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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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.PowerSystemSecurity State Estimation Powersystem compensationincludingsvs and FACTS Load Forecasting VoltageStability
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The book is very comprehensive,well organised,up-to-dateand (above all) lucid and easy to follow for self-study.lt is ampiy illustratedw1h solved examplesfor everyconceptand technique.
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Modern PorroerSYstem Third Edition
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 research papers 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
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
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.
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
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.
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
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
r.8
29
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
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
5. Characteristics and Performance of power Transmission Lines
6.1 6.2
IIYJEIETLLCJ
lntrotluction 184 NetworkModel Formulation I85
)20
LJ7
242
7. Optimal System Operation
95
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
Introduction 128 Short TransmissionLine 129 Medium TransmissionLine i37 The Long Transmission Line-Rigorous Solution I 39 Interpretationof the Long Line Equations 143 FerrantiEffect 150 TunedPowerLines 151 The Equivalent Circuit of a Long Line 152 Power Flow througha TransmissionLine I58 Me th o d so l ' V o l ra g eC o n trol 173 Problems 180 References 183
6. Load Flow Studies
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 Introduction g5 4.2 Single-phaseSolutionof Balanced Three-phaseNetworks 95 4.3 One-LineDiagram and Impedanceor ReactanceDiagram 98 4.4 Per Unit (PU) System 99 4.5 ComplexPower 105 4.6 SynchronousMachine 108 4.7 Representation of Loads I2I Problems 125 References 127
5.1 5.2 5.3 5.4 5.5 5.6 5.1 5.8 5.9 5 .1 0
.
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
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 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9
ll.
Introduction 369 SymmetricalComponentTransformation370 Phase Shift in Star-DeltaTransformers 377 SequenceImpedancesof TransmissionLines 379 SequenceImpedancesand SequenceNetwork of Power Systern 381 SequenceImpedancesand Networks of SynchronousMachine 381 SequenceImpedancesof TransmissionLines 385 SequenceImpedancesand Networks of Transformers 386 Constructionof SequenceNetworks of a Power System 389 Problems 393 References 396
Unsymmetrical Fault Analysis
Problems 506 References 508 13. Power System Security 13.1 13.2 13.3 13.4 13.5 13.6
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 12.2 12.3 12.4 I2.5 12.6 12.7 I2.8
Introduction 433 Dynamics of a SynchronousMachine 435 Power Angle Equation 440 Node Elimination Technique 444 SimpleSystems 451 Steady State Stability 454 Transient Stability 459 Fq'-ralArea Criterion 461
Introduction 510 SystemStateClassification 512 SecurityAnalysis 512 ContingencyAnalysis 516 SensitivityFactors 520 Power System Voltage Stability 524 References 529
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
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 17.2 17.3 11.4
In tro d u c ti o n 5 9 1 Comparisonof Angle and Voltage Stability 592 ReactivePower Flow and Voltage Collapse 593 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
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.
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 llfenislable -The 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 developcdcorrntrieshas 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
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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.
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.
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
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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 |
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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",.:''
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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
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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,?" --
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(iiD the consumermay be asked to use shunt capacitorsfor improving the
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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.
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"
3' staggering the offrce timings 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
1'1
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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% L
_ Average demand Installedcapacity
..
Plant usef(tctor Actual energyproduced _ _ --_ (kWh) -. . -,:- -plant capacity(kw) x Time (in hours)th" plunrh^
I ' b;i"
il;;ti""
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
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
InPut dematrd=
= 30 x 1.000= Rs 30,000
i
Annual running chargesdue to kwh consumed = 1000x 0.8 x 48 x 52 x 0.10 = Rs 1. 99. 680
Introduction
WI I
t
Total charges/annum = Rs 2,36,gg0 Max. demand correspondingto tariff(a) j 950 kVA Annual running chargesfor kWh consumed
=950x0.8x48x52x0.10 = Rs t,89,696 Total= Rs 2,20,096 Therefore, tariff (a) is economical.
.'.600 +0.03 or
l"'
4E-too-o.r3dE= o
dP
dP
dE=3m dP dE=dPxt
From triangles ADF and ABC,
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
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
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 unit. load from the base
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
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.
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
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.
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. 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'
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 structure of 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
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
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.
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.
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.
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
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
where W = dischargem3ls through turbine p = densiry 1000 kg/m3 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 away from 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,
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. Controlrods
Fuelrods_
* Existing capacity (small hydro) is 1341 MW as on June 200I. Total estimated potentialis 15000 MW. Waterintake
W
iiiriociucrion
ModernPo*", system An"tysis
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
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.
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
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
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
Introduction
Modern Power System Anatysis
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
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").
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.
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
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 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 -E n 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
lntroduction
Modem Pow'er systemAnatysis
ffiffi|
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
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. 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
HE FI
Introduction power to the load centres,transmissionlines of up to 110 were constructed.
and peak load in the regions of the country with projectedenergy requirement year 2011-12
[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
420 890 1095 2720
7503 t6372 27074 71060
\'./ 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 basic ancl heavy inclustriesand thus there was a need to step up power generation.The total installedcapacity which was around 3420 MW at the end of tn" First Five year Plan became 5700 MW at the end of the SecondFive year plan. The introduction of 230 kv transmissionvoltage came up in Tarnil
DieseI
398
=2700 MW renewable
Total
r4704 28640 42240 101630
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 BIIEL Today singrauli' at cornmissioned was first 500 MW turbo-generator in the manufacturers equipment plant power is consideredone of the major world. T.7
ENERGY CONSERVATION
of energy' we should resort Energy conservationis the cheapestnew source (discussedearlier), and to various conservationmeasuressuch as cogeneration
,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.
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
Modernpo*", Syster Anulyri,
I nt r oduct ion
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.
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
W
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
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 ntphasisis 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.
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.
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
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.
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.
rytrfi\ -----r-----
powerSvstemAnaivsis tutodern
A large central computer is used in central load despatch centres for cc
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
'.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.
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.
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 l. Nagrath,I.J. and D.P. Kothari, Electric Machines,Tata McGraw-Hill. New Delhi. 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).
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.
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
illffilf?;r
L*",,;i;;.";;r""i"J
lines'rhis transmission ii overhead
-r- --'i+r- +ha caricq line narameters,
#;;t;
ffiffir"##r'"*
1'e' lnductance
and resistance'
oi.itiuu,.datongthelineandthevtogether
th" ,"ri., imPedance of the line' i"; 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 2,2
DEFINITION
OF INDUCTANCE
given by Voltage induced in a circuit is
, =VY
(2.r)
the circuit in weber-turns(Wb-T)' ", ,ti" flux linkagesof
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
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
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! t'l
-ir-."
r1/
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)
/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)
\rrr') \r") FromEqs.(2.9) and(2.10),we obtain =)!H,. t
AT/m
2Tr"
(2.rr)
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.
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
*Throughout the book ln denotes natural logarithm (base e), while log denotes logarithm to base 10.
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 rl
(2.Zra)
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
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
li$Bfr f--
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.
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 permeabilityof 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.
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.
\^
Dt
)in
, D i + 4 ' u ! - + . . . +1 ,ln rl. D,z
+... + In The above equation can be reorganizedas
)i = 2xro'[[r,
But,
h + +h n [ * . +
I,m]+..+
+ (/, ln D r + I r h D z + . . . + I i k D , + . . + I n ^ O , ) - (1, + Iz +... + In-). I,
lnductanec
Substiiuti'g for /n in the secondterm of Eq. (2.29)and simplifying, we have f/
)i' = 2 * to-tlI t, ml * L,h L_.+...+1, rna Dir'-""'Diz""''''^'rti * I,l n = e+ . ..* l i l n at-
U
Dt
, (
^, \
D,2
+...+I^t" wb-r/m +) 2.7
INDUCTANCE
(2.30)
We are now ready to study the inductanceof transmissionlines 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
Wb_T/m
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
(2.3r)
, _Lu,,_ \+ 12+...+L,
_A
=-
.*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
(D,rD,r. . . D,,.. . D,n)r'"
(Dnr,.. . Dnj,... Dr^,17r/ntn
[ ( D n . .D . u . . .D r n ) . . . ( D i t . . n , t . 4.
Htm (2.33)
(Dnt... Dni ... D rn)]r,n'
o
2
^^' 6 - J - 1 |. ." .' *' l'rn t ) Diz, D,^, )
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))
r' r ' , ' ! - + 1-", l---n- 1
f.. k,...i
Llnes
I a /1 lA\t^cr Applying Eq. (2-30) ro filamenr i of conductorA, weobtain its flux tintages u,
zxfir !
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. approachln = t o. Also for the sake of symmetry, denoting ,.{^ D,,,-wi have
. ^i T---^--! ur I rallsmlssfon
,Frl a s raqi or rruagn a a
-
L('
l i = z x 1 o rI t r n * *
anrl
I
,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
"
of
rnl denominaror isdefin"o u,;;;w;;:;;:;,1;::; I!:::::1li:*Ti:r{. of conduc torA,and,s "r';#;;;":::"*
!:{Kr,"-li:t:t:y? GMD is also called
auureviate; ;,
geometric mean radius (GMR). "J In terms of the above symbols, we can write Eq (2.33) as
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 shownin 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
Fig. 2.8
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 L^^ = 0.461 los " 0.367 8'8 = 0.85 mH/
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
Fig. 2.10 Cross-sectionalview of a three-phaseline with unsymmetricalspacing
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 j * o - r r , r , Dt,)
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 " D, uzr
/
c
b
(l)
i'-) \7
*o-rrc'l
() - 1 u
r c t l I , , h ! r , , t n( D n D r D l t ) t / 3 \
4
,
T2
+1.ln (Dt2D2.)3t)t/3
F19.2.12 Effect of transpositionon Inducedvoltage of a telephoneline
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,
) 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
The harmonic line currentsare troublesomein two wavs: (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 alwaysbe present 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+ Iu* Ir=J[o
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.
64
I
Modern Power System Analysis
I
lnductanceand Resistanceof TransmissionLines
t a
b
c
,r"n.'f(')
(. 1
(:;
--. 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 15 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^
Vn = 0 .9 4 2 x 1 06 = 100 V (b) From Eq. (2.30), the flux linkages of the conductor a are (
r
I
r
\
)o=2x ro-?[ r,rnl* ru rn**r rn* | wu-r" r'o D 2D)
\ 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
\)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
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
o Section 1
Section2
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 Self GMD in section1 of phase a (i.e., conductorsa and a/) rs 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) the over parallel circuit each of conductors of rotation Becauseof the cyclic The section. transposition each in same the remains also D. transpositioncycle, reader should verify this for sections2 and 3 in Fig. 2.15. The inductancePer Phaseis = (r')t''qtt3hrt6
\0,
o
Section 3
L = 2 x 1 0 - 7l o
Ds
'|l'ltl"tlt)u = z x ,ot ,n't'u (r')t'' qr/3hrt6
Fig. 2.15 Arrangementof conductorsof a double-circuitthree-phaseline
=2x10-7h
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
D',n
( 2,,u(q')"'[a)"' ] rvpnur.rrn(2.4s)
[' \r') \q) The self inductanceof each circuit is givenby (2)'''p Lr = 2 x 10- 7ln
r
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)
di
where M is the mutual inductancebetweenthe 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
(}
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'
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
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 same reactancevalue 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
(2.s0)
R = resistanceat temperature0"C ao = temperaturecoefficient of the conductorat 0"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
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
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.
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. c'
I oi-----z.sm ---j C) Io c ) l1om-l 4 m
b
I
'
10. Kimbark, E.W., Electrical Transmission of Power and Signals, John Wiley, New York, 1949.
.)
Paper I l.
Fig. P-2.11
a
b
() -i- 1m i'1m
b/
a/ l
t I
-lI
l
1m
l
>l< I
Reichman.J.,
'Bundled
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.
(_)
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.
r.___zsm___l ')
New Delhi, 1997. 4. Stevenson,W.D., Elements of Power System Analvsis,4th edn, Mccraw-Hill, York, 1982.
1. Woodruff. L.F., Principles of Electric Pov,er Trun.snissiorr,John Wiley & Sons, New York, 1947.
l-
I
Electric'ttl Transrnission and Distributiort Book, Westinghouse Electric and Manufacturing Co., East Pittsburgh, Pennsylvania, 1964. 2. Waddicor, H., Principles of Electric Power Transmission, 5th edn, Chapman and Hall, London, 1964. 3. Nagrath, I.J. and D.P. Kothari, Electric Machines, 2nd edn, Tata McGraw-Hill,
5. Edison Electric Institute, EHV Transmission Line Reference Book, 1968. 6. Thc Aluminium Association, Aluminium Electrical Conductor Handboo,t. New York, 1971.
4m
b
l.
a) \..,_/ 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.
Conductor Voltage Gradient Calculations," AIEE Trans.
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
Figure 3.2 showsa group of parallelchargedconductors.It is assumedthat the conductorsare far removed from the ground and are sufficiently removed from 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 eachconductorremainsuniformly distributedaroundits peripheryand length. 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
10-12 F/m. Relative
(3.1)
i,o - --
Fig. 3.2 A group of parallelcharged conductors
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
--J
-
l
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,
Since
, l ; t ( n " h*D" ^ + )
(3.3)
Qr,= - qu, we have
-
In (D / (ror)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
Fig. 3.3 Cross-sectionalview of a two_wireline
V , = ' 0t)
| 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
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.
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)
(3.11)
!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 zTTk r
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)
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 , )
(\ 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,
')
- t e.zt"+l (3.16b) uzt utz)
f
(3.1e)
Adding Eqs. (3.18) and (3.19),we get
For the second section of the transposition cycle
zTrK\
, t j r" ,* )
vot # vo, =
V n ti
(3.16c)
(3.20)
As per Eq. (3.12) for balancedthree-phasevoltages
For the third sectionof the transpositioncycle
vob=|-.(r"rln -&r-* q6rtn-!-i e,tt"+l Dr,) zi'!\ Dy
,,6 *( r,/r^\
2r t r ( qt - r q") r n t f %)
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 ))
)
"'
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
E4
ModernPowerSYstemAnalYsis
|
we get 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.
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 irk wab
h
\
F/m line-to-line
(3.26a)
,
It__
and Cn= -
ln
F i g . 3 . 8 Electricfield of two long, parallel,oppositelycharged conductors
Capacitance
-
of a Single'Phase
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
(3.24)
2nk D
F/m to neutral
41+tO't4h21stt2
(3.26b)
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 --o ron@"httu')'!t
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
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 .
= 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,^(###)]
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 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
100x8
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''' Drr, = ( jh)''' (D"pop,o)r,t = [(i, grjh)r,t)t,, "n ln the first section of the transpositioncycle self GMD is D,o = ?f rf )rt4 = ?f)''' D
pF /km
1 n rf z * o s * s t o fr o o) 3 - l ' / 3 [
3.8
\0.86s/ |
D'l' = QAt'' Dr, = (At''
= 0.018I 1F/km Q C , = 3 1 4 x 0 ' 0 1 8 1x 1 0 - 6 = 5.68 x 10-6 Ulkm Chargingcurrent/phase =
x 5.6g x 10{ = 0.361
D, = (D,oDroD,,)''3= l?3f Arl3fitz Now
Cn=
2d(
2nk
"t#* Charging current/conductor= 0'361 L
= 0.1805A/km
4 ltk
F/m
ai
,'lt I
FThis result Example 3.3. 3.9
obviously checks with the fundamentally derived expressionin
LsL 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 phaseto 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).
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.
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 of Electrical Energy,3rd cdn, Hodder and Stoughton, 1970. 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.
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 '
See Section 10.3.
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 I
o
-fl
ft __'l
l
L
r
T2 .tF
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 \v/B
- > 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
*Per
A
(4.2a)
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.
I
tir.n' l .,rvq..il
MooernHowersystem Analysis
I
Vi
Baseimpedance z, = Y-4-: "
ohms
(VA )"
IB
(4.2b)
If the actual impedanceis Z (ohms), its per unit value is given by
'er unit imPedanceZ (Pu) =
Z (oltrns)x (MVA)' @
Z(ohms)x (kVA), ,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)
Z(ohms)x (kVA)u
o
(4.7)
x (kv)2, GD3 _ 1,ooo ^L_: onms (MVA)' G"Ab
(4.8)
Jr 1rv;u
Baseimpedance7o "
_
of a Transformer
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 gure4. 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. the transformer in the ratio of (vA)a and voltage bases on the two sides of transformation,i.e. V r s- ! Vt, a
(4.11a)
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),
l'ooox(MVA)u
(4.10)
-{_J--->
(kV)?x 1,000
BasecurrentIr-
n^n
to (MVA)a' n'*' andtV base when MVA baseis changecltrom (MVA)r, oto 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,
(4.e)
12
transformer of single-phase (a) '-' Representation neglected) impedance (mignetizing
l'ooox(kv)u J3rB transformer circuitof single-phase (b) Perunitequivalent Fig. 4.7
-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), we get Vz(pu)= Vr(pu) - I,(pu)Zr(pu)_ Ir(pu)Z,(pu) (4.13)
Now or
Ir
12 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)
(4.r4)
where
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.rablesimplification 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
l-
zru
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.
+=+=, I'o
zz(pu)= ' +-++o'1' zzu zB
[,.t'ff.
(4.1s)
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:
Flepresentation of power System eomponents
Transmission line:
2o.5x3o GT2
Transform er T,: TransformerTt:
Generator2:
Transformer Z :
16x30
= 0.44
0.209 x
0.22 x ff =0.++
Transformet Tr:
1 . 6x 3 0
u t7- = o'396
1 . 2x 3 0
tazr Generator3:
= 0.564
-l-!,?I l_0_= 0 . 4 1 8
aiT Generator1:
I iliffij IExample 4.I, we now calculatethe pu values of the reactancesof transfonners and generatorsas per relation (4.10):
Generator2:
The reactancediagramof the system is shown in Fig. 4.g.
(ll)'
U000._--J--
0.44
I
0 . 4 1 3r *
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
(10'5i1 =0.3e6
= o'936
0.56x 30 = 0'437 AZI-
JU
0.43sx
Generator1:
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 Reactancecriagramof the systemof Fig. 4.5 (roadsneglected) 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; ";
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 or
t S l= ( p 2 + e \ t , ,
@.17)
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 - voltamperesreactive (VAR) = 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
In case of a series RC load carrying current I P _I2R O - IzX, qreactivepower absorbedis negative) 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
Then
Ir =llil I (6p- A .S = ,'5 lvLl lILl I 0
- Ji tvLtvLtcosd + iJT t
[email protected])
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.
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 \Xp>{
(4.22b)
o --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.
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
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
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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.
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.
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-- -'-/ 6666
l
I
E1
Fig.4.22Synchronousmachineconnectedtoinfinitebus l/ol cos dof the phasor Io on V' It rneansthat since lV,l is fixed, the projection Phasordiagramscorresponding varied' is remains constant,whiie the excitation in Fig' 4'23' T\e phasor presented are to high, medium and low excitations unity power factor case'It is obvious diagramof Fig. 4.23(b)colrespondsto the frot the phasor diagram that for this excitation
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 bus' and of the power exchangebetweenthe synchronousmachine and the 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
(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
^
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# . . . . . . . ' @ , ^
^
l L ^
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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
lnA _ - rl,lx,
sin (90" + 0)
sin 6
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.
t
_
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^
-
1!1
of Fqwer SystemComponents Ftepreseniation
ModernPower System Analysis
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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
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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
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
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) 2XdXq
si n 26
(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.
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.
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.
I
9o.s Leading
,
0.5
tl
, -,---
1.0
>- Reactivepower(pu)lagging
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.
.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
)L
ASa= i
Full load (MVA) = I pu, 0.9 pf lagging Load current = generatorcurrent
Power Representation
Current
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"
Here the load current is given by Eq. (4.17), i.e.
= 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. (iii) Constant
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 pu
Synchronous reactance X," = +# = 1.344 - ' r - pu (24)z
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=
r.99
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) sin d= 0.5065
or
6 = 30.4"
or
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
l - - Y Y -I
t.59149,5"-rlo" .i1.344 =0.9-j0.024
In=
= 0.9 l-1.5" (a) pf = cos 1.5" = 1; unity pf
-.*fF-., [y^
which gives 6 - 49.5"
60 MW 0.9pf tagging
Fig. p_4.1
Reactivepower 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
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.
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 .5M W, 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
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.
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 Also the following identity holds for ABCD constants: AD-BC=l
(sr) /< 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'..
SHORT TRANSMISSION
LINE
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:
=I l z f l v o l Lr,J Lo rJLr-.J lyrl
(5'3)
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 )
-{u ) r l 2 ( 5 . 4 ) * z t v R | i l ( R c o sQ ^ + X s i nfw It2
. rl/ 1il21n2+ x2 r l ?JlJr cos/o = tvnt r * + t / A Qp f 4#rin , I / i " "L lVRl tvRP J
-;- to Appenclix B.
For overhead transmissionlines, shunt admittanceis mainly capacitive susceptance (iwcl) as the line conductance(also called leakance)is always negligible.
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 + 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|,'%' lvRl
(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
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 overhead transmissionline 20 km long. The line resistanceand inductanceare 0.0195 ohm 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 5l =
l V o l + l 1 l ( R c o sQ * + X s i n / ^ ) = 1 0 ,0 0 0+ 7 0 7 (0.39x 0.701 + 3.96 x 0.707,\y = 72.175kV
Voltage regulation= pfTL-- l9- x roo :Zt.j7vo 10 (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
(i)
(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 = 549( 0. 911- j0'412)- 707( 0. 107 - j0. 70'7) = 0.29 + j273'7 Note that the real part of 0.29 appearsdue the approximationin (i) Ignoring it, we have I, = j273.7 A
' or (c)
'Y ^L
:ltl -loxlooo 3wxc
C-81 P'F Efficiency of transmission;
= l0.9Vo Case (a)
lyst- 10 = 0.109 t0
!
-,L-
Fig. 5.4
(a) This is the case of a short line with I = Ia= 1, given by
From Eq. (5.5),
l
l-
X = 3I4 x 0.63 x 10-3x Z0 = 3.96 e
l 1 l= - - 5 0 0 0 = 7 0 7 A 1 0 x 0 .7 0 i
133
ll. I
273'7
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 connectedwindings arJ0.06 and 0.36 ohms. what is the voltage at the bus at the power station end?
6.6/33kV
33/6.6kV
Fig. 5.5 Solution lt is convenient here to employ the per unit method. Let us choose, BaseMVA = 5 Base kV = 6.6 on low voltageside = 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
=
(O.OU = (0.02+ /0.12) O/phase + 7O.36) f
( 0 . 0 2 +j 0 . 1 2 ) x 5 ( 0 . 5 + 1 3 . 7 5 ) x 5 _ 6.q2 Q'2
= (0.0046+ j0.030) pu 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)
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. I
I
2,000 kw I at 0.8 pf *Vs
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)
Modern PowerSystem Anatysis
'F6, I
Squaring(i) and (ii), adding and simplifying, we get lvrl2 lll2 = lVnlzlll2 + zlvRl lll2 (l1lR cos
5.3 /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 V5 | = l 1 l c o s d s = 2,000 l V 5l =
2000 :3 . 44 kY
7 2 5 x 0 .8
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 )
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 + Vs=
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
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
'r i4 -'t ', I ('* | f /
1
\
/
z l,u.r ., ll -" | |
= (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 tto 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
= 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.
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 drv, -d-T =
Ea
i''
= rZv'
(5.16)
This is a linear differentiai equation whose general solution can be written as follows: V*=CpI**Cre-1x
(s.r7)
where
7= ,lW
Cr=
*
(vn+zJn)
1
2Un-
(s.18)
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.
(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
dI.
Substituting the valueof + from Eq. (5.15),we ger dx d2v
,r=
I. = VoL( " " 2 " \ "*
2
-.t-" 2
/
/
+
r"^' 2( , ( e : J : : ) 2
)* ," ( "\ )
e1' + e-t' \ 2 )
)
These can be rewritten after intloducing hyperbolic functions, as Vr= Vn cosh 1r + I^2, sinh 1r
Characteristics and Performance of PowerTransmission Lines ffi -
sinh 7/- |.+*4* 3!
(5,22)
..=Jyz(H+)
)!
\^'
\(s.28a)
6.)
This seriesconvergesrapidly for values of 7t usually encountered for power -r,Lqrrrrrnwi-o.l.r
I,= Io cosh rr + V- -l-.i
lines
and
can
he convenie-nflw
^l-^i^-.^
--
expressionsfor ABCD constanis art when x = l, Vr= V, Ir= Is
':::i;:l;;t =l;::,, Hl
(s.23)
Bx,z (t.+)
Here A=D -cosh 7/ B = Z, sinh 7/
(s.24)
\ 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 _=
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\
c = J-sinh :r/ Z, *In
YZ 2
A=D=l*
Z
INTERPRETATION
1
;(e"
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 Z,lRlr.,,ritgx+n V^, = I V R + *lVn-Z,lnl
|
I
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
cosh 7/=r ++* #*. .=(t.+)
(5.30)
dz= I (Vn- InZ,) The instantaneous voltagev*(t) canbe writtenfrom Eq. (5.30)as ,o' ,i@t+r,,.+o,) l
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.
= 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,
-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=
oflight velocitv
(s.42)
of wave along the line would be The actual velocity of the propagation light' somewhatless than the velocityof
7= (yz)1/2= Qu,C(r+ juL))rt2
- iu (LQ'''(r- t
' =
., = 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 For a single-phasetransmissionline L=
C_
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"
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
At no-load Vs= AVn' ls = CVa
A and C are comPutedas follows: /AL = -
l*
tl
Y Z= l +
)
= 0.91+ j0.028
l
*x 2
0 . 1 8 81 1 6 2 ' , 6 "
.r14E-f
Modern Power Svstem Analvsis Characteristics. and Performanceof Power TransmissionLines I 14g^ tsimplifying, we obtain the maximum permissiblefrequency as f = 57.9 Hz
l A l = 0 .9 1
C = Y(l + YZ/6) = = 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% ,r, Now
2
1 ^ = 1 + -; t', .i2.8 x 10-6 x (0.125+ 70.4) z
,l
th of the real part. lzll can be
,A,=
1 . 0 6 gx l 0 - 3
;
Vn I
Incident vcrltage-
?cl n 2
vR-zclR
Reflected voltage -
220= 0.88 250 I A-l*;xi1.l2xl0-3x
a = 0 . 1 6 3x t O - 3 ;f =
V,,
- 220/J3 = 63.5110" kV (to neurral) 2
r-0.936 0 . 5 6x l 0 6
/ = 3 3 8k m ;;,,
= ( 0. 163+ i1. 068) x t O - - -r 0+ j{J
For open circuit 1n= 0
l A l = | - 0 . 5 6x l r J 6 P= 0 . 9 3 6 p_
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
(a) At the receiving-end;
= (1 - 0._56x t0-6P1+ j0.t75 x r0-6P
approxirnatedas
solution
y- 1yz)t'2= (2.8 x 10-6x 0.42 /.(g0 + 72.6))t/2 = 1 . 0 8x 1 0 - 3l g I . 3
1 A = l + L Y Z
S in c eth c i rn a g i n a ry p a rt w i l l b c l essthan,
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. (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.
-
)
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 &=
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)
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 neglectedin comparisohto yc,tl. Thus 15 - jVruCl Now
Vn= Vs - Is QwLl) = V, + V,tj
= vs0+ Jctt2)
lncreasino/ -\.-
CLlz
(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
(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
;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
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
1- --!r_.
,,1_ - | (t. It,t,)
z,
.Jru. ,
fi:I=filH;':"! l'lt,^t
. cosh 7/ L1,_, I z, ___.. JLl^J
(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.
( 5. 48)
According to exact solurion of a long line [refer to Eq. (5.23)].
= y, the velocity of light
THE EOUIVALENT
t(-tU2) Y ganh1il2)
_ _ ___{l
Lr,J +!v,2,) (r*ir,r,)lL,_) Lr,(,
2nrfJ LC
5.8
.o,nh..,tt2
For a zr-networkshown in Fig. 5.15 [refer ro Eq. (5.13)].
llsl = llal
,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/ =
For exact equivalence,we must have Z/= Z, sin h 7/ f * *YtZ=cosh 2 From Eq. (5.50)
7/
Z' = ,l-:. sinn1/ = Z, tlnh'' : -( z[*init rr 1 vY tJyz 7t )
(s.4e)
(s.s0) (s.51)
(5.52)
11-"'' -sln!-24 is thc fitct.r by which thc scrics 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.
Lv,2
(s.53)
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
equivalent- n- network reduces to that of nominal-n-.
Is= In= 0.764/_-36.9kA Sending-end powerfactor= cos (4.9"+ 36.9"_ 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
+^ 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( 4 r nl 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
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
- 155 |
ModernPowerSystem4nalysis
-
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
= 128.5172.7'
20.49" = 0.886 - j0.331
e4'oss4l-20.49" = 0.946 l-
gosh 7/ = 0.938+ iO.O2= 0.938 11.2"
c= y(w!Z') = o.oo1tetr+ ]x to{ lrgo xr3t.2172.3" \ 6 ) 6
sinh 7/ = 0.052 + 70.35 = 0.354 181.5"
E
L..= 1-
"
l3nnn"
: -t-----
.,^
= soz.2I l-
Y 10-'190" A = D = c o s hf l = O : 9 3 81 L . 2 "
t
lY
Z+ | yf 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 'fhe
= 52.58MW resultsaretabulatedbelow: Short line uppntximatktn
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
= 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 )
* ?313 x 0.t3 x 0.984
Nominal-r
Exact
Approximation
(s.28h) 238 kV o.I3 116.5"kA 0.984leading 52. 7M W
237.23kV 0.12861t5.3" kA 0.987leading 52. 15M W
?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.
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 Similerly,
ay
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'cos(B- a) v - a tBt lrl lyslly*l .
/1 Qn= ff
A \
(s.s6)
-
(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 Substitutingfor 1o in Eq. (5.54)'we get
(s.60)
,n - ^ sin1/ 6t
lAl .__."
sin(/- a) lliivni'?
(5.61) (s.62)
similarly, the real and reactive powers at sending-endare
currentscan, however,t" .*pr"rsed in terms of Receiving-and sending-end receiving- and sending-endvoltages[see Eq. (5'1)] as
-* 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#
lvsllv'l p,' = I ?l v,P cos(o- o)lBl J
tBicos(0*A
p r = Qs
at
(5'63)
tvsttvRt' .'ur,t ' . , sin (p - a) (/1+ b) l*l If |
6.e) 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
: (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* e, v o _ ry_srn \ 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)
(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 reducethe line reactance.This is accomplishedby adding series capacitorsin the line. This idea will be pursuedfurtherin chapter 12. Seriescapacitors woulclof c
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 So z = 2 5 + 7 1 5 P u 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.
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, -
) =
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)
f2o=
lvtllv,l -Tcos
d'
tv12- Qs = - 0.638Pu X
Reactivepower loss* in the cable is
QL= Qs- en _ 2es _ 1.276pu Total load on station1 = (15 + i5) + (5 + j0.638) = 20 + j5.638
__>
Sn=5-l SDI= 15 +/5 (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
Reactivc powcr loss can also be cornputedas l/l2X=
:t +(9.!1q)1 o.os= t.z7 pu. " I
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"
#,
o - !5x]75200
(iii)
sin (75"- d) -
ij;
x Q75)2sin(75"-5")
0=378sin(75"-A-302
- ,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).
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
From Eq. (5.61)
275^x275 cos(75o- 22")- 9 85 * (2712cos70o 200 200 - 227.6- 109.9= 117.7D{W
Pn= "
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 5 = 28.46"
or 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
1 5 0+ / 0
sin 70'
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).
(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 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
(ii)
_ 0.00l45tyRl2 150 = 1.375 lynt (1 _ (0.002912lv^121u2 Solving the quadratic and retaining the higher value of lv^|, we obtain lVal = 244.9 kV
sR=4/r-r. +P 4r- b) lfl rv^r' n=l+lvs(t(r-o)-+3 4/r+o
(c)
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
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 shown in 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
cos(f - a)Mw B-ltv;2
Fig. 5.21 Sending-endcircle diagram
(s.7e)
=u':;=o i,i,=',!"i
in the polar coordinates or in terms of rectangular coordinales. Horizontal coordinate of the centre
=
+
(s.80)
The corresponding receiving- and sending-end circle diagrams have been clrawn
in Figs5.22 and5.23.
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:
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.
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
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
(5.84)
Since the real power demandedby the loacl must be delivered 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-endvoltage 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 balance equationat 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= ^ == l v * l ' (lYsl - lVol) for (QD> QsR) Q^= o 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).
VAR grenerator
'of
lIc
Fig. 5.28
Static capacitor bank
kA
course, sincct{tis spccificcl withina buntl,Ql rury vrry withil a corresponding band.
: 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)
.C__ I-
(lvRl - IEGD/0. ,. ^ KA
If inductors are employed instead,vARs fed into the line are
Q{3-phase)=-'F''tuo* XL
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
i e c_ = 3Jt v R3l 4 oG i l
(s.87)
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.
J5;E-
= 3 W ( - l Y R rr-r c l ) J3 ( _jxsJl ) = jlVpt(tE6t _ IVRt)lXsMVA 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
il]ffi'ir'$ t:L::"::ch Control
in the svstem *r,"i"' a taigeu.oun,-.orvAR
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.
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 rVs = 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.
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.
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
!.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
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.
,,1€?il -T-
uooern power system hnalysis
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 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 mF 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
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.
characteristicsand Performanceof power TransmissionLines
183
NCES REFERE 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.
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.
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).
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 *fn, ., - Yrz _ !* 0
-ln (lzo*ln * hz * lzq,
-\z
0
- lzt
- !z+
(y:o * yrg
-t23
(b) Equivalent circuit
-lu
*rzt*yy, -J z +
-
( 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+ Yzz= lzo * ltz t Ytt= ):o* ln* Yu= lqo * lzq * Ytz= Yzt = - lni
Sor
ln (c) Power network of Fig. 6.1 (b) lumped and redrawn
lzt + lzq lzz*
Fig. 6.1 Samplefour-bussystem
lzq
ly YZt = YZZ= - lZt Soz
v1
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
(6.s)
or, in matrix form "Inus= Isus Vnus
(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)
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.
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
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.
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
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^
(6.11)
Similarly, the current relation for Fig. 6.4b is ir, *
jr, = lrrv^
(6.r2)
I
7
---------+ o-------------
\.e - Branch - - Link e =9 m=5 b=m-1=5-'l=4=n l=e-b=5
yr"=E -E"
(a) lmpedanceform
(b) Admittance form
Fig. 6.4 Representation of a networkelement 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
E,
(b) Co- tree
Fig. 6.3 Treeand cotreeof the orientedconnectedgraphof Fig.6.2 cotree.The number of links / of a connectedgraph with e elementsis I = e - b = e - m + l (6.10) Note that a tree (and therefbre, cotree) of a graph is not unique.
Jr,=
-
!r.,=
|/7r,
Yrs€rs
Also A set of unconnectedelements is defined as a primitive network. The performanceequationsof a primitive networkerre given below: In impedance form
V+E=ZI
(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. 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= VtVta= Vt Vn=VtVts= VqVtg= Vtor, in matrix form
*Another
<
v = AV,'J'
Vn
(6.15)
v2 v2 v2 vr
useful set of network variables are the / link (loop) currents which constitute a zero redundancyset of network variables [6].
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
_
l
1
0
0 1 0 0 1
1 0
branches
links
(6.r7)
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 = 0 if the ith element is not incident to the kth node SubstitutingEq. (6.16) into Eq. (6.14), we get 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.
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 - )r:
( yzo* yn
- lzt
* lzt * t-zq, '- ,. .t'23
-
()'lo f- .yr
-
* lzt * Jy,
!'tn
}'l
_!zc
0
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 0
)ro 0
YA=
0 0 0 0 !n 0 -.Yrr
0 0
0
0 lzo 0 J n 0 0
0 l
- jzz -lrz -
y lzz 0
!24
0
0
l
jqo -
ft+
0 0
0
lzq
n
0
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
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
(6.2sb)
n
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
(6.27)
(iv)
LOAD FLOW PROBLEM
The complex power injected by the sourceinto the ith bus of a power sysrem IS
(6.24)
(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.
: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 andrv,r p6;) ' 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 x = d_ependent or state vector of dimension 2n (2n unspecifiedvariables) ) = vector of independentvariablesof dimension 2n (2n independentvariabreswhich are specified.i prrori)
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.
-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.
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 n
.I
i:2
n
Po,- D i:2
Po,;(Pr= 0). Equations (6.37)can be solvedexplicirly
(non-iteratively) for 62,61,..., 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)
k:1
et=- 'u,'
n
Pr =
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 Pi =lVil
Wffi
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
iP,ts
'l.o lVzl=
2 .S.= I + i^ I Uz ,,-. -
Fig. 6.6 Four-buslosslesssamplesystem Real demand
1 2 3 4
Por = Poz = Poz = Poq =
1.0 7.0 2.0 2.0
Reactive demand
Qot = Qoz = Qoz = Qoq =
0.5 0.4 1.0 7.0
Real generation
Pcl ='Pcz = 4'0 Pct=o Pcq=o
Reactive generatrcn
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:: slnce ffie"_bviyus 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+.
'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 _ r.+r+
_ L,> = u.JJ4 pu
(viii)
Now, let us find the line flows. Equation (5.6g) can be written in the form
( l z l =X , 0 = 9 0 " )
lvi]-lvkl Pik = - Pki sin ({ - 6o)
x,o
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 )
(ii) (iii )
(iv) = (iv), (ii), (iii) we 0, and solving and Taking bus 1 as a referencebus, i.". 4 get --O.0ll rad = 4.4I" 4. 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
where P* is the real power flow from bus j to bus k.
sin-1.23" pn = - pz,= -r/ = 0.492pu \ r - q)+0 . 1 5 sin(d, 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.,
Qq = - l 0 c o s 5 .1 1 " - 6 .6 6 7cos 9.52o+ 16.667
@
OI'
Qr = 0'07Pu Qz = 0'22Pu Qz = 0'732Pu 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
"
(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
.2Oftril I
ModernPowerSystemAnalysis
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)
t^ .
v,= *l '':jq
,
(6.40)
-l
- t Yit' vr'l = 2'3""' n Y" I vr* ilt I" k*t I I
(6.4r)
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.
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.
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 updatedvaluesof 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 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
Sri = Vk (V*k- Vf) yfo+ VoVfyi,l,
(6.47)
a 1y.r,1i y,rv,,(,*t) ],r,',) D y,kvk,,,I " ,u.ro, Lr, I f!, )
2' The revisedvalueof {.is obtainedfrom Eq. (6.45) immediately following 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
(6.48) where
(6.4e)
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, /<).
.
P-i?t'rtt _ r^1(. ;r +. t=)_-_ ,u The algorithm for pe buses remains unchanged.
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 approachthe 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
- 'r^'iI
l -- - - - - 11 -0' -. 5 r+ -i '0- . -2 -1' .- 0' \ 4( - z- +' rj-6/ ) - (\ - 0v .' v6v6v6+' j 2 ) _ ( _ t+ r! fll Yzz t -iO I
') |
ModernPowerSystemAnalysis
fei.fii,',1 I
-
Load Flow Studies
4.246- jrr.04 = 1.019+ 70.046pu - jLr 3.666
1 - -
"
Ytt
_
1 Yzt
(vl)*
T#-frL+re - r.04 (- 1 + 13)
- e2+ i6) (to4 + io)
- (- 0.666+ j2)(t + 70)_ (_ 1+ I
= t( +'z^?e;t - itt'lzs = , (r.rstz+ 70.033e) ( 3 . 6 6 6 _ j t t )J-
- (- 0.666+ j2) (1.019+ ;0.046)- (- 2 * i6) |
- 70.087 = ''!'-,-1"'.9?' = 1.028 pu - jrr 3.666
- r^rr\l
j3)(t. io)jlr)
= 1.84658o= 0.032 rad or 612 .' . v) = 1. 04( cos 6) * j sin dj) = 1.04 (0.99948+ j0.0322) = L03946 + 70.03351
19 + 70.046) I - (- 2 + j6) (1.028- jo.o87) | )
( o , v' tl ==- l - j ' 1 - = i Q -t yY"v' . v . - yYszv; _ v t -_ vYrovl ,,0I
,l
1 114tr
[ - r _ 7 0 . s- (- t + i3) 1'04
-
3.666-in L ,f;;
2.99r- j9.2s31.025- 70.0093pu 3-je
- (- 0.666+ j2) (1.03s46 + 70.03351) _ (_ z +
io)f
r.0317In 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)
- 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
ModernPowerSystem Analysis 0 .2 5 < Q z < 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
I aar{ Lwqv
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 , - (- 0.666+ .i2)- (- r + i 3 ) ] (* 2 + i6)1.04 - 4'246- jlr-,? = t.05.59 + io.o341 3 . 6 6 6- j t l (
- 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,
]
- 70.0893pu 1.0347
- (- I +,r3) (r'0s0e + i0'0341) -1
3- je
/ ^- \0
I
\ux, )
l
)' are the derivatives whcrefg)',( !f, )'. .f9 ' ' of It with respectto d"r
( Dr, ) \ / [ ,", ,/ x1, x2, ..., x, evaluat edat ( x! , *1, . . . , *0, ) . Neglecting higher order terins we can write Eq. (6.55) in matrix fbrm
- (- 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+ 70.0341)- (- 2 + j6) _ 2 . 8 tt 2 - j n . 7 0 9 3 . 6 6 6- j r r
Considera set of n non-linearalgebraicequations 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
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) 46^ AlV^
I ous jmtn
i t h b u snlp , l = l T:tT' L----J t-- ---- ----rH''l -
I l
(6.62c) mth bus
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. 2 3
Yi^= G* + jB,* Vi= e, + jf,
(a* i ib)
I l 2
= (Gi- + jBi*) @* + jk\
I
Y
case2
f,z
Hzs
Hzn
Nza
Jzz
Lzz
Jzs
Jz+
Lz+
Hsz
Nsz Hss
Hqz
Nn
Haa
Naz.
H+s
Jqz
Lqz
Jcc
Lu
J+s
Nsq
Hss
Hss
J
IA A<\ \\J
' \'J,'
J ti = Pi - Giilvil' Li;= Qi-
5
Nzz
o
I,,==t- n, - Biirvirz Mii= Pi + G,,lV,lz
4
Hzz
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
c0
(6.67)
[:_]
Hss H5a
t Jacobian (Evaluatedat trial values of variables)
lalv4ll
l]ql-l
i_it_l t
Corrections in variables
,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 valuesare 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 violates the 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
2 .0
1.0
Unspecified
0 .0
0 .0
0.5
l 5
0 .6
0.0
demand PD
demand
specification power generation O6 UnspecifiedV'=1.94 * ;g (slack bus)
1.0 Qct=?
Unspecified (PO hus) lvll= 1.04 (PV bus)
C o n tro l l a b l crc a c ti v op o w o r s o urccl s avai l abl cnt btts3 w i th the ctl nstrai nt 0
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 - z . g 4 r - j 1 1 . 7 &- r 2 . r 3 l - 7 5 . 9 6 " 1 , snes en."= 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.
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
tQT= 1 - (- o'e6)- t'e6
I
The changesin variablesat the end of the first iteration are obtainedas follows:
0P, 061 aP, 061, }Qz
flows
The following matrix shows the real part of line flows
Aror- -1.5 - (0.12)= - !.62
0P, 06, 0P, 06, aQ,
0P, alv2l 7Pu 0lv2l 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 Q o \= Q\ + Q rt = 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
-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 flows
'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
- 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 .7 91)
(lbr each PQ bus)
(6.69)
222 |
Modern Porgl Jyllem Anatysis
I
(specified)2- @i2+ f,') = 0 (for each PV bus) (6.70) Using the NR method,the linearisedequationsin the rft iteration of iterative processcan be written as
Load Flow Studies
l(l
AQ AIVP F o ri *
Jr
J2
J3
J4
Js
J6
Decoupled
j - Jqij = Gij ei fi
Juj=
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.
.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
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:
'*i -::,:-,t
I I
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
gt'r ioi)
(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
-'
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
rcqotlvc
power mismatchesas follows: (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.
( 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 .0 4)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 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
1.(6.60a) = PV bus ------
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)
; Afr/,R
[alY"(')ll - [24.5r] Lz.rz5r
Ltu,a, l
e+ Flg.6. 13
--
<'-
re
fr=
\-
1 --
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.
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.
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 assessment and 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 system was discussedin Section 5.10. Though the same general conclusionshold 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
','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 byrn < 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) I z= l( Vz- 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). 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
Fig. 6.17 Regulatingtransformerfor controlof voltagephaseangle
I
ao.Ot,o* a,r.,.r
Modernpower SystemAnalysis
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
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
-(2- j6) 1T G z+ j6)
+ 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.
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.
(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 1
vi -710"
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
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.
.?38i I I
Modern Power Svstem Anatvsis Z-r----rvz -- .ffi() | ir o -y=_j5.0
y=-j5 0
\
3
vel
Fig. P-6.5
I ' 3
',\ , =-is'o ,lJ
,l
Real
Reactive
Real
demand
demand
Seneratrcn
Fig. P-6.7 Three-bus samplesystemcontaining a regulating transformer
Reactive generation
Por = 1.0 ',r2 = 0 Po: = 1.0
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: Pc t = 1 .0 p 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
6.8 Calculate V3for the systemof Fig. 6.5 for the first iteration, using the data of Example6.4. 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
1.02 1.01 Unspecified Unspecified
Unspecified Unspecified - 1.0 - 0.2
Bus type
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,
New Delhi. 1988.
' _ -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.
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.
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' load flow solution.Economy of operationis naturallypredominantin determiningallocationof generationto eachstationfor varioussystemload levels.The first problem in power system
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 Fig.7.1
(vw)min
(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.
|
. .
-
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
o o
Etr o E E ()
(MW) max 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 in the inverseform Pc;i= a, + {),(lC)i + 1,QC)', + ... Optimal
(7.8)
i l- P o = O
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>
(MW)min
\' G D
(7.3)
-""] it (Pci)-^[f "",
(7.e)
where X is the Lagrange multiplier. Minimization is achieved by the condition
of, =o dPo, dC' i
or
dPc,
(7.1o)
- ) i i = 1 , 2 , . . . ,k
Operation
[,et us assulnethat it is known a priltri which generutorsitre t
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
(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.
( 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
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 -
l c c
-dcz-= o.2,pcz+3o.o 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
@
^
45 i= a>
E P 40 t E
O c
J
c
Plantoutput,MW Fig.7.3
Incrementalfuel cost versus plant output,as found in Example7.1
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 saving in 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" ", - - 721.875Rs/hr
[ 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
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.
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.
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
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), Fz(2),..., Fz(9).Of these FzQ) = min tt6(0) + Ft(9)1, VzG) + Ft(8)l'
Unit number Load range
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.
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.
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
(7.13)
No. of cycles 'down' time), Mean time to repair (mean 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 .
If hy^dro.generationis available in the system, it could be brought on line in a matter of minutes to take up load.
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 ---tlr Up. Stale \K =
\/
.
rz\
i + I )t
!l--
-----1--l-:1:---
^t
L1^^
urs PIUDaUITILy Ur urtr systelll
l-^:-^
2-
tLl
Utrltr$ ttl ulIS Slaltr
is pt =
{r,r;(ul),{*
ot(down)
(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.
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.
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
I I I I I I
1 1
I I I I I I
l 0
l'l'
0
l 0 0
0 0 0
*Unit was started due to start-up considerations.
7.5
OPTIMUM
GENERATION
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-
;-^ vYuvrr luuv
l^--^^ rvirDsr
the overall cost of generation
c =
L^ s
4rE .1uuuurrttrg
,\-rci(Pc')
n^,-
t()f,
ry
Ine
ODJgCtfVg fS tO mirufiUze
(7.7)
at any time under equality constraint of meeting the load demand with transmissionloss, i.e.
k
- P, - P I - = DP", i:l
(7.re)
0
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
Equation (7.23) can also be written in the alternative form ( I C) i- An- Q TL) if i = 1, 2, . . . , k
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 PGto Po , # *: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)
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)
(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.
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 ^ f t o_, Ap I
*
*B^,
= ftl?lPo,"a^^'o')
{in pu) = B^n (in Mw-t)
x Base MVA
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, = atPo'+ b'
Substinrting APL|&G' and dcildPci Eq. (7.22), we have
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)
dP'
<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, -
B nPbr 100MW, Pr = 10 MW, i.e.
(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)
Pc t = 1 2 8 ' 5 7M W Pc z = 1 2 5 M w
= - 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
n v sn^r ravnr tr r
t ' E irc5 . \r
? 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.
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
*' (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 ,
P
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
;!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
----t''
'
i
i
ffi
-
'
b
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
Za = 0.Ol + 70.04pu Zo = 0.015 + 70.06 pu 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) - ^O : - l4oi ot- = t an- t+! o2: t an2. 2 . r 6 : - l4o 9t cos (or- ot) = cos 0o = 1
*For
more accuratemethodsand exactexpressionfor 0P,./0P6i,references122,231 may be consulted.
The plant power factors are pfi = cos (6.05" + l4') = 0.9393
i
OPtimal System Operation
Mociern PowerSvstemAna[,sis
ffiffi
ffi ffi$ffi l
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 [see
e -f
- 0.02224pu
and A 4 - Llvllvjlly,,lcos(9,, * 6i - 4)= 0 for each pv bus j:7
e0.2174)L0.7826X0.015) + 0.01x(0.217a)2 + 0.01x(0.7826)2 D _ 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.
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 [Eq. (7.a3)l I
100
0'01597 = 8.t.,= 0.01597x 1o-2Mw-l 100
'
0'0M06 = 0.00406 x lo-z Mw-l 100
OPTIMAL
f (x, y) =
l-ty,tl
Inequality
\
;
(7.47)
I
.= lr, jforeachrouusf
LOAD FLOW SOLUTION
Power Flow without
foreachr0 bus | foreachpV busi
| _tn.\7qql lEq.Q.a,
(7.46)
where the vector of dependentvariables is
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. Optimal
(7.4s)
It is to be noteclthat at the ith bus Pt= Pci- Poi
0.02224 LLL+ h - 0.02224 x lo2 Mw-l Drr =
7.6
(7.44)
j:r
= 0.01597 pu
Br.t =
(7.43)
3 Q, + Ltvinvjlllzulsin (0ul
q2 + 0.01x (0.782q2 0.0| 5 x (0.782q2 + 0.01x (0.217 (1.051)2 x(0.946)2
Bzz
tUilvjily,,tcos(0,, j:1
Constraints
The objective function to be minimized is ihe operating cost
(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
-r [fle l
l
w
ODleQtlVe l
L
l
L
tltnefinn l
g
L
l
mrrcf l
E
S
i_^1,-A^ r
a
c
d^^ K
D
u
_r 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
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) where ) is the vector of Lagrange multipliers of same dimension asf (x, u, p) The necessaryconditionsto minimize the unconstrainedLagrangianfunction 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 solution; the expressionsfor the elementsof Jacobianare given in Eqs. (6.64) 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
r,
I
Step 4
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'
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 unew= l.l.^6*
is
C = Pr(lVl, 6)
Make an initial guess for u, the control variables.
L,il
( 7. 57\
where L,u = - o"V-C,
(7.s8)
Here A,u is a step in the negative direction of the gradient. The step size is adjustedby the positive scalar o..
#*ff"
f
uooernPowerSvstem Analvsis 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, ** 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. e.g.
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
-:-
e.g.
lUn,in < lVl < lYl -.o ofl a PQ bus
(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
--L
du ou'+ a"*La"l'r:o =
Often, the upper and lower limits on dependentvariables are specifiedas rmir,SxSr*u^
'Ihe
)
^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.
'
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--_
\t.lr)
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
(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
.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)='
(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 .
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)
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€)
_ \m - ' t l
A D Tm VG
)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
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.
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 PGr=-100+50(IQt-2Aqi
Z, = 0.03+ /0.I2 pu Calculatethe loss formula coefficients6f the systemin per unit and in reciprocalmegawatts, if the baseis 100MVA Ref bus
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.
Flg. P-7.7
Samplesystemfor probtemp-7.7
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 Books 1. Billinton,R., Power SystemReliability Evaluation,Gqrdon and Breach,New york, t970. 2 . Billirrton, R., R.J. Ringlee and A.J. Wood. Power SystemReliabitity Calculations, The MIT Press,Boston,Mass, 1973. 5 . Kusic, G.L., computer Aided Power system Analysis,prentice-Hall, Nerv Jersey, 1986. 4 . Kirchmayer,L.K., Economicoperation of Power systems,John wiley, New york, I9)6.
5 . Kirchmayer, L.K., Economic control of Interconnectedsystems,wiley, New york, 1959. Knight, u.G., Power systemsEngineeringand Mathematics,pergamonpress,New York. 1972.
/. r\cuenswanoer, J.}(-, Modern power systems, International rext York, 1971.
w
Book co., New
8' singh, c. and R. Billinton, system Reliabitity, Modelling and Evaluation, Hutchinson of London,1977. 9. sullvan, R.L., power systempianning, McGraw-Hil, New york, 1977. 10' Wood, A'J' and B.F' Wollenberg,Power Generation,operation and Control, Znd edn., Wiley, New york, 1996 11' Mahalanabis,A.K., D.p. Kothari and s.I. Ahson, computer Aided power system Analysis and control, Tata McGraw-Hill, New Delhi. r9gg. 12. Bergen,A.R., power system Anarysis,prentice-Hall, rnc., Ncw Jersey,19g6. 13' Billinton, R. and R.N. Allan, Reliability Evaluation of power System,plenum Press,New york, 1984. 14. Sterling,M.J.H., power SystemControl,I.E.E., England, 197g. 15. Khatib, H., "Economicsof power systemsReriability,', Technicopy.r9g0. 16. warwick, K. A.E. Kwue and R. Aggarwar (Eds), A.r. Techniquesin power System.s, IEE, UK, 1997. 17. Momoh, J.A., Electic power SystemApplicationsof Optimization,Marcel Dekker, Inc., New York., 2001. 18' Yong-HuaSong (Ed.), Modern optimization Techniques power in Systems,Kluwer Academic Publishers,London, 1999. 19' Debs, A.S., Modern power systemscontror and operation, KAp, New york, 1988. 20. Berrie,T,W., Power SystemEconomics,IEE,London, 19g3. 2r. Berrie, T.w., Electricity, Economicsand pranning, rEE, London, rp92. Don o-o r ql,Er;,
22. Meyer, w.s. and v.D. Albertson, ..Improved Loss Formula compuration by optimally ordered EriminationTechniques',,IEEE TransmpAs, 1971,90: 716. 23. Hill, E.F. and w.D. stevenson,J.R., ..A New Method of Determining Loss Coefficients", IEEE Trans. pAS, July 196g, g7: 154g. 24' Agarwal, S'K. and I.J' Nagrath,"optimal Schcdulingof HydrothermalSystems,,, Proc. IEEE, 1972, 199: 169. 25' Aytb, A'K' and A.D. Patton, "Optimal Thermal Generating Unit Commitment,,, IEEE Trans.,July-Aug 1971,pAS_90: 1752. 26' Dopazo, J'F. et al., "An optimization Technique for Real and Reactive power Allocation", Proc, IEEE, Nov 1967. 1g77. 27. Happ, H.H., "optimal power Dispatch-A Comprehensive survey,,, IEEE Trans. 1977,PAS-96: 841. 28. Harker, 8.c., "A primer on Loss Formula", AIEE Trans, r95g, pt ilr,77: 1434. 29' IEEE CommitfeeReport, "Economy-security Functions power in -systemopoutionS".
IEEE
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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.
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,
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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. /.J5 - /4U. 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. 49.
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.
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.
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
.
W Turbine
Modern power system Analys,s 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
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) *,
= 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) and (g.4), we ger AYr(s)=- k&cApc(") + krAF(s) (8.5) Ayp(s)= kzAyd,s)+ koAyug) (8.6)
a y u ( g = - k s ol r U n
(8.7)
EliminatingAyr(s) andAyo(s), we can write AYu(s)-
k'ktk'AP' (s)- k,krAF(s) ') (oo '' t "'tr ,/ \
-lor,<,r-*^or",].i#)
(8.8)
controt /o
o\
:-
.
r .
t_
K2
K.,=
riyLr.Lru' \o.o., rs rcpfesenleo ln tne ronn of a block diagram in Fig. 9.3. Ks9 1+ fsss
.r.
+y
= l " = tlme constant of speed , rs governor ;-; r-KqkS
4Y5(s)
4F(s)
Flg. 8.3 ,Blockdiagramrepresentation of speed governorsystem 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. Turbine
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 stbam valve 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
= speedregulationof the governor
- gainof speed governor
t
-
where
n= klc
E
1 E^,,^ri^-
(a) Two-stage steamturbine
AYg(s)-FAPds) (b) Turbinetransferfunctionmodel Flg. 8.4
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)-
4Fis; -
AP,G) -4PoG) B*-'- s
Flg. 8.6 Blockdiagrammodelof load frequencycontrol (isolatedpowersystem) (8.i2) Steady States Analysis 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
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 ^ /P' (in PuMWunit change ai in frequency). Now
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 load demandremainingfixed (i.e. Apo= 0). The sready +)with state changein frequency is obtained as follows. (Or"
fi roa (J L
8.rog .c
102 101 100 0
*For
at li\
dA^t | | \r,, ruu-loLoao
(ii) 60% Load
Percent Load
Flg. 8.7 Steady qharacteristicof a speed governor "*-l?39-frequency system
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
'=(#)b :
#:
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 K rrK,= 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
af_
0.04x 50 200
x
(i)
af 0.05x 50 400 600-x (ii), (i) we get and in EquatingAf 231 MW (loadon generatorr)
(ii)
v -
-/A
600-x=
JOy
trltf
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
+L^ LfIs
^L^-^^+^--i^+in r,Il
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
Trr4T,
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. 8.6 K*\ 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, = 20, Apo = 0.01 pu e, 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
Firstorderapproximatiorl
(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
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
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 -1
+I
plus integralload frequencycontrol Fig. 8.10 Proportional
I
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=
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 Af l"t"^dy state , 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 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.
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.
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 representing integral 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
where q'q
- poweranglesof equivalentmachinesof thetwo areas.
(8.26)
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)
With referenceto Eq. (8.12), the incrementalpower balance equationfor area 1 can be written as
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 fho T -^l-^a rrrw lsl/l4vv
fL*lo4nl l oDfrnurrm rrr
^ufr
E^ LY.
/a ta\ \v.L9),
AF1(s)
-iE=
--n7ri"l Fig. 8. 15
= tYr:J cos({L - E): [S]ti z: ar2rrz rzr LL "
A F (s ) = IAP 6 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
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)
(8.34)
For the control area 2, Ap6", r(s) is given by tEq. (g.Zg)l - : gr r r , apt i", z( s)= [ 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) Similarly, for the control are a 2, ACE2 is expressedas ACEr(s) = APti".z(s)+ b2AF,(s)
(8.37)
( 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.
Modern Power 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
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
u I f|:-
-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
Y 4
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
sg2
( 8. 48) t sg2
ModernPower SystemAnatysis
'3i4',"1 T
For block 7 it=2iTtzxt-2iTr2xa
(8.4e)
is= brx, + x.i
(8.5O-)
'-- co","".t"" " 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 I Tpst, 0
2
7 8 9
7
0
- 1
1
Tt
Ttr
o
- 1 Trst
8
9
4
5
0
0
o
o
0
0 0
o
o
0
0 0
-bLoo Tprt
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
K p Tprz
-
0 0
-ss1
Br = |
Ttz
1
0
RzTrsz -2ilr2 0 0 0 b 2 0
[I o o TI
o o
o
l o oo o o + aco) 'O-
I L
,;T
Kprt 0 Tprt
0
z Tprz
--1-
o
i
'
0
Tprz
1 7,, I
0
0 0
0
0 0
0 1
0 0 0 0 0 0
TreZ
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 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.
00.l I
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
3
Y 'tPsl
1 Rr4er A _
Kirxs=- Kir IeCn,Ar
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
/= -
Ikl
(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
ulr, = wl r5rr=
x6rr= lv2
(8.s8)
b ? o0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0
0 0 0 0 0 0
0 0
-arzbz 0 0 0 0 0 0 0 0 0 0 0 0
4 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
(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].
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
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
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
(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.
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
(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
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
ModernPowerSystemAnalysis
i,i2[,:*.1 4
and voltageControl Automatic Generation
-
x - (x1 x2)'
- aF(s)' 1' af (t)dr: liq, * '4F(s): hm/F(")] I n,n,, tf^461dv JO JO s-0 s+0 ,t S
ut=-ktxt
L
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.
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
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.
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:
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
= - 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)
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
I.t
whcre i, = steadystatecurrent
- Jrv'sin(d- c) * E' lzl tzl
(e.2)
Sincetransmissionline resistanceis small. 0 - 9C,.
.
F i g .9.1
DC otT- set curnent
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*= _
tan-l +)
ir = transient current [it is such that t(0) = t(0) + L(0) = 0 being an inductive circuit; it decayscorrespondingiothe tim6 constantiRl.
Synrnretricalshort circuit current
9.2
A
Jiv rzr
cosa+
JTv 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).
.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
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
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.
X,J
where l1l = steady state current (rms) !//l = transientcurrent (rms) excluding DC component lltl = subtransientcurrent (rms) excluding DC component Xa = direct axis synchronousreactance
'ii#I:
Fault
Modern Power Svstenn Anelveie
Xtd= direct axis transientreactance X'j = direct axis subtransientreactance
pon ii'ragn-tir rutlrarion
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 constants with 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
rhe valuestf reactances j? normallyrie within f:T:jr:lltlq certain predictable",l"ilirion), 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/
0.6-1.5 0 .4 - 1.0 0.2-0.5 0 .1 3 -0.35 _ x,d 0.02-0.2 0 .0 0 3-0.01s
Synchronous motors*
capacitor)
r.5-r2.5 0.95-1.5 0.3-0.6 0.18-0.38 0.17-0.37 0.025-0.16 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 reactance.
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, =
s
I
PowerSystemAnalysis
Reactance of Tr= ;WY4 )
SymmetricalFault Analysis
|I
337
= (0.93+ j05s) + (71.6)+ (0.744+ i0.99) + (t1.0) = I.674+ j4.14= 4.43176.8"pu
= 71.6pu
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
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
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
- 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).
25 x 1,000 = 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,
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.
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 tI ; .6'f,1,'. 'dtI-. - 'ltrd-. -t 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 I'J= 1,312.2 (.- 0.6226- j6.8906)= (- 816.2_ jg,O4L8)A
short circuit (sc) current computation through the Thevenin Theorem
I (b) Prefaultequivalentcircuit
Fi9.9"12
(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
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
V" jXrn + Zt
(e.r2)
Current causedby fault in generatorcircuit
AI, - -
(a)
x'd^
(xhs+x + xl^
G
f
(e.13)
(b)
G
AI^
Xle +X (xlh,+x + xi
Postfault currents and voltages are I{= I" + alr I{=Postfault voltage vf where Av = -ix^tf
fffl:liiwith
(e.14)
obtainedas follows by superposition:
I" + AI^ (in rhedirectionof AI^)
(9.15)
vo + ( - , r xn, I f ) = v" + Av -= eJ6) is the voltage of thefaultpoint F/ 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 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
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
,Qrrrnmalrinal
Modern Power SystemAnatysts 9'3 for computationof postfault The circuit model for the systemof Example conditionis shownin Fig' 9'14'
_ 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" + AI^= (- 0.623 76.891)P u already which are the same (and shoutd be) as calculated through th9 Thevenin alternatively 9.3 we have thus solved Example method for large powerful a is in4eecl, theorem ond ,up.rposition. This, networks. 9.5
SELECTION OF CIRCUIT BREAKERS
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.
(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 location of fault and generating capacity (also current irms) is then calculatedby 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 breaker the presenceof DC off-set current' must be capable of subtransient r'rsing by computed interrurpting is it. An exception is an LG (line-to-ground)f.ault Symmetricalcurrentto be intcrrupted 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 the multiplying by for accounted locations is must be tried our to obtain the highestst nava requiring interrupted obtain the current to be Lp"ur"a below: SC computations.This is ilustrated by the examplesthat symmetrical SC current by a factor as tabulated follow. Circuit Breaker SPeed 8 cyclesor slower 5 cycles 3 cyclcs 2 cycles
Multiplying Factor 1.0 1.1 1.2 1.4
motors during a short t'In some recent attempts,currentscontributedby induction circuit have been accountedfor.
Iii"'n"r.; I Three6.6 kv generatorsA, B and c, eachof I0o/oleakage reactanceand MVA ratings 40, 50 and 25, respectively are interconnected electrically, as shown in tThis will be explained in Chapter 1l.
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
(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 Reactor C reactance =
1
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
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.
,^o
Moclern Power Svstem AnalYsis
I I
= - jt.37463 pu tt = -*= j0.13s60 lo*
!o;E
14
D':
jo 1,), 7
I
r - ---rnTl
4 +I
_ 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 = 1.0
(
\.. ) V ? t
9.6
ALGORITHM
FOR SHORT CIRCUIT STUDIES
.
Ill I
Fig. 9.19 Systentaticreductionof the network of Fig' 9'18
So far we have carriedout short circuit calculationsfor simple systemswhose I n t his scct ionwc ext cndour st udy t o l l i tssi vct t ct wor kscanbe easilyr educed.
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
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 systemshown schematicallyin 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
(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
rf I, :
-I', f
(e.22)
:
(e.18)
Let us assume that the rth bus is faulted through a fault impedance Zf . The postfault bus voltage vector will be given by
e.2D
0 : /-
V { u r = V B u s+ A V
-'l '-
Substituting Eq. (9.22) in Eq. (g.20), we have for rhe rth bus AV, = - ZrJf By step 4, the voltage at the nh 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
(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 ows:
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.
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
,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 vrz= L L v: z
r,',
r
By Inventing = 1.0 -
oi 0 .-0 7:1: 0 x 1 . 0 = 0 . 4 6 9 8 p u
/nus = Yrus Vsus
j0.1356
or
7
v{= v\ - 1* vf Zoo
Y"u"
Vsus= [Y"ur]-t /eus = Znus/eus
Zsvs= [Yuur]-t 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. or
- : : _ : 1 . 0= 0 . 4 5 2 1 = 1 . 0- ' i0.0743 pu j0.r3s6
vi = o'o
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
- v{ q .I irq = ' u'l zz+
_ o.4z4v-o jo'l
-
o - - -0.46e8 =
- 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
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
.f
Modern PowerSystemAnalysis
356 |
Now let It = 0 and 12= 1. It similarly fbllows that Ztz=Vt=4= Zn
l7
41
Zvus= 6l l+
Becauseof the above computationalprocedure,the Zru, matrix is referredto as the 'open-circuitimpedancematrix'. Z"vs Building
Algo4ithm
SymmetricalFault Analysis
!, f5l
, 4r,
the dimension of Zsu5 goes up by one). This is type-2 modification. 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.
Type-2
Modification
Zo is addedfrom new bus ft to the old bus7 as in Fig. 9.25.It follows from this figure that
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
'o= -12
or
Type-Z modification
= Zr,I*+ ZiJr+ Zlzlz+... + Zii ei* I) + ... + Zinln
,,= lr^ h(z*
Z l z l z+ . . . + 2 , , 1 , + . . . +Z i , l n + ( Z i i + Z ) l k
* 1rr]a lr,,
rzuz,,>)r, "h
+ + lr^- ^i{zu
Consequently 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)
Rearranging,
.
(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
(\1Ir+ Zi2Iz+ "' + ZlnI)
Now
Vo= Zdo + V,
V*=4lt+
1
(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 +
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
(e.44)
or
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)
Step 3: Add branch ,rt = 0.1 (from bus 3 (new) to bus I (old)); type-Z modification
lo.2s o.zs 0.2s1
Step 4: 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)
0
(ii)
zsvs=',lZ:11, 31i]
V 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 vl
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 l | 0 . 1+ 0 . 1 4 5+80 . 2 4 5 8 - 2 x O . l O 4 2 ) 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 Fig. 9.28
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#
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)
symmetry of the given power These two voltages are equal becauseof the network (c) From Eq. (9.29)
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
JIc) | i-,,
(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)
V: 2,, + Zl
Y o - : j0.175 . _ 1 - -- - j s . j l
zn
(b) As per Eq. (9.26) rr.f v i -
vl' - --Zt--yu
zrr+zl'r
But
,/'0.,
Fig. 9.29
1f -
- - i2.86 As per Eq. (9.32) ,r r Gr-
\,
or
I\t=t|t= ro.186-o) fr
and
0.1
F
Jt -
=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
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
u = 1 OO si n(100n f + 15.)
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: TransformerT,: Transfbrmer Tr: Line:
20 MVA, 3.8 kV, l5%o subtransientreactance 25 MVA, 11/33 ky, gVoreactance 20 MVA, 33/3.3 kV, I \Vo reactance 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.
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 0.8 lagging ."u"'tun". is supptying a passiveload of 400 kW at 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 off-set currenr in rhe breaker, iui tt. maximum possible DC of the breaker' rating current (c) the momentary kVA' (d) the current,o U. intemrpted by the breakerand the intemrpting and (e) the sustainedshort circuit current in the breaker' 50 MVA at 1 1 k v , 0 . 8 9 . 8 The systemshown in Fig' P-9'8 is delivering 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
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
(Iault close to breaker).
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
inductive reactanceof the reactor required. system 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.
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.
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.
'r-[b-l Q9
I
111110kV 0.5 pu reactance 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
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.
above.Thus Vo-- Vot * b-
YbL-
Vr= VrI *
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
Voz I
Voo
(10.s)
Vro
( r0.7)
vbz
Vrz *
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 Vo ,Vr= & 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],
(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 'Lt
loo
(r0.1s)
r rl
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
voz= ! fr"+ ozvu+o%) 3 Vuo=
I
;
V,+ Vr,* r,)
(10.16)
Vao
V""
(10.17) (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
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
(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)
= 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.
I t^ (lo + IR + J
/c) = 0
(ii) (iii)
FromEq. (10.2) Icr = 141162"A Icz = 4.651128"A Ico=oA
Im= 141282"A Inz= 4'6518"A Iao=oA
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)
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
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) (viii)
! J
The reader should verify theseby calculatrng Io', andl*2from Eqs. (vii) and (viii) and comparing the results with Eqs. (iv) and (v).
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".
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 X" = sell'reactance o1'eachline 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
V6
vc
lr= l"+ lo+1"
-(-
Fig.10. 9
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" V, r
--
A (I/, -
or or Now
v,
x^x,x. x_x*x,
= J
I// / n-
(10.36)
Ib I"
zIo
(10.37)
v ! ) : zuIs -
,r//' s -
(10.38)
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 o o 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 Zr:
j(X, - X,r) : positive sequence impedance
Zz: j(X, - X^) : negative sequence impedance Zo: i(X, + 2X)
: zero ,tequence impedance
(10.43) (r0.44) (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.
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
3S2 I --]-
_
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^'. 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 betweenthe equivalent circuits of varioussequences*.
theshortcircuit occurslirrnr kradcclconditiorts,thc voltagt:behindappropriltc reactance(subtransient,transient or synchr:onous)constitutesthe positive 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
d-n>_ -u \
t6 --> -----b
L-----
____l_" Fig' 10'11 Three-phase synchronous generatorwithgroundedneutral
b
I
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 Sincea synchronousmachineis clesignedwith symmetricalwindings, it induces emfsof positivesequenceonly, i.e. no negatrveor zerosequence voltagesare inc lt rc ei n d i t' Wh c n th c tttl tc l ti nc u rr i 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)
= jXa (if steady state value is of interest) (10.48) If the machine short circuit takes place from unloaded conditions, the terminaivoltageconstitutesthe positivesequencevoltage; on the other hand.if *'fhis can be shown to be so by synchronousmachine theory,[5].
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
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 to the Voo = - Z{oo
(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
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 dependson soil conditions, it is esseniialto make some simplifying assumptionsio obtain analytical results. The zero sequenceimpedanceof transmissionlines usuallv
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).
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.
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
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.
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 cgcrrc ri ttosru p p l i c stw o rrto to rso v 0r' r,r transnri ssi on l i nc w i Lhtral tsl ol l rrcl ' s at both ends as shownin 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
,-66X-L-g ,.I-6XU j0'164 i oo8o5
.
-,
r- "xfi i 00805
) Eara
i
l + --,1
iio6e
i o 3 4 5 :- r P -----r--'---
In
9
Fig. 10.23 Positivesequencenetworkfor Example10.3 Referencebus
p ' u910' (-t,) 'r' .
IYA
) +l I .r:t!
ioz:), d
I Enrt(
(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.
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
' 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
: 0.082
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
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 are:
\7
Fig. P- 10. 9 Note: Zero sequencereactanceof each line is 250Voof its positive sequencereactance.
j,,W
voo"inpo*",,syrt"r Rn"tu"i,
10.10 Considerthe circuit shownin 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.
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.
*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 assumedthat 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 fault point 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
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
(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 .
4+
Zo)+ 3zfllor= Eo Eo
(zr * zz -t Z) +3zr
The voltage of line b to ground under fault condition is 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.
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 passivepositive sequencenetwork (with voltage generatorsshort circuited).
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
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 -
l I u -Ir t ' lI, : _i
Vr= IoZf
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 )
UnsymmetricalFault Analysis--_1
t
F ro m Eq . (1 1 .1 5 ),w e g e t Iu ' - -
!
L
I SQt*
or
- jJi J ' - u n, o
-
(l 1.23)
Voo= Vot * 3Zf Ino
(\ 1. 1 . 1 8 )
Z "+ 4 t,+*+2222 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 a power system.The fault may in generalhave an impedanceZf as shown. tr a
a-
Y/u=9 b-
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
( 11 . 1 9 )
i", + Io, + I"o =o|
V,,= V, = Zf (lt, + Ir) = 37rf1,,r, The symmetricalcomponentsof voltagesare given by
l!:'1 [r' aa2 "']lr.l o
l ! : ' l =*,l_l l 1"v,,,) I
l l v Iu
(r 1.20)
(b)
(rr.21)
rll,ur) (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 I ,. = l-
Eo
@
En
v,,t= V.z= Llv,, + (a + r11V,,1 3"'
voo- vot=
+
(11.24)
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
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 ubst 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
o
o
z;' ll E,,- 2,r,,, +3zfI,,nl
z o t ) L o )L r , o _ J
(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
E o - Z rl o t= Z z (Io r+ Io s)
Fig. 11.12 Sequencenetworksof synchronous generator groundedthrough neutralimpedance
or
'I u^ =o -E o - ( ,-l
zt+z'\'
," 1"'
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_ xn ----l'FJf[\-: | : """ /
Ea
(iii)
Comparing (ii) and (iii), it is easy to seethat llol Lc> llol3L
.--\"/ ) +))
b c
Fig' i i.i i
3lE"l
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.
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
- 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 i ti 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 earthed through 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
Iy (fault current for LG fault) - Io = 3lo, =
Fig. 11. 16 one- r inediagr amof t hesyst emof Exam pr 11. e 3
3E X t" o+ X z" ql Zo" ,
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
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 Il t o" l - |la 2l a^
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
'
l
'
I)L-j0.447J L-j0.136_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.
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 xam 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
ModernPower SystemAnalysis
4!4" 1 I
UnsymmetricalFault Analysis
Voltage of the healthy phase, V o = 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
Fig.11.22 Connection of sequence networksfor two conductors open
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
Z-trl
: 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.
/ 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
(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)
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 ue
( 11 . 3 e )
s ub s ti tu ti n gE q .(1 1 .3 9 ) i n E q . (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) 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 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-,
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
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,
'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:
orl 0 " r r0l 0
e 0
-r4.M6 2.024 0
f 0 2.024 -14.446 0
o 6
0 0 0 -0.584
lnverting the three matricesaboverendersthe fbllowing threebus impedance matrices
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) = - 0.202pl
0 0 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
3x1 j 0J7636+ j 0.r7636+ j 0.07061 = j7.086pu The fault curent with LG faulton bus /'is
il= e
Irr
__3x1 .i0.t82gg+ iOlAZgg+/O"O?Gi
(i)
) j0.436s9
= 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
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
A single line to ground fault (on phasea) occurs on the bus I of the system of Fig. 11.27.Find
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 0.15
0.15
0.05
0.05
(a)
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 7 zO-BUS
Reference bus
l-0.0s 0.051
-
;
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
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 Zt_svs= j[0.15]
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
""
[o'ls o'3s] Lo.rt.l
O) Vfvr = Vo r-, = Zt-rr lfr-t = 1. 0- 70. 105x - j3. 92 = 0. 588; Vot ; = |
(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 =
= -70.105x - j3.92 = 0.412 V{-z= - Zr-r, Ifr-,
.:= to - (- 0.412))t3o" j0.1s r.37- i2.38
rLc=
- - j0.045x - j3.92 = _ 0.176 vfo-t= - zurrlL,
IIo-c= 0 (seeFig' 1I'29) If o_c= et37 - j2.38)+ (1.37- i2.38) = _ j4.76
- - j0.045x - j3.92 = _ 0.176 = - 7O.005 x - j3.92 = - 0.02 I{-rz= yvrz (VI_r- Vfr_r)
(1 - o's88)t-33"
,o; ---1.37-i2.38
vtr;= - zr-trfr-t
vt-z=- Zuy IL,
l
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
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
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 eachgeneratorand feeder are given below:
Positive sequence Negative sequence Zero sequence
rI'6
Generator (per unit) jo.2 i0.15 j 0.08
Feeder (ohms/phase) 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 generatorterminals. 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 r ---'. so rl ^vos. u \^ ^€ '. r/ t ll \vt rr rvrv 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.
fauft X'= Xz= 2oo/o Xo = 5o/o Xn = 2'5o/o
Yr6i* F i g .P - l 1 . 8 n.g
the A double line-to-ground fault occurs on lines b and c at point F in of c phase in system of Fig. i-tf.q. Find the subffansientcurrent are machines Both machine 1, assuming prefault currents to be zero. of x//= xz=lUvo and xo= 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' The reactancesof the neutral grounding reactors are 5Vo on the kVA base of the machines.
_r_rr*"1_
i^
,_L
vn6l -:
F l g .P ' 1 1 . 9 11.10 A synchronousmachine1 generating1 pu voltageis connectedthrough in a Y/Y transformerof reactance0.1 pu to two transmissionlines YN a through parallel. The other ends of the lines are connected transformerof reactance0.1 pu to a machine2 generating 1 pu voltage' 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)
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
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
oo (i.e. neutral isolated)
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:
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
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.
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. ,q 6L-_'
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.
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.
'
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 udie 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
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
Power SystemStability
'.: -: il
MociernPower Sysiemnnaiysis
{36 i
It immediatelyfollows that 2GH = GH MJ-s/elect rad M = (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**
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
9-6 7-4 4-3
machine and electricalpowersin a synchronous Fig.12.1 Flowof mechanical 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 .
lnrltar ,vvrvr
L-J
2-4
- Ln v- . 2. \,\.l .l l-l l,l =l l y* : * *
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
D t -
,
-
mechanical ptlwer inPut l n electrical power outPttt t n
MW MW; statorcopPerlossis assumed
negligible. Rewriting Eq. (12.4) s2a
/ )\2 x 1o-6) ul;) u.r, ff
where * 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.
-
'ttt
ol
- P^ P,Mw
0, = angle in rad (elect) - pP" nod'\, = = ,P,,Mi;-
(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\-
' 26 , , - d: = M P^-PeMw ot-
(r2.7)
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 (12.8)
I +Sl
_
Gryrt",o= system base
From Eq. (12.6) d rg , _ d 2 6 ,Jt2 dt2 Hence Eq. (12.5) can be written in terms of d as
-
"f (12.r0)
a;
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 = P*trf dtz
(IZ.1I)
P"
(r2.16"
whcrc Prr=P*r+ Pn Pr=P"l * Prz H"q=Hr+ H,
(r2.r7)
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
R O MI/I\/IVA
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 electricalload of 50 MW, tind rotor acceleration,neglecting rrlechanicaland electrical losses.
(c) If the accelerationcalculatedin part (b) is maintainedfor 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
I=Ia+ Is
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
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) Equation(12.19)during the transientmodifiesto (12.2r) Et =V + jxt lo+ jXnln =V + jXq(I I) + jXotlo ' (12.22) = (Y + jxp + j(X'a - Xq)Id 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
x'a = xq
- r 2 o x 5 o+ ' z 8 . t z 5 x 0.2 4
= 1505.625rpm I2.3
(r2.20)
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
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.
Fig. 12.2 Phasor diagram-salient pole machine
(r2.23)
r"orrr r
,firZlll
atysis _
Power Svstem Stabilitv
equation (12.22)now becomes
\
E,=V+jXnI = V + jXotl
t
synchronousreactance)
---
-
2
o
(r2.24)
The machine mod also applies to a cylindrical rotor machine where
[ECAC,+'
T*
@
-D'
O
X,l = X*/(transient
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
(IZ.ZB)
} t z = n 1 2 ,: . J = 0 P,=P^*sin 6
(12.29a)
j0.5
I
I
I lE'' | P^^,' = lE'' - x ''
where
(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)
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.
(r2.31)
lh | lY^ llv,I Llrl Lv,, Yn rrrJL%l
Equivalent
Sinceno sourceis connectedat the bus3 It =o or
or
Y r r V r +Y r r V r + Y r r V r = O
vz=-?rr- ?r, Yrt Y.,
(r2.32)
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" Yr, Yr, -(.^tt
)''
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 L 0.646 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
ll.0 + 11.0x.t1.35 _ j1.35x.r1.l+ "11.1x i 1.0 - j3.e35
-!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 From power angle equation
(
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
_t v- -t-l l v I s t n o = p " X < 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 ) = 0 .8 9 2+ j 0 .6 - 1.075133.9"
I 900
0 33.90
A
Fig. 12.13 Poweranglecurue,s-
(a) Systemhealrhy F,.\,-, p^u* = lv )-lEt | -- 1x 1.075= - ., t't9 P U x,, c,5 P, = I'79 sin d (b) One line shorredin the middle:
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
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
\
H"q d26 -* ., =Pn7rI (lt-
where
6=4-
HrH,
,.- - P,) )
(r2.4r)
P,
(r2.42)
6.
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
The swing equation Eq. (12.41) and the power angle equation F;q- (12.aa) have the same form as for a single machine connectedto infinite bus. Thus a two-machine systemis equivalentto a single machineconnectedto infinite bus. Becauseof this, the single-machine(connectedto intinite bus) qYstemwould be 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
(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)
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
Mp, +[#], =o
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. 12.14whose clynamicsis describect by equations = P^
M = H 7 and
p, = !4)!
P " M W; Eq. (12.8) ln Pu sYstem
x,t
)
,in 6'= p^u*sin d
d dt
The system stability to small changesis determined from the characteristic equation
STEADY STATE STABILITY
M* dl
P
where
- l34I elect deg/s2
(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
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
po*e,systmst"uititv lEllvl c o sd ^ < o X
Il -a- ee 1|
or
(12.4e) 4>90" Themaximumpowerthatcanbe transmitted withoutlossof stabili
( r2.s0) and is given by P^u*
lEnvl
(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 correspondingcharacteristic 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 steadystatestability limit 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
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) =
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 = 1.0pr) throughtransformersand a line of total reactanceof 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 '
max
= 0 . 5 o r4 = 3 0 o
')*
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 The systemis shown in Fig. 12.16.Let the voltage of the infinite bus be taken as reference.
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 E = Vt + jXdI = 1.2 l0 + j0.5
rr . 2l g
I
L
I
Now
J
0 = 73.87" Vt = 1.2/.73.87"= 0.332+ j t . 1 5 2
r _ 0.332+.ir.rs2_r = 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.
- 1.0
.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
EsE
=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.
I2.7
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
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.
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.
Considerthe swing equation d26
AF
=
I
*@^-
M=! rf
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
ln pu system
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
(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
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
be possible to find angle d2 such that rg condition is finally reached when 41 own in Fig.l2.2L Under this condition, hat - 6t= 6. = 6^o= T
6 b f % ?'i
i
6
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
mechanicalinput Fig. 12.21 Limitingcase of transientstabilitywith increased suddenlY availablefor A, is lessthan A1' Any turther increase in P^, meansthat the area beyond point c and the so that the excesskinetic energy causes d to increase with the system power, decelerating power changes over to accelerating use of the equal by shown consequentlybecoming uistable. It has thus been input mechanical in 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
trs#ffi -{i..fuu.xdr
r._r_..._ ruooern A 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
(12.59)
4nu*=T- d; and
(r2.60)
P*= Pr* sin 6o
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
uoo"rnpo*"r systemAnatvsis
ffil
W
For the systemto be stable, A2= A1,which yields
cos{. = !^
(5,^^" \ -tniu(- d) + cos 4o"* Prn*
,, /__\ tt \--l
,.774
(r2.61)
Pm
During the period the fault is persisting,the swing equation is
l_Lr__l
|
I L
where
4, = critical clearing angle SubstitutingEqs. (1259) and (12.60) in Eq. (12.61), we get 4r = cos-t [(r, _ Z6l sin do_ cos 6o]
)
r
l
bus
IVVO"
(a)
(r2.62)
d,2d
= rf P^: P, = o d,r, 1r:
I Infinite
-l
IVVO"
(12.63)
Integrating twice
6 = -,rf- P*tz + $ 2H
0 " , = # P ; 2 " ,1 6 0
(b)
Fi1.12.25 Singlemachinetiedto infinitebus throughtwo par:allel lines (12.64)
where /cr = critical clearing time 4, = critical clearing angle From Eq. (12.6a)
2 H ( 6 , -, 4 ) TrfP*
(r2.6s)
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 infinitebus 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 by
P"n= g:+ ,\d
-T
rt7
sin d= pmaxrsin d
Fig.12.26
Equalareacriterionappliedto the openingof one of the two linesin parallel
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.
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
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)
- sin-r (:t_) 4,,* =T
,.a
V
postfault(1 line) P"11;,
maxIII
(r2.66) ./
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"< {,
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. trin 1n 1A
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) = 1.0 (6", - 0.45) = 6c,- 0.45 4ou*=r-sin-l
dr*
Az= !{r,n-p^)d,6
f, Prefault bus is
2.41 I 6.,
r2.41
= - 1 . 5 c o ds _ d l | 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
(i)
1.0= 1.69sin ,fr
6,,
= | ( 1 . 5s i n 6 - 1 ) d d J
operation
or
do = 0.633 rad
The positive sequencereactancediagram during fault is IL Durtng fault presentedin Fig. 12.37a.
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
lE'l=1.2
and
A-z =
V=1.OlOo
dmax f J
| 1 . 2s i n d d d - 1 . 0( 2 . 1 5 5- 4 )
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- -=
fir. Postfault
lal! i.+Z+
operation
sin vd -= 0.495 uur v.a/, sin (faulty
6
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.
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.
,'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^lt 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.
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
(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 r M
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.
^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
(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"
t
l 0.1
l
t l l 0.3 0.2 f (s) --
l i 0.4
i 0.5
_ r l
0.6
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
c rr r' rr ri n e r Vn
vn ur r' n / a
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
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. 0 .8 8 ouu, 0. 05 0 . 8 8 0. 05+ 2 .0 0 0.05uus 0 . 1 0 2.00 0 . i 5 Z .U U o. 20 2 .0 0 0.25 2 .0 0 0 . 3 0 2 .0 0 0 . 3 5 2 .0 0 o. 40 2 .0 0 0.45 2 .C 0 0. 50 T
Pr,=P,rr.,*,tin5 pr,= 0.9- pu pu
0.368 0 .3 6 8 0 .3 6 8 0.41 0 .4 1
0 .3 6 0 .8 2
0.493 0 .5 6 0.s91 0.597 0.561 0.494 0.41 0.337
0 .9 8 6 I.t2 T.I9 r.t9 I.t2 0 .9 8 9 0.82 0.615
0 .9 0.324
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
Fault Cleared in 6.25 Cycles Time to clearfault= ua?t = 0.125 s s0
-
A6
6
deg
deg
2.57
2.57
2.767 0.767 1.96 2.s8 2.58 1.96 0.79 0.71 2.0
5.33 4.56
_ _ -
2.60 0.02 2.56 4.52 5.31 4.60 2.6
21.64 21.64 21.64 24.21 24.21 24.21 29.54 34.10 36.70 37.72 34.16 29.64 24.33 19.73 17.13
0 0+ ouu, 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 A
A ?
u.4J
2.44 0.88 0. 88 0.88 2.O0 2.00 2.00 2. 00 2. 00 2.00 ' \ N
Z.W
6
deg 0. 368 0. 368 0. 368 0.41 0.524 0.680 0.767 0. 78 0. 734 0.613 0.430
0.9 0.324 0.36r o.46t 1.36 1.53 1.56 1.46 r.22 0. 86
u. z-1-1
u.4c)0
-
0.0 0.576 0. 288 0.539 0.439 4.46 0.63 0.66 0. 56 0.327 0.04 u.4-J4
0.50 2.oo T2.TO MULTIMACHINE
deg
.
-
6
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/
2r.& 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
STABITITY
From what has been discussedso far, the following steps easily 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.
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).
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.
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
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)"
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 wlr, wlll,
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, Y s s =Y r s * Y s q * Y s r * 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 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 6.6598_j44.617 -1.4488 +j8.8538 0 + j7.1428 -1.4488+ j8.8538 8.9769 + j57.2972 0
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,
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 = Y
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 (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
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.
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
= *I*\:|+,tt'i"l iL'),*,
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., ch 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
SOME FACTORS AFFECTING
TRANSIENT
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
|
STABILITY
We have seenin this chapterthat the two-machine system can 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
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
'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. Seriescornpensation becomesmore 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-phase fault. 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,
. i::
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
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
Fast
Valuing
or Bypass
Valuing
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
lr /' - fXt r n 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
-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
o'1 P Negative sequence
H = 4.167MJA{VA
Zero sequence
Flg.12-48 Connectionof sequencenetworkswith faultedline switchedoff
- 4.63 x 10a sec2lelectricaldegree 1,1= JU1 8 0x 5 0 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 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
P,u,o
sec d=
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
0.726
1. 5 1. 5 1.5 1. 5 1.5 r.5
0. 565 0. 052 - 0.55 - 0. 984- 0 . 6 5 10.497
0.0 0.0 0.0 0.0 0.0
1.0 1.0 1.0 1.0 1.0
0.85 0.078 0.827 r . 48 0.98 0. 146
0.15 0.922 r.827 2. 48 1.98 0. 254
.5.4P,,
ifi
elec deg elec deg
1.2 N A
1.2 3.6
6 elec deg 41.8 41.8 41.8 43.0
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. 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 that rhe
PowerSystemStability
\4gqe!_lg',gf tyq]g1_Analysis
I
300 !
I
I 250I
tI
i
I
L
\ I
r pote ii switchoff i
2ooi- / / l I
6q) g)
o E o
0.40 0.45 0.50 0.55
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 0.894 0.781 0.62 0.433 0.2s9 0.133 0.079 0.107 0.214 0.38 0.566 0.738 0.867 0.946 0.983 0.997
- 0.44 - 2.4 -0.34 -1.8 - 0.17 - 0.9 0.068 0.4' 0.35 r.9 0.61 3.3 0.8 4.3 0.881 4.8 0.839 4.5 0.678 3.7 0.43 2.3 0 .1 6 0.9 - 0.11 - 0.6 - 0.3 - 1.6 - 0.42 - 2.3 - 0.48 - 2.6 - 0.s - 2.7
r. 0 5 r . 5
I
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)
5'4P,, elec deg
Ab elec deg
b elec deg
0.827 0.682 0.s64
0.0 0.448 0.224 0.436
1.2 2.4
r.2 3.6
4t . 8 0 41.8 41.8 43.0
0.98s 0.726 0.98s 0.784 0.985 0.848 0.98s 0.908 0.985 0.956
0.7Ls 0.77 0.834 0.893 0.940
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
0.988 1.485
0.485 - 2.6
5.5
81.1
oory
0.05 0.075--0.10 0.15 o.20 o.25 0.30 4325---+ 0.35
I
l'r, (pu)
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 \ '
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
(faultcleared) Reclosure
| ' /
A
I I
,s:itth
-
Single pole switch off
Table 12.10 swing curvecalculation-single pole switching I',uu^
2.8 0.1 - 2.6
I
- 2.7 - 2.7 - 2.7 -2.7
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
t.44 1.34 t.l1 0.932 0.6s 0.39 0.2 0.119 0.161 0.322 0.57 0.84 1.11 1.3 t.42 1.48 1.5
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)
1.5
(Contd....)
Fig. 12.51 Swingcurvefor singlepoleswitchingwith reclosure
506 |
I ina
Modernpo@is
PROB TEIlIS
from its prefault position, determine the maximum load that could be transferredwithout loss of stability.
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.
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 ri rmr 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
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
A synchr t ) nous gener at orr epr esent cd by a volt ageof l. l5 pu in ser ies with a transientreactanceis c
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 Transient Energy Function Method, Prentice-Hall, New Jersy, 1992.
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+.
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.
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 oycdin 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.
j5! 2 ,1
I srr
M o d e rnP o @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
o o o o
SYSTEM STATE CLASSIFICATION
SECURITY ANALYSIS
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
= -
L
3
pb. 6 q
EE
o, c o (!6
8E
L
q)
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z
BE
r: .9
qF
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o o E o o_ o o. o-
=E;.
o E o
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o_o x o--c.Y
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abg
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g)'E c c F O (g(J
=
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d H x
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9 c
P c
'oE9 .b
.
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 c o o o
9 c
at
=
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.=
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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 t i 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 MV AR 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 tcrestecl 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 ulti 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.
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
| +80.e+13.21
| +s,s* 1tz.t
':: )ll
tl;f Itl_qq.z_j1z.s
':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
-;'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 | 1 . 0 6 t 0 "1'l_ l -!.6 + ^ _ I 41.O043t-6.1" 4s.9_j7.6 <_i __22.4_ j2.6 _22 .3 +j0.7 <_ 1t.s L --i-1-I {nze+iz1.7l { t.s - 1t.t I --- V_21.5_j4.8 -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 explainthe 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,contingencyanalysis 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 g cs)or nrul ti pl c ccpri pnrcnt fi ri l urccvc^ts (failure of multiple unit or lines or their combination) one after another until all "credible outages"are considered.For each outage,all lines ancl voltagesin 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
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 processors or 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, 4t
= Change in MW power flow on hne I when a changein generation, AP", takes place at the ith bus
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 /
Now, for checking the /th line flow, we may write
j, = ff * 0r; APo, \ri LPotl E,lau,
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,,,= *
f i = power flow on /th line before the failure or precontingency power flow
(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
(13.8)
?,= 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.
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 n e l- 2) 1. 0001- 0. 3331- 0. 2685 - 0. 2094 0. 3735 0. 2091 0. 0 --0.4542 v =3 (line 0.4545 0 2-3) L -J) J \rr\rw -
= = = =
4 5 6 7
0.4476 (hne 2-4) -0.3634 0.3636 0.4443 0.0 (line 2-5) -0.1819 0.1818 0.2222 0.2835 (line 3-4) 0.5451 -0.5451 -0.6662 0.7161 (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.
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 of generation "ii.G 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 phenomena and 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
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-secondsFiq. 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^
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
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
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
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
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
n'F EA/-'n-r vr I nv
i,
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.
,{^*,:^^^ \rlvv.ltvgD.
(ii) Betterand probabilisticloadmodelling. (iii) Developtechniquesandmodelsfor studyof non-lineardynamics of largr siz.esystems.For example,new methodsto obtainnetworkequivalentr suitablefor voltagestabilityanalysis.
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.
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.
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
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
/\( r|1+/. J4,(, .\ 5 )
J =7'V
lJ).
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 anotirer related 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 values of the vector y. Since the variable r is assumedto be zero m ean,o n e ma 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)
gradoJ = 0
It is easyto check (see,e.g. [1]) that Eq. (1a.7)leadsto the following result.
j' = Hf,
'notmal equation' and may be solved explicitly This equation is called the 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)
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
whcre
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
\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
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)
(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
(lr/21) y,
^ *=[ L- ellr) y,
- (r}/Zt) y,
(2ot2r) y,
(t0/2t) yr
I
(r/2\ y, )
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 estimation error ls zero on an averase. x =kr The covariance of the error of estimation is therefore given by P, = KRK
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.
I
536 | I
Modern Power SystemAnalysis
An lrylfgduction to_StateEstimation of PowerSystems
I
(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
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
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
useful result in the sensethat it shows a mechanismfor improving on the initial estimafeby making use of the available nleasurements. Having obtainednew 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 ma te.i o b tai 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
I S3T
i ,00"," ) eq.(+..227
I
/
c"l::'1: -,=n) \,* ! -_I\,,ll ./ :-
No
\-
-''!'
ls .-2a.,
r Yes
\ Stcp
rig. r+.1
',: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).
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 )2
l, N
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
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
!=hlxl+ r with the components of the non-linearfunctiongivenby
(14.31)
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 Yi 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 , | yil I vjl I IUI
|
/
C
a
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' (r4.33)
(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
'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
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 Vf 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
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 z= Bx + r-
(r4.39)
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.
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. Vi 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
\
,
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 m p 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.
[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.
| 545 T-
\" Internal system
Boundary system
Flg. 14. 4
,/
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 t i ma ti o ni n th e p !e s e n c eo f b a d d ata.
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 test X@).Once the valuc of c is determined,it is simple to cary out the 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)p ot 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
,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
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. I4.8
NETWORK OBSERVABILITY MEASUREMENTS
I
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. UbsefvaDlllty
\
( 1 4 . s1 b )
Ct = l, f ot l; lo, S ai = 0, f or lilo, > a'
I
- -r--
I
l---:-
^
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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
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
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topology,
lhrs
kceps
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.
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In real-time environmentthe state estimatorconsistsof different modules such 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
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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)
100 r00 100
Mrz Mtt Mzz
Accuracy (MW)
o (pu)
+ 8 +4 + 0.8
0.02 0.01 0.002
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,
d2 given the
Mzr
Full scale (MW)
100 100
Meter Standard Deviation ( o ) in full scale
Reading
1 4
32 -26
Meter
(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.
'l I
I j
t
-
ModernPowerSystemAnalysls
;554.;l
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
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 O8 -2 1 5 . 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
ii
I
_T-I 555
3'79r-3799. 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.
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.
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
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.
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
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,-
I
, ,
ModernPower System AnalYsis rL X
I
=degree of compensation = 25 to J5Vo (recommended)
fc
' 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)
i'r ;
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 e rl 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
Power
I
transformer I
I -? {-rf '-- l
neaitr i I
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.
i] l.:
Fixed caoacitor
I n"r.to, f =..'x',rnc.I
)tl II
u
v
v
v
I
I 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.
| ;
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
/
Fig. 15.4 Thyristorswitchedcapacitor(TSC)
As suchthey are applied as a discretlyvariablereactivepower source,where this type of 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)
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
Damping reactor
-
TCR.FC (1)
Hannorric generation
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 o rcl 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-TSC (3)
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)
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
I n t c r n a l l yh i g h (large pu TCR) Requires significant filtering l/2 cycle
hrtcrnallyvcry low Resonancemay necessiratetuning reactors I cycle
Poor (FC ('iluscs transientovervoltages in response to step disturbances)
Cun bc lrcutral. S a n r ca s i n ( 2 ) (Capacitorscan be switchedout to minimise trattsicntovcr-voltages)
T5.7 COMPARISON BETWEEN STATCOM AND SVC It
ll Capacitor
Fig. 15.5 A combinedTCR/TSCcompensator
a
Neutral
mqw "'*J
he
nnterl
trhr roqtL
ri -r r
tLhr o rv
rnrnvrl r, l-l,a' ll
l;-^^lrlrt/q.r
^*^-,-f:^\,Pgr4trlrE
rdlrBc
^r ul
-r- ule
r t t v -l
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
.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) The rapid developmentof power electronicstechnologyprovidesexciting opportunities to developnew powersystemequipment for betterutilizationof
* 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
'
-^'.'^PrJw9r
^-,^f^^J^^ E^Urr4rrBtr
O=vI=
:^ l5
v(v-vo) X
563 l
ModernPo
I
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,
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
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)
/
Synchronous
Series
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
,
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
rr$il#4
Modern
l
l"r'.
-
r'
i:
I
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
Load flow control
V control
X
XXX
XX
x x
XXX
Transient
Oscillation
stability
Damping
xxx xxx
XX
XXX
XX
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
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
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 I0-
Schemes for Ac Transmission svstems". IEEE Trans. pAs 96 6). l8-10.
Nature of forecast Very short term
A few seconds to several minutes
Short term
Half an hour to a ferv hours
rg77. pp lglg-
12. Kimbark.
"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.
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
Lead time
Medium term Long term
A few days to a t'ew' rveeks A t'ew months to a few years
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
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)
(r6.2)
Technique LoadForecasting
Modern PowerSystem Analysis 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 E {bkz - ya(k)k+ tdkl - 0
(16.4a) (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)
kt5f,*A I
it 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 g5 from 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.
--/
,
,
i16
I t
the data
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 /
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 consecutiveweeks 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 the tn" uurrage and the porynomial rerms. consider for example ;ilri'.o,',; supplied power the active curve shown in Fig. tZ.Z wtrich givei the variation of that tlrg daily load observed is It weeks. t*o of period a over by a power utility fluctuations' random some variations are repetitive from day to day except for those of the from It is also seen that the curve for sundays differ significantly ior ihe curve the out that week days in view that Sundaysare holidays. I,,]rTt the mid till Sunday one of entire weekly period starting fiom, say, the mid night with waveform periodic night of the next Sunday behaves as a clistinctty variations' superposedrandom 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
[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 )
(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.
(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.
(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)) 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'+
GQG-DG',
to use the solution of Eq. (16.18a)for the vector x (k+ d)to get the result
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) (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
= 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.
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
12.=1 9
I
4
l
L
I
o l c o
E
l l
y(k)=h'(k)x+
'1C
E
i3
14
where
n f L
15
16
17
18
Comment
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.
(16.20b)
h'(k)= [yr(ft)yz(k)... yu&)] and x = far a2... ayl.
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. Techniques
e(k)
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
on'line
(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
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
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.
@
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. 12. Mehra, R.K., "On-line Identification of to Kalmann Filtering", IEEE Trans. Vol. AC-16, lg7l, pp lZ_21.
77
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.
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
REACTIVE
POWER FLOW AND VOLTAGE
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
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
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 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 0.9 pf lead
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 [l].
-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 changeover occurs 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 i n flrrv l r a rvwwr l^.',o^^* /..^.,+^Ll^ --,--\ 0 and ^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.
where
S = complex power at load bus 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
(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
Ers6= E cos5l !9 + 2Y]
or
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)
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)
(r7.1r) dZ dV Application of this criterion gives value of z";. \ (iii) Ratio of source to load reactance is very important dnd for voltage stability
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 stage of voltage stability is given by
Criteria
> Ers,
cntenon 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 (lU -
0.75
0
.
Pr", is the maxlmum powertransferat upf
1.0
X J
Stability
..it.rion: (E'=generatorvoltage; V=load voltage). Using this crite-
rion, the voltage stability limit is reachedwhen
cos d{#.#}+sindffi-}=o
(r7.e)
(17.r2)
X load
a indicates the off-nominal tap ratio of the OLTC transformer at the load end. T7.5
VOLTAGE
STABILITY
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,
.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 "'?'J"6;ffiffi;;;g."riuulirv s 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' Generators
and their
excitation
controls
AVR, load It is necessary to consider the droop chatacteristics of the should controls 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
*=f(X,n and a set of algebraicequations I (X, V) = YxV (Xo' Ve)' with a set of known initial conditions X = system state vector where = but voltage vector Y 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
(r7.r3) (r7.r4)
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 maY refer to [3]' *For dctailcd accountof HVDC, the reader
'
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
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
ModernPo
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=-
Krt V2
Qn"t= Qnua t
.'. Here,
Qn r= v' -
[-ve sign denotesinductive reactivepowerinjection.l
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,
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.
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 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. .
. . t ?
i -
d -
t
! l ! -
F
L -
- - - , - - - - ^
L - -
AppnNDrx A
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.
2t, .)
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. 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. 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
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)
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.
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
(A-2)
-
.
flTY'
t I
A
0x.= 0 The unit 'tector e the.rest of the componentbare zero, i.e.
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.
*tv AD",y,Aytx
0
€k:
A
0 1 kth component 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
(A_3)
i:l
Also, it is interesting to note that 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
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). '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
For each square matrix, there exists a determinant which is formed by taking the determinant of the elements of the 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
aoJA=
l3 lz l-1 t-l lI 2 1 lI 3
l-t
l 1 l2
lr
l 2 l-1
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
cqrnc
o,rAcr
qrc
ctz= anbn + arrb2
AB+BA
i-?
then,
lhe
r
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
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
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
,l[ 3 ol=fo 0l [ r 0J 0J 0l L0
L-l
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
Multiplication of any matrix by a unit matrix results in the original maffix, 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 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^
(A-14)
I f A is given by Eq. (A-9), then from Eqs. (A-10), (A-1I), (A-17), we get
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)
It can be written as a scalarfunction of a vector variable x, i.e. y - f(x) where x is an n-dimension vector,
0*,
0*, 0fz 0*,
0*,
0*,
o f a }fz
A scalarfunctionof n scalarvariablesis definedas
0x
(A-1e)
af- af; ..
o*n ?fz 0rn
(A-zs)
af; o*n
(A-20)
(A-26)
In general, a scalarfunction could be a function of sever al 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^
f (x, u, p)
(A-23)
LJm\rrUrP)J d Let us now find -. ox
DERTVATNTES OF SCALAR AND \ZECTOR FUNCTIONS
s
^
i
.
According to Eq. (A-24), we can write
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*, oJ orn -
(A-24)
6-2g)
. Accordingro Eq. (A-24), we can write
In general, a vector function is a function of several vector variables, e.g.
y-f(x,u,p)
(A-27)
la"l
Lar,J
(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.
.+ 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, passiveand 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)
r
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
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 rAz + 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:
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:
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
C
AC, A
1 [usingEq. (B-5)] AC 1 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=
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).
(Zss(Zno - Zo))''' Z^o
(zso(z^o- zo))t''
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 3rd edn., B.I. Publishers,New Delhi. 1970.
t;l
1'l['t I = l'L3 ,l1,,I
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 Upon back substituting,that is first solvingfor x2 and then for 11, we get
o-+
-
-
I _
I Y
(2)xr+(3)xz+(4)xt=)
1
( 3 ) x r+ ( 4 ) x r + ( 7 ) 4 = 1 4
(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)
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:
lL o ?t : r) Ll xl z_l l- Ll t , Jr x2= --7:-2
Consider the following system of linear equations: (Z)xt + Q)x2 + (3)x, = $
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 - + (*)rt = + (1)x, + (tr)*r+(*)r, _ (1)x,+ (I)*t
(c_'7\
J 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
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=
Thus for heading 3, z
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 neededto normalize
h 6 9 t4
lt-
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.
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].
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
NCES REFERE
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
= 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^)
Trans.,Nov. 1967,No. ll, PAS-86: 1449.
(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^ #=
- ( a ^ e i +b , f , ) =J i ^
tvrOtlcItI rOWgf l r '
I
For the case of m = i. we have
= l%l exp(- j6,)ilY,pl exp(i0*) lvplexp(7d')
expQfi) *06, -i*" 06, = - jtvitexp(- j6,)i ty,niexp(j4) tV1,t k:l
k:l
+ lV,lzlY,,lexp (7d,,)
+ jlV,l exp (- j6 ) (tytil exp Q?,,)tV;texp (id)) = - j(Pi - jQ) + jlV,l'(G,, + jB,,)
oPi'
-alv)
av,l ,et_
P i - G ' i t v i l 2= J '
avil
Now differentiate Eq. (D-1) partially with respect to lV^l (m r i). We have
a4
0Q, = L (- j6) (lYi*l exp (j0,^) exp Qd^)) ' exp " a l v ^ l lvil
it*
Case 7
lvil = Qi - Biilvil, = L t
m* i Hi^= Li*=
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
(a* * jb^) = (Gi^ + jBi^) @* + jk) Case 2
a ^ e , +b , f i = N i ^
(D-7)
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
Ao'
lv,l-i "", lv,l alvil ' " awil
(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)
\v'l = Pt * GiilViP = Nii
Ni*=-
!a! rl vr *^tl=
+
The above results are'summattzed below:
Multiplyingby lV^l on bothsides, aPt
lQ)
It followsfromEq. (D-5)that
= - Qli- Biilv;12= H"
ao. = ;t
= \Ft-
(D-3)
From Eq. (D.3), we can write
#
q - " "
f.lr.$
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.
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 0u
du
\*o-0
(E-e)
In Eq. (E-9), di=
Qi.^u*
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)+ and
of*u*{, -u^u*)+
oT*,n1r-in-z)
(E-6)
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.
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.
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.
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.
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 A k t ' y l' c i l tu rc $ 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, f^.-fe
rilur!
l^^-r.l^-^
--l
luL;aLtuil, ailu
su
un,
r
1
1.() tmprove
t
|
!t',
availaDtlty.
t.'
tr aISo lntegrates
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.
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 rrvr yt zu rr rt vr irruavrr n r raon f < '
horra rrqvw
frnia 6vuw
rrh up
iur
^o.rroy\ryvvr
*l^^+. pr4rrLD
(^rlr-D^ ( J . II l-l
^-l^uluttt
a^:-.--^-.^ tu lrrrpl'uvc
rLule
-r--^ Ixant
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.
l,@E+l
ModernPo@is
I
| -^^
Analog Inputs (i) (ii) (iii) (iv)
Pressures,flows, electrical parameters,etc. Analog input of 0-10 V DC Thermocoupleinputs 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 System 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.
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.
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
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 't
t
I
Matrix
To generate an identity matrix and store it in variabte K give the following command
[-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
9
3
1l
abs(x) gives 5 If applied to a matrix, it results in a matrix storing absolute values of all the elementSof a given matrix.
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 s ul 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
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.
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.
flH i U ccrlrnn i".Sl oircrctc '. S furtmsaroUos
Dbcret
Unu
, H nmr'oer i .fi sgn*aryrtan
Fqdin t l$hr
iHs*'
Ma['
:. fl sar:er f,l crrtrd9y*entooDox
Nmlle
Sign* t SJ,s{emt 1 :l
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.
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
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 transferfunction 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.
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.
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
'
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
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
% 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 . % % yo
% t5
elementno I connectedI lFromlTo I IB u s n Ioe u s nI o
y d a t a = [1 2 3 4 5
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 ) )) ;
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 d e 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 ) ;
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 )
end 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 ) e nd 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 end 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 ) ; end 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 ) ;
Hffi -
MooernpowersystemA_nalysis
W
lFitlilE4lq. ffi.
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 ;
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 5 . 8 8 2 2 8 - j * 2 3 . 5 0 5 1 a 1 ; % B u st y p e s a r e j n i t i a l j s e d i n t y p e a r r a y t o code1 whichst andsf or PQ %bus. %code2 stands for PV bus type=ones(n,1); % w h e nQ l ' i m j t s a r e e x c e e d e df o r a p V b u s B u s t y p e i s c h a n g e dt o p Q %temporari1y Lan
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% i s t e m p o r a r i l y c h a n g e df r o m p Q t o p V . O t h e r w i s e i t i s z e r o ( ns, 1 ) ; typechanged=zero % s i n c e m a x a n d m i n Q l ' i m i t s a r e c h e c k e do n l y f o r p v b u s e s , % m a x& m i n Q l i m l t s f o r o t h e r t y p e s o f b u s e s c a n b e s e t t o a n y v a l u e s . % here we have set them to zeros for convenience Q l i m i t m a x = z e r o(sn , 1 ) ; Q 1i m i t m i n = z e r o (sn , l ) ; V m a g fxi e d = z e r o (sn , 1 ) ;
ModernPowersystemAnalysis
W
I
t I t o t o t a l n o . o f e q u a t i o n s( e q c o u n t ) for ceq=1:eqcount, for ccol=1:eqcount, v s J \ e e y / t s J r v v v
pause u p d a t e =ni v( J a c o b )* m is m a t c h ;' noofeq=L; r \ u J J v e v r u u J \ v w w r l l r ,
b m =mi a g( Y( a s s o e q b u(sc e q ) ,a s s o c obl u s( c c o :)l ) * V ( a ss o c o lb u s( c c o l) ) ) ; e i = r e a l ( V( a s s o e q b u(sc e q )) ) ; f i = i m a g( V( a s s o e q b u( sc e q )) ) ; i f a s s o e q v a r ( c e=q=) ' P" ' & a s s o c ovl a i ( c c o 1) = = ' d ' , i f a s s o e q b u( sc e q )- = a s s o c obl u s( c c o l) , H = a m * f-i b m * e i; else 1=1 - Q( a ss o e q b u(sc e q )) i m a g( Y( a s s o e q b u( sc e q ) ,a s s o c obl u s( c e q ) ). . . * a b s( V( a s s o e q b u( sc e q ) ))^2) ; end J a c o b( c e q ,c c o l) = H end i f a s s o e q v a r ( c e=q=) ' P ' & a s s o c ovl a r ( c c o )1= = ' V ', ' if a s s o e q b u(sc e q )- = a s s o c o bl u s( c c o l ), N=am*ei+bm*fi; el se (assoeqbus (ceq)) +real(Y(assoeqbus (ceq), assocolbus(ceq)) . . . N=P * a b s( V( a s s o e q b u(sc e q ) ) )^2); end J a c o b( c e q ,c c o l) = 1 1 end i f a s s o e q v a r ( c e q ) = =& ' Qa' s s o c o l v a r ( c c o 1 ) = = ' d ' , i f a s s o e q b u( sc e q )- = a s s o c obl u s( c c o l) , J=am*ei+bm*fi; el se ( cse q) , a s s o c obl u s( c e q )) J = P( a ss o e q b u(sc e q) ) r e a l( Y( a s s o e q b u * a b s( V( a s s o e q b u( sc e q ) ) )^2) ; end J a c o b ( c e q , c c )o=l J end i f a s s o e q v a r ( c e q ) = =& ' Qa' s s o c o l v a r ( c c o 1 ) = = ' V ' , l u s( c c o l) , 1f a s s o e q b u( sc e q )- - a s s o c o b L = a m * f -i b m * e i' else ce L = Q ( a s s o e q b( u s q )) - . . . imrn/V/'cc^o^h,,.t/^o^\ 'e.^^^'1 h , , . / ^ ^ ^ \ \ * . h . / \ ,q w J \ r \ q 1 ^ ^V^Uh U , , J- \/ ^L ^g ^r \{ \,\ ,' \ D \ . J .J .v. E rfffuv\r\.uJJvsqvuJ\lsY/rqJJvuvlpuJ\LsY/r/, ll
if type(i;==1 n e w c hni a n g V = u p d a(tneo o f e q ); n e w a n g V = a neg( V 1( i ) ) + n e w c hni a n g V ; n e w c hni m a g V = u p d a t e ( n o o f e Q + l ) *(aVb( is) ) ; j nmagV; (V('i ) ) +newch newmagV=abs V( j ) = p o la r T o r e c t( n e w m a gnVe, w a n g V * 1 8 0 / p;i ) noofeq=noofeq+l; else n e w c hni a n g V = u p d a(t ne o o f e q ); V( ' i ) ) + n e w c tnr ia n g V n e w a n g V = a neg( 1 ; V ( i ) = p o la r T o r e c t( a b s( V ( i ) ) , n e w a n g V * l 8 O /)p;i noofeq=noofeq+l; end end % A l I t h e f o l I o w in g v a r i a b le s / a r r a y s a r e c l e a r e d f r o m % m e m o r y .T h i s i s b e c a u s et h e ' i r d i m e n s i o n sm a y c h a n g ed u e t o % b u s s w i t c h e da n d o n c e u p d a t e sa r e c a l c u l a t e d , t h e v a r i a b l e s 9rare of no use as they are being reformulated at the % end of each iterat'ion \ c l e a r m i s m a t c hJ a c o b u p d a t e a s s o e q v a ra s s o e q b u sa s s o c o l v a fa s s o c o l b u s ; d i f f = m 'ni ( a b s( a b s( V( 2 : n ) ) - a b s( V p r e v( 2 : n ) ) ) ) ; n o o f i t e r = n o o f it e r + 1 ; end
Ex. G.8 (Table 7.1) % MATLAP B r o g r a mf o r o p t ' i m u ml o a d i n g o f g e n e r a t o r s % T h e d a t a a r e f r o m E x a m p l e7 . 1 % r t f i n d s l a m d ab y t h e a l g o r i t h m g i v e n o n t h e s a m ep d 9 € , o n c e t h e d e m a n d rk is spec'ified % W e h a v e t a k e n t h e d e m a n da s 2 3 1 . 2 5 M W c o r r o s p o n d i n gt o t h e l a s t b u t o n e % r o w o f T a b l e 7 . 1 a n d c a l c u l a t e dl a m d aa n d t h e l o a d s h a r l n g % n is no of generators n=2 % Pd stands for load demand. % a l p h a a n d b e t a a r r a y s d e n o t ea l p h a b e t a c o e f f j c i e n t s % for given generators.
Ll,
end J a c o b( c e q ,c c o l) = L en0 end en0 %New U p d a t ev e c t o r i s c a l c u l a t e d f r o m I n v e r s e o f t h e J a c o b i a n
Pd=?31.25 a l p h a =[ 0 . 2 0 0.251 b e t a =[ 4 0
301 % i n i t i a l g u e s sf o r l a m d a I amda=20
Hffi
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I amdaprev=l amda % t o l e r a n c ei s e p s a n d i n c r e m e ni tn l a m d ai s d e l t a l a m d a eps=1 de'tl al am d a = 0 .2 5
m u m| ' rm tt s o t e a c h g e n e r a t in g u n %ar es t ore d i n a rra y s P g m i na n d P g m a x . % I n r e a l l i f e l a r g e s c a l e p r o b l e m s ,w e c a n f i r s t i n i t i a l s e t h e P g m a x % a r r a yt o i n f u s i n gf o r I o o p a n d % P g n i n a r r a y t o z e r o u s i n g P g m i n = z e r o s ( n , 1 c) o m m a n d % L a t e r w e c a n c h a n g et h e l ' i m i t s i n d i v ' i d u a l l y Pgmax=[125 I25] P g m i n = 1 2Z00 ) P g - - 1 O 0 * o (nne,s1 ) w h il e a b s( s u m ( P S- e) d ; ' g t t for i=1:n, P s ( i) = ( 1a m d a - b e t(ai ) ) / a l p h a( i ) ; i f P g ( i) > P s m a x)( i P g ( i) = P s m a x)( ;i end if Ps(i).Pgmln(i) Pg(i)=Pgmin(i); end
m 1 n" l n f i cost=0; for j=0:n, f o r i = 0 :n ,
m'tn'tmum
for I =0:n, unit = [0 0 0 0]; % H e r e w e e l i m n a t e s t r a i g h t a w a y t h o s e c o m b i n a t i o n sw h i c h % d o n t m a k eu p t h e generation % n M Wd e m a n do r s u c h c o m b i n a t i o n sw h e r em a x i m u m % on lndividual % g e n e r a t i o n i s e x c e e d i n gt h e m a x i m u cma p a c i t y o f a n y o f t h e % generators &)I < . . . i f ( ' i + j + k + ' l ) = = n& i c P g m a x ( 1&) i < P g m a x ( 2 & ) kcPgmax(3 (4) Pgmax if i-=0 unit(1,1)=i; % F t n d o u t t h e c o s t o f g e n e r a t i n gt h e s e u n i t s a n d % qdd 'it up to total cost c o s t = c o s t + 05.* a lp h a( I ) * i * i + b e t a( I ) * i ; end i f 3-=g u n it ( L , 2 ) = i ; c o s t = c o s t + 0 . 5 * a l p h a ( 2 ) * j * j + b e t a ( 2 ) * i; \ end i f k-=0 u n it ( 1 , 3 )= k ; c o s t = c o s t + 0 . 5 * ap lh r ( 3 ) * k * k + b e t a ( 3 ) * k; end i f l-=0 unit(1,4)=l; c o s t = c o s t + 0 . 5 * ap1h a ( a ) * l* l + b e t a ( 4 ) * l; end % If the total cost i s coming out to be I ess than of the cost in % minimum % p r e v i o u s c o m b i n a t ' i o ntsh e n m a k em i n e q u a l t o c o s t a n d % c u n i t ( s t a n d f o r c o m m i t t e du n i t s ) e q u a l t o u n ' i t s % c o m m i t t e di n t h i s i t e r a t i o n t ( d e n o t e db Y v a r i a b l e u n i t s ) if cost < min
end i f ( s u m ( P g ) - P)d. 0 l a m d a p r e va =m l da; l a m d a =al m d a + d e t al 1a m d a ; el se I a m d a p r e vI = a m d ;a l a m d a =al m d a - d et la l a m d a ; end end d i s p ( ' T h e f i n a l v a l u e o f L a m d ai s ' ) 1amdaprev d i s p ( ' T h e d i s t r i b u t i o n o f I o a d s h a r e d b y t w o u nj t s ' i s ' ) Pg
Ex. G.9 (Table 7.2) u n i t c o mmi ttment by B rute Forcemethod % M A T LAPBro g ra m fo r o p ti mu m f t T h ed a t a f o r t h l s p r o g r a m c o r r e s p o n dt so T a b l e7 . 2
cun'it- un'iti mi n=nnc.f : r r r
cI e a r ; % a l p h aa n d b e t a a r r a y sd e n o t ea l p h ab e t a c o e f f i c i e n t sf o r g i v e n g e n e r a t o r s alpha=10.1 7 .76 02 . 0 0 2 . 5 0 1 ' ; b e t a = 1 2 32. 5 6 . 5 3 0 . 03 2 . 0 1 ' ; P g m i n =1[ 1I 1 ] ' ; Pgmax=ll? 12 12 I2l':' n= 9 % n denotestotal MWto be commi tted
.
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end end , d i s p ( ' c u n i t d i s p l a y t h e n o o f c o m m j t t e du n i t s o n e a c h o f the four generators') d i s p ( ' I f c u n i t f o r a p a r t i c u ) a r g e n e r a t o r i s 0 j t m e a n st h e u n i t i s n o t comm d i s p ( ' T h e t o t a l n o o f u n i t s t o b e c o m mt ti e d a r e ' ) cunit
Ex
G.lO
E
for i=1:n, (i , : ) *Pg-B( i , i ) *Pg( i ) ; si gma=B
i, i ) ) ; P g( i ) = ( 1 - ( b e t a( i ) / l a m d a-) ( 2 * si s n a ) )/ ( a l p h a( i ) / l a m d a + z *(B I
if Pg(i)'Pgmax(j) (i ) ; Pg(i ) =Pgmax end if Pg(i)
(Ex. 7.A)
clear % MATLAB P r o g r a mf o r o p t i m u ml o a d i n g o f g e n e r a t o r s % T h i s p r o g r a mf i n d s t h e o p t i m a l l o a d i n g o f g e n e r a t o r si n c l u d i n g % penalty factors % I t i m p l e m e n t st h e a l g o r i t h m g i v e n j u s t b e f o r e E x a m p l e7 . 4 . % T h e d a t a f o r t h i s p r o g r a ma r e t a k e n f r o m E x a m p l e7 . 4 % H e r e w e g i v e d e m a n dP d a n d a l p h a , b e t a a n d B - c o e f f i c i e n t s t l , { ec a l c u l a t e l o a d s h a r e d b y e a c h g e n e r a t o r % n is no of generators n=2 ? Pd stands for I oad demand % alpha and beta arrays denotealpha beta coefficjents for given % generators P d = 2 3 .70 4 ; a l p h 6 =[ 0 . 0 2 0
0.0a1; beta= [16 2oli' s I amda I i n i t i a l g u e s for I amda=20; I amdaprev= I amda; % t o l e r a n c e i s e p s a n d i n c r e m e n ti n l a m d a j s d e l t a l a m d a eps=1; d e lt a l a m d a = O . 2 5 ; % t h e m i n i m u ma n d m a x i m u m l i m i t s o f e a c h g e n e r a t i n gu n i t % a r e s t o r e d i n a r r a y s P g m i na n d P g m a x . % I n a c t u a l l a r g e s c a l e p r o b l e m s ,w e c a n f i r s t i n i t i a l i s e t h e P g m a xa r r a y % to inf using for loop % a n d P g m i na r r a y t o z e r o u s i n g P g m i n = z e r o s ( n , 1c)o m m a n d % L a t e r w e s h o u l d c a n c h a n g et h e l i m i t s i n d i v ' i d u a 1 1 y P q m a x = f 2 0 02 0 0 . l : P g m i n = [ 00 ] ; B = [ 0 . 0 0 10 0 0l; Pg=zeros(n,1); n o o fi t e r = 0 ; PL=0; Pg=zeros(n,1);
end PL=Pg'*B*Pg;
( s u m ( P g ) - P d - P)L< 0 1amdaprev=l amda; I amda=lamda+del tal amda; el se I amdaprev=1 amda; I amda=lamda-deltal amda; end noofi ter=noofi ter+1 ; Pg; end d i s p( ' T h e n o o f i t e r a t i o n s r e q u i r e d a r e ' ) n o o f it e r 'lamda ' d i s p( ' T h e f i n a l v a l u e o f is ) 1amdaprev d i s p( ' T h e opt 'im a1 I oading of gener at or si ncluding penal t y f act o, r s'is' ) Pg d i s p( ' T h e I o s s e sa r e ' ) PL if
Ex. G.II % I r i t h i s example the parameters f o r a l l t h e b l o c k s f o r t h e s y s t e m i n T h i s program has to be run prior to the % Fis. 8.8 are initialized s' i mul ati on b o t h f o r F i g s . G . 3 a n d G . 4 . Tsg= .4 Tt=0.5 Tps= 20 K ps= 100 R=3 K sg= 1g K t= 0.I K i= 0 . 0 9
po*b,syrt"r Rn"trrri, Moo"rn
SS1
Conetant
Ksg .s+1 TraneferFcn TraneferFcn 1
lffiw ffiflI
. Kps Tps.s+ 1
Fig. G.3 Firstorderapproximationfor load frequencycontrol Ex. G.12
I curren t busno indicat esm axim um no. of busesaddedunt il now currentbusno=0
for count=1:elements, [ r o w sc o 1s ] = s j z e ( z b u s ) , f rom=zprmi ar y( count2) to= z prm i ar y( count , 3) , valu e=zprmi ar (ycount4) % n e w b uvsa r i a b l ei n d i c a t e st h e m a x i m u m of the two buses % newbusbus mayor maynot a l r e a d y b e a p a r t o f e x i s t i n g z b u s ( f ax r om t, o) newbus=m
The system of Fig. 8.10 is simulated using simulink as was done in E x am pleG.l l . Ksg Tsg.s+ 1 TransferFcn TransferFcnl
I
Kps Tps.s+ 1 TransferFcn2
Flg. G.4 Proportional plusIntegralloadfrequency control Ex. G.13 (Example 9.8) % P r o g r a mf o r b u i ] d i r r g o f Z b u s b y a d d i t ' i o n o f b r a n c ho r l . i n k % Z p r i m a r y = [ e l e m e n t n of r o m t o v a l u e % % yo
--l % % H e r ec a r e s h o u l db e t a k e nt h a t t o b e g i nw i t h a n e l e m e n ti s a d d e dto % r ef er en c ea n d b o th fro ma n d to n o d e ss houl dnot be new nodes clear zpri mary= [ 1 I 0 0.25 2 2 I 0.1 3 3 1 0.1 4 2 0 0.25 s 2 3 0.11 [ el em ent sco 1u mn s=] s i z e (z p ri m a ry ) t T o b e g i nw i t h z b u sm a t r i xi s a n u l l m a t r i x
% r e f v a r i a b l e i n d i c a t e s t h e m i n i m u mo f t h e t w o b u s e s % & not necessarjly the reference bus % r e f b u s m u s t a l w a y se x i s t i n t h e e x i s t i n g z b u s ref=min (from, to) % Modjfication of typel % A n e w e l e m e n t i s a d d e d f r o m new bus t o r ef er encebus i f n e w b u s> c u r r e n t b u s n o& r e f ==0 z b u s = [ z b u sz e r o s ( r o w s1, ) z e r o s ( 1 , c o l s )v a l u e l currentbusno=newbu s \ conti nue end % Modification of type? % A new element is addedfrom newbus to old bus other than reference bus i f n e w b u s> c u r r e n t b u s n o& r e f - = 0 z b u s = [ z b u sz b u s (: , r e f ) z b u s( r e f , : ) v a l u e + z b u (sr e f , r e f ) ] currentbu sno=newbu s continue end % M o d fi i c a i t o n o f t y p e 3 % A n e w e l e m e n t i s a d d e d b e t w e e na n o l d b u s a n d r e f e r e n c e b u s i f n e w b u s< = c u r r e n t b u s n o& r e f = = O z b u s = z b u s - 1 / ( z b (unse w b u sn,e w b u s ) + vuael )* z b u s( : , n e w b u s*)z b u s( n e w b u s: ,) c o n t in u e end % M o d ' i f i c a t j o no f t y p e 4 % A n e w e l e m e n t i s a d d e db e t w e n t w o o l d b u s e s i f newbus<= currentbusno & ref -=0 z b u s = z b u-s L / ( v a l u e + z b u (sf r o m ,f r o m ) + z b u(st o , t o ) 2*zbus(from,to))*((zbus(:,from)-zbus(:,to))*((zbus(from,;|-zbuS(tor:)))) c o n t in u e end end
Rnatysis uodginPowerSystem
Wl
Ex. G.14 (Example 12,10) j=j+1
t Program f o r t r a n s ' i e n ts t a b i ' l i t y o f s i n g l e m a c h i n ec o n n e c t etdo l n f i n i t e %bus this program s i m u l a t e sE x a m p l1e2 . 1 0u s j n g p o i n t b y p o i n t m e t h o d clear t-0 tf=0 t f i n a l= 0 . 5 t c = O .1 2 5 t s t e P = 9 .9 5 l4=?.5?/( 180*50) j=2 d e 1t a = 2 1 . . 6 4 * p/ i1 8 0 d d e lt a = 0 t i m e ( 1 )= 9 ang(1)=2I.64 Pm=0.9 Pnaxbf=2.44 Pmaxdf=0.88 Pmaxaf=2.00 w h iI e t < t f i n a l , 'if (t==tf ), P a mn' iu s = 0 . 9 - P m a x b f *ns(i d e l t a ) P a p lu s = O . 9 - P m a x d f * s('di ne l t a ) us) /2 Paav=( Pam'i nus+Pap1 ' Pa=Paav end i f (t==tc), P a mni u s = 0 . 9 - P m a x d f *ns(i d e lt a ) P a p lu s = 0 .9 - P m a x a f *jsn ( d e lt a ) Paav=(Pamjnus+Paplus)/2 Pa=Paav end if(t>tf & ttc), P a = P m - P m a x a fn* (sdi e lt a ) end t, Pa d d e lt a = c i d etla + ( t s t e p * t s t e p * p a / M ) d e l t a - ( d e lt a * 1 8 0 / p j+ d d etl a ) * p i / 1 8 0 d e ' tl a d e g = d et1a * 1 8 0 / P ' i t=t+tstep pause t i m e ( i) = 1 ang(i)=deltadeg
end
a x i s ( [ 00 . 6 0 1 6 0 ] )
Ex. G.15 Here the earlier Example G.l4 is solved again using SIMULINK. Before running simulation shown in Fig. G.5 integrater t has to be initialized to prefault value of 4 i.e. 4. Thit can be done by double-clicking on integrater 1 block and changing the initial value from 0 to 6s(in radius). Also double click the switch block and changethe threshold value from 0 to the fault clearing time (in sec.). Ex. G.16 (Ex. 12.11) v" Th'tsprogramsimulates transi ent stab'i1i ty of multimachine systems 12. 11 % Thedata i s f r om Exam p'le clear alI format l ong % S t e p1 I n ' i t i a l i s a t i o n w i t h l o a d f l o w a n dm a c h ' i ndea t a f = 5 0 ; t s t e p = 9 . 961=. [ 1 29 J ,; 3 .[ 2 5 2 . 1 0 ] '; Pgnetterm= Q g n e t t e r m =0 [. 6 9 8 60 . 3 1 1 0 ];' X g = [ 0 . 0 607. 1 0 ] ' ; % Note the .use of .' operator here % T h i s d o e s a t r a n s p o s ew j t h o u t t a k i n g t h e c o n j u g a t e o f e a c h e l e m e n t yg=[po'larTorect(1.08 3 ., 2 3 5 ) p o l a r f o r e c t ( 1 . 0 2 , 7 . 1 6 ) ]. ' % n is no of generatorsother than sl ack bus n=2i % S t e p2 V 0 c o n i = c o r(i V 0 ) ; I g 0 = c o n j( ( P g n e t t e m + j * Q g n e t t e r m.)/ U O ); E d a s h 0 = V 0"+(j X g .* I g 0 ) ; P g 0 = r e a('E l d a s h 0 . * c o n( jI g 0 )) ; x 1 r = a n g le ( E d a s h 0;) % lnit'ial isation of state vector Pg r=Pg0; % Pg_ro1usl.=PgO; xZ r=[0 0],; xtdot r=[0 0] ,; x2dot r=[o o] ,; x l d o t r p l r t l = [ 0 0 J' i x 2 d o t r p lr t l = J 0 0 J ' i
SStep3 e e h a v en o t r e a l l y c a l c u l a t e dY B u sB u t - s n e % H e r ei n t h i s e x a m p lw write seParate % can Program. yB usdf= [ -0.0681+i*5.1661 0 1f986-j*35-6301 -j*11.236 0 0 j . 2737fi - 0 . 0 6 8 l + j * 51. 6 6 1 0. 1362-*6 0
o f
d]
.
o
E
c
.g o P E o
() o c c o
. c ) o
=(\' o) E
c at,
IU
.E o (u
E c
L L
g
e O
o
o
C
,
o ' o / o ts L
= F
_o (U o c
Y B uspf= [ - 0 . 0 9 0 l +*j6 . 0 9 7 5 -0.2214+j*7.6289 I .3932-j*13.8731 -0.2?I++j*7.6289 0 0. 5+j*7. 7898 - 0 . 0 9 0 l +*i 6 . 0 9 7 5 0 . 1 5 9 1 = j * 6 . 1 1 6; 8 1 0 %Step4 % S e t t h e v a ] u e sf o r i n i t i a l t i m e t ( o c c u r a n coef f a u l t ) a n d % t c i s t i m e a t w h i c hf a u l t i s c l e a r e d t = 0 ;t c = 0 . 0 8t;f i n a l=1. 0 ; r=1; pl 1=Edash0 ; E dash_r =Edash0 ; Edash_rus w h i l et < t f i n a l , GeneratorPowersusing appropriateYBus %Step5 Compute % t h e Y B u sc h o s e ni n t h e f o l l o w i n g s t e p i s s e t a c c o r d i n gt o t h e c u r r e n t %time i f t < = t c Y B u s = Y B u sedl fs;e Y B u s = Y B u s p f ; e n d % N ote t hat her ewe obt ain t he cur r ent s iniect ed at gener at orbus r ow of t he Ybuswjt h t he ve\ t or of % by mu lt ip'lyingt he cor r espond'ing % v o l t a g e sb e h i n dt h e t r a n s i e n t r e a c t a n c e sT. h i s s h o u l da l s o j n c l u d e % slack bus voltage % a n d h e n c e t h e e n t r Y 1 a P P e a r sj n t h e b u s v o l t a g e v e c t o r i n a d d i t i o n t o % generator bus vo'ltages I=YBus(2:m+1,:)*[1 E d a s hr l ; P g _ r = r e a 1( E d a s h _ r . * c o n(iI ) ) ;
.9 o c (g
.s o o
L
F
u? (5 6 IL
-1,
(I' L
o) o c
%Step 6 computexldot-r and xZdot-r x ldot_r=x2_r; for k=1:m, x Z d o t - r ( k , 1) = ( p i * f / H ( k ) ) * ( P g O( k ) - P g - r ( k ) ) ; end % S t e p7 C o m p u t ef i r s t s t a t e e s t i m a t e s f o r t = t ( r + 1 ) x 1_rp1usI =xl_r+xldot-r*tsteP ; xZ_rpl us1=x2_r+x2dot_r*tsteP ; % S t e p8 C o m p u t ef i r s t e s t i m a t e s o f E ' - r + 1 E d a s hr p l u s 1 = a b s ( E d a s h. *0()c o s ( x 1 - r p 1u s l ) + i * s i n ( x l - r p l u s l ) ) ; % S t e p9 C o m p u t eP g f o r t = t ( r + 1 ) I=YBus(2:m+1,:)*[1 us1l ; Edash_rp'l P g _ r p 'ul s 1 = r e a l( E d a s h _ r pu1s 1 . * c o n i( I ) ) ; State-denivatives'at t=t(r+1) % S t e p1 0 C o m p u t e
Mod"rnpo*",. sy.t.r An"ryri,
W
Iffi
xldot_rplrtl=J0 J,i x Z dot _r prrl l = 1 0 0 J, i for k=1:m, xldot_rpu l s l( k ,1 )= x 2 _ r pu' sl l ( k , l ) ; x 2 d o t r. p l u s l ( k ;l ) = p i* f / H ( k )* ( p q g( % s t e p1 1 c o m p u taev e r a g ev a r u e so f s t a t e d e r i v a t i v e s x I dotav_r=(xldot_r+x1dot_rp'l usl)/ ?.0; x 2dot av _r= (x 2 d o t_ r+ x Z d o t_urps 1l ) /2 .0 ; %s t epL2 C o mp u te fj n a l Sta te e s tt' m a te fsbr t= t(r+ 1) x l_r p l us1= x l _ r+ x l d o ta v _ r* ts te ; p x 2_r plus lix Z r+ x 2 d o ta v _ r* ts te; p % s t e p1 3 C o m p u ft ien a l e s t i m a t ef o r E d a s ha t t = t ( r + 1 ) Edash_ru p s11 = a b(sE d a s h .0*)( c o s( x l _ r p lu s1 )+ j * si n( x t _ r p 1u s l ) ); % S t e p1 4 P r i n t S t a t eV e c t o r x?_r=x?_rpl usl; x l _ r = x l _ r pu1s l ; E das h _ r= E d a s hu_sIrp; ' l %S t ep15 time(r)=1; for k=l:m, a n g( r , k )=( x t _ r ( k *) 1 8 0/)p i ; end t=t+tstep; r-r+1; end plot(time,ang)
'Fl 88
N
E
o o > 6 II J 6
E E i l $ n E
o o o Il f
U'
o o a o .tt
o
tt
sE 3
-8 o
l
.= l ct
o c
.9 U'
c, 6 F o
.= .c, ct
G'
Ex. G. 17 (Example
E =
IZ,II)
%Examp7 l e2 . 1 1i E s o l v e du s i n g S I M U L I N K % T h ec o d eg i v e nb e l o ws h o u l db e r u n p r i o r t o s i m u ' l a t i osnh o w n in Fig.G.6. cl ear alI globa1 n ryyr yngg globalPmfHE global r t d d tr % c o n v e rs j ofa n c to r ra d /d egree 91obal Y bf Yd f y a f f=50; ngg=2; r=5; nbus=r; rtd=180/pi; d t r = p i/ 1 8 0 i 9o
Gen. gendata=[ 1
2 3
Ra Xd' 0 0 0 0.067 0 0.100
H inf 12.00 9.00 l;
q (J dt IL
iltl
Modern Power svstem Anatvsts
ydf'I * 3 5 .6 3 0 1 5. 798 6 -J 0 - 0 . 0 6 8 1 +* 5 i .1 6 6 1 Y af =[ 1.3932-j*13.8731 -0.22L4+j*7 .6289 - 0 . 0 9 0 1 +* 6 j .0975
0 -j*11.236 0 -0.2 2 I4 + j * 7 .6?89 0.5+j*7.7898 0
-
Appendix G I
-0.0681+J*5.1661 0 0 . 1 3 6 2 - j * 6 . 2 7 3i 7 1 -0.0901+* j6 .0975 0 0.1591-j*6.116 ; 81
f c t = i n p u t ( ' f a u l t c l e a r i n gt i m e f c t = ' ) ;
Complex Yaf (2,1) Product21
&Dampi ng factors d a m p 2 =0O; . dam p 3 =. 00;
E2
%Ini t i a l g e n e ra to rA n ge' ls d2- 0, 3 3 7 7 * rtd l d3=0.31845*rtd; %Initial Powers Pn2=3.25; Pm3=2 . 10; %G en e ra toIn r te rn a l V o l ta g e s El.= 1. 0 ; E 2 =.10 3 ; E3=1.02; % M a c h i nI e nertia Constants; H 2 = g e n daa( 2 t , 4 )t 1 1 3 = g e n d(a3t,a4 ); & M a cn he l X d '; , 3 ); 1 6 6 1 = g e n d (a2t a , 3 ); 1 6 6 1 = g e n d (a3t a Note : For the simulation for multimachine stability the'two summation boxes sum 2 & sum 3 give the net acceleating powers Po2 and Por. T}iregains of the gain blocks Gl, G2 G3 and G4 are set equal to pi*fAlZ, dampZ,pi*f/Fl3 and damp3, respectively. The accelerating power P" is then integrated twice for each mahcine to give the rotor angles 4 ^d 4. Th" initial conditions for integrator biocks integrater i, integrator 2,integrator 3 anciintegrator 4 are set to 0, d2/r+d, 0 and d3lrtd, respectively.The gain blocks G5 and G6 convert the angles $ and d, into degrees and hence their gains are set to rtd. The electrical power P"2ts calculatedby using two subsystems1 and 2.The detailed diagram for subsystem 1 is shown in Fig. G.7. Subsystem I gives two outputs (i) complex voltage Etl6 and (ii) current of generator 12 which is equal to
E
Flg. G.7 Multimachine transientStability: Subsystem1
vodernpowersysterRnalysis
ffi
E/-61 *Yar(2,1)+ Erlq*Yq{z,z) + ErlErYar(2,3).The switchesareusedto switch betweenprefault and post fault conditionsfor eachmachineand their thresholdvaluesare adjustedto fct, i.e. the fault clearingtime.
cl ear 'l % T h i s Program fi nds t h e r e d u c e d matri x for stabi i ty s t u d ie s % w h i c h el imnates the I oad I b u s e s a n d r e t a i n s o n l y t h e generator b u s e s % lIgll Y l l Y L ? . Yln Yln+l YLn+Z. Y 1 n + m l IV 1 I % lIszl YZn+l YZn+Z. Y l n + m l IV 2 . YZn iIY21. Y?r
% l . l
I
v o l . l I Y o l l g n l = l y n l YnZ % lrLll IY n + 1 2 yo lrL2l I I % l . l I v o l . l I % lILml I Yn+m1
t
.
Ynn
Yfull= _j*5 0 7*5 -j*7.5 j*2.5 0 -j*12.51 j*5 j*?.5 (Yfull) [row columns]=sjze n=2 Y A A = Y f u l (l 1 : n , 1 : n ) YAB;Yfull (1:n, n+1;columns) Y B A = Y f u1l ( n + 1 :r o w s ,n + L :c o lu m n s ) Y B B = Y f u1l ( n + 1 :r o w s ,n + 1: c o 1u m s ) % T h i s g i v e s t h e r e d u c e dm a t r i x Y r e d u c e d = Y A A - Y AnBv*(iY B B*)Y B A
Y n n + 1 Y n n + 2Y n n + m Yn+Lm
rs to Problems CHAPTER 2
l
Vn V n+ 1 V n+ 1
2.1 Lirt =
-1x 2
ro-7 x *Vlr,:
-,lt-+n,e|-,,,)+arlr?]
2.2 0.658ohmlkm F loR w^ 2r r 2.4 260.3Vlkm 2.3 L=
Yn+mn+m
Vn*m
- .,:I 2.5 He n = 3 n d, AT/mz(directedupwards) 2.6X=
(Xt-Xn)(X2-Xn)
\ * x2-zxn
2.7 0.00067m}J/krlt,0.0314V/r
Chapter 4 4.1 12 kV
ModrrnPo*"r. syrt"t An"lv.i.
W Chapter 5
5.1 (a) 992.75kW (b) No solutionpossible 5 . 2 A t - 0 . 9 1 I . 5 o ,B ' = 2 3 9 . 9 1 6 6 . 3C " ,' = 0 . 0 0 1 1 0 2 . 6 ,D t = 0 . 8 5 1 I . 9 6
IE (b) j0.3049 j0.1694
(b) 165.44ky,0.2441-28.3" kA, 0.808lagging,56.49 MW (c) 7O.8Vo
j0.1948
(d) 28.L5Vo
(b) l,IT4 A 5.4 (a) 273.5MVA (c) 467.7MVA 5.5 133.92kV , 23.12MW 5.6 202.2kV 5.7 At x = 0, irr = 0.3L4cosQ,st-21.7"), irz=0.117 cos (utt + 109"), - 9.3o),irz = 0.t12 cos (,*t + 96.6) At x = 200km, it = 0.327cos(c..r/ 5.8 135.817.8okV, 0.138115.6o kA, 0.99 leading,55.66 MW, 89.8Vo, 373.L1- t.5o,3,338km, 1,66,900km/sec Y' 5.g v - 12g.3172.60, - 0.000511g9.5". 2 5.10 7.12",pfr = 0.7 lagging,pfz= 0.74 \agging 5.11 47.56MVAR lagging 5.I2 10.97kV, 0.98leading,- 0.27Vo,86.2Vo 5.13 51.16kV, 38.87MVAR leading,40 MW 5.I4 238.5kV, P,+ jQ, = 53 - jI0, pf - 0.983leading 5"15 17.39MVAR leading,3.54MW Chapter 6 6.1 For this networktreeis shownin Fig. 6.3a;Ais givenby Eq. (6.17).The matrix is not unique.It dependsuponthe orientationof the elements. 6.2 V) = 0.9721- 8.15"
6.3 V) = 1.26 l:74.66"
6.a@)
j0.3134 0.807_ js.6s 0.645_ j4,517 0.968- j6.776 0.968_ j6.776 0.880_ j6.160 (c)
t
z
}
6.5 Pn = - 0.598pu, PB = 0.2 pu, pzs= 0.796pu Qn= Qzr = 0.036 pu, Qn= Qn= 0.004 pu, ezz= ezz= 0.064 pu 6.6 (a) Pn = - 0.58 pu, P* = 0.214pu, pzl = 0.792pu Qn = - 0.165pu, ezt _ 0.243 pu, en = 0.204pu Qy = - 0.188pv, Qzt = 0.479pu, en= - 0.321pu (b) Pn = - 0.333Pu, Pzz= 0.664pu, Pr: = 0.333pu ' , B = Q z r= 0 . 0 1 1p u , o n = Q z r= 0 ' 0 1 1 p u Q ezt= en=g.g44 pu
6.7 (a)(i)
-ito s/s3" i5 f-iro.rors j5.o5o5 js I lI -j10 js ;s.osos -ilo tul js | j5 lst87" I _j10 I js L js J
t 2 3 4 --:) 6 7 I 9
I 0 0 0 0 - l l 0 - l
U
I 0 0
U
0 I 0
U Qr
0 I
0 1 -l 0 0 1 1 0 0 0 l 1 0 l 0
4
fh\/i\
D
-n
n
n A^A
\.r,, \i,, i 12: v.iJVU pu, ts13= U.LUZ pu,
L
pzZ = 0.794 pu
Qn= 0.087pu, Qzt = - 0.014pu Qn= QzF 0.004 pu, Qzz= Qsz= 0.064pu //::\
I)
\u,, f 12:-
t\
js
.oE
u.ooJpu,
Pn = 0.287pu, Pzt= 0.711pu Qn = 0.047 Pu,Qn= 0.008 ptr, Qzt = 0.051pu 6.8 Vrt = I.025 - y0.095= l.AZ9/.- 5.3. pu
_jl0j
powerslrstemnnatysii Mooern
4 Chapter 7
RL.,-iiiis;E: lit+i(Atlldt
I--
7.1 Rs 22.5/hr 7.2 (a) Pcr = 140.9MW,
8.1 Load on
159.1MW
123MW, Load on G2 = 277 MW, 50.77H).. 2
S4VlIl$ =
7 . 3 (i) Gen A will sharemore load than Gen B (ii) Gen A andGen B will shareload of po each (iii) Gen B will sharemore load than Gen A. 7.4 Pcr = 148MW, Pez,= L42.9MW, Pct = 109.1MW 7.5 (dcldPd=0.175Pc+23 7 . 6 (a) Pq = 138.89MW, Pcz = 150 MW, Po = 269.6MW 0) Pcr = 310.8MW, Pcz = 55.4 MW (c) For part (a): Cr = Rs 6,465.141hr Forpart (b): Cr = Rs 7,708.15/hr 7 . 7 Bn = 0.03387pu or 0.03387x ITa MW-l Brz = 9.6073x 10-5pu or 9.6073x 10-7MW-t Bzz= 0.02370pu or 0.02370x 10-2MW-r. 7 . 8 EconomicallyoptimumUC Load (MW)
8.2 Af (t) = - 0.029 - 0.04ea.58'cos (1.254t + 137.g") 8.3 1/(50K) sec 9.4 aPtie. | -
1 r 1 1 1 1
1 1 1 1 0 1
System is found to be unstable. Chapter
9 . r it
1 1 0 1 0 0
1 0 0 0 0 0
Unit Number
t 2 3 4 04 4-8 8-12 I2_16 l6-20 20-24
9 = 3.14 sin (314t - 66.) + 2.97e sor,;"* = 5A
9.2 (a) 81" (b) - 9' 9.3 (1) Ie = 2.386kA, .IB- 1.75kA (ii) IA = 4.373 kA, = IB |.ZS ta 9.4 8.87 kA, 4.93 kA 9.5 26.96kA 9.6 6.97kA 9.7 (a) 0.9277kA (b) 1.312kA (c) 1.4843kA (d) 1.0205 kv, 53.03MVA (e) 0.1959kA 9 . 8 8 . 3 1 9k A 9.9 2.39pu 9 . 1 0 132.1,47.9; 136.9,45.6 gJt 0.6 pu 9 , 1 2{ - - j8,006pu, 1{: = - 74.004pu Chapter 10
Optimal and secureUC Period
L
80s51
Unit Number
20 l4 6 l4 4 10
t O OO f 'zt t t 0. 9t
8.5 APt i. , r ( s)= -
1 2 3 4
0-4 4-8 8-r2 t2-16 r6-20 20-24
ll R2)
_ an (K iz+D0 / K p,1* K ipl +l / Rt)+ (K, r
1 1 1 1 1
1 1 1 ' 11 0 1 0 0 . 1 , 10 1 0 0
r t 0 0
7.9 Total operating cost (both units in service for24 hrs) = Rs 1,47,960 Total operating cost (unit tr put off in light load period) = Rs 1,45,g40
10.1 (i) 1.7321270" (ii) zl0" (iii) 1.7321150.(iv) tl2r0" lO.2 IA= jl.l6 pu, Vtn = 1.171109.5" pu O<2 ./ 4.< Ao --' vBC -- nv.zirz_oJ.+pu, YVCB = U.9952_Il3.lo
ll
pu.
19 3 Vor= l97.Vl- 3.3" V, VozJ 20.Zll51.lo V, V"o= 21.61110.63" V 10.4 lor = 19.23_1. 3V A, Ioz = Ig.23lISV A,1oo= 0 A
1 0 . 5 1^ ,r , = 2 7 - 8 7 , / 3 f f
A
t , - - '- t ? L. J/ 1--
^AZ
A A o . l oI A l,
a-'.ZJ
r
lA0
=
n U
a A
Iobr= 16.1A, Iab2= 7.5/_-75"A, Iob1=l.,S/lS e, 10.6 Io = 16.16+ j1.335A, Iu= _ 9.24_"1\O.AO e, Ic - - 6.93+ j9.32A, lVN,l= lVosl= 40.75y t0.7 1,500.2w
Il
ModernPowerSystemAnal'sis
ffi4 Chapter 11
Index
11.1 - i6.56 kA, lV6,l- 12.83 kV, lvobl = 6.61 kV, ly"rl = 6'61 kV = I = -2./1 (b) V*= Vo, = 0'816 pu, 116l= llrl = 5'69 pu 11.3 (i) - i6.2.5 (ii) - 4.33 (iii) 6'01 (iv) /5 pu In order of decreasingmagnitude of line currentsthe faults can be listed AS
(a) LG (b) LLG (c) 3-Phase(d) LL Il.4 0.1936ohm,0.581ohm,- 4.33pu, j5 pu 11.5(a)3.51pu(b)Vt,=I.|91_159.5"po,V,=I,681|29.8"pt1 (c) 0.726Pu 1 1 . 6I u = - I r = - 2 . 8 8 7 P u = j10.02 kA ||.7 (a) Iv = _ 5.79+ j5.01kA, 18= 5.79+ j5.01kA, ,IG (b) /n - - IY = - 6'111kA, Ic= 0 Io^ = - 73'51Pu 11.8 In, = 0 Ib^= - jl.Z Piu Ius = - 72.08pu I,^ = - jl.Z Pu IrB = 72.08pu 1 1 . 9 5 , 2 6 6A 11.10j2.0 pt 11.11 / = - j6.732Pt, Io(A)= - i4.779 pu, /a(A) = - i0.255 po, 1" (A) - - j0'255 prt l|lz 0.42 ptt,- j9'256Pu 11.13 -ill.l52 pu, -i2.478pu, -Jl -239pn '1.14 4.737Pu, 1 Pu r 1.15 f, = - i12.547pu, Ifrr(b)= - i0.0962 pu Chapter 12 deg l2.I 4.19 MJA4VA, 0.0547MJ-sec/elec 12.3 40'4 MJiTVIVA 12.2 4.315 MJA4VA 12.4 140.1MW, 130.63MW, 175.67MW
r2.5 72.54MW !2.7 L27.3MW
12.64 = 58'
12.8 53'. We need to know the inertia constant M to determine /.' 12.10 70.54",0.1725sec 12.9 The system is stable t2.t2 63.36" l2.Il The system is unstable 12.14 The systemis stable stable 12.13 The system is 12.15 The system is unstablefor both three pole and single pole switching
ABCD constants 61,7 for varioussimple networks 618 in power flow equations lS9 measurementof 621 of networks in seriesand parallel 620 AC calculatingboard lB4 Accelerationfactor 207 Accelerating power 462 Acid rain 16 Adjoint matrix 609 Admittancematrix (seeBus admittance matrix) Advanced gas reactor (AGR) 19 Algorithm for building the Zru, 355 for load flow solution by GS method 205 by NR method 214 by FDLF 223 for optimum generation scheduling 262 for optimal hydrothermal scheduling 280 for optimal loading of generators on a bus 246 for optimal load flow solution 273 for short circuit current computation 343 for short circuit studies 349 for staticstateestimationof power systems 540 for transient stability, analysis of large system 4gS Alternator (sea Synchronousmachines) Aluminum conductor steel reinforced (ACSR) conductors 52 A
*aa
Arvc
a^-+-^1
Lrrllll(I
^-^-
Ellul
/
a
^F\
\f\\-.8,
4A
f,
JU+
Armature leakage reactance 331 Armature reaction 109, 110, 330 Attenuationconstant 143 Augmented cost function 279 Automatic generationconirol (see Load frequencycontrol) Automaticvoltagccontroi 318
B-coefficients 261.267 Bad data detection 547 Bad data detection 547 identification 547 suppression 548 treatment 546 Baseload 3 Basevalue 99 Bharat Heavy Electricals Lrd (BHEL) 14,31 Boiling water reactor 19 Breaking resistors for improving stabiliry 499 Branch 190 Breakers (see Circuit breakers) Brown, H. E. 368 Bundledconductors, \ capacitivereactance 92 inductive reactance 68 Bus generator 198 Ioad 198 PQ 198 PV 198 reference 198 slack Bus admittancematrix 188 formulationof 187. 189 Bus impedancematrix building algorithm 355 for unsymmetrical fault analysis 416 crra*^+-l^^l ri nr r D Jrrrrtlsttrucu
f^..fr
---!---'
latuil. al|alysls
Bus incidence matrix
-Er
JJ I
I92
Capacitance u4ruuliluonoy memoo oI modrlred geometric mean distances 9l effect of earth on 83 effect of non-uniform charge distribution 79 effect of strandedconductors 79 line-to-line 78 line-to-neutral 79 ^^f^--1^-i---
t
-t
tndex of parallel circuit three-phase lines 88 of three-phaselines: with equilateral spacing 80 with unsymmetrical spacing 81 a two-wire line (see also Reactance) 78 Central Electricity Authority (CEA) Zg Charpcteristic (surge) impedance l4l qf .linesand cables 145 Chaiging current 76, 180 Circle diagrams 167 Circuit breakers 327, 329 autoreclose 460 rated intemrpting capacity of 344 rated momentary current of 344 selection of 344 Circulating currents 389 Cogeneration 15 Coherent group 291,303 Compact storagescheme 189,628 Comparisonof angle and voltage stability 592 Compensation series 779,498, ,Sg shunt 179,498,562 Compensator combined TCR and TSC .564 shunt 562 static synchronousseries(SSCC) 571 Complex power 105 Compositeconductors capacitance 9l inductance 54 Computational flow chart for load flow solution using FDLF 227 GS method 20S NR method 224 Conductors ACSR 52 Bundled 52,68 expandedACSR 52 iypes 5 i Conductance 45 Constraints equality 272, 632 inequality 632 on control variables 274 on dependentvariables 274
.
for hydro-thermal scheduling problems 277 Contingency analysis 51
rankins 520 screening 520 selection 515 ' direct method 515 indirect method 515 Contingency analysis by dc model 520 by distribution factors 521 modelling for 5I4 Contingency evaluation (seecontingencyanalysis) Control area 291,303,306 by transformers 23I integral 304 isochronous 305 of voltage profile 230 of WAfiS and VARS along a transmission line 230 optimal 310 parameters 199,272 proportioiral pft"s-fuBral 303 suboptimal 318 Control area concept 303 Control strategy 303 Control variables 199, 632 Control vector 199 Controller interline power flow (IpFC) 571 interphasepower (IPC) 572 unified power flow (UPFC) 571 Converters 567 Ccxrrdinatisn equations 2fr Corona 52,68 Cost function 251,270 i,t^e-^^
\-ull(i('
1nn
L>V
Covariance matrix 535 Critical clearing angle and time 467 Current distribution factors 265 Current limiting reactors 346 DAC (Distribution Automation and Control System) 634 Damper winding 438 DC network analyser 39 Data acquisition systems(DAS) 636
Digiral LF controllers 322 Direct axis reactance llg Disturbanceparameters 271, 272 Distribution factors generation_shift SZ0 Iine-outage 520 Diversity factor 4 Dommel, H. W. 270 Doubling effect 3Zg Double line-to-ground(LLG) fault 404 Driving point (self) admittance lg7 Dual variables 279,632 Dynamic programming applied to unit commitment 251 Economic dispatch(see also optimum generation scheduling) 306 Electricity Board Zgl Elements 189 Energy conservation 3l Energy control centre 637 !MS.(Energymanagement system) 634 Energy sources conventional 13 hydroelectricpower generation l7 nuclearpower stations lg thermal power stations 13 biofuels 28 biomass 28 gas turbines 16 geotbermal power plants 24 magnetohydrodynam i c (MHDI generation 23 OTEC 27 renewable Zs solar energy 26 wave energy 27 wind power stations 25 Energy storage Zg fuel cells 29 hydrogen Zg pumpedstorageplanf lg secondarybatteries 2g Equal area criterion 461 Equal incremental fuel cost criterion 246
the medium length line nominal_ zr representation l3g nominal-Trepresentation 137 short transmission line l}g synchronous generator llg Error amplifier 3lg Error signat 319 Estimation methods least squares 532 weighted least-squares 533 Estimation of periodic components 5g l Exact coordination equation 26I Expectation 532 External system equivalencing 545 Facrors affecting srability 496 FACTS controllers 569 Failure rate 255 Fast breeder reactor 22 Fast decoupled load flow 223 Fasr valving 49,9 Faults balanced(see symmetrical fault analysis) calculationusing Zsus 351,417 open conductor 414 unbalanced series type 397 shunt type 397 Ferranri effect 150 Field erciretion l0g Field rriading 32g Flat voltagestarr 205,209,21g Flexible AC transmissionsystems (F-ACTS) 566 Fluidized-bed boiler 15 Flux linkages external 49 internal 46 of an isolated current carrying conductor 46 ball speedgovernor Zgz ,Fly rorecastlngmethodology 577 Fortescue,C.L. 370,3t6 Frequencybias 309 Fuel cost ofgenerators 243
@
!no.* 50
Full load ri,iection techniquefor ; ,, improving systemstability 500 Fusion 23 Gausselimination 623
of single phasetwo wire line of three-phaselines 59 Inequality constraints on control variables 274
constants) Generator(see Synchronousgenerator) Generatorload model 296 Generatoroperatingcost 243 Geometricmean distance(GMD) mutual 55 self 55 Geometric mean radius (GMR) 55 Governors, characteristics 298 dead band 32I free operation 301 model with GRC 32I Gradient method 272 Graph connected 189 linear 189 oriented 189 . theory 189 Greenhouseeffect 16 Growth of installed capacity in India 30 Growth of power systemsin India 29
Inertia constant 436 Infinite bus Ll2 Infinite line 145 Inner product 607 Input-output curve of a unit 243 Interrupting capacity of breakers 345 Interconnectedpower systems 10 Interference with communication circuits 6l-63 Irrverse of matrix 6L3 Iterative algorithm for NR method for load flow studies 218
HVDC 572 Hydraulic'amplifier 292 Hydroelectricpower generation l7 Hyperbolic'functions 142, 143 Ideal transformer 99 Illconditioning 229, 545 Impedance driving point 187 transfer 187 Impedancediagram 98 Incidencematrix (seeBus incidence matrix) Incident waves I44 lncrementalfuel cost 244 Incrementaltransmissionloss 260 Inductionmotors 122,344 lnductance definition 45 due to internalflux 46 mutual 46 of compositeconductorlines 54 of double circuit threc-phaseline 66
Jacobianmatrix 2I4,272 elementsof 2I4,629 Kalman filtering approach 583 application to short term load forecasting 583 Kimbark, E.W. 508 Kinetic energy 435 Kirchhoff's current law (KCL) 186. Kirchmayer, Leon K. 325 Kron, G. 265 Kuhn Tucker theorem (theorY) 632 Lagrangian function 279, 632 Lagrange multiplier 219, 632 Laplace transformation 294 Leakage reactance 331 Leakance 129 532 Lcast squaresestimation the basic solution 532 Linear graph (see GraPh) Line charging current (see capacitance) Link 190 Line-to-line (LLXault 402 Line compensation L-. -^-:^^ uy ugrrrrJ
^^-^^l+^-ualP4r/rturJ
1?O /'lOQ r t 7, 1/v
by shunt capacitors 179,498 Load angle(see also power angle; torque angle) 110 bus 198 oontro I I compcnsation 557
delta connected 374 representation 122 symmerrical (balanced) lO7 unbalanced 394 Loading capability 557 Load factor 3 Load flow problem 196 Load flow methods comparison of 229 continuation 599 decoupled 222 FDLF 223 Gauss 205 Gauss-Seidel204 Newton-Raphson 213 decoupled 223 rectangularversion Z2I Load forecasting 9 Load forecastingtechniques estimation of averageand trend terms 577 estimation of periodic components 581 incorporation of weather variables 5g3 time series approach 592 autoregressivemodels Sgz autoregressivemoving-average models 582 Load frequencyconrrol (LFC) and economic dispatchcontrol 305 decentralizedcontrol 323 digital 322 single area case 29I two area 307 with cRC 320 Load management 33 Load prediction on-line techniques586 A Dflt Al\llYl.l1
A
^^)^r^ rrluugllt
Eoz Joo
non-dynamicmodel 596 time varying model 5g6 ' long term 587 using econometricmodels 5g7 Long transmissionline equations l4Z cvaluntionof ABCD con$tsnts l42 hypcrbolic form of l4Z
Lossesas function of generation 260 assumptionsin calculating 266,269 equation for 267 Maintenance 32 Magnetohydrodynamic (MHD) generation 23 Magnetomotiveforce around closed path 47 MATLAB examples 651 introduction to 640 programs 652 programming 640 Matrices additionof 610 associativelaw 6l l commutativelaw 611 equalityof 610 multiplicationof 611 associativelaw 6lI distributivelaw 611 singular and nonsingular (seealso matrix) 610 Matrix '607 definition diagonal 608 element of 607 identity 609 null 608 symmetric 609 transposeof 609 unit 508 zero (seea/so Bus admittancematrix: Bus impedancematrix; motri^^^\ rrElr rvgJrf
zAo Ul.,O
Medium transmissionline 137 Method of images g3 Methods of voltage control 173 Mini hydel plants 18 Micro hydel plants lg Model generatorload 42 of speedgoverningsystem 2g3 turbine Z9S Modil'icdEuler's methodfor stability analysis 495
Modification of Z
BUS
for network
changbs360 Mutual (transfer) admittance Mutual coupling 46
I87
National Thermal Power Corporation (NTPC) 4r Negative sequencecomponents 370 Negative sequenceimpedance 383 of passiveelements 383 machines 386 of synchronous table 334 of transformers 390 lines 390 of transmission Negative sequencenetworks in fault analysis 399 examples 409 of powersystems 386 of unloadedgenerators 388 of Zr* elements 417 Network branch 190 model formulation for SLFE 185 node 90 observability 545, 549 primitive 191 Nodes 190 Node eliminationtechnique for stability study 444 Nominal- t circuit of medium transmissionline 138 Nominal-T representationof medium transmissionlinc 137 Non-linearprogrammingproblem 632 Non-linear programming technique 279 Nuclearreactors 2I Objectivefunction 270, 27I, 275 Off-nominal turn (tap) ratio or off-nominal tap setting 234 One linc diugrtnr 98 On-line -+^+^ ^.+1.-^+i^Jlctlly\,DLlllldllutl
J1a
techniquesfor non-stationary load prediction 586 ARMA models 5E6 rime r-ar1-ingmcxlel -s36 Apaaror a 370 310 l cstrrdla Wn ft'dmrl hrsd fltrrr' srrludon :?0
Optimal (two area) load frequency control 310 Optimal ordering I89, 627 Cptimal operation of generatorson a bus
of transformer 390 of transmissionlines 389 Positive sequencenetworks 398 in fault studies
Reactance
Optimal operating strategy 242 Optimal reactive power flow problem 272 Optimal scheduling of hydrothermal , system 276 Optimal unit commitment 250 Optimal security constrainedunit commitment 256 Optimum generationscheduling 259 Optimum real power dispatch (sZe optimum generationscheduling)
of Zsu5 elements 417 Power accelerating 46? angle 437 _ apparent 106 by symmetric components 374 complex 105 factor 105, 112, Il5 flow through a transmissionline 158 in three-phasecircuits Il2 maximum transmitted I59 real and reactive (seealso Reactor power; Real power ) 106 Power anglecurve 1I5,444 Power system compensation 556 engineer 39 state estimation 538 structure 10 studies 39 Primitive impedance (admittance) matrix I92 Primitive network (seeNetwork) I9l Propagationconstant l4l Proportional plus integral control 303 Proximity effect 7 L Pumped storage scheme 18
inductive 67 quadratureaxis Ll7 subtransient 332 synchronous 111,332 transient 332 Reactancediagram 98 Reactive load forecast 587 Reclosure 474 Referencebus 198 Reflected waves 144 Regulating transformers(see Transformer, regulating) Regulatingconstant(R) 294 Regulation of voltage 130 Reheat turbine model 295 Reliability considerationsfor economic dispatch 253 Reluctancepower . 118 Repair rate 255 \ Reserve 254 Resistanceof lines 7O RotatingVAR generator 176 Rigid limits 275
Parameterestimation 552 Parametersof overhead lines 45 Patton,A.D. 254,256 Penalty factor 260,275 Penalty function 270, 275 Percentvalue 99 Per unit system changeof base 101 definitioir of 99 selectionof base 100 Performance chart of synchronous machines 119 Performance index GI) 316 Phaseconstants 143 Phase shift in star-delta transformers 377 Planning commission 4L z'-equivalent for long line 152 for medium length line 138 zrmodel for transformer with off-nominal turns ratio 234 Plantcapacitytactor 4 Plnntuse factor 4 Photovoltaic generation 2l Pilct vulve 292 Point-by-pointsolutionof swing equation 481 Positive sequencecomponents 370 Positive sequenceimpedance of pa-ssiveelements i8l nrchirc,s 3E3 of syfrrwus table 334
direct axis 118. 121 leakage,of transformers 101 of bundledconductors
Qurdrrturcaxis synchronous reactance I17 Rcuctivcpowcr 106 control of flow: by machine excitation ll3, I14 injection of 175 machincs
---
,r^:-^
^,,:rl^
....1r--.^-
l--,..l
t - c t a u u n s n t pw i l , u v u r r u B cr c v s l sign of l06 Real power 106 f'ormulas for 159, 160 in sr nchrtrn(lus nt&('hin('s I l-5. I l9 Real dme computer control of Ww'q $-r-Silems 6Y
144
LI J
generator ll7 Salient pole synchrono-us Saturatedreactor 563 SCADA 634 Scalar product 607 Sctrlnrand v0ctorfunctions 614 derivativesof 614 Securityanalysis 256 Scnsitivityluctors generation shift factors 520 line outagedistribution factors 520 Seqtienceimpedances lirr synchronous machines 3ti5 ol passiveelements 383 of transformers 386 of transmission lincs 3E5 table lbr typical valuesof 334 Sequenceimpedances and sequencenetworks ul ptr$ E'rs\stL'nl iss of syrrchrowus mar.+tiffi 3t5 of transformers 390
i
lndex Sequencenetviork connectionsfor unsymmetrical faults 4OO,403,4A5 Sequen,:enetworks constructionof sequencenetworks Short circuit current limiters 499 Short circuit megavoltamperes 345 Short circuit ratio (SCR) 498 Short term load forecasting estimating of stochastic component 583 innovation model approach 588 using Kalman filtering approach 583 using the periodic load model 581 Short transmission line equations 12,9 Short transmissionline equivalent circuit 129 Shunt capacitors effect upon voltage regulation 132 for power factor correction 133 for voltage control 175 Simulink 640 basics 648 example 648 Single line-ro-ground(LG) fault 399 Skin effect 71 Slack bus 198 Soft limit 275 Solid LG fault 407 Sparsematrix 188 Sparsity 188, I 89, 623 Speed changer 293 Speed governing system 293 Speed regulationconstant(seeregulation constant) Spinning reserve 245, 254 Stabiliry analysis for multimachinesystem 497 definition 433 dynamic 434 study 461 steady state 454 transient, defined 434 clearing angle 466 clearing time 466 critical clearingangle 467 critical clearing time 467 equal area criterion 461 point-by-pointmerhod 4gl
lndex
some factois affecting voltage 591 Stability limit steady srate 456
496
Start up cost 250 Statcbm 562 State electricity boards 303 State estimation external system equivalencing 545 of power systems 538 the injections oniy algorithm 540 the line only algorithm 543 nonlinear measurements 535 problemsof illconditioning 545 stochasticcomponent 583 Statevariablemodel 3lz Statevariables 199,312, 494 Static load flow equarions(SLFE) 198 Static reservecapacity 254 Static VAR generator 175, 565 Static VAR sysrem 562,565 State-vector 198 Steadystaresrabiliry 454Step-by-stepfofmulation of bus impedancematrix addition of a branch 357 additionof a link 359 Stiffnessof sy,nchroneus machine 455 Steepestdescentmethod (see Gradient method) Suboptimalconrrol 318 Substation bulkpowcr 13 distribution l3 Superconductingmachinel4 Surge impbdance (see also characteristic impedance) 145 SVC (Static var compensator) 562.565 Swing bus (seeSlack bus) Swing curve 459,495 determinationusing modified Euler's marhn-| tllvLtlvu
,{ n< a7J
step-by-stepdeterminationof Swing equation 438, 480, 494 state variable form 495 Switching single pole 500 three pole 500 Symmetric marrix 609
481
Symmetrical comPonets definition of 369 of unsymmetricalPhasors 372 power in terms of 374
transformer (TCPST) 572 Thyristor controlled reactor (TCR) 563 Thyristor controlledvoltage regulator
algorithm for large sYstemsustng zBUs350 selection of circuit breakers 344 short circuit current computation throughthe Thevenin's theorem 341 shortcircuit of a synchronousmachine on load 339 on no load 330 transienton a transmissionline 320 Symmetricalshort circuit current 329, 331 Synchronisingpower coefficient I 18, 308, 455 Synchronousreactance33 I Synchronousmachine armaturereaction 108 construction 108 dynamicsof 435 e q u i v a l e nct i r c u i t l l l excitation 113 inertia constantsof 435 leakagereactance 331 ioad angleof I 10 (performance)chart ll9 -' operating n(qr. of values of sequenceimpedances of 385 per unit reactance,table 334 phasordiagrams 112
TSC) 564 Tie line power 307 Tolerance 206 Torque angle(see also Power angle) 438 Tracking stateestimationof power systems 544 Transformers control (seeregulatingbelow) 231 off-norninal tapes 234 of lOl per unit representation phaseshift in Y-A 377 phase shifting 232 polarity markings 377 regulating 232 equivalentcircuit of 234 for magnitudecontrol 232 for phaseangle control 233 sequenceimpedancesof 385 TCUL 177.232 \ Triangularisationand back substitution 623 Transient reactance 332 Transientstability definition of 433 digital computersolutionmethod 610 equalareacriterion 461 Euler modified method 495 multigeneratorcase487 numcricalsolutionof swing cquation for singlegeneratorcase 480 Transmissioncapability 12 Transmissionlevel l3 Transmissionloss (seeLossesas function of plant generation)'
T circuit of medium length line 137 Taylor seriesexpansion 213 TCUL (tap changingunder load) (see transformers) Thevenin'stheorem in calculationof three-phasefault currents 341, 35O in caleulationof unsymmetricalfault currents 416 Three phasecircuits power in lO7 voltageand current in 107 Three phasefaults (see Faults) Thyristor controlled braking resistor 577. Thyristor controlled phase shifting
t.T.rl!.f D \ \ r L v r\,,
{?t J t -
Transmissionloss formula 261 derivationof 264 Transpositionof transmissionlines tn lv
lvrsal lsar n r vnva
nqnqnif
qnne
60
fl, I
to balanceinductance 6l Tree 190 Tuned power lines 15t Turbine fast valving 499 Turbine model 295 Turbine speedgoverning system 292 Two-area load frequency control 307
Index Two reaction theory
Il7
Unit comrnitment 250 branch and bound technique 250, 252 dynamic programmingmethod Z5l reliability (security) considerations 253 securityconstrained 256 stait-up consideration Z5g Unit matrix 609 Unsyrirmetricalfaults (seealso Faults) analysis u,singZsu5 rhbthod 416 symmetriCal component analysis of 398 Use of computersand microprocessors in power systems', 39 VAR (see also Reactive power) Variance 582 Vector column 607 control (seecontrol vector) disturbance 27l equation 271 null 605 of fixed parameters Ig9 orthogonal 607 row 607 state (see State vector) sum 606 unit 606 Velocity of propagation 146 Voltage collapse 528, 592 prevention of 526,593, 600 control by transformers i'77 control by VAR injecrion I75 eff'ectof capacitorson l3Z, 175
equations: for long line 142 for medium transmissionlines 137-t39 for short transmission lines l/g stability 524, 592 methods for improVing 5Zg Voltage controlled bus 198 Voltage regulation 130 Voltage stability 591 analysis 592, 5gl criteria of 596 with IdVDC links 599 Watts (see qlso Real power) 106 Wavelength 145 Waves, incident and reflected 144, I45 Weighted least squaresestimation 533 Weighting marrix 534 Y"r, (see Bus admittance matrix) Y-A transfonners 96 phaseshift in 377 single-phaseequivalentof 96 zero sequencenetwor[s of 397 Z^.,^ (seeBus impedancematrix) BUS
./'
Zero sequencecomponents 371 Zero sequenceimpedance of circuit elements 384 of synchronousmachines 384 table' 334 of transformers 387 of transmissionlines 3g5 Zero sequencenetworks in fault studies 399 of delta connectedloads 389 of transformers 387 of unloadedgenerators 384 of Y-connecredloads 397
