modern power systems analysis d p kothari i j nagrath

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In simple anguage,he book providesa modern ntroductiono powersystemoperation, ontroland analysis.

Key Features of the Third

New chaptersadded on

) .PowerSystemSecurity

) State Estimation

) Powersystemcompensationncluding vs and FACTS) Load Forecasting

) VoltageStability

New appendiceson :

> MATLAB nd SIMULINK emonstratingheiruse n problem olving.) Real ime computer ontrolof powersystems.

From the Reviewen.,

The book s verycomprehensive, el lorganised, p-to-date nd (aboveall) ucidand easy o follow or self-study. t s ampiy llustratedw1h solvedexamples or everyconceptand technique.

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Modern PorroerSYstem

Third Edition



About the Authors

D P Kothari is vice chancellor,vIT University,vellore. Earlier,he wasProfessor,Centre for Energy Studies,and Depufy Director (Administration)Indian nstituteof Technology, elhi. He hasuiro t."n the Headof the centrefor Energy Studies 1995-97)and Principal l gg7-g8),VisvesvarayaRegional

Engineeringcollege,Nagpur.Earlier lflaz-s: and 19g9),he was a visitingfellow at RMIT, Melbourne,Australia. He obtained his BE, ME and phDdegrees rom BITS, Pilani. A fellow of the Institution Engineers India),prof.Kothari has published/presented 50 papers in national and internationaljournals/conferences.He has authored/co-authoredmore than 15 books,including Power systemEngineering, Electric Machines, 2/e, power systemTransients, Theory and problems of Electric Machines, 2/e., and. BasicElectrical Engineering.His research nterests nclude power systemcontrol,optimisation, eliabilityand energyconservation.

I J Nagrath is Adjunct Professor,BITS Pilani and retired as professorofElectricalEngineering nd DeputyDirectorof Birla Instituteof Technologyand Science,Pilani. He obtainedhis BE in Electrical Engineering rom theuniversity of Rajasthann 1951

and MS from the Unive.rity of Wi"sconsinn1956' He has co-authored everalsuccessful ooks which include ElectricMachines 2/e, Power systemEngineering, signals and systems and.systems:Modelling and Analyns. He has also puulistred,"rr.ui researchpapers inprestigious ationaland nternationalournats.

Modern Power SystemAnalysis

Third Edition

D P Kothari

Vice Chancellor

VIT University

VelloreFormer Director-Incharge, IIT Delhi

Former Principal, VRCE,Nagpur

I J Nagrath

Adjunct Professor, and Former Deputy Director,Birla Ins1i1y7" f Technologt and Science

Pilani

Tata McGraw Hill Educationprivate LimitedNEWDELHI

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Preface to the Third Edition

Since he appearance f the secondedition n 1989, he overallenergysituationhas changed considerablyand this has generatedgreat interest in non-conventionaland renewable

energysources, nergyconservationand manage-ment,powerreformsand estructuringand distributedarrddispersed eneration.Chapter has been therefore,enlargedand completelyrewritten. In addition,the influencesof environmentalconstraintsare alsodiscussed.

The present edition, like the earlier two, is designed or a two-semestercourseat the undergraduateevel or for first-semester ost-graduate tudy.

Modernpower systems ave grown largerand spread ver arger geographi-cal areawith many interconnectionsetweenneighbouring ystems.Optimalplanning,operationand control of such arge-scale ystems equireadvancedcomputer-basedechniquesmany of which areexplained n the student-orientedand reader-friendlymannerby meansof numericalexamples hroughout hi sbook. Electric utility engineerswill also be benefittedby the book as it willprepare hem more adequatelyo face the new challenges. he style of writing

is amenableo self-study.Ihe

wide rangeof topics acilitates ersarile electionof chaptersand sections br completion n the semesterime frame.

Highlights of this edition are the five new chapters.Chapter13 deals withpower system security. Contingency analysis and sensitivity factors aredescribed.An analytical ramework s developedo control bulk power systemsin sucha way that security s enhanced. verything eemso havea propensityto fail. Powersystems reno exception.Powersystem ecurity ractices ry tocontrol and operatepower systems n a defensivepostureso that the effectsofthese nevitable failures are minimized.

Chapter14 s an introduction o the use of stateestimationn electric powersystems.We have selected east SquaresEstimation o give basic solution.External system equivalencingand treatmentof 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) hasalways motivated the power engineers o develop compensatory ystems.Therefore,Chapter 15 addresses ompensationn power systems.Both seriesand shuntcompensationof linqs havebeen horoughlydiscussed.ConceptsofSVS, STATCOM and FACTS havc-beenbriefly introduced.

Chapter 16 covers the important topic of load forecasting technique.Knowing load is absolutely ssentialor solvingany power systemproblem.

Chapter 17 dealswith the mportant problemof voltagestability.Mathemati-cal formulation, analysis, state-of-art, future trends and challenges arediscussed.



Wl Prerace o rne hlrd Edrtion

MATLAB andSIMULINK, dealprogramsor powersystem nalysisre

includedn thisbookasanappendix longwith 18solved xamplesllustrating

theiruse n solvin tive tem problems.The help rendered

by Shri Sunil Bhat of VNIT, Nagpur n writing this appendix s thankfully

acknowledged.

Tata McGraw-Hill and the authors would like to thank the following

reviewers of this edition: Prof. J.D. Sharma, IT Roorkee; Prof. S.N. Tiwari,

MNNIT Allahabad; Dr. M.R. Mohan, Anna University, Chennai;Prof. M.K.Deshmukh, ITS,Pilani;Dr. H.R. Seedhar, EC, Chandigarh; rof.P.R.Bijwe

andDr. SanjayRoy, IIT Delhi.

While revising the text, we have had the benefit of valuable advice and

suggestionsrom manyprofessors,students ndpractising engineerswho used

the earlier editions of this book. All these ndividuals have influenced this

edition.We express ur thanks and appreciationo them.We hope his support/

responsewould continue n the future also.

D P Kors[m

I J Nlcn+rn

Preface to the First

Mathematicalmodellingand solutionon digital computerss the only practicalapproach o systemsanalysisand planning studies or a modern day power

system with its large size, complex and integratednature.The stage has,therefore,been reachedwhere an undergraduatemust be trained n the latesttechniques f analysis f large-scale owersystems. similarneedalsoexistsin the ndustrywherea practisingpowersystem ngineers constantlyacedwiththe challengeof the rapidly advancing ield. This book hasbedndesignedo fulfilthis need by integrating he basicprinciplesof power systemanalysis llustratedthrough he simplestsystemstructurewith analysis echniquesor practicalsizesystems.n this book arge-scaleystem nalysisollows as a natural xtensionof the basicprinciples.The form and evelof someof thewell-known echniquesare presented n such a manner that undergraduates an easily grasp andappreciate hem.

The book is designed or a two-semester ourseat the undergraduateevel.With a judicious choice of advanced opics,some nstitutionsmay also frnd it

useful for a first course or postgraduates.The reader s expected o have a prior grounding n circuit theoryandelectricalmachines.He should also have been exposed o Laplace transform, ineardifferential equations,optimisation techniquesand a first course in controltheory.Matrix analysis s applied hroughout he book. However,a knowledgeof simple matrix operations would suffice and these are summarised n anappendix br quick reference.

The digital computerbeingan indispensableool for power systemanalysis,computationalalgorithms or various systemstudiessuchas oad flow, fault levelanalysis,stability, etc. have been ncludedat appropriate laces n the book. Itis suggestedhat where computer acilitiesexist,students houldbe encouragedto build computerprograms or these studiesusing the algorithmsprovided.Further, the students can be asked to pool the various programs or moreadvanced nd sophisticated

tudies, .g.optimalscheduling.An importantnovelfeatureof thebook s the nclusionof the atestand practicallyuseful opics ikeunit commitment, generation eliability, optimal thermal scheduling,optimalhydro-thermalschedulingand decoupledoad flow in a text which is primarilymeant or undergraduates.

The introductory chapter contains a discussionon various methods ofelectricalenergygeneration nd heir techno-economicomparison. glimpse sgiven nto the uture of electricalenergy.The reader s alsoexposedo the ndianpower scenariowith facts and figures.

Chapters and3 give the transmissionine parameters nd heseare ncludedfor the sakeof completness f the ext. Chapter on the representationf powersystem omponents ives he steady tatemodelsof the synchronous achine ndthe circuit modelsof compositepower systems long with the per unit method.



Wpreface ro rhe Frrst Edition

Chapter5 dealswith the performanceof transmissionines. The load flowproblem s introduced ight at this stage hrough he simple wo-bus systemandbasicconcepts f watt and varcontrol are llustrated.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 andmodern control approacheshave 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 emphasises he concepts of various types <lf stabil ity in a power

system. In particular the concepts of transient stability is well illustrated throughthe equal area criterion. The classical numerical solution technique of the swing

equation as well as the algorithm for large system stability are advanced.Every concept and technique presented is well supported through examples

employing mainly a two-bus structure while sometimes three- and four-bus

il lustrations wherever necessary have also been used. A large number ofunsolved problems with their answers are included at the end of each chapter.

These have been so selected that apart from providing a drill they help the

reader develop a deeper nsight and illustrate some points beyond what is directlycovered by the text.

The internal organisation of various chapters is flexible and permits the

teacher to adapt them to the particular needs of the class and curriculum. Ifdesired, some of the advanced level topics could be bypassed without loss ofcontinuity. The style of writing is specially adapted o self-study. Exploiting this

fact a teacher will have enough time at his disposal to extend the coverage ofthis book to suit his particular syllabus and to include tutorial work on thenumerous examples suggested n the text.

The authors are indebted to their colleagues at the Birla Institute ofTechnology and Science, Pilani and the Indian Institute of Technology, Delhi

fo r the encouragement and various useful suggestions hey received from themwhile writing this book. They are grateful to the authorities of the Birla lnstituteof Technology and Science, Pilani and the Indian Institute of Technology, Delhifor providing facilities necessary for writing the book. The authors welcomeany constructive criticism of the book and will be grateful for any appraisal by

the readers.

I J NlcRArH

D P KorHlnr

A Perspective I

Structure f Power Systems I0ConventionalSourcesof Electric Energy I3RenewableEnergy Sources 25Energy Storage 28

Growth of Power Systems n India 29

EnergyConservbtion 3I

Deregulation 33

Distributedand DispersedGeneration 34Environmental Aspectsof Electric Energy Generation 35PowerSystemEngineers ndPower SystemStudies 39Use of Computersand Microprocessors 39ProblemsFacing Indian Power ndustry and ts Choices 40References 43

2. Inductance and Resistanceof Transmission Lines

1 .

vn

I

1 . 1

1 . 21 . 3r .41 .51 .61 . 7r .81 . 91 . 1 01 . 1 1T . I 21 . 1 3

2 . 12. 22.3

2. 42.52. 62. 72.82.92 . 1 02 . l I2 . r 2

Introduction 45

Definition of Inductance 45Flux Linkages of an Isolated

Current-CtrryingConductor 46Inductanceof a Single-Phase wo-Wire Line 50ConductorTypes 5I

Flux Linkages of one Conductor n a Group 53Inductanceof CompositeConductorLines 54Inductanceof Three-Phase ines 59Double-CircuitThree-PhaseLines6BundledConductors 68

Resistance 70

Skin Effect and Proximity Effect 7IProblems 72

References 75

45

3. Capacitance of Transmission Lines

3.1 Introduction 76

3.2 Electric Field of a Long StraightConductor 76

Contents

Preface to First Edition

Introduction

76



fW . contents .

3.3 PotentialDiff'erencebetween wo Conductorsof a Group of ParallelConductors 77

3.4 Capacitance f a Two-Wire Line 783.5 Capacitance f a Three-phaseLine

with EquilateralSpacing B0

6.4 Load Flow Problem 196

6.5 Gauss-SeidelMethod 204

6.6 Newton-RaphsonNR) Method 213

6.7 DecoupledLoad Flow Methods 222

6.9 Controlof Voltage Profile 230

Problems 236D ^ t - - ^ - - - . ) 2 0

I I Y J E I E T L L C J LJ 7

7. Optimal System Operation 2427.I Introduction 242

1.2 Optimal Operation of Generators n a Bus Bar 243

7.3 Optimal Unit Commitment(UC) 250

7.4 ReliabilityConsiderations 253

1.5 OptimumGenerationScheduling 259

7.6 Optimal Load Flow Solution 270

7.7 OptimalScheduling f Hydrothermal ystem 27 6

Problems 284

References 286

8. Automatic Generation and Voltage Control 291'l

8.1 Introduction 290

8.2 Load FrequencyControl (SingleArea Case) 291

8.3 Load FrequencyControland

EconomicDespatchControl 305

Two-Area Load FreqlrencyControl 307

Optimal (Two-Area) Load FrequencyControl 3I0

AutomaticVoltageControl 318

Load FrequencyControl with Generation

Rate Constraints GRCs) 320

SpeedGovernorDead-Band nd ts Effect on AGC 321

Digital LF Controllers 322

DecentralizedControl 323

Prohlents 324

References 325

9. Symmetrical Fault Analysis 327


9.2 Transienton a Transmission ine 328

9.3 ShortCircuit of a Synchronous achine

(On No Load) 330

9.4 ShortCircuit of a LoadedSynchronousMachine 339

9.5 Selection f Circuit Breakers 344

UnsymmetricalSpacingBI3.7 Effect of Earth on TransmissionLine capacitance g3

"

o l t - t l - - l a r . ^ / r .

J.o rvleln(Jo or \r lvl l -, (vlool l led) yl

3.9 BundledConductors 92

Problems 93References 94

4. Representation,of Power System Components

4. 1 Introduct ion g5

4.2 Single-phase olutionof BalancedThree-phase etworks 95

4.3 One-LineDiagramand mpedance rReactanceDiagram 98

4.4 Per Unit (PU) System 994. 5 ComplexPower 10 54.6 Synchronous achine 1084.7 Representation

f Loads I2IProblems 125

References 127

5. Characteristics and Performance of power

Transmission Lines

5.1 Introduct ion 12 85.2 ShortTransmission ine 1295.3 MediumTransmission ine i375.4 The LongTransmissionine-Rigorous Solut ion I 395.5 Interpretationof the Long Line Equations 1435.6 FerrantiEffect 150

5.1 TunedPowerLines 15 1

5.8 The EquivalentCircuit of a Long Line 1525.9 PowerFlow througha Transmission ine I585.10 Methods l 'Vol rageContro l 17 3

Problems 180References 183

6. Load Flow Studies

6. 1 lntrot luct ion 18 4

6.2 NetworkModelFormulat ion I8 5

95

128

8. 4

8 . 5

8 . 6

8 . 7

8 . 8

8 .9

8 . 1 0

t84



rffi#q confenfsI

9.6 Algorithm for ShortCircuit Studies 3499.7 Zsus Formulation 355

Problems 363

References 368

Symmetrical Com

10.1 Introduction 36910.2 SymmetricalComponentTransformation 7010.3 PhaseShift in Star-DeltaTransformers 37710.4 Sequencempedances f TransmissionLines 379

10.5 Sequencempedancesnd Sequence etworkof PowerSystern 381

10.6 Sequencempedancesnd NetworksofSynchronous achine 381

10.7 Sequencempedances f TransmissionLines 38510.8 Sequencempedancesnd Networks

of Transformers 386

10.9 Construction f Sequence etworksofa Power System 389

Problems 393

References 396

ll. Unsymmetrical Fault Analysis

11.1 Introduction 39711.2 SymmetricalComponentAnalysis of

UnsymmetricalFaults39 8

, 11.3 SingleLine-To-GroundLG) Fault 3gg11.4 Line-To-Line LL) Fault 40211.5 Double Line-To-GroundLLG) Fault 40411.6 Open Conductor aults 41411.1 Bus ImpedanceMatrix Method For Analysis

of Unsymmetrical huntFaults 416Problems 427

References 432

12. Power System Stability


12.2 Dynamicsof a Synchronous12.3 Power Angle Equation 44012.4 Node Elimination Technique

I2.5 SimpleSystems 45112.6 Steady StateStability 454

12.7 TransientStability 459

I2.8 Fq'-ralArea Criterion 461

Machine 435

444

12.10MultimachineStabilitv 487

Problems 506

References 508

13. Power System Security

13.1 Introduction 51013.2 SystemStateClassification 51213.3 SecurityAnalysis 51213.4 Contingency nalysis 51613.5 SensitivityFactors 52013.6 Power System Voltage Stability 524

References 529

14. An Introduction to state Estimation of Power systems 531

l4.l Introduction 531

I4.2 Least SquaresEstimation:The BasicSolution 532

14.3 StaticStateEstimationof PowerSystems 538

I4.4 Tracking State Estimation of Power Systems 54414.5 SomeComputationalConsiderations54414.6 ExternalSystemEquivalencing 545I4.7 Treatmentof Bad Dara 54614.8 Network observabilityandPseudo-Measurements 4914.9 Application of Power SystemStateEstimation 550

Problems 552

References 5.13

397

433

55 05. Compensation in Power Systems


15.2 LoadingCapability 55715.3 LoadCompensation55 7

15.4 Line Compensation 558

15.5 SeriesCompensation 559

15.6 ShuntCornpensators562

I5.7 Comparison etweenSTATCOM and SVC 56515.8 FlexibleAC Transmission ystems FACTS) 56615.9 Principleand Operationof Converrers 56715.10FactsControllers 569

References 574



16. Load Forecasting Technique

16.1 Introduction 57516.2 ForecastingMethodology 577

timationof Averageand Trend Terms 577Estimationof PeriodicComponents 581Estimation f y., ft):Time SeriesApproach 582Estimationof Stochastic omponent:Kalman Filtering Approach 583Long-TermLoadPredictions singEconometricModels 587ReactiveLoad Forecast 587References 589

Voltage Stability

11.1 Int roduct ion 59 117.2 Comparisonof Angle and Voltage Stability 59217.3 ReactivePowerFlow and VoltageCollapse 59311.4 MathematicalFormulation f

Voltage StabilityProblem 59311.5 Voltage StabilityAnalysis 597

17.6 Prevention f VoltageCollapse 600ll.1 State-of-the-Art,FutureTrends and Challenses 601References 603

Appendix A: Introduction to Vector and Matrix Algebra

Appendix B: Generalized Circuit Constants

Appendix C: Triangular Factorization and Optimal Ordering

Appendix D: Elements of Power System Jacobian Matrix

Appendix E: Kuhn-Tucker Theorem

Appendix F: Real-time Computer Control of power Systems

Appendix G: Introduction to MATLAB and SIMULINK

Answers to Problems

Index

I. T A PERSPECTIVE

Electric energy is an essential ngredient for the industrial and all-rounddevelopment f any country. t is a coveted orm of energy, ecauset can begenerated entrally n bulk and ransmitted conomically ver long distances.

Further, t can be adaptedeasily and efficiently to domesticand industrialapplications,particularly for lighting purposesand rnechanicalwork*, e.g.drives.The per capitaconsumption f electricalenergy s a reliable ndicatorof a country'sstateof development-figures or 2006 are615 kwh for Indiaand 5600 kWh for UK and 15000kwh for USA.

Conventionally, lectricenergy s obtainedby conversioniom fossil fuels(coal,oil, naturalgas),and nuclearandhydrosources. eatenergy eleased yburning ossil uelsor by fissionof nuclearmaterial s converted o electricityby f irst convert ing ea tenergy o th e mechanicalor m through thermocycleand then convertingmechanical nergy through generatorso the electricalform. Thermocycle s basicallya low efficiencyprocess-highestefficienciesfor modern argesize plants angeup to 40o/o, hile smallerplantsmay haveconsiderably lower efficiencies.The

earth has fixed non-replenishable e-sourcesof fossil fuels and nuclear materials,with certain countriesover-endowedby natureand othersdeficient.Hydro energy, hough eplenishable,salso imited in termsof power.The world's increasing owerrequirements anonly be partially met by hydro sources.Furthermore, cologicaland biologicalfactorsplacea stringent imit on the use of hydro sourcesor power production.(The USA has already developedaround 50Voof its hydro potential andhardlyany furtherexpansions plannedbecause f ecological onsiderations.)

x Electricity is a very inefficient agent or heatingpurposes, ecauset is generatedbythe low efficiency thermocycle rom heat energy. Electricity is used fo r heatingpurposes or only very specialapplications,say an electricfurnace.

16.4

1 6 . 5

16 .6

16.7

r6.8

17 . 59 1

60 5

617

623

629

632

634640



Introduct ion

with the ever ncreasing er capitaenergyconsumption nd exponentially__ _ _ _ - - ^ D v^ ,v t6J v u r r J L u r l p [ r u l t i l t l u g x p o n g n l l a

rising population, technologistsalready r* the end of the earth,s ncnon-

l lfenislablefuel reso.urces*.

-Theoil crisis of the 1970shas dramatically

intensepollution in their programmes of energy

generating stations are more easily amenable o

centralizedone-pointmeasures an be adopted.

development.Bulk power

control of pollution since

drawnattentiono this fact. n fact,we canno lontor generationof electricity. In terms of bulk electric energy generation,adistinct shift is takingplaceacross he world in favour of coalLJin particular

*varying estimatcsav ebccnpu t orth or rescrvcs l 'oi l ,ga san dcoalancl issionablernaterials'At the projected onsumption ates,oil and gasesare no t expected o lastmuch beyond50 years;severalcountrieswill face seriousshortages f coal after2200A'D' while fissionablematerials may carry us well beyond the middle of the nextcentury. Theseestimates,however, cannot be regardedas highly dependable.

Cufiailment of enerry consumption

The energyconsumptionof most developcdcorrntries as alreacly eachecllevel, which this planetcannotafford. There s, n fact, a need o find waysandmeansof reducing his evel.The developingcountries,on the otherhand,haveto intensify heirefforts o raise heir evel of energyproduction o providebasicamenities to their teeming millions. of course,-n doing ,o th"y need toconstantlydraw upon the experiencesof the developedcountries and guardagainst bsoleteechnology.

rntensification of effofts to develop alternative sources ofenerw including unconventional sources like solan tidalenergy, etc.

Distanthopesare pitchedon fusion energybut the scientific and technologicaladvances ave a long way to go in this regard.Fusion when harnessed ouldprovide an inexhaustible ourceof energy.A break-throughn the conversionfrom solar to electric energy could pr*io" anotheranswer o the world,ssteeplyrising energyneeds.

Recyclingr of nuclear wastes

Fastbreeder eactor echnology s expected o provide he answer or extendingnuclear energy

esources o last much longer.D e velopm ent an d applicati on of an ttpollu tion techn ologries

In this regard, the developing countries already have the example of thedeveloped countries whereby they can avoid going through the phasesof

consumption n a worldwidebasis.This figure s expectedo rise as oil supply

for industrial usesbecomesmore stringent.Transportation an be expected o

go electric in a big way in the long run, when non-conventional nergy

resources re we[ developedor a breakthrough n fusion is achieved.

To understandsome of the problems that the power industry faces let usbriefly review some of the characteristic featuresof generationand transmis-

sion.Electricity, unlike water and gas, cannotbe storedeconomically (except

in very small quantities-in batteries),and he electricutility can exercise ittle

control over the load (power demand) at any time. The power system must,

therefore,be capableof matching the output from generators o the demand at

any ime at a specified oltage nd requency. he difficulty encounteredn this

task can be imagined from the fact that load variationsover a day comprises

three components-a steady component known as base load; a varying

componentwhosedaily patterndependsupon he ime of day; weather, season,

a popular estival, etc.;and a purely randomly varying componentof relatively

small amplitude.Figure 1.1 showsa typical daily load curve. The characteris-

tics of a daily load curve on a grossbasisare ndicatedby peak load and the

time of its occurrenceand load factor defined as

averageoad = less han unitymaximum (peak) oad

Fig. 1. 1 Typical daily oa dcurve

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.

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mterconnection,rgreajlyids n ackinguF n,a factors ter in&viJJp i of the.station.staff.excessPowerof a plantaudng ight"toaa eriodss evacuatedhroughong i

Tariff structures aybe suchas o influencehe oadcurveand o improvedistance igh voltageransmissionlo"r, hl"

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I.Ah igh loadfac torhe lps indrawngmoreenergy* 'nu, 'u"n ,n , .u* I ' , ivrl,; ur (xawrrg more energy lrom a given installation. IAs ndividuaroadentrisaveJir o*n

"r,#u","'.i.,ii,;;ilnffi | Tt:"9:"lT:f:,_Tj_.:llT:11,:::."1:,1.j:ij:::.,j:j:*T:jeneralhavea time dJveniry,whichwhen 61il ;&;';;"irili,jij Io:rynd..o_l heunitsproducednd herefore n h€ uel chargesnd hewages

power.'"*-- Prur rwervcs

] 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 DDiversity Factor i i;;;;.i." and distribution) lossesare conespondingly ncreased. The

rhiss dennedsheumf ndividualaximumemandsnhe onsumers, ::m:r,:_".*'Ji:[1t z%""l,i""irT3H::Tgl""rii:'J],f;fi:_'J""1trividedyr,"*i-u--i;#"#UH:ffiTTfi,1"ff":"il:ffi:I :#:*'::i,:"ff"Hf,.j,f.',.d1:$:"*Yil'.:fff:fii*,:.;iff::::*1":i""1".:1t:i1;ft;,"."'.::"'?;3:":ffii"T,"-jT,ffiH:tr;'3f:lxl#1rJ,ffi?,1?Til'jl[,"ff"tJ'trJ;li"tH:s,?"i:"::19'c,.r"d

ransmissionrant.f au he emands"r-"

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i.e. nitvivenitvu"to',r,"'iotJ;';";;;il.;;;;;;ilTffi? i lilm

tharee,lfe:"T:l:_^,:Tl it1:Y_T^:*ore.Luckily,rhe actor s muchhigher rranunity,"il*t,

f-;;#; Itlil a pf penaltyclausemay be mposed n the consumer.

loads. , (iiD the consumermay be asked o use shuntcapacitors or improving theA high diversity factor could be obtained bv; I po*er factor of his installations.1' Giving incentives o farmers and/or some ndustries o use electricity n

th e nightor l ighr oa d periods.ru ruEirr ur UBI|L roao pef loos.

2 using daylight saving as in many other counfies. Llg4"1'1

L------3' staggering

the offrce timings A factory to be set up is to have a fixed load of 760 kw gt 0.8 pt. The4' Having different time zones in the country like USA, Australia, etc. L electricrty board offeri to supplJ,energy at the following alb;ate rates:5' Having two-part tariff in which consumer has to pay an amount (a) Lv supply at Rs 32ftvA max demand/annum + 10 paise/tWh

dependent on the maximum demand he makes, plus u"h.g;

fo."u"t

(b) HV supply at Rs 30/kvA max demand/annum + l0 paise/kwh.unit of energyconsumed.sometimesconsumer i chargedo? tt" u"si, i rne lrv switchgear costs Rs 60/kvA and swirchgear osses at full loadof kVA demand nstead of kW to penalize to"O. of to'* lo*", tin"tor. I amount to 5qa- Intercst depreciation charges br the snitchgear arc l29o of the

other factors used frequently are:plant capacity foctor

enuy 3re: 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

@ased n instarelptant capaciiyy-

Loss in switchgear= 5%

_Average

demand 950Installedcapacity .. InPut dematrd= j- = 1000 kvA

Plant use (tctorI

"ostof switchgear 60 x 1000= Rs 60,000

_ _ --_ ActualenergyproducedkWh)-. . -,:- -- ' Annualcharges n degeciation 0.12x 60,000= Rs 7,200

plantcapacitykw) x Time inhours)h" plunrh^ b;i" il;;ti""Annual fixed chargesdue to maximum demandcorresponding o tariff (b)

Tariffs= 30 x 1.000 Rs 30,000

The cost of electricpower s normally givenby the expression a + D x kW Annual running chargesdueto kwh consumed+ c x kWh) per annum,where4 is a rixea clarge f_ ,f," oiifif,'ina"p".a"* = 1000 0.8x 48 x 52 x 0.10f the power output;b depends n themaximumdemandon tir" syrie- ano i

= Rs 1.99.680



t

Total charges/annum Rs 2,36,gg0

Max. demandcorrespondingo tariff(a) 950 kVA

Annual running charges or kWh consumed

= 9 5 0 x 0 . 8 x 4 8 x 5 2 x 0 . 1 0

= Rs t,89,696

Total= Rs 2,20,096Therefore, tariff (a) is economical.

B0 Hours afoo

Fig. 1.2 Loadduration urveAnnual 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)

+ 0.t3(219x 10 7 E)

For minimum cost, 4Q- = 0dP

A region has a maximum demandof 500 MW at a road factor of 50vo.TheIoad durationcurve can be assumed o be a triangle.The utility has to meetthis load by settingup a generatingsystem,which is partly hydro and partrythermal. The costsare as under:

Hydro plant: Rs 600 per kw per annumand operatingexpenses t 3pper kWh.

Thermal plant: Rs 300 per kw per annumand operatingexpenses t r3p

Determinehe:ffily:f hydroprT!, rheenergy eneratednnually yeach, and overall generation os t per kWh.Solution

Total energy generated er year = 500 x 1000 x 0.5 x g760

- 21 9 x 10 ' kw hFigure 1.2 shows the load duration

curve.Sinceoperating ostof hydroplantis low, the base oad would be suppliedfrom the hydro plant and peak load fromthe thermal plant.

Ler the hydro capacity be p kW andthe energy generaredby hydro plant E

kWh/year.Thermal capacity= (5,00,000_ p) kWThermal energy = (2lg x107_ E) kwhAnnual cost of hydro plant

= 6 0 0 P + 0 . 0 3 E

I500,000 P

Introduction WII

. '.600+0.03

or

l"'

4E-too- .r3dE odP dP

d E = 3 m dP

d E = d P x t

From triangles ADF and ABC,

5,00,000-P_ 3000

5,00,000 8760

P = 328,say 330MW

Capacityof thermalplant= 170MW

Energy generatedby thermal plant = 170x3000x1000

= 255 x106kwh

Energy generatedby hydro plant = 1935x i06 kwh

Total annual cost = Rs 340.20 x 106/year

overall enerationost###P

x100

= 15.53paise/kWh

l 50.50

installed apacity

Installed apacity += 30 MW0.5

A generating tationhas a maximum demandof 25 MW, a load factor of 6OVo,

a plant capacity factor of 5OVo, nd a plant use factor of 72Vo.Find (a) the

daily energy produced, (b) thereservecapacity of the plant, and (c) the

maximum energy that could be produceddaily if the plant, while running as

per schedule,were fully loaded.

Solution

Load factor = average emand

maximumdemand

0.60=average emand

25Average demand= 15 MW

average emandPlant capacity factor=

;#;;..0".,,,



Reservecapacityof the plant = instalredcapacity- maximum demand= 3 0 - 2 5 = 5 M W

Daily energyproduced flver&g€emand 24 = 15 x 24=360 MWh

Energy orrespondingo installed apacity er day= 2 4 x 3 0 _ 7 2 0 M W h

axlmum energy t be produced

_ actualenergyproduced n a day

plant use actor

=:9

=5ooMWh/day0.72

From a load durationcurve, the folrowing data are obtained:Maximum demand n the sysrem s 20 Mw. The load suppliedby the two

units s 14 MW and 10 MW. Unit No. 1 (baseunit) works for l00Voof thetime, and Unit No. 2 (peak oad unit) only for 45vo of the time. The energygenera tedbyun i tis 1x 108un i t s ,and tha tbyun i ti s7 .5 x 106un i t s .F indthe oad factor, plant capacity actor and plant use actor of eachunit, and theload factor of the total plant.

Solution

Annual load factor for Unit 1 = 1 x 1 0 8 x 1 0 0:81.54Vo14,0008760

The maximumdemand n Unit 2 is 6 MW.

Annual load factor or Unit 2 = 7 . 5 x 1 0 610 0= 14.27Vo

6000 8760

Load actorof Unit 2 for the time t takes he oad

7 . 5 x 1 0 6 x 1 0 0

6000x0.45x8760

= 3I.7 7o

Since no reserve s availableat Unit No. 1, its capacity factor is the

same s the load actor, .e . 81.54vo. ls o sinceunit I ha s been unningthroughout he year, the plant use factor equals he plant capacity actori .e.81 .54Vo.

Annual plant capacity 'actorof Unit z = lPgx100

l o x g 7 6 o x l o o= 8 '5 6 7 o

7 . 5 x 1 0 6 x 1 0 0

Introduction NI

The annual oad factor of the total plant = 1 . 0 7 5 x 1 0 E x 1 0 0= 6135%o20,000 8760

CommentsThe various plant factors, the capacity of baseand peak load

units can thus be found out from the load duration curve. The load factor of

than that of the base oad unit, and thus the

cosf of power generationfrom the peak load unit is much higher than that

from the base oad unit.

i;;;";-l- ' - . - - * - "i

There are threeconsumersof electricityhaving different load requirementsat

different times.Consumer t has a maximum demandof 5 kW at 6 p.m. and

a demandof 3 kW at 7 p.m. and a daily load factor of 20Vo.Consumer2 has

a maximum demandof 5 kW at 11 a.m.' a load of 2 kW at 7 p'm' and an

average oad of 1200 w. consumer 3 has an average oad of I kw and his

maximum demand s 3 kW at 7 p.m. Determine: a) the diversity actor, (b )

the load factor and average oad of eachconsumer,and (c) the average oad

and oad factorof the combined oad.

Solution(a ) Consumer M D 5 K W

a t 6 p m

M D 5 K W

at 11 amM D 3 K W

a t T p m

Maximum demandof the system s 8 kW at 7 p'm'

sum of the ndividual maximumdernands 5 + 5 + 3 = 13 kw

DiversitY actor = 13/8= 7.625

Consumer2

Consumer3

3 k wa t T p m

2 k w

a t T p m

LF

ZOVo

Average oad

1\2 kWAverage load

1 k w

(b) Consumer Average oad 0'2 x 5 = I

Consumer Average oa d 1. 2kW ,

Consumer Average oad I kW,

(c ) Combinedaverage oad = I + l'2 +

kW, LF= 20Vo

L F =l ' 2 * 1 0 0 0 - 2 4 V o

5

IL F=

5 x

1 00 3 3 . 3 Vo

l = i . 2 k W

Combined oad factor

Load Forecasting

As power plantplanning and construction equirea gestation eriod of four to

eight yearsor even onger for the present ay superpower stations, nergy anrl

load demand nrecastingplaysa crucial ole in power systemstudies.

= + x 1 0 0 4 0 V o

Plant use factor of Unit 2 =1 0 x 0 . 4 5 x 8 7 6 0 x 1 0 0

= 19.027o

ower



ffiil,ftffi| Modern Syslem natysisI

This necessitatesong range orecasting.while sophisticatedmethodsexist in literature 5, 16, 28], the simple extrapolationquite adequateor long range orecasting. ince weatherhas ainfluenceon residential han the industrialcomponent, t mayprepare orecast n constituentparts to obtain total. Both power

uru ractorsrnvolved re ng an involvedprocess equiring experienceand high analytical ability.

Yearly forecastsare basedon previous year's loading for the period underconsiderationupdatedby factorssuch as general oad increases,major loadsand weather rends.

In short-term oad forecasting,hour-by-hour predictionsare made for the

decade f the 21st century t would be nparing2,00,000Mw-a stupendoustask ndeed.This, in turn, would requirea corresponding evelopmeni n coalresources.

T.2 STRUCTURE OF POWER SYSTEMS

Generating tations,ransmissioninesand he distributionsystems re he maincomponents f an electricpower system.Generating tationsand a distributionsystemare connectedhrough ransmissionines,which alsoconnectone power

* 38Vo f the total power

of electricity n India wasless than 200 billion kWh

required n India is for industrial consumption.Generationaround 530 billion kWh in 2000-2001 A.D. compared toin 1986-87.

system gtid,area) o another.A distribution ystemconnects ll the

a part icular re a o the transmissionines.

For economical nd echnologicaleasonswhich will be discussed

probabilistic

technique s

much more

be better to

and energy

loads in

in detail

electricallyconnected reasor regionalgrids (also calledpowerpools).Each

areaor regionalgrid operates echnicallyand economically ndependently,but

theseare eventually nterconnected* o form a national grid (which may even

form an internationalgrid) so that eacharea s contractually ied to other areas

in respect o certaingeneration nd schedulingeatures.ndia s now heading

for a nationalgrid.

The siting of hydro stations s determinedby the natural water power

sources. he choiceof site or coal ired hermalstationss more lexible.The

following two alternatives re possible.

l. power starionsmay be built close o coal nines calledpit headstations)

and electric energy is evacuatedover transmission ines to the load

centres.

Z. power stationsmay be built close to the load ceutresand coal is

transported o them from the minesby rail road'

In practice, owever,power station itingwill depend ponmany actors-

technical, economical and environmental.As it is considerablycheaper to

transport bulk electric energy over extra high voltage (EHV) transmission

lines than to transportequivalentquantities f coal over rail roqd, he recent

trends in India (as well as abroad) s to build super (large) thermal power

stations near coal mines. Bulk power can be transmitted to fairly long

distancesover transmission ines of 4001765 V and above.However, the

country'scoal resources re ocatedmainly n the eastern elt and somecoal

fired stationswill continue o be sited n distantwesternand southern egions.

As nuclearstationsare no t constrained y the problemsof fuel transport

and air pollution, a greater flexibility exists in their siting, so that these

stations re ocated lose o loadcentreswhile avoidinghigh densitypollution

areas o reduce he risks, however emote,of radioactivity eakage.

*Interconnectionhas the economic advantageof reducing the reservegeneration

capacity n eacharea.Under conditionsof suddenncrease n load or lossof generation

in one area, t is immediately possible o borrow power from adjoining nterconnected

areas.nterconnectionausesargercurrentso flow on transmissioninesunder aulty

condition with a consequent increase in capacity of circuit breakers. Also, the

centres. t providescapacity savingsby seasonal xchangeof power between areas

having opposingwinter and summer requirements. t permits capacity savingsfrom

time zonesand random diversity. It facilitates ransmissionof off-peak power. It also

gives the flexibility to meet unexpectedemergency oads'

lntroduct ion



In India, asof now, abou 75voof electricpower used s generatedn thermalplants includingnuclear). 3vo rommostlyhydrostationsandZvo.come rom

:^:yft.sand.others.oal is the uer or mostof the sream lants, he rest

substation, here he reduction s to a rangeof 33 to 132kV, depending n the

transmissionine voltage.Some ndustriesmay require power at thesevoltage

level.

The next stepdown n voltage s at the distributionsubstation.Normally, two

distributionvoltage evels are employed:

l. The primary or feedervoltage 11 kV)

2. The secondaryor consumervoltage (440 V three phase/230V singlephase).

The distribution system, fed from the distribution transformer stations,

suppliespower to ttre domesticor industrialand commercial consumers.

Thus, the power system operatesat various voltage levels separatedby

transformer.Figure 1.3 depictsschematicallyhe structureof a power system.

Though the distribution systemdesign,planning and operation are subjects

of great mportance,we are compelled, or reasonsof space, o exclude hem

from the scopeof this book.

1.3 CONVENTIONAL SOURCES OF ELECTRIC ENERGY

Thermal(coal, oil, nuclear) and hydro generations are the main conventional

sources of electric energy. The necessityto conserve fosqil fuels has forced

scientistsand technologistsacross the world to search for unconventional

sourcesof electric energy. Someof the sourcesbeing explored are solar, wind

and tidal sources.The conventionaland someof the unconventionalsourcesand

techniquesof energygenerationare briefly surveyedherewith a stresson future

trends, particularly with reference o the Indian electric energy scenario-

Ttrermal Power Stations-Steam/Gas-based

The heatreleased uring the combustionof coal, oil or gas s used n a boiler

to raise steam. n India heat generation s mostly coal basedexcept n small

sizes, becauseof limited indigenous production of oil. Therefore, we shall

discussonly coal-fired boilers for raising steam o be used n a turbine for

electric generation.The chemical energy stored in coal is transformed into electric energy in

thermal power plants. The heat releasedby the combustionof coal produces

steam n a boiler at high pressureand temperature,which when passed hrough

a steam urbine gives off someof its internal energyas mechanicalenergy. The

axial-flow type of turbine is normally usedwith severalcylinders on the same

shaft.The steam urbine actsas a prime mover and drives the electric generator

(alternator).A simple schematicdiagramof a coal fired thermalplant is shown

in Fig. 1.4.

The efficiency of the overall conversionprocess s poor and ts maximum

value s about4OVo ecause f the high heat ossesn the combustiongasesand

a OGenerating

tations.qi-aji, '-qff-9-a,

at 11kV- 25kv

Tie ines oother ystems

Largeconsumers

Small onsumers

Fig. 1.3 schematic diagram depicting power system structure

Transmissionevel(220kv - 765 kV)



E r t ^ - r ^ - - h - . -^ - ^

and the arge quantity of heat ejected o the condenserwhich has to be givenoff in cooling towers or into a streamlake in the caseof direct condensercooling' The steampower stationoperateson the Rankinecycle, modified to

\vv'yvrDrwrr ur r'.lr. r. u Inecnanlcal energy) can be increased byusing steam at the highest possible pressure and temperature. with steam

Ah Step-upuE transformer

10-30kv /

turbinesof this size, additional ncrease n efficiency is obtainedby reheatingthe steam after it has been partially expanded by an ext;;; i"ui"r. rn"reheatedsteam s then returned o the turbine where it is expanded hrough the

final statesof bleedins.To take advantageof the principle of economy of scale(which applies to

units of all sizes), he present rend s to go in foilarger sizesof units. Largerunits can be installedat much ower cost per kilowatt. Th"y are also cheaperto opcrate because of higher efficiency. Th"y require io*", labour andmaintenanceexpenditure.According to chaman Kashkari [3] there may be asavingof as high as l|vo in capitalcost per kilowatt by going up from a 100to 250 MW unit size and an additional saving in fuel cost of ubout gvo perkwh. Since arger units consume ess fuer pJr kwh, they produce ress air,thermaland wastepollution, and this is a significantadvantagen our concernfor environment' The only trouble in the cai of a large unit is the tremendousshock to the system when outageof such a large capacityunit occurs. Thisshock can be tolerateclso long as this unit sizeloes not exceed r}vo

of theon-line capacity of a large grid.

rntroduction Effi

perhaps ncrease unit sizes to several GWs which would result in better

generatingeconomy.

Air and thermal pollution is always present n a coal fired steamplant. The

COz,SOX, etc.) are emitted via the exhaustgasesand thermal pollution is due

to the rejected heat transferredfrom the condenser o cooling water. Cooling

towers are used in situations where the stream/lake cannot withstand the

thermal burdenwithout excessive emperature ise.The problem of air pollution

can be minimized through scrubbers and elecmo-staticprecipitators and byresortingto minimum emission dispatch [32] and Clean Air Act has already

beenpassed n 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 s being developed and

perfected.The fluidized-bedboiler is undergoingextensivedevelopmentand s

being preferred due to its lower pollutant level and better efficiency. Direct

ignition of pulverizedcoal is being introducedbut initial oil firing support s

needed.

Cogeneration

Considering the tremendousamount of waste heat generated n tlbrmal power

generation, t is advisable o save fuel by the simultaneous enerationof

electricity and steam(or hot water) for industrialuse or spaceheating.Now

called cogeneration, uch systemshave ong beencommon, here and abroad.

Currently, there s renewed nterest n thesebecause f the overall increase n

energy efficiencieswhich are claimed to be as high as 65Vo.

Cogeneration of steam and power is highly energy efficient and is

particularlysuitable or chemicals,paper, extiles, ood, ertilizer and petroleum

refining industries.Thus these ndustriescan solveenergyshortageproblem n

a big way. Further, hey will not have o dependon thegrid power which is not

so reliable.Of course hey can sell the extra power to the government or use

in deficientareas.They may aiso seil power to the neighbouring ndustries,a

conceptcalled wheelingPower.As on 3I.12.2000, otal co-generation otential n India is 19,500MW

-and

actual achievements 273 MW as per MNES (Ministry of Non-Conventional

Energy Sources,Governmentof 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 mechanicalpower and the wasteheat from the power

generation rovides heprocess eat. n thebottomingcycle, uel first produces

process heat and the waste heat from the process6ss then used to produce

power.

Stack

Coolirrgower

-Condenser

mi l l

Burner

Preheatedair Forced

draft fan

Flg. 1.4 schematic iagram f a coar iredsteamprant

In India, in 1970s the first 500 Mw superthermalunit had beencommissioned at Trombay. Bharat Heavy Electricals Limited (BHEL) hasproducedseveral urbogenerator etsof 500 MW capacity.Today;smaximumgeneratorunit size is (nearly 1200 Mw) limited by the permissiblecurrentcjensitiesused n rotor and stator windines. Efforts are on to develoo srDer.

H



-Coal-fired plants share environmental problems with some other types of

fossil-fuel plants; these nclude "acid rain" and the ,,greenhouse,,effect.

Gas Turbines

With increasing availability of natural gasuangladesh)primemoversbasedon gas turbineshave been developedon thelines similar to those used in aircraft. Gas combustion generateshightemperatures and pressures, so that the efficiency of the las turbine iscomparable to that of steam urbine. Additional advantage s that exhaustgasfrom the turbine still has sufficient heat content, which is used to raise steamto run a conventional steam turbine coupled to a generator. This is calledcombined-cyclegas-turbine CCGT) plant. The schernaticdiagramof such aplant is drawn n Fig. 1.5.

Steam

Fig.1.5 CCGTpower tation

CCGT plant has a fast start of 2-3 min for the gas turbine and about20 minutes for the steam turbine. Local storage tanks Jr gur

"ui-u"ured in

caseof gas supply intemrption. The unit can take up to ITVooverload for shortperiodsof time to take care of any emergency.

CCGT unit produces55voof CO2 producedby a coal/oil-firedplant. Unitsare now available for a fully automatedoperation for 24h or to meet the peakdemands.

In Delhi (India) a CCGT unit6f 34Mw is installed at IndraprasthapowerStation.

There are culrently many installationsusing gas turbines n the world with100 Mw generators.A 6 x 30 MW gas turbine stationhas alreadybeenputup in Delhi. A gas urbine unit can alsobe usedas synchrono.r, ornp"nsatorto help maintain flat voltage profile in the system.

I

The oldest and cheapestmethod of power generations that of utilizing thepotential energy of water. The energy is obtainedalmost free of nrnning costand is completely pollution free. Of course, t involves high capital cost

requires a long gestationperiod of about five to eight years as compared tofour to six years or steamplants.Hydroelectricstationsare designed,mostly,as multipurpose projects such as river flood control, storageof irrigation anddrinking water, and navigation. A simple block diagram of a hydro plant isgiven in Fig. 1.6. The vertical difference

between he upper reservoir and tailrace is called the head.

Surge hamberHeadworks

Spillway

Valve house

ReservoirPen stock

Powerhouse

Tailrace ond

Fig. 1.6 A typical ayout for a storage ype hydro plant

Hydro plants are of different types suchas run-of-river (use of water as itcomes), pondage (medium head) type, and reservoir (high head) type. Thereservoir type plants are the ones which are employed for bulk powergeneration.Often, cascaded lantsare alsoconstructed,.e., on the sa.mewaterstream where the dischargeof one plant becomes he inflow of a downs6eamplant.

The utilization of energy in tidal flows in channetshas long been thesubject of researeh;Ttrsteehnical and economicdifficulties still prevail. Some

of the major sites under investigation are: Bhavnagar,Navalakhi (Kutch),Diamond Harbour and GangaSagar. The basin in Kandala (Gujrat) has beenestimated to have a capacity of 600 MW. There are of course intense sitingproblems of the basin. Total potential is around 9000 IvftV out of which 900MW is being planned.

A tidal power station has been constructedon thenorthern France where the tidal height range s 9.2 mestimated o be 18.000m3/sec.

Different types of turbines such as Pelton.Francis and Kaplan are used forstorage,pondageand run-of-river plants, respectively.Hydroelectricplantsare

La Rance estuary inand the tidal flow is

Generator

Modern ower



W - system narvsist -

p = g p W H W

where

W = dischargem3ls through turbine

p = densiry1000kg/m3

11= head m)

8 = 9.81mlszProblemspeculiar o hydro plant which inhibit expansionare:

1. Silting-reportedly Bhakra deadstoragehas silted fully in 30 years2. Seepage

3. Ecologicaldamage o region

4. Displacement f humanhabitation rom areasbehind he dam which willfill up and becomea lake.

5. These annotprovidebase oad,mustbe used or peak.shaving ndenergysaving n coordinationwith thermalplants.

India alsohasa tremendous otential(5000MW) of having argenumberofmicro (< 1 Mw), mini (< 1-5 Mw),

and, mall (< 15 Mw) Mrl plants inHimalayan region, Himachal, up, uttaranchal and JK which must be fullyexploited o generate heapand clean power or villages situated ar away romthe grid power*. At present500 MW capacity s und"r construction.

In areaswheresufficienthydro generations not available,peak oad may behandled by meansof pumped storage.This consistsof un ,rpp". and lowerreservoirs and reversible urbine-generator ets,which cun ulio be used asmotor-pump sets.The upperreservoirhas enoughstorage or aboutsix hoursof full load generation.Sucha plant actsas a conventionalhydro plant duringthe peak oad period, when productioncostsare the highest.The urbines aredrivenby water from the upper eservoir n the usualmanner.During the ightload period, water in the lower reservoir s pumped back into the ipper oneso as to be ready for use in the next cycle of the peak ioad p.rioo. rn"

generators n this period change o synchronousmotor action and drive theturbineswhich now work as pumps.The electricpower is supplied o the setsfrom the general power network or adjoining thermal plant. The overallefficiency of the sets s normarly as high ut 60-7oEo.The pumped sroragescheme, n fact, is analogous o the chargingand dischargingor u battery.Ithas the added advantage hat the synchronousmachin", tu1 be used assynchronouscondensers or vAR compensationof the power network, ifrequired. n-a way, from the point of view of the thermal sectorof the system,

* Existing capacity (small hydro) is 1341 MW as on June 200I. Total estimatedpotential s 15000MW.

daily load demandcurve.Someof the existingpumpedstorageplantsare I100 MW Srisailem n Ap

and 80 MW at Bhira in Maharashtra.

Nuclear Power Stations

With the end of coal reservesn sight in the not too distant uture, the immediatepracticalalternativesourceof large scaleelectricenergygeneration s nuclearenergy. n fact, the developed ountrieshave already

switchedover in a big wayto the use of nuclear energy for power generation. n India, at present, hissourceaccounts or only 3Vo f the total power generationwith nuclearstationsat Tarapur (Maharashtra),Kota (Rajasthan),Kalpakkam Tamil Nadu), Narora(UP) and Kakrapar (Gujarat). Several other nuclear power plants will becommissioned y 20I2.In future, t is likely thatmoreandmore power will begenerated sing this important resource(it is planned o raise nuclear powergenerationo 10,000MW by rhe year 2010).

When Uranium-235 s bombardedwith neutrons, ission eaction akesplacereleasingneutronsand heatenergy.Theseneutrons hen participate n the chainreactionof fissioning more atomsof 235U.

In order that the freshly releasedneutronsbe able to fission the uranium atoms, heir speedsmust be ieduced oa critical value- Therefore, or the reaction to be sustained, uclear fuel rods

mustbe embedded n neutronspeed educingagents like graphite,hqavywater,etc.) called moderators.For eactioncontrol, rods madeof n'eutron-absorbingmaterial (boron-steel)are usedwhich, when inserted nto the reactor vessel,control the amount of neutron flux thereby controlling the rate of reaction.However, his ratecanbe controlledoniy within a narrow ange.The schemadc,diagramof a nuclearpowerplant s shown n Fig. 1.7.The heit releasedby the'uclear reaction is transportedo a heat exchanger ia primary coolant (coz,water,etc.). Steam s then generatedn the heatexchanger,which is used n aconventionalmanner o generate lectric energyby meansof a steam urbine.Various types of reactorsare being used n practice or power plant pu{poses,viz., advancedgas reactor (AGR), boiling water reactor (BwR), und h"uuywater moderated eactor.etc.

Waterntake

Controlods

Fue l rods_



WModernPo*", systemAn"tysis

CANDU reactor-Natural uranium in cixide orm), pressurized eavywatermoderated-is adopted in India. Its schematic diagram is shown in Fig.1 . 8 .

Containment

Fig. 1.8 CANDU eactor-pressurizedeavywater noderated-adoptednIndia

The associatedmerits and problemsof nuclear power plants as comparedto conventional hermal plants are mentionedbelow.

Merits

1. A nuclearpower plant is totally free of air pollution.

2. It requires inle fuel in termsof volume and weight, and hereforeposesno transportation roblems and may be sited, independentlyof nuclear

i i i r iociucr ion -

require that they be normally located away from populatedareas.

Demerits

Nuclear reactors produce radioactive fuel waste, the disposal

posesserious environmentalhazards.

The rate of nuclear eactioncan be lowered only by a small margin, so

that the load on a nuclear power plant can only be permitted to be

marginally reducedbelow its full load value. Nuclear power stations

must, therefore, be realiably connected o a power network, as tripping

of the lines connectingthe station can be quite serious and may required

shutting down of the reactor with all its consequences.

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 storageschemementionedearlier.

The greatestdanger n a fission reactor s in the caseof loss of coolant

in an accident.Even with the control rods fully loweredquickly called

scrarn operation, the fission does continueand its after-heatmay cause

vaporizing and dispersalof radioactivematerial.

The world uraniumresources re quite limited, and at the present ate may

not last much beyond50 years.However, there s a redeeming eqture. Duringthe fission of

235U,some of the neutrons are absorbedby lhe more abundant

uranium sotope238U

lenricheduraniumcontainsonly about3Voof 23sUwhile

most of its is 238U)converting t to plutonium ("nU), which in itself is a

fissionablematerial and can be extracted rom the reactor uel wasteby a fuel

reprocessingplant. Plutonium would then be used in the next generation

reactors (fast breeder reactors-FBRs), thereby considerablyextending the

life of nuclear uels.The FBR technology s being ntenselydevelopedas t

will extend the availability of nuclear fuels at predicted rates of energy

consumption o severalcenturies.

Figure 1.9 shows the schematicdiagram of an FBR. It is essential hat for

breeding operation, conversionratio (fissile material generated/fissilematerial

consumed) has to be more than unity. This is achieved by fast moving

neutronsso that no moderator s 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 necessary hat the

coolant should possessgood thermal propertiesand hence iquid sodium is

used.The fuel for an FBR consists of 20Voplutonium phts 8Vouranium oxide.

The coolant, liquid sodium, .ldaves he reactor at 650"C at atmospheric

pressure.The heat so transported s led to a secondarysodiumcircuit which

transfers it to a heat exchanger o generatesteam at 540'C.

2.

3 .

4.

t Modprn pnrrrar erro lam Anal . , ^ l^r r t y y v r r .



r v r r v r v y g l g t t l n t t d t v s t s

with a breeder reactor the release of plutonium, an extremely toxicmaterial, would make the environmentalconsiderationsmost stringent.

An experimental ast breeder est reacror FBTR) (40 MW) has-been

builtat Kalpakkam longside nucrear owerplant.FBR technology,

"*f..l"Jconventionalthermalplants.

- Core

Coolant

Containment

Fig. 1.9 Fastbreedereactor FBR)

An important advantage f FBR technology s that it can alsouse thorium(as fertile material)which gets converted o t33U

which is fissionable.Thisholds great promise or India as we have one of the world's largestdepositsof thoriym-about 450000 ons in form of sanddunes n Keralu una along theGopalpfurChatrapurcoastof Orissa.We havemerely 1 per cent of the world's

suited or India,with poorqualitycoal, nadequareydro potentiaiilentifulreserves f uranium 70,000 ons)and horium,and manyyearsof nuclearengineeringexperience.The presentcost of nuclearwlm coal-ttred power plant, can be further reduced by standardisingpl4ntdesignand shifting from heavy wate,reactor o light waterreactor echnology.

Typical power densities1MWm3) in fission reactorcores are: gas cooled0.53,high temperaturegas cooled 7.75,heavywarer 1g.0,boiling iut., Zg.O,pressurizedwater 54.75, fast breeder eactor 760.0.

Fusion

Energy is produced n this processby the combinationof two light nuclei toform a single heavier one under sustainedconditions of exiemely hightemperaturesin millions of degreecentigrade).Fusion is futuristic. Genera-tion of electricity via fusion would solve the long-tenn energyneedsof theworld with minimum environmental problems. A .o--"i.iul reactor isexpectedby 2010 AD. Consideringradioactivewastes, he impact of fusionreactorswould be much less than the fission reactors.

In caseof successn fusion technologysometime n the distant uture or abreakthroughn the pollution-free solarenergy,FBRs would becomeobsolete.However, there is an intense need today to develop FBR technologyas an

insuranceagainst ailure to deverop hese wo technologies. \In the past few years, serious doubts have been raised.aboutthe safetyclaims of nuclearpower plants.Therehavebeenas many as 150 neardisasternuclear accidents from the Three-mile accident in USA to the recentChernobylaccident n the former USSR.There s a fear.thatall this may purthe nuclearenergydevelopment n reverse ear. f this happens herecould beserious energy crisis in the third world countries which have pitched theirhopeson nuclearenergy to meet their burgeoningenergyneeds.France with78Voof its power requirement rom nuclearsources)and Canadaare possiblythe two countrieswith a fairty cleanrecordof nucleargeneration.ndia needsto watch carefully their design,constructionand operatingstrategies s it iscommitted to go in a big way for nuclear generationand hopes to achieve acapacity of 10,000MW by z0ro AD. As p.er ndian nuclear scientists,our

heavywater-based lants are most safe.But we must adoptmore conservativestrategiesn design,constructionand operationof nuclearplants.

World scientistshave to adoptof different eactionsafetystrategy-may beto discover additives to automatically nhibit feaction beyond cr;ii"at ratherthanby mechanically nsertedcontrol rods which havepossibilitiesof severalprimary failure events.

Magnetohydrodynamic (MHD) Generation

In thermal generation of electric energy, the heat released by the fuel isconverted to rotational mechanical energy by means of a thermocvcle. The

Introduction w



ryModernower ystemnatysis

mechanicalenergy s then used to rotate he electric generator.Thus twostagesof energy conversion are involved in which the heat to mechanicalenergy conversionhas inherently ow efficiency. Also, the rotating machineha s it s associatedossesand maintenance roblems. n MHD technology,

cornbustionof fuel without the need or mechanicalmoving parts.

In a MHD generator, lectrically conductinggas at a very high temperature

is passed n a strong magnetic leld, thereby generatingelectricity. Hightemperature is needed to iontze the gas, so that it has good eiectrical

conductivity.The conducting as s obtainedby burning a fuel and njectinga seeding materials such as potassium carbonate in the products ofcombustion. The principle of MHD power generation s illustrated in Fig.1.10.Abotrt 50Vo fficiency can be achieved f the MHD generator s operatedin tandem with a conventionalsteamplant.

Gas flow

at 2,500 C

Strong magnetic

f ield

Fig.1.10 Thepr incip lef MH Dpower enerat ion

Though the technological easibility of MHD generation as been estab-

lished, ts economic 'easibilitys yc t to be demonstrated.ndia had starteda

researchand 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 naturalgas as fuel had been n

operation for someyears. n fact with the developmentof CCGT (combined

cycle gas turbine) plant, MHD developmenthas been put on the shelf.

Geothermal Power Plants

In a geothermal power plant, heat deep inside the earth act as a source of

power. There has been some use of geothermalenergy n the form of steam

coming from underground n the USA, Italy, New Zealand,Mexico, Japan,

Philippines and some other countries. n India, feasibility studies of 1 MW

station at Puggy valley in Ladakh is being carried out. Another geothermal

field has been ocatedat Chumantang. here are a number of hot springs n

India, but the total exploitableenergypotentialseems o be very little.

Ttre present nstalled geothermal plant capacity in the world is about 500

MW and the total estimated apacity s immenseprovidedheatgeneratedn the

I

volcanic regionscan be utilized. Since he pressureand temperatures re low,

the efficiency s even ess han the conventionalossil fuelled plants,but the

capital costs are ess and the fuel is available ree of cost.

I.4 RENEWABLE ENERGY SOURCES

To protectenvironmentand for sustainable evelopment, he importanceof

renewableenergysources annotbe overemphasized.t is an established nd

acceptedact that renewableand non-conventionalorms of energywill play

an increasingly mportant role in the future as they are cleanerand easier to

use and environmentally enignand are bound o becomeeconomicallymoreviable with increased se .

Becauseof the limited availability of coal, there is considerablenterna-

tional effort into the developmentof alternative/new/non-conventionaUrenew-

able/cleansourcesof energy. Most of the new sources someof them n fact

have been known and used for centuries now!) are nothing but the

manifestation f solar energy, e.g., wind, sea waves, ocean hermal energy

conversion(OTEC) etc. In this section, we shall discuss he possibilitiesand

potentialitiesof various methods of using solar energy.

Wind Power

Winds are essentially reatedby the solar heating of the atmosphere. everal

attemptshave been madesince 1940 to use wind to generateelectric

energyand development s still going on. However, technoeconomic easibility has

yet to be satisfactorily stablished.

Wind as a power source s attractivebecauset is plentiful, nexhaustible

and non-polluting.Fnrther, it does not impose extra heat burden on the

environment.Unlbrtunately, t is non-steady nd undependable. ontrol

equipmenthas beendevised to start he wind power plant whenever he wind

speed eaches 0 kmftr. Methods have also been ound to generate onstant

frequencypower with varying wind speeds nd consequently aryingspeeds

of wind mill propellers. Wind power may prove practical for small power

needs n isolatedsites.But for maximum flexibility, it should be used in

conjunctionwith other methodsof power generation o ensure ontinuity.

For wind power generation, here are three types of operations:

1. Small, 0.5-10 kW for isolatedsinglepremises2. Medium, 10-100 kW for comrnunities i

3. Large, 1.5 MW for connection o the grid.

The theoretical ower in a wind stream s given by

P = 0. 5 pAV3W

densityof ai r (1201g/m' at NTP)

mean air velocity (m/s) and

p =

V _

where

A = sweptarea rn").

Introduct ion



2. Rural grid systems re ikely to be 'weak,in these reas. ince

retatrvely ow voitagesupplies e.g.33 kV).3. There are alwaysperiods without wind.In India, wind power plants have been installed in Gujarat, orissa,

Maharashtra nd Tamil Nadu, where wind blows at speeds f 30 kmftr duringsummer'On the whole, the wind power potentialof India has beenestimatedto be substantialand is around 45000 Mw. The installed capacity as onDec. 2000 is 1267 Mw, the bulk of which is in Tamil Nadu- (60%). Theconespondingworld 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 about600 W/rn2 but the actualvalue varies considerably.t has the advantage fbeing ree of cost,non-exhaustiblend completelypollution-free.On the otherhand, t has severalcrrawbacks-energydensitypei unit area s very row, it isavailable for only a part of the day, and cl,oud and,hazy atmosphericconditions greatly reduce the energy received. Therefore,harnessing solar

energy or electricitygeneration,hallengingechnological roblemsexist, hemost mportant being that of the collection and concentrationof solar energyand ts conversion o the electrical orm through efficient and comparativelyeconomicalmeans.

Totalsolarenergyotentialn India s 5 x lOls kwh/yr.Up ro 31. t2.2000.462000solar cookers,55 x10am2solar hermai systemcollectorarea,47 MWof SPV power, 270 community lights, 278000 solar lanterns PV domesticlighting units),640 TV (solar),39000PV street ights and3370 warer pumps

MW of grid connected solar power plants were in operation.As per oneestimate 36], solarpower will overtakewind in 2040 andwould become heworld's overall largest source of electricity by 2050.

Direct Conversion to Electricity (Photovoltaic Generation)

This technology onvertssolarenergy o theelectrical orm by meansof siliconwafer photoelectric ells known as"Solar Cells". Their theoreticalefficiency isabout25Vobut the practical value s only about I5Vo.But that doesnot matteras solar energy s basically free of cost. The chief problem is the cost andmaintenance f solar cells.With the ikelihoodof a breakthroughn the largescaleproductionof cheap solar cells with amorphoussilicon, this technologymay competewith conventionalmethodsof electricity generation, articularlyas conventionaluels becomescarce.

Solar energy could, at the most, supplementup to 5-r0vo of the totalenergydemand. t has beenestimated hat o produce1012 wh per year, thenecessary ellswould occupy about0.l%o f US landareaasagainst ighwayswhich occupy 1.57o in I975) assuming 07o efficiencyand a daily insolation

of 4 kWh/m'. .\In all solar hermalschentes, torages necessary ecause f the fluctuating

nature of sun's energy.This is equally rue with many otherunconventionalsources s well as sources ike wind. Fluctuat ing ourceswith f luctuat ingloads complicate till further the electricitysupply.

Wave Energy

The energyconientof sea waves s very high. In India, with several undredsof kilometersof coast ine, a vastsourceof energy s available. he power inthe wave is proportional o the squareof the anrplitudeand to the period ofthe motion. Therefore, he ong period - 10 s), argeamplitude - 2m) wavesare of considerable interest for power generaticln,with energy fluxes

commonly averaging etween50 and 70 kW/m width of oncomingwave.Though the engineeringproblemsassociated ith wave-powerare formidable,the amountof energy hat can be harnesseds largeand development ork isin progress alsosee he sectionon HydroelectricPowerGeneration, age 17).Sea wave power estimatedpoterrtial s 20000 MW.

Ocean Thermal Energy Conversion (OTEC)

The ocean s the world's largest solar coilector. Temperaturedifference of2O"Cbetween ,varrn, olar absorbingsurfacewater and cooler bottorn'

water

At present, two technologiesare being developed or conversion of solarenergy o the electrical orm.-'Inone technology,collectorswith concentratorsareemployed o achieve emperaturesigh enough 700'C) to operatea heatengrneat reasonableefficiency to generateelectricity. However, there areconsiderable ngineeringdifficulties in building a single racking bowi with adiarneter xceeding30 m to generate erhaps200 kw. The scheme nvolveslargeand intricate structuresnvoiving

lug" capital outlay and as of today sf'ar from being competitive with

"otru"titionalJlectricity generation.

Th e solar power tower [15] generates team or electrici typrocluct ion.]'here s a 10 MW installationof such a tower by the SouthernCaliforniaEdisonCo' in USA using 1818plane nirrors,each m x 7 m reflectingdirectracl iat iono th c raisecl oi ler.

Electricitymay be generatedrom a Solar pond by using a special .lowtemperature' eat enginecoupled o an electricgenerator. solarpond at EinBorek n Israelprocluces steady150 kW fiorn 0.74 hectare t a busbarcostof about$ O. O/kwh.

Solarpower potential s unlimited,however, otal capacityof about 2000MW is being planned.



ffiffi| Modem ow'erystem natysis

can occlrr.This can providea continually eplenished toreof thermalwhich is in principle available br conversion to other energy forms.refers to the conversionof someof this thermal energy nto work and

lntroduct ion

solar.The most widely usedstorage attery s the lead acid battery. nventedby Plante n 1860.Sodiutt t-sulphurattery 20 0 Wh/kg) an d othercolrbina-tions of materials re a-lso eingdevelopedo get more outputand storage erunit weisht.

Fuel Cells

A fuel cel lconverts hemical nerryof a fuel nt o electrici ty l irect ly,with nointermediate otnbustioncycle. In the fuel cell, hyclrogens supplied o thenegativeelectrode nd oxygen (or air) to the positive.Hydrogenand oxygenare combined to give water and electricity.The porous electrodesallowhydrogen ons to pass.The main reason ;rhy fuel cells are not in wide use stheir cost (> $ 2000/kW). Global electricity generating apacity rom full cellswil l grow ro m us t 75 Mw in 2001 o 15000MW bv 2010.US .Germanv ndJapan may take lead for this.

Hydrogen Energy Systems

Hydrogen can be used as a medium for energy transmission nd storage.Electrolysiss a well-establishedommercial rocess ieldingpurehydrogen.Ht can be converted ery efficientlyback o electi'icity y rneans f fuel ceils.Also the useof hydrogena.s uel for aircraftand automcbiles ouldencouraseits large scaleproduction,storageand distriburion.

1"6 GROWTH OF POWER SYSTEII{S IN INDIA

India is fairly rich in natural esourcesike coal and ignite; while sorneoilreserves avebeendiscovered o far. ntenseexploration s beingundertakeriin vitrious egit lns f thc country. ndiaha s mmensewaterpower .csourcesalsoof whichonly around25To av eso arbeenuti l iseci,.e.,oniy25000 \,IWhas so far beencommissioned p to the end of 9th plan.As per a recent eportof t lreCE A (CcntlalFlectrici t ,v uthority), he otalpotent ial f h1,dro oweris 84,040 v{Wat ('L't%oad actor.As regards uclear ower, ndia s cleflcientin uranium, ut has ich deposits f thorir-imvhich an be uti l ised t a futureclatc n l 'astbrccclor ci. tctor.s.ince ndepcndcncc,hc coull try ha s nndetremendous rogressn the development f electricenergyand oday t has he

largest systemamong he developingcountries.When lndia attained ndependence,he installecl apacitywas as low as

1400 MW in the early stages f the growth of power system, he major portionof generationwas through thermal stations, ut due to economical easons.hydro developmenteceivedattentionn areas ike Kerala,Tamil Nadu. UttarPradeshand Punjab.

In the beginningof the First Five Year Plan (1951-56), he rotal nstalledcapacitywas around2300 MV/ (560MW hydro, 1004MW thermal,149 MWthrough oil stations and 587 MW throughnon-utilities).For transporting his

energyOTECthence

50,000Mw .

A proposedplant usingsea emperaturedifferencewould be situated 25 kmcastol 'Mianii (USA), where he temperature l i l ' l 'eronces 17.5"C.

Biofuels

The material of plants and animals is called biomass, which may betransformed by chemical and biological processes o produce intermediatebiofuels sttch as methanegas, ethanol liquid or charcoal solid. Biomass isburnt to provide heat for cooking, comfort heat (spaceheat), crop drying,tactory processes nd raisingsteam or electricity productionand transport. nIndia potent ia l' t t l io-Energys 17000MW and hat br agr icul tunr l i rstc sabout 6000 MW. Thereare about 2000 community biogasplants and tamilysize biogasplants are 3.1 x 106.Total biomasspower harnessed o far is22 2 MW .

Renewableenergyprogrammes re specially designedo meet the growingenergy needs in the rural areas for prornoting decentralizedand hybriddcvelopment t. las to stemgrowingmigrat ionof rural populat ion o urban

areas n searchof better iving conditions. t would be through his integrationof energy conservation fforts with renewableenergyprogrammes hat Indiawould be able to achievea smooth ransition from fossil fuel economy tosustainable enewableenergybasedeconomy and bring "Energy for ali" forec;uitable nd environrnentalriendlysustainableevelopment.

1. 5 ENERGY STORAGE

'l 'hereis a lol ol problenr n storingclectrici ty n largc quantit ies.Enclgy

wliich can be converted nto electricitycan be stored n a number of ways.Storage f any nature s l rowever er y cost ly arrclts cconomicsmust beworked out properly. Variousoptionsavailable are: pLrmped torage,c:onl-pressed ir, heat,hydrogengas,secondary atteries,lywheels and supercon-duct ingcoi ls.

As already mentioned, gas turbines are normally used for meeting peakloads but are very expensive.A significant amount of storage capable ofinstantancoussewould be betterway of meetingsuchpeak oads,and so farthe most mportantway is to havea pumpedstorage lantas discussed arlier.Other methods are discuss-edelow very briefly.

Secondary Batteries

Large scalebattery use s almost ruled out and they will be used for batterypowered vehiclesand local fluctuating energy sources uchas wind mills or

HEIntroductionFI



power to the load

were constructed.

centres, ransmission ines of up to 110 regionsof the country with projectedenergy equirementand peak oad in the

year 2011-12 [19]'io ororrcrt crcncreri At the

During the Fourth Five Plan, ndia startedgeneratingnuclearpower'

Tarapur \uclear Plant 2 x 210 MW units were comrnissionedn April-May

. This stationuses wo boiling water reactorsof American design.By

commissionedbY 2012.

The growth of generatingcapacityso

2012 A.D. are given in Table 1'1'

far and future projectionfor 2011-

Tabte1. 1 Growth f Instal ledapacity n lndia ln MW )

Year Hydrtt NuclearThermal DieseI Total

Northern egion

308528

(49674).,.

MW* 9

Westernegion

299075

(46825)

1970-7t1978-791984-852000-01

39 8

=2700 MW

renewable

r47042864042240

101630

6383l 1378t427125141

42089 0

10952720

7503t63722707471060

\'./

Fig. 1.11 Ma pof India howingiv e egional rojectednergy equirementn

MkWh ndpark oa d n MW or year2011-12'

The emphasisduring he SecondPlan (1956-61) was on the developmentof

basic ancl heavy inclustries nd thus there was a need to step up power

generation.The total nstalledcapacitywhich was around3420MW at the end

of tn" First Five year Planbecame5700 MW at the end of the SecondFiveyear plan. The introductionof 230 kv transmission oltagecame up in Tarnil

Pattern of utlization of electrical energy in 1997-98 was: Domestic

{O.6g\o,commercial .91 o, nigation 30.54Vo,ndustry35'22Vo nd others s

6.657o.It is expected o remainmore or lesssame n 2004-05'

To be self-sufficient in power' BHEL has plants spreadout all over the

countryancl hese urn ou t an entire angeof powerequipment, iz ' turbo sets'

hydro sets, urbines or nuclearplants, iigft pi".ture boilers,power transform--

ers, switch gears,etc. Each plant specializesn a rangeof equipment'BHEL's

first 500 MW turbo-generatorwas cornmissionedat singrauli' Today BIIEL

is consideredone of the major power plant equipmentmanufacturersn the

world.

T.7 ENERGY CONSERVATION

Energyconservations the cheapest ew sourceof energy'we should esort

to variousconservationmeasures uch as cogenerationdiscussed arlier),and

,r32 I Modernpower SvstemAnalvsis lntroduct ion



lu

use energy efficient motors o avoid wastefulelectricuses.We can achieveconsiderable lectricalpower savingsby reducingunnecessary igh lightinglevels,oversizedmotors,etc.A 9 W cornpact luorescentamp (CFL) may beused nsteadof 40 w fluorescent ube or 60 w lamp, all having the same

Load Management

As mentionedearlierby various'load management' chemes. t is possible o

shift demanrlaway frorn peak hours (Section .1.). A more direct method

would be the control of the load either throughrnodified tariff structure hat

encourage

schedules r direct electricalcontrol of appliancen the form of remote imer

controlled on/off switches with the least inconvenience o the customer.

Various systems for load rnanagementare described n Ref. [27]. Ripple

control has been tried in Europe. Remote kWh meter reading by carrier

sysrems s being tried. Most of the potential for load control lies in thedomestic sector. Power companies are now planning the introduction of

system-wide oad managementschemes.

1.8 DEREGULATION

For over one hundredyears, he electricpower ndustryworldwide operatedas

a regulated ndustry. n any area herewas only one company oI government

agency(mostly state-owned) hat produced, ransmitted,distributedand sold

electricpower and services.Deregulationas a conceptcame n early 1990s. t

brought n changesdesignecio enc<.rutageompetition.

Restructuring nvolves disassemblyof the power industry and reassembly

into another orm or functional organisation.Privatisation startedsale by a

government f its state-owned lectricutility assets, nd operatingeconomy,to private companies. n somecases, eregulationwas driven by privatization

needs.The state wants to sell its electric utility investmentand change he

rules (deregulation) o make the electric ndustry more palatable or potential

investors, hus raising the price it could expect rom the sale. Open access s

nothingbut a common way for a govenlment o encourage ompetition n the

electric industry and tackle monopoly.The consumer s assuredof good

quality power supply at competitive price.

The structure or deregulation s evolved in terms of Genco (Generation

Company),Transco Transrnission ompany)and ISO (Independent ystem

Operator). t i s expectedhat the optimal bidding will help Genco o maximize

its payoffs. The consumersare given choice to buy energy from different retail

energy supplierswho in turn buy the energy rom Genco n a power market.

(independentpower producer, IPP).

The restructuringof the electricity supply ndustry hat norrnally accompa-

nies the introduction of competiiion providesa fertile ground for the growth

of embeddedgeneration, .e. generation hat is connected o the distribut-icn

system ather han to the transmission ystetn.

The earliest eforms n power industrieswere nitiated in Chile. They were

followed by England, the USA, etc. Now India is also implementing the

restructuring. ot of researchs neededo clearlyunderstandhe power system

operation under deregulation. The focus of, research s now shifting towards

a year.Everyoneshouldbe madeaware hroughprint or electronicmediahowconsumptionevelscanbe reducedwithout any essentialoweringof comfort.Rate estructuringcan have ncentives n this regard.There s no conscious-nesson energy accountabil i tyet etndno sense f urgencyas in developedcountries.

Transmissionand distribution ossesshoulclnot exceed2OVo. hi s can beachievedby employing series/shunt ompensation, ower factor improvementmethods, static var compensators,HVDC option and FACTS (flexible actechnology) devices/controllers.

Gas turbirre combined with steam turbine is ernployed for peak loadshaving. This is more efficient than normal steamturbine and has a quickautomated starl and shut doivn. It improves the load factor of the steamstaflon.

Energy storage can play an important role where there is time or ratemismatchbetweensupplyand demandof energy.This hasbeen discussednSection1.5. Pumpedstorage hyclro)schemehas beenconsicleredn Section1 . 3 .

Industry

In India where most areashave arge number of sunny days hot water forbatharrdkitchen by solarwaterheaterss becomingcommon or commercialbui ldings,hotelsevenhospitals.

In India where vast egionsare deficient in electricsupply and,aresubjectedto long hours of power sheddingmostly random, he use of small diesel/petrolgenerators nd invertersare very conmon in commercialand domesticuse.Theseare highly wastefulenergydevices.By properplannedmaintelance thedowntime of existing large stationscan be cut down. Plant utilization factorsof existingplantsmustbe mproved.Maintenancemust be on schedule atherthan an elner-qency.Maintenancemanpower training shouldbe placed on warfoot ing. These actionswil l also improve th e load factor of most powerstations,which would indirectlycontribute o energyconservation.

W Modernpo*", Syster Anulyri, Int roduct ion



finding the optimal bidding methodswhich take into account ocal optimaldispatch, evenueadequacy nd marketuncertainties.

India has now enactedhe ElectricityRegulatoryComrnission'sAct, 1998and the Electricity (Laws) AmendmentAct, 1998.These aws enablesettinguo of

State Electricity RegulatoryComrnissions SERC) at srate evel.'fhe

mainpurposeof CERC is to promoteefficiency,economy and competition n bulkelectricity supply. orissa, Haryana,Andhra Pradesh,etc. have started heprocess f restructuring he power sector n their respective tates.

1.9 DISTRIBUTED AND DISPERSED GENERATION

DistributedGenerationDG) entailsusing nanysrnallgenerators f 2-50 MWoutput, nstal led t variousstrategic oints hroughouthe area, o that eachprovidespower to a small numberof consumers earby.Thesemay be solar,mini/micro hydel or wind turbine units,highly efficient gas turbines,smallcombincdcycle pl i tnts,sincc hcsearo the rnostccon<lnricalhoiccs.

Dispersedgeneration eferes o useof still smaller generatingunits, of lessthan 500 kW outputand often sized o serve ndividualhomesor businesses.Micro gas urbines, uel cells,diesel,and small wind and solarPV seneratorsmake up this category.

Dispersed enerationhas been used or clecades s an emergency ackuppower source.Most of theseunitsare usedonly fbr reliability

reinfbrcement.Now-a-days nvertersare being increasinglyused n domesticsectoras anemergency upplyduring blackouts.

The distributed/dispersed enerators an be stand alone/autonomousrgrid connecteddependingupon the requirement.

At the time of writing this (200i) therestill is and will probablyalwaysbesome economy of scale f-avouring arge generators.But the margin ofeconomydecreasedonsiderablyn last 10 years 23].Even f thepower tselfct lsts bit rt t t l rc hi tnccn(r 'a ltat ion owcr, here s no nccd<t f ransrnissionlines, and perhapsa reducedneed br distribution equipmentas well. Anothermaior advantageof dispersed ene.rations its modularity, porlability andrelocatabil i ty. ispersed eneratorsls o nclude wo new typesof tbssi l ue lunits-fuel cells and microgas urbines.

The main challenge oday s to upgrade he existing technologies nd toproniotedeveloprnent, emonstrat ion,cal ing up and cornmercial izat ionfnew and emerging technologies or widespreadadaptation. n the rural sectormain thrust areasare biomassbriquetting,biomass-basedogeneration, tc. Insolar PV (Photovoltaic), arge size solar cells/modules asedon crystallinesilicon hin films need o be developed. olarcellsefficiency s to be mprovedto 15%oo be of useat commercialevel.Otherareas redeveloprnentf higheificiency inverters.Urban and ndustrialwastesare used or variousenergyapplicationsncluding power generationwhich was around 17 Mw in 2002.

There are already32 million improvedchulhas.f growingenergyneeds n the

rural areasare met by decentralised nd hybrid ener-qy ystems distributed/

dispersed eneration),his can stemgrowing migrationof rural population o

urban areas n searchof better iving conditions. hus, India will be able to

able-energybased econolny iind bring "Energy for all" for equitable,

environment-friendly, nd sustainabie evelopment.

1.10 ENVIRONMENT/\L ASPECTS OF ELECTRIC ENER,GY

GENERATION

As far as environmental nd health isks involved n nuclearplantsof various

kinds are concerned,hese ave already'been iscussedn Section1.3. The

problerns elated o large rydroplantshavealsobeen welledupon n Section

1.3.Therefore,we shallnow focus our attention n fossil uel plant ncluding

gas-based lants.

Conversion of clne lornr ol' energy or another o electrical tortn has

unwantedsideeffectsand the pollutantsgeneratedn the processhave o be

disposed off. Pollutants know no geographical boundary, as result the

pollution issuehas becomea nightmarishproblemand strong national and

internationalpressuregroupshave sprung up and they are having a definite

impacton the development f energy esources. overnmentalwarenessas

created umerousegislation t nationaland nternationalevels,w[ich powerengineers ave o be fully conversantwith in practiceof their professionand

survey an d planning of large power projects.Lengthy, t ime consuming

procedures t governrnentevel,PIL (public nterest itigation)and demonstra-

tive protestshave delayedseveralprojects n several ountries.This has led

to favouringof small-size rojectsan d redevelopmentf exist ing sites.Bu t

with the increasing ap n electricdernand ndproduction, ur country has o

move forward fbr several arge thermal,hydro and nuclearpower projects.

Entphasiss rc ing ai don cor]scrvi l t ior tssucs.urtu i l tnentf t ransnt issi t tn

losses, theft, subsidizedpower supplies and above all on sustainable

devektpnrenlwittr uppntpriata technolog-)' hercver easible. It has to be

particularly assured hat no irreversible damage s caused o environment

which wouid affect the living conditions of the futuregenerations.rreversible

damagesike ozone ayerholesandglobalwarming aused y increasen CO2in the atmosphere re alreadyshowing up.

Atmospheric Pollution

We shall treathere only pollutrorras caused y thermalplants using coal as

feedstock.Certain issuesconcerning this have already been highlighted in

Section 1.3. The fossil fuel basedgenerating lants onn the backboneof

power generation in our country and also giobally as other options (like

nuclear and even hydro) have even strongerhazardsassociatedwith them.

f f i f f i | r r ^ r ^ - - n ^ . . . ^ - ^ . , - r - - a - - r . - - ,w_ tviouern ow-eruystemAnaiysts lntroduct ion



Also t shouldbe understoodhat pollution n large cities ike Delhi is causedmore by vehicrtlar raffic and their emission. n Delhi of course nderprasthaand Badarpurpower stations ontribute heir share n certainareas.

Problematicpollutants n emissionof coal-basedgeneratingplants are.

Oxides of Carhon (CO, COt)

CO is a very toxicpollutant ut t getsconvertedo CO'.,n the openatmosphere

(if available)surroundinghe plant. On the otherhandCO2 has been dentified

developing ountries.

Ifydrocarbons

During the oxidationprocess n cornbustion harnbercertain light weight

hydrocarbon may beformed. Tire compounds are a major source of

photochemicalreaction hat adds to depleti,rnof ozone ayer.

Particulates (fIY ash)

Dust content is particularly high in the Indian coal. Particulatescome out of

the stack in the form of fly ash. t comprises ine particles of carbon, ash and

other inert materials. n high concentrations,hesecausepoor visibility and

respiratorydiseases.

Concentrationof pollutantscan be reducedby dispersalover a wider area

by use of high stacks.Precipitators canbe used o removeparticlesas he flue

gases ise up the stack. f in the stacka verticalwire is strung n the middle

and charged to a high negativepotential, t emits electrons.These electrons

are capturedby the gas molecules herebybecomingnegative ons. These ons

accelerate owards the walls, get neutralizedon hitting the'walls and theparticlesdrop down the walls. Precipitators ave high efficiency up to 99Vo or

large particles,but they have poor performance or particles of size ess than

0.1 pm in diameter.The efficiency of precipitatorss high with reasonable

sulphurcontent n flue gases ut drops or'low sulphur ontentcoals;99Vo or

37o sulphur and 83Vo or 0.5Vosulphur.

Fabric filters in form of bag lnuses have also been employed and are

located before the flue gasesenter the stack.

Thermal Pollution

Steam fronr low-pressure urbine has to be liquefied in a condenser and

reduced to lowest possible temperature o maximize the thermodynamic

efficiency.The bestefficiencyof steam-cycle racticallyachievable s about4\Vo.It means hat60Vo f the heat n steam t hecycleendmust be removed'

This is achievedby following two methods'

1. Once throughcirculation hrough condenser ooling tubesof seaor river

water whereavailable.This raises he temperature f water in these wo

sourcesand threatenssea and river life around n sea and downstream

in river. ThesE, re serious environmentalobjections and many times

cannot be overruled ard also theremay be legislation against t.

2. Cooling tov,ersCool water is circulated ottnd he condenser ube to

remove heat from the exhaust steam n order to condense t. The

a

a

o

a

2

NO.r, nitrogenoxides

CO

coz. Certain hydrocarbons

o Particulates

Though the account hat follows will be general, t needs o be mentionedhere hat Indian coal has comparatively ow sulphur content but a very highash content which in some coals may be as high as 53Vo.

A brief accountof various pollutants, heir likely impact and methods ofabatements re presentedas follows.

Oxides of Sulphur (SOr)

Most of the sulphur present in the fossil fuel is oxidized to SO2 in thecombustion hamberbefore being emittedby the chimney. n atmospheretgets urther oxidized to HrSOo and metallic sulphateswhich are the majorsourceof concernas thesecan causeacid rain, impaired visibility, damage o

buildings and vegetation. Sulphate concenffationsof 9 -10 LElm3 of airaggravate sthma, ung and heart disease.t may also be noted that althoughsulphur doesnot accumulate n air, it does so in soil.

Sulphur emission an be controlledby :

o IJse of fuel with less than IVo sulphur;generallynot a feasiblesolution.

o LJseof chemical reaction to remove sulphur in the form of sulphuricacid, from combustionproductsby lirnestonescrubbersor fluidized bedcombustion.

. Removing sulphur rom 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, nitrogen xides, s a majorconcern s a pollutant.t is solublein water and so has adverseaff'ecton human health as t enters he lungs oninhaling and combining with moisture converts to nitrous and nitric acids,which dannge he ungs. At ievelsof 25-100partsper million NO, can causeacutebronchitis and pneumonia.

Emissionof NO_, an be controlledby fitting advancedechnology urnerswhich can assuremore completecombustion,hereby educing heseoxidesfrom being emitted.These can also be removed rom the combustionproductsby absorption rocess y certainsolvents oing on to the stock.

Gfrfudffi-ffii Mociern owerSysteqAnaiysis

I

sffintrcCuction EEF



circulatingwater gets hot in the process. t is pumped o cooling owerand s sprayed hrough nozzlesnto a rising volume of air. Someof thewaterevaporatesrovidingcooling.Th e atentheatof water s 2 x 10 6J/kg and coolingcan occur ast,Bu t this has he disaclvantagef raising

unoes t raoteJevels n th c sul r f t l und lng reas .course he waterevaporatedmust be macleup in the systemby adctingfresh water rom the source.

Closed cooling owerswherecondenr;atelows through ubcsanclair isblown in these ubes voids he humidityproblembut at a very high cost. nIndia only v,et

towersare being used.Electromagnetic Radiation from Overhead Lines

Biological effectsof electromagneticadiation rom power lines and evencables n close proximity of buildings have recently attractedattentionandhave alsocaused omeconcern.Power frequency 50 or 60 Hz) and even heirharmonics are not consideredharmful. Investigationscarried out in certainadvanced countrieshave so far proved inconclusive.The electrical andelectronicsengineers,while being awareof this controversy,must know thatmany otherenvironmental gentsare moving around hat can cause ar greaterharm to humanhealth han does electromagneticadiation.

As a pieceof information t may be quoted hat directly underan overheadline of 400 kV , th e electric ield strengths 11000V/ m an d magneric lu xdensity (dependingon current) may be as much as 40 ptT. Electric fieldstrength n the rangeof 10000-15000v/m is considered afe.

Visual and Audible Impacts

These environmentalproblems are causedby the following factors.

l. Right of way acquiresand underneath. ot a seriousproblern n Indiaat present.Could be a problem in future.

2. Lines converging t a largesubstat ion ar th e beautyof th e anclscapearound.Underground ablesas alternative re too expensive proposi-tion except n congesteclity areas.

3' Radio interferenceRI) has to be taken nto accountand countered vvarlous means.

4. Phenomenon f corona (a sort of electric dischargearound he hightension ine) producesa hissingnoisewhich is aucliblewhenhabitationis in close proximity. At the to'wers great attention must be paid totightness of joints, avoidanceof sharp edgesand use of earth screenshielding o lirnit audible noise o acceptableevels.

5' Workers inside a power plant are subjected o various kinds of noise(particularlynear the turbines)and vibration of floor. To reduce hisuoise to tolerable evel foundations and vibration filters have to bedesignedproperlyand simulation studiescarried out. The worker nlustbe given regularmedicalexaminations nd soundmedicaladvice.

T.TT POWER SYSTEMENGINEERSAND POWERSYSTEM STUDIES

The power systemengineerof the first decadeof the twenty-first century has

abreastof the recentscientific advancesand the latest echniques.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

constructionasksof greatmagnitudeboth n generation nd ransmission.He

has o 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 ike India, he has to additionally ace the perennial

problemof power shortages nd to evolve strategiesor energyconservation

and load management.

For planning he operation,mprovement nd expansion f a power system,

a power systemengineerneeds oad flow studies,short circuit studies,and

stabilitystudies.He has o know the principlesof economic oad despatch nd

load frequencycontrol. All theseproblemsare dealt with in the next few

chapters after some basic concepts n the theory of transmission ines are

discussed.The solutions o these problemsand the enormouscontribution

made by digital cornputerso solve he planningand operational roblemsof

power systems s also nvestigated.

I . I2 USE OF COMPUTERS AN D MICR.OPROCESSOiTS

Jlhe rst rnethosirl solvingvarious owcrsystem roblenis er eAC and DC

networkanalysers evelopedn early 1930s.AC analysers ere used or load

florv and stability studieswhereasDC were preferred or short-circuit tudies.

AnaloguecompLrtersere developedn 1940s nd wereused n conjunc-

tion with AC networkanalyser o solvevariousproblems or off--linestudies.

In 1950s many analoguedevices were developed o control the on-line

tunctions such as genelation --ontrol,i'equency nd tie-line controt.

The 1950salso saw the adventof digital computerswhich were first used

to solve a. oad flow problem n 1956.Power systemstudiesby computers

gave greater lexibility, accuracy, peedand economy.Till 1970s, herewas

a widespreaduse of computers n systemanalysis.With the entry of micro-processorsn the arena, ow, besidesmain ramecompLlters, ini, micro and

personal omputers reall increasingly eingused o carry out various power

systern studies and solve power system problems for off-line and on-line

applications.

Off-line applications nclude research, routine evaluation of system

performanceand data assimilationand retrieval. t is mainly used or planning

and arralysingsome new aspects of the system. On-line and real time

applications nclude data-loggingand the monitoring of the system state.

rytrfi\ tutodernowerSvstem naivsis----r----- '.1g.r.,'""



A large central computer is used in central load despatchcentres forcc<ln<lmicnd securc ontrolof ' largc ntegratedystems.Microprocessorsnc lcomputersnstalled n generating tations ontrolvarious ocalprocessesuchas startingup of a generatorrom the cold state, tc. Table 1.2depicts he ime

microprocessors.ome of theseproblemsare tackled n this book.

Table1. 2

Tirne scale Control Problems

Fseveralsuper hermal stationssuchas at Singrauli (Uttar Pradesh), arakka

(WestBengal),Korba (MadhyaPradesh), arnagundamAndhraPradesh) nd

Neyveli (Tamil Nadu), Chandrapur Maharashtra)all in coal mining areas,

2000 MW*. Manv more super hermal

plants would be built in future. Intensivework must be conductedon boiler

furnaces to burn coal with high ash content. Nationai Thennal Power

Corporation NTPC) is in chargeof these arge scalegeneration rojects.

Hydro power will continue to remaincheaper han the other types for the

next decade.As mentioned earlier, India has so far developedonly around

l87o of its estimated otal hydro potentialof 89000MW. The utilization ofthis perennialsourceof energywould involve massive nvestmentsn dams,

channelsand generation-transrnissionystem.The CentralElectricity Author-

ity, the PlanningCommissionand he Ministry of Power arecoordinating o

work out a perspectiveplan to developall hydroelectricsources y the end of

this century to be executed by the National Hydro Power Corporation

(NHPC). NTPC has also started ecentlydevelopment f hydro plants.

Nuclear energyassumes pecialsignificance n energyplanning n India.

Because f limited coal reserves nd ts poor quality, ndia hasno choice bu t

to keep going on with its nuclearenergy plans. According to the Atomic

EnergyCommission, ndia's nuclearpower generationwill increaseo 10000

MW by year2010.Everythingseemso be set or a take off in nuclearpowel'

productionusing the country's thoriumreservesn breeder eactors.

In India, concerted efforts to develop solar energy and other non-

conventional ources f energyneed o be emphasized,o that he growing

clemanclan be met and depleting bssil uel resourcesmay be conserved.To

meet he energy equirement,t is expectedha t the coal productionwill have

to be ipclcascdo q)orc ha n 15 0 ri l l ion otts tt 200'+2005 ts cott tpltrcd o

18 0 mil l ion tonnes n 1988.

A number of 400 kV lines are operatingsuccessfully ince 1980s as

mentioned lreacly. hi s was he irsi step n working owards nationalgrid.

There s a need n future to go in for even higher voltages 800 kV). It is

expecred ha t by the year 2Ol1-12,5400 ck t krn of 800 kV lines and 48000

ck t kn i gf 40 0 kV lineswould be n operat ion. ls o l inesma y be sericsan d

shuntcompensatedo carry hugeblocksof power with greaterstability. There

is a need or constructingHVDC (High VoltageDC) links n thecountry sinceDC linescan carry considerablymorepower at the samevoltageand require

fewer conductors.A 400 kV Singrauli-Vindhyachal of 500 MW capacity

first HVDC back-to-back cheme asbeencommissioned y NPTC (National

power Transmission Corporation) followed by first point-to-point bulk

EHVDC transmission f 1500MW at-+

500 kV over a distance f 91-5km

from Rihand o Delhi, PowerGrid recently ommissionedn 14'Feb.2003 a

'k NTPC has also built sevengas-basedombinedcyclepower stations uch as Anta

an d Auraiya.

Mill iseconds

2 s 5 minutes

10 min-few hours- do-

few hours-l week

I month6 months

I yr - 10 years

Relayingand systemvoltage control andexcitationcontrol

AGC (Automatic generationconrrol)

ED (Economicdespatch)

Securityanalysis

UC (Unit commitment)

Ma intcrranccchedLrlng

Systern lanning

(modification/extension)

1.13 PROBLEMS FACING INDIAN POWER INDUSTR.Y

AND ITS CHOICES

The electricity requilements of .[ndia have giowntremendously anC thedemand has been running ahead of supplyl Electricity generation and

t ransmissionroccsscsn Indiaar c vcry nef f ic icntn c<l lnpar isoni t l r hoseof somedeveloped ountries. s per one estimate,n India generating apacityis utilized on an averageor 360t)hoursout of 8760hclurs n a year, ,vhilenJapan t is rrsedbr 5 00 hours. l ' the ut i l izat ionactorcouldbe ncreascd,tshouldbe possible o avoidpower cuts.The transmissionoss n 1997-98ona nationalbasiswa s 23.68Voonsist ing f both technical ossesn transmis-sion l ines ancl ransfonners,nd also non-technicalosses aused y energytheftsand metersno t being read correctiy. t should be possible o achieveconsiderableavingby leducing his oss o 1570 y the end of theTenthFiveYear Plan by r-rsingwell known ways and nreansand by adootingsoundcommercial practices.Further, evcry attempt should be made to improve

system oad factors by flattening he load curve by giving proper tariffincentives and taking other administrativem.easures. s per the CentralElectricity Authority's (CEA) sixteenthannualpower survey of India report,the all India oad actorup to 1998-99was of the order of 78Vo.In uture tis l ikely to be 7I7o.By 200i,5.07 lakhof vi l lages 86Vo) av ebeen lectri f iedand 117 akh of pumpsets ave been energized.

Assuminga very modestaverageannualenergygrowth of 5Vo, ndia'selectricalenergy equirement n the year 2010 will be enormouslyhigh. Adifficult and challenging ask of planning,engineering nd constructing ewpower stat ionss imrninent o rneet his situat ion. he governnlent as bLri l t

t



2000 MW Talcher-Kolar + 500 kV HVDC bipoleenablingexcesspower from East o flow to South.HVDC line is expectedby Z0ll-I2.

At the time of writing, the whole energy sce

real time controlof power system. t may alsobe pointedout that his book willalso help n training and preparing he argenumber of professionalsrained incomputeraidedpower systemoperationand control that would be required tohandle v

REFERENES

Books

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2. Eilgerd,O.1., Basic Electric Power Engineering,Reading,Mass., 1977.3. Kashkari,C., EnergyResources, emandand Conservation ith SpecialReference

to India, Tata McGraw-Hill, New Delhi, 1975.4. Parikh,Kirit, .sacond ndia studies-Energy, Macmillian,New Delhi, 1976.5. Sull ivan,R.L, Power System lanning,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.,ElectricPowerStations,ols and I, Chapman ndHall,London, 1944.8. Central Electricity GeneratingBoard, Modern Power Station Practice, 2nd edn,

' Pergamon, 976.

9t Golding,E.W., The Generationo.f' lectricitlt b1t WindPower,Ctnpman and Hall,London,1976. i10. McMillan, J.T., et. al., Energy Resoorces nd Supp\t, Wiley, London, 1976.I L Bennet,D.J., The Elements J'NuclearPoveer, ongman,1972.12. Berkowitz,D.A:,. Power Generation nd Environmental hange,M.I.T. press,

Cambridge,Mass., 1972.

13. Steinberg, .J. and T.H. Smith, Econonr-loading f PowerPlantsand ElectricSystems, iley, New York, 1943.

14. Power SystemPlanning and Operations:Future Problemsand ResearchNeeds.EPRIEL-377-SR, ebruary1977.

15. Twidell,J.w. and A.D. weir, Renewuble nergyResources, . and F. N, spon,London.1986.

16. Mahalanabis, .K., D.P. Kothariand S.l. Ahson, ConxputeridedPower,S),srenrAnalysisand Control, Tata McGraw-Hill,New Delhi, 1988.

17. RobertNoyes Ed.), Cogeneration f Steamand ElectricPower,NoyesDali Corp.,usA, 1978.

18. weedy,B.M. and B.J.cory, ElectricPower svstems,4th dn ,wiley, New york,

1998.

19. cEA 12 Annual surveyof PowerReport,Aug. 1985; 4th Report,March l99l;16thElectricPower Surveyof India,Sept 2000.

20. Kothari,D.P. and D.K. sharma(Eds),EnergyEn.gineering. heoryand practice,

S. Chand,2000.

21. Kothari, D.P. and I.J. Nagrath, Basic Electrical Engineering, nd edn, TataMcGraw-Hill,New Delhi, 2002. Ch. 15).

transmissionsystem thus7000 ckt km of + 500 kV

is so clouded with

future. However, certain trends that will decide the future developmentsofelectric power industry are clear.

Generally,unit sizewill go furtherup from 500 MW. A higher voltage 76511200kV) will comeeventuallyat the transmissionevel. There s little chancefor six-phase ransmission ecomingpopular hough hereare ew such ines nUSA. More of HVDC lines will do-. in operation.As populhtionhas alreadytouched the 1000 million mark in India, we may see a trend to go towardunderground ransmissionn urbanareas.

Public sector nvestment n power has increased rom Rs 2600 million inthe First Plan to Rs 242330million in the SevenrhPlan (1985 90). Shortfallin the Sixth Plan has been around 26Vo.There have been serious powershortages nd generationand availabilityof power in turn have agged oomuch from the industrial, agricultural and domestic requiremeni. Hugeamountsof funds (of the order of Rs. 1893200million) will be required f wehave o achievepower surpluspositionby the time we reach he terminalyearto the XI Plan (201I-2012). Otherwiseachievinga rargetof 975 billion unitsof electric power will remain an utopian dream.

Power grid is planning creationof transmissionhighways to conserveRight-of-way.Strongnationalgrid is being developedn phasedmanner. n

20Ol the interregionalcapacitywas 5000 MW. It is Lxpecred hat by 2OlI-12,it will be 30000Mw. Huge investment s planned o the tune of us $ 20billion in the coming decade.presenrfigures for HVDC is 3136 ck t km,800 kV is 950 ck t km, 400 kV is 45500ck t krn and.220/132 v is 215000ckt km. State-of-theart technologieswhich are, being used n India currentlyare HVDC bipole, HVDC back-to-back, svc (static var compensator),FACTs (Flexible AC Transmissions)devices etc. Improved o and M(Operation and Maintenance) echnologieswhich are being used tgda arehotline maintenance,emergency estorationsystem, thermovision scanning,etc.

Becauseof power shortages,many of the industries,particularly power-

intensiveones,have nstalled heir own captivepower plants.* Curcently 0Voof electricity generatedn lndia comes rom the captivepower plantsand thisis bound to go up in the future. Consortiumof industrial .onru-.rs shouldbeencouraged o put up coal-based aptiveplants. Import should be liberalizedto support this activity.

x Captive diesel plants (and small diesel sets or commercial and domesticuses)arevery uneconomical rom a national point of view. Apart from being lower efficiencyplants they use diesel which shouldbe conserved or transportationsector.

ffiffi N4odernowerSystem lqtysis ,

20. Kothari, D.P. and D.K. Sharma Eds),EnergyEngineering. heoryand practice,



S. Chand, 2000.

21. Kothari, D,P. and I.J. Nagrath, Basic Electrical Engineering,2nd edn, TataMcGraw-Hill, New Delhi, 2002.(Ch. l5).

22. Wehenkel,L.A. Automatic Learning Techniquesn Power Systems,Norwell MA:Kiuwer, 997.

23. Philipson,L and H. Lee Willis, (Jnderstanding lectric Utilities and Deregulation,Marcel Dekker nc, NY. 1999.

Papers

24 . Kusko, A., 'A Predictionof Power SystemDevelopment, 968-2030', IEEESpectrum, pl . 1968, 5.

25. Fink, L. and K. Carlsen,OperatingunderS/ress nd Strain',IEEESpectrum,Mar.r978.

26. Talukdar,s.N., et. al., 'Methodsfor Assessing nergyManagement ptions', IEEETrans., an.1981,PAS-100, o. I, 273.

27. Morgen, M.G. and S.N. Talukdar, 'Electric Power Load Management:someTechnological,Economic,Regularityand Social Issues',Proc. IEEE, Feb. L979,v o l .67 , no .2 ,241 .

28. Sachdev, .S.,'LoadForecasting-Bibliography,EEE Trans., AS-96,1977,697.29. Spom, P., 'Our Environment-Options on the Way into the Future', ibid May

1 9 7 7 , 4 9 .

30. Kothari D.P., Energy ProblemsFacing the Third World, Seminar to the Bio-Physics Workshop,8 Oct., 1986 Trieste Italy

31 . Kothari,D.P,'Energy systemPlanningand Energy conservation', resented tY Y I V N n t i n n n l f n n v o n t i n n n ^ F l l l t r N I ^ r ' , n - l h i E - k l O Q Os v J t t r L , r \ w w y v t t l l , I w u , I 7 Q L ,

32. Kothari, D.P. et. a1., Minimizationof Air Pollution due to Thermal Plants'. ,I1E

(India),Feb. 1977,57, 65.

33. Kothari, D.P, andJ. Nanda, PowerSupplyScenarion India' 'RetrospectsandProspects',Proc. NPC Cong.,on Captive Power Generation,New Delhi, Mar.1986 .

34. NationalSolarEnergyConvention, rganised y SESI,1-3Dec. 1988,Hyderabad.

35. Kothari, D.P., "Mini and Micro FlydropowerSystems n India", invited chapter nthe book, EnergyResources nd Technology,ScientificPublishers,1992,pp 147-1 5 8 .

36. PowerLine, vol .5. no . 9, June2001.

37. United Nations.'Electricity Costsatxd Tariffs: A GeneralStudy; 1972.38. Shikha,T.S. Bhatti and D.P. Kothari, Wind as an Eco-friendly nergySource o

meet the Electricity Needsof sAARC Region", Proc. Int. conf. (icME 2001),BUET, Dhaka,Bangladesh. ec. 2001,pp 11-16.

39. Bansal,R. C., D.P. Kothari & T. S. Bhatti, "O n Someof the Design Aspects fWind EnergyConversionSystems", nt. J. of Energy Conversion nd Managment,Vol . 43, 16, Nov. 2002,pp.2175-2187.

40. l). P. Kothari and Amit Arora, "Fuel Cells in Transporation-Beyondatteries",Proc.Nut. Conf. on Transportation ystems,IT Delhi,April 2002,pp. 173-176.

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ComputerSimulation nd Growtft",EJEISA,UK, Vo l 3, 1 Sep 2001, 14-28.

j^a

2.I INTRODUCTION

The four parameterswhich affect the performanceof a transmission ine as an

e l e me n t o f a p ow e r Sy s t e ma r e i n d u ct a n c e , c a p a c i ta n c e , r e s i s t a n c ea n d, .rL..++ ^nrrrrnrAnce hich is normally due o leakale over line

illffilf?; L*",,;i;;.";;r""i"J ii overheadransmissionines'hisr -r- -- 'i+r- +ha caricq l ine narameters, 1'e' lnductance an d resistance'

ffiffir"##r'"* #;;t; oi.itiuu,.d tonghe ineand hevogether

i"; th" ,"ri., imPedanceof the line'

Inductar,c. i, ;;f; the most dominant ine parameter rom a power system

e n g i n e e r , s v i ew p o i n t . As w e s ha l l s e e i n l a t e rc h a p t e r s , i t is t h e i n d u c t i v e

reactancewhich limits the transmission apacityof a line'

2,2 DEFINITION OF INDUCTANCE

Voltage induced n a circuit is given by

, = V Y

", ,ti" flux linkages f the circuit n weber-turnsWb-T)'

This can be written in the form

dr b d i , d i . ,e = , - : L - : v

dt dr dr

anceof the circ'lit in henrys' which in

nearmagneticcircuit, 'e'' a circuit

Ies vary linearly with current such that

(2 . r )

(2.2)

+f ,,il Modern o**, Syrtm An"lyri, lnductancend Resistancef Transmissionines ffffi-



L = ! HI

or

Fig. 2.1 Flux inkages ue o internallux(cross-sectionaliew)

where

H, = magnetic field intensity (AT/m)

/y = current enclosed A)

By symmetry, H, is.constantand s in direction of ds all alongi he circular

path. Therefore, rom Eq. (2.8) we have

(2.e)

(2 .10)

(2.rr)

(2.r2)

M r n =) t ,

,rLIz

The voltagedrop in circuit 1 due to current in circuit 2 is

V, = jwMnlz = 7t \12 V

A= LI e.4)where ) and I arc the rms values of flux linkagesand current respectively.Theseare of course n phase.

Replacing

i +Eq. (2.1)by

ir,we get the steadystate

AC volrage dropdue to alternating lux linkagesas

Y= jwLI = jt^r) V e.5)-On similar ines, hemutual nductancebetweenwo circuits s definedas he

flux linkagesof one circuit due to current in another, .e.,

(2.6)

(2.7)The conceptof mutual nductances requiredwhile considering he coupling

between arallel inesand the nfluenceof power ineson telephone ines.

2.3 FLUX LINKAGES OF AN ISOTATED CURRENT.CARRYTNG coNqucroR

Transmissionines arecomposed f parallel contluctorswhich, for all practicalpurposes, an be considereds nfinitely long.Let us irst developexpressionsfor flux linkagesof a long isolatedcurrent-canyingcylindricat conductorwithreturn path ying at nfinity. This system orms a single-turn ircuit, flux linkingwhich is in the orm of circular inesconcentric o the conductor.The total luxcan be divided into two parts, hat which is internal o the conductor and theflux external o the conductor.Sucha division s helpful as the internal luxprogressively inks a smalleramountof currentaswe proceednwards towardsthe centreof the conductor,while the external lux always inks the total currentinside he conductor.

Flux Linkages due to Internal Flux

Figure 2.1 shows the cross-sectionali,ew of a long cylindrical conductorcarrying current 1.

The mmf round a concentricclosedcircular path of radius y internal to theconductoras shown n the figure is

2rryH,=1,A s s r r m i n o r r n i f o r m c r r r r e n f d c n s i f v *

, " = ( - ) t : [ 4 )\ r r r ' ) \ r " )

FromEqs. 2.9)and 2.10),we obtain

H,.=)!- AT/mt

2 T r "

The flux density By, y metres rom the centreof the conductors s

r ntIBu=pHu= + Wb/m2

z ztr-wherep is the permeability of the conductor.Considernow an infinitesimal tubular elementof thicknessdy and lengthone

metre. The flux in the tubular element dd = Bu dy webers inks the fractional

trrrn (Iril - yzly'l resulting in flux linkagesof

(2.3)

*For power frequency of 50 Hz, it is quite reasonable o assumeuniform current

density. The effect of non-uniform current density s considered ater in this chapter

while treating resistance.

,---f\-]y, '.rt t, -? 5

t l l

i r ' i \ \i + - - - - r - l - l -1 t i !\ \. I t' lt . r t t -

- i r - . " r1 /

{nr.ds=I y (Ampere's aw ) (2,8)

lnductancend Resistancef rransmissionines ffi.$ModernPqwerSystemAnalysis



Integrating,we get the total nternal lux li n

^^,=I#fdy:f fwat^ (2.14)

For a relative permeability lf,, = | (non-magneticconductor), 1t = 4n xl0-'[Vm. therefore

f -

an d

^r n =T*10-/ wb-T/m

Lint= ]xto-7 rVmz

Flux Linkage due to Flux Between Two Points Externalto Conductor

Figure2.2 shows wo pointsP, and Prat distances , and Drftoma conductorwhich carriesa cunent of 1 amperes.As the conductor s far removed rom thereturn current path, the magnetic ield external o the conductor s concentriccircles around the conductorand therefore all the flux betweenP, and Pr lineswithin the concentriccylindrical surfacespassing

hrough P, andP2.

Fig.2.2 Flux inkagesue o luxbetween xternal oints

Magnetic field intensityat distancey from the conductor s

t-The flux dd contained n the tubular elementof thickness dy is

dd = +dy Wb/m engthof conductor

The flux dQbeing external o the conductor inks all the current in the conductor

which togetherwith 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 conductordue to flux betweenpoints

P, and Pr is

p D " , , f n

\," = |' tn' dy - -t"- I ln "2 wb-T/m

,1Dt2 n.v 2r Dr

where ln stands or natural ogarithm*.

SinceFr=I, F = 4t x10-7

(2.r3)

(2.1s)

(2.16)

The

points

or

- n) rz= 2 x l }- t l ln =L wb/m

Dr

inductanceof the conductorcontributedbv

P, and Pr is then

Lrz = 2 x I0-1 n -? fV*Dl

L,n = 0.461 los.D' mH,/kmL L " D l

(2.r7)

the flux included between

(2 .18)

(2 .1e)

(2.20)

external lux are

H.,=I

AT/m' Z r ry

Flux Linkages due to Flux up to an External Point

Let the externalpoint be at distanceD from the centreof the conductor.Flux

linkagesof the conductor due to external lux (from the surfaceof the conductor

up to theexternalpoint) s obtained rom Eq. (2.17)by substitutin D t = r and

Dz = D, i 'e',

) . * , = 2 x 7 0 - 1 l l n

D

r

Total flux linkages of the conductor due to internal and

) = ) i n , * ) " * ,

= I x1o -7+Zx ro - : . I n22 r

*Throughout the book ln denotes natural logarithm

logarithm to base 10.

PI, P2

(base e), while log denotes

W Modern ower ystgmAnalysis - - lnductancendFlesistanaef Transmissionines li$Bfrf--



- . ^ - t + r n 2 )2 x 1 0 - , / [ - 4r )

x ro_71" *J_r,

I-et ,t - ,r-r/4 = 0.7788r

\ = 2 x ro4rh + wb-T/mrl

Inductance of the conductor due to flux up to an external

(2.Zra)

point is therefore

L= 2x 1o-7 I w^ (z.zrb)rHere r' can be regardedas the radius of a fictitious conductor with no

internal inductance ut the same otal inductanceas the actual conductor.

2.4 INDUCTANCE OF A SINGLE.PHASE TWO.WIRE LINE

Considera simple wo-wire ine composed f solid round conductors arryingcurrents1, and 1, as shown n Fig. 2.3.|n a single-phaseine,

1 1 + r = Q

Iz = - I t

D - r z

D

D + r z

Fig. 2.3 Single-phasewo-wirein eand he magneticielddue o currentnconductor only

It is important to note that the effect of earth's presenceon magnetic ieldgeometry* s insignificant.This is so becausehe relative permeabilityof earthis about he sameas hat of air and ts electrical onductivitv is relativelvsmall.

*The electric field geometry will, however, be very much affected as we shall seelater while dealing with capacitance.

To startwith, let us consider he flux linkagesof the circuit causedby current

in conductor 1 only. We make three observations n regard to these flux

linkages:

1. External flux from 11 o (D - ,) links all the current It in conductor 1.

2. External flux from (D - r) to (D + rr) links a current whose magnitude

progressively educes rom Irto zeroalong this distance,becauseof the

effect of negative current flowing in conductor 2.

3. Flux beyond (D + 12) inks a net cunent of zero.

For calculating the total inductance due to current in conductor 1, a

simplifying assumptionwill now be made. If D is much greater han rt and 12(which is normally the case or overhead ines), t canbe assumed hat he flux

from (D - r) to the centre of conductor 2 links all the current ^Ir and the flux

from the centre of conductor2 t o (D + rr) links zero current*.

Basedon the above assumption, he flux linkagesof the circuit causedby

current n conductor 1 as per Eq. (2.2Ia) are

(2.22a)

The inductance of the conductor due to current in conductor 1 only is then

) r = 2 x 1 0 - 71 , n - Lr\

Lt= 2 x 10-7" +f' 1

(2.22b)

Similarly, the inductanceof the circuit due to current n conductor 2 is

L z = 2 x 1 0 - 7 hD . '

r, 2(2'23)

r lo inc rha c r rncr r rns i f inn fhenre rn fhe f l r r x l inkaoes and l i kewise fhe ind t tc fancesv u r r r E ru v u s r v r r ! a r v v ^ v ^ ^ ^ t

of the circuit caused y current n eachconductor cnsidered eparatelymay be

added o obtain the total circuit inductance.Therefore, or the completecircuit

L = L t + 4 = 4 x 1 0 - ' l n

If r/r= r'z= /; then

L= 4 x 10- ' l n D/ / Wm

L - 0.92t log Dlr' mHlkm

FVm (2.24)

Transmission lines are nfinitely long compared o D in practical situations

and therefore the end effects in the above derivation have been neglected.

2.5 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 n practiceare always stranded xcept

*Kimbark [l9] has shown hat the resultsbasedon this assumption re fairly

accurate venwhen D is not much arger than 11and 12.

(2.25a)

(2.zsb)

52 | Vodern PowerSystemAnalysisI

fo, .r.ry small cross-sectionalareas.Strandedconductors are composed ofwire, parallel,

l n e l r r n + a n a ^ ^ ^ - f a t ^ ^ l ^ . ^ - - - ^ t r - ^ - - - ^ : - - ! - , r . | - ^r r rLrL.vtclt tutt ct ' t tu nt 'st l ' lat lug ut I lal lsrnlsslon Unes ! 5.* "

l*2.6 FIUX LINI{AGES OF ONE CONDUCTOR



strands of electrically in with alternate ayers spiralled in

opposite direction to prevent unwinding. The total number of strands M) in

concentricallystrandedcableswith total annular space illed with strands of

uniform diameter rD s givenby

N = 3 x ' - 3 x + l (2.26a)

wherex is the number of layers wherein he single central strand s counted as

the first layer.The overall diameter D) of a stranded onductor s

P - ( 2 x - r ) d (2.26b)Aluminium is now the most commonlyemployedconductormaterial. t has

the advdntages f being cheaperand lighter than copper though with lessconductivity and tensile strength.Low density and low conductivity result in

larger overall conductordiameter,which offers another ncidentaladvantage n

high voltage lines. Increased diameter esults in reduced electrical stress at

conductor surface or a given voltageso that the line is corona ree. The low

tensile strength f aluminium conductorss made up by providingcentral

strandsof high tensilestrength teel.Sucha conductor s known as alurninium

conductor steel einforced ACSR) and s most commonly used n overhead

transmissionines. Figure 2.4 shows he cross-sect ionaliew of an ACSR

conductorwrth 24 strandsof aluminium and 7 strandsof steel.

Steel trands

Aluminiumstrands

Fig.2.4 Cross-sect ionalie wof ACSR-7 teel trands, 4 aluminiumtrands

In extra high voltage (EHV) transmissionine, expandedACSR conductors

are used. Theseare provided with paperor hessianbetweenvarious ayers ofstrandsso as o increase he overallconductordiameter n an attempt o reduce

electrical stressat conductor surfaceand prevent corona. The most effective

way of constructing orona-free HV lines s to provideseveral onductors er

phase in suitable geometrical configuration. These are known as bundled

conductorsand are a common practicenow for EHV lines.

N A GROUP

As shown n Fig. 2.5,considera group of n pnallel roundconductorscarryingphasorcurrents Ip 12,-, I, vvhose sum equals zero. I)istances of these

an expression for the total flux linkages of thel t u l t , . . t u n . us oDtam

ith conductor of the groupconsidering lux up to the point P only.

Fig.2. 5 Arbitraryroup f n paral lelound onductorsarrying urrents

The flux linkages f ith conductor ue o itsown current , self inkages)aregiven by [see F,q. (2.21)]

on(,

3

2

-4I

(2.27)

The flux linkages of conductor i due to current n conductor7 1rlf"r to Eq.(2 .17) l s

) i i= 2 x 10-7, h! Wb-T/mr i

5,,= 2 x l;-il,fn a Wb-T/mDij

)i = Xir + )iz + ... + )i i * . . . * )i n

= 2 x rca r, ' * + 4u !- +. . .+ ,\ ^ D t

'D , z

(2.28)

whereDu is the distanceof ith conductor rom 7th conductorcarrying current1r.From F,q. (2.27) and by repeateduse of Eq. (2.28), rhe rotal flux linkagesof conductor i due to flux up to point P are

, D iln

rl.

+...+ In

The above equationcan be reorganizedas

)i = 2xro'[[r, h+ +hn[* .+ I ,m]+. .+

+ (/, ln

But, I, - - (1, +

D r + I r h D z + . . . + i k D, + . .+ I n^ O , )

Iz +... + In-).

Substiiuti'g for /n in the second erm of Eq. 2.29)and simplifying, we have

l nduc tanec anr l F laa io ranaa ^ i T - - - ^ - - ! . f. ., r s rq r rug ur I ra l l sm lss fon L lnes k , . . . i<E r- I

-- -

a / 1 l A \ t ^ c r IApplying Eq. (2-30)ro filamenr of conductorA, weobtain ts



f/

)i = 2 * to-tl ,ml * L,h L_.+...+1,nL ( ' D i r ' - " " ' D i z " " ' ' ' ' ^ ' r t i

u, flux tintages

zxfir ! h+a6J-a . . . * rn l * . . . * rn I

, (-2xto-74l t " ] - * 6-J-1. . . * ln t

), \ Dir,^^ '

Diz,| " ' ' 'r

D,^, )

= 2 x lo-716(P,r:D,r,...D,t

Wb_T/m(D,rD,r.D,,. D,n)r'The inductance-of ilament f is then

L i = ] -! : 2nx70-t,n(Dn'"' D,, ' ' D,^')1/^ '

* I , ln =e+. ..*l i ln at -

.. . * I , - r1nDr - t1 )' 4 ) )

In order to account or total flux linkagesof conductor i, let the point p now

recede o infinity. The terms suchas n D1/Dn,etc. approachn t = o. Also forthe sakeof symmetry,denoting ,.{^ D,,,-wi have

l i= z x 1orI t rn** rr ' , ' ! -+1, ln 1U Dt

'D,2

-" ---Dii

+...+I^t"+)

wb-r/m

(D, rD, r . . .D , ,D '*FVm (2.3r)/ n

(2.30)

2.7 INDUCTANCE OF COMPOSITE CONDUCTOR LINES

We are now ready o study the inductanceof transmission ines composedofcomposite onductors. uture 2.6 showssucha single-phaseine comprising

compositeconductorsA and B with A havingn paraliel filamentsand B havingmt parallel ilaments.Though the nductance f each ilamentwill be somewhatdifferent their esistances ill be equal f conductor iameters rechosen o beunttorm), t is sufficientlyaccurate o assum.ehat the current s equallydividedamong he filamentsof eachcompositeconductor.Thus, each ilament of A i staken to carry a current I/n, while each filament of conductor B carries thereturn current of - Ihnt.

oCompositeonductor

Fig.2. 6 single-phasein econsist ingf tw o compositeonductors

Theaveragenductancef the ilaments f compositeonductor is

L u u , =L t + L 2 + 4 + " ' + L n

Since onductor is co#posed f n filaments lectricallyn parallel, tsinductances

, _Lu , ,_ \ + 1 2 + . . . + L ,_A =-

*ru"rorr:rr"rh".*pr"f,rionor ir"#i"t indu*anceromEq. 2.31)i"E;. 3'::r',

Le=2 x 10-7 n

l ( D n , . .Drj , . .Dr . , ) . . . (D^ ,. .p r j ,. . .D , ^ , ) . . .

(Dnr,. Dnj, .. Dr^,17r/ntn

[ (Dn . . . u . .D rn ) . . . (D i t . . n , t .4 .

Htm (2.33)

(Dnt . . . Dn i .. D rn) ] r , n '

11?x-.::ir:l :f,he,arsumenrf the ogarithmnEq. 2.33)s rhem,nth

t:?,T,:!"^:.]-tl-*'ofonductortim'ri"L.il"il"#il#:ili:ffi;

l:#1:# *: jr:lr* ryn2.thntofnz ,;d;;;._,i; f ffiilil:;yflr,""1i) T:l

setof nproducterm ertainso"

fii";;;;rrj"",i,sts of

I!:::::1li:*Ti:r{.rnl denominarorsdefin"o, ;;w;;:;;:;,1;::;

!:{Kr,"-li:t:t:y?ofconducorA,andsauureviate;,

"J"r';#;;;":::"*

oo

2o

l I

r )m

GMD is also called geometricmean radius (GMR).lf

In terms of the above symbols,we can write Eq (2.33) as

56 ,1 todern FowerSystemAnalysisI

Inductance nd Resistance f Transmission ines



Lt=2x 10-7h - I i lmDr.t

(2.34a)

'^^Hn^

Note the similarity of the above relation with Eq. (2.22b),which gives the

inductance f oneconductorof a single-phaseine for the specialcaseof two

solid, round conductors. n Eq. (2.22b) r\ is the self GMD of a single

conductor andD is the mutual GMD of two single conductors.

The inductance f the composite conductor B is determined n a similarmanner, and he total inductanceof the line is

L = L e + L n (2.3s)

A conductor s composedof seven dentical copper strands, each having a

radiusr, as shown n Fig. 2.7. Frnd the self GMD of the conductor.

-- D^^= 2..fir

Fig.2.7 Cross-sectionf a seven-strandonductor

Solution The self GMD of the sevenstrandconductor s the 49th root of the49 distances. hus

D, = (V)7 (D1zD'ruDrp r)u (zr)u )t'on

Substituting he values of various distances,

p..- ((tJ.7788r)72212 3 x 22 f x 22rx 2r x 2r)u)t 'o'

srngre ayer oI alurrunlumconductor hown n Fig. 2.8 s 5.04cm. The diameterof eachstrands 1.6gcm.Determine he 50 Hz reactance t I rn spacing; eglect he effectof the centralstrandof steeland advance easons or the same.

Solution The conductivity of steelbeingmuch poorer han hat of aluminium

and the internal nductanceof steelstrandsbeing p-times that of aluminiumstrands, he currentconductedby thecentralstrandsof steelcan be assumed obe zero.

Diameterof steelstrand= 5.04 2 x 1.68= 1.68cm .Thus, all strandsare of the samediameter, say d. For the arrangement of

strandsas given n Fig. 2.8a,

D t z = D r c = d

D r t = D n = J l d

D u = 2 d

D r = (l(+)- dTil,eilf')

(b ) Line composedof two ACSR conductors

(a) Cross-section f ACSR conductor

, _2r(3(0.7788)) t t7 2. t77s -6U+e Fig.2. 8

I

5t I ModernPowerSvstemAnalvsis

Substitutingd' = 0.7188dand simplifying

lnductance nd Resistance i Transmission ines

The self GMD fbr side A is



D ,= l . l 55d = 1 . 1 5 5 1 .68 1 . 9 3 m

D ^ = D s i n c eD > >d

L= 0.46t to e1P - 0.789mH/km" 1.93

Loop nductance 2 x 0.789 = 1.578mHlkm

Loop reactance 1.578x 314 x 10-3 - 0.495ohms/lcm

;;;,;,;I

D,A = ((DnD nD n)(DztDzzDn)(D3tDtrDtr))''eHere.

D, , = Do c= Da t= 2. 5 x 10-3 0.7788m

Substitutinghe values of various nterdistances nd self distancesn D16, we

get

D,A= (2.5 x 10-3x 0.778U3x 4a x 8\tte

= 0.367m

D,B= (( 5 x 10-3 0.778q2 4') t 'o= 0.125m

Substitutinghe values of D^, D6 andDr, in Eq. (2.25b),we get the various

inductances s

L^ = 0.461 os8' 8 = 0.635mHlkm^ " 0.367

L. = 0.461 toe8'8 = 0.85 mH/<mD -

0J25

L = Lt + Ln = 1.485 mH/km

If the conductors n this problem are each composed of seven dentical

strands s n Example2.1, the problemcanbe solved by writing t[e conductcr

self distances s

Di i= 2 '177r t

where r, is the stranci aciius.

2.8 INDUCTANCE OF THREE.PHASE LINES

So far we have considered only single-phaseines. The basic equations

developedan ,however, e easilyadapledo the calculat ion f the nductance

of three-phaseines. Figure 2.10 shows he conductorsof a three-phaseine

with unsymmetrical pacing.

Similarly,

The arrangement f conductorsof a single-phaseransmissionine is shown nFig.2.9, wherein he orwardcircuit is composed f threesolid wires 2.5 mmin radius and the return circuit of two-wires of radius 5 mm placedsymmetricallywith respect o the forward circuit. Find the inductanceof eachside of the line and that of the complete ine.

Solution The mutual GMD betweensidesA and B is

D.= ((DMD$) (Dz+Dz) 1D3aD3))t t6

( ) z 4 m

l i4 m s ( t l II " ' " ' t

SideA SideB

Fig.2. 9 Arrangementf conductorsor Example .3

From the figure it is obvious hat

Dtq= D z q = D z s - D.u= JO am

D r s = D y = 1 0 m

D^ = (682 100) l /6 8. 8 m

Dzsn

Fig. 2.10 Cross-sectional iew of a three-phaseine wi th unsymmetrical pacing

UgdernPowerSystemA

Assume hat there s no neutralwire, so that

lnductancend Resistance f Transmissionines

But, 1, I I, = - /n, hence



I r , + r + I r - 0

Unsymmetricalspacing auseshe lux linkagesand herefore he inductance feachphase o be different esultingn unbalancedeceiving-end oltages ven

en senolng-e tages and llne currents are balanced. AIso voltages will beinduced in adjacentcommunication lines even when line currents are balanced.This problem is tackled by exchanging the positions of the conductors at regularintervais aiong the line such that each conductor occupies the original positionof every other conductor over an equal distance. Such an exchange of conductorpositions is called transposition. A complete transposition cycle is shown inFig.2.11. This alrangementcauseseach conductor to have th e same averageinductance over the transposition cycle. Over the length of one transpositioncycle, the total flux linkages and hence net voltage induced in a nearbytelephone line is zero.

b1

Ao=2 x 10-7 ok(D"D ' t rD t ' ) ' ' '

r'a

D"o (DtzDnDrr)t'' equivalent quilateralpacing

Lo= 2x t0-7 + = 2 x fO-?n{ : r FVmf ' o D ,

This is the same elation as Eq. (2.34a)where Dn,= D"o, the mutual GMD

between he three-phaseonductors.f ro = 11, r6t we have

L o = L O = L ,

It is not he present ractice o transposehe power inesat regular ntervals.

However,an nterchangen the positionof the conductorss madeat switching

stations o balance he inductanceof the phases.For all practical purposes he

dissynrnrctryan bc neglectcd nd he nductancc f an untransposecline can

be taken equal to that of a transposedine.

If ttre spacing s equilateral, hen

D " o = D

and

L o = 2 x l 0 - 7 mI f m nra

If ro = 11, r,-, it follows from F,q. (2.37) that

L o = L t r = L ,

Fig.2.11 A completeransposit ionycle

To find the averagenductance f eachconductorof a transposedine, theflux linkages of the conductorare found for each position it occupies n theiransposeciycie.appiying Eq. (2.s0) to conciuctor of Fig. z.lI, for section1 of th e transposit ionyclewhereina is in posit ion1, b is in posit ion2 andc is in posit ion3, we get

)ur 2 x to- r f, , r " I I , ,m]- * l In 1 j*o- r r , r ,\ " ' ' r , " D r ,

(D t , )

For the second section

- ( | | \

Forher,ira".* :=

'x to-?[r" + * 16nb:

. '' t"],;i

wuv-

)o3 z x ro-71"h !t 16 n*+ r.,nl- l *o-rr-\ f',, " D, uzr /

Average flux linkages of conductor a are

(2.36)

(2.37)

Show that over the ength of one transposition ycle of a power ine, the total

f lux l inkages f a nearby elephonein e ar e z,ero,tr r balancedhree-phase

currents.

b

(l )c

i'-)\7

c'l

r c t l I , , h ! r , , t n\ 4 ,

()T2

- 1 u

(DnDrD l t ) t / 3+ 1 . l n

(D t2D2 . )3 t ) t / 3F19.2.12 Effectof transposi t ion n Induced ol tageof a telephone in e

62 | Modern Power Svstem Anatrrqic.

solution Referring to Fig. 2.r2, the flux linkagesof the conductor r, of thetelephoneine are

lnductance nd Resistance f Transmission ines

(ii) third and multipleof third harmoniccurrenrsunderhealthy ondirion,



) t 2 = 2 x to t l " L n5 * r 6 tn + *1 .1 n + lw b _ r im ( 2 .3 s )( " D o z

- o - -D u , D , , )

" v ' ' t

The net flux linkagesof the telephone ine are

) , = ) , t - ) t z

= 2 x to-r(" hD: ,

* 16 n? *r " ,"+) wb-rim 2.40)rheemf nduced" ,n! .,"p'#ir" *

"3Jti,

"-

D" )

E, = Zr f ) , \ lmfjnder balanced oad conditions, ), is not very large because there is acancellation o a greatextentof the flux linkagesdue to Io, 16and 1r. Suchcancellation oesno t akeplacewith harmonr.

"u.r.ntswhich aremultiplesof

three and are therefore n phase.Consequently, hese requencies, f present.ma y be very roublesome.

lf the power line is f'ully transposedwith respect o the telephone ine

), ,= )" (I ) 4, (tr) ),,(III)3

where ,r(I), )/r(II) ancl ,,(III) are he flux linkages f the elephoneine r, inthe three ransposition ections f the power ine.

Writing for ),,(l), ,\/2(II) nd ),r([l) by repeared seof Eq . (2.3g),we have

A , t = 2 x l 0 - / ( t n + 1 6 + 1 , , ) l n(Dn tDu tD , . i t t 3

Similarly,

= 2 x 1 0 4 ( 1 , + . I b + 1 , ) l n

(Dn2Db2D, , r ) t , ,

)r = 2 x 10-711 ,It ,+ I , ) l n (D"2Db2D, ) r t : (DorDarD,r)r , ,

If Io + Iu + I, = 0, ), = 0, i.e. voltage nduced n the telephone oop is zeroover one transpositioncycle of the power line.

It may be notedhere ha t he condition Iu+ Iu+ I, = 0 i s not satisfied or-(i) rpwer frequencyL-G (line-to-ground ault) currents.where

I o + I u * I r = J [ o

, " ^ * + 1 , l n* . 1

t " * Wb-T/m 2.38)

Similarly,

(2.4r)

where

1 , , (3 )+ oQ)+ I , (3 )= 31(3 )

The harmonic ine currents re troublesomen two wavs:(i ) Inducedernf s proportional o the frequency.

(ii) Higher frequencies ome within the audible ange.Thus there s need o avoid the presence f suchharmoniccurrentson power

Iine from considerationsf the performance f nearby elephoneines.It has beenshownabove hat voltage nduced n a telephoneine runningparallel to a power line is reduced o zero f the power line is transposed ndprovided it carries balancedcurrents. It was also shown that ptwer linetranspositions ineffective n reducing he nduced elephoneine uoitug. whenpower line currents are unbalancedor when they contain third harmonics.Power ine transpositionaparrfrombeing neffective ntroducesmechanicalandinsulationproblems. t is, therefore,easier o eliminate nducedvoltagesbytransposinghe telephoneine instead.n fact, he reader an easilyverify thateven when the power line currents are unbalancedor when t-h"y cgntainharmonics, he voltage nducedover complete ransposition ycle (called abarrel) of a telephoneine s zero.Some nduced oltagewill alwaysbe presenton a telephone ine runningparallel to a power line becausen actu4lpractice

transpositions nevercompletelysymmetrical. herefore,when the lines runparallelovera considerableength, t is a goodpractice o transpose oth powerand telephone lines. The two transpositioncyeles ^re staggered. nd thetelephone ine is transposed ver shorter engthscompared o the power line.

rc-7

i,."rp,"u If

A three-phase,0 Hz,15 km long ine has our No. 4/0 wires 1 cm dia) spacedhorizontally 1.5 m apart n a plane.The wires n orderare carryingcurrents1o,Iu and I, and the fourth wire, which is a neutral, carries ,".o iu.."nt. Thecurrentsare:

Io = -3 0 + 75 0 A

Iu = -25 + j5 5 A

I" = 55 - j105 AThe line is untransposed.

(a) From the undamental onsideration,ind the flux linkagesof the neutral.Also find the voltage nduced n the neutralwire.

(b) Find the voltage drop in each of the three-phase ires.

I64 I ModernPowerSystemAnalysis

t

a b c n

lnductance nd Resistance f Transmission ines

In 2



(:; (. 1 ,r"n.'f(') (:)

t-r.s n.'-. t--1.5 m l*- r.s 1l

Fig.2.13 Arrangementf conductorsor Example .5

Solution (a ) From Frg.2.I3,

Don= 4.5 m, Dbn= 3 m , Dr n = 1.5 mFlux linkages of the neutral wire n are

- ( I 1 r \) - = 2 x 1 0 - 7 1 - l n

' + 1 ^ l n ' * l l n 'l w b - T / m

\ " D o n"

D r n D , n )

Substitutinghe valuesof D,,n,Dg, and D,.n,andsimplifying, we get

An = - 2 x lO-' 0.51 I, + 1.1 u + 0.405 1") Wb-TimSince , = - (Io+ I) (this s easily checked rom the given values),

2,, = - 2 x l0-' (1.1051o 0.695I) Wb-T/mThe voltage nduced n the neutral wire is then

Vn = w),nx 15 x 10 3V

=- j314 x 15 x 10 3x Z x 10-7(1 05Io+0.695) y

Vn = - j0.942 1.105 o + 0.695 ) V

Substituting he valuesof 1, and 16,and simplifying

Vn = 0.942x 10 6 = 100 V

(b) From Eq. (2.30), he flux linkagesof the conductor a are

( r I r \

)o=2x ro-?[,rnl* ru n**r rn | wu-r-\ r 'o " D 2D )

The voltage rop/metren phase canbe writtenas

a v , , = 2 x r 0 - 7 i w ( , - r nI

+ 1 ^ n l * 1 - l n I ) t r , o\ . " f ' o

' 'D

c2 D )

Since r= - (lr,+ Iu), and urther ince o= rb= rr= r, the expressionorAVo canbe written in simplified form

Avo= 2 x to-ty,(r , rn!* romz) v/ m

Similarly, voltage drop/petre of phasesb and c can be written as

AVo= x r;a iulbo_#

AV ,= 2 t o t / , ( 1 , n + 1 , "+ )

Using matrix notation, we can present he result n compact orm

) , r = 2 x I o - 7 , r n t - l 6 t ) = 2 x\ D t D r )

ln Dl r l

ln 2

voltage p o s e 4 r s c a ulated below:

Av o= j2xra- lx r4 rs x rd(6 !9 e (-30+ 50)+0.6e3(-25+is5))

= - (348.6 j204) V

Flux linkagesof conductor 7,

\ , 2= 2 x 10-71 lnD 'D^

A single-phase 0 Hz power ine is supported n a horizontalcross-ann. hespacing between the conductors s 3 m. A telephone ine is supportedsymmetricallybelow the power line as shown n Fig. 2.14. Find the mutualinductancebetween he wo circuits and he voltagenducedper kilometre n thetelephone ine if the current n the power line is 100A. Assume the telephoneIine current to be zero.

Solution Flux linkaees of conductor Z,

ro-1 hD'

D1or

Fig.2.14 Power nd elephoneines or Example .6

Total flux linkage of the telephonecircuit

D2

Dr

\)d/, -Yt'

\-l-l I

" l*-O.e

) , = \ , r \ r z = 4 x 1 0 - ' l l n

66i

Modernpower SystemAnalysisI

A4pt= x tl-i n ! Ul^

lnductaneend Resistancef Transmissionines

eachother.(Thereader an ry otherconfigurationso verity that thesewill lead

the equivalent quilateral pacing s



Dl to low D..) Applying the methodof GMD,

D rn (DnbDbrD.n)' ' t (2.42)

a andb in section of the

transpositionYcle

= (DpDp)rr+ 7DP)tt2

Dt,=mutual GMD between hasesb an d c in sect ion1 of th e

transpositionYcle

= (DP)r;zD,o=mutual GMD between hases an d a in sect ion of th e

r- transPositionYcle

= (2Dh)v2

Hence D" o 2rt6Drt2pr/3htt6 (2'43)

It uray be noteclhere hat D"u t'eurainshe sallle n each sectionof the

transposition ycle,as he conductorsof eachparallelcircuit rotatecyclically,

so do D,,h, DbrandD,.,,. he reader s advised o verify this for sections2 and

3 o f th e transposit ionycle n Fig' 2'15'

Self GMD in section1 of phasea (i.e.,conductors and a/) rs

Drr,= (r'qy'q)t 'o= (r'q)'/z

SerrGMD"t *;:;=',*;;';:^':'ffi'1'are

respectivervi

Drr= (y 'qr 'q) ' ' ' = (r 'q) ' ' t

Equivalent self GMD D, = (Dr,,DrbD,,)rt3

= (r')t''qtt3hrt6 (2.44)

Dr= Q.12+ 2\ t /2 = (5 .2 I ) t / 2

D2 = (L92 + Z\r /2 = (7.61)t /2

vot= o.g2rbs (Jsf ) t "= 0.0758mH/km' -

\ 5 2 1Voltage nduced n the telephone ircuit V,= jttMr,I

lV, l= 31 4 x 0.0758x l0-3 x 10 0 _ 2.3,/9 Vlkm

2.9 DOUBI"E.CIRCUIT THREE.PHASE LINES

It is commonpractice o build double-circuithree-phaseinesso as o increasetransmissioneliability at somewhatenhanced ost. From the point of view ofrrower ransfer ro m on e en d of th e l ine to th e other (seeSec. 12.3), t isdesirableo build the wo lineswith as ow an nductance/phasespossible. norder to achieve his, self GMD (D") shouldbe made high and mutual GMD(D') should be made ow. Therefore, he individual conductors f a phase

shouldbe kept as ar apartaspossible for high self GMD), while the distancebetween phases e kept as ow as permissible for low mutual GMD).Figure2.15shows he hree ections f the ranspositionycleof two parallel

circuit three-phaseines with vertical spacing (it is a very commonly usedconfiguration).

c b ' b

Section1

Becauseof the cyclic rotation of conductors f eachparallel circuit over the

transposition ycle,D. also emains he samen each ransposition ection. he

readershouldverify this for sections2 and 3 in Fig. 2.15.

The inductancePer Phases

L= 2x 10-7 oD ' ,n

Ds

= z x ,ot ,n't'u|l'ltl"tlt)u

( r ' ) t ' ' q r /3hr t6ection2

Fig. 2.15 Arrangementof conductors f a double-ci rcui three-phasein e

It may be noted here that conductors a and a' in parallel compose phase aand sirnilarly b and b'compose phaseb and, and c'compose phasec. in orderto achieve high D" the conductors of two phases are placed diametricallyopposite to each other and those of the third phase are horizontally opposite to

\ 0 ,

oSection 3

o

= 2 x 1 0 - 7 h

The self inductance f eachcircuit

(2,,u(q')"' a)"' ] rvpnur.rrn 2.4s)[ ' \ r ' ) \ q )

is given y

(2) ' ' 'p

rr= 2 x 10-7 n

lnductancend Resisiance f Transmissionines

8-10 times he conductor's iameter,rrespective f the numberof conductorsn no\,v be written asEquation 2.45) can



(2.47)

in the bundle.

dit t , d

Configuration f bundledconductors

=,(t,

+ M )

whereM is the mutual nductancebetween he two circuits. i.e.

M = z x 1 o - 7( z \ ' ' '\ q )

This is a well known result fo r the two coupled curcuits connected in parallel(a t similar polarity ends).

/ \rf h >> ,[ L

l- l unaM '--+0,i.e. hemutuarmpedanceetweenhe ucuits\ q )

becomes zero. Under this condition

" I J2DL = I x 1 0 - ' l n " ' ;

The GMD method, hough applied above to a particular configurationof adouble circuit, is valid for any configurationas long as the circuits areelectricallyparallel.

While the GMD method s valid.forfully transposed ines, it is commonlyapplied or untransposedines and s quite u.rrrui" for practical purposes.

2 1n El rT l . t r r r r : rn AAr?s . rv sv l r r - r r r .q.Lr tetJM,rUUl-L,t(S

It is economical 0 transmit argechunksof power over long dista'cesbyemployingEHV lines.However, the ine voltages hat can be usedare severelylimited by the phenomenon f corona. orona,ln fact, s the resultof ionizationof the atmospherewhena certain ield intensity(about 3,000kv/m at NTp) isreached'Coronadischargecauses ommunication nterferenceand associatedpower oss which can be severe n bad weatherconditions.Critical line voltagefor formation of corona can be raised considerablyby the use of bundledconductors-a group of two or more conductorup". phase.This increase n

crit ical orona oltages clepenclcntn number f concluct<lrsn th cgroup, hcclearancebetweenthem and the distancebetween the groups forming theseparatehases*. eichman[11] hasshown hat the spacingof conductorsna bundle affectsvortagegradient and the optimum spacing s of the order of

L=![, ",u ^ c[.2 *2xro-7rn(t) '^l ,r.ou,

r- \ d /- \\ . /

-

Fig. 2 .16

f " = o . a m l ,, ( ) I ou ' ,

-] s = 0. 4m" "

be) I On '

---1-)d \_..

d l

I

(}

4 m! .

'/^' :' 6/

- s = 0 .

" \ '

-The bundle usually comprises wo, threeor four conductorsarranged n configura-

tions llustrated n Fig' 2.16.The currentwill not divideequally among he conductorsof the bundle unlessconductorswithin the bundleare ruly transposed. he GMDmethod s still fairly accurate or all practicalpurposes.

l -_- d =7 m---*]*- d =7 m--- >l:I

Fig.2.17 Bundled onductorhree-phasein e

Further, because of increased self GMD- line inductance is reduced

considerablywith the incidentaladvantageof increased ransmissioncapacity

of the line.

Find the inductive reactancen ohms per kilometer at 50 Hz of a three-phase

bundledconductor ine with two conductors er phaseas shown n Fig. z.ri.

All the conductors re ACSR with radii of 1.725cm'

Even hough he power inesarenot normally ransposedexceptwhen hey

enter and leave a switching station), it is sufficiently accurate to assume

complete ransposition of the bundlesas well as of the conductorswithin the

bundle)so that the methodof GMD can be applied'

The mutual GMD betweenbundlesof phasesa and b

D o b = @ @ + s ) ( d - i d ) r t 4

Mutual GMD betweenbundlesof phasesb and c

Dh, = D,,,, bY sYmmetrY)Mutual GMD betweenbundlesof phasesc uruJ

D,n = Qd (2d + s ) (2d - t)Zd)tt+

D,o= (DrPaPro)t ' t

= @f @ + s)z(d s)2(2a i(Zd - t))t t"

= GQ)6Q .q 26.0)2 14.4)(13.6))', ' ' ,2

= 8 . 8 1m

The more the number of conductors n a bundle, the more is the self GMD'

70 I n l l a r l a r n D n r e , o ' e r r a + ^ - ^ - ^ r - , ^ .

D.,= (r 'sr ' r ) ' 'o= rk)r /z= (0.77gg 1.725 l0-2x 0.4\t /z= 0.073m

lnductance nd Resistance f Transmissionines

A = cross-sectional rea, n2



Inductive eactance er phaseThe effectiveresistance iven by Eq. (2.48) s equalro the DC resistanceof theconductor given by Eq. (2.49) only if the current distribution is uniformthroughouthe conductor.

r small changesn temperature,he resistancencreaseswith temperaturein accordancewith the relationship

R, = R (1 + oor) (2.s0)where R = resistance t temperature "C

ao = temperature oefficientof the conductorat 0"C

Equation (2.50) can be used to find the resistanceRo at a temperature /2, ifresistanceRr1 at temperaturetl is known

8.8r0.073

= 0.301ohm/kmIn mostcases't is sufficientlyaccurateo use he centre o centredista'ces

betweenbundles ather han mutual GMD betweenbundles or compu ingD"n.with this approximation,we have for the example n hand

D r n = e x T x l 4 y r r t - g . g 2 m

Xr = 3I 4 x 0.461x 10-3 os8'82

" 0.073= 0.301ohmlkm

Thus he approximatemethodyielcls lmost he same eactance alueas heexactmethod' t is instructive o compare he nductive reactance f a bundledconductorin ewith an equivalent onheurist ic asis) ingle onductorine.Fo rthe example n hand, the equivalent ine will have d = 7 m and conductordiameter for same otal cross-sectionalrea)as JT x 1.i25 cm

Xr = 31 4x 0.461x l0-3 6n(7 x7 xl4)t /3

0.7799 J2 xl.725x 10-3

= 0.531ohm/kmThis is 7-6'4rvo igher han he colresponding alue or a bundledconductor

l l t l A A c o l r a n . l " - ^ l - r ^ l - - . r ri i i iv' -\i di iuduJ l iuir i t€c out' iower reactance of a bundled conductor l ineincreasests transmissionapacity.

2.TT RESISTANCE

Though the contribution of line resistance o series ine impedancecan beneglectedn most cases' t is the main sourceof line power loss.Thus whileconsideringransmissionine economy, he presence f line resistance ustbeconsidered.

The effectiveAC resistances givenby

O _ average ower oss n conductorn wattsohms (2.48)

where 1is the rms current n the conductoi n amperes.Ohmic or DC resistances given by the formula

n l

Rn - '" ohlns' A

p = resistivityof the conductor,ohm_m/ = length,m

2.T2 SKIN EFFECT AND PROXIMITY EFFECT

The distribution of current throughout the cross-sectionof a conductor isuniform only when DC is passing hrough it. on the contrary when AC isflowing througha conductor, he current s non-uniformlydistributedover thecross-sectionn a manner hat the currentdensity s higherat the surfaceof theconductorcompared o the currentdensity at its

centre.This effect becornesInorepronounced s requencys increased. his phenonlenons cil led st irrqffect.It causesargerpower loss or a given rms AC than he oss when theSairr€ aiueof DC is flowing ihrough he conciuctor. onsequently,he effectiveconductoresistances more br AC then br DC . A quali tat ive xplanation fthe phenomenons as ollows.

Imagine a solid rottnd conductor (a round shape s considered forconvenience nly) to becomposed f annular ilaments f equal ross-sectionalarea. The flux linking the filaments progressively decreases s we movetowards he outer ilaments br the simplereason hat he lux insidea filamentdoesno t link it. The inductive eactance f the nraginary ilaments hereforedecreases utwardswith the result that the outer filamentsconductmore ACthan the inner filaments (filaments being parallel). With the increase of

frequency he non-uniformity of inductive reactance f the filamentsbecomesmore pronounced,so also the non-uniformity of currentdistribution.For largesolid conductors the skin effect is quite significant even at 50 Hz. Theanalytical study of skin effect requires he use of Bessel's unctions and isbeyond the scopeof this book.

Apart fronl the skin effect, non-uniformityof currentdistribution s alsocaused y proximity eJJ'ect.onsider a two-wire line as shown n Fig. 2.1g.Each ine conductor an be divided nto sections f equalcross-sectionatrea(say hreesections).Pairsaat, bbt and, ct can form three oops n parallel. The

(2.s )

where

(2.49)

I72 | ModernPowerSystemAnaiysis

t

flux linking loop aat (and herefore ts inductance) s the least and t increases

somewhat or loops bbt and ccl. Thus the density of AC flowing through the

Inductance nd Resistancef Transmissionines

b



conductorss highest t he nneredges au') of theconductors nd s the east

at the outer edges cc').This type of non-uniform AC current distribution

Decomes more pronounceo as me olstance Detween conouctors ls reouceo. LlKe

skin effect, the non-uniformity of current distribution causedby proximity effect

also increases he effective conductor resistance.For normal spacing of

overhead lines, this effect is always of a negligible order. However, for

underground ableswhere onductors re ocated lose o eachother,proximity

etfect causes n appreciablencreasen effectiveconductor esistance.

Fig. 2 .18

Both skin and proximity effects depend upon conductor size, fiequency,

distance between conductors and permeability of conductor material.

PROBEVI

Derive the formula for the internal inductance n H/m of a hollow

r;onductor aving nside adius , andoutside adius , andalsodetermine

the expressionor the nductancen H/rn of a single-phaseine consisting

of the hollow conductors described above with conductors spaced a

distanceD apart.

Calculate he 50 Hz inductive reactance t I m spacing n ohms/km of a

cable consisting f 12 equal strandsaround a nonconducting ore. The

diameterof eachstrand s 0.25cm and he outsidediameterof the cable

is 1.25 m.

A concentriccable consistsof two thin-walled tubesof mean radii r and

It respectively. erive an expressionor the inductance f the cableper

unit length.A single-phase0 H z circuit comprises wo single-coreead-sheathed

cables aid sideby side; f the centres f the cablesare 0.5 m apart and

each sheath as a mean diameterof 7.5 cm, estimate he longitudinal

voltage inducedper km of sheathwhen the circuit carries a current of

80 0A.

2.5 Two longparallel onductors arrycurrents f + 1 and 1. What s themagneticield ntensity t a pointP, shownn Fig. P-2.5?

Fig.p-2.5

2.6 Two three-phaseinesconnectedn parallelhaveselt'-reactancesf X, andX2. If the mutual reactance etween hem is Xp, what is the effective

reactance etween he two endsof the line?2.7 A single-phase 0 Hz power ine s supported n a horizonral ross-arm.

The spacing between conductors s 2.5 m. A telephone ine is alsosupported n a horizontalcross-armn the samehorizontalplane as thepower ine. The condttctors f th e telephrlncin e ar e of sol id copperspaced0.6 m between centres.The distance between the nearestconductors f the two lines s 20 m. Find the mutual nductance etweenthe circuitsand he voltage erkilometre nduced n the eiephoneine for150 A current lowing over the power ine.

2.8 A telephoneine runsparallel o an untrasposedhree-phaseransmissionline, as shown n Fig. P-2.8.The power ine carriesbalanced urrent of400 A per phase.Find the mutual nductancebetween he circuits andcalculate he 50 Hz voltage nduced n the telephoneine ptsr

m.a b c h b

( r ( r ( , , .l L

f ^ 5 m - - - + + + - - 5 m - - - f - 1 5 m - - - * 1 r n . _ _

Fig. P-2.8 Telephone in e paral lel o a power in e

2.9 A 500 kV line has a bundling arrangement of two conductors per phaseas shown in Fic. P-2.9.

Fig.P-2.9 50 0kV, hree-phaseundled onductorin e

Compute he reactance er phaseof this line at 50 Hz Eachconductorcarries50Voof the phase urrent.Assume ull transposition.

2 . 1

2 .?

2 .3

2 .4

lgq e 1-tgrygf.!yg!U_An4ygp

2.10 An overheadine 50 kms n length s to be constructed f conductors .56cm in diameter, or single-phaseransmission. he ine reactancemust not

Inductance nd Resistance f Transmission ines

REFERECES



exceed31.4 ohms.Find the maximumpermissible pacing.

2.11 n Fig. P-2.1 whichdepicts wo three-phaseircuitson a steel ower here

cal cenre tlnes.

three-phase ircuit be transposedby replacing a by b and then by c, sothat the reactances f the three-phases re equaland he GMD methodofreactance alculations an be used. Each circuit remainson its own sideof the tower.Let theselfGMD of a singleconductorbe 1 cm. Conductorsa and at and other correspondingphase conductorsare connected n

parallel. Find the reactance er phaseof the system.

b b '

o c )l - 1o - - l

r.___zsm___l') .)

Fig. P-2.11

)-.12 A double-ci rcui t three-phaseine is shown in Fig. P-2.I2. Th e conductors

a, a/l b, bt and c, c/ belong to the same phase respectively. The radius ofeach conductor is 1.5 cm. Find the inductance of the double-circuit l ine in

mH/km/phase.

l t

I- l - 1 m

I

b /

I4 m

Il-4 m

II

i-----z.sm---j

l. Electric't t l Transrnission and Distributiort Book, Westinghouse Electric an d

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,

New Delhi, 1997.4. Stevenson,W.D., Elements of Power System Analvsis,4th edn, Mccraw-Hil l , New

York, 1982.

5. Edison Electric Institute, EHV Transmission Line Reference Book, 1968.

6. Thc Aluminium Association, Aluminium Electrical Conductor Handboo,t. New

York, 1971.

1. Woodruff. L.F., Principles of Electric Pov,er Trun.snissiorr, ohn Wiley & Sons,

Ne w York, 1947.

8. Gross, C.A., Power System Analysis, Wiley, New York, 1979.

9. Weedy, B.M. an d B.J. Cory Electric Power Systems,4thedn, Wiley, New york,

1998.

10. Kimbark, E.W., Electrical Transmission of Power and Signals, John Wiley, New

York, 1949.

Paper

I l . Reichman. J.,'Bundled

Conductor Vol tage Gradient Calculat ions, "AIEE Trans.

1959, Pr I I I . 78: 598.

c '

C)

a) [r..,_/ \. _./1 m - l - , r - l

Fig. P-2.12 Arrangementf conductorsor a double-circuithree-phasein e

2.13 A three-phaseine with equilateral pacingof 3 m is to be rebuilt withhorizontalspacing .Dn = ZDn - ZDrr).The conducrors re to be fullytransposed.Find the spacingbetween adjacentconductorssuch that thenew line has the same nductanceas the original line.

2.14 Find the self GMD of threearrangernentsf bundledconductorsshown nFig. 2.16 n termsof the otalcross-sect ionalre aA of concluctorssame

in each case)and the distanced between hem.

a /

l l

>l<

I

a b

(_) ()

i ' 1 m- i - 1 m

Capacitance f Transmission ines

t



V, "= 6eav6- - -q-dv VtL

|.',

) , , ,r r L / t ' \ V

Fig. 3. 1 Electric ield of a lc lng traight onductor

As the potential difference is independent of the path, we choose the path of

integration as PrPP2 shown in thick line. Since the path PP2lies along an

equipo',entral,Vrris obtained simply by integrating along PyP, t.e.

3. 1 INTRODUCTION

The capacitanceogetherwith conductanceorms the shunt admittanceof atransmissionine.As mentioned arlier he conductances the resultof leakageover the surfaceof insulatorsand s of negligible order.When an alternating

voltage s applied o the ine, he in e capacitancerawsa leadingsinusoidalcurrentcalled the chargingcurrent which is drawnevenwhen the line is opencircuitedat the far end. The ine capacitance eingproportional o its length, hecharging urrent s negligibleor l ines es s ha n10 0km long.Fo r longer inesthe capacitance ecomesncreasinglymportantand has o be accountedor .

3.2 ELECTRIC FIELD OF A LONG STRAIGHT CONDUCTOR

Imaginean nfinitely long straight onductor ar removed rom other conductors(includingearth) carryinga unifbrrnchargeof 4 coulomb/metre ength.Bysymmetry, he equipotential urfaces ill be concentric ylinders,while the inesof electrostaticstresswill be radial.The electricfield intensitv at a distancevfrom the axis of the conductors

,= Q v/^2nky

where t is the permittivity* of the medium.

As shown n Fig. 3.1 considerwo pcl intsP, andP, located t cl istances,and Dr respectively from the conductor axis. The potential difference Vn(betweenP, and Pr) is given by

* In SI units the pennitt ivi ty of free space is ko = 8.85 x 10-12 F/m. Relativepermittivity for air is ft, = klko = 1.

( 3 . 1 )

3.3 POTENTIAL DIFFERENCE BETWEEN TWO CONDUCTORS

OF A GROUP OF PARALLEL CONDUCTORS

Figure3.2 showsa group of parallelcharged onductors.t is assumedhat he

conductorsare ar removed rom the groundand are sufficiently removed rom

eachother-,.e. he conductor adii aremuchsmaller han he distances etween

them.The spacing ommonly used n overhead ower ransmissioninesalways

meets heseassumptions. urther, heseassumptionsmply that the chargeon

eachconductor emainsuniformly distributed round ts peripheryand ength.

vrz=l:, ' trav:ht"?u

i,o ---

nn

Fig. 3. 2 A group of paral lel hargedconductors

7S'l

Modern owerSystemAnalysis

The potentialdifference etween ny two conductors f the groupcan hen beobtained y adding he contributions f the ndividualcharged onductors:by

l ' . - ^Capacitance f Transmission ines | 79

r-or



repeatedapplication of Eq. (3.1). so, the potential difference betweenconductorsa and b (voltagedrop from a to b) is

v (3.2)

Each erm n Eq. (3.2) s the potentialdrop from a to b causedby chargeonone of the conductorsof the group. Expressionson similar lines could bewritten for voltage drop betweenany two conductorsof the group.

If thecharges ary sinusoidally, o do the voltages this s the case or ACtransmissionine), the expression f Eq. (3.2) still applieswith charges/metrelengthand voltages egarded s phasorquantities. quation 3.2) s thus validfor instantaneous uantitiesand for sinusoidalquantitiesas well, wherein allcharges nd voltagesarephasors.

3.4 CAPACITANCE OF A TWO.WIRE LINE

Consider two-wire ine shown n Fig. 3.3 excited rom a single-phaseource.The ine developsequalandopposite inusoidal harges n the wo conductorswhich can be represented s phaso$ QoNd qb so that eo = _ eu.

Cot0.0121

trtF/km (3.4b)lo g (D / (ror)t tz)

*lcln tn+.

qon*;. n,h++..

qn" +)

The associated ine charging current is

I,= ju.Co6VnuAllvn

(3.4c)

(3 .5)

(3.6)

(3.7)

(a ) Line- to- l inecapaci tance

l---r )oi . . - -

Cnn",

C . n = C b r = 2 C " a

(b ) Line- to-neut ralcapaci tance

Fi g . 3. 4

As shown in Figs. 3.a @) and (b) the line-to-line capacitance an beequivalently onsidered s wo equal capacitancess senes. he voltageacrossthe ines dividesequallybetween he capacitancesuch hat he neutralpoint nis at the groundpotential. he capacitance f each ine to neutral s thengivenby

l*- -- --Jl - l

F ig. 3. 3 Cross-sectionalie w of a two_wire in e

The potential difference Vo6 can be written in terms of the contributionsmade by qo and q6 by use of Eq. (3.2) with associatedassumptions (i.e. Dl r islarge and ground is lar away). Thus,

V , = 0t) (3 .3)

Since Qr,= - qu , we have

vob=+mL27rk rorb

The line capacitance Cnh s then

C,= Co,= Cb,= 2Cou=,!#A

pflkm

"O___l [__{b

pFkm

, l ; t ( n " h D" ^+ )

The assumptionsnherentn the abovederivat ion re :(i) The chargeon the surfaceof eachconductor s assumed o be uniformly

distributed,but this is strictly not correct.

If non-uniforrnityof chargedistribution s taken nto account, hen

0.0242C,

rcn(+( 4-, )" ' )[ 2 r\ 4 r ' ) )

Qo

Vot

lf D/2r >> , the above expression educes o that of Eq. (3.6)and he errorcausedby the assumptionof uniform chargedistribution s negligible.

(ii) The cross-section f both the conductors s assumedo be circular,while

in actual practice stranded onductorsare used.The use of the radius of the

circumscribing ircle for a stranded onductor ausesnsignificanterror.

"a b-

In (D / (ror) t tz1F/m length of line Q.aa)

q0| Modernpb*gl Jysteq 4!gly..!s

3.5 CAPACITANCE OF A THREE-PHASE LINE WITHEOUILATERAL SPACING

s 1 ^ - 4

a

es Lg-For air medium (k , = l) ,

T '

,,=#ffi p,Flkm



Figure3. 5showsa three-phasein e composed f three clent ical oncluctorsf

(3 .e)

(3.0)

Since here lr e no othcr charges n th e vicinity, thc sunr of 'c l rar_{es n th e threeconductors s zero. Thus q6 * Qr=- Qu ,which when substi tuted n Eq . (3.10)vields

A\_-/b

Fig. 3. 5 cross-section f a three-phase in ewith equi lateral pacing

Using Eq. (3.2) we can write the expressions fo r Vu,,and V,,..as

vub=*(0"

6 2 * t1,,nt

, r, ,,+)

(3.8)

vn ,=i ! * (n "6P - q1 , tn#* r , r j )AddingEqs. (3.8)an d (3.9),we ge t

v,,t, v,,,, 'oolrr, ,

, D+ (q ul ,1 ,t, ,

; ]

V,,h Vn, !+ r"2zTTk r

With balanced three-phase voltages appliedphasor diagrarn ol ' I ,r ig. 3.(r that

Vo u Vor= 3Von

Substi tut ing or (Vot,+ V,,,) trom Eq . (3.12)

v^_ - 4o tn227Tk r

The capacitance of line to neutral immediately follows as

/- Qo 2 Tlc" Vo n ln (D/r)

( 3 .11 )

to the line, it follows from the

i n Eq . (3 .11) ,we ger

(3.r2)

(3.13)

(3 . 4a )

1o (line charging) = ju,CnVnn

(3 .14b)

\ D

(3.1s)

Vac,/

Vab Vac=2 rt co s 30 " V"n= 3 V " n

\V'o

IF

vb

Vr

Fig.3. 6 phasordiagram f barancedhree-phaseortages

3.6 CAPACITANCE OF A THREE.PHASE LINE WITHUNSYMMETRICAL SPACING

For the first,sectionof the transposition ycle

vob +(eotrn!*qur tnf + e,r rn} ) (3. r6a)zTvc\ r-'

Dr z Dr , )\

Vca1

v",

a/

,''h1 " ,t "

, " III

I

II tvun

n " .

6 D

30

! b

. l

E2 ' I ModernPowerSystemAnalysist Capacitancef Transmissionines I m

t



* l r " h o - 3 L + a a tn L( 3 .18 )

(3 .1e)

D rn = (D nDnDT)tl3

Fig. 3.7 Cross-section f a three-phase(ful ly ransposed)

l ine wi th asymmetrical pacing

For the second section of the transposition cycle

- e.zt" l (3.16b)z T r K \ f u z t u t z )

For the third section f the transposition ycle

vob=|- . ( r "r ln-&r-* q6rtn-!- e,t t"+l (3.16c)

z i ' ! \ D y D r , )

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, -120. Thrce more equations can be written equating to zero the

summation of all l ine charges n 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.

With the usual spacing of conductors sufficient accuracy is obtained by

assuming

(3.20)

As per Eq. (3.12) for balanced hree-phase oltages

Vnt Vor= '3Vnn

and also (qu + qr) = - qo

Use of these elationshipsn Eq. (3.20) eads o

l/ ..Qo

lnD"n

o n - Z n k - " r

The capacitance f line to neutralof the transposedine

C. = 3s-: -2nk F/m to neutral" Von ln (D"o r)

Forairmediumo;=

-'uut--n = -- p,F/km to neutral" lo g (D,o r)

It is obvious hat for equilateral pacingD,, = D, the above(approximate)formula gives the exact resdlt presentedearlier.

The line charging current or a three-phaseine in phasor orm is

Io Qine charging) = jut,,Vnn Alkm (3.23)

3.7 EFFECT OF EARTH ON TRANSMISSION LINECAPACITANCE

In calculating he capacitance f transrnissionines, he presence f earthwasignored,so far. The effect of earth on capacitance an be conveniently akeninto accountby the methodof images.

Method of Images

The electric field of transmissionine conductorsmustconforrn o the presenceof the earthbelow. The earth or this purposemay be assumedo be a perfectty

r / - - l ( ^ r ^ D " n , t j'o,=

lf i ln,tn;*n, r"

,* )

Adding Eqs. (3.18)an d (3.19),we ge t

vo t # vo , =*( ,, 6 2r t r (qt -r q")rn

t')

r , / r ^ \ f-

% )

(3.2r)

is then givehby

(3.22a)

(3.22b)Q a t = Q a 2 = Q a 3 = Q " i Q u t = Q a z = Qu z = Q o ,

4 c t = Q , ' 2 = 4 , 3 = 4 r ' (3 . t7 )

This assumptionof equaicharge/unit ength of a ine n the threesectionsof the

transpositioncycle requires,on the other hand, three diff'erentvalues of Vnu

designated s Vo61, o62andVoo, n the three sections. he solutioncan be

considerablysimplifiedby taking Vou s he average f these hreevoltages, .e.

v ,q ,@vg)=! {v , , u r+ vour+ vooz)J

or ' I l- ( DtzD?tDy1*o"rn r'

) nb=6 q1aln'

n[ ,, )

'"' \ Dt2D23D3t)

( DrrDr^D^, 1+ q ^ l n l - : l l

\ DnDnD3t ))

E4 | ModernPowerSYstemAnalYsis

conducting horizontal shdet of infinite extent which thereforeacts like an 2h D (3.2s)'l

ubstituting the valuesof different chargesand simplifying' we get



equipotential urface.and unit

is such that it has a zero potentialplane midway between he conductorsas

shown n Fig. 3.8. If a conductingsheetof infinite dimensionss placed at the

zero potentialplane, the electric field remains undisturbed.Further, if the

conductor carrying charge -q is now removed, the electric field above the

conductingsheetstays ntact, while that below it vanishes.Using thesewell

known results n reverse,we may equivalently eplace he presence f ground

below a chargedconductorby a fictitious conductorhavingequaland opposite

chargeand ocatedas ar belowthe surfaceof groundas heoverhead onductor

above it-such a fictitious conductor s the mirror image of the overhead

conductor.This methodof creating he sameelectric ield as n the presence f

earth s known as the methodof imagesoriginally suggested y Lord Kelvin.

Zero potentialplane ground)

Fig. 3 .8 Electric ield of tw o long,paral lel ,opposi tely hargedconductors

Capacitance of a Single'Phase Line

Consider single-phasein e shown n Fig. 3.9.lt is requiredo calculatets

capacitanceaking he presence f earth nto acoount y the methodof images

describedabove.The equation or the voltage drop Vo6as determinedby the

two charged onductors and, , and heir images a'and b' canbe writtenas

follows:

lmage charge

F i g ' 3 . 9 S i ng | e - p h a s e t r a n s m is s i o n I i n e wi t h i m a g e s

It immediatelY follows that

wa b - F/m line-to-line (3.26a)

V o b =r(4hz + D2)'tz

Radius

F/m to neutral (3.26b)

rnrk

l--t ,

h \ ,

It__

h

III

/'7

irk

and

C n = -2nk

Dln

vub *1, "m2+nrrni* e,,,"gt#Y

*q,,tnGFfUl

41+tO' t4h21st t2

(3.24)

86 ,* ModernpowerSvstemAnalvsis

images.with conductora in position1, b in position2, andc in position 3,

Capacitaneef Transmission ines

The equation for the averagevalue of the phasor %. ir found in a similarmanner.Proceedingon the lines of Sec. 3.6 and using Vou Vo, = 3Von and

Qo Qt * Qc = 0, we ultimately obtained the following expression or the



lnr - ln

h,

(3.27)

Similar equations or Vo6 an be written for the secondand third sectionsof thetranspositionycle. f the fairly occurate ssumption f constant harge er unitlengthof theconductor hroughout he ransmissioncycle is made, he average

valueof Vou or the three sectionsof the cycle is given by

capacitanceo neutral.

F/m to neutral (3.29a)

1

2 7rl(

tnD'u - rr( rn"n"\'),)!t\r ( (hrlhhs)' ' ' )

0.0242C, pPttcmo neutral (3.29b)

lonD'n- ron@"httu')'!t-or

- -o(4h24 ) t t 3

(3.28)

ComparingEqs. (3.22a) and (3.29a), t is evident hat the effect of earth is toincrease he capacitanceof a line. If the conductorsare high above earttrcompared o the distancesamong hem, the effect of earthon the capacitanceof three-phaseines can be neglected.

Calculate the capacitance o neutrallkm of a single-phaseine composed ofNo. 2 single strand conductors radius = 0.328cm) spaced3 m apart and,7.5m above the ground. compare the results obtainedby Eqs. (3.6), (3.7) and

(3.26b).Solution (1) Neglectingthe presenceof earth tEq. (3.6)l

^ 0.0242 n,C , = 3 1 f i t k m

log -

o.oi+z=ff i=o '00817 P'Flkm-

0.328

By the rigorous elationship(Eq. (3.7)]

cno.0242

where D, = (DnDnDrr)"t

"'(+.(#-')"')Since+

=915, the effect of non-uniformityof chargedistribution s almost

negligible.

C" = 0.00817 pFkm

Flg. 3.10 Three-phasein e with images

r r l

_uu I Modern owel€ysteryr_Anslygls-

(2) Consideringhe eff'ect f earth ndneglecting on-uniformity f chargedistributionEq. 3.26b)]

0.0242

Capacitance f Transmission ines | 89-----------l

b c a



c'- o:o?? = o'0082 'Fkm 2.9s3

Note: The presenceof earth increaseshe capacitanceby approximately3parts n 800.

d = 8 m

- h = 6 m

Fig.3.11 Cross-sect ionf a double-circuithree-phasein e

Solution As in Sec. 3.6,assume hat he chargeper conductoron eachphase

is equal n all the three sectionsof the transpositioncycle. For section of the

transposition ycle

V,,n(l)=

(3.30)

r ( l * (

30 0

,Jto4 o32BJLo4

/4h") )

- 897

0.0242

tog(440/0.525)

Exampe 3. 2

A three-phase 0 Hz transmission ine has lat horizontal spacingwith 3.5 mbetweenadjacentconductors.The conductorsare No. 2/0 hard-drawn sevenstrand opper outsideconductordiameter= 1.05cm). The voltageof the lineis 110 kV. Find the capacitance o neutral and the charging current perkilometreof line.Solution D"o= (3.5 x 3. 5x

'7)t 't= 4. 4 m

v l _ 1 0 6n" -u,Cn

-314.o.oos%

= 0.384 x 106O/km to neutral

chargingunent +-(l 19l ]e) x -000

Xn 0.384x 06

= 0.1I Aftm

For section I of the transpositioncycle

v"bGr),lAln.("i*r"#)+c, rn; '"*)

+ a , ( r n j . , " : ) ] ( 3 3 r )

For section II of the transpositioncycle

vturrr)*lr. ('" .^i)+a,rnI.

^I)

+a.(rn;.r"f)f "r ,Average value of Vo6over the transposition ycle is given by

v,*av) iL*[n.'nttrj+* , (###)]

z*l*(" *' ' ;)+c,(rn;'"f)*n.(,n+r";)]

The six conductors f a double-circuit hree-phaseine havingan overall radiusof 0.865 x 10-2m are arrangedas shown n Fig. 3.11. Find the capacitivereactance o neutral and charging current per kilometre per conductor at11 0 V, 50 Hz.

90 | Modernpower SystemAnalysisCapacitancef Transmission ines I gf

-l ' . h = 6 m

=fi,^-,,)h(ffi)"' (3 .33)



i 'B' jh

,' f 'd

Capacitance o neutral per conductor = 2dc

vou* o,= von=fien"- Qt -q,)tn(ffi)' ' ' (3.3s)

3von=*r"( :#) Fig.3.12

3.8 METHOD OF GMD (MODTFTED)

A comparison of various expressions or inductance and capacitance oftransmissionines [e.g.Eqs. 2.22b)and (3.6)] bringsour rhe acr ha t the rwoare sirni lar xcept hat n inductancexpressions e have o us e he ict i t iousconductor adius rt = 0.7788r,while in the expressionsor capacitanceactualconductor adius r is used.This fact suggestshat the methodof GDM wouldbe applicablen the calculationsor capacitance swell provided t is modifiedby using the outer conductor adius or finding D,, the self geometric meandistance.

Exarnple .3 can be convenient ly olvedas underby using he rnodif ledGMD method.

For the first section of the transposition ycle mutual GMD is

Dub= ((ts) (ts))t ta l iglt tz

Db' = Qil'''

Drr, = ( jh)'''

D"n

(D"pop,o)r,t = [(i,grjh)r,t)t,,

ln the first sectionof the transpositionycle self GMD is

D,o = ?f rf )rt4= ?f)'''

D'l' = QAt''

Dr, = (At''D, = (D,oDroD,,)''3= l?3f Arl3fitz

2d ( 2nk

(3.36)

(3.37)

Now h= 6 m; d = 8 m; -/ = 8 m. Referring o Fig. 3.I2, we can write

f , . . ? . , , , 1 2 1 ' 1 / 2

r = l f / ) - + ( o - h \ " 1J i mL \ 2 ) \ 2 ) )

f = ( j '+ h2)r t2 l0 m

g = (72 + 42)rt2 J65 -

Conductor adius overall)= 0.865 x 10-2mSubstituting he values or variousdistances. e have

4 rx7 x 8.85 10-12 106 1000pF km

Total capacitanceo neutral for two conductors n parallel

C n =4r k

1n fz * o s * s to f o o) 3 - l ' / 3[ 1 0 0 x 8 \ 0 . 8 6 s / |

= 0.018 1F/km

QC,= 31 4x 0 '0181 10-6

= 5.68 x 10-6 Ulkm

Chargingcurrent/phase

"t#*x 5.6g x 10{ = 0.361

Chargingcurrent/conductor 0'361

L

cn

4 ltk= 0.1805A/km

Now C n =

F/m

ai , ' l t I

FThis result obviously checkswith the fundamentallyderived expressionnExample3.3. LsL

PROBEMS



3.9 BUNDTED CONDUCTORS

:::f:i::t"^tll:;;;:i;g;:;fi H#,",iHffi,.$l;:i,h*j bundledconductor ine is shown n Fig. 3.13.The conductorsof any onebundle are in parallel, and t is assumed hat the chargeper bundle dividesequallyamong he conductors f the bundleas Drr> r/. Also Drr- d x D*+ d x D12for he same eason. he resultsobtained-withheseurru'rnptions refairly accurate for usual spacings.Thus if the charge on phase a is qo, theconductorsa and a'havea charge

of qolz each;similarly the charge s equallydivided for phasesb andc.

[ ? 1 F - q - - i F d +aQI O,, oQ

I eu, "6 iec/____DP ---->f-

Dzs --------

l' Dy -- '..l

Fig' 3'13 Cross-sectionf a bundledconductor hree-phaseransmissionin e

Now, writing an equ'ation or the voltage from conductor a to condu ctor b,we get

vob=*lotn,(rn4.*ro,

*o.Sq"lr"-u *rn4. )-\

D, Dt,)

or

/r t

+ . q n tn g .n " t n D r r \,, h,* l r ,

,n ' . rro ,. ,

Dt z Dz t

appliedvoltage s barancedhree-phase,0 Hz. Take he voltageof phasea as referencephasor.All conductorshave the same adii. Also find thechargingcurrent of phasea. Neglect the effect of grouno.

Fig.p-3.1

3'2 A three-phaseouble-circuit ine s shown n Fig. p-3.2.Thediameterofeachconductor s 2.0 cm. The line is transffio and carriesbalancedload' Find the capacitance er phase o neutral of the ine.

Q c '

Qu,

aQ

o Q

T2 m

,fI

I2 m

_t

Considering he ine to be ransposed ndproceedingn the usualmanner, hef inal esul twi l l be

',=^ffi

p,Fkmtoeutral

whereDo, = (DnDnD3)In

It is obvious rom Eq.e.aD that the methodof modifiedGMD is equallyvalid n this case as t shouldbe).

3'3 A three-phase, 0 Hz overhead ine has regularly transposed onductorsequilaterally paced4 m apart The"upu.Itun.e

of sucha line is 0.01tFkm'Recalculate the capacitanc. "i kilometre to neutral when theconductorsare in the samehorizontaipranewith successive pacingof4 m and are regularly transposed.

3'4 consider he,500 kv , three-phaseundledconductor ine as shown nFig' P-2'9' Find the capacitivi reactanceo neutral n ohms/kmat 50 Hz.3. 5 A three-phaserernsnri.ssionin c ha s r,rat, orizontar pacingwith 2 rnbetween djacent onductors.The radiusof eachconductors 0.25cm. Ata certain nstant the chargeson the centreconductorand on one of theoutsideconductorsare identicaland voltagedrop between hese denti-cally charged onductors s 775v.xegtecithe effectof ground,and indthe value of the identical charge n coulomblkm at the nstantspecified.3'6 Find the 50 Hz susceptanceo neutralper kilometre of a double-circuit

threephase ine with transposition s shown n Fig. p_3.6.Given D = Jmand radiusof eachof the six conductorss 1.3gcm.

(3.3e)

(3.40)

94 i - Modern o*", Syrt"r An"lysi.t

6 6 6 5 6 5o - o-*_ D >)<--D 'l-- o-J



l'- +--

3 . 8

. P-3.6 Doublecircui t hree-phase ine wi th f lat spacing

3.7 .{ singleconductorpower cable has a conductorof No. 2 solid copper(radius= 0.328 cm). Paper nsulation separatinghe conductor rom theconcentric lead sheath has a thickness of 2.5 mm and a relativepermittivity of 3.8. The thicknessof the lead sheath s 2 mm. Find the

capacitivereactance er kilometre between he inner conductorand thelead sheath.

Find the capacitance f phase o neutralper kilometre of a three-phaseline having conductors f 2 cm diameterplacedat the cornersof a trianglewith sides5 m, 6 m and 7 m respectively. ssume hat the ine is fullytransposedand carriesbalanced oad.

Derive an expression or the capacitance er metre length between wolong parallel conductors, ach of radius,r,with axes separated y adistanceD, whereD ,, r, the insulatingmedium being air. Calculate hemaximum potential ifference ermissible etweenhe conductors,f theelectric ield strength etween hem s not to exceed25 kY lcm, r being0.3 cm and D = 35 cm.

REFERECES

Books

l . Stevenson,w.D., Elementsof Power SystemAnalysis,4th edn, McGraw-Hil l , New

York, 1982.

Cotton, H., and H. Barber, The Transmission and Distribution of Electrical

Energy,3rd cdn, Hodder and Stoughton, 1970.

Starr, A.T., Generation, Transmission and Utilization of Electric Power, Pitman,

1962.

Papers

Parton, J.E. and A. Wright, "Electric StressesAssociated with Bundle Conduc-

tors", International Journal of ELectrical Engineering Education 1965,3 :357.

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 representingall the three phases

becomes oo complicated or a systemof practical ize,so much so that t mayno longer convey the information it is intended o convey. It is much morepractical o represent power systemby meansof simple symbbls or each

component esulting n what is calleda one-linediagram.Per unit system leads to great simplification of three-phasenetworks

involving ransformers.n impedance iagram rawnon a pe runit basis oes

not require deal transfbrrnerso be included n it.

An important lementof a powersystem s thesynchronous achine,whichgreatly nfluences he systembehaviourduringboth steady tateand ransientconditions. he synchronousmachinemodel n steady tate s presentedn thischapter.The transient model of the machine will be presentedn Chapter9.

4.2 Single-Phase Solution of Balanced Three-Phase Networks

The solution of a three-phase etwork under balanced onditions s easilycarriedout by solving he single-phase etworkcorrespondingo the ref'erence

phase. igure4.1 showsa simple,balancedhree-phaseetwork.The generatorand load neutrals are therefore at the samepotential,so that In = 0. Thus theneutral mpedance Zn does not affect network behaviour.For the referencephasea

En = (Zc+ ZL)I '

3 . 9

2 .

3 .

AT.

5 .

( 4 . 1 )

The currentsand voltages n the other phases ave the samemagnitudebut

are progressively hifted n phaseby 120".Equation 4.1) conespondso the

single-phase etwork of Fig. 4.2 whose solutioncompletelydetermines hesolutionof the three-phase etwork.

Modernpower SystemAlelygis

Ia

\;

\ e , t L_-l - .*-'- In=o

Representationf PgryerSystemComponents

If the transformers YIA connectedas n-Fig. 4.4a, the delta sidehas o bereplacedby an equivalent tar connection s showndottedso as to obtain he

-l

,",lrJ'ZL



*c) zntrb

16 Z1

Ic

single-phaseequivalent of Fig. 4.4b. An important fact has, however, to be

AN m0 lrne culTent /A

have a certainphaseangle shift- from the starside values Vorand Io(90" forthe phase labelling shown). In the single-phaseequivalent (Vew, I) arerespectively n phasewith (Von, o). Sinceboth phasevoltage and line currentshift through the samephaseangle from star o delta side, the transformerperphase mpedanceand power flow arepreservedn the single-phase quivalent.In most analytical

studies,we are merely nterestedn the magnitudeof voltagesand currentsso that the single-phase quivalentof Fig. 4.4b is an acceptableproposition. Whereverproper phaseanglesof currentsand voltagesare needed,correction can be easily applier after obtaining he solution through a single-phase ransformerequivalent.

(a) Y/A ransformer ithequivalenttarconnection

!:Fig. 4.1 Balancedthree-phasenetwork

Ea

Fig' 4. 2 Single-phase quivalent f a balanced hree-phase etworkof Fig.4. 1

Z6

€ l >

Ia

->, A

Ia

(a)Three-phase /y ransformer

(b) Single-phase quivalentof 3-phasey/y transformer

n _

(b) Single-phaseequivalentof Y/A transformer

Fig.4.4

It may be noted here that irrespective of the type of connection, thetransformation ratio of the single-phaseequivalentof a three-phase ransformeris the same as the line-to-line transformation atio.

'See Section 10.3.

Fig. 4. 3

tro IYlii

ModernPowerSystemAnalysisI

4.3 ONE.IINE DIAGRAM AND IMPEDANCE OR REACTANCEDIAGRAM

A one-linediagram f a powersystem shows the main

Representationf PowerSystemComponentst

99

The impedancediagramon single-phase asis or useunderbalanced perating

condit ions an be easilydrawn ro m th e one-l ine iagram.Fo r th esystemof

Fie. 4.5 the mpedance iagram s drarvn n Frg.4.6. Single-phaseranstbrmer

are shown as ideal transformerswith transformer impedances



Y r

iv--

qr

, Fig. 4. 5 One-l ine epresentation f a simple power system

Generator o. 1 30 MVA, 0. S V,X' = 1. 6ohms1 Generator o.2 : 15 MVA, .6kV ,X' = 1. 2ohms

Generator o.3: 25 MVA, .6kV ,X' = 0.56 hm sTransformer1 3phase) 15MVA, 3/ 1 kV ,X = 15.2 hm s erphase n high ension id eTransformer r( 3phase): 5MVA,3316.2V,X= 16 ohms erphase n rignension id eTransmissionine: 0. 5ohms/phaseLoadA : 15 MW,11 kV ,0. 9 agging ower actorLoadB: 40 MW, .6 kV ,0.85 laggingower ac tor

Note: Generatorsre speci f iedn hree-phase VA , ine-to-l ineol tage ndpe rphasereactance equivalenttar). ransformersre speci f iedn hree-phase VA , ine-to-l inetransformationatio, nd pe rphase equivalenttar) mpedance n on eside.Loadsar especifiedn hree-phase W, ine-to-lineoltage nd power actor.

I- r_-1 --i--tI--ntrn^'"1'"rr' - I

'-litr-d-'r,n'rq J-'rn,A,n-?fi-d\ r- i , I

connectionsand

showrrdepending n the information equired n a systemstudy,e.g. circuitbreakersneednot be shown n a oad flow studybut are a must or a piotectionstudy. Power systemnetworksare represented y one-linediagramsusingsuitable symbols fbr generators,motors, transformersand loads. It is aconvenientpractical way of network representationather than drawing theactual three-phasediagram which may indeed be quite cumbersome and

confusing for a practical size power network. Generator and transformerconnections-star, delta,and neutralgroundingare ndicatedby symbolsdrawnby the side of the representation f these elements.Circuit breakersarerepresented s rectangularblocks.Figure 4.5 shows he one-linediagram of asimple power system.The reactance ata of the elementsare given below thediagram.

T2.t F

f t - f ll F t la t '

__'lr-YA

r l

- >Fr-F. + : : - - ,F_..- :-J ; - f f i

G e n l 'I

T r a n s f o r m e r l l L i n e' r ' T r a n s f o r m e r T 2 - t C ; / C " ' S

LoadA Loadg

equivalents

havebeenneglected. his s a fairly good approximation or mostpower system

studies.The generators re epresented svoltagesourceswith series esistance

an d nductive eactancesynchronous achinemodelwil l be discussedn Sec.

4.6). The transmissionine is representedy a zi-model to be discussedn

Chapter5). Loadsareassumedo be passiveno t nvolving rotatingmachines)

and are represented y resistanceand inductive reactance n series.Neutral

grounding mpedances o not appear n the diagramasbalancedconditionsare

assumed.

Three voltage evels 6.6. I 1 ancl33 kV ) ar e present n this system. he

throughout his book.

4.4 PER UNIT (PU) SYSTEM \

It is usual to expressvoltage, current, voltamperesand impedance of an

electricalcircuit n perunit (o r percentage)f base r ret'erencealuesof these

quantities.The per unit* value of any quantity s defined as:

th eactual alue n an Y nits

the baseor reference alue n the sameunits

The per unit method s particularly convenientn power systemsas he various

sectionsof a power systemare connected hrough translormersand have

different voltage evels.

Consider irst a single-phaseystem.Le t

BasevoltamPerss (VA)s VA

Basevoltage Vu V

Then

(4.2a)

*Percent value= per unit value

Per cent value is not convenient

computations.

T1

I(( 'l

L

Y o

A -

Basecurrent u= -[4)e\/v B

A

x 100.for use

Fig. 4.6 lmpectance iagram of the power system of Fig. 4.5

as the factor of 100 has to be carried in

t i r . n ' l. , rvq.. i l MooernHower ystemAnalysisI

I

(4.e)(4.2b)

Z(ohms)x (kVA),

Basempedance, = Y-4-: Viohms" I B (VA)"

If the actual mpedance s Z (ohms), ts per unit value s given by

Z (oltrns) (MVA)''er unit imPedanceZ (Pu)=

@



, r t r t 2 - , i n ^ n

z(pu) = J--- Z(ohms) (vA)'

ZB V;

For a power system,practical choiceof basevalues are:

Basemegavoltamperes (MVA)B

or

Basekilovoltamperes= (kVA)B

Basekilovolrs = (kv)a

Basecurrent 1,1,000 (MVA)u

(kV)a

when MVA base s changeclrom (MVA)r, oto o (MVA)a' n'*' andtV base

s changed rom (kv)r, oro J (kV)n,new'he nt* p"t unit impedance rom Eq'

.4.9) s given bY

Z(pu)n"*z(pu)o.,.ffi"ffi

ohms

(4.3)

(4.4)

(4.s)

(4.6)

(4.7)

(4 .10)

(4 .11a)

Basempedance- - 1'ooq] GV)r,u_ __Ia

_ GV)3 _ 1,000 (kv)1

(MVA)B (kvA)8

Per Unit Representation of a Transformer

It has been said n Section4.2 thata three-phaseransformer orming part of

a three-phasesystemcan be represented y a single-phase ransformer in

obtainingper phase olutionof the system'The deltaconnectedwinding of the

transformer s replacedby an"quiuaient

starso that the transformation atio of

the equivalentsingle-phaseransformer s always he ine-to-line voltageratio

of the three-Phaseransformer'

F igure4.Tarcpresentsasingle-phasetransfgrmlr intermsofpr imaryand

secondary eakagi reactancesZp artdZ, andan ideal transformer of ratio 1 : a'

The magnetizing mpedance s neglected.Let us choose-a oltamperebase of

(vA)a and voltage baseson the two sidesof the transformer in the ratio of

transformation, .e.

V r s- !Vt, a

(a) Representationf single-phaseransformer'- '

(mignetizingmpedanceeglected)

(b )Perunitequivalentircuit fsingle-phaseransformer

Fig. 4. 7

Z(ohms)x (kVA)u

(kV)? 1,000In a three-phase ystem ather han obtaining he per unit valuesusingper phasebasequantities, he per unit valuescan be obtaineddirectlyUy u.ing three-phasebasequantities.Let

Three-phase asemegavoltamperes (MVA,)BLine-to-line basekilovolts = (kV)BAssuming starconnection equivalentstarcan always be found),

Per unit impedance (pu) - Z (ohms) (MVA),

(kv)"

Base urrentr- l 'ooox(MVA)u oJr 1rv;u

Base mpedance o - l 'ooox(kv)u" J 3 r B

_ GD3 _ 1,ooo(kv)2, ^L_-(MVA)' G"Ab

: onms

-{_J--->

zs 12

(4.8)

-W I todern Power Srrstem -A-nalr-rsie- vr v . v r . ; , r r r q t y s t r

I

Therefore = a (as (VA)a is common)I' 2 8

Representationf Power SystemUomponents [,.t'ff.l-

zz(pu)=-++o'1' zzu zB zru

(4.1 b)

( 4 . 1 1 c )Vr n z Vz a

. La D - --

1, , 1, ,



From Fig. 4.7awe ca n write

V z = ( V t - I f l p ) a - l r Z , (4.12)

We shallconvertEq . (4.12) nto per unit form

Vz(pu) zn= [Vr(pu)Vw - I r(pu)I6Zo(pu)Zru]a

-Ir(pu)IruZ,(pu)Z*

Dividing by vzn throughout and using base elations (4. rra, b, c), we get

Vz(pu)= Vr(pu) I,(pu)Zr(pu)_ Ir(pu)Z,(pu) (4.13)

N o w+ = + = ,Ir 12

orI' o I' u

I r ( p u ) =1 2 ( p u ) = ( p u )

Equation 4.13)can thereforebe writtenas

Vz@u) Vr(pu)- (pu)Z(pu)

Z(pu)=Z,,(pu)+ Z,(pu)

Equation (4.I4) can be represented y the simple equivalentcircuit of Fig.4'7b whichdoesno t requirean deal ransfurmet. oni ia.rablesimplif icat ionhas hereforebeenachievedby the per unit methodwith a commonvoltamperebaseand voltagebaseson the two sides n the ratio of transformation.

Z(pu) canbe determined irectly rom the equivalentmpedance n primaryor secondaryside of a transformerby using he appropriaie mpedancl base.

On primary side:

Z r = Z p + Z , / a 2

Z(pu)+: ! -+L*I z r u Z r o z , o " a 2

But a2Ztn = Zzr

Zr(pu)= Zo(pu)+ Z,(pu) Z(pu)

On secondaryide:

zz= Z, + o2zo

where

(4.r4)

Thus he per unit mpedance f a transformers the samewhethercomputed

from primary or secondary ide so long as the voltagebaseson the two sides

are in the ratio of transformation equivalentper phase atio of a three-phase

transformerwhich is the sameas the ratio of line-to-line voltage rating).

The pu transformer mpedanceof a three-phaseransformer s conveniently

obtainedby direct use of three-phaseMVA baseand line-to-line kV base n

relation(4.9).Any other mpedance n eitherside of a transformer s converted

to pu value ust like Zo or Zr.

Per Unit Impedance Diagram of a Power System

From a one-linediagramof a power systemwe can directlydraw the mpedance

diagramby following the stepsgiven below:

1. Choosean appropriate ommonMVA (or kVA) base or the system.

2. Consider he system o be divided into a number of sections by the

transfbrmers.Choosean appropriate V base n one of the sections.

CalculatekV basesof other sections n the ratio of transformation.

3. Calculate er unit valuesof voltages nd mpedancesn eqchsectionand

connect hem up as pe r he opologyof the one-l ine iagram. he result

is the single-phase er unit impedancediagram.

The above stepsare illustratedby the fallowing examples.

j Example 1

Obtain he per unit impedance reactance) iagram of thepower system of Fig.

4.5.

Solution The per phase mpedance iagramof the powersystemof Fig. 4.5

has been drawn in Fig. 4.6. We shall make some further simplifying

assumptions.

1. Line capacitance nd resistance re neglected o hat t i s represented s

a series eactanceonly.

2. We shall assume hat the impedance iagram s meant or short circuit

studies.Currentdrawnby static oadsundershortcircuit conditionscan

be neglected. oads A and B are herefore gnored.

Let us convert all reactanceso per unit form. Choosea common hree-phase

MVA baseof 30 and a voltagebaseof 33 kV line-to-lineon the transmission

line. Then he voltagebase n thecircuit of generator1 s 11 kV line-to-line and

that n the circuits of generators and 3 is 6.2 kV.

The per unit reactances f various components re calculatedbelow:

(4 .1s)

Transmissionine: 2o.5x3o

GT2

Transformr T,: -l-!,? l_0_

= 0.564

= 0.418

Flepresentationf power Systemeomponents I iliffijI-

Example4.I, we now calculate he pu valuesof the reactances f transfonnersand generators s per relation (4.10):



Transformer Z : 0.209 x

TransformerTt:

Generator :

Generator2:

Generator3:

1 6 x 3 0

aiT= 0.44

1. 6 30

u t7- = o'396

1.2 30

tazr = o'9360.56 30

AZI-= 0'437

TransformetTr:

Generator1:

0.22 ff=0.++

0.43s(10'5i1

=0.3e6( l ) '

G e n e r a t o r 2 : 0 .413 * * ( 6 ' 6 1 2 - =0 . 9 3 6I) 6 .2 ) '

Generator : 0.3214 * i9 x(6'6)1

= 0.431/.r 6.21'

obviously these values are the sameas obtainedalready n Exampre4.r.

4.5 Complex Power

Considera single-phaseoad fed from a sourceas in Fig. 4.9. Let

v - t v t 1 6

r_ t n

t( 6 _

0 )

The reactance iagramof the system s shown n Fig. 4.g.

{-)frL_/-X-fX-)<1J U - _ "U 0 0 0 . _ - - J - -

64 0.44 I i

Fig.4.9 Reactanceriagramf the system f Fig.4.5 (roads eglected)Et ' Ez and E, are per unit valuesof voltages o which the generators re

iJ"',ltl3;3llllrlt""nashortircuittudy,heseil be akeni t /.,"pu (no

volta_eef I I kV in the circuit of generatorthe circuit of generators2 and 3

";,r; i;

Example4.2

The reactancc ata of gencrators nd transtbrmers re usuallyspecified n puor per cent) values,basedon equiprnent atings rather than in actual ohmicalues sgiven n Exampl 4.7; *iit" rhe ransmirr;;; hne rnpedancesraybe

;"";:,[J:Ti]#?l-et

us c-sotveJxarnpre.1b;assuminghe buowing

Transformer ,: 0.209TransformerT): 0.220GeneratorGr: 0.435GeneratorGr : 0.413GeneratorG3: 0.3214

With a baseMVA of 30 . baseI md b:ise oltage f 6. 1 k\ ; in

Source

Fig.4. 9 Complex ower lo w n a single-phaseoa d

When d is positive, he current ags behind voltage.This is a convenientchoiceof signof 0 n power systemswhere oadshavemostly aggingpower

factors.Complex power flow in the directionof current indicated s siven bv

S = V I *

= l V l l l l 0

= lV l l 1l cos d+ j l v l l1 ls in 0= P + ie @.17)or

(a)

t S l ( p 2 + e \ t , ,

ModernPowersystemAnatysisI

Here

S - complexpower (VA, kVA, MVA)

lSl = apparent ower (VA, kVA, MVA); it signifies rating of

nepresentationtpower

systemeompqlents l.i; fiiT-As per Eq.(4.19), Kirchhoff's current aw applies o complex power (also

appliesseparately o real and reactivepowers).In a series RL load carrying current {

V = l ( R + j x r )



I

P = lV l l1 l cos 0 - real (active)power (watts,kW, MW)

Q = lVl l1l sin 0 = reactivepower

- voltamperes eactive (VAR)

= kilovoltamperes eactive(kVAR)

= megavoltamperes eactive (MVAR)It irnmediately ollows from Eq. (4.17) thatQ, the eactivepower, s positive

for laggingcurrent (laggingpower factor oad) andnegative or leadingcunent(leadingpower actor oad).With the directionof current ndicated n Fig. 4.9,.9= P + iQ is suppliedby the sourceand is absorbedby the load.

Eqr-ration4.17)can be representedy the phasor iagramof Fig.4.10 where

, n0 = Lan'' i1 = positive for lagging current

P

:gative br leadingcurrent

P = I"R = active power absorbed y load

Q - IzXr = r-eactive ower absorbedby load

In caseof a seriesRC load carrying current I

P _ I 2 R

O- - IzX, qreactive ower absorbeds negative)

Consider now a balanced hree-phase oad represented n the form of anequivalentstar as shown n Fig. 4.L2.The three-phase omplexpower fed intoload is given by

S = 3vpl-t= 3 lvpt 6pl; : JT lvrlzOrti (4.20)If

I r = l l i l I ( 6 p - A

Then .S= ,'5 lvLl lILl I 0

- Ji tvLtvLt osd+ JT t 0 = P + i Q @ . 2 1 )

Here

Fig. 4.12 Complex ower ed to three-phaseoad

ts l = Ji tvrt tr l

P - Ji tvLl ILt osde = Ji lvLl ILt ind

d - power factor angle

lf vL, he ine voltage, s expressedn kv; and y,the line current n amperes,s is in kvA; and if the line current s in kiloamperes,s is in MvA.

(4 .18)

(4.1e)

Fig.4.10 Phasorepresentat ionf complex ower lagging f oad)

If two (o r more) oadsare n parallelas n Fig.4.ll

s = v F _ v ( i + i )

f Yr i . . r ' r ' ! -pr+ pr) (e t+ez)

In terms of load impedanceZ,

f, _ v , _ l vL l l 6Pr L -z J i z

Substitutingor I, in Eq. (4.20)

= rotor speed synchronous peed) n rpm

= numberof poles

N

P



" - V'l'r -r

tr V,. s in kV, ,S s now given n MVA. Load

calculated rom

, _ l v r l ' _ l v l

"-

Jt --T1O

4.6 SYNCHRONOUSMACHINE

The synchronousmachine s the most mportant elementof a power system. t

convertsmechanicalpower into electrical form and feeds t into the power

network or, in the caseof a motor, t draws electricalpower rom the network

and converts it into the mechanical orm. The machine excitation which is

controllable determines he low of VARs into or out of the machine.Books on

electrical machines 11-51may be consulted for a detailed account of the

synchronousmachine.We shall presenthere a simplified circuit model of the

machinewhich with suitablemodif icat ions herever ecessaryunder ransient

conditions) will be adopted hroughout his book.Figure 4. 13 shows he schematic ross-sectional iagramof a three-phase

synchronousgenerator alternator) aving a two pole structure.The stator has

a balanced hree-phasewinding-aat, bbt and cct. The winding shown is a

concentratedone, while the winding in an actualmachine s distributed across

the statorperiphery.The rotor shown s a cylindrical" one (roundrotor or non-

salient pole rotor) with rotor winding excited by the DC source. The rotor

winding is so arrangedon rotor periphery that the field excitation produces

nearlysinusoidallydistributed lux/pole (d) in the air gap.As the rotor rotates,

three-phase emfs are produced n stator winding. Since the machine is a

balanced ne and balancedoading will be considered,t can be modelledonper phasebasis for the reference hase a.

T n o m q n h i n c r r r i f h m n r a f h q n f r r r n n n l e c f h p q l r n r r e A p f i n p A c f n r n f r r r o r o n p a f cv

l / v rvo ,L l l v qu v v v sv r l ll vu JLr uvLur v rvyvo lo

electrically or everypair of poles.The frequencyof inducedemf is givenby

f =ff i nzwhere

.High-speed urbo-generators ave cylindrical rotors and ow sppedhydro-generators

have salient pole rotors.

winding

Fig.4.13 Schematic iagram f a round otor ynchronousenerator

On no load the voltage EJ nduced n the referencephasea lags 90" behind

dywhichproduces t and s proportional o dyif the magneticcircuit is assumed

to be unsaturated. his phasor elationships indicated n Fig. 4.14. Obviously

the terminal vclltage V, = Er

Il -

E f=V tFig.4.14 Phasor elat ionshipetweenuandE,

As balancedsteady oad is drawn from the three-phase tator winding, the

stator currentsproduce synchronously otating flux Q/poIe (in the direction of

rotation of the rotor). This flux, called armature reaction flux, is therefore

stationarywith respect o field flux Qy.It ntuitively fbllows that Qo s in phase

with phase c current 1o which causes t. Since the magnetic circuit has been

(4.22a)

impedance if required anbe

(4.22b)

Qr

Field winding -..

1)xr\Xp>{

o--F

vlI-T-

:t-F

I

lot

-t-I

\,. \'\r \lQ

NN

,liO* | Modernpowsr Syglem_Anatygis

urrurnJd o be unsaturated,he superposition rinciple s applicable o that heresultantair gap flux is given by the phasorsum

d' = d1+ Q, , @,3)Furtherassuminghat the armatureeakage eactancend resistance re

Rapr"."n,",ion f po*"r' Syrr"r Cornp.on"n,, NEffit -

The circuit of Fig. 4.L6 can be easilymodified o include he effectofarmatureeakageeactancend esistancethese reseries ffects)o give hecompleteircuitmodel f thesynchronousenerators n Fig.4.I7. The otal



Phasordiagram

and voltagesas

re eml whtch equals the termin tage V,. + x l = X s l s c synchronous reactance of the machine.Equation 4.24)now becomes

V, = Et - jlo X, - IoRa

Fig.4.16

This model of the synchronousmachinecan be further modified to accountfor the effect of magneticsaturationwhere the principle of super-positiondoesnot hold.

Fig-4.17 circuitmodelof round otor ynchronousenerator

Armature resistanceRn is invariably neglected n power system studies.Therefore, n the placeof the circuit model of Fig. 4.I7, the simplified circuitmodelof Fig. 4.18will be used hroughouthisbook.The correspondinghasordiagram is given in Fig. 4.i9. The fieici induceciemi Ey eacis he terminal

voltageby the torque(load) angle d This, in fact, is the-condition

for acrivepower to flow out of the generator.The magnitudeof power delivereddependsupon sin d

In the motoring operationof a synchronousmachine, he current1,, eversesas shown n Fig. 4.20, so that Eq. @.25)modifies o

Ef = V, - jIoX, (4.26)

which is represented y the phasordiagramof Fig. 4.2I.It may be noted hatV, now leads lby d, This in fact is the condition or power to flow into motorterminals.

under oaded(balanced)conditionsshowing fluxes, currentsphasorss drawn n Fig. 4.15. (4.2s)

j lX"= - E"

Fig.4.1S phasordiagram f synchronousenerator

Here

d - power factor angle

6 - angle by which Et leads v, called load angre or torque angleWe shall see n Sec.5.10 ha t dmainly determineshe power delivered y

the generatorand the magnitude f E, (i.e. excitation)determines he VARsdeliveredby it.

Becauseof the assumedinearity of the magneticcircuit, voltagephasorE,Eo and v, areproportional o flux phasors dr, doand d, respectively; urthei,voltage phasors ag 90' behind lux phasori. It thereforeeasily follows fromFig.4.15 that phasorAB =- Eois proportionalto (o (and herefore o)and s90' leadin d" @r 1,).With the directionof phasorAB indicatedon the diagram

AB = lo Xo

where X" ir the constantof propotionality.I n t e r r n c n F t h p q l r n r r o r l o f i - i r i ^ - ^ s v l i - ^ ^ r r - - - - - ^ r r - 4 t - - r r tquvvv uvruul rurr vL , l ra, wE ual t urr€ut ly w l- l [ c ) u le l0 i low lng

expression or voltageswithout the need of invoking flux phasors.V, = Ef - jloXo (4.24)

where

Ef = uolrage nducedby field flux Q, alone= uo load emf

The circuit model of Eq . (4.24) is drawn in Fig. 4.16 wherein X, isinterpretedas inductive reactancewhich accounts or the effect of armaturereaction hereby avoiding heneedof resorting o additionof fluxes l&[email protected])1.

&

the'i I ModernPowerSystemAnalysis

The flow of reactivepower ancl erminal voltageof a synchronousmachine

is mainly controllecl y meansof its excitation.This is discussedn detail n

Section5.10.Voltageand eactivepower low are oftenautomatically egulated

- Representationf Power ystem omponentsFIIS':

lvtl ll,,lcos d = constant active oweroutput



uy Y U I L 4 S \ / I v S u r q l v r o \ u v vv v v e ^ v ^ r vr

and by automatic ap changingdeviceson transformers.

F ig .4 .22Synchronousmach ineconnected to in f i n it ebus

It rneans hat since V,l is fixed, the projection /ol cos dof the phasor o on V'

remainsconstant,whiie the excitation s varied'Phasordiagramscorresponding

to high, mediumand low excitationsare presentedn Fig' 4'23' T\e phasor

diagramof Fig. 4.23(b)colrespondso the unitypower actorcase' t is obvious

frot the phasordiagram hat for this excitation

lEJ l os 5= lV)

E1

-l it"x'---"' /

0

(a) Overexcited

c

I

|- - - ' - / 6 6 6 6

l x "I

( ;

It

4.18 Simpl i f ied

round roto

generator

E1

Fig. i rcui tmodel of

synchronous

Fig. 4.20 MotoringoPeration f

sYnchronousmachine

Ia

(b)Normal exci tat ion

(c) Underexcited

Phas o rd i ag ramsc fs y nc h ronous gene ra to r f eed i ngc ons tan tpower as excitation s varied

j IJ 'f '

1I

JV t ' ,

Fig.4.21 Phasor iagram f motoring

oPerat ion

Normally, a synchronous enerator peratesn parallelwith othergeneratorsconnectedo the powersystem. or simplicityof operationwe shallconsider

generator onnected o an inJinitebus as shown in Fig' 4'22' As infinite bus

means argesystemwhosevoltage nd requencyemainconstantndependent

of the power exchange etween he synchronousmachineand the bus' and

independent f the excitationof the synchronousmachine.

considernow a synchronous eneratoreeding onstant ctivepower nto an

infinite b's bar. As the machineexcitation s varied,armature urrent n and ts

angle g, t.e. power factor, change n such a manner as to keep

Fi1.4.23

This is defined snormalexcit t t t i t t rt ortheovetexciterlas e Fig' a'23a)' 'e '

|,8,.1os 6>|v),1, lagsbehind V, so thatthe generatoreedspositive eactive

p o w e r i n t o t he b u s ( o r d r a w s n e g a ti v e r e a c t i v e p o we r f r o mt h e b u s ) ' Fo r t h

Fig.4.19 Phasor iagram f synchro-

nousgenerator

., I"

;iiiiI Modern ower ystemAnalysist

underexcited case Fig. 4.23c), i.e. lErl cos 6 < lV), 1o eads V, so that thegeneratoreeds egative eactive ower nto the bus or drawspositive eactive

^ # . . . . . . . ' @ , ^ ^ l L ^ "j

a ^ { - A l 4 ^ l i ^ / \ ^ ^ A a A u t _ ^ ^ -

Representationf PowerSystemComponents I iilis'.$T.._

or

ln , ll1,l os : E: si n (4.27)



Figure 4.24 shows he overexcitedand underexcitedcasesof synchronousmotor (connectedo infinite bus) with constantpower drawn from the infinitebus. n the overexcitedcase, o leads Vu .e. the motor draws negative eactivepower (or suppliespositive reactivepower); while in the underexcited ase .Iolags V, r.e. the motor draws positive reactive power (or suppliesnegativereactivepower).

E1

(a) Overexcited

V1

--

(b) Underexcited

Fig.4.24 Phasordiagramsof synchronousmotor drawing constantpower asexci tat ion s varied

From the above discussionwe can draw the generalconclusion hat a

synchronousmachine (generatingor motoring) while operating at constantpower suppliespositive reactivepower into the bus bar (o r draws negativer c q e t t v c n r t r x / c r f r n r n f h c h r r c h q r \ r r r h a n n r r c r c v n i f a r l A n r r n r l a r o w n i f p r l m q n l " i - o

I v vv v s U Y v r r v r r v Y v l v l \ v l L v v . / l l l u l l u v t v n v l t v u l l l 4 v l l l t l v

on the oih.. hand, feeds negativereactivepower into the bus bar (or drawspositive eactive ower from the bus bar).

Considernow the power deliveredby a synchronous eneratoro an nfinite

bus.From Fig. 4.19 hi s power s

P = lVtl llol cos 0

The aboveexpression.can e written in a more useful form from the phasor

geometry.From Fig. 4.19

lnA _ rl,lx,sin 90"+ 0)

-sin 6

(4.2e)

(4.28)

The plot of P versus shown n Fig. 4.25, s called the power angle curve.

The maximum power that can be deliveredoccurs at 6 = 90" and is given by

For P ) P** or for 6> 90' the generator alls out of step.This problem(the

stability) will be discussed t length in Chapter12.

Fig.4.25 Power ngle urveof a synchrcnousenerator

Power Factor and Power Control

While Figs 4.23 and4.24 illustrate how a synchronousmachinepower factor

changeswith excitation or fixed power exchange,hesedo not give us a clue

regarding he quantitative alues 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

, infinite bus bar) (4.30)Figure 4.26 shows he phasor diagram for a generatordelivering constant

power to infinite bus but with varying excitation.As lEtl sin dremainsconstant,

the ip of phasorErmovesalonga line parallel o y, as excitation s varied.The

directionof phasor1o s always90o agging I"X, and ts magnitude s obtained

from (l1olX5)/X5. igurc4.27 shows he caseof limiting excitationwith d= 90".

For excitation lower than this value the generatff becomesunstable.

ModernPowerSystemAnalysis Ftepreseniationf Fqwer SystemComponents|

1!1

Salient Pole Synctrronous Generator

A salientpole

from a round

synchronousmachine,as shown n Fig.

rotor machine by constructional eatures

4.29, s dist inguished

of field poles which



/.{1\3

Iaz

Effectof varyingexcitationof generator deliveringconstantpower

to inf ini tebu s ba r

employed n machinescoupled o hydroelectric urbineswhich are inherently

slow-speedones so that ttre synchronousmachinehas rnultiple pole pairs as

different from machinescoupled to high-speedsteamturbines (3,000/1,500

rpm) which havea two- or four-polestructure. alientpolemachineanalysiss

made hrough the two-reaction heory outlinedbelow.

Directaxis

I

' l

Fig. 4.29 Sal lentpole synchronousmachine 4-polestructure)

In a round rotor machine, armatLlre urrent n phase wi th f ield induced em f

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 in-

phase or quadrature current is the same (X,, + X1 = Xr), In a salient polel - , - - - 1 - - - ^ ^ - :- L ^ - - . f t i ^ + L ^ l ^ ^ ^ + ^ l ^ - ^ + L ^

machrne at gap ls non-unllorTn arong IULOI'ljcrlPilury. rL ls Lllc rtrilsL .lruug trrtr

axis of main poles called direct axis) and s the largestalong he axis of theinterpolarregion (called quadrature oxis).Armaturecurrent n quadraturewith

El produces lux along he direct axis and the eluctanceof flux pathbeing ow

(becauseof small air gap), it produces arger flux linkagesper ampere and

hence he machinepresentsargerarmature eaction eactance , (calleddirect

axis reactance)o the flow of quadrature omponent l of armature urrent 1o.

On the other hand,armaturecurrent n phasewith { produces lux along the

quadrature xis and he reluctance f the flux pathbeinghigh (because f large

62

Fig. 4.26

/ , "

V1

Fig. 4.27 Case of l imi t ing xci tat ion f generatordel ivering onstantpower

to inf ini tebus bar

Similar phasor diagrams can be drawn for synchronous motor as well for

constant input power (o r 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 excita-

tion. 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 i s shown in Fig. 4.28. It is seen from this diagranr that as d increases,

current magnitude increases and power t'actor improves. lt wil l be shtlwn in

Section 5.10 that as dchanges, there is no significant change in the flow of

reactive Power'

Locus f Er -E r ' ,--4,

, - - ' l \ - - - j l u z X ", n' l/-

Operationof synchronous eneratorwi th variablepower an d fixed

excitation

Fig. 4.28

4 r i5 I I r l , l - r -

ffil Mod"rn o*r. syrt"t Rn"tyri,

interpolarair gap), t producessmaller lux linkages per ampereand hence he

machine presents smaller armature reaction reactance Xu (guadratureaxis

reactancea X) to the flow of inphase component Io of armature current /o.

Since a salientpole machineoffers different reactanceso the flow of Il and

Representationf power

Resultant



1ocornponentsof armature current Io, a circuit model cannot be drawn. The

phasordiagramof a salientpole generators shown n Fig. 4.30.It can be easily

drawn by following the stepsgiven below:

Fig.4.30 Phasor iagram f sal ient ol esynchronousenerator

1. Draw % *d Io at angle 02. Draw IoRo.Draw CQ = .il,X,t(L to 1,,)

3. Make lCPl- llol Xq and draw the ine OP which gives he direction of Ey

phasor

4. Draw a I from Q to the extended ine OP such that OA = Ef

It can be shownby the above heory that the power outputof a salientpole

generator s given by

lv,l' xo xn)si n26 (4.31)

2XdXq

The first term is the same as for a round rotor machine with X, = Xa and

constitutes he major part in power transfer. The second erm is quite small

(about I0-20Vo)compared o the first term and s known as reluctancepower.P versus d is plotted in Fig. 4.31. It is noticed that the maximum power

output occursat 6 < 90' (about70'). Furt1t"r34

(change n power per unit d 5 '

change in power angle for small changes in power angle), called the

synchronizing ower.cofficient,in the operating egion (r< 70') is larger n

a salientpole niachin.: han in a round rotor machine.

Fig.4.31 powerangre urve or sarient oregenerator

In this book we shall neglect the effect of sariencyand take

X ' = X ' t

in all typesof power systemstudiesconsidered.During a machine transient, he direct axis reactancechangeswith time

acquiring he following distinct valuesduring the complete ransieht.

X/=

subtransientdirect axis reactanceXh = transientdirect axis reactanceX,r = steadystatedirect axis reactanceThe significanceand use of these hreevaluesof direct axis reactancewill

be elaboratedn Chapter9.

Operating Chart of a Synchronous Generator

while selectinga large generator,besides ated MVA and power factor, thegreatestallowable stator and rotor currentsmust also be consideredas theyinfluencemechanicalstresses nd temperature ise. Such imiting parameters nthe operationare broughtout by meansof an operatingchart or- erformancechart.

For simplicity of analysis, he saturation ffects,saliency,and resistance ueignored and an unsaturatedvalue of synchronous eactance s considered.ConsiderFig. 4.32, the phasor diagram of a cylindrical rotor machine. Thelocus of constantllolx,V) and henceMVA is a circle centered tM. The locusof constantEtl (excitation) s alsoa circle centered t O. As Mp is proportionalto MVA,QP is proportional o MVAR and Me to MW, all to the same scalewhich is obtainedas follows.

lv , lE, l, = - 1 ; - s i n d +

I \ , / l n r l o r n D n r r r a r Q r r c i a m A n a l r r c i cr Y r v u ! r r r I v Y Y v r v ) ' s r v r r | , rr r q r t u r u

- 2.O uexcitation



Fig.4.32 Phasor iagram f synchronousenerator

Fo r zeroexcitat ion..e. E.l = 0

- i IoXr '= Y,

or

Io = V,lX,

i.e. lo l =lV)lXr leadingat 90" to OM which correspondso VARs/phase.

Consider ow the chartshown n Fig. 4.33 which s drawn or a synchronous

machine avingXt = 1.43pu. For zeroexcitation, he current s 1 01I.43 0.J

pu, so that he length MO conesponds o reactivepower of 0.7 pu, fixing bothactive and reactivepower scales.

With centre at 0 a number of semicircles are drawn with radii equal to

differentpu MVA loadings.Circles of per unit excitation are drawnfrom centre

M with 1.0pu excitation orrespondingo the ixed terminalvoltageOM Lines

may also be drawn from 0 conesponding o various power factors but for

clarity only 0.85 pf lagging ine is shown.The operational imits are ixed as

fbllows.

Taking 1.0 per unit active power as the rnaximum allowable powel', a

horizontalimirl ine ubc s drawn hroughb at 1.0pu . It is assumedhat he

machine s rated o gire 1.0 per unit activepowerat power factor 0.85 agging

and his tixespoint c. Limitation of the statorcurrent o the corresponding alue

requireshe imit-line to becomea circular arccd aboutcentre0. At pointd the

rotor heatingbecomesmore mportantand the arc de is fixed by the maximumexcitation urrentallowable, n this caseassumedo be lEtl = 2.40 pu (i.e.2.4

t imes y, l).The remaining imit is decided y lossof synchronism t eading

power factors. The theoretical irnit is the line perpendicular o MO at M (i.e.

d= 90o),but in practicea safety margin s brought n to permit a further small

increasen load belore nstabil i ty.n Fig. 4.33,a 0. 1 pu margin s employed

and is shownby the curve afg which is drawn in the following way.

Fig.4.33 operat ing hart or rarge ynchronousenerator,

Considera point h onthe theoreticalimit on the ETl 1.0pu excitationsarc,the pcrwerMh is reducedby 0.1 pu to Mk; the op"ruiingpointmust,however,

still be on rheon the desired imiting curve. This is repeated or otherexcitationsgiving thecurveafg. The completeworking area. hown shaded.s gfabcde. { workingpoint placedwithin this areaat oncedefines he MVA, Mw, MVAR, current,power actorand excitation.The oad angle6 canbe measured sshown n thefigure.

4.7 REPRESENTATION OF LOADS

Loaddrawnby consutnerss the oughest arametero assesscientifically. hemagniiudeof the oad, n iact, changes ontinuouslyso hat he oad orecasting

problern s truly a statisticalone. A typical daily load curve is shown inFig' 1.1. The loads are generally composed of industrial and domesticcomponents. n industrial oad consistsmainly of large hree-phasenductionnlotorswith sulficient oad constancy nd predictable uty .y.lr, whereas hedomestic cladmainly consistsof lighting, heating and many single_phasedevices se d n a randomwa y by houscholders.he design n6 upJrot ionofpower systems oth economicallyandelectricallyaregreatly nfluencedby ttrpnatureand magnitudeof loads.

. 0.85 pf agging

i\E

a-o(E

C]

(Uo

c)F

0. 7

\I

, t l , - , - - -M 9 o . s

Leading

0 .5 1 .0

>- Reactive ower pu) agging

lN

ILocus 1,X"

(circle entreM)

. s a I._IaZ I ModernPowerSvstemAnatr-rsic

In representation f loads or various systemstudiessuch as load flow andstabilitystudies,t is essentialo know the variation f real and reactive owerwith variation of voltage.Normatly in such studies he load is of compositenaturewith both industrialand domesticcomponents. typical compositionof

Representationf pJlfZJ.t

Sotution

BaseMVA = 645, _phase

Base V = 24, ine-to_line



Load volt)L

ASa=i

= 1 pu

Synchronouseactance, = +# = 1.344 u ( 2 4 ) z- ' r -

Full load (MVA) = I pu, 0.9 pf lagging

Load current= generator urrent

Io = 7 pu , 0.9 pf lagging

= 0. 9 - 7 0.436pu

(a ) Excitationemf (seeFig. 4.Ig)

Ef = V, + j XJ "

= 1 1 0 " + j 1 . 3 4 4 0 . 9 - 0 . 4 3 6 )

= 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) Reactivepower drawn by load

Q = VJ,, sin r/

= 1 x I x 0.436= 0.436pu or 0.436x M5

= 281 MVAR

The generatorof Example 4.3 is carrying full load at rated voltage but itsexcitationemf is (i ) increased y 20vo and ii) reducedby 20vo.

Calculaten eachcase(a ) load pf

(b) reactivepower drawn by load(c) load angle 6

Solution

Full load, 1 x 0 . 9 = 0 . 9 p u

r.99

InductionmotorsSynchronousmotorsLighting andheating

55-757o

5:75Vo

20-30VoThough it is alwaysbetter to consider he P-V and Q-V characteristics f

each of these loads for simulation, the analytic treatment would be verycumbersomeand complicated.In most of the analytical work one of thefollowing threeways of load representations used.

(i) Constant Power Representation

This is used n load flow studies.Both the specifiedMW and MVAR are akento be constant.

(ii) Constant Current Representation

Here the oad current s given by Eq. (4.17), .e.

I = P : i Q - t n V { <

- " ' l ( 6 - 0 )

where V = lVl 16and 0= tan-l QlP is the power factor angle. t is knownasconstant urrent epresentationecause

he magnitudeof current s regardedas constant n the .study.

(iii) Constant Impedance Representation

This is quite olten used n stability studies.The load specified n MW andMVAR at nominalvoltage s used o compure he oad mpedance Eq. (4.?2b)).Thus I "

z = ! : - w * - l v l 2 tI

- P = J O -P - J Q : T

which then is regarded as constant throughout the study.

f l---- .---l

l Fvattr t l i a e II

F ^ s r . . l s r v zr vI

t - - TA synchronous enerators rated 645 MVA , 24 kv,0.9 pf lagging. t hasasyrrchronouseactance.z o. The generator s feeding full load-at 0.9 pflaggingat ratedvoltage.Calculate:

(a ) Excitationemf (E1) andpower angle6(b ) Reactive ower drawnby the load

Carry out calculations n pu form and convert the result to actual values.P _

Ef=

lM I ftrodernPower SystemAnalysis

(i) Et is

V , = 7

d by 20Vo t same eal oad.Now

(b )

or Q = 0.024x 645= 15.2MVAR

tation of Power

x 0. 9x si n 1. 5= 0.024



As pe r Eq . (a.28)

lE l l v , lP = ' ' s i n d

x,

0.9= (2'388xr) ,i ' d\ 1.344

sin d= 0.5065

6= 30.4"

= 0.89 j0 .79 = 1 .183 - 4L2"

(a) pf = cos 4I.2" = 0.75 agging

(b) ReactivepowerdrawnbY oad

Q = lV,lllol in /

= 1 x 1 . 1 8 3 x 0 . 6 5 9

= 0.78 pu or 502.8MVAR

(ii) E, decreased y 20o/o r

Ef= 1.99 0. 8 = 1 .59

Substituting n Eq. (i )

o e = L 2 \L ) , i n 6\ 1.344

whichgives

649.5"

I n =t.59149,5"-rlo"

4'r Figure P-4'l shows the schematic diagram of a radial transmissionsystem'The ratings

and reactances f the variouscomponentsare shownrherein.A load of 60 MW at 0.9-power factorragging s tapped rom the66 kv substationwhich is to be mainraineoalt oo kv. calcurate rheterminalvoltageof the synchronousmachine.Representhe transmissionline and the transforrnersby series eactances ntu.

PROBEIvI

11t220v

V1 100MVAX = 10o/o

Fig.p_4.1

4.2 Draw the pu inrpedancediagram br the powersystemshown in nig. e-4.2.Neglect esistance,ndusea baseof ioo vrre , 220 v in 50 () rine.The ratings of the generator,motor and transformersare

Generator 0 MVA, 25 kV, Xu = 20VoMotor 50 MVA, I I kV , X,t = 30VoY-Iltransformer,40 MVA, 33 y_220 y kV, X = I5VoY- l transformer, 0 MVA, ll L_220 y kV , X = l5*o

(i )

or

or

2.388130.4"-110"jr.344

220t66 V

dh 160v

_-__-EF_-----l___> 60 MW5 F

|0.9pf agging

100MVA V2X = Bo/o

- 1 f1 M ) Y\ ' )

/I

U-vGi-r+1,-__iF--r-F50o-.*fF-.,

l_. l l - - Y Y - [ y ^

2

Fig. p-4.2

A synchronousgenerator s rated 60 MVA, 1r kv. It has a resistanceRo = 0-l pu and xo

7r.65 pu . It is feeding nto an infinite bus bar ar11 kV deliveringa cirrrent3.15kA at 0.g ff hgging.

(a) Determine E, and angte d(b) Draw a phasor diagram or this operation.(c) Bus bar voltage fails to 10 kv while the mechanicalpower input togeneratorand its excitationremainsunchanged.wtrat is the value

and pf of the current delivered o the bus. In this caseassume he

. i1.344

= 0 . 9 - j 0 . 0 2 4

= 0. 9 -1.5"

= 1; unitypf(a ) pf = co s 1.5"

4.3

I126 | ModernPowerSystemAnalysisT

generatoresistanceo be negligible.

4.4 A 250 MVA, 16kV ratedgenerators feeding nto an nfinite bus bar at15 kV. The generatorasa synchronouseactancef 1.62pu.lt is foundthat the machineexcitationand mechanicalpower input are adiusted ogive E, = 24 kY and powerangle 6 = 30o.

- Representationoflgygr_gyg!"_!L_qg[p!g$q

REFERE.ICES

Books



(a) Determine he ine currentand active and reactivepowers ed to thebu s bars.

(b) The mechanicalpower input to the generator s increasedby 20Vofrom that in part (a) but its excitation s not changed.Find the newline currentand power factor.

(c) With reference o part (a) current is to be reducedby 20Voatthesame power factor by adjusting mechanical power input to the

generatorand ts excitation.DetermineEy, 6 and mechanicalpowerrnput.

(d) With the reduced urrent as n part (c), the power s to be deliveredto bus barsat unity pf, what are ttrecorresponding aluesof El andd and also he rnechanical ower nput to the generator.

4.5 The generator f Problern .4 is feeding150 MVA at 0.85pf lagging oinfinite bus bar at 15 kV.

(a) DetermineEy and d for the above operation.What are P and Q fedto the bus bars?

(b) Now E, is reduced y l0o/okeepingmechanicalnput to generatorsame, ind new dand Q delivered.

(c) Et is now maintainedas in part (a) but mechanicalpower input togenerator s adjusted ill Q = 0. Find new d and P.

(d) For the value of Eyin part (a) what is the maximum Qthat can bedelivered o bus bar.What is the corresponding and {,? Sketch hephasordiagram or eachpart.

Answers

4 . 1 1 2 k V

4.3 (a ) 26.8 kV (l ine),42.3" eading

(c ) i. i3 l -28.8" kA; 0.876 ag

4. 4 (a ) 0.5l l"l- 25.6"kA; 108MW , 51.15MVAR(b ) 6.14 kA, 0.908 agging

(c) 1.578, 3 .5o ,3 .3 MW

(d) 18 .37 v , '35 .5 " , 6 MW

4. 5 (a ) 25.28kV, 20.2 ' ,127.5MW , 79.05MVAR

(b ) 33.9",54 14 MVAR

(c ) 41.1",150.4MW

(d ) 184.45,MVAR,53.6", 7 0.787pu

l ' Nagrath, 'J. and D.P- Kothari, ElectricMachines, nd edn Tata McGraw-Hill,Ne w Delhi , 7997.

2' van E' Mablekos, lectricMachineTheoryor PowerEngineers,HarperlnoRaw,Ne w York, 1980.

3. Delroro, v., Electric Machines and power systems, rentice_Hall,Inc., NewJ e rs ey ,1985 .

4' Kothari,D.P. and I.J. Nagrath, Theoryand.Problemsof Electric Machines, 2ndEdn, Tata McGraw-Hill, New De\hi, 2002.

5. Kothari, D.p. and I.J. Nagrath, Basic Electicar Engineering,2nd Edn., TataMcGraw-Hill,New Delhi. 2002.

Paper

6' IEEE CornitteeReport, The Effectof Frequency nd Voltageon power SystemLoad", Presented t IEEE winter po,ver Meeting,New york, 1966.

ehara-cteristicsnd Perforr"nancef povrerTransmission ines

The following nomenclature as beenadoptedn this chapter:

z = series mpedance/unit ength/phase

y = shuntadmittance/unit ength/phaseo neutral



5. 1 INTRODUCTION

This chapter deals primarily with the characteristicsand performance oftransmissionines.A problemof major mportancen powersystemss the lowof load over transmission ines such that the voltage at various nodes ismaintainedwithin specified imits. While this general nterconnected ystemproblem will be dealt with in Chapter6, attention is presently ocussedonperformanceof a single transmission ine so as to give the reader a clearunderstanding f the principle involved.

Transmissioninesarenormallyoperatedwith a balanced hree-phaseoad;the analysiscan hereforeproceedon a per phasebasis.A transmissionine ona per phasebasiscan be regardedas a two-port network, wherein he sending-end voltage Vr and current 15are related o the receiving-endvoltage Vo andcurrent 1o hroughABCD constants"as

;=;l_*#:,T.":Jff::C = cepacitance/unitength/phaseo neutral

/ = transmissionine length

Z = zl = total series mpedance/phase

Y= ll = total shunt admittance/phaseo neutral

Note: Subscript S tandsor a sending-end uantityand subscriptR stands ora receiving-endquantity

5.2 SHORT TRANSMISSION LINE

For short lines of length 100 km or less, he total 50 Hz shunt admittance*QwCl) s smallenough o be negligible resulting n the simpleequivalentcircuitof F ig. 5.1.

Fig.5. 1 Equivalentircuit f a short in e

This being a simple series circuit, the relationshipbetween sending-endreceiving-endvoltagesand currentscan be immediatelywritten as:

l y r l I l z f lvo l

Lr,JLo JLr-.J (5'3)

The phasordiagram or the short line is shown n Fig. 5.2 for the laggingcurrent case.From this figure we can write

lV5l= l(ly^l cos /o + lllR)z + (lV^l sin dn + lllxyzlr/2

l v5 = [ tvRP+l t2 Rz f ) *z tvR| i l (RcosQ^ + X si n u) r l2 (5 .4)

(s )

/ < t \

These constants can be determined easily for short and medium-length linesby suitable approximations lumping the line impedance and shunt admittance.

For long lines exact analysis has to be carried out by considering thedistribution of resistance, nductance and capacitance parameters and the ABCDconstantsof the line are determined therefrom. Equations for power flow on aline and receiving- and sending-end circle diagrams will also be developed inthis chapter so that various types of end conditions can be handled.

*R"f'..to AppenclixB.

lyr l lA Bl Iy* l

L1,J Lc p lL roJ

Also the following identityholds or ABCD constants:

A D - B C = l

It 2

tvRP

-; - -For overhead ransmissionines, shunt admittances mainly capacitivesusceptance

(iwcl) as the line conductance also called leakance) s always negligible.

= tvntf r* ?JlJrcoso + 4#rin QpL l V R l/ A , I / i - "

-{w

. r l /

r lJ

1il2 n2 x2

The last term is usually of negligible order.

Characteristicsnd Performance f PowerTransmission ines J"#,-1

l/ l R co s ^+l1l X sind^ x 10 0IyR l

In the abovederivation,Q*has been considered ositive for a lagging oad.

(s.7)

Expanding binomially and retaining first order terms, we get | n x tin



(s.8)

(5.e)

(s.s)The above equation s quite accurate for the normal load range.

Fig. 5.2 Phasor iagram f a short ine or lagging urrent

Voltage Regulation

Voltage regulationof a transmission ine is definedas the rise n voltage at the

receiving-end,expressed spercentage f full load voltage,when full load at a

specifiedpower factor s thrown off, i.e.

Per cent egutation

''^?)-:'Yu'x

100 (5.6)lvRLl

where lVool magnitudeof no load receiving-endvoltage

lVprl= magnitudeof full load receiving-endvoltage

Ivs=l frff

cos +ff

sin )'''

l V 5 = l V ^ l + l X ( R c o so + X s i n / o )

(for leading oad)

Voltage regulationbecomesnegative i.e. oad voltage s more thanno load

voltage),when in Eq. (5.8)

X sin Qo> R cos /p, or tan </o leading) >+X

It also follows from Eq. (5.8) that for zerovoltageregulation

per cenr egularion !!4 9"-t t:l 4-r x 100lvRl

ta n n=X

=cot d

i .e . , / ^ ( tead ing=[-

e

where d is the angle of the transmission ine impedance. This is, however, an

approximatecondition.The exact condition for zero regulation is determinedas

follows:

Fig. 5.3 Phasor iagram nder ero egulat ionondit ion

Figure 5.3 shows the phasor diagram under conditions of zero voltageregulation, .e.

lV5 = lVpl

or OC= OA

sn /AOD _AD - AClz -_ llllzl

oA lyR zlvRl

(at a specifiedpower factor)

For short ine, lV^61 lVsl, Vs l = lVpl

Per cent egutat ion'u1|, '%'

lvRl

',

or IAOD= sin-r !4z l vR l

lt follows rom thegeometry f angles t A, that or zerovoltaeeegulation,

/p (leading) =

Characteristicsnd Performance f PowerTransmission ines 133

or ne w valueof lV r = 11.09kV

Figure5.4 shows he equivalent ircuit of the ine with a capacitive eactance

placed in parallel with the load.

R + j x



l l !

l - - , L - - - l

Fig. 5 .4

Assuming cosy''^

now to be the power factorof load and capacitive reactance

taken together, we can write

(11.09 10 ) x 10 3 l1 n (R co s d^ + X sin dn )

Since he capacitanceoes not draw any real power,we have

5000l / o l =

10x cos ^

Solving Eqs. (i ) and (ii), we ge t

co s dn = 0'911 agging

and

l l a l= 5 49A

Now

I c = I n - I

= 549(0.911 j0 '412) 707(0.107 j0.70'7)

= 0.29+ j273'7

Note that the realpartof 0.29 appears ue he approximationn (i ) Ignoring t,

we have

I, = j273.7 A

' Y : l t l - l o x l o o o' ^ L3 w x c l l . I 2 7 3 ' 7

or C - 8 1 P 'F(c) Efficiency of transmission;

(5 .10)

From the above discussion it is seen that the voltage regulation 6f a line isheavily dependent upon load power factor. voltage regulation improves(decreases)as the power factor of a lagging load is increasedand it becomeszero at a leading power factor given by Eq. (5.10).

A single-phase 0 Hz generator uppliesan inductive oad of 5,000 kw at apower factor of 0'707 agging by means of an overhead ransmission ine20 km long. The line resistance nd nductanceare0.0195ohm and 0.63 mHper km' The voltageat the receiving-ends required o be kept constantat 10kv .

Find (a) the sending-end oltageand voltage regulationof the line; (b) thevalue of the capacitorso be placed n parailet viittr the load such that theregulation s reduced o 50voof that obtained n part (a); and (c) compare thetransmissionefficiency in parrs (a) and (b).Solution The line constants

reR = 0.0195 20 = 0.39 f)

X = 3I 4 x 0.63x 10-3x Z0 = 3.96 e

(a) This is the caseof a short ine with I = Ia= 1, given by

l 1 l - - 5 0 0 0 = 7 0 7 A10x0.70i

From Eq . (5.5),

l V 5 = l V o l + 1 l R c o sQ * + X s i n / ^ )

= 10,000 707(0.39 0.701+ 3.96x 0.707, \

=72.175 V

Voltage egulation=pfTL-- l9 - x roo :Zt.j7vo

10

(b ) Voltage regularion esired= ?+t = l0.9Vo

ly s - 10

(i )

(ii)

t 0= 0.109

Case (a )

Case b)

characteristicsnd Pedormance f PowerTransmission ines LI{Sf*

Per unit transformer mpedance,

5000r/ = - g7.7%o 5000 (549)20.39 10-3

- '

/u Lrr6f.LJ prdvrtg , uapacltor ln parallel wlth the load, thereceiving-end ower factor mproves from 0.707 iug to 0.911 ag), the line

Zr=(0 .02+0 .12 )x5

_ (0 .5+13 .75 )x5

6.q2 Q' 2

= (0.0046+ j0.030)pu



current educes from 707 A to 549 A) , the line voitage egulationdecreases(one half the previous value) and the transmission

"ffi"i"nJyi-proves (from

96'2 to 97 7vo)'Adding capacitorsn parallelwith load s a powerfulmethodof improving the performanceof a transmissionsystemand will be discussedfurther towards he end of this chapter.

=f

(O.OU7O.36) (0.02+ /0.12)O/phase

A substation s shown n Fig.5.5 receives MVA at 6 kv , 0.g5 aggingpowerfactor on the ow voltage sideof a transforner from a power station hrough acablehavingper phase esistance nd eactance f 8 and2.5 ohms, espectively.Identical 6.6/33 kV transfoffnersare installed at each end of the line. The6'6 kV sideof the transfonnerss deltaconnectedwhile the 33 kV side s starconnected. he resistance nd eactance f the star connectedwindings are 0.5and3'75 ohms, espectivelyand or the deltaconnectedwindingsarJ0.06 and0.36 ohms.what is the voltageat the bus at the power stationend?

6.6/33 V 33/6.6 V

Fig.5. 5

Solution lt is convenienthere o employ the per unit method.Let us choose,BaseMVA = 5

BasekV = 6.6 on low voltageside

= 33 on high voltageside

Cabie mpeciance (8 + jZ.S) e/phase

_ (8+ r2 . s ) xs(33)'

: = (0'037 io'0r15) u

Equivalentstar mpedanceof 6.6 kv winding of the transformer

Total series mpedance (0.037+ j0.0115)+ 2(0.0046 j0.030)

= (0.046+ j0.072) pu

Given: Load MVA = 1 pu

Loadvol tage + = 0.91 pu6. 6

Load current= -.1- = 1.1 pu0.91

Using Eq. (5.5),we ge t

lV s = 0.91 + 1.1(0.046 0.85 + 0.072x 0.527)

= 0.995pu

= 0.995 x 6.6 - 6.57 kV (line-to-line)

Input to a single-phasehort ine shown n Fig. 5.6 s 2,000kw at 0.8 aggingpower actor.The ine hasa series mpedance f (0.4 + 0.a) ohms. f the oadvoltage s 3 kV, find the load and receiving-end ower f'actor.Also tind thesupplyvoltage.

2,000w Iat 0.8pf *Vs

lassinsL

Fig.5. 6

Solution It is a problem with mixed-endconditions-load voltage and nput

power are specified.The exact solution is outlinedbelow:Sending-endactive/reactive ower = receiving-end ctive/reactive ower +

activekeactive ine lossesFor activepower

lys l1 l co s ds = lVRl l l cos /a + l1 l2p (i )

For reactivepower

l ys l / l s in g55= Vp l 1 l s in Qo + l t l 2X ( i i )

I

I3 k v

I__1

'F6, I ModernPowerSystemAnatysis

Squaring i) and (ii), addingand simplifying,we get

lvrl2 ll 2 = lVnlz l l2 + zlvRl ll 2 (l1lRco s /o+ tItX sh /n ) + tlta @2 + f) (i i i )

the numericalvaluesgiven

Characteristicsnd Performancef Power ransmissioninesl. l3?.,-{I-

5.3 MEDIUM TRANSMISSION I,INE

For linesmore han 100km long,charging urrents ue o shuntadmittance100 km to 250 km leneth. it is

sufficiently accurateto lump all the line admittanceat the receiving-end



l Z l 2 = R 2 + ) = 0 . 3 2

lys l 1 l2 ,oo-ox1o3

- 2 ,500 1030. 8

lVs l1l cos /, - 2,000 x 103

lys l l l l s in /5 = 2 ,500 103 0.6 = 1 ,500 10 3

From Eqs. (i ) and (ii), we get

,n 2 0 0 0 x 1 0 30.4 l tPl 1 l o s P o = F

l1lsin /o1 5 0 0 x 1 0 30 . 4 t I f

3000

Substituting ll the known values n Eq. (iii), we have

(2,500 103;2 (3,000)' il 2 2 x 3,000|210.4*29W"19!p'{lt

L3000

+0.4x1s00xlq1r0.4112l+ o.zz t a3000 J

Simplifying, we get

0.32 Vf - 11.8 106 l l2+ 6.25x 1012 0

whichuponsolutionyields

t | _ 7 2 5 A

Substitutingor l1l in Eq . (iv), we ger

cos Qp= 0.82

Load Pn = lVRl ll cos /a = 3,000 x 725 x 0.82

= 1,790kW

Now

lV 5 = l1l cos ds = 2,000

2000

resulting n the equivalentdiagramshown in Fig. 5.7.

Starting frorn fundamentalcircuit equations, t is fairly straightforward to

write the transmission ine equationsn the ABCD constant orm given below:

[:]l'*;'llVl

Fig. 5.7 Mediumine, ocalizedoad-endapacitance

Nominal-f Representation

If all the shuntcapacitances lumped at the middle of the line, it leads to the

nominal-Zcircuit shown n Fig. 5.8.

Fig.5. 8 Mediumine,nominal-Tepresentat ion

For the nominal-Z circuit, the following circuit equations an be written:

Vc = Vn + Io(Zl2)

Is = In + VrY = In -r Wo+ IR(Z|L)Y

Vs = Vc + it (ZiZ)

Substituting or Vg and 1, in the last equation,we get

( s .1 )

(iv)

(v)

vs= vn+ I^(zt2) (zD)[r^(t +)+

YvR)

= vn( , . { )+ rnz( t . f )l V 5 =

725x0.8:3.44 kY

Rearranging he results,we get the following equations

(s .12+

2

l .- r

Characteristics nd Performance f Power Tralq4lqslon Lines [l$kt

MVA at 0.8 lagging power factor to a balanced oad at 132 kV. The line

conductorsare spaced quilaterally3 m apart. The conductor esistance s 0.11

ohmlkm and its effective diameter s 1.6 cm. Neglect eakance.



Nominal- zrRepresentation

In this methoc he total line capacitances divided nto twoare lumped at the sending- and receiving-ends esultingrepresentations shown n Fie. 5.9.

equalpartswhichin the nominal-zr

= 0.0094pFlkrrr

Fig.S. 9 Mediumine,nominal_7representat ion

From Fig. 5.9,we have

_ t lI s = I n + - V o Y +

2 V s y

R = 0.11x 25 0 = 27.5 C)

X - ZrfL = 2rx 50 x I.24 x 10-3 250= 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 U

r^ = #9 l-- i6.g"o = 109.3 -369"A"13xl32

vo (perphase) (I32/d, ) 10" = 76.2 l0 kv

/ 1 \

vs=[ ++YZ l vR + I *\ 2 ) "

r 1I +: x 7.38 o-a 190"x10r.2134.2"\ to.z\ 2 )+ 101.2 74.2"x 109.3 10-3 -36.9"

76.2 2 .85 1&.2 ' + 11.06 37.3"

82.26 j7.48- 825 15.2"

82.6xJl - 143kv

1 + 0.0374 l&.2" = 0.964 70.01

rv^orrineo oad)ffid:;#

= 148.3v

t z l

Voltage regulation= W#2 x 100 - l2.3vo

5.4 THE LONG TRANSMISSION LINE-RIGOROUS SOLUTION

For lines over 250 km, the fact that the parameters f a line are not lumped but

distributeduniformally throughout ts length, must beconsidered.

V s =

1 s =

vo eo ,)vov>zvn(r.*

r* *tvoy

+t ; t r r - ( t+ |vz)

vov t+ t^vz ) .+ ( ,+ )vz)

f / 1 \

I( '*4 z l,u.r|

'r -' t ',/ | ., l -" | (5.13)

LrU+;r t )[ t * r " ) ]L1nrnominal-zandnominal-rrwith he aboveconstants re

ther.The readershouldverify this fact by applyingstar_eitherone.

at

t o

to

lV5 (line)=

I + L Y Z =2

d tha

each

ion t

l noted

3nt o e

ormati<

J b e

vale

nsfc

ul d t

lurvatrans

ho u

equ

:a tr

It sl

not

delt

Finally, we have

Using the

regulationnominai-z- method, find

of a 250 km, three-phase,the sending-end oltage50 Hz, transmission ine

and voltage

delivering 25

-149 1 mooetn o*", Sv sisI

characteristics nd Performance f PowerTransmission ines

C 1 7 e ) r ' - C z l e - ) ' - 2 1 . ,

9 t , r , -C, , - ' , ' ,Z, Z,

(5.1e)



Fig.5.10 Schematic iagram f a long in e

Figure 5.10 shows one phaseand the neutral return (of zero impedance)ofa transmission ine. Let dx be an elemental section of the line at u dirtunr.

"rom the receiving-endhaving a series mpedancezdx and a shunt admittanceyd-r.The rise in voltage* to neutral over the elementalsection n the direction

of increasing "r is dV". We can write the following differential relationshipsacross he elementalsection:

dVx = Irzdx o, Y- = ZI,

d,lx= v*ldx o,!1'

= yvx

It may be noticed that the kind of connection (e.g. T or r) assumed or theelemental section, does not affect these irst order differential relations.

DifferentiatingEq. (5.14)with respecr o -tr,we obtain

drv , d I.-d-T =i' '

Substitutinghe valueof + fromEq. (5.15),we gerdx

d 2 v

Ea= rZv'

This is a linear differentiai equationwhose general solution can be writtenas follows:

where

V * = C p I * * C r e - 1 x

7= ,lWand C, andC, arearbitrary onstantso be evaluated.

Differentiating q. (5.17)with respecto x:

*-_-Here V' is the complex expressionof the rms voltage, whosemagnitude and phase

vary with distance along the line.

(5.20)

The constantsC, and C2 may be evaluatedby using the end conditions, i.e.when .r = 0, Vr= Vn and 1r= In. Substituting hesevalues n Eqs. (5.17) and

(5.19)gives

Vn = Cr + Cz

r^= (c t - cz)Lc

which upon solving yield

(s.2r)

(s.14)

(s.1s)

(5 .16)

(s.r7)

(s .18)

, r=*

(vn+ zJn)

1Cr =

2Un-ZJ* )

with 9r ant c, as determined bove,Eqs. (5.r7) and (5.19)yield thesolutionor V.-and 1. as

,,= Yn+/o),,.(h.? ),-,.,. _(bITk),,. w*),-,.

Here Z, is called he charqcteristic mpedanceof the line and 7is called thepropagatton constant.

Knowing vp, In and the parametersof the line, using Eq. (5.21) complexnumber rms values of I/, and , at any distancex along the line can be easilyfound out.

A more convenient orm of expressionbr voltageand current s obtainedbyintroducing hyperbolic functions.RearrangingEq. (5.21),we get

( o 7 * + o - 7 ' \ / +v*=vn t+ l+ r^2, (e :J: : )\ 2 /

" ' (2 )

I. = VoL(" *

- . t - ")* ," ( e1 ' + e-t '

\" " 2 " \ 2 ) " \ 2 )

Thesecan be rewritten after ntloducing hyperbolic unctions,as

Vr= Vn cosh 1r + I^2, sinh 1r (5,22)

I,= Io cosh rr + V- -l-.i

Characteristicsnd Performancef PowerTransmissionines fi-

sinh/ | .+*4* ..=Jyz(H+) (s.28a)3 ! ) ! \ ^ ' 6 . )

\

This seriesconverges apidly for valuesof 7t usually encounteredor powerl ines and can he convenie-nflw qrrrrrnwi-o.l .r ^l-^i^-.^ -r,L- --

whenx = l, Vr = V, Ir = Is

expressions or ABCD constanisart



(s.23)Hl l;::,,:::i;:l;;t

HereA = D - c o s h

7 /

B = Z, sinh 7/

c = J-sinh :r/Z,

A = D = l * Y Z

2

Bx, (t .+)

c x Y r * V \\ 6 )

The aboveapproximation is computationally convenientand quite accurate orlines up to 400/500 km.

Method 3

cosh(o/ + i7t)- tat"iot +-'-?t'-iot 1

Z_=

;(e"tp t +

sinh o/ + jpl) - , a t " i 0 t _e :de- i p t

*Incase vs 1s is known fvn Inl can be easily oundby inverting Eq. (5.23).

Hl=l-i,:l[:] (s.2s)

Evaluation of ABCD Constants

The ABCD constantsof a long line can be evaluated iorn the resultsgiven inF4. 6.24). It must be noted that 7 -Jw is in generala complex numberandcan be expressedas

^ 7 - a + j p (s.26)The hyperbolic function of.complex numbers nvolved in evaluating ABCDconstants an be computedby any one of the threemethodsgiven uJtow.

Method I

cosh (cr./+ j1l) = cosh ul cosgl+ 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 numbersas n Eq. (5.27) can belooked tp in standard tables.

Method 2

(s.24)

:){r",

tpt -

(s.28b)

e-"1 -Bt1

(s.2e)t-d l-Bt1

(5.30)

5.5 INTERPRETATION OF THE LONG LINE EOUATIONS

As alreadysaid n Eq. (5.26), 7is a comprexnumberwhich can be expressedas

7 = a + j p

The real part a is called the attenuationconstantand the imaginary partBiscalled the phase constant.Now v, of Eq. (5.2r) can be writtin as

^

V, = IVR+Z, lRl r . , , r i tgx+n

* lVn-Z , ln l o - -c , x , - t t / , x -e2) | 2 | | z I 'where

Qr= I (Izn I&,)

dz = I (Vn- InZ,)

The nstantaneousoltagev*(t) canbe written rom Eq. (5.30)as

v* (t ) = xd Ol!-+3/tl ,o ' i@t+r,,.+o,)

L t 2 l

cosh/ ++* #*. .=(t+)

144 | nrodern Power System Analysis

, nlV^- Zr I *l ^-* -j (at-/ tx+t 1

+Jll tr le-*ett ' ' - tLt-rh)I

(5.31)

The nstantaneousoltageconsists f two termseachof which is a function

of two variables-time and distance.Thus theyrepresentwo travelling waves,

characteristicsnd Performance f PowerTransmission ines

- A t ( f+ 40

S e n d i n g e n d x = /Vx = Vr l * Vx Z (s.32)



Fig. 5.12 Reflected ave

If the load impedance Zr = + = 2,, i.e. the line is terminated n its. IR

characteristic mpedance,he reflected voltagewave is zero (vn- zJn= 0).A line terminated n its characteristic mpedances called the infinite line.

The incident wave under his condition cannotdistinguishbetweena termina-tion and an infinite continuationof the line.

Powersystemengineers ormally caIIZrthe surge mpedance.t hasa valueof about 400 ohms for an overhead ine and its phaseangle normally variesfrom 0" to - 15o.For undergroundcablesZ. is roughly one-tenthof the valuefor overhead ines.The term surge mpedances, however,used h connectionwith surges due o ightningor switching)or transmissionines.where he inesloss can be neglected uch hat

( ; , . , 1r l / 2 i / 1

z,= z,=l: i : r )

(;)"'. ,, purccsist ' 'cc

SLtrge mpetlance Loading (SIL) of a transmission line is.defined as thepower delivered by a line to purely resistive load equal in value to the surgeimpedance of the line. Thus for a line having 400 ohms surge impedance,

Now

At any instant of time t, v.rl s sinusoidallydistributedalong the distance

from the receiving-endwithamplitude ncreasingexponentiallywith distance,

as shown n Fig. 5.11 (a > 0 for a line havingresistance).

Envelopeox

u-rAf

B

S e n d i n g e n d x = /- x = 0 R e c e i v i n g

en d

-'- Directionoftravel l ingwave

Fig.5.11 lncident av e

After t-imeAt, the distribution advancesn distancephaseby (u'Atlfl. Thus

thiswave s travelling owarclshe receiving-end nd s the ncidentv'ave'Line

losses ause ts arnplitucleo decrease xponentiallyn gcling }onr tht: sendirlg

to the receiving-end.

No w

"',,JIlh-?\,* "or{'l

gx+h)

u,r= lt+tlu^ cosaz0x+ z)

(5.33)

(s.34)

After time At the voltagedistribr-rtionetards n distancephaseby (uAtl4.

This s the reflecterlwave ravelling trom the receiving-end o the sending-end

with amplitudedecreasingxponentiallyn going rom the receiving-endo the

sending-end,s shown n Fig' 5.12.

At any point along he line, the voltage s the sum of incident and reflectedvoltagewavespresentat the point tEq. (5.32)1.The same s true of current

waves.Expressions or incident and reflectedcurrent waves can be similarly

writtendown by proceedingrom Eq. (5.21).If Z" is pure resistance, urrent

waves an be simply obtained rom voltagewavesby dividingby Zr .

where Vol s the ine-to-line eceiving-end oltage n kV . Sometirnes,t is

found convenient o express ine loading n per unit of SIL, i.e. as the ratio ofthe power transmitted o surge mpedance oading.

At any time the voltageand current vary harmonicallyalong the linc withrespect o x, the space oordinate.A completevoltageor currentcycle along heline corresponds o a change of 2r rad in the angular argument Bx. Thecorresponding ine length s defined as the wavelength.

It 0 i.s expressedn radlm,

SIL= JT -y!- tv, I x t00okwJ3 x aoo

lvo x looo

= 2. 5 lyRl2 w (s.3s)

-iAd

IModernPo*g!Sy$e!l Anelysis

) - Z n / g m ( s . 3 6 )

Now for a typicalpower ransmissionine

g (shunt onductancelunitength)= 0

' =*fu=

velocitvf ight

The actual velocity of the propagation

somewhates s ha n he velocityof l ight '

., = 1" lo t = 6,000 m

147

(s.42)

of wave along the line would be

7= (yz)1/2 Qu,C(r+ juL))rt2



" - 5 0 -

Practicalransmissioninesare much shorterhan his (usuallya few

hundred ilometre ). It needs o be pointed out heye hat the vyaveslrau'tt n

F i g s , 5 . 1 1 a n d 5 . ] ' 2 a r e f o r i t l u s t r a t i o n o n l y a n d d o n o t p e r t a i n n a r e a l

powertransmission ine'

A three-phase 0 Hz transmissionine is 400 km long' The voltageat the

sending.en ts 220kV. The ine parameters re r = o.!25 ohnr/km,x = 0.4 ohm/

tm an i y = 2.8 x 10-6 nho/km'

Find the following:

( i )Thesending-enclcurrentandrebeiving-endvol tagewhenthereisno- load

Solution The total line parameters re:

R - 0.125x 400 = 50.0 f)

X = O.4 x 400 = 160'0 fl

Y = 2.8 x 10-6x 400 lg)" - l'12 x IA-3 lxf U

Z= R + i X= ( 5 0 ' 0 j 1 6 0 ' 0 )= 1 6 8 '01 7 2 ' 6 " Q

YZ = l.l2 x 1O-390 " x 16 8 172'6"

= 0 .1881162 '6 "

(i) At no-loadVs = AVn' s = CV a

A and C are comPutedas follows:

l l

A = l *t

Y Z= l + * x 0 . 1 8 8 1 6 2 ' , 6 "/ L -

) 2

= 0.91+ j0.028

- iu (LQ''' r- t i)' ' '

7= a+ jg = ju\Lc),, ,(t- #)

Now time for a phasechangeof 2n is 1f s, where/= cul2r s the frequency

in cycles/s.During this time the wave travelsa distanceequal o ). i.e. one

wavelensth.

r ( C \ l t za - - l - l

2 \ L )

0 = a (Lq'' '

C _2ffi0

\Velocity of propagation f wave, , = -4=: f^ m/s

'"'

r i f

which is a well known result.

For a losslessransmissionine (R = 0, G = 0) ,

,= 7yz)''' - iu(LC)tlz

such that e. = A, 0 = . (Lq'''

) - 2 1 1 0 - , ? n = , = :1 , , - m

and

v = fA = ll(LC)rlz m|s

For a single-phaseransmission ine

L= l to ,nD

2 r r '

(s.37)

(s.38)

(5.3e)

(s.40)

(s.41)

ln D/ r

v = 4 ( P o l ^ D 2 * o ) t / 2

'

I t ; , . t -' l t "nG )

Sincer and rt arequite close o eachother,when og is taken, t is sufficiently

accurate o assume hat lnq, = h D/r.

. 14E f ModernPowerSvstemAnalvsis

lA l= 0.91

C = Y(l + YZ/6)=

= 1.09 10-3

Characteristics.nd Performance f PowerTransmission ines I 14g^t-

simplifying, we obtain the maximum permissible requencyas

f = 57.9Hz

If in Example



' lvnhn'#:#

= 242 Y

11 5= lcl lVRl= 1.09x 10-3x y * 10 3= 15 2A"J3

It is to be noted that under no-load conditions, the receiving-endvoltage(242 kV) is more than the sending-end oltage.This phenomenons known asthe Ferrantielfect and s discussed t length in Sec.5.6.(ii) Maximum permissibleno-loadreceiving-endvoltage= 235 kv.

, r , l V " l 2 2 0^ , l % l : f f i =0 . e 3 6

Now

1A = l + L Y Z

2

1 ^= 1 + -; t' , .i2.8x 10-6x (0.125+ 70.4)z

= (1 - 0._56 t0-6P1 j0.t75 x r0-6P

,lth of the real part. zll can beince hc rnaginary ar twi l l bc ess han,

approxirnated s

5.5 the ine is opencircuitedwith a receiving-endvoltag of 220kV, find the rms value and phaseangleof the following:

(a) The incident and reflectedvoltages o neutralat the receiving-end.(b) The incident and reflected voltages to neutral at 200 km from the

receiving-end.

(c) The resultantvoltageat 200 km from the receiving-end.Note: Use the receiving-end ine to neutral voltageas reference.

solution From Example 5.5, we have following line parameters:

r = 0.725 Qlkm; x = 0.4 Olkm; y = j2.g x 10{ Uncm

z = (0.125+ j0.4) Olkm = 0.42 172.6 CI/km

y- 1yz)t '2 (2.8 x 10-6 0.42 /.(g0 + 72.6))t /2

= 1 .08 10-3 gI .3

= (0.163 i1.068)x tO--r 0+ j{ J

a= 0 .163 tO-3 ;f= 1 .06g l0 -3

(a ) At the receiving-end;

For open circuit 1n= 0 ;

IncidentvcrltageVn I ?cl n -

2

- 220/J3= 63.51

2

Reflected voltage - vR-zclR-

)

= 63.5110 " kV (t o neurral)

At 200 km from the receiving-end:

l / l

Incident vol tage= :!u+url t lx l2 l r:2oo m

= 63.51ex p (0.163 l0-3 x 200)

x ex p 01.068x 10-3 200)

lA l= | - 0 .56 l rJ6P= 0.936

p _ r-0.936

0 .56 0 6

/= 338 m

= 0.88

V,,

VR

10" kV (to neurral)

;;,, ,A,=220

250I

A - l * ; x i 1 . l 2 x l 0 - 3 x

Neglecting the imaginary part, we can write

(b )

L ( t o + 1 6 0 , . I )J0 \ s0 )

tA l 1 +" r.r2 1o-3160&=

0.88

ffi0'.I Modern o*e, System natysisf

= 65.62112.2" kV (t o neutral)

Reflected oltageYs-"-'ur-itt 'l

2 l ' , , ,n ,u-

characteristicsndperformanceof powerTransmissionines rsr

/ v ^ , \-

I i.u"'J and urnshroughpositivengle r (represenredy phasor B);

while the reflected voltage wave decreases n magnitude exponentiaily

-Y-E--u,

= 6I.47 l-12.2" kV (to neutral)It is apparent rom the geometryof this figure that the resultant hasor voltageVs is such ha t



(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"

Resultantine-to-linevoltageat 200 km

= 124.2 x J3 - 215.1kV

5.6 FERRANTI EFFECT

As has been llustratedn Exarnple .5, he eff'ectof the ine capacitances tocause the no-load receiving-endvoltage to be more than the sending-endvoltage.The effectbecomesmorepronounced s he ine length ncreases. hisphenomenons known as the Ferranti.effect.A general explanation of thiseffect is advanced elow:

Substituting = / and In = 0 (no-load) n Eq. (5.21), we have

t / -7 5Vn

,at4gt * (< A2 \\ J . - t J I

- lncreasino\.-

Locus of V5wlth

DVpfor I = 0

QF) lVol > lysl.A simple explanationof the Ferranti effect on an approximatebasiscan be

advanced y lumping the nductanceand capacitance arameters f the line. Asshown n Fig.5.14 the capacitances lumpld at the eceiving-end f the line.

F ig.5.14

Here Ir = ,V,

(-+-.\ juCt )

Sincec is small compared o L, uLl can be neglected n comparisoh o yc,tl.Thus

Vn,-at --ifl

2 Now

Magnitudeof voltage ise _ lvrltJ CLf

15 - jVruCl

Vn= Vs - Is QwLl) = V, + V,tj CLlz= vs0+ Jct t2)

=olv.rt+

(s.M)

(s.4s)

Increasing

F i g . 5.13

The above equationshows hat at I = 0, the ncident (E,o)and reflected E o)voltage waves are both equal to V^/2. With reference to Fig. 5.13, as Iincreases,he incidentvoltage wave increases xponential lyn magnitude

where v = 7/J LC. i1the

velocity of propagationof the electromagneticwavealong the line, which is nearly equal to trr" velocity of light.

5.7 TUNED POWER LINES

Equation(5-23)characterizeshe performanceof a long line. For an overheadline shuntconductanceG is alwaysnegligibleand t is sufficientlyaccurate oneglect ine resistanceR as well. with this approximation

7- Jyz = jalLC

cosh 7/ = cosh wlJTC = cos Lt,lrc

\En= Ero= Vpl2

;fiTiI uooern

owerystem narysis

sinh ?/ = sinh jalJ-LC = j sin wtJLC

HenceEq. (5.23) simplifies to

cos lJ LC jZ, sin tJET .r sina'tJLC cosutE

Z,

eharacteristicsnd performanceof powerTr@

t -Z = Z, sinhl / =

= ! .o,nh..,tt2

_1 =Y

ganh1il2)(-tU )_ _ ___{ l



(5.48)

N o w f a l J t C = h r , n = I , 2 , 3 , . . .

lV5l= lVpl

llsl = llal

i.e. the receiving-end voltage and current are numerically equal to thecolresponding ending-end alues, o hat here s no voltagedrop on oad.Sucha line is called a tuned line.For 50 Hz, the length of line for tuning is

1- --!r_.2nrfJ LC

Since ll^frc = y, the velocity of light

Fig.5.15 Equivarent-zetwork f a transmissionin e

For a zr-network hown n Fig. 5.15 [refer ro Eq. (5.13)].

,,1_ (t. It,t,) z,

.Jru.

,Lr,JLr,(,!v,2,)r*ir,r,)lL,_)

According o exactsolurionof a long line [refer to Eq. (5.23)].

fi:I ilH;':"! l'lt,^t 1, , I z,

___.. cosh/JLl^

For exactequivalence,we musthave

Z/= Z, sin h 7/

f * *YtZ=cosh 7/2From Eq. (5.50)

Z' =,l-:.sinn1/ = Z,

tlnh ' : z[*init rr 1vY tJyz-(

7t )11-"'' -sln!-24 is thc fitct.r by which thc scrics rupcduncel'the no'ri'al-z

must be multiplied to obtain thez parameterof the equivalent-a Substituting7 from Eq . (5.50) n Eq . (5.51),we ge r

I

1 *;

YtZ, sitrh 7/ = cosh fl

r=+@))=i^,^,?rx.. .

= 3,000 m, 6,000 m,...

(s.47)

(s.4e)

(s.s0)

(s.51)

(5 .52)

It is too long a distanceof transmissionrorn the point of view of cost and

efficiency (note that line resistancewas neglected n the above analysis).Fora given ine, engthand freouency uning can be aehieved y increa-sing orC, i.e. by adding series nductances r shunt capacitances t severalplacesalong he ine length.The method s impracticaland uneconomicalor powerfrequencyines ancl s adopted or tclephonywhere higher recluenciesreemployed.

A methodof tuning power lineswhich is beingpresentlyexperimentedwith,usesseriescapacitors o cancel he effect of the line inductanceand shuntinductors o neutralize ine capacitance. long line is divided into severalsectionswhich are ndividuatly tuned.However, so far the practicalmethodofimproving ine regulationand power ransfercapacity s to add series apacitorsto reduce ine inductance; huntcapacitors nderheavy oad conditions;andshunt nductorsunder ight or no-loadconditions.

5.8 THE EOUIVALENT CIRCUIT OF A LONG LINE

So far as the end conditionsareconcerned,he exact equivalentcircuit of atransmissionine can be establishedn the form of a T- or zr-network.The parameters f the equivalentnetworkare easilyobtainedby comparing heperfbrmance quationsof a z--network nd a transmission ine in termsof endquantities. L v , -

2 (s.53)

Wl Modern owerSystem nqlysisI

d ( t a n h f l l 2 \ .rnus I is thc lactor by which thc shunt aclmittancc nr ol ' th e

\ i l / z )nominal-n-mustbe multiplied to obtain the shunt parameter Ytl2) of the

equivalent-.

/ 1 \ l

Note hat Ytl | + +Y' Z' I j- sinh 7/ is a consistent quation n termsof the

VR=+20- lz710kv

V J

(a) Short ine approximation:

Vs= t27 + 0.164 ._36.9 x 131.2 72.3



\ 4 ) 2 ,

above alues f Y/and Z/ .

For a line of medium lengthtuth-l!/2 - 1 and

sinh 7/ = 1 s o that thefll2 1t

equivalent- n- network reduces to that of nominal-n-.

Fig. 5.16' Equivalent-T etworkof a transmission in e

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 ^ ^ . 1 { ' l ^ , r , o t r r . l i o o o - . 1 i o f } r o r e f ' n r o r r n i r r p r c q l l . , p m n l n r r e r l< f l l IWP LV LT L\J l \ - ra t . l l l L rW JLUUM ql l u l o t l l t v l v l v l v t (4 l r l v v rL tqr rJ v r^ lH rvJ vs'

Z= 10-' 190" U

Leceiving-endoad s 50 MW at ?20kV,0.8pf lagging.

1o= ,=-+ - l-36.9" = 0.164 -36.9" kAJ3 x220x0.8

A 50 Hz transmissionine 300 km long has a total series mpedance f 40 +

j125 ohmsand a total shuntadmittanceof 10-3mho. The receiving-end oad s

50 MW at220 kV with 0.8 aggingpower actor.Find the sending-end oltage,

current,power and power actor using

(a) short ine approximation,

(b ) nominal-zrmethod,

(c ) exact ransmissionine equation Bq.(5.27)1,

(d ) approximationEq. (5.28b)].

Compare he resultsand comment.

Solution Z = 40 + iI25 - 131.2 172.3" Q

= 14514.9

lYsltin" 25L2kV

Is = In = 0.764 _-36.9 A

Sending-endower actor= cos 4.9"+ 36.9" 41.g.)= 0.745aggrng

Sending-endower JT x 251.2 0.764 0.745= 53.2MW

x@)Nominal-trmethod:

A = D = l + ! y z = . I '

2I + _x l0-r lg0" x l3 l .2 / .72.3"

= 1 + 0.0656 162.3= 0.93g _L2"

B = Z = I 3 L . 2 1 7 2 . 3 "

c = v ( r + ! v z ) = + f z\ 4 / 4

= 0.00 190"+ I . t0-6 Jg0 . x 131.2 72.3"4= 0.001 .90

7s= 0'9381r.2 x r27+ r3r.2 172.3"x 0.164L-36.9"= 719.7 7.2 + 21.5135.4" 73j.4 /.6.2

lVsll in" 238kV

1s 0.001Z9 0 x 12 7+ 0.93gZl.Z x 0.164 ._ 36,9"

= 0.127 9A + 0.154_35.7"= 0.13 .16.5"Sending-endf = co s 16.5. 6.2) = 0.9g4eacl ing

Sending-endpower=T * 23g 0.13x 0.gg4

= 52.7MW(c) Exact ransmissionine equarionsEq. 5.29)).

fl = al + jpt =JyZ

The

Exirmple .7

155 | - Modern owerSystem4nalysis

= Jro- t lgo"* l3 t .z l7z .3" = 0.0554 i0 .3577

= 0 .362 8L2"

cosh al + i 0l ) = !k"' lgt + e*t l-Bt1

'gl =a.5'fi1(radians)- '6.4,g"

eo.0s54e0.49") 1.057zo.49"= 0.99+ j0.37

characteristicsnd Performancef PowerTransmission ines

= 0.938 ll.2 (alreadycalculatedn part (b))

B = z ( t * ! Z \ = Z + | y f\ 6 ) 6

-{ - x 1o-3go " x (13l.D2 1144.6.6

= 731.2 72.3 +

= 131.2 72.3"+ 2.87 -125.4"



e4'oss4 -20.49" = 0.946 l- 20.49"= 0.886- j0.331

gosh7/ = 0.938+ iO.O2 0.938 11.2"

sinh 7/=

0.052+ 70.35=

0.354 181.5"

E l3nnn" .,^L..= 1- : -t----- = soz.2I l - 8.85'" l Y Y 1 0 - ' 1 9 0 "

A= D =coshf l =O :938 L.2"

t=7::: ,lr ul"u"'

t- 88s" 0354sr s'

Vs= 0.93811.2"x 127 10"+ 128.2172.65" 0.164 -36.9"-'

= 119.131.2" 2L03 135.75"

= 136.97 6.2"kV

lVshin. 81,23--W

C = Lsinh .y/ f x 0.354 181.5"z, 362.211- .85"

= 9.77 x 10 4 190.4" .\

Is=9.J7 x lO-a190.4" x 12 7+ 0.938 11.2" x0.164 l -36.9

= 0.124 190.4"+ 0.154" 35.7"

= 0.1286 115.3"kA

Sending-end f = co s (15.3' 6.2" - 9.1")= 0.987 eading

Sending-end ower = Jt x 237.23x 0.1286x 0.987

= 52.15MW

(d ) Approximation 5.28b):

A '=D = | +|

Yz)

= 128.5 72.7'

c= y(w!Z') = o.oo1et r+\ 6 )

= 0.001 9O "

]x to{ lrgo xr3t.2172.3"6

Vs= 0.93811.2"x 12 7 10 "+ 128.5 72.7'x 0.164

= ll9.I3 11.2"+ 21.07 35.8" 136.2 it4.82

= 137 16.2" kV

I Ys in" 237 3 kY

/s = 0.13 116.5" same s calculatedn part b) )

Sending-endf = cos(16.5' 6.2 = 10.3") 0.984 eading

Sending-endower Jl * ?313x 0.t3 x 0.984

= 52.58MW'fhe

results re abulatedelow:

I -36.9"

No w

Short line

uppntximatktn

Nominal-r Exact Approximation

(s.28h)l ys lhn" 5I .2 kv

I, 0J& l-36.9" kA

p.f, 0.745 lagging

P" 53.2 MW

238 kV 237.23 Vo.I3 116.5" A 0.1286 t5.3" kA0.984eading 0.987eading52.7MW 52.15MW

?37.3 V0.131t6.5"kA0.984 eading52.58MW

Comments

We find from the above example hat the resultsobtainedby the nominal-zrmethodand the approximation (5.28b) are practically the sameand are veryclose o those obtainedby exactcalculations part (c)). On the otherhand theresults obtained by'the short line approximationare in considerableerror.Therefore,tlr a line of this length about300knr), t i s sufficientlyaccurateouse the nominal- (or approximation 5.28b))which results n considerablesaving n computational flort.

15El ModernPowerSystemAnalysis

5.9 POWER FLOW THROUGH A TRANSMISSION LINE

So far he ransmissionine performance quationwaspresentedn the form of

voltase and current relationshipsbetween sending-and eceiving-ends.Since

loadsare moreoften expressedn terms of real (wattslkW)and reactive VARs/

kVAR) power, t is conveniento deal with transmissionine equationsn the

form of sending- and receiving-endcomplex power and voltages.While the

characteristicsnd performanceof power

Transmission ines

= ul'll^'t(/r d) l-4v^t21p - ay

t, =I?l,r,ftur- u)-lll*lf t@+6)

sn=vRt0 ti+lvstt13--l!lrv^tz{1-*t]

Similerly, (s.se)



problem of flow of power in a general network will be treated n the next

chapter, he principles involved are llustratedhere through a single transmis-

sion ine (2-nodel2-bus ystem) s shown n Fig. 5.17.

Sp = Pp +lQP

Fig.5.17 A two-bus ystem

Let us take the receiving-endoltageas areferencephasor Vn=lVRl 10")

and et the sending-end oltage ead t by an angle 6 (Vs = lVsl16). Tlp- angle

d is known as the torque angle whose significance has been explained in

Chapter4 and will further be taken up in Chapter 12 while dealing with the

problem of stability.

The complex power eaving he receiving-endand entering he sending-end

of the ransmissionine canbeexpresseds(onper phase asis)Sn= Pn+ .iQn

- Vn n

Ss = Ps + iQ s= YsI;

Receiving- nd sending-endurrents an, however, " .*pr"rsed in termsof

receiving-and sending-end oltages seeEq. (5'1)] as

(s.s6)

(s.s7)

S s = iQs

In the aboveequations o and s, areper phase omplex voltamperes,while v*and vr areexpressedn per phasevolts. f yRand 7, are expressedn kv line,then the three-phaseeceiving-endcomplexpor., is given by

so 3-phaseA)= t {l#%+191 t( s -r ,- l4l tvot2to6,, ,, ,-/ -LJt *J T Bl -'"'-lEl

3L\tl-a) F

s^ 3-phaseVA)T# 4p - , -l*lvop 1p a1 (s.60)

Let A, B, D, the transmission ine constants,be written as

A=

lAl la, B=

lBl lP, D= lDl lo (sinceA = D)

Therefore,we can write

r 1 r, _ 0 ) _ l A l r v " ra _ 0 )^=

l ; ltv5t(

rBl( '

r,=i+l ,t(a+ - -l+lvRt-p

Substitutingor 1o n Eq. (5.54)'wege t

This indeed s the sameas Eq. (5.59).The same esultholds or sr . Thus wesee that Eqs. (5.58) and (5.59) give rhe three-phaseMVA if vs ina vo *"expressedn kV line.

If Eq' (5'58) is expressedn real and maginaryparts,we can write the realand reactivepowers at the receiving_endas

r* = JI]l-v^]cos/ - a - l{ l rv^r' os B- a) (5.61) t B tv - /

l r l "

/ 1 l y s l l y * l . , n ^ l A l . _ _ . "Qn=ff

sin1/ 6tlliivni'?

sin /- a) (s.62)

similarly, the real and reactivepowers at sending-end re

p, = I?l v, P co so- o) -lvsllv'l

' lB l JtB icos(0*A (5 '63 )

Qs = l*l 'ur,tsin p a)tvsttvRt

prI |

. ' . ,l - s r t l

( / 1+ b) 6 .e )

It is easy o see rom Eq. (5.61) that he receivedpowerpo will be maximumat

6 = 0

such that

Po max) 'u:'l l^'-t Attv'Pv)

:.

IBt lB-c,os/1_ a) (5.65)

The corresponding en @t max po ) is

o = -l4LYrytn = - f f f : - s i n ( 0 - u )

/ E E A \

( J . J + /

(5.s5)

t,r=ivr,=

-* vR

, r=*r , -* ro

I

f60|

ModernPo*e, Sy.t"r Analysir_

Thus the load must draw this much leading MVAR in order to receive hemaximum ea l power.

Considernow the specialcaseof a short ine with a series mpedanceZ. Now

A - D = I l 0 : B = Z = l z l l e

Substitutinghese n Eqs. (5.61) o (5.64),we get the simplified esults orthe short ine as

t -Equation5.72) anbe urther implified yassumingos 6= r,sincedisnormallysmall*. hus

Let lvtl - lvRl= lAv1, he magnitudeof voltage rop across he rransmissionline.



n^=#tavl (s.74)

The above short ine equationwill also apply for a long line when the lineis replacedby its equivalent-r(o r nominal- ) and he shunt admittances relumped with the receiving-end oad and sending-endgeneration. n fact, thistechnique s always used n the load flow problem to be treated n the nextchapter.

From Eq.(5.66),he maximum eceiving-end ower s received,when 6= 0

so hatP^ (rnax)lv]\Yol Ytc's //

tz t t z l

po- Y r i l l o l cos (g -6 \ - l vRPcosolz l tz l

go = lv ' ) lY* l -s in(d - b) -

lv* l ', ino

l z l t z lfor the eceiving-endnd or the sending-end

ps= I:1.o,e-tl1' lunlo,@+6) l z l tz l

e,_ry_srn_ u:'lI.'sind*oolz l l z l

- \

Now cos d= RllZl,

Ppmax)'tr',|,o'-l'r-t o

several mportanr onclusions ha t easily ollow from Eqs. (5.71) to (5.74)are enumerated elow:

I' For R = 0 (which is a valid approximation or a transmission ine) thereal power ransferredo the receiving-ends proportional o sin 6(= 6forsmall valuesof d ), while the reactiu. po*", is proportional to themagnitudeof the voltage drop across he ine.

2. The real power received is maximum for 6 = 90o and has a varuelvsllvRvx. of course, d is restricted to varues weil below 90o fromconsiderationsf stability to be discussedn Chapte 12 .

3. Maximum real power transferred for a given line (fixed X) can beincreased y raising ts voltage evel. t is from this considerationha tvoltage evelsare beingprogressivelypushed p to transmit argerchunksbf power over ronger distances wananted ty l*g; ;irtlln"rutirrgstations.

For very ong inesvoltage evel cannotbe aisedbeyond he imits placed' 4'1'present-dayigh voltage echnoiogy. o increase ower ransmittedn

suclt citscs, lt c only choicc is to reduce he line reactance. hi s isaccomplished y adding seriescapacitorsn the line. This idea will bepursued urther n chapter 12. Series apacitors oulclof c<lursencrcos!.the severityof line over voltagesunderswitchingconditions.

4. As said n 1 above, he vARs (lagging eactive ower) derivered y a ineis proportional o the ine voltagedrop and s independent f d Therefore,in a transmission ystem f the vARs demandof the load is large, thevoltage profile at that point t-endso sag rather sharply. To maintain adesired oltageprofile, he vARs demanJ f the oad mustbe met ocallyby employing positive vAR generators condensers). hi s will bediscussed t length n

Sec.5.10.A somewhatmore accurateyet approximate esult expressing ine voltagedrop in terms of active and rlactive powers can be written directly fromEq . (5.5), .e .

lAVl= l ln lR co s Q + llolX sn Q

lvRl*small

dis necessaryromconsiderationsf systemtability hichwill bediscussedat length n Chapter12.

Normally the resistanceof a transmission ine is small compared to itsreactance since t i s necessaryo maintain a high efficiency of transmission),so that 0 = tan-t xlR = 90"; wherez = R + jx.The receiving-end qs. (5.66)and (5.67)can then be approximated s

(5.66)

(s.67)

(s.68)

(s.6e)

(5.70)

(s.1r)

(s.12)

i

r iIII

162 |"odern

power SvstemAnalvsis

=R!ry XQn

lvRl

This esulteduceso thatof Eq . 5.74)f R = 0.

characteristicsnd Performance f PowerTransmission ines

Case a): Cable mpedance 70.05pu .

Sincecable esistances zero, here s no realpower oss n the cable.Hence

Pc r t Pcz= Por * Poz = 40 pu

(s.75)

j Example .8 Pot= Pcz = 20 pu

The voltage of bus 2 is taken as ref'erence,.e. V, 10" and voltageof bus



An interconnectorable inks generating tations and 2 asshown n Fig. 5.18.Th e desired oltageprof i le s f lat, .e. lVrl=lVzl = 1 pu . The otaldemands tthe two busesare

S p r = 1 5 + 7 5 P u

Soz=25 + 71 5 Pu

The station oads are equalizedby the flow of power in the cable.Estimatethe torqueangleand the stationpower actors: a) for cablez = 0 + 70.05pu ,and (b) for cable Z - 0.005 + 70.05pu. It is given rhat generator G, cangeneratea maximum of 20.0 pu real power.

1 is V1 16r. Further, or flat voltage profile lVll = lV2l - LReal power flow from bus I to bus 2 is obtained from

recognizing hat sinceR = 0, 0= 90".Hence

D _ D l v t l l v lPs= Pn = -f si n6,

- l x l) = -SlI l d,

0.05

or 4 = I4.5"

v r= I l L 4 ' 5 "

FromEq. (5.69)

Eq . (s.68) by

Solution The powers at the various points insystemare defined n Fig. 5.18(a).

the fundamental two-bus)

@I

Sn.' = P61+ Q61

l v l t 51

Jp2 = I-p2+ J\Jp2

I

^ ' l v , l ' l v , l l v l -Q t=

Xcosdt

-

":

-, : x 0.e68 0.638 uU,\,J U.UJ

(a )From Eq. (5.67)

fl v t l l v , l

2o = -Tcos d'

Reactivepower oss* n the cable s

c) tv12

X- - Qs = - 0.638 u

SDI = 15 +/ 5

__>

S n = 5 - l

(c )

Fig.5.18 Two-bus ystem

{ zo.ro i16.12

V 2 = . 0 o "

QL= Qs en _ 2e s _ 1.276 u

Total oadon station = (15 + i5) + (5 + j0.638)

= 20 + j5.638

Power actorat starion = cos [run-' t'f:t') = 0.963 agging\ 2 0 )

Total oadon station - (25 + jls) - (5 - j0.638)

= 2 0 + i 1 5 . 6 3 8

(b )

G)

1 5 + j 5

Q - . - D - . r i Av u 1 - , D 1 ' l \ 1 D 1

25 + 1S

Reactivcpowcr loss can also be cornputedas /l2X= :t +(9.!1q)1"

o.os=.z7pu.I

164 I toctern power Svstem Analvsis

Power actorar stat ion2 = co S (ron*r 15'638)

= O.zgs agging\ 2 0 )

Th e stat ionoads, oa d demands, nd in e f lows ar eshown n Fig.5.1g(b).

ir-;-{ z i, r" p""

t""rv.

Case b) : Cable mpedance 0.005+70.05 = 0.0502 B4.3 pu. In rhis casethe cableresistanceauses eal power oss which is not known a priori. The real

characteristicsnd Performance fpower

Transmission ines 1656r= 14.4"

Substitutingr n Eqs. ii), (iii) and iv), we ger

Qct= 5.13,Qc 2 16.12, Gz 20.10

It may be noted that the realGz(Pcz - 20.10).

^ ' - - ^ D '" ' ^ '

power loss of 0. pu is supplied by

The above presented roblem is a two-bus oad flow problem. Explicit



load flow is thus not obvious as was in the case of R = 0. we specify thegenerationat station I as

Pc r = 20 Pu

The consideration or fixing this generation s economic as we shall see inChapter7.

The generationat station2 will be 20 pu plus the cable oss.The unknownvariables n the problem are

P62, 6p Qcp Qcz

solution s alwayspossible n a two-bus case.The readershould ry the casewhen

Q c z =7 1 0an d V2 l ' -

The general oad flow problem will be takenup in Chapter6. It will be seenthat explicit solution s not possible n the generalcaseand terative echniqueshave to be resorted o.

A 275 kV transmissionine has he following line constants:

A = 0.85 15": B - 200 175"

(a) Determine he power at unity power factor that can be received f thevoltage profile at eachend is to be maintaine at 275 ky .

\

(b) What type and atingof compensationquipmentwouldbe required f theload is 150 MW at unity power factorwith the samevoltageprofile asin part (a).

(c ) With the oadas n part b), what r,vould c th c receiving-endoltage fthe compensation quipments not installed?

Solution (a ) Given lV5l= lVal= 215 kY; e,= 5o,C = J5". Since he poweris receivedat unity power factor,

Q n = o

Substituting hesevalues n Eq. (5.62),we can write

Let us now examineas to how many systemequationscan be formed.From Eqs. 5.68)an d (5.69)

Pc t Po r , =#cos

d- V#*s@+ 61 )

5 =-= cos 84.3" ^-= cos (84.3"+ 4) (i)U.U)UZ U.U)UZ

ecr-eot= s=f f r t^ t - t r r , \ i ,s in(d+4)

Qcr - S= ;j - sin 84.3" =+-sin (84.3" Ur) (ii)0.0502 0.0502

FromEqs. 5.66) nd 5.67)

P o z - P - P - l v t l l v z l l v ' PGz =Pn= -f f co s d- 6r ) - -1 : -cose

2s - Pc2=*;rcos

(84.3" A, l#rcos

84.3o (ii i)

eor-ecz=n=Uffiri, (o- 6t>,Yl!rr",l z l

15 Qcz=#,

si n 84.3" 4)#,

si n84.3" (v )

Thuswe have our equations, qs . i) to (iv), in four unknowns 52, 51 , 61,Q62.Even hough hesearenon-linear lgebraic quations, olution s possiblein this case.SolvingEq. (i) for d,, we have

o - !5x]75- sin(75"- d)200

0 = 3 7 8 s i n ( 7 5 " - A - 3 0 2

whichgives6- ) ) "

From Eq . (5.61)

ij ;x Q75)2in 75"-5" )

Pn =275^x275

cos 75 o 22") 985

* (2712cos70o" 200 200

- 227.6 109.9 117.7 {W

150=215zq7cos

(75o 5) -200

16 6 | todern powerSvstemAnalvsisI

(b ) No w lV5l= lVpl= 27 5 kV

Powerdernandedy load = 15 0MW at UP F

P n = P R = 1 5 0 M W ;Q o = 0

0 = !:y-*t si n 75'- .5 ) *tvot2 si n70' (i i)200 200

FromEq. (ii), we get

s in (75 " - A=0 .00291VR1

characteristicsnd Performancef PowerTransmissionines Ifil

15 0= 1.375 yn t (1 _ (0.002912v^121u20.00l45tyRl2

0.85 .^- , )

,*x (275)'cos 0o



1 5 0 = 3 78co s 75 " 4 - 1 10

or 5 = 28.46"

FromEq . (5.62)

Solving the quadraticand retaining the higher value of lv^|, we obtain

lVal= 244.9kV

Note: The second and lower value solution of lVol though feasible, is

impractical as t corresponds o abnormally ow voltage and efficiency.It is to be observed rom the resultsof this problem that arger power can

be transmitted over a line with a fixed voltage profile by installingcompensation quipmentat the receiving-end apableof feedingpositive VARsinto the ine.

Circle Diagrams

It has been shown above that-the flow of active and reactivepower over atransmissionine canbe handledcomputationally.t will now be shown that theIocusof complexsending- nd eceiving-end ower s a circle.Sincecirclesareconvenient o draw, the circle diagramsare a usefulaid to visualize he loadflow problem over a single transmission.

The expressionsor complexnumbereceiving-andsending-endowersarereproduced elow from Eqs. (5.58)and (5.59).

en="#

sin (75. 28.46") 0'85x e75)2sin 70o

200

- 274.46 302= - 2l.56 MVAR

Thus n order o maintain2T5kV at areceiving-end, n= -27.56 MVAR mustbe drawnalongwith the real powerof Po = 150 MW. The oad being 150MWat unity power actor, .e. Qo = 0, compensation quipmentmust be installedat the receiving-end.With referenceo Fig. 5.19, we have

- 2 7 . 5 6 + Q c = 0

or Qc = + 27.56MVAR

i.e. the compensation quipmentnnust eed positive VARs into the line. Seesubsection .10 for a

more detailedexplanation.15 0 j27.56 15 0 /0

sR=-lfl rv^r'/r- . +P 4r- b)

n=l+lvs((r-o) -+3 4/r+

(s.s8)

(5.5e)

(s.16)

F ig .5.19

(c) Sinceno compensation quipments provided

P n = 1 5 0 M W ,Q n = 0Now,

The units or Sp and S, are MVA (three-pha,se) ith voltages n KV line.As per the above equations,So and ,9, are each composedof two phasorcomponenfs-6nea constant hasor nd he othera phasor f fixed magnitudebut variableangle. The loci lor S^ and S, would, therefore,be circles drawnfrom the tip of constantphasorsas centres.

It follows from Eq. (5.58) that the centreof receiving-end ircle is located

at the tip of the.phasor.

l V 5= 2 7 5 k V , l V a l =

Substitut inghis data n Eqs. (5.61)an d (5.62),we have

15 0t': \ ! '"os

(75" A - ggtv^t2 co s 0"200 200

-l+lvRP(r - a)IB l

' \

in polar coordinates r i n termsof rectangular oordinates,Horizontal coordinateof the centre

(i)

=-l|lrv-f cos//- a)MW (s.77)

e. - '1 , ,,

.I

.d6Eif _ ModernpowerSystemAnalysis .,.,,

Verticalcoordinateof the centre

= -i{- ltv*t2si n tJ - a) MVARl B l

The radiusof the receiving-endcircle is

t y s l l y R l MVA

tBl

characteristicsnd performanceof powerTransmissionines t69

Vertical coordinateof the centre

= l+ilvrt2 in F a) MVARt B t

The radius f the sending-endircle s

(5.78) ( s .81 )

The sending-endircle diagram s shown n Fig. 5.2I.Thecenrre s locatedby



l A l l t 2

lal l "l

Reference in efo r angle6

Fig. 5.20 Receiving-end irc lediagram

F'or constant lVol, the centre Co rernains ixed and concentric circles resultfor varying l7rl. However, fo r the case of constant ll{l and varying lvol thecentrcso1'c i rc lesmovc along th e in e OCoaru),have acli i n accordance o l tzrllvRtABt.

Similarly, it follows from Eq. (5.59) that the centre of the sending-end circleis located at the tip of the phasor

drawing OC, at angle rt? a) from the positive MW-axis. From the centre the

sending-end ircle is drawn with a .uoi.rr${e(same as in the case oflB t\

receiving-end).he operatingpoint N is locatedby measuring he torqueangled(asread rom the recefving-end irclediagram) n ttredirection ndicated rom

th c re'f i 'rcncc'inc.

l4l,u,,'(0.-a)I B l

in the polar coordinates or in terms of rectangular coordinales.Horizontal coordinate of the centre

C5

Roforoncoino- fo rangle

Radius YsllVnl

lB l

Phasor 5=P5+ie5

1--Qs

+

Fig. 5.21 Sending-end irc lediagram

i,i,=',!"iu':;=o

The corresponding receiving- and sending-end circle diagrams have been clrawnin Figs5.22and .23.

(s.7e)

MVAR MVAR

=B-ltv;2

os f - a)Mw (s.80)

characteristics nd Performance f powerTransmission ines [giMt -

Resistance= 0.035 Olkm per phaseInductance= 1.1 mHlkm per phase

Capacitance= 0.012 pFlkm per phase

If the line is supplied at 275 kV, determine the MVA rat ing of a shunt

the eceiving-endhen he ine s delivering o oad.Usenominal-rmethod.Solution

R = 0 . 0 3 5 x 4 0 0 = 1 4 Q



Pa =oK

Qn= KM

Fig.5-22 Receiving-endircre iagramor a short ine

MVAR

Ps =oLQ s = L N

L

Fig.5.23 Sending-endirclediagramor a stror.tine

oJl: useof circlediagramss $ustrated ymeans f the woexamplesiven

A 50 Hz, three-phasg, 75 kY,400 km transmissionine has the followingparameters:

X = 31 4x 1. 1 l0-3x 40 0= 138.2O

Z = 14 + 7138 138.1184.2"Q

Y= 314x 0.012 10-6 400 lg0" - 1.507 10-3 _W U

A= ( t+Lvz\ = 1 + - l - x t 507 10-3 r3g. i r i4 .z"\ 2 ) 2

= (0.896 70.0106) 0.896 j.l '

B = Z - 1 3 8 . 7 1 8 4 . 2 "

lV5 = 27 5kV , lVpl= 275kV

Radius f receiving-endirclelysllyRl-275x275 - 545.2MVA

tBl 138.7Location f the cenre of receiving-endircle,

l4:l1,, 275x275x0.896= 488.5MVA

I B Iil l

=

138J

= 'rdd') rl l\

l@ - a) = 84.2" 0.7"= 83.5"

i , \ /A P

_ > M W

488.5MVA

.tuou.rrro

Cp

Fi1.5.24 Circle iagramor Example .1 0

From the circle diagramof Fig. 5.24, + 55 MVAR mustbe drawn from thereceiving-endof the line in order to maintain a voltageof 275 kV. Thus ratingof shuntreactor needed= 55 MVA.

lvnl2

tzl

lvs l2

lzl

55 MVAR

Example .1 1

X12,:jl Modern owerSystemAnalysis

A - 0.93 11.5", B = Il 5 l l7"

If the receiving-endvoltage s 2'r-5 V, determine:

(a) The sending-end oltage equired f a load of 250MW at 0.g5 agging pf

Characteristicsnd Performance f PowerTransmission ines | ,i?3-;-t-(a ) Locate OP conespondingo the receiving-endoad of 250 MW at 0.85

laggingpf (+ 31.8). Then

, ^ o r^ l ys l l yR l 275 lvs l

l Vs l = 3 5 5 . 5 V

(b) Given lV5 = 295 kV.

t".l?"



is being deliveredat the receiving-end.

(b) The maximum power that can be delivered f the sending-endvoltage sheld at 295 kV.

(c) The additionalMVA that has to be provided at the receivihg-endwhendelivering 400 MVA at 0.8 lagging pt the suppty voltage beingmaintainedat 295 kV.

Solution In Fig. 5.25 the centreof the receiving-end ircle is locatedat

l4li t' - 2t5x275x0'93611.6 VA

l B lR '

t t 5

cos- l 0.85= 31.8"

l @- a ) =7 7 o - 1 .5 " 7 5 . 5 "

Radiusof circle diagram - 705.4 MVA1 1 5

Drawing the receiving-endcircle (seeFig. 5.25) and the line C^Q parallel to

the MW-axis, we read

PR-o = RQ = 55 6 MW

(c) Locate OPt conesponding o 400 MVA at 0.8 laggingpf (+ 36.8"). Draw

P/S parallel to MVAR-axis to cut the circle drawn in part (b) at S. For the

specifiedvoltage profile, the line load should be O^S. herefore, additional

MVA to be drawn from the line is

P/S = 295 MVAR or 295 MVA leading

5.10 METHODS OF VOLTAGE CONTROL

Practicallyeach equipmentused n power systemare rated or a cprtainvoltage

with a permissible band of voltagevariations.Voltage at various busesmust,

therefore,be controlled within a specifiedregulation figure. This article will

discuss he two methodsby meansof whieh voltage at a bus can be controlled.

lvslt6 lvRltj

MVAR

I

a) Ip"-.io, P^ io* |

Fig.5.26 A two-bus ystem

Consider he two-bussystem hownn Fig. 5.26 (akeadyexemplified n Sec.

5.9). For the sake of simplicity let the line be characterizedby a series

reactance i.e. it has negligibleresistance). urther, since he torqueangle d is

smallunderpracticalconditions,eal and eactive owersdelivered y the ine

for fixed sending-end oltage Vrl and a specified eceiving-encloltage { | can

be-writtenas below from Eqs. (5.71) and (5.73).

Fig. 5.25 Circlediagram or ExampleS.11(5.82)

ef, li, ,,u, rv,ilX

Equation 5.83) uponquadratic olution*can also be written as

rr{ r=}vrt

.+tys (1 4xesn vrtzlt/z

Since he real power demandedby the loaclmust be deliveredby the line,Pn= Po

varying real power demandp, is met by consequent hanges n the rorque

nsmtssionines"ffieReactive Power Injection5.83)

(5.84)

It follows from the above discussion hat in order to keep the receiving-endvoltageat a specified alue {1 , a fixed amountof VARs tai I *;;; drawn

Qn, a local VAR generator(controlled reactivepower source/compensatingequipment)must be usedas shown n Fig. 5.27. le vAR balance quation t

from the line-. To accomplish his underconditions

the receiving-end s now

Oi * Qc= Qo



angle d.

It is, however' -tobe noted hat the received eactivepower of the line mustremain ixed at esnasgivenby Eq. (5.g3) or fixed rv , I and specified 4r. il"line would, therefore,operatewith specified

eceiving-end oltage or only onevalueof Qo given by

Qo = QsnPractical loads are generally agging in nature and are such that the vARdemandQn mayexceedet*.rt easily ollows from Eq. (5.g3) hat or or ; ot-the receiving-end oltagemust change rom the specifiedvalue n' i somevalue Tol to meet the demandedVARs. Thus

Q ^ = o ^ =l v * l

'Jo =

Qn =;i

(lYsl lVol) or (QD> QsR)

The modified lVol s thengiven by

rvlqllvrt

-+ty3

(1 4xeRltvrtr, , (s.85)

Fluctuations n Qo ue absorbedby the local vAR generator o6 such hatthe vARs drawn from the line remain fixed at esn.The receiving-endvoltagewould thus remain

{1ed

at l4l (this of

"ourr"lrrumesa fixed sending_end

voltage lVrl). L,ocalVAR compensationcan, in fact, be made automaticbyusing he signal from the VAR meter nstalledat the receiving-endof the ine.

Fig.s-zr use of rocar AR generator t the roadbus

Trryo ypes of vAR generatorsare employed n practice-static type androtating type.Theseare discussedbelow.

Static VAR grenerator

crrmparison f Eqs. (5.84)ancl 5.85) rcvcals hu t r^r. n. - n - . ,-),s!,.,receiving-end oltage s r{r, but ior bo = Oo ; A:,"'

YD - vR - vp ' tttt:

tvo < t4 lThus a VAR demand arger han Qf is met by a consequentall in receiving-

11d ":t!q,"from the specified alue.similarly, if the vAR demand s less han

Q"*, t fo l lows ha t

tyRt tr4lIndeed,under light load conditions, he charging

"upu.lanceof the line may

cause the VAR demandto become negative resuliing in the receiving-endvoltageexceeding he sending-end oltage (this is the Ferrantieffect alreadyillustraredn Section5.6).

In order to regulate he ine voltageunder varying demandsof VARs, the twomethods iscussed elowareemployed.

'Negativesign in the quadratic solution is rejected becauseotherwise the solution

would not match the specified eceiving-endvoltage which is only slightly less thanthe sending-end ortage the difference s ress thai nqo).

It is nothing but a bank of three-phasestatic capacitors and/or inductors. Withreferenceo Fig. 5.28, f lV^l s in line kV,and Xg is the per phasecapacitive reac_tanceof the capacitorbank on an equiva_Ient star basis, the expression for theVARs fed into the line can be derived as

under.

l I c

Fig. 5.28 Static capacitorbank

r,=iH kA

'o fcourse,incc {t is spccif iccli thina buntl,Ql ruryvrry withil a corresponding

band.

: iii,',1,1 Modern owersystemAnarysis

iQcG-Phase)-ry (- IF )J3

- i 3x# .HMVA

tigure 5'29 showsa synchronousmotor connected o the receiving-endbusars and runningat no load. since the motoro.u*, n.grigible real power from

$",.t":::H:,.,:o,^fl :T-? T*lyinpr,use.

",; th" yn.r,ronouseachnce

:j 3:^t1o1wtr,icrr, usium"aohaverr,r*,n ;r;:r*::':il;

t v PQsQ-Phase)- MVAR

XC

If inductorsare employed nstead,vARs fed into the line are

I- _ (lvRl IEGD/0. ,. C_J5;E- KA

(s.86)



Q{3-phase)=- 'F ' ' t uo*XL

Underheavy oad conditions,whenpositiveVARs are needed, apacitorbanksare employed; while under light load conditions, when negativevARs are

needed,nductorbanksare switchedon.The following observations an be rnade or.static vAR generators.

(i ) Capacitorand inductor bankscan be switchedon in steps.However,stepless smooth)VAR control can now be achievedusing SCR (SiliconControlledRectifier) circuitrv.

iec= 3tvRl4oG lJ 3

= 3W(-lYRr-

rc l )J3 ( _jxsJl)

= jlVpt(tE6t _ IVRt)lXsMVA

ec= tVRt EGt _ tVRt)lXs VAR (5.8g)It immediately follows from the above relationshipthat the machine feedsositive vARs into the line when rEGt> tv^r (werexcited case)and injectsegarive VARs if lEGlcontinuouslyadjustableby adjustingmachine

"Jtutionwhich controls tE6l.n contrast o statrcvAR generators,he ollowing observations remade nespectof rotatingVAR generators. .,

(i) Thesecan provide both positive and negativevARs which are continu_usly adjustable.

(ii) vAR inje*ion ar a given excirarion s resss. .

vol age. s I zode eas andrE6r r ^ rti,Lt

tlt""r""J"-:ff i::: {.:::maller eductionn Qc compared o the 0.r. or staticcapacitors.From rhe observarionsua"

ibr_""in ,.rp..t of ,tuti.

""d;;;;;ingvARenerators,t seemshat rotatingvAR g.n"ruro* would be preferred.However,conomicconsiderat ions,nstal l . t ionan d 'rai 'r.rrun.. problernsimit their

il]ffi'ir'$t:L::"::chbusesn thesvstemr,"i"' a taige .oun,-.orAR

Control by Transformers

The vAR injectionmethoddiscussed bove acks he lexibility andeconomyf voltage control by transformer tap changing.The transformer tap changings obviously rimited to a narrow range of voltage control. If the vortageorrectionneeded xceedshis range, ap changings used n conjunctionwithhe VAR injectionmethod.

Receiving-endvoltagewhich tends o sagowing to vARs demanded y theoad, can be raised by simultaneousry.t,uogir; th. taps of sending_andeceiving-end ransformers. uch tap changes iust"bemade ,on_road,and cane done either manua'y or automaiically,-theoo*io.,o"r being ca'ed a Taphanging Under Load ifCUf_l transformer.

(s.87)

(ii) Since Qg is proportional o the squareof terminal voltage,capacitorbank, their effectivenessendsto decrease s theunder ull load conditions.

(iii) If the systemvoltagecontainsappreciableharmonics, he fifth being themost troublesome,he capacitorsmay be overloadedconsiderably.

(iv) capacitors act as short circuit when switchedon.

(v ) There is a possibility of seriesresonancewith the line incluctanceparticuia.riyat harmonic requencies.

Rotating VAR grenerator

It is nothingbut a synchronousmotor running at no-loadand havingexcitationadjustable ve r a wide range. t feedsposit iveVARs into th e line uncleroverexcited onditionsand 'eeds egativeVARs whenunderexcited. machinethus running is called a synchronouscondenser.

lvnl

for a given

voltage sags

Fig.5.29 Rotat ing AR generat ion

z = R + j x

,tig;'J Modern owerSystem nalysisI

Considerheoperation f a transmissionine with a tap changing ransformer

at eachend as shown n Fig. 5.30. Let /5and r^ be the fractionsof the nominal

transformation atios, i.e. the tap ratio/nominal ratio. For example, a trans-

kV input has rr - I2lll =

Characteristicsnd Performance f Power Transmission ines fi.l7!J,I

l.{Vl which is to be compensated. hus merely tap settingas a method of

voltage drop compensation vould give rise to excessively arge tap setting if

compensationexceeds ertain imits. Thus, if the tap setting dictatedby Eq.

setting range (usually not more than + 20Vo), t would be necessary o

simultaneously nject VARs at the receiving-end n order to maintain the

desiredvoltage evel.



1 : fsnl

Fig.5.30 Transmissionin ewith apchangingransformert eachen d

With reference o Fig. 5.30 et the mpedances f the transformerbe lumped

tn Z alongwith the line impedance.To compepsateor voltage n the line and

transformers, et the transformer taps be set at off nominal values, rr and ro.

With reference o the circuit shown. we have

t rnrVs= t^nrVo+ IZ (s.8e)

From Eq. (5.75)the voltage drop ref'erred o the high voltageside s given by

tAvl= !I,!jIQs-ton,rlVol

lAVl - tsn, lTsl - ton2lVol

Compensation of Transmission Lines

The perfonnanceof long EHV AC transmissionsystemscan be improved by

reactive compensationof seriesor shunt (parallel) ype. Seriescapacitorsand

shunt reactors are used to reduce artificially the series eactanceand shuntsusceptance f lines and thus hey act as the ine compensators. ompensation

of lines results n improving the system stability (Ch. 12) and voltageconffol,

in increasing he efficiency of power transmission, acilitating ine energization

and reducing temporaryand transient overvoltages.

Series compensation educes.he series mpedanceof the ine which causes

voltagedrop 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 herefore he also affected by change n the value of.Z, but

thesechanges are small in comparison to the change n B as B = Z for the

nominal -rr and equalsZ (sinh 4ll) for the equivalent zr. .,

The voltage drop AV due o series compensation s given by

AV = 1Rcos S, + I(X,. X.) sin ,!, (s.e4)Here X, = capacitive eactanceof the seriescapacitorbank per phaseancl

X, is thc total incluctive cactance f the ine/phasc.n practice,X. may be so

selected hat the factor (XL - X.) sin Q, becomesnegativeand equals (in

magnitude) R cos /, so that AV becomes zero. The ratio X=IXL is called"compensation actor" and when expressedas a percentages known as the"percentage ompensation".

The extent of effect of compensationdependson the number, ocation and

circuit arrangements f series capacitor and shunt reactor stations.Whileplanning long-distance ines, besides the average degree of compensation

required, t is required o find out the most appropriateocationof the reactors

and capacitor banks, the optimum connection scheme and the number ofintermediate stations.For finding the operatingconditionsalong the line, the

ABCD constantsof the portionsof line on eachside of the capacitorbank,and

ABCD constants of the bank may be first found out and then equivalent

constants f the series ombinationof line-capacitor-line an henbe arrivedat

by using the formulae given in Appendix B.

In India, in statesike UP, series ompensations quite mportantsincesuper

thermal plants are ocated (east) several hundred kilometers from load centres(west) and large chunks of power must be transmittedover long distances.

Seriescapacitorsalso help in balancing he voltagedrop of two parallel ines.

Now

(s.e0)

(s.e1)

In order that the voltage on the HV side of the two transformers e of the

sameorder and he tap settingof each ransformerbe the minimum, we choose

t s t n = 1 ( 5 . 9 2 )

Substitut inEn= llttin Eq . (5.91)an d eorganising, e obtain

t rn r l v r l - t on r l vo l+RPR+xQR

t* n r l Vo l

.r( , RP R xg o ) _ n2 lvRl

" [ ' "r"rWW )- " , W

(s.93)

For completevoltage drop compensation,he 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. quation (5.90) indicates hat /^ tends o increase* he voltage

-Thisis so because n < 1 increases he line current / and hencevoltage drop.

r-hOt" uodern Power SvstemAnalysis

When seriescompensations used, here are chances f sustainedovervoltage

to the ground at the seriescapacitor terminals.This overvoltage can be the

power limiting criterion at high degreeof compensation.A spark gap with a

high speed contactor is used to protect the capacitors under overvoltage

trons.

Under ight load or no-loadconditions,chargingcurrentshouldbe kept less

than the rated full-load currentof the line. The charging current is approxi-

matelygiven by BrltA whereB. is the total capacitivesusceptance f the line

and Vl is the ratedvoltage o neutral. f the total nductivesusceptances Br due

Characteristicsnd Performance f PowerTransmissionines

kW at a eadingpower factor. At what valueof P is the voltage egulatior

zero when the power factor of the load s (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= 150 165"C I h a s a t t h eoa r

end a transformerhaving a series mpedanceZr = 100 167" Q. The loar

form of

r81

ff]=[1



to several nductors connected shunt compensation) rom line to neutral at

appropriateplaces along the line, then the chargingcurrent would be

and evaluate hese constants.

5.3 A three-phase verhead ine 200 km long has esistance 0.16 Qlkrn anconductordiameterof 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 he ine is delivering full load of 50 MW at 132kV and 0.8 lagginpf, (c) efficiency of transmission,and (d) the receiving-end voltagregulation.

5.4 A short230 kV transmission ine with a reactance f 18O/phasesupplira load at 0.85 lagging power factor. For a line current of 1,000 A ttreceiving- and sending-end oltagesare to be maintainedat 230 k\

Calculate (a) rating of synchronous capacitor required, (b) the loacurrent,(c) the load MVA. Power drawn by the synchronous apacitt

may be neglected. .\

5.5 A 40 MVA generatingstation s connected o a 'three-phaseline havin

Z = 300 175" Q Y = 0.0025 19tr U.

The power at the generatingstation s 40 MVA at unity power factor aa voltage of L20 kV. There s a load of 10 MW at unity power factor athe mid pointof the ine.Calculate he voltageand oad at the distantencof the line. Use nominal-T circuit for the line.

5.6 The generalized ircuit constantsof a transmissionine are

A - 0 . 9 3 + 7 0 . 0 1 6

B = 2 0 + j I 4 0

The load at the receiving-end s 60 MVA, 50 H4 0.8 power factor

lagging. The voltage at the supply end is 22OkV. Calculate the loadvoltage.

Find the ncident and reflectedcurrents or the ine of Problem5.3 at the

receiving-endand 200 km from the receiving-end.If the ine of Problem5.6 s 200 km longand delivers50 MW at22OkY

and0.8 power actor agging,determinehe sending-end oltage, urrent,power factor and power. Compute the efficiency of transmission,

characteristicmpedance,wavelength, nd velocityof propagation.

For Example 5.7 find the parameters f the equivalent-ncircuit for theline.

",]l';l(s.es)

Reductionof the chargingcurrent s by the factor of (1 - Br lBc) and81lBg is

the shunt compensation actor.Shunt compensation t no-load also keeps he

receiving end voltage within limits which would otherwise be quite high

becauseof the Ferranti Effect. Thus reactorsshouldbe introduced as oad is

removed,f<lr ropervoltage ontrol'

As mentionedearlier, he shuntcapacitorsare usedacrossan inductive oad

so as to provide part'of the reactive VARs requiredby the load to keep the

voltage within desirable imits. Similarly, the shunt eactorsare kept across

capacitive oadsor in light loadconditions,as discussed bove, o absorbsome

of the leading VARs for achievingvoltage control. Capacitorsare connected

eithcrcl irect lyo a bu sor through crt iarywincl ing f the main ransformerndare placed along the line to minimise lossesand the voltage drop.

Ii may be noted hat for the samevoltageboost, he reactivepower capacity

of a shunt capacitrtr s greater han that of a seriescapacitor.The shunt

capacitor mproves thepf of the oad while the series apacitorhas hardly any

impact on the pf. Series capacitors are more effective for long lines for

irnprovement f systemstability.

Thus, we see hat n both seriesand shuntcompensation f long transmission

lines it is possible o transmit arge amountsof power efficiently with a flat

voltage profile. Proper type of compensationshould be provided in proper

quantityat appropriate laces o achieve he desiredvoltagecontrol.The reader

is enceuraged o read the details about the Static Var Systems (SVS) in

References , 8 and 16. For complete reatmenton'compensation',the reader

may refer to ChaPter15.

PROB.EMS

A three-phase oltageof 11 kV is applied o a line having R = 10 f) and

X = 12 ft p"t conductor.At the end of the line is a balanced oad of P

I ,he,=Bc -Br) lvl= BclV [r+)

5.7

5 . 8

5.95 . 1

,,1€? l uooern power system hnalysis-T-

5.10 An interconnectorcable having a reactanceof 6 O links generating

stations1 and 2 as shown n Fig. 5.18a.The desired oltageprofile is lVtl

= lVzl= 22 kY. The oadsat the two-busbarsare 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,275V, 400 km transmissionine has following

parameters per phase).

Resistance 0.035 Qlkm

characteristicsnd Performance f powerTransmission ines 183

REFERECES

Books

l. Tron'smission ine ReferenceBook-345kV and Above,Electric Power ResearchInstitute,Palo Alto calif, 1975.

2. Mccombe, J. and F.J.Haigh, overhead-Iinepractice,Macdonalel,London, 1966.3. Stevenson, .D., Elements f Power Sy.stem nalysis,4thedn,McGraw-Hill. New

York, 1982.



, Inductance=1mFl /km

Capacitance 0.01 p,Flkm

If the line is suppliedat 275 kV, determine he MVA rating of a shuntreactorhaving negligible osses hat would be required o maintain 275

kV at the receiving-end,when he ine is deliveringno-load.Use nominal-

zr method.

5.12 A,three-phaseeederhaving resistancef 3 Q and areactance f 10 f)

suppliesa load of 2.0 MW at 0.85 lagging power factor. The receiving-

end voltage is maintained at 11 kV by means of a static condenser

drawing2.1 MVAR from the ine. Calculate he sending-end oltageand

power actor.What s the voltage egulationandefficiency of the feeder?

5.13 A three-phaseverheadine has esistance nd eactance f 5 and 20 Q,

respectively. he load at the receiving-ends 30 MW, 0.85 power factor

laggingat 33 kV. Find the voltage at the sending-end.What will be the

kVAR rating of thecompensating quipment nsertedat the receiving-end

so as o maintain a voltageof 33 kV at eachend?Find also he maximum

load that can be transmitted.

5.I4 Constructa receiving-endpower circle diagram or the line of Example

5.7.Locate the point corespondingto the oadof 50 MW at 220 kV with

0.8 agging power factor.Draw the circle passing hrough he load point.

Measure he radiusanddetermineherefrom Vrl.Also draw the sending-

end circle and determine therefrorn the sending-endpower and power

factor.

5.15 A three-phase verhead ine has resistanceand reactance er phaseof 5

and25 f), respectively.The load at the receiving-end s 15 MW, 33 kV,

0.8 power factor lagging. Find the capacity of the compensation

equipment eeded o deliver his load with a sending-end oltage 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) nstalled, if the receiving-endvoltage is permitted to drop to

28kV.

4. Arrillaga, J., High VohageDirect Curuent Transmission, F,E Power EngineeringSeries6, PeterPeregrinus td., London, 1983.

5. Kirnbark,E.w., Direct current Transmission, ol. 1, wiley, New york, 1971.

6. IJhlmann,E., Power Transmissionby Direct current, Springer-verlag,Berlin-Heidelberg,1975.

7. Miller, T.J.E., ReactivePower control in Electric systems,wiley, New york

t982.

8. Mathur, R.M. (Ed.), Static Compensatorsor ReactivePub.,Winnipeg,1984.

9. Desphande,M.V., Electrical Power System Design,Delhi, 1984.

Power Control, Context

Tata McGraw-Hill. New

Papers

10 . Dunlop,R.D., R. Gutmanan d D.p. Marchenko,Analyt icalDevelopment fLoadabilityCharacteristicsor EHV and UHV Transmission ines", IEEE Trans.PAS,1979,98:606.

11 . "EHVTransmission",special ssue), EEE Trans, une 1966,No.6, pAS-g5.12. Goodrich,R.D., "A UniversalPower circle Diagram", AIEE Trans., 1951, 7o:

2042.

Indulkar, c.s. Parmod Kumar and D.p. Kothari, "sensitivity Analysis of aMulticonductorTransmission ine", Proc. IEEE, March 19g2,70: 299.Indulkar, c.s., Parmod Kumar and D.P.Kothari, "some studies on carrierPropagationn overhead ransmissionLines", IEEE Trans.on pAS,No. 4, 19g3,102:942.

Bijwe, P.R., D.P. Kothari, J. Nanda and K.s. Lingamurthy, "optimal voltageControl Using ConstantSensitivityMatrix", ElectricPowerSystemResearch,Oct.1986, : 195.

Kothari, D.P., et al. "Microprocessorsontrolled static var s5istems",proc. Int.conf. Modelling & Simulation,Gorakhpur,Dec. 1985,2: 139.

1 3 .

14 .

1 5 .

1 6 .

6.2 NETWORK MODEL FORMULATION

The load flow problem has, n fact, been already ntroduced n Chapter5 with

the help of a f'undamentalystent, .e. a two-busproblem(seeExample5.8)'

Eor4 lq4d flqW llq4ypfufg4l life power systemcomprising a large number

of buses, t is necessary o proceed systematicallyby first formulating the

network model of the sYstern.

A power systemcomprises everalbuseswhich are nterconnected y rneans

of transmissionines.Power s injected nto a bus from generators'while the

rom it. Of course, heremay be buseswith only generators



6. 1 INTRODUCTION

With the backgroundof the previouschapters,we are now ready to study the

operationaleatures f a composite owersystem.Symmetrical teady tate s'

in fact, the most mportantmodeof operationof a power system. hreemajor

problems encountered n this mode of operation are listed beiow in their

hierarchicalorder.

1. Load flow problem

2. Optimal oa dscheduling roblem

3. Systems ontrolProblemThis chapter s devoted o the load flow problem, while the other two

problemswill be treateil n later chapters. t-radlow study n power systetn

parlance s the steadystatesolution of the porversystemnetwork. The main

information obtained rom this study comprises he magnitudesand phase

angles of load bus voltages, eactive powers at generator uses, ea l and

reactinepower low on transmissionines,othervariablesbeingspecified.This

information s essentialor the continuousmonitoringof the currentstateof the

systernand or analyzinghe effectiveness f alternative lans or future system

expansion o meet ncreasedoad demand.

Before the adventof digital computers, he AC calculatingboard was the

only meansof carryingou t load flow studies.Thesestudieswere, herefore,tediousand ime consuming.With the availability of fast and arge size digital

computers, ll kinds of power system tudies,nciuding oad low, can now be

carried out conveniently. n fact, some of the advanced evel sophisticated

studieswhich were almost mpossible o carry out on the AC calculatingboard

have now become possible.The AC calculatingboard has been rendered

obsoleteor all practicalpurposes.

loadsare tapped

and no-loads,and there may be others with only loads and no generators.

Further,VAR generatorsmay also be connectedo somebuses.The surplus

power at some of the buses s transportedia transmission ines to buses

deficient n power.Figure6.1ashows he one-line iagramof a four-bus ystem

with generatorsand loads at each bus. To arrive at the network model of a

po*"i system, t is sufficientlyaccurate o represent short ine by a series

impedanceand a long line by a nominal-zrmodel. (equivalent-7T ay be used

foi very long lines).Often, ine resistancemay be neglectedwith a small oss

in accuracybut a great dealof saving n computationime.

For systematicanalysis, t is convenient to regard loads as negative

generators nd ump togetherhe generator nd oadpowersat the buses. hus

at the ith bus, the net complexpower injected nto the bus s given by

S ;= P i + j Q i = ( Pc i - Po ) + j ( Q c i - Qo )

where he corrrplex owersuppliedby the generatorss

Sc i = Po t + Qa iancl he ccltnplexpower drawnby the loads s

Sp i= Por+ Qo i

The real ancl eactivepowers njected nttl thc itlt bus arc thcn

P i = P o i - P ^ i = 1 ,2 , " ' ' f l ( 6 ' 1 )

Q i = Q c i - Q o i

Figure6.lb shows he networkmodel of the sample ower system repared

on the above ines.The equivalent ower source t eachbus s represented y

a shaded ircle.The equivalent ower sourceat the th bus njectscurrentJr into

the bus. It may be observedha t the structure f a power system s such hat

al l the Sources re alwaysconnectedo a commonground node.

Th e networkmodelof Fig.6. lb ha sbeen edrawnn Fig. 6. lc af ier urnping

the shuntadmittances t the buses.Besides he groundnode, t has our other

nociesbuses)at which the current rom the sourcess injected nto the network.

The ine admittance etween odes andk is depicted y !ip= Jri'Further, he

mutual admittancebetween ines is assumedo be zero'

.Linetransformersare represented y

tion by seriesand shunt impedances,

a series mpedance or for accurate epresenta-

i.e. invertedL-network).

apitying Kirchhoff s current aw (KCL) at nodes ,2,3 and ,respectively,we get he following fbur equations:

Jt = Vrlro + (Vr - V) ln * (Vr - Vr ) ),:Jz = Vz)zo (Vz- Vr ) n + (Vz_ V) yz t + (Vz_ Vq )yz q

s= vztgo+ (y : - v) ln + (V t _ V) lz z + (Vt_ Vq )yz qJq= Vayqo (Vq Vr ) zc * (Vq_ V) yzq

Rearrangingand writing in matrix form, we get

(yrc * tn - l n - \ z



* f n ,0

. , ( l z o * l n- Yrz

*hz

*lzq,

- lzt - !z+

_ !*(y:o * yr g

- t 2 3* r z t * y y ,

- l u

0 -Jz+ -lzq

Equation (6.3) can be recognized,o be of the standard orm

(b )Equivalentircuit

(c) Power network of Fig. 6.1 (b) umped and redrawn

Fig. 6.1 Sample our-bus ystem

Y l t= Y r c = - l n i Yv = Yqt - ) r + = 0

Y z q = Y + z = - l z q i Y y = y q z = -l y

Each admittancey,, (i = r,2,3, 4) is calred theserfadmittance or drivingpoint admittance)of node and equals he algebraic um of all the admittancesterminaringon the node.Each ofi-diagonal erm v* (i, k = r, 2, 3, 4) is themutual admittance (transfer admittance) betweennoim i and ft and equals henegativeof the sum of all admittances onnected irectlybetween hesenodes.Further, Yr*= Yri.

using index notation,Eq. (6.a) can be written n compact orm asn

J i= Dyp vp i = r ,2 , . . . , l

k = l

or, n matrix form

"Inus Isus Vnus

(Yqa yz q

* Yz+)

ComparingEqs. (6.3) and (6.4),we can write

(6.4)

Y r r = ) r o * l n + l n

Yzz= lz o * lt z t lz t + lz q

Y t t = ) : o * l n * l z z * l z q

Yu = lq o * lz q * ly

Ytz= Yz t = -ln i YZ t = YZ Z= -

lZ t

(a) One-l inediagram

Fig. 6. 1 Sample four bu s system

(6.s)

(6.6

Sor Soz

v1

,?-blrssystem.Furthermore, yi* = 0 if buses i d k are not connected(e.9. YA = 0). since in a power network each bus is connected nly to a fewotherbuses usually o two or threebuses), he ys.y5of a rargenetwork is very

urhanp V J^-^+^^ aL^ - - ^^ :'vrrvrv -BUS \re,'ulsD urc llralnx or DUS admlttanceand is known as busadmittancematrix. The dimensionof the y"u5 matrix is (n x n) where n is the

numberof buses'[The otalnumberof noclesi e m = n + | including he ground(reference) ode.l

As seenabove, yru, is a symmetric except when phase shifting

transformersare nvolved, so that onlyn v + t )

terms are to be stored for an2

1.,.^L{"

involved algorithms.Furthermore, he impedancematrix is a fuil matrix.**The bus mpedancematrix, however, s most useful for short circuit studies

as will be seen n Chapters10 and 11.Note: In the samplesystem of Fig. 6.1. the

arDrrrary manner, although in more sophisticatedstudies of large powersystems, t has been shown that certain ordering of nodes produfes fasterconvergence nd solutions.Appendix c dealswith the topicsof sparsity ndoptimal ordering.



sparse' 'e' it has a large numberof zeroelementi.-itrougtr tf,i, prop"rty is notevident in a small system ike the sample system under consideration, n asystem containinghundredsof buses, he sparsity may be as high

as 90vo.Tinney and associates[22] at Bonnevile Power Authority were the first toexploit the sparsity eatureof zsu5 n greatlyreducing numericalcomputationsin load flow studiesand in minimizing the memory requiredas only non-zerotermsneedbe stored.

Equation (6.6) can also be written in the form

6.3 FORMATION OF fsus BY SINGULARTRANSFORMATION

Graphnus= Zausleus

where

Zsu5 (bus impedancematrix) = fsLsfbr a networkof fbur buses fbur inclependentodes)

(6.7)

(6.8)

-flmmetricrnus yields symmetric Zsus.The diagonalerements f Zuu, are

calleddriving point imped,ancesf the nodes,and th! off-diagonalelementsarecalled transfer impedancesof the nodes. Zsus need not be obtained byinverting rnus.while y"u, is a sparsematrix,-Auris a full matrix. .e., zeroelementsof I'ru, becomenon-zero n the .o,,.rp"o"niing Zsu, erements.

It is t' be stressed ere that yBus/zBusconstitutemodelsof the passiveportionsof the power network.

Bus admittancematrix is often used n sorving road flow problem. t ha lgainedwidespreadapplicationowing to its simpticity of data preparationandthe easewith which the bus admittance

matrix can be formed ano moOfied fornetworkchanges-addition of lines, egulating ransformers, tc. seeExamples6'2 and6'7)' of course,sparsity s one of its greatestadvantages s t heavilyreducescomputer'memory and time requirernents. n contrast to this, the

zss" can be referred o ground or slack bus. In the former case, t is usuallynecessary 'o creal'eat least one strong artificial tie to ground to avoid numericaldifficultieswhen obtainingZsg5,becausen absence f this, rr * is ilr-conditionedorevensingular' A large shuntadmittance nsertedat the slack bus most simply achievesthe desiredresult[20].**The

disadvantages f the conventional impedancematrix may be overcome bymaking use of LU factorsof the admittancematrix and bf employingcompacrsroragescheme' Piecewise methods or tearing techniques(dialoptics) have recently beenapplied o overcome he disadvantagesf excessiv, rorug. requirementsIlg].**For

convenience,direction is so assignedas to coincide with the assu'redpositive direction of the elementcurrent.

i90 l Modern ower ystemnatysis

O r @

Primitive Network



oFig.6. 2 Linear raph f he circuitn Fig.6.1c

here that each source and the shunt admittance connected across it arerepresentedy a singleelement. n fact, this combinationrepresentshe mostgeneralnetwork elementand is describedunder the subheading primitiveNetwork".

A connecteds'bgraph containing all the nodes of a graph but having noclosedpaths s calleda tree.The elements f a tree are called branchesor ffeebranches.The number of branches b that form a tree are given by

b=m * 1 - n ( n u mb e r o f u s e s ) ( 6 . 9 )

Thoseelements f the graphhat are not ncluded n the treeare called inks (orlink branches) and they form a subgraph,not necessarilyconnected,called

- Branch- - Link

e = 9

m = 5

b = m - 1 = 5 - ' l = 4 = n

l = e - b = 5

@

(a) Tree

(b) Co- reeFig.6.3 Treeand cotreeof the oriented onnected raphof Fig.6.2

cotree.The numberof links / of a connected raph with e elements s

I = e - b = e - m + l

Note that a tree (and therefbre,cotree)of a graph s not unique.

vr , = Er - E,

where E, and E" are the voltages of the element nodes r and s, respectively.It may be remembered here that for steady state AC performan.., ull element

variables (vr* E, 8", irr,7r.,) are phasors and element parameters (zrr, .rr") ar€complex numbers.

The voltage relation for Fig. 6.4a can be written as

v r r * € r r= Z r r i ^

Similarly, the current relation for Fig. 6.4b is

ir , * j r , = l r r v^

(a ) lmpedance orm (b) Admittance form

( 6 . 1 1 )

(6.r2)

rn7

o--------------------+

\. e

@

!r. ,= | /7r ,

A set of unconnectedelements s defined asperformance quations f a primitive networkerre

In impedance orm

I

y r " = E -E "

a primitive network. Thegivenbelow:

Fig.6. 4 Representat ionf a network lement

The orms of Figs.6.4a and, are equivalentwherein heparallelsource urrenrin admittance orm is related o the seriesvoltage in impedanceorm by

Jr ,=-

Yrs€rsAlso

(6.10)

V + E = Z I(6 .13 )

'f{W';lModernPower vstemAnatvsis

In admittanceorm

I + J = W (6.14)

Load FiowStucjiesi flH:<

v = AV,'J' I(6.16)

where the bus incidence matrix A isHere V and.Iare he elementvoltageand currentvectors espectively, nd l andE are he sourcevectors.z and Y are eferred to as he primitive iadmittancematrices, espectively.'Thesere relatedas Z = Y-r.If there s nomutual couplingbetweenelements,Z and Y arediagonal where the diagonalentries are the impedances/admittances f the network elements and arereciprocal.

NeturorkVariables

1 0 0 0

0 1 0 0

0 0 1 0

e

I

2

3 branches



in Bus Frame of Reference

Linear network graphhelps n the systematicassemblyof a network model. The

main problem in deriving mathematical models for large and complex powernetworks s to selecta minimum or zeroredundancy linearly independent) etof current or voltagevariableswhich is sufficient to give the nformation aboutall element voltages and currents. One set of such variables is the b treevoltages.* t may easily be seen by using topological easoning ha t thesevariables constitutea non-redundantset. The knowledge of b tree voltagesallows us to compute all element voltages and therefore, all bus currentsassumingall elementadmittancesbeing known.

Considera tree graphshown in Fig. 6.3a where he ground node s chosenas the referencenode. This is the most appropriate ree choice for a powernetwork.With this choice, he b treebranchvoltagesbecome denticalwith thebus voltagesas the tree branchesare ncident to the ground node.

Bus Incidence Matrix

For the specificsystemof Fig. 6.3a,we obtain the following relations etweenthe nine elementvoltagesand he four bus (i.e. treebranch)voltagesV1, V2,V3and Va.

Vut = Vt

V tz = V z

Vut = Vt

Vuq= Vq

V t s = V t -

Vta= Vt -

V n = V t -

V t s = V q -

V t g = V t -

or, in matrix form

*Anotheruseful set of network variables are the / link (loop) currents which

constitutea zero edundancyset of network variables[6].

0 0 0 1

0 0 1 _ l

0 - l 0l - l 0

0 - 1 1- l 0 1 0

This matrix is rectangular and thereforesingular.as per the following rules:

4

)

67

links

8

9

(6.r7)

Its elementsa,oare ound

and oriented away from the ftth

to but oriented towards he ftth

aik = 1 if tth element s incident tonode (bus)

= - | if tth element s incident

node

= 0 if the ith element s not incident to the kth nodeSubstitutingEq. (6.16) nto Eq. (6.14),we get

I + J = y A V s u 5

Premultiplyingby Ar,

A r I + A r J = A r y A V " u ,

F,ach omponentof the n-dimensionalvector ATI s theelement urrents eaving he nodes7,2, ..., n.

Therefore, he application*of the KCL must result inA r I =o

(6 .20)Similarly, eachcomponentof the vector ATJ

"unbe

recognizedas healgebraicsumof all source urrentsinjectednto nodes ,2, ...,n. These omponents retherefore he bus currents**.Hencewe can write

*Fornode , Arl gives

ir o+ ir , - l : r= 0

-The eader houlci erify this for another ode.*Fornode , AT"/ ives

j61 =currentnjectednto bus 1becausetherelementsonnectedo bus haveno sources.

(6 . 8 )

(6.1e)algebraicsum of the

Vn

v2

v2v2

vr

(6.15)

i ,i l9:4iitl Modern Power Svstem Analvsis.

-J

- - - - - - - " - - ' f -- -

IArJ = Jsus

Equation6.19)hen s simplifiedo

,/eus ATYAV",,,,

- - J - - - . ^ ^ _ - . - - r r ^ ' v r r r r g v l ' v a

the samenodalcurrentequation s (6.6).The bus admittancematrix can thenbe obtained rom the singular ransformationof the primitive Y, i.e.

Yeus= ATYA

Load Flow Studies I ftt^

I

v - l T v , rr B U S = A IA

(6.2r)

(6.22)

(6.23)

0

-! ' t n

A computerprogramme an be developed o write the bus ncidencematrix

* y r : )(yzo yn

- Jtz* lz t * t-zq,

- lzt

- ) r :,.

()'lo f- .yr'-.t'23 -

}'l



A from the interconnection ata of the directedelements f the power systetn.Standardmatrix transpose nd multiplicationsubroutines an then be used ocompute Yu* fiorn Eq. (6.23).

*"*'^-"r

, Example .1 |I

Find the Y6u" using singular ransformation or the systemof Fig. 6.2.

Solution

Y -

Using A fiom

* lzt * Jy,

0 _ ! z c

The elements f this matrix, of course, greewith thosepreviouslycalculatedin Eq . (6.3)

)r o

Figure6.5 shows he one-linediagramof a simple our-bussystem.Table 6.1gives he ine mpedancesdentifiedby the buseson which these erminate.Theshuntadmittance t all the busess assumed egligible.

(a ) Find Yuu. assuming ha t he ine showndotted s no t connected.(b) Whatmodificationsneed o be carriedout n Yuu, if the ine showndotted

is connected.

Fig. 6. 5 Sample ystemor Example .2

Table 6. 1

!zo

-Vro

!qo

lt n

v- ,J Z )

!n

Y.-n

ln IJ1'Eq .

)ro

0

0

0

0

0

!n

0

- . Y r r

0

0

0

jq o

- ft+

0

0

lzq

0

& p u

0.050 . 1 00 . 1 50 . 1 00.05

x p u

0 .150.300.450.E00 .15

(6.17),we ge t

0 0

lzo 0

0 J n

0 0

0 l y

- jzz lzz- l r z 0

-!24 0

0 l n

Line,

bus to bus

t- 21- 32-32434

Y A =

trrrr.F,b"F

'f|/0.!:l Modern Power Svstem Anatvcist

Table 6. 2

f.siffi

where V, s the voltage at the ith bus with respect o groundand /, s the source

current injected nto the bus.

The load low problem s handledmore convenientlyby useof "/, ather han

,I,t. Therefore, aking the complex conjugateof Eq. (6.24), we have

(6.25a)

Substituting or J, = Y*Vr from Eq. (6.5),we can write

Line

r-2G p u

2. 0

1 . 02.0

B , p u

- 6. 0-

J . \ ,

- 2. 0- 3 .0- 6. 0

2-32434 t

k : lsolution (a) From Table 6.1,Table6.2 is obtained rom which yuu, for thesystemcan be written A



as

addedbetweenbuses7 and2.

Ytz,n"* = Yrz,on - (2 - j6) = Yzr.o"*Xrr,n" * = ytt ,ora+ (2 - j6 ) (iv)

Y2z,n"* = Yzz,ot,t (2 - j6)

Modified Y"u, is written below

6.4 LOAD FLOW PROBLEM

The complexpower njectedby the source nto the ith bus of a power sysremIS

Qi (reactivepower) = - Irn

In polar form

V; = ll/,1si6t

Y r* = lY'oleio'r

Real and reactive powers can now be expressed asn

P, (realpower)= lvil D lvkl lYiklcos (0,0

k : li = r ,

P i - jQ i=Vf L Yi l ,Vp ik : l

Equating real and imaginary parts

IP; (real power) = Re ]Vi

t

i = l 2 n (6.2sb)

(6.26a)

(6.26b)

(6.27)

6,) ;

(6.28)

6t - 6');

2, . . . , f l

Qi Qeactivepower) - - lvil D lvkl lYiklsin (e,1+ 6t -

k : l

i = 1 , 2 , . . . , f l

Equations(6.27) and (6.28) represent2n power low equationsat n busesof

a power system (n real power flow equationsand n reactive power flow

equations).Each bus is characterized y four variables; P; , Qi, l7,l and 6i

resulting n 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

considerations llow a power systemanalyst o fix a priori two variablesat

each bus. The solution for the remaining2n bus variables s rendereddifficult

by the fact that Eqs. (6.27) and (6.28) are non-linearalgebraicequations bus

voltagesare nvolved n product orm and sine and cosine erms are present)

and herefore, xplicit solution s notpossible. olution anonly be obtained y

iterative numerical echniques.(6.24)

:flg.,0iil toaern power system Anatvsis

Dependingupon which two variablesare specified a priori, the busesareclassifiednto threecategories.

(I) PQ Bus

r rr rrrrorJyw vr LruD,Lrrc ucr puwers ri ano ai arc known (po, and Qpi arc known

l."l 1:1t"t:i:.t"g ancl , and gile specified).he*tnn*i, ars /,tand

6, .A pure oa d bu s (n o generat ing*if ty-at th e bus, .e., pcr=eci=0) is a

Pp bus.

(2) PV Bus/Generator Bus/voltagre controlled Bus

state variables. These adjustable independent variables are called controlparameters.Vector J can then be partitioned into a vector u of confrolparameters nd a vector p of fixed parameters.

Control parametersmay be voltagemagnitudesat pV buses, eal powersp,, etc.The vector p includesall the remaining parameterswhich are uncontrollable.

For SLFE solution to have practical significance,all the stateand control

(6.30)



At this type of bus P' andeo, are known a priori and v,r and. ,(hence p6;)are specified.The unknowns are e, (henceeo,) and 6,.

' ' \

(3) Slack Bus/Swing Bus/Reference Bus

In a load low studyreal and reactivepowers(i.e.complexpower) cannotbe fixed a priori at alr the buses as the net complex power flow into thenetwork s not known in advance,he systempower oss beingunknown till theload low study s complete.t is , therefore, .."rrury to haveonebus (i.e. heslack bus) at which complex power is unspecified so that it supplies thedifference n the total

system oad plus lossesand the sum of the complexpowersspecifiedat the rcrnainingbuses. y the same easoninghe slackbusmustbe ageneratorbus.The complexpowerallocated o this bus s determinedas part of the solution. In order that the variations n real and reactivepowersof the slack bus during the iterative processbe a snrall percentageof itsgeneratingcapacity, the bus connectecl o the largest generating station isnormally selectedas the slack bus. Further, for corivenience he slack bus isnumbered s bu s 1.

Equations 6.27) and (6.28) are referred to as stutic load ftow equations(SLFE)' By transposingall the variableson one side. heseequationscan bewritten n the vector form

f ( x , y ) - o

,f = vector function of dimension2n

x = d_ependentr statevector of dimension2n(2n unspecifiedvariables)

variables must lie within specified practical limits. These limits, which aredictatedby specifications f power systemhardwareand operatingconstraints,

are describedbelow:(i) Voltage magnirude V,l must satisfy the inequality

lv, l^tn< lvi l ( lv, l_* (6.31)

The power system equipment s designed o operateat fixed voltages withallowable variationsof t (5 - l})Vo of the rated values.(ii) Certain of the 6,s statevariables)must satisfy he inequality constraint

16,- 6ft1 l6i- 6rln,o (6.32)

This constraint imits the maximum permissiblepower angle of transmissionline connectingbuses i and ft and is imposed by considerationsof systemstability (seeChapter12).

\ -

(iii) owing to physical imitations of p and/ore generation ources,po, and

Qci Ne constrainedas follows:(6 .33)

(6.34)

(6.35)

(6.36)

Pc,, ^in 1 Pc, S Pc,. ,n*

Qci, ^rn 1 Qci S Qc,, ^u**

(6.2e)

It is, of course,obvious hat he total generation f realand eactivepower mustequal the total load demandplus losses, .e.

D Po,= t Po ,+ ,l t

D Q c i = l Q o t +Q ri i

where Ptand Qpare system eal and reactivepower oss, espectively.Optimalsharing of active and reactive power generationbetween sources will be

discussedn Chapter7.

-Voltageata PV bus can be maintainedconstantonly if conrollable esource is

available at the bus and the reactive generation equired s within prescribed imits.

where

) = vectorof independent ariablesof dimension2n(2n independent ariabreswhich are specified. prrori)

ffi#ilfl,.l.i| Mod".n o*"t sv"ttt An"lvtitI

"Ihe load low problem can now be fully defined as follows:

Assumea certainnominal bus load configuration.Specify P6i+ iQci at all

the pQbuses this specifiesP, + iQi at thesebuses);specify Pcr (this specifies

P,) and lV,lat all the PV buses.Also specify lVll and 6, (= 0) at the slack bus'

T.L,,- r- . ,-- iolr lac nf thp rrer-fnr u ere snecif ie.d The 2n SLFE can now be solved

(iteratively) to determine the values of the 2n vanables of the vector x

comprisingvoltagesand anglesat the PQ buses, eactivepowers and anglesat

fhepV busesand active and reactivepowersat the slack bus. The next logical

step s to comPuteine flows.'

So far we havepresented,he methodsof assemblinga Yeusmatrix and oad

genpral orm with

Wffi

n n

Pr =.I

Po,- D Po,; Pr= 0) . Equations6.37) anbe solved xplicirlyi : 2 i : 2

(non-iteratively) for 62,61, .., d, which, when substitutedn Eq. (6.3g),yields

madehaveo..ouiLii*i;:;ffi"(;:;'iLT,",",T:,:T:J"#il-,1simultaneously ut can be solvedsequentially[solutionof Eq . (6.3g) followsimmediatelyupon simurtaneousorutionof Eq . (6.37)).Since he sorution snon-iterative and the dimension is retlucecr o (rr-l) from Zrt, it iscomputationallyhighly economical.



flow equationsand havedefined he load flow problern n its

definitionsof various ypesof buses. t has beendemonstratedhat oad flow

equations,being essentiallynon-linearalgebraicequations,have to be solved

through iterative numerical techniques. Section 6.5 presents some of the

algorithmswhich are used or load flow solutionsof acceptable ccuracy or

systemsof practicalsize.

At the cost of solution accuracy, t is possible to linearize load flo-w

equationsby making suitableassumptions nd approximationsso that f'astand

eiplicit solutionsbecomepossible.Such echniques ave value particularly for

planning studies,where load flow solutionshave to be carried out repeatedly

but a high degreeof accuracy s not needed.

An Approximate Load Flow Solution

Let us make he following assumptions nd approximationsn the load flow

Eqs. 6.27)an d (6.28). '

(i) Line resistanceseingsmaiiare reglecie,Cshiintconductancef overhead

lines s alwaysnegligible),.e. P7 , the activepower oss of the system s

zero.Thus n Eqs. (6'21) and (6.28) 1it = 90' and 1ii - - 90o

(ii) (6, - 6r) s small (< r/6) so that sin (6, 6o)= (6r 6r). This is justified

from considerationsof stability (see Chapter 72)'

(i i i ) AII buses ther ha n he slackbu s numbered sbu s 1) arePV buses,.e .

voltage magnitucles t all the buses ncluding the slack bus are specified.

Equations 6.27)and (6.28) hen reduce o

P i = l V i l l vk l Yik l 6i 6r) ; = 2,3, . . . , n (6.37)

n

et=- 'u, '

E

rvkr ly ikros 6,-6u) rv, r2y, , r ;= r ,2, . . . , (6.39)

Since lv,ls are specified,Eq . (6.37) represents se t of linear algebraic

equationi n 6,s ,vhichare n - l) in numberas 6, is specifiedat the slackbus

(6, = 0). The nth equationcorresponding o slack bus (n = l) is redundantas

the reat power injectedat this bus is now fully specifiedas

consider the four-bussamplesystemof Fig. 6.6 wherein ine reactances reindicated n pu. Line resistancesre considerldnegligible.The magnitudeof allthe four bus vortages re specified o be r.0 pu. itJuu, powersLe specifiedin th e tablebelow:

53=- 2 +7O,--r J

j 0 .15

jo.2

iP,tslVzl=

l.o

.S.= I + i^,,-. - I Uz

Fig.6.6 Four-busossless ample ystem

2

Real

demandReactive

demand

Realgeneration

Reactive

generatrcn

1234

Por = 1. 0 Qo t = 0 .5Po z = 7. 0 Qo z = 0. 4Po z= 2. 0 Qo z = 1. 0Poq = 2.0 Qoq = 7.0

061 (unspecified)

Q62 (unspecified)

O63 (unspecified)

06a (unspecified)

P c l = ' -

Pcz = 4'0P c t = o

P c q = o\-,/--r

k : 1Figure 6.6 indicatesbus njections fbr the data specified n the table.

As bus voltagesare specified,all the busesmust havecontrollable e sources.

Il_r::t:: "_bviyusrom the data harbuses3 and 4 haveonrye sources. urther,slnce ffie system s assumeciossless, he real power generationat bus I isknown a priori to be

Pct = Por * Poz * pot * poo_ pcz = 2.0 puTherefore,we have7 unknownsnstead f 2 x 4 = 8 unknowns. n the presentproblem the unknown stateand control variablesare {, e, 60, ct, ecz, eczand Qc+.

'yili.\Modern owerSystem nalysis

I

Though the reali63sesare zero, he presence f the reactive osses equiresthat the total reactivegenerationmust be more than the total reactive demand

(2.9 pu).From he datagiven,Yru5can be writtenas ollows:

4 4

er=D eo , -D ep i

_ r .+r+ _ L, > = u.JJ4 pu (vi i i )Now, let us find the line flows. Equation(5.6g) can be written in the form

( l z l =X , 0 = 9 0 " )

Pi k = - Pk ilvi]-lvkl

si n ({ - 6o )x,o

where



Using the above Y"u, and bus powers as shown in Fig. 6.6, approximate

load flow Eqs. (6.37)are expressed s(all voltagemagnitudesare equal o 1.0

pu)

2 + j0.ET

1 30.891 /O . 4

0.891 7O.04

,38570 .01 5 0.49270 .01 8

0.385 ,p .01 S

1.s02i0 . 11 3

1.502 10 . 13

2

t, jD.4

1.103 j0.0s2

4 1.103 p. 09 2

Pz=3 = 5 (6- 6) + 10 6- 4) + 6.6676- 6q)

Pt =- 2 - 6 .6676- 4) + l 0 (4 - 6)P + = - 2 - 1 0 ( 4 - 4 ) + 6 . 6 6 76 q : 6 )

P* is the real power flow from bus j to bus k.

pn = - pz ,+ si n d, q) -

sin-1.23"

= 0.492 u0 . 1 5 \ r - r / 0 . 1 :

Pt z = - Pz r -1 - si n (4 - 6) = - $n 4'41oL' L

0. 2 02= - 0 '385 Pu (ix)

Pru=- Pqt= + s in ( { - 6o ) 10 s in 5 .11o = 0.g91pu0.1

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 )

e i* =W - lv i - l l vk lcos , { - 6o) "

Xik Xikwhere Q* ir the reactivepower flow from

bus i to bus ft.

Q p = Q , z r = + -I

_1 ,0.2 i. ,

cos d,_ hl = 0.015 u

@

* l i't ',t,

Taking bus 1 as a reference us, i.". 4 = 0, and solving (ii), (iii) and (iv), we

get

4.--O.0ll rad = 4.4I"

4 = - 0 . 0 7 4 r a d = - 4 . 2 3 ' ( v )

6 q = - 0 . 0 8 9 r a d = - 5 . 1 1 '

Substituting 6s in Eqs. (6.38),we have

Qr= -

5 cos 4.4I"-

6.667 cos4.23"-

10 cos 5.11' + 21.667Qz = - 5 cos 4.41" - 10 cos 8.64" 6.667 cos 9.52o+ 21.667

Qz = - 6.667 cos4.23o 10 cos 8.64' + 16.667

Qq = - l0 cos5.11" 6.667co s9.52o 16.667

OI'

Qr = 0'07Pu

Qz = 0'22Pu

Qz = 0'732Pu

Q+= 0.132PuReactive ower generation t the four busesare

Qa=

Qt + 0.5=

0.57puQcz = Qz + 0.4 = 0.62pt r

Qa = Qs + 1.0= 1.L32 u

Qc+= Q++ 1.0= 1.132 u

n:n l |Fig. 6.7 Load low solutionor he our'-busystem

Qrc Qu=# #

cosd1,O = 0.018u

I

(ii)

(iii )

(iv)

(vi)

(vii),

.2Oftril Modern owerSystem nalysisI

eA = e+t = + - -1- .o , (6r 64) 0.04pu*+'0. 1 0. 1

Reactiveower lowson other inescan be similarlycalculated.

Generationsnd oaddemands t all thebuses ndall the line flows are

6. 5 GAUSS.SEIDELMETHOD

The Gauss-Seidel GS) method s an iterative algorithm for solving a set of

non-linearalgebraic equations.To start with, a solution vector is assumed,

practical experiencen a physical situation.One of the

I en*.

carriedout at the end of a complete teration, he processs known as he Gaussiterativemethod. t is much slower to convergeand may sometimes ail to doso .

Algorithm for Load Flow Solution

Presentlywe shall continue o consider he casewhereall buses ther han heslack arePQ buses.The stepsof a computational lgorithm'aregiven below:

1. With the load profile known at each bus (i.e. P^ and 0p; known),allocate*Po, and.Q5; to all generatingstations.



based n guidance rom

equations s then used o obtain the revised value of a particular variableby

substituting n it the presentvalues of the remainingvariables. The solution

vector s immecliately pdatecln respectof this variable.The process s then

repeatedor all the variables hereby completingone iteration. The iterative

processs then repeatedill the solution vectorconvergeswithin prescribed

accuracy.The convergences quite sensitive o the starting values assumed'

Fortunately,n a oad flow studya startingvectorclose o the final solutioncan

be easily dentifiedwith previousexperience.

To explain how the GS method s applied o obtain the load flow solution,

let it be assumedhat all busesother than he slackbus are PB buses.We shall

see ater hat the methodcanbe easily adoptedo includePV busesas well. The

slackbus voltage being specified, here are n - 1) busvoltagesstartingvalues

of whosemagnitudesand anglesare assumed. hesevalues are then updated

throughan iterativeprocess.During the courseof any iteration,the revised

voltageat the ith bus is obtainedas follows:

While active andreactivegenerations re allocated o the slackbus, these

arepermitted

o vary during iterative computation.This is necessaryasvoltagemagnitudeandangle are specified t this bus (only two variablescan be specified at any bus).

With this step,bus iniections P, + Q) are known at all buses ther hanthe slackbus.

2. Assembly of bus' admittance matrix rsus: with the line and shuntadmittance ata stored n the computer,Yru, is assembled y using therule for self and mutual admittances Sec. 6.2). Alternatively yru, isassembled sing Eq. (6.23)where nput data are n the form of primitivematrix Y and singularconnectionmatrix A.

3. Iterative computationof bus voltages V;; L = 2, 3,..., n): To start theiterationsa set of initial voltagevalues s assumed.Since, n a powersystem he voltage preads not too wide. t is normalpract ice o use a

flat voltagestart,** i... initialiy ali voltagesare set equal o (r + 70)except he voltageof theslack bus which s fixed. t shouldbe noted hat(n - l) equations6.41) n cornplex umbers re o be solved terat ivelyfor findin1 @ - 1) complex voltages V2,V3, .., V,. If complexnumberoperat ions lr eDot availablen a computer. qs (6.41)ca nbe convertedrnto 2(n - 1) equations n 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 byperforming in advanceall the arithmetic operations hat do not changewith the iterations.

Define

i = 2 , 3 , . . . , t (6.43)

'Activeand reactive generationallocationsare made on econorfc considerations

discussed n Chapter 7..*A

flat voltagestart means hat to start he teration set he voltagemagnitudes ndanglesat all busesother than the PV busesequal o (i + l0). The slackbus angle sconveniently aken as zero. The voltage magnitudesat the PV busesand slackbus areset equal to the specifiedvalues.

J, = (P, - jQ)lvi [from Eq. (6.25a)]

From Eq . (6.5)

, [ Iv , = * l L , _v , o v o l,,'

L T:i ISubstitut ingbr J, from Eq . (6.39) nt o (6.40)

t^ . ,- l

v,=*l'':jq - t Yit'r'l

"=2'3""' n

Y" I vr* ilt II k * t I

(6.3e)

(6.40)

(6.4r)

The voltages substituted n the right hand side of Eq . (6.41) are the most

recentlycalcuiated updated)values or the corresponding uses.During each

iteration oltagesat buses = 2,3, ..., n aresequentially pdated hroughuse

of Eq . (6.41). Vr , the slack bus voltagebeing fixed is not required o be

updated. terationsare repeatedill no bus voltagemagnitudechangesby more

than a prescribedvalue during an iteration.The computationprocess s then

said o converge o a solution.

206 | Modernpower SvstemAnatvsis

(6.44)

I*r l .d

flows on the lines erminatingat the slackbus.

Acceleration of convergence

acceleration factor. For the tth(r + l)th iteration is given by

up by the use of thebus, the acceleratedvalue of voltage at the

y(r+r)(accelerated) V,Q'* a(v.G+r) V,Ql,where a i s a real numbercalled the accelerationfactor. A suitablevalue of afor any system can be obtained by trial load flow sfudies. A generally

1 7 ( r + l ) -A i

' S- r t ( ,

n

v i ' - t B , o v o , , * t , -f r , o v l , , = 2 , 3 , . . . , n

( V l " ) i l t k = i r l

(6.4s)

The iterative process s continued till the change n magnitude of bus



recommended alue s a = 1.6.A wrong choice of o. may indeedslow downconvergence r evencause he method to diverge.

This concludes he load flow analysis or thJ caseof pe busesonry.

Algorithm Modification when pv Buses are arso present

At the PVbuses,p andrv]rarespecifiedande ancrdare the unknowns o bedetcnnincd.'l'hererirre,he valuesof e and d are to be updated n every GSiteration through appropriatebus equations.This is accomplished n thefollowing steps or the ith pV bus.

l. From Eq . (6.26b)

f " )ei =- Im j yr *

D,y,ovof

L f t : l )

The revised value ofei

is obtainedfrom the above equation bysubstitutingmost updatedvaluesof voltageson rhe right handside. In

fact, for the (.r + 1)th iteration one can write from the aboveequationI

g.(r+t) -Ln],r, ',)

-i y,rv,,(,*t)a 1y.r,1- y,kvk,,,I,u.ro,I L r , f ! , "

)2' The revised alueof {.is obtainedrom Eq. (6.45) mmediatelyollowing

step 1. Thus

6Q+r) ay!+r)

=Ansle"t fei]l- - i Bovo(,+r)D B,ovo,,,l6.s1)

L t1 ' ' ' ) * , r : k : ,+r Jwhere

. ^ ( r + t ) - P - i ? t ' r t tr1 . ; . =__

,u_

6 .52 )

The algorithm for pe buses remains unchanged.

voltage, av.('*r)|, between two consecutiveterations s less than acertain tolerance or all bus voltages, .e.

IAV.G*r)l 1y.t+r) V,e)l < 6, i = 2, 3, ..., n

4.

5 .

computation of slack bus power: substitution of all bus voltagescomputed n step 3 along with V, in Eq. (6.25b)yields Sf = pr- jey

computtt t ion f l ine .f lows:This is thc as tstep n th c oacl low analysiswherein he power lows on the various inesof thenetworkare computed.Consider he line connectingbuses andk. The line and transformersateachend can be represented y a circuit with seriesadmittance y* andtwo shuntadmittances1l;ro nd )no as shown in Fig. 6.8.

Bu s Bu s

(6.46)

(6.47)

(6.48)

(6.4e)

l*io sm

Fig. 6.8 7i'-representationf a lineand ransformersconnected etweenwo buses

The current fed by bus i into the line can be expressed s

Iit = Iitt + Iirc= (V i - 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 isSr i= Vk V*k- Vf) yfo+ VoVfyi,l,

The power oss n the (t - t)th line is thesumof the power flows determinedfrorn Eqs. (6.48) and (6.49). Total transmissionoss can be computedbysununingall the line llows (i.e. 5';a Sri fbr all i, /<).

7 Voltagemagnitudeimits Vr min and Vi max'lor -(J o

b n"u"iiu" pow"t limitsQi min and Q;max or PV buses

l'fiffi

demandat any bus must be in the range e^rn - eû*.If ai any stageduring hecomputatibn,Q at anybusgoesoutside treJeimits, t is fixedAt Qîn or Qô, as the case may be. and thedropped, .e. the bus is now treated ike a pe bus. Thus step 1 abovebranches ut to step 3 below.

3. ff 9.('+r)a 0;,6;n, se t e,Q*r) er,r^n and treatbus f as a pe bus.compute4-(r+1) nd y(r+t) from Eqs. (6.52) and (6.45), espectively. f

Otu.. : , ')Qt , , se tO-(r+l) ei,^*and treatbu s as a pebus. Compure

1 Primit ive matrix

2 Bus ncidencematrix

3 Slack us vol tagelY1l , 1)

4 Real us powersPi for= 2,3'4, " " 'n

5 Reactive us powersQi, or = m + 1" -{fObuses)

Vf I or = 2,....,m PV buses)

tg.Yr,.r.i"g Eq. 6.2

Make nit ial ssumptions io or = m + 1,"',n and O;0or = 2!: 'm

= t 1,"n B;p'for= 1'2"



F i g . 6 . 9 F | o w c h a r t f o r l o a d f l o w s o | u t i o n b y t h e G a u s s . S e i d e |iterativemethod using YBUS

4.Q+r) 6 y(r+l) from Eqs. (6.52)and (6.45), respecrively.

Now all thecomputationalstepsaresummarizedn thedetailed low chartof Fig. 6.9 which serves as a basis for the reader to write his own

computer rogramme. t is assumedhat out of n buses,he first is slackas usual , hen2,3, . . . , m arePVbuses nd he remainingm + l, . . . , ftare PQ buses.

For the sample ystemof Fig. 6.5 the generators re connected t all the fourbuses,while loadsare at buses2 and3. Valuesof real and reactivepowers arelisted n Table 6.3. Al l busesother han he slack are pe typd.

Assuminga flat voltagestart, )nd the voltagesand bus anglesat the threebusesat the end of the first GS iteration.

Solution

Table 6. 3 lnput data

Bu s

I 6.4xampl

Retnarks

Slack bu s

PQ bu s

Pp bu s

Pp bu s

The l"ur for the samplesystem asbeen alculated arlier n Example6.2b(i.e. the dotted ine is assumed o be connected).n order to approach heaccuracy of a digital computer, the computationsgiven below have been

performedon an electroniccalculator.T ) - - - - - - r - - - , , i-Dusvoltages at rne eno or tne llrst rteratronare calculated using Eq. (6.a5).

I2J

4

Pp PU

0. 5- 1 .0

0. 3

0u pu

- 0. 20 .5

- 0 . 1

Vt' Pu

r. u 10 "

v t z = + { W - Y z t v t - Y " v ! - r^'iI

Computehe parameters ; for ^ .and-.;; i: 1 2,-'...,nexcept = ) rom Eqs 6'43)and (6'44)

Set teration ount = 0

l 1 0 . 5 + i 0 . 2' )

- - - - -1 - ' - r - ' - - 1 . 0 4-z + j6 ) - ( - 0 . 6 6 6 j 2 ) _ ( _ t+ f l l Y z z I t - i O

- ' - ' \ - ' r - / \ v ' v v v ' r !|

fei.fii,',1 Modern owerSystemAnalysis -I

4.246 jrr.04= 1.019 70.046 u

3.666 jL r

(vl) T#-frL+re -e2+ 6)to4 o)

- (- 0.666 j2) t + 70 )_ (_ 1+r)j3) t. io) l

+'z^?e;t

1 - -" Ytt

_ 1Yzt

- r.04(- 1 + 13)

LoadFlowStudies

I

- (- 0.666 j2) (1.019 (- 2 *



= ( - itt'lzsJ= , (r.rstz+70.033e) 3 . 6 6 6 _ j t t ) -

or 61 2 1.84658o 0.032ra d

.' . v) = 1.04 co s 6) * j si n dj )

= 1.04 0.99948+ j0.0322)

= L03946+ 70.03351

, ( o

v l = - l - j ' 1 - = i Q t- y . v . - y _ v t _ v , , 0 I t =

1 114tr- Y"v' Yszv;Yrovll

[ - r _ 7 0 . s 3.666-in L ,f;;

- (- t + i3 ) 1'04

- (- 0.666 j2) 1.03s4670.03351)(_ z + io)f

I [ 0 . 3 + o . r il tf:;

- (- I + i3)(1.03e470.0335)

- (- 2 + j6 ) r.03t7 yo.08e3z)IJ

_ 2.967r j8.9962

3 - i g= 0 ' 9 9 8 5 - 7 O ' 0 0 3 1

",,il1;,;llii,""rrirhe

permissibteimits on ez (reactive ower njection) re

- Yo, J

;0.046) i6 ) |

= ''!'-,-1"'.9?' = 1.02870.087u3.666 jr r

I- (- 2 + j6) (1.028 jo.o87) |

)

- 1.025 70.0093 u

- r^rr\l

19+ 70.046)

2.99r j9.2s33 - j e

In Example6.4, let bu s 2 b e a PV bus no w 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), i e", (Note fz= 0, i.e. Vl - I.04 + i0)

n,=_[J

,:,,,:.

A,,i".r!:,,.,,?,:;:::+ (- 0.666+ j2) + (- I + i3)l )

= - Im {- 0.0693 j0.20791 0.2079 u

' O" = 0'2079 u

FromEq . (6.51)

r.0317

ModernPowerSystemAnalysis

0.25 <Qz

< i. 0 pu

It is clear, ha tother data remaining he same, he calculatede2 (= 0.2079) snow less han he Qz, ^in.Hence e, it set equal o ez, _in, .e.

Qz = 0'25 Pu

a- p -- - u v r v r v r v , tf

2tv q L t l l v

longerremain iied at I.04 pu. The valueof V, ar the end of the first iterationis calculated s follows. (Note VL = t + 70 bf virrue of a flat start.)

I a a r { E l a . . , O r . , - r : ^ ^Lwqv I t vyy \ ) t ,U (J lU l i

6.6 NEWTON-RAPHSON (NR) METHOD

The Newton-Raphsonmethod is a powerful method of solving non-linearalgebraicequations. t works faster and is sure to converge n most casesas

vL = -( 'r ,^!?, - yztvt yztv? yrovl)z

Y r ,[

( Y 3 ) -- z t | - z r J - ' "

" )



_ [ o . s - j o . z s

3 .6 6 6j r l L t - r ,

(* 2 + i6)1.04 (- 0.666 .i2) (-

- 4'246- jlr-,?= t.05.59 io.o341

3 . 6 6 6 j t l

(

vj = : l':-:* -.Y,,v,Y,,vJ" r, ,[ (Y,') '

[_t;u.t _ (_ r + j3) oa3 . 6 6 6 - j r r L - ; o *

Considera set of n non-linearalgebraicequationsf i ( \ , x2 , . . , xn ) 0 ; i = I , 2 , - . . , n

r + i 3 ) ] Assur.cnitialvulucs f u'know'sas *l : *'), ..., r"r.Le t J.r(1, g, ..., J_rlbe he corrections,hichon beingaddedo the nitialguess, ivelhe actualsolution. herefore

J t@\+ ut l ,*ur+Axl, . . , ,x0 ,+Axf ;1 g; i = 1,2 , . . ,n (6 .54)Expandinghese quationsn Taylorseries roundhe nitialguess,we have

f;(x01, or ,

(6.ss)

Neglectinghigherorder terinswe can write Eq. (6.55) n matrix fbrm

(6.s3)

- (- 0.666 2X1.0s5e 70.0 41 ) (- 2 + j6 )]

,,r).[[*) ' a*0,(#)' Axt

/ ^ - \ 0 I

*[ -l!t] o- i l* n,rn-, rdere,,nso\ux, ) l

whcrefg) ' , (f , ) ' . . f9 ) '\ d"r / ( Dr, )

' '[ ,", ,/

are the derivatives f It with respect o

x1 ,x2, . . . , x, evaluated t (x! , *1 , . . . , *0, ) .

vL ;;W,#- Yo,v,Yo,u|v*v))

=+,t"=#

- (- +,r3)r'0s0ei0'0341)

_ 2 . 8 t 2 j n . 7 0 9

3.666 jr r

/ , .0630 j9 .42M _3 - j e

1.0347 70.0893u

-1

- (- 2 + j6) r.034i j0.08e3)I

1.0775j0.0923 u

f'''I ' r l

a'l'

Llxi

Ax:

or in vector matrix form

(6.56a)

2ll I Modern Power Svstem Analvsis

- l( , + .f ,A* , * 0 (6.56b)

,f is k,o*n as he Jacobianmatrix(obtained y differentiatinghe function

vector with especto x andevaluatingt at r01.Equation6.56b) anbe

writtenas: l _ J

Approximate aluesof corrections -r0 can be obtained rom Eq (6.57).These

being a set of linear algebraic equations can be solved efficiently by

triangularization nd back substitution see Appendix C) .

Updated aluesof x are then

tl =" 0 + Ax I

(6.63a)

m

It is o be mmediately bservedhat he Jacobian lementsorrespondingo the



or, in general, for the (r + 1)th iteration

" ( r + l ) _ r ( r ) + A x Q ) (6 .s8)Iterations re continued ill Eq. (6.53) s satisfiedo any desired ccuracy,.e.

at Alvml = 0. We

(6.62c)

mth bus

First,assumehatall busesarePQ buses.At anyPQbus the oad low solution

rnustsatisfy he following non-linearalgebraic quations

l J iG" )l < e (a spec i f i eda lue) ; i = 1,2, . . . , n

NR Algorithm for Load flow Solution

fip= Pi (specified) - Pi (calculated) = APi

.f ie= Qi(speci f ied)- Qi @alculated) AQ i

(6.5e)

(6.60a)

(6.60b)

(6 .61a)

(6 .61b)

J'ip lV, 6) = I' i (sPccificcl) Pi = 0

fiq (1v1, 6) = Qi (specified) - Qi = o

whereexpressionsor P, andQ, aregiven n Eqs. 6.21)and (6.28).For a trial

set of variablesV;1, ;, he vectorof residuals0 of Eq. (6.57)comespondso

while the vector of corrections y'xo corresponds to

alvil, a, i

Equation6.51)or obtainingheapproximateorrectionsector anbewritten

for the load flow case as If the mth br-rss also a

can now wnte

II

f th usl, lP i

aQ ii thbuslp, l= l T: tT ' I

L- --J t-- ---- - - - - rH' ' l - l46 ^

AlV^

I(6.62a)

jmtnus

ith bus residualsandmth buscorrectionsare a 2 x 2 matrixenclosed n the box

in Eq. (6.62a)where i and m Ne both PQ buses.Sinceat the slackbus (busnumber ), Prand Qr areunspecifiedand lV,l,

Q are ixed, there are no equationscorresponding o Eq. (6.60) at this bus.Hence he slack bus doesnot enter he Jacobian n Eq. (6.62a).

consider now the presence f PV buses. f the ith bus s a pv bus, e, isunspecifiedso that there s no equationcorresponding o Eq. (6.60b) for thisbus.Therclbrc, the Jacobian leurents f the lth bus becomea sinele rowpertainingo AP,, i.e.

r thbusFl=(6.62b)

PV bus, Vrl becomes ixed so hmthbus

2L6 | rtrodernPower SystemAnalysis

Also if the ith bus s a PQ bus while the mth bus s a PV bus, we can then write

(6.62d)

l ztzt

Ninr= N^ i = 0

l , ^ = J ^ ; = O ( 6 . 6 6 )

corresponding o a parricularvecror of variables tqlv2lq64lval6lr, the



It is convenient or numerical solution to normalize he voltage corrections

alv^ l

l v ^ l

asa consequencef which, he correspondingacobianlements ecome

vector of residuars aP2 aez ah ap4 ae4 Apr], and the Jacobian 6 x 6in this example)

are computed. Equation rc.an is then solved bytriangularization and back substitutionprocedure to obtain the vector of

f n v t A t r l | 1 rcorrections

I oo,# oo1464+ /,6rl . correctionsre henaddedoL lv2l

J alVor I

update he vector of variables.

2(Pa) 3(PVl

(6.63b)

Expressions or elementsof the Jacobian in normalized orm) of the loadflow Eqs. (6.60aand b) are derived n Appendix D and are given below:

Case I m t l

H,^= Li^= a^fi - b*et 6.64)

Nr*=- J i r r= a.er+ bJ t

Yi^= G* + jB,*

Vi = e, + jf ,

(a* i ib) = (Gi- + jBi*) @* + jk\

case2

I,,==t-n, - Biirvirz

Mii= Pi + G,,lV,lz I A A < \\\J' \ ' J , '

J ti = Pi - Giilvil'

L i ;= Qi - Bi i lv i lz

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

Il 2IY

o

f , zJc0

Fig.6.10 Sampleive-bus etwork

-'-* BusNo.

2 3 4 5

Jacobian(Evaluated t tr ialvaluesof variables)

[:_]l a l v 4 l ll]ql-l

i_it_lt

Correctionsin variables

Hzz Nzz Hzs Hzn Nza

Jzz Lzz Jzs Jz+ Lz+

Hsz Nsz Hss Hss

Hqz Nn Haa Naz. H+s

Jqz Lqz Jcc Lu J+s

Hss H5a Nsq Hss

t

(6.67)

,re I rrrrodcrn ower Svstem AnalvsisLL V I-T--

Iterative Algorithm

Omittingprogramming etails, he iterativealgorithm br the solutionof the

load low problemby the NR method s as follows:

L W'ith voittge anclangle (usually f'= €I) at s

absence of anY tltherat atl PQ busesand d at all PV buses' n the

information flat voltagestart s recommended'

2. ComputeAP,(for PV and PB buses) nd AQ,(for aII PQ buses) rom

Eqs. (6.60aandb). If all the valuesare ess han the prescribed oletance,

stop the iterations,calculate P, and Q, and print the entire solution

including ine flows.

Lcad Ftor; Studies I i:ig,:

-Find the oad flow solutionusing the NR method.Use a toleranceof 0.01 forpowermismatch.

Solution Using the nominal-rcmodel or transmissionines, "u, for the givensystem s obtainedas follows:

eacn tn e

1,sen . "=I

^ -z .g4r - j11.7& r2 . r3 l -75 .96"eso.oz+jo.og

Eachoff-diagonal erm = - 2.94I + jll.764

Eachself term = 2l(2.941- j11.764)+ j0.011

= 5.882 j23.528= 24.23 -75.95"



3. If the convergence riterion is not satisfied,evaluate elementsof the

Jacobian singEqs. (6.64)an d (6.65).4. SolveEq. (6.67) or correctionsof voltageanglesand magnitudes'

5. Update voltage angles and magnitudesby adding the corresponding

changeso th eprevious aluesan dreturn o step2'

Note: 1. In step2, if thereare limits on the controllable B sources t P V

buses,Q is computedeach ime and f it violates he limits, it is madeequal o

the limiting value and he correspondingPV bus is made a PQ bus in that

iteration. f in the subsequentomputation,Q doescome within the prescribed

limits, the bus is switchedback to a PV bus.

2. If there are imits on the voltageof a PQ bus and if any of these imits

is violated,the correspondingPp bus s madea PV bus in that terationwith

voltage fixed at the limiting value.

Consider he three-bus ystemof Fig. 6.11.Eachof the three ineshas a series

impedanceof 0.02 + 1O.OSu and a total shuntadmittanceof 70'02 pu' The

specifiecl uantities t the _b"!91j9!qulatedbelow:

R e a l l o a d R e a c t i v e l o a d R e a l p o w e r R e a c t i v e V o l t a g e

0 +, O

2

Fig.6.11 Three-busystemor Example .6

Pz = lVzl vl lY2l cos 0r1 6r $) + lVrlz yrrl cos0zz lvzl \llYr r l os Qzt+q- 6)

Pt = lVl lvl l\rl cos dr ,+ 6r 6r )+ lVrl v2 l \zl x cos ez + h- 6r)+ lV,rlzyrrl cos0r,

Qz =- ivz i ivl tyz l s in gr+ dt - h) - lvz lzy22 l sn 4z_ lv2 llv3l Y.,.lsin (0r,+ bz 6)

Substitutingivenandassumedalues f different uantities,e get hevaluestr fPowersts

PB. -0.23pu

To start terationchoose4= t +70 and 4 = 0. From Eqs. 6.27)and (6.2g),we ge t

Bus

2. 0 1 .0 Unspecified UnspecifiedV'=1.94* ;g(slackbus)

0. 0 Unspecified

(P O hus)

0. 6 Q c t = ? l v l l = 1 . 0 4

(P V bus)

Contro l lablccact ivo oworsourccs avai lablc t bt ts3 with he ct lnst ra int

0 < Q c t S 1 . 5 P u

demand

PD

demand generation power specification

Qo PG generation O6

0. 0 1 .0.5

0. 05

3 1.04O"1. 5 7O.6

O } n r ' l i l a d a - n D a r r r a r C r r a l a m A n a h r a i at t v . l r v r u u E r r r r V Y Y E r \ ) y o l g l t r n r r q r y o r o

t P3 0.12 u

Qor= 0 '96 Pu

Power residualsas per Eq. (6.61) are

- rt (calcu

(- 0.23)= 0.73

Aror- -1.5 (0.12) - ! .62

tQT= 1 (- o'e6) t 'e6The changes n variablesat the end of the first iteration are obtainedas ollows:

Load FtowStudies i zzrt i ^ - - i .

S z = 0 . 5 + j 1 . 0 0 | -

S r = - 1 . 5 - j 0 . 1 5

Transmissionoss= 0.031

Line flows

The the eal partof line flows

0.1913r2E000.839861E'00- l

0.0 0.6s4697 00-0.673847E00

following matrix shows

I o o

I-0.r8422eE00L-0.826213^800



0P, 0P, 0P,

06, 061 alv2l0P, aP, 7Pu

06, 061, 0lv2laQ, }Qz aQ,

06, 06: 'av |

Jacobian elements can be evaluated by differentiating the expressionsgiven

above r> r Pr, Py Qz with respectto 6r , d1an d lVrl and substi tut ing th e given

and assumed values at the start of iteraticln. The chanses in variables are

obtained as

The0.0 J

the maginaryartof line lows-0.5994&E00 _0.r9178zE00]

0.0 0.39604s800-0.37sr6s800 0.0 I

Rectangular Power-Mismatch Version

This version se se, ancl.f th e rear ancr maginary parts of th e v. l tagesresl lect ivcly, s variables'fh e

numberof equations nd variabless greaterthan hartir r Eq . (6.6r), by th enurnber ttpi buses. inceat pv buses ,and.f; canvary but ,,' +.f,' - lviP, a voltage-magnitudequarecl isn-latchquationis required br eachPV bus. with sparsityprogramming, his increasen orderis of hardly any significance. ndeed

each teration s rnarginally aster han forEq ' (6'67) since herear e no time-consunringin eancl osine erms. t nray,however,be noted hat even the polar versionavoids heseas far as possible

9fusing rectangulararirhemetic in constructingEqs (6.64) and (6.65).

However' hc rcct: t t tgtrl l t rcrsion ccnls o bc sl ighi iy .r , rcl iable ut aster nconvergencehan he polar version.

The otal numberof non-linearpower low equations onsideredn thiscasear c f ixed ur dcqual2 (trl). These bllow fr.o'i Eqs. (6.26a) nd (6.26b)an dare

following matrixshows

I o.oI

l|0.60s274800L0.224742E00

I t d j l l - 2 4 . 4 i- t 2 . 2 3 s . 6 4 - 1 - r [ 0 . i 3 f[ - 0 . 0 2 3 - 1

| ^ r I | . ^ ^ ^ ^ ^ - l | - ^ l | ^ ^ . - . 1I A a i l : l - t / . . 2 5 / . 4 . e ) - J . u ) l | - t . o z l : l - u . u o ) 4 |

[nrv,r ' - ]-uu 3.0szz.s4) r.qol ooar l

la ) l Ia i - l l - ^4 I t -0 .1 002 .3I [ 0023.1

I a ] : l I l * l ^ 4 l : l o l * l - 0 0 6 s 4 1 : l - 0 0 6 s 4 1Itv,t ' jL 'y,roJnrv.,r ' .1' i I oosoJ r.08eJ

We can now calculate usingEq. (6.28)]

Qt t= 0.a671

Qo\= Q\ + Qr t = 0.4677+ 0. 6 = 1.0677

which s within l imits.

If' the sanre problem is solved using a digital computer, the solutionconvergesn three terations. he final resultsare given below:

Vz = 1.081 - 0.024ra d

Vt = I.M l- 0.0655 ad

Qu = - 0. I5 + 0 .6 = 0.45 wi th in in r i t s )

Sr = 1.031 j( - 0.791)

P, (specifi"d) I{ e,(epG11,f,B,t) + fihrG* -t epB,o} 0k: l

i = l , 2 , . . . ,n

i = slack(s)

nr t .

B; (specif i"d) )lf,koG,o

- ftB,t1- e;(. f*Gi* eoB,oy]}gk =l

(6.68)

(lbr each PQ bus) (6.69)

222 | Modern Porgl Jyllem AnatysisI

l( l (specified)2 @i2+f,')

= 0 (for each PV bus) (6.70)

Using the NR method, he linearisedequationsn the rft iterationof iterativeprocess an be written as

Jr J2

Load FlowStudies

necessarymotivation n developing he decoupledoad flow (DLF) method, nwhich P-6 and Q-V problemsare solvecl eparately.

Decoupled Newton Methods

F o r * j

Juj= - Jqij = Gij ei fi

Jzi j= J t , j=- Bi jei+ Gi l f i 6 .72)

In any conventionalNewton method,half of the elementsof the Jacobeanmatrix represent he weak coupling referred o above, and thereforemay beignored' Any suchapproximation educes he true quadraticconvergenceogeometricone, but-thereare compensating omputationalbenents.i largenumber of decoupled algorithms have been dlveloped in the literature.However,only the most popular decoupledNewton veision s presented ere

AQ

AIVP

J3 J4

Js J6



Jsij= Jo,j = O

F o r = , r

t;^"':::-::::!-

tlr,,,-!'u,u-',) J:,

Jsii= 2" i J6ii = 2fi

a, and b, are he componentsof the current louring into node i, i.e.,

a, + b, =frCo * jBi)@r+ f* )

k: l

(6.74)

Steps n solution procedureare similar to the polar coordinatescase,exceptthat the initial estimates f real and maginarypartsof the voltagesat the PQbuses re madeand he corrections equiredare obtained n each terationusing

The correctionsare hen applied o e andf and he calculationsare repeatedillconvergences achieved.A detailedflow chart describing he procedure orload low analysis singNR method s given n Fig. 6.12.

F E F F A A ' ? h I E h ' A A

O. I L'|aUL,UTL|1L' L('AL' .F LL'VV IVI.EI.tsI(-,L'Ii

An important characteristic f any practicalelectricpower transmission ystemoperating n steady tate s the strong nterdependenceetween eal powersandbus voltages ngles nd between eactive owersand voltagemagnitudes. hi sinterestingpropertyof weak coupling betweenP-6 and Q-V variable.s ave he

In Eq. (6.67), he elementso be neglected re submatrices[1v]and [,/]. Theresultingdecoupled inear Newton equations ecome

tApl = lHl lA6l

I^Qt tLt 4!1L t v t J

where t canbe shown hatgU= Lij = lvil lvjl (CU in

(6.73)

(6.7s)

H i i = - 8 , , l V i P -

Lii= - 8, , lV, lz+

6u B,i cos {r)i * j

(Eq. (6.6s))

(Eq. (6.6s))

Qi

Qi

(6.76)

(6.77)

(6.78)

(6.7e)

(6.80)

Equations 6.76) and 6.77) can be constructed nd solvedsimultaneouslywith each other at each teration, updatingthe [H] and [r] matrices n eachiteration using Eqs (6.78) to (6.80). A betteropproo.h is to conducteachiteration by frst solving Eq. (6.76) for 44 *o use the updared 6 inconstructingand then solving Eq. (6.77) or Alvl. This will result n fasterconvcl'geucehan n the sirnultaneous ode.

The main advantage f the DecoupredLoadFlow (DLF) ascompared o theNR method s its reducedmemory requirementsn storing the Jacotean.Thereis not much of an advantagerom the point of view of speed ince he time periteration of the DLF is almost the same as that of NR method and it alwaystakesmore numberof iterationsto convergebecause f the approximation.

Fast Decoupled Load Flow (FDLF)

Furtherphysicallyustifiablesimplificationsmaybe carriedout o achieve omespeed advantagewithout much loss in accuracyof solution using the DLFmodel described n the previous subsection.This effort culminated n thedevelopmenr f the FastDecoupledLoad Frow (FDLF) merhodby B. stott in1974l2ll. The assumptions hich arevalid n normalpowerryr,"n1operationare made as follows:

'* iI vodern Power SvstemAnalvsis

- : : , : - , t

I

I zas-

With theseassumptions,he entriesof the [1{ and [L] submatrices ill becomeconsiderablysimplified and are given by

IAduanceterationountI r = r + 1

Determine -1

.'. comPuteAefr) 2n6

Lfiv)

and ?=i: -' ' i,,,11,,u": ,::(6 .82)

(6.83)

(6.84)

(6 .8s)

c-mpute ano--lprint ine lows,l

power oss, Ivoltages, I

elci ----.]

Matrices /fl and [L] are squiuemafricswith dimension(npe+ npy) and nrnrespectively.

Equations(6.76) and (6.77) can now be written as

LAPI ltyjt vjl Bfi AA

tAQl= [%t v j tB,, i j ,[#]



Are al lmaxA within

tolerance2

(6.86)

(6.87)

Determine axchangenpowermax APr,AO

where 8t,,, B(are elementsof[-

B] matrix.

Further decouplingand ogical simplificationof the FDLF algorithm s achievedby :

1. Omitting from [B/] the representation f those network elements thatpredominantly affect reactive power flows, i.e., shunt reactancesandtransformeroff-nominal in-phase aps;

2. Neglecting rom [B//] the angleshifting effects of phaseshifters;

3. Dividing eachof the Eqs. (6.84)and (6.85)by lv,l and setting Vr l= I pu

in the equations;

4. Ignoring series esistancen calculating he elements f tdll which thenbecomes he dc approximationpower flow matrix.

With the above modifications, the resultant simplified FDLF equationsbecome

I astnode')

Compute

el,),1v112

-' ls-\

gt'r ioi)

IAP| ly l I = [B'] [L6]

tAQl tr 4 - LB" t^lytl

Fig.6 .12

In Eqs. (6.86)and (6.87),both [B/] and Bttf are eal, sparse nd have thestructuresof [I{ and [L], respectively.Since they contain only admittances,

they are constantand need to be invertedonly once at the beginningof thestudy. If phase shifters are not present,both [B'l and lB"] are always

symmetrical, and their constant sparseupper riangular factorsare calculatedand stored only once at the beginningof the solution.

Equations 6.86)and (6.87) are solvedalternatively.alwaysmploying hemost recent voltage values. One iteration implies one solution for [4fl, toupdate d] and henone solut ion or IA lVl] to update Vl ] to be called -dandl - V i t e r a f i o n S e n a r a f e . c r l n v e . r s e n ce t c e t c a r e a n n l i e d f h r t h e r c q l qn A rcqot lvc

" - r * ^ * ' - - - ^ ^ ' - ^ o - ^

power mismatches s ollows:

max [APf I €pi max lAQl S eo

where eoand €e are the tolerances.

A flow chart giving FDLF algorithm s presentedn Fig.6.13.

cos 6, : 1;

s i n{ r : 0

G,, sin 6,,< B4;

Qi < Bi i l v i l z

ComputeLlVi,12 1V,cnd12- %

l2

and

(6 .81) (6.88)

,bZC Modern owerSvstemAnalvsis

Consider he three-bussystemof Example6.6. Use (a) DecoupledNR method

(a) Decoupled NR method:

Equations o be solved are (seeEqs. (6.76) and (6.77)).Substitutingelevantvalues n Eqs. (6.78)- (6.80) we get

Hzz= 0'96 + 23.508 24.47

Hzt= Hzz= 1.04 - Brz)= - 1.04 x 11.764 - 12.23

LoadFtowStudies I ##

(Start

Read LF data and form Ysr.



H t t = - Q t - B n 0 . 0 q 2

= [- Bs r t\P -42 tv3t Bz z \P ] - Bn (r.0a1.2

= - f 1.764 (1.04)2 1t .764 x 1.04+ ( l .oq2

x 2 x 2 3 . 5 0 82 5 . 8 9

Calculate 6i.1= 6[+ 66[

(i )

(ii)

Lzz= Qz Bzz= t + 23.508 24.508

[ 0.731_ [ 24.47 r2.2311\6t\1

l-t.oz)-

L-n.23 zs.ssJ[64trl

tLQzllz4.srl+uJ"'l' L 't;(-i)l

SolvingEq. (i) we getL6;') = - 0.0082 0.0401 - 0.002

Aa{tr= - 0.018 0.08 - 0.062

Qz= - lvzl vtl lYzl sin 0r, + 6r 6r) - lvzlzV2zlsin82

lvzl lh l llrrl sin (9zt+ 6, - q)

= - 1.04 12.13 in 104.04 0 - 0.115)24.23

sin(-75.95) 1.04 12.13 in (104.04 .115 3.55")

- - 1 2 . 2 4 + 2 3 . 5 0 5 - 1 2 . 3 9

Qz= - l '125

A r - l - 1 / 1 1 a < \ ^ 1 ? r Et - t v 2 - r -

\ - r . r L J ) = L . I L J

Substituting n Eq. (ii)

f o r i = 1 , 2 , . . . , n , i - s

1. 6.60a)= PVbus

------X

- ' - -I

r c E q . l

. ,n, I: u s - l

; Afr/,R

Calculate libus poweral l ine lo v

an d print

-- re \-

<'-fr= 1 --

l slacker andflows

e+

Solveor A6f usingEq.(6.86)=1,2, . . . ,, * s

Lz.rz5r[24.5r]alY"( ') l l

L u,a,Flg.6.13

;'..ng Modernower ystem nalysis -

AN9

l= 0.086

lvftt - lvlo)t+atv|'tt 1.086 u

Q(rl canbe similarly calculated sing Eq. (6.28).

The matrix equations or the solutionof load flow by FDLF methodare [see

Eqs. (6.86)an d (6.87)l

lui:.

completean teration.This is becauseof the sparsityof the network matrix and

the simplicity of the solution echniques. onsequently,his method requiresless ime per iteration, With the NR method, he elementsof the Jacobianare

to be computed n each teration, o the ime s considerablyonger.For typicallarge systems, he time per teration n the NR rnethod s roughly equivalent o7 times that of the GS method [20]. The time per iteration n both thesemethodsincreases lmost directly as the number of busesof the network.

The rate of convergenceof the GS method is slow (linear convergence

characteristic),equiring a considerablygreaternumberof iterations to obtaina solution han he NR methodwhich hasquadraticconvergence haracteristics

and is the best among all methods from the standpointof convergence. naddition, the number of iterations or the GS method ncreasesdirectly as the

-Brr1l f',f-8,,)Lz4"

(iii)



number of busesof the network, whereas he numberof iterations for the NRmethod remains practically constant, independentof system size. The NRmethod needs 3 to 5 iterationsto reach an acceptablesolution for a large

system. n the GS method and other methods,convergences affected by thechoice of slack bus and the presenceof seriescapacitor,but the sensitiviry ofthe NR method s minimal to these actors which causepoor convergence.

Therefore, for large systems he NR method is faster,more accurateandmore reliable than the GS method or any other known method.In fact, it works

for any size and kind of pro6lemand is able to solve a wider variety of ill-

conditioned problems t23). Its programming logic is considerably morecomplex and t has he disadvantage f requiring a largecomputei memory evenwhen a compact storage scheme s used for the Jacobian and admittancematrices. In fact, it can be made even faster by adopting the scheme of

optimally renumberedbuses.The method s probably best suited for optimalload flow studies Chapter7) because f its high accuracywhich is restrictedonly by round-off errors.

The chief advantage f the GS method s the ease f programmingand most

efficientutilizationof core menrory. t is, however, estrictedn use of smallsize system becauseof its doubtful convergence nd longer time needed or

solution of large power networks.

Thus the NR method s decidecllymore suitable han he GS method for allbut very small systems.

For FDLF, the convergences geometric, wo to five iterations are normally

required for practical accuracies, nd it is more reliable than the formal NRmethod. This is due to the fact that the elementsof [81 and [Btt] are fixed

approximation o the tangents f the defining unctions LP/lVl andL,QAV , and

are not sensitive o any 'humps' in the ciefining unctions.

fi LP/lVl and A^QIIV arecalculatedefficiently, hen he speed or iterations

of the FDLF is nearly five times hat of the ormal NR or about wo-thirds that

of the GS method.Storage equirements re around60 percentof the formal

NR, but slightly more than the decoupledNR method.

and

lffil=r-Bzztatrt')tl (iv)

fil6411oqrl23.508.1la4"l

I o.tz I| -t.oz I -|

-l -

| 1.04 IL: - 1'5571

Solving Eq. (v) we get

46;r) -

A6t''=

6')lz.rzsl

alvtl=

l v t t =

- 0.003

- 0.068

- 0.003ad; {t l

[23.508]atvitl

0.09

1.09 u

Now Q3 can be calculated.

These values are used to computebu s

iteration.Using the values of 'LAPAVll and

solved alternatively,using the most recent

within the specified imits.

0.068 rad

power mismatchesor the next

lAQAl\l the above equationsare

values, ill the solution converges

(v)

6.8 COMPARISON OF IOAD FLOW METHODS

In this section,GS and NR methodsare comparedwhen both use liu5 as the

network model. It is experienced hat the GS method works well when

programmed sing rectangular oordinates,whereasNR requiresmore memory

when rectangularcoordinatesare used.Hence,polar coordinatesare preferred

for the NR method.

ilai:i;'l ,odern powersystemAnatvsisI

Changes n systemconfigurations can be easily taken into account andthoughadjusted olutionsake many more teration.s,achone of them akesless time and hence he overall solution time is still low.

system,buseswith generators reusuallymadePV (i.e. voltagecontrol) buses.Load flow solution hengives he voltage evelsat the oad buses.f some of

lines for specifiedvoltage imits cannotmeet he reactive oad demand reactiveline flow from bus 1o bus t s proportionalo lAvl = lvil - lvkD.Thissituation

The FDLF canbe employed n optimizationstudiesand s speciallyused orthe oad bus voltageswork out to be ess han he specified ower voltage imit,i t is indicative of the fact that the reer-f ivc n^rr/ar ff^.,, ^-*^^ie, ^r4-^-^--:--:-

estudies,as in contingencyevaluationenhancement nalysis.

for system security assessment nd

Note: When a seriesof load flow calculationsare performed, the finalvaluesof bus voltagesn eachcaseare normally usedas he nitial voltagesofthe next case.This reduceshe numberof iterations, articularlywhen hereareminor changesn system onditions.



tAvt tEthttv!= -ffi O,

tv ';t=Erhl+#l

n,

=tvi t *hn,

Flg.6.1S

Since we are consideringa voltagerise of a few percent, V(l canbe furtherapproximated s

lV|l= lV,l+ffiO,

Thus the VAR injection of +jQc causes he voltageat the jth bus to riseapproximately

by (XhllViDQ.. The voltagesat other load buseswili also riseowing to this injection o a varyingbu t smallerextent.

Control by Transformers

Apart from beingVAR generators,ransformersrovidea convenientmeansofcontrolling real power' and reactivepower flow along a transmissionine. As

6.9 CONTROL OF VOLTAGE PROFILE

Control by Generators

Control of voltageat the receivingbus n the fundamentalwo-bus systemwasdiscussed n Section5.10.Though the samegeneralconclusionshold for aninterconnected ystem,t is important to discuss his problem n greaterdetail.

At a bus with generation,voltage can be conveniently controlled byadjustinggenerator xcitation.This is illustratedby meansof Fig. 6.14 wherethe equivalent generatorat the ith bus is modelled ty a synch.onou, reactance(resistances assumed egligible)and voltagebehindsynchronouseactance.timmediately follows upon applicationof Eqs. (5.71) and (5.73) that

P ai + Qe t6iVilt

t i

Xei

P = |v,l -Epl.;;,' -'-i,xo,

eci=#r- ,vit+tEcit)

W i t h / P t i A \ - - , { l r / l , / f ^ : , - ^ - L - - r r - - r , f t

YYrtrr\rGi - iVci l an o i 't l r i Lo i glven bY-ihe ioaci i jow soiution- these velrres

(6.8e)

(6.e0)(6.e1)

','d&r1'lModernPower SvstemA-nalvsis- l

has already been clarified, real power is controlled by means of shifting thephase of voltage, and reactive power by changing its magnitude. Voltage

magnitude an be changedby transforrners rovided with tap changing under

Ioad TCUL) gear.Transformers pecially designed o adjustvoltagemagnitude

smail values are calleo re rnxers.

Figure6.16 showsa regulating ransformer or control of voltage nagnitude,

which is achievedby adding n-phase oostingvoltage n the ine. Figure 6.17a

shows a regulating transformer which shifts voltage phase angle with no

appreciablechange n its magnitude.This is achieved by adding a voltage n

series with the line at 90" phaseangle to the corresponding ine to neutral

voltage as illustrated by meansof the phasor diagram of Fig. 6.17b. Here

= - jJT)

Since he transformer s assumed o be deal, complex power output rom itequalscomplex power input, i.e.

51 V, / i= c rv l for

For the transmission ine

I\ = ! @V 1- V2 )

or

Ir = dl'a lo:lzyV, cfyVz

Also

(6.e4)



VLn= (Von tV6) Q Von= dVon 6.92)

where = (1- jJ- l t )=71-tan-t J1 t

since is small.

The presence f regulating ransformersn lines modifies he l/uur matrix

thcrcby n<i l i fy ing he oacl low solut ion.Consicler l ine, connect ingwo

buses,having a regulating transformerwith off'-nominal turns (tap) ratio a

includedat one end as shown n Fig. 6.18a. t is quite accurate o neglect he

small mpedance f the rr:gulatingranslonucr,.c . i t is rcgardcd s an iclcal

device.Figure 6.18bgives the corresponding ircuit 4epresentationith line

representedy a series dmittance.

-_o A/

Iz = l (Vz- oV,) = - WV r + lV z (6.es)Equations 6.94)and (6.95) cannotbe represented y a bilateral network.

The I matrix representationan be written down as follows liom Eqs. (6.94)

an d (6.95).

n. ,%n

C...,%n.___l"r<_tc,

(b )

(a )

J

I( V " n + t V " r ) = a V " ,

/ ^ ic raal nr rmhor\

Flg.6.16 Regulat ingransformeror control f voltagemagnitudeFig.6.17 Regulatingransformeror control f voltage haseangle

I ModernpowerSystemAnalysisI

Bus,1Bq s

neffitingtransformer

(a )

ao.O,o* a,r.,.r __f#

(i ) V/V\ = l .M or a - l/1.04(ii) V/V\ = d3 or d = s-i3o

Solution (i) With regulating transformerin line 3-4, the elementsof the6.2 are modifiedas under.orrespondingubmatrix n (v ) of Exa

sz v2aV'l <L I

. - - - . - O-- f__1--* -<- I

l''r Y 12

(b)

Fig.6.18 Linewith regulatingransformer nd ts circuit epresentation

3 4-r.92+ s.777

3 - j e

5 1 6j10.s47

r.92js.777

j2 )

3



The entries f f matrixof Eq. (6.96)would thenbe used n writing, the rru,matrixof the complete power network.

For a voltageregulatingtransformera is real, .e. d = e, therefore,Eqs.(6.94)and (6.95) can be represenred y the zr-networkof Fig. 6.19.

Fis.6.1e:jiiil:':3;:i;ffiTH#*:

with rr-nominarap ettinsr

If the line shown n Fig. 6.18a s represented y a zr-networkwith shuntadmittance u at eachend,additional huntadmittancelol2ysappearst bus 1and yo at bus 2.

The above derivations also apply for a transformer with off-nomin al tapsetting' where a = (k[nuJ(kD,r:up, a real value.

The four-bus systemof Fig. 6.5 is now modified to include a regulatingtransformern line 3-4 near bus 3. Find the modified rru5 gf the systemfor-With

off-nominafup **ing transformerst eachend of the ine as shown nFig.6.20,we canwrite.

t c

Fig. 6.20 Off-nominalransformerst both ineends a., a, real)

I t -Y l f v i l - [ I i - l

L- , y) lv t - Ut )

ly . i =[o, o I [ - ' 4 . l l r ll t /q o I f4 lLv ; l - l o q )Lv , ) ' L , ; l : [ v " , ) ] , 1

Substituting (ii) in (i) and solving we get

| 4t-"!-',vf

[q 1: '"1L-otory 4y ) LV2 ltz )

- - l o?v -aqzt f'=l-^*, d; )

Note: Solve it for the case when ao &2 are complex.

(6.e6)

j2 )

j6 )

a

J

r- j3)-(0.666

-(2 j6)1T

Gz +

3

4

j6)

4

i3"e2 +

3 - j e3- e

3.e

(i) .

(i i

/ i i i \

Thus

(iv)

of Example6.2 is

i n (Modifiedsubmatrix (v )

4

.3r13

5.887

. l

236 | ModernpowerSystemAnqtysisI

6.10 CONCLUSION

In this chapter, erhaps he most mportantpowersystem tudy,viz. oad flowhas been ntroducedand discussedn detail. mportantmethodsavailablehave

methods s the best,becausehe behaviourof different oad flow methods sdictated by the types and sizesof the problems to be solved as well as theprecisedetails o implementation.Choiceof a particularmethod n any givensituation s normally a compromisebetween he variouscriteriaof goodnessofthe load flow methods. t would not be ncorrect o say hat among he existingmethodsno single oad flow methodmeetsall the desirable equirements f anideal load flow method;high speed, ow storage, eliability for ill-conditionedsystems,versatility n handlingvariousadjustments

Load FlowStudies

Bu s I is slackbus with Vr = 1.0 10"Pz+ iQz = -5.96 + j1.46

lVTl= 1102

Assume:

\ = 1.02/0" and vi -710"

1 2 3t- '

oottooott.-o.oz

.7orr:"

Fig. P-6.2

6.3 For the systemof Fig. P-6.3 ind the voltageat the receivingbus at theend of the first iteration.Load is 2 + 70.8pu. voltage at the sending end



nd simplicity in program-

ming. Fortunately,not all the desirable eaturesof a load flow method areneededn al l s i tuat ions.

Inspite of a large number of load flow methodsavailable, t is easy to seethat only the NR and the FDLF load flow methodsare he most mportantonesfbr generalpurposeoad flow analysis.The FDLF method s clearly superior othe NR method rom the point of view of speedas well as storage.Yet, the NRmethod s stili in usebecause f its high versatility,accuracyand eliability andas such s widely being used or a variety of systemoptimizationcalculations;it gives sensitivityanalyses nd can be used n moderndynamic-response ndoutage-assessmentalculations. f course ewermethodswouldcontinue o bedevelopedwhich would eitherreduce he computationrequirements or largesystemsor which are more amenableo on-line implementation.

PROBEVI

For the powersystemshown n Fig. P-6.1,obtain hebus ncidencematrixA. Take ground as reference. s this matrix unique?Explain.

(-)

Fig. p-6='!

F<rr he network shown n Fig. P-6.2,obtain the complex bus bar voltageat bus 2 at the end of the first iteration..use the GS method. Lineimpedances hown in Fig. P-6.2 are n pu. Given:

(slack)is 1 + /0

pu.Line admittance s 1.0

-74.0 pu. Transformerreactances 70.4 pu. off-nominal turns ratio is lll.04. Use the GS

technique. ssumeVn = ll0.

c>+J=---F"dFig. P-6.3

6.4 (a) Find the bus ncidencematrix A for the four-bussysrem n Fig. p-6.4.

Take ground as a reference.

(b) Find the primitive admittancematrix for the system. t is givbn rhat allthe lines are charactenzed y a series mpedanceof 0.1 + j0.7 akrnand a shunt admittanceof 70.35x 10-5 O/km. Lines are rated at220

kv.(c) Find the bus admittancematrix for the system.Use the basevalues

22OkV and 100MVA. Expressall impedances nd admittancesn perunit.6.r

6. 5

1

1 1 0 m

3

Fig. P-6.4

Consider he three-bussystemof Fig. P-6.5. The pu line reactancesareindicatedon the figure; the ine resistances re negligible.The magnitudeof all the three-busvoltagesare specified o be 1.0 pu. The bus powersare specified n the following table.

6.2

.?38 I ModernPowerSvstemAnatvsisI

3

Fig. P-6.5

.ffi Z-r----rvz-- () r o-- |

y = _ j 5 . 0

', \

,l J, = is'o

\ /v e l 1 3

Fig.P-6.7 Three-busample ystem ontainingregulat ingransformer

y = - j 5 0 , l

Real

demand

Reactive

demand

Real

Seneratrcn

Reactive

generation

I Por= 1.0 = 0.6



6.8 CalculateV3for the systemof Fig. 6.5 for thefirst iteration,using he dataof Example6.4. Start he algorithmwith calculations t bus 3 rather han

at bus 2.

6.9 For the sample systemof Example 6.4 with bus I as slack, use thefbl lowingnrcthudso obtairr load lo w solut ion.(a ) Gauss-Seidel sing luur, with acceleration actor of 1.6 and

tolerancesof 0.0001 for the real and imaginary componentsofvoltuge.

(b) Newton-Raphsonsing BUS,with tolerances f 0.01pu for changesin the real and reactivebus powers.

\Note: This problem requires he use of the digital computer.

6.10 Performa load low study or the systemof Problem6.4.The bus powerand voltage specificationsare given in Table P-6.10.

Table P-6.10

Bu s power, uBu s Voltagemagnitude, u Bu s yp e

Qot PGr ? ect (unspecified)' ',r2 = 0 Qoz= 0 Pcz = 1.4 pn, lunspecified)3 Po:= 1.0 Qot = l.O Pa = 0 go. lunsp"cified)

Carry out the completeapproximateoad flow solution.Mark generations,loaddemands nd in c f-lows n th c one-l inediagram.

6.6 (a) RepeatProblem6.5 with bus voltagespecificationshanged sbelow:lV t l= 1 . 0 0 u

lV2l= 1.04pu

lV3l= 0.96pu

Your resultsshould show that no significant changeoccurs in realpower flows, but the reactive flows changeappreciably as e issensitiveo voltage.

(b)ResolveProblem6.5 assuminghat the real generations scheduled sfollows:

Pc t = 1. 0pu , Pc z = 1. 0pu , Pc t = 0

The realdemand emainsunchanged nd he desired oltageprofile isflat, i.e. lvrl = lv2l = lv3l= 1.0 pu. In this case he resultswill showthat he react ive lows ar c essential ly nchangccl,ut th c rcal f lgwsarechanged.

6.7 Consider he three-bus ystemof problem 6.5. As shown n Fig. p-6.7where a regulating transforrner RT) is now introclucedn the ljne l-2near bus 1. Other systemdata emain as that of Problem6.5. Considertwo cases:

(i) RT is a magnitude egulatorwith a rario = VrlVl = 0.99,(ii) RT is a phaseangle

egr-rlaroraving a ratio= V/Vi = ",3"a) Find ou t the modified )/ur5 matrix.(b) Solve the load flow equations n cases i) and (ii). compare the

load flow picture with the one n Problem6.5.The readershouldverify that n case i) only the reactive low will change;whereasin case ii) the changeswill occur in the real power flow.

Compute he unspecif ied usvoltages, ll bus powers nd all l ine lows.Assumeunlimited Q sources.Use the NR method.

REFERECES

I2a-)4

Unspecified0.95- 2. O- 1 .0

Unspecified

Unspecified- 1 . 0- 0. 2

1.021 . 0 1

UnspecifiedUnspecified

Slack

PV

PQ

PQ

Books

l . Mahalanabis, A.K., D.P. Kothari and S.I.

Analysis and Control, Tata McGraw-Hill,

Ahson, Computer Aided Power System

Ne w Delhi . 1988.

, 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, he Iowa state universityPress,Ames, owa, 1923.

9' Brown, H'8., sorution

'2A0" ' l Modprn pnr r ra r e r ra ra* A^^r - . - , -- r____ . - . - - - ' , , , v ' r v r eyor r r i l l A1 a tys ls

2' weedy, B.M. andB.J.cory, Erectricar ower-systents,4th d. ,Johnwiley, NewYork, 1998.

Gross,C.A., power SystemAnalysis,2nd 8d..,Sterling,M.J.H., power SystemControl, IEE.

Power Frow Argorithmswith Referenceo Indianpower systems,, proc. of IIsymp.on Powerplant Dynamics ncrcontror,Hyderabad, 4_i6 Feb. rg:/g, zrg.

26. Tinney, W.F. and W.S. Mayer, .Solution

Triangular actorization',EEE Trans.on Auto contor, August 1973, or . AC_lg,J J J .

27 Saro, N. and w.F Tinney, 'Techniquesfo r ExproitingSparsity

AdmittanceMatrix', IEEE Trans. pAS, I)ecember L963,g2, 44.28' sachdev,M.s. and r.K.p. Medicherla, 'A

secondorder Load FlowIEEE Trans., pAS, Jan/Feb1977, 96, Lgg.

29. Iwamoto, S.andy. Tamura, A Fast Load Flow Method' IEEE Trans.,pAS. Sept/Oct.197g, 97, 15g6.'Exact

John Wiley, New york, 19g6.England,1978.

of Network

Technique',

Retaining Non linearitv'.



of Large Networksby Matix Methods, John w1rey, New, York, 1975.

l0' Knight' rJ'G', Power systemEngineeringand Mathematics, ergamonpress,NewYork 1972.

11. Shipley,R.B., Introduction to Matrices and power systems,John w1ey, NewYork, 1976.

12' Hupp,H.H., Diakopticsand 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, .J. and D.p. Kothari, power system Engineering, TataMcGraw_Hilr,

New Delhi, 1994.l5' Bergen,A.R., power systemAnarysis,rentice-Hail,Engrewood riffs, N.J. r9g6.l6' Anillaga' J and N'R' watson, computerModelting of Electrical power systems,

2n dedn.Wiley, N.y. 2001.

Papers

17. Happ, H.H., 'Diakoptics-Thesolutionof systemproblemsby Tearing, proc. ofthe IEEE, Juty t974, 930.

l8' Laughton'M'A', 'DecontpositionTechniquesn powersystemNetworkLoadFlow

Analysi.s si'g the Nodar mpcdanceMatrix,, proc. IEE, 196g,r15, s3g19' stott, B., 'Decouprecr

NewronLoad FIow,, IEEE Trans., 1972,pAS_91,1955.20 . stott,B., 'Reviewof Load-Frow arcurationMethod, proc. IEEE, Jury 1974,916.

21 ' stott,B., and o. Alsac, FastDecoupled oadFlow,, IEEE Trans.,1974,pAS_93:

859 .

22' Tinney' w'F" and J'w. walker, 'Directsolutionsof sparseNetwork Equations y

Optimally Ordered TriangularFactorizaiions,, proc. IEEE. Nnvernhe" 1oA155 :1801 ' v "

23' Tinney, w.F. and c.E. Hart, 'powerFlow sorutionby Newton,s Method,, ,EEETrans.,November 967,No . ll , pAS_g6:1449.

24' ward, J.B. andH.w. Hare, DigitarcomputerSorution f powerproblems,,AIEETrans., une 1956,pt III, 75 : 39g.

30. Roy, L.,SecondOrder Load Flow,, proc. ofDermstadt,Yol. 2, August lg7g, 7Il.

31' Iwamoto,s' andY. Tamura, ALoad Flow calculationMethod or Ill-conditioned

PowerSystems', EEE TranspAS, April lggl , 100, 1736.32. Happ, H.H. and c.c. young, 'Tearing

Argorithms or Large scare NetworkProblems', EEE Trans pA,S,Nov/Dec Ig71, gO,2639.

33. Dopazo, .F.O.A.Kiltin and A.M. Sarson, stochasticLoadFlow ,, .IEEE

Trans.PAS, 975 ,94 ,2gg .

34' Nanda' J'D'P' Kothari and s.c. srivastava, Some ImportantobservationsonFDLF'Algorithm", proc. of the IEEE, May 19g7, pp.732_33. \

35. Nanda, ., p.R., Bijwe, D.p. Kothari and D.L. stenoy, .,secondorder Decoupred

LoadFlow', ErectricMachines nd power systems,or r2 , No . 5, 19g7,pp . 301_312.

36' Nanda '' D'P' KothariandS.c. srivastva, A Novelsecondorder FastDecoupledLoad Flow Method n Polarcoordinatcs", Electic Machinesand power,s.ysrems,Vol. 14 , No . 5, 19g9, p 339_351

37 . Das, D. , H.s. Nagi and D.p. Kothari, .,ANovel 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 Simpleand EfficientMethod fo r Load

Flow sorurionof RadiarDistributionNerworks',, nr . J. of EpES, r995, pp. 335_346.

the Sixth pSCC Conf.

OptimalSystem Operation A3

pariances caiied heunii

commitment' UC) probiemand he seconds calleci

the loadscheduling' LS) problem.One must first solve he UC problembeforeproceedingwith the LS problem.

Throughout this chapter we shall concern ourselves with an existing

installation,so that the economicconsiderations re hat of operating running)

costand not the capitai outiay.

7.2 OPTIMAL OPERATION OF GENERATORS ON A BUS BAR

Before we tackle the unit commitmentproblem, we shall consider he optimal

operationof generatorson a bus bar.

Generator Operating Cost

The major componentof generatoroperating cost s the fuel input/hour, while



7. 1 INTRODUCTION

Th e optimal systemoperat ion, n general, nvolved th e considerat ion feconomy f operation, ystemsecurity,emissions t certain ossil-fuelplants,optimal eleases f waterat hydro generation, tc.All theseconsiderationsmaymake or confl ict ing equirementsndusually compromise as o he made oroptimal ystem peration.n this chapterwe cor..;iderhe economyof operationonly, alsocalled he ec'omonic i.spcttchroblem.

The mainaim n theeconomic ispatch roblerns to rninimize hc totalcost

of generatingeal power (productioncost) at variousstations while satisfyingthe oads nd he ossesn the ransmissioninks.Fo r sirnpl icitywe consiclerhepresence f thermalplantsonly in the beginning. n the ater part of this chapterwe will consider hepresence f hydro plantswhich operate n conjunctionwiththermalplants.While there s negligibleoperating ost at a hydro plant, hereis a imitat ion l availabi l i ty l 'watcr' ve ra pcriodof t inrewhich nrust c usedto savemaximum fuel at the thermal plants.

In the oad low problem sdetailed n Chapter , two variables respecifiedat eachbus and he solution s then obtained or the rernainingvariables. hespecified ariablesare eal andreactivepowersat PQ buses, eal powersandvoltagemagnitudes t PV buses, nd voltagemagnitude nd angleat the slack

bus.The additionalvariableso be specified or load low solution are he ap

settings f regulating ransformers.f thespecified ariablesare allowed o vary

in a regionconstraineciy pract icai onsicierat ionsupperanci ower imits onactiveand eactivegenerations,us voltage imits,and angeof transformerap

settings),here esultsan nfinite numberof load low solutions,eachpertaining

to on ese tof values 1'specif ied ariables. he best 'choice n sorne ense f

thevalues f specified ariableseads o the best'load low solution.Economy

of operations naturallypredominantn determining llocationof generationo

each tation or varioussystem oad evels.The irst problem n powersystem

maintenanceoutributes nly to a small extent.The uel cost s meaningtul n

case of thermal and nuclear stations,but for hydro stationswhere the energy

storages'apparentlyfree', the operatingcost as such s not meaningful. A

suitablemeaningwill be attachedo the cost of hydro storedenergy n Section

7.7 of this chapter.Presentlywe shall concentrateon fuel fired stations.

o (vw)min (MW)max

Power utput, W '- --

Fig.7.1 Input-output urve of a generatrng ni t

The input-output curve of a unit* can be expressed n a mill ion kilocalories

per hour or directly in terms of rupeesper hour versusoutput in megawatts. The

cost curve can be determined experimentai ly. A typical curve is shown in

Fig. 7.1 where (MW)o'in is the minimum loading limit below which it is

uneconomical (o r may be technically infeasible) to operate the unit and

(MW)n,u, is th e max imunr output l imi t . The inpLl t -output curve ha s

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

lIII

t -L \

- al)

; Pt r o. c 86 o8 b =e 5g Eo

O 6f

LL

A unit consists of a boiler, turbine and generator.

_ Q ;

2

The slopeof the cost curve, .".43 is called the ncremental ue l cost lQ,dPo,

and s expressedn units of rupeesper megawatt our (Rs/lvIWh).A typicalplot

of incrementalue l costversuspower output s sho.wnn Fig. 7.2.If the cost

curve s approximated s a quadraticas n Eq. (7.1), we have

| . . - - A - - - ! - - A - ^ r - - ^ : -

.L4+ | MOOern rower DySIeIrl Arlaly

I

Ci(Pc) Rs/hourat outPutPc,

where he suffix i stands or the unit number. t generatlysuffices to fit a second

degree olynomial, .e.

Considerations f spinning eserve, o be explained ater n this section, equire

that

D Po,, *) Po

marsin. .e. Eq. (7.6) must be a strict inequality.

Since he operatingcost s insensitive o reactive oadingof a generator, he

rnanner n which the reactive oad of the station s sharedamongvarious on-

line generators oesnot afl'ect he operating conomy'.

The question hat has now to be answereds: 'What is the optimal manner

in which the load demand Po must be sharedby the generators n the bus?'

This is answeredby minimizing the operatingcost

k

c = D ci(pci)f: l

b,Pc, + d, Rs/hour (7.r)

(7.2)

(7.6)

(7.7)(lc)i= aiP"t + bt



under he equality onstraint f meetinghe oaddemand,.e.

(f - \ D l

L' G i - P o = O

i: t

wherek = the number of generators n the bus.

Further, the loading of each generator s constrainedby the inequality

constraintof Eq . (7.5).

Since Ci(Pc) is non-linearand C, is independentof P6t (i+ i), this is a

separable on-linearprogrammingproblem. \If it is assumedat present, hat the inequality constraintof Eq. Q.q is not

effective, he problem can be solvedby the method of Lagrangemultipliers.

Define the Lagrangianas

(7 .3)

(7.4)

of the ith generator

Equat ion 7.10) ca n

dC^ \

dPoo

(7 .8)

(7.e)

( 7 . 1 o )

(units:Rs/TvIWh),

be written as

(M Wmi n

(M Wma xPower utput,MW

Fig.7.2 Incrementaluel costversus ower utput or he unitwhose

input-outputurve s shownn Fig.7. 1

i.e. a linear relationship.For better accuracy ncremental fuel cost may be

expressedby a number of short line segments(piecewise lineanzation).

Altcrnativcly,wc can it a polynomialof suitable egree o representC curve

in the inverse orm

Pc;i= a, + {),(lC)i + 1,QC)', + . . .

Optimal Operation

[,et us assulnehat t is known a pri l t r i whichgenerutorstr e <l t ln to ntccta

p:.rrt icularoa d clenrandn the statton. bvtously

f -

where X is the Lagrangemultiplier.

Minimization is achievedby the condition

o f , = odPo,

dCo r

' i- ) i i = 1 , 2 , . . . ,

dPc,

where lc i is the incremental os t

d4,,a Iurctio'ol'.gencra,il;,":':t

:":d4( i l - d r - -

lII

-. I(J

ooo c

E B, d >

o oE t roEE()

i t (Pci)-^[f"",

-""]

DPr,,,, ' ',,* P,

where Pci, ,r.,,"*s the rated real power capacityof the ith generatorand Po is

the otal power demandon the station.Further, he oad on eachgenerator s to

irc constrainedwithin lower and upper imits, i.e.

Pcr, .i n 1 Po, 1 Po,, rn.*, = L, 2, "' , k (7.s)

*Theeffect of reactive loading on generator osses s of negligible order.

(7 . )

Computer solution fbr optimal loa<ling of generatorscan be obtainediteratively as follows:

1. Choose a trial value of ), i.e. IC = (IC)o.

2. Solve for P" , (i = 1, 2, ..., k) from Eq. (7.3).

3. If ItPc,- Po l < e(a specifiedvalue), he optimal solution s reached.

Otherwise,

I

n - . : - ^ r A - ^ ^ r - - - ^ . . L - ^ -

costof the plant colresponds o that of unit 2 alone.When rhe plant oad is 40Mw, eachunit operatesat its minimum bound, .e.2o Mw wiitr plant \ = Rs35/I4Wh.

When dczldPcz= Rs 44/MWh,

0.25PG2+ 0 - 44

or pnt = JI- = 56 MW0.25

The total plant output is then (56 + 20) = 76 MW. From this point onwards,the valuesof plant oad sharedby the two unitsare found by assuming ariousvaluesof \. The resultsare displayed n Table7.1.



4. Increment lC) bv A (1"), I:fl Po,- ,r l < 0 or decrement/c ) by A(tr)

if [D Pc, - Pr] r 0 and repeat rom step2. This step s possiblebecauseP.-, s monotonicallyncreasing unctionof (1g).

consider now the effecr of the inequality constraint (7.5). As (1c) isincreasedor decreasedn the terativeprocess,f a particulargeneratoroadingP", reaches he imit PGi,^o or P6;, min, its loading from now on is held fixedat this value and the balance load is then shared between the remaininggeneratorson equal incremental cost basis. The fact that this operation isoptimal can be shownby rhe Kuhn-Tucker heory(seeAppendix n;.

Incremental fuel costs n rupeesper MWh for a plant consistingof two unitsare:

dt

i * - o . 2 o p c t + 4 0 . 01

-dcz-= o.2,pcz+3o.odPo,

Assume that both units are operatingat all times,and otal oad varies rom 40MW to 250 MW, and he maximum and minimum loadson eachunit are o beI25 and 20 MW, respectively.How will the load be sharedbetween he twounitsas the system oad vanesover the full range?What are he correspondingvaluesof the plant ncrementalcosts?Solution At light loads,unit t has the higher ncremental uel cost and will,therefore,operare t ts lower limit of zo Mw, for which dcrldpcr is Rs 44 perMWh.When th eourput f unit 2is20 MW , dczldpcz=Rs 35 p" i UWt. Thus,with an increase n the plantoutput, he additional oad shouldbe bornebv unit

Table 7-1 Outputof each unit and plantoutput or variousvaluesof) for Example .1

Plant ),

RszMWh

354450556061.2565

40.076.0

130.0175 .0220.0\231.25250.0

Figure 7.3 shows he plot of the plant .trversusplant output. t is seen romTable 7.7 that at .\ = 61.25,unit 2 is operatingat its upper imit and herefore,the additional load must now be taken by unit 1, which then determines heplant ).

60

+t - -l c cI

8 5 0o

@ ^

i = 4 5a >

E P 4 0t -EOc J c

Plant utput, W

Fig.7.3 Incrementalue l cost versusplantoutput,as found in Example7. 1

Unit I Unit 2

Pcz, NfWPlant Output

Pcl, MW

t :Hzut" l.Z.48: Mocjern ower Systemnnaiysis

To find the oad sharingbetween he units for a plantoutput of say 150 MW,we find from the curve of Fig. 7.3, that the corresponding lant X is Rs 52,22

per MWh. Optimum schedules or each unit for 150 MW plant load can now

be found as

Net saving aused y optimumschedulings772.5 721.875 50.625 s/lr

Totalyearly aving ssumingontinuousperation

This saving ustifies he need for optimal

installed or controlling the unit loadings

load sharing and the ddvices o be

automaticallv.

0 . 2 P a * 4 0 - 5 2 . 2 2 ;

0.25PG2 30= 52.22;

Pcr + Pcz= 150MW

Proceeding on the above ines, unit outputs for variousplant outputs are

computedand havebeenplotted n Fig. 7.4. Optimum oad sharing or anyplant

load can be directlv read from this fieure.

Pc t = 6 1 ' 1 1MW

Pc z = 88'89MW

Let the two units of the svstem studied

curves.

in Example7.1 have he ollowing cost



100 150 200 250

Plantoutput,MW -_-->

uni t versus plant output or Example7. 1

For the plant describedn ExampleT.l find the saving n fuel cost n rupeesper

hour for the optimal scheduling f a total oad of 130 MW ascompared o equal

distribution of the same oad between he two units.

Solution Example 7.I reveals hatunit I should ake up a load of 50 MW and

unit 2 should supply 80 MW. If eachunit supplies65 MW, the ncrease n cost

for unit 1 is

1 6 5 r n n n . , ^ \ r n r r . ' t n Z

| \U.L I ' r : rt *U) |J I l

nr= (U. I fn t

Jso

Similarly, for unit 2,

J*co.rs"",+ 30) por=0.25PGz+oro;1"

Cr = 0.lPto, + 40Pc + 120 Rs/hr

Cz= 0.l25Pzcz+ 30Po,+ 100 Rsftrr

220 MW

Monday

12l l1 2 6

(noon) PM (night) AM

Time ----

Fig. 7.5 Daily oad cycle

Let us assume daily load cycle asgiven n Fig. 7.5. Also assumehat a cost

of Rs 400 is incurred n taking eitherunit off the ine and returning t to serviceafter 12hours.Consider he 24 hour period rom 6 a.m. onemorning 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 service or 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 oad s 220MW, referring o Table 7.1

of Example 7.1, we ge t the optimum schedule s

Pcr= 100 MW' Pc z = 12 0MW

Total fuel cost for this period is

[ 0 . 1 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 ] 1 2

= Rs. 1,27,440

I tzsI

= 10 0

E r uo

E 5 0f

250

t 200I

I

3 15 0

E 10 0o

J

500 5 0

Flg.7.4 Output of each

16 5; l f \ r 1 r l Fn ^ a D ^ t L -

t ' t U I ' 6 y l l = l l L . J l \ S / I [

" ' l so

Sunday

- - 721.875 s/hr

ltgtiFl Modernower vstem nAVsis- t

If both units operate n the light loadperiod (76 MW from 6 p.m. to 6 a.m.)also, then from the same able, we get the optimal scheduleas

Pcr = 20 MW, Pcz = 56 MW

Total fuel cost for this period is then(0.1x 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 hourperiod s R s I,65,024.

If only one of the units s run during he ight load period, t is easily verifiedthat t is economical o run unit 2 and o put off unit 1. Then he total fuel costduringhisperiot:tXf :

762+ 3ox 76+ 100)x rz = Rs37,224

t

-

Dynamic Programming Method

In a practicalproblem, the UC table s to be arrived at for the complete oadcycle. If the oad s assumed o increasen small but finite size stensldvnaminprograrrurung can be used to advantage or computing the uc table,wherein it is not necessary o solve the coordinationequations;while at thesame time the unit combinations to be tried are much reduced n number. Forthesereasons, nly the Dp approachwill be advancedhere.

The total numberof units available, heir individual cost characteristics ndthe load cycleon the stationare assumedo be known a priori.Further, it shallbe assumed hat the load on each unit of combination of units changei'insuitably small but uniform stepsof size /MW (e.g. I MW).

Startingarbitrarily with any two units, the



Total fuel cost for this case= L,27,440+ 37,224

= Rs 1,64,664

Total operatingcost for this casewill be the total fuel costplus the start-up costof unit l , i .e .

1,64,6& + 40 0 = Rs 1,65,064

Comparing his with the earliercase, t is clear that t is economical o run boththe units.

It is easy o see hat if the start-upcost is Rs 200, then it is economical orun only 2 in the light load period and to put off unit 1.

7.3 OPTTMAL UNrT COMMTTMENT (UC)

As is evident, t is not economical o run all the units availableall the time. Todetermine the units of a plant that should operate for a particular load is theproblemof unit commitment (UC). This problem is of importance br thermalplantsas for other types of generation uch as hydro; their operatingcost andstart-up imes are negligible so that their on-off status s not important.

A simple but sub-optimal approach o the problem is to impose priorityordering,wherein the most efficient unit is loaded first to be'followed by thelessefficient units in order as the Ioad ncreases.

A straightforward but highly time-consuming way of finding the mosteconomicalcombinationof units to meeta particular oad demand, s to try allpossible ombinationsof units that can supply this load; to divide the loadoptimally among the units of each combination by use of the coordination

equaiions, o as o finci the most economicaioperatingcostof the combination;then, o determine he combinationwhich has he eastoperating ost among allthese.Considerable omputationalsavingcan be achievedby usingbranch andbound or a dynamic programming method for comparing the economics ofcombinationss certaincombinationsee tnot be triedat al l .

most conomical combination sdetermined or all the discrete oad levels of the combined output of the twounits. At each oad level the most economicanswermay be to run either unitor both units with a certain oad sharingbetween he two. The most economicalcost curve in discrete orm for the two units thus obtained,can be viewed asthe cost curveof a single equivalentunit. The third unit is now addedand theprocedure epeatedo find the cost curveof the threecombinedunits. t may benoted that in this procedure he operatingcombinationsof third and first, alsothird and secondare not required to be worked out resulting in considerablesaving in computationaleffort. The process s repeated, ill all availableunitsare exhausted.The advantageof this approach s that having oitiined thegPli-u.| way,of loading ft units, it is quite easy o determine he Jptimal mannerof 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 generatingy 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 applicationof DP results n the following recursive elation

FN@)= TnVn9) * Fu-r @ y)| (7.r2)Using the above ecursive elation,we can easilydetermine he combination

of units, yielding minimum operatingcosts or loads anging n convenientstepsfrom the minimum.permissible load of the smallestunit to the sum of thecanae i f ies n f q l l q r ro i io l - lo r r - i+o i - +L ;^ - - ^^^ - ^ .1^ - 1 -1 r .

$vs^rqurv uurrD. rrr Lrr lD

PruuttJsure total

nunlmum oDerat ing eostand the load sharedby each unit of the optimal combination are ;il.u,i"determined or each oad level.

The use of DP for solving the UC problem s best llustratedby meansof anexample.Considera samplesystemhaving four thermalgeneratingunits withparameters isted n Table 7.2.It is required o determinr h. most-economicalunits to be committed or a load of 9 MW. Let the oad changes e n stepsofI MW.

. ModernPow

Table7.2 Generating nit parametersor the samplesystem

Capacity (MW) Cost curve pararneters (d = 0)

Unit No.

The effect of step size could be altogether eliminated, if the branch andbound technique [30] is employed. The answer to the above problem using

branch and bound s the same n termsof units to be committed, ,e. units 1 and

2, but with a load sharing of 7 34 MW and 1.66 MW, respectivelyand a totalgraung cost oI I<s z,y.zL /J/nour.

In fact the best scheme is to restrict the use of the DP method to obtain the

UC table for various discrete load levels; while the load sharing among

committed units is then decidedby use of the coordinationEq. (7.10).

For the example under consideration, he UC table is prepared n steps- f I

MW. By combining the load range over which the unit commitment does not

change, he overall result can be telescoped n the form of Table 7.3.

Table 7.3 Status*of units or minimum

1234

NowFt@) = ft@)

= LorP'ot+ btPcr

1 .01. 01. 01 .0

12.012.012.012.0

0.771.602.002.50

23.526.530.032.0



operating ost (Unit commitmenttable or the samplesystem)

Load range

Unit number

r 2 3 4

l-56-r3t4-181948

*l = unit running;0 = unit not running.\

' The UC table s preparedonceand for all for a given set of units. As the load

cycle on the station changes, t would only mean changes n starting and

stopping of units with the basic UC table remaining unchanged.Using the UC tableand ncreasing oad n steps, he most economicalstation

operatingcost s calculated or the complete ange of stationcapacityby 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, f necessary) an then be fitted to this data or later use n economic

load sharing among generatingstations.

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

powersystems

has assumedgreat mportance. Every eleciric utilify is normaily

under obligation to provide to its consumersa certain degreeof continuigl and

quality of service e.g. voltage and requency n a specified ange).Therefore,

economy and reliability (security) must be properly coordinated n arriving at

the operationalunit commitmentdecision. n this section,we will seehow the

purely economic UC decision must be modified through considerations of

reliability.

fr(9) f{9)== ollss x 92 + 23.5x9 = Rs 242.685lhour

From the recursive relation (7.12), computation s made for F2(0), Fz(l),

Fz(2),.. , Fz(9).Of these

FzQ) = min tt6(0) + Ft(9)1, VzG) + Ft(8)l'

VzQ)+ Ft(7)1, VzQ) + Fr(6)l' Vz@)+ F1(5)l'

t6(5)+ Fr(4)1, zG) Ft(3)1, zT + Fr(2)1,

tfr(s) F,(1)1,zg) + 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 he recursive elation 7.12),we now computeFl(O),F:(1), ...' F3(9).

Of these

Fse)=min {t6(0) Fr(9)f,6(1)+ Fl8)1, ..'[6(9)+ rr(0)]]

= [6(0)+ FzQ)l= Rs239.565ftour

Proceedingsimilarly, 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.

It must be pointed out here, hat the optimai iiC tabie is inclependentof 'u\enumberingof units, which couldbe completelyarbitrary.To verify, the reader

rnaysolvethe above problem onceagainby choosinga different unit numbering

scheme.

If a higheraccuracy s desired, he stepsizecould be reduced e.g.*

t*r,

with a considerable ncrease n computation ime and requiredstoragecapacity.

0 00 01 01 1

In order to meet the load demand under contingency of failure (forcedoutage) of a generatoror its derating causedby a minor defect, static reservecapacity is always provided at a generatingstation so that the total installedcapacity exceeds he yearly peak oad by a certainmargin. This is a planning

In arriving at the economicUC decisionat any particular ime, the constrainttaken into accountwas merely the fact that the total capacityon line was atleast equal to the load. The margin, if any, betweenthe capacity of unitscommitted and oad was incidental. If under actualoperationone or more of theunits were to fail perchance randomoutage), t may not be possible o meet theload requirements.To start a spare standby) hermal unit* and to bring it onsteam o takeup the load will take severalhours (2-8 hours),so that the loadcannot be met for intolerably long periods of time. Therefore, to meet

periods are a random phenomenon with operatingperiods being much longer

than repair periods. When a unit has been operating or a long time, the random

phenomenon an be described y the following parameters.

Mean time to failure (mean up' time),

Mean time to repair (mean down' time),

Et , (down)

Z (down)No. of cycles

No. of cycles( 7 .13 )

(7.r4)

Mean cycle time = f (up) + Z(down)



contingencies,he capacityof units on line (running)must have a definitemargin over the oad requirements t all times.This margin which is known asthe spinning reserveensurescontinuity by meeting the load demandup to acertainextentof probable oss of generation apacity.While rules of thumbhave beenused,basedon pastexperience o determine he system'sspinningreserveat any time, Patton's analytical approach o this problem is the mostpromising.

Since the probability of unit outage ncreaseswith operating ime and sincea unit which is to provide the spinning reserveat a particular time has to bestartedseveral oursahead, he problem of securityof supplyhas o be treatedin totality over a period of one day. Furthefinore, he loads are never knownwith complete certainty. Also, the spinning reserve has to be provided at

suitable generatingstations of the system and not necessarily at everygenerating tation.This indeed s a complexproblem.A simplified treatmentofthe problem s presented elow:

f1(down) f2 down) f3 down)

Time-------->

Fig. 7.6 Random nit performanceecord eglectingcheduled utages

A unit during its useful life spanundergoes lternateperiodsof operationandrepair as shown in Fig. 7.6. The lengths of individual operating and repair

z(up) z(down)

Z(down)

(7.rs)

(7 .16)

Inverseof these imes 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

engineeringudgement must be exercised n arriving at these estimates.The

failure rates are affected by preventive maintenance and the repair rates are

scnsitive o size, compositionand skill of repair teams. .r

By ratio definition of probability, we can write the probabiliiy of a unit being

in 'up' or 'down' statesat any time as

r(up) _ l.tp (up) =

P (down) =

p + A

)Z(up) Z(down) tr+ A

Obviously,

p (up) + p(down) = 1

p (up) andp (down) in Eqs. 7.15) and (7.16)are also ermedas availability

and unavailability, respectively.

When ft units are operating, he system statechangesbecauseof random

outages.Failure of a unit can be regardedas an event ndependent f the state

of otherunits. If a particularsystemstate is defined asX, units n 'down'state

1 r , ' ( t - - - t - l r \ / . r z\ ! l - - - - - - - 1 - - l - : 1 : - - - ^ t L 1 ^ ^ l - ^ : - ^ 2 - t L l

an0 I j ln Up. Stale \K = i + I )t urs PIUDaUITILy Ur urtr systel l l Utr l t r$ t t l ulIS Slaltr

is

pt ={r,r;(ul),{*

ot(down).

If hy^dro.generations available n the system, t could be brought on line in amatter of minutes to take up load.

(7.r7)

gW ModernPower ystemAnalysisI

Patton's Security FunctionA breach of system security is defined as some intolerable or undesirablecondition.The only breachof securityconsidered ere s insufficientgeneration

probability that the availablegenerationcapacity (sum of capacitiescbmmitted) at a particular hour is less than the system load at thatdefinedas [25]

S = Ep,r,

where

of units

time, is

( 7 . 1 8 )

p, = probabilityof systembeing n state [seeEq. (7.17)]

r, = probability that system state causesbreachof syStem

WWt

the sample system or the loadcurve of Fig. 7.7

Unit number

t 2 3 4

ABCDEF

100000

I

I

0

I

0

0



securlty.When system oad s deterministic i.e. known with completecertainty),r, = 1if available capacity is less than load and 0 otherwise. S indeed, is aquantitativeestimateof system nsecurity.

Though theoreticallyEq. (7.18) must be summed over all possiblesystemstates(this in fact can be very large), from a practical point of view the sumneeds o be caried out over states eflecting a relatively small numberof uniqson forced outage,e.g. stateswith more than two units out may be neglected as

the probability of their occurrencewill be too low;

Security Constrained Optimal Unit Commitment

Once he units to be committed at a particular load level are known from purelyeconomic

considerations,he security unction S s computed asper Eq. (7.18).This figure should not exceeda certain maximum'tolerable insecurity level(MTIL). MTIL for a givcn system s a management ecisionwhich is guided

by past experience. f the value of S exceeds MTIL, the economic unitcommitment schedules modified by bringing n the next most economicalunitas per the UC table. S is then recalculatedand checked. The process scontinued till ^t< MTIL. As the economicUC table has some nherentspinningreserve, arely more than one iteration is found to be necessary.

For illustration, econsider he fbur unit exampleof Sec.7.3. Let the dailyload curve for the systembe as ndicated n Fig. 7.7. The economicallyoptimal

UC fbr this load curve is immediately obtainedby use of the previouslypreparedUC table (seeTable 7.3) and s given in TabIe 7.4.

Let us now check f the aboveoptimal UC table s secure n every period of

the ioaci curve.For the minimum load of 5 MW (periodE of Fig. 7.7) according o optimal

UC Table 7.4., only unit 1 is to be operated.Assuming dentical ailure rate )of l/year and repair ratepr, f 99/year or all the four units, et us check f thesystem s secure crr heperiodE. Further assumehe systemMTIL to be 0.005.Unit I can be only in two possiblestates-operatingor on forced outage.

Therefore,

where

Time in hours ------------>

Flg.7.7 Daily oadcurye

2S =

,?pi t i : p1r1* p2r2

Pr= P(up)ff i

= 0.99, rr = 0 (unit = 12MW> 5 MW)

pz = p(down)=p + ^

= 0.01 rz = t (with unit I down load

demandcannot be met)

HenceS= 0.99x 0 + 0.01x 1 = 0.01> 0.005 MTIL)

Thus unit I alone supplying5 MW load fails to satisfy ne prescribedsecurity riterion. n order o obtainoptimalandyet secure C, t is necessaryto run henextmosteconomicalnit, .e.unit2 (Table .3) alongwith unit1.

0 L0

(noon)

ifl5 #Wl Mooernower ystemnarysis

With both units I and2 operating, security function is contributed only by

the statewhen both the units are on forced outage.The stateswith both unitsoperatingor eitherone failed can meet the load demand of 5 MW and so donot contribute o the security unction. Therefore,

S = p (down; x p (down) x 1 = 0.0001

This combination units 1 and 2both committed)does meet the prescribedMTIL of 0.005, .e. ^S MTIL.

Proceeding imilarly and checkingsecurity unctions or periodsA, B, c, Dand F, we obtain the following optimal and secureUC table for the samplesystem or the load curve given in Fig. 7.7.

Table 7.5 Optimal and secure UC table

Unit number

Period

E#nft#

= Rs 4,463.216(start-upcost = 0)Clearly Caseb results n overall economy.Therefore, he optimal and secure

UC table or this load cycle s modified asunder,with due consideration o theoverall cost.

Unit numberPeriod

I

I

I

I

I

I

I

I

I

I

1 l

1 0

l 'l ' 0

l 0

A

B

C

D



I

* Unit was started due to security considerations.

Start-up Considerations

The UC tableas obtainedabove s secureand economicallyoptimal over eachindividual period

of the oad curve. Sucha tablemay require that certainunitshave o be startedandstoppedmore thanonce.Therefore,start-upcostmustbetaken into consideration rom the point of view of overall economy.Forexample,unit 3 has to be stoppedand restarted wice during the cycle. Wemust, therefore,examinewhether or not it will bemore economical to avoid onerestarting by continuing to run the unit in period C.

Casea When unit 3 is not operating n periodC.Total fuel cost for periodsB, c and D as obtainedby most economic oadsharingare as under (detailedcomputation s avoided)

= 1,690.756 1,075.356 1.690.756 Rs 4.456.869

Start-upcost of unit 3 = Rs 50.000 (say)

Iotal operatrng ost= Rs 4,506.868

Caseb When all threeunits are running n periodC, i.e. unit 3 is not stoppeclat the end of period B.

Totaloperat ing osts= I,690.756+ 1,0g1.704 1,690.756

I I0 00 0

*Unit was starteddue to start-upconsiderations.

7.5 OPTIMUM GENERATION SCHEDUTING

From the unit commitment table of a given plant, the fuel costcurve of the plantcan be determined n the orm of a polynomial of suitabledegreeby the methodof least squares it. If the transmission ossesare neglected, he total systemload can be optimally divided among the variousgeneratingplants using theequal ncrementalcost criterion of Eq . (2.10). t is , howrurr, on."alistic toneglect transmission osses particularly when long distance ransmission ofpower s involved.

A modern electric utility serves over a vast area of relatively low loaddensity.The transmissionossesmay vary from 5 to ISVoof the total load, 4'dtherefore,t is essentialo account or lcsseswhile developingan economic oaddispatch olicy. t is obvious hat when osses represent,we canno onger usethe simple 'equal

incrementalcost' criterion.To illustrate he point, consider atwo-bussystemwith identicalgenerators t eachbus (i.e. the same C curves).Assume hat the load is locatednear plant I and plant 2 has o deliver powervia a lossy ine. Equal incrementalcost criterionwould dictate hat each plantshouldcarry half the total load; while it is obvious n this case hat the ptunt1 should carry a greatershareof the oad demand hereby educing ransmissio'losses.

In this section,we shall nvestigatehow the load shouldbe sharedamongr r q r i r r r r a n l q n f c . r r l r o - l ; - ^ l ^ - - ^ ^ L^ s n ^ , - r y

r$rrvsu l.rquLor vYuvrr uuv rvirDsr 4rE .1uuuurrttrg t()f, Ine ODJgCtfVg fS tO mirufiUzethe overall cost of generation

c =,\-rci(Pc')

at any time under equality constraint of meeting the load demand withtransmissionoss, .e.

EF

I

10100

t<

ABCDEF

I00000

(7.7)

- P I - = 0

where

k = tatalrnumber of generating plants

Pci= generation f lth plant

Pp = sum of load demandbt all buses system oad demand)

Pr= total system ransmissionoss

To solve he problem,we write the Lagrangianas

k

DP", - P,i: l

(7.re)

(7.20)t=tr,(Pc)-^[t""Po-".]i: l

Equation (7.23) can also be written in the alternative orm(IC) i - An - QTL)i f = 1,2, . . . , k (7.2s)

This equation s referred o as the exact coordination equation.Thus it is elear tha-t o solve the optimum lead sehedulingproblem; it is

necessary o compute ITL for each plant, and thereforewe must determine thefunctionaldependence f transmissionoss on realpowersof generatingplants.There are severalmethods,approximateand exact, for developing a transmis-sion loss model. A full treatmentof these s beyond the scopeof this book.One of the most important, simple but approximate,methodsof expressingtransmission oss as a function of generatorpowers s through B-coeffrcients.This method is reasonably adequate or treatment of loss coordination ineconomic schedulingof load betweenplants. The general form of the lossformula (derived

ater in this section)using B-coefficients s



(7.23)

It will be shown ater n this section hat, f the power factor of load at eachbus s assumed o remain constant, he system oss P, can be shown to be afunction of active power generationat eachplant, i.e.

Pr = Pt(Pcp P52, . . . , P51r) (7.2r)

Thus n the optimization roblemposedabove,Pc i Q = I,2, ..., k) are he onlycontrol variables.

For optimum real power dispatch, ')

(7.22)

RearrangingEq. (7 22) and recognizing that changing the output of only one

plant can affect the cost at only that plant, we have

AL =d C, - \ r , 1 P t -

oPo, dPGt-^+ A

#*:O'i = r ' 2 ' "" k

= ) or#Li=

), i = r ,2, . . ,

where

L i =Q -APL iAPGi )

(7.24)

is called he penaltyfactor of the ith plant.

The Lagrangianmultiplier ) is in rupees er megawatt-hour,when uel costis in rupees per hour. Equation (7.23) impiies that minimuqr ftrei eost isobtained,when he ncremental uel costof eachplant multiplied by its penaltyfactor s the same or all the plants.

The k + 1) variables P6r, P62,..., Pct, )) canbe obtained rom k optimaldispatchF,q. (7.23) together with the power balanceEq,. (1.19). The parrialderivative PLIAPGi s referred o as the incremental ransmission oss(ITL),,

associated ith the lth generatingplant.

P. =II

pG^B*,pGnm:7 n:I

where

PG^, PGr= real power generationat m, nth plants

B^n= loss coefficientswhich are constantsundercertainassumedoperatingconditions

If P6"s re n megawatts,B*n are n reciprocalof megawatts*. emputations,ofcourse,may be carriedout in per unit. Also, B*r= Bn^. ,:

Equation (7.26) for transmissionoss may be written in the rnatrix form as

(7.26)

(7.27)r= PIBPI

Where

It may be noted that B is a symmetric matrix.For a three plant system,we can write the expression or loss as

PL= Bn4, + Bzz4, + Bzz4, + 28rrpcrpcz+ ZBnpGzpG3

+ 2BrrPorpo, e.ZS)wiih the systemDowerossmorieiasper

Eq. e.z6), we c.an ow writeAp ^ f t o_, I

*=

ftl?lPo,"a^^'o')*B^,

{in pu) = B^n (in Mw-t) x Base MVA

ffif Moderno*ersystem natysis

It may be noted hat n the aboveexpressionother ermsare ndependent f Po,and are, therefore, eft out.

Simplifying Eq. (7.29) and recognizing that B, = Bir, we can write

a P k-+ =D zBijpcj)Fo, j:l

Assuming quadraticplant cost curves as

Ci(Poi )=

*o,4,+b,p.,+ d,

k k4. Calculate , =If pciBijpcj.

j:lJ=l

5. Check f power balanceequation ?. o) is satistied

l s llLPot

- PD pr.l' t (a specifiedalue)l n * t I

If yes, stop.Otherwise,go to step6.(7.30a)

6. Increase by A) (a suitable

decrease by AA

stepsize)' tt

[*

pc ,^

-

")

- po

< 0 o r

(a suitable



We obtain the incremenlal ost as

dc,

dP '= atPo'+ b'

Substinrting APL|&G' and dcildPci from above in

(7.30b)

the coordination

repeat rom step 3.

step ize);f (8", p,) > 0 ,

Eq . (7.22),we have

k

a iPc i+b ,+ S lzn , , r o , : ^ (1 .3 I )J: l

Collecting all terrnsof P, and solving for Po, we obtain

k

(a i + 2M,,) Pci=-

)D ZBijpGjbi

+ ^' - l

J* T

t-r &

n ' t -

l - a - )' 2 8 , , P

^) L " " i i ' G i

]: I

, ,* r B = ;

i - 7 ' 2 ' " ' ' k ( 7 ' 3 2 )

)

Example7,4'

A two-bus system s shown n Fig. 7.8. If 100 Mw is transmited from plant1 to the load, a transmission oss of 10 MW is incurred.Fin( the requiredgenerationor each plant and he power receivedby loadwhentie system ,\ isRs 25llvlWh.

The incremental uel costsof the two plants are given below:

dCA*

- o'ozPcr 16'oRs,Mwh

:f:- o'o4PG2+o.oRs/lvIWh

Pc,

For any particularvalue of \, Eq . (7.32) can be solved iteratively byassuming nitial valuesof P6,s(a onvenient hoice s P", = 0; i = l, 2, ..., k) .Iterations are stoppedwhen Po,s convergewithin specified accuracy.

Equation (7.32) along with the Dower balance F,q,. 7.19) for a pa-rtieular

ioaci ciemanci o are soiveci teratively on the following lines:1. Initially choose = )0 .

2 . As s u me tG i= ; I = 1 ,2 , . . ,k .

3. Solve Eq. (7.32) teratively or Po,s.

6 ^ r - - r : - -outuuon

Therefore

Hence

Fig. 7.8 A two-bus ystemor Example .4

since the ioad is at bus za\one,

p",v,.r,!not have any effect ortFr.

B z z = a n d B n = 0 = B z ,

Pl-=

For Po, -

BnPbr

100MW, Pr = 10MW, i.e.

(i)

" l Mod"rnPo*"r.Syrt"r An"lyri,I

10= Bn (100)2

Brr= 0'001MW-r

Equation7.31) or plant1 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 ( i i i )

Substituting he valuesof B-coefficientsand ) - 25, we get

Pc t = 128'57MW

Pcz= 125Mw

The transmissionpower loss s

Pr = 0.001x (128.57)'= 16.53MW

At plant 2 the load increases rom 37.59 MW to 125 MWcoordination.The saving at plant 2 is

due to loss

= - Rs 2,032.431hr

The net savingachievedby coordinating osseswhile scheduling he receivedload of 237.04MW is

r'

2,937.69 2,032.43 Rs g,OtS.Ztm,

Derivation of Transmission Loss Forrnula

An accuratemethod of obtaining a general formulafor transmission oss hasbeengiven by Kron This, however, s quite complicated.



and the load is

Po = Pct * Pcr- Pt- = 128.57+ I25 - 16.53 237.04 MW

Consider he system of Example7 4 with a load of 237 04 MW at bus 2. Find

the optimum load distributionbetween he two plants for (a) when lossesare

included but not coordinated,and (b) when lossesare also coordinated.Also

find the savings n rupeesper hour when losses are coordinated.

Solution Case a If the transmission oss is not coordinated, the optimum

schedules re obtainedby equating he ncremental uel costsat the two plants.Thus

0 .02P-+ 1 6 = 0 . O 4 P c z + 2 0

The powerdeliveredo the oad s

Pct * Pcz= 0.001P4r 237.04

SolvingBqs. i) and ii ) for P61nnclP6.2, e get

Pc r= 275.18 W; an dPrr2 37.59MW

Caseb This case s already olved n Example7.4. Optimumplant oadings

with loss coordinationare

Pct= 128,57MW; Pcz = 125 MW

Losscoordination auseshe oad on plant I to reduce rom 275.18 MW to

128.57MW. Therefore,savingat plant I due to loss coordination s

(i )

(i i

Kgi:",+16)dPc,o.MPl,rr6Pctli),'r',

= Rs 2,937.691hr

[4]. The aim of thisarticle is to give a simpler derivation by making certain assumptions.

Figure 7.9 (c) depicts the caseof two generatingplants connected o anarbitrary number of loads through a transmissionnetwork. One line within thenetwork is designated s branchp.

Im,agine hat the total load current 1, is supplied by plant 1 only, as inFig. 7.9a.Let the current n line p & Irr.Define

\ (7.33)

(c )

Flg. 7.9 Schematicdiagramshowing wo plantsconnected hrougha power network o a number of loads

S i r n i i n r i v w i t h n i e n t ) q i n r n c c t r n n i r r i n o t h e f n t e l l n q d n n r a n t t 'E i c ? o k \ r ' a ^ a -

r^_-^ ,vs^rv rr r

\rr5 .

, . rv ) , wv v( l t ldefine

I^ nMrz= i= e.34)

t D

Mo1 wrd Mp2 arecalled curcent distribution factors. The values of currentdistribution factors depend upon the impedances of the lines and theirinterconnectionand are independentof the current Ip.

fW Modern owerSygtemAnatysist

When both generators t and 2 are supplying current into the network as nFig.7.9(c), applying theprincipleof superposition he current n the line p canbe expressed s

where 1ot and Io2 are he currentssupplied by plants I and 2, respectively.At this stage et us makecertainsimplifying assumptions utlined below:(1) All load currentshave he samephaseanglewith respect o a common

refere{ce. To understandhe implication of this assumption onsider he loadcurrent at the ith bus. It can be written as

VDil (6t- d) = lloil l1i

where {. is the phaseangleof the bus voltage and /, is the aggingphaseangleof the oad.Since { and divary only through a narrow range at various buses,

it is reasonableo assume hat 0, s the same or all load currentsat all times.

p

Substituting or llrl2 fromEq. (7.37),and l1o,l and llurl from Eq. (7.38),'we

obtain

D2D rGr

rLu|rn,"-

-1v31"*fr)p

*Wl,r,rMpzRplVllv2lcos /, cosQ, 7

P

* P3''

tvzP("o,ritT

M32RP (7 3s)

e, = lstrp( Re



(2) Ratio X/R is the same or all network branches.These wo assumptionseadus to the conclusion hat Ip1and , fFig.7 .9(a))

have he samephaseangle andso have Ioz and 1o [Fig. 7.9(b)], such that thecurrent distribution factors Mr, and Mr, are real rather than complex.

Let, . Ict = llcrl lo, and cz = 1162lo2

where a, and 02 arephaseanglesof 1", and or, respectiveiywith respectto the common reference.

From Eq, (7.35),we can write

llrl2 - (Moll6l cos a1+ Mpzllo2lcos oz)2a (Mrll6lstn o;

Mr2lls2lsnoz)2 (7.36)

Expanding the simplifying the aboveequation,we getll,,l2= Mzrrllorlz+ tutf,zllczlz 2MolMrzllctl llGzlcosa1 - oz)

l l . r l= = '? ' : I I .J- -&z-r$ lvt lcos/ ,

' vLJl lvr lcosf "

Equation73e)"T,o:

;:;::::;:,PczB,z + 4,8,,

8 . . -n-WGoshfDmS,no

f f i D M " M " R o(7.40)

Bn = T.M' , rR.lV , l ' ( cos6) ' 7

YL Y

The terms Bs, Bp and 82, are called loss cofficients or B-cofficients.lfvoltagesare ine to line kV with resistancesn ohms, he unitsof B-coefficientsare in MW-I. Further, with Po, and Po, expressedn MW, P, will also be inMK

The\,bove results can be extended o the general case of ft plants withtransrnishur loss expressedas

k k

P, =Df PG^B^.PG.m:l n:l

where

cos (a,, - on )

Brz

Now

(7.37)

(7.38)

where Pot and Po, are he three-phaseeal power outputs of plants I and 2 atpower factors of cos (t, and cos Q2, nd yl and V2are the bus voltages at theplants.

ff Ro is the resistanceof branchp, the total transmission oss is given by*

(7.41)

(7.42)The

generalexpression or the power system with t plants is expressedas

P F3' T,*3,R,t.-r4r-lu|*ryr -1y f 1*r6f L' 'nt"P ' '

tVr, l2cosff iLu '

8 r,,, =l v - l l v , l cosQ-cosQ-

It can be recognized as

cos(a, - on )

lcosQ*cosQ,

Pcn

i lV,

Pc^

lV *

P'o4oo* zDpG^B^npGn

m,n:l

;!l

Pr = trB, * .. .*

i?ffil ruodernower ystem natysisI

The ollowing assumptionsncluding hosementioned lready re necessary,

if B-coefficientsare to be treated as constantsas total load and load sharing

between lantsvary. Theseassumptions re :

1, All load currents maintain a constant atio to the total current.

2. Voltage magnitudesat all plantsremain constant.

3. Ratio of reactive o real power, .e. power factor at eachplant remains

constant.

4. Voltage phaseangles at plant buses emain fixed. This is equivalent o

assuming that the plant currents maintain constant phase angle with

respect o the common reference, ince sourcepower factorsare assumed

constant as per assumption3 above.

In spite of the number of assumptionsmade, t is fortunate hat treating B-

coefficientsas constants, ields reasonablyaccurate esults, when the coeffi-

cients are calculated for some average operating conditions.Major system

..-#

r !nrr!-na!ffi i

- i i

l r"Y

Refbusv =1 0"pu

l ,oY

I ro"o

l b - - -1 ' '

- - - - t ' ' b



changes equire recalculationof the coefficients.

Losses s a function of plantoutputs an be expressed y othermethods*, ut

the simplicity of loss equations s the chief advantageof the B-coefficients

method.

Accounting for transmission losses results in considerable operating

economy.Furthermore, his considerations equally mportant n future system

planningand, n particular, with regard o the location of plants and building

of new transmission ines:

Figure7.10shows a systemhaving trrvo lants 1 and 2 connected o buses1 and

2, respectively. There are two loads and a network of four branches.The

referenceus with a voltage f l.0l0o pu is shownon the diagram.The branch

cunentsand mpedancesare:

I o =2 - 7 0 . 5 u

Iu= 1.6 j0.4 Pu

Zo = 0.015+ 70.06pu

Zo = 0.015+ 70.06pu

I , = 7 - j 0 . 2 5 P u

Id = 3. 6 - 70.9Pu

Z, = 0.OI + 70.04pu

Za = 0.Ol + 70.04pu

Calculate he oss formula coefficientsof the system n pu and n reciprocal

mesawat f s i f t he hase i s 100 MV A^ ^ - D -

*Formore accuratemethods nd exactexpression or 0P,./0P6i, eferences 22,231

may be consulted.

Fig.7.10 Sample ystem f Example .6

Solution As all load currentsmaintaina constant atio to the total current,wehave

rd _ 3.6_ o. g_ o. t8z6

I, + Id 4. 6 j l . l5

r- i0.25: -" - - - - -0.2174

4. 6 j1.1sI"

I , + l d

Mor= L, Mt r - - 0.2174,Mr r = 0.2774,Mu = 0.7826

M,,2= 0, Mnz= 0.7826, Mrz = 0.2174,Mrtz= 0,7826Since he sourcecurrentsare known, the voltagesat the sourcebusescan be

calculated.However, n a practicalsize networka load flow study has o bemade to find power factorsat the buses,bus voltagesand phaseangles.The bus voltagesat the plants are

Vr = 1.0 + (2 - j0.5) (0.015+ 70.06)

= 1.06+ jO.I725 = 1.06616.05"pu

Vz = 7 + (1.6 jO.4) 0.015 70.06)

= 1.048+ jO.O9= 1.051 4.9" pu

The current phaseanglesat the plants are (1, = Io, 12= 16r Ir)

ot = tan-t+! . :- l4oi o2: tan-r -^ O9t : - l4 o2 2 . 6

cos (or- ot) = cos 0o = 1

The plant power factorsare

pfi = cos (6.05" + l4') = 0.9393

ffiffi MociernowerSvstem na[,sis

Pfz = cos (4.9 + 14") 0.946The osscoefficientsre Bq. Q.a\l

- 0.02224 u

0.0 5x (0.7 2q2 + 0.0 x (0.2172 + 0.01 (0.7 2q2

(1.051)2(0.946)2

= 0.01597u

D _ e0.2174)L0.7826X0.015)0.01x 0.217a)2 0.01 (0.7826)2D

2

=

rJ66 Lotl x 0.9393 0.946

Bzz

OPtimalSystem Operation ffiffi$ffi

p l

c = f ci(Pci)

e -f tUilvj i ly, , tcos(0,,j: 1

3Q,+ Ltvinvjlllzulsin 0u

j: r

and

A4 -

Llvl lvj l ly, , lcos(9,,* 6i -

4) =0 fo r

eachpv

bu sj: 7

(7.43)

(7.44)

(7.4s)

(7.7)

subject to the load flow equations[see



= 0.00406 u

For a baseof 100MVA, theseoss coefficientsmustbe dividedby 100 oobtain heir valuesn unitsof reciprocalmegawatts,.e.

h 0.02224Dr r =

LLL+- 0.02224 x lo 2 Mw-l

10 0

8.t.,0'01597 = 0.01597 1o-2Mw-l

100

Br.t=0'0M06 = 0.00406x lo-z Mw-l

100

7.6 OPTIMAL LOAD FLOW SOLUTION

The problem of optimal real power dispatchhas been treated n the earliersection using the approximate oss formula. This section presents he moregeneral problem clf real attd reactive povrer flow so as to minirnize theinstantaneous peratingcosts. t i s a static optimizationproblem with a scalarobjective unction (also called cost function).

The solution echnique iven here was irst given by Dommel and Tinney[34]. t is based n load low solutionby theNR method,a first ordergradientadjustmentalgorithm or minimizing the objective unction and use of penaltyfunctions o account or inequality constraints n clepenclentariables.Theproblemof unconstrainedptirnal oad flow is first tackJed. ater he nequality

constraintsare ntroduced, irst on control variablesand then on dependentvariables.

Optimal Power Flow without Inequality Constraints

The objective unction o be minimized is ihe operating cost

] =

t r4n

4 ia. slack us

4 l

Oforeach e bus

It is to be notecl hat at the ith bus

P t= Pc i - Po i

Q i= Q c i - Quwhere Po, and ep; are oad demandsat bus i.Equarions 7.43), (7.44) and 7.45) can be expressedn vector orm

[Eq. (7.a3)l I'f (x,y) =

| _tn.7qqlforeach0 bus

|\

lEq.Q.a, foreach Vbus;

where the vector of dependent ariables s

l-t , l I.= l r ,

j fo reach rouus f

Ld, for each V bus_Jand he vectorof independentariabless

,1,,,j"each V us

f n f h a q l - ^ . r o f ^* , , . t ^+:^ - - rqvvvv r t - l r u r u l a l L l u l l , [ f l e OD l e Qt l V e t l t n e f i n n m r r c f i _ ^ 1 , - A ^ d ^ ^ _ r -

p o w e r . r r r u D l l l l w l L l L l g L l l E S r a c K D u S

The vector of independent ariabresy can be partitioned nto two parts_avectoru o f control variableswhich are to be variea to achieveoptimum valueof the objective function anda vector p of fixed or disturbange r unconhollable

(7.46)

(7.47)

(7.48a)

(7.48b)

ffi@ ModernPowerSvstem nalvsis

puru-"t"rs.Control arameters*aybe voltagemagnitudesn PVbuses, 6t

buseswith controllablepower, etc.The optimizationproblem**can now be restatedas

min C (x' u)

at

(7.4e)

subject o equalitYconstraints

.f (x , u, p) = 0 (7'50)

To solve the optimizationproblem, define the Lagrangian unction as

L (x , u, p)= C (x , u7 + Ar f x, u, P) (7'51)

where ) is the vectorof Lagrangemultipliers of samedimensionas (x, u, p)

The necessary onditions o minimize the unconstrained agrangianunction

are (see Appendix A for differentiation of matrix functions).

af ,= 0c*ly1 ' )_o0x 0x L}x J (7.s2)

feasiblesolutionpoint (a set of valuesof x which satisfiesEq. (7.5a) for givenu andp; it indeed s the oad flow solution) n the direction of steepest esceht(negative gradient) to a new feasible solution point with a lower value ofobjeetlve funstion. By repeating the-.semoves in *rc dkestisn +f the negadvegradient, the minimum will finally be reached.

The computational procedure or the gradientmethod with relevant details isgiven below:

Step I Make an initial guess or u, the control variables.

Stey 2 - Jind a feasible oad flow solution from Eq. (7.54) by the NR iterativemethod. The methodsuccessivelymproves he Solutionx as follows.

*(r +r)

,(r) + Ax

where A-r is obtainedby solving the set of linear equations(6.56b) reproducedbelow:



0L= 0c*ly1 ' )_o0u 0u Ldu-J

, a ru;

= (x,u,P)= o

Equation (7.54) is obviously the same as the equality

(7.s3)

(7.s4)

constraints.'The

tS^a L as needed n Eqs. (i.52) and (7.53)are ratherexpressionsor

; 0u

involved***. t may howeverbe observed y comparisonwith Eq. (6.56a) ha t

Y= Jacobianmatrix [same as employed in the NR method of load flow0x

solution; the expressionsor the elementsof Jacobianare given in Eqs. (6.64)

an d (6.65)1.

Equations 7.52),0.53) and 7.54) are non-linearalgebraicequations nd

can only be solved teratively.A simpleyet efficient iterationscheme,hat can

by employed, s the steepestdescentmethod (also called gradient method).

-Slackbus voltage and regulating transformer ap setting may be employed as

additional control variables.Dopazo et all26ltuse Qo, as control variable on buses'

with reactive Powercontrol'**rr *L^ . . , .*^- .pal nnrrrcr lncc ic to he minimized- the obiect ive funct ion is

lI LrMJ D lv r r l l vs r rv

C = Pr ( l V l , 6)

Since in this case he net injected real powers are fixed, the minimization of the real

injectedpower P, at the slack bus is equivalent o minimization of total system oss,

This is known as optimal reactivepower flow problem'***The

original pup". of Dommel and Tinney t34l may be consulted or details'

rheend esurts"^-,

,;; $; Iti.[l]'*u,,on orx andhe acobianatri,...Step3 Solve Eq. (7.52) or

(7.s5)I

Step 4 I ns er t ) from Eq. (7.55) into Eq. (7.53),and compute the gradient

l#r,"',rl]4"- r (*('),)

r , ^ - - r r - l

\ = - i { \ ' l a cL\dxl J 0x

(7.s6)

It may be noted that for computing the gradient, he JacobianJ - + is already0x

known from the load low solution step2 above).

step 5 rf v -c equalszero within prescribedolerance, he minimum hasbeen reached.Otherwise,

Step 6 Find a new set of control variables

where

unew= l. l .^6* L,i l

L,u = - o"V-C,

Here A,u is a step n the negative direction of the gradient. The step size isadjustedby the positive scalar o..

Y.c, oc*l 9L1'0u L0u

(7.57\

(7.s8)

#*ff"f

uooernPower vstemnalvsis

Steps1 through 5 are straightforward and pose no computational problems.

Step 6 is the critical part of the algorithm, where the choice of a is veryimportant.Too smalla value of a guaranteeshe convergencebut slows downthe rate of convergence; too high a value causes oscillations around the

Inequality Constraints on Control Variables

Though in the earlier discussion, he control variables are assumed o beunconstrai""o,

1.j":T.1",::tutt

are, n fact, always contrained,

(7.se)

e.g. Pc,, ,nin Po, S Pct, **

These nequality constraintson control variablescan be easily handled. f the

correctionAu,inBq. (7 57) causes ito exceed ne of the imits,a, s setequalto the correspondingimit, i.e.

to the constraint limits, when these limits are violated. The penalty function

method s valid in this case,because heseconstraintsare seldom rigid limitsin the strict sense, ut are n fact, soft limits (e.g. vl < 1.0 on a pebus really

ryv-should not exceed1.0 too much and lvl = 1.01may still be

The penalty method calls for augmentationof the objective function so thatthenew objective function becomes

C t = C ( x , u ) *f U t

where the penalty W, is introduced for each violated inequality constraint. Asuitable penalty function is defined as

w, = {7i@i- xi,^o)2 whenever i ) xi,rnax

'[ 71G,-xi ,^ i )z i wheneverxr(ry,min

Q'64)

(7.63)



where Tiis Treal positive number which controlsdegreeof penalty and s calledthe penalty factor.

Xmln

Fig.7.11 Penaltyunction

A plot of the proposedpenalty unction s shown n Fig. 7.11,which clearlyindicateshow the rigid limits are replacedby soft limits.The necessary onditions (7.52) and (7.53) would now be modified as givenbelow,while the conditions(7.54), .e. oad flow equations, emainunchanged.

(7.6s)

(7.66)

if u,,oro Au , ) ui,^^

7f u,,oro Au, 1ui,^in

otherwise

(7.60)

(7.6r)

After a control variable reachesany of the limits, its component n thegradient should continueto be computed n later iterations, as he variable maycome within limits at some ater stage.

In accordance with the Kuhn-Tucker theorem (see Appendix E), thenecessary onditions or minimization of I, under constraint (7.59) arc:

0L: 00u,

of .oou,o r , o0r,

-

Thereforernow, in step5 of the computationalalgorithm, the gradientvectorhas to satisfy he optimality condition 7.61).

Inequality Constraints on Dependent Variables

Often, the upper and ower limits on dependent ariables are specifiedas

r m i r , S x S r * u ^

e.g. lUn,in < lVl < lYl -. o ofl a PQ bus (7.62)

Such inequality constraintscan be conveniently handled by the penalty

function method. The objective function is augmented by penalties forinequality constraintsviolations. This forces he solution to lie sufficiently close

^ t r ZU W :'Ihe

vecto obtainedrom Eq. (7.64)wouldcontainonlyonenon-zero.0x

term orrespondingo thedependentariable ;; while#

= 0 as hepenalty

functions n dependentariables re ndependentf thecontrolvariables.

7f u,,*n<u i <ui ,^^ ,

if u, - ui,^*

ui: ui.^u*

a x _ a c , \ - a w j , f a f l ' ,T- = __l_ )d x o x 4 a r * L a " i ) - o

l

a x _ A C , s d w ; , f A f 1 ' ,:-= -- L

)d u o u ' + a " * L a " l ' r : o

'ModernPowerSystemAnalysis

By choosinga higher value fot 1,,the penalty unction can be madesteeperso hat hesolution ies closer o therigiA fimits; the convergence, owever,will

become oorer.A good schemes to startwith a low value of 7 and o increase

imization process, f the solution exceedsa certain tolerance

limit.

This sectionhas shown that the NR method of load flow can be extended o

yield the optimal load flow solution hat s feasiblewith respect o all relevant

inequalityconstraints.These solutionsare often required for systemplanning

and operation.

7.7 OPTIMAL SCHEDULING OF HYDROTHERMAL SYSTEM

The previous sectionshave dealtwith the problem of optimal schedulingof a

Dower systemwith thermal plants only. Optimal operating policy in this casedetermined at any instant without reference o operationat

Mathematical Formulation

For a certainperiod of operation7 (one year,one month or one day, depending.upon the requirement), t is assumed hat (i) storageof hydro reservoir at the

reservoir(after accounting or irrigation use) and oad demandon the systemare known as unctions of time with completecertainty(deterministiccase). Theproblem s to determine q(t),,the water discharge rate) so as to minimize thecost of thermal generation.

rTCr =

Jo CPor(t ) )dt

under the following constraints:(i) Meeting the load demand

Pcr(r) * Pca t) - Pr(t)- PoG)= 0; te 10,71 (7.68)

(7.67)



can be completely

other times. This, indeed, is the static optimization problem. Operation of a

systemhaving both hydro and thermalplants is, however, far more complex as

hydro plants have negligible operatingcost, but are required to operateunder

constraintsof water available or hydro generation n a given period of time.

The problem thus belongs to the realm of dynamic optimization. The problem

of minimizingthe operatingcostof a hydrothermal sYstem anbe viewed asone

of minimizing the fuel cost of thermal plants under the constraint of water

availability (storageand inflow) for hydro generation over a given period of

operation.J (water nflow)

Fig. 7.12 Fundamentalydrothermalystem

For the sakeof simplicity and understanding,he problem formulation and

solution techniqueare llustrated through a simplified hydrothermal systemof_ _ - - - - ^ _ 1 - - _ Q

Fig. 7 I2. This systemconsistsof one hydro and one thermaipiant suppiyingpower 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 transmissionossaccountedor by the oss ormula of Eq. (7.26),

This is called the power balance equation.(ii) Water availability

X (T)- x, (o) l' t@ at+ lrqgl t=oJO JO _

whereJ(t) is the water nflow (rate),X'(t) water storage, nd X/(0) , Y (T) arcspecified water storagesat the beginning and at the end of the optimizationinterval.(iii) The hydro generation Pcrlt) is a function of hydro dischaige and waterstorage or head), .e.

Pcn( r )= f (X ' ( t ) , q ( t ) ) (7.70)

The problem can be handled conveniently by discretization.The optimizationinterval Z is subdivided nto M subintervals achof time length47. Over eachsubinterval it is assumed that all the variables remain fixed in value. Theproblem is now posedas

: u^r^4-r:....*r"$^-r'(pt)n^(ELrfrrrt,under he following constraints:

(i) Power balanceequation

Pt *Pt r -P I -P t = 0

where

(7.72t

Ptr = thermal generation n the mth interval

Ptn = hydro generation n the zth interval

PI =transmission loss n the rnth interval

(7.6e)

\ t . l r )

ffiffifj- ModernPowerSystem nalysisn

n t n m t 2 | . r n n m I D t n m t 2= D77\r57 ) t LDTHTGH -f Dpy\r6p )

PI = load demand n the mth interval

(ii) Water continuity equation

y^ _ yt(m_r)_ f 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

Y - X^-r - J* + q^ = 0; m = I,2, , . . , M (7,73)

whereY = X/*IAT = storage n dischargeunits.

InEqs. (7.73), Xo and XM are he specified storagesat the beginningandend the

Optimalystem peration ffiffi

l -e = water head correction factor to account for head variation withstorage

In the aboveproblem formulation, it is convenient to choosewater dischargesin all subintervalsexcept one as independentvariables,while hydro genera-tions, thermal generationsand water storages n all subintervalsare treated asdependentvariables. The fact, that water discharge in one of the subintervalsis a dependent ariable, s shownbelow:Adding Eq. (7.73) for m = l, 2, ..., M leads o the following equation,knownas water availability equation

xM "o-D J^+la^ = o (7.7s)m m

p = non:effective discharge(water discharge needed o run h dro



of optimization interval.(iii) Hydro generation n any subinterval

Ptn = ho{I + o.5e Y + Y) l (7.74)

where

ho=9.8rx ro- rh to

fto = basic water head (head corresponding to dead storage)

*P3 , = 9.81 x 1o-2hk@^ -, p ) Mw

where

(q^ - p) = effective discharge n m3ls

hry,= average head in the mth interval

Now

LT(X^ +X^-r)

(7.76)

Becauseof this equation, only (M - l) qs can be specified ndependentlyandthe remaining one can then be determined rom this equationand s, therefore,a dependent ariable.For convenience, l is chosenas adependent ariable, orwhich we can write

M

qt = xo - xM DJ^-Dn^

Solution Technique

The problem is solved here using non-linear programming technique in

conjunction with the first order gradient method. The Lagrangian L isformulated by augmenting the cost function of Eq. (7.7L) with equaliryconstraints f Eqs. (7 72)- (7.74) hroughLagrangemultipliers dual variables)\i \i'and )i. Thus,

.c, D tc(%r) xT 4r+ 4, - ry- ff i + M (y - y-t-r* +

qr) ^T p1, h, r + 0.5e(y it11* @^ p)rj (777)The dual variablesare obtainedby equating to zero the partial derivatives

of the Lagrangianwith respect o the dependentvariablesyielding the followingequations

A P A r , r D m t / r t . \rJ)e ut/ \ I CT) ,z I n Uf t | ^

= - - - A r l l --

l - u7Pt dPt '[- 7Pt ) 0 /78)

[The reader may compare this equation with Eq. (7.23)]

can be expressed* s

(q* - p)

l f f i= 7ro*2A

where

A = draa. f cross-sectionof the reservoir at the given storage

h'o= basic water head (head corresponding o dead storage,

hk= hL l l + o .Se(x '+ X"- t ) l

where

AT

Ahto

.

Now

4n= ho {! + o.Se(x^ x^-t)l @^ P)

where

ho= 9.87 l0-3hto #r, M-^r['-ffi)=' (7.7e)

ffil Modern ower ystem nalysisa

( +) =)7,-Ar*' \r p.sh"r(q*p) l )i*t 1o.5hoI ax^ )^**t/

* U

(q^ '-DI - o

and using Eq. (7.73) n Eq. (7.77),we get

(pt-) = )rr- t\n" fl + 0.5ezy + Jt - zq t+d)= 0 (7.81)laq' )

The dual variables or any subinterval may be obtainedas 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.'

(7.80).The gradientvector is given by the partial derivatives of the Lagrangian with

(7.80)

ientlyby augmentinghe cost unctionith p"nulty unctionrur air";;;i"Sec. .6.

Themethod utlinedaboves quitegeneral nd canbedirectlyextendedoa slzstemhaving multihas the disadvantage

hydro and multi-ther+nal plan+s.The method, however,of large memory requirement, since the independent

variables,dependent ariablesand gradients eed to be storedsimultaneously.A modified techniqueknown as decomposition 24) overcomeshis difficulty.In this techniqueoptimization is carried out over each subintervaland thecompletecycle of iteration is repeated,f the water availability equationdoes.not check at the end of the cvcle.

Consider the fundamental hydrothermal system shown in Fig. 7.I2. The



respect o the independent ariables.Thus

(+) =\f.- ^Zh" r + 0.5e 2Y-t + J^ - 2q^+ p) } (7.82)\o q )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

subintervalsexcept he first.

Obtain the values of dependent variables

(7.73), (7.74), (7.72) and (7.76).

variables q* (m*I) for all

3. Obtain the dual variables f, \f, )i @ = 1) and )rr using Eqs. (7.78),

(739), (7.80)an d (7.81).

4. Obtain the gradient ectorusingEq. (7.82)and check f all its elements

are equal to zero within a specifiedaccuracy. f so, optimum is reached.

If not, go to step 5.

5. Obtain new values of control variables using the first order gradient

method, .e.

ek*=q;a-++) ;m=r (7 .83)\oq* )

where cr is a positive scalar. Repeat from step 2

In the solution techniquepresentedabove, f some of the control variables

(waterdischarges) ross he upper or lower bounds, hese are made equal to

their respectivebounded values. For these control variables, step 4 above

is checked n accordancewith the Kuhn-Tucker conditions (7.61) given in

Sec. .6 .

Y, Ptu, F\r, qt using Eqs.

objective s to find the optimal generation chedule or a typical dav, whereinload varies n three stepsof eight hourseachas 7 Mw, 10 Mw and 5 Mw,respectively.There s no water inflow into the reservoir of the hydro plant. Theinitial water storage n the reservoir is 100 m3/s and the final water storageshould be 60m3/s, .e. the total water available or hydro generationduring theday is 40 m3/s.

Basic head s 20 m. Water head correction actor e is given to.be 0.005.Assume or simplicity that the reservoir s rectangularso that e doesnot ehangewith water storage. Let the non-effective water discharge be assumed as2 m3/s.Incrementalue l cost of the thermalplant is

dc - r.opcr + 25.0Rsft'dPcr

Further, ransmissionossesmay be neglected.The aboveproblemhas beenspeciallyconstructed ratherovelsimplified) to

illustrate the optimal hydrothermal schedulingalgorithm, which is otherwisecomputationally nvolved and the solution has to be worked on the digitalcomputer.Stepsof one complete terationwill be given here.

Since hereare hreesubintervals,he control variables e q2and q3.Let usassume heir initial values to be

q2 = 75 m3ls

15 m2ls

The value of water diseharge n the first subintervalcan be immediateiy foundout using Eq. (7.76), .e.

er = LO O 60 - (1 5 + 15 ) = 10 m3ls

It is given that X0 = 100 m3/sand X3 = 60 m3/s.From Eq. (7.73)

W Modern o

Xt = f + Jr - gt = 90 m3ls

f = X r + 1 2 - q 2 = 7 5 m 3 / s

The valuesof hydro generationsn the subintervalsan be obtained singEq. 7.7q as ollows:

Pb , t=9 ,81 10 -3 20 [ l + 0.5x 0 .005 xr+ X0 ; ) {q ' - p )

= 0.1962 I + 25 x 104 x 190}x 8

= 2.315MW

4n =0 .1962{ I + 2 5 x lOa x 165} 13

= 3.602MW

PZa=0 .1962l + 25 x 10ax 135} 13

= 3.411MWThe thermalgenerationsn the hree ntervalsare hen

xt, x3 )l {o.Shoeqt p)l _

x?, ll - ^Z {0.5hoe q, _ p)l _Substitutingvarious values, we get

) t r=8.q14 29.685 0.5x0.1962x .005 8l _ 31.3980. 5x0.1962 0.005x 13 )

= 8.1574

)t , = 9.1574 31.398 0.5 0.tg6}x 0.005 t3 )- 26.589 0.5x 0.1962 0.005 t3) = 7.7877

UsingEq. (7.82),he gradient ectors

(#)= ^7- f t n"{1 + 0.5 0.0052x eo 2 x 15 z) l

S! 1O.Sn"e, p)l = 0

)3, 1o.5ho,qt - p)] = o



pL r = pL- pl, = J - 2.515 4.685MW

4r = Pto- PL, = 10 3.602 6.398MW

Fcr=p| , - 4" = J - 3 .41I= 1.589 W

FromEq. (7.78),we havevalues f )i as

dc(P€)_ \m

A D m- ' t l

V G T

or

Calculating ),

Also from Eq.

From Eq . (7.81

\l'=

A\hn i + o.5e?)(0

il- 29.685 0.1962 l + 25

- ,8 .474

FromEq. (7.E0)or m = 1 and2, we have

^ l- z q ' + o l

x loa (2oo 2o + 2)l

) T = P [ , + 2 5

for all the three subintervals,we have

[^i [2e 85]

l r ? l = l r r . l s a lLr? fzo.ssrJ(7.79), we can write

lril hil lzs.68s1| ^3| = |^?l: I r:ea I for treosslessase

Lri Ld Lzesag

)

= 8.1574 31.398 0.1962{l + 25 x 10a x l52l= - 0.3437

( ar . l - r3lrf )-

A:2-strn,l + 0.5e2X2 f - zqt+ p)I

lf:i=[] ,'l-:::Ijflililand from Eq. {7.76)

4'r*= 10 0 - 60 - (15.172+ 14.510)= 10.31gm3lsThe above computation brings us to the starting point of the next iteration.

Iterations are carried out till the gradientvector U"comeszero within specifiedtolerance.

- 7.7877 26.589 0.1962 I + 25 x 10 a x t22}

= 0.9799

If the tolerancebr gradientvector s 0.1, hen optimal conditionsarenot yetsatisfied,since he gradient vector s not zero, .i. (< 0.1); hence he seconditerationwill have o be carried out startingwith the following new valuesofthe control variablesobtainedfrom Eq. (7-g3)

lnk_l=la[,oLl#lLq:"*J=;i;l-1+l

Loq"Let us take a = 0.5, then

ffil uodern ower lrstemnalvsis

PROBEii/iS

7.1 For Example 7.1 calculate he extra cost incurred n Rsftr, if a load of220 MW is scheduled s Pct= Pcz = 110 MW.

7.2 A constant oad of 300 Mw is supplied by two 200 Mw generators,and 2, for which the respective ncremental fuel costs are

dcr

Po ,=o ' l O PG l+2 0 '0

dcz

dPo,- o'lzPc2 + 15'o

with powers Pc in MW and costs c in Rsar. Determine (a) the most

economicaldivision of loadbetween he generators, nd (b) the saving nRs/day thereby obtained compared to equal load sharing between

Hffiffi

miiiions of iriiocaiories perhour can be expressedas a function of poweroutput Poin megawattsby the equation

0.00014+ O.$ft + r2.0po+150Find the expression or ineremental uel eost n rupeesper megawatthouras a function of power output in megawaffs.AIso find a good linearapproximationto the incremental fuel cost as a function of Fo.Given: Fuel cost is Rs Zhmltion kilocalories.

7.6 For the systemof Example 7.4, the system ) is Rs 26a4wh. Assumefurther the fuel costs at no load to be Rs 250 and Rs 350 per hr,respectively for plants I and2.(a) For this value of system ),, what are the values of p61, po, and,

received oad for optimum operation.(b) For the above value of received

oad, what are the optimum valuesof Pot and Por, if system losses are accounted for but notcoordinated.



machines.

7.3 Figure P-7.3 shows he incremental uel cost curvesof generatorsA andB. How would a load (i) more than ZPo, (ii) equal o 2p6, and (iii) lessthan ZPo be sharedbetweenA and B if both generatorsare running.

(MW)mtn P6

Flg. P-7.3

7.4 Consider he following three C curves

P G r = - 1 0 0 + 5 0 ( I Q t - 2 A q i

Pcz= - 15 0 + 60 (IQz - 2. 5 Aq Z

Pct= - 8. 0+ 4a Qq 3 - 1. 8 Aq i

where ICs are n Rs/IVIWhand P6s are in MW.The total load at a certain hour of the day is 400 MW.

transmission oss and develop a computer programme forgeneration chedulingwithin and accuracyof + 0.05 MW.

Note: All P6s must be real positive.

Neglect

optimum

(c) Total fuel costs n RsArr or parrs (a) and (b).

7.7 FigureP-7.7showsa systemhaving wo plants I and2 connectedo buses1 and 2, respectively. There are two loads and a network of threebranches. he bus 1 is the reference us with voltageof 1.0 I 0" pu. Thebranch currentsand impedancesare

Io=2 - 7 0 . 5 p u

L = 1 6 - i O 4 n r ro

1,= 1.8 i0.45 pu

Zo= 0.06+ j0..24 uZt = 0.03+ J0.12 u

Z, = 0.03+ /0.I2 pu

Calculatehe oss ormulacoefficients f the systemn per unit and nreciprocalmegawatts,f the bases 100MVA

Ref bus

Flg. P-7.7 Samplesystem or probtemp-7.7

Wuoo"rnpo*", syrt"r Anutyri,

7.8 Fot the powerplantof the llustrative exampleused n Section7.3. obtain

the economicallyoptimumunit commitment for the daily load cycle givenin Fig. P-7.8.

Correct he schedule o meetsecurity requirements.

8 1 2 16

i "E , o

0

w. r\cuenswanoer, J.}(-, Modern power systems, International

rext Book co., NewYork, 1971.

8' singh, c. and R. Billinton, system Reliabitity, Modelling and Evaluation,Hutchinsonf London, 977.

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, 199611' Mahalanabis, .K., D.p. Kothari and s.I. Ahson, computerAided power system

Analysisand control, Tata McGraw-Hill, New Delhi. r9gg.12 . Bergen, .R., power systemAnarysis, rentice-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., "Economics f power systemsReriabil ity,',Technicopy.9g0.16. warwick,

K. A.E. Kwue and R. Aggarwar (Eds), A.r. Techniques n powerSystem.s,EE, UK, 1997.17. Momoh,J.A., Electic power System pplications



20 24

Time nhours --------'

Flg. P-7.8 Daily oadcurve or problemp-7.9

7.9 RepeatExample 7.3 with a load of 220 Mw from 6 a.m. to 6 p.m. and40 MW from 6 p.-. to 6 a.m.

7.10 Reformulate he optimathydrothermalschedulingproblemconsidering heinequality constraints on the thermal generation and water storageemploying penaity functions. Find out the necessaryequations andgradientvector to solve the problem.

REFERECES

Books

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t970.

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I 9 ) 6 .

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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,

ma l

< | Q a n Q ^ - i n D I f ^ + L ^ - j . . / \ - . : - -tvv r r . u. r q r rs v . l . nu l ua t l l, \ _ rp l , l l l l a l

Review, "Int. J. EPES,Vol. 20, No.52. Kulkarni,P.S.,A.G. Kothari and D.p.

Dispatchusing mprovedBpNN", /nf.3t - 4_7

53. Aryu, L.D., S.C.Chaudeand D.P. Kothari, "EconomicDespatchAccountingLineFlow Constraintsusing FunctionalLink Network," Int. J. of Electrical Machine &Power Systems, 8, l, Jan 2000, pp. 55-6g.

54. Ahmad, A. and D.P. Kothari, "A practical Model for GeneratorSchedulingwith rransmissionconstraints", Int. J. of EMps, vol. 2g,

Approach for

No. 2, Nov.

ThermaiGeneradng'tjnirCommitment-A7, Oct. 1998,pp . 443-451.Kothari, "Combined Econonic and EmissionJ. of EMPS,Vol. 28, No. l, Jan 2000,pp.

Maintenance

No. 6, June2000, pp. 501 514.

55. Dhillon, J.s. and D.P. Kothari, "The surrogate worth rrade offMutliobjective Thermal power Dispatch hobelm". EpsR, vol. 56,2000,pp . 103-110.

56. Son, S. andD.P. Kothari, "Large ScaleThermalGeneratingUnit Commitment:ANew Model", in TheNext Genera\ionof Electric Power (Jnit CommitmentModels,

edited by B.F. Hobbset. al. KAp, ,Boston,2001, pp. Zll-225-



and II", 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 usingConstantSensiiivityMatrix", ElectricPowerSystemResearch, ol.II, No. 3, Dec. 1986,pp. 195-203.

42. Nanda,J., D.P. Kothariand K.S. Lingamurthy, Economic-emission oad Dispatchthroughcoal Programming echniques",EEE Trans.on Energyconversion,vol.3, No. 1, March 1988,pp. 26-32.

43. Nanda, J.D.P. Kothari and s.c. srivastava,"A New optimal power DispatchAlgorithm using Fletcher's QP Method", Proc. IEE, pte, vol. 136. no. 3, May1989 , p. 153 -161 .

44 . Dhillon, J.S.,S.C. Parti and D.P. Kothari, "stochasticEconomicEmissionLoad

Dispatch", nt. J. of Electric Power systemResearch,Yol. 26, No. 3, 1993,pp.17 9 183.

45 . Dhillon, J.S.,S.C.Partiand D.P.Kothari, MultiobjectiveOptimalThermalPowerDispatch", nt. J. of EPES,Vol. 16, No.6, Dec. L994, p.383-389.

46. Kothari, D.P. and Aijaz Ahmad, "An Expert system Approach to the unitcommitment hoblem", Energy conversion and Management", ol. 36, No. 4,Apri l 1995,pp . 257-261.

47 Sen,Subir, D.P. Kothari and F.A Talukdar, "EnvironmentallyFriendly ThermalPower Dispatch An Approach", nt. J. of EnergySources,Vol. 19, no. 4, May1997,pp.397-408.

48. Kothari, D.P. and A. Ahmad, "Fuzzy Dynamic ProgrammingBased optimalGeneratorMaintenanceSchedulingncorporatingLoad Forecasting",n Advancesin Intelligent systems,edited by F.c. Morabito, IoS press,ohmsha, 1997, pp./.J5 /4U.

49. Aijaz Ahmad and D.P. Kothari, "A Review of Recent Advances in GeneratorMaintenance 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.

57. Dhillon, J.s., s.c. Parti and D.p. \ Kothari ,,FuzzyDecision Making in Multi

objective Long;term scheduling of Hydrothermalsystem,,, 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 DispatchbyFuzzy Logic based Searching Weightage Pattern," Electric power SystemsResearch,Vol. 63, 2002, pp. 149-160.

59. Dhillion, J.s., S.c. Parti and D.p. Kothari, "FuzzyDecision-mgkingn stochasticMultiobjective short-termHydrothermalscheduling,"Ip,Bproc.tcTD, vol. l4g, z,March 2fi02, pp l9i-200.

60' Kothari, D.P., Applicationof Neural Netwdrks o PowerSystems Invited paper),Proc. Int. Conf., ICIT 2000,Jan. 2000, pp. 62l_626.

R1.,

caused y momentaryhargen generaforpeecl,I'r.r.tnr*,-i;;?t;qffi; ;;excitation oltagecontrols renon-interactiveor smallchanges nd can bemodelledndanalysedndependently.urthermore,xcitationoltage ontrolsF : t c l : t c f i n r r i n r r r h i n h t h c - , r i ^ r f i r r r o n , r n . . r , r h r ^ 6 ^ , r r i - +^ - ^ . 1 : - r L ^ e ^$ r L - - ^ - ^ - - ^ -r r r v v r r r v r r L r r v r r r c r J v r r r l t t w v r J r r J r - ( l r r r u r l L U u l t L t r l c u r5 l l l a ! ul u l c ; g g i r t c f a l o r

field; while the power frequency ontrol is slow actingwith major time constantcontributedby the turbineand generatormomentof inertia-this time constantis much arger han ha tof thegeneratorield.Thus, he ransientsn excitationvoltagecontrol vanish much faster and do not affect the dynamics of powerfrequencycontrol.



8. T.INTRODUCTION

Power ystem perat ion onsidcrcdo ar wasunder ondit ions f stcadyoad.However, both active and reactivepower demandsare never steadyand theycontinually change with the rising or falling trend. Steam input to turbo-generatorsor water nput to hydro-generators) ust, herefore,be continuouslyregulated o match the active power demand, ailing which the machinespeedwill vary with consequent change in frequency whieh may be highlyundesirable*maximumpermissible hange n power fiequency s t 0.5 Hz).Also the excitation of generatorsmust be continuously egulated o match he

reactive power demandwith reuctivegeneration, therwise he voltagesatvarious systembuses may go beyond the prescribed imits. In modern argeinterconnectedsystems, manual regulation is not feasible and thereforeautomaticgenerationand voltage egulationequipment s installedon eachgenerator.Figure 8.1 gives the schematicdiagram of load frequency andexcitationvoltage regulatorsof a turbo-generator. he controllersare set or aparticularoperatirrg ondition and they take care of small changes n loaddenrandwithout fiequencyand voltageexceedinghe prescribed imits.Withthe passageof time, as the change in lcad demand becomes large, thecontrcl lersmustbe reset i thernianuallyor automatical ly.

It has beenshown n previouschapters hat for small changesactivepoweris dependent n internalmachineangle 6 and s inderrendent f bus voltage:whiie bus voitage s dependenton machine excitation (thereforeon reactive- " -

Changen frequencyauses hangen speed f the consumers' lantaffectingproduction rocesses.urther,t is necessaryo maintain etwork requency onstantso hat he powerstationsun satisfactorilyn parallel,he variousmotorsoperatingon the system un at the desired peed, orrect ime s obtained rom synchronousclocks n the system, nd he entertainingevicesunctionproperly.

Fig. 8.1 schematic iagram f load requency nd excitationvoltage egulatorsf a turbo-generator

Change n load demandcan be identified as: (i) slow varying changes nmeandemand,and (ii) fast randomvariationsaround he mean.The regulatorsmustbe dusignedo be nsensitiveo thst andom hanges, therwisehe systemwill be prone to hunting resulting in excessivewear and tear of rotatinsmachinesand control equipment.

8.2 LOAD FREOUENCY CONTROL (STNGLE AREA CASE)

Let us consider he problemof controlling he poweroutputof the generatorsof a closely knit electricareaso as o maintz,in hescheduledrequency.All thegeneratorsn such an areaconstitutea coherentgroupso thatall the generators^ - ^ ^ l - I ^ l - - - - . - l ^ - - - - ^ L - - - . r - - ^ . _ _ _ : - , . r .speeoip anci iow riowii ogetiiernarntarnrnghelr eiarrve owerangies.Suchan area s defined as a control area. Tire boundariesof a coqtrol area willgenerally oincide with thatof an individual ElectricityBoardCompany.

To understand he load fiequencycontrol problem, et us considera singleturbo-generator ystemsupplyingan isolated oad.

IP+JQ

W Modern power system Analys,s

Turbine Speed Governing System

Figure8.2 showsschematicallyhe speedgoverningsystemof a steam urbine.The system onsists f the following components:

Steam

Speed changer

turbine.Its downward movementopens he upperpilot valve so that more steemis admitted to the turbine under steadyconditions (hencemore steadypower

. The reverse

Model of Speed Governing System

Assume that the system is initially operating under steady conditions-thelinkage mechanismstationaryand pilot valve closed,stearnvalve opened by adefinite magnitude,turbinerunning at constantspeedwith turbin" po*"r outputbalancing the generator oad. Let the operating conditions be characteizedby

"f"= system requency (speed)

P'c = generatoroutput = turbine output (neglecting generator loss)

.IE= steam valve setting

We shall obtain a linear incremental model around these operating



Mainpiston

AI

rHydraul icampli f ier(speed control mechanism)

Fig.8,2 Turbine peedgoverningystem

Reprinted ith permissionf McGraw-HiltookCo.,New York, rom Olle . Elgerd:ElectricEnergy SystemTheory:An lntroduction,1g71,p. 322.

(i) FIy ball speedgovernor:This is the heartof the systemwhich senseshechangen speed frequency).As the speedncreaseshe fly balls move outwardsand he point B on linkage mechanismmovesdownwards.The reversehappenswhen the speeddecreases.

G) Hydraulic amplifier: It comprises a pilot valve and main pistonalrangement. ow power level pilot valve movement s converted nto highpower evelpiston valvemovement.This is necessaryn order to openor closethe steamvalve againsthigh pressuresteam.

(xl) Lintcagemechanism:ABC is a rigid link pivoted at B and cDE isanother igid link pivoted at D. This link mechanism rovides a movement othe control valve in proportion to change n speed. t also provides a feedback

,,fr9rnhe steamvalve movement(link 4).

conditions.

Let the point A on the linkage mechanismbe moved downwardsby a smallamountAye.It is a commandwhich causes he turbine power output to changeand can therefore be written as

Aye= kcAPc

-

Pilotvalue

oilHigh

pressure (8 .1)

(8.2)

where APc is the commanded ncrease n power. \The command signal AP, (i.e. Ayi sets nto rnotion a bequenceof events-the pilot valve moves upwards,high pressureoil flows on to the top of the mainpiston moving it downwards; the steamvalve openingconsequently ncreases,the turbine generatorspeed ncreases,

.e. the frequencygoesup. Let us modeltheseevents mathematically.

Two factors contribute to the movement of C:

(i) Ayecontributer-[?J Aya or - krAyo(i.e. upwards)of - ktKcApc\ r l l

(ii) Increase n frequency ff causes he fly balls to move outwards so thatB moves downwards by a proportional amount k'z Af. The consequent

movemen of Cwith A remaining fixed at Ayo- . (+) orO, - + kAf

(i.e. downwards)

The net movement of C is therefore

AYc=- k tkcAPc+kA fThe movementof D, Ayp, is the amountby which the pilot valve opens. t iscontributedby Ayg and AyB and can be written as

Ayo=(h) Ayc+(;h) *,

= ktayc + koAys (g.3)The movement

ay.o-d,ependingpon ts sign opensone of the ports of the pilotvalve admitting high pressure'o' into thJ"ynnJ.ithereby

moving the mainpiston and opening the steamvalve by ayr. certain ustifiable simprifyingassumptions,which ean be rnadeat this .tugl, ur",(i) Inertial reaction orces of main pistoi and steamvalve are negligiblecompared o the forcesexertecl n the iirton by high pressureoil.(ii) Because of (i) above, the rate of oil admitted to the cylinder isproportional to port openingAyo.The volume of oil admitted o the cylinder is thus proportional to the timeintegralo,f ayo. The movementay"i.s obtainedby dividing the oil volume bythe areaof the cross-sectionf the-piston.Thus

Avn= krfoeayrlat

It can be verified from the schematicdiagramthat a positive movemen ayo,causesnegative upward)movement ayulccounting for the

(8.4)

controt E1

E ^ , , ^ r i ^ - / o o \ : - . r . - triyLr.Lru' \o.o., rs rcpfesenleo ln tne ronn of a block diagram in Fig.

9.3.

4Y5(s)

4F(s)

Flg. 8.3 ,Block iagram epresentationf speedgovernor ystem

The speed governing system of a hydro-turbine is more involved. Anadditional feedback loop provides temporary droop compensation o prevent

instability.This is necessitated y the targe nertia or the penstoct gut" whichregulatesthe rate of water input to the turbine. Modelling of a hyjro-turbineregulatingsystem s beyond he scopeof this book.

Ks9

1+ sss



n"gutiu" ,ign usedin Eq . (8.4).

Taking he Laplaceransform f Eqs. (g.2), g.3)and g.4),we gerAYr(s)=- k&cApc(")+ krAF(s)

Ayp(s)= kzAyd,s) koAyug)

ayu (g= -ks l rUnEliminatingAyr(s) andAyo(s),we can write

AYu(s)

k'ktk'AP' (s) k,krAF(s)

(o o ' t ')

\ "'tr ,/

-lor,<,r-*^or",].i#)

(8.5)

(8.6)

(8.7)

(8 .8 )

where

n=kl ct_K2

= speed egulationof the governor

K.,=+y

- gainof speedovernor

. r. l ", rs=;-; = tlme constantof speedgovernor K q k S r- -

Steam valve-=-&

Turbine Model

Let us now relate the dynamic responseof a steamturbine in tenns of changesin power ouFut to changesn steamvalve opening ^4yr. Figureg.4ashows atwo stage steam turbine with a reheatunit. The dynamic *ponr" is targelyinfluencedby two factors, (i) entrainedsteambetwein the inlet stbamvalve andfirst stageof the turbine, (ii) the storage ction n the reheaterwhich causes heoutput of the low pressurestage o lag behind hat of the high pressurestage.fttus,

the turbine transfer unction is characterized y two time constants.Foreaseof analysis t will be assumed

ere hat the turbinl can be modelled to haveSsingle equivalent ime constant.Figure 8.4bshowsthe transfer unction modelof a sream urbine.Typicaly the time constant{ lies'in the rangeo.i ro z.ssec.

AYg(s)-FAPds)

(b) Turbineransferunctionmodel

Flg. 8.4

(a) Two-stageteam urbine

#ph-Si rrrroarrno*", s),rt"r An"ly.i,I

Generator Load Model

The ncrementn powernput o thegeneratbr-loadystemsAPG _ APD

whele AP6 = AP,, incremental turbineincremental oss to be negligible) and App is the load increment.

This ncrement n power input to the syrtem s accounted or in two ways:(i) Rate of increase of stored kinetic energy in the generator rotor. At

scheduled requency (fo ), the stored energy is

Wk, = H x p, kW = sec (kilojoules)

whereP, is the kW rating of the turbo-generator andH is defined as ts inertiaconstant.

The kinetic energybeing proportional to squareof speed frequency), hekinetic energy at a frequency of (f " + Arf ) is given by

AutomatlcGeneration nd VoltageControl I

=tAP6g)_Po(,)r.[#j (s.13)

2H

Bf"= pow€r system ime constant

Kp,+

=power ystemain

Equation (8.13) can be representedn block diagram form as n Fig. g.5.

laeo(s)^Po(s)6---ffioro,



=nr,(r.T)

Rate of change of kinetic energy is therefore

$rr*"rfffrr"n(ii) As the frequencychanges, he motor load changesbeing sensitiveto

speed, he rate of changeof load with respect o frequ"n.y, i.e. arot\ycan be

regarded as nearly constant for small changes n frequency Af ard can beexpressed s

Flg.8.5 Block iagramepresentationf generator-loadodel

complete Block Diagrram Representation of Load FrequenryControl of an Isolated Power System

(8.e)

(8 .10)

( 8 . 1 1 )

positivoor a

@PDl? f lA f=BA f

where he constantB can be determined mpirically,B ispredominantly otor oad.Writing he powerbalance quation, e have

A P c - a P ^ = T H P ' d (r= - f . ] * <of l+B Af

Dividing hroughoutby , and earanging,we get

AP(s)=trPn15;

AP6(s)

Flg. 8.6 Blockdiagrammodelof oad requency ontrol(isolated owersystem)

Steady States Analysis

The model of Fig. 8.6 shows that there are wo important incremental nputs to

the load frequency control system- APc, the change n speedchangersetting;

and APo, the change n load demand. Let us consider,,.4,.simpleituatiqn in

AP6 $ u ) - AP ; q ; u ) = 1 d/ A ' ^ ' n ' 7 ' - - - \

f dt

(Afi + B(ptt) af (8.i2)

Taking the l,aplace transforrn,we can write AF(s) as

4Fis; - AP,G) -4PoG)B * - ' - s

Modern

which the sneer l .hqnrro' hoo . . g.- .^) ^- .- .

r.han.,oo 't::; ;-::--::^'(rr cr rr^tr(r ucttrng \7'e' af

c= o) and the load demand

:,:il?:3l i: T: : a2re g ,, ;,2;; *;:r;;ffi; *'#ff:Tiil:teadyhangensystemrequen-cyorasudd.nhung",fi;ffi"ffi;ti'l;

anaount, (, e.Apog):+)is obtained s ollows:

aF@)l*,(s):o - AP^

^f

. r L^ ^L^- - ^ Ir'E .1uuy'cquauonglves tne steadystatechanges n frequencycausedbychanges n load demand. Speed regulation R is-naturally so adjusted thatchanges n frequency are small (of the order of 5vo from no load to ruu load).

Therefore, he inear ncremental elation(g.16)icanbe applied rom no load tofull load' with this understanding,Fig. 8.7 shows the linear relationshipbetween requencyand load for free governoroperation with speedchangersetto give a scheduledrequencyof r00% at full toao.The .droop,

or slopeof this(

relationship s -l I'l

-\ B+( t /R) )

Power systemparameter B is generaily much smalrer* than r/R (a typicalvalue s B = 0.01 pu Mwalz and l/R = U3) so that B canbe neglected ncomparison.Equation(8.16) then simplifiesto

rhe roop"r,,fl", fjfli;], curvesspeedgovernor egulation.

(8 .17)thusmainlydetermined y R, the



K I ( = 1^ s o r r , . I

It is also rccognized hat Ko, =

in frequency).Now

4=-(#6)o,.

7 / B , w h e r e B - Y ^

ai

/P' (inPuMWunit change

(8 .16 )

fi roa(JL

8. og.c

102 atl i \ d A ^ t | |

\r,, ruu-loLoao

(i i)60% Load01

10 00

Percent Load

Flg. 8.7 Steady"*-l?39-frequency

qharacteristicof a speedgovernor ystem

MW. let the change in load

= 50 Hz). Then

ap,=_*"r: (r^;)o",

Decreasen systemoad BAf= (uffi)*,

Of course, he contribution of decreasen system oad is much less than theincrease n generation.For typical values of B and R quoted earlierAPo = 0.971 APo

Decreasen system oad = 0.029 ApDconsider now the steady effect of changing speed changer setting

(Or"<rl-+)with

load demand emaining ixed (i.e. Apo= 0). The sready

statechange n frequency s obtainedas follows.

*For250 MW machine with an operating load of 125

be i%o or IVo change n frequency (scheduledrequency

a-:?:r?: :2.5 NNVtHzaf 0. s

:#:

o'olPuMwgz=(#)b

WuodernPower ystem nalysis

I

AF@lap,{s):o:

r t v (t t sgt t f t ^ps A D

x u ' c( 8 . 1 8 )

s1+T,rs ) ( l *4s)

I4,flr*uoyro,":

IAP',:g

xl-

_ t( 1

K

\Il

r l

p, R

AP,

* zors) KseKt

KreKrKp,

+ K.sKtKps R(8 .1e)

(8.20)

IfKrrK,=l

Ar= | \rc"" \ B + l l R )

If the speedchanger setting s changedby AP, while the load demand

changes y APo, the steady requencychanges obtainedby superposition,.e.

( 8 .21 )

According to Eq. (8.2I) the frequencychangecausedby load demandcan be

Ar= (".

ru ) 'o " - APo)

Autor"ticG"n"r"tion ndVolt"g"ConttolF

Two generators rated 200 MW and 400 MW are operating in parallel. The

droop characteristics f their governorsare4Vo and 5Vo, espectively rom no

load to full load. Assuming that the generators re operatingat 50 Hz at no

load, how would a load of 600 MW be shared etween hem?What will be the

system frequency at this load? Assume free governor operation.

Repeat he problem f both governorshave a droop of 4Vo.

Solution Since the generators are in parallel, they will operateat the same

frequency at steady oad.

Let load on generator 1 (200 MW) = x MW

and load on generator 2 (400 MW) = (600 - x) MW

Reduction n frequency=

AfNow

a f _ 0.04 50



compensated y changing he settingof the speedchanger, .e.

APc- APo, for Af = Q

Figure 8,7 depicts two load frequency plots-one to give scheduled

frequencyat I00Vo ated oad and he other o give the same requencyat 6O7o

rated load.

A 100 MVA synchronous enerator perates n full load at at frequencyof 50Hz. The load is suddenly educed o 50 MW. Due to time lag in governor

system, he steam alve begins o closeafter 0.4 seconds.Determine he change

in frequency hat occurs n this time.

Given H = 5 kW-sec/kVA of generatorcapacity.

Solution Kinetic energy stored n rotatingparts of generatorand turbine

= 5 x 10 0 x 1.000= 5 x 10 5kW-sec

Excesspower input to generatorbefore he steamvalve

begins o close= 50 MW

Excess energy nput to rotating parts n 0.4 sec

= 50 x 1,000x 0.4 = 20,000kW-sec

Storedkinetic energy oo (frequency)2

Frequencyat the end of 0.4 sec

=5o I soo,ooozo,ooo)t"=5r rfz

\ 500,000 )

x

af6 0 0 - x

EquatingAf in (i) andv -

6 0 0 - x =

System requency 50 - 0'0-1150x 231= 47 69 Hz'

200

It is observed here that due to difference in droop characteristics of

governors,generator getsoverloadedwhile generator is underloaded.

It easily follows from above hat f both governorshavea droop of.4Vo, hey

will share he load as 200 MW and 400 MW respectively, .e. they are oaded

corresponding to their ratings. This indeed is desirable from operational

considerations.

Dynamic Response

To obtain the dynamicresponse iving the change n frequencyas function of

the time for a stepchange n load, we must obtain the Laplace nverseof Eq.

(8.14). The characteristic quationbeing of third order,dynamicresponse anr ' r | 1 - ! - - - I f - , - - - ^ ^ ^ t C : ^ - - - * ^ - : ^ ^ 1 ^ ^ ^ ^ t I ^ . - , ^ , , - + L ^ ^ L ^ - ^ ^ + ^ - - i ^ +i n

Onfy Dg ODIalneU luf A SPtrUfffU ll|'l l l l( 'l lua1'I Ua1DE. II(rwsYsIr LfIs r,I l<ll4ivLsllDrlv

equation can be approximated as first order by examining the relativemagnitudesof the ime constantsnvolved.Typical valuesof the time constants

of load frequencycontrol systemare rdlatedas

200

0.05 50

400(ii), we get

231 MW (loadon generator

- / A t r l t f / 1 - l ^ -JOy lvlw (IUau ull Btrrltrriltur

(i)

(ii)

r)

L)

T r r 4 T , < T o ,

Typically* t, = 0.4 sec, Tt = 0.5 sec and

Flg' 8.8 Firstorderapproximaterockdiagram f roadfrequency ontrot f an isolated rea

Irning Tro= T, =reduced to thlt of F'ig.

0:Iuld

K* \ =1), the block diagramof Fig. 8.6 is8.8, from which we can write

Dynamic esponse_ofhangen requencyor a stepchange n oad(APo= 0.01pu, 4s = 0.4 sec,| = 0.5 sLc, o.= 2b sec, " = 100,

R = 3 )

The plot of change n frequencyversus time for first order approximadongiven above

^rirst

Time (sec)------->

-1tI

Io

Firstorderapproximatiorl



AF(s)l*r(s):o =

Ar,)=ft{' -*,[-,,a[n#)]], g22)TakingR = 3, Kp,= llB = 100,e, = 20, Apo = 0.01pu

Af (t )= - 0.029 I - ,-t:tt ',

Aflrt"udystare - 0.029 Hz

-to,

.-.APo

(1 + Kp s R) + Zp.s s

- - "o{1:- =xaP,, l +^+ro '1L R 4 , J

the exact responseare shown in Fig. a.g. orderapproximation s obviously a poor approximation.

Gontrol Area Concept

So far we have considered the simplified case of a single turbo-generatorsupplying an isolated load. Considernow a practical systemwith e number ofgeneratingstationsaird oads. t is possibleto divide an extendedpower system(say, national grid) into subareasmay be, StateElectricity Boards) in whichthe generatorsare tightly coupled ogether so as to form a coherent group, i.e.all the generators respond in unison to changes n load o, ,p"rJ changersettings.Such a coherentarea s called a control area in which the frequencyis assumed o be the same hroughout n static as well as dynamic conditions.For purposesof developing a suitable control strategy, a control area can bereduced to a single speedgovernor, turbo-generatorand load system. All thecontrol strategiesdiscussedso far are, therefore, applibable o an independentcontrol area.

Proportional Plus fntegral Control

It is seen from the above discussion hat with the speedgoverning sysreminstalled on each machine, the steady oad frequency charartitirti" fi agivenspeedchangersettinghas considerable roop, e.g. for the systembeing used orthe illustration above, the steadystate- roop in fieo=ueney ill be 2.9 Hz [see

Eq. (8.23b)l from no load to tull load (l pu load). System frequencyspecificationsare rather stringentand, therefore,so much change n frequencycannot be tolerated. In fact, it is expected hat the steadychange n frequencywill be zero. While steadystate requencycan be broughtback o the scheduled

(8.23a)

(8.23b)

"Fora 250 MW machine uoted arlier,nertiaconstanr

, = 4 : . 2 * 5 = = 2 o s e c Bfo 0.01x0

Il = SkW-seclkVA

ffil Modernower ystem nalysI

vaiue by adjus'ring peed hangersetting, he systemcould undergo intolerable

dynamic frequency changeswith changes n load. It leads to the naturalsuggestion that the speed changer setting be adjusted automatically by

monitoring the frequency changes.For this purpose, a signal from Af is fed

througfan integrator to the s diagram

configurationshown n Fig. 8.10.The systemnow modifies to a proportional

plus integral controller, which, as s well known from control theory, gives zero

steadystateerror, i.e. Af lrt""d",,ut,= 0.

Integralcontroller

APp(s)

AP6(s)

l+t-r8-I t - +

t l

APe(s)

AF(s)

AutomaticGenerationand VoltageControlt-

in ihe above schemeACE being zero unciersteaciyconditions*, 4 logical

designcriterion is the minimization of II,CZ dr for a step disturbance. Thisintegral is indeed the time error of a synchronouselectric clock run from thepower supply. Infact, modern powersystems keep Eaekofintegra+e4tinae errsrall the time. A corrective action (manual adjustment apc, the speed changersetting) s taken by a large (preassigned)station n the areaas soon as the timeerror exceedsa prescribedvalue.

The dynamics of the proportional plus integral controller can be studiednumerically only, the systembeing of fourth order-the order of the systemhasincreasedby one with the addition of the integral loop. The dynamic responseof the proportional plus integral controller with Ki = 0.09 for a step loaddisturbanceof 0.01 pu obtained through digital computer are plotted in Fig.8.11. For the sakeof comparison he dynamic responsewithout integral control

action s also plotted on the same igure.



Frequencyensor

Fig. 8.10 Proportionallus ntegraload requency ontrol

The signal APr(s) generated y the ntegralcontrol must be of oppositesign

to /F(s) which accounts or negative sign in the block for integral controller.

Now

( l * f , rs ) ( l+4s)

RKo,s(l+ rs)(l+ 4s )

obviousry

+ {'s)(1 + 4sXl f zo's)R Ko ' KiR s)

Af l"t"^dytate ,so/F(s) : o

In contrast o Eq. (8.16) we find that the steadystatechange n frequency

has beenreduced o zero by the additio4 of the integral controller. This can be

arguedout physically as well. Af reachessteadystate (a constantvalue) only

rr.,lrsrr Ap^ - Ap- = .ons-fant Becarrs-e of fhe intes!'atins actiOn Of thew l M l u r c - Hr D - v v u u l q r ! .

controller, his is only possible f Af = 0.In central oad frequency control of a given control area, he change (error)

in frequency s known asArea Contol Error (ACE). The additionalsignal ed

back in the modified control schemepresented above is the integral of ACE.

Kn,AF(s1=

(r %"s).* +). Ko,

"+

-1+II

torx

Flg. 8.11 Dynamicesponse f load requencyontroller ith and withoutintegral ontrol ction APo = 0.01pu, 4s = 0.4 sec, Ir = 0.5sec, ps= 20 sec,Kp .= 100,B - B, Ki= 0.-09)

8.3 IOAD FREOUENCY CONTROL AND ECONOMICDESPATCH CONTROL

Load freouencv control with inteorel eonfrnl ler qnhierrAe ?a?^ craolrr ora+oI

_ _ J _ _ _ _ _ _ _ _ , . _ _ _ _ - - - _ O 'v u l v r v o t v s u J D l 4 l g

frequencyelTor and a fast dynamicresponse,but it exercises o control over therelative loadings of various generatingstations(i.e. economicdespatch) of thecontrol area.For example, f a suddensmall increase n load (say, 17o)occurs

'Sucha control is known as isochronous control, but it has its time (integral of

frequency) error though steady frequency error is zero.

(8.24)

(8.25)

1i..l1r_::ltrolarea, the road

-frequencyconrior ,changes

the speed changerDcrurgsor tne governors of all generating units of the area so that, together,theseunitsmatch he oad and the frequenry returns p the scheduledvalue (thisaction akesplace n a few seconds).However, in the,process f this change heIoadingsof u@units change in a manner independent ofeconomi@ In fact, someunits in the pro""r, may evenget overloaded. ome control over loadingof individual unitscafi beLxercisedby adjusting he gain factors (K,) includeJin the signal representing ntegral ofthe area cogtrol error as fed to individual unitr. However, this is notsatisfactory.

"fnceot

Automaticf

_T---

commandsignai generated'oy he centraieconomic despatch omputer.Figure8'12 gives he schematicdiagramof both thesecontrolsior two typi.ut units ofa control area.The signal to change he speedchan3ersetting s lonstructed inaccordancewith economicdespatch rror, [po (desired) pJactual)]. suitabrymodified by the signal representingntegral ncg at that nstantof time. Thesignal P6 (desired) s computed by the central economicdespatch omputer(CEDC) and is transmitted o the local econornic despatch ontroller (EDC)installed at eachstation.The system husoperateswith economicdesfatch erroronly 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 numberof load frequencycontrol areasnterconnectedby meansof tie lines. Without

lossof generality weshall considera two-areacaseconnectedby a single tie line as ilusnated inFig. 8.13.



lr. EDC - Economicdespatch controllerCEDC - Central economic despatchcomputer

Flg. 8-12 Control area load frequency and economicdespatch 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.

Speed

Fig.B.i3 Tw o nterconnectedontror reas singreie ine)

The control objective now is to regulate he frequencyof eachareaand tosrnnultaneouslyegulate he tie line poweras per

inter-areapowercontracts.Asin the caseof frequency,proportionalplus integral controllerwill be installedso as to give zero steady stateerror in tie line power flow as compared o thecontractedpower,

It is convenientlyhssumed hat eachcontrol area canberepresented y anequivalent urbine,generatorand governorsystem.Symbolsusedwith suffix Irefer to area7 and those with suffix 2 refer to area 2.

In an isolatedcontrol areacase he incrementalpower (apc _ apo) wasaccounted or by the rate of increaseof storedkinetic energyand ncrease narea oad caused y increase n fregueircy.since a tie line t *rport, power inor out of an area, this fact must be accounted or in the incremental powerbalanceequationof each area.

Power transportedout of area 1 is .eivenbv

Ptie,r = ''rrl''l sin({ - qX,,

where

q'q - powerangles f equivalent achines f the wo areas.

(8.26)

I

308 | ModefnPower SystemAnalysisI

For incrementalchangesn { and 6r, the ncre.mentalie line power can be

expressedsAP,i,, (pu)= Tp(Afi - 462)

where

T, ='Y:t'-Yf

cos f - E)- synchronizing oefficient

PrrXrz

Since ncrementalpoweranglesare ntegralsof incremental requencies,we

ca n write Eq. (8.27)as

AP,i,, = 2*.(l AfrdtI Urat)

where Afi nd Af,, arc incremental frequency changes of areas 1

respectively.

Similarly the incremental ie line power out of area2 i s given by

aPt;",z = 2ilzr([ yrat -[

ayrat)

(8.27)

(8.28)

and 2,

(8.2e)

Automatic enerationndVortageontrorFi l

I APti".r(s)

Fig. 8.14

The corresponding block diagram is shown in Fig. g.15.

+

APti",r(s)

AF1(s) -iE= --n7ri"l

Fig.8.15



where

rzrtYr:J

co s{ - E): [S]t i z: ar2rrz (s.30)L L

P r z x z rL "

\ P r r )

With reference o Eq. (8.12), he incremental ower balanceequation or

area 1 can be written as

APo, APor + *w)+ nrz|r* AP,,", tJr" or

It rnay be noted hat all quantit ies ther han iequencyar e in pe r unit in

Eq . 8 . 3 l ) .

Taking he -aplace ransf 'orm f Eq. (8.31)an d eorganizing, e ge t

AF(s) = IAP61G) APr, (s) - APt i " , ,1r ; ]" t$-

$.32)I + 4,, t , !

where as definedearlier seeEq. (8.13)]

Kp31 I/81

Tpil = LHr/BJ"

Comparedo Eq. (8.13)of the isolatedcontrol areacase, he only change s

the appearancel the signal APri"J s) as shown n Fig. 8.14.' - l ' ^ L i - - f h o T - ^ l - ^ a f * o n o f n r m ^ f E ^ / a t a \ t h a c i o n o l , 4 P / " \ i c n l r f o i n e r lI4ArrrS rr rw ls l / l4vv Ll4 l lDrurr r r ur LY . \v .L9) , l l rv or6rrs^

", t ie . I \ . r /

AS

AP,i.,1(s)ffroor(s)

- /4 (s)l

(8 .31)

(8 .33)

(8.34)

For the control area2, Ap6", r(s) is given by tEq. (g.Zg)l

apt i " ,z(s)=-:grrr ,

[AFr(s) 4F , (s) ] (g:35)

which is also ndicatedUy ,i. block diagramof Fig. 8.15. \Let us now turn our attention o ACE (areacontrol error; in the presenceof

a tie line. In the case of an isolatedcontrol area,ACE is the change n areafrequencywhich when used n integral control loop forced the steady statefrequency lror to zero. n order ha t the steadystate ie line power error in a

two-area ontrolbe madezeroanotherntegral ontrol oop one or eacharea)mustbe introduced o integrate he ncremental ie line powersignaland feedit back o th espeed hanger. hi s s aeeomplishedy a singlentegrat ing loekby redefningACE as a linearcombination f incrementalrequenry nd ie linepower.Thus, br control area

ACEI = APu" . r+ brAf ,

where the constantb, is called area requency bias.Equation (8.36) can be expressedn the Laplace transformas

ACEl(s) = APo., r(s)+ b1AF1g)

Similarly, for the control area 2, ACE2 is expressedas

ACEr(s) = APti".z(s)+ b2AF,(s)

Combining he basic block diagramsof the two control areas orrespondingto Fig. 8.6, with AP5rg) and Apr2(s) generatedby integralsof respectiveACEs (obtained hrough signals epresenting hanges n tie line power and ocalfrequencybias)and employing he block diagramsof Figs.g.t+ to g.15,weeasily obtain the compositeblock diagramof Fig. g.16.

(8.36)

(8.37)

(8.38)

WIU&| ModernPower svstem Analvsis

Let the stepchangesn loadsAPo, and APrrbe simultaneously pplied ncontrol areas 1 and 2, respectively.When steadyconditionsare reached, heoutput signalsof all integratingblocks will becomeconstantand in order forthis to be so, their input signalsmust become zero.We have, therefore, romFie .8 .16

APu",, + b Afr= O finputof ntegratinglockKtL)

-\ , r )

APti", + brAfr= o finpotof ntegrating lockK'z)

-\ r l

Af r - Afz =o finpurot ntegratinglock-'4'\- \ s )

FromEqs. 8.28)and 8.29)

A P n " , ,= - T r , - . I . = c o n s t a n tA P . i " , z , T z t ; a r 2

HenceEqs. 8.39) (8.41) resimultaneouslyatisfiednly or

(8.39a)

(8.3eb)

(8.40)

(8.41)

.Yo(U-o!togo y ,

E F8 6

:pEo 6=ac)- ( d

E 9g t u

Eoo

a\

.gq<.1

trJo

oiraI

il

S AN

oo-

* l io l . a



(8.42)

and

Thus, under steadycondition change n the tie line power and frequency ofeach area is zero. This has been achieved by integration of ACEs in thefeedback oops of eacharea.

Dynamic response s difficult to obtain by the transfer unction approach(asused n the single areacase)becauseof the complexity of blocks-and multi-input (APop APor) andmulti-output APri",1,Ap6",2, Af r Afr) situation.

more organizedand moreconvenientlycarried out analysis s through the statespaceapproach a tirnedomainapproach). ormulationof thestatespacemodelfor the two-areasystemwill be illustrated n Sec.8.5.

The resultsof the two-areasystem APri", change n tie line power and,Af,change n frequency)obtainedhroughdigital computerstudy areshown n theform of a dotted ine in Figs.8.18 and 8.19. The two areas re assumed o beidentical with systemparameters iven by

Trs= 0.4 sec, 7r = 0.5 sec, ?r, = 20 sec

K o r =1 0 0 , = 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 n this section to design an optimal loadfrequencycontroller for a two-a3ea ystem. n accordance ith modern controlterminology APcr arrdAP62 will be referred to as control inputs q andu2.lnthe conventionalapproach l anduzwere providedby the ntegral of ACEs. In

Q c le . >

( g ( )o oF AO E5 ** o

b 6t r ( ' )E gH'.s

* 35 u= oa ' 5' 8 9o o .* , nE O -o oo o

@

d

<;l!

oooG'

6

ooEoE

(D()(UCL.nooor\

ai

citlr

tt)

.9d

A P r i " , r =A P , : " , 2 = 0

A f i = A f z = 0

t

l +

5 l

- l d

uI |:-

v'itlr

itZ I rrrrodernower ystem nalysis-

moderncontrol theoryapproachur and u2 wtll be createdby a linear

combinationf all the system tatesfull stateeedback).or ormt'latinghestatevariable nodel or this purpose he conventionaleedbackoopsare

resented v a se block as shown n

Fig. 8.17.Statevariables re definedas he outputsof all blocksbavingeither

an ntegratoror ar irne, onstanf.. e immediatelynotice hat the systern asnine

statevariables.

-1-+-r+--i, -f-.'f--\

Optimalcase (ful lstate eedback)

'With integralcontrolaction

AutomaticGeneration nd Voltage ontrot M&f*

ComparingFigs.8.16 and8.17,

xt = Aft

.r2 AP,;1

xq = Af.

x5 = AP52

XS= JACE t

t, = JACE, dt

Itt-

- l

t'-2

-3

+I

II

o

o

X

(

t t 1 = APg ,

w 1= AP"

For block 1

x1 + T.rr i , =L P

. 1h l

-4 l

' p s l

For block 2

x .2+ T i l i z = x t

u2 = /)Pa

w2 = APp,

K^ t (x z - h- w)

,Kp r t

, - Kp r t - , Kp r t* f x z - ; - x t - ; - w t ( 8 . 4 3 )

t psl t ptl t ptl



change n ie in epower ue o step oa d 0.01 u)changen area1

,I1 8

1

8. 'Fig.

AL

I

N

Iox

IL

r;+-.1 I-21.--+-_'--';-;;7-1=a.-1-1=--1

/' 8' - - - ' 12 14 16 18 20

/ Time sec)-----

or *z=+-r**n

For block 3

t r+ { , s r i : = -L r r+ r ,' R , r r

or * t=^h

r,-t* ,**, ,

For block 4

X n *+

or i q=

For block 5

x s t

o r i s =

For block 6

x s *

(8.445

(8.45)

(8.46)

(8.47)

with integralcontrolaction

Optimal ase (ful lstate eedback)

Fig. 8.19 Change n frequency f area 1 due to step oa d (0.01pu )

change n a.rea

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 U V *1. d t '

Torz*+= Krrz(xs + ar2x7 - wz)

I Knrz at?K or 2 Ko*2' \ A ' 1 - - . { <

- T- - y ' - a - _ - W ^

Tprz-

Tps2''

Tps2'

Tpszz

7,2i5 - x6

l 1Y I - V

4 < t 4 l

Ttzr

T,zu

. lI ,szx6

-; x4 + u2I\2

or i o= # *o - * *u'2 t sg2 t sg2

(8.48)

'3i4',"1 ModernPower SystemAnatysisT

For block 7

i t = 2 i T t z x t - 2 i T r 2 x aFor block 8

is = brx, + x. i

For block 9

(8.4e)

(8.5O-)

(8 .s )

(8.s2)

i9 = b2xa- anx t

The nine equations 8.43) o (8.51)canbe organized n the ollowing vector

matrix form

where

x _ l x r

u = f u t

w = l w t

while the matrices

* = A x + B u + F w

x2 ... xg)r= statevector

u2fT= control vector

w2fT= clisturbance ector

A, B and F are defined below:

"

'-- co","".t""

constructedas under from the statevariables x, and -rnonly.

u t = - K i r x s = - K i r I e C n , A r

uz=- K i {s=- K iz la.Cerar

ln the optimal control scheme he control inputs u, anduz aregeneratedbymeansof feedbacks rom all the nine stateswith feedbackconstants o bedeterminedn accordancewith an optimality criterion.

Examinationof Eq. (8.52) everals hat our model s not n the standardormemployed n optimal control theory.The standard orm is

i = A x + B u

which does not contain the disturbance term Fw present n Eq. (g.52).Furthermore,a constantdisturbance ector p would drive some

of the systemstates nd he control vectorz to constantsteadyvalues;while the cost unctionemployed n optimal control requires hat the systemstateand control vectorshave zero steady statevalues or the cost function to havea minimum.



A _

2 3 4 5Y' tPsl

0 0 0Tprt

- 1 1 o oTt Ttr

o - 1 o oTrst

0 0 - 1 K p ' zTprz Tprz

o o i o --1-Ttz

I

I

Tpst,

0

1

Rr4er

0

0

0

2 rrz

bL

0

7 8 9

- b L o oTprt

0 0 0

0 0 0

atzKprz0 0

Tprz

0 0 0

6

7

8

9

o o - 1 0RzTrsz

0 0 -2i l r2 0

0 0 0 0

0 0 b 2 0

[ o oI o o o

I TBr = |

-ss1

l o o o o +I a c o )L 'O -

1

7,,I

TreZ

0

0

0

0

0

1-a tz

0 0

0 0

0 0

0 0

-Kp r t

0Tprt

0 0 . lII

0 0 lI

J

,;T

(8.s7b)

For a constantdisturbance ector w, the steadystate s reachedwhen

* = 0

in Eq . 8.52);which hen ives

0=A.r r "+ Burr+ Fw (8.s3)Defining and z as hesumof transient ndsteady tateerms,we canwrite

,x = x' * Ir " (8.54)

n = ut * z' , (8.55)

Substituting and z from Eqs. (8.54) and (8.55) n Eq. (8.52),we havei' = A (r/ + x"r) + B(at + usr)+ Fw

By virtue of relationship 8.53),we get

*' = Ax t + Bu t (g.56)

This representssystemmodel in terms of excursionof state and conholvectors iom their respective teadystatevalues.

For full state feedback, he control vector z is constructedby a linearcombinationof all states..e.

u = - Kx ( 8 . 5 7 a )

whereK is the feedback matrix.

Now

t t t+ I t r r=- l ( ( r / + r r " )For a stable systemboth r/ and ut go to zero, therefore

ur, = _ Kx*

Hence

tt/= - Ikl


Examinationof Fig. 8.17 easily reveals he steadystatevalues of stateandcontrol variables

or constantvaluesof disturbancenputs w, andwr.TheseareI l r r =X 4" r= /7 r " = 0

AutomaticGeneration

b ? o0 0

0 0

0 0

0 0

0 0

0 0 0 0 0 0

0 0 0 0 0 0

4 0 0 - a n b z 0 00 0 0 0 0 0

0 0 0 0 0 0

4 0 0

0 0 0

0 0 0-arzbz 0 0

0 0 0

0 0 0

Q+a?)o o0 1 0

0 0 1

0

0

0

0

0

0

u l r , = wl

r5r r= x6rr= lv 2

uzr, = wz

(8.s8)

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:

utrr= kttxtr, + .. . + ft t8r8", kt*sr, = wl

r,t2ss k2txlr, + ... + kzgxgr., kz*gr, = wz (8.se)

= symmetric matrix

R - kI = symmetricmatrix



(8.63)

for whichthe system emainsstable.the system ynamics ith foedbacks

The feedback natrix K in Eq. (8.57b) s to be determinedso that a certainperformance index (PI) is minimized in transferring the system from anarbitrary nitial state x' (0) to origin in infinitie tirne (i.e. x' (- ) = 0) . AconvenientPI has he quadratic orm

' Pr=;l l '.'' Qx ' u' r Ru'dt

The manices Q arrdR are defined for the problem in hand through thefol lowing design onsiclerat ions:

(i) Excursionsof ACEs about the steadyvalues(r,t + brx\; - arrxt,+bzx,q)are minimized.The steadyvalues of ACEs are of coursezero.

(ii) Excursions f JnCg dr about he steady alues(xts,xte) re nrinimized.

The steacly aluesof JeCg dt are,of course, onstants.

(i i i ) Excursions 1'thecontt 'olvector(ut1,ut2) about th e steadyvaluear erninirnized. he steady alueof the controlvector s, of course, constant.

' This nrinimizat ions intendedo indirect lyimit the controleffbrtwithinthe physicalcapability of components. or example, he steamvalvecatmotbe openedmore than a certainvalue without causing he boilerpresisureo drop severely.

With the above easoning,we can write the PI as

K = R-rBrS

The acceptablesolutionof K is thatSubstitutingEq. (8.57b) n Eq. (8.56),definedbv

i' = (A - BIgx, (g.64)Fo l stabilityal l thc cigenvalues f the matrix (A - Bn shouldhavenegative

real parts.

For illustrationwe consider wo identicalcontrolareaswith the followingsyste|ll arameters:

4r* = 0'4 scc; T'r= 0.5 sec; 7'r*= 20 sec

/l = 3: (n * = l/ lJ = 10 0

b = O.425;Ki = 0.09;up = I; 2i ln = 0.05

f, [ 0.52tt6 l. l4l9 0.683 0.00460.021| -0.01000.74370.gggg0.00001^ =L-o.t l046-0.o2t l-0.01000.5286 t.t4rg 0.68130.74370.0000.gggsl

(8.60)

pr= * fU-+ + h,.r,,)2 (- tt,2xt, brxta)z (.r,? ,,])2 J t t '

+ kfu'l u,|11t

From the PI of Eq . (8.51),Q md R can be recognizedas

(8 .61)

'*ReferNagrath and Gopal [5].

iiii'f:l Modernowerystemn4gs

As the control areasextend over vast geographical egions, here are twoways of obtaining ull state nformation in eacharea or control purposes.

(i) Transport the state nformation of the distant area over communicationchannels. his s, of course,expensive.

8.6 AUTOMATIC VOLTAGE CONTROL

c= vR.f vr'_ -

The error initiates the corrective action of adjusting the alternator excitation.Error wave form is suppressed arrier modulated, tt" carrier frequency beingthe system requencyof 50 Hz.

Change in voltagecausedby load

Load change

Fig. 8.21 Brock iagram f arternatorortage egurator

tG1+Iers

skrt



Figure 8.20 gives the schematicdiagram of an automatic voltage regulator ofa generator. t basicallyconsistsof a main exciterwhich excites the alternatorfield to control the outputvoltage.The exciter ield is automaticallycontrolledthrough error e = vr"r - vr, suitably amplified through voltage and poweramplifiers. It is a type-0 systemwhich requiresa constanterror e for aspecifiedvoltage at generator erminals.The block diagramof the system s given in

Fig. 8.20 Schematic iagramof alternator oltage egulator cheme

Fig. 8.21.The functionof important components nd heir transfer unctions sgivenbelow:

Potential transformer: It gives a sample of terminal voltage v..Dffirencing device; It gives the actuating error

cheme

Error amplifier: It demodulatesand amplifies the error signal. Its gain is Kr.scR power amplffier and exciter fierd: It provides the n"."rriry poweramplification to the signal for controlling thl exciter n"ro.- arr*;"g ,rr"amplifier time constant o be small enoughio be neglected, he ovelail fansferfunction of these wo is

K,

l* T"rs

where T"y s the exciter field time constant.

Alternator; Its field is excited by the main exciter voltage vu. Under no roadit producesa voltageproportional o field current.The no load transfer unctionis

Ks

7 * T * s

where

T*= generator ield time constant.

The load causesa voltage drop which is a complex function of direct andquadrature xis currents.The effect s only schematically eBresentedv hlockG.. The exact oad model of the alternators beyond

,t" ,iop" ;rhtJ;;:stabitizing transformer: T4*d

-lqare large enough time constants o impair

the system'sdynamic response. tjs weil known that the dynami. r"rpoor" ofa control systemcan be improved by the internal derivative feedback oop. Thederivative feedback in this system is provided by means of a stabi zingtransformer excited by the exciter output voltage vE. The output of the

LoAD

Potential

I

320' lModernPowerSystemAnalysis

I

stabiliz,ingransformers fcclncgativclyat he nput erminals f thc SCRpower

amplifier. The transferunction of the stabilizing ransfo"mer s derivedbelow.

Since he secondarys connectedat the input ternfnals of an amplifier, it can

be assumed o draw zero current. Now

d tvr = Rr ., + LrJilL' d t

' r r= MYdt

Taking the Laplace ransform,we get

%, s)_

VuG) R, * s,Lt

sK",

sMlRt

l * I r s

sM

1+ , sAccurate state rariablemodelsof loadedalternatoraroundan operatingpoint

n literatureusing which optimal voltageregulationschemes an

Automatic enerationndVoltage ontrol Jffif-----_-----lE

The banded aluesmposedhy the imitersareselectedo resffict hegeneration

rate by l}Vo per minute.I

I.g 9t",

u' t +/_+(

-t*9r"'--l

Fig.8.22 Governor odelwithGR C

The GRCs esult n larger deviations n ACEs as he rateat which generation

can cha-ngen the area is constrainedby the limits imposed.Therefore, the

Al-



are available

be devised.This is, of course,beyond he scopeof this book.

8.7 LOAD FREOUENCY CONTROL WITH GENERATION

RATE CONSTRAINTS (GRCs)

The <-raclrcquency ontrolproblcmdiscussedo ar doesno t consiclerhe effect

of the restrictionson the rate of changeof power generation. n power systems

havingsteam lants, owergenerat ionanchange nly at a specif ied aximum

rate. The generation ate(fiom

saf'ety onsiderations 1 the equipment)or

reheatunits is quit low. Most of the reheatunits have a generatiol rate around

3%olmin.Some have a generation ate between 5 to 7jo/o/min. f these

constraints rc not consirlcrcd, ystert ts l ikely to c:ha.scargc tt tottrclt trry

disturbances, hrs results n unduewear and tear of the controller.Several

methocls av ebeenproposeclo considerhe effectof GRCs or the clesign f

automatic eneration ontrollers.When GRC s considered,he systeln ynamic

rnodelbecomes on-l inear nd inearcontrol echniquesannotbe applied or

th e optimizat ion f the control ler ett ing.

If the generation atesdenotedby P", are ncluded n the statevec:tor,he

systerm rder will be altered. nsteadof augntenting hem, while solving the

stareequations,t may be verified at each step if the GRCs are viclated.

Another way of consicieringGRCs for both areas s to arjri iinriiers io ihe

governors 15 , 17] as shown n Fig. 8.22, r.e., he maximum rate of valveopeningor closingspeeds restricted y the imiters.Here2" , tr,r,, S he power

rate limit irnposedby valve or gate control. In this model

lAYEl . - - u ,nr (8 .6s)

duration for which the power needs o be imported increases onsiderablyas

cornpared o the case wheregeneration ate s not constrained.With GRCs, Rshouldbe selectedwith care so as o give the best dynamicresponse.n hydro-

thennalsystem,he generation ate n the hydro areanorrnally emainsbelowthe safe imit and therefore GRCs for all the hydro plantscan be.ignored.

8.8 SPEED GOVERNOR DEAD-BAND AND ITS EFFECT

ON AG C

The eff'ectof the speedgovernordead-bands that for a given position of thegovernor control valves, an increase/decreasen speedcan occur before thepositionof the valvechanges. he governor ead-band anmaterially ffect he

system response. n AGC studies, the dead-bandeff'ect ndeed can besignificant, ince elativcly small signalsareunder considerations.

TlLe peed overnor haracterrist ic.hough on-l irrear,asbeen pproxinraaed

by linear characteristics n earlier analysis.Further, there is another non-iinearity introduced y the dead-bandn the governoroperation.Mechanical

f'rictionand backlashand also valve overlaps n hydraulic relayscause hegovernordead-band. ur to this, though he nput signal ncreases,he speedgovernormay not irnmediately reactuntil the nput reaches particularvalue.Similar a.ctionakesplacewhen he nput signaldecreases.hus he governor

dead-band s defined as the total rnagnitudeof sustained peed hangewithin

which there s no change n valve position.The limiting value of dead-band sspecifiedas 0.06Vo. t was shown by Concordia et. al [18] that one of theeffects of governordead-band s to increase he apparentsteady-state peedregulation R.

lFFf Modrrn o*., svrt.t Analuri,

The effect of the dead-bandmay be included in the speedgovernor control

loop block diagram s shown n Fig. 8.23.Consideringhe worst case orthe

dead-band, i.e., the system starts responding after the whole dead-band straversed) nd examining he dead-band lock in Fig. 8.23,thefollowing setof

ly define the behaviourolthe dead.band [9]-

Speed governor

Dead-band

Flg.8.23 Dead-bandn speed-governorontroloo p

u(r+1)= 7(r)1: _ x, 1 dead-band

Discrete-Time Control Model

The continuous-time ynamic systems describedby a set of linear differentialequations

x = A x + B u + f p (8.67)where f u, P are state,conhol and disturbancevectors respectively and A,Band f are constantmatricesassociated ith the above vectors.

The discrete-time ehaviourof thecontinuous-time ystems modelledby thesystemof first order inear differenceequations:

x ( k + 1 ) =Qx(k )+ V u ( k ) + p &) (8.68)

where x(k), u(k) andp(k) are the state,control and disturbancevectors and arespecif ied t t= kr , ft = 0, 1,2,. . . etc.an d ri s the sampling eriod.6, t l ,nd7 Te the state, control and disturbance transition matrices and they areevaluatedusing the following relations.



"(r+1)-

"(r+l)_ dead-band; if x('+l) - ,(r) I g

-

"(r+1).tf Xr*l _ xt < 0

(r is the step in the computation)

Reference 20] considers he effect of governordead-bandnonlinearity by using

the describing unction approach 11] and ncluding the inearisedequationsn

the statespacemodel.

The presence f governordead-bandmakes he dynamic esponse scillatory.

It has been seen [9J that the governordead-band oes not intluence the

selectionof integral controller gain settings n the presenceof GRCs. In the

presence f GRC and deadband even or small oad perturbation,he system

becomes ighly non-linearand hence he optimizationproblembecomes ather

complex.

8.9 DIGITAL LF CONTROLLERS

In recentyears, ncreasinglymore attention s being paid to the questionof

digital implementationof the automaticgeneration ontrol algorithrns.This is

mainly due to the facts that digital control turns out to be more accurate andr c l iqh lc nnrnnae f in q ize less cens i f i ve to nn ise end d r i f t nnd more f lex ih le T tr v ^ r E v ^ v t

may also be implemented n a time shared ashion by using the computersystemsn load despatch entre, f so desired.The ACE, a signal which is used

for AGC is available n the discrete orm, i.e., here occurssampling operation; between he systemand the controller. Unlike the continuous-timesystem, he

control vector n the discretemode is constrained o remain constantbetween

(8.66) d= eAT

{ = ( { r _ l n - t r

j = ( e A r - D A - t f

where A, B and, are the constantmatricesassociatedwith r, ,,LO p vectorsin the conespondingcontinuous-time ynamic system.The matrix y'r can beevaluatedusingvariouswell-documentedapproachesike Sylvestor'sexpansiontheorem,seriesexpansion echniqueetc. The optimal digital load frequency

controllerdesignproblem s discussedn detail n Ref [7].

8.10 DECENTRALIZED CONTROL

In view of the large size of a modern power system, t is virtually impossibleto implementeither he classicalor the modernLFC algorithm n a centralizedmanner. n Fig. 8.24, a decentralizedontrolscheme s shown.x, is used o findout the vector u, while x, alone s employed o find out u". Thus.

Flg. 8.24 Decentralizedontrol

i,i2[,:*.1 Modern ower ystem nalysis -4

x - (x 1 x2)'

u t = - k t x t

u 2 - - k z x z

been shown possible using the m odal control principle. Decentralized or

hierarchical mplementationof the optimal LFC algorithmsseemso have been

studiedmore widely for the stochastic ase sincethe real oad disturbances re

truely stochastic.A simple approach s discussed n Ref. [7].It may by noted hat other techniques f model simplification are available

in the literatureon alternative ools to decentralizedcontrol.These nclude the

method of "aggregation", "singular perturbation", "moment matching" and

other techniques 9] for finding lower order models of a given large scale

system.

PROBE/IS

Automatic enerationndvoltageControl ff i

In,n,, tf^461dv

aF(s)1' af (t)dr liq, * '4F(s): hm/F(")]

L JO ,t JO s- 0 S s+ 0

8.4 For the two area oad frequencycontrol of Fie. 8.16 assume hat nte

controllerblocks are eplaced y gainblocks, .e. ACEI and ACE are fed

to the respective speedchangers hrough gains - K, and- Ko. Derive an

expression or the steadyvaluesof change n frequency and tie line power

for simultaneouslyappliedunit step oad disturbance nputs in the two

areas.

8.5 For the two area oad frequencycontrol employing integral of area control

error in eacharea (Fig. 8.16), obtain an expression or AP6"$) for unit

step disturbance n one of the areas.Assume both areas o be identical.

Comment upon the stability of the system for parameter values given

below:

4e = 0'4 sec; Z, = 0'5 sec;

K p r =1 0 0 ; = 3 ; K i = l ; b

Zp. = 20 sec

= 0.425



Two generatorsated 200 MW and 400 MW are operating n parallel.

The droop characteristics f their governors are47o and5Vo espectively

from no oad o full load.The speed hangersare so set hat he generators

operateat 50 Hz sharing he full load of 600 MW in the ratio of their

ratings. f the oad educeso 400 MW, how will it be shared mong he

generators nd whatwill the s)/stem requencybe? Assume ree governor

operatlon.

The speed hangers f thegovernorsare resetso that he oad of 400 MW

is sharedamong he generators t 50 Hz in the ratioof their ratings. What

are he no load requenciesf the generators?

Consider he block diagrammodel of lcad frequencycontrolgiven 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 parameters iveu below. Given AP, - 0.01 pu

T"q= 0.4 + 0.5 = 0.9 sec; 70 , = 20 sec

K r r K , = 1 ; p r = 1 0 0 ; = 3

Coinparewith the exact esponseiven n Fig. 8.9.

For the load frequency control with proportionalplus integral controllero c o l r n ' r n . i - T i i c e 1 n n h f a i n e n A s n r A c c i n n f n r t h a c f e n r l r r c f r f p e r r n r i nc l J o r l v Y Y l l l l ( L L 6 . v . r v t v u L a r r r

cycles, .". f'41t)d r; for a urrit stepAPr. What s thecorrespondingimet ^ "

, 1 ,

l i r n l *m

error n seconds with respecto 50 Hz).lCommenton the dependence f

error in cycles upon the integral controller gain K,.

a r 2 = I ; 2 t T r , = 0 . 0 5

lHint: Apply Routh's stability criterion

the system.l

to the characteristicequationof

8 . 1

8 .2

REFERECES

Books

l . Elgcrd, O.1., Elccu'ic Energv.Sv,s/clrT'lrcorv: An l t t tnxlut 'l ion. 2nd cdn. McCraw-

Hil l , New York, 1982.

2. Weedy, B.M. and B.J. Cory Electric Pow'er Systems,4th edn, Wiley, New York,

I 998 .

Cohn, N., Control of Generation and Power Flou, on Interconnected Systents,

Wi ley, New York, i971.

Wood, A.J., and B.F. Woolenberg, Power Generation, Operation an d Control,2nd

edn Wi ley, New York, 1996.

Nagarth, I.J. and M. Gopal, Control Systems Engineering, 3rd edn. New Delhi,

200 .

Handschin, E. (Ed.), Real Time Control of Electric Power Systems, Elsevier, New

York 1972.

Mahalanabis, A.K., D.P. Kothari and S.I Ahson, Computer Aided Power Systent

Analysis and Control, Tata McGraw-Hill, New Delhi, 1988.

Kirclrrnayer,L.K., Economic Control of lnterconnected Systems,Wiley, New York,

t959.

Jamshidi, M., Inrge Scale System.s:Modelling and Control, North Holland, N.Y.,

1 9 8 3 .

a1

4.

5 .

6.

7 .

8 .

9.

8 . 3

10. Singh, M.G. and A. Titli, SystemsDecomposition,Optimizationand ControlPergamonPress,Oxford, 197g.

I I' Sil jak,D'D., Non-LinearSystems: he Parametcr Analysis ntl Design, Wiley,N.Y. 1969.

.t,apers

12. Elgerd,o.I. and c.E..Fosha, The MegawattFrequency ontrol problem: A NewApproachvia optimal control Theory", IEEE Trans.,April 1970,No. 4, pAS g9:556.

13' Bhatti, T'S., C.S Indulkar and D.P. Kothari, "Parameteroptimization of powerSystems or Stochastic oad Demands"Proc. IFAC. Bangalore,December19g6.

l4' Kothari,M'L., P.S.Satsangi nd J. Nanda, sampled-Data utomaticGenerationControl of Interconnected eheatThermal SystemsConsideringGenerationRateConstraints",EEE Trans.,May 19g1,pAS_100;2334.

15' Nanda,J', M.L. Kothari and P.S.Satsangi, Automatic GenerationControl of anInterconnectedHydro-thermal ystem n continuous and DiscreteModes consid-eringGeneration ate onstraints'EE proc., prD, No. l, January 9g3,130 17 .

16' IEEE committee Report, DynamicModels or SteamandHydro-turbinesn power

9.1 INTRODUCTION



systemstudies" IEEE Trans.,Nov/Dec. rg73, pAS-92, 1904.l7' Hiyama,T', "Optimizationof Discrete-type oad FrequencyRegulatorsConsider-

ing Generation-Rateonstraints"proc.lE4 Nov. g2, r2g, pt c, 2g5.I8 . concordia, ., L.K. Kirchmayer nd E.A. Szyonanski,.Effect

of speedGovernorDead-bandon Tie Line Power and FrequencyControl performance,,AIEE Trans.Aug. 1957, 6,429.

19' Nanda,J', M.L. Kothari and P.S.Satsangi,Automatic Controlof ReheatThermalSystemConsideringGenerationRateConstraintand CovernorDead-band,,. .I.E.(India), un e 1983,63,245.

20 . Tripathy,s.9., G.s. Hope an d o.p. Marik, , ,optimisatiorrof Load-frcqucncyC<lntrolParametersor Powersystemswith ReheatSteamTurbinesand GovernorDead-bandNonlinearity", proc. IEE, January gg2, rzg, pt c, No. r, r0.

21 . Kothari ,M.L., J. Nanda, .p . Kothari nd D. Das,.,Discrete-modeGC of a.two_area Reheat Thermal system with New Area control Error,,, IEEE Trans. onPower System,Vol. 4, May 19g9,730

22' Daq D. J. Nanda, M.L. Kothari and D.p. Kothari, ,.AGCof a Hydro_Thermal

systemwith New ACE considering RC", Int. J. EMps,1g, No. 5, rggo, 46r.23' Das, D', M.L' Kothari, D.P. Kothari and J. Nanda, "Variable StructureControl

strategy o AGC of an Intcrconncctcd cheatThermalsystcm,,,prctc.IEE, r3g,pt D, 1991 , 79.

Jalleli, Van Slycik et. al.. "lJndersfandingAutonnaticGenerationControl,,, EEETrans.on P.S.,Vo l 07, 3 Aug. 92, 1106_1122.Kothari,M.L., J. Nanda,D.p. Kothari and D. Das, ,,Discrete

Mode AGC of a twoArea ReheatThermalSystemwith a NACE consideringGRC,,,J.LE. (rndia), vol.72, Feb. 1992,pp Zg7-303.

Bakken, B.H. and e.s. Grande,"AGC n a Deregulatedpower system,,, EEETrans.on PowerSystems, 3, 4, Nov. 199g,pp . 1401_1406.

24 .

25.

26 .

So far we have dealt with the steadystatebehaviourof power systemundernormal operating conditions and its dynamic behaviour under small scaleperturbations.This chapter s devoted o abnormalsystem behaviour underconditionsof symmetricalshortcircuit (symmetrical hree-phase.fault*).Suchconditions are caused n the systemaccidentallythrough insulation failure ofequipment or flashover of lines initiated by a lightning stroke or throughaccidental aulty operation.The systemmustbe protected gainst low of heavyshortcircuit currents which can cause effnanent amage o major equipment)

by disconnecting he faulty part of the systemby means of circuit breakersoperated by protective relaying. For proper choice of circuit breakers andprotectiverelaying, we must estimate he magnitudeof currents that would flowunder short circuit conditions-this is the scopeof fault analysis(study).

The majority of system aults are not three-phaseaults but faults nvolvingone ine to groundor occasionally wo ines o ground.Theseare unsymmetricalfaults requiring special ools like symmetricalcomponentsand form the subjectof study of the next two chapters.Though the symmetrical faults are rare, thesymmetrical fault analysis must be carried out, as this type of fault generallyleads to most severe fault current flow against which the system must beprotected.Symmetrical ault analysis s, of course,simpler to carry out.

A power network comprisessynchronousgenerators, fansfonners, ines andloads. Though the operating conditions at the time of fault are important, theloads can be neglectedduring fault, as voltagesdip very low so that currentsdrawn by loads can be neglected n comparison o fault currents.

*Symmetrical fault

impedance.

may be a solid three-phase short circuit or may involve are

325 | ModernPowerSystemAnalysist

The synchronousgeneratorduring short circuit has a characteristic ime-

varyingbehaviour. n the eventof a short circuit, the flux per pole undergoes

dynamic changewith associatedransientsn damperand ield windings.Thereactanceof the circuit model of the machine changes n the first few cycles

from a low subtransient eaetanec o 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 reactance alue is used or the

circuit model of synchronousgeneratorsor short circuit analysis.

In a m odern large interconnectedpower system, heavy currents flowing

during a fault must be interruptedmuch before the steady stateconditionsare

established. urthermore, rom the considerations f mechanical orces hat act

on circuit breakercomponents, he maximum current that a breakerhas o carry

momentarilymust also be determined. or selectinga circuit breakerwe must,

therefore, determine the initial current that flows on occulTenceof a short

circuit and also the current in the transient hat flows at the time of circuit

intemrption.

9.2 TRANSIENT ON A TRANSMISSION LINE

tffiffiI

42V= --*sin (cr,rf a_ A

l z l

ir = transientcurrent [it is such that t(0) = t(0) + L(0) = 0 being aninductivecircuit; t decayscorrespondingiohe tim6 constantRl.

= - i,(6)e-$tL)t

= 9Y s in d- a )g - . (R tDttz l

Thus short circuit current s given by

z =(Rz Jr\tt"(t: tan-l+)

Synrnretricalhort

circuit current

DC otT- setcurnent

A plot of i* i, and'i = i, + i, is shown n Fig. 9.2.rnpower system errninology,

(e.1)



Let us consider the short circuit transient on a transmission ine. Certain

simplifying assumptions re madeat this stage.

(i ) The l ine s led I'rorn constant oltagc oLrrcctltecasewhcn the ine is

fed from a realist ic ynchronons a.chrne i l l t re reated n Sec.9.3).

(i i ) Short circuit takesplacewhen he l ine is unloaded th e caseof short

circuit on a loaded ine will be treated ater n this chapter).

(iii) Line capacitances negligible nd he ine can be representedy a umped

RZ series ircuit.

, , F . Lr+V\'\-'

. lv = JI vsin (o,t *) rV)

Ii _

Fig.9. 1

With the above ssumptionshe ine canbe representecly the circuit noclel

of Fig. 9.1 The short circuit s assumedo takeplaceat t = 0. The parameter

<rcontrols he nstanton the voltagewavewhen shortcircuit occLrrs.t is known

from circuit theory hat he currentafter shortcircuit s composed f two parts,

1. t r .

t - - I " + I . t

whcre

i, = steadystatecurrent

the sinusoidalsteadystatecurrent is called the symmetricalshort circuitcurrent and the unidirectional transient component s called the DC off-setcurrent, which causeshe otal shortcircurit urrent o be unsymmetrical ill thetransient ecays.

It easily follows fiom F'ig.

currcnt ,,,,,,orrespondso theshort ime is neglected,

9.2 that the maximum momenro) short circuit

f irstpenk.f the lecayf trnnsienturrentn his

(e.2)

(e.3)

- Jrv'sin(d - c)*

E'

lzl tzlSince ransmissionine resistances small.0 - 9C,.

. Jiv JTvIm *=rz r

cosa+rz l

This has the maximum possiblevalue for o. = 0, i.e. short circuit occurringwhen the voltagewave s going throughzero.Thus

i,n,nlrnu* ossible)'# e.4)

= twice the maxirnumof symmetrical hort circuit current

(doubling effect)

For the selectionof circuit breakers.momentary hortcircuit current s takencorrespondingo its maxirnumpossiblevalue a sat'e hoice).

.

ffiffif Modern ower ystem nalysis

T h e n ev f n t r e c f i nn i c t u r h q f i c f h c r . r r r r c n f f n h c i n f e r r r r n f p r l ? t A a h q o h a a ne l / l vs

pointed out earlier, modern day circuit breakers are designed to intemrpt the

cunent n the first few cycles (five cycles or less).With reference o Fig, 9.2

it means hat when he current s intemrpted, he 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 arge size), the symmetrical short

circuit current alone s calculated.This figure is then ncreased y an empirical

rnultiplying factor to account or the DC off-set current. Details are given in

Sec. .5.

b,a#&

reactancewhen combined with the leakage eactanceXi of the machine s calledsynchronous eactanceX4 (direct axis synchronous eactancen the case of

salientpole machines).Armature resistance eing small canbe neglected.Thene s snown n rrg.

on per phasebasis.

(a) Steady tateshortcircuitmodelof asynchronous achine

(b)Approximateircuitmodel uringsubtransienteriod f short ircuit

X1



Fig.9. 2 Waveformf a short ircuiturrentn a transmissionin e

9.3 SHORT CTRCUTTOF A SYNCHRONOUS MACHTNE (ON

NO LOAD)

Under steady state short circuit conditions, he armature eaction of asynchronouseneratorroduces demagnetizinglux. In terms f a circuit his

(c)Approximateircuit odeluringtransienteriod fshort ircuit

Fig.9. 3

Consider now the sudden short circuit (three-phase)of a synchronousgenerator initially operating under open circuit conditions. The machineundergoesa transient n all the three phase inally ending up in steadystateconditions describedabove.The circuit breakermust, of course, ntemrpt thecurrent much before steady conditions are reached. Immediately upon shortcircuit, the DC off-set currentsappear n all the three phases,each with adifferent magnitude since the point on the voltage wave at which short circuitoccurs s different for eachphase.These DC off-set currentsare accounted orseparately on an empirical basis and, therefore, for short circuit studies, weneed to concentrate our attention on syimmetrical (sinusoidal) short circuitcurrent only.Immediately in the event of a short circuit, the symmetrical.short

circuit current s limited only by the eakage eaitanceof the machine.Since heair gap lux cannotchange nstantaneouslytheoremof constantlux linkages),

to counter the demagnetizationof the armature short circuit current, currentsappear n the field winding as well as n the damperwinding in a direction tohelp the main flux. These currents decay n accordancewith the winding timeconstants.The time constant of the damper winding which has low leakage

inductance s much ess han that of the field winding, which has high leakage

MocjernPowerSysiemnnaiysis

inductance. hus during the nitial partof the short circuit, the damperand field

windings ave ransfurnrerurrents nduced n themso hat n thecircr,ritmodelthcir reactances--X,of field winding and Xa* of damperwinding-appear in

parallelxwith Xo as shorvn n Fig. 9.3b. As the danqpqlruadlag cullqqls 4!qfirst to die out, Xr* effectively becomesopen circuited and at a later stage X1

becomes pencircuited.The rnachine eactancehuschaugesrom the parallel

combinationof Xo, Xy andXu. during the initial period of the short circuit to

X,,and Xrinparal lel (Fig.9.3c) n the rniddleper: iod f the shortcircuit , and

finally to X,, n steady tate Fig.9.3a).The reactance resentedy the machine

in the initial period of the short circuit, .e.

X. -r - 1_ - : X'jL( 1 1 x , , + U X J + l l x d , , ) "

(e.5)

is called the subtrunsient eoctutxc:e>f he nrachine.While the reactance

effective after the darnperwinding currents have died out, i.e.

X' , t= X, + (X, , l l X, ) (e.6)

is called he transient eactance f the machirre.Of course, he eactanceunder

steacly onditions s the synchronous eactance f the machine.Obviousiy Xf7<

symmerricalaultAnalysis I'lt,5ffit

SteadystatecurrentamPlitude

(b )Envelopef synchronousachineymmetricalhort ircuiturrent

F ig.9. 4

If we examine he oscillograrn f the shortcircuit currentof a synchronous

machineafter he DC ott-setcuitentshave been ettrovedrom it, we will tind

the currentwaveshapeas given in Fig. 9.4a.The envelopeof the currentwave

I

I

0)

Tlme



X'd Xu.The machine husoffersa time-varying eactance hich changes ront

Xttoto Xtaand inally to Xn .

Subtransienteriod

Steady state period

Extrapolation fsteadyvalrre

Extrapolation f transientenvelope

(a) Symmetrical hort circuitarmaturecurrent n synchronousmachine

Fig. 9. 4 (Contd.)

Time

iActualenvelope

I

b

lI

III

E aq)

f()' o 0t

a

o

oEEa

*Unity turn ratio is assumedhere.

shape s plotted n Fig. 9.4b,The shortcircuit currentcan be divided rtto hree

periods-initial subtransientperiod when the current is large as tire tnachine

offers subtransienteactance, he middle transientperiod where the machine

offers transient eactance, nd inally the steadystateperiod w\n the machine

ofterssynchronouseactance. :

If the transientenvelope s extrapolatedbackwards n tinre, the difference

betwecn he tlansicrrt nc lsubtransiert tnvelopess the cunent Ai/' (corre-

sponding o the clamperwinding current) which decays ast according to the

clamperwinding time constant.Similarly, he differenceAi/ between he steadystate nd ransicnt nvelopes ecaysn accordance it h the ield im e constant.

In termsof the oscillogram,hecunentsand eactances iscussed bove,we

can wrlte

lEs l

Y ,

tltt= 32.: Y+J2 X,J

where

l1l = steadystatecurrent (rms)

!//l = transient urrent rms)excludingDC component

lltl = subtransient urrent(rms) excluding DC component

Xa = direct axis synchronous eactance

l I l =

l l l =

oat;

\t z.

obt;

\ I L

_ l E 8 l

xtd

(9.7a)

(e.7b)

(9.7c)

Modern Power Svstenn Anelveie

Xtd=direct axis transient eactance

X'j = direct axis subtransienteactancelErl = pe r phase o loadvoltage rms)

Oa,Ob,Oc= interceptsshorvn LEigs- 9Aa and, _The intercept Ob for finding transient reactancecan be determined

accurately by means of a logarithmic plot. Both Ai, and al decavexponentiallyas

Aitt = Ai( exP - t/q,)

Ait = Ai6 exp 1_ t/r7)

where r4, and rf are espectivelydamper,and field winding time constantswithTd* 4 ry At time / 2' r4*, Aitt practicalry dies out and we can write

lo g (Ait t+ At,)1,, , - log Ai' = - Aint,/ ryTr*

tl c-a b*+

gO)

Fault

pon ii'ragn-tir rutlrarion

f:T:jr:lltlqj?

",l"ilirion),rhevaluesf reactances

ormally ie withincertain predictable limits for different types of machines.Tabie 9.r givestypical valuesof machine eactances hich can be userj n fault calculations ndin stability studies.

Normally both generatorand motor subtransient eactancesare used todeterminethe momentary current flowing on occurrenceof a short circuit. Todecide the intemrpting capacity of circuit breakers,except those which openinstantarreously, ubtransient eactance s used for generatorsand transientreactance or synchronousmotors. As we shall see ater the transient eactancesare used or stability studies.

The machinemodel to be employedwhen the shortcircuit takesplace fromloadedconditionswill be explained n Sec. 9.4.

The method of computing short circuit currents is illustrated throughexamples ivenbelow.

'ii#I:

n

s

For the radial network



Fig. 9. 5

Ai, l , :o : Aitoexp(-r ,t )1,:o Ai,o ob

Table9. 1 Typical aluesof synchronous achineeactances(All valuesexpressedn pu of ratedMVA)

o

o r f

Type of

machineTurbo-alternator Salient pole

(Turbine (Hydroelectric)

generator)

Synchronous

compensator

(Condenser/

capacitor)

Synchronous

motors*

X, (or X,)X

.t

xdxti

x2xoru

l .00--2.00 .9 -1 . s0.12-0.350.r-0.25

_ x,d0.04-0.140.003-0.008

0 .6 -1 .50.4-1.00.2-0.50.13-0.35_ x,d0.02-0.2

0.003-0.01s

r.5-r2.50.95-1.50.3-0.60.18-0.380.17-0.370.025-0.160.004-0.01

0 . 8 - 1 . 1 00.65-0.80.3-0.35

0.18-0.20.19-0.350.05-0.070.003-0.012

ro = AC resistanceof the armaturewinding per phase.* High-speedunits tend to have ow reactanceand low speedunits high reactance.

shown in Fig. 9.6, a three-phase ault occurs at F.Determine he fault current and the line voltage at l l kv bus under faultconditions.

1OMV A15% eactance

11 V

10MV A \12.5oheactance \

ne 3okm ,

=(0.27 o.3e a/ km

r NO2: 5 MVA,8oh eaclance

riO.0B) / km

z xn caote/

F

Fig. 9. 6 Radialnetwork or Exampleg. 1

Solution Select a system base of 100 MVA.V6ltage basesare: I I kV-in generators, 33 kV fo r overhead line and 6.6 kVfor cable.

Reactance f G, =

Reactanceof G2 -

Reactanceof Z, =

PowerSystemAnalysis

Reactancef Tr= ;WY4 = 71.6pu)

overheadine mpedanceZ (in ohms) MVA""'"

(kvBur"2

30x (0 .27+ j0 .36 )x100

Q' 2

- (0.744 70.99)u

cablempedanc"3(9-1!t,rJr0,q,21tlq = (0.93 70.55) u

(6.6) '

Circuit model of the system or fault calculations s shown n Fig. 9.7. Since

the system s on no load prior to occurrence f the fault, thevoltagesof the two

generators re identical (in phaseand magnitude) and are equal to 1 pu. The

generator ircuit can thus be replacedby a single voltage source n serieswiththe parallel combination of generator eactances s shown.

11kV bus

I

Symmetrical aultAnalysis|

337I

= (0.93 j05s) + (71.6) (0.744 i0.99) + (t1.0)

= I.674+ j4.14= 4.43176.8"pu

Voltage t 11kV bu s= 4.43167.8"x 0.196 - 70-8"

= 0;88 I :T ptt = ft88 x 11 = 9;68 kV

A 25 MVA, 11 kV generatorwith Xl = 20Vo s connected through a

transformer, ine and a transfbrmer to a bus that supplies hree dentical motors

as shown in Fig. 9.8. Each motor has Xj = 25Voand Xl = 3OVo n a base of

5 MVA, 6.6 kV. The three-phaseating of the step-upransformer s 25 MVA,

11/66 kV with a leakage reactance of l0o/o and that of the step-down

transformer s 25 MVA, 6616.6 V with a leakage eactance f l0%o.The bus

voltageat the motors s 6.6 kV when a three-pha.seault occursat the point F.

For the specified ault, calculate

(a) the subtransient urrent n the fault,

(b) the subtransient urrent n the breaker 8,



Fig. 9. 7

Totalmpedance(j1.5 l j1.25) (t1.0) (0.744 i0.99)+ (i1.6)+

(0.93 70.55)

-1 .674 j 4 .82 5.1170 .8 "pu

Is c='' ?

tt = 0'196 - 70'8"Pu5.r170.8"

18u." to*.10; = 8,750AJ3 x6.6

Is c= 0.196 8,750 1,715

Total irnpedanceetween and11 kV bus

i

55)

I t

Il

j | .0 (o.744 i0.99) i1.6 (0.93 70 .

6 6 d . I | , 6 6 6 ' I IT1 Line T2 Cable

(c) the momentarycurrent n breaker B, and

(d) the current o b e interrupted y breakerB in five cycles.

Given:Reactance f the ransmissionine = l5%o n a baseof 25 MVA, 66

kV. Assurne hat the system s operatingon no load when the fdul" occurs.

Flg.9. 8

Sotution Choosea systembaseof 25 MVA.

For a generator oltagebaseof 11 kV, line voltagebase s 66 kV and motor

voltagebase s 6.6 kV.

(a) For eachmotor

X',j*= j0.25 x

Line, transtbrmersand generator eactances re alreadygiven on proper base

values.

The circuit model of the system or fault calculations s given in Fig. 9.9a.The systembeing nitially on no oad, hegenerator nd motor inducedemfs are

identical.The circuit can hereforebe reduced o that of Fig. 9.9b and then to

Fis. 9.9c.No w

+= i1.25u

fiffirel Modernower ystem natysis!

Isc=3><- l-+=+ - - j4 .22puj1.25 j0.55 r

Basecun'ent n 6.5 kV circui, - 25x 1,000 = 2.187 A

Issc=,;; r4rrlt]frf*o(b) From Fig. 9.9c, current throughcircuit breaker B is

I sc (B)2x-++ . ]_ : - i 3 . 4 2 j1.25 j0.55 r -. -

110' jo.2+

= 3.42x 2,187 7,479.5

j0.15 j0.1

110"

110 '

Ftll?9 ;< to"

momentarycurreni ihrough breaker B -- 1.6 x 7,4i9.5

- 17,967A

(d) To compute the current to be intemrpted by the breaker, motor

subtransientreactance (X!j = j0.25) is now replaced by transient reactance

(X a = /0.3O).

XI (motor)= 70.3x

The reactances f the circuitof Fig. 9.9c now modify to that of Fig. 9.9d.

Current (symmetrical) o be ntemrpted by the breaker as shown by arrow)

1 " 1= 2 x ^

+ = 3 . 1 5 1 5 p uj l .s jO.ss

Allowance s made or the DC off-set value by multiplying with a factor of 1.1(Sec.9.5). Therefore,the current o be interrupted s

1. 1x 3.1515 2 .187= 7.581A

9.4 SHORT CIRCUIT OF A LOADED SYNCHRONOUS

MACHINE

25

T= J r . ) pu



(b )i0.55

Fig.9. 9

(c) For finding momentary current hrough the breaker,we must add theDC off-set current o the symmetricalsubtransient urrent obtained n part (b).Rather han calculatingthe DC off-setcurrent,allowance s made or it on anempiricalbasis.As explained n Sec.9.5,

rcuit reaker) (c)

i0.55

In the previous article on the short circuit of a synchronousmachine, it was

aBsumedhat the machine was operating at no load prior to the occurrence of

shortcircuit. The analysisof short circuit on a loadedsynchronousmachine s

complicatedand is beyond he scopeof this book. We shall, howevbr, present

here the methods of computing short circuit current when short circuit occurs

under oadedconditions.r l : - - - ^ f \ 1 / . | ^ L ^ . - , ^ + L ^ ^ i - ^ . , i + * ^ A ^ l ^ flr rBtlc >. Lv Slluws Llrg urrUurt lrlu(lEl ur a

synchronous eneratoroperating nder steadycon-

ditions supplying a load current /" to the bus at aterminal voltage of V ". E, is the nduced emf under

loadedcondition andXa s the direct axis synchro-

nous eactanceof the machine.When short circuit

occursat the terminals of this machine, the circuit

model to be used for computing short circuit

current is given in Fig. 9.11a for subtransient

current,and n Fig. 9.1lb for transient urrent.The

inducedemfs to be used n thesemodels are given

bY

E,l= v" + i lTt j

EL- V'+ i l"Xto

The voltageE!is known as the voltage behind the subtransient eactanceand

the voltage E!is known as the voltagebehind the transient eactance.Infact,

if 1o s zero (no loadcase),EJ= Etr= Er, the no load voltage, n which case

the circuit model reduces o that discussed n Sec. 9.3.

Fig.9.10 Circuitmodel fa loadedmachine

(e.8)

(e.e)

340| Modern o*s1_qqe!l inslygsI

t/ o

F ig .9.11

Synchronousmotorshave nternalemfsand reactances imilar to that of ageneratorexcept that the current direction is reversed.During short circuitconditions these can be replaced by similar circuit moclelseicept that thevoltagebehind subtransient/transienteactances eiven bv

E'lr=v" - jI"xU

E'*= v"-

jI"4Wheneverwe aredealingwith shortcircuitof an nterconnectedystem,he

synchronousmachines (generatorsand motors) are replacedby their corre-sponding i rcui tmocle ls avingvol tage ehincl rrht ransientt ransient )eac-tancen serieswith subtransienttransient)eactance.he restof th enetwork

(a ) Ci rcui tmodel forcomput ingsubtransient urrent

(b) Circuitmodel or computingtransientcurrent

_Symmetrical FaultArralysis

solution Aii reactances re given on a baseof 25 MVA andvoiiages.

I 36.9"pu

I 36.9"

appropriatgi

:t

(e .10)

(e .11)

Prefaultvoltage V" = J' 9 = 0.9636 l0 pul 1

Load = 15 NfW, 0.8 pflEading

= l: = 0.6pu ,0.8 pf leading25

prefaultcurrenl " = _9{__ _ 136.9. 0.77g30.96360. 8

Voltage ehindsubtransienteactancegenerator)

E", 0.9636 tr + j0.45 x 0.1783

- 0.753670.28

u

Voltage ehind ubtransienteactancemotor)

El,, 0.9636_ r -i0.15 x 0.7783_ 36.9"

= 1.0336 .70.0933u



beingpassiveentains nchanged.

II Example .3r^_._ . ...-: .._ ._

A synchronousgeneratoran d a synchronousmotor each rated 25 MVA, I I kVhaving l5Vo subtransientreactance are connected through transfbrmers and aline as shown in F'ig.9.12a. The transfbrmersare ratecl25 MVA. ll l66kV and66lll kV with leakagereactanceof l\Vo each.The line has

a reactanceof lTToon a baseof 25 MVA, 66 kv. The motor is drawing 15 Mw at 0.9 power factorleadingand a terminal voltage of 10.6kV when a symmetrical three-phase aultoccurs at the motor terminals. Find the subtransient culrent in the generator,motor and fault.

t" j0.1 j0.1 j0.1

t ; .6'f ,1, '. 'dt I-. - ' l t rd-.I

. 1 i0 .15' )

r+ l

I

(b) Prefault quivalent ircuit

Gen t Tt'| | ' ! L ine) l r ;

(a ) One- l ineiagramor ho systom f Example 3

F-ti

II

(c) Equivalent ircuit uring ault

F i 9 . 9 " 1 2

Current n fault

Th eplefurul t .equivalenti rcui ts shownn Fig.9. l2b.Unclerhrr l tecl.ncl i -t i on l r i g . . 2c )

., , 0.7536+ 0.21t00I ' ; , - - " - = 0 . 6 2 2 6 _ j 1 . 6 7 4 6 p u" i 0.45

1 .03 i6 ro oc)?I',l,=

i 0 .1s

Base

Now

IJ= I :i + 1 , , , ,=_g.5653 u

cttrret (gen/moto= 44q1 = 1. 3 2. 2J 3 x l l

I' J 1,312.00.6226 j1.6746) (816.4 jL, tgt.4)A

I'J= 1,312.2.- 0.6226 j6.8906) (- 816.2 jg,O4L8)

1 t - - j t t , 2 3 g A

short circuit (sc) current computation through theThevenin Theorem

An alternatemethod ol ' cornputing short circuitapplicationof the Thevenin heorem.This method s

currents s through thefasterand easily adopted

I

342 | Modern Po*er SystemAnalysisI

to ,yrtl-atic computation or large networks. While the method s perfectly

gcncrerl,t rs l lustratcd crc t lrrougha sinrplc xanrplc.

Consider a synchronous enerator eeding a synchronousmotor over a line.Figure9.I3a shows he circuit model of the systemunder conditions of steady

As a first step he circuit model is replacedby the one shown n Fig. 9.13b,

wherein hesynchronousmachinesare epresented y their transient eactances(o r subtransienteactancesf subtransienturrents re of interest) n serieswith

voltagesbehind ransient eactances. his changedoes not disturb he prefault

current I" andprefaultvoltage V" (at F).

As seen rom FG the Theveninequivalent ircuit of Fig. 9.13b s drawn n

Fig. 9.13c. t comprises refaultvoltageV" rn serieswith the passive hevenin

impedancenetwork. t is noticed that the prefaultcurrent 1" doesnot appear n

the passivc hcvcnin rnpcdancc ctwork. t is thcretore <lbe rcmcnrbcrcdhat

this current must be accounted or by superposition fter the SC solution s

obtained hroughuseof the Theveninequivalent.Consider ow a ault at 1, ' hloughan rnpedancel . l i igure 9.13d hows he

Thevenin equivalent of the system feeding the fault impedance. We can

immediate ly ri te

AI^ +X(xlh,+x + xi

Xle

(e.14)

Postfaultcurrentsand voltagesare obtained s ollows by superposition:I{ = I" + al r

I{=- I" + AI^ (in rh edirect ion f AI^) (9.15)Postfaultvoltage

vf-=

vo + (- , rxn, I f ) = v" + AveJ6)whereAv = -ix^tf is the voltageof the aultpointF/ on the Theveninpassive

fffl:liiwithrespectothe eferenceus t-;;;..d by the low

-offaurt

ince he prefaultcurrent flowing out oft curent out of F. s independeniof load

on s summarizedn the following four



, l ' - - V "'

jXrn + Zt

Current causedby fault in generatorcircuit

Step I: Obtain steadystatesolution of loaded ystemstep 2: Replacereactances f synchronousmachinestransient values. Short circuit uit

"rrsources.The

Theveninnetwork.'Step

2 at the fault point by negariveofies with the fault impedanc".6o_pur"rterest.

rareobtainedby addingresultsof Steps

The following assumptionsan be saferymade n SC computationseading:ation:

ragnitudesre I pu.tre zero.

to actualconditions as under normalnity.

ruc cnanges ln current caused by shortcircuit are quite large, of the order of 10_20

(load flow study).by their subtransienUresult is the passive

x'd^fI, -

- (xhs+x + xl^

(b ) Ga) G

tion 2.

Let us illustrate the above method b,Fi9'e't+

recalculatinghe resultsof Example9.3.

F is th e ault oint nthe passive heveninnetwork

(c) (d)

Fig.9.13 Computat ionf SC current y he Thevenin quivalent

(e.r2)

(e.13)

ModernPowerSystemAnatysts

The circuit model or the systemof Example9'3 for computationof postfault

condit ion s shown n Fig' 9'14'

_ 0.9636x 0.60= - 78.565 u

Change n generator urrentdue to fault'

AI^=- l8.s6sio '+ - - i2 '141u,B

_ r " . " "_ j0 .60

Change n motor currentdue to fault'

At^=- 78.565 j.$*i-- i6'424 tt

To these changeswe add ttre prefault currentto obtain the subtransient

current in machines. hus

I 'l= I" + AI r - (0.623 j1.67$ Pu

In = - I" + AI^= (- 0.623 76.891) u

, Q r r r n m a l r i n a l E ^ . . t + A - - r . - - ! -

If sc MVA (explained elow) s more han 500, the above ultiplyingiactorsare ncreasedby 0.1 each.The multiplying

factor for air breakersated 600 vor lower is 1.25.The current that a circuit breakercan ntemr

rng voltage over a certain range, i.e.

Amperesat operatingvoltage

Rated ntemrpting MVA (three-phase)apacity

= '6ty(t i f le)lratedx 11(l ine)lrated

inremrptingunentwhere V(line) is in kV and 1 (line) is kA.

Thus, insteadof_computing he sc current to be intemrpted,we cbmputethree-phaseSC MVA to be intemrpted,where

r --' '



which are the same and shoutdbe) as calculatedalready

we have thus solved Example 9.3 alternatively through th9 Thevenin

theorem ond ,up.rposition. This, in4eecl, s a powerful method for large

networks.

9.5 SELECTION OF CIRCUIT BREAKERS

. Iwoot thcc i rcu i t b rcakcr ra t i ngswh ic l r rcc ;u i rc thcc trn lp t l t t t ti ono fSCcur ren t

are: rated momentary urrent and rated symmetrical nterruptirtgc:nrrent'

Syrnmetrical SC current is obtained by using subtransient eactances or

synchronous nachines.Momcntary current irms) is then calculatedby

multiplyingthe symmetrical o-"nory currentby a factorof 1'6 o account or

the presence f DC off-setcurrent'

Symmetrical urrent o be intcrrupteds computed y r'rsing ubtransient

reactancesor synchronouseneratorsnd ransient eactancesor synchronous

motors-induction motorsare neglected*.The DC off-set value o be added o

obtain the current to be interrupted is accountedfor by multiplying the

symmetrical SC currentby a factor as tabulatedbelow:

Multiplying Factor

SC MVA (3-phase)_ Jt x prefault ine voltage n kVx SC current n kA.

If voltage and current are in per unit valueson a three-phaseasis

SC MV A (3-phase)= ly lp,. .roul , 11116 (MVA)uur. (e.r7)

Circuit Breaker SPeed

8 cyclesor slower

5 cycles3 cyclcs

2 cycles

attempts, urrentscontributed

accounted or.

1 . 0

1 . 11 . 2

1 . 4

by induction motors during a short

O h v i o r r s l r r r q f o A l \ / \ / A i - + ^ * , - - . : - - - -- 'L 'Yru'uJrJ 'i c i iuu lvtvA iniei rupi i i lg capacl ty of a ci rcui t breaker is to bernurc hl n (o r cclual o) th c sc MV A requiredo be ntemupted.

For the selectionof a circuit breaker or a particular ocation,we must find.the maximum possible SC MVA to be intemrptedwith respect o type andlocation of fault and generatingcapacity (also synchronousnotorl oad)connected o the system.A three-phaseault though are s generally he onewhich gives the highestSC MVA and a circuit breakermust be capableofinterrurptingt. An exception is an LG (line-to-ground).ault close to asynchronous enerator*.n a simplesystem he ault ocationwhich gives thehighestsc MVA may be obvious bu t in a largesystemvarious possiblelocationsmust be tried our to obtain the highestst nava requiring Lp"ur"aSC computations.This is ilustrated by the exampleshat follow.

Iii"'n".;

IThree6.6 kv generators , B and c, each f I0o/oeakage eactance ndMVAratings40, 50 and 25, respectivelyare nterconnected lectrically, as shown n

t'In some recent

circuit have been

tThis will be explained n Chapter1 .

15:!"'l .";: I

{d46..t| Modern owerSystem nalysis-TFig. f.i5, by a tie bar throughcurent timiting reactors,eachof I2Vo eactance

baiecl po n he rat ingofthe machine o which t is connected. threc-phase

feeder s supplied rom the bus bar of generatorA at a line voltageof 6'6 kV'^ f n 1 a

Q/phase.Estimate he maximumMVA thatcan be fed into a symmetricalshort

circuit at the far end of the feeder.

currents an hen be calculatedby the circuitmodel of Fig. g.l6acorresponding

to Fig. 9.13d.Th e circuit s easily educedo rhatof Fig.9.16b,where7-(0.069 + j0.138)+ j0. r2s il 00.15+ jo.22| j0.44)

= 0.069 j0.226= 0.236173

SC MVA = Volf = V"('+') = + pu (sinceVo = 1 pu)\ Z ) Z

Fig.9.15

Sotution Chooseas base50 MVA, 6.6 kV '

Feeder mPedance

(MVA)Ba."

= 212MVA0.236

Consider he 4-bussystemof Fig. 9.17.Buses1 and 2 aregeneratorusesand3 and 4 are oad buses.The generators re rated l l kv , 100 MVA, with

I

Z

50

Tie bar



=%

=(o.o6e+/0 .138)pu

GenA reactanceo'1[50 = 0.125Pu

40

GenB reactance 0.1 Pu

GenCreac tance=0 .1425

Reactor reactan."o' t 'I tn

40

Reactor reactance0.12Pu

transient eactance f l07o each.Both the transformers re 1ll110 kV, 100MVA with a leakage eactanceof 5Vo.The reactances f the ines to a baseof100 MVA, 110kv are ndicatedon the figure.obtain the shortcircuit solutionfor a three-phase olid fault on bus 4 (loadbus).

Assumeprefault voltages to be 1 pu and prefault currents o be zero.

(G)

Fig.9.17 Four-busystem f Example .5

Solution Changes n voltages and currentscausedby a short circuit can becalculated rom the circuit model of Fig. 9.18.Fault current1/ is calculatedbysystematic etwork eductionas n Fig. 9.19,

Reactor C reactance =0.12 50

= 0.2pu

= 0 .15 u

- 0.24pu

(0.069 i0.138)

(a )

j0.2

j0.12 jo.24

1

yo = 1Z0o(

Fig. 9.16

(b)

lo* !o;E

,^o I Moclern Power Svstem AnalYsis

I

1 4

tt = -*= = - jt.37463puj0.13s60

t,= rsx i: i3;:: = j3837oruj0.37638

12 = , x {: i: : : : = - j3.53 62 u '' j0.37638

Let us no w compute he voltage hangescl r buses ,2and 3. From Fig.9 . 9 b , w c g i v c

AVr- 0 - ( /0.15) - j3.8370r)= - 0.57555 u

AV , = 0 - (iO.l.s) - . i3.53762) - 0.53064puNow

jo 1, ,7

D': _ jo.2[ 0t , i0.1

A ,l , zs

2

t, ,t ,

10.15

II r - ---rnTl

4+ ib . rs Izt

\i ro 'rso)

r -f - i.0.11I

t - - - ' A t i A ' -

l, I F ' t t I

V , ' = + l V t = 0 . 4 2 4 4 . 5u

V ) , l + A V z = 0 . 4 6 9 3 6 p u



i 0 1 ' ' : , , , t , . ( ^. -

ti o t s

l l4 T

( t v l = t o ( a )*

i r t

(c )

l ' ' l

I 2 i I 0 6 ' )( r l r

io.t \P' ' r ) , }o ' rs

-l(b ) (. )v l = .o

i rfr '

(, - l

, -J 10.04166t*)

- i( ) V ? 1 . 0\. .t .

I l lI

(e )

+I

,-j 0.'uuuo

l( ) v?= .o

Tr

. v ,J vJ/ l- t=

,o f tr .o. t= J0'17964u

AVy= 0 - [( /0.15)- j3.83701) Q0.15)/0.17964)l

= - 0.54fi(r0u

Vt t= I - 0 .54860 0.4514 u

v lo= o

The determination f currents n theremainingines s left as an exercise otltcr r'ldcr.

Short circuit study s completewith the computation f SC MVA at bus4.(SC MVA)^ = 7 37463x 10 0= 737.463MV A

It is obvious hat the heuristicnetwork eductionprocedureadoptedabove snot practical for a real power networkof evenmoderatesize. It is, therefore,essentialo adopta suitablealgorithm br carryingou t shortcircuit study on adigital computer.This is discussedn Sec.9.6.

9.6 ALGORITHM FOR SHORT CIRCUIT STUDIES

So far we havecarriedout shortcircuitcalculationsor simplesystemswhosel l i tssivctctworksanbe easi ly educed.n th isscct ionwc extcnd ur study o

No w

Fig. 9.19 Systentat iceductionof th e networkof Fig' 9'18

i*.3€0 l ModernPowerSystemAnalysis

large ,tyrt"-r. In orcler o apply the four stepsof short circuit computation

developedearlier o largesystems,t is necessary o evolve a systematic eneral

algorithm so that a digital computercan be used.

Gen 2

Fig. 9.20 n-bus ystemundersteadyoad

Consideran n-bussystem hownschematicallyn Fig.9.20 operating t steady

load. The first step towards short ciicuit computation s to obtain prefaultvoltagesat all buses andcurrents n all lines througha load flow study. Let us

indicate he prefaultbus voltagevector as

_ Symmetr.ic, . . . . . ,*", , ;al aul tAnat l rs is

4V = Z"urJfwhere

.7 -' l '-

ot nI

iI

= bus mpedance atrixof the e.2DZnn) passive hevenin etwork

(e.20)

(e.22)

u// = bus current njection vectorSince he network s injectedwith current 1/ only at the rth bus,we have

0

0:

r f , fI , : - I '

/ -

network ._ - - . r r . vv \ , r r r v u ( r r vu r raegcs oI ml s

Now



rth bus

(e.23)

(e.24)

Let us assume that the rth bus is faulted through a fault impedance Zf . The

postfault bus voltage vector will be given by

V{ur= VBus AV

:

SubstitutingEq. (9.22) in Eq. (g.20), we have for rheAV, = - ZrJf

By step4, the voltageat the nh bus under ault is

v != vor+avo,-vor - Z,J fHowever, his voltagemustequal

Vd= 7f 1f

We have from Eqs. (9.23) and (g.24)

zftf - vo,_ z,Jf

or f=V:

Zr, + Zf

At the rth bus fromEqs 9.20) nd g.22))

AV, = - Z,Jf

v{ = v? - Z,Jf, = 1,2, . . . ,substitutingor // from Eq. 9.25).we have

vI= vf - :zl';rv!

z *+ L

(e.18)

(e.1e)

where AV is the vector of changesn bus voltages aused y the fault.As'step 2, we drawn the passiveThevenin network of the system with

generatorseplacedby transient/subtransienteactances ith their emfs shorted

(Fie.9.21).

Fig. 9.21 Networkof th e system of Fig. 9.20 or computing hanges n

bus voltagescaused by the fault

(e.2s)

(e.26)

(e.27)

I rcaiimmetrica! ault Analysis | -?53

First of all the bus admittancematrix for the network of Fig. 9.18 s formed

as ol lows:

-r - - j28.333

352 i rtllodern owerSystemAnalysis

F o r i = r i n E q . ( 9 . 2 7 )

(e.28)

In the above elationshipV,o'r, the prefault bus voltagesare assumed o be

known from a load flow study. Zuu, matix of the short-circuitstudy network

of Fig. 9.21 canbe obtained y the nversionof its furr5matrix as n Example

9. 6 or the Zru, bui lding algorithmpresentedn Section9.7.It shouldbe

observedhere that the SC study network of Fig. 9.2I is different from the

correspondingoad flow study network by the fact'that the shunt branches

correspondingo the generator eactances o not appearn the oad low study

network.Further, n formulating he SC studynetwork, he oad mpedances re

ignored, these being very much larger than the impedancesof lines and

ginerators. Of coursesynchronousmotors must be included n Zuur tormula-

tion for the SC study.Postfaultcurrents n lines are given by

f u=yu vri- vt ) Q.29)

For calculation of postfaultgeneratorcurrent, examineFigs. 9.22(a)and (b).

From the oad flow study(Fie. 9.22(a))

- lY r . > = Y r t = _ - 1 7 5 . 0 0 0L Lt

j 0 . 2

- l

Y3 = Yy =;r *

- i6.667

- 1Yrq= Yq=

J,3 . l= i10 .000

Yzz=: -++ -+++ :^=- j28 .333

j0.15 j0.1s j j. r j0.2

Yzt=Yn= += j10.000j0 .1

- l

I-t--- r



Fi1.9.22

f" , = &#; (prefault eneratorutput Pc i+ Qci) (9.30)v,u

E'Gi = V, + jXt"/t)", (9'31)

From the SC study, Vf ,is obtained. It then follows from Fig. 9.22(b) that

Prefault generator output = PGr+ iQci Yzq=Yn =F;

- i6.667

Y - . =I

+I = - i 1 6 . 6 6 733 -

j 0 . 1 5 j o . 1

Ytq= Yqt= 0.000

v - I - IYqq=

,-

*

,.""

-- i76'667

(b )a)

By inversion we get Z",tt as

rrc,=tfr!

i r;rnr;;.;I

(e.32)

j0.0s97

j0.0903

j0.0780

jo.o719

j0.0719

j0.0780

j0.13s6

j0.0743

Now, the postfaultbus voltages an be obtained sing Eq. (9.27) as

To illustrate he algorithmdiscussed bove,we shall recomputehe shortcircuit

solution or Example9.5 which was solvedearlier using he networkreduction

technioue.

V { = Vo Z to VPr r Z o o a

,2.tr!A | ^/ lndarn Pnrrrar Qrratarn AnalrrcicJ J ' I l Y l V V V l l l I V l t v l V t v l v l l l , r l r s r t v r e

I

Th e prlfault condit ion eingno load, V0r= Voz= V03- Voo=1p u - r 1.000J - _ -

2,,{orZzz)

m a l r i a a l| | t g l t t v q l

E a tqu

1.00j0.0903

= - j I L074 I97 u

r .o - 99199x 10=0 .4248pui0 .1356

vrz= : - ?': r, ' ,L z o o r

i 0 . 0 7 1 0= 1. 0 o - :: x 1 .0 = 0 .4698 pu

j0 .1356

7

v{ =v\ - 1* vfZoo

i0.0743= 1 .0 ' - : : _ : 1 .0 0 .4521 uj0 . r3s6

vi = o'o

Using Eq. (9.25)we can obtain he ault currentas

7r = .r-^0.9^o=-, j7.3j463pu

By Inventing Y"u"

/nus= YrusVsus

or Vsus [Y"ur]-t eus = Znus eus

or Zsvs= [Yuur]-t

The sparsityof fsu, may be retained by using an efficient inversion technique

[1] and nodal impedancematrix can then be calculated directly from the

factorizedadmittancematrix. This is beyond the scopeof this book.

Current Iniection Technique

Equation(9.33) can be written in the expanded orm

V1 211\ * Z tz lz 1 .. . ZnIn (e.34)

(e.33)



j0.13s6

These alues greewith those btained arlier n Example .5.Let usalso

calculatehe short ircuit urrentn lines1-3,1-2, -4,2-4 an d2-3.

,r v, r v{ 0.4248-0.4521[ 1 2 = : - - j o ' 1 8 2P u

z. n i0.15

- r v, r-v {

0.4248-0.4698t r rz= - '_ - '2 - : - _ j0 .225 uzr z . i0 .2

ryr -v {

_ o.4z4v-o, t4

- i4.248 u4q jo' l

. r u'l - v{ 0.46e8oI i q = ' -

q - - - = - j 3 . 1 3 2 p uzz+ i0'15

-- r v{ -v^ 0.46980.4521I-Izt= iutt =

TO-= - ./0.177 u

For the exampleon hand his methodmay appearmore involved compared

to the heurist icnetwork eductionmethodemployed n Example9.5. This,

however, is a systematicmethod and can be easily adopted on the digital

computer for practicalnetworksof large size.Further, another mportant feature

of the method is that having computed Zu's, we can at once obtain all the

required shortcircuit data or a fault on any bus. For example, n this particular

system, he fault current for a fault on bus I (or bus 2) will be

v2:22111+ 22212 . . . + zzn ln

(e.35)

V, : Zntl l * 2,,21r .. + Z,,nl , ,

It immediately follows from Eq. (9.34) that

z - v ' l" i i

-I r l , r : r z . . . : r n =o

I 1 - r 0

Also Zi, - Ziii (Znvs is a symmetrical matrix).

As per Eq. (9.35) f a unit current s injected t bus(node)7, while theother

buses rekept oponcircuited, he busvoltages ield the valuesof theTthcolumn

of Zuur. However, no organizedcomputerizable echniquesare possible for

finding the bus voltages.The techniquehad utility in AC Network Analyzers'where the bus voltagescould be read by a voltmeter.

Consider he network of Fig. 9.23(a) with three busesone of which is a

reference.Evaluate Zsus.

Sotution Inject a unit currentat bus I keepingbus 2 open circuit, i.e., Ir = I.

and r= 0 as in Fig. 9.22(b).Calculatingvoltages t buses and 2, wehave

Z t t = V t = 7

Z z t = V z = 4

.f

356 | ModernPowerSystemAnalysis

Now et It = 0 and12= 1. It similarly fbllows that

Z t z = V t = 4 = Z n

Becauseof the abovecomputationalprocedure, he Zru, matrix isreferred o as the 'open-circuit

impedancematrix'.

Z"vs Building Algo4ithm

It is a step-by-step rogrammable echniquewhich proceedsbranchby branch.It has the advantage hat any modification of the network does not require

complcte ebuilding f Z"ur.Consider hat Zrur has been formulatedupto a certain stageand anotherbranch s now added. hen

Zrur (old)Zo:branch impedance

Zsus (new)

Symmetrical ault Analysis !, f5l

, 4r,the dimensionof Zsu5 goes up by one). This is type-2 modification.

3. Zuconnects n old bus o the eference ranch i.e.,a new oop s formed

dimensionof 2o,," doesnot change). his is type-3modification.

4. Zuconnectswo old buses i.e.,new loop is formedbut the dimensionof

Zuu, doesnot change).This rs type-4 rnodittcation.

5. Zu connects wo new buses Zeus remainsunaffected n this case).This

situationcan be avoidedby suitable numbering of busesand from now

onwardswill be ignored.

Notation: i, j-old buses; -lsfelence bus; k-new bus.

Type- 1 Modification

Figure 9.24 shows a passive(linear) n-bus network in which branch with

impedance2,, is added o the new bus k and the referencebus r. Now

Hence

V*=ZJ*

Z r i = Z * = 0 ; = 1 , 2 . . . . . t

Zm = Zu

l7 41Zvus=

l+ 6 l



Upon adding a new branch,one of the following situations s presented.

Fig. 9.23 Currentnject ion ethod f computing ru.

26 is added rom a new bus to the referencebus (i.e. a new branch sadded and the dimension of Zry5 goes up by one). This is type-Imodificution.

Fig. 9.24 Type-1modification

Type-2 Modification

Zo s addedrom new bus ft to the old bus7 as n Fig. 9.25.It follows from this

figure that

s,,"old)Zsvs (new) - (e.36)

t .

Passive inear

n-bus network

35S;,1 MorJern Powcr Srrctarn anatrroio

1O _

n

Passive inearn-bus network

Fig. 9.25 Type-Z modification

Vo= Zdo + V,

= Zr,I*+ZiJr+ Zlzlz .. . + Zi i ei * I) + .. .+ Zinln

Rearranging,

V * = 4 l t + Z l z l z . . . + 2 , , 1 , + . . . +Z i , l n + ( Z i i + Z ) l k

Consequently

Zrj

zzjZsvs(old) I : (e.38)

o lzyziz...zi, zi i+ z, )lI*

lr

Eriminatein he3":';,,iiy';;;?::"i"ti.I" a;y>;i:ation

(e' )'

1or 'o = -1 2 (\1Ir+ Zi2Iz+ "' + ZlnI) (9'39)

NowV; = 2;111+Z,r I , + "' + Z 'nIn Z;/r (e.40)

Zt i

SubstitutingEq. (9.40) n Eq. (9.39)

,,=lr^ h(z* 1 r]alr,,"h

rz z,,>)r,

+ +lr^- ^i{zu

2,,1)r, (s.4t)



(e.42)

Zzj

:

Znj

(e.37)

Equation 9.37) can be written in matrix orm as

. |t"fZsLr5new) znvs(old ;+;l : ltz't"Zv)

" j jT "u l z , , ljj + zb

IYpe-3 Modification

zu connectsan old bus (l) to the referencebus (r) as in Fig. 9.26. This casefollows from Fig.9.25 by connectingbus t to the reference us r, i.e. by settingv * = o '

Type- Modification

zo connectswo old buses s n Fig. 9.27 Equationsanbewrittenas ollows

for all the networkbuses.

Fig. 9.27 TYPe-4modification

Vi = Z;11, Z, l r+ "' + Zr i (I i + /) + ZU Qi -

Similar equations ollow for other buses.

J

Passivelinearn-buSnetwork

( l ;+ p

Fig. 9.26 Type-3 modification/o )+ ...+ZiJ"(9.43)

360 | Modern power System Analysis

I he voltages of the buses i and j are, however, constrained by the equation(Fig. 9.27)

Symmetrical aultAnalysis I fOt

Step I: Add branch 2,, - 0.25 (from bus I (new) to bus r)

StepZBUS t}.25l

(i)

Add branch Zzt = 0.1 (from bus 2 (new) to bus I (old)); type-Z

V, = Z, r lo+ Vi

o r 21111 \212+ .. . + Z it (1,+ I )* Zi i ( I i - I ) + . . . +

(e.44)

zi,In= Z t l *+ Z, r l , + Z , r l "+ .. . + Z lU,+ I

earTangtng_ I

0 = ( Z i t - 2 1 ) r + . . . + ( Z t i - Z ) I i + e u - Z ) I i

+...+ (Z* - 21,) In + (Zu * Zii * Zii - Zi - Z) 11, e.45)

Collecting equations imilar to Eq. (9.43)and Eq. (9.45)we can writevl

V

vn

0

(9.46)

Eliminating 1o n Eq. (9.46) on lines similar to whar was donemodif icat ion,t follows ha t

zsvs=',lZ:11,31i] (ii)Step 3: Add branch ,r t = 0.1 (from bus 3 (new) to bus I (old)); type-Z

modification

lo.2s o.zs 0.2s1zsvs

|o.zs o.3s o.2s

|(ii i)

lo.2s 0.2s 0.3s..l

Step4: Add branch zz, (from bus 2 (old) to bus r); type-3 modification

10.2s o.zs 0.25'l [0.25-lZet)s

|o.zs 0.3s 0.2s |o.ts to.zs .3 s .251

Lo.2so.2s o3slo'3so

" Ln.trj

0.10420.14s8-l



l'"Z,,

with the useof fbur relarionships qs (9.36),(9.37), 9.42)and (9.47)busimpedancematrix can be built by a step-by-step rocedure bringing in onebranch at a t in le) as i l l t rst ratc ' tln l ixrrrnple .8 . This pror.cdrrr-c

-hcinga

mechanicalone can be easilycomputerized.When the networkundergoes hanges,he modiflcationprocecluresan be

cttrploycclo rcvisc hc bLrsrtrpcdancenatrixul 'thc nctwol 'k. hc opeling o1a line (Ztt) s equivalent o addinga branch n parallel o it with impedance- 2, , (seeExample9.8).

j Example .8

Fo r the 3-busnetwork

III

!,--0.25,'

(--IlI

Ref bus

F i g . 9.28

[0.14s8= I o.ro+z o.r4s8 o. lo42

fo.tott o.ro42o.24s8l

Step5: Acldbranch z- r= 0.1 (frombus2 (olct) o brrs31old)): type-4

modification

[o. l4s8 0.1042 0.l4s8l

zsvs lo.'oo, 0.1458o.to42l-| 0 .1 0 .14580 .2458 -2xO. lO4210.l4s8 O.tO42 0.24s8

[- 01042.l

=I

0.0417| [-o.to+z 0.0411 -0.0417]

[-o.o+tz-l

lo. r3e70.1103 0.12501

=|0.1ro3 o.I3e7 0.1250

I|0.1250 0.1250 0.17sol

Oltenirtg tinu l ine 3-2):This s equivalento connecting n ntpeclance0. I

betweenbus 3 (old) and bus 2 (old) i.e. type-4modification.

Zsus= uur(olcl)(-01)+ol?5

Z131,snew) = Zsuts old) -

lzit - z1t) .. (zi, - zi))

z b + z i i * Z i i - z z i j

Zsus

Ii,J .

I

shownn Fig.9.28buildr 9 l zI

' i r l

lL-,tr li -, -- 6"11-.

0. 13

o .t

filf.lid, ModernPower SystemAnatysis- r - r - ^ r r ^ . . t r Ana* ra io f f i r

:)ymmelr lual raul l rurqryo'sr**

Now

I- | 0.1042 0.14s8 0.1042l; (same s n step4)

0.1458 0.1042 0.2458

For the power systemshown n Fig. 9.29 the pu reactances re shown therein.For a solid 3-phase ault on bus 3, calculate he following

(a) Fault current

(b) V\ andvt

(c ) I t , r , I ' \ , ux l Il ,(d ) 1fr, and f,

Assumerefault oltageo be pu.

0.2 0-.09

a nd i l= (FZE- )=0 . 286i 2 - [ ^z u )

These wo voltagesare equal because f the symmetry of the given power

network(c) From Eq. (9.29)

and

[ 0.0147-lt l

|-0.0147

| 10.0147 0.0147 .0s001

L o.osoo_J

(d)

Iri = Y,j vl - vl)

Irtz= (0.286 - 0.286;gj0.1

I \ t= t | t=f r

ro.186-o)

- - i2.86

As per Eq. (9.32)



1

But

| f f r

Jt -

or 1f -

(b) As per Eq. (9.26)

rr.fv

i -

\,//

o '\ ,/'0.,

\ /\.. ,/. ,

l l3 ' '

Fig. 9.29

, r E' or-vrfr Gr

ixtic+ xr

E'Gr 1 Pu (Prefault o oad)

J I - F0 .1

c) | i-,,

IL.=1-0'286--=- j2 '86

Gt -io.2+io.os

SimilarlY

I fcz= i2.86

Solution The Theveninpassivenetwork for this system s drawn n Fig. 9.28with its Zru, given n Eq. (iv) of Example9.8.(a ) As pe r Eq . (9.25)

PROBEMS

A transmissionine of inductance .1 H ancl esistance ohms s suddenly

shortcircuitedat t =0 at the bar end as shown n Fig' P-9'1' Write the

expressionor short circuit current (r). Find approximately he value of

the first current maximum (maximum momentary current)'

[Hint: Assume hat the first currentntaximumoccursat the same ime as

the first current maximum of the symmgtricl.:nonttrcuit current')

V:

2, , + Zl

Yo- : ._1 - - - j s . j lzn j0.175

vl - --Zt--yu'

z r r + z l ' r

9 . 1

564,'l rr]logernowerSystemAnalvsisI

errrnrnorrinal Farrl t Analvsis IgeS

v y r r r r l v t " v E ' ' | - - " - " - -J

I

9 .3

i 0 . 1H SO

I' 64-A-'v^/v\

i

fl u=1OOin 100 + 15. )

Fig.p-9.1

9 2 (a ) What should he nstantof short circuit be in Fig. p-9.1 so that heDC off-set current s zero?

(b) what should he instantof short circuit be in Fig. p-9.1 so rhar heDC off-set current s maximum?

For the systemof Fig. 9.8 (Example9.2) find the symmetricalcurrents obe interruptedby circuit breakersA and B for a fault ar (i) p and (ii) e.For the system n Fig. p-9.4 the ratings of the variouscomponenrsare:

Fig.P-9.5

As y n c h r o n o u s g e n er a t o f r a t e d 5 00 k VA , 4 4 0 v , 0 . l pu s u b t r a n s i e n t

."u"'tun". is supptying a passive oad of 400 kW at 0.8 lagging powel

factor.Calculate fr" in'itiui symmetricalrms current for a three-phase ault

at generator erminals.

A generator-transformer nit is connected o a line through a circuit

breaker.The unit ratings are:

Generator: 10MVA, 6. 6 kV ; X,,d= 0. 1pu, Xi= o,2o pu an d

X,r = 0'80 Ptt

T r a n s f o r me r : 10 MVA, 6 . 9 1 3 3 k V , r ea c t a n c e 0 . 0 8 p u

The systems operating o loadat a ine voltageof 30 kv , when a three-

phase ault occurson ih " tin" just beyond he circuit breaker'Find

(a) the initial symmetrical rms current in the breaker'

9. 6

9.7

9.4

Generator:

Motor:

25 MVA, 12.4 kV, l\Vo subtransienteactance

20 MVA, 3.8 kV, l5%osubtransienteactance

TransformerT,: 25 MVA, 11/33 ky , gVoreactance

TransfbrmerTr: 20 MVA, 33/3.3 kV, I \Vo reactance

Line:



iui tt. maximum possibleDC off-set currenr n rhe breaker,

(c) the momentarycurrent rating of the breaker'

(d) the current,o U. intemrptedby the breakerand he intemrpting kVA'

20 ohms reactance

The system s loadedso that he motor is drawing15 Mw at 0.9 loadingpower factor, the motor terminal voltage being 3.1 kv. Find thesubtransient urrent n generatorand motor for a fault at generatorbus.lHint: Assumea suitable oltage ase br the generator. he voltagebasefor transformers, ine and motor would then be given by the transforma_t i0n i l t i< ls. rl r cxanrplc,f 'wc choosc crrcrator.o l tagc as cas I I kv ,the ine voltagebase s 33 kv and motor voltagebase s 3.3 kv . per unitreactancesre calculated ccordingly.]

T,1 T 2

Fig. p-9.4

Two synchronousmotors are connected o the bus of a large systemthrougha short transmissionine as shown in Fig. p-9.5. The ratings ofvanouscomponents re :

Motors (each): 1 MVA, 440 v,0.1 pu transient eactanceLine:

0.05ohm reacranceLargesystem: Shortcircuit MVA at its bus at 440 V is g.when the motors are operatingat 440 v, calculate the short circuitcuffent (symmetrical) ed into a three-phaseault at motor bus.

and

(e ) the sustained hortcircuit current n the breaker'

Th e systemshown n Fig' P-9'8 s delivering 0 MV A at

l a g g i r r g p 0 w 0r l . i t c t t tr i l l t t l l tb r r s w h i c h r t r i r y be r e g i r r d e c lParticulars f varioussystemcomponents re :

Generator; 60 MVA, 12 kV' X,/ = 0'35 Pu

Transtbrmerseach): 80 MVA, 12166 V' X = 0'08 pu

Line: Reactance 2 ohms' resistanc:negligible'

Calculate he syrnmetrical urrent hat the circuitbreakersA and B will

be calledupon to interrupt n the eventof a three-phaseault occurring at

F near he circuit breaker B'

9 . 811 kv , 0 .8

il s inf ini te.

9.5

Line

Fig. P-9.8

*Sf-,S tgdern power SvstemAnatvsist . -

9'9 A two generatorstaticlnuppliesa f'eederhrougha busas shown n Fig.P-9'9' Additional power is fed to the bus throJgh a transfoilner from a

largesystemwhich may be regardedas nfinite. A reactorX is included

:,'jT:^"1^:t::1u1.',f"'Ternd,1"-!T o imit heSC upruringapacityf

4v<rl\er rr ru JJJ rvtyA (Iault close to breaker). Find theinductivereactanceof the reactorrequired.system data are:

-bus 2 of the system.Buses 1 and 2 are connected hrough a transformer

and a transmission ine. Per unitreactances f the

variouscomponents are:

Generator connectedo bus bar 1) 0.25

Transmissionine 0.28

The power network can be represented i a geneator with a reactance

(unknown) in series.

With the generator n no load and with 1.0 pu voltage at each bus

under operatingcondition,a three-phase hort circuit occurring on bus 1

causesa currentof 5.0 pu to flow into the fault. Determine he equivalent

reactanceof the power network.

9.12Consider he 3-bussystemof Fig. P-9.I2. The generators re 100 MVA,

with transient eactance jVo each. Both the transforners are 100 MVA

with a leakage eactance f 5%t.The reactanceof each of the lines to abaseof 100MVA, 110KV is 707o.Obtain he shortcircuit solution or

a three-phase olid shortcircuit on bus 3.

Assume prefault voltages o be 1 pu and prefault currents o be zero.

GeneratorGr:

Generator Gr:

TransformerTr:

Transformer T2:

Assume hat all reactancesregiven on appropriate oltagebases. hoosea baseof 100 MVA.

ti Tz

25 MVA, 757o eactance

50 MVA, 20Voreactance

10 0 MVA;8Vo reacrance

40 MVA; l\Vo reactance.

Fig. p_9.9G)



9'10 For the three-phase ower networkshown n Fig. p-9.10,the variouscomponentsare:

T111 / 10 V

^T" Tf l rn kv\-x-x-x-"

2

the ratings of

j 0 . 1

Fig.p-9.10

Generators Gr : 100 MVA, 0.30 pu reactanceGz: 60 MVA, 0.18 pu reactance

Transformers each):50 MVA, 0.10 pu reactanceInductivereactorX: 0.20 pu on a base of 100 MVALines (each):80 ohms (reactive);neglect esistance.

with th enetwork nit iai ly unroaded nd a rine vortageof 11 0 kv , asymmetrical short circuit occurs at mid point F of rine r-.

calculate the short circuit MvA to be intemrpted by the circuitbreakersA and B at the endsof the line. what would thesevaluesbe, ifthe reactorX were eliminated?Comment.

Fault

Fig. P-9.12

9.13 In the system onfiguration f Fig. P-9.12, hesystemmpedance ata are

given below:

Transient reactance f each generator= 0.15 pu

Leakagereactance f each ransformer= 0.05 pu

Ztz =

i0.1,zp -

i0.12,223 70.08

PuFor a solid 3-phase ault on bus 3, find all bus voltagesand sc currents

in each component.

365 j ModernPowerSystemAnalysist

9.14 For the ault (solid) ocationshown n Fig. P-9.I4, find the sc currents nl ines 1.2 and 1.3. Prefault ystem s on no-loadwith 1 pu voltagean dprefaultcurrents arezero.Use Zuu, method and compute ts elementsby

the current njection technique.

'r-[b- 111 10 V 0.5pu reactance

-lQ9 0.1pu eactance

F ig. P-9.14

REF RECE

10.1 INTRODUCTION

In our work so far, we have considered both normal an d abnormal (short

c i rcui t) operett ions f power systernLlncler ornpletelybalanced (symmetrical)

l



Books

l . BIrrwtt.H.L).. Solttt iotr t . f .ar,g,e etwrtrk ty Mutr i r Mct lutds, Wi lcy, Ncw York,

tL)'7 .

2. f . l c t rct tswar tdu' ,. i t . , Mttt i t : t 't t ' tnv 'u ' 5 'vs ' /el l , l , r r tcr r tat i tt r taji 'cxtbook

eorr rplny,

New York, 1971.'3 .

Stagg, G.W. and A.H. El-Abiad. Computer Methods in Power SystemsAnalysis,McCraw-Hi l l B ook Co. , Ncw York, 1968.

4. Anderson, P.M., Analysis of Faulted Power Systems,Iowa State Press,Ames, Iowa,

1973.

( ' f r r t 'kc.l - , . . ' in ' t t i t Arrtt l . t , , t ' i , t. f 'A l t t rnul i t t ,qCtrrrcrt l l 't twrr S\,.r '/r,r l .r.ol . I,

Ne w York. 1943.

Stcvcrrson,W.D. Jr., Eletrrcnt.sf Powcr SystemsAnuly-sis,4th dn, Mc( i rz

Nc w Y ork , l 9 t l 2 .

Paper

"7. Brown. H.E. et al. "Digital Calculation o[ Three-PhaseShort Circuits by Matrix

Methods", AIEE Trani., 1960, 79 : 1277.

col t t l i t i o t ts .LJ l r r l c r uc l t o l tc t 'a t io t t l tc systcrl r ru l tcdanccsn caclr phasc ar eidentical and the thrce-phase ol tages an d currents hroughout th e system ar e

col l tplctc ly h l t l i tncc t l. .c . thcy huvc crpr i r lnurgrr i tu t l csn c l rch phusc un d tr cprtrgrcs.sivcly isplaced n ti rne phase by 120' (phaseu leads/laesphase b by120 ' an d phase r leads / l l gsphi tsec by 120") . n a b: r l l ncedsys tenr, na lys is

can proceed on a single-phasebasis. The knowledge of voltage and current inone 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 dueto unsymmetrical faul t , e.g. l ine-to-ground faul t or l ine-to-l ine faul t. These

f 'aul tsatrc, t t lhct, ol ' lnot 'econlnrol l occurrence+ hal t th e syrnntetric:al three-

phase) aul t. Systern operation may also become unbalancedwhen loads ar eunbalanced as in th e presenceof lar-eesingle-phase oads. Analysis under

unbalancetl conditions has to be carried ou t on a three-phase basis. Alterna-

tively, a nlore convenient method of analyzing unbalanced operation is through

symtnetricalcomponents where th e three-phase ol tages and currents) which

may be unbalanced are transfbrmed into three sets of balanced voltages (and

* Typical relative requencies f occurrence f differentkinds of faults in a power

syst(: l l tir r ordcr ol ' dccrcasing cvcrity)are:

- r .

6 .

Three-phase 3L) faults

Double line-to-ground LLG) faults

I)outrlc i rrc l -1,) i rul ts

Single ine-to-groundLG ) faul ts

5V o

lj%o

| 5t/o

7jVo

currents)calledsymmetricalcomponents. ortunately, n sucha transformation

the impedancespresentedby various power system elements(synchronous

generators,ransformers,ines) to symmetricalcomponentsaredecoupled rom

each other resulting n independentsystem networks for each component

anced set). This is the basic reason or the simolicitv of the svmmetrical

componentmethodof analysis.

TO.2 SYMMETRICAL COMPONENT TRANSFORMATION

A set of three balanced oltages phasors)Vo, V6, V" is charactertzed y equal

magnitudesand nterphase ifferencesof 120'. The set s said o have a phase

sequence bc (positivesequence)f VulagsVoby l2O" andV. lags Vuby I20".

The threephasors an^then e expressedn terms of the reference hasorVoas

Vo = Vo, V6 = a"Va, V, = aVo

where the complex number operator cr s defined as

sL - air20"It has the following properties

symmetrica!ompo1gnlg ! .a?.,{fi-i

above. hus

Vo-- Vot * Voz Voo (10.s)

b - Y b L - v b z

Vr= VrI * Vrz * Vro ( 0.7)

The three phasor sequences positive, negative and zero) are called the

symmetrical omponents f the originalphasorset Vo, V6,V,. The additionof

symmetricalcomponentsas per Eqs. (10.5) to (10.7) to generateVo, Vr, V, is

indicated y the phasordiagramof Fig. 10.1.

V61=crV61 V6fo?V11

, * 2 : e i 2 4 o ' : e - i l A o " _ *

(o ' )* : o

a 3 l( 10 .1 )



Fig.10.1 Graphicalddit ion f the symmetricalomponentso obtainthe setof phasorsV" , V6 ,V" unbalancedn general)

Let us now expressEqs. (10.5) o (10.7) n termsof reference hasors Voy

Vo2and Voe.Thus

V o = V o t * V o z l V o o

Vu = a .2Vor+ aVor * Voo

V" = c-Vol+ o2v o r * Voo

These equations can be expressed in the matrix form

l + a l a 2 : 0

(r0.2)

If the phasesequences acb (.negative equence),hen

Vo = Vo ,Vu= tuVo, r= &Vo

Thus a set of balancedphasors s fu'lly characterizedby its referencephasor

(say V,) and its phasesequencepositiveor negative).Suffix 1 s commonly used o indicatepositivesequence. set of (balanced)

positive sequence hasors s written as

Vo1, V61 &Vu1, Vrr = aVot

Similarly, suffix 2 is used to indicatenegativesequence.A set of (balanced)

negativesequence hasors s written as

Vo2, V62= dVn2, Vrz= Q'Voz (10.3)

A set of threevoltages phasors) qual n magnitudeand having he samephase

is said to have zero sequence. hus a set of zero sequencephasors s written

AS

Vng, V6g = Vo1, Vr1 = Vo1 (10.4)

Considernow a set of three voltages phasors)Vo, V6, V, which in generalmaybe unbalanced.According to Fortesque's heorem* the three phasorscan be

(10 .8 )

(10.e)

(10.10)

* The theorem s a generalone and applies o the case of n phasors 6],

ffi#d : ModernPo*etsytttt Analysi,__r_-

andIo = il'

whereI' = A-rI'

(10.19)

(10.20)

(10.21)

(r0.22)(r0.23)

(ro.24)

(ro.2s)

Vp = AV,

1,"1v, =

l'u |= vectorof originalPhasors

Lv ,

Iu' l% = | Voz | = vector of symmetrical components

Lu"'.1[ ' I r - l

A 1 " o nI

I a o 2 t - l

We can wrire Eq . (10.12)as

( 1 0 . 1 1 )

(10.12)

(10 .13 )

l ' l [ r , 'I r = l I u

l ; a n a, = l I o ,

IL I , ) L r , o _ i

Of course A and A-r are he sameas given earlier.In expanded orm the relations (10.19) and (10.20) can be expressedas

follows:

(i) Constructionof current phasors rom their symmetricalcomponents:

Io = Iot * Ioz * Io o

I a= o 2 l o t t d o z r l o o

I r= do t+ az l o r * I oo(ii) Obtaining symmetrical components of current phasors:

1I o t= * e"+ du+ o2 l r )

;t , r .= ( Io+ az lu+ aI , )



(r0.26)

V, = 4-'V,ComputinE ,{r and utilizing relations (10.1), we get

[ t d o ' ]

o - ' = * l r c r 2 I' L t r r lIn expanded orm we can write Eq. (10.14)as

IVo r

,<V"+c-Vu+ 2V S

vo z ! f r"+ ozvu+o%)3I

Vuo=;

V ,+ Vr , * r , )

(10.14)

( r0 .1s)

(10.16)

(10.17)

(10 .18 )

i

; .I r o = , Q o + u + I r )

Certainobservations an now be made egardinga three-phase ystemwithneutral return as shown in Fig. 10.2.

Equations 10.16) o (10.18) give the necessaryelationships or obtainingsymmetricalcomponents f the original phasors,while Eqs. (10.5) o (10.7)give the relationships or obtaining original phasors frorn the symmetricalcomponents.

The symmetricalcomponent transformations hough given above n termsofvoltageshold for any set of phasorsand hereforeautomaticallyapply for a set

of currents.Thus

Fig. 10.2 Three-phaseystemwithneutraleturn

The sum of the three line voltages will always be zero.Therefore, the zerosequence omponentof line voltages s always zero, .e.

ln

Vao V""

vobotrr**

vu,+v"o) o (10.27)

On the otherhand, he sum of phasevoltages line to neutral)'may not be zero

so that theirzero sequence omponentVn, may exist.

Since the sum of the three line currentsequals he current in the neutralwire. we have

(10.28)

i.e. the current n the neutral s three imes the zerc sequenceine current. f the

neutral connection s severed.

I o o = ! U " + 1 6 + r " ) = ! +

I o o = l r , = o3

t.e. in the absence of a neutral connection the

always zero.

B --_____}_----

Flg.10.3

Solution Io + I" * Is = Q

or10130" + l5l- 60o+ Ic = O

Ic = - 16.2+ j8.0 = 18 1154" A

(ro.2e)

zero sequence line current is

FromPower Invariance

We shall now show that the symmetrical component ransformation s power

invariant, which means that the sum of powers of the three symmetrical

componentsequals he three-phase ower.

Total complex power in a three-phase ircuit is given by

S = f p $ = 4 1 , + V u t t +V , t ! (10.30)

Eqs. 10.24)o (10.26)1

Itr= :0Ol3O" + 751(-60"+ l2O"5

= 10.35 j9.3 = 14142"A+ I8l(154" + 240"))

(i )

151(- 60o+ 240")+ I8l(154' + 120"))

4.651248" (ii)

IIez= ;Q0130' +

J

= --1.7 j4.3 =

_\



( 10 .31 )

or

s = [Av,,]'t'lAl,l.

= v! 'qr'q.I

Now

[ ro ? ' T t

o 1 6 . = l ta" t l l

a

[r r I JLa,. . s=31y t i ,=341 . ,

(10.33)

A delta connected alanced esistive load is connectedacross an unbalanced

three-phase upply as shown in Fig. 10.3. With currents n lines A and B

specified, find the sym metrical componentsof line currents. Also find the

symmetrical componentsof delta currents. Do you notice any relationship

betweerisymmetricalcomponentsof line and delta currents ? Comment.

/c ) = 0 (ii i)I

Iao= t^ (lo + IR +J

FromEq . (10.2)

Im= 141282"A

Inz= 4'6518"A

I a o = o A r ll -1

0 0ln ' . r l : r | o r o l - t , ( 1 0 . 3 2 )c - l l L 0 0 l l

Ic r = 141162"

Icz = 4.651128"A

I c o = o A

Check:

Ia= Iat * I ,rz* I,to= 8'65 + j5 = 10130"

Convertingdelta oad nto equivalent tar,we can redrawFig. 10.3as n Fig.

10.4.la

= 3V^1, + 3V"r!),+ 3V"oI),

= sumof symmetrical omponentowers

I Example10.1|T

$,$i.M Modernpowersystem anatysisI

Deltacurrents reobtained s ollows

Qrrmmatriaal l-amnnnanla tf f if f i

Positiveand negativesequence oltagesand currentsundergoa phaseshift inpassing hrough a star-delta transformer which depends upon the labelling ofterminals.Before considering his phaseshift, we need to discuss he standardpolarity marking of a single-phase ransformer as shown in Fig. 10.5. Thetransformer endsmarked with a dot have the samepolarity. Therefore, voltage

Vun, is in phasewith voltage V..,. Assuming that the small amount ofmagnetizingcurrent can be neglected, heprirnary current 1r, entering he dotted

end cancels he demagnetizing ampere-turns f the secondarycurrent 1, so thatI, and12with directionsof flow as ndicated n the diagram are n phase. f thedirectionof 1, is reversed, 1, and 1, will be in phase opposition.

veB=Io U^-

Ir)

I tB= AB)RtUo-

ru)

Similarly,

r n c=er - I r )5

Ice= frr - Io)5

Substituting he valuesof Io, Iu and Ir,

Ien=!

Oozzo"r5l- 60")

rnc=lOsz-

60" rllr54")

rcn= - rol3o")

we have

6186 A

= lO.5l- 41.5"A

8.31173"



!W2154"The symmetricalcomponents f delta currents are

1Iem= ;(6186" + IO.5l (- 4I .5 '+ l2O")+ 8.31(173"+ 240"))( iv)

J

= 8172" A

II,qaz= 6186" + 1051(- 41.5" 240") 8.31(173" 120")) v)J

= 2.71218"

Ieno=0

Incr, Ircz, IBC,,lsn1, ga2and 1.oo can be found by using Eq. (10.2).

Comparing Eqs. (i) and (iv), and (ii) and (v), the following relationship

betweensymmetricalcomponentsof line and delta currents are immediately

observed:

t 'IeBr=+130" (vi i )

V J

Ienz= \ z-zo" (viii)! J

The reader shouldverify theseby calculatrng Io', and *2from Eqs. (vii) and(viii) and comparing he resultswith Eqs. (iv) and (v).

Flg. 10.5 Polaritymarking f a single-phaseransformer\

Consider now a star/delta transformer with terminal labelling as ndicated inFig. 10.6 a). Windings shown parallel o eachother are magneticallycoupled.Assume that the transformer is excited with positive sequence oltages andcarries positive sequencecurrents. With the polarity marks shown, we canimmediatelydraw the phasordiagramof Fig. 10.7.The following interrelation-ship between he voltages on the two sidesof the transformer s immediatelyobserved rom the phasordiagram

VtBt= x Vabr 3V, -r - phase ransformationatio (10.34)

As per Eq. (10.34), he positive sequenceine voltageson starside ead thecorresponding oltageson the delta side by 30" (The same esult wo,'ld applyto line-to-neutralvoltages on the two sides).The same also applies or linecurrents.

If the delta side s connectedas n Fig. 10.6(b) he phaseshrft reverses the

readershoulddraw the phasor diagram); he delta side quantities ead he starsidequantitiesby 30".

(vi)

A __--______

ar System Analys'is Symmetricat omponentg| 3?9I

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

l-/\

o a

(a ) Star sidequanti t iesea d delta side quanti t ies y 30 o

(b) Deltaside quantities ead star side quantitiesby 30o

Fig. 10.6 Labelling f star/deltaransformervcea Vcrcz Vesz

Fig. 10.8 Negative equenceoltages n a star/deltaransformer

IO.4 SEOUENCE IMPEDANCES OF TRANSMISSION LINES



Fig. 10.7 Positive equenceoltages n a star/deltaransformer

Instead, f the transformerof Fig. 10.6(a) is now excited by negativesequence oltagesand currents, he voltage phasordiagramwill be as in Fig.10.8.The phaseshift in comparisono the positive sequence asenow reverses,i.e., the star side quantities ag the delta side quantitiesby 30'. The result for

Fig. 10.6(b)also correspondinglyeverses.It shall from now onwardsbe assumed hat a star/delta transformer is so

labelled that the po,sitivesequencequantities on the HV side lead their

Figure 10.9 shows the circuit of a fully transposed ine carrying unUalancecl

currents.The return path for 1,, s sufficiently away for the mutual effect to be

ignored. Let

X" = sell 'reactance1'eachin e

X. = mutual reactanceof any line pairThe fbllowing KVL equations an be written down from Fig. 10.9.

Vo V'o= jXJo + jX*Iu + .ixmlc

l r = l " + l o+1"

-(-

V6

Vec'l

vc

Fig.10.9

Modern PgwgfSy$gln Snalysis Svmmetrienl cnmnonent( tflIlFffi

T

t, Z1 12 22o-------{__]- -o

"----f--}-

---o

4 - r[: jxJo

l', - V!: iXJ"

or in rnatrix form

+ xh + x*["

+ xmlb+ xJ"

(10.3s)

(10.36)

(10.37)

(10.38)

(10.3e)

(10.40)

(a) Positivesequencenetwork

(b) Nagative equencenetwork

ls Zso---)--f---_l.-

@)Zerosequencenetwork

^ a

Ib

I"

vIvi

vb

v"I// -- - / n -

v ! ) :r// - , ' s -

V,r

= J x^x,x.x_x*x,

or A (I/, -

or v,

Now

A-I ZA :

0

X , -X*

0

Fig.10 .10

The decouplingbetween sequence etworksof a fully transposedransmis-

sion holds also in 3-phasesynchronousmachinesand 3-phase ransformers.This fact leads to considerable simplications in the use of symmetrical

componentsnethodn unsymmetricalaultanalysis.

In caseof three static unbalanced mpedances, oupling appearsbetween

sequence etworks and the method s no more helpful than a straight orward3-phaso nalysis.

10.5 SEQUENCE MPEDANCES AND SEOUENCENETWORKOF POWER SYSTEM

zIo

zuIs

A-IZAI,

jx,

JX^jx^

jx^

jx,jx^

l*, ,:J

L :



(r0.42)

(10.43)

(r0.44)

(10.4s)

(or negative)

z t 0 00 2 2 0

0 0 z o

wherein

Zr: j(X, - X,r) : positive sequence impedance

Zz: j(X, - X^) : negative sequence impedance

Zo: i(X, + 2X) : zero ,tequence impedance

We conclude that a fully transposed transmission has:/ t ; \ o ^ r r o l ^ ^ o . i + i " o ^ - l - ^ ^ ^ + i - - ^ : ^ - ^ ) ^ - ^ ^-\ r , , vyuar pvrr l ryu ( t r l r l u t / t s4Lrvg s t rqu{r r ruE rrup€ual lut is .

(ii) zero sequencempedancemuch larger than the positive

Power system slernsnfs-transmission lines, transformersand.synchronousnlachines-have a three-phase ymmetrybecause f which when'currentsof aparticularsequenceare passed hrough these elements,voltage drops of thesame sequence ppear, .e. the elementspossessonly self impedancos-o

sequence urrcnts. Each eleurentcan therelbre be representedby tlreedecoupled equence etworks (on single-phase asis) pertaining to positive,negative nd zerosequences,espectively.

MFs are nvolvedonly n a positivesequence etwork of synchronousmachines. or finding a particularsequenceimpedance,he element n question s subjected o currentsand voltagesof thatsequencenly.With the elementoperating nder hese onditions,he sequence

impedance an be determined analyticallyor through experimentalest results.

With the knowledge of sequence etworks of elements,completepositivel'

negativeand zero sequencenetworks of any power systern an be assembled.

As will be explained in the next chapter. these networks are suitably

interconnected o simulate different unsymmetrical faults. The sequence

currents nd voltagesduring the fault are hen calculated rom which actual

far"rlt urrentsand voltases can be found.

10.6 sEQuENCg I;IpTDANCES AND NETWoRKs oF

SYNCHRONOUS MACHINE

Figure10.11depicts n unloaded ynchronouslachine generatorr motor)

grounded hrougha reactor(impedanceZ).8o, E6andE, are he nducedemfs

Thus Eq. (10.37)can be written as

| 41 l'tr( Ix"-x^

l " l - l ' t: ' I oL r r o J l r r t J o

0 o ll-r,x, - x,,, o ll , | (lo.4l)

o x, *zx^)j,l

l[ /' .lll ,lLr,

sequencempedance it is approximately 2.5 times).It is further observed hat the sequencecircuit equations (10.42) are in

decouplecl bnn, i.c. thcrre rc no rnutual scqucncc nducternccs. )quation(10.42)can be representedn network form as in Fig. 10.10.

3S2 I _ Modern owerSystem natysis-]-

of the three phases.when a fault (not shown in the figure) takes placeatmachineerminals, urrents ,, , ,,and . flow in th e ines.whenever he aultinvolvesground,current n= In+ Iu +

^I" lows to ne'tral from ground"i^I^'.

negative and zero sequenceculrents, respectively. Because of windingsymmetry currents of a particular sequenceproduce voltage drops of thatsequence nly.Therefore, here s a no coupling between he equivalentcircuitsof various equences*.

uence voltage.Hrvvvvs yyrrrr r( rLr l l 4rr .1IyJIs ( \_napler i l) , we must

know the equivalentcircuitspresentecly the *u.hin" to the flow of positive,

Symmetrigel_qg4pe!e$s

the hortcircuit occursirrnrkradccl ondit iorts,hc voltagt: ehindappropri l tc

reactancesubtransient,ransient or synchr:onous) onstitutes he positive

Figure lO.IZa shows the three-phase osi t ive sequencenetwork ntodel of a

synchronousmachine.Z, doesnot appear n the model as Iu = 0 for positive

sequenceurrents.Since t is a balancednetwork it can be represented y the

single-phaseetwork model of Fig. 10.12b or purposesof analysis.The

reference us for a positive sequence etwork is at neutral potential.Further,

since o current lows from ground o neutral, he neutral s at ground potential.

la t> a

(I

II

(a)Three-phasemodel

la

-*)e "

.n'-----( t +\=\'-..

t:.6 d-n>_- u\ t6

-- > -----b

L----- ____l_"Fig'10'11 Three-phaseynchronousenerator it hgrounded eutral

ln't, b

lc't

Reference us

(b ) S ingle-phase odel



Fig. 10.12 Posit iveequence etwork f synchronous achine

With referenceo Fig. 10.12b, he positive sequence oltageof terminal c

with respect o the referencebus is given by

V , , l = E , , - Z l l , , l (10.4e)

Negative Sequence Impedance and Network

It hasalreadybeensaid ha ta synchronousnachine aszeronegative equence

inducedvoltages.With the flow of negativesequence urrents n the statora

rotating ield is createdwhich rotates n the oppositedirection to that of the

positivesequenceield and, herefore,at doublesynchronous peedwith respect

to rotor. Currents at double he stator requencyare herefore nduced n rotor

field and damper winding. In sweeping over the rotor surface, he negative

sequence mf is alternately resentedwith reluctances f direct andquadrature

axes. The negative sequence mpedance presented by the machine with

consideration iven to the damperwindings, is often defined as

Positive Sequence Impedance and Network

Since synchronousmachine s clesigned ith symmetricalwindings, t inducesemfsof positivesequence nly, .e. no negatrve r zero equence oltages reincl t rcedn it ' Whcn th c tt t l tcl t incurr ics osi t ivc cqucnccurr .cntsl ly, th isntodeof operation s the balancedmoclediscussedailength in Chapter9. Thearmature eaction field caused by positive sequencecurrents rotates at'synchronouspeed n the salneclirection s the ,otu., i.e., t is stationarywithrespect o field excitation. The machine equivalently offers a direct axisreactance hose value reduces rom subtransicnt eactance X,a) to transientreactanceXtr) and inally to steadystate synchronous) eactanJe Xa),as theshortcircuit transientprogressesn time. If armatureresistance s assumednegligible,he positive sequencempedanceof the machine s

21= jXtj (i f I cycle transients of interest)

= jX'a Gf 3-4 cycle transient s of interest)

= jXa (if steadystatevalue is of interest) xt: + x,!

Z . t = ; l Z 2 l < l Z r l2

(10.46)

(10.47)

(10.48)

If the machine short circuit takes place from unloaded conditions, theterminai oltage onstituteshe positive equence oltage; n the otherhand. f Negativesequence etworkmodelsof a synchronousmachine, n a three-

phase nd single-phaseasisare shown n Figs. 10.13a ndb, respectively. he

referencebus is of courseat neutral potential which is the same as ground

potential.

(10.s0)

*'fhis can be shown to be so by synchronousmachine theory, 5].

fl$4.:rl MooernowLrSystem narysist -

From Fig' 10.13b he negativesequence oltageof terminal a with respecrto referencebus is

Voz= - Zzloz

symmetricatomponentsFjSiffi

Zs = 3Zra Zo s (10.53)in order or it to have hesame oltage rom a to reference us.Thereferencebus

here s, of course, t ground otential.From Fig. 10.14bzero sequencevoltage of point c with respect o the

Voo= - Z{oo (10.54) .

Order of Values of Sequence Impedances of aSynchronous Generator

Typical valuesof sequencempedances f a turbo-generatorated5 MVA, 6.6kV, 3;000rpm are:

Zr = lZ%o subtransient)

Zr = 20Vo transient)

Zr = 7l0Vo synchronous)

Zz = I2Vo

Zo = 5Vo

For typical valuesof positive, negativeand zero sequence eactances f asynchronousmachine efer to Tablu 9.1.

(10.s1)

la za

(b) Single-phase model

Flg.10.13 Negativeequence etwork f a synchronous achine

Zero Sequence Impedance and Network

we state once again that no zero sequencevoltages are induced in asynchronousmachine.The flow of zero sequence urrents creates hree mmfswhich are in time phase but are distributed in space phase by 120". Thetesultant air gap field caused by zero sequencecurrents is therefore zero.Hence, he rotor windings present eakage eactanceonly to the flow of zerosequence urrents(Zos< Zz < Z).



IO.7 SEOUENCE IMPEDANCES OF TRANSMISSIO\ LINES

A fully transposed hree-phaseine is completely symmetrical and thereforetheper phase mpedanceoffered by it is independent f the phase sequenceof abalanced etof currents.n otherwords, he mpedancesffired by ii to positiveand negativesequence ulrents

are dentical.The expression oi its per phaseinductive reactanceaccountingfor both self and mutual linkages as beenderived in Chapter2.

When only zero sequence urrents low in a transmissionine, the currentsin each phaseare identical n both magnitudeand phaseangle. part of thesecurrents return via the ground, while the rest return throu h the overheadground wires. The ground wires being groundedat several owers, the returncurrents n the groundwiresare not necessarily niformalong the entire ength.The flow of zerosequenceurrents hrough he transrnissionln"r, groundwiresand ground createsa magnetic ield patternwhich is very different from thatcausedby the flow of positiveor negativesequence urrentswhere he currentsh , r . r o ^ ^ L ^ ^ ^ - J : f f ^ - ^ - ^ ^ ^ f t n or rqYv (r

Prrd'Ds' urrrsr tr I lutr UI IL V an( l tne fetufn Cuffent lS ZefO. T\e ZefOsequence mpedance of a transmission ine also accounts or the ground

impedance zo = z.to+ 3zri.since the ground mpedance eavily depends nsoil conditions, t is esseniial o make somesimplifying assumptionso obtainanalytical results.The zero sequencempedanceof transmissionines usuallv

/66

Reference bus

(a)Three-phasemodel

Ia o-a

(b )S ing le -phase mode l

1o.14 Zero equence etwork f a synchronous achine

Zero sequence etworkmodelson a three-and single-phase asisare shownin Figs. 10.14a nd b. In Fig. r0.l4a, the cu'ent flowing in the inpedance nbetweenneutraland ground s In = 3lno.The zerosequence oltageof t"r-inula with respect o ground, he referencebus. s therefore

Vr1= - 3znioo Zorlno=,r, (32, + Z0)loo e0.52)wh11eZo, is the zerosequencempedanceper phaseof the machine.

Since he single-phaseero sequenceetworkof Fig. 10.14b arriesonlyperphasezero sequence urrent, ts total zero sequencempedancemust be

3Sd I ModernPowerSystemAnatysis

rangesromZ to 3.5 imes he posit ive equencempedance*. hi s rat io s onthe higherside or doublecircuit ines without groundwires.

10.8 SEOUENCE MPEDANCES AND NETWORKS OF

It is well known that ahnosl al l presentday installationshave three-phasetransformerssince they entail lower initial cost, have smaller space equire-mentsand higherefficiency.

The positive sequence eries mpedanceof a transformerequals ts leakageimpedance.Sincea transformer s a staticdevice, he leakage mpedancedoesnot changewith alterationof phasesequence f balancedapplied voltages.Thetransformernegativesequencempedance s also thereforeequal o its leakagereactance.Thus. for a transformer

Zt= Zz= Z r "uk ug "

Symmetria

ter" -"*"- -"-t--t"

transformers

the zero'sequenceetworksof various ypesof transformer

importantobservations re made:

currentonly if there s cunent florv on the secondary ide.

(ii) Zero sequence urrents an flow in the legsof a starconnectiononly if

the star point is groundeclwhich provicies he necessary eturn path for

zero sequence ulrents.This fact is illustratedby Figs. 10.15aand b.

Beforeconsidering

connections,hree

( 0.55)

-': o

__----, a

{->

(a) Groundedstar

Assumingsuch ransformer onnectionshat zerosequence unentscan lowon both sides,a transformeroffers a zerosequencempedancewhich may differslightly from the corresponding ositiveand negativesequence alues. t is,however,normal practice o assume hat the series mpedances f all sequencesare equal regardlessof the type of transformer.

The zero sequencemagnetizing urrent s somewhat igher n a core typethan in a shell type transformer. This difference does not matter as the

r ' l

/co=

0 --f-'

/oo=

(a )Ungroundedta r

Fig. 10.15 Flowof zero equence urrentsn a starconnection

o a delta

/ao= 0



magnetizingcurent of a transformer s always neglected n short circuitanalysis.

Above a certain rating (1,000 kvA) the reactanceand impedanceof atransformerarealmostequal and are hereforenot distinguished.

*We can easily compare he forward path positive and zero sequence mpedances fa transmissionin e with ground eturnpath nfini telyaway.Assume hateach ine hasa self inductance, and mutual inductance M between any two lines (completelysymmetricalcase).The voltagedrop in line a causedby positive sequence urrents s

VAnt= uL lnr+ uMI , , r+ aMI , .1

-[uL+ (& + a) aMllo, = a(L-tu|)Ior

Positive sequence eactance= a(L- fuI)The voltage drop in line a causedby zero sequence urrents s

VAno= aLloo+ utMluo+ uMlrs

= a(L + 2M)InoZero sequenceeactance= w(L + 2W

Obviously,zerosequenceeactances muchmore hanpositivesequenceeactance.This resul thas already ee nderived n Eq . (10.45).

(iii) No zero sequencecurrentscan flow in the lines connecte.d

connection sno returnpath s availableor these ulrents.Zero sequence

currents can, howevet, flow in the legs of a delta-such currents are

caused y the presence f zerosequenceoltagesn the deltaconnection.

This fact s i l l t rst ratedY Fig. 10.16'

/ a o = 0 a

Fig.10.16 Flowof zeroSequenceurrentsi' , delta onnection

Let us now considervarious ypes of transformerconnections.

Case ; Y-Y transformerbank with any one neutrttlgrounded.

If any one of the two neutrals f a Y-Y ransformers ungrounded, ero sequence

currents annot low in theungrounded tarandconsequently,hese annot low

in the groundedstar.Hence,an opencircuit exists n thezero sequence etwork

betweenH and L, i.e. between he two parts of the systemconnectedby the

transformer s shown n Fig. 10.17.

Zg

---tll d---L

Y-Ytransformeran kwithon eneutral roundednd ts zerosequence etwork

Case2: Y-Y transforrnerbank both neutralsgrounded

When both the neutralsof a Y-Y transformerare grounded,a path through hetransformerexists for zero sequencecurrents n both windings via the twogroundedneutrals.Hence, n the zerosequence etwork 11andL are onnectedby the zero sequencernpedanceof the transformeras shown

n Fig. 10.1g.Case3: Y-A transformerbank with grounded y neutral

Referencebus

ffi

(seeFig. 10.19). f the star neutral s grounded hrough 2,, an impedance3Zn

appearsn series with Z, in the sequence etwork-

Casb 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 sequencecurent can flow in the transformer

windings.The zerosequence etwork then modifies o that shown n Fig. 10.20.

Reference us

"ifr6LH 2 ;

Fig. 10.20 Y- l transformeran kwithungroundedta ran d ts

zerosequence etwork

Case5: A-A transformer bank

Sincea delta circuit provides no return path, the zero sequence urrents cannot

flow in or out of A-A transformer;however, it can circulate in the delta

{ L

----------o

L



o---4TdL=-oH Z s L

Fig. 10.18 Y-Ytransformeran kwithsequenceetwork

neutrals rounded nd ts zero

Referencebus

windings*. Theretore, here is an open circuit between H anE L and Zo is

connectedo the reference us on both ends o account or any circulating zero

sequence urrent in the two deltas (seeFrg. 10.2I).

Fig. 10.21 A-Atranstormeran kan d tszerosequence etwork

10.9 CONSTRUCTIONOF SEOUENCENETWORKS OF A

POWER SYSTEM

In the previous sections he sequencenetworks for various power system

elements-synchronous machines, ransformersand lines-have been given.

Using thesq,complete sequencenetworks of a power system can be easily

constructed.To start with, the positive sequence etwork is constructedby

*Such circulating currentswould exist only if zero sequence oltagesare somehow

induced in either delta winding.

d-E-;1 ---_.

^--r--I

Fig.10.19 Y-1 ransformeran kwithgrounded neutralan d ts zerosequence etwork

If the neutralof starside s grounded, ero sequence urrentscan low in star

because path is available o ground and the balancing zerosequence urrentscan low in delta.Of courseno zerosequenceurrents an flow in the ine onthe deltaside.The zerosequence etwork must herefbrehavea path from theline 1{ on the starside through the zero sequencempedance of the transformer


examination f the one-linediagrarnof the system. t is to be noted hat positivesequenceoltages representn synchronous achinesgcneratorsndmotors)only. The transition i'orr-r ositil 'e sequence etwork to negativesequencenetwork s straightforward, ince the positiveand negativesequencernped-

,) \ r r r r v Dq r t u L t o t r J l t r t t t l v t J r r , r r r v L r l t l J v r r a r r

neclessaryn positivesequence etwork to obtain negatit'esequence etworkin respect of synchronousmachines. Each machine is representedbyneg;ttive equencempedance,he negative equence oltagebeingzero.

The referencebus or positive and negativesequence etworks s the systemneutral.Any impedance onnected etween neutraland ground s not ncludedin thesesequence etworksas neither of these sequence urrentscan flow insuchan impedance.

Zero sequencesubnetworks or various parts of a system can be easilycombined o form completezero sequence etwork.No voltagesources representn the zerosequence etwork. Any impedance ncluded in generator r

transformerneutralbecomeshree irnes ts value n a zero sequence etwork.Special care needs o be taken of transforners n respect of zero sequencenetwork.Zero sequence etworks of all possible ransformer connections avebeen dealt with in the preceding section.

The procedure for drawing sequence networks is illustrated through the

following examples.

- -:---- '

ffiMVA base n all other circuits and the following voltagebases.

Transmissionine voltasebase= 11 x 121 - 123.2 Y10.8

Reactance f motor I = 0.25 x = 0.345 u

Motorvoltagebase 123.2 ++ = 11 kV121

The reactancesf transformers,ine andmotorsareconvertedo pu values

on appropriateases s ollows:

Transformereactance0.1x =. f++) = 0.0s05 u30 \ 11

Line reaclloo x 25:ance=f f i

=o .164pu

2 s / 1 0 \ 2- x t - |1 5 \ l l l

The

Reactance f motor 2 - 0.25 x ?1, rf -Lo)t = 0.69 pu

7 . 5 \ l l /

requiredpositivesequence etwork s presentedn Fig. 10.23.



f-"--rExample 0. 21t _ _ _ _ _ _ : _ - l

A 25 MVA, 1l kV, three-phaseenerator as a subtransienteactance f 20Vo.

Tl rcgcrrcr i t toruppl icswo rr totors v0r ' r , rransnr issioninc wiLh ral ts lo l l r rc l 'sat both endsas shown n the one-linediagramof Fig. I0.22. The motorshaverated nptrtsof 15 and 7.5 MVA, both 10 kV with 25Vo ubtransienteactance.

The hree-phaseransformers re both rated30 MVA, I0.8/L2I kV, connectionA-Y with leakage eactance f 70Vo ach.The series eactance f the line is100 ohms. Draw the positive and negativesequence etworks of the systemwith reactances arked n per unit.

e

Referencebus

IEs ( )

+ l

ioz: ,

I

Enrt( )

+ lI

.r: !

i o345 : r

i ) Eara

l +

--,1

i i o6e

t

p u910'' (-t, 'r '

. ( 2 ) ,q \ - / r \'

Motor ,,1

I

I t srt - o

IYA

. P -----r--'--- 9

d ,-66X-L-g ,.I-6XU -, "xfi r- - Ini oo8o5 j0'164 i 00805

Fig. 10.23 Posit iveequence etworkor Example 0.3

Reference us

i 0.0805 j4164 loo8o5

Fig. 10.24 Negativesequencenetwork or Example10.3

Since all the negative sequence reactances of the system are equal to the

positive sequence eactances, he negative sequencenetwork is identical to the

Fig. 10.22

Assume that the negative sequence reactance

subtransienteactance.Omit resistances.Select

generator ircuit.

of each machine is

generator rating as

equal o its

base n the

' rti..;;:1 . ::i.l

3,42'l l-4odern ov"'ei' vstemAnalvsis

positivesequence etwork but for the omissionof voltage sources.The negativesequenceetwork is drawn in Fig. 10.24.

For the power systemwhoseone-line diagram s shown in Fig. 10.25,sketchthe zero sequence etwork.

F ig.10 .25

Solution The zero sequence etwork is drawn in Fig. 10.26.

Reference bus

T21

32n

symmetricatompblents ffi------_-1

Zerosequenceeactancef motor - 0.06" +. (i+)'

= 0.164 u

Reactancef current imiting reactors''.1.i3t = 0.516pu(1 )' ,

Reactancef current imiting reactorncludedn zero sequenceetwork'= 3 x 0.516 - 1 .548 u

Zerosequenceeactance f transmissionine = ry:21(r23.D2

= 0.494pu

Thezero sequenceetwork s shown n Fig. 10.27.

j1-548

io06

Reference us

j1.i l8

jo164



Zrt Z6(line) 272

Fig. 10.26 Zero sequence network of the system presented in Fig. 10.25

i - - i - - - -- _ - * - ;

1Example10.4|*-"1

Draw the zero sequence etrvork for the systemdescribed n Example 10.2.Assumezerosequenceeactancesor the generator ndmotorsof 0.06per unit.Current imiting reactors f 2.5 ohms eachare connected n the neutralof thegeneratorand motor No. 2. The zero sequence eactanceof the transmissionline is 300 ohms.

Solution The zero sequence eactanceof the transfonner is equal to itspositivesequenceeactance.Hence

Transformerzerosequence eactance 0.0805pu

Generator ero seqllenceeactances: 0.06 pu

Zero sequenceeactance f motor 1 0.06 x

Zost

Fig. 1O.27 Zero sequencenetworkof Example10.5

PROBEIvIS10.1 Compute he following in polar orm

(i) o2-t 1ii; I- a -"?

(iii) 3 d + 4cy+ 2 (iv) ja

10.2 Thlee identical esistorsare star connected nd rated2,500 V, 750 kVA.This thre.e-phasenit of resistors s connected o the I sideof a A-Ytransformer.The following are the voltagesat the resistor oad

lVo6l 2,000 Y; lVu,l= 2,900 V; lV,ol= 2,500V

Choosebaseas 2,500 V, 750 kVA and determine he ine voltagesandcurrents n per unit on the delta side of the transforrner. t may be assumedthat the load neutral is not connected o the neutral of the transformer

secondary.

10.3 Determine he symmelrical components f three voltages

Vo= 2001ff, Vt = 2001245" and V, = 2001105' y

26(l ine)

: 0.082

PowerSystemAnalysis

10.4 A single-phaseesistiveoad of 100kVA is connected cross ines bc of

a balanced upplyof 3 kV. Compute he symmetrical omponents f the

line currents.

10.5 A delta connected esistive oad is connectedacrossa balanced hree-

pnase upply

250 )B

Fig.P-10.5 Phase equence BC

of 400 V as shown n Fig. P-10.5.Find the symmetricalcomponentsof

line cunentsand deltacurrents.10.6 Three resistances f 10, 15 and 2O ohms are connected n star acrossa

three-phase upply of 200 V per phase as shown in Fig. P-10.6. The

supply neutral s earthedwhile the load neutral s isolated.Find the

currents n each oad branchand the voltageof load neutral aboveearth.

Use the methodof symmetricalcomponents'la

"-"*' " Generator: 25 MVA, 11kV, Xtt = ZOVo

Three-phaseransformer each):20 MVA, ll Y1220Y kV, X = I5Vo

The negativesequence eactance f each synchronousmachine s equal

machine s 8Vo.Assume that the zero sequence eactances f lines are

25OVo f their positive sequenceeactances.

X = 5o/oat machine,l rating at machine 2 rating

Fig.P-10.8

For the power systemof Fig. P-10.9draw the positive, negativeand zero

sequence etworks.The generators nd ransformersare ated as follows:r0.9



I r

t B c 1 t b

I-te

_L__ _l1--ls

F ig.P-10.6

10.7 The voltagesat the terminalsof a balancedoad consistingof three20

oh m X-connectedesistors rc 2OO4O",00 1255.5'and 20 0 llsf V.

Find the line currents from the symmetrical componentsof the line

voltagesf theneutralof the oad s isolated.What elationexistsbetween

the syrnrnetricalcomponentsof the line and phase voltages'/ Find the

power expanded in three 20 ohm resistors from the symmetrical

components f currentsand voitages.

10.8.Draw the positive, negativeand zero sequencempedancenetworks for

the power systemof Fig. P-10.8.Choosea baseof 50 MVA, 220 kV in the 50 0 transmissionines,and

mark all reactancesn pu. The ratingsof the generators nd transformers

are:

Generator:2 5 MVA, 11 kV, Xt t=0.2, X2 = 0.15,Xo = 0.03pu

Generator: 15 MVA, 11 kV, Xt =0.2, Xz= 0.15,X0 = 0.05pu

Synchronous otor 3: 25 MVA, 11 kV, Xt = 0-2, Xz= 0-2,Xo = 0.1 pu

Transformer : 25 MVA, I 1 Lll20 Y kV, X = IIVo

2: 12.5MVA, 11 LlI20 Y kV, X = l07o3: 10 MVA, I2O Ylll Y kV, X - IjVo

Choosea baseof 50 MVA, I I kV in the circuit of generator .

Fig.P-10.9

Note: Zero sequence eactance f each ine is 250Vo f its positive

sequenceeactance.

'0.:-"t

T1

\7

j,,W voo"inpo*",,syrt"r Rn"tu"i,

10.10Consider he circuit shown n Fig. P-10.10.Suppose

Vo, = L00 l0

vbn = 60 160

X r = 1 2 Q

X o b = X b r = X * = 5 d )

Fig.P-10.10

16, nd 1, without using symmetricalcomponent.

16,and 1" using symmetrical component.

REFERECES

II.I INTRODUCTION

Chapter 9 was devoted to the treatment of symmetrical (three-phase) aults in

a powersystem. ince he system emains alanced uringsuch aults,analysis

could convenientlyproceedon a single-phase asis. n this chapter, we shall

(a) Calculate Io,

(b) Calculate Io,



Books

l. wagncr,c.F. andR.D.Evans, ymmetrit:al omponenfs,McGraw-Hill,Ncw york

, 1933.

2.Clarke,E., Circuit Analysisof Alternating Current Power Systems, ol. 1. Wiley,Nc w York, 1943.

3. Austin Stigant,5., MasterEquations nd Tables or SymmetricalComponent ault

Studies,Macdonald, ondon, 1964.

4. stcvenson, .D., Elcments J' owerSy.stemnalysis,4thedn,McGraw-Hill,New

York, 1982.

5. Nagrath, .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

Polyphase etworks',AIEE, 1918,37: 1O27.

deal with unsymmetrical aults. Various types of unsymmetrical faults that

occur n power systems re:

Shunt Wpe Faults

(i) Single ine-to-ground(LG) fault

(ii) Line-to-line (LL) fault(iii) Double line-to-ground(LLG) fault

Series Type Faults

(i) Openconductor one or two conductorsopen) ault.

It was stated n Chapter 9, that a three-phase3L) fault being the most severe

must be used o calculate he rupturing capacity of circuit breakers,even though

this type of fault has a low frequency of occurrence, when compared to the

unsymmetrical aults isted above.There are, however, situationswhen an LG

fault can causegreater fault current than a three-phase ault (this may be so

when the fault location is close to large generatingunits). Apart fiom tliS,

unsymmetrical ault analysis is important for relay setting, single-phase

switching and systemstability studies Chapter12).The probability of two or more simultaneous aults (cross-country aults) on

a power system is remote and is therefore gnored in system design for

abnormalconditions.

*g9,S. Modernpower SystemAnatysis

The method of symmetricar clmponents resented n chapter 10 , is apowerful tool for study of unsymmetrical aults and witl be fully exploited nthis chapter.

UNSYMMETRICAL FAULTS

Considera general power network shown in Fig. 11.1. It is assumed hat ashund ype ault occursat point F in the system,s a resultof which currenmIo, 16,, flow out of the system,and vo, v6, v"are voltagesof lines a, b, c withrespect o ground.

Fig.11.1 A general owernetwork

Unsymmetrical ault Analysis

Fig. 11 2 Sequenceetworks s seen Fig. 11.3 Thevenin quivalentsf hefrom he aultpointF sequence etWorkss seen

(a )Vat

(b )(b )

(c)(c)

lao

Vao

_l



Let us also assume that the system is operating at no load before theoccurrence of a fault. Therefore, the positive sequencevoltages of allsynchronous achineswill bc dentical nd will equal he prefaultvoltageat F.Let this voltagebe labelledas Eo.

As seen rom F, the power systemwill presentpositive,negative and zerosequence etworks,which are schematically epresented y Figs. Il.2a, b and

c. The reference us is indicatedby a thick line and he point F is identifiedoneachsequence etwork. Sequence oltagesat F and sequence urrents lowingout of the networks at F are also shown on the sequence etworks. Figures11.3a,b, andc respectively, ive theTheveninequivalents f the hreesequencenetworks.

Recognizing hat voltageEo s presentonly in the positivesequence etworkand hat here s no couplingbetween equence etworks, he sequence oltagesat F can be expressed n terms of sequence urrentsand Thevenin sequenceimpedances s

from he aultpointF

Dependingupon the type of fault, the sequence unentsand voltagesare

constrained, eading to a particular connection of sequencenetworks. The

sequencecurents and voltages and fault currents and voltages can then be

easilycomputed.We shall now consider he various ypesof faults enumeratedearlier.

11.3 STNGIE LINE-TO-GROUND (tG) FAULT

Figure 11.4showsa line-to-groundault at F in a powersystem hrougha fault

impedanceZI. The phasesare so labelled hat the fault occurs on phase a.

Fig. 11.4 Single ine-to-groundLG ) aul t at F

1r",1 t"1 t, 0 ol|-r",'l

l r " , l = l o l - l 2 2 o l l , * lLv"o)Lo l Io o zo ) l r "o )( 1 1 . 1 )

At the ault point F, the currentsout of the power systemand he ine to groundvoltagesare constrainedas follows:

UnsvmmetricalaultAnatvsisfffiH

r = E o'a t(z r* zz* z) +3zr

Fault current /o is then given byu=o

I r = 0

vo = zllo

The symmetrical componentsof the fault currents are

Expressineq ,r':l; ,';',.*1;,ij;.*.ar componenrs,ehaveVot * Voz * Voo= ZfIo = 3zflor

As per Eqs. (11.5) and (11.6) all sequence urrentsare equaland the sumof sequence oltagesequals3zf lot Theiefore, heseequationssuggest seriesconnectionof sequencenetworks through an impedane 3zf ut rrio"*n in Figs.11.5a nc i .

(z r* zz* zo )+3 2The above resultscan also be obtaineddirectly from Eqs. (11.5) and (11.6)

by using Vop Vo2 nd Voe rom Eq. (11.1).Thus

(Eo- IolZ) + (- IorZr) + (- I"&d = 3Zf lot

I (4+ 4+ Zo )+ 3zf l l o r= Eo

Ior =(zr * zz -t Z) +3zr

The voltage of line b to ground under fault condition is

Vu= &Vor+ aVo2* Voo

=o,2n' +)* -',+).-"+)Substituting or /o from Eq. (11.8) and reorganizing,we ger

3rl2r + Z2(&- a)+ zo(&- D

(rr.2)

1 .3 )

(11 .4 )

(11 .5 )

(11.6)

. ( 11 .7 )

Eo



la,t=az t"o=t h

l"s= ,,1=t t,

(Z r *22+Zo)+3Zt

be similarly obtained.

( 11 .e )Va = Eo

The expression for V, can

/a \(b )

Fig' 11'5 Connectionf sequence etworkor a singleine-to-groundLG ) ault

In termsof the Theveninequivalentof sequence etworks,we can write fromFig. 11.5b.

Fault Occurring Under Loaded Conditions

When a fault occursunderbalancedoad conditions,positivesequence urrentsalone flow in power system before the occurrence of the fault. Therefore,negative nd zerosequenceetworksare hesanleaswithout oacl.The positivesequence etwork must of course carry the load current.To account for loadcurrent, the synchronousmachines in the positive sequencenetwork arereplacedby subtransient,ransientor synchronous eactancesdependinguponthe time after the occulTence f fault, when currentsare o be determined) andvoltagesbehind appropriatereactances.This changedoesnot disturb the flowof prefault positive sequence urrents(seeChapter9). This positive sequencenetwork would then be used n the sequence etworkconhectionof Fig. 11.5afor computing sequence urrentsunder fault

In case he positivesequence etwork is replacedby its Theveninequivalentas n Fig. 11.5b, he Theveninvoltageequals he prefaultvoltageVi atthe aultpoint F (under loadedconditions).The Thevenin mpedances th6 impedancebetweenF and the reference us of the passivepositivesequenceetwork(withvoltagegeneratorsshortcircuited).

i '4ot ', ,1 ModernPo@is

fni, i, illustratecl y a two machine ystem n Fig. 11.6. t is seen iom thisfigure that while the prefault currents flow in the actual positive sequencenetwork<lfFig. 11.6a,he samcdo not exist n its Theveninequivalent etwork

of Fig' 11.6b.Therefore,when the Theveninequivalentof positivesequencenetwork is used fot calculatlng fault eurre+ts, he positive sequeneeeurrentswithin the network are thosedue to fault aloneand we must superimpose nthese the prefault currents.Of course, he positive sequence urrent nto thefault is directly the correctanswer, he prefaultcurrent nto the fault beingzero.

(c)

Fig. 11.6 Posit iveequence etwork nd tsThevenin quivalenteforeoccurrencef a fault

The above emarksare valid for the positive sequenienetwork, ndependentof the type of fault.

t7.4 LrNE-TO-LrNE (LL) FAULT

(b )

from whlch we get

Io2= - Io l

1.,0= 0

The symmetrical components of voltages at F under fault are

Writing the

3voz= v, + (t r + ,l ) vo ull Iu

from which we get

3(va- voz) (a - &1zrtu iJi y' to

Now

Iv,,f , t a o'f l ' 'Il r " , l = i l t 2 " l l r ,

LV"o ) Lr 1 I lL " , z r ra )

first two equations, e have

3vot= vo + (a + &) vu - &zf tu

( 1 1 . 1 1 )

( 1 1 . 1 2 )

( 1 1 . 1 3 )

( 11 . 1 4 )

16 = (& - a) Io t (: Ioz= - Io t , 1oo= 0) .r



Figure I 1.7 showsa line-to-line ault at F i n a power systemon phases andc througha fault impedanceZf The phases an alwaysbe relabelled,such hatthe fault is on phasesb md c.

b

l6

Fig. 11.7 Line-to-l ine Lt) faul t hrough mpedanceZ/

fault can be expressedasThe currents and voltagesat the

[ 1 " : oI

I o =l I u

lI ,: -I r

The symmetrical omponents f

Il ,t '

_ithe

Vo- Vr= IoZf

fault currentsare

( 1 1 . 1 0 )

= - j J t I " r ( 11 .15 )

Substitut ing6 rom Eq . (11.15) n Eq. (11.14),we ge t

Vot Voz=Zf Iot ( 1 1 . 1 6 )

Equations11.11)an d (11.16) uggest aral lel onnection f posit ivean d

negative equence etworks hrougha series ntpedance as shown n Figs.11.8a nd b. Since loo= 0 as pe r Eq.(11.12),the zero sequence etwork s

unconnected.

Vaz

_1laz

Fig. 11.8 Connectionof sequencenetworks or a l ine-to-l ineLL ) aul t

In terms of the Thvenin equivalents, we get from Fig. 11.8b

F

la t

(11 . 1 7 )

t

FromEq . (11.15), e ge t

- jJi n,

4 + 2 2 + Z l

\ .

Knowing ,-trwe arl alsqld{e ,,, and, orfromwhichvsltages t he ault,

I ' - - ! - J ' - u ou L

Z , * 2 " + z ft + 2 2 +

(1 .18 )

be found.

If the fault occurs rom loadedconditions, he positive sequence etworkcanbe modified on the ines of the aterportion of Sec. 1r.3.

11.5 DOUBLE LrNE-TO-cRouND (Lrc) FAULT

Figure 11.9 showsa doubre ine-to-groundault at F in a power system. hefault may in generalhave an impedanceZf as shown.

tra - aY / u = 9

b -- , . . \ .

_.__ |r" l

t' o z! t'"o ___,__r:]tr"

Fig. 11.9 Double ine-to-ground.LLG) ault through mpedanceZl

The current and voltage (to ground) conditions at the fault are expressed as

Unsymmetrical ault Analysis I SQt*-_1

or

Voo= Vo t * 3Z f In o ( 1 .23)

From Eqs. (11.19), l l .22a) an d (11.23),we can draw the connection f

sequenceetworksas shown n Figs. 11.10a nd b. The readermay verify this

by writing mesh and nodal equations or these igures.

(b )

Fig. 11.10 Connection f sequencenetworks or a double ine-to-ground(LLG) fault

IIVssI

+IVaz

: 1laz

Ea

Z1



( 1 .19 )

( r 1 .20)

I o = o lot

i" , + Io , + I"o =o |

V,,= V, = Zf (lt, + Ir) = 37rf ,,r,

The symmetricalcomponentsof voltagesare given by

l ! : '1 [r a "' ] l r . ll ! : ' l= l a2 o l lvu"v,, ,)l_ l

I r l l ,ur)fiom which i t fol lows rhat

v,,t= V.z= Llv,, + (a + r11V,,13 " '

v,,o=1""

+ 2vu)

From Eqs. (11.22a) nd (t I.Z2b)

voo-vot=tr ,

- ,r - &1 vu = vt = 3zfloo

(rr.21)

( I l .22a)

(rr.22b\

ln terms of the Thevenin equivalents, we can write from Fig. 11.10b

I . =E o

, l - @

En (11 .24)zt+ z2(zo 3z I ) (2 2 zo+ 3z t

The above esult can be obtainedanalyticallyas follows:

Subst i rut ingor Vut , Vuz and V,* in termsof E, , in Eq . (11.1) an d

premultiplyingboth sidesby Z-t (inverse f sequencempedancematrix), weget

l t ; 'o l [ r , , -zt rut

II o z;' o lln,-2,r", II o o z;' l lE,, 2,r,,,3z f,,nl'=17'

;, ilt?l';"1I o o z o t ) L o ) r , o _ J(rr.2s)

From Eq . (11.22a), e have

Eo - Z t lo t=- Zz loz

Substitut ing loz= - (Ior + Io ) fseeEq . (11.19)]

Eo - Zr lot= Zz(Ior Ios)

or

I ^ =E o - ( z t + z ' \ '' u o -, - l , " 1 " '

Substituting his valueofIoo n Eq. (rr.26) and simplifying, we finally get

f" l

=

If the fault takesplace rom loadedconditions,hepositive sequence etworkwill be modifiedas discussedn Sec. 11.3.

r- -- ----- -l

406t'l Modern powe1_syglgll4!g!yg!s

1-.gltiitiUyingboth sidesby row marrix tl 1 1l and using Eqs. (11.19)and

(1I.20), we ger be more than the three-phaseault current.

Fig. 11.12 Sequence etworks f synchronouseneratorroundedhroughneutralmpedance

(d) Write expression or neutralgrounding eactance,current s less han the three-phaseault current.

Solution (a) Figure 11.12 givesthe se-quencenetworksof the generator.As stated

earlier voltage source is included in thepositivesequence etwork only.

(b) Connectionof sequence etworks or asolid LG fault (ZI = 0) is shown in Fig.11.13, rom which we can write the faultcurrentas

\O<

lx,?il .

Negative

such hat the LG fault



ie+tflP.f;;!utrl,,r|

Figure 11.11 showsa synchronous eneratorwhoseneutral s grounded hrougha reactancaXn.The generatorhas balancedemfs and sequence eactances 1,X., and Xu such that X, = Xz > Xo.

xnEa

----l'FJf[\--| "" " .-- \" /: : / )+)- )

b

c

Fig' i i. i i Syne ronous enerator roundedhrough eutral eactance

(a) Draw the sequence etworksof the generator s seen rom the terminals.(b ) Derive expressionor fault current or a solid line-to-ground ault on

phasea.

ll)rc - 3 lE , l

zx t+ & , *3 X , ,(i)

(c) If the neutral s solidly grounded

llolLG

==

3lE" l

2 \ + x cFor a solid three-phaseault (seeFig.

IEo l _ 3 l E " ll I ) r r = ; : : t

Comparing ii) and (iii), it is easy o

llol Lc > l lo l3L

An important observation s madehere that,when the generator neutral is solidlygrounded,LG fault is more severe han a 3I_fault. It is so because, Xo * Xr = X, ingenerator.However, or a line Xo D Xt = Xz,

so that for a fault on a line sufficiently awayfrorngenerator, L fault will be ntoreseverethanan LG fault.

lqs= 131.

F ig . 11 .13 LG fau t t

nE " ( ) I

f t l

\ ll

Xt'4 I

7 iL _ _

Fig. 11.14 Three-phase

fauft

(ii)

1 . 1 4 )

(iii)

see hat

usqelrl@(d) with generatorneu,tralgrounded hrough reactance, omparingEqs. i)

and (iii), we have for LG faurt current to be less than 3L fault

3 lE , l24 + Xo+3X,

2X , + Xo + 3Xn> 3X l

*^ , I(xr - xo )J

Two 11 kV, 20 MVA, three-phase, tarconnected enerators peraten parallelassht rwn n Fig' l l '15, th c posi t ive, egat ivc nd-zeroequenceeactancesfeachbeing, espectively,0.1g, 0.r5,7o.10 pu . Th e starpoint of on e of thegeneratorss isolatedand hat of the other s earthed hrougha 2.0 ohm resistor.A single line-to-ground ault occursat the terminalsof one of the generators.Estimate(i) the faurt current, (ii) current n groundingresistor, and (iii) thevoltageacrossgrounding esistor.

408

(iv)

i;;;ili",i:rl

- a . 1 . - . 4 s ^ & g . . . - ' r } . } . * ' * . 3 a - - - } l ! - } - } . } ' Q t p t |

Unsymmetrical aultAnalysis l' i

tltrffi3 x 1 f

J

= 0.965 = 6.13 V

For hesystemof Example10.3 he one-linediagram s redrawn n Fig. I 1.16.Ona baseof 25 MVA and I 1 kV in generator ircuit, the positive,nega"tiuendzerosequence etworksof the systemhavebeendrawn already n Figs. 10.23,10.24and 10.27-Before the occurrenceof a solid LG at bus g, the motors areloaded o draw 15 and 7.5 MW at 10 kV, 0.8 eading power factor. fiprefaultcurrent s neglected, alculate he fault currentand subtransient urrent n all

(a ) Ir =f -:

J - ' A A A r : A A 4 E , . n rj 0.0e 0.07 + 0. 1 0.9e 0.99 j0.26s

= 2.827 j0.756

(b ) Cunent n the grounding esisror= If = 2.g27_ j0.756

llrl = 2.926 x -:L- = 3.07 kA" J 3 x l l

(c ) Voltageacross rounding esisror= * e.g2i - j0.756)12r

- 0.932 j0.249



F ig . 1 .15

Solution (Note:Al l valuesare given n per unit.)Since the two identicargeneratorsoperate n paralrer,

Xr.o= = i0.0g,Xr"q= -'# = j0.075

Since he starpoint cf the second enerators isolated,ts zerosequencereactanceoes ot comento picture.Therefore,

Z o " q = 7 O . 1 03 R n = j 0 . 1 0 + 3 x4 : + = 0 . 9 9 + 0 . 1( l l ) ' , r -

Fo r an LG faul t, using Eq . (11.1g), we ge r

3E

j 0 . 18

Iy (fault current or LG fault) - Io = 3lo, =Xt"o+ Xz"qlZo",

partsof the system.What voltage behind subtransient

sequencenetwork il prefault current

Xo= 0.06pu Tz e

(H+----2 . 5 A - d ) ( | Xo = 300 fl

Fig.11.16 one-r ine iagram f hesystem f Exampre1. 3

solution The sequence etworks given in Figs. 10.23, 10.24and r0.2i areconnectedn Fig. 11.17 o simulatea solidLG fault at bus g (seeFig. 11.16).[ f n re for r l t n r r f fan fo ara nan lo^+^. i

vur rv r r lD qr v r rvS l vvL( / \ I

E'l= E',!,t= E',1,2 vj (prefaultvoltage at g)

=1+=o.eoeu

reactancesmust be used in a posi t iveis to be accounted fbr'/

. l ' '.E"o(

)fl02:l

( io'szsdt rb-6T--___

410 I Modern owerSystem nalysisI

I

l- io.t+z

------+i0.136-i0.136

*_-- - j0.311

.447

i 0.608 11.712

I -p.oo,

Fig. 11.17 Connection f the sequencenetworksof Example11.3.

Subtransienturrents re shown n he diagramn pu ora solid

--'>

0.99 i0.607

. UnsymmetricalauttAnatysir mblT-

Zz= Zr = j0 .16pu

Fiom hescquenceetwork onnection

Vff _ J' o r

7 3 7 ; 4

= q'901=- jo.Mlpuj2.032

Ioz= Ioo= Ior = - j0.447 pu

Faultcurrent 3loo= 3 x (- j0.447)= - jL341 pu

The componentof Io, flowing towards g from the generatorside s

j0.447 !: ?:= = - 70.136uj0.7ss

and ts componentlowing towardsg from the motorsside s

- jo.Ml i,?s?-s=- j0.311uj0.7ss

Similarly, the componentof Io2from the generatorside is - j0.136 pu andits componentrom the motors side s -70.311.Al l of Iosflows owards fromtnotor2.

Fault currents rom the generator owards g are



l ine-to-groundaultat g

The positive sequence etwork can now be easily replacedby its Thevenin

e q u i v a l e n tss h o w n n F ig . 11 .18 .

Reference busT

r

Now

I r, 1 [ r I rl [-ro.r30l -i0.2721l l l . l l l l

I r u l = l a z G r l l - i o . r r 6l = l 7 0 .1 3 6p uLr .J lo u2 rJL o j L jo . r36 j

and to g fiom motorsare

[ t . . ] [ t I l l f - ro . l t t1 [ - . r l ,06elI t o l - a 2 a r l r o . s ul : l - i o . r 3 6 l p ul " l l ^ l l " l l ' l 'L/,J la e" I)L-j0.447JL- j0.136_l

The positiveand negativesequence omponents f the transmissioninecunents are shifted -90" and +90o respectively, from the correspondingcomponentson the generatorside of Tr, i.e.

Posit ive equenceurrent= - j(-jO.136) - - 0.136 puNegative sequence urrent - j(- j0.I36) = 0.136 puZero sequence urrent = 0 ('.' thereare no zero secluenceurrents

on the transmissionine, seeFrg. l.17)Line a current on the transmission ine

= - Q . 1 3 6 + 0 . 1 3 6 + 0 = 0

Iu and I, can be similarly calculated.

4l:iN Modernpower SystemAnatvsis

Let us now calculate he voltagesbehind subtransienteactanceso be usedif the load currentsare accounted or. The per unit motor currents are:

25x0 .909x0 .8 136.86= 0.66+ j0.495pu

25x0 .909x0 .8= 0.4125 36.86'=0.33+ j0.248pu

Total cuffent drawn by both motors = 0.99 + j0.743 puThe voltages behindsubtransienteactancesare calculatedbelow:

Motor:r'^'=llt- :';:: "ri;Yilii.il"

Motor:E!"2= .l:t-,';:::iir:tfi il r"

Generator,E' { = 0.909+ j0.525 x I.2375136.86"

= 0.52+ j0.52 = 0.735145. pu

It may be noted hat with thesevoltagesbehindsubtransienteactailces,heThevcnin equivalentcircuit will still be the same as that of Fig. 11.19.Therefore, in calculating fault currents taking into account prefault loadingcondition, we need not calculate EIy E/ft and E(. Using the Thevenineqtrivalent pproach,we can irst calculatecurrents auiedby fault to which theload currentscan thenbe added.

UnsymmetricataultAnglysislll4#hF

Ft u at a l -

x*q* Xr", i0.09+ i0.075

UsingEq . (11.15),we have

1y fault current) = Io = -i

Now

= _ 6.06

3X- j6 .06 ) : -10 .496

Vot = Voz = Eo - Iorxleq = 1'0 - (- j6.06)

= 0.455

V o o = - I ^ / ' o - g

Voltage of the healthy phase,

Vo= Vo t * Vnz * Voo= 0.91

u0.0e)

(" ' /oo = 0)

3orExample 11.2,assume hat the grounded enerators solidly grounded.Find

.he ault current n eachphaseand voltageof the healthy phase or a doubleine-to-ground ault on terminals of the generator.Assumesolid fault (Zf - 01.)olution Using Eq. (1I.24) and substituting he valuesof Zp* Zr"rand Z*,ro mExample l .2,we ge t (noteZf =0 , Z0"q=j0.1)

t + 7 0I _r a l -

' ' r -- - ; 7 < ?

10.075x 0.10



Thus, the actual value of positive sequencecurrent from the generator. ^ - - . ^ - l ^ r l - - f ^ - - l r ! -

ruw:lrus ule Iault ls

0.99+ -j0.743 -j0.136= 0.99+ j0.607

and heactual alue f positive equenceurrentrom hemotors o the ault

l s ' -0.99 -j0.743 -j0.311= - 0.99 -jr.054

In this problem, because f large zero sequenceeactance, oad current scomparablewith (in fact, more than) the fault current. In a large practicalsystem,however, the reversewill be the case,so that it is normal practice oneglect oad current without causingan appreciableerror.

For Example I1.2, assumehat the grounded generators solidly grounded.Findf } r o f ^ " 1 + n r r * a * # ^ - J , , ^ l r ^ - ^ ^ f e L ^ l - ^ ^ t r r ^ - - - r - - - - r r . r a .r r rv rcr lr rL t- tursrr l auLr vurLi l$tr ul t t l9 lr€art l ly pl lase IO f a l fne-to- l lne laUlt On

terminalsof the generators. ssume solid fault (Zf - 0).Solution For the LL fault, using Eq. (11.I7) and substituting he valuesofX,"uand Xr"u rom Example11.2,we get

iio.oe.: Yj: J: :l: j0.075 0. i0

Vot= Vo2= Vo1= Eo- Ior Z r .q= 1 - (- j7 .53) ( /O.09)

= 0.323

_ V o z -

Zr.,

Voo

Zo.,

0.323I _r a 2 -

t -ta 0 -

- j4.306j0.07s

0.323 .^A^= t 1 / 1

j0 .10

Iu= rl lot + alo, + Ioo

= (-0.5 -/0.866) (-j7.53)+ (-O.5+ _i0.866) 4.306) j3.23

= - 10.248 j4.842 - 11.3341154.74'

Ir = e l , r , + oz lor* In o

= (-0.s + 70.866) -j7.s3) + (-0.5 -j0.866) (j4.306)+ j3.23

4!4" 1 ModernPowerSystemAnalysisI

- 10.248+ j4.842 - 71.334125.28"

Voltage of the healthy phase,

Vo= 3Vat= 3 x 0.323= 0.969

11.6 OPEN CONDUCTOR FAULTS

An open conductor fault is in serieswith the line. Line currents and series

voltagesbetweenbrokenendsof the conductorsare required o be determined,

lcc i c '

Fig. 11.19 Currentsnd voltagesn openconductorault

Figure 11.19 shows currentsandvoltages n an open conductor ault. The ends

of the system on the sides of the fault are identified as F, F', while the

conductor nds are dentifiedas ua ,bb and cc'. The setof series urrents nd

voltagesat the fault are

l - l I f v . l

Unsymmetrical aultAnalysis|

,415

Posit ive equence

network

Negativesequencenetwork

Fig. 11.20 Sequence networks or open conductor aull at FF/

[n terms of symtnctrical cotnponents, we can wrlte

Zero equencenetwork

Vonl * Voo,2* Vno,g = O



l ' , l ' " " ' lr - l t , l : v : l v . . , lp

l - , 1 ' ' P Io o ' l

L / , L V , , ,

The symmetricalcomponents f currents and voltagesare

[ . ' 1v,,,,,,1t. = | ^. : v" l v^-,"" l " ' l | " " ' l

L/.,, lVou'o

The sequence etworkscan be drawn for the power systemas seen rom FF/

and are schematicallyshown n Fig. ll.2O. These are o be suitablyconnected

dependingon the type of fault (one or two conductorsopen).

Two Conductors Open

Figure lI.2l represents he fault at FF /with conductors

currentsand voltages due to this fault are expressed s

V oo '= 0

1 6 = I r - Q

b and c open. The

(rr.27)

( 11 .28 )

F I F '

I

I

I

c " c 'I

Fig 11.21 Two conduc tors pe n

' t D

I o t = o 2 = n o -+ 1 "

Fig.11.22 Connectionf sequencenetworksor wo conductorsopen

Onc Con-luctor Open

1V o o l = V o o r 2 = V o o r y = * V o o ,

J

I o t * I n + I o o = 0

Equations 11.33) an d (1I.34) suggest paral lel

networksasshown n Fie. 1I.24.

tt'6 il Modern ower ystem natysist

Equations 11.29)an d (11.30)suggest seriesconnection f sequence

networksas shown n Fig. II.22. Sequenceurrelrtsand voltages an now becomputed.

For one conductoropen as n Fig. 11.23, he circuit conditions equire

V b b , = V r r , = O ( 1 1 . 3 1 )

Io = O (11.32)

In termsof symmetrical omponentshese onditionscan be expressed s

( 11 .33 )

(11 .34 )

connection of sequence

prefault voltage Vf_. of bus r in series with the passivepositive sequence

network al l voltage ources hortcircuited).Sincenegative nd zerosequence

prefault voltages are zero,both these are passivenetworks only.

Referencebusfor passivepositivesequencenetwork

Fig.11.25 Connection f sequence etworksor LG faulton he r th bus (positiveequence etworkrepresentedy itsThevenin quivalent)

[--l



la

F

a

t c ' , c /, c + -

i

Fig. 11.23 On e conductoropen

la o

Fig. 11.24 Connectionof sequence

networks or one conductoropen

II.7 BUS IMPEDANCE MATRIX METHOD FOR ANALYSIS OFUNSYMMETRICAL SHUNT FAULTS

Bus impedancemethod of fault analysis, given for symmetrical faults in

Chapter , canbe easilyextendedo the caseof unsymmetrical aults.Considerfbr example an LG fault on the rth bus of a n-bus system. The connectionofsequence etworks o simulate he fault is shown n Fig. I1.25. The positivesequence etwork has been replacedhere by its Thevenin equivalent, ,e.

It may be noted hat subscripta has beendropped n sequence urrents and

voltages, while integer subscript s introduced for bus identification. Super-

scripts o and respectively, indicate prefault and postfault values.

For the passivepositivesequence etwork

Vr-"us= Zr-nus r-"utwhere

V t - u u s posit ivesequenceus voltagevector (11.36)

Zr-nus

and

- positive sequence us mpedancematrix

/ 1 1 2 ? \\ L r . J t )

bu scunent nject ion ector (l1.38)

( 1 1 . 3 5 )

Z-t r l

: lZt-nn )

[/'-'l t t . ' | = pos i t iveequencel : l

l

t lI V""'z I

L-'i-- ' ryl

Jr-sus=

t4l8 | todern power SvstemAnalvsisI

Thus the passive ositiveseguence etworkpresents n impedanceZr_ r, to theposit ive equenceurrent {_r.

For the negativesequence etwork

Vz-uus Zz_sus z_nus (11.41)

The negativesequence etwork s injectedwith current lfr_, at the rth busonly. Therefore,

nly at the

(1 .3e )

sequence

(11.40)

As per the sequence etwork connection,current IJr_, s injectedofaulted rth bus of the positive sequence etwork, we have therefore

subst i tut ing q. (11.39) n Eq. (11.35),we can wri re th e posi t ivevoltageat the rth bus of the passivepositive sequence etwork as

V,-, '= - Zr-rrl f r-,

0

0

-,i{ (1r.42)Jz-sus

Fromthe Sequenceetworkconnection f Fig. 11.25,we can now

- rf t 2 - r

vro-, (rr.47)

2r-,,* zz-,, Zo-r,+3zf

Zl-rr, Zr-r, and Zo-n

other types of faults can be simila*Seomputed using

in place f Zr , Zrand Zoin Eqs. 1I.7), (11.17)an d

write

ue

(11.24) ith E, - Vi-,.

Postfaultequenceoltagest anybuscannowbecomputedy superposing

on prefault bus voltage, he voltage developedowing to the injection of

appropriate equence urrentat bus r'.

Foi passive ositive^sequenceetwork, he voltagedeveloped t bus owing

to the injection of - IIr-, at bus r is

Vt- r=- Zr - ,J f r - ,

Hence ostfaultpositive equenceoltage t bus I is givenby

V l - ,= V i - , - Z r - , , f r - , ; = l ' 7 ' " ' ' t t

where

prefaultpositivesequence oltageat bus i

Zr-,, = irth component f Zt-"ut

Since he prefault negativesequence us voltages

negative equence us voltages re given by

V f ' - ' = 0 + V z - r

-- -

are zero, the postfault

( 1 1 . 4 8 )

( r.49)

( r . s0 )



0

The negative sequencevoltage at the rth bus is then given by

Y r , = - z r r r l f , I

Thus, he negative equence etworkoffersan mpedance r_rrto the negativesequenceurrent tr_,

Sirnilarly, br the zero scqucnce etwork

Vu-uu,= Zo-susJ,,-",r, ( 1 1 4 4 )

Jo-sus

0

0

-r{ u- r

0(r 1.46)

Zrr_,.,.o the zero

( l 1 .43 )

( .4s)

and Vo_, = - Zs_,.rlfs_,.,

That is, the zero s.equencenetwork off'ers zrn intpeclancesequence cuffent l '-*r.

zr - , r l ' f r - ,

where

l r - , , = i r t l t co l l l po l l c l t tol ' Z t - t , , t

Similarly, the postfault zero sequencebus voltages are given by

Vd- ' = - Zu- ' ' l fu - ' ' ; j = l' 2' "' ' tI

where

= irth component tl f Z9 - sLr.

With postfaultsequenceoltages nown at the buses, equenceurrents n

l inesca n be comPr'rteds:

For ine uv , havingsequence drnittarlcesl -ur,Jz',u, nd yo-r,

f , - r r = l t - u , ( V f t - u v [ - r )

I f r - r r= !2 - , , r ,V t - , , V5- r , )

I i , r -ur= Jo-u, Vio-, - Vfo-r l

Knowing sequenceoltages ndcurrents, hase oltages ndcurrents anbe

easilycomputedby the use of the symmetrical omponentransformationVr, = AV"

Ir, = AI ,

( l . s1 )

( r.s2)

'i[di;,lModernpowSrJy$ern_4nglysis

It appears t first, as f this method s more aborious han computing aultcurrents rom Thevenin impedancesof the sequencenetworks,as it requirescomputationof bus mpedancematricesof all the threesequence etworks. tmust,however, e pointedout here hat once he bus mpedancematriceq avebeenassembled,ault analysiscan be convenientlycarriedout for all the buses,which, in fact, s the aim of a fault study.Moreover, bus impedancematricescan be easilymodified o account or changesn powernetworkconfiguration.

For Example 10.3,positive, negativeand zero sequence etworkshave beendrawn in Figs. 70.23, 10.24 and 10:27.Using the bus impedancemethodoffault analysis, ind fault currents or a solid LG fault at (i) bus e and (ii) busI Also find bus voltages and line currents n case (i). Assume the prefaultcurrents o be zero and the prefault voltages o be 1 pu.

Solution Figure 11.26 shows the connectionof the sequencenetworks ofFigs. 10.23, 10:24and 10.27 or a solid LG fault at bus e.

I PositiveequenceE"*

anetwork

10.345

Unsymmetricalault narvsisliffiffi

l _+_+^- =_ t j .422r-aa=

io.z io.o'os

Yr-re-Yr-a"=to;;t

: jr2-4zz

Yr-n=t_,,=#*

.# - -i78.s1e

Yr-"f-*h :

i6.0s1

Y . *I

+I = - i 1 6 . 7 6 9- t-88 j0.085 j0.345 j0.69

vr I -BUS_ vI2_B US _

vr o-dd-

d e f

-t7.422 12.422 0t2.422 - 18.519 6.097

0 6.097 - 1 .5 9

0 0 12.422

I

00

12.422

-16.769

I

j1.608= _ j0.62I

1



Negativesequencenetwork iO'345

@ @

i 0.6e

i0.0805

Zero sequencenetwork

JU.4v+ /U.UUUs v/

Fig. 11.26 Connectionf the sequenceetworks f Example 1 6 or an LG

fault t bu s e

Refer o Fig. 11.26 o find the elements f the bus admittancematricesof thethreesequence efworks.as follows:

@{@

@ @ v - Y , - - - = - i 1 4 . 4 4 6' o - e e - ' t t - t t -j 0 . 0 8 0 5

'j 0 . 4 9 4

1Y r - r r =

j t h = - i o ' 5 8 4

Yo-ar= 'o

-. 1Yo-,r=jofu

= 2.024

Yo-fs= 0.0

o6

0f0

d e

-0.621 0

vI O_BUS0 -r4.M6 2.024 0

0 2.024 -14.446 0

0 0 0 -0.584

lnverting the threematricesabove enders he fbllowing threebus mpedance

matrices

orl

" r r l

Analysis

Th e fault currentwith LG fault on

i l=

The fault curent with

0

0.07061

0.00989

0

bu sc is

3 x 1j 0J 7636 j 0.r7 36 j 0.07 61

LG faulton bus'i s

_ _ 3 x 1.i0.t82gg iOlAZgg /O"O?Gi

0

0.00989

0.07061

0

0

0

0

r .71233

= - j7.086pu

a)

(i)

Irr

= - 76.871 u

Bu s voltages nd line currents n case i) ca nEqr. (11.49)-(t I .sZ). Given below is a samplevoltageat bus f and current n line efFrorrrEq . (11.49)

j0.436s9

(ii)

easi ly be computcdusingcalculationfor computing

- - 0.417 u

VL" = - Zu,,Ifo-, = - j0.0706 (- j2.362)- - 0.167 u

Vfz-r=- 4-r" l l r1 =- j0.08558 x( - j2.362)

= - 0.202pl

V f u s = - Z u r J f u " = O

Using Eq. (11.52), the currents in various

cornputed s ollows:

II+= Yr-p I+- vI+,

- - j6.097(0.728 0.s84)

= _ 70.88If ,-0"= Yr4" (V f ,- a

- Vf ,-r)

= - j12.422 0.703 0.584) - jL482

I,,t= If t-f" * If t-,t,=- 70.88 + (- ,t l .482)

- - i2.362

which is the sameas obtainedearlier seeEq. (i)l where If, = 3lut.

= -

parts of Fig. 1I.26 can be



VI-a= Vi_a- Zr_0"- I fr_"

= t. 0 - j0. t2s7s(- 7'0ttb) = 0.703 u

\ " 3 )

Vft- t=Vi- t - Z, tu - If , ,

= r. o jo.ns4j(-r,orru

)= o.zzs ,

VJ r- "= V"r-" - Z r_"o- l Jr_ "

= 1. 0 i0.17638- j2.363)= 0.584 u

vI-r= vi-s - z,-r"-I{-,

= 1. 0 j0.08558- j2.363) 0.798 u

vfz-f= - zr-.fJfr-"

= - ./0.11547(- j2.362)= - 0.272 u

Vfo-r= Zo-f"Ifo-,- j0.00989 (- j2.362)

= _ 0.023 u

IJrsf YFcf (vf '-, vf '-)

- j12.422 (-0.798 0.728)= - i0.88

Notice thatas per Fig. 11.26, t was required o b e 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.29Ipu

Ittn (a) = IJr-f" * It)-r, + {*

= - j0.88 + (- 70.88)+ (- j0.291)

- - jz.os

Sirnilarly,othercurrentscan be computed.

A single ine to ground fault (on phasea) occurson the bus I of the systemof

Fig. 11.27.Find

qU;l Modern owerSystem natysis

Ftg. 11.27

(a) Current in the fault.

(b) sc current on the transmission ine in all the threephases. :(c) SC current n phasea of the generator.

(d) Voltage of the healrhyphasesof the bus 1.Given: Rating of each machine 1200 kvA, 600 v with x, = x, = rTvo,xo = 5vo.Each three-phaseransformer s rated rz0o kvA, 600 v - nlgrooV-Y with leakageeactance f SVo, he reactances f the ransmissionine are

xr = Xz = 20voandXo = 40voon a baseof 1200kVA, 3300 V. The reactancesof the neutral grounding reactors are5Voon the kVA and voltage baseof themachine.

Note: Use Z"u, method.

Solution Figure 11.28shows the passivepositive sequence etwork of thesystem f Fig.l1.27. This also epresentshe negativesequence etwork for thesystem.Bus impedancematricesare computedbelow:

I

UnsymmetrlcalaultAnalysis f,r'4l#T

i t-0.105 0.0451zr-ws Lo.o+so.1o5.J:

z-.sus

sequencenetwork of the system s dr aw n in Fig. II.29 and its bus

matnx ls ted below.

(i )

Tero

0 .15

0.05

0 . 1 5

0.05

0.05 0. 4

Fig.11.29

Bus 1 to ref'erence usZo-sus= i [0.05]

Bus 2 to bus I

0.05

l-0.0s

Referenceus

l-0.05 0.051Zo-nus/Lo.os 0.451

Bus 2 to referencebus

7zO-BUS -

0.051 ; l-0.051o.4s l



'Lo.os

.1[0.04s 0.0051

Zo-susJZIO.OO50.0451

0.05

Bus I to referencebus

Zt_svs= j[0.15]

Bu s 2 to Bus I

0. 2 0.05

F lg. 11 .28VO

fr_,

Zr-8r , .= fo '15, t - r lO. tS

Bus2 to referenceus

zLBr , .= [o '15rs "/Lo.rs

o 5 t

;;;l

Reference bus

or

As pe rEq . (11.47)

I I - t =Z; t r * Zz_ t r Zo_r , 3Zl

unloadedbefore fault)ut V", = I Pu (sYstem

Then

-j1.0= - j3.92pu

,0 .105+0.105+0.045

I f r - t = l t r ; = f- ^ 2 P u

(a) Fault current, I\ = 3If ,-, =

O) Vfvr = Vo r-, = Zt-rr lfr-t

= 1. 0 70.105 - j3.92 = 0.588;Vot ; = |

_ l l _ l to .os0.4s10.45 0.0s 0.451

(ii)

0.1s1 r l-O.tst

o35J-*""

Lo.rt.lo'ls o'3s]

Vf t-r= Vor-r- Zr-rJfr-ri Vora = 1.0 system nloaded eforeault)= 1.0 j0.M5 x - j3.92= 0.g24

vtr;=- zr-trfr-t= -70.105 - j3.92= 0.412

V{-z=- Zr-r, Ifr-,

- - j0.045 - j3.92 = _ 0.176

vfo-t - zurrlL,

- - j0.045 - j3.92 = _ 0.176

vt-z= Zuy IL,

= - 7O.005 - j3.92 = - 0.02

I{-rz= yvrz (VI_r Vfr_r)

= - 1-(0 .588 0.824)=l . r8

j0.2

Ifr-rr= !z-n U{-t - Vfr-r)

= * . r - 0 .412 0.176) i1 .180 . 2 '

If o_r,= lo_tz Vfu, - Vfur)

r l(c) 4-c =

,o;(1 - o's88) -33"

- - - 1 . 37 - i2 . 38

rLc= .:= to - (- 0.412)) 3o"j0.1s

r.37 i2.38

IIo-c= 0 (seeFig' 1I'29)

If o_c=et37 - j2.38)+ (1.37 i2.38)

= _ j4.76

Currentn phases andt:c f thegenerator an be similarlycalculated.

(d) Vfo-r= ZVf r + Vf -, + Vfur

= 0.588 1240" 0.4L2 1120" 0.176= - 0.264 j0.866= 0.905 - 107"

VIr-t= Vf -t + Vfr-,+ Vfu'

= 0.588 l20' - 0.4121240" 0.176

- - 0.264+ i0.866= 0.905 1107"



=;,

(- 0'176 o.o2o) io.3s

I ro_tz= l.18 + 71.18 j0.39_ j2.75

7 f , - - i l t a . / 1 A I r o r . ' 1 -ro / 7 ^ f r . .n ^^ D_ t .z_

J r. ru 4-av _rJ r. ro z_ Lv + Jv .Jy

= _ j07g

If,-rz= l.18 lI20 + 71.1gZ4 V + il.3g= 0.79

PROBEIVIS

11.1A 25 MVA, 11 kv geaerator as a x"o= 0'2 pu ' Its negativeand zero

sequenceeactances re espectively .3 and 0.1 pu. The neutralof the

generatoi s solidly grounded.Determine he subtransient urrent n the

generatorand the line-to-linevoltages or subtransient onditionswhen

an LG f'ault occurs at the generator erminals.Assume hat before the

occuffenceof the fault, the generator s operatingat no load at rated

voltage. gnore resistances.

11.2RepeatProblem 11.1 or (a ) an LL fault; and (b ) an LLG fault.

11.3A synchronous enerators rated25 MVA, 11 kV. It is star-connected

with the neutralpoint solidly grounded.The generators operatingat no

load at ratedvoltage. ts reactances re Xt' = Xz = 0.20 and Xo = 0'08

r r^ r^ i - r^+^ + l^^ - . ,m, -o+einol orr l r f ronc icnt l ine et r r re.nfs for (i ) SinSlepu. \ -aruula lLE Ll ls DJr luuv l r rv4r ouvuera -- - -o--

line-to-groundfault; (ii) double line fault; (iii) double line-to-ground

fault; and (iv) symmetrical three-phaseault. Compare hese currents

and comment.

ll.4 For the generator f Problem 1.3,calculatehe valueof reactanceo be

included n the generator eutraland ground, so that ine-to-ground ault

11'5 Two 25 MVA, 11 kv synchronousgeneratorsare connected o acommon bus bar which suppliesa feeder.The star point of one of thegeneratorss grounded hrougha resistance f 1.0 ohm,while that of theother generator s isolated.A line-to-ground fault occursat the far endof the feeder.Determine: a) the fault current; (b) the voltage o groundof the sound phasesof the feederat the fault point; and r.l

"orilg"bi

the star point of the groundedgeneratorwith iespect o ground.The mpedanceso sequenceurrentsof eachgenerator nd eederare

given below:

lW:il MqdernpowerSyst€mAnatysis

currentequals he three-phaseault current. What will be the value ofth egrounding esistanceo achicvc hc samcconcli t ion, l

with the reactance value (as calculated above) included betweenneutral and ground, calculate th e double l ine

faul t current ancl rlsodouble line-to-ground faul

t*t. sub-^

prelault urrentgnored.7. 5MV A3.3/0,6v

X = 10o/o

fauft X'= Xz= 2oo/oXo = 5o/o

Xn = 2'5o/o

Yr6i*

F ig.P- l1 .8

n.g A double ine-to-ground ault occurson lines b and c at point F in the

systemof Fig. i-tf.q. Find the subffansient urrent in phase c of

machine 1, assumingprefault currents o be zero. Both machinesarerated7,200 vA,600 v with reactancesf x//= xz=lUvo an d xo =

5% . lac5 hrcc-phirscrtnslorutcr s rutcd1.200 VA . 60 0 V-A/3.300

V-ts with leakage eactanceof 5Vo.The reactances f the transmission

l ine ar eX,=X,-=20vo andXo=4oTo n a base f 1,200 VA , 3,300V'

The reactances f the neutral groundingreactors are 5Vo on the kVA

baseof the machines.

Positive sequence

Negativesequence

Zero sequence

Generator

(perunit)

jo.2

i 0 . 1 5

j0.08

Feeder

(ohms/phase)

j0.4

j0.4

j0.8rI'6 Determine the fault currents n each phase ollowing a double line-to-

ground short circuit at the terminals of a star-connected ynchronousgeneratoroperating nitially on an open circuit voltageor i.o pu. Thepositive, negative and zero sequence eactanceof the generator are,respectively,70.35, 0.25 and 0.20, and its star point is isolated from



ground.

11'7 A three-phase ynchronous eneratorhas positive,negativeand zerosequenceeactanceser phase espectively, f 1.0,0.g and0.4 ohm.Thewinding resistances re negligible.The phasesequence f the generatoris RYB with a no oa dvoltage f I I kV betweenines.A shortcircuitoccursbetween ines I and B and earth at the generator erminals.

Calculatesequence urrentsn phaseR and current n the earth eturncircuit , (a) if th e gencrator eutral s sol idly earthecl;nc l b) il th egeneratorneutral is isolated.

Use R phasevoltageas reference.11.8 A generatorsuppliesa group of identical motors as shown in Fig.

P-11'8. The motors are rated600 V, 9O%o fficiencyat full load unitypower factor with sum of their output ratings being i rrrrW.The motorsa f e s h a r i n o e n r r q l l r r o l ^ o . l ^ € . / t l \ t r r r ^ t - ^ . ^ r - - - ' .s rvsu \^'r l vrvv il r laL€u v'r 'age, u.6 power tactorlagging and90voefficiencywheaan LG fault occurson the ow voltageside of the transformer.

Specify completely he sequence etworks to simulate he fault so asto include the effect of prefault current. The group of motors can betreated s a singleequivalent otor.

_r_rr*1_

xl

0.35

Xz xo

0.25

0.30 0.20

0.05

0.04

0.80

i^ vn6l,_L

-:

F lg .P '11 .9

11.10 A synchronous achine1 generating pu voltage s connectedhrough

a Y/Y transformerof reactance .1 pu to two transmission ines in

parallel. The other ends of the lines are connectedthrough a YN

transformerof reactance .1 pu to a machine2 generating1 pu voltage'

For both transformers X, = Xz = Xo'

Calculate he current ed into a double ine-to-ground ault on the line'side

terminals of the transformer fed from machine2.

The star point of

machine I and of the two transiormers. re soiiriiy grourrded.The

reactancesof the machines and lines referred to a common base are

Machine 1

Machine 2

Line (each) 0.40 0.40

Modernpower

ll'lliltr"::,i"il"ti .1T:-Toow.er

erwork ith wogeneratorsonnecredil"lili1:' : ::, :,il.::9,,.*il;; ; ".il:i.H lffi

,:i:T"tl'

."."11'::::1.r:

nninier l; *'' ;ffi# Hffi ff#,f

i,:il".5.T1",ffi,1,iJl.:^::l:l;,-i""i;ffi*"#,J'#::,irfrfl:f{",,":t'Thepot+-: q"geEyeno ero eque;r1:#;:"#viltr'om vv'rv\,rerr,-" n pEr fiit *:"""*"t-;;;il#,Positive Negative Zero

Generator 0.15 0.15 0.0gGeneraror 0.25 0.25 oo (i.e. neutral solated)Each rransformer 0.15 0.15 0.15Inf ini te bu s 0.15 0.15 0.05Line 0.20 0.20 0.40

l.] P_-....*,theequence etworks of the power ;;_b) with borhgenerarors nd nfinite bus op"rutinf

"ar.o pu voltageono road,a rine-to-grouncrfaurt occursat .ne of the terminarsof thetar-connectedwinding

of the transformerA. caiculate the currentsflowing (i) in the fauli; and (ii) through rhe transformerA.6L-_' ,q\--7 | )r I

I_ L I, x l J | l

I t z 2 , H" . . '\ Y-_ t

F ig .p -11 .11

rl'r2 A star connected ynchronous

'I lnc rr rnmat r inal Farr l t Anahraio lL. 'A+\ l

Generator: Xr = Xz = 0.1 pui Xo = 0.05 pu

X, (Brounding eactance)= 0.02 pu

Transformer: Xr: Xr:Xt = 0J?u

X, (grounding reactance)= 0.04 pu

Form the positive,negativeand zero sequence us mpedancematrices.For a solid LG fault at bus 1, calculate the fault current and itscontributions rom the generatorand ffansformer.

1 , L 2_f6l-Y €fff

rTft-YA

Fig.P-11 .13

Hint: Notice that the line reactancesare not given. Therefore it isconvenient o obtain Zt, svs directly rather than by inverting Ir, sus.Also ro,

"usit singular and zs, BUS annotbe obtained rom it. In such

situations he method of unit current njection outlined below can beused.

For a two-buscase

li;,1=17,',7,',)l';,1



enerator eedsbus bar r of a powerystem. Bus bar I rs connected o bus bar 2 thro'gh a star/crertalt'itnsl0t'ttlcl'lt scrics with a transmissionline. The power networkonnected o bus bar 2 can be,

"quiuutently'

epresented y astar_onnectedgeneratorwith equar posit ive incr ncg.t ivc sc(rr.r0rccsrcactances. lr starpornts re solidryconnectedo ground.The per unitequencee:lct*nces f v'rious corrlponents re given berow:

pol;itiveNc,,gutivt: Zt:ro

Generator 0.20 0.l5 0.05Transfbrmer OJ Z 0.12 0.12TransmissionLine 0.30 0.30 0.50PowerNerwork X X 0.10

Under no load condition with 1.0 go voltage at each bus bar, a currentf 4'0 pu is fed to a three-phase short circ-uit on bus bar Z.Deitrmhethe positive sequence reactance X of the equivarent generator of the,, . power network.

For the same initial conditions, find the faurt current for single line_o-ground fault on bus bar l.

Inject ingunit currentat bus 1 (i.e. Ir = tr, 2= 0) , we ge t

Z n = V t

Zzt = Vz

Sirn i lar lynjcct ing rui t urrcut t bu s2 1i .c. r = 0, lz = l ) , we ge t

Z tz = Vl

7:tz= Vz

Zou5could husbc dircct ly obtained y this technique.

l l . l4 Consider he 2-bus systemof Example11.3.Assume ha t a sol id LLfault occurs n busf Determinehe ault currentand voltage to ground)of the healtlryphase.

11.15 Write a computer programme o be employed or studying a solid LGfault on bus 2 of the sy'stem hown n Fig. 9.17.our aim is to find thefault current and all bus voltagesand the line currents following thefault. use the

impedance data given in Example 9.5. Assume alltransformers o be YlA type with their neutrals (on HV side) solidlygrounded.

Assume that the positive and negative sequence, eactancesof the

generatorsare equal,while their zero sequence eactances one-fourth

of their positivesequenceeactance. he zerosequenceeactances f the

lines are to be taken as 2.5 times their positive sequenceeactances. etall prefault voltages= 1 pu.

REFERECES

Books

1. Stevenson,W.D., Elementsof Power SystemAnalysis,4th edn., McGraw-Hill,

New York, 1982.

2. Elgerd, O.I., Electric Energy Systems Theory: An Introduction, 2nd edn.,

McGraw-Hill, New York, 1982.

3. Gross,C.A., PowerSystem nalysis, Wiley, New York, 1979.

4. Ncuenswander,.R., Modern Power Systems,nternational extbookCo., Ncw

York, 1971.

5. Bergan,A.R. and V. Vittal, Power SystemAnalysis,2nd edn.,PearsonEducation

Asia, Delhi, 2000.

6. Soman, S.A, S.A. Khapardeand ShubhaPandit, ComputationalMethods or

Large SparsePower Systems nalysis, KAP, Boston,2002.

Papers

Short

12

T2.T INTRODUCTION

The stabilityof an nterconnectedowersystems ts ability to return o normal

or stableoperationafter having been subjected o some orm of disturbance.

Conversely, nstability means a condition denoting loss of synchronism or

falling out of step.Stabilityconsiderations avebeen ecognizedas an essential

part of power systemplanning for a long time. With interconnectedsystems

continually growing in sizeand extendingover vastgeographical egions, t is

becoming ncreasinglymore difficult to maintainsynchrortism dtweenvarious

parts of a power system. '.

The dynamicsof a powersystemare characterisedy its basic eaturesgiven

i



7. Brown, H.E. and C.E. Person, Circuit Studies f Large Systems y the

ImpedanceMatrix Method",Proc. PICA, 1967, . 335.

8. Smith, D.R., "Digital Simulationof SimultaneousUrrbalancesnvolving Open

and Faul tcdConductors",EEE Trans.PnS, 1970,1826.

t j E l t w .

1. Synchronous ie exhibits the typical behaviour hat as power transfer s

gradually ncreaseda maximum limit is reachedbeyond which the system

cannotstay n synchronism,.e., t falls out of step.2. The system s basicallya spring-inertiaoscillatory systemwith inertia on

the mechanicalsideand springaction providedby the synchronous ie wherein

power transfer is proportional to sin d or d (for small E, 6 being the relative

internal angleof machines).

3. Becauseof power transfer being proportional to sin d, the equation

determining system dynamics is nonlinear for disturbances causing large

variations n angle d, Stability phenomenonpeculiar o non-linear systemsas

distinguished rom linear systems s thereforeexhibited by power systems

(stable up to a certain magnitude of disturbanceand unstable for larger

disturbances).Accordi^rglypower systemstability problemsare classified nto three basic

types*-steady state,dynamic and transient.

*There are no universally accepted precise definitions of this terminology. For adefinition of some important tenns related to power system stability, refer to IEEE

StandardDictionary of Electrical and ElectronicTerms, IEEE, New York, 19i2.

' l

434 'l Modernpower SystemAnalysis

Th"t study of steady state stability is basically concerned with thedeterminationf the upper imit of machine oadingsbeiore osing synchronism,provided he oading s increased radually.

Dynamic instabilityis more probable than steady state instability. Smalldisturbances are esntinuaHy oeeurring irr a po*.. system r"#"ti"* i"

loadings, hangesn turbine speeds, tc.)which are small enoughnot to causethe system o losesynchronismbut do excite he system nto the"itateof naturaloscillations.The system s said to be dynamicallystable f the oscillationsdonot acquiremore han certainamplitudeand die out quickly (i.e., he system swell-damped). n a dynamically unstablesystem, he oscillation amplitude slargeand thesepersist or a long time (i.e., he system s underda-p"a;. rni,kind of instability behaviourconstitutesa serious hreat o systemsecurityandcreatesvery difficult operatingconditions.Dynamic stability can be signifi-cantly mproved through the use of power systemstabilizers.Dynamic systemstudyhas to be carriedout for 5-10 s and sometimes p to 30 s. computersimulation s the only effectivemeansof studyingdynamic stability problems.

The samesimulationprogrammes re,of course, ppiicable o transient tabilitystudies.

Following a sudden disturbanceon a power system rotor speeds, otorangulardifferencesand power transferundergo ast changeswhosemagnitudesaredependent pon he severityof disturbance. or a argedisturbanc",

"hung..n angulardifferencesmay be so large as o .:ause he machines o fall out ofstep' This type of instability is known as transient instability and is a fastphenomenonsuallyoccurringwithin I s fbr a generator lose o the cause fdisturbance.here s a argerangeof disturbanceiwhich may occur on a power

D n r r r n ' Q r r a + ^ - O r ^ L : l : r - . I - ^ -

machineswhich are not separatedby lines of high reactanceare lumpedtogetherand consideredas one equivalent machine.Thus a multimachine

system can often be reduced to aq equlyalg4t fery lq4qhrue system- IfSynchronlsmi lost, ttre n'achinei f eaitr gioup stay ogetheralthoughthey goout of step with othergroups.Qualitative behaviourof machines n an actualsystems usually thatof a two machinesystem.Because f its simplicity, thetwo machinesystem s extremelyuseful in describing he generalconceptsofpowersystemstabilityand he nfluenceof various 'actors n stability. t willbe seen n this chapter bata two machinesystem an be regarded s a singlemachine ystemconnected.tonfinite system.

Stabilitystudyof a multimachine ystemmustnecessarily e carriedou t ona digital computer.

I2.2 DYNAMICS OF A SYNCHRONOUS MACHINE

The kinetic energyof the rotor at synchronousmachine s

JJ,^ x 10-6MJ

where

Bu t

-/ = rotor momentof inertia in kg-m2

aro,= synchronous peed n rad (mech)/s

u.r,n= rotor speed in rad (elect)/s

K E = 12



system, ut a fault on a heavily oacleclin ewhichrequires pening hc ip c t< lclear he ault is usuallyof greatest oncern.The tripping of a loadJdgeneratoror the abruptdroppingof a large load may also cause nstability.

The effectof shortcircuits (faults), he most severe ypeof disturbanceowhich a power system s subjected,must be determined n nearly all stability

studies'During a fault, electrical power from nearby generators s reduceddrastically,while power rom remotegeneratorss scarcelyaf1'ecte4.n somecases' he systemmay be stableeven with a sustained ault, whereasothersystemswill be stable only if the fault is cleared with sufficient rapidity.Whether he system s stableon occurrence f a fault depends ot only on thesystemtself, but alsoon the typeof fault, ocationof fauit, rapidity of clearingand methodof clearing, .e., whetherclearedby the sequentialopeningof twoor morebreakersor by simultaneous peningand whetheror not the faulted ineis reclosed.The transientstability imit is atmostalways ower than the steadystate imit, but unlike the atter, t may exhibitdifferentvaluesdepending n thenature,ocation and magnitudeof disturbance.

Modernpower systemshavemany nterconnected eneratingstations,eachwith several enerators nd many oads.The machineJlocateda-t ny one point

ln a system ormally act n unison. t is, therefore, ommonpractice n stability

whorc

We shall

P = nuurbelo1'rnaclrineoles

= momentof inertia in MJ-s/elect ad

detine the inertia constantH such that

M = J ( ? \ ' u , x 1 0 - 6\ P /

' ,

G H = K E =! u % M J2

KE+(t(?)'c.,.ro-.)*

- L M ,2 ' ,

G = machine ating (base) n MVA (3-phase)

H = inertictconstant n MJ/I4VA or MW-s/MVA

'. : -: il{36 i MociernPowerSysiemnnaiysis

I

It immediately ollows that

M =2GH

= GHMJ-s/elect ad

(ts lt f

= ffi14J-s/electdegree

180-f

M is alsocalled he nert iaconstant.

Taking G as base, he inertia constant n pu is

M (pu) = + s2lelect adnf

(r2.r)

(r2.2)

H ) , ,=ffi

s'lelect egree

The inertia constant H has a characteristic alue or a range of values for

each classof machines. able 12.1 ists some ypical inertia constants.

, Table 12.1 Typical nert ia onstants f synchronous achines*

Type oJ Mat:hine Intertia Con.slunt H

Stored Energy in MW Sec per MVA**

Turbine Generator

Condensing

Non-Condensing

Water wheelGencrzttor

1,800 pm3,000 pm3,000 pni

9- 67- 44-3

PowerSystemStability I437

I

e Swing Equation

;ure12.1 shows the torque,speedand flow of mechanicaland electrical

wers n a synchronousmachine. t is assumed hat the windage, riction and

n-loss torque is negligible. The differential equation governing the rotornamicscan then be written as

- d 'o^J -:; t =T^- r" Nt

ot -

'here

0*T*

T,

= angle n rad (mech)

= turbine torque in Nm; it acquires

machine= electromagnetic orquedeveloped

for a motoring machine

o _ >t m

( r2 .3)

a negativevalue for a motoring

in Nm; it acquires egative alue

Fig.12.1 Flowof mechanicalndelectricalowersn a synchronous achine

While the rotor undergoes ynamics as per Eq. (12'3), the rotor' speed

changes y insignificant magnitude or the time period of interest(1s) [Sec.

0.i. Equation (12.3) can thereforebe converted nto its more convenient. - L - . . , . . - , = * : * * r ! : a * n r ^ r ( ' n e r r . l



Slow-speed< 200 rpm)

High-speed > 200 rpm)

Synchlorrousorrdcrrsc r '+4

LargeSma l l

Synclrrcnous otor wi th load varying i 'ortr

1.0 o 5.0 andhighcr or hcavy 'lywhccls

It is observedrom Table 12.1 h'at he value of H is considerably igher or

steam urbogeneratorhzrnbr watc:rwheelgenerator. hirty to sixtyper centof

the total inertiaof a steam urbogenerator nit is that of the priine mover,

whereasonly 4 -I5Vo of the nertia of a hydroelectricgeneratingunit is that of

the waterwheel,ncluding water.

L- J

2-4

t . 2 5l .00

2.00

* Rcprinted with permission of the Westinghous Electric CorporationElectrical Transmission and Distribution Reference Book.

+* Where range s given, the first figure applies to the smaller MVA sizes.*tc+ Hydrogen-Cooled,25er cent ess.

from

mechanical pt lwer inPut

electrical power outPttt

t n r c n r a i n c n n s t : t n t a t t h g S V n C h f O n O U Sl n r l t a r T n r m n v 2 \ \ l l l l l l ll y l l l c l t r l t r l J U U U U L v l v l l r q r t l ! v r r u r q r r r, v v r v r t v r r l t vJ -

I

peed ur,.).Multiplying both sidesof Eq . ( 12.3)by u),,^'we can write

J6"n,

':t);"

x lo-('

- P^ P, .Mwd t t

where

MWMW; stator opPer

( t 2 . 4 )

loss s assumedl n

t n

p' t t t -

Dt - , -

negligible.

RewritingEq. (12.4)

/ ) \ 2 s 2 a

ul;) u.r, 1o-6)ff

- P^ P,Mw

where 0, = angle n rad (elect)

n o d ' \ , = , - p ( 1 2 . 5 )ol Mi;- = P,,- P"

43S I Modernpower SvstemAnalvsis

- -

It is moreconveniento measurehe angularpositionof the rotor with respectto a synchronouslyotating frame of reference.Let

o =(a;,?;['

;"':?: f#u ar di sP1 cemen from synchronou s y

(r2.6)

(r2.7)

(called torque angle/powerangle)From Eq. (12.6)

drg, _ d2 6

,Jt2 dt 2

Hence Eq. (12.5)can be written n termsof d as

, , d 2 6 'M - : = P ^ - P e M w

ot -

With M as defined n Eq. (12.1),we can write

GH d2 d

f V - P * - P " M wDividing throughoutby G, the MVA rating of the machine,a

M(pu) = P*- P,;dt '

in pu of machine ating as base

where

(12.8)

(r2.9)

(12.r0)

M(Pu)+

_ PowerSystem tabitity _ I +S l-

Grnu"h= machine rating (base)

Gryrt",o system base

Equation (12.11) can then be written as

+-" (Yry*l = p- p.) '""Gsystem- / dt' )

\"t c'

Grrr,.*

= machine nertia constant n systembase

Machines Swinging Coherently

Consider he swing equationsof two machinesor a commonsystembase.

or { " " " ' o ' ! = ' -P

"fdl

=P* - P"PUn sYstemas e

where Hryr,"* H-u"ht'+*l*"\..Gt""'n

/

Hr d261

"fa;

=P*r - P"t Pu

H2 d262

"fA;

= P*2- P"zPu

Since he machine otorsswing together coherently r in unison)

6 , = [ r = [

( t : . r2)

( r2. r3)

( r2 .14)

. (12. rs )



lr j

H d z bor ^ = P*- P, pu (IZ.1I )

n f d t "

ThisequationEq . (12.1D)l}q.(12.1 )) , s cal led the tving cluat ioy nd tdescribeshe otordynamics br a synchronous achine generating/motoring).It is a second-order ifferential equationwhere he darnping erm (proportionalto d6ldt) s absent ecause f the assumption f a losslessmachine nc l he actthat the torqueof damper winding has been gnored.This assumptioneads opessimistic esultsn transientstabilityanalysis-dampinghelps to stabilize hesystem.Dampingmust of coursebe consideredn a dynamicstabilitystudy.Since heelectricalowerP, depends pon he sineof angled(seeEq. 12.29)),the swing equations a non-linear second-order ifferential equation.

Multimachine System

In a multimachine ystema commonsystembasemust be chosen.Let

AddingEq s (12.14) nd (12.15)

H" q d2 6

tr f dt z

= P*- P"

whcrcP r r = P * r + P n

P r = P " l * P r z

H " q = H r + H ,

ffimel"L2rl1

( r2.16"

( r2 . r7)

The two machinesswinging coherentlyare hus reduced o a singlemachineasin Eq. (12.16).The equivalent nertia n Eq. (12.17) canbe wrirtenas

H"q, = Hl ,nu.h Gl âch/Grystem Hz û"h G2 -u"h/G.yr,.. (12.18)

The above esultsare easilyextendableo any numberof machines wingingcoherently.

A 50 Hz, four pole turbogeneratorated 100 MVA, 11 kv has an inertiag6nsf an f nf R O MI / I \ / IVA

Find the storedenergy n the rotor at synchronousspeed.If the mechanicalnput s suclcJenlyaised o 80 MW fbr an electricaloadof 50 MW, tind rotor acceleration, eglectingrrlechanical nd electrical

losses.(c) If the accelerationalculatedn part (b) is maintainedor 10 cycles, ind

the changen torqueangleand otor speed n revolutionsper minuteat theend of this period.

)olution

(a) Storedenergy= GH = 100 x g = g00 MJ

( b ) , = 8 0 - 5 0 = 3 0 M W = M4dt '

r, CH 800. 1 - -

180/ 180 50

4 d 2 6+' j 6, 2

= 3o

or

r ) c

(c) 10cycres"lT

- 3375elect eg/s2

Changen d= !{ZZI.S) x (0.2)2 6.75 electdegrees

3 3 7 ' 5 .o 1Ê !

(a )

(b )

t4/s'I torJern Power Srrctarn Anarrraio--

4MJ-r/elect deg

teryer-Svele-qtsqllv--

3. Effect of voltageregulating oop during the transient s ignored, as a

consequencehe generated achine mf remains onstant. hi s assumption lso

leadso pessimist icesults-voltageregulator elps o stabi l izehe system.

Before the swingequationcan be solved, t is necessaryo determine he

dependencef the electricalpowel otttput P,,)upon the rotor angle.

Simplified Machine Model

For a nonsalient ole machine, he per phase nduced emf-terminalvoltage

equation ndersteady onditions s

where

E =V + jXolu+ jXol, , ;X,r) X, r

I = I a + I s

( r2 . Ie)(r2.20)

(12.2r)

'(12.22)

andusualsymbolsare used.

Under ransient ondition

X a X ' a 1 X a

but

X'o = Xn since he main fleld is on the d-axis

Xtd< Xo ; but the differences less han n Eq ' (I2.I9)

Equation 12.19)during the transientmodifies o

Et =V + jxt lo+ jXnln

=V + jXq(I - I) + jXotlo

= (Y + jxp + j(X'a - Xq)Id

The phasordiagramcolresponding o Eql (12.21) and (12.22) s drawn in

Fig. 12.2.



= 60 x2x360J

= z6'12r {PnVs

.' . Rotorspeed t th e en d of l0 cvcles

- r2ox5o+'z8. tz5 0.24

= 1505.625pm

I2.3 POWER ANGLE EQUATION

In solving he s'uvingquation Eq (12.10)),certain implifyingassumpticns reusually made.Theseare:

1. Mechanicalpower nput to the machine P*) remains onstantduring heperiod of electromechanicalransientof interest. n other words, t means hatthe effectof the turbine governing oop is ignoredbeing much slower than hespeedof the transient.This assumptioneads o pessimistic esult-governingloop helps o stabil izehe sysrem.

'2.Rotor speed hangesare nsignificant-these have alreadybeen gnored

in formulatingthe swing equation.

Sinceunder ransientcondition,X'a 1X, but Xn remainsalmost unaffected,

it is fairly val id to assttmeha t

x'a = xq ( r2.23)

Fig. 12.2 Phasor diagram-sal ient pole machine

,firZlll r"orrr r atysis

equation 12.22)now becomes

E , = V + j X n I

= V + jXotl (r2.24)The machine modalso applies to a cylindrical rotor machine where

-- - - - D '

X,l = X*/(transientsynchronouseactance)

The simplifiedmachineof Fig. 12.3will be used n all stabilitv studies.

P.ower Angle'Cunre

For the purposes f stabilitystudies El1, transientemf of generatormotor,remains constantor is the independentvariable determined by the voltageregulating oop but v, the generatordetermined erminalvoltage s a dependentvariable'Therefore,he nodes buses) f the stabilitystudy networkpertain othe ernf erminal n the machinemodelasshown n rig. 12.4,while he machinereactanceXu) is absorbedn the systemnetworkas clifferent rom a load flowstt lcly' t t rther,he oirt ls othcr hrrniu'gc ;ynchronous

Fig. 12.3 Simpl i f iedmachinemodel

Fig. 12.5 Two-bus tability tudynetwork

For the 2-bussystemof Fig. 12.5

f Y , , Y"1Ynus I - j ' - : ' l ; Y,z=Yzr

LY^ Yr,[ I

Complex power nto bus is given by

P i + Q i - E l f

At bus I

Pr + jQr - Er ' (YyE1)* + E, (YpEil*

Bu t

(r2.2s)

(r2.26)

E ' t = l E i l l 6 ; E / = l E ' z l1 4 .

Y, u= Gr t + jBt i Yr z= lYrzl 0 p

Since n solutionof the swing equation nly real power s involved,we havefrom Eq. (12.26)

PowerSvstemStabilitv [ECAC,+'T*

\ t 2

o @

O f r



i i t i t t i r.s) il l bc .cpiacetlby equivalent static admittances connected n shunt between transmissionnetwork busesand the referencebus). This is so because oad voltagesvaryclt lr ing stabil i tystut ly in a loacl lo w stucly,

he.seemainconstantwithinanalrow bancl).

,/

System etwork

I r t= l E l i 2 G t r + lEr t iE i i i y , r l c os , t i , h 0 ,2 ) u 2 . 2 7 )

A similar equationwill hold at bus 2.

Let

l E l l z G , = P ,

lEt ' l lEzt lY7l = Pn,*

4 - 6 = 6an d Qn = x/ 2 -1

Then Eq. (12.27)can be written as

Pr = P, * P,n"*sin (6 - 1); Power Angle Equation (IZ.ZB)

For a purely reactivenetwork

Gt t = 0 (. '. P. = 0); lossless etwork

} t z = n 1 2 , . J = 0

P , = P ^ * s i n 6

Fig. 12.4

(12.29a)

where P^^,= lE'' I lE'' | '' - x '

simplified power angle equation

Y l aZt

where X = transfer eactance etween nodes i.e., betweenE{'Ihe

graphical plot of power angle equation (Eq.(12.29)) is

Fig. 12.6.

p " lI

DI max

(Ps6+APe).-....t-----*

Peo

-Generator

Fig.12.6 Powerangle urve

The swing equation(Eq. (12.10)) can now be written as

j0.5

i0.5

lE' l t61 o"

(b)

F |g .12 .7As imp |esys temw i thi t s reac tanced iag ram

0. 5Xr z=0 '25+ 0 '1+ -

= 0. 6

(r2.29b)andEi)

shown n



+=- P^ Pn'u* in dpu

7rI dt- Consider owamorecomplicatedasewherein 3-phaseaultoccurs t the

midpointof oneof the ines n whichcasehe eactanceiagram ecomeshat

of Fig. 12.8 a) .

Star-Delta Conversion

Convertinghe starat the bus 3 to delta, he network ransformso that of

Fig. 12.8(b) herein

(,

j0.25

at(U

(12.30)

which, asalreadystated, s a non-linearsecond-order ifferentialequationwith

no damping.

T2.4 NODE ELIMINATION TECHNIOUE

In stability studies, t has been ndicated that the buses o be consideredare

thosewhich areexcitedby the nternalmachine oltages transient mf's) and

not the oad buseswhich are excitedby the terminalvoltagesof the generators.

Therefore, n Y"u, formulation for the stability study, the load buses must be

eliminated. Three methods are available for bus elimination. These are

illustratedby the simple systemof Fig. 12.7(a)whose reactancediagram is

drawn n Fig. I2.7(b).In this simplesituat ion, us 3 getseasilyel iminated y

parallel combinationof the lines.Thus

o

(a)

lE/lt61ttr

446 L Modern owerSystem natysis

lE/lt6

CD

(c )

Fig. 12.8

v _ 0.25 0.35+ 0.35x 0. 5+ 0. 5x0.25

= 1 . 5 5

This method for a complex network, however, cannot be mechainzed orpreparinga computerprogramme.

l u towerSystemStability,

- _

This method obviously is cumbersome o apply for a network of even small

complexity and cannot be computenzed.

Node Elimination Technique

Formulate the bus admittances or the 3-bus system of

network is redrawn in Fig. 12.9 wherein instead of

admittances re shown. For this network.

Fig. 12.8(a).This

reactance branch.

(r ra@ o

ruus

The bus 3 is to be eliminated.

In general or a 3-bus system

Fig.12.9



Thevenin's Equivalent

With ref 'crcncco Fig. l2.t i(a), ho Thcvcnirr's cluivalcntbr th c networkportion to the left of terminalsa b as drawn in Fig. 12.8(c)wherein bus t hasbeenmodified to 1/.

(r2.31)

(r2.32)

of Eq. (12.31),

f ' ' I fr ' Y" "lIu'Ilh |

=lY^

Y, vullv,

Llrl Lv,,Yn rrrJL%l

Sinceno sources connected t thebus3

I t = o

or Y r rV r+Yr rV r+Yr rV r=O

o r v z = - ? r r - ? r ,Yrt Y.,

Substituting this value of V3 n the remaining two equations

thereby eliminating Vy

It =Y,Vt * YrzVz+ Yt tVt

=(" , - Y,rYr,

) u +(y,"!+)v"( . ^ t t

Y r , ' [ ' t ' Y r , ) ' '

, r 0.25vr h =

0 2 5 +0 - , 5lEt l l 5

= 0.417 Et I 16

0.35x0.25xrh =

035+025- o '146

Xt2 =0 .146+ 0.5 = 0 .646*

*This value is different from that obtainedby stardelta ransformationaslonger Et l I { in fact t is 0.417 Et l 16 .

Itlow

V* is no

In compact orm

Ynusreduced)lt, i, ' l, l, '1lY'r, Y'r,

Y ' t l = Y " - r y'33

Y'12= Y'21= Ytz -

v t , = y r r - Y r t Y tt 22 - 'z zyT

In general, in eliminating node n

Yo^(old)Y,,(old)

Yo, new) = Yry old)

Applying Eq. ( 12.34) o the example n hand

l-t.gztYsu5reduced); 1 0.646L

It then follows that

X , t = : - = 1 . 5 4 8 (= 1 . 5 5 )o.646

to be 1. 0pu .

Solution

l E r l l E ^ l I . Z x I=f f i=m=o '49Pu

(2) Equivalent circuit with capacitive reactor s shown n Fig. 12.71 (a).

j1.o jo.l i0.25 io.1 i1.op'e65

lEnl 1.2 -i1.0= lEml 1.0

(a) (b)

F lg .12 .11

Converting star to delta, he network of Fig. 12.11(a)s reduced to that of

Fig. 12.11(b)where

7X( t rans fe r ) - / 1.35X/1 .1+/1 .1X(_J1 .0 )+( - / 1 .0 )XJ1 .35 . .j1.0

= j0.965

YrtYt,

Ytt

(12.33)

Q23aa)

(r2.34b)

(r2.34c)

(r2.3s)



In th e systenr hown n Fig. 12,10, three-phasetat iccapacit ive eactor f

reactance1 pu per phase s connected hrough a switch at motor bus bar.

Calculate he imit of steady tatepower with and without reactorswitch closed.

Recalculate he power limit with capacitivereactor replacedby an inductive

reactorof the samevalue.

. 1 .2x1Steady tate ower imit =

ff i= 1-244 u

(3) With capacitiveeactanceeplaced y inductive eactance, e getequivalentircuitof Fig. 12.12. onvertingte r o delta,we have

trasfer eactance f

i1.35 i1.1

Fig.12.12

_j1.35

.r1.l "11.1xl. 0 + 11.0.t1.35

i 1.0

- j3.e35

thethe

Fig.12 .10

7X(transfer)

-!5o' l ""atttt"-t,

Steady tate ower imit - ': ' ! ] = 0.304pu3.935

Example 2. 3

The generatorof Fig. 12.7(a)s delivering1.0pu power to the nfinite bus (lVl= 1'0 pu), with the generator erminalvoltageof v,r = 1.0 pu. calculate thegeneratoremf behind ransient eactance. ind the maximu- io*". thatcan betransferredunder the following conditions:

(a ) System ealthy(b) One line shorted 3_phase)n rhe middle(c) One line open..Plot all the threepower anglecurves.

Solution

L,et Vt = lV, l la = | a

From power angleequation

t v t l v_ - - - s tn o=p"X

( t " t )

[ 0 2 5 + o J J s r n< t= I

or rr = 20.5"

Current nto infinitebus.

PowerSystemStability hiffi

As already alculatedn this section,

Xn = l '55

= !".!;!! = 0.6e4u

or P, = 0.694 sin d (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 .85

P*. *= t i t ^o ] t = r .2650.85

or P" = I.265 sin 6 (iii)

The plot of the threepower angle curves (Eqs. (i), (ii) and (iii)) is drawn in

Fig. 12.13.Under healthycondition, the system s operatedwith P,, = P, = 1.0

pu and 6o= 33.9", .e., at the point P on the poweranglecurve 1.79 sin d As

one ine is shorted n the middle, Po, emains ixed at 1.0pu (governing system

act instantaneously)and is further assumed o remain fixed throughout the

transient governingaction s slow), while theoperating oint instantly shifts o

Q on thc curvc 0.694 in dat d= 33.9".Noticc hutbccuusc f machine nert iu,

the rotor angleca nnot change uddenly.

1 . 7 9



1.265

Pn=1O

0.694

0

0.694sin 6

33.90I

90 0 A

Fig.12.13 Power ngle urue,s-

I2.5 SIMPLE SYSTEMS

Machine Connected to Infinite Bus

Figure 12.14 s the circuit modelof a singlemachine onnectedo infinitebus

througha line of reacthnce r.In this simplecase

l V , l l a - l V l l 0 "

jx

| 120.5- 0i0.3s

= l + j 0 . 1 8 = 1 . 0 1 61 1 0 . 3 "Voltagebehind ransient eactance,

E t = t l t r + . j 0 . 6 ( l + 7 0 . 1 g )

= 0.892 j0.6 - 1.075133.9"

(a ) System ealrhy

p^u* =lv )- lEt | - 1x 1.075- ., F,.\ -,

x, ,-

c ,5= t ' t 9 PU

P, = I '79 si nd(b) One line shorredn the middle:

I _

(i )

I

Xtransf.er=X'a* X,

From Eq. (12.29b)

,, ='4Usin d=

p.u*sindXt urrrf..

The dynamicsof this sysremare describecln Eq. (12.11)as

-#ft= P^-P"Pu

(r2.36)

lE/lt6

Fi1.12.14 Machine onnectedo inf inite us

Two Machine System

The caseof two finite machinesconnected hrougha line (X") is illustrated nFig. 12.15whereone of the machinesmustbe generating und h" othermustbe motoring.Understeadycondition,before he systemgoes nto dynamicsand

t -

PowerSystemStability I lt5'

d 2 6 " . ( P - r - P " r \( r - - n - )

and it=" f

l2?:)= "tlH')

SubtractingEq.(12.39b) rom Eq. (12.39a)

d2@,;6)=^r( ' : j r ! , ] , . - - P,)dt z

.J

\ HrH, )

H"q d26o r - *

. , = P n - P ,7r (lt-

w h e r e 6 = 4 - 6 .

rr - HtH,

"eq Hl + Hz

The electrical power interchange is given by expression

p"= ,E!4!- ,in6' X'0, x, + xd2

(r2.39b)

(r2.40)

(r2.4r)

(r2.42)

(r2.43)

(12.44)

The swing equationEq. (12.41) and the power angleequationF;q- (12.aa)

have the same orm as for a single machineconnected o infinite bus. Thus a

two-machinesystems equivalent o a singlemachineconnectedo infinite bus.

Because f this, the single-machineconnected o intinite bus)qYstemwould be

studiedextensively n this chapter.



P r l = - P r z = P "

The swing equationsor the two machinescan now be written as

t L _ _ r ( P^ r _ " r )_ _ " p . - + \d t z

" ' if f , ) : u t t r r , l

Fig. 12.15 Two-machine vstem

P * t = - P * z = P . (12.38a)

the mechanicalnput/outputof the two machiness assumed o remainconstantat these values throughout the dynamics (governor action assumedslow).During steadystateor in dynamic condition, he electricalpower outputof thegeneratormust be absorbed y the motor (networkbeing lossless).Thus at alltime

(12.38b)

(12.39a)

In the systemof Example 12.3, he generatorhas an inertiaconstantof 4 MJ/

MVA, write the swingequationuponoccurrence f the ault.What is the nitial

angularacceleration?f this acceleration an be assumedo remain constant or

Lt = 0.05s, find the rotor angleat the end of this time interval and the new

acceleration.

Solution

Swing equationupon occurrence f fault

H d ' 6 _ o D

1 g 0 f v- r m - ' e

4 d,26 ,

* " t # = l - 0 ' 6 9 4s i n

t4 = z2so1 0.6e4ind;.d t "

,.4#jirf vooernpowersysremAnatysisI

Initial rotor angle do= 33.9"(calculatedn Example 12.3)

a 2 l;l = 2250 l - 0.694si n 33.9")dt ' l, n+

#l *= 0; rotorspeed annot hange uddenly

Cl l r: o +

A, (i n A, t = 0.05s) x 1379 (0.05)2

7"

+ 4, 6= 33.9 1.7" 35.6"

I

2

t .

6t =

a26l. , | = 2250 l - 0.694sin 35.6")

d r llr = 0 .05s

- l34I elect deg/s2

Observe hat as he rotor angle ncreases,he electricalpower outputof thegenerator ncreases nd so the accelerationof the rotor reduces.

12.6 STEADY STATE STABILITY

The steadystatestability imit of a particularcircuit of a power system sdefineclas the maximutnpower that can be transmitted ri fhe receivinoen;without oss of synchronism.

v'r6 vrru

L ltc..:']i;E!!";1&t

ffiLinearizing about the operatingpoint Qo (P"0, 4) *" can write

LP,=(* ) . o ,

The excursionsof A d are then describedby

no9i+' - P^ (P,o aP,') L,P,d, r

Md'+'*dr

(r2.47)

where

The system

equation

P ddt

stability to small changes s determined rom the characteristic

Mp, [#], =o

whosc two roots are

f / t \ t r , . 1 ( \ f {

P = + l - \ u 1 t 0 o ) o -



Consider he sirnplesystemof Fig. 12.14whoseclynamicss describectyequations

M* = P^ P" MW ; Eq . (12.8)d l

M H 7

ln Pu sYstem

and p, = !4)! ),in 6'=p^u*sin d

x,tFor determination f steady tatestability, he direct axis reactanceX.r)ant,voltagebehindX4 are used n the aboveequahons.

The plot of Eq. (12.46) s given in Fig. 12.6.Le t the systembe operaringwith steadypower transferof P^ = P^with torqueangle d as ndicated n the

figure.Assumea small ncrementAP in the electricpower with the nput fromthe prime mover emaining ixed at p*(governor r.rforrr" is slow compared o

(12.4s)

(12.46'

L M I

As long as (0P/0 0o it positive, he rootsare purely imaginaryand conjugate

and the systembehaviour s oscillatory about do.Line resistance nd damperwindingsof machine,which havebeen gnored n the abovemodelling,cause

the systemoscillations to decay.The system s thereforestable or a small

increment n power so long as

(aP,/aa|> o (12.48)

When (0 P/AD, is negative, he roots are real, one positive and the other

negativebut of equalmagnitude. he torqueangle hereforencreaseswithout

bound upon occurrcncc l a small powcr incretrtentdisturbancc) nd th e

synchronism s soon lost. The system s thereforeunstable or

@Pe/aDo < 0

@p/A[ois known as synchronizing officienr. This is alsocalled stffiess

(electrical) of synchronousmachine.

Assuming El and lV l to remainconstant,he system s unstable,f

l E l l v lc o s ^ < o

X

po*e,systmst"uititv b{dffir-

o r 4 > 9 0 "The

maximum ower hatcanbe ransmittedithout ossof stabili

(12.4e)

and s given by( 2.s0)

(12.sr)E n v l

If the system s operatingbelow the limit of steadystability condition (Eq.12'48), it may continue to oscillate for a long time if the iamping is low.Persistentoscillations are a threat to system security. The study oi systemdamping s rhe studyof dynamicalstability.

The above procedure is also applicable for complex systems whereingovernor action and excitationcontrol are also accounted or. The describing

di f ferent ia l quat ions l inear izeclbout he opcrat ing oint .Concl i t iepbrsteadystatestability is then determined rom the corresponding haracteristicequation(which now is of order higher than two)

It was assumed n the above account hat the internal rnachincvoltagelEl remainsconstant i.e., excitation s held constant).The result s that asloading increases, he terminal voltage lv,l dips heavily which cannot betoleratcdn practice. hereforc,we mustconsiclerhe steady tate tability imitby assuming that excitation is adjusted for every load increase to keeplv,l constanr.This is how the systemwill be operaiedpractically. t may beunderstocd hat we are still not considering he effect of automaticexcitationcontrol.

I ae1 L2xrl - -e | _ ____ cos 30"

L06 Jro" 1. 8

= 0.577MW (pu)/electadH 4

M(pu)=o*ro

=t rx5o

s?/crcctat r

From characteristic quation

-+ i(WiY")u == 4.76

P^u*

P=tr[(*),,"1*)'

FrequencYof oscillations=

(ii) For 807o oading

4.76 railsec

4'76 -0.758Hz2r

s i n o + = 0 . 8 o r6 = 5 3 . 1 "P-u*

rqa) - r 'Zxrco s 3.1"\ 05 ) r r , 1 .8

= 0'4 MW (Pu)/electad



steady statestabil i ty imit with lv,l anrt vl constant s consiclerednExample12.6.

A synchronous enerator f reactance .20pu is connectedo an infinite busbar (l Vl = 1.0 pr) through ransformers nda ine of total eactance f 0.60 pu.The generator o loadvoltage s I .20 pu and ts inertiaconstants H = 4 MW-silvIVA. The resistance nd machinedampingmay be assumed egligible.Thesystem requencys 50 Hz .

Calculate he frequencyof naturaloscillations f the generator s loaded o(1)50Vo nd (ii) 80Vo f its maximumpower imit.

Solution

(i) For 50Vo oading

p=!, (q+k)* =* i3s6

Frequencyof oscillations= 3.96 radlsec

Psm do i t

' max

? q 6

') *

Find the steady state power limit of a system consisting of a generator

equivalent reactance0.50 pu connected o an infinite bus through a series

rcactancef 1. 0pu .Th e terminal oltage f th egenerators heldat 1.20pu an d

the voltageof the infinite bus is 1.0 pu.

Solution

The system s shown n Fig. 12.16.Let the voltageof the nfinite bus be taken

as reference.= 0 . 5 o r4 = 3 0 o

Then

Now

V = 7 . 0 - f f ,

I _

lE4t6

E = Vt + jXdI = 1.2 l0 + j0.5

or

Now

0 = 73.87"

Vt = 1.2 .73.87"= 0.332

r _ 0.332+.ir .rs2_r=

Xa = O' 5

IVt =1.219

m Analysis

Vt = ! .2 l0

1 . 2 1 0 - 7 . 0

jI

V = 1.0100

Flg.12.16

.E= l. g l0 _ 0. 5= (t .g co s e_ 0.5)+ 71.g in 0Steady tate orver imit s reached henE hasanangle f 6= 90o,.e., ts realpart s zero.Thus,

1 . 8 c o s - 0 . 5 = 0

r. 2 g - 1. 0I

LIJ

PowerSrrstemStahilitu EsE

A knowledgeof steadystatestability limit is important for various reasons.A

system can be operated above ts transient stability limit but not above its

steady-tatelimit. Nowrwith increased fault el,earingspeedsjt is possible to

make the transient imit closely approach he steady state imit.

As is clear rom Eq. (12.51), he methodsof improving steadystate stability

limit of a systemare to reduceX and ncrease ither or both lEl and I Vl. If the

transmission ines are of sufficiently high reactance, he stability limit can be

raisedby using wo parallel lineswhich incidently also ncreases he reliability

of the system.Seriescapacitorsare sometimesemployed n lines to get better

voltage regulation and to raise the stability limit by decreasing the line

reactance.Higher excitation voltages and quick excitation system are also

employed to improve the stability limit.

I2.7 TRANSIENT STABITITY

It has been shown in Sec. L2.4 that the dynamics of a single synchrono,rs

machine connected o infinite bus bars s governed by the nonlineardifferential

equation

+ j t .152

,, d' 6M

iF=P^- P"

where P, = P-* sin d

- _ d 2 6or M -- + - P*- P** sind

ot -(r2.s2)



t . t52+ j0.669

E -0.332 jr .r52+70.5 1.152 j0.668)- 0.002 j1.728= 1.728 90 .

Steady tatepower imit is givenby

p ^ u l E l l V l 1 . 7 2 8 x L*=

V;-+V= -- i5 = l '152Pu

If instead,he generatormf s held ixed at a valueof r.2pu, the steadystate ower imit wouldbe

P*"*i#

=o'8 uIt is observedhat egulatinghegeneratormf o hold he erminalgenerator,'oltageat r. 2 pu raiseshepowerli. it frorn0.gpr'ro r.r52pu; this s how

thevoltage egulatingoophelpsn powersystem tab'ity.

As said earlier, his equation s 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 ntegrals.For smalldisturbance say,gradual oading), the equationcan be linearized see Sec. 12.6) eading to the

concept of steady state stability where a unique criterion of stability

(APrlAd>0) could be established.No generalizedcriteria are available* for

determiningsystem stability with large disturbances called ransient stability).

The practical approach to the transient stability problem is therefore to list all

important severedisturbances along with their possible locations to which the

systern s likely to be subjected according to the experienceand udgement ofthe power system analyst. Numerical solution of the swing equation (or

equations or a multimachine case) s then obtained n the presenceof 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

system s stableand the oscillation of daround he equilibrium point will decaytRecent literature gives methods of determining transient stability through

Liapunov and Popov's stability criteria, b:rt hesehave not beenof partical use so far.

ffiffif Modernower ystem natysisI

and inally die out. As alreadypointedout in the ntroduction,mportant severedistulbances re a short circuit or a sudden oss of load.

For ease f analysiscertainassumptions nd simplifications re alwaysmade(someof thesehave alreadybeenmade n arriving at the swing equation (Eq./1 /l < . t \ \ a r r 1 1 -

consequencespon accuracyof results.1. Transmission line as well as synchronous machine resistance are

ignored.This leads o pessimistic esult as resistancentroducesdamping termin the swingequationwhich helpsstability. n Example 2.11, ine iesistancehas beentaken nto account.

2. Damping erm contributedby synchronousmachinedamperwindings isignored.This also leadsto pessimistic esults for the transientstability limit.

3. Rotor speed s assumed o be synchronous. n fact it varies insignifi-cantly during the courseof the stability transient.

4. Mechanical nput to machine s assumed o remain constantdurins thetransient, .e., regulatingactionof the generator oop is ignored.This leais topessimistic esults.

5. Voltage behind transient eactances assumed o remain constant, .e.,action of voltage egulating oop s ignored. t also eads o pessimistic esults.

6. Shunt capacitances re not difficult to account or in a stability study.Where ignored, no greatly significant error is caused.

7. Loads are modelled as constant admittances.This is a reasonablvaccurate epresentation.

Note: Since otor speedandhence requencyvary insignificantly, he networkparameters emain fixed during a stability study.

A digital computer programme o compute the transient following sudden

Power ','stemtabilitY ffiffi

p"rrnon.ntly till clearedmanually.Since n the majority of faults the first

ieclosure will be successful, he chances of system stability are greatly

enhancedby using autoreclose reakers.

Fig.12.17

In the caseof a perrnanentault, this systemcompletely alls apart.This will

not be the case n a multimachine system.The steps isted, n fact, apply to a

systemof any size.

1. From prefault loading, determine he voltagebehind transient eactance

and the torque angle 16of the machine with reference o the infinite bus.

2. For the specified ault, determine he power transferequation Pr(A during

^ault. In this system P" = 0 for a three-phase ault' ''

From the swing equation starting with fi as obtained n step 1, calculate

das a function of time using a numerical techniqueof solving thetnon-

linear differential equation.

onceagain determineP, (A and solve further

5 .



disturbanceaanbe suitablymodified to include the effect of governlr actionand excitationcontrol.

Upon occulTenceof a severedisturbance,say a short circuit, the powertransfer betweenmachines s greatly reduced, causing the machine torqueangles to swing relatively. The circuit breakersnear the fault disconnect the

After clearanceof the

for d (r). In this case,P"(A = 0 as when the fault is cleared, he system

getsCisconnected.

After the transmissionine is switchedon, again ind P" (0 and continue

to calculate d (r).

If 6 (t) goes through a maximum value and starts o reduce, he system s

regardedas stable. t is unstable f d(r) continues o increase.Calculation

is ceasedafter a suitable ength of time.

An importantnumericalmethodof calculatingd(t) from the swing equation

will be giurn in Section12.9.For the singlemachine nfinite bus bar system,

stability can be conveniently determinedby the equal areacriterion presented

in the following section.

I2.8 EOUAL AREA CRITERION

In a systemwhere one machine s swingingwith respect o an infinite bus, tis possible to study transient stability by meansof a simple criterion, without

resorting to the numerical solution of a swing equation.

4.

5 .

6.

Consider he swing equation

&r t ," = acceleratingowerd2 6 I

AF=

*@^-P '1

M = !rf

ln pu system (r2.s3)

Fig. 12.1g protof 6 vs tforstabre nd unstabreystems

lf the system s unstabledcontinues o increase ndefinitely with time and themachine oses synchronism.on the other hand, if the system s stable, 6(t)performs oscillations (nonsinusoidal)whose amplitude decreases n actualpraeticebecause f darnping erms (not ncluded n the swing equation).Thesetwo situationsare shown in nig. 12.1g.since the system

Multiplying both sidesof the swing equation*[t#),

we get

The stability criterion for power systemsstatedabovecan be converted ntc

a simple and easily applicable orm for a single machine nfinite bus system.

2P"d6

M d t

Ifrtegrating,we have

(r2.s6)

where do s the initial rotor angle before t begins o swing due to disturbance.From Eqs. (12.55) and (12.56), he condition for stability can be written as

6

o r [ r "ad-o

6,,

The condition of stability can therefore be statedas: the system s stable f the

( r2.s7)



s non_linear, henatureof its response160l is no t unique and it may exhibit instability n afashion different from that ndicated n Fig. rz.rg,depending

upon the natureand severity of disturbance.However. experience ndicates hai the response6!'l j" a power systemgenerally alls in the two broadcategoriesas shown inthe figure' It can easilybe visualizednow (this has alsobeenstatedearlier) thatfor a stablesystem, ndicationof stabilitywill be given by observation f thefirst swing wheredwill go to a maximumand will Jturt o reduce. his factcanbe statedas a stability criterion, that the system s stable f at some time

d 6= o

d t

and is unstable, f

d 6- - > 0<l t

for a sufticiently long time (more than 1 s will genera'y do).

( r2.s4)

(12.ss)

areaunderPo(accelerating ower) -dcurve reduceso zero at somevalue of

d In otherwords, he positive(accelerating) reaunder Po- 6curve mustequal

the negative (decelerating)areaand hence he name equal area' criterion of

lstability.To illustrate the equal area criterion of stability, we now consider several

types of disturbanceshat may occur in a single machine nfinite bus bar

system.

Sudden Change in Mechanical Input

Figure 12.t9 shows he transientmodel of a singlemachine ied to infinite bus

bar. The electricalpower transmitted s given by

-->

Pm

Infinitebus bar

lv l r0o

F ig. 12 .19

ffifftl| Modern owervstem narys,s

l E t l l v lP, =

u, # si n d- P* * sind

,n d T ^ e

Under steadyoperating condition

P.o = Pro= P** sin do

Flg. 12.20 P" - 6diagramor suddenncreasen mechanicalnput ogeneratorf F ig.12.19

This is indicatedby the point ainthe Pr - 6 diagramof Fig. 12.20.

Let the mechanical input to the rotor be suddenly increased to Pn (by

opening the steamvalve). The accelerating ower 1o = P*t - P, causes'the

6 b f i 6 2%

?' i(' s

Q <( 4

4n

A r = ) ( P n - P " ) d 6

A z = i < r , - P ^ ) d 66l

be possible to find angle d2such that

rg condition is finally reached when 41

own in Fig.l2.2L Under this condition,

hat

6 . = 6^o= T - 6 t= n'-sin-l+:

(12.58)



rotor speed o increase u> a,,r)and so does the rotor angle. At angle 6r,,Po= P*r- Pr(= P-* sin 4) = O state oint atb)but the otor anglecontinues

to increaseas t., ur.Po now becomesnegative decelerating),he rotor speedbegins o reducebut the anglecontinues o increase ill at angle 6., a= ur onceagain (statepoint at c. At c), the-decelerating reaA, equals he accelerating

bc

areaA, (areas re shaded),.e.,J ,, Od= 0. Since he rotor is decelerating,6o

the speed educes elow ur and he rotor anglebegins o reduce.The statepointnow traverses he P, - 6 curve n the oppositedirection as ndicatedby arrowsin Fig. 12.20.It is easilyseen hat the systemoscillatesabout he new steadystatepoint b (6= 4) with angleexcursionup to 6 *d 4.on the two sides.Theseoscillationsare similar to the simple harmonic motion of an nertia-springsystemexcept hat theseare not sinusoidal.

As the oscillations decay out becauseof inherent system damping (not

modelled),.the systemsettles o the new steadystate where

P^t = P, = Prn.* sin dl

Fig. 12.21 Limiting aseof transient tability ith mechanicalnput

suddenlYncreased

Any turther ncreas in P^, means ha t he areaavailableor A, is less han A1 '

so that the excesskinetic energy causesd to increase eyond point c and the

decelerating power changesover to acceleratingpower, with the system

consequentlybecominguistable. It has thusbeenshown by use of the equal

areacriterion that here-is n upper imit to suddenncreasen mechanicalnput

(P^r- Po,s), or the system n question o remain stable'' 'ii

',,uy'ulsobe not"i from Fig. 12.21that he systemwill remain stableeven

though the rotor may oscillate beyo-nd-{^=.90"'

so long as the equal area

criterion is met. The condition of d = 90" is meant for use in steady state

stability only and doesnot apply to the transientstability case'

t rs#ff i r._r_.. ._{i..fuu.xdr ruooernFower uvslem Anarvsrs

Effect of Clearing Time on Stability

Let the systemof Fig. 12.22be operatingwith mechanical nput P^ at a steadyangleof d0 Pn,= P") as shown by the point a on the Pr - 6 cliagram f Fig.12.23.If a 3-phase ault occursat the point P of the outgoingradial ine, theelectricaloutput of the generator nstantly educes o zero, .e,, p, = 0 and hestatepoint drops to b. The acceleration reaA, begins o increaseand so doesthe otor anglewhile the statepoint movesalongbc. At time /. correspondingto angle 6, the faulted line is clearedby the openingof the line circuit breaker.The values of /, attd 4 are respectively known as clearing time and, learingangle. The systemonce again becomes ealthyand transmits p, = p,ou,.sin di.e. the state point shifts to d on the original P, - d curve. The rotor nowdecelerates nd the deceleratingareaA, begins while the statepoint movesalongde.

F19.12.22

If anangle fi canbe found such hat A2= Ap the system s found to be stable.The systern inally settlesdown to the steadyoperatingpoint a rn an oscillatorymannerbecauseof inherentdamping.

| /'-\- l l \

|-t___j co )l \ - /

l':fiffn#;i;Power a

t rresponding to a clearing angle can be

established nly by numerical ntegration except n this simplecase.The equal

areacriterion therefore gives only qualitative answer o systemstability as the

time whgn the breakershould be opened s hard to establish.

Pe

D' max

I d", 6r"*+

Crit ical learingangle

Fig. 12.24 Critical learing ngle

As the clearing of the faulty line is delayed,A, increases nd so\oes d, to

find A2 = Ar till 6r = 6^ as shown n Fig. 12.24.For a clearing ime (or angle)

larger han this value, he systemwould be unstable s A, < Ar The maximum

aiiowabievaiue of the clearing ime and angle or the system o remain stabie

are known respectively as critical clearing time and angle.

Pm



Pe

D, max

Pm

6 e l f . "

P"=o-.--/ i(3-phase ault) Clearing

61

angle

Fig. 12.23

For this simple case (P, = 0 during fault), explicit relationships or 6,

(critical) and t" (critical) are establishedbelow. All angles are in radians.

It is easily seen rom Fig. 12.24 that

4 n u * = T - d ;

and P*= Pr* sin 6o

Now

A t = (P^ --0) d 6= P^ (4 , - 6)

A z = (P** sin d- P^) d6

( 12.59)

(r2.60)

uct

Jh

6 ^^

J6,,

and

= P.u* (cos d, - cos d-*) - P* (6^o - 6"i)

ffil uoo"rnpo*"r system natvsis

For the system o be stable, A2= A1,which yields

cos{. = !^ (5,^^"Prn*

\ -tniu( d) + cos 4o"*

where 4, = critical clearing angleSubstitutingEqs. (1259) and (12.60) n Eq. (12.61),we get

4r = cos-t [(r, _ Z6l sin do_ cos 6o]During the period the fault is persisting, he swing equation s

d,2 rf

d,r,=

1r:P^: P, = o

Integrating twice

P*tz + $

/cr = critical clearing time

4, = critical clearing angleFrom Eq. (12.6a)

6 = -,rf-2H

0 " ,= #P ;2 " , 6 0

(r2.61)

(r2.62)

(12.63)

(12.64)

( r2.6s)

where

where d, is given by the expression f Eq, (12.62)An explicit elat ionship'rrr lctenninirtg

2H(6, , 4)TrfP*

,, /__\ l_Lr__l I Infinite, .774 ) | I -l bus

Pm t t \ - - lIVVO "

L r l

IVVO"

(b )

Fi1.12.25 Singlemachineie d o inf inite us hroughwo par:al lelines

Both thesecurvesare plotted n Fig. 12.26,whereinP-u*n ( P_u*ras (Yo * Xr)> (Ya + Xr ll X).The system s operating nitially with a steadypower transferPr= P^ at a torqueangle 4 on curve I.

Immediately on switching off line 2, the electrical operating point shifts tocurve II (point b). Acceleratingenergycorrespondingo areaA, is put into rotorfollowed by decelerating energy for 6 > q. Assuming that an area A2corresponding o deceleratingenergy(energyout of rotor) can be found suchthat At = Az, the system will be stable and will finally operate at ccorresponding o a new, otor angle 6, > 60" his is so because single lineoffers larger reactanceand larger rotor angle is needed o transfer he samesteady ower.

W

(a)



., , porisiblcn thiscaseas luringthe faultedcondition p" = o and so trre ,wing equation can be integrated nclosed orm.

This will not be the case n mosi other situations.Sudden Loss of One of parallel Lines

consider ow a singlemachine ie d o nfinitebu s hrough wo paral lel inesasin Fig. 12.25a. ircuit model of the sysrems given in Fig. r2.25b.Le t us study the transientstabilityof the ,yir"rn when one of the lines ssuddenlyswitchedoff with the systemoperatingat a steadyroad. Before

switching ff , poweranglecurve s givenby

P " r =l E ' l l v l

xa i xt l lx2si n d= Pm*l si nd

Immediatelyon switching off line 2, powerangrecurve is given by

P"n= g:+ sin d= pmaxrsin d,\d -T rt7

Fig.12.26 Equalareacriterion ppliedo the opening f oneof thewolines n paral lel

/

," (both ines n)

ffiffi-4l Mod"rno*rr. urt* nrryr',

4 = 4 o * _ T _ 6 ,

which is the samecondition as n the previous example.

Sudden Short Circuit on One of parallel Lines

Case a: Short circuit at one end of line

Let us now assumehe disturbance o be a short circuit at the generatorend ofline 2 of a doublecircuit line as shown n Fig. 12.27a.We shall assume hefault to be a three-phase ne.

Power Sy-t-- St"blllry-

via the healthy line (through higher linereactanceX2 in place of Xl ll Xz)7; ith power angle curve

sln sin d

obviously, P-o[ ( P-"*r. The rotor now starts to decelerateas shown inFig. 12.28.The system will be stable f a deceleratingareaA, can be foundequal to acceleratingarea A, before d reaches he maximum allowable value

4o*.At areaA, dependsuponclearing ime /. (correspondingo clearingangle

{), clearing time must be less than a certain value (critical clearing time) forthe system o be stable. t is to be observed hat the equalareacriterion helpsto determine critical clearing angle and not critical clearing time. Criticalclearing time can be obtained by numerical solution of the swing equation(discussedn Section 12.8).

P"y,prefault (2 lines)

P6n1, ostfault (1 line)

X2



? t 6

Ffg. 12.28 Equalareacriterion ppliedo the systemof Fig.12.24a,I systemnormal, l faultapplied,ll faulted ine solated.

It also easily ollows that larger nitial loading(P.) increases , for a givenclearingangle(and ime) and thereforequicker ault clearing would be neededto maintain stableoperation.

Case b: Short circuit away from line ends

When the fault occurs away from line ends say n the middle of a line), thereis somepower flow during the fault thoughconsiderably educed,as differentfrom case a where Pen= 0. Circuit model of the system during fault is nowshown in Fig. 12.29a.This circuit reduces o that of Fig. 12.29c hrough onedelta-starand one star-delta conversion. nstead,node elimination techniqueof

Section 12.3 couldbe employedprofitably.The power anglecurveduring faultis thereforegiven by

P" t=| E l l v l

s in d= Pmaxr rin d'1r'II

, (b )

F19.12.27Shoft ircuit t on een dof he in e

Before the occurrence f a fault, the power angle curve is given by'- -

p" t= ,)4,'rlr,,,a,ind= p_*,sinxi + xltx2

which is plotted n Fig. 12.25.Upon occulrenceof a three-phaseault at the generatorend of line 2 (see

Fig. I2.24a), he generator ets solatedrom the powersystem or purposes fpower flow as shownby Fig. 12.27b.Thus during the period the fauit lasts,

The rotor thereforeaccelerate, i;t:;i"s dincreases.synchronism will be

lost unless he fault is cleared n time.The circuit breakers at the two ends of the faulted line open at time tc

(correspondingo angle 4), the clearing time, disconnecting he faulted line.

ffiil Modern Power svstem Analvsis+ t * F ; 4 - l ' ! ' v e v " ' ' - ' - v ' v , - ' - " ' '

"' - ' , -' -

I

ls#z#i

systemoperation s shown in Fig. 12.30,wherein t is possible o 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. increases ill it has a value 4n*' t6"

-ooi*,,- ollnrvohlenr stahilitv This caseof critical clearineangle s shownin Fig. 12.3L

Pe

Fig. 12.31 Faulton middleof one ineof the systemof Fig. t2.l4a, caseof

critical learingangle

Annlvins eoualareaeriterion to thecaseof critical clearingangleof Fig. 12.31'

we can wnte

dntn'

(c )

Fig. 12.29

X6,t

(b)

Xr

Pr1,prefault 2 l ines)

P6111,ostfault (1 line)

xc @



4,

j (P^- 4n*u sinf ldd= J {r^*r sind-P^ )d6

60 6,,

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 clearingangle

dis less

Pe

Fig. 12.30 Faulton middle f one ineof the system f Fig.12.24a

with d"< {,

where

4,,* =T - sin-r (: t_)V maxIII ./

Integrating,we get

l6*(P^a+ Pmaxrr os d)

|* (P'*,,1 cos d +

1 6 o

or

(r2.66)

= Q

P^ (6", - 6) * P.u*u (cos '[. - cos do)

I P* (6** - 6"r)* P-om (cos fi*- cos 4J = 0

,.a

P"'Prefault 2 ines)

P"11;,ostfault 1 ine)

P"11,uring ault

t

cos {r = :otd",*4naxtn

- PmaxII

critical clearing ngle anbecalculatedrom Eq. 12.67) bove. heanglesn

:lt::t1|on

are n radians. heequationmooiiiesasbelowf the angJesreln oegrees.

cos {.- ft

r.(6 ^i* - do - Pmaxrrosdo * prnu*rucosd,ou*

Pmaxltr - Prnaxn

Case c: Reclosure

If the circuit breakersof line 2 are eclosedsuccessfully i.e., the fault was atransientone and thereforevanishedon clearing the faurty line), the powertransferonce again becomes

P"N = P"r= p*u*I sin dSince reclosure restores power transfer, the chances of stable operati'nimprove. A caseof stable

operation s indicatedby Fig. 12.32.For critical clearing angle

(12.67)

4 = 4r* = 1T sin-l 1p_/p*.*r;tucr

6r c

J @r,- Pmaxrrin 0 dd =J (p.*m sin d_ pm) d,6

606r,

dru,t

+ J (P,*r si n d_ p^ ) d66.-

i0.s

Give the system of Fig.P as shown.

12,33 where a three-phasc ault is applied at rhe point

Infinitebus

vFlloo

Flg.12.33

Find the critical clearing angle for clearing the fault with simultaneousopeningof the breakers I and 2. T\e reactancevalues of'various componentsare

indicatedon the diagram.The generators delivering 1.0 pu power at the instanrpreceding he fault.

Solution

With reference o Fig. 12.31, hree separate ower anglecurvesare involved.

f. Normal operation (prefault)

Xr=0.2s+f f i+0.05

=0.522 u



(Clearingangle) \(Angle f reclosure)

Fig- 12-32 Faurt n middre f a rineof the system f Fig. 12.27awhere rrj tr, + r; T = time betw,een learing ancl eclosure.

p,t=rysind:ffirino

= 2. 3 sin d (i l

Prefault operatingpower angle is given by

1. 0= 2.3 sin 6

or 6o=25.8" = 0.45 radians

IL During fault

It is clear from Fig. 12.31 hat no power is ffansferredduring fault, i.e.,

,0.:"o

= o

t r i n 1n 1A

iffi Modern owerSystem natysis

rrr. Post fault operation (fault cleared by openingr the faultedIiae)

UE

= - 1.5 cos2.41 cos 6, ) - (2.41 6")

= 1.5cos 6",+ 6r, I.293

Setting A = Az and solving

6r, 0.45= 1.5cos 6r ,+ 6r, 1.293

or cos {, . = 0.84311.5 0.562

or 4, = 55.8"

The correspondingoweranglediagrams re shown n Fig. 12.35.

Find the critical clearingangle or the systemshown n Fig. 12.36 or a three-

phase ault at the point P. The generator s delivering 1.0 pu power under

prefault conditions.

n l.2xl.0Perrr=

ffsin d= 1.5 in 6 (iii)

Pe

Pn=1 O

66=0.45ad 6^rr=2.41 dd

Fig. 12.35

T T t o - o o i * " * - ^ * : ^ ^ : L l ^ ^ - - l ^ C f ^ - - - - - - ^ .rrrw urour.rLurr psl l luDDlulc al i lBrtr Omax l()f afea Al = A2 (Sge flg.

given by

i0.1 i 0.1s

lnfinitebus

lvF1.otoo

.10.15 jo.15

Flg. 12.36

Solution

12.35)s

lF,l=1.2 u



4 o u * = r - s i n - l I = 2 . 4 L r a d i a n s

1.5Applying equal area criterion for critical clearing angle {Ar = P^ (6", - 6)

= 1.0 (6", - 0.45)= 6c, 0.45

dr*

A z =! { r , n - p ^ ) d , 66,,

2 .41I

= | (1 .5 in6 - 1 ) ddJ6.,

r 2 . 4 1

= - 1 . 5 c o s_ d l|6",

f, Prefault operation Transfer reactancebetweengeneratorand infinite

bus is

& = 0.25 0.17+0 .15+0 .28+0 .15= 0.71

P- , =r 'Zx l s in d= 1 .69 inc'0.71

(i)

2

6

The operating power angle is given by

1.0= 1.69 in ,fr

or do = 0.633 rad

IL Durtng fault The positive sequence eactancediagram during fault is

presented n Fig. 12.37a.

j0.25

J 00 0 L / 00 0 L----------r00 0 -

j 0 .15 j0.14 j0.14 j0.'15

jo .17

+

0 '

) E1=t.z V=1.0

(a) Positiveequenceeactance iagram uring ault

j0.25 j0.145 j0.145',j0.17

lE'l=1.2 V=1.OlOo

(b) Network fterddlta-staronversion

l9l=1'z V=1.0100

(c)Networkfter tar-deltaonversion

. Ftg. 12.32

Power ystem tabilit-v Mi#ffir*Perrr=U! sin d = r'2 sin 6l

With reference o Fig. 12.30and Eq. (12.66),we have

(iii)

To find the cqitical clearing angle, areasA1 and A, arc to be equated.

6",

At = l. o (6,,- 0.633) ,J

o.+e5in d dd60

and

dmaxf

Az = | 1 .2 in ddd- 1 .0 2.155 4)J

6cr

Now

A t = A zor

6r ,= 0.633 --

o

2.155

= [ t. Z rin 6 d6 2.t55 + 6, ,J '6cr

or - 0.633 0.495 osolo' = - 1.2 os ol"tt -2.155

J.63

0.495 in d dd

J



Converting delta to star*, the reactancenetwork is changed o that of Fig.12.37(b). Further, upon converting star to delta, we obtain the reactancenetwork of Fig..r2.37(c).The transferreactance s given 6y

(0.2s 0.145).072s(0.1450.17) .0725(0.25 0.145)

X u =(0.14s0.17)

0.075

_ 2.424

p - = lal! sin d= 0.495 in 6 eI-

i .+Z+uu r v - v.a/,

Postfault operation (faulty line switched off)

Xr l =0 .25 + 0.15+ 0.28+ 0.15 + 0.17= 1.0

*Node elirnination techniquewould be'used for complex network.

fir.

(i i)

lo.orr la.,

or - 0.633+ 0.495cos 6,, 0.399= 0.661+ 1.2 cos 6", 2.155

or cos 6r , = 0.655

U 6r, = 49.I"

A generatoroperating at 50 Hz delivers 1 pu power to an infinite bus througha transmission circuit in which resistance s ignored. A fault takes placereducing he maximumpower ransferableo 0.5 pu whereas efore he fault,this power was 2.0 pu and after he clearance f the fault, t is 1.5 pu. By theuse of equal areacriterion, determine he critical clearingangle.

Solution

All the three power angle curves are shown n Fig. 12.30.

,'ffi| Mod"rn o*.. sEl!"-nAn"lytit

Ilere P-"*r =2.0 pu, Pmaxl= 0.5 pu and Pmaxrrr 1.5 pu

Initial loading P^ = 1.0 pu

Applying Eq.

cos {,-

( p \

6r,ro=r intffiJ

E 7 - s i n l1 : 2 . 4 ! r a d

1. 5

(r2.67)

1.0(2.410.523)0. 5 os .523 1.5 os .41= o ?? ?1.5-0.5

6r ,= 70'3"

T2.9 NUMERICAT SOTUTION OF SWING EOUATION

In most practical systems,after machine umping has beendone, here are still

more than two machines to be considered rom the point of view of system

stability. Therefore, there is no choice but to solve thp swing equation of each

machineby a numerical techniqueon the digital computer.Even in the caseof

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! . . r r - - r r - - - - - - - l - - - - - : - - - - - - - L l ^ , z F L ^ - ^ | - ^ - L l ^ + i ^ ^ + ^ l * ^ + L ^ l -

numerlca[y mrougn swulg equauulr. t rt trIc aIU ssvtrI i l r JuPurDtrui lL('( l l l lELl luLlD

now available for the solution of the swing equation including the powerful

Runge-Kutta method.He.rewe shall treat he point-by-point method of solution

r>2 n-1

Discrete olution

n

Continuousolution

un-|/2

u13/2un-I/T-+tsn4l2

n-2 p3l2 n-'l r>112 n

6n-z

tAf

- tAf

$n-i

2.The angular rotor velocity u= d6ldt (over and above synchronousvelocity

t



which is a conventional,approximatemethod like all numerical methodsbut a

well tried and proven one.We shall llustrate the point-by-point method for one

machine tied to infinite bus bar. The procedure s, however,generaland can be

applied-to every machine of a multimachine system.

Consider the swing equation

d2 6 1 --

;T=

; e * - P^ * s i n d ) :Po l M ;

(* - 9H orinpusystem = +)\ 7t iTf)

The solution c(r) is obtainedat discrete ntervals of time with interval spread

of At uniform throughout.Acceleratingpower and change n speedwhich are

continuousfunctions of time are discretrzedas below:

1. The acceleratingpower Po computed at the beginning of an interval is

assumed o remain constant rom the middle of the preceding nterval to the

middle of the interval being consideredas shown in Fig. t2.38.

n-2 n-1 n

Fig. 12.38 Point-by-pointsolution of swing equation

In Fig. L2.38, he numbering on tl\t axis pertains o 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 change n velocit! (a= d6ldt),

causedby the Pa@-r),assumed onstantover At from (n-312) to (n-ll2) is

J-

Af

wn-'2- wn-3t2= Lt /M) Pa@-r)

The change n d during the (n-l)th interval is

L6r-t= 6r-1 6n-2= A'tun4'2

and during the nth interval

L6r- 6n- 6n-t= / \ tun-112

(12.6e)

(12.70a)

(12.70b)

,ir Yirtvt r Ar l r. rl b r )

SubtractingEq. (12.70a\ from Eq. (12.70b) and using Eq. (12.69),we get

L'6,= A6,-t +

Using his,we canwrite

(12.7r)

6n= 6n-t+ L,6nG2.72)

The processof computation s now repeated o obtain Pa61, L6r*tand d*t. The

time solution n discrete orm is thuscarried out over the desired ength of time,

normally 0.5 s. Continuousform of solution is obtainedby drawing a ;mooth

curve through discrete values as shown in Fig. 12.38. Greateraccuracy of

solution can be achi.eved y reducing the time duration of intervals.

The occurrence or removal of a fault or initiation of any switching event

causes discontinuity n accelerating ower Po.lf such a discontinuityoccurs

at the beginning of an interval, then the averageof the values of Po before and

after the discontinuitymust be used.Thus, n computing he ncrementof angle

occurring durirrg the first interval after a fault is applied at t = 0, Eq. (I2.7I)

becomes

7, ,6 , (Ar ) t *Pao+M 2

where Pos* is the acceleratingpower immediately after occurrenceof fault.

Immediately before the fault the system is in steady state, so that Poo-= 0 and

ds s 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) he value llP"6-r>-

+ Po6_9*), wherePa@_r)-s the accelerating ower immediately beforeclearing

and Po6_r)+ s that immediatelyafter clearing the fault. If the discontinuity

occurs at ihe miciciie of an intervai, no speciai proceciure s neecled.The

increment of angle during such an interval is calculated, as usual, from the

value of Po at the beginning of the interval.

(A r )2 DM

r a( .n - I )

powerSystemStabilitv[i{8il;r

Q ^ t , , 1 ; ^ ^ E ^ f ^ - ^ ^ - * 1 , , + L ^ ^ + ^ - L . . ^ + ^ - - ^ r l . ^ l - ^ - t ^ - - - r - - - r - a -\ rvtu.,v,t nsluls ws Lall aPPt.y ult t stEP-Uy-slt tP lI IculUU, Wtr l l t rC( l t( ) Calculate

the inertia constant M and the power angle equations under prefault andpostfaultconditions.

Base MVA = 20

IneRia coflstant, Mepu\ =18 0

1.0 L52180 50

I I Prefault

= 2.8 x 10+ s2le\ect egree

& = 0 . 3 5 +0 ' 2 = 0 . 4 5' 2

Pd= Pr.*r sin d

! , . l x t . r= -'.-;;sin d = 2.M sin 5 (i)

Prefault power transf'er= + = 0.9 pu20

Initial power angle is given by

2 . 4 4 s i n 4 = 0 . 9

or 6o= 21.64"\

II During fault A positive sequence reactancediagram is shown in Fig.12.39a.Converting star o delta,we obtain the networkof Fig. 12.39b, n which

,, 0.35 0. i + A.2x0. i + 0.35x0.2 1 A-trtr= -

0l-= I..Z) pu



The procedureof calculating solution of swing equation s illustrated n the

following example.

A 20 MVA, 50 Hz generatordelivers 18 MW over a double circuit line to an

infinite bus. The generatorhas kinetic energy of 2.52 MJA4VA at rated speed.

The generator ransient reactance s X/o = 0.35 pu. Each transmissioncircuit

has R = 0 and a reactance of 0.2 pu on a 20 MVA bgq-e.E/l = 1.1 pu and

infinite bus voltage V = 7.0 10". A three-phase hort circuit occurs at the mid

point of one of the transmission ines. Plot swing curves with fault cleared by

simrrltaneous peningof 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 periodof 0.5 s if the fault is sustained.

P.u = Pmaxtt sin d

- 1'1x r; n d = 0.88 in 61.25

(ii)

Fig. 12.39

lil. Postfault With the faulted line switched off,

J;=:::,J;2-055

1 . 1 x 1 . I= : : : t ' - ' si n d = 2.0 sin d

Let us choose Al = 0.05 s

The recursive relationshi'ps for step-by-step swingreproducedbelow.

Pa(n_r)=P^ - P* * si n 4_ r

L6n= L6n-t(Lt)z

o 'M

' a(n-l)

6n= 6n-t + A,6n

curve calculation are

(iv)

(v )

(vi)

Since here s a discontinuityn P, andhencen Po, the average alue f po

mustbe used or the irst interval.

P"(0-)= 0 pu andPo 0* )= 0. 9 0.88sin 2I.64"= 0.576 u

Po(ouu.,us"l9t#ZQ = 0.288puL

Sustained Fault

Calculations are carried out in Table 12.2 in accordancewith the recursiverelationship iv), (v) and (vi) above.The second olumn of the table showsP-*the maximum power that can be transferredat time r given in the first column.Pn * in the caseof a sustained ault undergoes sudden change at t = 0* andremainsconstant hereafter.The procedureof calculations s illustrated belowby calculatingthe row corresponding o t = 0.15 s.

which it is obvious that the systern s unstable'

fault clearedat 2.5 cycles

I t l l t l l l i io 0. 1 0. 2 0. 3 0. 4

_ r l

f (s) --0. 5 0. 6

Fig. 12.40 Swing uryesor Example 2.10or a sustainedaultan d or

clearingn 2. 5an d 6.25cYcles

! A A n ^ i - . l - . , ^ a i a * r r a t i n n c n { c r r r i n n n r r n / a f n r q t r c t a i n a r l f a t t l ta a D l e a Z . Z f U l l l l ' - U y - P u l l l t t / v l l l P u t q r r v r r o vr e r ' r r r V vu '

/f = 0.05s

t;100oE@

o 8 0o)c(5

o

i o oP



(0 ' lsec) 31 .59"P."* = 0.88

sin d (0.1 s) = 0.524

P, (0.1 s) = P,,,u* in 6 (0.1 s) = 0.88 x 0.524 - 0.46I

P, (0.1 s) = 0.9 - 0.46I - 0.439

( At\2

YP, (0.1s)= 8.929 0.439 3.92M

6 (0.1s )= Ad (0.1 ) + qL P, (0.1s)M U \

= J.38" 3.92" 11.33"

d (0 .15 )= d (0 .1 ) + Ad (0.15 )

= 31.59" 11.30' 42.89"

t P^u sin 6 P"=Prrr,*sin6P,,= 0'9- P,

sec pu Pu Pu

a 6 6

deg deg

2.440.88

0.880 .88

0 .880.880.88

0.880 .88

0 .880 .880.88

0.3680.368

0. 00.576

2.574 .8 r

3.922.68r.450.550 .18

0.426

1.302.87

2.57

7.381 1 . 3 01 3 . 9 815.4315 .98

r 6 . r 616 .58

r7 .8820.75

2r .642r.6474.2131.59

42.8956.8772.3088.28

r04.44

r2r.02138.901s9.65

0+o^u,0.050 .10

0 . 50.200.25

0.300.35

0.400.450.50

0324

0.3610.46r0.5980.7360.838

0.879

0.852

0 .1540 .578

0.368

0.410.5240.6800.8370.953

0.999

0.968

0.8560.657

0.288

0.5390.439

0.3010 .1630.060.021

0.048

0 .1450.32r

I Modernpo@isf -

F.at t l t F laae^ 'J : - o E n-- - t - -- su.a vtcrete.t tlt A,i, fryCIeS

Time to clear f'ault= 2.5 = 0.05 s50

P-u^ suddenly to 2.0 at t = 0.05-. Since the

be assumed o remain constant r-om 0.025 s to 0.075 s. Th e rest of th eprtrcedures the sameancl ompletecalculations re shown n Table 12.3.Theswing curve is plotted in Fig. 12.40 from which we find that the generatorundergoes maximum swing of 37.5" bu t is stableas c5 inai ly begins odecrease.

Table12'3 Computat ionsf swingcuryes or autt learedat2.scvcles(0.05 ), At = 0.05

6

deg

P,,,,,^

pu

.sin.5 Pr,=P,rr.,*,tin5 pr,= 0.9- pu

pu pu

A 6

deg

0

0 .

ouu,

0.05

0.05+

0.05uus

0 . 1 0

0 . i 5

o.20

0.25

2.44 0.368

0.88 0.368

0.368

0.88 0 .41

2.00 0.41

0. 0

0.576

0.288

0.54

0.08

0.31- 0.086

- 0.22- 0.29- 0.29

0. 9

0.324

0.36

0.82

0.986

I . t 2

T . I 9

r . t 9

2.57 2.57

21 .64

21.64

21.64

24.21

24 .21

24.21

29.54

34.10

36.70

2.00

Z.UU

2.00

2.00

0.493

0.56

0.s91

0.597

2.767 5.33- 0.767 4.56- 1.96 2.60- 2.s8 0.02- 2.58 -

l t r

progressively greater clearing time till the torque angle d increases withoutbound. In this example, however,we can first find the critical clearing angleusing Eq. (12.67) and then read the critical clearing time from the swing curvecorresponding o the sustained ault case.The values obtainedare:

Critical c learing angle = 118.62

Critical clearing ime = 0.38 s

Table 12.4 Computationsf swingcurve or faultcleared t6.25 ycles 0.125s), f = 0.05

P,no

pu

sin 6 P"=P^ *sin6 Po= 0.9- P" 4 6 6

deg deg

0 2.440+ 0.88

ouu,0.05 0.880 .10 0 .880 .15 2 .O00.20 2.000.25 2.00

0.30 2.00

0.35 2.00

0.40 2.00A A ? ' \ N

u.4J Z .W

0.50 2.oo

0.3680.368

0.3680.410.5240.6800.7670.780.734

0 .6130.430

u. z-1-1

0. 90.324

0 .36ro.46t1 .36

1 . 5 31 .561.46

r.220.86

u.4c)0

0. 00.576

0.2880.5390.439

- 4.46- 0.63- 0.66- 0.56- 0.327

0.04u.4-J4

2.574 .813.92

- 4 . 1 0- 5 .66- 5 .89- 5.08- 2.92

0.35

-r .6

. 2r .&21.64

2.57 zt .U7.38 24.2r

11 .30 3t .597.20 42.89r.54 50.09

- 4.35 51 .63- 9.43 47.28

- 12.35 37.85- 12.00 25.50- u. -J l_J. )u

5.37



0.30

0.35

o.40

0.45

0.50

-0.22- 0.089

0.08

0.225

I . t 20.989

0.82

0.615

37.72

34.1629.64

24.33

19.73

17 .13

2.00

2.00

2.00

2.C0

0.561

0.494

0.41

0.337

2.56

- 1.96 _ 4 .52- 0 .79 _ 5 .31

0 .71 - 4.60

2. 0 - 2. 6

Fault Cleared in 6.25 Cycles

Time to clear ault= ua?t= 0.125ss0

T2.TO MULTIMACHINE STABITITY

From what has been discussed o far, the following stepseasily follow for

determiningmultimachine tability.

1. From the prefault load flow data determineE/ovoltage behind transient. reactance or all generators.This establishes eneratoremf 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 ransient eactances.

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 generatoroutputs from power angle equations(generalized orms of Eq. (12.27))and solve swing equationsstep,by

step(point-by-point method).

4ffi | Modernpower SystemAnalysis

5. Keep repeating the above step for post fault mode and after linereclosuremode.

6. Examine d(r) plots of all generatorsand establish the answer to thestability question.

The above stgpsare llustratedin the following example.

A 50 Hz, 220 kV transmission ine has two generatorsand an infinite bus asshown in Fig. 12.4I. The transformerand ine data are given in Table I2.5. Athree-phase ault occursas shown.The prefault load flow solution is presentedin Table 12.6. Find the swing equation br each generatorcluring the faultperiod.

oVz=1.0328.2350

o(

l-liSe

).5+j(

F1.O217.16o

)vs=1 .o l l z22.490

6)

) v i

[5,

t-I'

i6,u

l,,m

220kV,100MV Abase

Bus to bus Series Z HaIf line charging

Line 4-5

Line5- 1Line 4-I

Trans;2-4

Trans:3-5

0 .018

0.0040.007

0 . 1 r0.02350.04o.0220.04

0 . 1 1 30.0980.041

Tabfe12.6 Bu s dataan dprefaultoad-f lowalues n pu on 220kV ,

10 0MV Abase

S.No. Voltage

and Polar

Bus Form

No.

Bus

Upe

Voltage Generation Load

Real Imaginary

e f

| 1.010" Slack

2 t.0318.35" PV

3 t .0217.16 PV

4 1.017414.32"PQ

5 1 . 0 1 1 2 1 2 . 6 9 " Q

0. 0 - 3.8083

0.1475 3.25

0 .1271 2 .10

0.167 0

0.0439 0

-0.2199 0 0

0.6986 0 0

0 . 3 1 1 0 0 0

1.0 1.0 0.44

0 0 .5 0 .16

1 .00

t.0194

1 . 0 1 2 1

1 .0146

1 .0102

Solution Before determining swing equations, we have to find transient

internal voltages.

The current into the network at bus 2 baseci n the <iata n Tabie i2.6 is

r -_Pr- iQ, _ 3 .25 - i 0 .6986

'zv : r .o3 l - 8 .23519"



o

Fig. 12.41

Data are given below for the two generatorson a 100 MVA base.Gen 1 500 MVA, 25 kV, XJ = 0.067 pu, H = 12 MJAyIVAGen 2 300 MVA, 20 kV, X, j = 0.10 pu, H = 9 MJA4VA

Plot the swing curves or the machinesat buses2 and3 for the above ault

which is clearedby simultaneous pening of the circuit breakersat the endsofthe faulted line at (i ) 0.275s and (ii) 0.0g s.

E{= (1.0194 j0.1475)3.2s j0.6986

x 0.06719Vr.03 -8.23519'

= 1.0340929 j0.3632368

= 1.096033319.354398" 1.0960o.337l tad I

El = I.0 l0' (slackbus)

E4 Q.0I2r + j0.1271)2 .1- 0 .31

x 0. 19O "r .021 - 7 .15811 '

= 1.0166979 j0.335177 1.0705 I8'2459"

- 1.07110.31845 ad

The loads at buses4 and 5 are represented y the admittances alculated as

follows:

yrr.=''9

.!.:,0! 0.9661jo.4zsr)6 (1.0174)"

;;'#;;"J###::il::H,:T:|, :::,::::^'l:,::,1,:nt

reactanlesf themachines e wlr,e wll l ,herefore'now designateas buses2 and3, the fictitious internalnodesbetweenthe internal voltagesand the transient eactances f the machines.Thus we get

Y-^=2 2 - @ = - i r r . 2 3 6

Yzq= 11.236= yq z

Ytt =

io'04+io]= - i7'143

Yz s j7.I43 = ys t

Y u = Y t q + y q t * y q s +* Yzq

= 0.9660877 j0.4250785+ 4.245 j24.2571 1.4488i8 .8538+0 .041 rO .113_ r t . 235g

_ 6.6598977 j44.6179

Yss=Yrs * Ysq* Ysr* 9 *U r ,

* r " ,2 2

' ' 35

- 0.4889 j0.1565 r.4488 j8.8538 7.039r j41.335+ /O.1 3+ j0.098 j7.t42}

I a,grr,owerSystem tability _ .

During Fault Bus Matrix

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 augmented refault Y".r. rnatrix.Reduced ault

matrix (to the generator nternal nodes) s obtained by eliminating the new 4th

row and column (node5) using the relationship

Y*iqn"*1= Y*j@tat - Yt n(oltt)ynj(old

)/Yrn @ta)

The reduced aulted matrix ()'eusduring ault) (3 x 3) is given n Table I2.8,

which clearly depicts hat bus 2 decouples rom the other busesduring the fault

and that bus 3 is directly connected o bus 1, showing that the fault at bus 4

reduces o zero the power pumped nto the system rom the generatorat bus 2

and renders he secondgeneratorat bus 3 to -eive ts power radially to bus 1.

Table 12.8 Elements f Yrus (during ault) and Ysus post ault)

for Ex. 12.11, dmittancesn pu .

Reduced during fault Yu^

Bus

v 0.5 0.16

prefaulr"r"

*Trl rr.off(0'488e i0'1s647)

Load admittances,al

Bo, Bo,') )

I

2a

J

s.7986-j35.6301

0- 0 .0681+5.1661

0- jrr.236

0

- 0.0681 j5 .166 l

0

0.1362'-6.2737

Reducedpost Jault Yurt

I

2-J

r.3932 jr3.873r- 0.2214 j7.6289- 0.0901 j6.0975

- 0.2214 j7.6289 - 0.0901+ j6.O975

0. s j7.7898 0

0 0.1591 j6 .1168



_ 8.97695s js7.297202The complete ugmentedrefault zuu,matrix s show .iri'Table12.1.

Table 2. T TheaugmentedrefaurtusadmittanceatrixorEx . 12.11,admittancesn pu

B u s

Post Fault Bus Matrix

Once the fault is cleared by removing the line, simultaneouslyopening the

circuit breakersat the either ends of the ine betweenbuses4 and 5, the prefault

Y"u5 has o be modified again. This is done by substituting Yqs= Ysq 0 and

subtracting he seriesadmittanceof line 4-5 and the capacitivesusceptance f

half the line from elements Yooand Ytt.

Y++lporrfault)Y++(prefault)Yqs- 84512

= 6.65989 j44.6179 1.448+ 78.853 j0.113

= 5.2III - j35.8771

Similarly, Ysr(oo*,rault) 7.528I - i48.5563

The reducedpost

faultY"u5 is shown n the lower half to Table 12.8. t

maybe noted that 0 element appears in 2nd and 3rd rorvs. This shows that,

2

3

4

rr.284_j6s.4730

0- j7.1428

0

0 + j7.1428

4.245 + j24.257

j11.23s9

0

6.6598_j44.617

-1.4488+ j8.8538

0 _i1r.23sg

0 6-4.245+ j24.257 jl l .2359

-7.039+ j41.355 0

-7.039 +j4 r .35s

0

j7.t428

-1.4488

+j8.85388.9769

+ j57.2972

{92 | ModernpowerSystemAnatysis

- L . . ^ : ^ ^ l r - . r r - -

lrrryr'ruarry'lle generarors an0 Z are not Interconnectedwhen line 4-5 isremoved.

During Fault Power Angle Equation

P r z = 0

P,3= Re [YrrEr,El* El * \F(] ; s ince y= 0

= E{ 2 Gn + lEi l lEi l l r r , l co s 6z r Lzt)

= (1 .071)20. 362) I x 1 .071x5.1665 os d3 90.755")

P" 3 0.1561 5.531 in h - 0 .755)

Postfault Power Angle Equations

p"z= lE/PG22 lElt lEll ly2Llos (dr,_ 0zr)

= 1.0962 0.5005 I x 1.096 7.6321 os { - 9I.662")

= 0.6012 8.365 in (d , _ I.662)

p,3= tE { 2q3 + El l tElt \l co s 6r ,_ 0rr)

- 7.0712 0.1591 1 x 1.07rx 6.09g os dr 90.g466")

= 0.1823 6.5282 in (d ,_ 0.9466")

Swing Equations-During Fault

PowerSystemStanitity I 491,1

It rnay be noted hat in the above swing equations,P,, ntay be written in

general as follows:

Pn = Pr , - Pr . P,r,n* in (6 - 7)

Solution of Swing Equation

The aboveswingequations during fault followed by post ault) can be solved

by the poinrby-point method presentedearlier or by the Euler's method

presented n the later part of this section.The plots of E and 4 are given in

Fig. 12.42 for a clearing ime of 0.21 s and n Fig. 12.43 or a clearing ime

of 0.08 s. For the case (i), the machine 2 is unstable,while the machine 3 i s

stable but it oscillateswherein the oscillationsare expected o decay f effect

of damperwinding s considered. or the case ii), bothmachines re stablebut

the machine2 has arge angularswings.

#=ff (P^z-P,z)=

= t:9/e.2s - o) elecrdeg/sz

1 2

Machine s referencelnfinite us)



(0.275saul t leared t)

F\g.12.42 Swing urves or machines and 3 of Example 2. 1 or

clearing t 0.275 .

If the fault is a transientone and the ine is reclosed,power angleand swing

equationsareneeded or the period after eclosure.Thesecan be computed romthe reduced Ysusmatrix after line reclosure.

d ,q_ 180a ' r = k

(Pn -P" t )

= t* : /e .r - {0 .1s61 5.531 in 6t -0 .7ss) } l

9

=Y

l.g43g - 5.531 in d, - 0.755") l lect ,eg/sz

Swing Equations-Postfault

#={ fp .2s-10.60r+ 8.365 in d) t .662") }Jrec t eg/s2

dtd, 18 0-* =*12.10-{0.1823 + 6.5282 in(d3 0.g466.)}lelecteg/szdt ' 9

494|

Modernpower SystemAnalysis

50 0Machine 1 is reference lnfini tebus)

Machine2

i l l t l_ L _ _ I _ | _0 .8 0 .9 1 .0

Fault learedafter4 cycles

Fig. 12.43 Swing curves fo r machines 2 and 3 ofExample2.11orclearingt0.0g

Gonsideration of Automatic voltage Regulator (AVR) andSpeed Governor Loops

PowerSystemStabilitvl,,4PFr

. i *= #(P"or - Pco), = 1,2 , . . ,mr7.2

Initial statevector (upon occurrenceof fault) is

xoLk= r= lEot

(t2.74)

(Eo*) using

rotor angle

10 0

at)c)oL

q,

o)!

o(tc

x " z k =

The state orm of swing equations Eq. (12.74)) can be solved by the many

available integration algorithms (modified Euler's method is a convenient

choice).

Computational Algorithm for Obtaining Swing Currzes Using

Modified Euler's Method

l. Carry ou t a load flow studY

voltagesand powers.

(prior to disturbance) using specified

2. Compute voltage behind transient reactances f generators

Eq . (9.31).This f ixes generator.mf magnitudes nd nit ial

lreferenceslack bus voltag" Y?).Compute, Ysu5(during fault, post fault, line reclosed).

Set time count r = 0.

Compute generatorpower outputsusing appropriatg s"us with the help

of the geniral form of Eq. (12.27).This givet Pg},for / 1 /').

Note: After the occurrenceof the fault, the period s divided into uniform

discrete ime intervals At ) so that t ime is counted s /(0), (l), ......A

typicalvalueol ' lt is 0.05 s.

a

J .

4.

5 .

6. Compute( i [ ; ' } , i \7 ' ,k

7. Corttputchc ' i ls l tutc

- - 1,2, . . . , ml f iom Eqs. 12 .74) .

cst i r r t i r lc 'siu ' t = , { t+l ) pr r



state variable Formulation of swing Equations

The swingequationor the hh generators

( r2.73)

For the multimachinecase, t is moreconvenient o organiseEq. (12.73) sstate variable orm. Define

x r k = 6r = lE* '

xz*= 6 t

i t * = x z t

+ = +(p 'oo -p" ) ; - r ,2 , . . ,f id t ' H k '

u

, [ , 0 * r , , t l o + i [ , 0 ) t

I = | 2 . . . . . r r t

,ff') - *V)+ *$'o)t

8. Cortrputehe irstestimatesf E^('+t)

BQ+D E? lcosx,(i*r)+7 si n *\lf '))

g. ComputeP8;'); (appropriate "u5and Eq. (12.72))'

10 . compute[t ;{ in ') , i t :o*t ') ,= r,2, . . . ,mf fromEqs. 12.74).

11. Computehe averagealuesof statederivatives

i [ i , ) urr= [ iu,c l +; l [* t) ]k = 1 , 2 , . . . , f f i

iL'),*,*I*\:|+ tt'i"

Then

1 ' t ALL. Lompute tne lrnal state estimates for | = t\r+t)-

*;Iu" = *([)+ ill) uus,t

,&*r) ,([l + if).^,ratk = 7'2' '' fr t

Compute the final estimate or Eo at t = r('*l) using

BQ+t)= l4llcos xf;+r) + 7 sin *f1r)

14 . Print (",9*t), ; :o*D; k = I,2, ..., m15. Test for time limit (time for which swing curve s to be plotted), .e.,

checki f r> rnnur.f not, r- r+ r an d repeatromstep5 above.Otherwise print results and stop.

The swing curves of all the machines are plotted. If the rotor angle of amachine (or a group of machines) with r"rp".t to other machines increaseswithout bound, such a machine (o r grouf of machines) s unstable andeventually alls out of step.

The computationalalgorithm given abovecan be easilymodified to include

simulation of voltage regulator, field excitation response,saturation of fluxpaths and governor action

Stability Study of Large Systems

To limit the computermemory and the time requirementsand for the sake ofcomputationalefficiency,a largemulti-machinesystem s divided nto a studysubsystemand an externalsystem.The study subsystems modelled n detailwhereasapproximatemodelling is carried out for the externalsubsystem.Thetotal study is rendered y ihe modern echniqueof dynamicequivalencing.nthe externalsubsysteln,utnber<lf nachiness drastically ecluced singvariousmethods-coherency based nethods eing most popurarand widely usedbyvanouspowerut i l i t ies n th e world.

Power System Stabitity|

491;,,

factors which affect transient stability and therefrom draw th" .on"llsions,regardingmethodsof improving the transientstability lirnit of a systemandmaking it as close to the steadystate limit as possible.

For the caseof one machineconnected o infinite bus, t is easi ly seen rom

the angle through which it swings in a given time interval offering thereby amethod of improving stability but this cannotbe employed n practice becauseof economic reasonsand for the reason of slowing down the responseof thespeed governor loop (which can even become oscillatory) apart from anexcessive otor weight.

With reference o Fig. 12.30, t is easily seen hat for a given clearing angle,the accelerating area decreasesbut the decelerating area increases as themaximum power limit of the various power angle curves is raised, therebyadding to the transient stability limit of the system.The maximum steady powerof a system can be increasedby raising the voltage profile of the system andby reducing the transfer eactance. heseconclusionsalong with the varioustransientstability casesstudied,suggest he following method of improving the

transientstability limit of a power system.

1. Increaseof systemvoltages,use of AVR.

2. Use of high speedexcitationsystems.

3. Reduction n system ransf-ereactance.

4. Use of high speed eclosingbreakers see Fig. 12.32).Mo&rn tendencyis to employ single-poleoperation of reclosingcircuit breakers.

When a fault takesplaceon a system, the voltagesat all busesare reduced.At generator terminals, theseare sensedby the automatic voltage regulatorswhich help restoregeneratorerminal voltagesby actingwithin the excitationsystem.Modern exciter systems aving solid statecontrolsquickly respond obus voltage reduction and canachieve rom one-half o one and one-h'alfcycles



I2.T7 SOME FACTORS AFFECTING TRANSIENT STABILITY

We have seen n this chapter hat he two-machinesystem an be equivalentlyreduced to a single machine connectedto infinite bus bar. The qualitativeconclusionsregarding system stability drawn from a two-machine or anequivalent one-machine nfinite bus system can be easily extended to amultimachine system. n the last article we have studied ihe algorithm fordetermining he stabilityof a multimachrne ystem.

It has been seen hat transientstability is greatly affectedby the type andlocationof a fault, so thata powersystemanalystmust at the very outsetof astability study decideon these wo factors. In ou,

"*u-pleswe have selected

a 3-phase fault which is generallymore severe rom point of view of powertransfer' Given the type of fault and its location let us now consider other

(l /2-l ]) gain incrit ical clearing imes br three-phaseaultson the HT bus

of the generator ransformer.

Reducing transfer reactance s another important practical method ofincreasingstability limit. Incidentally this also raisessystemvoltageprofile.The reactance of a transmission ine can be decreased i) by reducing theconductorspacing,and (ii) by increasingconductordiameter seeEq. (2.37)).Usually, however, the conductorspacing s controlledby other featuressuchaslightning protection and minimum clearance o prevent he arc from one phasemoving to another phase.The conductordiametercan be increased y usingmaterial of low conductivity or by hollow cores. However, norrnally, theconductor configuration is fixed by economic considerationsquite apart fromstability. The use of bundled conductors s, of course,an effectivemeans ofreducing series reactance.

Compensation for line reactanceby series capacitors s an effective andeconomical method of increasing stability limit specially for transmission

'4lI I Modern power SystemAnalysis

distancesof more than 350 km. The ciegreeof seriescompensation,however,accentuateshe problems of protective relaying, normal voltageprofiles, andovervoltages rrring ine-to-ground aults.Seriescornpensation ecomesmoreeffective and economical f part of it isof compensationupon the occurrenceSwitehed

serieseapaeitorssimultaneotand raise the transientstability limit tclimit. Switching shuntcapaciiorson olstability limits (see Example 12.2) but the MVA rating of shunt capacitorsrequired is three to six times the rating of switched seriescapacitors or thesame increase n stability limit. Thus series capacitors are preferred unlessshunt elementsare required for olher pu{poses,siy, control of voltageprofile.

Increasingthe number of parallel lines between ransmissionpoints is quiteoften used to reduce transferreactance. t adds at the same ime to reliabilityof the transmissionsystem.Aclditional ine circuits are not likely to proveeconomicalunit I aftetall feasible mprovementshave beencarried out in thefirst two circuits.

As the majority of faultsare ransient n nature, apid switchingand solationof unhealthy lines followed by reclosing has been shown earliei to be a great

help n improving he stabilitymarginr. h. moderncircuit breaker echnologyhas now made t possible br line clearing to be done as fast as n two cycles.Further, a greatmajority of transient faults a'e line-to-ground in nature. It isnatural hat methodshavebeendeveloped or selectivesinglepole openingandreclosing which further aid the stability limits. With ,"f"r".,"" to Fi;. lz.r7, ifa transientLG fault is assumedo occur on the generatorbus, t is immedi ately

::.1,:l_.,ol,lnt the fault herewill now be a definiteamountof power rransfer,

n d d r + + ^ s ^ ^ L L - - - -ab uurereni rrom zero power transfer for the case of a three-phaseault. Alsowhen the circuit breakerpole corresponding o the faulty line is opened, heother wo lines (healthyones) emain ntactso thatconsidlrablepower ransfercontinues o take placevia these ines in comparison o the caseof three-pole

II : . - ^ I r .

I .1. ': / t l l t t , i : :

stzeof rotor reduces nertiaconstant, owering thereby he stability margin.Theloss n stability margin is made up by such featuresas lower reactance ines,fastercircuit breakers nd fastercxcitation systenrs s tliscussetlalreacly, nda faster system valving to be discussed ater in this article.

A stage has now been reached in technology whereby the methods of

irnprovinE=stability; discussetlabove, have been pushed to their limits, e.g.,clearing times of circuit breakers have been brought down to virnrallyirreduciblevaluesof the order of two cycles.With the trend to reducemachineinertias there is a constant need to determine availability, feasibility andapplicability of new methods or maintainingand/or mproving systemstability.A brief account of someof the recent methodsof maintaining stability is givenbelow:

HVDC Links

Increaseduse of HVDC links ernploying thyristors would alleviate stabilityproblems.A dc link is asynchronous, .e., the two ac system at either end donot have to be controlled in phaseor even be at exactly the same frequency as

they do for an ac link, and the power transmittedcan be readily controlled.There s n o risk of a fault in one systemcausing oss of stability in the othersystem.

Breakingr Resistors

For improving stability where clearing is delayedor a large tn)a i, suddenlylost, a resistive load called a breaking resistor is connected at or near thegeneratorbus. This load compensatesor at leastsonreof the reduction of loadon the generators nd so reduces he acceleration. uring a f-ault, he resistorsare applied to the terminals of the generators hrough circuit breakersby meansof an elaboratecontrol scheme. he control scheme etermines he amountofresistanceo be appliedand ts duration.The breaking esistors emain on for



switching when the power ransferon fault clearingwill be reduced o zero. tis, therelbre,easy to see why the singlepole swiiching and reclosingaids instability problem and s widely adopted.These acts arelilustratedby meansofExample 12'12. Even when the stability margins are sufficient,single poleswitching is adopted to prevent large swings and consequentvortage dips.Singlepole switching and reclosing s, of course,expensiu. n t..*s of relayingand introduces he as.socjatec!roblernsof overvoltages ausedby singlepoleopening owing to line capacitances.Methods are available to nullify thesecapacitivecoupling effects.

Recent Trends

Recent trends in design of large alternators end towards lower short circuit

ratio (scR = r/x), which is achieved by reducing machine air gap withconsequentsavings n machine mmf, size,weight and cost. Reduction n the

a matter of cycles b<lthduring fault clearingand after system voltage isrestored.

Short Circuit Current Limiters

Theseare generally used to limit the short circuit duty of distribution lines.Thesemay also be used n long transmission ines to modify favourably thetransfer mpedanceduring fault conditionsso that the voltage profile of thesystems somewhat mproved, hereby aising he system oad level durins thefault.

Turbine Fast Valuing or Bypass Valuing

The two methods ust discussedabove are an attemptat replacing the sysremload so as to increase the electrical output

of the generator during faultconditions.Another recent method of improving the stability of a unit is todecreasehe mechanical nput power to the turbine.This can be accornplished

by rneansof fast valving. where the differeneebetween mechanical nput andreducedelectrical output of a generatorunder a fault, as sensedby a controlscheme, nitiates the closing of a turbine valve to reduce the power input.Briefly, during a fast valving operation, he interceptor valves are rapidly shut(in 0.1 to 0.2 sec) and immediately reopened.This procedure ncreases hecritical switching time lons enoush

stable or faults with stuck-breakerclearing imes. The schemehas beenput touse n somestat ionsn th e USA.

FUII Load Rejection Technique

Fast valving combined with high-speed learing time will suffice to maintainstability in most of the cases.However, thereare still situationswherestabilitvis difficult to maintain. n suchcases,he normal procedure s to automaticallytrip the unit off the line. This, however, causesseveral hours of delay beforethe unit can be put back into operation.The loss of a major unit for this lengthof tirne can seriouslyeopardize he remainingsystem.

To remedy thesesituations,a full load rejection schemecould be utilizedafter the unit is separated rom the system. To do this, the unit has to be

equipped with a large steam bypasssystem.After the systemhas recoveredfrom the shock caused by the fault, the unit could be resynchronizedandreloaded.The main disadvantageof this method is the extra cost of a largebypasssystem.

The systetnshown n Fig. 12.44 s loaded o I pu . Calculate he swing curveand ascertain ystern tability for:

(i ) LG fault hreepoleswitching ollowecl y reclosure.in c ouncl ealthy.(ii) LG fault singlepole switching bllowedby reclosure,

r--- - --lJ I-, t0.15 |7 lI 0 ' 1 p 0' 3 0. 1 |L --z -rfd]-ui-r

i-dTL_r6TT\_ I

(b) Negativesequence network

Fig. 12.45

For an LG fault at P the sequence etworkswill be connectedn series-asshown n Fig. 12.46.A star-delta ransformation educesFig. 12.38 o that ofFig. 12.47 rom which we have the transfer reactance

Xr2(LGaulQ 0. 4+ 0. 4+ nOII ' 'O = 1.450.246

I

I. - t .

lE l=1 '2 )I

t T- t

I- t-

; . i ) l _ Cao

'l]o go '

|=

l xo 0.0e15| | p 1 ' o | | IL----.-' I i-rT60 t;,'r";*u1n""

networl



ine foundhealthy.Switching occurs at 3.75 cycles (0.075 sec) and reclosureoccllrs at 16.25cycles (0.325 sec).All values shown n the figure are in pu. -

l . r . l= . ,,?_ Xr= .1

r - rd 600' 4

Fig.12.46 Connectionf sequence etworksor an LG ault

f-( , r

Fig.12.47 Transfermpedanceor an LG fault

When the circuit breakerpoles corresponding o the faulted ine are opened(it correspondso a single-lineopen ault) the connectionof sequence etworksis shciwn n Fig. 12.48. From the reduced network of Fig. 12.49 he rransferreactancewith faulted line switched off is

P o' 4

H = 4.167 -' .

Xr o3Oj ?*f -h'Ir-!.&-f!X t = 0 ' 1 5 ' L ( /

xo= .t AY-l

Fig. 12.44

Solution The sequence etworksof the systemare drawn and suitably educedin Figs. I2.45a,b and c.

I l ' Xn' l r

tI r / - f t r f \ r ___J

Y.^ (fett l ter l l ine nnen\ - 0 A t A A1 r (l A - 1 . .^l l \ - * - - - - v. - | v . a L-r

V . . + _ L . . / - L

Under healthy conditions transfer reactance s easily obtained from thepositive sequence etwork of Fig. 12.45a as

Xrr(line healthy)= 0.8

Zerosequence

Flg.12-48 Connectionf sequencenetworks ith aulted ine switched ff

lEl=1'z l v l=1 'o

Fig. 12.49 Reduced etworkof Fig. 12.49 iving ransfer eactance

I .o.ro

Pettt = 0

Now

PrN= Pd = 1. 5 si n d

46,- A6n-, @LP^,- ,,M

' a ( n - l )

H = 4.167MJA{VA' 1 P

1,1= JU- - 4.63 x 10a sec2lelectrical egree18 0 50

Taking At = 0.05 sec

(at)' 5. 44.63xI0-4

Time when single/threepole switching occurs

= 0.075 sec (during middle of At) .,Time when reclosing occurs = 0.325 (during middle of at)

Table12.9 swingcuryecalculat ion-threeol eswitchingPower angle eguations

PreJault

p, = rE vrsin d=

X, ,

7 .2x1- S l I l d = l . ) s t n b

0. 8

L

se c

P,u,o P,,(pu)

1. 5 0.6670.827 0.667

.5.4P,, ifielec deg elec deg

6 P "

(pu)6

elec degr .0 0. 00.552 0.448

4 1 . 8

4 1 . 8

0

o*

Negativesequence

P P/ o:4



Initial oad= 1.0puInitial orque ngles givenby

or

During ault

1 = 1.5 sin 5o

6o= 47.8"

0.827 0.682

0. 0 0.7260. 00.00. 00. 0

1. 5 0.5651. 5 0.0521. 5 - 0.551. 5 - 0.9841. 5 - 0 .651

r. 5 0 .497

0.075---+

0 . 1 00 .150.20o.250.30

0.325--+

0.3s0.400.45

0.500.s5

0.600.65

0.2240.564 0.436

0. 0 1. 00. 0 1. 00. 0 1. 00. 0 1 .00. 0 1 .0

0.85 0 .150.078 0.922

- 0.827 r.827- r .48 2.48- 0.98 1 .98

0.146 0.254

1 . 2 4 1 . 83. 6 43.0

9. 0 46.6r4.4 55.619.8 70.02s.2 89.830 .6 I 15 .0

31.436.446.359.710.47 1 . 8

ouu,0.05 1.2

N A

Fl . Z x l", ,=lf f

sind= 0.827in

I)uring single pole switching

Perrr+# sin d= 0.985in

5. 45. 45. 45. 45. 4

0. 85. 0

9. 913.410.7

t . 4

r45.6177.O2r3 .4259.7

3r9.4389.8461.6

The swing curve is plotted in Fig. 12.50 rom which it is obvious hat rhe

\4gqe!_lg',gf tyq]g1_Analysis PowerSystem tability|

505;I

0.40 l .s 0 .998 1.5 - 0. 5 - 2. 70 .45 1 .5 1 .0 1 .5 - 0 .5 - 2 .70 .50 1 .5 1 .0 1 .5 - 0 . s - 2 .7 - 2 .6 89 .50.55 1 .5 0.99851.5 - 0. 5 -2 .7 86.9

2.8 86.60.1 89 .4

r I , r l - - , r - r ] . . . r ro '5 1 .0

Time (sec)

Fig. 12.50 swingcurue or hreepoleswitching it h eclosure

Table12.10 swingcurvecalculat ion-singleol eswitching

- 10 .3 73 .7- 12.r 63.4- 1 3 . 0 5 1 . 3- 12 .6 38 .3- 10.7 25.7- 7. 4 15.0- 3 . r 7 .6

1. 7 4.56. 2 6.29.9 r2 .4

12.2 22.3l3 . l 34 .5

12.5 47.610.9 60.18.6 7 .O6. 0 79.63.3 \ 95 .6' gg .g

The swing curve s plotted n Fig. 12.51from which it follows that he sysremis stable.

Single pole switch off

/'/' Reclosurefault leared)| ' /

I300

II

250i

2oo-II

tII

L

6q)

g)

oE

o

L

(.)q)

J)to

\Iir pote i

switchoff i

/ / l

MACHINE NSTABLE

0 .65 1 .50 .70 1 .50 .75 1 .50.80 1.50.85 1 .50.90 1 .50.95 1 .51 .00 1 .5r 05 r.51 . 1 0 1 . 51 . 1 5 1 . 51 .20 1 .5

r .25 1 .51 .30 1.51 .35 1 .5r .40 1 .51 .45 1 .51 .50 1 .5

0.96 t .44 - 0.44 - 2. 40 . 8 9 4 1 . 3 4 - 0 . 3 4 - 1 . 8

0 . 7 8 1 t . l 1 - 0 .17 - 0 . 90.62 0.932 0.068 0.4'0 .433 0 .6s 0.35 r .90 .2s9 0 .39 0.61 3.30 .133 0 .2 0 .8 4 .30 .079 0 .119 0 .881 4 .80 .107 0 .161 0 .839 4 .50.214 0.322 0.678 3. 70.38 0 .57 0 .43 2.30.566 0.84 0 . 6 0.9

0 . 7 3 8 1 . 1 1 - 0 . 1 1 - 0 .60.867 1 .3 - 0 .3 - 1. 60.946 t .42 - 0.42 - 2. 30.983 1 .48 - 0 .48 - 2 .60.997 1. 5 - 0. s - 2. 7

1 50

100

A

I



i - l - - r

0 5

Time (sec)

Fig. 12.51 Swingcurve or single ol eswitching ith eclosure

I

SE C

I',uu^

(pu),s: i t th

BO

60

II

o)oo)o)

oC)L

ooo)

ra

t,,,(pu)

l'r, 5'4P,,

(pu) elec deg

A belec deg

b

elec deg

0

o ,oory

0.050.075---

0 . 1 00 . 1 5o.20o.250.30

4325---+0.35

1 .5 0 .661 1 .00.827 0.667 0.s52

0.827 0.682 0.s64

0. 00.4480.224 1. 20.436 2. 4

0.285 1. 50.230 r. 20 .166 0 .90.107 0.60.060 0. 3

0.485 - 2. 6

4 .804 1 . 8

r. 2 41.83. 6 43.0

5. 1 46.66. 3 5r .77. 2 58.01. 8 65.28.1 73 .0

5 . 5 8 1 . 1

(Contd... .)

0.98s 0.7260.98s 0.7840.985 0.8480.98s 0.9080.985 0.956

1.5 0 .988

0 .7Ls0.770.8340.8930.940

1 .485

MACHINE TABLE

t 2 . l

t 2 .3

12.4

506 | Modern o@is

PROBEIlIS

A two-pole, 50 Hz, 11 kv turboalternatorhas a rating of 100 Mw,powel factor0.85 agging.The rotor has a momentof inertia of a 10,000

12.2 Two turboalternatorswith ratings given below are nterconnectedvia ashort transmissionine.

Machine 1: 4 poIe, 50 Hz, 60 MW, power factor 0.g0moment of inertia 30,000 kg-rn,

lagging,

Machine 2 pole, 50 Hz, 80 MW, power factor 0.85 lagging,moment of inertia 10,000kg--'

Calculate he nertia constantof the single equivalentmachine on a baseof 200 MVA.

Power stationt has our identical generatorsetseach ated g0 MVA andeach having an inertia constant7 MJA4VA; while power station 2 hasthree sets eachrated 200 MVA, 3 MJA4VA. The stations are locatldclose ogether o be regarded sa singleequivalentmachine or stabilitystudies.Calculate he nertia constantof the equivalentmachine on 100MVA base.

A 50 Hz transmissionin e 50 0 km long with constants iven below ie sup two large power areas

I ina

from its prefault position, determine he maximum load that could betransferredwithout loss of stability.

I2.8 A synchronousgenerator s feeding250 MW to a large 5O Hz networkover a doublecircuit transmissionine. The maximum steadystateDower

that can be transmitted over the line with both circuits in operation is500 MW and is 350 MW with any one of the circuits.

A solid three-phase ault occurring at the network-end of one of the linescausest to trip. Estimate the critical clearing angle n which the circuitbreakersmust trip so that synchronism s not lost.

What further information is needed o estimate the critical clearing time?

12.9 A synchronousgeneratorrepresented y a voltage sourceof 1.05pu inserieswith a transient reactanceof 70.15pu and in inertia constantF/ =4.0 sec, s connected o an infinite inertia system hrough a transmissionline. The line has a series eactance f70.30 pu, while the nfinite inertiasystem s represented y a voltagesourceof 1.0 pu in series with atransient eactanceof 70.20 pu.

The generator s transmitting an acti're power of 1.0 pu when a three-phase ault occurs at its terminals. f the fault is cleared n 100 millisec,determine if the system will remain stable by calculating the swingcurve.

12.10 For Problem 12.9 ind the critical clearing ime from the swin! currre ora sustainedault.

l2. l l A synchrt )nouseneratorepresentcdy a vol tage f l. l5 pu in ser ieswith a transient eactances c<lnnectedo a large power systemwithvol t t tgc1. 0 pu throt lgh powcr rrc lwork.Th c cquivalcnt larrs ienttransf'er eactanceX betweenvoltagesources s 70.50pu.

After the occurrence f a three-phaseo grounclault on on eof th e inesof the power network, two of the line circuit breakersA and

F i n d t h e s f c : t r l v s l z f e c f e h i l i t rr l i m i t i f l l , / | - l l / | - t n l l t , \ / / . . ^ - ^ +^ - + r. - - ' . , , J r r r J r r r r r r r , , , r , t t _ | vR t

_ , \ r v A V \ L t , | t : l t d l l l r .

What will the steadystatestability limit be if line capacitance s alsoneglected?What will the steadystatestability imit be if line resistanceis alsonegloctcd ' / ornnrcl t t n t l rc rcsul ts.

R = 0.11 f ) /km

C = 0.009 lFlkm

L - 1.45mH/km

G = 0



t 2 .5

1 2 . 6

t 2 .1

B operatesequentiallyas follows with corresponding ransient transferreactancegiven therein.

(i) Short circuit occurs at 6 = 30", A opens nstantaneouslyo make X= 3. 0 pu .

(ii) At 6 = 60o, A recloses,X = 6.0 pu.(i i i ) At 5=75o, A reopens.(iv) At d = 90o, B also opens o clear the fault making X = 0.60 pu

Check f the systenrwil l operate tably.

12.12 A 50 Hz synchronousgeneratorwith inertia constantH = 2.5 sec anda transient eactanceof 0,20 pu feeds 0.80 pu active power into aninfinite bus (voltage I pu) at 0.8 agging power lactor via a network withan equivalentreactanceof 0.25 pu.

A three-phaseault is sustained or 150 millisec acrossgeneratorterminals.Determine hrough swing curyecalculation he orque angle6,250 millisec, after fault initiation.

A power deficient area receives 50 MW over a tie line from anotherarea.The maximumsteady tatecapacityof the ie ine s 100MW. Findthe allowable sudden load that can be switched on without loss ofstability.

A synchronousmotor is drawing 30vo of the maximum steady statepower from an infinite bus bar. If the load on motor is suddenlyincreased y 100 per cent,would the synchronism e lost? If not, whatis the maximum excursionof torque angle about he new steadystater<ltorposition.

The transfer reactancesbetween a generator and an infinite bus baroper i l t ingrf 20 0 kV t rnder ar ious ondi t ions n fhe nterconnectorro :

Pretault

During faultPostfhult

S0 0 per phase

m O per phase2ffi {) per phase

12.13 A 50 Hz, 500 MVA,400 kV generator(with transformer) s connectedto a 400 kV infinite bus bar through an interconnector.The generatorhasF1 2.5 MJA4VA, voltagebehind transient eactanceof 450 kV andis loaded 460 MW. The transfer reactancesbetween generatorand bus

PowerSystemStabitityI

5(D

12. Kundur, P., Power SystemStability and Control, McGraw-Hill, New York, 1994.

13. Chakrabarti,A., D.P. Kothari and A.K. Mukhopadhyay,PerformanceOperation

and Cqntrol of EHV Power TransmissionSystems,Wheeler Publishing, New

Delh i , 1995.

14. Padiyar,K.R., Povter System Stability and Control, Znd

lcatlons, Hy15. Sauer,P.W. and M.A. Pai, Power SystemDynamicsand Stabiliry, Prentice-Hall,

New Jersey,1998.

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

Rejection", EEE Trans, L972, PAS 9I 1624.t

17. Kimbark, E.W., "Improvement of Power SystemStability", IEEE Trans., 1969,

PAS-88: 73.

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 1400MW BrakingResistor", EEE Trans.,1975,94:602.

Nanda,J., D.P. Kothari, P.R. Bijwe and D.L. Shenoy, A New Approach for

Dynamic EquivalentsUsing Distribution FactorsBasedon a Moment Concept",

Proc. IEEE Int. Conf. on Computers,Systemsand Signal Processi4g,Bangalore,

Dec. 10-12, 1984. \

Dillon, T.S., 'Dynamic Modelling and Control of Large Scale System", Int.

Journal of Electric Power and Energy Systems, an. 1982,4: 29.

Fatel,R. , T.S. Bhatti and D.P. Kothari, Improvement f PowerSystemTransient

stability using Fast valving: A Review", Int. J. of Electric Power components

and Systems,Vol. 29, Oct 2001, 927-938.

Patel,R., T.s. Bhatti and D.P. Kothari, "MATLAB/simulink Basedrransient

Stability Analysis of a Multimachine Power System, JEEE, Vol. 39, no. 4, Oct.

During fault

Postfault

1.0 u

0.75pu

Calculate he swing curveusing intervals of 0.05 sec and assuming hatthe fault,is clearedat 0.15 sec.

I2.I4 Plot swing curvesand checksystemstability for the fault shown on thesystemof Example 12.10 or fault clearing by simultaneousopeningofbreakersat the endsof the faulted line at three cycles and eight cyclesafter the fault occurs.Also plot the swing curye over a period of 0.6 secif the fault is sustained. or the generatorassumeH = 3.5 pu, G = 1 puand carry out the computations in per unit.

12.15Solve Example 12.10 or a LLG fault.

REFERECES

Books

1 Stevenson,W.D., Elementsof Power SystemAnalysis,

New York, 1982.

Elgerd, O.I., Electic Energy Systems Theory: An

McGraw-Hill, New York, 1982.

4th edn., McGraw-Hill,

iniroduciion, 2nd edn.,

The Iowa. Anderson,P.M. and A.A. Fund, Power SystemControl and Stability,

StateUniversity Press,Ames, owa, 1977.

20 .

2r .

22 .

23 .



2002,pp 339-355.

Patel R., T.S. Bhatti and D.P. Kothari, "A Novel scheme of Fast valving

Control", IEEE Power EngineeringReview,Oct.2002, pp. 4446.

Patel,R., T.S. Bhani and D.P. Kothari, "Improvementof PowersystemTransient

stability by coordinated operationof Fast valving and Braking Resistor", To

appear n IEE proceeriings-Gen., rans and Distribution.

4. stagg, G.w. and A.H. o-Abiad, computer Methods n Power system Analysis,Chaps and 10 ,McGraw-Hi l lBook Co., New York, 1968.

5. Crary, S.8., Power SystemStability, Vol. I (Steady State Stability), Vol. II(TransientStability), Wiley, New York, 1945-1947.

Kimbark,E.W., PowerSystem tability, Vols 1, 2 and3, Wiley, New york, 1948,

Veuikorz,Y.A., TransientPhenomenan Electrical Power System translatedrom

the Russian),Mir Publishers,Moscow, 1971.

8. Byerly, R.T. and E.w. Kimbark (Eds.), stability of l^arge Electric power

Systems,EEE Press,New York, 1974.

9. Neuenswander, .R., Modern Power Systems,International ext 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 TransientStability Analysis using the

Transient Energy Function Method, Prentice-Hall,New Jersy, 1992.

. AL+ .

25 .6.

7 .

13.1 INTRODUCTION

In Chapter7, we have beenprimarily concernedwith the economicaloperationof a power system' An equally important factor in the operation of a powersystem s the desire to maintain systemsecurity. System security involvespractices uitablydesigned o keep he systemoperatingwhen components ail.Besides conomizing n fuel costand minimizrngemissionof gases co , cor,Nox, sor), the power systern houldbe operationally .secure,,.

An operation_ally "secure" power system s one with low probability of, systernblack out(col lapse) r equipment amage.f th epro."r, uf cascadingailures ontinses.the systernas a whole or its tnajor parts may completely collapse.This isnormally referred to as systemblackout. All these aspectsrequire securityconst ra inedowersystem pt imizat ionSCO).

PowerSystemSecurityI

Sll

Most of the security related functions deal with static "snapshots" of thepower system.They have o be executed t intervals ompatiblewith th e rateof changeof system tate. hi s quasi-stat ic pproachs, to a largeextent, heonly practical approachat present,since dynamic analysisand optimization areconslder4blymole {!fficu!! 4nd cqmpurallo44lly 1aqte ime corrsulurg,

System security can be said to comprise of three major functions that arecarriedout in an energycontrol centre: i) systemmonitoring, ii) contingencyanalysis,and (iii) comectiveaction analysis.

Systemmonitoringsupplies he power systemoperators r dispatcherswithpertinentup-to-datenformationon the conditionsof the power systemon realtime basisas oad andgeneration hange.Telemetrysystems neasure,monitorand ransmit he data,voltages, urrents,current lows and he statusof circuitbreakers nd switches n every substat ionn a transrnissionetwork.Further,other critical and important information such as frequency,generator outputsand transformer ap positionscan also be telemetered. igital computers n acontrol centre hen process he telemetered ata and place hem in a data baseform and inform the operatorsn caseof an overloador out of limit voltage.

Important data are also displayedon large size monitors.Alarms or warningsma y be given if required.

Stateestimation Chapter14) s normally used n suchsystems o combinetelemetered ata to give the best estimate in statistical ense)of the currelttsystemconditionor "state".Such systemsotten work with supervi$ory ontrolsystems o help operators ontrol circuit breakersand operateswitchesand tapsremotely.Thesesystems ogetherare called SCADA (supervisory ontrol anddataacquisit ion) ystelns.

The second ma-ior security function is contingency analysis. Modernoperat ion omputers av econtingency nalysis rograms tored n them.Theseloresee ossiblcsystetn roubles outages) efore hey occur.They studyoutageevents and alert the operators o any potential overloadsor serious voltagevio lat i t lns. or exalnple,he si rnplesti r rmof cont ingencynalysis an be pu t



Since security and economy normally have conflicting requirements, t isinappfopriateo treat hem separately. he fina.laim of economy s the securitylunction of the utility company.The energymanagement ystem(EMS) is tooperate the system at minimum cost, with the guaranteedalleviation ofemergency onditions.The emergency onditionwill dependon the severityoft io lat ions f operat ingi rn i ts branch' lows nd bu svol tage imi ts) .Th e mostsevere iolations esult iom contingencies.n irnportant art of securitystudy,therefbre,movesaround he power system'sability to withstanrj he effectsofcontingencies.A particular systemstate s said to be secureonly with referenceto one or more specific contingency cases, and a given set of quantitiesmonitored or violation. Most power systemsare operatedn such a way thatany singlecontingencywill not eaveother.o-pon"nts heavily

overloaded, othat cascadingailures are avoided.

together with a standard LF program as studied in Chapter 6, along withprocedures o set up the load flow dafa for eachoutage o be studiedby theLF plogram. This allows the systemoperators o locatedef'ensive peratingstateswhere no singlecontingencyeventwill generate verloadsand/orvoltageviolat ion:; .This analysis hu s evolves operat ingconstraintswhich may becntpioycdn thc iD (ccot ro ln icispatch) nd UC (uni tcornnr i t rncl r t )rograrrr .Thus contingency nalysis arricsou t ornergcncydenti l ' icat ionnc l what if ' 'simulat ions.

The third major security unction, corrective actionanalysis,permits theoperator o change he operationof the power systemf a contingencyanalysisprogram predicts a seriousproblem in the event of the occurrenceof a certainoutage.Thus this providespreventiveand post-contingency ontrol. A simple

example of corrective action is the shifting of generation rom one station toanother.This ma y result n change n power f lows an d causinga change nloading on overloaded ines.

j5!2,1 ModernPo@is

Thresehree trnctions ogetherconsistoi a very compiex set of toois that heipin the secure perat ion l' a power system.

T3.2 SYSTEM STATE CLASSIFICATION

Dyliacco [13] and urtherclarifiedby Fink andCarlsen 23l in order o definerelevant EMS (EnergyManagementSystem) unctions.Stott et. al [15] havealso presenteda more practical static security evel diagram (seeFig. 13.1) byincorporatingcorrectivelysecure Level 2) andcorrectableemergency Level4)security evels.

In the Fig. 13.1, arrowed linep represent nvoluntary transitions betweenLevels 1 to 5 due to contingencies..The emoval of violations from Level 4normally requiresEMS directed"corrective rescheduling"or "remedial action"bringing the system o Level 3, from where it can return to either Level I or2 by further EMS, directed "preventive rescheduling" depending upon thedesired operationalsecurityobjectives.

Levels I and 2 represent ormal power systemoperation.Level t has the

ideal securitybut is too conservativeand costly. The power systemsurvivesanyof the credible contingencieswithout relying on any post-contingencycorrectiveaction.Level2 is more economical,but depends n post-contingency orrectiverescheduling to alleviate violations without loss of load, within a specifiedperiod of time. Post-contingency perating imits might be different from theirpre-contingencyvalues.

13.3 SECURITY ANALYSIS

System securitycan be broken down into two major functions that are carriedout in an operations ontrol centre: (i ) securityassessment, nd (ii) securitycontrol. The tormer gives the security level of the systemoperatingstate.The

I srr

=o-3

o-ocoo

oooo

ooL

o-o

p b .6 q

o

oo)

oo

oo

o

L

f()oU)

C)

(E

o

Eooo

o

oo_

at

.d,lr

P c 9 cE O = O

d H d 8

occ)o)c

oo

9 c

6 H

g >. F O E=-EfiE 8 E

z E C g

g#EE( ' 5 o 3

xo 9 .' E b

.= g,

g)'Ec c

F O( g ( J

b o d8 b b: 5 8d d ' 5( l ) - . o

. = 9 =

los i5 c o(t ) .=

HE ' =e o y 9 c

F . O

o

og,c

co(J

E E8 E( ! 6

B Er: .9q Fo oo o _

E; .^Et-E. ; E b

abg

o,c

oL

q)

o.oozEo

o.o-=o

Eoo

- o =

g H €' = O -

o _ o xo - - c . Y

! ie= 9 8



latter determines he appropriatesecurity constrainedscheduling required tooptimally attain he targetsecurity evel.The security functions n an EMS can be executed n 'real

time' and 'study'

modes. Real time application functions have a particular need for computingspeedand reliability.

'fhestatic securit.vevel of a power system s characterised y the presence

or rrtherwiseof emergencyoperatingconditions (limit violations) in its actual(pre-contingency)or potential (post-contingency)operating states.Systemsecurity assessments the processby which any such violations are detected.

Sy tem e.: €s lt€rrt nvolves two func tions

(i) systemmonitoringand (ii) contingencyanalysis.Systemmonitoring providesthe operatorof the power systemwith pertinentup-to-date nformation on thecurrentcondition:; lf the power system. n its simplest orm, this us t detectsviolations n the actualsystemoperatingstate.Contingencyanalysis s much

=eeo-o

8 s 9o o ' : =L - A

, ^ k a t , ^ i :* ' u o * 9o c f r o ) c ,> o x > oA z C . A i lJ 3 ( D J 3

o

0)o)

Eo)o-o([()q)L

C)

r ' / ^ Y

ea

d 8

6-L

oc)a

o

: : 0( l ) -> o

_9 )

- Ea dg E

,.sfe ;l Modern power SvstemAnalvsis

Only a small proportionof work on optimal power flow (OPF)has aken ntoaccount he security constraints.The most successful pplicationshavebeen othe security constrainedMW dispatch OPF sub-problem. The contingency-constrained oltageivarreschedulingproblem,asof the writing of this text, stillremains to be solved to a satisf desree.

The total number of contingencyconstraints mposed on SCO is enormous.The SCO or contingency constrainedOPF problem is solved with or withoutfirst optimizing with respect o the basecase precontingency)constraints.Thegeneralprocedureadopted s as follows:

(i) Contingency analysis s carried out and cases with violations or nearviolations are dentified.

(ii) The SCO problem s solved.(iii) The rescheduling in Step 1 might have created new violations, and

therefore step 1 shouldbe repeated ill no violations exist.Hence,SCO represents potentially massiveadditional computingeffort.An excellent comprehensiveoverview of various available methods spresented y Stottet. al [15].

There is still great potential for further improvement in power systemsecurity control. Better problem formulations, theory, computer solutionmethodsand implementation echniquesare required.

T3.4 CONTINGENCYANALYSIS

In the pastmany widespreadblackoutshave occurred n interconnectedpowersystems.Therefore, it is necessary o ensure that power systemsshould beoperatec!mosf economic:r! ly uc h that povrer s cle! i . rereclel iably.Reliableoperation mplies that there s adequatepower generationand the samecan betransmittedeliably to the oads.Most power systems re designedwith enoughredundancyso that they can withstandall rnajor ailure events.Here we shallst trdy hc possiblcconsccprcnccsrncl crncdialact ions cquircdby

I sraiower SystemSecurity _

SLACKBUSf + o * y s +s y s

1.02421-5 .0"3

1.O6tO" 4 1 . 0 2 3 6 t - 5 . 3 "--.>40.7 1.2 _ 39.5_i3.0:_ + 18.9 y5.2 _16.9 13.2-<

,f 8. ey8.6,

+Ja.a--/e.8 i 6 . 3 - j 2 .3

I z+.2js.a ----/--zt.s-i5.9t

'L?:'-q''---'t- az.s

1.0474t -2

1.0610" 1

!G-)) ,zo 1t o

t4 0 +i30

Fig. 13.2 Base Case AC

SLACKBUS

*o o it o

Line low or sample bussystem

l+-u-" ' '.i;n;l--+s 53.7i7.2 -* 5

i1 .o17gt -6 .2"Y

*S4.9 +17.3

:63 .1 + 1O.2

{ + s * y t s

1.01071-5.9" 3

->48.6 + 5.2 -46.7 - j5.3<- + 38.5 10.6 -38.4 -i1.1:-

+--

3a? /'e

f + o + 7 s

4 1.00682 6.6"

{ r.o 7a. s

i r .s ;t . r-<-- 61.6 i8.9 5

1.O114t-6.4"

loo*lt o

AC LoadFlow Linebetween and4 is open)

+81 8 i 5.5

t- eo.o

1.0468t -2.

rl r { 2s+ilo\

l J . /

t4 0+i30

Fig. 13.3 Postoutage



two nrainfa i lureevents; ine.orr tagesncl enerat ingrni t a i lures.To explain he probleln rief ly,we considerhe ive-bus ystem f Ref 'erence

L 0J .Th e base ase oad low results br th eexample re given n t ig. 13.2and,sl t r rw r f ' low<t f24.7MW an d3.6 MVAR on the ine }o m bu s2 to hu s3. L.cttl s lsst l l l lcha tat prcscnl , e lr conly ntcrestecln the MW loading f the ine.Let us examinewhat will happen f the ine from bus2 to bus 4 were o open*.' l ' l rc

rcsul t ingin c l ' lows rtc l o l tagcs lc shown n l r ig . 13.3. t nraybe notcdthat the flow on the line 2-3 has increased o 31.5MW and that most of theother line flows are also changed. t may also be noted fhat bus voltagemagnitudesalso get aff'ected, articularly at bus 4, the change s almost2To essfrom 1.0236 o 1.0068pu. Suppose he ine from bus 2 to bus 5 were to open.Figure 13.4shows he resulting lows and voltages.Now the inaximumchange

t:fgl.g.j",jus

5 which is almost 107o ess.

xSimulationof line outages more complex han a generator utage, ince ineoutage esultsn a changen system onfigurations.

t a s r s I ao 7s

- 52.si 13.0

SLACKBUS

( c )

| +s,s 1tz.t

': : )l l:t't'

- 5

0.8994t-16.?

12 0 i1 0 'foo yr

between and is open)

| 0.se35/ 6.s" o11 '06 t0 " l ' 54 .4+ i 1 i . 3s 2 .1 - i 9 .7 | - s 1 .4 i 7 .3 -s 1 .1_1_ r

tl;f+ 8 0 . e + 1 3 . 2 1

I t l _ q q . z _ j 1 z . s

Fig. 13.4 Postoutage C LcadFlow Line

5LAUK HU S

e t4 s+ l15

| 1 .oo61ts .7" 311 . 0 6 t 0 " l ^ Il_ -!.6 +1t.s _ 4s.9 j7.6 <_i __22.4 2.6

{nz e+ i z 1 .7 l L - - - i -1 - II - - - V_21.5_ j4 .8

GiveAlarmsignal

;'518 | rr,roderno@ist-

l + o * 1 s

4 1 . O 0 4 3 t - 6 . 1 "_22 .3 + 0.7 <_

{ t. s - 1t.t

-25.2- j4 .4 +

I ---ti ' ' '"

1- rzo .g1ts. t l( - ' - _-l l r

2..]--u+ _+ _53.6 16.81.o24s t -3 .7 .

iY

l ,zo 1l o

<- -52.5 -16.5

f - z s - 7 s . s5

o 9956/_-71"

{ o o * i r o

lostgenerat ionsGiveAlarmsignal

Fig. 13.5 Postoutage C LoadFlow Generator outage,picked p by generator )

Figure 13.5 s an example f generator utage nc l s selectedo explain hetact that generatoroutages an also result n changes n line flows and busvoltages. n the exampleshowr in Fig. 13.5al l the generation os t from bus 2is picked up on the generatorat bus 1. Had therebeen more thanZgenerarorsin the samplesystemsay at bus 3 also, t was possible he loss of lenerationon bus 2 is made up by an increase n generation at buses 1 and 3. Thedifferences n Iine flows anclbus voltages rouldshow how the ost gener.ationis sharedby the remaining units is quite significant.

It is important o know which line or unit outageswill render ine flows orvoltages o cross he irnjts.To find the eff'ects f outages, ontingency nalysistechniquesare empioyeci.Contingency analysis models single failure events( i 'e ' one- l inc utagcsiro l lc uni toutugcs) r nrul t ip lc cpr ipnrcnti r i lurc vc^ts(failure of multiple unit or lines or their combination)one afteranotheruntil all"credible outages"are considered.For eachoutage,all lines anclvoltages n the

- - ' t A - . .

1;no*rifii.iuior".""-i;{offi 'l

l'r'.,o' ''l''''



netrvorkare checkedagainst heir respectiveimits. Figure 13.6depictsa flowchart llustratinga simple method for carryingout a contingencyanalysis.One of the important problems is the selectionof "all credible outages,,.

Execution ime to analyse everal housand utagess typically I min based ncomputeran danalyt icalechnology sof 2000.An erpproxirnatenodel uclrasDC load flow ma y be used to achievespeedysolut ion if voltage s aisorequired, hen full AC load flow analysishas to be carried out.

_ _ l,rr\=_ ve s jciventarnviolation?>*-t_-qJ-T-

,*" I, N oI

' - - - iAil

--\

FIg. 13.6 A simple echniqueor contingencynalysis

620j;:f uodernPowerSystemAnatysisI

^ a ? f r E r r a r E ? r t r E l t

IJ.C DI I IYDI I IVI I Y .FAUI-L' I (s

A .security nalysis rogram s run in a load dispatch entrevery quickly to helpthe operators.This can be attemptedby carrying out an approximate analysisand using a computer systemhaving multiple processorsor vector processors

y anarysts. ne s may uate an equrvalentshouldbe used or neighboursconnected hrough ie-linos. We can eliminateallnon-violation casesand run complete exact program for "critical" casesonly.This can be achievedby using techniques uchas "contingencyselection"or"contingency screening", or "contingency ranking". Thus it will be easy towarn the operation staff in advance o enable hem to take correctiveaction fone or more outageswiil result n seriousoverloadsor any violations.One ofthe simplest ways to present a quick calculation of possible overloads s toemploy network (linear) sensitivity factors.These actors give the approximatechange n line f lows for changes n generat ionn the systemand can becalculated rom the DC load flow. They are mainly of two types:

1. Generationshift factors

2. Line outage distribution factorsBriefly we shall now describe the useof those actors without deriving them.

Reference 7] gives their deri iation.

The generation shift factorsl cr.,; re defined as:

limits and those violating their limit can be informed to the operator for

necessary ontrol action.

The generationshift sensitivity factors are inear estimatesof the change n

line flow with a change n power at a bus. Thus, the effects of simultaneous

principle of superposition.

Let us assume hat the oss of the ith generators to be made up by governor

action on all generatorsof the interconnectedsystemand pick up in proportion

to their maximum MW ratings.Thus, the proportionof generationpick up from

unit k (k * i) would be

where,

4t = Change n MW power flow on hne I whention, AP", takes place at the ith bus

Here, t is assumed hat LPotis fully compensated y an equaland oppositechange n generationat the slack (reference)bus, with all other generators

where

Pn,,,,,u, maximum MW rating for rnth generator

g*i= proportionality factor for pick up on kth unit when ith unit fails.

Now, for checking the /th line flow, we may write

j, = ff * 0r; APo, -E,lau,

\ri LPotl

In Eq. (13.8) t is assumedhat no unit will violate ts maximdm limit. For

unit limit violation, algorithm can easily be modified.

Similarly the line outagedistribution factorscan be used or checking if the

line overloadswhen solllc of the lines are lost.

The line outage distribution factor is defined as:

d, , ,= *J i

( 13 .4 )

a changen genera-

(13 .7 )

(13 .8 )

(13.e)



remaining ixed at their original power generations. he factor al,then gives hesensitivityof the /th line flow to a change n generation t ith bus.Let us nowstudy the outage of a large generating unit and assume that all the lostgeneration (Pod would be supplied by the slack bus generation.Then

APo,- -P1i

where

dt,i = line outagedistributionoutageof ith line.

Aft = change n MW flow on /th iine'

fi- precontingency ine flow on ith line

lf precontingency ine flows on lines / and i, tlrepower flow on line / with line

i out can be found out employing "d" factors.

( 13 .10 )

factor when monitoring /th line atter an

(13.s)

and he new power flow on each ine could be calculated sing a precalculatedset of " d' factors as given below.

ft = f i * dti APc, for all lines V / (13.6)

where, ft- power flow on /th line after the failure of ith generator

f i= power flow on /th line before the failure

orprecontingency

power flow

Here,

fi ^d foi =precontingency or preoutage lows on lines / and i respectively

?,= ff *d,, , f ,o

fr = power low on /th line with ith line out.

iii l Modernoa

Thus one can check quickiy by precaiculating 'd'factors ali the lines for

overloading for the outage of a particular line. This can be repeated br the

outage of each ine one by one and overloadscan be found out for corrective

action.

It may be noted hat 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 usmgtelemetry systemsor with stateestimation techniques. f the network undergoes

any significant structural change, he sensitivity factors must be updated.

Find the generationshift factors and the line outagedistribution factors or 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 Matrix or Five-busSampleSystem Bus 1 as a reference)

0.05057 0.03772 0.04029 0.4714

0.03772 0.08914 0.07886 0.05143

0.04029 0.07886 0.09514 0.05857

0.04714 0.05143 0.05857 0.13095

Table 13.2 Generat ion hif tDistribut ionactor or Five-bus ystem

J srr

j= l j=2 j=3 j=4 j=5 j=6 j = 7

(line 1-2)(line-3)(Iine2-3)(line2-4)(line -5)(line -4)(line4-5)

( l ine -2 ) 0. 0 1.00010.3331 0.2685 -0.2094 0.3735 0.2091

= 3 (line2-3) -0.4542 0.4545 0 0.4476 0.3488 -0.6226 -0.3488 J \ r r \ r w L - J ) -v

= 4 (hne 2-4) -0.3634 0.3636 0.4443 0.0 0.4418 0.6642 -O.44r8

= 5 ( l ine 2-5) -0 .1819 0.1818 0.2222 0.2835 0.0 0 '3321 1. 0

= 6 ( l ine 3-4) 0.5451 -0.5451 -0.6662 0.7161 0.5580 0.0 -0.5580

= 7 ( l ine4-5) 0.18 i6 -0.1818 -0.2222 0.2835 1.0002 -0.3321 0' 0

It has been found that if we calculate the line flows by the sensitivity

methods, hey come out to be reasonably lose o the valuescalculatedby the

full AC load flows. However, the calculations carried out by sensitivity

methodsare aster han thosemadeby full AC load flow methodsand herefore

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 contingencyanalysis.

The simplestAC securityanalysisproceduremerely needs o run an AC load

flow analysis or eachpossible unit, line and transformeroutage.One normally

does ranking or shortlisting of most likely bad caseswhich are ikely to result

in an overload or voltage imit violation and other casesneed not be analysed.

Any good P1(performance ndex can be selected) s used or rankirig. One such

P/ is

0

0

0

0

0.

( 1 3 . 1 1 )

For large n, PI will be a small number f all line flows are within limit, and

will be large f one or more lines are overloaded.

For rr = I exactcalculations iur be done br P1 .P1 ablecan be ordered rom

largestvalue to least.Suitablenumber of candidates hen can be chosen for



Bus Bus 2 furtheranalysis 7] .If voltagesare to be included, hen he following PI can be employed.

l = 1 ( l i n e 1 - 2 )

/ = 2 ( l i n e 1 - 3 )

/ = 3 ( l i n e 2 - 3 )

I = 4 ( l i n e 2 - 4 )

/ = 5 ( l i n e 2 - 5 )

/ = 6 ( l i n e 3 - 4 )

I = l ( l i n e 4 - 5 )

- 0.8428- o.t572

o.0714

0.0571

0.0286- 0.0857- 0.0285

(13.12)

Here, Alvil is the difference between he voltage magnitude as obtainedat the

end of the lPlQ FDLF algorithm Alvl-u* it the value ixed by the utility.

Largest vaiue oi Pi is piaceciat the top. The security arraiysis rray rrow -ue

started or the desirednumbel of cases own the ranking ist.

$ummary and Further Reading:

Reference[25] has discussed he concept for screeningcontingencies.Such

contingencyselection/screeningechniques onn the foundation or many real-

time computer security analysisalgorithms.

Reference[15] gives a broad overview of securityassessment nd containan

;;;;t]r",bibliography covering the literature on"security assessmenrup to

Reference [11] gives an excellent bibliography on voltage stability. Thistopic is discussedbriefly in the next section.

13.6 POWER SYSTEM VOITAGE STABILITY

Power transmissiol tuglgility has traditionally been Iimited by either rotorangle (synchronous) stability or by thermal loaoing capabilities. The blackoutproblem has been inked with transient stability. L-uckily this ;roblem is nownot that serious becauseof fast short circuii clearing; po*Lrru excitationsystems, and other special stability controls. ElectriJ *rnpuni"s are nowrequired to squeeze he maximum possible power through

"ii.Gnetworks

owing to various constraintsn the constructionof generationand transmissionfacilities.

voltage (load) stability, however, is now a main issue in planning andoperating electric power systemsand s a factor reading o limit po-w; transfers.

voltage stability is concernedwith the ability of a power system to maintainacceptable voltagesat all buses n the systemunder normal conditions and afterbeing subjected o a disturbance.A power system s said o have entereda stateof voltage instability when a disturbante resurts in a pro!."rriu" anduncontrollable decline n voltageInadequatereactive.-power upport from generatorsand transmission inesleads to voltage.instabilityot uoitug" collapse,which have resulted in several

major system failures n the world. Th"y -",(i) south Florida, usA, systemdisturbanc of 17 May r9g5, (transient, 4sec)

(ii) French systemdisturbances f Decembe 19, r97gand Januar 12, rgg7,(longer term).

(iii) swedish systemdisturbanceof December27, rgg3 (longer term, 55 sec)(iv) Japanese Tokyo)

l#GrffiEIE!F}ILfr

stability, the dynamics mainly involves the loads and the means or voltagecontrol. Ref [11] provides a comprehensive ist of books, reports, workshops

and technical papers related to voltage stability and security.

Definitions: [2]

A power system at a given operating state s small-disturbance voltage stableif, following any small disturbance,voltagesnear loads are dentical or closeto the pre-disturbancevalues. The conceptof small-disturbancevoltage stabilityis related to steady-statestability (Chapter 12) and can be analysedusing small-signal (linearised)model of the system.

A power systemat a given operatingstateand subject to a given disturbancets voltage stable f voltages near oads approachpost-disturbance quilibriumvaluei. The concept of voltage stability is related o the ffansientstability of apower system.The analysis of voltage stability normally requiressimulation ofthe system modelledby non-linear diffdrential-algebraicequations.

A power systemat a given operatingstateand subject to a given disturbanceundergoes voltage collapse if post-disturbanceequilibrium voltages are belowacceptable limits. Voltage collapse may be total (blackout) or partial. The

voltage instability and collapse may occur n a time frame of a second. n thiscase the term transient voltage stability is used. Sometimes it may take up totens of minutes n which case he term long-term voltage stability is used.

The term voltage security means he ability of a system, not only to operatestably, but also to remain stable following any reasonably crediblebontingencyor adversesystem change such as load increases 2].

Voltage stability involves dynamics, but load flow based static analysismethods are generally used for quick and approximate analysis.

Figure 13.7depicts how voltage stability can be classified nto transientandlong-term time frame l2l.

Transient oltage stability Longer-termvoltagestability

Inductionmotor dynamics Increase n load/power ransfer



systemdisturbanceof July 23, 1gg7irorg", term,z0min)

(v) NREB grid disturbance n 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, insecurefor hours)

Hence, a full understandingof voltage stability phenomenaand designingmitigation schemes o preventvoltase nstabilitv is nf o,raqr'or,,o+^,.+:r:a:^^

', ;,ilir; ;,:;; ffiri#il;";;agestability.lf phenomena.Becadseof this, voltage'ent

engineers.Voltage instabilitvandrterchangeably.by many ."r"ur"h..r.

voltage instability or co[apse is a fasterdynamic-process. s opposed o angle

Generator/excitation vnamics LTC ransf& Distvolt.Reg.

Primemover ontrol Load diversity /thermostat

Mech.switchedcapacitors/reactors Excitationimiting Gas urbine tart-up

Under voltage oadshedding

10 0

Time-seconds

Protective elaying ncludingoverloadprotection

Fiq. 13.7 Vol tage stabi l i tV henomenaan d time responses

;;;''',h.;A#";rvr^rvlv vv'Yvr Pr'lurs tu a large system

over long transmission lines. Voltage':::i:y,"::,*:,','31^^,?::^s,112itiu

nd oto."ungl"stabilitys basicauy

Effective counter Measures to prevent or contain voltageInstability

(i) Generator erminal voltageshourdbe raised.(ii) Generator ransformer ap value may be increased.

(iii) Q-injection should be carried out at an appropriate location.(iv) Load-end

oLTc (on-load ap changer)shouldbe suitablyused.

wffi;Ewe

desirable.

Unitypower actor

Nosepoint knee)

t\

Loaw of V6 and Pr"r;same or const Z loadfor other ypes of load,Vdegradations faster.

PplP6zy

0.2 0.4 0 .6 0.8 1 .0 1. 2 1.4 1 .6

Fig. 13.8 PV curveswithdifferentoad power actors

Only the operating points above the critical points represent satisfactoryoperating conditions. At the

'knee'of the V-P curve, the voltage drops rapidly

with an increase n load demand. Power-flow solution fails to convefgebeyondthis limit indicating instability. Operation at or near thel stability limit isimpractical and a satisfactory operating condition is ensured by permitting

sufficient "power margin".

SPfti*f uooern po@isI

Voltage stability problems normaliy occur in heavily stressed systems.voltage stability and rotor angle (or synchronous) stability are more or lessinterlinked' Rotor angle stability,.a3-.wex

as voltage stability is affected byreactivepower control. voltage stability is concern"o *itt, load areasand oad

Seneratorstability.In a large nter-connected ystem,voltagecoilapseof a loadarea s possible without lois of synchroni* or any generators.The slower forms of voltage instability are often analysedas steady-stateproblems' 'snapshots'

in time following an outageor during load buildup aresimulated' n addition to post-disturbanclload floivs, two other oad flow basedmethodsare widely used:P-V curves and e-v curves.pV curves are used orespecially or radialsystems.

/ curves (Fig. 13.g),eV curves (Fig.'ig. 13.8) and methods o quantify noseputed.Power flow analysis cletermines

as anousvstemarametersnd""lT:ir'#n#'f |HI'I,:IJHI#voltage exist for each value of load. The uppe. on" indicatesstable ;;i;;;whereas ower one is the unacceptablevalue (multiple load flow). At limitingl!ug"

of voltage stability i.e. at nose point single oad flow solution exists.Nearer the nose point, lesser s the staLility maigin,

1400

1200

1000P> P, > P, , iNMW

j+Allowable range



(v) For under voltage conditions,strategic oad sheddingshould be resortedto'

System reinforcement may be carried out by instpiling new transmissionlines betweengenerationand oad centres.series and shuntcompensationmaybe carriedout and svcs (staticvar compensation)may be nstalled.Generationreschedulingand s_tarting-upf gas ,rrui-"* ,.y i" .,*.i.,t out.Practicalaspectsof Q-flow problems eading to voltage collapse n EHv lines:

(i ) Fo r lons l ine.s uri fh 'r-^^nrrnu^r L--^^-"-!' s'vv'Lr.,'e' uus's, recelvlng_end or road voltagesincrease for light load conditions and decrease for heavy load conditions.

(ii) For radial transmission lines, if any loss of a line takes place, reactancegoes up, I2x loss increases resulting in increase in voltage drop. This

shouldbe suitablycompensatedy local e injection.of course hisinvolves ost. f there s a shortage f rocalq ,ourrr, then mport of e

Systemcharacteristics

Capacitorcharacteristics

0. 9 0.951.0 1.05

Vi npu

Fig. 13.9 Systemand shuntcapacitor teady-state -V characteristics,capacitorMVAr shown at ratedvoltage

ffuo. Boo

Voltage Collapse

Voltagecollapse s the processby which the seguence f eventsaccompanyingvoltage instability le_adso unacceptablevoltag6 profile in a significant part ofthe power system. It may be manifested in ieverat different"ways. Voltagecollapse may be characterisedas follows:

(iii) The voltage collapsegenerallymanifests tself as a slow decayof voltage.It is the result of an accumulative process involving the actions andinteractions of many devices,controlJ, and protective iyrt"*r. The timeframe of collapse n suchcaseswould be of the order of severalminutes.Voltage collapse is strongly influenced by system conditions andcharacteristics.

(iv) Reactivecompensation anbe mademost effectiveby the udicious choiceof a mixture of shunt capacitors, static var system and possiblysynetuonous ondensers.

Methods of Improvlng Voltage Stabllity

voltage stability can be improved by adopting the following rneans:(i) Enhancing he localised eactivepower support SVC) is more efl,ective

and C-banks are lnore economical. ra-tS devices or synchronouscondensermay also be used,

(ii) Compensating the line length reduces net reactanceand power flowincreases.

(iii) Additional transmissioninc muy be crectccl.t also mproveseliability.(iv) Enhancing excitation of generator, system

M

;(v) Post-disturbanceMwA4vAR margins should be ffanslated to pre-

disturbance perating imits thatoperators anmonitor.(vi) Training in voltage stability basis (a training simulator) for control centre

and power plant operators should be i

SUIvIMARY

Power system security (including voltage stability) is likey to challengeplanneqs,analysts; esearchersand operators or the foreseeable uture. As loadgrows, and as new transmission lines and new generations would beincreasingly difficult to build or add, more and more utilities will face thesecurity challenge.

Deregulation and socio-economic ffends compounded by technologicaldevelopmentshave increased he likelihood of voltage nstability.

Luckily many creative persons are working tirelessly to find new methodsand innovative solutions to meet this challenge.

REFERECES

Books

L l.J. Narguth ndD.P. Kothari, PttnterSystem ngincering,'fataMc0raw.Hill, New

Dclhi , 1994.

2, C.W, Taylor, Power SystemVoltag,eStabil iry,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, ReactivePowerControl n ElectricSyslens, ohn Wiley and

Sons,New York, 1982.



voltage improves and e issupplied o the system.

(v) HVDC tie may be usedbetween egional grids.(vi) By resorting to strategic oad shedding, voltage goes up as the reactive

. burden s reduced.

Future Trends and Challenges

(i \ Ontirnql oif i- - n 'F EA/- 'n- ,{^* , :^^^\-./ urrrrr6 vr I nv r i, \rlvv.l tvgD.

(ii) Betterandprobabilisticoadmodelling.(iii) Developechniquesndmodelsor studyof non-linear ynamics f largr

siz.e ystems. or example, ew methodso obtainnetworkequivalentr

suitableor voltagestabilityanalysis.

5. A. chakrabarti, D.P. Kothari and A.K. Mukhopadhyay,Perforurutnce, peration

and Control of EHV Power TransmissionSystems,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, JohnWiley, New York, 1996.

E. JohnJ. 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

11. v. Ajjarapu and B. Lee, "Bibliography on voltage stability", IEEE Trans. onPower 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, "TheAdaptive 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 n North America", IEEE

Trans. on Power Systems, ol. 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 ssue of Proc. IEEE, February2000.

17. P.R. Bijwe, D.P. Kothari and L.D. Arya, "Alleviation of Line Overloads and

voltage violations by corrective Rescheduling", EE proc. c, vol. 140, No. 4,

July 1993, pp 249-255.

18. P.R.Bijwe, D.P. Kothari and L.D. Arya, "OverloadRankingof Line Outageswithpostourage 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 ContingencyLine 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 StabilityLirnit Enhancementby 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

Overloads nder Line OutageConditionTaking Bus VoltageLimits 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 sttressand Strain", IEEE spectrum,

March 1978,pp . 48-50.

t4

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 stateestimation is to obtain the best possible values of the bus voltage magnitudes

and angles by processing the available network data. Two modifications are,

however, ntroducednow in order to achieve a higher degreeof accuracyof the

solution at the cost of someadditional c omputations. First, it is recognised that

the numerical values of the data to be processedfor the state estimation are

generally noisy due to the errors present.Second, t is noted that there are a

larger number of variables n the system (e.9. P, Q line flows) which can be

measuredbut arc not utilised in the load flow analysis.Thus, the process

involves imperfect measurements hat are redundant and the process of

estimating the system states s based on a statistical criterion that estimates he

true value of the statevariables to minimize or maximize the selectedcriterion.

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)



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, nd Analysis

Methods", Proc. IEEE, Vol. 88, No. 2, Feb. 2000, pp 208-227.

values of a function.

Most state estimation programs in practical use are formulated as

overdeterminedsystemsof non-linear equationsand solved as weighted least-

squares WLS) problems.

Stateestimatorsmay be both static and dynamic. Both have been developed

for power systems.This chapterwill introduce the basicprinciples of a static-

state estimator.

ln a power system, the state variables are the v oltage magnrtudesand phase

angles at the buses.The inputs to an estimator are imperfect (noisy) power

systemmeasurements. he estimator s designed o give the "best estimate"of

the system voltage and phaseangleskeeping in m ind that thereare errors in the

measuredquantities and that there may be redundant measurements. he outputdata are then used at the energy control centres for carrying out several

:: if.-j l:tor,01-li,r:, ysteT^studicsuchas cconorniclisplrchChaprcr ),

st rLur l -y analysts (Lnapter lJ) .

r4-2 LEASTsouARES ESTIMATION:THE BAsrc soluTroNI7 l Ie I

A l r r

ns wilf be seen ater n Sectionr4.3, th e probremof power systemstateestirnation is a special case of the more general problem of estimation of a

random vector x from the numerical valuesof anotirer elatedrandom vector ywith relatively little statistical nformation being availablefor both x and y. Insuch cases, he method of least-squared-error stimationmay be utilised withgood results and has accordingly been widely employed.

Assume that x is a vector of n random variabres x1tx2, ..., x' that y isanothervector of m (> n) random variables !1, J2, ..., J^ and both are relatedas

whereH is a known matrixof dimensionmx n ancl is a zero mean ancklmvariableof the samedin-rensions y. The vector x representshe variables obe estimated,while the vector

y represents he variables whose numericalvalues re avai lable. quat ion 4.l) suggestsha t he measurementector islinearly related o the unknown vector x and in addition is corruptedby thevector r (error vector).

Th e problem s basical ly o obtain hc bestpossible alueof th e vector xfrom the given valuesof the vectory. Since he variable r is assumed o be zeromean,on e ma y ak e he expectat ionf Eq. (14.1)ancl et th e relat ion

! = H x + r

f = Hx

( r4 .1)

(r4.2)whcre I , , = cxpcctccl alueol ' x iurdy, respec:tively.

This shows hat he oad flow methodsof chapter 6 could be used o estimatethe mean valuesof the bus voltages.Howeu"r, n. woulcl ike to estimate heactual valuesof bus voltages ather han their

averages.One possibleway of obtaining he bestpossible stimate f the vectorx from

An lntroductiono State Estimation f PowerSystems

J = 7 ' V ( 1 4 . 5 )

From Eqs. (14.31)and (14.4),one gets he following expressionor the index:

J = y t ! - y ' H * . - i < ' H ' y + * .' H t H * .

For minimizing J = f$), we must satisfy the tollowing condition.

gradoJ = 0

It is easy o check see, .g . 1] ) that Eq. (1a.7) eads o the ol lowing result .

H t H * , - H ' y - O ( 14 .8 )

This equation is called the'notmal equation' and may be solved explicitly

for the LSE of the vector i as

*. = (H,Il)-t Ht y (14.e)

E;il;il]. i

ln orcler o illustrate he methodof LSE, let us consider he simpleproblem

ofestimating wo random variables x, and .rz by using the data for a thret:

dimensional ector y.

( 4.6)

\ r + . t )

l- ' 0.1A s s u m e H = 1 0 l l

L r r lThe matrix Ht F1 s then given bY

u' u =1 2II un a ts nverses

L l 2 )

(H'm-'f '':' -:':1

l- u3 21 3

/ ( \\ r + . J , ,



j' = Hf , (r4.3)

y lies n the use of the methodof leastsquare stimation LSE).To develop hisrneth<td, ssllme hat i rcpresetttshe desirecl stirnate f ,r so that y given bythe equation

represents he estimateof the vectory. The effor ! of the estimationof y is thengiven bv

(r4.4)Th e estimate is defineclo irc th c I-sE if it is cornputccl y rninirnizing heestimation index J given by

Weighted LSE

mI ..

ln qgrs l l l l | : . i l t r

E i l v€ i l !r l D9 . \ !+ ,2 ) ls u l re r r l r - r s r r su r \ , ( r i ) r r ru \ r r \ . r r r r (uJ r1 ,61 i r

squares stimateand is obtainedby minimising the index function that puts

equalweightage o the efforsof estimationof all components f the vector y.

It is often desirable o put.dift'erentweightages n the ditt'erent omponents f

y sincesome of the measurementsmay be more reliableand accurate han

theothersand theseshould be given more mportance.T'o achieve his we define

the estimation index as

It is easy to form the vector Hty and combining this with the itrverseof

(H'H), the following estimate f x is obtained.

f r r t l a t - . / 1 t 1 \ / r-l

^ | \ L t r ) / r - \ r r t ) \ / 2 - Y z ) |* =L-( t 3)yr*(2/3)yr1(r /3)v- ,

rvtouernPower

where W is a real symmetric weighting matrix of dimension m x ru. This isoften chosen as a diagonal matrix for simplicity.

It is relatively straightforward o extend he method of LSE to the weightedform of "I and to derive the foilowing form of the normal equation.

H | W H ? _ H ' W y _ 0 G 4 . I l a )This leads to the desired weighted least squaresestimate(WLSE)

I = l 'wf

*. = (H,WH)-t H,W yThis pertains to minimization as the hessian

definite.

SomeProperties:

Rewrit ing Eq . (14. i lb) as

i = k y

k = (H'WH)-' HtW

(14 .11b )2H|WH is a non-negative

(I4.I2a).

(r4.rzb)

/ 1 4 1 n \\ r . i . r vi,

(14.r3)

(14.r4)

( 1 4 . 1 5 a )

where

Here the matrix k dependson the value of H and the choice of w.using Eqs. (14.1)and (r4.lzb) it is easy o get the relationas follows.

r=f::;,7, ,,,wH)+kr

Or X =r + kr

a nd E{ i } =E { x l

ln IJq. (14.14) it is assumed hat the error r is statistically independentofcolumns of H and the vector r has a zeromean. An estimate hat satisfiedEq . (l 4)4) is cal ledan unbiased stimate. hi s mplies ha t he estimation rrorls zero on an averase.

x = k r

The covarianceof the error of estimation s therefore given by=

An ntroductiono State stimationf Power ystems l" 3iiSl:-T--

[o.r It lw = l I IL o.tj

The matrir HLWH ts

H,wHl''' o'tl

10.11.lJand the matrix HtW is obtainedas

H' |w=fo ' t o t - lL 0 1 0 . 1 j

Theweightedeast quaresstimatef thevector is thenobtaineds fromEq. la .11b) )

^ [ (lr/21)y, - (r}/Zt) y, (t0/2t) yrI =

L- ellr) y, (2ot2r), (r/2\y, )

If this result is compared with the result in Example I4.1, the effect ofintroducingthe weighting on the estimate s apparent.Note that the choice ofW in this case suggests he data for y2 is consideredmore valuable and thisresults n the componentsof x being more heavily dependenton )2.

The matrix ft is in this case ound to be (Eq. (1a.12b))

rc=lrr/21ro/2r o/zrf

L- rlzt 20/21 r/21If the covariance of the measurementerror is assumed o be R = L the

covariance f the estimationerror is obtainedas (Ref. Eq . (14.15c))

P,=(1tr47)|*:_

.::1L-67 r34 )

The choice of W above yields unacceptably arge estimationerror variances.



(14.1sb)P, KRK

whereR is the covarianee 1 he error vector r. Note that the covariance " isa lneasureof the accuracyof the estirnaiionand a smaiier trace of this matrixindicatesa betterestimate.Eq . (14.15b)suggestsha t the best possible hoiceof the weighting matrix is to set w - R-1.The optimum value of the errorcovariance'matrix s then given by

P, - (HtR-rH)- l ( 1 4 1 5 c )

Assume hat n the Example I4.1, we want to obtain he WLSE of the variablex by choosing he following weighting matnx

t---*--**

1Et<ample14;2

Let us now choose he weightingmatrix W = I. The matrix Kis then obtainedAS

* = l ' ' 3- 1 / 3 / 3 f

L- U3 2/3 1/3JThe error covariance matrix is then given by

D / r / n \ [ 6 - 3 ]r - x = \ L t > )

L _ E 6 JThe error variances are now seen o be much smaller as is to be expected.

Non-linear Measurements

The case of special interest to the power system stateestimation problemcorrespondso the non-linearmeasurementmodel.

I

536 | Modern Power SystemAnalysisII

! = h ( x ) + r ( 14 .16 )

whereh(x) represents n rz dimensionalvector of nonlinear functionsof the

variable . It is assumedhat the components f the vector h(x) arecontinuous

in their arguments and therefore may be differentiated r.vith respect to the

componentsf x. The

problem s to extend he methodof least squaresn order

to estimate he vector x from the data or the vector y with these wo variables

being related hroughEq . (14.16).

To mimic our treatmentof the linear measurement ase, assume hat i

representshe desired estimateso that the estimateol the measLlrement could

be obtained using the relation

I = h( i )

This yields he error of est imation f the vectol y

j = y _ h ( 3 )

In ordcr to obtain he WLSE of ,,r,we nrustcho<lsehe index of estimation

J as fol lows.

J = U - h ( f i ) l ' W l J h ( i ) l ( 14 .18 )

The necessary ondition for the index ,I to havea minimum at .x, s given by

Eq. (14 .1e) .

t y - h ( x ) l H ( i ) = o (r4.re)

where H (fr) is the Jacobian f h (x ) evaluated t i. In general his non-linear

equation can not be solved for the desired estimate, . n way out of this

difficulty is to make use of the linearisa-tion echnique.Let us assume hat an

a priori estimate xu of the vector x is available (say from the load flow

solution).

Using Taylor seriesapproximation,we get

(1.4.17a)

(r4.r7b)

(14.20)! = h ( x d + H u & - x 1 ) +where 1/o stands or the Jacobianevaluated at x = x9 and the noise term r is

An lrylfgductiono_State stimationf PowerSystems I S3TI

useful result n the sense hat it showsa mechanism or improving on the initialestimafeby making use of the availablenleasurements. aving obtainednew

estimate , theprocess f linearisation s repeated smany times as desiredandthis leads to the {qllq*tng 4erylle {qryq qf the qqb]ro4 of thq 4on-linear

estimation roblem.i ( t + 1)= ( t )+ K(t ) {y h t t ( / ) l }

where he matrix K(D is deflned as

K(t) - LH/ wHil-' al w

The index / represents he iteration number and H lrepresents he value of the

Jacobianevaluatedat x = i (l). Usualiy the iterative process s terminatedwhenever he norm of the differenceof two successive aluesof the estimate

i 1t + l) - .i (/ ) reachesa pre-selecredhreshold evel.A flowchart for implementing the iterative algorithm is shown in Fig. 14.7.

A major source f computat ionn the algorithm ie s n the need o update heJacobian t every stageof iteration.As discussed arlier n Chapter6 (seeEqs.(6.86)

and (6.87)) t i s oftenpossible o reduce he computations y holding hevalueof H a constant,possibly after / exceeds2 or 3. T,ris is in general,permissible n view of the fact that the change n estimate ends to be rathersmall after a couple of iterations.

(r4.23)

(r4.24)



r11.22)

now assurnedo include he effectsof the higher order terms n the'Taylor

series.Equation 14.20) an be rewrit tenas :

A ) ' = y - h ( x o ) =H s A x + r ( 1 4 . 2 I )

where Ay is thep^-rturbedmeasurementand Ax is the perturbed valueof the

vector r. An \\LSE of .r is then ea^silv btaineda-s iscussedearlier and this

leads to the desired expression or the lineaized solurion of the non-linear

est jmat. ionroblem.

.i . = .rrr [H,, ' I+'HQ]- H()' ]l ' {- r - h t. t , , ,)}

I r is nor ikelv ha t he est imate i obtainedro m Eq . (11.22 s going o be

t - r f r tuci r use si l t . 'c . i r t tener . r i . t l t r ' . r i , t ' r t t t ' t c 'st i r lute . l ' , . rl )J\ - I l r r t h ' c lose t i r thf(Dt-irnal due rf rh e \ect()rx. Hrt,* 'ever, ,q . 14.22)prot ' idesus wirh a very

i ,00"," ) eq .+..227/

Ic"l::'1:-,=n\, ! -_I

\,, l l-''!'

./ ls: - \ -

. - 2 a . ,No

r Yes

\Stcp

r ig . r+.1

',:83--E-,I todern powerSystem nalvsisI

Considerhe simplecaseo1'ascalarvariable and assunleha t the relationship

_x s glven Dy

Y = x 3+ r

The JacobianH, is easily obtained n this caseand the terative algorithm takesthe explicit form

i (t + 1)= i (t )+ t3 (Dl-z y - ti (Dl3}

wherewe have used W = 1.Let the correct value of x/ be eclual o 2 and assume hat due to the effect

of r the measured alue of y is found to be 8.5. Also, assume hat the initialestimate (0) is taken o be equal to 1. The table below gives the resultsof thefirst few iterations.

, (t)

1 . 03. 5

2.56

2 . t 6

It is apparenthat he algorithmwould yield he correctsolutionafterseveraliterations.

14.3 STATIC STATE ESTIMATION OF POWER SYSTEMS

lrol-lr2l

As notedearlier, for a systemwith N buses, he statevector x may be definedas thc 2N - | vccl t l r l l ' l l t c N -- | vol lagcanglcs 2, . . , 6" and hc N vol tage

magnitudes 1, v2, ..., v". The load flow data,dependingon type of bus, aregenerallycomrpted by noiseand the problem s that of processingan adequate

An tntroductiono State Estimation f Power Systems | 539_r-It is thus apparent hat the problem of estimationof the power systemstate

'is a non-linearproblenn nd may be solvedusingeither the batchpr:ocessingr

sequentialprocessing ormula [seeSection3.3 of Reference1]. Also, if the

svstem s assumedo havereacheda steady-stateondition. he voltageangles

problem is then a staticproblem and the methodsof Sec. 14.2may be used. f

so desired.To developexplicit solutions, t is necessary o start by noting the

exact forms of the model equations or the componentsof the vector y (k).

Let P, arfi Qi denote he active and reactivepower injections of ith bus.

These are related to the componentsof the statevector through the following

equations.

I Y j | | Vjl lYij I cos (- 6t + 6, * 4ti) (r4.2s)

(r4.26)

l v iP Y, , lcos , ,

(r4.27)

I viP I Y,, sin Iij

(14.28)

l, N Qn .. . Q7,J 1, 7s,

b i n lv 1 l , . . . . , yN I /

(r4.2e)

N

P,=Dj - -r

0

I2a_)

N

e , = - D | % l I v j l l Y i j l s i n - 6 , + 5 t + 0 ; )j : 1

where I IU I repre5entshe uragnitudeand l/U- represents he angle of theadmittanceof the ine connecting he lth and7th buses.The active and reactive

components f thepower low from the th to the7th bus, on the otherhand are

given by the tbllowing relations.

P,j= | vi l I Vj l I Yi j co s (d , - 6i * 0,) -

Q , j = l % l I V j l l Y i j l s i n ( d ' 6 i * 0 t ) -

Let us assume hat fhe vector y has the general orm

J = lP t .. . PN Qt .. . Qr u Prz . . Pu -

)2



set of availabledata n order to estimate he statevector. The readily availabledata may not provide enough redundancy (the large geographical area overwhich the system s spread ften prohibits he elemetering f all the availabletnedsurementso the centralcomputing station).The redundancy actor,definedas the ratio m./nshould have a value in the range 1.5 to 2.8 in order that thecomputed alueof the statemay have he desired ccuracy. t may be necessaryto irleh-rdehe data or the power flows in both the directionsof some of the tielines in order to increase he redundancy factor. In fact, some 'psuedo

measurements' hich representhe computedvaluesof such quantitiesas theactivean d eactive nject ions t some emote usesma y alsobe ncluderl n th evector y (k).

The Jacobian H will theh have the form

Ht Hz

H3 H4

Hs H6

H7 Hs

/ru-r o

o IN

where /" is the iclentity matrixof dimension N, H, is the N x (N - l; submatrix

of the partial derivatives of the active power injections wrt 6's, H2 is the

H _ (14.30)

54q f Modern oT

N x N sub-matrixof the partial-derivativesof the active power injections wrt| 7l' and so on. Jacobian H will also be a sparsematrix since I is a sparsematrix.

Two specialcasesof interest are those corresponding o the use of only theactive and reactive niections and the use of onl

flows in the vector y. In the first case, here area total of 2N components ofy compared to the 2N - 1 componentsof the state x. There is thus almost noredundancy of measurements.However, this case s very close to the case ofload flow analysis and therefore provides a good measure of the relative

strengthsof the methodsof load flow and stateestimation. n the second case,it is possible to ensurea good enoughredundancy f there are enough tie linesin the system.One can obtain two measurements sing two separatemetersatthe two ends of a single tie-line. Since these two data should have equalmagnitudes but opposite signs, this arrangementalso provides with a readycheckof metermalfunctioning. hereare otheradvantagesf this arrangementas will be discussedater.

The Injections Only Algorithm

In this case, the model equationhas the form

! = h l x l + r

with the componentsf the non-linearunctiongivenby

Qa32a)N

- D t v * _ i l l v j l l y N _ i , ; l s i n6 , - 6 i +7 i i )j : 1

i = N + 1 , N + 2 . . . , 2 N (r4.32b)

The elements of the sub-matrices Hp H2, H, and Ho are then determinedeasily as follows.

An tntroductiono State Estimation f Power Systems I 541'I

Equation 14.33)may be used o determine he Jacobianat any specifiedvalue

of the systemstatevector.The injectionsonly stateestimationalgorithm s then

obtained directly from the resultsof sec 14.2.Since the problem is non-linear,

it is convenient o employ the iterativealgorithmgiven in Eq. (14.22).

ine. the submatricesH., andH^ becomenull

with the result that the linearised model equationrnay be approximatedas:

^ fH , o - l

rrwe arrition,J";,1:',o*^::::

'

f Ay,1 lArr l l ro lAv

l^'r 'r. j '=

Lor"l,=

L;

( r4.34)

( 14 .35 )

(r4.36)

(14 .31 )

then, Eq . (14.34)may be rewritten n the decoupledorm as he ollowing two

separate quations or the two partitionedcomponentsof the statevector

A l p = H1 Ax 5 + ro

Alq= H4 Axr, + rn

Basedon these wo equations,we obtain he ollowing nearlydecoupledstateestimationalgorithms.

i u Q + r ) = x d 0 + t H i0 w p H tU ) l - ' H , D t o - h r l i U ) l }

where he subscripts andq are used o indicate he partitionsof the weighting

matrix W and he non-linear function h (.) which correspond o the vectors yp

*rd lerespectively. As mentioned earlier, f the covariancesR*11d

Rn of the

effors r, and r(t are assumedknown, one should select Wo= R o' and Wn =

R q " i'Notethat Eqs. (14.37) and (14.38) are not truly decoupledbecause he

partitions of the non-linear function dependon the estimateof the entire state

h , [ x ) =Dj : 1

. l = 0 , 1 , 2 , . . . \ ( 14 .37 )

l V i l l V j l l Y i j l c o s 6 , - 6 i + 0 i ) , = I , . . . , N i, (/ + 1)= i, Q)+ tHl u) w, Hq j)l-' nl u) wo ur - hq i (j)ll

j = 0 | 2 ( 1 4 . 3 8 )



Ht ( i , ) = lV i l lV j l lY i j I s i n ( r { . 6 t+ 0t )

i = 1 , 2 , . . . ,N ,

j = I , 2 , " ' , N - 1

Hz Q, ) = lv i l lYi j I cos(d , 6i * 0; ) i = 7,2, . . . , N,

j = 1 , 2 , " ' , N 'r f / r ! \ | r t I I t r | | r , | / C a - n \ . n A

I a 3 \ t t J ) =- | y i l I v j l I I U I C O S \ O i - . O j + A i j ) I = I , Z ) . . . , l Y,

j = I , 2 , " ' , N - 1

H q ( i , ) = l V i l l Y i j l s i n { - 6 i + 0 , ) = I , 2 , . . . , N ,

j = I , 2 , " ' , N '

(r4.33)

vector. It may be possible to assume hat vi U) = I for all i and 7 while'

Eq. (14.37) s being used n order to estimate he angle part of the statevector.

similarly one may assume 6i U) = 0o for al l i and 7 while using Eq. (14.38)

in order to estimate he_yllltagepart of the statevector. Suchapproximations

allow the two equations o be completely decoupledbut may not yield very good

solutions.A betterway to decouple he wo equationswould be to use he load

flow sglutions or x, and x6 as therr suppose<iiyonsnnt vaiues n Eq. (14.37)

and Eq. (14.38) respectively. There are several forms of fast decoupled

estimationalgorithmsbasedon suchconsiderationsseee.g. 13], l4l).A flow

chart for one schemeof fast decoupledstateestimation in shown n Fig. I4.2.

542 | ModernPo*er System AnalysisAn Introductiono State Estimation f Power Systems | 543

1_-

* 1 1

Fig.14 .3

I

j Example14.4

Application of the LSE then yields the following expressions or the

estimates f the perturbationsn the threestatevariablesaround heir chosen

initial values:

- AP , - AP ,

= 0.78AQ z - 0.26AQ r

AV ' = g3 9 AQ ' + 0'14 AQ r

These equations houldbe used n order o translate he measured aluesof

th c pcrturhat ionsn the act iveancl eact ive ower nject ionsnt o he st inrates

of the perturbations f the statevariables.

It is interest ingo note ha t or lhe sirnple xample, art i t ionsHranrl Htare

null matrices o hat hedecoupied tate st imators re he sameas hosegiven

above.

The Line Only Algorithm

This algorithm has beendeveloped n order o avoid the need or solving a non-

l inearest imation roblem.which as seen arl ier, equires om eapproximationor other. n the ine tlow only algorithm, he data or the active and eactive ie

I , o , r l l i 1 i *1 t ) - * ( j ) l l = , '

Ai,

AT,

lso n , ^ t

, 2 a n 2>

t\ u -\\\r---'

1"",I

, \(3'9)

Fig. 14.2



In order to illustratean applicationof the njections only algorithm,consider he

"imple 2-bus systemshown n Fig. 14.3.Assumingoss lessine , 0 , j=90" . A lso et Yr r= j zz=2and yn= yzt=

. The power relations n this casewould be

P t= - | Y r | V2 l Y r r l s i n EP z = V t l l V z l I Y , r l s i n 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 = Y z z l V r l ' - l y t 2 l l y r | | v r l c o s6If we choosehe ni t ia lvaluesVf l=lV; l= 1, 6o2=0", he corresponding

power values areP f = Pf - 0, Qf = Qi = 1. The value of the Jacobianmatrix

evaluatedat the above nominal values of the variables urns out to be

line flows areprocessedn order to generate he vectorof the voltage difference

across te tie-lines.Le t z denote his vector.A model equation or this vector

may be expressed s

z= Bx + r - ( r4 .39)

whereB is the node-elernentncidence natrixand r- is the vectorof the errors

in the voltage el-ata. ince this is a linear equation.one may use the WLSE

technique o generatehe estimateas

*, = 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) s that he vector zis not directly measurable ut needs o be generatedrorn the tie line flow data.

r++| Moctern ower SystemAnalysisI

V,tdenotes he voltageacross he ine connecting he ith ancl heTthbuses, hefol lowinerelat ionholds.

Vi i= Zi i [@i i - - Qi ; tV7- Vi Y, i ]

HereZ,.,stands or th e mpedance f the l ine.This shows hat the vectorz is related o the vectorsx and

ashion and one ffidy use the notation

z = g (x, y) 04.42)In viewof thisnon-linearelation, cl. 14.40)maybeexpressecln the orm

i = LB ,WnyrB, W (i , y) (14.43)This,beinga non-linearelationship,an not be solvedexcept hrough

numerical pproachiterativeolution). he terativeorm of Eq. (14.43)s

i (j + r ) = lBt wBll H w s [i (i), y],

- /= 0 , 1 , 2 , . .

(r4.44)

Note that the original problemof estimation of x from the data or z is a linearproblemso that th e solut ion iven by Eq . (14.40) s th e opr inra lsol rr t ion.However, he data or z need o be generated sing he non-linear ransforma_

tion in Eq ' (14.42), which in turn has necessitateclhe use of iterative Eq.(14.44)' Compared to the injections only iterative algorithm, the pr-ese1talgorithmhas he advantage f a constantgain matrix [8, W B]-t gr I4z. hisresult n a considerableomputat ionalimpli f icat ion. he concept f decoupledestimations easilyextencledo th e caseof the irre lows [15].

T4.4 TRACKING STATE ESTIMATION OF POWERSYSTEMS t16l

Tracking he state st imation f a givenpowersystems mportant or real irnemonitoringof the system. s is well known, the voltages f all real systemvaryrandomly with time and should therefore be considered to be stochasticprocesses'

t is thus necessaryo make us e of the sequentialest imationtechniquesof Ref. [1] in order to obtain the stateestimateat any given time

An lntroductiono stateEstimationf powersystems | 545T-

Network Obsenrability [17]

Consider he staticWLSE formula Eq.(14. l lb)l which serves s th e srart ingpoint fbr al l the algorithrns. nverse of information matrix Mn,, = Ht wHshouldexistotherwise here s no stateestimate. his will happen f rank of f/ls equal o n (no. ot statevartables).Since on e ca n always choosea non-singular l4l,so if lt1 has a rank n, the power network is said to be observable.

Problem of lll-conditioning

Even if the given power system is an observable system in terms of themeasurements elected or the stateestimationpurposes, here s no guaranteethat the required inversion of the information matrix will exist. Duringmultiplications of the matrices, here s somesmall but definiteerror introduceddue to the finite word length and quantisation.Whether or no t these errorscreate ill-conditioning of the information matrix may be determined from aknowledgeof the condition number of the matrix. This number s defined as herat ioof th e argest nd he smallest igenvalues f the nforrnat ionnatrix.Th emaftix M becomesmore and more ill-conditionedas its condition numberincreasesn magnitude.Somedetailed esultson powersystemstateestimationusing Cholesky factorization echniquesmay be fbund in tl8l. Factorizationhelps o reduce ll-conditioningbu t may not reduce he computationalburden.A techrriqueo reducecomputationalburden s describecln Ref. []91.

14.6 EXTERNAL SYSTEM EOUTVAIENCTNG [20]

One of the widely practicedmethodsused or computational implification is todivide the given system nto three subsystems s shown n Fig. 14.4.One ofthese s referred o as the 'internal'

subsystem nd consistsof thosebuses nwhich we are really interested. he secondsubsystem onsistsof thosebuseswhich are not of direct interest to us and is referred to as the 'external,

subsystem'Finally, the buseswhich provide inks between hese nternal andexternalsubsystems onstitute he third subsystemeferred o as he 'boundary'

(14.41)



point. The power relations n Eqs. (14.25) and,74.26) are still valid bur mustbe rewrittenafter indicating hat the voltage rnagnitudes nd anglesare newfunctions of the discrete ime index ft.

I4.5 SOME COMPUTATIONAL CONSIDERATIONS

B r r f h t h e q t a f i n qn d f h e t r q o l z l n o o c f i m o f i n - ^ 1 - ^ * i + L * ^ a i ^ , r ^ 6 1 ^ - r i - r L ^ - - - ^ - - r : - -r r u v r \ r r 1 6 v r ) r r r r r q l r u l r C t r 5 \ J l t L l l l l . l J P I E i ) t ' f l t t r l t ll l t l l € p I C L : C U l l l g

sectionsare computationally ntensive,particularly for large power networkswhich may havenlore han200 mportantbuses. t is, therefore, ery mportantto pay at tent ion o suchcomputat ionalssuesas i l lcondi t ioning, omputerstorage nd inne equirements. owever,we need o first consider he questionof existenceof a solutionof the stateestimationproblem.

subsystem. For any given power network, the identification of the threesubsystemsmay be done either n a naturalor in an artiflcial way.

,/

-t,

/ \ ,

\ ", /

Internal

system

Boundary

system

Flg. 14.4

fl6 -l ModernPoI

To illustrate he simplification of the stateestimationalgorithm, consider he

linearisedmeasurement quation or the njectionsonly case.Since he system

is partitioned into three subsystems,his equation may be written as

A y i H,, H,, o A * ,

A * ,

A * ,

It may be noted hat the internal me ..rrerrrertvector Ayt is not completely

inclepenclentf the external subsystemstate Axu since Ay, dependson the

boundary subsystetn state Axr, and Ax6 dependson Axr.

An lntroductiono state Estimationof Power systemsI

t{7

Bad Data Detection [231

A convenient tool fcr detectingthe presenceof one or more bad data in the

vector y at an y given point of t ime is basedon the'Chi SquareTest '. To

appreciate his, flrst nc:te hat the trrethod -ll least square ensures ha t the

rflLW Ta 1r (t)l- /ltgr haritrminimum

value when x - i. Since he variable r is random, he minimum value ,I.in is

also a random quantity. Quite often, r may be assumed o be a Gaussian

variable and then .I*1nwould follow a chi squaredistribution with L = m - n

degreesof freedom. It turns out that the mean of ./,o1ns equal to L and its

variance s equal to 2L. This implies that if all the data processed or state

estimatiorr re reliable, hen he computedvalueof ../r,r,nhould be close o the

averagevalue (=L). On the other hand, if one or ntore of the data for )' tre

unreliable, hen he assumptiorrs f the leastsquares stimationare violated and

the computed value of J*1nwill deviate significantly from I.

It is thus possible o developa reiiable schetrreor the detectionof bad data

in y by computing he valueof [y - h(i)l 'w ly - h(r)] , i being he estimate

obtainedon the basisof the concerned . If the scalar o obtainedexceeds ome

thresholdTi - c L, c: being a suitablenuutber,we conclude hat the vector yincludessomebad data. Note that the data or the. ornponent ;, i - 1,2, ...,

nr wil l be consiclereclac l f i t deviates ro m th e nean f r' ,by more han 3e,.

whereo, is thestandard eviationof r;). Caremustbe exercisedwhile choosing

the valueof threshold arameter .If it is close o 1, he estmaf producemany'talse

alal1s' ancl l it is tt-rt-rarge, he testworrld ail to detectnlany bad data.

To select an appropriatevalue of c, we may start by choosing the

signif icanceevel d of the est by the relat ion'

I ' lJ (x ) > cLlJ (.r) bl lows ch i squale istribut ion)= 6/

We rnay select, br example, = 0.05 which correspondso a 5Vo alsealarm

sitrrl t ign. t is thcn possible o f ind th e valr.ref c hy making us eof th e table

X@).Once th e valuc of c is determined, t is simple o cary out the test

whetheror not .l(x\ exceeds L.Identification of Bad Data [231

Ayr,ay"

Hu, Huu Ho,

0 Hra H""

, o l

, " ]

(r4.4s)

(r4.46)

(14.47)

( 4.48)

Ayr= Ayt, i+ Aynr,+ Ayu, + ru

where Ayo" representshe njections ntc the boundary buses rom the external

buses,Ay66is the injection from the boundary busesznd Ay6, is the injection

fiom the nternalbuses. t is assurnedha t the term Ayu, (= Ht,, Axr), may be

approximatedas n A*, where A It estimated rom the relation

H=

Ayt,nlAxl,The conrponent Ayu" may be estimated if the terms Ayy, and Ayuo ne

computedas H,,, Ax,and H1,1,x1, and hen subtracted rom the measured alue

of Ayo. This woulc! esult n part of Eq. (14.45) to be rewritten as

l^;,)=l',;;,,;:;)l*,1. ;]

whcrc Ht,, = Ht,t,t U pcrprcscrntshc cff-ccfvc Jacohil tn f the hottncl i t ry

subsysternhat accounts or the eff'ectsof the external subsysternon the

boundary ubsystem. quation 14.48)has a lower dimension han he original

tncasurclncrt tc lLr l t ionnd worr l r l l tcrc lorc r tvolvcrcss ct l t t t l l tr ( . t l io l ls.lt c

conceptof externalsystemequivalencingmay be employed with the line or

rnixedCata ituat ions lso.

14.7 TREATMENT Ol.- BAD DATA 127,221



Once the presenceof bad data is detected, t is iraperative that these be

identified so that hey could be removed rom thevectorof measurements efore

it is processed.One way of doing this is to evaluate he componentsof the

measurementesidual , = y, - h, (.x\, = 1,2,..., m. If we assLlmeha t th e

residualshave the Gaussian istributionwith zeromean and the varLance ,

then he magnitucie f the resiciuai , shouici ie in the range 3o i ( )i ( 3or with

95Voconfidenceevel. Ihus, if any one of tlte cotnputed esidual urns out to

be significantly larger n magnitude han three imes ts standarddevia,tion,hen

correspondingdata s taken o be a bad data. f this happenS or more than one

con)pot rct r(l '_y,hcn hccon' lponcnturvinglhc argcst csidual s assuntedo

be hc bad dataan d s removed ro m y. Th e estimation lgorithm s re-runwith

the remaining data and the bad data detectionand identification tests are

The ability to detectand dentify bad measurementss extremelyvaluable o a

load dispatchcentre.One or more of the datamay be affectedby rnalfunction-

ing of either he measuringnstruments r the data ransmission ystemor both.

Transt lucersnay havebecnwilcd incorrcct lyor t l to .rattsduccrtscl l rt taybe

malfunctioning so that it simply no longer gives accurate eadings.

If such aulty daia are ncluded n the vector Ay, the estimationalgorithm

will yielcl unreliableestimates f the state. t is therefbre mportant o develop

techniquesor detecting he presenceof faulty data n the measurement ector

at any given point of time, to identify the fauity dataand eliminate these rom

the vector y befble it is processcdbr stateestinration. t is alstl lttpoltant ornodify he estirnationalgorithms n a way that will permit tnore eliablestate

est imat ionn the p!esence f bad data.

,548 ,! Modern owerSystem nalysisI

performed again o find out if thereare additionalbad data n the measurementset. As we will see ater bad rneasurement lataare detected,eliminatedandreplacedby pseudoor calculated alues.

of Bad Data

The procedures described so far in this section are quite tedious and timeconsuming and rnay not be utilized to removeall the bad data which may bepresent n the vectory at a given point of time. t may often be desirable n theother hand to modify the estimationalgorithms n a way that will minimise theinfluence of the bad dataon the estimatesof the state vector. This would bepossible if the estimation ndex J(x) is chosen o be a non-quadratic unction.The reason hat the LSE algorithm does not perform very well in the presenceof bad data is the fact that because f the quadraticnatureof J(x), the indexassumesa large value or a data hat s too far removed rom its expectedvalue.To avoid this overemphasis n the effoneousdataanclat the same ime to retainthe analytical tractabilityof the quadratic: erlbrmance ndex, et us choose

/ ( i ) = s ' t ) w sG) (14.49a)

where 8(t) is a non-linear unction of the residual !. There may be severalpossiblechoices or this f'unction.A convenient orm is the so-called quadratic

flat' form. In this case, he componentsof the function g (y) are definedby thefollowing relation.

gi (j ) = li , fo r j , lo , 1 a,

= ai , f tx' l , /o,> u,

where a, is a pre-selected onstant hreshold evel. Obviously, the perform-ance ndcx ./("r)ma y bc cxpressed s

m

I( i) = Dy, c; ); - I. - - I

and eachcomponenthas a quadraticnature for small values of the residualbuthas a constantmagnitude

(r4.49b)

(14.s0)

An tntroduct iono State Estimation f PowerSystems I549.

t -

The main advantageof the choice of the form (14.49) for the estimation

index s that t is st i l l a quadrat icn the funct iong(. i) and so the LSE theory

may be mimicked in order o derive the following iterative ormuia for the state

estimate.

Hr(D CTWCH Q)Tfl (D WCtfiDI

(74.5ra)

wherehe matrixC is diagonal nd ts elements recomputed s

Ct = l, fo t l ; lo , S ai

= 0, fo r l i lo,> a'

(14.s b)

Comparing this solutionwith that given in Eq. (14.22), t is seen hat the

main effect of the particular hoiceof the estimationndex n Eq. (14.49) s to

ensure hat he dataproducing esiduals n excess f the hresholdevel will no t

change he estimate.This is achievedby the productionof a null value for the

matrix C for large valuesof the residual.

I4.8 NETWORK OBSERVABILITY AND PSEUDO-

MEASUREMENTS

A minimum amount of data is necessary or State Estimation (SE) to be

effective.A more analyticalway of determiningwhether a given data s enough

for SE s called observabilityanalysis. t forms an ntegralpart olany real time

state estimator.The ability to perform state estimation depends on whether

.suff icientmeasurementsre well distributed hroughout he system.When

sufficient measurementsare available SE can obtain the state vector of the

whole system. n this case he network is observable. s explainedearlier n

Sec. 14.5 his is true when the rank of measurententacobian lratrix is equal

to the number of unknown state variables. The rank of the measurement

Jacobian matrix is dependenton the locations and types of available

rneasurementss well as on the network topology.An auxiliary problem in stateestimation s where to add additional data or



or residualmagnitudes n excessof the threshold.Figure 14.5 showsa typicai variation of J, (.r) for the quadratic and the non-quadraticchoices.

\

pseudomeasurementso apowersystem n order o improve he accuracy f the

calculatedstate i.e. to improve observability.The additional measurements

representa cost for the physicat transducers, emote terminal or telemetry

sy.stem, nd software data processing n the central computer.Selectionof

pseudomeasurements,illing of missing data,providing appropriateweightage

are the functions of the observability analysisalgorithm.I I - - r - - I l - - - : - ^ C ^ ^ t ^ - i - ^ + l ^ - T f ^ - . ' ^ i " ^ + L ^ ^ ^ - ^ . . ' ^ - . '

UbsefvaDll l ty Ca n De CIICCKCU UUtrr lB l i lulul lzi t tLltr l l . Il 4IrJ Prv\rr r .rvtrr r .r r r lr YvrJ

small or zero during factoization, the gain matrix may be singular, and the

systemmay not be observable.

To finil the value ol an injection without nteasuringt, we tlrust know the

power systembeyond he measurementsurrentlybeingmade.For example,we

pormally know the generatedMWs and MV ARs at generators through

telemetry channels(i.e. thesemeasurementswould generallybe known to theFig. 14.5

i ,t,

stateestimator). f these hannelsare out, we canperhapscommunicatewith theoperatorsn the plantcontrol room by telephone nd ask for thesevaluesandenter hem manually.Similarly, if we requirea load-busMW and MVAR fora pseudomeasurement' e could use past records hat show the relationshipbetweenan individual load and the total qyrtemload. We canestimate he tstalsystetn oad quite accurately y f)nding the

total power being generated ndestimating the line losses.Further, if we have ust had a telernetry ailure, wecould use he mostrecently stimated alues iom the estimator a.ssgmingha ti', ' s run periodic:ally) spseudo lreasureftlents.hus, f required,we can give

. the stateestimatorwith a reasonable alue to use as a pseudomeasurement tany bus in the system.

Pseudomeasurementsncrease he data edundancyof SE. If this approachis adapted,care must be taken in assigning weights to various types ofmeasurements.echniquesha t ca n he useclo cleternrinehe rnctcror .pseu4gmeasurementocationsbr obtaininga completeobservabilityof the sysrem reavailable in Ref. [251.A review of the principal observabilityanalysisandmeterplaccrncnt lg ' r i thrnss avui rahrcn I (cr ' . 261.

T4.9 APPLICATION OF POWER SYSTEM STATEESTIMATION

In real-timeenvironmenthe stateestimator onsists f differentmodulessuchas network opologyprocessor, bservabilityanalysis, tateestimationand baddata processing.The network topology processor s required for ail powersystemanalysis.A conventional etwork opologyprogramuses ircuit breakerstatus nformationandnetworkconnectivitydata o determine he connectivityof the network.

Figtrrc 1r4.6s it schcrnaticl iagtarn howing he nfo.rnat ion lo w betweenthe varior-rsunctions o be performed n an operationscontrol centrecompurersystem. The systemgets nformation from remote terminal unit (RTU) that

cncodc llcasLlrcl l lcntl 'ul lsduccrr-rtputsnc lopcnccl/closccltaLusnlbmrationinto digital signalswhich are sent to the operationcentre over communicationscircuits. Control centre

,l' I

I

An introductiono stateEstimationf powqrsystems I FSIt -

o

oo

oLcoo

Eoa

U)L

o=oo

qsf

E;ll-

Io(!

(Eo

(L

II

II

II

I

I

il_

I

n lL U I> i

I

IIII

I

,t ll 6 ci o !

LLI .

L

- - 9F G. Y t= t h

dr x

o ga

aL

n ^v v

a. n F

A - L

q ' = . o

o 0 - ( E

I

o

oEoL

fa(!

o)Eo)o

trot-

FatrJ

]U

Fn

II

t , -L

oaU)o

X H; o -ii' >,z 8

F

o1fo

Eoo

U)

U)

U ' E

b A !l ol mI



can also transmit commands such as raiseflower togenerators and open/close o circuit breakers and switches. The analogmeasurements f generator utput would be rJirectlyusedby the AGC program(Chapter8) . However, es tof the datawill be processed y the stateestimatorbefbre being used br other unctionssuch as OLF (Optimal LoaclFlow) etc.

Before running the SE, we must know how the transmission ines arer - n n n e n f p r l f n f l r o l n ^ ' l ^ ^ J r - - - - - - -L\ / L r ru r \ / cu ( r r rL rl i c r r r s r .u . ( r l . ,usc l i l . e . ne lworK topo logy , lh rs kceps 6nchangingan dhcnce .h c urrert t .clenteteredbreaker'/switchtertus us tbe usedto restructurehe electrical ystemmodel.This is called the network r,tpop.tgyprogram or systemstatusproce,\.toror netwc;rkconfigurator.

C O eN ' E C

b 8 gE E . EE E E6 E oF O

o

il r l ModernoI

The output of the stateestimator .e. lvl, 6, P,j, Q,jtogether.with latestmodel

form the basis or the economic dispatch (ED) or minimum emission dispatch

(MED), contingencyanalysisprogram etc.

Further Readin

The weighted least-squares pproach o problems of static state estimation npower systems was introduced by Schweppe 11969-741. It was earlier

originated in the aerospace ndustry. 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 stateof the art

in stateestimation algorithms basedon this modelling approach were published

by Boseand Clements 27] and Wu [28]. Reviewsof externalsystemmodelling

are available n 1291. generalised tateestimator with integrated state,stutus

and parameterestimation capabilitieshas recently beenproposedby Alsac et al

[30]. The new role of stateestimationand otheradvanced nalytical unctions

in competitive energy markets was discussedn Ref. [31].A comprehensive

bibliography on SE from 1968-89 is available n Ref. [32].

PROBEMS

14.1 For Ex. 6.6 if the power injected at busesare given as Sr = 1.031-

j0.791,Sz = 0. 5 + 71.0an d 'S 3 - 1. 5 j0.15 pu . Consider Wt= Wz=

Wz = l. Bus I is a referencebus. Using flat start, ind the estimates f

lV,l and {. Tolerance= 0.0001.

lAns: Vl = l/0", V',= t.04223

Final values: Vt - 1.04./.0", V2

l- 3.736"1.

14.2 For samplesystemshown n Fig.

have the followine characteristics.

Meter Full scale (MW) Accuracy (MW) o (pu)

F ig .P.14.2

Given a single line as shown in Fig. p 14.3, two measurements reavailable.using DC load flow, calculate he best estimateof the powerflowing through he line.

14.3

4= 0 rad.

/1 |

( ,r-1-- t t----- [l y0.1pu1

Mt2(1ooMVAbase) Mzt l -2

Fig.P. 14.3

Meter Full scale

(MW)

Meter Standard

Deviation (o )in full scale

Meter

Reading

(Mw)10.4297",V\ = 0.9982q -2.1864";

= 1.080215-1.356, Va 1.03831

P. 14.2, ssumehat he hreemeters

Mrz

Mz r

10 0

10 0

1

432

- 2 6

REFERENES

An lntroduction o State Estimation f Power



Mrz

M t t

Mz z

10 0r0 0

10 0

+ 8

+ 4

+ 0. 8

0.02

0.01

0.002

the phase angles 4 *d d2 given the

Books

l. Mahalanabis, .K., D.P. Kothari and S.L Ahson, ComputerAidetl pr;wer SystemAnalysis 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 i.rtinrution n Eler:tric:Power Sysrems Gcnerali.reclpprct-at:h, Kluwcr AcademicPublishers, oston,1999.

4. Kusic,G'L., Computer-Aided ower Systems nalysis,Prentice-Hall, .J. 19g6.5. Wood, A'J. and B.F. Wollenberg,Power Generation,Operationand Control. Znd

Ed., JohnWiley, NY , 1996.

Calculate the best estimate forf , ' 1 1 , . . . , : n m n n f t ,| 1 - r l \ . , W | | l 5 | I lU { lDL l l t / l I U l l L , )

Meter Meusured value (MW)

Mtz

Mrz

Mt ,

70.0

4.0

30.5

'lIIj

t -

;554.;l Modern owerSystem nalysls

6. Grainger,J.J. and W.D. Stevenson,Power SystemAnalysis,McGraw-Hill, NY',

1994.

7. Deautsch,R., EstimationTheory, Prentice-Hall nc' NJ, 1965

g. Lawson,C.L. and R.J. Hanson, SolvingLeast SquaresProblens, Prentice-Hall.

inc.,NJ., i974.

g. Sorenson, .W., ParameterEstimation,Mercel Dekker,NY, 1980'

Papers

10. Schweppe, .C.,J. Wildes,D. Rom, "Power SystemStaticStateEstimation,Parts

l, ll and ll l", IEEE Trans',Vol. PAS-89,1970,pp 120-135'

ll. Larson,R.E., et. al., "StateEstimation n Power Systems", Parts I and II, ibid-,

pp 345-359.

lZ. Schweppe, .C . and E.J. Handschin, static StateEstimation n Electric Power

System",Proc. of the IEEE, 62, 1975,pp 972-982'

13 . Horisbcrgcr, .P.,J.C. Richarcl nd C. Rossicr, A Fast DccouplcdStaticStatc

Estimator or ElectricPower Systems",IEEE Trans. Vol. PAS-95, Jan/Feb1976,

pp 2O8-215.

Monticell i,A. and A. Garcia, Fast DecoupledEstimators", EEE Trans' PowerSys/,5, May 1990,pP 556-564.

Dopazo,J.F. et. al., "State Calculationof Power Systems rom Line Flow

Measurcments,arts an d l l ", IEEE Trans.,89 , pp. 1698-1708, 1, 1972,pp

1 4 5 - 1 5 1 .

16 . Dcbs, A.S. anclR.E. Larson,"A DynamicEstimator or Tracking the Stateof a

Power System", EEE Trans' 89 , 1970,pp 1670-1678'

iZ. K1u-pholz,G.R.et. al., PowerSystemObservability: PracticalAlgorithmUsing

Nctwork Topology", EEE Trans.99' 1980'pp 1534-1542'

lg. Sirnoes-Costa, and V.H. Quintana, A RobustNumericalTechnique or Power

SystemStateEstimation", EEE Trans.100, 1981,pp 691-698'

19 . Simgcs-Costa, an d V.H. Quintana,An Orthogonal ow Processing lgori thm

fo r Power System SequentialState Estimation", IEEE Trans., 100, 1981' pp

3 '79r-3799.

20. Debs, A.S., "Estimationof External Network Equivalents rom Internal System

1 4 .

1 5 .

An Introduct iono State Estrmation f Power Systems I 55 5_T--

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 PowerNetworks", IEEE

Proc., Special ssueon Computers n Power SvstemOperations,Vol. 75, No. 12,

Dee 1987;aP 76ff7=1ffi2:

28. Wu, F.F., "Power System StateEstimation:A Survey", Int: J. Elec. Power andEnergy 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 SecurityAnalysis", Int. J. Elec. Power and Energy Syst.,Vol. 5, Oct

1983,pp 222-235.

30. Alsac, O., et. aI., "Genetalized StateEstimation",IEEE Trans. on Power Systems,

Vol . 13, No. 3, Aug. 1998,pp 1069-1075.

31. Shirmohammadi, . et. al., "Transmission ispatchand CongestionManagement

in the EmergingEnergy Market Structures", EEE Trans.Power System.,Vol 13,

No. 4, Nov 1998,pp 1466-1474.

32 . Coufto,M.B. et. aI, "Bibliography on Power System State Estimation (1968-

1989)", EEE Trans.Power Sysr., ol .7, No.3, Aug. 1990,pp 950-961.



Data", IEEE Trans.,94, 1974,pp 1260-1268'

21 . Garcia. ., A. Monticell iand P. Abreu,"FastDecoupledStateEstimationand Bad

Dara Processing", EEE 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-.t 2 r .^-r :6 rJ r or nl "Flqd T)cre Detecf ion and ldentif icat ion" . Int. J. EleC. POWer,4J . r \ V 6 l r r r t L t . r . eL . @ e . ,

Vol. 12 , No . 2, April 1990,PP 94-103'

24 . Me-Jjl,H.M. andF.C. Schweppe,Bad Data Suppressionn Power SystemState

Estimation", EEE Trans.PAS-90, 1971'pp 2718-2725'

25. Mafaakher,F., et. al, "Optimum Metering Design Using Fast Decoupted

Estimator", EEE Trans. PAS-98, 7979,pp 62-68.

ii

I

15.1 INTRODUCTION

For reductionof cost and improved reliability, most of the world,s electricpower systems ontinue o be ntercorof diversityof loads,availabilityof so rto loads at minimum cost and pollrderegulatedelectric serviceenvironmeto the competitiveenvironmentof reli

Now-a-days, reater emands ave beenplacedon the ransmission etwork,and thesedemandswill continue o rise because f the increasingnumber ofnonuti l i tygenerators nd greater ompetit ionamonguti l i t ics he'rselves. t isnot easy to acquire new rights of way. Increaseddemandson transmission,absence of long-term planning, and the need to provide open access toenerating ompanies nd customers averesulted n lesssecurityand reducedquality of supply.

compensation n power systemss, therefore,essentialo alleviatesomeoftheseproblems'series/shunt ompensation as been n use or past

Compensationn lglver Systems |55?

I

can be connectecin the system n two ways, n seriesand rr shunt at ihe line

ends (or even n the midPoint)'

Apart from the well-known technologiesof compensation, he latest

technologyof Flexible AC TransmissionSystem FACTS) will be introduced

towards he 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

clearance.

There s a possibility of convertinga single-circuit to a double-circuit ine to

increasehe oadingcaPabil i tY.

Dieletric Limitations From insulationpoint of view, many ines are designed

very conservatively.For a given nominal voltage rating it is often possible to

increase ormaloperating oltages y l07o(i.e.400 kV-

440kV). One should,

however,ensurehat dynamicand ransient vervoltages rewithin limits. [See

Chapter 13 of Ref. 71 .

Stability Issues. here are certain stability issues hat limit the tlansmission

capability. These include steady-statestability, transient stability, dynamic

stability, frequencycollapse,voltagecollapseand subsynchronousesonance.

Several oocl ooks , , 2, 6, 7 8) ar eavailable n these opics.Th e FACTS

technology can certainly be used to overcome any of the stability limits, in

which case he final limits would be thermaland dielectric'

15.3 LOAD COMPENSATION

Load compensation s the managementof reactive pcwer to improve power

quality i.e. V profile and pf. Here the reactive power flow is controlled by

installing shunt compensatingdevices (capacitors/reactors)t the load end



many yearsto achieve his objective.In a power system,given the insignificant erectricarstorage, he powergenerat ion nd oa d mttstbalance t al l t imes.To someextent, he electricalsystem s self'-regulating.f generations less han oacl, oltageancl requencyd'op, and thereby reducing the road. However, there is only a few percentmargin for suchself-regulation.f ,,^r+--^ :- :_,-

support,henoadncreaseith on"se;:;iilTi: ,:t#'j|i;l:ffL;system ollapse' l ternat ively,f there sinadequateeactive ower, he systemrnay havcvoltage ollapse.

This chapter s devotecl o the stuclyof various methods f compensatingpowersystems nd variousypes lfcompensatingrevices,ai lccr, , ,rrj",rru,urr,to alleviate he problemsof powersystemoutlinedabove.Thesecompensators

bringing aboutproper balancebetweengenerated nd consurnedeactivepower.

This is most effective in improving the power transfer capabilityof the system

ancl ts voltagestabil i ty. t is desirable ot h economically nd technical ly o

operate he systemnearuriity power actor.This is why someutilities impose

a penalty on low pf loads. Ye t another way of irnproving the system

performa-nees to operate t undernearbalancedconditionsso as to reduce he

ho* of legative sequencecurrents thereby increasing he system's load

capabil i tyan d reducingpower oss'

A t ransmissionin e ha s hrcc r i t ica l oadings i) naturaloading ii ) steady-

stare tability imit an d (i i i ) thermal imit loading.Fo r a compensatedine the

natural oading s the lowest and before he thermal loading imit is reached,steady-state tability limit is arrived.

Ideal voltageprofile for a transmissionine is flat, which canonlyby loading the line with its surge mpedance oading while this

esJI

T5.4 LINE COMPENSATIONcompensated, t will behave as a purely resistive element and would. causeseries resonance even at fundamental frequency. The location of seriescapacitors s decidedby economical actorsand severityof fault currents.Seriescapacitor reduces ine reactance hereby evel of fault currents.

on on vanous lssues n in series and shunt

compensators ow follows.

15.5 SERIES COMPENSATION

A capacitor in series rvith a line gives control over the effective reactancebetween ine ends.This effectivereactances eiven bv

X r = X - X ,

where

Xl = line reactance

Xc = capacrtor eactance

It is easy o see ha t capacitor cduccs he cffect ivc in e rcactance*.This results n improvement n performance f the systemas

below.(i ) Voltage drop n the line reduces getscompensated).e. minimization ofend-voltage ariations.

(ii) Prevents oltagecollapse.

(i i i ) Steady-stt t teower transfer ncreases;t is inversclyproport ionalo Xl .(iv) As a result of (ii) rransientstability imit increasbs.

The benefitsof the series apacitor ompensator reassociated ith a problem.The capacitivereactanceXg fbrms a series esonantcircuit with the total seriesreactance

X = Xt * X*.n * Xoun,

The natural frequency of oscillation of this circuit is given by.

f ^ -|

rL 2 rJ rc

be achieved

may not beachievable, he characteristics f the ine can be modiso that

(i) Ferranti effect is minimized.

(ii) Underexcitedoperationof synchronous eneratorss 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

. Capacitorsand inductors

. Active voltage source (synchronousgenerator)

When a number of capacitors are connected in parallel to get the desiredcapacitance,t i s known as a bankof capacitors, imilarlya Uant<f incluctors.A bank of capacitorsand/or nductorscan be adjusted n

stepsby switching(mechanical). .

Capacitors and inductors as such are passive line compensators,whilesynchronousgenerator s an active compensator.When solid-statedevices areused for switching off capacitorsand inductors, this is regardecl s activecompensation.

Before proceeding o give a detailedaccountof line compensator, e shallbrief ly discuss ot h shuntan dseries ompensation.

Shunt compensation is more or less like load compensationwith all theadvantagesassociatedwith it and discussedn Section 15.3. t needs o bepointed out here that shunt capacitors/inductors can not be distributeduniformally along the line. Theseare normally connectedat the end of the lineand/orat midpoint of th e l ine.

Shunt capacitors raise the load pf which greatly increases the powertransmittedover the ine as t is not required o carry the reactivepower. Thereis a limit to which transmittedpower



can be increasedby shuntcompensationas t would require very iarge sizecapacitorbank, which would be mpractical.For increasing power transmittedover the line other and bettermeanscan beadopted.For example, eries ompensation,igher transmissionoltage,HVDCetc.

When switched capacitorsare employed or compensation, heseshculd bedisconnectedrnmediately nder ight oacl oncl i t ionso avoicl r.cessiveoltagerise and ferroresonance n presenceof transformers.

The purposeof seriescompensations to cancel part of the series nductivereactanceof the line using seriescapacitors.This helps in (i) increaseofmaximum power transfer (ii)

reduction n power angle for a given amount ofpower transfer (iii) increased oading. From practical point of view, it is

' 2 n

where l= system frequency

xReactive voltage drops of a seriesreactanceadded n a line is I2xIt is positive if X is inductive and negative f X is capacitive. So a seriescapacitivereactance educes he reactancevoltage drop of the line, which is an alternative wavof saying that

X ' t = \ - X , -iititi

I

f c < fwhich is subharmonicoscillation.

Even though seriescompensationhas often been found to be cost-effective

compared to shunt compensation,but sustainedoscillations below the funda-

mental system frequency can cause the phenomenon, referred to aS

subsynchronousesonance SSR) irst observed n 1937,but got world-wide

attentiononly in the 1970s,after wo turbine-generatorshaft failures occurred

at the Majave Generating station in Southern Nevada. Theoretical studies

pointed out that interaction betweena series capacitor-compensatedine,

oscillating at subharmonic frequency,and torsional mechanicaloscillation of

turbine-generatorset can result n negativedamping with consequentmutual

reinforcementof the two oscillations. ubsynchronousesonances oftennot a

major problem, and low cost countermeasures nd protective measures an beapplied. Some of the correctivemeasuresare:

(i) Detecting the low levels of subharmoniccurrents on the line by use of

sensitive elays, which at a certain evel of currents riggers the action to

bypass he seriescapacitors.

(ii) Modulation of generator field current to provide increased positive

dampingat subharmonic requency'

Series ncluctors reneecleclor line compensation nder ight loadconditions o

counter the excessivevoltage rise (Ferranti effect).

As the line load and, in particular the reactive power flow over the line

varies, here s need o vary the compensationor an acceptable oltageprofile.

The mechanical switching arrangement or adjusting the capacitanceof the

capacitorbank in serieswith the line is shown in Fig. 15.1. Capacitancesvariecl y opening he switchos f individual apacitances it h th ecapacitance

C1, being startedby a bypassswitch. This is a step-u'isearrangement. he

I , ,

' technology, the capacitanceof the series capacitancebank can be controlledmuch more effectively; both stepwiseand smoothcontrol.This is demonsffatedby the schematic iagramof Fig. 15.2wherein he capacitor s shuntedby two

nstors ln antl

current in positive half cycle and the other in negativehalf cycle.In each half cycle when the thyristor is fired (at an adjustable angle), it

conductscurrent for the rest of the half cycle till naturalcurrent zeto.Duringthe off-time of the thyristor current s conductedby the capacitorand capacitorvoltage s vr. During on-time of the thyristor capacitor s short circuited i.e.v, = 0 and current s conductedby the thyristor. The sameprocess s repeatedin the other half cycle. This means hat v, can be controlled for any given i,which is equivalent of reducing the capacitanceas C = vJi.By this schemecapacitance an be controlled smoothly by adjusting he firing angle.

l - -+{ - -_ -__-a , .Cur renu imi t i ns| | r' reactor^ o ' l l " " i

l-1<-------r lC r i--.

i

t7,c----]

Fig. 15.2

Thyristors are now available o carry largecurrent and to withstand (duringoff-time) large voltage encounteredn power systems.The latest device calleda Gate Turn Off (GTO) thyristor has the capability that by suitable firingcircuit, angle (time) at which it goes on and off can both be controlled.

This meanswider range and iner control over capacitance. imilarly confrolis possible ver series eactor n the l ine.

All controlled.".?nd uncontrolled (fixed) series compensators require a

ModernPowerSystemAnalYsis

rL=degree of compensation

X

= 25 to J5Vo recommended)



whole bank can also be bypassedby the starting switch under any emergent

conclitionson the line. As the switches n serieswith capacitorare current

carrying suitablecircuit breakingarrangement re necessary.However, breaker

switchedcapacitorsn seriesaregenerallyavoided hesedays the,capacitors

either fixed or thvristor switched.

protectivearrangement.Protectioncan be providedexternallyeither by voltagearrester or other voltage limiting device or an approximate bypass switcharrangement. n no case the VI rating of the thyristors should be exceeded.

Dependingon (i) kind of solid-state evice o be used ii) capacitorand/orreactor compensation and (iii) switched (step-wise) or smooth (stepless)control, severalcompensatron chemes ave beendevisedand are n use.Someof the more common compensationschemesare as under.

(i) Thyristor Controlled SeriesCappcitor (TCSC)

(ii) Thryristor\Switched SeriesCapacitor (TSSC)

(iii) Fllyristor controlled Reactor with Fixed capacitor (TCR + FC)(iv) GTO thyristor Coniiolled Series Capacitor (GCSC)

i'r;

*562i I Modern Power SystemAnalysisI

(v) Thyristor Controlied reactor (TCR)

Capacitorand/or reactor series compensatoract to modify line impedance.

An altcrnat ivo pproachs to introducc control lable oltagcsourcc n series

with the ine. This scheme s known as static synchronousseriescompensator

(SSSC). SSSC has the capabitity to induce both capacitive and inductive

voltage ln senes wrtn [lne, wrdenrngme operatmg eglon o

It can be used for power flow control both increasing or decreasing eactiveflow on the line. Further this schemegives betterstability and s more effective

in dampingout electromechanical scillations.

Though various types of compensators an provide highly effective power

flow control, their operating characteristicsand 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 maintenanceree operation mpedance

modifying (capacitorsand/or reactors)schemes re superior.The specifickind

of compensator to be employed is very much dependent on a particular

application.

15.6 SHUNT COMPENSATORS

As alreadyexplained n Sec.15.4and n Ch. 5 (Sec.5.10) shuntcompensators

are connected in shunt at various system nodes (major substations) and

sometimes t mid-point of lines.Theseserve he purposesof voltagecontroland

load stabilization. As a result of installation of shunt compensators n the

system,he nearbygenerators perate t nearunity pf and voltageemergencies

mostly clono t arise.The two kinds of compensatorsn use are:

(1) Static var compensators SVC): Theseare banks of capacitors (some-

t imes ncluctorsls o gr us eunder ight oad condi t ions)

(ii) STATCOM: staticsynchronous ompensator

(iii) Sync ronouscondenser: It is a synchronousmotor running at no-load

and having excitation adjustableover a wide range. It f'eedspositiveVARs into the line under overexcited conditions and negative VARS

when underexcited. For detailsseeSec. 5.10.)

calledstaticvar switchesor systems. t means ha t terminologywise

SV C = SV S

and we will use these nterchangeably.

Basic SVC Configrurations (or Desigrns)

Thyristors n antiparallelcan be used o switch on a capacitor/reactor ni t instepwisecontrol. When the circuitary is designecl o adjust the firing angle,capacitor/reactorni t actsas continuously ariable n the power circuit .

Capacitor or capacitor and incluctor bank can be varied stepwise orcontinuouslyby thyristorcontrol.Several mportantSVS configurationshavebeendevisedand are applied n shunt ine compensation. ome of the staticcompensators chemesare discussedn what follows.

(i) Sutu.rutcd euctlr

This is a multi-core reactor with the phase windings so arrangedas to cancelthe principalharmonics. t i s consiclered s a constant oltageieactivesource.It is almost maintenance ree but not very flexible with reipect to operatingcharacteristics.

(ii)'fhyristor-coilrroll,cd reuctor (T'CR)

A thyristor-controlled-reactorFig. 15.3) compensator onsists f a combina-tion of six ptrlse r twelvepluse hyristor-control lecleactprs it h a f ixe<t huntcapacitorbank. The reactivepower is changectby adjusting he thyristor firingangle. TCRs are characterisedby continuous control, no transients andgencrat ion f harmonics'k.he controlsystenr onsists

f voltage an dcurrent)

Power I

- - Line oi.rt

transformer i ] .:



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 operaf ion s explained n

Se c .15 . lO .

Statia VAR Compensator (SVC)

These omprise apacitor ank ixed or switched controlled)or fixed capacito

bank and switched reactor bank in parallel.Thesecompensatorsdraw reactiv

(leadingor lagging)power* from the line thereby egulatingvoltage, mprov

*A rcr rct l r rcccor)ncctcd in shunt to t inc at vol tage V draws react ive power Vzl ]It is negative (leading) if reactance is capacitive and positive (lagging) if reactance is

induct ive.

I-? {-rf -- lneaitr i

I

Fixedcaoacitor

I

) t lII

I

I n"r.to,f =..'x',rnc.-I

u v v v

I

I Neutral

Fig. 15.3 Thyristor ontrol ledeactor TCR)with ixedcapacitor*Though ) -connected TCR's are

since t is better configuration.usedhere, it is better o use A-connectedTCR,s

Modern PowerSystemAnalysis

measuringdevices, a controller for error-signal conditioning, a Iinearizing

circuit and one or more synchronising circuits.(iii) Thyristor switcherl capacitor (fSC)

It consistsof only a thyristor-switched capacitor bank which is split into a

numDer unrts equal ratrngs o achrevea stepwisecon

| ;

fe"

system (e.9. line faults, load rejection etc) TSC/TCR combinations arecharacterisedy continuous ontrdl,no transients,ow generations f harmon-ics, low losses, edundancy, lexible control and operation.

i n Tab le15 .1 .

Table 15.1 Comparison f Static Var Generators0 d 6 '

/

Type ofVar

Generator

TCR.FC

( 1 )TSC-(TSR)

(2 )TCR-TSC

(3 )

Dampingreactor

- Fig. 15.4 Thyristor witched apacitorTSC)

As such hey are appliedas a discretlyvariable eactivepower source,where

this typeof voltagesupport s deemed dequate. ll switching akesplacewhen

the voltageacross he thyristor valve is zero, hus providing almost transient

tree switching.Disconnection s eff'ected y suppressinghe firing plus to the

thyristors, which will block when the current reaches zero. TSCs are,

charetcLorisccly stcp wisc control, no transients, cry low ht lrnronics, ow

losses,edundancy nd flexibility.

(iv) CombinedTCR and TSC Compensator

A combinedTSC and TCR (Fig. 15.5) s the optimum solution n majority of

cases.With this,continuousvariable eactivepower s obtained irroughout he

cot t t l ; lc tcronl lo l r lngc. Fru ' t l tc l rnorci r l l cont ro l o1 'bot l r induct ive andcapacitivepartsof the cornpensators obtained.This is a very advantageous

VI and VQ

characteristics

Loss Vs varoutput.

Hannorr i c

generation

Max. theoret.

delay

Trrns i r :n l

behaviour

undersystent

voltagr:

Max comp. current

is proportional to

systemvoltage.

Max cap. var output

decreaseswith the

squareof the voltage

decrease.

High lossesat zerooutput. Losses

decrease moothly

with cap. output,

increirse it h

inductive output

In tc rna l l y ig h

(large pu TCR)

Requiressignificant

fi ltering

l/2 cycle

Poor(FC ('i luscs

transientover-

voltages n response

Max. Comp. current s Same as inproportional o system (1) or (2 )

vol tage.

Max. cap. var output

decreaseswith the

square of the voltage

decrease.

Low lossesat zero Low lossesatouput. Losses ncrease zero output.step-likewith cap. Losses ncreaseoutput step-like with

cup. output,

smoothly with

ind. outputhrtcrnal ly cr y ow Internal ly owResonancemay (small punecessirateuning TCR) Filteringreactors requiredI cycle I cvcle

Cu n bc l rcut ra l . Sanrcas n (2 )(Capacitors an be

switchedout to minimise



---{Neutral

l lCapacitor

T5.7 COMPARISON BETWEEN STATCOM AND SVC

I t m q w h e n n t e r l t h o t i - t h o n n r , - , ' l l ; - ^ ^ - ^ * ^ - , - f : ^ - ^ r - r - r t t" ' * J r r rqL rr r Lrrv r rv l r l la l l r l r t /q. r \ ,Pgr4t r l rE rdl rBc ul ule v -l

characteristicand functional compensation apabilityof theSTATCOM and hesVC aie similar l2l. However, the basic operating principles of theSTATCOM, which, with a converterbasedvar generator,unctions as a shunt-connectedsynchronousvoltagesource,are basicallydifferentfrom thoseof the

SVC, sinceSVC functions as a shunt-connected, ontrolled eactiveadmittance.This basic operational difference renders the STATCOM to have overall

disturbances

to step disturbances) trattsicntovcr-volt ges)

a

Neutral

Fig. 15.5 A combined CR/TSC ompensator

tto .t lvlooernPowerSystenrAnalystst

superior functional characteristics, etterperformance,and greaterapplication

flexibility ascompared o 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 betweenseries and shunt compensation:

(i) Series capacitorsare nherently self regulating and a control system s notrequired.

(ii) For the same performance,series capacitorsare often less costly than

SVCs and osses re very low.

(iii) For voltage stability, series capacitors lower the critical or coliapsevoltage.

(iv) Seriescapacitors ossess dequate imc-ovellt-radapability.

(v) Series capacitorsand switched series capacitorscan be used to controiloading of paralled ines to minimise activeand reactive osses.

Disadvantages f series ompensation:

(i) Series capacitorsare line connectedand compensation s removed for

outages and capacitors n parallel lines may be overloaded.

(ii) During tru.,vy oading, he voltage on oneside of the seriescapacitormaybe out-of ange.

(iii) Shunt reactorsmay be needed or light load compensation.

(iv) Subsynchronousesonance ay cal l for expensive ountermeasures.

Advantagesof SVC

(i) SVCs controlvoltagedirectly.

(ii) SVCs control emporaryovervoltagesapidly.

Disadvantages f SVC

(i) SVCs have imited ovcrload capability.

(ii) SVCs are expensive.

The bestdesignperhapss a combinationof seriesand shuntcompensation.

*FACTS echnology avebeen roposed nd mplemented.ACTSdevices a1be effectively used for power flow conffol, load sharing among parallelcorridors, voltage regulation, enhancementof transient stability anOmitlgation

enablea line tocarry power closer to its thermalrating. Mechanical switchinghas to be supplementedby rapid responsepower electronics. t may be noted

that FACTS is an enabling technology, and not a one-on-onesubstitute formechanicalswitches.

FACTS employ high speed hyristors for switching in or out transmissionline cornponents such as capacitors, reactors or phase shifting transformer forsome desirable performanceof the systems.The FACTS technology is not asinglehigh-power controller,but rather a collectionof controllers,wtrictr can beapplied ndividually or in coordination with others o conffol one or more of thesystemparameters.

Before proceeding to give an account of some of the important FACTScontrollers the principle of operationof a switching converterwill be explained,which forms the heart of thesecontrollers.

15.9 PRINCIPLE AND OPERATION OF CON\ZERTERS

Controllable reactive power can be generatedby dc to ac switching converterswhich are switched in synchronismwith the line voltage with whicn tfr" reactivepower s exchanged.A switchingpower converterconsistsof an array of solid-stateswitcheswhich connect he nput terminals to the output terminals. It hasno internal storageand so the instantaneousnput and output power are equal.Further the input and output terminations are complementar|, that is, if the inputis terminated by a voltage source (charged capacitor or battery), output ii acuffent source (which meansa voltage sourcehaving an inductive impedance)and vice versa. Thus, the converter can be voltage sourced (shunted by a

capacitoror battery) or current sourced (shuntedby an inductor).Single line diagram of the basic voltage sourcedconverter scheme for

reactivepower generation s drawn in Fig. 15.6.For reactivepower flow bus



Because of higher initial and operating costs, synchronous condensersare

normally not competitivewith SVCs. Technically, ynchronous ondensers re

better than SVCs in voltage-weak networks. Following a drop in network

voltage, the increase n condenser eactivepower output is ilrslantaneous.Most

synchrono.us ondenserapplicationsare now associatedwith HVDC installa-

l lons.

15.8 FLEXTBLEAC TRANSMTSSTON YSTEMS (FACTS)

The rapid development f power electronicsechnologyprovidesexciting

opportunitieso develop ewpowersystem quipmentor better tilization f

voltage V and converter terminal voltage V, are n phase.Then on per phase basis i

r = v - v 4x ' i

' F h ar a a ^ + ; " ^ - ^ ' . ' ^ - ^ - , ^ f ^ ^ J ^ ^ : ^

ru v l v c v L t v l yP r J w 9 r E ^ U r r 4 r r B t r l5

'O = v I =

v ( v - v o )X

563 l ModernPoI

System bus V---__i--'i ' Coupl ingransformer

I

. J ,X

:l J,Transformereakage eactance

Vd.

Fig. 15.6 Static eact ive ower enerator

The switching circuit s capable of adjusting Vo, he 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 on-

time of the solid-state switches. It is to be noted that transformer leakage

reactance s.quite small (0.1-0.15 pu), which means hat a amall differenceofvoltage (V-Vo) causes he required ,1and Q flow. Thus the converter acts ike

a static synchronous condenser or var generator).

A typical converter ircuit s shown n Fig. 15.7. t is a 3-phasewo-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

synchronismwith the bus voltagewaves.

[ 1 '

K

-

capacitor s zero. Also at dc (zero fiequency) the capacitor doesnot supply any

react ivepower. Thereibre, he capacitorvoltage cloesnot changeand the

capacitorestablishesonly a voltage level for the converter.The switching

causeshe converter o interconnect he 3-phase ines so that reactivecurrent

can flow between thent.

The converterdraws a small amountof real power to provide for the nternalloss(in switching). f it is required o feed reai power to the bus, he capacitor

is replace,lby a storagebattery. For this the circuit switching has to be

modified ro create a phase difference dbetween Vsand Vwith Vsleading V'

The aboveexplainedconverter s connected n shuntwith the ine. On sirnilar

lines a convertercan be constructedwith its terminals n serieswith the line.

It has to carry the line current and provide a suitablemagnitude (may also be

phase) oltage n serieswith the line. In such a connection t would act as an

impedancemodifier of the line.

15.10 FACTS CONTROLLERS

The developmentof FACTS controllershas followed two different approaches.

The first approachemploys reactive mpedancesor a tap changing fansformer

with thyristor switches as controlled elements, he second approachemploys

self-commutatedstatic convertersas controlled voltage sources.

ln general,FACTS controllerscan be divided into tour categories.

(i) series (ii) shunt (iii) combined series-seriesiv ) combined series-shunt

controllers.

The generalsymbol for a FACTS controller s given in Fig. 15.8(a).which

showsa thyristor arrow inside a box. The seriescontroller of Fig. 15.8bcould

be a variable mpedance,suchas capacitor, eactor,etc. or a power electronics

basedvariable source.Al l seriescontrollers nject voltage n serieswith the

line. If the voltage s in phasequadraturewith the ine, the series ontroller only

supplies r consumes ariable eactive ower. Any otherphase elationshipwill

involve real power also.Tlte shunt controllers of Fig. 15.8c may be variable impedance,variable

sourceor a combination of these.All shunt controllers nject current into the

Vo" Vot Vo,



Va"

C

Fig.15.7 Three-phase,wo-level ix-pulse ridge

systemat the point of connection.Combined series-series ontrollers of Fig.

15.8dcould be a combinationof separate eriescontrollerswhich areconffolled

in a coordinatedmanneror it could be a unified controller.

Combined series-shuntcontrollers are either controlled in a coordinated

ma-nner s in Fig. 15.8e or a unified Power Flow Controller with series and

shuntelementsas n Fig. 15.8f.For unified controller, herecan be a real power

exchangebetween the seriesand shunt controllers via the dc power link.

Storage ourcesuchas acapacitor, attery,superconducting agnet,or any

other sourceof energy can be added n parallel through an electronic nterface

to replenish the converter's dc storageas shown dotted in Fig. 15.8(b).

A

ff.*J ModernowerSystem nalysis

controller with 'storage is much more effective for conffolling the systemdynamics than the corresponding controller without storage.

Compensationnpower ystems tffi*ffilT-

Llne Line

(a)GeneralsymbolforFACTS ontrollerFC)

l u n " l[,--TT- I

lFclT-

(c )Shunt ontroller

Line

ac l ines

Flg. 15.8 DifferentFACTS ontrollers

The group of FACTS controllers employing switching converter-based

synchronousvoltage sources nclude the STATic synchronousCOMpensator(STATCOM), the static synchronousseriescompensator SSCC), the unifiedpower flow controller (UPFC) and the latest, the Interline Power Flow

(a )/

(b )

Fig. 15.9 (a) STATCOM asedon voltage-sourcednd (b )current-sourcedconverters.

. A combination of STATCOM and any energy source to supply or absorb

powei is called static synchronousgenerator SSG). Energy sourcemay be a

battery, flywheel, superconducting magnet, arge dc storagecapacitor, another

rectifi erlinverter etc.

Statlc Synchronous Series Compensator (SSCC)

It is a seriesconnectedcontroller. Though it is like STATCOM, bu(its output

voltage s in serieswith the line. It thus controls he voltageacross he ine and

hence ts impedance.

Interline Power FIow Controller (IPFC)

This is a recently ntroduced controller 12,3]. It is a cornbinationof two or

more staticsynchronous eries onlpensators hich arecoupledvia a common

dc link to facilitate bi-directional flow of real power between he ac terminals

of the SSSCs, and are controlled to provide independent reactive series

compensationfor the control of real power flow in each line and maintain thedesired distribution of reactivepower flow among he lines. Thus t managesa

comprehensiveoverall real and reactive power management or a multi-line

(d) Unified eries-series ontroller

!-.I i t

FCH Coordinatedcontrol

(e )Coordinatederiesand shunt ontroller

(f) Unified eries-shunt ontroller

dc power

l ink

<-oc



Controller (IPFC).

STATCOM

STATCOM is a static synchronousgeneratoroperatedas a shunt-connectedstatic var compensatorwhose capacitive or inductive output current can be^ - - - ^ - - ^ l l - l ! t a ,i;orrrr-oiieo noepenoentoi rhe ac system voltage. The STATCOM, like itsconventional countetpart, he SVC, controls transmissionvoltage by reactiveshunt compensation. t can be basedon a voltage-sourcedoi curtent-sourcedconverter. Figure 15.9 shows a one-line diagram of srATCoM basedon avoltage-sourced onverterand a current sourcedconverter.Normally a voltage-

source converter is preferred for most converter-basedFACTS controllers.STATCOM can bc designed o be an active filter to absorb systemharmonics.

transmission ystem.

Unified Power FIow Controller (UPFC)

This controller is connectedas shown in Fig. 15.10. t is a combinationof

STATCOM and SSSC which are coupled via a iommon dc lin-k to -allow

bi-directional flow of real power between he series output terminals of the

SSSCand the shunt output terminals of the STATCOM. These are controlled

fo provide concurrent real and reactive series line compensation without an

externalenergy source.The.UPFC,by meansof angularly unconsffained eries

voltage njection, s able to control, concurrently/simultaneously r selectively,the transmission line voltage, impedance, and angle or, alternatively, the real

rr$il#4 ModernI

and eactive ine flows. The UPFC may alsoprovide independentlycontrollabteshunt eactive compensation.

STATCOM

Fig.1S.10 Unif ied owerFlowControi ler pF C

Thyri s t or- co n tro I e d p h a se-sh ifti n g Tra n sfo rm er (TCp s r)

This controller is also called Thyristor-controlled Phase Angle Regulator(TCPAR). A phase hifting transformercontrolledby thyristor switches o givea rapidly variablephaseangle.

Thyristor- Con troll e d VoI tdgre R egu tra or flCVR)

A thyristor controlled transformer which can provide variable in-phase voltagewith continuouscontrol.

Interphase Power Controller (IpC)

A series-connectedontrollerof activeandreactivepower consisting, n eachphase,of inductive and capacitive branchessubjected to separatelyphase-shiftedvoltages.The active and reactive power can be set inpedendentlybyadjusting he phaseshifts and/or the branch mpedances,using mechanicalorelectronicswitches.

Thyristor Controlled Braking Resistor ICBR)

It is a shunt-connectedhyristor-switchedresistor, which is controlled to aidstabilization of a power system or to minimise power acceleration of a

l l " r ' .- r' i:

stability and cnnffol line flows. Voltage source converter based (self-commutated) HVDC system may have the same features as those ofSTATCOM or UPFC. This systemalso 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 comparative erformance f major FACTScontroller

Typeof FACTS

Controller

Load

flow control

Transient

stability

Oscillation

Damping

V

control

SVC/STATCOMTCSC

SSSCTCPARUPFC

X

XX

XX X

XX X

XX X

XX X

x

xX X

XX X

xx xxx x

XX X

X

XX X

XX

XX

XX

XX

XX X

* a;v-slrong inf luence; xx-average inf luence; x-smail infruence

sutyffvtARy

Since the 1970s,energy cost, environmental estrictions, ight-of-*ay difficul-ties, along with other legislative sociai and cost problems huu. postponedtheconstructionof both new generationand transmissionsystems n India as wellas most of other countries. Recently,because f adoption of power reforms orrestructuring or deregulation, competitive electric energy markets are beingdevelopedby mandatingopen accessransmission ervices.

In the late 1980s, he vision of FACTS was ormulated. n this variouspowerelectronicsbased ontrollers (compensators)egulate power flow and ransmis-sion voltage and through fast control aciion, mitigate dynamic disturbances.

Due to FACTS, transmission ine capacitywas enhanced. wo typesof FACTScontrollers were developed. One employed conventional thyristor-switchedcapacitorsand react<lrs, nd quadrature ap-changing ransformers uch as SVC



generatingunit during a disturbance.

Thyristor-controlled Voltage Limiter FCVL)

A thyristor-switched etal-oxide aristor Mov) used o limit theacrossts terminals urins ransient onditions

'TIVDC

It may be noted that normally HVDC and FACTS are compleme tarytechnologies.he role of HVDC, for economiceasons,s to interconnect c

system?,wherereliableac nterconnection ould be too expensive. VDC

voltage

and TCSC. The secondcategory was of self-commutatedswitching convertersas synchronousvoltage sources,e.g. STATCOM, SSSC,UPFC and pFC. Thetwo groups of FACTS controllers have quite different operatingand perform-ance characteristics.The second group usesself-commutateddc to ac converter.The converter, supported by a de power supply or energy storagede.,,icecanalso exchange eal power with the ac systembesidesconmolling eactivepowerindependently.

The increasing use of FACTS controllers in future is guaranteed.Whatbenefits are required for a given systemwould be a principal ustification for

the choice of a FACTS controller. Its final form and operation will, ofcourse,depend not only on the successfuldevelopmentof the necessary ontrol and

r:5?4i I Modernpower SystemAnalysisI

communieation eehnologies nd protocols,but alsoon the final structureof ihreevolving newly restructuredpower systems.

REFERECES

Books

I Chakrabarti,A', D.P. Kothari and A.K. Mukhopadhyay,Performance Operationand Control of EHV Power TransmissionSystems.Wheeler,New Delhi, 1995.

2. Hingorani, N.G. and Laszlo Gyugyi, (Jnderstanding FACTS. EEE press, NewYork, 2000.

3. Song,Y.H' andA.T. Johns,FlexibleAC TransmissionSystems,IEE,bndon, 1999.4. Miller, T'J.E., ReactivePower Control in Electric Systems, ohn 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 systemvortage stability, McGraw-Hill, singapore, 1994.7. Nagrath, .J and D.P. Kothari, power S),stem ngineering,TataMcGraw_Hill,

NewDelhi, 1994.

8. Indulkar,C.S. and D.P. Kothari, Power SystemTransientsA StatisticalApproach,Prentice-Hall f India, New Delhi, 1996.

9. Mathur, R.M' and R.K. Verma, Thyristor-BasedFACTI; Controllers or ElectricalTransmissionSystems. ohn Wiley, New york, 2002.

Papers

I0- Edris- A-, "FACTS Technologv Development: An Update".\EEE Pott 'er Engirteer-ing Rerieu'. \'ol. 20. lVlarch 2@1i. pp f9.

1l - I l iceto F- and E. Cinieri , "Comparative Analysis of Series and Shunt Compensation

Schemes or Ac Transmission svstems". IEEE Trans. pAs 96 6). rg77. pp lglg-l8-10.

12 . Kimbark. E.W.. "Hou, to Impror,e S.r'srem Stabi l i r- i , u, i thout RiskingSuhs-rtchnrnous R('s(')nitncc"'.IEEE lr.-rrs. P.-tJ-. 96 t-i,I. Sept/Oc.t 1977. pp 160.S-

1 9 .

16.1 INTRODUCTION

Load forecasting plays an important role in power system planning, operationand control. Forecastingmeans estimating active load at various load busesaheadof actual oad occurrence.Planningand operationalapplicationsof loadforecasting requires a certain

'leadtime' also called forecasting iritervals.

Nature of forecasts,ead times and applicationsare summarised n Table 16.1"

Table 16.1

Nature of

forecast

Lead time Application

Very short term

Short term

A few seconds to

several minutes

Half an hour toa ferv hours

Generation,distribution schedules,contingency analysis for system

secunfy

Allocation of spinning reserve;operarionilI planning ild unitcommitment; maintenancescheduling



13. CIGRE/ 'wG 38-01. srrzric/a r Conrp<,r-rdrorr.IGRE/.prrr-is.

l{ . Ptrv l r. ). . "Llsc 'ofHDVC an dF.ACTS". EE E procc'edings.vol .pp 235-245.

15 . Kinrbark,E.W. "A New Look At Shunt Compensation.,EEE102.No. l . Ja-n 983. n 212-2!8

i 986

.\8. 2, Feb. 2000,

Trans. Vol PAS-

Medium term

Long term

A few days to

a t'ew' veeks

A t'ew months to

a few years

Planning for seasonal peak-

wrnter. summer

Planning generation

growth

A good t'orecast eflecting current and future r.rends,temperedwi.Ji goodjudgement, is the key to all planning, indeed.o financial success. he accuracyof a forecast s crucial to any electric utility, since t determines he iming andcharacteristicsof major system additions.A forecast hat is too low can resultin low revenue rom sales o neighbouringutilitiesor even n loadcurtailment.

Forecasts that are too high can result in severe financial problems due toexcessive nvestment in a plant that is not fully utilized or operatedat low

capacity actors.No forecastobtained rom analytical procecluresan be strictly

r"ii.d upon the judgement of the forecaster, which plays a crucial role in

ariving at an acceptableforecast.

Choosing a forecasting technique for use in establishing future load

requirements is a nontrivial task in itself. ing on nature o

variations,one particularmethcd may be superior to another.The two approaches o load forecasting namely total load approach and

component approachhave their own merits and demerits.Total load approach

has the merit that it is much smootherand indicative of overall growth trends

and easy to apply. On the other hand, he merit of the componentapproach s

that abnormal conditions in growth trends of a certain component can be

detected, hus preventingmisleading orecastconclusions.There s a continuing

need,however, to improve the methodology for forecastingpower demand more

accurately.

The aim of the presentchapter s to give brief expositionsof some of the

techniques that have been developed'in order to deal with the various load

forecastingproblems. All of theseare basedon the assumptionhat the actual

load suppliedby agiven systemmatches he demandsat all points of time (i.e.,

there has not been any outagesor any deliberate sheddingof load). It is then

possible o make a statisticalanalysisof previous load data n order to se t up

a suitablemodel of the demandpattern.Once this hasbeendone, t is generally

possible to utilize the identified load model for making a prediction of the

estirnateddemand for the selected ead time. A major part of the forecasting

task is thus concernedwith that of identifying the bestpossiblemodel for the

past oad behaviour.This is bestachieved y decomposinghe oad demandat

any given point of tirne into a number of distinct components.The load is

dependenton the industrial, commercial and agricultural activities as well as the

weathercondition of the system/area.he weathersensitive omponentdepends

on temperarure,cloudiness,wind velocity, visibility and precipitation. Recall

the brief discussions n Ch. 1 regarding the nature of the daily load curve which

has been shown to havea constantpari corresponding o the base oad and other

variable parts. For the sakeof load forecasting, a simple decompositionmay

serveas a cdnvenientstartingpoint. Let y(k) represent he total load demand

LoadForecastingechnique Llif,iffiI

T6.2 FORECASTING METHODOLOGY

Forecasting techniquesmay be divided into three broad classes.Techniquesmay be basedon extrapolation or on correlation or on a combination of both.

rrnlnlstlc,

Extrapolation

Extrapolation echniques nvolve fitting trend curves to basic historical dataadjusted to reflect the growth trend itself. With a trend curve the forecast isobtained by evaluating the trend curve function at the desired future point.Although a very simple procedure, it produces reasonableresults in someinstances.Such a technique s called a deterministicextrapolationsince andomerrors in the data or in analytical model are not accounted or.Standard analytical functions used in trend curve fitting are [3].

(i) Straight line

(ii) Parabola'

(iii)S-curve(iv) Exponential

(v) Gempertz

! = a + b x

! = a + b x + c * 2

! = a + b x + c i + d x 3! = c e &

! = In-r 7a + ced''1

' The most corlmon curve-fitting technique for fitting coefficients andexponents a4) of a function in a given forecast s the methodof leastsquares.If the uncertainty of extrapolated results is to be quantified using statisticalentities such as mean and variance, he basic techniquebecomesprobabilisticextrapolation. With regression analysis the best estimate of the modeldescribing the trend can be obtained and used to forecast he trend.

Correlation

Correlation techniques of forecasting relate system loads to various demo-

graphic and economic factors. This approach s advantageous n forcing theforecaster to understand clearly the interrelationship between load growthpatterns and other measurable actors. The disadvantage s the need to forecastdemographic and economic factors,



(either or the whole or a part of the system)at the discrete ime k = l, 2,3,

....It is generallypossible o decompose (k) into two parts of the form

y(k)= ya(k) y"(k)

where the subscript d indicates the deterministic part and the subscript s

indicatesthe stochasticpart of the demand. f k is considered o be the present

time, then y(k + j), j > 0 would represbnta future load demandwith the index

7 being the lead ime. For a chosenvalue of the indexT, he forecastingproblem

is then he sameas he problemof estimating he valueof y(k +/) by processing

adequatedata br the past oad dernand.

which can be more difficult than forecastingsystem load. Typically, such factors as population, employment, buildingpermits, business,weatherdata and the like are used n correlation techniques.

No one forecasting method is effective in all situations. Forecastingtechniques must be used as tools to aid the planner; good judgement a:rdexperiencecan never be completely replaced.

16.3 ESTIMATION OF AVERAGE AND TREND TERMS

The simplest possible orm of the deterministic part of y(k) is givenby

( 16 .1 )

ya &) = !-a + bk + e(k) (r6.2)


where larepresentsthe averageor the mean value of yd(k), bk represents he'trend' term that grows linearly with k and e(k) represents he error of

modelling the complete load using the averageand the trend terms only. The

question s oneof estimating he valuesof the two unknownmodelparameters

la aldb !o ensurea good model.As seen n Ch, 14, when little orlo st1listical

information s available egarding he error erm, he methodof LSE is helpful.

If this method s to be used orestimating yo andb,the estimationndex "/isdefined using the relation

J - E{ez(D} (16 .3 )

where E(.) represents he expectationoperation. Substituting for e(A) from

Eq. (16.2) and making use of the first order necessaryconditionsfor the index

J to have its minimum value with respect o ya md b, it is found that the

following conditionsmust be satisfied 2) .

Since the expectationoperationdoesnot affect the constantquantities, t is

easy to solve these two equationsn order to get the desiredrelations.

E {y a - ya&)+ bk I= 0

E {bkz ya(k)k tdkl - 0

ta= E{yd&) l - b {E (k ) }

b - lE{ya&)kl yo E{kllt4{k2l

(16.4a)

(16.4b)

(16.5a)(16.5b)

(16.6a)

(16.6b)

If y(k) is assumedo be stationary statisticsare not time dependent) ne may

involve the ergodic hypothesisandreplace he expectatiorioperationby the time

averaging fonnula. Thus, if a total of N data are assumed o be avai.labLeor

determining the ime averages,he wo relationsmay be equivalentlyexpressed

as follows.

LoadForecastingechnique kt5f,*AI

ln order to illustrate the nature of results obtainable rom Eqs' (16'6a) and

(16.6b),consider he clatashown in the graphsof Fig. 16'l which give the

iopurerion in millionT. Th= cash values of the agricirtturaland the jndust+ial

ooiput, in millions r.rf upeesand the amountof electricalenergy consumpdon

(toaAdemand) n MWs 1n Punjab over a period of sevenyears starting from

1968.A total of g5 datahavebeen generated rom the graphsby sampling the

graphsat intervals of 30 days.Thesehave beensubstitutedn Eqs' (16'6a) and

if O.OUIn order to compute the avetageand the trend coefficients of the four

variables.The resultsare given in Table 16'2'

Table 16.2

Variable Average Trend Cofficient

itIIl

it

T

I

l

Population

Industrial output

Agricultural output

Load demand

13 million

Rs 397million

Rs 420.9million

855.8MW

o.20.540.78r .34

- peprJlationnmill ions

lndustryn mill ions f ruPees

-- - Agriculture il l ions f rupees

- Load emandn MW- - -

/ /t , ,



These two relationsmay be fruitfully employed in order to estimate he average

and the trend coefficient for any given load data.

Note that Eqs. (16.6a) and (16.6b)are not very accurate n case he load

data behavesas a non-stationaryprocesssince the ergodic hypothesisdoesnot

holcl or suchcases.t may still be possible o assume hat he dataover a finite

window is stationaryand the entire set of data may then be consideredas the

juxtaposition of a number of stationaryblocks, each having slightly different

statistics.Equations 16.6a)and 16.6b)may thenbe repeateclve r the different

blocks in order to compute the averageand the trend coefficient for each

window of data.

i 1 6

I t

the data

ffi4 Modern owerSystemAnalysisI

600

72

-t--tr in"-ert-

the load model maY be assumed o be

y(k)- fu3,+e(k)i: l

where he coefficient b,needto be estimated rom the past oad d

model above is obviously a non-linear function of the time index

need L coefficients to be estimated. A much simpler approach

modelling of the load is to introduce an exponential form

{II,

III

;:400

3

120

(16.8a)

/<and wouldto non-linear

(16.8b)k (hours) _---

192 216 240 264 288 312 336

k (hours)--------

Fig.16.2 Hourly oad behaviour f Delhiover wo consecutive eeks

Caution

The 85 data, used n Example 16 . , are generally not adequate or makingstatistical caiculations so that the vaiues given ubour may not be entirelyadequate. n addition, the statistical characteristicsof the set of variablesconcernedmay have changed (i.e., the data may in fact be non-stationary)andthis alsomav inffoducesomeerror in the results.Finally, the graphs n Fig. 16.1are actually based on half yearly data obtained from the planning commissiondocumentand an interpolationprocesshas beenemployed in order to generatethe monthly data. This may add someunspecified

errors to the data which willalso affect the accuracyof the estimates.

Prediction of ya &+j)

y(k)= c ex p [bk]+ e(k)

which involves only two unknown coefficients.Besides educing he number of

unknowns, th" exponential model has the additional advantageof being readily

transformed into u linru, form. All that is required is to take the natural log of

the given data. In either case, he method of LSE is easily extended to estimate

the model parameters rom the given historical data.

|6,4ESTIMATIoNoFPERIoDIGcoMPoNENTS

The deterministic part of the load may contain some periodic components in

;ilri'.o,',; tn" uurrage and the porynomial rerms. consider for example the

curve shown in Fig. tZ.Z wtrich givei the variation of the active power supplied

by a power utility over a period of t*o weeks. It is observed hat tlrg daily load

variations are repetitive from day to day except for somerandom fluctuations'

It is also seen hat the curve for sundays differ significantly from those of the

week days n view that Sundaysare holidays.I,,]rTt

out that the curve ior ihe

entire weekly period starting fiom, say, the mid night of one Sunday till the mid

night of the next Sunday behaves as a clistinctty periodic waveform with

superposed andom variations'

If it is assumed hat the load data are being sampledat an hourly interval,

then there are atotal of 168 oad data in one period so that the load pattern may

be expressed n terms of a Fourier series with thefundamental frequency ul

i.f"g'.q" ut to Z7rl\68rads.Arsuitablemodel for the oad y(k) is then given by

y(k) = y + [a, sin iuk + b, cos iuk] + e(k) (15'9)

600

tIi.

400

3E 2ooo

J

01



Once the model for the deterministic component of the load has beendetermined, t is simple to make the prediction of its future value.For the simplemodel in Eq. (76-2), the desired prediction is computedusing the relation

v ' (k + i ) = T , + h (k + i \4 \ r / J A " U ' - J l ( i6 .7)

More General Forms of Models

Before leavingthis section, t may be pointed out that the load model may begeneralized y including secondand higher order terms on the right hand side

ti= l

where L represents he total number of harmonics present and a; and b; are the

amplitudesof respectively he sinusoidal and the cosinusoidalcomponents'only

rlnrrrinqnt ',ormonicsnee-do be included in the model'( l t J l l l l l l g rr ! l rq^ r^ \

Once the harmonic load model is identified, it is simple to make a prediction

of the future load ya(k + 7) using the relation

9a&+ /) = h ' (k+ ) i (k )(16.10)

'eifrz#dModernPo@is

!6.5 ESTIMATION OF YS(^ft): iME SERIESAPPROACH

Auto-regressive Models

The sequence,(ft) s said o satisfy nAR model f ordern i.e.it is [AR(n)],if it can be expresseds

n

y,(k)=Do,r,(k

- i) + w(k)i_ l

lie inside the unit circle in the e-plane.The problem in estimating he value of n is refer-red n qc tha ntnhtaw )^s

i,&) = -D a,y, (k- i)

LoadForecastingechnique HffiT--h v 2

15(ft) -t a,y, (k - r) +D b, w(k- ) + w(k) (16.14)j:l

Estimation of two structuralparametersn and rn as well asmodel parametersap bi and the variance d of the noise term w(k) is required. Moie complex

i:l

( 1 6 . 1 1 )

can be represented.The identification problem is solved off-line. The

acceptable load model is then utilized on-line for obtaining on-line loadforecasts.ARMA model can easily be modified to incorporate the temperature,rainfall, wind velocity and humidity data [2]. In some cases, t is desirable toshow the dependenceof the load demand on the weather variables n an explicitmanner. The time seriesmodels are easily generalized in order to reflect thedependence f the load demandon one or more of the weathervariables.

16.6 ESTIMATION OF STOCHASTIC COMPONENT:

KALMAN FILTERING APPROACH

The time series approachhas been widely employed in dealing with the loadforecasting problem in view of the relative simplicity of the model forms.However, this method tends o ignore the statistical nformation about the loaddata which may often be available and may lead to improved load forecasts futilized properly. In ARMA model, the model identificationproblem is not thatsimple. These difficulties may be avoided in some situations f the Kalmanfiltering techniques are utilized. , '

Application to Short-term Forecasting

An application of the Kalman filtering algorithm to the load forecastingproblem has been first suggested y Toyada et. al. [11] for the very short-termand short-termsituations.For the atter case, or example, t i s possible o makeuse of intuitive reasoningso suggest hat an acceptablemodel or load demandwould have the form

v,(k)= y'(k)+ v(li) ( 16 .15 )where y,(ft) is the observedvalue of the stochastic oad at time ft, y,(k) is thetrue value of this load and u(fr) s the error in the observedoad. In addition.the dynamics of the true load may be expressedas



i: l

The variance d of w(k) is then estimatedusing the relation

N

n 2 - ( 1 l n n F - Z r r t \ r ' l ' )

/ , - r c\ t r )

k: l

( 16 .16 )

where a(k) represents he increment of the load demand at time k and u1@)represents 'disturbance erm which accounts or the stochastic erturbations ny,(k).The incrementai ioaci tseif is assumeri o remain constanton an averageat every time point and s modelled by the equation

z(k +l ) = z(k)+ ur(k)

where he term uz&) represents stochastic isturbanceerm.

(16.r2)

( 6 . 3)

y,(k +I ) = y{k) + z(k)+ u1&)

Auto-Regressive Moving-Average Models

In some cases, he AR model may not be adequate o representthe

observedIoad

behaviour unless he order n of the model is madevery hrgfi. In sucha caseARMA (n, m) model is used.

(16.r7)

In order to make use of the Kalman filtering techniques,the noise termsu(k), uz&) and u(k) are assumed o be zero mean ndependentwhite Gaussiansequences.Also, the model equations are rewritten in the form

x (ft+l) = Fx(k)+ Gu(k)

a-.,.".m

to use the solution of Eq . (16.18a) or the vector x (k+ d)to get the result

f ;g+ d) = pd i ( tc t tc ) (16.199)

In order to be able to make use of this algorithm for generatingthe forecastof thg load v-(k + d\. i t 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 accuracyof the metersemployed. However, it is very unlikelythat the value of the covariance Q(k) will be known to start with and willtherefore have to be obtained by some means. An adaptive version of theKalman filtering algorithm may be utilized in order to estimate the noisestatisticsalongwith the statevector x(k) tZ). Now let it be assumed hat bothR(k) and Q(k) are known quantities. Let it also be assumed that the initial

estimate (0/0) and the covarianceP"(0/0) are known. Basedon thesea prioriinformation, it is possible to utilize Eq. (16.19a)-(16.19e) ecursively toprocess he data for yr(l), yr(z), ..., yr(k) to generate he filtered estimate ,(klk). Once this is available,Eq. (16.19g)may be utilized to ger,eratehe desiredload forecast.

To illustrate the nature of the results obtainable through the allorithm justdiscussed, he data for the short term load behaviour for Delhi have beenprocessed. total of 1030datacollectedat the nterval of 15 minuteshave bee.nprocessed. t has been assumed hat, in view of the short time interval overwhich the total data set lies, the deterministic part of the load may be assumedto be a constantmean term. Using the sample averageFormula (16.7) (with

b = 0), we get y = 220 Mw. The data for yr(k) have then been generated bysubtracting the mean value from the measured oad data.

To process hese stochasticdata, the following a priori information have

beenused:

(16.18a)

v(k

where he vectorsx(ft)andu(ft) are definedas

x(k)= ly,(k)= (Df andu(k) = fur(k) uz&))r

The matrices G and ' are henobtainedromEqs. 16.15)-(16.17)asilyand have he ollowingvalues.

"= 1 1-l ,= 1 o1,r= tlLo u' Lo 1l ' Lo l

Based on model (16.18), t is possible to make use of the Kalman filteringalgorithmto obtain heminimum varianceestimate f the vectorx (k ) basedonthe datay,(k): {y,(1), ),,(2) .. y,(k)}. This algorithmconsisrsof the followingequations.

i (k/k)= i (ktt< 1) + K" (k) ty"(ft)

i(ktk -1)= F i((k-L)/(k-r))

h'ft(k/k-I)l (16.19a)

k

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',

where,

Q(k) = covariance f u(k)

R(ft) = covariance f y(ft)

ft(klk) = filtered estimateof x(ft)

ft (k/k-L) = singlesteppretlictionof x(ft)

K.r(k) = filter gain vector of samedimension as a(ft)

Pr(k/k) = filtering error covariance

(16.19b)

(16.19c)

(16 .19d )

(16.19e)

R(k)= 3.74,



The resultsof applicationof the prodietionAlgonthnn (16.19) are shown inFig. 16.3. t i s noted hat the elror of 15 minutesahead oad prediction s around8 MW which is about 3Voof the average oad and less than2Voof the dailymaximum load.

Pr(k/k-l) = predictionerror covariance

FronrEq . (16.18b) bta in he predict ion ((k+1) lk)

From this the one stepahead oad fbrecast s obtained as

Y " ( f t + ) = h ' i ( ( k + l ) l k ) (16.1e0It may be noted that filtering implies removal of disturbanceor stochastic

term with zero mean.

It is alsopossible o obtaina multi-step aheadpredictionof the load from the

multi-step aheadprediction of the vector x (k). For example, if the prediction

i (oro)[;]

1.Actual oad2. 15 mts.prediction rror

200

18 0

i 3 1 4 1 5 1 6 1 7 1 8

Fig.16.3

Comment

Application of Kalman filtering and prediction techniques s often hamperedbythe non-availability of the requiredstatevariablemodel of the concerned oaddata' For the few casesdiscussedn this section,a partof the model has beenobtainable from physical considerations.The part that has not been availableinclude the state and output noise variancesand the data for the initiai stateestimate and the conesponding covariance. In a general load forecastingsituation, none of the model parametersmay be available to start with and itwould be necessaryo make use of system dentification techniques n order toobtain the required statemodel. It has been shown that the Gauss-Markov

model described by Eq. (16.18) in over-parameterisedrom the modelidentification point of view in thescnse hat the data or y,(k) dr: not permit theestimation of ali the parametersof this model. It has been shown in Ref. [12]that a suitable model that s identifiable and s equivalent o the Gauss-Markov

I - , - r6 rsH s

ECONOMETRTC MODELS

If the oad orecastsre equiredor planning urposes,t is necessaryo selectthe lead time to lie in thrro aJeulyears-In sucLcases-the load demandshould be decomposed n a manner that reflects the dependence

of the oad on the various segmentsof theeconomy of the concerned egion. Forexample, he total load demand y(ft) may be decomposedas

M

y(k )=D" ,y , (k )+ e(k )i: l

whereaiare the regressioncoefficients,y,(k) are he choseneconomicvariablesand e(k) represents he error of modelling. A relatively simple procedure is torewrite the model equation in the tamiliar vector notation

y(k )=h ' (k )x+ e(k) (16.20b)

where h'(k)= [yr(ft)yz(k) .. yu&)] andx = fa r a2... ayl.

The regressioncoefficients may then be estimatedusing the convenient least

squaresalgorithm. The load forecastsare then possible through the simplerelation

i( k + 1)= i '(k) i t (k +I lk)\

(16.21)

where i (k) is the estimate of the coefficient vector basedon the data availabletill the tth samplingpoint ana r6 + Uk) s the one-step-aheadredictionof theregressionvector h(k).

16.8 REACTTVE LOAD FORECAST

Reactive loads are not easy to forecast as compared to active loads, sincereactive oads are made up of not only reactivecomponentsof loads,but alsoof transmissionand distribution networksand compensatingVAR devicessuchas SVC, FACTs etc. Therefore, past datamay not yield the correct forecast asreactive oad varies with variations in network configuration during varyingoperating conditions. Use of active load forecast with power factor predictionmay result in somewhat satisfactory results.Of course, here

3

E Ic lO IE IO Io lE(D

o

(16.2Oa)

1 6 =

1 2 . = 19 lL I

I o lc lo l

4 E'1C

En f L



model for state estimationpurposes s the innovation model usecl or estimationof the stochasticcomponent.

on'line Techniques f,or Non-stationanl Load predietion

Most practical load data behaveas non-stationaryand it is therefore mponantto consider the questionof adapting he techniques liscusseclo f,ar o the non-stationarysituation.Ref. [2] has discussedhe threemodels or this purposeviz.(i) ARIMA Models, (ii) Time varying model and (iii) Non-dynamic models.

also, only veryrecent past data (few minutes/hours) may be used, thus assuming steady-statenetwork configuration. Forecasted reactive loads are adapted with currentreactive requirementsof network including var compensationdevices.Suchf - - - | | a

rorecastsare neeoed or secunty anaiysls,voitageireactivepower schedulingetc. If control action s insufficient, sffucturalmodifications have to be carriedout, i.e., new generatingunits, new lines or new var compensatingdevicesnormally have to be installed.

mitul ModrrnPo*"rSyrt"r An"turi.

su uttilARy

Load forecasting s the basic step n power systemplanning.A reasonablyself-containedaccount of the various echniques or the oad prediction of a modern

predictionproblems.Applications of time series,Gauss-Markovand nnovationmodels n setting up a suitabledynamicmodel for the stochastic art of the oaddatahavebeen discussed. he time seriesmodel dentification problem has alsobeen dealt with through the least squaresestimation techniquesdeveloped n

Chapter 14.

In an interconnectedpower system, oad forecastsare usually needed at allthe important load buses.A great deal of attention has in recent years beengiven to the question of setting up the demand models for the individualappliances and their impact on the aggregateddemand. It may often benecessaryo make useof non-linear orms of load models and he questionofidentificationof the non-linear models of different forms is an mportant issue.

Finally, a point may be made that no particular method or approach willwork 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 isexhaustive.The intent has been to introduce some ideas currentlv used in

forecasting ystem oad requirements.

Future Trends

Forecastingelectricity loadshad reacheda comfortable stateof performance nthe yearspreceding he recent waves of industry restructuring.As discussed n

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, orecasting is becoming the rnost crucialactivity in a deregulated market.

li':l;{lrJ.

**and integrationwith conventional ime-seriesmethodsn order o provide amorepreciseorecasting.

REVIEWQUESTIONS

L6.7 Which method of load forecasting would you suggest or long term andwhy?

16.2 Which method of load forecastingwould you suggest or very short termand why?

16.3 What purposedoes medium term forecasting erve?

16.4 How is the forecaster'sknowledgeand ntuition considered uperior o anyload forecasting method? Should a forecaster intervene to modify aforecast,when, why and how?

16.5 Why and what are the non-stationarycomponentsof load changesduringvery short, short, medium and long terms?

REFTRTCES

Books

1. Nagrath .J. and D.P. Kothari, Power System ngineering,TataMcGraw-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., ElectricolPower Systemslanning,Macmillan ndiaLtd., New Delhi,

1998.

5. Pabla, .5., Electric Power Distribution,4thEdn, Tata McGraw-Hill,New Delhi.

1997.

6. Wang, X and J.R. McDonald (Eds), Modern Power SystemPlanning, McGraw-



(iii) Information is'becoming commercially sensitive and increasingly trade

secret.

(iv) Distributed, embeddedand dispersedgenerationrhay increase.

A recentpaper [7] takesa selective ook at someof the forecasting ssueswhich

ar enow associateclit h decision-makingn a competit ivemarket.Forecast ing

loads and prices in the wholesalemarkets are mutually intertwined activities.

Models based on simulated artificial agents may eventually become as

importanton supply side as artificial neural networks have already become

Hill, Singapore,1994.

Papers

7. Bunn, D.W., "Forecasting oads andPriees n CompetitivePowerMark-ets"- rac-

of the IEEE, Vol. 88, No. 2, Feb 2000, pp 163-169.

8. Dash,P.K. et. al., "Fuz,zyNeural Network and Fuzzy ExpertSystem or Load

Forecasting", roc. IEE, Yol. 143,No. l, 1996,106-114.

9. Ramanathan, . ef. a/., "Short-run Forecasts f Electricity Loadsand Peaks". lnt

J, Forecasring, ol . 13, 1997,pp 161-174.

@Mod"rnpo*", syrt"r An"ly.i.

10. Mohammed,A. et. al., "Short-term Load Demand Modelling and ForecastingAReview," 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 LoadForecasting", EEE Trans.,Vol. pAS-89, 1970, pp l67g-16gg.

12. Mehra,R.K., "On-line Identificationof

to KalmannFiltering", IEEE Trans. Vol. AC-16, lg7l, pp lZ_21.

77

T7.T INTRODUCTION

Voltage control and stability problems are very much familiar to the electricutility industry but are now receiving specialattentionby every power system

analyst and researcher.With growing size alongwith economic and environ-mental pressures, the possible threat of voltage instability is becomingincreasingly pronounced n power systemnetworks. In recent years, voltagiinstability has been responsible for severalmajor network collap$s in NewYork, France,Florida, Belgium, Swedenand Japan [4, 5]. Researchworkers,R and D organizations and utilities throughout the world, are busy inunderstanding,analyzingand developing newer and newer strategies o cope upwith the menaceof voltage instability/collapse.

Voltage stability* covers a wide range of phenomena.Because of this,voltage stability meansdifferent things to different engineers.Voltage stabilityis sometimesalso called load stability. The terms voltage instability and voltagecollapse are often used nterchangeably.The voltage instability is a dynamic

process wherein contrast to rotor angle (synchronous) stability, voltagedynamics mainly involves loads and the means for voltage control. Voltagecollapse s alsodefinedas a processby which voltage nstability eads o verylow voltage profile in a significant part of the system.Voltage nstability limit



is not directly correlated to the network maximum power transfer limit.A CIGRE Task Force [25] has proposed he following definitions or voltage

stability.

Small-disturban-ce voltage stability

A power systemat a given operatingstate s small-disturbancevoltage stableif, following any small disturbance, oltagesnear oadsdo not changeor remain

*The problemof voltagestability hasalreadybeenbriefly rackledn Ch. 13. Here

it is againdiscussedn greaterdetailsby devotinga full chapter.

close to the pre-disturbance alues. The concept of small-disturbance oltage

stability is related to steady-state tabiiity and can be analysed using small-

signal (linearised)model of the system.

Voltage Stability

to a certain disturbance, he voltagesnear oads approach he 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

systemmodelled by non-linear differential-algebraicequations.

Voltage Collapse

Following voltage nstability, a power system undergoes oltagecollapse f the

post-disturbanceequilibrium voltagesnear loads are below acceptable imits.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 he purposeof quick and approximate analysis.

T7.2 COMPARISON OF ANGTE AND VOLTAGE STABILITY

The problern of rotor angle (synchronous)stability (covered n Ch. 12) is well

understood and documented t3l. However, with power system becoming

overstressedon accourrtof economic and resource constraint on addition of

generation, transfofiners, ransmission ines and allied equipment, he voltage

instability has become a seriousproblem. Therefore, voltage stability studies

have attracted he attention of researchers nd plannersworldwide and is an

active area of research.

Real power is related to rotor angle nstability. Similarly reactive power is

central to voltage nstability analyses.Deficit or excess eactivepower leadsto

voltage instability either locally or globally and any increase n loadings maylead to voltage collapse.

Voltage Stability Studies

lEq!tl#*l , rc'r

considered. any of the ndicesused o assessoltagestabilityare relatedoNR load flow study. Detailsof static and dynamicvoltage stability will beconsideredurthern Section 7.5.

Some Counter Measures

n counter measures o avord voltage r lty are:

(i) generator terminal voltage increase (only limited control possible)(ii) increase of generator transformer tap

(iii) reactive power injection at appropriate ocations

(iv) load-end OLTC blocking

(v) strategic oad shedding(on occurrenceof undervoltage)

Counter measures to prevent voltage collapse will be taken up in

Section17.6.

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 hat can occur are

listed and explained below.(1) Long Transmission Lines.' In power systems, ong 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 transmission ines. Any loss of an EHV

line in the network causes an enhancement n system reactance.Under

certain conditions the increase n reactive power delivered by the line(s)

to the oad for a given drop n voltage, s less han the increase n reactive

power required by the load for the same voltage drop. In such a case a

small increase n load causes he system o reach a voltageunstable state.

(iii) .Sftortageof Local Reactive Power: There may occur a disorganised

combination of outage and maintenance schedule hat may cause ocalisedreactivepower shortage eadingto voltagecontrol 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



The voltage stability can be studied either on static (slow time frame) or

dynamic (over long time) considerations.Depending on the nature of distur-

bance and system/subsystem ynamicsvoltage stability may be regardeda slow

or fast phenomenon.

Static Voltage 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 analysis ool wherein dynamics s not explicitly

drop.

I7.4 MATHEMATICAL FORMULATION OF VOLTAGEcrrrr, lltE l.? Trnl't l it tD/^t T Eriltrit .l l.l.EDllrr I I r.ClLr.Cr!.GrlVl

The slower forms of voltage instability are normally analysed as steady state

problems using power flow sirnulation as the primary study method. "Snap-

shots" 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{methods reoftenused;pv curvesandve curves. seealsosec. 13.6)Thesetwo methods ive steady-stateoadability imits which are elated o vohagsstability. conventional load flow programs can be used for approximateanalysis.

P-V. curvesare useful for conceptualanalysis of voltage stability and for

The model that will be employedhere to judge voltage stability is basedona single line performance. The voltage performun"" of this simile system squalitatively similar to that of a practical system with many voltage sources,Ioadsand the network of transmissionines.

Consider he radial two bus systemof Fig. 17.1.This is the samediagrarnas that of Fig. 5.26 except hat symbols are simplified. Here Eis 75 and yisvn and E and v are magnitudeswith E leading v by d, Line angle" : tunliXlR and lzl = X.

Fig.17.1

In termsof P and e,the system oaclenclvoltagecan be expressed s [l].

V

-'<- Locus of V66 and Pr",

Noseof he curve

0.9 pf lead

0.8pfla g

Fig.17.2 PV curves or variouspower actors

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 s indicated by the singularityof the Jacobianof Load Flow equations

and level of voltage instability is assessed y the minimum singular value.

Certain esults hat are of significance or voltage stabilityare as under.o Voltage stability limit is reached when

It is seen ro m Eq . (17.1) ha t Vi s a double-valuedunction i.e. thas tw osolutions) of P for a particular pf which determinese in terms of p. The pVcurves 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 lowervoltage lies in the unstablevoltageoperation zone.

fhechangeover

occurs atv".t (critical) and Pro*. The locusof v.r,-p^u* points for variJus pfs is drawnin dotted ine in the figure. Any attempt o iniiease the load abov"

"-*causes

a reversal of voltage and load. Reducing voltage causesan increasingcurrentto be drawn by the load. In turn the arger reactive ine drop causes he voltage

( r7.2)

where S = complex power at load bus

Yrt= load bus admittance

', V = load bus voltage

Nearer he magnitude n Eq. (17.2) to unity, lesser he stability margin.o The loading limit of

'atransmission ine can.be determined rom

lsl = v"3 x"; ( r7.3)

X".i is the critical system eactance eyond which voltage stability is lost. It can

be expressed s

, - z z x - z : - _ i r - 1 - :, , _ < \

L z- i ' l r zox -E ) ' -4x2G2+o \ l e7 . t )

F zX"n= * ( - t an Q+sec Q)

2 P

We have so far consideredhow the PV characteristicswith constant oad

(r7.4)



to dip further. This being unstableoperationcauses he system o suffer voltagecollapse.This is also brought out by the fact that in upperpart of the curve

i n f l r a l ^ . ' ,o - ^^* / . . ^ . ,+^L l^ - - , - - \ dP^rr lr rv rvwwr panL \u'Dr.rurc pdtL),, ,

> U (feduclng lOad meansg V

o = EY.o, -11

X X

ff .0 and

d V

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 oreactivepower injections (+ve or -ve).

Consideronce again he simple radial systemof Fig. 17.1.For p flow it issufficiently accurate o assume X > R i.e. 0 = 90".It then follows that

reducing voltage and vice-versa). t may be noted here that the type of loadassumed n Fig' 17.2 is constant mpedance. n practical syste-.

-tir"type of

loads aremixed or predomirrantly onstantpower type such hat systemvoltagedegraclationis nore and voltagelnstability occursmuch prior to the theoreticalpower limit.

(17.s)

or

Taking

a

V" - EV cos 6+ QX = 0 ( r7.6)

(17.7)

The QV characteristic on normalized basrs etf**. VIE) for various valuesof P/P^ are plotted in Fig. 17.3.The system s voltage stable n the regionwhere

dQldv is positive, while the voltage stability limit is reachedat d,eldV= 0 which may also be termed as the critical operating point.

derivativewrt V gives

dQ = Ecos -2V

d V X

various aluesof plp^^r.

The limiting value of the reactive power transfer atvoltage stability is given by

oD, max

Pr", is hemaxlmumpower ransfer t upf

1 . 0

0.75

Unstable. . PIP^",=9. 5operation

0 0. 2 0. 6 0. 8 1. 0VIE

Fig- 17.3 QV characteristicsor he systemof Fig. 17.1 or

the limiting stageof

Q,n^ t-"o '26 ( 17 .8 )

The inferencesdrawn from the simple radial system qualitatively apply to apracticalsize system.Other factors hat contribute o systemvoltagecollapseare:strengthof transmissionsystem,power transfer evels, load chaiacteristics,generator eactivepower limits and characteristicsof reactive power compen-satingdevices.

Using ihe decoupiing principle 1.".dP

= 0, we setd V

+=."'r[#.+]or I sc= co sloO+11

L d V X J

or Ers6 E cos5l !9 + 2Y]L d V X J

Voltage tability s achievedwhen

E cos (# .+)

> Ers, (short ircuitMVA of power ource)

(17.10), . . \ dz(lU - cntenon

d V

voltage instability occurs when the system Z is such that

d v M- = o o o r - = udZ dV

Application of this criterion gives value of(iii) Ratio of source to load reactance is

stability

Xsourcea o2

X load

z";.very important

( r7.1r)

\dnd for voltage

(17.r2)

a indicates he off-nominal tap ratio of the OLTC transformerat the load end.

T7.5 VOLTAGE STABILITY ANALYSIS

The voltage stability analysis for a given system state involves examiningfollowing two aspects.

(i) Proximity to voltage nstabitity: Distance to instability may be measuredin termsof physicalquantities, uch as oad evel,rea power flow througha critical



Other Criteria of Voltage Stability

(i) 34 ..it.rion: (E'=generator oltage; V=load voltage). Using this crite-

rion, the voltagestability imit is reachedwhen

cos

{#.#}+sindff i }=o ( r7.e)

interface, and reactive power reserye.posiible contingenciessuchas a line outage, oss of a generatingunit or.a reactiu. po*"rio*..mustbe given due consideration.

(ii) iuiechanismof voitage nstabiiity: How and why doesvoltage instabititytakeplace? What are the main factors leading to instability?'ulhat are thevoltage-weakareas?What are he most effectiveways to improv"

"of,ug,tability?

.ffiiffi| . Modern owerSystem nalysis

The static analysis techniquespermit examination of a wide range of system

conditions and can descriUe he nature of the problem and give the main

contributing factors.Dynamic analysis is useful for detailed study of specific

voltage coliapsesituations,coordination of protectionand controls, and testing

of remedial rneasures.Dynamic simulations further tell us whether and how the

steady-state quilibrium point wr

Modelling Requirements of various Power system components

Loads

Load modelling is very critical in voltage stability analysis. Detailed

subiransmission ystem epresentationn a voltage-weakareamay be required'

This may includeiransformerULTC action,reactivepower compensation'and

voltage regulators

It is essential.o consider he voltage and frequency dependenceof loads'

Induction motors shouldalso be modelled'

Generators and their excitation controls

It is necessary o consider the droop chatacteristics of the AVR, load

compensation,SVSs (staticvar system),AGC, protection and controls should

also be modelledappropriately4l .

Dynamic AnalYsis

'lhegeneral tructuc of the systcm moclel or voltage stability analysis s

similar to that for transientstability analysis.overall system equationsmay be

cxprefihcd s

* = f ( X , nand a set of algebraicequations

I (X, V) = YxV

with a se tof knowninitial conditions (Xo' Ve)'

where X = systemstatevector

Y= but voltage vector

1= current njeition vector

Static AnalYsis

The staticapproach aptures.snapshotsf system onditions t variousdme

framesalong the time-domain trajectory. At eachof these ime frames' X in

ii h?lrJdil. frame. hus,heoverau ystemquationsedPceopurely

algebraicquationsllowingh.t..Ytt:f ,t'1tit Tlv:i: ::*:1,::t"t' vlp qnrt o"'?'J"6;ffiffi;;;g."riuulirv isdetermrned-ycomputin.vPand Q

curvesat selected oad buses.Special techniquesusing static analysishavb been

reported n literature.Methodsbasedon VQ sensitivity such as eigenvalue(or

modal) analysis have been devised. These methods give stability-related

information from a system-wideperspective and also identify areasof potential

problems 13-151.

Proximity to InstabilitY

proximity to small-disturbance voltage instability is determined by increasing

il;^";ruiion in stepsuntil,the systembecomesunstableor the oad flow fails

to converge.Refs. t16-181discussspecial echniques or determining he point

of voltage collapse and proximity to volage instability'

The Continuation Power-flow Analysis

The Jacobianmatril becomessingular at the voltage stability limit. A! a result'

conventional toad-flow algorith*, rnuy have convergenceproblemsat operating

conditions near the stability limit. The continuation power-flow analysis

overcomes his problem by reformulating the load-flow equationsso that they

remain well-conditioned at all possible loading conditions. This allows the

solutionof load-flow problemforioth upper *d lo*"t portionsof the P-V

*fi"t:lltinuation-methodof power-flow analysis s ,obrrrt and flexible and

with convergence ifficulties'However'

suming.Hencethe better approach s to

flow irethod (NR/FDLF) and continua-

:ase,LF is solved using a conventional

ns or successivelyncreasing oad evels

Hereafter, the continuation method is

(r7.r3)

(r7.r4)



Ilv = rletwork node admittance matrix'

Equations17.13)an d 17'14)ca n be s

oi tt e numerical integration methods

analysis rethoclslescribedn Ch ' 6' T

,ninutes.As the special models repl

ieading to voltage collapsehave bee

differentialequations s considerably

models.Stiffness s alsocalledsynchrc

ns. Normally, the continuationmethod is

I exactly at and past the critical point'

Voltage StabilitY with HVDC Links

High voltagedirect current* (HVDC).litt-tl:1r:t.:d

for extremely ong distance

f f a n s m i s s i c l n an c l f t r r a s y n c h r o no u s i n t e r c o n n e c ti o n s . An H VD C l i nk c a n b e

*For dctai lcdaccountof HVDC, the readermaY refer to [3]'

f00 | Modern owerSystem natysis

either a back-to-back rectifier/inverter link or can includetransmission. Multi-terfninal HVDC links are also feasible.

The technology has come to such a level that HVDCconnected even at voltage-weakpoints in power systems.present unfavourable "load" characterisfics o fhe nnrr/rrconverter consumes eactivepower equal to 50-60vo of the dc power.

HVDC-related voltagecontrol (voltage stability and fundamental requencytemporary over voltages)may be studied using a transient stability program.Transient stability is often nterrelatedwith voltagestability. Ref. t2i .onJia.r,this problem in greaterdetail.

17.6 PREVENTION OF VOLTAGE COLLAPSE

(i) Application of reactivepower-compensatingdevices.Adequate stability margins should be ensured by proper selection ofcompensation chemesn terms of their size, atingr-und ocations.

(ii) control of network voltage and generator reactive outputSeveral utilities in rhe world such as EDF (France),ENEL (Italy) are

developingspecialschemes or control of network voltages and reactivepower.

(iii) Coordination of protections/controls

Adequatecoordination houldbe ensuredbetweenequipmentprotections/controls basedon dynamic simulation studies.Tripping of equipment toavoid an overloaded ondition should be the last alternative.Controlledsystems€paration nd adaptive or irrtelligentcontrol could also be used.

(iv) Control of transfurmer tap chan.gers

T'apchangers anbe controlled,either ocally or centrally, so as o reducethe risk of voltagecollapse.Microprocessor-basedOLTC controls offeralmostunlimited lexibility for implementingULTC control strategies oas to take advantage f the load characteristics.

(v) Under voltage load shedding

For unplanned or extreme situations, it may be necessary to useundervoltage oad-shedding chemes.This is similar to under ir"q1r"rr.yload shedding,which s a common practice o dealwith extremesituations-

I' VoltaseStability __{-tr*g

monitoringand analysis o identifypotentialvoltagestabilityproblems

and appropriateemedialmeasuresreextremely elpful.

T7.7 STATE-OF-THE.ART, FUTURE TRENDS AND

CHALLENGES

The presentday transmissionnetworks aregetting more and more stressed ue

to economic andenvironmentalconstraints.The trend is to operate he existing

networks optimally close o their oadability imit. This consequentlymeans hat

the system operation s also near voltage stability limit (nosepoint) and there

is increasedpossibility of voltage instability and even collapse.

Off-line and on-line techniquesof determining state of voltagestability and

when t enters he unstablestate,provide the tools for systemplanningand real

time control. Energy management ystem EMS) provide a variety of measured

and computer processeddata. This is helpful to system operators n taking

critical decisions inter alia reactive power management and control. In this

regard autornationand specializedsoftware elieve the operatorof good part of

the burden of system managementbut it does add to the complexity of the

systemoperation.

Voltage stability analysis and techniques have been pushed forward byseveral researchersand several of these are in commercial use as outlined in

this chapter.As it is still hot topic, considerable esearcheffort is beingdevoted

to it. ,

Pw et al. l8l considered an exponential ype voltage dependent oad model

and a new index called condition number or staticvoltagestabilityprediction.

Eigenvalue analyseshas been used o find critical group of buses esponsible

for voltage collapse. Some researchers 26] have also investigatedaspectsof

bifurcations(local, Hopf, global) and chaos and their implications on power

systemvoltage stability. FACTS devicescan be effectively used or controlling

the occurrenceof dynamic bifurcations and chaos by proper choice of error

signalan d control lergains.

Tokyo Electric Power Co. has developed a pP-based controller for

coordinatedcontrol of capacitor bank switching and network transformer ap

ctranging.HVDC power control s used o improve stability.

More systematicapproach s still required for optimal siting and sizing of

FACTS devices.The availability of FACTS controllers allow operation lose to

long distance dc

terminals can beHVDC links mav



resulting from generationdeficiency.

Strategic oad sheddingprovides cheapestway of preventing widespreadVOl taSg COl lanse Lnar l sher ld ino cn l ,a -oo o l rn , , r , r L^ r ^ : - ^^ ) ^^ - - --___r uvrrvrrrvJ orr \_rr l l \ . t ug ugsrBlr tru su as t( )differentiate berween faults, transient voltage dips unJ lo* voltageconditions leading o voltage collapse.

(vi) Operators' role

Operatorsmust be able to recognise voltage stabiiity-relatedsymptomsand take required

emedialactions o prevenrvoltagecollapse.On-line

the thermal imit of the lines without jeopardizingsecurity. The reactivepower

compensationclose to the load centresand at the critical buses s essential or

overcoming voltage instability. Better and probabilistic load modelling [11]

shouldbe tried. It will be worthwhile developing echniquesand models or

studyof non-lineardynamicsof large sizesystems. his 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 assistoperators n applying C-banksso that

a00.l,*l ModernPot

generators peratenear upf. The design of suitableprotective measures n the

eventof voltage nstability s necessary.

So far, computed PV curvesare he most widely used method of estimating

voltagesecurity, providing MW margin type indices.Post-disturbanceMW or

MVAr margins should be translated to predisturbanceoperating limits that

operators can monitor. Both control centre and power.plant operators should be

trained n the basicsof voltagestability. For operator raining simulator [10] a

real-timedynamic model of thepower system hat nterfaceswith EMS controlssuch as AGC is of greathelp.

Voltage stability s likely to challengeutility plannersand operators or the

foreseable future. As load grows and as new transmission and load areagenerationbecome ncreasinglydifficult to build, rnore and more utilities will

face he voltage stability'challenge.Fortunately, many creative researchers nd

plannersare working on new analysismethods and an innovative solutions o

the voltage stability challenge.

A load bus is composedof induction motor where the nominal reactive poweris I pu. The shunt compensations K,n. Find the reactivepower sensitivityat

the bus wrt change n voltage.

Solution

Qrcot= Qno V2 [given] '

Qcomp= Krt V2

Qn"t= Qnua t Qcomp

Qn r= v' - Krn vz lQoo^ ='. Here,

[-ve sign denotesnductivereactive ower njection.l

1.0givenl

dQn"t= 2v-2v K..,

dvi' ,

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 . 4 p u .

I . .

AQ = reactivepower variation(i.e. the size of the compensator)

Srr.= systemshort circuit capacity

Then AV =

AQ = AVSsk

= 1 ( 0 . 0 5 x 5 0 0 0 )

= + 250 MVAR

The capacityof the staticvAR compensators +250MVAR.

REFERECES

Books

1. Chakrabarti, ., D.P. Kothari and A.K. Mukhopadhyay,Perfurmance,Operationand Control of EHV Power Transmission Systerts,Wheeler Publishing, NewDelhi, 1995.

2. Taylor,c.w., Power system voltage stabil ity,McGraw-Hill, New york, 1994.3. Nagrath, .J. and D.P. Kothari, Power SystemEngineering, Tata McGraw-Hill,

New Delhi, 1994.

4. Kunclur, ., Powcr Sy,stem tubility tu l 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 Issueon Voltage Stability and Collapse,', Int. J. ofElectrical Power and Energy systems,voi. i5, no. 4, August 1993.Pai, M.A. and M.G.O, Grady,"VoltageCollpaseAnalysiswith ReactiveGenerationand voltage

trg

d,.

8 .



dv

Find the capacity of a static VAR compensator o be installed at a

x 5Vovoltage luctuation.The short circuit capacity s 5000 MVA.

Solution For the switching of static shunt compensator,

Example:,17.2

bus withl l .

Dependentcons,traints",J. of Elect Machines and power systems,Vol . 17 ,No . 6, 1989, p 379-390.

cIGRE/ Task Forcc 38-06-0l,"Expcrt SystemsApplied to voltage and varControl / / ,1991.

"operator Training simulator", EpRI Final Report EL-7244, May 1991, preparedby EMPROS Systems nternational.

Xu , W and Y Mansour, VoltageStabil ityusingGenericDynamicLoad Models",IEEE Trans. on Power Systems,Vol 9, No l, Feb 1994,pp 479493.

9.

1 0 .

1 5 .

ModernPot

. . . t ? i - d - t ! l ! - F L - - - - , - - - - ^ L - -

IZ. VerTna,l.lt., L.IJ. Arya ano u.r. I\.oman, 'YoltageJlaDrtrtyDnnansEulenry

ReactivePower Loss Minimization", JIE (I), Vol. 76, May 1995, pp.4449.

13. IEEE, special Publication90 TH 0358-2 PWR, "Voltage Stabil ity of Power

Systems:Concepts,Analytical Tools, and Industry Experience" 1990.

14. Flatabo, ., R. Ogncdal nd T. Carlsen, Vol tagcStabi l i tyCondi t ionin a Power

Tiansmisiion System calculated by Sensilivity Methods", IEEE Tians. VoI.

PWRS-5,No. 5, Nov 1990,pp 12-86-93.

Gao,8., G.K. Morisonand P. Kundur, Voltagc Stabil ityEvaluationUsingModal

Analysis", IEEE Trans. Vol. PWRS-7, No. 4, Nov. 1992, pp 1529-1542.

Cutsem,T. Van, "A Method to Compute Rbactive Power Margins wrt Voltage

Collapse", EEE Trans.,Vol. PWRS-6,No . 2, Feb 1991,pp 145-156.

Ajjarapu,V. and C. Christy, "The continuationPower Flow: A Tool for Steady

StateVoltageStabil ityAnalysis", EEE PICA Conf.Proc., May 1991,pp 304-311.

Ldf, A-P, T. Sined, G. Andersonand D.J. Hill, "Fast Calculationof a Voltage

StabilityIndex", IEEE Trans., Vol. PWRS-7, No. 1, Feb 1992,pp 54-64.

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, p 126-129.

Bijwe, P.R,,S.M. Kelapure,D,P. Kothari and K.K. Saxena, Oscil latoryStabil ity

Limit Enhancement y Adaptive Control Rescheduling",nt J. of Electrical Powerand EnergySystems, ol. 21, No. 7, 1999,pp 507-514.

Arya, L.D., S,C, Chaubean d D.P. Kothari , "Linc Switching or Al leviat ing

OverloadsunderLine OutageConditiontaking 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 o Voltage

SccurityAnalysis and Enhancement", nt J. of EP and,ES., Vol. 22, No. 7, Oct.

2000, pp 483486.

Arya, L.D., S.C., Chaubeand D.P. Kothari, "ReactivePower Optimizationusing

Static Stability Index (VSD", Int J. of Electric Power Componentsand Systems,

Vol. 29 , No. 7, July 2001,pp 615-628.

Arya, L.D., S.C. Chaube and D.P. Kothari, "Line Outage Ranking for Voitage

Limit Violations witti Conective ReschedulingAvoiding Masking",Int. J. of EP

and ES,Vol. 23, No. 8, Nov. 2001,pp 837-846.

CIGRE Task Foice 38-02-10, "Cigre Technical Brochure: Modelling of Voltage

CollapsencludihgDynamicPhenomena", lecta, No. 147,April 1993,pp' 7l-77.

Mark J. Laufenbergand M.A. Pai, "Hopf bifurcationcontrol in power system with

T6,

1 7 .

1 8 .

19.

'20.

24 .

25.

26.

2 t ,

. )

23.

AppnNDrx A

In this appendix, our aim is to presentdefinitions andelementaryoperationsof

vectors and matrices necessary or power system analysis.

VECTORS

A vectorx is defined san ordered etof numbersrealor complex),.e.

, (A-1)

xp ...t xn are known as he componentsof the vector r. Thus the vector x is

a n-dimensionalcolumn vector. Sometimes ransposedorm 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 s zero, i.e.

(A-2)



staticvar compensators",Electrical Power and Energy Systems,Vol. 19, No. 5,

1997,pp 339-347.

The sum vector i has each of its components quai o unity, .e.

A

. flTY'

tI

0x.=0

The multiplication of two vectors x and y of samedimensions results in avery important product known as inner or j9g!g! pfodqcij.e.

*t v AD",y,Aytxi: l

Also, it is interesting to note that

xT x= lx 12

cos d 4 "tY l x l l y l ,

wtere Q is angle betweenvectors, lxl and lyl are the geometric lengths ofvectorsx and y, respectively.Two non-zerovectorsare said to be orthJgonral,if

* ty= o (A-6)

The unit 'tector ethe.rest of the componentbare zero, .e.

0

0

1

0

0

kth component

Some Fundamental Vector Operations

A€ k :

(A_3)

(A-4)

(A-5)

Two vectors x and y areknown as equal f, and only if, .yk= !*for k = r,2,. . . , n . Then we sa v

x = y

The product of a vector by a scalar is carried out by multiplying eachcomponent of the vector by that scalar, .e.

If a vector y is to be added o or subtractedrom anothervectorx of the samedimension, theneachcomponentof the resulting vector will consistof

additionor subtractionof the correspondingcomponentsof the vectors x and vr,i.e.

MATRICES

Definitions

Matrix ' '

'\n m x n (ot m' n) matrix is an ordered ectangular rrayof elementswhichmaybe real numbers, omplexnumbers,unctions r operators. hematrix

(A-7)

is a rectangular array of mn elements.

.o,t

!"notesthe (i, i)th element, .e. the element ocated n the ith row and the

7th column. The matrix A has m rows ano n coiumnsand s said to be of orderm x n .

When m = rt, i.e. the numberof rows is equal o that



T h e f n l l n u r i n c n r n n A r f i o o ^ * ^ ^ - - + : ^ ^ l ^ l ^ r ^ r L - - - ' ' -Y v r r r 6 y r v P v r !a v D 4 r v 4 y I - ,u U < r U I t r L Uu r c v e L : t 0 r a r g e Dr a :

x * ! = y + x

x + ( v + ' z ) = ( x + y ) + z

ar @zx) (afiz)x

(ar+ e )x - dF+ a2x

of columns, the matrixis 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 columnvector. An I x n matix, i.e. a matrix having only one row is called a rorevector.

*Sometimesinner product is also representedby the following alterrative forms

x . y, ( x, y ) , ( x , y ) .

. ,1r\!r

's.t

frffi-H Modern ower vstem nalvsis

Diagronalmatrix

A diagonalmatrix is a squarematrix whose elements off the main diagonal area l l z e r o s a i j = 0 f o r i + j ) .

NUII matrix

If all the elements of the squarematrix are zero, the matrix is a nuII or zeromatrix.

(A-8)

Unit (identity). 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 diagonalmatrix with all diagonal elementsequal to 2. This matrix is known as a scalarmaffix.

T

de t A)= tA t= 213

2l l- l 2l , l-1 3l

'1,+l- - 1 I 4l* 1 2l= 2 ( 8 ) + ( ' 6 ) + ( - 5 ) = 5

(A-e)

(A-10)

Transpose of a matrix

'l.h-.etransposeof matrixA denotedby At is the matrix formed by interchanglng

the rows and columns of A.

Note that (Ar)r = A

Symmetrtc matrix

A squarematrix is symmetric, if it is equal to its transpose, .e.

A T = ANotice that the matrix A of Eq. (A-9) is a symmeffic matrix.

Minor

The mino, Mij of an n x n rnatrix is the determinant of (n - l) x (n - 1) matrixformed by deleting the ith row and the 7th column of the n x n matrix.

Cofactor

The cofactor AU of element a,, of the matrix A is defined as

aU = Gl)'*t M,i

Adjoint matrix

The adjoint matrix of a squarematrix A is found by replacingeachelementauof matrix A by its cofactor A,, and then transposing.

For example, f A is given by Eq. (A-9), then



= 3 x 3 s c a l a r m a t r i x

Determinant of a matrix

For each squarematrix, there exists a determinant which is formed by taking

the determinantof the elementsof the matrix.

l -1- l

l 2

ll:Lll;l- ^ J

= 4 x 4 u n i t m a t r i x

l-tl 1l 2

l rl 2l - 1

1

3

l3lzt-Il-I

a o J A =

MotJel l r Powor S

6

7

-5

- \ l_;l) l

(A-11)

A squarematrix is called singular, if its associateddeterminant is zero, and

non-singular, f its associated eterminant s non-zero.

ELEMENTARY MATR.IX OPERATIONS

Eguality of matrices

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 r = 7 , 2 , . . . , f f i , . i = 1 , 2 , . . . , f l

Then we write

A = B

Multiplication of a matrix by a scalar

A matrix s multiplied by a scalar a if all the mn elementsare multiplied by

a. i .e.

(A -12 )

Addition (or suhtraction) of matrices

To add(or subtract) wo matrices of the sameolder (samenumber of rows, and

samenumberof columns),simply add (or subtract) he correspondingelementsnf the f rx rn rnef r ; cce i e r r rhen f rx rn rnqfr i / -ec A enA I l nf lhe cqrnc o, rAcr qr c

added,a new matrix C resultssuch that

C - A + B ;

whose rythelernentequals

ci j= ai j * bi j

Appendix I ,6tlI -

15 - 1 - lC = A + B = l ' I

L2 2)

Additionand subtraction re definedonly for matrices f the sameorder.The bllowingawshold br addition:

(+ fhe cammatatLve n+y:A + B -B + ,{ :

(ii) The associative aw: A +- (B + C) = (A + B) + C

Further( A t B ) r = A r + Br

Matrix Multiplication

The product of two matrices A x B is defined if A has the samenumber ofcolumns as the number of rows in B. The natrices are then said to beconform.able.Ifamatrix A is of order mx n and B is an n x q matrix, theproduct C = AB will be an m x q matrix. The elementc,, of the product is givenbv

,l

r ,cii =

)_rai*0*ik: l

Thus the elementscu are obtainedby multiplying the elementsof the ith rowofA with the correspondingelementsof theTth column of B and hen summingtheseelementproductp.

For example

where

c t t = a t t b t ,+ a r r b ^

c t z= anbn + ar rb2

c z l = uz lb t ,+ a r r b 2 1

czz= aztbtz+ crrb2

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, A are not, in general,equal.Thus, t is important o note ha t n generalmatrix multiplication s not

(A-13)

1""arz l lbn b,r l_

[ ' ' rcnl

La u azz)Lbn brrJ Lr ^ czr l



l - - - - - - _ -__- i r 1r ,ExampleI Il : r

I

i - ? O - r f ) - 1 1

I A A = i- - l z - r l " - L o 3 . 1then,

commutat ive,.e .

A B + B A

The associariveand distributive larr-shold for matrix mulriplicarion (wtrenthe 'aryprrtixiateerpernirnysaredef;tv.4)- re.

Asscrica'rv &rrt': 4{B [ = A(BC) = -tBC

Distibutive law:. A(B + C) - AB + AC

tffiWl ModernoI

[ 1- 1 3 ]

Lo z r lFind AB and BA.

A and B areconformable A has two columnsand B has two rows), thus we

have

l- 1 -1 3l| | r--1 0t

4 8 = 1 2 4 9 l t B A = l , , '" - L ;; ; l

" ^ - l - 4 1 )

A matrix remainsunaffected, f 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

A 0 = 0 A = 0

Also

A - A = 0

Note that equationAB = 0 does not mean hat either A or B necessarilyhas

to lre a null matrix, e.g.

[ r , l [ 3 o l = f o 0 lL0 0J -l 0l Lo 0J

Multiplication of any matrix by a unit matrix results n the original maffix,

l . e .

A I - I A = AThe transposeof the product of two matrices is the product of their

transposes n reverse order, i.e.

6Dr = BrAr

multiplication assistsn the solution of simultaneous

I,j].qffil*l

l.,i95I3n

orn

D " * t =c i i = 1 , 2 , . . . ,m

i:l

Using the rules of matrix multiplication defined above. Eqs (A-14) canlewritten in the compact notation as

A x = c (A-1s)where

It is clear that the vector:mntrix Eq. (A-15) is a useful shorthandrepresentationof the set of linear algebraic equations A-14).

Division does not exist as such in matrix algebra. However, f A is a squarenon-singularmatrix, its inverse (A-l) is defined by the relation

A _ I A _ A A _ I _ I (A-16)

The conventional method for obtaining an inverse s to use the followingrelation

A- ; _ ad jA

det A

It is easy o prove that the inverse s unique

The following are the important properties charactenzing the inverse:

(AB)-r = 3-t4-l

(A-tf = (Ar)-r (A-18)

Matrix Inversion

(A-17)



The conceptof matrix

linear algebraicequations.Consider such a Set of equations

a t f t + anxz+ . . . + abJ n = ct

aL lx t+ azzxz+ .. + azr t r= c2

:

Q^lXl * C^*z+ . . . + A-rJn = C^

14-r1-t 4

I f AA-14) is given by Eq. (A-9), then from Eqs. (A-10), (A-1I), (A-17), we get

ffd-ffi.:| ModernowerSystem natysis

A-1+:if : i _;l:f:i;,:_ilet tL-r -s 5j L ; -; l.J

SCALAR AI\rD VECTOFFUNCTI O]NIS

A scalar unctionof n scalar ariabless definedas

y ! f \ r , x2 , . . . , xn)

It can be written as a scalar unction of a vector variablex, i.e.y - f(x)

where x is an n-dimensionvector,

(A-1e)

(A-20)

In general,a scalar unction could be a functionof severy - f ( x , u , p )

where x, u and p are vectorsof various dimensions.A vector function is defined as

y - f ( x , u , p )

DERTVATNTES OF SCALAR AND \ZECTOR FUNCTIONS

(A-23)

A derivafit'e of a scalar funcritrn 1A-lt)) s,ith re-sper.r r) u \.eL-R)r .ariable -r isdefind as

al vector variables,e.g.(A-2r)

In general, a vector function is a function of severalvector variables, e.g.

- a f

It may be noted that ihe derivative of a scaiar function with respect to a vectorof dimension n is a vector of the samedimension.

The derivative of a vector function (A-22) with respect o a vector variabler is defined as

o fa0x

0*,

}fz0*,af-0*,

0*,

0fz0*,af;0*,

(A-zs)

o*n

?fz0rn

.. af ;o*n

Consider ow a scalar unctiondefinedast - ff@, u, p)

= 21fi(x,u, p) + Lzfz(*, , p)

Let us incf { . According o Eq. (A-24),a^

(A-26)

(A-27)

+ ... + 2*f*(4, u, ) 6-2g)

we can write

f (x , u, p) (A-2e)

(A-24), we can writeet us now find

L J m \ r r U r P ) J

d s ^ i .-. According o Eq .ox



0r,af

0*,

oJ

orn -

A t0x

(A-24)l a " lLar,

fijfilftf.f Modern ower'system natystsI

}ft 0f, af^ AppnNDrx B

We can represent,as we saw in Chapter5, a three-phase ransmission ine* bya circuit with two input terminals (sending-end,where power enters) and twooutput terminals(receiving-end,where power exits). This two-terminal paircircuit is passive (since t does not contain any electric energy sources), inear.(impedancesof its elements are independentof the amount of current flowingthrough them), and bilateral (impedancesbeing independent of direction ofcurrent flowing). It can be shown that such a two-terminal pair network can be

representedby an equivalent T- or zr-network.Consider he unsymmetrical T-network of Fig. B-1, which is equivalenr o

the general two-terminal pair network.

0*, 0*,Ofi }fz

0r, 0r,

0*,af^0*,

)r

^2

0*n 0r, o*n

REFERECES

l, Shiplcy, R,8,, Intoduction to Matric:r:sand

r976.

2. Hadley,G., Linear Algebra,Addison-Wesley

3. Bellman,R., Introduction to Matrix Analysis,

1960.

Power Sy,rtems, iley, New

Pub.Co. Inc., Reading,Mass.,

McGraw-HillBook Co.. New

VJI UJ2

(A-30)

York,

1961.

York,

Flg. B-1

Fo r Fig. B-1,1 s -

or

I s=

Unsymmetrical-circuitquivalento a generalnair networkr - " " - - - ' - " _

the ollowing circuitequationsanbe writtenI^+ Y(Vo+ ^Zr)

Y V * + ( l + YZr ) I *

two-terminal

(B-1)

s lp

Vg

Z1 22

Y Vp



* : V* + I^Zr+ IrZ,

V^ + I^2, + ZrWn+ I^2,, + I^YZ.Z, .+iqY

*A transformer is similarly

terminals.

representedby a circuit with two input and two ouput

tlCtf I Modern orelsyslern inal,sis

or

V5= (1 + YZr)V* + (2, + Z" + yZrZr)Io

EquationsB-1)and(B-2)canbe simplified y lettingA- l + Y Z , B = 2 . , + Z r + y Z r Z ,

C = Y D = l + y Z ,

A D - B C = 1 (B-s)

Fig. B-2:#:::representation

of a two-rerminarai rnetworkusing

ABCD CONSTANTS FOR VARIOUS SIMPLE NETWORKS

We havealreadyobtained theABCD constants.ofan unsymmetricalT-network.The ABCD constants f unsymmetricala-network shown n Fig. B-3 may be-

(B-2)

(B-3)

using these,Eqs. (B-1) and (B-2)can be written in matrix form as

lyrl lA Bllv*1

Lr,.JLt olj^l ,t-0,This equation s the sameas Eq. (5.1) and s valid for any linear,passiveandbilateral wo-terminal pair network. The constantsA, B, C andD arecalled thegeneralizedcircuit constantsor the ABCD constantsof the network, and theycan be calculated for any such two-terminal pair network.

It may be noted that ABCD constantsof a two-terminal pair network arecomplex numbers in general,and always satisfy the following relationship

Also, for any symmetrical network the constantsA and, areequal. From Eq.(B-4) it is clear

thatA andD are dimensionless,while B has the dimensionsofimpedance ohms) and c has the dimensionsof admittance mhos).The ABCD constantsare extensivelyused in power system analysis.A

general wo-terminal pair network s often representeds in Fig. B_2.

Fig. B-3 Unsymmetricalr-circuit

A series mpedanceoften representsshort ransmission ines and transform-ers. The ABCD constants for such a circuit (as shown in Fig. B-4) canimmediately be determinedby inspection of Eqs. (B-1) and (B-2), as follows:

A = 1

B = Z

C = 0

D = l

Fig. B- 4 Series mpedance

Another simple circuit of Fig. B-5 consistingof simple shuntadmittancecanbe shown to possess he following ABCD constants

A = 7

B = 0

C = Y

D = I

/E l_a \\s-u,,

(B-7)



obtained n a similar mannerand are given below:

A = l + Y r Z

B = Z,C = Y r + Y r + Z Y r y z

D = I + Y r Z

(B-6)

Fig. B- 5 Shuntadmittance r

It may be noted that whenever ABCD constants re computed, t should bechecked that the relation AD-BC = 1 is satisfied.For examole, using Eq.(B-8) we get

A D - B C = 1 x 1 - 0 x Y = 1

:fflfuii'l Modern ower:System natysis

II ABCDconstants f a circuit aregiven, ts equivalent - or a-circuitcanbe determinedy solvingEq. (B-3) or (8-6) respectively,or the valuesofseriesandshunt ranches. or the equivalentn-circuitof Fig. B-3, we liuu.

(B-e)

, r =#

ABCD CONSTANTS OF NETWORKS IN SERIES AND FANEr,r,Ur,

Whenever a power system consists of series and parallel combinations ofnetworks, whoseABCD constantsare known, the overall ABCD constants orthe systemmay be determined o analyze the overall clperationof the system.

Fig. 8-6 Networksn series

Consider he two networks n series,as shown n Fig. 8-6. This combinationcan be reduced o a single equivalentnetwork as follows:For the first network. we hal'e

t ,

, , Append8B l , ;621+I

tr = (ArBr+ ArBr)/(8, Br )

B= BrBz l (Bt+Br )

C= (Cr * Cr) + (A r - Ar ) (Dz- D)l(Br * Ur )

(B-13)

Fig. B-7 Networksn parallel

Measurement of ABCD Gonstants

The generuIized circuit constants rnay be computed for a transmission linewhich is being designed rom a knowledge of the system mpedance/admittanceparametersusing expressionssuch as those clevelopedabove. If\the line isalready built, the generalizedcircuit constantscan be measuredby making afew ordinary testson the line. Using Eq. (B-4), theseconstantscan easily beshown o be ratiosof either voltageor currentat the sending-endo voltageorcurrent ai the receiving-endof the network with the receiving-end openor short-circuited. When the network is a transformer, generator, or circuit havinglumped parameters,voltageand current measurements t both ends of the linecan be made, and the phase angles betweenthe sending and receiving-endquantities can be found out. Thus rhe ABCD constantscan be determineC.

It is possible, also, to measure he rnagnitudesof the required voltagesandcurrents sinrultaneously at both ends of a transmission ine, but there is nosimplemethod to find the difference n phaseanglebetween he quantitiesat thetwo ends of the line. Phase ifference s necessary ecauseheABCD constantsare complex. By measuring wo impedancesat eachend of a transmission ine,however', he generalized ircuit constants an be computed.

[u,l_ o,",

'l[u,l

Lr, 1. , o, lt, )For the secondnetwork, we can write

I

y*

I IA, Brll vo ]

I t = t | | |l r r ) l_c , D r J l r * l

From Eqs. (B-10) an d (B-11), we can write

[u' l lo' u'11o, ' l [u^]-

(B -10 )

( B - 1 1 )



The following impedancesare to be measured:7 - o o - ' l . i - - o - . l i * ^ o ' l ^ - ^ ^ . . ' i + L - ^ ^ ^ : . , : - - ^ - l ^ - ^ - ^ : - ^ - . - - r

"so- rvrrulrS-vr:\r uuPv\r(rrrlvs wrlrl r9L9rvlttE-trlru uP(,il-ull'uulteo

Zs s= sending-endmpedancewith receiving-endhortcircuited

zno = receiving-endmpedancewith sending-end pen-circuiteci

Zns = receiving-endmpedancewith sending-end hort-circuitedThe impedancesmeasured rom the sending:end an be determined n terms

of the ABCD constantsas follows:

L+ 1., n,ll.c, or)j^

f A ,A, + B tC . , AtBz qD2l [y* I= ll crAz + D1C2 CrB' -t DrD, Jl t o l

If two networksare connectedn parallelas shown n Fig. B-7, the ABCDconstants f thecombinednetworkca-n e oundout similarly with somesimplemanipulationsof matrix algebra.The results

arepresented

below:

From Eq. @- ), with 1R= 0,

Z s o =V s l l s =A l C

and with VR= 0,

Vn = DVs + BIt

In= CVs+ AIt

From Eq. (8-16), with Is = 0,

Z n o = VR | I R =D /C

and when Vs = 0,

Zns= VRl lR= BIA

When the impedances remeasured rom the receiving-end, he direction of

current flow is reversedand hence he signs of all current terms n Eq. (5.25).

We can thereforerewrite this equation as

AppENDIX C

We know that the nodal maffix Yu.r, and its associatedJacobian are very

sparse,whereas their inverse matrices are full. For large power systems the'sparsity

of thesematricesmay be as high as 98Voand must be exploited.Apart

fiom reducing storageand time of computation,sparsityutilization limits theround-off computational errors. In fact, straight-foru'ard application cf the

iterativeprocedure or systemstudies ike load flow is not possible or large

systemsunless the sparsity of the Jacobian s dealt with effectivet).

GAUSS ELIMINATION

One of the recent techniquesof solving a set of linear algebraic equations,

called triangular factorizatiott, replaces the use of matrix inverse which is

highly inefficient for large sparsesystems. n triangularization the elements of

.u.h ,o* below the main diagonal are madezero and the diagonalelement of

eachrow is normalized as soon as the processingof that row is completed- It

,is possible to proceed

columnwise but it is computationally nefficient and is

theref,re no t used.After triangularization he solution s easily obtainedby

hack sub.stitution. he technique s illustrated n the examplebelow'

(B -14 )

(B-16)

(B -17 )

(B -18 )

Solving Eqs. B-14), B-15), (B-17) and (B-18) we can obtain he valuesof

the ABCD constantsn terms of the measured mpedancesas follows.

AD-BC

C _

[usingEq. (B-5)]AC

I C 1

A C , A A 2

AC

o=(ffi)'''By substitutinghis value of A in Eqs. (B-14),

value of C so obtained n Eq. (B-17), we get

( z\ 1l/2r > - . 7 , - '

I= L R S l z * o - z n r )

(Zss(Zno- Zo))' ' '

Z^o

(B - e)

(B-18) and substituting the

(B-20)

(B-21)



Consider the linear vector-maffix equation

p = (B-22)(zso(z^o zo))t' '

REFERECE

The Transmi,ssionnd Distribution of Electrical Energy,

New Delhi. 1970.

l'L3

1' l[ ' t I

,l1,,It;l. Cotton,H. and H, Barber,

3rd edn.,B.I. Publishers,

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

[ r + [ ' ' - l + ll z i l t = t o _ + ll o ? : r l l _- l , rL t ) L x z l L t J

Upon back substituting,hat is first solving or x2 and hen or 11,we get

o - + 1x 2 = - - 7 : - - j

2

- - I _ I Y - I _ 1 1 _ 3 \ - 5

Check

br + xz=2(*) - t r=

t3x,+ 5rz-3(+) s(f) = o

Thus,we havedemonstratedheuseof thebasicGauss limination ndback

substitution rocedureor a simplesystem, ut the sameprocedureppliesoany general ystem f linear algebraic quations,.e.

A x = b (c-1)

An added advantage f row processing elimination of row elementsbelow

the main diagonal and normalization of the self-element) is that it is easily

amenable o the useof low storage ompactstorageschemes-avoiding storage

of zero elements.

GAUSS ELIMINATION USING TABLE OF FACTORS

Where repeatedsolution of vector-matrixPq. (C-1j with cons ant Abut varying

values of vector D is required, t is computationallyadvantageous o split the

matrix A into triangular f'actor (termed as'Table

of factors' or'LU

decomposition') using the Gauss elimination technique. If the matrix A is

Consider the following systemof linear equations:

(Z)xt + Q)x2 + (3)x, = $

( 2 ) x r + ( 3 ) x z + ( 4 ) x t = )

( 3 ) x r ( 4 ) x r + ( 7 ) 4 = 1 4

(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

reducedsetsof equationsare as follows:

( l )xr+ ( t )*z+ (*) r t = *

(Z)xt+ (3)x2+ (4)xt = )

(3)xr+ (4)xz+ (7)x3= 14

(1 ) . r , ( i l r r + ( | ) x t -z

(2)xz+ (1)x, = l

( 3 ) x r + ( 4 ) x 2 + ( 7 ) 4 = 1 4

(1 )x ,+t r ) r r+ * ) " , -3

(1 )x2+* ) r , +

( 3 ) x r + ( 4 ) x 2 + Q ) x t = 1 4

( l )xr+ ( t ) r r+ (J)x, = 3

( l )x2+ *) " r ' = +,i

( t ) , r + ( * ) r , - s

(1)x, (t)rr+ (*)", = J

( t)xr+ t)* t - +

(*)rt = +

(c-3)

(c-4)

(c-s)

(c-6)

(c_'7\



sparse,so s the table of factors which can be compactly stored herebynot only

reducing core storage equirements,but also the computational effort. Gauss

elimination using the table of factors s illustrated in the following example.

(1)x, (tr)*r+ *)r,

(1)x, (I)*t (c-8)

x 3 =

These steps are referred to as 'elimination' operations.The solution r may

be immediately determinedby'back substitution' operationusing Eq. (C-8).

_ J

32

I

Thesolution or a new set of valuesor b canbe easilyobtained y usinga ableof factors repared y a carefulexaminationf eqs.1c-l) to (c-g).wecanwrite he ableof factorsF asbelow or the examplen hand.

fn

fzt

The row elements of F below the diagonal are the multipliersof thenormalizedrows required or the elimination of the row element,e.g. zz =

*,the multiplier of normalizedrow 2 l&q. (c-6)l to eliminate the element(3,2),i'e- (])x2. The diagonalelementsof F are he mulripliers needed o normalize

the rows after the row elimination has beencompleted,e.g. fzz=! , the actor

by which tow 2 of Eq. (C-4) must be multiplied to normalize the row. Theelementsof F above the diagonal can be immecliately written down byinspectionof Eq. (C-8). These are needed or the back substitutionprocess.

In rapidly solvingEq. (C-1) by useof the ableof factorsF, succesiive tepsappearas columns (left to right) in Table C.l below:

Table C. l

- 5 ( ;X;) : iand for heading3, 3

lt = fszlz

= +x+)In fact, operation(C-10) representsow normalizationand (C-11) represents

elimination and back substitution procedures.

Optimal Ordering

In power systemstudies, he matrix A is quite sparse o that thenumber of non-zero operations and non-zero storagerequired in Gauss elimination is verysensitive o the sequence n which the rows are processed.The row sequencethat leads o the east number of non-zerooperations s not, in general, he sameas the one which yields least storage requirement. It is believed that theabsoluteoptimum sequence f ordering he rows of a arge network matrix (this

is equivalent o renumbering of buses) s too complicatedand time consumingto be of any practical value.Therefore,somesimple yet effectiveschemeshave

been evolved to achieve near optimal ordering with respect o both the criteria.Some of the schemesof near optimal ordering the sparsematrices,which arefully symmetrical or at least symmetric in the pattern of non-zero off-diagonalterms, are describedbelow [4].

Scheme I

Number the matrix rows in the order of the fewest non-zero erms n each row.If more than one unnumbered row has the same number of non-zero terms.number these n anv order.

Scheme 2

Number the rows in the order of the fewest non-zero erms n a row at each step

of elimination. This scheme equires updating the count of non-zero erms aftereach step.

Scheme 3

Number the rows in order of the fewestnon-zerooff-diagonal erms generatedin the renraining ows at eachstepof elimination.This scheme lso nvolves an

ft , ft z

fu fzz

32

r l 3' ) . r a

L

n l lL T T (c-e)

1 , 35/2

I

I

x

I

I

1

The heading ow (i, ) of Table C-1 representshe successive liminationandback substitution teps.Thus,

representsnormalization of row 1represents limination of element 2, l)

h69t 4

l , I39

T4

2 , 1 2 , 2 3 , 3 , 23 3 3 33 3t2 3/2 3t2

14 t4 5 5 t4

1 , 2a

33t )

1

2 , 3aJ

31

7 , 72 , L2,2 represents onnalizationof row 23, L; 3,2 represent limination of elements 3, 1) an,J 3,2) respectively3. 3 represents ormalizationof row 32,3 represents limination of element 2,3)

by back substitutionr, 3; 7, 2 reprelenteliminationof elements 1, 3) and (r, z) respectively yback substitution.

The solution vector at any stageof development s denotedby

Ut lz y3Jr= y

The modificationof solution vector fiom column to column (left to right) is

l t = l t - f n ) z



updatingprocedure.

I he chotceot scheme s a trade-otf betweenspeedof executionand thenumber of times the result is to be used.For Newton's method of load flowsolution, scheme 2 seems o be the best. The efficiency of scheme3 is notsufficiently established o offset the increased ime required or its execution.

carriedout for the headin (i , j) as per the operationsdefined below:

!i = f i i j i t t j = i (c-10)

(c-11)i= !r - f i / i lt i + i

Thus for heading3, z

Mffi# Modernot -

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 argely upon conserving

computer storage ucmg the nu oI non-zerocomputatlons.'l'o ettect

these deas on the computer,elimination of lower triangle elements s carried

out a row at a time using the concept of compact working row. The non-zeromodified upper triangle elements and mismatches are stored in a compact and

convenientway. Back substitutionprogresses ackwards hrough the compact

upper triangle table. A properly programmed compact storage scheme results n

considerablesaving of computer time during matrix operations.

Naturally, there are as many compact working rows and upper triangle

storageschemesas there are programmers.One possiblescheme or a general

matrix stores the non-zero elements of successive ows 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

schemes re given in [2].

REFERECES

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.

Tinney,W.F. and J.W. Walker, "Direct Solutionsof SparseNetwork Equationsby

Optintally OrderedTriangular Factorizations", Proc. IEEE, Nov. 1967,.55: 1801.

Tinney,W.F. and C.E. Hart, "Power Flow Solution by Newton's Method", IEEE

Trans.,Nov. 1967,No. l l , PAS-86: 1449.

AppBNDrx I)

Expressionso be used n evaluatinghe elements f theJacobianmatrix of apowersystem rederivedbelow:

From Eq. (6.25b)

fr*rrk: r

A \exp (- i6,)L lY,/ exp (i?il lVll exp (7dn) (D-l)

k:r

Differentiating partially with respect o 6* (m * i)

+- i+ =Tvil exp -r4) (Yi^l expQ0,^) v^texpj5^))06^

-a6^

P, jiQ,= ',

l . = lvil

3 .

4.-- j(ei - jf) (a^ + jb^) (D-2)

where

Y,^= G,* + jB,^

V i= € i+ j f i '

(a^ * jb*) = (c* * jBi) @^+ jf^)

Although the polar form of the NR method s beingused, ectangularcomplexarithmetics employedor numerical valuation s t is faster.



From Eq. (D-2),we can write

# = ( a J i - b ^ e ) = H i ^

#=- (aêi+ , f , )= i^

t v rOt lc I t rOWgf

For the case of m = i. we have

* -i* = - jtvitexp- j6,)i ty,ni xp j4) tV1,txpQfi)06 , " 06 , k: l

+ jlV,l exp (- j6 ) (tytil expQ?,,)tV;t xp (id))

l r 'q - " "

I f . l r . $

= l%lexp - j6,)ilY,pl exp i0*) lvplexp 7d')k:l

+ lV, lz lY,, l xp 7d,,)

= \F t - lQ ) + i t * (D-s) - j(Pi - jQ) + jlV,l'(G,, + jB,,)

From Eq. (D.3), we can write

(D-3)

#= - Qli- Bi iv;12 H"

ao.; t

= P i- G' i t v i l2 J '

Now differentiateEq. (D-1) partially with respect o lV^l (m r i). We have

a4 0Q,-- = lvil exp(- j6) (lYi*l exp j0,^) exp Qd^))a l v ) " a l v ^ l

' L

Multiplyingby lV^l on bothsides,

aPt I AO.a t v ^ t v ^ t - i f f i w ^ t

- lvil exp(- j6) lY,^l exp j1i)'tV*l exp (i6^)

= (ei- jf,) (a^ + jb*) (D-4)

It follows from Eq. (D-4) that

! ! r r * t= aê,+ , f i= Niâ v^l

-!,?:,tv^l= a,,fi bê, L,â l v . l

Now for the caseof m = i. we have

aP t - j-P.- ex p jilf tyat xp j0*) ve texpjQ,)a l v i l " 7 l v i l , . .

o : ,

+ lV,lexp (- j6) lyi exp (/4)

It follows romEq. (D-5) hat

oPi-

' av,l\v'l = Pt * GiilViP = Nii

, e t _ , = L t

avillvil = Qi - Biilvil

The above results are'summattzedbelow:

Case 7

Case 2

m * i

Hi^= Li*= aJ, - b*e,

N i *=- J i *= d*€ i+ b"f t

Yi*= Gt^ + jBt*

V i= e i + j f i

(a* * jb^) = (Gi^ + jBi^) @* + jk)

m =

H,=- Q i - B i i l v i l z

Ni i= P,+ G,, lV, lz

Ji i - Pi- Gii lvr

L i i= Qr - B i i l v i l z

(D-6)

(D-7)

(D-8)

REFERECES

Tinney, W.F. and C.E. Hart "PowerFlow Solutionby Newton'sMethod", IEEE



Q0,,) xp

Multiplyingby lV,l on bothsides

an Ao'- tl v , l - i " " ,

l v , la lv i l ' " aw i l

Trans.,Nov 1967,No. 11, PAS-86:1449.

Van Ness, J.E., "Iteration Methods or Digital Load Flow Studies", Trans.AIEE,

Au g 1959,78A: 583.

AppBNDrx E

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)

subject to equality constraints

g ( x , u , P ) = o

and to the inequality constraints

u - u ^ u 1 0

u^1- - u < 0

The Kuhn-Tuckerheorem] gives he necessaryonditionsor the minimum,

assuming onvexity or the unctions E-1)-(E-4), s

A.C = 0 (gradientwith respect o u, x, )) (E-5)

where .C is the Lagrangian formed as

-C=f (x , u) + )Tg(x ,u ,p )+ o f *u* { , -uû* )+ oT* ,n1r - i n -z )

and (E-6)

(E-7)

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

AL0x Ax

\ * o - 0 (E-e)

In Eq. (E-9),

oL = 0f +rq)'d u 0 u ' \ a u l

It is evident that a computed from Eq. (E-9) at any feasible solution, with

) from Eq. (E-8) is identical with negativegradient, .e.

di= Qi.û* tf Ui- ui ,*o ) 0

ai= - Ai,-io if ,r, ,t,- ui ) 0

a L , .# = g ( x , u , p ) = oOA

d= -

K=

negativeof gradientwith respect o u

if ur, ,n;o< ui < ui, ^

t f u , = u i , ^ *

tf u, - ili, în

(E-10)

(E-11)

(E-12)

(E- l )

(E-2)

(E-3)

(E-4)

At the optimum,

state that

a must also satisfy the exclusion equations(E-7), which

, \

ff ,i, ,rt, < ui < ui, ,n"*

lf u, = ui, ,iru*

Q i = - Q , r n i n S 0 l f u i = 4 i , *i n

which can be rewritten in ternts of the gradientusing Eq. (E-11) as follows:

o x= o0u,

o f , . o0u,

of , roout

d i = 0

di = di , -* 2 0



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 constraintscase.

REFERECE

1. Kuhn, H.W. and A.W. Tucker, "Nonlinear Programming", Proceedings of the

Second.BerkeleySymposiumon Mathematical Statisticsand Probability, Univer-

sity of CaliforniaPress,Berkeley,1951.

AppnNDrx F

In developedcountries he ocus s shifting in the power sector rom the creationof additional capacity to better capacity utilization through more effectivemanagement and efficient technology. This applies equally to developingcountrieswhere this focus will result n reduction n need or capacityaddition.

Immediate and near future priorities now are better plant management,higher availability, improved load management, educed transmission osses,revamps of distribution system, improved billing and collection, energyefficiency, energy audit and energy management.All this would enable anelectric power system o generate,ransmit and distributeelectric energyat thelowest possibleeconomicand ecologicalcost.

These objectives can only be met by use of information technology (IT)enabled services in power systems management and control. Emphasis istherefore, being laid on computer control and information transmission andexchange.

The operations nvolved in power systems equire geographicallydispersedand functionally complex monitoring and control system. The monitory andsupervisorycontrol that is constantlydevelopingand undergoing mprovementin it s control capabil i ty s schematical lyresentedn Fig. F.1 which is easi lyseen o be distributed in nature.

Starting from the top, control system unctions

EMS Energy ManagementSystem It exercisesoverall control over thetotal system.

scADA Supervisory control and Data Acquisition system -

Fig. F.l Real imemonitoring nd controllingf an electricpowersystem.

SCADA refers to a system that enablesan electricity utitity ro remotelymonitor, coordinate,control and operate ransmissionand distribution compo-nents,equipmentand devices n a real-time mode from a remote ocation withacquisitionof data or analysisand planning rom one con[ol location.Thus,the purposeof SCADA is to allow operators o observeand control the powersystem. The specific tasksof SCADA are:

o Data acquisition,which providesmeasurementsnd status nformation o

operators.. Trending plots and measurementson selected ime scales.. Supervisorycontrol, which enablesoperators o remotely co:rtrcl devices

such as circuit breakersand relays.Capabilityof SCADA system s to allow operatorso controlcircuitbreakers

ancldisconnect witches nd change ransformerapsand phase-shifter ositionremotely. It also allows operators o monitor he generationand high-voltagetransmission systemsand to take action to ccrrect overloads or out-of-limitvoltages. It monitors all status points such as switchgear position (open orclosed), substation oads and voltages, capacitorbanks, tie-line flows andinterchangeschedules. t detects through telemetry the failures and errors inbilateral communication links between the digital computer and the remote

equipment.The mostcritical functions,mentionedabove,are scanned very fewseconds.Other noncritical operations,such as the recording of the load,foiecastingof load,unit start-upsand shut-downs re carriedout on an hourlvbasis.

Most low-priority programs(thoserun less requently) may be executedondemandby the operator or study purposesor to initialize thepower system.An



It coversgenerationanci ransmission ystem.

DAC Distribution Automation and Control System - It oversees hedistributionsystem ncludingconnectedoads.

Automation, monitoring and real-time control have alwaysbeen apart of scADA system. with enhancedemphasis on IT in powersystems,scADA has beenreceiving a lot of attention lately.

operator may also change he digital computercode in the execution f aparameterchanges n the system. For example, he MWmin capability of ageneratingunit may change f one of its throttlevalues s temporarily emovedfor maintenance,so the unit's share of regulating power must accordingly bedecreasedby the code.The computer softwarecompilers and datahandlersaredesigned o be versatile and readily acceptoperator nputs.

DA C is a lower level versionof SCADA applicablen distribut ion ystem(including loads), which of course draws power from the transmission/subtransmissionevels. Obviously then there is no clear cut demarcationbetweenDAC and SCADA.

In a distributionnetwork, computerisationcan help manage oad, maintainquality, etectheftand ampering nd hus educe ystemosses. ornputeri-sation alsohelps n centralisationof data collection. At a central oad dispatch'centre, data such as culTent, voltage, power factor and breaker status are

telemetered nd displayed.This gives the operatoran overall view of the entiredistribution network. This enableshim to have effective control on the entirenetwork and issue instructions for optimising flow in the event of feederoverload or voltage deviation. This is carried out through switching inlout ofshunt capacitors,synchronouscondensersand load management.This wouldhelp in achievingbetter voltage profile, loss reduction, mproved reliability,quick detectionof fault and restorationof service.

At a systems level, SCADA can provide status and measurements ordistribution eedersat the substation.Dictributionautomationequipmentcanmonitor selectionalising evices ike switches,ntemrptersand fuses. t can alsooperate 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 ofdistribution control centres.

SCADA can be used extensively fbr compilation of extensive data andmanagementof distribution systems.Pilferage points too can be zeroed n on,as the flow of power can be closely scrutinised.Here again, trippings due tohuman effors can be avoided.Modern meteringsystemsusing electronicmeters,automatic meter readers(AMRs), remote meteripg and spot billing can go along way in helping electric utility. These systbmscan bring in additionalrevenuesand also reduce he time lag betweenbilling and collection.

Distribution automation hrough SCADA systemsdirectly leads o increasedreliability of power or consumersand ower operatingcosts or the utility. Itresults n fbrecasting ccurartelenrand nd supplymanagement,aster estora-tion of power in case of a tailure and ahernative routing of power in anemergency,

A kt 'y 'c i l turcll l ' thcst rystcl l lss hc rur lo lc l ontnr l i rc i l i ty l r l t l l lows nstcrexecutionof decisions.Manual errorsand oversightsare eliminated.Besidesonline and cal-time nlbnnution, he systenr rovidesperiodic eports hat help nthe analysisof performanceof the power system. Distributi'onautomationcombinesdistributionnetwork monitoring functionswith geographicalmapping,

IAn energycontrolcentremanageshese asksandprovides ptimaloperationof the system.A typicalcontrolcentre anperform he following functions:

(i ) Short,medium nd ong-termoad orecastingLF)(ii) System lanning SP)

(iii) Unit commitmentUC) and maintenancechedulingMS)

(v) Stateestimation(SE)

(vi) Economic dispatch(ED)

(vii) Load frequencycontrol (LFC)

The above monitoring and control functions are performed in the hierarchicalorder classified according o time scales.The functions perfotmed in the controlcentre are based on the availability of a large information base and requireextensive software for data acquisition and processing.

At the generation level, the philosophy of 'distributedconffol' has

dramatically reduced he cabling cost within a plant and has the potential ofreplacing traditional control rooms with distributedCRT/keyboard stations.

Data acquisition systems provide a supporting role to the applicationsoftware in a control centre.The data acquisition system(DAS) collects rawdata from selectedpoints in the power system and converts these data intoengineering units. The data are checked for limit violations and statuschangesand are sent to the data base for processing by the application software. Thereal-time data baseprovidesstructured nformationso that applicationprogramsneeding the information have direct and efficient access o it. ,

The Man-Machine interface provides a link between the operator and thesoftware/lrardwareused to control/monitor the power system. The interfacegenerally is a colour graphic display system.The control processors nterfacewith the control interfaceof the display system.The DAS and Man-Machineinterface support the following functions:

(i) Load/GenerationDispatching(ii) Display and CRT control

( i i i ) Dat t BaseMaintcnancc( iv) AlarrnHandl ing

(v ) Supervisory ontrol(v i I l ) rograrnnr ingbnct ions

(vi i) DaLa ogging

(vi i i ) Event ogging

(ix) Real-time Network Analysis

With the introduction of higher size generating units, the monitoring



f ^ . - f e l ^ ^ - r . l ^ - ^ - - l r 1 t | ! t ' , t. '

rilur! luL;aLtuil, ailu su un, 1.() tmprove availaDtlty. tr aISo lntegrates load

management,oad despatchand ntelligent metering.

Data Acquisition Systems and Man-Machine Interface

The use of computers nowadays encompasses ll phases of power systemoperation:planning, orecasting,scheduling,security assessment, nd control.

r r t z r t r i r a r n a nf < ' h o r r a f r n i a r r h i - ^ o . r r o - * l ^ ^ + . ^ l - ^ I - ^ - l ^ - a ^ : - . - - ^ - . ^ r L - - r - - ^rvyurrv ruv rr ro rrqvw 6vuw up ur y\ryvvr pr4rrLD (rrD(J . I l l u lut t t tu l r r rp l 'uvc ule Ixant

performance, now all the utilities have installed DAS in their generating unitsof sizes 200 MW and above.The DAS in a thermal power stationcollects thefollowing- inputs from various locations in the plant and converts thern intoengineering'units.

l,@E+l ModernPo@isI

Analog Inputs

(i) Pressures,lows, electrical parameters,etc.(ii) Analog input of 0-10 V DC

(iii) Thermocouplenputs(iv) RTD input

Digital Inpurs:

(i) Contractoutputs(ii) Valve position, pressureand limit switches

All these process nputs are brought from the field through cables to theterminals. The computer processes he information and ,uppli", to the Man-Machine interface o perform the following functions.

(i) Display on CRT screen(ii) Graphicdisptay of plant sub-systems

(iii) Data ogging(iv) Alarm generation(v) Event ogging

(vi) Trendingof analoguevariables(vii) Performance alculation

(viii) Generationof controlsignals

Some of the above unctions are briefly discussedas follows.

The DAS softwarecontainsprogramso calculateperiodically the efficiencyof various equipment ike boiler, turbine, generator,condenser, ans, heaters,

etc.

| -^^a a D J t ,

"

power systemengineerswho are adopting the ow-cost and relatively powerfulcomputing devices n implementing their distributedDAS and control systems.

Computer control brings in powerful algorithms with the following advan-

and so in raw materials

and modifiability, (iv)effectiveness.

paurly ururzauon rn generailon, (u) savings n energydue to increasedoperationalefficiency, (iii) flexibiliiy

reduction in human drudgery, (v) improved operator

Intelligent database rocessorswill becomemore cornmon n power systemssince the search, retrieval and updating activity can be speededup. Newfunctional concepts from the field of Artificial Intelligence (AI) will beintegrated with power system monitoring, automatic restoration of powernetworks, and real-time control.

Personal omputers'(PCs) re being used n a wide rangeof power systemoperationsncludingpower station ontrol, oaclmanagement, CAOe systems,protection, operator training, maintenance functions, administrative dataprocessing,generatorexcitationcontrol and control of distributionnetworks. ITenabledsystems hus not only monitor and control the grid, but also improveoperational efficiencies and play a key part in maintaining the security of

thepower system.

RFFERECES

I. Power Line Maga7ine,yol .7, No . l, October2002,pp 65_71.2. A.K' Mahalanabis, .P. Kothari and S.I. Ahson, ComputerAided power System

Analysis and Control, TMH, New Delhi, 199g.3. IEEE Tutorial course, Fundamentalsof supervisory control system, l9gr.4. IEEE Tutorial course, Energy control centre Design, 19g3.



AppBNDrx G

MATLAB has been developed by MathWorks Inc. It is a powerful softwarepackage used for high performance scientific numerical computation, dataanalysisand visualization.MATLAB stands or MATrix LABoratory. Thecombination of analysis capabilities, flexibility, reliability and powerfulgraphics makes MATLAB the main software package for power systemengineers.This is because nlike otherprogramming anguageswhereyou haveto declare natricesand operateon them with their indices, MATLAB providesmatrix as one of the basic elements. t providesbasic operations,as we will seelater,like addition, subtraction,multiplication by use of simple mathematicaloperators. Also, we need not declare the type and size of any variable in

advance. t is dynamicallydecideddependingon what value we assign o it. But

MATLAB is case sensitiveand so we have to be careful about the case ofvariables while using them in our prograrns.

MATLAB givesan interact ive nvironmentwith hundreds f rel iablean daccuratebuilrin functions.These unctionshelp in providing the solutions o avariety of mathematical problems including matrix algebra, inear systems,

differentialequations, ptimization,non-linear ystems nd manyother ypesofscientific and technicalcomputations.The most mportant featureof MATLABis its programming capability, which supportsboth types of programming-

object orientedand structuredprogrammingand s very easy o learn and useand allows user developed unctions. It facilitates access o FORTRAN and Ccodesby meansof external nterfaces.There are severaloptional toolboxes or

F;;ff$:

You can startMATLAB by doubleclicking on MATLAB icon on your Desktopof your computeror by clicking on StartMenu followed by 'programs'

and thenciicking appropriateprogram group such as 'MATLAB

Release12,. you willvisualizea screenshown n Fis. G.1.

F lg.G. l

The command prompt (characterisedby the symbol >>)'is the one afterwhich you type the commands.The main menu contains he submenussuch asEile, 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 thedesiredM'file. This will open up a MATLAB File Editor/Debugger windowwhere you can enter your program and save t for later use.You can run thisprogram by typing its name n front of command prompt. Let us now learn somebasiccommands.

Matrix Initialization

A matrix can be initialized by typing its name followed by = sign and anopening squarebracket after which the user supplies he valuesand closes thesquare brackets. Each element is separated rom the other by one or morespaces r tabs.Each row of matrix is separatedrom the other by pressingEnterkey at the end of each row or by giving semicolon at its

"na.potiowing

examples llustrate his.

Click on this to changethe cunent directory

Commandrompt

Simulink browser



simulating specializeel roblems of rJifferentareas anrder-tensionso link upNIATLAB and other programs. SIMULINK is a program build on top ofMATLAB environment,which along with its specializedproducts,enhances hepower of MATLAB for scientificsimulationsand visualizations.

For a detailed description of commands,capabilities,MATLAB functionsand many other useful features, the reader is referred to MATLAB lJser's

Guide/lvlanual.

7 2 -9

- 3 2 -5 1

The above operation canalso be achievedby typing

64?,..1 ModernPo@sist

If we do not give a semicolon at the end of closing square brackets,

MATLAB displays he value of matrices n the command winCow. If you do

not want MATLAB to disptay the results, ust type semicolon(;) at the end

of the statement. That is why you rvill find that in our programmes,whenever

we want o display valueof variables ayvoltages,we have ust typed he nameof the variable without semicolon at the end, so that user will see he values

during the program.After the program s run with no values of the variables

being displayed, f the user wants to see he valuesof any of thesevariables nthe program he can simply type the variable narne and see he values.

Common Matrix Operations

First let us declaresome matrices

Addition

Adds matricesA and B and stores hem n matrix C

Subtraction

Subtractsmatrix B from matrix A and stores he result.in D

Multiplication

Multiplies two conformablematricesA and B and stores he result n E

Inverse

This calculates he inverse of matrix A by calculating the co-factorsand stores

the result n F.

Transpose

Single quote ( /) operator s used to obtain transposeof matrix. In case the

matrix elernent s complex, it stores he conjugateof the element while takingthe transpose,e.g.

t em*,

This storesdeterminantof matrix A in H.

VaIues

obtains igen alues f rnatrixA andstores lrem n K.

G.2 SPECIAL MATRICES AND PRE.DEFINED VARIABLES

AND SOME USEFUL OPERATORS

MATLAB has some preinitialised variables. These variablescan directly

used n programs.

pi

be

This gives the value of n

convertsdegrees d rnto radians and stores t in variable r.

inf

You can specify a variable to have value as oo.

iandj

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.

The abovestatement defines complex power S.

A word of caution here s that if we use i and variablesas loop counters

then we cannot use them for defining complex numbers.Hence you may find

that n someof the programswe have used I and.i1 as oop variables nstead

of i and 7.

eps



or

> > G t = [ 1 3 ; 4 5 ; 6 3 ] '

also stores he transposeof matrix given in squarebrackets n 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. i

I't

tI

This variabie is preinitiiized to 2-s2.

Identity Matrix

To generatean identity matrix and store it in variabte K give the following

command

A= [1 5

? 65 4

#ffif Moderno*erSystem nalysis

So K after this becomes

K = [ 1 0 0

0 1 0

0 0 1 l

Zeros Matrix

generates 3 x 2 matrixwhoseall elements re zeroand, tores hem n L.

Ones Matrix

generates 3 x 2 matrixwhoseall elements reoneand stores hem n M.

: (Colon) operator

This is an important operator which can extract a submatrix from a givenmatrix. Matrix A is given as below

7 8

9 1 03 1

e 3 r 2 l

This commandextractsall columns of matrix A corresponding o row Z andstores them n B.S o B b e c o m e s 1 2 6 9 l , } l t

Now try this command.

The abovecofirmandextractsall the rows corresponding o column 3 of matrixA and stores hem in matrix C.So

C becomesl 79

3

1 l

Now try this command

[-ait+

" (.. oPERATOR)

This operation unlike complete matrix multiplication, multiplies element of one

matdx WitlreOnrespond'rng;iementof other matrixlaving same ndex. However

in latter caseboth the matricesmust have equal dimensions.

We have used this operator n calculatingcomplex 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 obtained s [- 0.0774+ 0.385Li 0.7404+ 0.4852t 0.6106+ 0.0198i]

Note here, hat f the result is complex MATLAB automaticallyassigns in the

resultwithout 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 elementsof a given matrix A.

min

This function can be used n two forms

(a) For comparing two scalarquantities

if either a or b is complex number, then its absolutevalue is taken for

comparison(b) For finding minimum amongsta matrix or ilray

e.g. f A = [6 -3 ;2 -5]

>> min (A) results in - 5

abs

If applied to'a scalar, it gives the absolute positive value of that element



This command extracts a submatrix such that it contains all the elementscorresponding o row number 2 to 4 and column number 1 to 3.

S o D = [ 2 6 9

5 4 3

9 3 1 l

(magnitude).For example,

> > x = 3 + j * 4

abs(x)gives 5

If applied to a matrix, it results n a matrix storing absolutevalues of all the

elementSof a given matrix.

G,5 CONTROL STRUCTURES

IF Statement

The generalrom of the F statementsI F e x p r e s s i o n

s t a t e m e n t s

E L S Ee x p r e s s i o ns t a t e m e n t sE L S E I F

s ta temen tsEND

Expression s a logical expressionesul t ing n an answer, t rue, ( l ) orfalse'(0).Th e logicalexpressionan consistof

(i ) an expression ontaining elationaloperators abulatedalong with theirmean ingsn Tab leG. l .

Table G. 1

Relational Operator Meaning

Greater than

Greater than or Equal toLess than

Less than or Equal toEqual to

No t equal o

(i i ) or many ogicalexpressionsombincdwith Logicaloperators.Variouslogical operatorsand their meanings are given in Table G.2.

Table G. 2

Logical Operator Meaning

ANDOR

NOT

FOR Loops

This repeatsa block of statements recletermined umberof times.

&

t - ' _ _I 041/

comes o an end when k reaches r exceeds he final valuec. For example,f o r i - 1 : 1 : 1 0 ,

a ( i ) - 1

e n d

This initializes every elementof a to 1. If increment s of 1,as n this case, henthe incrementpart may as well be omitted and the above oop could be writtenAS

f o r - 1 : 1 0 ,

a ( i ) = 1

e n d

While Loop

This loop repeatsa block of statementsill the condition given n the oop is truew h i l e e x p r e s s i o n

s ta tements

e n d

For example,

j - 1w h i l e i < = 1 0

a ( i ) = 1

i = i + 1 ;

e n d'

This loop makes irst ten elementsof array a equal to l.

break statement

This staternentllowsone o exit prerr'aturelyrom a for or while loop.

G.6 HOW TO RUN THE PROGRAMS GIVEN IN THISAPPENDIX?

1. Copy theseprograms nto the work subfolderunderMATLAB foider.

2. Just ype the nameof the programwithout'.m' extension nd theprogramwil l run.

3. If you wanL o copy them n some other older say c:\power, hen aftercopying hose iles in c:\power. hange he work folder o e \power. ou



The most common{&m of FOR loop used sf o r k = a : b t c ,

s ta tements

end

where k is the loop variable which is initialised to value of initial variable a.If the final value (i'e. c) is no t reached,he statementsn the body for the loop

cando this by clicking on toolbarcontaining hreeperiods .. which is onthe right side to the Current Directory on the top rilht corner.

4. You can seeor edit theseprogramsby going throughFite - open menuand opening the appropriate ile. However do not save hose programs,unless you are sure that you want the changesyou have made to these

orisinal files.

ffiffi| Modern ower ystemAnalysis

5. You canseewhich re hevariableslready efined y ypingwhos nfront 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 softwarepackagedevelopedby MathWorks Inc. which is one

of the most widely usedsoftware n academiaand industry or modeling and

simulating dynamical systems.t can be used or modeling inear and nonlinear

systems,either n continuous ime frame or sampled ime frame or even a hybrid

of the two. It provides a very easy drag-drop type Graphical user nterface to

build the models in block diagram form. It has many built-in block-library

componentsha t you canuse o modelcomplex systems.f these uilt-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 simulatinga system.You can try

to construct the models given in the examples.

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 thatyou see n 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 nput,

integrate it and observe he output. The stepsare outlined as below.

1. Click on the Sources n the SIMULINK library browser window.

2. You are able to see various sources hat SIMULINK provides. Scroll

Fig. G. 2

Similarly click on continuous library icon. You can now see various

built-in blocks such as derivative, integrator, transfer-function, state-spaceetc. Select ntegratorblock and drag-drop t in your model window.

Now click on sinks and drag-dropscopeblock into your model. This isone of the most common blocks used for displaying the values of theblocks.

7. Now join output of sine-wave source o input of Integratorbloc(. This can'be done n two ways. Either you click the eft button and drag mouse romoutput of sine-wave source to input of Integrator block and leave left

button or otherwise click on right button and drag the mouse to formconnectionfrom input of integrator block to output of sine,wavesource.

8. Now in the main menu, click on Simulation and click Start. Thesimulation runs and stops after the time specifiedby giving readyprompt

at bottom left'corner.

9. Now double-click on scope to see he output. s something wrong? The

f l Hi U ccrlrnni".Sloircrctc

'. S furtmsaroUos

Unu, H nmr'oer

i .fi sgn*aryrtan

i H s * ':. fl sar:er

f,l crrtrd y*enooDox

Dbcret

Fqdin t l$hr

Ma['

Nmlle

Sign* t SJ,s{emt

Si*r

1

:l

5 .

6 .



down and you will see a Sine Wave sources con.

3. Click on this sources con and without releasing he mouse button drag

and drop it in your model window which is currently named asuntitled'.

4. If you double-click on this source, you will be able to see Block

parametersor sine wave which includes amplitude, requency,phase,etc.

Let not change heseparametersight now. So click on cancel to go back.

result is a sine wave of magnitude 2. Is there something wrong withSIMULINK 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 specifyany initial condition for output of the ntegrator,simulink assumest to be0 and calculates he constant. So it calculates

ff i Modern o

- cos 0+ C = 0 at r = 0 givinl C = 1. So he equationor output ecomes- cos 0 + l. Thusnaturally,t starts rom 0 at t = 0 and eachests peakvalue f 2 at 0= n, .e .3.I4.

10. To rectifv this error. double click on Inte block and in the ini tial

conditionsnter 1 which hould e heoutput f theblockat = 0. Nowrun the simulationagainand see or yourself hat he result s correct.

Some Commonly used Blocks

1. Integrator We have alreadydescribed ts use n the aboveexample.

2. Transfer unction Using this block you can simulate a transfer unction

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 denominatorof the expression n ascendingorder of s which are o be

enclosed nside squarebrackets and separatedby.a space.

3. Sum You can ind this block underMath in Simulink block. By default,

it has wo inputs with both plus signs.You can modify it to have equired

number of inputs to be surnmed up by specifying a string of + or- dependingupon the nputs. So f there are3 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 negativefeedback.

4. Gain This block is also ound underMath in Simdlink block. t 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 rrput being numbered 1. When the nput number

2 equals or exceeds he thresholdvalue specified n 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 availableunderSignals and systems

block in Simulink. The Mux block combines ts inputs nto a singleoutput

and s mostly used to form a vector out of input scalarquantities.Demux

block does the reverse hing. It splits the vector quantity into multiple

scalaroutputs.

7. Scope We have already described ts use. However; if you are plotting

a large number of points, click on properties toolbar and select Data

History tab. Then uncheck he Limit datapoints box so that all points are

,*rr-rr r^tin Sources lockunderSimulink.We havemade seof thisblock nstability studies o provide constant mechanical nput.

generatea sine wave of any amplitude, requency and phase.

The reader is encouraged to work out, the examples given in thisAppendix to gain greater nsight into the software.

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 doesnot returnany values as output parameters. It is just a set of Matlab statementswhich is stored n a file so that one canexecute his set by just typing the

file name n front of the command prompt,eg. prograrnmes n G2 to Gl8are script files.

(ii) Function m-files - This file accepts nput agrumentsand return values asoutput parameters. It can work with variables which belong to theworkspace 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 structureof function m-file is given below:

(a) Function definition line - This is the first line of a function. It specifiesfunction name, number and order of input variables.

Its syntax is- function [output variables]= function-name (list of inputvariables)

(b) First line of help - Whenever help is requestedon this funciton or lookfor is executedMATLAB displays this line.

Itssyntax is -

Vounction-name

help(c) Help text - Whenever help is requestedon the function-name help textis displayed by Matlab in addition to first help line.

Its syntax is - Vo unction-name (input variables)

(d) Body of the function - This consistsof codesacting on the set of inputvariables to produce he output variables.

Tlcc r nqn fhrrs r lcwelnn hic /hc r nrr rn r t rnt r rqrnc end f i ' rnnf innc qnA qAA fham fn fhaLarvrII uv ulv



plotted. Also when the rvaveform doesnot appear smooth, n the general

tab of properties oolbar, selectsample ime instead of decimation n the

sample time box and enter a suitable value like le-3 (10-3).The scope

then uses he value at sample-time nterval to plot.

8. Clock This block is used to supply time as a source input and is

available n Sources block under Simulink.

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 usingMATLAB/SIMULINK to encourage the reader o solve more power system

problems using MATLAB.

ffil uodern owerSystefnnalysis

Ex G.I

% This func t ionYoTh eargument

% This func t ion

c onv e r t s po1 rt o th i s f u n c t i o n

ha s been used

to rec tangu l a rc oo rd i na tesi s 1 ) Magn i t ude ? )Ang l e s . in deg rees

append i

f unc t i on rec t=po l a rT o rec ta ,b )

rec t=a* c osb* p i / 1S0 )+ j * a * s i b * p i/ l g0 ) ;

function s used n solutionof many of the examples olved ater.

Ex G.2 (Example S.O)

% Thls i l l us t ra tes the F e r ran t i e f fec t% I t s i mu l a tes he e f f ec t by v a ry i ng the l eng th o f l i ne f rom% zero(rece iv ingend) to 5000kmn steps of 10 km% an dp l o t s the s end i ngend v o l t age phas o r% This c o r res pondso F i g . 5 .13 an d da ta f rom Example .6cl ea r

VR=220e3/sqr t3) ;a1 h a = 0 . 1 6 3 e - 3 ;

be ta=1068e -3 ;

I =5000 ;k = 1 ;

f o r i = 0 : 1 0 : 1 ,

V S = ( V R / z ) * e x p( a ' l p h a * i ) * e x p( j * b e t a * j ) + ( V R /Z ) * e x p ( - a l p h a *i ) * e x p ( j * b e t a* j ) ;

X (k ) rea l VS);Y ( k ) = i m a g ( V S ) ;

' k =k +1 ;

end

p 1 t ( X , )

Ex. G.3 (Example 8.7)

% This P rog raml l us t ra tes th e use of d i f f e ren t l i ne mode l s s% i n Examp' le.7c l e a rf=50

| =300

7=4Q+j * lZ5

Y= e -3

PR=50e6/3

YR=?20e3/sqrt (3 ) )p f1oad=0 .8

' Appendix ffiT -

% vary ing l i ne l eng ths f rom 10 to 300 km n s teps o f 10 km% and c omparehemas a f unc t i on o f l i ne l eng ths .% The methods re

% l ) Sho r t L i ne app rox i ma t i om

Nomi na l -P Ime thod

% 3) Ex ac t rans m i s s i oni ne e q u a t i o n s% 4) App rox i ma t i onf exac t equat ionsi =1 .

f o r I = 1 0 : 1 0 : 3 0 0 ,

% S h o r t I i n e a p p r o x i m a t i on

Vs _s ho r t l ' i e i ) =Yq+2*1)*I R

I s s h o r t l i n e ( i) = 1 P

spf_short l ne ) =cos ang1 vs_short l ne i ) angle s_short l ne ) ) ) )Spower_short l ' ie i ) =rea l Vs short l ne )

*conj( s_short l ne ) ) )

%Nomial PI method

4=1+y* 1) * (z * 1 )/ZD= A

B=z* l

C=y * t ( + y* 1) * (z * 1 )/ a )

Vs_nomialpi ( ) =A*VR+B*IR

I s_nom'ia' l i (i ) =C*VR+D*lRS p f _ n o m i n a l pi ( i ) = c o s ( a n g 1 e( V s _ n o m i n a l p' i ( i ) - a n g t e ( I s _ n om i n a l p i ( i ) ) ) )

Spower_nomia' l i ( ) =real (Vs_nomial pi ( )*conj ( s_norpia pN.i ) )

% Exac t r ansmjss ion i ne Equa t j ons

Lc=sqrt(z/y)

garnma=sqrty*z)

Vs_exac t ( i )=cosh(gamma* l*V R + Zc*sinh ganrma*1)* IRI s _ex ac t i )= ( I l l c ) * s i nh (gamma* 1) + c os h (gamma* l*I RSpf_exac t ) =cos ang1 Vs_exac t ) -ang1 s_exac t ) ) ) )Spower_exacti ) =real (Vs_eiact i ) *conj (Is_exlct (i ) ) )

% Approx imat ion f aboveexac t equat ions

4 = 1 + ( y * l* ( z * t ) ZD= Ag = ( z * 1) * ( t + ( y * l * ( z * t ) / 6 )C= y* l ) * (1+ (y * l * ( z * 1 )/ 6)Vs_aPProxi ) =A*VR+B*IR

I s_approx ) =[* !P+P*1P

Spf_approxi ) =cos ang1 Vs_approx ) -ang e s_approxi ) ) ) )Spower_appnox( i=rea1 Vs_approx'i ) *con j ( Is_approx( i) ) )p o i n t ( j ) = i



IR=PR/ (VR* p f1ad )z =7 /1y=Yl1

% Now e calcu la te the s end i ng -end o l t age , s end . i ng -endu r ren t% a n ds e n d i n g - e n df . by f o l l o w i n g m e t h o d so r v a r i o u s

l e n g t h soF I i ne

i = i + L

en d

% The reader can uncommentny of th e four p lo t s ta tementsgiven below% by remov ing he percenlages ign agains t that s ta tement% for ex . in the plot s ta tementuncommentedelow

% t p l o t s the s end i ngend v o ' l t ages or s ho r t l i ne mode l n red

% by nominalp i -mode1 n green , by exac t parametersmoder% an dby approx . pi model n bluepl o t (po ' i n t , bs Vs _s ho r t lne) ' r ' , poi nt , abs Vs _nomia . l p i ,

,( V s _ e x a c t ) ,b ' ,

Doi t, ab s Vs_approx),,k )

, po in t , abs . . .

, po i t, abs . . .

Ex G.4 (Eiample 6,2)

% This program orms yBUSby Singular Trans format ion& The da ta for t h i s p rog ram s a p r i m i t i v e admi t t anc ema t r i x y% wh i c h s to be g i v en i n the fo i l ow i ng fo rma t an d s to red in ydata.% G r o u n ds g i v e n a s b u s no 0 .% If t he e l emen t s no t mu tua l ] y c oup l edw i t h an y o the r e l emen t ,% the( ' the ent ry correspondi g to 4t h an d 5t h co i um nof ydata% has ' I o be z e ro

i n b l a c k

',po i t, abs. . .

%

% elemento I connectedI y (se l f ) lMutua l l y I v (mutua l )? | F romI To I t coup ledo l% |BusnoIBusnoI%

y da ta= [ I

2

3

4

5

% fo rm pr i m i t i v e

% to s ta r t wi th

t 2 L / ( 0 . 0 5 + ; * 9 . 1 5 ;I 3 t / ( 0 . 1 + j * 0 . 3)2 3 t / ( 0 . 1 s + j * 0 . 4 s )? 4 1 / 0 .10+ j *9 .39 ;3 4 1/ O05+j 0 .15)

y mat r i x rom hi s da taand i n i t i a l i z e i t t o z e r o .

0000o l ;

00000

%

%

%

%

%

%

%

%

%

%yo

%

el ements=maxydata , 1)ypr i mi ve=zerose1 mentse1 ments

% Processydata matr . i x owwjse o form

element o I connectedI

l F r o m l T oIBusnoeusno

y ( s e l )

1 2 L/ O Os+j0 .15) 01 3 r / ( o . i + j * 0 . 3 ) 02 3 t / ( 0 . r 5 + j * 0 . 4 5 ) 02 4 t (o 19+30.30) 0? , A 1 l l n n c ' : + ^ 1 r \v r r / \ v . s 3 r . . ; " U . I C , f U

lMu tua ly I y (mutua l)Icoupl d to

0000o j ;

ypr imi i ve

coupl d wi h an y other el ementi n 5t h co lum nof ydata above)th e i th ro w i s madeequal to y (mutua. l)

n o w i t h w h i c h t h e l e m e n t s m u t u a l l yc o u p l e d

fo r i = 1 : e l e m e n t s ,y p r im i i ve i ,

i ) y da tai , 4)% A l s o ' i f t h e e l e m e n t s mu tua l l yeo whos e l emen tno i s i nd i c a ted% the c o r res pond i ngo l umn o in

t5

yda ta=[12345

i f ( y d a t a ( i , S ) = g ;j i s i he e l e m e n t

ffi_f--

buses=max(max(yda ta2 , ) ) ,maxyda ta3 ,A=zerose' l me ns bu es

f o r i =1 :e lemen ts% y d a t a ( i , ? ) g i v e s t he , f r o m '

b u s no .% fh e ent ry corrospondingo columncorrospondingo this bu s% n A m a t r i x i s made i f t h i s

i s n o t g r o u n d u ii f y d a t a ( i , 2 1 = g

A ( i , y d a t a ( i , Z ; ; = 1 .

end

% y d a t a ( i , 3 ) g i v e s h e ' t o ' b u s n o .% The en t ry c o r ros pond i ngo column or res pond i ngo th is bus% n A m a t r i x i s m a d e i f t h i s i s n o t g r o u n d u s

i f y da ta ' i, 3 ; -= 9A ( i , y d a t a ( i , 3 ; ; = - i

en d

end

YBUS=A ' * y p r i m i t i v e * A

Ex G.5 (data same as Example 6.2)

This program orms yBUSby,adding on e erement t a t . imeTh e da ta fo r t h i s p rog ram s a pr imi t i ve admi t t anc ema t r i x ywh i c h i s to be given i n th e fo l l ow i ng fo rma t an d s to red in y da taG round s g i v en as bus no . 0If t he e l emen t s not mutual l y c oup l edw i t h an y o the r e l emen t ,then th e ent ry corresponding o 4t h an d 5t h coiumn f ydataha s to be zeroTh e da ta must be a r ranged n ascending rde r o f e l emen tno .



j =y da ta j , 5)

Ymu tua i; ' da ta ( i , 6 )

y p r i m i t i v e ( i , j ) = y m u t u a len d

end

% FormBu s incidencematr i x A f rom ydata

% Fol1owjng ta temen t i v es max i mumo. o f e l emen ts y c a l c u l a t i ng the% max imumut o f a ll e l emen ts deno ted y : ) o f f i r s t c o l umn f y aa tae l ements=maxydata : , ) ) \

% th is g i v es no . o f bus eswh i c h s no th i ngbu t t he max i mumnt ry% ou t o f ?nd an d 3 rd column f y da ta wh i c h s ' f rom '

and , t o 'ao l r rn ,

b u s e s = m a xm axy d a t az, Z ) ) , m a xy d a t a 3 ) ;

ffi,W uooernPower ystemnavsis

YBUS=zerosbuses,uses)

f o r row=1 :e l emen ts ,

% if y da ta ( row,5 ) s z e ro tha t means he c o r res pond i ng l emen t

% is not mutua l l y c oup l edw i t h any o the r e l emen t

nppencix Hffi-

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 ) ;YBUSi 1 k ) = YBUSi 1 ,k)

YBU Sk , 1 ) = YBU Si 1 , )YBUS(II,I) = YBtll(j!,L)YBUSl , j 1) = YBUSj 1 , ) ;Y B U S ( i 1 , . | )Y B U S ( i 1 , 1 )

YBUS. | 1) = YBUSj 1 , . |;Y B U S ( j 1 , k ) Y B U S ( j 1 , k )

Y B U S ( k , j l ) Y B U S ( j 1 , k )

en di f i 1 ==0 & 1 -=0 & k - =0& l - =0

Y B U S ( j 1 , j 1 )Y B U S ( i 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 ) ;

YBU S( ll ) = YBU S( 1 , . | )y s n ( ? , 2 ) ;

Y B U S ( k , l )Y B U S ( k , l )y s m ( 2 , 2 ) ;

YBU S( lk ) = YBU S( k , lY 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 )en di f i 1 - =0& l ==Q& k - =0& l - =0

Y B U S ( i 1 ,1 ) = Y B U S ( i 1 , ' i 1 )y s m ( 1 , 1 ) ;

YBU S( k , k )YBU S( k , k )y s n ( 2 , 2 ) ;

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 ( k , 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 ) ;

YBUSl , ' i1) = YBUSi 1, . | ;en dj f i 1 - =0& 1 - =Q& k ==0 l - =0

YBU Si 1 , 1 ) = YBU Si 1 , 1 ) + y s m( 1 , l . )

Y B U S ( j 1 , i 1 )Y B U S ( i 1 , i 1 )y s m ( 1 , 1 ) ;

Y B U S ( l , l ) Y B U S ( 1 , . | )y s m ( 2 , 2 ) ;YBU Si 1 , 1 ) = YBU Si 1 , i1 ) - y s m( 1 , l . )

Y B U S ( j 1 , j 1 )Y B U S ( i 1 , i 1 )

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 ) ;en d

+ y s m ( 1 , 2 ) ;

+ v s m ( 1 . 2 ) :

jf y da ta ( row,5 )== 0

i 1 = y d a t a ( r o w , 2 ) ;

j 1 = y d a t a ( r o w , 3 ) ;

i f i 1 - = 0 & l - = 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 ) ;

en d

i f i l = = 0 & j l - = 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 ) ;

en d

i f i l - = 0 & j l = = 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 ) ;

en d

en d

eo i f y da ta ( row,5 ) s NOT er o tha t means he c o r res pond i ng l emen t% i s mu tua l l yc o u p l e dw i t h e l e m e n t i v e n n y d a t a ( r o w , 5 )

i f y d a t a ( r o w , 5 )= g

i 1 = y d a t a ( r o w , 2 ) ;

j 1 =y da ta ( row,3 );% mutua l w i t h ives the e l emen tno w i t h wh i c h t he c u r ren t e l emen t

% s mutual l y coupledw j t h k and 1 g l v e the bus nos between

% which he mu tua l l y c oup l ede l emen t s c onnec tedmu tua l. i 11=y da ta1 ,5 )

;q=ydatamutual i h ,2) ;1 y da tamu tua l i h , 3 ) ;

zs I=I /ydata(row,4) ;

zs?=L/ydatamutual ' i th,4) ;zm = l / y da ta ( row,6 ) ;

756=zs 1 z m

zn zs?l;ysm='iv zsm)

% Fol ' l owingf b lock g ives the nec es s a ry od i f i c a t i ons n YBUS

% whennoneo f t he bus es s re fe renc e g round )bus .i f i l - = 0 & j l - = 0 & k - = 0 & l - = 0

- y s m ( 1 , 2 ) ;

- y s m ( 1 , 2 ) ;



i f i 1 - = 0 & j l - = 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 ) ;

Y B U S ( ' i 1 , i 1 )Y B U S ( 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 S ( i, j 1 ) = Y B U S ( i, j 1 ) - y s m ( 1 , 1 ) ;

Y B U S ( j 1 , i 1 ) Y B U S ( i, j 1 )

WMooern owersystem _nalysis

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 ) ;YBUSk, 1 ) = YBUSi 1, )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 )

end

YBUS

Ex G.6

% T h i s ' i s p r o g r a mor g a u s s i e d e rL o a dr o w .Thedata sc lea rn=4V = [ 1 . 0 4 . 0 41 t ]

Y = [ 3 - i * 9 - 2 + i * 6 - l + 5 * 3 0- 2 + i * 6 3 6 6 6 _ j * 1 1 _ 0 . 6 6 6 +* 2 _ t + 3 * 3- 1 + ; * 3 - 0 . 6 6 6 +* 2 3 . 6 6 6 -* 1 1 _ Z + i * 6

0 - 1+ 3 ' * 3 - 2 + t r * 6 3 - j " 9 1t y p e =o n e sn ,1 )t ypechanged=zerosn,1 Ql mi max=zerosn ,1Ql mi mi =zerosn, )Vmagf ied=zerosn 1tYpe(2)=2

Ql mi ma x2)=1 g

Q1mi mi 2 )=9.Vmagf ied(2) 1.04d i f=10 ; oof i e r=1Vprev=V;wh i le (d i f f >O.0000 1| n o o f i t e r ==1 ) ,ab s V)ab s Vprev)

%pauseVprev=V;P = [ i n f . 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 ) ] ;f o r i =Z i n ,

i f t y p e ( i ) ==2l t y p e c h a n g e d ( i ; ==1 ,

.Q,mimi(i ,

f rom Example .5

lFt ilE4lq.f .Appendix NilFffi

T-typechangedi ) =1t

e lsetype i ) =2 ;

en d

en d

sumyv=0;

f o r k = L : n ,

i f ( i - = k )

sumyv=sumyv+Yi, k) V(k ) ;en d

en d

v i ) = 1 /Y i, i ) )* ( (p i ) - j * Q ( i ) ) c on j ( v i ) ) - s umy v );

if t y pe ( i == 2 & t y pec hanged ( i) -=1 ,

V( i =pe1 rT o rec t (Vmag f i x ed ( i) , ang1 (V ( i ) * 1g0 /p i ;en d

en d

di f f=max ab s ab s U Zzn)) -abs Vprev Z:n) ) ) ;noof i er=noof i er+ l ;en d

V


% Program

c l e a r ;

% n s tands

fo r load flow by Newton-Raphsonethod.

fo r number f busesn = 3 ;% Y ,vo l t ages t t h o s e u s e s r e n i t j a l j s e dv = [ 1 . 0 4 . 0 1 . 0 4 ] ;% Y i s YBu sY=[ 5 .88228- j ' t 23 .505142.9427+j*Lt . t6t6

-2 9427j*Lt .767 5 88228_j*23505L4

-? 9427*L l . 7 7 -2 .9427* tt .76 t6%Busypesare jn i t i a l j sed in t ypear ray o%bus.%code stands fo r PV bu st y p e = o n e s ( n , 1 ) ;

%when l ' imj ts ar e exceeded or a pV bu s Bu s type is changedo pQ%temporariL a n a ' l o m o n f i n f # . r ^ o a h r n a n ' l .i a - ^ + . ^ 1 i -' e " , L , , e I vr u J p s L r s r v s u r ) J t r L L u r r l r c d s g r r5 D us s t a t u s

% i s t empora r i l y pQ pV .O the rw i s e

-Z 9427* 7 I t 67-2 9427* II .7 7

5 .88228- j *23 .5051a1;code1 whichstands or PQ



",llilil;:il;l#ii#ll,llQ ( j =Qlim i mi n i ) ;

e l s e

Q ( ' i ) = Q l i m i t m a x ( i ) ;en d

t Y P ei ) = 1 ;

c hangedrom to it is z e rotypechanged=zerosn, 1)% s ince maxand mi n Q l ' i m i t s a re checked nl y fo r pv buses ,% max& min Q l iml ts fo r o the r t y pes of bus esca n be se t t o any v a l ues .% here we have se t them o zeros fo r convenienceQl mi max=zerosn, 1)

Q1mi m in=z e rosn, ) ;Vmagf i ed=zerosn, 1)

W ModernPower ystemAnalysisI

t I to to ta l no. of equat ionseqcoun t )f or ceq=1:eqcoun t ,

fo r cco l=1 :eqcoun t ,

v s J \ e e y / t s J r v v v r \ u J J v e v r u u J \ v w w r l l r ,

bm=i agY assoeqbusceq), ssocol us cco: l) )*V as ocol us ccol) ) ) ;

ei =rea l V assoeqbusceq)) ) ;

fi = i magV as s oeqbusc eq )) ) ;

i a s s o e q v a r ( c e q )= ' P" ' & a s s o c o la i ( c c o1= = ' d ',

i f assoeqbusc eq ) =assocolbus ccol) ,H=am*f ibm*ei;

e l s e

11-Q as oeqbusc eq ) mag Y as s oeqbusc eq ) , ssocol us c eq ) ) .* abs V as s oeqbusc eq )) ) ^2 )

en d

Jacob c eq , c ol =H

en d

i f a s s o e q v a r ( c e q)= ' P ' & a s s o c o la r ( c c o 1= = ' V ','i f assoeqbusceq) =assocolbu s ccol) ,

N=am* e i +bm* f i ;e l se

N= Passoeqbusceq))+real Y assoeqbusceq),assocol us ceq)) . .* abs V assoeqbusc eq )) ) ^2 )

en d

J ac obceq, co l =11

en d

i f a s s o e q v a r ( ce q ) = = ' Q 'a s s o c o l v a r ( c c o 1 ) = = ' d ' ,

i f as s oeqbusc eq ) =as s oc o lbus ccol) ,J =am* e i +bm* f i ;

e l s e

J= Pas oeqbusc eq ) rea l Y as s oeqbusce q) ,as s oc o lus ceq))*abs V as s oeqbusceq) )^2 )

en d

J ac ob (c eq , c c o l= J

endi f a s s o e q v a r ( ce q ) = = ' Q 'a s s o c o l v a r ( c c o 1 ) = = ' V ' ,

1 assoeqbusc eq ) -as s oc o lbus ccol) ,L=am*f ibm*ei

e l s e

L=Q (as s oeqbusc eq )) . . .i m r n / V / ' c c ^ o^ h , , . t / ^ o ^ \ ' e . ^ ^ ^ ' 1 h , , . / ^ ^ ^ \ \ * . h . / \ , 1 . . . ^ ^ ^ h , , - / ^ ^ ^ \ \ \ ' \ D\ .r f f f u v \ r \ . u J J v sq v u J \ l s Y / r q JJ v u v l p u J \ L s Y/ r / , q w J \ r \ q J J v E V U U J \ L g r { , ,

l l L l ,

end

pause

update=i v Jacob) mismatch '

noofeq=L;

i f t y p e ( i ; ==1newchi angV=updatenoofeq);

newangV=ang1V ) )+newchiangVnewchi magV=update(noofeQ+l )*absV i ) ) ;newmagV=absV 'i ) ) +newchnmagV;

V j ) =pol rTorec t newmagV,ewangV*180/p i )noofeq=noofeq+l

e lse

newchi angV=updatenoofeq);newangV=ang1V ' i) )+newctr iangV

V( i =po l rT o rec t abs V( i) ) , newangV* l 8O /p i;noofeq=noofeq+l

en d

en d

% Al the fo l I ow i g var i ab les /arrays ar e c l eared f rom% memory. hi s i s because he ' i r d imens ionsma y change ue to

% bus swi tched nd onc e upda tesa re c a l c u l a ted , t he v a r i ab l es9r are of no use as they are be ing reformulatedat the% end of each terat ' i on \c l ea r m i s ma tc h ac ob upda teassoeqvar ssoeqbus ssocolvafassocolbus ;d i f =m ' i abs abs V 2: n ) abs Vprev2: n ) ) ) ;noof er=noof i er+1

en d

Ex. G.8 (Table 7.1)

% MATLABrogram or opt ' imumoading of generators

% The data are f rom Example .1% r t f i nds l amdaby the a l go r i t hm g i v en on t he same d9€ ,onc e he demandr k s spec ' i f i ed

% Wehav e ak en th e demand s 231 .25MWor ros pond i ngo the l as t but on e% row o f T ab l e7 . 1 a n d c a l c ul a t e d a m d a n d h e l oad s h a r l n g% n is no o f gene ra to rs

n= 2

% Pd stands or load demand.

% a l pha and be ta a r ray s deno tea l phabe ta c oe f f j c i en t s% fo r g i v en gene ra to rs .



J ac obc eq , co l =L

en0

en d

en 0

% New pda te ec to r i s ca lcu la ted f rom Inv e rs e o f t he J ac ob i an

Pd=?31.25

al pha=[0 .200.251

beta= 40301

% n i t i a l g u e s s o r l a m d a

I amda=20

m'tn'tmum

%are tored n arrays% n r e a l i f e l a rge

% ar ray o inf us ing

ffil Moderno

I amdaprev=lmd a% ole rances epsand ncrementn l amdas de l t a lamdaeps=1de' lal amda=0.25

Hffi

m 1 n" l n f icos t=0 ;f o r j = 0 : n ,

f o r i =0 : ,

f o r I = 0 : n ,

u n i t = [ 0 0 0 0 ] ;

% Here we e l imnate s t ra i gh taway hos e c omb i na t i ons hich% dont makeup th e

% n MW emand r such combinat ionswheremax imumenerat ion

% on l nd i v i dua l

% gene ra t i on s exceedinghe max i mumapac i ty o f any o f t he

% generators

i f ( ' i + j +k + ' l )==n& icPgmax(1) i <Pgmax (2 ) kcPgmax(3) < . . .

Pgmax4)

i f i - = 0

u n i t ( 1 , 1 ) = i ;

% F tnd ou t t he c os t o f gene ra t i ng hes e un i t s and

% qd d it up to to ta l cos t

cos t=cos t+0. *a lpha )* i * i +be ta )

*i ;en d

i f 3 -=gu n i ( 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 \en d

i f k-=0

u n i ( 1 , 3 ) k ;

c os t=c os t+0 .5 * a lh r (3 ) * k * k +be ta (3 ) * k

en d

i f l -=0

u n i t ( 1 , 4 ) = l ;

c os t=c os t+0 .5 * a1ha (a ) * l l +be ta (4 ) * l

en d

% If the total cost i s comi g out to be I ess than

% m i n i mumf t h e c o s t i n

% prev i ousc omb i na t ' i onshen makemi n equa l o c os t and

% cuni t (s tand for commi t teduni ts) equal to un ' i ts

% c ommi t t edn th i s i t e r a t i o n

t ( d e n o t e d Y v a r i ab l e un i t s )

i f c o s t < m i n

c u n ' i t u n ' i t im i n = n n c . f :rr r . | | e v v v t

e l s e

mu m rmt s ot each generat i g unPgmin nd Pgmax.

s c a l e p r o b l e m s , e c a n f i r s t i n i t i a l s e the Pgmax

fo r I oop and

L a m d as ' )

I oad s ha redby two un t s ' i s ' )

% Pgninarray to zero us ing Pgmin=zeros(n,1)ommand% La te rw e c a n c h a n g e h e l ' i m i t s i nd i v ' i dua l l yPgmax=[12525 ]

Pgmin=1200)

Pg - -1O 0* onesn ,1 )

wh i e abs s um(PS)ed ; ' g t t

f o r i = 1 : n ,

Ps ( i = 1amda-be tai ) / al ph a i ) ;i f P g ( i > P s m a x ( i

P g ( i = P s m a x ( i;en d

i f P s ( i ) . P g m l n ( i )

P g ( i ) = P g m i n ( i ) ;

en d

end

i f ( s u m ( P g ) - P d. 0l amdaprev = lmda;

lamda=lmda+dela1amda;

e l s e

I amdaprev=amda

lamda=lmda-dela l amda;

end

e nd

d i s p ( ' T h e i n a l v a l u e o f

1amdaprev

d i s p ( ' T h ed i s t r i b u t i o n o f

Pg

Ex. G.9 (Table 7.2)

% MATLABrogramor opt imumni t commit tmenty Brute Forcemethod

f t The a ta or th l s p rogramor respondso Tab le7. 2

c ear ;



c os = 0 ;

en d

en d

en d

% a lpha ndbetaar raysdenote lphabetacoef f ic ients or g i ven genera to rs

a l p h a = 1 0 . 7 7. 6 0 . 0 02 . 5 0 1 ' ;b e t a = 1 2 3 . 56 . 53 0 . 03 2 . 0 1 ' ;P g m i n = [ 1I 1 ] ' ;Pgmax=ll?2 12 I2l ': '

n= 9% n denotesotal MWo be commite d en d

,f,f6.'l Modern o@

en d

E

f o r i = 1 : n ,s gma=Bi , : *Pg-B , )*P g ) ;

en d

, d i s p ( ' c u n i t d i s p l a y h e n o o f c ommj t t ed n i t s o n e a c ho ft o r s ' )

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 s 0 j t m e a n s

fo r I amda

the four genera-

the un i t i s n o tcomm

di sp ( The to ta l no of un i ts to be commited are ' )c u n i t

Ex G.lO (Ex. 7.A)

c l ea r

% MATLAB rogram or opt imumoading of generators

% This p rog ram i nds the op t i ma l l oadi ng o f gene ra to rs nc l ud i ng% penal y factors

% I t i mp l emen tshe a l go r i t hmg i v en jus t be fo re Ex amp l e .4.% Th e data for th i s program re taken f rom Example .4% Herewe give demand d and a l pha , be ta and B -c oe f f i c i en t st l , {e a lcu la te load shared by each generator

% n is no of generators

n= 2

? Pd s tands for I oad demand% alpha and beta arrays denotea lpha beta coef f i c jents for g iven

% generators

Pd=23704 ;

al ph6=[0.0200 .0a1 ;

beta=162o l '

I i n i t i a l g u e s sIamda=20;I amdaprev=amda;

% to l e ranc e is eps and nc remen t n lamda s de l t a l amdae p s = 1 ;

de l a l amda=O .25 ;

% the mi n i mum nd max imumi m i t s of each gene ra t i ng ni t%are s tored in arrays Pgmin nd Pgmax.

% In ac tua l large s c a l e p rob l ems ,we can f i rs t i n i t i a l i se the Pgmax r ray% to in f us i ng f o r loop

% and Pgmin r ray t o zero us ing Pgmi n=z e ros (n ,1 )ommand% Later we should can change he l imi ts i nd i v ' i dua11yPqmax=f20000. l :

Pgmi n= [0 ] ;

Pg i ) = 1- beta i )/ l amda)(2*s sna)) (a l ph a i ) l amda+z*Bi, i ) ) ;I

i f P g ( i ) ' P g m a x ( j )

Pg i ) =Pgmaxi ) ;

en di f P g ( i ) < P g m i n ( i )

P s ( i ) = P s m i n ( i ) ;

en d

en d

PL=Pg ' * B * Pg ;

if ( s um(Pg) -Pd -PL)<01amdaprev=lmda;

I amda=l mda+delal amda;

e l se

I amdaprev=1mda;

I amda=l mda-delal amda;

en d

noofi er=noofi ter+1;Pg ;

en d

di sp Th e

noof i er

d i s p T h e

1amdaprev

d i s p Th e

Pg

d i s p Th e

PL

Ex. G.II

% r i th i s% F is . 8 .8s' imult i onTsg=4Tt=0.5Tps=20Kps=100

no o f i t e ra t i ons requ i red a re ' )

f i nal val ue of' l amda

i s )

opt ' ima1oadi g of generatorsnc ludi g penaly facto, rs is ')

I osses re ' )

example he parameters

a re i n i t i a l i z ed T h i s

bo th f o r F i gs . G.3 and

fo r al l the blocks fo r the sys tem nprogramhas to be run prior to th e

G . 4 .



B = [ 0 . 0 0 1

0 0 l ;

P g = z e r o s ( n , 1 ) ;

noof e r=0 ;

PL=0 ;

P g = z e r o s ( n , 1 ) ;

R= 3Ksg=gKt=0.K i=0 .09

Tranefer cn TraneferFcn 1

SS1 Moo"rno*b,syrt"r Rn"trrri, .

Ksg KpsTps.s 1

Conetant

Fig. G.3 Firstorderapproximationor load requency ontrol

Ex. G.12

The system of Fig. 8.10 is simulated using simulink as wasExampleG. l l .

Flg.G.4 Proportionallus ntegraload requencyontrol


% Prog ram or bu i ] d i r rg of Zbus by add i t ' i ono f b ranc ho r l . i nk% Zpr imary=[e lementnoro m to va lue

. s + 1

done in

%

%yo

% -- l% Here ar e shou ld e aken ha t to beg inw i th an% reference nd both froman d o nodes houldno tc l e arzprimary=

1 I 0 0 . 2 52 2 I 0 . 1

e l emen t s added

be new nodes

to

Ksg

Tsg.s 1Kps

Tps.s+ 1Transfer cn TransferFcnl I TransferFcn2

lffiwffiflI

I currentbusnondicatesmaximumo. of buses dded nt i l no wcurrentbusno=0

fo r coun t=1 :e lements ,

[ rows o1 ]=s j ze (zbus)

f rom=zpr iar ycount , )to=zpr imarycount,3)va lue=zpr i ar ycount , )

% newbusar iab le nd ica tes% newbus us ma yor mayno t

newbus=maxf rom,o)

th e maximumf th e tw o busesal ready be a part of exi st i ng zbus

bu s o reference us

% re f v a r i ab l e i nd i c a tes the m i n i mum f t he two bus es% & not necessar j l y the re ference bu s

% re f bus mus t a l way sex i s t i n t he ex is t ing z busref=min from, o)

% Modj f i cat ion o f t y pe l

% A ne welement s added romi f newbus currentbusno& ref

zbus=[zbus e ros ( rows , )

z e r o s ( 1 , c o l s ) a l u e l

cu rentbusno=newbu

cont inu e

end

ne w==0

% Modi f i cat ion of type?

% A new e lement i s added rom newbu s to o ld busbus

i f newbus currentbusno re f -= 0

zbus=[zbus bus ( , re f )

zbus ref , : va lue+zbusref , re f ) ]cu rentbusno=newbu

c on t inue

\

other than reference

en d

% Modi i ca i on of type3

% A new element s addedbetweenan old bus and re ference bu si f newbus =currentbusno ref==O

z bus =z bus -1 / (z busnewbus ,ewbus )+v a le ) z bus( :, newbus )zbus(newbus ,c on t inue

en d

% Mod' i f i cat jono f t y pe4% A new element s addedbetwen wo old busesi f newbus<= currentbusno& ref -= 0

zbus=zbus (va lue+zbusf rom, -



3 3 1 0 . 14 2 0 0 . 2 5s 2 3 0 . 1 1

[e l ements o1 mns]size(zpr imary)

t To beg inwi th zbusmat r i x s a nu l l mat r i x

L/ rom)+zbusto, to)2 * z b u s ( f r o m , to ) ) * ( ( z b u s ( : , f r o m ) - zb u s ( : , t o ) ) * ( ( z b u s( f r o m , ; | - z b u S ( t o r : ) ) ) )

c on t inue

en d

en d

Wl uodginPower ystem natysis

Ex. G.14(Example 12,10)

t Programor t rans ' i en t t ab i ' l i t y o f s ing lemach ineonnectedo ln f i n i t e%bus h i s p rogramimu la tes xample2 .10 s jngpo in t by po in tmethodc leart -0

t f=0

t fi na l=0 .5

t c =O . 25t s teP=9 . 5

l4=?5? ( 180*50)j = 2

de1 a=21. .64*p i18 0ddel a= 0

t ime 1 )=9

a n g ( 1 ) = 2 I . 6 4

Pm=0 .9

Pnaxbf=2.44

Pmaxdf=0.88

Pmaxaf=2.00

wh i e t<t f i na l

' i f ( t = = t f) ,Pam' ius=0.9-Pmaxbf*s ide l a)

Papl s=O.9-Pmaxdf*s ' inde l a)

Paav=Pam'ius+Pap1s 2' Pa=Paav

en d

i f ( t = = t c ) ,

Pami us=0.9-Pmaxdf*s ide l a)

Papl s=0. -Pmaxaf*sn de l a)

Paav=Pamj us+P plus) 2Pa=Paav

en d

i f ( t > t f & t < t c ) ,

Pa=Pm-Pmaxdf*s ide l a)en d

i f ( t > t c ) ,

Pa=Pm-Pmaxaf*s ide l a)

end

t , P a

ddel a=c ide la+ ts tep* ts tep*pa/M)

de l a- (de l a * 180 /p j dde la) * pi/ 180

de ' l adeg=de1a*180/P ' i

j = j + 1

en d

a x i s ( [ 0 . 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 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 change he threshold value rom 0 to the fault clearing time

(i n sec.).

Ex. G.16 (Ex. 12.11)

v" Th'tsprogram imul tes transient stab'i ty of mu limachi e systems

% Thedata is f rom Examp' le2.11

c lear a l

format ong%Step In ' i t i a l i sa t i onwi th load f low andmach ' i neat a

f =5 0 ; t s t e p =9 . 9 1 .= [ 1 2 J ,Pg n e t t e r m= [.2 52 .10 ] '

Qgnet te rm=[.6986 .3110 ] 'X g = [ 0 . 0 6 7. 1 0 ] ' ;% Note the .use of . ' operator here

% This does a t ransposewj thout tak ing the conjugateof each e lement

y g = [ p o ' l a r T o r e c t ( 1 . 0 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 s l ack bu s

n=2 i

%Step

V0c on i =c o r iV0 ) ;

Ig0=conj (Pgnet tem+j *Qgnet term)UO )Edash0=V0+j(Xg. Ig0)

Pg0=rea ' lEdas h0 . * c on jI g0 ) ;

x 1 r=ang l (Edas h0 )

% l n i t ' i a l i s a t i on o f s ta te vec tor

Pg r=Pg0;

% Pg_ro1us l .=PgO;

x Z r = [ 0 0 ] , ;;



t= t+ts tep

paus e

t i m e ( i = 1a n g ( i ) = d e l t a d e g

x t d o t r = [ 0 0 ]

x 2do t r= [o o ] ,;

x l do t rp l tl = [0 0J ix2dot rp l t l= J0 0J i

SStep3% Here n this exampleehave o t rea l l y ca l cu la ted Bus u t -sne

% canwrite seParate rogram.yBusdf=[1f986-j*35-6301 0 -0.0681+i*5.1661

- j * 1 1 . 2 3 6 0-0 .068 l+ j *5 .66 1

YBuspf=I .3932- j *13. 8731

-0.2?I++j*76289-0 .090 l+ i6 .0975

%Step

0

-0.2214+j*762890.5+j 7.7898

0

0. 1362- j6.2737f

-0 .090 l+ j6 .0975

00 .1591=j *6 .11681

of

d]

. oEc

.go

P

Eo()occo

. c )o

.Eo(u

E Lc Lg o

e o ,O Co ' o / -

L

ots

F=_o(U

o

c.9oc(gL

F

u?(5

6IL

.so

o

% Set the va ]ues or in i t i a l t ime t (occurancef faul t ) and

% c i s t ime at wh ich aul t i s c lea red

t=0; c=0 .08 ;f i na l 1 0 ;

r = 1 ;Edash_r=Edash0Edash_rpls =Edash0

w h i l e < t f i n a l ,%Step Computeenerator owers sing appropriateYBus

% he YBus hosenn the fo l l ow ings tep s se t accord ingo the cur ren t

% imeif t <= tc YBus=YBusd f ;l se YBus=YBusp f ;end% Note ha t herewe obtain he currents niected at generator us

% by mul t ip ' ly ing he correspond' ingow of the Ybuswj th the ve\ tor of

% vo l t ages eh indhe t rans ien t eac tances.h i s shou ld l so jnc lude

% slack bus v o l t age

% and hence he ent rY 1 aPPears

% generator bu s vo ' l tages

I = Y B u s ( 2 : m + 1 , : ) * [ 1

Edas h l ;

Pg_r=rea1Edash_r. *coniI) ) ;

j n t he bus v o l t age v ec to r i n add i t i on t o

%Step6 compute ldot-r an d xZdot-r

x dot_r=x2_r;

f o r k = 1 : m ,

xZdot -r (k, 1 = pi * f /H (k ) )* (PgOk) Pg-r k) ;

en d

%Step7 Computei rs t s ta te es t imates for t=t ( r+1)

x 1_rp1 s =xl_r+x dot-r*tsteP ;

xZ_rpl us1=x2_r+x2dot_r*tsteP

%Step8 Computei rs t es t imates of E ' - r+1

Edas h p l us 1=abs (Edas h0)* (c os x 1 - rp1 s l )+ i * s i n (x l - rp l us l ) ) ;

%Step9 Compute g for t=t ( r+1)

(\'=o)

E

cat,

IU



I = Y B u s ( 2 : m + 1 , : ) * [ 1Edash_rp'ls1l ;

Pg_rp ' l s1=real Edash_rp1s1.*coni I) ) ;

% S tep10 Compu te ta te -den i v a t i v es 'a t t = t ( r+1 )

(IL

o)oc

-1,

W Mod"rn o*",.sy.t.r An"ryri,

x l d o t _ r p l r t l =J 0J , ixZdot_rplr l=10 0J if o r k =1 : m ,

I

x ldo t_ rp l s l k ,1) x2_rp ' lus l k , ) ;x2do t .p lus l k ; )=p i f/ H k) (pqg

%step 1 computeveragearues f s ta te der i va t i vesx dotav_r=x dot_r+x1dot_rp'lsl ) ?. 0

x2dotav_r=x2dot_r+xZdot_rp1usl )/2.0%stepL2 Computejnal State est t 'matesbr t=t (r+1)x _r p us =xl_r+xldotav_r* tstep;x2_rpl s l ixZ r+x2dotav_r* tstep;%step 3 Computeina l es t imate or Edash t t= t ( r+1)Edash_rp1s =absEdash0)* (cos x l_ rp l s ) j si n x t_ rp1 s l)%Step 4 Pr in t Sta teVecto r

x?_r=x?_rplsl ;x l_ r=x l_ rp1s l;Edash_r=Edash_rp ' ls ;

%Step 5t i m e ( r ) = 1 ;f o r k = l : m ,

an gr , k )= x t_ r (k )180)/p i ;

en dt=t+tstep;r - r+1 ;en dp lo t ( t ime,ang)

Ex. G. 17 (Example IZ,II)

%Example2 .11 E so lved s ingSIMULINK% Th e od e i v e n e low h o u l d e runp r i o r t o s i mu ' l a t i o nh o w nn F ig .G.6 .c l ear alg l o b a 1 r y y rg l o b a l P m f H E n g gglobal rt d dt r %conversjonactor

rad/degree91 ba l Yb f Yd f ya f

f=50 ;ngg=2;r =5 ;n b u s = r ;

r t d=180 /p i ;d t r=p i

E E i l $ n E

'Fl8 8

tt

Es3-8 .=

o l lct

o

c

.9U'c,6Fo.=.c,ctG'

E

=

q(J

dtIL

o

oa

o.

o

N

oo

oIf

U'

Eo

o>6

IJ

6



/ 180 i

9o Gen.

g e n d a t a = [ 1

23

R a X d ' H0 0 i n f0 0 . 0 6 7 1 2 . 0 00 0 . 1 0 0 9 . 0 0

l ;

iltl Modern owervstem natvsts

yd f ' I

5. 7986-J35.6301

0-0 .0681+i5.1661

- AppendixE

I

Complex

Product 1

E2

0- j * 1 1 . 2 3 6

0

-0 .0681+J*5 .1661

00.1362- j *6 .27371

Yaf=1 . 3 9 3 2 - j * 1 3 . 8 7 3 1-0.22L4+j*76289

-0 .0901+j6.0975

-0 22I4+j*76?89 -0.0901+j6 0970 .5+ j *7 .7898 0

0 0.1591- j *6 .11681

fc t= i npu t ( ' f au l t c l e a r i n g i me c t = ' ) ;

&Dampig factorsdamp2=O.;damp3=00;

%I i i al generator ng' lsd2-0,3377*rtd ld3=0 .31845* r td ;

%In i t i a l PowersPn2=3.25;

Pm3=210 ;

%Generatornternal Vol tagesEl .= 0 ;E2 =103 ;E 3 = 1 . 0 2 ;

%Mach inener t i a Constan ts ;H2=gendat(2 ,4 )113=gendata3,4)

&Mach le Xd '1661=gendata2 ,3 )

1661=gendata3 ,3 )

Note : For the simulation or multimachine stability the'two summation boxes

sum 2 & sum 3 give the net acceleatingpowers Po2 and Por. T}iregains of the

gain blocks Gl, G2 G3 and G4 are set equal to pi*fAlZ, dampZ,pi*f/Fl3 anddamp3, respectively. The acceleratingpower 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 anci ntegrator 4 are set

to 0, d2/r+d, 0 and d3lrtd, respectively.The gain blocks G5 and G6 convert the

Flg. G.7 MultimachineransientStability Subsystem

Yaf 2,1)



angles $ and d, into degreesand hence heir gains are set to rtd. The electricalpower P"2ts calculatedby using two subsystems1 and 2.The detailed diagramfor 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

ffivodern owerysterRnalysis

E/-61 *Yar 2,1)+ Erlq*Yq{z,z) + ErlErYar 2,3).The switches reused oswitch between refault andpost ault conditions or eachmachineand theirthreshold aluesare adjustedo fct, i.e. the fault clearing ime.

cl ear

% T h i s

% w h i c h% l I g l l% lIszl% l . lv o l . l

Y ol l g n l% l r L l ly o

l r L 2 l% l . lv o l . l

% l I L m l

II

= l y n lIYn+12

III

IIYn+m1

Y f u l l = _ j * 5 0 7 *50 - j * 7 . 5 j * 2 . 5

j * 5 j * ? . 5 - j * 1 2 . 5 1[ r o w c o l umns ]=s j z eY fu l l )

n=2

Y A A = Y f u l l 1 : n , 1 : n )

Y A B ; Y f u l l 1 : n , n+1 ;c o l umns )YBA=Y fu l (n+1 : ows , +L : o lumns )YBB=Yfu l (n+1: ows, +1 co1 ms)

% This g ives the reducedmatr i x

Yreduced=YAA-YAB* iv YBB)YB A

matri x for stabili ty

and re ta i ns on l y t heY l n+ l YLn+Z.

YZn+l YZn+Z.

Y n n + 1 Y n n + 2

Program fi nds

e l mna tes theIY l l YL?

iIY21. Y?r

reduced

I bus es. Y l n

. YZ n

. Yn n

th e

I oad

s tud i s

generatorY1n+ml IV1Yln+ml IV2

t l

buses

2.1 Lirt =

2.2 0.658 hmlkm

2.3 L= F loR w^2 r r

2.4 260.3Vlkm

2.5 He= - .,:I

,AT/mz directed

pwards) 3nd

2.6X=(X t -Xn ) (X2 -Xn )

\ * x2 zxn

2.7 0.00067m}J/krlt,0.0314V/r<Tl2.8 0.W44/-t4C, ml{lkm, 0.553 .l40 Vlkm2. 9 0.346 hmtkm 2.rc 1.48m

2.ll 0.191 hmlkm/phase.12 0.455mlllkm/phase2.132.38m

2.14 (i) 0.557dtt2Att4 (ii) 0.633d2t3A16 er) 0.746d3t4Ar/8

Chapter 3

3.1 n^ _ 2 rk lVl(ln r/ 2D)130"_!n DlLr)l_30o) -,u2h(D/r)ln(r/2nffi t'tm,

Io = 2rfqo/90" A0.0204p,FlkrrrO OlO? t tF l1r 'm rn na' l .-^ r

/ r^ , rut r Lv uvuLl4 l

5,53x 10{ mhoslkm

ro-7*Vlr,:

-,lt-+n,e|,,,)+arlr?]

rs to Problems

3.3 0.0096pFlkm

3.5 3.08x iO-scoulomb/km3.7 8.54x 103 hmslkrn

CHAPTER 2

YnZ Y n n + m

Yn+Lm

Yn+mn+m

Vn

Vn 1

Vn 1

V n * m

1- x2

3 .2

3"4

3 .6



8.72x l0-3 1fi/xm 3.9 71.24 v3 .8

Chapter 4

4. 1 12kV

WModrrnPo*"r.syrt"t An"lv.i.

Chapter 5

5.1 (a) 992.75 W (b) No solution ossible5.2 At -0 .91I .5o,B '=239.9166.3" ,'= 0 .001102.6 ,Dt =0.851I .96

(b) 165.44 y,0.2441-28.3"kA, 0.808 agging, 6.49MW(c ) 7O.8Vo (d)

28.L5Vo5. 4 (a ) 273.5MVA (b ) l,IT4 A (c ) 467.7MV A5.5 133.92 V 23.12MW 5.6 202.2 V

5.7 At x = 0, ir r = 0.3L4 osQ,st21.7"), irz=0.117 cos (utt+ 109"),

At x = 200km, it = 0.327 os c..r/ 9.3o),rz = 0.t12 cos(,*t+ 96.6)5.8 135.817.8o V, 0.138115.6oA, 0.99 leading,55.66 MW, 89.8Vo,

373.L1 t.5o,3,338 m, 1,66,900m/sec

5.g v - 12g.3172.60,' - 0.000511g9.5".

2

5.10 7.12", fr = 0.7 agging, fz= 0.74 agging

5.11 47.56MVAR lagging

5.I2 10.97 V, 0.98 eading,0.27Vo,86.2Vo5.13 51.16 V, 38.87MVAR leading, 0 MW

5.I4 238.5 V, P,+ jQ, = 53 - jI0, pf - 0.983 eading

5"15 17.39MVAR leading, .5 4MW

Chapter 6

6.1 For hisnetworkree s shown n Fig.6.3a;A sgivenby Eq. (6.17). hematrix s not unique.t depends pon he orientation f the elements.

6. 2 V) = 0.9721-8.15" 6.3 V) = 1.26 :74.66"

6.a@)

IE

(b)

j0.3049

j0.1694

j0.1948

j0.31340.807 js.6s

0.645 j4,517

0.968 j6.776

0.968 j6.776

0.880 j6.160

6.5 Pn = - 0.598pu , PB = 0.2pu , pz s 0.796 uQn= Qzr = 0.036pu, Qn= Qn= 0.004pu, ezz=ezz= 0.064pu

6.6 (a) Pn = - 0.58pu, P* = 0.214pu, pzl = 0.792puQn = - 0.165pu ,ez t _ 0.243pu ,en = 0.204 uQy = - 0.188pv , Qz t= 0.479pu ,en= - 0.321pu

(b ) Pn = - 0.333Pu , Pzz=0.664 u, Pr : = - 0.333 u' on = Qzr= 0 '011pu,QB = Qzr= 0.011 u,ezt= en=g.g44 pu

(c )

t z } 4

I l-ito s/s3" i5 I

I tu l lst87" -ilo js IJ L js js _j l0 j

f-iro.rors j5.o5o5 js

6.7 a) i) | ;s.osos -j10 jsL j5 js _j10

t

2

3

4--:-

U

I

0

0

I

0

0

0

U

0

I

0

U

Q r

0

I

f h \ / i \ D - n < n n - . . n n A ^ A\.r,, i, , i 12: v.iJVUpu , ts13= U.LUZ pu , pzZ = 0.794 pu

Qn= 0.087pu, Qzt= - 0.014puQn= QzF 0.004 pu, Qzz= Qsz= 0.064pu

/ / : : \ I ) t \ . oE\ u , , f 12 : - u .ooJ pu ,

Pn = 0.287pu, Pzt= 0.711pu



)

6

7

I

9

0 0 1- l 0 0

l - 1 0

0 - l 0

- l 0 l

Qn = 0.047Pu,Qn= 0.008 ptr,Qzt = 0.051pu6.8 Vrt = I.025 y0.095= l.AZ9/.- 5.3. pu

- l

1

0

1

0

4Mooernower lrstemnatysii

Chapter 7

7.1 Rs 22.5/hr

7.2 (a) Pcr = 140.9MW, 159.1MW

7.3

7. 4

7 .5

7.6

7.8

7.7

S4VlIl$ =

(i) Gen A will sharemore oad thanGen B(ii) Gen A andGen B will

share oad ofpo

each(iii) Gen B will sharemore oad thanGen A.Pcr = 148MW, Pez, L42.9MW, Pct = 109.1MW( d c l d P d = 0 . 1 7 5 P c + 2 3

(a) Pq = 138.89MW, Pcz = 150MW, Po = 269.6MW

0) Pcr = 310.8MW, Pcz = 55.4MW(c) For part (a): Cr = Rs 6,465.141hr

Forpart (b): Cr = Rs 7,708.15/hr

Bn = 0.03387 u or 0.03387 ITa MW-lBrz = 9.6073 10-5 u or 9.6073x 10-7MW-tBzz= 0.02370 u or 0.02370x 10-2MW-r.

Economically ptimumUC

Load (MW) Unit Number

1 2 3 4

0-44-8

8-r2t2-16r6-2020-24

OptimalandsecureUC

Period Unit Number

t 2 3 4

044-88-12I2_16l6-20

1 1 1 1

r 1 1 0

1 1 0 0

1 1 1 0

1 0 0 0

1 1 0 0

20

l4

6

l4

4

10

1 1 1 11 ' 1 1 01 1 0 0 .1 1 , 11 1 0 0

RL.,-iiis;E:lit+i(Atldt

I--

8.1 Load on 123MW, Loadon G2= 277 MW, 50.77H)..

2

8. 2 Af (t ) = - 0.029- 0.04ea.58'os (1.254t + 137.g")

8.3 1/(50K) sec9.4 aPtie. |

-

_ ll R2)an (K iz+D0 / Kp,1 Kipl +l / Rt)+ (K, r

8. 5 APt i . , r (s) - tOOf 'zt t t0.9t L80s51

System s found to be unstable.

Chapter 9

9.r

9. 29. 3

it = 3.14 sin (314t - 66.) + 2.97e sor,;"* = 5A

(a ) 81" (b) - 9'(1) e = 2.386kA, .IB

9.4 8.87kA, 4.93kA

9. 6 6.97kA

9.7 (a) 0.9277 A (b) 1.312 A (c)(e )0.1959 A

8.319 A

1.75kA (ii) IA = 4.373kA, IB = |.ZS a9.5 26.96 A

1.4843 A (d) 1.0205 v, 53.03MVA

9. 9 2.39 u.8

9.10

9 , 1 2

132.1, 7.9;136.9, 5.6 gJ t 0. 6pu{ - - j8,006pu, 1{: = - 74.004 u

Chapter 10

10.1 (i) 1.7321270"ii ) zl0" (i i i) 1 7321150.iv ) tl2r0"lO.2 IA= jl.l6 pu , Vtn = 1.171109.5"u

ll - n O< 2 . / 4 . < A o - - - Y 'vBC - v.zi rz_- oJ.+- pu , VC B= U.9952_Il3. lo pu .193 Vor= l97.Vl- 3.3" V, Voz 20.Zll51.lo V, V"o= 21.61110.63"10.4 or = 19.23_1.V A, Ioz = Ig.23lISV A,1oo= 0 A1 0 . 5 1 , , = 2 7 - 8 7 , / 3 f f A t , - - ' t ? . / - A A o . l o A r n a

^r ^A Z - L J 1 - a - ' . Z J Il , lA 0 = U A

Iobr= 16.1A, Iab2 7.5/_-75"A, Iob1=.,S/lS e,10.6 o = 16.16 j1.335 , Iu= _ 9.24_"1\O.AOe,



20-24

7.9 Total operatingcost (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

r t 0 0 Ic - - 6.93+ j9.32A, lVN,l= lVosl 40.75t0.7 1,500.2

Il

ffi4 Modern ower ystem nal'sis

Chapter 11

11. 1 - i6.56 kA , lV6,l 12.83kV, lvobl= 6.61 kV, ly"rl = 6'61 kV

= I = -2./1

(b ) V* = Vo , = 0'816pu, 116l l lr l = 5'69 pu

11.3 (i ) - i6.2.5(i i) - 4.33 (i i i ) 6'01 (iv) - /5 pu

In order of decreasingmagnitudeof line currents he faults can be listedAS

(a) LG (b) LLG (c) 3-Phased) LL

Il.4 0.1936 hm,0.581 hm, 4.33pu , 5 pu

1 1 . 5 ( a ) 3 . 5 1 p u ( b ) V t , = I . | 9 1 _ 1 5 9 . 5 " p o , V , = I , 6 8 1 | 2 9 . 8 " p t 1

(c ) 0.726Pu1 1 . 6 u = - I r = - 2 . 8 8 7 P u

||.7 (a ) v = _ 5.79+ j5.01kA , 18= 5.79+ j5.01kA , ,I G j10.02kA

(b) /n - - IY = - 6'111 A, Ic= 0

11.8 In , = 0 Io^ = - 73'51Pu

Ius= - 72.08pu Ib^= - jl.Z Piu

IrB= 72.08pu I,^ = - jl.Z Pu

11.95,266

11.102. 0 pt

11.1 / = - j6.732Pt , Io(A)= - i4.779pu ,

/a(A) = - i0.255 po, 1"(A) - - j0'255 prt

l| lz 0.42ptt, j9'256Pu

11.13 -il l. l52 pu , i2.478pu , -J l -239 n'1.14

4.737Pu ,1Pu

r 1.15 f, = - i12.547 u, Ifrr(b) - i0.0962pu

Chapter 12

l2.I 4.19MJA4VA,0.0547MJ-sec/eleceg

12.24.315MJA4VA 12.340'4 MJiTVIVA

12.4 140.1MW, 130.63 W, 175.67MW

r2.572.54MW

!2.7 L27.3MW

12.64 = 58 '

12.8 53'. We need to know the inertia constant M to determine /.'

ABCD constants 61,7for various

simplenetworks 618in power low equations lS9measurement f 621of networks n seriesand parallel 620

AC calculating oard lB4Accelerationfactor 207Acceleratingpower 462Acid rain 16Adjoint matrix 609Admittancematrix (seeBus admittance

matrix)

Advancedgas eactor AGR) 19Algorithm

for building the Zru, 355for load flow solution

by GS method 205by NR method 214by FDLF 223

for optimum generation

scheduling 262for optimal hydrothermal

scheduling 280

for optimal loadingof generators

on a bus 246for optimal load flow solution 273for shortcircuit current

computation 343for short circuit studies 349for staticstateestimation f

power systems 540for transientstability, analysis

of largesystem 4gSAlternator (seaSynchronousmachines)Aluminum conductorsteel einforced

(ACSR) conductors 52A * a a a ^ - + - ^ 1 ^ - ^ - / a ^ F \ 4 A f ,

Arvc Lrrl l l l (I El lul \f\\-.8, JU+

Armature leakage reactance 331Armature eaction 109,110,330

Index

12.9 The system s stable 12.1070.54", .1725 ec

B-coefficients 261.267

Bad data detection 547Bad data

detection 547identification 547suppression 548treatment 546

Base oad 3Basevalue 99BharatHeavy Electricals

Lr d (BHEL) 14,31Boiling water reactor 19Breaking esistors or

improving stabil iry 499Branch 190

Breakers (see Circuit breakers)Brown,H. E. 368Bundledconductors, \

capacitivereactance92inductive eactance 68

Bu s

generator 198Ioad 19 8

PQ 19 8

PV 19 8reference 198slack

Bus admittancematrix 188formulationof 187.189

Bus mpedancematrixbuilding algorithm 355for unsymmetrical ault analysis 416i n c r r a * ^ + - l ^ ^ l f ^ . . f r - - - ! - - - ' - E rrr r DJrr r r t ls t t rucu latu i l . a l |a lys ls JJ I

Bus ncidencematrix I92

Capacitance^ ^ f ^ - - 1 ^ - i - - - t - tu4ruuliluon y memoooI modrlred

geometricmeandistances 9leffect of earth on 83



Attenuationconstant 143Augmentedcost function 279Automatic generation onirol (seeLoad

frequencycontrol)Automatic ol tagc ontroi 31 8

l2.Il The system s unstable

12.13 The system is stable

t2.t2 63.36"

12.14The systems stable

effectof non-uniform

chargedistribution 79effectof strandedconductors 79l ine-to-l ine 78line-to-neutral 79

12.15The system is unstable or both three pole and single pole switching

tndex

of parallel circuit three-phasel ines 88

of three-phaseines: withequilateralspacing 80

with unsymmetricalspacing 81a two-wire ine(seealso Reactance) 78

CentralElectricity Authority (CEA) Zg

Charpcteristic (surge) impedance l4lqf .linesand cables 145Chaiging current 76, 180Circle diagrams 167Circuit breakers 327, 329

autoreclose 460rated intemrpting capacityof 344ratedmomentarycurrentof 344selectionof 344

Circulating currents 389Cogeneration 15Coherentgroup 291,303Compactstoragescheme 189,628Comparisonof angleand voltage

stability 592Compensation

series 779,498, ,Sgshunt 179,498,562

Compensator

combinedTCR and TSC .564shunt 562

staticsynchronous eries SSCC) 571Complex power 105Compositeconductors

capacitance 9l

inductance 54Computational flow chart for load flow

solution using

FDLF 227GS method 20SNR method 224

Conductors

ACSR 52

Bundled 52,68expandedACSR 52iypes 5

Conductance 45Constraints

for hydro-thermal scheduling. problems 277

Contingency

analysis 51rankins 520screening 520selection 51 5

' directmethod 515

indirectmethod 515Contingencyanalysis

by dc model 520by distribution factors 521modelling for 5I4

Contingencyevaluation(seecontingencyanalysis)

Control

area 291,303,306

by transformers 23I

integral 304isochronous 305of voltage profile 230of WAfiS and VARS alonga

transmission line 230optimal 310parameters 199,272proportioiral pft"s-fuBral 303suboptimal 318

Control areaconcept 303Control strategy 303

Control variables 199, 632Control vector 199Controller

interline power flow (IpFC) 571interphasepower (IPC) 572unified power flow (UPFC) 571

Converters 567

Ccxrrdinatisn equations 2frCorona 52,68

Cost function 251,270i , t ^ e - ^ ^ 1 n n\ -u l l ( i ( ' L> V

Covariancematrix 535Critical clearingangle and ime 467Current distribution factors 265Current limiting reactors 346

DAC (Distribution Automationand

Digiral LF controllers 322Direct axis reactance llgDisturbanceparameters 271,

272Distribution factorsgeneration_shift SZ0Iine-outage 520

Diversity factor 4Dommel, H. W. 270Doubling effect 3ZgDouble line-to-ground LLG) fault 404Driving point (self) admittance lg7Dual variables 279,632Dynamic programming

appliedto unit commitment 251

Economicdispatch seealso optimumgenerationscheduling) 306

Electricity Board ZglElements 189Energy conservation 3lEnergy control centre 637

!MS.(Energymanagement system) 634Energy sources

conventional 13hydroelectricpower generation l7nuclearpowerstations lgthermalpower stations 13

biofuels 28biomass 28gas turbines 16geotbermalpower plants 24

magnetohydrodynam c (MHDIgeneration 23

OTEC 27renewable Zssolar energy 26wave energy 27wind power stations 25

Energy storage Zgfuel cells 29hydrogen Zg

the medium length inenominal_zr representation l3gnominal-Trepresentation 137

short transmission line l}gsynchronousgenerator llg

Error amplifier 3lgError signat 319Estimationmethods

least squares 532weighted least-squares 533

Estimationof periodic components 5gExact coordinationequation 26IExpectation 532Externalsystemequivalencing 545

Facrorsaffecting srability 496FACTS controllers 569Failure rate

255Fast breeder reactor 22Fast decoupled oad flow 223Fasr valving 49,9Faults

balanced seesymmetrical aultanalysis)

calculation singZsus 351,417

openconductor 414unbalanced

series type 397shunt ype 397

Ferranrieffect 150Field erciretion l0g

Field rriading 32gFlat voltagestarr 205,209,21gFlexible AC transmission ystems

(F-ACTS) 566Fluidized-bedboiler 15Flux linkages

external 49internal 46of an solatedcurrentcarrying

conductor



equality 272, 632

inequality 632on control variables 274on dependent ariables 274

Control System) 634Damper winding 438DC network analyser 39Data acquisitionsystems DAS) 636

pumpedstorage lanf lgsecondarybatteries 2g

Equal areacriterion 461Equal incremental fuel cost criterion 246

46

,Flyball speed overnor Zgz

rorecastlngmethodology577

Fortescue,C.L. 370,3t6Frequencybias 309

Fuel costofgenerators 243

@ !no.*Full load ri,iection echnique or ; ,,

improving systemstability 500

Fusion 23

Gausselimination 623

constants)

Generator seeSynchronous enerator)

Generatoroad model 296

Generatoroperatingcost 243

Geometricmeandistance GMD)

mutual 55

self 55

Geometricmeanradius(GMR) 55

Governors,characteristics 298

deadband 32I

free operation 301

model with GRC 32I

Gradientmethod 272

Graph

connected 189

l inear 18 9

oriented 189

. theory 189

Greenhouse ffect 16

Growth of installedcapacity n India 30

Growth of power systemsn India 29

HVDC 57 2

Hydraulic'amplifier 292

Hydroelectric owergeneration l7

Hyperbolic'functions 142, 143

Ideal transformer 99

Illconditioning 229, 545

Impedance

drivingpoint 187

transfer 187Impedance iagram 98

Incidencematrix (seeBus incidence

matrix)

Incidentwaves I44

lncremental uel cost 244

Incremental ransmissionoss 260

Inductionmotors 122,344

lnductance

definition 45

of single phase wo wire line 50

of three-phaseines 59

Inequality constraints

on control variables 274

Inertia constant 436

Infinite bus Ll2

Infinite line 145

Inner product 607

Input-outputcurve of a unit 243

Interruptingcapacityof breakers 345

Interconnected ower systems 10

Interference with communication

circuits 6l-63

Irrverseof matrix 6L3

Iterative algorithm for NR methodfor load

flow studies 218

Jacobianmatrix 2I4,272

elements f 2I4,629

Kalman filtering approach 583

application to short term loadforecasting 583

Kimbark, E.W. 508

Kinetic energy 435

Kirchhoff's current aw (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 reactance331

Leakance 129

Lcastsquares stimation 532

the basic solution 532

Linear graph (seeGraPh)

Line charging current (seecapacitance)

Link 190

Line-to-line (LLXault 402

Line compensationL - . - ^ - : ^ ^ ^ - ^ ^ l + ^ - - 1 ? O / ' l OQuy ugrr r rJ ualP4r / r turJ r t 7, 1/v

by shuntcapacitors 179,498

Loadsee power angle;

delta connected 374

representation 122symmerrical balanced) lO7unbalanced 394

Loading capability 557Load factor 3Load flow problem 196Load flow methods

comparisonof 229continuation 599decoupled 222FDLF 223

Gauss 205

Gauss-Seidel204

Newton-Raphson 213decoupled 223rectangularversion Z2I

Load forecasting 9Load forecasting echniques

estimationof averageandtrend terms 577

estimationof periodic

components 581incorporationof weathervariables 5g3time seriesapproach 592

autoregressivemodels Sgzautoregressive oving-average

models582Load frequencyconrrol (LFC) and

economicdispatchcontrol 305decentralizedcontrol 323digi tal 32 2

single areacase 29Itwo area 307with cRC 320

Load management 33Load prediction

on-line techniques586A D f l t A ^ ^ ) ^ r ^ E o z

A l \ l l Y l . l 1 r r l uug l l t Joo

non-dynamicmodel 596time varyingmodel 5g6

'long term

generation 260assumptionsn calculating 266,269equation for 267

Maintenance 32Magnetohydrodynamic(MHD)generation 23

Magnetomotive orce aroundclosedpath 47

MATLAB

examples 651introduction to 640programs 652programming 640

Matrices

additionof 610associativeaw 6l lcommutative aw 611

equalityof 610multiplicationof 611associativeaw 6lIdistributive aw 611

singularand nonsingular(seealso matrix) 610

Matrix

definition'607

diagonal 608elementof 607identity 609null 608

symmetric 609transposeof 609

uni t 50 8zero (seea/soBus admittancematrix:Bus impedancematrix;

m o t r i ^ ^ ^ \ z A orrElr rvgJrf Ul.,O

Medium transmissionine 137Methodof images g3

Methodsof voltagecontrol 173Mini hydelplants 18Micro hydelplants lgModel

Lossesas unctionof



due o internal lux 46

mutual 46

of composite onductorines 54

of doublecircuit hrec-phaseine 66

angle also

torque angle) 110

bus 198

oontro I I

compcnsation 557

587usingeconometricmodels5g7

Long transmissionine equations l4Zcvaluntion f ABCD con$tsnts l4 2hypcrbolic ormof l4Z

generatoroad 42of speedgoverningsystem 2g3turbine Z9 S

Modil 'icdEuler'smethod or stabil ityanalysis 495

Modificationof Z for networkBU S

changbs 60

Mutual (transfer) admittance I87

Mutual coupling 46

NationalThermal Power Corporation

(NTPC) 4r

Negativesequence omponents 370Negativesequencempedance 383

of passive lements 383

of synchronous achines 386

table 334

of transformers 390

of transmissionines 390

Negativesequence etworks

in fault analysis 399

examples 409

of powersystems 386

of unloaded enerators 388

of Zr* elements 417

Network

branch 190

model ormulation or SLFE 185

node 90

observability 545, 549primit ive 19 1

Nodes 19 0

Node elimination echnique or stability

study 444

Nominal- t circuit of medium

transmission ine 138

Nominal-T representation f medium

transmissioninc 137

Non-linearprogrammingproblem 632

Non-linearprogramming echnique 279

Nuclear eactors 2I

Objective unction 270, 27I, 275

Off-nominal urn (tap) ratio or off-nominal

tap setting 234

One lincdiugrtnr 98

On-l ine- + ^ + ^ ^ . + 1 . - ^ + i ^- < A AJ l c t l l y \ , D L l l l l d l l u t l J1 a

techniquesor non-stationary

load prediction 586

ARMA models 5E6

Optimal (two area) oad requency

control 310

Optimal ordering I89, 627

Cptimal operation of generatorson a bus

Optimal operatingstrategy 242

Optimal reactivepower low

problem 272

Optimal scheduling of hydrothermal ,system 276

Optimal unit commitment 250

Optimal securityconstrained nit

commitment 256

Optimumgeneration cheduling 259

Optimum real power dispatch sZe

optimum generationscheduling)

Parameterestimation 552

Parametersof overhead ines 45

Patton,A.D. 254,256

Penalty actor 260,275

Penalty unction 270,275

Percentvalue 99Per unit system

change f base 101

definitioir of 99

selection f base 100

Performancechart of synchronous

machines 11 9

Performance ndex GI) 316

Phaseconstants 143

Phaseshift in star-delta

transformers 377

Planningcommission 4L

z'-equivalent

for long ine 152

for medium ength ine 138zrmodel for transformerwith off-nominal

turns ratio 234

Plantcapacity actor 4

Plnntus e actor 4

Photovoltaicgeneration 2l

Pilct vulve 29 2

Point-by-point olutionof swing

equation 48 1

Positive sequencecomponents 370

of transformer

of transmission

Positive sequence

in fault studies

39 0

lines 389

networks

39 8

of Zsu5elements 417

Power

accelerating 46?

angle 437_apparent 106

by symmetric components 374

complex 105

factor 105, 112, Il5

flow through a transmissionine 158

in three-phaseircuits Il2

maximum transmitted I59

real and reactive(seealso Reactor

power;Realpower 106

Poweranglecurve 1I5,444

Power system

compensation 556

engineer 39

state estimation 538structure 10

studies 39

Primitive impedance admittance)

matrix I92

Primitive network (seeNetwork) I9l

Propagationconstant l4l

Proportionalplus integralcontrol 303

Proximity effect 7 L

Pumpedstoragescheme 18

Qurdrrturcaxissynchronous

reactance I1 7

Rcuctivc owcr 10 6

control of flow: by machine

excitation ll3, I14

injectionof 175

<l l ' ynchronous achincs

with ovcr cxcitutiou l 14

with underexci tat ion 11 4- - l - - r ! - - - , r ^ : - ^ ^ , , : r l ^ . . . . 1 r - - . ^ - l - - , . . l 1 4 4

t -ctauunsntpwi l ,u vurruBc cvsl L J

s i gn o f l 06

Real power 10 6

f'ormulas for 159, 160

directaxis 118.12 1

leakage,of transformers 101

of bundledconductors

inductive 67

quadratureaxis Ll7

subtransient 332

synchronous111,332transient 332

Reactance iagram 98

Reactive oad forecast 587

Reclosure 474

Reference us 198

Reflectedwaves 144

Regulating ransformerssee Transformer,

regulating)

Regulating onstantR) 294

Regulationof voltage 130

Reheat urbine model 295

Reliability considerationsor economic

dispatch 253

Reluctance ower . 118Repair rate 255

Reserve 254 \

Resistance f lines 7O

RotatingVAR generator 176

Rigid l imits 275

Salientpole synchrono-usenerator ll7

Saturated eactor 563

SCADA 634

Scalar product 607

Sctrlnr ndv0ctor unctions 61 4derivativesof 614

Securityanalysis 256

Scnsi t iv i ty uctorsgenerationshift factors 520line outagedistribution actors 520

Seqtiencempedances

l i rr synchronous achines 3ti5ol passive lements 38 3

of transformers 386

of transmissionincs 3E 5table br typicalvalues f 334

Sequencempedancesnd

Reactance



rime r-ar1-ingmcxlel -s36

Apaaror a 370

Wn l cstrrdla 310

ft'dmrl hrsd fltrrr' srrludon :?0

Positivesequencempedance

of pa-ssive lements i8l

of syfrrwus nrchirc,s 3E3

table 334

in sr nchrtrn(lus nt&('hin('s I -5. I l9

Real dme computer control of

Ww'q $-r-Silems 6Y

sequence etworks

ul ptr$E'rs\stL'nl iss

of syrrchrowus mar.+tiffi 3t5

of transformers 390

i

lndex lndex

Sequencenetviorkconnections orunsymmetrical aults 4OO, 03,4A5

Sequen,:enetworksconstructionof sequence etworks

Short circuit current imiters 499Short circuit megavoltamperes 345Short circuit ratio (SCR) 498

Short term load forecastingestimatingof stochastic

component 583innovationmodel approach 588using Kalman filtering

approach 58 3using the periodic oad model 581

Short transmission ine equations 12,9Short transmissionine equivalent

ci rcui t 12 9

Shunt capacitors

effect upon voltage egulation 132for power factor correction 133fo r voltagecontrol 175

Simulink 640basics 648

example 648Single line-ro-ground LG) fault 399Skin effect 71

Slack bu s 19 8

Soft l imi t 27 5Solid LG fault 407Sparsematrix 18 8Sparsi ty 188, 89 ,62 3Speedchanger 293Speedgoverningsystem 293Speed egulationconstant see egulation

constant)

Spinning reserve 245,254Stabi l i ry

analysis or multimachine ystem 497definition 433dynamic 434study 461

steady state 454

transient,defined 434clearing angle 466clearing time 466

some factois affecting 496vol tage 59 1

Stability imit

steadysrate 456

Start up cost 250Statcbm 562Stateelectricity boards 303

Stateestimationexternalsystemequivalencing 545of powersystems 538

the njectionsoniy algorithm 540the line only algorithm 543

nonlinearmeasurements535problems f i l lconditioning 545stochastic omponent 583

Statevariablemodel 3lzStatevariables 199,312, 494Static oad low equarions SLFE) 198Static reservecapacity 254StaticVAR generator 175,565StaticVAR sysrem 562,565

State-vector 19 8Steadystare rabil iry 454-Step-by-stepofmulationof bus

impedancematrixadditionof a branch 357additionof a link 359

Stiffness f sy,nchroneus achine 455Steepest escentmethod see Gradient

method)

Suboptimal onrrol 318Substation

bulk powcr 13distribution l3

Superconductingmachine4

Surge mpbdance seealso characteristicimpedance) 14 5

SVC (Static var compensator) 562.565Swing bus seeSlack bus)Swingcurve 459,495

determination sing modifiedEuler'sm arhn- | , { n<t l l vL t l vu a7 J

step-by-stepetermination f 481Swingequation 438, 480, 494

statevariable orm 495

SymmetricalcomPonets

definitionof 369

of unsymmetricalPhasors 372

power n terms of 374

algorithm for large sYstems stng

zBUs 50

selectionof circuit breakers 344

shortcircuit current computation

through he Thevenin's

theorem 341

shortcircuitof a synchronous achine

on load 339

on no oad 330

transient n a transmissionine 320

Symmetrical hort circuit

current 329, 331

Synchronisingower

coefficient I 18, 308, 455

Synchronouseactance 3

Synchronous achine

armature eaction 108

construction 108

dynamicsof 435

equiva lenti rcu i t ll l

exci tat ion 113

inertiaconstants f 435

leakage eactance 331

ioadangleof I 10

operating performance) hart ll9-'

n(qr. of valuesof sequencempedances

of 38 5

pe r unit reactance,able 334

phasor iagrams 11 2

T circuitof medium ength ine 137

Taylor series xpansion 213TCUL (tapchangingunder oad) see

transformers)

Thevenin's heorem

in calculationof three-phaseault

currents 341, 35O

in caleulation f unsymmetricalault

currents 41 6

Threephase ircuits

power in lO7

transformer (TCPST) 572

Thyristor controlled reactor (TCR) 563

Thyristor controlledvoltage egulatort.T.rl!.f D \ {?t\ r L v r\,, J t -

TSC) 56 4

Tie line power 307

Tolerance 206

Torque angle seealsoPower angle) 438Tracking stateestimationof power

systems 544

Transformers

control (see egulatingbelow) 231

off-norninal tapes 234

per unit representationf lOl

phase hift n Y-A 377

phaseshifting 232

polarity markings 377

regulat ing 23 2

equivalent ircuit of 234

fo r magnitude ontrol 232

for phaseanglecontrol 233

sequencempedancesof 385TCUL 177.232

Triangularisation nd back\

substi tut ion 62 3

Transient eactance 332

Transientstabil ity

definitionof 433

digitalcomputer olutionmethod 610

equalarea ri terion 46 1

Eulermodifiedmethod 495

mult igeneratorcase48 7

numcrical olut ion f swingcquation

for singlegenerator ase 480

Transmissioncapabi l i ty12

Transmissionevel l3Transmissionoss seeLossesas function

of plantgeneration)

Transmissionoss ormula 261

derivationof 264

Transposition f transmissionines 60t n l r a l a n n a n q n q n i f q n n e fl ,lv vs l s r rvv

to balancenductance 6l

Tree 190

Tunedpower ines 15t



critical clearingangle 467critical clearing time 467equal areacriterion 461point-by-pointmerhod 4g l

Switching

singlepole 500threepole 500

Symmetricmarrix 609

voltageand current n 107

Threephase aults (seeFaults)

Thyristorcontrolledbraking esistor 577.

Thyristorcontrolledphaseshifting

Turbine ast valving 499

Turbine model 295

Turbine speedgoverningsystem 292

Two-area load frequencycontrol 307

Two reaction theory Il7

Unit comrnitment 250branch and bound technique 250,

252

dynamicprogrammingmethod Z5lreliability (security)

considerations253securityconstrained 256

stait-up consideration Z5gUnit matrix 609Unsyrirmetrical aults (seealso Faults)

analysisu,singZsu5 rhbthod 416

symmetriCal component analysisof 39 8

Use of computersand microprocessorsin power systems',39

VAR (see also Reactive power)Variance 582Vector

column 607

control (seecontrol vector)disturbance 27lequation 271

nul l 60 5

of fixed parameters Ig9orthogonal 607

row 607state see Statevector)sum 606

unit 606Velocity of propagation 146Voltage

collapse 528, 592preventionof 526,593, 600

control by transformers i'77control by VAR injecrion I75eff'ect f capacitors n l3Z, 175

equations: or long line 142for medium transmission ines13 7 t3 9

for short transmission ines l/g

stability 524, 592methods or improVing 5Zg

Voltagecontrolledbus 198

Voltage regulation 130Voltage stabil ity 591

analysis 592,5glcriteria of 596with IdVDC links 599

Watts (see qlso Real power) 106Wavelength 145Waves, ncidentand reflected 144, I45Weighted eastsquaresestimation 533Weighting marrix 534

Y"r, (seeBus admittancematrix)

Y-A transfonners 96phaseshift in 377single-phase

quivalentof 96zero sequence etwor[s of 397Z^.,^ (seeBus mpedancematrix)

BUS

./'Zero sequence omponents 371Zero sequencempedance

of circuit elements 384of synchronousmachines 384

table' 334of transformers 387of transmissionines 3g5

Zero sequencenetworksin fault studies 399of deltaconnectedoads 389

of transformers 387of unloadedgenerators 384of Y-connecredoads 397

Index

modern power systems analysis d p kothari i j nagrath

Documents

power system engineering

power reforms

power systemsare

economics of power

power system transients

power system security

modern power systemshave

system operation