6.2 NETWORK MODEL FORMULATION 6.1 INTRODUCTION The load flow problem has,in fact, been alreadyintroducedin Chapter5 with the help of a f'undamentasl ystent,i.e. a two-busproblem (seeExample 5.8)' With the backgroundof the previous chapters,we are now ready to study the Eor4 lq4d flqW llq4ypfufg4l life power systemcomprising a large number operationalfeaturesof a compositepower system.Symmetricalsteadystateis' of buses, it is necessaryto proceed systematicallyby first formulating the in fact, the most importantmode of operationof a power system.Three major problems encounteredin this mode of operation are listed beiow in their network model of the sYstern. hierarchicalorder. A power systemcomprisesseveralbuseswhich areinterconnectedby rneans 1. Load flow problem of transmissionlines. Power is injected into a bus from generatorsw' hile the 2. Optimalload schedulingproblem loadsare tappedfrom it. Of course,there may be buseswith only generators and no-loads,and there may be others with only loads and no generators. 3. SystemscontrolProblem Further,VAR generatorsmay also be connectedto some buses.The surplus This chapteris devotedto the load flow problem, while the other two power at some of the busesis transportedvia transmissionlines to buses problems will be treateilin later chapters.Lt-radllow study in power systetn deficientin power.Figure6.1ashowsthe one-linediagramof a four-bussystem parlance is the steadystatesolution of the porversystemnetwork. The main with generatorsand loads at each bus. To arrive at the network model of a information obtained from this study comprisesthe magnitudesand phase po*\"i system,it is sufficientlyaccurateto representa short line by a series angles of load bus voltages,reactive powers at generatorbuses,real and impedanceand a long line by a nominal-zrmodel. (equivalent-7Tmay be used reactinepower flow on transmissionlines, othervariablesbeing specified.This foi very long lines). Often,line resistancemay be neglectedwith a small loss information is essentiaflor the continuousmonitoring of the currentstateof the in accuracybut a greatdeal of saving in computationtime. systernandfor analyzingthe effectivenessof alternativeplansfor future system expansionto meet increasedload demand. For systematic analysis,it is convenient to regard loads as negative Before the adventof digital computers,the AC calculatingboard was the generatorsand lump togetherthe generatorandloadpowersat the buses.Thus only meansof carrying out load flow studies.Thesestudieswere, therefore, at the ith bus, the net complexpower injectedinto the bus is given by tedious and time consuming.With the availability of fast and large size digital computers,all kinds of power systemstudies,inciudingload flow, can now be S;= Pi + jQi= (Pci- Po)+ j(Qci- Qo) carried out conveniently.In fact, some of the advancedlevel sophisticated studieswhich were almostimpossibleto carry out on the AC calculatingboard where the corrrplexpower suppliedby the generatorsis have now become possible.The AC calculatingboard has been rendered obsoletefor all practicalpurposes. Sci= Pot+iQai anclthe ccltnplexpower drawn by the loadsis Spi= Por+ iQoi The real anclreactivepowersinjectedinttl thc itlt bus arc thcn (6'1) Pi= Poi- P^ i = 1,2, \"'' fl Qi= Qci- Qoi Figure 6.lb showsthenetworkmodel of the samplepower systemprepared on the abovelines.The equivalentpower sourceat eachbus is representedby a shadedcircle. The equivalentpower sourceat the ith bus injectscurrentJr into the bus. It may be observedthat the structureof a power systemis such that all the Sourcesare alwaysconnectedto a commonground node. The networkmodelof Fig. 6.lb hasbeenredrawnin Fig. 6.lc afierlurnping the shunt admittancesat the buses.Besidesthe groundnode,it hasfour other nocies(buses)at which the currentfrom the sourcesis injectedinto the network. The line admittancebetweennodesi andk is depictedby !ip= Jri'Further,the mutual admittancebetweenlines is assumedto be zero' .Line transformers are representedby a seriesimpedance(or for accuraterepresenta- tion by series and shunt impedances,i.e. invertedL-network).
apitying Kirchhoff s currentlaw (KCL) at nodesr,2,3 and4,respectively, we getthe following fbur equations: Jt = Vrlro + (Vr - V) ln * (Vr - Vr) ),: Jz= Vz)zo+ (Vz- Vr) ln + (Vz_ V) yzt + (Vz_ Vq)yzq s= vztgo+ (y: - v) ln + (Vt _ V) lzz + (Vt_ Vq) yzq Jq= Vayqo+ (Vq- Vr) lzc * (Vq _ V) yzq Rearrangingand writing in matrix form, we get (yrc* tn -ln -\\z 0 *fn, - Y. ,rz (lzo*ln - lzt - !z+ (b) Equivalenctircuit _ !* * hz * lzq, (y:o * yrg -lu *rzt*yy, -t23 0 -Jz+ - lzq (Yqa* yzq * Yz+) Equation (6.3) can be recognizedt,o be of the standardform (6.4) ComparingEqs. (6.3) and (6.4), we can write (c) Power network of Fig. 6.1 (b) lumped and redrawn Yrr=)ro* ln+ ln Yzz= lzo * ltz t lzt + lzq Fig. 6.1 Samplefour-bussystem Ytt= ):o* ln* lzz* lzq Yu= lqo * lzq * ly Y l t = Y r c = - l n i Y v = Y q t= - ) r + = 0 Ytz= Yzt = - lni YZt = YZZ= - lZt Yzq=Y+z=-lzqiYy= yqz=- ly Sor Soz FnpmteueourrgiEmttnhuaatieatnacirlvah,daermaYdoindmrfi*gtmtti=ahottinenattYcnastrechnui.e)ecmoe(nftoyronfa,do,an(edslil.ef=Eeairdaaramc,n2dhdi,mt3otea,ifqtint-uadc4anei)alcsscgeitsooh) nncbenaeaaeltllrwigceeeetrdemebtdnrhdaneivriocse*siemcu(rtilmf,yaikbodaefm=ntadwitlrltfe,attehna2necn,tedah3(dee,omsq4reu)dintatriosalisvdnittenchhsgeee.s v1 using index notation,Eq. (6.a) can be written in compact form as (a) One-linediagram Fig. 6.1 Sample four bus system n (6.s) J i = D y p v p ii = r , 2 , . . . ,f l k=l or, in matrix form \"Inus= Isus Vnus (6.6
u'rvhrarnvp rv -VBUS \\Jr^e-^,+'^u^ lsD auLr^c l-l-ra^^lnx: or DUS admlttance and is known as bus 1.,.^L{\" admittancematrix. The dimensionof the y\"u5 matrix is (n x n) where n is the ainsvTwohlivlelebbdueaslsgimeoeprnietihdnmaCnsc.hFeamuprattehtrreisxr1m,0hooarwen,dethv1ee1ri.,mispmedoasnt cuesmefautlrfioxrisshaofruticl irmcuaittrisxt.u**dies numberof buses'[The totalnumberof noclesii e m = n + | includingthe ground Note: In the samplesystem of Fig. 6.1. the (reference)node.l ocsapoyrDstnitmvrerermaarlgsroye, irmntdceaheranainnnsgdeb.rse, oealnluthtsioohunogswh.An pitnhpaemtnodceri xertcasiodnpeoharildssetwirciiatnhtgetdohfsetnut oodpdieiecsssopofrf oldasuprgfaeersspiftoayawsnteedrr As seenabove, yru, is a symmetric except when phase shifting 6.3 FORMATION OF fsus BY SINGULAR transformersare involved, so that only nv+ t) terms are to be stored for an TRANSFORMATION 2 tiesTesn,o(e?xyeivptrnh-psl.miaob9dntelroeal.esesrrimdYnsynbet tsteAuf'icahl'iyoesenone=sewd'ndtssaibe0ttp(aasem)uhsai.tssnsumar.ssusFitdsnoioaianuiatgcelrylcrlleisfhltyaesaehdauttyraeoei.nngnstrsdduettm[wea2rrinneoemo2pudro]moemolsfiwa,kriobznetteysfehiBrmiurr*tbnoheo5iuzef=eenisnitnbnzwes0eugegasovsrr,mtreotiiehkfhleaepseeelt)blePlem,aytusmhocsrpseeewhyeeamsdnbeysrtoustusreirci..simtyA-yiyini5tudsrrugmoetonhcfnqaudooauuygernmkirtirrbntraeyeceetrdafrgoc,iwaaricent,esesneaspndoerlihcodreontioonwegtltpmylrhhoya\"cenprrotatktiouoynofsiintsnrnaia-ss9e,tzvitinf0coeneetvtnrorweoaooystd. Graph Equation (6.6) can also be written in the form Vnus= Zausleus (6.7) where Zsu5 (bus impedancematrix) = fsLs (6.8) fbr a networkof fbur buses(fbur inclependenntodes) ieccnlaaev-llfmleleelmrdedt imndnttrrgesiaovtrnrifninscuIgf'serpur.rnow, iuimhnbsitepl eyicemioedpymlade\"nesudcn,,seaoyisnsnmoc-afmezoesstefhrptoterahiincrensZtonehmsdoeuedas.seto.r.Ts,i,x,ha.Z,ren-psAdu\"doutsiah\"rnng!i sieoionenfgfdaa-ldZfenuisaroleulgt m,ombeneearanetlrtoemsiobxleef.tmain.Zietensu.ne.,utdsz, aebaryrreoe It is t' be stressedhere that yBus/zBusconstitute models of the passive portionsof thepower network. tednhvieefecf*ind*ezcTesssuhsisslnteia\"irgereuyscdllaw'aroenrh'sceAburneeltlaoar[lbre'2egtf0aeea]ir.tsnrhelienudagntsZottasodggnmr5oe,iubttsneatdcnroaconuergisnsesailneartrcaitfkebicdsbiaaeultnstc.thiIeeoenfstotltahhceigskr,foboruurr*mnsdmeirotsoscitalarss-viecmoo,ipidntlydniisauticmouhnseieeuridvacoelalrsyl rtn6gheea'2edtiBnweuaueaconsdesrdkeswa6ccwdi'hd7oimtae)mh'nsitpowpgtuafrehtenseicacc-roahde'umamdtrphdesapeimettlri,biooiscxunrpaysiatosiaorafsdonnliimftnodtyeewiittsnstiiman,ourngeenscgeeteorumdoeliifqaatnsiutttirsnissirxogeigmtrrcrrvneaapieanntnintcsgbetifsetrsoyo.trfIaamonoddrfemvcfrdaloseoan,nedwttattaarcpagpn.r(seroesotsmaeptboseaolEeirOttamxhhftaiii.eesomI,tadnpavthfhlionealeyrdsl asmcpahp*ke*lii*mneFgdeotu'roPsdeioiesovcafeedrLwvcUaiosnmeftaaemcgttheeoetsrhosdofoidsfthsatehdoevrcaaotndnetmvaaerigitnntetagisoonftnceeaecmlxhicmanetipqrsieusxdieavas,nn,(drdcobiearfmulogeapm.trtriipcxelsqom)uyhiianrayegvmbceeoermneotcvpsIeealgnrcc]trl.osymrobereabegyne convenience, direction is so assigned as to coincide with the assu'red positive direction of the element current.
i90 l ModernpoweSr ystemAnatysis Or@ Primitive Network o vr, = Er- E, Fig.6.2 Lineargraphof thecircuitin Fig.6.1c where E, and E\" are the voltages of the element nodes r and s, respectively. here that each source and the shunt admittance connected across it are It may be remembered here that for steady state AC performan.., ull element representedby a single element. In fact, this combination representsthe most variables (vr* E, 8\", irr,7r.,) are phasors and element parameters (zrr, .rr\") ar€ generalnetwork element and is describedunder the subheading\"primitive complex numbers. Network\". The voltage relation for Fig. 6.4a can be written as A connecteds'bgraph containing all the nodes of a graph but having no vrr* €rr= Zrri^ (6.11) closedpathsis calleda tree. The elementsof a tree are called branchesor ffee branches.The number of branches b that form a tree are given by Similarly, the current relation for Fig. 6.4b is b = m * 1 - n ( n u m b e r o bf u s e s ) ir, * jr, = lrrv^ (6.r2) (6.9) Thoseelementsof the graphthat are not includedin the tree are called links (or I link branches) and they form a subgraph,not necessarilyconnected,called c@ rn yr\"=E -E\" vo------------------7--+ - Branch \\.e - - Link e =9 (a) lmpedanceform (b) Admittance form m=5 b=m-1=5-'l=4=n Fig.6.4 Representationf a networkelement l=e-b=5 iTnhaedfomrimttsanocf eFfiogrsm. 6i.s4arealantdeb,datoretehqeusievarileesnvtwolhtaegreeiinnthimeppaerdaallneclsefoourrmcebcyurrenr @ Jr,= - Yrs€rs (a) Tree (b)Co-tree Also Fig. 6.3Treeandcotreeof the orientedconnectedgraphof Fig.6.2 cotree.The numberof links / of a connectedgraph with e elementsis !r.,= |/7r, I=e-b=e-m+l (6.10) A set of unconnectedelements is defined as a primitive network. The performanceequationsof a primitive networkerregi enbelow : Note that a tree (and therefbre,cotree)of a graph is not unique. In impedanceform v V+E=ZI (6.13)
'f{W';l ModernPowersvstemAnatvsis In admittanceform Load Fiow Stucjies < i flH: I+J=W (6.14) v = AV,'J' I where the bus incidence matrix A is (6.16) Here V and.Iarethe elementvoltageandcurrentvectorsrespectively,andrl and e branches E are the sourcevectors.z and Y arereferred to as the primitive i links admittancematrices,respectively.'Thesaere relatedas Z = Y-r.If there is no I 1000 mutual coupling betweenelements,Z and Y arediagonal where the diagonal 20 1 0 0 entries are the impedances/admittancesof the network elements and are 30 0 1 0 reciprocal. 40 0 0 1 ) 001_l NeturorkVariables in Bus Frame of Reference 6 0 - l1 0 7 l - l0 0 Linear network graph helps in the systematicassemblyof a network model. The 8 0 - 10 1 (6.r7) main problem in deriving mathematical models for large and complex power 9 - l0 1 0 networks is to selecta minimum or zero redundancy(linearly independent)set of current or voltagevariableswhich is sufficient to give the information about This matrix is rectangular and therefore singular. Its elementsa,oare found all element voltages and currents. One set of such variables is the b tree per the following rules: voltages.*It may easily be seen by using topologicalreasoningthat these as variables constitutea non-redundantset. The knowledge of b tree voltages allows us to compute all element voltages and therefore, all bus currents aik = 1 if tth element is incident to and oriented away from the ftth assumingall elementadmittancesbeing known. node (bus) Considera tree graph shown in Fig. 6.3a wherethe ground node is chosen = - | if tth element is incident to but oriented towardsthe ftth as the referencenode. This is the most appropriatetree choice for a power node network. With this choice,the b tree branchvoltagesbecomeidentical with the bus voltagesas the tree branchesare incident to the ground node. Bus Incidence Matrix For the specificsystemof Fig. 6.3a, we obtain the following relationstetween = 0 if the ith element is not incident to the kth node ( 6 .l 8 ) the nine elementvoltagesand the four bus (i.e. treebranch)voltagesV1,V2,V3 SubstitutingEq. (6.16) into Eq. (6.14),we get and Va. I+J=yAVsu5 Vut = Vt Premultiplyingby Ar, Vtz= Vz ArI+ArJ=AryAV\"u, (6.1e) Vut = Vt F,achcomponentof the n-dimensionalvector ATI is the algebraicsum of elementcurrentsleaving the nodes7,2, ..., n. Vuq= Vq the V t s = V t - Vn (6.15) Therefore,the application* of the KCL must result in V t a = V t - v2 ArI =o (6.20) V n = V t - v2 tsShu*iemFmroeiolrafforarlyel,ltheseaocubhruccsoeccmuuprrrroeennnettsns*tion*.fjHetchetenecvdeiencwttoeonrcoAadTneJw\"surri,nt2eb,e..r.e,nc.oTgnhiezseedcaosmthpeoanlgeenbtsraariec V t s = V q - v2 V t g = V t - vr nodel, Arl gives or, in matrix form iro+ ir, - l:r= 0 *Another *-*TFhoerreadershoulcvi erify this for anothernode. useful set of network variables are the / link (loop) currents which nodel, AT\"/gives j61 =currentinjectedinto bus 1 constitute a zeroredundancyset of network variables [6]. becausoetherelementcsonnectetdo bus t haveno sources.
i,il9:4iitl Modern Power S v s t e m- - - - - - A n a l v s i s- - \" - - ' f Load Flow Studies I ftt^ J --- -. I I ArJ = Jsus (6.2r) (6.22) Equation(6.19)thenis simplifiedto rvB U S =- Al T vI,Ar ,/eus= ATYAV\",,,, --J---.^^_-.--rr^'vrrrrgvl'va * yr:) ( yzo* yn - lzt 0 - Jtz * lzt * t-zq, the samenodalcurrentequationas (6.6). The bus admittancematrix can then - )r: - !'tn be obtainedfrom the singulartransformationof the primitive Y, i.e. 0 - }'l Yeus= ATYA (6.23) A computerprogrammecanbe developedto write the bus incidencematrix '- .,t.'23 ()'lo f- .yr A from the interconnectiondataof the directed elementsof the power systetn. * lzt * Jy, Standardmatrix transposeand multiplication subroutinescan then be usedto compute Yu* fiorn Eq. (6.23). _!zc *\"*'^-\"r The elementsof this matrix, of course,agreewith thosepreviouslycalculated , Example6.1 | in Eq. (6.3) I Find the Y6u\" using singulartransformationfor the systemof Fig. 6.2. Solution Figure 6.5 showsthe one-linediagramof a simple four-bussystem.Table 6.1 gives the line impedancesidentifiedby the buseson which theseterminate.The )ro shuntadmittanceat all the busesis assumednegligible. !zo (a) Find Yuu. assumingthat the line shown dottedis not connected. -Vro (b) What modificationsneedto be carriedout in Yuu, if the line shown dotted is connected. !qo ltn Y- v-, JZ) !n Y.-n 3IJ1' ln Fig. 6.5 Samplesystemfor Example6.2 UsingA fiom Eq.(6.17),we get )ro 0 0 0 Line, Table 6.1 xpu 0 lzo 0 0 bus to bus 0 0Jn 0 &pu 0.15 t-2 0.30 0 00 jqo 1-3 0.05 0.45 2-3 0.10 0.E0 YA= 0 0 l y - ft+ 24 0.15 0.15 34 0.10 0 - jzz lzz 0 0.05 !n -lrz 0 0 0 - !24 0 lzq -.Yrr 0 l n 0
trrrr.F,b\"iFl Modern Power Svstem Anatvcis f.siffi -'ft|/0.!:l Table 6.2 where V,is the voltage at the ith bus with respectto ground and ,/, is the source current injectedinto the bus. Line Gpu B,pu The load flow problem is handledmore convenientlyby useof \"/,ratherthan r-2 2.0 - 6.0 ,I,t. Therefore,taking the complex conjugateof Eq. (6.24), we have - (6.25a) J.\\, tSubstitutingfor J, = Y*Vr from Eq. (6.5),we can write 2-3 - 2.0 24 1 . 0 - 3.0 34 2.0 - 6.0 solution (a) From Table 6.1, Table6.2 is obtainedfrom which yuu, for the k:l systemcan be written as A n (6.2sb) P i - j Q i = V f L Y i l , V p ii = l 2 (6.26a) k:l Equating real and imaginary parts I tP; (real power) = Re ]Vi Qi (reactivepower) = - Irn (6.26b) In polar form V; = ll/,1si6t Y r* = lY'oleio'r Real and reactive powers can now be expressedas addedbetweenbuses7 and2. n lvkl lYiklcos (0,0+ 6t - 6'); (6.27) Ytz,n\"* = Yrz,on - (2 - j6) = Yzr.o\"* i = r , 2, ..., fl Xrr,n\"* = ytt,ora+ (2 - j6) P, (realpower)= lvil D k:l Y2zn, \"* = Yzz,ot,+t (2 - j6) Modified Y\"u, is written below (iv) Qi Qeactivepower) - - lvil D lvkl lYiklsin (e,1+ 6t - 6,); k:l i=1,2,...,fl (6.28) 6.4 LOAD FLOW PROBLEM Equations(6.27) and (6.28) represent2n power flow equationsat n busesof a power system (n real power flow equationsand n reactive power flow The complex power injected by the sourceinto the ith bus of a power sysrem equations).Each bus is characterizedby four variables; P;, Qi, l7,l and 6i resulting in a total of 4n variables.Equations(6.27) and (6.28) can be solved IS for 2n variables if the remaining 2n variables are specified. Practical considerationsallow a power systemanalystto fix a priori two variablesat (6.24) eachbus.The solution for the remaining2n bus variablesis rendereddifficult by the fact that Eqs. (6.27) and (6.28) are non-linearalgebraicequations(bus voltagesare involved in product form and sine and cosinetermsare present) and thereforee, xplicit solutionis not possibleS. olutioncanonly be obtainedby iterative numericaltechniques.
:flg.,0iil toaern power system Anatvsis Dependingupon which two variablesare specified a priori, the busesare state variables. These adjustable independent variables are called control classifiedinto three categories. parameters. Vector J can then be partitioned into a vector u of confrol parametersand a vector p of fixed parameters. (I) PQ Bus r rr rrrro rJyw vr LruD,Lrrc ucr puwers ri ano ai arc known (po, and Qpi arc known g ancpl , 6l.,\".lA 1p:u1rte\"lto:ia:.dt\"b s(no gene andegile specified)T. he*tnn*i, arsfl/,tand (6.30) r a t i n gf * i f t y -at the bus,i.e., pcr= eci=0) is a u Pp bus. Control parametersmay be voltagemagnitudesat pV buses,real powersp,, etc. The vector p includes all the remaining parameterswhich are uncontrollable. (2) PV Bus/Generator Bus/voltagre controlled Bus For SLFE solution to have practical significance,all the state and control aArtetshpisetcyipfieedo.fTbhuesuPn'knaonwdnesoa, raeree,kn(hoewnnceaepori,o) rianadn6d,r.v',r andp. ,'(h\\ ence p6;) variables must lie within specified practical limits. These limits, which are (3) Slack Bus/Swing Bus/Reference Bus dictatedby specificationsof power systemhardwareandoperatingconstraints, are describedbelow: (i) Voltage magnirudelV,l must satisfy the inequality lv,l^tn< lvil ( lv,l_* (6.31) The power system equipment is designedto operateat fixed voltages with allowable variationsof t (5- l})Vo of the rated values. (ii) Certain of the 6,s(statevariables)must satisfythe inequality constraint nngoampdslnboouefsoileeufaranmpwfttmsIcefewdhanitebkrxreafrboeraelatelboesrlnsotrdykusliwaeclfnoiapsssedgagatacs)eehiendckstnlcapeuenoafbitrbldcetfsptiouiohuktwoyreawesanresildchiusdtoa1ioidtatstccwoti.yarusotbhott,untrnahutmtidhahi.nlecsnyIpelego.srnrrTlacymseebtthrdsohlhtapnuaetevreeelladsca.ecIamketibnioxcnntrubicemloidsptonsuerhn,oparaegs,tatnewhldbths.eteiaeFvueeatxpcchessurrspltteeuyeerioiptvtfssscshvhworetl.Boeatelerpeuooceryrmornnis,e,anatwsfpsetshot.hlpeoti.elseoro\"eeswcrbrnccscaslroeeasaoui(namfrirtirirndm.aegleiyopevdedtresl.eshteestrcossnnoeextaoobritgalehhampseaetsnailuoihslnpnnocvmwapdelgbeneeteetruruiohoxanrerisntcnefptgaifiketesonltcshsonhnbgwtlduewoaietuvaesepwcsestgtcprkliaen)apn(oeltriibctcotmeoi.miooeuaklwsflpnsi.nbnettttlinhehiuhthiersssosexeseedt 16,- 6ft1S l6i- 6rln,o (6.32) This constraintlimits the maximum permissiblepower angle of transmission line connectingbuses i and ft and is imposed by consideratio\\nsof s-ystem stability (seeChapter12). (iii) owing to physical limitations of p and/ore generationsources,po, and Qci Ne constrainedas follows: Pc,, ^in 1 Pc, S Pc,. ,n* (6.33) Qci, ^rn 1 Qci S Qc,, ^u** (6.34) It is, of course,obviousthatthe total generationof realandreactivepower must equal the total load demandplus losses,i.e. D P o , = t P o , +P , (6.35) Equations(6.27) and (6.28) are referred lt (SLFE)' By transposingall the variableson written in the vector form to as stutic load ftow equations D Qci=l Qot+Qr (6.36) one side. theseequationscan be ii f(x,y)-o (6.2e) where Ptand Qpare systemreal andreactivepower loss,respectively.Optimal sharing of active and reactive power generationbetween sources will be where discussedin Chapter7. ,f = vector function of dimension2n x = d_ependenotr statevector of dimension2n (2n unspecifiedvariables) ) = v(2enctionrdoefpiennddeepnetvnadreianbtvraersiawbhliecshoaf rdeimspeencsiifoiend2.in prrori) -Voltage ata PV bus can be maintained constantonly if conrollable esource is available at the bus and the reactive generationrequired is within prescribed limits.
ffi#ilfl,.l.i| Mod\".nPo*\"t sv\"ttt An\"lvtit Wffi I \"Ihe loadflow problem cannow be fully defined as follows: nn Assumea certainnominal bus load configuration. Specify P6i+ iQci at all Pr = Po,- D Po,;(Pr= 0). Equation(s6.37)canbe solvedexplicirly the pQbuses(this specifiesP, + iQi at thesebuses);specify Pcr (this specifies i.:I2 i:2 P,) and lV,lat all the PV buses.Also specify lVll and 6, (= 0) at the slack bus' (non-iteratively) for 62,61,..., d, which, when substitutedin Eq. (6.3g), yields T.L,,- r- .,--iolrlac nf thp rrer-fnr u ere snecifie.d The 2n SLFE can now be solved madehaveo..ouiLii,*\"i(;;::;;f'fiiLT,\",\",T:,:T:J\"#il-,1 (iteratively) to determine the values of the 2n vanables of the vector x cinsmooimmmnu-peiltudteatiraanatteiteoiovlnyueauslalplyybnohudnigtshctihlmayenuebrcdteoaimnsnooeemlonvusiecisdaoosln.reuiqstuioenrneottfilauElclqye[.sc(or6tlo.u3t7io()rn)r.-oSlf)inEcqfer.ot(hm6e.3sZgo)rrtfu,otliiltoonwiisss comprising voltagesand anglesat the PQ buses,reactivepowers and anglesat fhepV busesand active and reactivepowers at the slack bus. The next logical stepis to comPuteline flows.' tiicnnhodetnihfcsoeaiudttereadrbbitulnhesepbvuefoo.lrLoutiawrn-g:ebeurseasssreiassmtappnelcceeisfsaiyersdettecomobnoesfirdF.e0igrlp.du6n..e6itgJwliughuieb, rlee.ipTnohliweneemrraseLgaencittsuapdneecoceifsfaaierleld So far we havepresentedt,he methodsof assemblinga Yeusmatrix and load flow equationsandhave definedthe load flow problernin its genpralform with definitionsof varioustypes of buses.It has been demonstratedthat load flow equations,being essentiallynon-linear algebraicequations,have to be solved through iterative numerical techniques. Section 6.5 presents some of the algorithmswhich are usedfor load flow solutionsof acceptableaccuracyfor systemsof practicalsize. -5-r3 = - 2 +7O, At the cost of solution accuracy,it is possible to linearize load flo-w J equationsby making suitableassumptionsand approximationsso that f'astand eiplicit solutionsbecomepossible.Such techniqueshavevalue particularly for planning studies,where load flow solutionshave to be carried out repeatedly but a high degreeof accuracyis not needed. j0.15 An Approximate Load Flow Solution jo.2 Let us makethe following assumptionsand approximationsin the load flow ' iP,ts Eqs.(6.27) and (6.28). lVzl='l.o 2 (i) Line resistancebseingsmaiiarerreglecie,(Cshiintconductanceof overhead ,.,S-..= -I + Ii^Uz lines is alwaysnegligible),i.e. P7, theactivepo wer loss of the systemi s Fig. 6.6 Four-buslosslesssamplesystem zero.Thusin Eqs. (6'21) and (6.28) 1it = 90' and 1ii- - 90o' (ii) (6, - 6r) is small (< r/6) so that sin (6,- 6o)= (6r- 6r). This is justified Real Reactive Real Reactive demand generation generatrcn from considerationsof stability (see Chapter 72)' demand 1 Por = 1.0 Pcl ='- 061 (unspecified) (iii) AII busesotherthantheslackbus (numberedasbus 1) arePV buses,i.e. 2 Poz = 7.0 Qot = 0.5 Pcz = 4'0 Q62 (unspecified) voltage magnituclesat all the busesincluding the slack bus are specified. 3 Poz= 2.0 Qoz= 0.4 Pct=o O63 (unspecified) 4 Poq = 2.0 Qoz= 1.0 Pcq=o 06a (unspecified) Equations(6.27) and (6.28)then reduceto Qoq= 7.0 n l v k l l Y i k(l 6 i- 6 r);i = 2,3, ..., n (6.37) P i = l V i l \\- ,/--r kIsAlnl_snorFc:bw:eiutgn:fsu:fairveeo\"p_sl6rtbyai.o6vsgriteyiienusmdtsaofiricrsoebametsaepssteshbcueuimdfsi eaei ndtcajl,ieaor hclslatsitorlhebnesussbsf,eubt hssr3eetsharmeenadudl4saptthhaoaawsvvpeeeer occgoniefnirneyteedr ori nalsl atoitbouhlnreecateetasbb.sFluoeusu.r trIhceei srs,. k:1 n r v k r l y ickor s( 6 , - 6 u+) r v , r 2r y , ,ir ;= r , 2 , . . . ,n ( 6 . 3 9 ) et=- 'u,' E Pct = Por * Poz* pot * poo_ pcz = 2.0 pu Since lv,ls are specified, Eq. (6.37) representsa set of linear algebraic Therefore,we have7 unknownsinsteadof 2 x 4 = 8 unkno wns.In t he present equationiin 6,sr,vhichare(n - l) in numberas 6, is specifiedat the slackbus problem the unknown 60,ect, ecz, ecz (6, = 0). The nth equation correspondingto slack bus (n = l) is redundantas and Qc+. stateand control variablesare {, e, the reat power injected at this bus is now fully specifiedas
'yili.\\ ModernPowerSystemAnalysis I Though the reali63sesare zero,the presenceof the reactivelossesrequires that the total reactive generationmust be more than the total reactivedemand 44 (2.9 pu). er=D eo,-D epi Fromthedatagiven, Yru5canbe writtenasfollows: _ r.+r+ _ L,> = u.JJ4 pu (viii) Now, let us find the line flows. Equation (5.6g) can be written in the form ( l z l =X , 0 = 9 0 \" ) Pik = - Pki- lvi]-lvkl sin ({ - 6o) x,o Using the above Y\"u, and bus powers as shown in Fig. 6.6, approximate where P* is the real power flow from bus j to bus k. load flow Eqs. (6.37) areexpressedas (all voltagemagnitudesareequal to 1.0 pn = - pz,= +0 . 1 5 sin(\\ dr ,- q- r)/- s i n - 1 . 2 3=\" 0 . 4 9 2p u pu) 0.1: P z =3 = 5 ( 6 - 6 ) + 1 0( 6 - 4 ) + 6 . 6 6 7( 6 - 6 q ) (ii) Ptz = - PLz' Lr= -1- sin (4 - 6) = - $n 4'41o = - 0'385 Pu (ix) P t = - 2 - 6 . 6 6(76 - 4 ) + l 0 ( 4 - 6 ) (iii) 0.2 02 P + = - 2 - 1 0 ( 4 - 4 ) + 6 . 6 6(76 q : 6 ) (iv) Taking bus 1 as a referencebus, i.\". 4 = 0, and solving (ii), (iii) and (iv), we P r u = - P q t = +0 . 1 s i n ( { - 6 o )= 1 0 s i n 5 . 1 1 o = 0 . g 9 1p u get Real power flows on other lines can be similarly calculated. 4. --O.0ll rad = 4.4I\" For reactive power flow, Eq. (5.69) can be written in the general form (lZl=X,0=90o) 4=-0.074rad=-4.23' (v) 6q=-0.089rad=-5.11' ei* =W - lvi-llvkl cos(,{- 6o) \" Xik Xik Substituting6s in Eqs. (6.38), we have Qr = - 5 cos4.4I\" - 6.667cos4.23\"- 10 cos 5.11' + 21.667 where Q* ir the reactivepower flow from bus i to bus ft. Qz = - 5 cos 4.41\" - 10 cos 8.64\" - 6.667cos 9.52o+ 21.667 Qz = - 6.667cos4.23o- 10 cos 8.64' + 16.667 Q p =_Q1, z r0=.2+ - I cos(d,_ hl = 0.015pu Qq = - l 0 c o s5 .1 1 \"- 6 .6 6 7cos 9.52o+ 16.667 i., OI' (vi) @ , 3 8 5- 7 0 .0 1 5 * l i't ',t, (vii), Qr = 0'07Pu 2 + j0.ET 3 Qz = 0'22Pu 1 0.492-70.018 1 . s 0 2- i 0 . 1 1 3 Qz = 0'732Pu Q+= 0.132Pu 0 . 8 9 1+ / O .0 4 Reactivepower generationat the four busesare Qa = Qt + 0.5= 0.57pu 0 . 3 8 5+ , p . 0 1 S Qcz= Qz+ 0.4= 0.62ptr Qa = Qs+ 1.0= 1.L32pu 0.891-7O.04 1.103- j0.0s2 1.502+ 10.113 Qc+= Q++ 1.0= 1.132pu 1 . 1 0 3+ p . 0 9 2 n : n 4 2 t, jD.4 l| Fig. 6.7 Loadflowsolutionfor the four'-busystem Q r c= Q u= # # c o s( d 1 ,O = 0 . 0 1p8u I
.2Oftril ModernPowerSystemAnalysis I en*. I eA = e* ++'t = +0 . - -1- .o, (6r- 64=) 0.04pu 1 0.1 carriedout at the end of a completeiteration,the processis known as the Gauss Reactivepowerflows on otherlinescanbe similarlycalculated. iterative method.It is much slower to convergeand may sometimesfail to do Generationasndload demandsat all the busesandall the line flows are so. Algorithm for Load Flow Solution 6.5 GAUSS.SEIDELMETHOD Presentlywe shall continueto considerthe casewhereall busesotherthan the slack are PQ buses.The stepsof a computationaal lgorithm'aregiven below: The Gauss-Seidel(GS) method is an iterative algorithm for solving a set of 1. With the load profile known at each bus (i.e. P^ and 0p; known), non-linear algebraic equations.To start with, a solution vector is assumed, allocate*Po, and.Q5; to all generatingstations. basedon guidancefrom practicalexperiencein a physical situation. One of the While active and reactivegenerationsareallocatedto the slackbus, these are permitted to vary during iterative computation.This is necessaryas equationsis then used to obtain the revised value of a particular variableby voltagemagnitudeandanglearespecifiedat this bus (only two variables can be specified at any bus). substitutingin it the presentvalues of the remaining variables. The solution vectoris immecliatelyupdateciln respectof this variable.The processis then repeatedfor all the variablesthereby completing one iteration. The iterative processis then repeatedtill the solution vector convergeswithin prescribed With this step,bus iniections(P, + jQ) areknown at all busesother than the slackbus. accuracy.The convergenceis quite sensitiveto the starting values assumed' Fortunately,in a load flow study a startingvectorcloseto the final solutioncan 2. Assembly of bus' admittance matrix rsus: with the line and shunt be easilyidentifiedwith previousexperience. admittancedata storedin the computer,Yru, is assembledby using the rule for self and mutual admittances(Sec. 6.2). Alternatively yru, is To explain how the GS method is appliedto obtain the load flow solution, assembledusing Eq. (6.23) where input data are in the form of primitive let it be assumedthat all busesother than the slackbus are PB buses.We shall matrix Y and singularconnectionmatrix A. seelaterthat the methodcanbe easily adoptedto includePV busesaswell. The 3. Iterative computationof bus voltages(V;; L = 2, 3,..., n): To start the slackbus voltage being specified,there are(n - 1) bus voltages startingvalues iterationsa set of initial voltage valuesis assumed.Since,in a power of whosemagnitudesand anglesare assumed.Thesevalues are then updated throughan iterativeprocess.During the courseof any iteration, the revised systemthe voltagespreadis not too wide. it is normalpracticero use a voltageat the ith bus is obtainedas follows: flat voltage start,** i... initialiy ali voltagesare set equal to (r + 70) exceptthe voltageof the slack buswhich is fixed. It shouldbe notedthat J, = (P, - jQ)lvi [from Eq. (6.25a)] (6.3e) (n - l) equations(6.41)in cornplexnumbersareto be solvediteratively for findin1 @ - 1) complexvoltages V2,V3,..., V,. If complexnumber From Eq. (6.5) operationszlreDot availablein a computerE. qs (6.41)can be converted v ,[ I , rnto 2(n - 1) equationsin real unknowns(ei,fr or lV,l, 5) by writing , =,*,L'l L _f v , o vo l (6.40) (6.4r) T:i I Vr = €i + ifi = lV,l ei6' (6.42) t ^ . , - lv,=*Yl\" Substitutingfbr J, from Eq. (6.39)into (6.40) A significant reduction in the computer time can be achieved by performing in advanceall the arithmetic operationsthat do not change '':jq - itlt Y ivt'rl' =2'3\"\"' n with the iterations. I vr* I \" Define I k*t I i = 2,3, ...,ft (6.43) The voltages substitutedin the right hand side of Eq. (6.41) are the most 'Active and reactive generationallocations are made on econorfc considerations recentlycalcuiated(updated)valuesfor the correspondingbuses.During each iterationvoltagesat busesi = 2,3, ..., n aresequentiallyupdatedthroughuse dis.c*uAssedin Chapter7. of Eq. (6.41). Vr, the slack bus voltagebeing fixed is not requiredto be flat voltage start meansthat to start the iteration set the voltage magnitudesand updated.Iterationsarerepeatedtill no bus voltagemagnitudechangesby more than a prescribedvalue during an iteration. The computation processis then anglesat all busesother than the PV busesequal to (i + l0). The slack bus angle is conveniently taken as zero. The voltage magnitudesat the PV busesand slack bus are saidto convergeto a solution. setequal to the specifiedvalues.
206 | Modernpower SvstemAnatvsis I*rl.d (6.44) flows on the lines terminatingat the slackbus. Acceleration of convergence 1v7i( r'+ l ) - Ai 'S n acceleration factor. For the tth up by the use of the (r + l)th iteration is given by bus, the acceleratedvalue of voltage at the (Vl\") - t B , o- v or t,( ,, * t , - f r , o v l , , i = 2 , 3 , . . . , n y(r+r)(accelerated=) V,Q'* a(v.G+r)- V,Ql, ilt k=irl crfwoeohrcneovarmeenrmyageeisnsnycdaseetoredervmaaellvcuneaeunnmiscbbaaeeurs=coeabt1hltl.aee6idmn. Aetehdtewhbaoryocdncttgeorilcadehlirvaoletoiciroagendefoa.ffclotoow.rm. aAsyfusidnuidieteaseb. dAlesvlagolewunedeoorafwllnay (6.4s) This concludesthe load flow analysisfor thJ case of pe busesonry. The iterative processis continued till the change in magnitude of bus voltage, lav.('*r)|, between two consecutiveiterations is less than a certain tolerancefor all bus voltages,i.e. IAV.G*r)_l 1y.t+r)_ V,e)l < 6,; i = 2, 3, ..., n (6.46) 4. computation of slack bus power: substitution of all bus voltages Algorithm Modification when pv Buses are arso present computedin step3 along with V, in Eq. (6.25b)yields Sf = pr- jey fidAoteetllrtotcahwnteinionPinngVctsbhdtur.eo'slpu'ehsgsfeoh,rpreartphiarpenrerdtio,thrhpver]pirvaaVatreleubbseuupssseo.cfeifqeieudaaatninoddndesa.TarhenitcsordibsaereauctphcdeoamutenpdkliinnsohewevdneisnrtyoGthbSee 5 . computtttionofline .flows:This is thc laststepin thc loaclflow analysis l. From Eq. (6.26b) whereinthe powerflows on the variouslinesof the network arecomputed. Considerthe line connectingbusesi andk. The line and transformersat eachend can be representedby a circuit with seriesadmittancey* and two shunt admittances1l;roand )no as shown in Fig. 6.8. B u si B u sk ei =- Im f yr * \" ) j D, y,ovof l*io sm L ft:l ) fsTauhcbet,sftroietruvittsihneegdm(v.roa+slut1eu)ptohdfaitteeerdai vtiaioslnueoosnboetafcinvaeondlwtafrrgioteemsfornothmrehteahrbeigoahvbteohveaeqneudqasutiiadotneio.bnIny I g . ( r + t=) -Ln ]I, r , ' , ) - Li y,r v , , ( , a* t )1 y . r , f1D!-, y , k\"v k , ,, ,uI. r o, , ) Fig. 6.8 7i'-representatoiofna lineandtransformers r c o n n e c t e db e t w e e nt w o b u s e s 2' The revisedvalueof {.is obtainedfrom Eq. (6.45)immediatelyfollowing The current fed by bus i into the line can be expressedas step 1. Thus Iit = Iitt + Iirc= (Vi - V) !it,+ V,y,oo (6.47) 6Q+r)_ ay!+r) The power fed into the line from bus i is. (6.48) S* = Pir* jQiF Vi lfr= Vi(Vf - Vr\\ yft+ V!,*y,f, =Ansl\"et Lfe( ti1]l'-' ' ) - ,ir :r Bovo(,-+krD): , B , o v o , (, ,6l . s 1 ) Similarly, the power fed into the line from bus k is * +r J where Sri= Vk(V*k- Vf) yfo+ VoVfyi,l, (6.4e) Thepower lossin the (t - t)th line is thesumof thepower flows determined . r^1(. ;r +. t=)_-_P - i ? t ' r t t _ 6.52) frorn Eqs. (6.48) and (6.49). Total transmissionloss can be computed by ,u sununingall the line llows (i.e. 5';a+ Sri fbr all i, /<). The algorithm for pe buses remains unchanged.
l'fiffi 1 PrimitiveY matrix 2 BusincidencematrixA 3 S l a c kb u sv o l t a g e( l Y 1 l6, 1 ) demandat any bus must be in the range e^rn - e^u*.If ai any stage during thecomputatibn,Q at anybusgoesoutsidettreJetimits, it is fixed 4 R eeaaclbtuivsebpuoswpeorwsPeirfVsoQfriIi=,ffoo2rri,I3==' 42m,,.\".+\".'.n1,m\" ( P- {Vf Ob u buses ) At Q^in or Q^o, as the case may be. and the 5 R ses) dropped,i.e. the bus is now treatedlike a pe bus. Thus step 1 above branchesout to step 3 below. 7 VoltagemagnitudelimitsIVrI min andIViI max'lor'-(J o b n\"u\"iiu\" pow\"t limitsQi min and Q;maxfor PV buses 3. ff 9.('+r)a 0;,6;n, set e,Q*r)- er,r^n and treat bus f as a pe bus. compute 4-(r+1)and y(r+t) from Eqs. (6.52) and (6.45),respectively.If tg. Yr,.r.i\"g Eq.(6.2 Otu..:,')Qt,^ , set O-(r+l=) ei,^*and treatbus i asa pebus. Compure 4.Q+r)-6 y(r+l) from Eqs. (6.52) and (6.45), respecrively. MakeinitialassumptionsViofor I = m + 1,\"',n and O;0for i = 2!:'m Now all the computationalstepsaresummarizedin thedetailedflow chart Compu tethe parametersA ;f or i= ^ t 1,\"n 4. a3n)adnBd; p( 6' f oir = 1'2\" of Fig. 6.9 which serves as a basis for the reader to write his own -.;; i: 1, 2 , - ' . . . ,(ne x c e p tk = i) fro m E qs (6' '44) computerprogramme.It is assumedthat out of n buses,the first is slack as usual,t hen2, 3, . . . , m ar ePVbusesand t he r em ainingm + l, . . . , f t Set iterationcountr = 0 arePQ buses. IExampel 6.4 For the samplesystemof Fig. 6.5 the generatorsare connectedat all the four buses,while loadsare at buses2 and3. Valuesof real and reactivepowers are listed in Table 6.3. All busesotherthan the slack are pe typd. Assuminga flat voltage start,f)nd the voltagesand bus anglesat the three busesat the end of the first GS iteration. Solution Table 6.3 lnput data Bus Pp PU 0u pu Vt' Pu Retnarks I - 0.2 r.u 10\" Slackbus 0.5 PQ bus 2 0.5 Pp bus J - 1.0 - 0.1 Pp bus 4 0.3 The l\"ur for the samplesystemhasbeencalculatedearlierin Example6.2b (i.e. the dotted line is assumedto be connected).In order to approachthe accuracy of a digital computer, the computationsgiven below have been performed on an electronic calculator. -TD) - -u- s-v- o- r l-t-ag- es a,t r, ine eno or tne llrst rteratron are calculated using Eq. (6.a5). vtz=+{W-Yztvt-Y\"v! -'r^'iI Fig.6.9F|owchartforloadflowso|utionbytheGauss.Seide| -- - -l - - 11-0' -. 5 r+ -i '0- . -2 1- ' .- 0' \\ 4( - z- +' ') iterativemethod using YBUS rj-6/ ) - (\\ - 0v .' v6v6v6+' rj!2 ) _ ( _ t+ f l l Yz z I t -i O |
fei.fii,',1 ModernPowerSystemAnalysis - Load FlowStudies I 4.246- jrr.04 3.666- jLr = 1 . 0 1 9+ 7 0 . 0 4 6p u 1\" - - Ytt ( v l )* T#-frL+re - e2+i6()to+4io) _1 - r.04(- 1 + 13) - (- 0 . 6 6 6+ j2)(t + 70)_ (_ 1+ j 3 )( t . r) Yzt i o )j l I =t( + ' z^?-ei;ttt' lz sJ = , (r.rstz+70.033e) - (- 0.666+ j2) (1.019+ ;0.046)- (- 2 * i6) | ( 3.666_jt t )- = ''!'-,-1\"'.9?' = 1.02-870.08p7u or 61=2 1.84658o= 0.032 rad 3.66-6 jrr .' . v) = 1. 04( cos 6) * j sin dj) -r^rr\\l = 1.04(0.99948+ j0.0322) = L03946+ 70.03351 19+ 70.046) ==- , ( o' 1 - = i Q -t , lv' tl l- j yY. v\".v-' yY_ sv zt v_- v;Y r,o, 0vIl I 1 114tr )| - (- 2+ j6) (1.028- jo.o87) - [ - r _ 7 0 . s- 3.666-in L ,f;; (- t + i3) 1'04 2.99r- j9.2s3- 1.025- 70.0093pu - (- 0.666+ j2)(1.03s4+670.0335_1)(_ z + io)f 3-je r . 0 3 1 7- In Example6.4,let bus 2 be a PV bus now with lV2l= 1.04pu. Once again - Yov, J assuming a flat voltage start, find Qz, 6., V3, V4 at the end of the first GS iteration. =I [0.3+io.r- (- I + i3) (1.03e+4 70.0335) il tf:; Given:0.2 < Q, < l. - (- 2 + j6)(r.03t7_ yo.08e3Iz) n=,:_[J.,:A,,,,^:,.irJ\"!^,.:.,,,:?:;,:!:^:From Eq. (6.5),ii e\", (Note fz= 0, i.e. Vl - I.04 + i0) J + (- 0.666+ j2) + (- I + i3)l ) _ 2.967r_ j8.9962 = - Im {- 0.0693- j0.20791- 0.2079pu =0'9985-7O'0031 3-ig ' O\" = 0'2079Pu permissibtelimits on ez (reactivepowerinjection)are FromEq. (6.51) \",,il1;,;llii,\"\"rrirhe
ModernPowerSystemAnalysis 0.25< Qz < i.0 pu LI waqavr { Ela.., Or.,-r:^^ I tvyy \\)t,U(JlUli It is clear,that other data remainingthe same,the calculatede2 (= 0.2079)is now lessthanthe Qz, ^in.Hence e, it set equalto ez, _in,i.e. 6.6 NEWTON-RAPHSON (NR) METHOD Qz = 0'25 Pu The Newton-Raphsonmethod is a powerful method of solving non-linear algebraic equations.It works faster and is sure to convergein most casesas a - p --- uvrvrvrv, tf vqLt llv longer remainfiied at I.04 pu. The valueof 2t V, ar the end of the first iteration is calculatedas follows. (Note VL = t + 70 bf virrue of a flat start.) v' zL= l-( 'r, ^!?, - -yzzt t|v-t -yzzr tvJ?- -y' \"rovl) Yr, [ (Y3)- \") 3 . 6 6-6_j r[loL. s - jo. z s t -r, Considera set of n non-linearalgebraicequations (* 2 + i6)1.0-4 (- 0.666+ .i2)- (- r + i 3 ) ] f i ( \\ , x 2 , . . . , x n )= 0 ; i = I , 2 , - . . , n (6.s3) - 4'246- jlr-,? = t.05.5+9 io.o341 Assur.cinitialvulucsof u'know's as *l: *'), ..., r\"r.Let J.r(1J, g, ..., J_rl be the correctionsw,hichon beingaddedto theinitialguessg, ivelhe actual 3 . 6 6 6- j t l solutionT. herefore ( J t @ \\ + f u t l , * u r + A x l , . . , ,x 0 , +A x f ; 1= g ; i = 1 , 2 , . . . ,n ( 6 . 5 4 ) Expandingtheseequationsin Taylorseriesaroundtheinitial guessw, e have v\"j = :r,, l[':(-Y:*,')' -.Y,,v-,Y,,vJ f;(x01,xor, ,,r).[[*)' a*0.(,#)' Ax+t 3 . 6 6 6 - j r[r_Ltr;u- ;.to _ (*_ r + j3)toa / ^- \\0 I *[ -l!t] o-i l* n,rn-o, rderer ,,n- so (6.ss) - (- 0.666+ j2X1.0s5e+ 70.0341)- (- 2 + j6) \\ux, ) ] l _ 2 . 8 tt 2 - j n . 7 0 9 1.0347- 70.0893pu whcrefg)',!(f, )''. '. f 9 )' 3 . 6 6 6- j r r \\ d\"r / ( Dr, ) [ ,\", are the derivativesof It with respectto ,/ x1,x2, ..., x, evaluat edat ( x! , *1, . . . , *0, ) . vL=;;W,# - Yo,-vY, o,-uv| *v)) Neglectinghigherorder terinswe can write Eq. (6.55)in matrix fbrm =+,t\"=# - (- I +,r3()r'0s0+ei0'0341) f'l'' a'l' -1 I'rl Llxi (6.56a) - (- 2 + j6)(r.034-i j0.08e3I) / , . 0 6 3 0- j9 .42M _ 1.077+5 j0.0923pu 3- je Ax: or in vector matrix form
2ll I Modern Power Svstem Analvsis (6.56b) - l(, + .f,A*, * 0 ,f is k,o*n asthe Jacobianmatrix (obtainedby differentiatingthe function m vectofr withrespectto x andevaluatinigt at r01.Equation(6.56b)canbe writtenas (6.63a) :l_J It is to be immediatelyobservedthattheJacobianelementscorrespondingto the ith bus residuals andmth buscorrectionsare a 2 x 2 matrix enclosedin the box Approximatevaluesof corrections/-r0 canbe obtainedfrom Eq (6.57).These in Eq. (6.62a) where i and m Ne both PQ buses. being a set of linear algebraic equations can be solved efficiently by triangularizationand back substitution(seeAppendix C). Sinceat the slackbus (busnumber l), Prand Qr areunspecifiedand lV,l, Q are fixed, there are no equationscorrespondingto Eq. (6.60) at this bus. Updatedvaluesof x are then Hencethe slack bus doesnot enter the Jacobianin Eq. (6.62a). tl =\"0 + AxI consider now the presenceof PV buses.If the ith bus is a pv bus, e, is unspecifiedso that there is no equation correspondingto Eq. (6.60b) for this or, in general, for the (r + 1)th iteration bus. Therclbrc, the Jacobianeleurentsof the lth bus becomea sinele row pertainingto AP,, i.e. (6.s8) \"(r+l)_r(r)+AxQ) Iterationsare continuedtill Eq. (6.53)is satisfiedto any desiredaccuracy,i.e. l J i G \" ) l < e ( a s p e c i f i e dv a l u e ) ; i = 1 , 2 , . . . , n (6.5e) NR Algorithm for Load flow Solution First,assumethatall busesarePQ buses.At anyPQbus the loadflow solution rnustsatisfythe following non-linearalgebraicequations J'ip(lV, 6) = I'i (sPccificcl)- Pi = 0 (6.60a) fiq (1v1, 6) = Qi (specified) - Qi = o (6.60b) whereexpressionfsor P, andQ, aregiven in Eqs.(6.21) and (6.28).For a trial set of variableslV;1,6;,the vectorof residuals/0 of Eq. (6.57)comespondsto fip= Pi (specified) - Pi (calculated) = APi (6.61a) r t h b u s Fl = (6.62b) .fie= Qi(specified) - Qi @alculated)- AQi (6.61b) while the vector of corrections y'xo corresponds to alvil, a,i Equation(6.51)for obtainingtheapproximatceorrectionvsectorcanbewritten for the load flow case as If the mth br-rsis also a PV bus, lVrl becomesfixed so that Alvml = 0. We cannow wnte mthbus I (6.62a) i t h b uL-sn-l-p-,Jl = lt-- ---- ----rHT'':lt T ' - I (6.62c) I ,lPi l mth bus f t hb u s l aQi 46^ I t no u s AlV^ jm
2L6 | rtrodernPowerSystemAnalysis l ztz Also if the ith busis a PQ buswhile the mth busis a PV bus, we can thenwrite t (6.66) Ninr= N^i = 0 l,^=J^;=O (6.62d) It is convenientfor numerical solution to normalizethe voltage corrections correspondingto a parricularvecror of variables tqlv2lq64lval6lr, the vector of residuars[aP2 aez ah ap4 ae4 Apr], and the Jacobian(6 x 6 alv^l in this example) are computed. Equation rc.an is then solved by lv^l triangularization and back substitution procedure to obtain the vector of asa consequencoef which,thecorrespondinJgacobiaenlementbsecome correctionLIfsoo,#nl vv2tl Atrl| 1r ooJ146a4+l V o r /,6rl . correctionasrethenaddedto I updatethe vector of variables. 2(Pa) 3(PVl (6.63b) Expressionsfor elementsof the Jacobian(in normalized form) of the load flow Eqs. (6.60aand b) are derived in Appendix D and are given below: Case I mtl H,^= Li^= a^fi - b*et 6.64) Fig. 6.10 Samplefive-busnetwork -'-* BusNo. Nr*=- Jirr= a.er+ bJt 2345 Yi^= G* + jB,* Vi= e, + jf, Hzz Nzz Hzs Hzn Nza (a* i ib) = (Gi- + jBi*) @* + jk\\ I Jzz Lzz Jzs Jz+ Lz+ l2 case2 Hsz Nsz Hss Hss I,,==t-n, - Biirvirz I Mii= Pi + G,,lV,lz IA A<\\ Y (6.67) Jti = Pi - Giilvil' \\\\J ' \\'J,' o [:_] f,z lalv4ll J l]ql-l c0 i_tit_l Hqz Nn Haa Naz. H+s Corrections in variables Jqz Lqz Jcc Lu J+s Li;= Qi- Biilvilz Hss H5a Nsq Hss An important observation can be made in respect of the Jacobian by t 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 Jacobian (Evaluatedat trialvalues of variables)
-TL,Lr-Ve- I rrrrodcrnPower Svstem Analvsis Lcad Ftor; Studies I i:ig,: I - Iterative Algorithm Find the load flow solutionusing the NR method.Use a toleranceof 0.01 for Omitting programmingdetails,the iterativealgorithm fbr the solutionof the powermismatch. loadflow problemby the NR methodis as follows: Solution Using the nominal-rcmodel for transmissionlines,I\"u, for the given L W'ith voittge anclangle (usually f'= €I) at s systemis obtained as follows: at atl PQ busesand d at all PV buses' In the absence of anY tlther eacn ltne information flat voltage start is recommended' I -z.g4r - j11.7&- r2.r3 l-75.96\" 1 , sneens. \" =o. oz+jo. og^ 2. Compute AP,(for PV and PB buses)and AQ,(for aII PQ buses)from Eqs. (6.60aandb). If all the valuesarelessthan the prescribedtoletance, Eachoff-diagonalterm = - 2.94I + jll.764 stop the iterations,calculate P, and Q, and print the entire solution Each self term = 2l(2.941 - j11.764)+ j0.011 includingline flows. = 5.882- j23.528= 24.23 l-75.95\" 3. If the convergencecriterion is not satisfied, evaluate elementsof the 0 +,O JacobianusingEqs. (6.64) and (6.65). 2 4. Solve Eq. (6.67)for correctionsof voltageanglesand magnitudes' 5. Update voltage angles and magnitudesby adding the corresponding changesto the previousvaluesand returnto step2' Note: 1. In step2, if there are limits on the controllable B sourcesat PV and if it violatesthe limits, it is madeequalto buses,Q is computedeachtime espondingPV bus is made a PQ bus in that the limiting value andthe corr iteration.If in the subsequenct omputation,Q doescome within the prescribed 3 1 . 0 4l O \" limits, the bus is switchedback to a PV bus. 1 . 5+ 7 O . 6 2. If there are limits on the voltage of a PQ bus and if any of theselimits Fig. 6.11 Three-busystemfor Example6.6 is violated, the correspondingPp bus is madea PV bus in that iteration with voltage fixed at the limiting value. Considerthe three-bussystemof Fig. 6.11.Eachof the threelineshasa series To startiterationchoose4= t +70 and,4 = 0. From Eqs.(6.27) and (6.2g), impedanceof 0.02 + 1O.OSpu and a total shunt admittance of 70'02 pu' The we get specifiecql uantitiesat the_b\"!91j9!qulated below: Pz = lVzllvl lY2l cos(0r1+ 6r- $) + lVrlzlyrrl cos0zz+ lvzl l\\l RealloadReactiveloadRealpowerReactiveVoltage l Y r r lc o s( Q z t +q - 6 ) Bus demand demand generation power specification Pt = lVl lvl l\\rl cos(dr,+ 6r- 6r)+ lVrllv2ll\\zl x cos(ez + h - 6r)+ lV,rlzlyrrl cos0r, PD Qo PG generation O6 Q z= - i v z i i v l t y z l s i n( g r + d t - h ) - l v z l zl y 2 2xl s n 4 z _ l v 2 l 2 . 0 1 . 0 Unspecified UnspecifiedV'=1.94* ;g lv3l lY.,.lsin (0r, + bz- 6) (slackbus) Substitutinggivenandassumevdaluesof differentquantitiesw, e getthevalues 0.0 0.0 0.5 1 . 0 Unspecified trf Powersrts (PO hus) PB=. -0.23pu l5 0.6 0.0 Qct=? lvll= 1.04 (PV bus) C o n tro l l a b l crc a c ti v op o w o rs o urccl s avai l abl cnt btts3 w i th thectl nstrai nt 0<Qct S 1.5Pu
O}nr'l ilada-n Darrrar Crralam Anahraia Load FtowStudies ttv.l rvruuErrr r VYYEr \\)yolgltr nrrqryoro Sz=0.5+j1.00 Sr=-1.5-j0.15 |i zz-rti^--i. t P3= 0.12pu Transmissionloss= 0.031 Qor=- 0'96 Pu (calcu Power residualsas per Eq. (6.61) are - rt (- 0.23)= 0.73 Line flows Aror- -1.5 - (0.12)= - !.62 The following matrix shows therealpartof line flows tQT= 1- (- o'e6-) t'e6 I oo 0.1913r2E000.839861E'00-l The changesin variablesat the end of the first iterationare obtainedasfollows: LI--00..r882462221e3E^0800-00.6703.8047E00 0.6s4697E00| 0.0 J 0P, 0P, 0P, The following matrix shows theimaginarypartof line flows 06, 061 alv2l II o.o -0.5994&E00 _0.r9178zE00] 0P, aP, 7Pu 06, 061, 0lv2l l|0.60s274800 0.0 0 . 3 9 6 0 4 s 8 0 0I aQ, }Qz : '' aaQv |, L0.224742E0-00.37sr6s800 0.0 I 06, 06 Jacobian elements can be evaluated by differentiating the expressionsgiven Rectangular Power-Mismatch Version above fr>r Pr, Py Qz with respectto 6r, d1and lVrl and substituting the given and assumed values at the start of iteraticln. The chanses in variables are 9.hiEiTtrfsshfeo;qhcoraws'ieafusnl(eqlnh6stevvuhvai'ceen6iaarrtrrdg7erirs,vytlbd)yiicrorbsetealrbinuyncnnEruat,tocyesqa,st,esae.ne'vti(csgdhg+a6heunet.r.hP,fil6ir,afaaaeV'ircrbnt)a-aa,belcrerbnluve.livfsycres,nih''etP.ntfhoew.hh,Itemnheeatitidehnmevneutproueieesrcolm-dtpnalaciaeanbrbograreeaensvcrcr<in-shetoomyutcirfnttpsrpnaeseriiirgrmootaqirnngnutabuisaigrtocguaaiuvntnistmodiniieneossiemadgasqnrrsEinS.nusynatcqaaihgpnnlsrrc,eadgcteohsriec(vnsteim6aslsaaai.tniir6lsonspleiy4fafncvtaf)erb-aertlrlsbaahaematnuteessscisdsershepv.iteInhqo(g.ts6laeusrton.ns,aea6raridgtafab5inoeoetyl)rdeer.ns,r obtained as I t d j l l - 2 4 . 4 i - t 2 . 2 3 s . 6 4 - 1 - r [ 0 . i 3[ -f0 . 0 2 3 - 1 | A a^i r I| .t ^/ .^.^2 5 / . 4 . e ) - J^. ^u-)ll | - t .-o^zl l : l -| u .^u^o. )- 4. 1 | I l:l- | [ n r v , r ' -L]- u u 3 . 0 s z z . s 4L) r . q o lI o o a r l l a ) l I a i - l l - ^ 4 I t - 0 .[1 0 0 2I. 3[ 0 0 2 3 . 1 However'thc rcct:tttgtrllvtrcrsionsccnlsto bc slighiiyt.r, r c l i a b l eb u t f a s t e ri n I a]l:l I l*l ^4 l:lol*l-006s41:l-006s41 convergencethan the polar version. Itv,t'jL'y,rolJnrv.,r'L.1'i I oosoJI r.08eJ The total numbero f non-linearpower flow equationsconsideredin this case We can now calculatefusingEq. (6.28)] arc fixed iurdcqual2 (trl). Theselbllow fr.o'i Eqs. (6.26a)and (6.26b)and Qtt= 0.a671 are Q o \\= Q\\ + Q rt = 0 .4677+ 0.6 = 1.0677 P, (specifi\"d)-I{ e,(epG1-1,f,B,t) + fihrG* -t epB,o}: 0 (6.68) whichis within limits. (6.69) k:l If' the sanreproblem is solved using a digital computer, the solution convergesin threeiterations.The final resultsare given below: i = l , 2 , . . . ,n i = slack(s) Vz= 1.081l- 0.024 rad n Vt = I.M l- 0.0655rad B; (specifi\"d)-)r t . lf,koG,o - ftB,t1- e;(.f*Gi* eoB,oy=]}g Q u = - 0 . I 5 + 0 . 6 = 0 . 4 5( w i t h i nl i n r i t s ) k=l S r = 1 .0 3 1* j (- 0 .7 91) (lbr each PQ bus)
222 | Modern Porgl Jyllem Anatysis I l(l (specified)2- @i2+ f,') = 0 (for eachPV bus) (6.70) Load FlowStudies Using the NR method,the linearisedequationsin the rft iteration of iterative necessarymotivationin developing the decoupledload flow (DLF) method, in processcan be written as which P-6 andQ-V problemsare solveclseparately. Jr J2 Decoupled Newton Methods J3 J4 AQ Js J6 HntgimIgn1ueonamowetoarmbernilxeeve.ydretrcre'rioAo,pcfonnronevyndlseeyeesn,ctnutbhoitcotueuhhntpme-aaltepohwlNdpseeetrraaowpelkgxoatooipcrmnreouiatulcmhatpoimroleimdnsntehgrpceoheroddaenuu,vfshepecaarleterblisefndteNdhogecetefnoowttmrhdatuopelbvenuoeeqtvvlaloueeetpa,miisoaedienndornaadnttlibtisnihscoecefnptorheretnehenfvsoeteserliJer.tngieamterecaanoldhatycbuereergbtreoaeee.n AIVP Fori* j Juj= - Jqij = Gij ei fi 6.72) Jzij= Jt,j=- Bijei+ Gilfi Jsij= Jo,j = O In Eq. d( 6e.c6o7u)p, tlheedlei nl eema reNnet swt ot obneenqeugal et icotnesdbaerceosmu bem a t r i c e s[ 1 v ] a n d resulting [,/]. The For i=,r tApl = lHl lA6l (6.76) t ; ^ \".'-: : : : : ! - tlr,,, (6.73) I^Qt- tLtl4!1 (6.77) -!'u,ul,)-i' J:, (6.78) Ltvt J Jsii= 2\"i J6ii = 2fi whereit canbe shownthat a, and b, arethe componentsof the currentflouring into node i, i.e., gU = Lij = lvil lvjl (CUsin 6u- B,i cos {r) i*j a,+ jb,= frCo * jBi)@r+jf*) (6.74) Hii= - 8,,lViP - Qi (Eq.(6.6s)) (6.7e) k:l Lii= - 8,, lV,lz+ Qi (Eq. (6.6s)) (6.80) Stepsin solutionprocedureare similar to the polar coordinatescase,except Equations(6.76) and(6.77) can be constructedand solvedsimultaneously that the initial estimatesof real and imaginary parts of the voltagesat the PQ cciiwttooeeitnnrrhaavsttctieiroloua'ngnccethubiunsyocgitnehatfhgrensardEtnatqithnsseeot(anlh6vcse.ihn7osgi8litvre)niErntuaoqglt.tiEo(a6(nnq6.,e.8.u7o(06p6u).d).sm7Aaf7ooti)drnbfegoe.rtt4ht4eAerlo[vH*plo.]prTaohnuoidss.he[wri]tshillemtorauectpsrodiucnaletdrsieunidncftaee6saaticcenhhr busesare madeandthe correctionsrequiredareobtainedin eachiterationusing tiiNtsaeRknrTeaomsthtioemmenmtouhrocaoefhidnntiohsuafemdiatvsDbnareLanerFdtdoavufigascienteoeatdfalrmmatghtoeeiefosmrDntoosmtehrtcyeotohrsuceeaopqpmnruoevierideneLartmosgoaetefhbdnvaeFtitseclioonawfwusoNst(ofeRDorsiLfpnmFtegh)eetehtdahesasoipJcndapocacrmeonotpxhdteiaemiarttenaiamd.tTltiweoohnapet.yherseer (6.7s) The correctionsarethen applied to e andf and the calculationsare repeatedtill convergenceis achieved.A detailed flow chart describingthe procedurefor load flow analysisusing NR methodis given in Fig. 6.12. FE FFAA'?hIEh 'AA Fast Decoupled Load Flow (FDLF) O. I L'|aUL,UTL|1L' L('AL' .F LL'VV IVI.EI.tsI(-,L'Ii An important characteristicof any practicalelectricpower transmissionsystem amds1Frpe9uoeev7rdmtee4ehdllaloe2adprdlpdlem.ehvaseayTscnsnhrfitriooecbaflaeglaotldeslhwywisjenusuiFts:mhtathiopsfeuittaDitpobmernleecswuvsochiiuomhipucplshloelisadfsisLruceobainvastaiedoalcinFcdtscirmoiounnawrn.ayT(ocbFhryeiDmsocLfaaeFlsrfp)rfoooimelruwdteteoicorruhunrtoyltmuords,bia\"niynnca1gBhtoei.tephdsveeitenorsDatottLtmihiFoneen operatingin steadystateis the stronginterdependencbeetweenreal powersand busvoltagesanglesandbetweenreactivepowersandvoltagemagnitudesT. his interestingpropertyof weak coupling betweenP-6 and Q-V variable.sgave the
'*i I vodern Power SvstemAnalvsis I zas- I Aduanceiterationcount -::,:-,t I r=r+1 With theseassumptionst,he entriesof the [1{ and [L] submatriceswill become considerablysimplified and are given by I and ?=i:=-''i,,,11::,,u,:\": (6.82) (6.83) Matrices [/fl and [L] are squiuemafricswith dimension(npe+ npy)andnrn respectively. Equations(6.76) and (6.77) can now be written as D e t e r m i n eJ - 1 c-mpute ano--l LAPI= ltyjtlvjl Bfi AA (6.84) .'.comPuteAefr) 2n6 printlineflows,l Lfiv) powerloss, I v oltag es, I tAQl= [%t tvjtB,,ij, elci ----.] [#] (6.8s) Are all where 8t,,, B(are elementsof [- B] matrix. maxA within Furtherdecouplingand logical simplificationof the FDLF algorithmis achieved tolerance by: 2 Determinme ax 1. Omitting from [B/] the representationof those network elements that changein power predominantly affect reactive power flows, i.e., shunt reactancesand max APr,AO transformer off-nominal in-phasetaps; 2. Neglectingfrom [B//] the angle shifting effects of phaseshifters; 3. Dividing eachof the Eqs. (6.84)and (6.85) by lv,l andsettinglVrl= I pu in the equations; Compute I last')node 4. Ignoring seriesresistancein calculatingthe elementsof tdll which then becomesthe dc approximationpower flow matrix. el,),1v112 Compute LlVi,12= 1Vs, cnd12-l%tl2 With the above modifications, the resultant simplified FDLF equations -' ls -\\ become gt'rioi) IAP| lyl I = [B'] [L6] (6.86) tAQl tr4I - LB\"I t^lytl (6.87) In Eqs. (6.86)and (6.87),both [B/] and lBttf arereal, sparseand have the structuresof [I{ and [L], respectively.Since they contain only admittances, they are constantand need to be invertedonly once at the beginningof the study. If phase shifters are not present, both [B'l and lB\"] are always symmetrical, and their constant sparseupper triangular factorsare calculated and storedonly once at the beginningof the solution. F i g .6 . 1 2 Equations(6.86) and (6.87) are solvedalternatively.alwaysemploying the cos 6, : 1; most recent voltage values. One iteration implies one solution for [4fl, to s i n{ r : 0 G,, sin 6,,< B4; update[d] andthenonesolutionfor IA lVl] to update[Vl] to becalled l-dand and Q i < B i i l v i l z l-V iterafion S\" -ern* a^ r*a' -f e . c- -r l^n^v' e- ^. ros-e^n c e tcetc are annlied fhr the rcql qnA rcqotlvc power mismatchesas follows: (6.81) max [APf I €pi max lAQl S eo (6.88) where eoand €e are the tolerances. A flow chart giving FDLF algorithmis presentedin Fig.6.13.
,bZCI ModernpowerSvstemAnalvsis LoadFtowStudies I ## Considerthe three-bussystemof Example6.6. Use (a) DecoupledNR method (Star)t Read LF data and form Ysr. (a) Decoupled NR method: Equationsto be solved are (seeEqs. (6.76) and (6.77)). Substitutingrelevantvaluesin Eqs. (6.78) - (6.80) we get H z z = 0 ' 9 6 + 2 3 .5 0 8- 24.47 Hzt= Hzz= 1.04(- Brz)= - 1.04 x 11.764= - 12.23 Htt=- Qt- Bn0.0q2 = [- Bsr t\\P - 42 tv3t- Bzzt\\P] - Bn (r.0a1.2 = - f 1 .7 6 4x (1 .0 4)2- 1t.764 x 1.04+ (l .oq2 x 2 x 2 3 . 5 0 8- 2 5 . 8 9 Lzz= Qz- Bzz= t + 23.508- 24.508 [ 0.73-_1 [ 24.47_r2.2311\\6t\\1 (i) l-t.oz) L-n.23 zs.ssJ[64trl tLQz-l lz4' .srLf+l'tu;(J-i)\"lJ'l (ii) Solvefor A6f usingEq. ( 6 . 8 6i)= 1 , 2 , . .n. ,, i * s SolvingEq. (i) we get Calculate 6i.1= 6[+ 66[ L6;') = - 0.0082- 0.0401- - 0.002 fori=1,2,...,n,i-s Aa{tr= - 0.018- 0.08- - 0.062 1.(6.60a) C a l c u l a tles l ia c k = PVbus bus power ai nd Qz= - lvzl lvtl lYzl sin(0r,+ 6r- 6r) - lvzlzV2zlsin 82 ------ lvzl lhl llrrl sin (9zt+ 6, - q) all linefloowvs X = - 1.04x 12.13sin(104.04+ 0 - 0.115-)24.23 and print sin(-75.95-) 1.04x 12.13sin (104.04+ .115- 3.55\") --12.24+23.505-12.39 Qz=- l'125 -'-- I tA- rt -vl 2 - - 1r - / 1r . r1La <J \\) = ^L . I1L? Jr E -- re \\- rcEq.l \\- .,n, I <'- fr= 1 -- Substitutingin Eq. (ii) :us- l ; Afr/,R Lz.rz5- r[24.5rL[]atluY,\"(al'),ll e+ F l g .6 . 1 3
;'..ngI ModerPn oweSr ystemAnalysis - lui:. AN 9 l=0.086 completean iteration.This is becauseof the sparsityof the network matrix and lvftt - lvlo)t+atv|'t=t 1.086pu the simplicity of the solution techniquesC. onsequentlyt,his method requires less time per iteration, With the NR method,the elementsof the Jacobianare Q(rl canbe similarly calculatedusing Eq. (6.28). to be computedin eachiteration,so the time is considerablylonger.For typical large systems,the time per iteration in theNR rnethodis roughly equivalentto The matrix equationsfor the solutionof load flow by FDLF methodare [see 7 times that of the GS method [20]. The time per iterationin both thesemethods Eqs. (6.86)and (6.87)l increasesalmost directly as the number of busesof the network. -Brr1al f',f (iii) The rate of convergenceof the GS method is slow (linear convergence characteristic),requiring a considerablygreaternumberof iterations to obtain -8,),Lz4l\" a solutionthanthe NR methodwhich hasquadratic convergencecharacteristics and is the best among all methods from the standpointof convergence.In and (iv) addition, the number of iterationsfor the GS method increasesdirectly as the number of busesof the network, whereasthe numberof iterations for the NR lffil= r-Bztaztrt')tl method remains practically constant, independentof system size. The NR method needs 3 to 5 iterations to reach an acceptablesolution for a large I - -ot..otzz I - fil6411oqrl (v) system.In the GS method and other methods,convergenceis affected by the | It 23.508.1lla4\" choice of slack bus and the presenceof seriescapacitor,but the sensitiviry of : l- the NR method is minimal to thesefactors which causepoor convergence. | -1 .0 4 1 1' 5 I Therefore, for large systemsthe NR method is faster, more accurateand | 57 more reliable than the GS method or any other known method. In fact, it works L for any size and kind of pro6lem and is able to solvea wider variety of ill- conditioned problems t23). Its programming logic is considerably more Solving Eq. (v) we get complex and it has the disadvantageof requiring a largecomputei memory even when a compact storage scheme is used for the Jacobian and admittance 46;r) - - 0.003 matrices. In fact, it can be made even faster by adopting the scheme of optimally renumberedbuses.The method is probably best suited for optimal A6t'' = - 0.068 0.068 rad load flow studies(Chapter7) becauseof its high accuracywhich is restricted only by round-off errors. 6')- - 0.003rad; {tl The chief advantageof the GS methodis the easeof programming and most lz.rzs-l [23.508]tatvitl efficient utilization of core menrory.It is, however,restrictedin use of small alvtl= 0.09 size system becauseof its doubtful convergenceand longer time needed for solution of large power networks. l v t t =1 . 0 9p u Thus the NR method is decidecllymore suitablethan the GS method for all Now Q3can be calculated. but very small systems. These valuesare used to computebus power mismatchesfor the next iteration. Using the values of 'LAPAVll and lAQAl\\l the above equationsare For FDLF, the convergenceis geometric,two to five iterations arenormally solved alternatively,using the most recent values,till the solution converges required for practical accuracies,and it is more reliable than the formal NR method. This is due to the fact that the elementsof [81 and [Btt] are fixed within the specifiedlimits. approximationto the tangentsof the definingfunctions LP/lVl andL,QAVl, and are not sensitiveto any 'humps' in the ciefiningiunctions. 6.8 COMPARISON OF IOAD FLOW METHODS fi LP/lVl andA^QIIVI are calculatedefficiently, then the speedfor iterations In this section,GS and NR methodsare comparedwhen both use liu5 as the of the FDLF is nearly five timesthat of theformal NR or abouttwo-thirds that network model. It is experiencedthat the GS method works well when of the GS method. Storagerequirementsare around 60 percent of the formal programmedusing rectangularcoordinates,whereasNR requiresmore memory NR, but slightly more than the decoupledNR method. when rectangularcoordinatesare used.Hence, polar coordinatesare preferred for the NR method.
ilai:i;'l ,odern powersystemAnatvsis I Changes in systemconfigurations can be easily taken into account and system,buseswith generatorsareusuallymadePV (i.e. voltagecontrol) buses. though adjustedsolutionstake many more iteration.se,achone of them takes Load flow solution then gives the voltagelevels at the load buses.If some of less time and hencethe overall solution time is still low. the load bus voltageswork out to be lessthan the specifiedlower voltage limit, The FDLF canbe employedin optimization studiesandis speciallyusedfor it is indicative of the fact that the reer-fivc n^rr/ar ff^.,, ^-*^^ie, ^r4-^-^--:--:- e studies, as in contingencyevaluation for system security assessmentand lines for specifiedvoltagelimits cannotmeetthe reactiveload demand(reactive enhancementanalysis. line flow from bus i1o busft is proportionatlo lAvl = lvil - lvkD.Thissituation Note: When a seriesof load flow calculationsare performed, the final valuesof bus voltagesin eachcaseare normally usedas the initial voltagesof the next case.This reducesthe number of iterations,particularly whenthereare minor changesin systemconditions. 6.9 CONTROL OF VOLTAGE PROFILE Control by Generators Control of voltageat the receiving bus in the fundamentaltwo-bus systemwas tAvt- tEth- tv!= -ffi O, discussedin Section5.10. Though the samegeneralconclusionshold for an interconnectedsystem,it is important to discussthis problem in greaterdetail. t v ' ; t =l E r h l + n, #l At a bus with generation, voltage can be conveniently controlled by adjustinggeneratorexcitation.This is illustrated by meansof Fig. 6.14 where =tvit* the equivalent generatorat the ith bus is modelled ty a synch.onou, reactance hn, (resistanceis assumednegligible)and voltagebehindsynchronousreactanceI.t immediately follows upon application of Eqs. (5.71) and (5.73) that Pai+ jQet l V i l t6i Xei ti F l g .6 . 1 S Pci = |v,l -Epl.;;,' -'-i, (6.8e) Since we are considering a voltage rise of a few percent, lV(l canbe further approximatedas xo, lV|l= lV,l+ffiO, eci= (6.e1) #r-,vit+tEcit) (6.e0) Thus the VAR injection of +jQc causesthe voltage at the jth bus to rise approximatelyby (XhllViDQ.. The voltagesat other load buseswili also rise WYiYt hrtrr\\/ PrGi -t iVi Acil \\ a- -n, {o lir't/llri L,o/ fi ^g:l,v- ^e-n Lb-Y- -rihr -e- rioac,i ifjtow soiution- these velrres owing to this injectionto a varying but smallerextent. Control by Transformers Apart from beingVAR generatorst,ransformerpsrovidea convenienmt eansof controlling real power' and reactivepower flow along a transmissionline. As
','d&r1'l ModernPower SvstemA-nalvsis -l has alreadybeenclarified, real power is controlled by meansof shifting the Sincethe transformerls assumedto be ideal, complex power output from it equalscomplex power input, i.e. phase of voltage, and reactive power by changing its magnitude. Voltage 5 1= V , / i = c r v l f magnitudecan be changedby transforrnersprovided with tap changing under or Ioad(TCUL) gear.Transformersspeciallydesignedto adjustvoltagemagnitude smail values are calleo re rnxers. Figure 6.16 showsa regulatingtransformerfor control of voltagernagnitude, For the transmissionline which is achievedby addingin-phaseboostingvoltagein the line. Figure 6.17a I\\= ! @V1- V2) or shows a regulating transformer which shifts voltage phase angle with no Ir = dl'a lo:lzyV,- cfyVz appreciablechangein its magnitude.This is achievedby adding a voltage in series with the line at 90\" phaseangle to the correspondingline to neutral (6.e4) voltage as illustrated by meansof the phasor diagram of Fig. 6.17b. Here VLn= (Von* tV6) = Q - jJT) Von= dVon Also 6.92) Iz= l(Vz- oV,) = - WVr + lVz (6.es) wher ea = (1 -j J -l t)= 7 1 -ta n -t J 1 t Equations(6.94) and (6.95) cannotbe representedby a bilateralnetwork. sincer is small. The I matrix representationcan be writtendown as follows liom Eqs. (6.94) The presenceof regulatingtransformersin lines modifies the l/uur matrix and (6.95). t hc r c byrn < i l i fy i n gth e l o a c lfl o w s o l u ti on.C onsi cl ear l i ne, connecti ngtw o buses,having a regulating transformerwith off'-nominal turns (tap) ratio a n.,%n includedat one end as shownin Fig. 6.18a.It is quite accurateto neglectthe small impedanceof the rr:gulatingtranslonucr,i.c. it is rcgardcdas an iclcal device.Figure 6.18bgives the correspondingcircuit 4epresentatiown ith line representedby a seriesadmittance. -_o A/ C...,%n .___Jl\"r<_tc, (a) IJ (V\"n+tV\"r)=aV\", /^ ic raal nr rmhor\\ Flg.6.16 Regulatintgransformefor r controol f voltagemagnitude (b) Fig. 6.17 Regulatingtransformefror controlof voltagephaseangle
I Modernpower SystemAnalysis ao.Ot,o* a,r.,.r __ f# I Bus,1 (i) V/V\\ = l.M or a - l/1.04 B q s2 (ii) V/V\\ = d3 or d = s-i3o neffiting Solution (i) With regulating transformer in line 3-4, the elementsof the correspondingsubmatrixin (v) of Exa 6.2 aremodified as under. transformer (a) aV'l sz v2 3 34 . ---.- O--f__1--*-<- j2) 516 -r.92 <L II j10.s47 +js.777 l''r Y 12 r.92 3-je js.777 (b) Modified submatrixi n ((v) of Example6.2 is Fig. 6.18 Linewith regulatingtransformear nd its circuitrepresentation 4a 4 (6.e6) J .3r13 5.887 Theentriesof f matrix of Eq. (6.96)would thenbe usedin writing, the rru, r- j3) 3. 3-je matrix of the completepower network. 3 -(0.66-6j2) ei3\"e2 + j6) For a voltageregulating transformer a is real, i.e. d = e, therefore,Eqs. (6.94)and (6.95) can be represenredby the zr-networkof Fig. 6.19. -(2- j6) 4 1T Gz+ j6) 3- je Fis.6.1e withorr-nomtianpasrettinosr tc :jiiil:':3;:i;ffiTH#*: Fig. 6.20 Off-nominatlransformerast bothlineends(a., a, real) If the line shown in Fig. 6.18a is representedby a zr-networkwith shunt I t -Ylfvil -[Ii- l admittanceyu at eachend, additionalshuntadmittancelol2ysappearast bus 1 L - y ) l v t) - Ut) and yo at bus 2. , (i) . (ii) The above derivations also apply for a transformer with off-nomin al tap lLyv. i;1l=- l[oo , oI [ - ' 4 . lIl r ll t / q o ,I)f ]4,l1 setting' where a = (k[nuJ(kD,r:up, a real value. q) L v , ) ' L , ;ol : [ v\" Substituting (ii) in (i) and solving we get | 4t -\"!-',v[fq1:['\"1 /iii\\ L-otory 4y ) LVJ2 ltz ) Thus (iv) The four-bus system of Fig. 6.5 is now modified to include a regulating transformerin line 3-4 near bus 3. Find the modified rru5 gf the system for '-=- l l - o?v -aqztf -With ^*, d; ) off-nominaf,up **ing transformerast eachend of the line as shownin Note: Solve it for the case when ao &2 are complex. Fig.6.20,we canwrite.
.l Modernpower SystemAnqtysis Load FlowStudies 236 | I 6.10 CONCLUSION Bus I is slackbuswith Vr = 1.0 10\" Pz+ iQz = -5.96 + j1.46 In this chapter,perhapsthe mostimportantpower systemstudy,viz. load flow lVTl= 1102 has beenintroducedand discussedin detail. Important methodsavailablehave Assume: methodsis the best,becausethe behaviourof differentload flow methodsis dictated by the types and sizes of the problems to be solved as well as the \\ = 1.02/0\" and vi -710\" precise details to implementation.Choice of a particular method in any given situation is normally a compromisebetweenthe variouscriteria of goodnessof t1-' tt2.-o.oz 3 the load flow methods.It would not be incorrect to say that amongthe existing methodsno singleload flow methodmeetsall the desirablerequirementsof an oottooo .7orr:\" ideal load flow method; high speed,low storage,reliability for ill-conditioned systems,versatility in handling variousadjustmentsand simplicity in program- Fig. P-6.2 ming. Fortunately,not all the desirablefeatures of a load flow method are neededin a l l s i tu a ti o n s . 6.3 For the systemof Fig. P-6.3 find the voltageat the receivingbus at the end of the first iteration.Load is 2 + 70.8pu. voltage at the sendingend Inspite of a large number of load flow methods available,it is easy to see (slack) is 1 + /0 pu. Line admittanceis 1.0 - 74.0 pu. Transformer that only the NR and the FDLF load flow methodsare the most important ones reactanceis 70.4 pu. off-nominal turns ratio is lll.04. Use the GS fbr generalpurposeload flow analysis.The FDLF methodis clearly superior to techniqueA. ssume Vn = ll0. the NR methodfrom the point of view of speedas well as storage.Yet, the NR method is stili in usebecauseof its high versatility, accuracyandreliability and c>+J=---F\"d as suchis widely being usedfor a variety of systemoptimizationcalculations; it gives sensitivityanalysesand can be usedin modern dynamic-responseand Fig. P-6.3 outage-assessmecnatlculationsO. f coursenewermethodswould continueto be developed which would either reducethe computation requirementsfor large 6.4 (a) Find the bus incidencematrix A for the four-bussysremin Fig. p-6.4. systemsor which are more amenableto on-line implementation. Take ground as a reference. PROTBEIVSI (b) Find the primitive admittancematrix for the system.It is givbn rhat all the lines are charactenzedby a seriesimpedanceof 0.1 + j0.7 akrn 6 . r For the powersystemshownin Fig. P-6.1,obtainthebus incidencematrix and a shunt admittanceof 70.35x 10-5 O/km. Lines are rated at220 A. Take ground as reference.Is this matrix unique?Explain. kv. (-) (c) Find the bus admittancematrix for the system.Use the basevalues 22OkV and 100MVA. Expressall impedancesandadmittancesin per unit. 1 1 1 0k m Fig. p-6='! 3 6 . 2 F<rrthe network shown in Fig. P-6.2,obtain the complex bus bar voltage Fig. P-6.4 at bus 2 at the end of the first iteration..use the GS method. Line impedancesshown in Fig. P-6.2 are in pu. Given: 6 . 5 Consider the three-bussystem of Fig. P-6.5. The pu line reactancesare indicatedon the figure; the line resistancesare negligible.The magnitude of all the three-busvoltagesare specifiedto be 1.0 pu. The bus powers are specifiedin the following table.
.?38i I ModernPower Svstem Anatvsis I --.ff(i)ir o --Z-r----rv|z y=_j5.0 y=-j5 0 ,l ',\\ =-is'o ,lJ , \\ 3 vel / Fig. P-6.5 13 Real Reactive Real Reactive Fig. P-6.7 Three-busamplesystemcontaininag r e g u l a t i ntgr a n s f o r m e r demand demand Seneratrcn generation I Por= 1.0 Qot = 0.6 PGr= ? ect (unspecified) 6.8 CalculateV3for the systemof Fig. 6.5 for the first iteration,using the data Pcz= 1.4 pn, lunspecified) of Example6.4. Startthe algorithmwith calculationsat bus 3 ratherthan ' ',r2 = 0 Qoz= 0 Pa = 0 go. lunsp\"cified) at bus 2. 3 Po:= 1.0 Qot = l.O Carry out the completeapproximateload flow solution.Mark generations, 6.9 For the sample systemof Example 6.4 with bus I as slack, use the load demandsand linc f-lowson thc one-linediagram. fbllowing nrcthudsto obtairra load flow solution. 6.6 (a) RepeatProblem6.5 with busvoltagespecificationcshangedasbelow: (a) Gauss-Seideul sing luur, with accelerationfactor of 1.6 and l V t l= 1 . 0 0p u tolerancesof 0.0001 for the real and imaginary componentsof lV2l= 1.04pu voltuge. lV3l= 0.96 pu (b) Newton-RaphsonusingIBUS,with tolerancesof 0.01pu for changes Your results should show that no significant changeoccurs in real in the real and reactivebus powers. \\ power flows, but the reactive flows change appreciably as e is sensitiveto voltage. Note: This problem requiresthe use of the digital computer. (b) ResolveProblem6.5 assumingthat the real generationis scheduledas 6.10 Performa load flow studyfor the systemof Problem6.4. The buspower follows: and voltage specificationsare given in Table P-6.10. Pc t = 1 .0p u , P c z = 1 .0pu, P ct = 0 Table P-6.10 The real demandremainsunchangedand the desiredvoltageprofile is Bus power,pu flat, i.e. lvrl = lv2l = lv3l = 1.0 pu. In this casethe resultswill show that the reactiveflows arc essentiallyunchangccbl,ut thc rcal flgws Bus Voltagemagnitudep, u Bus type arechanged. I Unspecified Unspecified 1.02 Slack 1.01 PV 6.7 Considerthe three-bussystemof problem 6.5. As shown in Fig. p-6.7 2 0.95 Unspecified Unspecified PQ where a regulating transforrner(RT) is now introclucedin the ljne l-2 Unspecified PQ near bus 1. Other systemdataremain as that of Problem6.5. Consider -a ) - 2 . O - 1.0 two cases: 4 - 1.0 - 0.2 (i) RT is a magnituderegulatorwith a rario = VrlVl = 0.99, Computethe unspecifiedbusvoltagesa, ll bus powersandall line flows. (ii) RT is a phaseangleregr-rlarohraving a ratio = V/Vi = Assumeunlimited Q sources.Use the NR method. (a) Find out the modified )/ur5 matrix. \",3\" (b) Solve the load flow equationsin cases(i) and (ii). compare the REFERNECES load flow picture with the one in Problem6.5. The readershould Books verify that in case(i) only the reactiveflow will change;whereas l. Mahalanabis, A.K., D.P. Kothari and S.I. Ahson, Computer Aided Power System Analysis and Control, Tata McGraw-Hill, New Delhi. 1988. in case(ii) the changeswill occur in the real power flow.
' M. - .o-d- -p' ,r,n p, nvr'rrrvar r eeyr roarrrar i*l l AA1^t^art-y. -s,l-s _ -2rA_0_\"_'_l 2' weedy, B.M. andB.J.cory, Erectricarpower-systents,4thEd., Johnwiley, York, 1998. New Gross,C.A., power SystemAnalysis,2nd 8d..,John Wiley, New york, 19g6. Power Frow w.Aaerngr pdolrWaitnh.tSmD.syMwnaiatyhmeRirc,es.afSenorcelurnctocioentntoroInr,dHiaydneproawbeards1, y4s_tie6mFse,b, .prrgo:c/.g,ozf II Sterling,M.J.H., power SystemControl, IEE. England,1978. symp.on Po rg. 26. Tinney, W.F , u.r., L,teclrrc Lnergy System Theom: An Introduction, 2nd Ed, McGraw_ TriangularFactorization'I,EEE Trans.on Auto contor, August 1973,yor. AC_lg, Hill, New York. 19g2. JJJ. 6. stagg, G.w. and A.H. EI-Abiad, computer Methods in power system 27' Saro, N. and w.F Tinney, 'Techniquesfor Exproiting Sparsity of Network McGraw-Hill,New york, 196g. Anarysis, 7' Rose' D'J' and R'A willough (Eds), AdmittanceMatrix', IEEE Trans. pAS, I)ecember L963,g2, 44. Plenum,New york, 1972. sparse Matrices and their Applications, 28' sachdev,M.s. and r.K.p. Medicherla, 'A second order IEEE Trans., pAS, Jan/Feb1977, 96, Lgg. Load Flow Technique', 8' Anderson,P'M., Analysisof Faulted Power systemsT, he Iowa state university 29. Iwamoto, S.andy. Tamura,,A Fast Load Flow Press,Ames, Iowa, 1923. ' IEEE Trans., pAS. Sept/Oct.197g, 97, 15g6. Method Retaining Non linearitv'. 9' Brown, H'8., sorution of Large Networks by Matix Methods, John w1rey, New 30. Roy, L., 'Exact Second Order Load Flow,, proc. of the Sixth pSCC Conf. , York, 1975. Dermstadt,Yol. 2, August lg7g, 7Il. l0' Knight' rJ'G', Power systemEngineeringand Mathematics,pergamonpress,New 31' Iwamoto, s' and Y. Tamura,'A Load Flow calculationMethodfor Ill-conditioned York 1972. PowerSystems',IEEE TranspAS, April lggl , 100, 1736. 11. Shipley,R.B., Introduction to Matrices and power 32. Happ, H.H. and c.c. young, 'Tearing Argorithmsfor York, 1976. systems,John w1ey, New Problems', IEEE Trans pA,S,Nov/Dec Ig71, gO,2639. Large scare Network 12' A2Hnrurdpilplea,dgHna.,H,WJ..,ilaDenyida, kNNo.epRwt.icwysoaartnskdo, n2N,0ec0tow1mo. rpkust,eArcMadoedmeiliicngproefsEs,rNeecwtricyaorrpko, w1e9r7s1lt.stems, 33. Dopazo,J.F.O.A.Kiltin andA.M. Sarson,,stochasticLoad Flows,, .IEEE Trans. l3' P A S ,1 9 7 5 , 9 4 , 2 g g . 14' Nagrath,I.J. and D.p. Kothari, power system Engineering, TataMcGraw_Hilr, 34' Nanda' J'D'P' Kothari and s.c. srivastava,\"Some Importantobservationson New Delhi, 1994. 35. the IEEE, May 19g7, pp.732_33. \\ FDLF'Algorithm\", proc. of Kothari and D.L. stenoy, .,secondorder Decoupred l5' 2ABnnedirlgleaedgnan,A'. W.JRia.l,enpdyo,NwN'Re.yr' .swy2as0tt0seo1mn.A, ncaormypsuistp,erreMntoidceel-tHinagiloE, fnEglreecwtroicoadcl rpioffws,eNr s.Jy.srt9egm6s., Nanda,J., p.R., Bijwe, D.p. l6' LoadFlow', ErectricMachinesand power systemsv,or r2, No. 5, 19g7,pp. 301_ 312. 36' NandaJ'' D'P' KothariandS.c. srivastva,\"A Novelsecondorder FastDecoupled Load Flow Method in Polar coordinatcs\", Electic Machinesand power,s.ysrems, Papers Vol. 14, No. 5, 19g9,pp 339_351 17. Happ, H.H., 'Diakoptics-The solutionof system problemsby Tearing,, proc. of 37. F3DDD4laiaos6sswt.,r,DisDb'ou,.,rtuDioHr'inP.oNs'n.eKotNfowtaRohgraakidrsiai\"aan,nrpdDdriDosAc.t.pr.iIK.bEauKEltoai,otmphnNta.c\"re,Air,vw..Sr,oA.irmkrN4spr'ol,,e,vInanernol.d.MJ4E.e.fotJfhfiucoErideypnfEro9tSMr9,4esr.t9ohp9lovp5id.n, fpgzopgrR.rL_3aoz3dag5iadg_r. the IEEE, Juty t974, 930. 38' l8' 8sssALtt5tanooo9uatttttt.gl,,,yBhBBst..io.,,.,snau'R''nDMsedie''vgAoci.'oe,tuwAh'pDolesrfeeNaLcccoNoor,daen'Fadwtapr-rIsoFmotsrDnopiLtewciooccdaTnoaadeurnccpFcuhlIeeMronadwiLatqiot,our,aineIMxdEsi,nFe,EtlppEohrowooTwdc,r,e,,.aIrpIEnsErsEyoE.sEc,,t.e1TI1m9Er9a7EN6n2Eegs,pt,,.wrA,J11oSu59rr_,k7y9Ls411o3,9p,1ag7A9d45FS,59l_o.19w63.: 19' 20. 21' 22' Tinney' w'F\" and J'w. walker, 'Direct solutionsof sparseNetwork Equationsby Optimally Ordered Triangular Factorizaiions,, proc. IEEE. Nnvernhe\" '1voA1 55:1801 \" 23' Tinney, w.F. and c.E. Hart, 'power Flow sorution by Newton,s Method,, T r a n s . ,N o v e m b e r1 9 6 7 ,N o . ll, pAS_g6:1449. ,EEE 24' ward, J.B. andH.w. Hare,'DigitarcomputerSorutionof power problems,,AIEE Trans.,June 1956,pt III, 75: 39g.
OptimalSystem Operation A3 parianceis caiiedthe 'unii commitment'(UC) probiemandthesecondis calleci the 'load scheduling'(LS) problem.One must first solvethe UC problem before proceedingwith the LS problem. Throughout this chapter we shall concern ourselves with an existing installation,so that the economicconsiderationsarethat of operating(running) costandnot the capitaioutiay. 7.2 OPTIMAL OPERATION OF GENERATORS ON A BUS BAR Before we tackle the unit commitmentproblem, we shall considerthe optimal operationof generatorson a bus bar. 7.1 INTRODUCTION Generator Operating Cost The optimal systemoperation,in general,involved the considerationof The major componentof generatoroperatingcost is the fuel input/hour, while economyof operation,systemsecurity,emissionsat certainfossil-fuelplants, maintenancceoutributesonly to a small extent.The fuel costis meaningtulin optimalreleasesof waterat hydro generation,etc.All theseconsiderationsmay caseof thermal and nuclear stations,but for hydro stationswhere the energy makefor conflictingrequirementasndusuallya compromisehasfo he madefor storageis 'apparentlyfree', the operatingcost as such is not meaningful.A optimalsystemoperationl.n this chapterwe cor..;idetrhe economyof operation suitablemeaningwill be attachedto the cost of hydrostoredenergyin Section only, alsocalled the ec'omonicdi.spcttcphroblem. 7.7 of this chapter.Presentlywe shall concentrateon fuel fired stations. Themainaim in theeconomicdispatchproblernis to rninimizethc totalcost l of generatingreal power (production cost) at variousstations while satisfying I theloadsandthelossesin thetransmissiolninks.For sirnplicitywe consicletrhe I presenceof thermalplantsonly in the beginning.In the later part of this chapter tI - we will considerthepresenceof hydro plantswhich operatein conjunctionwith thermalplants.While thereis negligibleoperatingcost at a hydro plant,there L\\ is a limitationol availabilityol'watcr'overa pcriodof tinrewhich nrustbc used to savemaximum fuel at the thermal plants. - al) In theloadflow problemasdetailedin Chapter6, two variablesarespecified ;t r Po at eachbus and the solutionis then obtainedfor the rernainingvariables.The .c8 specifiedvariablesarereal and reactivepowersat PQ buses,real powersand 6o voltagemagnitudesat PV buses,and voltagemagnitudeand angleat the slack bus.The additionalvariablesto be specifiedfor load flow solutionare the tap 8b= settingsof regulatingtransformersI.f the specifiedvariablesare allowed to vary e5 in a regionconstrainecbiy practicaiconsicieration(uspperanciiower iimits on activeandreactivegenerationsb,us voltagelimits,andrangeof transformetrap gE settings)t,hereresultsan infinite numberof loadflow solutions,eachpertaining o to onesetof valueso1'specifiedvariablesT. he 'best'choicein sornesenseof thevaluesof specifiedvariablesleadsto the 'best'loadflow solution.Economy O6 of operationis naturallypredominantin determiningallocationof generationto eachstationfor varioussystemload levels.The first problem in power system f LL o (vw)min (MW)max Fig.7.1 PoweroutputM, W-'- -- Input-outpuct urve of a generatrngunit The input-output curve of a unit* can be expressedin a million kilocalories per hour or directly in terms of rupeesper hour versusoutput in megawatts. The cost curve can be determined experimentaily. A typical curve is shown in Fig. 7.1 where (MW)o'in is the minimum loading limit below which it is uneconomical (or may be technically infeasible) to operate the unit and (MW)n,u, is the maximunr output limit. The inpLlt-output curve has discontinuities at steam valve openings which have not been indicated in the figure. By fitting a suitable degree polynomial, an analytical expression for operating cost can be written as A unit consistsof a boiler, turbine and generator.
| M. . O O e r n r o w e r- - DA - y- - S I! - - e I r l A r l a l yA - ^ r - - ^ : - Considerationsof spinningreserve,to be explainedlaterin this section,require .L4+ | Ci(Pc) Rs/hourat outPutPc, I that wherethe suffix i standsfor the unit number.It generatly suffices to fit a second D Po,,^*) Po (7.6) degreepolynomial,i.e. _ Q ; b,Pc, + d, Rs/hour (7.r) marsin. i.e. Eq. (7.6) must be a strict inequality. 2 Sincethe operatingcost is insensitiveto reactive loading of a generator,the rnannerin which the reactiveload of the station is sharedamong various on- The slopeof the costcurve, i.\".43 is called the incrementaJl uel cost(lQ, dPo, line generatorsdoesnot afl'ectthe operatingeconomy'. The questionthat has now to be answeredis: 'What is the optimal manner andis expressedin unitsof rupeesper megawatthour (Rs/lvIWh).A typical plot in which the load demand Po must be sharedby the generatorson the bus?' of incrementalfuel cost versuspower outputis sho.wnin Fig. 7.2.If the cost This is answeredby minimizing the operatingcost curve is approximatedas a quadraticas in Eq. (7.1), we have k (lc)i= aiP\"t + bt (7.2) c = D ci(pci) (7.7) f:l undertheequalityconstrainot f meetingtheload demandi,.e. l ( I f- \\ ' GD i l- P o = O (7.8) I L -i. I i:t (J where k = the number of generatorson the bus. Further, the loading of each generatoris constrainedby the inequality o oo c constraintof Eq. (7.5). EB Since Ci(Pc) is non-linear and C, is independentof P6t (i+ i), this is a ,d> separablenon-linear programmingproblem. \\ oo If it is assumedat present,that the inequality constraintof Eq. Q.q is not Etr o E (E) effective,the problem can be solvedby the methodof Lagrangemultipliers. ( M W) m i n (MW) max Define the Lagrangianas PoweroutputM, W (Pc i ) - ^ [\"f\" - \"\" ] (7.e) Fig.7.2 Incrementaful elcostversuspoweroutputfor the unitwhose i tf - , input-outpucut rveis shownin Fig.7.1 i.e. a linear relationship.For better accuracyincremental fuel cost may be where X is the Lagrangemultiplier. Minimization is achievedby the condition expressedby a number of short line segments (piecewise lineanzation). Altcrnativcly,wc canfit a polynomial of suitabledegreeto representIC curve of, =o in the inverseform dPo, Pc;i= a, + {),(lC)i + 1,QC)', + ... (7.3) o r dC' i - ) i i = 1 , 2 , . . . ,k (7.1o) dPc, Optimal Operation [,et us assulnethatit is known a priltri which generutorsitre t<lrtln to ntccta where lci is the incrementacl ost of the ith generator (units: Rs/TvIWh), p:.rrticulalor ad clenranodn the stattonU. bvtously d4,, a Iurctio'ol'.gencra,il;,\":':t Equation(7.10) can be writtenas DPr,,,,'',,)* P, (7.4) where Pci, ,r.,,i\"s* the rated real power capacityof the ith generatorand Po is :\": dC^ \\ ( 7 .r l ) the total power demandon the station.Further,the load on eachgeneratoris to dPoo d4(il- dr-- irc constrainedwithin lower and upper limits, i.e. *The of reactive loading on generatorlossesis of negligible order. Pcr, .in 1 Po, 1 Po,, rn.*,I = L, 2, \"', k effect (7.s)
n-.:-^r A-^^r--- ^ .. L -^- Computer solution fbr optimal loa<ling of generatorscan be obtained cost of the plant colrespondsto that of unit 2 alone. When rhe plant load is 40 iteratively as follows: Mw, each unit operatesat its minimum bound,i.e.2o Mw wiitr plant \\ = Rs 35/I4Wh. When dczldPcz= Rs 44/MWh, 0.25PG2+30 - 44 1. Choose a trial value of ), i.e. IC = (IC)o. or pnt = JI- = 56 MW 2. Solve for P\", (i = 1, 2, ..., k) from Eq. (7.3). 0.25 3. If ItPc,- Pol < e(a specifiedvalue),the optimal solution is reached. The total plant output is then (56 + 20) = 76 MW. From this point onwards, the valuesof plant load sharedby the two units are found by assumingvarious Otherwise, valuesof \\. The resultsare displayedin Table 7.1. 4. Increment(lC) bv A (1\"), t fI:l I Table 7-1 Outputof each unit and plantoutput for variousvaluesof Po,- ,rl < 0 or decremen(t/c) by A(tr) ) for Example7.1 if [D Pc, - Pr] r 0 and repeatfrom step2. This stepis possiblebecause P.-, is monotonicallyincreasingfunction of (1g). Plant ), Unit I Unit 2 Pcl, MW Pcz, NfW consider now the effecr of the inequality constraint (7.5). As (1c) is RszMWh Plant Output increasedor decreasedin the iterative process,if a particulargeneratorloading 35 40.0 P\", reachesthe limit PGi,^o or P6;, min, its loading from now on is held fixed 44 76.0 at this value and the balance load is then shared between the remaining 50 130.0 generators on equal incremental cost basis. The fact that this operation is 55 175.0 optimal can be shownby rhe Kuhn-Tuckertheory (seeAppendix n;. 60 220.0\\ 61.25 231.25 65 250.0 Incremental fuel costsin rupeesper MWh for a plant consistingof two units Figure 7.3 showsthe plot of the plant .trversusplant output. It is seenfrom are: Table 7.7 that at .\\ = 61.25,unit 2 is operatingat its upper limit andtherefore, the additional load must now be taken by unit 1, which then determinesthe dt plant ). i * -Go1 . 2 o p c t + 4 0 . 0 60 -dcz-= o.2,pcz+3o.o + dPo, t-- lcc Assume that bothunits areoperatingat all times,andtotal load variesfrom 40 MW to 250 MW, andthe maximum and minimum loadson eachunit areto be I I25 and 20 MW, respectively.How will the load be sharedbetween the two units as the systemload vanesover the full range?What arethe corresponding 850 valuesof the plant incrementalcosts? Solution At light loads,unit t has the higher incrementalfuel cost and will, o therefore,operareat its lower limit of zo Mw, for which dcrldpcr is Rs 44 per MWh. When theourputof unit 2is20 MW, dczldpcz= Rs 35 p\"i UWt. Thus, @^ with an increasein the plant output, the additionalload shouldbe bornebv unit ia=> 4 5 EP 40 t- E O cJc PlantoutputM, W Fig.7.3 Incrementafluel cost versus plantoutput,as found in Example7.1
t:Hzut\"l MocjernPower Systemnnaiysis .Z.48i: To find the load sharingbetweenthe units for a plant outputof say 150MW, Net savingcausedby optimumschedulinigs 772.5- 721.875= 50.625Rs/lr we find from the curve of Fig. 7.3, that the correspondingplant X is Rs 52,22 Totalyearlysavingassumingcontinuouosperation per MWh. Optimum schedulesfor each unit for 150 MW plant load can now This savingjustifies the need for optimalload sharing and the ddvices to be be found as installedfor controlling the unit loadingsautomaticallv. 0.2Pa*40-52.22; P c t = 6 1 ' 1 1M W 0.25PG+2 30= 52.22; Pcz = 88'89MW P c r + P c z = 1 5 0M W Proceedingon the above lines, unit outputs for various plant outputs are Let the two units of the svstem studiedin Example7.1 havethefollowing cost computedand havebeenplottedin Fig. 7.4. Optimum load sharingfor any plant curves. load can be directlv read from this fieure. Cr = 0.lPto, + 40Pc + 120Rs/hr I tzs Cz= 0.l25Pzcz+ 30Po,+ 100 Rsftrr I = 100 250 220 MW Eru t 200 o I E50 I 3 150 f Eo 1 0 0 050 100 150 200 250 J Plantoutput,MW -_--> 50 Sunday Monday Flg.7.4 Output of each unit versus plant outputfor Example7.1 12 l l (noon) PM 12 6 Time----- (night) AM Fig. 7.5 Daily load cycle For the plant describedin ExampleT.l find the savingin fuel cost in rupeesper Let us assumea daily load cycle asgivenin Fig. 7.5. Also assumethat a cost hour for the optimal schedulingof a total load of 130MW ascomparedto equal of Rs 400 is incurredin taking eitherunit off the line and returningit to service distribution of the sameload betweenthe two units. after 12 hours.Considerthe 24 hour period from 6 a.m. one morning to 6 a.m. Solution Example 7.I revealsthat unit I shouldtake up a load of 50 MW and the next morning. Now, we want to find out whether it would be more unit 2 should supply 80 MW. If eachunit supplies65 MW, the increasein cost economical to keepboth the units in servicefor this 24hour period or to remove for unit 1 is one of the units from service for the 12 hours of light load. J|1s6o\\5'Ur.nLnI ' n . ,^\\rn (rUr..I'f tnnt Z 165 Fn^ a D^tL- For the twelve-hourperiod when theload is 220 MW, referringto Table 7.1 r: t *U)|JIl nr ; lf\\r1 rl llL.J l\\S/I[ of Example7.1, we get the optimum scheduleas r = = \" ' l s ot ' t U I ' 6 y l l Pcr= 100 MW' Pcz= 120MW Similarly, for unit 2, Total fuel cost for this period is J*co.rs\",\" + 30d) por=(0r.25PGz:o+ro;1\" [ 0 . 1x 1 0 0 2 + 4 0 x 1 0 0 + 1 2 0 + 0 . 1 2 5 x 1 2 0 2+ 3 0 x 7 2 0 + 1 0 0 ]x 1 2 - - 721.875Rs/hr = Rs. 1,27,440
-l-ttgtiFl Modernpowesr vstemnnAVsis If both units operatein the light load period (76 MW from 6 p.m. to 6 a.m.) Dynamic Programming Method t- also, then from the same table, we get the optimal scheduleas In a practicalproblem, the UC table is to be arrived at for the complete load Pcr = 20 MW, Pcz = 56 MW cycle. If the load is assumedto increasein small but finite size stensldvnamin prograrrurung Total fuel cost for this period is then can be used to advantagefor computing the uc table, (0.1 x 20' + 40 x 20 + 120+ 0.125x 5 6 + 3 0 x 5 6 + 1 0 0 ) x 1 2 wherein it is not necessaryto solve the coordination equations;while at the same time the unit combinations to be tried are much reduced in = Rs 37,584 these reasons,only the Dp approachwill be advancedhere. number. For Thus the total fuel cost when the units are operating throughout the 24 hour bt heeTal ohsaesdut omctyaecldlnet huoamnt btthheeer oslftoauatnidoi tnosanarveeaaaislcashbuulmen,eitthdteooifrbicenodkminvboi diwnunaatali ocpnorsiootfrcih.uFanuri tarstchtceehrr,ai snittgi cessahi ' inandl l periodis Rs I,65,024. suitably small but uniform stepsof size /MW (e.g. I MW). Starting arbitrarily with any If only oneof the units is run duringthe light load period,it is easily verified determinedfor all the discrete two units, the most iconomical combinationis that it is economicalto run unit 2 andto put off unit 1. Then the total fuel cost load levels of the combined output of the two duringthisperiot:tXf ': units. At eachload level the most economicanswermay be to run either unit 762+ 3ox 76+ 100)x rz = Rs37,224 or both units with a certain load sharingbetweenthe two. The most economical cost curve in discreteform for the two units thus obtained,can be viewed as Total fuel cost for this case= L,27,440+ 37,224 pnthrooeteccedodtshutarceturirenvpetehoaisfteapdrtsooicnfeigndldeuertehqteuhicevoaoslpet cenurtaurvtnienitog. fcTothhmeebtthhinirraedteiuocnonimstoibsfintnheoidrwdunaaidntsdd.efIitdrsamtn,adyatlbhseeo third and secondare not required to be worked out resulting in considerable = Rs 1,64,664 saving in computationaleffort. The processis repeated,till all available units are exhausted.The advantageof this approachis that having oitiined the Total operatingcost for this casewill be the total fuel costplus the start-upcost of unit l, i.e. 1,64,6& + 400 = Rs 1,65,064 Comparingthis with the earlier case,it is clear that it is economicalto run both the units. It is easyto see that if the start-upcost is Rs 200, then it is economical to run only 2 in the light load period and to put off unit 1. gPli-u.| way,of loading ft units, it is quite easyro determinethe Jptimal manner 7.3 OPTTMAL UNrT COMMTTMENT (UC) of loading (ft + 1) units. Let a cost function F\" (x) be defined as follows: As is evident,it is not economicalto run all the units availableall the time. To F,y (x) = the minimum cost in Rs/hr of generating.r MW by N units, determine the units of a plant that should operate for a particular load is the problem of unit commitment (UC). This problem is of importancefbr thermal fN 0) = cost of generatingy MW by the Nth unit plants as for other types of generationsuch as hydro; their operatingcost and start-uptimes are negligible so that their on-off statusis not important. F*-{x - y) - the minimum cosr of generating(.r - y) Mw by the remain_ ing (1/- t) units A simple but sub-optimal approachto the problem is to impose priority ordering,wherein the most efficient unit is loaded first to be'followed by the Now the applicationof DP results in the following recursiverelation lessefficient units in order as the Ioad increases. FN@)= TnVn9) * Fu-r @- y)| (7.r2) A straightforward but highly time-consuming way of finding the most economicalcombination of units to meet a particular load demand,is to try all Using the aboverecursiverelation, we can easilydeterminethe combination possiblecombinationsof units that can supply this load; to divide the load of units, yielding minimum operating costsfor loadsranging in convenient steps optimally among the units of each combination by use of the coordination from the minimum.permissible load of the smallestunit to the sum of the equaiions,so asto finci the most economicaioperatingcost of the combination; canaeifies nf q l l $qvr rso^iriqoul -rlvo ruru- ri +rDo. ri -rr +L Lr r;l^D --^^^-^ .u1r^e- 1-1 r . then,to determinethe combinationwhich hasthe leastoperatingcost among all PruuttJ s total nunlmum oDerating eost these.Considerablecomputationalsavingcan be achievedby using branch and bound or a dynamic programming method for comparing the economics of and the load sharedby each unit of the optimal combination are ;il.u,i\" determinedfor eachload level. combinationsas certaincombinationsneetnot be tried at all. The useof DP for solving the UC problemis bestillustratedby meansof an example. Considera samplesystemhaving four thermal generatingunits with parameterslisted in Table 7.2.It is requiredto determinr th. most-economical units to be committedfor I MW. a load of 9 MW. Let the load changesbe in stepsof
. ModernPow Table 7.2 Generatingunit parametersfor the samplesystem The effect of step size could be altogether eliminated, if the branch and bound technique [30] is employed. The answer to the above problem using Capacity (MW) Cost curve pararneters (d = 0) branch and bound is the samein terms of units to be committed, i,e. units 1 and 2, but with a load sharing of 7.34 MW and 1.66 MW, respectively and a total Unit No. graung cost oI I<s z,y.zL /J/nour. 1 1.0 12.0 0.77 23.5 26.5 In fact the best scheme is to restrict the use of the DP method to obtain the 2 1.0 12.0 1.60 30.0 UC table for various discrete load levels; while the load sharing among 32.0 committed units is then decidedby use of the coordinationEq. (7.10). 3 1.0 12.0 2.00 For the example under consideration,the UC table is preparedin steps-of I 4 1.0 12.0 2.50 MW. By combining the load range over which the unit commitment does not change,the overall result can be telescopedin the form of Table 7.3. Now Ft@) = ft@) fr(9) = f{9)= LorP'ot+ btPcr Table 7.3 Status*of units for minimum operatingcost (Unit commitment = ollss x 92 + 23.5x9 = Rs 242.685lhour tablefor the samplesystem) From the recursiverelation (7.12), computationis made for F2(0), Fz(l), Fz(2),..., Fz(9).Of these Load range Unit number FzQ)= min tt6(0) + Ft(9)1, VzG) + Ft(8)l' r234 VzQ) + Ft(7)1, VzQ) + Fr(6)l' Vz@)+ F1(5)l' l-5 00 6-r3 00 t6(5)+ Fr(4)1V, zG)+ Ft(3)1V, zT + Fr(2)1, t4-18 10 1948 11 tfr(s)+ F,(1)1v, zg) + Fr(O)l) *l = unit running;0 = unit not running. \\ On computing term-by-term and compdng, we get ' The UC table is preparedonce and for all for a given set of units. As the load cycle on the station changes,it would only mean changesin starting and FzQ) = Vz(2)+ Ft(1)) = Rs 239.5651how stopping of units with the basic UC table remaining unchanged. Similarly, we can calculate Fz(8),Fz(1), ..., Fz(l), Fz(O). Using the UC table andincreasingload in steps,the mosteconomicalstation Usingthe recursiverelation(7.12),we now computeFl(O),F:(1), ...' F3(9). operatingcost is calculatedfor the completerangeof stationcapacity by using the coordination equations.The result is the overall station cost characteristic Of these in the form of a set of data points. A quadratic equation (or higher order equation,if necessary)can then be fitted to this data for later use in economic Fse)=min {t6(0)+ Fr(9)ft,6(1)+ Fl8)1,...'[6(9)+ rr(0)]] load sharing among generatingstations. = [6(0)+ FzQ)l= Rs239.565ftour Proceedingsimilarly, we get FoQ) = [f4(0)+ Fr(9)] = Rs 239.565lhour 7.4 RELIABILITY CONSIDERATIONS Examinationof Fr(9), Fz(9),Fl(9) and Fa(9) leads to the conclusion that With the increasing dependence of industry, agriculture and day-to-day optimum units to be lommitted for a 9 MW load are 1 and 2 sharing the load household comfort upon the continuity of electric supply, the reliability of ur Z l,tW and 2 MW, respectively with a minimum operating cost of Rs power systemshas assumedgreatimportance. Every eleciric utilify is normaily under obligation to provide to its consumersa certain degreeof continuigl and 239.565/hour. tabie is inclependentof 'u\\e quality of service(e.g. voltageand frequencyin a specifiedrange).Therefore, It must be pointed out here, that the optimai iiC economy and reliability (security) must be properly coordinatedin arriving at the operationalunit commitmentdecision.In this section,we will seehow the numberingof units, which could be completely arbitrary.To verify, the reader purely economic UC decision must be modified through considerations of rnaysolvethe above problem onceagainby choosinga different unit numbering reliability. scheme. If a higheraccuracyis desired,the stepsize could be reduced(e.g. * t*r, with a considerableincreasein computationtime and required storagecapacity.
In order to meet the load demand under contingency of failure (forced periods are a random phenomenonwith operating periods being much longer outage) of a generator or its derating causedby a minor defect, static reserve than repair periods. When a unit hasbeen operatingfor a long time, the random capacity is always provided at a generatingstation so that the total installed phenomenoncan be describedby the following parameters. capacity exceedsthe yearly peak load by a certain margin. This is a planning Mean time to failure (mean 'up' time), In arriving at the economicUC decisionat any particulartime, the constraint No. of cycles (7.13) taken into accountwas merely the fact that the total capacity on line was at Mean time to repair (mean 'down' time), (7.r4) least equal to the load. The margin, if any, between the capacity of units committed andload was incidental. If under actual operationone or more of the Z (down) Et, (down) units were to fail perchance(random outage),it may not be possible to meet the No. of cycles load requirements.To start a spare(standby)thermal unit* and to bring it on steam to take up the load will take severalhours (2-8 hours), so that the load Mean cycle time = f (up) + Z(down) cannot be met for intolerably long periods of time. Therefore, to meet Inverseof thesetimes can be defined as rates[1], i.e. contingencies,the capacity of units on line (running) must have a definite margin over the load requirementsat all times. This margin which is known as Failure rate, A = IIT (up) (failures/year) the spinning reserveensurescontinuity by meeting the load demandup to a certain extentof probableloss of generationcapacity.While rules of thumb Repair rate, trt= llT (down) (repairs/year) have been used,basedon past experienceto determinethe system's spinning reserveat any time, Patton's analytical approachto this problem is the most Failure and repair rates are to be estimated from the past data of units (or promising. other similar units elsewhere) by use of Eqs. (7.13) and (7.I4). Sound engineeringjudgement must be exercised in arriving at theseestimates.The Since the probability of unit outageincreaseswith operatingtime and since a unit which is to provide the spinning reserveat a particular time has to be failure rates are affected by preventive maintenanceand the repair rates are startedseveralhours ahead,the problem of securityof supply hasto be treated in totality over a period of one day. Furthefinore,the loads are never known scnsitiveto size, compositionand skill of repair teams. .r with complete certainty. Also, the spinning reserve has to be provided at suitable generating stations of the system and not necessarily at every By ratio definition of probability, we can write the probabiliiy of a unit being generatingstation.This indeedis a complexproblem. A simplified treatmentof in 'up' or 'down' statesat any time as the problem is presentedbelow: p (up) = r(up) _ l.t (7.rs) z(up)* z(down) p + A f1(down) f2(down) P (down) = Z(down) ) (7.16) Z(up)* Z(down) tr+ A f3(down) Obviously, p (up) + p(down) = 1 p (up) andp (down) in Eqs.(7.15) and (7.16)are alsotermedas availability andunavailability, respectively. Time--------> When ft units are operating,the system state changesbecauseof random Fig. 7.6 Randomunitperformancreecordneglectingscheduledoutages outages.Failure of a unit can be regardedas an event independentof the state A unit during its useful life spanundergoesalternateperiods of operationand of other units. If a particularsystemstatei is definedasX, units in 'down' state repair as shown in Fig. 7.6. The lengths of individual operating and repair 1 r, ' ( t ---t- lr = \\ / i +. Ir z \\ !ul -r-s -----1--l-:1:--- U^ t r uL 1r^t^r s y s t e l l l Ul - ^t: r- ^l t r $ t2 t- l ut LllI S S l a l t r an0 I j ln Up. Stale \\K ) t PIUDaUITILy is pt = ot(down) (7.r7) {r,r;(ul),{* . If hy^dro.generationis available in matter of minutes to take up load. the system, it could be brought on line in a
gW ModernPowersystemAnalysis WWt I the sample systemfor the load curveof Fig. 7.7 Patton's Security Function Unit number A breach of system security is defined as some intolerable or undesirable condition.The only breachof securityconsideredhere is insufficientgeneration t234 A I1 probability that the available generationcapacity (sum of capacitiesof units B I0 C 00 cbmmitted) at a particular hour is less than the system load at that time, is D I0 E 00 defined as [25] F 00 S = Ep,r, (7.18) where p, = probabilityof systembeingin state i [seeEq. (7.17)] r, = probability that system state i causesbreach of syStem securlty. When systemload is deterministic(i.e. known with completecertainty),r, = 1 if available capacity is less than load and 0 otherwise. S indeed, is a quantitative estimateof systeminsecurity. Though theoreticallyEq. (7.18) must be summed over all possiblesystem states(this in fact can be very large), from a practical point of view the sum needsto be caried out over statesreflecting a relatively small numberof uniqs on forced outage,e.g. stateswith more than two units out may be neglectedas the probability of their occurrencewill be too low; Security Constrained Optimal Unit Commitment 0L 0 Time in hours ------------> (noon) Oncethe units to be committed at a particular load level areknown from purely Flg.7.7 Dailyloadcurye economicconsiderationst,he securityfunction S is computedasper Eq. (7.18). This figure should not exceeda certain maximum'tolerable insecurity level Therefore, 2 (MTIL). MTIL for a givcn systemis a managemendt ecisionwhich is guided where by past experience.If the value of S exceeds MTIL, the economic unit S = piti : p1r1* p2r2 commitment scheduleis modified by bringing in the next most economicalunit ,? as per the UC table. S is then recalculatedand checked. The process is continued till ^t < MTIL. As the economic UC table has someinherentspinning Pr= P(up=) ffi = 0.99, rr = 0 (unitI = 12MW> 5 MW) reserve,rarely more than one iteration is found to be necessary. pz= p(down)= p + ^ = 0.01, rz = t (with unit I down load For illustration,reconsiderthe fbur unit exampleof Sec.7.3. Let the daily demand cannot be met) load curve for the systembe as indicatedin Fig. 7.7. The economicallyoptimal UC fbr this load curve is immediately obtainedby use of the previously Hence preparedUC table (seeTable 7.3) and is given in TabIe 7.4. S= 0.99x 0 + 0.01x 1 = 0.01> 0.005(MTIL) Let us now checkif the aboveoptimal UC table is securein every period of the ioacicurve. Thus unit I alonesupplying5 MW load fails to satisfytne prescribed securitycriterion.In orderto obtainoptimalandyet secureUC, it is necessary For the minimum load of 5 MW (period E of Fig. 7.7) accordingto optimal to runthenextmosteconomicaulnit,i.e.unit2 (Table7.3) alongwith unit1. UC Table 7.4., only unit 1 is to be operated.Assuming identical failure rate ) of l/year and repair rate pr,of 99/yearfor all the four units, let us check if the systemis securefcrrtheperiod E. Further assumethe systemMTIL to be 0.005. Unit I can be only in two possiblestates-operatingor on forced outage.
ifl5\"#Wl MooernpowerSystemRnarysis E#nft# With both units I and2 operating, security function is contributed only by = Rs 4,463.216(start-upcost = 0) the statewhen both the units are on forced outage.The stateswith both units Clearly Caseb resultsin overall economy.Therefore,the optimal and secure operatingor either one failed can meet the load demand of 5 MW and so do UC tablefor this load cycle is modified as under,with due considerationto the not contributeto the security function. Therefore, overall cost. S = p (down; x p (down) x 1 = 0.0001 Unit number This combination(units 1 and 2both committed)does meet the prescribed Period MTIL of 0.005, i.e. ^S< MTIL. A II 1l Proceedingsimilarly and checking securityfunctions for periodsA, B, c, D and F, we obtain the following optimal and secureUC table for the sample systemfor the load curve given in Fig. 7.7. Table 7.5 Optimal and secure UC table B II 10 C I I l'l' 0 Unit number D II l0 Period E I I 00 F II 00 A II *Unit was starteddue to start-upconsiderations. B 10 C 00 7.5 OPTIMUM GENERATION SCHEDUTING D 10 E I t< 0 0 From the unit commitment table of a given plant, the fuel cost curve of the plant F 00 can be determinedin the form of a polynomial of suitabledegreeby the method of least squaresfit. If the transmissionlossesare neglected,the total system * Unit was started due to security considerations. load can be optimally divided among the various generatingplants using the equal incrementalcost criterion of Eq. (2.10).It is, howrurr, on.\"alistic to Start-up Considerations neglect transmissionlossesparticularly when long distancetransmission of poweris involved. The UC table as obtainedabove is secureand economically optimal over each individual period of the load curve. Sucha tablemay require that certainunits A modern electric utility serves over a vast area of relatively low load haveto be startedandstoppedmore thanonce.Therefore,start-upcostmust be density.The transmissionlossesmay vary from 5 to ISVoof the total load, 4'd taken into considerationfrom the point of view of overall economy. For therefore,it is essentialto accountfor lcsseswhile developingan economicload example, unit 3 has to be stopped and restartedtwice during the cycle. We dispatchpolicy. lt is obviousthatwhen losse must, therefore,examinewhether or not it will be more economical to avoid one the simple 'equal incrementalcost' criterion. sarepresent,we canno longer use restarting by continuing to run the unit in period C. To illustratethepoint, consider a Casea When unit 3 is not operatingin period C. two-bus systemwith identical generatorsat eachbus (i.e. the sameIC curves). Total fuel cost for periods B, c and D as obtainedby most economic load Assumethat the load is locatednearplant I and plant 2 hasto deliver power sharingare as under (detailedcomputationis avoided) via a lossy line. Equal incrementalcost criterion would dictatethat each plant should carry half the total load; while it is obvious in this casethat the ptunt = 1,690.756+ 1,075.356+ 1.690.756= Rs 4.456.869 1 should carry a greatershareof the load demandtherebyreducing transmissio' losses. Start-upcost of unit 3 = Rs 50.000 (say) In this section, we shall investigatehow the load should be sharedamong Iotal operatrngcost= Rs 4,506.868 rrr$q rrirrvr rsr au nl.lrqqnuf cLor .vr rYl ruo v- rr lu; -u^v rl ^v-i-r^D^ sr 4rE .1uuuurrttLr^g s nt(^),f-, ry ODJgCtfVg fS tO mirufiUze Caseb When all threeunits are running in periodC, i.e. unit 3 is not stoppecl at the end of period B. Ine Total operatingcosts= I,690.756+ 1,0g1.704+ 1,690.756 the overall cost of generation c = (7.7) ,\\-rci(Pc') at any time under equality constraint of meeting the load demand with transmissionloss, i.e.
k (7.re) Equation (7.23) can also be written in the alternativeform DP\", - P,- P I - = 0 (IC)i- An- QTL)ifi = 1,2, ..., k (7.2s) where i:l This equationis referredto as the exact coordination equation. k = tatalrnumber of generating plants Thus it is elear tha-tto solve the optimum lead sehedulingproblem; it is necessaryto compute ITL for eachplant, and therefore we must determine the Pci= generationof lth plant functional dependenceof transmissionloss on real powersof generatingplants. Pp = sum of load demandbt all buses(systemload demand) There are severalmethods, approximate and exact, for developing a transmis- Pr= total systemtransmissionloss sion loss model. A full treatmentof theseis beyond the scopeof this book. To solve the problem, we write the Lagrangianas One of the most important, simple but approximate,methodsof expressing transmissionloss as a function of generatorpowers is through B-coeffrcients. t=tr,(Pc)-^[t\",-\"Po-\".] (7.20) This method is reasonably adequatefor treatment of loss coordination in i:l economic scheduling of load betweenplants. The general form of the loss formula (derivedlater in this section)using B-coefficientsis It will be shownlaterin this sectionthat,if the power factor of load at each P. = II pG^B*,pGn (7.26) bus is assumedto remain constant, the systemloss P, can be shown to be a m:7 n:I function of active power generationat eachplant, i.e. Pr= Pt(Pcp P52, ..., P51r) (7.2r) where Thusin the optimizationproblem posedabove,Pci Q = I,2, ..., k) arethe only PG^, PGr= real power generationat m, nth plants control variables. B^n= loss coefficients which are constantsunder certainassumed For optimum real power dispatch, ') operatingconditions A L = d C , - ^\\+r A, 1 P t - i = r' 2' \"\" k (7.22) If P6\"sarein megawatts,B*n arein reciprocalof megawatts*C. emputations,of oPo, dPGt- #*:O' ,: course,may be carriedout in per unit. Also, B*r= Bn^. RearrangingEq. (7.22) and recognizing that changing the output of only one Equation (7.26) for transmissionloss may be written in the rnatrix form as plant can affect the cost at only that plant, we have Pr= PIBPI (7.27) Where = ) or #Li= ) , i = r , 2 ,. . . ,k (7.23) where It may be noted that B is a symmetric matrix. For a three plant system,we can write the expressionfor loss as Li= (7.24) Q-APL iAPGi) PL= Bn4, + Bzz4, + Bzz4, + 28rrpcrpcz+ ZBnpGzpG3 is calledthe penaltyfactor of the ith plant. TheLagrangianmultiplier ) is in rupeesper megawatt-hour,when fuel cost + 2BrrPorpo, e.ZS) is in rupees per hour. Equation (7.23) impiies that minimuqr ftrei eost is wiih the systemDoweriossmorieiasper Eq. e.z6), we c.annow write obtained,when the incrementalfuel cost of eachplant multiplied by its penalty factoris the samefor all the plants. Ap =f^tl?f tlPo_o, ,\"a^I^'o') The(k + 1) variables(P6r, P62,..., Pct, )) canbe obtainedfrom k optimal * dispatchF,q. (7.23) together with the power balance Eq,. (1.19). The parrial derivative)PLIAPGiis referredto as the incrementaltransmission loss (ITL),, *B^, {in pu) = B^n (in Mw-t) x Base MVA associatedwith the lth generatingplant.
ffif ModerPn o*ersystemAnatysis kk 4. Calculatep, =If pciBijpcj. j:l J=l It may be notedthatin the aboveexpressionother termsareindependentof Po, 5. Check if power balanceequation(?.t o) is satistied and are, therefore,left out. l -sPD- prl.l' t (a specifiedvalue) Simplifying Eq. (7.29) and recognizing that B, = Bir, we can write lLPot ln*t I -a+ P=Dk zBijpcj (7.30a) If yes, stop. Otherwise, go to step6. )Fo, j:l Assuming quadraticplant cost curves as 6. Increase) by A) (a suitablestepsize)' tt pc,- ^ - <0or [* \") Ci(Poi)=*o,4,+ b,p.,+ d, decrease) by AA (a suitablestepsize);if (8\", - po- p, We obtain the incremenlalcost as repeatfrom step 3. )>0, dc, = atPo'+ b' (7.30b) dP' Example7,4 ' Substinrting APL|&G' and dcildPci from above in the coordination Eq. (7.22), we have k (1.3I) A two-bus systemis shown in Fig. 7.8. If 100 Mw is transmited from plant 1 to the load, a transmissionloss of 10 MW is incurred. Fin( the required aiPci+b,+ Slzn,,ro,:^ generationfor eachplant and the power receivedby load when tie system ,\\ is Rs 25llvlWh. J:l Collecting all terrnsof P, and solving for Po, we obtain k The incrementalfuel costsof the two plants are given below: (ai + 2M,,) Pci= - )D ZBijpG-jbi +^ dC - o'ozPc+r 16'oRs,Mwh '-l A* J*T r t- :f: - o'o4PG22+o.oRs/lvIWh & n ' t '28,,P - ^l - a - ) ) L\"\"ii'Gi ]:I Pc, ,, i-7'2'\"''k (7'32) ) *rB=; For any particularvalue of \\, Eq. (7.32) can be solved iteratively by Fig. 7.8 A two-bussystemfor Example7.4 assuminginitial valuesof P6,s(aconvenientchoiceis P\", = 0; i = l, 2, ..., k). Iterations are stoppedwhen Po,s converge within specifiedaccuracy. o6 ^ur -tuu- ro: - n- since the ioad is at bus z a\\one,p\", v,.r,!not have any effect ort Fr. Therefore Equation (7.32) along with the Dower balanceF,q,.(7.19) for a pa-rtieular ioaci ciemancPi o aresoiveciiteratively on the following lines: 1. Initially choose) = )0. B z z =0 a n d B n = 0 = B z , 2 . A s s u m eI t G i =0 ; I = 1 , 2 , . . . ,k . Hence Pl-= BnPbr For Po, - 100MW, Pr = 10MW, i.e. 3. Solve Eq. (7.32)iterativelyfor Po,s. (i)
\" l Mod\"rnPo*\"r.Syrt\"r An\"lyri, 10= Bn (100)2 I Brr= 0'001MW-r At plant 2 the load increasesfrom 37.59 MW to 125 MW due to loss Equation(7.31)for plant 1 becomes coordination.The saving at plant 2 is A.A2P;1+2^;B1P;1 +2^aBr2P62= ^-76 (ii) = - Rs 2,032.431hr (iii) and for plant 2 The net savingachievedby coordinatinglosseswhile schedulingthe received load of 237.04MW is r' 0 . 0 4 P c 2 + 2 A B r r P \" r + Z ) B u P c r =) - 2 0 Substituting the valuesof B-coefficientsand ) - 25, we get 2,937.69- 2,032.43= Rs g,OtS.Ztm, Pc t = 1 2 8 ' 5 7M W Pcz= 125Mw Derivation of Transmission Loss Forrnula The transmissionpower loss is An accuratemethod of obtaining a general formula for transmissionloss has Pr = 0.001x (128.57)'= 16.53MW been given by Kron [4]. This, however,is quite complicated.The aim of this article is to give a simpler derivation by making certain assumptions. and the load is Po = Pct * Pcr- Pt- = 128.57+ I25 - 16.53- 237.04 MW Figure 7.9 (c) depicts the caseof two generatingplants connectedto an arbitrary number of loads through a transmissionnetwork. One line within the network is designatedas branch p. Im,aginethat the total load current 1, is supplied by plant 1 only, as in Fig. 7.9a.Let the current in line p & Irr.Define Considerthe systemof Example 7.4 with a load of 237.04 MW at bus 2. Find \\ (7.33) the optimum load distributionbetweenthe two plants for (a) when lossesare includedbut not coordinated,and (b) when lossesare also coordinated.Also find the savings in rupeesper hour when lossesare coordinated. Solution Case a If the transmission loss is not coordinated, the optimum schedulesare obtainedby equatingthe incrementalfuel costsat the two plants. Thus 0 . 0 2 P -+ 1 6 = 0 . O 4 P c z + 2 0 (i) The powerdeliveredto theloadis Pct * Pcz= 0.001P4+r 237.04 (ii) SolvingBqs.(i) and(ii) for P61nnclP6.2w, e get Pcr= 275.18MW; andPrr2= 37.59MW Caseb This caseis alreadysolvedin Example7.4. Optimumplant loadings (c) with loss coordinationare Flg. 7.9 Schematic diagramshowingtwo plants connectedthrough Pct= 128,57MW; Pcz = 125 MW a power networkto a number of loads Losscoordinationcausesthe load on plant I to reducefrom 275.18MW to Sirniinriv with nient ) qinrnc ctrnnirrino t h e f n t e l l n q d nv sn^rravnr tr r t 'E irc5 . ?, . rov )k,\\ rw' av v^ (al-t l 128.57MW. Therefore,saving at plant I due to loss coordinationis r^_-^, \\ r define Kgi:\",+ 16)dP: co,.MP+l,rr6Pctli),'r', I^n e.34) = Rs 2,937.691hr Mrz= it=D Mo1 wrd Mp2 arecalled curcent distribution factors. The values of current distribution factors depend upon the impedances of the lines and their interconnection and are independentof the current Ip.
fW ModernpowerSygtemAnatysis e, = lstrp( Re p t Substitutingfor llrl2 fromEq. (7.37),and l1o,l and llurl from Eq. (7.38),'we When both generators t and 2 aresupplying current into the network as in obtain Fig.7.9(c), applying the principleof superpositionthe currentin the line p can be expressedas where 1ot and Io2 arethe currentssupplied by plants I and 2, respectively. D- rDG2 r At this stagelet us make certainsimplifying assumptionsoutlined below: \"- (1) All load currentshavethe samephaseangle with respectto a common 1v31\"*frrL) pu|rn, refere{ce. To understandthe implication of this assumptionconsider the load *Wl,r,rMpzRp P current at the ith bus. It can be written as lVllv2lcos /, cosQ, 7 VDilI (6t- d) = lloil l1i *' P3' M32RP (7'3s) where {. is the phaseangleof thebus voltage and /, is the lagging phaseangle tvzP(\"od,ritT of the load. Since { and divary only through a narrow rangeat various buses, it is reasonableto assumethat 0,is the same for all load currentsat all times. Equatio(n7 3e)\"T,o: + 4,8,, ;:;::::;:,PczB,z (2) Ratio X/R is the samefor all network branches. Thesetwo assumptionsleadus to the conclusionthat Ip1 andI, fFig.7 .9(a)) 8 '.n. -- have the samephaseangleand so have Ioz and 1o [Fig. 7.9(b)], such that the WGoshfDp mS,no current distribution factors Mr, and Mr, are real rather than complex. Brz f f i D M \" M \" R o (7.40) Let, . Ict = llcrl lo, and lcz = 1162llo2 where a, and 02 arephaseanglesof 1\", and lor, respectiveiywith respect to the common reference. From Eq, (7.35),we canwrite llrl2 - (Moll6l cos a1+ Mpzllo2lcos oz)2a (Mrll6lstn o; Bn = T . M ' ,YrLRY. l V , l ' ( c o s6 ) ' 7 Mr2lls2lsnoz)2 (7.36) Expanding the simplifying the aboveequation, we get The terms Bs, Bp and 82, arecalled loss cofficients or B-cofficients.lf ll,,l2= Mzrrllorlz+ tutf,zllczl+z 2MolMrzllctl llGzlcos(a1 - oz) (7.37) voltagesareline to line kV with resistancesin ohms,the units of B-coefficients are in MW-I. Further,with Po, and Po, expressedin MW, P, will also be in Now l sl .rr l = ='?' :' I I .vJL- -&z- (7.38) MK $lvtlc o s / , Jllvrlcosf\" The\\,bove results can be extendedto the generalcase of ft plants with where Pot and Po, arethe three-phasereal power outputs of plants I and 2 at transrnishur loss expressedas power factors of cos (t, and cos Q2,and yl and V2are the bus voltages at the plants. kk (7.41) ff Ro is the resistanceof branchp, the total transmissionloss is given by* P, =Df PG^B^.PG. m:l n:l where 'The 8 r,,, = cos (a,, - on) (7.42) general expression for the power system with t plants is expressedas l v - l l v , l c o s Q -c o s Q - P'r - F3' LT' ,*'3nt,\"RP' ,' t.'-trV4r,rl12-clous|f*firLyu' 1yf 1*r6f It can be recognizedas Pc^ Pcn cos(a, - on) P r =f t r B , * . . . * P'o4oo* zDpG^B^npGn lV* ilV, lcosQ*cosQ, m,n:l ;!l
i?ffil ruoderpnowersystemnnatysis r !nrr!-na! ..-# ffi i I -ii Thefollowing assumptionsincludingthosementionedalreadyarenecessary, lr\" lb-- - if B-coefficientsare to be treatedas constantsas total load and load sharing betweenplants vary. Theseassumptionsare: Y 1'' 1, All load currents maintain a constantratio to the total current. ----t'' b 2. Voltage magnitudesat all plantsremain constant. Refbus v =110\"pu 3. Ratio of reactiveto real power,i.e. power factor at eachplant remains constant. l,o 4. Voltage phase anglesat plant busesremain fixed. This is equivalent to Y assuming that the plant currents maintain constant phase angle with respectto the common reference,sincesourcepower factorsare assumed I ro\"or constantas per assumption3 above. Fig.7.10 Samplesystemof Example7.6 In spite of the number of assumptionsmade, it is fortunatethat treating B- coefficientsas constants,yields reasonablyaccurateresults,when the coeffi- Solution As all load currentsmaintain a constantratio to the total current,we cients are calculated for some averageoperating conditions. Major system have changesrequire recalculationof the coefficients. rd _ 3.6_jo.g _ o.t8z6 Lossesas a function of plant outputscan be expressedby othermethods*,but I , + I d 4. 6- jl. l5 the simplicity of loss equationsis the chief advantageof the B-coefficients method. I\" : 4-r.-6-\"ji-01-..-21-5s - 0 . 2 1 7 4 I,+ld Accounting for transmission losses results in considerable operating economy.Furthermore,this considerationis equally importantin future system planning and, in particular, with regardto the location of plants and building of new transmissionlines: Mor= L, Mtr - - 0.2174,Mrr = 0.2774,Mu = 0.7826 Figure7.10 showsa systemhaving trrvoplants 1 and 2 connectedto buses1 and M,,2= 0, Mnz= 0.7826, Mrz = 0.2174,Mrtz= 0,7826 2, respectively.There are two loads and a network of four branches.The Since the sourcecurrentsare known, the voltagesat the sourcebusescanbe calculated.However,in a practical size networka load flow study has to be referencebuswith a voltageof l.0l0o pu is shownon thediagram.The branch made to find power factors at the buses,bus voltagesand phaseangles. The bus voltagesat the plants are cunentsand impedancesare: Vr = 1.0 + (2 - j0.5) (0.015+ 70.06) I o = 2 - 7 0 . 5P u I,=7 - j0.25Pu = 1.06+ jO.I725 = 1.06616.05\" pu I u = 1 . 6- j 0 . 4 P u Id = 3.6 - 70.9Pu Zo= 0.015+ 70.06pu Z, = 0.OI + 70.04pu Vz= 7 + ( 1. 6 - jO . 4) ( 0. 015+ 70. 06) Zo= 0.015 + 70.06pu Za = 0.Ol + 70.04pu = 1.048+ jO.O9= 1.05114.9\" pu Calculatethe loss formula coefficientsof the systemin pu and in reciprocal The current phaseanglesat the plants are (1, = Io, 12= 16r Ir) m^ ^ ^e- Ds- a w a t f s i f t h e h a s e i s 1 0 0 M V A o-t = t an- t+2! . :- l24oi o2: t a.n- r -^O - l4o 96t : *For accuratemethodsand exactexpressionfor 0P,./0P6i,references122,231 cos(or- ot) = cos 0o = 1 more The plant power factorsare may be consulted. pfi = cos (6.05\" + l4') = 0.9393
ffiffi MocierPn owerSvstemAna[,sis OPtimalSystem Operation ffffii$ffi p l Pfz = cos(4.9 + 14\")- 0.946 The losscoefficientsare lBq. Q.a\\l c = f ci(Pci) (7.7) subject to the load flow equations [see - 0.02224pu e -f tUilvjily,,tcos-(0,, (7.43) (7.44) Bzz 0.0| 5x (0.782q2+ 0.01x (0.21q72 + 0.01x (0.782q2 j:1 ( 1 . 0 5 1x) 2( 0 . 9 4 6 ) 2 3 = 0.01597pu Q,+ jL:rtvinvjlllzulsin(0ul and DDt 2 =_ e0.2174)L0.7826X0.0+150).01x(0.217a)+2 0.01x(0.7826)2 4 - A * 6i - 4)= 0 for eachpv bus (7.4s) rJ66. Lotl x 0.9393x 0.946 Llvllvjlly,,lcos(9,, j:7 = 0.00406pu It is to be noteclthat at the ith bus For a baseof 100MVA, theseloss coefficientsmust be dividedby 100to Pt= Pci- Poi obtaintheir valuesin unitsof reciprocaml egawattsi.,e. Qi= Qci- Qu Dhr r = 0.02L2L2L4+ - 0 . 0 2 2 2 4 x l o 2 M w - l where Po, and ep; areload demandsat bus i. (7.46) 100 Equarions(7.43), (7.44) and(7.45) can be expressedin vector form 8.t.,= 0'01597 = 0.01597x 1o-2Mw-l ' [Eq. (7.a3)l foreachr0 I \\ 100 (x,y) = f | _tn\\.7qql bus (7.47) | ; Br.t= 0'0M06 = 0.00406x lo-z Mw-l lEq.Q.a, foreachpV busi 100 where the vector of dependentvariables is 7.6 OPTIMAL LOAD FLOW SOLUTION l-ty,tl I (7.48a) The problem of optimal real power dispatch has been treated in the earlier .= lr, jforeachrouusf section using the approximateloss formula. This section presentsthe more general problem clf real attd reactive povrer flow so as to minirnize the Ld, for eachpV bus_J instantaneousoperatingcosts.It is a static optimization problem with a scalar objectivefunction (also called cost function). andthe vectorof independenvtariablesis The solutiontechniquegiven here was first given by Dommel and Tinney t r 4In [34]. It is basedon load flow solutionby the NR method,a first ordergradient a. slackbus adjustmentalgorithmfor minimizing the objectivefunction and use of penalty 4i functions to accountfor inequality constraintson clepenclenvtariables.The 4l problem of unconstrainedoptirnalload flow is first tackJed.Later the inequality ]= O] foreachpe bus (7.48b) constraintsare introduced,first on control variables and then on dependent variables. ,1,,t,\"j eachPVbus Optimal Power Flow without Inequality Constraints f n f h a qqvl -v^v. vr o fr^t -*l,r,u. tr^u+l a: ^l L- l u l l , -r tltnefinn mrrcf i_^1,-A^ d^^ _r - [fle ODleQtlVe The objectivefunctionto be minimized is ihe operatingcost power.rrruDllllwlLlLlgLllESracKDuS ovfetcThtheoeroubvjeoecfcttcoivorenotffruoinnlcdvteiaoprnieaanbndledesanwtvvhaeicrcithaobrarpreesotyfofcibxaeendvbaoerripedaaisrtttouitriaobcnahneigdeevinoetrooupnttwicmoonuphmaorlvltaasb_lulaee
ffi@ ModernPowerSvstemAnalvsis puru-\"t\"rs. Controlparametersm*aybe voltagemagnitudeson PV busesP, 6t at buseswith controllablepower, etc. feasiblesolution point (a setof valuesof x which satisfiesEq. (7.5a) for given u andp; it indeedis the load flow solution)in the direction of steepestdesceht The optimization problem** can now be restatedas (negative gradient) to a new feasible solution point with a lower value of objeetlve funstion. By repeating the-.semoves in *rc dkestisn +f the negadve min C (x' u) (7.4e) gradient, the minimum will finally be reached. subjectto equalitYconstraints The computational procedurefor the gradientmethod with relevant details is given below: .f (x, u, p) = 0 (7'50) To solve the optimizationproblem, define the Lagrangianfunction as L (x, u, p)= C (x, u7+ Arf(x, u, P) (7'51) Step I Make an initial guessfor u, the control variables. where ) is the vector of Lagrangemultipliers of samedimensionasf (x, u, p) Smteetyh2od- . TJihned a feasibleload flow solution from Eq. (7.54) by the NR iterative The necessaryconditionsto minimize the unconstrainedLagrangianfunction method successivelyimprovesthe Solutionx as follows. are (see Appendix A for differentiation of matrix functions). * (r+r)- ,(r) + Ax a f ,= 0 c* l y 1 ' ) _ o (7.s2) where A-r is obtained by solving the set of linear equations(6.56b) reproduced below: 0x 0x L}x J 0L= 0c*ly1' )_o (7.s3) l#r,\"',rl]4-\"- r (*(y')), 0u 0u Ldu-J , ar =f (x,u,P)= o (7.s4) u; rheendresurt\"s^-J,,;; $; Iti.[l]'*u,,on orx andtheracobiamn atri,... Equation (7.54) is obviously the same as the equality constraints.'The Step3 Solve Eq. (7.52)for tS ^a L as neededin Eqs. (i.52) and (7.53) are rather r (, { \\ '^ - - r lr - l ac expressionsfor ; 0u \\ = - i (7.s5) involved***.It may howeverbe observedby comparisonwith Eq. (6.56a)that L\\dxl J 0x I Y= Jacobianmatrix [same as employed in the NR method of load flow Insert ) from Eq. (7.55) into Eq. (7.53), and compute the gradient 0x Step 4 solution; the expressionsfor the elementsof Jacobianare given in Eqs. (6.64) and (6.65)1. Y.c=, oc*l 9L1t' (7.s6) Equations(7.52),0.53) and(7.54) are non-linearalgebraicequationsand 0 u L 0 u) can only be solvediteratively. A simple yet efficient iteration scheme,that can by employed, is the steepestdescentmethod (also called gradient method). It may be noted that for computing the gradient,the JacobianJ - +0x is already known from the load flow solution(step2 above). -Slack bus voltage and regulating transformertap setting may be employed as additional control variables.Dopazo et all26ltuse Qo, as control variable on buses' step 5 rf v -c equalszero within prescribedtolerance,the minimum has been reached.Otherwise, Step 6 Find a new set of control variables with reactive Powercontrol' (7.57\\ **rr *L^ ..,.*^- the obiective function is unew= l.l.^6* L,il lI LrMJ Dlvrrl l.vpsar l nnrrrcr lncc ic to he minimized- rv C = Pr(lVl, 6) where Since in this casethe net injected real powers are fixed, the minimization of the real L,u = - o\"V-C, (7.s8) injected power P, at the slack bus is equivalentto minimization of total systemloss, Here A,u is a stepin the negative direction of the gradient. The step size is Th*is**iTshkenoorwigninaasl optimal reactivepower flow problem' be consultedfor details' adjustedby the positive scalar o.. pup\". of Dommel and Tinney t34l may
#*ff\" uooernPoweSr vstemAnalvsis f to the constraint limits, when these limits are violated. The penalty function Steps1 through 5 are straightforward and pose no computational problems. methodis valid in this case,becausetheseconstraints are seldom rigid limits Step 6 is the critical part of the algorithm, where the choice of a is very in the strict sense,but are in fact, soft limits (e.g.lvl < 1.0 on a pebus really ry v-should not exceed1.0 too much and lvl = 1.01 may still be important.Too small a value of a guaranteetshe convergencebut slows down the rate of convergence; too high a value causes oscillations around the Inequality Constraints on Control Variables The penalty method calls for augmentationof the objective function so that thenew objective function becomes Though in the earlier discussion, the control variables are assumedto be C t = C ( x , u )f*U t (7.63) unconstrai\"\"o, are, in fact, always contrained, 1.j\":T.1\",::tutt (7.se) where the penalty W, is introduced for each violated inequality constraint. A suitablepenalty function is defined as e.g. Pc,,,nin1 Po, S Pct, ** w', = - xi,^o)2 i wheneverxi ) xi,rnax Q'64) Theseinequality constraintson control variablescan be easily handled.If the {7i@i wheneverxr(ry,min correctionAu,inBq. (7,57)causesuito exceedone of the limits, a, is setequal [ 71G,-xi,^i)zi to the correspondinglimit, i.e. where Tiis Treal positive number which controlsdegreeof penalty and is called if u,,oro* Au, ) ui,^^ the penalty factor. 7f u,,oro* Au, 1ui,^in otherwise (7.60) After a control variable reachesany of the limits, its componentin the gradient should continue to be computedin later iterations, asthe variable may come within limits at somelater stage. In accordancewith the Kuhn-Tucker theorem (see Appendix E), the necessaryconditionsfor minimization of I, under constraint (7.59) arc: 0L :0 7fu,,*n<ui <ui,^^, Xmln 0u, if u, - ui,^* ui: ui.^u* Fig.7.11 Penaltyfunction oouf ,.-o (7.6r) A plot of the proposedpenaltyfunction is shownin Fig. 7.11,which clearly or,o indicateshow the rigid limits are replacedby soft limits. 0r, - The necessaryconditions (7.52) and (7.53) would now be modified as given below,while the conditions(7.54),i.e. load flow equations,remainunchanged. Thereforernow, in step 5 of the computationalalgorithm, the gradientvector a x _= a__cl_ , \\ 4-) a w j ,f a f l ', (7.6s) has to satisfythe optimalitycondition(7.61). Td- x ox a r * La \" i ) -o l Inequality Constraints on Dependent Variables a x_ A C ,--L sd w ; , fAf 1 ', - :- u= ou' +) a \" * La \"l 'r (7.66) Often, the upper and lower limits on dependentvariables are specifiedas d : o rmir,SxSr*u^ 'Ihevecto U^ t rWZ : e.g. lUn,in < lVl < lYl -.o ofl a PQ bus (7.62) obtainedfrom Eq. (7.64)wouldcontainonly onenon-zero. 0x Such inequality constraintscan be conveniently handled by the penalty termcorrespondintgo thedependenvtariablex;; while # = 0 asthepenalty function method. The objective function is augmented by penalties for functionson dependenvt ariablesareindependenotf thecontrolvariables. inequality constraintsviolations. This forces the solution to lie sufficiently close
' ModernPowerSystemAnalysis Mathematical Formulation By choosinga higher value fot 1,,the penalty function can be madesteeper For a certainperiod of operation7 (one year,onemonth or one day, depending. so thatthesolutionlies closerto therigiA fimits; the convergenceh, owever,will upon the requirement),it is assumedthat (i) storageof hydro reservoir at the becomepoorer.A good schemeis to startwith a low value of 7j andto increase reservoir (after accountingfor irrigation use) and load demandon the system imization process,if the solution exceedsa certain tolerance areknown asfunctions of time with completecertainty (deterministiccase).The problem is to determineq(t),,the water discharge(rate) so as to minimize the limit. cost of thermal generation. This sectionhas shown that the NR method of load flow can be extendedto rT yield the optimal load flow solutionthat is feasiblewith respectto all relevant inequality constraints.These solutionsare often required for systemplanning and operation. 7.7 OPTIMAL SCHEDULING OF HYDROTHERMAL SYSTEM Cr= (Por(t))dt (7.67) JoC The previous sectionshave dealt with the problem of optimal schedulingof a under the following constraints: (7.68) Dower system with thermal plants only. Optimal operating policy in this case (i) Meeting the load demand can be completely determined at any instant without referenceto operation at other times. This, indeed, is the static optimization problem. Operation of a Pcr(r) * Pca(t) - Pr(t) - PoG)= 0; te 10,71 systemhaving both hydro and thermalplants is, however, far more complex as This is called the power balance equation. hydro plants have negligible operatingcost, but are required to operateunder (ii) Water availability constraintsof water available for hydro generationin a given period of time. The problem thus belongs to the realm of dynamic optimization. The problem X (T)- x, (o)-l' t@ at+ lrqgdl t=o (7.6e) of minimizingthe operatingcost of a hydrothermal sYstemcanbe viewed asone JO JO_ of minimizing the fuel cost of thermal plants under the constraint of water whereJ(t) is the water inflow (rate),X'(t) water storage,and X/(0) , Y (T) arc availability (storage and inflow) for hydro generation over a given period of specified water storagesat the beginning and at the end of the optimization operation. interval. (iii) The hydro generation Pcrlt) is a function of hydro dischaige and water storage(or head), i.e. J (waterinflow) Pcn(r)=f(X'(t),q(t)) (7.70) The problem can be handled conveniently by discretization.The optimization interval Z is subdividedinto M subintervalseachof time length 47. Over each subinterval it is assumed that all the variables remain fixed in value. The problem is now posedas Fig. 7.12 Fundamentahlydrothermaslystem : u^r^4-r:....*r\"$^--nr'^(p(Et)L,rfrrrt, \\t.lr) (7.72t For the sakeof simplicity and understanding,the problem formulation and under the following constraints: (i) Power balanceequation solution technique- _ _ - - - - ^ _ 1 - - _ are illustrated through a simplified hydrothermal systemof Q Pt *Ptr-PI-Pt =0 Fig. 7.I2. This systemconsistsof one hydro and one thermai piant suppiying where power to a centralized load and is referred to as a fundamental system. Ptr = thermal generationin the mth interval Optimization will be carried out with real power generation as control Ptn = hydro generationin the zth interval variable,with transmissionlossaccountedfor by the loss formula of Eq. (7.26), PI =transmission loss in the rnth interval
ffiffifj- ModernPowerSystemAnalysis n = n tnm t2 | .rn nm -fI D tnm t2 ) t LDTHTGH OptimaSl ystemOperation D77\\r57 Dpy\\r6p ) ffiffi PI = load demandin the mth interval l- e = water head correction factor to account for head variation with (ii) Water continuity equation storage y^ _ yt(m_r)_ f AT + q^ AT = 0 p = non:effective discharge(water discharge neededto run h dro X ^ = water storageat the end of the mth interval In the aboveproblem formulation, it is convenient to choosewater discharges J* -- water inflow (rate) in the mth interval in all subintervalsexcept one as independentvariables,while hydro genera- tions, thermal generations and water storagesin all subintervalsare treated as q^ = water discharge (rate) in the ruth interval dependentvariables. The fact, that water discharge in one of the subintervals is a dependentvariable, is shown below: The above equation can be written as (7,73) Y - X^-r - J* + q^ = 0; m = I,2, ,..,M Adding Eq. (7.73) for m = l, 2, ..., M leadsto the following equation,known as water availability equation whereY = X/*IAT = storagein dischargeunits. xM- J^+la^ = o (7.7s) InEqs. (7.73), Xo and XM are the specified storagesat the beginning and \" o - Dm m end of the optimization interval. (iii) Hydro generation in any subinterval can be expressed*as Becauseof this equation, only (M - l) qs can be specifiedindependentlyand the remaining one can then be determinedfrom this equationand is, therefore, Ptn = ho{I + o.5e(Y + Y)l (q* - p) (7.74) a dependenvt ariable.For convenienceq, l is chosenasa dependenvt ariable, for where which we can write ho=9.8rx ro-rhto M fto = basic water head (head corresponding to dead storage) qt = xo - xM* DJ^ -Dn^ (7.76) * Solution Technique P3, = 9.81 x 1o-2hk@^ -,p ) Mw The problem is solved here using non-linear programming technique in conjunction with the first order gradient method. The Lagrangian L is where formulated by augmenting the cost function of Eq. (7.7L) with equaliry constraintsof Eqs. (7.72)- (7.74)throughLagrangemultipliers(dual variables) (q^ - p) = effective discharge in m3ls hry,= averagehead in the mth interval \\i \\i'and )i. Thus, Now lffi= 7ro* LT(X^ +X^-r) .c=, D tc(%r)- xT(4r+ 4, - ry- ffi + M (y - y-t-r* + where 2A qr)* ^Ttp1,- h,(r + 0.5e(y* it11* @^- p)rj (7.77) A = draa.of cross-sectionof the reservoir at the given storage h'o = basic water head (head correspondingto dead storage, The dual variablesare obtainedby equating to zero the partial derivatives hk= hLll + o.Se(x'+ X\"-t)l of the Lagrangianwith respectto the dependentvariablesyielding the following equations where AT Dmt Now C- -T) . 7Pt dPt '[- 7Pt )rAJ)PeuAtr/\\,I r / Ur tf . -t \\ where Ahto - A ,rzl I n |^ = l- l-u 0 /78) 4n= ho {! + o.Se(x^+ x^-t)l (7.7e) h o = 9 .8 7x l 0 -3 h to @^- P) [The reader may compare this equation with Eq. (7.23)] #r,-M-^r['-ffi)='
ffil ModerPn owesr ystemAnalysis a (-+) /)^* *U * = )7,- Ar*'- \\r p.sh\"r(q-*p)l- )i*t 1o.5eho ientlyby augmentintghe costfunctioniith p\"nulty functionrur air\";;;i\" I\\ ax^ t (q^*' - DI - o S e c .7 . 6 . (7.80) The methodoutlinedaboveis quitegeneralandcanbe directlyextendedto andusingEq. (7.73)in Eq. (7.77),we get a slzstemhaving multi hydro and multi-ther+nal plan+s. The method, however, has the disadvantage of large memory requirement, since the independent (pt-) = )rr- t\\n\" fl + 0.5e(zy + Jt - zqt+d)= 0 (7.81) variables,dependentvariablesand gradientsneed to be storedsimultaneously. A modified techniqueknown as decomposition[24) overcomesthis difficulty. laq') In this techniqueoptimization is carried out over each subintervaland the completecycle of iteration is repeated,if the water availability equation does. The dual variablesfor any subintervalmay be obtainedas follows: not check at the end of the cvcle. (i) Obtain { from Eq. (7.78). Consider the fundamental hydrothermal system shown in Fig. 7.I2. The (ii) Obtain )! from Eq. (7.79). objective is to find the optimal generationschedulefor a typical dav, wherein load varies in three stepsof eight hourseachas 7 Mw, 10 Mw and 5 Mw, (iii) Obtain )1, from Eq. (7.81) and other values of ry (m * 1) from Eq. respectively.There is no water inflow into the reservoir of the hydro plant. The ' (7.80). initial water storage in the reservoir is 100 m3/s and the final water storage should be 60m3/s,i.e. the total water availablefor hydro generationduring the The gradient vector is given by the partial derivatives of the Lagrangian with day is 40 m3/s. respectto the independentvariables.Thus Basic head is 20 m. Water head correctionfactor e is given to.be 0.005. Assumefor simplicity that the reservoiris rectangularso that e doesnot ehange (+) =\\f.- ^Zh\"{r + 0.5e(2Y-t + J^ - 2q^+ p)} (7.82) with water storage. Let the non-effective water discharge be assumed as 2 m3/s.Incrementafluel cost of the thermalplant is \\oq )m+r dc - r.opcr + 25.0Rsft' For optimality the gradient vector should be zero if there are no inequality dPcr constraintson the control variables. Further, transmissionlossesmay be neglected. The aboveproblem has beenspeciallyconstructed(ratherovelsimplified) to Algorithm illustrate the optimal hydrothermal schedulingalgorithm, which is otherwise Assume an initial set of independent variables q* (m*I) for all computationallyinvolved and the solution has to be worked on the digital subintervalsexceptthe first. computer.Stepsof one complete iteration will be given here. Obtain the values of dependent variables Y, Ptu, F\\r, qt usingEqs. (7.73), (7.74),(7.72) and (7.76). Sincethereare three subintervals,the control variablesre q2andq3.Let us assumetheir initial values to be 3. Obtain the dual variables)f, \\f, )i @ = 1) and )rr using Eqs. (7.78), (739), (7.80)and (7.81). q2= 75 m3ls 4. Obtain the gradientvectorusing Eq. (7.82) and checkif all its elements 15 m2ls are equal to zerowithin a specified accuracy.If so, optimum is reached. If not, go to step5. The value of water diseharge in the first subinterval can be immediateiy found out using Eq. (7.76),i.e. 5. Obtain new values of control variables using the first order gradient method,i.e. er = LOO- 60 - (15 + 15) = 10 m3ls It is given that X0 = 100 m3/sand X3 = 60 m3/s. e k * = q ; a{\\+-o+q *) ;) m = r (7.83) From Eq. (7.73) where cr is a positive scalar.Repeat from step 2 In the solution techniquepresentedabove,if some of the control variables (water discharges)crossthe upper or lower bounds, theseare made equal to their respectivebounded values. For these control variables, step 4 above is checkedin accordancewith the Kuhn-Tucker conditions (7.61) given in S ec .7. 6 .
W Modernpo Xt = f + Jr - gt = 90 m3ls f=Xr+12-q2=75m3/s xt,- x3- )l {o.Sho(eqt- p)l _ S! 1O.Sn@\"e,- p)l = 0 x?,- ll - ^Z {0.5hoe(q,_ p)l _ )3, 1o.5ho(,qt - p)] = o The valuesof hydro generationisn the subintervalscan be obtainedusing Eq.(7.7q asfollows: Substituting various values, we get ) t r = 8 . q 1 4- 2 9 . 6 8 5( 0 . 5 x 0 . 1 9 6 2 x0 . 0 0 5x 8 l _ 3 1 . 3 9 8{ 0 . 5x P b , t = 9 , 8 1x 1 0 - 3x 2 0 [ l + 0 . 5x 0 . 0 0 51 x r+ X 0 ; ) { q ' - p ) 0.1962x 0.005x 13) = 0.1962{I + 25 x 104 x 190}x 8 = 8.1574 = 2 . 3 1 5M W )t, = 9.1574- 31.398(0.5x 0.tg6}x 0.005x t3) 4 n = 0 . 1 9 6 2{ I + 2 5 x l O a x 1 6 5 }x 1 3 - 26.589(0.5 x 0.1962x 0.005x t3) = 7.7877 = 3.602MW UsingEq. (7.82),the gradientvectoris P Z a = 0 . 1 9 6 2{ l + 2 5 x 1 0 a x 1 3 5 }x 1 3 ( # ) = ^ 7 - f t n \" { 1 + 0 . 5x 0 . 0 0(52x e o- 2 x 1 5+ z ) l = 3.411MW = 8.1574- 31.398x 0.1962,{l + 25 x 10a x l52l The thermalgenerationsin the threeintervalsarethen = - 0.3437 pLr = pL- pl, = J - 2.515= 4.685MW ( ar.l - Ar 3: 2 - s t r nf,l + 0.5e(2X2+ f - zqt+ p)I 4r = Pto- PL, = 10- 3.602= 6.398MW F c r = p | , - 4 \" = J - 3 . 4 1 I= 1 . 5 8 9M W lrf )- FromEq. (7.78),we havevaluesof )i as dc(P€) -_' t l\\m - 7.7877- 26.589x 0.1962{I + 25 x 10a x t22} AV GD Tm = 0.9799 or ) T = P [ , + 2 5 tishteaetrIiafcstoftiihoneentdrtw,ooslililenvl rachaeranitavchbeeefletbgosrroabgbderatiaecdianinerternvideet vfdcretooocmurtotisEsritsqna.o0rtt(.i71nz-,gegthwr3oei)t,nhi.oi.tpht(ei<mfoa0lll.oc1ow)n;inhdgeitninocenewstahvreeasnlueoectsoyonefdt Calculating ), for all the three subintervals, we have [^iI [2e685] Ll rr?j?l =lrr.lsa l fzo.ssrJ Llnqk:_\"L*lJ=;i=l;al-[1,o+Ll l#l Also from Eq. (7.79), we can write L o q \") lril hil lzs.68s1 Let us take a = 0.5, then | ^3| = | ^?l: LI Izr.:eseaaJIgfor ttrelosslescsase lf:i=[[]-,'l-:::Ifjllilil LriI LdI and from Eq. {7.76) From Eq. (7.81) 4'r*= 100 - 60 - (15.172+ 14.510)= 10.31gm3ls \\l' = A\\hnii + o.5e(?)(0+ il - z^ ql ' + o l The above computation brings us to the starting point of the next iteration. Iterations are carried out till the gradientvector U\"comeszero within specified - 29.685x 0.1962{l + 25 x loa (2oo- 2o + 2)l tolerance. -,8.474 FromEq. (7.E0)for m = 1 and2, we have
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