E4 | ModernPowerSYstemAnalYsis ubstituting the valuesof different chargesand simplifying' we get conducting horizontal shdet of infinite extent which therefore acts like an 'l rn 2hD (3.2s) equipotentiasl urface. Vob= rk r(4hz + D2)'tz and unit is such that it has a zero potential plane midway betweenthe conductorsas Radiusr shown in Fig. 3.8. If a conductingsheetof infinite dimensionsis placed at the zero potential plane, the electric field remains undisturbed.Further, if the conductor carrying charge -q is now removed, the electric field above the conducting sheetstaysintact, while that below it vanishes.Using these well known resultsin reverse,we may equivalentlyreplacethe presenceof ground below a chargedconductorby a fictitious conductorhavingequaland opposite chargeandlocatedasfar below the surfaceof groundastheoverheadconductor above it-such a fictitious conductoris the mirror image of the overhead conductor.This methodof creatingthe sameelectric field asin the presenceof earth is known as the method of imagesoriginally suggestedby Lord Kelvin. lt-- ,/'7 h I I I lmage charge Zero potential Fig'3.9Sing|e-phasetransmissionIinewithimages plane(ground) It immediatelY follows that irk F/m line-to-line (3.26a) wab - h\\, and I Cn= - 2nk F/m to neutral (3.26b) D t__ ln F i g . 3 . 8 Electricfield of two long, parallel,oppositelychargedconductors 41+tO't4h21stt2 Capacitance of a Single'Phase Line Considera single-phasleine shownin Fig. 3.9.lt is requiredto calculateits capacitancetaking the presenceof earthinto acoountby the methodof images describedabove.The equationfor the voltage drop Vo6as determinedby the two chargedconductorsa andb, , andtheir images a'and b' canbe written as follows: vub= *1, \" m2 +nrrin* e,,t,\"gt#Y *q,,tnGFfUl (3.24)
86 ,*| M o d e r np o w e r S v s t e mA n a l v s i s Capacitaneeof TransmissionLines images.with conductora in position1, b in position2, and c in position 3, The equation for the averagevalue of the phasor %. ir found in a similar manner.Proceedingon the lines of Sec. 3.6 and using Vou* Vo, = 3Von and 1 ln r - ln h, Qo* Qt * Qc = 0, we ultimately obtained the following expression for the 2 7rl( capacitanceto neutral. (3.27) tn D'u - rr( rn\"n\"\\'),)!t\\ F/m to neutral (3.29a) Similar equationsfor Vo6can be written for the secondand third sectionsof the r ( (hrlhhs)''' ) transpositiocnycle.If the fairly occurateassumptionof constanct hargeper unit length of theconductorthroughoutthe transmissioncycle is made,the average C, 0.0242 pPttcmto neutral (3.29b) value of Voufor the three sectionsof the cycle is given by D'n - r- -oon@\"httu')'!t -l-oon r (4h24)tt3 (3.28) ComparingEqs. (3.22a) and(3.29a),it is evidentthat the effect of earth is to increase the capacitanceof a line. If the conductors are high above earttr comparedto the distancesamong them, the effect of earth on the capacitance of three-phaselines can be neglected. where D, = (DnDnDrr)\"t Calculate the capacitanceto neutrallkm of a single-phaseline composed of No. 2 single strandconductors(radius = 0.328 cm) spaced3 m apart and,7.5 m above the ground. compare the results obtainedby Eqs. (3.6), (3.7) and (3.26b). Solution (1) Neglecting the presenceof earth tEq. (3.6)l C^ , = 301.0f i2t4k2m n, log - =ffoi=.oo'i0+0z817 P'Flkm - 0.328 By the rigorousrelationship[(Eq. (3.7)] cn o.0242 \"'(+.(#-')\"') Since =915, the effect of non-uniformity of chargedistributionis almost + negligible. C\" = 0.00817 pFkm Flg. 3.10 Three-phaseline with images
rrl Modernpowel€ysteryr_Anslygls Capacitanceof TransmissionLin--e--s-------l | 89 _- uu I bca (2) Considerintgheeff'ectof earthandneglectingnon-uniformityof charge distribution[Eq.(3.26b)] 0.0242 r(l*( /4h\")) ,Jto4 300 - 897 d=8m- o32BJLo4 c\"'- o:o?? = o'0082P'Fkm 2.9s3 Note: The presenceof earth increasesthe capacitanceby approximately 3 partsin 800. Example 3.2 - h=6m A three-phase50 Hz transmissionline has flat horizontal spacingwith 3.5 m Fig.3.11 Cross-sectioonf a double-circutiht ree-phasleine betweenadjacentconductors.The conductorsare No. 2/0 hard-drawn seven Solution As in Sec.3.6, assumethat the chargeper conductoron eachphase is equal in all the three sectionsof the transpositioncycle. For section/ of the strandcopper(outsideconductor diameter= 1.05 cm). The voltage of the line transpositioncycle is 110 kV. Find the capacitanceto neutral and the charging current per V,,n(l)= z*l*(\"i*'';)+c,(r.n'\"; f) kilometreof line. D\"o= (3.5 x 3.5 x '7)t't= 4.4 m Solution 0.0242 * n . ( ,jn+ r \" ; ) ] (3.30) tog(440/0.525) For sectionII of the transpositioncycle nv \" - l _ -1 0 6 u,Cn 314.o.oos% v\"bG=r),lAln.(\"i* r\"#)+c,(rn;.'\" *) = 0.384 x 106O/km to neutral chargingcunen-t +X-n (l 19l]e) x l-000 +a,(rnj.,\":)] (33r) 0 . 3 8 4 x1 0 6 For sectionIII of the transpositioncycle = 0.1I Aftm vt(urr=r)*lr. ('\"j .^ i)+a(,rIn. ^I) The six conductorsof a double-circuitthree-phaseline having an overall radius +a.(rn;.r\"f)f,\"r, of 0.865 x 10-2m are arrangedas shown in Fig. 3.11. Find the capacitive reactanceto neutral and charging current per kilometre per conductor at Average value of Vo6over the transpositioncycle is given by 110k V, 5 0 H z . v,(*asv=) iL*[n(.t'ntrj+*r,^(###)]
90 | Modernpower SystemAnalysis =fi,^-,,)h(ffi)\"' (3.33) Capacitanceof TransmissionLines I gf l'. h=6m - i'B' jh ,'f 'd vou*vo,=3vonf=ien\"- Qt-q,)tn(ffi)''' (3.3s) 3von=*r\"(:#) (3.36) Fig.3.12 Capacitanceto neutral per conductor = 2dc 3.8 METHOD OF GMD (MODTFTED) Total capacitanceto neutral for two conductorsin parallel A comparison of various expressionsfor inductance and capacitance of transmissionlines [e.g.Eqs.(2.22b)and (3.6)] bringsour rhefacr that the rwo Cn= 4rk (3.37) are sirnilarexceptthat in inductancexpressionws e haveto usethe tictitious conductorradius rt = 0.7788r,while in the expressionsfor capacitanceactual Now h= 6 m; d = 8 m; -/ = 8 m. Referringto Fig. 3.I2, we can write conductorradius r is used.This fact suggeststhat the methodof GDM would be applicablein the calculationsfor capacitanceaswell providedit is modified r = f , .. ? - . , ,, 12 1' 1/ 21: J i m by using the outer conductorradiusfor finding D,, the self geometric mean distance. l f / ) + (o - h \\ \" L\\2) \\ 2 )) Exarnple3.3 can be convenientlysolvedas underby usingthe rnodifled GMD method. f = (j ' + h 2 )rt 2= l 0 m For the first section of the transpositioncycle mutual GMD is g = (72 + 42)rt2=J65 - Dub= ((ts) (ts))tta_ liglttz Db' = Qil''' Conductorradius(overall)= 0.865 x 10-2m Drr, = ( jh)''' Substitutingthe valuesfor variousdistancesw. e have cn 4 r x 7 x 8 . 8 5x 1 0 - 1 2x 1 0 6x 1 0 0 0p F / k m D (D\"pop,o)r,t = [(i,grjh)r,t)t,, \"n 1 nrf z * o s * s t o fr o o) 3 - l ' / 3 ln the first sectionof the transpositioncycle self GMD is [ 100x8 \\0.86s/ | D,o = ?f rf )rt4 = ?f)''' = 0.018I 1F/km Q C , = 3 1 4 x 0 ' 0 1 8 1x 1 0 - 6 D'l' = QAt'' = 5.68 x 10-6 Ulkm Dr, = (At'' D, = (D,oDroD,,)''3= l?3f Arl3fitz Chargingcurrent/phas=e x 5.6g x 10{ = 0.361 Now 2d( 2nk Cn= \"t#* Chargingcurrent/conductor= 0'361 L = 0.1805A/km 4 ltk F/m
Fa i , ' l t I This result obviously checkswith Example3.3. the fundamentallyderived expressionin LsL 3.9 BUNDTED CONDUCTORS PROIBEMS dcf+ebAaoiquvidrnnuibldydxdaueullneldacDydctafao1lcoermr2uerdsfraoopcainnrhotaetgannhpftsddeaoheurresasaca'bculhtlmooseaaunrlevl,nriadeanedluaneasccdpsc.itsaohiotcnarsisi.roTnhsgfghoeasetwoh.srfnTesehquisbnuomuuslFlnzetsiiddefgotla.ehbtcha3atahet.s1i;tnDsch3eihrem.draT>ci-rhlwghaeearitlrcrhoy/gton.hteAhneppedlssheuceoarchustbDareorurrurgnras'-erodniiflsspedatedqinoxqioyvnu,iDsaodatlnhe*rlyeese ::::f:i::t\"^tjl;l:;;:i;g;:;fiIH#,\",iHTf,f.i$l;:i,h*jcaahpaapsrlgireeindfegvroceultnracrgeeenpithsoabfsaoprrha.aAnsclleeacdt.ohNnredeeug-clpetohcratstshheae5,v0eefHftehzce.tTsoaafkmgeertorhauednvioio..lAtalsgoe of phase find the a[Q?I1O , , o QF -Iqe- -ui , Fd+ Fig.p-3.1 DP ---->f- \"6 iec/_____ 3'2 A three-phasdeouble-circuitline is shownin Fig. p-3.2.Thediameterof Dzs--------- each conductoris 2.0 cm. The line is transffio and carriesbalanced load' Find the capacitanceper phaseto neutral of l' Dy -- '..l the line. Fig' 3'13 Cross-sectionof a bundled conductorthree-phasetransmissionline TaQ Qc' Now, writing an equ'ationfor the voltage from conductor a to condu ctor b, 2m we get ,f o Q Qu, I I vob=*lot-n, ,(rn4.*ro 2m *o.S--q\\ \"lor\"-u*rn4. ) _t D, Dt,) or v,,h= r t/ , n+' .rr.oq n, .t,n gD. tnz\" t n D rr\\ (3.3e) 3'3 4ceAtFoqmnktuhdmiarluae'nRctedete-orapcrahrsaleaallycrrsseeuepgil5,anau0ctleetaHhdrezlt4yhsotemravamcenaraeshpphpeaaooacrrdsiittlzeTianodnhen.cetha.p\"aiu\"psiprrauken.iglIeotuumwlaniert.lhetyroetsrfuatocsncusnecpeshoussatiervaledicsnl owpenahidcseiunn0cgtt.hoo0erf1s ,*lr, D z t) Consideringthe line to be transposedandproceedingin the u s u a l m a n n e r ,t h e final resultwill be ',= ^ffi p,Fkmtnoeutral (3.40) 3'4 atFtcoabAhFchnaiueroniecdglettltdnesvwy'hresriaaPtpdcertihaldedh-heeue2iieeaacnneu'a5-sor9risogpnd0ten'ofhleFhsjdiafHetndtiauhancec,ez5sceendoate0twsntnrioctd0tuethidchrhtessrehoukeonncancvcftestrctd,iraeotcpnhauhpartarintcsaedaihlrs.tinecgcrsosspeicnhesrito7xeisetasitvsiro7.oo-crcTiiipg5ntanorhniheevolcneetai.aanhndxharunscaeuesatdcetbrdgcacssoauvittenuhurloeonpn,csolcrlrdoeettoswaitraimltrosfeehtngk,hebcdenie1inaoollce.okdcnr3ouFmmreihdgnztoifcrgeucfaodpaoe.tcmtunlrncbptietctno.tdhe_atoroueto3oraswffcr.hinpgln6teaimodnras.eGordtescionasuio/ivtnnknanhu0emtsdgeob.nsw2,snlsapaeD5eeihten-tci5ohccodd=m0iwiferff2Jcii.nHntnemAuhtrddziiietnn-t.. 3.5 whereDo, = (DnDnD3)In 3'6 It is obviousfrom Eq.e.aD that the methodof modified GMD is equally valid in this case(as it shouldbe).
94 i - ModernPo*\", Syrt\"r An\"lysi. t 666565 l'- o - +-- o-*_ D >)<--D 'l-- o-J . P-3.6 Doublecircuitthree-phaseline with flat spacing 3.7 .{ singleconductorpower cable has a conductorof No. 2 solid copper 4.I INTRODUCTION (radius = 0.328 cm). Paperinsulation separatingthe conductorfrom the concentric lead sheath has a thickness of 2.5 mm and a relative A complete diagram of a power system representingall the three phases permittivity of 3.8. The thicknessof the lead sheathis 2 mm. Find the becomestoo complicatedfor a systemof practicalsize,so much so thatit may capacitivereactanceper kilometre betweenthe inner conductor and the no longer convey the information it is intendedto convey. It is much more lead sheath. practicalto representa power systemby meansof simple symbblsfor each componentresultingin what is called a one-linediagram. 3 . 8 Find the capacitanceof phase to neutral per kilometre of a three-phase line having conductorsof 2 cm diameterplacedat the cornersof a triangle Per unit system leads to great simplification of three-phasenetworks with sides5 m, 6 m and 7 m respectively.Assumethat the line is fully involvingtransformersA. n impedancediagramdrawnon a per unit basisdoes transposedand carriesbalancedload. not requireideal transfbrrnersto be includedin it. 3 . 9 Derive an expressionfor the capacitanceper metre length betweentwo An importantelementof a powersystemis the synchronousmachine,which long parallel conductors,each of radius,r, with axes separatedby a greatlyinfluencesthe systembehaviourduringboth steadystateand transient distanceD, whereD ,, r, the insulating medium being air. Calculatethe conditions.The synchronousmachinemodelin steadystateis presentedin this maximum potentialdifferencepermissiblebetweenthe conductorsi,f the chapter.The transientmodel of the machinewill be presentedin Chapter9. electriclield strengthbetweenthem is not to exceed25 kY lcm, r being 0.3 cm andD = 35 cm. REFERNECES 4.2 Single-Phase Solution of Balanced Three-Phase Networks Books The solution of a three-phasenetwork under balancedconditionsis easily carriedout by solvingthe single-phasenetworkcorrespondingto the ref'erence l . Stevenson,w.D., Elementsof Power SystemAnalysis,4th edn, McGraw-Hill, New phase.Figure4.1 showsa simple,balancedthree-phasneetwork.The generator York, 1982. and load neutralsare thereforeat the samepotential,so that In = 0. Thus the neutralimpedance Zn does not affect network behaviour. For the reference 2 . Cotton, H., and H. Barber, The Transmission and Distribution of Electrical phase a Energy,3rd cdn, Hodder and Stoughton, 1970. En= (Zc+ ZL)I' (4.1) 3 . Starr, A.T., Generation, Transmission and Utilization of Electric Power, Pitman, 1962. Papers TA. Parton, J.E. and A. Wright, \"Electric StressesAssociated with Bundle Conduc- tors\", International Journal of ELectrical Engineering Education 1965,3 :357. 5 . Stevens, R.A. and D.M. German, \"The Capacitance and Inductance of Overhead Transmission Lines\", International Journal of Electrical Engineering Education, 1 9 6 5 , 2: 7 1 . The currentsand voltagesin the other phaseshave the samemagnitudebut are progressivelyshifted in phaseby 120\".Equation(4.1) conespondsto the single-phasenetwork of Fig. 4.2 whose solutioncompletelydeterminesthe solution of the three-phasenetwork.
Modernpower SystemAlelygis \\; Ia -l Representationof PgryerSystemComponents L_-l - .*- ,\", \\*ec, ) 't- If the transformeris YIA connectedas in-Fig. 4.4a, the delta side hasto be trb zn In=o lrJ'ZL replacedby an equivalentstar connectionas showndottedso as to obtainthe single-phaseequivalentof Fig. 4.4b. An important fact has, however, to be 16 Z1 AN m0 lrne culTent /A !: Ic have a certain phaseangle shift- from the star side values Vorand Io(90\" for Fig. 4.1 Balanced three-phasenetwork the phase labelling shown). In the single-phaseequivalent (Vew, I) are respectivelyin phasewith (Von,/o). Sinceboth phasevoltage and line current Z6 shift through the samephaseangle from starto delta side, the transformerper Ea phaseimpedanceandpower flow arepreservedin the single-phaseequivalent. In most analytical studies,we are merely interestedin the magnitudeof voltages Fig' 4.2 Single-phaseequivalentof a balancedthree-phasenetwork of Fig.4.1 and currents so that the single-phaseequivalentof Fig. 4.4b is an acceptable proposition. Wherever proper phaseanglesof currentsand voltagesare needed, correction can be easily applier after obtainingthe solution through a single- phasetransformerequivalent. (a) Y/A transformewr ithequivalensttarconnection €l> ->, A Ia Ia (a)Three-phaseY/y transformer n_ (b) Single-phaseequivalentof 3-phase y/y transformer (b) Single-phase equivalentof Y/A transformer Fig. 4.3 Fig.4.4 It may be noted here that irrespective of the type of connection, the transformation ratio of the single-phaseequivalentof a three-phasetransformer is the sameas the line-to-line transformationratio. ' See Section10.3.
tYroli I ModernPowerSystemAnalysis Representatioonf PowerSystemComponents 99 iI t 4.3 ONE.IINE DIAGRAM AND IMPEDANCE OR REACTANCE The impedancediagramon single-phasebasisfor useunder balancedoperating conditionscan be easilydrawnfrom the one-linediagram.For the systemof DIAGRAM Fie. 4.5 the impedancediagramis drarvnin Frg.4.6. Single-phasetranstbrmer equivalents are shown as ideal transformerswith transformer impedances A one-linediagramof a powersystem shows the main connectionsand showrrdependingon the informationrequiredin a systemstudy,e.g. circuit havebeenneglected.This is a fairly goodapproximationfor most power system breakersneednot be shownin a load flow study but are a must for a piotection studies.The generatorsarerepresentedasvoltagesourceswith seriesresistance study. Power system networksare representedby one-line diagramsusing andinductivereactanc(esynchronoums achinemodelwill be discussedin Sec. suitable symbols fbr generators,motors, transformersand loads. It is a 4.6). The transmissionline is representedby a zi-model(to be discussedin convenient practical way of network representationrather than drawing the Chapter5). Loadsareassumedto be passive(not involving rotatingmachines) actual three-phasediagram which may indeed be quite cumbersome and and are representedby resistanceand inductive reactancein series.Neutral confusing for a practical size power network. Generator and transformer grounding impedancesdo not appearin the diagramasbalancedconditionsare connections-star, delta,andneutralgroundingareindicatedby symbolsdrawn by the side of the representationof these elements.Circuit breakers are assumed. representedas rectangularblocks. Figure 4.5 showsthe one-line diagram of a Three voltagelevels(6.6. I 1 ancl33 kV) are presentin this system.The simple power system.The reactancedata of the elementsare given below the diagram. T1 T2 Yr ((I ' Y o ft -fl .tF tl i l __'l laFt ' v-- L r rl - Y A q throughoutthis book. A- r 4.4 PER UNIT (PU) SYSTEM \\ , Fig. 4.5 One-linerepresentationof a simple power system It is usual to expressvoltage, current, voltamperesand impedance of an electricalcircuit in perunit (or percentageo) f baseor ret'erencevaluesof these G e n e r a t oNr o .1 : 3 0 M V A ,1 0 . Sk V ,X ' = 1 . 6o h m s quantities.The per unit* value of any quantityis defined as: 1 G e n e r a t oNr o .2 : 1 5 M V A ,6 . 6k V ,X ' = 1 . 2o h m s theactualvaluein anYunits G e n e r a t oNr o .3 : 2 5 M V A ,6 . 6k V ,X ' = 0 . 5 6o h m s Transforme1r 1(3phase): 15MVA,33/11 kV,X = 15.2ohmsperphaseon hightensionside the baseor referencevaluein the sameunits TransformeTr r(3phase):15MVA,3316.2kV,X= 16 ohmsperphaseon frigntensionside The per unit methodis particularly convenientin power systemsasthe various Transmissiolnine2: 0.5ohms/phase sections of a power system are connectedthrough translormersand have LoadA : 15 MW,11 kV,0.9laggingpowerfactor different voltagelevels. L o a dB : 4 0 M W ,6 . 6k V ,0 . 8 5 l a g g i npgo w e rf a c t o r Considerfirst a single-phasesystem.Let Note: Generatorasrespecifiedin three-phaseMVA,line-to-linveoltageandper phase Base voltamPerss= (VA)s VA reactance(equivalensttar).Transformerasre specifiedin three-phaseMVA,line-to-line transformatiornatio,and perphase(equivalensttar)impedanceon oneside.Loadsare Basevoltage= Vu V specifiedinthree-phaseMW,line-to-linveoltageand powerfactor. I- r_-1 --i--t I--ntrn^'\"1'\"rr'f- I '-litr-d-'r,n'rq J-'rn,A,n-?fi-d\\1 r- i , I Then Basecurrent/u= -[4)e A (4.2a) \\v/B - e > F r - F . r o l+l ' :: - - , F' r _' T.r a. -n :s f o r m e r - J; - f fi \" ' S *Per cent value = per unit value x 100. a n s f r m e r Line Per cent value is not convenientfor useas the factor of 100 has to be carried in G nl' T T 2- - t C ; /C computations. I LoadA Loadg Fig. 4.6 lmpectancediagram of the power system of Fig. 4.5
I .t,i rr.vnq' ..lil MooernHowersystem Analysis I Baseimpedancez,\" = Y-4-: Vi 'er unit imPedanceZ (Pu)= Z (oltrns)x (MVA)' (4.e) IB (VA)\" @ ohms (4.2b) (4.3) If the actual impedanceis Z (ohms),its per unit value is given by Z(ohms)x (kVA), (4.4) ,r trt2 -, i n^n z(pu) = J--- Z(ohms)x (vA)' when MVA baseis changectlrom (MVA)r, ototo (MVA)a' n'*' andtV base ZB V; changedfrom (kv)r, orotJ (kV)n, newt' he nt* p\"t unit impedancefrom Eq' For a power system,practical choice of base values are: s Base megavoltamperes= (MVA)B .4.9)is given bY Z(pu=)zn(\"p*uf)foi\".,f.fi (4.10) or Base kilovoltamperes= (kVA)B Per Unit Representation of a Transformer Basekilovolrs = (kv)a Iaotttrhbaehttnhaaesriseqnfobeuinre-imvgpeahpenlaeersisrnsaeptirdssheyiaipnnsslgtaeeSlcsemeeo-pcdlcubthaitoyainonsaneb4tnroe.a2\"fqnrtutehshifpeuaorsatreamyisesetenthnertritmsseetea'daT-trlpyhwsheoaaadtyhsseeasttltrhtianatehcgnloelsiennft-oernpar-ehmtnocaset-sfelroiefndrotwemrramivannotdisniloiftngoangrgrpmaeoaterifroatrtthooiinoeff of the three-Phasetransformer' Basecurrent 1, 1,000x (MVA)u (kV)a sT(vehAcFeo)iamngduaaargrnenyd4ele.tvaTizokailantraggcgiipemrrebpeaaescsdeteaansnntocsceneasisstZhipnneegatgwlretldeo-Zcp,stheiadadne.sLdseeaottnfruaitdshneecsahftlgrotarroanmsnselfsro-faiornvmrotmeelterramrimnospftoheraferpetirrboaiamt1isoae: roaoyf'fand transformation,i.e. Baseimpedancez,-u_- _1_'oIoaq] GV)r _ GV)3 _ 1,000x (kv)1 ohms (4.s) (MVA)B (kvA)8 Per unit impedancez (pu) - Z (ohms)x (MVA), (4.6) V r s- ! Vt, a (kv)\" (4.11a) Z(ohms)x (kVA)u (kV)?x 1,000 -{_J---> zs 12 In a three-phasesystemratherthan obtaining the per unit valuesusing per phase ('-a' )(Rmeipgrneesteiznitnagitmioopnfesdinagnlceen-pehgalestcertaends)former basequantities,the per unit valuescan be obtained directly Uy u.ing three- p h a s eb a s e q u a n t i t i e s .L e t Three-phasebasemegavoltamperes= (MVA,)B Line-to-line basekilovolts = (kV)B Assuming star connection(equivalentstar can always be found), B a s ec u r r e nItr - l'ooox(MVA)u (4.7) Jr 1rv;u o Baseimpedance7o\" - l'ooox(kv)u J3rB _ GD3 _ 1 , o oxo( k v ):2 , o^ Ln_m-s (4.8) (b)Perunitequivalenctircuiot fsingle-phasteransformer (MVA)' G\"Ab Fig. 4.7
-W I todern PowYe- r Sv r rrstv . v r . e; m ,-Ar r -nr q t aly s trr -rsie Therefore I (4.1i b) Representationof Power SystemUomponents [l-,.t'ff. (4.11c) = a (as (VA)a is common) z z (' p u+zz)-u=+ +zBo ' 1 ' zru (4.12) 'I2 8 Vrn z - Vza . LaD -- 1,, 1,, From Fig. 4.7awe can write Thusthe per unit impedanceof a transformeris the samewhethercomputed from primary or secondaryside so long as the voltage baseson the two sides Vz=(Vt-Iflp)a-lrZ, are in the ratio of transformation(equivalentper phaseratio of a three-phase transformerwhich is the sameas the ratio of line-to-line voltage rating). We shallconvertEq. (4.12) into per unit form The pu transformerimpedanceof a three-phasetransformeris conveniently Vz(pu)Vzn= [Vr(pu)Vw - I r(pu)I6Zo(pu)Zru]a obtainedby direct use of three-phaseMVA base and line-to-line kV base in -Ir(pu)IruZ,(pu)Z* relation (4.9). Any other impedanceon eithersideof a transformeris converted to pu valuejust like Zo or Zr. Dividing by vzn throughout and using baserelations (4. rra, b, c), we get Per Unit Impedance Diagram of a Power System Vz(pu)= Vr(pu) - I,(pu)Zr(pu)_ Ir(pu)Z,(pu) (4.13) From a one-linediagramof a power systemwe can directly draw the impedance Now +=+=, diagramby following the stepsgiven below: or Ir 12 1. Choosean appropriatecommon MVA (or kVA) basefor the system. I'o I'u 2. Considerthe systemto be divided into a number of sectionsby the I r ( p u ) = 1 2 ( p u ) =1 ( p u ) transfbrmers.Choose an appropriatekV base in one of the sections. CalculatekV basesof other sectionsin the ratio of transformation. Equation(4.13)can thereforebe written as 3. Calculateper unit valuesof voltagesandimpedancesin eqchsectionand connectthemup as per the topologyof the one-linediagram.The result Vz@u)= Vr(pu)- (pu)Z(pu) (4.r4) is the single-phaseper unit impedancediagram. The above stepsare illustrated by the fallowing examples. where Z(pu)=Z,,(pu)+ Z,(pu) j Example4 1 Equation (4.I4) can be representedby the simple equivalentcircuit of Fig. 4'7b which doesnot requirean idealtransfurmetC. oniia.rablesimplification Obtain the per unit impedance(reactance)diagramof the power system of Fig. hasthereforebeenachievedby theper unit method with a commonvoltampere 4.5. baseand voltagebaseson the two sidesin the ratio of transformation. Solution The per phaseimpedancediagramof the powersystemof Fig. 4.5 has been drawn in Fig. 4.6. We shall make some further simplifying Z(pu) can be determineddirectlyfrom theequivalentimpedanceon primary assumptions. or secondaryside of a transformerby using the appropriaieimpedancl base. 1. Line capacitanceand resistanceare neglectedso that it is representedas On primary side: a seriesreactanceonly. Zr=Zp+ Z,/a2 2. We shall assumethat the impedancediagramis meantfor short circuit studies.Currentdrawn by staticloadsundershortcircuit conditionscan Z ( p' =u+) : ! - + L * I (4.1s) be neglected.Loads A and B are thereforeignored. zru Zro z,o\"a2 But a2Ztn = Zzr Let us convertall reactancesto per unit form. Choosea commonthree-phase Zr(pu)= Zo(pu)+ Z,(pu)- Z(pu) MVA baseof 30 and a voltagebaseof 33 kV line-to-lineon the transmission On secondarsyide: line. Then the voltagebasein the circuit of generator1 is 11kV line-to-line and that in the circuits of generators2 and3 is 6.2 kV. zz= Z, + o2zo The per unit reactancesof various componentsare calculatedbelow:
Flepresentationof power System eomponents I iliffij I- T r a n s m i s s i ol i nn e : 2o.5x3o= 0.564 Example4.I, we now calculatethe pu valuesof the reactancesof transfonners Transformer T,: GT2 and generatorsas per relation (4.10): -l-!,?I l_0_= 0 . 4 1 8 Transformer Z : 0.209 x TransformerTt: 16x3 0= 0.44 Transformet Tr: 0.22x ff =0.++ Generator1: aiT 0.43sx (10'5i1=0.3e6 Generator1: 1 . 6x 3 0 = o'39 6 (ll)' ut 7 - Generator2: 1 . 2x 3 0 = o ' 9 3 6 tazr Generator2: 0 . 4 1 3r * I ) *(6'612- =0.936 6.2)' Generator3: 0.56x 30 = 0'437 AZI- The reactancediagramof the systemis shown in Fig. 4.g. G e n e r a t o r3 : 0.3214 * i/9.r x (6'6)1 = 0.431 6.21' {J-)UfrL_/--X_-f\"X-)<U1 0 0 0 . _ - - J - - i obviously thesevalues are the sameas obtained alreadyin Exampre4.r. 64 0.44 I 4.5 Complex Power Consider a single-phaseload fed from a sourceas in Fig. 4.9. Let v -tvt16 r _tn t (6_0) Fig.4.9 Reactancecriagramof the systemof Fig.4.5 (roadsneglected) iJ\"E',tl'tEl3z;3anlldllErl,t\"a\"rienpaesr hunoitrctviarcluueisstotuf vdoytlth,aegeswsetoil which the g enerator sare Source berakeuni t /.,\"pu (no Example4.2 ;v(T\"oah\"rl;eu:p,ere[esJraac:scTegtnaiiv]tn#)ecvn?cdailnal-uetEeatsxo,afbmgaespenleduc4ros.a7nrtc;oe*-riqssiatuo\"niptdvrrrntherJeaexntnrtarsaartnbtnisrnpmmg4rsie.er1rrras;ab;tr;;heheaunrssteshuiaurannlmlpyienisndpragahecncetrcuifbeaieunsl doroiahnwympbiineuc g (a) Fig.4.9 Complexpowerflowin a single-phasloead When d is positive, the currentlags behind voltage.This is a convenient choiceof sign of 0 in power systemswhere loadshavemostlylaggingpower factors. TransformerT,: 0.209 Complex power flow in the direction of current indicatedis siven bv TransformerT): 0.220 GeneratorGr: 0.435 S=VI* GeneratorGr: 0.413 GeneratorG3: 0.3214 =lVlllll0 = lVl l1lcos d+ jlvl l1lsin 0= P + ie @.17) or With a baseMVA of 30. basevolta_eoef II kV in the circuit of generator t S l= ( p 2 + e \\ t , , md b:isevoltageof 6.1 k\\; in the circuit of generators2 and I 3 ,r; i; \";
ModernPowersystem Anatysis nepresentatioont powersystemeompqlents Tl.i-; ffii I As per Eq.(4.19), Kirchhoff's currentlaw appliesto complex power (also Here appliesseparatelyto real and reactivepowers). S - complexpower (VA, kVA, MVA) In a series RL load carrying current { lSl = apparentpower (VA, kVA, MVA); it signifies rating of V=l(R+jxr) I P = lVl l1l cos 0 - real (active)power (watts,kW, MW) P = I\"R = activepower absorbedby load Q = lVl l1l sin 0 = reactivepower Q - IzXr = r-eactivepower absorbedby load - voltamperesreactive (VAR) In caseof a seriesRC load carrying current I P _I2R = kilovoltamperesreactive(kVAR) O - - IzX, qreactivepower absorbedis negative) = megavoltamperesreactive (MVAR) Consider now a balanced three-phaseload representedin the form of an equivalentstar as shownin Fig. 4.L2. The three-phasecomplexpower fed into It irnmediatelyfollows from Eq. (4.17) thatQ, thereactivepower, is positive load is given by for laggingcurrent (laggingpower factor load) andnegativefor leadingcunent (leadingpowerfactor load).With the directionof currentindicatedin Fig. 4.9, S = 3vpl-t= 3 lvptl6pl; : JT lvrlzOrti (4.20) .9= P + iQ is supplied by the sourceand is absorbedby the load. If Eqr-ratio(n4.17)canbe representebdy the phasordiagramof Fig.4.10 where 0= , in1 = positive for lagging current (4.18) Ir =llil I (6p- A Lan'' .S= ,'5 lvLl lILl I 0 P Then - Ji tvLtvLtcosd+ iJT t :gativefbr leadingcurrent [email protected]) Fig.4.10 Phasor epresentatioonf complexpower(laggingpf load) Fig. 4.12 Complexpowerfed to three-phasleoad If two (or more) loadsarein parallelas in Fig. 4.ll tsl = Ji tvrtttrl P - Ji tvLltILtcosd s=vF_v(i+i) e = Ji lvLltILtsind Here f Y r i . . r ' r('p! -r + p r+)j ( e t +e z ) (4.1e) d - power factor angle lf vL, the line voltage,is expressedin kv; andIy,the line currentin amperes, s is in kvA; and if the line currentis in kiloamperes,s is in MvA.
In terms of load impedanceZ, N = rotor speed(synchronousspeed)in rpm P = numberof poles r,f L_- v , _ l v L l l 6 P z Jiz Substitutingfor I, in Eq. (4.20) r\" -- V'l' (4.22a) winding Fieldwinding -.. r tr V,.is in kV, ,Sis now given in MVA. Load impedancZe if requiredcanbe 1)xr o calculatedfrom \\Xp>{ --F vl \", _- lvr l ' -_T 1l Ov l (4.22b) I Jt- -T- 4.6 SYNCHRONOUSMACHINE :t -F I lot The synchronousmachineis the most important elementof a power system.It -t- converts mechanical power into electrical form and feeds it into the power '. \\ '\\r I network or, in the caseof a motor, it draws electricalpower from the network NN \\lQ\\, and converts it into the mechanicalform. The machine excitation which is controllable determinesthe flow of VARs into or out of the machine.Books on electrical machines 11-51may be consulted for a detailed account of the synchronousmachine.We shall presenthere a simplified circuit model of the machinewhich with suitablemodificationswherevernecessar(yundertransient conditions) will be adoptedthroughoutthis book. Fig.4.13 Schematicdiagramof a roundrotorsynchronougsenerator Figure 4. 13 showsthe schematiccross-sectionadl iagramof a three-phase On no load the voltage EJinducedin the referencephasea lags 90\" behind synchronousgenerator(alternator)having a two pole structure.The statorhas dywhich producesit and is proportionalto dyif themagneticcircuit is assumed to be unsaturatedT. his phasorrelationshipis indicatedin Fig. 4.14. Obviously a balanced three-phasewinding-aat, bbt and cct. The winding shown is a the terminal vclltage V, = Er concentratedone, while the winding in an actualmachineis distributed across Qr the statorperiphery.The rotor shownis a cylindrical\" one(roundrotor or non- I salient pole rotor) with rotor winding excited by the DC source. The rotor l - Ef=Vt winding is so arrangedon rotor periphery that the field excitation produces Fig.4.14 Phasor elationshbipetweenfuandE, As balancedsteady load is drawn from the three-phasestator winding, the nearly sinusoidally distributedflux/pole (d) in the air gap.As the rotor rotates, statorcurrentsproduce synchronouslyrotating flux Q/poIe (in the direction of rotation of the rotor). This flux, called armature reaction flux, is therefore three-phaseemfs are produced in stator winding. Since the machine is a stationarywith respectto field flux Qy.It intuitively fbllows that Qois in phase with phase c current 1o which causesit. Since the magnetic circuit has been balancedone and balancedloadingwill be consideredi,t can be modelledon per phasebasis for the referencephase a. Tn o mqnhinc rrrifh mnra fhqn f r r r nv nnlec fLhl lpv qqul rvn r r e v sAvpr lflilnvpuA JcLf nr r nufvrLr ur or v rr voynvpoal of c l/vrvo, v electricallyfor everypair of poles.The frequencyof inducedemf is given by f =ffinz where . High-speedturbo-generatorshave cylindrical rotors and Iow sppedhydro-generators have salient pole rotors.
,liO* | Modernpowsr Syglem_Anatygis urrurnJdto be unsaturatedth, e superpositionprincipleis applicableso that the Rapr\".\"n,\",ionof po*\"r' Syrr\"r Cornp.on\"n,, NEffi resultantair gap flux is given by the phasorsum t- The circuit of Fig. 4.L6 canbe easilymodifiedto includethe effectof d' = d1+ Q,, armatureleakagereactancaendresistanc(etheseareserieseffects)to givethe @,3) Furtherassumingthat the armatureleakagereactanceand resistanceare completecircuitmodelof thesynchronougseneratoarsin Fig.4.I7. Thetotal re eml whtch equals the termin tage V,. +xl=Xslsc synchronous reactance of the machine. Phasordiagram under loaded(balanced)conditions showing fluxes, currents Equation(4.24) now becomes and voltagesas phasorsis drawn in Fig. 4.15. V, = Et - jlo X, - IoRa (4.2s) jlX\"= - E\" Fig. 4.1S phasordiagramof synchronougsenerator Fi g.4. 16 Here This model of the synchronousmachinecan be further modified to account d - power factor angle for the effect of magnetic saturation where the principle of super-positiondoes 6 - angle by which Et leads v, called load angre or torque angle not hold. We shall seein Sec.5.10that dmainly determinesthe power deliveredby & the generatorand the magnitudeof E, (i.e. excitation)determinesthe VARs deliveredby it. Fig-4.17 circuitmodelof roundrotorsynchronougsenerator Becauseof the assumedlinearity of the magneticcircuit, voltage phasorE, Armature resistanceRn is invariably neglectedin power system studies. Eo and v, areproportionalto flux phasors dr, doand d, respectively;furthei, Therefore,in the place of the circuit model of Fig. 4.I7, the simplified circuit voltage phasorslag 90' behindflux phasori. It thereforeeasily follows from Fig.4.15 that phasorAB =- Eois proportionalto (o (andtherefore Io)andis 90' leading d\" @r 1,).With thedirection of phasorAB indicatedon the diagram AB = jlo Xo where X\" ir the constantof propotionality. modelof Fig. 4.18will be usedthroughout hisbook.The correspondingphasor I n t e r r n c n F t h p qqul rvnvrvr o ur lvorfui -uilrriu^r-r v^ Ls ,vl r a , wE ualt li-^^rr-- ---^rr- 4t-- r rt diagram is tghievetonrqinueF(ilgo.a4d.)i9a.ngTlehedfiTehiciis,inindufcaecct,ieismtiheE-ycioenadciitsiothnefotreramcirnivael urr€utly wl-l[c) ule l0ilowlng voltage by expressionfor voltageswithout the need of invoking flux phasors. V, = Ef - jloXo (4.24) power to flow out of the generator.The magnitudeof power delivereddepends where upon sin d Ef = uolrage induced by field flux Q, alone In the motoring operationof a synchronousmachine,the current1,,reverses = uo load emf as shownin Fig. 4.20, so that Eq. @.25)modifiesro The circuit model of Eq. (4.24) is drawn in Fig. 4.16 wherein X, is interpretedas inductive reactancewhich accountsfor the effect of armature Ef = V, - jIoX, (4.26) reactiontherebyavoidingtheneedof resortingto additionof fluxes l&[email protected])1. which is representedby the phasordiagramof Fig. 4.2I.It may be notedthat V, now leadslby d,This in fact is the conditionfor power to flow into motor terminals.
the'i I ModernPowerSystemAnalysis - RepresentatioofnPowerSystemComponents FIIS': The flow of reactivepower anclterminal voltage of a synchronousmachine lvtl ll,,lcos d = constan=t activepoweroutput is mainly controlleclby meansof its excitation.This is discussedin detail in Section5.10.Voltageandreactivepower flow areoften automaticallyregulated uy YUIL4S\\/ IvSurqlvro \\uvv vvve^v^r vr andby automatictap changingdeviceson transformers. -- -'-/ 6666 Fig.4.22Synchronousmachineconnectedtoinfinitebus | Irtdteoimarhngaeirgiaanhnms,scomtohfneaFsdtitgisau.innm4tc.,wa2enh3lVdi(i,bell)oitcshwoefilxereeexxsdccp,iittoaathnttieidoostpnnorissotahjvereaecurtpiineoriedntys'lP/peohonlacwtesoeodsirnrfddaiFoacfigtgort'arhc4mea'2pssch3eoa''IrTtsro\\iesersIopopboohvnnaidosiVuno'sgr lx\" frot the phasordiagramthat for this excitation I l E J lc o s 5 = l V ) E1 ( ; E1 It -l it\"x' Fig. 4 . 1 8 S i m p l i f i e dcircuit model of Fig. 4.19 Phasordiagramof synchro- ---\"' / round rotoI synchronous n o u sg e n e r a t o r 0 generator ., I\" Ia (a) Overexcited 1E f ' jIJ' (b)Normalexcitation I Fig. 4.20 MotoringoPerationof J s Y n c h r o n o u sm a c h i n e Vt', Fig. 4.21 Phasordiagramof motoring oPeration Normally, a synchronousgeneratoroperatesin parallel with other generators (c) Underexcited connectedto the powersystem.For simplicity of operationwe shall considera generatorconnectedto an inJinitebus as shown in Fig' 4'22' As infinite bus Fi1.4.23 Phasordiagramscfsynchronousgeneratorfeedingconstant meansa largesystemwhosevoltageandfrequencyremainconstanitndependent power as excitationis varied of the power exchangebetweenthe synchronousmachine and the bus' and independenot f the excitationof the synchronousmachine. This is definedas nloargmsbael ehxincdittVtt,isttortFthoarttthheeogveenteerxactioterfercleadssep(Fosigit'iav'e2r3eaa)c'it'iev'e |,8,.c1os 6>|v),1, considernow a synchronousgeneratofreedingconstantactivepower into an infinite b's bar. As the machineexcitationis varied,armaturecurrentIn andits powerintothebus(ordrawsnegativereactivepowerfromthebus)'Forthe angle g, t.e. power factor, changein such a manner as to keep
^ #.......'@,^ ^ lL^ j a ^{ -A l4 ^ li^/\\^^ AaAu t _^ ^- \" ;iiii I ModernPowersystemAnalysis Representationf PowerSystemComponents T.I.i_ilis'.$ t or underexcitedcase(Fig. 4.23c), i.e. lErl cos 6 < lV), 1oleads V, so that the l1,cl ose: ln , ls i n6 (4.27) generatorfeedsnegativereactivepowerinto thebus (or drawspositivereactive E: Figure 4.24 shows the overexcitedand underexcitedcasesof synchronous (4.28) motor (connectedto infinite bus) with constantpower drawn from the infinite bus. In the overexcitedcase,Io leads Vu i.e. the motor draws negativereactive The plot of P versus{ shownin Fig. 4.25,is called thepower angle curve. power (or suppliespositive reactivepower); while in the underexcitedcase .Io The maximum power that can be deliveredoccursat 6 = 90\" and is given by lags V, r.e. the motor draws positive reactivepower (or suppliesnegative reactivepower). (4.2e) For P ) P** or for 6> 90' the generatorfalls out of step.This problem (the stability) will be discussedat length in Chapter 12. E1 (a) Overexcited -- V1 Fig. 4.25 Poweranglecurveof a synchrcnougsenerator Power Factor and Power Control Fig.4.24 (b) Underexcited While Figs 4.23 and4.24 illustrate how a synchronousmachinepower factor changeswith excitationfor fixed power exchange,thesedo not give us a clue Phasordiagramsof synchronousmotor drawingconstantpower as regardingthe quantitativevaluesof llnl and d This can easily be accomplished excitationis varied by recognizing from Eq. (4.27) that From the above discussionwe can draw the general conclusionthat a lEll sin 6 -llolX, cos d synchronousmachine (generatingor motoring) while operating at constant power suppliespositive reactivepower into the bus bar (or draws negative = PX\" = constant (for constantexchangeof power to # rcqettvcI v nrtrx/cr frnrn fhc hrrc hqr\\ rrrhan nrrcrcvnifarl An rrnrlarownifprl mqnl\"i-o vsU Yv rrvrr v Y vlvl\\vlLvv. / lll ulluvtvnvltvu lyrl vv lll4vllltlv , on the oih.. hand, feeds negativereactivepower into the bus bar (or draws infinite bus bar) (4.30) positivereactivepower from the bus bar). Figure 4.26 shows the phasor diagram for a generatordelivering constant Considernow the power deliveredby a synchronousgeneratorto an infinite power to infinite busbut with varying excitation.As lEtl sin dremainsconstant, thetip of phasorErmovesalonga line parallelto y, asexcitationis varied.The bus.From Fig. 4.19 this power is P = lVtl llol cos 0 direction of phasor1ois always90o laggingjI\"X, andits magnitudeis obtained The aboveexpression.canbe written in a more useful form from the phasor from (l1olX5)/X5F. igurc 4.27 showsthe caseof limiting excitationwith d= 90\". geometry.From Fig. 4.19 For excitation lower than this value the generatff becomesunstable. lnA _- rl,lx, sin(90\"+ 0) sin6
ModernPowerSystemAnalysis Ftepreseniatioonf Fqwer SystemComponents 1!1 Salient Pole Synctrronous Generator | A salientpole synchronousmachine,as shownin Fig. 4.29, is distinguished from a round rotor machine by constructionalfeaturesof field poles which /.{1 62 employed in machinescoupled to hydroelectricturbines which are inherently Iaz slow-speedones so that ttre synchronousmachinehas rnultiple pole pairs as \\3 different from machinescoupled to high-speedsteam turbines (3,000/1,500 rpm) which havea two- or four-polestructureS. alientpole machineanalysisis Fig. 4.26 Effect of varying excitationof generator deliveringconstant power madethrough the two-reactiontheory outlined below. to infinitebus bar Directaxis /,\" I V1 'l Fig. 4.27 Case of limitingexcitationof generatordeliveringconstantpower Fig. 4.29 Sallent pole synchronousmachine(4-polestructure) to infinitebus bar Similar phasor diagrams can be drawn for synchronous motor as well for In a round rotor machine, armatLlrecurrent in phase with field induced emf constant input power (or constant load if copper and iron lossesare neglected Ey or in quadrature (at 90\") to S, produces the same flux linkages per arnpere and mechanical ioss is combined with load). ai the air gap is uniform so that the armature reaction reactance offered to in- Another important operating condition is variable power and fixed excita- phase or quadrature current is the same (X,, + X1 = Xr), In a salient pole tion. In this case lV,l and lE1trare fixed, while d and active power vary in machrne at gap ls non-unllorTn arl -o, -ng I-U- 1L- -OI'l-j^c^ r- l: P- L i^l-u- .ry. frtL ils^ +LLl^lc rl ^t^r^il+sL .^ ll r^ u- ^ug +trL r^ tr accordance with Eq. (a.28). The corresponding phasor diagram for two values axis of main poles(called direct axis) and is the largestalong the axis of the of d is shown in Fig. 4.28. It is seen from this diagranr that as d increases, interpolarregion (calledquadrature oxis).Armaturecurrentin quadraturewith current magnitude increases and power t'actor improves. lt will be shtlwn in El producesflux along the direct axis and thereluctanceof flux path being low Section 5.10 that as dchanges, there is no significant change in the flow of (becauseof small air gap), it produces larger flux linkages per ampere and reactive Power' hencethe machinepresentslarger armaturereactionreactanceX, (called direct axis reactance)to the flow of quadraturecomponentIl of armaturecurrent 1o. Locusof Er - On the other hand, armaturecurrent in phasewith { producesflux along the Er', quadratureaxis andthe reluctanceof the flux pathbeing high (becauseof large --4, ,--' l\\---jluzX\" , n' l/- Fig. 4.28 Operationof synchronousgeneratorwith variablepower and fixed excitation
4 r i5 I I rl,l-r- ffil Mod\"rnPo*r. syrt\"t Rn\"tyri, Representatioonf power Resultant interpolarair gap), it producessmallerflux linkages per ampereand hencethe machine presentssmaller armaturereaction reactance Xu (guadratureaxis reactancea X) to the flow of inphase component Io of armaturecurrent /o. Sincea salientpole machineoffers different reactancesto the flow of Il and 1ocornponentsof armature current Io, a circuit model cannot be drawn. The phasordiagramof a salientpole generatoris shown in Fig. 4.30.It can be easily drawn by following the stepsgiven below: Fig.4.30 Phasordiagramof salienpt olesynchronougsenerator Fig. 4.31 powerangrecurvefor sarienpt oregenerator 1. Draw % *d Io at angle 0 In this book we shall neglect the effect of sariencyand take 2. Draw IoRo.Draw CQ = .il,X,t(L to 1,,) X'= X't 3. Make lCPl- llol Xq anddraw the line OP which givesthe direction of Ey in all types of power systemstudiesconsidered. phasor During a machine transient, the direct axis reactancechangeswith time 4. Draw a I from Q to the extendedline OP such that OA = Ef acquiringthe following distinct valuesduring the completetransieht. It can be shownby the abovetheory that the power output of a salientpole X/ = subtransientdirect axis reactance Xh = transientdirect axis reactance generatoris given by X,r = steadystatedirect axis reactance , = - 1lv;,-lEs,iln d + lv,l'(xo- xn) (4.31) The significanceand use of thesethree valuesof direct axis reactancewill be elaboratedin Chapter 9. 2XdXq sin26 Operating Chart of a Synchronous Generator The first term is the same as for a round rotor machine with X, = Xa and constitutesthe major part in power transfer.The secondterm is quite small while selectinga large generator,besidesrated MVA and power factor, the (about I0-20Vo) comparedto the first term and is known as reluctancepower. greatestallowable stator and rotor currentsmust also be consideredas they influence mechanical stressesand temperaturerise. Such timiting parametersin P versus d is plotted in Fig. 4.31. It is noticed that the maximum power the operationare brought out by meansof an operatingchart orp- erformance chart. output occursat 6< 90' (about 70')'. Furt1t\"r34 (changein power per unit d5' For simplicity of analysis,the saturationeffects,saliency,andresistanceiue ignored and an unsaturatedvalue of synchronousreactanceis considered. change in power angle for small changes in power angle), called the Consider Fig. 4.32, the phasor diagram of a cylindrical rotor machine. The locus of constantllolx,V) and henceMVA is a circle centeredat M. The locus synchronizingpower.cofficient,in the operatingregion (r< 70') is larger in of constantlEtl (excitation)is alsoa circle centeredat O. As Mp is proportional to MVA,QP is proportionalto MVAR andMe to MW, all to the same scale which is obtainedas follows. a salientpole niachin.:than in a round rotor machine.
Ir\\Y, /rl nv rul o! rr rnr DI nvrYr rYa vr r vQ)r'rsc ri avmr r | , r Ar rnqarl tr rucriuc l - 2.Opu excitation N . 0.85pf lagging E I a i\\ Locus11I,X\" (circlecentreM) -o Fig.4.32 Phasordiagramof synchronougsenerator (E C]' For zeroexcitation.i.e. lE.l = 0 - iIoXr'= Y, (U o or Io = jV,lX, c) F i.e. llol =lV)lXr leadingat 90\" to OM which correspondsto VARs/phase. Considernow the chartshownin Fig. 4.33 which is drawn for a synchronous 0.7 9o.s , \\ M machinehavingXt = 1.43pu. For zeroexcitation,the currentis 1.01I.43- 0.J 0.5 1.0 I pu, so that the length MO conespondsto reactivepower of 0.7 pu, fixing both tl , -,--- activeand reactivepower scales. >- Reactivepower(pu)lagging Leading With centre at 0 a number of semicirclesare drawn with radii equal to differentpu MVA loadings.Circlesof per unit excitationaredrawn from centre Fig.4.33 operatingchartfor rargesynchronougsenerato,r M with 1.0pu excitationcorrespondingto thefixed terminalvoltageOM. Lines Considera point h onthe theoreticalimit on the lETl= 1.0pu excitationsarc, may also be drawn from 0 conespondingto various power factors but for the pcrwerMh is reducedby 0.1 pu to Mk; theop\"ruiingpoint must,however, clarity only 0.85 pf laggingline is shown.The operationalimits arefixed as still be on rhe fbllows. on the desiredlimiting curve. This is repeatedfor other excitationsgiving the curve afg. The completeworking area.shown shaded.is gfabcde.,{ working Taking 1.0 per unit active power as the rnaximum allowable powel',a point placedwithin this areaat oncedefinesthe MVA, Mw, MVAR, current, horizontalimirline ubc is drawnthroughb at 1.0 pu. It is assumedthat the power factorandexcitation.The loadangle 6 canbe measuredasshownin the machineis ratedto gire 1.0 per unit activepowerat power factor0.85 lagging figure. andthis tixes point c. Limitation of the statorcurrentto the correspondingvalue requiresthe limit-line to becomea circular arccd aboutcentre0. At point d the 4.7 REPRESENTATION OF LOADS rotor heatingbecomesmore importantand the arcde is fixed by the maximum Load drawnby consutnersis thetoughespt arameterto assesscientifically.The excitationcurrentallowable,in this caseassumedto be lEtl = 2.40 pu (i.e.2.4 magniiudeof the ioad, in iact, changescontinuouslyso that the load forecasting times ly,l). The remaininglimit is decidedby lossof synchronismat leading problern is truly a statisticalone. A typical daily load curve is shown in power factors.The theoreticallirnit is the line perpendicularto MO at M (i.e. Fig' 1.1. The loads are generally composed of industrial and domestic d= 90o),but in practicea safetymarginis broughtin to permit a further small componentsA. n industrialload consistsmainly of largethree-phaseinduction increasein load beloreinstability.ln Fig. 4.33,a 0.1 pu margin is employed nlotorswith sulficient load constancyand predictableduty .y.lr, whereasthe and is shownby the curve afg which is drawn in the following way. domestic lcladmainly consistsof lighting, heating and many single_phase devicesusedin a randomway by houscholdersT.he designun6upJrotionof power systemsboth economically andelectrically aregreatlyinfluencedby ttrp natureand magnitudeof loads.
._. sIaaZ I ModernPower SvstemAnatr-rsic Representatioofnp Sotution I BaseMVA = 645,3_phase In representationof loads for various systemstudiessuch as load flow and BasekV = 24,line-to_line JlfZJ.t stabilitystudies,it is essentiatlo know the variationof real and reactivepower with variation of voltage.Normatly in such studiesthe load is of composite Load volt ASa= )L = 1 pu naturewith both industrialand domesticcomponentsA. typical compositionof i Induction motors 55-757o SynchronouresactancXe\", = +#( 2 4 ) z = 1.-3' 4r -4pu Full load (MVA) = I pu, 0.9 pf lagging Synchronousmotors 5:75Vo Load current = generatorcurrent Io= 7 pu, 0.9pf lagging Lighting andheating 20-30Vo = 0.9- 7 0.436pu Though it is always better to consider the P-V and Q-V characteristicsof (a) Excitationemf (seeFig. 4.Ig) each of these loads for simulation, the analytic treatment would be very Ef = V,+ j XJ\" cumbersomeand complicated. In most of the analytical work one of the = 1 1 0 \" + j 1 . 3 4 4( 0 . 9 - j 0 . 4 3 6 ) following three ways of load representationis used. = 1 . 5 8 6- j l . 2 l = 1 9 9 1 3 7 . 1 \" (i) Constant Power Representation E, (actual) = 1.99 x 24 = 47.76 kV (line) 6= 3j.1\" (leading) This is usedin load flow studies.Both the specifiedMW and MVAR aretaken to be constant. (b) Reactivepower drawn by load Q = VJ,,sin r/ (ii) Constant Current Representation = 1 x I x 0.436= 0.436pu or 0.436x M5 = 281 MVAR Here the load currentis given by Eq. (4.17),i.e. I' = P :iQ- t- n\" ' l ( 6 - 0 ) V{< where V = lVl 16and 0= tan-l QlP is the power factor angle.It is known asconstantcurrentrepresentationbecausethe magnitudeof current is regarded as constantin the .study. (iii) Constant Impedance Representation This is quite olten usedin stability studies.The load specifiedin MW and MVAR at nominal voltageis usedro compuretheload impedance(Eq. (4.I?2b\")). Thus z = !I: - Pw=* J O - l v l 2- t P-JQ:T which then is regarded as constant throughout the study. f l- --- .---l The generatorof Example 4.3 is carrying full load at rated voltage but excitationemf is (i) increasedby 20vo and(ii) reducedby 20vo. l FF^vsar .t.tl srrtv li ae I its zrv I I t-- Calculatein eachcase T (a) load pf A synchronousgeneratoris rated 645 MVA , 24 kv,0.9 pf lagging.It has a (b) reactivepower drawn by load syrrchronourseactancel.z o. The generatoris feeding full load-at 0.9 pf laggingat ratedvoltage.Calculate: (c) load angle 6 (a) Excitationemf (E1) andpower angle6 Solution (b) Reactivepower drawnby the load Full load, P _ 1 x 0 . 9 = 0 . 9 p u Carry out calculationsin pu form and convert the result to actual values. Ef= r.99
lM I ftrodernPower SystemAnalysis tation of Power (i) Et is V,=7 (b) x 0.9x sin 1.5= 0.024 d by 20Voat samereal load.Now or Q = 0.024x 645= 15.2MVAR As perEq. (a.28) P= l E', l l v', l s i n d (i) PROBIEIvSI x, 0.9= (\\ 2'13.8384x4r)) ,i' d 4'r lt6rFsiehn6yirgeesmukrtaeevirnneimnad.'slPATvtuh-hob4leeols'tlaatrtraadgasttenioihonosfongff6woswt0rhsahrenMnitcdhesWheryrsenbisasacycchtth0tsroaoee.9nnmrb-ciopeeaeuostsomswirmcefeadaatrihicncafetahrgaacvirnnitaancoermeer.iRoorsoauoeangsfplttgcruoaeio.nos.mgreakpinsvdott.ithnaaeelpcntpatrtrealsacadnnurfsserrmaomstmiheissossrtwhihiooeennn or sin d= 0.5065 or 6= 30.4\" 2.388130.4\"-110\" 11t220kv 220t66kV 1 6 0k v 60 MW jr.344 V1 100MVA _-__-EF_d--h---l___> = 0 . 8 9- j 0 . 7 9 = 1 . 1 8 3l - 4 L 2 \" X = 10o/o (a) pf = cos 4I.2\" = 0.75lagging 5 F | 0.9pf tagging (b) Reactivepower drawnbY load 100MVA V2 X = Bo/o Q = lV,lllolsin / =1x1.183x0.659 Fig. p_4.1 = 0.78pu or 502.8MVAR 4.2 T4Dh.r2ea.wNraetthginelgepscutorifenstrhipseetadganencneceae,rdnaitdaougrs,rmaeoamtbofbarrsaetnhodef tpiroaoonwsvefrrorsremyse,tre2sm2a0rsekhvowinn5in0 nig. e- () rine. (ii) E, decreasedby 20o/oor E f = 1 . 9 9x 0 . 8 = 1 . 5 9 Generator40 MVA, 25 kV, Xu = 20Vo Substitutingin Eq. (i) Motor 50 MVA, I I kV, X,t = 30Vo Y-Iltransformer,40 MVA, 33 y_220 y kV, X = I5Vo Y-l transformer,30 MVA, ll L_220 y kV, X = l5*o oe=(L2\\L),in 6 U-vGi-r+1,-__iF-,-5r-0Fo -.*fF-., l_- 1 f 1\\M ) 'Y ) . l l - - Y Y -I 2 / \\ 1.344) [y^ I whichgives 6- 49.5\" Fig. p-4.2 t.59149,5\"-rlo\" 4 . 3 R1A(a1o)sk=yVDn0ec-dthleerrplomiuvnieonarueninsdEgg,exaoancne7idrrararrtone.6rgn5ittse3pr.dau1t.5eIktdAi6s0afteM0eV.dgAinf,fgi1hnrgtogkvian.ng.Iitnfhinaistea resistance bus bar ar In= .i1.344 =0.9-j0.024 = 0.9l-1.5\" (b) Draw a phasor diagramfor this operation. (a) pf = cos 1.5\"= 1;unitypf (c) agBneudnsepbrfaartoovfroatlhntaedgcietusfrareeilxnscttidtoaet1liio0vnekvrreemdwtaohiinltehseuthnbecumhsa.eIncnghetahdni.swicctaralapstoewiasestrshiuenmpvueattlhutoee
I ModernPower SystemAnalysis - Representationoflgygr_gyg!\"_!L_qg[p!g$q generatorresistanceto be negligible. 126 | T 4.4 A 250 MVA, 16 kV ratedgeneratoris feedinginto an infinite bus bar at REFERI.EICES 15 kV. The generatohrasa synchronourseactanceof 1.62pu.lt is found that the machineexcitationand mechanicalpower input are adiustedto Books give E, = 24 kY andpowerangle 6 = 30o. (a) Determinethe line current and active and reactivepowersfed to the l' Nagrath,I'J. and D.P- Kothari, Electric Machines,2nd edn Tata McGraw-Hill, bus bars. New Delhi, 7997. (b) The mechanicalpower input to the generatoris increasedby 20Vo 2' van E' MablekosE, lectricMachineTheoryfor PowerEngineers,HarperlnoRaw, from that in part (a) but its excitation is not changed.Find the new New York, 1980. line currentand power factor. 3. Delroro, v., Electric Machines and power systems,prentice_Hall,Inc., New (c) With referenceto part (a) current is to be reducedby 20Voat the Jersey,1985. same power factor by adjusting mechanical power input to the generatorand its excitation.Determine Ey, 6 and mechanicalpower 4' Kothari, D.P. and I.J. Nagrath, Theory and.Problemsof Electric Machines, 2nd Edn, Tata McGraw-Hill, New De\\hi, 2002. rnput. 5. Kothari, D.p. and I.J. Nagrath, Basic Electicar Engineering, 2nd Edn., Tata (d) With the reducedcurrent as in part (c), the power is to be delivered McGraw-Hill,New Delhi. 2002. to bus barsat unity pf, what are ttrecorrespondingvaluesof El and Paper d and also the rnechanicapl ower input to the generator. 6' IEEE CornitteeReport,\"The Effect of F requencyand Voltageon pow er Sys t em Load\", Presentedat IEEE winter po,ver Meeting,New york, 1966. 4.5 The generatorof Problern4.4 is feeding 150MVA at 0.85 pf laggingto infinite bus bar at 15 kV. (a) DetermineEy and d for the above operation.What are P and Q fed to the bus bars? (b) Now E, is reducedby l0o/okeepingmechanicalinput to generator same,find new dand Q delivered. (c) Et is now maintainedas in part (a) but mechanicalpower input to generatoris adjustedtill Q = 0. Find new d and P. (d) For the valueof Eyin part (a) what is the maximum Qthat can be deliveredto busbar.What is the corresponding6and {,? Sketchthe phasor diagramfor eachpart. Answers 4.1 12kV 4.3 (a) 26.8 kV (line),42.3\"leading (c ) i .i 3 l -2 8 .8 \" k A ; 0 .8 7 6l a g 4.4 (a) 0.5ll\"l- 25.6\"kA; 108 MW, 51.15MVAR (b) 6.14 kA, 0.908lagging ( c ) 1 . 5 7 8 ,1 3 . 5 o5, 3 . 3M W ( d ) 1 8 . 3 7k v , ' 3 5 . 5 \" ,9 6 M W 4. 5 (a ) 2 5 .2 8k V , 2 0 .2 ' ,1 2 7 .5M W, 79.05MV A R (b) 33.9\",54 14MVAR (c ) 4 1 .1 \" , 1 5 0 .4MW (d) 184.45,MVAR,53.6\",- 7 0.787 pu
5.1 INTRODUCTION ehara-cteristicasnd Perforr\"nancoef povrerTransmissionLines This chapter deals primarily with the characteristicsand performance of The following nomenclaturehas beenadoptedin this chapter: transmissionlines.A problemof majorimportancein powersystemsis the flow z = seriesimpedance/unitlength/phase of load over transmissionlines such that the voltage at various nodes is y = shunt admittance/unitlength/phaseto neutral maintainedwithin specified limits. While this general interconnectedsystem problem will be dealt with in Chapter 6, attention is presently focussedon ;=;l_*#:,T.\":Jff:: performanceof a single transmissionline so as to give the reader a clear understandingof the principle involved. C = cepacitance/unitlength/phaseto neutral / = transmissionline length Transmissionlinesare normallyoperatedwith a balancedthree-phaseload; Z = zl = total seriesimpedance/phase the analysiscan thereforeproceedon a per phasebasis.A transmissionline on Y = ll = total shunt admittance/phaseto neutral a per phasebasiscan be regardedas a two-port network, whereinthe sending- Note: Subscript,Sstandsfor a sending-endquantityand subscriptR standsfor a receiving-endquantity end voltage Vr and current 15are relatedto the receiving-endvoltage Vo and current 1othroughABCD constants\"as 5.2 SHORT TRANSMISSION LINE For short lines of length 100 km or less,the total 50 Hz shunt admittance* QwCl) is small enoughto be negligible resultingin the simple equivalentcircuit of Fig. 5.1. Fig.5.1 Equivalenctircuiot f a shortline lyrl lA BlIy*l ( sr ) This being a simple series circuit, the relationshipbetween sending-end L1,J Lc plLroJ receiving-endvoltagesand currents can be immediately written as: /< t\\ Also the following identityholdsfor ABCD constants: lyrl =JLI l zflvol - . J (5'3) AD-BC=l Lr, orJLr These constants can be determined easily for short and medium-length lines The phasordiagramfor the short line is shownin Fig. 5.2 for the lagging by suitable approximations lumping the line impedance and shunt admittance. current case.From this figure we can write For long lines exact analysis has to be carried out by considering the distribution of resistance, inductance and capacitance parameters and theABCD lV5l= l(ly^l cos /o + lllR)z + (lV^l sin dn + lllxyzlr/2 constants of the line are determined therefrom. Equations for power flow on a line and receiving- and sending-end circle diagrams will also be developed in l v 5l = [ t v R P +l l t 2( R z+ f ) * z t v R | i l ( R c o sQ ^ + X s i nwf-u{ ) r l 2 ( 5 . 4 ) this chapter so that various types of end conditions can be handled. 1il21n2+ x2 . r lI/t 2 tvRP J r l= tv\" Lnf tr * ?lJVlJRrlcos/oA+ 4,#I r/ in- \" Qpt *R\"f'.. -;- - to Appenclix B. For overheadtransmissionlines, shunt admittanceis mainly capacitivesusceptance (iwcl) as the line conductance(also called leakance)is always negligible.
Characteristicasnd Performanceof PowerTra--n1smissionLines J\"#, The last term is usually of negligible order. l/l R cos/^+l1l X s i nd ^ x 100 (s.7) IyRl In the abovederivation, Q*has beenconsideredpositive for a lagging load. Expandingbinomially and retaining first order terms, we get per cenrregularion- |!!4 9\"-t!t:l n x tin4-r x 100 (s.8) lvRl Ivsl=vl ^rfr. ff co/s^+ff sinf^)''' (for leadingload) l V 5l = l V ^ l + l X ( R c o s / o + X s i n / o ) The above equation is quite accurate for the normal load range. (s.s) Voltage regulationbecomesnegative(i.e. load voltage is more than no load voltage),when in Eq. (5.8) X sin Qo> R cos /p, or tan </o(leading) >+ X It also follows from Eq. (5.8) that for zerovoltage regulation tan/n= X =cot d i . e . , / ^ ( t e a d i n g =[ - e (5.e) where d is the angle of the transmission line impedance. This is, however, an approximatecondition. The exact condition for zero regulation is determinedas follows: Fig. 5.2 Phasordiagramof a shortlinefor laggingcurrent Voltage Regulation Voltage regulationof a transmissionline is definedas the rise in voltage at the Fig.5.3 Phasordiagramunderzeroregulationcondition receiving-end,expressedaspercentageof full load voltage,when full load at a specifiedpower factoris thrown off, i.e. Figure 5.3 shows the phasor diagram under conditions of zero voltage regulation, i.e. Per centregutation= ''^?)-:'Yu'x 100 (5.6) lvRLl lV5I = lVpl or OC= OA where lVool= magnitudeof no load receiving-endvoltage sn /AOD _ AD - AClz -_ llllzl lVprl= magnitudeof full load receiving-endvoltage oA lyR| zlvRl (at a specified power factor) For short line, lV^61= lVsl,lVsl = lVpl Per centregutation= 'u1|,'%' lvRl
', Characteristicasnd Performanceof PowerTransmissionLines 133 or new valueof lVr | = 11.09kV or IAOD= sin-Ur !4 Figure 5.4 showsthe equivalentcircuit of the line with a capacitivereactance zlvRl placed in parallel with the load. lt follows from thegeometryof anglesat A, thatfor zerovoltaeeregulation, R+jx /p (leading) = (5.10) From the above discussion it is seen that the voltage regulation 6f a line is ll! heavily dependent upon load (decreases) as the power factor power factor. voltage regulation improves l- -,L- - -l of a lagging load is increased and it becomes Fig. 5.4 zero at a leading power factor given by Eq. (5.10). Assuming cos y''^now to be the power factor of load and capacitive reactance taken together, we can write (11.09- 10) x 103= l1n| ( R cos d^+ X sin dn) (i) kppA2eov0rws.kkimenmrg'llfoeaTn-chpgteho. arTvsoohelfe5ta0l0ign'H7eeaz0rt7egtlsheaiensgertgaeriancntecgoeiabvrsniyundpgmipn-eeldianeundsciasstanornefincqadeuanuirrceoetdvi0vte.oe0rlh1boe9eaa5kddoeothprfamt5nca,os0nmn0ds0ist0aks.nw6iot3antmlain1tHea0 Since the capacitancedoesnot draw any real power, we have 5000 (ii) l/ol= 10x cos/^ Solving Eqs.(i) and (ii), we get Find (a) the sending-endvoltage and voltage regulationof the line; (b) the cosdn= 0'911lagging the value of the capacitorsto be p lacedin parailet viittr th e load such that the and regulation is reducedto 50voof that obtainedin part (a); and (c) compare transmissionefficiency in parrs (a) and (b). llal= 549A Solution The line constantsare Now R = 0 .0 1 9 5x 2 0 = 0.39 f) Ic= In- I = 549( 0. 911- j0'412)- 707( 0. 107 - j0. 70'7) X = 3I4 x 0.63 x 10-3x Z0 = 3.96 e (a) This is the caseof a short line with I = Ia= 1, given by = 0.29+ j273'7 Note that the realpart of 0.29 appearsduethe approximationin (i) Ignoring it, l 1 l= - - 5 0 0 0 = 7 0 7 A we have 10x0.70i I, = j273.7 A From Eq. (5.5), l V 5l = l V o l + l 1 l ( R c o sQ * + X s i n / ^ ) ' 'Y^ L : l t l - l o x l o o o = 1 0 ,0 0 0+ 7 0 7 (0.39x 0.701+ 3.96 x 0.707,y\\ = 72.175kV 3wxc ll. I 273'7 Voltageregulation= pfTL-- l9- x roo :Zt.j7vo or C-81 P'F 10 (c) Efficiency of transmission; (b) Voltage regulariondesired= ?+t = l0.9Vo Case (a) l y st - 10 = 0.109 t0
Case(b) characteristicsand Pedormanceof PowerTransmissionLines fL*I{S Per unit transformerimpedance, r'/ = 5000 - g' 7.7%o Z r = ( 0 . 0 2 +j 0 . 1 2 ) x 5_ ( 0 . 5 + 1 3 . 7 5 ) x 5 5000+ (549)2x 0.39x 10-3 6.q2 Q'2 9c(roeu6ncr'2ereeithvnoaitnl9rfeg7td-'h7euevncopdep)rs'eoA(fvw/driuooedmuriLnsfrrag76vc0fac.tLu7laouJpAreiap)mrtcadopinvtrrod5togr4tvshi9nee,Aspt(u)raf,aarrpontaahmslcelmleto0lliirswn.7seii0tlohnv7noli\"poiufatfaagria\"dgiltl\"oieenslrJ0eay.gw9pult1oihl-a1wptlretaiohorgevfn)ued,llsoetmahcd(ere,fetrlhaoitnhomseedes = (0.0046+ j0.030) pu of improving the performanceof a transmissionsystemand will be discussed further towardsthe end of this chapter. Total seriesimpedance= (0.037+ j0.0115)+ 2(0.0046+ j0.030) = (0.046+ j0.072) pu Given: Load MVA = 1 pu cfAaacbstuloebrhosatnvaittnhi ogenpaloeswrsphvhooawlstnaergi neesFsisiigdta.e5no.cf5eaarnetdrcareeni vasefcots5arnnMecrVeofArfo8maat na6dpk2ov.w,5e0or.hgsmt5asl at,irgoengstiphnergocputoigvwehelayr. Loadvolt age= +6. 6 = 0. 91 pu 0c6Iad.o'n36end6nkn3tVoei'7chca5stmeliodsd6he...Twm6oh/hfs3eat,3rhrteeekssiVstiprsaetthtancrenatsicnvfvoeesonafllyonntaafedfgnnrredeseiasfarosctdratteahtrhelneteacibnedcousoefstnlattanhallteeectcdoshtnteeaandtrpewcoecohwatneicelnedhrweteschitnnteaeddt3diionwo3ngfiknsetVdhnaiedrnsJ?gil0disn.e0aei6r.seTas0hnt.a5edr Load current= -.1- = 1.1 pu 0.91 Using Eq. (5.5),we get lVs| = 0. 91 + 1. 1( 0. 046x 0. 85 + 0. 072x 0. 527) = 0.995pu = 0.995 x 6.6 - 6.57kV (line-to-line) 6.6/33kV 33/6.6kV Input to a single-phaseshortline shownin Fig. 5.6 is 2,000kw at 0.8 lagging power factor.The line hasa seriesimpedanceof (0.4 + j0.a) ohms.If the load Fig.5.5 voltage is 3 kV, find the load and receiving-endpower f'actor.Also tind the supplyvoltage. Solution lt is convenienthereto employ the per unit method.Let us choose, 2,00k0w I II BaseMVA = 5 at0.8pf *Vs 3kv BasekV = 6.6 on low voltageside = 33 on high voltageside lassins I L __1 Fig.5.6 Cabieimpeciance= (8 + jZ.S) e/phase Solution It is a problem with mixed-endconditions-load voltage and input power are specified.The exact solution is outlined below: _ ( 8 +r2.s ):x s= (0'037+ Sending-endactive/reactivepower = receiving-endactive/reactivepower + (33)' io'0r15P) u activekeactiveline losses For activepower Equivalentstarimpedanceof 6.6 kv winding of the transformer = (O.O+U7O.36=) (0.02+ /0.12)O/phase l ys I l1l cos ds= lVRllll cos / a + l1l2p (i) f For reactivepower l y sI l / l s i n g 5 5 =l V p ll 1 l s i n Q o + l t l 2 X (ii)
'F6, I ModernPowerSystemAnatysis Characteristiacnsd Performancoef PowerTransmissioLninesIl-. l3?.,-{ 5.3 MEDIUM TRANSMISSION I,INE Squaring(i) and (ii), adding and simplifying, we get (iii) For linesmorethan 100km long,chargingcurrentsdueto shuntadmittance lvrl2 lll2 = lVnlzlll2 + zlvRllll2 (l1lR cos /o 100 km to 250 km leneth. it is + tItX sh /n) + tlta @2+ f) sufficiently accurate to lump all the line admittanceat the receiving-end resulting in the equivalentdiagramshown in Fig. 5.7. the numerical values given (iv) l Z l 2 =( R 2 +f ) = 0 . 3 2 Starting frorn fundamentalcircuit equations,it is fairly straightforward to l y s l l 1 l - 2 , o o - o x 1 o 3- 2 , 5 0 0x 1 0 3 [ : ]= l ' * ; ' l l V lwrite the transmissionline equationsin the ABCD constantform given below: 0.8 ( s . 11 ) lVsI l1l cos /, - 2,000 x 103 Fig.5.7 Mediumline,localizedload-endcapacitance l y s l l l l s i n / 5 = 2 , 5 0 0x 1 0 3x 0 . 6 = 1 , 5 0 0x 1 0 3 From Eqs.(i) and (ii), we get Nominal-f Representation If all the shunt capacitanceis lumped at the middle of the line, it leads to the ,l n1cl o s P o = F2 0 0 0 x 1 0 3- 0 . 4 l t P nominal-Zcircuit shownin Fig. 5.8. l1lsin /o 1 5 0 0 x 1 0- 30 . 4 t I f (v) Fig. 5.8 Mediumline,nominal-Trepresentation 3000 For the nominal-Z circuit, the following circuit equationscan be written: Substitutingall the known valuesin Eq. (iii), we have Vc= Vn+ Io(Zl2) Is = In + VrY = In -r Wo+ IR(Z|L)Y (2,500x 103;=2 (3,000)'til2+ 2 x 3,000t|21L0.4*29W\"1390!00p'{lt Vs = Vc + it (ZiZ) x1 s 0 0 xlq1r 0 . 4 112 Substitutingfor Vg and 1, in the last equation,we get + 0 . 4 3000 Jl+ o . z zt t t a vs= vn+ I^(zt2)+ (zD)[r^(t. +)+ YvR) Simplifying, we get = v n ( , . t { ) + rn z ( t . t f ) 0 . 3 2 V f - 1 1 . 8x 1 0 6l l l 2 + 6 . 2 5 x 1 0 1 2= 0 which uponsolutionyields t|_725 A Substitutingfor l1l in Eq. (iv), we ger cos Qp= 0.82 Load Pn = lVRllll cos /a = 3,000 x 725 x 0.82 = 1,790kW Now lV5| = l1l cos ds = 2,000 l V 5l = 2000 : 3 . 4 4 k Y 725x0.8
Rearrangingthe results,we get the following equations l. -r Characteristicsand Performanceof PowerTralq4lqslon Lines [l$k t MVA at 0.8 lagging power factor to a balancedload at 132 kV. The line conductorsare spacedequilaterally3 m apart.The conductorresistanceis 0.11 ohmlkm and its effective diameteris 1.6 cm. Neglect leakance. (s.12 + 2 Nominal- zrRepresentation = 0.0094 pFlkrrr In this methoc the total line capacitanceis divided into two equal parts which R = 0.11x 250 = 27.5 C) are lumped at the sending- and receiving-endsresulting in the nominal- zr X - Zr f L = 2r x 50 x I . 24 x 10- 3x 250= 97. 4 Q representationas shown in Fie. 5.9. Z= R + jX = 27.5 + j97^4 = I0I.2 174.T Q Y= jutl = 314 x 0.0094 x 10{ x 250 lg0\" = 7.38 x 104 lW U Fig. S.9 Mediumline,nominal_r7erpresentation r^ = #\"193xl32 l--i6.g\"o = 109.3I -369\"A From Fig. 5.9, we have vo (perphase)= (I32/d, ) 10\" = 76.2 l0 kv I_s = I n + - Vt o Y + l v s / t + 1 Y \\ l vR +z I * 2Vsy =[\\ + Z \" 2 ) r i +: 1 x 7 . 3 8x io-a 190\"x10r.2134.2\"\\to.z I V s = v o+ e o* , ) v o v >=zv n ( r . * \\2) + 101.2174.2\"x 109.3x 10-3l-36.9\" 1s=r** tvoy+ t;trr-(t+|vz) 7 6 . 2+ 2 . 8 5l 1 & . 2 ' + 1 1 . 0 61 3 7 . 3 \" 82.26+ j7.48 - 825 15.2\" v o v( t+ t ^ v z ) .+ ( ,+ ) v z ) lV5| (line)= 82.6xJl - 143kv Finally, we have I + L Y Z = 1 + 0.0374ll&.2\" = 0.964+ 70.01 2 I('*i4 z l,u.rf / 1 \\ rv^(orrinneoload=) = 148k.3v 'r -'t ', / ffid:;# | | .,ll -\" | (5.13) tzl LrU+;rt) [t*r\")]L1nr Voltage regulation= W#2 x 100 - l2.3vo ndItoestlth:leaoqltuuurulardJvnvnasbatslfeecl3onrnmtotatoetidie<oatnhctathoot ntehoietmhr.eiTnrhoaenl-reze.aanddenrsohmoiunldavl-errrwifyitthhtihsefaacbt obvyeacpopnlysitnagnststaarre_ 5.4 THE LONG TRANSMISSION LINE-RIGOROUS SOLUTION Using the nominai-z- method, find the sending-endvoltage and voltage regulation of a 250 km, three-phase,50 Hz, transmissionline delivering 25 For lines over 250 km, the fact that the parametersof a line arenot lumped but distributeduniformally throughoutits length, must be considered.
-149 1 mooetnPo*\", Sv sis characteristicsand Performanceof Power TransmissionLines I C17e)r'-Czle-)'-21., 9t ,r, -C, ,-',', (5.1e) Z, Z, Fig. 5.10 Schematicdiagramof a longline (5.20) Figure 5.10 shows one phaseand the neutral return (of zero impedance)of The constantsC, and C2 may be evaluatedby using the end conditions, i.e. a transmission line. Let dx be an elemental section of the line at u dirtunr. when .r = 0, Vr= Vn and 1r= In. Substitutingthesevaluesin Eqs. (5.17) and from the receiving-end having a seriesimpedance zdx and a shunt admittanc\"e (5.19)gives yd-r.The rise in voltage* to neutral over the elementalsectionin the direction of increasing \"r is dV\". We can write the following differential relationships Vn= Cr + Cz acrossthe elementalsection: r^=! (ct- cz) dVx= Irzdx o, Y- = ZI, (s.14) Lc which upon solving yield d,lx= v*ldx o, !1' = yvx (s.1s) ,r= (vn+zJn) * It may be noticed that the kind of connection (e.g. T or r) assumedfor the elemental section, does not affect thesefirst order differential relations. 1 Differentiating Eq. (5.14) with respecrro -tr,we obtain Cr= ZJ*) 2Un- -ddr v-T, = d I . i'' sowluitthionf9orr ant c, as determinedabove,Eqs. (5.r7) and (5.19)yield the V.-and 1. as Substitutingthe valueof +dx from Eq. (5.15),we ger ,,=(Yn+/o)*,(,h. .? ),-,. ,.-_(bITk),,._(w*),-,. d2v = rZv' (5.16) (s.2r) Ea Here Z, is called the charqcteristic impedanceof the line and 7is called the propagatton constant. This is a linear differentiai equation whose general solution can be written as follows: Knowing vp, In and the parametersof the line, using Eq. (5.21) complex number rms values of I/, andI, at any distancex along the line can be easily V*=CpI**Cre-1x (s.r7) found out. where A moreconvenientform of expressionfbr voltageandcurrentis obtainedby introducing hyperbolic functions. RearrangingEq. (5.21), we get 7= ,lW (s.18) and C, andC, are arbitraryconstantsto be evaluated. v * = v n t( o7 +* + l +o-7'\\ r\"^' 2(/ , +( e : J : : ) DifferentiatingEq. (5.17)with respecto x: \\ 2/ 2) *-_- I .\" = V\"o2L\"(\\ \" * -.t-\" )* , \"\" ( e 1' +e - t ' \\ 2 ) \\ 2 ) Here V' is the complex expressionof the rms voltage, whose magnitude and phase vary with distance along the line.
Characteristicasnd Performancoef PowerTransmissioLninesffi - Thesecan be rewritten afterintloducing hyperbolicfunctions, as sin7h/- |.+*4* Vr= Vn cosh 1r + I^2, sinh 1r (5,22) 3! )! ..=Jyz(H+) \\( s . 2 8 a ) \\^' 6.) I,= Io cosh rr + V- -l-.i This seriesconvergesrapidly for qvrarrlrurnewsio-of .l7.rt us^lu-^ai^ll-y.^enc-or,uL-nte--redfor power lines and can he convenie-nflw expressionsfor ABCD constanis art H=l l;::,',:::i;:l;;twhenx = l, Vr= V, Ir= Is A=D=l* YZ 2 (s.23) Bx,z (t.+) (s.28b) Here c x Y(r*V\\ A=D -cosh 7/ (s.24) \\ 6) B = Z, sinh 7/ The aboveapproximation is computationally convenient and quite accuratefor lines up to 400/500 km. c = J-sinh :r/ Z, Method 3 *In case[vs 1s] is known, fvn Inl can be easilyfound by inverting Eq. (5.23). cosh(o/ + i7t) - tat\"iot +-'-?t'-io_t = 1 tpt + e-\"1 l-Bt1 Z H=l l-i-,:l[:] (s.2s) ;(e\" sinh(o/ + jpl) - ,at\"i0t _e:de-ipt (s.2e) : ){r\", tpt - t-d l-Bt1 Evaluation of ABCD Constants 5.5 INTERPRETATION OF THE LONG LINE EOUATIONS The ABCD constantsof a long line can be evaluatedfiorn the resultsgiven in As alreadysaid in Eq. (5.26), 7is a comprexnumber which can be expressed F4. 6.24). It must be noted that 7 -Jw is in generala complex number and as can be expressedas ^ 7- a+ jp (s.26) 7= a+ jp cTohneshtaynptesrcbaonlibcefucnocmtipounteodf.bcyomanpyleoxnneuomf btheersthinrveoelmveedthiondesvgaivlueantiunJgtoAwB.CD The real part a is called the attenuationconstantand the imaginary ^part Bis called the phase constant.Now v, of Eq. (5.2r) can be writtin as V^, = Z,lRlr.,,ritgx+n |I V R +2 | * l V| n - Zz, l n l I 'o - - c , x , - t t / , x - e 2 ) Method I where cosh (cr./+ j1l) = cosh ul cos gl+ j sinh a/ sn pt (5.27) Qr= I (Izn+ I&,) (5.30) sinh (a/ + jQl) = sinh al cos gl + j cosh a/ sn pt dz= I (Vn- InZ,) The instantaneouvsoltagev*(t) canbe writtenfrom Eq. (5.30)as Note that sinh, cosh, sin and cos of real numbersas in Eq. (5.27) can be looked tp in standard tables. Method 2 v*(t)= xd Ol!- +3/tl ,o',i@t+r,,.+o,) Lt2l cos7h=/ r++* #*. .=(.t+)
144 | nrodern Power System Analysis characteristicas nd Performanceof PowerTransmissionLines +, Jnl llVt r^ - -Z*r Ie* lt t^'-'*- t L- j (ta-tr- /ht x)+ t' 1I (5.31) le The instantaneouvsoltageconsistsof two termseachof which is a function - At(f+40 of two variables-time and distance.Thus they represent wo travelling waves, Vx= Vrl * VxZ (s.32) Sendingendx=/ Now \"',,- JIlh-?\\,* \"or{'l+gx+h) (5.33) At any instant of time t, v.rl is sinusoidallydistributed along the distance Fig. 5.12 Reflectedwave from the receiving-endwith amplitudeincreasingexponentiallywith distance, as shownin Fig. 5.11 (a > 0 for a line havingresistance). Envelop eox If the load impedance Zr = + = 2,, i.e. the line is terminatedin its u-rAf -x=0Receiving . IR B end characteristic impedance,the reflected voltage wave is zero (vn- zJn= 0). -'- ' Direction of A line terminatedin its characteristicimpedanceis called the infinite line. Sendingendx=/ t r a v e l l i n gw a v e The incident wave under this condition cannotdistinguishbetweena termina- tion and an infinite continuationof the line. F i g .5 .1 1 l nci denwt ave Power systemengineersnormally caIIZrthe surgeimpedance.It hasa value After t-imeAt, the distribution advancesin distancephaseby (u'Atlfl. Thus of about 400 ohms for an overheadline and its phaseangle normally varies thiswave is travellingtowarclsthe receiving-endandis the incidentv'ave'Line from 0\" to - 15o.For undergroundcablesZ. is roughly one-tenthof the value lossescauseits arnplitucleto decreaseexponentiallyin gclingl}onr tht: sendirlg for overheadlines.The term surgeimpedanceis, however, usedih connection to the receiving-end. with surges(dueto lightningor switching)or transmissionlines.wherethe lines loss can be neglectedsuch that Now z,=z,= ( rl/2 i/1 u,r=Elt+tlu^ co(sa-z0x+dz) (s.34) ;,.,1 (;)\"'. ,,purcrcsist''cc After time At the voltage distribr-rtionretardsin distancephaseby (uAtl4. l:i:r) This is the reflecterlwavetravelling trom the receiving-endto the sending-end with amplitudedecreasingexponentiallyin goingfrom the receiving-endto the SLtrgeImpetlance Loading (SIL) of a transmission line is.defined as the sending-enda,s shownin Fig' 5.12. power delivered by a line to purely resistive load equal in value to the surge impedance of the line. Thus for a line having 400 ohms surge impedance, At any point alongthe line, the voltageis the sum of incident and reflected voltagewaves presentat the point tEq. (5.32)1.The sameis true of current SIL= JT -y!- tlv,oI x lto00ookw waves.Expressionsfor incident and reflectedcurrent waves can be similarly J3 xaoo written down by proceedingfrom Eq. (5.21).If Z\" is pure resistancec, urrent wavescan be simply obtainedfrom voltagewavesby dividingby Zr. = 2.5 lyRl2kw (s.3s) where lVol is the line-to-linereceiving-endvoltagein kV. Sometirnesi,t is found convenientto expressline loading in per unit of SIL, i.e. as the ratio of the power transmittedto surge impedanceloading. At any time the voltage and current vary harmonicallyalong the linc with respectto x, the spacecoordinate.A completevoltageor currentcycle alongthe line correspondsto a change of 2r rad in the angular argument Bx. The correspondingline length is defined as the wavelength. It 0 i.sexpressedin radlm,
-iAd ModernPo*g!Sy$e!l Anelysis 147 I )-Zn/gm (s.36) ' = velociotvflight (s.42) Now for a typicalpowertransmissiolnine *fu= g (shuntconductancelunleitngth)= 0 The actual velocity of the propagation of wave along the line would be somewhaltessthan the velocityof light' 7= (yz)1/2= Qu,C(r+ juL))rt2 ., = 1\" lot = 6,000km \"- 50- - iu (LQ'''(r- t i)''' Practicaltransmissionlines are much sho rterthan this (usuallya few hundredkilometres). It needsto be pointed out heyethat the vyavestlrau'ttin 7= a+ jg = ju\\Lc),,,(t-t#) Figs,5.11and5.]'2areforitlustrationonlyanddonotpertainnareal power transmissionline' a - - rl ( l C \\ l t z (s.37) - 2\\L) 0 = a (Lq''' (s.38) Asenthdrieneg-.epnhdatsse52200HkVz .tTrahneslimneispsaioranlmineeteisrsa4r0e0r km long' The volta geat the = o.!25 ohnr/km,x = 0.4 ohm/ Now time for a phasechangeof 2n is 1f s, where/= cul2r is the frequency in cycles/s.During this time the wave travelsa distanceequalto ). i.e. one tm ani y = 2.8 x 10-6rnho/km' wavelensth. Find the following: \\' (5.3e) (i)Thesending-enclcurrentandrebeiving-endvoltagewhenthereisno-load Velocity of propagationof wave, , = -r4i =f: f^ m/s \"' which is a well known result. (s.40) Solution The total line parametersare: For a losslesstransmissionline (R = 0, G = 0), R - 0.125x 400 = 50.0 f) ,= 7yz)'''- iu(LC)tlz X = O.4x 400 = 160'0 fl such that e. = A, 0 = . (Lq''' Y = 2.8 x 10-6x 400 lg)\" - l'12 x IA-3 lxf U Z = R + i X = ( 5 0 ' 0+ j 1 6 0 ' 0 )= 1 6 8 ' 01 7 2 ' 6 \" Q )-2110- ,?n=,=: 1,,- m YZ = l.l2 x 1O-3/90\" x 168 172'6\" and (s.41) v = fA = ll(LC)rlz m|s = 0.1881162'6\" For a single-phasetransmissionline L= lto ,n D (i) At no-load 2r r' Vs= AVn'ls = CVa 2ffi0 C_ ln D/r v' = 4 2*o ' A and C arecomPutedas follows: (Pol^D 'lt\"nG )t / 2 tl Y Z= l + It;,.t- ) /AL =- l * l 0 . 1 8 81 1 6 2 ' , 6 \" *x )2 Since r and rt arequite closeto eachother, when log is taken,it is sufficiently = 0.91+ j0.028 accurateto assumethat ln q, = h D/r.
.r14E-f ModernPowerSvstemAnalvsis lAl= 0.91 Characteristicasn. d Performanceof PowerTransmissionLines I 14g^ t- simplifying, we obtain the maximum permissiblefrequency as C = Y(l + YZ/6)= f = 57.9Hz = 1.09x 10-3I ' lvnh=n'#:# = 242kY If in Example 5.5 the line is opencircuited with a receiving-endvoltage of 220 kV, find the rms value and phaseangle of the following: 1 1l5= l c l l V R l = 1.09x 10-3xry * 103= 152A \"J3 (a) The incident and reflectedvoltagesto neutral at the receiving-end. (b) The incident and reflected voltages to neutral at 200 km from the It is to be noted that under no-load conditions, the receiving-endvoltage (242 kV) is more than the sending-endvoltage. This phenomenonis known as receiving-end. the Ferrantielfect and is discussedat length in Sec.5.6. (ii) Maximum permissibleno-loadreceiving-endvoltage= 235 kv. (c) The resultantvoltageat 200 km from the receiving-end. Note: Use the receiving-endline to neutral voltage as reference. ^, r ,,=llV%\" l l :2f2f0i = 0 . e 3 6 solution From Example 5.5, we have following line parameters: r = 0.725 Qlkm; x = 0.4 Olkm; y = j2.g x 10{ Uncm z = (0.125+ j0.4) Olkm = 0.42 172.6 CI/km Now y- 1yz)t'2= (2.8 x 10-6x 0.42 /.(g0 + 72.6))t/2 = 1 . 0 8x 1 0 - 3l g I . 3 1 A=l+LYZ = ( 0. 163+ i1. 068) x t O - --r 0+ j{J 2 = 1+ -1; t ^ .i2.8x 10-6x (0.125+ 70.4) a = 0 . 1 6 3x t O - 3 ;f = 1 . 0 6 gx l 0 - 3 z ', (a) At the receiving-end; = (1 - 0._56x t0-6P1+ j0.t75 x r0-6P For open circuit 1n= 0 ; S in c eth c i rn a g i n a rpy a rtw i l l b c l essthan, th of the real part. lzll can be Incidentvcrltage- Vn I ?cl n - V,, ,l 2 approxirnatedas l A l = | - 0 . 5 6x l r J 6 P= 0 . 9 3 6 - 220/J3 2 p _ r-0.936 = 63.5110\" kV (to neurral) 0 . 5 6x l 0 6 Reflected voltage - vR-zclR - VR / = 3 3 8k m ) ;;,, ,A,= 220= 0 . 8 8 = 63.5110\" kV (to neurral) 250 (b) At 200 km from the receiving-end: I L( to+160,.I) l/l A-l*;xi1.l2xl0-3x J0\\ s0) Incident voltage= :!u+urltlxl Neglecting the imaginary part, we can write 2 lr:2oorm tA=l 1- +\" r . r 2x 1 o -x3 1 6 0x 0.88 = 63. 51exp ( 0. 163x l0- 3 x 200) &= x exp 01.068 x 10-3x 200)
ffi0'.I ModernPo*e, SystemAnatysis characteristaicnsdperformanocfepowerTransmissLioinesI rsr f = 65.62112.2\" kV (to neutral) - / v ^u\",'J\\ andturnsrhrougahpositiveanglepr (represenrbeydphasoor B); I i. Reflectedvoltage- Ys-\"-'ur-itt'l 2 l' ,,,n,u- while the reflected voltage wave decreasesin magnitude exponentiaily -Y-E---u, = 6I.47 l-12.2\" kV (to neutral) It is apparentfrom the geometryof this figure that the resultantphasor voltage (c) Resultantvoltage at 200 km from the receiving-end Vs QF) is suchthat lVol > lysl. - 65.62 112.2\" + 61.47 I - 12.2\" sahdovAawnsnciimnedpFbleyigel.u5xmp.1lpa4inntaghtteihocenaoinpf dathucecittaFanencrcreeiasannltudi mecfapfepldcatcaoittnathnaecneipaeapcrpearimovxienimtge-aresotnefbdtoahfseitslhicneaeln.inAbese. = 124.2+ j0.877 = 124.2 10.4\" Resultantline-to-linevoltageat 200 km = 124.2 x J3 - 215.1 kV 5.6 FERRANTI EFFECT As has beenillustratedin Exarnple5.5, the eff'ectof the line capacitanceis to Fig.5. 14 cause the no-load receiving-end voltage to be more than the sending-end voltage.The effectbecomesmorepronouncedas the line lengthincreasesT. his V, phenomenonis known as the Ferranti.effect. A general explanationof this (-+-.Here Ir= , effect is advancedbelow: \\ juCt ) Substitutingx = / and In = 0 (no-load)in Eq. (5.21), we have Since c is small comparedto L, uLl can be neglectedin comparisohto yc,tl. Thus Vn Vn 15 - jVruCl ,at4gt * ,-at t7 /5 -- --ifl (< A2\\ Now Vn= Vs- Is QwLl) = V, + V,tj CLlz 2 \\J.-tJ I = vs0+ Jctt2) - -\\l.n-creasino/ (s.M) Magnitudeof voltagerise _ lvrltJ CLf Locus of V5wlth/ =olv.rt+ (s.4s) D where v = 7/J LC. i1 the velocity of propagationof the electromagneticwave Vpfor I = 0 along the line, which is nearly equal to trr\" velocity of light. \\ 5.7 TUNED POWER LINES En= Ero= Vpl2 I n c r e a s i n g/ nlEineqegulasethcioutlnnint(ec5o-r2ne3ds)uisccthataannrcaceceRtGearisiszewasltewhlale.ypwseintrhefogrtlmihgiaisbnalecpeaponrfodaxitliomisnagstiuloifnnfeic.ieFnotrlyaancocvuerrahteetaod 7- Jyz = jalLC Fig. 5.13 cosh 7/ = coshjwlJTC = cos Lt,lrc The aboveequationshowsthat at I = 0, the incident (E,o)andreflected(E o) voltage waves are both equal to V^/2. With reference to Fig. 5.13, as I increasest,he incident voltage wave increasesexponentiallyin magnitude
;fiTiI uooernpowesr ystemRnarysis eharacteristicsand performanceof power Tr@ I Z = Z, sinhl/ = sinh ?/ = sinh jalJ-LC = j sin wtJLC t- HenceEq. (5.23) simplifies to cosulJ LC jZ, sinutJE = ! .o,nh..,tt2 cosutE rT . _1_ _ = g a n h 1 i l 2t () - t U2 ) Z, sina'tJ LC __ Y_ { l Nowif alJtC = hr, n = I,2,3, ... Fig. 5.15 Equivarent-nzetworkof a transmissiornine lV5l= lVpl For a zr-netwoLr,ksr,h1,o-_JwL|n in(rtF,.i+(gI.,t!5,.v1t5,,)[2ref,e(r)rro*Eziqr.,(,5r.1,3.))J]l.rLu, ,._ ) llsl = llal (5.48) i.e. the receiving-end voltage and current are numerically equal to the Accordingto exactsolurionof a long line [refer to Eq. (5.23)]. (s.4e) colrespondingsending-endvalues,so that thereis no voltagedrop on load. Such a line is called a tuned line. Lf1i,_:,=IIfzi,lH;'___... :c\"o!s7hl/ J'Lltl^J,^t For 50 Hz, the length of line for tuning is For exactequivalence,we must have 1- --!r_. 2nrfJ LC Since ll^frc = y, the velocity of light (s.47) Z/= Z, sin h 7/ r =+ @ ) ), i=^ , ^ , ? r x . . . (s.s0) = 3,000km,6,000km,... f * *2Y t Z = c o s h 7 / (s.51) From Eq. (5.50) It is too long a distanceof transmissionfrorn the point of view of cost and efficiency (note that line resistancewas neglectedin the above analysis).For Z' = ,vl -Y: . s i n n1 / = Z, tlnh'' : z-([*init rr 1 (5.52) a given line, length and freouencytuning can be aehievedby increa-singL or tJyz 7t ) C, i.e. by adding seriesinductancesor shunt capacitancesat severalplaces along the line length.The methodis impracticaland uneconomicaflor power 11-\"'' -sln!-24 is thc fitct.r by which thc scricsirupcdunceol'the no'ri'al-z frequencylines ancl is adoptedfor tclephonywhere higher frecluencieasre employed. must be multiplied to obtain thez parameterof the equivalent-a Substituting 7 from Eq. (5.50)in Eq. (5.51),we ger A methodof tuning power lines which is being presentlyexperimentedwith, uses seriescapacitorsto cancel the effect of the line inductanceand shunt I inductorsto neutralizeline capacitanceA. long line is divided into several sectionswhich are individuatly tuned.However, so far the practical method of 1 * ; YtZ, sitrh 7/ = cosh fl improving line regulationand powertransfercapacityis to add seriescapacitors to reduceline inductance;shuntcapacitorsunder heavy load conditions;and shuntinductorsunder light or no-loadconditions. 5.8 THE EOUIVALENT CIRCUIT OF A LONG LINE So far as the end conditions are concerned,the exact equivalentcircuit of a Lv,- (s.53) transmissionline can be establishedin the form of a T- or zr-network. 2 The parametersof the equivalentnetwork areeasily obtainedby comparingthe perfbrmanceequationsof a z--networkand a transmissionline in terms of end quantities.
Wl ModernPowerSystemAnqlysis I dr n u s (tanhfll2\\. thc lactor by which thc shunt aclmittanccanr ol' the V R =+ 2 0 - l z 7 1 0k v I is VJ \\ il/z ) (a) Shortline approximation: nominal-n-mustbe multiplied to obtain the shunt parameter(Ytl2) of the e q u i v a l e nzt -. /1\\l Vs= t27 + 0.164/._36.9 x 131.2172.3 = 14514.9 Note thatYtl | + +Y' Z' I: j- sinh 7/ is a consistenet quationin termsof the \\4)2, lYsltin=\" 25L2kV abovevaluesof Y/and Z/. Is= In= 0.764/_-36.9kA Sending-enpdowerfactor= cos(4.9\"+ 36.9\"_ 41.g.) For a line of medium length tuth-l!/2 - 1 and sinh 7/ = 1 so that the = 0.745laggrng fll2 1t equivalent- n-network reduces to that of nominal-n-. Sending-epnodwer_ JT x 251.2x 0.764x 0.745 = 53.2MW x@)Nominal-trmethod: A=D=l+ !y z= .I + _I x ' 2 l0-r lg0\" x l3l .2 /.72.3\" Fig. 5.16' Equivalent-Tnetworkof a transmissionline = 1 + 0.0656L162.3= 0.93g/_L2\" Equivalent-T network parameters of a transmission line are obtained on B=Z=I3L.2 172.3\" similar lines. The equivalent-T network is shown in Fig. 5.16. As we shall see in Chapter 6 equivalent-r (or nominal-r) network is easily c = v ( r + ! v z ) =y + l f z \\4/4 ^.1^^+^l +^ l^^.1 {l l'Ll ^r W, r , oJ Lt rUr U. l Mi o o qol-l.u1 lioot fl l}trvol vr el vfl'vntr o r( 4r nl rilr r pvr vcrqLltlq. ,r r J pmnlnrrerl vs' <fllIWPLVLT L\\J l\\-rat.l vr^lHrvJ Exirmple5.7 = 0.00| 190\"+ 4I . t0-61Jg0. x 131.2L72.3\" = 0.001/.90 A 50 Hz transmissionline 300 km long hasa total seriesimpedanceof 40 + j125 ohmsand a total shuntadmittanceof 10-3mho. The receiving-endload is 7s= 0'9381r.2 x r27+ r3r.2 172.3\"x 0.164L-36.9\" 50 MW at220 kV with 0.8 laggingpowerfactor.Find the sending-endvoltage, = 719.7/7.2 + 21.5135.4\"= 73j.4 /.6.2 current,power and power factor using lVsllin=\" 238kV (a) shortline approximation, (b) nominal-zrmethod, 1s= 0.001Z90 x 127+ 0.93gZl.Z x 0.164/._ 36,9\" (c) exacttransmissionline equationlBq. (5.27)1, = 0.127l9A + 0.154l_35.7\"= 0.13 /.16.5\" (d) approximation[Eq. (5.28b)]. Comparethe resultsand comment. Sending-epnfd= cos(16.5_. 6.2) = 0.9g4leacling Solution Z = 40 + iI25 - 131.2 172.3\" Q Sending-endpoweJrT= * 23gx 0.13x 0.gg4 Z= 10-' 190\" U = 52.7MW The Leceiving-elnodadis 50 MW at ?20kV,0.8 pf lagging. (c) Exacttransmissiolnine equarion(sEq.(5.29)). 1o= J,3=-x+220x0- .8 l-36.9\" = 0.164 l-36.9\" kA fl = al + jpt =JyZ
- 155 | - ModernPowerSystem4nalysis characteristicsand Performanceof PowerTransmissionLines = 0.938 ll.2 (alreadycalculatedin part (b)) = J r o - t l g o \" * l 3 t . z l 7 z . 3 \" = 0 . 0 5 5 4+ i 0 . 3 5 7 7 = 0 . 3 6 21 8 L 2 \" Z+ | yf 6 cosh(al + i 0l) = !k\"' lgt + e*t l-Bt1 B= z(t*!Z\\= 'gl = a.5'fi1(radians)-r'6.4,g\" \\ 6) eo.0s514e0.49\")= 1.057lzo.49\"= 0.99+ j0.37 = 731.2172.3 + -{- x 1 o - 3l g o \" x (13l.D2 1144.6. 6 e4'oss4l-20.49\" = 0.946 l- 20.49\"= 0.886 - j0.331 = 131.2172.3\"+ 2.87l-125.4\" = 128.5172.7' gosh7/ = 0.938+ iO.O2= 0.938 11.2\" c= y(w!Z') ) = o.oo1tetr+ ]6x to{ lrgo xr3t.2172.3\" sinh 7/ = 0.052+ 70.35= 0.354 181.5\" \\6 E l3nnn\" = .s,o^z.2I = 0.001l9O\" L .\". = 1- : -t----- l- 8.85' Vs= 0.93811.2\"x 127 10\"+ 128.5172.7'x 0.164I -36.9\" Y 10-'190\" lY A = D = c o s hf l = O : 9 3 81 L . 2 \" = ll9.I3 11.2\"+ 21.07135.8\"- 136.2+ it4.82 t t- 88sx\" 0354tsr s' = 13716.2\" kV =7::: ,lr ul\"u\"' IYsIrin\"= 237.3 kY /s = 0.13116.5\"(sameascalculateidn part(b)) Now Vs= 0.93811.2\"x 127 10\" + 128.2172.65\"x 0.164 l-36.9\"-' Sending-enpdf = cos(16.5'- 6.2 = 10.3\")= 0.984leading = 1 1 9 . 1 31 1 . 2 \"+ 2 L 0 3 1 3 5 . 7 5 \" = 136.9716.2\"kV Sending-enpdower= Jl * ?313x 0.t3 x 0.984 lVshin.= 81,23--W = 52.58MW 'fhe resultsaretabulatedbelow: C = Lsinh .y/ f x 0.354 181.5\" Short line Nominal-r Exact Approximation z, 3 6 2 . 2 1 1 -8 . 8 5 \" uppntximatktn 238kV 2 3 7 . 2 3k V (s . 2 8 h ) = 9.77x 104 190.4\" .\\ l y s l h n \"2 5 I . 2 k v o.I3 116.5\"kA 0.12861t5.3\" kA I, 0J& l-36.9\" kA 0 . 9 8 4l e a d i n g 0 . 9 8 7l e a d i n g ? 3 7 . 3k V Is = 9 .J 7 x l O -a1 9 0 .4 \" x 127+ 0.938 11.2\" x0.164 l -36.9 p.f, 0.745 lagging 52. 7M W 52. 15M W 0. 131t 6. 5\"kA P\" 53.2 MW 0.984leading 5 2 . 5 8M W = 0.124190.4\"+ 0.154\"- 35.7\" = 0.1286 115.3\"kA Comments Sending-endpf = cos (15.3'- 6.2\" - 9.1\") = 0.987leading We find from the above examplethat the resultsobtainedby the nominal-zr Sending-endpower = Jt x 237.23x 0.1286 x 0.987 method and the approximation (5.28b) are practically the sameand are very close to thoseobtainedby exact calculations(part (c)). On the other hand the = 52.15MW results obtained by'the short line approximation are in considerableerror. Therefore,ftlr a line of this length(about300 knr), it is sufficientlyaccurateto (d) Approximation(5.28b): use the nominal-r (or approximation(5.28b))which resultsin considerable savingin computationael flort. A'=D = | + | Yz )
1 5 E lI M o d e r nP o w e rS y s t e mA n a l y s i s characteristicsand performanceof power TransmissionLines 5.9 POWER FLOW THROUGH A TRANSMISSION LINE sn=lvRt0t ti+lvstt13a- -l!lrv^tz{1-*t] = 'ul'll^t'(/r-d-) l-4v^t2l1p - ay So far the transmissionline performanceequationwas presentedin the form of voltase and current relationshipsbetween sending-andreceiving-ends.Since Similerly, t, =I?l,r,ftur- u)-lll*lf t@+6) (s.se) loadsaremore often expressedin terms of real (wattslkW) and reactive(VARs/ kVAR) power, it is conveniento deal with transmissionline equationsin the taIhnnedtnhvetrhaeabrtohevreeexeep-qrpuehasatsisoeerndienscspeoeiavrinpndghs-ae,sneadvrceoopltmesr.pIpflehxyaRpsoeacrn.o,dm7pi,slaegxrievveeonxltpabrmyepseseredisn,wkhvilelinve*, form of sending- and receiving-endcomplex power and voltages.While the problem of flow of power in a general network will be treated in the next chapter,the principles involved are illustrated here through a single transmis- sion line (2-nodel2-bussystem)as shown in Fig. 5.17. so(3-phavsA'e-/)= -t {Ll#Jt%*+J1T9lB1l.t(-'s\"'-rlE,-ll4l tvo3tx2to6L,,\\,t,l-a,)iF S s = iQs Sp = Pp +lQP s^(3-phaMseVA=) T# 4p - , -l*lvopt1p- a1 (s.60) F i g .5 .1 7 A tw o-bussystem This indeedis the sameas Eq. (5.59).The sameresultholds for sr. Thuswe see that Eqs. (5.58) and (5.59) give rhe three-phaseMVA if vs ina vo *\" Let us take the receiving-endvoltageas areferencephasor(Vn=lVRl 10\") expressedin kV line. and let the sending-endvoltagelead it by an angle 6 (Vs = lVsl 16). Tlp- angle If Eq' (5'58) is expressedin real and imaginaryparts,we can write the real d is known as the torque angle whose significance has been explained in and reactivepowers at the receiving_endas Chapter4 and will further be taken up in Chapter 12 while dealing with the problemof stability. r*\" = JI]tlB-vt ^]co(sv/ - a- / - ll{rll \"rv^rc' os(B- a) (5.61) The complex power leavingthe receiving-endand enteringthe sending-end of thetransmissiolnine canbeexpresseads(onperphasebasis) Q/ 1 n=flfy s l l y * l s.in1, n/- 6^ t- l A l v.n_i_'?. \"s Sn= Pn+ .iQn-- Vnfn /E E A\\ l li i i n( /- a) (s.62) (5'63) (J.J+/ Ss= Ps+ iQs= YsI; (5.s5) similarly, the real and reactive powers at sending-endare Receiving-and sending-endcurrentscan, however,t\" .*pr\"rsed in termsof p', = lIB?ll v,JP cos(o- o)- lvsllv'l tBicos(0*A receiving- and sending-endvoltages[seeEq. (5'1)] as t,r=ivr,= -* vR (s.s6) tvsttvR't ( / 1 + b ) Qp rs = Il*fl | .'u' .r, ,t sin(p- a)- l-srtl 6.e) ,r=*r, - (s.s7) It is easyto seefrom Eq. (5.61) that the receivedpower po will be maximum *ro at Let A, B, D, the transmissionJine constants,be written as 6=0 A = lAl la, B = lBl lP, D = lDl lo (sinceA = D) such that Therefore,we can write ' u : ' l l ^ '- t A t t v ' Pc , o s( / 1_ : t^= r1 rt v 5tt( , _ 0 ) _ rl BAll r v4\" ar _ 0 ) P o( m a. x=) IBt lB- av) ) (5.65) 'l ;(l' The corresponding en @t max po ) is r , =i + lv , t ( a +6 -, - l + l v Rtt- p oA^n== -l4-Lf fYf :r!yt-.s i n ( 0 - u ) Substitutingfor 1o in Eq. (5.54)'we get
I ModernPo*e, Sy.t\"r Analysir _ f60 | Thus the load must draw this much leading MVAR in order to receivethe Equation(5.72)canbefurthersimpl ifiedby a ssumincgos 6 = t- maximumreal power. normallysmall*.Thus r,sincedis Considernow the specialcaseof a short line with a seriesimpedanceZ. Now A-D=I l0:B=Z=lzlle Substitutingthesein Eqs. (5.61)to (5.64),we get the simplifiedresultsfor Let lvtl - lvRl= lAv1,the magnitudeof voltagedrop acrossthe rransmission the shortline as line. po- Yrillol cos(g-6\\-lvRPcoso (5.66) n ^ =# t a v l (s.74) lzl tzl g o = l v ' ) l Y * l - s(i nd - b ) - l v * l ' , i n o (s.67) severalimportanrconclusionsthat easilyfollow from Eqs. (5.71) to (5.74) lzl tzl are enumeratedbelow: for thereceiving-enadndfor thesending-end I' msrFemoaargal pRnlloivtw=uadelue0retor(sfawonthhfsiecfedhvrori)sel,tdaatwgovhetaihdlleeridoretphacaepecpirvrreooinasxgcsim-tthieuaen.tdliipiosnonepf*.or\"o,rpaoisrttriaopnnrosampltooisrtssioiionnn6alil(n=teo)6ttfhhoeer 2. cTlvohsnellsvriRdeveaxrl a.ptooiofwnecsorof ursertscaeeb,ivildeitdyistisorbemesatdrxiiicsmtceuudmstsofeodvirnarC6ueh=sapw9teet0ilo1ab2ne.dlohwas90aovfarorume ps =l I :1 . o ,e - tl 1'lu n. ol ,@ + 6 ) (s.68) 3. bsviMnftoaaclttxrpaieioomganweusselmee.rdvbeorylevsearaalrriepsroboinnweggieniertgsrptrdvraooisngltstraafeengsrcerseeilvsedevwfleoyaplr.nulaatsnhigtseeidvfdureotpnymtolitnlth*regais;n(csfiomx;eniirdtstllaildnrXeg\"r)reuartctciirahornguntnhbkaest \" l zl tzl evo,_ ry_srnd_'u:'lI.s'- in(\\ d*o (s.6e) lzl lzl The above short line equationwill also apply for a long line when the line is replacedby its equivalent-r (or nominal-r) and the shunt admittancesare lumped with the receiving-endload and sending-endgeneration.In fact, this techniqueis always used in the load flow problem to be treatedin the next chapter. From Eq.(5.66),the maximum receiving-endpower is received,when 6 = 0 ' 4ptasFh'ucu1oecrc'rpssolvturemeecevrspideyteflsriulnosictrthnsyt-hoeg,dtefdlalbtirlhnyciiynneigoeaschndvholvodyavolieptnlcarttghaegvsogreoeei1llcetrt2aceive.gecSiseslhcecsnaratuoiopneniaonsdrccogeeaityrdtbposT.ueawrocrsciaieintintictsohcthehrreisdewneblagolieicnsunyoeelepocn.onrldoeTwdifathteichcoir<settnrallisuaidnm.rnecsisateeismnT.wpcihitlrltialcsecodbeiisnesd!. sothatP^ (rnax=) lv]\\Yol Ytc's // tzt tzl Now cos d= RllZl, Pp(ma=x)'tr',|,o-'l'r-t o (5.70) 4. dbdviiAnsioyesslpsatcsarieuroagtmesrpieaddspovniperlnoostrdailomyo1ttfainiinallsgaeegbselntoipaopogvrtonttoehshts,fhtyieiihnltaselieitvtn,SepevtemheAoevcviRinAfo.v5stlARtt.aht1R(-gele0asegn.vgdeddAgrsneiRtonoempsgrsaaareadntnoageJdorcmrissaft(iattvcihnhneodeepdnerolodospwfeaheetndahnrsrmed)pedeullryoenss.arto)tiT.dbTvfoeehdisrmimesTldeaahbwtrienylgoirtleealac,failotnibnhlrleaeyee, Normally the resistanceof a transmissionline is small compared to its reactance(since it is necessaryto maintain a high efficiency of transmission), so that 0 = tan-t xlR = 90\"; wherez = R + jx.The receiving-endEqs. (5.66) and (5.67) can then be approximatedas (s.1r) A 5sin.o5m)te,eirw.meh.satomf oarcetiavcecaunradterylaecttiavpeprpooxwimearstecreasnublteexwprritetessnindgiiriencetlvyoflrtaogme i drop Eq. ( ri I I I (s.12) lAVl= llnlR cos Q + llolX sn Q *small lvRl dis necessarfyromco nsidera tionosf sys temstab ilitywhichwil l bediscu ssed lengthin Chapter12. at
162 | power SvstemAnalvsis characteristicsand Performanceof PowerTransmissionLines \"odern = R!ry* XQn (s.75) Case(a): Cable impedance= 70.05 pu. Sincecableresistanceis zero,thereis no realpower lossin the cable.Hence lvRl Pcr t Pcz= Por * Poz= 40 pu Thisresulrt educetso thatof Eq.(5.74)if R = 0. j Example5.8 Pot= Pcz= 20 pu An interconnectocrablelinks generatingstations1 and2 asshownin Fig. 5.18. The voltage of bus 2 is taken as ref'erencei,.e. V, 10\" and voltageof bus The desiredvoltageprofileis flat, i.e.lVrl=lVzl = 1 pu. The totaldemandsat 1 is V1 16r. Further,for flat voltage profile lVll = lV2l- L the two busesare Real power flow from bus I to bus 2 is obtained from Eq. (s.68) by Spr=15+75Pu recognizingthat sinceR = 0, 0= 90\". Soz=25+ 715Pu Hence The stationloads are equalizedby the flow of power in the cable.Estimate the torqueangleand the stationpower factors:(a) for cablez = 0 + 70.05pu, DPs_=DPn = -lfv t l l v l 6, and (b) for cable Z - 0.005 + 70.05 pu. It is given rhat generator G, can sin generatea maximum of 20.0 pu real power. - -Sl lxIl l d, Solution The powers at the various points in the fundamental(two-bus) systemare definedin Fig. 5.18(a). )= 0.05 @ or 4 = I4.5\" Sn.' = P61+jQ61 vr= I lL4'5\" I FromEq. (5.69) lvlt51 I ^' lv,l' lv,llvl - Qt= cosdt X vQu-1. -- ,DD-1. r i A' l \\ 1 D 1 Jp2 = I-p2+ J\\Jp2 - -,: x 0.e68= 0.638pu c) \" :U,\\,J U.UJ From Eq. (5.67) (a) f2o= -l vTtcl lovs, l d' t v 1 2- - Qs= - 0 . 6 3 8P u X Reactivepower loss*in the cable is __> QL= Qs- en _ 2es _ 1.276pu SDI= 15+/5 Sn=5-l Totalload on station1 = (15 + i5) + (5 + j0.638) G) (b) = 20+ j5.638 15+j5 { zo.ro+ i16.12 Powerfactorat starionI = cos [\\run-' t'f:t') = 0.963lagging 20 ) V 2 =1 . 0l o \" Totalload on station2 - (25 + jls) - (5 - j0.638) =20+i15.638 25 +j1S (c) Reactivc powcr loss can also be cornputedas l/l2X= :t + ( 9 .! 1 q )\"1o . o st=.z 7 pu. Fig. 5.18 Two-bussystem I
164 I toctern power Svstem,Analvsis characteristicsand Performanceof power TransmissionLines 165 Powerfactorar station2 = coS (ron*r 15'638') = O.zgslagging 6r= 14.4\" \\20) Substitutingdr in Eqs.(ii), (iii) and(iv), we ger The stationloads,load demandsa, nd line flows areshownin Fig.5.1g(b). Q c t = 5 . 1 3 ,Q c 2= 1 6 . 1 2P, G z= 2 0 . 1 0 Case(b): Cableimpedance= 0.005+70.05 = ir-;-t{ iz pi,ur. \"Inrpr\"h\"irst\"c\"arvse. ^'- - ^D' \"'^' 0.0502lB4.3 It may be noted that the real power loss of 0.i pu is supplied by Gz(Pcz - 20.10). the cable resistancecausesreal power losswhich is not known a priori. The real The above presentedproblem is a two-bus load flow problem. Explicit load flow is thus not obvious as was in the case of R = 0. we specify the solution is alwayspossiblein a two-bus case.The readershouldtry the case when generationat station I as Q c z = 7 1 0a n dl V 2 l= ' - Pcr= 20 Pu The generalload flow problem will be takenup in Chapter6. It will be seen The considerationfor fixing this generationis economic as we shall see in that explicit solutionis not possiblein the generalcaseanditerative techniques Chapter7. have to be resortedto. The generationat station 2 will be 20 pu plus the cableloss.The unknown variablesin the problem are P62, 6p Qcp Qcz Let us now examineas to how many system equationscan be formed. From Eqs.(5.68)and (5.69) P c t- Po=r ', = d- V#*s@+ 61) A 275 kV transmissionline has the following line constants: #cos A = 0.85 15\": B - 200 175\" (a) Determ inethe power at unity power factor that can be r ece\\ ive di f the voltage profile at each end is to be maintained at 275 ky . 5= U-=. cos 84.3\"- U^-.=U)UZ cos (84.3\"+ 4) (i) U ) U Z (b) What type andratingof compensationequipmentwould be requiredif the e c r - e o t =e s =f f r t ^ t - t r r , \\ i , s i n ( d +4 ) load is 150 MW at unity power factor with the samevoltageprofile as in part (a). Qcr - S= ;0j -. 0 50 sin 84.3 \"- =0+. 0- s5i0n2 (84.3\"* Ur) (ii) (c) With the loadas in part (b), what r,vouldbc thc receiving-endvoltageif 2 the compensationequipmentis not installed? F r o mE q s .( 5 . 6 6 )a n d ( 5 . 6 7 ) Solution (a) Given lV5l= lVal= 215 kY; e,= 5o,C = J5\". Sincethe power is receivedat unity power factor, P o z - PG z-=PP n-=l v- ftf l l v z cl o s( d - 6 r ) - -l 1v:'-Pc o s e Qn= o Substitutingthesevaluesin Eq. (5.62), we can write 2s- Pc2= (84.3-\" A,l- 84.3o (iii) o - !5x]75- sin(75\"- d)- x Q75)2sin(75\"-5\") 200 ij; *;rcos #rcos 0=378sin(75\"-A-302 6- ))\" e o r -e c z =e n =U f f i r i , (o- 6 t->, Y l ! r r \" , whichgives lzl 15- Qcz=#, sin(84.3-\" 4) #, sin84.3\" (v) From Eq. (5.61) Thuswe havefour equationsE, qs.(i) to (iv), in four unknownsp52, 51e, 61, P\" n= 2 7 5 ^ x 2 7c5o s( 7 5 o- 2 2 \" )- 85 * (2712cos70o Q62.Eventhoughtheseare non-linearalgebraicequationss, olutionis possible 200 9200 in this case.SolvingEq. (i) for d,, we have - 227.6- 109.9= 117.7D{W
166 | todern powerSvstemAnalvsis characteristicsand Performanceof PowerTransmissionLines Ifil I (b) Now lV5l= lVpl = 275 kV Powerdernandedby load = 150MW at UPF 0 = !:y-*t sin(75'- . 5 -) *2t v0o0t 2 sin70' (ii) 200 FromEq. (ii), we get P n = P R = 1 5 0 M W ;Q o = 0 sin(75\"- A=0.00291VR1 1 5 0 = 2 1 5 z q 7 c o s ( 7 5 o- 5 ) - 0 . 8 5 x (.2^7--5, ))'cos70o 150= 1.375lynt (1 _ (0.002912lv^121u_2 0.00l45tyRl2 200 ,* Solving the quadraticand retaining the higher value of lv^|, we obtain 1 5 0 = 3 7 8c o s( 7 5 \"- 4 - 1 1 0 or 5 = 28.46\" lVal= 244.9kV FromEq. (5.62) Note: The second and lower value solution of lVol though feasible, is impractical as it correspondsto abnormally low voltage and efficiency. en= sin (75.- 28.46\"-) 0'85x e75)2 sin 70o \"# 200 It is to be observedfrom the results of this problem that larger power can be transmitted over a line with a fixed voltage profile by installing - 274.46- 302= - 2l.56 MVAR compensationequipmentat the receiving-endcapableof feedingpositive VARs into the line. Thusin orderto maintain2T5kV at a receiving-ende, n= -27.56 MVAR must be drawnalongwith thereal powerof Po = 150MW. The loadbeing 150MW Circle Diagrams at unity power factor,i.e. Qo = 0, compensationequipmentmust be installed at the receiving-endW. ith referenceto Fig. 5.19, we have It has been shown above that-the flow of active and reactivepower over a transmissionline can be handledcomputationally.It will now be shown that the - 2 7 . 5 6 +Q c = 0 Iocusof complexsending-andreceiving-endpoweris a circle.Sincecirclesare convenientto draw, the circle diagramsarea useful aid to visualizethe load or Qc = + 27.56MVAR flow problem over a single transmission. i.e. the compensationequipmentnnustfeed positive VARs into the line. See The expressionsfor complexnumberreceiving-and sending-endpowers are subsection5.10 for a more detailedexplanation. reproducedbelow from Eqs. (5.58) and (5.59). 1 5 0- j 2 7 . 5 6 1 5 0+ / 0 sR=lf-l rv^4r/'r-r. +P 4r- b) (s.s8) F i g .5 . 1 9 n=l+lvs(t(r-o)-+3 4/r+o (5.5e) (c) Sinceno compensationequipmentis provided The units for Sp and S, are MVA (three-pha,sew) ith voltagesin KV line. As per the above equations,So and ,9, are each composedof two phasor P n = 1 5 0 M W ,Q n = 0 componenfs-6ne a constantphasorand the othera phasorof fixed magnitude but variable angle. The loci lor S^ and S, would, therefore,be circles drawn from the tip of constantphasorsas centres. It follows from Eq. (5.58) that the centre of receiving-endcircle is located at the tip of the.phasor. Now, -l+l vR' \\ Pt(r - l V 5l = 2 7 5 k V , l V a l =? IBl a) (s.16) Substitutingthis datain Eqs. (5.61)and (5.62),we have in polar coordinatesor in terms of rectangularcoordinates, 150= t':\\!'\"os ( 7 5 \"- A- ggtv^t2 c o s7 0 \" (i) Horizontal coordinateof the centre 200 200 = -l|lrv-f co(s//- a)MW (s.77)
e.-'1,, , . I characteristicsand performanceof power TransmissionLines .d6Eif _ ModernpowerSystemAnalysis Vertical coordinateof the centre .,.,, t69 (5.78) = tl+Bitlvrt2sin(F- a) MVAR Verticalcoordinateof the centre The radiusof thesending-encdircleis = -i{-ltv*t2sin(tJ- a) MVAR (s.81) lBl The radiusof the receiving-endcircle is tysllyRlMVA tBl dTrhaewsinegndOinCg, -aetnadcnigrclelertd?i-agar)afmroismsthhoewpnoisnitFivieg.M5W.2-Ia.Txhise. cFernomrrethiselcoecnattreedtbhey sending-endcircle is drawn with a .uo i.rr ${e(sa m\\ e as in the case of lBt receiving-end).The operatingpoint N is located by d(as readfrom the recefving-endcircle diagram)in measuringthe torque angle ttredirectionindicatedfrom thc re'fi'rcncIci'nc. MVAR MVAR C5 ll Aal llul \"t 2l Roforoncolino - forangled RadiuslYsllVnl lBl Phasor55 = P5+ie5 Referenceline for angle6 Fig. 5.20 Receiving-endcircle diagram 1-- F'or constant lVol, the centre Co rernains fixed and concentric circles result +Qs for varying l7rl. However, for the case of constant ll{l and varying lvol the centrcso1'circlesmovc along the line OCoaru),haveraclii in accordanceto ltzrl Fig. 5.21 Sending-endcircle diagram lvRtABt. Similarly, it follows from Eq. (5.59) that the centre of the sending-end circle is located at the tip of the phasor l4l,u,,t'(0.- a) (s.7e) i,i,==',u!\"':i ;=o (s.80) IB l The corresponding receiving- and sending-end circle diagrams have been clrawn in the polar coordinates or in terms of rectangular coordinales. in Figs5.22and5.23. Horizontal coordinate of the centre = B-ltv;2cos(f - a)Mw
characteristicsand Performanceof power TransmissionLines [giM t- Resistance= 0.035 Olkm per phase Inductance= 1.1 mHlkm per phase Capacitance= 0.012 pFlkm per phase If the line is supplied at 275 kV, determine the MVA rating of a shunt lvnl2 Pa=oK thereceiving-enwdhenthelineis deliveringno load.Usenominalz- rmethod. Qn=KM Solution tzl R=0.035x400=14Q X = 314x 1.1x l0-3x 400= 138.2O Z = 14+ 7138- 138.1184.2\"Q Y= 314x 0.012x 10-6x 400 lg0\" - 1.507x 10-3/_W U Fig.5-22 Receiving-encdircrediagramfor a shortrine A = ( t + L v z \\ = 1 + - l - x t . 5 0 7x 1 0 - 3x r 3 g . i l r i 4 . z \" MVAR \\2)2 = (0.896+ 70.0106=) 0.896lj.l' B=Z-138.7 184.2\" lV5| = 275kV, lVpl= 275kV Radiusof receiving-encdircle- lysllyRl-275x275 - 545.2MVA tBl 138.7 Locationof the cenre of receiving-encdircle, Ps=oL l 4 : l 1 i l , , = 275x275x0.896== 4'rd8d8').5Mrll VA Qs=LN IB I l 138J \\ lvsl2 l@ - a) = 84.2\"- 0.7\"= 83.5\" lzl i,\\/A P 55 MVAR _ >MW L 488.5MVA Fig. 5.23 Sending-endcirclediagramfor a stror.ltine .tuou.rrro oJl: useof circlediagramiss i$ustratedby meansof therwoexamplesgiven Cp A 50 Hz, three-phasg,275 kY,400 km transmissionline has the following parameters: Fi1.5.24 Circlediagramfor Example5.10 From the circle diagramof Fig. 5.24,+ 55 MVAR mustbe drawn from the receiving-endof the line in order to maintain a voltageof 275 kV. Thus rating of shuntreactor needed= 55 MVA.
X12,:jl ModernpowerSystemAnalysis Characteristicasnd Performanceof PowerTransmissionLines t|-,i?3-;- E x a m p l e5 . 1 1 (a) Locate OP conespondingto the receiving-endload of 250 MW at 0.85 laggingpf (+ 31.8). Then , ^ or^ lysllyRl 275lvsl A - 0.93 11.5\", B = Il5 ll7\" l V s l = 3 5 5 . 5k V If the receiving-endvoltageis 2'r-5kV, determine: (b) Given lV5| = 295 kV. (a) The sending-endvoltagerequired if a load of 250MW at 0.g5 lagging pf Radiusof circle diagram- t\".l?\" - 705.4MVA is being deliveredat the receiving-end. 115 (b) The maximum power that can be deliveredif the sending-endvoltageis Drawing the receiving-endcircle (seeFig. 5.25) and the line C^Q parallel to held at 295 kV. the MW-axis, we read (c) The additionalMVA that has to be provided at the receivihg-endwhen delivering 400 MVA at 0.8 lagging pt the suppty voltage being PR-o = RQ = 556 MW maintained at 295 kV. (c) Locate OPt conespondingto 400 MVA at 0.8 lagging pf (+ 36.8\"). Draw Solution In Fig. 5.25 the centre of the receiving-endcircle is located at P/S parallel to MVAR-axis to cut the circle drawn in part (b) at S. For the specified voltage profile, the line load should be O^S.Therefore, additional l4li Rt'' -- 2t5x275x0'93- 611.M6 VA MVA to be drawn from the line is lBl tt5 P/S= 295 MVAR or 295 MVA leading c o s -l 0 .8 5= 3 1 .8 \" l @ - a ) = 7 7 o - 1 . 5 \"= 7 5 . 5 \" 5.10 METHODS OF VOLTAGE CONTROL MVAR Practicallyeachequipmentusedin power systemareratedfor a cprtainvoltage with a permissible band of voltage variations.Voltage at various busesmust, I therefore,be controlled within a specifiedregulation figure. This article will discussthe two methodsby meansof whieh voltage at a bus can be controlled. lvslt6 lvRltj a) p\"-.io, I P^iio* | Fig.5.26 A two-bussystem Considerthe two-bus systemshownin Fig. 5.26 (akeadyexemplifiedin Sec. 5.9). For the sake of simplicity let the line be characterizedby a series reactance(i.e. it has negligible resistance)F. urther, sincethe torqueangle d is smallunderpracticalconditions,real andreactivepowersdeliveredby the line for fixed sending-endvoltagelVrl and a specifiedreceiving-encvl oltagel{ | can be-writtenas below from Eqs.(5.71) and (5.73). (5.82) Fig. 5.25 Circlediagram for ExampleS.11
e'f=, 'li, ,,u,r- rv,ril nsmtssioLninesl\"ffie X (5.83) Reactive Power Injection Equation(5.83) upon quadraticsolution*can also be written as It follows from the above discussionthat in order to keep the receiving-end rr{r= . +tys| (1- 4xesnAvrtzlt/z (5.84) voltageat a specifiedvalue l{1, a fixed amountof VARs tai I *;;; drawn }vrt from the line-. To accomplishthis underconditions Since the real power demandedby the loaclmust be deliveredby the line, Qn, a local VAR generator (controlled reactive power source/compensating Pn= Po equipment)must be usedas shown in Fig. 5.27.fle the receiving-endis now vAR balanceequationat varying real power demandp, is met by consequenct hangesin the rorque Oi * Qc=Qo angle d. Fluctuationsin Qo ue absorbedby lvrienamelIutaweiisoon,fufhilxQdoe,wodtehgaveitvereeer'fsno-ntrboaeybs,oegpinveoertanetbdeywthEiathqt t.sh(p5ee.rgcei3cfi)eefiodvrreefdcixreeeiavdcinrtvigv,-eeI panondwvdoeslprtaoegfceitfhfioeerdlroin4nerl.ymiolu\"nset the vARs drawn from the line remain the local vAR generator o6 suchthat fixed at esn.The receiving-endvoltage vuwosoiltnuaglgdtehthelVusrsilg)r.neaLml,ofarcionam{l V1teAhdeRVacAt oRl4mlpme(entthseiarsitnioosfnt\"acolaluenrd,rai\"nltrtrfhuaemcrt,eescbeeivaminfaigxd-eeednadsuoetofnmtdhiaentgilcin_beey.nd Qo = Qsn Practical loads are generally lagging in nature and are such that the vAR demandQn may exceedet*.rt easilyfollows the receiving-endvoltagemust changefrom from Eq. (5.g3)that for or; ot- the specifiedvalue 'n'i some value lTol to meet the demandedVARs. Thus Q'J^o== Qo ^n== l v * l (' lYsl- lVol)for (QD> QsR) ;i The modified lVolis thengiven by rvlq=l - +ty3 (1- 4xeRltv)rrt,, (s.85) lvrt Fig. s-zr use of rocarvAR generatoar t the roadbus crrmparisonof Eqs. (5.84)ancl(5.85) rcvcalsthut r^r. nY.D -- vnR -- .v,p-)' ,s,t!t,t.t,: receiving-endvoltageis r{r, butior bo= Oo; A:,\"' Trryotypes of vAR generatorsare employed in practice-static tvo| < t4l rotating type. Theseare discussedbelow. type and Thus a VAR demandlarger than Qf is met by a consequentfall in receiving- Static VAR grenerator 11d ,\":i tt!qf o,\"l from the specifiedvalue. similarly, if the vAR demandis lessthan It is nothing but a bank of three-phase Q\"* ows that static capacitors and/or inductors. With l referenceto Fig. 5.28,if lV^l is in line kV, and Xg is the per phasecapacitive reac_ tyR>t tr4l tanceof the capacitorbank on an equiva_ Ient star basis, the expression for the cInadueseedth,uendVeArRligdhet mloaanddctoondbieticoonms,ethneegcahtaivrgeinregs\"uulpiiun.glainncetheofrethceeilvininegm-eanyd VARs fed into the line can be derived as lIc voltage exceeding the sending-endvoltage under. Static capacitor bank illustraredin Section5.6). (this is the Ferranti effect already r,=iH kA In order to regulatethe line voltage under varying demandsof VARs, the two Fig. 5.28 m e t h o d sd i s c u s s e db e l o w a r e e m p l o y e d . 'Negative sign in the quadraticsolution is rejected becauseotherwise the solution would not match the specifiedreceiving-end voltage which is only slightly less than 'of the sending-endvortage(the differenceis ress thai nqo). c o u r s es,i n c ct { t i s spccificcwlithina buntlQ, l ruryvrry withil a corresponding band.
: iii,',1,1 ModernpowersystemAnarysis iQcG-Phase3)r-y (- IF) :$bja\"3r,Fs.:itag^\"nu:t:dr1e:rHou5n'1:2n,w9.i,nts:groh,ao,itc^win,frsolu!ra:lsTosaiy-und?cm. hsTir\"nto*aoclnyehoutahisnvem;emportroro,,truorc*rsoo,\"enn.,un.*e;;,rc;ttn:ehr.dg*\"tro:,i:gy't:ihnbiell.;erre,rrceoeainlvpionorugwe-seetarncfdrhobmunsce J3 (s.86) .I-C__ (lvRl- IEGD/0. K,. ^A -i3x J5;E- #.HMVA tvP QsQ-Phase)-+ MVAR XC If inductors are employed instead,vARs fed into the line are Q{3-phase)=-'F''tuo* (s.87) i e c_= 3Jt v R3l 4 oG i l XL Under heavy load conditions,whenpositive VARs are needed,capacitorbanks = 3W(- l Y R rr-r c l ) are employed; while under light load conditions, when negative vARs are J3 ( _jxsJl ) needed,inductor banksare switchedon. The following observationscan be rnadefor.static vAR generators. = jlVpt(tE6t _ IVRt)lXsMVA (i) Capacitor and inductor banks can be switched on in steps.However, ec= tVRt(EGt _ tVRt)lXMs VAR (5.8g) stepless(smooth) VAR control can now be achievedusing SCR (Silicon Controlled Rectifier) circuitrv. It immediately follows from wthheenarbEoGvet>retvla^trio(wnsehreipxtchiatetdthceamsea)cahnidneinfjeeecdtss positive vARs into the line (ii) Since Qg is proportionalto the squareof terminal voltage,for a given negarive VARs capacitorbank, their effectivenesstends to decreaseas the voltage sags if lEGl under full load conditions. rceosnrpnteinccuotoonuftrsraolysttaatotdinjsugtsaVttarAcbRvleAbgRyeangdeejrnuaestotriarnstgo. mrsa,thcehfinoell\"oJwtuintigonobswerhvicahtiocno.s,natrreomlsatEde6iln. (iii) If the systemvoltagecontainsappreciableharmonics,rhe fifth being the most troublesome,the capacitorsmay be overloadedconsiderably. (i) Thesecan provide both positive and negativevARs ously adjustable. (iv) capacitors act as short circuit when switched on. which are continu_ (v) There is a possibility of series resonancewith the line incluctance iegl]ce(fionfFii)en'riororsavvm'mm$tAooiRtalrrrc:shalcLl,ieegtoi:nro:snerj\"beeeAs.:sedi*:dsceimuoerhIcnsvrrttzahai oarraotiinbrdtooiraenunnoJstssg-iQaeun,eitvaacsisnie\"tnsecgansolvtleimbA.hxtarRpiec_onai\"snr\"drga(vea.rrEidsnnni ot6\"tdorne,ru.i'-trsrmrhp*aoer.rri*v.e',t.r\"s0^w.irrs.\"roorrts'.uiufa,onlL,drt.tutt..bsaltt.eitp\"i.agpr\"torrJ\"eiec\"\"ub\"d\"fcl-.e;eao:;r;frpru;fne;aisnidnc:l i.:g,Himt:-oo.iorwt{sv.rve.t:AhAv::eeRRri r, particuia.riyat harmonic frequencies. Rotating VAR grenerator It is nothing but a synchronousmotor running at no-load andhaving excitation Control by Transformers adjustableover a wide range.It feeds positive VARs into the line uncler overexcitedconditionsandf'eedsnegativeVARs when underexcitedA. machine thus running is called a synchronouscondenser. lvnl Fig.5.29 RotatingVAR generation tcioThsofheroverVoebvAlcvtAaRitoiRgouenisnniclnjyeeojceenrtticdimroteoinioldteebnmxdymectetetrhoateohndadosst.dfhnodiarsimsrrrceaournwstgaserpea,dtcnaahgbpaeocnvohgefailnnavggco.ikTlntshagtIeghseeturfacslnoeesnxdftiiobrnorimlcli.toeyInrfajtuanthnpdeccethcivoaoonnnrwgtoaiinmtghgey Cbrl oeehacaRdednoi,evgncicineenagigenv-iieUtnbhnngeedd-reterrmnaar diLansvonseofaudoldatrab'miygfyCeeorUwsrsifhm.Sa_iuculuthclotthartmaent ananepsidoifcscuoahtsrolmarl yysne,a.g-trtg,e.huoseonwogi uioinrs*g;ti o\"tb.ot,ehomv.\"AraRtdasebpes,doi neongmf_crasoanea' edndded,adinnbadgyT_ctaaahpnned
,tig;'J ModernPowerSystemAnalysis Characteristicsand Performanceof Power TransmissionLines fi.l7!J, I I Considertheoperationof a transmissionline with a tapchangingtransformer l.{Vl which is to be compensatedT. hus merely tap settingas a method of voltage drop compensationrvould give rise to excessivelylarge tap setting if at eachend as shown in Fig. 5.30. Let /5and r^ be the fractionsof the nominal compensationexceedscertain limits. Thus, if the tap setting dictatedby Eq. transformationratios, i.e. the tap ratio/nominal ratio. For example, a trans- kV inputhas rr - I2lll = setting range (usually not more than + 20Vo), it would be necessaryto simultaneously inject VARs at the receiving-end in order to maintain the z=R+jx desiredvoltage level. Compensation of Transmission Lines The perfonnance of long EHV AC transmissionsystemscan be improved by reactivecompensationof seriesor shunt (parallel) type. Seriescapacitorsand shunt reactors are used to reduce artificially the seriesreactanceand shunt 1 : fsnl susceptanceof lines and thus they act as the line compensatorsC. ompensation of lines results in improving the systemstability (Ch. 12) and voltageconffol, Fig.5.30 Transmissiolninewithtapchangingtransformeart eachend in increasingthe efficiencyof power transmission,facilitating line energization andreducing temporaryand transientovervoltages. With referenceto Fig. 5.30 let the impedancesof the transformerbe lumped Seriescompensationreducest.he seriesimpedanceof the line which causes tn Z alongwith the line impedance.To compepsatefor voltagein the line and voltagedrop and is the most important factor in finding the maximum power transformers,let the transformer taps be set at off nominal values, rr and ro. transmissioncapability of a line (Eq. (5.70)). A, C and D constants are With referenceto the circuit shown. we have functions of Z andtherefore the also affected by changein the value of.Z, but trnrVs= t^nrVo+ IZ (s.8e) these changes are small in comparison to the change in B as B = Z for the nominal -rr andequalsZ (sinh 4ll) for the equivalent zr. ., From Eq. (5.75) the voltage drop ref'erredto the high voltage side is given by The voltage drop AV dueto seriescompensationis given by tAvl= !I,!jIQs- (s.e0) AV = 1Rcos S, + I(X,.- X.) sin ,!, (s.e4) ton,rlVol Here X, = capacitivereactanceof the seriescapacitorbank per phaseancl lAVl - tsn, lTsl - ton2lVol Now X, is thc total incluctivercactanceof the line/phascI.n practice,X. may be so selectedthat the factor (XL - X.) sin Q, becomesnegative and equals (in RPR+xQR (s.e1) magnitude) R cos /, so that AV becomes zero. The ratio X=IXL is called trnrlvrl- tonrlvol+ \"compensationfactor\" and when expressedas a percentageis known as the t* n r l V o l \"percentagecompensation\". In order that the voltage on the HV side of the two transformersbe of the The extent of effect of compensationdependson the number,location and sameorder andthe tap settingof eachtransformerbe the minimum, we choose circuit arrangementsof series capacitor and shunt reactor stations.While tstn= 1 (5.92) planning long-distancelines, besides the average degree of compensation SubstitutinEtn= llttin Eq. (5.91)and reorganisingw, e obtain required,it is requiredto find out the most appropriatelocationof the reactors .r( , RPR+xgo ) _ n2 lvRl (s.93) and capacitor banks, the optimum connection scheme and the number of \"[' \"r\"rWW )- \", W intermediatestations.For finding the operatingconditions along the line, the ABCD constantsof the portions of line on eachside of the capacitorbank, and For completevoltage drop compensation,the right hand side of Eq. (5.93) ABCD constants of the bank may be first found out and then equivalent shouldbe unity. constantsof the seriescombinationof line-capacitor-linecan thenbe arrived at It is obvious from Fig. 5.30 that rr > 1 and tn 1 I for voltage drop by using the formulae given in Appendix B. compensationE. quation (5.90) indicatesthat /^ tends to increase*the voltage In India, in stateslike UP, seriescompensationis quiteimportantsincesuper thermal plants are located (east) several hundred kilometers from load centres -This 1 increasesthe line current / and hence voltage drop. (west) and large chunks of power must be transmittedover long distances. is so becausefn < Seriescapacitorsalso help in balancing the voltage drop of two parallel lines.
r-hOtl\" uodern Power SvstemAnalysis Characteristicasnd Performanceof PowerTransmissionLines r81 When seriescompensationis used,there are chancesof sustainedovervoltage kW at a leadingpower factor.At what valueof P is the voltageregulatior to the ground at the seriescapacitor terminals. This overvoltage can be the zerowhen the power factor of the load is (a) 0.707, (b) 0.85? power limiting criterion at high degree of compensation.A spark gap with a 5 . 2 A l o n g l i n e w i t h A = D = 0 . 9 l 1 . 5 \" a n d B = 1 5 0 1 6 5 \"C I h a s a t t h el o a r high speed contactor is used to protect the capacitors under overvoltage end a transformerhaving a seriesimpedanceZr = 100 167\" Q. The loar trons. ff]=[1form of Under light load or no-loadconditions,chargingcurrentshouldbe kept less \",]l';land evaluatethese constants. than the rated full-load current of the line. The charging current is approxi- matelygiven by BrltA whereB. is the total capacitivesusceptanceof the line 5.3 A three-phaseoverheadline 200 km long hasresistance= 0.16 Qlkrn an andlVl is the ratedvoltageto neutral.If the total inductive susceptanceis Br due conductordiameter of 2 cm with spacing4 m, 5 m and 6 m transpose( to several inductors connected(shunt compensation)from line to neutral at Find: (a) the ABCD constantsusing Eq. (5.28b), (b) the V,, 1,, pf,,'I appropriateplaces along the line, then the charging current would be whenthe line is delivering full load of 50 MW at 132kV and 0.8 laggin pf, (c) efficiency of transmission, and (d) the receiving-end voltag I,he,(=Bc-Br) lvl= BclVf[r (s.es) regulation. +) Reductionof the chargingcurrentis by the factor of (1 - Br lBc) and81lBg is the shunt compensationfactor. Shunt compensationat no-load also keepsthe receiving end voltage within limits which would otherwise be quite high 5.4 A short230 kV transmissionline with a reactanceof 18 O/phasesupplir becauseof the Ferranti Effect. Thus reactorsshould be introduced as load is a load at 0.85 lagging power factor. For a line current of 1,000 A tt removed,f<lpr ropervoltagecontrol' receiving-and sending-endvoltagesare to be maintainedat 230 k\\ As mentionedearlier,the shuntcapacitorsareusedacrossan inductive load Calculate (a) rating of synchronouscapacitor required, (b) the loa so as to provide part'of the reactive VARs requiredby the load to keep the current, (c) the load MVA. Power drawn by the synchronouscapacitt voltage within desirablelimits. Similarly, the shuntreactorsare kept across capacitiveloadsor in light load conditions,as discussedabove,to absorbsome maybe neglected. .\\ of the leading VARs for achievingvoltage control. Capacitorsare connected eithcrclirectlyto a busor throughtcrtiarywinclingof the main transformear nd 5.5 A 40 MVA generatingstationis connectedto a 'three-phaseline havin Z = 300 175\" Q Y = 0.0025 19tr U. are placed along the line to minimise lossesand the voltage drop. The power at the generatingstation is 40 MVA at unity power factor a Ii may be noted that for the samevoltageboost,the reactivepower capacity a voltageof L20 kV. There is a load of 10 MW at unity power factor a themid point of the line. Calculatethe voltageand load at the distantenc of a shunt capacitrtris greaterthan that of a seriescapacitor.The shunt of the line. Use nominal-T circuit for the line. capacitorimproves thepf of the load while the seriescapacitorhas hardly any impact on the pf. Series capacitors are more effective for long lines for 5.6 The generalizedcircuit constantsof a transmissionline are irnprovementof systemstability. A-0.93+70.016 Thus, we seethat in both seriesand shuntcompensationof long transmission B=20+ jI40 lines it is possibleto transmitlarge amountsof power efficiently with a flat The load at the receiving-end is 60 MVA, 50 H4 0.8 power factor voltage profile. Proper type of compensationshould be provided in proper lagging. The voltage at the supply end is 22OkV. Calculate the load quantity at appropriateplacesto achievethe desiredvoltagecontrol. The reader voltage. details about the Static Var Systems(SVS) in is enceuragedto read the completetreatmenton 'compensation',the reader 5 . 7 Find the incident and reflectedcurrentsfor the line of Problem5.3 at the References7, 8 and 16. For receiving-endand 200 km from the receiving-end. may refer to ChaPter15. 5 . 8 If the line of Problem5.6 is 200 km long anddelivers50 MW at22OkY and0.8 powertactor lagging,determinethe sending-endvoltage,current, PROB!.EMS power factor and power. Compute the efficiency of transmission, characteristiicmpedance,wavelength,andvelocity of propagation. 5 . 1A three-phasevoltageof 11 kV is appliedto a line having R = 10 f) and X = 12 ft p\"t conductor.At the end of the line is a balancedload of P 5 . 9 For Example 5.7 find the parametersof the equivalent-ncircuit for the line.
-,T,1- €?il uooern power system hnalysis characteristicsand Performanceof power TransmissionLines 183 5.10 An interconnectorcable having a reactanceof 6 O links generating REFERNECES stations1 and 2 asshownin Fig. 5.18a.The desiredvoltageprofile is lVtl = lVzl= 22 kY. The loadsat the two-busbarsare40 MW at 0.8 lagging Books power factor and20 MW at 0.6 lagging power factor, respectively.The l. Tron'smissionLine ReferenceBook-345 kV and Above, Electric Power Research torqueangle and the stationpower factors. Institute,Palo Alto calif, 1975. 5.11 A 50 Hz, three-phase,275kV, 400 km transmissionline has following 2. Mccombe, J. and F.J. Haigh, overhead-Iinepractice, Macdonalel,London, 1966. parameters(per phase). 3. Stevensonw, .D., Elementsof Power Sy.stemAnalysis,4thedn, McGraw-Hill. New Resistance= 0.035 Qlkm York, 1982. , Inductance=1mFl/km 4. Arrillaga, J., High Vohage Direct Curuent Transmission,IF,E Power Engineering Capacitance= 0.01 p,Flkm Series6, Peter PeregrinusLtd., London, 1983. 5. Kirnbark,E.w., Direct current Transmission,vol. 1, wiley, New york, 1971. If the line is suppliedat 275 kV, determinethe MVA rating of a shunt 6. IJhlmann,E., Power Transmissionby Direct current, Springer-verlag, Berlin- reactorhaving negligible lossesthat would be required to maintain 275 kV at the receiving-end,whenthe line is deliveringno-load.Use nominal- Heidelberg,1975. zr method. 7. Miller, T.J.E., ReactivePower control in Electric systems,wiley, New york 5.12 A,three-phasfeeederhavinga resistanceof 3 Q and areactanceof 10 f) t982. suppliesa load of 2.0 MW at 0.85 lagging power factor. The receiving- 8. Mathur, R.M. (Ed.), Static Compensatorsfor ReactivePower Control, Context end voltage is maintained at 11 kV by means of a static condenser drawing2.1 MVAR from theline. Calculatethe sending-endvoltageand Pub., Winnipeg,1984. powerfactor. What is the voltageregulationandefficiency of the feeder? 9. DesphandeM, .V., Electrical Power System Design, Tata McGraw-Hill. New 5.13 A three-phaseoverheadline hasresistanceandreactanceof 5 and 20 Q, Delhi, 1984. respectivelyT. he load at the receiving-endis 30 MW, 0.85 power factor lagging at 33 kV. Find the voltage at the sending-endW. hat will be the Papers kVAR rating of thecompensatingequipmentinsertedat the receiving-end so as to maintain a voltageof 33 kV at eachend?Find also the maximum 10. Dunlop, R.D., R. Gutman and D.p. Marchenko,\"AnalyticalDevelopmenot f load that can be transmitted. Loadability Characteristicsfor EHV and UHV TransmissionLines\", IEEE Trans. P A S ,1 9 7 9 , 9 8 : 6 0 6 . 5.I4 Constructa receiving-endpower circle diagramfor the line of Example 5.7.Locate the point coresponding to the loadof 50 MW at 220 kV with 11. \"EHV Transmission\"(,specialIssue),IEEE Trans,June 1966,No.6, pAS-g5. 0.8 lagging power factor. Draw the circle passingthroughthe load point. 12. Goodrich,R.D., \"A Universal Power circle Diagram\", AIEE Trans., 1951,7o: Measurethe radiusand determinetherefromlVrl.Also draw the sending- end circle and determine therefrorn the sending-endpower and power 2042. 1 3 . Indulkar, c.s. Parmod Kumar and D.p. Kothari, \"sensitivity Analysis of a factor. MulticonductorTransmissionLine\", Proc. IEEE, March 19g2,70: 299. 5.15 A three-phaseoverheadline has resistanceand reactanceper phaseof 5 14. Indulkar, c.s., Parmod Kumar and D.P.Kothari, \"some studies on carrier and25 f), respectively.The load at the receiving-endis 15 MW, 33 kV, 0.8 power factor lagging. Find the capacity of the compensation Propagationin overheadrransmissionLines\", IEEE Trans.on pAS,No. 4, 19g3, equipmentneededto deliver this load with a sending-endvoltageof 33 102: 942. 1 5 . Bijwe, P.R., D.P. Kothari, J. Nanda and K.s. Lingamurthy, \"optimal voltage kv. Control Using ConstantSensitivityMatrix\", Electric PowerSystemResearch,Oct. 1 9 8 6 ,3 : 1 9 5 . Calculate the extra load of 0.8 lagging power factor which can be 1 6 . Kothari, D.P., et al. \"Microprocessorscontrolled static var s5istems\",proc. Int. delivered with the compensatingequipment (of capacity as calculated conf. Modelling & Simulation,Gorakhpur,Dec. 1985,2: 139. above) installed, if the receiving-end voltage is permitted to drop to 28kV.
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