41 INTRODUCTION A complete dtagram ron et System representing all the three phases becomes too complicated for System of practical size. go Much so that it may ao longer convey the information it 'S Intended to convey, It is much more practical to Tepresent a power System by means of simple symbols for each Somponent resulting in what is called a on ng diagram. Per unit System leads to great simplification of three-pkase networks involving transformers. An impedance diagram drawn on a Per unit basis does Tot require ideal transformers to be included in it An important element of 4 poWer system is the syachronous machine, which Sreatly influences the system behaviour during boch steady state and transient Smnditions. The synchronous machine model in steady state is presented in this Stapter. The transient model of the machine will be presented in Chapter 9. 42 Single-Phase Solution of Balanced Three-Phase Networks The solution of a three-phase network under balanced ee sel y Settied out by solving the single-phase network corresponding to the referen Phase, Figure 4.1 shows a simple, balanced fueeaiat oe a cea i tential, so that /, = s Hout neutrals ae Heeler atari ee eens impedance loes: Phase g (4.1) Eq= Zo + Zo ame magnitade but x hases have the same magni gt currents, and voltages in ee ie Ea uation (4.1) corresponds to ite *® progressively shifted in Ne ‘whose solution completely determines the network of Fig. 4- ‘Ohution of the three-phase network.bad) Modem Power System Analysis i h : <\ Ze 4 \B a % ve/, ty Inve N - n i Zz a Ee 2g qi a Fig. 4.1 Balanced three-phase network a N ° Fig. 4.2 Single-phase equivalent of a balanced three-phase network of Fig. 4 Consider now the case where a three-phase transformer forms pat ds three-phase system. If the twansformer is Y/Y connected as shown in Fig ah in the single-phase equivalent of the three-phase circuit it can be obvioal represented by a single-phase transformer (as in Fig. 4.3b) with primay# secondary pertaining to phase a of the three-phase transformer. rsha ~ifgs —___Fepreseniaton Power: System: Compononts [a If the fea obfortmer is Y7 connec iy Fig, 44a, the delta side replaced by an equivalent star connection ay shown do single-phase equivalent of Fig. 4.4b. An Important fact has, however, to he observed here. On the delta side the Voltage to neutril Y, have a certain phase angle shift’ from the the phase labelling shown), In the sin, respectively in phase with (V,,. 1) shift through the same phase angle from st phase impedance and power flow are § to be ted 80 a8 to obtain the aw atid line current 1, Slat side values V,, and J, (90" for sle-phase equivalent (Vay, Ay) are both phase voltage and fine current far to delta side, the transformer per Preserved in the single-phase equivalent Tn most analytical studies, we are merely interested in the magnitude of volta iges and currents so that the Single-phase equivalent of Fig. 4b is an acceptable Proposition. Wherever proper phase angles of currents and voltages are needed, Correction can be easily applier after obtaining the solution through a single- phase transformer equivalent. 8 = +1 . ls ~ NaiVv3 L Von Ny a No Ne n a Ie z ——ie a y ts | aA o— de as 0 equivalent of YIA transformer Fig. 4.4 ion, the i tive of the type of connection, acai Sa neaetae au of a three-phase transformer a a same as oe peuine transformation ratio.a =, 58. Modern Power System Analysis 4.3 ONE-LINE DIAGRAM AND IMPEDANCE OR REACTANo, DIAGRAM se A one-line diagram of a power system shows the main conection arrangements of components, Any particular component may or m; shown depending on the information required in a system study, cite breakers need not be shown in a load flow study but are a must fora Protection study. Power system networks are represented by one-line diagrams suituble symbols for generators, motors, transformers and loads, |, j, 5 convenient practical way of network representation rather than drawing the actual three-phase diagram which may indeed be quite cumbersome ay4 confusing for a practical size power network. Generator and transfon connections—star, delta, and neutral grounding are indicated by symbols drayyy by the side of the representation of these elements. Circuit breakers are Tepresented as rectangular blocks. Figure 4.5 shows the one-line diagram of , simple power system. The reactance data of the elements are given below the diagram. 8 arg EY Ot bye Using Ta ON ¥ OH { YY va (3) ¥ * ; ~8 Fig. 4.5 One-line representation of a simple power system Generator No. 1:30 MVA, 10.5kV, X"=1.8 ohms Generator No. 2: 15 NVA, 6.6 kV, X’=1.2 ohms Generator No. 3:25 MVA, 6.6 kV, X" = 0.58 ohme Transformer T; (3 phase): 1SMVA, 33/11 KV, X= 15.2 ohms par phase on high tension side Transformer F, (2phasel: 15MVA, 33/6.2 KV, X=1€ onms per phase on high tensor si Trarsmission line: 20.5 onmeiphase Load A: 18 MW, 11kV, 0.9 legging power factor Load B: 40 MW, 6.6 kV, 0.85 lagging power factor ‘Note: Generators are specified in three-phase MVA, line-to-line voltage and per pss? reactance (equivelent star). Transformers are spacttied in three-ohase MVA, line-to-in® transformation ratio, and per phase (equivalent star) impedance on cne side. Loads 3° Specified in three-phase MW, line-to-line voltage and power factor. ld me pe Fig. 4.6 impedance diagram of the power system of Fig. 4.5Phase basis awn from. the ine under titlanced opening conditions can be easily sly dr One-line diagram, Fup the system of Fig. 4.5 the impedance « STAM is drawn in’ Big. d 6, § equivalents are shown as ides transformers with qe gs se tnstormer have been neglected, This is a "ly'2006 appro imation Tapes ome studies. TRE Renerttors are Fepresented us voltage equre ee : ee and inductive reactance Synchronous machine model will he diseu fon line is represented by a mode! i model (to be Chapter 5). Loads are a: Ssumed t0 be passive (not involy ig rotating machines) and are represented BY resistance ane inductive reactance in series, Neutral grounding impedances do Not appear in the d ‘grim as balanced conditions are assumed. Three voltage levels (6.6, 11 and 33 kV) are present in this system. The analysis would Proceed by transforming all voltages and impedances to any selected voltage level, say that of the transmission line (33 kV). The voltages mation and all impedances by the square of ratio of transformetion. This is a very cumbersome procedure | for a large network with Several voltage levels. The per unit method discussed : : below is the’ most convenient for PoWer system analysis and will be used throughout this book. 4.4 PER UNIT (PU) SYSTEM It is usual to express voltage. current, voltamperes and impedance of an electrical circuit in per unit (o- percentage) of base or reference values of these lantities. The per unit’ value of any quantity is defined as: the actual value in any units the base or reference value in the same units Phe per unit method is particularly convenient in power systems as the various Sections of a power system are connected through transformers and have | different voltage levels. Consider first a single-phase system, Let Base voltamperes = (VA), VA Base voltage = Vy V ‘Then 2 _ Way y (42a) Base current |, Vp i 100, ai eee factor of 100 has to be carried in “computations.100 | Modern Power System Analysis My Ya ohms Base impedance Zp = “A * WA), Ifthe actual impedance is Z Coline), ils per unit value is given by Z — Hohms)*(VA),y ipuy = = Tg, vy For a power system, practical choice of base values are: Base megavoltamperes = (MV A), or Base kilovoltamperes = (kVA), Base kilovolts = (kV)g, 1,000«(MVA), _ (KVA)y Base current /, = : = a = (kY), (kv), Base impedance Z, = 000% UV). 5 (KY), 1,00 2 = = OO oes (MVA), (kWA), Per unit impedsnee Z (pu) = Zichms) (MVA) «&Y); = Mohs) (kVA), (kV 1,000 ‘Three-phase base megavol = on (equivalent star can always be found), Base: current |, = 1000%(MVA), °° Fray, 4 Base impedance Zp = et la = KM 1.00004) -On TRE MN _ Ropresontation OF Power g *"Y8Om Cainponanta Per unit Impedance 2 (pu) © Zlohins) ly (kV); ae Aohing) (KYA), (KV). ¢1,000) changed from (MVA), OUR) y ys When MVA base is is chinged from (kV yn it © (MVAY)y sen, and KV binge (4.9) is given by he new por unit impedance trom Iq (MVA} i, ny (MYA), Z (Whew = 2 (pu aia (4.10) Per Unit Representation of 4 Transformer Whas been said in Se 4.2 that # threesphase ttanstomner forming part of 4 three-phase system can be Tepresented by a single-phase nsformer in obtaining per phase solution Of the system. The delta conn winding of the transformer is replaced by an equivalent star so that the (ransformation catio of the equivalent Single-phase transformer is always the line-turline voltage ratio of the three-phase transformer, Figure 4.7a represents a wale-phase transformer in terms of Primary ane Secondary leukage reaclances Z, and Z, and an ideal transformer of ratio | +a, The magnetizing impedance is neglected, Let us choose a voltampere base of (VA), and voltage bases on the two sides of the Gansformer in the transformation, i.e. io of Ma aa (Atta) Van a 1p omab voan Fag ee v, f ie -ahase transformer: Re earn iste oo £ ‘ Nowy pu) | vio) ua) [at (b) Per unit eqquivatont circuit of single-phase transfmmar Fig. 47> J Modern Power System Anal) (Git _Notom fower Sytem anny Therefore, fa = a (as (VA)y is common) np) 28 Mn Von Ziy= + Zp= Zip i an a (illo, From Fig. 4.7a we can write = (VY, 42) a- bs We shall convert Eq. (4.12) into per unit form (419) V{PUVz_9= [Vi(pu) Vig = 2)(pu)lipZ,(pu)Z, gla ADU hygZ,(PUZyy Dividing by V5, throughout and using base relations (4.14, , o), we get Vx(pu)= V,(pu) — J(pwZ,(pu) — ,(pv)Z,(pa) (4.13) Now epee as boots or in hs lbp 4 (pu) = L(puj = 7 (pu) Equation (4.13) can therefore be written as V2(pu) = Vj (pu) — Kpu)Z(puy 414) where Apu) = Z,(pu) + Z(pu) Equation (4.14) can be represented by the simple equivalent circuit of Fig. 4.7b which does not require an ideal transformer, Considerable simpiification has therefore been achieved ‘by the per unit method with a common voltampere base and voltage bases on the two sides in the ratio of transformation. . Apu) can be determined directly from the equivalent impedance on primary or secondary side of a transformer by using the appropriate impedance base Z=2,+ 24a 2, ZA % (pu) = oa =a yp Sy Zn Zn hn a? = E fou) = Zp) + Zipu) = 2 pu) (4.15) On secondary side:or Thus the PEr-Unit impedy from prims ALY OF secon are in the ratio of try Om the tiwry Transformer which jg * Phase ratio Of w three-phase The pu transformer impedance Of & three Phase tansfoemen ee ait obtained by direct use of sper hse MVA base and Line arene telation (4.9). any oO de ofa her impe 2,0 Z, etO-line KV base ig dance on forte % pu value just like Cither side of «tray t ls converted From a one-line diagram of diagram by following the 1 4 power system wee ‘an direcily draw the impedance Steps given below common MVA (or kVA) base ot the system © be divided into a number of sections an appropriate KV base in one of other sections in th 3, Calculate per unit values of voltages and impedances in Sonnect them up as per the topology of the one-line dh is the single-phase per unit impedance diagram, The above steps are illustrated by the fallow ‘Example 41 | Obtain the per unit impedance (reactance) diagram of the power system of Fig. aie phase impedance diagram of the power system of Fig. 4,5 agram has: been ee Be. 45. We shall make some further simplifying ce are neglected wo that iis represented ay and resistance are 1. ‘Line capacitance is atl ig impedance diagram is meant for short circuit _ Dade ae ‘on by static loads under short ciscuit cxbitions can neglected. Losds A and Mare therefore ignored oe Pe conven all reactances to per unit fons. Choose 4 commoq epee roe e cpcaseab hier ALE” lige acca pepe of 30 by the the sections, Choose Calculate kV bases of '¢ ratio of transformation. h section and ran. The result ing examples, voltage base in the cirewit af generabor 1 is 11 k¥ lime-tline anid Then the voltage a 2 and Fis 6,2 bY. criss of varie conpacs ue calculi bloTransinission line: — 0.564 Transformer 7; = 0418 Transformer 7: = 044 Generator 1: = 0,396 Generator 2; 0.936 Generator 3; 0.437 The reactance diagram of the system is shown in Fig. 4.8. 0418 0.564 0.44 0396 = 0339 oua7 + : ls BO [ Ex Bs Fig. 4.8 Reactance diagram of the system of Fig. 4.5 (oads neglected) £,, Ey and E; are per unit values of voltages to which the generators ee excited. Quite often in a short circuit study, these will be taken as 1 20° pa (ae load conditinn). a ‘The reactance data of generators and transformers are Sly (or per cent) values, based on equipment ratings rather than in oye values as given in Example 4.1; while the transmission line ee es given in actual values. Let us re-solve Example 4.1 by assuming (pu values of reactances - A sireuit of 300° 30, base voltage of 11 kV in the lteult °° y ee the circuit of generators 2 and2 R ——Textesttiaion of Power System Comptnants is Example 4.1, we alege / Paponens __ i633 and generators ar aulate the pu vainey op the reactance SENETALOES as per reation Ghat ance’ of transformers Transformer 7; Transformer Ty Generator |: Generator 2: Generator 3: 0.5214 , 30 Obviously these values are the same as obtained already in Example 4,1 4.5 Complex Power load Zed from a source as in Fig. 49. Let 2b T= 2 (8-% hang (0) Fig. 4.9 Complex power flow in 2 single-phase foad Whea i i nt lags behind voltage, This is a convenient sicies afcign ot Jin power aysoas where loads have moaly lagging powec pice 0 on rection of curtent indicated is given by in the direction of current ‘Complex power flow in Sew =Wile @ ’ =IVI Ui cos 0+ AVI Ii! sin #= P+ jO (47) yt sl=(P? + 9"106 4 Modern Power System Analysis Here 5 = complex power (VA, kVA, MVA) |S| = apparent power (VA, kVA, MVA); it signifies rating of equipments (generators. transformers) P=IVI II cos @= real (active) power (watts, kW, MW) Q=IVIIT sin @= reactive power = voltamperes reactive (VAR) = kiloveltamperes reactive (kVAR) = megavoltamperes reactive (MVAR) mi Eq. (4.17) that Q, the reactive power, is positive ower factor load) and negative for leading current (leading power factor load). ‘With the direction of current indicated in Fig. 49, S=P+jQis supplied by the source and is absorbed by the load. Equation (4.17) can be represented by the phasor diagram of Fig. 4.10 whee It immediately foliows fro for lagging current (lagging Pp 6= tan" £ = positive for lagging current (4.18) = negative for leading current Reactive 4 power P Real power Fig. 4.10 Phasor representation of complex power (lagging pi load) [f two (or more) loads are in parellel as in Fig. 4.11 s=vi= VR +h) VI} + Via «ill = 5, + $= (Pit Pd + HQ Q;) pts ‘ fy | ’ H \ Fig. 4.11 Two loads in parallelce tee tnwtths hms Le t As per Eq. (4.19), Kirchhoff's cument | ; applies separately to real and re. oe eT e ak ors) Ina series RL load carrying currey I V=I(R 4 jX,) S= Vf PR + PX P=TR = active power absorbed by loud Q = 1°X, = reactive power absorbed by load In case of a se RE jo arrying current 7, P=PR Q@ =— UX ¢ (reactive power absorbed is: negative " f NOW a balanced three-p presented in the form of an 5 star as shown in Fig, 4 c-phase complex power fed into 9 ren by ! S=3V,/, =31V I y el me it 18) qi ° Then $= J3 Wu, Zo P+ 10 (4.21) Fig. 4.12 Complex power fed to three-phase load isiev3 Wil! P= VB Wl ly) cos 8 Q= 3 WAll sin o where @ = power factor angle IY, the line voltage, is expressed in KY; and J), the Line current in amperes, ‘Sis in KYA; and if the line curreat is in kiloamperes, 5 is in MVAModem Power System Analysis }_—____Medatn Power System An $a Tn terms of load impedance Z, |Vileop V3Z Substituting for J, in Eg. (4,20) p= Mal? (4225 If Vj is in kV, Sis now given in MYA. Load impedance Z if required cay 5, calculated from 4.6 SYNCHRONOUS MACHINE The synchronous machine is the most important element of a power system. converts mechanical power into electrical form and feeds it into the power network or, in the case of a motor, it draws electrical power from the network and converts it into the mechanical form. The machine excitation which controllable determines the flow of VARs into or out of the machine, Books or electrical machines [1-5] may be consulted for a detailed account of the synchronous machine. We shall present here a simplified circuit model of the machine which with suitable modifications wherever necessary (under (ransiett conditions) will be adopted throughout this book, Figure 4.13 shows the schematic cross-sectional diagram of a three-pit* synchronous generator (alternator) having a two pole structure. The stator bit 4 balanced three-phase winding—aa'‘, bb! and cc! The winding shown 8? concentrated one, while the winding in an actual machine is distributed the siator periphery. The rotor shown is a cylindrical” one (round rotor or salient pole rotor) with rotor winding excited by the DC source. Tre winding is so arranged on rotor periphery that the field excitation prol? ‘nearly sinusoidally distributed flux/pole (¢) in the air gap. As the reir three-phase emifs are produced in stator winding. Since the macie * balanced one and balanced loading will be considered, it ean be model asis for the reference phase a. eps ‘with more than two poles, the above defined structure" ‘for every pair of poles, The trequency of induced ent 8 2 f= ty HeN= rotor speed is p YAchronoUs = number of po} peed) in rp les on ‘Stator (armature) windy Fiele winding Stator 120 120° Fig. 4.13 Schematic diagram of a round rotar synchronous gererator On no load the voltage E, induced in the feferetce phase a lags 96° Sehind @ which produces it and is proportional to @ if the magnetic circuit is assumed to be unsaturated. This phasor zelationship is indicated in Fig. 4.14. Obvious!y the terminal voltage V,= Ej. ore . . Fajen 1 ulema Fig. 4.14 Phasor relationship between o/and E, i from the three-phase stator winding, the een co dea evaing fn dole (inthe diection of Pema ak the\rovor, This hx, called amarure reccton (Lx, i tele Boa aon ae oe peld x dy Icinutivey follows ta: ois im phase es which causes it. Since the magnetic circuit has been a isModem Power System Analysis assumed to be unsaturated, the superposition principle is app! resultant air gap flux is given by the phasor sum $= %+ Cy (423) Further assuming that the armature leakage reactance and resistanes ar negligible, , induces the armature emf which equals the terminal Voltage y. Phasor diagram under loaded (balanced) conditions showing fluxes, curreqg and voltages as phasors is drawn in Fig. 4.15. . 50 that the Wie a sp £8 \ i (MeXe= —Es \ So Fig. 4.15 Phasor diagram of ‘synchronous generator Here @= power factor angle ea nee by which E leads V, called load angle or torque angle Pe 2 Se 5.10 that @ mainly determines the power delivered by moe a magnitude of E, (i.e. excitation) determines the VARs Because of the assumed linearit : y of the magnetic circuit Bs eee Y, are oe to flux phasors &, 4, and 4, aa ae f ae ee 2 ee {Tux phasors. It therefore easily follows fom MA? leading dr). With tence as, and therefore) of pl indi oe yhasor AB indicated on the «itr where X, is the constant of tonalit i. Propotionality, san fori tote definition oF X,, we can directly write the o0wilt ‘expression iges without the need of invoking flux. phasors. WE HX, es where Ey = voltage induced = 00 load emf el aest 10 PIeSenitation of Power Systam Components i iu 8 4.16 can’ be easily Modified to include the effect of Astance (these are series effect 8) 10 give the ig. 4.17. The total of the machine. complete circuit model of the sync Besianes (Xi, 4X) = Xie ealed hronows generaior as in Equation (4.24) now becomes TEs eee iy X,~1,R, (4.25) Fig. 4.18 Circuit model of round rotor synch Onous Generator (resistance and leakage reactance neglected) This model of the synchronous m for the effect of magi not hold. aachine car be further modified to scconnt etic saturation where the principle of superposition does X 5 y Fig. 4.17 Circuit model of round rotor synchronous generator ted in. power system studies. ig. 4.17, the simplified circuit Armature resistance A, is ee : ‘Therefore, in the place of the circuit model of Fig. 4 eerie rie a be used throughout this book. The corresponding phasor ver i . iven in Fig. 4.19. The field induced emf leads the terminal ane ve the tomque Cloud) angle & This in fac, isthe condition for active To en or afte generator The magnitude of power deiered deperas power upon sin & operation of « syehronus favhibe, the eument reverses aoa ceripie, 420, v0 that £4, (4.25) meliies (0 i a ee (42 ote he moted that je repuoseated sor diagram of Fig. 4.21. may be no ops aoe athe condion or power to low ito tne V, now leads Ey ‘terminals.2 _Madern Power System Analysis The flow of reactive power and terminal vol is mainly controtle Section 5.10, Voltage and 1 by voltage regulators (see d) by automatic tip changi ection & Fig. 418 Simplified circuit model of round rotor synchronous generator x be Fig. 4.20 Motoring operation of synchronous machine Normally, 4 synchronous generator operates in parallel with other connected to the power system, For simplicity of operation we shall consider + genesator connected to an iafinite bus as shown in Fig, meaty a large system whose voltage and frequency remain constant indepe™!"! of the power exchange between the synchronous. mac! independent of the excitation of the synchronous machine. Consider now a synchronous generator feeding constant nciive-power Me™ infinite bas bar. Asthe machine excitation is varied, armature current 3" angle @ i.e. power factor, change in such a mannet as to keep by means of its excitation, This is ¢ jive power flow are often 1 cussed in dstuil in comatically regulated ting on the field cisevits of generator, devices on transformers, gy Md, : i i 6 “ty Fig. 4.19 Phasor diagram of synchro- nous generator Fig. 4.21 Phaisor diagram of motoi ‘operation ner tO 22. As infinite P 2 and the bussSHUN = active power ouput M Unfit bs Fig. 4.22 Synchronous machine connected to infinite tus It mens that since |V,1 is f; Ned, the projection If,| cos 4 remain the e\citation is varied, Phasor diagrams corresponding + medium and low excitations are presented in Fig. 4.23. The phasor gran of Fig, 4.23(b) corresponds to the from the phasor diagram that for this exe IE of the phasor J, en ¥, IS Constant, While unity power factor case, Ibis obvious | ‘tition cos 6= IVI he (al Cverexctied jLX (b) Normal excitation Eres jl rie 5 (o) Undarexcitod % c constant diagrams of syrchronous generator feeding Phaso! Fig, 4.23 power as exctalion § varied verexcited case (Fi mF wwisatinin. For the overe islet This et i th the generator eds Lape Ij ens 6> 1k. aes bein tive power ftom “into the bus (or dra power intoedenilethad cio underexeited case (Fig. 4.23¢), ie. \EAcos 5 < Wi 1, leads V, so ‘ Seherator feeds negative reactive power into the bus (or draws positive i Power from the bus), Figure 4.24 shows the overexcited and underexcited cases of S¥nChrongys Motor (Connected to infinite bus) with constant power drawn from the bus. In the overexcited case, /, leads V, i.e. the motor draws hat the re ‘tive infinite ative reictiy Power (or supplies positive reactive power); while in the undereacited case ) lags Vj, i.e. the motor draws positive reactive power (or supplies nepative reactive power). Z to liek: ws Yn & (a) Overexcited “I F ——# HlaXs rf oe “e h, (6) Underexcited Fig. 4.24 —Phasor diagrams of synchronous motor drawing constant power 8s excitation is varied From the above discussion we can draw the general conclusion that ® synchronous machine (generating or motoring) while Operating at const Power supplies positive reactive power into the bus bar (or draws segalite reactive power from the bus bar) when overexcited. An underexcited muctil® on the other hand, feeds negative reactive power into the bus bar (or drws Positive reactive power from the bus bar), ee Consider now the power delivered by a synchronous generator to an intial bus. From Fig, 4.19 this power is P=IVA If, cos 0 I The above expression can be written in w more usetul form from the pe? geometry, From Fig. 4,19wer Syst istem Components __| Ja or : (4:27) fi: x sind (4.28) The plot of P versus 4 shown in Fig, Ik 25, is called the power angle curve. The maximum power that can be del ivered Occurs at 6= 90° and is given by iz) 1¥ man = (4.29) P For P > Pia: or for &> 90° the generator falls out of step. This problem (the stability) will be discussed at length in Chapter 12. Pa Pra Fig. 4.25 Power angie curve of a synchronous generator Power Factor and Power Control : mus machine power factor Ti d 4.24 illustrate how a synchronous mac! p a, Ie atioe for fixed power exchange. these do not give us a clue ine de quantitative values of l,l and & This can easily be accomplished re by recognizing from Eq. (4.27) that Ej sin 5=U,1X, c0s 8 constant (for constant exchange of power to 4.30) infinite bus bar) a é . Figure hasor diggram for 4 generator deliverin — power 4.26 shows the Pi ying excitation. Assn 9 remais esa a ee eal Hel to V, as excitation is varied. : Ejmoves along 1 Pa it magnitude is obained ee 27 shows the case of limiting excital ae ee yen this value the generator becomes un ‘For excitation lower| Modem Power System Analysis a Ex En lay oa | y Y | Mea Xe Na | H4aX, Pp oh (Erisin = PX b= iM i iPalte ' b The Vi S Jai Fig. 4.26 Effect of varying excitation of generator delivering constant Poway to infinite bus bar th Ere. Ia, Vy Fig. 4.27 Case of limiting excitation of generator delivering constant power to infinite bus bar Similar phasor diagrams can be drawn for synchronous motor as well ‘or Constant input power (or constant load if copper and iron losses are neglected and mechanical loss is combined with load). Another important operating condition is variable power and fixed excita- tion, In this case IV,l and 1E/| ate fixed, while 6 and active pow accordance with Eq. (4.28), The i responding phasor diagt for two values of 4 is shown in Fig, 4.28. It is seen from this diagram that as 0 increases, current magnitude increases and power factor improves, It will be shown it Section 5.10 that as 6 changes, there is no significant change in the flow of reactive power, Locus of & —= 5 ‘a dhaXs Fig. 4.28 Operation of synchronous generator with variable power and suet excitationeT Representation of Power System Components __ _| a7 Salient Pole Synchronous Generator A salient pole synchronous machine from a round rotor Project with « larg employed in a5 Shown machine by constructional ¢ interpolar air Sap. This type machines coupled to hydroelectric turbines which are inherently nes So that the synchronous machine has multiple pole pairs as eed steam turbines (3,000/1.500 eaure, Salient pole machine anslysis is in Fig. 4.29, is distinguished features of field poles which of construction is: commonly lined below. Direct axis, | Quadrature a axis bf ey e%4 [on 'y bi a , s & \ bb ts & Same by a Fig. 4.29 Sallent pole synchronous machine (4-pole structure) i field induced emt ine, amature current in phase with Ina round rotor machine, a ces the same flux linkages per ampere See ee he cceamid Ractok wectnes chee et E SEE | f ‘ the air gap is uniform so X,= X,). In a salient pole phase current is the same Oat y= X, a Sete eS in ee aera ft iis (called direct axis) and is the ES CE eat “ ae is). Armature current in q ‘ eee oie -¢ oF flux path being low nces the direct axis and the reluctanc ne area it produces larger flux linkages pel eee Lew aoe ae ee aoa rection reactance X,, e ee ids rare 1, of armatut ler " component 1, ue a i il roduces flux along ee saree a ed (because of large d we of the flux axis and the reluctama Power System Analysis interpolar air gap), j = kts ae Mt produces smaller flux linkages per ampere and ce ee ‘Jents smaller armature reaction reactance xX, (quade — San < a) the flow of inphase compon rm es ce a Salient pole machine offers different reactances to the flow of J, on i mponents of armature current /,,, a circuit model ou lasor diagram ofa Salient pole generator is she rawn by following the steps given below: tL, of an mine |, of armature curren, | cannot be drawn, ‘Th G own in Fig, 4.30.11 can be easity laoXq PA = ilelX -%) Fig. 4.30 Phasor diagram of salient pole synchronous generator 1. Draw V, and J, at angle 6 2. Draw 1,R,. Draw CO= jl, X;(4 © |.) 3. Make ICPI = |J,| X, and draw the line OP which gives the direction of £ phasor 4, Draw a | from @Q to the extended line OP sucha that OA = Ey Te can be shown by the above theory that the power output of a salient pole generator is given by WIlEd WP (=X) (430) P= % sin 6+ SOI sin 26 ‘The first term is the same as for a round rotor machine with X, = Xuan i is quite S! constitutes the major part in power transfer. ‘The second term is qui : (about fea eoageared to the first term and is known as reluctance ies P versus Sis plotted in Fig. 4.31. It is noticed that the maximum Po : ower per ust change in power i abled the i : for small changes in power angle). a mchroncin 9 Sams in the operating Yegion (6-< 70") ae" ‘a salient pole machi than in a round rotor machine. dP 7 output occurs at &< 90° (about 70°). Further’, (change inppe SB ~~ Siem: Components Rosuitant — fae Ne | 90° eames 2 =i Fig. 4.. 19: 4:31 Power angle curve tor salient pole generator In this book we shall neglect the effect of saliency and take aks in all types of power system studies considered. During fa machine transient, the direct axis reactance changes with time ig the following distinct values during the complete trénsient. ubtransient direct axis reactance _ Xf = transient direct axis reactance -X, = steady state direct axis reactance The significance and use of these three values of direct axis reactance will ‘be elaborated in Chapter 9. Operating Chart of a Synchronous Generator While selecting a large generalor, besides rated MVA and power factor, the sreatest allowable stator and rotor currents must also be considered as they 4 anil stesses and ternpercture rise. Such limiting parameters in ‘operation are brought out by means of an operating chart or performance ii ition simplicity of analysis, the saturation ¢ and an unsaturated vay of syne effects, saliency, and resistance are hronous reactance 1s cae ‘of a cylindsical rotor mactine, en MVA isa circle ame UM. The locus comtered at 0. As MP is ; SO and MQ wo MW, all 10 the same saleModern Power System Analysis Locus of constant (é (cirete canta (3) s Loous [f,|.X, » (circle centre AN) Fig. 4.32 Phasor diagram of synchronous generator For zero excitation, ie, lel =0 ~ HT, X= or y hx, te. 17,1 = IV,UX, leading at 90° to OM which corresponds to VARSiphase Consider now the chart shown in Fig. 4.33 which is drawn for a sys machine having Xj = 1.43 pu. For zero excitation, the current is. 1.0.4. pu, So that the length MO corresponds to reactive power of 0.7 pu, fixing both active and reactive power scales. With centre at 0 a number of semicircles are drawn with radii equal to different pu MVA loadings. Circles of per unit excitation are drawn from cent M with 1.0 pu excitation corresponding to the fixed terminal voltage OM. Lines may also be drawn from 0 corresponding to various power fuctors bat fe clarity only 0.85 pf lagging line is shown. The operational limits are fixed follows. Taking 1.0 per unit active power as the maximum allowable powss * horizontal limit-line abc is drawn through b at 1.0 pu, It is assumed dat & machine is rated to give 1.0 per unit active power at power factor 0.85 lig and this fixes point c. Limitation of the stator current to the eotresponding be fequires the limit-line to become a circular arc ef about centre (), At poitt ! = rotor heating becomes more important and the are de is fixed by the nase excitation current allowable, in this case assumed (o be Lé)| = 2.40 pa ie times |V,1). The remaining limit is decided by loss of synchiopism 3 7 - power factors, The theoretical limit is the line perpendicular to MO a’ or 5= 90°), but in practice a safery margin is brought in to permit a further 2 increase in load before instability, In Fig. 4.83, a 0.1 pa margin is ea and is shown by the curve afy which is drawn in the following way runes: 0.85 pt lapping 3 10), = e z u a 08 8 si 2.4 pu exctaton 5 t 2 06 \ o4 5 10py 02 excitation opin An | e | mM 905 9 05 ey | Leading <— ~ Reactive power (py) lagging Fig. 4.33 Operating chart for large syrctronous gensrater Consider a point A on the theoretical limit on the t LO pu excitations are, power Mh is reduced by 0.1 pu to Mf: the opetating point must, however, Still be on the same II ere and kis projected to f which is the required po! on the desired limiting curye. This is repeated for other excitations giving the curve afg. The comy “a, shown shaded, is gfabede. A working point placed within this area at once defines the MVA, MW. MVAR, curreat, power factor and excitation. The load angle écan be measured as shown ia the figure. 4.7 REPRESENTATION OF LOADS + is the toughest parameter to assess scientifically, The eee problem is truly a statistical one. A typical daily load curve is ben u Fig. 1.1. The louds are generally composed of industrial and cori ‘components, An industrial Joad consists mainly of lange thse ee ae motos with sufficient load constancy and predictable diy cele, whereas moe) sinly consists of lighting, heating and many single-phase co Bets wiay by householders. The design and operation i + sysiems both economically and electrically are greatly influenced by power sy: nature and magnitude of lo122 | Modern Power System Analysis - ‘ In representation of loads for various system studies such as load flow an s stability studies, it is essential to know the variation of real and reactive power with variation of voltage, Normally in such studies the load is of cc nature with beth industria! and domestic components. A typical composit | Toad at a bus may be | Induction motors % Synehrorious motors 5-15% L ing and heating 20-30% 5 Though it is always better to consider the PV and Q-V characteristics y cach of these loads for simulation, the analytic treatment would be ver Sumbersome and complicated. In most of the analytical work one of d following three ways of load representation is used. (i) Constant Power ‘Representation This is used in lead flow studies. Both the specified MW and MVAR are taken to be constant, (a) (i) Constant Current Representation Here the load current is ven by Eg. (4.17), ie P- Te a = 5-8 7 | whew V=IV.Zéand 6= tan " QIP is the power Factor an as constant cutreat representation because the m1 | 4s constant in the study It is known agmituds of current is regarded (ti) Constant impedance Representation OR This is quite often used i MVAR at nominal vole, ‘Thus Sail iQ” F—-jo ~y which then is rg; wparc Hed as constunt throughout the study Example 4,3 A sanchtonais generator i rated 645 MV, 24 py synchrovows reactings 1.24 a 0.9 pr ts feeding ‘full load at 0.9 Ff The generato At rated Voltage, Cateutate AG) Escitation emt (Ey) aad power angle: (0) Reactive power drawn by the load Canry out calculations in pu fort a eonve tthe result to actual values.@Ntalion oF py ‘Wer Syatem Components Ses ‘yale Component fa Base MVA = 645, 3.ppnse j Base KV = 24, teins teat voluge= 24. 24 Synchronous reactance y — 12-645 | y NOUS Teactance Y — GaP lupe ONE OF the | Full load (Mya) - PU, 0.9 pF tae Load caren I] | { R are taken = 095 0.436 ou | (2) Excitation emt (see Fig. 4,10) tis known F; (wetual) = 1.99 x 24 = 47.76 Ry (ine) | | is regarded $= 37.1" (leading) (6) Reactive pawer drawn by load = Vi, sin ¢ 1 MW. and = 1X 1 x 0.436 = 0.436 pi or 0.436 x 645 4 422) if = 281 MVAR [Example a4 | BS Sceraion of Enanspie 4.3 is eimying fll Toad ot twp i j Eruttion emf is (i) increased by 20% and (ii) reduced by 20%, ; —— in each case : ead py PY state power drawn by load 7 ) lod angle Full oud, p= 1 x 09 = 0.9 pu y= 1.99Modern Power System Analysis _ =1 @) Ey is increased by 20% at same real load. Now E)= 1.99 x 1.2 = 2.388 As per Bq, (4.28) = 0.89 = 0.79 = 1.183 2 41.2° (a) pf = cos 41.2°= 0.75 lagging (b) Reactive power drawn by load Q=1V)lU,! sin 3 = 1 1.183 x 0.659 = 0.78 pu or 5028 MVAR (i) E, decreased by 20% or 199 x 0.8 = 159 Substituting in Eq. (i) 0.9= ( which gives— Representation of Bower System Companeris 125 (by f : 4 Pape sin 1.5 = 024 or O 0024 % 645 = 15.2 my AR (i PROBLEMS Pog ov rae i "5 the schemutic diagram of a midial transmission 1 tings aad reactances of the various components ate show ] oe A Lonel of 60 MW a 0.9 neyes actor Iiguing is tapped ie A SUbstation es ( ii tie Which is ta be Maintained at 6) € te mie era ss tl ite i ain : at V. Calculate the ichroniows machine. Represe nt the transmissi¢n es only Azo HOR | | ~ saw ie Bf ageing VY 190m OMA X= 10% x Fig. P-4.4 i 42 Draw the pu impedance di 4.2 + base of 100 MVA, 220k in 50.2 line, ; ‘The ratings of the generator, motor and tr Generator 40 MVA, 25 kV, X" = 204 Motor 50 MVA, 11 kV, X"" = 30% ¥-¥ transformer, 40 MVA. 33 Y-220 Y kV, X= [8% ¥-A transformer, 30 MVA, 11 A-220 ¥ kV, ¥ = 15% m for the power system shown in Fig, P. plect resistance, and use ansformers are My G6 js Fig. P-4.2 i y. Mas a resistance 43. synchronous generator is rated 60. MVA, 11 kV. Ie hi Daren R= 0.4 pu and X, = 1.65 pw. It is feeding into an infiite Stati Fae eving. TRY delivering a curren 3.15 KA at 0.9 pf lagging 8) Determine £; and angle 4 1s 1%) Draw a phasor diagram for this operation, — Aaatoue (s) Bus bar yoituge falls w 10 kV nie te ae : ee ; cl is unchange z eneracor and its excitation renin ae tnd ph of the current delivered 6 te bus 101iSS ee ae BS, i | ____Modem Power System Analysis generator resistance to be negligible 4.4 A 250 MVA, 16 KV generator is feeding into an infinite bys 15. kV. The generator yynichrondus reactarice of 1.62 pu. Itig that the machine ion and mechanical power input are a ¢ E = 24 kV and power angle 6 = 30° “(a) Determine the line current and active and react bus bars. (6) The mechanical power input to the generator is increase from that in part (a) but its excitation is not changed. Find the » cit line current and power factor, ) With reference to part (a) current is ta be reduced by 20% same power factor by adjusting mechanical power input to generator and its excitation. Determine £, 4 and mechanical powe input (d) With the reduced current as in part (c), the power is to be d to bus bars at unity pf, what are the corresponding values of E,and Sand also the mechanical power input to the generator. 4.5 The generator of Problem 4.4 is feeding 150 MVA at 0.85 pf infinite bus bar at 15 kV. ta) Determine £; and 6 for the above operation. What are P and to the bus bars? {b) Now £; is reduced by 10% keeping mechanical input to generator same, find new Sand @ delivered. (c) &; is now maintained as in part (a) but mechanical power inp generator is adjusted till @ = 0. Find new Sand P (d) For the value of £,in part (a) what is the maximum @ that delivered to bus bar. What is the corresponding Sand /,? Ske phasor diagram for each part Answers: 4.1 12 kV 43 (a) 26.8 KV (line), 42.3° leading (©) 113 2-288" kA; 0.876 lag 4.4 (a) 0.S11°Z— 25.6" kA: 108 MW, 51.75 MVAR (b) 6.14 kA, 0.908 lagging (©) 1,578, 13.5%, 93.3 MW (d) 1837 KY, 35.5%, 96 MW 4,5 (a) 25.28 kV, 20.2%, 127.5 MW, 79.05 MVAR (b) 33.9%, 54.14 MVAR (©) 411°, 1504 MW (d) 184.45 MVAR, 53.6%, — 7 0.787 pu—— Representation 6 PB —s fh I Power System Components | 7 _ REFERENCES : ue ey DAP. ‘Kothari, Electrie Machines, 2nd) edn Tata MCraw-Hit 2. Van E. Mablekos, Electric Mac w York, 1930, 3. DelToro. V. Jersey, 1985. $. Kothari, DP. and 1J. Nagrath, Theory and Problemy of Electric Machines, 2nd Edn, Tats McGraw-Hill, New: Delhi, 2002. 5. Kothari, DP. and Li, Nagrath, Basic Electrical Biginee McGraw-Hill, New Delhi, 2002 hine Theory for Power Engineers, Harper and Raw. Electric Machines and Power Systems. Prentice-Hall, Inc. New fe 2nd Edn, Tata Paper 1 Voltage on Power System IEEE Comitice Report, “The Effect of Frequency 4 Saas SLYESE Winter Power Meeting. New York, 1966.