WW lH’lanHé I l U ’ ’ WI 120 186 HTHS COMPARATIVE STUDY OF SIMULTANEOUS FAULTS Thesis for the Degree of M. S. MICHIGAN STATE COLLEGE A. S. M. Nuruliah "i954 This is to certify that the thesis entitled "Comparative Study of Simultaneous Faults" presented by Abu Ssdeque Iaioha nmed Nurul lah has been accepted towards fulfillment of the requirements for ___I‘L-_§_:_degree in E E Major profgssorU ‘ Date 1583' 27. 19 54 0-169 Comparative Study of SIMU TANEOUS FAULTS investigated by CLARKE COMPONENTS in comparison with SYIIJ '31". I C AL C 01.11” I ITSII‘IS by A. S. M. Nurullah A Thesis Submitted to he School of Graduate Studies of Michigan State College of Agriculture and Applied Science in partial fulfilment of the requirement for the degree of M. S. Department of Electrical Engineering 195A THESIS mmnwmmmmmm The author eXpresses his deep heart- felt gratitude to Dr. J. Strelzoff, Professor of Electrical Engineering at Michigan State College, for his guidance and valuable counsels every now and then. Also, the help rendered by M. Rahman, a graduate student, in checking the computations and to D. W. Plezia, Barbara Walsh and Sue Ball in certain aspects, cannot be left unaccounted. 331289 l. 2. 3. 7. Contents Introduction. Chapter I. Symmetrical Components Chapter II. Clarke Components Chapter III. Solution by Clarke Components Chapter IV. Merits of Clarke Components Appendix. Table I, II, and III Bibliography. Pox gs F—Z '3-\6 “1-49 (a\~ 6‘2. 63~ 6L 67- G8 INTRODUCTION This thesis is mainly intended to investigate two simultaneous faults that cccured in the power system by Clarke Components. Supposing the problem constitutes a paper mill which buys power from a big system to run a motor besides their own station capacity is thrown out because of a line to ground fault which occured in the 'systemf with a simultaneous opening of a breaker in the motor circuit. (shown in Fig. 3.1) The origin of this new method of attack came to the mind of Dr. J. Strelzoff, Professor of Electrical Engineering at Michigan State College, during his work in The Consumer Power Company, Jackson, Michigan. Though he solved the prob- lem.by the usual Symmetrical Components method, he had an idea that it might be possible to solve the same sort of problem with ease by using Clarke Components. Clarke Come ponents were defined in 1938 by Miss Edith Clarke of the General Electric Company and have not been put to much use as yet. The author was entrusted to solve the problem stated above by Clarke Components as well as to draw a com! parison with the Symmetrical Components. In view of this, the Symmetrical Components method of solving the problem with a review of a few basic relations involved was used as well as the Clarke method, to show the simplicity of the Clarke method. The reference books used have been listed in the bibliography and their numbers occur as exponents through- out the thesis. An appendix is provided at the back for ready reference of vitally important relations. Chapter I SYMMETRICAL COIPONENT History: The first paper indicating the possibilities of re- solving an unbalanced system of currents into positive and negative sequence components of current was published in 1912 by L. G. Stokvis. A second paper dealing with third harmonic voltages in alternators was presented under the sponsorship or Andre Bondel at a meeting of the French Academy of Science in lth. It is interesting to note that positive and negative sequence currents as they are now known were a by-product of Stokvis's main endeavor, which was to find a means of determining the magnitude of the third harmonic voltage produced by unbalanced line to line loads. A more detailed treatment of the resolution into positive and negative sequence currents of the unbalanced currents in the three phase ungrounded system was given2 in 1915. In 1918, Dr. C. L. Fortescue presented before the American Institute of Electrical Engineers a paper3 which introduced the concept of zero sequence currents and voltages and provided a general method for the solution of unbalanced polyphase systems. In this paper Dr. Fortescue proved that, " a system of n vectors or quantities may be resolved when n is prime into n different symmetrical groups or systems, one of which consists of n equal vectors and the remaining (n-l) systems consists of n equi-spaced vectors which with the first mentioned groups of equal vectors forms an equal number of symmetrical n-phase systems...." The method of symmetrical components is a general one, applicable to any polyphase system. Review of a few basic relations: In three-phase power systems, sinusoidal currents and voltages of fundamental frequency are represented for the purpose of calculation by vectors revolving at an angular velocity,co=23fradians per second. The components which replace them must therefore be sinusoidal quantities of the same frequency, represented by vectors revolving at the same angular velocity. Since the angles between vectors, Va, Vb, and Vc, and the somponents which are to replace them can.be represented in the same vector diagram, with any current or voltage vector revolving at the same rate as reference vector. Any three co-planar vectors Va’ Vb, and Vc’ can be expressed in terms of three new vectors V1, V2, and V3, by three simultaneous linear equations with constant coefficients. Thus: v3... r"n V1+ C12 V2+°13 V3 (1) vb z 021 v1. 022 v2... 023 v3 (2) vczc31 v1.032 V2+C33 v3 (3) Where the choice of coefficient is arbitrary, except for the restriction that the determinant made up of co- efficient must not be zero. The purpose of expressing the original vectors in terms of three known vectors is to simiplify calculation and there- by to gain a better understanding of a given problem. With this thought in mind, two conditions should be satisfied in selecting systems of components to replace three phase current and voltage vectors. (1) Calculations should be simiplified by the use of the chosen systems of components. This is possible only if the impedences (or admittances) associated with the components of current or voltages can be obtained readily by calculation or test. (2) The systems of components chosen should have physical significance and be an aid in determining power system per- formance. A system of three symmetrical vectors out of many possible systems is one in which three vectors are equal in magnitude and displaced from each other by equal angles. If Va, Vb, and Vb are a set of voltage or current vectors, referring to phase a, b, and c reapectively, of a three phase system, the three systems of three symmetrical vectors replacing Vfi, V5, and Vb are: 1. A system of three vectors equal in magnitude dis- placed from each other by 120°, with the component of phase b lagging the component of phase a by 1200 and the component of phase c lagging the component of phase b by 120° as in Fig. 1(a). 2. A system of three vectors equal in magnitude dis- placed from each other by 120°, with the component of phase b lagging the component of phase a by 2h00, and the com- ponent of phase c lagging the component of phase b by 2h0° as in Fig. l (b). 3. A system of three vectors equal in magnitude dis- placed from each other by 0° or 360° as in Fig. l (c). Vb V0 v 2 Vane: Vbo: Vco '\ / ‘: Von ch \, Van. v... (a) U (0 Figure l. (a) positive, (b) negative and, (c) zero-sequence of vectors. Three given vectors VA, V5 and V0 are expressed in terms of their symmetrical components by the equations: Vat: Von A. Van. - Vac (h) V»: V». + \(m. + \lbo (5) Va: Vu + ch '*‘ v°° ((3) Phase A as reference. Vb. -= «Him 5 Va .- OLVm Vex-c OkVaz I V“. = 0&le Vb: . V00 3 VCO .- Vac Substituting these relations in (h) - (6), the results Vfi: Vm + Va'). 4- \Iao (7) Vb = OKzVOq a. OkVoa + \loo (8) Vc : 0\\I0L\ 4 &Val 4 V050 (9) Again VOA = e (Va. 4. 0AM. + OGVC) (10) Van: 1:; (Va-t (RA/be 00/1) (11) V10 = *3 (VOL-t: \lb+ Va) (12) Any three phase systems can be represented by sequence networks. (This includes the equivalent sequence networks of generator, transformer; and transmission line etc.) The method used by symmetrical somponent method in solving two simultaneous faults on the same phase-~a line to ground and an open wire fault is presented here.li - C - Solution by the equivalent wye method. “I“ *1 D L“ 1M. ‘ 1 b The equivalent circuit for replacing a Y 1$¢ :hc single fault in the positive-sequence network is a single impedance connected in shunt ( or series) with the positive-sequence network at the point of fault. The equivalent circuit for a dnuble fault is a 3-terminal network. The 3- terminals are the two points of flault and the zero-potential bus. The simplest forms of 3-termina1 networks are the wye and the delta. The equivalent wye circuit is derived by first replacing the negative and the zero-sequence networks by their equivalent wye and then combining these two into an equivalent wye Which is then connected at the corresponding points of the positive-sequence network. From.then the computation is similar to that for a simgle fault. Extract from Dr. Strelzoff's work. Zero-sequence Wye Negative-sequence Wye The values of Co, Do, SO, C2, D2, and 32 may be obatined by standard method of reduction of a network to its equivalent wye. For the zero-sequence and the negative-sequence networks reduced to their equivalent wye, we can write new the follow- ing equations: (a) For the zero-sequence network: O-VXQ: IxO (00+ SQ)+ Iyoso (l) Solving the above equations for the currents 1x0: Iyo: 50 340: “Va“ L—OIS" + V¥0- boo Doc (19-) 110: VXO=§2 ~V10£9i§3 oo .900 (28.) Doo~ COD0+ C030 1- D080 (b) For the negative-sequence network, the following two equations may be written: O-ng .—. 1x2 (02 *r 32) + 13,232 (3) O-Vy2= 1x2 'tIy2 (D2 32) (A) Solving the above equations for the currents 1x2, Iygz D11 2'). X- 2 __ sz 51 V C2+ $2. l ._ .__.. ~ 2 ._.._.___.... ‘8. Y .921 \! ~D22 (t' ) where D22=02D2-r6282+.D232 In general this problem involves 12 unknowns; they are Ix0, 1x1? 1x23 VXO! VXI: ViZ: Iyoa Iyl: Iy2: Vjo: vyl’ V32. To solve it analytically we have the h equations 1, 2, 3,‘& h and the 6 equations determined by the respective fault conditions. There equations are: (a) For the line-to-ground fault on phase a: Vxfi == 0 be ‘l' be 1' O IXC —\- ch = O (b) For the open circuit on phase a: Iya: O V;{b"' Vyb = O VXC " Vyc = O In terms of symmetrical components, these equations are: oni-Vxl-VV:2==O (5) xyoilyhlyggo (6) IXO 1 Iyo =-. Ix]. + Iyl (7) 1x2 -\- Iy2 = Ix], + Iyl (8) VxO " Vyo = Vxl" Vyl (9) vxg — v3.2 = vxl- vyl (i0) Hence we have 12 unknowns and only 10 equations. By eliminating negative- and zero-sequence currents and voltages from.the 10 known equations, 2 equations containing only the 2 positive-sequence currents and the 2 positive-sequence voltages at points of fault, x and y remain. These equations 10 will be of the form similar to those of equations (1) and (2), or (3) and (h). vxl = 11x1* mlyl (11) vy1= nle . klyl (12) For these equations to correspond to a set of equations involving an equavalent wye, it is necessary that me=n. Equations (11) and (12) are derived as follows: From equations (1) and (2) and (9) Exo - Eyor-Exl - Eyl= - Ixc (00+ 30) - IyOSO +Ixoso+ I'yO (Dos-So)= - IxOCo+'IyODO From equation (7) 111" L571 ' 1370 ’ Ixo Hence EXl'Eyl::' C0 (le+ Iyl‘IyO)‘*Iyo DO::’ cOle‘colyl‘tlyO (Cod-Do) (13) Using equations (3) and (h) and (8) Exg-EyzzEXl-Eylz -IX2 (02+32) - Iy232+1x282+ly2 (D2+Sg) - = ~ CgIX2+ DZIyE I2:2 = Ix1+ Iyl'Iy2 Also from equation (6) -Iy2==IyO+Iy1 Hence Exl-Eylz- -02 (le +Iy1) + (Cg +112) 13,2: -021n-021y1-(02+D2) EXl-Byl= -CZIfl-(2CZ+D2) 1371'“? +D2)Iyo (11+) Comparing equations (13) and (1h), we write: Solving for IyO l| Let 00+ D0 = Z0 024:- D2-‘-' Z2 Then 110 x C°“C2. C = xx + 1 iii 20 +¥2 7" ¥o+izz (15) Substituting the value of IyO from equation (15) in equation (1h) which for this purpose is rewritten as follows: Exl-Eylz quay-13,102 4: 13,2 (62+ D2) ='-leC2-Iy102-(Iy0-+Iy1)(02+-D2) .- ’leCZ’Iyl (02+ Z$437032 E)“ .- EY1 : “IX\C1 i 17\QC1*21)-5£1(C° (1)1)“ flake—€22; h”): £5 +i1 £01?)- EY" E)“ + 1w {Cue} 55‘.qu ] + ENE; +932+ ziQCO-C2‘22)](16) zo +11 3° + *2. Using equations (5), (l) and (3) (1x14'Iy1'1y0)(co*‘SO)'*IyOSO'*(le-tIyI'IYZ) (02+ 32) + 15232 le (CO - 30+ Cg * $2)+ Iyl(Co*SO+C2*32)- ll IyOCO'IYZCZ = Ix1 (00+ so+ca+sz)+ Iyl (eo.so+02+32)-Iyoco +(Iyl IyO) 02 : le (CO+SO+C2+32)+Iy1 (00*SO*2C2*32)’ yO (Co-Cg) Substituting equation (15) for IyO C2) Exl': IX]. (CD‘SO" C2‘SZ) 4 Iyl (C0*SO+202+32)-Z (‘———Z I,“ Eta-*2). (C0 C2) (Co {2.21) K )2. (C ~6 in: IX\Y_C°*SO+C1+S2“ Pfiz)‘ *37\[<°+S°*2C2+52 (1'?) 201‘21 (Co-C2) (Co “iflj Substituting the value of Exl from equation (17) in the equation (16) _ (C EV! — lxsxco—k- so + Cl—x- 51 20—55 (4)1+ 1j\[co+So+lC2—’r g ~ (_C£~c2)(c° (2:22)] (3+2. 2 20+£ z°+zL + 2 (C .. 1X\[C?_~+; 'L 0 :2)]-" 1Y1[ [CZ+£1+\¥1(CQ~C1~ZI)] 0? EV) 7' X-choa— So+lCz+ 91+ EZ~ -fiaw zcec )- b+ZL EDA-21 }°+22 ~ M‘t (ii—221 0+3; 10+ZZ 2. Zd=2C2+D2- rifl— + 3.5M 0*21 Kai-3:1 i 7. = _ 2 + Dz-t- 2621a+2C222+12C° ~‘izCz ZO+ZL 26+ {\— 2. 2'1 : a. _.__.3___ + 32 + 3C2Z0+ C22; +7221. 20+Zz z°+ Z1 2 —222+D220+D \ 2z1+ (2201‘ Czig‘i- (221*- ZZCO Zd z Clio-\- Co z: (92* Cz)-Z'° + (D + C 2 .. L 20“- 21. + z 1) 1 £2 ED 1' 21 Zd = (23504-03551 + 223° (23) 2c 2 _‘ M1 30‘? %1 20+ 351 2° + $2- 25.. (o +S°+2Q2+52~M {2(C°~C;_) 20*21 + “—— Analytical solution. 30 + I 1 First step. To determine the values of positive- sequence fault currents: le’ IVl’ ( or fault voltages Vil’ V§l). By analytical computations, we equate the following two sets of equations: VXI‘; VX-le (61+ Sl)'Iylsl (228.) v3,1 . vy-les1 43,1 (SI-x D1) (22b) And vxlz (zc+zs)1x1* zsxyl (23a) \S Vy1= 2313a. (2.3+ zdnyl (23b) Where Vx and V? are the voltages existing under normal opera- tions at the fault points, and Cl, 81’ and D1 are determined in the same manner as 02, 82, and b . (If the negitive react- ances of the generators are the same as the positive-sequence values, then 012:02 etc.) Solving the above 2 sets of equations simultaneously results in vi==(cl+sl+zc+zs)1xl +(SI+ZS)Iy1 Vy, (51+ZS) le+ (31+ D1+ZS+Zd)ly1 The above 2 equations on solution yield the values of 1x1 and Iyl which substituted into (23a,b) give the corresponding values of Vxl’ vyl’ Knowing the values of 1:1 and Iyl and using equation 15, the value of Iyo is then found. Next, equation (7) yields the value of 1*0, and equation (6) the value of Iyg, and equation (8), the value of Ix2° Knowing the values of Ix2’ Iyg, IyO and Ixoand using equations (1), (2), (3), a (h), the corresponding fault voltages are evaluated. The current distribution in the positive-sequence net- work is obtained by superposition of the following values: (a) The currents for normal operation before the faults occur 0 (b) The currents due to the fault voltages V31, V&1. (c) The currents due to the fault currents 1x1, Iylo The current distribution in the negative - and zero-se- quence networks are obtained by the superposition of the fol- \6 lowing values: (a) (b) The currents due to the fault voltages. (ng, V32, The currents due to the fault currents. (132, I 2, or I ) Y xO’ IyO Having found the current distributions in the positive-, negative-, and zero-sequence networks, the currents in the phases a, b, c are found by the usual procedure of symmetrical components method. Chapter II CLARKE COMEONENTS History: Clarke Components9 or (X, (g, Oeomponents of current are not new. Components of current answering to the description of OK , Q), 0 although not so named were used in a method developed by Dr. W. W. Lewis, and published in 1917, to determine system currents and voltages during line to ground faults. In figure 2‘of Dr. Lewis' paper, which is similar to Figure 2.1 of this chapter, phase currents are represented by arrows both in direction and magnitude, the number of arrows indicating relative magnitudes of currents in each circuit. A :; a N1: 1) “C Figure 2.1 Phase currents represented by arrows in direction and magnitude, number of arrows showing relative magnitudes of currents in any circuit. Currents in the Y-A transformer bank and in the line to the right of the fault are 0 currents; currents in the transmission line to the left of the fault are X currents; currents in second\ac5transformer bank and in the line at generator terminals are (gourrents; currents in the generator are O\currents. In the method as used before symmetrical components were applied to unsymmetrical short circuits, each component of phase current met its respective impedences, but calculations were made with phase voltages and currents, not with component networks, and therefore it was time consuming if many circuits operated at different voltages were to be considered. In problems involving unsymmetrical three—phase circuits, and in particular circuits with two of the phases symmetrical with respect to the third phase, the use of the components of current which flow in one phase and divided equally between the other two phases is a logical develOpment. Such components, as yet unnamed, were usedh in 1931, in paper,u!5 both of which deal with transient conditions in rotating machines where development is materially simiplified by their use. Two papers have developed exclusively to these components. In one paper they are called 0(, (3, 0 components and the system modified symmetrical components. In the other paper, entitled, "Two-Phase Co-ordinates of a Three- Phase System," by Dr. E. W. Kimbark, the components are called x, y, and 2. Comparing these two set: x and!‘ com- ponents are identical; y and 6 components differ only in sign; 2 components of voltage are 0 (zero-sequence) com— ponents of voltage, 2 components of currents are twice 0 (zero-sequence) components of currents and z impedences are one half 0 (zero-sequence) impedences. \9 Clarke Components8 were introduced in 1938 by Edith Clarke, but their usefulness in the solution of the three phase circuit (unbalanced) is not yet fully appreciated. The paper by Camburn and Gross and another paper by Duesterhoeft are welcome because they indicate an increase of interest in Clarke Components. There are still promising Opportunities, however, for further work in exploring the application of these components to both steady state and transient problems. Definition of Clarke Components: With phase a as a reference phase in a Three-Phase System. (1) C(components in phase b and c are equal; they are opposite in sign and of half the magnitude of the component of phase a. (2) 6 components in phase b and c are equal in magnitude and opposite in sign; in phase a they are zero. (3) ()(Zero)components are equal in three phases. Relation between phase currents and voltages and their 0‘, (3, 0 components. Va \(b \(C \ -59: -—‘,; o( (3 O {.31. -T l O \ ‘ fl) ‘i-Vk F\ 0 ‘fi ’VA : -‘ 5 VB 1 g \ V9 SO, Voq : VOL +\(O (1) (2) (3) The required condition i.e determinant must not be zero, \ Vb = ~§Vd~ '* LEVQ +Vo Vc= --\5_V& ~§V© +Vo is satisfied. Again solving (1) - (3)V¢=%(VQ~V§N‘)U+) VQ-t KVb-“Vc) (5) V0: '% (Va+Vb+Vc)( 6) The corresponding current equations are I“ I“ *10 (7) 1b;- —}Ql*+§2—I@ + 30 (8) lg =-5.11*_%i@ +10 (9) and 10k = %L1a— 1211.35.) (10) 1e= i=3 (lb—IQ) (11) lo .- 3; (lmlbdc) (12) 0(, 6, 0 one line diagrams: When components of phase currents and voltages instead of phase quantities are used in calculations, each set of components is conviently represented by a seperate one line disgram or component network for which the components of currents and voltage in the three phases can be obtainede To draw component networks, it is necessary to determine: 1. references for the components of voltage, 2. components of generated voltage and, 3. impedenees offered to the components of current or admittances associated with the components of voltage. :20 2) Generatelev 6, and O voltages: In a synchronous machine with generated voltages Ea’ Eb, and Ec’ the Ed! E5, and E0 voltages obtained by sub- stituting Ea, Eb and EC for va, VB and v0 in (k) - (6) becomes: 7‘ E4 = ’3‘ (Es— FMMR<) \ (13) E@= ff?" (Rb — EC) (ILL) E0 = “3* (smegma (15) If the generated voltages are balanced Eb: 03%“ Vic: 0-ha a“: fig (16) E@ = ~3Ea E0 = O and currents in a balanced system: In a symmetrical system Operating under balanced con- ditions, the currents in phases b and c at any point of the system are IbzraZIa; Io: a I a. Substituting these values for 1b and Ic in (10) - (12), 1*: I“ 1@::_31m (16a) ‘10:: 0 Equations (16) and (16a) show that generated voltages and load currents are present in both the c( and 6 networks of a symmetrical system.during normal Operation. Because two networks must be considered instead of one,c(, Q, 0 components are not as convinient as symmetrical components 22. for the study of symmetrical systems during normal Operation or during three phase faults. fie OK, © and 0 networks: * I- ixnga Em r 2 w “.1 bub , R 'F‘ on: b0 evil “35km Ea. Z d‘ «deo‘lk ‘ 35:9—————q ; ._+ fiI‘x F wEb Xb=rgi+r1§x€ I Zeopoumam (M, _ g 2% \Deiouhoxl bub ‘) K“ 2 0- moJCme Q-“ekAnoflK Q3 GD Figure 2.2 (a) Flow of 0(and @ currents in balanced system with equal self impedences in three phases and balanced applied voltages. (b) &_network for system (a). (c) 6 network for system (a). (d) O-network for system (a). Figure 2. shows a symmetrical three-phase system.with balanced applied voltages and equal self impedences z in the three phases. I“ , flowing in phase a and returning one half in each of phases b and c flows into a loop circuit.‘ The voltage applied to this loop, as shown in the Fig. 2.2(a) is Ea—(-Ea/2) = “BifEa' The d-loop impedence for a symmetrical JKB three phase circuit of equal self-impedence Z in the three phases is 31 z. The current I“ in phase a is :1 ; 1°“- ‘s-‘E‘i— = 5;- <17) -:§£ The impedence met by ins Z. The equivalent circuit for phase a in thecKsystem is shown in Fig. 2.2(b), with the applied voltage Ba and the self-impedence Z. In this equivalent circuit, voltages are referred to neutral, base voltage is line to neutral voltage, and base current is line current. Since thetX currents and voltages in phases b and c at any point in the system are -1/2 those Of phase a at the same point, it is unnecessary to have additions equivalent circuits for these phases. The equivalent circuit for phase a in the 0( system will be called the 0( network. @ currents, flowing in phase b and returning in phase c, flow in a loop circuit. The voltage applied to this loop, as shown in Fig. 2.2(a), is -j[§ Ea. The @ loop impedence for the symmetrical three-phase circuit of equal self-impedences Z in the three phases is 22. The 6 current flowing in phase b in the direction indicated by arrow is 916. Therefore 3 - 3R” . _ - gIB:‘3g—§Eq’ omo\1@_ was The impedence met by IQ is Z. The equivalent circuit for the 9 system is shown in Fig. 2.2(c) with the applied voltage -j Ea and the self-impedence Z. In this equivalent circuit, which will be called the 6 network, voltages are referred to neutral, base voltage is line to neutral voltage, and base current is line current. The @ voltages and currents in 24 phases b and c are the voltages and currents in the Q3 net- work multiplied by g and—g , respectively. The @ network does not give directly the G voltages and currents in either phase b or phase c. This slight disadvantage is more than Offset by the convenience of having the same line to neutral voltage and line current as base quantities in the @ as in the (X and 0 networks. With a path for O currents through the circuit of equal self-impedences Z in the three phases, the impedence met by IO is Z. The 0 network for the system of Fig.2.2(a) is shown in.Fig. 2.2(d). d\, @ and 0 equivalent circuits to replace the various equipment, machines and transmission circuits of a three phase power system in the o(, 6 and 0 networks can be determined when the o(, CBand.O self and mutual impedences of the circuits are known. (X, G and O impedences, just as positive, negative and zero-sequence impedenees can be Obtained by calculation or test. Before developing equivalent for use in the¢fi ,(3, and 0 networks, relations between symmetrical components and 0b Qand O compo:':ents will be established. 0<, éBand 0 components of voltages and current in terms of symmetrical components of voltage and current: Symmetrical Components in matrix form; FVQ \ i )1 fiVa{) (18) Vb :: Q} a \ VA: NC a at \_J _ )qu 25' Clarke Components in.m£trii form \(a \ O \ \(K -1. 3 Vb '~‘ 2 g \ V9 (19) \ VC A: .‘g% ‘ \(o Similarly for current, Symmetrical Components in matrix form 1“ 1 P \ \ \ F Ian .= x 1" °‘ 0‘ x 1m. (20) i ICJ L 0‘ Q). | L 10m and Clarke Components ‘ Ia} "\ O \ I“ 1 2 -5- 3 20 b 2 Ci \ 1G ( ) L IQJ L-Ai “I; \ Io From the relations shown above, Symmetrical Components and Clarke Components can be related to each other. Form (18) VCL: V0“ +\lao. +\/O\O and form (19) Va : VaL+Vo So Vd+Vo= er+VQI-+VGO V0 1. Vac (Com be \n'SquXSeA from ® Md @) Va: Va\+Vaz 26 Now from (18) \lb -.- 03V“ -\- aVaz 4. VOLO from (19) vb: —35Va+§\1© +Vo SO ’ 03Voq+ QVOL2= ‘éVok" {EVQ’ 1- _- 3 OK - «5 5—5;- a= 4;: +35% new + as W) = ~12- New?“ 0% 9V? z {3% \Iou—t sgVQZ \(g, = 45 {Val - \lcQ) In similar way relations of two components of current can be found. I“ = 104 ~+ 102. I@ s -'3(Im-'Sa2) (21) ls . “lac VA: VGV‘V \(Q2 \(9: —3(\/ou—\)au) (22) Va : \(O~0 Conclusion: «components are positve plus negative components, @components are positive minus negative components turned backward by 90° and 0 components are same. 2'7 Symmetrical Components of voltage and current in terms of 0(, @ , 0 components of voltage and current. Solving simultaneous equations of (21) Van: tins-3*?) Val: “‘i (VOL-file) Vac: V0 (23) I“: i {Ii + 31(3) 100.: {(3% ~ J :9) Iqo= 10 OK, @and 0 self and Mutual Impedences and their relation with Symmetrical Component: A P IQ -> Q ~ 1bfi~i _i I c —? V: '0 Vat val Vb, Vc’ In: 1q+ lu+1c = 3140 L l Ill/7,3,77’rrT’I7ll’lIII],I/Ifif(I'llfITII](I]r(’,]’, Fig. 2.3 Symmetrical Components: The Symmetrical Component of voltage drop in an un- symmetrical three phase series circuit without internal voltages are expressed in terms of the symmetrical components A cwunmxk ofAflowing in the circuit and self and mutual impedences of the sequence networks. The basic assumptions that have considered are: (1) effect of saturation is negligible (2) linear relations between currents and voltages assumed. V5, V5, and Vb are phase voltages referred to ground at iaand vg', Vb' and V3? are phase voltages referred to ground at Q.. .28 So , “jog: Va~V0~ w. Iouzau+ l02—2,“). "" :°‘°£°~° 43b: Vb- Vb] :- Ql—K-CMan -* alalib1+ 1a°£b° é4> Uc : VC _ Vol : ogIouiu "* Nimitz “’t XWE“ New Vac: Vac—V03 = Jgklh+13b+13c) g \900 2 ng-Xc“ Q2a|+alibt+c1203 + ICAZK_EOQ.+ QZbl-t (113:2) + IQO (fag ‘5 lzbo-b ICQ)] 1 19cm : \lou ~V0q 2. 13 (13¢... amt”, 03.1%) 19°“ Ionian 4" lazirqz't 1&0 Iago avb= lmim-r dLSQsz1+ alaeibo Q3= hit-$0 (25) QIUC a :Q\Z(\ + QIQ: Zc2+ O‘IIMECo Qq“ 0‘ " ”0‘ = i 1 z z 2 ’L 3 GHQ 04 4 Inn-t- a) ‘1' 101 Vial" qzb2‘* @263) + I“ \Zko 4’ azbo* Q12(°)] Similarly, / Um = val ~ vm = ‘% {wot-£- 039nm: Qua} = L 1’ ”' 3i1a\ ki‘\+ Qib‘ "V Q ZCI) + leCZQx‘¥ ibz“ 2C2) 4' 14° Q ides cilia-‘- align with positive-sequence currents only flowing in the circuit, Ia2 and IaO are zero, and equation (25) becomes 1. mac : $q\ Eai+o~£b\+azcg 3 (1.90“ = :a‘ gl" Zb|+ 1C; 3 __ (26) “’07.: :al ion-- ai‘h‘q-Qli‘g 3 29 Equations (26) show that, with positive sequence currents only flowing in the circuit, voltage drops of all three sequences will occur between F’and qunless the co-efficients of Ia‘are zero. .Likewise, with only negative sequence currents or'zero sequence currents flowing in an unsymmetrical circuit, voltage drops of all three sequences may be obtained. The equations of (25) can be rewritten in much more compact form. / _. 19m : V0q~ V04 =- lquvamEu +10w§£lo / qu : V09. — V00 2 10:. 22. + 10.1221 + Iao 1.10 (27) I Uao : VONO —-\I0Lo : 10H Zot filalzoz *10to‘ioo Where Zn :43 iat+Zm+Im ) _. self impedence to positive sequence currents. Zu.= self impedence to negative sequence currents. ’l i (5500. + 1102+ 10. ) (iao + Zea ‘1' Eco) .L 3 Zoo: J2 . _ self impedence to zero sequence "currents. in = A37 (Em—salibydzq ratio of the positive sequence ) =vcltage drop produced by Ia2 to 1820 .L f ' (28) ' 3K Gr+aitvufia) ratio of the negative sequence =pvoltage drop produced by Ia,to 131 o .. l 3 Kiqo+aibo+aliu) ratio of the positive sequence =voltage drop produced by IaO to I80. 2; .L ol :- 3 1b, 'Iao So, Ial: Ia2: and lab of equations (27) in terms ofXK, @, 0 components are: I‘M : 5541,“ 31(3) I“? i (labile) XQO = 10 (23) 3\ The equation (27) becomes [52 (lowfilgflim Efiuak‘319fl2n + loam (4 190m 10K &i\\—\-Z‘1) .+ 32: 31.9, £ZH-1\Z)+ qum L '2 «9&2: é 10g K174 +i21> —\ 32—. 5 16 (Zn—112) 4* It: {10 _ L -“ (UGO - 2 XAQZokfi-Ioz —\' 1i A 3-(3 kZo\—‘ioz)+lo£oo From (23) again 1900 : DC) ‘21 : U01“? 19(12 MA 198,: *3L190u ~30?) '= A (1.9a; -- 190k\) a .- 5 an“ an 22¢ 222] - a 3M2 Mz- 2.2-222] * Whoa») u - l- - ~ 6 _ 2 31.4 £21” +121 In 10.1% lileiiu* 121-?2‘-1l2 '\ IOS<£2°~3$°> a, = l . 1 Id KEN-k in) + i2 119 3K10\‘291)'\ I. I“ The equations can be expressed in compact as follows: 604 = Vol‘VoL = Iafla*1{5§k@*3b he 196’: \‘©_Vé =KOKEQ’O‘A' 1E£QQ+1o XQO ”Do = \lo—vo' = Lloai'quoqstIoZN (29) °" “5% Via £49 1“ 1; q 2 29.4 gee 39° I? L ‘70 L1 5L ioQ 20‘ To 32, Where £4“ 2 '5‘ (Zn + in“ 12“? 227‘) Zea = e {Zn-r'izz‘zu-Zz‘ ) a0& 2 200 A@ :: 332'\‘I\\*121'* 12“ 2‘1) . O 190‘: —~\% \{\\~Izz*¥tz“ Z2‘) (3 ) 200‘: 1i (£e\ +102.) 2046 : (Zlo’r 721°) 2°C” 3% KIM—z») 290: ‘J (A0 " 110) Self and mutual impedances of the sequence networks in terms of 0k, @, 0 self and mutual impedences: Proceeding in a manner analogous to that used to determine (30) or by solving (30) for 211, 222 etc, 2“ : ”Elie-I + lee -'skhe-1‘et)] in = EL. lied * EMMA (its 53%)] 2&0 = £06 7.1:. : 3i- x1401“ 2%? +3 (kdcrk igdll 2..“ i [mesa new we) (31) Zko: J2 (3&9 4.31930) Zv\ 2 (7:0,; .. bio?) 220:: ,li K2$O _. 52226) 261: (5304 «31%) Symmetrical circuit—equal positive and negative sequence impedences: In a circuit with equal positive and negative sequence self-impedences and no mutual impedances oftained from (30) are 33 ad; EQQ :: in: 71! ZOO :Zoo 3 £0 (32) 29{@=¥@o(: 2403204:Z@02Zo@ : 0 When there are no mutual impedances between the sequence networks, the positive negative and zero sequence self impedances are customarily indicated by 21, 22 and 20 respectively instead of 211, 222, and 200‘ Unsymmetrical static circuit: If 211-4222, 210: 202 and 233:201' (30) becomes 2,“ = in + 31(zzvtzn.) ZQQz z“ ~15: (zznln) Zoo T- 2200 (33) 2‘9: 29% __. JJiQ-X'2\~Zt2) {do = 2104 = kzto* Z16) 29° : nice} 2 -.3Kz\o“120) becomes z“: z\\"\"1\7_ EQQ: an“ 1‘2- 2'00 = 2 ‘ ° (3L0 2‘? = 3534 :Zeo=3§oc: o 246 2 2264 c 21\° Symmetrical Components can also be represented in terms of Clarkes Components. From (33) In 2.. 3522: J2" Lima-559?) :00 2‘06 . Zn : i Kiii" 131‘“, ‘ £9?) (35) in. 2 1): LI“ ”SI-Lo ~19?) 1 = um +3 in») = iou'fim 1,. = A; m: an.) ==ifas-fi° Vai 77/,(l/1L77/lTiff—I—f/r—I’rI/IIr17f77’IiII7/77—T 4———-—- 31.0 3 lat’lbfi-XQ Fig. 2.h Let Fig. 2.h represent a general three phase satic circuit composed of bilateral circuit elements without internal voltages between points P and Q , with a return path for O currents. With phase voltages atI’and Q referred to ground or to a neutral conductor at? P and Q, respectively, the voltage drops va, vb and vc in phases a, b, and c in the direction of flow are 36 \3Q =1 Va ~VaI—a laiaq* lbickb —\- lczocc m, = vbwb’ = $10.14!: + lbibb * Libs .. "BC -.-. VC _Vcl 2 szom‘)‘ szbc + Iczc‘ (33) Equations (38) are general equations expressing phase voltage drOps in terms of phase currents after all other currents in the circuit have been eliminated. For example, in a three phase transmission circuit with a neutral conductor or ground wires, Zaa’ Zab“zba etc., may include the effects of neutral conductor or ground wires. Replacing Ia’ Ib and Ic in.(38) by'd”.@3 0 components given by (7)-(9), va, vb and v0 are eXpressed in terms of Ix, Iéand Id. Substituting these equations for v3, vb and Vc A and cameras ' in (ll.)--(6),’,\v',(l,vé5 and v0 to the corresponding co-efficients in (29), the o(, 5, 0 self and mutual impedences in terms of phase impedences are 2“: ESIXAM 35% .. (Zab+zac« 3:15)] 269 = “‘2‘ [Qibb + Icy-J. ibc)] e6~ L 2 - 3I£aa +be+ ¥¢¢ +2Q¥qb+¥ac+1bc)] 35.! = -~‘- e zgaqfi [icc—zhb+2(IO~L-§ac)] (39) 354 O: =- J. 3[1241‘ zbb‘zcc+ (245,1- 14¢“2zbc)] _ J. - fi[( 25" ‘ 55cc "’r Zak-zaWJJ Two phases with equal self impedances and equal mutual impedences with the third phase: Iet be:=ch and 230::Zab. Equations (39) tnen becomes 3'7 a .- ~[a—- We?” z@@= %[Q;bb+;€cc-l;bc)3 Zoo : A3— [Ian + lib!) “" 1 (‘1 Z043". sz)] £99 2 39.1 1&0 : 10%:0 ((+0) £40 :1 10.1 H g. [1M - ibb—i (Zab- ibCDJ Symmetrical circuit: With all self impedences equal to Zaa and all mutual impedences equal to Zab’ (39) or (hO) becomes 3““ ‘ 1% -‘ Zea-Zoe. 16o? zm+ alga-b (12.1) 2:19 = 194 =- 2(5e =Z°Q 7- 2210: 25%! = O Unsymmetrical three phase self impedance circuit with finite 0 self impedance: In a three phase series circuit between P and Q, let the self impedences of phases a, b, and c be Za’ Z and Zc’ b respectively and with no mutual impedance between phases. The 0k, 6, 0 self and mutual impedences can be obtained by replacing Zaa’ be and 206 in (39) bg Za, Zb and Zc respectively, and equating all mutual impedences between phases to zero. Then, 2),; : g3- (Ea + Ewb; z‘) 2% ._. 2. a; 1‘: (A2) 20. a 2"Vzb +1: '3 38 r 1(Ea~§€44) (:12 cent.) :90 3 lie, : zb-ZC f2? Two phases with equal self impedances: Let Zb=Zc, then (11(2) becomes a 1 a? 35M 3 {21+ ‘5?) 2@(b: zb z0°: “5(iq512b) (1+3) Ease.- 11w :ZQ6:£OQ:¢ 2.15: 2 Zen! 2 23- (gag-1b) : 2<£4-zd4) Symmetrical self.impedence circuit: Let Zae 2b: 202 Z. From (1L2) or (1.3-3) 2““ g@@ 2 2.9: 2 Equivalent circuits to replace an actual circuit in the 0(, Q, 0 network: Synchronous machine with equal positive and negative sequence impedances: From (32), the dand Qself impedances are equal to Z, and the 0 self impedanees to 20. There are no mutual impedances between.the<&, Q, and 0 networks. With balanced generated voltages in the machine, the generated voltage in the X_ network from (16) is Ea; in theenetwork it is -an. The O(, Q’ and 0 equivalent circuits for a synchronous machine with balanced generated voltages and equal positive and 319 negative impedances are chown in Fig. 2.5. Points T are the terminals of the machine to which the equivalentC¥ , 6 and O circuits for the rest of the system are to be connected. So in a three phase power system consisting of symmetrical circuit with equal positive and negative sequence impedances, the one line impedance diagrams for theci and C?systems are the same as the positive sequence impedance diagram. Generated C(voltages are positive sequence generated voltages; thec% network, is the same as the positive sequence network. The 6 network differs from.the positive sequence network only in its generated voltages, which are positive sequence voltages multiplied by -j. 2cm botwfio—l \oub zen. pow bws 7:063. pow low: Ea. oL-mcrufd‘n‘ 0_T\-Qhwo‘1K 1‘ 3750 7' JM ._. 10k '1' 45:0 T Fig. 2.5 Symmetrical circuit with unequal positive and negative sequence impedances: From (30) with Zl#=22 and all sequence mutual impedances zero, 3w gaze +9. (Zea) got: 2. (an) ZA@ 1 — £9“ = J 32er 12) 2J0 ‘ 10$ 2 29¢, ; £6920 40 When the positive and negative sequence self impedances of a circuit are unequal and there are no sequence mutual impedances, theo( and @self impedances are average of the positive and negative sequence impedences. There is no mu- tual coupling with the 0 network; but the 0(and anetworks are coupled through non-reciprocal mutual impedences. Be- cause of this non-reciprocal coupling between theo< and 6 networks in rotating machines in which 217*— 22, the o( , 6, 0 components are not convenient for determoning fundamental frequency currents and voltages in systems in Which the positive and negative sequence impedences cannot be assumed equal. However, if there is but one machine or group machines in which Zlqfi 22, (29) can be rewritten to give a reciprocal mutual coupling between the 0(network and a modified (3 net- work, from which an equivalent circuit can be obtained. Modified 5 network: Substituting Z4, = 2’“: 29,0: 30$ 2 Q from (21.3) in (29), the 0(,(@ , and 0 components of voltage drop in the circuit in the direction of current flow are 19d : Kdzobl + 192.1% : Id 32- (z\+ zz)+ AIQE KEV-11') “9e- Lfiw ~ 191% - 43W: (11-32)th (few W) «96: Lab Rewriting vxand V6 in (ES) in terms of (~19 ), with.—Zg@ replacing Zed , vac Liane Hex-249) Ia 1.” {-IQBELM (W is ( Z14- 19.04 (Irwin reg... Iiia+ (.19) gig) cg x9) (- swim) —~ 64-392 M I) U 4| 1901: like: “-*‘ ngitdg : 30‘ @4010“ ig) + (34 Pig) Lg W ~ 1g C“ 1900* 3%(‘31-(n) = Li idQ’ +1g(-’ZQQ) (w) . ‘ = iii-”(4% ~iu® +CXa+T®1dQ In (is) and (LY), the mutual impedances between thecm network and a modified @ network are reciprocal. In (k6), v? has been retained, but 19 has been replaced by (-IQ ) flowing in the direction assumed as positive for I . In (h7), Ig,Ihas‘boen retained but vQ has been replaced by -v9 measured in the direction of VQ . Equivalent circuits for synchronous machine with 217E 22 fqo ‘DO‘tC/‘AM 5W3 ‘0? Ok’YWL—wa1k T fiat Vac é [Zu+£z+3(¥.-i‘=2) 1 L ———+ & _3%(Z|-ZI 4 a -i [2, + '1'er (1:46)] Id-T(‘1Q) \a -19 T Q: 9) ——+ 1‘15 fife Tb I 2% $30in be is seam“ Fig. 2.6 (a). Current in the@ network is -IQ . Fig. 2.6 (a) satisfies the equation forCX and gicomponents of voltage drop in the direction of current flow given by 42 Euro @o’cmfi'oel ewe 5hr 0k Tatum. Ea -&[}\*zz‘$ (iFLJJ %[Z‘+z,+3(lrifij ' ———v I? 19—4 —= T@ 3K“ {—Vg) J. éfi0h56%tevdkm1 bvn (or P 7mkumfifi. Fig. 2.6 (b) In Fig. 2.6 (b) and equation (14.7), voltages in the? network are negative @ voltages, Giving a modified (3net- work in which currents are @ currents. The generated voltages in the modified 6 network of Fig. 2.6 (b) becomes -(-an)= an as indicated. The points TA and T@ are the terminals of the synchronous machine to which the o( and @ networks, re- spectively, for the system exclusive of synchronous machine are to be connected, after all impedences in the @ network have been multiplied by -1. If the impedances in the @ net- work include resistances, negative resistances will be pre- sent in the network, capacitive reactances will become induc- tive reactances and vice versa. The modification of (5’ net- work presents no difficulties in_a analytic solution. Zero network is the same as in Fig. 2.5. Connections of 0( , é , 0 networks to represent an unsymmetrical circuit and a short circuit; (no in- tervening A - Y transformer bank) Direct connections of the 0(, 5, and 0 networks to 43 satisfy the fault conditions of Table I (Appendix) are shown in Fig. 2.7 'Eqw<fifiewbeac3ue Inm€fifimm£ed<3ub gsefihxedheanua . .35“ .3 5 P Q F 9 Q Case I(a)-Three phase fault Ea Fig. 2.7 The fault is at F, the unsymmetrical circuit between P and Q. Mutual impedances between the component networks because of the unsymmetrical circuit are not indicated, but they may be present. For simplicity, two synchronous machines only are shown, but the system, exclusive of the unsymmetrical cir- cuit, may be any symmetrical three phase system.with equal positive and negative-sequence impecanees. Ea and Ea! are the generated voltages in phase a of the two machines. Case I(b) is similar to I(a) except that F in the 0 network is shorted to the zero potential bus for the 0 network. Case II(b) and II(c) involve modified @ network. The fault is (a) Line to ground - a and ground (b) Line to ground - b and ground (c) Line to ground - c and ground i’ 44 ‘ I 2440 eotcmhal Qwo {or 0(th ‘ fl 1‘93th (BM «Soy dW Ea E0: F P G. vjazzjo in.» Vowel law. go. o M F P a Ru Zqo SeLb- WPQAMQS 2V0 ‘ dUw'xéaé. ‘33 1 Au 05¢“, mpamM p Q The-Misha 531 F Val) 21. I ,i’m Goiw-HaLQ—wa ‘SEGY (3M '3“ $5.3? ‘GJOitnmfig'aJ‘ (Bum fa», QMUIYK F P a 3“ (5 $1.00 WRMM mayuea b3 ‘5- an (3* %M& VOUtcL Mapua 53 r; "swarms-m a. {3 warm; Case II(a) Case II(b) Fig. 2.7 Other cases are not shown here because of limited Space. Open Conductors: ‘ I r a , I 'F... I: | :F Ib TF l lb 6 iF I l , I | l J}. i 41 =0 i ° f ——+——‘ . I I n I ' ' I (a) (b) Fig. 2.8 Fig. 2.8 (a) and (b) show one and two open conductors, respectively. Let v and I' with different subscripts repre- sent voltage drOp between F' and F respectively. Here v is a series voltage drOp and I' used to indicate voltages to 4E? ground at a fault and currents flowing from the phases into a fault. Table II (appendix) gives equations expressing re- lations between v“ , v@ and vb , the components of voltage drop between F' and F, and between 14 4, I 1 and Io ', the @ components of line current flowing from F' to F. Conductor a open: Conductor b open: Connection of the 04 , Q and 0 networks to represent open conductors: Fig. 2.8 (a) and (b) represent cases I.(a) and II(a) in Table II. it“ pa‘iw’n-‘ol So.- dx~h¢kwoiK gum gawk“); ”(Wok “dunk fiat E." 2| 2:, m “—4 HI: -L.’ E—ZVQ-fl" .95. ._m_.. 120 lfo/ Zero ‘poiwfiol {CY 0-h1—‘Zwafik 2a.“) Po§wm by" O hckaTK '1‘er Pfiw’dal for 6 "*me , Zero fbtemfiol 166* 6 “WK 4% as; as :fio Z, in I i\ F’ F 2| I 5% fie T 4|" . Yb “’1? r’ F ”19:0 (0) (d) Fig. 2.8 44.. Fig. 2.8(c) and (d) show direct connections of theci and 0 networks to satisfy Fig. 2.8(a) and (b) respectively. For simplicity two generators only are indicated. The sys- tem could be any system with equal positive and negative se- quence impedances. Direct connection of theo) (50) Va, = V4 + 1901 Eliminating Vb, IO, IO', v0 and WE from simultaneous equations of (23) and (2h), “L and Mg’ interms of Id and Igfare obtained and may be written Vd~I= 16‘ ( So +13Do+3£§)+ 1&1K2C0* Eng.) 5\ Vok a 1d Q°2v3'he+3£) -\- (Isl ‘11,)(“310 ( -) F (16. + 3b.) (~30) F L W Vat] fl LA :‘L’ VA (25) .. i-(Se+3bo+i'” . a“. The equivalent Y shown in figure 2.9(d) satisfies equation (35). It can be used to re lace the fault and open conductor in thé aknetwork. For a fault through 0 impedences, 2f: 0. If this equivalent circuit is substituded in the d.network, between F' and F, the currents and voltages throughout the system can be determined. The<§currents and voltages are unaffected by the fault and open conductor. In the 0 network: from 49. g—a . I___ C Isl BO f 19- Id) 1" 2x4 (58') I Va: ~V¢l*&(§i{) Knowing the oLcomponents, with these relations and the 0 network, the 0 components of current and voltages throughout the system can be determined. The phase voltages and currents at any point are obtained by substituting the x, Q: 0 components in (l) - (3) and (7) - (9), respectively. The effect ofaw(transformer bank: CI ——_————_ I I I I I I I Fig. 2 H8 ,r In The choice of the reference circuit is arbitrary. With the circuit C as reference, components of voltage and current at D can be related with C. Components at D are indicated by primed symbols. 30l=1<°us Vor- Va" 19’4“ 1' “t ’ ““d (55) Line reactance: As do 9, O impedences are not available in reference books, symmetrical impedences are calculated for stranded copper conductor on gmd. basis, and they are found out from later's relation with former. Chapter III Solution by Clarke Components Problem: _ o, 000 KW“ were, 2 0 Y A A 'Y 50,000 KVA a} 37, 500 KVA \B'KV fl 40M “0 Kw " g ‘ \5‘6KV Ii 30W saw 43 . Id: \00% "we: X4; 1.5% . A Y X=X:1°% Y A X=X:zs.. \ 1 6° ZsoooKVA [SIOOOK‘O’A \ a. . / Kat /o 'XSsc/o 3:8/0 X.‘ 5/. QWW 5:0” ‘1. Cigflunxszr No. 2. gvacimn 6k CwA-vdcd‘fs I H0 11“: 15—441.“? NWLAMY 5,000 KVA \ unfavt> x,=5#b= X1 3 66 Kw: L4L46ng—mL9 5112 5k CJMAMfi’fidY ‘. 4/0 Qt?" ® SJOOOKVA Xv: 3%:XL ?(o: l'S‘x Motm’f to Yufix ikL— w/Cs Fig. 301 50 5\ Basis assumptions: 1. The fault currents are to be calculated using transient reactances. . 2. A base of 50,000 kva for calculation. 3. All resistances can be neglected. h. 100% synchronous impedence is used as reference. 5. The mutual impedence of the transmission line is neglected. ] l 6. Saturation of transformer is neglected. Terminology: Xlerositive sequence reactance X2=-Negative sequence reactance. Xb==Zero sequence reactance. Eal’e’3 voltage to neutral of phase a at ‘1 generator or motor 1,2,3. Generator equivalent circuit: G1.=x,_ 10/ '1 zau = 2@@ ° 10% X0:- 670 2:0 3 X0 2 ‘70 jun Wok/M bu; {m @otwiiak \ws Ea (i- meiwo‘rK _5g;{ $-Y\ekbooYK 30.0 T 13“?” a T ow an Wee/yaw; to) Q9 1’ 0— nziwom -——sa§w- Fig; ./ 3.2 (9 G2 : 501000 : ___. x X, 2537 $00 333/ 1. X t: “50/000 : ° 5 37.500 6'€‘% 2J4» :- ¥G<§ 3 7“ 3 7‘2, 3o 3 x6 Ki: 75' x $0,000 1'00]; 37,500 6'2, 2m awai (Ema 2‘60 GotMQM gal d‘ MdWOVK "35.2 B ._. thIK 13m. i-r 333/, 53.3% T 9) £360 @owa—Q. 03% Q3) O~ 73de m ‘T e-ee% ‘ 50,0 , F1 . 3.3 x‘ ~ GMS’OOOOO : 60%; X1 g M Y = l‘5 2&3? . $804M 63» 5,000 : ‘5/° Euro Gown-R49» (EM E: 0L?” 1K -ng 6 ~‘h0kwo‘6K 40% T . “r at... rammed, “/0 O~TtekuaoYK J T .afr‘ \5% Transmission lines: Reactance of llOkv line For 4/ocopper or conductor xov— ‘4‘5‘1 ohms Yum. M we: {we bur met-N: Maw Wat-r X4?“ “SW = 53 ('310 4: '320+ '404) = "548 ohms yum “Ah. Xv: x. = 14“ m = mam—r '3°7= -%45 0W“ ‘5’“ ‘AG $2= 'SALLAO xso’cm 7‘ ‘470 -.-. 1°“: {Q9 “0 xuo )uO X0: 53- (Zam zbtfc) = {a E In: Kb: is] :: Xad-XQ- ~23- [Xi bw 54,—1 Xak [WT M," x‘k V138,] ‘ '4Q'l 1 1.89- ax.343 2 l-S okrns MM 2.6 x QOXS‘OAJOO A = m%, XO‘ “0 )l \‘0 x \O 5‘2, 2°69 Gotw (Bub 20(3) GoLe/A‘QLOJ. QM EOE- d~ deoflK "SEQ? B _ ”$07K as. i, 333/0 53.3% T W9 in“ @oievxfilafl. 633% Q?) 0— ‘hejwmfk (554% A T ‘ 50,0 , pi, - 3.3 x“ 6 5,00000:€OA’ X2. 3 M Ac 0 . Yo : [5 5&0—39 : 5. olefin-1 63% 5,000 ‘ /° Zara 6)on (bug E2 0“?” 1K -sfig e ~“kao1K 60% ”V . '1' 5Com - (5)0in (3w. 50/: O~ “nehpoYK 1—_nsm____..r \5% Transmi s 3 ion line 8 : Reactance of llOkv line For 4/ocopper or conductor 5 new (>ka You widt— 450(4 bar Mush» Mg” ugh X41941 18%) = 33 ('310 + '320 -\ '404.) = "548 ohms pumi- Xa H M. KW K2 = 1(a+ 7Q: ¢Ml+'307~. 4545 ohms $4“- ><\= x2: 154:1“ ”“999 = \41o= 53a: fee “0 xuo M0 X0: .33 (1a.; {bi-{CD : za [cf-{(12 219: 35c] :- Xa+xe~ 23 [Xd' M1421 Xdk ‘5“, ‘4'& X4 Swag/J ‘— '4°H ‘t a-S‘i- 2x343 2 9mg ohms WM 2.6 . X0. x 40% 50,000 # = \Hfl HO¥ \\O X\O ‘.‘5s N 53 Reactance of 66 kv line For 11/0 copper conductor x“: '46” ()ka hum. Xa: % K“ \W \otyt-t MW 3052‘,» Xd. be; 205%) ~ EQ‘RH +~27q»t—3g4) : 307 ohms amm- Xx= X2: Xodem -4‘i‘l-t-307:-304 ohms ramm- x‘z X1 . ~§04x4oxsaooo 68x 66.x \0 ’4 39'970 de=zQQ : Xu-“Xz: 39‘97" 10" Yq+x - —2- e 22(de ¥W\OS'*-rio\‘6mlob%*x¢§"7~°b%) ‘Ac‘h R-S‘i-r-QX.301 2 3.65 ohm am Wow, -‘ 550. Xo~ sz-e'ax4oxso,ooo . q fl : 2° ééxéé X\O 4 /° Transformersé on "new base \. 25,000 KVA . X-- \07, X‘ ”10% 2" 15, 000 KVA x_ \670 X? 870 50:0 0 3. 20, 000 KVA X : q-fiiggoaz 22-5/o )(2 4°]. 4 ' ‘9)000 KVA X : SO'OOO ._.; ° a 8 264/. 5. 5 000 KVA §~ . x: 5 £2.22; 507 ‘ 5/° 5,000 ‘ $2a=7zb=lz~ci 2‘1": £W22°°:z 54 Equivalent circuits of the system: ok-network Zero-potential bus 510% |4% 22.5% 30° 33.3% /° F": ' |€>°/o 27.7% ’5? ”WK. 26-77. ‘0‘ Zero-potential bus 1. E2 1; . E« 1-59 m ~ i 3.33, 3-437 5.35 a Lil-\O (9 Sample calculation: '.5c '. 31 -243 . z = 3 X5 4 ~— - -—"——'"" = . a 356 *‘5.3£«3.43‘( 3\-'35‘] ‘8 ‘8‘ A (C) a”? all EA. . EJ- 1‘1 3.\3‘ F’F 30l4q $035 - m Fig. 3.5 Q9 5‘5 Zero potential bus \ . . # E“ % 3°38 F I F 5048 £01“ Fivl we Fig. 305 O-network X010 @‘kMM (BUD 10% 111% 11-5 ‘7. (:70 6' CLO/o rm ‘ M [£70 3\-$% (0.57. 297% 507. t5], «1) Sample calculation: r? w_ 5541”,!“ 3:53 {5.47 . id“ W : ' s-Boq ‘—\. = 145+ Zia-Pica 3 ‘-5+S.3‘H:]-4" ‘ZamD 6%kewjbkkl GBMD $39 H5 3'3 4; Agrmmr\___ . ‘. l j.30U--\-J906f 3,36 j.239+—j.0665j.30 j.0735+ 3.65: j.72 Fig. 3.6 ‘Z%0 thevdL&L dowb Ezfio themiiak.63uo UL-notwork RCQ‘FBDO :- .3 51-88 “Do =-d)72 Fig. 3.8 SQ: ‘37 Substituting Y—network of (Kin original d network. 2.59/60 630+WM QM ‘ Ea: ii}; 33-25 - 3.71 ‘ 5.48 W . ml) 3m. 1. “.13 4* E0. E3: Fig. 3 09 Considering figure 3.8 base voltage 13.8kv de" Vd=d4z Va L4 (Sm-31>.) + 1,2(1(°+3Do) 7. VOL Iok (So-+396) + (L‘- IJ) (‘D") H 01‘ \~. 3\.23 Id - 351-881; 449. .- 3 \.23 I.“ "t 3.71(I‘-L() So Miriam ' Considering figure 3.9 (b) 33.21. —3-u i '48 - ejvvmv IVNHWN——————1 4—__ , 4—- 33" SGT “-2 'JLS “23 1 ji-ZZ‘ _ 33 KG 15.? Fig. 3.10 O-network Em 630th GM 303$ 3'3 ——Jumm» awnL—- ififfiél+ie "’ ho-Ig) Fig. 3.11 L, : Align) = 3.4; 16': “Ink = —'3\.12 (Xe-149’) = 3-6\-_';\.n —- 3 \-83 5 network is not affected by the fault and does not contribute anything in the calculation and solution. Hence it is not included here. So far the effect of A—Y transformer bank is not considered. All the relations developed are based on Y#Y bank. Proceeding left of the fault F-' and F. P (10¢ '—‘ 41:00 ‘, (EGOD ‘= <14)( 10 (MM “’t a“? ‘58 5‘) At G1 Y A IQQCJ:O “IACDJ : b Rugs—3.31 EQCD>= {3-81 Io‘rdec-lo’ : O‘Shaa _ _X4 (3 ’ ‘ ~ 1b- 3- + 319*13 :~ '8GCH«‘<‘>7—5 L22 .— —:§\.°\T:‘: 1¢= ~ 5:; - [3165+S; 2 -3c¢,x3.gq-gl.aa 2 «3.441 At G2 V/ [5 I (C :0 (3' 3 -1.kcb)=o ( 2 \ ' 1d c) J S 19(3):3\-5. 1&2 “*io - A C\ Ib= — 3:; .. {Elem = «cur-slew = w» 1e = - £25 - E: 19 *3. : .. -eLex;\\-S+3.e\ = ~5-S°\ At M Y A 1<5(C)-'O “S-cLCD)‘; C) iaCc)= 3.(: Ken): :34. 1a: 3.4+ S: 3 ‘8‘.83 l _._I. (3 - - ’ ~b~ €+1£QJ*1:: °$Qtfiyé+5h63= ‘11-‘55. LC: . l.‘ - ‘3 “%X@*lz = -'€C€Yj-L +:\\~€3 : 3-3\ In the transmission line (1) ~j.87—-j1.22 =-j2.2 (2) 31.5 4: j.6l (3) 3.6* j1.83 32.11 32oh3 ll 60 . - .'s -.\ - 46'! c ..3. 9 c. —’ f b @ ® ‘* “L b ~3 m3 “-9 r a. ‘— —. —'- 4— b 4—— -S\.12 -sa.2 32.. 5.06. 3C —r Eb“ ——-9 £8“ f I» C b€£ ® \€~[ —v >——? Fig. 3.12 Phase currents. 6| Chapter IV Merits of Clarke Components System. In power systems in which the positive and negative sequence impedences of rotating machines can be assumed equal, use of Clarke Components entails far less work in analytic solutions than the use of Symmetrical Components. They provide simpler equivalent circuit because of the mutual impedences between component networks resulting from unsymmetrical static circuits, if not zero, are receprocal. This is not the case with Symmetrical Components. When symmetrical components are to be used, the self and mutual impedences in the sequence network can often be derived from Clarke Components self and mutual impedences more simply than direct determination. Clarke Components provide a point of view which is helpful in vis- ualizing a problem even if it will eventually be solved by Symmetrical Components. In an unsymmetrical circuit, where the impedences of two phases are equal, or two phases are symmetrical with reSpeet to the third phase, Clarke Components give almost an immediate solution to many simple problems which require appreciable time by other methods. Discussion on Symmetrical Somponents System and Clarke Components System: It has been shown that there are twelve equations in the case of symmetrical components connecting the twelve 62, unknown components of current and voltage of phase a at the two fault points-~three for each fault point and two for each of the three sequence networks. Eight unknown negative and Zero sequence components are eliminated, leaving four equations in terms of the four positive sequence components. Two of these four equations will be in terms of the negative and zero-sequence system impedences. The other two equations involve positive sequence quantities only. The twelve un- knowns mentioned before are to be solved by ten available simultaneous equations and as a result, for other two left, an equivalent circuit has to be drawn. Val :' KIa1+ mIAl VA1= nlel + 11A1 constants k, m, n, and l are then evaluated. On the other hand, in Clarke Components, we need nine equations as @-network is unaffected by the fault in question and these equations are straight forward. The e- volution of constants in Yecircuit to represent the fault is quite sinple. Besides, the component networks are tied by simple relations. From these facts and properties dis- cussed in Chapter 2, it can be concluded that Clarke Com- ponents are distinctively advantageous because they are less time consuming. Appendix Table I Short Circuit on Three Phase System Relations between (X , 5 , and 0 components of voltages and currents at the fault. CaSe ije 0‘: 0350565 inVOlved WOA'LCM {m eats 0% Cowpcnnonb: 0;, [Qault CeerOT‘ c V“ . \IoLtovan- ad: Faulk. wt,“ u'th:u1k\?w 1e) WWW «Hove VoL=o Ale-‘0 1020 Q3) a,\o, c and arm Va=o ,‘ V0.20, voeo “B (a) Lv'm‘w fiwomé ax cued» catch/ea Va -.~. - v0 19 =0; 3.13210 w Lone. 153th b GMJ~ %7mo\ VoL=BV@£-1Vo Id:‘%-9/Id~-30 at) \e‘w to 757““ C we ‘me v, = JEN? +1“ r.‘ - Io. - fi- 9' 1“: ~16 - ' ' A V 2 m (a) LW‘B qu. baud“: (5 :9 Idrol 30:0 ' him ,.___ 0”) ”fauna... 0* Ma d- 5 1““530 L=° \Auvu. ' OK M C. V ‘ ‘19 (Q) 01.. 3 xdzfifc’ 3°~° Q (a) 30‘” WVTM b’c’mfi’YMCK V9.70; V¢:1VO 142-10 Q9) we um. row; 095. M WM“ Vol 2 31%; Vet ”W R = -519” 23‘, ‘ , )Ouu 7M :_V‘ KC) ‘lum Wh‘vmfif QC aLfi \h % ', Var-Va azfifp *Zlo (.4 Appendix Table II One or Two Open Conductors in a Three-Phase Circuit. Relations between at, Q , 0 components of series voltages v across openings and line current I' flowing through Openings. WOW 63 Volkafizo Cow M 5t W Wvaes . Wm ~ - across ope/M25 W “PM. 1% Q ”(2.20; 1901:2130 x&1;_16, [0,) \o bxengé 3-1-1: tall-519511; =0 KCC) C. 1901.“ %Q_ ““130 KAI—afil@f—Zl“’:° 1(a) b owl c 19,, = .130 I)": 252;; IQ, 2 o 1 l 1 I 1 EU») (km-Mb 190L«fil9(b-7—13=O &:T§:‘X° II(c) mwc use-Event” I,’:.%:_:,’ 3 Appendix Table III 2115 Function of Operator a(= e3) (1:. \L\_20°= -o.s+30,8c,g 03‘: \L7:46= *0-9—30.8€Q (X3 2 \L3_6061\-O+$O Ct 2 0‘: ‘1‘}.5= ~O-S-t 50.8gg ‘0“ ”3—0. :95 *lotee —(f 2 \LQO 2 O'S+J©.%££ \+(l~\- Q'Lzo “+0.1 : --\ 0“a? = O+$\-132=5[_30° 03-0\~ O-lfl-‘HZ =55}; \—0\ = L’s—Seats: szfo' \—-OQ‘ 2 \-5+30.3L5=I§L§P Om 2 —\.S+3o.sa=fiL‘§° 03-4 = -\.s~ _. soseerfifon‘ 65 Appendix Symmetrical Component impedences in: e {a ~ its ~s mew 112 i “5; [204.2 -\ flee-\- S (h@"}@"‘)] £0..- :00 .. ,L. . 20—- zi‘t‘weeucaedafl 22\ ~ Jiissalak“ 2%.? ‘3 (fd%+;é%‘0] 75m : 17'- (hb + ‘3 its) 201: Kzoi ”.55EGQ33 £20 3 2 (£410 “:3 z~@°) £01.: (£Dul ‘i' 3 £0?) Clarke Component impedences: £40K 2 ii (In *- £\’L*' 3E2\ * £22) ‘L "' in. “£‘) 52(ch = LKZu‘t 121 ZOO ‘3 100 dee = 331 (id—111. *171‘7‘5'1) 20‘: “.3 xi k1" ritl'*¥ll “£19 ZOOK‘ i£10[+¥01) 240 : (*‘O+’£‘103 2°C” ‘3 31(1°1~ E02,) 29° ‘ ".5 (ito’ $1.) 6‘. l. 3. 5. 7. 9. 91 Bibliography Circuit Analysis of A-C Powcr Systems Volume I and Volume II by Edith Clarke Analysis of Unbalanced Three-phase System by L. G. Strokvis Electric World, volume 65, May 1, 1915 pp 1111-1115 Methods of Symmetrical Co-ordinates Applied to the Solution of Polyphase Networks b y C. L. Fortescue, A.I.E.E. Trans. volume 37, part II, 1918, pp 1027-llu0 Short Circuit Currents on Grounded Neutral System by W. W. Lewis Gen. Elec. Rev. June, 1917 pp. SZE-SZ? Circuit Breaker Recovery Voltage by R. E. Park and W. F. Skeats A. I. E. E. Trans, vol. 50 , March, 931 p. 20h Over Voltage Caused by Unbalanced Short Circuit by Edith Clarke, C. N. Weygandt and C. Concor- dia, A. I. E. E. Trans. vol. 58. 1939 Two-Phase Co-ordinates of a Three-Phase Circuit by E. W. Kimbark, A. I. E. E. Trans. vol 58,1939 Discussion: Analysis of Synchronous Machine Short Circuit (by R. D. Camburn and E. T. B. Gross) by E. w. Kimbark, A. I. E. E. TRAHS. vol. 69, Part II, p. 678 The Principle of Synchronous machines by W. A. Lewis, Chapter 10 and 11 10. Electrical Transmission and Distribution Reference Book - Westinghouse £8 lllllfllli 1 3