modern-power-systems-analysis-d-p-kothari-i-j-nagrath-

ffil uodernPoweSr lrstemAnalvsis Hffiffi PROiBEii/iS miiiions of iriiocaiories per hour can be expressedas a function of power output Poin megawattsby the equation 7.1 For Example 7.1 calculatethe extra cost incurred in Rsftr, if a load of 220 MW is scheduledas Pct= Pcz = 110 MW. 0.00014+ O.$ft + r2.0po+150 7.2 A constantload of 300 Mw is supplied by two 200 Mw generators,I Find the expressionfor inerementalfuel eost in rupeesper megawatt hour and 2, for which the respective incremental fuel costs are as a function of power output in megawaffs.AIso find a good linear dcr = o ' l O P G l+ 2 0 ' 0 approximation to the incremental fuel cost as a function of Fo. Po, Given: Fuel cost is Rs Zhmltion kilocalories. 7.6 For the systemof Example 7.4, the system ) is Rs 26a4wh. Assume dcz - o'lzPc2 + 15'o further the fuel costs at no load to be Rs 250 and Rs 350 per hr, dPo, respectively for plants I and2. (a) For this value of system ),, what are the values of p61, po, and, with powers Pc in MW and costs c in Rsar. Determine (a) the most economicaldivision of load betweenthe generatorsa, nd (b) the savingin received load for optimum operation. Rs/day thereby obtained compared to equal load sharing between (b) For the above value of received load, what are the optimum values machines. 7.3 Figure P-7.3 showsthe incrementalfuel cost curvesof generatorsA and of Pot and Por, if system losses are accounted for but not B. How would a load (i) more than ZPo, (ii) equalto 2p6, and (iii) less coordinated. than ZPo be sharedbetween A and B if both generatorsare running. (c) Total fuel costs in RsArrfor parrs (a) and (b). 7.7 FigureP-7.7showsa systemhavingtwo plants I and 2 connectedto buses (MW)mtn P6 1 and 2, respectively. There are two loads and a network of three branches.The bus 1 is the referencebus with voltageof 1.0 I 0\" pu. The Flg. P-7.3 branch currents and impedancesare Io=2 -70.5 pu 7.4 Considerthe following threeIC curves PGr=-100+50(IQt-2Aqi -Lo= 1 6 - i O 4 n r r Pcz= - 150+ 60 (IQz - 2.5 AqZ 1,= 1.8- i0.45 pu Pct= - 8.0+ 4a Qq3 - 1.8 Aqi Zo= 0.06+ j0..24pu Zt = 0.03+ J0.12pu where ICs are in Rs/IVIWhand P6s are in MW. Z, = 0.03+ /0.I2 pu The total load at a certain hour of the day is 400 MW. Neglect Calculatethelossformulacoefficients6f the systemin per unit and in transmission loss and develop a computer programme for optimum reciprocaml egawattsif, the baseis 100MVA generationschedulingwithin and accuracyof + 0.05 MW. Note: All P6s must be real positive. Refbus Flg. P-7.7 Samplesystemfor probtemp-7.7

uoo\"rnpo*\", syrt\"r Anutyri, w W /. r\\cuenswanoer, J.}(-, Modern power systems, International rext Book co., New York, 1971. 7.8 Fot thepower plant of theillustrative exampleusedin Section7.3. obtain the economicallyoptimum unit commitment for the daily load cycle given 8' singh, c. and R. Billinton, system Reliabitity, Modelling and Evaluation, in Fig. P-7.8. Hutchinsonof London,1977. Correct the scheduleto meet security requirements. 9. sullvan, R.L., power systempianning, McGraw-Hil, New york, 1977. i\" 10' Wood, A'J' and B.F' Wollenberg,Power Generation,operation edn.,Wiley, New york, 1996 and Control, Znd E,o 11' Mahalanabis,A.K., D.p. Kothari and s.I. Ahson, computer Aided power system Analysis and control, Tata McGraw-Hill, New Delhi. r9gg. 12. Bergen,A.R., power system Anarysisp, rentice-Hall,rnc., Ncw 13' Billinton, R. and R.N. Allan, Reliability Evaluation of power Jersey,19g6. Press,New york, 1984. System,plenum 0 14. Sterling,M.J.H., power SystemControl,I.E.E., England,197g. 8 1 2 16 20 15. Khatib,H., \"Economicsof power systemsReriability,',Technicopy.r9g0. Timein hours---------' 16. warwick, K. A.E. Kwue and R. Aggarwar (Eds), A.r. Techniquesin power Flg. P-7.8 Dailyloadcurvefor problemp-7.9 System.sIE, E, UK, 1997. 24 17. Momoh, J.A., Electic power SystemApplicationsof Optimization,MarcelDekker, Inc., New York., 2001. 7.9 RepeatExample 7.3 with a load of 220 Mw from 6 a.m. to 6 p.m. and 18' Yong-HuaSong(Ed.),Modern optimization Techniquesin power Systems,Kluwer 40 MW from 6 p.-. to 6 a.m. Academic Publishers,London, 1999. 7.10 Reformulate the optimat hydrothermal schedulingproblem considering the 19' Debs, A.S., Modern power systemscontror and operation, KAp, New york, inequality constraints on the thermal generation and water storage 1988. employing penaity functions. Find out the necessaryequations and gradientvector to solve the problem. 20. Berrie,T,W., Power SystemEconomics,IEE,London, 19g3. 2r. Berrie,T.w., Electricity, Economicsand pranning, rEE, London, rp92. Dr o nq lo, -Eor ; , 22. CHoMpoieltelyi,mfefiErac,.ileFlwyn..otsas.rn\"d,dIeaErnwEedd.EDEvT..rDirmas. tniensAva.lpebt iAneosSrntTo,sneJo,cunJhl.,yRn..i.1Iq,m9u.6.peAgrso,'Ngv, ,I7eeE:dwE1LE5oM4sTegsrt.ahFnoosdrmmopufAlaDs ,ceot1em9r7mp1ui,nr9ai0nti:gon7L1ob6sys. 23. REFERNECES 24' Agarwal,S'K. and I.J' Nagrath,\"optimal Schcdulingof HydrothermaSl ystems,,, Proc. IEEE, 1972, 199: 169. Books 25' Aytb, A'K' and A.D. Patton,\"Optimal Thermal GeneratingUnit Commitment,,, 1. Billinton,R., Power SystemReliabilityEvaluation,GqrdonandBreach,New york, IEEE Trans.,July-Aug 1971,pAS_90:1752. t970. 26' Dopazo,J'F. et al., \"An optimization Techniquefor Real and Reactive power 2 . Billirrton, R., R.J. Ringleeand A.J. Wood. Power SystemReliabitity Calculations, Allocation\", Proc, IEEE, Nov 1967. 1g77. The MIT Press,Boston,Mass, 1973. 27. Happ, H.H., \"optimal power Dispatch-A Comprehensivesurvey,,, IEEE Trans. 5 . Kusic, G.L., computer Aided Power system Analysis,prentice-Hall,Nerv Jersey, 1 9 7 7 ,P A S - 9 6 : 8 4 1 . 1986. 28. Harker,8.c., \"A primer on Loss Formula\", AIEE Trans, r95g, pt ilr,77: 1434. 4 . Kirchmayer,L.K., Economicoperation of Power systemsJ, ohnwiley, New york, 29' IEEE CommitfeeReport, \"Economy-security Functionsin power -systemopou- I9)6. tionS\". IEEE ,snprinl Ptthlirntinn 1, 5J vf \\rrrJvn o7 awo7 . ( DE \\Yr rYDr \\ , lL\\re-w IorK, Iyl). w 5 . Kirchmayer, L.K., Economiccontrol of Interconnectedsystems,wiley, New york, 1959. 30. Kothari, D.p., \"optimal HydrothermalSchedulingand Unit commitment,. ph. D. Knight, u.G., Power systemsEngineeringand Mathematicsp, ergamonpress,New York. 1972. Thesis,B.I.T.S, Pilani, 1975. 31' Kothari, D.P. and I.J. Nagrath,\"security ConstrainedEconomicThermal Generat- ing Unit Commitment,, J.I.E. (India), Dec. 197g,59: 156. 32' Nagrath,I.J. and D.P. Kothari,\"optimal StochasticSchedulingof CascadedHydro- thermalSystems\",J.I.E. (India), June 1976, 56: 264.

ffi ModernpowersvstemAnalvsis mal J2J2. rDv^D\"wl Lr r^w- t r t JI. v^ rr c^ll. t ../\\-r:-^l t1^-t-^l ^f D --^ri--^ n------ Fr ----rr rnnh ar <| vQvar rn. uQ. r q^ r- ri s nD If ^+L^-j ../\\-.:- -t ThermaiGeneradng'tjnir Commitment-A r.,tPtlrrr.u \\.VtlLfUl Ul I\\|jAL;|IYtr fUWEf lfQnS, v.l . nuluatl l, fIUW , IEDL IyO6, \\_rpl,llllal P A S .8 7 : 4 0 . Review, \"Int. J. EPES, Vol. 20, No. 7, Oct. 1998,pp. 443-451. 34. Dommel,H.w. and w.F. Trinney, \"optimal power Flow solution\", IEEE Trans., 52. Kulkarni, P.S.,A.G. Kothari and D.p. Kothari, \"Combined Econonic and Emission October1968,PAS. 87: 1866. Dispatchusing ImprovedBpNN\", 3t - 4_7 /nf. J. of EMPS,Vol. 28, No. l, Jan 2000,pp. 35. Sasson,A.M. and H.M. Itlerrill, \"SonneApplicationsof OptimizationTechniques 53. Aryu, L.D., S.C. Chaudeand D.P. Kothari, \"EconomicDespatchAccountingLine to Power SystemProblems\",Proc. IEEE, July 1974,62: 959. Flow Constraintsusing Functional Link Network,\" Int. J. of Electrical Machine & Power Systems,28, l, Jan 2000, pp. 55-6g. 36. wu, F. et al., \"A Two-stageApproachto solving optimal power Flows\", proc. 1979 PICA Conf. pp. 126-136. 37. Nanda, J., P.R. Bijwe and D.P. Kothari, \"Application of ProgressiveOptimality 54. Ahmad, A. and D.P. Kothari, \"A practical Model for Generator Maintenance Algorithm to Optimal Hydrothermal Scheduling ConsideringDeterministic and Schedulingwith rransmission constraints\", Int. J. of EMps, StochasticData\", International Jountal of Electrical Power and Energy Systems, 2000,pp. 501-514. vol. 2g, No. 6, June January1986,8: 61. 55. Dhillon, J.s. and D.P. Kothari, \"The surrogate worth rrade off Approach for 38. Kothari, D.P. et al., \"Some Aspects of Optimal MaintenanceScheduling of Mutliobjective Thermal power Dispatch hobelm\". EpsR, vol. 56, No. 2, Nov. 2000,pp. 103-110. GeneratinglJnits\", \"/./.E (India), August 1985, 66: 41. 39' Kothari, D.P. and R.K. Gupta, \"Optimal StochasticLoad FIow Studies\", J.I.E. 56. Son, S. and D.P. Kothari, \"Large ScaleThermalGeneratingUnit Commitment:A New Model\", in TheNext Genera\\ion of Electric Power (Jnit CommitmentModels, (India),August 1978,p. 34. 40. \"Description and Bibliography of Major Economy-security Functions-Part I, il, 57. edited by B.F. Hobbset. al. KAp, ,\\BKoosthtoanri,,20,,0F1u,zpzpy. Zll-225- Making in Multi and III\", IEEE committeeReport,IEEE Trans. Jan 1981.pAS-r00, zlr-235. Dhillon, J.s., s.c. Parti and D.p. Decision 41. Bijwe, P.R., D.P., Kothari, J. Nanda, and K.S. Lingamurthy,.,Optimal Voltage objective Long;term scheduling of Hydrothermalsystem,,, rnt. t. oy erns, vol. 23, No. l, Jan 2001,pp. lg-29. Control using ConstantSensiiivityMatrix\", Electric Power SystemResearch,Vol. II, No. 3, Dec. 1986,pp. 195-203. 58. Brar, Y.s., J.s. Dhillon and D.p. Kothari, \"Multi-objective Load Dispatch by Fuzzy Logic based Searching Weightage Pattern,\" Electric power Systems 42. Nanda,J., D.P. Kothariand K.S. Lingamurthy,\"Economic-emissionLoad Dispatch Research,Vol. 63, 2002, pp. 149-160. throughcoal ProgrammingTechniques\",IEEE Trans.on Energyconversion, vol. 3, No. 1, March 1988,pp. 26-32. 59. Dhillion, J.s., S.c. Parti and D.p. Kothari, \"Fuzzy Decision-mgkingin stochastic Multiobjective short-term Hydrothermal scheduling,\" Ip,B proc.tcTD, vol. l4g, z, 43. Nanda, J.D.P. Kothari and s.c. srivastava, \"A New optimal power Dispatch March 2fi02,pp l9i-200. Algorithm using Fletcher's QP Method\", Proc. IEE, pte, vol. 136. no. 3, May 1 9 8 9 ,p p . 1 5 3 - 1 6 1 . 60' Kothari, D.P., Applicationof Neural Netwdrksto Power Systems(Invited paper), Proc. Int. Conf., ICIT 2000, Jan. 2000, pp. 62l_626. 44. Dhillon, J.S., S.C. Parti and D.P. Kothari, \"stochasticEconomicEmissionLoad Dispatch\",Int. J. of Electric Power systemResearch,Yol. 26, No. 3, 1993, pp. 1 7 9- 1 8 3 . 45. Dhillon,J.S.,S.C.Partiand D.P. Kothari,\"MultiobjectiveOptimalThermalPower Dispatch\",Int. J. of EPES,Vol. 16, No.6, Dec. L994,pp.383-389. 46. Kothari, D.P. and Aijaz Ahmad, \"An Expert system Approach to the unit commitment hoblem\", Energy conversion and Management\",vol. 36, No. 4, April 1995,pp. 257-261. 47 - Sen, Subir, D.P. Kothari and F.A Talukdar, \"EnvironmentallyFriendly Thermal PowerDispatch- An Approach\", Int. J. of EnergySources,Vol. 19, no. 4, May 1 9 9 7 ,p p . 3 9 7 - 4 0 8 . 48. Kothari, D.P. and A. Ahmad, \"Fuzzy Dynamic ProgrammingBased optimal GeneratorMaintenanceSchedulingIncorporatingLoad Forecasting\",in Advances in Intelligent systems,edited by F.c. Morabito, IoS press, ohmsha, 1997, pp. /.J5- /4U. 49. Aijaz Ahmad and D.P. Kothari, \"A Review of Recent Advances in Generator Maintenance scheduling\", Electric Machines and Power systems, yol 26, No. 4, 1998, pp. 373-387. 50. Sen S., and D.P. Kothari, \"Evaluation of Benefit of Inter-Area Energy Exchange of Indian Power System Based on Multi-Area Unit Commitment Approach\", Int. J. of EMPS, Vol.26, No. 8, Oct. 1998, pp. 801-813.

R causedby momentarcyhargein generafosrpeectlI,'r.r.tnr*,-i;;?t;qffi; ;; excitationvoltagecontrolsarenon-interactivfeor smallchangesand can be 1., modelledandanalyseidndependentlFyu. rthermore,xcitationvoltageeontrolis F : t c l : t c f i n r r ri nr r rvrvr rhrirnvhr r tLhr cr v -r r, rr ci ^r rJ v r frirrl trtrwo vnr,Jr nr r.J. rr -, (rlhr rr r u^ 6r l^L,Ur ruil-t+L^t r-l^c.u1 :- rL^e ^$rL- -^-^--^- r5 llla! ul ulc; ggirtcfalor field; while the power frequencycontrol is slow actingwith major time constant contributedby the turbineandgeneratormomentof inertia-this time constant is muchlargerthanthatof thegeneratortield. Thus,the transientsin excitation voltagecontrol vanish much faster and do not affect the dynamics of power frequencycontrol. 8.T .INTRODUCTION I Powersystemoperationconsidcrcdso far wasunderconditionsof stcadyload. P+JQ However, both active and reactivepower demandsare never steadyand they continually change with the rising or falling trend. Steam input to turbo- Fig. 8.1 schematicdiagramof loadfrequencyandexcitation generators(or waterinput to hydro-generatorsm) ust,therefore,be continuously voltageregulatorsof a turbo-generator regulatedto match the active power demand,failing which the machinespeed will vary with consequentchange in frequency whieh may be highly Changein load demandcan be identified as: (i) slow varying changesin undesirable*(maximum permissiblechangein power fiequency is t 0.5 Hz). meandemand,and (ii) fast randomvariationsaroundthe mean.The regulators Also the excitation of generatorsmust be continuouslyregulatedto match the mustbe dusignedto be insensitiveto thstrandomchangeso, therwisethe system reactive power demand with reuctive generation,otherwise the voltagesat will be prone to hunting resulting in excessivewear and tear of rotatins various systembusesmay go beyond the prescribedlimits. In modern large machinesand control equipment. interconnected systems, manual regulation is not feasible and therefore automaticgenerationand voltageregulationequipmentis installedon each 8.2 LOAD FREOUENCY CONTROL (STNGLE AREA CASE) generator.Figure 8.1 gives the schematicdiagram of load frequency and excitationvoltageregulatorsof a turbo-generatorT. he controllers are setfor a Let us considerthe problemof controlling the power outputof the generators particularoperatirrgcondition and they take care of small changesin load of a closely knit electricareaso asto maintz,inthescheduledfrequency.All the denrandwithout fiequencyand voltageexceedingthe prescribedlimits. With generatorsin such an areaconstitutea coherentgroup so thatall the generators the passageof time, as the change in lcad demand becomes large, the s^ - p^ ^ el eoiip ancisiow- I ^ l - - - - . riowii- l ^ - - - - ^ tL o- - get- . r -i-ie^ r. _na_ _ : r-n, tarnrngt. rhe. lrreiarrvepower angies.Such contrcllersmustbe reseteithernianuallyor automatically. an areais defined as a control area. Tire boundariesof a coqtrol area will generallycoincide with that of an individual ElectricityBoard Company. It has beenshownin previous chaptersthat for small changesactivepower is dependenot n internalmachineangle 6 and is inderrendenot f bus voltage: To understandthe load fiequencycontrol problem,let us consider a single whiie bus voitageis dependenton machine excitation (therefore on reactive turbo-generatorsystemsupplyingan isolated load. -\"- Changein frequenccyauseschangein speedof the consumersp' lantaffecting productionprocesseFs.urtheri,t is necessartyo maintainnetworkfrequencyconstant so thatthe powerstationsrun satisfactoriliyn parallel,the variousmotorsoperating on the systemrun at the desiredspeedc, orrecttime is obtainedfrom synchronous clocksin the systema, ndtheentertainingdevicesfunctionproperly.

W Modern power system Analys,s turbine. Its downward movementopensthe upperpilot valve so that more steem Turbine Speed Governing System is admitted to the turbine under steadyconditions (hencemore steadypower Figure 8.2 showsschematicallythe speedgoverningsystemof a steamturbine. The systemconsistsof the following components: . The reverse Steam Model of Speed Governing System Speed changer Assume that the system is initially operating under steady conditions-the linkage mechanism stationary and pilot valve closed, stearnvalve opened by a definite magnitude, turbine running at constantspeedwith turbin\" po*\"r output balancing the generatorload. Let the operating conditions be characteizedby \"f\" = systemfrequency (speed) P'c = generatoroutput = turbine output (neglecting generator loss) .IE= steam valve setting We shall obtain a linear incremental model around these operating conditions. Pilot Let the point A on the linkage mechanism be moved downwards by a small amountAye.It is a commandwhich causesthe turbinepower output to change --t-\\ and can therefore be written as value Main Aye= kcAPc (8.1) piston High pressure oil A rI Hydraulic amplifier where APc is the commandedincreasein power. \\ (speed control mechanism) The command signal AP, (i.e. Ayi sets into rnotion a bequenceof events- the pilot valve moves upwards,high pressureoil flows on to the top of the main piston moving it downwards; the steamvalve opening consequentlyincreases, the turbine generatorspeedincreases,i.e. the frequencygoesup. Let us model Fig.8,2 Turbinespeedgoverningsystem theseevents mathematically. Reprintewd ithpermissioonf McGraw-HiBltookCo.,NewYork,from Ollel. Elgerd: Two factors contribute to the movement of C: ElectricEnergy SystemTheory:An lntroduction,1g71,p. 322. (i) Ayecontributer- [\\ ?r Jl l Aya or - krAyo(i.e. upwards)of - ktKcApc (i) FIy ball speedgovernor: This is the heartof the systemwhich sensesthe changein speed(frequency).As the speedincreasesthe fly balls move outwards (ii) Increase in frequency ff causesthe fly balls to move outwards so that and the point B on linkage mechanism moves downwards.The reversehappens B moves downwards by a proportional amount k'z Af. The consequent when the speeddecreases. movement of Cwith A remaining fixed at Ayo- . (+) orO, - + kAf G) Hydraulic amplifier: It comprises a pilot valve and main piston alrangement.Low power level pilot valve movement is converted into high (i.e. downwards) power level piston valve movement.This is necessaryin order to openor close the steamvalve againsthigh pressuresteam. The net movement of C is therefore (xl) Lintcagemechanism:ABC is a rigid link pivoted at B and cDE is AYc=- ktkcAPc+kAf (8.2) anotherrigid link pivoted at D. This link mechanismprovides a movementto the control valve in proportion to changein speed.It also provides a feedback The movementof D, Ayp, is the amountby which the pilot valve opens.It is ,,fr9rn the steamvalve movement (link 4). contributedby Ayg and AyB and can be written as Ayo=(h) Ayc+(;h) *,

= ktayc + koAys controt E 1 (g.3) rEi^y,L, ^rr.iL^ -ru' \\/ oo.oo \\., r: -s rcpfesenl.eor . tne ronn of - t pcapvTrosiaohsms(l(petviiuoi)poe)mmnarItaonrBpiaeodvetendinemrocdtatmniaioatlosutetlipto,snnwhreegtepehnaaooihfiyconcrfi.hgtrtgoicoh(oe-tiephd)napsee,fnereaonesbpxbristnceeeoesgenrvarutsndAeemraioe,ycndvf'oltuogohae.pnm'elaovitantnehrittaneiohbttsiyeipsitrshits.oioJatgtfunoyn\"griyoo.lnb,ipalnyenucJandhre.\"disir,tgsotmhtahneeiietnprateoerjmuebfdsvsytttashiofuleivmarpetebohoolaveeriltir.nsecsgoyinmflteihtnpghedrleiiegmfpyriibailniolisengt diagram in Fig. ln a block 9.3. Ks9 4Y5(s) 1+fsss tinhteeTaghrreaelavooo,flfuatmyhoee.coTrfohoseislm-asodevmcetimitoteeondnf tttoahyet\"h-i.epsicsyotolbinntad.Tienhreuidssbtyhudsivpidroinpgortthioenoaill to the time 4F(s) volume by Flg. 8.3 ,Blockdiagramrepresentatioonf speedgovernorsystem Avn= krfoeayrlat (8.4) The speed governing system of a hydro-turbine is more involved. An additional feedback loop provides temporary droop compensationto prevent cItacuasnesbneegveartiivfiee(dufprwomardth)me socvheemmeantticadyiualgcrcaomunthtiantga instability. This is necessitatedby the targeinertia or the penstoct gut\" which in Eq. (8.4). regulates the rate of water input to the turbine. Modelling of a hyjro-turbine positive movement ayo, regulating systemis beyond the scopeof this book. for the n\"gutiu\" ,ign used Takingthe Laplacetransformof Eqs. (g.2),(g.3)and(g.4),we ger Turbine Model AYr(s)=- k&cApc(\")+ krAF(s) (8.5) Let us now relate the dynamic responseof a steam turbine in tenns of changes Ayp(s)= kzAyd,s)+ koAyug) (8.6) in power ouFut to changesin steamvalve opening ^4yr. Figure g.4a shows a two stage steam turbine with a reheat unit. The dynamic *ponr\" is targely a y u ( g = - k sol r U n (8.7) influenced by two factors, (i) entrainedsteambetwein the inlet stbamvalve and 'fofitrutsuttpsus, tttahogefettohuferbtlhionewetutprrarbeninssesfeu, (rrifeiu)snttahcgetieostntooirslaacgghebaaercahticiontnedirtnihzatehtdeboyrfetthhweeoahttiiegmrhwehpcircoehnsscstauaurnestsest.saFtghoeer. EliminatingAyr(s) andAyo(s), we can write AYu(s)- k'ktk'AP' (s)- k,krAF(s) easeof analysisit will be assumedherethat the turbinl can be modelled to have ') (oo'' t Ssingle equivalent time constant.Figure 8.4b shows the transferfunction model of a sreamturbine. Typicaly the time constant{ lies'in the rangeo.i ro z.s \\ \"'tr ,/ sec. -lor,<,r-*^or\",].i#) -=-&Steam valve (8.8) where n = k l cKt_2 = speedregulationof the governor K.,= - gainof speedgovernor (a)Two-stagesteamturbine +y AYg(s)-FAPds) .,rr.-s= l\" = tlme constantof srp- -eedgovernor (b) Turbinetransferfunctionmodel ;-; Flg. 8.4 KqkS

#ph-Si rrrroarprno*\", s),rt\"r An\"ly.i, AutomatlcGenerationand Voltage Control I =tAP6g)a_Po(,)r.[#j I Generator Load Model (s.13) Theincremenitn powerinputto thegeneratbr-loasdystemis 2H APG _ APD Bf\" = pow€r systemtime constant whele AP6 = AP,, incremental turbine Kp,= + =powersystemgain incremental loss to be negligible) and App is the load increment. Equation (8.13) can be representedin block diagram form as in Fig. g.5. This incrementin power input to the syrtem is accountedfor in two ways: (i) Rate of increase of stored kinetic energy in the generator rotor. At ^Po(s1)6la-e-o-(fsf)ioro, scheduledfrequency (fo ), the stored energy is Wk, = H x p, kW = sec (kilojoules) whereP, is the kW rating of the turbo-generator andH is defined as its inertia constant. The kinetic energy being proportional to squareof speed(frequency),the kinetic energy at a frequency of (f \" + Arf ) is given by =nr,(r.T) (8.e) Flg.8.5 Blockdiagramrepresentatioofngenerator-lomadodel Rate of change of kinetic energy is therefore complete Block Diagrram Representation of Load Frequenry Control of an Isolated Power System $rr*=\"frffrr\"n (8.10) (ii) As the frequency changes,the motor load changesbeing sensitive to AP(s)=trPn15; speed,the rate of changeof load with respectto frequ\"n.y, i.e. arot\\ycan be regarded as nearly constant for small changes in frequency Af ard can be expressedas @PDl?flAf=BAf (8.11) wherethe constantB can be determinedempirically,B is positivofor a predominantlmy otorload. AP6(s) Writing the powerbalanceequationw, e have A P c - a P, r^== T-Hf .P] *' d <( o f l +B A f Dividingthroughoutbyp, andrearanging,we get Flg. 8.6 Blockdiagrammodelof loadfrequencycontrol (isolatedpowersystem) AP6$u)- AP;q;u)= 1d (/ AAf'i^ +' Bn('p7t't-)-a- \\f (8.i2) f Steady States Analysis dt The model of Fig. 8.6 shows that therearetwo important incrementalinputs to Taking the l,aplace transforrn,we can write AF(s) as the load frequency control system- APc, the change in speedchangersetting; and APo, the change in load demand. Let us consider,,.4,.simplesituatiqn in 4Fis; - AP,G) -4PoG) B*-'- s

Modern which the sneerl .hqnrro' 'h(roro c. .r g.- .^) u^ -c.t-t.r n g c = o) and the load tbfTccouehhhlgtlaaew.rirnnrv'LelEeggo^eefeeaoan.ss^dr1fLisiern'n^uce,-wtu-hqfh^rlyieueoet'hdeqaclinuudqncleuteedyhnaadeiafcusrrmnyeionduaqancrlnnugoreddeelavs.enmdemrcSessfyapontotalertlnfanefe(rlddrore0siefren0teltegagh%ga,toeuFidovaloiaynegtrstrf.d(iuntogea8lon.lrt1.reot7o6oRcfp)ahsi5ecoaihvsra.noao-Tngnwtfbhiaeroseeotsnumiat.nwrhdpaeirnfptlrolhoylelioieqsnlpodspue,afoeeoardoenrratdmcsodrcylejohuncrlpuaaasouetntuIileooogsldofnaeeasrttddhhsdhbtaei)oiysp.tt rr^t r(r \\7'e' af :,:il?;:3l i:T::: a2rreego,*, ;, 2;;;;*;:r;;ffi; *' #ff:Tiil:steadcyhangiensystemfrequen-focyrasudd.n.hung\"ff,i;ffi\"ffi;ti'l; r.han.,oo 't::; ;-::--::^ demand anaoun*t, (, e.Apog):+)is obtainedasfollows: aF@)l*,(s)::o - AP^ ( 'l relationshi-p is -l ^f I \\ B+(t/R) ) cvoamlPupeoawisreisrBosny=s.Et0eq.mu0a1ptapioruanmM(8ew.t1ea6rlz)Bthieasnngdseimnl/eRprlai=fiiley sUmt3ou)cshostmhaatlrBer*cthaannber/nReg(aletcytpeidcianl sperehdgedovreornoo\"rrrpe,g,uflla\"t,ionf.jfli;], (8.17) curvisethusmainly determinedby R, the K Ir ^ s o r r , ( = 1 ap,=_*\"r: (r^;)o\", .I It is also rccognizedthat Ko,= 7 / B , w h e r e B - Y ^/P' (in PuMWunit change Decreaisnesystemload= BAf= (uffi)*, ai in frequency).Now iOncf recoausersine,tgheenceoranttiroinb.uFtioorntyopf icdaelcvreaaluseeisnosf yBstaenmdloRadquisotmeduecharlleiesrs than the APo = 0.971 APo 4=-(#6)o,. (8.16) Decreasein systemload = 0.029 ApD consider now the steady effect of changing speed changer setting (Or\"<rl- +)with load demandremainingfixed (i.e. Apo= 0). The sready fi roa state changein frequencyis obtainedas follows. (J at *For 250 MW machine with an operating load of 125 MW. \\rli\\ ,, ruudA^t -loL| oao| for IVo changein frequency (scheduledfrequency = 50 L be i%o let the change in load (ii)60% Load Hz). Then 8.rog a-:?:r?: .c af 0. 102 101 100 :2.5 NNVtHz 0 s Percent Load ' = ( # ) b: Flg. 8.7 Steady qharacteristicof a speed #: o'olPuMwgz govern\"o*r-sly?s3t9e-mfrequency

uodernPowersystemAnalysis Autor\"ticG\"n\"r\"tionandVolt\"g\"Conttol F W I trt stg t tvf (t ^ p s AD AF@lap,{s):o: xu'c (8.18) ( 1 +T , r s ) ( l *4 sx) l-* zors)+ KseKKt p,/R s (8.1e) I4,flr*uoyro,\"_: t KreKrKp, \\lIlAP, Two generators rated 200 MW and 400 MW are operating in parallel. The K .sK tK ps/ droop characteristicsof their governors are4Vo and5Vo,respectivelyfrom no I AP',:g ( 1+ - load to full load. Assuming that the generatorsare operatingat 50 Hz at no load, how would a load of 600 MW be sharedbetweenthem?What will be the Rr system frequency at this load? Assume free governor operation. If Repeatthe problem if both governorshave a droop of 4Vo. K rrK,= l Solution Since the generatorsare in parallel, they will operate at the same A\" r =( B | l \\r c \" (8.20) frequency at steadyload. Let load on generator1 (200 MW) = x MW \\ +l R) and load on generator 2 (400 MW) = (600 - x) MW Reductionin frequency= Af If the speedchanger setting is changedby AP, while the load demand changesby APo, the steadyfrequencychangeis obtainedby superposition,i.e. Ar= ( ru) 'o\" - APo) (8.21) Now According to Eq. (8.2I) t\"h.e frequencychangecausedby load demandcan be a f _ 0.04x 50 (i) compensatedby changingthe settingof the speedchanger,i.e. x 200 APc- APo, for Af = Q af 0.05x 50 (ii) Figure 8,7 depicts two load frequency plots-one to give scheduled 6 0 0 - x 400 frequencyat I00Vorated load and the other to give the samefrequencyat 6O7o EquatingAf in (i) and (ii), we get rated load. v - 231MW (loadon generator) 600- x= -/A trltf /1 ^-l u^ -ll Btrrltrriltur L) JOy lvlw (IUau Systemfrequenc' y= 50 - 0'0-1150 x 231 = 47.69 Hz 200 A 100MVA synchronousgeneratoroperateson full load at at frequencyof 50 Hz. The load is suddenlyreducedto 50 MW. Due to time lag in governor It is observed here that due to difference in droop characteristics of system,the steamvalve beginsto closeafter0.4 seconds.Determinethe change governors,generatorI getsoverloadedwhile generator2 is underloaded. in frequencythat occurs in this time. Given H = 5 kW-sec/kVA of generatorcapacity. It easily follows from abovethat if both governorshave a droop of.4Vo,they Solution Kinetic energy storedin rotating parts of generatorand turbine will sharethe load as 200 MW and 400 MW respectively,i.e. they are loaded corresponding to their ratings. This indeed is desirable from operational = 5 x 100 x 1.000= 5 x 105kW-sec considerations. Excesspower input to generatorbefore the steam valve Dynamic Response beginsto close= 50 MW To obtain the dynamic responsegiving the changein frequencyas function of Excessenergyinput to rotating parts in 0.4 sec = 50 x 1,000x 0.4 = 20,000kW-sec the time for a stepchangein load, we must obtain the Laplaceinverse of Eq. Storedkinetic energy oo (frequency)2 (8.14). The characteristicequationbeing of third order, dynamic responsecan Frequencyat the end of 0.4 sec r' r | 1-!-- - I f-,- - -^^^tC: ^ ---*^-:^^1 U^ ^a^1^DE. tIII(^r.w-s, ^Y,s,I^r- +LLfI^s r^,LIl^<-ll^4^iv+L^s-l-lDi ^r+lvi n = 5ox I soo,o+oozo,oo)to\"= 5r rfz Onfy Dg ODIalneU luf A SPtrUfffU ll|'llll('llua1'I \\ 500,000 ) equation can be approximated as first order by examining the relative magnitudesof the time constantsinvolved.Typical valuesof the time constants of load frequencycontrol systemare rdlatedas

Trr4T, <To, Time (sec)-------> Typically* t, = 0.4 sec, Tt = 0.5 sec and t -1 Firstorderapproximatiorl I Io Flg' 8.8 Firstorderapproximatberockdiagramof road Dynamicresponse_cohf angein frequencyfor a stepch ang ein load frequencycontrotof an isolatedarea (APo= 0.01pu, 4s = 0.4sec, | = 0.5sLc,Io. = 2bse c,(\" = 100, R= 3) Irning =ofTF, '=ig.08:.I8u,lfdromK*w\\ hic=h1w),e reduced to Tro the block diagramof Fig. 8.6 is The plot of change in frequency versus time for thlt can write given above and the exact responseare shown first order ap^prriorsxtimadon to, approximationis obviously a poor approximation. in Fig. a.g. order (1+ KpslR)+ AF(s)l*r(s):o = - Z p . s.\"-.APs o Gontrol Area Concept - - \"o{1:- =xaP, So far we have considered the simplified case of a single turbo-generator , l, + ^R+4r o, J' 1 supplying an isolated load. Considernow a practical systemwith e number of generatingstationsaird loads.It is possible to divide an extendedpower system L (say, national grid) into subareas(may be, StateElectricity Boards) in which the generatorsare tightly coupledtogether so as to form a coherent group, i.e. A(r,)-=ft{' -*,[-,,a[n#)]*], g 22) all the generators respond in unison to changes in load o, ,p\"rJ changer settings.Such a coherentareais called a control area in which the frequency TakingR = 3, Kp,= llB = 100,e, = 20,Apo = 0.01pu (8.23a) is assumedto be the same throughout in static as well as dynamic conditions. Af (t)= - 0.029(I - ,-t:tt', (8.23b) For purposes of developing a suitable control strategy, a control area can be reduced to a single speedgovernor, turbo-generator and load system. All the Aflrt\"udystare= - 0.029 Hz control strategiesdiscussedso far are, therefore, applibable to an independent control area. Proportional Plus fntegral Control It is seen from the above discussionthat with the speedgoverning sysrem installed on each machine, the steadyload frequency charartitirti\" fi agiven speedchangersetting has considerabledroop, e.g. for the systembeing usedfor the illustration above, the steadystate-droop in Eq. (8.23b)l from no load to tull load (l fieo=ueneywill be 2.9 Hz [see pu load). System frequency specifications are rather stringentand, therefore, so much changein frequency \"For a 250 MW machinequotedearlier,inertiaconstanrIl = SkW-seclkVA cannot be tolerated. In fact, it is expected that the steadychangein frequency ', = 4B:fo. 2 *05.015x=0 = 2 o s e c will be zero. While steadystatefrequencycan be brought back io the scheduled

ffil ModernPowesr ystemAnalys AutomaticGenerationand VoltageControl t- I in ihe above scheme ACE being zero uncier steaciyconditions*, 4 logical vaiueby adjus'ringspeedchangersetting,the systemcould undergo intolerable design criterion is the minimization of II,CZ dr for a step disturbance. This integral is indeed the time error of a synchronouselectric clock run from the dynamic frequency changeswith changes in load. It leads to the natural power supply. Infact, modern powersystems keep Eaekofintegra+e4tinae errsr all the time. A corrective action (manual adjustment apc, the speed changer suggestion that the speed changer setting be adjusted automatically by setting) is taken by a large (preassigned)station in the areaas soon as the time error exceedsa prescribed value. monitoring the frequency changes.For this purpose, a signal from Af is fed The dynamics of the proportional plus integral controller can be studied througfan integrator to the s diagram numerically only, the systembeing of fourth order-the order of the system has increasedby one with the addition of the integral loop. The dynamic response configuration shown in Fig. 8.10. The systemnow modifies to a proportional of the proportional plus integral controller with Ki = 0.09 for a step load disturbanceof 0.01 pu obtained through digital computer are plotted in Fig. plus integral controller, which, as is well known from control theory, gives zero 8.11.For the sake of comparisonthe dynamic responsewithout integral control action is also plotted on the samefigure. steadystateerror, i.e. Af lrt\"\"d\",,ut,= 0. + -1 Integral APe(s) APp(s) controller I AF(s) l+t-r8- I tor I I t-+ x tl AP6(s) Frequencsyensor Fig. 8.10 Proportionapllusintegral oadfrequencycontrol The signal APr(s) generatedby the integral control must be of oppositesign to /F(s) which accountsfor negative sign in the block for integral controller. Now AF(s1= Kn, ( r+% \" s()*.* + ) . ( l * f , r Ko, + 4 s ) s)(l \" +RKo,s(l+{rs)(l+ 4s) (8.24) Flg. 8.11 Dynamicresponseof loadfrequencycontrollewr ith and without + {'s)(1 + 4sXl f zo's)R* Ko'(KiR f s) integralcontrolaction(APo = 0.01pu, 4s = 0.4 sec, Ir = 0.5 sec, Ips= 20 sec,Kp.= 100,B - B,Ki= 0.-09) obviousry Af l\"t\"^dsytat=e so/F(s) : o (8.25) , In contrastto Eq. (8.16) we find that the steadystatechangein frequency 8.3 IOAD FREOUENCY CONTROL AND ECONOMIC hasbeenreducedto zeroby the additio4 of the integral controller. This can be DESPATCH CONTROL arguedout physically as well. Af reachessteady state (a constant value) only Load f r e o u e n c v_ _ J c_ _o_n_ _t r_ o_ _l w, . i_t_h_ i_n- -t- e_ Oo r e l eonfrnller qnhierrAe ?a?^ craolrr ora+o rwrl.M,llrsrr uAr cp-^ - HAr pD - -= v.ov unusl -qfra! .nt Becarrs-e of fhe intes!'atins actiOn Of the 'vu lvrv otvsuJ I Dl4lg controller,this is only possibleif Af = 0. In central load frequency control of a given control area, the change (error) frequencyelTor and a fast dynamic response,but it exercisesno control over the in frequencyis known asArea Contol Error (ACE). The additional signalfed relative loadings of various generatingstations(i.e. economicdespatch) of the back in the modified control schemepresentedabove is the integral of ACE. control area.For example, if a suddensmall increasein load (say, 17o) occurs 'Such a control is known as isochronous control, but it has its time (integral of frequency) error though steady frequency error is zero.

f Automatic _T--- D1ic..rlu1rrg_s::oltrrotnl e gaorevae,rtnhoersrooaf da-llfregqeuneenractyincgounnriiotsr ,changesthe speed changer aac8co'c1cmo2onmrgtdriavoanelndsacsrtehewi geaint.shTachiehgeceeosmnniegaorntmaiactdilectidoad'geocrshyatpahmanetgoccefheetbnhroterrtoahsrpi,teh[ecpeeoosdnec(ocdhmoeansni cti3rrdeoeedlrsss)-iepotarpttiJtncwahgccoi sot utmlyaoppln)iu.s]u.ttert usru.cuFntieiitgtadsuboirrneyf of the area so that, together, theseunits match the load and the frequenry returnstp the scheduledvalue (this tbgeIahoyecceatotaidonodaivonnjrueemtgasraslitko@ioecnaofsgdpgteuhltarde@ocS.gleuoainenminirtreasfoacfrecotwnaotsrrssoef(lceKood,v)neditnrosIcln)ol.uHaifndadodeciwicntJv,hegiindsavouoenfamtrghi,lneeeinduiusintvnnihgiiidettnarsu,a.piamlnrrlHoeautcophnnewreniesteepsssovrercfeoinanrt\"tfh,\"diinribes,tgehpcinimLehstxnaeaedngiysregrceaneilnvsttoheoeofentdf (smCigoEndDaifClieP)d6abn(yddetihsseirtersadign)nissamlcirtoetmepdrpetuosteetnhdtebinylogitcnhateel gecrceaonlntnroacrlgneiccaodtneotshmpaaitctincdhsectsaopnnattrotocflhltceimor m(eE.pDuTCthee)r installed at eachstation. The systemthus operateswith economicdesfatch satisfactory. only for very short periods of time beforJ it is readjusted. error \"fnceot 8.4 TWO-AREA LOAD FREOUENCY CONTROL An extendedpower system can be divided into a number of load frequency control areasinterconnectedby meansof tie lines. Without shall considera two-area caseconnectedby a single tie loss of generality we Fig. 8.13. line aslilusnated in Speed Fig.B.i3 Two interconnectecdontroar reas(singretie rine) lr. EDC - Economic despatch controller isnrntThnheuelctacanoseneotorufosflrloreyebqgjueueclatnitvceeyt,hnpeorotwipeioslirnttieoonpreaoglwpuelluarsatesintphteeergfirrnaetlqecuro-eannrtecroyalpoleof rweweaircllchobanerteirnaasacttnasdl.lAetsod CEDC - Central economic despatchcomputer so as to give zero steady stateerror in tie line power flow as comparedto the contractedpower, Flg. 8-12 Control area load frequency and economicdespatch control Reprinted (with modification) with permission of McGraw-Hill Book Company, reeqfueItirvitsaolceaonrntetvuaer7bniianenend,tgltyehhnosessreuamwtoietrhadtnhsdaugtffoei xvaec2rhncroeorfnseytrrstootlemaarr.eeSaayc2ma. bnobles uresperdewsiethntseudbffyixaIn New York from Olle I. Elgerd: Electric Energy SystemsTheory: An Introd.uction, In an isolatedcontrol areacasethe incrementalpower (apc _ apo) was I971,p. 345. accountedfor by the rate of increaseof storedkinetic energyand increasein areaload causedby increasein fregueircy.since a tie line t *rport, power in or out of an area, this fact must be accountedfor in the incremental power balanceequationof each area. Power transportedout of area 1 is .eivenbv Ptie,r = ''rrl''l sin({ - q (8.26) X,, where q'q - poweranglesof equivalenmt achinesof thetwo areas.

I ModefnPower SystemAnalysis AutomatiGc eneratioanndVortagecontror Fil 308 | I APti\".r(s) I For incrementalchangesin { and 6r, the incre.mentatlie line power can be expresseads AP,i,,r(pu)= Tp(Afi - 462) (8.27) where T, = 'Y:t'-Yfcos(f - E) - synchronizingcoefficient PrrXrz Fig. 8.14 Since incrementalpower anglesare integralsof incrementalfrequencies,we The corresponding block diagram is shown in Fig. g.15. can write Eq. (8.27)as + AP,i,,r = 2*.(l Afrd-t I Urat) (8.28) APti\",r(s) where Afi nd Af,, arc incremental frequency changesof areas 1 and 2, respectively. -iE= --n7ri\"l Similarly the incrementaltie line power out of area2 is given by AF1(s) aPt;\",z = 2ilzr([ yrat - [ ayrat) (8.2e) Fig.8.15 where For the control area2, Ap6\", r(s) is given by tEq. (g.Zg)l rLzLr= t Y r : J c o s( {L - E\" ) : [S]ti z: ar2rrz (s.30) a p t i \" , z ( s )= -:grrr, [ AFr ( s)- 4F, ( s) ] Przxzr \\Prr) (g:35) With referenceto Eq. (8.12), the incrementalpower balanceequationfor which is alsoindicatedUy ,i. block diagramof Fig. 8.15. \\ Let us now turn our attentionto ACE (areacontrol error; in the presenceof area 1 can be written as a tie line. In the case of an isolatedcontrol area,ACE is the changein area APo,- APo=r +Jr\" *owr )+ nrz|r* AP,,\",t (8.31) frequency which when used in integral control loop forced the steady state frequencyelror to zero. In orderthat the steadystatetie line power error in a It rnay be notedthat all quantitiesother thanfiequencyare in per unit in two-areacontrolbe madezeroanotherintegralcontrol loop(onefor eacharea) Eq.(8.3l). must be introducedto integratethe incrementaltie line powersignal and feed it backto thespeedchangerT. his is aeeomplishebdy a singleintegratingbloek Taking the l-aplacetransf'ormof Eq. (8.31)andreorganizingw, e get by redefiningACE as a linearcombinationof incrementaflrequenryand tie line power.Thus,fbr control areaI A F (s ) = IAP 6 1 G-) A P r,(s) - APti\",,1r;] $.32) \" t$I +- 4,,t,! where as definedearlier [seeEq. (8.13)] Kp31= I/81 ACEI = APu\".r+ brAf, (8.36) Tpil = LHr/BJ\" (8.33) where the constant b, is called areafrequency bias. Equation (8.36) can be expressedin the Laplace transform as Comparedto Eq. (8.13)of the isolatedcontrol areacase,the only changeis ACEl(s) = APo.,r(s)+ b1AF1g) (8.37) the appearancoel the signal APri\"J(s) as shown in Fig. 8.14. Similarly, for the control area 2, ACE2 is expressedas '-l'^Li-- frhr row Tl s l-/ l^4lv- ^va fL*lo4nl loDfrnurrmr r r ^ufr EL Y^ . /a ta\\ tlhl rav oc iro6nr rosl^ ,4P /\"\\ ic nlrfoinerl ACEr(s) = APti\".z(s)+ b2AF,(s) (8.38) I4ArrrS \\v.L9), \", tie.I\\.r/ AS Combining the basic block diagramsof the two control areascorresponding to Fig. 8.6, with AP5rg) and Apr2(s) generatedby integralsof respective AP,i.,1=(sff)roor(s) - /4 (s)l (8.34) ACEs (obtainedthrough signalsrepresentingchangesin tie line power and local frequencybias) and employingthe block diagramsof Figs. g.t+ to g.15,we easily obtain the compositeblock diagramof Fig. g.16.

WIU&| ModernPower svstem Analvsis Let the stepchangesin loadsAPo, andAPrrbe simultaneouslyappliedin control areas 1 and 2, respectively.When steadyconditions are reached,the output signals of all integratingblocks will becomeconstantand in order for this to be so, their input signalsmust become zero. We have, therefore,from F i e .8 . 1 6 APu\",, + brAfr-= O f\\inputof integratingblock- KtL) (8.39a) ,r) torJ APti\",, + brAfr-= o f\\inpotof integratingblock- K'z) (8.3eb) .(oYU rl roai -o !t Afr - Afz- =o finpuro\\tintegratinbgslock -'4'\\ ) (8.40) I il go FromEqs.(8.28)and(8.29) SA oy, A P n \" ,=, - T r , - . I . = c o n s t a n t (8.41) oE AP.i\",z, Tzt; ar2 ooN- EF HenceEqs.(8.39)- (8.41)aresimultaneousslyatisfiedonlyfor o a(\\ :8p6 E o APri\",r=AP,:\",2=0 (8.42) oo .g o6 and A f i = A f z = 0 q =a c) G' Thus, under steadycondition changein the tie line power and frequency of <.1 -(d 6 each area is zero. This has been achieved by integration of ACEs in the feedbackloops of eacharea. *li E9 o o Dynamic responseis difficult to obtain by the transferfunction approach(as v'ito l . a gtu Eo used in the single areacase)becauseof the complexity of blocks-and multi- Qe .c> l input (APop APor) andmulti-output(APri\",1,Ap6\",2, Afr Afr) situation.A lr (g() E more organizedand moreconvenientlycarried out analysisis through the state spaceapproach(a tirnedomainapproach).Formulationof the statespacemodel tt) oo ((D) for the two-area systemwill be illustrated in Sec. 8.5. .9 FA (U The resultsof the two-areasystem(APri\", changein tie line power and,Af, C.nL changein frequency)obtainedthroughdigital computerstudyare shown in the d OE o form of a dotted line in Figs. 8.18 and 8.19.The two areasare assumedto be o identical with systemparametersgiven by uIf|:- *5 o* o b6 r\\ Trs= 0.4 sec,7r = 0.5 sec, ?r, = 20 sec tr(') ai Eg cit lr H'.s t *3 -ld =5 uo 'a8' 95 l+ oo. 5l E* , On - oo oo @ d <; l! K o r =1 0 0 ,R = 3 , b = 0 . 4 2 5 ,& = 0 . 0 9 , 2 f l r 2 = 0 . 0 5 8.5 OPTTMAL (TWO-AREA) LOAD FREOUENCY CONTROL Modern control theory is applied in this section to design an optimal load frequencycontroller for a two-a3easystem.In accordancewith modern control terminology APcr arrdAP62 will be referred to as control inputs q and u2.ln the conventionalapproachul anduzwere providedby the integral of ACEs. In

AutomaticGenerationand Voltagecontrot fM*& itZ I rrrrodePr-nowerSystemAnalysis ComparingFigs.8.16 and 8.17, (8.43) (8.445 moderncontrol theory approachur and u2 wtll be createdby a linear xt = Aft xq = Af. XS= JACEit combinatioonf all thesystemstates(full statefeedbackF).orformt'latingthe (8.45) statevariablernodelfor this purposethe conventionafleedbackloops are (8.46) block as shownin .r2- AP,;1 x5 = AP52 t, = JACE, dt (8.47) resentedbv a se tt1= APg, u2 = /)Pa (8.48) Fig. 8.17.Statevariablesaredefinedas the outputsof all blocks baving either w1= AP\" w2 = APp, an integratoror artirne,constanf.W. e immediately noticethat the systernhasnine state variables. For block 1 -1-+-r+--i, -f-.'f-- x1 + T.rri, = K^t(xz- h - w) LP + \\ . 1 *,f Kpr tx, -z Kp rt - , - ; - Kwpt r t I Optimalcase (fullstatefeedback) xt I hl - 4l - ; - ' With integralcontrolaction I 'psl t psl t ptl t ptl -l For block 2 o x.2+ Tiliz= xt o o r * z =- + - r * * n IX t( t'-2 -3 t- 1,I F i g . 8 .1' 8 changein tie linepowerdueto stepload(0.01pu)changein area1 For block 3 -21.--+-_'--';-;;7-1=a.-1-1=--1 tr+ { , s r i ': = -Lrr+r, r R,r 12 14 16 18 20 A /r' ;+-8.1' - I o r * t=- ^h r,-t* ,* * ,, --' For block 4 L/ Time(sec)----- I N with integralcontrolaction oI x IL X +n * Torz*+= Krrz(xs + ar2x7 - wz) Optimalcase (fullstate feedback) or iq= I Knrz at?K or2 Ko*2 Tpsz z ' \\ A-' 1 - - . {< '' - T -- - y ' - a - _ - W ^ ' Tps2 Tprz Tps2 For block 5 Fig. 8.19 Changein frequencyof area 1 due to step load (0.01 pu) x s t 7,2i5 - x6 change in a.rea1 o r i s = l 1rY I - V 4 lu Before presenting the optimal design, we must formulate the state model. 4<t This is achieved below by writing the differential equations ciescribing each Ttz T,z individual block of Fig. 8.17 in terms of state variables (note that differential For block 6 equations are written by replacing s U'V *d1.t ' x s * I ,sz.x6- -; l x4 + u2 I\\2 or io=-#*o-**u '2t sg2 t sg2

'3i4',\"1 ModernPower SystemAnatysis T '-- co\",\"\".t\"\" c\" onstructedas under from the statevariables x, and -rnonly. For block 7 ut=- Kirxs=- Kir IeCn,Ar it=2iTtzxt-2iTr2xa (8.4e) For block 8 is= brx, + x.i (8.5O-) For block 9 uz=- Ki{s=- Kiz la.Cerar i9= b2xa- anxt ( 8 . s1 ) ln the optimal control schemethe control inputs u, anduz aregeneratedby meansof feedbacksfrom all the nine stateswith feedbackconstantsto be The nine equations(8.43)to (8.51)canbe organizedin the following vector determinedin accordancewith an optimality criterion. matrix form *=Ax+Bu+Fw (8.s2) Examinationof Eq. (8.52)reveralsthat our model is not in the standardform where employedin optimal control theory. The standardform is x _ l x r x2 ... xg)r= statevector i=Ax+Bu u = f u t u2fT= control vector which does not contain the disturbance term Fw present in Eq. (g.52). Furthermore,a constantdisturbancevector p would drive someof the system w = l w t w2fT= clisturbancevector statesand the control vectorz to constantsteadyvalues;while the cost function while the matricesA, B and F are defined below: employedin optimal control requiresthat the systemstateand control vectors have zero steady statevaluesfor the cost function to havea minimum. I2 3 4 5 7 89 I 'YtPsl 0 0 For a constant disturbancevector w, the steadystateis reachedwhen Tpst, Tprt 0 -bLoo o o *=0 0 -1 1 o o Tprt in Eq.(8.52);whichthengives Tt Ttr 0 00 0 = A . r r \"+ B u r r + F w (8.s3) 1o -1 0 00 Definingx andz asthesumof transienat ndsteadystateterms,we canwrite x = x', * Ir\" (8.54) Rr4er Trst atzKprz 00 A_ 0 00-1Kp'z n = ut * z', (8.55) Tprz Tprz Tprz 6 0 Substitutingr and z from Eqs. (8.54) and (8.55)in Eq. (8.52),we have 7 o o i o --1- 1 0 00 8 0 Ttz 7,, i' = A (r/ + x\"r) + B(at + usr)+ Fw 9 2 irrz 0 00 By virtue of relationship(8.53),we get oo-10 I bL 0 00 RzTrsz TreZ 1 00 *' = Axt + But (g.56) 0 -atz 0 0 0 0 -2ilr2 0 0 This representssystemmodel in terms of excursion of state and conhol vectorsfiom their respectivesteadystate values. 0 0 0 00 00b200 For full state feedback,the control vector z is constructedby a linear combinationof all states.i.e. [oo I oo o I -Tss1 00.l u=- Kx (8.57a) Br = | oo oo+ I where K is the feedback matrix. Now l o 0 0 Il IL a c' Oo -) I ttt+ Itrr=- l( (r/+ rr\") J Kprt For a stable systemboth r/ andut go to zero, therefore -0 ur, = _ Kx* ,;T Tprt Hence tt /= - Ikl (8.s7b)

M o d e r n P o w e r S y s t e mA n a l y s i s AutomaticGeneration Examinationof Fig. 8.17 easily revealsthe steadystatevaluesof stateand b ? o0 0 0 0 4 0 0 control variablesfor constantvaluesof disturbanceinputs w, andwr.Theseare 0 00 0 0 0 0 0 0 Ilrr=X4\"r= /7r\" = 0 0 00 0 0 0 0 0 0 ulr, = wl -arzbz 0 0 0 00 0 00 0 00 r5rr= x6rr= lv2 (8.s8) 0 00 0 00 0 00 uzr, = wz 4 0 0 - a n b z 0 0 Q+a?)o o Igr, = COnstant 0 00 0 00 0 1 0 0 00 0 00 0 01 I9r, = Constant = symmetric matrix The values of xr* and xe* depend upon the feedback constants and can be determined from the following steady state equations: R - kI = symmetricmatrix utrr= kttxtr, + ... + ftt8r8\",* kt*sr, = wl r,t2ss= k2txlr, + ... + kzgxgr.*, kz*gr, = wz (8.se) The feedbackrnatrix K in Eq. (8.57b) is to be determinedso that a certain K = R-rBrS (8.63) performance index (PI) is minimized in transferring the system from an arbitrary initial state x' (0) to origin in infinitie tirne (i.e. x' (-) = 0). A The acceptablesolution of K is that for which the systemremainsstable. convenientPI hasthe quadraticform SubstitutingEq. (8.57b)in Eq. (8.56), the systemdynamicswith foedbackis definedbv ' Pr= ;ll '.'' Qx+' u'r Ru'dt (8.60) The manices Q arrdR are defined for the problem in hand through the i' = (A - BIgx, (g.64) following designconsiclerations: Fol stability all thc cigenvaluesof the matrix (A - Bn shouldhavenegative (i) Excursionsof ACEs about the steadyvalues (r,t + brx\\; - arrxt,+ bzx,q) real parts. are minimized.The steadyvalues of ACEs are of coursezero. For illustration we considertwo identicalcontrol areaswith the following (ii) Excursionsof JnCg dr about the steadyvalues (xts,xte)are nrinimized. syste|llparameters: The steaclyvaluesof JeCg dt are,of course,constants. 4r* = 0'4 scc; T'r= 0.5 sec; 7'r* = 20 sec (iii) Excursionso1'the contt'olvector (ut1,ut2) about the steadyvalue are /l = 3: (n* = l/lJ = 100 rninirnizedT. he steadyvalueof thecontrolvectoris, of course,a constant. b = O . 425;Ki = 0. 09;up = I ; 2iln = 0. 05 ' This nrinimizationis intendedto indirectlylimit the controleffbrtwithin the physical capability of components.For example, the steamvalve catmot be openedmore than a certainvalue without causingthe boiler presisurteo drop severely. With the abovereasoning,we can write the PI as pr= *2 JftUt -' + + h,.r,,)2+ (- tt,2xt+, brxta)+z (.r,?+ ,,]) ^f , = [ 0.52tt6 l.l4l9 0.68l3 - 0.0046-0.021| -0.0100-0.7437 0.gggg0.00001 L-o.tl046-0.o2tl-0.0100 0.5286 t.t4rg 0.6813 0.74370.0000.gggsl + kfu'l+ u,|1a1t (8.61) From the PI of Eq. (8.51), Q md R can be recognizedas '*Refer Nagrath and Gopal [5].

iiii'f:l Modernpowerrystemin4gs As the control areasextend over vast geographicalregions, there are two c= vR.f- vr '_- ways of obtainingfull stateinformation in eachareafor control purposes. The error initiates the corrective action of adjusting the alternator excitation. (i) Transport the state information of the distant area over communication Error wave form is suppressedcarrier modulated, tt\" carrier frequency being channels.This is, of course,expensive. the systemfrequency of 50 Hz. Change in voltage Load change caused by load tG 1+Iers skrt 8.6 AUTOMATIC VOLTAGE CONTROL Fig. 8.21 Brockdiagramof arternatovrortagereguratosrcheme Figure 8.20 gives the schematic diagram of an automatic voltage regulator of a generator.It basicallyconsistsof a main exciter which excites the alternator faaEsummcrnrRpopclrtliiifpfoiiaeconmrawotpteifiomlrintfhieeaetrmcos:oepnttlhfswIftteioaednrseitsimatgonnobaddelueflsaoxmtrceaitsclealorennfnditeroaorumdllgi:nphgliiofIitetbhsepl trnhoeeevxgiecdleiretrceostrretdsnhi,\"egtrhnonea.-\"lo..a\"vIrrtesrri*lra;gy\"iagl ifnapnisoswfK,ererrr\".r field to control the outputvoltage.The exciterfield is automaticallycontrolled through error e = vr\"r - vr, suitably amplified through voltage and power K, amplifiers. It is a type-0 systemwhich requiresa constant error e for aspecified l* T\"rs voltage at generatorterminals.The block diagram of the systemis given in where T\"yis the exciter field time constant. L Alternator; Its field is excited by the main exciter voltage vu. Under no road o it producesa voltageproportional to field current.The no load transferfunction A is D Ks Potential 7*T*s Fig. 8.20 Schematicdiagramof alternatovr oltageregulatosr cheme Fig. 8.21.The function of important componentsand their transferfunctionsis where givenbelow: Potential transformer: It gives a sample of terminal voltage v.. T*= generatorfield time constant. Dffirencing device; It gives the actuating error The load causesa voltage drop which is a complex function of direct and quadratureaxis currents.The effect is only schematicallyreBresentedhv hlock G.. The exact load model of the alternatoris beyond ,t\" ,iop\" ;rhtJ;;: dstahtecaerobisnviytatirsztoiitnvel egsmyf'tsesrtadeenydmsnbfcaaoamcrnmkibceinerre:imstThpp4iosr*ondsvsyeesd.I-ttlqjebsmyatwrhiseeeiliplanrkrtogenevroniewdanenloddtuehbgrayihvt attmhitmieveeedayncfenoseanodmsfbtaiaa.ncrt\"kssrlttopaooobpiomiy.rz\"pTianhoiegrf transformer excited by the exciter output voltage vE. The output of the

I ModernPowerSystemAnalysis AutomatiGc eneratioanndVoltageC---o---n_-t-r--o-ll Jffif 320' l E I The bandedvaluesimposedhy the limitersareselectedto resffictthegeneration stabiliz,intgransformeirs fcclncgativclyat the inputterminalsof thc SCR power amplifier. The transferfunction of the stabilizingtransfo\"meris derivedbelow. rate by l}Vo per minute. Since the secondaryis connectedat the input ternfnals of an amplifier, it can I be assumedto draw zero current. Now dt I.g 9t\", vr = Rr i., + Lr' JdilLt -t*9r\"'--l 'rr= MY _u+' t( + / A dt l- Taking the Laplacetransform,we get %,(s) _ sM sMlRt VuG) R, * s,Lt l * I r s sK\", Fig.8.22 Governomr odelwith GRC 1 +{ , s The GRCsresultin larger deviationsin ACEs as the rateat which generation can cha-ngein the area is constrainedby the limits imposed. Therefore, the Accurate staterrariablemodels of loadedalternatoraround an operatingpoint duration for which the power needsto be imported increasesconsiderably as are available in literatureusing which optimal voltage regulation schemescan cornparedto the casewhere generationrate is not constrained.With GRCs, R be devised.This is, of course,beyondthe scopeof this book. shouldbe selectedwith care so as to give the best dynamic responseI.n hydro- thennal system,the generationrate in the hydro areanorrnallyremainsbelow 8.7 LOAD FREOUENCY CONTROL WITH GENERATION the safe limit and therefore GRCs for all the hydro plants can be.ignored. RATE CONSTRAINTS (GRCs) 8.8 SPEED GOVERNOR DEAD-BAND AND ITS EFFECT The l<-rafcrlcquencycontrolproblcmdiscussedsofar doesnot consicletrhe effect ON AGC of the restrictionson the rate of changeof power generation.In power systems havingsteamplantsp, owergenerationcanchangeonly at a specifiedmaximum The eff'ectof the speedgovernor dead-bandis that for a given position of the governor control valves, an increase/decreasien speedcan occur before the rate. The generationrate (fiom saf'etyconsiderationso1 the equipment)for positionof thevalvechangesT. he governordead-bandcanmateriallyaffect the reheatunits is quit low. Most of the reheatunits have a generatiol rate around system response.ln AGC studies, the dead-band eff'ect indeed can be 3%olmin.Some have a generationrate between 5 to 7jo/o/min.If these significant,sincerelativcly small signalsare under considerations. constraintsarc not consirlcrcd,systerttis likely to c:ha.slcargc tttottrclttrry T l L es p e e dg o v e r n o rc h a r a c t e r r i s tti hc .o u g hn o n - l i r r e a rh,a sb e e na p p r o x i n r a a e d disturbancesT, hrs resultsin undue wear and tear of the controller. Several by linear characteristicsin earlier analysis. Further, there is another non- methoclshavebeenproposectlo considerthe effectof GRCs for the clesignof iinearity introducedby the dead-bandin the governoroperation.Mechanical f'riction and backlashand also valve overlapsin hydraulic relayscause the automaticgenerationcontrollers.WhenGRC is consideredt,hesystelndynamic governordead-bandD. ur to this, thoughthe input signalincreasest,he speed rnodelbecomesnon-linearand linearcontroltechniquescannotbe appliedfor governor may not irnmediately reactuntil the input reachesa particular value. Similar a.ctiontakesplace when the input signaldecreasesT.hus the governor the optimizationof the controllersetting. dead-bandis defined as the total rnagnitudeof sustainedspeedchangewithin If the generationratesdenotedby P\", areincludedin the statevec:tor,the which thereis no changein valve position.The limiting value of dead-bandis specifiedas 0.06Vo.It was shown by Concordia et. al [18] that one of the systermorder will be altered.Insteadof augntentingthem, while solving the effects of governor dead-bandis to increasethe apparentsteady-statespeed stare equations,it may be verified at each step if the GRCs are viclated. regulationR. Another way of consicieringGRCs for both areas is to arjri iinriiers io ihe governors[15, 17] as shown in Fig. 8.22, r.e.,the maximum rate of valve openingor closingspeedis restrictedby the limiters.Here 2\", tr,r,,iSthe power rate limit irnposedby valve or gate control. In this model l A Y E l . - -g u , n r (8.6s)

lFFf ModrrnPo*., svrt.t Analuri, Discrete-Time Control Model The effect of the dead-bandmay be included in the speedgovernor control loop block diagramas shownin Fig. 8.23.Consideringthe worst caseforthe dead-band,(i.e., the system starts responding after the whole dead-bandis traverseda) ndexaminingthe dead-bandblock in Fig. 8.23,the following setof ly define the behaviourolthe dead.band [9]- The continuous-timedynamic systemis describedby a setof linear differential equations x=Ax+Bu+ fp (8.67) where f u, P are state, conhol and disturbancevectors respectively and A,B and f are constantmatrices associatedwith the above vectors. The discrete-timebehaviourof thecontinuous-timesystemis modelled by the systemof first order linear differenceequations: Speed governor x(k+1)=Qx(k)+Vu(k)+jp&) (8.68) Dead-band where x(k), u(k) andp(k) are the state,control and disturbancevectors and are Flg.8.23 Dead-banidn speed-governcoor ntroloop specifiedat t= kr, ft = 0, 1,2,... etc.and ris the samplingperiod. 6, tl,nd 7 Te the state, control and disturbance transition matrices and they are u(r+1)= 7(r)1: _ x, 1 dead-band evaluatedusing the following relations. \"(r+1) d= eAT - _ dead-band; if x('+l) - ,(r) I g (8.66) \"(r+l) {=({r_ln-tr - tf Xr*l _ xt < 0 \"(r+1). (r is the step in the computation) j=(eAr-DA-tf Reference[20] considersthe effect of governor dead-bandnonlinearity by using where A, B andI, are the constantmatrices associatedwith r, ,,LO p vectors the describingfunction approach[11] and including the linearisedequationsin in the conespondingcontinuous-timedynamic system.The matrix y'r can be the statespacemodel. evaluatedusing various well-documentedapproacheslike Sylvestor's expansion theorem, seriesexpansiontechniqueetc. The optimal digital load frequency Thepresenceof governordead-bandmakesthe dynamicresponseoscillatory. controllerdesignproblem is discussedin detail in Ref [7]. It has been seen [9J that the governor dead-banddoes not intluence the 8.10 DECENTRALIZED CONTROL selectionof integral controller gain settingsin the presenceof GRCs. In the presenceof GRC and deadband even for small load perturbation,the system In view of the large size of a modernpower system,it is virtually impossible becomeshighly non-linearand hencethe optimizationproblembecomesrather to implementeither the classicalor the modern LFC algorithmin a centralized manner.ln Fig. 8.24,a decentralizedcontrol schemeis shown.x, is usedto find complex. out the vector u, while x, aloneis employedto find out u\". Thus. 8.9 DIGITAL LF CONTROLLERS In recentyears,increasinglymore attentionis being paid to the questionof digital implementationof the automatic generationcontrol algorithrns.This is mainly due to the facts that digital control turns out to be more accurate and rcliqhlc nnrnnaef in qize less censifive to nnise end drift nnd more flexihle Tt r v^rEv^vt may also be implementedin a time sharedfashion by using the computer systemsin load despatchcentre,if so desired.The ACE, a signal which is used for AGC is availablein the discreteform, i.e., there occurssampling operation ; betweenthe systemand the controller. Unlike the continuous-timesystem,the controlvector in the discretemode is constrainedto remain constantbetween Flg. 8.24 Decentralizecdontrol

i,i2[,:*.1 ModernPowerSystemAnalysis - AutomatiGc eneratioanndvoltageControl ffi 4 x - (x1 x2)' ut=-ktxt u2-- kzxz LI n , n , , t f^4 6 1 d v- aF(s)' 1' af (t)dr: liq, * '4F(s:) hm/F(\")] JO ,t JO s-0 S s+0 been shown possible using the modal control principle. Decentralized or 8.4 For the two areaload frequencycontrol of Fie. 8.16 assumethat inte hierarchicalimplementationof the optimal LFC algorithmsseemsto have been controllerblocks arereplacedby gainblocks, i.e. ACEI and ACE are fed studiedmore widely for the stochasticcasesince the real load disturbancesare to the respective speedchangersthrough gains - K, and- Ko. Derive an expressionfor the steadyvaluesof changein frequency and tie line power truely stochastic.A simple approachis discussedin Ref. [7]. for simultaneouslyapplied unit stepload disturbanceinputs in the two It may by noted that other techniquesof model simplification are available areas. in the literatureon alternativetools to decentralizedcontrol.Theseinclude the 8.5 For the two areaload frequencycontrol employing integral of areacontrol method of \"aggregation\", \"singular perturbation\", \"moment matching\" and error in each area (Fig. 8.16),obtain an expressionfor AP6\"$) for unit other techniques[9] for finding lower order models of a given large scale step disturbance in one of the areas.Assume both areasto be identical. system. Comment upon the stability of the system for parameter values given below: PROIBEI/IS 4e = 0'4 sec; Z, = 0'5 sec; Zp. = 20 sec 8 . 1 Two generatorsrated 200 MW and 400 MW are operatingin parallel. K p r = 1 0 0 ;R = 3 ; K i = l ; b = 0.425 The droop characteristicsof their governors are47o and5Vorespectively from no load to full load.The speedchangersareso setthatthe generators ar2= I;2tTr, = 0.05 operateat 50 Hz sharingthe full load of 600 MW in the ratio of their ratings.If the loadreducesto 400 MW, how will it be sharedamong the lHint: Apply Routh's stability criterion to the characteristicequation of generatorsand what will the s)/stemfrequencybe?Assumefree governor the system.l operatlon. REFERNECES The speedchangersof the governorsare resetso that the load of 400 MW Books is sharedamongthe generatorsat 50 Hz in the ratio of their ratings. What are the no load frequencieosf the generators? l. Elgcrd, O.1., Elccu'ic Energv.Sv,s/clrT'lrcorv: An ltttnxlut'lion. 2nd cdn. McCraw- 8 . 2 Considerthe block diagrammodelof lcad frequencycontrol given in Fig. Hill, New York, 1982. 8.6.Make the following approximatron. (1 + Z.rs) (1 + Z,s) =- t + (7rg+ T,),s= 1 + Z\"c.r 2. Weedy, B.M. and B.J. Cory Electric Pow'er Systems,4th edn, Wiley, New York, Solve for Af (l) with parametersgiveu below. Given AP, - 0.01 pu I998. T\"q= 0.4 + 0.5 = 0.9 sec; 70, = 20 sec a Cohn, N., Control of Generation and Power Flou, on Interconnected Systents, 1 Wiley, New York, i971. 4. Wood, A.J., and B.F. Woolenberg, Power Generation, Operation and Control,2nd K r r K , = 1 ;K p r = 1 0 0R; = 3 edn Wiley, New York, 1996. Coinparewith the exactresponsegiven in Fig. 8.9. 5 . Nagarth, I.J. and M. Gopal, Control SystemsEngineering, 3rd edn. New Delhi, 8 . 3For the load frequencycontrol with proportionalplus integral controller 2 0 0l . oc olrn'rn.i- Tiic e 1n nhfain en AsnrAccinn fnr tha cfenrlrr cfrfp errnr in 6 . Handschin, E. (Ed.), Real Time Control of Electric Power Systems,Elsevier, New clJ orlvYYll ll( L L6. v. rvt vuLarrr York 1972. cycles,i.\". ft '^4\"1t)d r; for a urrit stepAPr. Whatis thecorrespondingtime 7 . Mahalanabis, A.K., D.P. Kothari and S.I Ahson, Computer Aided Power Systent ,1, lirnl*m Analysis and Control, Tata McGraw-Hill, New Delhi, 1988. error in seconds(with respect o 50 Hz).lCommenton the dependenceof 8 . Kirclrrnayer,L.K., Economic Control of lnterconnected Systems,Wiley, New York, error in cycles upon the integral controller gain K,. t959. 9 . Jamshidi, M., Inrge Scale System.s:Modelling and Control, North Holland, N.Y., 1983.

10. Singh, M.G. and A. Titli, SystemsDecomposition, Optimization and Control 9.1 INTRODUCTION PergamonPress,Oxford, 197g. So far we have dealt with the steadystatebehaviourof power system under I I' Siljak,D'D., Non-LinearSystems:The Parametcr Analysisantl Design, Wiley, normal operating conditions and its dynamic behaviour under small scale N.Y. 1969. perturbations.This chapter is devotedto abnormal system behaviour under conditions of symmetricalshort circuit (symmetricalthree-phase.fault*).Such .t,apers conditions are causedin the system accidentally through insulation failure of equipment or flashover of lines initiated by a lightning stroke or through 12. Elgerd, o.I. and c.E..Fosha,\"The Megawatt Frequencycontrol problem: A New accidentalfaulty operation.The systemmust be protectedagainstflow of heavy Approachvia optimal control Theory\", IEEE Trans.,April 1970,No. 4, pAS g9: shortcircuit currents(which can causepeffnanentdamageto major equipment) 556. by disconnectingthe faulty part of the systemby means of circuit breakers operated by protective relaying. For proper choice of circuit breakers and 13' Bhatti, T'S., C.S Indulkar and D.P. Kothari, \"Parameteroptimization of power protective relaying, we must estimatethe magnitudeof currents that would flow Systemsfor StochasticLoad Demands\" Proc. IFAC. Bangalore,December 19g6. under short circuit conditions-this is the scopeof fault analysis (study). l4' Kothari,M'L., P.S.Satsangai nd J. Nanda,\"sampled-DataAutomaticGeneration The majority of systemfaults arenot three-phasefaults but faults involving Control of InterconnectedReheatThermal SystemsConsideringGenerationRate oneline to groundor occasionallytwo linesto ground.Theseareunsymmetrical Constraints\"I,EEE Trans.,May 19g1,pAS_100;2334. faults requiring specialtools like symmetricalcomponentsand form the subject of study of the next two chapters.Though the symmetrical faults are rare, the 15' Nanda, J', M.L. Kothari and P.S. Satsangi,\"Automatic GenerationControl of an symmetrical fault analysis must be carried out, as this type of fault generally InterconnectedHydro-thermalsystem in continuous and DiscreteModes consid- leads to most severe fault current flow against which the system must be protected.Symmetrical fault analysisis, of course,simpler to carry out. 16' IeErEinEgGcoemnmeriattteioenRRaepteocrot,n'DsytrnaainmtiscI'MEEodperloscfo.,rpSrDte,amNaon. dl,HJyadnruoa-truyr1b9inge3s,i1n3p0o:w1e7r. system studies\" IEEE Trans.,Nov/Dec. rg73, pAS-92, 1904. A power network comprisessynchronousgenerators,ffansfonners, lines and loads. Though the operating conditions at the time of fault are important, the l7' Hiyama, T', \"Optimization of Discrete-typeLoad FrequencyRegulatorsConsider- loads can be neglectedduring fault, as voltagesdip very low so that currents ing Generation-Rateconstraints\"proc.lE4 Nov. g2, r2g, pt c, 2g5. drawn by loads can be neglectedin comparisonto fault currents. I8. concordia,c., L.K. KirchmayerandE.A. Szyonanski.,.Effectof speedGovernor *Symmetrical fault may be a solid three-phase short circuit or may involve are Dead-bandon Tie Line Power and FrequencyControl performance,,AIEE Trans. impedance. A u g . 1 9 5 7 ,7 6 , 4 2 9 . 19' Nanda,J', M.L. Kothari and P.S.Satsangi,\"Automatic Control of ReheatThermal SystemConsideringGenerationRateConstraintand CovernorDead-band,,.J.I.E. (India),June 1983, 63,245. 20. Tripathy,s.9., G.s. Hope and o.p. Marik, ,,optimisatiororf Load-frcqucncy C<lntrolParametersfor Power systemswith ReheatSteamTurbinesand Governor Dead-bandNonlinearity\", proc. IEE, Januaryrgg2, rzg, pt c, No. r, r0. 21. Kothari,M.L., J. NandaD, .p. KothariandD. Das,.,Discrete-moAdGe C of a.two_ area Reheat Thermal system with New Area control Error,,, IEEE Trans. on Power System,Vol. 4, May 19g9,730 22' Daq D. J. Nanda, M.L. Kothari and D.p. Kothari, ,.AGC of a Hydro_Thermal systemwith New ACE consideringGRC\", Int. J. EMps,1g, No. 5, rggo, 46r. 23' Das, D', M.L' Kothari, D.P. Kothari and J. Nanda, \"Variable StructureControl strategy to AGC of an IntcrconncctcdRcheatThermal systcm,,, prctc. IEE, r3g, p t D , 1 9 9 1 ,5 7 9 . 24. Jalleli, Van Slycik et. al.. \"lJndersfandingAutonnaticGenerationControl,,, IEEE Trans.on P.S., Vol 07, 3 Aug. 92, 1106_1122. 2 5 . Kothari, M.L., J. Nanda,D.p. Kothari and D. Das, ,,DiscreteMode AGC of a two Area ReheatThermal Systemwith a NACE consideringGRC,,,J.LE. (rndia), vol. 72, Feb. 1992,pp Zg7-303. 2 6 . Bakken, B.H. and e.s. Grande,\"AGC in a Deregulatedpower system,,, IEEE Trans.on Power Systems,13,4, Nov. 199g,pp. 1401_1406.

325 | ModernPowerSystemAnalysis tffiffi t I The synchronousgeneratorduring short circuit has a characteristictime- = -4-*2siVn lzl varying behaviour.In the eventof a shortcircuit, the flux per pole undergoes (cr,r+f a_ A dynamic changewith associatedtransientsin damperand field windings.The reactanceof the circuit model of the machine changesin the first few cycles z = (Rz+ Jr\\tt\"(t: tan-l +) from a low subtransientreaetanecto a higher transient value, finally settling at a s'iitt higher synchronous (steady state) value. Depending upon the arc ir = transientcurrent [it is such that t(0) = t(0) + L(0) = 0 being an inductivecircuit; it decayscorrespondingiothe tim6 constantiRl. intemrption time of circuit breakers,a suitable reactancevalue is usedfor the circuit model of synchronousgeneratorsfor short circuit analysis. = - i,(6)e-$tL)t In a modern large interconnectedpower system, heavy currents flowing during a fault must be interruptedmuch before the steadystateconditionsare establishedF. urthermore,from the considerationsof mechanicalforcesthat act = 9tYz l s i n( d - a ) g - . ( R t D t on circuit breakercomponents,the maximum current that a breakerhasto carry Thus short circuit currentis given by momentarilymust also be determined.For selectinga circuit breakerwe must, therefore, determine the initial current that flows on occulTenceof a short circuit and also the current in the transientthat flows at the time of circuit (e.1) intemrption. S y n r n r e t r i c asl h o r t DC otT- setcurnent circuit current 9.2 TRANSIENT ON A TRANSMISSION LINE A plot of i* i, and'i = i, + i, is shown in Fig. 9.2.rnpower systemterrninology, Let us consider the short circuit transient on a transmission line. Certain the sinusoidalsteadystatecurrent is called the symmetricalshort circuit current and the unidirectional transient componentis called the DC off-set simplifying assumptionsare madeat this stage. current,which causesthetotal short circurict urrentto be unsymmetricaltill the transiendt ecays. (i) The line is led I'rorna constanvt oltagcsoLrrcc(tltecasewhcn the line is fed from a realisticsynchrononms a.chrnwe ill tre treatedin Sec.9.3). It easily follows fiom F'ig.9.2 that the maximum momenro) short circuit currcnt i,,,,,c,orrespondtso the firstpenkI.f theclecaoyf trnnsiencturrenitn this (ii) Short circuit takesplacewhen the line is unloaded(the caseof short shorttime is neglected, circuit on a loadedline will be treatedlater in this chapter). (iii) Line capacitanceis negligibleandthe line canbe representedby a lumped - J r v ' s i n( d - E' lzl tzl RZ seriescircuit. L c)* (e.2) v = JI vsin (o,t+ *) , ,F. Sincetransmissionline resistanceis small. 0 - 9C,. . lr+V\\'\\-' I.m*= Jiv cosa+ JTv rV) rzr rzl I (e.3) i_ F i g .9 . 1 This has the maximum possiblevalue for o. = 0, i.e. short circuit occurring when the voltagewave is going throughzero.Thus With the aboveassumptiontsheline canbe representebcyl thecircuitrnoclel i,n,nlrnu*possib=le)'# e.4) of Fig. 9.1. The short circuit is assumedto take placeat t = 0. The parameter <rcontrolstheinstanton thevoltagewavewhenshortcircuit occLrrsI.t is known = twice the maxirnum of symmetricalshort circuit current from circuit theorythat the currentaftershortcircuit is composedof two parts, (doubling effect) 1.tr. For the selectionof circuit breakers.momentaryshortcircuit currentis taken correspondingto its maxirnumpossiblevalue(a sat'echoice). t-- I\"+ I.t whcre i, = steadystatecurrent .w

ffiffif ModernPoweSr ystemAnalysis b,a#& The nevf ntrecfinn ic turhqf ic fhc r.rrrrcnf fn hc inferrrrnfprl?t Aa hqo haan reactancewhencombined with the leakagereactanceXi of the machine is called el/lvs synchronousreactanceX4 (direct axis synchronousreactancein the case of salientpole machines).Armature resistancebeing small canbe neglected.The pointed out earlier, modern day circuit breakers are designed to intemrpt the nels snownln rrg. cunent in the first few cycles (five cycles or less).With referenceto Fig, 9.2 on per phasebasis. it meansthat when the currentis intemrpted,the DC off-set (i,) hasnot yet died the value of the DC off-set at the time of intemrption (this would be highly complex in a network of even moderatelylarge size), the symmetrical short circuit current aloneis calculated.This figure is then increasedby an empirical rnultiplying factor to accountfor the DC off-set current. Details are given in S ec .9. 5 . ( a ) S t e a d ys t a t es h o r tc i r c u i tm o d e l (b)Approximatecircuitmodelduring of a synchronoums achine subtransienpteriodof shortcircuit X1 Fig.9.2 Waveforomf a shorct ircuict urrenotn a transmissiloine (c)Approximacteircuimt odedl uring transienpteriodofshorct ircuit 9.3 SHORT CTRCUTTOF A SYNCHRONOUS MACHTNE (ON NO LOAD) Fig.9.3 Under steady state short circuit conditions,the armaturereaction of a Consider now the sudden short circuit (three-phase)of a synchronous synchronougseneratoprroducesa demagnetizinfglux. In termsof a circuitthis generator initially operating under open circuit conditions. The machine undergoesa transientin all the three phasefinally ending up in steady state conditions describedabove. The circuit breakermust, of course,intemrpt the current much before steady conditions are reached. Immediately upon short circuit, the DC off-set currents appearin all the three phases,each with a different magnitude since the point on the voltage wave at which short circuit occurs is different for each phase.TheseDC off-set currents are accountedfor separately on an empirical basis and, therefore, for short circuit studies, we need to concentrateour attention on syimmetrical (sinusoidal) short circuit current only.Immediately in the eventof a shortcircuit, the symmetrical.short circuit current is limited only by the leakagereaitance of the machine.Sincethe air gap flux cannotchangeinstantaneously(theoremof constantflux linkages), to counter the demagnetization of the armature short circuit current, currents appearin the field winding as well as in the damper winding in a direction to help the main flux. These currents decayin accordancewith the winding time constants.The time constant of the damper winding which has low leakage inductanceis much less than that of the field winding, which hashigh leakage

MocjernPowerSysiemnnaiysis symmerricaFlaultAnalysis I'lt,5ffit inductanceT. hus during the initial part of the shortcircuit, the damperand field I windingshavetransfurnrecr urrentsinducedin themso thatin thecircr,ritmodel thcir reactances--X,of field winding and Xa* of damper winding-appear in I parallelxwith Xo as shorvnin Fig. 9.3b. As the danqpqlruadlag cullqqls 4!q first to die out, Xr* effectivelybecomesopen circuited and at a later stageX1 0) becomesopencircuited.The rnachinereactancethuschaugesfrom the parallel combinationof Xo, Xy and Xu. during the initial period of the short circuit to Steadystatecurrent amPlitude X,,and Xrinparallel (Fig.9.3c) in the rniddleper:iodof the shortcircuit, and finally to X,,in steadystate(Fig. 9.3a).The reactancepresentedby the machine in the initial period of the shortcircuit, i.e. -r- 1_- : (e.5) Tlme ( XJ+ X.L 1 1 x , , + U l l x d , , ) X\"'j is called the subtrunsientreoctutxc:<e>fthe nrachine.While the reactance (b)Envelopoef synchronomusachinseymmetricsahl orct ircuicturrent Fig.9. 4 effective after the darnperwinding currentshave died out, i.e. X',t= X, + (X,,ll X,) (e.6) If we examinethe oscillograrnof the shortcircuit currentof a synchronous machineafterthe DC ott-set cuitentshavebeenrettrovedtrom it, we will tind is called the transientreactanceof the machirre.Of course,the leactanceunder the currentwave shapeas given in Fig. 9.4a.The envelopeof the current wave steaclyconditionsis the synchronousreactanceof the machine.Obviousiy Xf7< shapeis plottedin Fig. 9.4b,The shortcircuit currentcanbe divided irttothree X'd< Xu.The machinethusoffersa time-varyingreactancewhich changesfront periods-initial subtransientperiod when the current is large as tire tnachine Xttoto Xtaandfinally to Xn. offers subtransientreactance,the middle transient period where the machine offers transientreactance,and finally the steadystateperiod w\\n the machine I ofters synchronousreactance. : Subtransienoteriod b If the transientenvelope is extrapolatedbackwardsin tinre, the difference betwecnthe tlansicrrtanclsubtransierettnvelopesis the cunent Ai/' (corre- l Steady state period sponding to the clamperwinding current) which decaysfast according to the I clamperwinding time constant.Similarly, the differenceAi/ betweenthe steady I state1ndtransicnet nvelopesdecaysin accordancwe ith thefield time constant. I I Ea q) In termsof the oscillogram,thecunentsandreactancesdiscussedabove,we ()f, can wrlte 'g o 0 lEsl Y, t Time lIl = oa (9.7a) t; a \\t z. o o lll = ob _ l E 8 l (e.7b) t; E i xtd E \\IL a E x t r a p o l a t i o no f steadyvalrre Actualenvelope tlt t= 32 . : Y+ (9.7c) J2 X,J Extrapolationof transientenvelope where (a) Symmetricalshortcircuitarmaturecurrent in synchronousmachine l1l = steadystatecurrent (rms) Fig. 9.4 (Contd.) !//l = transientcurrent(rms) excludingDC component lltl = subtransientcurrent (rms) excluding DC component *Unity turn ratio is assumedhere. Xa = direct axis synchronousreactance

Modern Power Svstenn Anelveie Fault 'ii#I: Xtd= direct axis transientreactance pon ii'ragn-tir rutlrarion n X'j = direct axis subtransientreactance lErl = per phaseno load voltage(rms) ictfny:eTprsit:ctajarai:nblllivltpialtrqyleudsetiscuotdafji?bemlsea.\"cl,ihlm\"iiniltiesrrieofaoncr)t,adrnihfcfeeersvweanhltiuctheypcstaef nsrboeefaucmsteaarcjnihncinfeanesuosl.trTmcaaablcileulylra9iet.irowngsiiatvhneidsn s Oa,Ob,Oc = interceptsshorvn iLEigs- 9Aa andb, _ The intercept Ob for finding transient reactance can be determined raiddnereesaecttNceauidtroansmerentmaitcdnhreaferefoelotlriyhorneustsbestymaomlnybtocrh,simplhuittrgeibyonentngsrntaoateucnurrydassamipicteeoauosnrcr.ttraioreteynrnsadto.cffAmtloascownirwtciconeeurgiisstshoubaunbrlseltoesracadekcneefusolrarsirret,eenegnrxtecrtcehneeeaeopcrfttrataatathnonsosrchssieeoeasnrnawttdrrcheeiirtaccruhcaustniaetos.ndpiTceeteoonnst accurately by means of a logarithmic plot. Both Ai, and al decav exponentiallyas The machinemodel to be employedwhen the short circuit takesplace from loadedconditionswill be explainedin Sec.9.4. Aitt = Ai( exP(- t/q,) The method of computing examplesgiven below. short circuit currents is illustrated through Ait = Ai6 exp 1_ t/r7) where r4, and rf arerespectivelydamper,and field winding time constantswith For the radial network shown in Fig. 9.6, a three-phasefault occurs at F. Td* 4 ry At time / 2' r4*, Aitt practicalry dies out and we can write Determinethe fault current and the line voltage at l l kv bus under fault log (Aitt+ At,)1,, , -Trl* log Ai' = - Aint,/ ry conditions. t 1 OM V A -l c 15%reactance ab 1 1k V *+ g oO) o rf Fig. 9.5 10MVA \\ 12.5ohreactance \\ ne: 3o km,z = (0.27+jo.3e a/ km Ai,l,:o : Aito exp(-r/ ,t )1,:o: Ai,o: ob r NO 2: 5 MVA,8ohreaclance Table 9.1 Typicalvaluesof synchronousmachinereactances riO.0Bo) / km (All valuesexpressedin pu of ratedMVA) F z xn caote/ Type of Turbo-alternator Salient pole Synchronous Synchronous Fig. 9.6 Radial networkfor Exampleg.1 machine (Turbine (Hydroelectric) compensator motors* (Condenser/ generator) Solution Select a system base of 100 MVA. capacitor) V6ltage basesare: I I kV-in generators, 33 kV for overhead line and 6.6 kV for cable. X, (or X,) l.00--2.0 0.6-1.5 r.5-r2.5 0.8-1.10 0.9-1.s 0.65-0.8 X^ 0.12-0.35 0.4-1.0 0.95-1.5 0.3-0.35 .t 0.r-0.25 0.18-0.2 _ x,d 0.2-0.5 0.3-0.6 0.19-0.35 Reactanceof G, = xd 0.04-0.14 0.05-0.07 xti 0.003-0.008 0.13-0.35 0.18-0.38 0.003-0.012 Reactanceof G2 - x2 _ x,d 0.17-0.37 xo 0.02-0.2 0.025-0.16 ru 0.003-0.01s 0.004-0.01 ro = AC resistanceof the armaturewinding per phase. Reactanceof Z, = * High-speedunits tend to have low reactanceand low speedunits high reactance.

PowerSystemAnalysis I SymmetricaFl aultAnalysis | 337 I Reactancoef Tr= ;WY4) = 71.6pu = (0.93+ j05s) + (71.6+) (0.744+ i0.99) + (t1.0) = I.674+ j4.14= 4.43176.8\"pu overheadline impedance- Z (in ohms)x MVA\"\"'\" Voltageat 11kV bus= 4.43 167.8\"x 0.196l- 70-8\" (kvBur)\"2 = 0;88 I :T ptt = ft88 x 11 = 9;68 kV 30x(0.27+j0.36)x100 Q'2 - (0.744+ 70.99p) u cableimpedanc=\" 3(9-1!t,rJr0q,q2,1tlq = (0.93+ 70.55p) u A 25 MVA, 11 kV generatorwith Xl = 20Vois connected through a transformer, line and a transfbrmer to a bus that suppliesthree identical motors (6.6)' as shown in Fig. 9.8. Each motor has Xj = 25VoandXl = 3OVoon a base of 5 MVA, 6.6 kV. The three-phaserating of the step-uptransformeris 25 MVA, Circuit model of the systemfor fault calculationsis shownin Fig. 9.7. Since 11/66 kV with a leakage reactance of l0o/o and that of the step-down the systemis on no load prior to occurrenceof the fault, the voltagesof the two transformeris 25 MVA, 6616.6kV with a leakagereactanceof l0%o.The bus generatorsare identical (in phaseand magnitude) and are equal to 1 pu. The voltage at the motors is 6.6 kV when a three-pha.sfeault occursat the point F. generatorcircuit can thus be replacedby a single voltagesourcein serieswith For the specifiedfault, calculate the parallel combination of generatorreactancesas shown. (a) the subtransientcurrent in the fault, 11kV bus (b) the subtransientcurrent jn the breaker.8, ( o . 7 4 4+ i 0 . 9 9 ) (0.93+705. 5) (c) the momentarycurrentin breaker B, and .I | , (d) the currentto be interruptedby breakerB in five cycles. Line Given: Reactanceof the transmissionline = l5%oon a baseof 25 MVA, 66 Ij|.0 i1.6 ' II I t Cable kV. Assurnethat the systemis operatingon no load when the fdul\" occurs. 66d 666 T1 T2 i l Fig. 9.7 Flg.9.8 Totalimpedanc=e (j1.5ll j1.25)+ (t1.0)+ (0.744+ i0.99)+ (i1.6)+ Sotution Choosea systembaseof 25 MVA. (0.93+ 70.55) For a generatorvoltagebaseof 11 kV, line voltagebaseis 66 kV and motor - 1 . 6 7 4+ j 4 . 8 2= 5 . 11 7 0 . 8 \" p u voltagebaseis 6.6 kV. (a) For eachmotor Isc= ''? tt = 0'196I - 7 0 ' 8 \"P u 5.r170.8\" X',j*= j0.25 x + = i1.2p5u 18u.=\" to*.10; = 8,750A Line, transtbrmersand generatorreactancesare alreadygiven on proper base J3 x6.6 values. Isc= 0.196x 8,750= 1,715A The circuit model of the systemfor fault calculationsis given in Fig. 9.9a. The systembeinginitially on no load,thegeneratorand motor inducedemfs are Total irnpedancbeetweenF and11 kV bus identical. The circuit can thereforebe reducedto that of Fig. 9.9b and then to Fis. 9.9c.Now

fiffirel ModerpnoweSr ystemAnatysis momentarycurreni ihrough breaker B -- 1.6 x 7,4i9.5 ! I s c = 3 > <- - l - + = + - - jr4 . 2 2 p u - 17,967A j1.25 j0.55 Basecun'entin 6.5 kV circui, - 25x 1,000 = 2.187 A (d) To compute the current to be intemrpted by the breaker, motor subtransientreactance (X!j = j0.25) is now replaced by transient reactance Issco=,;;* r4rrlt]frf* o (X a = /0.3O). (b) From Fig. 9.9c, current through circuit breaker B is XI (motor)= 70.3x 25 = Jr.) pu T I s c (' B- )2 x -++. ] _ : - i 3r -..4-2 The reactancesof the circuit of Fig. 9.9c now modify to that of Fig. 9.9d. j1.25 j0 . 5 5 Current (symmetrical)to be intemrpted by the breaker(as shown by arrow) = 3.42x 2,187= 7,479.5A =2x j 1^ \" 1 =3.1515pu + l.s jO.ss 110\" Allowance is madefor the DC off-set value by multiplying with a factor of 1.1 (Sec.9.5). Therefore, the currentto be interruptedis 1 1 0+' jo.2 j0.15 j0.1 110' 1 . 1 x 3 . 1 5 1 5x 2 . 1 8 7= 7 . 5 8 1A F tll?9 ;< to\" 9.4 SHORT CIRCUIT OF A LOADED SYNCHRONOUS MACHINE In the previous article on the short circuit of a synchronousmachine, it was aBsumedthat the machine was operating at no load prior to the occurrence of shortcircuit. The analysisof short circuit on a loadedsynchronousmachine is complicatedand is beyond the scopeof this book. We shall,howevbr, present here the methods of computing short circuit current when short circuit occurs underloadedconditions. lrrl :r- B- - ^tlc f>\\ . 1L/v. | S^ Ll^lu. -w, ^s +LLlr^g u^ ir-r^U. , iu+rt l*r^lAu^(llEl u^ fr a synchronousgeneratoroperatingunder steadycon- (b) i0.55 ditions supplying a load current /\" to the bus at a i0.55 terminal voltage of V \". E, is the induced emf under loadedcondition andXa is the direct axis synchro- nousreactanceof the machine.When short circuit occursat the terminals of this machine, the circuit model to be used for computing short circuit Fig. 9.10 Circuitmodelof current is given in Fig. 9.11a for subtransient a loaded current,and in Fig. 9.1lb for transientcurrent.The machine inducedemfs to be usedin thesemodels are given rcuitbreaker) (c) bY Fig.9.9 E,l= v\" + ilTtj (e.8) (c) For finding momentary currentthrough the breaker,we must add the EL- V'+ il\"Xto (e.e) DC off-set currentto the symmetricalsubtransienct urrent obtainedin part (b). Ratherthan calculating the DC off-set current,allowanceis madefor it on an The voltageE!is known as the voltage behind the subtransientreactance and empiricalbasis.As explainedin Sec.9.5, the voltage E!is known as the voltagebehind the transientreactance.Infact, if 1ois zero (no load case),EJ= Etr= Er, the no load voltage,in which case the circuit model reducesto that discussedin Sec.9.3.

340 | Modernpo*s1_qqe!l inslygs _Symmetrical FaultArralysis I solution Aii reactancesare given on a base of 25 MVA and appropriatgi t/o voiiages. :t Prefaultvoltage V\" = J'9 = 0.9636 l0 pu l1 Load = 15 NfW, 0.8 pflEading ( a ) C i r c u i tm o d e l f o r c o m p u t i n g (b) Circuitmodelfor computing = l2:5 = 0.6 pu, 0.8pf leading s u b t r a n s i e n tc u r r e n t transientcurrent F i g .9 . 1 1 Synchronousmotorshave internalemfs and reactancessimilar to that of a prefaultcurrenlI\" = _9{__ _ 136.9.= 0.77g3I 36.9\"pu generatorexcept that the current direction is reversed. During short circuit 0.963x6 0.8 conditions these can be replaced by similar circuit moclelseicept that the voltagebehind subtransient/transienrteactanceis eiven bv Voltagebehindsubtransienret actance(generator) E'lr=v\" - jI\"xU (e.10) E\",- 0.9636I tr + j0.45 x 0.1783I 36.9\" E'*= v\" - jI\"4 (e.11) - 0.7536+ 70.28pu Voltagebehindsubtransiernetactanc(emotor) Wheneverwe aredealingwith shortcircuit of an interconnectedsystem,the synchronousmachines (generatorsand motors) are replaced by their corre- El,,- 0.9636/_ tr -i0.15 x 0.7783/_ 36.9\" s po n d i n gc i rc u i tm o c l e l sh a v i n gv o l tagebehi ncsl rrhtransi en(ttransi entr)eac- tancein serieswith subtransien(ttransientr)eactanceT. he restof the network = 1.033-6 .70.093p3u b e i n gp a s s i v er e n t a i n su n c h a n g e d . T h ep l e furult.equ iv acl ei rnctu i ti s show ni n Fi g . 9 . l 2 b .U n cl efrh r rl t e c l. t i o n( l r i g .9 .l 2 c ) n c l i- IIr^E_.x_, a m p l e9 . 3 . ....-:\".._. ._. I. ', ;, 0 . 7 5 3 6 +\"i 0 . 2 1-t 0=00 - , - . 6 2 2 6 _ j 1 . 6 7 4 6 p u \" i0.45 A synchronousgeneratorand a synchronousmotor each rated 25 MVA, I I kV having l5Vo subtransient reactance are connected through transfbrmers and a I',l,= 1 . 0 3 i 6- r o o c ) ?1 line as shown in F'ig.9.12a. The transfbrmers are ratecl25 MVA. lll66kV and i0.1s 66lll kV with leakagereactance of l\\Vo each.The line has a reactanceof lTTo on a baseof 25 MVA, 66 kv. The motor is drawing 15 Mw at 0.9 power factor Currentin fault leadingand a terminal voltage of 10.6 kV when a symmetrical three-phasefault occurs at the motor terminals. Find the subtransient culrent in the generator, I J = I : i + 1 , , , , = _j g . 5 6 5 3p u motor and fault. Gen t Tt' Basecttrrnet (gen/moIto= 44q1 = 1.312.2A |) | '! Line J3xll lr; Now ( a ) O n e - l i n ed i a g r a mf o r t h o s y s t o mo f E x a m p l e9 3 I'J- 1,312.(00.622-6 j1.6746=) (816.4_ jL,tgt.4) A I'J= 1,312.2(.-0.6226- j6.8906)= (- 816.2_ jg,O4L8A) t ;t \" . j0.1 , ' . j0.1 - j0.1 -t F 1t--jtt,23gA 6'f,1 'dtI-. 'ltrd-. i I I I '. ), 1 i 0 . 1 5 + lr short circuit (sc) current computation through the Thevenin Theorem I An alternate method ol' cornputing short circuit currentsis through the applicationof the Thevenintheorem.This methodis fasterand easily adopted (b) Prefaultequivalentcircuit (c) Equivalentcircuitduringfault Fi9.9\"12

I ModernPo*er SystemAnalysis 342 | I to ,yrtl-atic computationfor large networks.While the method is perfectly gcncrerli,t rs illustratcdhcrc tlrrougha sinrplccxanrplc. Considera synchronousgeneratorfeedinga synchronousmotor over a line. Xle +X (xlh,+x + xi Figure 9.I3a showsthe circuit model of the systemunder conditionsof steady AI^ (e.14) As a first step the circuit model is replacedby the one shown in Fig. 9.13b, Postfault currents and voltagesare obtainedasfollows by superposition: whereinthe synchronousmachinesarerepresentedby their transientreactances I{= I\" + alr (or subtransienrteactanceisf subtransiencturrentsareof interest)in serieswith voltagesbehindtransientreactancesT. his changedoesnot disturb the prefault I{=- I\" + AI^ (in rhedirectionof AI^) (9.15) current I\" andprefault voltage V\" (at F). Postfaultvoltage As seenfrom FG the Theveninequivalentcircuit of Fig. 9.13bis drawn in whereAv = -ixr^etfsvfp-i=es cttvhotoetv+hoe(lt-ra,ergxfeeno,rfIetfh)ne=bcfaeuvu\"sltcp+to-iAn;;vt;F./.donbthyetThheefvloenwin-opafesfJasi6uv)ert Fig. 9.13c.It comprisesprefaultvoltageV\" rn serieswith the passiveThevenin impedancenetwork.It is noticed that the prefaultcurrent 1\" doesnot appearin fffl:liiwith thepassivcThcvcninirnpcdanccnctwork.It is thcretore t<lbe rcmcnrbcrcdthat this current must be accountedfor by superpositionafter the SC solution is incetheprefault current flowing out of obtainedthroughuseof the Theveninequivalent. t curent out of F.is independeniof load Considernow a lault at 1,'thloughan irnpedancZel .liigure 9.13dshowsthe on is summarizedin the following four Thevenin equivalent of the system feeding the fault impedance.We can im m e d i a te lwy ri te ', l ' - - V \" (e.r2) Step I: nORveaebtlpwtualeoainsrck.es.Streehaaodcryttascntiacrcteeussoitof luusiyttino\"cnrhrorfolsnoooauudrsmecdeasscy.hTsithneeemsrb(elyosauthdltefliisor wstuhsbetturpadanyss)s.ieivneU jXrn + Zt step 2: transient Current causedby fault in generatorcircuit Thevenin AI, - x'd^ f (e.13) 'Step 2 at the fault point by negariveof - (xhs+x + xl^ ies with the fault impedanc\". 6o_pur\" rterest. rareobtainedby adding resultsof Steps The following assumptionscan be saferymadein SC computationsreading :ation: ragnitudesare I pu. tre zero. (a) G (b) G to actual conditions as under normal nity. ruc cnanges ln current caused by short circuit are quite large, of the order of 10_20 (c) (d) tion 2. Fi9'e't+ F is thefaultpointon the passiveThevenin Fig.9.13 Computatioonf SC currenbt y theTheveninequivalent re Let us illustrate the above method b, network calculatingthe resultsof Example9.3.

M o d e r n P o w e r S y s t e mA n a t y s t s ,Qrrrnmalrinal E^..t+ A --r. --!- The circuit modelfor the systemof Example9'3 for computationof postfault If sc MVA (explainedbelow) is morethan500, the abovemultiplyingiactors conditionis shownin Fig' 9'14' are increasedby 0.1 each.The multiplying factor for or lower is 1.25. air breakersrated 600 v _ 0.9636x70.60= - 78.565pu The current that a circuit breakercan intemr rng voltage over a certain range, i.e. Changein generatorcurrentdue to fault' Amperesat operatingvoltage A, ,IB^_= - rl\"8. \". s\" _6 st io'+ - - i2'14P1 u j0.60 Changein motor currentdue to fault' At^=- 78.56x5 j.$*i --- i6'424Ptt To these changeswe add ttre prefault current to obtain the subtransient Rated intemrpting MVA (three-phasec) apacity current in machines.Thus = '6ty(tifle)lrated x 11(line)lratedinremrptincgunent I'l= I\" + AIr - (0.623- j1.67$ Pu where V(line) is in kV and 1 (line) is kA. In = - I\" + AI^= (- 0.623- 76.891)P u Thus, insteadof_computingthe sc current to be intemrrpt-e-'d,w' e cbmpute three-phaseSC MVA to be intemrpted,where which are the same(and shoutdbe) as calculatedalready we have thus solved Example 9.3 alternatively through th9 Thevenin SC MVA (3-phase)_ Jt x prefault line voltagein kV for large x SC currentin kA. theorem ond ,up.rposition. This, i n4eecl,is a powerful method networks. If voltage and current are in per unit valueson a three-phasbeasis 9.5 SELECTION OF CIRCUIT BREAKERS SC MVA (3-phase) = lylp,..rouxl, 1111x6 (MVA)uur. (e.r7) .Iwootthccircuitbrcakcrratingswhiclrrcc;uircthcctrnlptlttttionofSCcu r r e n O' -h' Lv i'oYrrr us l'rur J r J ' ircqifiouAu ll \\v/t\\ /v AA ii -n+i^e*i,r- u- .p: -i i i l g - c- a- p a c l t y o f a c i r c u i t b r e a k e r is to be Ssmayyrunerlnt:cimrphalreytoetirnndigcomatuhlosSermnCsaeycnmchtumairnrreyeectnrsuitc.Mriarseol-nomo-tbc\"ntnaaotniarndyreyrdacbtcueyurdrrruesesnynimtntbigrmymesasutfr)baictcirastaolnirntsohtifeeer1nnr'tu6rcpetaoatilcracttcugaclcnao:cntueernsdrtfefboonyrrt' trnurcthln (or cclualto) thc sc MVA requiredto be intemupted. trmosheybeoamtStacpomyitrrnamesen-sttmicrhenieceendtsatcuocrlecroiSuctsfriCaoyrDcelnncuCncurtmhrrotrerooefofntn-nobtsttrobeoesuytagsbicnraeueetnerfniarerneercrtugntacoplttre'rtorceuartdsaspendtitsaed*bd.tTiarushaclaencctosoDeiumdeCnbnpteoterulefdotfae-fwsocde:brt atyvbnayrc'lerusfmseoi nturoglsstbiyupenlbycatihrdnardgonensdtohiteouenst hsiwcltnohiyoghtcenneFhiaccnrmoetrhheirusaorcgttroxnthspienivmectodoeinftsuusogeMmsgfitlttaheVhe.puecenAoAlthetissonirysagmnasinohbtetaodfeelxyermsac*gSteb.I.cSenACpeinrCtcaieMotouhMrsbniVartiveVmtAibiiesoAnpr-uegpltaeasochanskabanbyepusdLesreatfGfatoaiceinnrucimtt(aielytlrtaihmc(pntaheualeralofiprs-tartuttobigeuocgredl-ushetgswlylraaoraynoikrctsrclhueoeathenrctimrrsimdaeoougt)sfnnvi.spoewoaatennuurhbc,eiwslioerttcrtnuaecohcolslmalogtytpyposiutvpohaerselesebtlstoslfooaeiiatnbnhnoddleadefe). locations must be tried our to obtain the highestst nava requiring Lp\"ur\"a SC computations.This is ilustrated by the examplesthat follow. Circuit Breaker SPeed Multiplying Factor I i i \" ' n \"r . ; I 8 cyclesor slower 1.0 5 cycles 1.1 Three6.6 kv generat orsA, B andc, eachof I0o/oleakagereactanceandMVA 3 cyclcs 1.2 ratings40, 50 and25, respectivelyareinterconnectedelectrically, as shown in 2 cycles 1.4 t'In some recent attempts,currentscontributedby induction motors during a short tThis will be explainedin Chapter 1l. circuit have been accountedfor.

15:!\"'l .\";: I ModernPowerSystemAnalysis- T{d46..t| currentscan thenbe calculatedby the circuit model of Fig. g.l6acorresponding Fig. f.i5, by a tie bar throughcurent timiting reactors,eachof I2Voreactance to Fig. 9.13d.The circuit is easilyreducedro rhat of Fig. 9.16b,where baieclupon the ratingofthe machineto which it is connectedA. threc-phase feederis suppliedfrom the bus bar of generatorA at a line voltage of 6'6 kV' 7- ( 0. 069 + j0. 138) + j0. r 2s il 00. 15+ jo. 22| j0. 44) ^f n 1a = 0.069+ j0.226= 0.236173 Q/phase.Estimatethe maximum MVA that can be fed into a symmetricalshort circuit at the far end of the feeder. SCMVA = Volf = V\"('+') = pu (sinceVo = 1 pu) \\Z) +Z I (MVA)Ba.\" Z 50 = 2 1 2M V A 0.236 Tie bar Fig.9.15 Sotution Chooseas base50 MVA, 6.6 kV' Considerthe 4-bussystemof Fig. 9.17.Buses1 and 2 l aregeneratobr use sand FeederimPedance 3 and 4 are load buses.The generatorsare rated l kv, 100 MVA, with = =(o.o6e+/0.138)pu transientreactanceof l07o each.Both the transformersare 1ll110 kV, 100 % MVA with a leakagereactanceof 5Vo.The reactancesof the lines to a baseof 100 MVA, 110kv areindicated on the figure.obtain the shortcircuit solution o'1[50 GenA reactanc-e 40 = 0.125Pu for a three-phasesolid fault on bus 4 (load bus). GenB reactanc=e 0.1 Pu Assumeprefault voltages to be 1 pu and prefault currentsto be zero. (G) G e n C r e a c t a n c e = 0* . 1425 = 0.2pu 1 ReactoAr reactan.\"= o't''I tn = 0 . 1 5p u 40 ReactoBr reactanc=e 0.12Pu Reactor C reactance = 0 . 1 2x 5 0 - 0.24pu j0.2 Fig.9.17 Four-bus ystemof Exampleg.5 Solution Changesin voltages and currentscausedby a short circuit can be j0.12 jo.24 calculatedfrom the circuit model of Fig. 9.18.Fault current 1/ is calculatedby systematicnetworkreductionas in Fig. 9.19, ( 0 . 0 6 9+ i 0 . 1 3 8 ) yo = 1Z0o( (a) (b) Fig. 9.16

,^o I Moclern Power Svstem AnalYsis I tt = -j*0=. 1 3 s = - jt.37463pu 60 lo* !o;E 14 jo 1,), D': _[ jo.2 i0.1 t,= rsx ji0:.3i736;3:8: =- j3837opru 7 0t , I r - ---rnTl Izt A, 1-2=r',' x j{0:.3i:7:6::38 = - j3.53762pu l,zs 4 +I ib.rs 2 Let us now computethe voltagechangesfclr busesl,2and 3. From Fig. 9.l9b, wc givc t-,ti,, 1 0 . 1 5 AVr - 0 - ( / 0. 15)( - j3. 8370r ) = - 0. 57555pu \\ AV, = 0 - (iO.l.s)(- .i3.53762)= - 0.53064pu i ro'rs Now o) V , ' = l + l V t = 0 . 4 2 4 4 .p5u r -f- t-- - ' iA.0t .i 1A1' - I l, V ) ,= l + A V z = 0 . 4 6 9 3 6 p u II 2Fi I' t0t 6 ' I i0 1'':, , , t , .( .^t-s ( rl )r t i o io.t\\P ''r),}o'rs /. l-t= v , J- v J = J0'1796p4u ,of tr.o.t ll -l (a) 4T (b) (. )vl =t.o ir ( tvl=to Now * fr' irt AVy= 0 - [(/0.15()- j3.83701+) Q0.15()/0.17964)l = - 0.54fi(r0pu V t t = I - 0 . 5 4 8 6=0 0 . 4 5 1 4p u (c) (e) vlo= o l''l + The determinationof currentsin the remaininglinesis left as an exerciseto I tltcrrr'ldcr. ( ,-j;0.''uuuo l Short circuit studyis completewith the computationof SC MVA at bus 4. ,-l ( ) v?=t.o (SC MVA)^ = 7.37463x 100= 737.463MVA , -J10.04166 Tr It is obviousthat the heuristic networkreductionprocedureadoptedaboveis tt*) not practical for a real power network of evenmoderatesize. It is, therefore, essentialto adopta suitablealgorithmfbr carryingout shortcircuit study on a -i digital computer.This is discussedin Sec.9.6. (\\.. ) V ?= 1 . 0 9.6 ALGORITHM FOR SHORT CIRCUIT STUDIES t. So far we havecarriedout shortcircuitcalculationsfor simplesystemswhose l l i tssi vct t ct wor kscanbe easilyr educedI.n t hisscct ionwc ext cndour st udyt o Ill I Fig. 9.19 Systentaticreductionof the networkof Fig' 9'18

i*.3€0-l ModernPower System Analysis large ,tyrt\"-r. In orclerto apply the four stepsof short circuit computation _ S- ,y.m. . .m. ,e*t\"r,.,i;caFl aultAnat lr sis developedearlierto largesystemsi,t is necessaryto evolve a systematicgeneral algorithm so that a digital computercan be used. _--.rr.vv\\, rrrv u(rr vurraegcs oI mls network. 4V = Z\"urJf (e.20) e.2D Now where .7 -'l '- otn I Gen 2 i = busimpedancematrixof the I passiveTheveninnetwork Fig. 9.20 n-bussystemundersteadyload Consideran n-bussystemshownschematicallyin Fig.9.20 operatingat steady Znn) load. The first step towards short ciicuit computation is to obtain prefault voltagesat all busesand currentsin all lines througha load flow study. Let us u// = bus current injection vector indicatethe prefault bus voltagevector as Sincethe network is injectedwith current- 1/ only at the rth bus, we have (e.18) 0 0 : /- rf -I', f (e.22) I, : : Let us assume that the rth bus is faulted through a fault impedance Zf . The SubstitutingEq. (9.22) in Eq. (g.20), we have for rhe rth bus AV, = - ZrJf postfault bus voltage vector will be given by By step4, the voltageat the nh bus underfault is V { u r = V B u s+ A V (e.1e) v!= vor+avo,- vor- Z,Jf where AV is the vector of changesin bus voltagescausedby the fault. (e.23) As'step 2, we drawn the passiveThevenin network of the system with However,this voltagemustequal (e.24) Vd= 7f 1f generatorsreplacedby transient/subtransienret actanceswith their emfs shorted (e.2s) (Fie.9.21). We have from Eqs. (9.23) and (g.24) zftf - vo,_ z,Jf (e.26) Fig. 9.21 Networkof the system of Fig. 9.20 for computingchanges in (e.27) bus voltagescaused by the fault or f= V: Zr, + Zf At the rth bus(fromEqs(9.20)and(g.22)) AV, = - Z,Jf v{= v?- Z,Jf,i = 1,2, ..., substitutinfgor // from Eq.(9.25).we have vI= vf- z:*z+lL';rv!

SiimmetricaF! ault ,Analysis I| -rc?a53 352 i rtllodernPower System Analysis Fori=rinEq.(9.27) First of all the bus admittancematrix for the network of Fig. 9.18 is formed asfollows: (e.28) - - r I I -t- 1 -r I - - j28.333 In the aboverelationship V,o'r, the prefault bus voltagesare assumedto be -l YtrL. > = YL t r t = _ - 1 7 5 . 0 0 0 known from a load flow study. Zuu, matix of the short-circuitstudy network j0.2 of Fig. 9.21 canbe obtainedby the inversionof its furr5matrix as in Example 9.6 or the Zru, building algorithmpresentedin Section9.7.It should be -l - i6.667 observedhere that the SC study network of Fig. 9.2I is different from the correspondingload flow study network by the fact'that the shunt branches Y3= Yy = ;r* correspondingto the generatoreactancesdo not appearin the load flow study network.Further,in formulatingthe SC study network, the load impedancesare -1 ignored, these being very much larger than the impedancesof lines and ginerators. Of coursesynchronousmotors must be included in Zuur tormula- Yrq= Yq= = i10.000 J,3.l tion for the SC study. Y z z =j.0:.-1+5+j-0+.1+s\"+j:j^.r =- j28.333 j0.2 Postfault currentsin lines are given by f u= yu(vri- vt) Q.29) Y z t =Y n = +j0=.1 j 1 0 . 0 0 0 For calculation of postfault generatorcurrent, examine Figs. 9.22(a) and (b). From the load flow study(Fie. 9.22(a)) -l - i6.667 Prefault generatoroutput = PGr+ iQci Yzq=Yn= F; 'Y3-3.-= j I 1 5+' I =-i16.667 0. jo.1 Ytq= Yqt= 0.000 vYqq- = I -* I -- i76'667 ,- ,.\"\" (a) (b) Fi1.9.22 f\", = &#; (prefaulgt eneratoorutput= Pci + iQci) (9.30) v,u By inversion we get Z\",tt as E'Gi = V, + jXt\"/t)\", (9'31) j0.0s97 j0.0719 From the SC study, Vf,is obtained. It then follows from Fig. 9.22(b) that j0.0903 j0.0780 j0.0780 j0.13s6 rrc,=tfr! (e.32) jo.o719 j0.0743 iir;rnr;;.I; Now, the postfaultbus voltagescan be obtainedusingEq. (9.27) as To illustratethe algorithmdiscussedabove,we shall recomputethe shortcircuit V {r = r V Zo - Z t o VaP solutionfor Example9.5 which was solvedearlier usingthe network reduction o o technioue.

,2.tr!A | ^/lndarn Pnrrrar Qrratarn Analrrcic malriaal E qau t | | tglt tvql JJ' I lYlVVVlll I Vltvl Vtvlvlll , rlrsrtvre I -r 1.000 1.00 = - j I L 0 7 4 I 9 7p u The prlfault conditionbeing no load, V0r= Voz= V03- Voo=1pu ,J - _ - r . o - i 9 9 1 9x9 1 0= 0 . 4 2 4 8 p u 2,,{orZzz) j0.0903 i0.1356 vrLz=L v:z - o?'o: r r,', By Inventing Y\"u\" = 1.0- oi 0 ..-10 73:1:506x 1.0 = 0.4698 pu /nus= YrusVsus j0 v{=v\\ - 7 or Vsus= [Y\"ur]-t /eus = Znus/eus (e.33) 1 * v f or Zsvs= [Yuur]-t Zoo The sparsityof fsu, may be retainedby using an efficient inversion technique = 1 . 0- 'ji00-..:r0:37_s4:63 1 . 0= 0 . 4 5 2 1p u [1] and nodal impedancematrix can then be calculated directly from the factorizedadmittancematrix. This is beyond the scopeof this book. vi = o'o Current Iniection Technique UsingEq. (9.25)we canobtainthe fault currentas Equation (9.33) can be written in the expandedform 7r= j.0r.-1^30s.96^=o, =j-7.3j463pu V 1: 2 1 1 \\ * Z t z l z 1 . . .* Z n I n (e.34) Thesevaluesagreewith thoseobtainedearlierin Example9.5.Let us also v 2 : 2 2 1 1 1 + 2 2 2 1 2+ . . . + z z n l n calculatteheshortcircuitcurrenitn lines1-3,1-2,l-4,2-4 and2-3. V, : Zntll * 2,,21r* ...+ Z,,nl,, [,1r 2 =-v:,-r -v{ 0.4248-0.4521- j o ' 1 8 2P u It immediately follows from Eq. (9.34) that z.n i0.15 z -- v'l -t rrr z = v- ',_r --v{' 2 - : - 0.4248-0.4698_ j 0 . 2 2 5p u I r l , r : r z: . . . : rn=o (e.35) zrz .i0.2 \"ii I1-r0 ,r rt 4- yr -v{ _ o.4z4v-o - i4.248pu AlsoZi, - Ziii (Znvs is a symmetrical matrix). 4q jo'l As per Eq. (9.35)if a unit currentis injectedat bus (node)7, while the other - j3.132pu .I irq u'l - vq{ - - -0=.46e-8o buseserekept oponcircuited,the bus voltagesyield the valuesof theTthcolumn of Zuur. However, no organized computerizabletechniquesare possible for = ' - zz+ i0'15 finding the bus voltages.The techniquehad utility in AC Network Analyzers 'where the bus voltagescould be read by a voltmeter. I---Irzt=tviu{ t-tv^ = 0.4698-0.4521= - ./0.177Pu Considerthe network of Fig. 9.23(a) with three busesone of which is a TO- reference.Evaluate Zsus. For the exampleon handthis method may appearmore involved compared Sotution Inject a unit currentat bus I keepingbus 2 opencircuit, i.e., Ir = I. to the heuristicnetworkreductionmethodemployedin Example9.5. This, andIr= 0 as in Fig. 9.22(b).Calculatingvoltagesat busesI and 2, wehave however, is a systematicmethod and can be easily adopted on the digital computer for practicalnetworksof large size.Further, anotherimportant feature Ztt=Vt=7 of the method is that having computed Zu's, we can at once obtain all the required shortcircuit datafor a fault on anybus.For example,in this particular Zzt=Vz=4 system,the fault current for a fault on bus I (or bus 2) will be

.f M o d e r n P o w e rS y s t e mA n a l y s i s SymmetricaFl ault Analysis !, f5l 356 | Now let It = 0 and12= 1. It similarly fbllows that , 4r, the dimensionof Zsu5 goesup by one). This is type-2modification. Ztz=Vt=4= Zn 3. Zuconnectsan old busto thereferencebranch(i.e.,a new loop is formed Z v u sll=7+ 41 dimensionof 2o,,\" doesnot change).This is type-3modification. 6l 4. Zuconnectstwo old buses(i.e.,new loop is formed but the dimensionof Zuu, does not change).This rs type-4 rnodittcation. referBreecdatousaesotfheth'eoapbeonv-ceicrcoumitpimutpaetidoannalcpermoacetrdixu'r.e,the Zru, matrix is 5. Zu connectstwo new buses(Zeus remains unaffectedin this case).This Z\"vs Building Algo4ithm situation can be avoidedby suitable numbering of busesand from now onwardswill be ignored. It is a step-by-stepprogrammabletechniquewhich proceedsbranchby branch. It has the advantagethat any modification of the network does not require Notation: i, j-old buses;r-lsfelence bus; k-new bus. complcterebuildingof Z\"ur. Type- 1 Modification Considerthat Zrur has been formulatedupto a certain stageand another branchis now added.Then Figure 9.24 shows a passive(linear) n-bus network in which branch with impedance2,, is addedto the new bus k and the referencebus r. Now Zrur (old) Zo:branch impedance Zsus (new) V*=ZJ* Upon adding a new branch,one of the following situationsis presented. Z r i = Z * = 0 ; i = 1 , 2 . . . . .t t Zm= Zu Hence Zsvs (new) - s,,\"(old) (e.36) Fig. 9.23 Currentinjectionmethodof computingZru. Passivelinear n-bus network t . 26 is added from a new bus to the referencebus (i.e. a new branch is added and the dimension of Zry5 goes up by one). This is type-I Fig. 9.24 , Type-1modification modificution. Type-2 Modification Zo is addedfrom new bus ft to the old bus7 as in Fig. 9.25.It follows from this figure that

35S;,1 MorJern Powcr Srrctarn anatrroio 1O_ Zrj lr n zzj Zsvs(old) I : (e.38) P a s s i v el i n e a r oI lzyziz...zi,I zii + z,)lI* n-bus network Erimin{aitnethe3\":';,,iiy';;;?::\"i\"t i.I\" (e3'I)' a;y>;i:ation or 'o= -12 1 (\\1Ir+ Zi2Iz+ \"' + ZlnI) (9'39) Fig. 9.25 Type-Z modification Vo=Zdo+ V, Now V; = 2; 111+Z, r I , + \"' + Z'nI n+ Z; / r (e.40) = Zr,I*+ZiJr+ Zlzlz+... + Zii ei* I) + ...+ Zinln SubstitutingEq. (9.40) in Eq. (9.39) V * = 4 l t + Z l z l z+ . . . + 2 , , 1 , + . . . +Z i , l n + ( Z i i + Z ) l k Rearranging, ,,=lr^- h(z* 1rr]*alr,,- rzuz,,>)r, Consequently + +l r ^ - \"h 2,,1)r, (s.4t) Zti ^i{zu Zzj : Equation(9.37)can be written in matrix form as Znj (e.37) ZsLr(5new=) znvs(ol-d ;+;l . |t\"f (e.42) : ltz't\"Zv) zjj + zb \"jjT\"ulz,,l IYpe-3 Modification Type- Modification zu connectsan old bus (l) to the referencebus (r) as in Fig. 9.26. This case zo connecttswo old busesasin Fig.9.27. Equationscanbewrittenasfollows follows from Fig.9.25 by for all the networkbuses. connectingbus ft to thereferencebus r, v*=o' i.e. by setting (l;+ lp Passive linear J n-buS network Fig. 9.26 Type-3 modification Fig. 9.27 TYPe-4modification Vi= Z;11+, Z,lr+ \"' + Zri (Ii + /) + ZU Qi- /o)+ ...+ZiJ\"(9.43) Similar equationsfollow for other buses.

360 | Modern power System Analysis SymmetricaFl aultAnalysis I fOt I he voltages of the buses i and j are, however, constrained by the equation Step I: Add branch 2,, - 0.25 (from bus I (new) to bus r) (Fig. 9.27) V,= Z,rlo+ Vi (e.44) ZBUS= t}.25l (i) o r 2 1 1 1+1 \\ 2 1 2 + . . . + Z i t ( 1 , + I ) * Z i i ( I i - I ) + . . . + zi,In Step Add branch Zzt = 0.1 (from bus 2 (new) to bus I (old)); type-Z = Ztl*+ Z,rl,+ Z,rl\"+ ... + ZlU,+ I _I earTangtng 0 =(Zit- 21)Ir+ ... + (Zti- Z) Ii+ eu- Z) Ii zsvs=',3lZ1:i1] 1, (ii) +...+ (Z* - 21,) In + (Zu * Zii * Zii - Zi - Z) 11, e.45) Step 3: Add branch ,rt = 0.1 (from bus 3 (new) to bus I (old)); type-Z Collecting equationssimilar to Eq. (9.43) and Eq. (9.45) we can write vl modification V lo.2s o.zs 0.2s1 (iii) zsvs= | o.zs o.3s o0..32ss| ..l vn (9.46) lo.2s 0.2s Step4: Add branch zz, (from bus 2 (old) to bus r); type-3 modification 0 10.2s o.zs 0.25'l [0.25-l Eliminating 1oin Eq. (9.46) on lines similar to whar was done Z e t=) s| o . z s 0.3s 0 . 2 s| o'3+so | o.tsI to.zs0.3s0.251 modification,it followsthat \" Lo.2 so.2s o3sl Ln.trj Z131,(snew) = Zsuts(old) - zb+zii*Zii-zzij l'\" [0.14s8 0.10420.14s8-l = I o.ro+z o.r4s8 o.lo42| lzit - z1t)... (zi, - zi)) lZ,, fo.tott o.ro42o.24s8l with the useof fbur relarionshipsEqs (9.36), (9.37),(9.42) and (9.47) bus impedancematrix can be built by a Step5: Acldbranchiz-r= 0.1 (frombus 2 (olct)to brrs31old)): type-4 bra n c h a t a ti n l e ) a s i l l trs tra tc 'itnl step-by-stepprocedure(bringin-ghicni nogne l i xrrrnpl e9.8. Thi s pror.cdrrr-c a modification [o.l4s8 0.1042 0.l4s8l mechanicalonecan be easilycomputerized. zsv=s l o.'oo, 0 . 1 4 5 8o . t o 4 2 l - When the networkundergoeschanges,the modiflcationproceclurescan be 10.l4s8 O.tO42 0.24s)8| 0 . 1+ 0 . 1 4 5+80 . 2 4 5 8 - 2 x O . l O 4 2 cttrploycctlo rcviscthc bLrsirtrpcdancrenatrixul'thc nctwol'k.Thc opeling o1 a line (Ztt) is equivalentto adding a branchin parallelto it with impedance [-0 1042.l 0.0411 -0.0417] - 2,, (seeExample9.8). = I 0.0417| [-o.to+z j Example9.8 [-o.o+tz-l For the 3-busnetworkshownin Fig.9.28buildZsus = lo.r3e7 0.1103 0.12501 |0.1ro3 o.I3e7 0.1250 II r 'ir9l l z |0.1250 0.1250 0 . 1 7 sIo l I I 6\"11-lI. I Oltenirtgo tinu (line 3-2): This is equivalentto connectingan intpeclanc-e0.I 0 . 2 5(l!,,--' -- L-t,rili -, -- o.t i,Jor. betweenbus 3 (old) and bus 2 (old) i.e. type-4 modification. 0.1 3 II Zsus=Zuur(olc- l)(-01)+ol?5# l I Ref busr Fig. 9.28

filf.lid, ModernPower System Anatysis -r-r-^r rra^ .u.ltlr rAunr qar*yroa'iso ffir :)ymmelrlual r** [ 0.0147-l t- 0.0147l | 10.0 14 7- 0.01470.0s001 Now o.oso| L o_J - I 0.14s8 0.1042l; (sameas in step4) i' li=2 (-F[ ^Z E - ) = 0 . 2 8 6 | 0.1042 and zu) 0.1458 0.1042 0.2458| Thesetwo voltagesare equal becauseof the symmetry of the given power network (c) From Eq. (9.29) Iri = Y,j(vl - vl) For the power systemshownin Fig. 9.29 the pu reactancesare shown therein. Irtz=+j0(.01.'286 - 0.286=;g For a solid 3-phasefault on bus 3, calculate the following and I \\ t = t | t =f r r o . 1 8 6 - o ) (a) Fault current - - i2.86 (b) V\\ and vt (d) As perEq. (9.32) (c) It,r, I'\\, uxl Il, (d) 1fr, andIf, Assumeprefaulvt oltageto be I pu. | f f r0.2 0-.09 1 0.1 r, r r- E'or-vrf G ixtic+ ixr c) | iJ-I,-, F \\, // But E'Gr= 1 Pu (Prefaulnt o load) o '' \\ \\ \\... ,/'0., IGL .t=- 1 - 0 ' 2 8 6 - -= - j2'86 / io.2+io.os l , ,/ 3' SimilarlY 'l Fig. 9.29 Ifcz= i2.86 Solution The Thevenin passivenetwork for this systemis drawn in Fig. 9.28 PROBLEMS with its Zru, givenin Eq. (iv) of Example9.8. (a) As per Eq. (9.25) Jt - V: 9 . 1 Aeshxtporarretncssisrmciouinsiftsoeirodsanlihtnoter=ot c0f iirncadutuittchcteaubnrraceer0net.1ni(Hdr)a.asFnsicnhlrdoewaspinspitnraonFxciimeg5'oaPthe-m9ly'st1hi'seWsvuraidtleudeethnoelfy 2,, + Zl or 1f - zYno - :j.0_.117-5-- - j s . j l the first current maximum (maximum momentary current)' sametime as t[hHeinfitr:stAcsusurrmenetthmaatxtihmeufmirstocf uthrreensytnmtamxgimtriucml.:noocncursttarct uthite current') (b) As per Eq. (9.26) rv r. f- vl' - --Zt--yu i - zrr+zl'r

564,'l rr]logerpnower SystemAnalvsis errrnrnorrinal F' | a- - \"r r l t A n- \" - a- l v s i s I geS I vyrrrrlvt\"vE' J I 'i 0.1H v^S/Ov\\ 64-A-' Ii fl u= 1 OsOi n(100n f + 15.) Fig. p-9.1 Fig.P-9.5 9'2 (a) What should the instantof short circuit be in Fig. p-9.1 so that the 9 . 6 A.f\"ausc\"ty'otnur.cnCh\".arliocsunlsoautuepstpfgrty\"eiinnng'eitairuapi tsaoysmfsrimaveteeltordiac5da0lor0mf k4sV0cA0ur,kr4eWn4t0faovtr,00a..8tlhplraeugesg-upinhbgat rspaeofnawuselitle n t DC off-set current is zero? at generatorterminals. (b) what shouldthe instantof short circuit be in Fig. p-9.1 so rharthe 9 . 7 A generator-transformerunit is connectedto a line through a circuit DC off-set currentis maximum? breaker.The unit ratings are: 9 . 3 For the systemof Fig. 9.8 (Example9.2) find the symmetricalcurrentsto Generator: 10 MVA, 6.6kV; X,,d= 0.1pu, Xi= o,2o pu and be interruptedby circuit breakersA and B for a fault ar (i) p and (ii) e. X,r= 0'80 Ptt 9 . 4 For the system in Fig. p-9.4 the ratings of the various componenrsare: Generator: 25 MVA, 12.4 kV, l\\Vo subtransienrteactance Motor: 20 MVA, 3.8 kV, l5%osubtransienrteactance Transformer:10MVA,6.9133kV,reactance0.08pu TransformerT,: 25 MVA, 11/33 ky, gVoreactance The systemis operatingno load at a line voltageof 30 kv, when a three- phasetault occurson ih\" tin\" just beyondthe circuit breaker'Find Transfbrmer Tr: 20 MVA, 33/3.3 kV, I \\Vo reactance (a) the initial symmetrical rms current in the breaker' Line: 20 ohmsreactance iui tt. maximum possibleDC off-set currenrin rhe breaker, The systemis loadedso thatthe motor is drawing 15 Mw at 0.9 loading (c) the momentarycurrent rating of the breaker' kVA' power factor, the motor terminal voltage being 3.1 kv. Find the (d) the current,o U. intemrpted by the breakerandthe intemrpting subtransientcurrent in generatorand motor for a fault at generatorbus. lHint: Assumea suitablevoltagebasefbr thegeneratorT. he voltagebase and for transformers,line and motor would then be given by the transforma_ (e) the sustainedshort circuit currentin the breaker' ti 0 n ri l ti < l sF. rl r c x a n rp l ci,f' w c chooscgcrrcratovro. l tagcbascas I I kv, The systemshown in Fig' P-9'8 is delivering50 MVA at 1 1 k v , 0 . 8 the line voltagebaseis 33 kv and motor voltagebaseis 3.3 kv. per unit 9.8 l a g g i r r g p 0 w 0 r l . i t c t t t r i l l t t l l t b r r s w h i c h r t r i r y b e r e g iilrsr dinefcinlite. reactanceas re calculatedaccordingly.] Particularsof varioussystemcomponentsare: 60 MVA, 12 kV' X,/= 0'35 Pu Generator; Transtbrmers(each): 80 MVA, 12166kV' X = 0'08 pu Line: Reactance12 ohms' resistanc:negligible' T, 1 T 2 Calculatethe syrnmetrica lcurrentthat the circuit breakersA and B will be calledupon to interrupt in the event of a three-phasefault occurring at Fig. p-9.4 F nearthe circuit breaker B' 9 . 5 Two synchronousmotors are connectedto the bus of a large system Line through a short transmissionline as shown in Fig. p-9.5. The ratings of vanouscomponentsare: Motors (each): 1 MVA, 440 v,0.1 pu transientreactance Line: 0.05 ohm reacrance Largesystem: Shortcircuit MVA at its bus at 440 V is g. Fig. P-9.8 when the motors are operatingat 440 v, calculate the short circuit cuffent (symmetrical)fed into a three-phasefault at motor bus.

*Sf-,S tgdern power Svstem Anatvsis t . - 9'9 lPAa-r9tgw'e9os' Aygsdetdneiemtiroawnthoailrcsphtoamwtiaceylrnsbiusepfrpeeldgieatsoradtefh'deeaebsduiensrtfhtinhriortoeuJ.ggAhhaarbetuarscatanossrfsXohiloinsweinrnfcirnloumFdiegad. - :,'jT:^\"1^:t::1u1.',f\"'Taenrd,1\"-!T to limit theSCruprurincgapacitoyf bus 2 of the system.Buses 1 and 2 areconnectedthrough a transformer and a transmissionline. Per unit reactancesof the various componentsare: Generator(connectedto bus bar 1) 0.25 inductive 4v<rl\\er rr ru JJJ rvtyA (Iault close to breaker). Find the reactanceof the reactor required. system data are: Transmissionline 0.28 GeneratorGr: 25 MVA, 757oreactance The power network can be representedbi a geneator with a reactance (unknown) in series. Generator Gr: 50 MVA, 20Voreactance TransformerTr: 100MVA;8Vo reacrance With the generatoron no load and with 1.0 pu voltage at each bus under operatingcondition, a three-phaseshort circuit occurring on bus 1 Transformer T2: 40 MVA; l\\Vo reactance. causesa current of 5.0 pu to flow into the fault. Determinethe equivalent reactanceof the power network. Assumethat all reactancesaregiven on appropriatevoltagebases.choose a baseof 100 MVA. ti Tz 9.12Considerthe 3-bussystemof Fig. P-9.I2. The generatorsare 100 MVA, with transientreactanceIjVo each.Both the transforners are 100 MVA with a leakagereactanceof 5%t.The reactanceof eachof the lines to a baseof 100 MVA, 110KV is 707o.Obtain the shortcircuit solutionfor a three-phasesolid shortcircuit on bus 3. Assume prefault voltagesto be 1 pu and prefault currents to be zero. Fig. p_9.9 G) T1 1 1 /1 1 0k V f l rn kv\\-x^-Tx-\"x-T\"'2 9'10 For the three-phasepower network shown in Fig. p-9.10, the ratings of the various componentsare: j0.1 Fig.p-9.10 Fault Generators Gr: 100 MVA, 0.30 pu reactance Fig.P-9.12 Gz: 60 MVA, 0.18 pu reactance 9.13 In the systemconfigurationof Fig. P-9.12,thesystemimpedancedataare given below: Transformers(each):50 MVA, 0.10 pu reactance Transient reactanceof each generator= 0.15 pu Leakage reactanceof each transformer= 0.05 pu InductivereactorX: 0.20 pu on a base of 100 MVA Ztz = i0.1, zp - i0.12, 223= 70.08Pu For a solid 3-phasefault on bus 3, find all bus voltagesand sc currents Lines (each):80 ohms (reactive);neglectresistance. in eachcomponent. with the network initiaily unroadedand a rine v o r t a g eo f 110 kv, a symmetrical short circuit occurs at mid point F of rine r-. tbhreeareckaaelcrcstuAolraXatnewdtehBreeaest hltihmoeritencnaidtrecsduo?iftCtMhoemvAlmineent.otw. bheatinwteomulrdpttheedsbeyvatlhueescbirec,uiift

365 j ModernPowerSystemAnalysis t 9.14 For the fault (solid)locationshownin Fig. P-9.I4, find the sc currentsin ll lines 1.2 and 1.3.Prefaultsystemis on no-loadwith 1 pu voltageand I prefault currentsarezero.Use Zuu, method and computeits elementsby the currentinjection technique. 'r-[b- 111110kV 0.5pu reactance -l 0.1pureactance Q9 Fig. P-9.14 10.1 INTRODUCTION REFERENCES In our work so far, we have considered both normal and abnormal (short circuit) operettionsof power systern Llnclercornpletely balanced (symmetrical) Books c o l t t l i t i o t t s .L J l r r l c rs u c l t o l t c t ' a t i o t t l t c s y s t c r l ri r u l t c d a n c c si n c a c l r p h a s c a r e l . B I r r w t t . H . L ) . . S o l t t t i o t rt t . fI . a r , g , eN e t w r t r k l t y M u t r i r M c t l u t d s , W i l c y , N c w Y o r k , identical and the thrce-phasevoltages and currents throughout the system are c o l l t p l c t c l y h l t l i t n c c t l .i . c . t h c y h u v c c r p r i r l n u r g r r i t u t l c si n c l r c h p h u s c u n d t r c t L ) ' 75 . prtrgrcs.sivclydisplaced in tirne phase by 120' (phase u leads/laesphase b by 1 2 0 ' a n d p h a s e i r l e a d s / l l g sp h i t s ec b y 1 2 0 \" ) . I n a b : r l l n c e d s y s t e n r ,l n a l y s i s 2 . f . l c t r c t t s w a r t d uj'.,i t . , Mtttit:t'tti'tnv'u' 'i'cxtbook eorrrplny, can proceed on a single-phasebasis. The knowledge of voltage and current in 5 ' v s ' / e l l , l ,i r r t c r r t a t i t t r t a j one phase is sufficient to completely determine voltages and currents in the other two phases. Real and reetctive powers are simply three times the New York, 1971. corresponding per phase values. '3. Stagg, G.W. and A.H. El-Abiad. Computer Methods in Power SystemsAnalysis, Unbalanced system operation can result in an otherwise balanced system due McCraw-Hill Book Co., Ncw York, 1968. to unsymmetrical fault, e.g. line-to-ground fault or line-to-line fault. These f'aultsatrc,itt lhct, ol' lnot'econlnroll occurrence+thalt the syrnntetric:al(three- 4. Anderson, P.M., Analysis of Faulted Power Systems,Iowa State Press,Ames, Iowa, phase)fault. Systern operation may also become unbalancedwhen loads are unbalanced as in the presenceof lar-eesingle-phaseloads. Analysis under 1973. unbalancetl conditions has to be carried out on a three-phase basis. Alterna- tively, a nlore convenient method of analyzing unbalanced operation is through - r . ( ' f r r t ' k c .l - , . . ( ' i n ' t t i t A r r t t l . t , , to' i.,ft' A l t t r n u l i t t , qC t r r r c r t l l ' t t w r r S \\ , . r ' / r , r l .Vr .o l . I , symtnetrical components where the three-phasevoltages (and currents) which may be unbalanced are transfbrmed into three sets of balanced voltages (and New York. 1943. 6 . S t c v c r r s o n ,W . D . J r . , E l e t r r c n t .osf P o w c r S y s t e m sA n u l y - s i s , 4 t hc d n , M c ( i r z Ncw York, l9tl2. Paper \"7. Brown. H.E. et al. \"Digital Calculation o[ Three-Phase Short Circuits by Matrix Methods\", AIEE Trani., 1960, 79 : 1277. * Typical relative frequenciesof occurrenceof differentkinds of faults in a power syst(:ll(tirr ordcr ol' dccrcasingscvcrity)are: Three-phase(3L) faults 5Vo Double line-to-ground(LLG) faults lj%o I)outrlc lirrc(l-1,)lirults | 5t/o Single line-to-ground(LG) faults 7jVo

symmetricac!ompo1gnlg ! .a?.,{fi-i currents)calledsymmetricalcomponents.Fortunately,in sucha transformation aboveT. hus the impedancespresentedby various power system elements (synchronous generatorst,ransformersl,ines) to symmetricalcomponentsaredecoupledfrom Vo-- Vot * VozI Voo (10.s) each other resulting in independentsystem networks for each component b- YbL- vbz ancedset).This is the basic reasonfor the simolicitv of the svmmetrical componentmethodof analysis. Vr= VrI * Vrz * Vro (r0.7) TO.2 SYMMETRICAL COMPONENT TRANSFORMATION The three phasor sequences(positive, negative and zero) are called the symmetricacl omponentsof the original phasorset Vo,V6,V,. The addition of A set of three balancedvoltages(phasors)Vo, V6, V\" is charactertzedby equal symmetricalcomponentsas per Eqs. (10.5) to (10.7) to generateVo, Vr, V, is magnitudesand interphasedifferencesof 120'. The set is saidto have a phase sequenceabc (positivesequence)if VulagsVoby l2O\" and V. lags Vuby I20\". indicatedby the phasordiagramof Fig. 10.1. The threephasorscan^thenbe expressedin terms of the referencephasor Voas Vo = Vo, V6= a\"Va, V, = aVo where the complex number operator cr is defined as sL - air20\" It has the following properties ,*2:ei24o':e-ilAo\" _* (o')* : o V61=crV61 V6fo?V11 a3:l l+ala2:0 (10.1) If the phasesequenceis acb (.negativesequence),then V o = V o ,V u = tu Vo ,Vr= & V o Thus a set of balancedphasorsis fu'lly characterizedby its referencephasor (say V,) and its phasesequence(positive or negative). Suffix 1 is commonlyusedto indicatepositive sequence.A setof (balanced) positive sequencephasorsis written as Vo1, V61- &Vu1, Vrr = aVot (r0.2) Similarly, suffix 2 is used to indicate negativesequence.A set of (balanced) Fig. 10.1 Graphicaal dditionof the symmetricaclomponenttso obtain theset of phasorsV\", V6,V\"(unbalanceidn general) negative sequencephasorsis written as Vo2, V62= dVn2, Vrz= Q'Voz (10.3) A set of three voltages(phasors)equalin magnitudeand havingthe samephase Let us now expressEqs. (10.5)to (10.7)in termsof referencephasors Voy is said to have zero sequence.Thus a set of zero sequencephasorsis written Vo2and Voe.Thus AS Vo= Vot* Vozl Voo (10.8) Vng, V6g = Vo1, Vr1 = Vo1 (10.4) Vu= a.2Vor+ aVor* Voo (10.e) Considernow a setof three voltages(phasors)Vo,V6, V, which in generalmay V\"= c-Vol+ o2vor* Voo (10.10) be unbalanced.According to Fortesque'stheorem* the three phasorscan be These equations can be expressed in the matrix form * The theorem is a generalone and appliesto the case of n phasors[6],

-ff_i#_d r_-: ModernPo*et sytttt Analysi, Io = il' (10.19) (10.20) (10.11) and where I' = A-rI' Vp = AV, (10.12) Lll I'Iu', l)l ; nIa, =[l Iro,,I'I Lr,o_ I r = a 1,\"1 i = v, l'u | = vectorof originalPhasors Of course A and A-r arethe same as given earlier. In expandedform the relations (10.19) and (10.20) can be expressedas Lv), follows: Iu'l (i) Constructionof current phasorsfrom their symmetricalcomponents: % = | Voz| = vector of symmetrical components Io= Iot * Ioz * Ioo (10.21) Lu\"'.1 Ia= o2lott dozr loo (r0.22) I r-l Ir= dot+ azlor* Ioo (r0.23) (ii) Obtaining symmetrical components of current phasors: A - 1[ '\" o nt -Il (10.13) I o t = 1 e\"+ du+ o2lr) (ro.24) Ia (10.14) o2 t * We can wrire Eq. (10.12)as ; (Io+ azlu+ aI,) (ro.2s) t,r.= i Qo+Iu+ Ir) V, = 4-'V, . ; (r0.26) ComputinE ,{r and utilizing relations (10.1), we get Iro=, [t d o'] Certain observationscan now be maderegardinga three-phasesystem with neutral return as shown in Fig. 10.2. o - ' = *' lLr t c r 2a I (r0.1s) r r l In expandedform we can write Eq. (10.14)as V o r= I c - V u +a 2 V S (10.16) Vao V\"\" ,<V\"+ voz= !3 fr\"+ ozvu+o%) (10.17) Vuo= I V,+ Vr,* r,) (10.18) ; Equations(10.16) to (10.18) give the necessaryrelationshipsfor obtaining ln symmetricalcomponentsof the original phasors,while Eqs. (10.5) to (10.7) Fig. 10.2 Three-phasseystemwith neutrarl eturn give the relationshipsfor obtaining original phasors frorn the symmetrical components. The symmetricalcomponent transformationsthough given abovein terms of The sum of the three line voltages will always be zero.Therefore, the zero sequencecomponentof line voltages is always zero, i.e. voltageshold for any set of phasorsandthereforeautomaticallyapply for a set of currents.Thus v o b=o trr** vu,+v\"o=) o (10.27)

On the otherhand,the sum of phasevoltages(line to neutral)'may not be zero _\\ so that theirzero sequencecomponent Vn, may exist. Since the sum of the three line currentsequalsthe current in the neutral wire. we have Ioo=!U\"+16+r\")=!+ (10.28) B --_____}_---- i.e. the currentin the neutralis three times the zercsequenceline current.If the Flg.10.3 neutral connectionis severed. Ioo=lr,=o (ro.2e) Solution Io + I\" * Is = Q 3 or 10130\" + l5l- 60o+ Ic = O t.e. in the absence of a neutral connection the zero sequence line current is I c= - 16. 2+ j8. 0 = 18 1154\" A always zero. Power Invariance From Eqs.(10.24t)o (10.26) Itr= 1 + 751(-60\"+ l2O\") + I8l(154\" + 240\")) We shall now show that the symmetrical componenttransformationis power :0Ol3O\" 5 invariant, which means that the sum of powers of the three symmetrical = 10.35+ j9.3 = 14142\"A (i) componentsequalsthe three-phasepower. I e z = I + 151(- 60o+ 240\")+ I8l(154' + 120\")) Total complex power in a three-phasecircuit is given by ;Q 0 1 3 0 ' J S = f p $ = 4 1 , + V u t t +V , t ! (10.30) = --1.7- j4.3 = 4.651248\"A (ii) or Iao= tI^ (lo + IR + /c) = 0 (iii) s = [Av,,]'t'lAl,l. J = v! 'qr'qtI. (10.31) FromEq. (10.2) Now Im= 141282\"A Icr = 141162A\" Inz= 4'6518\"A Icz = 4.651128\"A [r o? 'Tt r rl l-1 0 0l Iao=oA Ico=oA o16.=lta \"ItJl la n' .rl:r|o r ol-t, (10.32) Check: [r r La , c-llL00 ll Ia= Iat * I,rz* I,to= 8'65 + j5 = 10130\" .. s=31yti,=341., Convertingdeltaload into equivalentstar,we can redrawFig. 10.3as in Fig. 10.4. = 3V^1, + 3V\"r!),+ 3V\"oI), (10.33) la = sumof symmetricaclomponenptowers I-TE x a m p l e1 0 . 1 | A delta connectedbalancedresistive load is connectedacrossan unbalanced three-phasesupply as shown in Fig. 10.3. With currents in lines A and B specified, find the symmetrical componentsof line currents. Also find the symmetrical componentsof delta currents.Do you notice any relationship betweerisymmetricalcomponentsof line anddelta currents? Comment.

$,$i.M Modernpower system anatysis Qrrmmatriaal l-amnnnanla tffiffi I Delta currentsareobtainedas follows veB=Io U^- Ir) Positiveand negativesequencevoltagesand currents undergoa phaseshift in passingthrough a star-delta transformer which depends upon the labelling of I t B =z A B )-R tUo- ru) terminals.Before considering this phaseshift, we need to discussthe standard polarity marking of a single-phasetransformer as shown in Fig. 10.5. The Similarly, transformerendsmarked with a dot have the samepolarity. Therefore,voltage Vun, is in phasewith voltage V..,. Assuming that the small amount of r n c =I e r - I r ) magnetizingcurrentcan be neglected,the prirnary current 1r, enteringthe dotted 5 end cancelsthe demagnetizing ampere-turnsof the secondarycurrent 1, so that I, and12with directionsof flow asindicatedin the diagram arein phase.If the directionof 1, is reversed,1, and 1, will be in phase opposition. Ice=!frr- Io) 5 Substitutingthe valuesof Io, Iu and Ir, we have Ien= ! Oozzo-\"r5l- 60\")= 6186 A rnc= 60\"- rllr54\")= lO.5l- 41.5\"A lOsz- rcn=!W2154\" - rol3o\")= 8.31173A\" Flg. 10.5 Polaritymarkingof a single-phasetransformer\\ The symmetricalcomponentsof delta currentsare Considernow a star/delta transformerwith terminal labelling asindicated in Fig. 10.6(a). Windings shown parallelto eachother are magneticallycoupled. 1 l 2O\" )+ 8.31(173\"+ 240\" ))(i v) Assume that the transformer is excited with positive sequencevoltages and carries positive sequencecurrents. With the polarity marks shown, we can Iem= ;(6186\" + IO.5l(- 4I.5'+ immediatelydraw the phasordiagramof Fig. 10.7.The following interrelation- J ship betweenthe voltages on the two sidesof the transformeris immediately = 8172\" A observedfrom the phasor diagram I + 1051(- 41.5\"+ 240\")+ 8.31(173\"+ 120\"))(v) VtBt= x Vabrl3V, -r - phasetransformationratio (10.34) I , q a z =: 6 1 8 6 \" As per Eq. (10.34),the positive sequenceline voltages on star sidelead the J correspondingvoltageson the deltasideby 30\" (The sameresult wo,'ld apply to line-to-neutralvoltages on the two sides).The same also appliesfor line = 2 . 7 1 2 1 8A\" currents. Ieno=0 (vi) If the delta side is connectedasin Fig. 10.6(b)the phaseshrft reverses(the readershould draw the phasor diagram);the delta side quantitieslead the star Incr, Ircz, IBC,,lsn1,Iga2and 1.oocan be found by using Eq. (10.2). side quantitiesby 30\". Comparing Eqs. (i) and (iv), and (ii) and (v), the following relationship between symmetricalcomponentsof line and delta currentsare immediately observed: t' (vii) I e B r =+ 1 3 0 \" VJ Ienz= \\ z-zo\" (viii) !J Eqs. (vii) and The reader shouldverify theseby calculatrng Io', andl*2from (viii) and comparingthe resultswith Eqs. (iv) and (v).

ar SystemAnalys'is SymmetricaCt omponentg | 3?9 I l-A __--______ oa correspotxding positive sequence quantities on the LV side by 30'. The reverse /\\ is the case for negative sequence quantities wherein HV quantities lag the corresponding LV quantities by 30\". vsc2 (a) Star side quantitieslead delta side quantitiesby 30o (b) Delta side quantitieslead star side quantitiesby 30o vcea Vcrcz Vesz Fig. 10.6 Labellingof star/deltatransformer Fig. 10.8 Negativesequencevoltageson a star/deltatransformer Vec'l IO.4 SEOUENCE IMPEDANCES OF TRANSMISSION LINES Fig. 10.7 Positivesequencevoltageson a star/deltatransformer Figure 10.9 shows the circuit of a fully transposedline carrying unUalancecl Instead, if the transformerof Fig. 10.6(a) is now excited by negative currents. The return path for 1,,is sufficiently away for the mutual effect to be sequencevoltages and currents,the voltage phasor diagram will be as in Fig. ignored.Let 10.8.The phaseshift in comparisonto the positive sequencecasenow reverses, i.e., the star side quantitieslag the delta side quantitiesby 30'. The result for X\" = sell'reactancoe1'eachline Fig. 10.6(b)also correspondinglryeverses. It shall from now onwardsbe assumedthat a star/delta transformer is so X. = mutual reactanceof any line pair labelled that the po,sitivesequencequantities on the HV side lead their The fbllowing KVL equationscan be written down from Fig. 10.9. Vo- V'o= jXJo + jX*Iu + .ixmlc V6 vc lr= l\"+ lo+1\" -(- F i g .1 0 . 9

Modern PgwgfSy$gln Snalysis Svmmetrienl cnmnonent( tflIlFffi 4 - r[: jxJo + jxh +jx*[\" (10.3s) T l', - V!: iXJ\" +jxmlb + jxJ\" t, Z1 12 22 ls Zs or in rnatrix form o-------{__]- -o ---o o---)--f---_l.- \"----f--}- ^a (10.36) (a) Positivesequence (b) Nagativesequence @)Zero sequence network network network vb vI = J x^x,x. Ib v\" vi x_x*x, I\" --zIoV,r /I / / - (10.37) n - (10.38) (10.3e) F i g .1 0 . 1 0 or A (I/, - v ! ) : zuIs The decouplingbetween sequencenetworksof a fully transposedtransmis- or sion holds also in 3-phase synchronousmachines and 3-phasetransformers. Now v, - ,r/'/s-- A-IZAI, This fact leads to considerable simplications in the use of symmetrical componentrsnethodin unsymmetricaflaultanalysis. jx, jx^ In case of three static unbalancedimpedances,coupling appearsbetween A-I ZA : JX^ jx, (10.40) sequencenetworksand the method is no more helpful than a straight forward 3 - p h a s ou n a l y s i s . jx^ jx^ (lo.4l) : Jl*,-', 0 (r0.42) 10.5 SEQUENCEIMPEDANCES AND SEOUENCENETWORK L: X,-X* (10.43) OF POWER SYSTEM 0 Power system slernsnfs-transmission lines, transformers and.synchronous nlachines-have a three-phasesymmetrybecauseof which when'currentsof a Thus Eq. (10.37)can be written as particularsequenceare passedthrough these elements,voltage drops of the same sequenceappear, i.e. the elementspossessonly self impedancost-o | 41 l'tr(I Ix \"-x ^ 0 o ll-rI, sequencecurrcnts. Each eleurent can therelbre be representedby tlree decoupledsequencenetworks (on single-phasebasis) pertaining to positive, l \" l - l 'lt: ' I o x,- x,,, o llr, | negativeandzerosequencesr,espectivelyE. MFs areinvolvedonly in a positive sequencenetwork of synchronousmachines.For finding a particular sequence L r r o J l r r t JL o o x,*zx^)j,l impedancet,he elementin questionis subjectedto currentsand voltagesof that sequencoenly. With the elementoperatingundertheseconditionst,he sequence ll,,Iz t 0 0l[ /' .l impedancecan be determinedanalyticallyor through experimentaltest results. 0220 With the knowledge of sequencenetworks of elements,complete positivel' 0 0 z olLri, negativeand zero sequencenetworksof any power systerncan be assembled. As will be explained in the next chapter. these networks are suitably wherein interconnectedto simulate different unsymmetrical faults. The sequence currentsand voltagesduring the fault are then calculatedfrom which actual Zr: j(X, - X,r) : positive sequence impedance far\"rltcurrentsand voltases can be found. Zz: j(X, - X^) : negative sequence impedance (r0.44) Zo: i(X, + 2X) : zero ,tequence impedance (10.4s) We conclude that a fully transposed transmission has: /t;\\ voy^ur ar or l ^p^v ror.l ri +y ui \" o (^t-rll r l -u^t /^t^s +4 Li r v g s t r q u { r r r u E- - ^ : r^r-u^p)€^u- ^a^l l-u t i s . \\r,, (ii) zero sequenceimpedancemuch larger than the positive (or negative) 10.6 sEQuENCg I;IpTDANCES AND NETWoRKs oF SYNCHRONOUS MACHINE sequenceimpedance(it is approximately 2.5 times). It is further observedthat the sequencecircuit equations (10.42) are in Figure10.11depictsan unloadedsynchronourslachine(generatoor r motor) groundedthrougha reactor (impedanceZ).8o, E6andE, arethe inducedemfs decoupleclfbnn, i.c. thcrrearc no rnutual scqucnccinducternccsF. )quation (10.42) can be representedin network form as in Fig. 10.10.

--3]-S2 I _ ModernpowerSystemAnatysis Symmetrigel_qg4pe!e_$s of the three phases.when a fault (not shown in the figure) takes placeat theshortcircuit occurslirrnr kradcccl onditiortst,hc voltagt:behindappropriltc machineterminalsc, urrents1,,,I,,and /. flow in the lines.whe neverthefault reactance(subtransient,transient or synchr:onousc) onstitutesthe positive involvesground,currentIn= In+ Iu + ^I\"flows to ne'tral from ground \"i^I^'. uence voltage. Figure lO.IZa shows the three-phasepositive sequencenetwork ntodel of a Hrvvvvs yyrrrr r(rLrll 4rr.1IyJIs (\\_napler i l), we must synchronousmachine. Z, doesnot appearin the model as Iu = 0 for positive sequencecurrents.Sinceit is a balancednetwork it can be representedby the know the equivalentcircuits presentecbl y the *u.hin\" to the flow of positive, single-phasenetwork model of Fig. 10.12b for purposesof analysis.The negative and zero sequence culrents, respectively. Because of winding referencebus for a positive sequencenetwork is at neutral potential. Further, osseyf mqvuamerienotcureyossnceluyqr.ruTeehnnetscreeofsfo*ra.e,pthaertrieciuslaarnsoecqouuepnlicnegpbroedtwueceenvthoeltaegqeuidvarolepnstcoifrctuhiatst sinceno current flows from groundto neutral,the neutralis at groundpotential. la lat >a -* R e f e r e n c eb u s )e\" t6 ln't . n'- - - - - --> -----b (\\=\\'t-+.. t:.6 d--nu>_ \\ L----- ( ,b ____l_\" I I I lc't Fig' 10'11 Three-phasseynchronougseneratowr ithgroundedneutral ( a ) T h r e e - p h a s em o d e l ( b ) S i n g l e - p h a s em o d e l Positive Sequence Impedance and Network Fig. 10.12 Positivesequencenetworkof synchronoums achine 'aneiSsnrtmyiomcnndflcascteerothoaucfrfrseopeioyndnorpneosiectauirhs'tcsaiWprvttoieeioonhsennocediunisqfnsimutethhtelahdcencetcbhtsceaitoanallntauelcnlinsysleetc,cciedinll.iedecbrmse.uyniocrgortcnpiilonceoendsedswapiigsstiotiacvthshtuerisestviyses,vmooeectrdsqmuacuz.ieq,leeetiunrr.nieocccgs.eant,ehlcicwqtuciiuinrnsuerdersCnirnntc.hatcgesavtnsipoor,ttolginsetttalarailngy9rtydee,.wusTstaichhtarehieetss With referenceto Fig. 10.12b,the positive sequencevoltageof terminal c with respect to the referencebus is given by V,,l= E,,- Zll,,l (10.4e) Negative Sequence Impedance and Network respectto field excitation. The machine equivalently offers a direct axis It hasalreadybeensaidthata synchronousrnachinehaszeronegativesequence reactancewhose value reducesfrom subtransicntreactance(X,a) to transient inducedvoltages.With the flow of negative sequencecurrentsin the stator a rotating field is createdwhich rotates in the oppositedirection to that of the nsreheaogcrlittgacinibrccleeu(,Xtithtertr)paaonnssdiitefiivnnetapsllryeoqgtuoreessntsceeeaismdinypsettidamateen.(csIeyfonfacrthhmreoanmtuoaruecshr)reiensaeiscisttaanncJeei(sXaa)s,sausmtheed positivesequencefield and,therefore,at doublesynchronousspeedwith respect to rotor. Currents at doublethe statorfrequencyare thereforeinduced in rotor 21= jXtj (if I cycle transientis of interest) (10.46) field and damper winding. In sweeping over the rotor surface,the negative sequencemmf is alternatelypresentedwith reluctancesof direct andquadrature = jX'a Gf 3-4 cycle transientis of interest) (10.47) axes. The negative sequenceimpedance presentedby the machine with considerationgiven to the damper windings, is often defined as = jXa (if steadystatevalue is of interest) (10.48) xt: + x,! (10.s0) terImf itnhaeivmolatacgheincoensshtoitrut tceisrtchueipt otsaikteivsepselaqcueefnrcoemvolutangloea;odnedthceoontdhietirohnasn, dth.ife Z . t =j 2 ;lZ2l<lZrl *'fhis can be shown to be so by synchronousmachine theory,[5]. Negativesequencenetwork modelsof a synchronousmachine,on a three- phaseand single-phasbeasisareshownin Figs. 10.13aandb, respectivelyT. he referencebus is of course at neutral potential which is the same as ground potential.