PowerSystemAnalysis "This page is Intentionally Left Blank"Power System Analysis Prof. P.S.R.Murthy B.Sc.(Engg.)(Hons.)ME., Dr.- ING (Berlin),F.I.E.(India) LifeMember - ISTE (Formerly Principal O.U. College of Engineering & Dean, Faculty of Engineering, O.U. Hyderabad) Principal, Sree Datha Institute of Engineering and Science, Sheriguda, IbrahimPatnam,Hyderabad (AP). BSPBS Publications 4-4-309, Giriraj Lane,Sultan Bazar, Hyderabad - 500095 A.P. Phone: 040-23445688 Copyright2007, by Publisher Allrights reserved No part of this book or parts thereof may be reproduced, stored in a retrieval system ortransmittedinanylanguageorbyanymeans,electronic,mechanical, photocopying,recordingorotherwisewithoutthepriorwrittenpermissionof the publishers. Published by: SSPBS Publications 4-4-309, Giriraj Lane, Sultan Bazar, Hyderabad - 500 095 A. P. Phone: 040-23445688 e-mail: contactus@bspublications.net www.bspublications.net Printed at: AdithyaArt Printers Hyderabad. ISBN: 978-81-7800-161-6 Preface Power Systemanalysisisapre-requisitecourseforelectricalpower engineering students. InChapterI.introductoryconceptsaboutaPowersystem,networkmodels, faultsand analysis;theprimitive network andstabilityarepresented. Chapter 2 deals with the graph theory thatisrelevant tovariousincidence matrices required for network modelling areexplained. Chapter 3explainsthevariousincidencematricesandnetworkmatrices. Chapter 4discusses,step-by-stepmethodof buildingof networkmatrices. Chapter 5 dealswithpower flowstudies.Both Gauss-SeidelmethodandNewton-Raphson methodsareexplained.InNewton-RaphsonmethodboththeCartesioncoordinatesmethod andpolar coordinatesmethodsarediscussed. Inchapter 6shortcircuit analysisisexplainedPerunit quantityandpercentage values are defined.Analysisforsymmetricalfaultsisdiscussed.Theutilityof reactorsforbusbar and generator protectionis also explained. Unbalancedfaultanalysisispresentedinchapter7.Useof symmetricalcomponentsand networkconnectionsareexplained. Chapter8dealswiththepowersystemstabilityproblem.Steadystatestability.transient stability anddynamicstability arediscussed. Itisearnestlyhopedthatthisbookwillmeettherequirementsof studentsinthesubject powersystemanalysis. - Author "This page is Intentionally Left Blank"Acknowledgment MysincerethanksgotoMr.NikhilShah,Proprietorof BSPublicationsforhisconstant encouragement tometowriteand completethisbook on'Power System Analysis'. Mythanksalsogo toMr.M.Y.L.Narasimha Rao,mywellwisher fortaking great pains in transferring script material and proof materialbetween myresidence and the press with a great smiley face,dayandnight over thelast fewmonths. I thankMrs.Swarupa forMATLABassistance. -Author "This page is Intentionally Left Blank"Contents Preface .................................................................................................................. (vii) Acknowledgment................................................................................................... (ix) 1Introduction I. ITheElectricalPower System ..................................................................I 1.2NetworkModels ......................................................................................3 1.3Faults and Analysis..................................................................................3 1.4ThePrimitiveNetwork ............................................................................4 1.5PowerSystemStability...........................................................................5 2GraphTheory 2.1Introduction ............................................................................................. 6 2.2Definitions ................................................................................................6 2.3TreeandCo-Tree ....................................................................................8 2.5Cut-Set ....................................................................................................9 2.4BasicLoops.............................................................................................9 2.6BasicCut-Sets.......................................................................................10 vVorked Examples..'......II Prohlems.......IS Questions16 (xii)C o n t e n t ~ 3IncidenceMatrices 3.1Element NodeIncidence Matrix ............................................................18 3.2Bus Incidence Matrix............................................................................18 3.3Branch- PathIncidenceMatrixK.......................................................19 3.4BasicCut-Set IncidenceMatrix............................................................20 3.5BasicLoop Incidence Matrix .................................................................21 3.6NetworkPerformanceEquations ..........................................................22 3.7NetworkMatrices ..................................................................................24 3.8Bus Admittance Matrix andBusImpedance Matrix.............................25 3.9Branch AdmittanceandBranchImpedanceMatrices ...........................26 3.10LoopImpedance and Loop Admittance Matrices.................................28 3.11Bus Admittance Matrix by Direct Inspection ........................................ 29 Worked Examples....................... ....... ............. ................................33 Problems........................... : .....................................................51 Questions .......... ...........................................'"..........................52 4Buildingof Network Matrices 4.1PartialNetwork ......................................................................................53 4.2Addition ofa Branch ..............................................................................55 4.3Addition ofa Link ..................................................................................60 4.4Removalor ChangeinImpedanceof Elementswith MutualImpedance .................................................................................66 Worked Examples........."... 70 Problems... .............. :............ ..... 96 Questions.. ............. .. .............97 5PowerFlowStudies 5.1NecessityforPowerFlowStudies.......................................................98 5.2ConditionsforSuccessfulOperationof aPowerSystem ....................99 5.3ThePowerFlowEquations ...................................................................99 5.4Classification of Buses........................................................................101 5.5Bus Admittance Formation..................................................................102 5.6SystemModelforLoadFlowStudies................................................104 5.7Gauss-SeidelIterativeMethod............................................................105 5.8Gauss - SeidelIterativeMethodof LoadFlowSolution ....................106 5.9Newton-RaphsonMethod...................................................................109 5.9.1TheRectangular CoordinatesMethod ..................................110 5.9.2The Polar Coordinates Method............................................112 Contents(xiii) 5.10Sparsityof Network AdmittanceMatrices ..........................................115 5.11Triangular Decompostion ....................................................................116 5.12Optimal Ordering .................................................................................118 5.13DecoupledMethods .............................................................................119 5.14FastDecoupledMethods .....................................................................120 5.15LoadFlow SolutionUsing Z Bus .....................................................121 5.15.1BusImpedanceFormation...................................................121 5.15.2Addition of aLine to theReference Bus..............................122 5.15.3Addition of aRadialLine and New Bus...............................122 5.15.4Addition of aLoop Closing TwoExisting BusesintheSystem.............................................................123 5.15.5Gauss - SeidelMethodUsing Z-busfor LoadFlow Solution..............................................................124 5.16ConvergenceCharacteristics ...............................................................124 Worked Examples...........126 Problems..............161 Questions....................................... ........................175 6Short Circuit Analysis 6.1Per Unit Quantities ...............................................................................176 6.2Advantagesof PerUnitSystem..........................................................178 6.3ThreePhaseShortCircuits.................................................................178 6.4ReactanceDiagrams ............................................................................181 6.5Percentage Values................................................................................181 6.6Short Circuit KVA................................................................................182 6.7Importanceof ShortCircuitCurrents .................................................183 6.8Analysis ofR-L Circuit ........................................................................184 6.9ThreePhaseShort Circuit onUnloadedSynchronous Generator ......185 6.10Effect of LoadCurrentorPrefault Current ........................................185 6.11Reactors ...............................................................................................186 6.12Constructionof Reactors....................................................................186 6.13Classification of Reactors ....................................................................187 Worked Examples.......................... .. .. ...........189 Problems. ................................... .. ............216 Questions............................................................................2 I 6 (xiv)Contents 7UnbalancedFault Analysis 7. ITheOperator "'a'".................................................................................218 7.2SymmetricalComponentsof UnsymmetricalPhases .........................219 7.3PowerinSequenceComponents........................................................221 7.4UnitaryTransformationforPowerInvariance...................................222 7.5SequenceImpedances ......................................................................... 224 7.6Balanced Star ConnectedLoad...........................................................224 7.7TransmissionLines ..............................................................................226 7.8SequenceImpedancesof Transformer ...............................................227 7.9SequenceReactancesof SynchronousMachine ................................ 228 7.10SequenceNetworksof SynchronousMachines .................................228 7.10.1PositiveSequence Network ..................................................228 7.10.2NegativeSequence Network ................................................229 7.10.3ZeroSequenceNetwork .......................................................230 7.11UnsymmetricalFaults ..........................................................................231 7.12AssumptionsforSystemRepresentation............................................232 7.13UnsymmetricalFaults on anUnloaded Generator ...............................232 7.14Llne-to-Line Fault ................................................................................235 7.15Double Line toGroundFault ...............................................................238 7.16Single-Line toGround Fault withFault Impedance ............................241 7.17Line-to-Line Fault withFault Impedence ............................................242 7.18DoubleLine-to-GroundFault withFaultImpedence ..........................243 Worked Examples.............................. .............................. ..............245 Problems....................................................................... ........257 Questions . ......................................................................................257 8Power SystemStability 8.1Elementary Concepts ........................................................................... 259 8.2Illustrationof SteadyState Stability Concept.....................................260 8.3MethodsforImprocessingSteadyStateStabilityLimit .....................261 8.4SynchronizingPowerCoefficient .......................................................262 8.5Transient Stability ................................................................................262 8.6Stability ofa Single Machine Connected to lnfinite Bus....................262 8.7The SwingEquation............................................................................263 8.8Equal Area Criterion and Swing Equation........................................... 267 Contents(xv) 8.9Transient Stability Limit......................................................................269 8.10Frequency of Oscillations ....................................................................270 8.11Critical Clearing Time and Critical Clearing Angle ..............................272 8.12Fault on aDouble-Circuit Line ............................................................274 8.13Transient Stability WhenPower isTransmittedDuring theFault ......275 8.14Fault Clearance andReclosureinDouble-Circuit System ..................277 8.15SolutiontoSwing EquationStep-by-Step Method.............................277 8.16Factors Affecting TransientStability..................................................279 8.17Dynamic Stability................................................................................280 8.18Node Elimination Methods ...................................................................282 Worked Examples............. . , ......................... ........................285 Problems..... ,........................................................................303 Questions.......................................... ................................ ......304 ObjectiveQuestions .................................................................................305 AnswerstoObjectiveQuestions ............................................................ 317 Index...........................................................................................................319 "This page is Intentionally Left Blank"1 INTRODUCTION Power isanessentialpre-requisite fortheprogressof any country.The modernpower system hasfeaturesuniquetoitself.It isthelargestmanmadesysteminexistenceandisthemost complexsystem.Thepowerdemandismorethandoublingeverydecade. Planning,operationandcontrolof interconnectedpowersystemposesavarietyof challengingproblems,thesolutionof whichrequiresextensiveapplicationof mathematical methodsfromvariousbranches. Thomas AlvaEdison was the first to conceive anelectric power station and operate itin Newyork in1882. Since then, power generation originally confined tosteam engines expanded using(steamturbines)hydroelectricturbines,nuclearreactorsandothers. The inter connection of the various generating stations toload centers through EHV and UHVtransmissionlinesnecessitatedanalyticalmethodsforanalysingvarioussituationsthat ariseinoperationandcontrolof thesystem. Powersystemanalysisisthesubjectinthebranchof electricalpowerengineering whichdealswiththedeterminationof voltagesat variousbuses andthe currents that flowin the transmissionlines operating at different voltagelevels. 1.1The Electrical Power System The electrical power system isa complex network consisting of generators, loads, transmission lines,transformers,buses,circuit breakers etc.For the analysis of a power systeminoperation 2PowerSystemAnalysis a suitable modelisneeded.Thismodelbasically dependsupon the typeof problemonhand. Accordinglyit maybe algebraic equations,differentialequations, transfer functionsetc.The power systemisneverinsteady state astheloadskeepchanging continuously. However,itispossibletoconceiveaquasistaticstateduringwhichperiodtheloads could be considered constant.This period could be15to 30 minutes.Inthis state power flow equationsarenon-linear duetothepresenceof product termsof variablesandtrigonometric terms.The solution techniquesinvolvesnumerical(iterative)methods forsolving non-linear algebraicequations.Newton-Raphsonmethodisthemostcommonlyusedmathematical technique.Theanalysisof thesystemforsmallloadvariations,whereinspeedor frequency and voltage control maybe required to maintain the standard values, transfer function and state variable models are better suited to implement proportional, derivative and integralcontrollers oroptimalcontrollersusingKalman'sfeedbackcoefficients.Fortransientstabilitystudies involvingsuddenchangesinloadorcircuitconditionduetofaults,differentialequations describing energy balance over a few half-cycles of time period are required.For studying the steady stateperformancea number of matrixmodelsareneeded. ConsiderthepowerSystemshowninFig.1.1.Theequivalentcircuitforthepower systemcanberepresentedasinFig.1.2.Forstudyof faultcurrentstheequivalent circuitin Fig.1.2can be reduced to Fig.1.3upto the load terminals neglecting the shunt capacitances of the transmissionlineandmagnetizing reactancesof thetransformers. Generators Generators Sending endstep up transformer Sending end Transformer Transmission Line .;;>----+-+ Loads Receiving end step-down transformer Fig. 1.1 Transmission Lines Fig.1.2 Receiving end transformer Load Introduction TransmIssIon Lines Fig.1.3 3 Tran,former Whilethereactancesof transformersandlineswhicharestaticdonotchangeunder varyingconditionsof operation,themachinereactancesmaychangeandassumedifferent valuesfotdifferentsituations.Also,compositeloadscontaining3-phasemotors,I-phase motors,d-cmotors,rectifiers,lightingloads,heaters.weldingtransformersetc.,mayhave verydifferentmodelsdepending uponthe compositionof itsconstituents. Thecontrolof aturbogeneratorsettosuittothevaryingloadrequirementrequiresa model.For smallvariations,alinearizedmodelisconvenient tostudy.Such amodelcan be obtainedusingtransferfunctionconceptandcontrolcanbeachievedthroughclassicalor moderncontroltheory.This requiresmodeling of speed governor,turbo generator andpower systemitself asalltheseconstitutethecomponentsof afeedbackloopforcontrol.The ultimateobjectiveof powersystemcontrolistomaintaincontinuoussupplyof powerwith acceptable quality.Qualityisdefinedintermsof voltageandfrequency. 1.2Network Models Electricalpowernetworkconsistsof largenumberof transmissionlinesinterconnectedina fashionthatisdictatedbythedevelopmentof loadcenters.Thisinterconnectednetwork configurationexpandscontinuously.A systematicprocedureisneededtobuildamodelthat canbeconstantlyup-gradedwithincreasinginterconnections. NetworksolutionscanbecarriedoutusingOhm'slawandKirchoff'slaws. Eithere=Z. i ori =Y. e modelcanbeusedforsteadystatenetworksolution.Thus,itisrequiredtodevelopboth Z-bus and V-busmodelsfor thenetwork.Tobuild suchamodel,graphtheoryandincidence matriceswillbe quite convenient. 1.3Faults and Analysis Studyof thenetworkperformanceunderfaultconditionsrequiresanalysisof agenerally balancednetworkasanunbalancednetwork.Underbalancedoperation,allthethree-phase voltages are equalinmagnitude and displaced fromeach other mutually by1200 (elec.).It may be noted that unbalanced transmission line configuration is balanced in operation by transposition, balancing the electricalcharacteristics. 4PowerSystemAnalysis Under fault conditions, the three-phase voltages may not be equal inmagnitude and the phase angles toomay differ widely from1200 (elec.) evenif the transmission and distribution networksarebalanced.Thesituationchangesintoacaseof unbalancedexcitation. Network solution under these conditions can be obtained by using transformed variables throughdifferent component systemsinvolving theconceptof powerinvariance. Inthiscoursealltheseaspectswillbedealtwithinmodelingsothatatanadvanced level, analyzing anddeveloping of suitable controlstrategies could be easilyunderstoodusing thesemodelswherevernecessary. 1.4The Primitive Network Networkcomponentsarerepresentedeitherbytheirimpedanceparametersoradmittance parameters.Fig (1.4)represents theimpedance form,thevariables are currents and voltages. Everypower system element canbedescribedby aprimitive network.A primitivenetwork is asetof unconnectedelements. a 8 b Zab v1 eab ~ I iab Vb Fig. 1.4 aandb are the terminalsof anetwork element a-b.VaandVbarevoltages ata andb. Vabisthevoltageacrossthenetwork element a-b. eab isthesourcevoltageinserieswiththenetwork element a- b zabistheself impedanceof network element a-b. jabisthecurrentthroughthenetworkelement a-b. FromtheFig.( 1.4) wehavetherelation Vab + eab =zab iab..... (1.1) IntheadmittanceformthenetworkelementmayberepresentedasinFig.(1.5) . .lab +-(]) ab Yab Var ~ rVb lab ~ .+' lab.lab Vab=Va-Vb Fig.1.5 Introduction5 Yabistheself admittance of thenetwork element a-b jabisthe source current inparalielwith the network element a-b FromFig.( 1.5)we havetherelation iab + jab =Yab Vab..... ( 1.2) The series voltage in the impedance form and the parallel source current in the admittance formare relatedby the equation. -Lb=Yabeab..... (1.3) A set of unconnected elements that are depicted in Fig.( 1.4) or (1.5) constitute a primitive network.The performanceequationsfortheprimitivenetworksmaybeeitherintheform ..... (1.4) or in theform i+ j=[y]y..... (1.5) Ineqs.{lA)and(1.5)thematrices[z]or[y]containtheself impedancesorself admittances denoted by zab, abor Yab,ab'The off-diagonal elements may ina similar way contain themutualimpedances or mutual admittances denotedbyzab, cdor Yab,cdwhere abandcdare two different elements having mutual coupling.Ifthere is no mutual coupling, then the matrices [z]and [y]are diagonalmatrices.While in general[y]matrix canbe obtained byinverting the [z]matrix,whenthereisnomutualcoupling,elementsof [y]matrixareobtainedbytaking reciprocalsof theelementsof [z]matrix. 1.5Power System Stability Powersystemstabilityisawordusedinconnectionwithalternatingcurrentpowersystems denotingaconditionwherein,thevariousalternatorsinthesystemremaininsynchronous witheachother.Studyof thisaspectisveryimportant,asotherwise,duetoavarietyof changes,suchas,suddenloadlossorincrement,faultsonlines,shortcircuitsatdifferent locations,circuitopeningandreswitchingetc.,occuringinthesystemcontinuouslysome whereor othermaycreateblackouts. Study of simple power systems with single machine or a group of machines represented bya single machine, connected toinfinitebus gives aninsightintothestabilityproblem. At a firstlevel,study of these topicsisveryimportant forelectricalpower engineering students. 2GRAPH THEORY 2.1Introduction Graph theory hasmany applicationsinseveralfieldssuch as engineering, physical, social and biologicalsciences,linguistics etc.Anyphysicalsituation thatinvolves discrete objects with interrelationships canberepresentedbya graph.InElectrical Engineering Graph Theoryis used topredict thebehaviour of thenetwork inanalysis.However, for smaller networks node or meshanalysisismoreconvenient thantheuseof graphtheory.It maybementioned that Kirchoff wasthefirsttodeveloptheoryof treesforapplicationstoelectricalnetwork.The advent of high speed digital computers has made itpossible to use graph theory advantageously forlargernetworkanalysis.Inthischapterabrief accountof graphstheoryisgiventhatis relevanttopower transmissionnetworksandtheiranalysis. 2.2Definitions Elemellt of aGraph:Eachnetworkelementisreplacedbyalinesegmentoranarcwhile constructingagraphforanetwork.Eachlinesegmentorarciscailedanelement.Each potentialsourceisreplacedbyashortcircuit.Eachcurrentsourceisreplacedbyanopen circuit. Nodeor Vertex: The terminalof anelement iscalled a node or avertex. T.dge:Anelement of a graphiscalled anedge. Degree:Thenumber of edges connected to avertexor nodeiscalleditsdegree. GraphTheory7 Graph: Anelementissaidtobeincident onanode,if thenodeisaterminalof theelement. Nodescanbeincident tooneor moreelements.Thenetworkcanthusberepresentedbyan interconnectionof elements.Theactualinterconnectionsof theelementsgivesagraph. Rank: Therankof a graphisn - Iwherenisthenumber of nodesinthegraph. Sub Graph: Anysubset of elements of thegraphiscalleda subgraphA subgraphissaidto beproperif itconsistsof strictlylessthanalltheelements andnodesof thegraph. Path:Apathisdefinedasasubgraphof connectedelementsSLlchthatnotmorethantwo elementsareconnectedtoanyone node.If thereisapathbetweeneverypairof nodesthen the graphissaid tobe connected.Alternatively,a graphissaid tobe connectedif there exists atleastonepathbetweeneverypair of nodes. PlanarGraph:Agraphissaidtobeplanar,if itcanbedrawnwithout-outcrossoverof edges.Otherwiseitiscallednon-planar(Fig.2. J). (a)(b) Fig. 2.1(a)Planar Graph (b) Non-Planar Graph. Closed Pathor Loop: Theset of elementstraversedstarting fromonenodeandreturning to thesamenodeformaclosedpathorloop. OrientedGraph:Anorientedgraphisagraphwithdirectionmarkedforeachelement Fig. 2.2(a) shows the single line diagram of a simple power network consisting of generating stations.transmissionlinesandloads.Fig.2.2(b)showsthepositivesequencenetwork of thesysteminFig.2.2(a).TheorientedconnectedgraphisshowninFig.2.3forthe samesystem. r-------,3 (a)(b) Fig. 2.2 (a)Power system single-line diagram(b) Positive sequence network diagram PowerSystemAnalysis 5 ~ CD.---+-....... - - - 1 ~ - - - Q ) 4 Fig. 2.3Oriented connected graph. 2.3Tree and Co-Tree Tree: A treeisanorientedconnected subgraphof anorientedconnectedgraph containing all thenodesof thegraph,but,containingnoloops.Atreehas(n-I)brancheswherenisthe number of nodes of graph G.The branches of a tree are called twigs.The remaining branches of thegrapharecalledlinksor chords. Co-tree:Thelinksformasubgraph,not necessarily connectedcalledco-tree.Co-treeisthe complementof tree.Thereisaco-treeforeverytree. Foraconnectedgraphandsubgraph: I.There existsonly onepathbetween anypair of nodesonatree ')everyconnectedgraphhasatleastonetree 3.every treehastwoterminalnodes and 4.therankof atreeisn-Iandisequaltotherank of thegraph. Thenumber of nodesandthenumber of branchesinatreearerelatedby b =n-I..... (2.1 ) If eisthe totalnumber of elements then the number of links I of a connected graphwith hranchesbisgivenby I==e-h Hence,fromeq.(2.1).itcanbewrittenthat l=e-n+1 ..... (2.2) ..... (2.3) Atreeandthecorresponding co - tree of the graph for the system showninFig.2.3are indicatedinFig.2.4(a)andFig.2.4(b). 5 ._-...----..../ . ~~3 "-CD-..-/ 2..('4 / / @ n=number of nodes =4 e=number of elements =6 b=n-I=4-1=3 I=e-n+I=6-4+1=3 Fig. 2.4 (a)Tree for the systeminFig.2.3. GraphTheory9 5 CD---+-..... -+-...-: @ Fig. 2.4 (b)Co-tree for the systeminFig.2.3. 2.4Basic Loops 1\loop isobtained whenever a link isadded to a tree,whichisa closed path.As an example to thetreeinFig.2.4(a)if thelink6isadded,aloopcontaining theelements1-2-6isobtained. Loopswhich contain only one linkarecalledindependent loopsor hasic loops. It canbe observed that thenumber of basicloopsisequalto thenumber of linksgiven byequation(2.2)or (2.3).Fig.2.5showsthebasicloopsforthetreeinFig.2.4(a). 5 .. - .. - - - . CD GL ~ = - - + - - - ~ @ Fig.2.5Basic loops for the treeinFig.2.4(a). 2.5Cut-Set A Cut setisa minimalset of branchesK of a connected graph G,such that the removalof all Kbranchesdividesthegraphintotwoparts.It isalsotruethat theremovalof Kbranches reducestherankof Gbyone,providednoproper subset of thissetreduces therankof Gby onewhenitisremovedfromG. Consider thegraphinFig.2.6(a). I ._ - --e ._ - --e 3 (a)(b) Fig. 2.6 I ._ ---e _e I {, ...... .---3 (e) 10PowerSystemAnalysis Therankof thegraph= (no.of nodesn- 1)= 4- I= 3.If branches1 and3are removedtwosubgraphsareobtained asinFig.2.6(b).Thus1 and 3maybea cut-set.Also, if branches1,4and3areremovedthegraphisdividedintotwosubgraphsasshownin Fig.2.6(c)BranchesI,4,3mayalsobeacut-set.Inboththeabovecasestherankbothof the sub graphsis1 + 1 =2.It canbenoted that (I, 3) set isa sub-set of (I, 4,3) set.The cut set isa minimalset of branches of thegraph,removalof whichcutsthegraphintotwoparts. It separates nodesof thegraphsinto two graphs.Eachgroupisinone of the twosub graphs. 2.6BasiC Cut-Sets If each cut-set contains only one branch,then theseindependent cut-sets arecalledbasic cut-sets.Inorder tounderstandbasic cut-sets select a tree.Consider a twig bk of the tree.If the twig isremovedthetreeisseparatedinto twoparts.Allthelinkswhich gofromonepart of thisdisconnected tree totheother,together with thetwig bk constitutesa cut-set calledbasic cut-set.Theorientation of thebasic cut-set is chosen asto coincide with that of the branch of thetreedefiningthecut-set.Eachbasiccut-setcontainsatleastonebranchwithrespectto which the treeisdefinedwhichisnot containedintheother basiccut-set.For thisreason, the n -Ibasiccut-setsof a treearelinearlyindependent. Nowconsider thetreeinFig.2.4(a). Considernode(1)andbranchortwig1.Cut-setAcontainsthebranch1 andlinks 5 and6 andisoriented in the same way asbranchJ.In a similar way C cut-set cuts thebranch 3 andlinks 4 and5 and isoriented inthe same direction asbranch 3.Finally cut-set B cutting branch 2 and alsolinks 4,6 and5 isoriented asbranch 2 and the cutsets are shown in Fig.2.7. 85 - - - -"If'--@3:;..:----1 ~ ~ ~ 4 - - - . @~ Fig. 2.7Cut-set for the tree inFig.2.4(a). GraphTheory11 Worked Examples 2.1Forthenetworkshowninfigurebelow,drawthegraphandmarkatree. Howmanytreeswillthisgraphhave?Markthebasiccutsetsandbasic loops. 6 Solution: Assumethat bus(1)is thereferencebus 4 Number of nodesn =5 Number of elements e =6 Thegraphcanberedrawnas, 5 Fig.E.2.1 Fig.E.2.2 QDe-__________ ~ 5 . _ - - - - - - - - - - - - _ . 63 - - - - - - ~ ~ - - - - - - ~ - - - - - - ~ - - - - - - - - @ CD2 Fig.E.2.3 3 Tree:Aconnectedsubgraphcontainingallnodesof agraph,butnoclosedpathis called a tree. 12PowerSystemAnalysis 5 ...... 0 ././\ 6 ./"-;)..3 ./\ ./\ ././\ @2 Fig.E.2.4 Number of branchesn-1=5-1=4 Number of links= e-b = 6-4= 2 (Note:Number of links=Number of co-trees). D.... ___ c

Fig.E.2.S Thenumberof basiccutsets=no.of branches=4;thecutsets A,B,C,D,areshown infigure. 2.2Show the basic loops and basic cutsets for the graph shown below and verify anyrelationsthatexistbetweenthem. (Take1-2-3-4astree1). 8 Fig.E.2.6 GraphTheory Solution: ,.' Fig.E.2.7Tree and Co-tree for the graph. 13 If alinkisaddedtothetreealoopisformed,loopsthatcontainonlyonelinkarecalled basic loops. Branches,b=n-1=5-1=4 I=e-b=8-4=4 Thefourloops areshown inFig. Fig. E.2.SBasic cut sets A,B,C,D. Thenumberof basiccuts(4)=numberof branchesb(4). 2.3For thegraphgiveninfigurebelow,drawthetreeand thecorresponding co-tree.Choose a tree of your choiceand hence write the cut-set schedule. CDFig. E.2.9Oriented connected graph. 14PowerSystemAnalysis Solution: Fig. E.2.10Basic cut sets A,B,C,D. Thef-cutset schedule(fundamentalor basic) A:1,2 B:2,7,3,6 c:6,3,5 D:3,4 2.4Forthepowersystemsshowninfiguredrawthegraph,atreeandits co-tree. Fig. E.2.11 Solution: '2 Substituting thevalues =k1.03.)2 B.75+ (of 8.75 J+ 0(1.05)(-1.25) + 0.(-3.75) -1.03[(0)(-1.25) - (1.05)(-3.75)] + (0)[(1)( -1.6667) + (0)(-5.0)] - 1.03[(0)( -1.6667) - (1)( -5)] =0.07725 Mvargeneratedatbus2 Mvar injectionintobus2 + loadMvar 0.07725+ 0.2=0.27725p.u. 27.725Mvar 128 Thisiswithinthelimitsspecified. The voltageatbusiis OperationandControlinPowerSystems y(m+l)=~ [ P I- jQI_ ~y(m+l)_~VIm)] IyY (In )*L.. YIkkL.. YIkk IIIk ~ 1k=I+1 (2.9167 - j8.75) [ - 0.3 - 0.?7725_ (-1.25 + j3.75)(l.05 + jO.O) + (-1.6667 + j5.0)(1+ jO.O)] _1.03 - -,0.0 Vii)=1.01915- jO.032491 =1.0196673L -1.826 Anaccelerationfactorof 1.4isusedforbothrealandimaginaryparts. The accelerated voltagesisobtainedlIsing v ~=1.03 + 1.4(1.01915 -1.03) =1.01481 v;=0.0 + 1.4(-0.032491 - 0.0) =-0.0454874 Yil) (accelerated) =1.01481- jO.0454874 = 1.01583L - 2.56648 Thevoltageathus3isgivenby 6.6667 -- j20 =1.02093 - jO.0351381 PowerFlowAnalysis The acceleratedvalue of obtainedlIsing v',= 1.0 + 1.4(1.02093 - 1.0) = 1.029302 = 0 + 1.4( -0.0351384 - 0) = -0.0491933 Vjl)= 1.029302- jO.049933 =1.03048L - 2.736240 Thevoltagesattheendof thefirstiterationare VI=1.05+ jO.O = 1.01481- jO.0454874 Vjl)= 1.029302- jO.0491933 Check for convergence: Anaccuracyof 0.001istakenforconvergence [ 'l 0 I[, ll)[, IO) L1V2J=v1J- v1 =1.01481-1.03=-0.0152 [ "IO)[" ll)[" Ill) L1v2 =V2J- v2 =-0.0454874-0.0=-0.0454874 ["lUI[,,11[,,01 tWJ =V,J- V3J=1.029302--1.0=0.029302 129 [AVJIII=-0,0491933--0.0:--0.0491933 Themagnitudesof allthevoltagechanges aregreater than0.001. Iteration2 :Thereactivepower Q2 at bus2iscalculated asbefore togive 8(11= tan-I hE = tan-I [=-0.0454874J = -2.56648 2[vJII1.01481 = 1.03cos(-2.56648) = 1.02837 [ ,,)11_II'(II-' -V2J- V2sch sm82 -1.03sm(-2.56648) - -0.046122 [V2ne ..JI)-=1.02897 - jO.046122 = (1.02897)' (8.75)+- (-0.046122)" (8.75) +( -0.0461 22)[1.05( -\.25) + (0)( - 3. 75)J - (1.02897)[(0)(-\.25) - (1.05)( -3.75)1 + - (1.02897)[( --n.0491933)( -1.6667) --5)] co-0.0202933 130OperationandControlinPowerSystems Mvar tobegeneratedatbus2 =Net Mvar injectionintobus2 +loadMvar =- 0.0202933+0.2=0.1797067p.u.=17.97067Mvar Thisiswithinthespecifiedlimits.The voltagesare,therefore,thesame asbefore VI=1.05+ jO.O vii)= 1.02897 - jO.0.46122 Vjl)=1.029302- jO.0491933 ~ h eNew voltageatbus2isobtained as VO)_1[- 0.3 + jO.0202933] 2- 2.9 I 67 - j8.751.02827 + jO.046122 - (-1.25 + j3.75)(I- 05+ jO) - (--1.6667 + j5) . (\ .029302 - jO.0491933)] = 1.02486 - jO.0568268 The acceleratedvalue ofV?)isobtainedfrom " v ~= 1.02897 + 1.4(\ .02486 - 1.02897) = 1.023216 v ~=- -0.046 I 22 + 1.4( -0.0568268) -- (-0.046122 = -0.0611087) vi2)'=1.023216 - jO.0611 087 Thenew voltageatbus3iscalculatedas V(l)- I[-6.6 + jO.25] J- 6.6667 - j201.029302 + jO.0491933 -- ( - 5 + jI5)(1.05 + jO.O) - (-1.6667 + j5.0) (1.023216 - jO.0611)] =1.0226 - jO.0368715 The acceleratedvalue ofVi2)obtainedfrom v ~= 1.029302 + 1.4(1.0226 - 1.029302) = 1.02 v ~=(-0.0491933)+1.4(-0.0368715)+ (0.0491933) = -0.03194278 vf)= 1.02 - jO.03194278 The voltagesat theendof theseconditerationare V I=1.05+ jO.O PowerFlowAnalysi.\ vi2)=1.023216- jO.0611087 Vfl=1.02 - jO.03194278 131 Theprocedureisrepeatedtillconvergenceisobtainedattheendof thesixthiteration. The resultsare tahulatedinTableE5.1 (a) TableES.I(a)BusVoltage Iteration8usI8us 28us 3 01.05+jO1.03+jO1.0 + jO I1.05+ jO1.01481-jO.045481.029302 - jO.049193 21.05+- iO1.023216-,0.0611087I02 - jO.OJ 19428 3I05',0I 033l76-,0 04813X3,003508 1lOS +iOI 0227564 - ,0051 J2l)I 0124428,00341309 5I05+ ,01027726,00539141I0281748 - ,0.0363943 61.05+jO1'()29892 - ,0.050621.02030 I - jO.0338074 71.05+ jOI. 0284 78 -.iO05 101171.02412 -jO.034802 Lineflowfromhus1 tobus2 S'2.'V,(V;- V;)Yt2=0.228975+ jO.017396 Lineflowfrombus2tobus1 S2'=v2(v;- V;)Y;1=-0.22518- jO.OOS9178 Similarly, theother lineflowscanbe computedand are tabulatedinTable ES.I (b).the slack bus power obtainedbyadding theflowsinthelinesterminating at theslackbus,is Line 1-2 2-1 1-3 3-1 2-3 3-2 PI+ jQI= 0.228975+ jO.017396-t0.684006+ jO.225 =(0.912981+ jO.242396) Table E5.1(b)LineFlows PPower Flow Q +0.2289750.017396 - 0.2251830.0059178 0.683960.224 - 0.674565-0.195845 - 0.0741290.0554 0074610.054 132OperationandControlinPowerSystems E 5.2ConsiderthebussystemshowninFig.E5.2. 5 E 5.2A sixbuspower system. The followingisthe data: Lineimpedance(p.u.)RealImaginary 140.57000E-I0.845E-I 1-51.33000E-23.600/::-2 2-33.19999E-21750E-l 24I. 73000E-20.560E-I 2-(,30()()OOL-21500E-I 4-51 94000f:-20625E-l Scheduled generationandbusvoltages: Bus CodePAssumedbusvoltageGenerationLoad MWp.u.Mvar p.uMWp.u.Mvarp.u 11 05+jO0--- --- --- ---(specitied) 2--- 1.20.05--- ---3--- 1.20.05--- ---. ~ --- --- --- 1.40.05 :\--- --- --- 0.80.D3 6--- --- --- 0.70.02 (a)Taking bus - Iasslack bus andusing anaccelerating factor of 1.4,performload flowbyGauss - Seidelmethod.Takeprecisionindexas0.000 I. (b)Solvetheproblemalsousing Newton-Raphsonpolar coordinatemethod. PowerFlowAnalysis133 Solution: Thebusadmittance matrixisobtainedas: BusCodeAdmittance (p.u.) P-QRealImaginary 1-114.516310-32 57515 1-4-5486446813342 1-5-9.02987024.44174 2-27.329113-28.24106 2-3-1.0 II 0915.529494 2-5-503597016.301400 2-6-1.2820516410257 3-2-1.0110915.529404 3-31.011091-5.529404 4-1-5.4864468133420 4-410.016390-22727320 4-5-4.52994814.593900 5-1-90298702..J.441740 5-2-5.035970163014()0 5-4-4.52994814593900 5 _.5 18.59579()-55.33705() 6 - 21.2820516..J 1()257 6 - 61.282051-6.410254 Allthebusvoltages,y(O),areassumedtobeI+ jOexcept thespecifiedvoltage atbus I whichiskept fixedat1.05 + jO.The voltage equations for the fistGause-Seidel iteration are: V(I)'"_1_11'2-.iQ 2 2yIV(0)' 2l_2 .Y2'V,(0).- Y2,V;"I.y,VIO)'1 _hI,! .i (I)I[ p ~ ' - jQ-1 VCO" ..-- -----.-~YVIO)' III 134OperationandControlinPowerSystems y!l)=_I_[P5-jQ5_yY_yy(l)_Y_y(l)] )Yy (0)'51I512)44 555 Substituting thevalues.theequationforsolution are 1-.28.24100)X[I.2-jO.05Jl - 7.329113Jl-jO - (-I .011091+ j5.529404)x (I + jO) -(-5.03597 + jI6.3014)(1 + jO) - (I- 282051+ j16.30 14 )(1+ jO) = 1.016786 + jO.0557924 y(l)=(I_.5.52424) x [1.2 - jO.05] :1 1.0 I 109 IJI - jO _.(- 1.0 I 1091+ j5.529404) x (1.016786 + jO.0557924) =I .089511 + j0.3885233 y(l) =(I_.22.72732)X[-I.4+ jO.005] 410.01639J.1- jO - (-5.4862,46 + j8.133342) x (1.05 + jO.) - (-4.529948 + jl 4.5939)(1+ jO) = 0.992808 - jO.0658069 y!l)= ( __I ___.55.33705) x [- 0.8 +)18.59579J1- jO - (-9.02987 + j24.44174) x (1.05 + jO) - (-5.03597 + jl 6.30 14)(1.016786 + jO.0557929) - (-4.529948 + jl 4.5939)(0.992808 - jO.0658069) =1.028669 - jO.O 1879179 y(I)=(1-.6.410257JX[-0.7+jO.02] 61.282051J1 - jO - (-1.282051 - j6.4 I 0257) x (1.0 I 6786 + jO.0557924) =0.989904 - jO.0669962 The resultsof theseiterationsisgiveninTable5.3(a) TableES.3(a) It.NoBus 2Bus 3Bus4Bus5Bus 6 01+ jO.OI+-.10.0I+-jO0I+jO.OI+jO 0 I1.016789 + jO.05579241.089511+ jO38852330.992808 - jO06580691.02669-jO.0 18791790989901-.10.0669962 21.05306 + jO.1O 187351.014855 + .10.23233091.013552 - jO.05772131.0-l2189 + jO.O 1773221.041933 +.10.0192121 3I043568 + jO089733I054321+ jO32760351.021136 - jO03527271034181+jO.00258192I 014571-jO 02625271 41.047155 +.10.101896I02297 + .10.027635641.012207 - .10.05()05581.035391+ jO00526437I02209 +.100064356n 51040005+jO.093791103515+ jO.30508141.61576 _. jO042586920.033319 + jO.0036970561014416-jO.01319787 6I04212 + .10.09784311.027151+.1029013581.013044 -.100464654610.33985 + jO.0045044171.01821-jOOO1752973 71.040509 + jO.09634051.031063+ jO.29940831.014418 - jO.0453101I033845 ... jO004304541016182-.1000770664 1\1.041414+j00975181.028816 + jO2944651.013687 - jO045610 II033845 + jO.004558826I 017353 - jO0048398 91.040914 + jO.0970021.030042 + jO.29732871-014148 --jO044876291.033711+jO.0044 136471016743 -jO.0060342 101.041203+ jO.09728181.02935 +.1029732871.013881- jO.045111741.03381+ jO.004495542IOI7089-jO.00498989 II1.041036 + jO.0971641.029739 + jO.2965981.01403-.1 04498312I03374 +jO.0044395591.016877 -jO 00558081 121.041127 +-jO.09719981.029518 + jO29607841013943-.10045062121.033761+ jO004470961.016997 -- .10.00524855 131.041075+-.10.09714511.029642 + .10.2963 7151.019331-j0045014881.033749 + jO.OO44540021.016927 -jO 00543323 141.041104 + .10.09717771.02571+ jO.29620841.0013965 -jO.045042231.033756 +- .10.004463 713IOI6967-jO.00053283 136OperationandControlinPowerSystems Inthepolarform,allthevoltagesattheendof the14thiterationaregiven inTableE5.3(b). Table E5.3(b) BusVoltagemagnitude(p.u.)Phaseangle(0) II OS0 2()0456295.3326 3107133416.05058 41.014964-2.543515 5I0337652.473992 61.016981-3.001928 (b)Newton- Raphsonpolar coordinatesmethod Thebusadmittancematrixiswritteninpolar formas [19.7642L-71.60 3.95285L-I08.4 Ysus =3.95285L-108.49.2233IL-71.6 15.8114L - 108.4 5.27046L - 108.4 0 Note that Theinitialbus voltages are VI=1.05LOO viol=1.03LOO viOl=1.0LOo 15.8114L -108.41 5.27046L -108.4 21.0819L-71.6 Therealandreactivepowers atbus2arecalculatedasfollows: P2=IV1 VI Y11 Icos(oiO)- 01- 821)+ [vi Yn[ cos(- 822)+ 7(1.03)(1.05)(3.95285)cos(108.4)-+- (1.03)2(9.22331)cos(-108.4) +(1.03)2(9.22331)cos(71.6)+(1.03)(1.0)(5.27046)cos(--108.4) ~0.02575 PowerFlowAnalysis 137. O2 =IV2 VI Y21- 81 - 821)+ Iv} y221sin(- 822)+ IV2V3Y23 I sin (8iol=(1.03) (1.05) (3.95285) sin(-108.4) + (1.03)2(9.22331) sin (71.6) +(1.03)(1.0)(5.27046)sin(108.4) =0.07725 Generationof p. u Mvaratbus2 =0.2+ 0.07725 =0.27725=27.725Mvar Thisiswithinthelimitsspecitied.Therealandreactivepowers atbus3 arecalculated ina similar way. P3=-81 -831)+!Vj01V2 Y32! -82 -832)+ IVjO)2Y33Icos(-833) =(1.0)(1.05)(15.8114)cos(-108.4)+ (1.0)(1.03)(5.27046)cos(-108.4) +(1.0)2(21.0819)cos(71.6) =- 0.3 03 =-81 -831)+lvjO)V2 -82 -932)+ IVJ012 Y31 Isin(- 833) =(1.0)1.05(15.8114)sin(-108.4)+ (1.0)(1.03)(5.27046) sin(-108.4) + (1.0)2(21.0891)sin(71.6) =- 0.9 Thedifferencebetweenscheduledandcalculatedpowersare =-0.3 - 0.02575 = -0.32575 = -0.6 -;- (-0.3) = -0.3 =-0.25 - (-0.9) = -0.65 It maybenotedthat hasnotbeencomputedsincebus2 isvoltagecontrolledbus. since I, 1 andI 138OperationandControlinPowerSystems are greater thanthespecifiedlimit of 0.0 I,thenextiterationiscomputed. IterationI:Elements of theJacobianarecalculatedasfollows. oP2 =Iv,V(O)y"- 8(0)- 8,,) c8,-)-'- ,-.' =(1.03)(1.0)(5.27046)sin(-I 08.4())-5.15 ap2II . ((O)(0)) 282 =-v2 VIY21SIn\82 -81 -821 + IV2 -823) =- (1.03)(1.05)(3.95285)sin(108.4)+ (1.03)(!.O)(5.27046)sin(-108.4) =9.2056266 CP2II((O)(0)) -8,-823 Ou, J =0(1.03)(5.27046)cos(108.4) =- 1.7166724 cP}=IV(O)VYISin(8(O)-8)) "I::3I313_L CU3 =(0.0)(1.03)(5.27046)sin(-108.4) =- 5.15 oP,IV(O)VyI.(dO)I::8)IV(O)VYI.{I:: (0)dO)8) 00=31)1SIn\u)-UI- 31+3232SIn\U3-u2- 32 3 =- (1.0)(1.05)(15.8114)sin(-108.4) - 5.15 =20.9 IV2 Y32Icos(ojO)-oiO)-832) =2(1.0)(21.0819)cos(71.6)+ (1.05)(15.8114)cos(-108.4)+ (1.03)(5.27046)cos(-108.4) =6.366604 PowerFlowAnalysis 8Q3 :=:-IV(O)VyICOS(O(O)- 0(0)- e) cO,3I32,232 =(1.0)(1.03)(5.27046)cos(--108.4() =1.7166724 ?Q3 Iv(OIVI(dO)e) cO,:=:,IY cos u 3- UI- 31+ .' IV(O)VI \:(0)8) 32Y32 cosu2-u3- 32 =c(1.0)(1.05)(15.8114)cos(-108.4()- 1.7166724 =- 6.9667 IV2Y32Isin(oiO) -832) 139 =2(1.0)(21.0819)sin(71.6)+(1.05)(15.8114)sin(-108.4)+ (1.03)(5.27046)sin(-108.4(1) =19.1 Fromeqn.(5.70) r- 0.325751[9.20563 -0.3=-5.15 0.651.71667 - 5.15 20.9 - 6.9967 Following themethodof triangulationandbacksubstations l- 0.353861lI --0.3=-5.15 - 0.035386+ 1.71667 -0.5594420.96.36660t\b, -6.966719.1 r-0.353861[I - 0.482237=0 +0.710746_0 -0.55944 1 18.025.40623M, - 6.00632619.42012L{\IV l 140 Finally. Thus, OperationandControlinPowerSystems r -0.35386]11-0.55944 l-0.0267613=lOI0.3 0.550021.22202 = (0.55)/(21.22202)= 0.025917 =- 0.0267613- (0.3)(0.025917) =- 0.0345364rad '"- 1.98 Ml=- 0.035286- (-0.55944)(--0.034536)- (--0.18648)(0.025917) ==- 0.049874rad =- 2.8575 At theendof the firsiterationthebusvoltagesare VI=1.05LOo V 2 =1.03L2.85757 V3==1.025917L-1.9788 Therealandreactivepowersatbus2 arecomputed: pil)=(1.03)(1.05)(3.95285)[cos(-2.8575) - 0(-108.4) + (1.03) 2 (1.025917)(5.27046) cos[( -2.8575) - (-1.9788) - 108.4 -0.30009 =(1.03)(1 .05)(3.95285)[sin( -2.8575) - O( -108.4 0) + (1.03)2 (9.22331) sin[( -2.85757) - (-1.9788) - 108.4 0)] =0.043853 Generationof reactivepower atbus2 =0.2+ 0.043853=0.243856p.ll.Mvar =24.3856Mvar Thisiswithinthespecifiedlimits. Therealandreactivepowersatbus3arecomputedas pjl) =(1.025917)(1.05)(l5.8117)cos[(-1.09788) - 0 -108.4 0)] + (l.025917)(l.03)(5.27046)cos[(-1.0988) - (-2.8575) -108.4] + (1.025917)\21.0819) cos(71.60) =- -0.60407 PowerFlowAnalysis ==(1.025917)(1.05 )(15.8114) sin[( -1.977) - 108.4 0)] + (1.025917)(1.03)(5.27046) sinl( -1.9788) - (-2.8575) - 108.4 0)1 + (21.0819)sin(71.6o) = -0.224 Thedifferencesbetweenscheduledpowersandcalculatedpowersare = -0.3 - (-0.30009) = 0.00009 I)= -0.6 - (-0.60407) = 0.00407 ;\Q\I)co-0.25 - (-0.2224) =-0.0276 141 Eventhoughthefirsttwodifferences arewithinthelimitsthelastone,oil)isgreater than the specified limit 0.0 I.The next iterationis carried out ina similar manner.At the end of the seconditeration evenalsoisfound to bewithin the specified tolerance.The results are tabulatedintableE5.4(a) andE5.4(b) TableE5.4(a)Busvoltages IterationBusIBus2Bus 3 0I.05LOoI03LOoI.LOo II.05LOoI.03L-2.857571.025917 L-1.9788 2I.05LOo 1.03L-2.85171.024 76L -1. 94 7 TableE5.4(b)LineFlows LinePPower Flow Q . 1-202297n.O 16533 2-1- 022332o ()O49313 1-3O.683C)/)0.224 3-1- 0674565-0.0195845 2-3-0.0741260.0554 3-2007461- 0.054 142OperationandControlinPowerSystems E5.3ForthegivensamplepowersystemfindloadflowsolutionusingN-Rpolar coordinatesmethod,decoupledmethodandfastdecoupledmethod. 5 34 (a)Power system BusCodeLineimpedance ZpqLinecharging 1-20.02 +jO.24j 0.02 2-3004 + jO.02j 002 3-50.15 + jO.04j0.025 3-40.02+jO.06j 0.01 4-50.02 +jO.04i 0.01 5-10.08 +jO.02j 0.2 (b)Line-data BusCode(Slack)GenerationLoad I 11.724 - j24.27 2-10 + j20 30+ jO 40 + jO 5-1.724+ .14.31 I 2 3 4 5 MwMvarMW 00 () 502515 0045 0040 0050 . (c)Generationandloaddata 2 -10 +j20 3 0+ jO 4 0+ jO 10962 - j24.768-0.962 +j4.8080 +jO -0.962+j4.8086783-.121.944-5+jI5 O+jO-)1.11515-j34.98 0+ jO-0 82,.12.192-10+ .120 (d)Bus admittance matrix Fig.E 5.3 Mvar 0 10 20 15 25 5 -1.724+ j4.31 +.jO -0822 +j2.192 -10 +j20 12.546 - j26.447 PowerFlow Analysis Solution: The Residualor Mismatchvector foriterationno: Iis dp[2]= 0.04944 dp[3]=--0.041583 dp[4]= -0.067349 dp[5]=-0.047486 dQ[2]=-0.038605 dQ[3]=-0.046259 dQ[4]=-0.003703 dQ[5]=-0.058334 The New voltage vector afteriterationIis: Busno1 E: 1.000000F : 0.000000 Busno2E :1.984591F :- 0.008285 Busno3 E : 0.882096F :- 0.142226 Busno4E : 0.86991F :- 0.153423 Busno5E : 0.875810F :- 0.142707 Theresidualor mismatchvector foriterationno: 2is dp[2]=0.002406 dp[3]= -0.001177 dp[4]=--0.004219 dp[5]=--0.000953 dQ[2] =-0.001087 dQ[3]=-0.002261 dQ[4]=--0.000502 dQ[5]=-0.002888 The Newvoltage vector afteriteration2is: BusnoIE: 1.000000F : 0.000000 Bus no2E : 0.984357F :- 0.008219 Busno3 E: 0.880951F :- 0.142953 Busno 4E : 0.868709F :- 0.154322 Busno5 E: 0.874651F :- 0.143439 Theresidualor mismatchvector foriterationno: 3is 143 144OperationandControlinPowerSystems dp[2]=0.000005 dp[3]=-0.000001 dp[4]=-0.000013 dp[5]=-0.000001 dQ[2]=-0.000002 dQ[3]=-0.000005 dQ[4]=-0.000003 dQ[5]=-0.000007 The finalloadflowsolution(forallowableerror.OOOt): busno1 SlackP = 1.089093Q = 0.556063E = 1.000000F = 0.000000 busno2pqP = 0.349995Q = 0.150002E = 0.984357 busno3 pqP = -0.449999Q = -0.199995E = 0.880951 busno4pqP = -0.399987Q = -0.150003E = 0.868709 busno5 pqP = -0.500001Q = -0.249993E = 0.874651 Decoupledloadflowsolution(polar coordinatemethod) Theresidualormismatchvector foriterationno: 0is dp[2]=0.350000 dp[3]=-0.450000 dp[4]=-0.400000 dp[5]=-0.500000 dQ[2]=-0.190000 dQ[3]=-0.145000 dQ[4]=-0.130000 dQ[5]=-0.195000 Thenewvoltagevector afteriteration0: BusnoIE: 1.000000F : 0.000000 Busno2 E: 0.997385F :- 0.014700 Bus no3 E: 0.947017F:- 0.148655 Busno4 E : 0.941403F :- 0.161282 Bus no5 E: 0.943803F:- 0.150753 Theresidualor mismatchvector foriterationno: Iis dp[2] =0.005323 F =-0.008219 F =-0.1429531 F =-0.154322 F =-0.143439 PowerFlowAnalysis dp[3]=-0.008207 dp[4]=-0.004139 dp[5]=-0.019702 dQ[2]=-0.067713 dQ[3]= -0.112987 dQ[ 4]=-0.159696 dQ[5]=-0.210557 The new voltage vector after iteration1 : Bus no1 E :1.000000 F : 0.000000 Busno2E : 0.982082F :- 0.013556 Busno3E : 0.882750F :- 0.143760 Busno4E : 0.870666 F :- 0.154900 Busno5 E: 0.876161F :- 0.143484 Theresidualor mismatchvector foriterationno:2is dp[2]=0.149314 dp[3]=-0.017905 dp[4]=-0.002305 dp[5]=-0.006964 dQ[2]=-0.009525 dQ[3]=-0.009927 dQ[4]=-0.012938 dQ[5]=0.007721 The new voltage vector after iteration 2: Bus no1 E: 1.000000F : 0.000000 Bus no2 E: 0.981985F:- 0.007091 Busno3 E: 0.880269F :- 0.142767 Busno4E: 0.868132F:- 0.154172 Bus no5 E : 0.874339F :- 0.143109 Theresidualor mismatchvector foriterationno:3is dp[2]=0.000138 dp[3]=0.001304 dp[4]=0.004522 145 146 dp[ 5]=-0.006315 dQ[2]=0.066286 dQ[3]=0.006182 dQ[ 4]=-0.001652 dQ[5]=-0.002233 OperationandControlinPowerSystems Thenewvoltagevector afteriteration 3: BusnoIE: 1.000000F : 0.000000 Busno2E : 0.984866F :- 0.007075 Busno3E : 0.881111F::- 0.142710 Busno4E : 0.868848F :- 0.154159 Busno5 E : 0.874862F :- 0.143429 Theresidualor mismatchvector foriterationno:4is dp[2]=-0.031844 dp[3]=0.002894 dp[4]=-O.M0570 til dp[5]=0.001807 dQ[2]=-0.000046 dQ[3]=0.000463 dQ[ 4]=0.002409 dQ[5]=-0.003361 Thenewvoltagevector afteriteration4: BusnoIE: 1.000000F : 0.000000 Busno2E : 0.984866F :- 0.008460 Bus no3 E: 0.881121F:- 0.142985 Busno4E : 0.868849 F :- 0.1546330 Busno5 E : 0.874717F :- 0.143484 Theresidualor mismatchvector foriterationno:5is dp[2]=0.006789 dp[3]=-0.000528 dp[4]=-0.000217 dp[5]=-0.0000561 dQ[2J~ .--0.000059 PowerFlowAnalysis dQ[3 J =-0.000059 dQ[4]=-0.000635 dQ[5]=-0.000721 Thenewvoltagevector afteriteration5: Bus no1 E :1.000000F : 0.000000 Busno2E: 0.984246F :- 0.008169 Busno3E: 0.880907F :- 0.1.42947 Busno4E : 0.868671F :- 0.154323 Busno5 E: 0.874633F :- 0.143431 Theresidualor mismatchvector foriterationno: 6is dp[2]=0.000056 dp[3]=0.000010 dp[ 4]=0.000305 dp[5]=-0.000320 dQ[2]=0.003032 dQ[3]=-0.000186 dQ[4]=-0.000160 dQ[5]=-0.000267 Thenewvoltagevector afteriteration 6: Busno1 E :1.000000 F : 0.000000 Busno2 E: 0.984379F :- 0.008165 Busno 3E: 0.880954F :- 0.142941 Busno4 E: 0.868710F :- 0.154314 Busno5 E: 0.874655F :- 0.143441 Theresidualor mismatchvector foriterationno:7is dp[2]=- 0.001466 dp[3]=0.000 I 06 dp[4]=-0.000073 dp[5]=0.000156 dQ[2]=0.000033 dQ[3]=0.000005 dQ[41=0 . O O O I 5 ~ 147 148OperationandControlinPowerSystems dQ[5]=-0.000166 Thenewvoltage vector afteriteration7 : Busno1 E: 1.000000F : 0.000000 Busno2E : 0.954381F :- 0.008230 Busno3E : 0.880958F :- 0.142957 Busno4E : 0.868714F :- 0.154325 Busno5 E: 0.874651F:- 0.143442 Theresidualor mismatchvector foriterationno: 8is dp[2]=-0.000022 dp[3]=0.00000 I dp[4]=-0.000072 dp[5]=-0.000074 dQ[2]=-0.000656 dQ[3]=0.000037 dQ[4]=--0.000048 dQ[5]=-0.000074 Thenewvoltagevector afteriteration8 : Busno1 E :1.000000F : 0.000000 Bus no2E : 0.984352F :- 0.008231 Busno3 E: 0.880947F :- 0.142958 Busno4E : 0.868706F :- 0.154327 Busno5 E : 0.874647F :- 0.143440 Theresidualor mismatchvector foriteration no:9is dp[2]=0.000318 dp[3]=-0.000022 dp[ 4]=0.000023 dp[5]=-0.000041 dQ[2]=-0.000012 dQ[3]=-0.000000 dQr 4]=0.000036 dQ[5]=-0.000038 Thenewvoltagevector afteriteration 9: PowerFlowAnaly.flis Busno1 E :1.000000F : 0.000000 Busno 2E: 0.984352F :- 0.008217 Busno3E: 0.880946F :- 0.142954 Busno4E: 0.868705F :- 0.154324 Busno5E: 0.874648F :- 0.143440 Theresidualor mismatchvector foriterationno: lOis dp[2]=0.000001 dp[3]=-0.000001 dp[4]=0.000017 dp[S]=-0.000017 dQ[2]=0.000143 dQ[3]=-0.000008 dQ[4]= 0.000014 dQ[5]=-0.000020 Thenew voltagevector after iteration10: Bus no1 E:1.000000F : 0.000000 Busno 2E: 0.984658F :- 0.008216 Busno 3E: 0.880949F :- 0.142954 Busno 4E: 0.868707F :- 0.154324 Busno5E: 0.874648F :- 0.143440 Theresidualor mismatchvectorforiterationno: II is dp[2]=--0.000069 dp[3]=0.000005 dp[ 4]=-0.000006 dp[51=0.000011 dQ[2]=0.000004 dQ[3]=-0.000000 dQ[4]=0.000008 dQ[5]=-0.000009 The finalloadflowsolution after11iterations (for allowablearror.OOO I) 149 150OperationandControlinPowerSystems Thefinalloadflowsolution(for allowableerror.OOOl): Busno1SlackP =1.089043Q= 0.556088E =1.000000 Busno 2pqP = 0.350069Q= 0.150002E = 0.984658 Bus no3pqP = -0.450005Q=-0.199995E = 0.880949 Busno4pqP = -0.399994Q= -0.150003E = 0.868707 Busno5pqP = -0.500011Q= -0.249991E = 0.874648 Fastdecoupledloadflowsolution(polar coordinatemethod) Theresidualor mismatchvector foriterationno:Ois dp[2]=0.350000 dp[3]=-0.450000 dpf4]~0.400000 dp[5]=0.500000 dQ[21=0.190000 dQ[3]=-0.145000 dQ[4]=0.130000 dQ[5]=-0.195000 Thenewvoltagevector afteriteration0: Busno1 E :1.000000 F : 0.000000 Busno2E : 0.997563F :- 0.015222 Busno3E: 0.947912F:- 0.151220 Busno4E: 0.942331F :- 0.163946 Busno5 E: 0.944696F:- 0.153327 Theresidualor mismatchvector foriterationno: Iis dp[2]=0.004466 dp[3]=-0.000751 dp[ 41=.0.007299 dpr 5]~-0.012407 dQ[21=0.072548 dQ[3]==-0.118299 dQ[4]=0.162227 dQr5]=-0.218309 Thenewvoltagevector afteriteration1 : Busno1 E :1.000000F . 0.000000 F= 0.000000 F =-0.008216 F =-0.142954 F=-0.154324 F = -0.143440 PowerFlowAnalysis Busno2E; 0.981909F ;- 0.013636 Busno3E; 0.882397F ;- 0.143602 Busno4E; 0.869896F ;- 0.154684 Busno5E; 0.875752F ;- 0.1433 12 Theresidualor mismatchvectorforiterationno:2is drl21~O.I 5366 I dp[3]~-0.020063 dp[4]=0.005460 dpr5]=-0.009505 dQr2]=0.011198 dQ[31=---0.014792 dQ[4]=-0.000732 dQ[5]=-0.002874 Thenewvoltagevector afteriteration2: BusnoIE; 1.000000F ; 0.000000 Busno2E; 0.982004F ;- 0.007026 Busno3E; 0.8805 15F ;- O.142597 Busno 4E; 0.868400F ;- 0.153884 Busno5E; 0.874588F ;- 0.143038 Theresidualor mismatchvectorforiterationno:3is dp[2]=-0.000850 dp[3]=-0.002093 dp[4]=0.000155 dp[5]=-0.003219 dQ[2]=0.067612 dQ[3]=-0.007004 dQ[4]=-0.003236 dQ[5]=--0.004296 Thenewvoltagevector afteriteration3: BusnoIE; 1.000000F : 0.000000 Busno2E: 0.984926F :- 0.007086 Busno3E: 0.881246F :- O.I 42740 151 152OperationandControlinPowerSystems Busno4E: 0.869014F :- 0.154193 Bus no 5 E: 0.874928F :- 0.143458 Theresidualor mismatch vector foriterationno:4is dp[2]=-0.032384 dp[3]=0.003011 dp[4]=-0.001336 dp[5]=-0.002671 dQ[2]=-0.000966 dQ[3]=-0.000430 dQ[41=-0.000232 dQ[5]=-0.001698 Thenewvoltagevector afteriteration4: Busno1 E: 1.000000F : 0.000000 Busno2E : 0.984862F :- 0.008488 Busno3 E: 0.881119F :- 0.143053 Bus no4E : 0.868847F :- 0.154405 Busno5E: 0.874717F:- 0.143501 Theresidualor mismatchvector foriterationno:5is dp[2]=0.000433 dp[3]=0.000006 dp[ 4]=--0.000288 dp[5]=0.000450 dQ[2]=-0.014315 dQ[3]=-0.000936 dQ[4]=-0.000909 dQ[5]=-O.OOJ 265 Thenewvoltagevector afteriteration6: Bus noIE: 1.000000 F : 0.000000 Busno2E : 0.984230F :- 0.008463 Busno 3E: 0.881246F :- 0.143008 Busno4E : 0.869014F :- 0.154357 PowerFlowAnalysis Bus no5E : 0.874607F :- 0.143433 Theresidualor mismatch vector foriterationno:6is dp[2]=0.006981 dp[3]=-0.000528 dp[4]=0.000384 dp[5]=-0.000792 dQ[2]=0.000331 dQ[3]=0.000039 dQ[4]=-0.000155 dQ[5]=0.000247 Theresidualor mismatchvectorforiterationno:7 is dp[2]=-0.000144 dp[3]=-0.000050 dp[4]=0.000080 dp[5]=-0.000068 dQ[2]=0.003107 dQ[3]=-0.000162 dQ[4]=-0.000255 dQ[5]=-0.000375 Thenewvoltagevector afteriteration7: BusnoIE: 1.000000F : 0.000000 Busno2E : 0.984386F :- 0.008166 Busno3E : 0.880963F :- 0.142943 Busno4E: 0.868718F:- 0.154316 Busno5 E : 0.874656F:- 0.143442 Theresidualormismatchvector foriterationno:8is dp[2]=-0.001523 dp[3]=-0.000105 dp[4]=-0.000115 dp[5]=-0.000215 dQ[2]=0.000098 153 154 dQ[3]=-0.000024 dQrtl=-0.000037 dQ15]-0-0.000038 OperationandControlinPowerSystems Thenewvoltagevectorafteriteration8: BusnoIE: 1.000000F: 0.000000 Busno2E: 0.984380F:- 0.008233 Busno3E: 0.880957F :- O.I 4296 I Busno4E: 0.8687 I 4F :- O.I 54329 Busno,)E: 0.874651F :- 0.143442 Theresidualormismatchvectorforiterationno:9is dp[2]=-0.000045 dp[3]=0.000015 dp[ 4]=-0.000017 dp[5]=0.000008 dQ[2]~0.000679 dQl3]=0.00003 I dQ[4]= -0.000072 dQ[5]=-0.000 105 Thenewvoltagevector afteriteration9: BusnoIE: 1.000000 F : 0.000000 Busno2E : 0.984350F :- 0.008230 Busno3E: 0.880945F :- 0.142958 Busno4E: 0.868704F :- O.I 54326 Busno5E: 0.874646F :- O.I 43440 Theresidualor mismatchvector foriterationno:lOis dp[2]=0.000334 dp[3]=-0.000022 dpl41""0.000033 dp[5]=0.000056 dQI21=0.000028 dQI3 J=0.000007 dQr 41=--0.000007 PowerFlowAnalysis dQ[5]=0.000005 Thenewvoltagevector after iteration10: Busno1 E:1.000000 F: 0.000000 Busno2E: 0.984352F :- 0 .. 008216 Busno3E: 0.880946F :- 0.142953 Busno 4E: 0.898705F :- 0.154323 Busno5E: 0.874648F :- 0.143440 Theresidualor mismatchvector foriterationno:11is dpP1=--0.000013 dpl31~-0.000004 dp[4]=0.000003 dpr5]= -0.000000 dQr21=0.000149 dQ[31=-0.000007 dQ[4]=0.000020 dQ[5]=-0.000027 Thenewvoltagevector after iteration11: Busno1 E:1.000000F : 0.000000 Busno2E: 0.984358F:- 0.008216 Busno3E: 0.880949F :- 0.142954 Bus no 4E : 0.868707F :- 0.154324 Busno5E: 0.874648F :- 0.143440 Theresidualor mismatchvector foriterationno:12is dp[21-c-0.000074 dp[ 31=0.000005 dp[4]=-0.000009 dp[5]=-0.000014 dQr21=0.000008 dQr31: ~-0.000002 dQf41=-0.000001 dQr 5 J'--0.000000 155 .. 156OperationandControlinPowerSystems Theloadflowsolution BusnoISlackP =1.089040Q= 0.556076E =1.000000F=0.000000 Busno2pqP = 0.350074Q=0.150008E = 0.984358F=-0.008216 Busno3pqP = -0.450005Q' ~ - O .199995E = 0.880949F = -0.142954 Busno4pqP =--0.399991Q~-0.150001E ==0.868707F= -0.154324 Busno5pqP = -0.500014Q=-0.250000E = 0.874648F= -0.143440 E5.4ObtaintheloadflowsolutiontothesystemgiveninexampleE5.1lIsingZ-Bus.Use Gauss- Seidelmethod.Takeaccuracyforconvergenceas0.000 I. Solution: Thebusimpedancematrixisformedasindicatedinsection5.15.Theslackbusis takenasthereferencebus.Inthisexample.asinexample5.1busIischosenasthe slackbus. (i)Addelement1-2.Thisisadditionof anewbustothereferencebus Z=(2) BUS(2)[0.05 + jO.24[ (ii)Addelement1-3.r hisisalsoadditionof anewbustothereferencebus (2)(3) (2)0.08 + jO.240.0 + jO.O Zsus=(3)0.0 + jO.O0.02 + jO.06 (iii)Addelement2-3.Thisistheadditionof alinkbetweentwoexistingbuses2 and3. Z2-IOOp= Zloop_2= Z22- Z23= 0.08+jO.24 Z3_loop= Zloop_3=Z32- Z33= -(0.02+jO.06) Zloop-looP=Z22+ Z33- 2 Z23+ Z2323 =(0.08+jO.24)+(0.02+jO.06)(0.06+ .i0.18) =0.16+j0.48 (2) Zsus=(3) C Theloopisnow eliminated (2) 0.08 + jO.024 0.0 + jO.O 0.08 + jO.24 .Z2-loopZloop-2 Z22=Z 22- ---'--------'-Zloop-loop (3) (0 0+ jO0.08 + jO.24 0.02 + jO.06- (0.02 + jO.06) - (0.02 + jO.006)0.16+jO.48 PowerFlowAnalysis Similarly =(0.08 + jO.24) _(0.8 + jO.24)2 0.16 + j0.48 =0.04 + jO.12 =z'= [Z?'_Z2-loopZIOOP-3] _332_.'Z loop-loop =(0.0 + jO.O)_(0.8 + jO.24)( -0.02 - jO.06) 0.16+j0.48 = 0.0 1+ jO.03 =0.0175+ jO.0526 The Z- Busmatrix isthus Z_[0.04+jO.1210.01+jO.03]_ Bus- 0.0 I + jO.030.017 + jO.0525-[ 0.1265L71.565I 0.031623L71.5651 0.031623L71.5650.05534L71.S6SO The voltages at bus2 and3areassumedtobe via)= 1.03 + jO.O viol=1.0 + jO.O Assuming that thereactivepowerinjectedintobys2iszero, Q2 =0.0 Thebuscurrents and arecomputedas 1(0)=-0;3+ jO.O=-0.29126- '0.0 = 0.29126L1800 21.03 _ jO.OJ =-0.6 + !0.25= -0.6 _ jO.25= 0.65LI57.380 .1.0 + JO.O IterationI:The voltageatbus2iscomputedas VI+ Z22 + Z231(0) 1.05LOO+ (0.1265L71.5650(0.29126L180o+ (0.031623L71.565 )(0.65L157.38 1.02485 - jO.05045 1.02609L-2.8182 157 158OperationandControlinPowerSystems Thenewbuscurrent isnowcalculated. (I)IIII 1I= V2I V,eh_ 1 ! - Zl' I'VII)II -- 2I ..J =1.02609L - 2.8182x(1.03_I) = 0.0309084L _ 74.38320 0.1265L71.5651.02609 0(0)=Im[Vill =Im[I.02609L - 2.8182 JO.0309084L74.383 =0.03 = + =0.0 + 0.03= 0.03 III)=-0.3- jO.3= 0.29383LI82.891 8 2J.02609L _ 2.81820 Voltageatbus3isnow calculated VII)=V+ Z1\1)+ Z11(0) 3I32_3.3 =1.05LO.OO+ (0.031623L71.5650) (0.29383LI82.8320) + (0.05534':::71.565 )(0.65LI57.38 ) =(1.02389 - jO.036077) = 1.0245'::: - 2.018 1(1)=0.65LI57.38= 0.634437 LI55.36 31.0245L2.0 180 The voltages attheendof thefirstiterationare: VI=1.05LOO vii) = 1.02609L - 2.8182 VP)= 1.0245L - 2.018 Thedifferences in voltages are 8 Vjl)=(1.02485 - jO.05045) - 0.03 + jO.O) =-0.00515 - jO.05045 Vjl)= (1.02389 - jO.036077) - (1.0 + jO.O) =(0.02389 - jO.036077) PowerFlowAnalysis159 Boththerealandimaginarypartsare greater than thespecifiedlimit 0.00 I. Iteration2 : =VI+Z22Iil)+Z21 =1.02LOo+(0.1265L71.5650)(0.29383L 182.892)+ (0.031623L71.5650)(0.63447 L155.36) =1.02634 - jO.050465 =1.02758L-2.81495 1.02758L-2.81495 i1.03_I] 1.1265L - 71.565L1.02758 =0.01923L-74.38 L'10i') =Im[ vi2)r] =[m(1.02758L - 2.81495)(0.0 1913L74.38) =0.0186487 =oil)+ L'10i') =0.03+ 0.0 I 86487=0.0486487 -0.3 - jO.0486487 I - ----"----2- I .02758L2.81495 ==0.295763LI86.393 Vf) = 1.05LOo+ (0.31623L7 I .565 (O.295763LI 86.4 + 0.05534L71.565 )(0.634437 LI55.36 ) =0.65LI57.38=0.6343567LI55.4340 - 1.02466LI.9459 L'1 viI)=(1.02634 - jO.050465) - (1.02485 - jO.0504 I) =0.00149 - jO.000015 L'1Vjl)=(1.024 - jO.034793) - (1.02389 - jO.036077) =0.00011+ jO.OOI28 Astheaccuracyisstillnot enough,another iterationisrequired. Iteration3: =1.05LOo + (0.1265L71.5650)(0.295763LI86.4) + (0.031623L71.565 )(0.63487 LI 55.434 0) =1.0285187 - jO.051262 :c1.0298L _ 2.853 160OperationandControlinPowerSystems = J.0298L - 2.853= 0.001581L74.4180 - 0.1265L71.5651.0298 =0.00154456 Qi3)=0.0486487 + 0.001544 =0.0502 1\3)=-0.3- jO.0502=0.29537LI86.6470 - 0.0298L2.853 Vj31= 1.05LOo + (0.031623L71.5650) + (0.29537 LI86.6470) + (0.05534L71.565 )(0.634357 L155.434 0) = 1.024152 - jO.034817 ==1.02474L -1.9471 1(3)=- 0.65 - LI57.38= 0.6343LI55.4330 3J.02474LJ.9471 oil vi 2)= (1.0285187 - jO.051262) - (1.02634 - jO.050465) = 0.0021787- 0.000787 oilV12 I= (1.024152 - jO.034817) - (1.024 - jO.034793) = 0.000152 - jO.00002 Iteration4 " =1.02996L - 2.852 =0.0003159L - 74.417 = 0.0000867 = 0.0505 =0.29537 LI86.7 vi4)= 1.02416 - JO.034816=1.02475L -1.947 oil V?) = 0.000108 + jO.000016 oil Vj3)=0.00058 + jO.OOOOOI The finalvoltages are VI=1.05+ jO.O V2 =1.02996L-2.852 V3=1.02475L-1.947 The lineflowsmaybecalculated furtherif required. PowerFlowAnalysis161 Problems PS.IObtainaloadflowsolutionforthesystemshowninFig.PS.lLlse (i)GaLlss~Seidelmethod (ii)N-Rpolar coordinatesmethod Bus code p-qImpedance ZpqLine charges Ypq/s 1 ~ 20.02 +jO 20.0 2 ~ 30.01+ jO.0250.0 3 ~ 40.02 + jO.40.0 3 ~ 50.02 +0.050.0 ~ ~ 50.015 +jO04O.D 1-50015-t.lOO40.0 Valuesaregiveninp.Ll.ona base of IOOMva. 162PowerSystemAnalysis Thescheduledpowersareasfollows Bus Code (P)GenerationLoad MwMvarMWMvar 1 (slack bus)0000 280352515 30000 4004515 5005520 TakevoltageatbusIasI LOop.u. PS.2Repeat problemP5.1withline chargingcapacitance Y p/2 =jO.025for eachline PS.3ObtainthedecoupledandfastdecoupleloadflowsolutionforthesysteminP5.1and compare theresultswiththeexact solution. PS.4Forthe51bussystemshowninFig.P5.!, thesystemdataisgivenasfollowsinp.u. Performloadflowanalysisforthesystem LinedataResistanceReactanceCapacitance 2-30.02870.07470.0322 3-40.00280.00360.0015 3-60.06140.14000.0558 3-7002470.05600.0397 7-80.00980.02240.0091 8-9001900.04310.0174 9-100.01820.04130.0167 10- 110.02050.04680.0190 11-120.06600.01500.0060 12- 130.04550.06420.0058 13- 14011820.23600.0213 14 - 150.02140.27430.0267 15- 160.13360.05250.0059 16-170.05800.35320.0367 Contd. .... PowerFlowAnalysis163 LinedataResistanceReactanceCapacitance 17 - 180.15500.15320.0168 18- 19015500.36390.0350 19-200.16400.38150.0371 20 - 210.11360.30600.0300 20- 230.07810.20000.0210 23- 240.10330.26060.0282 12 - 250.08660.28470.0283 25- 260.01590.050800060 26 - 27008720.287000296 27 - 280.01360.04360.0045 28 - 290.01360.04360.0045 29 - 300.0125004000.0041 30 - 310.01360043600045 27 - 3100136()04360.0045 30 - 3200533()16360.0712 32 - 330.03110.100000420 32 - 340.0471o 15110.0650 30 - 510.06670.17650.0734 51- 33002300.06220.0256 35- 500.0240013260.0954 35- 3600266014180.1146 39 - 490.0168008990.0726 36 - 380.02520.13360.1078 38 - 1002000.110700794 38 - 470.02020.107600869 47 - 430.02500.13360.1078 42 - 430.02980.15840.1281 40-410.02540.14000.1008 Comd..... 164PowerSystemAnalysis LinedataResistanceReactanceCapacitance 41-- 430.03260.18070.1297 43- 450.02360.125201011 43- 440.0129007150.0513 45- 460.00540.02920.0236 44-10.03300.18180.1306 46-10.03430.20870.1686 1-490.01100.05970.1752 49 - 50000710.04000.0272 37 - 380.00140.00770.0246 47 - 390.02030.10930.0879 48 -20.04260.11000.0460 3 - 3500000005000.0000 7360.0000()04500.0000 t-I1- .170.00000.05000.0000 14 -- 470.00000.09000.0000 16- 390.00000.09000.0000 18-400.00000.04000.0000 20- 420.0000.0.08000.0000 24-B0.00000.09000.0000 27 - 450.00000.09000.0000 26- 440.00000.05000.0000 30 - 460.00000.04500.0000 1-340.00000.06300.0000 ! 21-20.00000.25000.0000 ! 4 -- 50.00000.20850.0000 I 19-410.00000.08000.0000 PowerFlowAnalysis165 Fig. P5.451BusPower System. BusP- QTAP 3 -351.0450 7 - 361.0450 11- 371.0500 14 - 471.0600 16 - 391.0600 18- 401.0900 19-411.0750 20 - 421.0600 24 -431.0750 30-461.0750 1 -341.0875 21-221.0600 5-41.0800 27 - 4510600 26 - 441.0750 166PowerSystemAnalysis BusData- VoltageandScheduledPowers BusnoVoltagemagnitudeVoltagephaseangleRealpowerReactive power (p.u.)(p.u.)(p.u.) 110800000000.00000.0000 2] 000000000- 0.5000- 0.2000 31.000000000- 09000-0.5000 41.00000.00000.00000.0000 51.00000.0000-0.11900.0000 61.00000.0000-0.1900- 0.1000 71.000000000- 0.3300-0.1800. 81000000000- 0 4400- 02400 9100000.0000- 0.2200-0.1200 10100000.0000-0.2100- 01200 II1.00000.0000- 0.3400-- 0.0500 121.000000000- 0.2400- 0.1360 13100000.0000- 0.1900- 0.1100 141.000000000- 0.1900- 0.0400 IS1.00000.00000.24000.0000 16100000.0000- 0.5400- 0.3000 17100000.0000- 0.4600- 0.2100 18100000.0000- 0.3700- 0.2200 191.00000.0000- 0.3100-0.0200 20100000.0000- 03400- 0.1600 211000000000000000.0000 22100000.0000- 01700- 0 0800 13100000.0000- 0.4200- 0.2300 2 ~1.000000000- 0.0800- 0.0200 251.00000.0000-0.1100- 0.0600 261.00000.0000- 0.2800- 0.1400 271.000000000- 0.7600- 0.2500 28IOOO() 00000- 08000- 0.3600 Contd. .... PowerFlowAnalysis167 BusnoVoltagemagnitudeVoltagephaseangleRealpowerReactive power (p.u.)(p.u.)(p.u.) 291.00000.0000- 0.2500- 0.1300 301.00000.0000- 0.47000.0000 311.00000.0000- 0.4200-0.1800 321.00000.0000- 0.3000- 0.1700 331.00000.00000.50000.0000 341.00000.0000- 05800- 0.2600 351.00000.00000.00000.0000 361.00000.00000.00000.0000 371.00000.00000.00000.0000 38I. 00000.0000170000.0000 391.00000.00000.00000.0000 401.00000.00000.00000.0000 411.00000.00000.00000.0000 421.00000.00000.00000.0000 431.0000000000.00000.0000 441.00000.0000175000.0000 451.00000.0000000000.0000 461.00000.00000.00000.0000 471.00000.00000.00000.0000 481.0000000000.55000.0000 491.00000.00003500000000 501.00000.00001.2000o.oono 511.00000.0000- 0.5000- 0.3000 168PowerSystemAnalysis BusNo.Voltage at VeBReactivepowerlimit 151.03000.1800 301.00000.0400 331.00000.4800 381.06000.9000 441050004500 481060002000 491.07000.5600 501.07001.500 P 5.4Thedatafora13machine,71bus,94linesystemisgiven.Obtaintheloadflow solution. Data: No.of buses71 No.of lines94 Basepower(MVA)200 No.of machines13 No.of shuntloads23 BUSNOGENERATIONLOADPOWER I- - 0.000 2000.00.00.0 3506.0150.00.00.0 40.00.00.00.0 50.00.00.00.0 6100.032.00000 70.00012.88.3 8300.0125.00.00.0 90.00.0185.0130.0 100.00.080.050.0 II0.00.0155.096.0 120.00.00.00.0 Contd..... PowerFlowAnalysis169 BUSNOGENERATIONLOADPOWER 130.00.0100.062.0 140.00.00.00.0 15180.0110.0000.0 160.00.073.045.5 170.00.036.022.4 180.00.016.09.0 190.00.032019.8 20000.027016.8 210.00.032.0In 220.00.00.000 23000.075.046.6 240.00.0000.0 250.000133.0825 26000.00000 27300075.00.000 280.00.030.020.0 29260.070.00.00.0 300.00.0120.00.0 310.00.0160.074.5 320.00.00.0994 330.00.00.000 340.00.0112.069.5 350.00.00.000 360.00.050.032.0 37000.0147092.0 38000.0935880 3925.030.00.00.0 400.00.00.00.0 410.00.0225.0123.0 420.00.00.00.0 430.00.00.00.0 Contd. .... 170PowerSystemAnalysis BUSNOGENERATIONLOADPOWER 44180.055.00.00.0 450.00.00.00.0 460.00.078.038.6 470.00.0234.0145.0 48340.0250.00.00.0 490.00.0295.0183.0 50000.040.024.6 510.00.02270142.0 520.00.00.00.0 530.00.00.00.0 540.00.0\08.068.0 550.00.025.548.0 560.00.00.00.0 57000055635.6 580.00.042027.0 590.00.057.027.4 600.00.00.00.0 610.00.00.00.0 620.00.040.027.0 630.00.033.220.6 64300.075.00.00.0 650.00.00.00.0 6696.025.00.00.0 670.00.014.06.5 6890.025.00.00.0 690.00.00.00.0 700.00.011.47.0 710.00.00.000 PowerFlowAnalysis171 LINE DATA LineNoFromBusToBusLineimpedance112Y chargeTurnsRatio 1980.00000.05700.00001.05 2970.32000.07800.0090100 3950.06600.16000.00471.00 49100.05200.12700.01401.00 510110.06600.16100.01801.00 67100.27000.07000.0070100 712110.00000.05300.00000.95 8II130.0600014800.03001.00 914130.00000.08000.00001.00 1013160.97000.23800.02701.00 1117150.00000.09200.00001.05 1276000000.22200.00001.05 13740.00000.08000.00001.00 14430.00000.03300.00001.05 15450.00000.16000.00001.00 164120.01600.07900.07101.00 1712140.01600.07900.07101.00 1817160.00000.08000.00000.95 19240.00000.06200.00001.00 204260.01900.09500.19300.00 21210.00000.03400.00001.05 2231260.03400.1670015001.00 2326250.00000.08000.00000.95 242523024000.52000.13001.00 25222300000008000.00000.95 2624220.00000.08400.00000.95 2722170.04800.25000.05051.00 28224001000.10200.33531.00 2923210.03660.14120.01401.00 3021200.72000.18600.00501.00 3120III014600.37400.01001.00 Contd. 171PowerSystemAnalysis LineNoFromBusToBusLineimpedance112Y chargeTurnsRatio 3219180.05900.15000.00401.00 3318160.0300007550.00801.00 3428270.00000.08100.00001.05 3530290.00000.06100.0000105 3632310.00000.0930000000.95 3731300.0000008000.00000.95 382832000510.05100.67061.00 393330.0130006400.05801.00 403147O.()1100.0790017701.00 412320.01580.15700.51001.00 423334000000.08000.0000095 4335330.00000.08400.00000.95 4435240.0062006120.21201.00 4534360.07900.20100.02201.00 4636370.1690043100.01101.00 4737380.0840018800.02101.00 484039000000.38000.00001.05 4940380.08900.21700.02501.00 5038410.10900.19600.22001.00 5141510.23500.60000.01601.00 5242410.00000.05300.00000.95 5345420.00000.08400.00000.95 5447490.21000.10300.92001.00 5549480.00000.04600.0000105 56495000170008400.0760100 5749420.03700.1950003901.00 585051000000.0530000000.95 5952500.00000.08400.00000.95 6050550.02900.15200.03001.00 6150530.01000.05200.03901.00 6253540:00000.08000.00000.95 635754002200.05400.00601.00 Contd. .... PowerFlowAnalysis173 LineNoFromBusToBusLineimpedance112Y chargeTurnsRatio 6455560.01600.08500.01701.00 655657000000.08000.00001.00 6657590.02800.07200.00701.00 6759580.04800.12400.01201.00 6860590.00000.08000.00001.00 6953600.03600.18400.37001.00 7045440.00000.12000.00001.05 7145460.03700.09000.01001.00 7246410.08300.15400.0170100 7346590.10700.19700.02101.00 7460610.01600.08300.01601.00 7561620.00000.08000.00000.95 76586200420010800.00201.00 7762630.03500.08900.0090100 7869680.00000.22200.00001.05 7969610.02300.11600.10401.00 8067660.00000.18800.00001.05 816564000000.06300.00001.05 8265560.02800.1-l400.02901.00 8365610.02300.11400.02401.00 8465670.02400.06000.09501.00 8567630.03900.09900.01001.00 8661420.02300.22930.06951.00 8757670.05500.291000070LOO 88457001840OA680001201.00 8970380.16500.42200.01101.00 9033710.05700.29600.05901.00 4171370.00000.08000.00000.95 9245410.15300.38800.10001.00 -n35430.0131013060.42931.00 ~ -9-l52520.01640.16320.53601.00 174PowerSystemAnalysis ShuntLoadData S.NoBusNoShuntLoad 12000- 0.4275 2130.000.1500 3200.000.0800 4240.00- 0.2700 5280.00- 0.3375 6310.000.2000 7320.00- 0.8700 8340.000.2250 9350.00- 0.3220 10360.000.1000 11370.000.3500 12380.000.2000 13410.000.2000 14430.00- 02170 15460.000.1000 16470000.3000 17500.000.1000 18510.000.1750 19520.00-0.2700 20540.000.1500 21570.000.1000 22590.000.0750 23210.000.0500 PowerFlowAnalysis Questions 5.1Explaintheimportance of loadflowstudies. 5.2Discussbreifly thebusclassification. 5.3Whatistheneedforaslack busor referencebus? Explain. 5.4Explain Gauss-Seidelmethodof loadflow solution. 175 5.5Discuss the method ofNewton-Raphson method ingeneral and explain its applicalibility forpowerflowsolution. 5.6Explain the Polar-Coordinates method of Newton-Raphsonload flowsolution. 5.7Give the Cartesian coordinates method or rectangular coordinates method of Newton-Raphsonloadflow solution. 5.8GivetheflowchartforQ.No.6. 5.9GivetheflowchartforQ.No.7. 5.10Explain sparsity and its application inpower flowstudies. 5.11How are generator buses areP,Vbusestreatedinloadflowstudies? 5.12Givethealgorithmfordecoupledloadflowstudies. 5.13Explainthefastdecoupedloadflowmethod. 5.14Comparethe Gauss-SeidelandNewton-Raphsonmethodforpower flowsolution. 5.15ComparetheNewton-Raphsonmethod,decoupledloadflowmethodandfast decoupedloadflowmethod. 6SHORTCIRCUITANALYSIS Electricalnetworksandmachinesaresubjecttovarioustypesof faultswhileinoperation. During the fault period, the current flowing is determined by the internal e.m.fs of the machines inthe network, andby the impedances of the network and machines.However, the impedances of machines may change their values from those that exist immediately after the fault occurrence todifferentvaluesduring thefaulttillthefaultiscleared.Thenetworkimpedancemayalso change,if thefaultisclearedbyswitching operations.Itis,therefore,necessarytocalculate the short-circuit current at differentinstants whenfaultsoccur.For suchfaultanalysisstudies andingeneralforpowersystemanalysisitisveryconvenienttouseperunitsystemand percentage values.Inthefollowing thissystemisexplained. 6.1Per Unit Quantities The per unit value of any quantity is the ratio of the actualvalueinany units to the chosen base quantifyof thesame dimensionsexpressedasadecimal. Actualvalueinanyunits Per unit quantity =.. base or reference value In the same umts Inpowersystemsthebasicquantitiesof importancearevoltage,current,impedance andpower.For allperunitcalculations abaseKVAor MVAandabase KVaretobechosen. Oncethebasevaluesorreferencevaluesarechosen.theotherquantitiescanbeohtainedas follows: Selecting the totalor 3-phaseKVAasbase KVA,fora 3-phase system ShortCircuit Analysis baseKVA Base currentinamperes=h ,,3[baseKV(Iine-to-line)] .[base KV(line-to-line)2x 1000] Base Impedance In ohms =h[] ,,3(base KV A)/3 B .d.h(baseKV (line-to-line)2 aseImpeance In 0ms= baseMVA d.h(baseKV(line-to-line)2x\000 BaseImpeanceIn 0m =------'------'----baseKVA Hence, wherebase KVAandbase MVAarethe totalor threephasev.tllles. If phase valuesareused baseKVA Base currentinamperes=-b-a-se-K-V-base voltage Base impedance inohm =----=-base current (baseK V)2x \000 baseK V A per phase .(base KV)2 BaseImpedance In ohm=bMVAh aseper pase Inalltheaboverelationsthepower factorisassumedunity,so that base power K W = baseK VA (actualimpedance inohm)xKVA Per unitimpedance =2 (baseK V)x1000 Now, Some times,itmaybe requiredtousetherelation (I .d.h)(Per unit impedance in ohms) (base KV)2x\000 actuaImpeanceIn 0m=. . : . . . . . . . - - - ~ - - - - - . : . . . . . . . : . . - - - - - - ' ' - - - -baseKVA 177 Veryoftenthevaluesareindifferentbasevalues.Inordertoconverttheperunit impedance fromgivenbase toanother base,the followingrelationcanbederivedeasily. Per unitimpedance onnewbase (new KVA base)(giVen KV base)2 Z-u =ZU newpgIvenp.given KV A basenew KV base 178PowerSystemAnalysis 6.2Advantages of Per Unit System 1.Whileperforming calculations, referring quantitiesfromone side of the transformer to the other side serious errorsmay be committed.This canbe avoidedbyusing per unitsystem. 2.Voltages,currentsandimpedancesexpressedinperunitdonot changewhenthey are referred from one side of transformer to the other side. This is a great advantage'. 3.Per unit impedances of electrical equipment of similar type usually lie within a narrow range,whentheequipment ratingsareusedasbasevalues. 4.Transformer connectionsdonotaffecttheperunitvalues. 5.Manufacturersusuallyspecify theimpedances of machines and transformers inper unitor percent of nameplate ratings. 6.3Three Phase Short Circuits Intheanalysisof symmetricalthree-phaseshortcircuitsthefollowingassumptionsare generally made. 1.Transformersarerepresentedbytheir leakagereactances.The magnetizing current, andcorefussesareneglected.Resistances,shuntadmittancesarenotconsidered. Star-delta phaseshiftsarealsoneglected. 2.Transmissionlinesarerepresentedbyseriesreactances.Resistancesandshunt admittances areneglected. 3.Synchronousmachinesarerepresentedbyconstantvoltagesourcesbehind subtransient reactances. Armature resistances, saliency and saturation areneglected. 4.All non-rotating impedance loads areneglected. 5.Induction motors are represented just as synchronous machines with constant voltage source behindareactance.Smaller motorloadsaregenerallyneglected. Per unit impedances of transformers : Consider a single-phase transformer with primary and secondary voltagesandcurrentsdenotedbyV I' V2andI I'12respectively. VI12 wehave,-=-V2 II V Baseimpedanceforprimary =_I II V2 Baseimpedanceforsecondary=G ZIIIZI Per unitimpedancereferredtoprimary =(VI/II)VI ShortCircuit Analysis [2Z2 Perunitimpedancereferredtosecondary =V-2 Again,actualimpedance referred to secondary =Z, Perunitimpedancereferredtosecondary ( Yy2,)2 =Per unitimpedancereferredtoprimary 179 Thus,theperunitimpedancereferredremainsthesameforatransformeroneither side. 6.4Reactance Diagrams [npowersystemanalysisitisnecessarytodrawanequivalentcircuit forthesystem.Thisis animpedancediagrams.However,inseveralstudies,includingshort-circuitanalysisitis sufficienttoconsideronlyreactancesneglectingresistances.Hence,wedrawreactance diagrams.For 3-phasebalancedsystems,itissimpler torepresent thesystembyasingleline diagram without losing the identify of the 3-phase system. Thus, single linereactance diagrams canbedrawnforcalculation. ThisisillustratedbythesystemshowninFig.6.1(a)&(b)andbyitssingleline reactancediagram . .cY ei Gene transformerLines (a) A power system Fig. 6.1 6.5Percentage Values (b)Equivalent single-line reactance diagram The reactances of generators, transformers and reactors aregenerally expressedinpercentage values topermit quick short circuit calculation. Percentagereactanceisdefinedas: [X %X=-xIOO y 180 where,I =fullload current V=phase voltage X=reactanceinohmsper phase PowerSystemAnalysis ShortcircuitcurrentIscinacircuit thencanbeexpressedas, _V=V. IxlOO Isc- XV. (%X) I.100 %X Percentage~ a c t a n c ecan expressedinterms of KVAandKV as following Fromequation Alternatively (%X) . V(%X)V2 X =----= -'--'----1.1 00100. V. I (%X)(KV)2 10 ==-----'--'----KVA KVA (%X) =X.10(KV)2 VV (%X)-- .--xIOOO 10001000 V 100.-.I 1000 Ashasbeenstatedalreadyinshortcircuit analysissincethereactanceXisgenerally greater thanthreetimes theresistance,resistancesareneglected. But, in case percentage resistance and therefore, percentage impedance values are required then,inasimilar manner we can define and IR %R =-x 100 V IZ %Z=-xIOO V withusualnotation. The percentage values of Rand Z also do not change with the side of the transformer or eithersideof the transformer theyremainconstant.Theohmicvaluesof R,XandZchange fromonesidetotheother sideof thetransformer. whenafaultoccursthepotentialfallstoavaluedeterminedbythefaultimpedance. Shortcircuitcurrentisexpressedintermof shortcircuitKVAbasedon thenormalsystem voltage at thepoint of fault. 6.6Short Circuit KVA Itisdefinedastheproductof normalsystemvoltageandshort circuit currentatthepointof fault expressedinKVA. ShortCircuit Analysis LetV =normalphase voltagein volts 1= fallload current inamperes at base KVA %X =percentagereactanceof thesystemexpressedonbaseKVA. Theshortcircuitcurrent, 100 JSc=I.%X The three phase or totalshort circuit KVA 3.VIsc3.V.I1003VI100 1000(%X) 10001000%X 100 Therefore short circuit KVA =Base KVAx(%X) 181 Inapowersystemoreveninasinglepowerstationdifferentequipmentmayhave differentratings.Calculationarerequiredtobeperformedwheredifferentcomponentsor unitsarerateddifferently.Thepercentagevaluesspecifiedonthenameplateswillbewith respect to their name plate ratings. Hence. it it necessary to select a common base KVA or MVA andalsoabaseKY.The followingaresome of theguideIinesforselectionof basevalues. 1.Rating of the largest plant or unit forbase MVAor KVA. 2.The totalcapacity ofa plant or system forbaseMVAor KVA. 3.Any arbitrary value. (BaseKVA). (%X)=.(% XatUOItKVA) onnewbaseUnitK VA Ifa transformer has8% reactance on 50KVAbase,its value at100 KVAbasewillbe (100) (%X)IOO KVA=50x8= 16% Similarly thereactance values change with voltagebase asper therelation where X I=reactanceatvoltageV I andX2 =reactance atvoltageV 2 Forshortcircuitanalysis,itisoftenconvenienttodrawthereactancediagrams indicating the values inper unit. 182PowerSystemAnalysis 6.7Importance of Short Circuit Currents Knowledgeof shortcircuitcurrentvaluesisnecessaryforthefollowingreasons. 1.Faultcurrentswhichareseveraltimeslargerthanthenormaloperatingcurrents producelargeelectromagneticforcesandtorqueswhichmayadverselyaffect the statorendwindings.Theforcesontheendwindingsdependonboththed.c.and a.c.componentsof statorcurrents. 2.The electro dynamic forceson the stator end windings may result in displacement of the coils against one another. This may result inloosening of the support or damage to theinsulationof thewindings. 3.Following ashortcircuit,itisalwaysrecommendedthat themechanicalbracing of theendwindingstocheckedforanypossibleloosening. 4.The electricaland mechanicalforcesthat develop due to asudden three phase short circuit are generallysevere when themachineisoperating under loaded condition. S.As the fault is cleared with in 3 cycles generally the heating efforts are not considerable. Shortcircuitsmayoccurinpowersystemsduetosystemovervoltagescausedby lightningorswitchingsurgesor duetoequipmentinsulationfailureor evenduetoinsulator contamination.Sometimesevenmechanicalcausesmaycreateshortcircuits.Otherwell known reasons include line-to-line, line-to-ground, or line-to-line faults on over head lines. The resultantshortcircuit has totheinterrupted withinfewcyclesby thecircuitbreaker. Itisabsolutelynecessarytoselectacircuitbreakerthatiscapableof operating successfullywhenmaximumfaultcurrentflowsatthecircuitvoltagethatprevailsatthat instant. Aninsight canbegainedwhenweconsider anR-L circuit connected to an alternating voltagesource,thecircuitbeingswitchedonthroughaswitch. 6.8Analysis ofR-L Circuit Consider the circuitintheFig.6.2. + s t=0' R L Fig.6.2 Let e= EmaxSin(rot+ a) when theswitch Sisclosedatt= 0+ di e= EmaxSin(rot + a) = R+ Ldt ShortCircuit Analysis aisdeterminedbythemagnitudeof voltagewhenthecircuitisclosed. The general solutionis where and E--[ -Rt1 i=maxSin(cot+a-8)-eLSin(a-8) IZI IZI=~ R 2 + co2L2 coL 8=Tan-I-R Thecurrentcontainstwocomponents: Emax a.c.component =IZT Sin (cot+a - 8) and -Rt EmaxeLe) d.c.component=IZTSm(a-If theswitchisclosedwhena- e = 1tor whena- 8= 0 thed.c.componentvanishes. 1t thed.c.componentisamaximumwhena - 8= '2 6.9Three Phase Short Circuit on Unloaded Synchronous Generator 183 If athreephaseshortcircuitoccursattheterminalsof asalientpolesynchronousweobtain typicaloscillogramsasshowninFig.6.3fortheshortcircuitcurrentsthethreephases. Fig.6.4 shows the alternating component of theshort circuit current when the d.c. component iseliminated.Thefastchangingsub-transientcomponentandtheslowlychanging transient components areshown at A and C.Figure 6.5shows theelectrical torque.The changing field currentisshowninFig.6.6. Fromtheoscillogramof a.c.componentthequantitiesx;;,x ~ ,xdandx ~canbe determined. If Visthelinetoneutralprefault voltage then thea.c.component. iac=~=I", the r.m.ssubtransient short circuit.Its durationis determinedbyTd'' the xl] V subtransientdirectaxistimeconstant.Thevalueof iacdecreasesto7whent>Td' ,d withTdasthedirect axistransient timeconstantwhent >Td V 184PowerSystemAnalysis Themaximumd.c.off-setcomponent thatoccursinanyphaseat(l =0is .k2V-IITA 1(t)-"L.-e d,c,maxxd whereT Aisthearmaturetimeconstant. Fig. 6.3Oscillograms of the armature currents after a short circuit. F J: ~----- ------- --- -- ___0 - ----- B i::r-~-0 u1 ~ 61 tsec.~ Fig. 6.4Alternating component of the short circuit armature current ShortCircuit Analysis o0.10020030 ---.....t 0.400.50 Fig. 6.5 Electrical torque onthree-phase termlnaJshort circuit. Field current after short circuit o,-_+-_-_-_-_....._-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_...;.. t----+ Fig. 6.6 Oscilogram of the field current after a short circuit. 6.10Effect of Load Current or Prefault Current 185 Considera3-phasesynchronousgeneratorsupplyingabalanced3-phaseload.Letathree phase fault occur at theload terminals.Before the fault occurs, a load current ILis flowing into theloadfromthegenerator.Let thevoltageatthefaultbev fandtheterminalvoltageof the generatorbeVt'Under faultconditions,thegeneratorreactanceisxd' The circuit inFig.6.7indicates the simulation of fault at the load terminals bya parallel switchS. E; = Vt + jXd'IL= VI'+(Xext+ jxd)IL whereE;isthesubtransient internalvoltage. 186PowerSystemAnalysis s Fault Fig. 6.7 For thetransientstate E ~=VI+ jxdIL =Vf + (Zext+ jxd ) IL E;orE ~areusedonlywhen thereisaprefault currentIL.OtherwiseEgthesteady state voltage in series with the direct axis synchronous reactance is to be used for all calculations. EgremainsthesameforallILvalues,anddependsonlyonthefieldcurrent.Everytime,of course,anewE;isrequiredtobecomputed. 6.11Reactors Wheneverfaultsoccurinpower systemlargecurrentsflow.Especially,if thefaultisadead short circuit at the terminals or bus barst?normouscurrents flow damaging the equipment and itscomponents.Tolimit theflowof largecurrentsunder therecircumstances current limiting reactorsareused.Thesereactorsarelargecoilscoveredforhighself-inductance. Theyarealsosolocatedthattheeffectof thefaultdoesnotaffectotherpartsof the systemandisthuslocalized.Fromtime to timenew generating unitsareadded toanexisting systemtoaugmentthecapacity.Whenthishappens,thefaultcurrentlevelincreasesandit maybecomenecessarytochangetheswitchgear.Withproperuseof reactorsadditionof generatingunitsdoesnotnecessitate changesinexisting switchgear. 6.12Construction of Reactors Thesereactorsarebuiltwithnonmagneticcoresothatsaturationof corewithconsequent reductionininductanceandincreasedshortcircuitcurrentsisavoided.Alternatively,itis possible touseiron core with air-gaps includedin the magnetic core so that saturation is avoided. 6.13Classification of Reactors (i)Generator reactors,(ii)Feederreactors,(iii)Bus-bar reactors ShortCircuit Analysis187 The above classification isbased on the location ofthe reactors. Reactors may be connected in serieswith thegenerator inserieswitheachfeederor tothebusbars. (i)Generator reactors Thereactorsarelocatedinserieswitheachof thegeneratorsasshownin Fig.6.8so that current flowingintoa faultF fromthegenerator islimited. Generators Bus Bars Fig. 6.8 Disadvantages (a)Intheeventof afaultoccuringonafeeder,thevoltageattheremaining healthy feeders also may loose synchronism requiring resynchronization later. (b)Thereisaconstantvoltagedropinthereactorsandalsopowerloss,even during normal operation. Since modern generators are designed to with stand deadshortcircuitattheirterminals,generator reactorsarenow-a-daysnot usedexcept foroldunitsinoperation. (ii)Feederreactors:Inthismethodof protectioneach isequippedwitha seriesreactorasshowninFig.6.9. Intheeventof afaultonanyfeederthefaultcurrent drawn isrestrictedby the reactor. Generators Reactors Bars JC' Fig. 6.9 188 F PowerSystemAnalysis Disadvantages:I.Voltagedropandpowerlossstilloccursinthereactor fora feeder fault.However, the voltage drop occurs only in that particular feeder reactor. 2.Feederreactorsdonotofferanyprotectionforbusbarfaults.Neverthless, bus-barfaultsoccurveryrarely. Asseries reactorsinhererbly create voltagedrop,systemvoltage regulationwill beimpaired.Hencetheyaretobeusedonlyinspecialcasesuchasforshort feedersof largecross-section. (iii)Bus bar reactors: Inboth the above methods, the reactors carry fullload current under normal operation.The consequent disadvantage of constant voltage drops andpowerlosscanbeavoidedbydividing thebusbarsintosectionsandinter connectthesectionsthroughprotectivereactors.Therearetwowaysof doing this. (a)Ring system : Inthismethodeachfeederisfedbyonegenerator.Verylittlepower flows acrossthereactorsduringnormaloperation.Hence,thevoltagedropand power loss arenegligible.If a fault occurs onanyfeeder,only the generator to which the feederisconnected willfeedthefaultando ~ h e rgenerators are requiredtofeedthefaultthroughthereactor. (b)Tie-bar system: Thisis animprovement over the ring system. This isshown inFig.6.11.Currentfedintoafaulthastopassthroughtworeactorsin seriesbetweensections. FF BVS Bars BVS bar Tie bar Generators F2 Peeders Fig.6.10Fig. 6.11 Anotheradvantageisthatadditionalgenerationmaybeconnectedtothe systemwithoutrequiringchangesintheexisting reactors. Theonlydisadvantageisthatthissystemsrequiresanadditionalbus-bar system,thetie-bar. ShortCircuit Analysis189 Worked Examples E 6.1Twogeneratorsratedat10MVA,11KVand15MVA,11KVrespectivelyare connectedinparallel toabus.The busbarsfeedtwomotorsrated 7.5MVA and 10MVArespectively.Theratedvoltageof themotorsis9KV.Thereactanceof eachgeneratoris12%andthatof eachmotoris15%ontheirownratings. Assume50MVA,10KVbase anddrawthereactance diagram. Solution: Thereactances of the generators and motorsarecalculated on 50MVA,10KVbase values. Reactanceof generatorI= XG1 = 12.C~r.( ~ ~ )= 72.6% (11)2(50) Reactanceof generator 2= XG2 = 1210.10= 48.4% Reactanceof motorI= XM1 = 15.( I ~r(~ . ~ )= 81 % Reactanceof motor 2= XM2 =15(190 r( ~ ~J = 60.75% ThereactancediagramisdrawnandshowninFig.E.6.1. Fig.E.6.1 E.6.2A 100MVA,13.8KV,3-phase generator has areactance of 20%. The generator is connectedtoa 3-phase transformer T Irated100MVA12.5KV 1110KVwith10% reactance.Theh.v.sideof thetransformer isconnectedtoatransmissionlineof reactance 100 ohm. The far end of the line is connected to a step down transformer T 2'made of three single-phasetransformerseachrated30MVA,60KV /10KV with10%reactancethegenerator suppliestwomotorsconnectedonthel.v.side T2as showninFig.E.6.2.Themotorsareratedat 25MVAand50MVAbothat 10KVwith15% reactance.Draw the reactance diagram showing all the values in perunit.Takegeneratorratingasbase. 190PowerSystemAnalysis Solution: Base MVA =100 BaseKV=13.8 110 BaseKVfortheline = 13.8x12.5= 121.44 ,J3 x66KV114.31 Line-to-line voltage ratio ofT2 =10KV:=----w-121.44xl0, Base voltage formotors =114.31=10.62KV %Xforgenerators=20%=0.2p.ll. (12.5J2100 %Xfortransformer T)=10x13.8x 100=8.2% %X fortransformer T2on,J3x 66:10KVand3x30 MVAbase =10% %X for T2on100MVA,and121.44 KV:10.62 KVis %XT2= 10xC ~ . ~ 2 rx C ~ ~ J= 9.85%=0.0985p.ll. (121.44J2 Base reactanceforline =100=147.47ohms 100 Reactanceof line =147.47=0.678p.ll. ( 10J2 (90J Reactanceof motor M)=lOx10.6225=31.92% =0.3192p.ll. ( 10J2 (90J Reactanceof motor M2= 10x10.6250= 15.96% Thereactance diagramisshowninFig.E.6.2. Fig.E.S.2 ShortCircuit Analysis191 E.6.3Obtaintheperunitrepresentationforthethree-phasepowersystemshownin Fig.E.6.3. Generator 1 :50MVA, Generator 2:25MVA, Generator 3: 35 MVA., Transformer T 1:30 MVA, Transformer T2: 25 MVA, Fig.E.6.3 10.5 KV;X = 1.8ohm 6.6KV;X= 1.2ohm 6.6 KV;X = 0.6ohm 11/66 KV,X = 15ohm