of the individual maximum dernands= 5 + 5 + 3 13 kw
(b)
100
Annual plant capacityf'actorof Unit z = lPgx
= 8'567o
loxg76oxloo
Plant use factor of Unit 2 =
7.5x106x100
= 19.027o
10x0.45x8760x100
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DiversitY factor = 13/8 = 7.625
ConsumerI Average load 0'2 x 5 = I kW,
Consumer2 Averageload 1.2 kW,
= 3I.7 I7o
Since no reserveis available at Unit No. 1, its capacity factor is the
sameas the load factor, i.e. 81.54vo.Also since unit I has been running
throughout the year, the plant use factor equals the plant capacity factor
i . e .8 1 . 5 4 V o .
N
Consumer3 Average load I kW,
(c)
LF= 20Vo
l'2*1000-24Vo
5
I
100-33.3Vo
LF=
5x
LF=
Combinedaverageload = I + l'2 + l = i . 2 k W
Combinedload factor = +
=40Vo
x100
Load Forecasting
As power plant planning and constructionrequire a gestationperiod of four to
eight yearsor even longer for the presentday superpower stations,energy anrl
load demandfnrecastingplays a crucial role in power systemstudies.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffiil,ftffi|
ModernpowerSyslemnnatysis
I
This necessitates
long range forecasting.while sophisticatedprobabilistic
methodsexist in literature [5, 16, 28], the simple extrapolationtechnique is
quite adequatefor long range forecasting.since weatherhas a much more
influence on residentialthan the industrial component,it may be better to
prepare forecast in constituentparts to obtain total. Both power and energy
uru ractors rnvolved re
ng an involved
processrequiring experienceand high analytical ability.
Yearly forecastsare basedon previous year's loading for the period under
considerationupdated by factors such as general load increases,major loads
and weathertrends.
In short-term load forecasting,hour-by-hour predictions are made for the
lntroduction
system(gtid,area) to another.A distributionsystem connectsall the loads in
a particularareato the transmissionlines.
For economicaland technologicalreasons(which will be discussedin detail
electricallyconnectedareasor regionalgrids (also called power pools). Each
area or regional grid operatestechnicallyand economically independently,but
these are eventuallyinterconnected*to form a national grid (which may even
form an internationalgrid) so that each areais contractually tied to other areas
in respectto certaingenerationand schedulingfeatures.India is now heading
for a national grid.
The siting of hydro stations is determinedby the natural water power
sources.The choiceof site for coal fired thermalstationsis more flexible. The
following two alternativesare possible.
l. power starionsmay be built closeto coal tnines (calledpit headstations)
and electric energy is evacuatedover transmission lines to the load
centres.
decadeof the 21st century it would be nparing2,00,000Mw-a stupendous
task indeed.This, in turn, would require a correspondingdevelopmeni in coal
resources.
T.2
STRUCTURE OF POWER SYSTEMS
Generatingstations,transmissionlines and the distributionsystemsare the main
componentsof an electric power system.Generatingstationsand a distribution
systemare connectedthrough transmissionlines, which alsoconnectone power
* 38Voof the total power required
in India is for industrial consumption. Generation
of electricity in India was around 530 billion kWh in 2000-2001 A.D. compared to
less than 200 billion kWh in 1986-87.
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Z. power stations may be built close to the load ceutres and coal is
transportedto them from the mines by rail road'
In practice,however,power stationsiting will dependupon many factorstechnical, economical and environmental. As it is considerablycheaper to
transport bulk electric energy over extra high voltage (EHV) transmission
lines than to transportequivalentquantitiesof coal over rail roqd, the recent
trends in India (as well as abroad) is to build super (large) thermal power
stations near coal mines. Bulk power can be transmitted to fairly long
distancesover transmissionlines of 4001765kV and above. However, the
country's coal resourcesare locatedmainly in the easternbelt and some coal
fired stationswill continueto be sitedin distantwesternand southernregions.
As nuclear stationsare not constrainedby the problemsof fuel transport
and air pollution, a greater flexibility exists in their siting, so that these
stationsare locatedclose to load centreswhile avoiding high densitypollution
areasto reducethe risks, however remote,of radioactivityleakage.
*Interconnectionhas the economic advantageof reducing the reserve generation
capacity in eacharea.Under conditionsof suddenincreasein load or loss of generation
in one area, it is immediately possible to borrow power from adjoining interconnected
areas.Interconnectioncauseslarger currentsto flow on transmissionlines under faulty
condition with a consequent increase in capacity of circuit breakers. Also, the
centres.It provides capacity savingsby seasonalexchangeof power between areas
having opposing winter and summer requirements.It permits capacity savings from
time zones and random diversity. It facilitates transmissionof off-peak power. It also
gives the flexibility to meet unexpectedemergencyloads'
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
In India, as of now, about 75voof electric power used
is generatedin thermal
plants(including nuclear).23vofrommostly hydro
stationsandZvo.comefrom
:^:yft.s
and.others.
coal is the fuer for most of the sreamplants,the rest
substation,wherethe reductionis to a rangeof 33 to 132 kV, dependingon the
transmissionline voltage. Some industriesmay require power at these voltage
level.
The next stepdownin voltage is at the distribution substation.Normally, two
distribution voltage levels are employed:
l. The primary or feeder voltage(11 kV)
2. The secondaryor consumer voltage (440 V three phase/230V single
phase).
The distribution system, fed from the distribution transformer stations,
supplies power to ttre domestic or industrial and commercial consumers.
Thus, the power system operatesat various voltage levels separatedby
transformer.Figure 1.3 depicts schematicallythe structureof a power system.
Though the distribution system design, planning and operation are subjects
of great importance,we are compelled,for reasonsof space,to exclude them
from the scope of this book.
1.3
a
.qi-aji,
Generating
stations
at 11kV - 25kv
'-qff-9-a,
O
Transmission
level
(220kv - 765 kV)
Tie linesto
othersystems
CONVENTIONAL
Thermal (coal, oil, nuclear) and hydro generations are the main conventional
sources of electric energy. The necessity to conserve fosqil fuels has forced
scientists and technologists across the world to search for unconventional
sources of electric energy. Some of the sourcesbeing explored are solar, wind
and tidal sources.The conventional and some of the unconventional sourcesand
techniquesof energy generation are briefly surveyedhere with a stresson future
trends, particularly with reference to the Indian electric energy scenarioTtrermal
Large
consumers
Smallconsumers
Fig. 1.3 schematic diagram depicting power system
structure
www.muslimengineer.info
SOURCES OF ELECTRIC ENERGY
Power Stations-Steam/Gas-based
The heat releasedduring the combustionof coal, oil or gas is used in a boiler
to raise steam. In India heat generationis mostly coal basedexcept in small
sizes, because of limited indigenous production of oil. Therefore, we shall
discuss only coal-fired boilers for raising steam to be used in a turbine for
electric generation.
The chemical energy stored in coal is transformed into electric energy in
thermal power plants. The heat releasedby the combustion of coal produces
steam in a boiler at high pressureand temperature,which when passedthrough
a steamturbine gives off some of its internal energy as mechanicalenergy. The
axial-flow type of turbine is normally used with several cylinders on the same
shaft. The steam turbine acts as a prime mover and drives the electric generator
(alternator). A simple schematic diagram of a coal fired thermal plant is shown
in Fig. 1.4.
The efficiency of the overall conversionprocessis poor and its maximum
value is about 4OVobecauseof the high heat lossesin the combustiongasesand
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
rt^-r^--
E
h-.-^-
rntroduction
^
and the large quantity of heat rejectedto the condenser
which has to be given
off in cooling towers or into a streamlake in the
case of direct condenser
cooling' The steam power station operateson the
Rankine cycle, modified to
\vv'yvrDrwrr ur r'.lr. r.u Inecnanlcal energy) can be increased
by
using steam at the highest possible pressure
and temperature. with steam
turbines of this size, additionalincreasein efficiency
is obtainedby reheating
the steam after it has been partially expanded by
an ext;;;
i"ui"r. rn"
reheated steam is then returned to the turbine where
it is expandedthrough the
final states of bleedins.
To take advantageof the principle of economy of
scale (which applies to
units of all sizes),the presenttrend is to go in foilarger
sizesof units. Larger
units can be installed at much lower cost per kilowatt.
Th"y are also cheaper
to opcrate because of higher efficiency. Th"y
require io*", labour and
maintenance expenditure. According to chaman Kashkari
[3] there may be a
saving of as high as l|vo in capitalcost per kilowatt
by going up from a 100
to 250 MW unit size and an additional saving in
fuel cost of ubout gvo per
kwh. Since larger units consumeless fuer pJr
kwh, they produce ress air,
thermal and waste pollution, and this is a significant
advantagein our concern
for environment' The only trouble in the cai of a
large unit is the tremendous
shock to the system when outage of such a large
capacity unit occurs. This
shock can be tolerateclso long as this unit sizeloes
not exceed r}vo of the
on-line capacity of a large grid.
Stack
Step-up
Ah
uE
transformer
10-30 kv /
Coolirrgtower
-Condenser
Burner
mill
Preheated
air
Forced
draft fan
Flg. 1.4 schematicdiagramof a coarfiredsteamprant
In India, in 1970s the first 500 Mw superthermal
unit had been
commissioned at Trombay. Bharat Heavy Electricals
Limited (BHEL) has
produced severalturbogeneratorsetsof 500 MW
capacity.Today;s maximum
generator unit size is (nearly 1200 Mw) limited
by the permissible current
cjensitiesused in rotor and stator windines. Efforts
are on to develoo srDer.
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Effi
perhaps increase unit sizes to several GWs which would result in better
generatingeconomy.
Air and thermal pollution is always present in a coal fired steam plant. The
COz, SOX, etc.) are emitted via the exhaust gasesand thermal pollution is due
to the rejected heat transferred from the condenserto cooling water. Cooling
towers are used in situations where the stream/lake cannot withstand the
thermal burden without excessivetemperaturerise. The problem of air pollution
can be minimized through scrubbers and elecmo-staticprecipitators and by
resorting to minimum emission dispatch [32] and Clean Air Act has already
been passedin Indian Parliament.
Fluidized-bed
Boiler
The main problem with coal in India is its high ash content (up to 4OVomax).
To solve this, Jtuidized bed combustion technology is being developed and
perfected.The fluidized-bed boiler is undergoingextensivedevelopmentand is
being preferred due to its lower pollutant level and better efficiency. Direct
ignition of pulverized coal is being introduced but initial oil firing support is
needed.
Cogeneration
Considering the tremendous amount of waste heat generatedin tlbrmal power
generation,it is advisable to save fuel by the simultaneousgenerationof
electricity and steam (or hot water) for industrial use or space heating. Now
called cogeneration,such systemshave long been common, here and abroad.
Currently, there is renewedinterest in thesebecauseof the overall increasein
energy efficiencies which are claimed to be as high as 65Vo.
Cogeneration of steam and power is highly energy efficient and is
particularly suitablefor chemicals,paper,textiles,food, fertilizer and petroleum
refining industries.Thus theseindustriescan solve energyshortageproblem in
a big way. Further, they will not have to dependon the grid power which is not
so reliable. Of course they can sell the extra power to the government for use
in deficient areas.They may aiso seil power to the neighbouring industries,a
concept called wheeling Power.
-and
As on 3I.12.2000, total co-generationpotential in India is 19,500MW
actual achievementis 273 MW as per MNES (Ministry of Non-Conventional
Energy Sources,Government of India) Annual Report 200H1.
There are two possible ways of cogenerationof heat and electricity: (i)
Topping cycle, (ii) Bottoming cycle. In the topping cycle, fuel is burnt to
produce electrical or mechanical power and the waste heat from the power
generationprovides the processheat.In the bottoming cycle, fuel first produces
process heat and the waste heat from the process6sis then used to produce
power.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
H
I
Coal-fired plants share environmental problems with some
other types of
fossil-fuel plants; these include "acid rain" and the ,,greenhouse,,
effect.
Gas Turbines
With increasing availability of natural gas
uangladesh) primemovers basedon gas turbines have been
developedon the
lines similar to those used in aircraft. Gas combustion
generates high
temperatures and pressures, so that the efficiency of the
las turbine is
comparable to that of steam turbine. Additional advantageis
that exhaust gas
from the turbine still has sufficient heat content, which is
used to raise steam
to run a conventional steam turbine coupled to a generator.
This is called
combined-cyclegas-turbine(CCGT) plant. The schernatic
diagram of such a
plant is drawn in Fig. 1.5.
The oldest and cheapestmethod of power generationis that of utilizing the
potential energy of water. The energy is obtained almost free of nrnning cost
and is completely pollution free. Of course, it involves high capital cost
requires a long gestation period of about five to eight years as compared to
four to six years for steamplants. Hydroelectric stationsare designed,mostly,
as multipurpose projects such as river flood control, storage of irrigation and
drinking water, and navigation. A simple block diagram of a hydro plant is
given in Fig. 1.6. The vertical difference between the upper reservoir and tail
race is called the head.
Headworks
Surgechamber
Spillway
Valve house
Reservoir
Pen stock
Powerhouse
Tailracepond
Fig. 1.6
A typical layout for a storage type hydro plant
Generator
Steam
Fig. 1.5 CCGTpowerstation
CCGT plant has a fast start of 2-3 min for the gas turbine
and about
20 minutes for the steam turbine. Local storage tanks
Jr gur
ured in
caseof gas supply intemrption. The unit can take up to ITVo "ui-u"
overload for short
periods of time to take care of any emergency.
CCGT unit produces 55vo of CO2 produced by a coal/oil-fired
plant. Units
are now available for a fully automatedoperation for 24h
or to meet the peak
demands.
In Delhi (India) a CCGT unit6f 34Mw is installed at
Indraprasthapower
Station.
There are culrently many installationsusing gas turbines
in the world with
100 Mw generators.A 6 x 30 MW gas turbine station has
already been put
up in Delhi. A gas turbine unit can also be used as synchrono.r,
.ornp"nsator
to help maintain flat voltage profile in the system.
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Hydro plants are of different types such as run-of-river (use of water as it
comes), pondage (medium head) type, and reservoir (high head) type. The
reservoir type plants are the ones which are employed for bulk power
generation.Often, cascadedplants are also constructed,i.e., on the sa.mewater
stream where the discharge of one plant becomesthe inflow of a downs6eam
plant.
The utilization of energy in tidal flows in channets has long been the
subject of researeh;Ttrsteehnical and economic difficulties still prevail. Some
of the major sites under investigation are: Bhavnagar,Navalakhi (Kutch),
Diamond Harbour and Ganga Sagar. The basin in Kandala (Gujrat) has been
estimated to have a capacity of 600 MW. There are of course intense siting
problems of the basin. Total potential is around 9000 IvftV out of which 900
MW is being planned.
A tidal power station has been constructed on the La Rance estuary in
northern France where the tidal height range is 9.2 m and the tidal flow is
estimated to be 18.000 m3/sec.
Different types of turbines such as Pelton. Francis and Kaplan are used for
storage,pondageand run-of-river plants, respectively.Hydroelectric plants are
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
W
t
-
-
ModernpowersystemAnarvsis
daily load demand curve.
Some of the existingpumpedstorageplantsare I100 MW Srisailem
in Ap
and 80 MW at Bhira in Maharashtra.
p=gpWHW
Nuclear Power Stations
W = dischargem3ls through turbine
p = densiry 1000 kg/m3
With the end of coal reservesin sight in the not too distant future,
the immediate
practical alternative sourceof large scale electric energy generation
is nuclear
energy.In fact, the developedcountries have already switched over
in a big way
to the use of nuclear energy for power generation.In India,
at present, this
sourceaccountsfor only 3Voof the total power generationwith
nuclear stations
at Tarapur (Maharashtra),Kota (Rajasthan),Kalpakkam (Tamil Nadu),
Narora
(UP) and Kakrapar (Gujarat). Several other nuclear power
plants will be
commissionedby 20I2.In future,it is likely that more and more power
will be
generatedusing this important resource (it is planned to
raise nuclear power
generationto 10,000MW by rhe year 2010).
When Uranium-235 is bombardedwith neutrons,fission reaction
takesplace
releasingneutrons and heat energy. These neutronsthen participate
in the chain
reaction of fissioning more atoms of 235U.In order that the
freshly released
neutronsbe able to fission the uranium atoms, their speedsmust
be ieduced to
a critical value- Therefore,for the reaction to be sustained,nuclear
fuel rods
must be embeddedin neutronspeedreducing agents(like graphite,
hqavy water,
etc.) called moderators.Forreaction control, rods made of n'eutron-absorbing
material (boron-steel) are used which, when inserted into the
reactor vessel,
control the amount of neutron flux thereby controlling the
rate of reaction.
However, this rate can be controlled oniy within a narrow range.
The schemadc,
diagramof a nuclear power plant is shown in Fig. 1.7. The heit
releasedby the
'uclear reaction is transportedto a heat exchangervia primary
coolant (coz,
water, etc.). Steam is then generatedin the heat exchanger,
which is used in a
conventionalmanner to generateelectric energy by meansof
a steam turbine.
Various types of reactorsare being used in practicefor power
plant pu{poses,
viz., advancedgas reactor (AGR), boiling water reactor (BwR),
und h"uuy
water moderated reactor.etc.
where
11= head (m)
8 = 9.81 mlsz
Problems peculiar to hydro plant which inhibit expansion are:
1. Silting-reportedly Bhakra dead storagehas silted fully in
30 years
2. Seepage
3. Ecological damageto region
4. Displacementof human habitation from areasbehind the dam
which will
fill up and become a lake.
5. Thesecannotprovide baseload, mustbe usedfor peak.shavingand
energy
saving in coordination with thermal plants.
India alsohas a tremendouspotential (5000MW) of having large number
of
micro (< 1 Mw), mini (< 1-5 Mw), and,small (< 15 Mw)
Mrl plants in
Himalayan region, Himachal, up, uttaranchal and JK which must
be fully
exploited to generatecheapand clean power for villages situated far
awayfrom
the grid power*. At present 500 MW capacityis und"r construction.
In areaswheresufficienthydro generationis not available,peak load
may be
handled by means of pumped storage. This consists of un ,rpp". and
lower
reservoirs and reversible turbine-generatorsets, which cun ulio
be used as
motor-pump sets.The upper reservoir has enough storagefor about
six hours
of full load generation.Such a plant acts as a conventional hydro plant
during
the peak load period, when production costs are the highest. The iurbines
are
driven by water from the upper reservoir in the usual manner. During
the light
load period, water in the lower reservoir is pumped back into the ipper
one
so as to be ready for use in the next cycle of the peak ioad p.rioo.
rn"
generatorsin this period change to synchronousmotor action
and drive the
turbineswhich now work as pumps. The electric power is supplied to
the sets
from the general power network or adjoining thermal plant. The
overall
efficiency of the sets is normarly as high ut 60-7oEo. The pumped srorage
scheme,in fact, is analogousto the charging and discharging or u battery.
It
has the added advantage that the synchronousmachin", tu1 be used
as
synchronous condensersfor vAR compensationof the power network, if
required. In-a way, from the point of view of the thermal sector of the system,
Controlrods
Fuelrods_
* Existing capacity (small
hydro) is 1341 MW as on June 200I. Total estimated
potentialis 15000 MW.
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Waterintake
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
iiiriociucrion
ModernPo*", system An"tysis
W
CANDU reactor-Natural uranium (in cixideform), pressurizedheavy water
moderated-is adopted in India. Its schematic diagram is shown in Fig.
1.8.
-
require that they be normally located away from populated areas.
Demerits
Nuclear reactors produce radioactive fuel waste, the disposal
poses serious environmentalhazards.
2. The rate of nuclear reaction can be lowered only by a small margin, so
that the load on a nuclear power plant can only be permitted to be
marginally reduced below its full load value. Nuclear power stations
must, therefore, be realiably connected to a power network, as tripping
of the lines connecting the station can be quite serious and may required
shutting down of the reactor with all its consequences.
3 . Because of relatively high capital cost as against running cost, the
nuclear plant should operate continuously as the base load station.
Wherever possible, it is preferable to support such a station with a
pumped storage schemementioned earlier.
4. The greatestdangerin a fission reactor is in the caseof loss of coolant
Containment
Fig. 1.8 CANDUreactor-pressurized
heavywater rnoderated-adopted
in
In d i a
The associatedmerits and problems of nuclear power plants as compared
to conventional thermal plants are mentioned below.
Merits
1. A nuclear power plant is totally free of air pollution.
2. It requireslinle fuel in terms of volume and weight, and thereforeposes .
no transportationproblems and may be sited, independentlyof nuclear
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in an accident. Even with the control rods fully lowered quickly called
scrarn operation, the fission does continue and its after-heatmay cause
vaporizing and dispersalof radioactive material.
The world uranium resourcesare quite limited, and at the presentrate may
not last much beyond 50 years.However, there is a redeemingfeqture. During
235U,some of the neutrons are absorbed by
the fission of
lhe more abundant
23sUwhile
238U
uranium isotope
lenriched uranium containsonly about 3Voof
238U)
converting it to plutonium ("nU), which in itself is a
most of its is
can be extractedfrom the reactor fuel waste by a fuel
and
material
fissionable
reprocessing plant. Plutonium would then be used in the next generation
reactors (fast breeder reactors-FBRs), thereby considerably extending the
life of nuclear fuels. The FBR technologyis being intenselydevelopedas it
will extend the availability of nuclear fuels at predicted rates of energy
consumption to several centuries.
Figure 1.9 shows the schematicdiagram of an FBR. It is essentialthat for
breeding operation, conversion ratio (fissile material generated/fissilematerial
consumed) has to be more than unity. This is achieved by fast moving
neutrons so that no moderatoris needed.The neutrons do slow down a little
through collisions with structural and fuel elements.The energy densitylkg of
fuel is very high and so the core is small. It is therefore necessarythat the
coolant should possessgood thermal properties and hence liquid sodium is
used. The fuel for an FBR consists of 20Voplutonium phts 8Vouranium oxide.
The coolant, liquid sodium, .ldaves the reactor at 650"C at atmospheric
pressure.The heat so transportedis led to a secondarysodium circuit which
transfers it to a heat exchanger to generate steam at 540'C.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t_
Modprn
rrtyyvrr.
pnrrrar
r
vrrvr
errolam
vyglgttl
Anal.,^l^
nttdtvsts
with a breeder reactor the release of plutonium, an extremely
toxic
material, would make the environmentalconsiderationsmost
stringent.
-been
An experimentalfast breeder test reacror (FBTR) (40 MW) has
built
at Kalpakkamalongsidea nucrearpowerplant.FBR technologyi,
"*f..l"J
conventional thermal plants.
suitedfor India, with poor quality coal,inadequare
hydro potentiaiilentiful
reservesof uranium(70,000tons) and thorium,and many yearsof nuclear
engineering experience. The present cost of nuclear
wlm coal-ttred power plant, can be further reduced by standardisingpl4nt
design and shifting from heavy wate,rreactorto light water reactortechnology.
Typical power densities 1MWm3) in fission reactor cores are: gas cooled
0.53, high temperaturegas cooled 7.75, heavy warer 1g.0, boiling iut.,
Zg.O,
pressurizedwater 54.75, fast breederreactor 760.0.
Fusion
-
Core
Coolant
Containment
Fig. 1.9 Fastbreederreactor(FBR)
An important advantageof FBR technologyis that it can also use thorium
(as fertile material) which gets convertedto t33U which is fissionable.
This
holds great promise for India as we have one of the world's largest
deposits
of thoriym-about 450000 tons in form of sand dunes in Keralu una along
the
GopalpfurChatrapurcoastof Orissa. We have merely 1 per cent of the world's
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Energy is produced in this processby the combination of two light nuclei
to
form a single heavier one under sustained conditions of exiemely
high
temperatures(in millions of degree centigrade).Fusion is futuristic. Generation of electricity via fusion would solve the long-tenn energy needs
of the
world with minimum environmental problems. A .o--"i.iul
reactor is
expectedby 2010 AD. Considering radioactive wastes, the impact of
fusion
reactors would be much less than the fission reactors.
In case of successin fusion technologysometime in the distant future
or a
breakthroughin the pollution-free solarenergy,FBRs would becomeobsolete.
However, there is an intense need today to develop FBR technology
as an
insuranceagainst failure to deverop these two technologies.
\
In the past few years, serious doubts have been raised.about the safety
claims of nuclear power plants. There have been as many as 150 near
disaster
nuclear accidents from the Three-mile accident in USA to the
recent
Chernobyl accident in the former USSR. There is a fear.that all this may pur
the nuclear energy developmentin reversegear. If this happensthere
could be
serious energy crisis in the third world countries which have pitched
their
hopes on nuclear energy to meet their burgeoningenergy needs.France (with
78Voof its power requirement from nuclear sources)and Canadaare possibly
the two countries with a fairty clean record of nuclear generation.India
needs
to watch carefully their design, constructionand operating strategiesas
it is
committed to go in a big way for nuclear generation and hopes to
achieve a
capacity of 10,000 MW by z0ro AD. As p.er Indian nuclear scientists,
our
heavy water-basedplants are most safe.But we must adopt more conservative
strategiesin design, constructionand operationof nuclearplants.
World scientistshave to adopt of different reaction safety strategy-may
be
to discover additives to automatically inhibit feaction beyond cr;ii"at rather
than by mechanically inserted control rods which have possibilitiesof several
primary failure events.
Magnetohydrodynamic
(MHD) Generation
In thermal generation of electric energy, the heat released by the fuel is
converted to rotational mechanical energy by means of a thermocvcle. The
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Introduction
Modern
Power
System
Anatysis
ry
mechanicalenergy is then used to rotate the electric generator.Thus two
stages of energy conversion are involved in which the heat to mechanical
energy conversion has inherently low efficiency. Also, the rotating machine
has its associatedlossesand maintenanceproblems.In MHD technology,
cornbustionof fuel without the need for mechanicalmoving parts.
In a MHD generator,electrically conducting gas at a very high temperature
is passed in a strong magnetic fleld, thereby generatingelectricity. High
temperature is needed to iontze the gas, so that it has good eiectrical
conductivity. The conductinggas is obtainedby burning a fuel and injecting
a seeding materials such as potassium carbonate in the products of
combustion. The principle of MHD power generation is illustrated in Fig.
1.10. Abotrt 50Voefficiency can be achievedif the MHD generatoris operated
in tandem with a conventional steam plant.
Gas flow
at 2,500'C
Strong magnetic
field
e MH Dpow ergenerati on
F i g .1 .1 0 T h e p ri n c i p lof
Though the technologicalfeasibility of MHD generationhas been established, its economicf'easibilityis yct to be demonstrated.lndia had starteda
research and developmentproject in collaboration with the former USSR to
install a pilot MHD plant based on coal and generating2 MW power. In
Russia, a 25 MW MHD plant which uses natural gas as fuel had been in
operation for some years. In fact with the developmentof CCGT (combined
cycle gas turbine) plant, MHD developmenthas been put on the shelf.
Geothermal
Power Plants
In a geothermal power plant, heat deep inside the earth act as a source of
power. There has been some use of geothermalenergy in the form of steam
coming from undergroundin the USA, Italy, New Zealand,Mexico, Japan,
Philippines and some other countries. In India, feasibility studies of 1 MW
station at Puggy valley in Ladakh is being carried out. Another geothermal
field has been located at Chumantang.There are a number of hot springsin
India, but the total exploitableenergy potential seemsto be very little.
Ttre present installed geothermal plant capacity in the world is about 500
MW and the total estimatedcapacityis immenseprovided heat generatedin the
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w
I
volcanic regionscan be utilized. Since the pressureand temperaturesare low,
the efficiency is even less than the conventionalfossil fuelled plants,but the
capital costs are less and the fuel is available free of cost.
I.4
RENEWABLE ENERGY SOURCES
To protect environmentand for sustainabledevelopment,the importance of
It is an establishedand
renewableenergysourcescannot be overemphasized.
acceptedtact that renewableand non-conventionalforms of energy will play
an increasingly important role in the future as they are cleanerand easier to
use and environmentallybenign and are bound to becomeeconomicallymore
viable with increaseduse.
Because of the limited availability of coal, there is considerableinternational effort into the development of alternative/new/non-conventionaUrenewable/cleansourcesof energy. Most of the new sources (some of them in fact
have been known and used for centuries now!) are nothing but the
manifestationof solar energy, e.g., wind, sea waves, ocean thermal energy
conversion (OTEC) etc. In this section, we shall discuss the possibilities and
potentialities of various methods of using solar energy.
Wind Power
Winds are essentiallycreated by the solar heating of the atmosphere.Several
attempts have been made since 1940 to use wind to generateelectric energy
and developmentis still going on. However, technoeconomicfeasibility has
yet to be satisfactorilyestablished.
Wind as a power source is attractivebecauseit is plentiful, inexhaustible
and non-polluting. Fnrther, it does not impose extra heat burden on the
Control
environment.Unlbrtunately, it is non-steadyand undependable.
equipment has been devised to start the wind power plant wheneverthe wind
speedreaches30 kmftr. Methods have also been found to generateconstant
frequencypower with varying wind speedsand consequentlyvarying speeds
of wind mill propellers. Wind power may prove practical for small power
needs in isolated sites. But for maximum flexibility, it should be used in
conjunction with other methods of power generationto ensurecontinuity.
For wind power generation, there are three types of operations:
1. Small, 0.5-10 kW for isolated single premises
i
2. Medium, 10-100 kW for comrnunities
3. Large, 1.5 MW for connectionto the grid.
The theoreticalpower in a wind streamis given by
P = 0.5 pAV3W
where
p = densityof air (1201 g/m' at NTP)
V _ mean air velocity (m/s) and
A = sweptarea (rn").
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Introduction
2. Rural grid systemsare likely to be 'weak, in theseareas.
since
retatrvely low voitage supplies(e.g. 33 kV).
3. There are always periods without wind.
In India, wind power plants have been installed in
Gujarat, orissa,
Maharashtraand Tamil Nadu, where wind blows at speeds
of 30 kmftr during
summer' On the whole, the wind power potential of India
has been estimated
to be substantial and is around 45000 Mw. The installed
capacity as on
Dec. 2000 is 1267 Mw, the bulk of which is in Tamil
Nadu- (60%). The
conesponding world figure is 14000 Mw, rhe bulk
of which is in Europe
(7UVo).
Solar Energy
The average incident solar energy received on earth's
surface is about
600 W/rn2 but the actual value varies considerably.It
has the advantageof
beingfree of cost,non-exhaustible
and completelypollution-free.On the other
hand,it has severalcrrawbacks-energy density pei unit
area is very row, it is
available for only a part of the day, and cl,oudy
and, hazy atmospheric
conditions greatly reduce the energy received. Therefore,
harnessing solar
energyfor electricitygeneration,challengingtechnologicalproblems
exist,the
most important being that of the collection and concentration
of solar energy
and its conversion to the electrical form through efficient
and comparatively
economicalmeans.
At present, two technologiesare being developedfor conversion
of solar
energyto the electrical form.-'In one technology,collectors
with concentrators
are employedto achievetemperatureshigh enough(700'C)
to operatea heat
engrne at reasonable efficiency to generate electricity.
However, there are
considerableengineeringdifficulties in building a single
tracking bowi with a
diarneterexceeding30 m to generateperhaps200 kw.
The schemeinvolves
large and intricate structuresinvoiving
lug" capital outlay and as of today is
f'ar from being competitive with
Jlectricity generation.
"otru"titional
The solar power tower [15] generates
steam for electricity procluction.
]'here is a 10 MW installationof such a tower by
the Southern California
EdisonCo' in USA using 1818plane rnirrors, eachi m x
7 m reflecting direct
racliationto thc raiseclboiler.
Electricity may be generatedfrom a Solar pond by using
a special .low
temperature'heat enginecoupledto an electricgenerator.A
solar pond at Ein
Borek in Israel proclucesa steady150 kW fiorn 0.74 hectare
at a busbarcost
of abo u t$ O.tO/k w h .
Solar power potential is unlimited, however, total capacityof
about 2000
MW is being planned.
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Tota l solar ener gypot ent ialin I ndia is 5 x lO ls kwh/ yr . Up r o 31. t 2. 2000.
462000solar cookers,55 x10am2solar thermai system collector area,47 MW
of SPV power, 270 community lights, 278000 solar lanterns(PV domestic
lighting units),640 TV (solar), 39000 PV streetlights and 3370 warer pumps
MW of grid connected solar power plants were in operation. As per one
estimate[36], solar power will overtakewind in 2040 and would become the
world's overall largest source of electricity by 2050.
Direct Conversion
to Electricity
(Photovoltaic
Generation)
This technologyconvertssolar energyto the electrical form by meansof silicon
wafer photoelectriccells known as "Solar Cells". Their theoreticalefficiency is
about 25Vobut the practical value is only about I5Vo. But that does not matter
as solar energy is basically free of cost. The chief problem is the cost and
maintenanceof solar cells. With the likelihood of a breakthroughin the large
scale production of cheap solar cells with amorphoussilicon, this technology
may competewith conventional methodsof electricity generation,particularly
as conventionalfuels becomescarce.
Solar energy could, at the most, supplementup to 5-r0vo of the total
energy demand.It has been estimatedthat to produce 1012kwh per year, the
necessarycells would occupy about0.l%oof US land areaas againsthighways
which occupy 1.57o(in I975) assumingI07o efficiency and a daily insolation
.\
of 4 kWh/m'.
In all solar thermalschentes,storageis necessarybecauseof the fluctuating
nature of sun's energy. This is equally true with many other unconventional
sourcesas well as sourceslike wind. Fluctuatingsourceswith fluctuating
loads complicatestill further the electricity supply.
Wave Energy
The energyconientof sea wavesis very high. In India, with severalhundreds
of kilometersof coast line, a vast sourceof energyis available.The power in
the wave is proportionalto the squareof the anrplitudeand to the period of
the motion. Therefore,rhe long period (- 10 s), large amplitude(- 2m) waves
are of considerable interest for power generaticln, with energy fluxes
commonly averagingbetween50 and 70 kW/m width of oncoming wave.
Though the engineeringproblems associatedwith wave-powerare formidable,
the amountof energythat can be harnessedis large and developmentwork is
in progress(alsoseethe sectionon HydroelectricPower Generation,page 17).
Sea wave power estimated poterrtial is 20000 MW.
Ocean Thermal Energy Conversion (OTEC)
The ocean is the world's largest solar coilector. Temperaturedifference of
2O"Cbetween\,varrn,solar absorbingsurfacewater and cooler 'bottorn' water
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modem
Pow'er
systemAnatysis
lntroduction
can occlrr.This can provide a continuallyreplenishedstoreof thermal energy
which is in principle available fbr conversion to other energy forms. OTEC
refers to the conversion of someof this thermal energy into work and thence
solar. The most widely used storagebatteryis the lead acid battery.invented
by Plantein 1860.Sodiuttt-sulphur
battery(200 Wh/kg) and other colrbinations of materialsare a-lsobeing developedto get more output and storageper
unit weisht.
ffiffi|
50,000 Mw.
A proposedplant using seaiemperaturedifferencewould be situated 25 km
cast ol'Mianii (USA), wherethe temperatureclil'l'eronce
is 17.5"C.
Biofuels
The material of plants and animals is called biomass, which may be
transformed by chemical and biological processesto produce intermediate
biofuels sttch as methane gas, ethanol liquid or charcoal solid. Biomass is
burnt to provide heat for cooking, comfort heat (space heat), crop drying,
tactory processesand raising steamfor electricity production and transport. In
I ndia p o te n ti aI'l ttlb i o -En e rg yi s 1 7 0 00MW and thatfbr agri cul tunrlw i rstci s
about 6000 MW. There are about 2000 community biogas plants and tamily
size biogas plants are 3.1 x 106. Total biomass power harnessedso far is
222 MW .
Renewableenergy programmesare specially designedto meet the growing
energy needs in the rural areas for prornoting decentralized and hybrid
dcvelopmentst.las to stem growing migration of rural populationto urban
areasin searchof better living conditions. It would be through this integration
of energy conservationefforts with renewable energy programmesthat India
would be able to achievea smooth transition from fossil fuel economy to
sustainablerenewableenergy basedeconomy and bring "Energy for ali" for
ec;uitableand environrnental
friendly sustainabledevelopment.
1.5
ENERGY STORAGE
'l'here
is a lol ol problenrin storing clectricity in largc quantities.Enclgy
wliich can be convertedinto electricity can be storedin a number of ways.
S t or ag eo f a n y n a tu rei s l ro w e v e rv ery costl y arrcli ts cconomi csmust be
worked out properly. Various options available are: pLrmpedstorage, c:onlpressedair, heat, hydrogengas,secondarybatteries,flywheels and superconduc t in gc o i l s .
As already mentioned, gas turbines are normally used for meeting peak
loads but are very expensive. A significant amount of storage capable of
instantancoususe would be betterway of meetingsuch peak loads, and so far
the most importantway is to have a pumped storageplant as discussedearlier.
Other methods are discuss-ed
below very briefly.
Secondary Batteries
Large scale battery use is almost ruled out and they will be used for battery
powered vehicles and local fluctuating energy sourcessuch as wind mills or
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Fuel Cells
A fuel cell convertschemicalenerry of a fuel into electricityclirectly,with no
intermediatecotnbustioncycle. In the fuel cell, hyclrogenis supplied to the
negative electrodeand oxygen (or air) to the positive. Hydrogenand oxygen
are combined to give water and electricity. The porous electrodesallow
hydrogen ions to pass.The main reason';rhy fuel cells are not in wide use is
their cost (> $ 2000/kW). Global electricity generatingcapacity from full cells
will grow from just 75 Mw in 2001 ro 15000MW bv 2010.US. Germanvand
Japan may take lead for this.
Hydrogen
Energy Systems
Hydrogen can be used as a medium for energy transmissionand storage.
Electrolysisis a well-established
commercialprocessyielding pure hydrogen.
Ht can be convertedvery efficiently back to electi'icityby rneansof fuel ceils.
Also the use of hydrogena.sfuel for aircraftand automcbilescould encourase
its large scaleproduction, storageand distriburion.
1"6
GROWTH OF POWER SYSTEII{S IN INDIA
India is fairly rich in natural resourceslike coal and lignite; while sorne oil
reserveshave been discoveredso far. intenseexplorationis being undertakeri
in vitriousregitlnsof thc country. India has immensewater power l.csources
also of which only around25Tohave so farbeen utiliseci,i.e.,oniy 25000 t\,IW
has so far beencommissionedup to the end of 9th plan. As per a recentreport
of tlre CEA (Ccntlal Flectricit,vAuthority),the total potentialof h1,dropower
is 84,040Iv{W at ('L't%load factor. As regardsnuclearpower, India is cleflcient
in uranium,but has rich depositsof thorir-imrvhichcan be utilisedat a future
clatc in l'ast brccclorrci.tctor.s.
Since indepcndcncc,thc coulltry has nnde
tremendousprogressin the developmentof electricenergyand todayit has the
largest systemamong the developingcountries.
When lndia attained independence,the installeclcapacity was as low as
1400 MW in the early stagesof the growth of power system,the major portion
of generationwas through thermal stations,but due to economicalreasons.
hydro developmentreceivedattentionin areaslike Kerala, Tamil Nadu. Uttar
Pradesh and Punjab.
In the beginningof the First Five Year Plan (1951-56), the rotal installed
capacity was around2300 MV/ (560 MW hydro, 1004MW thermal,149 MW
through oil stations and 587 MW through non-utilities). For transportingthis
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
HE
FI
Introduction
power to the load centres,transmissionlines of up to 110
were constructed.
regions of the country with projectedenergy requirement
year 2011-12
and peak load in the
[19]'
io ororrcrt crcncreri
nuclear power' At the
During the Fourth Five Plan, India startedgenerating
in April-May
comrnissioned
were
units
Tarapur i\uclear Plant 2 x 210 MW
design. By
American
of
reactors
water
. This station uses two boiling
Northern region
308528
(49674)
.,.
MW* 9
Westernregion
299075
(46825)
commissionedbY 2012.
projection for 2011The growth of generating capacity so far and future
1'1'
2012 A.D. are given in Table
(ln MW)
Tabte 1.1 Growthof Installedcapacityin lndia
Year
1970-7t
1978-79
1984-85
2000-01
Hydrtt
6383
l 1378
t4271
25141
Nuclear
Thermal
DieseI
Total
420
890
1095
2720
7503
t6372
27074
71060
398
r4704
28640
42240
101630
\'./
Fig. 1.11
in
energyrequirement
Mapof Indiashowingfive regionalprojected
year
2011-12'
for
MW
park
in
load
and
MkWh
The emphasisduring the SecondPlan (1956-61) was on the developmentof
power
basic ancl heavy inclustriesand thus there was a need to step up
the
end
at
MW
3420
around
was
which
capacity
installed
total
generation.The
Five
the
Second
year
of
end
the
at
MW
5700
became
Plan
Five
of tn" First
year plan. The introduction of 230 kv transmissionvoltage came up in Tarnil
=2700 MW
renewable
was: Domestic
Pattern of utlization of electrical energy in 1997-98
industry 35'22Voand othersis
{O.6g\o,commercial6.917o, inigation 30.54Vo,
in 2004-05'
6.657o.It is expectedto remain more or less same
spread out all over the
plants
To be self-sufficient in power' BHEL has
viz' turbo sets'equipment,
power
country ancltheseturn out an entirerangeof
transformpower
boilers,
pi".ture
tiigft
hydro sets,turbinesfor nuclearplants,
BHEL's
equipment'
of
range
a
in
ers, switch gears,etc. Each plant specializes
Today BIIEL
singrauli'
at
cornmissioned
was
first 500 MW turbo-generator
manufacturersin the
is consideredone of the major power plant equipment
world.
T.7
ENERGY CONSERVATION
of energy' we should resort
Energy conservationis the cheapestnew source
(discussedearlier), and
to various conservationmeasuressuch as cogeneration
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,r32 I
lntroduction
Modernpower Svstem Analvsis
use energy efficient motors to avoid wasteful electric uses.We can achieve
considerableelectricalpower savings by reducing unnecessaryhigh lighting
levels,oversizedmotors,etc. A 9 W cornpactfluorescentlamp (CFL) may be
used instead of 40 w fluorescenttube or 60 w lamp, all having the same
lu
a year.Everyoneshouldbe madeaware throughprint or electronicmedia how
consumptionlevels can be reducedwithout any essentiallowering of comfort.
Rate restructuringcan have incentivesin this regard.There is no consciousnesson energy accountability
yet etndno senseof urgencyas in developed
countries.
Transmissionand distributionlossesshoulclnot exceed2OVo.This can be
achievedby employing series/shuntcompensation,power factor improvement
methods, static var compensators,HVDC option and FACTS (flexible ac
technology) devices/controllers.
Gas turbirre combined with steam turbine is ernployed for peak load
shaving. This is more efficient than normal steam turbine and has a quick
automated starl and shut doivn. It improves the load factor of the steam
staflon.
Energy storage can play an important role where there is time or rate
mismatchbetween supplyand demandof energy.This has been discussedin
Section 1.5. Pumped storage(hyclro)schemehas been consicleredin Section
1.3.
Industry
In India where most areashave large number of sunny days hot water for
bath arrdkitchen by solarwater heatersis becomingcommon for commercial
buildings,hotels even hospitals.
In India where vastregionsare deficient in electric supply and,aresubjected
to long hours of power sheddingmostly random, the use of small diesel/petrol
generatorsand invertersare very conmon in commercialand domestic use.
Theseare highly wastefulenergydevices.By properplanned maintelance the
downtime of existing large stationscan be cut down. Plant utilization factors
of existingplants must be improved.Maintenancemust be on schedulerather
than an elner-qency.Maintenancemanpower training should be placed on war
footing. These actions will also improve the load factor of most power
stations,which would indirectly contribute to energy conservation.
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Load Management
'load management'schemes.It is possibleto
As mentionedearlier by various
shift demanrlaway frorn peak hours (Section I .1.). A more direct method
would be the control of the load either through rnodified tariff structurethat
encourage
schedulesor direct electrical control of appliancein the form of remote timer
controlled on/off switches with the least inconvenience io the customer.
Various systems for load rnanagementare described in Ref. [27]. Ripple
control has been tried in Europe. Remote kWh meter reading by carrier
sysrems is being tried. Most of the potential for load control lies in the
domestic sector. Power companies are now planning the introduction of
system-wideload managementschemes.
1.8
DEREGULATION
For over one hundred years,the electricpower industry worldwide operatedas
a regulated industry. In any area there was only one company oI government
agency (mostly state-owned)that produced,transmitted,distributed and sold
electric power and services.Deregulationas a conceptcame in early 1990s.It
competition.
brought in changesdesignecito enc<.rutage
power industry and reassembly
of
the
Restructuring involves disassembly
Privatisation started sale by a
organisation.
functional
or
into another form
governmentof its state-ownedelectric utility assets,and operatingeconomy,
to private companies.In some cases,deregulationwas driven by privatization
needs. The state wants to sell its electric utility investment and change the
rules (deregulation)to make the electric industry more palatable for potential
investors,thus raising the price it could expect from the sale. Open accessrs
nothing but a common way for a govenlmentto encouragecompetition in the
electric industry and tackle monopoly. The consumer is assuredof good
quality power supply at competitive price.
The structure for deregulation is evolved in terms of Genco (Generation
Company),Transco (TransrnissionCompany)and ISO (IndependentSystem
Operator).It is expectedthat the optimal bidding will help Genco to maximize
its payoffs. The consumersare given choice to buy energy from different retail
energy suppliers who in turn buy the energy from Genco in a power market.
(independentpower producer, IPP).
The restructuringof the electricity supply industry that norrnally accompanies the introduction of competiiion provides a fertile ground for the growth
of embeddedgeneration,i.e. generationthat is connectedto the distribut-icn
systemrather than to the transmissionsystetn.
The earliest reforms in power industrieswere initiated in Chile. They were
followed by England, the USA, etc. Now India is also implementing the
restructuring.Lot of researchis neededto clearlyunderstandthe power system
operation under deregulation. The focus of, researchis now shifting towards
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
W
Modernpo*", Syster Anulyri,
finding the optimal bidding methodswhich take into account local optimal
dispatch,revenueadequacyand market uncertainties.
India has now enactedthe Electricity RegulatoryComrnission'sAct, 1998
and the Electricity (Laws) AmendmentAct, 1998. These laws enablesetting
uo of
State Electricity RegulatoryComrnissions(SERC) at srate level. 'fhe main
purpose of CERC is to promote efficiency, economy and competition in bulk
electricity supply. orissa, Haryana, Andhra Pradesh,etc. have started the
processof restructuringthe power sector in their respectivestates.
1.9
DISTRIBUTED AND DISPERSED GENERATION
DistributedGeneration(DG) entailsusing lnany srnallgeneratorsof 2-50 MW
output,installedat variousstrategicpoints throughoutthe area,so that each
providespower to a small numberof consumersnearby.Thesemay be solar,
mini/micro hydel or wind turbine units, highly efficient gas turbines,small
combincdcycle plitnts,sincc thcsearo the rnost ccon<lnrical
choiccs.
Dispersedgenerationreferesto use of still smaller generatingunits, of less
than 500 kW output and often sized to serve individual homes or businesses.
Micro gas turbines,fuel cells, diesel,and small wind and solar PV senerators
make up this category.
Dispersedgenerationhas been used for clecadesas an emergencybackup
power source.Most of theseunits are used only fbr reliability reinfbrcement.
Now-a-daysinverters are being increasinglyused in domestic sector as an
emergencysupply during black outs.
The distributed/dispersedgeneratorscan be stand alone/autonomousor
grid connecteddepending upon the requirement.
At the time of writing this (200i) there still is and will probably alwaysbe
some economy of scale f-avouringlarge generators.But the margin of
economydecreased
considerablyin last 10 years [23]. Even if the power itself
c t ls t sa b i t rtttl rcth i tnc c n (r' asl ta ti o np o wcr,therei s no nccd < tftransrni ssi on
lines, and perhapsa reducedneed fbr distribution equipment as well. Another
maior advantageof dispersedgene.rationis its modularity, porlability and
relocatability.Dispersedgenerators
also include two new types of tbssil fuel
units-fuel cells and microgas turbines.
The main challenge today is to upgradethe existing technologiesand to
proniotedeveloprnent,demonstration,
scaling up and cornmercialization
of
new and emerging technologiesfor widespreadadaptation.In the rural sector
main thrust areasare biomassbriquetting,biomass-basedcogeneration,etc. In
solar PV (Photovoltaic),large size solar cells/modulesbased on crystalline
siliconthin films need to be developed.Solarcells efficiencyis to be improved
to 15%o
to be of use at commerciallevel.Otherareasare developrnent
of high
eificiency inverters.Urban and industrial wastesare used for variousenergy
applicationsincluding power generationwhich was around 17 Mw in 2002.
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I nt r oduct ion
There are already32 million improved chulhas.If growing energy needsin the
rural areasare met by decentralisedand hybrid ener-qysystems(distributed/
dispersedgeneration),this can stem growing migrationof rural populationto
urban areasin searchof better living conditions.Thus, India will be able to
able-energy based econolny iind bring "Energy for all" for equitable,
environment-friendly,and sustainabiedevelopment.
1.10 ENVIRONMENT/\L
GENERATION
ASPECTS OF ELECTRIC ENER,GY
As far as environmentaland health risks involvedin nuclearplants of various
kinds are concerned,thesehave already'beendiscussedin Section1.3. The
problernsrelatedto largelrydroplantshavealsobeendwelledupon in Section
1.3. Therefore,we shall now focus our attentionon fossil fuel plant including
gas-basedplants.
Conversion of clne lornr ol' energy or anotherto electrical tortn has
unwanted side effects and the pollutants generatedin the processhave to be
disposed off. Pollutants know no geographical boundary, as result the
pollution issue has become a nightmarish problem and strong national and
international pressuregroups have sprung up and they are having a definite
has
awareness
Governmental
impacton the developmentof energyresources.
creatednumerouslegislationat national and internationallevels, w[ich power
engineershave to be fully conversantwith in practiceof their professionand
survey and planning of large power projects.Lengthy, time consuming
proceduresat governrnentlevel, PIL (public interestlitigation) and demonstrative protestshave delayedseveralprojects in severalcountries.This has led
of existing sites.But
to favouring of small-sizeprojectsand redevelopment
with the increasinggap in electric dernandand production,our country has to
move forward fbr severallarge thermal, hydro and nuclear power projects.
cur t uilt nent
of t r ansnt issit t n
t
E ntph asisis lr cinglaid on cor ] scr vilt iorissucs.
losses, theft, subsidized power supplies and above all on sustainable
devektpnrenlwittr uppntpriata technolog-)'whercver feasible. It has to be
particularly assuredthat no irreversible damageis caused to environment
which wouid affect the living conditions of the future generations.Irreversible
damageslike ozonelayer holesand global warmingcausedby increasein CO2
in the atmosphereare alreadyshowing up.
Atmospheric
Pollution
We shall treat here only pollutrorras causedby thermalplants using coal as
feedstock. Certain issues concerning this have already been highlighted in
Section 1.3. The fossil fuel based generatingplants fonn the backbone of
power generation in our country and also giobally as other options (like
nuclear and even hydro) have even stronger hazardsassociatedwith them.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffiffi|
rr^r^--
^...^- ^.,-r-a--r.--,
tviouernnrow-er
uystem Anaiysts
w_
Also it should be understoodthat pollution in large cities like Delhi is caused
more by vehicrtlar traffic and their emission.In Delhi of course Inderprastha
and Badarpur power stationscontributetheir share in certain areas.
Problematic pollutants in emission of coal-basedgeneratingplants are.
a
2
o
NO.r, nitrogen oxides
CO
a
coz
a
. Certain hydrocarbons
o Particulates
Though the account that follows will be general, it needs to be mentioned
here that Indian coal has comparatively low sulphur content but a very high
ash content which in some coals may be as high as 53Vo.
A brief account of various pollutants, their likely impact and methods of
abatementsare presentedas follows.
Oxides of Sulphur
(SOr)
Most of the sulphur present in the fossil fuel is oxidized to SO2 in the
combustionchamberbefore being emittedby the chimney. In atmosphereit
gets further oxidized to HrSOo and metallic sulphateswhich are the major
sourceof concern as these can causeacid rain, impaired visibility, damageto
buildings and vegetation. Sulphate concenffations of 9 -10 LElm3 of air
aggravateasthma,lung and heart disease.It may also be noted that although
sulphur does not accumulatein air, it does so in soil.
Sulphur emissioncan be controlledby:
o IJse of fuel with less than IVo sulphur; generally not a feasible solution.
o LJseof chemical reaction to remove sulphur in the form of sulphuric
acid, from combustionproducts by lirnestonescrubbersor fluidized bed
combustion.
. Removing sulphurfrom the coal by gasificationor floatationprocesses.
It has been noticed that the byproduct sulphur could off-set the cost of
sulphur recovery plant.
Oxides of Nitrogen
(NO*)
Of theseNOz, nitrogenoxides,is a majorconcernas a pollutant.It is soluble
in water and so has adverseaff'ect on human health as it enters the lungs on
inhaling and combining with moisture converts to nitrous and nitric acids,
which danngethe lungs. At ievels of 25-100 parts per million NO, can cause
acutebronchitis and pneumonia.
Emissionof NO_,can be controlledby fitting advancedtechnologyburners
which can assuremore complete combustion,thereby reducing theseoxides
from being emitted. These can also be removedfrom the combustionproducts
by absorptionprocessby certain solventsgoing on to the stock.
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lntroduction
Oxides
of Carhon (CO, COt)
CO is a very toxic pollutantbut it getsconvertedto CO'.,in the openatmosphere
(if available) surroundingthe plant. On the other hand CO2 has been identified
developingcountries.
Ifydrocarbons
During the oxidation process in cornbustioncharnbercertain light weight
hydrocarbon may be formed. Tire compounds are a major source of
photochemical reaction that adds to depleti,rnof ozone layer.
Particulates
(fIY ash)
Dust content is particularly high in the Indian coal. Particulatescome out of
the stack in the form of fly ash. It comprisesfine particles of carbon, ash and
other inert materials.In high concentrations,these cause poor visibility and
respiratory diseases.
Concentration of pollutants can be reducedby dispersal over a wider area
by use of high stacks.Precipitators can be used to remove particles as the flue
gasesrise up the stack.If in the stack a vertical wire is strung in the middle
and charged to a high negative potential, it emits electrons. These electrons
are captured by the gas molecules therebybecomingnegative ions. These ions
accelerate towards the walls, get neutralized on hitting the'walls and the
particles drop down the walls. Precipitatorshave high efficiency up to 99Vofor
large particles, but they have poor performancefor particles of size less than
0.1 pm in diameter.The efficiency of precipitatorsis high with reasonable
sulphur contentin flue gasesbut drops for'low sulphurcontentcoals;99Vofor
37o sulphur and 83Vofor 0.5Vosulphur.
Fabric filters in form of bag lnuses have also been employed and are
located before the flue gases enter the stack.
Thermal
Pollution
Steam fronr low-pressureturbine has to be liquefied in a condenser and
reduced to lowest possible temperatureto maximize the thermodynamic
efficiency. The best efficiency of steam-cyclepracticallyachievableis about
4\Vo.It meansthat60Voof the heat in steamat the cycle end must be removed'
This is achievedby following two methods'
1. Once through circulation through condensercooling tubes of seaor river
water where available.This raises the temperatureof water in these two
sources and threatenssea and river life around in sea and downstream
in river. ThesE,are serious environmental objections and many times
cannot be overruled ard also there may be legislation againstit.
2. Cooling tov,ers Cool water is circulatedrottnd the condensertube to
remove heat from the exhaust steam in order to condenseit. The
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Gfrfud
ffi-ffii
lntrcCuction
MociernPowerSysteqAnaiysis
sffi
EEF
I
circulating water gets hot in the process.tt is pumped to cooling tower
and is sprayedthrough nozzlesinto a rising volume of air. Some of the
water evaporates
providingcooling.The latentheat of water is 2 x 106
J/kg and cooling can occur fast, But this has the disaclvantage
of raising
u n o e s t r a o t e Jt e v e l s l n t h c s u l r f t l u n d l n ga r e a s .
course the water evaporatedmust be macleup in the systemby adcting
fresh water from the source.
Closed cooling towers where condenr;ate
flows through tubcs anclair is
blown in thesetubesavoidsthe humidity problembut at a very high cost. In
India only v,et towers are being used.
Electromagnetic
Radiation from Overhead Lines
Biological effects of electromagneticradiation from power lines and even
cables in close proximity of buildings have recently attractedattentionand
have also causedsomeconcern. Power frequency(50 or 60 Hz) and even their
harmonics are not considered harmful. Investigations carried out in certain
advanced countries have so far proved inconclusive. The electrical and
electronics engineers,while being aware of this controversy, must know that
many other environmentalagentsare moving around that can causefar greater
harm to human health than does electromagneticradiation.
As a piece of information it may be quoted that directly under an overhead
line of 400 kV, the electricfield strengthis 11000 V/m and magnericflux
density (dependingon current) may be as much as 40 ptT. Electric field
strengthin the rangeof 10000-15000 v/m is consideredsafe.
Visual and Audible Impacts
These environmentalproblems are causedby the following factors.
l. Right of way acquiresland underneath.Not a seriousproblernin India
at present.Could be a problem in future.
2. Lines convergingat a large substationmar the beauty of the lanclscape
around. Undergroundcablesas alternativeare too expensivea proposition except in congestecl
city areas.
3' Radio interference(RI) has to be taken into account and counteredbv
varlous means.
4. Phenomenonof corona (a sort of electric dischargearound the high
tension line) producesa hissingnoise which is aucliblewhen habitation
is in close proximity. At the to'wers great attention must be paid to
tightness of joints, avoidance of sharp edges and use of earth screen
shielding to lirnit audible noise to acceptablelevels.
5' Workers inside a power plant are subjectedto various kinds of noise
(particularly near the turbines) and vibration of floor. To reduce this
uoise to tolerable level foundations and vibration filters have to be
designed properly and simulation studiescarried out. The worker nlust
be given regularmedical examinationsand sound medical advice.
www.muslimengineer.info
T.TT POWER SYSTEMENGINEERSAND POWER
SYSTEM STUDIES
The power system engineerof the first decadeof the twenty-first century has
abreastof the recent scientific advancesand the latest techniques.On the
planning side, he or she has to make decisions on how much electricity to
generate-where, when, and by using what fuel. He has to be involved in
constructiontasksof greatmagnitudeboth in generationand transmission.He
has to solve the problemsof planning and coordinatedoperationof a vast and
complex power network, so as to achieve a high degree of economy and
reliability. In a country like India, he has to additionally face the perennial
problem of power shortagesand to evolve strategiesfor energyconservation
and load management.
For planning the operation,improvementand expansionof a power system,
a power system engineerneeds load flow studies,short circuit studies, and
stability studies.He has to know the principlesof economicload despatchand
load frequency control. All these problems are dealt with in the next few
chapters after some basic concepts in the theory of transmission lines are
discussed.The solutions to these problems and the enormouscontribution
made by digital cornputersto solve the planning and operationalproblems of
power systemsis also investigated.
I.I2
USE OF COMPUTERS AND MICR.OPROCESSOiTS
Jlhef irst rnethoslirl solvingvariouspowcr systemprobleniswere AC and DC
network analysersdevelopedin early 1930s.AC analyserswere used for load
florv and stability studieswhereasDC were preferredfor short-circuitstudies.
were developedin 1940sand were used in conjuncAnalogue compLrters
tion with AC network analyserto solve variousproblemsfor off--linestudies.
In 1950s many analogue devices were developed to control the on-line
tunctions such as genelationr--ontrol,Ii'equencyand tie-line controt.
The 1950s also saw the advent of digital computerswhich were first used
to solve a. load flow problem in 1956. Power system studiesby computers
gave greater flexibility, accuracy,speedand economy.Till 1970s,there was
a widespreaduse of computersin systemanalysis.With the entry of microprocessorsin the arena,now, besidesmain frame compLlters,
mini, micro and
personalcomputersare all increasinglybeing used to carry out various power
systern studies and solve power system problems for off-line and on-line
applications.
Off-line applications include research, routine evaluation of system
performanceand data assimilationand retrieval. It is mainly usedfor planning
and arralysing some new aspects of the system. On-line and real time
applications include data-logging and the monitoring of the system state.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
rytrfi\
-----r-----
powerSvstemAnaivsis
tutodern
A large central computer is used in central load despatch centres for
cc<ln<lmic
and securccontrolof'largc integratedsystems.Microprocessors
ancl
computersinstalledin generatingstationscontrol various local processessuch
as startingup of a generatorfrom the cold state,etc. Table 1.2 depictsthe time
microprocessors.some of these problemsare tackled in this book.
T a b l e1.2
Tirne scale
Milliseconds
2 s -5 minutes
10 min-few hours
- dofew hours-l week
I m o n t h- 6 m o n t h s
I yr- 10 years
Control Problems
Relaying and system voltage control and
excitation control
AGC (Automatic generation conrrol)
ED (Economic despatch)
Securityanalysis
UC (Unit commitment)
Mai ntcrranccschedLrl
i ng
Systernplanning
(modification/extension)
1.13 PROBLEMS FACING INDIAN POWER INDUSTR.Y
AND ITS CHOICES
The electricity requilements of .[ndia have giown tremendously anC the
demand has been running ahead of supplyl Electricity generation and
t r ans m i s s i opnro c c s s cisn In d i a a rc v c ry i neffi ci cnti n c< l l npari son
w i tl r those
of somedevelopedcountries.As per one estimate,in India generatingcapacity
is utilized on an averagefor 360t) hours out of 8760 hclursin a year, r,vhilein
Japanit is rrsedlbr 5 t00 hours.ll' the utilizationlactor could be increascd,it
should be possibleto avoid power cuts.The transmissionloss in 1997-98 on
a national basiswas 23.68Voconsistingof both technicallossesin transmission lines ancltransfonners,
and also non-technicallossescausedby energy
thefts and meters not being read correctiy.It should be possibleto achieve
considerable
savingby leducingthis lossto 1570by the end of the Tenth Five
Year Plan by r-rsingwell known ways and nreans and by adooting sound
commercial practices.Further, evcry attempt should be made to improve
system load factors by flattening the load curve by giving proper tariff
incentives and taking other administrativem.easures.As per the Central
Electricity Authority's (CEA) sixteenthannual power survey of India report,
the all India load factor up to 1998-99was of the order of 78Vo.In future it
is likely to be 7I7o.By 200i,5.07 lakh of villages(86Vo)havebeenelectrified
and 117 lakh of pumpsetshave been energized.
Assuming a very modest averageannualenergy growth of 5Vo,India's
electrical energy requirementin the year 2010 will be enormouslyhigh. A
difficult and challengingtask of planning,engineeringand constructingnew
power stationsis imrninentto rneetthis situation.The governnlenthas bLrilt
www.muslimengineer.info
'.1g.r.,'""
F
severalsuper thermal stationssuch as at Singrauli (Uttar Pradesh),Farakka
(West Bengal),Korba (Madhya Pradesh),Rarnagundam(AndhraPradesh)and
Neyveli (Tamil Nadu), Chandrapur (Maharashtra)all in coal mining areas,
2000 MW*. Manv more super thermal
plants would be built in future. Intensive work must be conductedon boiler
furnaces to burn coal with high ash content. Nationai Thennal Power
Corporation(NTPC) is in chargeof theselarge scale generationprojects.
Hydro power will continue to remain cheaper than the other types for the
next decade.As mentioned earlier, India has so far developed only around
l87o of its estimatedtotal hydro potentialof 89000 MW. The utilization of
this perennialsource of energy would involve massiveinvestmentsin dams,
system. The Central Electricity Authorchannelsand generation-transrnission
ity, the Planning Commissionand the Ministry of Power are coordinating to
work out a perspectiveplan to develop all hydroelectric sourcesby the end of
this century to be executed by the National Hydro Power Corporation
(NHPC). NTPC has also startedrecently developmentof hydro plants.
Nuclear energy assumesspecialsignificancein energy planning in India.
Becauseof limited coal reservesand its poor quality, India has no choice but
to keep going on with its nuclear energy plans. According to the Atomic
Energy Commission,India's nuclearpower generationwill increaseto 10000
MW by year 2010. Everything seemsto be set for a take off in nuclearpowel'
production using the country's thorium reservesin breederreactors.
\other
nonIn India, concerted efforts to develop solar energy and
growing
the
conventionalsourcesof energy needto be emphasized,so that
can be met and depletingfbssil fuel resourcesmay be conserved.To
clemancl
meet the energyrequirement,it is expectedthat the coal productionwill have
-2005 lts cotttpltrcdto
to be ipclcascdto q)orc than .150nrillion totts itt 200'+
180 million tonnesin 1988.
A number of 400 kV lines are operating successfullysince 1980s as
mentionedalreacly.This was the firsi stepin working towardsa nationalgrid.
There is a need in future to go in for even higher voltages(800 kV). It is
expecredrhat by the year 2Ol1-12,5400 ckt krn of 800 kV lines and 48000
ckt kni gf 400 kV lines would be in operation.Also lines may be sericsand
shunt compensatedto carry huge blocks of power with greaterstability. There
is a needfor constructingHVDC (High Voltage DC) links in the country since
DC lines can carry considerablymore power at the samevoltageand require
fewer conductors. A 400 kV Singrauli-Vindhyachal of 500 MW capacity
first HVDC back-to-backschemehas beencommissionedby NPTC (National
power Transmission Corporation) followed by first point-to-point bulk
-+
EHVDC transmissionof 1500 MW at 500 kV over a distanceof 91-5 km
on 14'Feb.2003 a
from Rihandto Delhi, Power Grid recentlycommissioned
'k NTPC has also built seven gas-basedcombined cycle power stationssuch as Anta
and Auraiya.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t
2000 MW Talcher-Kolar + 500 kV HVDC bipole transmissionsystem
thus
enabling excesspower from East to flow to South. 7000 ckt km of +
500 kV
HVDC line is expected by Z0ll-I2.
At the time of writing, the whole energy sce
is so clouded with
future. However, certain trends that will decide the future developments
of
electric power industry are clear.
Generally,unit size will go further up from 500 MW. A higher voltage (7651
1200 kV) will come eventually at the transmissionlevel. There is little
chance
for six-phasetransmissionbecomingpopular though there are few suchlines
in
USA. More of HVDC lines will do-. in operation.As populhtion has already
touched the 1000 million mark in India, we may see a trend to go toward
undergroundtransmissionin urban areas.
Public sector investment in power has increasedfrom Rs 2600 million
in
the First Plan to Rs 242330 million in the SevenrhPlan (1985 -90). Shortfall
in the Sixth Plan has been around 26Vo. There have been serious power
shortagesand generationand availability of power in turn have lagged
too
much from the industrial, agricultural and domestic requiremeni.
Huge
amounts of funds (of the order of Rs. 1893200million)
will be required if we
have to achievepower surplusposition by the time we reach the terminal year
to the XI Plan (201I-2012). Otherwise achieving a rarget of 975 billion
units
of electric power will remain an utopian dream.
Power grid is planning creation of transmissionhighways to conserve
Right-of-way. Strong national grid is being developedin phasedmanner.
In
20Ol the interregional capacitywas 5000 MW. It is Lxpecredthat by 2OlI-12,
it will be 30000 Mw. Huge investmentis planned to the tune of
us $ 20
billion in the coming decade.presenr figures for HVDC is 3136 ckt
km,
800 kV is 950 ckt km, 400 kV is 45500 ckt krn and.220/132kv is 215000
ckt km. State-of-theart technologieswhich are, being used in India
currently
are HVDC bipole, HVDC back-to-back, svc (static var compensator),
FACTs (Flexible AC Transmissions) devices etc. Improved o and
M
(Operation and Maintenance) technologieswhich are being
used tgday are
hotline maintenance,emergencyrestoration system, thermovision scanning,
etc.
Because of power shortages,many of the industries, particularly powerintensive ones,have installed their own captive power plants.* Curcently
20Vo
of electricity generatedin lndia comesfrom the captive power plants
and this
is bound to go up in the future. Consortiumof industrial .onru-.rs
should be
encouragedto put up coal-basedcaptive plants. Import should be liberalized
to support this activity.
x Captive diesel plants (and small
diesel sets for commercial and domestic uses) are
very uneconomical from a national point of view. Apart from being lower
efficiency
plants they use diesel which should be conservedfor transportationsector.
www.muslimengineer.info
real time control of power system.It may also be pointed out that this book will
also help in training and preparingthe large number of professionalstrained in
computeraided power system operationand control that would be required to
handle v
CES
REFEREN
Books
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3rd edn, 1997.
2. Eilgerd, O.1., Basic Electric Power Engineering,Reading,Mass., 1977.
3. Kashkari,C., Energy Resources,Demandand Conservationwith SpecialReference
to India, Tata McGraw-Hill, New Delhi, 1975.
4. Parikh,Kirit, .sacondIndia studies-Energy, Macmillian, New Delhi, 1976.
5. Sullivan,R.L, Power SystemPlanning,McGraw-Hill, New york, 1977.
6. S. Krotzki, B.G.A. and W.A. Vopat, Power Station Engineeringand Economy,
McGraw-Hill, New York. 1960.
7 . Car,T.H.,Electric Power Stations,vols I and lI, Chapmanand Hall, London, 1944.
8. Central Electricity GeneratingBoard, Modern Power Station Practice, 2nd edn,
' Pergamon,L976.
9t Golding, E.W., The Generationo.f'Electricitlt b1t Wind Power, Ctnpman and Hall,
London,1976.
i
10. McMillan, J.T., et. al., Energy Resoorcesand Supp\t, Wiley, London, 1976.
I L Bennet,D.J., The ElementsoJ'NuclearPoveer,Longman,1972.
12. Berkowitz,D.A:,. Power Generationand Environmentalchange, M.I.T. press,
Cambridge,Mass., 1972.
13. Steinberg,M.J. and T.H. Smith, Econonr-loadingof Power Plants and Electric
Systems,Wiley, New York, 1943.
14. Power System Planning and Operations:Future Problemsand ResearchNeeds.
EPRI EL-377-SR,February1977.
15. Twidell, J.w. and A.D. weir, RenewubleEnergy Resources,
E. and F. N, spon,
London. 1986.
16. Mahalanabis,
A.K., D.P. Kothari and S.l. Ahson, Conxputer
Aided Power,S),srenr
Analysisand Control, Tata McGraw-Hill, New Delhi, 1988.
17. RobertNoyes (Ed.), Cogenerationof Steamand Electric Power, NoyesDali Corp.,
usA, 1978.
18. weedy, B.M. and B.J. cory, ElectricPower svstems,4thedn, wiley, New york,
1998.
19. cEA 12 Annual survey of Power Report,Aug. 1985; l4th Report,March l99l;
16th ElectricPower Survey of India, Sept 2000.
20. Kothari, D.P. and D.K. sharma (Eds), Energy En.gineering.Theory and practice,
S. Chand,2000.
21. Kothari, D.P. and I.J. Nagrath, Basic Electrical Engineering,2nd edn, Tata
McGraw-Hill, New Delhi, 2002. (Ch. 15).
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
powerSystemAlqtysis
N4odern
ffiffi
,
20. Kothari, D.P. and D.K. Sharma(Eds), Energy Engineering.Theory and practice,
S. Chand, 2000.
21. Kothari, D,P. and I.J. Nagrath, Basic Electrical Engineering, 2nd edn, Tata
McGraw-Hill, New Delhi, 2002. (Ch. l5).
22. Wehenkel,L.A. Automatic Learning Techniquesin Power Systems,Norwell MA:
Kiuwer, i997.
23. Philipson,L and H. Lee Willis, (JnderstandingElectric Utilities and Deregulation,
Marcel Dekker Inc, NY. 1999.
j
^a
Papers
24. Kusko, A., 'A Predictionof Power System Development,1968-2030', IEEE
Spectrum,Apl. 1968,75.
25. Fink, L. and K. Carlsen,'OperatingunderS/ressand Strain',IEEE Spectrum,Mar.
r978.
26. Talukdar, s.N., et. al., 'Methodsfor AssessingenergyManagementoptions', IEEE
Trans.,Jan. 1981,PAS-100,no. I, 273.
27. Morgen, M.G. and S.N. Talukdar, 'Electric Power Load Management: some
Technological,Economic, Regularity and Social Issues',Proc. IEEE, Feb. L979,
vol.67,no.2,241.
28. Sachdev,M.S.,'LoadForecasting-Bibliography,
IEEE Trans.,PAS-96, 1977,697.
29. Spom, P., 'Our Environment-Options on the Way into the Future', ibid May
1977,49.
30. Kothari D.P., Energy ProblemsFacing the Third World, Seminar to the BioPhysics Workshop,8 Oct., 1986 Trieste Italy
31. Kothari, D.P,'Energy system Planning and Energy conservation',presentedat
YYIV
Nntinnnl
fnnvontinn
sn
^F
vJ
llltr
ttrL,
NI^r',
r \w w
yn v- lt ht lil ,
E-k
I wu,
lOQO
I 7QL,
32. Kothari, D.P. et. a1., 'Minimization of Air Pollution due to Thermal Plants'. ,I1E
(India), Feb. 1977, 57, 65.
33. Kothari, D.P, and J. Nanda, 'Power Supply Scenarioin India' 'Retrospectsand
Prospects', Proc. NPC Cong., on Captive Power Generation,New Delhi, Mar.
1986.
34. NationalSolarEnergyConvention,Organisedby SESI, 1-3 Dec. 1988,Hyderabad.
35. Kothari, D.P., "Mini and Micro FlydropowerSystemsin India", invited chapterin
the book, Energy Resourcesand Technology,Scientific Publishers,1992, pp 147158.
36. PowerLine, vol.5. no. 9, June2001.
37. United Nations.'Electricity Costs atxd Tariffs: A General Study; 1972.
38. Shikha,T.S. Bhatti and D.P. Kothari,"Wind as an Eco-friendlyEnergy Sourceto
meet the Electricity Needs of sAARC Region", Proc. Int. conf. (icME 2001),
BUET, Dhaka, Bangladesh.Dec. 2001, pp 11-16.
39. Bansal,R. C., D.P. Kothari & T. S. Bhatti, "On Some of the Design Aspectsof
Wind Energy ConversionSystems",Int. J. of Energy Conversionand Managment,
Vol. 43, 16, Nov. 2002,pp.2175-2187.
40. l). P. Kothari and Amit Arora, "Fuel Cells in Transporation-BeyondBatteries",
Proc. Nut. Conf. on Transportation
Systems,IIT Delhi, April 2002, pp. 173-176.
41. Saxena,Anshu, D. P. Kothari et al, "Analysisof Multimedia and Hypermediafor
ComputerSimulationand Growtft", EJEISA, UK, Vol 3, 1 Sep 2001, 14-28.
www.muslimengineer.info
2.I
INTRODUCTION
line as an
the performance of a transmission
The four parameterswhich affect
e l e m e n t o f a p o w e r S y s t e m a r e i n d u c t a n c e , c a pdue
a c ito
t a leakale
n c e , r e sover
i s t aline
nceand
, .rL..++^^nrrrrnrAncewhich is normally
lines'rhis
transmission
ii overhead
illffilf?;r -r- L*",,;i;;.";;r""i"J
resistance'
and
--'i+r- +ha caricq line narameters, 1'e' lnductance
#;;t;
ffiffir"##r'"*
oi.itiuu,.datongthelineandthevtogether
th" ,"ri., imPedance of the line'
from a power system
the most dominant line parameter
Inductar,c. i, ;;f;
e n g i n e e r , s v i e w p o i n t . A s w e s h a l l s e e i n l a t eofr cahline'
apters,itistheinductive
capacity
reactancewhich limits the transmission
i";
2,2
DEFINITION
OF INDUCTANCE
Voltage induced in a circuit is
given by
(2.r)
, =VY
", ,ti" flux linkagesof
the circuit in weber-turns(Wb-T)'
This can be written in the form
drb di
, di.,
, - : L - : v
e =
dr
dt dr
(2.2)
in
anceof the circ'lit in henrys' which
near magneticcircuit, i'e'' a circuit
current such that
Ies vary linearly with
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpo**, Syrtm An"lyri,
+f ,,il
lnductance
and Resistance
of Transmission
Lines
-
ffffi
L = ! H
I
or
(2.3)
A = LI
,---f\-]y
t
,'.rtl
t, -?5
l
i
i r '
i+----r
e.4)
where ) and I arc the rms values of flux linkages and current respectively.
These are of course in phase.
i
I
!
t'l
-ir-."
r1/
1 t
\ \.
t.rtt-
Replacing
Eq. (2.1) by ir, we get the steady state AC volrage drop
i +
due to alternating flux linkages as
\ \
-l-l-
Y= jwLI = jt^r) V
e.5)
-On similar lines,the mutual inductancebetweentwo circuits is defined asthe
flux linkages of one circuit due to current in another,i.e.,
)t,
M r rL
n=
(2.6)
,
Iz
The voltage drop in circuit 1 due to current in circuit 2 is
V, = jwMnlz = 7tl\12 V
where
(2.7)
The conceptof mutual inductanceis requiredwhile consideringthe coupling
betweenparallel lines and the influence of power lines on telephonelines.
2.3
FLUX LINKAGES OF AN ISOTATED CURRENT.
CARRYTNG coNqucroR
due to Internal
Flux
=Iy (Ampere'slaw)
www.muslimengineer.info
/y = current enclosed(A)
By symmetry, H, is.constantand is in direction of ds all alongithe circular
path. Therefore,from Eq. (2.8) we have
2rryH,=1,
Figure 2.1 shows the cross-sectionalvi,ew of a long cylindrical conductor
carrying current 1.
The mmf round a concentricclosed circular path of radius y internal to the
conductor as shown in the figure is
{nr.ds
H, = magnetic field intensity (AT/m)
Assrrmino
Transmissionlines are composedof parallel contluctorswhich, for all practical
purposes,can be consideredas infinitely long. Let us first developexpressions
for flux linkages of a long isolatedcurrent-canying cylindricat conductor with
return path lying at infinity. This systemforms a single-turncircuit, flux linking
which is in the form of circular lines concentricto the conductor.The total flux
can be divided into two parts,that which is internal to the conductor and the
flux externalto the conductor.Such a division is helpful as the internal flux
progressivelylinks a smalleramountof current as we proceedinwards towards
the centreof the conductor,while the external flux alwayslinks the total current
inside the conductor.
Flux Linkages
Fig. 2.1 Flux linkagesdue to internalflux (cross-sectional
view)
(2,8)
rrniform
(2.e)
crrrrenf dcnsifv*
,' " = ( - ) t : [ 4 ),
(2.10)
=)!H,.
t
(2.rr)
\rrr')
\r")
FromEqs.(2.9) and(2.10),we obtain
AT/m
2Tr"
The flux density By, y metres from the centre of the conductorsis
r ntI
Bu=pHu=:+
Wb/m2
z ztr-
(2.r2)
where p is the permeability of the conductor.
Consider now an infinitesimal tubular elementof thicknessdy and length one
metre. The flux in the tubular element dd = Bu dy webers links the fractional
trrrn (Iril - yzly'l resulting in flux linkages of
*For power frequency of 50 Hz, it is quite reasonableto assume uniform current
density. The effect of non-uniform current density is consideredlater in this chapter
while treating resistance.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPqwer System Analysis
of rransmission
Lines
and Resistance
lnductance
(2.r3)
The flux dd containedin the tubular elementof thickness dy is
dd = +dy
Integrating,we get the total internal flux lin
^^, = I #f d y :ffw a t^
(2.14)
For a relative permeability lf,, = | (non-magnetic conductor), 1t = 4n x
l0-'[Vm. therefore
f
^rn =T*10-/
-
wb-T/m
(2.1s)
rVm
(2.16)
and
Lint= ]xto-7
z
Flux Linkage
to Conductor
due to Flux Between
Two Points External
ffi.$
t-
Wb/m lengthof conductor
The flux dQbeing external to the conductor links all the current in the conductor
which together with the return conductor at infinity forms a single return, such
that its flux linkages are given by
d ) = 1 x d 6' = F I d ,
2ny
Therefore, the total flux linkages of the conductor due to flux between points
P, and Pr is
p D " , , f
n
' tn'
\," = |
dy - -t"- I ln "2 wb-T/m
,1Dt2 n.v
2r
Dr
where ln standsfor natural logarithm*.
Since Fr=I, F = 4t x10-7
Figure 2.2 showstwo pointsP, and Prat distancesD, and Drftoma conductor
which carries a cunent of 1 amperes.As the conductoris far removed from the
return current path, the magnetic field external to the conductor is concentric
circles around the conductor and therefore all the flux between P, and Pr lines
within the concentriccylindrical surfacespassingthrough P, and P2.
-
n
(2.r7)
) r z= 2 x l}- t l ln =L wb/ m
Dr
The inductanceof the conductor contributedbv the flux included between
points P, and Pr is then
Lrz = 2 x I0-1 fn -?
Dl
(2.18)
fV*
or
(2.1e)
= 0.461 los.D' mH,/km
L,n
LL
"Dl
Flux Linkages
due to Flux up to an External
Point
Let the external point be at distanceD from the centre of the conductor.Flux
linkages of the conductor due to external flux (from the surface of the conductor
up to the externalpoint) is obtainedfrom Eq. (2.17)by substitutin
E D t = r and
Dz = D, i'e',
D
(2.20)
r
Total flux linkages of the conductor due to internal and externalflux are
).*,= 2x70-1 lln
)=
Fig.2.2 Fluxlinkagesdue to flux betweenexternalpoints PI, P2
)in,*
)"*,
= I x 1 o - 7 + Z x r o - : I. I n 2
2
r
Magnetic field intensity at distancey from the conductor is
I
H.,=
' Z r r y AT/m
www.muslimengineer.info
*Throughout the book ln denotes natural logarithm (base e), while log denotes
logarithm to base 10.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowersystgmAnalysis
W
- -
- .2^x- 1 0 - , / t[ +
=
- 4r n 2 )
r )
x ro_71
t" *J_r,
I-et
,t - ,r-r/4 = 0.7788r
\ = 2 x ro4rh + wb-T/m
(2.Zra)
rl
Inductance of the conductor due to flux up to an external point is therefore
L= 2x 1o-7
n Ir w^
(z.zrb)
Here r' can be regarded as the radius of a fictitious conductor with no
internal inductancebut the same total inductanceas the actual conductor.
2.4
INDUCTANCE
OF A SINGLE.PHASE
TWO.WIRE
LINE
Considera simple two-wire line composedof solid round conductorscarrying
currents1, and 1, as shown in Fig. 2.3.|n a single-phaseline,
11+ Ir= Q
Lines
of Transmission
andFlesistanae
lnductance
li$Bfr
f--
To start with, let us considerthe flux linkages of the circuit caused by current
in conductor 1 only. We make three observationsin regard to these flux
linkages:
1. External flux from 11to (D - ,) links all the current It in conductor 1.
2. External flux from (D - r) to (D + rr) links a current whose magnitude
progressivelyreducesfrom Irto zero along this distance,becauseof the
effect of negative current flowing in conductor 2.
3. Flux beyond (D + 12)links a net cunent of zero.
For calculating the total inductance due to current in conductor 1, a
simplifying assumption will now be made. If D is much greater than rt and 12
(which is normally the casefor overheadlines), it can be assumedthat the flux
from (D - r) to the centre of conductor 2 links all the current ^Ir and the flux
from the centre of conductor2 to (D + rr) links zero current*.
Based on the above assumption,the flux linkages of the circuit causedby
current in conductor 1 as per Eq. (2.2Ia) are
) r = 2 x 1 0 - 7 1 ,l n - L
r\
(2.22a)
The inductance of the conductor due to current in conductor 1 only is then
(2.22b)
l" +
Lt= 2 x 10-7
f'1
Iz= - It
Similarly, the inductanceof the circuit due to current in conductor 2 is
'
D
.
(2'23)
Lz=2x10-7h
r,2
rv lu o
inc
rrrE
rr uhv a cu srr vrr n
crrrnsifinn
r
f! ahr vev ^nv ^r^ e^ t r n
fhe flrrx
linkaoes
and likewise
fhe indttcfances
of the circuit causedby currentin each conductorccnsideredseparatelymay be
addedto obtain the total circuit inductance.Therefore,for the complete circuit
L= Lt+ 4=
D-rz
D
D+rz
Fig. 2.3 Single-phase
two-wirelineand the magneticfielddueto currentin
conductor1 only
*The electric field geometry will, however, be very much affected as we shall see
later while dealing with capacitance.
www.muslimengineer.info
FVm
(2.24)
If r/r= r'z= /; then
L= 4 x 10-'ln D// Wm
(2.25a)
(2.zsb)
L - 0.92t log Dlr' mHlkm
Transmission lines are infinitely long comparedto D in practical situations
and therefore the end effects in the above derivation have been neglected.
2.5
It is important to note that the effect of earth's presenceon magneticfield
geometry* is insignificant. This is so becausethe relative permeability of earth
is about the sameas that of air and its electricalconductivitv is relativelv small.
4 x 10-'ln
CONDUCTOR
TYPES
So far we have considered transmission lines consisting of single solid
cylindrical conductors for forward and return paths. To provide the necessary
flexibility for stringing, conductorsused in practiceare always strandedexcept
*Kimbark [l9] has shownthat the resultsbasedon this assumptionare fairly
accurateeven when D is not much larger than 11and 12.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
52
|
Vodern PowerSystemAnalysis
I
lnelr
rn+ana^
^^-f
at^^l^.^--rrrLrL.vtclttutt
ct'ttu nt'stl'latlug
fo, .r.ry small cross-sectionalareas.Stranded conductors are composed of
strands of wire, electrically in parallel, with alternate layers spiralled in
opposite direction to prevent unwinding. The total number of strands (M) in
concentrically stranded cables with total annular space filled with strands of
uniform diameter(rD is given by
(2.26a)
N=3x'-3x+l
where x is the number of layers wherein the single central strandis counted as
the first layer. The overall diameter(D) of a strandedconductoris
r-^-
--^:-
-!
-
I lallsrnlsslon
,
r
.
|
-^
!
Unes
l*
5.* "
2.6 FIUX LINI{AGES OF ONE CONDUCTOR IN A GROUP
As shown in Fig. 2.5, considera group of n pnallel round conductorscarrying
phasor currents Ip 12,-, I, vvhose sum equals zero. I)istances of these
lt ult,..t
un.
us oDtam
an expression for the total flux linkages of the ith conductor of the group
consideringflux up to the point P only.
o
(2.26b)
P-(2x-r)d
Aluminium is now the most commonlyemployedconductormaterial.It has
the advdntagesof being cheaperand lighter than copper though with less
conductivity and tensile strength.Low density and low conductivity result in
larger overall conductordiameter,which offers anotherincidental advantagein
high voltage lines. Increased diameter results in reduced electrical stress at
conductor surfacefor a given voltage so that the line is coronafree. The low
tensile strengthof aluminium conductorsis made up by providing central
strandsof high tensilestrengthsteel.Sucha conductoris known as alurninium
conductor steelreinforced (ACSR) and is most commonly used in overhead
transmissionlines. Figure 2.4 shows the cross-sectional
view of an ACSR
conductor wrth 24 strands of aluminium and 7 strands of steel.
^t
ut
n
(,
3
2
4
I
Fig. 2.5 Arbitrary
groupof n parallelroundconductors
carryingcurrents
The flux linkagesof ith conductordue to its own current1,(self linkages)are
given by [see F,q. (2.21)]
)ii= 2 x 10-7t, h!
Steelstrands
Wb-T/m
(2.27)
ri
The flux linkages of conductor i due to current in conductor 7 1rlf"r to Eq.
( 2 . 1 7 ) li s
5,,= 2 x l;-il,fn a
Wb-T/m
(2.28)
Dij
Aluminium
strands
where Du is the distanceof ith conductor from 7th conductorcarrying current
1r.From F,q. (2.27) and by repeateduse of Eq. (2.28), rhe rotal flux linkages
of conductor i due to flux up to point P are
)i = Xir + )iz + ... + )ii *... *
= 2 x rca(r, t'*
viewof ACSR-7steelstrands,24 aluminium
strands
Fig.2.4 Cross-sectional
In extra high voltage (EHV) transmissionline, expandedACSR conductors
are used. Theseare provided with paper or hessian between various layers of
strandsso as to increasethe overall conductordiameter in an attemptto reduce
electrical stressat conductor surfaceand prevent corona. The most effective
way of constructingcorona-freeEHV linesis to provide severalconductorsper
phase in suitable geometrical configuration. These are known as bundled
conductors and are a common practice now for EHV lines.
\^
, D i
+ 4
' u ! - + . . . +1 ,ln
D,z
rl.
+... + In
The above equation can be reorganizedas
)i = 2xro'[[r, h +
But,
www.muslimengineer.info
Dt
)in
+h n [ * . +
I,m]+..+
+ (/, ln D r + I r h D z + . . . + I i k D , + . . + I n ^ O , )
- (1, + Iz +... + In-).
I,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
lnductanec
Substiiuti'g for /n in the secondterm of Eq. (2.29)and
simplifying, we have
u,
f/
zxfir !
* I,l n = e+ . ..* l i l n at-
, (
^, \
In order to accountfor total flux linkages of conductor i,
let the point p now
recedeto infinity. The terms such as ln D1/Dn,etc. approach
ln t = o. Also for
the sake of symmetry, denoting ,.{^ D,,,-wi have
Dt
r' r ' , ' ! - + 1-", l---n- 1
D,2
+...+I^t" wb-r/m
+)
2.7
INDUCTANCE
(2.30)
We are now ready to study the inductanceof transmission
lines composedof
compositeconductors.Future 2.6 showssuch a single-phase
line comprising
compositeconductorsA and B with A having n paraliel filaments
and B having
mt parallelfilaments.Though the inductanceof each filament
will be somewhat
different(theirresistances
will be equalif conductordiametersare chosento be
unttorm), it is sufficiently accuratefo assum.e
that the currentis equally divided
among the filaments of each compositeconductor.Thus, each
filament of A is
taken to carry a current I/n, while each filament of conductor
B carries the
return current of - Ihnt.
o
^^'- J - 1 |. ." .' *' l'rn t )
6
Diz,
D,^, )
(D,rD,r. . . D,,.. . D,n)r'"
The inductance-offilament f is then
(Dn'"' D,,'"' D,^')1/^'
FVm
7/n
( D , r D , r . . . D , ,. D ' *
The averageinductanceof the filamentsof composite
conductorA is
Lt+L2+4+"'+Ln
Luu,=
: 2nx70-t,n
o
=-
.*pr"f,rionfor rir"#i"t indu*ance
fromEq.(2.31)i"E;. 3'::r',
*ru"rorr:rr"rh"
l ( D n ,. . .D rj , . . . D r . , ) . . . ( D ^.,. .p r j ,. . . D , ^ , ) . . .
L e = 2 x 1 0 - 7l n
11?x-.::ir:l
2l
o
I
(Dnr,.. . Dnj,... Dr^,17r/ntn
[ ( D n . .D
. u . . .D r n ) . . . ( D i t . . n , t .
4.
Htm (2.33)
rn)]r,n'
,he,arsumenr
of thelogarithm
in Eq. (2.33)is rhem,nth
:f
t:?,T,:!"^:
u.]-tl-*'ofconductor
Atim'iri"L.il
"il"#il#:ili:ffi;
r )
m
intofnzp,;d;;;._,i;ff ffiilil:;
jr:lr* n2.th
l:#1:#
producttermpertains
yflr,""1i) *: setof nry
to fii";;;;rrj"",i,sts
T:l
Composite
conductorA
Fig. 2.6 single-phase
lineconsisting
of two compositeconductors
www.muslimengineer.info
(2.3r)
, _Lu,,_ \+ 12+...+L,
_A
(Dnt... Dni ... D
2
Wb_T/m
Sinceconductora is co#posedof n filaments
electricallyin parallel,its
inductance
is
OF COMPOSITE CONDUCTOR LINES
o
Dir,
= 2 x lo-716 (P,r:D,r,...D,t
L i = ] -!
Dii
---
I
h+a6J-a...*rnl*...*rn I
- 2 x t o -47l t " ] - *
. . . * I , - r 1 n' D4r -)t 1))
U
f..
k,...i<Er
Llnes
I
a
/1
lA\t^cr
Applying Eq.
(2-30)
ro filamenr i of conductorA, weobtain
its flux tintages
Dir'-""'Diz""''''^'rti
l i = z x 1 o rI t r n * *
.
^i T---^--!
ur I rallsmlssfon
,Frl a
s raqi or rruagn a a
-
)i' = 2 * to-tlI t, ml * L,h L_.+...+1,
rna
L('
anrl
"
of
rnl denominaror
isdefin"o
u,;;;w;;:;;:;,1;::;
I!:::::1li:*Ti:r{. of conduc
torA,and,s
"r';#;;;":::"*
auureviate;
;,
!:{Kr,"-li:t:t:y?
GMD
is also called geometric mean radius (GMR).
"J
In terms of the above symbols, we can write
Eq (2.33) as
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
lf
56 ,1
todern FowerSystemAnalysis
Inductanceand Resistanceof TransmissionLines
I
L t = 2 x 1 0 - 7 h+ -
Dr.t
'^
Iilm
(2.34a)
^Hn^
Note the similarity of the above relation with Eq. (2.22b), which gives the
inductanceof one conductorof a single-phaseline for the specialcaseof two
solid, round conductors.In Eq. (2.22b) r\ is the self GMD of a single
conductor and D is the mutual GMD of two single conductors.
The inductanceof the composite conductor B is determined in a similar
manner, and the total inductance of the line is
(2.3s)
L= Le+ Ln
A conductor is composed of seven identical copper strands, each having a
radius r, as shown in Fig. 2.7. Frnd the self GMD of the conductor.
srngrelayer oI alurrunlum
conductorshownin Fig. 2.8 is 5.04 cm. The diameterof eachstrandis 1.6g cm.
Determinethe 50 Hz reactanceat I rn spacing;neglectthe effectof the central
strand of steel and advance reasonsfor the same.
Solution The conductivity of steelbeing much poorer than that of aluminium
and the internal inductance of steel strandsbeing p-times that of aluminium
strands,the current conductedby the central strandsof steel can be assumedto
be zero.
Diameterof steel strand= 5.04 -2 x 1.68= 1.68 cm.
Thus, all strands are of the same diameter, say d. For the arrangement of
strandsas given in Fig. 2.8a,
Dtz= Drc= d
Drt= Dn=
Jld
Du= 2d
Dr=
(l(+)- dTil,eilf')
-- D^^= 2..fir
Fig.2.7 Cross-section
of a seven-strand
conductor
Solution The self GMD of the sevenstrandconductor is the 49th root of the
49 distances.Thus
(a) Cross-sectionof ACSR conductor
D, = (V)7 (D1zD'ruD
rp r)u (zr)u )t'on
Substituting the values of various distances,
p..- ((tJ.7788r)7
(2212x 3 x 22f x 22rx 2r x 2r)u)t'o'
(b) Line composed of two ACSR conductors
, -_ 2 r(3 (0 .7 7 8 8))tt7_ 2.t77r
us
6U+e
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Fig. 2.8
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
5t
I
lnductanceand Resistanceoi TransmissionLines
ModernPower Svstem Analvsis
I
Substitutingd' = 0.7188dand simplifying
D , = l . l 5 5 d = 1 . 1 5 5x 1 . 6 8= 1 . 9 3c m
D ^ = D s i n c eD > >d
1P L = 0 .4 6 t to e
" 1.93 0.789mH /km
Loop inductance- 2 x 0.789 = 1.578 mHlkm
Loop reactance= 1.578 x 314 x 10-3 - 0.495 ohms/lcm
;;;,;,;
I
The arrangementof conductorsof a single-phasetransmissionline is shown in
Fig.2.9, whereinthe forward circuit is composedof three solid wires 2.5 mm
in radius and the return circuit of two-wires of radius 5 mm placed
symmetrically with respectto the forward circuit. Find the inductanceof each
side of the line and that of the complete line.
The self GMD fbr side A is
D,A = ((D nD nD n)(DztDzzDn)(D3tDtrDtr))''e
Here.
D,, = Doc = Dat = 2.5 x 10-3x 0.7788m
Substitutingthe values of various interdistancesand self distancesin D16, we
get
D,A= (2.5 x 10-3x 0.778U3x 4a x 8\tte
= 0. 367 m
Similarly,
D, B= ( ( 5 x 10- 3x 0. 778q2, 4') t 'o
= 0 . 1 2 5m
Substitutingthe values of D^, D6 andDr, in Eq. (2.25b),we get the various
inductancesas
8'8 =
0.635 mHlkm
0.367
8'8 =
0.85 mH/<m
L.D = 0.461 toe
0J25
L^^ = 0.461 los
"
Solution The mutual GMD between sides A and B is
L = Lt + Ln = 1.485 mH/km
((DMD$) (Dz+Dz) 1D3aD3))tt6
D.=
If the conductors in this problem are each composed of seven identical
strandsas in Example2.1, the problemcan be solved by writing t[e conductcr
self distancesas
Dii= 2'177rt
where r, is the stranci raciius.
2.8
()z
l
4m
4m
"s (' t"l ' It
I
Side A
i
Side B
INDUCTANCE
OF THREE.PHASE LINES
So far we have considered only single-phaselines. The basic equations
developedcan,however,be easilyadapledto the calculationof the inductance
lines. Figure 2.10 shows the conductorsof a three-phaseline
of three-phase
with unsymmetricalspacing.
Dn
Dzs
Fig. 2.9 Arrangement
of conductors
for Example2.3
From the figure it is obvious that
D t q = D z q = D z s - D . u = J O am
Drs=Dy = 10m
D ^ = (6 8 2x 1 0 0 )l /6= 8.8 m
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Fig. 2.10 Cross-sectionalview of a three-phaseline with unsymmetricalspacing
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Ugdern Power SystemA
Lines
of Transmission
lnductanceand Resistance
Assume that there is no neutralwire, so that
But, 1, I I, = - /n, hence
Ir,+Ir+ Ir-0
Unsymmetrical spacingcausesthe flux linkagesand thereforethe inductanceof
eachphaseto be differentresultingin unbalancedreceiving-endvoltageseven
en senolng-e
tages and llne currents are balanced. AIso voltages will be
induced in adjacent communication lines even when line currents are balanced.
This problem is tackled by exchanging the positions of the conductors at regular
intervais aiong the line such that each conductor occupies the original position
of every other conductor over an equal distance. Such an exchange of conductor
positions is called transposition. A complete transposition cycle is shown in
Fig.2.11. This alrangement causes each conductor to have the same average
inductance over the transposition cycle. Over the length of one transposition
cycle, the total flux linkages and hence net voltage induced in a nearby
telephone line is zero.
b
1
(D"D'trDt')'''
A o =2 x 1 0 - 7I o k
r'a
equilateral
spacing
D"o (DtzDnDrr)t'' - equivalent
L o = 2 x t 0 - 7h +
f
= 2 x f O - ?f n { : r F V m
'
o
D
(2.36)
,
This is the same relation as Eq. (2.34a) where Dn, = D"o, the mutual GMD
betweenthe three-phaseconductors.lf ro = 11,= r6t we have
Lo= LO= L,
It is not the presentpracticeto transposethe power lines at regularintervals.
However,an interchangein the positionof the conductorsis madeat switching
stations to balancethe inductanceof the phases.For all practical purposesthe
line can
dissynrnrctry
can bc neglectcdand the inductanccof an untransposecl
be taken equal to that of a transposedline.
If ttre spacingis equilateral, then
D"o= D
and
L o = 2 x l 0 - 7 mI
Fig. 2.11 A completetransposition
cycle
To find the averageinductanceof each conductorof a transposedline, the
flux linkages of the conductor are found for each position it occupies in the
iransposeci
cycie. appiying Eq. (2.s0) to conciuctora of Fig. z.lI, for section
1 of the transposition
cycle whereina is in position1, b is in position2 and
c is in position3, we get
) u r= 2 x t o - r,f, , r "*
\"
''r,
-(
|
I I", , m ] - * l
Dr,
( I n- 1
Dt,)
j*o-rr,r,
For the second section
Forthetr,ira
,".* :=
\
Average flux linkages of conductor a are
Lo= Ltr= L,
Show that over the length of one transpositioncycle of a power line, the total
flux linkagesof a nearby telephoneline are z,ero,ftrr balancedthree-phase
currents.
\
|
'x to-?[r"
k + * 16tn . '' t"
wuvb:
],;i
r" h !t
)o3= z x ro-71
f',,
(2.37)
fmn
ra
If ro = 11,= r,-, it follows from F,q. (2.37) that
16
tn*+ r.,nl-l *o-rr"
D,
uzr /
c
b
(l)
i'-)
\7
c'l
()
- 1 u
r c t l I , , h ! r , , t n( D n D r D l t ) t / 3
\
4
,
www.muslimengineer.info
T2
+1.ln
(Dt2D2.)3t)t/3
F19.2.12 Effectof transpositionon Inducedvoltage of a telephoneline
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
62
Modern Power Svstem Anatrrqic.
|
solution Referring to Fig. 2.r2, the flux linkages of the
conductor r, of the
telephoneline are
rc-7 , " ^ * + 1 ,
ln
*.1
t"*
Wb-T/m (2.38)
lnductanceand Resistanceof TransmissionLines
(ii) third and multiple of third harmonic currenrsunder healthy
.ondirion,
where
1 , , ( 3 ) +I o Q ) + I , ( 3 ) = 3 1 ( 3 )
Similarly,
The harmonic line currentsare troublesomein two wavs:
) t 2 = 2 x t o t l, " L n5
( "
Doz
* r- 6o -t- n + * 1 . 1 n + l w b"_v r i' 'm
t (2.3s)
Du,
D,, )
The net flux linkages of the telephoneline are
),=
),t-
)tz
= 2 x to-r(t" h D:, * 16tn? *r" -,"+) wb-rim(2.40)
" D" )
rhe emfinduced
," ,n! ,.,"p'#ir"t*
"3Jti,
E,= Zrf),\lm
fjnder balanced load conditions, ), is not very large
because there is a
cancellationto a great extent of the flux linkages due
to Io, 16 and 1r. Such
cancellationdoesnot takeplacewith harmonr.
which are multiplesof
"u.r.nts
three and are thereforein phase.Consequently,
these frequencies,if present.
may be very troublesome.
lf the power line is f'ully transposedwith respectto the
telephone line
),, =
)" (I)* 4, (tr)* ),,(III)
3
where ),r(I), )/r(II) ancl),,(III) are the flux linkagesof the
telephoneline r, in
the three transpositionsectionsof the power line.
Writing for ),,(l), ,\/2(II)and ),r([l) by repeareduseof
Eq. (2.3g), we have
Similarly,
A,t= 2 x l0-/(tn+ 16+ 1,,)ln
(DntDutD,.itt3
(i) Inducedernf is proportionalto the frequency.
(ii) Higher frequenciescome within the audiblerange.
Thus thereis need to avoid the presenceof suchharmonic currentson power
Iine from considerations
of the performanceof nearbytelephonelines.
It has been shown above that voltage inducedin a telephoneline running
parallel to a power line is reduced to zero if the power line is transposed
and
provided it carries balanced currents. It was also shown that ptwer
line
transpositionis ineffective in reducing the inducedtelephoneline uoitug.
when
power line currents are unbalanced or when they contain third
harmonics.
Power line transpositionaparrfrom being ineffectiveintroducesmechanical
and
insulation problems.It is, therefore,easierto eliminate induced voltages
by
transposingthe telephoneline instead.In fact, the readercan easily verify
that
even when the power line currents are unbalanced or when t-h"y
cgntain
harmonics,the voltage induced over complete transpositioncycle (called
a
barrel) of a telephoneline is zero.Someinducedvoltagewill always present
be
on a telephoneline running parallel to a power line becausein actu4l practice
transpositionis never completelysymmetrical.Therefore,when the
lines run
parallel over a considerablelength, it is a good practiceto transpose
both power
and telephone lines. The two transposition cyeles ^re staggered.
and the
telephoneline is transposedover shorter lengths comparedto the power
line.
i,."rp,"i- u fI
A three-phase,
50 Hz,15 km long line hasfour No. 4/0 wires (1 cm dia) spaced
horizontally 1.5 m apart in a plane. The wires in order are carrying currents
1o,
Iu and I, and the fourth wire, which is a neutral, carries ,".o iu.."nt.
The
currents are:
=2x104(1,+.Ib+1,)ln
(Dn2Db2D,,r)t,,
) r = 2 x 1 0 - 7 1 1+, I t , + I , )' l n ( D " 2 D b 2 D , ) r t :
(2.4r)
(D orD arD ,r)r,,
If Io + Iu + I, = 0, ), = 0, i.e. voltage induced in the telephone
loop is zero
over one transpositioncycle of the power line.
It may be noted here that the condition Iu+ Iu+ I, = 0 is not
satisfiedfor
(i) rpwer frequency L-G (line-to-ground fault) currents.
where
Io= -30 + 750 A
Iu= -25 + j55 A
I"= 55 - j105 A
The line is untransposed.
(a) From the fundamentalconsideration,find the flux linkages of
the neutral.
Also find the voltage induced in the neutral wire.
(b) Find the voltage drop in each of the three-phasewires.
Io+ Iu* Ir=J[o
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
64
I
Modern Power System Analysis
I
lnductanceand Resistanceof TransmissionLines
t
a
b
(:;
(. 1
c
,r"n.'f(')
--.
t-r.s n.' -t--1.5 m-l*-
n
I n2
(:)
ln Dlr l
-l
r.sn1
ln2
voltage
Fig. 2.13 Arrangement
of conductors
for Example2.5
= - (348.6
+ j204) V
D o n = 4 .5 m, D b n= 3 m, D rn = 1.5 m
Flux linkages of the neutral wire n are
- (
'I + 1 ^ l n
) - = 2 x 1 0 - 7 11 - l n
"
Don
\"
r \
1' * l l n '
lwb-T/m
Drn
D,n)
A single-phase50 Hz power line is supportedon a horizontal cross-ann.The
spacing between the conductors is 3 m. A telephone line is supported
symmetricallybelow the power line as shown in Fig. 2.14. Find the mutual
inductancebetweenthe two circuits and the voltageinducedper kilometre in the
telephoneline if the current in the power line is 100 A. Assume the telephone
Iine current to be zero.
Solution Flux linkaees of conductor Z,
Substitutingthe valuesof D,,n,Dg, and D,.n,and simplifying, we get
An= - 2 x lO-' 0.51 I, + 1.1 Iu + 0.405 1") Wb-Tim
Since I, = - (Io+ I) (this is easily checkedfrom the given values),
2,, = - 2 x l0-' (1.1051o+ 0.695I) Wb-T/m
The voltage induced in the neutral wire is then
Vn = j w ),n x 1 5 x 103 V
or
s e 4 r s c a ulated below:
Avo = j2xra-lx3r4x rs x rd(6 !9e (-30+j50)+0.6e3(-25+is5))
Solution (a) From Frg.2.I3,
=-
p o
j 3 1 4 x 1 5 x 103x Z x 10-7(1J05Io+ 0.695I)
Vn = - j 0 .9 4 2 (1 .105 Io + 0.695I)
y
V
) , r = 2 x I o - 7( , r n t - l 6 t )
Dt
\
Dr)
= 2 x ro-1I h D'
Flux linkages of conductor 7,
Substituting the valuesof 1, and 16,and simplifying
\,2= 2 x 10-71lnD'
D^
V n = 0 .9 4 2 x 1 0 6 = 100 V
(b) From Eq. (2.30), the flux linkages of the conductor a are
(
r
I
)o=2x ro-?[
r,rnl*
r
\
ru
rn**r rn* | wu-r"
D
2D)
r'o
\
The voltagedrop/metre
in phasea canbe writtenas
I
av,,=2xr0-7iw(,-rn
\."
f'o
+ 1' ^' l n l * 1 - l cn I )
D
tr,o
2D)
Since Ir= - (lr,+ Iu), and furthersince ro= rb= rr= r, the expressionfor
AVo can be written in simplified form
Avo= 2 x to-ty,(r, rn!*
romz) v/m
Similarly, voltage drop/petre of phases b and c can be written as
AVo=2 x r;a iulbo_
#
A V , = 2x t o t / , ( 1 ,f n2 + 1 , " + )
Using matrix notation, we can present the result in compact form
www.muslimengineer.info
\)d/
-
\-l-l
I
, '-Yt'
" l*-O.et
Fig.2.14 Powerand telephonelinesfor Example2.6
Total flux linkage of the telephonecircuit
),=\,r-\rz=4x10-'lln
D2
Dr
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
D1
66
Lines
of Transmission
and Resistance
lnductanee
Modernpower SystemAnalysis
iI
A4pt=4 x tl-i n !
Dl
will lead
eachother. (The readercan try other configurationsto verity that these
is
spacing
equilateral
equivalent
to low D..) Applying the methodof GMD, the
Ul^
D rn
(2.42)
(DnbDbrD.n)''t
a andb in sectionI of the
transpositioncYcle
= (DpDp)rr+- 7DP)tt2
D r = Q . 1 2+ 2 \ t / 2 = ( 5 . 2 I ) t / 2
D 2 = (L 9 2 + Z \r/2 = (7.61)t/2
(Jsf)t" =
v or t=
' o .g 2 r b s\ 5 2 1)
Dt,=mutual GMD betweenphasesb and c in section 1 of the
transpositioncYcle
0.0758mH /km
= (DP)r;z
D,o=mutual GMD betweenphasesc and a in section I of the
r- transPositioncYcle
Voltage inducedin the telephonecircuit V,= jttMr,I
lV,l = 314 x 0.0758x l0-3 x 100 _ 2.3,/9 Vlkm
= (2Dh)v2
2.9
DOUBI"E.CIRCUIT THREE.PHASE
LINES
Hence
It is commonpracticeto build double-circuitthree-phaselines so as to increase
transmissionreliability at somewhatenhancedcost. From the point of view of
rrowertransferfrom one end of the line to the other (see Sec. 12.3), it is
desirableto build the two lines with as low an inductance/phase
as possible.In
order to achievethis, self GMD (D") should be made high and mutual GMD
(D') should be made low. Therefore,the individual conductorsof a phase
should be kept as far apart as possible (for high self GMD), while the distance
between phasesbe kept as low as permissible (for low mutual GMD).
Figure2.15 showsthe threesectionsof the transpositioncycleof two parallel
circuit three-phaselines with vertical spacing (it is a very commonly used
configuration).
c
b
'
b
D"o
(2'43)
2rt6Drt2pr/3htt6
of the
It uray be noteclhere that D"u t'eurainsthe sallle in each section
cyclically,
rotate
circuit
parallel
transpositioncycle, as the conductorsof each
2 and
so do D,,h, Dbrand D,.,,.The reader is advisedto verify this for sections
2'15'
Fig'
cycle in
3 of the transposition
rs
Self GMD in section1 of phase a (i.e., conductorsa and a/)
Drr,= (r'qy'q)t'o= (r'q)'/z
SerrGMD *;:;='
"t
respectiverv
'i
,*;;';:^':'ffi'1'are
Dr r = ( y'qr 'q) ''' = ( r 'q) ''t
Equivalent self GMD D, = (Dr,,DrbD,,)rt3
(2.44)
= (r')t''qtt3hrt6
Becauseof the cyclic rotation of conductorsof eachparallel circuit over the
section.The
transpositioncycle,D. alsoremainsthe samein eachtransposition
2.15.
Fig.
in
3
and
2
sections
for
this
verify
reader should
The inductancePer Phaseis
o
Section 1
Section2
\0,
o
Section 3
Fig. 2.15 Arrangementof conductorsof a double-circuitthree-phaseline
It may be noted here that conductors a and a' in parallel compose phase a
and sirnilarly b and b'compose phase b and,c and c'compose phase c. in order
to achieve high D" the conductors of two phases are placed diametrically
opposite to each other and those of the third phase are horizontally opposite to
www.muslimengineer.info
L = 2 x 1 0 - 7l o
D',n
Ds
'|l'ltl"tlt)u
= z x ,ot ,n't'u
(r')t'' qr/3hrt6
= 2 x 1 0 - 7 h ( 2,,u(q')"'[a)"'
['
\r')
\q)
] rvpnur.rrn(2.4s)
The self inductanceof each circuit is givenby
Lr = 2 x 10- 7ln
(2)'''p
r
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Lines
and Resisianceof Transmission
lnductance
Equation (2.45) can
n no\,v be written as
L=!
[, ",u
=
,(t,
' ^ c[.2 *2xro-7rn
(t)'^l ,r.ou,
8-10 times the conductor'sdiameter,irrespectiveof the numberof conductors
in the bundle.
+ M)
where M is the mutual inductancebetweenthe
di
two circuits. i.e.
r-\
\./
-d
M = z x 1 o -^7( z \ ' ' '
This is a well known result for the two coupled
curcuits connected in parallel
(at similar polarity ends).
/ \
rf h >>o,[ L l-l unaM '--+0,
i.e.themutuarimpedance
between
thecucuits
\ q )
becomes zero. Under this condition
IJ2D
x 1 0 "- ' l n " ' ;
ElrTl.trrr
r:rn
svlrr-rrr.q.Lr
---1-)
\_..
d
,() I ou',
l-_-
- s = 0 .4 m ! .
-] s = 0.4m""
be) I On'
d =7 m---*]*-
'/^' :' 6 /
"\ '
d =7 m---
>l
:
line
three-phase
Fig.2.17 Bundledconductor
(2.47)
AAr?
tetJM,rUUl-L,t(S
It is economical10 transmit large chunksof power
over long dista'ces by
employing EHV lines. However, the line voltagesthat
can be usedare severely
limited by the phenomenonof corona.corona,ln fact,
is the result of ionization
of the atmospherewhen a certain field intensity (about
3,000 kv/m at NTp) is
reached'Corona dischargecausescommunication
interference and associated
power loss which can be severein bad weather
conditions. Critical line voltage
for formation of corona can be raised considerably
by the use of bundled
conductors-a group of two or more conductorup".
phase. This increasein
criticalcoronavoltageis clepenclcnt
on numberof concluct<lrs
in thc group,thc
clearance between them and the distance between
the groups forming the
separatephases*.Reichman [11] has shownthat the
spacingof conductorsin
a bundle affects vortagegradient and the optimum
spacing is of the order of
-
(}
f"=o.aml,
I
The GMD method, though applied above to
a particular configuration of a
double circuit, is valid for any configuration
as long as the circuits are
electricallyparallel.
While the GMD method is valid.for fully transposed
lines, it is commonly
applied for untransposedlines and is quite u.rrrui"
for practical purposes.
s2. r v1 n
I
/-\
F i g . 2 . 1 6 Configurationof bundled conductors
\ q )
L=I
dl
tt,d
The bundle usually comprises two, three or four
conductors arrangedin configurationsillustratedin Fig' 2.16. The current will not
divide equally amongthe conductors
of the bundle unlessconductors within the bundle
are ruly transposed.The GMD
method is still fairly accurate for all practical purposes.
Further, because of increased self GMD- line inductance is reduced
considerablywith the incidental advantageof increasedtransmissioncapacity
of the line.
Find the inductive reactancein ohms per kilometer at 50 Hz of a three-phase
bundled conductorline with two conductorsper phaseas shown in Fig. z.ri.
All the conductorsare ACSR with radii of 1.725 cm'
Even though the power lines are not normallytransposed(exceptwhen they
enter and leave a switching station), it is sufficiently accurate to assume
complete transposition(of the bundles as well as of the conductorswithin the
bundle) so that the method of GMD can be applied'
The mutual GMD between bundles of phasesa and b
Dob= @ @ + s) (d - i d)rt4
bundles of phasesb and c
between
GMD
Mutual
Dh, = D,,,, (bY sYmmetrY)
Mutual GMD betweenbundles of phasesc uruJa
D,n = Qd (2d + s ) (2d - t)Zd)tt+
D, o= ( Dr PaPr o) t 't
= @f@ + s)z(d- s)2(2a+ i(Zd - t))tt"
= GQ)6Q .q2 6.0)2(14.4)(13.6))','',2
= 8 . 8 1m
The more the number of conductors in a bundle, the more is the self GMD'
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
70
I
n llarlarn
Dnre,o'
erra+^-
^ -^r-,^.
D .,= (r' s r' r)' ' o =(rk)r/z = (0.77ggx 1.725x l 0-2
x 0.4\t/z
= 0 .0 7 3m
Inductivereactanceper phase
8.8r
0.073
= 0 .3 0 1 o h m/k m
In most cases'it is sufficiently accurateto usethe
centreto centredista'ces
betweenbundlesratherthan mutual GMD between
bundles for compu ting D"n.
with this approximation,we have for the example
in hand
lnductanceand Resistanceof Transmission
Lines
A = cross-sectionalarea, ln2
The effective resistancegiven by Eq. (2.48) is equal ro the DC resistanceof
the
conductor given by Eq. (2.49) only if the current distribution is uniform
throughoutthe conductor.
r small changesin temperature,the resistanceincreaseswith temperature
in accordancewith the relationship
R, = R ( 1 + oor )
where
ao = temperaturecoefficient of the conductorat 0"C
Drn=exTxl4yrrt-g.g2m
Equation (2.50) can be used to find the resistance Ro at a temperature /2,
if
resistanceRr1 at temperature tl is known
Xr= 3 I4 x 0 .4 6 1 x 10-3l os 8' 82
" 0.073
= 0.301 ohmlkm
Thus the approximatemethod yielclsalmostthe
samereactancevalue as the
exact method'It is instructive to comparethe inductive
reactanceof a bundled
conductorline with an equivalent(on heuristicbasis)
singleconductorline. For
the example in hand, the equivalent line will
have d = 7 m and conductor
diameter(for sametotal cross-sectional
area)as JT x 1.i25 cm
Xr= 3 1 4 x 0 .4 6 1 x l 0-3 6n
= 0 .5 3 1o h m/k m
(7 x7 xl 4)t/3
0.7799x J2 xl.725x 10-3
This is 7-6'4rvohigher than the colrespondingvalue
lltlA
iiiiv'
Ac
-\i
olran.l"
diiuduJ
-^l-r^l
liuirit€c
--.r
out'
r
iower
increasesits transmissioncapacity.
2.TT
for a bundledconductor
reactance of a bundled
conductor
line
RESISTANCE
Though the contribution of line resistanceto
series line impedance can be
neglectedin most cases'it is the main sourceof
line power loss. Thus while
consideringtransmissionline economy,the presence
of line resistancemust be
considered.
The effectiveAC resistanceis given by
O_
averagepower lossin conductorin watts
ohms (2.48)
where 1is the rms current in the conductoi in amperes.
Ohmic or DC resistanceis given by the formula
nl
R'n - ' ' "A o h l n s
where
p = resistivity of the conductor,ohm_m
/ = length,m
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(2.s0)
R = resistanceat temperature0"C
(2.49)
( 2 . s1 )
2.T2
SKIN EFFECT AND PROXIMITY
EFFECT
The distribution of current throughout the cross-sectionof a conductor
is
uniform only when DC is passing through it. on the contrary when AC
is
flowing through a conductor,the currentis non-uniformlydistributedover
the
cross-sectionin a mannerthat the current density is higher at the surface of
the
conductor compared to the current density at its centre.This effect becornes
Inore pronouncedas frequencyis increased.This phenonlenon
is cilled stirr
qffect.It causeslarger power loss for a given rms AC than the loss when
the
Sairr€vaiueof DC is flowing ihroughthe conciuctor.Consequently,
the effective
conductorresistance
is more fbr AC then fbr DC. A qualitativeexplanationof
the phenomenonis as follows.
Imagine a solid rottnd conductor (a round shape is considered
for
convenienceonly) to be composedof annularfilamentsof equalcross-sectional
area. The flux linking the filaments progressively decreasesas we move
towards the outer filaments fbr the simple reasonthat the flux inside a filament
does not link it. The inductive reactanceof the inraginaryfilaments therefore
decreasesoutwards with the result that the outer filaments conduct more
AC
than the inner filaments (filaments being parallel). With the increase
of
frequency the non-uniformity of inductive reactanceof the filaments becomes
more pronounced,so also the non-uniformity of current distribution. For large
solid conductors the skin effect is quite significant even at 50 Hz. The
analytical study of skin effect requires the use of Bessel's functions and
is
beyond the scope of this book.
Apart fronl the skin effect, non-uniformity of current distribution is also
causedby proximity eJJ'ect.
consider a two-wire line as shown in Fig. 2.1g.
Each line conductorcan be divided into sectionsof equalcross-sectionat
area
(say three sections).Pairs aat, bbt and,cct can form threeloops in parallel.
The
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
72
I
|
ModernPowerSystemAnaiysis
Inductanceand Resistance
of Transmission
Lines
t
flux linking loop aat (and thereforeits inductance)is the least and it increases
somewhatfor loops bbt and ccl. Thus the density of AC flowing through the
conductorsis highestat the inneredges(au') of the conductorsand is the least
at the outer edges (cc').This type of non-uniform AC current distribution
b
Decomes more pronounceo as me olstance Detween conouctors ls reouceo. LlKe
skin effect, the non-uniformity of current distribution caused by proximity effect
also increasesthe effective conductor resistance.For normal spacing of
overhead lines, this effect is always of a negligible order. However, for
undergroundcableswhereconductorsare locatedcloseto eachother,proximity
etfect causesan appreciableincreasein effective conductorresistance.
Fig. 2.18
Both skin and proximity effects depend upon conductor size, fiequency,
distance between conductors and permeability of conductor material.
LEIVIS
PROB
2 . 1 Derive the formula for the internal inductancein H/m of a hollow
r;onductorhavinginsideradiusr, and outsideradiusr, andalsodetermine
line consisting
the expressionfor the inductancein H/rn of a single-phase
of the hollow conductors described above with conductors spaced a
distance D apart.
2 . ? Calculate the 50 Hz inductive reactanceat I m spacing in ohms/km of a
cable consistingof 12 equal strandsaround a nonconductingcore. The
diameterof each strandis 0.25 cm and the outsidediameterof the cable
i s 1 .2 5c m .
2 . 3 A concentriccable consistsof two thin-walled tubes of mean radii r and
It respectively.Derive an expressionfor the inductanceof the cable per
unit length.
50 Hz circuit comprisestwo single-corelead-sheathed
2 . 4 A single-phase
cableslaid side by side; if the centresof the cablesare 0.5 m apart and
each sheathhas a mean diameterof 7.5 cm, estimatethe longitudinal
voltage induced per km of sheathwhen the circuit carries a current of
800 A.
2.5 Two long parallelconductorscarrycurrentsof + 1 and- 1. What is the
at a point P, shownin Fig. P-2.5?
magnetic
fieldintensity
www.muslimengineer.info
Fig. p-2.5
2.6 Two three-phaselinesconnectedin parallelhaveselt'-reactances
of X, and
X2. If the mutual reactancebetween them is Xp, what is the effective
reactancebetween the two ends of the line?
2.7 A single-phase50 Hz power line is supportedon a horizonralcross-arm.
The spacing between conductorsis 2.5 m. A telephoneline is also
supportedon a horizontalcross-armin the samehorizontalplane as the
power line. The condttctorsof the telephrlncline are of solid copper
spaced 0.6 m between centres. The distance between the nearest
conductorsof the two lines is 20 m. Find the mutual inductancebetween
the circuitsand the voltageper kilometreinducedin the teiephoneline for
150 A current flowing over the power line.
2.8 A telephoneline runs parallelto an untrasposedthree-phase
transmission
line, as shown in Fig. P-2.8.The power line carriesbalancedcurrent of
400 A per phase. Find the mutual inductancebetweenthe circuits and
calculatethe 50 Hz voltageinducedin the telephoneline ptsrkm.
a
( r
f^
b
c
( r
( ,
l
L
5m---+++--5m---f -
h
b
, .
15m
---*1rn.__
Fig. P-2.8 Telephoneline parallelto a power line
2.9 A 500 kV line has a bundling arrangement of two conductors per phase
as shown in Fic. P-2.9.
Fig. P-2.9 500 kV,three-phase
bundledconductor
line
Computethe reactanceper phaseof this line at 50 Hz Each conductor
carries50Voof the phasecurrent.Assume full transposition.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
lgq er1-tgrygf.!yg!U_An4ygp
Inductanceand Resistanceof TransmissionLines
2.10 An overheadline 50 kms in length is to be constructedof conductors2.56
cm in diameter,for single-phasetransmission.The line reactancemust not
exceed31.4 ohms.Find the maximum permissiblespacing.
2.11In Fig. P-2.1I which depictstwo three-phase
circuitson a steeltower there
NCES
REFERE
cal cenre tlnes.
three-phasecircuit be transposedby replacing a by b and then by c, so
that the reactancesof the three-phasesare equal and the GMD method of
reactancecalculationscan be used. Each circuit remains on its own side
of the tower. Let the self GMD of a single conductorbe 1 cm. Conductors
a and at and other corresponding phase conductors are connected in
parallel. Find the reactanceper phaseof the system.
I oi-----z.sm ---j C)
Io
c
)
1om-
I
r.___zsm___l
4 m
I
3. Nagrath, I.J. and D.P. Kothari, Electric Machines, 2nd edn, Tata McGraw-Hill,
New Delhi, 1997.
4. Stevenson,W.D., Elements of Power System Analvsis,4th edn, Mccraw-Hill, New
8. Gross, C.A., Power System Analysis, Wiley, New York, 1979.
9. Weedy, B.M. and B.J. Cory Electric Power Systems,4th edn, Wiley, New york,
1998.
-l
')
2. Waddicor, H., Principles of Electric Power Transmission, 5th edn, Chapman and
Hall, London, 1964.
1. Woodruff. L.F., Principles of Electric Pov,er Trun.snissiorr,John Wiley & Sons,
New York, 1947.
'
l- l-
10. Kimbark, E.W., Electrical Transmission of Power and Signals, John Wiley, New
York, 1949.
.)
Paper
I l.
Fig. P-2.11
Reichman.J.,
'Bundled
Conductor Voltage Gradient Calculations," AIEE Trans.
1959, Pr III. 78: 598.
)-.12 A double-circuit three-phaseline is shown in Fig. P-2.I2. The conductors
a, a/l b, bt and c, c/ belong to the same phase respectively. The radius of
each conductor is 1.5 cm. Find the inductance of the double-circuit line in
mH/km/phase.
a
(_)
b
l
I
i'1m
b/
a/
()
- i - 1 m -lI
t
l
1m
a)
\..,_/
l
>l<
I
1m -l- ,r
[r
\. _./
-l
Fig. P-2.12 Arrangement
of conductors
for a double-circuit
three-phase
line
2.13 A three-phaseline with equilateralspacingof 3 m is to be rebuilt with
horizontal spacing(.Dn = ZDn - ZDrr).The conducrorsare to be fully
transposed.Find the spacingbetween adjacentconductorssuch that the
new line has the sameinductanceas the original line.
2.14 Find the self GMD of three arrangernentsof bundledconductorsshown in
Fig. 2.16 in termsof the total cross-sectional
(same
areaA of concluctors
in each case)and the distance d between them.
www.muslimengineer.info
and
York, 1971.
4m
b
Electric'ttl Transrnission and Distributiort Book, Westinghouse Electric
Manufacturing Co., East Pittsburgh, Pennsylvania, 1964.
York, 1982.
5. Edison Electric Institute, EHV Transmission Line Reference Book, 1968.
6. Thc Aluminium Association, Aluminium Electrical Conductor Handboo,t. New
c'
b
l.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Capacitanceof TransmissionLines
t
V , " = 6 e a v: 6 - - - q - d v V
tL
r
Fig. 3.1
3. 1
IN T R O D U C T IO N
The capacitancetogetherwith conductanceforms the shunt admittanceof a
transmissionline. As mentionedearlierthe conductance
is the result of leakage
over the surface of insulatorsand is of negligible order. When an alternating
voltageis appliedto the line, the line capacitance
drawsa leading sinusoidal
current called the charging current which is drawn even when the line is open
circuited at the far end. The line capacitancebeing proportionalto its length,the
chargingcurrentis negligiblefor lines lessthan 100km long. For longerlines
the capacitancebecomesincreasinglyimportantand has to be accountedfor.
3.2
ELECTRIC FIELD OF A LONG STRAIGHT CONDUCTOR
Imagine an infinitely long straightconductorfar removedfrom other conductors
(including earth) carrying a unifbrrn charge of 4 coulomb/metrelength. By
symmetry,the equipotentialsurfaceswill be concentriccylinders,while the lines
of electrostaticstresswill be radial. The electric field intensitv at a distancev
from the axis of the conductoris
,=
r
|
.',
L
),,,
/t'\V
Electricfield of a lclngstraightconductor
path of
As the potential difference is independent of the path, we choose the
an
along
path
PP2lies
the
integration as PrPP2 shown in thick line. Since
t.e.
PyP,
along
integrating
by
equipo',entral,Vrris obtained simply
vrz=
l:,'trav:ht"?u
3.3
POTENTIAL DIFFERENCE BETWEEN TWO CONDUCTORS
OF A GROUP OF PARALLEL CONDUCTORS
the
Figure 3.2 showsa group of parallelchargedconductors.It is assumedthat
from
removed
are
sufficiently
and
ground
conductorsare far removed from the
eachother-,i.e. the conductorradii are much smallerthan the distancesbetween
them. The spacingcommonly usedin overheadpower transmissionlines always
meets theseassumptions.Further, these assumptionsimply that the charge on
length.
eachconductorremainsuniformly distributedaroundits peripheryand
nn
Q v/^
2nky
where ft is the permittivity* of the medium.
As shown in Fig. 3.1 considertwo pclintsP, and P, locatedat clistances
D,
and Dr respectively from the conductor axis. The potential difference Vn
(betweenP, and Pr) is given by
* In SI units the pennittivity of free
space is
permittivity for air is ft, = klko = 1.
ko = 8.85 x
www.muslimengineer.info
10-12 F/m. Relative
(3.1)
i,o - --
Fig. 3.2 A group of parallelcharged conductors
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
7S
'l
Modernpower SystemAnalysis
Capacitanceof TransmissionLines
The potentialdifferencebetweenany two conductorsof the group can then be
obtainedby adding the contributionsof the individual chargedconductors:by
repeated application of Eq. (3.1). so, the potential difference between
conductorsa and b (voltagedrop from a to b) is
*lcln tn+. qotn*;. n,h ++.. qn" +)
v
CAPACITANCE
0.0121
Cot
(3.4c)
The associatedline charging current is
I,= ju.Co6Vnu Allvn
(3.5)
[__{b
"O___l
(a)
Line-to-linecapacitance
l---r )o
OF A TWO.WIRE LINE
i
c",
Considera two-wire line shownin Fig. 3.3 excitedfrom a single-phasesource.
The line developsequaland oppositesinusoidalchargeson the two conductors
which can be representedas phaso$ QoNd qb so that
eo = _ eu.
--l*l
Fig. 3.3 Cross-sectionalview of a two_wireline
The potential difference Vo6 can be written in terms of the contributions
made by qo and q6 by use of Eq. (3.2) with associated assumptions (i.e. Dlr is
large and ground is lar away). Thus,
V
, =
'
0t)
Since
, l ; t ( n " h*D" ^ + )
(3.3)
Qr,= - qu, we have
-
In (D / (ro r)ttz1
-
Line-to-neutralcapacitance
Fig. 3.4
C,= Co,= Cb,= 2Cou=
,!#A
pflkm
(3.6)
The assumptions
inherentin the abovederivationare:
(i) The charge on the surfaceof each conductoris assumedto be uniformly
distributed, but this is strictly not correct.
If non-uniforrnityof chargedistributionis takeninto account,then
0.0242
\4r'
rorb
Qo
Vot
-
pFkm
(3.7)
r -c[ 2nLr(+ ( 4 - , ) " ' )
The line capacitance Cnh is then
"ab
.
C.n=Cbr=2C"a
(b)
C,
v o b =! + m L
27rk
.
Cnn
As shown in Figs. 3.a @) and (b) the line-to-line capacitancecan be
equivalentlyconsideredas two equal capacitances
is senes.The voltageacross
the lines dividesequallybetweenthe capacitances
suchthat the neutralpoint n
is at the groundpotential.The capacitanceof eachline to neutralis then given
by
--J l
-
| 79
r( 3. 4b)
trtF/km
log (D / (ror)ttz)
(3.2)
Each term in Eq. (3.2) is the potentialdrop from a to b causedby charge on
one of the conductors of the group. Expressionson similar lines could be
written for voltage drop between any two conductorsof the group.
If the chargesvary sinusoidally,so do the voltages(this is the casefor AC
transmissionline), the expressionof Eq. (3.2) still applies with charges/metre
length and voltagesregardedas phasorquantities.Equation(3.2) is thus valid
for instantaneousquantities and for sinusoidal quantities as well, wherein all
chargesand voltagesare phasors.
3.4
or
l' .-^
F/m length of line
www.muslimengineer.info
Q.aa)
)
)
lf D/2r >>1, the above expressionreducesto that of Eq. (3.6)and the error
causedby the assumptionof uniform chargedistribution is negligible.
(ii) The cross-sectionof both the conductorsis assumedto be circular, while
in actual practice strandedconductors are used. The use of the radius of the
circumscribingcircle for a strandedconductorcausesinsignificanterror.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
a
q0
3.5
s
1
^
-
4
Modern pb*gl Jysteq 4!gly..!s
|
es
CAPACITANCE OF A THREE-PHASE LINE WITH
EOUILATERAL SPACING
LgT '
For air medium (k, = l),
,,=#ffi p,Flkm
(3.14b)
Figure3.5 showsa three-phase
line composedof threeiclenticalconcluctors
of
1o (line charging) = ju,CnVnn
(3.1s)
\D
Vac,/
Vab+ Vac= 2 rt cos 30" V"n
=3V"n
A
\_-/b
a/
Fig. 3.5
,''h
cross-sectionof a three-phaseline with equilateralspacing
1",
t "
, " I
Using Eq. (3.2) we can write the expressions for Vu,,and V,,..as
6 2 * t1,,
tn , r, ,,
t
+)
vub=
*(0"
vn,=
i ! * ( n " 6 P -t q 1 , t n # * r ,^ r j )
I
I
(3.8)
Vca 1
\V'o
I
I
30
Itvun
(3.e)
IF
n".
Adding Eqs. (3.8) and (3.9),we get
v,,t,
* v,,,=
,'oolrr,,
,, D + (qu'l ,1,)t,,
; ]
v",
(3.r0)
zTTk
(3.11)
r
!b
Vr
Since there ilre no othcr chargesin the vicinity, thc sunr of'clrar_{eson
the three
conductors is zero. Thus q6 * Qr=Qu, which when substituted in Eq. (3.10)
vields
V,,h
I Vn,= !+ r" 2
vb
6 D
Fig. 3.6 phasordiagramof baranced
three-phase
vortages
3.6
CAPACITANCE OF A THREE.PHASE LINE WITH
UNSYMMETRICAL
SPACING
With balanced three-phase voltages applied to the line, it follows
from the
phasor diagrarn ol'I,rig. 3.(r that
Vou I
Vor= 3Von
(3.r2)
Substituting for (Vot,+ V,,,) trom Eq. (3.12) i n E q . ( 3 . 1 1 ) ,w e g e r
v^_- 4o tn2
27Tk
r
(3.13)
The capacitance of line to neutral immediately follows as
/-
"
Qo
Von
2 Tlc
ln (D/r)
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For the first,sectionof the transpositioncycle
v o b= + ( e o t r n ! * q u r -t'n f
(3.l 4a)
zTvc\
r
Dr z
+e,rrn})
Dr , )
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(\ 3 . r 6 a )
.
l
ModernPower SystemAnalysis
E2' I
t
Capacitanceof Transmission
Lines
I
m
t
(3.18)
*lr"ho-3L+aatnL
D rn = (D nDnDT)tl3
r'o,=
/ --l
Fig. 3.7
Cross-sectionof a three-phaseline with asymmetricalspacing
(fully transposed)
( ^ r ^ D " n'
tn;*n,
lfiln,
')
vot # vo, =
- t e.zt"+l
f
(3.1e)
Adding Eqs. (3.18) and (3.19),we get
For the second section of the transposition cycle
zTrK\
j
, t
r"
,* )
uzt
(3.16b)
utz)
,,6
*(
r,/r^\
2r t r ( qt - r q") r n t
f
%)
(3.20)
As per Eq. (3.12) for balancedthree-phasevoltages
For the third sectionof the transpositioncycle
V n ti
vob=|-.(r"rln -&r-* q6rtn-!-i e,tt"+l
Dr,)
zi'!\
Dy
(3.16c)
and also
Vor='3Vnn
(qu + qr) = - qo
Use of these relationshipsin Eq. (3.20) leadsro
If the voltage drop along the line is neglected, Vno is the same in each
transposition cycle. On similar lines three such equations can be written for
Vbr= Vut, I -120. Thrce more equations can be written equating to zero the
summation of all line charges in each section of the transposition cycle. From
these nine (independent) equations, it is possible to determine the nine unknoWn
charges. The rigorous solution though possible is too involved.
D"n
- Qo
l/ o^..
(3.2r)
n - Z n k ln
-" r
The capacitanceof line to neutral of the transposedline is then giveh by
With the usual spacing of conductors sufficient accuracy is obtained by
assuming
Forair medium'o;=
C." = 3s-: -2nk
F/m to neutral
Von ln (D"o/ r)
-'uut\-n =
-"
log (D,o /r)
Qat= Qa2= Qa3= Q"i Qut= Qaz= Quz= Qo,
(3.t7)
4ct= Q,'2= 4,3= 4r'
This assumptionof equaicharge/unitlength of a line in the three sectionsof the
transposition cycle requires,on the other hand, three diff'erent values of Vnu
designatedas Vo61,Vo62andVoo,in the three sections.The solution can be
considerablysimplified by taking Vouas the averageof thesethreevoltages,i.e.
J
or
'/ nb=
.1*o"
I l- 'n ( DtzD?tDy
rnI
'
6 q1aln'
[
,,
)
( DrrDr^D^,\1
+q^lnl-:ll
\ DnDnD3t ))
www.muslimengineer.info
"'
r'
)
\ Dt2D23D3t
)
(3.22b)
It is obvious that for equilateralspacing D,, = D, the above (approximate)
formula gives the exact resdlt presentedearlier.
The line charging current for a three-phaseline in phasor form is
Io Qine charging) = jut,,Vnn
3.7
v , q , @ v g ) =! { v , , u r + v o u r + v o o z )
p,F/km to neutral
(3.22a)
Alkm
(3.23)
EFFECT OF EARTH ON TRANSMISSION LINE
CAPACITANCE
In calculatingthe capacitanceof transrnissionlines,the presenceof earthwas
ignored, so far. The effect of earth on capacitancecan be conveniently taken
into account by the method of images.
Method of Images
The electric field of transmissionline conductorsmust conforrn to the presence
of the earth below. The earthfor this purposemay be assumedto be a perfectty
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
E4
ModernPowerSYstemAnalYsis
|
ubstituting the valuesof different chargesand simplifying'
conducting horizontal shdet of infinite extent which therefore acts like an
equipotentialsurface.
unit
and
as
conductors
the
between
plane
midway
potential
is such that it has a zero
at
the
placed
is
dimensions
infinite
of
sheet
conducting
shown in Fig. 3.8. If a
zero potential plane, the electric field remains undisturbed.Further, if the
conductor carrying charge -q is now removed, the electric field above the
conducting sheet stays intact, while that below it vanishes.Using these well
known resultsin reverse,we may equivalentlyreplacethe presenceof ground
below a chargedconductorby a fictitious conductor having equal and opposite
chargeand locatedasfar below the surfaceof ground as the overheadconductor
above it-such a fictitious conductor is the mirror image of the overhead
conductor.This method of creatingthe same electric field as in the presenceof
earth is known as the method of images originally suggestedby Lord Kelvin.
we get
2hD
'l
rn
r(4hz + D2)'tz
rk
Vob=
(3.2s)
Radiusr
h
l--
,/'7
t
II
I
Zero potential
p l a n e( g r o u n d )
lmage charge
Fig'3.9Sing|e-phasetransmissionIinewithimages
It immediatelY follows that
wab
h
\
F/m line-to-line
(3.26a)
2nk
D
F/m to neutral
(3.26b)
,
It__
and
Cn= -
ln
F i g . 3 . 8 Electricfield of two long, parallel,oppositelycharged conductors
Capacitance
irk
-
of a Single'Phase
41+tO't4h21stt2
Line
line shownin Fig. 3.9.lt is requiredto calculateits
Considera single-phase
capacitancetaking the presenceof earth into acoountby the methodof images
describedabove. The equationfor the voltage drop Vo6as determined by the
two chargedconductorsa and,b, and their images a'and b' canbe written as
follows:
t"gt#Y
=
m2 +nrrn
vub
i* e,,,
*1, "
*q,,tnGFfUl
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(3.24)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Capacitaneeof TransmissionLines
86 ,* |
Modernpower SvstemAnalvsis
images.with conductor a in position1, b in position2, and c in position 3,
1
2 7rl(
ln
r
- ln h,
The equation for the averagevalue of the phasor %. ir found in a similar
manner. Proceedingon the lines of Sec. 3.6 and using Vou* Vo, = 3Von and
Qo* Qt * Qc = 0, we ultimately obtained the following expression for the
capacitanceto neutral.
(3.27)
tn
Similar equationsfor Vo6can be written for the second and third sectionsof the
transposition
cycle.If the fairly occurateassumptionof constantchargeper unit
length of the conductor throughoutthe transmissioncycle is made,the average
value of Voufor the three sectionsof the cycle is given by
C,
D'u -
rr( rn"n"\'),)!t \
(3.29a)
( (hrlhhs)''' )
r
0.0242
D'n -lon
-o
-ron@"httu')'!t
-o
pPttcmto neutral (3.29b)
(4h24)tt3
r
(3.28)
F/m to neutral
Comparing Eqs. (3.22a) and (3.29a), it is evident that the effect of earth is to
increase the capacitanceof a line. If the conductors are high above earttr
comparedto the distancesamong them, the effect of earth on the capacitance
of three-phaselines can be neglected.
where D, = (DnDnDrr)"t
Calculate the capacitanceto neutrallkm of a single-phaseline composed of
No. 2 single strand conductors(radius = 0.328 cm) spaced3 m apart and,7.5
m above the ground. compare the results obtained by Eqs. (3.6), (3.7) and
(3.26b).
Solution (1) Neglecting the presenceof earth tEq. (3.6)l
0.0242 n,
^
C,=31fitkm
log -
o.oi+z
=ffi=o'00817
P'Flkm
-
0.328
By the rigorousrelationship[(Eq. (3.7)]
cn
Since
+
negligible.
o.0242
"'(+.(#-')"')
=915, the effect of non-uniformity of chargedistribution is almost
C" = 0.00817 pFkm
Flg. 3.10 Three-phaseline with images
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
r
r
l
Capacitanceof TransmissionLines
-----------l
Modernpowel€ysteryr_Anslygls
I
_uu
-
(2) Considering
the eff'ectof earthandneglectingnon-uniformityof charge
distribution[Eq. (3.26b)]
0.0242
,Jto4
r(l*(
/4h"))
300
- 897
o32BJLo4
b
c
|
89
a
d=8m-
o:o??
c'" 2.9s3= o'0082P'Fkm
Note: The presenceof earth increasesthe capacitanceby approximately 3
partsin 800.
- h=6m
Example 3.2
A three-phase50 Hz transmissionline has flat horizontal spacing with 3.5 m
betweenadjacentconductors. The conductorsare No. 2/0 hard-drawn seven
strandcopper (outsideconductor diameter= 1.05 cm). The voltage of the line
is 110 kV. Find the capacitance to neutral and the charging current per
kilometre of line.
Solution
D"o= (3.5 x 3.5 x '7)t't= 4.4 m
0.0242
tog(440/0.525)
Fig. 3.11 Cross-section
of a double-circuit
three-phase
line
Solution As in Sec. 3.6, assumethat the charge per conductoron each phase
is equal in all the three sectionsof the transpositioncycle. For section / of the
transpositioncycle
V,,n(l)=
.'"f)
z*l*("i*'';)+c,(rn;
*n.(,n
j+r";)]
(3.30)
For section II of the transpositioncycle
v -l
n"
_ -1
0
6
u,Cn 314.o.oos%
= 0.384 x 106O/km to neutral
charging
cunent- +-
(l 19l ]e) x l-000
Xn
0.384x106
= 0.1I Aftm
The six conductorsof a double-circuitthree-phaseline having an overall radius
of 0.865 x 10-2m are arrangedas shown in Fig. 3.11. Find the capacitive
reactanceto neutral and charging current per kilometre per conductor at
110k V, 5 0 H z .
www.muslimengineer.info
=
v"bGr)
,lAln.("i* r"#)+c,(rn;.'" *)
(33r)
+a,(rnj.,":)]
For section III of the transpositioncycle
=*lr. ('"j .^ a,(rn ^I)
(rrr)
vtu
i)+ I.
+a.(rn;.r"f)f,"r,
Average value of Vo6over the transpositioncycle is given by
(av
v,*
s)=iL*[n.'n
(ttrj+*r,^(###)]
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
90
Modernpower SystemAnalysis
|
=fi,^-,,)h(ffi)"'
Capacitanceof TransmissionLines
(3.33)
h=6m
l'.
I
-
gf
i'B' jh
,'f 'd
vou*vo,=3von=
q,)tn(ffi)'''
fien"- Qt-
(3.3s)
3von=*r"(:#)
Fig. 3. 12
2dc
Capacitanceto neutral per conductor =
(3.36)
Total capacitanceto neutral for two conductorsin parallel
4rk
Cn=
(3.37)
Now h= 6 m; d = 8 m; -/ = 8 m. Referring to Fig. 3.I2, we can write
f ,
..?
.
,
,,121'1/2
r = l f / ) - + ( o - h \ " 1: J i m
\ 2 ))
L\2)
=
(j ' + h 2 )rt 2= l 0 m
f
g = (72 + 42)rt2=J65 -
3.8
METHOD OF GMD (MODTFTED)
A comparison of various expressions for inductance and capacitance of
transmissionlines [e.g. Eqs. (2.22b)and (3.6)] brings our rhe facr that the rwo
are sirnilarexceptthat in inductance
expressions
we haveto use the tictitious
conductorradius rt = 0.7788r, while in the expressionsfor capacitanceactual
conductorradius r is used.This fact suggeststhat the method of GDM would
be applicablein the calculationsfor capacitanceas well providedit is modified
by using the outer conductor radius for finding D,, the self geometric mean
distance.
Exarnple3.3 can be convenientlysolved as under by using the rnodifled
GMD method.
For the first section of the transpositioncycle mutual GMD is
Dub= ((ts) (ts))tta_ liglttz
Db' = Qil'''
Conductorradius (overall) = 0.865 x 10-2m
Substitutingthe valuesfor various distances.we have
cn
4 r x 7 x 8 . 8 5x 1 0 - 1 2x 1 0 6x 1 0 0 0
Drr, = ( jh)'''
1 n rf z * o s * s t o fr o o) 3 - l ' / 3
[
100x8
(D"pop,o)r,t = [(i, grjh)r,t)t,,
"n
ln the first section of the transpositioncycle self GMD is
D
pF /km
D,o = ?f rf )rt4 = ?f)'''
\0.86s/ |
D'l' = QAt''
= 0.018I 1F/km
Q C , = 3 1 4 x 0 ' 0 1 8 1x 1 0 - 6
Dr, = (At''
D, = (D,oDroD,,)''3= l?3f Arl3fitz
= 5.68 x 10-6 Ulkm
Chargingcurrent/phase
=
x 5.6g x 10{ = 0.361
Now
Cn=
2d(
2nk
"t#*
Charging current/conductor= 0'361
L
= 0.1805A/km
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4 ltk
F/m
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ai
,'lt I
FThis result
Example 3.3.
3.9
obviously checks with the fundamentally derived
LsL
expressionin
PROB
IEMS
BUNDTED CONDUCTORS
A bundled conductorline is shown in Fig. 3.13.
The conductorsof any one
bundle are in parallel, and it is assumedthat
the charge per bundle divides
equally among the conductorsof the bundle as
Drr> r/. Also Drr- d x D*
+ d x D12for the samereason.The results obtained-with
theseurru'rnptionsare
fairly accurate for usual spacings. Thus if the
charge on phase a is qo, the
conductorsa and a'have a chargeof qolz each;
similarly the charge is equally
divided for phasesb and c.
F-q--i
[?1
aQ
I O,,
DP
oQ eu,
I
---->fDy
l'
Fig' 3'13
Fd+
Fig. p-3.1
'..l
3'2 A three-phase
double-circuitline is shown in Fig. p-3.2.The
diameterof
each conductoris 2.0 cm. The line
is transffio and carries balanced
load' Find the capacitanceper phase
to neutral of the line.
aQ
_
"6 iec/____
---------
Dzs
--
j;;:i;g;:;fiIH#,",iHffi
T,.$l;:i,h*j
:::f::i::t"^tll:;
applied voltage is barancedthree-phase,
50 Hz. Take the voltage of phase
a as referencephasor. All conductors
have the sameradii. Also find the
charging current of phase a. Neglect the
effect of grouno.
Cross-sectionof a bundled conductorthree-phase
transmissionline
Now, writing an equ'ation for the voltage from
conductor a to condu ctor b,
we get
T
,f o Q
I
I
vob=
*lotn,(rn4.*ro
-,
o-u* 4.
*o.Sq"lr"
rn )
--\
D,
Qc'
2m
Qu,
2m
_t
D t,)
or
/
= r t
v,,h
,*lr,
,n'.r
+ro
. q n,.,
t n gDtz
.n"tnDrr\
Dzt)
(3.3e)
Consideringthe line to be transposedand proceeding
in the usual manner,the
f inal r e s u l t w i l l b e
',=
^ffi
p,Fkmto
neutral
(3.40)
where Do, = (DnDnD3)In
It is obvious from Eq. e.aD that the method of
modified GMD is equally
valid in this case (as it shouldbe).
www.muslimengineer.info
3'3 A three-phase,50 Hz overheadline
has regularly transposedconductors
equilaterallyspaced 4 m apart The
of such a line is 0.01
"upu.Itun.e
tFkm'Recalculate the capacitanc.p"i
kilometre to neutral when the
conductorsare in the same horizontai prane
with successivespacing of
4 m and are regularly transposed.
3'4 consider the,500 kv, three-phase
bundled conductorline as shown in
Fig' P-2'9' Find the capacitivi reactance
to neutral in ohms/km at 50 Hz.
3.5 A three-phase
trernsnri.ssion
rinc has r,rat,horizontarspacingwith
2 rn
betweenadjacentconductors.The radius
of eachconductoris 0.25 cm. At
a certain instant the chargeson the centre
conductor and on one of the
outside conductors are identical and voltage
drop betweenthese identically chargedconductorsis 775 v.xegtecithe
effect of ground, and find
the value of the identical chargein coulomblkm
at the instant specified.
3'6 Find the 50 Hz susceptanceto neutral
per kilometre of a double-circuit
threephaseline with transpositionas shown
in Fig. p_3.6.Given D = Jm
and radius of each of the six conductors
is 1.3g cm.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
94
i
ModernPo*", Syrt"r An"lysi.
-
t
6
l'-
6
o - +--
o-*_
6
5
6
D >)<-- D 'l-- o-J
5
. P-3.6 Doublecircuit three-phaseline with flat spacing
3.7 .{ single conductorpower cable has a conductorof No. 2 solid copper
(radius = 0.328 cm). Paper insulation separatingthe conductor from the
concentric lead sheath has a thickness of 2.5 mm and a relative
permittivity of 3.8. The thickness of the lead sheath is 2 mm. Find the
capacitive reactanceper kilometre between the inner conductor and the
lead sheath.
3 . 8 Find the capacitanceof phase to neutral per kilometre of a three-phase
line having conductorsof 2 cm diameterplacedat the cornersof a triangle
with sides5 m, 6 m and 7 m respectively.Assume that the line is fully
transposedand carriesbalanced load.
3 . 9 Derive an expressionfor the capacitanceper metre length between two
long parallel conductors,each of radius,r, with axes separatedby a
distanceD, where D ,, r, the insulating medium being air. Calculatethe
maximum potentialdifferencepermissiblebetweenthe conductors,if the
electric lield strengthbetweenthem is not to exceed25 kY lcm, r being
0.3 cm and D = 35 cm.
NCES
REFERE
Books
l . Stevenson,w.D., Elements of Power SystemAnalysis,4th edn, McGraw-Hill, New
York, 1982.
2 . Cotton, H., and H. Barber,
The Transmission and Distribution
Energy,3rd cdn, Hodder and Stoughton, 1970.
of Electrical
3 . Starr, A.T., Generation, Transmission and Utilization of Electric Power, Pitman,
1962.
Papers
A
T.
Parton, J.E. and A. Wright, "Electric Stresses Associated with Bundle Conductors", International Journal of ELectrical Engineering Education 1965,3 :357.
5 . Stevens, R.A. and D.M. German, "The Capacitance and Inductance of Overhead
Transmission Lines", International Journal of Electrical Engineering Education,
1 9 6 5 , 2: 7 1 .
4.I
INTRODUCTION
A complete diagram of a power system representing all the three phases
becomestoo complicatedfor a systemof practicalsize, so much so that it may
no longer convey the information it is intended to convey. It is much more
practicalto representa power systemby meansof simple symbbls for each
componentresultingin what is called a one-linediagram.
Per unit system leads to great simplification of three-phasenetworks
involvingtransformers.
An impedancediagramdrawn on a per unit basisdoes
not requireideal transfbrrnersto be includedin it.
An importantelementof a power systemis the synchronousmachine,which
greatly influencesthe systembehaviourduring both steadystateand transient
conditions.The synchronousmachinemodel in steadystateis presentedin this
chapter.The transient model of the machine will be presentedin Chapter 9.
4.2
Single-Phase Solution of Balanced Three-Phase Networks
The solution of a three-phasenetwork under balancedconditions is easily
carriedout by solving the single-phasenetworkcorrespondingto the ref'erence
phase.Figure4.1 showsa simple,balancedthree-phase
network.The generator
and load neutrals are therefore at the samepotential, so that In = 0. Thus the
neutral impedance Zn does not affect network behaviour. For the reference
phase a
En= ( Zc+ ZL) I '
(4.1)
The currentsand voltages in the other phaseshave the samemagnitudebut
are progressivelyshifted in phaseby 120".Equation(4.1) conespondsto the
single-phasenetwork of Fig. 4.2 whose solution completely determinesthe
solution of the three-phasenetwork.
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpower SystemAlelygis
Representationof PgryerSystemComponents
Ia
\;
\e,
t
* c ) 'trb-
L_-l
zn
-l
-
.*In=o
If the transformeris YIA connectedas in-Fig. 4.4a, the delta side has to be
replacedby an equivalentstar connectionas shown dotted so as to obtain the
single-phaseequivalent of Fig. 4.4b. An important fact has, however, to be
,",
lrJ'ZL
AN m0
16
!:
Z1
Ic
Fig. 4.1
Balanced three-phase network
Z6
lrne culTent /A
have a certain phaseangle shift- from the star side values Vorand Io(90" for
the phase labelling shown). In the single-phaseequivalent (Vew, I) are
respectively in phasewith (Von,/o). Since both phasevoltage and line current
shift through the samephaseangle from star to delta side, the transformerper
phaseimpedanceand power flow are preservedin the single-phaseequivalent.
In most analytical studies,we are merely interestedin the magnitude of voltages
and currents so that the single-phaseequivalentof Fig. 4.4b is an acceptable
proposition. Wherever proper phaseangles of currents and voltages are needed,
correction can be easily applier after obtaining the solution through a singlephasetransformerequivalent.
Ea
Fig' 4.2
Single-phaseequivalentof a balancedthree-phasenetwork of Fig.
4.1
(a) Y/A transformerwithequivalentstarconnection
€
l
>
->,
Ia
Ia
A
n_
(b) Single-phase equivalent of Y/A transformer
(a) Three-phaseY/y transformer
Fig.4.4
It may be noted here that irrespective of the type of connection, the
transformation ratio of the single-phaseequivalent of a three-phasetransformer
is the same as the line-to-line transformationratio.
(b) Single-phaseequivalentof 3-phase y/y transformer
Fig. 4.3
'
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See Section 10.3.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
tro
Yli
4.3
I
ONE.IINE
DIAGRAM
AND IMPEDANCE
OR REACTANCE
DIAGRAM
A one-linediagramof a power system shows the main connectionsand
showrrdependingon the information required in a system study, e.g. circuit
breakersneednot be shown in a load flow study but are a must for a piotection
study. Power system networks are representedby one-line diagrams using
suitable symbols fbr generators,motors, transformersand loads. It is a
convenient practical way of network representationrather than drawing the
actual three-phasediagram which may indeed be quite cumbersome and
confusing for a practical size power network. Generator and transformer
connections-star, delta,and neutralgrounding are indicatedby symbols drawn
by the side of the representationof these elements.Circuit breakers are
representedas rectangularblocks. Figure 4.5 shows the one-line diagram of a
simple power system.The reactancedata of the elementsare given below the
diagram.
T1
(( ' Y o
I
ft
L
__'l
l
r
T2
.tF
-fl
have been neglected.This is a fairly good approximationfor most power system
studies.The generatorsare representedas voltagesourceswith seriesresistance
in Sec.
(synchronous
machinemodelwill be discussed
and inductivereactance
4.6). The transmissionline is representedby a zi-model(to be discussedin
Chapter5). Loads are assumedto be passive(not involving rotatingmachines)
and are representedby resistanceand inductive reactancein series.Neutral
grounding impedancesdo not appearin the diagram as balancedconditions are
assumed.
Three voltagelevels(6.6. I 1 ancl33 kV) are presentin this system.The
Y r
i
laFt ' t l
r-YA
l
99
t
The impedancediagramon single-phasebasisfor use under balancedoperating
conditionscan be easilydrawn from the one-linediagram.For the systemof
Fie. 4.5 the impedancediagramis drarvnin Frg. 4.6. Single-phasetranstbrmer
equivalents are shown as ideal transformers with transformer impedances
of Power SystemComponents
Representation
Modern Power System Analysis
iI
v--
q
r
A-
throughout this book.
PER UNIT (PU) SYSTEM
4.4
, Fig. 4.5 One-linerepresentationof a simple power system
G e n e r a t oN
r o . 1 : 3 0 M V A ,1 0 . Sk V ,X ' = 1 . 6o h m s
r o .2 : 1 5 M V A ,6 . 6k V ,X ' = 1 . 2o h m s
1 G e n e r a t oN
G e n e r a t oN
r o .3 : 2 5 M V A ,6 . 6k V ,X ' = 0 . 5 6o h m s
Transformer
11(3 phase): 15MVA,33/11 kV,X = 15.2ohmsper phaseon hightensionside
TransformerTr(3 phase):15 MVA,3316.2
kV,X= 16 ohmsper phaseon frigntensionside
Transmission
line:20.5ohms/phase
LoadA : 15 MW, 11 kV, 0.9 laggingpowerfactor
L o a dB : 4 0 M W ,6 . 6 k V ,0 . 8 5 l a g g i npgo w e rf a c t o r
Note: Generators
are specifiedin three-phase
MVA,line-to-line
voltageand per phase
reactance(equivalent
star).Transformers
are specifiedin three-phaseMVA, line-to-line
transformation
ratio,and per phase(equivalent
star)impedanceon one side.Loads are
specifiedin three-phaseMW,line-to-line
voltageand powerfactor.
--i--t
'-litr-d-'r,n'rq J-'rn,A,n-?fi-d\
I- r_-1
I--ntrn^'"1'"rr' f- I
1 r- i , I
\
It is usual to express voltage, current, voltamperes and impedance of an
electricalcircuit in per unit (or percentage)of baseor ret'erencevaluesof these
quantities. The per unit* value of any quantity is defined as:
the actualvaluein anYunits
the baseor referencevaluein the sameunits
The per unit method is particularly convenientin power systemsas the various
sections of a power system are connectedthrough translormersand have
different voltage levels.
Consider first a single-phasesystem.Let
Base voltamPerss= (VA)s VA
Basevoltage= Vu V
Then
Basecurrent/u = -[4)e
\/
A
(4.2a)
vB
- > F r - F .T r a n s f o r m e r l+l '
Genl'
I
Load A
Fig. 4.6
-J;-ffi
. -n :s f o r m e r T 2- - t C ; / C " ' S
' T .r a
:L :i n e- - , F' r _
Load g
lmpectancediagram of the power system of Fig. 4.5
www.muslimengineer.info
*Per
cent value = per unit value x 100.
Per cent value is not convenient for use as the factor of 100 has to be carried in
computations.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
tir.n' l
.,rvq..il
MooernHowersystem Analysis
I
Vi
Baseimpedance
z, = Y-4-:
"
ohms
(V A)"
IB
(4.2b)
'er unit imPedanceZ (Pu) =
Z (oltrns)x (MVA)'
@
Z(ohms)x (kVA),
If the actual impedanceis Z (ohms), its per unit value is given by
,r trt2
Z(ohms)x (vA)'
z(pu) = J--ZB
V;
(4.3)
For a power system,practical choice of base values are:
Base megavoltamperes= (MVA)B
Per Unit Representation
Base kilovolrs = (kv)a
1,000x (MVA)u
(4.4)
(kV)a
1'ooq] GV)r
- __Ia
Baseimpedance
z-,u_
GV)3 _ 1,000x (kv)1
ohms
(MVA)B
Per unit impedancez (pu) -
(kvA)8
Z (ohms)x (MVA),
(kv)"
(4.s)
(4.6)
o
(4.7)
x (kv)2,
GD3 _ 1,ooo
^L_: onms
(MVA)'
G"Ab
(4.8)
Baseimpedance7o "
_
V r s- !
(4.11a)
a
zs
In a three-phasesystemrather than obtaining the per unit valuesusing per phase
base quantities,the per unit valuescan be obtained directly Uy u.ing
threephase base quantities.Let
Three-phasebase megavoltamperes= (MVA,)B
Line-to-line base kilovolts = (kV)B
Assuming star connection (equivalentstar can always be found),
Jr 1rv;u
of a Transformer
-{_J--->
(kV)?x 1,000
l'ooox(MVA)u
(4.10)
transformer forming part of
It has been said in Section 4.2 that a three-phase
ty a single-phasetransformer in
a three-phase system can be represented
The deltaconnectedwinding of the
obtaining per phasesolution of the system'
so that the transformationratio of
transformer is replacedby an "quiuaient star
is always the line-to-line voltage ratio
the equivalent single-phasetransformer
of the three-Phasetransformer'
Fi gur e4. Tar cpr esent sasingle- phaset r ansf gr m lr int er m sof pr im
ar yand
andan ideal transformer of ratio 1 : a'
secondary leakagi reactancesZp artdZ,
Let us choose-avoltamperebase of
The magnetizing impedanceis neglected.
of the transformer in the ratio of
(vA)a and voltage bases on the two sides
transformation,i.e.
Vt,
Z(ohms)x (kVA)u
BasecurrentIr-
n^n
(MVA)r, ototo (MVA)a' n'*' andtV base
when MVA baseis changecltrom
nt* p"t unit impedancefrom Eq'
s changedfrom (kv)r, orotJ (kV)n, new'the
.4.9) is given bY
Base kilovoltamperes= (kVA)B
_
-, i
=z(pu)o.,.
Z(pu)n"*
ffi"ffi
or
Base current 1,
12
transformer
of single-phase
(a)
'-' Representation
neglected)
impedance
(mignetizing
l'ooox(kv)u
J3rB
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(4.e)
transformer
circuitof single-phase
(b) Perunitequivalent
Fig. 4.7
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
-W
todern
I
Power Srrstem -A-nalr-rsie
Y-
vr
v.vr.;
, rr rqt
ystr
Representationof Power System Uomponents
I
Therefore
'I 2 8
= a (as (VA)a is common)
Vrn z
1,,
LaD
.
Vza
-
--
1,,
(4.1i b)
(4.11c)
From Fig. 4.7a we can write
Vz=(Vt-Iflp)a-lrZ,
(4.12)
We shall convertEq. (4.12) into per unit form
Vz(pu)Vzn= [Vr(pu) Vw - I r(pu)I
6Zo(pu)Zru]a
-Ir(pu)IruZ,(pu)Z*
Dividing by vzn throughout and using base relations (4. rra, b, c),
Vz(pu)= Vr(pu) - I,(pu)Zr(pu)_ Ir(pu)Z,(pu)
Now
or
we get
(4.13)
+=+=,
Ir
12
I'o
I'u
I r ( p u ) = 1 2 ( p u ) =1 ( p u )
Equation (4.13) can thereforebe written as
Vz@u)= Vr(pu)- (pu)Z(pu)
where
(4.r4)
Z(pu)=Z,,(pu)+ Z,(pu)
Equation (4.I4) can be representedby the simple equivalent circuit
of Fig.
4'7b which doesnot requirean idealtransfurmet.Coniia.rable
simplification
has thereforebeenachievedby the per unit method with a common voltampere
base and voltagebaseson the two sidesin the ratio of transformation.
Z(pu) can be determineddirectlyfrom the equivalentimpedanceon primary
or secondaryside of a transformerby using the appropriaie impedancl
base.
On primary side:
Zr=Zp+ Z,/a2
Z ( p' =
u+
) :!-+L*I
zru
But
Zro
z,o"a2
a2Ztn = Zzr
Zr(pu)= Zo(pu)+ Z,(pu)- Z(pu)
On secondary
side:
zz= Z, + o2zo
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(4.1s)
[,.t'ff.
l-
zz(pu)=
' +-++o'1'
zru
zzu zB
Thus the per unit impedanceof a transformeris the samewhethercomputed
from primary or secondaryside so long as the voltage baseson the two sides
are in the ratio of transformation(equivalentper phaseratio of a three-phase
transformer which is the same as the ratio of line-to-line voltage rating).
The pu transformerimpedanceof a three-phasetransformeris conveniently
obtained by direct use of three-phaseMVA base and line-to-line kV base in
relation (4.9). Any other impedanceon either side of a transformeris converted
to pu value just like Zo or Zr.
Per Unit Impedance
Diagram of a Power System
From a one-line diagramof a power systemwe can directly draw the impedance
diagramby following the stepsgiven below:
1. Choosean appropriatecommon MVA (or kVA) basefor the system.
2. Consider the system to be divided into a number of sections by the
transfbrmers.Choose an appropriatekV base in one of the sections.
Calculate kV bases of other sectionsin the ratio of transformation.
3. Calculateper unit valuesof voltagesand impedancesin eqch sectionand
connectthem up as per the topologyof the one-linediagram.The result
is the single-phaseper unit impedancediagram.
The above stepsare illustrated by the fallowing examples.
j Example4 1
Obtain the per unit impedance(reactance)diagram of the power system of Fig.
4.5.
Solution The per phaseimpedancediagramof the power system of Fig. 4.5
has been drawn in Fig. 4.6. We shall make some further simplifying
assumptions.
1. Line capacitanceand resistanceare neglectedso that it is representedas
a series reactanceonly.
2. We shall assumethat the impedancediagram is meant for short circuit
studies.Current drawn by static loadsunder short circuit conditions can
be neglected.Loads A and B are thereforeignored.
Let us convert all reactancesto per unit form. Choosea common three-phase
MVA base of 30 and a voltagebaseof 33 kV line-to-lineon the transmission
line. Then the voltagebasein the circuit of generator1 is 11 kV line-to-line and
that in the circuits of generators2 and 3 is 6.2 kV.
The per unit reactancesof various componentsare calculatedbelow:
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Flepresentation
of power System eomponents
2o.5x3o
Transmission
line:
GT2
Transform
er T,:
I iliffij
IExample 4.I, we now calculatethe pu values of the reactancesof transfonners
and generatorsas per relation (4.10):
= 0.564
-l-!,?I l_0_= 0 . 4 1 8
16x30
TransformerTt:
= 0.44
aiT
Transformer Z :
0.209 x
Transformet Tr:
0.22
x ff =0.++
1 . 6x 3 0
Generator1:
u t7- = o'396
1 . 2x 3 0
Generator2:
tazr
The reactancediagramof the
(10'5i1
=0.3e6
(ll)'
= o'936
Generator2:
0.56x 30
= 0'437
AZI-
Generator3:
0.43sx
Generator1:
systemis shown in Fig. 4.g.
JU
U000._--J--
0.44
I
I)
*(6'612- =0.936
6.2)'
(6'6)1
= 0.431
0.3214 * i9 x
/.r
6.21'
obviously these values are the same as obtained already in Exampre 4.r.
Generator3:
{-)frL_/-X-fX-)<1
-_"
64
0 . 4 1 3r *
i
4.5
Complex Power
Consider a single-phaseload fed from a source as in Fig. 4.9. Let
v -tvt16
r _tn t (6_0)
Fig.4.9 Reactance
criagramof the systemof Fig. 4.5 (roads
neglected)
Et' Ez and E, are per unit values
of voltagesto which the generators
are
in ashort
iJ"',ltl3;3llllrlt""
circuitstudy,
these
wil beraken
ui t /.,"pu (no
Source
Example4.2
(a)
The reactanccdata of gencrators
and transtbrmersare usually specified
in pu
(or per cent) values, based on
equiprnentratings rather than in
actual ohmic
valuesasgiven in Example 4.7; *iit"
rhetransmirr;;; hne irnpedances
nray be
us
;"";:,[J:Ti]#?l-et
rc-sotve
rJxarnpre
4.1b; assuming
rherbuowing
TransformerT,: 0.209
TransformerT): 0.220
GeneratorGr: 0.435
GeneratorGr: 0.413
GeneratorG3: 0.3214
With a baseMVA of 30. base volta_ee
of I I kV in the circuit of generator
I md b:isevoltageof 6.1 k\; in the
circuit of generators2 and 3
,r; i;
";
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Fig. 4.9 Complexpowerflow in a single-phase
load
When d is positive, the current lags behind voltage.This is a convenient
choice of sign of 0 in power systemswhere loads have mostly lagging power
factors.
Complex power flow in the direction of current indicatedis siven bv
S=VI*
=lVlllll0
= l V l l 1 lc o s d + j l v l l 1 l s i n 0 = P + i e
@.17)
or
t S l= ( p 2 + e \ t , ,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
nepresentation
ot powersystemeompqlents
Modern Power system Anatysis
I
l.i; ffii
T-
As per Eq.(4.19), Kirchhoff's current law appliesto complex power (also
applies separatelyto real and reactive powers).
In a series RL load carrying current {
Here
S - complex power (VA, kVA, MVA)
V=l(R+jxr)
lSl = apparentpower (VA, kVA, MVA); it signifies rating of
I
P = I"R = active power absorbedby load
Q - IzXr = r-eactivepower absorbedby load
P = lVl l1l cos 0 - real (active) power (watts, kW, MW)
Q = lVl l1l sin 0 = reactivepower
In case of a series RC load carrying current I
- voltamperesreactive (VAR)
P _I2R
O - IzX, qreactivepower absorbedis negative)
= kilovoltamperesreactive (kVAR)
= megavoltamperesreactive (MVAR)
It irnmediatelyfollows from Eq. (4.17) that Q, thereactivepower, is positive
for lagging current (lagging power factor load) and negativefor leading cunent
(leading power factor load). With the direction of current indicated in Fig. 4.9,
.9= P + iQ is supplied by the source and is absorbedby the load.
(4.17)can be represented
Eqr-ration
by the phasordiagramof Fig.4.10 where
, i1
n
= positive for lagging current
0 = Lan''
(4.18)
P
:gativefbr leadingcurrent
Consider now a balanced three-phaseload representedin the form of an
equivalentstar as shown in Fig. 4.L2. The three-phasecomplex power fed into
load is given by
S = 3vpl-t= 3 lvptl6pl; : JT lvrl zOrti
(4.20)
If
Ir =llil I (6p- A
Then
.S = ,'5 lvLl lILl I
0
- Ji tvLtvLtcosd + iJT t
0=P+iQ@.21)
Fig. 4.10 Phasorrepresentation
of complexpower(laggingpf load)
If two (or more) loads are in parallel as in Fig. 4.ll
Fig. 4.12 Complexpowerfed to three-phase
load
s=vF_v(i+i)
Here
f Y r i . . r ' r(' p
! -r + p r+) j ( e t +e z )
(4.1e)
tsl = Ji tvrt ttrl
P - Ji tvLltILtcosd
e = Ji lvLltILtsin d
d - power factor angle
lf vL, the line voltage,is expressedin kv; andIy,the line currentin amperes,
s is in kvA; and if the line current is in kiloamperes, s is in MvA.
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
In terms of load impedanceZ,
N = rotor speed (synchronousspeed)in rpm
v, _lvLll6P
r, L _Jiz
z
Substitutingfor I, in Eq. (4.20)
P = numberof poles
f
winding
r" -- V'l'
r
(4.22a)
Field winding -..
tr V,.is in kV, ,Sis now given in MVA. Load impedance
Z if requiredcanbe
calculatedfrom
,_lvrl'_ lvl
" - Jt- -T1O
4.6
1)xr o
\Xp>{
(4.22b)
--F
vl
I
-T:t
-F
SYNCHRONOUSMACHINE
I
The synchronousmachineis the most important elementof a power system.It
converts mechanical power into electrical form and feeds it into the power
network or, in the caseof a motor, it draws electrical power from the network
and converts it into the mechanicalform. The machine excitation which is
controllable determinesthe flow of VARs into or out of the machine.Books on
electrical machines 11-51 may be consulted for a detailed account of the
synchronousmachine. We shall presenthere a simplified circuit model of the
(undertransient
machinewhich with suitablemodificationswherevernecessary
conditions) will be adoptedthroughout this book.
Figure 4. 13 shows the schematiccross-sectionaldiagramof a three-phase
synchronousgenerator(alternator)having a two pole structure.The stator has
a balanced three-phasewinding-aat, bbt and cct. The winding shown is a
concentratedone, while the winding in an actual machineis distributed across
the stator periphery. The rotor shown is a cylindrical" one (round rotor or nonsalient pole rotor) with rotor winding excited by the DC source. The rotor
winding is so arrangedon rotor periphery that the field excitation produces
nearly sinusoidally distributedflux/pole (d) in the air gap.As the rotor rotates,
three-phase emfs are produced in stator winding. Since the machine is a
balancedone and balancedloading will be considered,it can be modelled on
per phase basis for the referencephase a.
Tn
o
mqnhinc
rrrifh
mnra
fhqn
f r r r nv
nnlec
l/vrvo,
fhp
Lllv
qlrnrre
quv
vv
ApfinpA
svrlllvu
cfnrnfrrro
JLr uvLur
v
ronpafc
rvyvolo
electricallyfor every pair of poles.The frequencyof inducedemf is given by
f =ffi nz
where
.
High-speedturbo-generatorshave cylindrical rotors and Iow sppedhydro-generators
have salient pole rotors.
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lot
-t'. \
'\r \lQ I
\,
NN
generator
Fig. 4.13 Schematicdiagramof a roundrotorsynchronous
On no load the voltage EJ inducedin the referencephasea lags 90" behind
dywhich producesit and is proportionalto dyif the magneticcircuit is assumed
This phasorrelationshipis indicatedin Fig. 4.14. Obviously
to be unsaturated.
the terminal vclltage V, = Er
Qr
I
l
-
Ef=Vt
Fig.4.14 Phasorrelationship
betweenfuand E,
As balancedsteady load is drawn from the three-phasestator winding, the
stator currents produce synchronously rotating flux Q/poIe (in the direction of
rotation of the rotor). This flux, called armature reaction flux, is therefore
stationarywith respectto field flux Qy.It intuitively fbllows that Qois in phase
with phase c current 1o which causesit. Since the magnetic circuit has been
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern powsr Syglem_Anatygis
,liO* |
urrurnJdto be unsaturated,
the superpositionprincipleis applicableso that the
resultant air gap flux is given by the phasor sum
d' = d1+ Q,,
@,3)
Further assumingthat the armatureleakagereactanceand resistanceare
re eml whtch equals the termin
tage V,.
Phasor diagram under loaded (balanced) conditions showing fluxes, currents
and voltagesas phasorsis drawn in Fig. 4.15.
Rapr"."n,",ionof po*"r' Syrr"r Cornp.on"n,,
NEffi
t The circuit of Fig. 4.L6 can be easilymodifiedto include the effect of
armatureleakagereactance
andresistance
(theseareserieseffects)to give the
complete
circuitmodelof the synchronous
generator
as in Fig. 4.I7. Thetotal
+xl=Xslsc
synchronous reactance of the machine.
Equation(4.24) now becomes
V, = Et - jlo X, - IoRa
(4.2s)
jlX"= - E"
Fi g.4. 16
This model of the synchronousmachinecan be further modified to account
for the effect of magnetic saturation where the principle of super-position does
not hold.
Fig. 4.1S phasordiagramof synchronous
generator
Here
d - power factor angle
6 - angle by which Et leads v, called load angre or torque angle
We shall see in Sec.5.10 that dmainly determinesthe power deliveredby
the generatorand the magnitudeof E, (i.e. excitation)determinesthe VARs
deliveredby it.
Becauseof the assumedlinearity of the magneticcircuit, voltage phasorE,
Eo and v, are proportionalto flux phasors dr, doand d, respectively;furthei,
voltage phasors lag 90' behind flux phasori. It therefore easily follows from
Fig.4.15 that phasorAB =- Eois proportionalto (o (and therefore Io)and is
90' leading d" @r 1,). With the direction of phasorAB indicatedon the diagram
AB = jlo Xo
where X" ir the constantof propotionality.
In
terrnc
nF
thp
q ul rvnvrvr o
q
rlofi-iri^uvruulrurr
^s
vL
v
,lra,
wE
ualt
li-^^rr-urr€utly
---^rrwl-l[c)
4t-ule
expressionfor voltageswithout the need of invoking flux phasors.
V, = Ef - jloXo
r rt
l0ilowlng
(4.24)
where
Ef = uolrage induced by field flux Q, alone
= uo load emf
The circuit model of Eq. (4.24) is drawn in Fig. 4.16 wherein X, is
interpreted as inductive reactancewhich accountsfor the effect of armature
reactionthereby avoiding the needof resortingto addition of fluxes l&q.@.23)1.
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&
Fig- 4.17 circuitmodelof roundrotorsynchronous
generator
Armature resistance Rn is invariably neglectedin power system studies.
Therefore,in the place of the circuit model of Fig. 4.I7, the simplified circuit
model of Fig. 4.18 will be usedthroughoutthisbook.The corresponding
phasor
diagram is given in Fig. 4.i9. The fieici induceciemi Ey ieacisthe terminal
-condition
voltage by the torque (load) angle d This, in fact, is the
for acrive
power to flow out of the generator.The magnitudeof power delivered depends
upon sin d
In the motoring operationof a synchronousmachine,the current 1,,reverses
as shown in Fig. 4.20, so that Eq. @.25) modifiesro
Ef = V, - jIoX,
(4.26)
which is representedby the phasordiagramof Fig. 4.2I.It may be noted that
V, now leads lby d, This in fact is the condition for power to flow into motor
terminals.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Power System Analysis
the'i I
-
machine
The flow of reactivepower anclterminal voltage of a synchronous
detail in
in
discussed
is
This
excitation.
is mainly controlleclby meansof its
regulated
automatically
often
are
flow
power
Section5.10. Voltage and reactive
uy
IvSurqlvro
YUIL4S\/
\uvv
vvve^v^r
of PowerSystemComponents
Representation
lvtl ll,,lcos d = constant= activepoweroutput
FIIS':
vr
and by automatic tap changing devices on transformers.
-- -'-/ 6666
|
l
E1 (
Fig.4.22Synchronousmachineconnectedtoinfinitebus
projection l/ol cos dof the phasor Io on V'
It rneansthat since lV,l is fixed, the
varied' Phasordiagramscorresponding
remains constant,whiie the excitation is
presentedin Fig' 4'23' T\e phasor
are
to high, medium and low excitations
the unity power factor case'It is obvious
diagramof Fig. 4.23(b)colrespondsto
excitation
frot the phasor diagram that for this
I
x
"
l E J lc o s 5 = l V )
;
E1
I
t
Fig. 4 . 1 8 S i m p l i f i e dc ircuit model of
round rotoI synchronous
generator
it"x'
-l
---"'
Fig. 4.19 Phasordiagramof synchronousgenerator
., I"
/
0
(a) Overexcited
Ia
Ef'
jIJ'
1I
J
( b ) N o r m a le x c i t a t i o n
Vt',
Fig. 4.20 MotoringoPerationof
sYnchronousmachine
Fig. 4.21 Phasordiagramof motoring
oPeration
generators
Normally, a synchronousgeneratoroperatesin parallel with other
considera
we
shall
operation
of
simplicity
connectedto the power system.For
bus
infinite
As
4'22'
Fig'
in
as
shown
bus
inJinite
generatorconnectedto an
independent
constant
remain
frequency
and
voltage
meansa largesystemwhose
the bus' and
of the power exchangebetweenthe synchronousmachine and
machine.
synchronous
the
of
excitation
the
independentof
into an
consider now a synchronousgeneratorfeedingconstantactivepower
and its
In
infinite b's bar. As the machineexcitation is varied, armaturecurrent
keep
angle g, t.e. power factor, changein such a manner as to
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(c) Underexcited
Fi1.4.23 P h a s o r d i a g r a m s c f s y n c h r o n o u s g e n e r a t o r f e e d i n g c o n s t a n t
power as excitation is varied
case(Fig' a'23a)'i'e'
ovetexciterl
This is definedas normal excittttittrtForthe
generatorfeedspositive reactive
cos 6>|v),1, lags behind V, so thatthe
|,8,.1
powerintothebus(ordrawsnegativereactivepowerfromthebus)'Forthe
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
^
;iiii
# . . . . . . . ' @ , ^
^
l L ^
"
j
a
^
{
-
A
l 4
^
l i ^ / \ ^ ^
of PowerSystemComponents
Representation
ModernPowersystemAnalysis
I
A a A u
I iilis'.$
T.._
or
t
underexcited case (Fig. 4.23c), i.e. lErl cos 6 < lV), 1o leads V, so that the
generatorfeedsnegativereactivepower into the bus (or drawspositivereactive
Figure 4.24 shows the overexcitedand underexcitedcasesof synchronous
motor (connectedto infinite bus) with constantpower drawn from the infinite
bus. In the overexcited case,Io leads Vu i.e. the motor draws negativereactive
power (or suppliespositive reactive power); while in the underexcitedcase .Io
lags V, r.e. the motor draws positive reactive power (or suppliesnegative
reactivepower).
ln,l
l1,lcose: E: sin6
(4.27)
(4.28)
The plot of P versus { shown in Fig. 4.25, is called the power angle curve.
The maximum power that can be delivered occurs at 6 = 90" and is given by
(4.2e)
For P ) P** or for 6> 90' the generatorfalls out of step. This problem (the
stability) will be discussedat length in Chapter 12.
E1
(a) Overexcited
--
V1
generator
Fig. 4.25 Poweranglecurveof a synchrcnous
Power Factor and Power Control
(b) Underexcited
Fig.4.24
Phasor diagramsof synchronousmotor drawing constantpower as
excitationis varied
From the above discussionwe can draw the general conclusion that a
synchronousmachine (generating or motoring) while operating at constant
power suppliespositive reactive power into the bus bar (or draws negative
r c q e t t v cI v
nrtrx/cr
vv
frnrn
fhc
hrrc
hv sqU r \
rYrvr rhr var rn
n
r rYcvrl vcl \vv lnL vi vf .a r l
v
/
Al n
ll
ru rl l n
rlarownifprl
uvtvnvltvu
mqnl"i-o
lll4vllltlv
on the oih.. hand, feeds negative reactive power into the bus bar (or draws
positive reactivepower from the bus bar).
Considernow the power deliveredby a synchronousgeneratorto an infinite
bus. From Fig. 4.19 this power is
P = lVtl llol cos 0
The above expression.canbe written in a more useful form from the phasor
geometry.From Fig. 4.19
While Figs 4.23 and 4.24 illustrate how a synchronousmachine power factor
changeswith excitation for fixed power exchange,thesedo not give us a clue
regarding the quantitativevalues of llnl and d This can easily be accomplished
by recognizing from Eq. (4.27) that
lEll sin 6 -llolX, cos d
PX"
= constant (for constantexchangeof power to
= #
lyrl
(4.30)
infinite bus bar)
,
Figure 4.26 shows the phasor diagram for a generatordelivering constant
power to infinite bus but with varying excitation.As lEtl sin dremainsconstant,
the tip of phasor Ermoves along a line parallel to y, as excitation is varied. The
direction of phasor1ois always 90o lagging jI"X, andits magnitudeis obtained
from (l1olX5)/X5.Figurc 4.27 shows the caseof limiting excitation with d= 90".
For excitation lower than this value the generatff becomesunstable.
lnA _
- rl,lx,
sin (90" + 0)
sin 6
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t
_
^
^
-
1!1
of Fqwer SystemComponents
Ftepreseniation
ModernPower System Analysis
|
Salient Pole Synctrronous Generator
A salient pole synchronousmachine,as shown in Fig. 4.29, is distinguished
from a round rotor machine by constructional features of field poles which
/.{1
62
Iaz
\3
Fig. 4.26
Effect of varying excitationof generator deliveringconstant power
to infinitebus bar
employed in machinescoupled to hydroelectric turbines which are inherently
slow-speedones so that ttre synchronousmachine has rnultiple pole pairs as
different from machines coupled to high-speed steam turbines (3,000/1,500
rpm) which have a two- or four-polestructure.Salientpole machineanalysisis
made through the two-reaction theory outlined below.
Direct axis
I
/,"
V1
Fig. 4.27
Case of limitingexcitationof generator deliveringconstant power
to infinitebus bar
Similar phasor diagrams can be drawn for synchronous motor as well for
constant input power (or constant load if copper and iron losses are neglected
and mechanical ioss is combined with load).
Another important operating condition is variable power and fixed excitation. In this case lV,l and lE1trare fixed, while d and active power vary in
accordance with Eq. (a.28). The corresponding phasor diagram for two values
of d is shown in Fig. 4.28. It is seen from this diagranr that as d increases,
current magnitude increases and power t'actor improves. lt will be shtlwn in
Section 5.10 that as dchanges, there is no significant change in the flow of
reactive Power'
Locusof Er
, n'
Er',
--4,
,--' l\---jluzX"
l/-
'
Fig. 4.29 Sallent pole synchronousmachine (4-polestructure)
In a round rotor machine, armatLlrecurrent in phase with field induced emf
Ey or in quadrature (at 90") to S, produces the same flux linkages per arnpere
ai the air gap is uniform so that the armature reaction reactance offered to inphase or quadrature current is the same (X,, + X1 = Xr), In a salient pole
machrne
Fig. 4.28
Operationof synchronousgenerator with variablepower and fixed
excitation
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l
at
gap ls non-unllorTn
l - , -
- - 1 - -
- ^ ^ - : - L ^ - - .
arong IULOI'ljcrlPilury.
f t
i ^
+ L ^
l ^ ^ ^ +
^ l ^ - ^
+ L ^
rL ls Lllc rtrilsL .lruug trrtr
axis of main poles (called direct axis) and is the largestalong the axis of the
interpolar region (called quadrature oxis).Armature current in quadraturewith
El producesflux along the direct axis and the reluctanceof flux path being low
(becauseof small air gap), it produces larger flux linkages per ampere and
hencethe machinepresentslarger armaturereactionreactanceX, (called direct
axis reactance)to the flow of quadraturecomponentIl of armaturecurrent 1o.
On the other hand, armaturecurrent in phasewith { producesflux along the
quadratureaxis and the reluctanceof the flux path being high (becauseof large
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
4
ffil
Mod"rnPo*r. syrt"t Rn"tyri,
r
i5 I
I
rl,l-r-
Representation
of power
interpolar air gap), it producessmaller flux linkages per ampereand hencethe
machine presents smaller armature reaction reactance Xu (guadrature axis
reactancea X) to the flow of inphase component Io of armature current /o.
Since a salientpole machineoffers different reactancesto the flow of Il and
1ocornponentsof armature current Io, a circuit model cannot be drawn. The
phasordiagramof a salientpole generatoris shown in Fig. 4.30.It can be easily
drawn by following the stepsgiven below:
Resultant
Fig. 4.31 power angrecurvefor sarientporegenerator
In this book we shall neglect the effect of sariency and take
X'= X't
generator
Fig. 4.30 Phasordiagramof salientpolesynchronous
1. Draw % *d Io at angle 0
2. Draw IoRo.Draw CQ = .il,X,t(L to 1,,)
3. Make lCPl - llol Xq and draw the line OP which gives the direction of Ey
phasor
4. Draw a I from Q to the extended line OP such that OA = Ef
It can be shown by the abovetheory that the power output of a salient pole
generatoris given by
lv,llE,l
,=-1;-sind+
lv,l'(xo- xn) si n 26
2XdXq
(4.31)
The first term is the same as for a round rotor machine with X, = Xa and
constitutesthe major part in power transfer. The second term is quite small
(about I0-20Vo) comparedto the first term and is known as reluctancepower.
P versus d is plotted in Fig. 4.31. It is noticed that the maximum power
34 (change power per
unit
in
output occurs at 6 < 90' (about 70').
' Furt1t"r
d5'
change in power angle for small changes in power angle), called the
synchronizingpower.cofficient,in the operatingregion (r< 70') is larger in
a salient pole niachin.: than in a round rotor machine.
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in all types of power system studiesconsidered.
During a machine transient, the direct axis reactancechanges with time
acquiring the following distinct values during the completetransieht.
X/ = subtransientdirect axis reactance
Xh = transient direct axis reactance
X,r = steady state direct axis reactance
The significance and use of thesethree values of direct axis reactancewill
be elaboratedin Chapter 9.
Operating
Chart of a Synchronous
Generator
while selecting a large generator,besidesrated MVA and power factor, the
greatestallowable stator and rotor currents must also be consideredas they
influence mechanical stressesand temperaturerise. Such timiting parameters in
the operationare brought out by meansof an operating chart or-performance
chart.
For simplicity of analysis,the saturationeffects,saliency,and resistanceiue
ignored and an unsaturated value of synchronous reactanceis considered.
Consider Fig. 4.32, the phasor diagram of a cylindrical rotor machine. The
locus of constantllolx,V) and henceMVA is a circle centeredat M. The locus
of constantlEtl (excitation) is also a circle centeredat O. As Mp is proportional
to MVA,QP is proportional to MVAR and Me to MW, all to the same scale
which is obtained as follows.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Ir \Y, /rl n
v rul o
! rr rnr
D
c
I nvrYr rYa vr r vQ)r 'rsc ri avm
rr|,rA
r rnqarl tr rucri u
-
2.Opu excitation
0.85 pf lagging
.
E
i\
a
l
-o
(E
N
C]'
I
(U
o
Locus11,I X"
(circlecentreM)
c)
F
generator
Fig.4.32 Phasordiagramof synchronous
For zero excitation.i.e. lE.l = 0
- i Io X r' = Y ,
\
0.7
M
or
Io = jV,lX,
i.e. llol =lV)lXr leading at 90" to OM which correspondsto VARs/phase.
Considernow the chart shown in Fig. 4.33 which is drawn for a synchronous
machinehaving Xt = 1.43pu. For zero excitation,the currentis 1.01I.43- 0.J
pu, so that the length MO conespondsto reactivepower of 0.7 pu, fixing both
active and reactivepower scales.
With centre at 0 a number of semicircles are drawn with radii equal to
different pu MVA loadings.Circles of per unit excitation are drawn from centre
M with 1.0pu excitationcorrespondingto the fixed terminalvoltageOM. Lines
may also be drawn from 0 conesponding to various power factors but for
clarity only 0.85 pf lagging line is shown. The operationallimits are fixed as
fbllows.
Taking 1.0 per unit active power as the rnaximum allowable powel', a
horizontallimirline ubc is drawn throughb at 1.0 pu. It is assumedthat the
machineis ratedto gire 1.0 per unit active power at power factor 0.85 lagging
and this tixes point c. Limitation of the statorcurrentto the correspondingvalue
requiresthe limit-line to becomea circular arccd aboutcentre0. At point d the
rotor heatingbecomesmore important and the arc de is fixed by the maximum
excitationcurrent allowable,in this caseassumedto be lEtl = 2.40 pu (i.e.2.4
times ly,l). The remaininglimit is decidedby loss of synchronismat leading
power factors. The theoreticallirnit is the line perpendicularto MO at M (i.e.
d= 90o), but in practice a safety margin is brought in to permit a further small
increasein load belore instability.ln Fig. 4.33, a 0.1 pu margin is employed
and is shown by the curve afg which is drawn in the following way.
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I
,
9o.s
0.5
tl
, -,---
1.0
>- Reactivepower(pu)lagging
Leading
Fig. 4.33 operatingchartfor rargesynchronous
generator
,
Considera point h onthe theoreticallimit on the lETl= 1.0 pu excitations
arc,
the pcrwerMh is reducedby 0.1 pu to Mk; the op"ruiing point must,
however,
still be on rhe
on the desiredlimiting curve. This is repeatedfor other excitationsgiving
the
curve afg. The completeworking area.shown shaded.is gfabcde.
,{ working
point placed within this area at once defines the MVA, Mw, MVAR,
current,
power factor and excitation.The load angle 6 canbe measuredas
shown in the
figure.
4.7
REPRESENTATION
OF LOADS
Load drawnby consutnersis the toughestparameterto assess
scientifically.The
magniiudeof the ioad, in iact, changescontinuouslyso that the load forecasting
problern is truly a statistical one. A typical daily load curve
is shown in
Fig' 1.1. The loads are generally composed of industrial and domestic
components.An industrial load consistsmainly of largethree-phase
induction
nlotors with sulficient load constancyand predictableduty .y.lr, whereas
the
domestic lclad mainly consists of lighting, heating and many single_phase
devicesusedin a randomway by houscholders.
The designun6 upJrotionof
power systemsboth economically and electrically are greatly influenced
by ttrp
nature and magnitude of loads.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.sa
I
._IaZ
I
Modern Power SvstemAnatr-rsic
Representation
of p
In representationof loads for various systemstudies such as load flow and
stability studies,it is essentialto know the variationof real and reactivepower
with variation of voltage. Normatly in such studies the load is of composite
nature with both industrialand domesticcomponents.A typical composition
of
Induction motors
55-757o
Synchronousmotors
5:75Vo
Lighting and heating
20-30Vo
Though it is always better to consider the P-V and
Q-V characteristicsof
each of these loads for simulation, the analytic treatment would be
very
cumbersome and complicated. In most of the analytical work one
of the
following three ways of load representationis used.
(i) Constant
BasekV = 24, line-to_line
Load volt ASa=
(24)z
Load current = generatorcurrent
Io= 7 pu, 0.9 pf lagging
= 0.9 - 7 0.436 pu
(a) Excitation emf (see Fig. 4.Ig)
Representation
Ef = V,+ j XJ"
= 1 1 0 " + j 1 . 3 4 4( 0 . 9 - j 0 . 4 3 6 )
"'l(6-0)
I' = P : i Q - -t n
V{<
where V = lVl 16and 0= tan-l QlP is the power factor angle. It is known
as constantcurrent representationbecausethe magnitudeof current is regarded
as constantin the .study.
Impedance
= 1 . 5 8 6- j l . 2 l = 1 9 9 1 3 7 . 1 "
E, (actual) = 1.99 x 24 = 47.76 kV (line)
6= 3j. 1" ( leading)
(b) Reactive power drawn by load
Representation
Q = VJ,, sin r/
This is quite olten used in stability studies.The load specified in
MW and
MVAR at nominal voltageis used ro compurethe load impedance(Eq. (4.?2b)).
Thus
I "
z = !I: - P
w =* J O -
= 1.344
- ' r - pu
Full load (MVA) = I pu, 0.9 pf lagging
Here the load current is given by Eq. (4.17), i.e.
(iii) Constant
= 1 pu
Synchronous
reactance
X," = +#
Power Representation
Current
)L
i
This is usedin load flow studies.Both the specifiedMW and MVAR are
taken
to be constant.
(ii) Constant
JlfZJ.t
Sotution
BaseMV A = 645,3_phase
= 1 x I x 0.436 = 0.436 pu or 0.436 x
M5
= 281 MVAR
l v l 2- t
P-JQ:T
which then is regarded as constant throughout the study.
f ll Fvattrt
F^sr..lsrv
I
--li
.---l
a e
zrv
I
I
t
-
-
T
A synchronousgeneratoris rated 645 MVA , 24 kv,0.9 pf lagging. It has
a
syrrchronous
reactancel.z o. The generatoris feeding full load-at 0.9 pf
lagging at rated voltage.Calculate:
(a) Excitation emf (E1) andpower angle 6
(b) Reactivepower drawn by the load
Carry out calculationsin pu form and convert the result to actual values.
The generator of Example 4.3 is carrying full load
at rated voltage but its
excitationemf is (i) increasedby 20vo and (ii) reducedby
20vo.
Calculatein each case
(a) load pf
(b) reactive power drawn by load
(c) load angle 6
Solution
Full load, P _ 1 x 0 . 9 = 0 . 9 p u
Ef=
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r.99
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
lM
tation of Power
ftrodernPower System Analysis
I
(b)
V,=7
(i)
Et is
x 0.9 x sin 1.5= 0.024
or
d by 20Voat samereal load.Now
Q = 0.024x 645= 15.2MVAR
As per Eq. (a.28)
P=
l E', l l v', l
x,
0.9= (2'388xr) ,i' d
\ 1.344)
or
sin d= 0.5065
or
6 = 30.4"
2.388130.4"-110"
jr.344
= 0 . 8 9- j 0 . 7 9 = 1 . 1 8 3l -
PROB
IEIvIS
(i)
sind
4'r Figure P-4'l shows the schematic
diagram of a radial transmission
system' The ratings and reactancesof the various
componentsare shown
rherein.A load of 60 MW at 0.9-power factor
ragging is tappedfrom the
66 kv substation which is to be mainraineo
alt oo kv. calcurate rhe
terminal voltage of the synchronousmachine.
Representthe transmission
line and the transforrnersby series reactances
ontu. .
11t220kv
220t66kV
kv
4L2"
V1
(a) pf = cos 4I.2" = 0.75 lagging
(b)
=1x1.183x0.659
= 0.78 pu or 502.8 MVAR
E, decreasedby 20o/oor
E f = 1 . 9 9x 0 . 8 = 1 . 5 9
Substitutingin Eq. (i)
o e =( L 2 \ L ) , i n 6
\ 1.344 )
4.2 Draw the pu inrpedancediagram fbr the
power system shown in nig. e4.2. Neglectresistance,and usea baseof
ioo vrre , 220kv in 50 () rine.
The ratings of the generator,motor and transformers
are
Generator40 MVA, 25 kV, Xu = 20Vo
Motor 50 MVA, I I kV, X,t = 30Vo
Y-Iltransformer, 40 MVA, 33 y_220 y kV,
X = I5Vo
Y-l transformer,30 MVA, ll L_220 y
kV, X = l5*o
,50o
U-vGi-r+1,-__iF--r-F
.l
-.*fF-.,
l - - Y Y -I
[y^
which gives
6 - 49.5"
t.59149,5"-rlo"
.i1.344
=0.9-j0.024
In=
= 0.9 l-1.5"
(a) pf = cos 1.5" = 1; unity pf
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60 MW
0.9pf tagging
Fig. p_4.1
Reactive power drawn bY load
Q = lV,lllolsin /
(ii)
100MVA
X = 10o/o
160
dh
_-__-EF_-----l___>
5F
|
100MVA
V2
X = Bo/o
-1f
1 \M ) 'Y )
l_
2
/
I
Fig. p-4.2
4 . 3 A synchronousgenerator is rated 60 MVA,
1r kv. It has a resistance
Ro = 0-l pu and xo
7 r.65 pu. It is feedinginto an infinite bus bar ar
11 kV delivering a cirrrent 3.15 kA at 0.g
ff hgging.
(a) Determine E, and angte d
(b) Draw a phasor diagram for this operation.
(c) Bus bar voltage fails to 10 kv while
the mechanicalpower input to
generator and its excitation remains unchanged.
wtrat is the value
and pf of the current delivered to the bus.
In this case assumethe
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
126 |
ModernPower SystemAnalysis
T
generatorresistanceto be negligible.
4.4 A 250 MVA, 16 kV ratedgeneratoris feedinginto an infinite bus bar at
15 kV. The generator
hasa synchronous
reactance
of 1.62pu.lt is found
that the machine excitation and mechanicalpower input are adiustedto
give E, = 24 kY and power angle 6 = 30o.
(a) Determine the line current and active and reactivepowers fed to the
bus bars.
(b) The mechanicalpower input to the generatoris increasedby 20Vo
from that in part (a) but its excitation is not changed.Find the new
line current and power factor.
(c) With referenceto part (a) current is to be reduced by 20Voat the
same power factor by adjusting mechanical power input to the
generatorand its excitation. Determine Ey, 6 and mechanicalpower
rnput.
(d) With the reducedcurrent as in part (c), the power is to be delivered
to bus bars at unity pf, what are ttre correspondingvalues of El and
d and also the rnechanicalpower input to the generator.
4.5 The generatorof Problern4.4 is feeding 150 MVA at 0.85 pf lagging to
infinite bus bar at 15 kV.
(a) Determine Ey and d for the above operation.What are P and Q fed
to the bus bars?
(b) Now E, is reducedby l0o/okeeping mechanicalinput to generator
same, find new dand Q delivered.
(c) Et is now maintainedas in part (a) but mechanicalpower input to
generatoris adjustedtill Q = 0. Find new d and P.
(d) For the value of Eyin part (a) what is the maximum Qthat can be
deliveredto bus bar. What is the corresponding6and {,? Sketchthe
phasor diagram for each part.
Representationoflgygr_gyg!"_!L_qg[p!g$q
-
REFERE
I.ICES
Books
l' Nagrath,I'J. and D.P- Kothari, Electric Machines,2nd
edn Tata McGraw-Hill,
New Delhi, 7997.
2' van E' Mablekos,Electric MachineTheory PowerEngineers,Harperlno
for
Raw,
New York, 1980.
3. Delroro, v., Electric Machines and power systems,prentice_Hall,
Inc., New
Jersey,1985.
4' Kothari, D.P. and I.J. Nagrath, Theory and.Problems
of Electric Machines, 2nd
Edn, Tata McGraw-Hill, New De\hi, 2002.
5. Kothari, D.p. and I.J. Nagrath, Basic Electicar Engineering,
2nd Edn., Tata
McGraw-Hill, New Delhi. 2002.
Paper
6' IEEE CornitteeReport,"The Effect of Frequencyand Voltage
on power System
Load", Presentedat IEEE winter po,ver Meeting,New york,
1966.
Answers
4.1 12kV
4.3 (a) 26.8 kV (line),42.3" leading
(c ) i .i 3 l -2 8 .8 " k A; 0 .8 7 6l a g
4.4 (a) 0.5ll"l- 25.6"kA; 108 MW, 51.15MVAR
(b) 6.14 kA, 0.908lagging
( c ) 1 . 5 7 8 ,1 3 . 5 o5, 3 . 3 M W
( d ) 1 8 . 3 7k v , ' 3 5 . 5 " ,9 6 M W
4. 5 (a ) 2 5 .2 8 k V, 2 0 .2 ' ,1 2 7 .5MW , 79.05MV A R
(b) 33.9", 54 14 MVAR
(c ) 4 1 .1 " , 1 5 0 .4MW
(d) 184.45,MVAR,53.6",- 7 0.787 pu
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ehara-cteristics
and Perforr"nance
of povrerTransmissionLines
The following nomenclaturehas been adoptedin this chapter:
z = seriesimpedance/unitlength/phase
y = shunt admittance/unitlength/phaseto neutral
;=;l_*#:,T.":Jff::
C = cepacitance/unitlength/phaseto neutral
/ = transmissionline length
Z = zl = total series impedance/phase
Y = ll = total shunt admittance/phaseto neutral
Note: Subscript,Sstandsfor a sending-endquantityand subscriptR standsfor
a receiving-end quantity
5.1
INTRODUCTION
5.2
This chapter deals primarily with the characteristics and performance of
transmissionlines.A problemof major importancein power systemsis the flow
of load over transmissionlines such that the voltage at various nodes is
maintained within specified limits. While this general interconnectedsystem
problem will be dealt with in Chapter 6, attention is presently focussed on
performance of a single transmissionline so as to give the reader a clear
understandingof the principle involved.
Transmissionlines are normally operatedwith a balancedthree-phaseload;
the analysiscan thereforeproceedon a per phasebasis.A transmissionline on
a per phasebasis can be regardedas a two-port network, wherein the sendingend voltage Vr and current 15are related to the receiving-endvoltage Vo and
current 1o through ABCD constants"as
lyrl lA BlIy*l
L1,J Lc plLroJ
(sr)
SHORT TRANSMISSION
For short lines of length 100 km or less, the total 50 Hz shunt admittance*
QwCl) is small enoughto be negligible resultingin the simple equivalentcircuit
of Fig. 5.1.
Fig. 5.1 Equivalent
circuitof a shortline
This being a simple series circuit, the relationship between sending-end
receiving-end voltagesand currents can be immediately written as:
l y r l =I l
/< t\
These constants can be determined easily for short and medium-length lines
by suitable approximations lumping the line impedance and shunt admittance.
For long lines exact analysis has to be carried out by considering the
distribution of resistance, inductance and capacitance parameters and the ABCD
constants of the line are determined therefrom. Equations for power flow on a
line and receiving- and sending-end circle diagrams will also be developed in
this chapter so that various types of end conditions can be handled.
*R"f'..
zflvol
(5'3)
Lr,J Lo rJLr-.J
Also the following identity holds for ABCD constants:
AD-BC=l
LINE
The phasor diagram for the short line is shown in Fig. 5.2 for the lagging
current case.From this figure we can write
lV5l = l(ly^l cos /o + lllR)z + (lV^l sin dn + lllxyzlr/2
l v 5l = [ t v R P +l l t 2 ( R z+ f )
* z t v R | i l ( R c o sQ ^ + X s i nw
f-{u ) r l 2 ( 5 . 4 )
. r l I/ t 2
1il21n2+ x2 r l
?JlJr /o
= tvnt
/ A + 4#rin
, I / i - " Qpt
" Lf r * l V R lcos
tvRP
J
-;- -
to Appenclix B.
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For overhead transmissionlines, shunt admittanceis mainly capacitive susceptance
(iwcl) as the line conductance(also called leakance)is always negligible.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Lines J"#,
Characteristics
and Performanceof PowerTransmission
--1
l/l R cos/^+l1l X sind^
x 100
I yRl
The last term is usually of negligible order.
(s.7)
In the above derivation, Q*has been consideredpositive for a lagging load.
Expanding binomially and retaining first order terms, we get
Ivsl=lv^rfr.
ff
cos
/^ +
ff
per cenrregularion- |!!4 9"-t!t:l n x tin 4-r x 100
lvRl
sinf^)'''
l V 5l = l V ^ l + l X ( R c o s / o + X s i n / o )
The above equation is quite accurate for the normal load range.
(s.8)
(for leading load)
(s.s)
Voltage regulationbecomesnegative(i.e. load voltage is more than no load
voltage), when in Eq. (5.8)
X sin Qo> R cos /p, or tan </o(leading) >+
X
It also follows from Eq. (5.8) that for zero voltage regulation
tan/n=
i.e.,
X
/^(teading=
[-
=cot d
e
(5.e)
where d is the angle of the transmission line impedance. This is, however, an
approximate condition. The exact condition for zero regulation is determinedas
follows:
Fig. 5.2 Phasordiagramof a shortlinefor laggingcurrent
Voltage
Regulation
Voltage regulationof a transmissionline is defined as the rise in voltage at the
receiving-end,expressedas percentageof full load voltage, when full load at a
specifiedpower factor is thrown off, i.e.
Per cent regutation=
where
''^?)-:'Yu'x
lvRLl
100
lVool= magnitude of no load receiving-endvoltage
lVprl = magnitude of full load receiving-endvoltage
(at a specified power factor)
= lVsl, lVsl = lVpl
For short line, lV^61
Per cent regutation=
'u1|,'%'
(5.6)
condition
Fig. 5.3 Phasordiagramunderzeroregulation
Figure 5.3 shows the phasor diagram under conditions of zero voltage
regulation, i.e.
lV5I = lVpl
or
O C= O A
AD - AClz -_ llllzl
sn /AOD _
zlvRl
lyR|
oA
lvRl
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
and Performanceof PowerTransmissionLines
Characteristics
',
or
IAOD= sin-rU!4
zlvRl
lt follows from the geometryof anglesat A, that for zero
voltaeeregulation,
/p (leading) =
or
new value of lVr | = 11.09 kV
Figure 5.4 showsthe equivalentcircuit of the line with a capacitivereactance
placed in parallel with the load.
R+jx
(5.10)
From the above discussion it is seen that the voltage
regulation 6f a line is
heavily dependent upon load power factor. voltage
regulation improves
(decreases) as the power factor of a lagging
load is increased and it becomes
zero at a leading power factor given by Eq. (5.10).
A single-phase50 Hz generatorsuppliesan inductive load
of 5,000 kw at a
power factor of 0'707 lagging by means of an
overheadtransmissionline
20 km long. The line resistanceand inductanceare
0.0195ohm and 0.63 mH
per km' The voltageat the receiving-endis required
to be kept constantat 10
kv.
Find (a) the sending-endvoltage and voltage regulation
of the line; (b) the
value of the capacitorsto be placed in parailet viittr
the load such that the
regulation is reducedto 50vo of that obtained in part (a);
and (c) compare the
transmissionefficiency in parrs (a) and (b).
Solution The line constantsare
R = 0 .0 1 9 5x 2 0 = 0.39 f)
l
l V o l + l 1 l ( R c o sQ * + X s i n / ^ )
(b) Voltage regulariondesired= ?+t
-l
Assuming cos y''^now to be the power factor of load and capacitive reactance
taken together, we can write
(11.09- 10) x 103= l1n| ( R cos d^+ X sin dn)
Since the capacitancedoes not draw any real power, we have
l/ol=
5000
(ii)
10 x cos/^
Solving Eqs. (i) and (ii), we get
cos dn= 0'911 lagging
and
l l a l = 5 4 9A
Now
Ic= In- I
= 0.29 + j273'7
'
'Y
^L
0.707,\y
or
(c)
:ltl -loxlooo
3wxc
ll. I
273'7
C-81 P'F
Efficiency of transmission;
= l0.9Vo
Case (a)
lyst- 10
= 0.109
t0
www.muslimengineer.info
(i)
Note that the real part of 0.29 appearsdue the approximationin (i) Ignoring it,
we have
I, = j273.7 A
From Eq. (5.5),
Voltage regulation= pfTL-- l9- x roo :Zt.j7vo
10
-
= 549( 0. 911- j0'412)- 707( 0. 107 - j0. 70'7)
l 1 l= - - 5 0 0 0 = 7 0 7 A
1 0 x 0 .7 0 i
= 72.175kV
!
-,L-
Fig. 5.4
(a) This is the case of a short line with I = Ia=
1, given by
= 1 0 ,0 0 0+ 7 0 7(0.39x 0.701 + 3.96 x
l
l-
X = 3I4 x 0.63 x 10-3x Z0 = 3.96 e
l V 5l =
133
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
characteristicsand Pedormanceof Power TransmissionLines LI{S
Case (b)
f*
Per unit transformerimpedance,
r/=
'
5000
- g7.7%o
'
5000+ (549)2
x 0.39x 10-3
/u Lrr6f.LuJ prdvrtg
, uapacltor
ln parallel
wlth
Zr=
the load,
the
receiving-endpower factor improves (from 0.707 iug to
0.911 lag), the line
current reduces(from 707 A to 549 A), the line voitage
regulationdecreases
(one half the previous value) and the transmission
i-proves (from
"ffi"i"nJy
96'2 to 97'7vo)' Adding capacitorsin parallel with load
is a powerful method
of improving the performanceof a transmissionsystem and
will be discussed
further towards the end of this chapter.
A substationas shown in Fig. 5.5 receives5 MVA at 6
kv, 0.g5lagging power
factor on the low voltage side of a transforner from a power
stationthrough a
cablehaving per phaseresistanceandreactanceof 8 and
2.5 ohms,respectively.
Identical 6.6/33 kV transfoffnersare installed at each
end of the line. The
6'6 kV side of the transfonnersis delta connectedwhile the
33 kV side is star
connected.The resistanceand reactanceof the star connected
windings are 0.5
and 3'75 ohms, respectivelyand for the delta connected
windings arJ0.06 and
0.36 ohms. what is the voltage at the bus at the power
station end?
6.6/33kV
Fig. 5.5
BaseMVA = 5
= 33 on high voltage side
Cabie impeciance= (8 + jZ.S) e/phase
_ ( 8 + r 2 . s ): x s= (0'037+
io'0r15)Pu
(33)'
Equivalent star impedance of 6.6 kv winding of the transformer
f
Total seriesimpedance= (0.037 + j0.0115)+ 2(0.0046+ j0.030)
= (0.046 + j0.072) pu
Given: Load MVA = 1 pu
= 0. 91 pu
Loadvolt age= +
6. 6
Load current= -.1- = 1.1 pu
0.91
Using Eq. (5.5),we get
lVs| = 0. 91 + 1. 1( 0. 046x 0. 85 + 0. 072x 0. 527)
= 0.995 pu
= 0.995 x 6.6 - 6.57 kV (line-to-line)
choose,
I
I
2,000
kw I
at 0.8 pf *Vs
Base kV = 6.6 on low voltageside
=
= (0.0046+ j0.030) pu
Input to a single-phase
short line shown in Fig. 5.6 is 2,000kw at 0.8 lagging
power factor. The line has a seriesimpedanceof (0.4 + j0.a) ohms. If the load
voltage is 3 kV, find the load and receiving-endpower f'actor.Also tind the
supply voltage.
33/6.6kV
Solution lt is convenient here to employ the per unit method.
Let us
( 0 . 0 2 +j 0 . 1 2 ) x 5 ( 0 . 5 + 1 3 . 7 5 ) x 5
_
6.q2
Q'2
(O.OU
= (0.02+ /0.12) O/phase
+ 7O.36)
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lassins
3kv
L
__1
I
Fig. 5.6
Solution It is a problem with mixed-end conditions-load voltage and input
power are specified.The exact solution is outlined below:
Sending-endactive/reactivepower = receiving-endactive/reactivepower +
activekeactiveline losses
For active power
l ys I l1l cos ds= lVRllll cos / a + l1l2p
For reactivepower
( i)
l y s I l / l s i n g 5 5 =l V p ll 1 l s i n Q o + l t l 2 X
(ii)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern PowerSystem Anatysis
'F6, I
Squaring(i) and (ii), adding and simplifying, we get
5.3
lvrl2 lll2 = lVnlzlll2 + zlvRl lll2 (l1lR cos /o
+ tItX sh /n) + tlta @2 + f)
(iii)
For lines morethan 100 km long, chargingcurrentsdueto shuntadmittance
100 km to 250 km leneth. it is
line
admittance at the receiving-end
the
all
to
lump
accurate
sufficiently
resulting in the equivalent diagram shown in Fig. 5.7.
Starting frorn fundamental circuit equations,it is fairly straightforward to
write the transmissionline equationsin the ABCD constantform given below:
the numerical values given
l Z l 2 =( R 2 +f ) = 0 . 3 2
lysl l1l-
Lines l. l3?.,-{
of PowerTransmission
and Performance
Characteristics
IMEDIUM TRANSMISSION I,INE
2,oo-ox1o3
- 2 , 5 0 0x 1 0 3
0.8
[:]=l'*;'llVl
lVs I l1l cos /, - 2,000 x 103
( s . 11 )
l y s l l l l s i n / 5 = 2 , 5 0 0x 1 0 3x 0 . 6 = 1 , 5 0 0x 1 0 3
From Eqs. (i) and (ii), we get
2 0 0 0 x 1 0 3- 0 . 4 l t P
,n
l 1 cl o s P o = F
(iv)
Fig. 5.7 Mediumline,localizedload-endcapacitance
1 5 0 0 x 1 0- 30 . 4 t I f
3000
l1l sin /o
Substitutingall the known valuesin Eq. (iii), we have
p'{lt
(2,500x 103;2
= (3,000)'til2+ 2 x 3,000t|210.4*29W"19!
3000
L
1s00xlq1r0.4112
+0.4x
l+ o.zztt ta
3000
J
(v)
Nominal-f
Representation
If all the shunt capacitanceis lumped at the middle of the line, it leads to the
nominal-Z circuit shown in Fig. 5.8.
Simplifying, we get
0 . 3 2 V f - 1 1 . 8x 1 0 6l l l 2 + 6 . 2 5 x 1 0 1 2= 0
which upon solution yields
t|_725 A
Substitutingfor l1l in Eq. (iv), we ger
cos Qp= 0.82
Load Pn = lVRl lll cos /a = 3,000 x 725 x 0.82
= 1,790kW
Now
l V 5 | = l 1 l c o s d s = 2 ,000
l V 5l =
2000 :3 .44
kY
Fig. 5.8 Mediumline,nominal-Trepresentation
For the nominal-Z circuit, the following circuit equationscan be written:
Vc= Vn+ I o( Zl2)
Is = In + VrY = In -r Wo+
IR(Z|L)Y
Vs = Vc + it (ZiZ)
Substitutingfor Vg and 1, in the last equation, we get
YvR)
.
vs= vn+ I^ (zt2)+ (zD)
[r^(t +)+
= v n ( , . t { ) + rn z ( t . t f )
7 2 5 x 0 .8
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
l.
-r
Characteristicsand Performanceof Power Tralq4lqslon Lines [l$k
t
Rearrangingthe results, we get the following equations
+
MVA at 0.8 lagging power factor to a balancedload at 132 kV. The line
conductorsare spacedequilaterally3 m apart. The conductorresistanceis 0.11
ohmlkm and its effective diameteris 1.6 cm. Neglect leakance.
(s.12
2
= 0.0094 pFlkrrr
Nominal- zr Representation
In this methoc the total line capacitanceis divided into two
equal parts which
are lumped at the sending- and receiving-endsresulting
in the nominal- zr
representation
as shown in Fie. 5.9.
R = 0. 11 x 250 = 27. 5 C)
X - Zr f L = 2r x
50 x I . 24 x 10- 3x 250 = 97. 4 Q
Z= R + jX = 27.5 + j97^4 = I0I.2 174.T Q
Y = jutl
= 314 x 0.0094 x 10{ x 250 lg0"
= 7.38 x 104 lW
l--i6.g"o = 109.3I -369" A
r^ = #9
"13xl32
vo (per phase)= (I32/d, ) 10" = 76.2 l0 kv
Fig. S.9 Mediumline,nominal_7r
representation
From Fig. 5.9, we have
/
_
t
Is= In + -VoY +
tvoy
+
\
\
2 ) "
r
1
I i +: x 7.38x io-a 190"x10r.2134.2"\to.z
\
2
)
+ 101.2174.2" x 109.3x 10-3l-36.9"
t;trr-(t+|vz)
v o v( t + t ^ v z ) . + ( , +
1
v s =[ t + + Y Z l v R +z I *
l
2Vsy
V s = vo+ eo*,)vov>z
= vn(r.*
1s=r**
U
7 6 . 2+ 2 . 8 5l 1 & . 2 ' + 1 1 . 0 61 3 7 . 3 "
82.26+ j7.48 - 825 15.2"
lV5| (line)= 82.6xJl - 143 kv
)vz)
I + L Y Z = 1 + 0.0374ll&.2" = 0.964+ 70.01
2
Finally, we have
z l,u.r
'r i4
-'t ',
I ('*
., ll -" |
|
/
|
f /
1
\
=
(rine
noload)
rv^or
(5.13)
LrU+;rt) [t*r")]L1nr
It slhou
uldJ bt el noted
d tha
at nominal-zand nominal-rrwith the aboveconstantsare
not equ
vale3nt to e
each
t other. The readershould verify this fact by applying star_
lurva
delt:a tr
trans
nsfcormati<
ion to
t either one.
Using the nominai-z- method, find the sending-endvoltage
and voltage
regulation of a 250 km, three-phase,50 Hz, transmission
line delivering 25
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= 148.3
kv
ffid:;#
t
z
l
Voltage regulation = W#2
5.4
x 100 - l2.3vo
THE LONG TRANSMISSION LINE-RIGOROUS
SOLUTION
For lines over 250 km, the fact that the parametersof a line are not lumped but
distributeduniformally throughoutits length, must be considered.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
mooetnPo*", Sv
-149 1
sis
characteristicsand Performanceof Power TransmissionLines
I
C17e)r'-Czle-)'-21.,
9t ,r, -C, ,-',',
Z,
Z,
(5.1e)
Fig. 5.10 Schematicdiagramof a longline
(5.20)
Figure 5.10 shows one phase and the neutral return (of zero impedance)
of
a transmission line. Let dx be an elemental section of the line at u dirtunr.
"
from the receiving-end having a series impedance zdx and a shunt admittance
yd-r. The rise in voltage* to neutral over the elemental section in the
direction
of increasing "r is dV". We can write the following differential relationships
acrossthe elemental section:
dVx = Irzdx o, Y-
= ZI,
The constantsC, and C2 may be evaluatedby using the end conditions, i.e.
when .r = 0, Vr= Vn and 1r= In. Substitutingthese values in Eqs. (5.17) and
(5.19) gives
Vn= Cr + Cz
(ct- cz)
r^=!
Lc
(s.14)
which upon solving yield
d,lx= v*ldx o,
!1'
= yvx
(s.1s)
It may be noticed that the kind of connection (e.g. T or r) assumedfor the
elemental section, does not affect these first order differential relations.
Differentiating Eq. (5.14) with respecrro -tr,we obtain
Substituting
the valueof + from Eq. (5.15),we ger
dx
Ea
= rZv'
Cr=
*
(vn+zJn)
1
2Un-
(5.16)
This is a linear differentiai equation whose general solution can be written
as follows:
V*=CpI**Cre-1x
(s.r7)
7= ,lW
(s.18)
where
and C, and C, are arbitraryconstantsto be evaluated.
DifferentiatingEq. (5.17)with respectto x:
,,=(Yn+/o),,.
*(h.? ),-,.
,.-_(bITk),,._(w*),-,.
Here V' is the complex expressionof the rms voltage, whose magnitude and phase
vary with distance along the line.
www.muslimengineer.info
(s.2r)
Here Z, is called the charqcteristic impedanceof the line and 7is called the
propagatton constant.
Knowing vp, In and the parametersof the line, using Eq. (5.21) complex
number rms values of I/, and I, at any distance x along the line can be easily
found out.
A more convenientform of expressionfbr voltageand currentis obtainedby
introducing hyperbolic functions. Rearranging Eq. (5.21), we get
( o7* +
o-7'\
v*=vnt+l+
\
*-_-
ZJ*)
with 9r ant c, as determinedabove,Eqs. (5.r7) and (5.19) yield the
solutionfor V.-and 1. as
drv,
-d-T
= dI.
i''
d2v
,r=
I. = VoL(
"
" 2 " \ "*
2
-.t-"
2
/
/
+
r"^' 2( , ( e : J : : )
2
)* ," (
"\
)
e1' + e-t'
\
2
)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
)
Characteristics
and Performance
of PowerTransmission
Lines ffi
-
These can be rewritten after intloducing hyperbolic functions, as
Vr= Vn cosh 1r + I^2, sinh 1r
sinh
7/- |.+*4*
3!
(5,22)
..=Jyz(H+)
)!
\^'
This seriesconvergesrapidly for values of
I,= Io cosh rr + V- -l-.i
lines
and
can
7t usually encountered
for power
-r,L-
qrrrrrnwi-o.l.r
he convenie-nflw
\(s.28a)
6.)
^l-^i^-.^
--
expressionsfor ABCD constanis art
when x = l, Vr= V, Ir= Is
':::i;:l;;t
=l;::,,
Hl
(s.23)
Bx,z (t.+)
Here
(s.24)
c = J-sinh :r/
Z,
*In
\
6 )
The above approximation is computationally convenient and quite
accuratefor
lines up to 400/500 km.
Method
case [vs 1s] is known, fvn Inl can be easily found by inverting Eq. (5.23).
-:l
=l-i,
Hl
[:]
Evaluation
(s.2s)
of ABCD Constants
7- a+ jp
(s.26)
The hyperbolic function of.complex numbers involved in evaluating ABCD
constantscan be computedby any one of the three methods given uJtow.
Method
3
cosh (o/ + i7t) -
tat"iot +-'-?t'-iot
1
Z
;(e"
sinh (o/ + jpl) -
,at"i0t _e:de-ipt
5.5
The ABCD constantsof a long line can be evaluatedfiorn the results given
in
F4. 6.24). It must be noted that 7 -Jw
is in generala complex number and
can be expressedas
^
(s.28b)
c x Y(r*V\
A=D -cosh 7/
B = Z, sinh 7/
YZ
2
A=D=l*
INTERPRETATION
_=
tpt + e-"1 l-Bt1
(s.2e)
:
){r",
tpt - t-d l-Bt1
OF THE LONG LINE EOUATIONS
As alreadysaid in Eq. (5.26), 7is a comprexnumber which can be expressed
as
7= a+ jp
The real part a is called the attenuation constantand the imaginary part
^
Bis
called the phase constant.Now v, of Eq. (5.2r) can be writtin as
V^, = I V R +
|
I
Z,lRlr.,,ritgx+n
*lVn-Z,lnl
2
|
|
z
o--c,x,-tt/,x-e2)
I'
where
cosh (cr./+ j1l) = cosh ul cos gl+ j sinh a/ sn pt
(5.27)
sinh (a/ + jQl) = sinh al cos gl + j cosh a/ sn pt
Note that sinh, cosh, sin and cos of real numbers as in Eq. (5.27) can be
looked tp in standard tables.
Method
Qr= I (Izn+ I&,)
v* (t) = xd Ol!- + 3/tl
L
t
2
2
(5.30)
dz= I (Vn- InZ,)
The instantaneous
voltagev*(t) canbe writtenfrom Eq. (5.30)as
,o' ,i@t+r,,.+o,)
l
cosh
7/=r ++* #*. .=(t.+)
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
nrodern Power System Analysis
144 |
- ZrI
characteristics
and Performanceof PowerTransmissionLines
'1
*l ^-* -j(at-/tx+t
, nlV^
+Jlltrle-*ett''-tLt-rh)
(5.31)
I
voltageconsistsof two termseachof which is a function
The instantaneous
of two variables-time and distance.Thus they representtwo travelling waves,
Vx=
Vrl *
(s.32)
VxZ
- A t ( f+ 4 0
Sendingendx=/
Now
- JIlh-?\,*
"',,
gx+ h)
"or{'l+
(5.33)
At any instant of time t, v.rl is sinusoidally distributed along the distance
from the receiving-endwith amplitude increasingexponentially with distance,
as shown in Fig. 5.11 (a > 0 for a line having resistance).
Envelop eox
Fig. 5.12 Reflectedwave
If the load impedance Zr = +
IR
.
u-rAf
B
-x=0Receiving
end
-'- ' Direction of
travellingwave
Sendingendx=/
w ave
F i g . 5 .1 1 l nci dent
After t-imeAt, the distribution advancesin distancephaseby (u'Atlfl. Thus
this wave is travellingtowarclsthe receiving-endand is the incidentv'ave'Line
lossescauseits arnplitucleto decreaseexponentiallyin gclingl}onr tht: sendirlg
to the receiving-end.
Now
u,r=Elt+tlu^
- 0x+dz)
(az
cos
(s.34)
After time At the voltage distribr-rtionretardsin distancephaseby (uAtl4.
This is the reflecterlwave travelling trom the receiving-endto the sending-end
with amplitudedecreasingexponentiallyin going from the receiving-endto the
as shownin Fig' 5.12.
sending-end,
point
along the line, the voltage is the sum of incident and reflected
At any
voltage waves presentat the point tEq. (5.32)1.The same is true of current
waves.Expressionsfor incident and reflected current waves can be similarly
written down by proceedingfrom Eq. (5.21).If Z" is pure resistance,current
wavescan be simply obtainedfrom voltage wavesby dividingby Zr.
www.muslimengineer.info
= 2,, i.e. the line is terminated in its
characteristic impedance,the reflected voltage wave is zero (vn- zJn= 0).
A line terminatedin its characteristicimpedanceis called the infinite line.
The incident wave under this condition cannot distinguish betweena termination and an infinite continuation of the line.
Power systemengineersnormally caII Zrthe surge impedance.It hasa value
of about 400 ohms for an overhead line and its phase angle normally varies
from 0" to - 15o.For undergroundcables Z. is roughly one-tenthof the value
for overheadlines. The term surge impedanceis, however, used ih connection
with surges(dueto lightningor switching)or transmission
lines.wherethe lines
loss can be neglectedsuch that
z, = z,=
(
;,.,1
l:i:r)
rl/2
i/1
(;)"'. ,, purcrcsist''cc
SLtrge Impetlance Loading (SIL) of a transmission line is.defined as the
power delivered by a line to purely resistive load equal in value to the surge
impedance of the line. Thus for a line having 400 ohms surge impedance,
SIL= JT -y!tv, I x looo
lvo
t00okkw
J3 x aoo
= 2.5 lyRl2kw
(s.3s)
where lVol is the line-to-line receiving-endvoltagein kV. Sometirnes,it is
found convenientto expressline loading in per unit of SIL, i.e. as the ratio of
the power transmittedto surge impedanceloading.
At any time the voltage and current vary harmonically along the linc with
respectto x, the spacecoordinate.A completevoltageor currentcycle along the
line correspondsto a change of 2r rad in the angular argument Bx. The
correspondingline length is defined as the wavelength.
It 0 i.s expressedin radlm,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
-iAd
147
ModernPo*g!Sy$e!l Anelysis
I
(s.36)
)-Zn/gm
line
power
transmission
a
typical
for
Now
length)= 0
g (shuntconductancelunit
*fu=
(s.42)
oflight
velocitv
of wave along
The actual velocity of the propagation
of light'
somewhatless than the velocity
7= (yz)1/2= Qu,C(r+ juL))rt2
- iu (LQ'''(r- t
' =
the line would be
., = 1" lot = 6,000km
"50(usuallya few
lines are much shorterthan this
Practicaltransmission
tlrau'ttin
i)'''
pointed out heyethat the vyaves
hundredkilometres). It needsto be
Figs,5.11and5.]'2areforitlustrationonlyanddonotpertainnareal
power transmissionline'
7= a+ jg = ju\Lc),,,(t-t#)
r ( C\ltz
a - - l - l
(s.37)
0 = a (Lq'''
(s.38)
2\L)
Now time for a phasechangeof 2n is 1f s, where/= cul2r is the frequency
in cycles/s.During this time the wave travels a distanceequal to ). i.e. one
wavelensth.
\
'
(5.3e)
Velocity of propagationof wave, , = -4=: f^ m/s "'
ri f
at the
line is 400 km long' The voltage
A three-phase50 Hz transmission
=
ohm/
0.4
x
are r = o.!25 ohnr/km,
sending.end ts 220kV. The line parameters
tm ani y = 2.8 x 10-6rnho/km'
Find the following:
(i )Thesending- enclcur r ent andr ebeiving- endvolt agewhent her eisno- load
which is a well known result.
For a losslesstransmissionline (R = 0, G = 0),
,= 7yz)''' - iu(LC)tlz
such that e. = A, 0 = . (Lq'''
)-2110- ,?n=,=:
1,,- m
(s.40)
are:
Solution The total line parameters
R - 0.125 x 400 = 50.0 f)
X = O.4 x 400 = 160'0 fl
and
(s.41)
v = fA = ll(LC)rlz m|s
U
- l'12 x IA-3 lxf
Y = 2.8 x 10-6 x 400 lg)"
=
172'6" Q
Z = R + i X = ( 5 0 ' 0+ j 1 6 0 ' 0 ) 1 6 8 ' 0
YZ = l.l2 x 1O-3/90" x 168 172'6"
For a single-phasetransmissionline
L=
C_
lto ,n D
r'
2r
= 0.1881162'6"
2ffi0
(i)
l n D /r
v' = 4
(Pol^D
It;,.t-
2*o
'lt"nG
)t/2
'
)
Since r and rt arequite close to each other, when log is taken, it is sufficiently
q, =
h D/r.
accurateto assumethat ln
www.muslimengineer.info
At no-load
Vs= AVn' ls = CVa
A and C are comPutedas follows:
/AL =
-
l*
tl
Y Z= l +
l
*x
0 . 1 8 81 1 6 2 ' , 6 "
2
)
= 0.91+ j0.028
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.r14E-f
Modern Power Svstem Analvsis
Characteristics.
and Performanceof Power TransmissionLines
l Al = 0 .9 1
simplifying, we obtain the maximum permissiblefrequency as
C = Y(l + YZ/6) =
I 14g^
t-
f = 57.9 Hz
= 1.09x 10-3I
=
lvnhn'
#:#
'
= 242kY
1 1 l5= l c l l V R l = 1.09 x 10-3xry
* 103= 152A
"J3
It is to be noted that under no-load conditions, the receiving-end voltage
(242 kV) is more than the sending-endvoltage. This phenomenonis known as
the Ferranti elfect and is discussedat length in Sec. 5.6.
(ii) Maximum permissible no-load receiving-end voltage = 235 kv.
" l l :2f2f0i = 0 . e 3 6
^ ,= llV%
(b) The incident and reflected voltages to neutral at
200 km from the
receiving-end.
(c) The resultant voltage at 200 km from the receiving-end.
Note: Use the receiving-endline to neutral voltage as reference.
solution
From Example 5.5, we have following line parameters:
r = 0.725 Qlkm; x = 0.4 Olkm; y = j2.g x 10{ Uncm
z = (0.125 + j0.4) Olkm = 0.42 172.6 CI/km
,r,
Now
y- 1yz)t'2= (2.8 x 10-6x 0.42 /.(g0 + 72.6))t/2
1
A = l + L Y Z
= 1 . 0 8x 1 0 - 3l g I . 3
2
1 ^
= 1 + -; t', .i2.8 x 10-6 x (0.125+
70.4)
z
S in c eth c i rn a g i n a ry
p a rt w i l l b c l e ssthan,
approxirnatedas
,l
th of the real part. lzll can be
?cl n 2
V,,
vR-zclR
-
)
VR
= 63.51 10" kV (to neurral)
(b) At 200 km from the receiving-end:
L( to+160,.I)
s0)
J0\
Neglecting the imaginary part, we can write
tAl= 1- +" r.r2x1o-3
x 160
x
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;
Vn I
Incident vcrltage-
Reflected voltage -
220=
0.88
250
I
A-l*;xi1.l2xl0-3x
1 . 0 6 gx l 0 - 3
- 220/J3 = 63.5110" kV (to
neurral)
2
r-0.936
0 . 5 6x l 0 6
/ = 3 3 8k m
,A,=
a = 0 . 1 6 3x t O - 3 ;f =
For open circuit 1n= 0
l A l = | - 0 . 5 6x l r J 6 P= 0 . 9 3 6
p_
= ( 0. 163+ i1. 068) x t O - - -r 0+ j{J
(a) At the receiving-end;
= (1 - 0._56x t0-6P1+ j0.t75 x r0-6P
;;,,
If in Example 5.5 the line is open circuited with a receiving-endvoltag
e of 220
kV, find the rms value and phase angle of the following:
(a) The incident and reflected voltages to neutral at the receiving-end.
&=
0.88
l
/
l
Incident voltage = :!u+urltlxl
2
lr:2oorm
= 63. 51exp ( 0. 163x l0- 3 x 200)
x exp 01.068 x 10-3x 200)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffi0'.I
characteristics
andperformance
of powerTransmission
LinesI rsr
/ v ^ , \
u"'J andturnsrhrough
a positiveanglepr (represenred
by phasoroB);
I i.
ModernPo*e, SystemAnatysis
f
= 65.62 112.2" kV (to neutral)
Reflectedvoltage-
Ys-"-'ur-itt'l
while the reflected voltage wave decreasesin magnitude
exponentiaily
l' ,,,n,u-
2
-Y-E--u,
It is apparentfrom the geometry of this figure that the
resultantphasor voltage
Vs QF) is such that lVol > lysl.
A simple explanationof the Ferranti effect on an approximate
basis can be
advancedby lumping the inductanceand capacitanceparameters
of the line. As
shown in Fig.5.14 the capacitanceis lumpld at the ieceiving-end
of the line.
= 6I.47 l-12.2" kV (to neutral)
(c)
Resultantvoltage at 200 km from the receiving-end
- 65.62 112.2" + 61.47 I - 12.2"
= 124.2+ j0.877 = 124.2 10.4"
Resultantline-to-line voltage at 200 km
= 124.2 x J3 - 215.1 kV
5.6
FERRANTI EFFECT
As has beenillustratedin Exarnple5.5, the eff'ectof the line capacitanceis to
cause the no-load receiving-end voltage to be more than the sending-end
voltage.The effect becomesmore pronouncedas the line length increases.This
phenomenonis known as the Ferranti. effect. A general explanation of this
effect is advancedbelow:
Substitutingx = / and In = 0 (no-load)in Eq. (5.21), we have
t7 /5 --
Vn
,at4gt *
-
Vn
2
,-at
(< A2\
--ifl
\J.-tJ
I
Fig.5. 14
Here
V,
(-+-.
Ir= ,
\ juCt
)
Since c is small comparedto L, uLl can be neglected
in comparisohto yc,tl.
Thus
15 - jVruCl
Now
Vn= Vs - Is QwLl) = V, + V,tj
CLlz
= vs0+ Jctt2)
lncreasino/
-\.-
(s.M)
Magnitude of voltage rise _ lvrltJ CLf
Locus of V5 wlth /
=olv.rt+
D
Vpfor I = 0
\
En= Ero= Vpl2
Increasing/
Fig. 5.13
The above equationshows that at I = 0, the incident (E,o)and reflected (E o)
voltage waves are both equal to V^/2. With reference to Fig. 5.13, as I
increases,the incident voltage wave increasesexponentiallyin magnitude
www.muslimengineer.info
(s.4s)
where v = 7/J LC. i1
velocity of propagationof the electromagneticwave
the
along the line, which is nearly equal to trr" velocity of
light.
5.7
TUNED POWER LINES
Equation (5-23) characterizesthe performanceof a long
line. For an overhead
line shunt conductanceG is always negligible and it is
sufficiently accurateto
neglect line resistanceR as well. with this approximation
7- Jyz = jalLC
cosh 7/ = cosh jwlJTC
= cos Lt,lrc
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
;fiTi I
I
uooernpowersystemRnarysis
sinh ?/ = sinh jalJ-LC
eharacteristicsand performanceof power Tr@
= j sin wtJLC
t
Z = Z,
Hence Eq. (5.23) simplifies to
cosulJ LC
.
rT sina'tJ
LC
Z,
jZ, sinutJE
=!
_1
cosutE
For a zr-networkshown in Fig. 5.15
llsl = llal
i.e. the receiving-end voltage and current are numerically equal to the
colrespondingsending-endvalues,so that thereis no voltagedrop on load. Such
a line is called a tuned line.
For 50 Hz, the length of line for tuning is
[refer ro Eq. (5.13)].
According to exact solurion of a long line
z,
.Jru.
,
[refer to Eq. (5.23)].
fi:I=filH;':"! l'lt,^t
JLl^J
. cosh
7/
L1,_, I z, ___..
= y, the velocity of light
(s.47)
= 3,000km, 6,000km,...
It is too long a distance of transmissionfrorn the point of view of cost and
efficiency (note that line resistancewas neglectedin the above analysis).For
a given line, length and freouencytuning can be aehievedby increa-singL or
C, i.e. by adding series inductancesor shunt capacitancesat severalplaces
along the line length. The methodis impracticaland uneconomicalfor power
frequencylines ancl is adoptedfor tclephonywhere higher frecluencies
are
employed.
A methodof tuning power lines which is being presentlyexperimentedwith,
uses seriescapacitorsto cancel the effect of the line inductanceand shunt
inductors to neutralize line capacitance.A long line is divided into several
sectionswhich are individuatly tuned. However, so far the practical method of
improving line regulationand power transfercapacity is to add seriescapacitors
to reduceline inductance;shunt capacitorsunder heavy load conditions;and
shunt inductors under light or no-load conditions.
THE EOUIVALENT
t(-tU2)
Y ganh1il2)
_ _ ___{l
( 5. 48)
Lr,J +!v,2,)
(r*ir,r,)lL,_)
Lr,(,
2nrfJ LC
5.8
.o,nh..,tt2
,,1-_| (t. It,t,)
1- --!r_.
,i^,^,?rx...
r=+@))=
=
Fig. 5.15 Equivarent-z
networkof a transmission
rine
N o wi f a l J t C = h r , n = I , 2 , 3 , . . .
lV5l= lVpl
Since ll^frc
-
sinhl/ =
(s.4e)
For exact equivalence,we must have
Z/= Z, sin h 7/
f * *YtZ=cosh
2
From Eq. (5.50)
(s.s0)
7/
(s.51)
Z' = ,l-:. sinn1/ = Z, tlnh'' : -(
z[*init rr 1
vY
tJyz
7t )
11-"'' -sln!-24 is thc fitct.r by which thc scrics
(5.52)
irupcdunceol'the no'ri'al-z
must be multiplied to obtain the z parameter
of the equivalent-a Substituting
7 from Eq. (5.50) in Eq. (5.51),we ger
1*
I
; YtZ, sitrh 7/ = cosh fl
CIRCUIT OF A LONG LINE
So far as the end conditions are concerned,the exact equivalentcircuit of a
transmissionline can be establishedin the form of a T- or zr-network.
The parametersof the equivalentnetwork are easily obtainedby comparingthe
perfbrmanceequationsof a z--networkand a transmissionline in terms of end
quantities.
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Lv,-
(s.53)
2
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowerSystemAnqlysis
Wl
I
(tanhfll2\.
d
rnus
I is thc lactor by which thc shunt aclmittanccanr ol' the
\ il/z )
nominal-n-mustbe multiplied to obtain the shunt parameter (Ytl2) of the
z.
equivalent/
1
\
l
Note that Ytl | + +Y' Z' I : j\
4
)
2
,
abovevaluesof Y/and Z/.
V R =+ 2 0
VJ
- l z 7 1 0k v
(a) Short line approximation:
Vs= t27 + 0.164/._36.9 x 131.2
172.3
= 14514.9
sinh 7/ is a consistentequationin termsof the
= 25L2 kV
lYsltin"
sinh 7/ =
tuth-l!/2 1 so that the
1 and
For a line of medium length
1t
fll2
Is= In= 0.764/_-36.9kA
Sending-end
powerfactor= cos (4.9"+ 36.9"_
equivalent- n- network reduces to that of nominal-n-.
41.g.)
= 0.745laggrng
Sending-end
power_ JT x 251.2x 0.764x
0.745
= 53.2MW
x@)Nominal-trmethod:
Fig. 5.16' Equivalent-Tnetwork of a transmissionline
Equivalent-T network parameters of a transmission line are obtained on
similar lines. The equivalent-T network is shown in Fig. 5.16.
As we shall see in Chapter 6 equivalent-r (or nominal-r) network is easily
^.1^^+^l
<fllIWPLVLT
+^
L\J
l^^.1
l\-rat.l
{'l^,r,
llLrW
otrr.lioo
JLUUM
o l-l.u1
q
io
lot
f}roref'nro
lltvlvlvlvt
r( r4nl ri lr r p r c q l l . ,
v vrLtqrrJ
pmnlnrrerl
vr^lHrvJ
vs'
Exirmple5.7
A 50 Hz transmissionline 300 km long has a total seriesimpedanceof 40 +
j125 ohms and a total shunt admittanceof 10-3mho. The receiving-endload is
50 MW at220 kV with 0.8 lagging power factor.Find the sending-endvoltage,
current,power and power factor using
(a) short line approximation,
(b) nominal-zrmethod,
(c) exacttransmissionline equation lBq. (5.27)1,
(d) approximation[Eq. (5.28b)].
Comparethe resultsand comment.
Solution Z = 40 + iI25 - 131.2 172.3" Q
Z= 10-' 190" U
loadis 50 MW at ?20 kV,0.8 pf lagging.
The Leceiving-end
1o=
l-36.9" = 0.164 l-36.9" kA
,=-+
J3 x220x0.8
www.muslimengineer.info
I
'
A = D = l + ! y z = .I + _x
l0-r lg0" x l3l .2 /.72.3"
2
= 1 + 0.0656L162.3 = 0.93g
/_L2"
B=Z=I3L.2 172.3"
c = v ( r + ! v z ) =y + l f z
\
4
/
4
= 0.00| 190"+
I . t0-61J g0. x 131.2L72.3"
4
= 0.001/.90
7s= 0'938 1r.2 x r27 + r3r.2 172.3"
x 0.164L-36.9"
= 719.7/7.2 + 21.5135.4"=
73j.4 /.6.2
= 238 kV
lVsllin"
1s= 0.001Z90 x 127+ 0.93gZl.Z
x 0.164/._ 36,9"
= 0.127l9A + 0.154l_35.7"=
0.13 /.16.5"
Sending-end
pf = cos (16.5._ 6.2) = 0.9g4leacling
Sending-endpower=
JT * 23gx 0.13x 0.gg4
= 52.7 MW
(c) Exacttransmission
line equarions
(Eq. (5.29)).
fl = al + jpt =JyZ
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowerSystem4nalysis
- 155 | -
characteristicsand Performance
of Power TransmissionLines
= Jro-tlgo"*l3t.zl7z.3" = 0.0554
+ i0.3577
= 0.938 ll.2
(alreadycalculatedin part (b))
= 0 . 3 6 21 8 L 2 "
cosh(al + i 0l) = !k"'
'
B= z(t*!Z\=
\
6 )
lgt + e*t l-Bt1
= 731.2172.3 + -{- x 1o-3lgo" x (13l.D2 1144.6.
6
= 131.2172.3"+ 2.87 l-125.4"
gl = a.5'fi1 (radians)-r'6.4,g"
1e0.49")= 1.057lzo.49"= 0.99+ j0.37
eo.0s54
e4'oss4l-20.49" = 0.946 l-
= 128.5172.7'
20.49" = 0.886 - j0.331
gosh 7/ = 0.938+ iO.O2= 0.938 11.2"
c= y(w!Z') = o.oo1tetr+ ]x to{ lrgo xr3t.2172.3"
\
sinh 7/ = 0.052 + 70.35 = 0.354 181.5"
E : -t----l3nnn"
L..= 1-
"
.,^
= soz.2I l-
Y 10-'190"
A = D = c o s hf l = O : 9 3 81 L . 2 "
t
lY
Z+ | yf
6
6 )
6
= 0.001l9O"
8.85'
Vs= 0.93811.2"x 127 10" + 128.5172.7'x 0.164I -36.9"
= ll9.I3 11.2" + 21.07135.8"- 136.2+ it4.82
= 137 16.2" kV
t- 88s"x 0354tsr s'
=7::: ,lr ul"u"'
I YsIrin"= 237.3 kY
/s = 0.13 116.5"(sameas calculated
in part (b))
Now
Vs= 0.93811.2"x 127 10" + 128.2172.65"x 0.164 l-36.9"-'
= 119.13
1 1 . 2 "+ 2 L 0 3 1 3 5 . 7 5 "
= 136.9716.2" kV
lVs hin.= 81,23--W
f
x 0.354 181.5"
362.211-8.85"
C = Lsinh .y/
z,
= 9.77 x 104 190.4"
.\
Is = 9 .J 7 x l O -a 1 9 0 .4 " x 127+ 0.938 11.2" x0.164 l -36.9
Sending-end
pf = cos (16.5' - 6.2 = 10.3")= 0.984leading
Sending-end
power= Jl
* ?313 x 0.t3 x 0.984
= 52.58MW
'fhe
resultsaretabulatedbelow:
Short line
uppntximatktn
Nominal-r
Exact
(s.28h)
l y s l h n "2 5 I . 2 k v
I,
0J& l-36.9" kA
p.f,
0.745 lagging
P"
53.2 MW
238 kV
o.I3 116.5"kA
0.984leading
52. 7M W
237.23kV
0.12861t5.3" kA
0.987leading
52. 15M W
= 0.124 190.4" + 0.154" - 35.7"
= 0.1286 115.3"kA
Sending-endpf = cos (15.3' - 6.2" - 9.1") = 0.987leading
Sending-endpower = Jt
x 237.23 x 0.1286 x 0.987
= 52.15 MW
(d) Approximation(5.28b):
A'=D = | +
|
Yz
)
www.muslimengineer.info
Approximation
?37.3kV
0. 131t 6. 5" kA
0.984leading
52.58MW
Comments
We find from the above example that the results obtained by the nominal- zr
method and the approximation (5.28b) are practically the same and are very
close to those obtained by exact calculations(part (c)). On the other hand the
results obtained by'the short line approximation are in considerableerror.
Therefore,ftlr a line of this length(about300 knr), it is sufficientlyaccurateto
use the nominal-r (or approximation(5.28b)) which results in considerable
savingin computationaleflort.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
5.9
characteristicsand performanceof power Transmission
Lines
ModernPowerSystemAnalysis
15ElI
POWER FLOW THROUGH A TRANSMISSION
sn=lvRt
t0 ti+l vstt13-a-l!lrv^tz{1-*t]
LINE
So far the transmissionline performanceequationwas presentedin the form of
voltase and current relationshipsbetween sending-andreceiving-ends.Since
loadsare more often expressedin terms of real (wattslkW) and reactive (VARs/
kVAR) power, it is convenientto deal with transmissionline equationsin the
form of sending- and receiving-endcomplex power and voltages. While the
problem of flow of power in a general network will be treated in the next
chapter,the principles involved are illustrated here through a single transmission line (2-nodel2-bussystem)as shown in Fig. 5.17.
= 'ul'll^'t(/r-d)- l-4v^t2l1p - ay
Similerly,
t, = I?l,r,f tur - u) -lll*lf t@+6)
(s.se)
In the above equationsso and s, are per phasecomplex
voltamperes,while v*
and vr are expressedin per phasevolts. If yR and 7,
are expressedin kv line,
then the three-phasereceiving-end complex por., is given
by
x to6,, ,, ,i
so(3-phase
vA)
'-/= -t {l#%+191 .-'"'-lEl
t( s -r, - l4l tvot2
L\tl- a)F
Jt
*JT
lBl
L
3
Ss=
s^(3-phase
MVA)=
Sp = Pp +lQP
iQs
F i g .5 .1 7 A tw o-bussystem
Let us take the receiving-endvoltage as areferencephasor(Vn=lVRl 10")
and let the sending-endvoltagelead it by an angle 6 (Vs = lVsl 16). Tlp- angle
d is known as the torque angle whose significance has been explained in
Chapter 4 and will further be taken up in Chapter 12 while dealing with the
problem of stability.
The complex power leaving the receiving-endand entering the sending-end
as (on per phasebasis)
line canbe expressed
of the transmission
-=
Sn Pn + .iQn Vnfn
Ss= Ps + iQs = YsI;
/ E
E
r*" = JI]l-v^]cos
(/
- / - l{l rv^r'
v - a
cos(B- a)
"
tBt
-
(s.s7)
,r=*r,
*ro
Let A, B, D, the transmissionJine constants,be written as
A = lAl la, B = lBl lP, D = lDl lo (since A = D)
Therefore, we can write
t^=
r1r
r _0)
t v 5tt( , _ 0 ) _ l A l r v4" a
rBl
'l ;(l '
r,=i+lv,tt(a+6-,-l+lvRtt-p
,n - ^ - lAl .__."
sin1/
6t
sin(/- a)
lliivni'?
lvsllv'l
p,' = I ?l v,P
J cos(o- o)lBl
tBicos(0*A
(5.61)
(s.62)
(5'63)
tvsttvRt
at
'
p
r = l*l .'ur,t
' . , sin (p - a) Qs
(/1+ b)
6.e)
If |
l-srtl
It is easyto seefrom Eq. (5.61) that the receivedpower po
will be maximum
6=0
such that
'u:'ll^'-t Attv'P
Po(max)
c,os(/1_ a)
.=
v)
I Bt
lBThe corresponding en @t max po ) is
- f f f : !t- .s i n ( 0 - u )
oA^n== --l4LYry
Substitutingfor 1o in Eq. (5.54)'we get
www.muslimengineer.info
(s.60)
similarly, the real and reactive powers at sending-end
are
currentscan, however,t" .*pr"rsed in terms of
Receiving-and sending-end
receiving- and sending-endvoltages[see Eq. (5'1)] as
(s.s6)
lrl
lyslly*l .
/1
Qn=
ff
A \
-* vR
t,r=
ivr,=
4p - , -l*lvopt1p- a1
This indeedis the sameas Eq. (5.59).The sameresult
holds for sr. Thus we
see that Eqs. (5.58) and (5.59) give rhe three-phaseMVA
if vs ina vo *"
expressedin kV line.
If Eq' (5'58) is expressedin real and imaginary parts,
we can write the real
and reactive powers at the receiving_endas
(J.J+/
(5.s5)
T#
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
:
(5.65)
I
f60 |
Modern Po*e, Sy.t"r Analysir
_
Thus the load must draw this much leading MVAR in order to receive the
maximum real power.
Consider now the specialcaseof a short line with a seriesimpedanceZ. Now
A-D=I
l0:B=Z=lzlle
Substitutingthesein Eqs. (5.61) to (5.64), we get the simplified resultsfor
the short line as
p o - Y r i l l o l c o s ( g - 6 \ - l v R Pc o s o
lzl
tzl
g o = l v ' ) l Y * l - s (i nd - b ) - l v * l ' , i n o
lzl
tzl
for the receiving-end
andfor the sending-end
(s.67)
(s.68)
d_'u:'lI.'sin
- (d*
v o _ ry_srn
e,
\
o
lzl
lzl
(s.6e)
lzl
tzl
The above short line equation will also apply for a long line when the line
is replacedby its equivalent-r (or nominal-r) and the shunt admittancesare
lumped with the receiving-endload and sending-endgeneration.In fact, this
technique is always used in the load flow problem to be treated in the next
chapter.
From Eq.(5.66),the maximum receiving-endpower is received,when 6 = 0
sothatP^ (rnax)=
lv]\Yol Ytc's
tzt
tzl
//
Now cos d= RllZl,
= 'tr',|,o'-l'r-t o
Pp(max)
Let lvtl - lvRl= lAv1, the magnitudeof voltage
drop acrossthe rransmission
line.
n^=#tavl
(5.66)
ps=lI:1.o,e-tl1'lunl
.o,@+6)
"
t Equation(5.72)canbe furthersimplifiedby assuming
cos 6 = r,since dis
normallysmall*.Thus
(5.70)
Normally the resistanceof a transmission line is small compared to its
reactance(since it is necessaryto maintain a high efficiency of transmission),
so that 0 = tan-t xlR = 90"; where z = R + jx.The receiving-endEqs. (5.66)
and (5.67) can then be approximatedas
(s.1r)
(s.12)
lvRl
dis necessary
from considerations
of systemstabilitywhichwill be discussed
at length in Chapter12.
*small
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(s.74)
several importanrconclusionsthat easily follow
from Eqs. (5.71) to (5.74)
are enumeratedbelow:
I' For R = 0 (which is a valid approximation
for a transmissionline) the
real power transferredto the receiving-endis proportional
to sin 6 (= 6for
small values of d ), while the reactiu. po*",
is proportional to the
magnitude of the voltage drop acrossthe line.
2. The real power received is maximum for
6 = 90o and has a varue
lvsllvRvx. of course, d is restricted to varues
weil below 90o from
considerationsof stability to be discussedin
Chaptet 12.
3. Maximum real power transferred for a given
line (fixed X) can be
increasedby raising its voltage level. lt is from
this considerationthat
voltage levels are being progressivelypushedup
to transmit larger chunks
bf power over ronger distances wananted ty
l*g; ;irtlln"rutirrg
stations.
For very long lines voltage level cannotbe raised
beyond the limits placed
' 4'1'present-day
high voltagetechnoiogy.To increasepower
transmittedin
suclt citscs,tltc only choicc is to reduce
the line reactance.This is
accomplishedby adding series capacitorsin
the line. This idea will be
pursuedfurtherin chapter 12. Seriescapacitors
woulclof c<lurseincrcos!.
the severityof line over voltagesunder switching
conditions.
4. As said in 1 above,the vARs (lagging reactive
power) deriveredby a line
is proportionalto the line voltage drop and is independent
of d Therefore,
in a transmissionsystem if the vARs demand
of the load is large, the
voltage profile at that point t-endsto sag rather
sharply. To maintain a
desiredvoltageprofile, the vARs demanJof the load
must be met locally
by employing positive vAR generators(condensers).
This will be
discussedat lengthin Sec. 5.10.
A somewhat more accurateyet approximateresult
expressingiine voltage
drop in terms of active and rlactive powers can
be written directly from
E q. (5.5) , i. e.
lAVl= llnl R cos Q + llol X sn
Q
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
i
r i
I
I
I
power SvstemAnalvsis
162 |
characteristicsand Performanceof Power TransmissionLines
"odern
R!ry* XQn
lvRl
Thisresultreduces
to thatof Eq. (5.74)if R = 0.
=
(s.75)
Case (a): Cable impedance= 70.05 pu.
Since cableresistanceis zero, thereis no real power loss in the cable.Hence
Pcr t Pcz= Por * Poz = 40 pu
j Example5.8
Pot= Pcz= 20 pu
An interconnector
cablelinks generatingstations1 and 2 as shownin Fig. 5.18.
The desiredvoltageprofile is flat, i.e.lVrl=lVzl = 1 pu. The total demandsat
the two busesare
Spr=15+75Pu
The voltage of bus 2 is taken as ref'erence,i.e. V, 10" and voltageof bus
1 is V1 16r. Further, for flat voltage profile lVll = lV2l - L
Real power flow from bus I to bus 2 is obtained from Eq. (s.68) by
recognizingthat since R = 0, 0= 90".
Hence
lvtllvl
_D
D
Ps=
Pn = -f
sin 6,
Soz= 25 + 715 Pu
The station loads are equalizedby the flow of power in the cable. Estimate
the torqueangleand the stationpower factors:(a) for cable z = 0 + 70.05 pu,
and (b) for cable Z - 0.005 + 70.05 pu. It is given rhat generator G, can
generatea maximum of 20.0 pu real power.
-
) =
From Eq. (5.69)
@
Sn.' = P61+jQ61
I
^'
Qt=
lvlt51
vQu-1.
Jp2 = I-p2+ J\Jp2
lv,llvl
X
-,:
U,\,J
From Eq. (5.67)
c)
lv,l'
-
-- ,DD- 1. r i A' l \ 1 D 1
d,
0.05
4 = I4.5"
vr= I lL4'5"
or
Solution The powers at the various points in the fundamental(two-bus)
systemare defined in Fig. 5.18(a).
I
lxl
-SlIl
":
-
cosdt
x 0.e68= 0.638pu
U.UJ
(a)
lvtllv,l
f2o= -Tcos
d'
tv12- X
Qs = - 0.638Pu
Reactivepower loss* in the cable is
QL= Qs- en _ 2es _ 1.276pu
Total load on station1 = (15 + i5) + (5 + j0.638)
__>
Sn=5-l
SDI= 15 +/5
= 20 + j5.638
(b)
G)
{
zo.ro+ i16.12
V 2 =1 . 0l o "
Powerfactorat starionI = cos [run-' t'f:t') = 0.963lagging
\
2 0 )
Total load on station2 (25 + jls) (5 - j0.638)
=20+i15.638
15+j5
25 + j1S
(c)
Fig. 5.18 Two-bussystem
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Reactivc powcr loss can also be cornputedas l/l2X=
:t +(9.!1q)1
o.os=
t.z7 pu.
"
I
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
characteristicsand Performanceof power TransmissionLines
toctern power Svstem ,Analvsis
164 I
')
Powerfactor ar station2 = coS (ron*r 15'638 = O.zgslagging
\
2
0
)
The stationloads,load demands,and line flows are shownin Fig.5.1g(b).
ir-;-t { iz i, r"rp rt v.
"" "case
"r
Cableimpedance= 0.005 +70.05 = 0.0502 lB4.3 pu. In rhis
Case(b):
the cable resistancecausesreal power loss which is not known a priori. The real
load flow is thus not obvious as was in the case of R = 0. we specify the
generation at station I as
Pcr= 20 Pu
The considerationfor fixing this generation is economic as we shall see in
Chapter 7.
The generationat station 2 will be 20 pu plus the cable loss. The unknown
variables in the problem are
165
6r = 14.4"
Substitutingdr in Eqs.(ii), (iii) and (iv), we ger
Q c t = 5 . 1 3 ,Q c 2= 1 6 . 1 2P, G z= 2 0 . 1 0
^'-
-
^D'
"'^'
It may be noted that the real power loss of 0. i pu is supplied by
Gz(Pcz - 20.10).
The above presentedproblem is a two-bus load flow problem. Explicit
solution is always possiblein a two-bus case.The readershould try the case
when
Q c z = 7 1 0 a n d l V 2 l= ' The general load flow problem will be taken up in Chapter 6. It will be seen
that explicit solution is not possiblein the generalcaseand iterative techniques
have to be resortedto.
P62, 6p Qcp Qcz
Let us now examine as to how many system equationscan be formed.
From Eqs. (5.68)and (5.69)
= ', =
Pct- Por
#cos
d- V#*s@+
5 = -=
cos 84.3"U.U)UZ
A 275 kV transmissionline has the following line constants:
A = 0.85 15": B - 200 175"
61)
cos (84.3"+ 4)
^-=
U.U)UZ
(i)
e c r - e o t =e s = f f r t ^ t - t r r , \ i , s i n ( d +4 )
sin 84.3"- =+-sin
Qcr - S = ;j0.0502
0.0502
FromEqs.(5.66)and (5.67)
(84.3"* Ur)
(ii)
*;rcos
(84.3"- A,l -
#rcos
84.3o
- 4)
sin(84.3"
#,
sin (75"- d) -
ij;
x Q75)2sin(75"-5")
0=378sin(75"-A-302
sin84.3"
whichgives
6- ))"
(v)
Thus we havefour equations,Eqs. (i) to (iv), in four unknownsp52, 51,e61,
Q62.Eventhoughtheseare non-linearalgebraicequations,solutionis possible
in this case.Solving Eq. (i) for d,, we have
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o - !5x]75200
(iii)
- ,Yl!rr",
eor-ecz=en=Uffiri, (o- 6t>
lzl
15- Qcz=
#,
(c) With the load as in part (b), what r,vouldbc thc receiving-end
voltageif
the compensationequipmentis not installed?
Solution (a) Given lV5l= lVal = 215 kY; e, = 5o,C = J5". Since the power
is receivedat unity power factor,
Qn= o
Substitutingthesevalues in Eq. (5.62), we can write
P o z - PG z- =PP n -=l v- ftf l l v z cl o s( d - 6 r ) - -l 1v :'-Pc o s e
2s - Pc2=
(a) Determine the power at unity power factor that can be received if the
\
voltage profile at each end is to be maintained at 275 ky .
(b) What type andrating of compensation
equipmentwould be requiredif the
load is 150 MW at unity power factor with the samevoltage profile as
in part (a).
From Eq. (5.61)
275^x275
cos(75o- 22")- 9 85 * (2712cos70o
200
200
- 227.6- 109.9= 117.7D{W
Pn=
"
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
todern power SvstemAnalvsis
166 |
characteristicsand Performanceof Power TransmissionLines
I
(b) Now lV5l= lVpl = 275 kV
0 = !:y-*t sin (75'- .5)- *tvot2
200
200
From Eq. (ii), we get
sin(75"- A=0.00291VR1
Power dernanded
by load = 150 MW at UPF
P n = P R = 1 5 0 M W ;Q o = 0
150=
215zq7cos
( 7 5 o- 5 ) 200
0 . 8 5 . ^ -- , )
x (275)'cos70o
,*
1 5 0 = 3 7 8 c o s ( 7 5 "- 4 - 1 1 0
From Eq. (5.62)
sin (75. - 28.46")- 0'85x e75)2 sin 70o
200
"#
- 274.46- 302= - 2l.56 MVAR
en=
Thusin orderto maintain2T5kV at a receiving-end,en= -27.56 MVAR must
be drawn alongwith the real power of Po = 150 MW. The load being 150 MW
at unity power factor, i.e. Qo = 0, compensationequipmentmust be installed
at the receiving-end.With referenceto Fig. 5.19, we have
- 2 7 . 5 6 +Q c = 0
or
Qc = + 27.56 MVAR
i.e. the compensationequipment nnustfeed positive VARs into the line. See
subsection5.10 for a more detailedexplanation.
1 5 0- j 2 7 . 5 6
sin 70'
_ 0.00l45tyRl2
150 = 1.375 lynt (1 _ (0.002912lv^121u2
1 5 0+ / 0
Note: The second and lower value solution of lVol though feasible, is
impractical as it correspondsto abnormally low voltage and efficiency.
It is to be observedfrom the results of this problem that larger power can
be transmitted over a line with a fixed voltage profile by installing
compensationequipmentat the receiving-endcapableof feedingpositive VARs
into the line.
Circle Diagrams
It has been shown above that-the flow of active and reactive power over a
transmissionline can be handledcomputationally.It will now be shown that the
Iocusof complex sending-and receiving-endpower is a circle.Sincecircles are
convenientto draw, the circle diagramsare a useful aid to visualize the load
flow problem over a single transmission.
The expressionsfor complex numberreceiving-and sending-end
powers are
reproducedbelow from Eqs. (5.58) and (5.59).
sR=4/r-r. +P 4r- b)
lfl rv^r'
n=l+lvs(t(r-o)-+3 4/r+o
Since no compensationequipmentis provided
P n = 1 5 0 M W ,Q n = 0
Now,
l V 5l = 2 7 5 k V , l V a l =?
Substitutingthis data in Eqs. (5.61)and (5.62),we have
t':\!'"os
150=
(75"- A - ggtv^t2 cos70"
200
200
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(s.s8)
(5.5e)
The units for Sp and S, are MVA (three-pha,se)with voltages in KV line.
As per the above equations, So and ,9, are each composedof two phasor
componenfs-6ne a constantphasorand the other a phasorof fixed magnitude
but variable angle. The loci lor S^ and S, would, therefore,be circles drawn
from the tip of constantphasorsas centres.
It follows from Eq. (5.58) that the centre of receiving-endcircle is located
at the tip of the.phasor.
F i g .5 . 1 9
(c)
(ii)
Solving the quadratic and retaining the higher value of lv^|, we obtain
lVal = 244.9 kV
5 = 28.46"
or
Ifil
(i)
-l+l vRP
' \ t(r - a)
(s.16)
= -l|lrv-f cos
(//- a)MW
(s.77)
IB l
in polar coordinatesor in terms of rectangularcoordinates,
Horizontal coordinate of the centre
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
e.-'1,,
,
.
I
.d6Eif
Modernpower SystemAnalysis
_
.,.,,
characteristicsand performanceof power Transmission
Lines
t69
Vertical coordinate of the centre
Vertical coordinateof the centre
= l+ilvrt2sin(F- a) MVAR
= -i{-ltv*t2sin(tJ- a) MVAR
tB t
The radiusof the sending-end
circleis
lBl
The radius of the receiving-endcircle is
tysllyRlMVA
(5.78)
tBl
(s.81)
The sending-endcircle diagramis shownin Fig. 5.2I.Thecenrre
is located by
drawing OC, at angle rt?- a) from the positive MW-axis.
From the centre the
sending-endcircle is drawn with a .uoi.rr${e(same
as in the case of
\
lBt
receiving-end).The operatingpoint N is located by measuring
the torque angle
d(as read from the recefving-endcircle diagram) in ttredirection
indicated from
thc re'fi'rcncc'
Iinc.
MVAR
MVAR
C5
Roforoncolino
- for angled
lAll
t2
lallu"l
Radius lYsllVnl
lBl
Phasor55 = P5 +ie5
Referenceline
f o r a n g l e6
Fig. 5.20
1--
Receiving-endcircle diagram
Qs
F'or constant lVol, the centre Co rernains fixed and concentric
circles result
for varying l7rl. However, for the case of constant ll{l and
varying lvol the
centrcso1'circlesmovc along the line OCoaru),have raclii in accordance
to ltzrl
lvRtABt.
Similarly, it follows from Eq. (5.59) that the centre of the sending-end
circle
is located at the tip of the phasor
l4l,u,,'t(0.- a)
IB l
=u':;=o
i,i,=',!"i
Horizontal coordinate of the centre
cos(f - a)Mw
B-ltv;2
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Fig. 5.21 Sending-endcircle diagram
(s.7e)
in the polar coordinates or in terms of rectangular coordinales.
=
+
(s.80)
The corresponding receiving- and sending-end circle diagrams have
in Figs5.22 and5.23.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
been clrawn
characteristicsand Performanceof power TransmissionLines [giM
t
Resistance= 0.035 Olkm per phase
Inductance= 1.1 mHlkm per phase
Capacitance= 0.012 pFlkm per phase
If the line is supplied at 275 kV, determine the MVA rating of a shunt
lvnl2
tzl
Pa=oK
Qn= KM
Fig.5-22 Receiving-end
circrediagramfor a short rine
MVAR
Ps=oL
Qs=LN
thereceiving-end
whentheline is deliveringno load.Usenominal-zrmethod.
Solution
R=0.035x400=14Q
X = 314x 1.1x l0-3 x 400 = 138.2O
Z = 14 + 7138- 138.1184.2" Q
Y= 314 x 0.012x 10-6x 400 lg0" - 1.507x 10-3/_W U
A = ( t + L v z \ = 1 + - l - x t . 5 0 7x 1 0 - 3x r 3 g . i l r i 4 . z "
\
2
)
2
= (0.896+ 70.0106)
= 0.896 lj.l'
B=Z-138.7 184.2"
lV5| = 275 kV, lVpl= 275kV
lysllyRl -275x275
- 545.2MVA
Radiusof receiving-end
circletBl
138.7
Locationof the cenre of receiving-end
circle,
,, = 275x275x0.896== '488.5
r d d ' ) rMVA
ll
l4:l1 i l l
IB I
\
138J
l@ - a) = 84.2"- 0.7" = 83.5"
lvsl2
lzl
i,\/A P
55 MVAR
_
> M W
L
488.5 MVA
Fig. 5.23 Sending-endcirclediagramfor a stror.tline
oJl:
.tuou.rrro
useof circlediagrams
is i$ustratedby meansof therwo examplesgiven
Cp
Fi1.5.24 Circlediagramfor Example5.10
A 50 Hz, three-phasg,275 kY,400 km transmissionline has
the following
parameters:
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From the circle diagramof Fig. 5.24, + 55 MVAR must be drawn from the
receiving-endof the line in order to maintain a voltageof 275 kV. Thus rating
of shunt reactor needed= 55 MVA.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpower SystemAnalysis
X12,:jl
and Performanceof PowerTransmissionLines | ,i?3-;Characteristics
t-
(a) Locate OP conespondingto the receiving-endload of 250 MW at 0.85
laggingpf (+ 31.8). Then
Example5.11
, ^
A - 0.93 11.5", B = Il5 ll7"
If the receiving-end voltage is 2'r-5kV, determine:
(a) The sending-endvoltagerequired if a load of 250 MW at 0.g5 lagging pf
is being delivered at the receiving-end.
(b) The maximum power that can be delivered if the sending-endvoltage is
held at 295 kV.
(c) The additional MVA that has to be provided at the receivihg-end when
delivering 400 MVA at 0.8 lagging pt the suppty voltage being
maintained at 295 kV.
Solution
In Fig. 5.25 the centre of the receiving-endcircle is located at
2t5x275x0'93611.6
MVA
l4li Rt'' -lBl
tt5
or^
lysllyRl
275lvsl
l V s l = 3 5 5 . 5k V
(b) Given lV5 | = 295 kV.
t".l?"
- 705.4 MVA
115
Drawing the receiving-end circle (see Fig. 5.25) and the line C^Q parallel to
the MW-axis, we read
PR- o = RQ = 556 M W
Radius of circle diagram-
(c) Locate OPt conespondingto 400 MVA at 0.8 lagging pf (+ 36.8"). Draw
P/S parallel to MVAR-axis to cut the circle drawn in part (b) at S. For the
specified voltage profile, the line load should be O^S.Therefore, additional
MVA to be drawn from the line is
P/S = 295 MVAR or 295 MVA leading
c o s -l 0 .8 5= 3 1 .8 "
l@-
a ) = 7 7 o - 1 . 5 "= 7 5 . 5 "
MVAR
I
5.10
METHODS OF VOLTAGE CONTROL
Practically each equipmentusedin power systemare ratedfor a cprtain voltage
with a permissible band of voltage variations.Voltage at various buses must,
therefore,be controlled within a specified regulation figure. This article will
discussthe two methodsby meansof whieh voltage at a bus can be controlled.
lvslt6
a)
lvRltj
I
p"-.io,
P^iio* |
Fig. 5.26 A two-bussystem
Considerthe two-bus systemshown in Fig. 5.26 (akeady exemplifiedin Sec.
5.9). For the sake of simplicity let the line be characterizedby a series
reactance(i.e. it has negligible resistance).Further, since the torque angle d is
small under practicalconditions,real and reactivepowersdeliveredby the line
for fixed sending-endvoltage lVrl and a specifiedreceiving-enclvoltagel{ | can
be-written as below from Eqs. (5.71) and (5.73).
(5.82)
Fig. 5.25 Circle diagram for ExampleS.11
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
nsmtssion
Linesl"ffie
'li,
ef,
rl
' = X ,,u,r- rv,i
Equation (5.83) upon quadraticsolution*can also be written
rr{ r=
(5.83)
as
. +tys | (1 - 4xesnAvrtzlt/z
}vrt
Since the real power demandedby the loacl must be delivered
(5.84)
by the line,
Pn= Po
varying real power demandp, is met by consequent
changesin the rorque
angle d.
It is, however' -tobe noted that the receivedreactive power
of the line must
remainfixed at esnas given by Eq. (5.g3) for fixed rv,
I and specifiedr4r. il"
line would, therefore,operatewith specified receiving-end
voltage for only one
value of Qo given by
Qo = Qsn
Reactive Power Injection
It follows from the above discussion that in order to keep
the receiving-end
voltageat a specifiedvalue l{1, a fixed amountof VARs
drawn
tai I *;;;
from the line-. To accomplish this under conditions
Qn, a local VAR generator (controlled reactive power source/compensating
equipment)must be usedas shown in Fig. 5.27.fle vAR
balanceequationat
the receiving-endis now
Oi * Qc= Qo
Fluctuationsin Qo ue absorbedby the local vAR generator
o6 such that
the vARs drawn from the line remain fixed at
esn.The receiving-endvoltage
would thus remain
at l4l (this of
a fixed sending_end
{1ed
voltage lVrl). L,ocal VAR compensation"ourr"lrrumes
can, in fact, be made automatic by
using the signal from the VAR meter installedat the receiving-end
of the line.
Practical loads are generally lagging in nature and
are such that the vAR
demandQn may exceedet*.rt easily follows from Eq. (5.g3)
that for or; otthe receiving-endvoltagemust changefrom the specified
value 'n'i
some
value lTol to meet the demandedVARs. Thus
'Jo= o ^ == l v * l '
(lYsl - lVol) for (QD> QsR)
Q ^ = Qn
;i
The modified lVol is then given by
=
rvlql
lvrt
-
+ty3
(1 - 4xeRltvrt
)r,,
(s.85)
crrmparisonof Eqs. (5.84)ancl(5.85) rcvcalsthut r^r. n. - n - .,-),s,!,.,
YD- vR- vp' tttt:
receiving-endvoltageis r{r, butior
bo= Oo; A:,"'
tvo| < t4l
Thus a VAR demandlarger than Qf is met by a consequent
fall in receivingfrom the specifiedvalue. similarly, if the vAR demandis less
than
11d ":t!q,"
Q " *, it fo l l o w s th a t
tyRt> tr4l
Indeed, under light load conditions, the charging
of the line may
"upu.lance
cause the VAR demand to become negative resuliing
in the receiving-end
voltage exceeding the sending-endvoltage (this is the Ferranti
effect already
illustraredin Section5.6).
In order to regulate the line voltage under varying demandsof
VARs, the two
methodsdiscussedbelow are employed.
Fig. s-zr use of rocarvAR generatorat the roadbus
Trryotypes of vAR generators are employed in practice-static
type and
rotating type. These are discussedbelow.
Static
It is nothing but a bank of three-phase
static capacitors and/or inductors. With
referenceto Fig. 5.28,if lV^l is in line kV,
and Xg is the per phase capacitive reac_
tance of the capacitor bank on an equiva_
Ient star basis, the expression for the
VARs fed into the line can be derived as
under.
r,=iH
'Negative
sign in the quadratic solution is rejected because otherwise
the solution
would not match the specified receiving-end voltage which
is only slightly less than
the sending-endvortage(the differenceis ress thai nqo).
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VAR grenerator
lIc
Fig. 5.28
Static capacitor bank
kA
'of
course,
sincct{tis spccificcl
withina buntl,Ql rury vrry withil a corresponding
band.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
: iii,',1,1
ModernpowersystemAnarysis
t
Figure 5'29 shows a synchronous
motor connectedto the receiving-end
bus
bars and running at no load. since
the motor o.u*, n.grigible real power
from
3ry (- IF)
iQcG-Phase)J3
-i3x
#.HMVA
t v P
QsQ-Phase)-+
XC
MVAR
in pr,use. ; th" ,yn.r,ronous
reachnce
$",.t":::H:,.,:o,^fl
!:T-? T*ly
wtr,icrr
i, usium"a
tohave
",
;rr,r*,n
;r;:r*::':il;
:j 3:^t1o1
(s.86)
-
(lvRl IEGD/0.
,. ^
.C__
IKA
J5;E-
If inductors are employed instead,vARs fed into the line are
Q{3-phase)=-'F''tuo*
XL
Under heavy load conditions,when positive VARs are needed,capacitor
banks
are employed; while under light load conditions, when negative
vARs are
needed,inductor banks are switchedon.
The following observationscan be rnade for.static vAR generators.
(i) Capacitor and inductor banks can be switched on
in steps.However,
stepless(smooth) VAR control can now be achieved using SCR (Silicon
Controlled Rectifier) circuitrv.
(ii) Since Qg is proportionalto the squareof terminal
voltage,for a given
capacitor bank, their effectivenesstends to decreaseas the voltage
sags
under full load conditions.
(iii) If the system voltage containsappreciableharmonics,
rhe fifth being the
most troublesome,the capacitorsmay be overloadedconsiderably.
(iv) capacitors act as short circuit when switched on.
(v) There is a possibility of series resonancewith
the line incluctance
particuia.riyat harmonic frequencies.
Rotating
VAR grenerator
It is nothing but a synchronousmotor running at no-load and having
excitation
adjustableover a wide range. It feeds positive VARs into
the line uncler
overexcitedconditionsand f'eedsnegativeVARs when underexcited.
A machine
thus running is called a synchronouscondenser.
lvnl
Fig. 5.29 RotatingVAR generation
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i e c_ = 3Jt v R3l 4 oG i l
(s.87)
= 3 W ( - l Y R rr-r c l )
J3 ( _jxsJl )
= jlVpt(tE6t _ IVRt)lXs
MVA
ec= tVRt(EGt _ tVRt)lXs
MVAR
(5.8g)
It immediately follows from the
above relationship that the machine
feeds
positive vARs into the line when
rEGt> tv^r (werexcited case)
and
injects
negarive VARs if lEGl
continuously adjustableby adjusting
machine
which controls tE6l.
rn contrastto statrcvAR generators,
"Jtution
the following
observations
are made in
respect of rotating VAR generators.
.,
(i) These can provide both positive
and negativevARs which are
continu_
ously adjustable.
(ii) vAR inje*ion ar a given
..
excirarionis ressst
tlt""r""
volrage.As Irzo
r deJeases and(rE - rv^rr i,Lt
6r
J"-:ff i::: {.:::
smallerreductionin
Qc comparedto the .0.r. or static capacitors.
From rhe observarions-ua"
in ,.rp..t of ,tuti.
vAR
generators,it seemsthat rotating ibr_""
""d;;;;;ing
vAR g.n"ruro* would be preferred.
However,
economic considerations,
install.tion and 'rai'r.rrun.. problerns
limit their
buses
in the svstem
*r,"i"' a taigeu.oun,-.orvAR
il]ffi'ir'$ t:L::"::ch
Control
by Transformers
The vAR injectionmethoddiscussed
abovelacksthe flexibility and
economy
of voltage control by transformer
tap changing. The transformer
tap changing
is obviously rimited to a narrow
range of voltage control. If
the vortage
correctionneededexceedsthis range,
tap changingIs usedin conjunction
with
the VAR injection method.
Receiving-endvoltage which tends
to sagowing to vARs demanded
by the
load, can be raised by simultaneousry
.t,uogir; th. taps of sending_and
receiving-end transformers.Such tap
changesniust"bemade ,on_road,
and can
be done either manua'y or automaiically,-the
oo*io.,o"r being ca'ed a Tap
Changing Under Load
ifCUf_l transformer.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Characteristicsand Performanceof Power TransmissionLines fi.l7!J,
ModernPowerSystemAnalysis
,tig;'J
I
I
Considerthe operationof a transmissionline with a tap changingtransformer
at eachend as shown in Fig. 5.30. Let /5 and r^ be the fractions of the nominal
transformationratios, i.e. the tap ratio/nominal ratio. For example, a transkV input has rr - I2lll
l.{Vl which is to be compensated.Thus merely tap setting as a method of
voltage drop compensationrvould give rise to excessivelylarge tap setting if
compensationexceedscertain limits. Thus, if the tap setting dictated by Eq.
setting range (usually not more than + 20Vo), it would be necessary to
simultaneously inject VARs at the receiving-end in order to maintain the
desiredvoltage level.
=
z=R+jx
Compensation
1 : fsnl
at eachend
linewithtap changingtransformer
Fig.5.30 Transmission
With referenceto Fig. 5.30 let the impedancesof the transformerbe lumped
tn Z alongwith the line impedance.To compepsatefor voltage in the line and
transformers,let the transformer taps be set at off nominal values, rr and ro.
With referenceto the circuit shown. we have
(s.8e)
trn rV s = t^ n rVo + IZ
From Eq. (5.75) the voltage drop ref'erredto the high voltage side is given by
tAvl = !I,!jIQs-
(s.e0)
t on,rlVol
Now
lAVl - tsn, lTsl -
trnrlvrl- tonrlvol +
ton2lVol
RPR+xQR
(s.e1)
t* n r l V o l
In order that the voltage on the HV side of the two transformersbe of the
sameorder and the tap setting of eachtransformerbe the minimum, we choose
(5.92)
tstn= 1
SubstitutinEtn= llttin Eq. (5.91) and reorganising,we obtain
.r( ,
RPR+xgo ) _ n2 lvRl
"['
"r"rWW )- ", W
(s.93)
For complete voltage drop compensation,the right hand side of Eq. (5.93)
shouldbe unity.
It is obvious from Fig. 5.30 that rr > 1 and tn 1 I for voltage drop
compensation.Equation (5.90) indicatesthat /^ tends to increase*the voltage
-This
is so becausefn < 1 increasesthe line current / and hence voltage drop.
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of Transmission
Lines
The perfonnance of long EHV AC transmissionsystemscan be improved by
reactive compensationof series or shunt (parallel) type. Seriescapacitorsand
shunt reactors are used to reduce artificially the series reactanceand shunt
susceptanceof lines and thus they act as the line compensators.Compensation
of lines results in improving the system stability (Ch. 12) and voltage conffol,
in increasingthe efficiency of power transmission,facilitating line energization
and reducing temporaryand transient overvoltages.
Series compensationreduces.the seriesimpedanceof the line which causes
voltage drop and is the most important factor in finding the maximum power
transmission capability of a line (Eq. (5.70)). A, C and D constants are
functions of Z and therefore the also affected by change in the value of.Z, but
these changes are small in comparison to the change in B as B = Z for the
.,
for the equivalent zr.
nominal -rr and equalsZ (sinh 4ll)
The voltage drop AV due to series compensationis given by
AV = 1Rcos S, + I(X,.- X.) sin ,!,
(s.e4)
Here X, = capacitivereactanceof the seriescapacitor bank per phaseancl
X, is thc total incluctivercactanceof the line/phasc.In practice,X. may be so
selected that the factor (XL - X.) sin Q, becomes negative and equals (in
magnitude) R cos /, so that AV becomes zero. The ratio X=IXL is called
"compensation factor" and when expressedas a percentageis known as the
"percentagecompensation".
The extent of effect of compensationdependson the number, location and
circuit arrangementsof series capacitor and shunt reactor stations.While
planning long-distance lines, besides the average degree of compensation
required, it is requiredto find out the most appropriatelocation of the reactors
and capacitor banks, the optimum connection scheme and the number of
intermediate stations.For finding the operating conditions along the line, the
ABCD constantsof the portions of line on eachside of the capacitorbank, and
ABCD constants of the bank may be first found out and then equivalent
constantsof the seriescombination of line-capacitor-linecan then be arrived at
by using the formulae given in Appendix B.
In India, in stateslike UP, seriescompensationis quite importantsincesuper
thermal plants are located (east) several hundred kilometers from load centres
(west) and large chunks of power must be transmitted over long distances.
Seriescapacitors also help in balancing the voltage drop of two parallel lines.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
r-hOt"
l
Characteristics
and Performanceof PowerTransmissionLines
uodern Power SvstemAnalysis
When seriescompensationis used, there are chancesof sustainedovervoltage
to the ground at the series capacitor terminals. This overvoltage can be the
power limiting criterion at high degree of compensation.A spark gap with a
high speed contactor is used to protect the capacitors under overvoltage
trons.
Under light load or no-load conditions, charging current should be kept less
than the rated full-load current of the line. The charging current is approximately given by BrltA where B. is the total capacitive susceptanceof the line
and lVl is the rated voltageto neutral.If the total inductive susceptanceis Br due
to several inductors connected(shunt compensation)from line to neutral at
appropriateplaces along the line, then the charging current would be
(Bc- Br) lvl= BclVf[r
I,he,=
+)
(s.es)
Reduction of the charging current is by the factor of (1 - Br lBc) and 81lBg is
the shunt compensationfactor. Shunt compensationat no-load also keeps the
receiving end voltage within limits which would otherwise be quite high
becauseof the Ferranti Effect. Thus reactors should be introduced as load is
removed,f<lrproper voltagecontrol'
As mentioned earlier, the shunt capacitorsare used acrossan inductive load
so as to provide part'of the reactive VARs required by the load to keep the
voltage within desirablelimits. Similarly, the shunt reactors are kept across
capacitive loads or in light load conditions, as discussedabove, to absorbsome
of the leading VARs for achieving voltage control. Capacitors are connected
eithcrclirectlyto a bus or throughtcrtiarywinclingof the main transformerand
are placed along the line to minimise lossesand the voltage drop.
Ii may be noted that for the same voltage boost, the reactive power capacity
of a shunt capacitrtr is greater than that of a series capacitor. The shunt
capacitor improves the pf of the load while the seriescapacitor has hardly any
impact on the pf. Series capacitors are more effective for long lines for
irnprovementof systemstability.
Thus, we seethat in both seriesand shunt compensationof long transmission
lines it is possible to transmit large amounts of power efficiently with a flat
voltage profile. Proper type of compensation should be provided in proper
quantity at appropriateplacesto achieve the desiredvoltage control. The reader
is enceuragedto read the details about the Static Var Systems (SVS) in
'compensation',the reader
References7, 8 and 16. For complete treatmenton
may refer to ChaPter 15.
!.EMS
PROB
5 . 1 A three-phasevoltage of 11 kV is applied to a line having R = 10 f) and
X = 12 ft p"t conductor.At the end of the line is a balanced load of P
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r81
kW at a leading power factor. At what value of P is the voltage regulatior
zero when the power factor of the load is (a) 0.707, (b) 0.85?
5 . 2 A l o n g l i n e w i t h A = D = 0 . 9 l 1 . 5 " a n d B = 1 5 0 1 6 5 "C I h a s a t t h el o a r
end a transformerhaving a seriesimpedanceZr = 100 167" Q. The loar
form of
ff]=[1
",]l';l
and evaluatethese constants.
5.3 A three-phaseoverheadline 200 km long has resistance= 0.16 Qlkrn an
conductordiameter of 2 cm with spacing4 m, 5 m and 6 m transpose(
Find: (a) the ABCD constantsusing Eq. (5.28b), (b) the V,, 1,, pf,,'I
when the line is delivering full load of 50 MW at 132 kV and 0.8 laggin
pf, (c) efficiency of transmission, and (d) the receiving-end voltag
regulation.
5.4 A short 230 kV transmissionline with a reactanceof 18 O/phasesupplir
a load at 0.85 lagging power factor. For a line current of 1,000 A tt
receiving- and sending-endvoltages are to be maintainedat 230 k\
Calculate (a) rating of synchronous capacitor required, (b) the loa
current, (c) the load MVA. Power drawn by the synchronouscapacitt
.\
may be neglected.
'three-phase
5.5 A 40 MVA generating station is connectedto a
line havin
Z = 300 175" Q Y = 0.0025 19tr U.
The power at the generatingstation is 40 MVA at unity power factor a
a voltage of L20 kV. There is a load of 10 MW at unity power factor a
the mid point of the line. Calculatethe voltageand load at the distantenc
of the line. Use nominal-T circuit for the line.
5.6 The generalizedcircuit constantsof a transmissionline are
A-0.93+70.016
B=20+ jI40
The load at the receiving-end is 60 MVA, 50 H4 0.8 power factor
lagging. The voltage at the supply end is 22O kV. Calculate the load
voltage.
5 . 7 Find the incident and reflectedcurrentsfor the line of Problem 5.3 at the
receiving-endand 200 km from the receiving-end.
5 . 8 If the line of Problem 5.6 is 200 km long and delivers50 MW at22OkY
and0.8 power tactor lagging,determinethe sending-endvoltage,current,
power factor and power. Compute the efficiency of transmission,
impedance,wavelength,and velocity of propagation.
characteristic
5 . 9 For Example 5.7 find the parametersof the equivalent-n circuit for the
line.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
uooern power system hnalysis
characteristicsand Performanceof power TransmissionLines
5.10 An interconnector cable having a reactanceof 6 O links generating
stations1 and 2 as shown in Fig. 5.18a.The desiredvoltage profile is lVtl
= lVzl = 22 kY. The loadsat the two-bus bars are 40 MW at 0.8 lagging
power factor and 20 MW at 0.6 lagging power factor, respectively. The
NCES
REFERE
,,1€?il
-T-
torqueangle and the stationpower factors.
5.11 A 50 Hz, three-phase,275kV, 400 km transmissionline has following
parameters(per phase).
Resistance= 0.035 Qlkm
,
In d u c ta n c e = 1 m F l /k m
Capacitance= 0.01 p,Flkm
If the line is supplied at 275 kV, determine the MVA rating of a shunt
reactor having negligible lossesthat would be required to maintain 275
kV at the receiving-end,when the line is delivering no-load.Use nominalzr method.
of 3 Q and areactanceof 10 f)
feederhavinga resistance
5.12 A,three-phase
suppliesa load of 2.0 MW at 0.85 lagging power factor. The receivingend voltage is maintained at 11 kV by means of a static condenser
drawing2.1 MVAR from the line. Calculatethe sending-endvoltage and
power factor. What is the voltage regulation and efficiency of the feeder?
5.13 A three-phaseoverheadline has resistanceand reactanceof 5 and 20 Q,
respectively.The load at the receiving-endis 30 MW, 0.85 power factor
lagging at 33 kV. Find the voltage at the sending-end.What will be the
kVAR rating of the compensatingequipmentinsertedat the receiving-end
so as to maintain a voltageof 33 kV at each end?Find also the maximum
load that can be transmitted.
5.I4 Constructa receiving-endpower circle diagram for the line of Example
5.7. Locate the point coresponding to the load of 50 MW at 220 kV with
0.8 lagging power factor. Draw the circle passingthrough the load point.
Measurethe radiusand determinetherefromlVrl. Also draw the sendingend circle and determine therefrorn the sending-end power and power
factor.
5.15 A three-phaseoverheadline has resistanceand reactanceper phase of 5
and25 f), respectively.The load at the receiving-endis 15 MW, 33 kV,
0.8 power factor lagging. Find the capacity of the compensation
equipmentneededto deliver this load with a sending-endvoltage of 33
Books
l. Tron'smissionLine ReferenceBook-345 kV and Above, Electric Power Research
Institute, Palo Alto calif, 1975.
2. Mccombe, J. and F.J. Haigh, overhead-Iinepractice, Macdonalel,London, 1966.
3. Stevenson,w.D., Elementsof Power Sy.stem
Analysis,4thedn, McGraw-Hill. New
York, 1982.
4. Arrillaga, J., High Vohage Direct Curuent Transmission,IF,E Power Engineering
Series 6, Peter PeregrinusLtd., London, 1983.
5. Kirnbark, E.w., Direct current Transmission,vol. 1, wiley, New york, 1971.
6. IJhlmann,E., Power Transmissionby Direct current, Springer-verlag, BerlinHeidelberg,1975.
7. Miller, T.J.E., Reactive Power control in Electric systems,wiley, New york
t982.
8. Mathur, R.M. (Ed.), Static Compensatorsfor ReactivePower Control, Context
Pub., Winnipeg, 1984.
9. Desphande,M.V., Electrical Power System Design, Tata McGraw-Hill. New
Delhi, 1984.
Papers
10. Dunlop, R.D., R. Gutman and D.p. Marchenko,"Analytical Developmentof
Loadability Characteristicsfor EHV and UHV TransmissionLines", IEEE Trans.
PAS,1979,98: 606.
11. "EHV Transmission",
(special Issue), IEEE Trans,June 1966,No.6, pAS-g5.
12. Goodrich, R.D., "A Universal Power circle Diagram", AIEE Trans., 1951, 7o:
2042.
1 3 . Indulkar, c.s. Parmod Kumar and D.p. Kothari, "sensitivity Analysis of a
MulticonductorTransmissionLine", Proc. IEEE, March 19g2, 70: 299.
1 4 . Indulkar, c.s., Parmod Kumar and D.P.Kothari, "some studies on carrier
Propagationin overheadrransmission Lines", IEEE Trans.on pAS,No. 4, 19g3,
102: 942.
1 5 . Bijwe, P.R., D.P. Kothari, J. Nanda and K.s. Lingamurthy, "optimal voltage
Control Using ConstantSensitivityMatrix", Electric Power SystemResearch,Oct.
1 9 8 6 ,3 : 1 9 5 .
1 6 . Kothari, D.P., et al. "Microprocessorscontrolled static var s5istems",proc. Int.
conf. Modelling & Simulation,Gorakhpur,Dec. 1985, 2: 139.
kv.
Calculate the extra load of 0.8 lagging power factor which can be
delivered with the compensatingequipment (of capacity as calculated
above) installed, if the receiving-end voltage is permitted to drop to
28 kV.
www.muslimengineer.info
183
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
6.2 NETWORK MODEL FORMULATION
6.1
INTRODUCTION
With the backgroundof the previous chapters,we are now ready to study the
operationalfeaturesof a compositepower system.Symmetricalsteadystateis'
in fact, the most importantmode of operationof a power system.Three major
problems encounteredin this mode of operation are listed beiow in their
hierarchical order.
1. Load flow problem
2. Optimalload schedulingproblem
3. Systemscontrol Problem
This chapter is devotedto the load flow problem, while the other two
problems will be treateilin later chapters.Lt-radllow study in power systetn
parlance is the steady state solution of the porver system network. The main
information obtained from this study comprisesthe magnitudesand phase
angles of load bus voltages,reactive powers at generatorbuses, real and
reactine power flow on transmissionlines, other variablesbeing specified.This
information is essentialfor the continuousmonitoring of the current state of the
systernand for analyzingthe effectivenessof alternativeplans for future system
expansionto meet increasedload demand.
Before the advent of digital computers,the AC calculatingboard was the
only meansof carrying out load flow studies.These studieswere, therefore,
tedious and time consuming.With the availability of fast and large size digital
computers,all kinds of power systemstudies,inciuding load flow, can now be
carried out conveniently.In fact, some of the advancedlevel sophisticated
studieswhich were almostimpossibleto carry out on the AC calculating board
have now become possible.The AC calculating board has been rendered
obsoletefor all practicalpurposes.
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The load flow problem has,in fact, been alreadyintroduced in Chapter 5 with
the help of a f'undamentalsystent,i.e. a two-bus problem (see Example 5.8)'
life power systemcomprising a large number
Eor4 lq4d flqW llq4ypfufg4l
of buses, it is necessaryto proceed systematicallyby first formulating the
network model of the sYstern.
A power systemcomprisesseveralbuseswhich are interconnectedby rneans
of transmissionlines. Power is injected into a bus from generators'while the
loads are tappedfrom it. Of course,there may be buseswith only generators
and no-loads, and there may be others with only loads and no generators.
Further, VAR generatorsmay also be connectedto some buses.The surplus
power at some of the buses is transportedvia transmissionlines to buses
deficientin power. Figure6.1ashowsthe one-linediagramof a four-bussystem
with generatorsand loads at each bus. To arrive at the network model of a
po*"i system, it is sufficiently accurateto representa short line by a series
may be used
impedanceand a long line by a nominal-zr model. (equivalent-7T
foi very long lines). Often, line resistancemay be neglectedwith a small loss
in accuracybut a great deal of saving in computationtime.
For systematic analysis, it is convenient to regard loads as negative
generatorsand lump togetherthe generatorand load powersat the buses.Thus
at the ith bus, the net complex power injectedinto the bus is given by
S;= Pi + jQi= (Pci- Po)+ j(Qci- Qo)
where the corrrplexpower suppliedby the generatorsis
Sci= Pot+ iQai
anclthe ccltnplexpower drawn by the loads is
Spi= Por+ iQoi
The real anclreactivepowersinjected inttl thc itlt bus arc thcn
Pi= Poi- P^ i = 1,2, "'' fl
(6'1)
Qi= Qci- Qoi
Figure 6.lb showsthe network model of the samplepower systemprepared
on the above lines. The equivalentpower sourceat eachbus is representedby
a shadedcircle. The equivalentpower sourceat the ith bus injects currentJr into
the bus. It may be observedthat the structureof a power systemis such that
all the Sourcesare alwaysconnectedto a commonground node.
The networkmodel of Fig. 6.lb hasbeenredrawnin Fig. 6.lc afier lurnping
the shunt admittancesat the buses.Besidesthe groundnode, it has four other
nocies(buses)at which the currentfrom the sourcesis injected into the network.
The line admittancebetweennodes i and k is depictedby !ip= Jri'Further, the
mutual admittancebetweenlines is assumedto be zero'
.Line
transformers are representedby a seriesimpedance(or for accuraterepresentation by series and shunt impedances,i.e. invertedL-network).
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
apitying Kirchhoff s currentlaw (KCL) at nodesr,2,3 and4,respectively,
we getthe following fbur equations:
Jt = Vrlro + (Vr - V) ln * (Vr - Vr) ),:
Jz= Vz)zo+ (Vz- Vr) ln + (Vz_ V) yzt + (Vz_ Vq)yzq
s= vztgo+ (y: - v) ln + (Vt _ V) lzz + (Vt_ Vq) yzq
Jq= Vayqo+ (Vq - Vr) lzc * (Vq _ V) yzq
Rearrangingand writing in matrix form, we get
(yrc * tn
-ln
*fn,
.,
- Yrz
(lzo*ln
* hz * lzq,
_ !*
-t23
0
-J z +
-\z
0
- lzt
- !z+
(y:o * yrg
(b) Equivalent
circuit
-lu
*rzt*yy,
-
( Y q a* y z q
lzq
* Yz+)
Equation (6.3) can be recognized,to be of the standardform
(6.4)
Comparing Eqs. (6.3) and (6.4), we can write
Yrr=)ro* ln+ ln
Yzz= lzo * ltz t lzt + lzq
Ytt= ):o* ln*
lzz* lzq
Yu= lqo * lzq * ly
Ytz= Yzt = - lni YZt = YZZ= - lZt
Soz
Sor
v1
(c) Power network of Fig. 6.1 (b) lumped
and redrawn
Fig. 6.1 Samplefour-bussystem
Y l t = Y r c = - l n i Y v = Y q t= )r+ = 0
Yzq=Y+z=-lzqiYy= yqz=ly
Each admittance y,, (i = r,2,3, 4) is calred
the serf admittance (or driving
point admittance) of node i and equals the
algebraicsum of all the admittances
terminaring on the node. Each ofi-diagonal
ierm v* (i, k = r, 2, 3, 4) is the
mutual admittance (transfer admittance) between
noim i and ft and equals the
negativeof the sum of all admittancesconnected
directly between thesenodes.
Further, Yr* = Yri.
using index notation, Eq. (6.a) can be written
in compact form as
n
Ji= D
y p v p ii = r , 2 , . . . ,f l
k=l
(a) One-linediagram
Fig. 6.1
Sample four bus system
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(6.s)
or, in matrix form
"Inus= Isus Vnus
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(6.6
1.,.^L{"
'vrrvrv
urhanp
V
J^-^+^^
-BUS
\re,'ulsD
aL^
--
^^
:
urc llralnx
or
DUS admlttance
and
is known
as bus
admittancematrix. The dimensionof the y"u5 matrix
is (n x n) where n is the
numberof buses'[The total numberof noclesii e m =
n + | includingthe ground
(reference)node.l
As seen above, yru, is a symmetric
except when phase shifting
transformersare involved, so that only n v + t ) terms
are to
be stored for an
2
,?-blrs system. Furthermore, yi* = 0 if buses
i d
k are not connected
(e.9. YA = 0). since in a power network each
bus is connectedonly to a few
other buses(usually to two or three buses),the ys.y5
of a rargenetwork is very
sparse'i'e' it has a large number of zeroelementi.-itrougtr
tf,i, prop"rty is not
evident in a small system like the sample system
under consideration,in a
system containing hundreds of buses,the sparsity
may be as high as 90vo.
Tinney and associates[22] at Bonnevile Power
Authority were the first to
exploit the sparsityfeature of zsu5 in greatly reducing
numericalcomputations
in load flow studies and in minimizing the memory
required as only non-zero
terms needbe stored.
Equation (6.6) can also be written in the form
Vnus = Zausleus
(6.7)
where
Zsu5 (bus impedancematrix) = fsLs
fbr a network of fbur buses(fbur inclependent
nodes)
6.3 FORMATION OF fsus BY SINGULAR
TRANSFORMATION
Graph
(6.8)
rnus yields symmetric Zsus.The diagonal erements
-flmmetric
of Zuu, are
called driving point imped,ances
of the nodes,and th! off-diagonalelementsare
called transfer impedances of the nodes. Zsus need
not be obtained by
inverting rnus.while y"u, is a sparsematrix,-Auris
a full matrix. i.e., zero
elementsof I'ru, become non-zeroin the .o,,.rp"o"niing
Zsu, erements.
It is t' be stressedhere that yBus/zBusconstitute
models of the passive
portions of the power network.
Bus admittancematrix is often used in sorving road
flow problem. It hal
gained widespreadapplication owing to its simpticity
of data preparationand
the easewith which the bus admittancematrix can be
formed ano moOfied for
networkchanges-addition of lines,regulatingtransformers,
etc. (seeExamples
6'2 and6'7)' of course, sparsityis one of its greatest
advantagesas it heavily
reducescomputer'memory and time requirernents.
In contrast to this, the
www.muslimengineer.info
involved algorithms. Furthermore,the impedance
matrix is a fuil matrix.**
The bus impedancematrix, however, is most useful
for short circuit studies
as will be seenin Chapters10 and 11.
Note: In the sample system of Fig. 6.1. the
arDrrrary manner, although in more sophisticated
studies of large power
systems, it has been shown that certain ordering
of nodes produfes faster
convergenceand solutions.Appendix c dealswith
the topics of sparsityand
optimal ordering.
zss" can be referred to ground or slack bus. In
the former case, it is usually
necessary l'o creal'e at least one strong artificial
tie to ground to avoid numerical
difficulties when obtainingZsg5, becausein absence
of this, rr* is ilr-conditionedor
even singular' A large shunt admittanceinserted at
the slack bus most simply achieves
the desired result [20].
**The
disadvantagesof the conventional impedancematrix
may be overcome by
making use of LU factors of the admittancematrix and
bf employing compacr srorage
scheme' Piecewise methods or tearing techniques (dialoptics)
have recently been
applied to overcome the disadvantagesof excessiv,
,rorug. requirementsIlg].
***For
convenience, direction is so assigned as to coincide
with the assu'red
positive direction of the element current.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernpowerSystem
Anatysis
i90 l
O
r
@
Primitive Network
o
vr, = Er-
Fig. 6.2 Lineargraphof the circuitin Fig.6.1c
here that each source and the shunt admittance connected across it are
representedby a single element. In fact, this combination representsthe most
general network element and is describedunder the subheading "primitive
Network".
A connecteds'bgraph containing all the nodes of a graph but having no
closedpathsis called a tree. The elementsof a tree are called branchesor ffee
branches. The number of branches b that form a tree are given by
b=m * 1- n (numberob
f uses)
(6.9)
Thoseelementsof the graph that are not includedin the tree are called links (or
link branches) and they form a subgraph,not necessarily connected,called
c
rn
@
v
E,
where E, and E" are the voltages of the element
nodes r and s, respectively.
It may be remembered here that for steady state
AC performan.., ull element
variables (vr* E, 8", irr,7r.,) are phasors and
element parameters (zrr,
.rr") ar€
complex numbers.
The voltage relation for Fig. 6.4a can be written
as
vrr*
€rr=
Zrri^
ir, *
jr, = lrrv^
(6.r2)
I
7
---------+
o-------------
- Branch
- - Link
(6.11)
Similarly, the current relation for Fig. 6.4b is
yr"=E -E"
\.e
e =9
m=5
b=m-1=5-'l=4=n
l=e-b=5
(a) lmpedanceform
The forms of Figs. 6.4a and,b are equivalentwherein
the parallel sourcecurrenr
in admittanceform is related to the seriesvoltage
in impedanceform by
@
(a) Tree
(b) Admittance form
Fig. 6.4 Representation
of a networkelement
(b) Co- tree
Fig. 6.3 Treeand cotreeof the orientedconnectedgraphof Fig.6.2
I = e - b = e - m + l
-
!r.,=
|/7r,
Yrs€rs
Also
cotree.The number of links / of a connectedgraph with e elementsis
Note that a tree (and therefbre, cotree) of a graph is not unique.
Jr,=
(6.10)
A set of unconnectedelements is defined as a primitive
network. The
performanceequationsof a primitive networkerre
given below:
In impedance form
V+E=ZI
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(6.13)
'f{W';l
ModernPowersvstemAnatvsis
In admittanceform
I+J=W
Load Fiow Stucjies
(6.14)
Here V and.Iare the elementvoltage and currentvectorsrespectively,and rl and
E are the sourcevectors. z and Y arereferred to as the primitive i
admittancematrices,respectively.'These
are relatedas Z = Y-r.If there is no
mutual coupling betweenelements, Z and Y are diagonal where the diagonal
entries are the impedances/admittancesof the network elements and are
reciprocal.
NeturorkVariables
in Bus Frame of Reference
Linear network graph helps in the systematicassemblyof a network model. The
main problem in deriving mathematical models for large and complex power
networks is to selecta minimum or zero redundancy(linearly independent)set
of current or voltage variableswhich is sufficient to give the information about
all element voltages and currents. One set of such variables is the b tree
voltages.*It may easily be seen by using topological reasoning that these
variables constitute a non-redundantset. The knowledge of b tree voltages
allows us to compute all element voltages and therefore, all bus currents
assumingall element admittancesbeing known.
Consider a tree graph shown in Fig. 6.3a where the ground node is chosen
as the referencenode. This is the most appropriatetree choice for a power
network. With this choice, the b tree branch voltagesbecomeidentical with the
bus voltagesas the tree branchesare incident to the ground node.
flH:
i
I
( 6. 16)
where the bus incidence matrix A is
e
I
2
3
4
)
6
7
8
9
-
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
-
l
l
-
l
0
l
1
_
1
0
1 0
0
1
l
branches
(6.r7)
links
0
0
1
0
This matrix is rectangular and therefore
singular. Its elementsa,oare found
as per the following rules:
aik = 1 if tth element is incident to and oriented
away from the ftth
node (bus)
= - | if tth element is incident
to but oriented towards the ftth
node
Bus Incidence Matrix
For the specific systemof Fig. 6.3a, we obtain the following relations tetween
the nine elementvoltagesand the four bus (i.e. tree branch) voltages V1, V2,V3
and Va.
Vut = Vt
Vtz= Vz
Vut = Vt
Vuq= Vq
Vts= Vt-
<
v = AV,'J'
Vn
(6.15)
Vta= Vt -
v2
V n = V t - v2
V t s = V q - v2
V t g = V t - vr
or, in matrix form
*Another
useful set of network variables are the / link (loop) currents which
constitute a zero redundancyset of network variables [6].
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= 0 if the ith element is not incident
SubstitutingEq. (6.16) into Eq. (6.14), we
get
to the kth node
I+J=yAVsu5
Premultiplying by Ar,
( 6 .l 8 )
ArI+ArJ=AryAV"u,
(6.1e)
F,achcomponentof the n-dimensional
vector ATI is the algebraicsum of
the
elementcurrentsleaving the nodes 7,2, ...,
n.
Therefore, the application* of the KCL must
result in
ArI =o
(6.20)
Similarly, eachcomponentof the vector ATJ
recognizedas the algebraic
sum of all sourcecurrentsinjectedinto nodes
"unbe
r,2, ...,n. Thesecomponentsare
thereforethe bus currents**. Hence we can
write
*For
node l, Arl gives
iro+ ir, - l:r= 0
-Thereadershoulciverify this for anothernode.
**For
node l, AT"/gives
j61 =currentinjectedinto bus 1
because
otherelements
connected
to bus t haveno sources.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Power Svstem Analvsis
i,il9:4iitl
-
-
.
-
J
- - - -
-
- " - - ' f
ArJ = Jsus
Equation(6.19)thenis simplifiedto
,/eus= ATYAV",,,,
-
-
J
-
-
-
.
^
^
_
-
.
-
I ftt^
Load Flow Studies
- - -
I
I
(6.2r)
rvB U S
- lTv,r
=
A IA
(6.22)
-
r
r
^
'
v
r
r
r
r
g
v
l
'
v
a
the samenodal currentequationas (6.6). The bus admittancematrix can then
be obtainedfrom the singular transformationof the primitive Y, i.e.
Yeus = ATYA
(6.23)
A computerprogrammecan be developedto write the bus incidencematrix
A from the interconnectiondata of the directed elementsof the power systetn.
Standardmatrix transposeand multiplication subroutinescan then be usedto
compute Yu* fiorn Eq. (6.23).
*"*'^-"r
Example
6.1 |
,
0
* yr:)
- Jtz
( yzo* yn
- lzt
* lzt * t-zq,
- )r:
'- ,.
.t'23
0
_!zc
-
()'lo f- .yr
-
* lzt * Jy,
!'tn
}'l
The elementsof this matrix, of course,agreewith thosepreviouslycalculated
i n Eq. ( 6. 3)
I
Find the Y6u" using singular transformation for the system of Fig. 6.2.
Figure 6.5 showsthe one-line diagramof a simple four-bussystem.Table 6.1
gives the line impedancesidentified by the buseson which theseterminate.The
shunt admittanceat all the busesis assumednegligible.
(a) Find Yuu. assumingthat the line shown dotted is not connected.
(b) What modifications needto be carriedout in Yuu, if the line shown dotted
is connected.
Solution
)ro
!zo
-Vro
!qo
Y-
ltn
v-,
J Z )
!n
Y.-n
ln
Using A fiom Eq. (6.17),we get
)ro
0
YA=
0
0
0
0
0
0
0
0
0
lzo
0 J n
0
0
jqo
0
0
l
- jzz
-lrz
!n
0
!24
-.Yrr
0 l
y
lzz
0
-
ft+
0
0
0
lzq
n
0
www.muslimengineer.info
3
IJ1'
Fig. 6.5 Samplesystemfor Example6.2
Table 6.1
Line,
bus to bus
&pu
xpu
t-2
1-3
2-3
24
34
0.05
0.10
0.15
0.10
0.05
0.15
0.30
0.45
0.E0
0.15
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
trrrr.F,b"iFl
'f|/0.!:l
-t
f.siffi
Modern Power Svstem Anatvcis
Table 6.2
Line
r-2
Gpu
2.0
B,pu
- 6.0
1.0
2.0
- 2.0
- 3.0
- 6.0
-
2-3
24
34
where V, is the voltage at the ith bus with respectto ground and ,/, is the source
current injected into the bus.
The load flow problem is handled more convenientlyby use of "/, ratherthan
,I,t. Therefore,taking the complex conjugateof Eq. (6.24), we have
(6.25a)
J.\,
Substitutingfor J, =
solution (a) From Table 6.1, Table 6.2 is obtained from which yuu,
for the
systemcan be written as
t Y*Vr from Eq. (6.5), we can write
k:l
Pi- jQi=Vf
A
L
Y i l , V p ii = l 2
n
(6.2sb)
k:l
Equating real and imaginary parts
I
P; (real power) = Re ]Vi
(6.26a)
Qi (reactive power) = - Irn
(6.26b)
t
In polar form
V; = ll/,1si6t
Y r* = lY'oleio'r
Real and reactive powers can now be expressed as
n
addedbetweenbuses 7 and 2.
Ytz,n"* = Yrz,on - (2 - j6) = Yzr.o"*
Xrr,n"* = ytt,ora+ (2 - j6)
P, (real power) = lvil D
k:l
lvkl lYiklcos (0,0+ 6t - 6');
i = r , 2, ..., fl
Y2z,n"* = Yzz,ot,t+ (2 - j6)
Modified Y"u, is written below
Qi Qeactivepower)
- - lvil
D
lvkl lYiklsin (e,1+ 6t - 6,);
k:l
i=1,2,...,fl
6.4
LOAD FLOW PROBLEM
The complex power injected by the sourceinto the ith bus of a power sysrem
IS
(6.24)
www.muslimengineer.info
(6.27)
(iv)
(6.28)
Equations (6.27) and (6.28) represent2n power flow equationsat n buses of
a power system (n real power flow equationsand n reactive power flow
equations).Each bus is characterizedby four variables; P;, Qi, l7,l and 6i
resulting in a total of 4n variables. Equations(6.27) and (6.28) can be solved
for 2n variables if the remaining 2n variables are specified. Practical
considerationsallow a power systemanalystto fix a priori two variablesat
each bus. The solution for the remaining 2n bus variablesis rendered difficult
by the fact that Eqs. (6.27) and (6.28) are non-linear algebraicequations (bus
voltagesare involved in product form and sine and cosine terms are present)
and therefore,explicit solutionis not possible.Solutioncan only be obtainedby
iterative numerical techniques.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
:flg.,0iil
toaern power system Anatvsis
Dependingupon which two variablesare specified a priori,
the buses are
classifiedinto three categories.
(I)
PQ Bus
r rr rrrro rJyw vr LruD, Lrrc ucr puwers ri ano
ai arc known
(po, and
state variables. These adjustable independent variables are called control
parameters. Vector J can then be partitioned into a vector u of confrol
parametersand a vector p of fixed parameters.
arc known
Qpi
g anclp, andegile specified).
The*tnn*i, ars fl/,tand
l."lA 1:1t"t:i:.t"
pure load bus (no generating
6,.
f*ifty-at the bus,i.e., pcr= eci=0)
Pp bus.
(2)
PV Bus/Generator
Bus/voltagre
controlled
is a
Bus
At this type of bus P' and eo, are known a priori and
p6;)
rv,r
' and.p,(hence
' \
are specified.The unknowns are e, (hence
eo,) and 6,.
(3) Slack Bus/Swing
Bus/Reference
Bus
(6.30)
Control parametersmay be voltage magnitudesat pV buses,real powers p,, etc.
The vector p includes all the remaining parameterswhich are uncontrollable.
For SLFE solution to have practical significance,all the state and control
variables must lie within specified practical limits. These limits, which are
dictated by specificationsof power systemhardwareand operating constraints,
are describedbelow:
(i) Voltage magnirudelV,l must satisfy the inequality
lv,l^tn< lvil ( lv,l_*
(6.31)
The power system equipment is designed to operateat fixed voltages with
allowable variations of t (5 - l})Vo of the rated values.
(ii) Certain of the 6,s (state variables) must satisfy the inequality constraint
In a load flow study real and reactivepowers (i.e. complex
power) cannot
be fixed a priori at alr the buses as the net complex power
flow into the
network is not known in advance,the systempower loss
being unknown till the
load flow study is complete.It is, therefore,n.."rrury
to have one bus (i.e. the
slack bus) at which complex power is unspecified so
that it supplies the
difference in the total system load plus losses and the
sum of the complex
powersspecifiedat the rcrnainingbuses.By the same
reasoningthe slack bus
must be a generatorbus. The complex power allocatedto
this bus is determined
as part of the solution. In order that the variations in real
and reactive powers
of the slack bus during the iterative process be a snrall
percentage of its
generating capacity, the bus connecteclto the largest
generating station is
normally selectedas the slack bus. Further, for corivenience
the slack bus is
numberedas bus 1.
Equations(6.27) and (6.28) are referred to as stutic
load ftow equations
(SLFE)' By transposingall the variables on one
side. these equationscan be
written in the vector form
f(x,y)-o
(6.2e)
where
,f = vector function of dimension 2n
16,- 6ft1
S l6i- 6rln,o
(6.32)
This constraint limits the maximum permissible power angle of transmission
line connecting buses i and ft and is imposed by considerationsof system
\
stability (see Chapter 12).
(iii) owing to physical limitations of p and/ore generationsources,po, and
Qci Ne constrainedas follows:
Pc,, ^in 1 Pc, S Pc,. ,n*
(6.33)
Qci, ^rn 1 Qci S Qc,, ^u**
(6.34)
It is, of course,obvious that the total generationof real and reactive power must
equal the total load demandplus losses,i.e.
D P o , = t P o , +P ,
l
D
i
(6.35)
t
Qci=l
Qot+Qr
(6.36)
i
where Ptand Qpare systemreal and reactivepower loss,respectively.Optimal
sharing of active and reactive power generation between sources will be
discussedin Chapter 7.
x = d_ependent
or state vector of dimension 2n
(2n unspecifiedvariables)
) = vector of independentvariablesof dimension 2n
(2n independentvariabreswhich are specified.i prrori)
www.muslimengineer.info
-Voltage
ata PV bus can be maintained constant only if conrollable esource is
available at the bus and the reactive generation required is within prescribed limits.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Mod".nPo*"t sv"ttt An"lvtit
ffi#ilfl,.l.i|
I
"Ihe load flow problem can now be fully defined as follows:
Assume a certainnominal bus load configuration. Specify P6i+ iQci at all
the pQbuses (this specifiesP, + iQi at thesebuses);specify Pcr (this specifies
P,) and lV,l at all the PV buses.Also specify lVll and 6, (= 0) at the slack bus'
T.L,,- r-
.,--iolrlac nf thp rrer-fnr u ere snecifie.d The 2n SLFE can now be solved
(iteratively) to determine the values of the 2n vanables of the vector x
comprising voltagesand anglesat the PQ buses,reactive powers and angles at
fhepV busesand active and reactive powers at the slack bus. The next logical
step is to comPuteline flows.'
So far we have presented,the methods of assemblinga Yeusmatrix and load
flow equationsand have defined the load flow problern in its genpralform with
definitions of various types of buses.It has been demonstratedthat load flow
equations,being essentially non-linear algebraic equations,have to be solved
through iterative numerical techniques. Section 6.5 presents some of the
algorithmswhich are used for load flow solutionsof acceptableaccuracyfor
systemsof practicalsize.
At the cost of solution accuracy, it is possible to linearize load flo-w
equationsby making suitable assumptionsand approximationsso that f'astand
eiplicit solutionsbecomepossible.Such techniqueshave value particularly for
planning studies,where load flow solutions have to be carried out repeatedly
but a high degreeof accuracyis not needed.
Load Flow
An Approximate
n
l i- 6 r); i = 2,3, ..., n
l v k l l Y i k (6
,/--r
k:1
n
et=- 'u,'
Pr =
n
n
.I
Po,- D
i:2
i:2
d, which, when substitutedin Eq. (6.3g), yields
madehaveo..ouiLii*i;:;ffi
,"(;:;'iLT,",",T:,:T:J"#il-,1
simultaneouslybut can be solved sequentially
[solution of Eq. (6.3g) follows
immediately upon simurtaneoussorutionof Eq. (6.37)).
Since the sorutionis
non-iterative and the dimension is retlucecr
to (rr-l) from Zrt, it is
computationallyhighly economical.
consider the four-bus samplesystem of Fig. 6.6
wherein line reactancesare
indicatedin pu. Line resistancesare considerld negligible.
The magnitudeof all
the four bus vortagesare specifiedto be r.0 pu. itJuu,
powersLe specified
in the table below:
53=- 2 +7O,
--r J
j0.15
jo.2
(6.37)
r y , , ir ;= r , 2 , . . . ,n ( 6 . 3 9 )
) rv,r2
r v k r l y ickor s( 6 , - 6 u +
E
Since lv,ls are specified, Eq. (6.37) representsa set of linear algebraic
equationi in 6,sr,vhichare(n - l) in numberas 6, is specifiedat the slack bus
(6, = 0). The nth equation correspondingto slack bus (n = l) is redundant as
the reat power injected at this bus is now fully specified as
www.muslimengineer.info
Po,;(Pr= 0). Equations
(6.37)can be solvedexplicirly
(non-iteratively) for 62,61,...,
Solution
Let us make the following assumptionsand approximationsin the load flow
'
Eqs. (6.27) and (6.28).
of overhead
(shiintconductance
being smaii are rreglecie,C
(i) Line resistances
the
systemis
of
power
loss
active
the
P7,
i.e.
lines is alwaysnegligible),
- - 90o'
=
1ii
and
(6.28)
90'
1it
(6'21)
and
zero. Thus in Eqs.
(ii) (6, - 6r) is small (< r/6) so that sin (6, - 6o) = (6r- 6r). This is justified
from considerationsof stability (see Chapter 72)'
(iii) AII busesotherthanthe slackbus (numberedas bus 1) are PV buses,i.e.
voltage magnituclesat all the busesincluding the slack bus are specified.
Equations(6.27) and (6.28) then reduceto
P i = l V i l \-
Wffi
iP,ts
'l.o
lVzl=
2
.S.= I + i^
I Uz
,,-. -
Fig. 6.6 Four-buslosslesssamplesystem
1
2
3
4
Real
demand
Reactive
demand
Real
generation
Reactive
generatrcn
Por = 1.0
Poz = 7.0
Poz = 2.0
Poq = 2.0
Qot = 0.5
Qoz = 0.4
Qoz = 1.0
Qoq = 7.0
Pcl ='Pcz = 4'0
Pct=o
Pcq=o
061 (unspecified)
Q62 (unspecified)
O63 (unspecified)
06a (unspecified)
Figure 6.6 indicates bus injections fbr the data specified
in the table.
As bus voltages are specified,all the busesmust have
controllable e sources.
from the datarhar buses3 and 4 have onry sources.
Further,
e
Il_r::t::ffie"_bviyus
slnce
system is assumecilossless, the real power generation
at bus I is
known a priori to be
Pct = Por * Poz * pot * poo_ pcz = 2.0 pu
Therefore,we have 7 unknownsinsteadof 2 x 4 = 8
unknowns.In the present
problem the unknown stateand control variablesare
{, e, 60,ect, ecz, ecz
and Qc+.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'yili.\
ModernPowerSystemAnalysis
I
Though the reali63sesare zero,the presenceof the reactive lossesrequires
that the total reactive generationmust be more than the total reactive demand
4
(2.9 pu).
From the datagiven, Yru5can be written as follows:
4
e r = D e o , -D e p i
_ L,> = u.JJ4 pu
_ r.+r+
(viii)
Now, let us find the line flows. Equation (5.6g) can be written
( l z l =X , 0 = 9 0 " )
Pik = - Pki Using the above Y"u, and bus powers as shown in Fig. 6.6, approximate
load flow Eqs. (6.37) are expressedas (all voltage magnitudesare equal to 1.0
pu)
(6- 6q)
P z =3 = 5 ( 6 - 6 ) + 1 0( 6 - 4 ) + 6 . 6 6 7
P t = - 2 - 6 . 6 6 (76 - 4 ) + l 0 ( 4 - 6 )
P + = - 2 - 1 0 ( 4 - 4 ) + 6 . 6 6(76 q : 6 )
(iii )
4=-0.074rad=-4.23'
(v)
6q=-0.089rad=-5.11'
Substituting 6s in Eqs. (6.38), we have
Qr = - 5 cos 4.4I" - 6.667 cos 4.23" - 10 cos 5.11' + 21.667
Qz = - 5 cos 4.41" - 10 cos 8.64" - 6.667 cos 9.52o+ 21.667
Qz = - 6.667 cos 4.23o - 10 cos 8.64' + 16.667
Qq = - l 0 c o s 5 .1 1 " - 6 .6 6 7cos 9.52o+ 16.667
Q+ = 0.132Pu
Reactivepower generationat the four busesare
Qa = Qt + 0.5 = 0.57pu
Qcz = Qz + 0.4 = 0.62 ptr
Qa = Qs + 1.0= 1.L32pu
Qc+= Q++ 1.0= 1.132pu
sin ({ - 6o)
pn = - pz,=
-r/
\ r - q)+0 . 1 5 sin(d,
sin-1.23"
= 0.492pu
0.1:
4'41o
Pt z = - Pzr
= - 0'385 Pu ( ix)
L'L= - 1- sin ( 4 - 6) = - $n
0. 2
02
Pqt= +
s i n ( { - 6 o )= 1 0 s i n 5 . 1 1 o = 0 . g 9 1p u
0.1
Real power flows on other lines can be similarly calculated.
For reactive power flow, Eq. (5.69) can be written in the general
form
( l Z l =X , 0 = 9 0 o )
Pru=-
- lvi-llvkl cos(,{ - 6o)
ei* =W
Xik
Xik
where Q* ir the reactive power flow from bus i to bus ft.
I
Q p = _1,
Q,zr=+cos(d,_ hl = 0.015pu
0.2 i.,
(vi)
2 + j0.ET
1
0.891+/O.04
* l i't ',t,
, 3 8 5- 7 0 .0 1 5
0.492-70. 018
3
1 . s 0 2- i 0 . 1 1 3
0.385+,p. 01S
0.891-7O.04
1.502+ 10.113
1.103- j0.0s2
4
1 . 1 0 3+ p . 0 9 2
n:n l|
2
t, jD.4
(vii),
Fig. 6.7 Loadflow solutionfor the four'-bus
system
Q r c= Q u=
#
#
c o s( d 1 ,O = 0 . 0 1p8u
I
www.muslimengineer.info
"
@
OI'
Qr = 0'07Pu
Qz = 0'22Pu
Qz = 0'732Pu
x,o
where P* is the real power flow from bus j to bus k.
(ii)
(iv)
=
(iv),
(ii),
(iii)
we
0, and solving
and
Taking bus 1 as a referencebus, i.". 4
get
--O.0ll rad = 4.4I"
4.
lvi]-lvkl
in the form
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowerSystemAnalysis
.2Oftril
I
I en*.
+ ' = + - -1- .o, (6r - 64)= 0.04 pu
eA = *e+t
0.1 0.1
lines can be similarlycalculated.
other
on
power
flows
Reactive
and load demandsat all the busesand all the line flows are
Generations
carried out at the end of a completeiteration, the processis known as the Gauss
iterative method. It is much slower to convergeand may sometimesfail to do
so.
Algorithm
6.5 GAUSS.SEIDEL METHOD
The Gauss-Seidel(GS) method is an iterative algorithm for solving a set of
non-linear algebraic equations.To start with, a solution vector is assumed,
basedon guidancefrom practical experiencein a physical situation. One of the
equationsis then used to obtain the revised value of a particular variable by
substituting in it the present values of the remaining variables. The solution
vectoris immecliatelyupdateclin respectof this variable.The processis then
repeatedfor all the variablesthereby completing one iteration. The iterative
processis then repeatedtill the solution vector convergeswithin prescribed
accuracy.The convergenceis quite sensitive to the starting values assumed'
Fortunately,in a load flow study a starting vector closeto the final solution can
be easily identified with previousexperience.
To explain how the GS method is applied to obtain the load flow solution,
let it be assumedthat all busesother than the slack bus are PB buses.We shall
seelater that the methodcanbe easily adoptedto include PV buses as well. The
slackbus voltage being specified,there are(n - 1) bus voltages starting values
of whose magnitudesand angles are assumed.These values are then updated
through an iterativeprocess.During the courseof any iteration, the revised
voltage at the ith bus is obtained as follows:
(6.3e)
J, = (P, - jQ)lvi
[from Eq. (6.25a)]
From Eq. (6.5)
,[
I
v , =,,'
*lL,_
fv,ovol
L T:i I
Substitutingfbr J, from Eq. (6.39) into (6.40)
(6.40)
-l
t^ . - , Yit'vr' = 2'3""' n
v,= *l '':jq
Y" I
I
vr*
t
l
ilt
k*t
I
(6.4r)
I"
The voltages substitutedin the right hand side of Eq. (6.41) are the most
recently calcuiated (updated)values for the correspondingbuses. During each
iterationvoltagesat buses i = 2,3, ..., n aresequentiallyupdated throughuse
of Eq. (6.41). Vr, the slack bus voltage being fixed is not required to be
updated.Iterations are repeatedtill no bus voltage magnitudechangesby more
than a prescribed value during an iteration. The computation processis then
said to converge to a solution.
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for Load Flow Solution
Presentlywe shall continueto considerthe casewhere all busesother than the
slack are PQ buses.The stepsof a computationalalgorithm'aregiven below:
1. With the load profile known at each bus (i.e. P^ and 0p; known),
allocate* Po, and.Q5; to all generatingstations.
While active and reactive generationsare allocatedto the slack bus, these
are permitted to vary during iterative computation.This is necessaryas
voltagemagnitudeand angle are specifiedat this bus (only two variables
can be specified at any bus).
With this step,bus iniections(P, + jQ) are known at all busesother than
the slack bus.
2. Assembly of bus' admittance matrix rsus: with the line and shunt
admittancedata storedin the computer, Yru, is assembledby using the
rule for self and mutual admittances(Sec. 6.2). Alternatively yru, is
assembledusing Eq. (6.23) where input data are in the form of primitive
matrix Y and singularconnectionmatrix A.
3. Iterative computationof bus voltages(V;; L = 2, 3,..., n): To start the
iterationsa set of initial voltage values is assumed.Since, in a power
systemthe voltagespreadis not too wide. it is normalpracticero use a
flat voltage start,** i... initialiy ali voltagesare set equal to (r + 70)
exceptthe voltageof the slack bus which is fixed. It shouldbe noted that
(n - l) equations(6.41)in cornplexnumbersare to be solvediteratively
for findin1 @ - 1) complex voltages V2,V3, ..., V,. If complex number
operationszlreDot availablein a computer.Eqs (6.41)can be converted
rnto 2(n - 1) equationsin real unknowns(ei,fr or lV,l, 5) by writing
Vr = €i + ifi = lV,l ei6'
(6.42)
A significant reduction in the computer time can be achieved by
performing in advance all the arithmetic operationsthat do not change
with the iterations.
Define
i = 2,3, ...,ft
'Active
(6.43)
and reactive generation allocations are made on econorfc considerations
discussedin Chapter 7.
.*A
flat voltage start means that to start the iteration set the voltage magnitudesand
anglesat all busesother than the PV busesequal to (i + l0). The slack bus angle is
conveniently taken as zero. The voltage magnitudesat the PV buses and slack bus are
set equal to the specified values.
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Modernpower SvstemAnatvsis
206 |
I*rl.d
(6.44)
flows on the lines terminatingat the slack bus.
Acceleration
vi'
17(r+l) -
Ai
'S
n
ilt
k=irl
- t B , o- v or t,( ,, * t , -
(Vl")
r , o v l , ,i = 2 , 3 , . . . , n
f
(6.4s)
The iterative process is continued till the change in magnitude of bus
voltage, lav.('*r)|, between two consecutiveiterations is less than a
certain tolerancefor all bus voltages,i.e.
IAV.G*r)l_ 1y.t+r)_ V,e)l < 6, ; i = 2, 3, ..., n
(6.46)
4. computation of slack bus power: substitution of all bus voltages
computedin step 3 along with V, in Eq. (6.25b) yields Sf = pr- jey
5 . computtttionofline .flows:This is thc last step in thc loaclflow analysis
whereinthe power flows on the variouslines of the network are computed.
Considerthe line connectingbuses i and k. The line and transformersat
each end can be representedby a circuit with series admittance y* and
two shunt admittances1l;roand )no as shown in Fig. 6.8.
B u si
up by the use of the
acceleration factor. For the tth bus, the accelerated
value of voltage at the
(r + l)th iteration is given by
y(r+r)(accelerated)
= V,Q'* a(v.G+r)- V,Ql,
where a is a real number called the acceleration
factor. A suitablevalue of a
for any system can be obtained by trial load flow
sfudies. A generally
recommendedvalue is a = 1.6. A wrong choice of
o. may indeed slow down
convergenceor even cause the method to diverge.
This concludesthe load flow analysisfor thJ case pe
of
busesonry.
Algorithm
when pv Buses are arso present
Modification
At the PVbuses, p andrv]rarespecified and ancr
dare the unknownsto be
e
detcnnincd.'l'hererirre,
the valuesof e and d are to be updatedin every
GS
iteration through appropriate bus equations.This
is accomplishedin the
following stepsfor the ith pV bus.
l. From Eq. (6.26b)
Busk
l*io
of convergence
f
"
)
L
ft:l
)
ei = - Im j yr* D, y,ovof
sm
The revised value of ei is obtained from the
above equation by
substitutingmost updatedvalues of voltageson rhe
right hand side. In
fact, for the (.r + 1)th iteration one can write from
the above equation
I
g.(r+t)
= -Ln
Fig. 6.8 7i'-representation
of a line and transformers
connectedbetweentwo buses
The current fed by bus i into the line can be expressedas
Iit = Iitt + Iirc = (Vi - V) !it,+ V,y,oo
(6.47)
The power fed into the line from bus i is.
S* = Pir* jQiF Vi lfr= Vi(Vf - Vr\ yft+ V!,*y,f,
Similarly, the power fed into the line from bus k is
(6.48)
Sri = Vk (V*k- Vf) yfo+ VoVfyi,l,
(6.4e)
2' The revisedvalueof
{.is obtainedfrom Eq. (6.45) immediatelyfollowing
step 1. Thus
6Q+r)_ ay!+r)
= Ansle fei]l- - i Bovo(,+r)
- D B,ovo,,,l
(6.s1)
"t L ( t 1 ' ' ' ) * , r :r
k:,+r
J
where
The power loss in the (t - t)th line is the sum of the power flows determined
frorn Eqs. (6.48) and (6.49). Total transmissionloss can be computed by
sununingall the line llows (i.e. 5';a+ Sri fbr all i, /<).
www.muslimengineer.info
a 1y.r,1i y,rv,,(,*t)
],r,',)
D y,kvk,,,I
" ,u.ro,
Lr,
I
f!,
)
.
P-i?t'rtt
_
r^1(. ;r +. t=)_-_
,u
The algorithm for pe buses remains unchanged.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
6.52)
l'fiffi
1 PrimitiveY matrix
2 Bus incidencematrixA
, 1)
3 S l a c kb u s v o l t a g e( l Y 1 l 6
4 R e a lb u s p o w e r sP i f o ri = 2 , 3 ' 4 , " " ' n
=
5 Reactivebus powersQi,for I m + 1" -{fO buses)
Vf I for i = 2,....,m(PV buses)
o
7 VoltagemagnitudelimitsIVrI min andIVi I max'lor '-(J
b n"u"iiu" pow"t limitsQi min and Q; max for PV buses
tg.
Yr,. r.i"g Eq.(6.2
=
for i = 2!:'m
Make initialassumptionsViofor I m + 1,"',n and O;0
=
i = 1'2"
Computethe parametersA; for i ^ t 1,"n .and B;p'for
(exceptk = i ) from Eqs (6'43)and (6'44)
-.;; i: 1,2,-'...,n
Set iterationcount r = 0
demand at any bus must be in the range e^rn - e^u*.If ai any stage
during the computatibn,Q at anybusgoes outsidettreJetimits, it is fixed
At Q^in or Q^o, as the case may be. and the
dropped,i.e. the bus is now treatedlike a pe bus. Thus step 1 above
branchesout to step 3 below.
3. ff 9.('+r) a 0;,6;n, set e,Q*r) - er,r^n and treat bus f as a pe bus.
compute 4-(r+1)and y(r+t) from Eqs. (6.52) and (6.45),respectively.If
Otu..:,')Qt,^ , set O-(r+l)= ei,^*and treatbus i as a pebus. Compure
4.Q+r)-6 y(r+l) from Eqs. (6.52) and (6.45), respecrively.
Now all the computationalstepsare summarizedin the detailedflow chart
of Fig. 6.9 which serves as a basis for the reader to write his own
computerprogramme.It is assumedthat out of n buses,the first is slack
as usual,t hen2, 3, . . . , m ar ePVbusesand t he r em ainingm + l, . . . , f t
are PQ buses.
Example 6.4
I
For the samplesystemof Fig. 6.5 the generatorsare connectedat all the four
buses,while loads are at buses2 and3. Values of real and reactivepowers are
listed in Table 6.3. All busesother than the slack are pe typd.
Assuming a flat voltage start,f)nd the voltagesand bus anglesat the three
busesat the end of the first GS iteration.
Solution
Table 6.3
Bus
I
2
J
4
Pp PU
0u pu
0. 5
- 1.0
0.3
- 0.2
0.5
- 0.1
lnput data
Vt' Pu
Retnarks
r.u 10"
Slack bus
PQ bus
Pp bus
Pp bus
The l"ur for the samplesystemhasbeencalculatedearlierin Example 6.2b
(i.e. the dotted line is assumedto be connected).In order to approach
the
accuracy of a digital computer, the computations given below have been
performed on an electronic calculator.
T)--- ---r-, ,i
-Dus
voltages at rne eno or tne llrst rteratron are calculated using Eq. (6.a5).
vtz=+{W-Yztvt-Y"v!
Fig.6.9F|owchartforloadflowso|utionbytheGauss.Seide|
iterative method using YBUS
l
-- - - - - 11 -0' -. 5 r+ -i '0- . -2 -1' .- 0' \ 4( - z- +' rj-6/ ) - (\ - 0v .' v6v6v6+' j 2 ) _ ( _ t+
r!
fll
YzzI
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- 'r^'iI
t -iO
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
')
|
ModernPowerSystemAnalysis
fei.fii,',1
I
4.246- jrr.04 =
3.666- jLr
1 - -
"
Ytt
_
-
Load Flow Studies
1.019+ 70.046pu
(vl)*
1
Yzt
- e2+ i6)
(to4
+ io)
T#-frL+re
- r.04 (- 1 + 13)
- (- 0.666+ j2) (1.019+ ;0.046)- (- 2 * i6)
- (- 0.666+ j2)(t +
I
= t( +'z^?e;t
- itt'lzs = , (r.rstz+
70.033e)
(
3 . 6 6 6 _ j t t )J-
|
- 70.087
= ''!'-,-1"'.9?' = 1.028
pu
- jrr
3.666
= 1.84658o= 0.032 rad
or 612
.' . v) = 1. 04( cos 6) * j sin dj)
- r^rr\l
= 1.04 (0.99948+ j0.0322)
= L03946 +
19 + 70.046)
I
- (- 2 + j6) (1.028- jo.o87) |
)
r)
70)_ (_ 1+ j3)(t. io)jl
70.03351
( o
v' tl ==- l - j ' 1 - = i Q --t yY"v'
. v . -- yYszv;
_vt _
,,0I
- vYrovl
,
,l
1 114tr
[ - r _ 7 0 . s- (- t +
-
3.666-in L ,f;;
2.99r- j9.2s31.025- 70.0093pu
3-je
i3) 1'04
- (- 0.666+ j2) (1.03s46
+ 70.03351)
_ (_ z + io)f
r.0317-
In Example6.4,let bus 2 be a PV bus now with lV2l= 1.04pu. Once again
assuming a flat voltage start, find Qz, 6., V3, V4 at the end of the first GS
iteration.
Given: 0.2 < Q, < l.
From Eq. (6.5), ii e", (Note fz= 0, i.e. Vl - I.04 + i0)
.,:,,,:.
n,=_
::;:!::
^
^r!:,
J,^.,,?,:
: [J A,,i".
+ (- 0.666+ j2) + (- I + i3)l )
= - Im {- 0.0693- j0.20791- 0.2079pu
' O" = 0'2079Pu
FromEq. (6.51)
www.muslimengineer.info
- Yo,vJ
=
I
il
[0.3+io.r(- I + i3) (1.03e4
+ 70.0335)
tf:;
- (- 2 + j6)(r.03t7_ yo.08e3z)
I
J
_ 2.967r_ j8.9962 = 0 ' 9 9 8 5 - 7 O ' 0 0 3 1
3-ig
permissibtelimits on ez (reactivepower
injection)are
",,il1;,;llii,""rrirhe
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowerSystem Analysis
I aar{
Lwqv
0 .2 5 < Qz < i .0 p u
It is clear, that other data remainingthe same,the calculatede2 (= 0.2079)is
now less than the Qz, ^in.Hence e, it set equal to ez,
_in, i.e.
=
0'25
Qz
Pu
-
a
p
---
uvrvrvrv,
tf
2t
vqLt
Ela..,
I tvyy
Or.,-r:^^
\)t,U(JlUli
6.6 NEWTON-RAPHSON (NR) METHOD
The Newton-Raphson method is a powerful method of solving non-linear
algebraic equations.It works faster and is sure to convergein most
cases as
llv
longer remain fiied at I.04 pu. The value of V, ar the end of the first iteration
is calculatedas follows. (Note VL = t + 70 bf virrue of a flat start.)
'r,
vL
' z = l-(
Yr,
[
z t | - -yztv?
z r J - -yrovl)
'"
^!?, - -yztvt
(Y3)-
")
_[o.s-jo.zs
3 . 6 6-6j r l L t - r ,
Considera set of n non-linearalgebraicequations
- (- 0.666+ .i2)- (- r + i 3 ) ]
(* 2 + i6)1.04
- 4'246- jlr-,? = t.05.59
+
3 . 6 6 6- j t l
io.o341
(
- Y,,vJ
vj" = : l':-:* -.Y,,v,
r,, (Y,')'
[
3 . 6 6 6 - j r[_t;u.t
r Lr - ; o
vL=;;W,#
=
f;(x01,xor,
70.0341)- (- 2 + j6)
]
- 70.0893pu
1.0347
I
\ux, )
l
)' are the derivatives
whcrefg)',(
!f,
)'.
.f9
'
'
of It with respectto
d"r
Neglecting higher order terins we can write Eq. (6.55) in matrix fbrm
- (- I +,r3)
(r'0s0e
+ i0'0341)
-1
3- je
/ ^- \0
( Dr, )
\
/
[ ,", ,/
x1, x2, ..., x, evaluat edat ( x! , *1, . . . , *0, ) .
- (- 2 + j6) (r.034i- j0.08e3)
I
/,.0630- j9 .42M
,,r).[[*)' a*0,
.(#)' Axt
+
*[ -l!t] o-i l* n,rn-,order
- o (6.ss)
re,,ns
- Yo,v,
- Yo,u|
- v*v))
+,t"=#
(6.s3)
Assur.cinitialvulucsof u'know's as *l: *'), ..., r"r. Let J.r(1,Jg, ..., J_rl
be the corrections,
whichon beingaddedto the initial guess,givelhe actual
solution.Therefore
J t @ \ + f u t l , * u r + A x l , . . , , x 0 , + A x f ; 1= g ; i = 1 , 2 , . . . , n ( 6 . 5 4 )
Expandingtheseequationsin Taylor seriesaroundthe initial guess,we have
_ (_
* r + j3)toa
- (- 0.666+ j2X1.0s5e+
_ 2 . 8 tt 2 - j n . 7 0 9
3 . 6 6 6- j r r
f i ( \ , x 2 , . . . , x n )= 0 ; i = I , 2 , - . . , n
f'''l
I'rl
_ 1.0775
+ j0.0923pu
a'l'
Llxi
(6.56a)
Ax:
or in vector matrix form
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2ll
I
Modern Power Svstem Analvsis
-
l(, + .f,A*, * 0
(6.56b)
,f is k,o*n as the Jacobianmatrix (obtainedby differentiatingthe function
it at r01. Equation(6.56b)can be
vectorf withrespectto x and evaluating
writtenas
:l_J
m
Approximatevaluesof corrections/-r0 can be obtainedfrom Eq (6.57).These
being a set of linear algebraic equations can be solved efficiently by
triangularizationand back substitution(see Appendix C).
Updatedvaluesof x are then
tl ="0 + AxI
or, in general, for the (r + 1)th iteration
(6.s8)
"(r+l)_r(r)+AxQ)
Iterationsare continuedtill Eq. (6.53) is satisfiedto any desiredaccuracy,i.e.
(6.5e)
value); i = 1,2, ..., n
lJiG")l < e (a specified
NR Algorithm
for Load flow Solution
First, assumethat all busesarePQ buses.At any PQbus the load flow solution
rnust satisfythe following non-linearalgebraicequations
(6.60a)
J'ip (lV, 6) = I'i (sPccificcl) Pi = 0
(6.60b)
=
=
(specified)
o
Qi
Qi
fiq (1v1, 6)
for P, and Q, are given in Eqs. (6.21) and (6.28).For a trial
whereexpressions
set of variableslV;1,6;,the vector of residuals/0 of Eq. (6.57) comespondsto
- Pi (calculated) = APi
(6.61a)
fip= Pi (specified)
- Qi @alculated)- AQi
(6.61b)
.fie= Qi(specified)
( 6. 63a)
It is to be immediatelyobservedthat the Jacobianelementscorrespondingto the
ith bus residuals andmth bus correctionsare a 2 x 2 matrix enclosedin the box
in Eq. (6.62a) where i and m Ne both PQ buses.
Since at the slackbus (bus number l), Prand Qr are unspecifiedand lV,l,
Q are fixed, there are no equationscorrespondingto Eq. (6.60) at this bus.
Hence the slack bus does not enter the Jacobianin Eq. (6.62a).
consider now the presenceof PV buses.If the ith bus is a pv bus, e, is
unspecifiedso that there is no equation correspondingto Eq. (6.60b) for this
bus. Therclbrc, the Jacobianeleurentsof the lth bus become a sinele row
pertainingto AP,, i.e.
(6.62b)
rthbusF
l=
while the vector of corrections y'xo corresponds to
alvil, a,i
vectorcanbe written
corrections
Equation(6.51)for obtainingtheapproximate
for the load flow case as
If the mth br-rsis also a PV bus, lVrl becomesfixed so that Alvml = 0. We
can now wnte
mth bus
I
I
fthbusl
,lPi
aQi
(6.62a)
I
ous
AlV^ jmtn
46^
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i t h b u snlp , l = l
T:tT'
L----J t-- ---- ----rH''l
-
(6.62c)
I
l
mth bus
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
rtrodernPower SystemAnalysis
2L6 |
l ztz
t
Also if the ith bus is a PQ bus while the mth bus is a PV bus, we can then write
Ninr= N^i = 0
l,^=J^;=O
(6.66)
(6.62d)
It is convenientfor numerical solution to normalize the voltage corrections
corresponding to a parricular vecror of variables
tqlv2lq64lval6lr, the
vector of residuars[aP2 aez ah ap4 ae4 Apr], and the Jacobian(6
x 6
in this example) are computed. Equation
rc.an is then solved by
triangularization and back substitution procedure to obtain the
vector of
f
alv^l
lv^l
elements
become
Jacobian
asa consequence
of which,the corresponding
nvt
Atrl|
1r
correctionsoo,#
oo1464
/,6rl . corrections
arethenaddedto
J a+
I
lv2l
lVor
L
I
updatethe vector of variables.
2(Pa)
3(PVl
(6.63b)
Expressionsfor elementsof the Jacobian (in normalized form) of the load
flow Eqs. (6.60a and b) are derived in Appendix D and are given below:
Case I
m t l
H,^= Li^= a^fi - b*et
Nr*=-
Jirr= a.er+
6.64)
Fig. 6.10 Samplefive-busnetwork
bJt
-'-* BusNo.
Yi^= G* + jB,*
2
Vi= e, + jf,
(a* i ib)
I
l 2
= (Gi- + jBi*) @* + jk\
I
Y
case2
f,z
IA A<\
\\J
' \'J,'
Biilvilz
An important observation can be made in respect of the Jacobian by
examination of the Y"u5 matrix. If buses i and m are not connected, Yi^= 0 (Gi^
= Bi^ - 0). Hence from Eqs. (6.63) and (6.64), we can write
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5
Hzs
Hzn
Nza
Jzz
Lzz
Jzs
Jz+
Lz+
Hsz
Nsz Hss
Hqz
Nn
Haa
Naz.
H+s
Jqz
Lqz
Jcc
Lu
J+s
Hss H5a
Nsq
Hss
Hss
J
J ti = Pi - Giilvil'
Li;= Qi-
4
Nzz
o
I,,==t- n, - Biirvirz
Mii= Pi + G,,lV,lz
3
Hzz
c0
(6.67)
[:_]
t
Jacobian
(Evaluatedat trial values of variables)
lalv4ll
l]ql-l
i_it_l
t
Corrections
in variables
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,re
-T-LLV
rrrrodcrnPower Svstem Analvsis
II
Lcad Ftor; Studies
Iterative Algorithm
Omitting programmingdetails, the iterative algorithm fbr the solution of the
load flow problem by the NR method is as follows:
L W'ith voittge anclangle (usually f'= €I) at s
at atl PQ busesand d at all PV buses' In the absence of anY tlther
information flat voltage start is recommended'
2. Compute AP,(for PV and PB buses)and AQ,(for aII PQ buses)from
Eqs. (6.60a and b). If all the values are lessthan the prescribedtoletance,
stop the iterations,calculate P, and Q, and print the entire solution
including line flows.
3. If the convergencecriterion is not satisfied, evaluate elements of the
JacobianusingEqs. (6.64) and (6.65).
4. Solve Eq. (6.67)for correctionsof voltageangles and magnitudes'
5. Update voltage angles and magnitudes by adding the corresponding
changesto the previousvaluesand returnto step 2'
Note: 1. In step 2, if there are limits on the controllable B sourcesat PV
to
buses,Q is computedeach time and if it violatesthe limits, it is made equal
in
that
bus
PQ
a
made
is
bus
PV
corresponding
the
and
value
the limiting
iteration. If in the subsequentcomputation, Q does come within the prescribed
limits, the bus is switchedback to a PV bus.
2. If there are limits on the voltage of a PQ bus and if any of these limits
is violated, the correspondingPp bus is made a PV bus in that iteration with
voltage fixed at the limiting value.
Considerthe three-bussystemof Fig. 6.11. Eachof the three lines has a series
pu and a total shunt admittance of 70'02 pu' The
impedanceof 0.02 + 1O.OS
specifieclquantitiesat the _b"!91j9!qulated below:
RealloadReactiveloadRealpowerReactiveVoltage
Bus
Qo
generation
PG
specification
power
generation O6
1.0
Unspecified
UnspecifiedV'=1.94 * ;g
(slack bus)
0 .0
0 .0
0.5
1.0
Unspecified
(PO hus)
l 5
0 .6
0.0
Qct=?
lvll= 1.04
(PV bus)
demand
PD
demand
2 .0
C o n tro l l a b l crc a c ti v op o w o r s o u rccl s avai l abl cnt btts3 w i th the ctl nstrai nt
0<Qct S 1.5Pu
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I i:ig,:
Find the load flow solution using the NR method.Use a toleranceof 0.01 for
powermismatch.
Solution Using the nominal-rcmodel for transmissionlines, I"u, for the given
systemis obtained as follows:
eacn ltne
I
1 , snes
en."=
^ - z . g 4 r - j 1 1 . 7 &- r 2 . r 3 l - 7 5 . 9 6 "
o. oz+jo. og
Each off-diagonalterm = - 2.94I + jll.764
Each self term = 2l(2.941 - j11.764)+ j0.011
= 5.882- j23.528 = 24.23 l-75.95"
0 +,O
2
3
1 . 0 4l O "
1 . 5+ 7 O . 6
Fig. 6.11 Three-bus
systemfor Example6.6
To startiterationchoose 4=
we get
t +70 and ,4 = 0. From Eqs. (6.27) and (6.2g),
Pz = lVzllvl lY2l cos(0r1+ 6r- $) + lVrlzlyrrl cos0zz+ lvzl l\l
l Y r r lc o s( Q z t +q - 6 )
Pt = lVl lvl l\rl cos(dr, + 6r- 6r)+ lVrl lv2ll\zl x cos(ez +
h
- 6r)+ lV,rlzlyrrl cos 0r,
Q z = - i v z i i v l t y z l s i n( g r + d t - h ) - l v z l zl y 2 2xl s n 4 z _ l v 2 l
lv3l lY.,.lsin (0r, + bz- 6)
Substituting
givenandassumed
valuesof differentquantities,
we getthevalues
trf Powersrts
= -0.23pu
PB.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
O}nr'l
ttv.l
ilada-n
rvruuErrr
t
Darrrar
r VYYEr
Crralam
\)yolgltr
Anahraia
nrrqryoro
Load Ftow Studies
i zzr
-
ti^--i.
P3= 0.12pu
Sz=0.5+j1.00
Q o r =- 0 ' 9 6 P u
Sr=-1.5-j0.15
Power residualsas per Eq. (6.61) are
- rt (calcu
Transmissionloss = 0.031
Line
(- 0.23)= 0.73
flows
The following matrix shows the real part of line flows
Aror- -1.5 - (0.12)= - !.62
tQT= 1 - (- o'e6)- t'e6
I
The changesin variablesat the end of the first iteration are obtainedas follows:
0P,
06,
0P,
06,
aQ,
0P, 0P,
061 alv2l
aP,
7Pu
061, 0lv2l
}Qz aQ,
I
t/..25
[ n r v , r ' -L]- u u
s . 6 4 - 1 - r [ 0 . i 3[ -f 0 . 0 2 3 - 1
^ ^ - l
/.4.e) -J.u)l
|
|
- t .-o^ zl l : l -| u .^u^o. )- 4. 1
|
3 . 0 s z z . s 4L) r . q o lI o o a r l
l a ) l I a i - l l - ^ 4 I t - 0 .[1 0 0 2I. 3[ 0 0 2 3 . 1
I a ]l : l I l * l ^ 4 l : l o l * l - 0 0 6 s 4 1 : l - 0 0 6 s 4 1
L'i I oosoJI r.08eJ
Itv,t'j L'y,roJ
lnrv.,r'.1
We can now calculatefusing Eq. (6.28)]
Qtt= 0.a671
Qo \= Q \ + Qrt = 0 . 4677+ 0.6 = 1.0677
which is within limits.
If' the sanre problem is solved using a digital computer, the solution
convergesin three iterations.The final resultsare given below:
Vz= 1.081 lVt = I.M l-
o.o
flows
-0.5994&E00 _0.r9178zE00]
0.0
0.39604s800
I
l|0.60s274800
-0.37sr6s800
L0.224742E00
0.0
I
Rectangular
I Aai l:l-
0.1913r2E00 0.839861E'00-l
0.0
0.6s4697
E00|
I-0.r8422eE00
-0.673847E00
L-0.826213^800
0.0
J
I
Jacobian elements can be evaluated by differentiating the expressions given
above fr>r Pr, Py Qz with respectto 6r, d1 and lVrl and substituting the given
and assumed values at the start of iteraticln. The chanses in variables are
obtained as
-t2.23
oo
The following matrix shows the imaginarypartof line
'av|
06, 06:'
I| t d j^l rl -I 2 4
.4
i
|
. ^ ^ ^
|
0.024 rad
0.0655 rad
Power-Mismatch
Version
This versionusese, ancl.f
,the rear ancr imaginary parts of the v.ltages
resllectivcly,as variables''fhe number of equationsand
variablesis greater
than tharttirr Eq. (6.6r), by the nurnber<ttpi buses.Since pv
at
busese,and
.f; canvary but ,,' +.f,' - lviP, a voltage-magnitude
squareclmisn-latch
equation
is required tbr each PV bus. with sparsityprogramming,
this increasein order
is of hardly any significance.Indeed eachiterationis rnarginally
faster than for
Eq' (6'67) sincethereare no time-consunring
sine anclcosineterms.It nray,
however,be noted that even the polar versionavoids these
as far as possible
using rectangular arirhemetic in constructing Eqs (6.64)
9f
and (6.65).
However'thc rcct:tttgtrlltr
vcrsionsccnlsto bc slighiiy t.r, rcliablebut faster
in
convergencethan the polar version.
The total numberof non-linearpower flow equationsconsidered
in this case
arc fixed iurd cqual 2 (trl). Theselbllow fr.o'i Eqs. (6.26a)
and (6.26b)and
are
P, (specifi"d)-I{
- f,B,t) +
e,(epG11,
fihrG*
-t epB,o}: 0
k:l
i = l , 2 , . . . ,n
( 6. 68)
i = slack(s)
n
rt.
B; (specifi"d)-)
=g
lf,koG,o - ftB,t1- e;(.f*Gi** eoB,oy]}
k=l
Q u = - 0 . I 5 + 0 . 6 = 0 . 4 5( w i t h i nl i n r i t s )
S r = 1 .0 3 1* j (- 0 .791)
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(lbr each PQ bus)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(6.69)
222 |
Modern Porgl Jyllem Anatysis
I
(specified)2- @i2+ f,') = 0
l(l
(for each PV bus)
Using the NR method,the linearisedequationsin the rft iteration of iterative
processcan be written as
AQ
AIVP
F o ri *
Jr
J2
J3
J4
Js
J6
- Jqij = Gij ei
fi
Jzij= Jt,j=-
Bijei+ Gilfi
6.72)
Jsij= Jo,j = O
For i=,r
tlr,,,
-!'u,u-'
l,)i J:,
(6.73)
Jsii= 2"i J6ii = 2fi
a, and b, are the componentsof the current flouring into node i, i.e.,
jBi)@r+jf*)
frCo *
(6.74)
k:l
Steps in solution procedureare similar to the polar coordinatescase,except
that the initial estimatesof real and imaginary parts of the voltages at the PQ
busesare madeand the correctionsrequiredare obtainedin each iteration using
(6.7s)
The correctionsare then applied to e andf and the calculationsare repeatedtill
convergenceis achieved.A detailed flow chart describing the procedure for
load flow analysisusing NR methodis given in Fig. 6.12.
F E
O. I
F F A A ' ? h I E h
L'|aUL,UTL|1L'
In any conventionalNewton method, half of the elements
of the Jacobean
matrix representthe weak coupling referred to above, and
therefore may be
ignored' Any such approximationreducesthe true quadratic
convergenceto
geometric one, but-there are compensatingcomputational
benents.i large
number of decoupled algorithms have been dlveloped
in the literature.
However, only the most popular decoupledNewton veision
is presentedhere
t1el.
tApl = lHl lA6l
( 6. 76)
I^Qt- tLtl4!1
( 6. 77)
Ltvt J
whereit can be shownthat
gU = Lij = lvil lvjl (CUsin 6u- B,i
cos {r)
i * j
(6.78)
H i i = - 8 , , l V i P - Qi
(Eq. (6.6s))
(6.7e)
Lii= - 8,, lV,lz + Qi
(Eq. (6.6s))
(6.80)
Equations(6.76) and (6.77) can be constructedand solved
simultaneously
with each other at each iteration, updating the
[H] and [r] matrices in each
iteration using Eqs (6.78) to (6.80). A better opproo.h
is to conduct each
iteration by frst solving Eq. (6.76) for 44 *o use
the updared 6 in
constructingand then solving Eq. (6.77) for Alvl. This
will result in faster
convcl'geucethan in the sirnultaneousmode.
The main advantageof the DecoupredLoad Flow (DLF) as
comparedto the
NR method is its reducedmemory requirementsin storing
the Jacotean.There
is not much of an advantagefrom the point of view of speed
sincethe time per
iteration of the DLF is almost the same as that of NR method
and it always
takes more number of iterations to convergebecauseof the
approximation.
Fast Decoupled Load Flow (FDLF)
' A A
L('AL'
Newton Methods
In Eq. (6.67), the elementsto be neglectedare submatrices
[1v] and [,/]. The
resulting decoupledlinear Newton equationsbecome
t;^"':::
.-::::!-
a,+ jb,=
necessarymotivation in developing the decoupledload flow (DLF)
method, in
which P-6 and Q-V problems are solveclseparately.
Decoupled
j
Juj=
Load Flow Studies
(6.70)
.F LL'VV
IVI.EI.tsI(-,L'Ii
An important characteristicof any practicalelectric power transmissionsystem
operatingin steadystateis the strong interdependence
betweenreal powers and
bus voltagesanglesand betweenreactivepowersand voltagemagnitudes.This
interestingproperty of weak coupling betweenP-6 and Q-V variable.sgave the
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Furtherphysicallyjustifiablesimplificationsmay be carried
out to achievesome
speed advantagewithout much loss in accuracy of solution
using the DLF
model described in the previous subsection.This effort culminated
in the
developmenrof the Fast DecoupledLoad Frow (FDLF) merhod
by B. stott in
1974l2ll. The assumptionswhich are valid in normal power
ryr,"n1operation
are made as follows:
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'*i
I
I
-::,:-,t
I zas-
vodern Power SvstemAnalvsis
With theseassumptions,the entriesof the [1{ and [L] submatriceswill become
considerablysimplified and are given by
count
I Aduanceiteration
I
r= r+1
and
(6.82)
? =i:=-''i,,,11,,u"
:: ,::
(6.83)
Matrices [/fl and [L] are squiue mafrics with dimension (npe + npy) and nrn
respectively.
Equations (6.76) and (6.77) can now be written as
DetermineJ-1
.'. comPuteAefr) 2n6
Lfiv)
c-mpute ano--l
printlineflows,l
powerloss, I
voltages, I
elci ----.]
tolerance
2
max
Determine
changein power
max APr,AO
')
Compute
el,),1v112
ls
-\
1;
l-V
s i n{ r : 0
and
Qi < Biilvilz
(6.86)
(6.87)
In Eqs. (6.86) and (6.87), both [B/] and lBttf are real, sparseand have the
structuresof [I{ and [L], respectively. Since they contain only admittances,
they are constantand need to be invertedonly once at the beginning of the
study. If phase shifters are not present, both [B'l and lB"] are always
symmetrical, and their constant sparseupper triangular factors are calculated
and stored only once at the beginning of the solution.
Equations(6.86) and (6.87) are solvedalternatively.alwaysemploying the
most recent voltage values. One iteration implies one solution for [4fl, to
update[d] and thenone solutionfor IA lVl] to update[Vl] to be called l-dand
F i g .6 . 1 2
G,, sin 6,, < B4;
(6.8s)
IAP| lyl I = [B'] [L6]
tAQl tr4 I - LB" I t^lytl
Compute
LlVi,12= 1V,scnd12-l%tl2
cos 6, :
tAQl= [%t tvjtB,,ij,
[#]
4. Ignoring seriesresistancein calculatingthe elementsof tdll which then
becomesthe dc approximation power flow matrix.
With the above modifications, the resultant simplified FDLF equations
become
I last node
-'
(6.84)
where 8t,,, B(are elementsof [- B] matrix.
Further decouplingand logical simplificationof the FDLF algorithmis achieved
by:
1. Omitting from [B/] the representationof those network elements that
predominantly affect reactive power flows, i.e., shunt reactancesand
transformer off-nominal in-phase taps;
2. Neglecting from [B//] the angle shifting effects of phaseshifters;
3. Dividing eachof the Eqs. (6.84)and (6.85) by lv,l and settinglVrl = I pu
in the equations;
Are all
maxA within
gt'r ioi)
LAPI= ltyjt lvjl Bfi AA
iterafion
Senarafe.
"-r*^*'-
c- -r l^n^v' e- ^. ros-e^n c e
tcetc
are
annlied
fhr
the
rcql
qnA
(6.81)
max [APf I
€pi max lAQl S eo
where eo and €e are the tolerances.
A flow chart giving FDLF algorithmis presentedin Fig.6.13.
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rcqotlvc
power mismatchesas follows:
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
( 6. 88)
ModernpowerSvstemAnalvsis
,bZCI
LoadFtowStudies
I ##
(Start)
Considerthe three-bussystemof Example 6.6. Use (a) DecoupledNR method
Read LF data and form Ysr.
(a) Decoupled NR method:
Equationsto be solved are (see Eqs. (6.76) and (6.77)).
Substitutingrelevant valuesin Eqs. (6.78) - (6.80) we get
H z z = 0 ' 9 6 + 2 3 .5 0 8- 24.47
Hzt= Hzz= 1.04 (- Brz) = - 1.04 x 11.764= - 12.23
Htt=- Qt- Bn0.0q2
= [- Bsr t\P - 42 tv3t- Bzzt\P] - Bn (r.0a1.2
= - f 1 .7 6 4x (1 .04)2- 1t.764 x 1.04+ (l .oq2
x2x
2 3 . 5 0 8- 2 5 . 8 9
Lzz= Qz- Bzz= t + 23.508- 24.508
24.47_r2.2311\6t\1
[ 0.731
-_ [
l-t.oz) L- n.23 zs.ssJ[64trl
- lz4.srl
tLQzl
' f+uJ"'l
't;(-i)l
(i)
Solvefor A6f usingEq.
( 6 . 8 6i)= 1 , 2 , . .n. ,, i * s
(ii)
L
J
SolvingEq. (i) we get
L6;') = - 0.0082- 0.0401- - 0.002
Calculate 6i.1 = 6[+ 66[
fori=1,2,...,n,i-s
Aa{tr = - 0.018- 0.08- - 0.062
1.(6.60a)
= PV bus
Qz= - lvzl lvtl lYzl sin(0r, + 6r - 6r) - lvzlzV2zlsin 82
lvzl lhl llrrl sin (9zt+ 6, - q)
= - 1.04x 12.13sin (104.04+ 0 - 0.115)-24.23
------
Calculate
l slack
sli
bus power
er and
i
all line flows
flov
and print
X
sin(-75.95)- 1.04x 12.13sin (104.04+ .115- 3.55")
--12.24+23.505-12.39
- ' - -
Qz= - l'125
Ar-l
t-tv2-
-
1
r -
/
\-
1
1a<\
r.rLJ)
=
^
I
rcEq.l
.,n, I
1?rE
L.ILJ
:us- l
Substituting in Eq. (ii)
--
<'-
re
fr=
\-
1 --
; Afr/,R
[alY"(')ll
- [24.5r]
Lz.rz5r
Ltu,a,
l
e+
Flg.6. 13
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Analysis
PowerSystem
Modern
;'..ng I
AN 9 l= 0.086
= 1.086pu
lvftt - lvlo)t+atv|'tt
Q(rl canbe similarly calculatedusing Eq. (6.28).
The matrix equationsfor the solution of load flow by FDLF method are [see
Eqs. (6.86) and (6.87)l
-Brr1laf',f
-8,,)Lz4"
l
(iii)
and
lffil=
r-Bzzt
tatrt')tl
o.tz I
I -t.oz
|
I
|
-
(iv)
fil6411oqrl
23.508.1la4"
l
t l -
| -1.04 I
L: 1'5571
(v)
Solving Eq. (v) we get
46;r) - - 0.003
A6t'' = - 0.068
6')- - 0.003rad; {tl
0.068 rad
lz.rzsl- [23.508]tatvitl
alvtl= 0.09
l v t t = 1 . 0 9p u
Now Q3 can be calculated.
These values are used to compute bus power mismatchesfor the next
'LAPAVll and
iteration. Using the values of
lAQAl\l the above equationsare
solved alternatively, using the most recent values, till the solution converges
within the specifiedlimits.
6.8
COMPARISON
OF IOAD FLOW METHODS
In this section,GS and NR methodsare compared when both use liu5 as the
network model. It is experienced that the GS method works well when
programmedusing rectangularcoordinates,whereasNR requiresmore memory
when rectangularcoordinatesare used. Hence, polar coordinatesare preferred
for the NR method.
www.muslimengineer.info
lui:.
-
completean iteration. This is becauseof the sparsityof the network matrix and
the simplicity of the solution techniques.Consequently,this method requires
less time per iteration, With the NR method, the elementsof the Jacobianare
to be computedin eachiteration,so the time is considerablylonger. For typical
large systems,the time per iteration in the NR rnethodis roughly equivalent to
7 times that of the GS method [20]. The time per iteration in both these methods
increasesalmost directly as the number of buses of the network.
The rate of convergenceof the GS method is slow (linear convergence
characteristic),requiring a considerablygreaternumber of iterations to obtain
a solution than the NR methodwhich hasquadratic convergencecharacteristics
and is the best among all methods from the standpointof convergence.In
addition, the number of iterations for the GS method increasesdirectly as the
number of busesof the network, whereasthe number of iterations for the NR
method remains practically constant, independentof system size. The NR
method needs 3 to 5 iterations to reach an acceptablesolution for a large
system. In the GS method and other methods,convergenceis affected by the
choice of slack bus and the presenceof seriescapacitor,but the sensitiviry of
the NR method is minimal to these factors which causepoor convergence.
Therefore, for large systemsthe NR method is faster, more accurate and
more reliable than the GS method or any other known method. In fact, it works
for any size and kind of pro6lem and is able to solve a wider variety of illconditioned problems t23). Its programming logic is considerably more
complex and it has the disadvantageof requiring a large computei memory even
when a compact storage scheme is used for the Jacobian and admittance
matrices. In fact, it can be made even faster by adopting the scheme of
optimally renumberedbuses.The method is probably best suited for optimal
load flow studies(Chapter7) becauseof its high accuracywhich is restricted
only by round-off errors.
The chief advantageof the GS method is the easeof programming and most
efficient utilization of core menrory.It is, however,restrictedin use of small
size system becauseof its doubtful convergenceand longer time needed for
solution of large power networks.
Thus the NR method is decidecllymore suitablethan the GS method for all
but very small systems.
For FDLF, the convergenceis geometric,two to five iterations are normally
required for practical accuracies,and it is more reliable than the formal NR
method. This is due to the fact that the elementsof [81 and [Btt] are fixed
approximationto the tangentsof the defining functions LP/lVl and L,QAVl, and
are not sensitiveto any 'humps' in the ciefiningiunctions.
fi LP/lVl and A^QIIVI are calculatedefficiently, then the speedfor iterations
of the FDLF is nearly five times that of the formal NR or about two-thirds that
of the GS method. Storagerequirementsare around 60 percent of the formal
NR, but slightly more than the decoupledNR method.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ilai:i;'l
,odern powersystemAnatvsis
I
Changes in system configurations can be easily taken into account
and
though adjustedsolutionstake many more iteration.s,
each one of them takes
less time and hencethe overall solution time is still low.
The FDLF can be employedin optimization studiesand is specially
used for
e
studies, as in contingencyevaluation for system security
assessmentand
enhancementanalysis.
Note: When a series of load flow calculations are performed,
the final
values of bus voltagesin each case are normally used as the initial
voltagesof
the next case.This reducesthe number of iterations,particularly when
there are
minor changesin systemconditions.
6.9
CONTROL OF VOLTAGE
system,buseswith generatorsare usually made PV (i.e. voltagecontrol)
buses.
Load flow solution then gives the voltagelevels at the load buses.
If some of
the load bus voltages work out to be lessthan the specified lower voltage
limit,
it is indicative
of the fact that the reer-fivc n^rr/ar ff^.,, ^-*^^ie,
^r4-^-^--:--:-
lines for specifiedvoltage limits cannotmeet the reactiveload demand
(reactive
line flow from bus i1o bus ft is proportionalto lAvl = lvil - lvkD.This
situation
PROFILE
Control by Generators
Control of voltage at the receiving bus in the fundamentaltwo-bus
systemwas
discussedin Section 5.10. Though the same general conclusions
hold for an
interconnectedsystem,it is important to discussthis problem in greater
detail.
At a bus with generation, voltage can be conveniently controlled
by
adjusting generatorexcitation.This is illustrated by meansof Fig.
6.14 where
the equivalent generatorat the ith bus is modelled ty a synch.onou,
reactance
(resistanceis assumednegligible)and voltagebehind synchronous
reactance.It
immediately follows upon application of Eqs. (5.71) and (5.73) that
Pai+ jQet
- tv!= -ffi O,
tAvt- tEtht
tv';t=lErhl+ n,
#l
=tvit*
hn,
lVilt6i
ti
Xei
Flg.6. 1S
W
ith
/P
YYrtrr
\rGi
-t
iA
Pci = |v,l -Epl.;;,' -'-i,
xo,
(6.8e)
eci=
(6.e0)
iVcil
\ --,{
#r-,vit+tEcit)
lr/l
,/ f
^:,-^-
L-- rr--
r
, ft
ano i'tlri Loi glven bY-ihe ioaci ijow
soiution- these velrres
Since we are considering a voltage rise of a few percent, lV(l can
be further
approximatedas
lV|l= lV,l+
ffiO,
(6.e1)
Thus the VAR injection of +jQc causesthe voltage at the jth bus
to rise
approximately by (XhllViDQ.. The voltagesat other load buseswili
also rise
owing to this injection to a varying but smaller extent.
Control by Transformers
Apart from being VAR generators,transformers
provide a convenientmeansof
controlling real power' and reactive power flow along a transmission
line. As
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
','d&r1'l
Modern Power Svstem A-nalvsis
- l
has already been clarified, real power is controlled by means of shifting the
phase of voltage, and reactive power by changing its magnitude. Voltage
magnitudecan be changedby transforrnersprovided with tap changing under
Ioad (TCUL) gear.Transformersspecially designedto adjustvoltagemagnitude
smail values are calleo re
rnxers.
Figure 6.16 showsa regulatingtransformerfor control of voltagernagnitude,
which is achievedby adding in-phaseboosting voltage in the line. Figure 6.17a
shows a regulating transformer which shifts voltage phase angle with no
appreciablechange in its magnitude.This is achieved by adding a voltage in
series with the line at 90" phase angle to the corresponding line to neutral
voltage as illustrated by means of the phasor diagram of Fig. 6.17b. Here
VLn= (Von* tV6) = Q - jJT)
Since the transformerls assumedto be ideal, complex power output from it
equals complex power input, i.e.
5 1= V , / i = c r v l f
or
For the transmissionline
I\= ! @V1 - V2)
or
Ir = dl'a
lo:lzyV,- cfyVz
(6.e4)
Also
Von= dVon
6.92)
wher ea = (1 -j J -l t)= 7 1 -ta n -t J 1 t
since r is small.
The presenceof regulatingtransformersin lines modifies the l/uur matrix
t hc r c byr n < i l i fy i n gth e l o a c lfl o w s o l u ti on.C onsi cl era l i ne, connecti ngtw o
buses, having a regulating transformer with off'-nominal turns (tap) ratio a
includedat one end as shown in Fig. 6.18a.It is quite accurateto neglectthe
small impedanceof the rr:gulatingtranslonucr,i.c. it is rcgardcdas an iclcal
with line
device. Figure 6.18b gives the correspondingcircuit 4epresentation
representedby a seriesadmittance.
-_o
(6.es)
oV, ) = - WVr + lVz
Equations(6.94) and (6.95) cannot be representedby a bilateral network.
The I matrix representationcan be written down as follows liom Eqs. (6.94)
and (6.95).
I z= l( Vz-
n.,%n
A/
.___
Jl"r<_tc,
C...,%n
(a)
I
J
(V"n+tV"r)=aV",
/^ ic raal nr rmhor\
(b)
Flg. 6.16 Regulating
transformer
for controlof voltagemagnitude
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Fig. 6.17 Regulatingtransformerfor controlof voltagephaseangle
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ao.Ot,o* a,r.,.r
Modernpower SystemAnalysis
I
I
Bus,1
(i ) V/ V\ = l. M or a - l/ 1. 04
(ii) V/V\ = d3 or d = s-i3o
Bqs2
neffiting
transformer
aV'l
O--f__1--*-<-
l''r
f#
Solution (i) With regulating transformer in line 3-4, the elements of the
correspondingsubmatrixin (v) of Exa
6.2 are modified as under.
(a)
. ---.-
__
Y
v2
sz
<L
3
I
I
3
4
-r.92
516
j10.s47 + js.777
j2)
12
r.92
js.777 3 - j e
(b)
Fig. 6.18 Line with regulatingtransformerand its circuitrepresentation
Modified submatrixi n (v)
( of Example 6.2 is
4
a
J
(6.e6)
4
r- j3)
3 -(0.666
- j2)
The entriesof f matrix of Eq. (6.96)would then be usedin writing, the rru,
matrix of the complete power network.
For a voltage regulating transformer a is real, i.e. d = e, therefore, Eqs.
(6.94) and (6.95) can be represenredby the zr-networkof Fig. 6.19.
4
e i3"e2 + j6)
-(2- j6)
1T
G z+ j6)
.3r13
5.887
3.
3- je
3- je
t
Fis.6.1e
:jiiil:':3;:i;ffiTH#*:
withorr-nominar
tapsettins
or
c
Fig. 6.20 Off-nominaltransformers
at both line ends (a.,, a, real)
-Ylfvil-[Ii-l
I t
L - , y ) l v t) - U t )
[ - ' 4 . lIl r ll t / q o I f 4 l
lLyv. i;1l =- l[oo , qo)I L
v , ) ' L , ; ol : [ v " , ) ] , 1
If the line shown in Fig. 6.18a is representedby a zr-networkwith shunt
admittanceyu at eachend, additionalshunt admittancelol2ysappears
at bus 1
and yo at bus 2.
The above derivations also apply for a transformer with off-nomin al tap
setting' where a = (k[nuJ(kD,r:up, a real value.
(i) .
(ii )
Substituting (ii) in (i) and solving we get
| 4t
The four-bus system of Fig. 6.5 is now modified to include a regulating
transformerin line 3-4 near bus 3. Find the modified rru5 gf the system for
-With
off-nominaf,up **ing transformers
at eachend of the line as shownin
Fig. 6.20,we can write.
www.muslimengineer.info
Thus
L-otory
--l
'=
-"!-',vf
[q 1: ['"1
/iii\
J ltz )
4y ) LV2
o?v
-aqztf
l-^*,
d; )
Note: Solve it for the case when ao &2 are complex.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(iv)
.
l
236 |
Modernpower SystemAnqtysis
Load Flow Studies
I
6. 10
CO N C L U SION
In this chapter,perhapsthe most importantpower systemstudy,viz. load flow
has been introduced and discussedin detail. Important methodsavailable have
methodsis the best, becausethe behaviourof different load flow methods is
dictated by the types and sizes of the problems to be solved as well as the
precise details to implementation.Choice of a particular method in any given
situation is normally a compromisebetweenthe various criteria of goodnessof
the load flow methods.It would not be incorrect to say that among the existing
methods no single load flow method meetsall the desirablerequirementsof an
ideal load flow method; high speed,low storage,reliability for ill-conditioned
systems,versatility in handling various adjustmentsand simplicity in programming. Fortunately, not all the desirable features of a load flow method are
neededin a l l s i tu a ti o n s .
Inspite of a large number of load flow methods available,it is easy to see
that only the NR and the FDLF load flow methodsare the most important ones
fbr generalpurposeload flow analysis.The FDLF method is clearly superior to
the NR methodfrom the point of view of speedas well as storage.Yet, the NR
method is stili in use becauseof its high versatility, accuracyand reliability and
as such is widely being used for a variety of system optimization calculations;
it gives sensitivity analysesand can be used in modern dynamic-responseand
outage-assessment
calculations.Of coursenewer methodswould continueto be
developed which would either reduce the computation requirementsfor large
systems or which are more amenableto on-line implementation.
Bus I is slack bus with Vr = 1.0 10"
Pz+ iQz = -5.96 + j1.46
lVTl= 1102
Assume:
\ = 1.02/0" and vi -710"
1
t-'
2
tt.-o.oz
o ottooo
.7orr:"
3
Fig. P-6.2
6.3 For the systemof Fig. P-6.3 find the voltage at the receivingbus at the
end of the first iteration. Load is 2 + 70.8 pu. voltage at the sending end
(slack) is 1 + /0 pu. Line admittance is 1.0 - 74.0 pu. Transformer
reactanceis 70.4 pu. off-nominal turns ratio is lll.04. Use the GS
technique.Assume Vn = ll0.
c>+J=---F"d
Fig. P-6.3
6.4 (a) Find the bus incidencematrix A for the four-bus sysremin Fig. p-6.4.
Take ground as a reference.
(b) Find the primitive admittancematrix for the system.It is givbn rhat all
the lines are charactenzedby a seriesimpedanceof 0.1 + j0.7 akrn
and a shunt admittanceof 70.35 x 10-5 O/km. Lines are rated at220
kv.
PROB
TEIVIS
6 . r For the power systemshownin Fig. P-6.1,obtain the bus incidencematrix
A. Take ground as reference.Is this matrix unique? Explain.
(c) Find the bus admittancematrix for the system. Use the base values
22OkV and 100 MVA. Expressall impedancesand admittancesin per
unit.
1
(-)
1 1 0k m
3
Fig. P-6.4
Fig. p-6='!
6 . 5 Consider the three-bussystem of Fig. P-6.5. The pu line reactancesare
6 . 2 F<rrthe network shown in Fig. P-6.2,obtain the complex bus bar voltage
at bus 2 at the end of the first iteration..use the GS method. Line
impedancesshown in Fig. P-6.2 are in pu. Given:
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indicatedon the figure; the line resistancesare negligible.The magnitude
of all the three-busvoltages are specified to be 1.0 pu. The bus powers
are specified in the following table.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.?38i I
I
Modern Power Svstem Anatvsis
Z-r----rvz
-- .ffi()
|
ir o -y=_j5.0
',\
, =-is'o
,lJ
y = - j 5 0 ,l
3
Fig. P-6.5
I
'
3
Real
Reactive
Real
demand
demand
Seneratrcn
PGr= ?
Qot = 0.6
ect (unspecified)
Pcz = 1.4 pn, lunspecified)
Qoz= 0
go. lunsp"cified)
Pa = 0
Qot = l.O
Carry out the completeapproximateload flow solution.Mark generations,
load demandsand linc f-lowson thc one-linediagram.
6.6 (a) RepeatProblem6.5 with bus voltagespecifications
changedas below:
l V t l = 1 . 0 0p u
lV2l= 1.04pu
lV3l= 0.96 pu
Your results should show that no significant change occurs in real
power flows, but the reactive flows change appreciably as
e is
sensitiveto voltage.
(b) ResolveProblem6.5 assumingthat the real generationis scheduledas
follows:
P c t = 1 .0p u , P c z = 1 .0pu, P ct = 0
The real demandremainsunchangedand the desiredvoltage profile is
flat, i.e. lvrl = lv2l = lv3l = 1.0 pu. In this casethe resultswill show
that the reactiveflows arc essentiallyunchangccl,
but thc rcal flgws
are changed.
6.7 Consider the three-bussystemof problem 6.5. As shown in Fig. p-6.7
where a regulating transforrner(RT) is now introclucedin the ljne l-2
near bus 1. Other systemdata remain as that of Problem 6.5. Consider
two cases:
(i) RT is a magnituderegulatorwith a rario = VrlVl = 0.99,
(ii) RT is a phaseangle regr-rlaror
having a ratio = V/Vi =
",3"
(a) Find out the modified )/ur5 matrix.
(b) Solve the load flow equationsin cases(i) and (ii). compare the
load flow picture with the one in Problem 6.5. The reader should
verify that in case(i) only the reactiveflow will change;whereas
in case (ii) the changeswill occur in the real power flow.
/
13
Fig. P-6.7 Three-bus
samplesystemcontaining
a
regulating
transformer
Reactive
generation
Por = 1.0
',r2 = 0
Po: = 1.0
\
vel
6.8 Calculate V3for the systemof Fig. 6.5 for the first iteration, using the data
of Example 6.4. Startthe algorithmwith calculationsat bus 3 rather than
at bus 2.
6.9 For the sample system of Example 6.4 with bus I as slack, use the
fbllowing nrcthudsto obtairra load flow solution.
(a) Gauss-Seidelusing luur, with accelerationfactor of 1.6 and
tolerancesof 0.0001 for the real and imaginary components of
voltuge.
(b) Newton-Raphsonusing IBUS,with tolerancesof 0.01 pu for changes
in the real and reactivebus powers.
\
Note: This problem requiresthe use of the digital computer.
6.10 Perform a load flow studyfor the systemof Problem6.4. The bus power
and voltage specificationsare given in Table P-6.10.
Table P-6.10
Bus power,pu
B us
I
2
-)
4
a
Voltagemagnitude,
pu
Unspecified
0.95
- 2. O
- 1.0
Slack
PV
PQ
PQ
Computethe unspecifiedbus voltages,all bus powersand all line flows.
Assume unlimited Q sources.Use the NR method.
NCES
REFERE
Books
l.
Mahalanabis, A.K., D.P. Kothari and S.I. Ahson, Computer Aided Power System
Analysis and Control, Tata McGraw-Hill,
www.muslimengineer.info
1.02
1.01
Unspecified
Unspecified
Unspecified
Unspecified
- 1.0
- 0.2
Bus type
New Delhi. 1988.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'
_ -2rA
_0
_ "_' _l
.M- .o- d- -p' ,r,n
pnrrrar
, v'rvr
e yr roarrrar i*l l
e
A^^r-.-,A1 tatysls
2' weedy, B.M. and B.J. cory, Erectricarpowersystents,4thEd., John wiley, New
York, 1998.
Gross,C.A., power System Analysis,2nd
8d..,John Wiley, New york, 19g6.
Sterling, M.J.H., power SystemControl,
IEE. England,1978.
, u.r., L,teclrrc Lnergy System Theom: An Introduction,
2nd Ed, McGraw_
Hill, New York. 19g2.
6. stagg, G.w. and A.H. EI-Abiad, computer
Methods in power system Anarysis,
McGraw-Hill, New york, 196g.
7' Rose' D'J' and R'A willough (Eds),
sparse Matrices and their Applications,
Plenum,New york, 1972.
8' Anderson,P'M., Analysisof Faulted
Power systems,The Iowa state university
Press,Ames, Iowa, 1923.
9' Brown, H'8., sorution of Large Networks
by Matix Methods, John w1rey, New
, York, 1975.
l0' Knight' rJ'G', Power systemEngineering
and Mathematics,pergamonpress,New
York 1972.
11. Shipley, R.B., Introduction to Matrices
and power systems,John w1ey, New
York, 1976.
12' Hupp, H.H., Diakoptics and Networks,
Academic press,New york, 1971.
l3' Arrillaga, J. and N.R. watson, computer
Modeiling of Erectricarpower sltstems,
2nd edn, Wiley, New york, 2001.
14' Nagrath, I.J. and D.p. Kothari, power
system Engineering, Tata McGraw_Hilr,
New Delhi, 1994.
l5' Bergen,A.R., power systemAnarysis,prentice-Hail,
Engrewoodcriffs, N.J. r9g6.
l6' Anillaga' J and N'R' watson, computer
Modelting of Electrical power systems,
2nd edn. Wiley, N.y. 2001.
Papers
17. Happ, H.H., 'Diakoptics-The solution
of system problems by Tearing,, proc. of
the IEEE, Juty t974, 930.
l8' Laughton'M'A', 'Decontposition
Techniquesin power systemNetworkLoad
Flow
Analysi.susi'g the Nodar ImpcdanceMatrix,, proc.
IEE,
196g,
r15,
s3g
19' stott, B., 'Decouprecr
NewronLoad FIow,, IEEE Trans., 1972,pAS_91,
1955.
20. stott, B., 'Review of Load-Frowcarcuration
Method,, proc. IEEE, Jury 1974,916.
21' stott, B., and o. Alsac, 'FastDecoupled
Load Flow,, IEEE Trans., 1974,pAS_93:
859.
22' Tinney' w'F" and J'w. walker, 'Direct
solutions of sparse Network Equationsby
Optimally Ordered Triangular Factorizaiions,, proc.
IEEE. Nnvernhe" ' 1oA1
v
55:1801
"
23' Tinney, w.F. and c.E. Hart, 'power
Flow sorution by Newton,s Method,, ,EEE
Trans.,November1967,No. ll, pAS_g6:1449.
24' ward, J.B. and H.w. Hare, 'Digitar
computerSorutionof power problems,,AIEE
Trans.,June 1956,pt III, 75: 39g.
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Power Frow Argorithmswith Referenceto Indian power
systems,, proc. of II
symp.on Power plant Dynamicsancrcontror, Hyderabad,
14_i6 Feb. rg:/g, zrg.
26. Tinney, W.F. and W.S. Mayer, .Solution
TriangularFactorization',IEEE Trans.on Auto contor,
August 1973,yor. AC_lg,
JJJ.
27' Saro, N. and w.F Tinney, 'Techniquesfor
Exproiting Sparsity of Network
Admittance Matrix', IEEE Trans. pAS, I)ecember
L963, g2, 44.
28' sachdev, M.s. and r.K.p. Medicherla, 'A second
order Load Flow Technique',
IEEE Trans., pAS, Jan/Feb 1977, 96, Lgg.
29. Iwamoto, S.and y. Tamura, ,A Fast Load Flow
Method Retaining Non linearitv'.
' IEEE
Trans., pAS. Sept/Oct. 197g, 97, 15g6.
30. Roy, L., 'Exact Second Order Load Flow,, proc.
of the Sixth pSCC Conf.
Dermstadt,Yol. 2, August lg7g, 7Il.
31' Iwamoto, s' and Y. Tamura,'A Load Flow calculation
Method for Ill-conditioned
Power Systems',IEEE Trans pAS, April lggl 100,
1736.
,
32. Happ, H.H. and c.c. young, 'Tearing Argorithms
for Large scare Network
Problems', IEEE Trans pA,S,Nov/Dec Ig71, gO,
2639.
33. Dopazo,J.F.O.A.Kiltin and A.M. Sarson,,stochastic
Load Flows,, .IEEE Trans.
P A S ,1 9 7 5 , 9 4 , 2 g g .
34' Nanda' J'D'P' Kothari and s.c. srivastava,"Some
Important observationson
FDLF'Algorithm", proc. of the IEEE, May 19g7, pp.732_33.
\
35. Nanda,J., p.R., Bijwe, D.p. Kothari and D.L.
stenoy, .,secondorder Decoupred
Load Flow', ErectricMachinesand power systems,
vor r2, No. 5, 19g7,pp. 301_
312.
36' NandaJ'' D'P' Kothari and S.c. srivastva,"A
Novel secondorder Fast Decoupled
Load Flow Method in Polar coordinatcs", Electic
Machinesand power,s.ysrems,
Vol. 14, No. 5, 19g9,pp 339_351
37. Das, D., H.s. Nagi and D.p. Kothari, .,A Novel
Method for solving Radiar
DistributionNetworks", proc. IEE, ptc, v.r. r4r,
no. 4. Jury r994. pp. zgr_zgg.
38' Das, D', D'P' Kothari and A. Kalam. "A Simple
and Efficient Method for Load
Flow sorurion of RadiarDistributionNerworks',,
Inr. J. of EpES, r995, pp. 335_
346.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
OptimalSystem Operation
A3
the 'unii commitment' (UC) probiemand the secondis calleci
parianceis caiied
the 'load scheduling' (LS) problem.One must first solvethe UC problem before
proceedingwith the LS problem.
Throughout this chapter we shall concern ourselves with an existing
installation,so that the economicconsiderationsare that of operating(running)
cost and not the capitai outiay.
7.2
OPERATION OF GENERATORS ON A BUS BAR
OPTIMAL
Before we tackle the unit commitment problem, we shall consider the optimal
operation of generatorson a bus bar.
Generator Operating
7.1
INTRODUCTION
The optimal system operation,in general,involved the considerationof
economyof operation,systemsecurity,emissionsat certain fossil-fuel plants,
optimalreleasesof water at hydro generation,etc. All theseconsiderationsmay
makefor conflictingrequirements
and usuallya compromisehasfo he madefor
optimalsystemoperation.ln this chapterwe cor..;iderthe economyof operation
only, also called the ec'omonicdi.spcttch
problem.
The main aim in theeconomicdispatchproblernis to rninimize thc total cost
of generatingreal power (production cost) at various stations while satisfying
theloadsandthe lossesin thetransmission
links.For sirnplicitywe consicler
the
presenceof thermalplantsonly in the beginning.In the later part of this chapter
we will considerthe presenceof hydro plants which operatein conjunctionwith
thermalplants.While thereis negligible operatingcost at a hydro plant, there
is a limitationol availabilityol'watcr'over a pcriodof tinrewhich nrustbc used
to savemaximum fuel at the thermal plants.
In the load flow problemas detailedin Chapter6, two variablesare specified
at eachbus and the solutionis then obtainedfor the rernainingvariables.The
specifiedvariablesare real and reactivepowersat PQ buses,real powersand
voltagemagnitudesat PV buses,and voltagemagnitudeand angle at the slack
bus.The additionalvariablesto be specifiedfor load flow solution are the tap
settingsof regulatingtransformers.If the specifiedvariablesare allowed to vary
in a regionconstraineci
(upperanciiower iimits on
by practicaiconsicierations
activeand reactivegenerations,
bus voltagelimits, and rangeof transformertap
settings),there resultsan infinite numberof load flow solutions,eachpertaining
to one set of valueso1'specifiedvariables.The 'best'choice in sornesenseof
thevaluesof specifiedvariablesleadsto the 'best' loadflow solution.Economy
of operationis naturallypredominantin determiningallocationof generationto
eachstationfor varioussystemload levels.The first problem in power system
www.muslimengineer.info
Cost
The major component of generatoroperating cost is the fuel input/hour, while
maintenance
coutributesonly to a small extent.The fuel cost is meaningtulin
case of thermal and nuclear stations,but for hydro stations where the energy
storageis 'apparentlyfree', the operatingcost as such is not meaningful. A
suitablemeaningwill be attachedto the cost of hydro storedenergyin Section
7.7 of this chapter. Presentlywe shall concentrateon fuel fired stations.
l
I
II
t
L
-
\
al)
;
P
t r o
. c 8
6
o
8b=
e 5
g
o
O
E
6
f
LL
o
(vw)min
Fig.7.1
(MW)max
Poweroutput,MW -'- --
Input-outputcurve of a generatrngunit
The input-output curve of a unit* can be expressedin a million kilocalories
per hour or directly in terms of rupees per hour versus output in megawatts. The
cost curve can be determined experimentaily. A typical curve is shown in
Fig. 7.1 where (MW)o'in is the minimum loading limit below which it is
uneconomical (or may be technically infeasible) to operate the unit and
(MW)n,u, is the maximunr output limit. The inpLlt-output curve has
discontinuities at steam valve openings which have not been indicated in the
figure. By fitting a suitable degree polynomial, an analytical expression for
operating cost can be written as
A unit consists of a boiler, turbine and generator.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
|
. .
-
MOOern rower
.L4+ |
I
Ci(Pc)
-
A - - - ! - -
A - ^ r - - ^ : -
DySIeIrl Arlaly
Considerationsof spinning reserve,to be explained later in this section,require
that
Rs/hourat outPutPc,
b,Pc, + d, Rs/hour
_Q;
2
(7.r)
is called the incrementalJuel cost(lQ,
The slope of the cost curve, i.".43
dPo,
and is expressedin units of rupeesper megawatthour (Rs/lvIWh).A typical plot
of incrementalfuel cost versus power output is sho.wnin Fig. 7.2.If the cost
curve is approximatedas a quadraticas in Eq. (7.1), we have
(lc)i=
marsin. i.e. Eq. (7.6) must be a strict inequality.
Since the operatingcost is insensitiveto reactive loading of a generator,the
rnannerin which the reactive load of the station is sharedamong various online generatorsdoes not afl'ectthe operatingeconomy'.
'What is the optimal manner
The question that has now to be answeredis:
by
the generatorson the bus?'
in which the load demand Po must be shared
cost
the
operating
This is answeredby minimizing
k
c = D ci(pci)
(7.2)
aiP"t + bt
(7.6)
D Po,,^*) Po
where the suffix i standsfor the unit number.It generatly suffices to fit a second
degreepolynomial,i.e.
( 7. 7)
f:l
underthe equalityconstraintof meetingthe load demand,i.e.
(
l
I
-i.
(J
II
f
-
L
i:t
o
o
o c
where k = the number of generatorson the bus.
Further, the loading of each generator is constrained by the inequality
constraintof Eq. (7.5).
Since Ci(Pc) is non-linear and C, is independentof P6t (i+ i), this is a
separablenon-linear programming problem.
\
If it is assumedat present,that the inequality constraint of Eq. Q.q is not
effective,the problem can be solved by the method of Lagrangemultipliers.
Define the Lagrangian as
E
B
,d>
o o
Etr
o
E
E
()
(MW) max
(MW)min
Poweroutput,MW
f -
fuel costversuspoweroutputfor the unitwhose
Fig.7.2 Incremental
curve is shownin Fig.7.1
input-output
i.e. a linear relationship.For better accuracy incremental fuel cost may be
expressedby a number of short line segments (piecewise lineanzation).
Altcrnativcly,wc can fit a polynomial of suitabledegreeto representIC curve
Optimal
(7.3)
-""]
it (Pci)-^[f
"",
(7.e)
where X is the Lagrange multiplier.
Minimization is achieved by the condition
of, =o
dPo,
in the inverseform
Pc;i= a, + {),(lC)i + 1,QC)', + ...
(7.8)
\' G
D i l- P o = O
dC' i
or
dPc,
(7.1o)
- ) i i = 1 , 2 , . . . ,k
Operation
[,et us assulnethat it is known a priltri which generutorsitre t<lrtln to ntccta
on the statton.Ubvtously
load clenrand
p:.rrticular
(7.4)
DPr,,, ) P,
where lci
,'',,*
is the rated real power capacity of the ith generatorand Po is
where Pci, ,r.,,"*
the total power demandon the station.Further,the load on each generatoris to
irc constrainedwithin lower and upper limits, i.e.
Pcr, .in 1 Po, 1 Po,, rn.*, I = L, 2, "', k
www.muslimengineer.info
(7.s)
is the incrementalcost of the ith generator (units: Rs/TvIWh),
d4,,
a Iurctio' ol'.gencra,il;,":':t
:":
d4(il- dr-*The
E q u a t i o n ( 7 . 1 0 ) c a n be written as
dC^
\
dPoo
effect of reactive loading on generatorlosses is of negligible order.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
( 7 .r l )
n-.:-^r
A-^^r---
^
..
L
-^-
cost of the plant colrespondsto that of unit 2 alone. When rhe plant load is
40
Mw, each unit operatesat its minimum bound, i.e.2o Mw wiitr plant \ = Rs
35/I4Wh.
Computer solution fbr optimal loa<ling of generators can be obtained
iteratively as follows:
1. Choose a trial value of ), i.e. IC = (IC)o.
2. Solve for P", (i = 1, 2, ..., k) from Eq. (7.3).
3. If ItPc,-
Pol < e(a specified value), the optimal solution is reached.
Otherwise,
4. Increment(lC) bv A (1"), t I: Po,- ,rl I < 0 or decrement(/c) by A(tr)
fl
if [D Pc, - Pr] r 0 and repeat from step 2. This step is possible because
P.-, is monotonicallyincreasingfunction of (1g).
consider now the effecr of the inequality constraint (7.5). As (1c) is
increasedor decreasedin the iterative process,if a particular generatorloading
P", reachesthe limit PGi,^o or P6;, min, its loading from now on is held fixed
at this value and the balance load is then shared between the remaining
generators on equal incremental cost basis. The fact that this operation is
optimal can be shown by rhe Kuhn-Tucker theory (seeAppendix n;.
Incremental fuel costs in rupeesper MWh for a plant consisting of two units
are:
When
dczldPcz= Rs 44/MWh,
0.25PG2+30 - 44
pnt = JI= 56 MW
0. 25
The total plant output is then (56 + 20) = 76 MW. From this point onwards,
the values of plant load sharedby the two units are found by assumingvarious
valuesof \. The results are displayedin Table 7.1.
or
Table 7-1 Outputof each unit and plant output for variousvaluesof
) for Example7.1
Plant ),
Unit I
RszMWh
Pcl, MW
Unit 2
Pcz, NfW
35
44
50
55
60
61.25
65
Plant Output
40.0
76.0
130. 0
175.0
220.0\
231.25
250.0
Figure 7.3 shows the plot of the plant .trversusplant output. It is seenfrom
Table 7.7 that at .\ = 61.25, unit 2 is operatingat its upper limit and therefore,
the additional load must now be taken by unit 1, which then determines
the
plant ).
dt
i * -G o
1 .2opct+40.0
60
+
t - -
-dcz-= o.2,pcz+3o.o
l c c
dPo,
I
8 5 0
Assume that both units are operating at all times, and total load varies from 40
MW to 250 MW, and the maximum and minimum loads on each unit are to be
I25 and 20 MW, respectively.How will the load be sharedbetween the two
units as the systemload vanes over the full range?What are the corresponding
values of the plant incrementalcosts?
Solution At light loads, unit t has the higher incrementalfuel cost and will,
therefore,operareat its lower limit of zo Mw, for which dcrldpcr is Rs 44 per
MWh. When the ourputof unit 2is20 MW, dczldpcz= Rs 35 p"i UWt. Thus,
with an increasein the plant output, the additionalload should be borne bv unit
o
@
E
P 40
t E
O
c
J
c
Plantoutput,MW
Fig.7.3
www.muslimengineer.info
^
45
i=
a>
Incrementalfuel cost versus plant output,as found in Example7.1
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t:Hzut"l
.Z.48:i
MocjernPower Systemnnaiysis
To find the load sharingbetweenthe units for a plant output of say 150 MW,
we find from the curve of Fig. 7.3, that the correspondingplant X is Rs 52,22
per MWh. Optimum schedulesfor each unit for 150 MW plant load can now
be found as
P c t = 6 1 ' 1 1M W
0.2Pa*40-52.22;
Pcz = 88'89 MW
0.25PG2
+ 30= 52.22;
Net savingcausedby optimumscheduling
is
= 50.625Rs/lr
772.5- 721.875
Totalyearlysavingassumingcontinuous
operation
This savingjustifies the need for optimal load sharing and the ddvices to be
installed for controlling the unit loadings automaticallv.
P c r + P c z = 1 5 0M W
Proceeding on the above lines, unit outputs for various plant outputs are
computedand have beenplottedin Fig. 7.4. Optimum load sharingfor any plant
load can be directlv read from this fieure.
I tzs
Let the two units of the svstem studiedin Example7.1 have the following cost
curves.
Cr = 0.lPto, + 40Pc + 120 Rs/hr
I
= 100
Cz= 0.l25Pzcz+ 30Po, + 100 Rsftrr
250
Eru
E 5 0
f
0
Flg.7.4
220 MW
tI 200
I
3 150
o
5
0
150
100
Plantoutput,MW -_-->
200
250
100
E
o
J
50
Monday
Sunday
Output of each unit versus plant output for Example 7.1
l
12
(noon)
PM
Time -----
Fig. 7.5
For the plant describedin ExampleT.l find the savingin fuel cost in rupeesper
hour for the optimal schedulingof a total load of 130 MW as comparedto equal
distribution of the same load between the two units.
Solution Example 7.I revealsthat unit I should take up a load of 50 MW and
unit 2 should supply 80 MW. If each unit supplies65 MW, the increasein cost
for unit 1 is
165rnnn
| \'U . L I ' r : r
Jso
t
.
,^\rn
*U)|JIl nr =
rr.'tnZ
(U.Ifnt
;
lf\r1
t'tUI'6yll
rl
"'
165
=
lso
Similarly, for unit 2,
Fn^ a
llL.J
D^tLl\S/I[
1
2
(night)
l
6
AM
Daily load cycle
Let us assumea daily load cycle as given in Fig. 7.5. Also assumethat a cost
of Rs 400 is incurredin taking either unit off the line and returning it to service
after 12 hours. Considerthe 24 hour period from 6 a.m. one morning to 6 a.m.
the next morning. Now, we want to find out whether it would be more
economical to keep both the units in servicefor this 24hour period or to remove
one of the units from service for the 12 hours of light load.
For the twelve-hourperiod when the load is 220 MW, referring to Table 7.1
of Example 7.1, we get the optimum scheduleas
Pcr = 100 M W' Pcz = 120 M W
Total fuel cost for this period is
+ 30)dpor=(0.r25PGz+
:oro;1"
J*co.rs"
",
[ 0 . 1x 1 0 0 2 + 4 0 x 1 0 0 + 1 2 0 + 0 . 1 2 5 x 1 2 0 2+ 3 0 x 7 2 0 + 1 0 0 ]x 1 2
= Rs. 1, 27, 440
- - 721.875Rs/hr
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ltgtiFl
--t
powersvstemnnAVsis
Modern
If both units operatein the light load period (76 MW from 6 p.m. to 6 a.m.)
also, then from the same table, we get the optimal scheduleas
Pcr = 20 MW, Pcz = 56 MW
Total fuel cost for this period is then
(0.1 x 20' + 40 x 20 + 120+ 0.125x 5 6 + 3 0 x 5 6 + 1 0 0 ) x 1 2
= Rs 37,584
Thus the total fuel cost when the units are operating throughout the 24 hour
period is Rs I,65,024.
If only one of the units is run during the light load period, it is easily verified
that it is economicalto run unit 2 andto put off unit 1. Then the total fuel cost
duringthisperiot:tXf ':
762+ 3o x 76+ 100)x rz = Rs 37,224
Total fuel cost for this case= L,27,440+ 37,224
= Rs 1,64,664
Total operatingcost for this casewill be the total fuel cost plus the start-up cost
of unit l , i .e .
1,64,6& + 400 = Rs 1,65,064
Comparingthis with the earlier case,it is clear that it is economicalto run both
the units.
It is easy to see that if the start-upcost is Rs 200, then it is economical to
run only 2 in the light load period and to put off unit 1.
7.3
OPTTMAL UNrT COMMTTMENT
(UC)
As is evident,it is not economicalto run all the units availableall the time. To
determine the units of a plant that should operate for a particular load is the
problem of unit commitment (UC). This problem is of importancefbr thermal
plants as for other types of generationsuch as hydro; their operating cost and
start-up times are negligible so that their on-off statusis not important.
A simple but sub-optimal approach to the problem is to impose priority
ordering, wherein the most efficient unit is loaded first to be'followed by the
less efficient units in order as the Ioad increases.
A straightforward but highly time-consuming way of finding the most
economicalcombination of units to meet a particular load demand,is to try all
possiblecombinationsof units that can supply this load; to divide the load
optimally among the units of each combination by use of the coordination
equaiions,so as to finci the most economicaioperatingcost of the combination;
then,to determinethe combinationwhich has the least operatingcost among all
these.Considerablecomputationalsavingcan be achievedby using branch and
bound or a dynamic programming method for comparing the economics of
combinationsas certaincombinationsneetnot be tried at all.
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t
Dynamic Programming Method
-
In a practical problem, the UC table is to be arrived at for
the complete load
cycle. If the load is assumedto increasein small but finite size stensldvnamin
prograrrurung
can be used to advantage for computing the uc table,
wherein it is not necessaryto solve the coordination equations;
while at the
same time the unit combinations to be tried are much reduced
in number. For
these reasons,only the Dp approachwill be advancedhere.
The total number of units available, their individual cost characteristics
and
the load cycle on the station are assumedto be known a priori.Further,
it shall
be assumedthat the load on each unit of combination
of units changei'in
suitably small but uniform steps of size /MW (e.g. I MW).
Starting arbitrarily with any two units, the most iconomical combination
is
determined for all the discrete load levels of the combined output
of the two
units. At eachload level the most economic answermay be
to run either unit
or both units with a certain load sharing betweenthe two. The
most economical
cost curve in discreteform for the two units thus obtained,
can be viewed as
the cost curve of a single equivalent unit. The third unit is now
added and the
procedurerepeatedto find the cost curve of the threecombined
units. It may be
noted that in this procedurethe operatingcombinationsof
third and first, also
third and secondare not required to be worked out resulting in
considerable
saving in computationaleffort. The processis repeated,till all
available units
are exhausted.The advantage of this approach is that having
oitiined the
gPli-u.| way,of loading ft units, it is quite easy ro determine the Jptimal manner
of loading (ft + 1) units.
Let a cost function F" (x) be defined as follows:
F,y (x) = the minimum cost in Rs/hr of generating .r MW by
N units,
fN 0) = cost of generating y MW by the Nth unit
F*-{x - y) - the minimum cosr of generating (.r - y) Mw by
the remain_
ing (1/ - t) units
Now the application of DP results in the following recursive relation
FN@)= TnVn9) * Fu-r @ - y)|
(7.r2)
Using the aboverecursiverelation, we can easily determinethe
combination
of units, yielding minimum operating costsfor loads ranging in
convenient steps
from the minimum.permissible load of the smallest unit to the
sum of the
canaeifies
qrroiiol-lo rr-i+o i- +L;^ --^^^-^
.
n f q l l $vs^rqurv
1^- 1-1 r
uurrD. rrr LrrlD PruuttJs .ure
total nunlmum
oDerating
eost
and the load sharedby each unit of the optimal combination are
;il.u,i"
determined for each load level.
The use of DP for solving the UC problem is best illustratedby
meansof an
example. Considera sample system having four thermal generating
units with
parameterslisted in Table 7.2.It is required to determinr th.
most-economical
units to be committed for a load of 9 MW. Let the load changes
be in steps of
I MW.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPow
.
Table 7.2 Generatingunit parametersfor the samplesystem
Cost curve pararneters (d = 0)
Capacity (MW)
Unit No.
1 2 .0
1 2 .0
1 2 .0
12.0
1.0
1.0
1.0
1.0
1
2
3
4
Now
0.77
1.60
2.00
2.50
23.5
26.5
30.0
32.0
Ft@) = ft@)
fr(9) = f{9)=
The effect of step size could be altogether eliminated, if the branch and
bound technique [30] is employed. The answer to the above problem using
branch and bound is the samein terms of units to be committed, i,e. units 1 and
2, but with a load sharing of 7 .34 MW and 1.66 MW, respectively and a total
graung cost oI I<s z,y.zL /J/nour.
In fact the best scheme is to restrict the use of the DP method to obtain the
UC table for various discrete load levels; while the load sharing among
committed units is then decidedby use of the coordination Eq. (7.10).
For the example under consideration,the UC table is prepared in steps-of I
MW. By combining the load range over which the unit commitment does not
change, the overall result can be telescopedin the form of Table 7.3.
Table 7.3 Status*of units for minimum
operatingcost (Unit commitment
table for the samplesystem)
LorP'ot+ btPcr
= ollss x 92 + 23.5 x9 = Rs 242.685lhour
From the recursive relation (7.12), computation is made for F2(0), Fz(l),
Unit number
Load range
Fz(2),..., Fz(9).Of these
FzQ) = min tt6(0) + Ft(9)1, VzG) + Ft(8)l'
r
l-5
6-r3
t 4- 18
1948
VzQ) + Ft(7)1, VzQ) + Fr(6)l' Vz@)+ F1(5)l'
t6(5)+ Fr(4)1,VzG)+ Ft(3)1,VzT + Fr(2)1,
tfr(s)+ F,(1)1,vzg) + Fr(O)l)
On computing term-by-term and compdng, we get
FzQ) = Vz(2) + Ft(1)) = Rs 239.5651how
Similarly, we can calculate Fz(8),Fz(1), ..., Fz(l), Fz(O).
Using the recursiverelation(7.12),we now computeFl(O), F:(1), ...' F3(9).
Of these
Fse)= min {t6(0)+ Fr(9)f,t6(1)+ Fl8)1, ...'[6(9)+ rr(0)]]
= [6(0)+ FzQ)l= Rs 239.565ftour
Proceeding similarly, we get
FoQ) = [f4(0) + Fr(9)] = Rs 239.565lhour
Examinationof Fr(9), Fz(9),Fl(9) and Fa(9) leads to the conclusion that
optimum units to be lommitted for a 9 MW load are 1 and 2 sharing the load
ur Z l,tW and 2 MW, respectively with a minimum operating cost of Rs
239.565/hour.
'u\e
It must be pointed out here, that the optimai iiC tabie is inclependentof
numberingof units, which could be completely arbitrary. To verify, the reader
rnay solvethe above problem onceagain by choosing a different unit numbering
scheme.
If a higher accuracyis desired,the step size could be reduced(e.g. * t*r,
with a considerableincreasein computation time and required storage capacity.
www.muslimengineer.info
2
3
4
0
0
1
1
0
0
0
1
*l = unit running;0 = unit not running.
\
' The UC table is preparedonce and for all for a given set of units. As the load
cycle on the station changes,it would only mean changes in starting and
stopping of units with the basic UC table remaining unchanged.
Using the UC table and increasingload in steps,the most economical station
operatingcost is calculatedfor the completerange of stationcapacity by using
the coordination equations. The result is the overall station cost characteristic
in the form of a set of data points. A quadratic equation (or higher order
equation,if necessary)can then be fitted to this data for later use in economic
load sharing among generating stations.
7.4
RELIABILITY
CONSIDERATIONS
With the increasing dependence of industry, agriculture and day-to-day
household comfort upon the continuity of electric supply, the reliability of
power systemshas assumedgreat importance. Every eleciric utilify is normaily
under obligation to provide to its consumersa certain degreeof continuigl and
quality of service (e.g. voltage and frequency in a specifiedrange). Therefore,
economy and reliability (security) must be properly coordinated in arriving at
the operationalunit commitment decision.In this section,we will seehow the
purely economic UC decision must be modified through considerations of
reliability.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
In order to meet the load demand under contingency of failure (forced
outage) of a generator or its derating caused by a minor defect, static
reserve
capacity is always provided at a generating station so that the total
installed
capacity exceedsthe yearly peak load by a certain margin. This is planning
a
In arriving at the economicUC decisionat any particular time, the
constraint
taken into account was merely the fact that the total capacity
on line was at
least equal to the load. The margin, if any, between the capacity
of units
committed and load was incidental. If under actual operation one
or more of the
units were to fail perchance(random outage), it may not be possible
to meet the
load requirements.To start a spare (standby) thermal unit* and to
bring it on
steam to take up the load will take severalhours (2-8 hours), so
that the load
cannot be met for intolerably long periods of time. Therefore,
to meet
contingencies,the capacity of units on line (running) must have
a definite
margin over the load requirementsat all times. This margin which is
known as
the spinning reserve ensurescontinuity by meeting the load demand
up to a
certain extent of probable loss of generationcapacity.While rules
of thumb
have been used, based on past experienceto determinethe system's
spinning
reserve at any time, Patton's analytical approach to this problem
is the most
promising.
Since the probability of unit outageincreaseswith operatingtime and
since
a unit which is to provide the spinning reserve at a particular time
has to be
startedseveralhours ahead,the problem of security of supply has
to be treated
in totality over a period of one day. Furthefinore, the loads are never
known
with complete certainty. Also, the spinning reserve has to be provided
at
suitable generating stations of the system and not necessarily
at every
generatingstation.This indeedis a complex problem. A simplified
treatmentof
the problem is presentedbelow:
f1(down)
f2 (down)
periods are a random phenomenon with operating periods being much longer
than repair periods. When a unit has been operating for a long time, the random
phenomenoncan be describedby the following parameters.
'up' time),
Mean time to failure (mean
Mean time
(7.13)
No. of cycles
to repair (mean 'down' time),
Et, (down)
Z (down)
(7.r4)
No. of cycles
Mean cycle time = f (up) + Z(down)
Inverseof these times can be defined as rates [1], i.e.
Failure rate, A = IIT (up) (failures/year)
Repair rate, trt= llT (down) (repairs/year)
Failure and repair rates are to be estimated from the past data of units (or
other similar units elsewhere) by use of Eqs. (7.13) and (7.I4). Sound
engineeringjudgement must be exercised in arriving at these estimates.The
failure rates are affected by preventive maintenance and the repair rates are
.r
scnsitiveto size, composition and skill of repair teams.
probabiliiy
probability,
we
can
write
the
of
a
unit being
By ratio definition of
in 'up' or 'down' statesat any time as
p (up) =
P (down) =
f3 (down)
r(up)
l.t
_
z(up)* z(down)
p+A
Z(down)
Z(up)* Z(down)
tr+ A
(7.rs)
)
(7.16)
Obviously,
p (up) + p(down) = 1
Time-------->
Fig. 7.6 Randomunit performance
recordneglectingscheduledoutages
A unit during its useful life span undergoesalternateperiods of operation
and
repair as shown in Fig. 7.6. The lengths of individual operating
and repair
.
p (up) andp (down) in Eqs. (7.15) and (7.16) are also termed as availability
and unavailability, respectively.
When ft units are operating, the system state changesbecauseof random
outages.Failure of a unit can be regardedas an event independentof the state
'down'
state
of other units. If a particular systemstate i is defined asX, units in
an0
1
r,
'
I j ln
(
t
---t-
lr
Up. Stale \K =
\/
.
rz\
i + I )t
!l--
-----1--l-:1:---
^t
L1^^
urs PIUDaUITILy Ur urtr systelll
2-
tLl
Utrltr$ ttl ulIS Slaltr
is
pt =
{r,r;(ul),{*
ot(down)
If hy^dro.generationis available in the system, it could be brought
on line in a
matter of minutes to take up load.
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l-^:-^
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(7.r7)
gW
WWt
ModernPowersystemAnalysis
I
Patton's Security Function
the sample system for the load
curve of Fig. 7.7
A breach of system security is defined as some intolerable or undesirable
condition.The only breachof securityconsideredhere is insufficient generation
Unit number
probability that the available generation capacity (sum of capacitiesof units
cbmmitted) at a particular hour is less than the system load at that time, is
defined as [25]
(7.18)
S = Ep,r,
t
2
A
B
C
D
E
F
where
p, = probability of system being in state i [see Eq. (7.17)]
3
4
I
I
0
I
0
0
1
0
0
0
0
0
r, = probability that system state i causesbreach of syStem
securlty.
When system load is deterministic(i.e. known with complete certainty), r, = 1
if available capacity is less than load and 0 otherwise. S indeed, is a
quantitative estimateof system insecurity.
Though theoretically Eq. (7.18) must be summed over all possible system
states (this in fact can be very large), from a practical point of view the sum
needsto be caried out over statesreflecting a relatively small number of uniqs
on forced outage, e.g. stateswith more than two units out may be neglected as
the probability of their occurrencewill be too low;
Security
0L
Constrained
Optimal
0
(noon)
Unit Commitment
Once the units to be committed at a particular load level are known from purely
economic considerations,the security function S is computed as per Eq. (7.18).
This figure should not exceed a certain maximum'tolerable insecurity level
(MTIL). MTIL for a givcn systemis a managementdecisionwhich is guided
by past experience.If the value of S exceeds MTIL, the economic unit
commitment scheduleis modified by bringing in the next most economicalunit
as per the UC table. S is then recalculated and checked. The process is
continued till ^t < MTIL. As the economic UC table has some inherent spinning
reserve,rarely more than one iteration is found to be necessary.
For illustration,reconsiderthe fbur unit exampleof Sec.7.3. Let the daily
load curve for the systembe as indicated in Fig. 7.7. The economicallyoptimal
UC fbr this load curve is immediately obtained by use of the previously
prepared UC table (see Table 7.3) and is given in TabIe 7.4.
Let us now check if the above optimal UC table is securein every period of
the ioaci curve.
For the minimum load of 5 MW (period E of Fig. 7.7) accordingto optimal
UC Table 7.4., only unit 1 is to be operated.Assuming identical failure rate )
of l/year and repair rate pr,of 99/year for all the four units, let us check if the
systemis securefcrrthe period E. Further assumethe systemMTIL to be 0.005.
Unit I can be only in two possiblestates-operating or on forced outage.
www.muslimengineer.info
Time in hours ------------>
Flg.7.7 Dailyloadcurye
Therefore,
2
S =
piti :
p1r1* p2r2
,?
where
=
Pr= P(up)
ffi
pz= p( down) =
= 0.99, rr = 0 (unitI = 12MW> 5 MW)
p+^
= 0.01,
rz = t
(with unit I down load
demand cannot be met)
Hence
S= 0.99x 0 + 0.01x 1 = 0.01> 0.005(MTIL)
Thus unit I alone supplying5 MW load fails to satisfytne prescribed
securitycriterion.In orderto obtainoptimalandyet secureUC, it is necessary
to run the next mosteconomical
unit, i.e. unit 2 (Table7.3) alongwith unit 1.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
powerSystem
Mooern
Rnarysis
ifl5"#Wl
E#nft#
With both units I and 2 operating, security function is contributed only by
the state when both the units are on forced outage.The stateswith both units
operatingor either one failed can meet the load demand of 5 MW and so do
not contributeto the security function. Therefore,
S = p (down; x p (down) x 1 = 0.0001
= Rs 4,463.216 (start-up cost = 0)
Clearly Caseb resultsin overall economy.Therefore,the optimal and secure
UC table for this load cycle is modified as under,with due considerationto the
overall cost.
This combination (units 1 and 2both committed) does meet the prescribed
MTIL of 0.005, i.e. ^S< MTIL.
Proceedingsimilarly and checking securityfunctions for periodsA, B, c, D
and F, we obtain the following optimal and secure UC table for the sample
systemfor the load curve given in Fig. 7.7.
Table 7.5
Unit number
Period
A
B
C
D
E
F
Optimal and secure UC table
Unit number
Period
A
B
C
D
E
F
I
I t<
1
0
1
0
0
I
0
0
0
0
0
* Unit was started due to security
considerations.
Start-up
Considerations
The UC table as obtainedabove is secureand economically optimal over each
individual period of the load curve. Such a table may require that certainunits
have to be startedand stoppedmore than once.Therefore,start-upcost must be
taken into consideration from the point of view of overall economy. For
example, unit 3 has to be stopped and restartedtwice during the cycle. We
must, therefore, examine whether or not it will be more economical to avoid one
restarting by continuing to run the unit in period C.
Casea When unit 3 is not operatingin period C.
Total fuel cost for periods B, c and D as obtained by most economic load
sharing are as under (detailed computation is avoided)
= 1,690.756+ 1,075.356+ 1.690.756
= Rs 4.456.869
Start-up cost of unit 3 = Rs
50.000 (say)
Iotal operatrngcost= Rs 4,506.868
Caseb When all threeunits are running in period C, i.e. unit 3 is not stoppecl
at the end of period B.
Total operatingcosts= I,690.756+ 1,0g1.704+ 1,690.756
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I
I
I
I
I
I
*Unit was started due to start-up
7.5
OPTIMUM
GENERATION
1
1
I
I
I
I
I
I
l
0
l'l'
0
l
0
0
0
0
0
considerations.
SCHEDUTING
From the unit commitment table of a given plant, the fuel cost curve of the plant
can be determinedin the form of a polynomial of suitabledegreeby the
method
of least squaresfit. If the transmissionlossesare neglected,the total system
load can be optimally divided among the various generatingplants using the
equal incremental cost criterion of Eq. (2.10). It is, howrurr, on."alistic
to
neglect transmission losses particularly when long distance transmission
of
power is involved.
A modern electric utility serves over a vast area of relatively low load
density.The transmissionlossesmay vary from 5 to ISVoof the total load, 4'd
therefore,it is essentialto accountfor lcsseswhile developingan economic load
dispatchpolicy. lt is obviousthat when lossesarepresent,we can no longer use
the simple 'equal incrementalcost' criterion. To illustrate the point, consider
a
two-bus system with identical generatorsat eachbus (i.e. the sameIC curves).
Assume that the load is located near plant I and plant 2 has to deliver power
via a lossy line. Equal incrementalcost criterion would dictate that each plant
should carry half the total load; while it is obvious in this casethat the ptunt
1 should carry a greatershareof the load demandthereby reducing transmissio'
losses.
In this section, we shall investigate how the load should be shared among
rr$rrvsu
rqrirrrra
nlqnfc
l.rquLor
.rrlro-
l;-^
vYuvrr uuv
l^--^^
rvirDsr
the overall cost of generation
c =
L^ s
4rE .1uuuurrttrg
n^,-
t()f,
ry
Ine
ODJgCtfVg fS tO mirufiUze
,\-rci(Pc')
(7.7)
at any time under equality constraint of meeting the load demand with
transmissionloss, i.e.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
k
- P, - P I - = 0
DP",
i:l
(7.re)
Equation (7.23) can also be written in the alternative form
( I C) i- An- Q TL) if i = 1, 2, . . . , k
where
k = tatalrnumber of generating plants
Pci= generationof lth plant
Pp = sum of load demand bt all buses (system load demand)
Pr= total systemtransmissionloss
To solve the problem, we write the Lagrangianas
t=tr,(Pc)-^[t""
,- Po-".]
(7.20)
i:l
This equation is referred to as the exact coordination equation.
Thus it is elear tha-t to solve the optimum lead seheduling problem; it is
necessaryto compute ITL for each plant, and therefore we must determine the
functional dependenceof transmissionloss on real powers of generatingplants.
There are several methods, approximate and exact, for developing a transmission loss model. A full treatmentof these is beyond the scopeof this book.
One of the most important, simple but approximate, methods of expressing
transmissionloss as a function of generatorpowers is through B-coeffrcients.
This method is reasonably adequatefor treatment of loss coordination in
economic scheduling of load between plants. The general form of the loss
formula (derived later in this section) using B-coefficients is
It will be shown later in this section that, if the power factor of load at each
bus is assumedto remain constant, the systemloss P, can be shown to be a
function of active power generation at each plant, i.e.
Pr= Pt(Pcp
(7.2r)
P52, ..., P51r)
Thus in the optimizationproblem posedabove,Pci Q = I,2, ..., k) are the only
control variables.
For optimum real power dispatch,
')
AL = dC,- \r ,1Pt ^+ A
d P G toPo,
# *:O'
i = r' 2' " " k
(7.22)
Rearranging Eq. (7 .22) and recognizing that changing the output of only one
plant can affect the cost at only that plant, we have
= ) or
#Li=
) , i = r , 2 , . . . ,k
Q-APL iAPGi)
(7.24)
is called the penalty factor of the ith plant.
The Lagrangianmultiplier ) is in rupeesper megawatt-hour,when fuel cost
is in rupees per hour. Equation (7.23) impiies that minimuqr ftrei eost is
obtained,when the incrementalfuel cost of eachplant multiplied by its penalty
factor is the same for all the plants.
The (k + 1) variables(P6r, P62,..., Pct, )) canbe obtainedfrom k optimal
dispatchF,q. (7.23) together with the power balance Eq,. (1.19). The parrial
derivative)PLIAPGi is referred to as the incrementaltransmission loss (ITL),,
associatedwith the lth generatingplant.
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P. = II
pG^B*,pGn
(7.26)
m:7 n:I
where
PG^, PGr= real power generationat m, nth plants
B^n= loss coefficients which are constantsunder certain assumed
operatingconditions
If P6"sare in megawatts,B*n are in reciprocal of megawatts*.Cemputations,of
,:
course,may be carried out in per unit. Also, B*r= Bn^.
Equation (7.26) for transmissionloss may be written in the rnatrix form as
Pr= PIBPI
(7.27)
Where
(7.23)
where
Li=
(7.2s)
It may be noted that B is a symmetric matrix.
For a three plant system, we can write the expressionfor loss as
PL= Bn4, + Bzz4, + Bzz4, + 28rrpcrpcz + ZBnpGzpG3
+ 2BrrPorpo,
e.ZS)
wiih the systemDowerioss morieias per Eq. e.z6), we c.annow write
I
= ^ f t o_,
* ftl?lPo,"a^^'o')
Ap
*B^,
{in pu) = B^n (in Mw-t)
x Base MVA
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern
Po*ersystemAnatysis
ffif
k
k
4. Calculatep, =If
j:l
It may be noted that in the aboveexpressionother termsare independentof Po,
and are, therefore, left out.
Simplifying Eq. (7.29) and recognizing that B, = Bir, we can write
a P k
-+
=D zBijpcj
)Fo,
(7.30a)
pciBijpcj.
J=l
5. Check if power balanceequation (?. t o) is satistied
l
s
lLPot
l
- PD- pr.l' t
(a specifiedvalue)
ln*t
I
If yes, stop. Otherwise, go to step 6.
j:l
6. Increase) by A) (a suitable step size)' tt
Assuming quadraticplant cost curves as
pc, [*
C i (P o i )=
decrease ) by AA (a suitable stepsize);if
We obtain the incremenlalcost as
dc, =
dP'
atPo'+ b'
from above in the coordination
k
a i P c i +b , + S l z n , , r o , : ^
(1.3I)
J:l
Collecting all terrnsof P, and solving for Po, we obtain
k
(ai + 2M,,) Pci= - )D
dC
A*
J*T
r
) > 0 ,
A two-bus system is shown in Fig. 7.8. If 100 Mw is transmited from
plant
1 to the load, a transmission loss of 10 MW is incurred. Fin( the
required
generationfor each plant and the power received by load when
tie system ,\ is
Rs 25llvlWh.
The incrementalfuel costs of the two plants are given below:
ZBijpGj-bi + ^
- o'ozPcr+ 16'oRs,Mwh
- o'o4PG2+
2o.oRs/lvIWh
t&
'28,,P
)
L""ii'Gi
n ' t -
- po- p,
'
Example7,4
'-l
^l - a )
(8",
repeatfrom step 3.
(7.30b)
Substinrting APL|&G' and dcildPci
Eq. (7.22), we have
<0or
")
b ,p .,+ d,
* o ,4 ,+
^ -
:f:
]:I
Pc,
,,
*rB=;
)
i-7'2'"''k
(7'32)
For any particularvalue of \, Eq. (7.32) can be solved iteratively by
assuminginitial valuesof P6,s(aconvenientchoiceis P", = 0; i = l, 2, ..., k).
Iterations are stoppedwhen Po,s converge within specified accuracy.
Equation (7.32) along with the Dower balance F,q,.(7.19) for a pa-rtieular
ioaci ciemanciPo are soiveciiteratively on the following lines:
1. Initially choose) = )0.
2 . A s s u m eI t G i =0 ; I = 1 , 2 , . . . , k .
3. Solve Eq. (7.32)iterativelyfor Po,s.
Fig. 7.8 A two-bussystemfor Example7.4
6^r-
-r:--
outuuon
Therefore
since the ioad is at bus z a\one,p", v,.r,!not have any effect ort
Fr.
B z z =0 a n d B n = 0 = B z ,
Hence
Pl-=
For Po, -
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B nPbr
100MW, Pr = 10 MW, i.e.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(i)
"
Mod"rnPo*"r.Syrt"r An"lyri,
l
I
10= Bn (100)2
At plant 2 the load increases from 37.59 MW to 125 MW due to loss
coordination.The saving at plant 2 is
Brr= 0'001MW-r
Equation(7.31) for plant 1 becomes
A.A2P;1+2^;B1P;1 +2^aBr2P62= ^-76
(ii)
and for plant 2
0 . 0 4 P c 2 + 2 A B r r P " r + Z ) B u P c r =) - 2 0
Substituting the values of B-coefficientsand ) - 25, we get
(iii)
P c t = 1 2 8 ' 5 7MW
Pcz= 125Mw
= - Rs 2,032.431hr
The net saving achieved by coordinatinglosseswhile schedulingthe received
r'
load of 237.04 MW is
2,937.69- 2,032.43= Rs g,OtS.Ztm,
Derivation
The transmissionpower loss is
Pr = 0.001 x (128.57)' = 16.53 MW
of Transmission
Loss Forrnula
An accurate method of obtaining a general formula for transmission loss has
been given by Kron [4]. This, however,is quite complicated.The aim of this
article is to give a simpler derivation by making certain assumptions.
Figure 7.9 (c) depicts the case of two generating plants connected to an
arbitrary number of loads through a transmission network. One line within the
network is designatedas branch p.
Im,agine that the total load current 1, is supplied by plant 1 only, as in
Fig. 7.9a. Let the current in line p & Irr.Define
and the load is
Po = Pct * Pcr- Pt- = 128.57+ I25 - 16.53 - 237.04 MW
Considerthe system of Example 7.4 with a load of 237.04 MW at bus 2. Find
the optimum load distribution between the two plants for (a) when losses are
included but not coordinated,and (b) when losses are also coordinated. Also
find the savings in rupeesper hour when losses are coordinated.
Solution Case a If the transmission loss is not coordinated, the optimum
schedulesare obtainedby equatingthe incremental fuel costsat the two plants.
Thus
(i)
0 . 0 2 P -+ 1 6 = 0 . O 4 P c z + 2 0
The powerdeliveredto the loadis
+ 237.04
Pct * Pcz= 0.001P4r
SolvingBqs.(i) and(ii) for P61nnclP6.2,we get
Pcr = 275.18MW; and Prr2= 37.59MW
\
(ii )
Caseb This caseis alreadysolvedin Example 7.4. Optimum plant loadings
with loss coordination are
Pct= 128,57MW; Pcz = 125 MW
Loss coordinationcausesthe load on plant I to reducefrom 275.18 MW to
128.57MW. Therefore,saving at plant I due to loss coordination is
Kgi:",+
: o.MPl,r
16)dPc,
+r6Pctli),'r',
(c)
Flg. 7.9 Schematic diagram showing two plants connectedthrough
a power network to a number of loads
Sirniinriv
with
nient
r^_-^,
)
qinrnc
ctrnnirrino
the
fntel
lnqd
vn sn^r ravnr tr r
www.muslimengineer.info
t'Eic
\r r5.
? ok\
,.rv),
r'a
wv
^av(ltl
define
I^n
Mrz= i=
tD
= Rs 2,937.691hr
(7.33)
e.34)
Mo1 wrd Mp2 are called curcent distribution factors. The values of current
distribution factors depend upon the impedances of the lines and their
interconnection and are independentof the current Ip.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernpowerSygtemAnatysis
fW
t
When both generators t and 2 are supplying current into the network as in
Fig.7.9(c), applying the principle of superpositionthe currentin the line p can
be expressedas
e, = lstrp( Re
p
Substitutingfor llrl2 fromEq. (7.37), and l1o,l and llurl from Eq. (7.38),'we
obtain
where 1ot and Io2 are the currentssupplied by plants I and 2, respectively.
At this stage let us make certain simplifying assumptionsoutlined below:
(1) All load currentshave the same phase angle with respectto a common
refere{ce. To understandthe implication of this assumptionconsider the load
current at the ith bus. It can be written as
VDil I (6t- d) = lloil l1i
where {. is the phaseangleof the bus voltage and /, is the lagging phase angle
of the load. Since { and divary only through a narrow range at various buses,
it is reasonableto assumethat 0,is the same for all load currentsat all times.
(2) Ratio X/R is the same for all network branches.
Thesetwo assumptionslead us to the conclusion that Ip1 andI, fFig.7 .9(a))
have the samephase angle and so have Ioz and 1o [Fig. 7.9(b)], such that the
current distribution factors Mr, and Mr, are real rather than complex.
Let,
Ict = llcrl lo, and lcz = 1162llo2
.
o;
Mr2lls2lsn oz)2
*Wl,r,rMpzRp
lVllv2lcos /, cosQ,
*'
(7 3e)
Equation
"T,o:
8 'n.. -
Expanding the simplifying the aboveequation, we get
ll,,l2= Mzrrllorlz + tutf,zllczlz
+ 2MolMrzllctl llGzlcos(a1 - oz)
Now
l sl .rr l = = ' ? '
$ l v tl c o s /,
:' I I .vL
J-
-&z-
Jl l vrl cosf"
(7.37)
(7.38)
where Pot and Po, are the three-phasereal power outputs of plants I and 2 at
power factors of cos (t, and cos Q2,and yl and V2 are the bus voltages at the
plants.
ff Ro is the resistanceof branchp, the total transmissionloss is given by*
'The
general expression for the power system with t plants is expressedas
P
'r -
Brz
Bn =
(7.36)
Pc^ Pcn cos(a, - on)
lV* ilV, l c o s Q * c o s Q ,
www.muslimengineer.info
P
7
P3'
M32RP
tvzP("o,
dritT
;:;::::;:,PczB,z
(7'3s)
+ 4,8,,
WGoshfDmS,no
p
ffiDM"M"Ro
(7.40)
T . M ' ,YrLRY.
7
The terms Bs, Bp and 82, are called loss cofficients or B-cofficients.lf
voltagesare line to line kV with resistancesin ohms, the units of B-coefficients
are in MW-I. Further, with Po, and Po, expressedin MW, P, will also be in
l V , l ' ( c o s6 ) '
MK
The\,bove results can be extendedto the general case of ft plants with
transrnishur loss expressedas
k
k
P, =Df
m:l
PG^B^.PG.
(7.41)
n:l
where
8 r,,, =
F3'
'nt"P' ' .-r4r-lu|*ry
'
T,*3,R,t.
1yf 1*r6f L'
tVr,l2
1cosffiLu'
rLu|rn,
1v31"*fr)
p
"-
where a, and 02 are phaseanglesof 1", and lor, respectiveiywith respect
to the common reference.
From Eq, (7.35), we can write
llrl2 - (Moll6l cos a1 + Mpzllo2lcos oz)2a (Mrll6lstn
2
rDGr
D-
cos (a,, - on)
(7.42)
l v - l l v , l c o s Q -c o s Q -
It can be recognized as
Pr=ftrB, * ...* P'o4oo * zDpG^B^npGn
m,n:l
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
;!l
..-#
i?ffil
powersystemnnatysis
ruodern
r !nrr!-na!
I
The following assumptionsincludingthosementionedalreadyare necessary,
if B-coefficients are to be treated as constantsas total load and load sharing
betweenplants vary. Theseassumptionsare:
1, All load currents maintain a constantratio to the total current.
2. Voltage magnitudes at all plants remain constant.
3. Ratio of reactive to real power, i.e. power factor at each plant remains
constant.
4. Voltage phase angles at plant buses remain fixed. This is equivalent to
assuming that the plant currents maintain constant phase angle with
respectto the common reference,since sourcepower factors are assumed
constant as per assumption3 above.
In spite of the number of assumptionsmade, it is fortunate that treating Bcoefficients as constants,yields reasonablyaccurateresults, when the coefficients are calculated for some average operating conditions. Major system
changesrequire recalculation of the coefficients.
Lossesas a function of plant outputscan be expressedby other methods*,but
the simplicity of loss equationsis the chief advantageof the B-coefficients
method.
Accounting for transmission losses results in considerable operating
economy.Furthermore,this considerationis equally important in future system
planning and, in particular, with regard to the location of plants and building
of new transmissionlines:
-
lb--
lr"
1
'
Y
i
ffi
i
-
'
b
----t''
Refbus
v =110"pu
l,o
Y
I ro"or
Fig. 7.10 Samplesystemof Example7.6
Solution As all load currentsmaintain a constantratio to the total current,we
have
rd
_ 3. 6_ jo. g _ o. t 8z6
I, + Id
4. 6- jl. l5
I"
I,+ld
: -r- "i0.25
---- -0.2174
4.6-j1.1s
Mor= L, Mtr - - 0.2174, Mrr = 0.2774,Mu = 0.7826
Figure7.10 shows a systemhaving trrvoplants 1 and 2 connectedto buses 1 and
2, respectively. There are two loads and a network of four branches. The
referencebus with a voltageof l.0l0o pu is shownon the diagram.The branch
cunents and impedancesare:
I,=7 - j0.25Pu
I o = 2 - 7 0 . 5P u
I u = 1 . 6- j 0 . 4 P u
Id = 3.6 - 70.9 Pu
Zo = 0.015 + 70.06pu
Z, = 0.OI + 70.04 pu
Zo = 0.015 + 70.06 pu
Za = 0.Ol + 70.04pu
Calculatethe loss formula coefficients of the systemin pu and in reciprocal
mesawatfs if the hase is 100 MVA
^^^-D-
M,,2= 0, Mnz = 0.7826, Mrz = 0.2174,Mrtz= 0,7826
Since the sourcecurrentsare known, the voltagesat the sourcebusescan be
calculated.However, in a practical size network a load flow study has to be
made to find power factors at the buses,bus voltagesand phase angles.
The bus voltages at the plants are
Vr = 1.0 + (2 - j0.5) (0.015+ 70.06)
= 1.06+ jO.I725 = 1.066 16.05" pu
Vz= 7 + ( 1. 6 - jO . 4) ( 0. 015+ 70. 06)
= 1.048+ jO.O9= 1.05114.9" pu
The current phaseanglesat the plants are (1, = Io, 12= 16r Ir)
ot- = t an- t+!
2.
:-
l4oi
o2: t an2
. r
cos (or- ot) = cos 0o = 1
*For
more accuratemethodsand exactexpressionfor 0P,./0P6i,references122,231
may be consulted.
www.muslimengineer.info
The plant power factors are
pfi = cos (6.05" + l4') = 0.9393
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
- ^O
6 :
9t
- l4o
i
OPtimal System Operation
Mociern
PowerSvstemAna[,sis
ffiffi
p
Pfz = cos (4.9 + 14")- 0.946
The losscoefficientsare lBq. Q.a\l
ci(Pci)
c =f
(7.7)
subject to the load flow equations
e -f
- 0.02224pu
A
4 - Llvllvjlly,,lcos(9,, * 6i - 4)= 0 for each pv bus
It is to be noteclthat at the ith bus
Pt= Pci- Poi
Qi= Qci- Qu
where Po, and ep; areload demands at
bus i.
Equarions(7.43), (7.44) and (7.45) can be
expressedin vector form
(7.a3)l
[Eq.
I
0'01597 =
8.t.,=
0.01597x 1o-2Mw-l
100
'
0'0M06 =
0.00406 x lo-z Mw-l
100
f (x, y) =
\
;
(7.47)
I
.= lr, jforeachrouusf
Constraints
The objective function to be minimized is ihe operating cost
www.muslimengineer.info
(7.46)
is
l-ty,tl
The problem of optimal real power dispatch has been treated in the earlier
section using the approximate loss formula. This section presentsthe more
general problem clf real attd reactive povrer flow so as to minirnize the
instantaneousoperatingcosts.It is a static optimization problem with a scalar
objective function (also called cost function).
The solutiontechniquegiven here was first given by Dommel and Tinney
[34]. It is basedon load flow solutionby the NR method,a first order gradient
adjustmentalgorithm for minimizing the objective function and use of penalty
functions to accountfor inequality constraintson clepenclent
variables.The
problem of unconstrainedoptirnalload flow is first tackJed.Later the inequality
constraints are introduced, first on control variables and then on dependent
variables.
Inequality
foreachr0 bus
|
foreachpV busi
| _tn.\7qql
lEq.Q.a,
where the vector of dependentvariables
LOAD FLOW SOLUTION
Power Flow without
(7.4s)
j:7
100
Optimal
(7.44)
and
0.02224
LLL+
h
- 0.02224 x lo2 Mw-l
Drr =
OPTIMAL
(7.43)
j:r
+ 0.01x(0.217a)2
+ 0.01x(0.7826)2
D =
_ e0.2174)L0.7826X0.015)
D
t2
rJ66 . Lotl x 0.9393
x 0.946
= 0.00406pu
For a baseof 100 MVA, theseloss coefficientsmust be dividedby 100 to
obtaintheir valuesin units of reciprocalmegawatts,
i.e.
7.6
tUilvjily,,tcos(0,,
3
Q, + Ltvinvjlllzulsin (0ul
= 0.01597
pu
Br.t =
[see
j:1
q2 + 0.01x (0.782q2
0.0| 5 x (0.782q2 + 0.01x (0.217
(1.051)2
x(0.946)2
Bzz
ffi
ffi$ffi
l
(7.48a)
Ld, for eachpV bus_J
andthe vectorof independent
variablesis
t r 4 In
a. slackbus
4i
4l
foreachpe bus
]=
O]
(7.48b)
t"eachPVbus
,1,,,j
fn
p
o
fha
w
e
qq vl -v^v. vr o
r
.
r
f^*,,.t^+:^rt-lrurulalLlull,
r
r
u
D
l
l
l
-r
[fle
ODleQtlVe
l
l
w
L
l
L
tltnefinn
l
g
L
l
mrrcf
l
E
S
i_^1,-A^
r
a
c
d^^
K
D
_r u
S
The vector of independentvariabresy can
be partitioned into two parts_a
vector u of control variableswhich are to
be variea to achieve optimum value
of the objective function and a vector p of fixed
or disturbangeor unconhollable
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowerSvstemAnalvsis
ffi@
may be voltagemagnitudeson PV buses,P6t
puru-"t"rs. Controlparameters*
at buseswith controllablepower, etc.
The optimization problem** can now be restatedas
(7.4e)
min C (x' u)
subject to equalitYconstraints
.f (x, u, p) = 0
(7'50)
To solve the optimization problem, define the Lagrangian function as
(7'51)
L (x, u, p)= C (x, u7+ Arf (x, u, P)
p)
where ) is the vector of Lagrange multipliers of same dimension asf (x, u,
function
Lagrangian
unconstrained
the
to
minimize
conditions
The necessary
are (see Appendix A for differentiation of matrix functions).
a f ,= 0 c* l y 1 ' ) _ o
(7.s2)
0 L = 0 c* l y 1 ' ) _ o
(7.s3)
a r = (x,u,P) = o
f
u;
(7.s4)
0x
0x
0u
0u
,
L}x J
feasible solution point (a set of valuesof x which satisfiesEq. (7.5a) for given
u andp; it indeed is the load flow solution) in the direction of steepestdesceht
(negative gradient) to a new feasible solution point with a lower value of
objeetlve funstion. By repeating the-.semoves in *rc dkestisn +f the negadve
gradient, the minimum will finally be reached.
The computational procedurefor the gradientmethod with relevant details is
given below:
Step I
Stey 2 - Jind a feasible load flow solution from Eq. (7.54) by the NR iterative
method. The method successivelyimproves the Solution x as follows.
(r
* +r) - ,(r) + Ax
where A-r is obtained by solving the set of linear equations (6.56b) reproduced
below:
-y)
l#r,"',rl]4" r (*('),
Ldu-J
Equation (7.54) is obviously the same as the equality constraints.'The
tS
as neededin Eqs. (i.52) and (7.53) are rather
^a L
;
0u
involved***.It may howeverbe observedby comparisonwith Eq. (6.56a)that
rhe endresurts J,;; $; Iti.[l]'*u,,on
"^-,
Step 3
expressionsfor
Y=
Jacobian matrix [same as employed in the NR method of load flow
0x
(6.64)
solution; the expressionsfor the elementsof Jacobianare given in Eqs.
and (6.65)1.
Equations(7.52),0.53) and (7.54) are non-linearalgebraicequationsand
can only be solvediteratively. A simple yet efficient iteration scheme,that can
by employed, is the steepestdescentmethod (also called gradient method).
-Slack
bus voltage and regulating transformer tap setting may be employed as
buses'
additional control variables.Dopazo et all26ltuse Qo, as control variable on
with reactive Power control'
**rr *L^ ..,.*^Dlvrrl
LrMJ
lI
.pal nnrrrcr lncc ic to he minimized-
lvsr
rv
the obiective
function
the real
Since in this case the net injected real powers are fixed, the minimization of
loss,
system
total
of
to
minimization
is
equivalent
bus
slack
the
at
injected power P,
This is known as optimal reactive power flow problem'
***The
original pup". of Dommel and Tinney t34l may be consulted for details'
www.muslimengineer.info
r,
I
Step 4
or x andtheracobian
matri,...
Solve Eq. (7.52) for
^- -rr-l
\ = - i( { \ ' l
L\dxl J
ac
0x
(7.s5)
Insert ) from Eq. (7.55) into Eq. (7.53), and compute the gradient
= oc *l 9L1't
Y.c,
0u
L 0 u)
(7.s6)
It may be noted that for computing the gradient,the JacobianJ - + is already
0x
known from the load flow solution (step 2 above).
step 5 rf v -c equals zero within prescribedtolerance,the minimum has
been reached. Otherwise,
Step 6
Find a new set of control variables
L,il
( 7. 57\
L,u = - o"V-C,
(7.s8)
unew= l.l.^6*
is
C = Pr(lVl, 6)
Make an initial guess for u, the control variables.
where
Here A,u is a step in the negative direction of the gradient. The step size is
adjustedby the positive scalar o..
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
#*ff"
uooernPowerSvstem
Analvsis
f
to the constraint limits, when these limits are violated. The penalty function
method is valid in this case, becausethese constraints are seldom rigid limits
in the strict sense,but are in fact, soft limits (e.g. lvl < 1.0 on a pebus really
Steps 1 through 5 are straightforward and pose no computational problems.
Step 6 is the critical part of the algorithm, where the choice of a is very
important.Too small a value of a guarantees
the convergencebut slows down
the rate of convergence; too high a value causes oscillations around the
ry
Inequality
The penalty method calls for augmentationof the objective function so that
the new objective function becomes
Constraints
on Control Variables
v-should not exceed 1.0 too much and lvl = 1.01 may still be
C t = C ( x , u )f*U t
Though in the earlier discussion, the control variables are assumedto be
unconstrai""o,
are, in fact, always contrained,
(7.se)
1.j":T.1",::tutt
Pc,, ,nin1 Po, S Pct, **
e.g.
These inequality constraintson control variablescan be easily handled.If the
correctionAu,inBq. (7,57) causesuito exceedone of the limits, a, is setequal
to the correspondinglimit, i.e.
if u,,oro* Au, ) ui,^^
7f u,,oro* Au, 1ui,^in
(7.63)
where the penalty W, is introduced for each violated inequality constraint. A
suitable penalty function is defined as
- xi,^o)2 i
w,' = {7i@i
[ 71G , - xi, ^i) zi
wheneverxi ) xi,rnax
whenever xr ( r y, m in
Q '64)
where Tiis
Treal positive number which controls degreeof penalty and is called
the penalty factor.
(7.60)
otherwise
After a control variable reaches any of the limits, its component in the
gradient should continue to be computed in later iterations, as the variable may
come within limits at some later stage.
In accordance with the Kuhn-Tucker theorem (see Appendix E), the
necessaryconditionsfor minimization of I, under constraint (7.59) arc:
Xmln
0L :0
0u,
7 f u ,,* n < ui < ui ,^^,
of .o
ou,
if u, - ui,^*
o r -, o
0r,
Fig. 7.11 Penaltyfunction
(7.6r)
u i : u i .^u*
A plot of the proposedpenalty function is shown in Fig. 7.11, which clearly
indicates how the rigid limits are replaced by soft limits.
The necessaryconditions (7.52) and (7.53) would now be modified as given
below, while the conditions (7.54), i.e. load flow equations,remain unchanged.
Thereforer now, in step 5 of the computationalalgorithm, the gradient vector
has to satisfy the optimality condition (7.61).
Inequality
Constraints
on Dependent
ax_
c , \ -) a w j , f a f l ' ,
__l_
= a
(7.6s)
ax_AC,sdw; ,fAf1',
(7.66)
T-
dx
Variables
o x 4l a r * L a " i ) - o
-:-
=
rmir,SxSr*u^
e.g.
lUn,in < lVl < lYl -.o ofl a PQ bus
'Ihe
(7.62)
Such inequality constraints can be conveniently handled by the penalty
function method. The objective function is augmented by penalties for
inequality constraintsviolations. This forces the solution to lie sufficiently close
www.muslimengineer.info
--L
du ou'+ a"*La"l'r:o
Often, the upper and lower limits on dependentvariables are specifiedas
)
^trZ
vecto
UW:
0x
obtainedfrom Eq. (7.64)would containonly one non-zero.
termcorresponding
to the dependent
variablex;; while
= 0 asthepenalty
#
functionson dependentvariablesare independent
of the control variables.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'
ModernPowerSystemAnalysis
By choosinga higher value fot 1,,the penalty function can be made steeper
that
the solutionlies closer to the rigiA fimits; the convergence,however,will
so
becomepoorer.A good schemeis to start with a low value of 7j and to increase
imization process,if the solution exceedsa certain tolerance
limit.
This sectionhas shown that the NR method of load flow can be extendedto
yield the optimal load flow solution that is feasible with respectto all relevant
inequality constraints.These solutions are often required for system planning
and operation.
7.7
OPTIMAL
SCHEDULING
OF HYDROTHERMAL
Mathematical
For a certainperiod of operation7 (one year,one month or one day, depending.
upon the requirement),it is assumedthat (i) storageof hydro reservoir at the
reservoir (after accounting for irrigation use) and load demandon the system
are known as functions of time with complete certainty (deterministic case). The
problem is to determine q(t),,the water discharge(rate) so as to minimize the
cost of thermal generation.
Cr =
SYSTEM
The previous sectionshave dealt with the problem of optimal schedulingof a
Dower system with thermal plants only. Optimal operating policy in this case
can be completely determined at any instant without reference to operation at
other times. This, indeed, is the static optimization problem. Operation of a
system having both hydro and thermal plants is, however, far more complex as
hydro plants have negligible operating cost, but are required to operate under
constraintsof water available for hydro generationin a given period of time.
The problem thus belongs to the realm of dynamic optimization. The problem
of minimizingthe operating cost of a hydrothermal sYstemcan be viewed as one
of minimizing the fuel cost of thermal plants under the constraint of water
availability (storage and inflow) for hydro generation over a given period of
operation.
J (waterinflow)
Formulation
rT
JoC
( Por ( t ) ) dt
(7.67)
under the following constraints:
(i) Meeting the load demand
Pcr(r) * Pca(t) - Pr(t) - PoG) = 0; te 10,71
(7.68)
This is called the power balance equation.
(ii) Water availability
X (T)- x, (o) -l' t@ at+ lrqgldt=o
JO
JO_
(7.6e)
where J(t) is the water inflow (rate), X'(t) water storage,and X/(0) , Y (T) arc
specified water storages at the beginning and at the end of the optimization
interval.
(iii) The hydro generation Pcrlt) is a function of hydro dischaige and water
storage(or head), i.e.
Pcn(r)=f(X'(t),q(t))
(7.70)
The problem can be handled conveniently by discretization. The optimization
interval Z is subdivided into M subintervalseach of time length 47. Over each
subinterval it is assumed that all the variables remain fixed in value. The
problem is now posed as
n^(EL,rfrrrt,
: u^r^4-r:....*r"$^-r'(pt)
hydrothermalsystem
Fig. 7.12 Fundamental
For the sake of simplicity and understanding,the problem formulation and
solution technique are illustrated through a simplified hydrothermal system of
Fig. 7 .I2. This systemconsistsof one hydro and one thermai piant suppiying
power to a centralized load and is referred to as a fundamental system.
Optimization will be carried out with real power generation as control
variable,with transmissionloss accountedfor by the loss formula of Eq. (7.26),
-__----^_1--_
under the following constraints:
(i) Power balance equation
Pt *Ptr-PI-Pt =0
where
Q
Ptr = thermal generation in the mth interval
Ptn = hydro generation in the zth interval
PI =transmission loss in the rnth interval
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\t.lr)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(7.72t
ModernPowerSystemAnalysis
ffiffifjn
=
n
tnm
D77\r57
t2
)
t
|
.rn
nm
LDTHTGH
D
tnm
-fI Dpy\r6p
t2
Optimal
SystemOperation
)
(ii) Water continuity equation
y^ _ yt(m_r)_ f
p = non:effective discharge (water discharge neededto run h dro
AT + q^ AT = 0
X ^ = water storage at the end of the mth interval
J* -- water inflow (rate) in the mth interval
q^ = water discharge (rate) in the ruth interval
The above equation can be written as
(7,73)
Y - X^ -r - J * + q ^ = 0 ; m = I,2, ,.., M
=
=
where Y
X/*IAT
storagein dischargeunits.
InEqs. (7.73), Xo and XM are the specified storagesat the beginning and
end of the optimization interval.
(iii) Hydro generation in any subinterval can be expressed*as
Ptn = ho{I + o.5e(Y + Y)l
(q* - p)
where
(7.74)
In the aboveproblem formulation, it is convenient to choosewater discharges
in all subintervalsexcept one as independentvariables, while hydro generations, thermal generations and water storagesin all subintervalsare treated as
dependentvariables. The fact, that water discharge in one of the subintervals
is a dependentvariable, is shown below:
Adding Eq. (7.73) for m = l, 2, ..., M leads to the following equation, known
as water availability equation
xM-
"o-D
J^ +la^ = o
m
Becauseof this equation, only (M - l) qs can be specifiedindependently and
the remaining one can then be determined from this equationand is, therefore,
a dependentvariable.For convenience,ql is chosenas a dependentvariable, for
which we can write
M
qt = xo - xM* DJ^ -Dn^
fto = basic water head (head corresponding to dead storage)
Solution
P3, = 9.81 x 1o-2hk@^ -,p ) Mw
where
(q^ - p) = effective discharge in m3ls
hry, = average head in the mth interval
(7.7s)
m
ho=9.8rx ro-rhto
*
ffiffi
l e = water head correction factor to account for head variation with
storage
PI = load demand in the mth interval
(7.76)
Technique
The problem is solved here using non-linear programming technique in
conjunction with the first order gradient method. The Lagrangian L is
formulated by augmenting the cost function of Eq. (7.7L) with equaliry
constraintsof Eqs. (7 .72)- (7.74) throughLagrangemultipliers (dual variables)
\i \i'and )i. Thus,
Now
L T (X ^ + X^ -r)
l ffi = 7 ro *
2A
where
A = draa.of cross-sectionof the reservoir at the given storage
h' o = basic water head (head correspondingto dead storage,
hk= hLll + o.Se(x'+ X"-t)l
where
AT
Ahto
.
Now
4n= ho {! + o.Se(x^+ x^-t)l @^ - P)
where
h o = 9 .8 7x l 0 -3 h to
www.muslimengineer.info
.c,=D tc(%r)- xT(4r+ 4, - ry- ffi + M (y - y-t-r* +
qr)* ^Ttp1,- h,(r + 0.5e(y* it11* @^- p)rj (7.77)
The dual variables are obtainedby equating to zero the partial derivatives
of the Lagrangian with respectto the dependentvariables yielding the following
equations
AP
rJ)e
7Pt
=
Ar,rDmt
ut/\I CT)
--
dPt
/
- A ,z
rl I n
l-
'[-
rt.
Uf -t
\
|
^
l-u
7Pt )
0 /78)
[The reader may compare this equation with Eq. (7.23)]
#r,-M-^r['-ffi)='
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(7.7e)
Modern
PowersystemAnalysis
ffil
a
(-+)
- p)l- )i*t 1o.5ho
= )7,- Ar*'- \r p.sh"r(q*
e
I\ ax^ /)^**t
*U
(q^*'- DI - o
(7.80)
and using Eq. (7.73) in Eq. (7.77), we get
(pt-) = )rr- t\n" fl + 0.5e(zy + Jt - zqt +d)= 0
laq' )
(7.81)
The dual variables for any subinterval may be obtained as follows:
(i) Obtain { from Eq. (7.78).
(ii) Obtain )! from Eq. (7.79).
(iii) Obtain )1, from Eq. (7.81) and other values of ry (m * 1) from Eq.
'
( 7. 8 0 ).
The gradient vector is given by the partial derivatives of the Lagrangian with
respectto the independentvariables. Thus
=\f.- ^Zh"{r + 0.5e(2Y-t + J^ - 2q^+ p)}
(+)
\oq
(7.82)
)m+r
For optimality the gradient vector should be zero if there are no inequality
constraintson the control variables.
Algorithm
Assume an initial set of independent variables q* (m*I)
subintervalsexcept the first.
for all
Obtain the values of dependent variables Y, Ptu, F\r, qt using Eqs.
(7.73), (7.74), (7.72) and (7.76).
3. Obtain the dual variables )f, \f, )i @ = 1) and )rr using Eqs. (7.78),
(739), (7.80) and (7.81).
4. Obtain the gradientvector using Eq. (7.82) and check if all its elements
are equal to zero within a specified accuracy.If so, optimum is reached.
If not, go to step 5.
5. Obtain new values of control variables using the first order gradient
method,i.e.
e k * = q ; a{ +
- +);m=r
(7.83)
\oq* )
where cr is a positive scalar. Repeat from step 2
In the solution techniquepresentedabove, if some of the control variables
(water discharges)cross the upper or lower bounds, these are made equal to
their respective bounded values. For these control variables, step 4 above
is checked in accordancewith the Kuhn-Tucker conditions (7.61) given in
S ec .7. 6 .
www.muslimengineer.info
iently by augmenting
the cost functioniith p"nulty functionrur air";;;i"
Sec.7.6.
The methodoutlinedaboveis quitegeneraland canbe directlyextendedto
a slzstemhaving multi hydro and multi-ther+nal plan+s. The method, however,
has the disadvantage of large memory requirement, since the independent
variables,dependentvariables and gradientsneed to be storedsimultaneously.
A modified techniqueknown as decomposition[24) overcomesthis difficulty.
In this technique optimization is carried out over each subinterval and the
complete cycle of iteration is repeated,if the water availability equation does.
not check at the end of the cvcle.
Consider the fundamental hydrothermal system shown in Fig. 7.I2. The
objective is to find the optimal generationschedulefor a typical dav, wherein
load varies in three steps of eight hours each as 7 Mw, 10 Mw and 5 Mw,
respectively. There is no water inflow into the reservoir of the hydro plant. The
initial water storage in the reservoir is 100 m3/s and the final water storage
should be 60m3/s,i.e. the total water available for hydro generationduring the
day is 40 m3/s.
Basic head is 20 m. Water head correction factor e is given to.be 0.005.
Assume for simplicity that the reservoir is rectangular so that e doesnot ehange
with water storage. Let the non-effective water discharge be assumed as
2 m3/s.Incrementalfuel cost of the thermalplant is
dc r.opcr + 25.0 Rsft'
dPcr
Further, transmissionlosses may be neglected.
The aboveproblem has been speciallyconstructed(ratherovelsimplified) to
illustrate the optimal hydrothermal schedulingalgorithm, which is otherwise
computationally involved and the solution has to be worked on the digital
computer. Stepsof one complete iteration will be given here.
Since there are three subintervals,the control variablesre q2 and q3. Let us
assumetheir initial values to be
q2 = 75 m 3ls
15 m2ls
The value of water diseharge in the first subinterval can be immediateiy found
out using Eq. (7.76), i.e.
er = LOO- 60 - (15 + 15) = 10 m3ls
It is given that X0 = 100 m3/s and X3 = 60 m3/s.
From Eq. (7.73)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpo
W
Xt = f
+ Jr - gt = 90 m3ls
xt, - x3- )l {o.Shoe
(qt- p)l _ S! 1O.Sn"e
@,- p)l = 0
x?,- ll - ^Z {0.5 hoe(q, _ p)l _ )3, 1o.5ho,
(qt - p)] = o
f=Xr+12-q2=75m3/s
The valuesof hydro generations
in the subintervalscan be obtainedusing
Eq. (7.7q as follows:
P b , t = 9 , 8 1x 1 0 - 3x 2 0 [ l + 0 . 5 x 0 . 0 0 51 x r+ X 0 ; ) { q ' - p )
= 0.1962{I + 25 x 104 x 190} x 8
= 2 . 3 1 5M W
4 n = 0 . 1 9 6 2{ I + 2 5 x l O a x 1 6 5 }x 1 3
= 3.602MW
P Z a = 0 . 1 9 6 2{ l + 2 5 x 1 0 a x 1 3 5 }x 1 3
= 3.411MW
The thermalgenerationsin the threeintervalsare then
pLr = pL- pl, = J - 2.515= 4.685MW
4r = Pto- PL, = 10 - 3.602= 6.398MW
F c r = p | , - 4 " = J - 3 . 4 1 I= 1 . 5 8 9M W
From Eq. (7.78),we havevaluesof )i as
dc(P€)
A
D Tm
VG
_
- ' t l \m
)T=P[,+25
or
Calculating ), for all the three subintervals, we have
[^i I
[2e685]
lr?l=lrr.lsal
j fzo.ssrJ
Lr?
Also from Eq. (7.79), we can write
lril
hil lzs.68s1
=
case
| ^3| | ^?l: I I r:ea I for ttrelossless
.sag
J
LriI LdI Lze
From Eq. (7.81)
\l'
= A\hnii + o.5e(?)(0+ il - z^ ql ' + o l
- 29.685x 0.1962{l + 25 x loa (2oo- 2o + 2)l
-,8.474
FromEq. (7.E0)for m = 1 and2, we have
www.muslimengineer.info
Substituting various values, we get
) t r = 8 . q 1 4 - 2 9 . 6 8 5( 0 . 5 x 0 . 1 9 6 2 x0 . 0 0 5x 8 l _ 3 1 . 3 9 8
{0.5x
0.1962x 0.005x 13)
= 8.1574
)t, = 9.1574- 31.398(0.5x 0.tg6} x 0.005x t3)
- 26.589(0.5 x 0.1962x 0.005x t3) =
7.7877
UsingEq. (7.82),the gradientvectoris
^ 7 - f t n " { 1 + 0 . 5x 0 . 0 0(52x e o- 2 x 1 5+ z ) l
(#)=
= 8.1574- 31.398x 0.1962,{l + 25 x 10a x
l52l
= - 0.3437
( ar.l - r3
A:2- strn,
fl + 0.5e(2X2+ f - zqt+ p)I
lrf )-
- 7.7877- 26.589x 0.1962 + 25 x
10a x t22}
{I
= 0.9799
If the tolerancefbr gradient vectoris 0.1, then optimal conditions
are not yet
satisfied, since the gradient vector is not zero,i.i. (< 0.1);
hence the second
iteration will have to be carried out starting with the following
new values of
the control variablesobtained from Eq. (7-g3)
lnk_l=la[,oLl#l
Lq:"*J=
L;i;l-1+l
L o q ")
Let us take a = 0.5, then
:flillil
lf:i=[[]-,'l-:::Ij
and from Eq. {7.76)
4'r*= 100 - 60 - (15.172+ 14.510)= 10.31gm3ls
The above computation brings us to the starting point of the
next iteration.
Iterations are carried out till the gradient vector U"comes zero
within specified
tolerance.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffil
Hffiffi
uodernPowerSlrstem
Analvsis
PROB
iEii/iS
7.1 For Example 7.1 calculate the extra cost incurred in Rsftr, if a load of
220 MW is scheduledas Pct= Pcz = 110 MW.
7.2 A constant load of 300 Mw is supplied by two 200 Mw generators,I
and 2, for which the respective incremental fuel costs are
dcr
= o ' l O P G l+ 2 0 ' 0
Po,
dcz o'lzPc2 + 15'o
dPo,
with powers Pc in MW and costs c in Rsar. Determine (a) the most
economicaldivision of load betweenthe generators,and (b) the saving in
Rs/day thereby obtained compared to equal load sharing between
machines.
7.3 Figure P-7.3 shows the incrementalfuel cost curves of generatorsA and
B. How would a load (i) more than ZPo, (ii) equal to 2p6, and (iii) less
than ZPo be shared between A and B if both generatorsare running.
miiiions of iriiocaiories per hour can be expressedas a function of power
output Poin megawattsby the equation
0.00014+ O.$ft + r2.0po+150
Find the expressionfor inerementalfuel eost in rupeesper megawatt hour
as a function of power output in megawaffs. AIso find a good linear
approximation to the incremental fuel cost as a function of Fo.
Given: Fuel cost is Rs Zhmltion kilocalories.
7.6 For the system of Example 7.4, the system ) is Rs 26a4wh. Assume
further the fuel costs at no load to be Rs 250 and Rs 350 per hr,
respectively for plants I and 2.
(a) For this value of system ),, what are the values of p61, po,
and,
received load for optimum operation.
(b) For the above value of received load, what are the optimum values
of Pot and Por, if system losses are accounted for but not
coordinated.
(c) Total fuel costs in RsArr for parrs (a) and (b).
7.7 FigureP-7.7 showsa systemhaving two plants I and 2 connectedto buses
1 and 2, respectively. There are two loads and a network of three
branches.The bus 1 is the referencebus with voltage of 1.0 I 0" pu. The
branch currents and impedancesare
Io=2 -70.5 pu
-Lo= 1 6 -
i O 4 n r r
1, = 1.8- i0.45 pu
Zo= 0.06+ j0..24pu
Zt = 0.03+ J0.12pu
(MW)mtn
P6
Flg. P-7.3
7.4 Consider the following three IC curves
Z, = 0.03+ /0.I2 pu
Calculatethe loss formula coefficients6f the systemin per unit and in
reciprocalmegawatts,
if the baseis 100MVA
Ref bus
PGr=-100+50(IQt-2Aqi
Pcz= - 150 + 60 (IQz - 2.5 AqZ
Pct= - 8.0+ 4a Qq3 - 1.8 Aqi
where ICs are in Rs/IVIWh and P6s are in MW.
The total load at a certain hour of the day is 400 MW. Neglect
transmission loss and develop a computer programme for optimum
generationschedulingwithin and accuracyof + 0.05 MW.
Note: All P6s must be real positive.
www.muslimengineer.info
Flg. P-7.7
Samplesystemfor probtemp-7.7
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
uoo"rnpo*", syrt"r Anutyri,
W
7.8 Fot the power plant of the illustrative exampleusedin Section 7.3. obtain
the economically optimum unit commitment for the daily load cycle given
in Fig. P-7.8.
Correct the scheduleto meet security requirements.
i"
E,o
0
8
1
2
16
20
24
Timein hours---------'
Flg. P-7.8 Dailyload curve for problemp-7.9
7.9 Repeat Example 7.3 with a load of 220 Mw from 6 a.m. to 6 p.m. and
40 MW from 6 p.-. to 6 a.m.
7.10 Reformulate the optimat hydrothermal schedulingproblem considering the
inequality constraints on the thermal generation and water storage
employing penaity functions. Find out the necessary equations and
gradient vector to solve the problem.
NCES
REFERE
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t970.
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I9)6.
5 . Kirchmayer, L.K., Economic control of Interconnectedsystems,wiley, New york,
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Knight, u.G., Power systemsEngineeringand Mathematics,pergamonpress,New
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rext
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w
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in
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Academic Publishers,London, 1999.
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Don
o-o
r ql,Er;,
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Proc. IEEE, 1972, 199: 169.
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Allocation", Proc, IEEE, Nov 1967. 1g77.
27. Happ, H.H., "optimal power Dispatch-A Comprehensive
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in
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IEEE
,snprinl
Ptthlirntinn
1, 5
J
fv\rr rJvn
oao (
7w7.w
D\rrD
LrE YYr\, l\ew
IorK,
Iyl).
30. Kothari, D.p., "optimal HydrothermalSchedulingand Unit
commitment,. ph. D.
Thesis,B.I.T.S, Pilani, 1975.
31' Kothari, D.P. and I.J. Nagrath,"security ConstrainedEconomic
Thermal Generating Unit Commitment,, J.I.E. (India), Dec. 197g, 59: 156.
32' Nagrath,I.J. and D.P. Kothari, "optimal StochasticScheduling
of CascadedHydrothermal Systems", J.I.E. (India), June 1976, 56: 264.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernpowersvstemAnalvsis
ffi
22
JJ.
D^"lL^rvDwrrwtrt
I
J.
^r
vr
^l
cl.t
../\-r:-^l
r.,tPtlrrr.u
t1^-t-^l
\.VtlLfUl
^f
Ul
D --^ri--^
I\|jAL;|IYtr
n-----fUWEf
Fr ----rr
fIUW
,
mal
rnnh
IEDL
ar
lfQnS,
< |
IyO6,
P A S .8 7 : 4 0 .
34. Dommel, H.w. and w.F. Trinney, "optimal power Flow solution", IEEE Trans.,
October1968,PAS. 87: 1866.
35. Sasson,A.M. and H.M. Itlerrill, "SonneApplications of Optimization Techniques
to Power SystemProblems", Proc. IEEE, July 1974,62: 959.
36. wu, F. et al., "A Two-stage Approach to solving optimal power Flows", proc.
1979 PICA Conf. pp. 126-136.
37. Nanda, J., P.R. Bijwe and D.P. Kothari, "Application of ProgressiveOptimality
Algorithm to Optimal Hydrothermal Scheduling ConsideringDeterministic and
StochasticData", International Jountal of Electrical Power and Energy Systems,
January1986, 8: 61.
38. Kothari, D.P. et al., "Some Aspects of Optimal Maintenance Scheduling of
GeneratinglJnits", "/./.E (India), August 1985, 66: 41.
39' Kothari, D.P. and R.K. Gupta, "Optimal StochasticLoad FIow Studies", J.I.E.
(India), August 1978,p. 34.
40. "Description and Bibliography of Major Economy-security Functions-Part I, il,
and III", IEEE committeeReport,IEEE Trans. Jan 1981. pAS-r00, zlr-235.
41. Bijwe, P.R., D.P., Kothari, J. Nanda, and K.S. Lingamurthy, .,Optimal Voltage
Control using ConstantSensiiivity Matrix", Electric Power SystemResearch,Vol.
II, No. 3, Dec. 1986,pp. 195-203.
42. Nanda,J., D.P. Kothari and K.S. Lingamurthy, "Economic-emissionLoad Dispatch
through coal ProgrammingTechniques",IEEE Trans.on Energy conversion, vol.
3, No. 1, March 1988,pp. 26-32.
43. Nanda, J.D.P. Kothari and s.c. srivastava, "A New optimal power Dispatch
Algorithm using Fletcher's QP Method", Proc. IEE, pte, vol. 136. no. 3, May
1 9 8 9 ,p p . 1 5 3 - 1 6 1 .
44. Dhillon, J.S., S.C. Parti and D.P. Kothari, "stochasticEconomicEmissionLoad
Dispatch", Int. J. of Electric Power system Research,Yol. 26, No. 3, 1993, pp.
1 7 9- 1 8 3 .
45. Dhillon, J.S.,S.C. Parti and D.P. Kothari,"MultiobjectiveOptimalThermalPower
Dispatch",Int. J. of EPES, Vol. 16, No.6, Dec. L994,pp.383-389.
46. Kothari, D.P. and Aijaz Ahmad, "An Expert system Approach to the unit
commitment hoblem", Energy conversion and Management",vol. 36, No. 4,
April 1995, pp. 257-261.
47 - Sen, Subir, D.P. Kothari and F.A Talukdar, "EnvironmentallyFriendly Thermal
Power Dispatch- An Approach", Int. J. of Energy Sources,Vol. 19, no. 4, May
1 9 9 7 ,p p . 3 9 7 - 4 0 8 .
48. Kothari, D.P. and A. Ahmad, "Fuzzy Dynamic Programming Based optimal
GeneratorMaintenanceSchedulingIncorporatingLoad Forecasting",in Advances
in Intelligent systems,edited by F.c. Morabito, IoS press, ohmsha, 1997, pp.
Qan
vvrr.
Q
u.r
^ r- ri s
q
n
D
v.l
.
If ^+L^-j
nuluatl
l,
../\-.:-t
\_rpl,llllal
Thermai Generadng'tjnir Commitment-A
Review, "Int. J. EPES, Vol. 20, No. 7, Oct. 1998,pp. 443-451.
52. Kulkarni, P.S., A.G. Kothari and D.p. Kothari, "Combined Econonic and Emission
Dispatchusing ImprovedBpNN", /nf. J. of EMPS, Vol. 28, No. l, Jan 2000,pp.
3t - 4_7
53. Aryu, L.D., S.C. Chaudeand D.P. Kothari, "EconomicDespatchAccounting Line
Flow Constraintsusing Functional Link Network," Int. J. of Electrical Machine &
Power Systems,28, l, Jan 2000, pp. 55-6g.
54. Ahmad, A. and D.P. Kothari, "A practical Model for Generator Maintenance
Scheduling with rransmission constraints", Int. J. of EMps, vol. 2g, No. 6, June
2000, pp. 501 -514.
55. Dhillon, J.s. and D.P. Kothari, "The surrogate worth rrade off Approach for
Mutliobjective Thermal power Dispatch hobelm". EpsR, vol. 56, No. 2, Nov.
2000, pp. 103-110.
56. Son, S. and D.P. Kothari, "Large ScaleThermal GeneratingUnit Commitment:A
New Model", in The Next Genera\ion of Electric Power (Jnit CommitmentModels,
edited by B.F. Hobbs et. al. KAp,
,Boston,2001, pp. Zll-22557. Dhillon, J.s., s.c. Parti and D.p. \ Kothari, ,,Fuzzy Decision Making in Multi
objective Long;term scheduling of Hydrothermal system,,, rnt. t. oy erns, vol.
23, No. l, Jan 2001,pp. lg-29.
58. Brar, Y.s., J.s. Dhillon and D.p. Kothari, "Multi-objective Load Dispatch by
Fuzzy Logic based Searching Weightage Pattern," Electric power Systems
Research,Vol. 63, 2002, pp. 149-160.
59. Dhillion, J.s., S.c. Parti and D.p. Kothari, "Fuzzy Decision-mgkingin stochastic
Multiobjective short-term Hydrothermal scheduling," Ip,B proc.tcTD, vol. l4g, z,
March 2fi02, pp l9i-200.
60' Kothari, D.P., Application of Neural Netwdrks to Power Systems(Invited paper),
Proc. Int. Conf., ICIT 2000, Jan. 2000, pp. 62l_626.
/.J5 - /4U.
49.
Aijaz Ahmad and D.P. Kothari, "A Review of Recent Advances in Generator
Maintenance scheduling", Electric Machines and Power systems, yol 26, No. 4,
1998, pp. 373-387.
50. Sen S., and D.P. Kothari, "Evaluation of Benefit of Inter-Area Energy Exchange
of Indian Power System Based on Multi-Area Unit Commitment Approach", Int.
J. of EMPS, Vol.26, No. 8, Oct. 1998, pp. 801-813.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
R
1.,
8.T
caused
by momentary
chargein generaforspeecl,
tI'r.r.tnr*,-i;;?t;qffi;
;;
excitationvoltagecontrolsare non-interactive
for smallchangesand can be
modelledandanalysed
independently.
Furthermore,
excitationvoltageeontrolis
F:tcl :tcfinrr
ri n
r r rvrvr rhrirnvhr r
tLhr cr v -r r, rr ci ^r rJ v r f i r r r o
rrlttw
n,rn..r,rhr ^6^,rri-+^-^.1
vrJrrJr-(lrrr urlLUultLtrlcu
:- rL^e ^$rL-^-^--^r5 llla! ul ulc; ggirtcfalor
field; while the power frequencycontrol is slow actingwith major time constant
contributedby the turbine and generatormomentof inertia-this time constant
is much larger than that of the generatortield. Thus,the transientsin excitation
voltage control vanish much faster and do not affect the dynamics of power
frequencycontrol.
.INTRODUCTION
Powersystemoperationconsidcrcdso far was underconditionsof stcadyload.
However, both active and reactive power demandsare never steady and they
continually change with the rising or falling trend. Steam input to turbogenerators(or water input to hydro-generators)
must, therefore,be continuously
regulatedto match the active power demand,failing which the machinespeed
will vary with consequent change in frequency whieh may be highly
undesirable*(maximum permissiblechangein power fiequency is t 0.5 Hz).
Also the excitation of generatorsmust be continuouslyregulated to match the
reactive power demand with reuctive generation,otherwise the voltagesat
various system buses may go beyond the prescribed limits. In modern large
interconnected systems, manual regulation is not feasible and therefore
automaticgenerationand voltage regulation equipment is installed on each
generator. Figure 8.1 gives the schematic diagram of load frequency and
excitation voltage regulatorsof a turbo-generator.The controllers are set for a
particularoperatirrgcondition and they take care of small changesin load
denrandwithout fiequency and voltageexceedingthe prescribedlimits. With
the passage of time, as the change in lcad demand becomes large, the
contrcllersmust be reseteithernianuallyor automatically.
It has been shown in previous chaptersthat for small changesactive power
is dependenton internalmachineangle 6 and is inderrendentof bus voltage:
whiie bus voitage is dependenton machine excitation (therefore on reactive
- " Changein frequency
causeschangein speedof the consumers'
plant affecting
productionprocesses.
Further,it is necessary
to maintainnetworkfrequencyconstant
so that the powerstationsrun satisfactorily
in parallel,the variousmotorsoperating
on the systemrun at the desiredspeed,correcttime is obtainedfrom synchronous
clocksin the system,andthe entertaining
devicesfunctionproperly.
www.muslimengineer.info
I
P+JQ
Fig. 8.1 schematicdiagramof loadfrequencyand excitation
voltageregulatorsof a turbo-generator
Change in load demand can be identified as: (i) slow varying changes in
meandemand,and (ii) fast random variations aroundthe mean. The regulators
mustbe dusignedto be insensitiveto thst randomchanges,otherwisethe system
will be prone to hunting resulting in excessivewear and tear of rotatins
machinesand control equipment.
8.2
LOAD FREOUENCY CONTROL (STNGLE AREA CASE)
Let us considerthe problemof controlling the power output of the generators
of a closely knit electric areaso as to maintz,inthe scheduledfrequency. All the
generatorsin such an areaconstitutea coherent group so that all the generators
speeoiip ancisiow riowii togetiierrnarntarnrng
thelr reiarrvepower angies.Such
an area is defined as a control area. Tire boundariesof a coqtrol area will
generallycoincide with that of an individual Electricity Board Company.
To understandthe load fiequency control problem, let us consider a single
turbo-generatorsystem supplying an isolated load.
^ - ^ ^ l
- I
^ l - - - - .
- l ^ - - - - ^
L - -
- . r -
-
^ . _ _ _ :
- ,
. r
.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern power system Analys,s
W
Turbine
Speed Governing
System
Figure 8.2 shows schematicallythe speedgoverningsystem of a steamturbine.
The systemconsistsof the following components:
Steam
turbine. Its downward movement opens the upper pilot valve so that more steem
is admitted to the turbine under steady conditions (hence more steady power
. The reverse
Model of Speed Governing
System
Assume that the system is initially operating under steady conditions-the
linkage mechanism stationary and pilot valve closed, stearnvalve opened by a
definite magnitude, turbine running at constant speedwith turbin" po*"r output
balancing the generator load. Let the operating conditions be characteizedby
=
"f" systemfrequency (speed)
Speed changer
P'c = generator output = turbine output (neglecting generator loss)
Pilot
value
.IE = steam valve setting
We shall obtain a linear incremental model around these operating
conditions.
Let the point A on the linkage mechanism be moved downwards by a small
amount Aye.It is a commandwhich causesthe turbine power output to change
and can therefore be written as
--t-\
High
pressure oil
Main
piston
A
I
rHydraulic amplifier
(speed control mechanism)
Fig.8,2 Turbinespeedgoverningsystem
Reprinted
with permission
of McGraw-Hilt
BookCo., New York,from Olle l. Elgerd:
Electric Energy System Theory: An lntroduction, 1g71, p. 322.
(i) FIy ball speedgovernor: This is the heart of the system which sensesthe
changein speed(frequency).As the speedincreasesthe fly balls move outwards
and the point B on linkage mechanism moves downwards. The reversehappens
when the speeddecreases.
G) Hydraulic amplifier: It comprises a pilot valve and main piston
alrangement.Low power level pilot valve movement is converted into high
power level piston valve movement. This is necessaryin order to open or close
the steamvalve againsthigh pressure steam.
(xl) Lintcage mechanism: ABC is a rigid link pivoted at B and cDE is
anotherrigid link pivoted at D. This link mechanismprovides a movementto
the control valve in proportion to change in speed.It also provides a feedback
the steamvalve movement (link 4).
,,fr9rn
www.muslimengineer.info
Aye= kcAPc
(8.1)
where APc is the commanded increase in power.
\
The command signal AP, (i.e. Ayi sets into rnotion a bequenceof eventsthe pilot valve moves upwards,high pressureoil flows on to the top of the main
piston moving it downwards; the steam valve opening consequently increases,
the turbine generatorspeedincreases,i.e. the frequencygoes up. Let us model
these events mathematically.
Two factors contribute to the movement of C:
(i) Ayecontributer -
[?J
\rll
Aya or - krAyo(i.e. upwards) of - ktKcApc
(ii) Increase in frequency ff causes the fly balls to move outwards so that
B moves downwards by a proportional amount k'z Af. The consequent
movement of Cwith A remaining fixed at Ayo - . (+)
orO, - + kAf
(i.e. downwards)
The net movement of C is therefore
AYc=- ktkcAPc+kAf
(8.2)
The movement of D, Ayp, is the amount by which the pilot valve opens. It is
contributedby Ayg and AyB and can be written as
Ayo=(h)
Ayc+(;h) *,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
controt
= ktayc + koAys
(g.3)
The movement ay.o-d,epending
upon its sign opensone of the ports of the pilot
valve admitting high pressure'o' into
thJ
moving the main
piston and opening the steam valve
"ynnJ.ithereby
by ayr.
certain justifiable simprifying
assumptions,which ean be rnade at this
.tugl, ur",
(i) Inertial reaction forces of main
pistoi and steam valve are negligible
compared to the forces exerteclon the
iirton by high pressureoil.
(ii) Because of (i) above, the
rate of oil admitted to the cylinder
is
proportional to port opening Ayo.
The volume of oil admitted to the cylinder
is thus proportional to the time
integral o,f ayo. The movementay"i.s
obtainedby dividing the oil volume by
the area of the cross-section
of the-piston.Thus
Avn= krfoeayrlat
(8.4)
It can be verified from the schematicdiagram
that a positive movement ayo,
causesnegative (upward)movement ayulccounting
for the n"gutiu" ,ign used
in Eq. (8.4).
Taking the Laplacetransformof Eqs. (g.2),(g.3)
AYr(s)=- k&cApc(") + krAF(s)
(8.5)
(8.6)
a y u ( g = - k s ol r U n
(8.7)
EliminatingAyr(s) andAyo(s), we can write
k'ktk'AP' (s)- k,krAF(s)
')
(oo '' t
"'tr ,/
\
-lor,<,r-*^or",].i#)
(8.8)
/o
o\
:-
.
r .
Ks9
1+ fsss
4F(s)
The speed governing system of a hydro-turbine is more
involved. An
additional feedback loop provides temporary droop compensation
to prevent
instability. This is necessitatedby the targe inertia or the penstoct
gut" which
regulates the rate of water input to the turbine. Modelling of
a hyjro-turbine
regulating system is beyond the scopeof this book.
Model
Let us now relate the dynamic responseof a steam turbine in tenns
of changes
in power ouFut to changesin steamvalve opening ^4yr. Figure g.4a
shows a
two stage steam turbine with a reheat unit. The dynamic
*ponr" is targely
influenced by two factors, (i) entrained steambetwein the inlet
stbamvalve and
first stageof the turbine, (ii) the storageaction in the reheaterwhich
causesthe
output of the low pressurestageto lag behind that of the high pressure
stage.
'fttus,
the turbine transfer function is characterizedby two time
constants.For
easeof analysisit will be assumedhere that the turbinl can be modelled
to have
Ssingle equivalent time constant.Figure 8.4b shows the transfer function model
of a sreamturbine. Typicaly the time constant lies'in the
range o.i ro z.s
{
sec.
-=-&
Steam valve
n= klc = speedregulationof the governor
t_
K2
+y
.r.
= l"
, rs
;-;
KqkS
- gainof speed
governor
= tlme constant of speedgovernor
r--
(a) Two-stage
steamturbine
AYg(s)-FAPds)
(b) Turbinetransferfunctionmodel
Flg. 8.4
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4Y5(s)
Flg. 8.3 ,Blockdiagramrepresentation
of speed governorsystem
where
K.,=
t
-
riyLr.Lru' \o.o., rs rcpfesenleo ln tne ronn of a block diagram
in Fig. 9.3.
Turbine
and (g.4), we ger
Ayp(s)= kzAyd,s)+ koAyug)
AYu(s)-
E
1
E^,,^ri^-
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
po*", s),rt"r An"ly.i,
rrrroarrn
#ph-Si
Automatlc Generationand Voltage Control
I
Generator Load Model
=tAP6g)_
aPo(,)r.[#j
The increment
in powerinputto the generatbr-load
systemis
APG _ APD
whele AP6 = AP,, incremental turbine
incremental loss to be negligible) and App is the load increment.
This increment in power input to the syrtem is accountedfor in two ways:
(i) Rate of increase of stored kinetic energy in the generator
rotor. At
scheduledfrequency (fo ), the stored energy is
Wk, = H x p, kW = sec (kilojoules)
where P, is the kW rating of the turbo-generator andH is defined as its inertia
constant.
The kinetic energy being proportional to square of speed(frequency),
the
kinetic energy at a frequency of (f " + Arf ) is given by
=nr,(r.T)
(s.13)
2H
= pow€r systemtime constant
Bf"
Kp,=
+
=powersystem
gain
Equation (8.13) can be representedin block diagram form as in Fig. g.5.
laeo(s)
^Po(s)
16---ffioro,
Flg.8.5 Blockdiagramrepresentation
of generator-load
model
(8.e)
Rate of change of kinetic energy is therefore
=fffrr"n
$rr*"r
I
complete Block Diagrram Representation of Load Frequenry
Control of an Isolated Power System
(8.10)
(ii) As the frequency changes,the motor load changesbeing
sensitive to
speed,the rate of changeof load with respectto frequ"n.y, i.e. arot\ycan
be
regarded as nearly constant for small changes in frequency Af ard can
be
expressedas
@PDl?flAf=BAf
(8.11)
wherethe constantB can be determinedempirically, B is positivofor a
predominantly
motorload.
Writing the powerbalanceequation,we have
AP(s)=trPn15;
AP6(s)
f . P] *' d (< o f l +B A f
APc- aP
, r^== T- H
Dividingthroughoutbyp, andrearanging,we get
/ A ' ^ +' B(ptt)
n ' 7 ' - -af
-\
AP;q;u)= 1d
(Afi
dt
f
Taking the l,aplace transforrn, we can write AF(s) as
AP6$u)-
Flg. 8.6 Blockdiagrammodelof loadfrequencycontrol
(isolatedpowersystem)
(8.i2)
Steady States Analysis
B*-'- s
The model of Fig. 8.6 shows that there are two important incrementalinputs to
the load frequency control system - APc, the change in speedchangersetting;
and APo, the change in load demand. Let us consider,,.4,.simplesituatiqn in
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
4Fis; -
AP,G) -4PoG)
Modern
which
the sneerl .hqnrro'
't::;
;-::--::^
hoo .. g.-.^) ^-.-.
'(rr
cr rr^tr(r ucttrng
.rL^
a2
rreego,,* ;, 2
\7'e'
af
= o) and the load
; ;;;*;:r;;ffi; *' #ff:Tiil:
:,:il?:
;3l i:inT:
:: frequen-cy
steady
change
system
fora sudd.n.hung",
ffi;ffi"ffi;ti'l;
r.han.,oo
anaount
*,
(, e.Apog):+)is
c
demand
obtainedas follows:
aF@)l*,(s):o
: -
AP^
^f
I
^L^--^
r'E .1uuy'
cquauon glves tne steadystate changesin frequency
causedby
changes in load demand. Speed regulation R
is-naturally so adjusted that
changesin frequency are small (of the order
of 5vo from no load to ruu load).
Therefore,the linear incrementalrelation (g.16)ican
be appliedfrom no load to
full load' with this understanding,Fig. 8.7
shows the linear relationship
betweenfrequency and load for free governor operation
with speedchanger set
to give a scheduledfrequency of r00% at full toao.
The .droop, or slopeof this
'l
(
I
relationship
- is -l
\ B+(t/R) )
Power system parameter B is generaily much smalrer*
than r/R (a typical
value is B = 0.01 pu Mwalz and l/R = U3)
so that B canbe neglected in
comparison.Equation (8.16) then simplifies to
(8.17)
rhedroop
fjfli;], curve
isthusmainly determinedby R, the
"r,,fl",
speedgovernor
regulation.
K
r^sorr,
ap,=_
*"r: (r^;)o",
I ( = 1. I
It is also rccognized that Ko, = 7 / B , w h e r e B - Y ^
in frequency). Now
ai
/P' (in PuMWunit change
4=-(#6)o,.
(8.16)
Decrease
in system
load= BAf= (uffi)*,
Of course,the contribution of decreasein system
load is much less than the
increase in generation. For typical values of B and
R quoted earlier
APo = 0.971 APo
Decreasein system load = 0.029 ApD
consider now the steady effect of changing
speed changer setting
(Or"<rl-
fi roa
+)with
load demandremainingfixed (i.e. Apo= 0). The
sready
state changein frequency is obtained as follows.
(J
L
8.rog
.c
102
101
100
0
*For
at
li\
dA^t
|
|
\r,, ruu-loLoao
(ii) 60% Load
Percent Load
250 MW machine with an operating load of 125
MW. let the change in load
be i%o for IVo change in frequency (scheduledfrequency =
50 Hz). Then
a-:?:r?: :2.5
NNVtHz
af
0.s
Flg. 8.7 Steady
qharacteristicof a speed
governor
"*-l?39-frequency
system
'=(#)b
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:
#:
o'ol PuMwgz
uodernPowersystemAnalysis
Autor"tic G"n"r"tionand Volt"g" Conttol
W
I
r t v (t ^ p s
ttsgttf
AF@lap,{s):o:
( 1 +T , r s ) ( l * 4 sxl) * zors)+ KseKtK p,/R
I
\
KreKrKp,
- lI
AP,
4,flr*uoyro,":_ t
r l
( 1 + K .sK tK ps/ R
I AP',:g
xu'c
s
(8.18)
(8.1e)
If
KrrK ,= l
(
Ar=
"
|
\rc"
(8.20)
\ B+llR)
If the speed changer setting is changedby AP, while the load demand
changesby APo, the steadyfrequencychangeis obtainedby superposition,i.e.
Ar = (
- APo)
ru) 'o"
F
AD
(8.21)
". frequency changecausedby load demandcan be
According to Eq. (8.2I) the
compensated
by changingthe settingof the speedchanger,i.e.
APc- APo, for Af = Q
Figure 8,7 depicts two load frequency plots-one to give scheduled
frequencyat I00Vorated load and the other to give the samefrequencyat 6O7o
rated load.
Two generators rated 200 MW and 400 MW are operating in parallel. The
droop characteristicsof their governors are4Vo and 5Vo,respectivelyfrom no
load to full load. Assuming that the generatorsare operating at 50 Hz at no
load, how would a load of 600 MW be sharedbetweenthem?What will be the
system frequency at this load? Assume free governor operation.
Repeat the problem if both governorshave a droop of 4Vo.
Solution Since the generators are in parallel, they will operate at the same
frequency at steadyload.
Let load on generator 1 (200 MW) = x MW
and load on generator 2 (400 MW) = (600 - x) MW
Reduction in frequency = Af
Now
a f _ 0.04x 50
(i)
200
x
af
0.05x 50
400
600-x
(ii),
we get
(i)
and
in
EquatingAf
231 MW (loadon generatorr)
(ii)
v -
-/A
trltf
6 0 0 - x = JOy lvlw
/1 ^-l
(IUau
^-
ull
Btrrltrriltur
L)
0'0-1150
x 231 = 47.69 Hz
200
It is observed here that due to difference in droop characteristics of
governors,generatorI gets overloadedwhile generator2 is underloaded.
It easily follows from above that if both governorshave a droop of.4Vo,they
will share the load as 200 MW and 400 MW respectively,i.e. they are loaded
corresponding to their ratings. This indeed is desirable from operational
considerations.
Systemfrequency
' = 50
A 100 MVA synchronousgeneratoroperateson full load at at frequencyof 50
Hz. The load is suddenly reducedto 50 MW. Due to time lag in governor
system,the steamvalve beginsto closeafter 0.4 seconds.Determinethe change
in frequencythat occurs in this time.
Given H = 5 kW-sec/kVA of generatorcapacity.
Solution Kinetic energy stored in rotating parts of generatorand turbine
= 5 x 100 x 1.000= 5 x 105 kW-sec
Excesspower input to generatorbefore the steam valve
beginsto close= 50 MW
Excess energy input to rotating parts in 0.4 sec
= 50 x 1,000 x 0.4 = 20,000 kW-sec
Stored kinetic energy oo (frequency)2
Frequency at the end of 0.4 sec
+ zo,ooo
= 5ox I soo,ooo
)t"= 5r rfz
\
500,000
Dynamic
Response
To obtain the dynamic responsegiving the change in frequency as function of
the time for a step changein load, we must obtain the Laplace inverse of Eq.
(8.14). The characteristicequationbeing of third order, dynamic responsecan
r'
Onfy
r
Dg
| 1-!-- - I
ODIalneU
f-,luf
A
-^^^tC: ^
SPtrUfffU
---*^-:^^1
ll|'llll('llua1'I
^^^^
Ua1DE.
tI^.-,^,,^II(rwsYsIr
)
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+L^
LfIs
^L^-^^+^--i^+in
r,Il<ll4ivLsllDrlv
equation can be approximated as first order by examining the relative
magnitudesof the time constantsinvolved. Typical valuesof the time constants
of load frequency control system are rdlated as
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Trr4T, <To,
Typically* t,
Time (sec)------->
= 0.4 sec, Tt = 0.5 sec and
t
-1
I
I
o
Flg' 8.8 Firstorderapproximate
brockdiagramof road
frequencycontrotof an isolatedarea
Irning Tro = T, = 0:
=1), the block diagram of Fig.
K*\
8.6 is
Iuld
reduced to thlt of F'ig. 8.8,
from which we can write
AF(s)l*r(s):o = -
to,
.-.APo
(1+ KpslR)+ Zp.s" s
, l, + ^ + r o ' 1
R4,J
-ft{' - *,[-,,a[n#)]]*, g 22)
Ar(,)=
TakingR = 3, Kp, = llB = 100,
e, = 20, Apo = 0.01 pu
Af (t) = - 0.029(I - ,-t:tt',
Aflrt"udystare= - 0.029 Hz
Dynamicresponse_of
changein frequencyfor a stepchangein load
(APo= 0.01pu, 4s = 0.4 sec, | = 0.5 sLc, Io. = 2b sec, (" = 100,
R= 3)
The plot of change in frequency versus time for first order approximadon
^rirst
given above and the exact response are shown in Fig. a.g.
order
approximation is obviously a poor approximation.
Gontrol Area Concept
- - "o{1:- =xaP,
L
(8.23a)
(8.23b)
So far we have considered the simplified case of a single turbo-generator
supplying an isolated load. Consider now a practical system with number
of
e
generating stations aird loads. It is possible to divide an extendedpower
system
(say, national grid) into subareas(may be, State Electricity Boards)
in which
the generatorsare tightly coupled together so as to form a coherent group,
i.e.
all the generators respond in unison to changes in load o, ,p"rJ changer
settings.Such a coherentareais called a control area in which the frequency
is assumedto be the same throughout in static as well as dynamic conditions.
For purposes of developing a suitable control strategy, a control area can
be
reduced to a single speed governor, turbo-generator and load system. All
the
control strategies discussedso far are, therefore, applibable to an independent
control area.
Proportional
"For a 250 MW machinequoted
earlier,inertiaconstanrIl = SkW-seclkVA
,' = 4 :o. 2 * 5
==2osec
Bf
0.01x
50
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Firstorderapproximatiorl
Plus fntegral
Control
It is seen from the above discussionthat with the speed governing sysrem
installed on each machine, the steady load frequency charartitirti" fi agiven
speedchanger setting has considerabledroop, e.g. for the system being used
for
the illustration above, the steadystate-droop in fieo=ueneywill be 2.9 Hz
[see
Eq. (8.23b)l from no load to tull load (l pu load). System frequency
specifications are rather stringent and, therefore, so much change in frequency
cannot be tolerated. In fact, it is expected that the steady change in frequency
will be zero. While steadystatefrequency can be brought back io the scheduled
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
PowersystemAnalys
Modern
ffil
AutomaticGenerationand VoltageControl
I
vaiue by adjus'ringspeedchangersetting,the systemcould under go intolerable
dynamic frequency changes with changes in load. It leads to the natural
suggestion that the speed changer setting be adjusted automatically by
monitoring the frequency changes.For this purpose, a signal from Af is fed
diagram
througfan integrator to the s
configuration shown in Fig. 8.10. The system now modifies to a proportional
plus integral controller, which, as is well known from control theory, gives zero
steady state error, i.e. Af lrt""d",,ut,= 0.
APe(s)
Integral
controller
APp(s)
l+t-r8I
I
t - +
t
AF(s)
l
AP6(s)
tin ihe above scheme ACE being zero uncier steaciyconditions*, 4 logical
design criterion is the minimization of II,CZ dr for a step disturbance. This
integral is indeed the time error of a synchronous electric clock run from the
power supply. Infact, modern powersystems keep Eaekofintegra+e4tinae errsr
all the time. A corrective action (manual adjustment apc, the speed changer
setting) is taken by a large (preassigned)station in the area as soon as the time
error exceeds a prescribed value.
The dynamics of the proportional plus integral controller can be studied
numerically only, the systembeing of fourth order-the order of the system has
increasedby one with the addition of the integral loop. The dynamic response
of the proportional plus integral controller with Ki = 0.09 for a step load
disturbance of 0.01 pu obtained through digital computer are plotted in Fig.
8.11. For the sake of comparisonthe dynamic responsewithout integral control
action is also plotted on the samefigure.
sensor
Frequency
+I -1
plus integralloadfrequencycontrol
Fig. 8.10 Proportional
The signal APr(s) generatedby the integral control must be of oppositesign
to /F(s) which accounts for negative sign in the block for integral controller.
Now
AF(s1=
I
t
o
r
x
Kn,
Ko,
(r+%"s).
(** +). ( l * f , r s ) ( l + 4 s )
RKo,s(l+{rs)(l+ 4s)
+ {'s)(1 + 4sXl f zo's)R* Ko'(KiR f s)
"+
(8.24)
Flg. 8.11 Dynamicresponseof loadfrequencycontroller
with and without
integralcontrolaction(APo = 0.01pu, 4s = 0.4 sec, Ir = 0.5
sec, Ips= 20 sec, Kp.= 100,B - B, Ki= 0.-09)
obviousry
(8.25)
so/F(s) : o
,
In contrast to Eq. (8.16) we find that the steady state changein frequency
has been reduced to zero by the additio4 of the integral controller. This can be
argued out physically as well. Af reaches steady state (a constant value) only
=
Af l"t"^dy
state
rr.,lrsrr
uAp^
rcwlMl
-
Ap-
HrD
=
-
.ons-fant
vvuulqr!.
Becarrs-e of
fhe
intes!'atins
actiOn
Of the
controller, this is only possible if Af = 0.
In central load frequency control of a given control area, the change (error)
in frequencyis known as Area Contol Error (ACE). The additional signal fed
back in the modified control schemepresented above is the integral of ACE.
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8.3
IOAD FREOUENCY CONTROL AND ECONOMIC
DESPATCH CONTROL
Load
freouencv
__J
I
control
________
with
,.___
inteorel
_---_O
eonfrnller
qnhierrAe
'vu
?a?^
lvrv
craolrr
otvsuJ
ora+o
Dl4lg
frequencyelTor and a fast dynamic response,but it exercisesno control over the
relative loadings of various generating stations (i.e. economic despatch) of the
control area. For example, if a sudden small increasein load (say, 17o) occurs
'Such
a control is known as isochronous control, but it has its time (integral of
frequency) error though steady frequency error is zero.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
f
Automatic
area, the road
conrior ,changesthe speed changer
1i..l1r_::ltrol
-frequency
Dcrurgs
or tne governors of all generating units of the area so that,
together,
theseunits match the load and the frequenry returns tp the scheduled
value (this
action takesplace in a few seconds).However, in the,process
of this change the
Ioadings of u@units
change in a manner independent of
economi@
In fact, some units in the pro""r, may even
get overloaded.Some control over loading of individual
units cafi be Lxercised
by adjustingthe gain factors (K,) includeJin the signal
representingintegral of
the area cogtrol error as fed to individual unitr.
However, this is not
satisfactory.
"fnceot
_T--command signai generated'oythe centrai economic despatch
computer.Figure
8'12 gives the schematicdiagram of both thesecontrolsior two
typi.ut units of
a control area.The signal to changethe speedchan3ersetting
is lonstructed in
accordancewith economic despatcherror,
[po (desired)- pJactual)]. suitabry
modified by the signal representingintegral ncg at that instant
of time. The
signal P6 (desired) is computed by the central economic despatch
computer
(CEDC) and is transmitted to the local econornic despatch
controller (EDC)
installed at each station. The system thus operateswith economic
desfatch error
only for very short periods of time beforJ it is readjusted.
8.4
TWO-AREA
LOAD FREOUENCY
CONTROL
An extendedpower system can be divided into a number
of load frequency
control areasinterconnectedby meansof tie lines. Without loss
of generality we
shall consider a two-area case connectedby a single tie line
aslilusnated in
Fi g. 8.13 .
Speed
Fig.B.i3 Two interconnected
controrareas(singre
tie rine)
lr.
EDC - Economic despatch controller
CEDC - Central economic despatch computer
Flg. 8-12 Control area load frequency and economic despatch
control
Reprinted (with modification) with permission of McGraw-Hill
Book Company,
New York from Olle I. Elgerd: Electric Energy SystemsTheory:
An Introd.uction,
I971,p. 345.
The control objective now is to regulatethe frequency of each
area and to
srnnultaneously
regulatethe tie line power as per inter-areapower contracts.
As
in the caseof frequency, proportional plus integral controller
will be installed
so as to give zero steady state error in tie line power flow as
compared to the
contractedpower,
It is convenientlyhssumedthat each control area canbe represented
by an
equivalentturbine, generatorand governorsystem.Symbols used
with suffix I
refer to area 7 and those with suffi x 2 refer to area 2.
In an isolated control area case the incremental power (apc _ apo)
was
accountedfor by the rate of increaseof stored kinetic energy and increase
in
areaload causedby increasein fregueircy.since a tie line t *rport, power
in
or out of an area, this fact must be accounted for in the incremental
power
balanceequation of each area.
Power transported out of area 1 is .eivenbv
Ptie,r =
''rrl''l
X,,
sin ({ - q
(8.26)
where
q'q
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- poweranglesof equivalentmachinesof thetwo
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
areas.
I
Automatic
Generation
and Vortagecontror
Modefn Power SystemAnalysis
308 |
I
expressed
as
AP,i,,r(pu)= Tp(Afi - 462)
where
T, =
'Y:t'-Yf
PrrXrz
(8.27)
cos(f - E) - synchronizingcoefficient
Since incrementalpower anglesare integrals of incrementalfrequencies,we
can write Eq. (8.27) as
AP,i,,r = 2*.(l
Afrdt-
aPt;", z = 2ilzr([ yrat -
[
(s.30)
\Prr)
nrz|r* AP,,",t
(8.31)
It rnay be noted that all quantitiesother than fiequencyare in per unit in
Eq.(8.3l).
Taking the l-aplacetransf'ormof Eq. (8.31) and reorganizing,we get
A P ti " ,,1r;]
$.32)
" t$I + 4,,t,!
where as definedearlier [seeEq. (8.13)]
Kp31= I/81
(8.33)
Tpil = LHr/BJ"
Comparedto Eq. (8.13) of the isolated control areacase,the only changeis
ol the signal APri"J(s) as shown in Fig. 8.14.
the appearance
fL*lo4nl l oDfrnurrm
rrr
^ufr
E^
LY.
/a ta\
\v.L9),
tl h
l rav
c iro6 nr rosl^
o
,4P
/"\
tie.I\.r/
",
ic
nlrfoinerl
AS
- /4 (s)l
=
AP,i.,1(s)
ffroor(s)
www.muslimengineer.info
-iE=
--n7ri"l
For the control area 2, Ap6", r(s) is given by tEq. (g.Zg)l
With referenceto Eq. (8.12), the incrementalpower balance equationfor
area 1 can be written as
fho T -^l-^a
rrrw lsl/l4vv
AF1(s)
Fig. 8. 15
= tYr:J cos({L - E): [S]ti z: ar2rrz
rzr
LL
"
AF (s ) = IA P6 1 G)- A P r,(s) -
APti",r(s)
(8.2e)
where
APo,- APor= + *w)+
Jr" or
The corresponding block diagram is shown in Fig. g.15.
+
ayrat)
Przxzr
Fig. 8.14
(8.28)
I Urat)
where Afi nd Af,, arc incremental frequency changes of areas 1 and 2,
respectively.
Similarly the incrementaltie line power out of area2 is given by
'-l'^Li-I4ArrrS
Fil
I APti".r(s)
For incrementalchangesin { and 6r, the incre.mentaltie line power can be
(8.34)
apt i", z( s)=
- : gr r r ,
[ AFr ( s)- 4F, ( s) ]
( g: 35)
which is also indicated Uy ,i. block diagramof Fig. 8.15.
\
Let us now turn our attentionto ACE (areacontrol error; in the presence
of
a tie line. In the case of an isolated control area, ACE is the change in area
frequency which when used in integral control loop forced the steady state
frequencyelror to zero. In order that the steadystatetie line power error
in a
two-areacontrol be made zeroanotherintegralcontrol loop (one for each area)
must be introducedto integratethe incrementaltie line power signal and
feed
it backto the speedchanger.This is aeeomplished
by a singleintegratingbloek
by redefining ACE as a linearcombinationof incrementalfrequenryand tie line
power. Thus, fbr control area I
ACEI = APu".r+ brAf,
( 8. 36)
where the constant b, is called areafrequency bias.
Equation (8.36) can be expressedin the Laplace transform as
ACEl(s) = APo., r(s) + b1AF1g)
(8.37)
Similarly, for the control are a 2, ACE2 is expressedas
ACEr(s) = APti".z(s)+ b2AF,(s)
( 8. 38)
Combining the basic block diagramsof the two control areascorresponding
to Fig. 8.6, with AP5rg) and Apr2(s) generatedby integrals of respective
ACEs (obtainedthrough signalsrepresentingchangesin tie line power and local
frequencybias) and employing the block diagramsof Figs. g.t+ to g.15, we
easily obtain the composite block diagram of Fig. g.16.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPower svstem Analvsis
WIU&|
Let the step changesin loads APo, and APrrbe simultaneouslyappliedin
control areas 1 and 2, respectively.When steady conditions are reached,the
output signals of all integratingblocks will become constantand in order for
this to be so, their input signalsmust become zero. We have, therefore, from
F i e .8 . 1 6
KtL)
APu",, + b rAfr=
block- O finputof integrating
\
,r)
(8.39a)
trJ
o
K'z)
APti",, + brAfr=
- o finpot of integratingblock\
r l
-'4'\
Afr - Afz- =o finpurot
block
\ integrating
s
(8.3eb)
ra
S
(8.40)
oy,
EF
N
o
8 6
a(\
I.=constant
ar2
(8.41)
A P r i " , r =A P , : " , 2 = 0
:pE
.g
q
o 6
c)
=a
- ( d
<.1
ol. a
v'it
lr
(8.42)
E
o
E 9
*li
HenceEqs.(8.39)- (8.41)aresimultaneously
satisfiedonly for
and
g
A
o-
FromEqs.(8.28)and(8.29)
APn",,
=-Tr, -.
AP.i",z, Tzt;
o
(U
-o
!t
o
oi
I
il
)
.Y
g t u
Q c l
e.>
(g()
o
o
o
o
F A
G'
o o
Afi=Afz=0
6
O E
Thus, under steady condition change in the tie line power and frequency of
each area is zero. This has been achieved by integration of ACEs in the
feedbackloops of each area.
Dynamic responseis difficult to obtain by the transfer function approach (as
used in the single area case)becauseof the complexity of blocks-and multiinput (APop APor) and multi-output(APri",1, Ap6",2, Afr Afr) situation.A
more organizedand more convenientlycarried out analysisis through the state
spaceapproach(a tirne domainapproach).Formulation of the statespacemodel
for the two-area system will be illustrated in Sec. 8.5.
The results of the two-areasystem(APri", changein tie line power and,Af,
changein frequency) obtainedthrough digital computerstudy are shown in the
form of a dotted line in Figs. 8.18 and 8.19. The two areasare assumedto be
identical with system parametersgiven by
Trs= 0.4 sec, 7r = 0.5 sec, ?r, = 20 sec
*5 o*
o
o
o
b 6
tr(')
E
E g
E
H'.s
tt)
(D
()
(U
* 3
5 u
= o
a
5
' 8' 9
.9
d
CL
.n
o
o
o
r\
ai
cit
lr
oo.
*,n
EOo o
o o
@
d
<;
t
l!
l+
5l
K o r =1 0 0 ,R = 3 , b = 0 . 4 2 5 , & = 0 . 0 9 , 2 f l r 2 = 0 . 0 5
8.5
OPTTMAL (TWO-AREA) LOAD FREOUENCY
u I f|:-
CONTROL
Modern control theory is applied in this section to design an optimal load
frequencycontroller for a two-a3easystem.In accordancewith modern control
terminology APcr arrdAP62 will be referred to as control inputs q and u2.ln
the conventional approachul anduzwere provided by the integral of ACEs. In
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
-ld
AutomaticGenerationand Voltagecontrot
M&
f*
SystemAnalysis
Power
rrrrodern
-
itZ I
moderncontrol theory approach ur and u2 wtll be createdby a linear
For formt'latingthe
of all the systemstates(full statefeedback).
combination
loops are
feedback
conventional
purpose
the
this
statevariablernodel for
ComparingFigs. 8.16 and 8.17,
block as shown in
resentedbv a se
baving either
blocks
of
all
outputs
as
the
defined
are
variables
State
Fig. 8.17.
We immediately notice that the systernhas nine
an integratoror artirne,constanf..
state variables.
xt = Aft
xq = Af.
XS= JACEit
.r2- AP,;1
x5 = AP52
t, = JACE, dt
tt1= APg,
u2 = /)Pa
w1= AP"
w2 = APp,
For block 1
-1-+-r+--i,
-f-.'f--
+I
L
\
I
I
x1 + T. r r i, = K ^ t ( x z -
-
l
o
o
X
( t'-2
-3
1
I
1
-
tt-
I
o
x
Kprt -,
Kprt
-;-wt
x, -z - ; - x t
t psl
x.2+ Tiliz=
'
With integralcontrolaction
or
t ptl
(8.43)
t ptl
,I
/
/'
' - I- - '
8
14
20
18
16
Time(sec)-----
with integral control action
t r + { , s r i ': = R- L r, r + rr ,
or
* t=-
r,-
^h
r
(8.45)
t* ,* * ,,
For block 4
IL
X +n *
Optimalcase (full state feedback)
( 8. 445
For block 3
-21.--+-_'--';-;;7-1=a.-1-1=--1
12
xt
* z =- + - r * * n
I
N
Kprt
,
*f
4 l
- w)
For block 2
r;+-.1
L
h l
'psl
F i g . 8 .1' 8 change in tie linepowerdue to step load(0.01pu) changein area1
A
.
Optimal case (full state feedback)
h
P
or
iq=
Krrz(xs + ar2x7 -
Torz*+=
Knrz
I
-
at?K or2
'\A'1--.{<
Tprz
wz)
Tps2
''
- T -- - y ' - a - _ - W ^
Tps2
Ko*2
'
Tpsz
z
( 8. 46)
For block 5
Fig. 8.19 Change in frequencyof area 1 due to step load (0.01 pu)
change in a.rea1
Before presenting the optimal design, we must formulate the state model.
This is achieved below by writing the differential equations ciescribing each
individual block of Fig. 8.17 in terms of state variables (note that differential
equations are written by replacing s UV
' *1.
dt'
x s t 7,2i5 - x6
or
is=
l
Ttz
r<
1
I - V
t
4
T,z
l
u
( 8. 47)
For block 6
.
l
x s * I ,szx6- -;
x4 + u2
I\2
or
io=-#*o-**u
'2t
www.muslimengineer.info
Y
4
sg2
( 8. 48)
t sg2
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPower SystemAnatysis
'3i4',"1
T
'-- co","".t""
For block 7
it=2iTtzxt-2iTr2xa
(8.4e)
is= brx, + x.i
(8.5O-)
"
constructed
as under from the state variables x, and -rnonly.
For block 8
ut=-
uz=- Ki{s=-
For block 9
( 8 . s1 )
i9= b2xa- anxt
The nine equations(8.43) to (8.51) can be organizedin the following vector
matrix form
(8.s2)
*=Ax+Bu+Fw
where
x _ l x r x2 ... xg)r = state vector
u = f u t u2fT= control vector
w = l w t w2fT = clisturbancevector
while the matrices A, B and F are defined below:
I
2
7
7
8
9
9
4
5
0
0
0
o
o
0
0 0
o
o
0
0 0
0
- 1
1
Tt
Ttr
1
Rr4er
o
- 1
Trst
-bLoo
Tprt
0
0
-
i=Ax+Bu
which does not contain the disturbance term Fw present in Eq. (g.52).
Furthermore,a constantdisturbancevector p would drive some of the system
statesand the control vector z to constantsteadyvalues;while the cost function
employedin optimal control requiresthat the systemstateand control vectors
have zero steady state values for the cost function to have a minimum.
For a constant disturbancevector w, the steady state is reachedwhen
* = 0
in Eq. (8.52);whichthengives
0 = A . r r "+ B u r r + F w
(8.s3)
Definingx and z as the sumof transientand steadystateterms,we can write
= ,
x
0
o
o
0
o
o
2 irrz
0
0
0
0
bL
0
1
i
-
0
0
'
o
--1-
0
z
Tprz
Ttz
1
0
RzTrsz
-2ilr2
0
0
0
b
2
0
[I o o TI
1
7,,
I
0
0
0
l o oo o o +
I
aco)
'O-
I
L
-
Kprt
0
Tprt
0 0
0
0 0
0
1
0 0
0 0
0 0
-atz
I
0 0l
x' * Ir"
(8.54)
n = ut * z',
(8.55)
Substitutingr and z from Eqs. (8.54) and (8.55) in Eq. (8.52), we have
0
TreZ
00.l
o o
0
Tprz
o
-ss1
Br = |
,;T
K
p
Tprz
Kiz la.Cerar
ln the optimal control schemethe control inputs u, anduz aregeneratedby
means of feedbacks from all the nine states with feedback constants to be
determinedin accordancewith an optimality criterion.
Examination of Eq. (8.52) reveralsthat our model is not in the standardform
employed in optimal control theory. The standardform is
atzKprz
0
6
8
3
Y
'tPsl
I
Tpst, Tprt
A _
Kirxs=- Kir IeCn,Ar
i' = A (r/ + x"r) + B(at + usr)+ Fw
By virtue of relationship(8.53),we get
*' = Axt + But
( g. 56)
This represents system model in terms of excursion of state and conhol
vectors fiom their respectivesteadystate values.
For full state feedback, the control vector z is constructed by a linear
combination of all states.i.e.
u=-
Kx
(8.57a)
where K is the feedback matrix.
Now
I
J
ttt+ Itrr=- l( (r/+ rr")
For a stable system both r/ and ut go to zero, therefore
ur, = _ Kx*
Hence
tt
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/= -
Ikl
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(8.s7b)
Modern Power SystemAnalysis
AutomaticGeneration
Examinationof Fig. 8.17 easily reveals the steady state values of state and
control variablesfor constantvaluesof disturbanceinputs w, andwr. These are
Ilrr=X4"r=
/7r" = 0
0 0 0
0 0 0
0
0
0
0 0
0 0
0 0
4
0
0
0 0
0 0
0 0
-arzbz 0 0
ulr, = wl
r5rr=
b ? o0
x6rr= lv2
(8.s8)
0 0 0
0 0 0
0
0
0 0
0 0
0
0
0 0
0 0
4 0 0 - a n b z 0 0 Q+a?)o o
0 0 0
0
0 0
0
1 0
0 0 0
0
0 0
0
0 1
uzr, = wz
Igr, = COnstant
I9r, = Constant
The values of xr* and xe* depend upon the feedback constants and can be
determined from the following steady state equations:
= symmetric matrix
R - kI = symmetric matrix
utrr= kttxtr, + ... + ftt8r8", * kt*sr, = wl
r,t2ss= k2txlr, + ... + kzgxgr.,
* kz*gr, = wz
(8.se)
The feedbackrnatrix K in Eq. (8.57b) is to be determined so that a certain
performance index (PI) is minimized in transferring the system from an
arbitrary initial state x' (0) to origin in infinitie tirne (i.e. x' (-) = 0). A
convenient PI has the quadratic form
'
Pr=
+ u'r Ru'dt
;ll '.'' Qx'
(8.60)
The manices Q arrd R are defined for the problem in hand through the
following designconsiclerations:
(i) Excursions of ACEs about the steadyvalues (r,t + brx\; - arrxt, + bzx,q)
are minimized. The steady values of ACEs are of course zero.
(ii) Excursionsof JnCg dr about the steadyvalues (xts, xte)are nrinimized.
The steaclyvaluesof JeCg dt are,of course,constants.
(iii) Excursionso1'the contt'ol vector (ut1,ut2) about the steadyvalue are
rninirnized.The steadyvalueof the controlvectoris, of course,a constant.
'
This nrinimizationis intendedto indirectlylimit the controleffbrt within
the physical capability of components.For example, the steam valve
catmot be openedmore than a certain value without causingthe boiler
presisure
to drop severely.
With the abovereasoning,we can write the PI as
From the PI of Eq. (8.51), Q md R can be recognizedas
www.muslimengineer.info
(8.63)
The acceptablesolution of K is that for which the systemremainsstable.
SubstitutingEq. (8.57b) in Eq. (8.56), the systemdynamicswith foedback
is
definedbv
i' = (A - BIgx,
(g.64)
Fol stability all thc cigenvaluesof the matrix (A - Bn shouldhave
negative
real parts.
For illustration we considertwo identical control areaswith the
following
syste|llparameters:
4r* = 0'4 scc; T'r= 0.5 sec; 7'r* = 20 sec
/l = 3: (n* = l/lJ = 100
b = O . 425;Ki = 0. 09; up = I ; 2iln
= 0. 05
0.52tt6 l.l4l9 0.68l3 - 0.0046-0.021| -0.0100-0.7437 0.gggg0.00001
^f, = [
L-o.tl046-0.o2tl-0.0100 0.5286 t.t4rg 0.6813 0.74370.0000
0.gggsl
pr= * fU-+ + h,.r,,)2+ (- tt,2xt,+ brxta)z
+ (.r,?+ ,,])
2Jtt'
+ kfu'l+ u,|11
at
K = R-rBrS
(8.61)
'*Refer
Nagrath and Gopal [5].
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
iiii'f:l
Modernpowerrystemin4gs
As the control areasextend over vast geographical regions, there are two
ways of obtaining full state information in each area for control purposes.
(i) Transport the state information of the distant area over communication
channels.This is, of course,expensive.
'_-
c= vR.f- vr
The error initiates the corrective action of adjusting
the alternator excitation.
Error wave form is suppressedcarrier modulated,
tt" carrier frequency being
the system frequency of 50 Hz.
Change in voltage
caused by load
Load change
tG
1+Iers
skrt
8.6
AUTOMATIC
VOLTAGE
CONTROL
Fig. 8.21 Brockdiagramof arternator
vortagereguratorscheme
Figure 8.20 gives the schematic diagram of an automatic voltage regulator
of
a generator.It basicallyconsistsof a main exciter which excites the alternator
field to control the output voltage. The exciter field is automaticallycontrolled
through error e = vr"r - vr, suitably amplified through voltage and power
amplifiers. It is a type-0 system which requiresa constant error e for aspecified
voltage at generatorterminals. The block diagram of the systemis given
in
Error amplifier: It demodulatesand amplifies the
error signal. Its gain is Kr.
scR power amplffier and exciter
fierd: It provides the n"."rriry power
amplification to the signal for controlling thl exciter
n"ro.- arr*;"g
,rr"
amplifier time constant to be small enoughio be neglected,
the ovelail fansfer
function of these two is
K,
l* T"rs
where T"yis the exciter field time constant.
L
o
A
D
Alternator; Its field is excited by the main exciter voltage
vu. Under no road
it producesa voltage proportional to field current.The
no load transferfunction
is
Ks
7*T*s
where
Potential
Fig. 8.20 Schematicdiagramof alternatorvoltageregulatorscheme
Fig. 8.21. The function of important componentsand their transferfunctions
is
given below:
Potential transformer: It gives a sample of terminal voltage v..
Dffirencing device; It gives the actuating error
www.muslimengineer.info
T*= generatorfield time constant.
The load causes a voltage drop which is a complex
function of direct and
quadratureaxis currents.The effect is only schematically
reBresented
hv hlock
G.. The exact load model of the alternatoris beyond ,t"
,iop" ;rhtJ;;:
stabitizing transformer: T4*d
are large enough time constantsto impair
the system's dynamic response.-lq
Itjs weil known that the dynami. r"rpoor" of
a control system can be improved by the internal derivative feedback
loop. The
derivative feedback in this system is provided by means
of a stabiyzing
transformer excited by the exciter output voltage vE. The
output of the
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Power SystemAnalysis
Automatic
Generation
and VoltageControl
------_-----l
transformeris fccl ncgativclyat the input terminalsof thc SCR power
stabiliz,ing
amplifier. The transferfunction of the stabilizing transfo"meris derived below.
Since the secondaryis connectedat the input ternfnals of an amplifier, it can
be assumedto draw zero current. Now
The bandedvaluesimposedhy the limitersareselectedto resffictthe generation
rate by l}Vo per minute.
I
I
320' l
Jffif
E
I
dt
vr = Rr i., + LrJilL
' d t
I.g 9t",
u't
_+(
'rr= MY
+/
A
l-
dt
-t*9r"'--l
Taking the Laplace transform, we get
%,(s) _ sM
VuG) R, * s,Lt
sMlRt
l*Irs
Fig.8.22 Governormodelwith GRC
sK",
1 +{ , s
Accurate staterrariablemodels of loaded alternator around an operating point
are available in literature using which optimal voltage regulation schemescan
be devised.This is, of course,beyondthe scopeof this book.
8.7
LOAD FREOUENCY CONTROL WITH GENERATION
RATE CONSTRAINTS (GRCs)
frcquencycontrolproblcmdiscussedso far doesnot consiclerthe effect
The l<-racl
of the restrictionson the rate of changeof power generation.In power systems
canchangeonly at a specifiedmaximum
havingsteamplants,powergeneration
(fiom
saf'etyconsiderationso1 the equipment)for
rate. The generationrate
reheat units is quit low. Most of the reheatunits have a generatiol rate around
3%olmin. Some have a generation rate between 5 to 7jo/o/min.If these
largc tttottrclttrry
constraintsarc not consirlcrcd,systerttis likely to c:ha.sc
Several
the
controller.
tear
of
and
wear
undue
disturbances,Thrs results in
of
clesign
for
the
of
GRCs
effect
the
to
consider
proposecl
been
methoclshave
dynamic
the
systeln
considered,
is
When
GRC
generation
controllers.
automatic
rnodelbecomesnon-linearand linearcontroltechniquescannotbe appliedfor
the optimizationof the controllersetting.
If the generationratesdenotedby P", are included in the statevec:tor,the
systermorder will be altered.Insteadof augntentingthem, while solving the
stare equations,it may be verified at each step if the GRCs are viclated.
Another way of consicieringGRCs for both areas is to arjri iinriiers io ihe
governors[15, 17] as shown in Fig. 8.22, r.e., the maximum rate of valve
openingor closingspeedis restrictedby the limiters.Here 2", tr,r,, iS the power
rate limit irnposedby valve or gate control. In this model
l A Y E l . - -g u , n r
www.muslimengineer.info
(8.6s)
The GRCs result in larger deviationsin ACEs as the rate at which generation
can cha-ngein the area is constrainedby the limits imposed. Therefore, the
duration for which the power needsto be imported increasesconsiderably as
cornparedto the case where generationrate is not constrained.With GRCs, R
should be selectedwith care so as to give the best dynamic response.In hydrothennal system,the generationrate in the hydro area norrnallyremainsbelow
the safe limit and therefore GRCs for all the hydro plants can be.ignored.
8.8
SPEED GOVERNOR DEAD-BAND
ON AGC
AND ITS EFFECT
The eff'ectof the speed governor dead-bandis that for a given position of the
governor control valves, an increase/decrease
in speed can occur before the
positionof the valve changes.The governordead-bandcan materiallyaffect the
system response.ln AGC studies, the dead-band eff'ect indeed can be
significant,sincerelativcly small signalsare under considerations.
TlLespeedgovernorcharacterristic.
thoughnon-lirrear,
hasbeenapproxinraaed
by linear characteristicsin earlier analysis. Further, there is another noniinearity introducedby the dead-bandin the governor operation.Mechanical
f'riction and backlashand also valve overlapsin hydraulic relays cause the
governor dead-band.Dur to this, though the input signal increases,the speed
governor may not irnmediately react until the input reachesa particular value.
Similar a.ctiontakesplace when the input signal decreases.
Thus the governor
dead-bandis defined as the total rnagnitudeof sustainedspeedchangewithin
which there is no change in valve position. The limiting value of dead-bandis
specifiedas 0.06Vo.It was shown by Concordia et. al [18] that one of the
effects of governor dead-bandis to increasethe apparentsteady-statespeed
regulation R.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModrrnPo*., svrt.t Analuri,
lFFf
The effect of the dead-bandmay be included in the speed governor control
loop block diagramas shown in Fig. 8.23.Consideringthe worst caseforthe
dead-band,(i.e., the system starts responding after the whole dead-band is
traversed)and examiningthe dead-bandblock in Fig. 8.23,the following set of
ly define the behaviourolthe dead.band [9]-
Discrete-Time Control Model
The continuous-timedynamic systemis describedby a set of linear differential
equations
x=Ax+Bu+ fp
(8.67)
where f u, P are state, conhol and disturbance vectors respectively and
A,B
and f are constantmatrices associatedwith the above vectors.
The discrete-timebehaviourof the continuous-timesystemis modelled by the
system of first order linear differenceequations:
x(k+1)=Qx(k)+Vu(k)+jp&)
Speed governor
controlloop
in speed-governor
Flg. 8.23 Dead-band
u(r+1)= 7(r) 1:
-
"(r+l)
_ x, 1 dead-band
"(r+1)
_ dead-band; if x('+l) - ,(r) I g
(8.66)
DIGITAL
d= eAT
{=({r_ln-tr
tf Xr*l _ xt < 0
"(r+1).
(r is the step in the computation)
j=(eAr-DA-tf
Reference[20] considersthe effect of governor dead-bandnonlinearity by using
the describingfunction approach[11] and including the linearised equationsin
the state spacemodel.
The presenceof governordead-bandmakesthe dynamicresponseoscillatory.
It has been seen [9J that the governor dead-banddoes not intluence the
selectionof integral controller gain settingsin the presenceof GRCs. In the
presenceof GRC and deadband even for small load perturbation,the system
becomeshighly non-linear and hencethe optimization problem becomesrather
complex.
8.9
LF CONTROLLERS
where A, B and,I are the constantmatrices associatedwith r, ,,LO p vectors
in the conespondingcontinuous-timedynamic system. The matrix y'r can be
evaluatedusing various well-documentedapproacheslike Sylvestor's expansion
theorem, series expansion techniqueetc. The optimal digital load frequency
controller designproblem is discussedin detail in Ref [7].
8.10 DECENTRALIZED
CONTROL
In view of the large size of a modern power system, it is virtually impossible
to implement either the classicalor the modern LFC algorithm in a centralized
manner.ln Fig. 8.24, a decentralizedcontrol schemeis shown.x, is usedto find
out the vector u, while x, aloneis employed to find out u". Thus.
In recent years,increasinglymore attentionis being paid to the questionof
digital implementation of the automatic generationcontrol algorithrns. This is
mainly due to the facts that digital control turns out to be more accurate and
rcliqhlc
r
v^rEv^vt
nnrnnaef
in qize
less censifive
to nnise
end drift
nnd more
flexihle
Tt
may also be implemented in a time shared fashion by using the computer
systemsin load despatchcentre,if so desired.The ACE, a signal which is used
for AGC is availablein the discreteform, i.e., there occurssampling operation
; betweenthe systemand the controller. Unlike the continuous-timesystem,the
control vector in the discretemode is constrainedto remain constant between
Flg. 8.24 Decentralized
control
www.muslimengineer.info
(8.68)
where x(k), u(k) and p(k) are the state,control and disturbancevectors and are
specifiedat t= kr, ft = 0, 1,2,... etc.and ris the samplingperiod.
6, tl,nd
7 Te the state, control and disturbance transition matrices and they are
evaluatedusing the following relations.
Dead-band
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowerSystemAnalysis
i,i2[,:*.1
4
and voltageControl
Automatic
Generation
-
x - (x1 x2)'
- aF(s)' 1' af (t)dr: liq, * '4F(s): hm/F(")]
IL n,n,, tf^461dv
JO
JO
s-0
s+0
,t
S
ut=-ktxt
kzxz
u2--
been shown possible using the modal control principle. Decentralized or
hierarchicalimplementationof the optimal LFC algorithmsseemsto have been
studied more widely for the stochasticcase since the real load disturbancesare
truely stochastic.A simple approachis discussedin Ref. [7].
It may by noted that other techniquesof model simplification are available
in the literatureon alternativetools to decentralizedcontrol. Theseinclude the
method of "aggregation", "singular perturbation", "moment matching" and
other techniques[9] for finding lower order models of a given large scale
system.
8.4 For the two area load frequencycontrol of Fie. 8.16 assumethat inte
controller blocks are replacedby gain blocks, i.e. ACEI and ACE are fed
to the respective speedchangersthrough gains - K, and - Ko. Derive an
expressionfor the steadyvalues of changein frequency and tie line power
for simultaneouslyapplied unit step load disturbanceinputs in the two
areas.
8.5 For the two area load frequencycontrol employing integral of area control
error in each area (Fig. 8.16), obtain an expressionfor AP6"$) for unit
step disturbance in one of the areas.Assume both areas to be identical.
Comment upon the stability of the system for parameter values given
below:
4e = 0'4 sec; Z, = 0'5 sec; Zp. = 20 sec
IEI/IS
PROB
K p r = 1 0 0 ;R = 3 ; K i = l ; b = 0.425
8 . 1 Two generatorsrated 200 MW and 400 MW are operating in parallel.
The droop characteristicsof their governors are47o and 5Vorespectively
from no load to full load. The speedchangersare so setthat the generators
operate at 50 Hz sharing the full load of 600 MW in the ratio of their
ratings.If the load reducesto 400 MW, how will it be sharedamong the
generatorsand what will the s)/stemfrequency be? Assumefree governor
operatlon.
The speedchangersof the governorsare resetso that the load of 400 MW
is sharedamong the generatorsat 50 Hz in the ratio of their ratings. What
of the generators?
are the no load frequencies
8 . 2 Consider the block diagrammodel of lcad frequencycontrol given in Fig.
8.6. Make the following approximatron.
(1 + Z.rs) (1 + Z,s) =- t + (7rg + T,),s= 1 + Z"c.r
Solve for Af (l) with parametersgiveu below. Given AP, - 0.01 pu
T"q= 0.4 + 0.5 = 0.9 sec; 70, = 20 sec
; =3
K r r K , = 1 ;K p r = 1 0 0 R
olrn'rn.iorlvYYll
T i iLc6 .
ll(
L
e
v.
1n
rvt
nhfain
vuLarrr
en
AsnrAccinn
lHint: Apply Routh's stability criterion to the characteristicequation of
the system.l
NCES
REFERE
Books
l. Elgcrd, O.1., Elccu'ic Energv.Sv,s/clrT'lrcorv: An ltttnxlut'lion. 2nd cdn. McCrawHill, New York, 1982.
2. Weedy, B.M. and B.J. Cory Electric Pow'er Systems,4th edn, Wiley, New York,
I998.
a
1
Cohn, N.,
Control of Generation and Power Flou, on Interconnected Systents,
Wiley, New York, i971.
4 . Wood, A.J., and B.F. Woolenberg, Power Generation, Operation and Control,2nd
edn Wiley, New York, 1996.
Coinparewith the exact responsegiven in Fig. 8.9.
oc
ar2= I;2tTr, = 0.05
5 . Nagarth, I.J. and M. Gopal, Control Systems Engineering, 3rd edn. New Delhi,
8 . 3 For the load frequency control with proportional plus integral controller
clJ
ffi
fnr
tha
cfenrlrr
cfrfp
errnr
in
r; for a urrit stepAPr. What is the correspondingtime
cycles,i.". f'41t)d
t ^ "
, 1 ,
lirnl*m
error in seconds(with respectto 50 Hz).lCommenton the dependenceof
error in cycles upon the integral controller gain K,.
2 0 0l .
6 . Handschin, E. (Ed.), Real Time Control of Electric Power Systems, Elsevier, New
York 1972.
7 . Mahalanabis, A.K., D.P. Kothari and S.I Ahson, Computer Aided Power Systent
Analysis and Control, Tata McGraw-Hill,
New Delhi,
1988.
8 . Kirclrrnayer, L.K., Economic Control of lnterconnected Systems,Wiley, New York,
t959.
9 . Jamshidi, M., Inrge Scale System.s:Modelling and Control, North Holland, N.Y.,
1983.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
10. Singh, M.G. and A. Titli, SystemsDecomposition,
Optimization and Control
PergamonPress, Oxford, 197g.
I I' Siljak, D'D., Non-LinearSystems:The Parametcr Analysis
antl Design, Wiley,
N.Y. 1969.
.t,apers
12. Elgerd, o.I. and c.E..Fosha,"The Megawatt Frequency
control problem: A New
Approachvia optimal control Theory", IEEE Trans.,April
1970,No. 4, pAS g9:
556.
13' Bhatti, T'S., C.S Indulkar and D.P. Kothari, "Parameter
optimization of power
Systemsfor StochasticLoad Demands" Proc. IFAC. Bangalore,
December 19g6.
l4' Kothari, M'L., P.S. Satsangiand J. Nanda,"sampled-Data
AutomaticGeneration
Control of InterconnectedReheatThermal Systems Considering
GenerationRate
Constraints",IEEE Trans.,May 19g1, pAS_100;2334.
15' Nanda, J', M.L. Kothari and P.S. Satsangi,"Automatic
GenerationControl of an
InterconnectedHydro-thermalsystem in continuous and
DiscreteModes considering GenerationRateconstraints'IEE proc., prD, No. l,
January19g3,130 : 17.
16' IEEE committee Report,'DynamicModels for Steamand
Hydro-turbinesin power
system studies" IEEE Trans., Nov/Dec. rg73, pAS-92, 1904.
l7' Hiyama, T', "Optimization of Discrete-typeLoad Frequency
RegulatorsConsidering Generation-Rateconstraints" proc.lE4 Nov. g2, r2g, pt
c, 2g5.
I8. concordia,c., L.K. Kirchmayerand E.A. Szyonanski,..Effect
of speed Governor
Dead-bandon Tie Line Power and Frequency Control performan
ce,, AIEE Trans.
A u g . 1 9 5 7 ,7 6 , 4 2 9 .
19' Nanda,J', M.L. Kothari and P.S. Satsangi,"Automatic
Control of ReheatThermal
SystemConsidering GenerationRate Constraint and Covernor
Dead-band,,.J.I.E.
(India),June 1983, 63,245.
20. Tripathy, s.9., G.s. Hope and o.p. Marik, ,,optimisatiorr
of Load-frcqucncy
C<lntrolParametersfor Power systemswith ReheatSteamTurbines
and Governor
Dead-bandNonlinearity", proc. IEE, January rgg2, rzg, pt
c, No. r, r0.
21. Kothari,M.L., J. Nanda,D.p. Kothariand D. Das,.,Discrete-mode
AGC of a.two_
area Reheat Thermal system with New Area control Error,,,
IEEE Trans. on
Power System,Vol. 4, May 19g9, 730
22' Daq D. J. Nanda, M.L. Kothari and D.p. Kothari, ,.AGC
of a Hydro_Thermal
systemwith New ACE consideringGRC", Int. J. EMps,1g, No.
5, rggo, 46r.
23' Das, D', M.L' Kothari, D.P. Kothari and J. Nanda, "Variable
Structure Control
strategy to AGC of an IntcrconncctcdRcheat Thermal systcm,,, prctc.
IEE, r3g,
p t D , 1 9 9 1 ,5 7 9 .
24. Jalleli, Van Slycik et. al.. "lJndersfandingAutonnaticGeneration
Control,,, IEEE
Trans.on P.S., Vol 07, 3 Aug. 92, 1106_1122.
2 5 . Kothari, M.L., J. Nanda,D.p. Kothari and D. Das, ,,Discrete
Mode AGC of a two
Area ReheatThermal Systemwith a NACE consideringGRC,,,
J.LE. (rndia), vol.
72, Feb. 1992,pp Zg7-303.
2 6 . Bakken, B.H. and e.s. Grande,"AGC in a Deregulatedpower
system,,, IEEE
Trans.on Power Systems,13, 4, Nov. 199g,pp. 1401_1406.
www.muslimengineer.info
9.1
INTRODUCTION
So far we have dealt with the steadystate behaviour of power system under
normal operating conditions and its dynamic behaviour under small scale
perturbations.This chapter is devoted to abnormal system behaviour under
conditions of symmetrical short circuit (symmetricalthree-phase.fault*).Such
conditions are caused in the system accidentally through insulation failure of
equipment or flashover of lines initiated by a lightning stroke or through
accidentalfaulty operation.The systemmust be protectedagainst flow of heavy
short circuit currents(which can causepeffnanentdamageto major equipment)
by disconnectingthe faulty part of the system by means of circuit breakers
operated by protective relaying. For proper choice of circuit breakers and
protective relaying, we must estimatethe magnitude of currents that would flow
under short circuit conditions-this is the scope of fault analysis (study).
The majority of systemfaults are not three-phasefaults but faults involving
one line to ground or occasionallytwo lines to ground.These are unsymmetrical
faults requiring special tools like symmetrical componentsand form the subject
of study of the next two chapters.Though the symmetrical faults are rare, the
symmetrical fault analysis must be carried out, as this type of fault generally
leads to most severe fault current flow against which the system must be
protected. Symmetrical fault analysisis, of course, simpler to carry out.
A power network comprises synchronousgenerators,ffansfonners, lines and
loads. Though the operating conditions at the time of fault are important, the
loads can be neglectedduring fault, as voltages dip very low so that currents
drawn by loads can be neglectedin comparisonto fault currents.
*Symmetrical fault may be a solid three-phase short
circuit or may involve are
impedance.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
tffiffi
ModernPowerSystemAnalysis
325 |
t
I
The synchronousgenerator during short circuit has a characteristictimevarying behaviour.In the event of a short circuit, the flux per pole undergoes
dynamic changewith associatedtransientsin damper and field windings.The
reactanceof the circuit model of the machine changesin the first few cycles
from a low subtransientreaetanecto a higher transient value, finally settling at
a s'iitt higher synchronous (steady state) value. Depending upon the arc
intemrption time of circuit breakers,a suitable reactancevalue is used for the
circuit model of synchronousgeneratorsfor short circuit analysis.
In a modern large interconnectedpower system, heavy currents flowing
during a fault must be interruptedmuch before the steady state conditions are
established.Furthermore,from the considerationsof mechanicalforcesthat act
on circuit breaker components,the maximum current that a breaker has to carry
momentarily must also be determined.For selecting a circuit breakerwe must,
therefore, determine the initial current that flows on occulTenceof a short
circuit and also the current in the transient that flows at the time of circuit
intemrption.
42V
= --*sin
(cr,rf+ a_
lzl
z = (Rz+ Jr\tt"(t:
A
tan-l
+)
ir = transient current [it is such that t(0) = t(0) + L(0) = 0 being an
inductive circuit; it decayscorrespondingiothe tim6 constantiRl.
= - i,(6)e-$tL)t
= 9Y
tzl
s i n( d - a ) g - . ( R t D t
Thus short circuit current is given by
(e.1)
Synrnretricalshort
circuit current
9.2
TRANSIENT
ON A TRANSMISSION
LINE
Let us consider the short circuit transient on a transmission line. Certain
simplifying assumptionsare madeat this stage.
(i) The line is led I'rorna constantvoltagcsoLrrcc(tlte casewhcn the line is
fed from a realisticsynchronons
ma.chrne
will tre treatedin Sec.9.3).
(ii) Short circuit takes place when the line is unloaded(the caseof short
circuit on a loaded line will be treatedlater in this chapter).
(iii) Line capacitanceis negligibleand the line can be represented
by a lumped
RZ seriescircuit.
L
, ,F.
r+V\'\-'
.l
v = JI vsin (o,t+ *) rV)
I
i
- Jrv'sin(d- c) * E'
lzl
tzl
by the circuit rnoclel
With the aboveassumptions
the line can be representecl
of Fig. 9.1. The short circuit is assumedto take place at t = 0. The parameter
It is known
<rcontrolsthe instanton the voltagewavewhen shortcircuit occLrrs.
two parts,
circuit
is
composed
of
current
after
short
from circuit theory that the
1.tr.
I.t
whcre
i, = steadystatecurrent
www.muslimengineer.info
(e.2)
Sincetransmissionline resistanceis small. 0 - 9C,.
.
F i g .9.1
t-- I"+
A plot of i* i, and'i = i, + i, is shown in Fig. 9.2.rnpower systemterrninology,
the sinusoidal steady state current is called the symmetrical short circuit
current and the unidirectional transient component is called the DC off-set
current, which causesthe total short circuritcurrentto be unsymmetricaltill the
transientdecays.
It easily follows fiom F'ig. 9.2 that the maximum momenro) short circuit
currcnt i,,,,,,
corresponds
to the firstpenk.If theclecay
of trnnsient
currentin this
short time is neglected,
Im* =
_
DC otT- set curnent
Jiv cosa+ JTv
rzr
rzl
(e.3)
This has the maximum possible value for o. = 0, i.e. short circuit occurring
when the voltage wave is going through zero.Thus
= '#
i,n,nlrnu*possible)
e.4)
= twice the maxirnum of symmetricalshort circuit current
(doubling effect)
For the selectionof circuit breakers.momentaryshort circuit currentis taken
correspondingto its maxirnumpossible value (a sat'echoice).
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.w
The
b,a#&
Modern
PowerSystem
Analysis
ffiffif
nevf
ntrecfinn
ic
turhqf
ic
fhc
r.rrrrcnf
fn
hc
i n f e r r r r ne fl / p
rl?t
lvs
Aa
hqo
haan
pointed out earlier, modern day circuit breakers are designed to intemrpt the
cunent in the first few cycles (five cycles or less). With referenceto Fig, 9.2
it meansthat when the current is intemrpted,the DC off-set (i,) has not yet died
the value of the DC off-set at the time of intemrption (this would be highly
complex in a network of even moderately large size), the symmetrical short
circuit current alone is calculated.This figure is then increasedby an empirical
rnultiplying factor to account for the DC off-set current. Details are given in
S ec .9. 5 .
reactancewhen combined with the leakagereactanceXi of the machine is called
synchronousreactance X4 (direct axis synchronousreactancein the case of
salient pole machines).Armature resistancebeing small can be neglected.The
ne ls snownln rrg.
on per phasebasis.
(a) Steadystateshortcircuitmodel
of a synchronous
machine
(b) Approximate
circuitmodelduring
periodof shortcircuit
subtransient
X1
(c)Approximate
circuit
modelduring
periodofshortcircuit
transient
Fig. 9.3
Fig.9.2 Waveform
of a shortcircuitcurrent
on a transmission
line
9.3
SHORT CTRCUTTOF A SYNCHRONOUS MACHTNE (ON
NO LOAD)
Under steady state short circuit conditions, the armature reaction of a
generator
producesa demagnetizing
synchronous
flux. In termsof a circuitthis
www.muslimengineer.info
Consider now the sudden short circuit (three-phase)of a synchronous
generator initially operating under open circuit conditions. The machine
undergoesa transientin all the three phasefinally ending up in steady state
conditions describedabove. The circuit breakermust, of course,intemrpt the
current much before steady conditions are reached. Immediately upon short
circuit, the DC off-set currents appear in all the three phases,each with a
different magnitude since the point on the voltage wave at which short circuit
occurs is different for each phase.These DC off-set currents are accounted for
separately on an empirical basis and, therefore, for short circuit studies, we
need to concentrate our attention on syimmetrical (sinusoidal) short circuit
current only.Immediately in the event of a short circuit, the symmetrical.short
circuit current is limited only by the leakagereaitance of the machine. Since the
air gap flux cannotchangeinstantaneously(theorem of constantflux linkages),
to counter the demagnetization of the armature short circuit current, currents
appearin the field winding as well as in the damper winding in a direction to
help the main flux. These currents decay in accordancewith the winding time
constants.The time constant of the damper winding which has low leakage
inductanceis much less than that of the field winding, which has high leakage
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
MocjernPower Sysiem nnaiysis
FaultAnalysis
symmerrical
inductance.Thus during the initial part of the short circuit, the damper and field
windingshavetransfurnrercurrentsinducedin them so that in the circr,ritmodel
thcir reactances--X,of field winding and Xa* of damper winding-appear in
parallelx with Xo as shorvn in Fig. 9.3b. As the danqpqlruadlag cullqqls 4!q
first to die out, Xr* effectively becomesopen circuited and at a later stage X1
becomesopen circuited.The rnachinereactancethus chaugesfrom the parallel
combinationof Xo, Xy and Xu. during the initial period of the short circuit to
X,,and Xrinparallel (Fig.9.3c) in the rniddleper:iodof the short circuit, and
finally to X,,in steadystate(Fig. 9.3a).The reactancepresentedby the machine
in the initial period of the short circuit, i.e.
1_:
X.L -r-(11x,,+UXJ+llxd,,)
is called the transientreactanceof the machirre.Of course,the leactanceunder
steaclyconditionsis the synchronousreactanceof the machine.Obviousiy Xf7<
X'd< Xu.The machinethus offers a time-varyingreactancewhich changesfront
Xttoto Xtaandfinally to Xn.
Subtransient
oeriod
b
l
I
Steady state period
I
I
I
0)
Steady state current amPlitude
Tlme
current
shortcircuit
symmetrical
machine
(b)Envelope
of synchronous
<>fthe nrachine.While the reactance
is called the subtrunsientreoctutxc:e
effective after the darnperwinding currents have died out, i.e.
(e.6)
X' ,t= X, + (X ,,l l X ,)
I
I
(e.5)
X'j
"
I'lt,5ffit
I
E a
q)
()
f,
Fig.9. 4
If we examinethe oscillograrnof the short circuit currentof a synchronous
machineafter the DC ott-set cuitentshave beenrettrovedtrom it, we will tind
the current wave shapeas given in Fig. 9.4a. The envelopeof the current wave
shapeis plottedin Fig. 9.4b, The shortcircuit currentcan be divided irtto three
periods-initial subtransientperiod when the current is large as tire tnachine
offers subtransientreactance,the middle transient period where the machine
offers transientreactance,and finally the steadystateperiod w\n the machine
ofters synchronousreactance.
:
If the transientenvelope is extrapolatedbackwards in tinre, the difference
envelopesis the cunent Ai/' (correbetwecnthe tlansicrrtanclsubtransiertt
sponding to the clamperwinding current) which decays fast according to the
clamperwinding time constant.Similarly, the difference Ai/ betweenthe steady
with the field time constant.
state1nd transicntenvelopesdecaysin accordance
discussedabove,we
reactances
and
cunents
the
oscillogram,
the
In termsof
can wrlte
'o0
g
Time
t
lIl =
a
oa
t;
\t z.
lEsl
(9.7a)
Y,
o
o
E
E
a
ob _ l E 8 l
t;
xtd
(e.7b)
tltt= 32.: Y+
(9.7c)
lll =
i
\IL
Actual envelope
Extrapolationof
steady valrre
J2
Extrapolationof transientenvelope
(a) Symmetricalshort circuit armature current in synchronousmachine
Fig. 9.4 (Contd.)
*Unity turn ratio is assumedhere.
www.muslimengineer.info
X,J
where
l1l = steady state current (rms)
!//l = transientcurrent (rms) excluding DC component
lltl = subtransientcurrent (rms) excluding DC component
Xa = direct axis synchronousreactance
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'ii#I:
Fault
Modern Power Svstenn Anelveie
Xtd= direct axis transientreactance
X'j = direct axis subtransientreactance
pon ii'ragn-tir rutlrarion
j? ",l"ilirion), rhe valuestf reactances
normallyrie within
lErl = per phaseno load voltage(rms)
Oa,Ob,Oc = interceptsshorvn iLEigs- 9Aa and,b_
The intercept Ob for finding transient reactance can
be determined
accurately by means of a logarithmic plot. Both Ai, and
al
decav
exponentially as
Aitt = Ai( exP (- t/q,)
Ait = Ai6 exp 1_ t/r7)
where r4, and rf arerespectively damper, and field winding time
constantswith
Td* 4 ry At time / 2' r4*, Aitt practicalry dies out and we can write
log (Aitt+ At,)1,
,
Trl*
n
, - log Ai' = - Aint ,/ ry
t
l c
a b
f:T:jr:lltlq
certain
predictable limits for different types of
machines. Tabie 9.r gives
typical valuesof machinereactanceswhich can be userj
in fault calculationsand
in stability studies.
Normally both generator and motor subtransient
reactancesare used to
determine the momentary current flowing on occurrence
of a short circuit. To
decide the intemrpting capacity of circuit breakers,
except those which open
instantarreously,subtransientreactanceis used for
generatorsand transient
reactancefor synchronousmotors. As we shall see later
the transient reactances
are used for stability studies.
The machinemodel to be employed when the short
circuit takesplace from
loaded conditionswill be explainedin Sec. 9.4.
The method of computing short circuit currents
is illustrated through
examplesgiven below.
*+
g
For the radial network shown in Fig. 9.6, a three-phase
fault occurs at F.
Determine the fault current and the line voltage
at l l kv bus under fault
conditions.
O)
o
o
r f
Fig. 9.5
1OMVA
15%reactance
10 MVA
12.5ohreactance \
\
1 1k V
ne : 3o km,z = (0.27+jo.3e a/ km
Ai,l,:o : Aito exp(-r/ ,t )1,:o: Ai,o : ob
r NO 2: 5 MVA,8ohreaclance
Table 9.1 Typicalvaluesof synchronousmachinereactances
(All valuesexpressedin pu of ratedMVA)
riO.0B)
o / km
F
z xn caote
/
Synchronous
Type of
machine
X, (or X,)
X^
.t
xd
xti
x2
xo
ru
Turbo-alternator Salient pole
(Turbine
(Hydroelectric)
generator)
l .0 0 --2 .0
0.9-1.s
0 .1 2 -0 .3 5
0.r-0.25
_ x,d
0.04-0.14
0.003-0.008
compensator
(Condenser/
Synchronous
motors*
capacitor)
0.6-1.5
r.5-r2.5
0 .4 -1.0
0.95-1.5
0.2-0.5
0.3-0.6
0 .1 3-0.35
0.18-0.38
_ x,d
0.17-0.37
0.02-0.2
0.025-0.16
0 .0 03-0.01s 0.004-0.01
0.8-1.10
0.65-0.8
0.3-0.35
0.18-0.2
0.19-0.35
0.05-0.07
0.003-0.012
ro = AC resistanceof the armature winding per phase.
* High-speed units tend to
have low reactanceand low speedunits high
www.muslimengineer.info
Fig. 9.6 Radial networkfor Example g.1
Solution
Select a system base of 100 MVA.
V6ltage bases are: I I kV-in generators, 33 kV for
overhead line and 6.6 kV
for cable.
Reactanceof G, =
Reactanceof G2 Reactanceof Z, =
reactance.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
s
I
PowerSystemAnalysis
Reactance
of Tr= ;WY4
)
SymmetricalFault Analysis
|I
337
= (0.93+ j05s) + (71.6)+ (0.744+ i0.99) + (t1.0)
= 71.6pu
= I.674+ j4.14= 4.43176.8"pu
Z (in ohms)x MVA""'"
overheadline impedance(kvBur"
)2
Voltageat 11 kV bus= 4.43 167.8"x 0.196l- 70-8"
= 0;88 I :T ptt = ft88 x 11 = 9;68 kV
30x(0.27+j0.36)x100
Q'2
- (0.744
+ 70.99)pu
q21tlq = (0.93
= 3(9-1!t,rJr0,q,
+ 70.55)pu
cableimpedanc"
(6 .6 )'
Circuit model of the systemfor fault calculationsis shownin Fig. 9.7. Since
the systemis on no load prior to occurrenceof the fault, the voltagesof the two
generatorsare identical (in phaseand magnitude) and are equal to 1 pu. The
generatorcircuit can thus be replacedby a single voltage sourcein serieswith
the parallel combination of generatorreactancesas shown.
11kV bus
j|.0 (o.744+i0.99) i1.6 (0.93+70.55)
,666' I
I
|
66d. I
Cable I t
T2
Line
T1
I
l
i
A 25 MVA, 11 kV generator with Xl = 20Vo is connected through a
transformer, line and a transfbrmer to a bus that suppliesthree identical motors
as shown in Fig. 9.8. Each motor has Xj = 25Voand Xl = 3OVoon a base of
5 MVA, 6.6 kV. The three-phaserating of the step-uptransformeris 25 MVA,
11/66 kV with a leakage reactance of l0o/o and that of the step-down
transformer is 25 MVA, 6616.6kV with a leakagereactanceof l0%o.The bus
fault occurs at the point F.
voltage at the motors is 6.6 kV when a three-pha.se
For the specified fault, calculate
(a) the subtransientcurrent in the fault,
(b) the subtransientcurrent jn the breaker .8,
(c) the momentary current in breaker B, and
(d) the current to be interruptedby breakerB in five cycles.
Given: Reactanceof the transmissionline = l5%oon a baseof 25 MVA, 66
kV. Assurne that the systemis operating on no load when the fdul" occurs.
Flg. 9.8
Fig. 9.7
= (j1.5ll j1.25)+ (t1.0)+ (0.744+ i0.99)+ (i1.6)+
Totalimpedance
(0.93+ 70.55)
- 1 . 6 7 4+ j 4 . 8 2= 5 . 1 1 7 0 . 8 " p u
''?
Isc=
tt = 0'196I - 70'8"Pu
5.r170.8"
to*.10; =
8,750A
18u."=
J3 x6.6
Isc= 0.196x 8,750= 1,715A
betweenF and11 kV bus
Total irnpedance
www.muslimengineer.info
Sotution Choosea systembaseof 25 MVA.
For a generatorvoltage baseof 11 kV, line voltagebaseis 66 kV and motor
voltage base is 6.6 kV.
(a) For each motor
X',j*= j0.25 x
+
= i1.25
pu
Line, transtbrmersand generatorreactancesare already given on proper base
values.
The circuit model of the systemfor fault calculationsis given in Fig. 9.9a.
The systembeing initially on no load, the generatorand motor inducedemfs are
identical. The circuit can thereforebe reducedto that of Fig. 9.9b and then to
Fis. 9.9c. Now
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
powerSystem
Modern
Anatysis
fiffirel
!
--l-+=+
Isc=3><
j1.25
Basecun'entin 6.5 kV circui, -
j0.55
momentarycurreni ihrough breaker B -- 1.6 x 7,4i9.5
- - jr 4 . 2 2 p u
25 x 1,000 =
- 17,967A
(d) To compute the current to be intemrpted by the breaker, motor
subtransient reactance (X!j = j0.25) is now replaced by transient reactance
(X a = /0.3O).
2.187 A
Issc=
o,;;* r4rrlt]frf* o
XI (motor)= 70.3x
(b) From Fig. 9.9c, current through circuit breaker B is
I s c (' B- ) 2 x - + + . ] _
j1.25
j0.55
The reactancesof the circuit of Fig. 9.9c now modify to that of Fig. 9.9d.
Current (symmetrical) to be intemrpted by the breaker(as shown by arrow)
1
^1 "
=3.1515pu
=2x
+
: - i 3r -.. 4-2
= 3.42x 2,187= 7,479.5
A
jl.s
110"
1 1 0+'
jo.2
j0.15
25 =
Jr.) pu
T
j0.1
110'
F tll?9
;< to"
jO.ss
Allowance is made for the DC off-set value by multiplying with a factor of 1.1
(Sec. 9.5). Therefore, the current to be interrupted is
1 . 1 x 3 . 1 5 1 5x 2 . 1 8 7= 7 . 5 8 1A
9.4
SHORT CIRCUIT OF A LOADED SYNCHRONOUS
MACHINE
In the previous article on the short circuit of a synchronousmachine, it was
aBsumedthat the machine was operating at no load prior to the occurrence of
short circuit. The analysisof short circuit on a loadedsynchronousmachine is
complicatedand is beyond the scope of this book. We shall, howevbr, present
here the methods of computing short circuit current when short circuit occurs
under loaded conditions.
rl:---^
lr rBtlc
(b)
i0.55
rcuitbreaker)
i0.55
(c)
Fig. 9.9
(c) For finding momentary current through the breaker, we must add the
DC off-set current to the symmetricalsubtransientcurrent obtainedin part (b).
Rather than calculating the DC off-set current, allowance is made for it on an
empiricalbasis.As explainedin Sec.9.5,
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f\
1/.|
>. Lv
^L^.-,^
Slluws
+L^
Llrg
^i-^.,i+
urrUurt
*^A^l
lrlu(lEl
^f
ur
a
synchronousgeneratoroperatingunder steadyconditions supplying a load current /" to the bus at a
terminal voltage of V ". E, is the induced emf under
loadedcondition andXa is the direct axis synchronous reactanceof the machine.When short circuit
occurs at the terminals of this machine, the circuit
model to be used for computing short circuit Fig.
9.10 Circuitmodelof
current is given in Fig. 9.11a for subtransient
a loaded
current,and in Fig. 9.1lb for transientcurrent. The
m achine
inducedemfs to be used in thesemodels are given
bY
E,l= v" + ilTtj
(e.8)
(e.e)
EL- V'+ il"Xto
The voltage E!is known as the voltage behind the subtransientreactance and
the voltage E!is known as the voltage behind the transient reactance.Infact,
if 1o is zero (no load case),EJ= Etr= Er, the no load voltage,in which case
the circuit model reducesto that discussedin Sec. 9.3.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
340
Modernpo*s1_qqe!l inslygs
|
_Symmetrical
I
Fault Arralysis
solution Aii reactancesare given on a base of 25 MVA and appropriatgi
voiiages.
:t
t/o
Prefaultvoltage V" = J'9
l1
= 0.9636 l0
pu
Load = 15 NfW, 0.8 pflEading
( a ) C i r c u i tm o d e l f o r c o m p u t i n g
subtransientcurrent
(b) Circuit model for computing
transient current
= l: = 0.6 pu, 0.8 pf leading
25
F i g .9 . 1 1
Synchronousmotors have internal emfs and reactancessimilar to that of a
generatorexcept that the current direction is reversed. During short circuit
conditions these can be replaced by similar circuit moclels eicept that the
voltage behind subtransient/transient
reactanceis eiven bv
E'lr= v" - jI"xU
E'*= v" - jI"4
prefaultcurrenlI" = _9{__
_ 136.9.= 0.77g3I 36.9"pu
0.9636
x 0.8
Voltagebehindsubtransient
reactance
(generator)
E",- 0.9636I tr + j0.45 x 0.1783I 36.9"
- 0.7536+ 70.28pu
(e.10)
(e.11)
Wheneverwe are dealing with shortcircuit of an interconnectedsystem,the
synchronousmachines (generatorsand motors) are replaced by their corres po n d i n gc i rc u i t m o c l e l sh a v i n g v o l tagebehi nclsrrhtransi ent
(transi ent)
reactancein serieswith subtransient
(transient)reactance.
The rest of the network
beingpassiverentainsunchanged.
I
Example9.3
Ir^_._
, . ...-.: . ._.
Voltagebehindsubtransient
reactance
(motor)
El,, - 0.9636/_ tr -i0.15 x 0.7783/_ 36.9"
- .70.0933
= 1.0336
pu
Theplefurult.equivalent
circuitis shownin Fig.9.l2b.Uncler
fhrrltecl
c.nclit i o n( l r i g .9 .l 2 c )
".._
A synchronousgenerator and a synchronousmotor each rated 25 MVA, I I kV
having l5Vo subtransient reactance are connected through transfbrmers and a
line as shown in F'ig. 9.12a. The transfbrmers are ratecl 25 MVA. lll66kV and
66lll kV with leakage reactance of l\Vo each. The line has a reactance of lTTo
on a base of 25 MVA, 66 kv. The motor is drawing 15 Mw at 0.9 power factor
leading and a terminal voltage of 10.6 kV when a symmetrical three-phasefault
occurs at the motor terminals. Find the subtransient culrent in the generator,
motor and fault.
Gen
|
I',l,=
1 . 0 3 i 6- r o o c ) ?1
i0.1s
Current in fault
I J = I : i + 1 , , , , = _j g . 5 6 5 3p u
Basecttrre
nt (gen/moto
I = 44q1
t Tt'
| '!
J3xll
Line
) lr;
= 1.312.2A
Now
(a) One-line
d i a g r a mf o r t h o s y s t o mo f E x a m p l e9 3
j0.1
. , , 0. 7536+"i0. 21t- 00
=0.6226_j1.6746pu
I';,-"
i 0.45
j0.1
j0.1
t"
F
t ; .6'f,1,'. 'dtI-. - 'ltrd-. -t
I
I'J= 1,312.2
(.- 0.6226- j6.8906)= (- 816.2_ jg,O4L8)A
i
1t--jtt,23gA
I
I
'. ), 1 i 0 . 1 5
r
+l
- j1.6746)= (816.4_ jL,tgt.4) A
I'J - 1,312.0
(0.6226
short circuit (sc) current computation through the
Thevenin Theorem
I
(b) Prefaultequivalentcircuit
Fi9.9"12
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(c) Equivalentcircuitduringfault
An alternate method ol' cornputing short circuit currents is through the
application of the Thevenin theorem.This method is faster and easily adopted
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
Modern Po*er SystemAnalysis
342 |
I
to ,yrtl-atic computation for large networks. While the method is perfectly
gcncrerl,
it rs illustratcdhcrc tlrrougha sinrplccxanrplc.
Consider a synchronousgeneratorfeeding a synchronousmotor over a line.
Figure 9.I3a showsthe circuit model of the systemunder conditions of steady
As a first step the circuit model is replacedby the one shown in Fig. 9.13b,
wherein the synchronousmachinesare representedby their transientreactances
(or subtransient
reactances
currentsare of interest)in serieswith
if subtransient
voltagesbehind transientreactances.This changedoes not disturb the prefault
current I" and prefault voltage V" (at F).
As seenfrom FG the Thevenin equivalentcircuit of Fig. 9.13b is drawn in
Fig. 9.13c.It comprisesprefault voltage V" rn serieswith the passiveThevenin
impedancenetwork. It is noticed that the prefault current 1" doesnot appearin
the passivcThcvcninirnpcdanccnctwork.It is thcretore t<lbe rcmcnrbcrcdthat
this current must be accountedfor by superpositionafter the SC solution is
obtainedthroughuse of the Thevenin equivalent.
Zl .liigure 9.13dshowsthe
Considernow a lault at 1,'thloughan irnpedance
Thevenin equivalent of the system feeding the fault impedance. We can
im m e d i a te lw
y ri te
V"
', l ' - jXrn + Zt
(e.r2)
Current causedby fault in generatorcircuit
AI, - -
x'd^
(xhs+x + xl^
f
(e.13)
AI^
Xle +X
(xlh,+x + xi
Postfault currents and voltages are
I{= I" + alr
I{=Postfault voltage
vf
where Av = -ix^tf
fffl:liiwith
G
(b)
G
obtainedas follows by superposition:
I" + AI^ (in rhedirectionof AI^)
vo + ( - , r xn, I f ) = v" + Av
-=
is the voltage of thefaultpoint F/
(9.15)
eJ6)
on the Thevenin
passive
-of
respect
to thereference
busct-;;;..d by theflow
faurt
incethe prefault current flowing
out of
t curent out of F.is independeniof
load
on is summarizedin the following
four
Step I: Obtain steady state solution
of loadedsystem (load flow study).
step 2: Replace reactancesof
synchronousmachines by their
subtransienU
transient values. Short circuit
uit
sources.The result is the passive
Thevenin network.
"rr
'Step
2 at the fault point by negarive
of
ies with the fault impedanc". 6o_pur"
rterest.
r areobtainedby adding results
of Steps
The following assumptionscan be
(a)
(e.14)
saferymadein SC computations
reading
:ation:
ragnitudesare I pu.
tre zero.
to actual conditions as under
normal
nity.
ruc cnanges ln current caused
by short
circuit are quite large, of the
order of 10_20
(c)
(d)
Fig.9.13 Computation
of SC currentby the Theveninequivalent
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tion 2.
Fi9'e't+
Let us illustrate the above method
b,
recalculatingthe results of Example
9.3.
F is the faultpointon
the passiveThevenin
network
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,Qrrrnmalrinal
Modern Power SystemAnatysts
9'3 for computationof
The circuit model for the systemof Example
conditionis shownin Fig' 9'14'
postfault
_ 0.9636x70.60 = - 78.565pu
E^..t+
A --r.
--!-
If sc MVA (explainedbelow) is more than 500, the
abovemultiplyingiactors
are increasedby 0.1 each. The multiplying factor for
air breakersrated 600 v
or lower is 1.25.
The current that a circuit breakercan intemr
rng voltage over a certain range, i.e.
Amperes at operating voltage
Change in generatorcurrent due to fault'
io'+ - - i2'141
Pu
, , B_ rl8.s6s
AI^="." "_ t j0.60
Change in motor current due to fault'
x j.$*i --- i6'424Ptt
At^=- 78.565
current to obtain the subtransient
To these changeswe add ttre prefault
current in machines.Thus
I'l= I" + AIr - (0.623 j1.67$ Pu
In = - I" + A I^ = (- 0.623 76.891)P u
calculated already
which are the same (and shoutd be) as
alternatively through th9 Thevenin
9.3
we have thus solved Example
is a powerful method for large
in4eecl,
theorem ond ,up.rposition. This,
networks.
9.5
Rated intemrpting MVA (three-phase)capacity
= '6ty(tifle)lrated
x 11(line)lrated
inremrpting
cunent
where V(line) is in kV and 1 (line) is kA.
Thus, instead of_computing the sc current to be
intemrpted,
' cbmpute
r --' we
three-phaseSC MVA to be intemrpted, where
SC MVA (3-phase)_ Jt x prefault line voltage in
kV
x SC currentin kA.
If voltage and current are in per unit values on
a three-phase
basis
SC MVA (3-phase) = lylp,..roul,
x 11116
x (MVA)uur.
SELECTION OF CIRCUIT BREAKERS
(e.r7)
O
' -h' Lv i' oYr rr u
s l'ru
r J r J ' ircq if i ouAu l \ / \ /
-.:l v t v AA ii -n+i^e*i,r- u
p i i i l g - c- a- p a c l t y o f a c i r c u i t b r e a k e r
is to be
. I w o o t t h c c i r c u i t b r c a k c r r a t i n g s w h i c l r r c c ; u i r c t h c c t r n l p t l t t t t i o n o f S C c u r r e n t rnurcthln (or cclualto) thc sc MVA required
to be intemupted.
symmetrical interruptirtg c:nrrent'
For the selectionof a circuit breakerfor a particular
are: rated momentarycurrent and rated
location,we must find.
using subtransient reactancesfor
the maximum possible SC MVA to be intemrpted
with
Syrnmetrical SC current is obtained by
respectto type and
by
location of fault and generating capacity (also
current irms) is then calculated
synchronous
synchronousrnachines.Momcntary
rnotorl load)
by a factor of 1'6 to accountfor
connectedto the system. A three-phasefault though
rare is generally the one
multiplying the symmetrical-o-"nory current
which gives the highest SC MVA and a circuit
breakermust be capable of
the presenceof DC off-set current'
subtransient
r'rsing
by
computed
interrurpting
is
it.
An
intcrrupted
exception
be
to
is
an
current
(line-to-ground)
LG
Symmetrical
f.ault close to a
for synchronous
synchronousgenerator*.In a simple systemthe fault location
generatorsand transientreactances
tor synchronous
reactances
which gives the
to
added
be
to
The DC off-set value
highest sc MVA may be obvious but in a large
system
motors-induction motors are neglected*.
various
possible
locations must be tried our to obtain the highestst
is accounted for by multiplying the
nava requiring Lp"ur"a
obtain the current to be interrupted
below:
SC computations.This is ilustrated by the examples
that follow.
symmetrical SC current by a factor as tabulated
Circuit Breaker SPeed
8 cyclesor slower
Iii"'n"r.; I
Multiplying Factor
1.0
5 cycles
3 cyclcs
2 cycles
1.1
1.2
t'In some recent attempts,currentscontributedby induction
circuit have been accountedfor.
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Three6.6 kv generatorsA, B and c, eachof I0o/oleakage
reactanceand MVA
ratings 40, 50 and 25, respectively are interconnected
electrically, as shown in
1.4
motors during a short
tThis will be explained
in Chapter 1l.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
15:!"'l
ModernPowerSystemAnalysis-
.";: I
{d46..t|
T
reactance
Fig. f.i5, by a tie bar through curent timiting reactors,each of I2Vo
threc-phase
A
connected.
is
it
which
to
baieclupon the rating ofthe machine
of 6'6 kV'
feeder is supplied from the bus bar of generatorA at a line voltage
^ f
n
1 a
short
Q/phase.Estimatethe maximum MVA that can be fed into a symmetrical
feeder.
the
of
end
far
circuit at the
currentscan then be calculatedby the circuit model of Fig. g.l6acorresponding
to Fig. 9.13d.The circuit is easilyreducedro rhat of Fig. 9.16b,
where
7- ( 0. 069 + j0. 138) + j0. r 2s il 00. 15+ jo. 22| j0. 44)
= 0.069+ j0.226 = 0.236173
SC MVA = Volf = V"('+') = + pu (sinceVo = 1 pu)
\Z)
Z
I
Z
(MVA)Ba."
50 =
2 1 2M V A
0.236
Tie bar
Fi g. 9.15
Sotution Chooseas base50 MVA, 6.6 kV'
Feeder imPedance
=
=(o.o6e+/0.138)pu
%
- o'1[50 = 0.125Pu
Gen A reactance
40
=
0.1
GenB reactance
Pu
Considerthe 4-bus systemof Fig. 9.17. Buses1 and 2 aregenerator
busesand
3 and 4 are load buses. The generatorsare rated l l kv, 100
MVA, with
transientreactanceof l07o each. Both the transformersare 1ll110
kV, 100
MVA with a leakagereactanceof 5Vo.The reactancesof the lines
to a base of
100 MVA, 110 kv are indicated on the figure. obtain the short
circuit solution
for a three-phasesolid fault on bus 4 (load bus).
Assume prefault voltages to be 1 pu and prefault currentsto
be zero.
(G)
G e n C r e a c t a n c e = 0* . 14 = 0.2 pu
25
o't''I tn = 0 . 1 5p u
ReactorA reactan."=
40
= 0.12Pu
ReactorB reactance
0.12x 50 - 0.24 pu
=
1
Reactor C reactance
j0.2
j0.12
Fig. 9.17 Four-bussystemof Exampleg.5
Solution Changesin voltages and currentscausedby a short
circuit can be
calculatedfrom the circuit model of Fig. 9.18. Fault current 1/ is calculated
by
systematicnetwork reduction as in Fig. 9.19,
jo.24
(0.069+i0.138)
yo = 1Z0o(
(b)
(a)
Fig. 9.16
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,^o
Moclern Power Svstem AnalYsis
I
I
= - jt.37463 pu
tt = -*=
j0.13s60
lo*
!o;E
14
D':
jo 1,),
7
I
r - ---rnTl
I
4+
_ jo.2
[ 0t ,
i0.1
t,= rsx i: i3;:: = - j3 837or
pu
j0.37638
A,
l,zs
ib.rs
2
Izt
t,- ti,,
12=
- r,
{: i:::: = - j3.53762pu
'' x j0.37638
10.15
Let us now computethe voltagechangesfclr busesl,2and 3. From Fig.
9 .l 9 b , w c g i v c
AVr - 0 - ( / 0. 15)( - j3. 8370r ) = - 0. 57555pu
AV, = 0 - (iO.l.s)(- .i3.53762)= - 0.53064pu
\
Now
i ro'rs
o)
r -fI
l,
, , t , . ( .^-
iots
l
4T
( tvl=to
*
irt
(a)
V ) ,= l +
F'tt
r l r
' ' r),} o' rs
i o.t\P
II
i0 1'':,
tl
V , ' = l + l V t = 0 . 4 2 4 4 .p5u
i.0.11
'AtiA'-
t-- -
(b)
(
2 i I 0 6 ' )I
.
/ l-t=
-l
(. )vl = t.o
ir
AVz=0.46936pu
v , J- v J
= J0'17964
pu
,of tr.o.t
N ow
fr'
AVy= 0 - [(/0.15)
(- j3.83701)
+ Q0.15)
(/0.17964)l
= - 0.54fi(r0
pu
V t t = I - 0 . 5 4 8 6=
0 0.4514
pu
(e)
(c)
l ' ' l
+
I
,-j;0.''uuuo
l
( ) v?= t.o
Tr
(
,-l
-J 10.04166
,
vlo= o
The determinationof currentsin the remaininglines is left as an exerciseto
tltcrrr'ldcr.
Short circuit study is completewith the computationof SC MVA at bus 4.
(SC MVA)^ = 7 .37463 x 100 = 737.463MVA
It is obvious that the heuristic networkreductionprocedureadoptedabove is
not practical for a real power network of even moderate size. It is, therefore,
essentialto adopt a suitablealgorithmfbr carryingout short circuit study on a
digital computer.This is discussedin Sec.9.6.
tt*)
-i
=
(
\.. ) V ? 1 . 0
t
9.6
ALGORITHM
FOR SHORT CIRCUIT STUDIES
.
Ill
I
Fig. 9.19 Systentaticreductionof the network of Fig' 9'18
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So far we have carriedout short circuit calculationsfor simple systemswhose
I n t his scct ionwc ext cndour st udyt o
l l i tssi vct t ct wor kscanbe easilyr educed.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Power System Analysis
i*.3€0-l
Symmetr.ic
- , . . . . . , *", , ; alFaultAnat lr sis
_
large ,tyrt"-r. In orcler to apply the four steps of short circuit computation
developedearlier to large systems,it is necessaryto evolve a systematicgeneral
algorithm so that a digital computercan be used.
_--.rr.vv\,
network.
rrrv u(rr vurraegcs oI mls
Now
4V = Z"urJf
(e.20)
where
.7
otn
-'l '-
I
i = busimpedance
matrix of the
I
Znn) passiveTheveninnetwork
Gen 2
Fig. 9.20 n-bussystemundersteadyload
u// = bus current injection vector
Since the network is injected with current -
Consideran n-bus systemshownschematicallyin Fig.9.20 operatingat steady
load. The first step towards short ciicuit computation is to obtain prefault
voltages at all buses and currentsin all lines through a load flow study. Let us
indicate the prefault bus voltagevector as
0
rf
I, :
(e.22)
-I', f
:
(e.18)
Substituting Eq. (9.22) in Eq. (g.20),
Let us assume that the rth bus is faulted through a fault impedance Zf . The
postfault bus voltage vector will be given by
(e.1e)
where AV is the vector of changesin bus voltagescausedby the fault.
As'step 2, we drawn the passiveThevenin network of the system with
generatorsreplacedby transient/subtransient
reactanceswith their emfs shorted
( F ie. 9 .2 1 ).
1/ only at the rth bus, we have
0
:
/-
V { u r = V B u s+ A V
e.2D
AV, = - ZrJf
By step 4, the voltage at the nh bus
we have for rhe rth bus
under fault is
v!= vor+avo,- vor- Z,Jf
However, this voltage must equal
Vd = 7f 1f
We have from Eqs. (9.23) and (g.24)
(e.23)
(e.24)
zftf - vo,_ z,Jf
or
f=
V:
Zr, + Zf
(e.2s)
At the rth bus (from Eqs (9.20)and(g.22))
AV, = - Z,Jf
Fig. 9.21
Network of the system of Fig. 9.20 for computingchanges in
bus voltages caused by the fault
v{= v?- Z,Jf, i = 1,2, ...,
substituting
for // from Eq. (9.25).we have
vI= vf - : zl';rv!
z*+L
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(e.26)
(e.27)
I -?53
rca
|
Siimmetrica!Fault ,Analysis
rtllodernPower System Analysis
352 i
First of all the bus admittancematrix for the network of Fig. 9.18 is formed
as fol l o ws:
Fori=rinEq.(9.27)
(e.28)
In the above relationship V,o'r, the prefault bus voltagesare assumedto be
known from a load flow study. Zuu, matix of the short-circuit study network
of Fig. 9.21 canbe obtainedby the inversionof its furr5matrix as in Example
9.6 or the Zru, building algorithm presentedin Section9.7.It should be
observed here that the SC study network of Fig. 9.2I is different from the
correspondingload flow study network by the fact'that the shunt branches
correspondingto the generatorreactancesdo not appearin the load flow study
network. Further, in formulating the SC study network, the load impedancesare
ignored, these being very much larger than the impedances of lines and
ginerators. Of course synchronousmotors must be included in Zuur tormulation for the SC study.
Postfault currents in lines are given by
f u= yu (vri- vt)
Q.29)
For calculation of postfault generatorcurrent, examine Figs. 9.22(a) and (b).
From the load flow study (Fie. 9.22(a))
Prefault generator output = PGr+ iQci
I
--r I
-t-
-l
Y trL. > = Y
L tr t = _ -
I
-r
- - j28.333
175.000
j0.2
Y3= Yy =
1
-l
- i6. 667
;r*
-1
Yrq= Yq=
J,3.l
= i10.000
Yzz=
. : - + +j0.1s
- + + + jj.r
:^
j0.15
"
j0. 1
-l
- i6. 667
F;
'Y3-3. -=
=- j28.333
j10.000
Y z t =Y n = + =
Yzq= Yn=
j0.2
I
I
=-i16.667
+'
j0.15 jo.1
Ytq= Yqt= 0.000
I
vYqq=
,-
*-
I
--
,.""
i76'667
(b)
(a)
Fi1.9.22
f", = &#;
v,u
(prefaultgenerator
output= Pci + iQci) (9.30)
E'Gi = V, + jXt"/t)",
(9'31)
From the SC study, Vf ,is obtained. It then follows from Fig. 9.22(b) that
rrc,=tfr!
(e.32)
iir;rnr;;.;
I
To illustrate the algorithm discussedabove,we shall recomputethe short circuit
solution for Example 9.5 which was solved earlier using the network reduction
technioue.
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By inversion we get Z",tt as
j0.0s97 j0.0719
j0.0903 j0.0780
j0.0780 j0.13s6
jo.o719 j0.0743
Now, the postfault bus voltagescan be obtainedusing Eq. (9.27) as
V {r = r V Zo - Zo t oo VaP
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,2.tr!A
JJ'
|
^/lndarn
lYlVVVlll
I
Pnrrrar
I Vltvl
Qrratarn
Vtvlvlll
Analrrcic
E qau t
malriaal
| | tglt
tvql
, rlrsrtvre
I
The prlfault conditionbeing no load, V0r= Voz= V03- Voo= 1pu
-r
, J
-
_
r . o - i 9 9 1 9 x9 1 0= 0 . 4 2 4 8 p u
1.000
2,,{or Zzz)
-
1.00 = - j I L 0 7 4 I 9 7
pu
j0.0903
i0.1356
- o?':o r
vrz=
L
L v:
z
r,',
By Inventing
= 1.0 -
oi 0 .-0 7:1: 0 x 1 . 0 = 0 . 4 6 9 8 p u
Y"u"
/nus = Yrus Vsus
j0.1356
7
or
Vsus= [Y"ur]-t /eus = Znus/eus
Zoo
or
Zsvs= [Yuur]-t
v{= v\ - 1* vf
- : : _ : 1 . 0= 0 . 4 5 2 1
= 1 . 0- ' i0.0743
pu
j0.r3s6
vi = o'o
The sparsity of fsu, may be retained by using an efficient inversion technique
[1] and nodal impedance matrix can then be calculated directly from the
factorized admittancematrix. This is beyond the scope of this book.
Current
Using Eq. (9.25)we can obtainthe fault currentas
Iniection
V 1: 2 1 1 \ * Z t z l z 1 . . . * Z n I n
0.4248-0.4521- j o ' 1 8 2
Pu
i0.15
-v{
-t rr r z = v,r
- ' _ - ' 2 - : - 0.4248-0.4698_ j 0 . 2 2 5p u
zrz
.i0.2
,r rt4-
4q
_ o.4z4v-o
jo'l
- v{
-o
q - - -0.46e8
.I irq = ' u'l
=
zz+
- i4.248pu
- j3.132pu
i0'15
(e.34)
v 2 : 2 2 1 1 1 + 2 2 2 1 2+ . . . + z z n l n
Thesevaluesagreewith thoseobtainedearlierin Example9.5.Let us also
the shortcircuitcurrentin lines1-3,1-2, l-4,2-4 and2-3.
calculate
yr -v{
Technique
Equation (9.33) can be written in the expandedform
=, j7.3j463pu
7r= .r-^0.9^o=j0.13s6
v,r -v{
[,r1 2 =- : z.n
(e.33)
V, : Zntll * 2,,21r* .. .+ Z,,nl,,
It immediately follows from Eq. (9.34) that
z -- v ' l
"ii
I r l , r : r z: . . . : r n = o
(e.35)
I1-r0
Also Zi, - Ziii (Znvs is a symmetrical matrix).
As per Eq. (9.35) if a unit currentis injectedat bus (node)7, while the other
busesere kept oponcircuited,the bus voltagesyield the valuesof theTth column
of Zuur. However, no organized computerizabletechniquesare possible for
finding the bus voltages.The technique had utility in AC Network Analyzers
'where the bus voltagescould be read by a voltmeter.
--r v{ -v^ = 0.4698-0.4521= I-Izt=tiutt
./0.177Pu
TOFor the exampleon hand this method may appearmore involved compared
to the heuristicnetworkreductionmethodemployed in Example9.5. This,
however, is a systematicmethod and can be easily adopted on the digital
computer for practical networks of large size. Further, another important feature
of the method is that having computed Zu's, we can at once obtain all the
required short circuit data for a fault on any bus. For example,in this particular
system, the fault current for a fault on bus I (or bus 2) will be
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Consider the network of Fig. 9.23(a) with three buses one of which is a
reference.Evaluate Zsus.
Sotution Inject a unit current at bus I keepingbus 2 open circuit, i.e., Ir = I.
andIr= 0 as in Fig. 9.22(b).Calculatingvoltagesat busesI and 2, wehave
Ztt=Vt=7
Zzt=Vz=4
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.f
Modern PowerSystemAnalysis
356 |
Now let It = 0 and 12= 1. It similarly fbllows that
Ztz=Vt=4= Zn
l7
!, f5l
, 4r,
the dimension of Zsu5 goes up by one). This is type-2 modification.
41
Zvus= 6l
l+
Becauseof the above computationalprocedure,the Zru, matrix is
referredto as the 'open-circuitimpedancematrix'.
Z"vs Building
SymmetricalFault Analysis
Algo4ithm
3. Zuconnectsan old bus to the referencebranch(i.e., a new loop is formed
dimensionof 2o,," doesnot change).This is type-3modification.
4. Zuconnectstwo old buses(i.e., new loop is formed but the dimensionof
Zuu, does not change).This rs type-4 rnodittcation.
5. Zu connectstwo new buses(Zeus remains unaffectedin this case). This
situation can be avoided by suitable numbering of busesand from now
onwardswill be ignored.
Notation: i, j-old buses; r-lsfelence bus; k-new bus.
Type- 1 Modification
It is a step-by-stepprogrammabletechniquewhich proceedsbranch by branch.
It has the advantagethat any modification of the network does not require
complcterebuildingof Z"ur.
Consider that Zrur has been formulated upto a certain stage and another
branchis now added.Then
Zrur (old)
Zo:branch impedance
Figure 9.24 shows a passive (linear) n-bus network in which branch with
impedance2,, is addedto the new bus k and the referencebus r. Now
V*= ZJ*
Z r i = Z * = 0 ; i = 1 , 2 . . . . .t t
Zsus (new)
Upon adding a new branch, one of the following situations is presented.
Zm = Zu
Hence
Zsvs (new) -
s,,"(old)
(e.36)
Passive linear
n-bus network
Fig. 9.24 , Type-1modification
Fig. 9.23 Currentinjection
methodof computingZru.
t . 26 is added from a new bus to the referencebus (i.e. a new branch is
added and the dimension of Zry5 goes up by one). This is type-I
modificution.
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Type-2
Modification
Zo is addedfrom new bus ft to the old bus7 as in Fig. 9.25.It follows from this
figure that
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
35S;,1
MorJern
Powcr
Srrctarn
anatrroio
lr
Zrj
1O_
zzj
n
(e.38)
Zsvs(old) I :
Passivelinear
n-bus network
oI
lzyziz...zi,I zii + z, )lI*
(e'3I )'
Eriminate
{ inthe
3":';,,iiy';;;?::"i"t i.I" a;y>;i:ation
Fig. 9.25
Type-Z modification
= Zr,I*+ ZiJr+ Zlzlz+... + Zii ei* I) + ... + Zinln
Rearranging,
(\1Ir+ Zi2Iz+ "' + ZlnI)
Z l z l z+ . . . + 2 , , 1 , + . . . +Z i , l n + ( Z i i + Z ) l k
,,= lr^ h(z*
* 1rr]a
lr,,
rzuz,,>)r,
"h
+ +
lr^- ^i{zu
Zti
2,,1)r, (s.4t)
Equation (9.37) can be written in matrix form as
Zzj
:
(9'39)
(e.40)
V; = 2; 111+Z, r I , + "' + Z'nI n+ Z; / r
SubstitutingEq. (9.40) in Eq. (9.39)
Consequently
.
(e.37)
zjj + zb
Type-
Modification
zu connectsan old bus (l) to the referencebus (r) as in Fig. 9.26. This
case
follows from Fig.9.25 by connectingbus ft to the referencebus r, i.e. by
setting
|t"f
: ltz't"Zv)
(new)= znvs(old
ZsLr5
;+;l
"jjT"ulz,, l
Znj
IYpe-3
1
Now
Vo= Zdo + V,
V*=4lt+
'o= -12
or
(e.42)
Modification
two old busesasin Fig. 9.27. Equationscanbe writtenas follows
zo connects
for all the networkbuses.
v*=o'
(l;+ lp
J
Passive
linear
n-buS
network
Fig. 9.27 TYPe-4modification
Fig. 9.26 Type-3 modification
"' + Zr i ( I i + / ) + ZU Q i- /o)+ ...+ ZiJ"(9.43)
Similar equationsfollow for other buses.
Vi= Z; 11,+ Z, lr +
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Modern power System Analysis
360 |
SymmetricalFault Analysis
I he voltages of the buses i and j are, however, constrained by the equation
(Fig. 9.27)
V ,= Z ,rl o + V i
or
(e.44)
2 1 1 1+1 \ 2 1 2 + . . . + Z i t ( 1 , + I ) *
Z i i ( I i - I ) + . . . + zi,In
= Z t l * + Z , r l , + Z , r l "+ . . . + Z l U , + I
_I
earTangtng
0 =(Zit- 21) Ir+ ... + (Zti- Z) Ii+ eu- Z) Ii
+...+ (Z* - 21,) In + (Zu * Zii * Zii - Zi - Z) 11,
Step I: Add branch 2,, - 0.25 (from bus I (new) to bus r)
ZBUS= t}.25l
e.45)
vl
(ii)
zsvs=',lZ:11,
31i]
Step 3: Add branch ,rt = 0.1 (from bus 3 (new) to bus I (old)); type-Z
modification
lo.2s o.zs 0.2s1
V
Step 4:
0
Eliminating 1o in Eq. (9.46) on lines similar to whar was done
modification,it follows that
l'"
(new) = Zsuts (old) Z131,s
zb+zii*Zii-zzij
lZ,,
lzit - z1t)... (zi, - zi))
with the useof fbur relarionshipsEqs (9.36), (9.37),(9.42) and (9.47) bus
impedancematrix can be built by a step-by-stepprocedure(bringing in one
-hci ng
bra n c h a t a ti n l e ) a s i l l trs tra tc 'itln l i xrrrnpl e9.8. Thi s pror.cdrrr-c
a
mechanicalone can be easily computerized.
When the network undergoeschanges,the modiflcationproceclurescan be
cttrploycclto rcviscthc bLrsirtrpcdance
rnatrixul'thc nctwol'k.Thc opeling o1
a line (Ztt) is equivalentto adding a branch in parallel to it with impedance
- 2,, (see Example9.8).
(iii)
zsvs= o.zs o.3s o.2s
|
|
lo.2s 0.2s 0.3s..l
(9.46)
vn
(i)
Add branch Zzt = 0.1 (from bus 2 (new) to bus I (old)); type-Z
Step
Collecting equationssimilar to Eq. (9.43) and Eq. (9.45) we can write
I fOt
Add branch zz, (from bus 2 (old) to bus r); type-3 modification
10.2s o.zs 0.25'l
= o.zs 0.3s 0.2s|
Zet)s
|
o'3s
+o
Lo.2so.2s o3sl
[0.25-l
| o.tsI to.zs0.3s0.251
" Ln.trj
[0.14s8 0.1042 0.14s8-l
= I o.ro+z o.r4s8 o.lo42|
fo.tott o.ro42o.24s8l
Step5: Acldbranchiz-r= 0.1 (from bus 2 (olct)to brrs31old)): type-4
modification
[o.l4s8 0.1042 0.l4s8l
= o.'oo, 0.1458o.to42lzsvs
| 0 . 1+ 0 . 1 4 5+80 . 2 4 5 8 - 2 x O . l O 4 2
l
)
10.l4s8 O.tO42 0.24s8
1042.l
[-0
= 0.0417 [-o.to+z 0.0411 -0.0417]
I
|
[-o.o+tz-l
j Example9.8
For the 3-busnetworkshownin Fig. 9.28build Zsus
I
I
I
!,-0.25l,'
(-I
l
r
z
'irl9 l
I
l
L-,trili -, -- 6"11-.
I
3
0.1
o.t
I
Ref bus r
I
i,J
or.
I
0.1103 0.12501
lo.r3e7
= 0.1ro3 o.I3e7 0.1250
|
I
| 0.1250 0.1250 0.17sol
- 0.I
Oltenirtgo tinu (line 3-2): This is equivalentto connectingan intpeclance
betweenbus 3 (old) and bus 2 (old) i.e. type-4 modification.
Zsus=Zuur(olcl)
(-01)+ol?5#
Fig. 9.28
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ModernPower System Anatysis
filf.lid,
-r-r-^r
:)ymmelrlual
[t 0.0147-l
-0.0147 l 10.0147
- 0.01470.0s001
|
|
L o.osoo_J
r^..tr
raull
Ana*raio
rurqryo's
ffir
r**
Now
I
- | 0.1042 0.14s8 0.1042l; (sameas in step 4)
0.1458 0.1042 0.2458|
and
i' li =2 (- F[ ^Z E - ) = 0 . 2 8 6
zu)
These two voltages are equal becauseof the
network
(c) From Eq. (9.29)
symmetry of the given power
Iri = Y,j(vl - vl)
For the power system shown in Fig. 9.29 the pu reactancesare shown therein.
For a solid 3-phasefault on bus 3, calculate the following
(a) Fault current
(b) V\ and vt
(c) It,r, I'\, uxl Il,
(d) 1fr, andIf,
Assumeprefaultvoltageto be I pu.
0.2
0-.09 1
|ffr
JI-
- - i2.86
(d)
//
o '' \
\
/
, ,/
\..
.
l
3 '
'l
Solution The Thevenin passivenetwork for this systemis drawn in Fig. 9.28
with its Zru, given in Eq. (iv) of Example 9.8.
(a) As per Eq. (9.25)
1f -
V:
2,, + Zl
Y o - : j0.175
. _ 1 - -- - j s . j l
zn
(b) As per Eq. (9.26)
rr.f
v
i -
But
,/'0.,
Fig. 9.29
or
As per Eq. (9.32)
,r
r Gr-
\,
Jt -
I\t=t|t= ro.186-o)
fr
and
0.1
F
c) | i-,,
=g
- 0.286;
Irtz=+(0.286
'
j0.1
E' or-vrf
ixtic + ixr
E'Gr= 1 Pu (Prefaultno load)
IL.=
Gt-
1-0'286--=- j2'86
io.2+io.os
SimilarlY
I f cz= i2. 86
LEMS
PROB
anclresistance5 ohms is suddenly
9 . 1 A transmissionline of inductance0.1 H
in Fig' P-9'1' Write the
as
shortcircuitedat t =0 at the bar end shown
the value of
approximately
Find
i(r).
expressionfor short circuit current
current)'
momentary
(maximum
the first current maximum
currentntaximum occurs at the sametime as
[Hint: Assumethat the first
ttrcuit current')
the first current maximum of the symmgtricl.:non
vl' - --Zt--yu
zrr+zl'r
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564,'l
power System Analvsis
rr]logern
I
errrnrnorrinal Farrlt
Analvsis
-"-'| --"
J
vyrrrrlvt"vE'
I
geS
I
I
fl
'i
0.1H
SO
64-A-' v^/v\
i
On(100n f + 15.)
u = 1 Osi
Fig. p-9.1
Fig. P-9.5
9 '2 (a) What should the instantof short circuit be in Fig. p-9.1
so that the
DC off-set current is zero?
(b) what should the instantof short circuit be in
Fig. p-9.1 so rhar the
DC off-set current is maximum?
9 . 3 For the system of Fig. 9.8 (Example 9.2) find the symmetrical currents
to
be interrupted by circuit breakersA and B for a fault ar (i) p
and (ii) e.
9 . 4 For the system in Fig. p-9.4 the ratings of the various componenrs
are:
Generator:
25 MVA, 12.4 kV, l\Vo subtransientreactance
Motor:
20 MVA, 3.8 kV, l5%o subtransientreactance
TransformerT,:
25 MVA, 11/33 ky, gVoreactance
Transfbrmer Tr:
20 MVA, 33/3.3 kV, I \Vo reactance
ansient
9 . 6 A s y n c h r o n o u s g e n e r a t o f r a t e d 5 0 0 k V A , 4 4 0 v , 0 . l p u s u b t rpowel
at 0.8 lagging
."u"'tun". is supptying a passiveload of 400 kW
a three-phasefault
for
factor. Calculate tfr" in'itiui symmetrical rms current
at generator terminals.
a line through a circuit
9 . 7 A generator-transformerunit is connected to
breaker.The unit ratings are:
pu, Xi= o,2o pu and
Generator: 10 MVA, 6.6 kV; X,,d= 0.1
=
0'80
X,r
Ptt
Transformer:10MVA,6.9133kV,reactance0.08pu
of 30 kv, when a threeThe systemis operatingno load at a line voltage
breaker'Find
phasetault occurson ih" tin" just beyondthe circuit
breaker'
the
in
current
rms
symmetrical
(a) the initial
DC off-set currenr in rhe breaker,
iui tt. maximum possible
rating of the breaker'
current
(c) the momentary
kVA'
(d) the current,o U. intemrpted by the breakerand the intemrpting
Line:
20 ohms reactance
The systemis loadedso that the motor is drawing 15 Mw at
0.9 loading
power factor, the motor terminal voltage being 3.1
kv. Find the
subtransientcurrent in generatorand motor for a fault at generator
bus.
lHint: Assumea suitablevoltagebasefbr the generator.The voltagebase
for transformers,line and motor would then be given by the transforma_
ti 0 n ri l ti < l sF. rl r c x a n rp l ci,f' w c chooscgcrrcrator.
vol tagcbascas I I kv,
the line voltage baseis 33 kv and motor voltagebaseis 3.3
kv. per unit
reactances
are calculatedaccordingly.]
T
,1
T2
Fig. p-9.4
9 . 5 Two synchronousmotors are connectedto the bus of
a large system
through a short transmissionline as shown in Fig. p-9.5. The
ratings of
vanouscomponentsare:
Motors (each):
1 MVA, 440 v,0.1 pu transientreactance
Line:
0.05 ohm reacrance
Large system:
Shortcircuit MVA at its bus at 440 V is g.
when the motors are operating at 440 v, calculate the short
circuit
cuffent (symmetrical) fed into a three-phasefault at motor bus.
www.muslimengineer.info
9.8
and
(e) the sustainedshort circuit current in the breaker'
MVA at 1 1 k v , 0 . 8
The systemshown in Fig' P-9'8 is delivering50
ils infinite.
laggirrgp0w0rl.itctttrillttlltbrrswhichrtriryberegirrdecl
Particularsof various systemcomponentsare:
60 MVA, 12 kV' X,/ = 0'35 Pu
Generator;
= 0'08 pu
Transtbrmers(each): 80 MVA, 12166kV' X
Reactance12 ohms' resistanc:negligible'
Line:
breakersA and B will
Calculatethe syrnmetricalcurrentthat the circuit
fault occurring at
three-phase
a
of
event
the
in
be called upon to interrupt
B'
F near the circuit breaker
Line
Fig. P-9.8
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
tgdern power Svstem Anatvsis
t
.
9'9 A two generatorstaticln
suppliesa f'eederthrougha bus as shown in Fig.
P-9'9' Additional power is fed to the bus throJgh
a transfoilner from a
large system which may be regardedas infinite.
A reactor X is included
*Sf-,S
and,1"-!T to limit theSCrupruring
capacity
of
:,'jT:^"1^:t::1u1.',f"'Ter
4v<rl\er
rr ru JJJ rvtyA
(Iault close to breaker).
inductive reactanceof the reactor required. system
bus 2 of the system.Buses 1 and 2 are connectedthrough a transformer
and a transmissionline. Per unit reactancesof the various components are:
Generator(connectedto bus bar 1) 0.25
Find the
data are:
Generator Gr:
25 MVA, 757o reactance
Generator Gr:
50 MVA, 20Vo reactance
Transformer Tr:
100 MVA;8Vo reacrance
Transformer T2:
40 MVA; l\Vo reactance.
Assumethat all reactancesare given on appropriate
voltagebases.choose
a baseof 100 MVA.
ti
-
Tz
0.28
Transmissionline
The power network can be representedbi a gene ator with a reactance
(unknown) in series.
With the generatoron no load and with 1.0 pu voltage at each bus
under operating condition, a three-phaseshort circuit occurring on bus 1
causesa current of 5.0 pu to flow into the fault. Determinethe equivalent
reactanceof the power network.
9.12 Considerthe 3-bus systemof Fig. P-9.I2. The generatorsare 100 MVA,
with transientreactanceIjVo each. Both the transforners are 100 MVA
with a leakage reactanceof 5%t. The reactanceof each of the lines to a
base of 100 MVA, 110 KV is 707o.Obtain the short circuit solution for
a three-phasesolid short circuit on bus 3.
Assume prefault voltagesto be 1 pu and prefault currents to be zero.
G)
Fig. p_9.9
T1
9'10 For the three-phasepower network shown
in Fig. p-9.10, the ratings of
the various componentsare:
^T" T
'2
f l rn kv\-x-x-x-"
1 1 /1 1 0k V
j0.1
Fault
Fig. p-9.10
Fig. P- 9. 12
Generators Gr:
100 MVA, 0.30 pu reactance
Gz: 60 MVA, 0.18 pu reactance
Transformers(each): 50 MVA, 0.10 pu reactance
Inductive reactor X: 0.20 pu on a base of 100
MVA
Lines (each): 80 ohms (reactive); neglect resistance.
with the network initiaily unroadedand a rine
vortageof 110 kv, a
symmetrical short circuit occurs at mid point
F of rine r-.
calculate the short circuit MvA to be intemrpted
by the circuit
breakersA and B at the ends of the line. what
would these values be, if
the reactor X were eliminated?Comment.
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9.13 In the systemconfigurationof Fig. P-9.12,the systemimpedancedata are
given below:
Transient reactanceof each generator= 0.15 pu
Leakage reactanceof each transformer = 0.05 pu
Ztz = i0.1, zp - i0.12, 223= 70.08 Pu
For a solid 3-phasefault on bus 3, find all bus voltagesand sc currents
in each component.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
365 j
ModernPower SystemAnalysis
t
ll
9.14 For the fault (solid) location shown in Fig. P-9.I4, find the sc currentsin
lines 1.2 and 1.3. Prefaultsystemis on no-loadwith 1 pu voltageand
prefault currents are zero. Use Zuu, method and compute its elementsby
the current injection technique.
I
'r-[b- 111110kV 0.5 pu reactance
-l
Q9
0.1 pu reactance
Fig. P-9.14
10.1
NCES
REFERE
B oo k s
l.
B I r r w t t . H . L ) . . S o l t t t i o t r t t . f I . a r , g , eN e t w r t r k l t y M u t r i r M c t l u t d s , W i l c y , N c w Y o r k ,
tL)'7
5.
2 . f . l c t r c t t s w a r t d uj'.,i t . ,
'i'cxtbook
M t t t i t : t ' t t i ' t n v ' u ' 5 ' v s ' / e l l , l ,i r r t c r r t a t i t t r t a j
eorrrplny,
New York, 1971.
'3.
Stagg, G.W. and A.H. El-Abiad. Computer Methods in Power SystemsAnalysis,
McCraw-Hill Book Co., Ncw York, 1968.
4. Anderson, P.M., Analysis of Faulted Power Systems,Iowa State Press,Ames, Iowa,
-r.
1973.
( ' f r r t ' k c .l - , . . ( ' i n ' t t i t A r r t t l . t , , to' i.,ft' A l t t r n u l i t t , qC t r r r c r t l l ' t t w r r S \ , . r ' / r , r l .Vr .o l . I ,
New York. 1943.
6 . S t c v c r r s o n ,W . D . J r . , E l e t r r c n t .os f P o w c r S y s t e m sA n u l y - s i s , 4 t hc d n , M c ( i r z
Ncw York, l9tl2.
P ap e r
"7. Brown. H.E. et al. "Digital Calculation o[ Three-Phase Short
Circuits by Matrix
INTRODUCTION
In our work so far, we have considered both normal and abnormal (short
circuit) operettionsof power systern Llnclercornpletely balanced (symmetrical)
c o l t t l i t i o t t s . L J l r r l c rs u c l t o l t c t ' a t i o t t l t c s y s t c r l r i r u l t c d a n c c si n c a c l r p h a s c a r e
identical and the thrce-phasevoltages and currents throughout the system are
c o l l t p l c t c l y h l t l i t n c c t l .i . c . t h c y h u v c c r p r i r l n u r g r r i t u t l c si n c l r c h p h u s c u n d t r c
prtrgrcs.sivclydisplaced in tirne phase by 120' (phase u leads/laesphase b by
1 2 0 ' a n d p h a s e i r l e a d s / l l g sp h i t s e c b y 1 2 0 " ) . I n a b : r l l n c e d s y s t e n r , l n a l y s i s
can proceed on a single-phase basis. The knowledge of voltage and current in
one phase is sufficient to completely determine voltages and currents in the
other two phases. Real and reetctive powers are simply three times the
corresponding per phase values.
Unbalanced system operation can result in an otherwise balanced system due
to unsymmetrical fault, e.g. line-to-ground fault or line-to-line fault. These
f'aultsatrc,itt lhct, ol' lnot'econlnroll occurrence+thalt the syrnntetric:al(threephase) fault. Systern operation may also become unbalanced when loads are
unbalanced as in the presence of lar-ee single-phase loads. Analysis under
unbalancetl conditions has to be carried out on a three-phase basis. Alternatively, a nlore convenient method of analyzing unbalanced operation is through
symtnetrical components where the three-phasevoltages (and currents) which
may be unbalanced are transfbrmed into three sets of balanced voltages (and
Methods", AIEE Trani., 1960, 79 : 1277.
* Typical relative frequenciesof occurrenceof different kinds of faults in a power
syst(:llt(irr ordcr ol' dccrcasingscvcrity)are:
Three-phase(3L) faults
5Vo
Double line-to-ground(LLG) faults
lj%o
I)outrlc lirrc (l-1,) lirults
| 5t/o
Single line-to-ground(LG) faults
7jVo
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
symmetrica!compo1gnlg
currents)called symmetricalcomponents.Fortunately, in sucha transformation
the impedancespresentedby various power system elements (synchronous
generators,transformers,lines) to symmetricalcomponentsare decoupledfrom
each other resulting in independentsystem networks for each component
anced set). This is the basic reason for the simolicitv of the svmmetrical
component method of analysis.
COMPONENT
TO.2 SYMMETRICAL
TRANSFORMATION
A set of three balancedvoltages (phasors)Vo, V6, V" is charactertzedby equal
magnitudes and interphasedifferencesof 120'. The set is said to have a phase
sequenceabc (positivesequence)if Vulags Voby l2O" and V. lags Vuby I20".
The three phasorscan^thenbe expressedin terms of the referencephasor Vo as
Vo = Vo, V6 = a"Va, V, = aVo
where the complex number operator cr is defined as
! .a?.,{fi-i
above.Thus
Vo-- Vot *
Voz I
YbL-
vbz
Vr= VrI *
Vrz *
b-
Voo
(10.s)
Vro
( r0.7)
The three phasor sequences(positive, negative and zero) are called the
symmetricalcomponentsof the original phasorset Vo, V6,V,. The addition of
symmetricalcomponentsas per Eqs. (10.5) to (10.7) to generate Vo, Vr, V, is
indicatedby the phasordiagramof Fig. 10.1.
sL - air20"
It has the following properties
,*2:ei24o':e-ilAo"
_*
V6fo?V11
V61=crV61
(o')* : o
(10.1)
a3:l
l+ala2:0
If the phase sequenceis acb (.negativesequence),then
V o = V o , V u = tu V o ,V r= & V o
Thus a set of balancedphasorsis fu'lly characterizedby its referencephasor
(say V,) and its phasesequence(positive or negative).
Suffix 1 is commonly usedto indicatepositive sequence.A set of (balanced)
positive sequencephasorsis written as
Vo1, V61- &Vu1, Vrr = aVot
(r0.2)
Similarly, suffix 2 is used to indicate negativesequence.A set of (balanced)
negative sequencephasorsis written as
Vo2, V62=
dVn2, Vrz=
Q'Voz
(10.3)
A set of three voltages(phasors)equal in magnitudeand having the samephase
is said to have zero sequence.Thus a set of zero sequencephasorsis written
AS
Vng, V6g =
Vo1, Vr1 = Vo1
(10.4)
Consider now a set of three voltages (phasors)Vo, V6, V, which in generalmay
be unbalanced.According to Fortesque'stheorem* the three phasorscan be
components
to obtain
additionof the symmetrical
Fig. 10.1 Graphical
in general)
the set of phasorsV", V6,V" (unbalanced
Let us now expressEqs. (10.5) to (10.7) in termsof referencephasors Voy
Vo2and Voe.Thus
(10.8)
Vo= Vot* Vozl Voo
Vu= a.2Vor+ aVor*
Voo
c-Vol+ o2vor*
Voo
V"=
These equations can be expressed in the matrix form
* The theorem is a general one and applies to the case of n phasors [6],
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(10.e)
(10.10)
ffi#d
-__r_:
ModernPo*et sytttt Analysi,
(10.11)
Io = il'
(10.19)
I' = A-rI'
(10.20)
and
where
Vp = AV,
v, =
l ''l
(10.12)
1,"1
= vectorof originalPhasors
Of course A and A-r are the same as given earlier.
In expanded form the relations (10.19) and (10.20) can be expressed as
follows:
(i) Constructionof current phasorsfrom their symmetricalcomponents:
l'u |
Lv,)
Iu'l
Io= Iot * Ioz *
% = | Voz | = vector of symmetrical components
Lu"'.1
['
A- 1 "
I r-l
o nt
I
Ia= o2lott
(10.13)
1
Iot= *
ComputinE ,{r
(10.14)
d
o']
o - ' = * l r c r 2a I
'Ltr rl
loo
(r0.1s)
e"+ du+ o2lr)
(ro.24)
(ro.2s)
t,r.=
and utilizing relations (10.1), we get
[t
dozr
;
(Io+ azlu+ aI,)
i
;
I r o = , Q o +I u + I r )
We can wrire Eq. (10.12) as
V, = 4-'V,
(10.21)
(r0.22)
(r0.23)
Ioo
Ir= dot+ azlor* Ioo
(ii) Obtaining symmetrical components of current phasors:
o2 t-l
Ia
[r,'I
Ir=lIu
I,=lIo,
l;ana
I
LI,)
Lr,o_i
.
(r0.26)
Certain observationscan now be made regarding a three-phasesystem with
neutral return as shown in Fig. 10.2.
In expandedform we can write Eq. (10.14) as
Vor=
I
c-Vu+a2VS
(10.16)
voz= ! fr"+ ozvu+o%)
3
(10.17)
Vuo=
,<V"+
I
;
V,+ Vr,* r,)
V""
(10.18)
Equations (10.16) to (10.18) give the necessaryrelationshipsfor obtaining
symmetricalcomponentsof the original phasors,while Eqs. (10.5) to (10.7)
give the relationships for obtaining original phasors frorn the symmetrical
components.
The symmetrical component transformationsthough given above in terms of
voltageshold for any set of phasorsand thereforeautomatically apply for a set
of currents.Thus
www.muslimengineer.info
Vao
ln
Fig. 10.2 Three-phase
systemwith neutralreturn
The sum of the three line voltages will always be zero. Therefore, the zero
sequencecomponent of line voltages is always zero, i.e.
=
vobo
trr**
vu,+ v"o)= o
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(10.27)
On the other hand, the sum of phasevoltages(line to neutral)'may not be zero
so that their zero sequencecomponent Vn, may exist.
Since the sum of the three line currentsequals the current in the neutral
wire. we have
Ioo=!U"+16+r")=!+
_\
B --_____}_----
(10.28)
i.e. the currentin the neutralis three times the zerc sequenceline current.If the
neutral connectionis severed.
Ioo=lr,=o
3
(ro.2e)
Flg. 10. 3
Solution
or
Io + I" * Is = Q
10130" + l5l- 60o+ Ic = O
I c= - 16. 2+ j8. 0 = 18 1154" A
t.e. in the absence of a neutral connection the zero sequence line current is
always zero.
From Eqs.(10.24)to (10.26)
Power Invariance
1
Itr= :0Ol3O" + 751(-60"+ l2O" ) + I8l(154" + 240"))
We shall now show that the symmetrical componenttransformationis power
invariant, which means that the sum of powers of the three symmetrical
componentsequalsthe three-phasepower.
Total complex power in a three-phasecircuit is given by
S = f p $ = 4 1 , + V u t t +V , t !
5
= 10.35+ j9.3 = 14142"A
I
Iez= ;Q0130' + 151(- 60o+ 240") + I8l(154' + 120"))
J
= --1.7- j4.3 = 4.651248"A
(10.30)
or
Iao=
s = [Av,,]'t'lAl,l.
= v! 'qr'q.
tI
(10.31)
..
= 3V^1, + 3V"r!), + 3V"oI),
(10.33)
I
t^ (lo + IR +
J
/c) = 0
(iii)
Icr = 141162"A
Icz = 4.651128"A
Ico=oA
Im= 141282"A
Inz= 4'6518"A
Iao=oA
Check:
Ia= Iat * I,rz* I,to= 8'65 + j5 = 10130"
Convertingdelta load into equivalentstar,we can redraw Fig. 10.3 as in Fig.
10.4.
la
= sumof symmetricalcomponent
powers
Example10.1 |
I-T
A delta connectedbalancedresistive load is connectedacross an unbalanced
three-phasesupply as shown in Fig. 10.3. With currents in lines A and B
specified, find the symmetrical componentsof line currents. Also find the
symmetrical componentsof delta currents. Do you notice any relationship
betweerisymmetricalcomponentsof line and delta currents ? Comment.
www.muslimengineer.info
(ii)
FromEq. (10.2)
Now
r rl l-1 0 0l
[r o? 'Tt
o16.=lta
a n' .rl:r|o r ol-t, (10.32)
"tll
I JLa, c - l l L 0 0 l l
[r r
s=31yti,=341.,
(i)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpower system anatysis
$,$i.M
I
Delta currentsareobtainedas follows
Qrrmmatriaal
veB= U^- Ir)
Io
ItB=zAB)R
tUo-
tffiffi
Positive and negativesequencevoltagesand currents undergo a phaseshift in
passing through a star-delta transformer which depends upon the labelling of
terminals.Before considering this phaseshift, we need to discussthe standard
polarity marking of a single-phasetransformer as shown in Fig. 10.5. The
transformer ends marked with a dot have the same polarity. Therefore, voltage
Vun, is in phase with voltage V..,. Assuming that the small amount of
magnetizingcurrent can be neglected,the prirnary current 1r, enteringthe dotted
end cancelsthe demagnetizing ampere-turnsof the secondarycurrent 1, so that
I, and12with directions of flow as indicatedin the diagram are in phase.If the
direction of 1, is reversed, 1, and 1, will be in phase opposition.
ru)
Similarly,
r n c =I e r - I r )
5
Ice=!frr-
l-amnnnanla
Io)
5
Substituting the values of Io, Iu and Ir, we have
-r5l- 60")= 6186 A
Ien= Oozzo"
!
rnc=
60"- rllr54") = lO.5l- 41.5"A
lOsz- rol3o")= 8 . 3 1 1 7 3A"
rcn=
!W2154"
Flg. 10.5 Polaritymarkingof a single-phase
transformer\
The symmetricalcomponentsof delta currents are
1
Ie m = ;(6 1 8 6 " + IO.5 l ( - 4I.5' +
l 2O" ) + 8.31(173"+ 240" ))(i v)
J
= 8172" A
I
I,qaz=:6186" + 1051(- 41.5"+ 240")+ 8.31(173"+ 120"))(v)
J
= 2 . 7 1 2 1 8A"
(vi)
Ie n o = 0
Incr, Ircz, IBC,,lsn1,Iga2 and 1.oo can be found by using Eq. (10.2).
Comparing Eqs. (i) and (iv), and (ii) and (v), the following relationship
between symmetrical componentsof line and delta currents are immediately
observed:
t
'
IeBr=+130"
V J
Ienz= \
z-zo"
(vii)
Consider now a star/delta transformer with terminal labelling as indicated in
Fig. 10.6 (a). Windings shown parallelto eachother are magneticallycoupled.
Assume that the transformer is excited with positive sequencevoltages and
carries positive sequencecurrents. With the polarity marks shown, we can
immediatelydraw the phasor diagramof Fig. 10.7.The following interrelationship betweenthe voltages on the two sidesof the transformer is immediately
observedfrom the phasor diagram
VtBt= x Vabrl3V, -r - phasetransformationratio (10.34)
As per Eq. (10.34), the positive sequenceline voltages on star side lead the
correspondingvoltageson the delta side by 30" (The same result wo,'ld apply
to line-to-neutral voltages on the two sides). The same also applies for line
currents.
If the delta side is connectedas in Fig. 10.6(b) the phaseshrft reverses(the
readershould draw the phasor diagram);the delta side quantitieslead the star
side quantities by 30".
(viii)
! J
The reader should verify theseby calculatrng Io', andl*2from Eqs. (vii) and
(viii) and comparing the results with Eqs. (iv) and (v).
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ar System Analys'is
SymmetricatComponentg
| 3?9
I
A
__--______
o a
l-
/\
correspotxding positive sequence quantities on the LV side by 30'. The reverse
is the case for negative sequence quantities wherein HV quantities lag the
corresponding LV quantities by 30".
vsc2
(a) Star side quantitieslead delta side quantitiesby 30o
(b) Delta side quantities lead star side quantities by 30o
Fig. 10.6 Labellingof star/deltatransformer
Vesz
Vcrcz
vcea
Fig. 10.8 Negativesequencevoltageson a star/deltatransformer
IO.4
SEOUENCE IMPEDANCES
OF TRANSMISSION
LINES
Figure 10.9 shows the circuit of a fully transposedline carrying unUalancecl
currents. The return path for 1,,is sufficiently away for the mutual effect to be
ignored. Let
o1'eachline
X" = sell'reactance
X. = mutual reactanceof any line pair
The fbllowing KVL equationscan be written down from Fig. 10.9.
Vo - V'o= jXJo + jX*Iu + .ixmlc
Vec'l
Fig. 10.7 Positivesequencevoltageson a star/deltatransformer
Instead, if the transformerof Fig. 10.6(a) is now excited by negative
sequencevoltages and currents,the voltage phasor diagram will be as in Fig.
10.8.The phaseshift in comparisonto the positive sequencecasenow reverses,
i.e., the star side quantitieslag the delta side quantities by 30'. The result for
Fig. 10.6(b)also correspondingly
reverses.
It shall from now onwards be assumed that a star/delta transformer is so
labelled that the po,sitive sequencequantities on the HV side lead their
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V6
vc
lr= l"+ lo+1"
-(-
Fig.10. 9
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern PgwgfSy$gln Snalysis
4 - r[: jxJo + jxh + jx*["
l', - V!: iXJ" + jxmlb + jxJ"
Svmmetrienl cnmnonent(
tflIlFffi
(10.3s)
T
t,
o-------{__]-
Z1
12
-o
or in rnatrix form
22
"----f--}-
---o
ls
Zs
o---)--f---_l.-
^a
vI
vi
vb
v"
(10.36)
zIo
(10.37)
A (I/, - v ! ) : zuIs
(10.38)
- - /I// V,
nr
or
x^x,x. Ib
x_x*x, I"
= J
or
Now
v, -
,r//' s - A-IZAI,
(10.3e)
jx,
A-I ZA :
JX^
jx^
- ',
l*,
:J
L:
jx^
jx,
jx^
(10.40)
I x"- x^
lL "r rlo-Jl l'rltr:t 'JLI oo
z
0
0
t 0
2 2
0 z
0
o
ll-r,I
x, - x,,,
o ll r, |
o
x, *zx^ )j,l
(l o.4l)
0 l[ /'
0
o
(r0.42)
.l
ll,,I
lLr,i
wherein
j(X, - X,r) : positive sequence impedance
(10.43)
Zz: j(X, - X^) : negative sequence impedance
(r0.44)
Zr:
Zo: i(X, + 2X)
: zero ,tequence impedance
(10.4s)
We conclude that a fully transposed transmission has:
/t;\
\r,,
o^rrol
vyuar
^^o.i+i"o
pvrrlryu
(^t-r ll r l
-^^^+i
ut/ts4Lrvg
(b) Nagativesequence
network
@)Zero sequence
network
F i g .1 0 . 1 0
The decouplingbetween sequencenetworks of a fully transposedtransmission holds also in 3-phase synchronousmachines and 3-phasetransformers.
This fact leads to considerable simplications in the use of symmetrical
rnethodin unsymmetricalfault analysis.
components
In case of three static unbalancedimpedances,coupling appearsbetween
sequencenetworks and the method is no more helpful than a straight forward
3-phasounalysis.
10.5 SEQUENCE IMPEDANCES AND SEOUENCE NETWORK
OF POWER SYSTEM
0
X,-X*
0
Thus Eq. (10.37) can be written as
| 41 l'tr(I
(a) Positive sequence
network
Power system slernsnfs-transmission lines, transformers and. synchronous
nlachines-have a three-phasesymmetrybecauseof which when'currentsof a
particular sequenceare passed through these elements, voltage drops of the
same sequenceappear, i.e. the elementspossessonly self impedancos-to
sequencecurrcnts. Each eleurent can therelbre be representedby tlree
decoupledsequencenetworks (on single-phasebasis) pertaining to positive,
negativeand zerosequences,
respectively.
EMFs are involvedonly in a positive
sequencenetwork of synchronousmachines.For finding a particular sequence
impedance,the elementin questionis subjectedto currentsand voltagesof that
only. With the elementoperatingundertheseconditions,the sequence
sequence
impedancecan be determined analytically or through experimentaltest results.
With the knowledge of sequencenetworks of elements,complete positivel'
negativeand zero sequencenetworks of any power systerncan be assembled.
As will be explained in the next chapter. these networks are suitably
interconnectedto simulate different unsymmetrical faults. The sequence
currentsand voltagesduring the fault are then calculatedfrom which actual
far"rltcurrentsand voltases can be found.
--^:^-^)^-^^strqu{rrruE
rrup€uallutis.
(ii) zero sequenceimpedancemuch larger than the positive (or negative)
sequenceimpedance(it is approximately 2.5 times).
It is further observedthat the sequencecircuit equations (10.42) are in
decoupleclfbnn, i.c. thcrrearc no rnutual scqucncc inducternccs.F)quation
(10.42) can be representedin network form as in Fig. 10.10.
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10.6
sEQuENCg I;IpTDANCES AND NETWoRKs
SYNCHRONOUS MACHINE
oF
rlachine (generator
or motor)
Figure 10.11depictsan unloadedsynchronous
groundedthrougha reactor (impedanceZ).8o, E6andE, are the inducedemfs
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernpowerSystemAnatysis
Symmetrigel_qg4pe!e$s
_
of the three phases.when a fault (not shown in the figure)
takes place at
machineterminals,currents1,,,I,,and /. flow in the lines.
whenever the fault
involvesground,currentIn= In+ Iu + ^I"flows to ne'tral
from ground
"i^I^'.
theshortcircuit occurslirrnr kradcclconditiorts,thc voltagt:behindappropriltc
reactance(subtransient,transient or synchr:onous)constitutesthe positive
3S2 I
--]-
_
Hrvvvvs
yyrrrr r(rLrll 4rr.1IyJIs (\_napler
i l),
we
must
know the equivalentcircuits presenteclby the *u.hin"
to the flow of positive,
negative and zero sequence culrents, respectively.
Because of winding
symmetry currents of a particular sequence produce voltage
drops of that
sequenceonly. Therefore,thereis a no coupling between
the equivalent circuits
of varioussequences*.
uence voltage.
Figure lO.IZa shows the three-phasepositive sequencenetwork ntodel of a
synchronousmachine. Z, doesnot appearin the model as Iu = 0 for positive
sequencecurrents. Since it is a balancednetwork it can be representedby the
single-phasenetwork model of Fig. 10.12b for purposesof analysis. The
referencebus for a positive sequencenetwork is at neutral potential. Further,
sinceno current flows from ground to neutral,the neutral is at ground potential.
lat
>
la
a
Referencebus
-*
)e"
.n
'----(
t +
\=\'-..
t:.6
L-----
d-n>_
-u
\
t6
--> -----b
____l_"
Sincea synchronousmachineis clesignedwith symmetricalwindings,
it induces
emfsof positivesequenceonly, i.e. no negatrveor zerosequence
voltagesare
inc lt r c ei n
d i t' Wh c n th c tttl tc l ti nc u rri csposi ti vcscqucncc
gl l y, thi s
curr.cnts
ntodeof operationis the balancedmoclediscussedailength
in Chapter 9. The
armature reaction field caused by positive sequence currents
rotates at
'synchronous
speedin the salneclirectionas the ,otu., i.e., it is stationary
with
respect to field excitation. The machine equivalently offers
a direct axis
reactancewhose value reducesfrom subtransicntreactance(X,a)
to transient
reactance(Xtr) and finally to steadystate (synchronous)reactanJe(Xa),
as the
short circuit transient progressesin time. If armature resistance
is assumed
negligible,the positive sequenceimpedanceof the machine
is
21= jXtj (if I cycle transientis of interest)
(10.46)
= jX'a Gf 3-4 cycle transientis of interest)
(10.47)
(10.48)
If the machine short circuit takes place from unloaded
conditions, the
terminaivoltageconstitutesthe positivesequencevoltage;
on the other hand.if
so by synchronousmachine theory,[5].
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I
I
lc't
( b ) S i n g l e - p h a s em o d e l
( a ) T h r e e - p h a s em o d e l
machine
Fig. 10.12 Positivesequencenetworkof synchronous
Positive Sequence Impedance and Network
= jXa (if steady state value is of interest)
b
I
Fig' 10'11 Three-phase
synchronous
generatorwithgroundedneutral
*'fhis can be shown to be
ln't
,
(
With referenceto Fig. 10.12b,the positive sequencevoltageof terminal c
with respect to the referencebus is given by
(10.4e)
V,,l= E,,- Zll,,l
Negative Sequence Impedance
and Network
It hasalreadybeensaidthat a synchronousrnachinehas zeronegativesequence
inducedvoltages. With the flow of negative sequencecurrentsin the stator a
rotating field is createdwhich rotates in the opposite direction to that of the
positivesequencefield and, therefore,at double synchronousspeedwith respect
to rotor. Currents at double the stator frequency are thereforeinduced in rotor
field and damper winding. In sweeping over the rotor surface, the negative
sequencemmf is alternatelypresentedwith reluctancesof direct and quadrature
axes. The negative sequence impedance presented by the machine with
considerationgiven to the damper windings, is often defined as
xt: + x,!
(10.s0)
;lZ2l<lZrl
2
Negative sequencenetwork models of a synchronousmachine,on a threebasisare shownin Figs. 10.13aandb, respectively.The
phaseand single-phase
referencebus is of course at neutral potential which is the same as ground
potential.
Z . t =j
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
fl$4.:rl
t
powLrSystem
Mooern
nnarysis
symmetricat
Components
-
From Fig' 10.13bthe negative sequencevoltage of terminal a with respecr
to referencebus is
Voz= - Zzloz
(10.s1)
FjSiffi
Zs= 3Zra Zos
(10.53)
in orderfor it to havethe samevoltagefrom a to referencebus.The reference
bushereis, of course,at groundpotential.
From Fig. 10.14b zero sequencevoltage of point c with respect
Voo = - Z{oo
to the
(10.54).
Order of Values of Sequence Impedances of a
Synchronous Generator
laz
a
Typical valuesof sequenceimpedancesof a turbo-generator
rated 5 MVA, 6.6
kV, 3;000 rpm are:
(b) Single-phase model
Zr = lZ%o(subtransient)
Flg. 10.13 Negativesequencenetworkof a synchronous
machine
Zero Sequence Impedance
Zr = 20Vo(transient)
and Network
we state once again that no zero sequence voltages are induced in a
synchronous machine. The flow of zero sequencecurrents creates three mmfs
which are in time phase but are distributed in space phase by 120".
The
tesultant air gap field caused by zero sequence currents is therefore zero.
Hence, the rotor windings present leakage reactanceonly to the flow of zero
sequencecurrents (Zos < Zz < Z).
Zr = 7l0Vo (synchronous)
Zz= I2Vo
Zo= 5Vo
For typical values of positive, negative and zero sequencereactances
of a
synchronousmachine refer to Tablu 9.1.
IO.7
/66
SEOUENCE IMPEDANCES
Reference bus
-a
Iao
(a)Three-phase model
(b)Single-phase model
1o.14 Zerosequencenetworkof a synchronous
machine
Zero sequencenetwork models on a three-and single-phasebasis are shown
in Figs. 10.14aand b. In Fig. r0.l4a, the cu'ent flowing in the inpedance
zn
betweenneutral and ground is In = 3lno.The zero sequencevoltage of t"r-inul
a with respect to ground, the referencebus. is therefore
Vr1= - 3znioo- Zorlno=,r, (32, + Z0)loo
e0.52)
wh11eZo, is the zero sequenceimpedanceper phase of the machine.
Sincethe single-phase
zero sequencenetworkof Fig. 10.14bcarriesonly per
phasezero sequencecurrent, its total zero sequenceimpedancemust be
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OF TRANSMISSIO\
LINES
A fully transposedthree-phaseline is completely symmetrical and
therefore the
per phase impedanceoffered by it is independentof the phase
sequenceof a
balancedsetof currents.In other words,the impedances
offired by ii to positive
and negative sequenceculrents are identical. The expressionfoi
its per phase
inductive reactanceaccounting for both self and mutual linkages
ias been
derived in Chapter 2.
When only zero sequencecurrents flow in a transmissionline,
the currents
in each phase are identical in both magnitude and phase angle. part
of these
currents return via the ground, while the rest return throu h the
overhead
ground wires. The ground wires being grounded at severaltowers,
the return
currentsin the ground wires are not necessarilyuniform along the entire
length.
The flow of zero sequencecurrentsthroughthe transrnissioniln"r, ground
wires
and ground createsa magnetic field pattern which is very different from
that
causedby the flow of positive or negativesequencecurrentswhere the currents
h,r.ro
rrqYv
^
(r
^L^^^
-J:ff^-^-^^
Prrd'Ds' urrrsrtrIlutr
^f
UI
t ^no
ILV
an(l
tne
fetufn
Cuffent
lS
ZefO.
T\e
ZefO
sequence impedance of a transmission line also accounts for the ground
impedance(zo = z.to+ 3zri.since the ground impedanceheavily depends
on
soil conditions, it is esseniialto make some simplifying assumptionsio obtain
analytical results. The zero sequenceimpedanceof transmissionlines usuallv
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
3Sd I
ModernPower SystemAnatysis
Symmetri
a
rangesfromZ to 3.5 timesthe positivesequenceimpedance*.This ratio is on
the higher side for double circuit lines without ground wires.
10.8 SEOUENCEIMPEDANCES AND NETWORKS OF
It is well known that ahnosl all present day installationshave three-phase
transformerssince they entail lower initial cost, have smaller spacerequirements and higher efficiency.
The positive sequenceseriesimpedanceof a transformerequalsits leakage
impedance.Since a transformeris a static device, the leakageimpedancedoes
not changewith alterationof phasesequenceof balancedapplied voltages.The
transformernegativesequenceimpedanceis also thereforeequal to its leakage
reactance.Thus. for a transformer
Zt= Zz= Zr"ukug"
( r0.55)
Assuming suchtransformerconnectionsthat zero sequencecunents can flow
on both sides,a transformeroffers a zerosequenceimpedancewhich may differ
slightly from the correspondingpositive and negativesequencevalues.It is,
however, normal practiceto assumethat the seriesimpedancesof all sequences
are equal regardlessof the type of transformer.
The zero sequencemagnetizingcurrent is somewhathigher in a core type
than in a shell type transformer. This difference does not matter as the
magnetizing curent of a transformer is always neglectedin short circuit
analysis.
Above a certain rating (1,000 kvA) the reactanceand impedance of a
transformerare almostequal and are thereforenot distinguished.
transformers
-"-t--t
ter" -"*"-
"
Beforeconsideringthe zero'sequencenetworks of various types of transformer
connections,three importantobservationsare made:
current only if there is cunent florv on the secondaryside.
(ii) Zero sequencecurrentscan flow in the legs of a star connectiononly if
the star point is groundeclwhich proviciesthe necessaryreturn path for
zero sequenceculrents.This fact is illustratedby Figs. 10.15aand b.
/ao= 0
-
':o
__----,
a
{.->
/co= 0 --f-'
r'l
/oo=o
(a) Grounded star
star
(a) Ungrounded
Fig. 10.15 Flowof zerosequencecurrentsin a starconnection
(iii) No zero sequencecurrents can flow in the lines connecte.dto a delta
connectionas no returnpath is availablefor theseculrents.Zero sequence
currents can, howevet, flow in the legs of a delta-such currents are
causedby the presenceof zerosequencevoltagesin the delta connection.
This f act is illt r st r at ed
bY Fig. 10. 16'
/ao= 0
a
*We can easily compare the forward path positive and zero sequenceimpedances
of
a transmissionline with groundreturnpath infinitely away.Assumethat eachline has
a self inductance,L and mutual inductance M between any two lines (completely
symmetrical case).The voltage drop in line a causedby positive sequencecurrents is
VAnt= uLlnr+ uMI,,r+ aMI,.1
-
[uL+ (& + a) aMllo, = a(L-tu|)Ior
Positive sequencereactance= a(L- fuI)
The voltage drop in line a causedby zero sequencecurrents is
VAno= aLloo+ utMluo+ uMlrs
= a(L + 2M)Ino
Zero sequencereactance= w(L + 2W
Obviously,zerosequencereactanceis much more than positive sequencereactance.
This result has alreadybeenderivedin Eq. (10.45).
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currentsii',a deltaconnection
Fig. 10.16 Flowof zeroSequence
Let us now consider various types of transformerconnections.
Case I; Y-Y transformerbank with any one neutrttl grounded.
If any one of the two neutralsof a Y-Y transformeris ungrounded,zero sequence
currentscannot flow in the ungroundedstar and consequently,thesecannotflow
in the grounded star.Hence,an open circuit existsin the zero sequencenetwork
between H and L, i.e. betweenthe two parts of the system connectedby the
transformeras shown in Fig. 10.17.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffi
(seeFig. 10.19). If the star neutral is groundedthrough 2,, an impedance3Zn
appearsin series with Z, in the sequencenetworkCasb
Zg
---tll
d---
L
Y-Ytransformer
bankwithone neutralgroundedand its zero
sequencenetwork
Y-A tran
rmer bank with ungroundedstar
This is the special case of Case 3 where the neutral is grounded through
Zn = oo. Thereiore no zero sequence curent can flow in the transformer
windings. The zero sequencenetwork then modifies to that shown in Fig. 10.20.
{ L
Case 2: Y-Y transforrner bank both neutrals grounded
Referencebus
When both the neutralsof a Y-Y transformerare grounded, a path through the
transformer exists for zero sequencecurrents in both windings via the two
groundedneutrals.Hence,in the zero sequencenetwork 11andL areconnected
by the zero sequenceirnpedanceof the transformeras shown in Fig. 10.1g.
Case 3: Y-A transformer bank with grounded y neutral
"ifr6L2 ;
H
Fig. 10.20
----------o
L
star and its
bankwith ungrounded
Y-l transformer
zerosequencenetwork
Case5: A-A transformer bank
Since a delta circuit provides no return path, the zero sequencecurrents cannot
flow in or out of A-A transformer; however, it can circulate in the delta
windings*. Theretore, there is an open circuit between H anE L and Zo is
connectedto the referencebus on both endsto accountfor any circulating zero
sequencecurrent in the two deltas (see Frg. 10.2I).
Reference bus
o---4TdL=-o
H
Z
s
L
Fig. 10.18 Y-Ytransformer
bankwith neutralsgroundedand its zero
sequence
network
a ---_.
-I
Reference bus
^--r-
d-E-;1
Fig. 10.19 Y-1 transformer
bankwith groundedyneutral and its zero
sequencenetwork
If the neutral of star side is grounded,zero sequencecurrentscan flow in star
becausea path is availableto ground and the balancing zero sequencecurrents
can flow in delta.Of courseno zero sequencecurrentscan flow in the line on
the delta side. The zero sequencenetwork must therefbre have a path from the
line 1{ on the star side through the zero sequenceimpedance of the transformer
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bankand its zerosequencenetwork
Fig. 10.21 A-Atranstormer
10.9 CONSTRUCTION OF SEOUENCENETWORKS OF A
POWER SYSTEM
In the previous sections the sequencenetworks for various power system
elements-synchronous machines,transformersand lines-have been given.
Using thesq, complete sequencenetworks of a power system can be easily
constructed.To start with, the positive sequencenetwork is constructed by
*Such circulating currents would exist only if zero sequencevoltages are somehow
induced in either delta winding.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Power SystemAnalysis
examinationof the one-linediagrarnof the system.It is to be noted that positive
sequence
voltagesarepresentin synchronous
machines(gcnerators
and motors)
only. The transitionli'orr-rpositil'e sequencenetwork to negative sequence
network is straightforward,Since the positive and negative sequenceirnped,) \rrrrvD
qrtu
Lt otrJltrt
ttlvt
Jrr,
rrrv
LrltlJ
been dealt with in the preceding section.
The procedure for drawing sequence networks is illustrated through the
following examples.
-
MVA base in all other circuits and the following voltagebases.
Transmissionline voltase base= 11 x
121 123.2kY
10.8
vrrarr
neclessary
in positive sequencenetwork to obtain negatit'e sequencenetwork
in respect of synchronous machines. Each machine is representedby
neg;ttivesequenceimpedance,the negativesequencevoltage being zero.
The referencebus for positive and negativesequencenetworks is the system
neutral.Any impedanceconnectedbetweena neutraland ground is not included
in these sequencenetworks as neither of these sequencecurrents can flow in
suchan impedance.
Zero sequencesubnetworks for various parts of a system can be easily
combinedto form completezero sequencenetwork. No voltage sourcesare
presentin the zero sequencenetwork. Any impedanceincluded in generatoror
transformerneutralbecomesthree tirnesits valuein a zero sequencenetwork.
Special care needs to be taken of transforners in respect of zero sequence
network. Zero sequencenetworks of all possibletransformer connectionshave
f-"--
ffi
Motor voltagebase= 123.2x ++ = 11 kV
121
line andmotorsareconvertedto pu values
of transformers,
The reactances
basesas follows:
on appropriate
= 0.1x =. f++) = 0.0s05pu
reactance
Transformer
3 0 \ 1 1/
loo x 25
Line reacl: a n c e =
ffi
=o.164pu
2s /10\2 =
|
0.345pu
Reactanceof motor I = 0.25 x - x t 15 \lll
Reactanceof motor 2 - 0.25 x ?1, rf -Lo)t = 0.69 pu
7.5 \ ll /
The requiredpositive sequencenetwork is presentedin Fig. 10.23.
-:----'
Referencebus
I
r Example10.2 1
t _ _ _ _ _ _ : _ - l
Es( )
+l
A 25 MVA, 1l kV, three-phase
generatorhas a subtransientreactanceof 20Vo.
T lr cgc rrc ri ttosru p p l i c stw o rrto to rso v0r' r,r
transnri ssi on
l i nc w i Lhtral tsl ol l rrcl ' s
at both ends as shown in the one-line diagramof Fig. I0.22. The motorshave
ratedinptrtsof 15 and 7.5 MVA, both 10 kV with 25Vosubtransientreactance.
Thethree-phase
transformersare both rated30 MVA, I0.8/L2I kV, connection
A-Y with leakagereactanceof 70Voeach.The seriesreactanceof the line is
100 ohms. Draw the positive and negative sequencenetworks of the system
with reactancesmarkedin per unit.
e
I' r tt - so
t
l +
--,1
iio6e
.
-,
r- "xfi
i 00805
P -----r--'---
In
9
Fig. 10.23 Positivesequencenetworkfor Example10.3
Referencebus
(2) ,
q' \ - / r \
Motor ,,1
I
Fig. 10.22
Assume that the negative sequence reactance of each machine is equalto its
subtransientreactance.Omit resistances. Select generator rating as basein the
generatorcircuit.
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,-66X-L-g ,.I-6XU
j0'164
i oo8o5
) Eara
i
i o 3 4 5 :- r
p
' u910'
(-t,) 'r'
.
IYA
)
+l
I
.r:t!
ioz:),
d
I
Enrt(
i 0.0805
j4164
loo8o5
Fig. 10.24 Negative sequencenetworkfor Example 10.3
Since all the negative sequence reactances of the system are equal to the
positive sequence reactances,the negative sequencenetwork is identical to the
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
' rti..;;:1 . ::i.l
3,42'l
symmetricat
compblents-------_-1
l-4odernPov"'ei'SvstemAnalvsis
positive sequencenetwork but for the omission of voltage sources.The negative
sequencenetwork is drawn in Fig. 10.24.
Zerosequence
reactance
of motor2 - 0.06
" +.
= 0.164pu
Reactance
of currentlimiting reactors=
For the power system whose one-line diagram is shown in Fig. 10.25,sketch
the zero sequencenetwork.
T2
T1
ffi
(i+)'
''.1.i3t =
0.516pu
(11)',
Reactance
of currentlimiting reactorincludedin zero sequence
network'
= 3 x 0 . 5 1 6- - 1 . 5 4 8p u
Zerosequence
reactance
of transmission
line = ry:21
(r23.D2
= 0.494pu
The zero sequence
networkis shownin Fig. 10.27.
Fig.10.25
Referencebus
Solution The zero sequencenetwork is drawn in Fig. 10.26.
j1-548
j1.il8
io06
jo164
Reference bus
32n
Zost
Fig. 1O.27 Zero sequencenetworkof Example 10.5
Zrt
Fig. 10.26
Z6(line)
26(line)
272
Zero sequence network of the system presented in Fig. 10.25
IEIvIS
PROB
i - - i - - - -
-
_ -
* - ;
10.4|
1Example
-*-"
1
Draw the zero sequencenetrvork for the system describedin Example 10.2.
Assumezero sequencereactancesfor the generatorand motors of 0.06 per unit.
Current limiting reactorsof 2.5 ohms eachare connectedin the neutral of the
generatorand motor No. 2. The zero sequencereactanceof the transmission
line is 300 ohms.
Solution The zero sequencereactance of the transfonner is equal to its
positive sequencereactance.Hence
Transformer zero sequencereactance: 0.0805 pu
Generatorzero seqllencereactances: 0.06 pu
Zero sequencereactanceof motor 1 : 0.06 x
10.1 Computethe following in polar form
(i) o2-t 1ii; I- a (iii) 3 d + 4cy+ 2 (iv) ja
"?
10.2 Thlee identical resistorsare star connectedand rated 2,500 V, 750 kVA.
This thre.e-phase
unit of resistors is connected to the I side of a A-Y
transformer.The following are the voltages at the resistor load
lVo6l= 2,000 Y; lVu,l = 2,900 V; lV,ol = 2,500 V
Choosebaseas 2,500 V, 750 kVA and determinethe line voltagesand
currents in per unit on the delta side of the transforrner.It may be assumed
that the load neutral is not connectedto the neutral of the transformer
secondary.
10.3 Determine the symmelrical componentsof three voltages
Vo= 2001ff, Vt = 2001245" and V, = 2001105' y
: 0.082
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Power SystemAnalysis
resistiveload of 100 kVA is connectedacrossiines bc of
10.4 A single-phase
a balancedsupply of 3 kV. Computethe symmetricalcomponentsof the
line currents.
10.5 A delta connectedresistive load is connectedacross a balanced threepnasesupply
'
"-"*'
2: "25 MVA, 11 kV, Xtt = ZOVo
Generator
Three-phasetransformer(each):20 MVA, ll Y1220Y kV, X = I5Vo
The negative sequencereactanceof each synchronousmachine is equal
machine is 8Vo.Assume that the zero sequencereactancesof lines are
25OVoof their positive sequencereactances.
250 f)
B
Fig. P-10.5 PhasesequenceABC
of 400 V as shown in Fig. P-10.5. Find the symmetrical componentsof
line cunents and delta currents.
10.6 Three resistancesof 10, 15 and 2O ohms are connectedin star acrossa
three-phasesupply of 200 V per phase as shown in Fig. P-10.6. The
supply neutral is earthedwhile the load neutral is isolated. Find the
currentsin each load branch and the voltage of load neutral above earth.
Use the method of symmetrical components'
la
I r
'0.:-"t
tB
I-
te
ls
c1
_L__
1--
tb
_l
Fi g. P -10.6
X = 5o/o
at machine,l rating
at machine 2 rating
Fig. P-10.8
r0.9For the power system of Fig. P-10.9draw the positive, negative and zero
sequencenetworks. The generatorsand transformersare rated as follows:
G ener at orl: 25 M VA, 11 kV, Xt t =0. 2, X2 = 0. 15,Xo = 0. 03 pu
2: 15 M VA, 11 kV, Xt =0. 2, Xz= 0. 15, X0 = 0. 05 pu
G ener at or
SynchronousMotor 3: 25 MVA, 11 kV, Xt = 0-2, Xz= 0-2, Xo = 0.1 pu
Transformerl: 25 MVA, I 1 Lll20 Y kV, X = IIVo
2: 12.5 MVA, 11 LlI20 Y kV, X = l07o
3: 10 MVA, I2O Ylll Y kV, X - IjVo
Choosea base of 50 MVA, I I kV in the circuit of generatorl.
T1
10.7 The voltagesat the terminalsof a balancedload consistingof three 20
V.
resistorsarc 2OO4O",100 1255.5'and 200 llsf
ohm X-connected
Find the line currents from the symmetrical components of the line
voltagesif the neutralof the load is isolated.What relationexistsbetween
the syrnrnetricalcomponentsof the line and phase voltages'/ Find the
power expanded in three 20 ohm resistors from the symmetrical
componentsof currentsand voitages.
10.8. Draw the positive, negative and zero sequenceimpedancenetworks for
the power systemof Fig. P-10.8.
Choosea baseof 50 MVA, 220 kV in the 50 0 transmissionlines, and
mark all reactancesin pu. The ratings of the generatorsand transformers
\7
Fig. P- 10. 9
Note: Zero sequencereactanceof each line is 250Voof its positive
sequencereactance.
are:
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
j,,W
voo"inpo*",,syrt"r Rn"tu"i,
10.10 Considerthe circuit shown in Fig. P-10.10.Suppose
Vo, = L00 l0
Xr=12 Q
=
vbn 60 160
Xob=Xbr=X*=5d)
F i g . P -10.10
(a) Calculate Io, 16,and 1, without using symmetrical component.
(b) Calculate Io, 16,and 1" using symmetrical component.
NCES
REFERE
Books
l. wagncr, c.F. and R.D. Evans,symmetrit:alcomponenfs,McGraw-Hill,Ncw york
, 1933.
2. Clarke,E., Circuit Analysisof Alternating Current Power Systems,Vol. 1. Wiley,
Ncw York, 1943.
3. Austin Stigant,5., Master Equationsand Tablesfor SymmetricalComponentFault
Studies,Macdonald,London, 1964.
4. stcvenson,w.D., ElcmentsoJ'Power Sy.stem
Analysis,4thedn,McGraw-Hill, New
York, 1982.
5. Nagrath, I.J. and D.P. Kothai, Electric Machines,2nd edn., Tata McGraw-Hill,
New Delhi, 1997.
Paper
6. Fortescue,C.L., "Method of SymmetricalCoordinatesAonlies to the Solution of
PolyphaseNetworks', AIEE, 1918,37: 1O27.
www.muslimengineer.info
II.I
INTRODUCTION
Chapter 9 was devoted to the treatment of symmetrical (three-phase)faults in
a power system.Sincethe systemremainsbalancedduring suchfaults,analysis
could conveniently proceed on a single-phasebasis. In this chapter, we shall
deal with unsymmetrical faults. Various types of unsymmetrical faults that
occur in power systemsare:
Shunt
Wpe Faults
(i) Single line-to-ground (LG) fault
(ii) Line-to-line (LL) fault
(iii) Double line-to-ground (LLG) fault
Series Type Faults
(i) Open conductor (one or two conductorsopen) fault.
It was statedin Chapter 9, that a three-phase(3L) fault being the most severe
must be usedto calculate the rupturing capacity of circuit breakers,even though
this type of fault has a low frequency of occurrence, when compared to the
unsymmetricalfaults listed above. There are, however, situationswhen an LG
fault can cause greater fault current than a three-phasefault (this may be so
when the fault location is close to large generating units). Apart fiom tliS,
unsymmetrical fault analysis is important for relay setting, single-phase
switching and system stability studies(Chapter 12).
The probability of two or more simultaneousfaults (cross-countryfaults) on
a power system is remote and is therefore ignored in system design for
abnormalconditions.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
*g9,S.I
Modernpower System Anatysis
UnsymmetricalFault Analysis
The method of symmetricarcclmponentspresentedin chapter 10, is a
powerful tool for study of unsymmetricalfaults and witl be fully exploited
in
this chapter.
Vat
UNSYMMETRICAL
(a)
FAULTS
Consider a general power network shown in Fig. 11.1. It is assumed
that a
shundtype fault occursat point F in the system,is a result of which
currenm
Io, 16,/, flow out of the system, and vo, v6, v"are voltagesof lines a,
b, c with
respectto ground.
(b)
(b)
Vao
(c)
(c)
_l
lao
as seen
Fig. 11.2 Sequencenetworks
fromthe faultpoint F
Fig. 11.1 A generalpowernetwork
Let us also assume that the system is operating at no load before the
occurrence of a fault. Therefore, the positive sequence voltages of all
synchronousmachineswill bc identicaland will equalthe prefaultvoltageat F.
Let this voltage be labelled as Eo.
As seenfrom F, the power systemwill presentpositive, negative and zero
sequencenetworks, which are schematicallyrepresentedby Figs. Il.2a, b and
c. The referencebus is indicatedby a thick line and the point F is identified on
each sequencenetwork. Sequencevoltagesat F and sequencecurrentsflowing
out of the networks at F are also shown on the sequencenetworks. Figures
11.3a,b, and c respectively,give the Thevenin equivalentsof the three sequence
networks.
Recognizingthat voltageEo is presentonly in the positive sequencenetwork
and that thereis no coupling betweensequencenetworks,the sequencevoltages
at F can be expressedin terms of sequencecurrents and Thevenin sequence
impedancesas
1r",1lt"1 lt, 0 ol|-r",'l
l r " , l = l o lo- l 2 2 o l l , * l
Lv"o)Lol Io
o
Fig. 11.3 Theveninequivalents
of the
sequencenetWorksas seen
fromthe faultpoint F
Dependingupon the type of fault, the sequencecunents and voltages are
constrained, leading to a particular connection of sequencenetworks. The
sequencecurents and voltages and fault currents and voltages can then be
easily computed.We shall now consider the varioustypesof faults enumerated
earlier.
11.3
STNGIE LINE-TO-GROUND
(tG) FAULT
Figure 11.4showsa line-to-groundfault at F in a power systemthrougha fault
impedanceZI. The phasesare so labelled that the fault occurs on phase a.
(11.1)
zo)lr"o)
Fig. 11.4 Single line-to-ground(LG) fault at F
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Unsvmmetrical
FaultAnatvsis
fffiH
At the fault point F, the currentsout of the power systemand
the line to ground
voltagesare constrainedas follows:
I u =o
(rr.2)
Ir= 0
11 . 3 )
vo = zllo
(11.4)
r'at
=
E
o
(zr * zz * z) +3zr
.
(11.7)
Fault current /o is then given by
(zr * zz * zo)+32
The above resultscan also be obtained directly from Eqs. (11.5) and (11.6)
by using Vop Vo2and Voefrom Eq. (11.1). Thus
The symmetrical components of the fault currents are
(Eo- IolZ) + (- IorZr) + (- I"&d = 3Zf lot
I(4+
Ior =
(11.5)
Expressine
Eq ,r':l; ,';',.*1;,ij;.*.ar
Vot * Voz *
componenrs,
wehave
Voo = ZfIo = 3zflor
4+
Zo)+ 3zfllor= Eo
Eo
(zr * zz -t Z) +3zr
The voltage of line b to ground under fault condition is
(11.6)
As per Eqs. (11.5) and (11.6) all sequencecurrentsare
equal and the sum
of sequencevoltagesequals3zf lot Theiefore, theseequations
suggesta series
connection of sequencenetworks through an impedanie
3zf ut rrio"*n in Figs.
11.5 aa n c ib .
Vu= &Vor+ aVo2* Voo
=o,2
(n'-^+)*4-',+).
(-"+)
Substituting for /o from Eq. (11.8) and reorganizing,we ger
Va= Eo
3rl2
zr + Z2(& - a)+ zo(& - D
(Zr*22+Zo)+3Zt
(11.e)
The expression for V, can be similarly obtained.
Fault Occurring
la,t=laz= t"o= h
t
l"s=I,,1=
t t,
/a\
(b)
Fig' 11'5 Connection
of sequencenetworkfor a singleline-to-ground
(LG)fault
In terms of the Thevenin equivalent of sequencenetworks,
we can write from
Fig. 11.5b.
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Under Loaded Conditions
When a fault occursunderbalancedload conditions,positive sequencecurrents
alone flow in power system before the occurrence of the fault. Therefore,
negativeand zero sequencenetworksare the sanleas without loacl.The positive
sequencenetwork must of course carry the load current.To account for load
current, the synchronous machines in the positive sequence network are
replacedby subtransient,transientor synchronousreactances(dependingupon
the time after the occulTenceof fault, when currents are to be determined) and
voltages behind appropriate reactances.This change does not disturb the flow
of prefault positive sequencecurrents (see Chapter9). This positive sequence
network would then be usedin the sequencenetwork conhectionof Fig. 11.5a
for computing sequencecurrentsunder fault
In casethe positive sequencenetwork is replacedby its Thevenin equivalent
as in Fig. 11.5b,the Theveninvoltage equalsthe prefaultvoltage Vi atthe fault
point F (under loaded conditions).The Thevenin impedanceis th6 impedance
betweenF and the referencebus of the passivepositivesequencenetwork (with
voltage generatorsshort circuited).
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
i'4ot',,1
ModernPo@is
fni, i, illustrateclby a two machinesystemin Fig. 11.6.It is seenfiom this
figure that while the prefault currents flow in the actual positive sequence
network <lfFig. 11.6a,the samcdo not exist in its Theveninequivalentnetwork
of Fig' 11.6b.Therefore,when the Theveninequivalentof positive sequence
network is used fot calculatlng fault eurre+ts, the positive sequeneeeurrents
within the network are those due to fault alone and we must superimposeon
these the prefault currents. Of course, the positive sequencecurrent into the
fault is directly the correct answer,the prefault current into the fault being zero.
from whlch we get
Io2= - Iol
(11.11)
1.,0= 0
(11.12)
The symmetrical components of voltages at F under fault are
I v,,f , [t a o'fl''
lr",l=ila
t1 2 "I l l r ,- z r r aI )
LV"o) Lr
(b)
(c)
Fig. 11.6 Positivesequencenetworkand its Theveninequivalent
before
occurrence
of a fault
The aboveremarksare valid for the positive sequenienetwork, independent
of the type of fault.
lL",
Writing the first two equations,we have
3vot = vo + ( a+ &) vu - &zf t u
3voz= v, + (tr + ,l) vo - ull Iu
from which we get
3(va- voz)- (a - &1zrtu- iJi y' to
LrNE-TO-LrNE (LL) FAULT
a) Iot (:
I o z = - I o t , 1 o o= 0 )
= - jJt I"r
Figure I 1.7 showsa line-to-linefault at F in a power systemon phasesb and
c through a fault impedanceZf . The phasescan always be relabelled,such that
the fault is on phasesb md c.
b
( 11 . 1 4 )
Now
16= (& -
t7.4
(11.13)
.r
( 11.15)
16from Eq. (11.15)in Eq. (11.14),we get
Substituting
Vot - Voz= Zf I ot
(11.16)
Equations(11.11) and (11.16)suggestparallelconnectionof positive and
negativesequencenetworksthrougha seriesintpedanceV as shown in Figs.
11.8aand b. Since loo= 0 as per Eq.(11.12),the zero sequencenetwork is
unconnected.
l6
Vaz
Fig. 11.7 Line-to-line(Lt) fault through impedance Z/
F
lat
The currents and voltages at the fault can be expressedas
_1
laz
[1":o I
Io=
I
l ,V o - V r = I o Z f
l I u -Ir t '
lI, :
_i
The symmetricalcomponentsof the fault currents are
(11.10)
Fig. 11.8
Connectionof sequencenetworksfor a line-to-line(LL) fault
In terms of the Thvenin equivalents, we get from Fig. 11.8b
( 11 . 1 7 )
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UnsymmetricalFault Analysis--_1
t
F ro m E q . (1 1 .1 5 ),w e g e t
or
- jJi
J ' - u n,
o
!L-
Iu
' - -
I SQt*
(l 1.23)
Voo= Vot * 3Zf Ino
(\ 1. 1 . 1 8 )
Z t,+*+2
222
"+
4
2+z+Zf l
Knowing l,-trwecarlcalsqld{ev,,, and,vorfromwhich vsltages
at thefault,
be found.
From Eqs. (11.19),(ll.22a) and (11.23),we can draw the connectionof
networksas shown in Figs. 11.10aand b. The readermay verify this
sequence
by writing mesh and nodal equationsfor these figures.
If the fault occursfrom loadedconditions,the positive sequence
network can
be modified on the lines of the later portion of Sec. 1r.3.
11.5
DOUBLE LrNE-TO-cRouND
(Lrc)
FAULT
Figure 11.9 showsa doubre line-to-groundfault at F in power
a
system.The
fault may in generalhave an impedanceZf as shown.
a-
tr
a
b-
-,..\.
Y/u=9
I
:1
Ea
Z1
_.__
|
r"l
t'o z! t'"o
| ___
_,__r:]tr"
The current and voltage (to ground) conditions at the
fault are expressed as
Io=o
l
i", + Io, + I"o =o|
V,,= V, = Zf (lt, + Ir) = 37rf1,,r,
The symmetricalcomponentsof voltagesare given by
l!:'1 [r a "']lr.l
l ! : ' l =*,l_l
l' a 2 o l l v Iu
1"v,,,)
I rll,ur)
( 11 . 1 9 )
(r 1.20)
(b)
(rr.21)
(Il .22a)
v,,o=
+ 2vu)
1""
(rr.22b\
tr,
- ,r- &1 vu= vt = 3zfloo
Fig. 11.10 Connectionof sequence networksfor a double line-to-ground
(LLG) fault
ln terms of the Thevenin equivalents, we can write from Fig. 11.10b
(11.24)
The above result can be obtainedanalyticallyas follows:
f or Vut , Vuz and V, * in t er m sof E, , in Eq. ( 11. 1) and
S ub st ir ut ing
premultiplyingboth sidesby Z-t (inverseof sequenceimpedancematrix), we
get
lt;'
I o z;'
ol[r,,-ztrut
o lln,-2,r",
I
I
'=17'
;, ilt?l
l';"1
I o
Io
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Eo
I ,. =
l-@
En
v,,t= V.z= Llv,, + (a + r11V,,1
3"'
voo- vot=
+
zt + z2(zo*3zI ) I (22+ zo+ 3zt )
fiom which it follows rhat
From Eqs. (11.22a)and (t I.Z2b)
Vss
I
laz
Fig. 11.9 Double line-to-ground.
(LLG) fault through impedanceZl
ot
II
Vaz
o
o
z;' ll E,,- 2,r,,,
+3zfI,,nl
z o t ) L o )L r , o _ J
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(rr.2s)
Modern powe1_syglgll 4!g!yg!s
406t'l
both sides by row marrix tl 1 1l and using Eqs. (11.19) and
1-.gltiitiUying
(1I.20), we ger
be more than the three-phasefault current.
\
From Eq. (11.22a),we have
O<
lx,
?
Eo- Ztlot=- Zzloz
i
loz= - (Ior + Io) fseeEq. (11.19)]
Substituting
l
.
Negative
Eo - Z rl o t= Z z (Io r+ Io s)
Fig. 11.12 Sequencenetworksof synchronous
generator
groundedthrough
neutralimpedance
or
'I u^ =o -E o - ( z t + z ' \ '
," 1"'
,-l
Substituting this value of Iooin Eq. (rr.26) and simplifying,
we finally get
=
f
"l
If the fault takesplacefrom loadedconditions,the positive sequence
network
will be modified as discussedin Sec. 11.3.
r---- ------ - l
i e+tflP.f
;;!utrl,,r|
(d) Write expressionfor neutral grounding reactance,such
that the LG fault
current is less than the three-phasefault current.
Solution (a) Figure 11.12 gives the sequencenetworks of the generator.As stated
earlier voltage source is included in the
positive sequencenetwork only.
(b) Connectionof sequencenetworks for a
solid LG fault (ZI = 0) is shown in Fig.
11.13, from which we can write the fault
current as
ll)rc -
Figure 11.11 showsa synchronousgeneratorwhoseneutral is grounded
through
a reactancaXn.The generatorhas balancedemfs and sequence
reactancesX1,
X., and Xu such that X, = Xz > Xo.
3 lE , l
zxt+&,*3X,,
(i)
(c) If the neutral is solidly grounded
llolLG= =
3lE"l
2\+xc
(ii)
For a solid three-phasefault (see Fig. 11 . 1 4 )
I
l I ) r r = ;E:o:l t_ 3 l E " l
xn
----l'FJf[\-: | : """ /
Ea
Comparing (ii) and (iii), it is easy to seethat
.--\"/
)
+))
llol Lc> llol3L
b
c
Fig' i i.i i
Synehronousgeneratorgroundedthroughneutralreactance
(a) Draw the sequencenetworks of the generatoras
seen from the terminals.
(b) Derive expressionfor fault current for a solid
line-to-groundfault on
phase a.
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(iii)
An important observationis made here that,
when the generator neutral is solidly
grounded,LG fault is more severethan a 3I_
fault. It is so because, Xo * Xr = X, in
generator.However, for a line Xo D Xt = Xz,
so that for a fault on a line sufficiently away
frorn generator,3L fault will be ntore severe
than an LG fault.
l q s =1 1 3 1 .
Fig. 11.13 LG fautt
n
E"()
I
ft
q
l
\
Xt'4
l
l
I
7L _ _ li
Fig. 11.14 Three-phase
fauft
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
- a . 1 . - . 4 s ^ & g . . . - ' r } . } . * ' * . 3 a - - - } l ! - } - } . } ' Q t p t |
408
usqelrl@
UnsymmetricalFault Analysis
(d) with generatorneu,tralgrounded through
reactance,comparing Eqs. (i)
and (iii), we have for LG faurt current to be less
than 3L fault
3 lE , l
(a) Ir=
If
J
-
3x1
:
j 0.0e+ j0.07s+ j0.1+ 0.9e 0.99+ j0.26s
- ' A A A
r
: A A 4 E ,
. n
r
llrl = 2.926 x -:L"
J3 xll
2X, + Xo + 3Xn> 3Xl
(c) Voltageacrossgrounding resisror=
= 3.07 kA
*1 2 r e.g2i - j0.756)
- 0.932- j0.249
(iv)
J
tltrffi
f
J
= 2.827 - j0.756
(b) Cunent in the grounding resisror= If = 2.g27 _ j0.756
24 + Xo+3X,
*^, I(xr - xo)
l'i
= 0.965
i;;;ili",i:rl
Two 11 kV, 20 MVA, three-phase,starconnectedgenerators
operatein parallel
a s s h trw ni n F i g ' l l ' 1 5 , th c p o s iti ve,negati vcand-zero
sequence
reactances
of
eachbeing,respectively,
j0.1g, j0.r5,7o.10 pu. The star point
of one of the
generatorsis isolatedand that of the otheris
earthedthrough a 2.0 ohm resistor.
A single line-to-groundfault occurs at the terminals
of one of the generators.
Estimate (i) the faurt current, (ii) current in grounding
resistor, and (iii) the
voltage acrossgrounding resistor.
= 6 . 1 3k V
For the systemof Example 10.3 the one-linediagramis redrawnin Fig.
I 1.16.
On a baseof 25 MVA and I 1 kV in generatorcircuit, the positive,nega"tiue
and
zerosequencenetworks of the systemhave been drawn alreadyin Figs.
10.23,
10.24and 10.27-Before the occurrenceof a solid LG at bus g, the
motors are
loadedto draw 15 and 7.5 MW at 10 kV, 0.8 leading power factor.
Ifiprefault
currentis neglected,calculate the fault current and subtransientcurrent
in all
partsof the system.
What voltage behind subtransient reactancesmust be used
in a positive
sequencenetwork il prefault current is to be accounted fbr'/
Xo = 0.06pu Tz e
(H+---d )( |
F i g .1 1 . 1 5
2.5A-
Xo = 300 fl
Solution (Note: All values are given in per unit.)
Since the two identicar generatorsoperatein paralrer,
Xr.o =
j0.18
= i0.0g,Xr"q=t-'#
= j0.075
Sincethe star point cf the secondgenerator
is isolated,its zero sequence
reactance
doesnot comeinto picture.Therefore,
Z o " q = 7 O .+130R n = j 0 . 1 0 + 3 x
solution The sequencenetworks given in Figs. 10.23, 10.24 and r0.2i
are
connectedin Fig. 11.17to simulatea solid LG fault at bus g (seeFig.
11.16).
[f
4:+
(ll)',
= 0 . 9 9 +ri-0 . 1
For an LG fault, using Eq. (11.1g), we ger
3E
Iy (fault current for LG fault) - Io = 3lo, =
Fig. 11. 16 one- r inediagr amof t hesyst emof Exam pr 11.
e 3
nreforrlt
n
r rrfrfvarnr fl D
o
vu
a rr va
q
nanlo^+^.i
rrvSlvvL(/\I
= vj (prefault voltage at g)
E'l= E',!,t= E',1,2
=
=
pu
1+ o.eoe
X t" o+ X z" ql Zo" ,
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ModernPowerSystemAnalysis
410 I
I
I
.l''.
E"o( )
dt
mbl
T-
Z z = Z r = j 0 . 1 6p u
Fiom thescquence
networkconnection
f
l02:l
(
Unsymmetrical
FauttAnatysir
.
' o rf
io'szs
rb-6T--___
_
Vf
J
737;4
= q'901 = - jo.Mlpu
--'>
j2.032
0.99+i0.607
Ioz= Ioo= Ior = - j0.447 pu
l- io.t+z
Fault current= 3loo= 3 x (- j0.447) = - jL341 pu
The componentof Io, flowing towards g from the generator side is
------+-i0.136
-i0.1 36
*_-- - j0.311
.447
j0.447
x !: ?:= = - 70.136pu
j0.7ss
and its componentflowing towards g from the motors side is
i 0.608
11.712
I -p.oo,
Fig. 11.17 Connectionof the sequence networksof Example 11.3.
currentsare shownon the diagramin pu for a solid
Subtransient
faultat g
line-to-ground
The positive sequencenetwork can now be easily replaced by its Thevenin
e q u i v a l e nats s h o w ni n F i g . 1 1 . 1 8 .
Reference bus
T
r
Now
- jo.Ml* i,?s?-s=
= - j0.311pu
j0.7ss
Similarly, the component of Io2from the generatorside is - j0.136 pu and
its componentfrom the motors side is -70.311. All of Iosflowstowardsg from
tnotor2.
Fault currentsfrom the generatortowards g are
I rl [-ro.r30ll-i0.2721
Il r,1l [ rl
.
l
l
l
l
6 l 7 0 . 1l 3p 6u
I r u l = l a z G r l l - i o . r rl =
Lr.J lo u2 rJL o j L jo.r36j
and to g fiom motors are
I
[t..] [t
I t o" l - |la 2l a^
l
L/,J la
e"
llf-ro.ltt1 [-.rl,06el
r l l - r o . s ul : l - i o . r 3 6 l p u
l
l
"
l
l
'
'
The positive and negative sequencecomponentsof the transmissionline
cunents are shifted -90" and +90o respectively, from the corresponding
componentson the generatorside of Tr, i.e.
Positivesequencecurrent = - j(-jO.136) - - 0.136 pu
Negative sequencecurrent - j(- j0.I36) = 0.136 pu
Zero sequencecurrent = 0
('.' there are no zero secluence
currents
on the transmissionline, see Frg. ll.17)
Line a current on the transmission line
=-Q.136+0.136+0=0
Iu and I, can be similarly calculated.
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l
I)L-j0.447J L-j0.136_l
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern power System Anatvsis
4l:iN
Unsymmetricat
FaultAnglysis
lll4#hF
Let us now calculatethe voltages behind subtransientreactancesto be used
if the load currents are accountedfor. The per unit motor currents are:
136.86 = 0.66+ j0.495pu
25x0.909x0.8
Vo= Vot *
("' /oo = 0)
V n z * V o o= 0 . 9 1
3orExample 11.2,assumethat the groundedgeneratoris solidly grounded.Find
.hefault current in each phase and voltage of the healthy phase for a double
ine-to-groundfault on terminals of the generator.Assume solid fault (Zf - 01.
)olution Using Eq. (1I.24) and substitutingthe values of Zp* Zr"rand Z*,
rom E x am ple1l. 2, we get ( not e Zf =0, Z0"q=j0. 1)
I
_
ral -
!-
t' '+r 7- 0
10. 075
^x i 0. 10
-io.oe.
{:
Yj:+ J::
l:
j0.075
j0.i0
Vot= Vo2= Vo1= Eo-
= 0.323
I
_
ra2-
tta 0 --
_Voz -
^"1+
nrr*a*#
rrrv frcrlrrL
t-tursrrl
^-J
,,^lr^-^
auLr vurLil$tr
^f
ul
eL^
ttl9
l-^^trr^--
lr€artlly
-r-
- - -
pllase
r
IOf
r.
a lfne-to-llne
r
a.
laUlt
On
terminals of the generators.Assume solid fault (Zf - 0).
Solution For the LL fault, using Eq. (11.I7) and substituting the values of
X,"u and Xr"u from Example 11.2,we get
0.323 - j4.306
Zr.,
j0.07s
Voo
0.323 =
j0. 10
Zo.,
Iu= rllot
For Example I1.2, assumethat the grounded generatoris solidly grounded. Find
- - ; 7 < ?
Ior Zr.q= 1 - (- j7.53) (/O.09)
'
-0.99 -j0.743 -j0.311= - 0.99 -jr.054
In this problem, becauseof large zero sequencereactance,load current is
comparable with (in fact, more than) the fault current. In a large practical
system,however, the reversewill be the case,so that it is normal practice to
neglect load current without causing an appreciableerror.
f}ro
u0.0e)
= 0.909 + j0.525 x I.2375136.86"
0.99 + -j0.743 -j0.136 = 0.99 + j0.607
andtheactualvalueof positivesequence
currentfrom the motorsto thefault
s
Iorxleq = 1'0 - (- j6.06)
Voltage of the healthy phase,
ruw:lrus ule Iault ls
l
3X-j6.06):-10.496
Now
Voo=- I^/'o-g
= 0.52 + j0.52 = 0.735145. pu
It may be noted that with thesevoltages behind subtransientreactailces,the
Thevcnin equivalent circuit will still be the same as that of Fig. 11.19.
Therefore, in calculating fault currents taking into account prefault loading
condition, we need not calculate EIy E/ft and E(. Using the Thevenin
eqtrivalentapproach,we can first calculate currentscauied by fault to which the
load currents can then be added.
Thus, the actual value of positive sequencecurrent from the generator
f^--lr
i0.09+ i0.075
= _ j6.06
= 0.455
Motor
2:E!"2
=l.l:t-,l';:::iir:tfi il r"
rl--
x*q* Xr",
Vot = Voz = Eo -
Motor
1:tr'^'
=lllt-r:';::="ri;Yilii.il
r"
.^--.^-l^
F
a
u
1y(fault current) = Io = -i
Total cuffent drawn by both motors = 0.99 + j0.743 pu
The voltages behind subtransientreactancesare calculatedbelow:
Generator,E'{
-
UsingEq. (11.15),we have
= 0.4125136.86'= 0.33+ j0.248pu
25x0.909x0.8
t
tal
.^ A^
/ 1
t 1
+ alo, + Ioo
= (-0.5 -/0.866) (-j7.53)+ (-O.5+ _i0.866)
U4.306)+ j3.23
= - 10.248+ j4.842 - 11.3341154.74'
Ir=
el,r, + ozlor*
Ino
= (-0.s + 70.866)(-j7.s3) + (-0.5 -j0.866) (j4.306)+ j3.23
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPower SystemAnalysis
4!4" 1
I
UnsymmetricalFault Analysis
Voltage of the healthy phase,
Vo = 3 Va t = 3 x 0 .3 2 3= 0.969
11.6
,415
|
- 10.248+ j4.842 - 71.334125.28"
Positivesequence
network
OPEN CONDUCTOR FAULTS
An open conductor fault is in series with the line. Line currents and series
voltagesbetween broken ends of the conductorsare required to be determined,
lc
Negativesequence
network
Zerosequence
network
cic'
fault
and voltagesin openconductor
Fig. 11.19 Currents
Figure 11.19 shows currentsand voltages in an open conductor fault. The ends
of the system on the sides of the fault are identified as F, F', while the
conductorends are identifiedas ua /, bb / and cc'. The set of seriescurrentsand
voltagesat the fault are
Fig. 11.20 Sequence networks for open conductor laull at FF/
[n terms of symtnctrical cotnponents, we can wrlte
Vonl *
l-l I
f v .l
l
'
,
I
l'""'l
l
t
,
l : v : l v .o. o, l' l
'r p
l-,1''P I
L / I,
Voo,2*
Vno,g = O
I o t = I o 2 = I n o -+ 1 "
L V ,), ,
The symmetrical componentsof currents and voltagesare
[ /.'I
1v,,,,,,1
t." = | I^. l: v":l v^-,"
I
" " ' l
l " ' l
|
L/.,,J
lVou'o)
The sequencenetworks can be drawn for the power system as seen from FF/
and are schematically shown in Fig. ll.2O. These are to be suitably connected
dependingon the type of fault (one or two conductorsopen).
Two Conductors
F IF'
Open
Figure lI.2l representsthe fault at FF /with conductors b and c open. The
currents and voltages due to this fault are expressedas
Voo'= 0
1 6 = I r - Q
(rr.27)
(11.28)
'tD
I
I
I
c " c '
I
F i g 1 1 . 2 1 T w o c o n d u c t o r so p e n
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Fig.11.22 Connection
of sequence
networksfor two conductors
open
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
tt'6 il
ModernPowerSystemAnatysis
t
Equations(11.29) and (11.30) suggesta seriesconnectionof sequence
networksas shown in Fig. II.22. Sequencecurrelrtsand voltagescan now be
computed.
prefault voltage Vf_. of bus r in series with the passive positive sequence
network (all voltagesourcesshort circuited).Sincenegativeand zero sequence
prefault voltages are zero, both these are passivenetworks only.
Onc Con-luctor Open
Reference bus
For one conductoropen as in Fig. 11.23,the circuit conditionsrequire
Vbb,=Vrr,=O
Io = O
for passive
positive
sequence
network
(11.31)
(11.32)
In terms of symmetricalcomponentstheseconditionscan be expressedas
Vool=
1
Voor2= Voory= *Voo,
(11.33)
J
(11.34)
Iot* In + Ioo=0
Equations(11.33) and (1I.34) suggesta parallel connection of sequence
networksas shown in Fie. 1I.24.
Fig. 11.25 Connectionof sequencenetworksfor LG fault
on the r th bus (positivesequencenetwork
by its Theveninequivalent)
represented
[--l
t
I
V""'z
I
l
It may be noted that subscripta has been droppedin sequencecurrents and
voltages, while integer subscript is introduced for bus identification. Superscripts o and /respectively, indicate prefault and postfault values.
For the passivepositivesequencenetwork
L-'i-- ' ryl
F
la
Vr-"us= Zr-nusJr-"ut
a
V t - u u s=
t
,c+-
(11.35)
where
c', c/
positivesequence
bus voltagevector (1 1.36)
lao
i
Fig. 11.23 One conductoropen
II.7
Fig. 11.24 Connectionof sequence
networks for one conductor
open
BUS IMPEDANCE MATRIX METHOD
UNSYMMETRICAL
SHUNT FAULTS
FOR ANALYSIS
: l- positive sequencebus impedance matrix
Zr-nus
Zt-nn )
OF
Bus impedance method of fault analysis, given for symmetrical faults in
Chapter9, canbe easilyextendedto the caseof unsymmetricalfaults.Consider
fbr example an LG fault on the rth bus of a n-bus system. The connectionof
sequencenetworks to simulatethe fault is shown in Fig. I1.25. The positive
sequencenetwork has been replacedhere by its Thevenin equivalent,i,e.
www.muslimengineer.info
Z-trl
/ 1 1 2 ?t \
\ L
r . J
)
and
[/'-'I
e cunent injectionvector
Jr-sus= l t t . ' | = p o s i t i vsee q u e n cbus
l : l
r
l
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(l1.38)
todern power SvstemAnalvsis
t4l8 |
I
As per the sequencenetwork connection,current - IJr_, is injected only at the
faulted rth bus of the positive sequencenetwork, we have therefore
From the Sequencenetwork connectionof Fig. 11.25,we can now write
--
vro-,
2r-,, * zz-,, I Zo-r,+3zf
rf
t2-r
(rr.47)
( 11 . 3 e )
other types of faults can be simila*Seomputed using
Zl-rr, Zr-r, and Zo-n in placeof Zr , Zr and Zoin Eqs. ( 1I . 7) , ( 11. 17) and
s ub s ti tu ti n gEq .(1 1 .3 9 ) i n Eq . (11.35),w e can w ri re the posi ti vesequence
voltage at the rth bus of the passivepositive sequencenetwork as
V,-,' = - Zr-rrlfr-,
( 11.40)
on prefault bus voltage, the voltage developed owing to the injection of
appropriatesequencecurrent at bus r'.
network,the voltagedevelopedat bus i owing
Foi passivepositive^sequence
to the injection of - IIr-, at bus r is
(11.48)
Vt-r=- Zr-,Jfr-,
ue
Thus the passivepositive seguencenetwork presentsan impedanceZr_ to the
r,
positivesequence
currentI{_r.
For the negativesequencenetwork
Vz-uus= Zz_susJz_nus
(11.41)
The negativesequencenetwork is injected with current lfr_, at the rth bus
only. Therefore,
0
0
Jz-sus=
-,i{,
(11.24)with E, - Vi-,.
by superposing
voltagesat anybuscannow be computed
Postfaultsequence
voltageat bus I is givenby
Hencepostfaultpositivesequence
V l - , = V i - , - Z r - , , f r - , ;i = l ' 7 ' " ' ' t t
( l r.49)
where
prefaultpositive sequencevoltageat bus i
Zr-,, = irth componentof Zt-"ut
Since the prefault negativesequencebus voltages are zero, the postfault
negativesequencebus voltagesare given by
Vf'-'=0+
(1r.42)
Vz-r
-- - zr-,rl'fr-,
(l r . s 0 )
where
0
l r - , , = i r t l t c o l l l p o l l c l t to l ' Z t - t , , t
The negative sequence voltage at the rth bus is then given by
Yr,=-
zr rrlf, I
Similarly, the postfault zero sequence bus voltages are given by
(l 1.43)
Thus, the negativesequencenetwork offers an impedanceZr_rrto the negative
sequencecurrentltr_,
Sirnilarly,fbr the zero scqucncenetwork
Vu-uu,= Zo-susJ,,-",r,
( 1 1. 4 4 )
0
0
Jo-sus=
- r^{u - r
( l l.4s)
0
Vo_, = - Zs_,.rlfs_,.,
(r 1.46)
That is, the zero s.equencenetwork off'ers zrn intpeclance Zrr_,.,.
to the zero
sequence cuffent l'-*r.
and
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Vd-' = - Zu-''lfu-''; j = l' 2' "'' tI
(i l . s1 )
where
= irth component tlf Z9- sLr.
With postfault sequencevoltagesknown at the buses,sequencecurrentsin
as:
linescan be comPr'rted
For line uv, having sequenceadrnittarlcesyl -ur, Jz',u,and yo-r,
f , - r r = l t - u , ( V f t - u- v [ - r )
(rr.s2)
I f r - r r = ! 2 - , , r (, V t - , ,- V 5 - r , )
I i, r - ur = Jo- u,( Vio- , - Vf o- r l
Knowing sequencevoltagesand currents,phasevoltagesandcurrentscanbe
easilycomputed by the useof the symmetricalcomponenttransformation
Vr, = AV"
Ir, = AI,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'i[di;,l
Modern powSr Jy$ern_4nglysis
Faultnnarvsis
Unsymmetrical
It appearsat first, as if this methodis more laboriousthan computingfault
currents from Thevenin impedancesof the sequencenetworks, as it requires
computation of bus impedancematrices of all the three sequencenetworks.It
must, however,be pointed out herethat once the bus impedancematriceqhave
been assembled,fault analysiscan be conveniently carried out for all the buses,
which, in fact, is the aim of a fault study. Moreover, bus impedancematrices
can be easilymodified to accountfor changesin power networkconfiguration.
l_+
Yr-aa=
_+^- =_ jtj.422
io.z io.o'os
Yr-re-Yr-a"=
to;;t
: jr2-4zz
Yt_,,=
Yr-n=
#*
.#
Yr-"f-For Example 10.3, positive, negative and zero sequencenetworks have been
drawn in Figs. 70.23, 10.24 and 10:27. Using the bus impedancemethod of
fault analysis,find fault currents for a solid LG fault at (i) bus e and (ii) bus
I Also find bus voltages and line currents in case (i). Assume the prefault
currents to be zero and the prefault voltages to be 1 pu.
Solution Figure 11.26 shows the connection of the sequencenetworks of
Figs. 10.23, 10:24 and 10.27 for a solid LG fault at bus e.
.
-Yt-88
Positive
sequence
j0.085
@
@
I
= _ j0.62I
j1.608
1
'vo - e e -- 'Yt t,- t t -- - -
Yr-rr=
Negative sequence
network
i 0.6e
iO'345
@
i0.0805
e
f
I
0
0
12.422
0
6.097
18.519
t2.422
12.422
6.097 - 18.519
0
-16.769
12.422
0
0
v
rv I - B U S _ I 2 _ B U S _
10.345
@
I
I
=-i16.769
+
j0.345 j0.69
*
d
-t7.422
rv o - d d -
a
network
- -i78.s1e
i6.0s1
*h:
E"*
I
liffiffi
j0.0805
=-i14.446
'
j0.494
1
jth=-io'584
Y o - a r =o ' o
-.1
Yo-,r=jofu = j2.024
Yo-fs= 0.0
Zero sequence
network
{
@
d
-0.621
@
JU.4v+ /U.UUUs
vI
_
O_BUS
v/
Fig. 11.26 Connection
of the sequencenetworksof Example11.6 for an LG
faultat bus e
Refer to Fig. 11.26to find the elementsof the bus admittancematricesof the
three sequencenefworks.as follows:
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orl
0
" r r0l
0
e
f
-r4.M6
0
2.024
-14.446
0
2.024
0
0
o
6
0
0
0
-0.584
lnverting the three matricesaboverendersthe fbllowing threebus impedance
matrices
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Analysis
- - 0.417pu
VL" = - Zu,,Ifo-, = - j0.0706x (- j2.362)
- - 0.167pu
0
0
0.07061 0.00989
0.00989 0.07061
0
0
Vfz-r=- 4-r"llr1 =-j0.08558 x(- j2.362)
0
0
= - 0.202pl
0
Vfus=-ZurJfu"=O
r.71233
Using Eq. (11.52), the currents in various parts of Fig. 1I.26 can be
cornputedas follows:
The fault currentwith LG fault on busc is
il=
e
3x1
j 0J7636+ j 0.r7636+ j 0.07061 = j7.086pu
(i)
The fault curent with LG faulton bus
/'is
Irr
__3x1
.i0.t82gg+ iOlAZgg+/O"O?Gi
= t.0 - j0.t2s7s(-' 7'0ttb = 0.703pu
\ " 3 ))
Vft-t= Vi-t - Z,
tu- If , .,
= r.o- jo.ns4j(-r, orru = o.zzsp,
)
V J r - "= V " r - " - Z r _ " o - l J
r_"
= 1.0- i0.17638
(- j2.363)= 0.584pu
vI-r= vi-s - z,-r"-I{-,
= 1.0- j0.08558(- j2.363)= 0.798pu
vfz-f= - zr-.fJfr-"
- - j6.097 (0.728- 0.s84)
= _ 70.88
)
j0.436s9
a
= - 7 6 . 8 7 1p u
(ii)
Bus voltagesand line currentsin case(i) can
easily be computcdusing
Eqr. (11.49)-(tI.sZ). Given below is a sample
calculation for computing
voltage at bus f and current in line ef
FrorrrEq. (11.49)
VI-a= Vi_a- Zr_0"- Ifr_"
II+= Yr-pUI+- vI+,)
If ,-0"= Yr4" (Vf ,-a -
Vf ,-r)
= - j12.422(0.703- 0.584)= - jL482
I,,t= If t-f" * If t-,t, = - 70.88 + (- ,tl .482)
- - i2.362
which is the same as obtainedearlier [seeEq. (i)l where If, = 3lut.
IJrsf = YFcf (vf '-, - vf '-)
- j12.422 (-0.798 - 0.728) = - i0.88
Notice that as per Fig. 11.26,it was requiredto be the sameas llr-s".
Iz-f" = Yr-r, (Vfr-r - Vf,-r)
= - j6. 097 ( - 0. 212+ 0. 417)= - 7O . 884
IL- tr.= nt-1a(Vtt-1- Vfrr,)
= - j2.024 (- 0.023+ 0.167) = - jO.29I pu
Ittn (a) = IJr-f" * It)-r, + {*
= - j0.88 + (- 70.88)+ (- j0.291)
- - jz.os
Sirnilarly,other currentscan be computed.
= - ./0.11547
x (- j2.362)= - 0.272pu
Vfo-r=- Zo-f"Ifo-,
=- j0.00989x (- j2.362)
= _ 0.023pu
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A single line to ground fault (on phasea) occurs on the bus I of the system of
Fig. 11.27.Find
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
qU;l
ModernPowerSystemAnatysis
FaultAnalysis-T
Unsymmetrlcal
f,r'4l#
i t-0.105 0.0451
zr-ws= rLo.o+s
Zz-.sus
o.1o5.J:
( i)
Tero sequencenetwork of the system is drawn in Fig. II.29 and its bus
ted below.
matnx ls
Referencebus
Ftg. 11.27
(a)
0.15
0.15
0.05
0.05
Current in the fault.
(b)
sc current on the transmissionline in all the three phases.
:
(c) SC current in phase a of the generator.
(d) Voltage of the healrhy phasesof the bus 1.
Given: Rating of each machine 1200 kvA, 600 v with x, = x, = rTvo,
xo = 5vo. Each three-phasetransformeris rated rz0o kvA, 600 v - nlgroo
V-Y with leakagereactanceof SVo,The reactancesof the transmissionline are
xr = Xz = 20vo and Xo = 40vo on a baseof 1200 kVA, 3300 V. The reactances
of the neutral grounding reactors are 5Vo on the kVA and voltage base of the
machine.
0.05
0.05
Fig. 11. 29
Bus 1 to ref'erencebus
Zo-sus= i [0.05]
Bus 2 to bus I
Note: Use Z"u, method.
Solution Figure 11.28 shows the passive positive sequencenetwork of the
systemof Fig.l1.27. This also representsthe negativesequencenetwork for the
system.Bus impedancematrices are computedbelow:
0.4
l-0.05 0.051
= /Lo.os
Zo-nus
0.451
Bus 2 to referencebus
l-0.0s 0.051
7
zO-BUS
Reference bus
-
;
l-0.051
_l
l _ l-t o . o so . 4 s l
'Lo.os0.4s1
0.45
+ 0.0s10.451
or
.1[0.04s 0.0051
= JZIO.OO5
Zo-sus
0.0451
0.05
0.2
0.05
II-t =
Flg. 11.28
Zt_svs= j[0.15]
Bus 2 to Bus I
Bus2 to reference
bus
VO
, 3Zl
Z ; t r * Z z _ t r* Z o _ r +
But V", = I Pu (sYstemunloaded before fault)
Then
-j1.0
Bus I to referencebus
o '.1t S5
Z r - 8, tr ,-. =r,lfO
As per Eq. (11.47)
fr_,
o l5t
;;;l
0.1s1
z L Brs
r , .-="/Lo.rs
,[o'15
o35J**
= - j3.92 pu
,0.105+0.105+0.045
Ifr-t=ltr;=f
-^2Pu
(a) Fault current, I\ = 3If ,-, =
r
l-O.tst
""
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[o'ls o'3s]
Lo.rt.l
O) Vfvr = Vo r-, = Zt-rr lfr-t
= 1. 0- 70. 105x - j3. 92 = 0. 588; Vot ; = |
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(ii)
Vf t-r= Vor-r- Zr-rJfr-ri Vora = 1.0 (systemunloaded
beforefault)
= 1.0- j0.M5 x - j3.92 = 0.g24
r
(c) 4-c =
l
(1 - o's88)t-33"
,o;
---1.37-i2.38
vtr;= - zr-trfr-t
= -70.105x - j3.92 = 0.412
.:= to - (- 0.412))t3o"
j0.1s
r.37- i2.38
rLc=
V{-z= - Zr-r, Ifr-,
- - j0.045x - j3.92 = _ 0.176
IIo-c= 0 (seeFig' 1I'29)
If o_c= et37 - j2.38)+ (1.37- i2.38)
= _ j4.76
vfo-t= - zurrlL,
- - j0.045x - j3.92 = _ 0.176
vt-z=- Zuy IL,
= - 7O.005
x - j3.92 = - 0.02
I{-rz= yvrz (VI_r- Vfr_r)
Currentin phasesb andt:cof the generatorcan be similarly calculated.
(d) Vfo-r= ZVfvr + Vf,-, + Vfur
= 0.588 1240"- 0.4L2 1120"- 0.176
= - 1- (0.588
- 0 . 8 2 4 ) =j l . r 8
j0.2
= - 0.264- j0.866= 0.905 l-
107"
VIr-t= Vf,-t + Vfr-, + Vfu'
= 0.588ll20' - 0.412 1240"- 0.176
Ifr-rr= !z-n U{-t - Vfr-r)
= *.r0 . 4 1 2+ 0 . 1 7 6 =
) i1.18
j0.2'
- - 0.264+ i0.866= 0.905 1107"
If o_r,= lo_tz(Vfu, - Vfur)
=
;,
Iro _ tz =j l .1 8 + 7 1 .1 8+ j 0.39_ j 2.75
. , - -_ i lr . tr a
-7 f D
u
_t.z J
LEIVIS
PROB
(- 0'176+ o.o2o)= io.3s
1 vA I r o _rr . ' 1 - ro
4. -/ a
J r. ro
/7^fr
z_tLv
11.1 A 25 MVA, 11 kv geaeratorhas a x"o= 0'2 pu' Its negative and zero
sequencereactancesare respectively0.3 and 0.1 pu. The neutral of the
generatoiis solidly grounded.Determine the subtransientcurrent in the
generatorand the line-to-line voltagesfor subtransientconditions when
an LG f'ault occurs at the generatorterminals. Assume that before the
occuffence of the fault, the generatoris operating at no load at rated
voltage. Ignore resistances.
11.2 RepeatProblem 11.1 for (a) an LL fault; and (b) an LLG fault.
11.3 A synchronousgeneratoris rated 25 MVA, 11 kV. It is star-connected
with the neutral point solidly grounded.The generatoris operating at no
=
load at rated voltage. Its reactancesare Xt' = Xz = 0.20 and Xo 0'08
. .n ^^
+ Jv.Jy
pu.
= _ j07g
If,-rz= jl.18 lI20 + 71.1glZ4V + il.3g
= j0.79
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ll.4
rr^r^i-r^+^
\-aruulalLE
+l^^
Llls
-.,m,-o+einol
DJrluuvlrrv4r
ou
r rvl rufer or an c i c n t
o
line
etrrre.nfs
for
(i)
S-l -e
-S-i-n- o
line-to-ground fault; (ii) double line fault; (iii) double line-to-ground
fault; and (iv) symmetrical three-phasefault. Compare these currents
and comment.
For the generatorof ProblemI 1.3,calculatethe valueof reactanceto be
included in the generatorneutraland ground, so thatline-to-groundfault
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Mqdernpower Syst€m Anatysis
lW:il
current equals the three-phasefault current. What will be the value
of
the groundingresistance
to achicvcthc samcconclition,l
sub-^
t*t.
ignored.
current
prelault
with the reactance value (as calculated above) included
between
neutral and ground, calculate the double line fault
current ancl rlso
double line-to-ground faul
7.5MVA
3.3/0,6kv
X = 10o/o
11'5 Two 25 MVA, 11 kv synchronousgenerators
are connected to a
common bus bar which suppliesa feeder. The star point
of one of the
generatorsis groundedthrough a resistanceof 1.0 ohm,
while that of the
other generatoris isolated.A line-to-ground fault occurs
at the far end
of the feeder. Determine:(a) the fault current; (b) the voltage
to ground
of the sound phasesof the feeder at the fault point; and
bi
r.l
"orilg"
the star point of the groundedgenerator with iespect to ground.
The impedancesto sequencecurrentsof eachgenerator
and feeder are
given below:
Positive sequence
Negative sequence
Zero sequence
rI'6
Generator
(per unit)
jo.2
Feeder
(ohms/phase)
i0.15
j 0.08
j0.4
j0.4
j0.8
Determine the fault currentsin each phase following a double
line-toground short circuit at the terminals of a star-connected
synchronous
generator operating initially on an open circuit voltage
or i.o pu. The
positive, negative and zero sequencereactance
of the generator are,
respectively,70.35,j0.25 andj0.20, and its star point is
isolated from
ground.
11'7 A three-phasesynchronousgeneratorhas positive,
negative and zero
sequencereactances
per phaserespectively,of 1.0,0.g and 0.4 ohm. The
winding resistancesare negligible. The phase sequenceof
the generator
is RYB with a no load voltageof I I kV betweenlines.
A short circuit
occurs between lines I and B and earth at the generator
terminals.
Calculate sequencecurrentsin phaseR and currentin the
earth return
circuit, (a) if the gencratorneutralis solidly earthecl;
ancl (b) il the
generator neutral is isolated.
Use R phase voltage as reference.
11.8 A generator supplies a group of identical motors
as shown in Fig.
P-11'8. The motors are rated 600 V, 9O%oefficiency at full
load unity
power factor with sum of their output ratings being i
rrrrW.The motors
afe
sharino
enrrqllrr
l
^€./t l\trrr
r
---'.
so rl ^vos. u
\ ^ ' r l v r v v i^ltr l- a^ .L^ €
u v'r'age, u.6 power tactor
lagging and 90vo efficiency whea an LG fault occurson
the low voltage
side of the transformer.
Specify completely the sequencenetworks to simulatethe
fault so as
to include the effect of prefault current. The group of motors
can be
treatedas a singleequivalentmotor.
www.muslimengineer.info
fauft X'= Xz= 2oo/o
Xo = 5o/o
Xn = 2'5o/o
Yr6i*
F i g .P - l 1 . 8
n.g
in the
A double line-to-ground fault occurs on lines b and c at point F
c of
phase
in
system of Fig. i-tf.q. Find the subffansientcurrent
are
machines
Both
machine 1, assuming prefault currents to be zero.
xo=
and
xz=lUvo
x//=
of
rated 7,200kvA,600 v with reactances
trtnslorutcris rutcd 1.200kVA. 600 V-A/3.300
5%. llac5 thrcc-phirsc
V-ts with leakage reactanceof 5Vo.The reactancesof the transmission
line areX,=X,-=20vo andXo= 4oToon a baseof 1,200kVA, 3,300V'
kVA
The reactancesof the neutral grounding reactors are 5Vo on the
base of the machines.
_r_rr*"1_
i^
,_L
vn6l
-:
F l g .P ' 1 1 . 9
through
11.10 A synchronousmachine1 generating1 pu voltageis connected
lines in
transmission
two
pu
to
0.1
reactance
of
transformer
a Y/Y
a YN
through
parallel. The other ends of the lines are connected
pu
voltage'
1
generating
transformerof reactance0.1 pu to a machine2
For both transformers X, = Xz = Xo'
fault on the line
Calculatethe current fed into a double line-to-ground
'2.
'side
terminals of the transformer fed from machine The star point of
The
machine I and of the two transiormers.are soiiriiy grourrded.
are
base
reactancesof the machines and lines referred to a common
xo
Xz
xl
0.05
0.25
0. 35
Machine 1
0.04
0.20
0.30
2
Machine
0. 80
0.40
0.40
Line (each)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpower
'
ll'll
,:i
;
;il;;
;
;".il:i
i.H
flffi
9,,.*
il"lilij.1:'
:" :i::,{:,il.:
f:
"
uninrinitefrusil; *'' ;ffi# Hffi ff
I lncrrrnmatrinal
Farrlt
Anahraio
lL.'A+\l
nerwork
withtwogenerators
connecred
iltr"::,i"il"ti .1T:-Toow.er
Xr = Xz = 0.1 pui Xo = 0.05 pu
X, (Brounding reactance)= 0.02 pu
= 0J?u
Transformer: Xr: Xr:Xt
Generator:
#,f
:T"tl': "."11'::::1.r:
i,:il".5.T1",ffi,1,iJl.:^::l:l;,-i""i;ffi*"#,J'#::,irfr
Thepot+-: q"geEye
unozeroseque;r1:#;:"#
fl:f{",,":t'
viltr'om vv'rv\,rerr,-"in pEr
fiit *:"
Positive
Zero
0.0g
GeneratorI
0.15
0.15
Generaror2
0.25
0.25
Each rransformer
oo (i.e. neutral isolated)
0.15
0.15
Infinite bus
0.15
0.15
0.15
0.05
0.20
0.20
0.40
Line
X, (grounding reactance)= 0.04 pu
Form the positive, negative and zero sequencebus impedancematrices.
For a solid LG fault at bus 1, calculate the fault current and its
contributionsfrom the generatorand ffansformer.
""*"t-;;;il#,
Negative
1
_f6l-Y
Hint: Notice that the line reactancesare not given. Therefore it is
convenientto obtain Zt, svs directly rather than by inverting Ir, sus.
Also ro,
it singular and zs, BUScannot be obtained from it. In such
"us
situations
the method of unit current injection outlined below can be
used.
For a two-bus case
-_t
li;,1=17,',7,',)l';,1
F i g .p - 1 1 . 1 1
rl'r2 A star connectedsynchronous
generatorfeeds bus bar r of
a power
system. Bus bar I rs connected
to bus bar 2 thro'gh a star/crerta
lt'itnsl0t'ttlcl'ilt scrics with a
transmissionline. The power
network
connectedto bus bar 2 can
be,
connectedgeneratorwith equar"quiuutently'representedby a star_
positive incr ncg.tivc sc(rr.r0rccs
rcactances.
Alr star porntsare solidry connected
to ground.The per unit
sequencere:lct*ncesof v'rious
corrlponentsare given berow:
Nc,,gutivt:
0.l5
Transfbrmer
0.05
OJZ
0.12
TransmissionLine
PowerNerwork
0.12
0.30
0.30
0.50
X
X
0.10
Under no load condition with
1.0 go voltage at each bus
bar, a current
of 4'0 pu is fed to a three-phase
short circ-uit on bus bar Z.Deitrmhe
the positive sequence reactance
X of the equivarent generator
of the
,, . power network.
For the same initial conditions,
find the faurt current for single
line_
to-ground fault on bus bar
l.
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Injectingunit currentat bus 1 (i.e. Ir = tr, !2= 0), we get
Zn= Vt
Zzt = Vz
S ir nilar lyinjcct ingr r uitcur r cutut bus 2 1i. c./ r = 0, lz = l) , we get
Ztz = Vl
7:tz= Vz
Zt:ro
0.20
2
F i g .P - 1 1 . 1 3
\--7
| )r I
I
- x lI_ J]L | l
,
Itz2,H
/ "\ . .Y' -
pol;itive
L
rTft-YA
sequencenetworks of the power
l.] P_-....*,the
;;_
(b)
with borh generarorsand infinite
bus op"rutinf
r.o pu voltage on
no road,a rine-to-grouncr
faurt occursat .ne of "a
the
terminarsof the
star-connectedwinding of the
transformerA. caiculate the
currents
flowing (i) in the fauli; and (ii)
through rhe transformer A.
6L-_' ,q
Generator
,
€fff
ll.l4
Zou5could thus bc dircctly obtainedby this technique.
Considerthe 2-bus systemof Example 11.3.Assumethat a solid LL
fault occurson busf Determinethe fault currentand voltage(to ground)
of the healtlry phase.
11.15 Write a computer programme to be employed for studying a solid LG
fault on bus 2 of the sy'stemshown in Fig. 9.17. our aim is to find the
fault current and all bus voltages and the line currents following the
fault. use the impedance data given in Example 9.5. Assume all
transformers to be YlA type with their neutrals (on HV side) solidly
grounded.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
il
Assume that the positive and negative sequence,reactances of the
generatorsare equal, while their zero sequencereactanceis one-fourth
of their positive sequencereactance.The zero sequencereactancesof the
lines are to be taken as 2.5 times their positive sequencereactances.Set
all prefault voltages = 1 pu.
12
NCES
REFERE
Books
1.
2.
3.
4.
5.
6.
Stevenson,W.D., Elementsof Power SystemAnalysis,4th edn., McGraw-Hill,
New York, 1982.
Elgerd, O.I., Electric Energy Systems Theory: An Introduction, 2nd edn.,
McGraw-Hill, New York, 1982.
Gross,C.A., Power SystemAnalysis, Wiley, New York, 1979.
J.R., Modern Power Systems,InternationalTextbook Co., Ncw
Ncuenswander,
York, 1971.
Bergan,A.R. and V. Vittal, Power SystemAnalysis,2nd edn.,PearsonEducation
Asia, Delhi, 2000.
Soman, S.A, S.A. Khapardeand Shubha Pandit, ComputationalMethods for
Large SparsePower SystemsAnalysis, KAP, Boston, 2002.
Papers
7.
8.
Brown, H.E. and C.E. Person,"Short Circuit Studiesof Large Systemsby the
ImpedanceMatrix Method", Proc. PICA, 1967,p. 335.
Smith, D.R., "Digital Simulation of SimultaneousUrrbalancesInvolving Open
IEEE Trans.PnS, 1970,1826.
and FaultcdConductors",
T2.T
INTRODUCTION
power systemis its ability to returnto normal
The stabilityof an interconnected
or stable operation after having been subjectedto some form of disturbance.
Conversely, instability means a condition denoting loss of synchronism or
falling out of step. Stability considerationshave beenrecognizedas an essential
part of power system planning for a long time. With interconnectedsystems
continually growing in size and extending over vast geographicalregions, it is
becomingincreasinglymore difficult to maintain synchrortismbdtween various
'.
parts of a power system.
by its basic featuresgiven
The dynamicsof a power system are characterised
tjElt
w.
1. Synchronoustie exhibits the typical behaviourthat as power transfer is
gradually increaseda maximum limit is reachedbeyond which the system
cannot stay in synchronism,i.e., it falls out of step.
2. The systemis basically a spring-inertiaoscillatory system with inertia on
the mechanicalside and spring action providedby the synchronoustie wherein
power transfer is proportional to sin d or d (for small E, 6 being the relative
internal angle of machines).
3. Because of power transfer being proportional to sin d, the equation
determining system dynamics is nonlinear for disturbances causing large
variations in angle d, Stability phenomenonpeculiar to non-linear systemsas
distinguished from linear systems is therefore exhibited by power systems
(stable up to a certain magnitude of disturbance and unstable for larger
disturbances).
Accordi^rglypower system stability problemsare classified into three basic
types*-steady state,dynamic and transient.
*There are no universally accepted precise definitions of this terminology. For a
definition of some important tenns related to power system stability, refer to IEEE
Standard Dictionary of Electrical and Electronic Terms, IEEE, New York, 19i2.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'
l
434 .'l
Modernpower SystemAnalysis
Th"t study of steady state stability is basically concerned
with the
determinationof the upperlimit of machineloadingsbeiore losing
synchronism,
providedthe loading is increasedgradually.
Dynamic instability is more probable than steady state instability.
Small
disturbances are esntinuaHy oeeurring irr a po*..
system r"#"ti"*
i"
loadings,changesin turbine speeds,etc.) which are small enough
not to cause
the systemto lose synchronismbut do excite the systeminto the"itate
of natural
oscillations.The system is said to be dynamically stable if the
oscillations do
not acquiremore than certainamplitudeand die out quickly (i.e.,
the systemis
well-damped).In a dynamically unstablesystem, the oscillation
amplitude is
large and thesepersist for a long time (i.e., the system is underda-p"a;.
rni,
kind of instability behaviour constitutesa seriousthreat to system
security and
createsvery difficult operating conditions. Dynamic stability
can be significantly improved through the use of power system stabilizers.Dynamic
system
study has to be carried out for 5-10 s and sometimesup to
30 s. computer
simulationis the only effective meansof studying dynamic stability
problems.
The samesimulationprogrammesare, of course,appiicableto transient
stability
s t udi e s .
Following a sudden disturbance on a power system rotor
speeds,rotor
angulardifferencesand power transferundergofast changeswhose
magnitudes
aredependentupon the severityof disturbance.For a large disturbanc",
in angulardifferencesmay be so large as to .:ausethe machines
to fall"hung..
out of
step' This type of instability is known as transient instability
and is a fast
phenomenonusuallyoccurringwithin I s fbr a generatorclose
to the causeof
disturbance.There is a large rangeof disturbanceiwhich may
occur on a power
system,but a fault on a heavilyloaclecl
line which requiresopeningthc lipc t<l
clear the fault is usually of greatestconcern.The tripping of
a loadJd generator
or the abrupt dropping of a large load may also causeinstability.
The effect of short circuits (faults), the most severetype of
disturbanceto
which a power systemis subjected,must be determinedin nearly
all stability
studies' During a fault, electrical power from nearby generators
is reduced
drastically,while power from remote generatorsis scarcelyaf1'ecte4.
ln some
cases'the system may be stable even with a sustainedfault,
whereas other
systems will be stable only if the fault is cleared with sufficient
rapidity.
Whetherthe systemis stableon occurrenceof a fault depends
not only on the
systemitself, but also on the type of fault, location of fauit, rapidity
of clearing
and method of clearing,i.e., whether clearedby the sequential
opening of two
or more breakersor by simultaneousopening and whether or
not the faulted line
is reclosed.The transientstability limit is atmost always lower than
the steady
statelimit, but unlike the latter,it may exhibit different values
dependingon the
nature,location and magnitudeof disturbance.
Modern power systemshave many interconnectedgenerating
stations,each
with severalgeneratorsand many loads.The machineJlocated
a-tany one point
ln a systemnormally act in unison.It is, therefore,common practice
in stability
www.muslimengineer.info
Dnrrrn'
Qrra+^-
Or^L:l:r-.
I
-^-
machines which are not separatedby lines of high reactance are lumped
togetherand consideredas one equivalent machine. Thus a multimachine
system can often be reduced to aq equlyalg4t fery
lq4qhrue system- If
Synchronlsmii lost, ttrern'achineiof eaitr gioup stay togetheralthough they
go
out of step with other groups.Qualitative behaviour of machinesin an
actual
systemis usually that of a two machine system.Becauseof its simplicity,
the
two machine system is extremelyuseful in describingthe general concepts
of
power system stability and the influence of varioust'actorson stability.
It will
be seenin this chaptertbata two machine systemcan be regardedas a single
machinesystemconnected.to
infinite system.
Stability study of a multimachinesystemmust necessarilybe carried out
on
a digital computer.
I2.2
DYNAMICS
OF A SYNCHRONOUS
MACHINE
The kinetic energy of the rotor at synchronousmachineis
K E = 1 JJ,^ x 10-6 MJ
2
where
-/ = rotor moment of inertia in kg-m2
aro, = synchronousspeedin rad (mech)/s
But
whorc
u.r,n= rotor speed in rad (elect)/s
P = nuurbelo1'rnaclrinepoles
KE=+(t(?)'c.,.
xro-.)*
-L M,
2
'
,
M = J ( ? \ ' u , x' ,1 0 - 6
\P/
= moment of inertia in MJ-s/elect rad
We shall detine the inertia constantH such that
G H = K E =! u % M J
2
G = machinerating (base) in MVA (3-phase)
H = inertict constant in MJ/I4VA or MW-s/MVA
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Power SystemStability
'.: -: il
MociernPower Sysiemnnaiysis
{36 i
It immediatelyfollows that
M =
=
2GH = GH
MJ-s/elect rad
(ts
lt f
(r2.r)
ffi
14J-s/electdegree
180-f
M (pu) = +
nf
= H
ffi
and electrical
;ure 12.1 shows the torque, speedand flow of mechanical
triction and
windage,
wersin a synchronousmachine.It is assumedthat the
the rotor
governing
n-loss torque is negligible. The differential equation
namicscan then be written as
- d ' o ^ =T^-
J -: ; t
ot-
M is also calledthe inertia constant.
Taking G as base, the inertia constantin pu is
(r2.2)
s2lelectrad
437
Equation
e Swing
I
II
(r2.3)
r " Nt
'here
0* = angle in rad (mech)
T* = turbine torque in Nm; it acquiresa negative value for a motoring
machine
T, = electromagnetictorque developedin Nm; it acquiresnegativevalue
),,
degree
s'lelect
for a motoring machine
The inertia constant H has a characteristicvalue or a range of values for
each classof machines.Table 12.1 lists some typical inertia constants.
o_>
t m
, Table 12.1 Typicalinertiaconstantsof synchronousmachines*
Type oJ Mat:hine
Turbine Generator
Condensing
Intertia
Con.slunt H
Stored Energy in MW Sec per MVA**
machine
and electricalpowersin a synchronous
Fig.12.1 Flowof mechanical
9-6
7-4
4-3
While the rotor undergoesdynamics as per Eq. (12'3), the rotor' speed
(1s) [Sec.
changesby insignificant magnitude for the time period of interest
0.i. Equation (12.3) can therefore be converted into its more convenient
1,800rpm
3,000rpm
3,000rpni
Non-Condensing
Water wheel Gencrzttor
Slow-speed(< 200 rpm)
High-speed(> 200 rpm)
S y n c h l o r r o uCso r r d c r r s c+r4' +
Large
Small
Motor with load varying li'ortr
Synclrrcnous
1.0 to 5.0 and highcr lor hcavy l'lywhccls
.
lnrltar
,vvrvr
L-J
2-4
Tnrm
tvrrlt
vJ
rl l !l c: a *l t nr l rt r^l- r J( U' nU UeUr r . l tL nv lrvcl l nr qrr at l i n
c! v nr r nu rsq rt r:rt n t a t t h g S V n C h f O n O U S
I
peed(ur,.). Multiplying both sidesof Eq. ( 12.3)by u),,^'we can write
t':t);"
x lo-(' - P^
J6"n,
t.25
l.00
P,. Mw
(t2.4)
dtt
where
2.00
It is observedfrom Table 12.1 th'atthe value of H is considerablyhigher for
thzrntbr watc:rwheelgenerator.Thirty to sixty per cent of
steamturbogenerator
of
a
steam turbogeneratorunit is that of the priine mover,
the total inertia
whereasonly 4 -I5Vo of the inertia of a hydroelectric generatingunit is that of
the waterwheel,including water.
p
'ttt
* Rcprinted with permission of the Westinghous Electric Corporation from
Electrical Transmission and Distribution Reference Book.
+* Where range is given, the first figure applies to the smaller MVA sizes.
*tc+ Hydrogen-Cooled,25per cent less.
-
D
t - , -
mechanical ptlwer inPut l n M W
electrical power outPttt t n MW; statorcopPerlossis assumed
negligible.
Rewriting Eq. (12.4)
s2a
/ )\2
x 1o-6)
ul;) u.r,
ff
where
www.muslimengineer.info
- Ln v- . 2. \,\.l .l l-l l,l =l l y* : * *
ol
- P^ P,Mw
0, = angle in rad (elect)
- pP"
nod'\, =
= ,P,,Mi;-
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(12.5)
Modernpower SvstemAnalvsis
43S I
t
- -
PowerSystemStabitity
_
It is more convenientto measurethe angularposition of the rotor with respect
to a synchronouslyrotating frame of reference.Let
o=
ul ar di sP1acement fro m sy nchronou sl y
(a;,?;[' ;"':?:
f#
(called torque angle/power angle)
Grnu"h= machine rating (base)
Equation (12.11) can then be written as
(r2.6)
+-" (Yry*l =(p-- p.)g'""
Gsystem\-
(r2.7)
'
26
, , - d: =
M
P^-PeMw
ot-
(12.8)
dt'
/
\
)
fV-P*-P"Mw
\..Gt""'n
(t:.r2)
(r2.r3)
/
= machine inertia constantin system base
(r2.9)
Machines
Swinging
Coherently
Consider the swing equationsof two machinesor a common systembase.
Dividing throughoutby G, the MVA rating of the machine,
a
Hr d261
M(pu)+ = P*- P,;
dt'
in pu of machine rating as base
Grrr,.*
Hryr,"*= H-u"h
**"
t'+*l
where
With M as definedin Eq. (12.1), we can write
GH d2d
c'
"t
{ " " " ' o ' ! = 'P*-- PP"PUin sYstem
base
dl
"f
or
Hence Eq. (12.5) can be written in terms of d as
I +Sl
_
Gryrt",o= system base
From Eq. (12.6)
d rg , _ d 2 6
,Jt2
dt2
-
a;
"f
(12.r0)
H2 d262
A;
where
= P*r - P"t Pu
(r2.14)
= P*2- P"zPu
. (12.rs)
"f
Since the machinerotors swing together(coherentlyor in unison)
6, = [r= [
=+
M(Pu)
lrj
A ddi ng Eqs ( 12. 14)and ( 12. 15)
or
H dzb
= P*^nf dt"
P, pu
This equation(Eq. (12.1D)l}q.(12.11)),is called thestvingecluatioyand it
describesthe rotor dynamicsfbr a synchronousmachine(generating/motoring).
It is a second-orderdifferential equationwhere the darnpingterm (proportional
to d6ldt)is absentbecauseof the assumptionof a losslessmachineanclthe fact
that the torqueof damper winding has been ignored. This assumptionleads to
pessimisticresultsin transientstability analysis-dampinghelps to stabilizethe
system.Dampingmust of coursebe consideredin a dynamic stability study.
Sincethe electricalpower P, dependsuponthe sine of angled(seeEq. (12.29)),
the swing equationis a non-linear second-orderdifferential equation.
Multimachine
System
In a multimachinesystema common systembase must be chosen.
Let
H"q d26
(IZ.1I)
= P*-
P"
(r2.16"
Prr=P*r+ Pn
Pr=P"l * Prz
H"q=Hr+ H,
(r2.r7)
trf
dtz
whcrc
The two machinesswinging coherentlyare thus reducedto a single machine as
in Eq. (12.16).The equivalentinertia in Eq. (12.17) can be wrirten as
H"q, = Hl ,nu.h Gl ^ach/Grystem* Hz ^u"h G2 -u"h/G.yr,..
(12.18)
The aboveresultsare easily extendableto any numberof machinesswinging
coherently.
ffimel"L2rl
1
A 50 Hz, four pole turbogeneratorrated 100 MVA, 11 kv has an inertia
g6nsf anf nf
www.muslimengineer.info
R O MI/I\/IVA
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
m-qtsqllv-teryer-Svele
t4/s'
--
I
torJern
Power Srrctarn Anarrraio
(a) Find the stored energy in the rotor at synchronous
speed.
(b) If the mechanicalinput is suclcJenly
raisedto 80 MW fbr an electrical
load
of 50 MW, tind rotor acceleration,neglecting rrlechanicaland electrical
losses.
(c) If the accelerationcalculatedin part (b) is maintained
for 10 cycles,find
the changein torque angle and rotor speedin revolutionsper minute
at the
end of this period.
)olution
(a) Storedenergy= GH = 100 x g = g00 MJ
( b )P , = 8 0 - 5 0 = 3 0 M W = M
4
dt'
r,
. 1
- -
CH
180/
800
1 8 0x 5 0
4
MJ-r/elect deg
Simplified Machine
Model
For a nonsalientpole machine, the per phaseinduced emf-terminalvoltage
equationundersteadyconditionsis
(r2.Ie)
E =V + jXolu+ jXol,,;X,r) X,r
where
andusual symbolsare used.
Undertransientcondition
Xa-X'a1Xa
X'o = Xn since the main fleld is on the d-axis
Xtd< Xo ; but the differenceis less than in Eq' (I2.I9)
or
c
- 337'5 electdeg/s2
(c) 10cycres
=
"lT
Changein d= !{ZZI.S) x (0.2)2= 6.75 elect degrees
= 60 x
337'5
2x360J
!
= z. o6 ' 112^r E
{PnVs
.'. Rotor speedat the end of l0 cvcles
Equation(12.19)during the transientmodifiesto
Et =V + jxt lo+ jXnln
=V + jXq(I - I) + jXotlo
= (Y + jxp + j(X'a - Xq)Id
x'a = xq
4
= 1505.625rpm
POWER ANGLE EQUATION
In solvingthe s'uving
equation(Eq (12.10)),certainsimplifying assumpticnsare
usually made. These are:
1. Mechanicalpower input to the machine (P*) remainsconstantduring the
period of electromechanicaltransientof interest. In other words, it means
that
the effect of the turbine governing loop is ignored being much slower than the
speedof the transient.This assumptionleads to pessimisticresult-governing
loop helpsto stabilizethe sysrem.
'2.
Rotor speedchangesare insignificant-these have alreadybeen ignored
in formulating the swing equation.
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(12.2r)
'
(12.22)
The phasor diagram colrespondingto Eql (12.21) and (12.22) is drawn in
Fig. 12.2.
Since under transient condition, X'a 1X, but Xn remainsalmost unaffected,
it is fairly valid to assttmethat
- r 2 o x 5 o+ ' z 8 . t z 5
x 0.2
I2.3
(r2.20)
I=Ia+ Is
but
4 d26
= 3o
+'j 6,2
r )
3. Effect of voltage regulating loop during the transientis ignored, as a
the generatedmachineemf remainsconstant.This assumptionalso
consequence
leadsto pessimisticresults-voltage regulatorhelps to stabilizethe system.
Before the swing equation can be solved,it is necessaryto determinethe
of the electrical powel otttput(P,,)upon the rotor angle.
dependence
Fig. 12.2 Phasor diagram-salient pole machine
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(r2.23)
r"orrr r
,firZlll
atysis
_
Power Svstem Stabilitv
equation (12.22)now becomes
\
E,=V+jXnI
(r2.24)
o
X,l = X*/(transient
O
= V + jXotl
The machine mod
also applies to a cylindrical rotor machine
where
---
synchronousreactance)
t
-
[ECAC,+'
T*
2
@
-D'
f
r
Fig. 12.5 Two-busstabilitystudy network
For the 2-bus system of Fig. 12.5
f Y,, Y"1
Y n u s= I - j ' - : ' l ; YI , z =Y z r
LY^
Fig. 12.3 Simplifiedmachine model
The simplified machine of Fig. 12.3 will be used
in all stabilitv studies.
Yr,[
Complex power into bus is given by
Pi+ jQi-Elf
At bus I
P.ower Angle'Cunre
Pr + jQr - Er' (YyE1)* + E, (YpEil*
For the purposesof stability studies lEl1, transient
emf of generatormotor,
remains constant or is the independent variable
determined by the voltage
regulatingloop but v, the generatordeterminedterminal
voltage is a dependent
variable'Therefore,the nodes(buses)of the stability
study network pertainto
the ernf terminalin the machinemodel as shown
in rig. 12.4, while the machine
reactance(Xu) is absorbedin the systemnetwork
as clifferent from a load flow
sttlcly'Fttrther,lhe loirtls(othcrthrrnliu'gcs;ynchronous
riitittir.s)
will bc r.cpiacetl
by equivalent static admittances (connected in
shunt between transmission
network buses and the referencebus). This is so
becauseload voltages vary
cltlringa stabilitystutly (in a loaclllow stucly,
the.se
remain constantwithin a
nalrow bancl).
(r2.2s)
(r2.26)
But
E't =lEil
l6;
Y, u = G r t + jBt i
E / = l E ' z l1 4 .
Yr z= lYr zl l0 p
Since in solutionof the swing equationonly real power is involved, we have
from Eq. (12.26)
I r t = l E l i 2 G t r + l E r tI i E i i i y , r l c o s ( , t i- , h - 0 , 2 ) u 2 . 2 7 )
A similar equationwill hold at bus 2.
Let
lEllzG, = P,
l E t' l lEztI lY7l = Pn, *
4- 6=6
,/
Qn = x/2 -1
Then Eq. (12.27) can be written as
Pr = P, * P,n"* sin (6 - 1); Power Angle Equation
For a purely reactive network
Gtt = 0 (.'. P. = 0); losslessnetwork
and
Systemnetwork
Fig. 12.4
} t z = n 1 2 ,: . J = 0
P,=P^*sin 6
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(IZ.ZB)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(12.29a)
j0.5
P^^,' = lE''
-
where
I lE'' | '
x
I
I
'
(r2.29b)
simplified power angle equation
X = transfer reactancebetween nodes (i.e., between E{ andEi)
graphical plot of power angle equation (Eq.(12.29)) is shown in
Fig. 12.6.
i0.5
where
'Ihe
p"l
I
ID max
(Ps6+APe).-....t-----*
-
Generator
Peo
1lo"
lE'lt6
(b)
F|g.12.7Asimp|esystemwithitsreactancediagram
Fig. 12.6 Poweranglecurve
0.5
X r z= 0 ' 2 5+ 0 ' 1 + -
The swing equation (Eq. (12.10)) can now be written as
Yl aZt
+=
- P^
Pn'u*sin dpu
(12.30)
7rI dt-
which, as already stated,is a non-linear second-orderdifferential equation with
no damping.
T2.4
NODE ELIMINATION
TECHNIOUE
In stability studies, it has been indicated that the busesto be considered are
thosewhich are excitedby the internal machinevoltages(transientemf's) and
not the load buseswhich are excited by the terminal voltagesof the generators.
Therefore, in Y"u, formulation for the stability study, the load buses must be
eliminated. Three methods are available for bus elimination. These are
illustrated by the simple system of Fig. 12.7(a) whose reactancediagram is
drawn in Fig. I2.7(b).In this simplesituation,bus 3 getseasilyeliminatedby
parallel combination of the lines. Thus
= 0.6
casewhereina 3-phasefaultoccursat the
Considernow a morecomplicated
diagrambecomesthat
midpointof oneof thelinesin whichcasethe reactance
of Fig. 12.8(a).
Star-Delta Conversion
that of
Convertingthe star at the bus 3 to delta,the networktransformsto
Fig. 12.8(b)wherein
o
(,
j0.25
- at(
U
1t t r
lE/lt6
(a)
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpower SystemAnatysis
446 L
l u- t _
,
This method obviously is cumbersometo apply for a network of even small
complexity and cannot be computenzed.
Power SystemStability
Node Elimination
lE/lt6
Technique
Formulate the bus admittances for the 3-bus system of Fig. 12.8(a).This
network is redrawn in Fig. 12.9 wherein instead of reactance branch.
admittancesare shown. For this network.
CD
(r
o
@
ra
Fig. 12. 9
(c)
Fig. 12.8
v
= 1.55
This method for a complex network, however, cannot be
mechainzed for
preparing a computerprogramme.
Thevenin's
ruus
_ 0.25x 0.35+ 0.35x 0.5 + 0.5 x0.25
The bus 3 is to be eliminated.
In general for a 3-bus system
Y"t"lIu'I
f''I=fr' Y,
vu
With ref'crcncclo Fig. l2.ti(a), tho Thcvcnirr'sccluivalcntlbr
thc network
portion to the left of terminals a b as drawn in Fig. 12.8(c)wherein
bus t has
been modified to 1/.
,vrr h =
0.25
lEtl l5
025+0-,5
= 0 .4 1 7| Et I 16
xrh =
0 .3 5 x 0 .2 5
035+025
- o'146
Itlow
Xt2 =0.146+ 0.5 = 0.646*
*This value is different
from that obtained by star delta transformation as V* is no
longer lEtl I { in fact it is 0.417 lEtl 16.
www.muslimengineer.info
(r2.31)
lh | lY^
llv,I
Llrl Lv,, Yn rrrJL%l
Equivalent
Sinceno sourceis connectedat the bus3
It =o
or
Y r r V r +Y r r V r + Y r r V r = O
or
vz=-?rr- ?r,
Yrt
(r2.32)
Y.,
Substituting this value of V3 in the remaining two equations of Eq. (12.31),
thereby eliminating Vy
It =Y,Vt * YrzVz+ YttVt
=(", - Y,rYr,
) u +(
' [ ' ty,"
' -!+)v"
-(.^tt
Yr,
Yr, )''
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
In compactform
to be 1.0 pu.
=lt,i,'
Ynus(reduced)
l,l,'1
Y'r,
lY'r,
(12.33)
Solution
)
Y ' t l = Y '33
"-ry
YrtYt,
Y'12= Y'21= Ytz -
Ytt
E^l
xI
=l fEfr lil =
m = o ' 4I9. ZP
u
Q23aa)
(r2.34b)
(2) Equivalent circuit with capacitive reactor is shown in Fig. 12.71 (a).
p'e65
j1.o jo.l i0.25
io.1 i1.o
(r2.34c)
t , =- y 'zz
rr-YrtYt
tv 22
yT
lEnl= 1.2
-i1.0
lEml= 1.0
=
In general, in eliminating node n
Yo,(new) = Yry(old)
Yo^(old)Y,,(old)
(r2.3s)
(b)
(a)
Flg.12.11
Applying Eq. ( 12.34) to the example in hand
l-t.gzt
=;
Ysu5(reduced)
1 0.646
L
It then follows that
X,t=:o.646
Converting star to delta, the network of Fig. 12.11(a)is reduced to that of
Fig. 12.11(b)where
7 X ( t r a n s f e r ) - / 1 . 3 5 X / 1 . 1 + / 1 . 1-j1.0
X(_J1.0)+(-/1.0)XJ1.35..
= 1 . 5 4 8 (= 1 . 5 5 )
static capacitivereactorof
In the systenrshown in Fig. 12,10,a three-phase
reactance 1 pu per phase is connectedthrough a switch at motor bus bar.
Calculatethe limit of steadystatepower with and without reactorswitch closed.
Recalculatethe power limit with capacitive reactor replaced by an inductive
reactor of the same value.
= j0.965
1 . 2 x 1=
.
1-244pu
Steadystatepowerlimit =
ffi
replacedby inductivereactance,we get the
(3) With capacitivereactance
sterto delta,we havethe
circuitof Fig. 12.12.Converting
equivalent
trasferreactanceof
i1.1
i1.35
Fig.12.12
7X(transfer)
F i g .1 2 . 1 0
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ll.0 + 11.0x.t1.35
_ j1.35x.r1.l+ "11.1x
i 1.0
- j3.e35
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
-!5o' l
""attt
t"-t,
PowerSystemStability
in this section,
As alreadycalculated
Xn = l'55
Steadystatepower timit -':'!]
= 0.304 pu
3.935
Example12.3
= !".!;!! = 0.6e4pu
The generatorof Fig. 12.7(a)is delivering 1.0 pu power
to the infinite bus (lVl
= 1'0 pu), with the generatorterminal voltage
of v,r = 1.0 pu. calculate the
generatoremf behind transientreactance.Find the maximuio*". that can be
transferredunder the following conditions:
(a) Systemhealthy
(b) One line shorted (3_phase)in rhe middle
(c) One line open.
. Plot all the three power angle curves.
Solution
L,et
Vt = lV,l la
_t v- -t-l l v I s t n o = p "
X
(
t"t
)
P, = 0.694 sin d
or
(ii)
(c) One line open:
It easily follows from Fig. 12.7(b) that
X r z= 0 . 2 5 + 0 . 1 + 0 . 5 = 0 . 8 5
P*.*=tit^o]t =r.265
0.85
P" = I.265 sin 6
or
(iii)
The plot of the three power angle curves (Eqs. (i), (ii) and (iii)) is drawn in
Fig. 12.13. Under healthy condition, the systemis operatedwith P,, = P, = 1.0
pu and 6o= 33.9",i.e., at the point P on the power anglecurve 1.79 sin d As
one line is shortedin the middle, Po,remains fixed at 1.0 pu (governing system
act instantaneously)and is further assumedto remain fixed throughout the
transient(governing action is slow), while the operatingpoint instantly shifts to
Q on thc curvc 0.694sin dat d= 33.9".Noticcthutbccuuscof machineinertiu,
the rotor angle can not changesuddenly.
= | la
From power angle equation
< t= I
[025+oJJsrn
or
hiffi
rr = 2 0 .5 "
Current into infinite bus.
1.79
lV,lla-lVll0"
I _
1.265
jx
| 1 2 0 . 5 -I l 0
Pn=1'O
i0.3s
0 . 6 9 4s i n 6
0.694
= l + j 0 . 1 8 = 1 . 0 1 61 1 0 . 3 "
Voltage behind transientreactance,
Et =tltr
+ . j 0 . 6x ( l + 7 0 . 1 g )
I
900
0
33.90
= 0 .8 9 2+ j 0 .6 - 1.075133.9"
A
Fig. 12.13 Poweranglecurue,s-
(a) Systemhealrhy
p^u* = lv )-lEt | -- 1x 1.075=
- t.,' tF,.\
,-,
9 P
U
x,,
c,5
P, = I'79 sin d
(b) One line shorredin the middle:
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I2.5
(i)
SIMPLE
Machine
SYSTEMS
Connected
to Infinite
Bus
Figure 12.14 is the circuit model of a single machineconnectedto infinite bus
through a line of reacthnceXr.In this simple case
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t
Power System Stability
I
Xtransf.er =X'a*
X,
,, ='4U
Xt urrrf..
sin d= p.u*sin d
(r2.36)
(r--n-)
="f. ( P - r - P " r \ "tlH')
it
l2?:)=
d26"
and
From Eq. (12.29b)
-
I lt5'
(r2.39b)
SubtractingEq. (12.39b)from Eq. (12.39a)
d2@,;6)=^r(':jr!,]
.J
The dynamicsof this sysremare describeclin Eq. (12.11) as
dtz
-#ft= P^-P"Pu
or
\
HrH,
H"q d26
-*
., =Pn7rI (lt-
P,
6=4-
6.
where
,.- - P,)
)
(r2.4r)
(r2.42)
HtH,
rr "eq
Hl + Hz
lE/lt6
(r2.40)
(r2.43)
The electrical power interchange is given by expression
p"
,in6
,E!4!' = X'0,
x, + xd2
Fi1.12.14 Machineconnected
to infinitebus
Two Machine System
The caseof two finite machines connectedthrough a line (X") is illustrated in
Fig. 12.15whereone of the machinesmust be generating und th" other must
be motoring.Under steadycondition, beforethe system goesinto dynamicsand
(12.aa)
The swing equation Eq. (12.41) and the power angle equation F;qThus a
bus.
have the same form as for a single machine connectedto infinite
bus.
infinite
to
two-machine systemis equivalentto a single machineconnected
be
would
qYstem
bus)
Becauseof this, the single-machine(connectedto intinite
studied extensively in this chapter.
In the system of Example 12.3,the generatorhas an inertia constant of 4 MJ/
MVA, write the swing equationupon occurrenceof the fault. What is the initial
angularacceleration?If this accelerationcan be assumedto remain constantfor
Lt = 0.05s, find the rotor angle at the end of this time interval and the new
acceleration.
Fig. 12.15 Two-machinesvstem
Solution
P*t=-P*z=P.
(12.38a)
the mechanicalinput/outputof the two machinesis assumedto remain constant
at these values throughout the dynamics (governor action assumed slow).
During steadystateor in dynamic condition, the electrical power output of the
generatormust be absorbedby the motor (network being lossless).Thus at all
time
Prl=-Prz= P"
(12.38b)
The swing equationsfor the two machinescan now be written as
tL _
- "_' ir ( P ^ r _P " r_) _ " ( p . - + \
dtz
ff, ):ut t rr, l
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(12.44)
+
Swing equation upon occurrenceof fault
H
d'6 _o
- r m - 'De
1 g 0 fd v
4
d,26
,
*"t#=l-0'694
s i n6
t4 = z2so(1- 0.6e4sin d;.
dt"
(12.39a)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
vooern powersysremAnatysis
,.4#jirf
ffi
I
Initial rotor angle do= 33.9" (calculatedin Example 12.3)
L ltc..:'
]i;E!!";1&t
Linearizing about the operatingpoint Qo (P"0, 4) *" can write
a 2s l
= 2 2 5 0 (l - 0.694 si n 33.9" )
;l
d t ' l ,- n +
LP,=(*).
o,
The excursionsof A d are then describedby
#l
C l l l r : o *+
= 0; rotor speedcannotchangesuddenly
+ aP,')= - L,P,
no9i+' - P^- (P,o
d,r
A , (i n A,t = 0 .0 5 s )= I x 1379x (0.05)2
*
M d'+'
dr
2
t .7"
(r2.47)
6 t = 6 + 4,6 = 33.9+ 1.7"= 35.6"
a26l
= 2250(l - 0.694sin 35.6")
.,|
d r ll r = 0 . 0 5 s
Observethat as the rotor angle increases,the electrical power output
of the
generatorincreasesand so the accelerationof the rotor reduces.
12.6
M*
dl
= P^
M = H
7
Mp,
+[#], =o
p, = !4)!
P " MW ; E q . (12.8)
ln Pu sYstem
x,t
)
,in 6'= p^u*sin d
(12.4s)
(12.46'
For determinationof steadystate stability, the direct axis reactance(X.r) ant,
voltagebehind X4 are usedin the above equahons.
The plot of Eq. (12.46)is given in Fig. 12.6.Let the system be operaring
with steadypower transferof P^ = P^with torque angle as indicated
in the
d
figure. Assume a small increment AP in the electric power with the input
from
the prime mover remainingfixed at p*(governor r.rforrr" is slow compared
to
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d
dt
The system stability to small changesis determined from the characteristic
equation
STEADY STATE STABILITY
The steady state stability limit of a particularcircuit of a power
system is
defineclas the maximutnpower that can be transmittedtri fhe receivinoen;
v'r6 vrru
without loss of synchronism.
Consider the sirnplesystemof Fig. 12.14whoseclynamicsis describect
by
equations
and
P
where
- l34I elect deg/s2
whosc two roots are
f
/ t \ t r
, . 1
( \
f {
P = + l - \ u 1 t 0 o ) ol L
M
I
As long as (0P/0 0o it positive, the roots are purely imaginary and conjugate
and the system behaviour is oscillatory about do.Line resistanceand damper
windings of machine,which have been ignored in the abovemodelling, cause
the system oscillations to decay. The system is therefore stable for a small
increment in power so long as
(aP,/aa| > o
(12.48)
When (0 P/AD, is negative, the roots are real, one positive and the other
negativebut of equal magnitude.The torqueangle thereforeincreaseswithout
bound upon occurrcnccol a small powcr incretrtent(disturbancc)and the
synchronismis soon lost. The systemis therefore unstablefor
@ Pe/aDo < 0
known as synchronizingcofficienr. This is also called stffiess
@p/A[ois
(electrical) of synchronousmachine.
Assuming lEl and lVl to remain constant,the systemis unstable,if
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
po*e,systmst"uititv
lEllvl
c o sd ^ < o
X
I a e 1|
l--e
or
(12.4e)
4>90"
Themaximumpowerthatcanbe transmitted
withoutlossof stabili
( r2.s0)
and is given by
P^u*
lEnvl
r-
_ _L_2
_ _x r c o s 3 0 "
1.8
L 06 Jro"
= 0.577MW (pu)/elect
rad
H
=
o*r o
From characteristicequation
M ( pu) =
(12.sr)
If the system is operating below the limit of steady stability
condition (Eq.
12'48), it may continue to oscillate for a long time if the
iamping is low.
Persistent oscillations are a threat to system security. The
study oi system
damping is rhe study of dynamicalstability.
The above procedure is also applicable for complex systems
wherein
governor action and excitation control are also accounted
for. The describing
dif f er e n ti a le q u a ti o ni s l i n e a ri z e cal bout the opcrati ngpoi nt.
C oncl i ti epfbr
steady state stability is then determinedfrom the corresponding
characteristic
equation (which now is of order higher than two)
It was assumedin the above account that the internal rnachinc
voltage
lEl remains constant(i.e., excitation is held constant).The result
is that as
loading increases,the terminal voltage lv,l dips heavily which
cannot be
toleratcdin practice.Thereforc,we must consiclerthe steadystate
stabilitylimit
by assuming that excitation is adjusted for every load increase
to keep
lv,l constanr.This is how the system will be operaiedpractically.
It may be
understocdthat we are still not considering the effect of automatic
excitation
control.
steady state stability limit with lv,l anrt lvl constantis consiclered
in
Example 12.6.
b{dffi
4
t r x5o
r at r
s?/ cr cct
P=tr[(*),,"1*)'
-+i(WiY")u
== i4.76
FrequencYof oscillations= 4.76 railsec
4'76 0. 758Hz
2r
(ii) For 807oloading
s i nd o= +
P-u*
rqa)
\ 05 )rr,
= 0 . 8 o r6 = 5 3 . 1 "
- r'Zxrcos53.1"
1.8
= 0'4 MW (Pu)/elect
rad
p =!, (q+k)* =* i3s6
Frequencyof oscillations = 3.96 radlsec
?q6
A synchronousgeneratorof reactance1.20pu is connectedto an infinite
bus
bar (l Vl = 1.0 pr) throughtransformersand a line of total reactance
of 0.60 pu.
The generatorno load voltageis I .20 pu and its inertia constantis H =
4 MWsilvIVA. The resistanceand machinedamping may be assumednegligible.
The
systemfrequencyis 50 Hz.
Calculate the frequency of natural oscillations if the generatoris loaded
to
(1) 50Voand (ii) 80Voof its maximum power limit.
Solution
(i) For 50Voloading
P
s m d o- i t
'
')*
Find the steady state power limit of a system consisting of a generator
equivalent reactance0.50 pu connectedto an infinite bus through a series
gf 1.0pu. The terminalvoltageof the generatoris held at 1.20pu and
rcactance
the voltage of the infinite bus is 1.0 pu.
Solution
= 0 . 5 o r4 = 3 0 o
max
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The systemis shown in Fig. 12.16.Let the voltage of the infinite bus be taken
as reference.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
m Analysis
Then
V = 7 . 0 / - f f , V t= ! . 2 l 0
Now
I_
Power SrrstemStahilitu
1.210-7.0
jI
Xa= O'5
I
lE4t6
Vt=1.219
V = 1.0100
Fl g. 12.16
r
I
E = Vt + jXdI = 1.2 l0 + j0.5 I r . 2l g - 1 . 0
L
J
.E = l.g l0 _ 0.5= (t.g cos e_ 0.5)+
71.gsin 0
Steadystateporverrimit is reached
whenE hasan angleof 6= 90o,i.e.,its real
part is zero.Thus,
1 . 8 c o 0s - 0 . 5 = 0
or
0 = 73.87"
Now
Vt = 1.2/.73.87"= 0.332+ j t . 1 5 2
t.t52 + j0.669
E -0.332+ jr.r52+ 70.5(1.152+ j0.668)
- 0.002+ j1.728= 1.728
I90.
Steadystatepowerlimit is givenby
1.728xL
p ^ u - l E l l V l = --i5
= l'152 Pu
*=
V;-+V
If instead,the generator
emf is held fixed at a valueof r.2pu, the
steady
statepower limit would be
=
P*"*
i#
A knowledgeof steady state stability limit is important for various reasons.A
system can be operated above its transient stability limit but not above its
steady-tatelimit. Nowrwith increased fault el,earing speedsjt is possible to
make the transient limit closely approach the steady state limit.
As is clear from Eq. (12.51),the methodsof improving steadystate stability
limit of a systemare to reduceX and increaseeither or both lEl and I Vl. If the
transmissionlines are of sufficiently high reactance, the stability limit can be
raisedby using two parallel lines which incidently also increasesthe reliability
of the system.Series capacitorsare sometimesemployed in lines to get better
voltage regulation and to raise the stability limit by decreasing the line
reactance.Higher excitation voltages and quick excitation system are also
employed to improve the stability limit.
I2.7
=o'8Pu
It is observedthatregulatingthegenerator
emf to hold the terminalgenerator
','oltage
at r.2 pu raisesthepowerli.it frorn0.g pr'ro
r.r52pu; thisis how
thevoltageregulatingloop helpsin power
systemstab'ity.
TRANSIENT
STABITITY
It has been shown in Sec. L2.4 that the dynamics of a single synchrono,rs
machine connectedto infinite bus bars is governed by the nonlinear differential
equation
,, d'6 = P ^ - P "
M
iF
where
r _ 0.332+.ir.rs2_r
=
EsE
P, = P-*
sin d
-_ d26
(r2.s2)
M --+ - P*- P** sind
otAs said earlier, this equation is known as the swing equation.No closed form
solution exists for swing equation except for the simple case P- = 0 (not a
practicalcase)which involves elliptical integrals.For small disturbance(say,
gradual loading), the equation can be linearized (see Sec. 12.6) leading to the
concept of steady state stability where a unique criterion of stability
(APrlAd>0) could be established.No generalizedcriteria are available* for
determining system stability with large disturbances(called transient stability).
The practical approach to the transient stability problem is therefore to list all
important severe disturbances along with their possible locations to which the
systernis likely to be subjected according to the experienceand judgement of
the power system analyst. Numerical solution of the swing equation (or
equationsfor a multimachine case) is then obtained in the presence of such
disturbancesgiving a plot of d vs. r called the swing curve. If d starts to
decrease after reaching a maximum value, it is normally assumed that the
systemis stableand the oscillation of daround the equilibrium point will decay
or
tRecent literature gives methods of determining transient stability through
Liapunov and Popov's stability criteria, b:rt thesehave not been of partical use so far.
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powerSystem
Modern
Anatysis
ffiffif
StabilitY
PowerS','stem
I
andfinally die out. As alreadypointed out in the introduction,important severe
distulbancesare a short circuit or a sudden loss of load.
For easeof analysiscertainassumptionsand simplificationsare always made
(some of thesehave already been made in arriving at the swing
equation (Eq.
/1 /l <.t\\
a rr 11consequences
upon accuracyof results.
1. Transmission line as well as synchronous machine resistance are
ignored. This leads to pessimisticresult as resistanceintroducesdamping
term
in the swing equationwhich helps stability. In Example I2.11, line iesistance
has been taken into account.
2. Damping term contributedby synchronousmachinedamperwindings is
ignored. This also leads to pessimisticresults for the transientstability limit.
3. Rotor speed is assumedto be synchronous.In fact it varies insignificantly during the course of the stability transient.
4. Mechanical input to machineis assumedto remain constantdurins the
transient,i.e., regulating action of the generatorloop is ignored. This leais
to
pessimisticresults.
5. Voltage behind transientreactanceis assumedto remain constant, i.e.,
action of voltage regulating loop is ignored. It also leadsto pessimisticresults.
6. Shunt capacitancesare not difficult to account for in a stability study.
Where ignored, no greatly significant error is caused.
7. Loads are modelled as constant admittances. This is a reasonablv
accuraterepresentation.
ffiffi
p"rrnon.ntly till cleared manually. Since in the majority of faults the first
ieclosure will be successful, the chances of system stability are greatly
enhancedby using autoreclosebreakers.
Fig.12.17
A digital computer programme to compute the transient following sudden
disturbance aan be suitably modified to include the effect of governlr action
and excitation control.
In the caseof a perrnanentfault, this system completely falls apart. This will
not be the case in a multimachine system.The stepslisted, in fact, apply to a
system of any size.
1. From prefault loading, determine the voltage behind transientreactance
of the machine with reference to the infinite bus.
and the torque angle 16o
2. For the specified fault, determine the power transfer equation Pr(A during
'
^ault. In this system P" = 0 for a three-phasefault' '
1, calculate
in
step
as
obtained
5 . From the swing equation starting with fi
thetnonsolving
of
technique
a
numerical
das a function of time using
linear differential equation.
4 . After clearanceof the fault, once again determine P, (A and solve further
for d (r). In this case,P"(A = 0 as when the fault is cleared,the system
Upon occulTenceof a severedisturbance, say a short circuit, the power
transfer between machines is greatly reduced, causing the machine torque
angles to swing relatively. The circuit breakers near the fault disconnect the
getsCisconnected.
5 . After the transmissionline is switched on, again find P" (0 and continue
to calculate d (r).
6 . If 6 (t) goes through a maximum value and starts to reduce,the system is
regardedas stable.It is unstableif d(r) continuesto increase.Calculation
is ceasedafter a suitable length of time.
An important numericalmethod of calculating d(t) from the swing equation
will be giurn in Section 12.9.For the single machineinfinite bus bar system,
stability can be conveniently determined by the equal area criterion presented
Note: Sincerotor speedand hencefrequency vary insignificantly, the network
parametersremain fixed during a stability study.
in the following section.
I2.8
EOUAL AREA CRITERION
In a systemwhere one machine is swinging with respectto an infinite bus, it
is possible to study transient stability by means of a simple criterion, without
resorting to the numerical solution of a swing equation.
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Considerthe swing equation
d26
AF
=
I
*@^-
The stability criterion for power systemsstatedabovecan be convertedintc
a simple and easily applicableform for a single machine infinite bus system.
P ' 1= &rt ,"= accelerating
power
M = ! ln pu system
rf
we get
Multiplying both sides of the swing equation *
[t#),
(r2.s3)
2P" d6
M d t
Ifrtegrating,we have
(r2.s6)
where dois the initial rotor angle before it begins to swing due to disturbance.
From Eqs. (12.55) and (12.56), the condition for stability can be written as
Fig. 12.1g protof 6 vs tfor stabreand unstabre
systems
lf the system is unstable dcontinuesto increase
indefinitely with time and the
machine loses synchronism.on the other hand,
if the system is stable, 6(t)
performs oscillations (nonsinusoidal)whose
amplitude decreasesin actual
praetice becauseof darnpingterms (not
included in the swing equation).These
two situations are shown in nig. 12.1g. since
the system is non_linear, the
nature of its response160l is not unique and
it may exhibit instability in a
fashion different from that indicated in Fig. rz.rg,depending
upon the nature
and severity of disturbance.However. experience
indicatesihai the response
6!'l j" a power system generallyfalls in the
two broad categoriesas shown in
the figure' It can easily be visualizednow (this
has alsobeenstatedearlier) that
for a stable system,indicationof stability will
be given by observationof the
first swing where dwill go to a maximumand
will Jturtto reduce.This fact can
be stated as a stability criterion, that the system
is stableif at some time
d6
=o
(r2.s4)
dt
and is unstable,if
-d-6 > 0
<lt
for a sufticiently long time (more than 1 s will
genera'y do).
(12.ss)
6
or
[r"ad
(r2.s7)
-o
6,,
The condition of stability can therefore be stated as: the system is stable if the
areaunderPo(acceleratingpower) -dcurve reducesto zero at some value of
d In other words, the positive (accelerating)areaunder Po- 6curve must equal
'equal area' criterion of
the negative (decelerating)area and hence the name
lstability.
To illustrate the equal area criterion of stability, we now consider several
types of disturbancesthat may occur in a single machine infinite bus bar
system.
Sudden Change in Mechanical
Input
Figure 12.t9 shows the transientmodel of a single machinetied to infinite bus
bar. The electrical power transmittedis given by
Infinite
bus bar
-->
Pm
lvlr0o
Fig. 12.19
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ffifftl|
,
Modern
Powersvstemnnarys,s
lEtllvl
P, = u, #
sin d- P** sind
ndT^e
Under steady operating condition
P.o = Pro= P** sin do
4n
Ar=)(Pn-P")d6
Az=i<r,-P^)d6
6l
be possible to find angle d2 such that
rg condition is finally reached when 41
own in Fig.l2.2L Under this condition,
hat
6. = 6^o= T
6 b f
%
?'i
i
6
- 6t=
n'-sin-l
+:
(12.58)
2
(' s
Q <(4
Flg. 12.20 P"- 6 diagramfor suddenincreasein mechanical
inputto
g e n e ra toorf F i g .1 2 .19
This is indicatedby the point ainthe Pr- 6 diagram of Fig. 12.20.
Let the mechanical input to the rotor be suddenly increased to Pn (by
opening the steam valve). The acceleratingpower 1o = P*t - P, causes'the
rotor speed to increase (u> a,,r)and so does the rotor angle. At angle 6r,,
Po= P*r- Pr(= P-* sin 4) = O (statepoint atb)but the rotor anglecontinues
to increaseas t.,) ur.Po now becomesnegative(decelerating),the rotor speed
begins to reducebut the anglecontinuesto increasetill at angle 6., a= ur once
again (statepoint at c. At c), the-deceleratingarea A, equals the accelerating
bc
areaA, (areasare shaded),j.e.,
=
J ,, Od 0. Since the rotor is decelerating,
6o
the speedreducesbelow ur andthe rotor angle begins to reduce.The statepoint
now traversesthe P, - 6 curvein the oppositedirection as indicated by arrows
in Fig. 12.20.It is easily seenthat the systemoscillatesabout the new steady
statepoint b (6= 4) with angle excursionup to 6 *d 4.on the two sides.
These oscillations are similar to the simple harmonic motion of an inertia-spring
system except that these are not sinusoidal.
As the oscillations decay out because of inherent system damping (not
modelled),.the systemsettlesto the new steady state where
P^t = P, = Prn.* sin dl
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mechanicalinput
Fig. 12.21 Limitingcase of transientstabilitywith
increased
suddenlY
availablefor A, is lessthan A1'
Any turther increase in P^, meansthat the area
increasebeyond point c and the
so that the excesskinetic energy causes d to
power, with the system
decelerating power changes over to accelerating
by use of the equal
shown
consequentlybecoming uistable. It has thus been
in mechanicalinput
increase
areacriterion that there-isan upper limit to sudden
stable'
remain
to
Po,s),for the systemin question
(P^r' 'ii
will remain stableeven
',,uy'ulso be not"i from Fig. 12.21thatthe system
as the equal area
long
so
beyo-nd
though the rotor may oscillate
-{^=.90"'
in steady state
use
for
=
is
meant
90"
d
of
criterion is met. The condition
case'
stability
transient
the
to
apply
stability only and does not
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
trs#ffi
-{i..fuu.xdr
r._r_..._
A
ruooern Fower
uvslem Anarvsrs
l':fiffn#;i;
Power a
Effect of Clearing Time on Stability
Let the system of Fig. 12.22 be operatingwith mechanicalinput P^ at a steady
angleof d0 (Pn,= P") as shown by the point a on the Pr- 6 cliagramof Fig.
12.23.If a 3-phasefault occurs at the point P of the outgoing radial line, the
electrical output of the generatorinstantly reducesto zero, i.e,, p, = 0 and the
statepoint drops to b. The accelerationareaA, begins to increaseand so does
the rotor angle while the statepoint movesalong bc. At time /. corresponding
to angle 6, the faulted line is cleared by the opening of the line circuit breaker.
The values of /, attd 4 are respectively known as clearing time and,clearing
angle. The system once again becomeshealthy and transmits p, = p,ou,.sin d
i.e. the state point shifts to d on the original P, - d curve. The rotor now
deceleratesand the decelerating area A, begins while the state point moves
along de.
-
|
l
- l|-t___j
l
rresponding to a clearing angle can be
t
establishedonly by numerical integration except in this simple case.The equal
area criterion therefore gives only qualitative answer to system stability as the
time whgn the breaker should be opened is hard to establish.
Pe
'D max
Pm
/'-\
\
co )
\-/
d",
I
6r"*
+
Criticalclearing
angle
Fig. 12.24 Criticalclearingangle
F19.12.22
If an angle fi canbe found such that A2= Ap the system is found to be stable.
The systernfinally settlesdown to the steadyoperatingpoint a rn an oscillatory
mannerbecauseof inherent damping.
Pe
,D max
As the clearing of the faulty line is delayed,A, increasesand so\oes d, to
find A2 = Ar till 6r = 6^ as shown in Fig. 12.24.For a clearing time (or angle)
larger than this value, the systemwould be unstableas A, < Ar The maximum
aiiowabie vaiue of the clearing time and angle for the systemto remain stabie
are known respectively as critical clearing time and angle.
For this simple case (P, = 0 during fault), explicit relationships for 6,
(critical) and t" (critical) are establishedbelow. All angles are in radians.
It is easily seen from Fig. 12.24 that
and
4nu*=T- d;
(12.59)
P*= Pr* sin 6o
(r2.60)
Now
uct
Pm
A t =
J
h
(P^ --0) d 6 = P^ (4, -
6)
and
6^^
6 e l f . "
P"=o
-.--/
(3-phasefault)
61
Az=
J (P** sin d- P^) d6
6,,
i
= P.u* (cos d, - cos d-*) - P* (6^o - 6"i)
Clearing
angle
Fig. 12.23
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uoo"rnpo*"r systemAnatvsis
ffil
W
For the systemto be stable, A2= A1,which yields
cos{. = !^
Prn*
where
,, /__\
tt \--l )
,.774
(5,^^"
\ -tniu(- d) + cos 4o"*
(r2.61)
Pm
l_Lr__l
|
I
L
4, = critical clearing angle
r
l
I Infinite
-l
bus
IVVO"
(a)
SubstitutingEqs. (1259) and (12.60) in Eq. (12.61),
we get
4r = cos-t [(r, _ Z6l sin do_ cos 6o]
During the period the fault is persisting,the
swing equation is
(r2.62)
d,2d
= rf P^: P, = o
d,r, 1r:
IVVO"
(12.63)
Integrating twice
6 = -,rf- P*tz + $
2H
(b)
Fi1.12.25 Singlemachinetiedto infinitebus throughtwo par:allel
lines
0 " , = # P ; 2 " ,1 6 0
(12.64)
where
/cr = critical clearing time
4, = critical clearing angle
From Eq. (12.6a)
2 H ( 6 , -, 4 )
(r2.6s)
TrfP*
where d, is given by the expressionof Eq, (12.62)
An explicitrelationshipI'rrrclctenninirtg
r., i, porisiblcin this caseas tluring
the faulted condition p" = o and so trre ,wing
equation can be integrated in
closedform. This will not be the casein mosi
other situations.
Both thesecurves are plotted in Fig. 12.26,wherein P-u*n ( P_u*ras (Yo * Xr)
> (Ya + Xr ll X).The systemis operatinginitially with a steadypower transfer
Pr= P^ at a torque angle 4 on curve I.
Immediately on switching off line 2, the electrical operating point shifts to
curve II (point b). Acceleratingenergycorrespondingto areaA, is put into rotor
followed by decelerating energy for 6 > q. Assuming that an area A2
correspondingfo deceleratingenergy (energy out of rotor) can be found such
that At = Az, the system will be stable and will finally operate at c
correspondingto a new,rotor angle 6, > 60"This is so becausea single line
offers larger reactanceand larger rotor angle is neededto transfer the same
steadypower.
,"
(both lines in)
/
Sudden Loss of One of parallel Lines
considernow a singlemachinetied to infinite
bus throughtwo parallellinesas
in Fig. 12.25a.circuit model of the sysremis given
in Fig. r2.25b.
Let us study the transientstability of the ,yir"rn
when one of the lines is
suddenly switched off with the system operating
at a steady road. Before
switchingoff, poweranglecurve is given by
P"r=
lE'llvl
sin d= Pm*l sin d
xa i xt llx2
Immediately on switching off line 2, power
angrecurve is given
P"n= g:+
,\d
-T
rt7
sin d= pmaxrsin d
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by
Fig.12.26
Equalareacriterionappliedto the openingof one of the two
linesin parallel
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Mod"rnpo*rr.surt*
ffiffi-4l
nrryr',
Power Sy-t-- St"blllry
via the healthy line (through higher line
reactance X2 in place of Xl ll Xz)7;with power angle curve
sln
4=4o*_T_6,
which is the samecondition as in the previous example.
Sudden Short Circuit on One of parallel
Case a: Short circuit
Lines
at one end of line
Let us now assumethe disturbanceto be a short circuit at the generatorend of
line 2 of a double circuit line as shown in Fig. 12.27a.We shall assumethe
fault to be a three-phaseone.
sin d
obviously, P-o[ ( P-"*r. The rotor now starts to decelerate as shown in
Fig. 12.28. The system will be stable if a decelerating areaA, can be found
equal to accelerating area A, before d reachesthe maximum allowable value
4o*.At areaA, dependsupon clearing time /. (correspondingto clearing angle
{), clearing time must be less than a certain value (critical clearing time) for
the system to be stable. It is to be observedthat the equal area criterion helps
to determine critical clearing angle and not critical clearing time. Critical
clearing time can be obtained by numerical solution of the swing equation
(discussedin Section 12.8).
P"y, prefault (2 lines)
P6n1,postfault (1 line)
X2
,
?
(b)
F19.12.27 Shoftcircuitat one end of the line
Before the occurrenceof a fault, the power angle curve is given by
'--
p"t' =
sind= p_*, sind
,)4,'rlr,,,a,
xi
+ xltx2
which is plotted in Fig. 12.25.
Upon occulrenceof a three-phasefault at the generatorend of line 2 (see
Fig. I2.24a), the generatorgetsisolatedfrom the power systemfor purposesof
power flow as shown by Fig. 12.27b.Thus during the period the fauit lasts,
The rotor therefore accelerate, .i;t:;i"s
dincreases. synchronism will be
lost unless the fault is cleared in time.
The circuit breakers at the two ends of the faulted line open at time tc
(correspondingto angle 4), the clearing time, disconnectingthe faulted line.
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t
6
Ffg. 12.28 Equalarea criterionappliedto the systemof Fig. 12.24a,
I systemnormal,ll faultapplied,lll faultedline isolated.
It also easily follows that larger initial loading (P.) increasesA, for a given
clearing angle (and time) and thereforequicker fault clearing would be needed
to maintain stable operation.
Case b: Short circuit
away from line ends
When the fault occurs away from line ends(say in the middle of a line), there
is some power flow during the fault thoughconsiderablyreduced,as different
from case a where Pen= 0. Circuit model of the system during fault is now
shown in Fig. 12.29a.This circuit reducesto that of Fig. 12.29cthrough one
delta-star and one star-delta conversion. Instead,node elimination technique of
Section 12.3 could be employed profitably. The power angle curve during fault
is therefore given by
P"t=
| Ellvl
'1r'II
s i n d = P m a xsr irn d
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffiil
ls#z#i
Modern Power svstem Analvsis
'!'vev"
+t*F;4-l
'
'
-
'-
v'
-'-"
v,
'
'
"
'-',
-'-
I
system operationis shown in Fig. 12.30,wherein it is possibleto find an area
A, equal-to A, for q. < 4nu*. At the clearing angle d. is increased, area
ai increat"t und to nna Az = Ar, 4. increasestill it has a value 4n*' t6"
-ooi*,,- ollnrvohlefnr stahilitv This caseof critical clearineangleis shown
in Fig. 12.3L
Pe
x,t
Pr1,prefault (2 lines)
X6
postfault (1 line)
P6111,
xc
G
@
(b)
Xr
Fig. 12.31 Faulton middleof one lineof the systemof Fig. t2.l4a, case of
criticalclearingangle
(c)
Annlvins eoual areaeriterion to the caseof critical clearing angle of Fig. 12.31'
we can wnte
Fig. 12.29
Pe
dntn'
4,
P"rand P,u as in Fig. 12.28 and Per as obtained above are all plotted in Fig.
I2.3O. Accelerating area A, corresponding to a given clearing angle d is less
j (P^- 4n*u sinfldd= J {r^*r sind- P^)d6
6,,
60
where
P"'Prefault (2 lines)
4,,* =T
,.a
V
postfault(1 line)
P"11;,
(r2.66)
- sin-r (:t_)
maxIII
./
Integrating, we get
l6*
(P^a + Pmaxrrcos d) | * (P'*,,1 cos d +
P"11,
duringfault
= Q
16o
or
P^ (6", - 6) * P.u*u (cos '[. - cos do)
I P* (6** - 6"r) * P-om (cos fi* - cos 4J = 0
Fig. 12.30 Faulton middleof one line of the systemof Fig. 12.24a
with d"< {,
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t
cos {r =
4naxtn
- PmaxII
:otd",*
(12.67)
critical clearinganglecanbe calculated
from Eq.(12.67)above.The anglesin
are in radians.The equationmooiiiesas belowif the
angJes
are
:lt::t1|on
ln oegrees.
cos {. -
ft
r.(6 ^i* - do) - Pmaxrr
cosdo * prnu*ru
cosd,ou*
Give the system of Fig. 12,33 where a three-phascfault is applied at rhe point
P as shown.
i0.s
Pmaxltr - Prnaxn
Infinite
bus
vFlloo
Case c: Reclosure
If the circuit breakersof line 2 arereclosed successfully
(i.e., the fault was a
transient one and therefore vanishedon clearing
the faurty line), the power
transfer once again becomes
P"N = P"r= p*u*I sin d
Since reclosure restores power transfer, the
chances of stable operati'n
improve. A case of stable operationis indicated
by Fig. 12.32.
For critical clearing angle
4 = 4r* = 1T- sin-l 1p_/p*.*r;
t
ucr
6rc
sin 0 dd =
pm) d,6
J @r,- Pmaxrr
J (p.*m sin d_
60
6r,
Flg. 12.33
Find the critical clearing angle for clearing the fault with simultaneousopening
of the breakers I and 2. T\e reactance values of'various components are
indicated on the diagram. The generatoris delivering 1.0 pu power at the instanr
preceding the fault.
Solution
With referenceto Fig. 12.31, three separatepower angle curves are involved.
f. Normal
dru,
t
+ J (P,*r sin d_ p^) d6
operation
(prefault)
Xr=0.2s+ffi+0.05
6.-
= 0.522pu
p,t=rysind:
ffirino
= 2.3 sin d
(il
Prefault operatingpower angle is given by
1.0 = 2.3 sin 6
or
IL During
6o =25.8" = 0.45 radians
fault
It is clear from Fig. 12.31 that no power is ffansferredduring fault, i.e.,
= o
(Clearing angle)
\
(Angleof reclosure)
,0.:"o
Fig- 12-32 Faurtin middreof a rineof
the systemof Fig. 12.27a
where trrj tr, + r; T = time betw,een
clearing anclreclosure.
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n 1A
هاجتإلا يمالسإلاt r i n- 1تالاصتإلاو
ةنجل روبلا
U
ModernpowerSystemAnatysis
iffi
E
= - 1.5 (cos2.41- cos 6,) - (2.41- 6")
rrr. Post fault operation (fault cleared by openingr the faulted
Iiae)
= 1.5cos 6", + 6r,- I.293
Setting A = Az and solving
6r, - 0.45 = 1.5cos 6r,+ 6r,- 1.293
l.2xl.0
n
Perrr=
ff
sin d= 1.5sin 6
(iii)
- 0.562
cos {,. = 0.84311.5
or
or
4, = 55.8"
poweranglediagramsare shownin Fig. 12.35.
The corresponding
Pe
Find the critical clearing angle for the system shown in Fig. 12.36 for a threephase fault at the point P. The generator is delivering 1.0 pu power under
prefault conditions.
Pn=1 'O
i 0. 1s
i0.15
lnfinite
bus
66=0.45
rad
Fig. 12.35
TTto
rrrw
-ooi*"*
urour.rLurr
-^*:^^:Ll^
psllluDDlulc
^--l^
alilBrtr
C
Omax
f^l()f
lvF1.otoo
lF,l=1.2Pu
6^rr=2.41rdd
jo.15
.10.15
----afea
^
Al
=
.
A2
given by
(Sge
flg.
Flg. 12.36
12.35)is
Solution
I
= 2.4Lradians
1.5
Applying equal area criterion for critical clearing angle
{
Ar = P^ (6", - 6)
4ou*=r-sin-l
= 1.0 (6", - 0.45) = 6c,- 0.45
dr*
Az=
!{r,n-p^)d,6
f, Prefault
bus is
2.41
I
6.,
Transfer reactance between generator and infinite
& = 0.25+ 0.17+
0 . 1 5 + 0 . 2 8 + 0 . 1=5
0.71
2
Pc'
- , = r ' Z x l s i n d = 1 . 6 9s i n 6
0. 71
power
angle
is given by
The operating
or
do = 0.633 rad
The positive sequencereactancediagram during fault is
IL Durtng fault
presentedin Fig. 12.37a.
r2.41
= - 1 . 5 c o ds _ d l
| 6",
www.muslimengineer.info
(i)
1.0= 1.69sin ,fr
6,,
= | ( 1 . 5s i n 6 - 1 ) d d
J
operation
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
PowerSystemStabilit-v
J
/ 000 L----------r \000
000 L
j0.25
0 '
Perrr=U!
'
l
Mi#ffi
r*
sin d = r'2 sin 6
(iii)
jo.17
j0.15
+
j0.14
j0.14
With referenceto Fig. 12.30and Eq. (12.66),we have
j0.'15
) E1=t.z
+
V=1.0
To find the cqitical clearing angle, areas A1 and A, arc to be equated.
6",
- ,
At = l.o (6,,- 0.633)
(a) Positivesequence
reactancediagramduringfault
j0.25
j0.145
j0.145
',
J
o.+e5
sin d dd
60
j0.17
and
dmax
f
lE'l=1.2
A-z = J | 1 . 2s i n d d d - 1 . 0( 2 . 1 5 5- 4 )
V=1.OlOo
6
- cr
Now
At =Az
(b) Networkafterddlta-star
conversion
or
J
6r, = 0.633---
0.495sin d dd
o.63J
l9l=1'z
V=1.0100
2.155
=
[ t.Z rin 6 d6 - 2.t55 + 6,,
'
J
6cr
(c)Network
afterstar-delta
conversion
.
or - 0.633+ 0.495cos olo' = - 1.2cos ol"tt -2.155
Ftg. 12.32
lo.orr
Converting delta to star*, the reactancenetwork is changed to that of Fig.
12.37(b). Further, upon converting star to delta, we obtain the reactance
network of Fig. .r2.37(c). The transfer reactanceis given 6y
(0.2s
0.072s
+ 0.145)
+ (0.145
+ 0.17)0.0725
+ (0.25
+ 0.145)
(0.14s
0.17)
+
Xu=
la.,
or - 0.633 + 0.495 cos 6,, - 0.399 = 0.661 + 1.2 cos 6", - 2.155
or cos 6r, = 0.655
U 6r, = 49.I"
0.075
_ 2.424
rpeI- -= lal!
i.+Z+
fir. Postfault
operation
uur
sin vd -= 0.495
v.a/,
sin 6
(faulty
line switched
(ii)
off)
X r l = 0 . 2 5 + 0 . 1 5+ 0 . 2 8 + 0 . 1 5 + 0 . 1 7= 1 . 0
*Node elirnination techniquewould
be'used for complex network.
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A generatoroperating at 50 Hz delivers 1 pu power to an infinite bus through
a transmission circuit in which resistance is ignored. A fault takes place
reducingthe maximum power transferableto 0.5 pu whereasbefore the fault,
this power was 2.0 pu and after the clearanceof the fault, it is 1.5 pu. By the
use of equal area criterion, determine the critical clearing angle.
Solution
All the three power angle curves are shown in Fig. 12.30.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,'ffi|
Mod"rnPo*.. sEl!"-nAn"lytit
Ilere
= 1.5 pu
P-"*r =2.0 pu, Pmaxl= 0.5 pu and Pmaxrrr
2.The angular rotor velocity u= d6ldt (over and above synchronous velocity
Initial loading P^ = 1.0 pu
(
p
\
6r,ro=
zrsinI
tffiJ
1 :2.4!rad
E7-sinl
1.5
Applying Eq. (r2.67)
Continuous
solution
U
- 0.523)
- 0.5cos0.523
1.0(2.41
+ 1.5cos2.41= o ???
cos {, 1.5-- 0.5
6r, = 70'3"
t
Af
n
n-1
r>2
Discretesolution
un-|/2
t
u13/2
un-I/T-+tsn4l2
T2.9
NUMERICAT
SOTUTION
OF SWING EOUATION
n-2
In most practical systems,after machine lumping has been done, there are still
more than two machines to be considered from the point of view of system
stability. Therefore, there is no choice but to solve thp swing equation of each
machine by a numerical techniqueon the digital computer. Even in the case of
a single machine tied to infinite bus bar, the critical clearing time cannot be
obtained from equal area criterion and we have to make this calculation
! . .rr--
numerlca[y
rr------
-l-
mrougn
----:-
-
swulg
-----Ll^,
equauulr.
zFL^-^
|
-^-Ll^+i^^+^l
*^+L^l-
t rttrIc aIU ssvtrIilr JuPurDtruilL('(l lllELlluLlD
now available for the solution of the swing equation including the powerful
Runge-Kutta method. He.rewe shall treat the point-by-point method of solution
which is a conventional, approximatemethod like all numerical methods but a
well tried and proven one. We shall illustrate the point-by-point method for one
machine tied to infinite bus bar. The procedure is, however, general and can be
applied-to every machine of a multimachine system.
Consider the swing equation
d26
1 -=
;T
;e*-P^*sind):
PolM;
(* - 9H orinpu systemM = +)
\
7t
iTf)
The solution c(r) is obtained at discrete intervals of time with interval spread
of At uniform throughout. Accelerating power and changein speed which are
continuous functions of time are discretrzedas below:
1. The acceleratingpower Po computed at the beginning of an interval is
assumed to remain constant from the middle of the preceding interval to the
middle of the interval being consideredas shown in Fig. t2.38.
www.muslimengineer.info
p3l2
n-'l
r>112
n
-t
Af
$n-i
6n-z
Jn-1
n-2
Fig. 12.38
n
Af
Point-by-point solution of swing equation
In Fig. L2.38, the numbering on tl\t axis pertains to the end of intervals. At
the end of the (n -l)th interval, the accelerationpower is
Pa (n_r)--Pm- P-* sh 4-r
Q2.68)
where d_1 has been previously calculated.The changein velocit! (a= d6ldt),
causedby the Pa@-r),assumedconstant over At from (n-312) to (n-ll2) is
w n - '2-
wn- 3t 2=( Lt / M ) Pa@- r )
The change in d during the (n-l)th interval is
L6r-t= 6r-1- 6n-2= A'tun4'2
(12.6e)
(12.70a)
and during the nth interval
L6r -
6n- 6n- t = / \ t un- 112
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(12.70b)
,ir Yirtvt r t Ar lr.rl b r )
power SystemStabilitv
[i{8il;r
SubtractingEq. (12.70a\ from Eq. (12.70b) and using Eq. (12.69), we get
L'6,= A6,-t +
(A r)2 D
M
r a(.n-I)
Q^t,,1;^^
\rvtu.,v,t
Using this, we can write
6n= 6n-t + L,6n
7 , ,, 6=,
(Ar)t
M * P a2o +
where Pos* is the acceleratingpower immediately after occurrence of fault.
Immediately before the fault the system is in steady state, so that Poo-= 0 and
ds is a known value. If the fault is cleared at the beginning of the nth interval,
in calculation for this interval one should use for Pa@-r)the value llP"6-r>+ Po6_9*), where Pa@_r)-is the acceleratingpower immediately before clearing
and Po6_r)+is that immediately after clearing the fault. If the discontinuity
occurs at ihe miciciie of an intervai, no speciai proceciure is neecled.The
increment of angle during such an interval is calculated, as usual, from the
value of Po at the beginning of the interval.
The procedure of calculating solution of swing equation is illustrated in the
following example.
A 20 MVA, 50 Hz generator delivers 18 MW over a double circuit line to an
infinite bus. The generatorhas kinetic energy of 2.52 MJA4VA at rated speed.
The generator transient reactanceis X/o = 0.35 pu. Each transmission circuit
has R = 0 and a reactance of 0.2 pu on a 20 MVA bgq-e.lE/l = 1.1 pu and
infinite bus voltage V = 7.0 10". A three-phaseshort circuit occurs at the mid
point of one of the transmissionlines. Plot swing curves with fault cleared by
simrrltaneousopening of breakersat both ends of the line at2.5 cycles and 6.25
cycles after the occuffence of fault. Also plot the swing curve over the period
of 0.5 s if the fault is sustained.
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^f^-^
ws
^-*1,,
+L^
Lall aPPt.y ultt
^+^-
L..
^+^-
stEP-Uy-slttP
-^rl.^l
lIIculUU,
- ^ -t
^-
- -r
---r-a-
Wtr lltrC(l t() Calculate
the inertia constant M and the power angle equations under prefault and
postfault conditions.
Base MVA = 20
(12.7r)
G2.72)
The processof computation is now repeatedto obtain Pa61, L6r*tand d*t. The
time solution in discrete form is thus carried out over the desiredlength of time,
normally 0.5 s. Continuous form of solution is obtained by drawing a ;mooth
curve through discrete values as shown in Fig. 12.38. Greater accuracy of
solution can be achi.evedby reducing the time duration of intervals.
The occurrence or removal of a fault or initiation of any switching event
causesa discontinuity in acceleratingpower Po.lf such a discontinuityoccurs
at the beginning of an interval, then the averageof the values of Po before and
after the discontinuity must be used.Thus, in computing the incrementof angle
occurring durirrg the first interval after a fault is applied at t = 0, Eq. (I2.7I)
becomes
E
nsluls
IneRia coflstant, Mepu\ =
1.0x L52
1 8 0x 5 0
1 8 0/
= 2.8 x 10+ s2le\ectdegree
I
I
Prefault
&' = 0 . 3 5 +
0'2 =0.45
2
Pd= Pr.*r sin d
!,.lxt . r =
= -'.-;;sin
2.M sin 5
d
(i)
Prefault power transf'er = + = 0.9 pu
20
Initial power angle is given by
2.44sin4=0.9
or
\
6o= 21.64"
II During fault A positive sequence reactancediagram is shown in Fig.
12.39a.Converting star to delta, we obtain the network of Fig. 12.39b, in which
2
, , = -0. 35x 0. i + A. 2x0. i + 0. 35x0.-=
trtr
0l
1 AI..Z) pu
P.u = Pmaxtt sin d
- 1'1x1 r;n d = 0.88sin 6
1.25
Fig. 12. 39
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(ii)
lil. Postfault
With the faulted line switched off,
-055
J;=:::,J;2
which it is obvious that the systernis unstable'
: : .: 1t ' x- '1s.i n dI = 2 . 0 s i n d
= 1
Let us choose Al = 0.05 s
The recursive relationshi'ps for step-by-step swing curve calculation are
reproducedbelow.
Pa(n_r)=P^ - P** sin 4_r
(iv)
L6n= L6n-t*
t
;100
o
E
@
(Lt)z
'
'oa(n-l)
M
6n= 6n-t + A,6n
(v)
o 8 0
o)
(vi)
ioo
P
c
(5
o
Sincethereis a discontinuity
in P, andhencein Po, the averagevalueof po
mustbe usedfor the first interval.
P"(0-)= 0 pu and Po (0*) = 0.9 - 0.88 sin 2I.64" = 0.576pu
= 9t#ZQ
Po(ouu.,us"l
fault cleared
at 2.5 cycles
= 0.288pu
L
I
o
Sustained Fault
Calculations are carried out in Table 12.2 in accordance with the recursive
relationship (iv), (v) and (vi) above. The secondcolumn of the table showsP-*
the maximum power that can be transferredat time r given in the first column.
Pn * in the case of a sustainedfault undergoesa sudden change at t = 0* and
remains constant thereafter.The procedure of calculations is illustrated below
by calculating the row correspondingto t = 0.15 s.
( 0 ' l s e c )= 3 1 . 5 9 "
P."* = 0.88
sin d (0.1 s) = 0.524
P, (0.1 s) = P,,,u*sin 6 (0.1 s) = 0.88 x 0.524 - 0.46I
P, (0.1 s) = 0.9 - 0.46I - 0.439
( At\2
YP,
M
(0.1 s) = 8.929x 0.439- 3.92
6 (0.1ss) = Ad (0.1s) + qL
M
P, (0.1s)
U
\
= J.38"+ 3.92"= 11.33"
d ( 0 . 1 5s )= d ( 0 . 1s ) + A d ( 0 . 1 5s )
= 31.59"+ 11.30'= 42.89"
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t
l
0.1
l
t
l
l
0.3
0.2
f (s) --
l
i
0.4
i _ r l
0.6
0.5
fault and for
Fig. 12.40 Swingcuryesfor Example12.10for a sustained
clearingin 2.5 and 6.25 cYcles
aaDle
!A A
aZ.Z
n^i-.
l-., ^aia*
fUllll'-Uy-Pulllt
rratinnc
t/vlllPutqrrvrro
n{
vr
ce rr r' rr ri n
r Vn
n
r r' n / a
vu
fnr
qtrctainarl
fattlt
/ f = 0. 05s
t
P^u
sec
pu
0+
o^u,
0.05
0.10
0 .t 5
0.20
0.25
0.30
0.35
0.40
0.45
0.50
2.44
0. 88
0. 88
0.88
0.88
0. 88
0. 88
0. 88
0.88
0.88
0.88
0.88
sin 6
Pu
0. 368
0. 368
0. 368
0. 41
0.524
0.680
0. 837
0.953
0.999
0.968
0. 856
0.657
a
P"=Prrr,*sin6 P,,= 0'9- P,
0324
0.361
0.46r
0. 598
0. 736
0. 838
0. 879
0. 852
0.154
0.578
deg
Pu
0. 0
0. 576
0. 288
0. 539
0.439
0. 301
0.163
0. 06
0.021
0.048
0.145
0.32r
6
2.57
4.8r
3. 92
2.68
r . 45
0. 55
0.18
0.426
1.30
2. 87
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
6
deg
2r.64
2r . 64
74.21
31.59
42.89
56.87
72. 30
r 6 . r 6 88. 28
1 6 . 5 8 r04.44
2.57
7.38
11.30
13.98
15.43
15.98
r 7 . 8 8 r2r.02
20.75 138. 90
1s9.65
I
f
ttlt
-F . a
su.a
Modern po@is
Flaae^'J
vtcrete.t
:tlt
ltr
o E n---t-A,i,
fryCIeS
Time to clear f'ault= 2.5 = 0.05 s
50
P-u^ suddenly
to 2.0 at t = 0.05-. Since the
be assumedto remain constant fr-om 0.025 s
to 0.075 s. The rest of the
prtrcedureis the sameanclcomplete calculations
are shown in Table 12.3.The
swing curve is plotted in Fig. 12.40 from which
we find that the generator
undergoesa maximum swing of 37.5" but is
stable as c5finaily begins to
decrease.
progressively greater clearing time till the torque angle d increases without
bound. In this example, however, we can first find the critical clearing angle
using Eq. (12.67) and then read the critical clearing time from the swing curve
corresponding to the sustainedfault case. The values obtained are:
Critical clearing angle = 118.62
Critical clearing time = 0.38 s
Table 12.4 Computations
of swing curve for fault clearedat
, = 0. 05s
6. 25cycles( 0. 125s)Af
Table 12'3 Computations
of swing curyesfor fauttcleared at2.s cvcles
(0 .0 5s ), At = 0.05s
P,no
sin 6
P"=P^ *sin6
4
Po= 0.9- P"
pu
P,,,,,^
.sin.5
pu
0
2.44 0.368
0.
0 .8 8 0 .3 6 8
ouu,
0 .3 6 8
0. 05 0 . 8 8 0 . 4 1
0. 05+ 2 .0 0 0 .4 1
0.05uus
0 . 1 0 2.00 0.493
0 . i 5 Z .U U 0 .5 6
o. 20 2 .0 0 0.s91
0.25 2 .0 0 0.597
0 . 3 0 2 .0 0 0 . 5 6 1
0 . 3 5 2 .0 0 0.494
o. 40 2 .0 0 0 . 4 1
0.45 2 .C 0 0.337
0. 50
T
Pr,=P,rr.,*,tin5 pr,= 0.9- pu
pu
0 .9
0.324
0 .3 6
0 .8 2
0 .9 8 6
I.t2
T.I9
r.t9
I.t2
0 .9 8 9
0.82
0.615
pu
0.0
0.576
0.288
0.54
0.08
0.31
- 0.086
- 0.22
- 0.29
- 0.29
- 0 .22
- 0.089
0 .08
0.225
A6
6
deg
deg
21.64
21.64
2.57
2 . 5 7 21.64
24.21
24.21
2.767
5.33 24.21
- 0.767
4.56 29.54
- 1.96
2.60 34.10
- 2.s8
0.02 36.70
- 2.58
- 2.56 37.72
- 1.96
_ 4 . 5 2 34.16
- 0.79
_ 5 . 3 1 29.64
- 4 . 6 0 24.33
0.71
2.0
2.6
19.73
17.13
Fault Cleared in 6.25 Cycles
Time to clearfault= ua?t = 0.125 s
s0
0
0+
ouu,
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
A
A ?
u.4J
0. 88
0.88
2.O0
2.00
2.00
2. 00
2. 00
2.00
' \ N
Z.W
deg
0. 368
0. 368
0. 368
0.41
0.524
0.680
0.767
0. 78
0. 734
0.613
0.430
0.36r
o.46t
1.36
1.53
1.56
1.46
r.22
0. 86
0.0
0.576
0. 288
0.539
0.439
- 4.46
- 0.63
- 0.66
- 0. 56
- 0.327
0.04
u. z-1-1
u.4c)0
u.4-J4
0.9
0.324
T2.TO MULTIMACHINE
6
deg
.
2r.&
2.57
2.57
4.81
7. 38
3. 92
11.30
- 4.10
7.20
- 5.66
r . 54
- 5.89 - 4.35
- 5. 08 - 9.43
- 2.92 - 12.35
0. 35 - 12.00
- u.t - J
- r . 6/
21.64
zt.U
24. 2r
3t.59
42.89
50. 09
51.63
47.28
37.85
25.50
l_J. ) u
5.37
0.50 2.oo
STABITITY
From what has been discussedso far, the following stepseasily follow for
determiningmultimachinestability.
1. From the prefault load flow data determine E/ovoltage behind transient
. reactancefor all generators.This establishesgeneratoremf magnitudes
lEll which remain constant during the study and initial rotor angle
6f = lEt. Also record prime mover inputs to generators,P*o - PoGk
2. Augment the load flow network by the generator transient reactances.
Shift network busesbehind the transient reactances.
3. End Inus for various network conditions*during fault, post fault
(faulted line cleared), after line reclosure.
4.
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2.44
0.88
6
For faulted mode, find generator outputs from power angle equations
(generalizedforms of Eq. (12.27)) and solve swing equationsstep,by
step (point-by-point method).
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern power SystemAnalysis
4ffi |
5. Keep repeating the above step for post fault mode and after line
reclosuremode.
6. Examine d(r) plots of all generators and establish the answer to the
stability question.
The above stgps are illustrated in the following example.
A 50 Hz, 220 kV transmissionline has two generatorsand an infinite bus as
shown in Fig. 12.4I. The transformer and line data are given in Table I2.5. A
three-phasefault occurs as shown. The prefault load flow solution is presented
in Table 12.6. Find the swing equation fbr each generatorcluring the fault
period.
o
o
Vz=1.0328.2350
) v i F1.O217.16o
([5, ) v s = 1 . o l l z
l-ltI'
-
22.490
i
Se
).5+j(
6)
i6,u
l,,m
220kV, 100M VAbase
0.1r
0.0235
0.04
o.022
0.04
0.018
0.004
0.007
Line 4-5
Line 5-1
Line 4-I
Trans;2-4
Trans:3-5
S.No.
and
Voltage
Polar
Bus
No.
Form
Bus
Upe
Voltage
Real
e
Slack
| 1.010"
PV
2 t.0318.35"
3 t.0217.16 PV
4 1.017414.32"PQ
5 1 . 0 1 1 2 1 2 . 6 9 "P Q
1.00
t.0194
1.0121
1.0146
1.0102
Data are given below for the two generatorson a 100 MVA base.
Gen 1 500 MVA, 25 kV, XJ = 0.067 pu, H = 12 MJAyIVA
Gen 2 300 MVA, 20 kV, X,j = 0.10 pu, H = 9 MJA4VA
Plot the swing curves for the machines at buses 2 and 3 for the above fault
which is cleared by simultaneousopening of the circuit breakersat the ends of
the faulted line at (i) 0.275 s and (ii) 0.0g s.
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Generation
Load
Imaginary
f
- 3. 8083 -0.2199 0
0
0. 0
0
0.6986 0
3. 25
0. 1475
0
0.3110 0
0.1271 2.10
1.
0
1.
0
0. 44
0
0.167
0
0.5 0.16
0
0.0439
Solution Before determining swing equations, we have to find transient
internal voltages.
The current into the network at bus 2 basecion the <iatain Tabie i2.6 is
Pr-iQ, _ 3.25-i0.6986
r.o3l- 8.23519"
v:
E{= (1.0194+ j0.1475)+
Fig. 12.41
0.113
0.098
0. 041
Tabfe 12.6 Bus data and prefaultload-flowvaluesin pu on 220 kV,
100M VAbase
'rz- _
o
HaIf line charging
Series Z
Bus to bus
3.2s-j0.6986
x 0.06719V
r.03l-8.23519'
= 1.0340929
+ j0.3632368
= 1.0960lo.337l tad I
= 1.0960333
119.354398"
El = I.0 l0' (slackbus)
2 . 1 - j 0 . 3 1r
x 0.1l9O"
E4 - Q.0I2r + j0.1271)+
r.021- 7.15811'
= 1.0166979+ j0.335177= 1.0705 lI8'2459"
- 1.071 10.31845 rad
The loads at buses 4 and 5 are representedby the admittancescalculated as
follows:
- jo.4zsr)
yr6r.= ''9 .!.:,0!(0.9661
(1.0174)"
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I_ a,grr,
.
PowerSystemStability
v
prefaulr
*Trl
"r"
0 . 5_ j 0 . 1 6
rr.off
During Fault Bus Matrix
- i0'1s647)
(0'488e
Load admittances, al
;;'#;;"J###::
reactanles
of the machines
we
we wlll,
wlr,
therefore'
il::H,:T:|,now :::,::::^'l:,::,1,:nt
designateas buses2 and3, the fictitious
internal nodes between
the internal voltages and the transient
reactancesof the machines.Thus
we get
-Y-^
2 2= - @ = - i r r . 2 3 6
Yzq= j11.236= yqz
Since the fault is near bus 4, it must be short circuited to ground. The Ynus
during the fault conditions would, therefore,be obtained by deleting 4th row
and 4th column from the above augmentedprefault Y".r. rnatrix. Reduced fault
matrix (to the generatorinternal nodes)is obtained by eliminating the new 4th
row and column (node 5) using the relationship
Y*iqn"*1= Y*j@tat -
The reducedfaulted matrix ()'eus during fault) (3 x 3) is given in Table I2.8,
which clearly depicts that bus 2 decouplesfrom the other busesduring the fault
and that bus 3 is directly connected to bus 1, showing that the fault at bus 4
reducesto zero the power pumped into the system from the generatorat bus 2
and renders the second generator at bus 3 to -eive its power radially to bus 1.
= - i7'143
Ytt =
io'04+io]
Yt n(oltt) ynj(old Yrn
)/
@ta)
Table 12.8 Elementsof Yrus (duringfault) and Ysus(post fault)
for Ex. 12.11,admittancesin pu.
Yzs= j7.I43 = yst
Reduced during fault Yu^
Y u = Y t q + y q t * y q s + Bo,
)
Bo,'
) *
Yzq
Bus
L
= 0.9660877
- j0.4250785
+
4.245- j24.2571+ 1.4488_
i 8 . 8 5 3 8 +j 0 . 0 4 1+ r O . 1 1 3 _ j r t . 2 3 5 g
I
2
a
J
s.7986-j35.6301
0
- 0.06817
+5 . 1 6 6 1
Ur,
9*
*' 'r3" ,5
2
2
- 0.4889- j0.1565+ r.4488- j8.8538
+ 7.039r_ j41.335
+ /O.113 + j0.098_ j7.t42}
_ 8.97695s_ js7.297202
The completeaugmented
prefaultlzuu,matrix is shown.iri'Table
12.1.
Table12.T Theaugmented
prefaurt
busadmittance
matrixfor Ex. 12.11,
admittances
in pu
Y s s =Y r s * Y s q * Y s r *
Bus
2
3
4
0
0
_i1r.23sg
0
6
-4.245+ j24.257 jll
.2359
-7.039+ j41.355
0
- 0.0681
+ j5.166l
0
0.1362'-j6.2737
Reduced post Jault Yurt
_ 6.6598977
_ j44.6179
rr.284_j6s.473
0
- jrr.236
0
0
- j7.1428
4.245 + j24.257 -7.039 +
j4r.35s
j11.23s9
0
0
j7.t428
0
6.6598_j44.617 -1.4488
+j8.8538
0 + j7.1428 -1.4488+ j8.8538
8.9769
+ j57.2972
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I
2
-J
- jr3.873r
r.3932
- 0.2214
+ j7.6289
- 0.0901
+ j6.0975
- 0.2214+ j7.6289
0.s- j7.7898
0
- 0.0901+ j6.O975
0
0 . 1 5 9 -1 j 6 . 1 1 6 8
Post Fault Bus Matrix
Once the fault is cleared by removing the line, simultaneouslyopening the
circuit breakersat the either ends of the line betweenbuses4 and 5, the prefault
Y"u5 has to be modified again. This is done by substituting Yqs= Ysq- 0 and
subtracting the seriesadmittance of line 4-5 and the capacitive susceptanceof
half the line from elements Yooand Ytt.
- Yqs- 84512
= Y++(prefault)
Y++lporrfault)
= 6. 65989- j44. 6179- 1. 448+ 78. 853- j0. 113
= 5.2III - j35.8771
Similarly, Ysr(oo*,
rault)= 7.528I i48.5563
The reducedpost fault Y"u5 is shown in the lower half to Table 12.8. It may
be noted that 0 element appears in 2nd and 3rd rorvs. This shows that,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern power System Anatysis
{92 |
-L..^:^^lr-.
PowerSystemStanitity
I 491,1
rr- -
tlle generarors I an0 Z are not Interconnectedwhen line 4-5 is
lrrryr'ruarry'
removed.
During Fault Power Angle Equation
Prz= 0
It rnay be noted that in the above swing equations,P,, ntay be written in
general as follows:
Pn= Pr, - Pr.- P,r,n*sin (6- 7)
Solution
P,3= Re [YrrEr,El** El* \F(];
sinceyy= 0
= E{2 Gn + lEil lEil lrr,l cos (6zr_ Lzt)
= ( 1 . 0 7 1 )(20 .1 3 6 2 +
) I x 1 . 0 7 1 x 5 . 1 6 6 5c o s( d 3_ 9 0 . 7 5 5 " )
P " 3= 0 . 1 5 6 1+ 5 . 5 3 1s i n ( h - 0 . 7 5 5 )
Postfault Power Angle Equations
p"z= lE/P G22+ lElt lEll ly2Llcos (dr, _ 0zr)
= 1.0962x 0.5005+ I x 1.096x 7.6321cos ({ - 9I.662")
= 0.6012+ 8.365sin (d, _ I.662)
p,3 = tE{ 2q3 + Ell tElt t\l cos ( 6r,_ 0rr)
- 7.0712x 0.1591+ 1 x 1.07rx 6.09gcos (dr - 90.g466")
of Swing Equation
The above swing equations(during fault followed by post fault) can be solved
by the poinrby-point method presented earlier or by the Euler's method
presented in the later part of this section. The plots of E and 4 are given in
Fig. 12.42 for a clearing time of 0.215 s and in Fig. 12.43for a clearing time
of 0.08 s. For the case (i), the machine 2 is unstable, while the machine 3 is
stable but it oscillates wherein the oscillations are expected to decay if effect
of damper winding is considered.For the case(ii), both machinesare stablebut
the machine 2 has large angular swings.
Machine1 is reference(lnfinitebus)
= 0.1823+ 6.5282sin (d, _ 0.9466")
Swing Equations-During
#=ff
Fault
(P^z-P,z)=
= t:9/ e.2s - o) elecr deg/sz
12
d , q _ 1 8 0f
(Pn-P"t)
a'r= k
= t*:/
9
e . r - { 0 . 1 s 6 1+ 5 . 5 3 1s i n ( 6 t - 0 . 7 s s ) } l
=
l.g43g - 5.531sin (d, - 0.755")lelectd,eg/sz
Y
(0.275sfaultclearedat)
Swing Equations-Postfault
p . 2 s - 1 0 . 6 0zr + 8 . 3 6 5s i n ( d ) - t . 6 6 2 " ) } eJ r e c d
t eg/s2
#={f
dtd, 180f
-*
=*12.10-{0.1823
dt'
9
+ 6.5282sin(d3_ 0.g466.)}lelect
deg/sz
www.muslimengineer.info
F\g.12.42 Swingcurvesfor machines2 and 3 of Example12.1for
clearingat 0.275s.
If the fault is a transient one and the line is reclosed,power angle and swing
equations are neededfor the period after reclosure.Thesecan be computed from
the reduced Ysus matrix after line reclosure.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
494
PowerSystemStabilitv
Modern power System Analysis
|
Machine 1 is reference (lnfinite bus)
500
.
(t2.74)
Initial state vector (upon occurrence of fault) is
xoLk=fr= lEot
M a c h i n e2
x " z k =0
at)
c)
o
The state form of swing equations (Eq. (12.74)) can be solved by the many
available integration algorithms (modified Euler's method is a convenient
choice).
L
q,
o)
!
o
(t
c
i * = # ( P " o r - P c o )k, = 1 , 2 ,. . . ,m
r7.2
l,,4PFr
Algorithm
Computational
Modified Euler's Method
100
i
l
l
t
l
_L__ I _
0.8
0.9
| _
1.0
Faultcleared
after4 cycles
Fig. 12.43
Swing curves for machines 2 and 3 of
Example
12.11forclearing
at 0.0gs
for Obtaining
Swing Currzes Using
Carry out a load flow studY (prior to disturbance) using specified
voltagesand powers.
2 . Compute voltage behind transient reactancesof generators(Eo*) using
emf magnitudesand initial rotor angle
Eq. (9.31).This fixes generator.
Y?).
voltag"
bus
lreference slack
Compute, Ysu5 (during fault, post fault, line reclosed).
J.
l.
a
4. Set time count r = 0.
5 . Compute generatorpower outputs using appropriatgts"us with the help
Gonsideration of Automatic voltage Regulator (AVR) and
Speed Governor Loops
of the geniral form of Eq. (12.27). This givet Pg},for / 1 /').
Note: After the occurrenceof the fault, the period is divided into uniform
discretetime intervals(At) so that time is countedas /(0),t(l), ......A
typical value ol' lt
is 0.05 s.
6. Computet(i[;'},i\7'),k - - 1 , 2 , . . . , m l f i o m E q s .( 1 2 . 7 4 ) .
liu' t = , {t +l)'pr r
thc I'ilslstutccst ir r t ir lc's
7. Corttputc
)t
, [ , 0 * r =, , t l o + i [ , 0 a
state variable Formulation of swing Equations
At
,ff') - *V)+ *$'o)
The swingequationfor the hh generatoris
+d t '
= + ( p ' ou o - p " ) ;k - r , 2 ,. . .f,f i
Hk'
(r2.73)
For the multimachine case,it is more convenientto organiseEq. (12.73)
is
state variableform. Define
xrk= 6r= lE*'
xz*= 6t
I = |.2. .... rrt
of E^('+t)
the first estimates
8. Cortrpute
BQ+D= E? lcos x,(i*r)+7 sin *\lf'))
g. ComputeP8;'); (appropriateY"u5and Eq. (12.72))'
10. compute[t;{in'),it:o*t'),k = r,2, ..., mf fromEqs.(12.74).
11. Computethe averagevaluesof statederivatives
i[i,), urr= ][iu,cl +;l[*t)]
k=1,2,...,ffi
Then
i t*= xzt
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= *I*\:|+,tt'i"l
iL'),*,
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Power System Stabitity
1't
LL.
A
Lompute tne lrnal state estimates for | = t\r+t)-
*;Iu" = *([)+ ill) uus,
at
,&*r)- ,([l + if).^,rat
k = 7'2'"'' frt
BQ+t) = l4llcos xf;+r) + 7 sin *f1r)
14. Print (",9*t),*;:o*D); k = I,2, ..., m
15. Test for time limit (time for which swing curve
is to be plotted), i.e.,
c h e c k i f r> rn n u r.If n o t, r- r+ r and repeatfromstep5
above.
Otherwise print results and stop.
The swing curves of all the machines are plotted.
If the rotor angle of a
machine (or a group of machines) with r"rp".t to other
machines increases
without bound, such a machine (or grouf of machines)
is unstable and
eventually falls out of step.
The computational algorithm given above can be easily
modified to include
simulation of voltage regulator, field excitation response,
saturation of flux
paths and governor action
Study of Large Systems
To limit the computer memory and the time requirements
and for the sake of
computationalefficiency, a large multi-machine system
is divided into a study
subsystemand an external system.The study subsystem
is modelled in detail
whereas approximate modelling is carried out for the
external subsystem.The
total study is renderedby ihe modern techniqueof
dynamic equivalencing.In
the externalsubsysteln,nutnber<lfrnachinesis drastically
reclucedusing various
methods-coherency based rnethodsbeing most popurar
and widely used by
v anous p o w e r u ti l i ti e si n th e w o rl d .
I2.T7
SOME FACTORS AFFECTING
TRANSIENT
STABILITY
We have seenin this chapterthat the two-machine
systemcan be equivalently
reduced to a single machine connected to infinite
bus bar. The qualitative
conclusions regarding system stability drawn from
a two-machine or an
equivalent one-machineinfinite bus system can
be easily extended to a
multimachine system. In the last article we have
studied ihe algorithm for
determiningthe stability of a multimachrnesystem.
It has been seen that transient stability is greatly
affected by the type and
location of a fault, so that a power systemanalyst
must at the very outset of a
stability study decide on these two factors. In ou,
we have selected
"*u-ples
a 3-phase fault which is generally more severe from
point of view of power
transfer' Given the type of fault and its location
let us now consider other
www.muslimengineer.info
491;,,
factors which affect transient stability and therefrom draw th" .on"llsions,
regarding methods of improving the transientstability lirnit of a system and
making it as close to the steady state limit as possible.
For the case of one machineconnected to infinite bus, it is easily seen from
Compute the final estimate for Eo at t = r('*l) using
Stability
|
the angle through which it swings in a given time interval offering thereby a
method of improving stability but this cannot be employed in practice because
of economic reasons and for the reason of slowing down the response of the
speed governor loop (which can even become oscillatory) apart from an
excessiverotor weight.
With referenceto Fig. 12.30,it is easily seenthat for a given clearing angle,
the accelerating area decreasesbut the decelerating area increases as the
maximum power limit of the various power angle curves is raised, thereby
adding to the transient stability limit of the system. The maximum steady power
of a system can be increasedby raising the voltage profile of the system and
by reducing the transfer reactance.These conclusionsalong with the various
transient stability casesstudied,suggestthe following method of improving the
transient stability limit of a power system.
1. Increaseof system voltages,use of AVR.
2. Use of high speedexcitation systems.
3. Reduction in systemtransf-erreactance.
4. Use of high speedreclosing breakers (see Fig. 12.32).Mo&rn tendency
is to employ single-pole operation of reclosing circuit breakers.
When a fault takes place on a system, the voltagesat all buses are reduced.
At generator terminals, these are sensed by the automatic voltage regulators
which help restoregeneratorterminal voltagesby acting within the excitation
system.Modern exciter systemshaving solid statecontrols quickly respondto
bus voltage reduction and can achieve from one-half to one and one-h'alfcycles
(l/2-l])
gain in critical clearingtimes fbr three-phase
taults on the HT bus
of the generator transformer.
Reducing transfer reactance is another important practical method of
increasing stability limit. Incidentally this also raises system voltage profile.
The reactance of a transmission line can be decreased(i) by reducing the
conductor spacing,and (ii) by increasingconductordiameter(see Eq. (2.37)).
Usually, however, the conductor spacing is controlled by other features such as
lightning protection and minimum clearanceto prevent the arc from one phase
moving to another phase.The conductor diametercan be increasedby using
material of low conductivity or by hollow cores. However, norrnally, the
conductor configuration is fixed by economic considerationsquite apart from
stability. The use of bundled conductorsis, of course,an effective means of
reducing series reactance.
Compensation for line reactance by series capacitors is an effective and
economical method of increasing stability limit specially for transmission
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'4lI
Modern power System Analysis
I
distancesof more than 350 km. The ciegreeof series
compensation,however,
accentuatesthe problems of protective relaying, normal
voltage profiles, and
overvoltagesdrrring line-to-groundfaults. Series
cornpensationbecomesmore
effective and economical if part of it is
of compensationupon the occurrencec
Switehed series eapaeitorssimultaneot
and raise the transient stability limit tc
limit. Switching shunt capaciiors on ol
stability limits (see Example 12.2) but the MVA
rating of shunt capacitors
required is three to six times the rating of switched
series capacitors for the
same increase in stability limit. Thus series capacitors
are preferred unless
shunt elements are required for olher pu{poses,siy,
control of voltage profile.
Increasing the number of parallel lines between transmission
points is quite
often used to reduce transfer reactance.It adds at the
same time to reliability
of the transmissionsystem. Aclditional line circuits
are not likely to prove
economical unit I aftet all feasible improvements have
been carried out in the
first two circuits.
As the majority of faults are transientrn nature,rapid switching
and isolation
of unhealthy lines followed by reclosing has been
shown earliei to be a great
help in improving the stability marginr. ih. modern
circuit breakertechnology
has now made it possible fbr line clearing to be done
as fast as in two cycles.
Further, a great majority of transient faults a'e line-to-ground
in nature. It is
natural that methodshave beendevelopedfor selective
single pole opening and
reclosing which further aid the stability limits. With
,"f"r".,"" to Fi;. lz.r7, if
a transient LG fault is assumedto occur on the generator
bus, it is immedi ately
ol,lnt the fault therewill now be a definite amount
of power rransfer,
::.1,:l_.,
ab
uurereni rrom zero power transfer for the case of a
three-phasefault. Also
when the circuit breaker pole corresponding to the faulty
line is opened, the
other two lines (healthyones)remain intact so that considlrable
power transfer
continuesto take place via theselines in comparison
to the caseof three-pole
switching when the power transferon fault clearing will
be reducedto zero. It
is, therelbre, easy to see why the single pole swiiching
and reclosing aids in
stability problem and is widely adopted.These facts
arelilustrated by means of
Example 12'12. Even when the stability margins are
sufficient, single pole
switching is adopted to prevent large swings and
consequent vortage dips.
Single pole switching and reclosingis, of course,expensiu.
in t..*s of relaying
and introducesthe as.socjatec!
problernsof overvoltagescausedby single pole
opening owing to line capacitances.Methods are
available to nullify these
capacitive coupling effects.
nd
dr++^s^^L
L----
I
I
:.-^Ir
I.1.':/tlltt,
stzeof rotor reducesinertia constant, Iowering thereby the stability margin. The
loss in stability margin is made up by such features as lower reactance lines,
fastercircuit breakersand faster cxcitation systenrsas tliscussetlalreacly,and
a faster system valving to be discussed later in this article.
A stage has now been reached in technology whereby the methods of
irnprovinE=stability; discussetl above, have been pushed to their limits, e.g.,
clearing times of circuit breakers have been brought down to virnrally
irreducible values of the order of two cycles. With the trend to reduce machine
inertias there is a constant need to determine availability, feasibility and
applicability of new methodsfor maintaining and/or improving system stability.
A brief account of some of the recent methods of maintaining stability is given
below:
HVDC Links
Increased use of HVDC links ernploying thyristors would alleviate stability
problems. A dc link is asynchronous,i.e., the two ac system at either end do
not have to be controlled in phase or even be at exactly the same frequency as
they do for an ac link, and the power transmitted can be readily controlled.
There is no risk of a fault in one system causing loss of stability in the other
system.
Breakingr
Resistors
For improving stability where clearing is delayed or a large tn)a i, suddenly
lost, a resistive load called a breaking resistor is connected at or near the
generatorbus. This load compensatesfor at least sonreof the reduction of load
on the generatorsand so reducesthe acceleration.During a f-ault,the resistors
are applied to the terminals of the generators through circuit breakers by means
of an elaboratecontrol scheme.The control schemedeterminesthe amount of
resistanceto be applied and its duration. The breakingresistorsremain on for
a matter of cycles b<lth during fault clearing and after system voltage is
restored.
Short
Circuit
Current
Limiters
These are generally used to limit the short circuit duty of distribution lines.
These may also be used in long transmission lines to modify favourably the
transferimpedance during fault conditions so that the voltage profile of the
systemis somewhatimproved, therebyraising the systemload level durins the
fault.
Recent Trends
Turbine
Recent trends in design of large alternators tend towards
lower short circuit
ratio (scR = r/x), which is achieved by reducing
machine air gap with
consequentsavings in machine mmf, size, weight
and cost. Reduction in the
The two methods just discussedabove are an attempt at replacing the sysrem
load so as to increase the electrical output of the generator during fault
conditions. Another recent method of improving the stability of a unit is to
decreasethe mechanical input power to the turbine. This can be accornplished
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.
i::
Fast
Valuing
or Bypass
Valuing
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
by rneansof fast valving. where the differenee between mechanicalinput and
reduced electrical output of a generator under a fault, as sensedby a control
scheme, initiates the closing of a turbine valve to reduce the power input.
Briefly, during a fast valving operation, the interceptor valves are rapidly shut
(in 0.1 to 0.2 sec) and immediately reopened. This procedure increasesthe
critical switching time lons enoush
stable for faults with stuck-breakerclearing times. The schemehas been put to
use in some stationsin the USA.
FUII Load Rejection
Technique
Fast valving combined with high-speedclearing time will suffice to maintain
stability in most of the cases.However, there are still situations where stabilitv
is difficult to maintain.In such cases,the normal procedureis to automatically
trip the unit off the line. This, however, causesseveral hours of delay before
the unit can be put back into operation.The loss of a major unit for this length
of tirne can seriouslyjeopardize the remaining system.
To remedy these situations, a full load rejection scheme could be utilized
after the unit is separated from the system. To do this, the unit has to be
equipped with a large steam bypass system. After the system has recovered
from the shock caused by the fault, the unit could be resynchronized and
reloaded. The main disadvantageof this method is the extra cost of a large
bypasssystem.
- --l
r--J
-,t0.15
7
0'1 p
I
L --z -rfd]-ui-r
0'3
0.1
i-dT L_r6TT\_
I
|
l
|
I
(b) Negative sequence network
t
;
ao
''
-t
Ti
.
)
l]o t
I_
l
go ''
|
|L----.-' |I p 1 ' o |
|
i-rT60 t;,'r";*u1n""
-t-
= C xo=0.0e15
l
I
networl
Fig. 12.45
For an LG fault at P the sequencenetworks will be connectedin series-as
shown in Fig. 12.46. A star-deltatransformationreducesFig. 12.38 to that of
Fig. 12.47 from which we have the transfer reactance
Xr2(LGfaulQ= 0.4+ 0.4+ nOII''O = 1.45
0.246
The systetnshown in Fig. 12.44is loaded to I pu. Calculate the swing curve
and ascertainsysternstability for:
(i) LG fault threepole switchingfolloweclhy reclosure.linc lounclhealthy.
(ii) LG fault singlepole switchingfbllowed by reclosure,line found healthy.
Switching occurs at 3.75 cycles (0.075 sec) and reclosure occllrs at 16.25
cycles (0.325 sec). All values shown in the figure are in pu. -
l . r . l =, ?
. ,_
H = 4.167
Xr=o3Oj
Xt=0'15
xo=o.t
I
I
.-t.
l E l = 1 ' 2(
)
I
r-rd 60
0'4
P
o'4
Fig.12.46 Connection
of sequencenetworksfor an LG fault
X r =0 . 1
-'.
?*f ( / -h'Ir-!.&-f!
'L
AY-l
f-
(,r
II
Xn
rl /' - f t r f \ r
'lr
t
___J
Fig. 12.44
Fig.12.47 Transferimpedancefor an LG fault
Solution The sequencenetworks of the systemare drawn and suitably reduced
in Figs. I2.45a, b and c.
When the circuit breakerpoles correspondingto the faulted line are opened
(it correspondsto a single-line open fault) the connectionof sequencenetworks
is shciwn in Fig. 12.48. From the reduced network of Fig. 12.49 the rransfer
reactance with faulted line switched off is
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
-Y.^
^ll
(fettlterl
\-*---
line
nnen\
--
0
A t|
v.-
A
A1
v.aL
-rr
(l A -_
V..+
I .o.ro
1 ..
L../-L
Under healthy conditions transfer reactance is easily obtained from the
positive sequencenetwork of Fig. 12.45 a as
Xrr(line healthy)= 0.8
Pettt = 0
PrN= Pd = 1.5 sin d
Now
46,- A6n-,+ @LP^,' a ( n - l ,,
)
M
Negative sequence
o'1 P
H = 4.167MJA{VA
Zero sequence
- 4.63 x 10a sec2lelectricaldegree
1,1= JU1 8 0x 5 0
Flg.12-48 Connectionof sequencenetworkswith faultedline switchedoff
Taking At = 0.05 sec
(at)'
l v l= 1 ' o
lEl=1'z
P
P/
o:4
4. 63xI 0- 4
Time when single/threepole switching occurs
Fig. 12.49 Reducednetworkof Fig. 12.49givingtransferreactance
= 0.075 sec (during middle of At)
Time when reclosing occurs = 0.325 (during middle of at)
L
PreJault
rE vr
sin d= -7 . 2 x 1 S l I l d =
X,,
0.8
Initial torqueangleis givenby
1 = 1.5 sin 5o
6o= 47.8"
Duringfault
F'",,=
I)uring
l.Zxl
lff
sin d = 0.827
sin
single pole switching
=
Perrr
+#
sin d = 0.985
sin
www.muslimengineer.info
P,u,o
sec
l.)
stn b
0
o*
ouu,
0.05
Initial load= 1.0 pu
or
.,
Table 12.9 swing curyecalculation-threepoleswitching
Power angle eguations
p,=
5.4
0.075---+
0.10
0.15
0 . 20
o.25
0 . 30
0.325--+
0 . 3s
0.40
0 . 45
0. 50
0. s5
0. 60
0.65
6
)
P "
(pu)
P,,
(pu)
1. 5
0. 667 r . 0
0.0
0. 827 0. 667 0. 552 0. 448
0.224
0.827 0.682 0. 564 0. 436
0.0
0. 0
0.0
0. 0
0. 0
1. 5
1. 5
1.5
1. 5
1.5
r.5
0.726
0.0
0.0
0.0
0.0
0.0
0. 565 0 . 8 5
0. 052 0.078
- 0 . 5 5 - 0.827
- 0. 984- r . 48
- 0 . 6 5 1- 0 . 9 8
0 . 4 9 7 0. 146
.5.4P,,
ifi
elec deg elec deg
1.2
N A
6
elec deg
1.2
3.6
41.8
41.8
41.8
43.0
1.0
1.0
1.0
1.0
1.0
5. 4
5. 4
5. 4
5. 4
5. 4
9. 0
r 4. 4
19. 8
2s. 2
30.6
46. 6
55. 6
70. 0
89. 8
I 15.0
0.15
0.922
r.827
2. 48
1.98
0. 254
0. 8
5. 0
9. 9
13.4
10.7
t.4
31.4
36.4
46.3
59.7
10.4
71.8
r45.6
177.O
2r3.4
259.7
3r9.4
389.8
461.6
The swing curve is plotted in Fig. 12.50 from which it is obvious
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
that rhe
PowerSystemStability
\4gqe!_lg',gf tyq]g1_Analysis
I
300 !
I
I
250
t I
II
i
L
\
I
r pote ii
switchoff i
2ooi- / / l
I
6q)
g)
o
E
o
l.s
1.5
1.5
1.5
0.998 1.5
1.0
1.5
1.0
1.5
0 . 9 9 8 51 . 5
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
0.96 t.44 - 0.44 - 2.4
0.894 1.34 -0.34 -1.8
0.781 t.l1 - 0.17 - 0.9
0.62 0.932 0.068 0.4'
0.433 0.6s
0.35
r.9
0.2s9 0.39
0.61
3.3
0.133 0.2
0.8
4.3
0.079 0.119 0.881 4.8
0.107 0.161 0.839 4.5
0.214 0.322 0.678 3.7
0.38 0.57
0.43
2.3
0.566 0.84
0 .1 6
0.9
0.738 1.11 - 0.11 - 0.6
- 0.3 - 1.6
0.867 1.3
0.946 t.42 - 0.42 - 2.3
0.983 1.48 - 0.48 - 2.6
- 0.s
- 2.7
0.997 1.5
1.10
1.15
1.20
r.25
1.30
1.35
r.40
1.45
1.50
MACHINEUNSTABLE
(.) 1 5 0
L
q)
J)
to
100
r
I
,r
o'5
l--,r- r
Time (sec)
]...r
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
/'/'
t,,,
SEC
(pu)
0
1.5 0.661 1.0
0.827 0.667 0.s52
o,
f
(pu)
oory
0.05 0.827 0.682 0.s64
0.075--0 . 1 0 0.98s 0.726 0 . 7 L s
0 . 1 5 0 . 9 8 s 0 . 7 8 4 0.77
o.20 0.985 0.848 0 . 8 3 4
o.25 0.98s 0.908 0.893
0 . 3 0 0 . 9 8 5 0 . 9 5 6 0.940
4325---+
0.35 1.5 0.988 1.485
I
l'r,
(pu)
5'4P,,
elec deg
Ab
elec deg
b
elec deg
1.2
2.4
r.2
3.6
4t . 8 0
41.8
41.8
43.0
0.285 1.5
0.230 r.2
0.166 0.9
0.107 0.6
0.060 0.3
5.1
6.3
7.2
1.8
8.1
46.6
5r.7
58.0
65.2
73.0
0.485 - 2.6
5.5
81.1
0.0
0.448
0.224
0.436
(faultcleared)
Reclosure
| ' /
A
I
I
,s:itth
73.7
63.4
51.3
38.3
25.7
15.0
7.6
4.5
6.2
r2.4
22.3
34.5
47.6
60.1
7r.O
79.6
95.6
gg.g
Single pole switch off
Table 12.10 swing curvecalculation-single
pole switching
I',uu^
- 10.3
- 12.r
- 13.0
- 12.6
- 10.7
- 7.4
- 3.r
1.7
6.2
9.9
12.2
l3.l
12.5
10.9
8.6
6.0
3.3 \
'
I
The swing curve is plotted in Fig. 12.51from which it follows that the sysrem
is stable.
r
1.0
Fig. 12.50 swing curuefor threepole switchingwith reclosure
I
2.8
0.1
- 2.6
- 2.7
- 2.7
- 2.7
-2.7
0.40
0.45
0.50
0.55
r. 0 5 r . 5
I
- 0.5
- 0.5
- 0.s
- 0.5
505;
|
86.6
89.4
89.5
86.9
o)
o
o)
o)
BO
MACHINESTABLE
60
o
C)
L
o
o
o)
ra
i-
l--r
05
Time (sec)
www.muslimengineer.info
(Contd....)
Fig. 12.51 Swingcurvefor singlepoleswitchingwith reclosure
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
506 |
I ina
Modernpo@is
PROB
TEIlIS
t 2 . l A two-pole, 50 Hz, 11 kv turboalternatorhas a rating of 100 Mw,
powel factor 0.85 lagging.The rotor has a momentof inertia of a 10,000
12.2 Two turboalternatorswith ratings given below are interconnectedvia a
short transmissionline.
Machine 1: 4 poIe, 50 Hz, 60 MW, power factor 0.g0 lagging,
moment of inertia 30,000 kg-rn,
Machine
2 pole, 50 Hz, 80 MW, power factor 0.85 lagging,
moment of inertia 10,000 kg--'
Calculate the inertia constantof the single equivalentmachine on a base
of 200 MVA.
t 2 . 3 Power station t has four identical generatorsetseachrated g0 MVA and
each having an inertia constant 7 MJA4VA; while power station 2 has
three sets each rated 200 MVA, 3 MJA4VA. The stations are locatld
close togetherto be regardedas a single equivalentmachinefor stability
studies. Calculate the inertia constant of the equivalent machine on 100
MVA base.
from its prefault position, determine the maximum load that could be
transferredwithout loss of stability.
I2.8 A synchronousgenerator is feeding 250 MW to a large 5O Hz network
over a double circuit transmissionline. The maximum steadystateDower
that can be transmitted over the line with both circuits in operation is
500 MW and is 350 MW with any one of the circuits.
A solid three-phasefault occurring at the network-end of one of the lines
causesit to trip. Estimate the critical clearing angle in which the circuit
breakersmust trip so that synchronismis not lost.
What further information is neededto estimate the critical clearing time?
12.9 A synchronousgenerator representedby a voltage source of 1.05 pu in
serieswith a transient reactanceof 70.15 pu and in inertia constant F/ =
4.0 sec, is connectedto an infinite inertia system through a transmission
line. The line has a series reactanceof70.30 pu, while the infinite inertia
system is representedby a voltage source of 1.0 pu in series with a
transient reactanceof 70.20 pu.
The generatoris transmitting an acti're power of 1.0 pu when a threephasefault occurs at its terminals. If the fault is clearedin 100 millisec,
determine if the system will remain stable by calculating the swing
curve.
12.4 A 50 Hz transmissionline 500 km long with constantsgiven below ties
up two large power areas
R = 0 .1 1 f)/k m
12.10 For Problem 12.9 find the critical clearing time from the swin! currrefor
a sustainedfault.
L - 1.45mH/km
G = 0
C = 0.009 lFlkm
F i n d t h e s, . f- -c' :. ,t,rJl v s l z f e c f e h i lrirtr rJ r lr ir m
r r rirtr
i, ,f l, rl ,, t/t | _-
| _-
ll/
| v
Rt
t, \nr vl l
l 2.l l
t,\/
A
V
/..^-^+^-+r
\Lt,|t:ltdlllr.
What will the steadystate stability limit be if line capacitanceis also
neglected?What will the steadystate stability limit be if line resistance
i s a l s o n e g l o c tc d C
' / o rn n rc l tton tl rc rcsul ts.
t 2 . 5 A power deficient area receives 50 MW over a tie line from another
area.The maximum steadystatecapacityof the tie line is 100 MW. Find
the allowable sudden load that can be switched on without loss of
stability.
1 2 . 6 A synchronous motor is drawing 30vo of the maximum steady state
power from an infinite bus bar. If the load on motor is suddenly
increasedby 100 per cent, would the synchronismbe lost? If not, what
is the maximum excursionof torque angle about the new steady state
r<ltorposition.
t 2 . 1 The transfer reactancesbetween a generator and an infinite bus bar
o p e ri l ti n gi rf 2 0 0 k V trn d e rv a r i ouscondi ti onson fhe i nterconnector
aro:
Pretault
S0 0 per phase
During fault
m
Postfhult
O per phase
2ffi {) per phase
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A synchr t ) nous
gener at orr epr esent cd
by a volt ageof l. l5 pu in ser ies
with a transientreactanceis c<lnnected
to a large power system with
volt t t gc 1. 0 pu t hr ot lgh l powcr r r clwor k.Thc cquivalcntt lar r sient
transf'erreactanceX betweenvoltage sourcesis 70.50 pu.
After the occurrenceof a three-phase
to grounclfault on one of the lines
of the power network, two of the line circuit breakersA and B operate
sequentially as follows with correspondingtransient transfer reactance
given therein.
(i) Short circuit occurs at 6 = 30", A opens instantaneouslyto make X
= 3.0 pu.
(ii) At 6 = 60o, A recloses, X = 6.0 pu.
(iii) At 5=75o, A reopens.
(iv) At d = 90o, B also opens to clear the fault making X = 0.60 pu
Check if the systenrwill operatestably.
12.12 A 50 Hz synchronous generatorwith inertia constant H = 2.5 sec and
a transientreactanceof 0,20 pu feeds 0.80 pu active power into an
infinite bus (voltage I pu) at 0.8 lagging power lactor via a network with
an equivalent reactance of 0.25 pu.
A three-phasefault is sustainedfor 150 millisec across generator
terminals.Determine through swing curye calculation the torque angle6,
250 millisec, after fault initiation.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
PowerSystemStabitity
12.13 A 50 Hz, 500 MVA,400 kV generator (with transformer) is connected
to a 400 kV infinite bus bar through an interconnector. The generator
has F1- 2.5 MJA4VA, voltage behind transientreactanceof 450 kV and
is loaded 460 MW. The transfer reactancesbetween generator and bus
During fault
Postfault
5(D
I
12. Kundur, P., Power SystemStability and Control, McGraw-Hill, New York, 1994.
13. Chakrabarti,A., D.P. Kothari and A.K. Mukhopadhyay, Performance Operation
and Cqntrol of EHV Power TransmissionSystems,Wheeler Publishing, New
Delhi, 1995.
14. Padiyar,K.R., Povter System
Stability
and Control,
Znd
lcatlons, Hy
1 . 0p u
0.75pu
15. Sauer,P.W. and M.A. Pai, Power SystemDynamics and Stabiliry, Prentice-Hall,
New Jersey,1998.
Calculate the swing curve using intervals of 0.05 sec and assuming that
the fault,is cleared at 0.15 sec.
I2.I4 Plot swing curves and check system stability for the fault shown on the
system of Example 12.10for fault clearing by simultaneousopening of
breakers at the ends of the faulted line at three cycles and eight cycles
after the fault occurs.Also plot the swing curye over a period of 0.6 sec
if the fault is sustained.For the generator assumeH = 3.5 pu, G = 1 pu
and carry out the computations in per unit.
12.15 Solve Example 12.10 for a LLG fault.
Papers
16. Cushing, E.W. et al., "Fast Valving as an Aid to Power System Transient
Stability and Prompt Resynchronisationand Rapid Reload After Full L,oad
t
Rejection", IEEE Trans, L972, PAS 9I 1624.
17. Kimbark, E.W., "Improvement of Power System Stability", IEEE Trans., 1969,
PAS-88:773.
18. Dharma Rao, N. "Routh-Hurwitz Condition and Lyapunov Methods for the
TransientStability Problem", Proc. IEE, 1969, 116: 533.
19. Shelton,M.L. et al., "BPA 1400 MW Braking Resistor",IEEE Trans., 1975,94:
602.
20. Nanda, J., D.P. Kothari, P.R. Bijwe and D.L. Shenoy,"A New Approach for
NCES
REFERE
Books
1
Stevenson,W.D., Elements of Power SystemAnalysis, 4th edn., McGraw-Hill,
New York, 1982.
Elgerd, O.I., Electic Energy Systems Theory: An iniroduciion,
McGraw-Hill, New York, 1982.
2nd edn.,
3.
Anderson, P.M. and A.A. Fund, Power SystemControl and Stability, The Iowa
State University Press,Ames, Iowa, 1977.
4.
stagg, G.w. and A.H. o-Abiad, computer Methods in Power system Analysis,
Chaps9 and 10, McGraw-HillBook Co., New York, 1968.
Crary, S.8., Power System Stability, Vol. I (Steady State Stability), Vol. II
(Transient Stability), Wiley, New York, 1945-1947.
Kimbark, E.W., Power SystemStability, Vols 1, 2 and 3, Wiley, New york, 1948,
5.
6.
7.
Veuikorz,Y.A., TransientPhenomenain Electrical Power System(translatedfrom
the Russian),Mir Publishers,Moscow, 1971.
8. Byerly, R.T. and E.w. Kimbark (Eds.), stability of l^arge Electric power
Systems,IEEE Press,New York, 1974.
9. Neuenswander,J.R., Modern Power Systems,InternationalText Book Co., 1971.
t0. Pai, M.A., Power SystemStability Annlysis by the Direct Method of Lyapunov.,
North-Holland, System and Control Services, Vol. 3, 1981.
I 1. Fouad, A.A and V. Vittal, Power System Transient Stability Analysis using the
Dynamic Equivalents Using Distribution Factors Based on a Moment Concept",
Proc. IEEE Int. Conf. on Computers, Systemsand Signal Processi4g, Bangalore,
\
Dec. 10-12, 1984.
'Dynamic
2 r . Dillon, T.S.,
Modelling and Control of Large Scale System", Int.
Journal of Electric Power and Energy Systems,Jan. 1982, 4: 29.
2 2 . Fatel,R., T.S. Bhatti and D.P. Kothari, "Improvementof PowerSystemTransient
stability using Fast valving: A Review", Int. J. of Electric Power components
and Systems,Vol. 29, Oct 2001, 927-938.
23. Patel, R., T.s. Bhatti and D.P. Kothari, "MATLAB/simulink Based rransient
Stability Analysis of a Multimachine Power System,IJEEE, Vol. 39, no. 4, Oct.
2002, pp 339-355.
Patel R., T.S. Bhatti and D.P. Kothari, "A Novel scheme of Fast valving
Control", IEEE Power Engineering Review, Oct.2002, pp. 4446.
25. Patel,R., T.S. Bhani and D.P. Kothari, "Improvementof Power system Transient
stability by coordinated operation of Fast valving and Braking Resistor", To
appearin IEE proceeriings-Gen.,Trans and Distribution.
. A
L+.
Transient Energy Function Method, Prentice-Hall, New Jersy, 1992.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Power System Security
13.1
INTRODUCTION
In Chapter7, we have beenprimarily concerned with
the economical operation
of a power system' An equally important factor in
the operation of a power
system is the desire to maintain system security.
System security involves
practicessuitably designedto keep the system
operating when componentsfail.
Besideseconomizingon fuel cost and minimizrngemission
of gases(co, cor,
Nox, sor), the power systernshould be operationally,.secure,,.
An operation_
ally "secure" power system is one with low probability
of, systern black out
(collapse)or equipmentdamage.If the pro."r,
uf cascadingfailurescontinses.
the systernas a whole or its tnajor parts may completely
collapse. This is
normally referred to as system blackout. All these
aspects require security
c ons t r a i n epdo w e r s y s te mo p ti mi z a ti on(S C O).
Since security and economy normally have conflicting
requirements, it is
inappfopriateto treat them separately.The fina.laim
of economy is the security
lunction of the utility company.The energy management
system (EMS) is to
operate the system at minimum cost, with the guaranteed
alleviation of
emergencyconditions.The emergencycondition will
dependon the severity of
t iolat io n so f o p e ra ti n gl i rn i ts(b ra n c hf ' l ow sand
bus vol tagel i mi ts).The most
severeviolationsresult fiom contingencies.
An irnportantpart of securitystudy,
therefbre,moves around the power system'sability
to withstanrjthe effects of
contingencies.A particular systemstateis said to be
secureonly with reference
to one or more specific contingency cases, and
a given set of quantities
monitoredfor violation. Most power systemsare operated
in such a way that
any singlecontingencywill not leaveother.o-pon"nts
heavily overloaded,so
that cascadingfailures are avoided.
www.muslimengineer.info
Sll
I
Most of the security related functions deal with static "snapshots" of the
power system.They have to be executedat intervalscompatiblewith the rate
of changeof systemstate.This quasi-staticapproachis, to a large extent,the
only practical approachat present,since dynamic analysisand optimization are
conslder4bly mole {!fficu!! 4nd cqmpurallo44lly 1aqtelime corrsulurg,
System security can be said to comprise of three major functions that are
carried out in an energycontrol centre: (i) systemmonitoring, (ii) contingency
analysis,and (iii) comectiveaction analysis.
System monitoring suppliesthe power systemoperatorsor dispatcherswith
pertinentup-to-dateinformation on the conditionsof the power system on real
time basisas load and generationchange.Telemetrysystemsrneasure,monitor
and transmit the data, voltages,currents,current flows and the statusof circuit
breakersand switchesin every substationin a transrnission
network. Further,
other critical and important information such as frequency, generator outputs
and transformertap positions can also be telemetered.Digital computers in a
control centre then processthe telemetereddata and place them in a data base
form and inform the operatorsin case of an overload or out of limit voltage.
Important data are also displayed on large size monitors. Alarms or warnings
may be given if required.
Stateestimation (Chapter 14) is normally used in such systemsto combine
telemetereddata to give the best estimate (in statisticalsense)of the curreltt
systemcondition or "state". Such systemsotten work with supervi$orycontrol
systemsto help operatorscontrol circuit breakersand operateswitches and taps
remotely. These systemstogetherare called SCADA (supervisorycontrol and
data acquisition)systelns.
The second ma-ior security function is contingency analysis. Modern
operationcomputershavecontingencyanalysisprogramsstoredin them.These
loreseepossiblcsystetntroubles(outages)beforethey occur.They study outage
events and alert the operators to any potential overloads or serious voltage
vi ol ati tl ns.For exalnple,t he sir nplestf ir r m of cont ingency
analysiscan be put
together with a standard LF program as studied in Chapter 6, along with
proceduresto set up the load flow dafa for each outageto be studied by the
LF plogram. This allows the system operatorsto locate def'ensiveoperating
stateswhere no single contingencyevent will generateoverloadsand/or voltage
violation:;.This analysis thus evolves operatingconstraintswhich may be
cntpi oyc din t hc liD ( ccot r olnicdispat ch)and UC ( unitcor nnr it r nclrpr
t ) ogr ar r r .
Thus contingencyanalysiscarricsout ornergcncyidentil'icationancl"what if''
simulations.
The third major security function, corrective action analysis, permits the
operatorto changethe operationof the power systemif a contingencyanalysis
program predicts a serious problem in the event of the occurrenceof a certain
outage.Thus this provides preventive and post-contingencycontrol. A simple
example of corrective action is the shifting of generationfrom one station to
another.This may result in change in power flows and causing a change in
loading on overloadedlines.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
j5! 2 ,1
I srr
Mo d e rnPo @ i s
Thresethreeftrnctionstogetherconsist oi a very compiex set of toois that heip
in the secureoperationol'a power system.
T3.2
Dyliacco [13] and further clarified by Fink and Carlsen l23l in order to define
relevant EMS (Energy ManagementSystem) functions. Stott et. al [15] have
also presenteda more practical static security level diagram (see Fig. 13.1) by
incorporating correctively secure(Level 2) andcorrectableemergency(Level4)
security levels.
In the Fig. 13.1, arrowed linep represent involuntary transitions between
Levels 1 to 5 due to contingencies..The removal of violations from Level 4
normally requiresEMS directed "corrective rescheduling" or "remedial action"
bringing the system to Level 3, from where it can return to either Level I or
2 by further EMS, directed "preventive rescheduling" depending upon the
desired operational security objectives.
Levels I and 2 representnormal power system operation.Level t has the
ideal security but is too conservativeand costly. The power system survives any
of the credible contingencieswithout relying on any post-contingency corrective
action. Level2 is more economical,but dependson post-contingencycorrective
rescheduling to alleviate violations without loss of load, within a specified
period of time. Post-contingencyoperating limits might be different from their
pre-contingencyvalues.
13.3
SECURITY ANALYSIS
=
-
L
pb.
6 q
EE
o,
c
o (!6
8E
L
q)
BE
o.
o r: .9
o
z
qF
- o =
gH€
o o
E o
o_
o
o.
o-
=E;.
o
E
o
o
o
o)
o
' = O -
o_o x
o--c.Y
^. ;EEt b- E
!ie
=98
abg
o
o
o
o
o
g,
c
c
o
(J
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.
System security can be broken down into two major functions that are carried
out in an operationscontrol centre: (i) security assessment,and (ii) security
control. The tormer gives the security level of the system operating state. The
latter determines the appropriate security constrained scheduling required to
optimally attain the target security level.
The security functions in an EMS can be executedin 'real time' and 'study'
modes. Real time application functions have a particular need for computing
speed and reliability.
'fhe
static securit.vlevel of a power systemis characterisedby the presence
or rrtherwise of emergency operating conditions (limit violations) in its actual
(pre-contingency) or potential (post-contingency)operating states. System
security assessmentis the process by which any such violations are detected.
o
o
oo
o
o -o o
o c
o
o o
3
o
SYSTEM STATE CLASSIFICATION
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System e.:^
s€sstlt€rrt involves two func tions:
(i) system monitoring and (ii) contingencyanalysis.Systemmonitoring provides
the operator of the power system with pertinent up-to-date information on the
current condition:;clf the power system. In its simplest form, this just detects
violations in the actual systemoperating state.Contingency analysisis much
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,.sfe ;l
Modern power Svstem Analvsis
Only a small proportion of work on optimal power flow (OPF) has taken into
account the security constraints.The most successfulapplications have beento
the security constrained MW dispatch OPF sub-problem. The contingencyconstrainedvoltageivar reschedulingproblem, as of the writing of this text, still
remains to be solved to a satisf
desree.
The total number of contingency constraintsimposed on SCO is enormous.
The SCO or contingency constrained OPF problem is solved with or without
first optimizing with respectto the base case(precontingency)constraints.The
general procedure adopted is as follows:
(i) Contingency analysis is carried out and cases with violations or near
violations are identified.
(ii) The SCO problem is solved.
(iii) The rescheduling in Step 1 might have created new violations, and
therefore step 1 should be repeatedtill no violations exist.
Hence, SCO representsa potentially massiveadditional computing effort.
An excellent comprehensiveoverview of various available methods is
presentedby Stott et. al [15].
There is still great potential for further improvement in power system
security control. Better problem formulations, theory, computer solution
methods and implementation techniquesare required.
T3.4
I_ srai
Power System Security
SLACKBUS
t +s*y ts
f+o*ys
1.02421-5.0" 3
1 . O 6 t O "1
--.>40.7+ j1.2 _ 39.5_i3.0:_
,l
,f88.e-y8.6
_
+ 1 8 . 9- y 5 . 2 _ 1 6 . 9+ 1 3 . 2 - <
-Ja.a--/e.8
+
i6.3-j2.3
--zt.s-i5.9
I z+.2*js.a
'L?:'-q''---'----/
t- az.s*
t
-*
-.i;n;l--+s
53.7-i7.2
'
*S4.9 +17.3
41.0236t-5.3"
1.0474t-2
'-"''
l+-u
5
i1.o17gt-6.2"
Y
*oo*ito
) , zo+1t o
!G-)
t40 +i30
Fig. 13.2
Base Case AC Lineflow for sample5 bus system
SLACKBUS
{+s*yts
f+o+7s
1 . 0 1 0 7 1 -5 . 9 " 3
1.0610" 1
->48.6 + j5.2 -46.7 - j5.3<- + 38.5-10.6
-38.4 -i1.1
-+ 3a?-/'er
+81.8-i 5.5
4 1.00682- 6.6"
:-
{ - r . o- 7 a. s
CONTINGENCYANALYSIS
In the past many widespreadblackouts have occurred in interconnectedpower
systems. Therefore, it is necessaryto ensure that power systems should be
operatec!mosf economic:r!lysuch that povrer is cle!i.rerecl
reliably. Reliable
operation implies that there is adequatepower generation and the same can be
transmittedreliably to the loads.Most power systemsare designedwith enough
redundancyso that they can withstand all rnajor tailure events.Here we shall
sttrdy thc possiblc consccprcnccs
:rnclrcrncdialactions rcquircd by two nrain
f ailur e ev e n ts ;l i n e .o rrta g easn c lg e n e ra ti ngrrni tfai l ures.
To explainthe problelnbriefly,we considerthe five-bussystemof Ref'erence
LI0J. The basecaseload llow resultslbr the exampleare given in t ig. 13.2and,
s lt r r wir f ' lo w< tf2 4 .7 MW a n d 3 .6 M VA R on the l i ne f}om bus 2 to hus 3. L.ct
t ls lls s t llll cth a t a t p rc s c n l w
, e i l rc o n l y i n tc restecl
i n the MW l oadi ngof the l i ne.
Let us examinewhat will happenif the line from bus 2 to bus 4 were to open*.
' l' lr c
r c s ul ti n gl i n c l ' l o w sa rtc vl o l ta g c su l c show ni n l ri g. 13.3.tt nray be notcd
that the flow on the line 2-3 has increasedto 31.5 MW and that most of the
other line flows are also changed. It may also be noted fhat bus voltage
magnitudesalso get aff'ected,particularly at bus 4, the change is almost2To less
from 1.0236to 1.0068pu. Supposethe line from bus 2 to bus 5 were to open.
Figure 13.4shows the resultingflows and voltages.Now the inaximum change
t:fgl.g.j
5 which is almost 107o less.
",jus
xSimulationof line outageis more complex than a generatoroutage,
since line
outageresultsin a changein systemconfigurations.
www.muslimengineer.info
i r . s- ; t . r
t- eo.o*
:63.1 + j1O.2
-<-- - 61.6 -i8.9
5
1.O114t-6.4"
1.0468t-2.
loo *lt o
rlr { 2s+ilo
t40+i30
\
lJ./
Fig. 13.3 PostoutageAC LoadFlow(Linebetween2 and4 is open)
SLACKBUS
I a o, 7 s
t a sr r t s
(c)
0.se35/- 6.s" o1
|
1 ' 0 6 t 0 "1 l ' 5 4 . 4+
i
1
i . 3- s 2 . 1 - i 9 . -7 | - s 1 . 4+ i 7 . 3 - s 1 . 1
_1__r
tl;f
Itl_qq.z_j1z.s
|
+80.e+13.21
| +s,s* 1tz.t
':: )ll
':t't'
- 52.s-i 13.0+
- 5
0.8994t-16.?
1 2 0+ i 1 0
'foo*yro
2 and5 is open)
Fig. 13.4 PostoutageAC Lcad Flow(Linebetween
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
|
-;'518
t-
rr,rodern
Po@is
5LAUK HUS
t 4 s+ l 1 5
e
l+o*1s
1 . o o 6 1_t s . 7 " 3 1
^ _
I
41.O043t-6.1"
+1t.s
4s.9_j7.6 <_i __22.4_ j2.6 _22 .3 +j0.7 <_
L--i-1-I
{ t.s - 1t.t
--- V_21.5_j4.8
|
1 . 0 6 t 0 "1'l_
l -!.6
{nze+iz1.7l
I
-25.2-j4.4 +
---t
I
i
'''"
1 - r z o .-g1 t s . t (l - ' - _ l
l
GiveAlarm signal
f-zs-7s.s
r
2..]--u+ _+ _53.6+16.8
1.o24st-3.7.
i
Y
l , z o+ 1 l o
<-
-52.5 -16.5
5
o 9956/_-7
.1"
{oo*iro
GiveAlarm signal
Fig. 13.5 PostoutageAC Load Flow(Generator
2 outage,lostgeneration
is
pickedup by generator1)
Figure 13.5is an exampleof generatoroutageanclis selectedto explain
the
tact that generatoroutagescan also result in changes in line flows and
bus
voltages.In the exampleshowr in Fig. 13.5 all the generationlost from
bus 2
is picked up on the generatorat bus 1. Had there been more thanZgenerarors
in the sample system say at bus 3 also, it was possible the loss of
leneration
on bus 2 is made up by an increase in generation at buses 1 and 3.
The
differencesin Iine flows anclbus voltagesvrould show how the lost gener.ation
is shared by the remaining units is quite significant.
It is important to know which line or unit outageswill render line flows or
voltagesto crossthe lirnjts. To find the eff'ectsof outages,contingency
analysis
techniques are empioyeci. Contingency analysis models single failure events
( i' e' one -l i n co u ta g c soiro l l c u n i t o u tu gcs)or nrul ti pl c
ccpri pnrcnt
fi ri l urccvc^ts
(failure of multiple unit or lines or their combination) one after
anotheruntil all
"credible outages"are considered.For each outage,all lines
anclvoltagesin the
netrvork are checkedagainsttheir respectivelimits. Figure 13.6 depictsa
flow
chart illustrating a simple method for carrying out a contingency analysis.
One of the important problems is the selection of "all credible outages,,.
Execution time to analyseseveralthousandoutagesis typically I min based
on
computerand analyticaltechnologyas of 2000.An erpproxirnate
rnodelsuclras
DC load flow may be used to achievespeedy solution if voltage is
aiso
required, then full AC load flow analysishas to be carried out.
'l
1;no*rifii.iuior".""-i;{offi
' ''l''''
l'r'.,o
_
_
l
--'t
A-..
,rr\=_
ves jciventarn
violation?
->*-t_-qJ
-T-
,*"
, N
'--o-i
I
Ail
I
--\
FIg. 13.6 A simpletechniquefor contingency
analysis
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
uodern PowerSystemAnatysis
620j;:f
I
^
a
?
frErrarE?rtrElt
IJ.C
DIIIYDIIIVII
Y .FAUI-L'I(s
A .securityanalysisprogramis run in a load dispatchcentre very quickly to help
the operators.This can be attempted by carrying out an approximate analysis
and using a computer system having multiple processorsor vector processors
y anarysts.I ne s
may
uate
an equrvalent
should be used for neighboursconnectedthrough tie-linos. We can eliminate all
non-violation casesand run complete exact program for "critical" casesonly.
This can be achievedby using techniquessuch as "contingency selection"or
"contingency screening", or "contingency ranking". Thus it will be easy to
warn the operation staff in advance to enable them to take corrective action if
one or more outageswiil result in seriousoverloadsor any violations. One of
the simplest ways to present a quick calculation of possible overloads is to
employ network (linear) sensitivity factors. Thesefactors give the approximate
change in line flows for changes in generationin the system and can be
calculated from the DC load flow. They are mainly of two types:
1. Generation shift factors
limits and those violating their limit can be informed to the operator for
necessarycontrol action.
The generation shift sensitivity factors are linear estimatesof the change in
line flow with a change in power at a bus. Thus, the effects of simultaneous
principle of superposition.
Let us assumethat the loss of the ith generatoris to be made up by governor
action on all generatorsof the interconnectedsystemand pick up in proportion
to their maximum MW ratings. Thus, the proportion of generationpick up from
unit k (k * i) would be
(13.7)
where
= maximum MW rating for rnth generator
Pn,,,,,u,^
2. Line outage distribution factors
g*i= proportionality factor for pick up on kth unit when ith unit fails.
Briefly we shall now describe the use of those factors without deriving them.
Reference [7] gives their deri iation.
The generation shift factorsl cr.,;are defined as:
(13.4)
where,
= Change in MW power flow on hne I when a changein generation, AP", takes place at the ith bus
4t
Here, it is assumedthat LPotis fully compensatedby an equal and opposite
change in generation at the slack (reference) bus, with all other generators
remaining fixed at their original power generations.The factor al,then gives the
sensitivityof the /th line flow to a changein generationat ith bus. Let us now
study the outage of a large generating unit and assume that all the lost
generation (Pod would be supplied by the slack bus generation. Then
APo, - -P1i
(13.s)
and the new power flow on each line could be calculatedusing a precalculated
set of " d' factors as given below.
ft = f i * dti APc,
where, ft
for all lines V /
j, = ff * 0r; APo, -
f i = power flow on /th line before the failure or precontingency
power flow
E,lau,
\ri LPotl
(13.8)
In Eq. (13.8) it is assumedthat no unit will violate its maximdm limit. For
unit limit violation, algorithm can easily be modified.
Similarly the line outage distribution factors can be used for checking if the
line overloadswhen solllc of the lines are lost.
The line outage distribution factor is defined as:
d,,,= *
(13.e)
Ji
where
dt,i = line outagedistribution factor when monitoring /th line atter an
outage of ith line.
Aft = change in MW flow on /th iine'
- precontingency line flow on ith line
fi
lf precontingency line flows on lines / and i, tlre power flow on line / with line
i out can be found out employing "d" factors.
(13.6)
- power flow on /th line after the failure of ith generator
www.muslimengineer.info
Now, for checking the /th line flow, we may write
?,= ff *d,,,f,o
(13.10)
Here,
fi
^d foi =precontingency or preoutageflows on lines / and i respectively
fr = power flow on /th line with ith line out.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
iii
J"srr
Po
Modern
l
a
'd' factors ali the lines for
Thus one can check quickiy by precaiculating
overloading for the outage of a particular line. This can be repeated fbr the
outage of each line one by one and overloads can be found out for corrective
action.
It may be noted that a line flow can be positive or negative. Hence we must
check / agarnst - Jt ^u* as well as h **. Llne tlows can be louncl out usmg
telemetry systemsor with stateestimation techniques.If the network undergoes
any significant structural change, the sensitivity factors must be updated.
Find the generationshift factors and the line outagedistribution factors for the
five-bus sample network discussedearlier.
Solution Table 13.1 gives the [x] matrix for the five bus sample system,
together with the generation shift distribution factors and the line outage
distribution factors are given in Tables I3.2 and 13.3 respectively.
Table 13.1 X Matrixfor Five-busSampleSystem(Bus 1 as a reference)
0
0
0.05057 0.03772 0.04029 0.4714
0
0.03772 0.08914 0.07886 0.05143
0
0.04029 0.07886 0.09514 0.05857
0.
0.04714 0.05143 0.05857 0.13095
Factorfor Five-busSystem
Table 13.2 Generation
Shift Distribution
Bus I
l=1(line1-2)
/=2(line1-3)
/=3(line2-3)
I=4(line2-4)
/=5(line2-5)
/=6(line3-4)
I=l (line4-5)
Bus 2
- 0.8428
- o.t572
o.0714
0.0571
0.0286
- 0.0857
- 0.0285
j=l
j=2
j=3
j=4
j=5
j=6
j = 7
(line 1-2)(linel-3)(Iine 2-3)(line2-4)(line2-5)(line3-4)(line4-5)
(l i ne l- 2)
1. 0001- 0. 3331- 0. 2685 - 0. 2094 0. 3735 0. 2091
0. 0
--0.4542
v
-= J3 \(line
0.4476
0.4545 0
L
-J)
2-3)
rr\rw
= 4 (hne 2-4) -0.3634 0.3636 0.4443 0.0
= 5 (line 2-5) -0.1819 0.1818 0.2222 0.2835
= 6 (line 3-4)
0.5451 -0.5451 -0.6662 0.7161
= 7 (line 4-5)
0.18i6 -0.1818 -0.2222-0.2835
0.3488 -0.6226 -0.3488
0.4418 0.6642 -O.44r8
0.0
0'3321 1.0
-0.5580
0.5580 0.0
1.0002 -0.3321 0'0
It has been found that if we calculate the line flows by the sensitivity
methods,they come out to be reasonablyclose to the valuescalculatedby the
full AC load flows. However, the calculations carried out by sensitivity
methods are faster than those made by full AC load flow methodsand therefore
are used for real time monitoring and control of power systems.However,
where reactive power flows are mainly required, a full AC load flow method
(NR/FDLF) is preferred for contingency analysis.
The simplest AC security analysisproceduremerely needsto run an AC load
flow analysis for each possible unit, line and transformer outage.One normally
does ranking or shortlisting of most likely bad caseswhich are likely to result
in an overload or voltage limit violation and other casesneed not be analysed.
Any good P1(performanceindex can be selected)is usedfor rankirig. One such
P/ is
(13.11)
For large n, PI will be a small numberif all line flows are within limit, and
will be large if one or more lines are overloaded.
For rr = I exact calculationsciur be done fbr P1. P1 tablecan be orderedfrom
largest value to least. Suitable number of candidatesthen can be chosen for
further analysis[7].
If voltagesare to be included, then the following PI can be employed.
(13.12)
Here, Alvil is the difference between the voltage magnitude as obtained at the
end of the lPlQ FDLF algorithm Alvl-u* it the value fixed by the utility.
-ue
Largest vaiue oi Pi is piaceciat the top. The security arraiysistrray rrow
startedfor the desired numbel of casesdown the ranking list.
$ummary and Further Reading:
Reference [25] has discussedthe concept for screening contingencies. Such
contingency selection/screeningtechniquesfonn the foundationfor many realtime computer security analysis algorithms.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
l#GrffiE
IE!F}ILfr
Reference [15] gives a broad overview of security
assessmentand contain an
bibliography covering the literature on"security
assessmenrup to
;;;;t]r",
Reference [11] gives an excellent bibliography
on voltage stability. This
topic is discussedbriefly in the next section.
13.6
POWER SYSTEM VOITAGE
STABILITY
Power transmissiol tuglgility has traditionally
been Iimited by either rotor
angle (synchronous) stability or by thermal loaoing
capabilities. The blackout
problem has been linked with transient stability.
L-uckily this ;roblem is now
not that serious because of fast short circuii
clearing; po*Lrru excitation
systems, and other special stability controls.
ElectriJ *rnpuni"s are now
required to squeezethe maximum possible power
through
networks
owing to various constraintsin the construction
"ii.G
of generation
and transmission
facilities.
voltage (load) stability, however, is now a
main issue in planning and
operating electric power systemsand is a factor
reading to limit po-w; transfers.
voltage stability is concerned with the ability
of a power system to maintain
acceptable voltages at all busesin the system under
normal conditions and after
being subjected to a disturbance.A power system
is said to have entereda state
of voltage instability when a disturbante
resurts in a pro!."rriu"
and
uncontrollable decline in voltage
Inadequate reactive.-powersupport from generators
and transmission lines
leads to voltage.instability ot uoitug" collapse,
which have resulted in several
major system failures in the world. Th"y -",
(i) south Florida, usA, system disturbanc
e of 17 May r9g5, (transient, 4
sec)
(ii) French systemdisturbancesof Decembe
r 19, r97g and January 12, rgg7,
(longer term).
(iii) swedish systemdisturbanceof December
27, rgg3 (longer term, 55 sec)
(iv) Japanese(Tokyo) system disturbance
of July 23, 1gg7irorg", term, z0
min)
(v) NREB grid disturbancein India in
19g4 and r9g7.
(vi) Belgium, Aug 4, 1992. (longer term,
4.5 min)
(vii) Baltimore, washington DC, usA,
5th July 1990 (longer rerm, insecure
for hours)
Hence, a full understandingof voltage stability
phenomenaand designing
mitigation schemesto prevent voltase instabilitv is nf o,raqr'or,,o+^,.+:r:a:^^
', ;,ilir; ;;,:;;
ffiri#il;";
;agestability.
lf phenomena.Becadseof this, voltage
'ent engineers.
Voltage instabilitv and
rterchangeably.by many ."r"ur"h..r.
voltage instability or co[apse is a faster dynamic-process.
As opposedto angle
www.muslimengineer.info
stability, the dynamics mainly involves the loads and the means for voltage
control. Ref [11] provides a comprehensivelist of books, reports, workshops
and technical papers related to voltage stability and security.
Definitions:
[2]
A power system at a given operating state is small-disturbance voltage stable
if, following any small disturbance, voltages near loads are identical or close
to the pre-disturbancevalues. The concept of small-disturbance voltage stability
is related to steady-statestability (Chapter 12) and can be analysedusing smallsignal (linearised) model of the system.
A power systemat a given operating stateand subject to a given disturbance
ts voltage stable if voltages near loads approach post-disturbanceequilibrium
valuei. The concept of voltage stability is related to the ffansient stability of a
power system. The analysis of voltage stability normally requires simulation of
the system modelled by non-linear diffdrential-algebraic equations.
A power systemat a given operating stateand subject to a given disturbance
undergoes voltage collapse if post-disturbanceequilibrium voltages are below
acceptable limits. Voltage collapse may be total (blackout) or partial. The
voltage instability and collapse may occur in a time frame of a second.In this
case the term transient voltage stability is used. Sometimes it may take up to
tens of minutes in which case the term long-term voltage stability is used.
The term voltage security means the ability of a system, not only to operate
stably, but also to remain stable following any reasonably crediblebontingency
or adverse system change such as load increases[2].
Voltage stability involves dynamics, but load flow based static analysis
methods are generally used for quick and approximate analysis.
Figure 13.7 depicts how voltage stability can be classified into transient and
long-term time frame l2l.
Transientvoltage stability
Longer-term voltage stability
Inductionmotor dynamics
Increase in load/powertransfer
LTCtransf& Distvolt.Reg.
Generator/excitationdvnamics
Primemovercontrol
Mech. switched capacitors/reactors
Load diversity /thermostat
Excitationlimiting
Gasturbinestart-up
Under voltageload shedding
Protective relaying includingoverload protection
100
Time-seconds-
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Fiq. 13.7 Voltage stabilitVphenomenaand time responses
SPfti*f
uooern po@is
I
Voltage stability problems normaliy occur
in heavily stressed systems.
voltage stability and rotor angle (or synchronous)
stability are more or less
interlinked' Rotor angle stability,
as voltage stability is affected by
.a3-.wex
reactive power control. voltage stability
is concern"o *itt, load areas and load
wffi;Ewe
desirable.
Unity powerfactor
;;;''',h.;A#";
rvr^rvlv vv'Yvr Pr'lurs tu a large system
over long transmission lines. Voltage
':::i:y,"::,*:,','31^
asvanous
svstem
parameters
and
""lT:ir'#n#'f
|HI'I,:IJHI#
voltage exist for each value of load.
The uppe. on" indicates stable ;;i;;;
whereas lower one is the unacceptable
value (multiple load flow). At limiting
of voltage stability i.e. at nose point single
l!ug"
ioad flow solution exists.
Nearer the nose point, lesser is the staLility
maigin,
Effective counter
Instability
Nosepoint(knee)
and ,oto."ungl"stabilityis basicauy
^,?::^s,112itiu
Seneratorstability.In a
large inter-connectedsystem,voltagecoilapse
of a load
area is possible without lois of synchroni*
or any generators.
The slower forms of voltage instability
are often analysed as steady-state
problems' 'snapshots' in time following
an outage or during load buildup are
simulated' In addition to post-disturbancl
load floivs, two other load flow based
methods are widely used: P-V curves and
e-v curves. pV curves are used for
especially for radial systems.
/ curves (Fig. 13.g),
eV curves (Fig.
'ig. 13.8) and methods
to quantify nose
puted. Power flow analysis cletermines
Measures to prevent
or contain
voltage
(i) Generator terminal voltage shourd
be raised.
(ii) Generator transformer tap value
may be increased.
(iii) Q-injection should be carried out
at an appropriate location.
(iv) Load-end oLTc (on-load tap
changer) should be suitably used.
(v) For under voltage conditions, strategic
load shedding should be resorted
to'
System reinforcement may be carried
out by instpiling new transmission
lines between generation and load centres.
series and shunt compensationmay
be carried out and svcs (static var compensation)
may be installed. Generation
rescheduling and s_tarting-upof gas ,rrui-"*
,.y i" .,*.i.,t out.
Practical aspectsof Q-flow problemsleading
to voltage collapsein EHv lines:
(i) For lons line.s urifh 'r-^^nrrnu^r L--^^uus's, recelvlng_end or road voltages
"-!' s'vv'Lr.,'e'
increase for light load conditions and
decrease for heavy load conditions.
(ii) For radial transmission lines,
if any loss of a line takes place, reactance
goes up, I2x loss increases resulting
in increase in voltage drop. This
should be suitably compensated
by local e injection.of course this
involvescost.If thereis a shortageof rocal q
,ourrr, thenimport of e
www.muslimengineer.info
t
\
Loaw of V6 and Pr"r;
same for const Z load
for other types of load,
V degradationis faster.
PplP6zy
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
Fig. 13.8 PV curveswith differentload powerfactors
Only the operating points above the critical points represent satisfactory
operating conditions. At the 'knee' of the V-P curve, the voltage drops rapidly
with an increasein load demand. Power-flow solution fails to convefgebeyond
this limit indicating instability. Operation at or near thel stability limit is
impractical and a satisfactory operating condition is ensured by permitting
sufficient "power margin".
1400
j+Allowable
range
1200
1000
P> P, > P,, iNMW
ffuo. Boo
Systemcharacteristics
Capacitor
characteristics
0 . 9 0 . 9 51 . 0 1 . 0 5
Vin pu
Fig. 13.9 Systemand shunt capacitorsteady-stateQ-V characteristics,
capacitorMVAr shown at rated voltage
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
M
Voltage
Collapse
;
Voltage collapse is the processby which the seguenceof events accompanying
voltage instability le_adsto unacceptable voltag6 profile in a significant part
of
the power system. It may be manifested in ieverat different"ways.
Voltage
collapse may be characterised as follows:
(v) Post-disturbanceMwA4vAR margins should be ffanslated to predisturbanceoperatinglimits that operatorscan monitor.
(vi) Training in voltage stability basis (a training simulator) for control centre
and power plant operators should be i
SUIvIMARY
(iii) The voltage collapse generallymanifests itself as a
slow decay of voltage.
It is the result of an accumulative process involving the actions
and
interactions of many devices, controlJ, and protective iyrt"*r. The
time
frame of collapse in such caseswould be of the order of several minutes.
Voltage collapse is strongly influenced by system conditions
and
characteristics.
(iv) Reactivecompensationcan be made most effective by the judicious
choice
of a mixture of shunt capacitors, static var system and possibly
synetuonouscondensers.
Methods
of Improvlng
Voltage
(i\
\-./
Trends and Challenges
Ontirnql
oifi-urrrrr6
NCES
REFERE
Stabllity
voltage stability can be improved by adopting the following rneans:
(i) Enhancing the localisedreactivepower support (SVC)
is more efl,ective
and C-banks are lnore economical. ra-tS
devices or synchronous
condenser may also be used,
(ii) Compensating the line length reduces net reactance and power
flow
increases.
(iii) Additional transmissionlinc muy be crectccl.It also
improves reliability.
(iv) Enhancing excitation of generator, system voltage improves
and e is
supplied to the system.
(v) HVDC tie may be used between regional grids.
(vi) By resorting to strategic load shedding, voltage goes
up as the reactive
burden is reduced.
.
Future
Power system security (including voltage stability) is likey to challenge
planneqs,analysts; researchersand operators for the foreseeablefuture. As load
grows, and as new transmission lines and new generations would be
increasingly difficult to build or add, more and more utilities will face the
security challenge.
Deregulation and socio-economic ffends compounded by technological
developmentshave increasedthe likelihood of voltage instability.
Luckily many creative persons are working tirelessly to find new methods
and innovative solutions to meet this challenge.
n'F
EA/-'n-r i, ,{^*,:^^^
vr
I nv
\rlvv.ltvgD.
(ii) Betterand probabilisticloadmodelling.
(iii) Developtechniquesandmodelsfor studyof non-lineardynamics
of largr
siz.esystems.For example,new methodsto obtainnetworkequivalentr
suitablefor voltagestabilityanalysis.
www.muslimengineer.info
Books
L l.J. Narguthand D.P. Kothari, PttnterSystemEngincering,'fataMc0raw.Hill, New
Dclhi, 1994.
2, C.W, Taylor, Power SystemVoltag,eStabiliry,McGraw-Hill, New york, 1994.
3. P. Kundur, Power SystemStability and Contol, Sections2.12, ll.2 and Chapter
14, McGraw-Hill, New York, 1994.
4, T,J.E,Miller, Editor, ReactivePower Control in ElectricSyslens,John Wiley and
Sons, New York, 1982.
5. A. chakrabarti, D.P. Kothari and A.K. Mukhopadhyay, Perforurutnce,operation
and Control of EHV Power Transmission Systems,Wheeler Publishing, New
Delhi. 1995.
6. T.V. Cutsemand C. Vournas, VoltageStability of Electric Power Syslerns,Kluwer
Academic Publishers,London, 1998.
7. A.J. Wood and W.F. Wollenberg, Power Generation,Operation, and Control, Znd
Edn, John Wiley, New York, 1996.
E. John J. Gratngerand W.D. Stevenson,Power SystemAnalysis,McGraw-Hill, New
York, 1994.
9. G.L. Kusic, Computer-AidedPower SystemsAnalysis,Prentice-Hall,New Jersey,
1986.
10. G.W. Stagg and A.H. El-Abiad, Computer Methods in Power System Analysis,
McGraw-Hill, New York, 1968.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Papers
t4
11. v. Ajjarapu and B. Lee, "Bibliography on voltage stability", IEEE Trans. on
Power Systems,Vol. 13, No. 1, February 1998, pp lI5-125,
12. L.D. Arya, "Security ConstrainedPower System Optimization", PhD thesis, IIT
Delhi, 1990.
13. T.E. Dyliacco, "The Adaptive Reliability Control System", IEEE Trans. on pAS,
Vol. PAS-86, May 1967, pp 517-531
(This is a key paper on system security and energy control system)
14. A.A. Fouad, "Dynamic Security AssessmentPractices in North America", IEEE
Trans. on Power Systems,Vol. 3, No. 3, 1988, pp 1310-1321.
15. B. Stott, O. Alsac and A.J. Monticelli, "security Analysis and Optimization", proc
IEEE, VoL 75, No. 12, Dec. 1987, pp 1623-1644.
16. Special issue of Proc. IEEE, February 2000.
17. P.R. Bijwe, D.P. Kothari and L.D. Arya, "Alleviation of Line Overloads and
voltage violations by corrective Rescheduling", IEE proc. c, vol. 140, No. 4,
July 1993, pp 249-255.
18. P.R. Bijwe, D.P. Kothari and L.D. Arya, "Overload Ranking of Line Outageswith
postourage generation rescheduling", Int. J. of Electric Machines and Power
Systems,Yol. 22, No. 5, 1994, pp 557-568.
19. L.D. Arya, D.P. Kothari et al, "Post Contingency Line Switching for Overload
Alleviation or Rotation", Int J. of EMPg Vol 23. No. 3, 1995, pp 345-352.
20. P.R. Bijwe, S.M. Kelapure, D.P. Kothari and K.K. Saxena,"Oscillatory Stability
Lirnit Enhancement by Adaptive Control Rescheduliig, Int. J. of Electric Power
and Energy Systems,Vol. 21, No. 7, 1999, pp 507-514.
21. L.D. Arya, S.C. Chaube and D.P. Kothari, "Line switching for Alleviating
Overloadsunder Line OutageCondition Taking Bus Voltage Limits into Account",
Int. J. of EPES, Yol. 22, No. 3, 2000, pp 213-ZZl.
22. P.R. Bijwe, D.P. Kothari and S. Kelapure, "An Effective Approach to Voltage
Security and Enhancement",'Int. J. of EPES, Yol. 22, No 7, 2000, pp 4g3-4g6.
23. L. Fink and K. carlsen, "operating under sttress and Strain", IEEE spectrum,
March 1978, pp. 48-50.
24. S.M. Kelapure, "Voltage Security Analysis and Enhancement", Ph.D. thesis, IIT
Delhi,2000.
25. G.C. Ejebe, et. al, "Fast Contingency Screening and Evaluation for Voltage
security Analysis", IEEE Trans. on Power systems,vol. 3, No. 4, Nov. l9gg, pp
1582-1590.
26. T. Van Cutsen, Voltage Instability: "Phenomena,Counter measures,and Analysis
Methods", Proc. IEEE, Vol. 88, No. 2, Feb. 2000, pp 208-227.
www.muslimengineer.info
T4.I
INTRODUCTION
State estimation plays a very important role in the monitoring and control of
modern power systems. As in case of load flow analysis, the aim of state
estimation is to obtain the best possible values of the bus voltage magnitudes
and angles by processing the available network data. Two modifications are,
however, introduced now in order to achieve a higher degreeof accuracy of the
solution at the cost of some additional computations. First, it is recognised that
the numerical values of the data to be processed for the state estimation are
generally noisy due to the errors present. Second, it is noted that there are a
larger number of variables in the system (e.9. P, Q line flows) which can be
measured but arc not utilised in the load flow analysis. Thus, the process
involves imperfect measurements that are redundant and the process of
estimating the system statesis based on a statistical criterion that estimates the
true value of the state variables to minimize or maximize the selected criterion.
A well known and commonly used criterion is that of minimizing the sum of the
squares of the differences between the estimated and "u1le" (i.e. measured)
values of a function.
Most state estimation programs in practical use are formulated as
overdetermined systems of non-linear equations and solved as weighted leastsquares(WLS) problems.
State estimatorsmay be both static and dynamic. Both have been developed
for power systems.This chapter will introduce the basic principles of a staticstate estimator.
ln a power system, the state variables are the voltage magnrtudes and phase
angles at the buses. The inputs to an estimator are imperfect (noisy) power
system measurements.The estimator is designedto give the "best estimate" of
the system voltage and phase angles keeping in mind that there are errors in the
measuredquantities and that there may be redundant measurements.The output
data are then used at the energy control centres for carrying out several
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
An lntroductionto State Estimationof Power Systems
or,01-li,r:,systeT^studics
suchas cconornicclisplrch(Chaprcr7),
::if.-jl:t
strLurl-y analysts (Lnapter
r4-2 LEAST souARES ESTIMATION:THE BAsrc
soluTroN
I7l- IeI
A
From Eqs. (14.31) and (14.4),one gets the following expressionfor the index:
( r4.6)
J =yt!- y'H*.- i<'H'y+ *.'HtH*.
For minimizing J = f$), we must satisfy the tollowing condition.
lrr
ns wilf be seen iater in Section r4.3, the probrem
of power system state
estirnation is a special case of the more general
problem of estimation of a
random vector x from the numerical values of
anotirerrelated random vector y
with relatively little statistical information being available
for both x and y. In
such cases,the method of least-squared-errorestimation
may be utilised with
good results and has accordingly been widely
employed.
Assume that x is a vector of n random variabres
x1tx2, ..., x' that y is
another vector of m (> n) random variables
!1, J2, ..., J^ and both are related
as
!=Hx+r
(r4.1)
where H is a known matrix of dimensionmx
n anclr is a zero mean rancklm
variable of the same din-rensionas y. The vector
x representsthe variablesto
be estimated,while the vector y representsthe
variables whose numerical
v aluesa re a v a i l a b l eE. q u a ti o n( l 4 .l ) suggests
thatthe measurement
vectory i s
linearly related to the unknown vector x and
in addition is corrupted by the
vector r (error vector).
The problem is basicallyto obtain thc best possible
value of the vector x
from the given valuesof the vector y. Since the
variable r is assumedto be zero
m ean,o n e m a y ta k e th e e x p e c ta ti o n
o f E q. (14.1)anclget the rel ati on
f = Hx
(r4.2)
whcre
(/\ r|1+/. J4,(, .\ 5 )
J =7'V
lJ).
gradoJ = 0
It is easyto check (see,e.g. [1]) that Eq. (1a.7)leadsto the following result.
j' = Hf,
the 'notmal equation' and may be solved explicitly
This equation is called
for the LSE of the vector i as
*. = (H,Il)-t
E;il;il]
r
.
ln orclerto illustrate the methodof LSE, let us considerthe simple problem of
estimating two random variables x, and .rz by using the data for a thret:
dimensionalvector y.
0.1
l-'
H=10 ll
Assume
Lrrl
The matrix Ht F1 is then given bY
I
is
u'u =12 I unaits inverse
Ll 2)
-:':1
=f ''':'
(H'm-'
2131
u3
l-
It is easy to form the vector Hty and combining this with the itrverse of
(H'H), the following estimateof x is obtained.
^
(r4.4)
www.muslimengineer.info
f
rrt
lat
-.
/1
t1\/
r
-l
\Ltr)/r-\rrt)\/2-Yz)
|
* =|
/
3
)
y
r
*
(
2
/
3
)
y
r
1
(
r
/
3
)
l v-,
L-(t
Weighted
mI
lnq
The estimate i is defineclto irc thc I-sE if it is
cornputcclby rninirnizingthe
estimation index J given by
(14.e)
Ht y
i
(r4.3)
representsthe estimateof the vector y. The effor
! of the estimation of y is then
given bv
(14.8)
HtH*,-H'y-O
I , , = cxpcctcclvalue ol' x iurd y, respec:tively.
This shows that the load flow methods of chapter 6
could be used to estimate
the mean valuesof the bus voltages.Howeu"r, tn.
woulcl like to estimatethe
actual values of bus voltagesrather than their averages.
One possibleway of obtainingthe bestpossibleestimate
of the vectorx from
y lies in the use of the method of leastsquareestimation
(LSE). To developthis
rneth<td,assllmethat i rcpresetttsthe desireclestirnate
of ,r so that y given by
the equation
\r+.t)
..
LSE
grsllll|:.iltr
Eilv€il
!rl D9.
\!+,2)
ls ulrerr
lr-rsrrsu
r\, (ri) rrru
\rr\.rrrr(uJ
r1,61ir
squaresestimate and is obtainedby minimising the index function that puts
equal weightage to the effors of estimation of all componentsof the vector y.
It is often desirableto put.dift'erentweightageson the ditt'erentcomponentsof
y since some of the measurementsmay be more reliable and accuratethan the
othersand these should be given more importance.T'o achievethis we define
the estimation index as
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
rvtouern Power
I = l'wf
An lntroduction
to StateEstimation
of PowerSystems-T--l" ,3iiSl:
/ 1 4 1 n \
\ r.i. r vi,
where W is a real symmetric weighting matrix of dimension m x ru. This
is
often chosen as a diagonal matrix for simplicity.
It is relatively straightforward to extend the method of LSE to the weighted
form of "I and to derive the foilowing form of the normal equation.
H|WH?_ H'Wy _0
[o.r
w =t l
i=ky
k = (H'WH)-' Ht W.
where
r
Or
and
E{i} =E{xl
(14.r3)
P, = KRK
(14.15a)
(14.1sb)
where R is the covarianeeo1 the error vector r. Note that the covariancep"
is
a lneasureof the accuracy of the estirnaiion and a smaiier trace
of this matrix
indicatesa better estimate.Eq. (14.15b) suggeststhat the best possible
choice
of the weighting matrix is to set w - R-1.The optimum value
of the error
covariance'matrix is then given by
P, - (HtR-rH)-l
H ' | w = f o 'ot o t - l
L0
(14 15c)
**
t---*-1Et<ample14;2
^
www.muslimengineer.info
(lr/21) y,
*=[
L- ellr) y,
- (r}/Zt) y,
(t0/2t) yr
(2ot2r)
y,
(r/2\ y, )
I
If this result is compared with the result in Example I4.1, the effect of
introducing the weighting on the estimate is apparent.Note that the choice of
W in this case suggeststhe data for y2 is consideredmore valuable and this
resultsin the components of x being more heavily dependenton
)2.
The matrix ft is in this case found to be (Eq. (1a.12b))
rc=lrr/21
L- rlzt
-ro/2r ro/zrf
20/21 r/21J
If the covariance of the measurementerror is assumedto be R = L the
covarianceof the estimation error is obtained as (Ref. Eq. (14.15c))
*:_
P,= (1tr47)
|L - 6 7 .::1
r34
)
The choice of W above yields unacceptablylarge estimation error variances.
Let us now choose the weighting matrix W = I. The matrix Kis then obtained
AS
-1/3t/3f
*=l''3
L- U3 2/3 1/3J
The error covariance matrix is then given by
6
-3]
rD- x = \/ rL/ tn>\ ) [
L_ E 6 J
The error variances are now seen to be much smaller as is to be expected.
Non-linear
Assumethat in the Example I4.1, we want to obtain the WLSE of the variable
x by choosing the following weighting matnx
1 0.1j
Theweightedleastsquares
estimateof the vectorx is thenobtainedas(from
E q .( l a . 1 1 b ) )
(14.r4)
ln IJq. (14.14) it is assumedthat the error r is statistically independent
of
columns of H and the vector r has a zero mean. An estimate that
satisfied
Eq. (l 4)4) is calledan unbiasedestimate.This implies that the
estimationerror
ls zero on an averase.
x =kr
The covariance of the error of estimation is therefore given by
10.1 1.lJ
and the matrix HtW is obtained as
(r4.rzb)
=f::;,7, ,,,wH)x + kr
X =r + kr
o'tl
H,wH-l'''
Here the matrix k dependson the value of H and the choice of w.
using Eqs. (14.1) and (r4.lzb) it is easy to get the relation as follows.
l
I
o.tj
L
G4.Ila)
(I4.I2a).
I
The matrir HLWH ts
This leads to the desired weighted least squares estimate (WLSE)
*. = (H,W H)-t H,W y
(14.11b)
This pertains to minimization as the hessian 2H|WH is a non-negative
definite.
Some Properties:
Rewriting Eq. (14.ilb) as
I
Measurements
The case of special interest to the power system state estimation problem
corresponds
to the non-linear measurementmodel.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
536 |
Modern Power SystemAnalysis
An lrylfgduction
to_StateEstimation
of PowerSystems
(14.16)
!=h(x)+ r
where h(x) representsan rz dimensional vector of nonlinear functions of the
variablex. It is assumedthat the componentsof the vector h(x) arecontinuous
in their arguments and therefore may be differentiated r.vith respect to the
componentsof x. The problem is to extendthe methodof least squaresin order
to estimate the vector x from the data for the vector y with these two variables
being relatedthrough Eq. (14.16).
To mimic our treatment of the linear measurementcase, assumethat i
representsthe desired estimate so that the estimateol the measLlrementy could
be obtained using the relation
useful result in the sensethat it shows a mechanismfor improving on the initial
estimafeby making use of the available nleasurements.
Having obtainednew
I
I
I = h(i)
(1.4.17a)
This yields the error of estimationof the vectol y
j =y_ h(3)
(r4.r7b)
In ordcr to obtain the WLSE of ,,r,we nrust cho<lsethe index of estimation
J as follows.
J=U-
h ( f i ) l ' W l J- h ( i ) l
(14.18)
The necessarycondition for the index ,I to have a minimum at .x,is given by
Eq. (14.1e).
ty-h(x)lH(i)=o
I
estimatei , the processof linearisationis repeatedas many times as desiredand
this leads to the {qllq*tng 4erylle {qryq qf the qqb]ro4 of thq 4on-linear
estimationproblem.
i ( t + 1 ) = i ( t )+ K ( t ) { y - h t t ( / ) l }
(r4.23)
where the matrix K(D is deflned as
K(t) - LH/ wHil-' al w
(r4.24)
The index / representsthe iteration number and H lrepresentsthe value of the
Jacobian evaluated at x = i (l). Usualiy the iterative process is terminated
wheneverthe norm of the differenceof two successivevalues of the estimate
i 1t + l) - .i (/) reachesa pre-selecredrhresholdlevel.
A flowchart for implementing the iterative algorithm is shown in Fig. 14.7.
A major sourceof computationin the algorithmlies in the need to updatethe
Jacobianat every stageof iteration.As discussedearlier in Chapter6 (seeEqs.
(6.86) and (6.87)) it is often possibleto reducethe computationsby holding rhe
value of H a constant,possibly after / exceeds2 or 3. T,ris is in general,
permissible in view of the fact that the change in estimate tends to be rather
small after a couple of iterations.
(r4.re)
where H (fr) is the Jacobianof h (x) evaluatedat i. In general this non-linear
equation can not be solved for the desired estimate, i. n way out of this
difficulty is to make use of the linearisa-tiontechnique. Let us assumethat an
a priori estimate xu of the vector x is available (say from the load flow
solution).
Using Taylor seriesapproximation,we get
! = h ( x d + H u & - x 1 ) +r
(14.20)
where 1/o standsfor the Jacobian evaluated at x = x9 and the noise term r is
now assurnedto include the effects of the higher order terms in the'Taylor
series.Equation (14.20)can be rewrittenas:
(14.2I)
A ) ' = y - h ( x o ) =H s A x + r
where Ay is the p^-rturbedmeasurementand Ax is the perturbed value of the
vector r. An \\LSE of .r is then ea^silvobtaineda-sdiscussedearlier and this
leads to the desired expressionfor the lineaized solurion of the non-linear
problem.
estjmat.ion
t
.i. = .rrr- [H,,' I+'HQ]- H()' ]l' {-r - h t.t,,,)}
r11.22)
I r is no r l i k e l v th a t th e e s ti m a te.i o b t ai nedfrom E q. (11.22r i s goi ngto be
t - r f I r t u c i r u s e s i l t . ' c . i r t t e n e r . r i . t l t r ' . r i , t ' r t t t ' t c ' s t i r l u t e . l ' , . rI l ) J \ - I l r r t h ' c l o s e t i r t h f
(Dt-irnalydue rf rhe \ect()r x. Hrt,*'ever,F,q.(14.22) prot'ides us wirh a very
www.muslimengineer.info
I S3T
i ,00"," ) eq.(+..227
I
/
c"l::'1:
-,=n) \,* ! -_I\,,ll
./
:-
No
\-
-''!'
ls
.-2a.,
r Yes
\
Stcp
rig. r+.1
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
',:83--E-,
I
| 539
_r-
todern powerSystemAnalvsis
An tntroductionto State Estimationof Power Systems
Considerthe simple caseo1'ascalar variablex and assunlethat the relationship
It is thus apparentthat the problem of estimation of the power system state
'is a non-linear problennand may be solved using either the batchpr:ocessingor
sequentialprocessingformula [see Section 3.3 of Reference 1]. Also, if the
condition. the voltageangles
svstem is assumedto have reacheda steady-state
I
_xls glven Dy
Y = x 3+ r
The JacobianH, is easily obtained in this caseand the iterative algorithm takes
the explicit form
i (t + 1)= i (t) + t3i (Dl-z {y - ti (Dl3}
where we have used W = 1.
Let the correct value of x/ be eclualto 2 and assumethat due to the effect
of r the measuredvalue of y is found to be 8.5. Also, assumethat the initial
estimatex (0) is taken to be equal to 1. The table below gives the results of the
first few iterations.
, (t)
0
I
2
_)
a
1.0
3.5
2.56
2.t6
It is apparentthat the algorithmwould yield the correctsolutionafter several
iterations.
14.3 STATIC STATE ESTIMATION OF POWER SYSTEMS
lrol-lr2l
As noted earlier, for a system with N buses,the state vector x may be defined
as t hc 2N - | v c c l tl rtl l ' l l tc N -- | v o l l agcangl cs62,..., 6" and thc N vol tage
magnitudes/1, v2, ..., v". The load flow data,dependingon type of bus, are
generally comrpted by noise and the problem is that of processing an adequate
set of available data in order to estimate the state vector. The readily available
data may not provide enough redundancy (the large geographical area over
which the system is spreadoften prohibits the telemeteringof all the available
tnedsurementsto the central computing station).The redundancyfactor, defined
as the ratio m./n should have a value in the range 1.5 to 2.8 in order that the
computedvalue of the statemay have the desiredaccuracy.It may be necessary
to irleh-rdethe data for the power flows in both the directions of some of the tie
lines in order to increase the redundancy factor. In fact, some 'psuedo
measurements'which representthe computedvalues of such quantitiesas the
active andreactiveinjectionsat some remotebusesmay alsobe includerlin the
vector y (k).
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problem is then a static problem and the methodsof Sec. 14.2 may be used. if
so desired. To develop explicit solutions, it is necessaryto start by noting the
exact forms of the model equations for the components of the vector y (k).
Let P, arfi Qi denote the active and reactive power injections of ith bus.
These are related to the components of the state vector through the following
equations.
N
P , =D
I Yj | | Vjl lYij I cos (- 6t + 6, *
4ti)
(r4.2s)
j --r
N
e,= -D
| % l I v j l l Y i j l s i n( - 6 , + 5 t + 0 ; )
(r4.26)
j:1
where I IU I repre5entsthe uragnitudeand l/U- representsthe angle of the
admittance of the line connecting the lth and7th buses.The active and reactive
componentsof the power flow from the ith to the7th bus, on the other hand are
given by the tbllowing relations.
P,j= | vil I Vjl I Yij I cos (d, - 6i * 0,) - l v i P l Y , , l c o s9 , ,
(r4.27)
Q , j = l % l I V j l l Y i j l s i n ( d ' 6 i * 0 t ) I viP I Y,,I sin Iij
(14.28)
Let us assume that fhe vector y has the general form
J = lPt . . . PN Q t . . . Q r u Pr z . . . Pu - l , N
)2
Qn
. . . Q 7 , J- 1 , 7 s ,
b i nl vl 1 l , . . . . ,I y N I l /
(r4.2e)
The Jacobian H will theh have the form
H_
Ht
Hz
H3
H4
Hs
H6
H7
Hs
(14.30)
/ru-r o
o
IN
(N - l; submatrix
where /" is the iclentity matrix of dimension N, H, is the N x
wrt
6's, H2 is the
power
injections
of the partial derivatives of the active
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
54q f
ModernPo
N x N sub-matrix of the partial-derivatives of the active power injections wrt
| 7l' and so on. Jacobian H will also be a sparsematrix since I is a sparse
matrix.
Two special casesof interest are those correspondingto the use of only the
active and reactive iniections and the use of onl
flows in the vector y. In the first case, there are a total of 2N components of
y compared to the 2N - 1 components of the state x. There is thus almost no
redundancy of measurements.However, this case is very close to the case of
load flow analysis and therefore provides a good measure of the relative
strengthsof the methods of load flow and state estimation.In the second case,
it is possible to ensurea good enough redundancy if there are enough tie lines
in the system. One can obtain two measurementsusing two separatemeters at
the two ends of a single tie-line. Since these two data should have equal
magnitudes but opposite signs, this arrangement also provides with a ready
check of meter malfunctioning.There are other advantagesof this arrangement
as will be discussedlater.
The Injections
I 541'
I
value
specified
any
at
the
Jacobian
determine
to
used
(14.33)
may
be
Equation
is
then
algorithm
state
estimation
only
injections
The
vector.
state
of the system
non-linear,
problem
is
the
Since
14.2.
of
sec
results
the
from
obtained directly
it is convenientto employ the iterative algorithm given in Eq. (14.22).
ine. the submatricesH., andH^ become null
equation rnay be approximatedas:
model
linearised
the
that
result
with the
An tntroductionto State Estimationof Power Systems
T
^
'
rol
Av=f A y , 1 '= l A r r l '= l j
l^'r'r.j' Lor"l, L;
then, Eq. (14.34) may be rewritten in the decoupledform as the following two
separateequationsfor the two partitionedcomponentsof the statevector
(14.35)
Alp= H1 Ax5+ ro
(14.31)
!=hlxl+ r
with the components
of the non-linearfunctiongivenby
Based on thesetwo equations,we obtain the following nearly decoupledstate
estimation algorithms.
i u Q + r ) = x d 0 + t H i 0 w p H tU ) l - ' H , ( D { t o - h r l i U ) l }
.l= 0, 1,2, ...
l V i l l V j l l Y i j l c o s( 6 , - 6 i + 0 i ) ,i = I , . . . , N
Qa32a)
N
t v * _ i l l v j l l y N _ i , ; l s i (n6 , - 6 i + 7 i i )
j:1
(r4.32b)
i=N+1,N+2...,2N
j = 0 | 2
i = 1 , 2 , . . . ,N ,
j=I,2,"',N-1
=
H z Q, i )
l v i l l Y i j I c o s (d, 6i * 0;) i = 7,2, ..., N ,
j = 1,2,"', N'
r f
/ r
! \
-
|
r t
I
I t r
|
| r ,
|
/ C
a
| yil I vjl I IUI COS \Oi-.Oj+
-
n
\
.
Aij) I =
n
I,
A
Z) ..., lY,
j=I,2,"',N-1
H q ( i ,j ) = l V i l l Y i j l s i n( { - 6 i + 0 , ) i = I , 2 , . . . , N ,
j = I,2, "', N'
'Note
i
that Eqs. (14.37) and (14.38) are not truly decoupledbecausethe
partitions of the non-linear function dependon the estimate of the entire state
vector. It may be possible to assume that vi U) = I for all i and 7 while'
Eq. (14.37) is being used in order to estimatethe angle part of the statevector.
similarly one may assume 6i U) = 0o for all i and 7 while using Eq. (14.38)
in order to estimate the_yllltage part of the state vector. Such approximations
allow the two equationsto be completely decoupledbut may not yield very good
solutions.A betterway to decouplethe two equationswould be to use the load
flow sglutions for x, and x6 as therr suppose<iiyconsnnt vaiues in Eq. (14.37)
and Eq. (14.38) respectively. There are several forms of fast decoupled
estimationalgorithmsbasedon suchconsiderations(seee.g. [13], tl4l).A flow
chart for one schemeof fast decoupled stateestimation in shown in Fig. I4.2.
(r4.33)
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(14.38)
where the subscriptsp andq are usedto indicatethe partitionsof the weighting
matrix W and the non-linear function h (.) which correspond to the vectors yp
*rd lerespectively. As mentioned earlier, if the covariances R* 11d Rn of the
effors r, and r(t are assumedknown, one should select Wo = R o' and Wn =
Rq"
The elements of the sub-matrices Hp H2, H, and Ho are then determined
easily as follows.
H t ( i , j ) = l V i l l V j l l Y i j I s i n ( r { .- 6 t+ 0 t )
\ (14.37)
i, (/ + 1)= i, Q) + tHl u) w, Hq (j)l-' nl u) wo ur- hq ti (j)ll
j:1
Ia3 \tt J)=
(r4.36)
Alq= H4 Axr , + r n
In this case, the model equation has the form
- D
(r4.34)
rrweparrition,J";,1:'
,,o*^::::
Only Algorithm
h , [ x ) =D
fH, o-l
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
542 |
Modern Po*er System Analysis
An Introductionto State Estimationof Power Systems
| 543
1_-
F i g .1 4 . 3
*11
Application of the LSE then yields the following expressions for the
estimatesof the perturbationsin the three state variablesaround their chosen
initial values:
A i , - AP, - AP,
I , o , r l l i 1 i * 1 t ) - * ( j ) l l = , 'j
on,^t
t\u
ls
,2
-\
an2
>
\\r---'
,
1"",
I
\
(3'9)
Fig. 14.2
AT,= 0.78 AQz - 0.26AQr
AV' = g39 AQ ' + 0'14 AQ r
These equationsshould be used in order to translatethe measuredvalues of
in t he act iveanclr eact ivepower inject ionsint o t heist inr at es
thc pcrturhat ions
of the perturbationsof the state variables.
It is interestingto note that for lhe sirnpleexample,partitionsHranrl Htare
null matricesso that the decoupiedstateestimatorsare the sameas thosegiven
above.
The Line Only Algorithm
I
j Example14.4
In order to illustrate an applicationof the injections only algorithm, consider the
"imple 2-bus system shown in Fig. 14.3.
A s s u m i n gl o s s l e s lsi n e , 0 , j = 9 0 " . A l s o l e t Y r r = j z z = 2 a n d y n = y z t =
. The power relations in this casewould be
Pt= - | Yr| | V2l lYrrl sin E
Pz=lVtl lVzl I Y,rlsin 6,
Q t = i r i l i i V r i ' - | y n l l y l | | V r l c o s$
Q z = l Y z z l V r l ' - l y t 2 l l y r | | v r l c o s6
I f we c h o o s eth e i n i ti a lv a l u e sl V f l = l V ; l = 1, 6o2= 0" , the correspondi ng
power values are P f = Pf - 0, Qf = Qi = 1. The value of the Jacobian matrix
evaluated at the above nominal values of the variables turns out to be
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This algorithm has been developed in order to avoid the need for solving a nonlinear estimationproblem.which as seenearlier,requiressome approximation
or other. ln the line tlow only algorithm,the data tor the active and reactive tie
line flows are processedin order to generatethe vector of the voltage difference
acrossfte tie-lines.Let z denote this vector.A model equationfor this vector
may be expressedas
(r4.39)
z= Bx + r-
where B is the node-elernentincidencernatrix and r- is the vector of the errors
in the voltage el-ata.Since this is a linear equation.one may use the WLSE
techniqueto generatethe estimate as
*, = lB, wBl-
| Bt wz
(r4.40)
where the weighting rnatrix may be set equai to the inverse of the covariance
of rrif this is known. The main problem with Eq. (14.40) is that the vector z
is not directly measurablebut needs to be generatedfrorn the tie line flow data.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
r++
Mocternpower System Analysis
|I
An lntroduction
to state Estimation
of powersystems
V,tdenotesthe voltage acrossthe line connectingthe ith ancltheTth buses,the
followine relation holds.
V i i = Z i i [@ i i -- j Qi ;tV 7- V i Y ,i ]
Here Z,.,standsfor the impedanceof the line.
This shows that the vector z is related to the vectorsx and
(14.41)
ashion and one ffidy use the notation
z = g (x, y)
04.42)
In view of this non-linear
relation,Ecl.(14.40)maybe expressecl
in theform
i = LB,Wnyr B,W I (i, y)
(14.43)
This, being a non-linearrelationship,
can not be solvedexceptthrougha
numericalapproach(iterativesolution).The iterativeform of Eq. (14.43)is
i (j + r) = lBt wBll H w s [i (i), y],
- /= 0 , 1 , 2 ,. . .
(r4.44)
Note that the original problem of estimation of x from the data for z is a linear
pr ob l e m s o th a t th e s o l u ti o ng i v e n by E q. (14.40) i s the opri nralsol rrti on.
However, the data for z needto be generatedusing the non-linear transforma_
tion in Eq' (14.42), which in turn has necessitateclthe use of iterative Eq.
(14.44)' Compared to the injections only iterative algorithm, the pr-ese1t
algorithm has the advantageof a constantgain matrix [8, W B]-t gr I4z.This
resultin a considerable
computational
simplification.The conceptof decoupled
estimationis easily extencled
to the case of the lirre flows [15].
Network Obsenrability [17]
Considerthe staticWLSE formula lEq.(14.llb)l which servesas the srarting
point fbr all the algorithrns.lnverse of information matrix Mn,, = Ht wH
should exist otherwisethere is no stateestimate.This will happenif rank of f/
ls equal to n (no. ot state vartables).Since one can always choose a nonsingular l4l, so if lt1 has a rank n, the power network is said to be observable.
Problem
TRACKING STATE ESTIMATION
SYSTEMS t16l
of lll-conditioning
Even if the given power system is an observable system in terms of the
measurementsselectedfor the state estimation purposes,there is no guarantee
that the required inversion of the information matrix will exist. During
multiplications of the matrices,there is some small but definite error introduced
due to the finite word length and quantisation.Whether or not these errors
create ill-conditioning of the information matrix may be determined from a
knowledge of the condition number of the matrix. This number is defined as the
ratio of the largestand the smallesteigenvaluesof the inforrnationrnatrix.The
maftix M becomesmore and more ill-conditioned as its condition number
increasesin magnitude.Somedetailedresultson power systemstateestimation
using Cholesky factorizationtechniquesmay be fbund in
tl8l. Factorization
helps to reduceill-conditioning but may not reducethe computationalburden.
A techrriqueto reduce computationalburden is describeclin Ref.
[]91.
14.6
T4.4
EXTERNAL
-t,
,/
Brrfh
SOME COMPUTATIONAL
the
qtafin
qnd
fhe
rt rr uq vorl \zr lrn1o6
o
f ilnr u- l r
v rc)frirm
r ror q
[20]
One of the widely practiced methods used for computationalsimplification is to
divide the given system into three subsystemsas shown in Fig. 14.4. One of
these is referredto as the 'internal' subsystemand consistsof those buses in
which we are really interested.The secondsubsystemconsistsof those buses
which are not of direct interest to us and is referred to as the 'external,
subsystem'Finally, the buseswhich provide links betweenthese internal and
external subsystemsconstitute the third subsystemreferredto as the 'boundary'
subsystem. For any given power network, the identification of the three
subsystemsmay be done either in a natural or in an artiflcial way.
/
I4.5
SYSTEM EOUTVAIENCTNG
OF POWER
Trackingthe stateestimationof a given power systemis importantfor real tirne
monitoring of the system.As is well known, the voltagesof all real systemvary
randomly with time and should therefore be considered to be stochastic
processes'It is thus necessaryto make use of the sequentialestimation
techniquesof Ref. [1] in order to obtain the state estimate at any given time
point. The power relationsin Eqs. (14.25) and,(74.26) are still valid bur must
be rewritten after indicating that the voltage rnagnitudesand angles are new
functions of the discrete time index ft.
\
,
CONSIDERATIONS
^1-^*i+L*^
ai^,r^61^-r
Ctr5\Jl tLllll.lJ
PIEi)t'f lttrlt
illl
rL^
tll€
---^--r:
-pICL:CUlllg
sections are computationally intensive, particularly for large power networks
which may have nlore than200 importantbuses.It is, therefore,very important
t o pa y a tte n ti o nfo s u c h c o mp u ta ti onali ssuesas i l l condi ti oni ng,computer
storageand tinnerequirements.However, we need to first considerthe question
of existenceof a solution of the state estimation problem.
\"
Internal
system
,/
Boundary
system
Flg. 14. 4
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| 545
T-
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
An lntroductionto state Estimationof Power systems
ModernPo
fl6 -l
I
To illustrate the simplification of the stateestimation algorithm, considerthe
linearisedmeasurementequationfor the injections only case.Since the system
is partitioned into three subsystems,this equation may be written as
Ayi
H,,
H,,
o
A*,
Ayr,
Hu,
Huu
Ho,
A*,
,ol
ay"
0
Hra
H""
A*,
,"]
(r4.4s)
It may be noted that the internal me ..rrerrrert vector Ayt is not completely
inclepenclentof the external subsystem state Axu since Ay, dependson the
boundary subsystetn state Axr, and Ax6 depends on Axr.
Ayr= Ayt,i + Aynr,+ Ayu, + ru
(r4.46)
where Ayo" representsthe injections intc the boundary buses from the external
buses,Ay66is the injection from the boundary buses znd Ay6, is the injection
fiom the internal buses.It is assurnedthat the term Ayu, (= Ht,, Axr), may be
approximatedas n A*, where A It estimated from the relation
(14.47)
H = Ayt,nlAxl,
The conrponent Ayu" may be estimated if the terms Ayy, and Ayuo ne
Ax1, and then subtractedfrom the measuredvalue
computedas H,,, Ax,and H1,1,
part of Eq. (14.45) to be rewritten as
in
woulc!
result
of Ayo. This
',;:;)l*,1
.[;]
l^;,)=l',;;,
( r4.48)
fhc cff-ccfivc Jacohiltn if the hottnclitry
whcrc Ht,, = H t,t,t U pcrprcscrnts
subsysternthat accounts for the eff'ects of the external subsysternon the
boundarysubsystem.Equation(14.48) has a lower dimensionthan the original
s. tc
c c l L rl ti o nu n d w o rrl rltl tcrcl orc i rtvol vcrl css ctl tttl l tr(.tl i ol lTl
t nc a s u rc l n c rtt
line
or
with
the
be
employed
may
conceptof external systemequivalencing
rnixedCatasituationsalso.
14.7
TREATMENT
Ol.- BAD DATA
127,221
The ability to detectand identify bad measurementsis extremely valuableto a
load dispatch centre. One or more of the data may be affected by rnalfunctioning of either the measuringinstrumentsor the data transmissionsystemor both.
itscll rttay be
rnay have becn wilcd incorrcctlyor tlto l.rattsduccr
Transtlucers
malfunctioning so that it simply no longer gives accuratereadings.
If such faulty daia are included in the vector Ay, the estimationalgorithm
will yielcl unreliableestimatesof the state.It is therefbreimportant to develop
techniquesfor detecting the presenceof faulty data in the measurementvector
at any given point of time, to identify the fauity data and eliminate these from
the vector y befble it is processcdlbr state estinration.[t is alstl ilttpoltant to
rnodify the estirnationalgorithmsin a way that will permit tnore reliable state
es ti ma ti o ni n th e p !e s e n c eo f b a d d ata.
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Bad Data Detection
t{7
I
[231
A convenient tool fcr detecting the presenceof one or more bad data in the
vector y at any given point of time is based on the'Chi Square Test'. To
appreciate this, flrst nc:te that the trrethod t-ll least square ensures that the
/ltgr haritrminimum
rflLW Ta: 1r (t)lvalue when x - i. Since the variable r is random,the minimum value ,I.in is
also a random quantity. Quite often, r may be assumed to be a Gaussian
variable and then .I*1nwould follow a chi squaredistribution with L = m n
degrees of freedom. It turns out that the mean of ./,o1nis equal to L and its
variance is equal to 2L. This implies that if all the data processedfor state
should be close to the
estimatiorrare reliable, then the computed value of ../r,r,n
averagevalue (=L). On the other hand, if one or ntore of the data for )' tre
unreliable, then the assumptiorrsof the least squaresestimation are violated and
the computed value of J*1nwill deviate significantly from I.
It is thus possibleto developa reiiable schetrrefor the detection of bad data
ly - h(r)], i being the estimate
in y by computingthe valueof [y - h(i)l'w
y.
scalarso obtainedexceedssome
the
If
the
concerned
of
obtainedon the basis
we conclude that the vector y
nuutber,
a
suitable
being
threshold Ti c L, c:
(Note
the.cornponenty;, i - 1,2, ...,
for
data
the
that
data.
includes some bad
of r', by more than t 3e,.
tnean
the
from
if
it
deviates
bacl
nr will be consiclerecl
while choosing
be
exercised
must
r;).
Care
of
deviation
where o, is the standard
producemany
maf
the
test
to
1,
is
it
close
c.If
parameter
the value of threshold
'talse alal1s' anclil it is tt-rt-r
large, the test worrld fail to detect nlany bad data.
To select an appropriatevalue of c, we may start by choosing the
significancelevel d of the test by the relation'
I'lJ (x) > cLlJ (.r) lbllows chi squaledistribution) = 6/
We rnay select,fbr example,d = 0.05 which correspondsto a 5Vofalse alarm
sitrrltign. [t is thcn possibleto find the valr.reof c hy making use of the table
out the test
X@).Once the valuc of c is determined,it is simple to cary
whether or not .l(x\ exceedscL.
of Bad Data [231
Identification
Once the presenceof bad data is detected,it is iraperative that these be
identified so that they could be removed from the vector of measurementsbefore
it is processed.One way of doing this is to evaluatethe components of the
residual !, = y, - h, (.x\, i = 1,2,..., m. If we assLlmethat the
measurement
residualshave the Gaussiandistribution with zero mean and the varLance4,
then the magnitucieof the resiciuaiy, shouiciiie in the range- 3o i ( )i ( 3or with
95Voconfidenceievel. Ihus, if any one of tlte cotnputedresidual turns out to
be significantly larger in magnitude than three times its standarddevia,tion,then
corresponding data is taken to be a bad data. If this happenSfor more than one
to
lur vinglt hc lar gcstr csidualis assunt ed
t hcn t hc con'lponcnt
con)pot r ct ol'_y,
r(
be thc bad dataand is removedfrom y. The estimationalgorithmis re-run with
the remaining data and the bad data detection and identification tests are
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
,548 ,!
performed again to find out if there are additional bad data in the measurement
set. As we will see later bad rneasurementclataare detected,eliminated and
replaced by pseudoor calculatedvalues.
estimate.
of Bad Data
Hr(D CTWCH Q)Tfl (D WCtfiDI
The procedures described so far in this section are quite tedious and time
consuming and rnay not be utilized to remove all the bad data which may be
presentin the vectory at a given point of time. It may often be desirableon the
other hand to modify the estimation algorithms in a way that will minimise the
influence of the bad data on the estimatesof the state vector. This would be
possible if the estimation index J(x) is chosen to be a non-quadratic function.
The reason that the LSE algorithm does not perform very well in the presence
of bad data is the fact that becauseof the quadraticnature of J(x), the index
assumesa large value for a data that is too far removed from its expectedvalue.
To avoid this overemphasison the effoneousdata anclat the same time to retain
the analytical tractability of the quadratic:perlbrmanceindex, let us choose
/(i) =s'(t) w sG)
(14.49a)
where 8(t) is a non-linearfunction of the residual !. There may be several
possiblechoicesfor this f'unction.A convenientform is the so-called'quadratic
flat' form. In this case,the componentsof the function g (y) are defined by the
following relation.
g i (j ) = l i ,
= ai,
fo r j ,l o , 1 a,
ftx ' l ,/o ,>
u,
(r4.49b)
where a, is a pre-selectedconstant threshold level. Obviously, the performance indcx ./("r)may bc cxpressedas
m
I(i) = Dy, c;)
; . - -
It -549.
The main advantage of the choice of the form (14.49) for the estimation
index is that it is still a quadraticin the function g(.i) and so the LSE theory
may be mimicked in order to derive the following iterative formuia for the state
to State Estimationof Power Systems
An tntroduction
ModernpowerSystemAnalysis
I
(14.s0)
I
I
and each componenthas a quadratic nature for small values of the residual but
has a constantmagnitudefor residual magnitudesin excessof the threshold.
Figure 14.5 shows a typicai variation of J, (.r) for the quadratic and the nonquadratic choices.
(74.5ra)
wherethe matrix C is diagonaland its elementsarecomputedas
= 0, f or lilo, >
I4.8
NETWORK OBSERVABILITY
MEASUREMENTS
Fig. 14.5
AND PSEUDO-
A minimum amount of data is necessaryfor State Estimation (SE) to be
effective. A more analytical way of determining whether a given data is enough
for SE is called observability analysis.It forms an integral part olany real time
state estimator.The ability to perform state estimation depends on whether
.sufficientmeasurementsare well distributedthroughoutthe system. When
sufficient measurementsare available SE can obtain the state vector of the
whole system. In this casethe network is observable.As explained earlier in
Sec. 14.5 this is true when the rank of measurententJacobiantlratrix is equal
to the number of unknown state variables. The rank of the measurement
Jacobian matrix is dependent on the locations and types of available
rneasurementsas well as on the network topology.
An auxiliary problem in state estimation is where to add additional data or
pseudomeasurementsto a power systemin orderto improvethe accuracyof the
calculated state i.e. to improve observability. The additional measurements
represent a cost for the physicat transducers,remote terminal or telemetry
sy.stem,and software data processing in the central computer. Selection of
pseudomeasurements,filling of missing data,providing appropriateweightage
are the functions of the observability analysis algorithm.
I
www.muslimengineer.info
a'
Comparing this solution with that given in Eq. (14.22), it is seen that the
main effect of the particularchoice of the estimationindex in Eq. (14.49) is to
ensurethat the data producingresidualsin excessof the thresholdlevel will not
changethe estimate. This is achieved by the production of a null value for the
matrix C for large values of the residual.
UbsefvaDlllty
\
( 1 4 . s1 b )
Ct = l, f ot l; lo, S ai
I
- -r--
I
l---:-
^
Can De CIICCKCU UUtrrlB
C^^t^-i-^+l^-
lilulullzittLltrll.
Tf
^-.'
^i"^+
Il 4IrJ Prv\rr
L^^^-^.
'^-.'
r.rvtrrr.rrrlr .YvrJ
small or zero during factoization, the gain matrix may be singular, and the
systemmay not be observable.
To finil the value ol an injection without nteasuringit, we tlrust know the
power system beyond the measurementscurrently being made.For example, we
pormally know the generated MWs and MV ARs at generators through
telemetry channels (i.e. these measurementswould generally be known to the
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
An introduction
to state Estimation
of powqr systems
i,
t,
stateestimator). If thesechannelsare out, we can perhapscommunicate
with the
operatorsin the plant control room by telephoneand ask for these
values and
enter them manually. Similarly, if we require a load-bus MW and
MVAR for
a pseudo measurement'we could use past recordsthat show the relationship
between an individual load and the total qyrtemload. We canestimate
the tstal
systetn load quite accuratelyby f)nding the total power being generated
and
estimating the line losses.Further, if we have just had a telernetry
failure, we
could use the most recentlyestimatedvaluesfiom the estimator(a.ssgming
that
i','is run periodic:ally)as pseudorlreasureftlents.
Thus, if required,we can give
. the state estimatorwith a reasonablevalue to use as a pseudomeasurement
at
any bus in the system.
Pseudomeasurements
increasethe data redundancyof SE. If this approach
is adapted, care must be taken in assigning weights to various
types of
measurements.
Techniquesthat can he useclto cleternrine
the rnctcror.pseu4g
measurementlocationsfbr obtaininga completeobservabilityof the sysrem
are
available in Ref. [251.A review of the principal observability
analysis and
m et erp l a c c rn c nat l g ' ri th rn si s a v u i ra hrci n I(cr' .
[261.
T4.9
APPLICATION
ESTIMATION
U ' E
L
bA
l
!
--9
F G
.Yt
o
l m
I
= t h
drx
tr
o
]U t-
o g
a
F
I
CnF
a
trJ
I
I
I
Io
o
1f
o
(!
(E
E
OF POWER SYSTEM STATE
a
o
(L
'=.o
I
I
L
n
v
^
v
I
r-nnnenfprl
L \ / fLl rror u lrn\^/'cl u (^ r^ Jr r L rl i c r r r s r . u . ( r rl -.-,-u-s- -c l- i- l . e . n e l w o r K
a
II
A
q
t , L
o
a
U)
o
o
o
o I
XH
;oii' >,
-
L
0-(E
I
I
I
I
,tl
l 6 ncl
L
f
a
(!
L
o
=
o
o
I
LUI
i LLI
o>
!
. i
o)
I
o)
o
lhrs
kceps
q
sf
E;
ll-
F
U)
C O e
C
N'E
b 8g
E E.E
E E E
6 E o
F O
o
6n
changingand hcncel.hccurrerttr.clentetered
breaker'/switch
stertus
must be used
to restructurethe electricalsystem model. This is called the network
tr,tpop.tgy
program or system status proce,\.tor or netwc;rk configurator.
www.muslimengineer.info
c
o
o
o
a
U)
I
i
l_
E
o
E
z 8
. n F
fn
topology,
o
o
o
E
I
,' Il
o
L
o
o
U)
In real-time environmentthe state estimatorconsistsof different
modulessuch
as network topologyprocessor,observabilityanalysis,stateestimation
and bad
data processing.The network topology processoris required
for ail power
systemanalysis.A conventionalnetwork topologyprogram usescircuit
breaker
statusinformation and network connectivity datato determinethe connectivity
of the network.
Figtrrc 1r4.6is it schcrnaticcliagtarnshowing the info.rnationflow between
the varior-rsfunctions to be performed in an operationscontrol centre
compurer
system. The system gets information from remote terminal unit (RTU)
that
cncodc lllcasLlrclllcnt
tl'ullsduccr0r-rtputs
anclopcnccl/closccl
staLusinlbmration
into digital signals which are sent to the operation centre over communications
circuits. Control centre can also transmit commands such as raiseflower
to
generators and open/closeto circuit breakers and switches.
The analog
measurementsof generatoroutput would be rJirectlyused by the AGC program
(Chapter 8). However, rest of the data will be processedby
the stateestimator
befbre being used fbr other functions such as OLF (Optimal Loacl
Flow) etc.
Before running the SE, we must know how the transmission lines
are
I
I FSI
t -
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
An lntroductionto State Estimationof Power
Po
Modern
ilr l
I
The output of the stateestimator i.e. lvl, 6, P,j, Q,jtogether.with latestmodel
form the basis for the economic dispatch (ED) or minimum emission dispatch
(MED), contingency analysis program etc.
Further
Readin
The weighted least-squaresapproach to problems of static state estimation in
power systems was introduced by Schweppe 11969-741. It was earlier
originated in the aerospaceindustry. Since 1970s, state estimators have been
installed on a regular basis in nern'energy (power system or load dispatch)
control centres and have proved quite helpful. Reviews of the state of the art
in stateestimation algorithms basedon this modelling approach were published
by Bose and Clements[27] and Wu [28]. Reviewsof externalsystemmodelling
are available in 1291.A generalisedstate estimator with integrated state,stutus
and parameterestimation capabilities has recently been proposed by Alsac et al
[30]. The new role of stateestimationand other advancedanalytical functions
in competitive energy markets was discussedin Ref. [31].A comprehensive
bibliography on SE from 1968-89 is available in Ref. [32].
F i g .P . 1 4 . 2
14.3Given a single line as shown in Fig. p 14.3, two measurementsare
available. using DC load flow, calculate the best estimate of the power
flowing through the line.
4= 0 rad.
/1
|
( ,r-1-r t t----1
LEMS
PROB
Mt2
[l
y0.1pu
(1ooMVAbase)
Mzt
l -
2
Fig. P. 14. 3
14.1 For Ex. 6.6 if the power injected at buses are given as Sr = 1.031 j0.791, Sz = 0.5 + 71.0and 'S3= - 1.5 - j0.15 pu. Consider Wt= Wz=
Wz = l. Bus I is a referencebus. Using flat start, find the estimatesof
lV,l and {. Tolerance= 0.0001.
lAns: Vl = l/0", V', = t.04223 10.4297",V\ = 0.9982q l-2.1864";
Meter
Mrz
l-1.356, Va - 1.03831
Final values: Vt - 1.04./.0", V2 = 1.080215
l- 3.736"1.
14.2 For samplesystemshown in Fig. P. 14.2,assumethat the threemeters
have the followine characteristics.
Meter
Full scale (MW)
Accuracy (MW)
o (pu)
Mrz
100
r00
100
+ 8
+4
+ 0.8
0.02
0.01
0.002
Mtt
Mzz
Calculate the best estimate for the phase angles 4 *d
f,'11,...,:
| 1-rl l\., W | | l5
| I lU{lDLll
nmnnf
t,
t/l I lUl lL,)
Meter
Meusured value (MW)
Mtz
70.0
4.0
30.5
Mrz
Mt,
www.muslimengineer.info
d2 given the
Mzr
Full scale
Meter Standard
Meter
(MW)
Deviation ( o )
in full scale
Reading
1
4
32
-26
100
100
(Mw)
REFEREN
CES
Books
l. Mahalanabis,A.K., D.P. Kothari and S.L Ahson, ComputerAidetl pr;wer System
Analysis and Control, Tata McGraw-Hill, New Delhi, 19gg.
2. Nagrath, I.J. and D.P. Kothari, Power SystemEngineering, Tata McGraw-Hill.
Ncw Dclhi, I q94.
3. Mtrnticclli' A., State li.rtinrution in Eler:tric:Power SysremsA Gcnerali.recl
Apprctat:h, Kluwcr Academic Publishers,Boston, 1999.
4. Kusic, G'L., Computer-AidedPower SystemsAnalysis,Prentice-Hall,N.J.
19g6.
5. Wood, A'J. and B.F. Wollenberg, Power Generation,Operation and Control. Znd
Ed., John Wiley, NY, 1996.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'l
I
I
j
t
-
ModernPowerSystemAnalysls
An Introductionto State Estrmationof Power Systems
6. Grainger, J.J. and W.D. Stevenson, Power System Analysis,McGraw-Hill, NY',
1994.
7. Deautsch,R., Estimation Theory, Prentice-Hall Inc' NJ, 1965
g. Lawson, C.L. and R.J. Hanson, Solving Least SquaresProblens, Prentice-Hall.
26. Clements, K.A., "Observability Methods and optimal Meter Placement", Int. J.
Elec. Power, Vol. 12, no. 2, April 1990,pp 89-93.
27. BoSe,A. and Clements,K.A., "Real-timeModelling of Power Networks", IEEE
Proc., Special Issue on Computersin Power Svstem Operations,Vol. 75, No. 12,
inc., NJ., i974.
g. Sorenson,H.W., ParameterEstimation,Mercel Dekker, NY, 1980'
Dee 1987;aP 76ff7=1ffi2:
28. Wu, F.F., "Power System State Estimation: A Survey", Int: J. Elec. Power and
Energy Syst.,Vol. 12, Jan 1990, pp 80-87.
29. Wu, F.F. and A. Monticelli, "A Critical Review on External Network Medelling
for On-line Security Analysis", Int. J. Elec. Power and Energy Syst.,Vol. 5, Oct
;554.;l
Papers
10. Schweppe,F.C., J. Wildes, D. Rom, "Power SystemStatic StateEstimation,Parts
l, ll and lll", IEEE Trans', Vol. PAS-89, 1970,pp 120-135'
ll. Larson, R.E., et. al., "State Estimation in Power Systems", Parts I and II, ibid-,
pp 345-359.
lZ. Schweppe,F.C. and E.J. Handschin,"static State Estimation in Electric Power
System", Proc. of the IEEE, 62, 1975, pp 972-982'
13. Horisbcrgcr,H.P., J.C. Richarcland C. Rossicr,"A Fast DccouplcdStatic Statc
Estimator for Electric Power Systems", IEEE Trans. Vol. PAS-95, Jan/Feb1976,
pp 2 O 8 -2 1 5 .
1983,pp 222-235.
30. Alsac, O., et. aI., "Genetalized State Estimation", IEEE Trans. on Power Systems,
Vol. 13, No. 3, Aug. 1998,pp 1069-1075.
D. et. al., "TransmissionDispatch and CongestionManagement
31. Shirmohammadi,
Market Structures",IEEE Trans. Power System.,Vol 13,
Energy
Emerging
in the
No. 4, Nov 1998,pp 1466-1474.
32. Coufto, M.B. et. aI, "Bibliography on Power System State Estimation (19681989)",IEEE Trans.Power Sysr.,Vol.7, No.3, Aug. 1990,pp 950-961.
1 4 . Monticelli, A. and A. Garcia, "Fast DecoupledEstimators", IEEE Trans' Power
Sys/, 5, May 1990,pP 556-564.
1 5 . Dopazo,J.F. et. al., "State Calculation of Power Systems from Line Flow
PartsI and ll", IEEE Trans.,89, pp. 1698-1708,91, 1972,pp
Measurcments,
145-151.
A.S. anclR.E. Larson, "A Dynamic Estimatorfor Tracking the Stateof a
Dcbs,
16.
Power System", IEEE Trans' 89, 1970,pp 1670-1678'
iZ. K1u-pholz, G.R. et. al.,"Power SystemObservability:A PracticalAlgorithm Using
Nctwork Topology", IEEE Trans. 99' 1980' pp 1534-1542'
A and V.H. Quintana,"A Robust Numerical Techniquefor Power
lg. Sirnoes-Costa,
SystemStateEstimation", IEEE Trans. 100, 1981, pp 691-698'
A and V.H. Quintana,"An OrthogonalRow ProcessingAlgorithm
19. Simgcs-Costa,
for Power System SequentialState Estimation", IEEE Trans., 100, 1981' pp
3'79r-3799.
ii
I
20. Debs, A.S., "Estimation of External Network Equivalentsfrom Internal System
Data", IEEE Trans., 94, 1974, pp 1260-1268'
21. Garcia.A., A. Monticelli and P. Abreu, "Fast DecoupledStateEstimationand Bad
Dara Processing",IEEE Trans. PAS-98, Sept/Oct 1979, pp 1645-1652.
22. Handschin,E. et. al., "Bad Data Analysis for Power System State Estimation",
IEEE Trans., PAS-94, 1975, pp 329-337.t2
4J.
r.^-r:6
r\V6lrrrt
rJ
r
Lt.r.
or @nle . ,
eL.
"Flqd T)cre Detecfion
and ldentification".
Int. J. EleC. POWer,
Vol. 12, No. 2, April 1990, PP 94-103'
24. Me-Jjl, H.M. and F.C. Schweppe,"Bad Data Suppressionin Power SystemState
Estimation",IEEE Trans. PAS-90, 1971' pp 2718-2725'
25. Mafaakher, F., et. al, "Optimum Metering Design Using Fast Decoupted
Estimator", IEEE Trans. PAS-98, 7979, pp 62-68.
www.muslimengineer.info
_T-I 555
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
in lglver Systems
Compensation
|I
55?
can be connecteciin the system in two ways, in series and irr shunt at ihe line
ends (or even in the midPoint)'
Apart from the well-known technologies of compensation, the latest
technology of Flexible AC TransmissionSystem (FACTS) will be introduced
towards the end of the chaPter.
15.2
LOADING
CAPABILITY
There are three kinds of limitations for loading capability of transmission
system:
(i) Thermal (ii) Dielectric (iii) Stability
Thermal capabitity of an overhead line is a function of the ambient
temperature, wind conditions, conditions of the conductor, and ground
15.1
INTRODUCTION
For reduction of cost and improved
reliability, most of the world,s electric
power systemscontinueto be intercor
of diversity of loads,availability of sor
to loads at minimum cost and pollr
deregulatedelectric service environme
to the competitive environmentof reli
Now-a-days,greaterdemandshave been
placedon the transmissionnetwork,
and these demandswill continue to
rise becauseof the increasingnumber
of
nonutility generatorsand greatercompetition
among utilitics the'rselves.It is
not easy to acquire new rights of way.
Increased demands on transmission,
absence of long-term planning, and
the need to provide open access
to
generatingcompaniesand customers
have resultedin less securityand reduced
quality of supply.
compensationin power systemsis, therefore,
essentialto alleviate some of
theseproblems'series/shuntcompensation
has beenin usefor past many years
to achieve this objective.
In a power system, given the insignificant
erectricarstorage, the power
generationand load mttst balance
at all times. To some extent,the electrical
system is self'-regulating.
If generationis less than roacl,voltage
anclfrequency
d'op, and thereby reducing the road.
However, there is only a few percent
margin for such self-regulation.lf ,,^r+--^ :- :_,-
support,
rhenloadincrease
withcon"se;:;iilTi: ,:t#'j|i;l:ffL;
systemcollapse'Alternatively,if there
isinadequatereactivepower,the system
rnay havc voltagecollapse.
This chapter is devoteclto the stucly
of various methods.f compensating
power systemsand varioustypesclf
compensating
crevices,
cailccr.,,,rrj",rru,urr,
to alleviatethe problemsof power system
outlined above.Thesecompensators
www.muslimengineer.info
clearance.
There is a possibility of converting a single-circuit to a double-circuit line to
increasethe loading caPabilitY.
Dieletric Limitations From insulationpoint of view, many lines are designed
very conservatively. For a given nominal voltage rating it is often possible to
increasenormal operatingvoltagesby l07o (i.e. 400 kV 440 kV). One should,
however,ensurethat dynamic and transientovervoltagesarewithin limits. [See
Chapter 13 of Ref. 71.
Stability Issues.There are certain stability issues that limit the tlansmission
capability. These include steady-statestability, transient stability, dynamic
stability, frequency collapse, voltage collapse and subsynchronousresonance.
Severalgooclbooks l, l, 2, 6,7 ,8) are availableon thesetopics.The FACTS
technology can certainly be used to overcome any of the stability limits, in
which casethe final limits would be thermal and dielectric'
15.3
LOAD COMPENSATION
Load compensation is the managementof reactive pcwer to improve power
quality i.e. V profile and pf. Here the reactive power flow is controlled by
at the load end
installing shunt compensating devices (capacitors/reactors)
reactive power.
consurned
and
generated
bringing about proper balancebetween
of the system
capability
power
transfer
the
This is most effective in improving
to
technically
and
economically
both
ancl its voltagestability. It is desirable
impose
utilities
why
some
is
This
power
factor.
operatethe systemnear uriity
a penalty on low pf loads. Yet another way of irnproving the system
performa-neeis to operate it under near balancedconditions so as to reduce the
ho* of legative sequence currents thereby increasing the system's load
capabilityand reducingpower loss'
line hast hr cccr it icalloadings( i) nat ur alloading( ii) st eadyA transm ission
line the
starestability limit and (iii) thermallimit loading.For a compensated
reached,
is
limit
loading
the
thermal
before
and
lowest
the
is
natural loading
steady-statestability limit is arrived.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
esJ
I
T5.4 LINE COMPENSATION
Ideal voltage profile for a transmissionline is flat, which can only be achieved
by loading the line with its surge impedance loading while this may not
be
achievable,the characteristicsof the line can be modi
so that
(i) Ferranti effect is minimized.
(ii) Underexcitedoperationof synchronousgeneratorsis not required.
(iii) The power transfer capability of the line is enhanced.Modifying
the
characteristicsof a line(s) is known as line compensation.
Various compensatingdevicesare:
o Capacitors
. Capacitors and inductors
compensated,it will behave as a purely resistive element and would. cause
series resonance even at fundamental frequency. The location of series
capacitorsis decidedby economical factors and severity of fault currents.Series
capacitor reducesline reactancethereby level of fault currents.
on on vanous lssues ln
in series and shunt
compensatorsnow follows.
15.5
A capacitor in series rvith a line gives control over the effective reactance
between line ends.This effective reactanceis eiven bv
Xr=X-X,
where
. Active voltage source (synchronousgenerator)
I
ii
tit i
When a number of capacitors are connected in parallel to get the desired
capacitance,it is known as a bank of capacitors,similarly a Uant<
of incluctors.
A bank of capacitors and/or inductors can be adjusted in stepsby switching
(mechanical).
.
Capacitors and inductors as such are passive line compensators,while
synchronous generator is an active compensator.When solid-statedevices are
used for switching off capacitorsand inductors, this is regardeclas
active
compensation.
Before proceedingto give a detailedaccount of line compensator,we shall
briefly discussboth shunt and seriescompensation.
Shunt compensation is more or less like load compensation with all the
advantagesassociatedwith it and discussedin Section 15.3. It needs to
be
pointed out here that shunt capacitors/inductors can not be distributed
uniformally along the line. These are normally connected at the end of the line
and/or at midpoint of the line.
Shunt capacitors raise the load pf which greatly increases the power
transmitted over the line as it is not required to carry the reactivepower. There
is a limit to which transmitted power can be increased by shunt compensation
as it would require very iarge size capacitorbank, which would be impractical.
For increasing power transmitted over the line other and better means can be
adopted.For example,seriescompensation,
higher transmissionvoltage,HVDC
etc.
When switched capacitors are employed for compensation,these shculd be
disconnected
irnmediatelyunderlight loaclconclitionsto avoicler.cessive
voltage
rise and ferroresonancein presenceof transformers.
The purpose of series compensationis to cancel part of the seriesinductive
reactanceof the line using series capacitors.This helps in (i) increase of
maximum power transfer (ii) reduction in power angle for a given amount of
power transfer (iii) increasedloading. From practical point of view, it is
www.muslimengineer.info
SERIES COMPENSATION
Xl = line reactance
Xc = capacrtor reactance
It is easy to see that capacitorrcduccsthe cffectivc line rcactance*.
This results in improvement in performanceof the system as below.
(i) Voltage drop in the line reduces(getscompensated)i.e. minimization of
end-voltagevariations.
(ii) Preventsvoltage collapse.
(iii) Steady-stttte
power transferincreases;it is inversclyproportionalto Xl.
(iv) As a result of (ii) rransientstability limit increasbs.
The benefitsof the seriescapacitorcompensatorare associatedwith a problem.
The capacitive reactanceXg fbrms a seriesresonantcircuit with the total series
reactance
X = Xt * X*.n * Xoun,
The natural frequency of oscillation of this circuit is given by.
rfL^ -
|
2rJrc
'2n
where
l=
system frequency
xReactive voltage drops of a series reactance
added in a line is I2x
It is positive if X is inductive and negative if X is capacitive. So a series capacitive
reactance reduces the reactance voltage drop of the line, which is an alternative wav
of saying that
X't= \-
X,-
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I
, ,
ModernPower System AnalYsis
rL
X
I
=degree of compensation
= 25 to J5Vo (recommended)
fc<f
which is subharmonicoscillation.
Even though series compensationhas often been found to be cost-effective
compared to shunt compensation,but sustained oscillations below the fundamental system frequency can cause the phenomenon, referred to aS
subsynchronousresonance(SSR) first observedin 1937, but got world-wide
attention only in the 1970s, after two turbine-generatorshaft failures occurred
at the Majave Generating station in Southern Nevada. Theoretical studies
pointed out that interaction between a series capacitor-compensatedline,
oscillating at subharmonic frequency, and torsional mechanical oscillation of
turbine-generator set can result in negative damping with consequentmutual
reinforcementof the two oscillations.Subsynchronousresonanceis often not a
major problem, and low cost countermeasuresand protective measurescan be
applied. Some of the corrective measuresare:
(i) Detecting the low levels of subharmonic currents on the line by use of
sensitiverelays, which at a certain level of currents triggers the action to
bypassthe series capacitors.
(ii) Modulation of generator field current to provide increased positive
damping at subharmonic frequency'
for line compensationunder light load conditionsto
Seriesincluctorsare neeclecl
counter the excessivevoltage rise (Ferranti effect).
As the line load and, in particular the reactive power flow over the line
varies, thereis need to vary the compensationfor an acceptablevoltage profile.
The mechanical switching arrangementfor adjusting the capacitance of the
capacitor bank in series with the line is shown in Fig. 15.1. Capacitanceis
with the capacitance
varieclby openingthe switchosof individualcapacitances
C1, being startedby a bypass switch. This is a step-u'isearrangement.The
whole bank can also be bypassedby the starting switch under any emergent
conclitions on the line. As the switches in series with capacitor are current
carrying suitablecircuit breaking arrangementare necessary.However, breaker
switched capacitorsin seriesare generallyavoided thesedays the,capacitoris
either fixed or thvristor switched.
' technology, the capacitanceof the series capacitance
bank can be controlled
much more effectively; both stepwiseand smooth control. This is demonsffated
by the schematicdiagramof Fig. 15.2 wherein the capacitoris shuntedby two
nstors ln antl
current in positive half cycle and the other in negative half cycle.
In each half cycle when the thyristor is fired (at an adjustable angle), it
conducts current for the rest of the half cycle till natural current zeto. During
the off-time of the thyristor current is conductedby the capacitor and capacitor
voltage is vr. During on-time of the thyristor capacitor is short circuited i.e.
v, = 0 and current is conductedby the thyristor. The sameprocess is repeated
in the other half cycle. This means that v, can be controlled for any given i,
which is equivalent of reducing the capacitanceas C = vJi.By this scheme
capacitancecan be controlled smoothly by adjusting the firing angle.
l--+{--_-__-a,.Currenuimitins
^ | o ' l l | "r' "reactor
i
l-1<-------r
---.
t
l
C r i
i
7,c-----]
Fig. 15. 2
Thyristors are now available to carry large current and to withstand (during
off-time) large voltage encounteredin power systems.The latest device called
a Gate Turn Off (GTO) thyristor has the capability that by suitable firing
circuit, angle (time) at which it goes on and off can both be controlled.
This meanswider range and finer control over capacitance.Similarly confrol
is possibleover seriesreactorin the line.
All controlled.".?nd uncontrolled (fixed) series compensators require a
protective arrangement.Protection can be provided externally either by voltage
arrester or other voltage limiting device or an approximate bypass switch
arrangement.In no case the VI rating of the thyristors should be exceeded.
Depending on (i) kind of solid-statedevice to be used (ii) capacitor and/or
reactor compensation and (iii) switched (step-wise) or smooth (stepless)
control, several compensatronschemeshave been devisedand are in use.Some
of the more common compensationschemesare as under.
(i) Thyristor Controlled Series Cappcitor (TCSC)
(ii) Thryristor\Switched Series Capacitor (TSSC)
(iii) Fllyristor controlled Reactor with Fixed capacitor (TCR + FC)
(iv) GTO thyristor Coniiolled Series Capacitor (GCSC)
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
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;
Modern Power System Analysis
*562i I
I
(v) Thyristor Controlied reactor (TCR)
Capacitor and/or reactor series compensatoract to modify line impedance.
An altcrnativoapproachis to introducca controllablevoltagc sourccin series
with the line. This scheme is known as static synchronous series compensator
(SSSC). SSSC has the capabitity to induce both capacitive and inductive
wrdenrngme operatmg reglon o
voltage ln senes wrtn [lne,
It can be used for power flow control both increasing or decreasingreactive
flow on the line. Further this schemegives better stability and is more effective
in damping out electromechanicaloscillations.
Though various types of compensatorscan provide highly effective power
flow control, their operating characteristics and compensating features are
different. These differences are related to their inherent attributes of their
control circuits; also they exhibit different loss characteristics.
From the point of view of almost maintenancefree operation impedance
modifying (capacitorsand/or reactors) schemesare superior. The specific kind
of compensator to be employed is very much dependent on a particular
application.
15.6
SHUNT COMPENSATORS
As alreadyexplainedin Sec. 15.4 and in Ch. 5 (Sec.5.10) shunt compensators
are connected in shunt at various system nodes (major substations) and
sometimesat mid-point of lines. Theseserveihe purposesof voltage control and
load stabilization. As a result of installation of shunt compensators in the
system,the nearbygeneratorsoperateat nearunity pf and voltage emergencies
mostly clo not arise. The two kinds of compensatorsin use are:
(1) Static var compensators (SVC): These are banks of capacitors (somea l s o fg r u s e u n d erl i ght l oad condi ti ons)
t im e s i n c l u c to rs
(ii) STATCOM: static synchronouscompensator
(iii) Synchronous condenser: It is a synchronousmotor running at no-load
and having excitation adjustableover a wide range. It f'eedspositive
VARs into the line under overexcited conditions and negative VARS
when underexcited.(For details see Sec. 5.10.)
It is to be pointed out here that SVC and STATCQM are stgtic var
generatorswhich are thyristor controlled. In this section SVC will be detailed
while STATCOM forrns a part of FACTS whose operafionis explainedin
called static var switchesor systems.It means that terminology wise
SVC = SVS
and we will use theseinterchangeably.
Basic SVC Configrurations
(or Desigrns)
Thyristors in antiparallelcan be used to switch on a capacitor/reactorunit
in
stepwise control. When the circuitary is designeclro adjust the firing angle,
capacitor/reactor
unit acts as continuouslyvariablein the power circuit.
Capacitor or capacitor and incluctor bank can be varied stepwise or
continuouslyby thyristor control. Several important SVS configurationshave
been devised and are applied in shunt line compensation.Some of the static
compensatorsschemesare discussedin what follows.
(i)
Sutu.rutcdreuctlr
This is a multi-core reactor with the phase windings so arrangedas to cancel
the principal harmonics.It is consicleredas a constantvoltageieactive source.
It is almost maintenance free but not very flexible with reipect to operating
characteristics.
(ii)'fhyristor-coilrroll,cd reuctor (T'CR)
A thyristor-controlled-reactor
(Fig. 15.3) compensatorconsistsof a combination of six ptrlseor twelve plusethyristor-controllecl
reactprswith a fixe<tshunt
capacitor bank. The reactive power is changectby adjustingthe thyristor firing
angle. TCRs are characterised by continuous control, no transients and
gencrationof harmonics'k.
The controlsystenrconsistsof voltage(and current)
- - Line oi.rt
Power
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S e c .1 5 . l O .
Statia
VAR Compensator
(SVC)
Thesecomprisecapacitorbank fixed or switched(controlled)or fixed capacito
bank and switched reactor bank in parallel. These compensatorsdraw reactiv
(leadingor lagging) power* from the line therebyregulating voltage, improv
*A rcrrctlrrcc cor)ncctcd in shunt to tinc at voltage V draws reactive power Vzl]
It is negative (leading) if reactance is capacitive and positive (lagging) if reactance is
inductive.
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Fixed
caoacitor
I n"r.to,
f =..'x',rnc.I
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II
u
v
v
v
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I Neutral
Fig. 15.3 Thyristorcontrolledreactor(TCR)with fixedcapacitor
*Though ) -connected
TCR's are used here, it is better to use A-connected
TCR,s
since it is better configuration.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
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Modern Power SystemAnalysis
fe"
system (e.9. line faults, load rejection etc) TSC/TCR combinations are
characterisedby continuouscontrdl, no transients,low generationsof harmonics, low losses, redundancy,flexible control and operation.
measuring devices, a controller for error-signal conditioning, a Iinearizing
circuit and one or more synchronising circuits.
(iii) Thyristor switcherl capacitor (fSC)
It consists of only a thyristor-switched capacitor bank which is split into a
num De o
r unrtso equal ratrngsto achrevea stepwisecon
in Table 15.1.
Table 15.1 Comparisonof Static Var Generators
'0d6'
Type of
Var
Generator
/
Damping
reactor
-
Fig. 15.4 Thyristorswitchedcapacitor(TSC)
As suchthey are applied as a discretlyvariablereactivepower source,where
this typeof voltagesupportis deemedadequate.All switchingtakesplace when
the voltage acrossthe thyristor valve is zero, thus providing almost transient
tree switching.Disconnectionis eff'ectedby suppressingthe firing plus to the
thyristors, which will block when the current reaches zero. TSCs are,
charetcLorisccl
by stcp wisc control, no transients,vcry low htlrnronics,low
losses,redundancyand flexibility.
(iv)
TSC-(TSR)
(2)
TCR-TSC
(3)
VI and VQ
Max comp. current
characteristics is proportional to
system voltage.
Max cap. var output
decreaseswith the
squareof the voltage
decrease.
Loss Vs var
High losses at zero
output.
output. Losses
decreasesmoothly
with cap. output,
increirsewith
inductive output
Max. Comp. current is
proportional to system
voltage.
Max. cap. var output
decreaseswith the
square of the voltage
decrease.
Low lossesat zero
ouput. Losses increase
step-like with cap.
output
Same as in
(1) or (2)
Hannorric
generation
hrtcrnallyvcry low
Resonancemay
necessiratetuning
reactors
I cycle
CombinedTCR and TSC Compensator
A combinedTSC and TCR (Fig. 15.5) is the optimum solution in majority of
cases.With this, continuousvariablereactivepower is obtainedtirroughoutthe
c ot t t l; lc tccr o n l l o l rl n g c . F ru ' tl tc l rn orcl i rl l control o1' botl r i nducti ve and
capacitiveparts of the cornpensatoris obtained.This is a very advantageous
Max. theoret.
delay
Trrnsir:nl
behaviour
under systent
voltagr:
disturbances
---{
Neutral
TCR.FC
(1)
I n t c r n a l l yh i g h
(large pu TCR)
Requires significant
filtering
l/2 cycle
Low losses at
zero output.
Losses increase
step-like with
cup. output,
smoothly with
ind. output
Internally low
(small pu
TCR) Filtering
required
I cvcle
Poor (FC ('iluscs
Cun bc lrcutral.
S a n r ca s i n ( 2 )
transientover(Capacitorscan be
voltages in response switchedout to minimise
to step disturbances) trattsicntovcr-voltages)
T5.7 COMPARISON BETWEEN STATCOM AND SVC
It
ll
Capacitor
Fig. 15.5 A combinedTCR/TSCcompensator
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characteristicand functional compensationcapability of the STATCOM and the
sVC aie similar l2l. However, the basic operating principles of the
STATCOM, which, with a converter based var generator,functions as a shuntconnectedsynchronousvoltage source,are basically different from those of the
SVC, since SVC functions as a shunt-connected,controlledreactive admittance.
This basic operational difference renders the STATCOM to have overall
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.t
lvlooernPower SystenrAnalysts
t
superior functional characteristics,better performance,and greater application
flexibility as comparedto SVC. The ability of the STATCOM to maintain full
capacitle output current at low system voltage also makes it rnore effective
than the SVC in improving the transient (first swing) stability.
Comparison between series and shunt compensation:
tto
(i) Series capacitorsare inherently self regulating and a control systemis not
required.
(ii) For the same performance, series capacitors are often less costly than
SVCs and lossesare very low.
(iii) For voltage stability, series capacitors lower the critical or coliapse
voltage.
(iv) Seriescapacitorspossessadequatetimc-ovellt-radcapability.
(v) Series capacitorsand switched series capacitors can be used to controi
loading of paralledlines to minimise active and reactive losses.
Disadvantagesof seriescompensation:
(i) Series capacitors are line connected and compensation is removed for
outages and capacitorsin parallel lines may be overloaded.
(ii) During tru.,vyioading, the voltage on one side of the seriescapacitormay
be out-of range.
(iii) Shunt reactorsmay be neededfor light load compensation.
(iv) Subsynchronous
resonancemay call for expensivecountermeasures.
Advantagesof SVC
(i) SVCs control voltagedirectly.
(ii) SVCs control temporaryovervoltagesrapidly.
Disadvantagesof SVC
(i) SVCs have limited ovcrload capability.
(ii) SVCs are expensive.
The best designperhapsis a combinationof seriesand shunt compensation.
Because of higher initial and operating costs, synchronous condensers are
normally not competitivewith SVCs. Technically,synchronouscondensersare
better than SVCs in voltage-weak networks. Following a drop in network
voltage, the increasein condenserreactive power output is ilrslantaneous.Most
synchrono.uscondenserapplications are now associatedwith HVDC installallons.
15.8 FLEXTBLE AC TRANSMTSSTONSYSTEMS (FACTS)
*
FACTStechnologyhavebeenproposedandimplemented.
FACTS devicesca1
be effectively used for power flow conffol, load sharing among parallel
corridors, voltage regulation, enhancement of transient stability anO mitlgation
enablea line to carry power closer to its thermal rating. Mechanical switching
has to be supplementedby rapid responsepower electronics.It may be noted
that FACTS is an enabling technology, and not a one-on-one substitute for
mechanicalswitches.
FACTS employ high speed thyristors for switching in or out transmission
line cornponents such as capacitors, reactors or phase shifting transformer for
some desirable performance of the systems. The FACTS technology is not a
single high-power controller, but rather a collection of controllers,wtrictr can be
applied individually or in coordination with others to conffol one or more of the
systemparameters.
Before proceeding to give an account of some of the important FACTS
controllers the principle of operation of a switching converter will be explained,
which forms the heart of these controllers.
15.9
PRINCIPLE
AND OPERATION
OF CON\ZERTERS
Controllable reactive power can be generated by dc to ac switching converters
which are switched in synchronism with the line voltage with whicn tfr" reactive
power is exchanged.A switching power converter consistsof an array of solidstate switches which connectthe input terminals to the output terminals. It has
no internal storage and so the instantaneousinput and output power are equal.
Further the input and output terminations are complementar|, that is, if the input
is terminated by a voltage source (charged capacitor or battery), output ii a
cuffent source (which meansa voltage source having an inductive impedance)
and vice versa. Thus, the converter can be voltage sourced (shunted by a
capacitor or battery) or current sourced (shunted by an inductor).
Single line diagram of the basic voltage sourced converter scheme for
reactive power generation is drawn in Fig. 15.6. For reactive power flow bus
voltage V and converter terminal voltage V, are in phase.
Then on per phase basis i
r=v-v4
x'i
'Fha
ruv
raa^+;"^
lvcvLtvly
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-^'.'^PrJw9r
^-,^f^^J^^
E^Urr4rrBtr
O=vI=
:^
l5
v(v-vo)
X
The rapid developmentof power electronicstechnologyprovidesexciting
opportunities
to developnew powersystemequipment
for betterutilizationof
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
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ModernPo
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System bus V
---__i-'i ' Coupling
transformer
I
X
.J,
:l J,Transformerleakagereactance
capacitoris zero. Also at dc (zero fiequency) the capacitor does not supply any
reactivepower. Thereibre, the capacitor voltage cloes not change and the
capacitor establishesonly a voltage level for the converter. The switching
causesthe converter to interconnectthe 3-phaselines so that reactive current
can flow between thent.
Vd.
Fig. 15.6 Staticreactivepowergenerator
The switching circuit is capable of adjusting Vo, the output voltage of the
converter. For Vo 1 V,1 lags V and Q drawn from the bus is inductive, while
for Vo > V, I leads V and Q drawn from the bus is leading. Reactive power
drawn can be easily and smoothly varied by adjusting Voby changing the ontime of the solid-state switches. It is to be noted that transformer leakage
reactanceis.quite small (0.1-0.15 pu), which meansthat a amall differenceof
voltage (V-Vo) causesthe required ,1and Q flow. Thus the converter acts like
a static synchronous condenser(or var generator).
A typical convertercircuit is shown in Fig. 15.7.It is a 3-phasetwo-level,
six-pulse H.bridge with a diode in antiparallel to each of the six thyristors
(Normally, GTO's are used). Timings of the triggering pulses ate in
synchronism with the bus voltage waves.
Vo"
Vot
Vo,
[1'
K
Va"
C
two-levelsix-pulsebridge
Fig. 15.7 Three-phase,
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The converter draws a small amount of real power to provide for the internal
loss (in switching). If it is requiredto feed reai power to the bus, the capacitor
is replace,l by a storage battery. For this the circuit switching has to be
modified ro create a phase difference dbetween Vsand Vwith Vsleading V'
The above explained converteris connectedin shunt with the line. On sirnilar
lines a converter can be constructedwith its terminals in serieswith the line.
It has to carry the line current and provide a suitable magnitude (may also be
phase)voltage in serieswith the line. In such a connectionit would act as an
impedancemodifier of the line.
15.10
FACTS CONTROLLERS
The development of FACTS controllershas followed two different approaches.
The first approach employs reactive impedancesor a tap changing ffansformer
with thyristor switches as controlled elements, the second approach employs
self-commutatedstatic converters as controlled voltage sources.
ln general,FACTS controllerscan be divided into tour categories.
(i) series (ii) shunt (iii) combined series-series(iv) combined series-shunt
controllers.
The generalsymbol for a FACTS controller is given in Fig. 15.8(a).which
shows a thyristor arrow inside a box. The series controller of Fig. 15.8b could
be a variable impedance, such as capacitor,reactor, etc. or a power electronics
based variable source. All series controllers inject voltage in serieswith the
line. If the voltage is in phasequadraturewith the line, the seriescontroller only
suppliesor consumesvariablereactivepower. Any other phaserelationshipwill
involve real power also.
Tlte shunt controllers of Fig. 15.8c may be variable impedance, variable
source or a combination of these. All shunt controllers inject current into the
system at the point of connection. Combined series-seriescontrollers of Fig.
15.8dcould be a combination of separateseriescontrollers which are conffolled
in a coordinatedmanner or it could be a unified controller.
Combined series-shunt controllers are either controlled in a coordinated
ma-nneras in Fig. 15.8e or a unified Power Flow Controller with series and
shunt elementsas in Fig. 15.8f. For unified controller, there can be a real power
exchangebetween the series and shunt controllers via the dc power link.
Storagesource such as a capacitor,battery,superconductingmagnet,or any
other source of energy can be added in parallel through an electronic interface
to replenish the converter's dc storage as shown dotted in Fig. 15.8 (b). A
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
in powerSystems
Compensation
ModernpowerSystem
Analysis
ff.*J
controller with 'storage is much more effective for conffolling the system
dynamics than the corresponding controller without storage.
tffi*ffil
T-
Line
Llne
Line
(a)General
symbolfor
FACTScontroller(FC)
(a)
l
u
n
[,--TTlFcl
-T-
"
I
l
(c) Shuntcontroller
<-oc
ac lines
(d) Unifiedseriesseriescontroller
Fig. 15.9 (a) STATCOMbasedon voltage-sourced
and (b) current-sourced
converters.
. A combination of STATCOM and any energy source to supply or absorb
powei is called static synchronous generator(SSG). Energy source may be a
battery, flywheel, superconducting magnet, large dc storage capacitor, another
rectifi erlinverter etc.
Statlc
!-.
I it
FCH
Coordinated
control
(e) Coordinated
series
and shuntcontroller
dc power
link
(f) Unifiedseriesshuntcontroller
(b)
/
Series
Synchronous
Compensator
(SSCC)
It is a seriesconnectedcontroller. Though it is like STATCOM, bu(its output
voltage is in serieswith the line. It thus controls the voltage acrossthe line and
hence its impedance.
Interline
Power
FIow
Controller
(IPFC)
The group of FACTS controllers employing switching converter-based
synchronous voltage sourcesinclude the STATic synchronous COMpensator
(STATCOM), the static synchronous series compensator (SSCC), the unified
power flow controller (UPFC) and the latest, the Interline Power Flow
Controller (IPFC).
This is a recently introduced controller 12,3]. It is a cornbinationof two or
more static synchronousseriesconlpensatorswhich are coupledvia a common
dc link to facilitate bi-directional flow of real power between the ac terminals
of the SSSCs, and are controlled to provide independent reactive series
compensation for the control of real power flow in each line and maintain the
desired distribution of reactive power flow among the lines. Thus it managesa
comprehensiveoverall real and reactive power managementfor a multi-line
transmissionsystem.
STATCOM
Unified
Flg. 15.8 DifferentFACTScontrollers
STATCOM is a static synchronousgeneratoroperatedas a shunt-connected
static var compensator whose capacitive or inductive output current can be
i;orrrr-oiieornoepenoentoi rhe ac system voltage. The STATCOM, like its
conventional countetpart, the SVC, controls transmission voltage by reactive
shunt compensation.It can be based on a voltage-sourcedoi curtent-sourced
converter. Figure 15.9 shows a one-line diagram of srATCoM based on a
voltage-sourcedconverterand a current sourcedconverter. Normally a voltagesource converter is preferred for most converter-basedFACTS controllers.
STATCOM can bc designedto be an active filter to absorb system harmonics.
^---^--^ll-l
!
t
a
,
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Power
FIow
Controller
(UPFC)
This controller is connectedas shown in Fig. 15.10. It is a combination of
STATCOM and SSSC which are coupled via a iommon dc lin-k to -allow
bi-directional flow of real power between the series output terminals of the
SSSC and the shunt output terminals of the STATCOM. These are controlled
fo provide concurrent real and reactive series line compensation without an
external energy source.The.UPFC, by meansof angularly unconsffainedseries
voltage injection, is able to control, concurrently/simultaneouslyor selectively,
the transmission line voltage, impedance, and angle or, alternatively, the real
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and reactive line flows. The UPFC may also provide independentlycontrollabte
shuntreactive compensation.
stability and cnnffol line flows. Voltage source converter based (selfcommutated) HVDC system may have the same features as those of
STATCOM or UPFC. This system also regulatesvoltage and provides system
A comparative perfonnance of major FACTS controllers in ac system is
given in Table 15.2 U4l.
Table 15.2 A comparativeperformanceof major FACTS controller
STATCOM
Fig. 1S.10 UnifiedpowerFlowControilerUpFC
Th yri s t or- co n tro I I e d p h a s e -sh ifti n g Tra n sfo rm er (TCp s r)
This controller is also called Thyristor-controlled Phase Angle Regulator
(TCPAR). A phaseshifting transformer controlled by thyristor switches give
to
a rapidly variable phase angle.
Th yristor-
Power
Controller
(IpC)
A series-connected
controller of active and reactive power consisting,in each
phase, of inductive and capacitive branches subjected to separately phaseshifted voltages. The active and reactive power can be set inpedendently
by
adjusting the phase shifts and/or the branch impedances,using mechanical
or
electronic switches.
Thyristor
SVC/STATCOM
TCSC
SSSC
TCPAR
UPFC
V
control
Transient
Oscillation
flow control
stability
Damping
X
XXX
XX
XX
x
x
xxx
xxx
XXX
XX
Load
XXX
XX
XXX
XX
X
XX
XXX
XXX
XXX
XXX
* a;v-slrong influence; xx-average
influence; x-smail infruence
Con troll e d VoI tdgre R egu trator flCVR)
A thyristor controlled transformer which can provide variable in-phase voltage
with continuouscontrol.
Interphase
Type of FACTS
Controller
Controlled
Braking
Resistor
ICBR)
It is a shunt-connectedthyristor-switched resistor, which is controlled to aid
stabilization of a power system or to minimise power acceleration of a
generatingunit during a disturbance.
Thyristor-controlled
Voltage Limiter
FCVL)
A thyristor-switched
metal-oxidevaristor (Mov) used to limit the voltage
acrossits terminalsdurins transientconditions
'TIVDC
It may be noted that normally HVDC and FACTS are complementary
technologies.
The role of HVDC, for economicreasons,is to interconnectac
system?,where
a reliableac interconnection
would be too expensive.HVDC
www.muslimengineer.info
sutyffvtARy
Since the 1970s,energy cost, environmentalrestrictions, right-of-*ay difficulties, along with other legislative sociai and cost problems huu. postponed the
construction of both new generation and transmission systemsin India as well
as most of other countries. Recently, becauseof adoption of power reforms or
restructuring or deregulation, competitive electric energy markets are being
developedby mandatingopen accesstransmissionservices.
In the late 1980s,the vision of FACTS was formulated. In this various power
electronics basedcontrollers (compensators)regulate power flow and transmission voltage and through fast control aciion, mitigate dynamic disturbances.
Due to FACTS, transmissionline capacity was enhanced.Two types of FACTS
controllers were developed. One employed conventional thyristor-switched
capacitorsand react<lrs,and quadraturetap-changingtransformerssuch as SVC
and TCSC. The second category was of self-commutatedswitching converters
as synchronousvoltage sources,e.g. STATCOM, SSSC, UPFC and IpFC. The
two groups of FACTS controllers have quite different operating and performance characteristics.The second group usesself-commutated dc to ac converter.
The converter, supported by a de power supply or energy storage de.,,ice can
also exchangereal power with the ac systembesidesconmollingreactive power
independently.
The increasing use of FACTS controllers in future is guaranteed.What
benefits are required for a given system would be a principal justification for
the choice of a FACTS controller. Its final form and operation will, ofcourse,
depend not only on the successful development of the necessarycontrol and
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
r:5?4i I
Modern power System Analysis
I
communieationteehnologiesand protocols, but also on the final structure of ihre
evolving newly restructuredpower systems.
REFERE
NCES
Books
I ' Chakrabarti,A', D.P. Kothari and A.K. Mukhopadhyay, Performance Operation
and Control of EHV Power TransmissionSystems.Wheeler, New Delhi, 1995.
2. Hingorani, N.G. and Laszlo Gyugyi, (Jnderstanding FACTS. IEEE press, New
York, 2000.
3. Song, Y.H' and A.T. Johns,Flexible AC TransmissionSystems,IEE,Lbndon, 1999.
4. Miller, T'J.E., ReactivePower Control in Electric Systems,John Wiley and Sons,
NY, lgg2.
5. Nagrath' I.J. and D.P. Kothari, Electric Machines, 2nd edn, Tata McGraw-Hill.
New Delhi, 1997.
6. Taylor, c.w., power system vortage stability, McGraw-Hill, singapore, 1994.
7. Nagrath,I.J and D.P. Kothari, power S),stemEngineering, TataMcGraw_Hill, New
Delhi, 1994.
8. Indulkar, C.S. and D.P. Kothari, Power SystemTransientsA StatisticalApproach,
Prentice-Hallof India, New Delhi, 1996.
9. Mathur, R.M' and R.K. Verma, Thyristor-BasedFACTI; Controllers
for Electrical
TransmissionSystems.John Wiley, New york, 2002.
16.1
INTRODUCTION
Load forecasting plays an important role in power system planning, operation
and control. Forecasting means estimating active load at various load buses
ahead of actual load occurrence.Planning and operational applicationsof load
forecasting requires a certain 'lead time' also called forecasting iritervals.
Nature of forecasts,lead times and applicationsare summarisedin Table 16.1"
Papers
Table 16.1
Edris- A-, "FACTS Technologv Development: An Update".\EEE Pott'er Engirteering Rerieu'. \'ol. 20. lVlarch 2@1i. pp f9.
1l - Iliceto F- and E. Cinieri, "Comparative Analysis of Series and Shunt Compensation
Nature of
forecast
Lead time
Very short term
A few seconds to
several minutes
12. Kimbark.
Short term
Half an hour to
a ferv hours
13. CIGRE/ 'wG 38-01. .srrzricl/ar Conrp<,r-rdrorr.
CIGRE/. prrr-is.i 986
l{. Ptrvlr.f).. "Llsc'ofHDVC and F.ACTS".IEEE procc'edings.
vol. .\8. 2, Feb. 2000,
pp 235-245.
15. Kinrbark,E.W. "A New Look At Shunt Compensation.,IEEE Trans. Vol PAS102. No. l. Ja-n1983.nn 212-2!8
Medium term
A few days to
a t'ew' rveeks
A t'ew months to
a few years
I0-
Schemes for Ac Transmission svstems". IEEE Trans. pAs 96
6).
l8-10.
rg77. pp lglg-
"Hou,
E.W..
to Impror,e S.r'srem Stabilir-i, u,ithout
Risking
Suhs-rtchnrnous R('s(')nitncc"'. IEEE lr.-rrs. P.-tJ-. 96 t-i,I. Sept/Oc.t 1977. pp 160.S19.
www.muslimengineer.info
Long term
Application
Generation, distribution schedules,
contingency analysis for system
secunfy
Allocation of spinning reserve;
operarionilI planning ild unit
commitment; maintenance scheduling
Planning for seasonal peakwrnter. summer
Planning generation
growth
A good t'orecastreflecting current and future r.rends,tempered wi.Ji good
judgement, is the key to all planning, indeed.to financial success.The accuracy
of a forecast is crucial to any electric utility, since it determinesthe timing and
characteristicsof major system additions. A forecast that is too low can result
in low revenuefrom salesto neighbouringutilities or even in load curtailment.
Forecasts that are too high can result in severe financial problems due to
excessive investment in a plant that is not fully utilized or operatedat low
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
LoadForecasting
Technique
Llif,iffi
I
capacity factors. No forecastobtainedfrom analytical proceclurescan be strictly
r"ii.d upon the judgement of the forecaster, which plays a crucial role in
ariving at an acceptable forecast.
Choosing a forecasting technique for use in establishing future load
ing on nature o
requirements is a nontrivial task in itself.
another.
to
superior
be
may
variations, one particular methcd
The two approaches to load forecasting namely total load approach and
component approach have their own merits and demerits.Total load approach
has the merit that it is much smoother and indicative of overall growth trends
and easy to apply. On the other hand, the merit of the component approach is
that abnormal conditions in growth trends of a certain component can be
detected,thus preventing misleadingforecastconclusions.There is a continuing
need, however, to improve the methodology for forecasting power demand more
accurately.
The aim of the present chapter is to give brief expositions of some of the
techniques that have been developed'in order to deal with the various load
forecasting problems. All of these are based on the assumptionthat the actual
load supplied by a given systemmatchesthe demandsat all points of time (i.e.,
there has not been any outagesor any deliberate sheddingof load). It is then
possible to make a statisticalanalysisof previous load data in order to set up
a suitable model of the demandpattern.Once this has been done, it is generally
possible to utilize the identified load model for making a prediction of the
estirnated demand for the selectedlead time. A major part of the forecasting
task is thus concerned with that of identifying the best possible model for the
past load behaviour.This is best achievedby decomposingthe load demand at
any given point of tirne into a number of distinct components.The load is
dependent on the industrial, commercial and agricultural activities as well as the
weathercondition of the system/area.The weather sensitivecomponentdepends
on temperarure,cloudiness, wind velocity, visibility and precipitation. Recall
the brief discussions in Ch. 1 regarding the nature of the daily load curve which
has been shown to have a constantpari correspondingto the baseload and other
variable parts. For the sake of load forecasting, a simple decomposition may
serve as a cdnvenient starting point. Let y(k) representthe total load demand
(either for the whole or a part of the system) at the discretetime k = l, 2,3,
....It is generallypossibleto decomposey(k) into two parts of the form
y(k)= ya(k)+ y"(k)
(16.1)
where the subscript d indicates the deterministic part and the subscript s
indicates the stochasticpart of the demand. If k is consideredto be the present
time, then y(k + j), j > 0 would represbnta future load demand with the index
problem
7 being the lead time. For a chosenvalue of the indexT, the forecasting
processing
y(k
+/)
by
of
value
the
of
estimating
is then the sameas the problem
adequatedata fbr the past load dernand.
T6.2
FORECASTING
METHODOLOGY
Forecasting techniques may be divided into three broad classes.Techniques
may be based on extrapolation or on correlation or on a combination of both.
rrnlnlstlc,
Extrapolation
Extrapolation techniquesinvolve fitting trend curves to basic historical data
adjusted to reflect the growth trend itself. With a trend curve the forecast is
obtained by evaluating the trend curve function at the desired future point.
Although a very simple procedure, it produces reasonable results in some
instances.Such a techniqueis called a deterministic extrapolation since random
errors in the data or in analytical model are not accounted for.
Standard analytical functions used in trend curve fitting are [3].
(i) Straight line
! = a+ bx
(ii) Parabola'
!=a+bx+c*2
(iii) S-curve
!=a+bx+ci+dx3
(iv) Exponential
!=ce&
(v) Gempertz
! = In-r 7a + ced''1
' The most corlmon curve-fitting technique for fitting coefficients and
exponents(a4) of a function in a given forecastis the method of least squares.
If the uncertainty of extrapolated results is to be quantified using statistical
entities such as mean and variance, the basic technique becomes probabilistic
extrapolation. With regression analysis the best estimate of the model
describing the trend can be obtained and used to forecast the trend.
Correlation
Correlation techniques of forecasting relate system loads to various demographic and economic factors. This approach is advantageousin forcing the
forecaster to understand clearly the interrelationship between load growth
patterns and other measurable factors. The disadvantageis the need to forecast
demographic and economic factors, which can be more difficult than forecasting
system load. Typically, such factors as population, employment, building
permits, business,weather data and the like are used in correlation techniques.
No one forecasting method is effective in all situations. Forecasting
techniques must be used as tools to aid the planner; good judgement a:rd
experience can never be completely replaced.
16.3
ESTIMATION
OF AVERAGE
AND TREND TERMS
The simplest possible form of the deterministic part of y(k) is given by
ya &) = !-a + bk + e(k)
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(r6.2)
Technique
LoadForecasting
Modern PowerSystem Analysis
kt5f,*A
I
it
where larepresents the averageor the mean value of yd(k), bk representsthe
'trend' term that grows linearly with k and e(k) representsthe error of
modelling the complete load using the average and the trend terms only. The
questionis one of estimating the valuesof the two unknown model parameters
la aldb !o ensurea good model. As seenin Ch, 14, when little orlo st1listical
information is availableregardingthe error term, the methodof LSE is helpful.
If this method is to be used forestimating yo and b,the estimationindex "/is
defined using the relation
J - E{ez(D}
(16.3)
where E(.) represents the expectation operation. Substituting for e(A) from
Eq. (16.2) and making use of the first order necessaryconditions for the index
J to have its minimum value with respect to ya md b, it is found that the
following conditions must be satisfiedl2).
E { y a - y a & )+ b k I = 0
(16.4a)
E {bkz - ya(k)k+ tdkl - 0
(16.4b)
Since the expectation operationdoes not affect the constant quantities, it is
easy to solve these two equationsin order to get the desired relations.
ta= E{yd&)l- b{E(k)}
b - lE{ya&)kl- yo E{kllt4{k2l
(16.5a)
I
I
l
it
(16'6a) and
ln order to illustrate the nature of results obtainablefrom Eqs'
give the
which
16'l
Fig.
of
(16.6b), consider the clatashown in the graphs
jndust+ial
the
and
agricirttural
values of the
iopurerion in millionT. Th= cash
consumpdon
ooiput, in millions r.rfrupees and the amount of electrical energy
from
starting
years
seven
of
(toaA demand) in MWs 1n Punjab over a period
the
sampling
by
graphs
the
from
g5
data have been generated
1968. A total of
(16'6a)
and
Eqs'
in
substituted
graphs at intervals of 30 days. These have been
and the trend coefficients of the four
if O.OUIin order to compute the avetage
variables.The results are given in Table 16'2'
Table 16.2
Trend Cofficient
Average
Variable
13 million
Rs 397 million
Rs 420.9million
855.8MW
Population
Industrial output
Agricultural output
Load demand
o.2
0.54
0. 78
r.34
(16.5b)
If y(k) is assumedto be stationary(statisticsare not time dependent)one may
involve the ergodic hypothesis and replace the expectatiori operation by the time
averaging fonnula. Thus, if a total of N data are assumedto be avai.labLefor
determining the time averages,the two relations may be equivalently expressed
as follows.
(16.6a)
(16.6b)
-
peprJlation
in millions
---
lndustryin millionsof ruPees
millionsof rupees
Agriculture
Loaddemandin MW
-
/
t
These two relations may be fruitfully employed in order to estimate the average
and the trend coefficient for any given load data.
Note that Eqs. (16.6a) and (16.6b) are not very accuratein case the load
data behavesas a non-stationaryprocesssince the ergodic hypothesis does not
holcl for suchcases.It may still be possibleto assumethat the data over a finite
window is stationary and the entire set of data may then be consideredas the
juxtaposition of a number of stationaryblocks, each having slightly different
over the different
statistics.Equations(16.6a)and (16.6b)may then be repeatecl
for each
coefficient
trend
the
blocks in order to compute the average and
window of data.
www.muslimengineer.info
--/
,
,
i16
I t
the data
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
T
I
l
ModernpowerSystemAnalysis
ffi4
I
{
I
600
tI
-t--tr
I
"-ertthe load model maY be assumedto be
I
400
;:
I
,II
in
3
y(k)-fu3,+e(k)
(16.8a)
i:l
72
past load d
where the coefficient b,needto be estimated from the
index /<and would
model above is obviously a non-linear function of the time
approach to non-linear
need L coefficients to be estimated. A much simpler
form
modelling of the load is to introduce an exponential
( 16. 8b)
y( k) = c exp [ bk] + e( k)
reducing the number of
which involves only two unknown coefficients. Besides
of being readily
advantage
additional
the
has
model
unknowns, th" exponential
natural log of
the
take
is
to
required
is
that
All
form.
transformed into u linru,
is easily extended to estimate
the given data. In either case,the method of LSE
data.
the model parametersfrom the given historical
120
k (hours)-_---
t
I
i.
600
400
3
2oo
E
o
J
0
1
192
216
240
264
k (hours)---------
288
312
336
Fig. 16.2 Hourlyload behaviourof Delhiover two consecutive
weeks
Caution
The 85 data, used in Example 16.r, are generally not adequate
for making
statistical caiculations so that the vaiues given ubour may
not be entirely
adequate. In addition, the statistical characteristics of the
set of variables
concernedmay have changed (i.e., the data may in fact be non-stationary)
and
this alsomav inffoduce some error in the results.Finally, the graphs
in Fig. 16.1
are actually based on half yearly data obtained from the planning
commission
document and an interpolation processhas been employed in
order to generate
the monthly data. This may add some unspecified errors to the data
which will
also affect the accuracy of the estimates.
Prediction
|6,4ESTIMATIoNoFPERIoDIGcoMPoNENTS
periodic components in
The deterministic part of the load may contain some
example the
tn" uurrage and the porynomial rerms. consider for
;ilri'.o,',;
supplied
power
of the active
curve shown in Fig. tZ.Z wtrich givei the variation
load
daily
tlrg
that
is observed
by a power utility over a period of t*o weeks. It
fluctuations'
random
some
variations are repetitive from day to day except for
from those of the
significantly
differ
sundays
for
curve
the
that
It is also seen
out that the curve ior ihe
week days in view that Sundaysare holidays. I,,]rTt
of one Sunday till the mid
entire weekly period starting fiom, say, the mid night
periodic waveform with
night of the next Sunday behaves as a clistinctty
superposedrandom variations'
at an hourly interval,
If it is assumed that the load data are being sampled
so that the load pattern may
then there are atotal of 168 load data in one period
fundamental frequency ul
be expressed in terms of a Fourier series with the
load y(k) is then given by
i.f"g'.q" ut to Z7rl\68rads. Arsuitable model for the
of ya &+j)
Once the model for the deterministic component of the load
has been
determined, it is simple to make the prediction of its future value.
For the simple
model in Eq. (76-2), the desired prediction is computed using the
relation
v'(k + i) = T, + h(k + i\
/ 4
\
r /
J A
" U ' -
J l
(i6.7)
More General Forms of Models
Before leaving this section, it may be pointed out that the load
model may be
generalizedby including second and higher order terms on
the right hand side
www.muslimengineer.info
[a, sin iuk + b, cos iuk] + e(k)
y(k) = y + t
(15'9)
i=l
present and a; and b; are the
where L represents the total number of harmonics
components' only
cosinusoidal
the
and
amplitudes of respectively the sinusoidal
model'
the
in
rlnrrrinqntt',ormonicsnee-dto be included
to make a prediction
Once the harmonic load model is identified, it is simple
relation
the
of the future load ya(k + 7) using
(ltJllllllgrr!
lrq^r^\
9 a & + / ) = h ' ( k+ l ) i ( k )
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(16.10)
'eifrz#d
!6.5
ModernPo@is
LoadForecasting
Technique
ESTIMATION OF YS(^ft):TiME SERIESAPPROACH
h
= -t
15(ft)
v
Hffi
-T--
2
a, y, (k - r) +D b, w(k- ) + w(k) (16.14)
j:l
i:l
Estimation of two structural parameters n and rn as well as model parameters
ap bi and the variance d of the noise term w(k) is required. Moie complex
can be represented.The identification problem is solved off-line. The
acceptable load model is then utilized on-line for obtaining on-line load
forecasts. ARMA model can easily be modified to incorporate the temperature,
rainfall, wind velocity and humidity data [2]. In some cases,it is desirable to
show the dependenceof the load demand on the weather variables in an explicit
manner. The time series models are easily generalized in order to reflect the
dependenceof the load demand on one or more of the weather variables.
Auto-regressive Models
The sequence
y,(ft)is saidto satisfyan AR modelof ordern i.e.it is
[AR(n)],
if it can be expressed
as
16.6
ESTIMATION OF STOCHASTIC COMPONENT:
KALMAN
FILTERING APPROACH
n
y,(k)=
Do,r,(k
- i) + w(k)
(16.11)
i_l
lie inside the unit circle in the e-plane.
The problem in estimating the value of n is refer-redrn qc tha ntnhtaw )^s
The time series approachhas been widely employed in dealing with the load
forecasting problem in view of the relative simplicity of the model forms.
However, this method tends to ignore the statisticalinformation about the load
data which may often be available and may lead to improved load forecasts if
utilized properly. In ARMA model, the model identification problem is not that
simple. These difficulties may be avoided in some situations if the Kalman
'
,
filtering techniques are utilized.
Application
to Short-term
Forecasting
An application of the Kalman filtering algorithm to the load forecasting
problem has been first suggestedby Toyada et. al. [11] for the very short-term
and short-term situations.For the latter case,for example,it is possible to make
use of intuitive reasoningsto suggestthat an acceptablemodel for load demand
would have the form
v,(k) = y'(k) + v(li)
i,&) = -D a,y, (k- i)
(16.r2)
i:l
The variance d of w(k) is then estimated using the
relation
N
n 2 --
Auto-Regressive
(1lnn F
-Zrrt
\ r'l' )
\tr )
/,-rc
k:l
Moving-Average
( r 6 .l 3 )
Models
In some cases,the AR model may not be adequate
to represent-the observed
Ioad behaviour unless the order n of the model is
made very hrgfi. In such a case
ARMA (n, m) model is used.
www.muslimengineer.info
(16.15)
where y,(ft) is the observedvalue of the stochasticload at time ft, y,(k) is the
true value of this load and u(fr) is the error in the observedload. In addition.
the dynamics of the true load may be expressedas
y,(k +I) = y{k) + z(k) + u1&)
(16.16)
where a(k) represents the increment of the load demand at time k and u1@)
representsa'disturbanceterm which accountsfor the stochasticperturbationsin
y,(k).The incrementai ioaciitseif is assumerito remain constanton an average
at every time point and is modelled by the equation
z(k +l) = z(k)+ ur(k)
wherethe term uz&) representsa stochasticdisturbanceterm.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(16.r7)
a-.,.".lm
In order to make use of the Kalman filtering techniques, the noise terms
u(k), uz&) and u(k) are assumedto be zero mean independent white Gaussian
sequences. Also, the model equations are rewritten in the form
x (ft+l) = Fx(k) + Gu(k)
(16.18a)
k
v(k
where the vectorsx(ft) and u(ft) are definedas
x(k) = ly,(k) = (Df and u(k) = fur(k) uz&))r
The matrices4 G andh' arethen obtainedfrom Eqs. (16.15)-(16.17)
easily
and have the followingvalues.
=[1 1o1,
c
[1
1-l,
"=L o u ' L o l ' r=Lltlo l
Based on model (16.18), it is possible to make use of the Kalman filtering
algorithm to obtain the minimum varianceestimateof the vector x (k) basedon
the data y,(k): {y,(1), ),,(2) ... y,(k)}. This algorithm consisrsof the following
equations.
i (k/k) = i (ktt<- 1) + K" (k) ty"(ft)- h'ft(k/k-I)l
i(ktk -1)= F i((k-L)/(k-r))
f;g+ d)= pdi(tcttc)
(16.199)
In order to be able to make use of this algorithm for generating the forecast
of thg load
v-(k + d\. it is necessarv
fhaf the nnisc
cfeficrinc qnr{ o^-o
nr}ra-
information be available. The value of R(ft) may often be estimated from a
knowledge of the accuracy of the meters employed. However, it is very unlikely
that the value of the covariance Q(k) will be known to start with and will
therefore have to be obtained by some means. An adaptive version of the
Kalman filtering algorithm may be utilized in order to estimate the noise
statistics alongwith the state vector x(k) tZ). Now let it be assumedthat both
R(k) and Q(k) are known quantities. Let it also be assumed that the initial
estimate i (0/0) and the covarianceP"(0/0) are known. Based on thesea priori
information, it is possible to utilize Eq. (16.19a)-(16.19e) recursively to
processthe data for yr(l), yr(z), ..., yr(k) to generatethe filtered estimate ,(kl
k). Once this is available, Eq. (16.19g) may be utilized to ger,eratethe desired
load forecast.
(16.19a)
(16.19b)
K,(k) = P,(k/k-l) hlh' P,(klk-I)h + R(k)l-r
P-(k/k)= V - K,(k) h') P,(k/k-l)
P,(k/k - l)= FP,(k-l/k-l)F'+
to use the solution of Eq. (16.18a)for the vector x (k+ d)to get the result
GQG-DG',
(16.19c)
(16.19d)
(16.19e)
where,
Q(k) = covarianceof u(k)
R(ft) = covarianceof y(ft)
ft(klk) = filtered estimateof x(ft)
ft (k/k-L) = single steppretliction of x(ft)
K.r(k) = filter gain vector of same dimension as a(ft)
Pr(k/k) = filtering error covariance
To illustrate the nature of the results obtainable through the allorithm just
discussed, the data for the short term load behaviour for Delhi have been
processed.A total of 1030 data collected at the interval of 15 minuteshave bee.n
processed.It has been assumedthat, in view of the short time interval over
which the total data set lies, the deterministic part of the load may be assumed
to be a constant mean term. Using the sample average Formula (16.7) (with
b = 0), we get y = 220 Mw. The data for yr(k) have then been generated by
subtracting the mean value from the measured load data.
To process these stochastic data, the following a priori information have
been used:
R(k)= 3.74,
Pr(k/k-l) = predictionerror covariance
F r onr E q. (1 6 .1 8 b )o b ta i nth e p re d i c ti o ni ((k+ 1)l k)
From this the one step aheadload fbrecast is obtained as
(16.1e0
Y " ( f t +I ) = h ' i ( ( k + l ) l k )
It may be noted that filtering implies removal of disturbance or stochastic
term with zero mean.
It is also possible to obtain a multi-step aheadprediction of the load from the
multi-step ahead prediction of the vector x (k). For example, if the prediction
www.muslimengineer.info
=
i (oro)
[;]
The results of application of the prodietion Algonthnn (16.19) are shown in
Fig. 16.3.It is noted that the elror of 15 minutes aheadload predictionis around
8 MW which is about 3Voof the averageload and less than 2Voof the daily
maximum load.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I-,-r6rsHs
ECONOMETRTC MODELS
1. Actualload
2. 15 mts. predictionerror
If theloadforecastsarerequiredfor planningpurposes,it is necessary
to select
the lead time to lie in
thrro aJeulyears-In sucLcasesthe load demand should be decomposedin a manner that reflects the dependence
of the load on the various segmentsof the economy of the concernedregion. For
example, the total load demand y(ft) may be decomposedas
3
200
E
c
O
I
l
O
I
E
o
E
M
I
y(k)=D",y,(k)+ e(k)
I
l
180
(D
o
where aiare the regressioncoefficients, y,(k) arethe choseneconomic variables
and e(k) represents the error of modelling. A relatively simple procedure is to
rewrite the model equation in the tamiliar vector notation
y(k)=h'(k)x+ e(k)
(16.20b)
12.=1
9
l
L
I
I
o l
4
'1C
c
o
E
l
l
E
i3
14
where
n f L
15
16
17
18
h'(k)= [yr(ft)yz(k)... yu&)] and x = far a2... ayl.
The regression coefficients may then be estimated using the convenient least
squares algorithm. The load forecasts are then possible through the simple
relation
F i g . 1 6.3
Comment
i(k + 1) = i'(k) it(k +Ilk)
Application of Kalman filtering and prediction techniques is often hampered by
the non-availability of the required state variable model of the concernedload
data' For the few casesdiscussedin this section, a part of the model has been
obtainable from physical considerations.The part that has not been available
include the state and output noise variances and the data for the initiai state
estimate and the conesponding covariance. In a general load forecasting
situation, none of the model parameters may be available to start with and it
would be necessaryto make use of system identification techniquesin order to
obtain the required state model. It has been shown that the Gauss-Markov
model described by Eq. (16.18) in over-parameterisedfrom the model
identification point of view in the scnse that the data for y,(k) dr: not permit the
estimation of ali the parametersof this model. It has been shown in Ref.
[12]
that a suitable model that is identifiable andis equivalent to the Gauss-Markov
model for state estimation purposesis the innovation model useclfor estimation
of the stochastic component.
on'line
Techniques
f,or Non-stationanl
Load predietion
Most practical load data behaveas non-stationaryand it is therefore imponant
to consider the questionof adaptingthe techniquescliscusseclso f,arto the nonstationarysituation. Ref. [2] has discussedthe threemodels for this purposeviz.
(i) ARIMA Models, (ii) Time varying model and (iii) Non-dynamic models.
www.muslimengineer.info
(16.2Oa)
i:l
1 6 =
\
(16.21)
where i (k) is the estimate of the coefficient vector based on the data available
till the ftth sampling point ana fr6 + Uk) is the one-step-aheadprediction of the
regressionvector h(k).
16.8
REACTTVE LOAD FORECAST
Reactive loads are not easy to forecast as compared to active loads, since
reactive loads are made up of not only reactive components of loads, but also
of transmissionand distribution networks and compensatingVAR devicessuch
as SVC, FACTs etc. Therefore, past data may not yield the correct forecast as
reactive load varies with variations in network configuration during varying
operating conditions. Use of active load forecast with power factor prediction
may result in somewhat satisfactory results. Of course, here also, only very
recent past data (few minutes/hours) may be used, thus assuming steady-state
network configuration. Forecasted reactive loads are adapted with current
reactive requirements of network including var compensation devices. Such
rorecasts are neeoed ior secunty anaiysls, voitageireactivepower scheduling
etc. If control action is insufficient, sffuctural modifications have to be carried
out, i.e., new generating units, new lines or new var compensatingdevices
normally have to be installed.
f-
-
-
|
|
a
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModrrnPo*"r Syrt"r An"turi.
mitul
sututttilARy
Load forecastingis the basic step in power systemplanning. A reasonablyselfcontainedaccount of the various techniquesfor the load prediction of a modern
prediction problems. Applications of time series,Gauss-Markovand innovation
models in setting up a suitable dynamic model for the stochasticpart of the load
datahave been discussed.The time seriesmodel identification problem has also
been dealt with through the least squaresestimation techniques developed in
Chapter 14.
In an interconnectedpower system, load forecasts are usually needed at all
the important load buses. A great deal of attention has in recent years been
given to the question of setting up the demand models for the individual
appliances and their impact on the aggregated demand. It may often be
necessaryto make use of non-linearforms of load models and the questionof
identification of the non-linear models of different forms is an important issue.
Finally, a point may be made that no particular method or approach will
work for all utilities. All methods are strung on a common thread, and that is
the judgement of the forecaster. In no way the material presented here is
exhaustive. The intent has been to introduce some ideas currentlv used in
forecastingsystem load requirements.
Future
Trends
Forecastingelectricity loads had reacheda comfortable stateof performance in
the yearspreceding the recent waves of industry restructuring. As discussedin
this chapter adaptive time-series techniques based on ARIMA, Kalman
Filtering, or spectral methods are sufficiently accurate in the short term for
operational purposes, achieving errors of l-2%o. However, the arrival of
competitive markets has been associated with the expectation of greater
consumerparticipation. Overall we can identify the following trends.
(i) Forecast errors have significant implications for profits, market shares,
and ultimately shareholdervalue.
(ii) Day ahead, weather-based,forecasting is becoming the rnost crucial
activity in a deregulated market.
(iii) Information is'becoming commercially sensitive and increasingly trade
secret.
(iv) Distributed, embeddedand dispersedgeneration rhay increase.
A recentpaper [7] takesa selectivelook at some of the forecastingissueswhich
are now associatecl
with decision-makingin a competitivemarket.Forecasting
loads and prices in the wholesale markets are mutually intertwined activities.
Models based on simulated artificial agents may eventually become as
important on supply side as artificial neural networks have already become
www.muslimengineer.info
li':l;{lrJ.
**
and integrationwith conventionaltime-seriesmethodsin order to provide a
more preciseforecasting.
REVIEWQUESTIONS
L6.7 Which method of load forecasting would you suggestfor long term and
why?
16.2 Which method of load forecasting would you suggestfor very short term
and why?
16.3 What purposedoes medium term forecastingserve?
16.4 How is the forecaster'sknowledge and intuition consideredsuperior to any
load forecasting method? Should a forecaster intervene to modify a
forecast,when, why and how?
16.5 Why and what are the non-stationarycomponentsof load changesduring
very short, short, medium and long terms?
NCES
REFTRT
Books
1. Nagrath I.J. and D.P. Kothari, Power SystemEngineering,Tata McGraw-Hill, New
Delhi, 1994.
2. Mahalanabis,A.K., D.P. Kothari and S.I. Ahson, Computer Aided Power System
Analysis and Control, Tata McGraw-Hill, New Delhi, 1988.
3. Sullivan, R.L., Power SystemPlanning, McGraw-Hill Book Co., New York, 1977.
4. Pabla,A.5., Electricol Power Systems
Planning,Macmillan lndia Ltd., New Delhi,
1998.
5. Pabla,4.5., Electric Power Distribution,4th Edn, Tata McGraw-Hill, New Delhi.
1997.
6. Wang, X and J.R. McDonald (Eds), Modern Power SystemPlanning, McGrawHill, Singapore,1994.
Papers
7. Bunn, D.W., "ForecastingLoads and Prieesin CompetitivePowerMark-ets"-Pracof the IEEE, Vol. 88, No. 2, Feb 2000, pp 163-169.
8. Dash,P.K. et. al., "Fuz,zy Neural Network and Fuzzy Expert System for Load
Forecasting",Proc. IEE, Yol. 143, No. l, 1996, 106-114.
9. Ramanathan,R. ef. a/., "Short-run Forecastsof Electricity Loads and Peaks". lnt
J, Forecasring,
Vol. 13, 1997,pp 161-174.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
@
Mod"rnpo*", syrt"r An"ly.i.
10. Mohammed,A. et. al., "Short-term Load Demand Modelling and ForecastingA
Review," IEEE Trans. SMC, Vol. SMC-12, No. 3, lggl, pp 370_3g2.
ll. Tyoda, J. et. al., "An Application of State Estimation to Short-term Load
Forecasting",IEEE Trans., Vol. pAS-89, 1970, pp l67g-16gg.
77
12. Mehra, R.K., "On-line Identification of
to Kalmann Filtering", IEEE Trans. Vol. AC-16, lg7l, pp lZ_21.
T7.T
INTRODUCTION
Voltage control and stability problems are very much familiar to the electric
utility industry but are now receiving special attention by every power system
analyst and researcher.With growing size alongwith economic and environmental pressures, the possible threat of voltage instability is becoming
increasingly pronounced in power system networks. In recent years, voltagi
instability has been responsible for several major network collap$s in New
York, France, Florida, Belgium, Sweden and Japan [4, 5]. Researchworkers,
R and D organizations and utilities throughout the world, are busy in
understanding, analyzing and developing newer and newer strategiesto cope up
with the menace of voltage instability/collapse.
Voltage stability* covers a wide range of phenomena. Because of this,
voltage stability meansdifferent things to different engineers.Voltage stability
is sometimes also called load stability. The terms voltage instability and voltage
collapse are often used interchangeably.The voltage instability is a dynamic
process wherein contrast to rotor angle (synchronous) stability, voltage
dynamics mainly involves loads and the means for voltage control. Voltage
collapse is also defined as a processby which voltage instability leads ro very
low voltage profile in a significant part of the system. Voltage instability limit
is not directly correlated to the network maximum power transfer limit.
A CIGRE Task Force [25] has proposedthe following definitions for voltage
stability.
Small-disturban-ce
voltage
stability
A power system at a given operating state is small-disturbancevoltage stable
if, following any small disturbance,voltagesnear loads do not changeor remain
*The problemof voltage stability hasalreadybeenbriefly rackled
in Ch. 13. Here
it is again discussedin greaterdetailsby devotinga full chapter.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
lEq!tl#*
l,rc'tr
close to the pre-disturbancevalues. The concept of small-disturbancevoltage
stability is related to steady-statestabiiity and can be analysed using smallsignal (linearised) model of the system.
voltagestabilityare relatedto
considered.
Many of the indicesusedto assess
NR load flow study. Detailsof static and dynamicvoltage stability will be
furtherin Section17.5.
considered
Voltage
Some Counter Measures
Stability
to a certain disturbance,the voltages near loads approachthe post-disturbance
equilibrium values.
The concept of voltage stability is related to transient stability of a power
system. The analysis of voltage stability normally requires simulation of the
system modelled by non-linear differential-algebraic equations.
Voltage
Collapse
Following voltage instability, a power system undergoesvoltage collapse if the
post-disturbanceequilibrium voltages near loads are below acceptablelimits.
Voltage collapse may be total (blackout) or partial.
Voltage security is the ability of a system, not only to operate stably, but
also to remain stable following credible contingencies or load increases.
Although voltage stability involves dynamics, power flow based static
analysis methods often serve the purpose of quick and approximate analysis.
T7.2
COMPARISON
OF ANGTE AND VOLTAGE
STABILITY
The problern of rotor angle (synchronous)stability (coveredin Ch. 12) is well
understood and documented t3l. However, with power system becoming
overstressedon accourrt of economic and resource constraint on addition of
generation, transfofiners, transmissionlines and allied equipment, the voltage
instability has become a serious problem. Therefore, voltage stability studies
have attracted the attention of researchersand planners worldwide and is an
active area of research.
Real power is related to rotor angle instability. Similarly reactive power is
central to voltage instability analyses.Deficit or excessreactive power leads to
voltage instability either locally or globally and any increasein loadings may
lead to voltage collapse.
Voltage
Stability
Studies
The voltage stability can be studied either on static (slow time frame) or
dynamic (over long time) considerations.Depending on the nature of disturbance and system/subsystemdynamics voltage stability may be regardeda slow
or fast phenomenon.
Static Voltage
lty are:
n counter measuresto avord voltage r
(i) generator terminal voltage increase (only limited control possible)
(ii) increase of generator transformer tap
(iii) reactive power injection at appropriate locations
(iv) load-end OLTC blocking
(v) strategic load shedding (on occurrenceof undervoltage)
Counter measures to prevent voltage collapse will be taken up in
Section 17.6.
T7,3
POWER FLOW AND VOLTAGE
REACTIVE
COLLAPSE
Certain situations in power system cause problems in reactive power flow
which lead to system voltage collapse.Some of the situations that can occur are
listed and explained below.
(1) Long Transmission Lines.' In power systems, long lines with voltage
uncontrolled buses at the receiving ends create major voltage problems
during light load or heavy load conditions.
(ii) Radial TransmissionLines: In a power system,most of the parallel EHV
networks are composedof radial transmissionlines. Any loss of an EHV
line in the network causes an enhancementin system reactance. Under
certain conditions the increasein reactive power delivered by the line(s)
to the load for a given drop in voltage, is less than the increasein reactive
power required by the load for the same voltage drop. In such a case a
small increasein load causesthe system to reach a voltage unstable state.
(iii) .Sftortageof Local Reactive Power: There may occur a disorganised
combination of outage and maintenance schedulethat may cause localised
reactive power shortageleading to voltage control problems.Any attempt
to import reactive power through long EHV lines will not be successful.
Under this condition, the bulk system can suffer a considerable voltage
drop.
I7.4
MATHEMATICAL
crrrr,lltE l.? Trnl't
rit .l l.l.EDllrr
I I
FORMULATION
OF VOLTAGE
littD/^t
T Erilt
r.ClLr.Cr!.GrlVl
Analysis
Load flow analysis reveals as to how system equilibrium values (such as
voltage and power flow) vary as various system parametersand controls are
changed.Power flow is a static analysistool wherein dynamicsis not explicitly
www.muslimengineer.info
The slower forms of voltage instability are normally analysed as steady state
problems using power flow sirnulation as the primary study method. "Snapshots" in time following an outage or during load build up are simulated.
Besides these post-disturbance power flows, two other power flow based
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
I 's-ls{
methodsareoftenused;pv curvesandve curves.(seealsosec. 13.6)
These
two methodsgive steady-state
loadabilitylimits which are relatedto vohags
V
-'<-
stability. conventional load flow programs can be used for
approxilmate
analysis.
P-V. curves are useful for conceptual analysis of voltage
0.9 pf lead
stability and for
The model that will be employed here to judge voltage stability
is based on
a single line performance. The voltage performun"" of this
simile system is
qualitatively similar to that of a practical system with
many voltage sources,
Ioads and the network of transmissionlines.
Considerthe radial two bus system of Fig. 17.1. This is the
same diagrarn
as that of Fig. 5.26 except that symbols are simplified. Here
Eis 75 and yis
vn and E and v are magnitudes with E leading v by d,
Line angle"p: tunli
XlR and lzl = X.
Locus of V66 and Pr",
Noseof the curve
0.8 pf
lag
Fig.17.2 PV curvesfor variouspowerfactors
As in the case of single line systerrr,,r, a general power system, voltage
instability occurs above certain bus loading and certain Q injections. This
condition is indicated by the singularity of the Jacobianof Load Flow equations
and level of voltage instability is assessedby the minimum singular value.
Certain results that are of significancefor voltage stability are as under.
o Voltage stability limit is reached when
(r7.2)
F i g . 1 7.1
In terms of P and e,the systemloaclenclvoltage can be expressed
as
where
-1-:
r
, - z z x - z : -- _i 'il r z o x - E ) ' - 4 x 2 G 2 + o \ l , , _ < \
L
z
e7.t)
It is seenfrom Eq. (17.1) that Vis a double-valuedfunction (i.e.
ithas two
solutions) of P for a particular pf which determines in terms
of p. The pV
e
curves for various values of pf are plotted in Fig. 17.2. For
each value of pf,
the higher voltage solution indicates stable voltage case,
while the lower
voltage lies in the unstable voltage operation zone. -fhe
changeoveroccurs at
v".t (critical) and Pro*. The locus of v.r,-p^u* points for variJus
pfs is drawn
in dotted line in the figure. Any attempt to iniiease the load abov"
causes
a reversal of voltage and load. Reducing voltage causes an
"-* current
increasing
to be drawn by the load. In turn the larger reactive line drop
causesthe voltage
to dip further. This being unstableoperation causesthe system
to suffer voltage
collapse. This is also brought out by the fact that in upper part
of the curve
ff.
d
P
n flrrv
l r a rvwwr
l^.',o^^*
/..^.,+^Ll^ --,--\
0 and i^rr
panL \u'Dr.rurc
pdtL)
> U (feduclng
lOad means
dV
g,,,V
reducing voltage and vice-versa).It may be noted here that
the type
-tir" of load
assumedin Fig' 17.2 is constantimpedance. In practical syste-.
type of
loads are mixed or predomirrantlyconstantpower type such that
system voltage
degraclationis inore and voltagelnstability occurs much prior
to the theoretical
power limit.
www.muslimengineer.info
S = complex power at load bus
[l].
Yrt= load bus admittance
',
V = load bus voltage
in Eq. (17.2) to unity, lesserthe stability margin.
Nearer the magnitude
'a
o The loading limit of transmissionline can.be determinedfrom
lsl = v"3lx";
(r7.3)
X".i is the critical system reactancebeyond which voltage stability is lost. It can
be expressedas
Fz
(-tan Q+sec Q)
X"n= *
2P
(r7.4)
We have so far consideredhow the PV characteristicswith constant load
power factor affect the voltage stability of a system. A more meaningful
charrcteristic for certain aspectsof voltage stability is the QV characteristic,
which brings out the sensitivity and variation of bus voltage with respect to
reactive power injections (+ve or -ve).
Consideronce again the simple radial systemof Fig. 17.1.For p flow it is
sufficiently accurate to assume X > R i.e. 0 = 90".It then follows that
o = EY.o,a- 11
X
X
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(17.s)
a
or
V" - EV cos 6+ QX = 0
Taking derivativewrt V gives
(r7.6)
d Q = E c o s6 - 2 V
d
V
X
Using ihe decoupiing principle 1.". dP = 0, we set
dV
+=."'r[#.+]
(17.7)
The QV characteristic on normalized basrs(etf**.
VIE) for various values
of P/P^ are plotted in Fig. 17.3.The system is voltage stable
in the region
where dQldv is positive, while the voltage stability limit is reached
at d,eldV
= 0 which may also be termed as the critical
operating point.
o
or
I s c = c odsl o O + 1 1
LdV
or
Ers6= E cos5l !9 + 2Y]
LdV
Voltagestability is achievedwhen
,D max
E cos , (#
Unstable. .
operation
P I P ^ " , =9 . 5
0.2
0.6
0.8
1.0
Xsource
a o2
(17.8)
Stability
..it.rion: (E'=generatorvoltage; V=load voltage). Using this crite-
rion, the voltage stability limit is reachedwhen
cos
d{#.#}+sindffi-}=o
www.muslimengineer.info
(r7.1r)
\
(iii) Ratio of source to load reactance is very important
dnd for voltage
stability
The inferencesdrawn from the simple radial system qualitatively
apply to a
practicalsize system.Other factors that contribute to system
voltage collapse
are: strengthof transmissionsystem,power transfer levels, load chaiacteristics,
generatorreactive power limits and characteristicsof reactive
power compensating devices.
(i) 34
(17.10)
dZ
dV
Application of this criterion gives value of z";.
VIE
= t-"o'26
Q,n^
of Voltage
(shortcircuitMVA of powersource)
v r - = uM
-d= o o o
The limiting value of the reactive power transfer at the limiting
stageof
voltage stability is given by
Criteria
> Ers,
dV
voltage instability occurs when the system Z is such that
Fig- 17.3 QV characteristics
for the systemof Fig. 17.1 tor
variousvaluesof plp^^r.
Other
+)
X J
,..\ dz cntenon
(lU -
0.75
0
.
Pr", is the maxlmum
powertransferat upf
1.0
X J
(17.r2)
X load
a indicates the off-nominal tap ratio of the OLTC transformer
T7.5
VOLTAGE
STABILITY
at the load end.
ANALYSIS
The voltage stability analysis for a given system state
involves examining
following two aspects.
(i) Proximity to voltage instabitity: Distance to instability
may be measured
in terms of physical quantities,such as load level, rea power
flow through
a critical interface, and reactive power reserye.posiible
contingencies
such as a line outage,loss of a generating unit or.a reactiu.
po*"rio*..
must be given due consideration.
(ii) iuiechanismof voitage instabiiity: How and why
does voltage instabitity
take place? What are the main factors leading to instability?'ulhat
are the
voltage-weak areas?What are the most effective ways
to improv"
stability?
"of,ug,
(r7.e)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.ffiiffi|
.
Analysis
ModernPowerSystem
of system
The static analysis techniques permit examination of a wide range
give
the main
and
problem
the
of
nature
the
conditions and can descriUe
specific
of
study
detailed
for
useful
is
analysis
contributing factors. Dynamic
testing
and
controls,
and
protection
of
coordination
voltage coliapse situations,
the
how
and
whether
us
tell
further
simulations
of remedial rneasures.Dynamic
wr
point
steady-stateequilibrium
Requirements
Modelling
of various
Power system
components
systemconditionsat variousdme
The staticapproachcaptures.snapshotsof
frames along the time-domain trajectory.
At each of these time frames' X in
to purely
redPce
frame.Thus,theoverausystemequations
ii h?lrJdil.
allowing'h.t..Ytt:f ,t'1tit Tlv:i: ::*:1,::t"t' vlp qnrtvo
equations
algebraic
andvQ
by computin
s
"'?'J"6;ffiffi;;;g."riuulirv
is determrned-
.vP
Loads
Detailed
Load modelling is very critical in voltage stability analysis.
required'
be
may
area
voltage-weak
a
in
representation
subiransmissionsystem
and
This may includeiransformer ULTC action, reactive power compensation'
voltage regulators
of loads'
It is essential.to consider the voltage and frequency dependence
Induction motors should also be modelled'
and their
Generators
excitation
controls
the AVR, load
It is necessary to consider the droop chatacteristics of
controls should
and
protection
AGC,
compensation,SVSs (static var system),
also be modelled appropriatelyl4l.
Dynamic
Static AnalYsis
AnalYsis
analysis is
generalstructurc of the systcm moclel for voltage stability
may be
equations
system
overall
similar to that for transient stability analysis.
cxprefihcdas
'lhe
(r7.r3)
*=f(X,n
and a set of algebraicequations
I (X, V) = YxV
with a set of known initial conditions
X = system state vector
where
Y
(r7.r4)
(Xo' Ve)'
= but voltage vector
1= current injeition vector
Ilv = rletwork node admittance matrix'
be s
Equations(17.13)and (17'14)can
methods
integration
oi tt e numerical
in Ch' 6' T
clescribed
analysisnrethocls
repl
models
special
the
,ninutes. As
bee
have
collapse
ieading to voltage
considerably
is
differential equations
models.Stiffnessis also called synchrc
www.muslimengineer.info
using static analysis havb been
curves at selected load buses.Special techniques
VQ sensitivity such as eigenvalue (or
reported in literature. Methods based on
methods give stability-related
modal) analysis have been devised. These
and also identify areasof potential
information from a system-wide perspective
problemst13-151.
Proximity
to InstabilitY
by increasing
proximity to small-disturbance voltage instability is determined
flow fails
load
the
or
unstable
in steps until,the system becomes
il;^";ruiion
the point
determining
for
techniques
to converge.Refs. t16-181 discussspecial
instability'
of voltage collapse and proximity to volage
The Continuation
Power-flow
Analysis
at the voltage stability limit. A! a result'
The Jacobian matril becomes singular
have convergenceproblems at operating
conventional toad-flow algorith*, rnuy
continuation power-flow analysis
The
conditions near the stability limit.
load-flow equationsso that they
the
overcomesthis problem by reformulating
the
possible loading conditions. This allows
remain well-conditioned at all
P-V
the
of
forioth upper *d lo*"t portions
solution of load-flow problem
*fi"t:lltinuation-method
and flexible and
of power-flow analysis is ,obrrrt
However'
difficulties'
with convergence
is to
approach
better
the
suming. Hence
continuaand
(NR/FDLF)
flow irethod
:ase, LF is solved using a conventional
levels
ns for successivelyincreasingload
is
method
Hereafter, the continuation
is
method
ns. Normally, the continuation
point'
critical
I exactly at and past the
Voltage
StabilitY
with
HVDC Links
for extremely long distance
(HVDC).litt
-tl:1r:t.:d
High voltage direct current*
ffansmissiclnanclftrrasynchronousinterconnections.AnHVDClinkcanbe
*For dctailcd accountof HVDC, the reader
maY refer to [3]'
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
'
ModernpowerSystemAnatysis
f00 |
either a back-to-back rectifier/inverter link or can include long
distance dc
transmission. Multi-terfninal HVDC links are also feasible.
The technology has come to such a level that HVDC terminals
can be
connected even at voltage-weak points in power systems.
HVDC links mav
present unfavourable "load" characterisficsfo fhe nnrr/rr
converter consumesreactive power equal to 50-60vo of the
dc power.
HVDC-related voltage control (voltage stability and fundamental
frequency
temporary over voltages)may be studied using a transient
stability program.
Transient stability is often interrelated with voltage stability.
Ref. t2i .onJia.r,
this problem in greaterdetail.
17.6
PREVENTION
OF VOLTAGE
COLLAPSE
(i) Application of reactive power-compensating devices.
Adequate stability margins should be ensured by proper selection
of
compensationschemesin terms of their size, ratingr-und locations.
(ii) control of network voltage and generator reactive
output
Several utilities in rhe world such as EDF (France), ENEL (Italy)
are
developing specialschemesfor control of network voltages and
reactive
power.
(iii) Coordination of protections/controls
Adequatecoordinationshould be ensuredbetweenequipment protections/
controls basedon dynamic simulation studies. Tripping of equipment
to
avoid an overloadedcondition should be the last alternative. Controlled
system s€parationand adaptive or irrtelligent control could also be
used.
(iv) Control of transfurmer tap chan.gers
T'apchangerscan be controlled, either locally or centrally, so as to reduce
the risk of voltage collapse. Microprocessor-basedOLTC controls
offer
almost unlimitedflexibility for implementingULTC control strategies
so
as to take advantageof the load characteristics.
(v) Under voltage load shedding
For unplanned or extreme situations, it may be necessary to
use
undervoltageload-sheddingschemes.This is similar to under ir"q1r"rr.y
load shedding,which is a common practiceto deal with extreme situationsresulting from generationdeficiency.
Strategic load sheddingprovides cheapestway of preventing widespread
VOltaSg
CO
l l a n s_e
_-___r
Lnarl
sherldino
cnl,a-oo
uvrrvrrrvJ
olrn,,r,r L^
orr\_rrll\.t
ug
r^^: -^^)
ugsrBlrtru
differentiate berween faults, transient voltage dips unJ lo*
conditions leading to voltage collapse.
(vi) Operators' role
^^
su
-as
-t()
voltage
Operators must be able to recognise voltage stabiiity-related symptoms
and take requiredremedial actions to prevenr voltage collapse.On-line
www.muslimengineer.info
VoltaseStability
I
__{-tr*g
monitoringand analysisto identify potentialvoltage stabilityproblems
are extremelyhelpful.
and appropriateremedialmeasures
T7.7 STATE-OF-THE.ART, FUTURE TRENDS AND
CHALLENGES
The present day transmission networks are getting more and more stresseddue
to economic and environmental constraints.The trend is to operatethe existing
networks optimally close to their loadability limit. This consequentlymeansthat
the system operation is also near voltage stability limit (nose point) and there
is increasedpossibility of voltage instability and even collapse.
Off-line and on-line techniquesof determining state of voltage stability and
when it entersthe unstable state,provide the tools for system planning and real
time control. Energy managementsystem (EMS) provide a variety of measured
and computer processed data. This is helpful to system operators in taking
critical decisions inter alia reactive power management and control. In this
regard autornationand specializedsoftware relieve the operator of good part of
the burden of system management but it does add to the complexity of the
systemoperation.
Voltage stability analysis and techniques have been pushed forward by
several researchers and several of these are in commercial use as outlined in
this chapter.As it is still hot topic, considerableresearcheffort is being devoted
,
to it.
Pw et al. l8l considered an exponential type voltage dependentload model
and a new index called condition number for staticvoltage stability prediction.
Eigenvalue analyseshas been used to find critical group of busesresponsible
for voltage collapse. Some researchers[26] have also investigated aspects of
bifurcations (local, Hopf, global) and chaos and their implications on power
systemvoltage stability. FACTS devicescan be effectively used for controlling
the occurrence of dynamic bifurcations and chaos by proper choice of error
signal and controllergains.
Tokyo Electric Power Co. has developed a pP-based controller for
coordinated control of capacitor bank switching and network transformer tap
ctranging.HVDC power control is used to improve stability.
More systematic approach is still required for optimal siting and sizing of
FACTS devices.The availability of FACTS controllers allow operationclose to
the thermal limit of the lines without jeopardizing security. The reactivepower
compensationclose to the load centresand at the critical busesis essentialfor
overcoming voltage instability. Better and probabilistic load modelling [11]
should be tried. It will be worthwhile developingtechniquesand models for
study of non-lineardynamics of large size systems.This may requireexploring
new methods to obtain network equivalents suitable for the voltage stability
analysis. AI is another approach to centralized reactive power and voltage
control. An expert system [9] could assistoperatorsin applying C-banksso that
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Po
a00.l,*l
I
. .
t
generatorsoperate near upf. The design of suitable protective measuresin the
event of voltage instability is necessary.
So far, computed PV curves are the most widely used method of estimating
voltage security, providing MW margin type indices. Post-disturbanceMW or
MVAr margins should be translated to predisturbance operating limits that
operators can monitor. Both control centre and power.plant operators should be
trained in the basics of voltage stability. For operator training simulator [10] a
real-time dynamic model of the power system that interfaceswith EMS controls
such as AGC is of great help.
Voltage stability is likely to challengeutility plannersand operatorsfor the
foreseable future. As load grows and as new transmission and load area
generationbecome increasingly difficult to build, rnore and more utilities will
face the voltage stability'challenge.Fortunately, many creative researchersand
plannersare working on new analysis methods and an innovative solutions to
the voltage stability challenge.
AQ = reactive power variation
(i.e. the size of the compensator)
Srr.= system short circuit capacity
Then
AV =
trg
d,.
AQ = AVSsk
=1(0.05x5000)
= + 250 MVAR
The capacityof the staticvAR compensator
is +250 MVAR.
REFERE
NCES
Books
A load bus is composed of induction motor where the nominal reactive power
is I pu. The shunt compensationis K,n. Find the reactive power sensitivityat
the bus wrt change in voltage.
Solution
Qrcot= Qno V2 [given] '
Qcomp=-
Qn"t= Qnua t
.'. Here,
[-ve sign denotesinductive
reactivepowerinjection.l
Krt V2
Qn r= v' -
Qcomp
Krn vz lQoo^ =
1.0givenl
dQn"t
= 2v-2v K..,
i',
dv
Sensitivity increasesor decreaseswith Krn as well as the magnitude of the
voltage.Say at V - 1.0 pu, Krn = 0.8
dQn t
-2-1.6=0.4pu.
dv
Example:,17.2
Find the capacity of a static VAR compensator to be installed at a bus with
x 5Vo voltage fluctuation. The short circuit capacity is 5000 MVA.
Solution For the switching of static shunt compensator,
www.muslimengineer.info
1. Chakrabarti,A., D.P. Kothari and A.K. Mukhopadhyay,Perfurmance,Operation
and Control of EHV Power Transmission Systerts,Wheeler Publishing, New
Delhi, 1995.
2. Taylor, c.w., Power system voltage stability,McGraw-Hill, New york, 1994.
3. Nagrath, I.J. and D.P. Kothari, Power SystemEngineering, Tata McGraw-Hill,
New Delhi, 1994.
4. Kunclur,P., Powcr Sy,stem
Stubilityatul Control, McGraw-Hill, New york, 1994.
5. Padiyar, K.R., Power System Dynamics: Stability and Control, John Wilev.
Singapore,1996.
6. Cutsem,T Van and CostasVournas, Voltage Stabitityof Electric power Systems,
Kluwer Int. Series, 1998.
Papers
7 . Concordia, C (Ed.), "special Issue on Voltage Stability and Collapse,', Int. J. of
Electrical Power and Energy systems, voi. i5, no. 4, August 1993.
8 . Pai, M.A. and M.G.O, Grady,"Voltage CollpaseAnalysis with ReactiveGeneration
and voltage Dependentcons,traints", J. of Elect Machines and power systems,
V o l . 1 7 , N o . 6 , 1 9 8 9 ,p p 3 7 9 - 3 9 0 .
9 . cIGRE/ Task Forcc 38-06-0l,"Expcrt Systems Applied to voltage and var
Control//,1991.
1 0 . "operator Training simulator", EpRI Final Report EL-7244, May 1991, prepared
by EMPROS SystemsInternational.
l l . Xu, W and Y Mansour,"Voltage Stability using GenericDynamic Load Models",
IEEE Trans. on Power Systems,Vol 9, No l, Feb 1994, pp 479493.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modern Po
t
'Yoltage JlaDrtrty DnnansEulenruy
IZ. VerTna,tl.lt., L.IJ. Arya ano u.r. I\.oman,
ReactivePower Loss Minimization", JIE (I), Vol. 76, May 1995, pp.4449.
13. IEEE, special Publication 90 TH 0358-2 PWR, "Voltage Stability of Power
.
. . t ?
i -
d -
t
! l ! -
F
L -
- - - , - - - - ^
L - -
AppnNDrx A
Systems:Concepts,Analytical Tools, and Industry Experience" 1990.
14. Flatabo,N., R. Ogncdaland T. Carlsen,"Voltagc StabilityCondition,in a Power
Tiansmisiion System calculated by Sensilivity Methods", IEEE Tians. VoI.
PWRS-5,No. 5, Nov 1990,pp 12-86-93.
1 5 . Gao, 8., G.K. Morison and P. Kundur, "Voltagc Stability EvaluationUsing Modal
Analysis", IEEE Trans. Vol. PWRS-7, No. 4, Nov. 1992, pp 1529-1542.
T6, Cutsem, T. Van, "A Method to Compute Rbactive Power Margins wrt Voltage
Collapse",IEEE Trans.,Vol. PWRS-6, No. 2, Feb 1991, pp 145-156.
1 7 . Ajjarapu, V. and C. Christy, "The continuation Power Flow: A Tool for Steady
StateVoltageStabilityAnalysis", IEEE PICA Conf.Proc., May 1991,pp 304-311.
1 8 . Ldf, A-P, T. Sined, G. Anderson and D.J. Hill, "Fast Calculationof a Voltage
Stability Index", IEEE Trans., Vol. PWRS-7, No. 1, Feb 1992, pp 54-64.
1 9 . Arya, L.D., S.C. Chaube and D.P. Itothari, "Line Outage Ranking based on
EstimatedLower Bound on Minimum Eigen Value of Load Flow Jacobian", JIE
(1),Vol. 79, Dec 1998,pp 126-129.
'20. Bijwe, P.R,, S.M. Kelapure,D,P. Kothari and K.K. Saxena,"Oscillatory Stability
Limit Enhancementby Adaptive Control Rescheduling",Int J. of Electrical Power
and Energy Systems,Vol. 21, No. 7, 1999, pp 507-514.
2 t , Arya, L.D., S,C, Chaube and D.P. Kothari, "Linc Switching for Alleviating
Overloadsunder Line OutageCondition taking Bus Voltage Limits into Account",
Int J. of Electric Power and Energy System,YoL 22, No. 3, 2000, pp 213-221.
In this appendix, our aim is to present definitions and elementary operationsof
vectors and matrices necessaryfor power system analysis.
VECTORS
A vectorx is definedas an orderedset of numbers(realor complex),i.e.
,
(A-1)
.)
Bijwc, P.R., D.P. Kothari and S. Kclapure, "An Efficient Approach to Voltage
Sccurity Analysis and Enhancement",Int J. of EP and,ES., Vol. 22, No. 7, Oct.
2000, pp 483486.
2 3 . Arya, L.D., S.C., Chaubeand D.P. Kothari, "Reactive Power Optimization using
Static Stability Index (VSD", Int J. of Electric Power Components and Systems,
Vol. 29, No. 7, July 2001, pp 615-628.
24. Arya, L.D., S.C. Chaube and D.P. Kothari, "Line Outage Ranking for Voitage
Limit Violations witti Conective Rescheduling Avoiding Masking", Int. J. of EP
and ES,Vol. 23, No. 8, Nov. 2001, pp 837-846.
xp ...t xn are known as the components of the vector r. Thus the vector x is
a n-dimensional column vector. Sometimestransposedform is found to be more
convenient and is written as the row vector.
rT A fx1, x2,..., xrf
Some Special Vectors
The null vector 0 is one whose each component is zero, i.e.
2 5 . CIGRE Task Foice 38-02-10, "Cigre Technical Brochure: Modelling of Voltage
CollapseIncludihgDynamicPhenomena",Electa, No. 147, April 1993,pp' 7l-77.
2 6 . Mark J. Laufenbergand M.A. Pai, "Hopf bifurcation control in power system with
static var compensators",Electrical Power and Energy Systems,Vol. 19, No. 5,
1997, pp 339-347.
equaito unity, i.e.
The sum vector i has each of its components
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(A-2)
-
.
flTY'
t
I
A
0x.= 0
The multiplication of two vectors x and y of same dimensions
results in a
very important product known as inner or j9g!g! pfodqcij.e.
The unit 'tector e
the.rest of the componentbare zero, i.e.
*tv AD",y,Aytx
0
(A_3)
i:l
Also, it is interesting to note that
€k:
0
1 kth component
A
0
0
Some Fundamental Vector Operations
Two vectors x and y are known as equal if, and only if, .yk=
!*for k = r,2,
..., n. Then we sav
xTx = lx 12
(A-4)
cos d' 4 "tY
lxllyl ,
(A-5)
wtere Q is angle between vectors, lxl and lyl are the geometric
lengths of
vectors x and y, respectively.Two non-zero vectors are
said to be orthJgonral,
if
*ty= o
(A-6)
MATRICES
x = y
The product of a vector by a scalar is carried out by multiplying each
component of the vector by that scalar, i.e.
Definitions
'
Matrix
'
'\n m x n (ot m' n) matrix is an orderedrectangulararray of
elementswhich
may be real numbers,complexnumbers,functionsor operators.
The matrix
(A-7)
If a vector y is to be addedto or subtractedfrom anothervector x of the same
dimension, then each component of the resulting vector will consist of addition
or subtraction of the correspondingcomponents of the vectors x and
vr,i.e.
The
f n l l n u r i nYcv r r r 6 y nr vr nPnvAr !r af ivoD
o 4 r v 4 y^ I* -^, u U^<- -r+U: I^t^rlL^ U
l ^u r c v e rL^: t 0 rr L
a -r g e -D- r' a :' -
x*!=y+x
x+(v+'z)=(x+y)+z
ar @zx)- (afiz)x
(ar+ e)x- dF+ a2x
www.muslimengineer.info
is a rectangular array of mn elements.
o,t
the (i, i)th element,i.e. the elementlocated in the ith row
and the
..
!"notes
7th column. The matrix A has m rows ano n coiumns and is said to be of order
mxn.
When m = rt, i.e. the number of rows is equal to that of columns,
the matrix
is said to be a square matrix of order n.
An m x 1 matrix, i.e. a matrix having only one column is
called a column
vector. An I x n matix, i.e. a matrix having only one
row is called a rore
vector.
*Sometimes
. ,1r\!r
inner product is also representedby the following alterr
ative forms
x . y, (x, y),(x,y).
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
's.t
ModernPowerSvstemAnalvsis
frffi-H
T
Diagronal matrix
A diagonal matrix is a squarematrix whose elements off the main diagonal are
a l l z e r o s( a i j = 0 f o r i + j ) .
(A-e)
2l
l-l 2l , l-1 3l
'1, +lr-t1 I 4l*
| 1 2l
det (A) = tAt = 213
=2(8)+('6)+(-5)=5
Transpose
NUII matrix
If all the elements of the square matrix are zero, the matrix is a nuII or zero
matrix.
(A-10)
of a matrix
'l.h-.e
transpose of matrixA denoted by At is the matrix formed by interchanglng
the rows and columns of A.
(Ar)r = A
Note that
Symmetrtc
matrix
A square matrix is symmetric, if it is equal to its transpose,i.e.
AT=A
Notice that the matrix A of Eq. (A-9) is a symmeffic matrix.
(A-8)
Unit
(identity).
Minor
The mino, Mij of an n x n rnatrix is the determinant of (n - l) x (n - 1) matrix
formed by deleting the ith row and the 7th column of the n x n matrix.
matrix
A unit matrix / is a diagonal matrix with all diagonal elementsequal to unity.
If a unit matrix is multiplied by a constant(),), the resulting matrix is a diagonal
matrix with all diagonal elements equal to 2. This matrix is known as a scalar
maffix.
Cofactor
The cofactor AU of element a,, of the matrix A is defined as
aU = Gl)'*t M,i
Adjoint
L -
^ J
=4x4unitmatrix
Determinant
ll:Lll;l
=3x3scalarmatrix
of a matrix
matrix
The adjoint matrix of a squarematrix A is found by replacing each element au
of matrix A by its cofactor A,, and then transposing.
For example, if A is given by Eq. (A-9), then
l3
lz
l-1
taoJA= - l
Il 12
lI 3
l-t
l 1
l2
lr
l 2
l-1
For each square matrix, there exists a determinant which is formed by taking
the determinant of the elements of the matrix.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
MotJellr Powor S
AppendixA
_;l
6
- \ l
7
-5
(A-11)
)l
A square matrix is called singular, if its associateddeterminant is zero, and
non-singular,if its associateddeterminantis non-zero.
15 -1-l
C = A + B = l '
I
L2 2)
Addition and subtractionare definedonly for matricesof the sameorder.
The fbllowinglawshold lbr addition:
(+ fhe cammatatLveln+y: A + B -B + ,{
(ii) The associative law: A +- (B + C) = (A + B) + C
Further
MATR.IX OPERATIONS
Matrix
Eguality
:
Ar + Br
(A tB)r=
ELEMENTARY
I ,6tl
I
-
Multiplication
of matrices
The product of two matrices A x B is defined if A has the same number of
columns as the number of rows in B. The natrices are then said to be
conform.able.If amatrix A is of order mx n and B is an n x q matrix, the
product C = AB will be an m x q matrix. The element c,, of the product is given
bv
Two matrices A(m x n) and B(m x n) are said to be equal, if the only if
a i = b i j f o ri = 7 , 2 , . . . , f f i , . i = 1 , 2 , . . . , f l
Then we write
A = B
Multiplication
of a matrix
cii =
by a scalar
r
,l
,
(A-13)
)_rai*0*i
k:l
A matrix is multiplied by a scalar a if all the mn elements are multiplied by
a. i.e.
(A-12)
Thus the elements cu are obtained by multiplying the elementsof the ith row
ofA with the correspondingelementsof theTth column of B and then summing
theseelementproductp.
For example
1""
(or suhtraction)
Addition
Lau
of matrices
the
frxrn
rnefr;cce
i e
rrrhen
frxrn
rnqfri/-ec
A
enA
added,a new matrix C results such that
C-A+
aij *
B;
l
bij
lhe
cqrnc
o,rAcr
qrc
ctz= anbn + arrb2
AB+BA
r
I
i-?
IA
nf
czz= azt bt z+ cr r b2
,ExampleI I
:
Il
ctt= attbt,+ arrb^
If the product AB is defined, the product BA may or may not be defined.
Even if BA is defined, the resulting products of AB and,BA are not, in general,
equal. Thus, it is important to note that in generalmatrix multiplication is not
commutative,i.e.
l------_-__-ir1
r
czrl
c z l = u z l b t ,+ a r r b 2 1
whose ryth elernent equals
cij=
azz)Lbn brrJ Lr^
cnl
where
To add (or subtract)two matrices of the sameolder (same number of rows, and
samenumber of columns), simply add (or subtract)the correspondingelements
nf
arzllbn b,rl_ r
[''
O-r
f)
-11
'A-=- il z - r l " - L o 3 . 1
then,
www.muslimengineer.info
The associariveand distributive larr-shold for matrix mulriplicarion (wtren
the 'aryprrtixiateerpernirnys.aredef;tv.4)- re.
Asscrica'rv &rrt': 4{B [
= A(BC) = -tBC
Distibutive law:. A(B + C) - AB + AC
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
tffiWl
Po
Modern
I,j].qffil*l
l.,i95I3n
I
or
n
D
i:l
[1
-1
"*t=
c i i i = 1 , 2 , . . . ,m
Using the rules of matrix multiplication defined above. Eqs (A-14) canle
written in the compact notation as
3]
Lo z rl
Ax=c
(A-1s)
where
Find AB and BA.
A and B areconformable (A has two columns and B has two rows), thus we
have
l-1 -1
3l
0t
ll ,- 4
'4"8 -=|L1 2; 4 9 |l t B" ^A =-r--1
,'1)
;;l
A matrix remainsunaffected,if a null matrix, defined by Eq. (A-8) is added
to it, i.e.
A + 0 = A
If a null matrix is multiplied to another matrix A, the result is a null matrix
A0= 0A = 0
Also
It is clear that the vector:mntrix Eq. (A-15) is a useful shorthand
representationof the set of linear algebraic equations (A-14).
Matrix Inversion
A - A = 0
Note that equationAB = 0 does not mean that either A or B necessarilyhas
to lre a null matrix, e.g.
A_IA_AA_I_I
[r ,l[ 3 ol=fo 0l
L0 0JL-l 0l
Division does not exist as such in matrix algebra. However, if A is a square
non-singular matrix, its inverse (A-l) is defined by the relation
(A-16)
The conventional method for obtaining an inverse is to use the following
relation
Lo 0J
Multiplication of any matrix by a unit matrix results in the original maffix,
A-; _
l.e.
AI-IA=A
The transpose of the product of two matrices is the product of their
transposesin reverse order, i.e.
6Dr = BrAr
The conceptof matrix multiplication assistsin the solution of simultaneous
linear algebraic equations. Consider such a Set of equations
atft + anxz+ ... + abJn = ct
adj A
det A
(A-17)
It is easy to prove that the inverse is unique
The following are the important properties charactenzing the inverse:
(AB)-r = 3-t4-l
(A-tf = (Ar)-r
(A-18)
14-r1-t= 4
a L l x t+ a z z x z + . . . + a z r t r = c 2
:
Q^lXl *
C^*z+
... + A-rJn = C^
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(A-14)
I f A is given by Eq. (A-9), then from Eqs. (A-10), (A-1I), (A-17), we get
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
powerSystemRnatysis
Modern
ffd-ffi.:|
-+:if
A-1
: -si _;l:f:i;
i,:_il
det
A t
5j L-;
L-r
SCALAR AI\rD VE CTOFFUNCTI
-;
l.J
It may be noted that ihe derivative of a scaiar function with respect
to a vector
of dimension n is a vector of the same dimension.
The derivative of a vector function (A-22) with respect to a vector
variable
r is defined as
O]NIS
0*,
y ! f\r,
x2, ..., xn)
0x
(A-1e)
It can be written as a scalarfunction of a vector
variable x, i.e.
y - f(x)
0*,
0*,
0fz
0*,
0*,
0*,
o f a }fz
A scalarfunctionof n scalarvariablesis definedas
af- af; ..
o*n
?fz
(A-zs)
0rn
af;
o*n
(A-20)
where x is an n-dimension vector,
(A-26)
In general, a scalarfunction could be a function of
several vector variables, e.g.
y-f(x,u,p)
(A-2r)
where x, u and p are vectors of various dimensions.
A vector function is defined as
Considernow a scalarfunctiondefinedas
t - ff@, u, p)
= 21fi(x, u, p) + Lzfz(*,u, p) + ... + 2*f*(4, u,
Let us fincf j{
a^
In general, a vector function is a function of several
DERTVATNTES OF SCALAR AND \ZECTOR
(A-23)
f (x, u, p)
LJm\rrUrP)J
d s ^ i .
Let us now find -.
According to Eq. (A-24), we can write
ox
FUNCTIONS
A derivafit'e of a scalar funcritrn
1A-lt)) s,ith re-sper.rtr) u \.eL-R)rr.ariable -r is
as
defind
-af
0r,
A t
0x
af
0*,
(A-24)
la"l
Lar,J
oJ
orn -
www.muslimengineer.info
)
6-2g)
. Accordingro Eq. (A-24), we can write
vector variables, e.g.
y-f(x,u,p)
(A-27)
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
(A-2e)
fijfilftf.f
ModernPower'systemAnatysts
I
}ft 0f,
af^
0*, 0*,
Ofi }fz
0r, 0r,
0*,
)r
0*,
^2
VJI
UJ2
0*n
0r,
AppnNDrx B
af^
o*n
(A-30)
NCES
REFERE
l, Shiplcy, R,8,, Intoduction to Matric:r:sand Power Sy,rtems,
Wiley, New York,
r976.
2. Hadley, G., Linear Algebra, Addison-WesleyPub. Co. Inc., Reading,Mass., 1961.
3. Bellman, R., Introduction to Matrix Analysis, McGraw-Hill Book Co.. New York,
1960.
We can represent,as we saw in Chapter 5, a three-phasetransmissionline* by
a circuit with two input terminals (sending-end,where power enters) and two
output terminals (receiving-end,where power exits). This two-terminal pair
circuit is passive (since it does not contain any electric energy sources),linear.
(impedances of its elements are independent of the amount of current flowing
through them), and bilateral (impedances being independent of direction of
current flowing). It can be shown that such a two-terminal pair network can be
representedby an equivalent T- or zr-network.
Consider the unsymmetrical T-network of Fig. B-1, which is equivalenr to
the general two-terminal pair network.
s
lp
22
Z1
Vg
Y
Vp
Flg. B-1 Unsymmetrical
T-circuit
equivalent
to a generaltwo-terminal
nair "network
---'-"_
r-"
For Fi g. B- 1, the following circuit equationscan be written
1 s - I^+ Y(Vo+ I^Zr)
or
(B-1)
Is= YV*+(l+ YZr)I*
*:
V*+ I^Zr+ IrZ,
V^ + I^2, + ZrWn+ I^2,, + I^YZ.Z,
*A transformer is similarly represented by a circuit with two input and two ouput
terminals.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
.+
iqY
Modernporelsyslern inal,sis
tlCtf I
or
V5= (1 + YZr) V* + (2, + Z" + yZrZr)Io
Equations(B-1) and (B-2) can be simplifiedby letting
A-l + YZ,
B= 2.,+Zr+ yZrZ,
C=Y
(B-2)
(B-3)
D=l+yZ,
using these, Eqs. (B-1) and (B-2) can be written in matrix form as
lyrl = lA
Bllv*1
Fig. B-3 Unsymmetrical
zr-circuit
,t-0,
Lr,.JLt olj^l
This equationis the sameas Eq. (5.1) and is valid for any linear, passive
and
bilateral two-terminal pair network. The constantsA, B, C andD arecalled the
generalized circuit constantsor the ABCD constantsof the network,
and they
can be calculated for any such two-terminal pair network.
It may be noted that ABCD constants of a two-terminal pair network are
complex numbers in general, and always satisfy the following relationship
AD-BC=1
(B-s)
A seriesimpedanceoften representsshort transmissionlines and transformers. The ABCD constants for such a circuit (as shown in Fig. B-4) can
immediately be determinedby inspection of Eqs. (B-1) and (B-2), as follows:
A= 1
B=Z
(B-7)
C=0
D = l
Also, for any symmetrical network the constantsA and,D areequal. From
Eq.
(B-4) it is clear that A and D are dimensionless,while B has the
dimensions of
impedance(ohms) and c has the dimensions of admittance (mhos).
The ABCD constants are extensively used in power system analysis.
A
generaltwo-terminal pair network is often representedas in Fig. B_2.
Fig. B-4 Seriesimpedance
Another simple circuit of Fig. B-5 consistingof simple shunt admittancecan
be shown to possessthe following ABCD constants
A= 7
Fig. B-2
:#:::representation
of a two-rerminar
pair networkusing
B=0
/El_a\
\s-u,,
C=Y
ABCD CONSTANTS
D = I
FOR VARIOUS SIMPLE NETWORKS
We havealready obtained theABCD constants.ofan unsymmetrical T-network.
The ABCD constantsof unsymmetricala-network shown in Fig. B-3
may be
obtained in a similar manner and are given below:
A=l+
YrZ
Fig. B-5 Shuntadmittance
B = Z,
C=Yr+Yr+ZYryz
D=I+ YrZ
(B-6)
It may be noted that whenever ABCD constantsare computed, it should be
checked that the relation AD-BC = 1 is satisfied. For examole, using Eq.
(B-8) we get
AD-BC=1x1-0xY=1
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r
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpower:Systemnnatysis
:fflfuii'l
II ABCD constantsof a circuit are given,its equivalentI- or a-circuitcan
be determinedby solving Eq. (B-3) or (8-6) respectively,for the valuesof
seriesand shuntbranches.For the equivalentn-circuitof Fig. B-3, we liuu.
t ,
Append8B
,,
l,;621+
I
tr= (ArBr+ ArBr)/(8,+ Br)
B = B r B z l ( B t +B r )
(B-13)
C= (Cr * Cr) + (Ar - Ar) (Dz- D)l(Br * Ur)
(B-e)
,r=#
ABCD CONSTANTS
OF NETWORKS
IN SERIES AND FANEr,r,Ur,
Whenever a power system consists of series and parallel combinations of
networks, whose ABCD constantsare known, the overall ABCD constantsfor
the system may be determined to analyze the overall clperationof the system.
Fig. B-7 Networksin parallel
Measurement
Fig. 8-6 Networksin series
Consider the two networks in series,as shown in Fig. 8-6. This combination
can be reducedto a single equivalent network as follows:
For the first network. we hal'e
'l
[u,l-_ |o, ", [u,l
Lr,J 1., o,)lt, )
(B-10)
For the secondnetwork, we can write
II
y*
It = It A,
lrr)
Brll vo]
| |
(B-11)
|
l_c, DrJlr*l
From Eqs. (B-10) and (B-11), we can write
u'11o,u'l[u^]
[u'l_lo'
L+I 1., n,ll.c, or)j^l
The generuIized circuit constants rnay be computed for a transmission line
which is being designedfrom a knowledge of the systemimpedance/admittance
parameters using expressionssuch as those clevelopedabove. If\the line is
already built, the generalizedcircuit constantscan be measured by making a
few ordinary tests on the line. Using Eq. (B-4), these constants can easily be
shown to be ratios of either voltage or currentat the sending-endto voltage or
current ai the receiving-end of the network with the receiving-end open or shortcircuited. When the network is a transformer, generator, or circuit having
lumped parameters,voltage and current measurementsat both ends of the line
can be made, and the phase angles between the sending and receiving-end
quantities can be found out. Thus rhe ABCD constantscan be determineC.
It is possible, also, to measurethe rnagnitudesof the required voltages and
currents sinrultaneously at both ends of a transmission line, but there is no
simple method to find the difference in phaseanglebetweenthe quantitiesat the
two ends of the line. Phasedifference is necessarybecausethe ABCD constants
are complex. By measuringtwo impedancesat each end of a transmissionline,
however',the generalizedcircuit constantscan be computed.
The following impedancesare to be measured:
7
f A,A,+ BtC.,
=l
A t B z+ q D 2 l [ y * I
l c rA z + D 1 C 2 CrB ' -t D rD , Jl t o l
If two networksare connectedin parallel as shown in Fig. B-7, the ABCD
constantsof the combinednetworkca-nbe found out similarly with somesimple
manipulations of matrix algebra. The results are presentedbelow:
www.muslimengineer.info
of ABCD Gonstants
"so
--
oo-'l.i-o-.l
rvrrulrS-vr:\r
i*^o'l^-^^
uuPv\r(rrrlvs
..'i+L
wrlrl
-^^^:.,:--
^-l
r9L9rvlttE-trlru
^-^-
^:-^-
. :--
uP(,il-ull'uulteo
r
Zss = sending-end
impedancewith receiving-end
short circuited
=
zno receiving-endimpedancewith sending-endcpen-circuiteci
Zns = receiving-endimpedancewith sending-endshort-circuited
The impedancesmeasuredfrom the sending:endcan be determinedin terms
of the ABCD constantsas follows:
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
From Eq. @- ), with 1R= 0,
Z s o = V s l l s =A l C
(B-14)
and with VR= 0,
AppENDIX C
When the impedancesare measuredfrom the receiving-end, the direction of
current flow is reversedand hence the signs of all current terms in Eq. (5.25).
We can therefore rewrite this equation as
(B-16)
Vn= DVs + BIt
In= CVs + AIt
From Eq. (8-16), with Is = 0,
(B-17)
Z n o = V R | I R =D / C
and when Vs = 0,
(B-18)
Zns= VRllR= BIA
Solving Eqs. (B-14), (B-15), (B-17) and (B-18) we can obtain the values of
the ABCD constantsin terms of the measuredimpedances as follows.
AD-BC
AC
I
1
[usingEq. (B-5)]
AC
C
1
AC, A
A2
o=(ffi)'''
GAUSS ELIMINATION
(B-1e)
By substitutingthis value of A in Eqs. (B-14), (B-18) and substituting the
value of C so obtained in Eq. (B-17), we get
1l /2
(
z \- '
r >=-L. 7
D
R S, l z * o - z n r ) I
(B-20)
C_
(B-21)
p=
(Zss(Zno - Zo))'''
Z^o
(zso(z^o- zo))t''
We know that the nodal maffix Yu.r, and its associated Jacobian are very
sparse,whereas their inverse matrices are full. For large power systems the
'sparsity
of these matricesmay be as high as 98Voand must be exploited. Apart
fiom reducing storageand time of computation, sparsity utilization limits the
round-off computational errors. In fact, straight-foru'ard application cf the
iterative proceduretor system studies like load flow is not possiblefor large
systems unless the sparsity of the Jacobian is dealt with effectivet).
One of the recent techniques of solving a set of linear algebraic equations,
called triangular factorizatiott, replaces the use of matrix inverse which is
highly inefficient for large sparse systems. In triangularization the elements of
.u.h ,o* below the main diagonal are made zero and the diagonal element of
each row is normalized as soon as the processingof that row is completed- It
, is possible to proceed columnwise but it is computationally inefficient and is
theref,re not used. After triangularizationthe solution is easily obtained by
hack sub.stitution.The techniqueis illustratedin the examplebelow'
(B-22)
Consider the linear vector-maffix equation
NCE
REFERE
and Distribution of Electrical Energy,
l . Cotton, H. and H, Barber, The Transmi,ssion
t;l
1'l['t I
=
l'L3 ,l1,,I
3rd edn., B.I. Publishers,New Delhi. 1970.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Procedure
1. Divide row 1 by the self-element of the row, in this case2.
2. Eliminate the element (2, 1) by multiplying the modified row 1, by
element (2, I) and subtract it from row 2.
3' Divide the modified row 2 by its self-element (f); and stop.
Following this procedure, we get the upper triangular equation as
[l r z+ li [l ' 't -= tlt +o _l + l
l o ? : r l l _l - l , r
J
L
t
)Lxzl Lt
Upon back substituting,that is first solvingfor x2 and then for 11, we get
o-+
1
x2= --7:-2
-
-
I _
I Y
Consider the following system of linear equations:
(Z)xt + Q)x2 + (3)x, = $
(2)xr+(3)xz+(4)xt=)
( 3 ) x r+ ( 4 ) x r + ( 7 ) 4 = 1 4
For computer solution, maximum efficiency is attained when elimination is
carried out by rows rather than the more tamiliar column order. The successive
reduced sets of equations are as follows:
(l)xr+ (t)*z+ (*)rt = *
(Z)xt+ (3)x2+ (4)xt = )
j
(c-3)
(3)xr + (4)xz+ (7)x3 = 14
-
I _
1 1 _ 3 \ -
5
Check
( 1 ) . r+, ( i l r r +
(|)xt -z
(c-4)
(2)xz+ (1)x, = l
b r + x z =2 ( * ) - t r = t
3x,+ 5rz-3(+)- s(f) = o
theuseof thebasicGausseliminationandback
Thus,we havedemonstrated
substitutionprocedurefor a simplesystem,but the sameprocedureappliesto
any generalsystemof linear algebraicequations,i.e.
(c-1)
Ax=b
An added advantageof row processing(elimination of row elementsbelow
the main diagonal and normalization of the self-element) is that it is easily
amenable to the useof low storagecompactstorageschemes-avoiding storage
of zero elements.
GAUSS ELIMINATION
(c-2)
USING TABLE OF FACTORS
Where repeatedsolution of vector-matrix Pq. (C-1j with cons tant Abut varying
values of vector D is required, it is computationally advantageousto split the
'Table of factors' or 'LU
matrix A into triangular f'actor (termed as
decomposition') using the Gauss elimination technique. If the matrix A is
sparse,so is the table of factors which can be compactly stored therebynot only
reducing core storagerequirements,but also the computational effort. Gauss
elimination using the table of factors is illustrated in the following example.
(3)xr+(4)x2+(7)4=14
(1)x,+
( t r ) r r +( * ) " , - 3
( 1 ) x 2(+* ) r , - +
(c-s)
(3)xr+(4)x2+Q)xt=14
( l ) x r+ ( t ) r r + ( J ) x , = 3
( l ) x 2 +( * ) " r ' = + , i
(c-6)
(t),r+(*)r,-s
(1)x,+ (t)rr+ (*)", = J
(t)xr+(t)*t - +
(c_'7\
(*)rt = +
(1)x,
+ (tr)*r+(*)r, _ J
(1)x,+ (I)*t
3
2
(c-8)
x3= I
'elimination'
operations. The solution r may
These steps are referred to as
'back substitution' operation using Eq. (C-8).
be immediately determined by
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
The solutionfor a new set of valuesfor b can be easily obtainedby using
a tableof factorspreparedby a carefulexaminationof eqs.
1c-l) to (c-g). we
can write the tableof factors F as below for the examplein hand.
ft,
ftz
fn
fu
fzz
fzt
r
'
)
.
l
r
a
3
lt=
- 5- ( ; X ; ) : i
and for heading3, 3
n
L
Tl
lt = fszlz
(c-e)
Tl
=(+x+)
=I
32
the rows after the row elimination has been completed, e.g.
.fzz= ! , the factor
by which tow 2 of Eq. (C-4) must be multiplied to normalize
the row. The
elementsof F above the diagonal can be immecliately written
down by
inspection of Eq. (C-8). These are neededfor the back substitution process.
In rapidly solving Eq. (C-1) by use of the table of factors F, succesiive
steps
appearas columns (left to right) in Table C.l below:
Table C.l
l, I
3
9
T4
2,1
3
3
14
2,2
3
3t2
t4
3,I
3
3/2
5
3,2
3
3t2
5t4
a
a
3
3t)
1
2,3
a
J
3
1
1,3
5/2
I
I
1,2
x
I
I
1
The headingrow (i, j) of Table C-1 representsthe successiveelimination
and
back substitutionsteps.Thus,
7,7
represents normalization of row 1
2,L
representselimination of element(2, l)
2,2
representsnonnalization of row 2
3, L; 3,2 representelimination of elements(3, 1) an,J(3,2) respectively
3. 3
representsnormalization of row 3
2,3
representselimination of element(2,3) by back substitution
r, 3; 7, 2 reprelent elimination of elements(1, 3) and (r, z) respectively
by
back substitution.
The solution vector at any stage of development is denoted by
Ut lz y3Jr = y
The modification of solution vector fiom column to column (left
to right) is
carried out for the heading (i, j) as per the operationsdefined below:
fiiji
ttj=i
(c-10)
!i = !r- fi /i
lt i + i
(c-11)
!i=
fn)z
L
The row elements of F below the diagonal are the multipliers
of the
normalized rows required for the elimination of the row element,
e.g.fzz = *,
the multiplier of normalized row 2 l&q. (c-6)l to eliminate the element (3,2),
i'e- (])x2. The diagonal elements of F are the mulripliers needed
to normalize
h
6
9
t4
lt-
Thus for heading 3, z
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In fact, operation (C-10) representsrow normalization and (C-11) represents
elimination and back substitution procedures.
Optimal
Ordering
In power systemstudies,the matrix A is quite sparseso that the number of nonzero operations and non-zero storage required in Gauss elimination is very
sensitive to the sequencein which the rows are processed.The row sequence
that leads to the least number of non-zero operationsis not, in general,the same
as the one which yields least storage requirement. It is believed that the
absoluteoptimum sequenceof ordering the rows of a large network matrix (this
is equivalent to renumbering of buses)is too complicated and time consuming
to be of any practical value. Therefore, some simple yet effective schemeshave
been evolved to achieve near optimal ordering with respectto both the criteria.
Some of the schemesof near optimal ordering the sparsematrices, which are
fully symmetrical or at least symmetric in the pattern of non-zero off-diagonal
terms, are described below [4].
Scheme
I
Number the matrix rows in the order of the fewest non-zero terms in each row.
If more than one unnumbered row has the same number of non-zero terms.
number these in anv order.
Scheme
2
Number the rows in the order of the fewest non-zero terms in a row at each step
of elimination. This scheme requires updating the count of non-zero terms after
each step.
Scheme
3
Number the rows in order of the fewest non-zero off-diagonal terms generated
in the renrainingrows at each stepof elimination.This schemealso involves an
updatingprocedure.
I he chotce ot scheme ts a trade-otf between speed of execution and the
number of times the result is to be used. For Newton's method of load flow
solution, scheme 2 seems to be the best. The efficiency of scheme 3 is not
sufficiently establishedto offset the increasedtime required for its execution.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
Modernpo
Mffi#
t
-
Scheme I is useful for problems requiring only a single solution with nq
iteration.
Compact
Storage Schemes
The usefulness of the Newton's method depends largely upon conserving
computer storage
ucmg the nu
oI non-zero computatlons.'l'o ettect
these ideas on the computer, elimination of lower triangle elements is carried
out a row at a time using the concept of compact working row. The non-zero
modified upper triangle elements and mismatches are stored in a compact and
convenient way. Back substitution progressesbackwards through the compact
upper triangle table. A properly programmed compact storage scheme results in
considerablesaving of computer time during matrix operations.
Naturally, there are as many compact working rows and upper triangle
storageschemesas there are programmers. One possible scheme for a general
matrix stores the non-zero elements of successiverows in a linear array. The
column location of these non-zero elements and the location where the next row
starts (row index) is stored separately. The details of this and various other
schemesare given in [2].
NCES
REFERE
AppBNDrx I)
Expressions
to be usedin evaluatingthe elementsof the Jacobianmatrix of a
power systemare derivedbelow:
From Eq. (6.25b)
P,- jiQ,=t', fr*rr
k:r
A
l . Singh, L.P., Advanced Power System Analysis and Dynamics, 2nd edn., Wiley
Eastern,New Delhi, 1986.
Agarwal, S.K., 'Optimal Power Flow Studies', Ph.D. Thesi.r, B.I.T.S.r Pilani,
r970.
3 . Tinney, W.F. and J.W. Walker, "Direct Solutionsof SparseNetwork Equationsby
Optintally Ordered Triangular Factorizations", Proc. IEEE, Nov. 1967,.55: 1801.
4. Tinney, W.F. and C.E. Hart, "Power Flow Solution by Newton's Method", IEEE
Trans.,Nov. 1967,No. ll, PAS-86: 1449.
= lvil exp (- i6,)L
lY,/ exp (i?il lVll exp (7dn)
\
(D-l)
k:r
Differentiating partially with respect to 6* (m *
i)
= Tvil exp(-r4) (Yi^l expQ0,^)tv^texp(j5^))
-i+
+06^ a6^
-- j(ei - jf) (a^ + jb^)
(D-2)
where
Y,^= G,* + jB,^
Vi= €i+ jfi '
(a^ * jb*) = (c* * jBi) @^ + jf^)
Although the polar form of the NR method is being used,rectangular
complexarithmeticis employedfor numericalevaluationas it is faster.
From Eq. (D-2), we can write
#=(aJi-b^e)=Hi^
#=
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- ( a ^ e i +b , f , ) =J i ^
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
tvrOtlcItI rOWgf
l r '
I
For the case of m = i. we have
*06, -i*" 06, = - jtvitexp(- j6,)i
k:l
= l%l exp(- j6,)ilY,pl exp(i0*) lvplexp(7d')
ty,niexp(j4) tV1,t
expQfi)
k:l
+ lV,lzlY,,lexp (7d,,)
+ jlV,l exp (- j6 ) (tytil exp Q?,,)tV;texp (id))
= - j(Pi - jQ) + jlV,l'(G,, + jB,,)
oPi'
;t
,et_
P i - G ' i t v i l 2= J '
Now differentiate Eq. (D-1) partially with respect to lV^l (m r i). We have
-alv)
av,l
avil
0Q, =
" alv^l
+
it*
lvil
L (- j6) (lYi*l exp (j0,^) exp Qd^))
' exp
\v'l = Pt * GiilViP = Nii
lvil = Qi - Biilvil, = L t
Case 7
m* i
Hi^= Li*=
Ni*=-
aJ, -
b*e,
Ji*= d*€i+ b"ft
I
Yi*= Gt^ + jBt*
- lvil exp (- j6) lY,^l exp (j1i)'tV*l exp (i6^)
= (ei- jf,) (a^ + jb*)
Vi= ei+ jfi
(D-4)
It follows from Eq. (D-4) that
(D-7)
(a* * jb^) = (Gi^ + jBi^) @* + jk)
Case 2
a ^ e , +b , f i = N i ^
!a! rl vr *^tl=
m = i
H,=-
Qi- Biilvilz
Nii= P, + G , , lV, lz
-!,?:
= a,,fi- b^e,= L,^
a l v .,tv^l
l
Jii - Pi-
j-P.-
alvil "7lvil
exp
, .e
. jilf
tyatexp(j0*)tvetexp(jQ,)
o:,
+ lV,l exp (- j6) lyi expQ0,,)exp (/4)
Multiplying by lV,l on both sides
lv,l-i
alvil '
(D-8)
Giilvr
Lii= Qr- Biilvilz
Now for the case of m = i. we have
an
--t
(D-6)
AO.
atv^tv^t-iffiw^t
aPt -
(D-s)
The above results are'summattzed below:
Multiplyingby lV^l on bothsides,
aPt
lQ)
It followsfromEq. (D-5)that
= - Qli- Biilv;12= H"
ao. =
a4
= \Ft-
(D-3)
From Eq. (D.3), we can write
#
q - " "
f.lr.$
Ao'
"",
NCES
REFERE
Tinney, W.F. and C.E. Hart "Power Flow Solution by Newton's Method", IEEE
Trans.,Nov 1967,No. 11, PAS-86:1449.
Van Ness, J.E., "Iteration Methods for Digital Load Flow Studies", Trans. AIEE,
Aug 1959,78A: 583.
" awil lv,l
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هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
AppBNDrx E
If z, violares a limit, it can either be upper or lower limit and not both
simultaneously.Thus, either inequality constraint (E-3) or (E-4) is active at a
time, that is, either ei.^^ of oi.minexists, but never both. Equation (E-5) can
be written as
AL
Ax
0x
oL = 0f +rq)'
'\aul \*o-0
du
0u
di=
Qi.^u*
(E-e)
In Eq. (E-9),
ai= - Ai,-io
The Kuhn-Tucker theorem makes it possible to solve the general non-linear
programming problem with several variables wherein the variables are also
constrained to satisfy certain equality and inequality constraints.
We can state the minimization problem with inequality constraints for the
control variables as
nln / (x' u)
(E-l)
subject to equality constraints
(E-2)
g(x,u,P)=o
and to the inequality constraints
(E-3)
(E-4)
u-u^u1 0
u^1-- u < 0
conditionsfor the minimum,
The Kuhn-Tuckertheorem[] givesthe necessary
assumingconvexityfor the functions(E-1)-(E-4),as
(E-5)
A.C = 0 (gradientwith respect to u, x, ))
where .C is the Lagrangian formed as
-C=f(x, u)+ )Tg(x,u,p)+
of*u*{, -u^u*)+
oT*,n1r-in-z)
(E-6)
and
tf Ui-
ui,*o ) 0
if ,r, ,t, - ui ) 0
a
L
,
.
#O A= g ( x , u , p ) = o
(E-10)
It is evident that a computed from Eq. (E-9) at any feasible solution, with
) from Eq. (E-8) is identical with negative gradient, i.e.
d =-
K
= negativeof gradientwith respectto u
(E-11)
At the optimum, a must also satisfy the exclusion equations (E-7), which
state that
,
\
ff ,i, ,rt, < ui < ui, ,n"*
di=
0
di=
di, -*
Qi=-
lf u, = ui, ,iru*
0
2
Q,rnin S
0
lf ui=
4i, *in
which can be rewritten in ternts of the gradient using Eq. (E-11) as follows:
o x= o
if ur, ,n;o< ui < ui, ^
of,.o
tf u,= ui,^*
0u,
0u,
of, ro
out
tf u, -
(E-12)
ili, ^in
(E-7)
NCE
REFERE
Equations (E-7) are known as exclusion equations.
The multipliers a,rr* and @,rri,,?re the dual variables associated with the
upper and lower limits on control variables. They are auxiliary variables similar
to the Lagrangian multipliers ) for the equality constraints case.
www.muslimengineer.info
1 . Kuhn, H.W. and A.W. Tucker, "Nonlinear Programming", Proceedings of the
Second.Berkeley Symposium on Mathematical Statistics and Probability, University of California Press,Berkeley, 1951.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
AppnNDrx F
Fig. F.l
In developedcountries the focus is shifting in the power sectorfrom the creation
of additional capacity to better capacity utilization through more effective
management and efficient technology. This applies equally to developing
countries where this focus will result in reduction in need for capacity addition.
Immediate and near future priorities now are better plant management,
higher availability, improved load management,reduced transmission losses,
revamps of distribution system, improved billing and collection, energy
efficiency, energy audit and energy management. All this would enable an
electric power system to generate,transmit and distribute electric energy at the
lowest possible economicand ecologicalcost.
These objectives can only be met by use of information technology (IT)
enabled services in power systems management and control. Emphasis is
therefore, being laid on computer control and information transmission and
exchange.
The operations involved in power systemsrequire geographically dispersed
and functionally complex monitoring and control system. The monitory and
supervisory control that is constantlydeveloping and undergoing improvement
in its control capability is schematicallypresentedin Fig. F.1 which is easily
seen to be distributed in nature.
Starting from the top, control system functions
EMS
Energy Management System - It exercisesoverall control over the
total system.
scADA Supervisory control and Data Acquisition system - It covers
generationancitransmissionsystem.
DAC
Distribution Automation and Control System - It oversees the
distribution systemincluding connectedloads.
Automation, monitoring and real-time control have always been a
part of scADA system. with enhanced emphasis on IT in power
systems, scADA has been receiving a lot of attention lately.
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Real time monitoringand controllingof an electricpowersystem.
SCADA refers to a system that enables an electricity utitity ro remotely
monitor, coordinate,control and operate transmissionand distribution components, equipment and devices in a real-time mode from a remote location with
acquisition of data for analysis and planning from one con[ol location. Thus,
the purpose of SCADA is to allow operatorsto observe and control the power
system. The specific tasks of SCADA are:
o Data acquisition,which provides measurements
and statusinformation to
operators.
. Trending plots and measurementson selectedtime scales.
. Supervisory control, which enablesoperatorsto remotely co:rtrcl devices
such as circuit breakers and relays.
Capability of SCADA system is to allow operatorsro control circuit breakers
ancldisconnectswitchesand changetransformertapsand phase-shifterposition
remotely. It also allows operatorsto monitor the generationand high-voltage
transmission systems and to take action to ccrrect overloads or out-of-limit
voltages. It monitors all status points such as switchgear position (open or
closed), substation loads and voltages, capacitor banks, tie-line flows and
interchange schedules.It detects through telemetry the failures and errors in
bilateral communication links between the digital computer and the remote
equipment.The most critical functions, mentionedabove, are scannedevery few
seconds. Other noncritical operations, such as the recording of the load,
foiecasting of load, unit start-upsand shut-downsare carried out on an hourlv
basis.
Most low-priority programs (those run less frequently) may be executed on
demand by the operatorfor study purposesor to initialize thepower system.An
operator may also change the digital computer code in the execution if a
parameterchangesin the system. For example,the MWmin capability of a
generatingunit may changeif one of its throttlevaluesis temporarilyremoved
for maintenance,so the unit's share of regulating power must accordingly be
decreasedby the code. The computer softwarecompilers and data handlers are
designed to be versatile and readily accept operator inputs.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
DAC is a lower level versionof SCADA applicablein distributionsystem
(including loads), which of course draws power from the transmission/
subtransmissionlevels. Obviously then there is no clear cut demarcation
betweenDAC and SCADA.
In a distribution network, computerisationcan help manage load, maintain
quality,detecttheftandtamperingandthusreducesystemlosses.Cornputerisation also helps in centralisation of data collection. At a central load dispatch
'centre, data such as culTent, voltage, power factor and breaker
status are
telemeteredand displayed. This gives the operator an overall view of the entire
distribution network. This enables him to have effective control on the entire
network and issue instructions for optimising flow in the event of feeder
overload or voltage deviation. This is carried out through switching inlout of
shunt capacitors,synchronous condensersand load management.This would
help in achieving better voltage profile, loss reduction, improved reliability,
quick detection of fault and restoration of service.
At a systems level, SCADA can provide status and measurementsfor
distribution feedersat the substation.Dictribution automation equipment can
monitor selectionalisingdeviceslike switches,intemrpters and fuses.It can also
operate switches for circuit reconfuration, control voltage, read customers'
meters, implement time-of-day pricing and switch customer equipment to
manage load. This equipment significantly improves the functionality of
distribution control centres.
SCADA can be used extensively fbr compilation of extensive data and
managementof distribution systems.Pilferage points too can be zeroedin on,
as the flow of power can be closely scrutinised.Here again, trippings due to
human effors can be avoided.Modern meteringsystemsusing electronicmeters,
automatic meter readers (AMRs), remote meteripg and spot billing can go a
long way in helping electric utility. These systbms can bring in additional
revenues and also reduce the time lag between billing and collection.
Distribution automation through SCADA systemsdirectly leads to increased
reliability of power for consumersand lower operating costs for the utility. It
results in fbrecastingaccurarte
clenrandand supply management,tasterrestoration of power in case of a tailure and ahernative routing of power in an
emergency,
rlll ' th c s tr
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$ y s tc l l l si s l h c ru r l ol clcontnrll i rci l i tyl l rl t l l l ow s l nstcr
execution of decisions.Manual errors and oversightsare eliminated. Besideson
line and rcal-timeinlbnnution, the systenrprovidesperiodic reportsthat help in
the analysis of performance of the power system. Distributi'on automation
combinesdistribution network monitoring functions with geographicalmapping,
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load
management,load despatchand intelligent metering.
Data Acquisition
Systems and Man-Machine
Interface
The use of computers nowadays encompassesall phases of power system
operation: planning, forecasting, scheduling, security assessment,and control.
www.muslimengineer.info
I
An energycontrolcentremanagesthesetasksandprovidesoptimaloperation
of the system.A typicalcontrolcentrecanperformthe following functions:
(i) Short,mediumandlong-termloadforecasting
(LF)
(ii) Systemplanning(SP)
(iii) Unit commitment(UC) and maintenance
scheduling(MS)
(v) State estimation (SE)
(vi) Economic dispatch (ED)
(vii) Load frequency control (LFC)
The above monitoring and control functions are performed in the hierarchical
order classified according to time scales.The functions perfotmed in the control
centre are based on the availability of a large information base and require
extensive software for data acquisition and processing.
At the generation level, the philosophy of 'distributed conffol' has
dramatically reduced the cabling cost within a plant and has the potential of
replacing traditional control rooms with distributed CRT/keyboard stations.
Data acquisition systems provide a supporting role to the application
software in a control centre. The data acquisition system (DAS) collects raw
data from selected points in the power system and converts these data into
engineering units. The data are checked for limit violations and status changes
and are sent to the data base for processing by the application software. The
real-time data baseprovides structuredinformation so that application programs
,
needing the information have direct and efficient accessto it.
The Man-Machine interface provides a link between the operator and the
software/lrardware used to control/monitor the power system. The interface
generally is a colour graphic display system. The control processorsinterface
with the control interface of the display system. The DAS and Man-Machine
interface support the following functions:
(i) Load/GenerationDispatching
(ii) Display and CRT control
(i i i ) Dat t BaseM aint cnancc
(i v) Alar r n Handling
(v) Supervisorycontrol
(vi I l) r ogr ar nnr ing
lbnct ions
(vii) DaLalogging
(vi i i ) Eventlogging
(ix) Real-time Network Analysis
With the introduction of higher size generating units, the monitoring
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performance, now all the utilities have installed DAS in their generating units
of sizes 200 MW and above. The DAS in a thermal power station collects the
following- inputs from various locations in the plant and converts thern into
engineering'units.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
l,@E+l
ModernPo@is
I
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Analog Inputs
(i) Pressures,flows, electrical parameters,etc.
(ii) Analog input of 0-10 V DC
(iii) Thermocoupleinputs
(iv) RTD input
Digital Inpurs:
(i) Contract outputs
(ii) Valve position, pressureand limit switches
All these process inputs are brought from the field through
cables to the
terminals. The computer processesthe information and ,uppli",
to the ManMachine interface to perform the following functions.
(i) Display on CRT screen
(ii) Graphic disptay of plant sub-systems
(iii) Data logging
(iv) Alarm generation
(v) Event logging
(vi) Trending of analogue variables
(vii) Performancecalculation
(viii) Generationof control signals
Some of the above functions are briefly discussedas follows.
a
aDJt,
"
power system engineerswho are adopting the low-cost and relatively powerful
computing devices in implementing their distributed DAS and control systems.
Computer control brings in powerful algorithms with the following advanpaurly ururzauon rn generailon, (u) savings in energy
and so in raw materials due to increased operational efficiency, (iii)
flexibiliiy
and modifiability, (iv) reduction in human drudgery, (v) improved operator
effectiveness.
Intelligent databaseprocessorswill becomemore cornmon in power systems
since the search, retrieval and updating activity can be speeded up.
New
functional concepts from the field of Artificial Intelligence (AI) will
be
integrated with power system monitoring, automatic restoration of power
networks, and real-time control.
Personalcomputers'(PCs)are being used in a wide range of power system
operationsincluding power stationcontrol, loaclmanagement,SCAOe systems,
protection, operator training, maintenance functions, administrative
data
processing,generatorexcitation control and control of distribution networks.
IT
enabled systems thus not only monitor and control the grid, but also improve
operational efficiencies and play a key part in maintaining the security
of the
power system.
RFFERE
NCES
I. Power Line Maga7ine,yol.7, No. l, October2002,pp 65_71.
2. A.K' Mahalanabis,D.P. Kothari and S.I. Ahson, ComputerAided power
Analysis and Control, TMH, New Delhi, 199g.
3. IEEE Tutorial course, Fundamentals of supervisory control system, l9gr.
4. IEEE Tutorial course, Energy control centre Design, 19g3.
The DAS softwarecontains programs to calculate periodically the efficiency
of various equipment like boiler, turbine, generator, condenser,fans,
heaters,
etc.
www.muslimengineer.info
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
System
F;;ff$:
AppBNDrx G
You can start MATLAB by double clicking on MATLAB icon on your Desktop
of your computer or by clicking on Start Menu followed by 'programs' and then
ciicking appropriate program group such as 'MATLAB Release 12,. you will
visualize a screenshown in Fis. G.1.
Click on this to change
the cunent directory
Commandprompt
Simulink browser
MATLAB has been developed by MathWorks Inc. It is a powerful software
package used for high performance scientific numerical computation, data
analysis and visualization.MATLAB stands for MATrix LABoratory. The
combination of analysis capabilities, flexibility, reliability and powerful
graphics makes MATLAB the main software package for power system
engineers.This is becauseunlike other programming languageswhere you have
to declare rnatricesand operateon them with their indices, MATLAB provides
matrix as one of the basic elements.It provides basic operations,as we will see
later,like addition, subtraction, multiplication by use of simple mathematical
operators. Also, we need not declare the type and size of any variable in
advance.It is dynamically decided dependingon what value we assignto it. But
MATLAB is case sensitive and so we have to be careful about the case of
variables while using them in our prograrns.
MATLAB gives an interactiveenvironmentwith hundredsof reliable and
accuratebuilrin functions. These functions help in providing the solutions to a
variety of mathematical problems including matrix algebra, linear systems,
differential equations,optimization,non-linearsystemsand many other typesof
scientific and technical computations.The most important featureof MATLAB
is its programming capability, which supports both types of programmingobject oriented and structuredprogramming and is very easy to learn and use
and allows user developedfunctions. It facilitates accessto FORTRAN and C
codes by means of external interfaces.There are several optional toolboxes for
simulating specializeelproblems of rJifferentareas anrder-tensionsto link up
NIATLAB and other programs. SIMULINK is a program build on top of
MATLAB environment,which along with its specializedproducts,enhancesthe
power of MATLAB for scientific simulationsand visualizations.
For a detailed description of commands, capabilities, MATLAB functions
and many other useful features, the reader is referred to MATLAB lJser's
Guide/lvlanual.
www.muslimengineer.info
Flg. G . l
The command prompt (characterisedby the symbol >>)'is the one after
which you type the commands.The main menu contains the submenussuch
as
Eile, Edit, Help, etc. If you want to start a new program file in MATLAB
(denotedby .m extension) one can click on File followed by new and
select the
desired M'file. This will open up a MATLAB File Editor/Debugger window
where you can enter your program and save it for later use. You can run this
program by typing its name in front of command prompt. Let us now learn some
basiccommands.
Matrix
Initialization
A matrix can be initialized by typing its name followed by = sign and an
opening squarebracket after which the user supplies the values and closes the
square brackets. Each element is separatedfrom the other by one or more
spacesor tabs.Each row of matrix is separatedfrom the other by pressing Enter
key at the end of each row or by giving semicolon at its
potiowing
"na.
examplesillustrate this.
7
2
-9
-3 2 -51
The above operation can also be achieved by typing
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t em*,
ModernPo@sis
64?,..1
t
If we do not give a semicolon at the end of closing square brackets,
MATLAB displaysthe value of matricesin the command winCow. If you do
not want MATLAB to disptay the results, just type semicolon(;) at the end
of the statement. That is why you rvill find that in our programmes,whenever
we want to display value of variablessay voltages,we havejust typed the name
of the variable without semicolon at the end, so that user will see the values
during the program. After the program is run with no values of the variables
being displayed, if the user wants to seethe values of any of these variables in
the program he can simply type the variable narne and see the values.
Common
Matrix
This stores determinant of matrix A in H.
VaIues
obtainseigenvaluesof rnatrixA and storestlremin K.
G.2
Operations
First let us declare some matrices
SPECIAL MATRICES AND PRE.DEFINED VARIABLES
AND SOME USEFUL OPERATORS
MATLAB has some preinitialised variables. These variables can directly be
used in programs.
pi
Addition
This gives the value of n
Adds matricesA and B and storesthem in matrix C
converts degrees d rnto radians and stores it in variable r.
Subtraction
inf
Subtracts matrix B from matrix A and stores the result.in D
You can specify a variable to have value as oo.
Multiplication
i andj
Multiplies two conformable matrices A and B and storesthe result in E
These are predefined variables whose value is equitl to sqrt (- 1). This is used
to define complex numbers. One can specify complex maffices as well.
Inverse
This calculatesthe inverse of matrix A by calculating the co-factors and stores
the result in F.
The above statement defines complex power S.
A word of caution here is that if we use i and j variables as loop counters
then we cannot use them for defining complex numbers. Hence you may find
that in some of the programswe have used i I and.i1 as loop variables instead
of i and 7.
Transpose
Single quote ( / ) operator is used to obtain transposeof matrix. In case the
matrix elernentis complex, it stores the conjugate of the element while taking
the transpose,e.g.
or
>>Gt=[1 3;45;63]'
also stores the transposeof matrix given in square brackets in matrix Gl
In case one does not want to take conjugate of elements while taking
transposeof complex matrix, one should use . 'operator instead of ' operator.
eps
This variabie is preinitiiized to 2-s2.
Identity
i
I
Matrix
To generate an identity matrix and store it in variabte K give the following
command
't
t
www.muslimengineer.info
I
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
[-ait+
Modernpo*er SystemAnalysis
#ffif
So K after this becomes
"
K=[1 0
0
0
0
1
0
0
1l
Zeros Matrix
generates
a 3 x 2 matrix whoseall elementsare zero and,storesthem in L.
Ones Matrix
generates
a 3 x 2 matrix whoseall elementsare one and storesthem in M.
: (Colon) operator
This is an important operator which can extract a submatrix from
a given
matrix. Matrix A is given as below
A= [1
?
5
6
5
4
7
8
9 1 0
3
1
e
3
r
2l
(.. oPERATOR)
This operation unlike complete matrix multiplication, multiplies element of one
eiement of other matrixlaving same index. However
matdx WitlreOnrespond'rng;
in latter case both the matrices must have equal dimensions.
We have used this operator in calculating complex powers,at the buses.
Say V = [0.845 + j*0.307 0.921 + j*0.248 0.966 + 7*0.410]
And I = [0.0654- j*0.432 0.876 - j*0.289 0.543 + j*0.210]'
Then the complex power S is calculated as
Here, conj is a built-in t'unctionwhich gives complex conjugateof its argument.
So S is obtainedas [- 0.0774+ 0.385Li 0.7404 + 0.4852t 0.6106+ 0.0198i]
Note here, that if the result is complex MATLAB automatically assignsi in the
result without any multiplication sign(*). But while giving the input as complex
number we have to use multiplication (*) along with i or 7.
G.4
COMMON
BUILT.IN
FUNCTIONS
sum
sum(A) gives the sum or total of all the elements of a given matrix A.
min
This command extracts all columns of matrix A correspondingto row and
Z
stores them in B.
SoBbecomes1269l,}lt
Now try this command.
The abovecofirmand extracts all the rows correspondingto column 3 of
matrix
A and stores them in matrix C.
So C becomes
l7
9
3
1l
This function can be used in two forms
(a) For comparing two scalar quantities
if either a or b is complex number, then its absolute value is taken for
comparison
(b) For finding minimum amongst a matrix or ilray
e . g .i f A = [ 6 - 3 ; 2 - 5 ]
>> min (A) results in - 5
Now try this command
abs
If applied to'a scalar, it gives the absolute positive value of that element
(magnitude).For example,
This command extracts a submatrix such that it contains all the elements
corresponding to row number 2 to 4 and column number 1 to 3.
>>x=3+j*4
SoD=[2
6
9
5
4
3
3
1l
9
www.muslimengineer.info
abs(x) gives 5
If applied to a matrix, it results in a matrix storing absolute values of all the
elementSof a given matrix.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
t -'__
I 041/
G,5
CONTROL STRUCTURES
comes to an end when k reachesor exceedsthe final value c. For example,
for i - 1:1:10,
a(i) - 1
end
This initializes every element of a to 1. If increment is of 1, as in this case,then
the increment part may as well be omitted and the above loop could be written
IF Statement
The generalfrom of the IF statementis
IF expression
statements
E L S Ee x p r e s s i o n
statements
ELSEIF
statements
END
AS
f o ri - 1 : 1 0 ,
a(i) = 1
end
E x pr es s io ni s a l o g i c a l e x p re s s i o nre sul ti ng i n
an answ er,true,(l ) or
'false'(0).
The logical expressioncan consist of
(i) an expressioncontaining relational operators
tabulatedalong with their
m e a n i n g si n T a b l eG . l .
Table G.1
Relational Operator
Meaning
Greater than
Greater than or Equal to
Less than
Less than or Equal to
Equal to
Not equal to
(ii) or many logical expressions
combincd with Logical operators.Various
logical operators and their meanings are given in
Table G.2.
While Loop
This loop repeatsa block of statementstill the condition given in the loop is true
while expression
statements
end
For example,
j - 1
while i <= 10
a(i) = 1
i = i + 1;
end
'
This loop makes first ten elementsof array a equal to l.
break statement
This staternent
allows one to exit prerr'aturelyfrom a for or while loop.
Table G.2
Logical Operator
Meaning
&
AND
OR
NOT
FOR Loops
This repeats a block of statementsprecletermined
number of times.
The most common {&m of FOR loop used is
f o r k = a : b t c ,
statements
end
where k is the loop variable which is initialised to
value of initial variable a.
If the final value (i'e. c) is not reached,the statements
in the body for the loop
www.muslimengineer.info
G.6
HOW TO RUN THE PROGRAMS GIVEN IN THIS
APPENDIX?
1. Copy theseprograms into the work subfolderunder MATLAB foider.
2. Justtype the name of the program without'.m' extensionand the program
will run.
3. If you wanLto copy them in some other folder say c:\power, then after
copying thosefiles in c:\power.changethe work folderto e :\power.you
can do this by clicking on toolbar containingthreeperiods ... which is on
the right side to the Current Directory on the top rilht corner.
4. You can see or edit theseprograms by going through Fite - open menu
and opening the appropriate file. However do not save those programs,
unless you are sure that you want the changes you have made to these
orisinal files.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ModernPowersystemAnalysis
ffiffi|
5. You canseewhicharethevariables
alreadydefinedby typingwhosin
front of command prompt. That is why you will normally find a clear
command at the beginning of our programs. This clears all the variables
defined so far from the memory, so that those variables do not interfere
G.7
SIMULINK
BASICS
SIMULINK is a software packagedeveloped by MathWorks Inc. which is one
of the most widely used softwarein academiaand industry for modeling and
simulating dynamical systems.It can be usedfor modeling linear and nonlinear
systems,either in continuoustime frame or sampledtime frame or even a hybrid
of the two. It provides a very easy drag-drop type Graphical user interface to
build the models in block diagram form. It has many built-in block-library
componentsthat you can useto model complex systems.If thesebuilt-in models
are not enough for you, SIMULINK allows you to have user defined blocks as
well. However, in this short appendix, we will try to cover some of the very
common blocks that one comesacrosswhile simulating a system.You can try
to construct the models given in the examples.
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How to Start?
You can start SIMULINK by simply clicking the simulink icon in the tools bar
or by typing Simulink in front of the MATLAB command prompt >>. This
opens up SIMULINK'hbrary browser, which should look similar to the one
shown in Fig. G2. There may be other tool boxes dependingupon the license
you have, The plus sign that you seein the right half of the window indicates
that there are more blocks availableunder the icon clicking on the (+) sign will
expand the library. Now for building up a new model click on.File and select
New Model. A blank model window is opened.Now all you have to do is to
select the block in the SIMULINK library browser and drop it on your model
window. Then connect them together and run the simulation. That is all.
An Example
Let us try to simulate a simple model where we take a sinusoidal input,
integrate it and observe the output. The steps are outlined as below.
1. Click on the Sources in the SIMULINK library browser window.
2. You are able to see various sources that SIMULINK provides. Scroll
down and you will see a Sine Wave sources icon.
3. Click on this sourcesicon and without releasing the mouse button drag
and drop it in your model window which is currently named as 'untitled'.
4. If you double-click on this source, you will be able to see Block
parametersfor sine wave which includes amplitude,frequency,phase,etc.
Let not change theseparametersright now. So click on cancel to go back.
www.muslimengineer.info
Si*r
Fig. G.2
5 . Similarly click on continuous library icon. You can now see various
built-in blocks such as derivative, integrator, transfer-function, statespaceetc. Select integrator block and drag-dropit in your model window.
6 . Now click on sinks and drag-drop scopeblock into your model. This is
one of the most common blocks used for displaying the values of the
blocks.
7. Now join output of sine-wave sourceto input of Integrator bloc(. This can
'
be done in two ways. Either you click the left button and drag mouse from
output of sine-wave source to input of Integrator block and leave left
button or otherwise click on right button and drag the mouse to form
connection from input of integrator block to output of sine,wave source.
8. Now in the main menu, click on Simulation and click Start. The
simulation runs and stops after the time specifiedby giving ready prompt
at bottom left'corner.
9. Now double-click on scope to see the output. Is something wrong? The
result is a sine wave of magnitude 2. Is there something wrong with
SIMULINK software?
No, we have in fact forgotten to specify the integration constAnt!
Integrationof sin ?is - cos d+ C. At 0=0, C - - 1. If we do not specify
any initial condition for output of the integrator,simulink assumesit to be
0 and calculates the constant. So it calculates
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffi
ModernPo
- cos 0+ C = 0 at r = 0 givinl C = 1. So the equationfor outputbecomes
- cos 0 + l. Thusnaturally,it startsfrom 0 at t = 0 andreachesits peak
valueof 2 at 0= n, i.e. 3.I4.
10. To rectifv this error. double click on Inte
,*rr-rr
r^t
in SourcesblockunderSimulink.We havemadeuseof this blockin
stability studies to provide constant mechanical input.
block and in the initial
conditionsenter- 1 whichshouldbe theoutputof theblockat t = 0. Now
run the simulationagainand seefor yourselfthat the resultis correct.
generatea sine wave of any amplitude, frequency and phase.
The reader is encouraged to work out, the examples given in this
Appendix to gain greater insight into the software.
Some Commonly used Blocks
1. Integrator We have already describedits use in the above example.
2. Transferfunction Using this block you can simulate a transfer function
of the form Z(s) = N(s)/D(s), where N(s) and D(s) are polynomials in s.
You can double click the block and enter the coefficients of s in numerator
and denominator of the expressionin ascendingorder of s which are to be
enclosed inside squarebrackets and separatedby.a space.
3. Sum You can find this block underMath in Simulink block. By default,
it has two inputs with both plus signs. You can modify it to have required
number of inputs to be surnmed up by specifying a string of + or
- depending upon the inputs. So if there are 3 inputs you can give the list
of signs as + - - . This will denoteone positive input and two inputs with
- signs which are often used to simuiate negative feedback.
4. Gain This block is alsofound under Math in Simdlink block. It is used
to simulate static eain. It can even have fractional values to act as
attenuator.
5. Switch This block is availableunder Non-linear block in Simulink. It
has 3 inputs with the top irrput being numbered 1. When the input number
2 equals or exceedsthe threshold value specified in the propertiesof this
block, it allows input number 1 to pass through, else it allows input
number 3 to pass through.
6. Mux and Demux Theseblocks are availableunder Signals and systems
block in Simulink. The Mux block combinesits inputs into a single output
and is mostly used to form a vector out of input scalar quantities.Demux
block does the reverse thing. It splits the vector quantity into multiple
scalar outputs.
7. Scope We have already describedits use. However; if you are plotting
a large number of points, click on properties toolbar and select Data
History tab. Then uncheckthe Limit data points box so that all points are
plotted. Also when the rvaveform does not appear smooth, in the general
tab of properties toolbar, select sample time instead of decimation in the
sample time box and enter a suitable value like le-3 (10-3). The scope
then uses the value at sample-time interval to plot.
8. Clock This block is used to supply time as a source input and is
available in Sources block under Simulink.
www.muslimengineer.info
G.8
SCRIPT AND FUNCTION
FILES
Types of m-files
There are two types of m-files used in Matlab programming:
(i) Script m-file - This file needs no input parametersand does not return
any values as output parameters. It is just a set of Matlab statements
which is stored in a file so that one can executethis set by just typing the
file name in front of the command prompt, eg. prograrnmesin G2 to Gl8
are script files.
(ii) Function m-files - This file acceptsinput agruments and return values as
output parameters. It can work with variables which belong to the
workspace as well as with the variables which are local to the functions.
These are useful for making your own function for a\ particular
application, eg. PolarTorect.m in Gl is a function m-file.
The basic structure of function m-file is given below:
(a) Function definition line - This is the first line of a function. It specifies
function name, number and order of input variables.
Its syntax is- function [output variables] = function-name (list of input
variables)
(b) First line of help - Whenever help is requested on this funciton or look
for is executedMATLAB displays this line.
Its syntax is - Vofunction-name help
(c) Help text - Whenever help is requestedon the function-name help text
is displayed by Matlab in addition to first help line.
Its syntax is - Vofunction-name (input variables)
(d) Body of the function - This consistsof codes acting on the set of input
variables to produce the output variables.
Tlccr
nqn
fhrrs
rlcwelnn
hic/hcr
nrrrn
rtrntrrqrnc
end
fi'rnnfinnc
qnA
qAA
fLarvrII
ham
fuvn
fulv
ha
existing library of functions and blocks.
G.9
SOME SAMPLE EXAMPLES
SOLVED BY MATLAB
In this section 18 solved examples of this book are solved again using
MATLAB/SIMULINK to encourage the reader to solve more power system
problems using MATLAB.
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
uodernpowerSystefn
nnalysis
ffil
Ex G.I
% T h i s f u n c t i o n c o n v e r t s p o 1a r t o r e c t a n g u l a r c o o r d i n a t e s
YoThe argumentt o t h i s f u n c t i o n i s 1 ) M a g n i t u d e
& ?)Angleis .in degrees
% This function has been used i
a p p e n dxi
f u n c t i o n r e c t = p o l a r T o r e c(ta , b )
i ( b * p i/ l g 0 ) ;
r e c t = a * c o s( b * p i / 1 S 0 ) + j * a * s n
function is used in solution of many of the examplessolved later.
Ex G.2 (Example S.O)
% Thls illustrates the Ferranti effect
% I t s i m u l a t e st h e e f f e c t b y v a r y i n g t h e l e n g t h o f l i n e f r o m
% z e r o ( r e c e i v i n g e n d ) t o 5 0 0 0 k mi n s t e p s o f 1 0 k m
% a n d p l o t s t h e s e n d i n ge n d v o l t a g e p h a s o r
% T h i s c o r r e s p o n d st o F i g . 5 . 1 3 a n d d a t a f r o m E x a m p l e5 . 6
cl ear
V R = 2 2 0 e 3 / s q(r3t ) ;
a 1p h a = 0 . 1 6 3 e - 3 ;
b e t a = 1. 0 6 8 e - 3 ;
I =5000;
k=1;
for i=0:10:1,
VS=(VR/z)*exp(a'lpha*i)*exp(j*beta*j)+(VR/Z)*exp(-alpha*i)*exp(j*beta*j);
X ( k )= r e a l( V S );
Y(k)=imag(VS);
' k=k+1;
end
p 1o t ( X ,Y )
Ex. G.3 (Example 8.7)
% T h i s P r o g r a mi l l u s t r a t e s t h e u s e o f d i f f e r e n t l i n e m o d e l sa s
% i n E x a m p ' l5e. 7
cl ear
f=50
| =300
7=4Q+j*lZ5
Y =I e - 3
PR=50e6/3
(sqrt (3) )
YR=?20e3/
p f 1o a d = 0 . 8
I R = P R / ( V R * pof a1 d )
z=7/1
y=Y
l1
% N o ww e c a l c u l a t e t h e s e n d i n g - e n dv o l t a g e , s e n d . i n g - e n cdu r r e n t
% a n d s e n d i n g - e npdf . b y f o l l o w i n g m e t h o d sf o r v a r i o u s l e n g t h s o F I i n e
www.muslimengineer.info
'
Appendix
G
% varying line lengths from 10 to 300 kmin steps of 10 km
% a n d c o m p a r et h e m a s a f u n c t i o n o f l i n e l e n g t h s .
% T h e m e t h o d sa r e
% l) Short Line approximatiom
N o m i n a l - P Im e t h o d
% 3 ) E x a c t t r a n s m i s s i o nl i n e e q u a t i o n s
% 4 ) A p p r o x i m a t i o no f e x a c t e q u a t i o n s
i =1.
for I =10:10:300,
ffi
T-
% Short Iine approximation
( 2 * 1) * I R
V s _ s h o r t l 'ni e( i ) = Y q +
Is shortline(i)=1P
s p f _ s h o r t l i n e( i ) = c o s( a n g 1e ( v s _ s h o r t li n e( i ) - a n g le ( I s _ s h o r t l i n e ( i ) ) ) )
S p o w e r _ s h o r t ln' ie( i ) = r e a l ( V s s h o r t l i n e( j ) * c o n j ( I s _ s h o r t l i n e( i ) ) )
%Nom
n ia l P I m e t h o d
4 = 1 +( y * 1) * ( z * 1 )/ Z
D=A
B=z*l
C = y * t* ( l + ( y * 1) * ( z * 1 )/ a )
*I
V s _ n o mnia lp i ( i ) = A * V R + BR
I s_nom'i
na'lpi (i ) =C*VR+D*l
R
Spf_nominalpi(i)=cos(ang1e(Vs_nominalp'i(i)-angte(Is_nominalpi(i))))
Spower_nomi
na'lpi ( i ) =real (Vs_nomi
nal pi ( i ) *conj ( I s_norpi
naI pN.i ) ) )
% E x a c t t r a n s m j s s i o nL i n e E q u a t j o n s
Lc=sqrt(z/y)
garnma=sqrt
(y*z)
V s _ e x a c t ( i ) = c o s h ( g a m m)a**Vl R + Z c * s i n h( g a n r m a)**1I R
I s _ e x a c t( i ) = ( I l l c ) * s i n h ( g a m m a *)1 + c o s h ( g a m m )a**IlR
S p f _ e x a c (t i ) = c o s( a n g 1e ( V s _ e x a c(t i ) - a n g 1e ( I s _ e x a c t( j ) ) ) )
(i ) =real (Vs_eiact (i ) *conj (Is_exlct (i ) ) )
Spower_exact
% A p p r o x i m a t i o no f a b o v e e x a c t e q u a t i o n s
4 = 1 + ( y * )l * ( z * t ) / Z
D=A
g = ( z * 1) * ( t + ( y * l ) * ( z * t ) / 6 )
C =( y * l ) * ( 1 + ( y * l ) * ( z * 1 )/ 6 )
(i ) =A*VR+B*IR
Vs_aPProx
I s _ a p p r o x( i ) = [ * ! P + P *1P
S p f _ a p p r o (xi ) = c o s( a n g 1e ( V s _ a p p r o(xI ) - a n gI e ( I s _ a p p r o (xi ) ) ) )
S p o w e r _ a p p n o x) (=i r e a 1( V s _ a p p r o(x' i) * c o n j ( I s _ a p p r o x ( i) ) )
point(j)=i
i=i+L
end
% T h e r e a d e r c a n u n c o m m e natn y o f t h e f o u r p l o t s t a t e m e n t sg i v e n b e l o w
% b y r e m o v i n g t h e p e r c e n l a g es i g n a g a i n s t t h a t s t a t e m e n t
% f o r e x . i n t h e p l o t s t a t e m e n t u n c o m m e n t ebde l o w
% i t p l o t s t h e s e n d i n ge n d v o ' l t a g e sf o r s h o r t l i n e m o d e l i n r e d
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffi
__f-% b y n o m i n a l p i - m o d e 1i n g r e e n , b y e x a c t p a r a m e t e r s
moderi n b l a c k
% a n db y a p p r o x . p i m o d e l i n b l u e
p l o t ( p o ' i n t a, b s( V s _ s h o r t il n e ) ' r ' p o i
n t , a b s( V s _ n o mn ia . l p )i , , ' , p o in t , a b s . . .
,
,
( V s _ e x a c t ) ,' b ' , D
p o in t , a b s( V s _ a p p r o, x, k) , )
b u s e s = m a x ( m a x ( y( 2d,a: )t)a, m a x( y d a t a( 3 ,: ) ) )
(e'lemen
A=zeros
t s , buses)
f o r i = 1 : e le m e n t s
% ydata(i,?) gives the ,from' bus no.
% f h e e n t r y c o r r o s p o n d i n gt o c o l u m n c o r r o s p o n d i n gt o
this bus
% i n A m a t r i x i s m a d e1 i f t h i s i s n o t g r o u n db u i
i f y d a t a ( i , 2 1- = g
A(i,ydata(i,Z;;=1.
end
% ydata(i,3) givesthe 'to' bus no.
% T h e e n t r y c o r r o s p o n d i n gt o c o l u m nc o r r e s p o n d i n gt o
this bus
% i n A m a t r i x i s m a d e1 i f t h i s i s n o t g r o u n db u s
, p o in t , a b s . . .
, p o in t , a b s . . .
Ex G.4 (Eiample
6,2)
% T h i s p r o g r a mf o r m s y B U Sb y S i n g u l a r T r a n s f o r m a t i o n
& T h e d a t a f o r t h i s p r o g r a mi s a p r i m i t i v e
a d m i t t a n c em a t r i x y
% whichis to be given in the foilowing format
and stored in ydata.
% G r o u n di s g i v e n a s b u s n o 0 .
% I f t h e e l e m e n ti s n o t m u t u a l ] y c o u p l e dw i t h
any other element,
% t h e ( ' t h e e n t r y c o r r e s p o n dni g t o 4 t h a n d
5 t h c o i u m no f y d a t a
% has'Io be zero
%
% e l e m e nnt o I c o n n e c t e dI
| F r o mI T o I
?
%
| B u s n oI B u s nIo
i f y d a t a( ' i, 3 ; - = 9
A(i,ydata(i,3;;=-i
end
end
YBUS=A'*yprimitive*A
y (self)
lMutually I
t c o u p l e dt o l
v(mutual)
Ex G.5 (data same as Example 6.2)
%
ydata=[ I
2
3
4
5
t
I
2
?
3
2
3
3
4
4
L/(0.05+;*9.15; 0
t / ( 0 . 1 + j * 0 . 3)
0
t/ (0.1s+j*0.4s) 0
1 /( 0 . 1 0 + j * 9 . 3 9 ; 0
1 / ( O. 0 5 +*j 0 .1 5 )
0
0
0
0
0
ol;
% f o r m p r i m i t i v e y m a t r i x f r o mt h i s d a t a a n d
initialize it to zero.
% to start with
e l e m e n t s = m (ayxd a t a( : , 1 ) )
y p r i m it i v e = z e r o s( e 1e m e n t se 1e m e n t s
,
)
% P r o c e s sy d a t a m a t r . i x r o w w j s et o f o r m y p r i m i
ti ve
for i=1:elements,
y p r i m it i v e( i , . i )= y d a t a( i , 4 )
% A l s o ' i f t h e e l e m e n ti s m u t u a l l y c o u p le d w i
th any other el ement
e o ( w h o s ee l e m e n tn o i s i n d i c a t e d
i n 5 t h c o l u m no f y d a t a a b o v e )
% t h e c o r r e s p o n d i n gc o l u m nn o i n t h e i t h r o w
i s m a d ee q u a l t o y ( m u t u a ). l
i f ( y d a t a ( i , S )- = g ;
j i s i h e e l e m e n tn o w i t h w h i c h j t h
e l e m e n ti s m u t u a l l y c o u p l e d
j = y d a t a( j , 5 )
Y m u t u a=i ; ' d a t a ( i , 6 )
yprimitive(i,j) = ymutual
end
end
% F o r m B u s ' i n c i d e n c em a t r i x A f r o m y d a t a
www.muslimengineer.info
% T h i s p r o g r a mf o r m s y B U Sb y , a d d i n g o n e e r e m e n t
at a t.ime
% T h e d a t a f o r t h i s p r o g r a mi s a p r i m i t i v e a d m i t t a n c e
matrix y
% which is to be given in the following format
and stored in ydata
% G r o u n di s g i v e n a s b u s n o . 0
% I f t h e e l e m e n ti s n o t m u t u a l l y c o u p l e dw i t h a n y
other element,
% t h e n t h e e n t r y c o r r e s p o n d i n gt o 4 t h a n d 5 t h c o i u m n
of ydata
% has to be zero
% T h e d a t a m u s t b e a r r a n g e di n a s c e n d i n go r d e r o f
e l e m e n tn o .
%
% element
no I connected
yo
%
t5
y d a t a = [1
2
3
4
5
I
lFromlTo
I
IB u s n Ioe u s nI o
1
1
2
2
2
3
3
4
v? ,
rA
y ( s e lf )
L / ( O. O s +*j 0 .1 5 )
r/(o.i+j*0.3)
t/(0.r5+j*0.45)
t / ( o. 1 9 +*30 . 3 0 )
1 lln
nc':+^
1r\
r/\v.s3r..;"U.IC,f
l M u t u al yl I
I coupled t o I
0
0
0
0
U
y ( m u t u a) l
0
0
0
0
oj;
% F o l 1 o w j n gs t a t e m e n tg i v e s m a x i m u m
n o . o f e l e m e n t sb y c a l c u l a t i n g t h e
% maximum
o u t o f a l l e l e m e n t s( d e n o t e db y : ) o f f i r s t c o l u m n
of yaata
\
e l e m e n t s = m a( yx d a t a( : , l ) )
% t h i s g i v e s n o . o f b u s e sw h i c h i s n o t h i n g b u t t h e m a x i m u m
entry
% o u t o f ? n d a n d 3 r d c o l u m no f y d a t a w h i c h i s ' f r o m ' a n d , t o '
aolrrn,
b u s e s = m (am
x a x( y d a t a( z , Z ) ) , m a x( y d a t a( : , 3 )
)) ;
هاجتإلا يمالسإلا- ةنجل روبلا تالاصتإلاو
ffi,W
uooernPowerSystem
nnavsis
nppencix
o
Y B U S = z e(rbouss e sb,u s e s; )
Y B U S ( k =, l Y
) B U S ( k -, l y) s m ( 2 , 2 ) ;
Y B U S ( l ,=
k )Y B U S ( k , l ) ;
i 1 ,k ) + y s m ( 1 , 2 ) ;
i 1, k ) = Y B U( S
Y B U( S
k ,i 1 ) = Y B U( iS1 ,k ) ;
Y B U( S
YBUS(II,I) = YBtll(j!,L) + v s m ( 1 . 2 ) :
for row=1:elements,
% i f y d a t a ( r o w , 5 ) i s z e r o t h a t m e a n st h e c o r r e s p o n d i n ge l e m e n t
% i s n o t m u t u a l l y c o u p l e dw i t h a n y o t h e r e l e m e n t
j f y d a t a ( r o w , 5 )= = 0
i 1 =ydata(row,2);
j1 =ydata(row,3);
if i1 -= 0 & jl -= 0
Y B U S ( i 1 , ' i 1= ) Y B U S ( j 1 , j 1+) y d a t a ( r o w , 4 ) ;
Y B U S ( i 1 , j l )= Y B U S ( i 1 , j 1 -) y d a t a ( r o w , 4 ) ;
Y B U S ( j 1 , i 1=) Y B U S ( i 1 , j l ) ;
Y B U S ( j 1 , j 1=) Y B U S ( j 1 , j 1+) y d a t a ( r o w , 4 ) ;
end
if il ==0 & jl -=Q
Y B U S ( i 1 , j 1=) Y B U S ( i 1 , i 1+) y d a t a ( r o w , 4 ) ;
end
if il -= 0 & jl ==Q
Y B U S ( j 1 , j 1=) Y B U S ( j 1 , j 1+) y d a t a ( r o w , 4 ) ;
end
end
e o i f y d a t a ( r o w , 5 )i s N O Tz e r o t h a t m e a n st h e c o r r e s p o n d i n ge l e m e n t
% i s m u t u a l l y c o u p l e dw i t h e l e m e n tg i v e n i n y d a t a ( r o w , 5 )
i f y d a t a ( r o w , 5 )- = g
i 1 =ydata(row,2);
j 1 = y d a t a ( r o w , 3; )
% m u t u a l w i t hg i v e s t h e e l e m e n tn o w i t h w h i c h t h e c u r r e n t e l e m e n t
% i s m u t u a l l y c o u p l e dw j t h k a n d 1 g l v e t h e b u s n o s b e t w e e n
% w h i c h t h e m u t u a l l y c o u p l e de l e m e n ti s c o n n e c t e d
m u t u a*l . i 1 1 = y d a t( ai 1 , 5 ) ;
; q = y d a t(am u t u awl i t h , 2 ) ;
1= y d a t a( m u t u awl it h , 3 ) ;
zsI=I/ydata(row,4);
zs?=L/ydata(mutualw'ith,4) ;
zm=l/ydata(row,6);
7 5 6 =[ z s 1 z m
zn zs?l;
ysm='i
nv (zsm);
% F o l ' l o w i n gi f b l o c k g i v e s t h e n e c e s s a r ym o d i f i c a t i o n s i n Y B U S
% w h e nn o n eo f t h e b u s e s i s r e f e r e n c e ( g r o u n d ) b u s .
if il -= 0 & jl -= 0 & k -=0 & l-=0
Y B U S ( ' i 1 , i 1= ) Y B U S i(,i i 1 ) + y s m ( 1 , 1 ) ;
Y B U S ( j 1 , j l )= Y B U S ( j 1 , j l )+ y s m ( 1 , 1 ) ;
Y B U S ( k ), = Y B U S ( k ,) + y s m ( 2 , 2 ) ;
Y B U S ( l , l )= Y B U S ,( ll ) + y s n ( Z , 2 ) ;
Y B U SI(,ij 1 ) = Y B U S 1( i, j 1 ) - y s m ( 1 , 1 ) ;
Y B U S ( j 1 , i 1=) Y B U S 1( i, j 1 ) ;
www.muslimengineer.info
Y B U(Sl , j 1 ) = Y B U(Sj 1 , I ) ;
Y B U S ( i 1 , . =| ) Y B U S ( i 1 , 1-) y s m ( 1 , 2 ) ;
Y B U(S. |, i 1 ) = Y B U(Sj 1 , . |) ;
Y B U S ( j 1 , k=) Y B U S ( j 1 , k-) y s m ( 1 , 2 ) ;
Y B U S ( k , j l )= Y B U S ( j 1 , ;k )
e nd
i f i 1 = = 0 & j 1 - = 0 & k - = 0& l - = 0
= )Y B U S ( i 1 , i 1
+ )y s m ( 1 , 1 ) ;
YBUS(j1,j1
Y B U S ( k ,=k )Y B U S ( k ,+k )y s m ( 2 , 2 ) ;
. | )y s n ( ? , 2 ) ;
Y B U S, (l )l = Y B U S ( 1 , +
Y B U S ( k ,=l ) Y B U S ( k , -l ) y s m ( 2 , 2 ) ;
Y B U S, (kl ) = Y B U S ( k) ,; l
Y B U S ( j 1 , =. | )Y B U S ( i 1 , +1 )y s m ( 1 , 2 ) ;
Y B U S ( . l , j=1 )Y B U S ( i l ,;l )
Y B U S ( j 1 , =k )Y B U S ( i 1 , -k )y s m ( 1 , ;2 )
Y B U S ( k , j=
1 )Y B U S ( i 1 ;, k )
end
i f i 1 - = 0 & j l = = Q& k - = 0& l - = 0
Y B U S ( ji 11 ), = Y B U S ( i 1 , ' +i 1y)s m ( 1 , 1 ) ;
k )y s n ( 2 , 2 ) ;
Y B U S ( k , k= )Y B U S ( k , +
Y B U S ( l , l=) Y B U S ( 1 ,+' l )y s n ( z , 2 ) ;
Y B U S ( k ,=l ) Y B U S ( k , -l ) y s m ( 2 , 2 ) ;
Y B U S ( l , k= )Y B U S ( k , l ) ;
Y B U S ( i 1 , =k )Y B U S ( i 1 , +k )y s m ( 1 , 2 ) ;
Y B U S ( k1 ,) i = Y B U S ( ' i l ;, k )
Y B U S ( i 1 , =. | )Y B U S ( i 1 , -1 )y s m ( 1 , 2 ) ;
i 1 , . )| ;
l , ' i1 ) = Y B U( S
Y B U( S
e nd
j f i 1 - = 0 & j 1 - = Q& k = = 0& l - = 0
i 1 ,i 1 ) + y s m ( 1 , l;. )
i 1 ,i 1 ) = Y B U( S
Y B U( S
= )Y B U S ( i 1 , i 1
+ )y s m ( 1 , 1 ) ;
YBUS(j1,i1
Y B U S ( l , l=) Y B U S ( 1 ,+. |y) s m ( 2 , 2 ) ;
i 1 , i 1 ) - y s m ( 1 , l;. )
i 1 ,j 1 ) = Y B U( S
Y B U( S
= )Y B U S ( i 1 , i; 1 )
YBUS(j1,j1
Y B U S ( j 1 , =1 )Y B U S ( j 1 , +1 )y s m ( I , 2 ) ;
Y B U S ( . l , j=1 )Y B U S ( i 1 ;, 1 )
Y B U S ( i 1 , =. | )Y B U S ( i 1 , -. | y) s m ( 1 , 2 ) ;
Y B U S ( l , j l=) Y B U S ( i 1 , 1 ) ;
e nd
if
i1-=0 & jl-=Q & k-=0 & l==0
Y B U S ( i 1 , i 1=) Y B U S ( i 1 , i 1+) y s m ( 1 , 1 ) ;
Y B U S ( j 1 , j l )= Y B U S ( j 1 , i 1+) y s m ( 1 , 1 ) ;
Y B U S ( k , k=) Y B U S ( k , k+) y s m ( 2 , 2 ) ;
Y B U S ( i 1 , j 1=) Y B U S ( i 1 , j 1 -) y s m ( 1 , 1 ) ;
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Appendixc
Y B U S ( j 1 , i =1 )y B U S ( i t , j t ) ;
Y B U S ( i 1 ,=k )Y B U S ( i 1 , +k )y s m ( 1 , 2 ) ;
k ,i 1 ) = Y B U( S
Y B U( S
i 1 ,k ) ;
Y B U S ( i 1 ,=k )y B U S ( i 1 , k_)y s m ( 1 , 2 ) ;
Y B U S ( k , i =l ) Y B U S ( j 1 ;, k )
NilFffi
T-
( i ) =1t
t ypechanged
else
t ype( i ) =2;
end
end
end
sumyv=0;
for k=L:n,
if(i -= k)
s u m y v = s u m y v(+i ,Yk ) * V( k ) ;
end
end
v ( i ) = ( 1 / Y( i , i ) ) * ( ( p ( i ) - j * Q ( i ) ) / c o n j ( v ( i ) ) - s u m y v;)
if type(i;==2 & typechanged(i)-=1,
V ( i ) = p e 1a r T o r e c t ( V m a g f i x e d) (, ia n g 1e ( V ( i ) ) * 1 g 0 / p i) ;
end
YBUS
Ex G.6
%This'is program
f o r g a u s ss i e d e rL o a df r o w . T h ed a t a i s f r o m E x a m p l e
6.5
clear
n=4
V = [ 1 . 0 41 . 0 4 1 t ]
Y=[3-i*9
-2+i*6
-1+;*3
0
-2+i*6
-l+5*3
3 6 6 6 _ j * 1 1 _ 0 . 6 6 6j *+2
- 0 . 6 6 6j *+ 2 3 . 6 6 6j -* 1 1
- 1+ 3 ' * 3
-2+tr*6
0
_t+3*3
_Z+i*6
3-j"91
t y p e = o n (ens, 1 )
t y p e c h a n g e d = z e( nr o,1s) Q li m it m a x = z e r(ons,1
i
Q li m it m in = z e r o(sn ,l )
V m agfxied= z e ro(n
s , 1)
t Y pe( 2) = 2
Q li m it m ax( 2 )= 1. g
Q 1i m it m in( 2 )= 9 .2
V m agfxi ed(2=) 1 .0 4
d i f f = 1 0 ;n o o f it e r = 1
Vprev=V;
w h i l e ( d i f f > O . 0 0 0 0| 1n o o f i t e r = = 1 ) ,
a b s( V )
abs( V pr ev )
%paus e
Vprev=V;
P = [ i n f0 . 5 - 1 0 . 3 ] ;
Q = [ i n f0 0 . 5 - 0 . 1 ] ;
$ = [ i n f + j * i n f ( 0 . 5 - j * 0 . 2 )( - 1 . 0 + j * 0 . 5 ) ( 0 . 1 - 1 * g . 1 ) ] ;
for i =Zin,
i f t y p e ( i ) = = 2l t y p e c h a n g e d ( i ; = = 1 ,
imi
tmi
n(i)),
).Q,
",llilil;:il;l#ii#ll,
ill
= Q il m i t m i ( i
Q ( j)
else
n
);
Q('i)=Qlimitmax(i);
end
t Y P e( i ) = 1 ;
www.muslimengineer.info
end
d i f f = m a x( a b s( a b s( U( Z z n )) - a b s ( V p r e v( Z : n ) ) ) ) ;
n o o f i t e r = n o o f it e r + l ;
end
V
Ex. G.7 (Example 6.6)
% Programf o r l o a d f l o w b y N e w t o n - R a p h s oMne t h o d .
cl ear;
% n s t a n d s f o r n u m b e ro f b u s e s
n=3;
% Y , v o l t a g e sa t t h o s eb u s e sa r e i n i t j a l j s e d
v = [ 1 . 0 41 . 0 1 . 0 4 ] ;
%Y is YBus
Y = [ 5 . 8 8 2 2 8 - j ' t 2 3 . 5 0 5- 2
1 .49 4 2 7 + j * L t . t 6 t 6 - Z. 9 4 2+7j * 7 I . t 6 76
-2 .9427
+j*Lt .7676 5.88228_j*23
.505L4 - 2 . 9427
+j* I I . 7676
- ? . 9 4 2+7j * L l . 76 76 - 2 . 9 4 2+7j * t t . 76 t 6
