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Transient analysis of six-phase synchronous machines Mozaffari, Said 1993

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TRANSIENT ANALYSISOF SIX-PHASE SYNCHRONOUS MACHINESbySAID MOZAFFARIB.Sc. (Civil E), United States International University, 1987B.A.Sc. (EE), University of British Columbia, 1990A MILS'S SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR 11:it, DEGREE OFMASTER OF APPLIED SCIENCEinTilt FACULTY OF GRADUATE STUDIESDEPARTMENT OF ELECTRICAL ENGINEERINGWe accept this thesis as conformingto the required standardTHE UNIVERSITY OF BRITISH COLUMBIAMarch 1993©Said Mozaffari, 1993In presenting this thesis in partial fulfilment of the requirements for an advanceddegree at the University of British Columbia, I agree that the Library shall make itfreely available for reference and study. I further agree that permission for extensivecopying of this thesis for scholarly purposes may be granted by the head of mydepartment or by his or her representatives. It is understood that copying orpublication of this thesis for financial gain shall not be allowed without my writtenpermission.(Signature) Department of  E6C-17 L•CcIt The University of British ColumbiaVancouver, CanadaDate  1-fai"ch e3, ict1.5DE-6 (2/88)ABSTRACTSix-phase synchronous machines have been used in the past for high power generation, andpresently are being used in high power electric drives and in ac and/or dc power supplies. Thisthesis presents a mathematical model for investigating the transient performance of a six-phasesynchronous machine, and develops a model for the simulation of such machines with the EMTP.The effect of the mutual leakage coupling between the two sets of three-phase stator windings areincluded in the development of a d-q equivalent circuit. The six-phase machine is constructedfrom a three-phase machine by splitting the stator windings into two equal three-phase sets. Thesix-phase machine reactances and resistances are derived by relating their values to the knownthree-phase machine values. The transient performance of the machine is tested by solving themachine equations for a case of a six-phase short-circuit at the terminals of the machine. TheEMTP model presented is based on representing the six-phase machine as two three-phasemachines in parallel. The EMTP model verification is achieved by comparing the EMTP resultsagainst an independent computer program, for the case of a six-phase short-circuit test at theterminals of the machine.iiTABLE OF CONTENTSAbstract^ iiList of TablesList of Figures^ viAcknowledgement viiDedication^ viii1. INTRODUCTION 11.1 History of Six-Phase Synchronous Machines 11.2 Objective of the Thesis 22. EQUIVALENT CIRCUIT OF A SIX-PHASE 4SYNCHRONOUS MACHINE2.1 Introduction 42.2 Differential Equations 52.3 Stator Mutual Leakage Inductances 92.4 Flux Equations 92.5 Transformation of the Machine Equations 112.6 Equivalent Circuit of the System 143. SIX-PHASE MACHINE PARAMETERS 173.1 Introduction 173.2 Synchronous Self-Reactance of a Three-Phase Machine 173.3 Stator Leakage Reactance of a Three-Phase Machine 17iii3.4^Relating Six-Phase Machine Parameters to a Three-Phase Machine^193.5^Stator Leakage Reactance of a Six-Phase Machine^213.6^Machine Parameters^ 234. TRANSIENT ANALYSIS^ 274.1^Introduction^ 274.2^Symmetrical Short-Circuit of a Unloaded Six-Phase Synchronous Machine 274.3^Solution of the Differential Equation^ 295. EMTP MODEL FOR THE SIX-PHASE 35SYNCHRONOUS MACHINE5.1^Introduction^ 355.2^Model of the Machine^ 355.3^Method of Current Injection 395.4^Numerical Instability Caused by Time Delays^ 425.5^Testing the EMTP Model^ 446.^CONCLUSIONS^ 476.1^Results in Thesis 476.2^Suggestion for Future Work^ 48REFERENCES^ 50APPENDIX A 52APPENDIX B^ 57APPENDIX C 59ivList of TablesTable 3.1^Three-Phase Synchronous Machine Parameters^ 24Table 3.2^Six-Phase Synchronous Machine Parameters^ 26Table 3.3^Leakage Reactances of the Six-Phase Synchronous Machine ^26vList of FiguresFigure 1.1^Basic Circuit of High-Power Drive with Six-Phase Synchronous Motor ^2Figure 2.1^Cross Section of Synchronous Machine^ 4Figure 2.2^Equivalent Circuit of the System, without the mutual leakage ^16coupling between ql,d2 and q2,d1Figure 3.1^Winding Distribution for a 5/6 Pitch Machine^ 20Figure 4.1^Computed Short Circuit Current in Direct and Quadrature Axises^31Figure 4.2^Short-Circuit Current in Windings a,b,c^ 32Figure 4.3^Short-Circuit Current in windings x,y,z 33Figure 4.4^Field Current After a Short-Circuit^ 34Figure 5.1 The EMTP Model Diagram^ 38Figure 5.2^Transformation to Phase Quantities 40Figure 5.3^Equivalent z-Transform Network^ 42Figure 5.4^Short-Circuit Armature Current in phase a and x ^ 45Figure 5.5^Steady-State Short-Circuit Current in Phase a 45Figure 5.6^Numerical Oscillations^ 46viACKNOWLEDGEMENTI would like to thank my wife Suzanne Belanger for her moral support and constantencouragement. I send my regards to my parents for their love, nurturing and support.I am indebted to my supervisor, Dr. H.W. Dommel, for his constant guidance and hisindispensable role in the successful completion of this thesis. I express my gratitude to Dr. M.Wvong and Dr. L.M. Wedepohl , for their careful examination of this thesis.I would also like to thank my good friends and fellow students in the Power Group of theElectrical Engineering Department at U.B.C, for their dedication to providing a friendly andcooperative environment for research. In particular, I like to thank A. Araujo, and Dr. N.Santiago for providing me with helpful comments and suggestions.viiDEDICATIONI would like to dedicate this work to my father Mohammad Hossein Mozaffari.viiiChapter 1INTRODUCTION1.1 History of Six-Phase Synchronous MachinesIn the late 1920's, the drive toward building larger generating units was hampered bylimitations in circuit breaker interrupting capacity and by the large size of reactors needed to limitthe fault currents. To overcome this problem, six-phase synchronous machines were built withtwo sets of three-phase windings [1], in which each set of three-phase stator windings wasconnected to its own station bus. The voltages of each set were equal and in phase. Suchgenerators, rated up to 175 MVA, were in use until the fault current problem was solved byconnecting each generator to its own step-up transformer, with switching accomplished on thehigh voltage side where the currents are lower.During the last few years, six-phase synchronous motors have been used in high powerelectric drives. In these inverter-fed synchronous motors, each of the three-phase stator windingsets is connected to its own inverter. Figure 1.1 shows the basic circuit diagram of such a twelve-pulse inverter-fed synchronous motor with two three-phase windings. This circuit configuration isvery beneficial for the reduction of harmonic losses and torque pulsations if the two sets ofwindings are displaced by 30° [2]. On the supply side, which consists of two bridge rectifiers fedby a three-winding transformer with Yz\ connection, giving a line shift of 30°, the 5th and 7thharmonic are canceled out. The two machine-side inverters each feed a set of three-phasewindings of the machine. The two three-phase sets are displaced by 30° and cancel out the fieldcaused by the 5th and 7th harmonics of the stator winding currents.12Figure 1.1: Basic Circuit of High-Power Drive with Six-Phase Synchronous Motor.A second application of six-phase synchronous machines is for generators where one three-phase set feeds an ac load, while the other set supplies power through a three-phase rectifier to adc load. This type of AC-DC generator has lower cost and lighter weight than a comparablethree-phase generator transformer-rectifier configuration [3]. Currently, these types of generatorsare being used as power supplies on aircrafts and ships. If the rectifier is replaced with an inverterin this configuration, the six-phase machine acts as a DC to AC conversion system. The weight ofsuch a system is less than that of a conventional DC to AC motor-generator set of the samecapacity. For example, in a practical system with a 140-kVA 200-V rating on the ac side and 150-kW 1500-V rating on the DC side, the weight of the six-phase system is about 70% of aconventional motor-generator set [3]. This kind of dc-to-ac conversion system is being used inelectric railway trains as ac power sources for the air-conditioning equipment.1.2 Objective of the ThesisThe main objective of this thesis is to derive a mathematical model for investigating thetransient performance of a six-phase synchronous machine with 30° displacement between thetwo sets of three-phase windings. With the help of this mathematical model, an equivalent modelis developed for the simulation of six-phase synchronous machines with theEMTP(Electromagnetic Transients Program). In order to verify this EMTP model, a comparisonwill be made between the EMTP results and results from an independent computer program, forthe case of a six-phase short-circuit test at the terminals of the machine.3In Chapter 2, the differential equations of a six-phase synchronous machine are first derivedin phase variables. Park's transformation [4] will then be used to transform the machine equationsto the d-q-O rotor reference frame. This model includes the stator mutual leakage reactances [5],which have been ignored in earlier works [6,7].Chapter 3 describes the calculation of the six-phase self and mutual reactances andresistances, which are needed for the simulation of transient phenomena. The six-phase machineparameters will be related to the better known three-phase machine parameters, based on turnratios and winding factors. The transient performance of such a machine in case of a six-phaseshort-circuit at the terminals is presented in Chapter 4.In Chapter 5, an EMTP model for six-phase synchronous machines is presented. In order toverify the validity of this model, the EMTP simulation results for a six-phase short circuit test arecompared with the results obtained in the previous chapter. In Chapter 6, general conclusions aredrawn, and recommendations are made for future research.rotor q-axisx-axisa-axisChapter 2EQUIVALENT CIRCUIT OF A SIX-PHASE SYNCHRONOUS MACHINE2.1 IntroductionIn this chapter, the differential equations for a six-phase synchronous machine and itsequivalent circuit model are derived. The synchronous machine is assumed to have six identicalwindings in the stator, in which they are divided into two symmetrical three-phase sets, called a-b-c and x-y-z. The three-phase system x-y-z is displaced with respect to the system a-b-c by 30°.The three rotor windings are the field winding f, that produces flux in the direct axis, and twoequivalent damper windings D and Q in the d- and q-axes. These nine windings are magneticallycoupled, and the magnetic coupling between the windings is a function of the rotor position.Figure 2.1 shows a cross section of the synchronous machine with its six stator and three rotorwindings.# z-axis^rotor d-axisFigure 2.1: Cross Section of Synchronous Machine452.2 Differential EquationsThe voltage-current-flux relationship of the 9-coil machine in Figure 2.1 is described by thefollowing equations:[v] = [R][ddt [x];[2] [L][d;f ) ID ,[v] = [va,vb,vc,vx,vy ,vz ,v f ,0,0 ;[11]={2a52b)11c,i1x,2y3/1"z,2 f,2D,2Q]T ;and[R] = diag[ra ,ra ,ra ,ra ,ra ,ra ,rf ,rD ,rQ ].The elements of the matrix [L] describing the linkage between the fluxes [X] and the currents [i]depend on the rotor position 13:[L] = [Lu (0 ^ (2.7)With all the coils having the same number of turns, or with rotor quantities referred to the statorside, the elements Lu of the matrix in equation 2.7 are given bellow [5] (the lower case "represents the leakage inductances).- Stator self inductances:▪ +L1Laa = ^11 – 1,2 cos2,8;^ (2.8.1)7"1"Lbb =^2• L1 – L2 cos2(fi– —31; (2.8.2)Lac = s + 11 -4 cos2(f3+ 1.2 );^ (2.8.3)3L,, = e + Li – 1,2 cos2G6-- 301; (2.8.4)s(2.1)(2.2)(2.3)(2.4)(2.5)(2.6)Lyy =L 5 +L1 –LZ cos2l/3-30° – 371;^(2.8.5)6Lzz = e + L, — L2 cos2(fi— 30°where(2.8.6)+ —231;.e s is the leakage inductance of a stator winding;L, is the constant component of the magnetizing inductance of a stator winding; andL2 is the amplitude of the second harmonic component of the magnetizing inductance.- Stator mutual inductances:1—^1 — L2Lab^—2.L2Lba; (2.9.1)2flcos^—31Lac = — — Li — L22 L.; (2.9.2)71-)=COS(213 + —237rLax =^+L, cos—6 —^cos(^7r^2r32,^—6 — —); (2.9.3)Lay = t ay + L,cos(Lr3 -1— L2 cos 7r2,3- 7 27r1 (2.9.4)— —3 ;La, =^az + Li cos —n- — —2 7r^27r— L2 COS(2)6— —+—);6^3 (2.9.5)7r)6^31Lb'^—2 LI — 1,2 cos2fi; (2.9.6)Lbx^bx^L, cos( 7r^27r — L COS 2flLr6 (2.9.7)—6^327r)  3Lby = "e by +^f=cos^—^COS(2g— 671-6 (2.9.8)23 71)-± 'Lb:^bz + Li cos( 7I^2 71" — L2 cos 7"C2/3--6 ; (2.9.9)6— + —327rL, =^+ L,cos(-7C + —)— L2 cos 7C210— 2g (2.9.10)36^3 6727z-Lc, = +^cos(— — — L2 cos 2)3-6); (2.9.11)6^3L = +L, cos 1,27- (2.9.12)cos(2fl —^_ 2 31 ;whereax , f ay , f a, are the mutual leakage inductances between phase a and phases x, y and z;bx , 'e by)"e bZ are the mutual leakage inductances between phase b and phases x, y and z; andare the mutual leakage inductances between phase c and phases x, y and z.- Rotor self inductances:Lff = t f + (Li + L2 ) ;^ (2.10.1)LDD =^+(LI + L2 ); (2.10.2)L„ = t + (L1 — L2 );^ (2.10.3)wheref is the leakage inductance of the field winding referred to the stator;.e r, is the leakage inductance of the d-axis damper winding referred to the stator; andQ is the leakage inductance of the q-axis damper winding referred to the stator.Rotor mutual inductances:Ln, = Lpf -7, L, + L2 ;^ (2.11.1)LiQ = LQt = LDQ = LQD = 0.0; and^ (2.11.2)- Stator-to-rotor mutual inductances:Laf = (L1 + L2 )sinf3;^ (2.12.1)Lbf = (Li + L2 ) sin(fl— 31;^ (2.12.2)Laf =( L, + L2 ) sin (fl+ —2371;^ (2.12.3)Lxf = (L1 + 4 ) sin (fi— , r);Lyf (L, + L2 ) sin (fl— 2,3r — Y-);=(L+ L2 )sin(fi—:- + 2-31;Lap = (L1 + 4 )sinLbD = ( L1 + 1,2) sin( — 23LCD = (L, + L2 ) sin(fl+LxD = (Li + L2 ) sin(13 r);LyD (L1 + L2 ) sin(fl-LZD = (Li + L2 ) sin (fi— 6+ y);LaQ = (Li — ) cos/3;2 7rLbQ = (L1 — L2 ) co?— —3--);LcQ^1=(L — 1,2 )cos(fi+ —237r);L„Q, =(11 — 1,2 )cos^—6 );2 71"L yo = (L1 — L2 ) cos(/3 — ^—3 );(2. 12.4)(2.12.5)(2.12.6)(2.12.7)(2.12.8)(2.12.9)(2.12.10)(2.12.11)(2.12.12)(2.12.13)(2.12.14)(2.12.15)(2.12.16)(2.12.17)97T 2 71"LZQ = (11 — ) COS^613– — + — .3 (2.12.18)2.3 Stator Mutual Leakage InductancesThe mutual leakage fluxes between the two sets of stator windings a-b-c and x-y-z do notcross the machine air gap and couple the three phases of the two sets of stators. The inductancematrix related to these fluxes between the two sets of three-phase windings, can be written as:t ax ,t ay f a,e 12] – ebx f by f bz^ (2.13)cy^czThe flux-current relationship equations for a two-layer stator winding with 30° difference inwinding for a six-phase machine can be formulated as:?t,ta = atfx^+ G aziz ;^ (2.14.1)X = L bxlx + byi y + e bzi z; and (2.14.2)sec – L cxix^cyiy + cziz^ (2.14.3)The two sets of stator windings a-b-c and x-y-z are uniformly distributed around the stator andtherefore, the leakage inductance matrix rp 12] is cyclic. This can be formulated as:L -= C by = cz;f ay = f bz^a,; andt az = fbx = cy2.4 Flux EquationsThe flux equations of the six-phase synchronous machine, written in matrix form, areshown in equation 2.15:[Aabc = [41 ][Iabc^[112 + £12^[Lri ][Ifiv ] ;^(2.15.1)[2xyz = [Ls2^+[L, 2 +^[lab,]+ [1, ,][IIDQ];^(2.15.2)where10[2.1,Q]=[L,J[la,]+[L,,2]`[I,,z]+[L,][1],d; (2.15.3)[Lsi ] =Lac,LbaLabLbbLacLbc (2.16.1)_L„ Lth L„[42]_L.Lam.L,„Lam,L,,L2,,LayLyzL„Laz(2.16.2)[ 1,12 ]= Lbx Lby Lbz (2.16.3)L,LafLayLL„L aQ[41 ] = Lbf LbD LbQ (2.16.4)L LCD LcQL L LxQ[42 ] , L yfL zfLyDL zDL yQLzQ; (2.16.5)Li. L JD Liu[1,„]= L Df LDD LDQ ; and (2.16.6)L Qf LQD LQQ(2.16.7)[ T] =^[T,^ 0^0^0 ^[T ]^0^x„(2.17) with112.5 Transformation of the Machine EquationsThe inductance elements in equations 2.9 through 2.13 are all functions of the rotor positionand this makes the solution of the equations difficult. However, these equations can be simplifiedby applying Park's coordinate transformation. This transformation projects the rotating fluxesonto the field axis, where they appear as stationary during steady-state operation. Thistransformation is identical for fluxes, voltages, and currents, and converts phase quantities a-b-cand x-y-z into dl, ql, 01 and d2, q2, 02 quantities. This transformation does not change thequantities on the field structure. The transformation matrix [T] can be written as:[I] = identiy matrixcosI3 cos p - —1■4„ 2^3 ^cost/(3 23sing sin(13 - 2-2 c ) sin(f3+ Lc )31^1 3^1; and^(2.18)cos`  —1 cos 13- —7c — —27c) COS 13 ± —71 + —27c )6^6 3^6 3sin(13- a6 ) sin(13 -6 —  —27c) sin(I3 + 1-c +3^6 31^1 1(2.19)12in which[T]1 = [T]r^ (2.20)Applying the transformation matrix to equations 2.1 (voltage equations) will result in:[Vdoi] = lidgod [cor]{2ood+ ydt [2401];^(2.21.1)[V dqO2] [Ra][' dg02] -1- [ a r][2 qd02] + —d ^dq02 ;^(2.21.2)[17iD2 ] , [Rr ][iiD2 + Tit [2.11V ];^(2.21.3)where0 o) 0[co r [Ti]ddt [T]ddt [ 7,2-1 –0) 0 0 ;0 0 0(2.21.4)co = —d fi;^ (2.21.5)dt[Ra ] =ra00r a0 -0 ; and0 0 rarf 0 0[Rd = 0 rD 00 0 Q(2.21.6)(2.21.7)Applying the transformation matrix to equations 2.15 (flux equations) will result in:[2 dqOI] – [ 71 Lsl][T] 1 [i dq01] [ TI][ L12^121-172] 1 [I iq02] + [1il1][ Lrl][i !EV ];^(2.22.1)[2 clq02] = [ T2][Ln][ T2]-1 [i cio2]±[T2][ 112 e 12 f [ nidq01] + [ T2 ][42 ][iim2 ]; (2.22.2)and[2„2 ] = [Lri ]T [T1 ]-1 [idol ]+ [L, ]T [7;][idg02 ] + [Lrr ][ipd.^(2.22.3)The products involving inductance matrices in the above equations are calculated as:13[TaLsa ] =+ -23 (LI - L2 )^0^00^ts+2(LI -1-L2 ) 00 0^S(2.23.1)(L — L2)^0^0[ 7142 I T1 = (2.23.2)-Nh + 1,2) 0o^o][ T2 ] 1 = (2.23.3) whereAil = ar cos(16-1 )-F ay cos(: + 37---t )+ t cos( 2 ;A:^315Al2 =t ax sin(:) +f ay sin( +^-1e az sin( 716 ^)(Li — L2 )^0^0o^+ 1.2) 13-  (LI ± )0 0^0[Ti][Lri =(2.23.4)(2.23.5)[12 ][i-s2^r = [TILA TI ;14[T2][LI2f[Ti] 1 = [711,1217'2r ;Elatun Td-1 ={[TileidT2Y[}T;[la42]=[1141];[L,]T [Ti ]-1 .{[Td[4jy;and[42]T[T2r ={[T2][42]} T(2.23.6)(2.23.7)(2.23.8)(2.23.9)(2.23.10)2.6 Equivalent Circuit of the SystemThe voltage equations of the six-phase synchronous machine, once transformed into d, qand 0 quantities, will be as follows:+ 3 (L, -^ - t „,)(io +io )-(-kt + t )-t i„+ ^,• (L, Lz )(ig +i)]+2^2 -^2 'dt^-V2 2ArJ, 1,2^-^2 "^"d[(^-3 (Li + L2))ia,+—2 kt - t ,)(ia,+id2 )+(kt + t^t^+ L2 )(„+i,,+if )]I(2.24.1)V =ri +o)[(t,+-3 (L+L,))id,+ —2 ktdt^2^Lmi,,+2(t..—e,)(i,, • )^+^-^+0](2.24.2)(t ,+-(Li - L2 ))i,2 +—2 kt^t " )(i +i,2)+(-kt + t )-^+ Aff (L, L2 )(ig2 +i2 )11-2 "/^(t ,+-23 (L,+ L2)id2+ 2 V - - t i(ia,+04-kt erc\•+1 )- t^,+ ^ (LI +1,2 )(id2 +iD +if )]2^q^Vj^1 I (2.24.3)-,1—^-.I—3P=ri"+6..)(t + 3 (L + D))ia2 +^( 21- --1°, \)(id' +id2)-(1- ( 1" +1- ) --t-V°' + V3( L' +1,1)( id' +i° +if +2^2 ^2^)-,1— --da^-3dt (e' +s (L - L))i" +^( 2t" ---e- ),(P' + l'1 ) - (1- (^2t" -ft')\ --t- \e+V3 ( L' - L2 )( io + P ) +[^2^2^1 ^ (2.24.4)vdl = ria, — [(Vd2^raid2dt15V0, , rai01 +dt[. sio,+(t^t ay – t JiceiV02 = rai02 +dt[t 102 +(t ax + t ay – tVf =rf i +—d [—(L +1.2)(idif dtd [ NdVD = " + L2 )(id,dt 1/2 d [VQ =rQ iQ +—dt 15^–L2 )(iql +ig2 ±/Q )+e Q/Q(2.24.5)(2.24.6)(2.24.7)(2.24.8)(2.24.9)1+id2 -ElD +1f -0 - F .e fif1-Fid2 -Fif -FiD ke DiDThe equivalent circuits of the six-phase synchronous machine corresponding to the equations 2.24are shown in figure 2.2.16Figure 2.2: Equivalent Circuit of the System, without the MutualLeakage Coupling between ql,d2 and q2,d1Chapter 3SIX-PHASE MACHINE PARAMETERS3.1 IntroductionIn this chapter, the six-phase synchronous machine reactances and resistances will bederived, assuming that the machine is constructed by splitting each 60° phase belt of a three-phase machine into two equal 30° phase belts, one for each three-phase winding set [8]. The six-phase machine parameters will be related to those of the three-phase machine, based on the turnsratios and winding factors.3.2 Synchronous Self-Reactances of a Three-Phase MachineFor a three-phase salient pole synchronous machine with N3 equivalent stator turns perphase, the reactances in the rotor reference frame can be written as [9]:Xd 3 = co-3 N2P : and3 ^3 dXq3 = w NP  2 3 q5wherePd is the d-axis equivalent permeance dependent on machine dimensions;P is the q-axis equivalent permeance dependent on machine dimensions;co is the base frequency; andN3 is the turns per coil of a three-phase machine.The subscript 3 is used to denote three-phase machine values.(3.1.1)(3.1.2)3.3 Stator Leakage Reactances of a Three-Phase MachineThe stator leakage reactances will be derived assuming that the stator winding has twolayers with sixty degree phase belts, and that the winding phase belt has either unity or 5/6 pitch.In both cases, the leakage reactances can be divided into two distinct components:1718(a) slot dependent leakage reactance;(b) all non-slot related leakage reactance including zigzag-leakage reactance, the belt-leakage reactance, and the coil end leakage reactance.From the above definitions, the leakage reactances of a three-phase machine can be written as:Xes. = Xe + XPsiot^ (3.2)whereXeslot is the slot dependent leakage reactance; andXe is the sum of all non-slot related leakage reactances.- Slot leakage reactancesThe slot leakage reactances are made up of three parts: the self reactance of the coil sides in thetop; the mutual reactances between top and bottom layer coil sides in the stator slots and the self-reactances of the coil sides in the bottom.X toot = Xer ± KS XeTB + X eB^ (3.3)whereX iT is the slot leakage reactance due to coil sides in the top;Xe.„ is the slot leakage reactance due to coil sides in the bottom;XeTB is the slot leakage reactance between top and bottom layer coil; andK is the coupling factor that is dependent on the winding pitch.The slot leakage reactance components in equation 3.3 are each defined as:A T 2 T PTX a = COIY 3 L." 3XpB CON3 2 L S3XtTB cDN32 L TB3(3.4.1)(3.4.2)(3.4.3)where19PT ,P, are the permeances in top half and bottom half of the slot;Pm is the permeance of mutual coupling between coils in the top and bottom half,L is the slot length;N3 is the turns per coil of a three-phase machine; andS3 is the number of stator slots per phase of the three-phase machine.For a three-phase winding, the effective value of the coupling factor Ks will vary linearly withpitch, so that for pitch values p between 2/3 and 1 [9]Ks =3p —1.^ (3.5)Substituting equation 3.5 into 3.3 results in the following slot leakage reactances:for a unity pitch ^ Xeslot = X€7, + 2 X trB ± X tB ; and^ (3.6.1)for a 5/6 pitch  ^Xesiot = XP7' +1 ' 5X/TB + X . (3.6.2)- Non-slot leakage reactancesThe non-slot leakage reactances are approximately proportional to the square of the equivalentstator turns and base frequency,XP = coKi N32 ,^ (3.7)whereKe is the proportionality factor dependent upon the end winding and slot dimensions and thenumber of turns per pole.3.4 Relating Six-Phase Machine Parameters to a Three-Phase MachineA six-phase machine can be constructed by splitting the 60 degrees phase belt of a three-phase machine into two equal portions each spanning 30 degrees. This technique is illustrated infigure 3.1 for a machine having six slots per pole.(3.8.1)(3.8.2)(3.8.3)N =6^2 K pd,K _ Kp6Kd6 pd Kp3Kd3S6S3220a-c-c-c-cbb-a-a-a-a -b-b-b(a)ax-c-c-z-zbb-a(b)6 -bMc turns0 -b tic turnsaabbax-b^-Y-vFigure 3.1: Winding Distributions for a 5/6 Pitch Machine a) Three-Phase Machineb) Six-Phase Machine With Split Phase BeltsIf the turns and winding factors of the two machines are the same, then the following relationshipsbetween the six- and the three-phase machine can be written as follows:whereN6 , N3 are the equivalent turns per phase of the six-phase and the original three-phase windings ,respectively;Kp6 ,Kp3 are the pitch factors of the six-phase and three-phase machine, respectively;Kd6 , Kd3 are the winding distribution factors of the six-phase and three-phase machine,respectively; andS6 is number of stator slots per phase of the six-phase machine.The magnetizing reactances Xq and Xd of a six-phase machine are defined as:X = co-3 N 2 P andq6 2 6 ql (3.9.1)212 Pw— N6 2^6 d • (3.9.2)From equations 3.8 and the impedance relationships of equations 3.1 and 3.9 , the followingequations relating six-phase parameters to their three-phase counterparts can be derived:Xq6 = X0-14 (Kpd )2 ;X d6 = Xd3 741 (Kpd )2 ;(3.10.1)(3.10.2)X tQ6 = X to 4-1- (K pd ) 2 ; (3.10.3)/^2XiD6 = XiD3 4 VC d—1)P•• (3.10.4)X if 6 = X tf 3 —14 (Kpd )2 ; (3.10.5)1^N2r^=r^ 1CpdQ6^Q3 4 ( (3.10.6)1^\ 2rD6 = rD3^Pd )   and (3.10.7)r^=r^-1-(K^)2f 6^f 3 4^pd^' (3.10.8)The stator resistances of a six-phase machine are directly proportional to the number of statorturns, as in the case of a three-phase machine. Therefore, for an even stator split, the statorresistance will be1ra6 2 r a3 • (3.11)3.5 Stator Leakage Reactance of a Six-Phase MachineAs in the case of a three-phase machine the stator leakage reactances of a six-phase machinecan be broken into two parts:22Xis = Xt6 + X  whereXtriot 6 is the slot dependent leakage reactance of a six-phase machine; andX16 is the sum of all non-slot related leakage reactances of a six-phase machine.The slot leakage reactance can be broken into three components so thatXts1ot6 = XiT6 Ks6X17736 + X06'(3.12)(3.13)For the same number of turns per pole of the three-phase machine, each component of equation3.13 can be defined as:XiT6 = ON 2 /,6 s6PXt B 6 = CON62 s62 ^X ^coN L TB€TB 6 — 6^ckl 6(3.14.1)(3.14.2)(3.14.3)whereKs6 is the coupling factor that is dependent on the winding pitch.As with the three-phase case, the coupling factors Ks6 vary linearly with the pitch factor pKs6 12p-10 for 5 5p<_ 1.6 (3.15)From equations 3.13 and 3.15, the slot leakage reactance for a unity pitch can be written as:X tslot6 XIT6 +2 XCTB6 + X CB6^ (3.16)and for a 5/6 pitch factorXtslot 6 X176 + XCB6'^ (3.17)The mutual leakage coupling between the two sets of windings is only caused by statorwinding from each three phase set occupying the same stator slot. Therefore, all mutual leakagereactances can be related to XiTB6 by coupling factors:23X Pax Kx XPTB6;^ (3.18.1)Xtay KyX€TB6; and (3.18.2)X taz = Kz Xtni.6 .^ (3.18.3)For a 5/6 pitch winding, the coupling factors can be determined by inspection of figure 3.1 asfollows:Kx =1;^KY = —1; and Kz = 0,^for 5P = 6 (3.19)From equations 3.4 , 3.7 , 3.8 and 3.14 , the following relationships between the six-phase andthree-phase leakage reactances can be derived:xt 6 = 3 Kpd ;1(^)2X PT6 + Xle136 = 2 (Xtr3 + XtB3 ); andXiTB 6 = 12 A 1TB 3 •(3.20.1)(3.20.2)(3.20.3)3.6 Machine ParametersThe six-phase synchronous machine parameters will be derived from a three-phase, four pole,salient pole, 125 kVA, 480 V synchronous machine whose parameters are given in table 3.1. Fora machine of this type, the slot component of the leakage reactance is 35% of the total leakagereactance [9] and thereforeXisolt 3 35= ^00 0.147) = 0.05145 f2 and1 (3.22.1)65^\X€3 = 00 0.147) = 0.0956 a1 (3.22.2)Xd = 4.0735 Q X ii3 = 0.1842 C2Xq =1.9612 Q Xt./. = 0.168 nXd' = 0.3041 52 rs= 0.0332 nx; = 0.236 Q rD = 0' 00826 QXq = 0.3557 fl rf = 0.00558 flXis = 0.147 fl rQ = 0. 00872 nX€Q, = 0.2354 f2Table 3.1: Three-Phase Machine ParametersFrom equation 3.6.2, for a 5/6 pitch, the slot leakage reactance was calculated as:Xis/0f 3 = Xin + 1.5 XiTB 3 + X a33 .24(3.23)Assuming a straight stator slot and neglecting tooth top fringing, the slot top and bottom mutualleakage reactances are related by:X €TB 3 = 0.3(X€T3 + X 03 ) .^ (3.24)By substituting equations 3.22 and 3.24 into equation 3.23, the following values of leakagereactances can be determined:X in + X tin -= 0.0355 f2 andX erB 3 = 0.0106 Q.The pitch and distribution factors for the sixty-degree phase belt are given as [9] :Kp3 = sin —PIT and2sin(30°)Kd = ^3^TC/6(3.25.1)(3.25.2)(3.26.1)(3.26.2)For a 5/6 pitch, the above factors are :Kp3 = 0.966 and^ (3.27.1)K d3 = 0.955 .^(3.27.2)25From all the above parameters for a three-phase machine, their six-phase counterparts can nowbe calculated.- Six-phase synchronous machine parametersThe value of the pitch factor Kp is the same as in the case of a three-phase machine. However, thevalue of the distribution factor will depend on the number of phase belts per pole and the numberof slots per phase belt:Kd6sin( 2qn sin(  n"2nqj(3.28)whereq is the number of phase belts per pole; andn is the number of slots per phase belt.For a 5/6 pitch the distribution factor is:Kd6 = 0. 989 .^ (3.29)Substituting the values of Kp3 ,Kd3 ,Kp6 and Kd6 into equation 3.8.2 will result inKpd = 1.0356.^ (3.30)Substituting the three-phase impedance values along with equation 3.30 into equation 3.10 willresult in the six-phase machine parameters that are shown in table 3.2. From equations 3.18, 3.19and 3.20, the leakage reactances and the mutual leakage coupling are calculated and their valuesare shown in table 3.3.X d6 = 1.0921 f2 X iD6 = 0.0493 flX q6 = 0.5258 f2 X4/6 = 0.0437 f2X d6. = 0.0815 f2 rs6^.= 0 0166 QX d6 = 0.0633 f2 rf6 = 0. 00149 Q= 0.0953 0 rD = 0. 0022 nXtQ6 = 0.0631 Q rQ^'= 0 0023 0Table 3.2: Six-Phase Machine ParametersXt6 = 0.0256 f2 Xta, = 0.0 f2X €TB 6^0.0053 f2 Xtskt6 = 0.01775 f2X tay = —0.0053 fl Xts6 = 0.0433 QXtax = 0.0053 f2 X€7. 6 ± X tB6 = 0.01775 QTable 3.3: Leakage Reactances of the Six-Phase Machine26Chapter 4TRANSIENT ANALYSIS4.1 IntroductionThe transient performance of synchronous machines is as important as its steady-statebehaviour. This is because the synchronous machine is part of the power system whose overallbehaviour and stability is greatly influenced by the dynamic characteristics of individual devices,and by their dynamic interaction. The sudden short-circuit test, in which all terminals of anunloaded synchronous machine are short-circuited simultaneously, is a well-established method ofchecking the transient characteristics of the machine. In this chapter, the differential equations ofthe six-phase synchronous machine, developed in chapter 2, will be solved for the case of a six-phase short-circuit, under the condition of constant field voltage and speed.4.2 Symmetrical Short - Circuit of an Unloaded Six-Phase Synchronous MachineThe differential equations for a synchronous machine under condition of constant speed canbe written in the following matrix format in the d-q reference frame:[v] = [R]p]^[-Lid^[L2 ][i] •^ (4.1)The elements of the matrix equation 4.1 are then defined as:[11]=[Vai ,Vgi ,Vd2 ,Vq2 ,V f ,VD ,Vd ;^ (4.2.1)[i] [idl iql id2 ,iq2 if iD 3 1dr .3^ (4.2.2)[R] = diag[r,„ra ,r„,r,„rf ,rp ,re ]; (4.2.3)270280^ IJI (t..-ti+(L, -4)^0^L1-L,€.+;(4 -L2)^ 02^2-t,+2(L,+/.2)^0 +1,2)^ -L -L,-4-L,^0(1 ° ,+1,,,,)-1,,2 --‘5 (e -C )- 2(L +L )2^°,^'^2 '^20000 0 01(1.,,-1,,,,)+;(4- L2 ) 1, + ;(1., -4)4/^+L^).-/2^"^'^" -1,-1(4 +4)0 -L -1,.., -L,-L,0 0 0 0 00 0 0 0 00 0 0 0 0and[4] = L,-L,0000_(4.3)^1, +(L, +4)^00^L,^-L,)-/ )+ 2(L, +L,)^2 " " 2 '^2000;(4-4))+(L +L2 " " 2^)^24(t,,,-t,,,,)+ 2(L, -4),e,+2(L+1,2 )^0t, +(L,-L2 )00-2-(4 -4)L,+L2^L,+L,^00^ - L,4+1.2^L,+L,^00^0^4-41,,+(L+L2 )^L,+L,^0L,+L,^1,,+(L+L,)^0o^0^to+(L-L,)[L, ]=(4.4)The elements of the inductance matrices [L 1 ] and [L2 ] with constant coefficients have alreadybeen defined in chapter 2. However, for each three-phase set, the addition of currents in all threephases will equal to zero. Therefore by knowing that i„ +iy +i2 = 0 or zZ = —(ix +iy ), the statormutual leakage inductances matrix of equation 2.13 can be simplified as :P12 ] =ay — C0 —e azt ay —t az^00ay^aze^'e azTo solve equation 2.1 for the currents it must be rearranged as follow:[L2 at [i] = -[R][i] +0 [L. ][i]± [v],^ (4.5)=^[R][d ± CD [L2^+ [1'2 ] I M'^(4.6)29c",[i]= {—[4]- I[R]+co[4]1[L1]l[i]+[1,2]-1[v].^(4.7)Since all six phases are short circuited, the voltages are:Vdl — Vd2 — Vql — Vq2 — VD — VQ — 0 '^ (4.8)Substituting equation 4.8 into 4.7 results in the following expression for the short circuitcurrents:',[i],{-[1,2 ]-1[R]+.[Li] 1[L1 ]}[i]+ [4]1[vi ].^(4.9)The initial value of the field current is calculated with id iq = 0 asVq(0) N^ withf\ 32m (f) (Li + L2 )Vq(0) — Vt' —P rated (RMS) •The value of if (0) is used to initialize vf ,Vf 0 = Rfif (0) •4.3 Solution of the Differential EquationEquation 4.9 is a set of first order differential equations that has the following generalform:di (t)n =f Vidt^n^iN) n= 1, ^ N,^(4.13)with known initial conditions i,,(0), n=1, ^N ("initial value problem"). With the parametersfrom chapter 3, these differential equations become:—d idi = —118.6id,+1795.910 +66.21d2 +1415.91q2 +2.4if +3.2iD +1015.4/2 —1101.3, (4.14.1)dt —dtiql= —2110.81d , — 106.610 —1736.8/d2 +76.9iq2 —1251.4if —1251.4/D +3.5/2 + 0.0, (4.14.2)30dt- id2 = 66.2id, +1415.9i0 –118.6id2 +1795.9iq2 +2.41f +3.2iD +1015.4iQ –1101.30, (4.14.3)—d iq2 = –1736.81d, +76.9ig, –2110.81d2 –106.61q2 -1251.41f –1251.4iD +3.5iQ +0.0, (4.14.4)dt dif = 40. 3id, – 2471.510 + 40.31d2 – 2471.51g2 – 10. Oif + 3. 8iD – 1562. 8iQ + 4552.1,^(4.14.5)dt —d in = 35.81d, –2191.7i0 +35.8id2 –2191.71q2 +2.6if –13.5iD –1385.8iQ –1166.1, and(4.14.6)dtdt -- = 4825.2id, +37.3ig, +4825.2id2 +37.3/q2 +3138.8if +3138.8iD –11.0ig, +0.0^(4.14.7)Even though it is known that the zero sequence currents are zero for a six-phase short circuit , itsequations were included for the completeness, as:—d i• °1 = –8055.55i01 , and^ (4.14.8)dt—d in, = –8055.55102 .^ (4.14.9)To solve the ordinary differential equations 4.14, the fourth-order Runge-Kutta method is chosen[10]. This method advances the solution over an interval by combining the information fromseveral forward Euler-type steps, and by using this information to match a fourth-order Taylorseries expansion to it. In the fourth-order Runge-Kutta method, the derivative is evaluated fourtimes: once at the initial point, twice at trial midpoints, and once at a trial endpoint. From thesederivatives, the final function value is calculated. The FORTRAN program "solve 1" inAppendix A was written to solve equations 4.14 for the case of a six-phase short circuit at theterminals of the machine. Figure 4.1 shows the computed wave forms for the direct andquadrature axis currents. From the results of this run it can be concluded that:id1 = id2 ; and^ (4.15.1)1ql = q2 •^ (4.15.2)uadrature axis current (A) Scale= 10**(3)1.00_00 —1 BB ^8.08^8.18^8.13^. 6.13^8.48^0.58^0.68^0.78^0.88Time (s) Scale 10**(-1)^CD Mic•-$11-...,:ssz.(b)2.0031However, in the transformation from d,q,0- quantities to phase quantities, the transformationmatrices [Ti ] and [7'2 ] are different. This causes the two sets of currents for windings a,b,c andx,y,z to be different. The two FORTRAN programs "phasel" and "phase2" in Appendix Atransform the dl,q1-quantities into a,b,c and d2,q2-quantities into x,y and z. Figure 4.2 and 4.3show the curves for the stator currents after the six-phase short circuit at the terminals of themachine.0 BBDirect axis current (A)^Scale 10**(3)- .813—2.00—3.00—4.88—5 000.00^8.18^8.20^0.38^8.48^8.50^8.60^8.70^8.88^Time (s) (a)^Scale 10**(-1)Figure 4.1: Computed Short Circuit Current: a) direct axis; b)quadrature axis.32Figure 4.2: Short Circuit Current in Windings a,b,c: a) Phase a; b) Phase b; c) Phase c.Figure 4.3: Short Circuit Current in Windings x,y,z: a)Phase x; b)Phase y; c)Phase z.0.08 0.48^8.50^8.60^0.70^0.800.18^8.28^8.38Time (s) (a) Scale 10**(-1) (4 111 e elre An. nfa.Time (s) (b) Scale 10**(3) Mi•rell.•■•■.1sva.0.58^0.68 ^82e0^8.66Scale 10**(3)8.00 8.18^8.20Time (s)0.30 8.48(c) ED Mit D. 71-an. tflGt.Phase current Ix (A) Scale 10**(3).ii Y \If I ‘I i Ii 1 f■i1 1II\ 1. 1/\\ !\iI^\(,\\ I^\\i' \\i,/^\ \\-2 080.80^8.10^8.20^8.30^0.48^0.58^0.68Phase current Iz (A) Scale 10**(3)._ r/, \ .,•.,\ie/) \it I^\'''\7 \\ I^\ \ I\'I11 1 ‘ /I t t \ / t.\ /t f \ t \ / .11 I„..\^1V332 HO1.8008- 1.08-2.00-3 003.082.081.088 88-2.08Field current If (A)^Scale 10**(3)8 007.006.005.084.883.002.001.088.18^8.28^8.38Time (s) 8.58^8.68^0.78Scale 10**(- 1 )0 088.80 8.48 0.88it Micr■Than. "Ma.34The short-circuit armature currents in figures 4.2 and 4.3 consist of a exponentially decayingunidirectional current and a superimposed sinusoidal current with decaying amplitude. The dccomponents of the armature currents all decrease to zero exponentially with the same timeconstant T. The initial values of these components depend upon the point in the cycle at which theshort-circuit occurs. The ac component can be separated into, steady, transient, and subtransientcomponents. The transient component with a long time constant 7'; and the subtransientcomponent with a very short time constant Td". Figure 4.4 shows the field current, which like thearmature current consists of dc and ac components. The dc component can be divided intosteady, transient, and subtransient components. The transient and subtransient dc componentsdecrease with the time constant Td , and Td . The ac component of field current decay with timeconstant T, which is the same as the dc component of the armature currents.Figure 4.4: Field Current After a Short-CircuitChapter 5EMTP MODEL FOR THE SIX-PHASE SYNCHRONOUS MACHINE5.1 IntroductionThe Electromagnetic Transients Program is a general purpose computer program whoseprincipal application is the simulation of transient effects in electric power systems. It wasdeveloped by H.W. Dommel in the late 1960's at the Bonneville Power Administration [11]. TheU.B.C. MicroTran version of EMTP has a model for a three-phase synchronous machine, inwhich the electrical part of the machine is modelled as an equivalent two-pole machine with sevenwindings [12]. These windings are: three armature windings; one field winding which producesflux in the direct axis; one winding in the quadrature axis to represent fluxes produced by eddycurrents; one winding in the direct axis and one winding in the quadrature axis to representdamper bar effects. In this chapter, an EMTP model for a six-phase, salient pole synchronousmachine, that includes the stator mutual leakage inductances, will be developed. This is achievedby paralleling two identical three-phase synchronous machines in such a way that the magneticcoupling between the two machines is included. The magnetic axes of the two three-phase sets aredisplaced by an angle of 30° . The windings of each set are uniformly distributed and have axesdisplaced 120° apart. In the first part of this chapter, the equations and the physical aspects of themodel will be discussed. In the second part, the EMTP model will be tested by simulating a six-phase short circuit test at the terminals of the machine. The results of this simulation are thencompared against the results obtained from chapter 4.5.2 Model of the MachineThe first proposal for including a multiple coil synchronous machine in the EMTP wasbased on the fact that the magnetic coupling between the two three-phase sets can be35411d2q240236approximated with equivalent voltages sources, which the program user can specify as TACS(Transients Analysis of Control System) variables, connected in series with the machine terminals[13]. This model was verified for several short-circuit tests. However, its implementation wasnumerically unstable. L. Bompa [13] suggested another EMTP model for the six-phasesynchronous machine in 1988, in which the magnetic coupling between the two three-phase setswas accounted for by injecting currents from one machine into the other and vice versa. In thismodel, the extra leakage inductance coupling effects between the two three-phase machines wereignored. In addition, both magnetic axes of the two machines were not displaced with respect toeach other.To develop a complete model, it must first be shown how the magnetic coupling and extraleakage inductances can be included. To do so, the voltage equations of 2.27 are partitioned asfollow:- Vd1Vql r[ Alli [ Al2]-1L[A21] [A22]]VolVd2Vq2V- O2^+—d^+ 1 ( i + L2 ))^dt^2coLe s +-23 (L, +L2 ))-+ 3 /,^-L2))^0ra +-61 1.e s -E(Li -L2 ))^0dt r +±-1 t0^ 0 dtwhere] = [An ](5.1)(5.2)The matrix of equation 5.2 represents a model of a three-phase synchronous machine. The factthat the two matrixes [A„ ] and [A22 ] are equal supports the idea of representing the six-phase37machine as two three-phase machines in parallel. Since the stator winding position is fixed in theEMTP machine model, the two sets of stator windings of the two machine in parallel can not beset 30° apart. However, a second approach is to keep the two sets of stator windings in the sameposition and rotate the rotor position of one machine with respect to another by 30°. The off-diagonal matrices [Al2 ] and [41 ] represent the extra magnetic couplings and leakage inductancesthat must be taken into account in paralleling the two three phase machines. The matrixes [A, 2 ]and [41 ] are found to be:oP(P„+/,)+t,P-P(4+/,,)+-1(t.—Pil2 dt 2 4) + 1(1. - € .- )] + :7, [(t., + t w ) - ,-] 0+ 4 )+ 1(c _^÷ i ,)+ t.]^ 00^ 0^ d(5.3)and2 +e" )—e^dt 2^2—L )+—^-)^—4 3i(L,-4)+1(1.—i.)1+:71t [!(e.+e.)—e.]^0^-[41= co[pL,+4)+ \fj.2  (e e,d+:7/ p2 „,+t,)_„]^_co[1(,„,+t,,)+,d,z(,_4)+1(i.,_ed^0^0^ 0^ 1(t-+t.-2€.)^(5.4)Based on the theoretical analysis of chapter 3, the above matrices can be simplified for a case of5/6 pitch windings. Using equations 3.18 and 3.19, the mutual leakage inductances between thetwo machines can be calculated as:= --t ay and^ (5.5.1)^az = 0. (5.5.2)Substituting equation 5.5 into the matrix equations 5.3 and 5.4 will result in the followingsimplified off-diagonal matrix [A, 2 ] :d [(L +L)+ (t -cit 2^-^2^- 03[(L2 , + L2 )± ^2 --4-3-(L1- +14-. (t _d 3(L - L2 ) +^-e380100^ 0(5.6)The rewriting of the voltage equations 2.27 with the above simplified matrices and the rearrangingof them, will show a common flux (I)„, which is created by the addition of the two currents id, d2in the direct axis and igi , iq2 in the quadrature axis. This common flux 4m in the direct andquadrature axis are defined as:[ A,^= [A,,] =,41) md 2^+ L2)(i dl^d2) and11) =^L2)(mq iqlq2) •(5.7.1)(5.7.2)The flux ,4),n which is created by currents coming from both machines represents the magneticcoupling effect between the two three-phase sets. This relationship is important because itsuggests the representation of this magnetic coupling between the two machines by injecting tothe terminals of each machine the current coming from the other machine. This is the basis of theEMTP model shown in figure 5.1.Figure 5.1: The EMTP Model Diagram39According to equation 5.7 and diagram 5.1, the armature resistances and self leakage inductancesdo not participate in the magnetic coupling phenomena between the two machines. Therefore,they are placed outside the machine. At the same time, the armature resistance R a and thearmature leakage reactance X1 , must be set to zero in the machine input data (card six of theEMTP synchronous machine data deck). Along with the armature resistance and self leakageinductance, the effect of the mutual leakage inductances between the two machine was calculatedand connected in series to the outside of the two machines. Also, it is assumed that thesynchronous machine maintains its synchronous speed, and therefore the mechanical behaviour ofthe machine need not be modelled.5.3 Method of Current InjectionTo inject the current from one machine to the other, the FORTRAN subroutine "source" inAppendix B was written and interfaced with the EMTP. This subroutine uses as input the valuesof the currents at the terminals of the machines at each time step. These values are put into avector "X" of length "LX" by the EMTP. The subroutine "source" will then use the values invector X, along with constant parameters appended to vector X, to generate six user-definedcurrent sources to be injected back into the two machines at each time step. The constantparameters which modifies the values of currents in vector X are necessary because the twomachines are rotated with respect to each other by 30°. The EMTP solves the machine equationsin d,q,0 quantities and only transforms them into phase quantities to connect it to the outsidenetwork. This is shown graphically in Figure 5.2. Therefore, when injecting the current as anextra load on machine 2, it must be rotated by —30° and then injected as idq02 . For the samereason the current must be rotated by +30° and then injected as ido, .1-1lc_(5.8)cosf3^cos((3 -120°) cos((3+120°) -sin (3^sin ((3 -120°) sin 03 +120°) (5.9)1/Nri^1/-5 1/5iqli01where[T1] 1 =40dqolTransformation i abc0Rotate by 13Transformationdqo2 Rotate by 13+30° xyzFigure 5.2: Transformation to Phase QuantitiesTo calculate the constant parameters matrix [DJ, the transformation matrix [T]-' is usedalong with the modified transformation matrix [CI ], according to equations 5.11. Thetransformation matrix [7] as defined in the EMTP is:_ • -'cll.^IQRotating the transformation equation of 5.9 by another 3 0, as required here, will result in thenew transformation matrix-^,cosQ3 -30°) cos(j3 -120° -30°) cos(3 +120° -30°) -[C,] = 3 sin ‘ (3 - 30°) sin(13 - 120° -30°) sin03 +120° - 30°)1/5^1/5^1/5(5.10)Matrix [Cd can be written as the product of two matrices. One matrix is the original1transformation matrix [7; and the other is a constant matrix [R] :41[CI][R]= [TIC, ]; andThe constant matrix [D1 ] from equation 5.11 becomes:(5.11.3)COS( 6 + 2^6cos(-51 + 2^2cos(-71 +-12(5.12)In an analogous way, the constant matrix [D2 ] for machine 2 becomes:( 6^2+ —1^cos it + —1 cos 57c + —1cos —7t 2 2^6^2[D, 7C(cos —2 2^—6 2+ —1  cost n + —1 cos(— + —57c) 16^257t(cos 6 2^—2 2+ —1 cos 7C)2)+ 2 cos( —6  l+ 2n^1) [D,cos(—5n ) 21+ — cos it + —1 cos It6^ +—6 2^2 27C ) 1^Sic) 21 ^1cos(—2 2+ — cos 6(— + — cos( 71 ) +—6 2 (5.13)The modification of the currents from one machine before injection into the other machine canthen be formulated as follow:lamodified laibmodifiedicmodified _ixmodifiediymodified= [4 ]= [D2iblxand_izmodified _ _iz(5.14)(5.15)This modification is the essential part of subroutine source.5.4 Numerical Instability Caused by Time DelaysThe EMTP model of the six-phase synchronous machine developed in the previous sectioncan cause numerical oscillations due to the time delay introduced by the interaction between thesubroutine Source and the main program EMTP. The delay is caused when the subroutinecalculates the current sources for the time step at time t from the results of the preceding timestep at (t-At). Therefore, the supplemental source equations (subroutine Source) and the powersystem equations (EMTP) are not solved simultaneously, and a time delay of At is introduced. Toshow how this time delay may cause numerical instability, consider a simple example of a"constant resistance" simulation using a one time-step lagging current generator [14]. Theequivalent circuit of such a system and its z-transform equivalent network is shown in figure 5.3.The current source ik is represented as rz, where z represents the time delay introduced by theinteraction between the source routine and the EMTP.42= g AK-1Figure 5.3: Equivalent z-Transform NetworkThe z-transform current response of the circuit in figure 5.3 is:^VIZI ^C (z)I =^, =^Z, i- L^p(z )^1+ rz '^2Z, Z,(5.16)43The stability of the system defined by equation 5.16 can be determined from the location of theclosed-loop poles in the z-plane, or the roots of the characteristic equationP(Z)= ld- rZZ I^2 =0.Zi Z2 (5.17)For the system to be stable, the closed-loop poles or the roots of the characteristic equation mustlie within the unit circle in the z-plane. Using the trapezoidal rule of integration, Z1 and Z2 aredefined as:2L z- 1=   andAt z + 1 (5.18.1)Z2 = R .^ (5.18.2)By substituting equation 5.18 into 5.17, the roots of the characteristic equation are found to be: r ( R 2 LR) -2^2LRAt)^At riR- 2L) + 2 LR At) Atz = (5.19)2r' At +RAt 2,-( 2L +R)At r(-2L R)•AtBy letting At ->0.0, the root of P(z) was calculated as:R - r R + rZ a =   ^ = 1 and^ (5.20.1)2r^2rR-r R+r^R (5.20.2)Zb -^2r^2rFor cases where R>r, the roots of equation 5.17 will be outside the unit circle, which meansinstability with a force free component of growing amplitude. To overcome this problem, the timedelay between the solution of the current generator and the EMTP should be eliminated.Therefore the power system equations should be solved simultaneously with the other equations.Research on this topic is beyond the objective of this thesis. It is the research topic of a Ph.D.thesis project at the University of British Columbia which addresses the problem of interfacingcontrol system equations with the power system equations in the EMTP [15].445.5 Testing the EMTP ModelIn order to test the EMTP six-phase synchronous machine model developed in the previoussection , a six-phase short circuit at the terminals of the machine is considered. The results fromthis run are then compared against the solution obtained from solving the differential equationswith the fourth order Runge-Kutta method (RK4), as shown in chapter 4. The synchronousmachine model of the EMTP accepts as data input either the transient and subtransient reactancesand time constants, which are then converted to self and mutual impedances, or directly the selfand mutual impedances of the windings. In this thesis, input was in the form of self and mutualimpedances, to insure that the data is identical with that of the Runge-Kutta solution. Thecomplete EMTP input data listing for modelling the six-phase synchronous machine and the six-phase short circuit is shown in Appendix C. The numerical simulation with the EMTP has beencarried out from t=Os to t=0.12s, with a step size of 50i,ts. The generator was open circuitedbefore the fault was applied at time zero. Figure 5.4 shows the comparisons between the EMTPmodel and the corresponding solution of the differential equations with the Runge-Kutta method,for the armature current for the first few cycles, in phase "a" and "x". The EMTP model gives,practically, the same results than those obtained from chapter 4. The fault current takes a longtime to reach steady state because of the small value of the resistances. Figure 5.5 shows thecomparison of the steady-state short circuit current in phase "a", for the two solution methods.Figure 5.5: Steady-State Short-Circuit Current in Phase "a".45300020001000Current 0(A)- 1000- 2000- 300046The problem of numerical instability caused by the time delay between the EMTP and thesubroutine Source, as discussed in section 5.4 , was a main concern in testing the EMTP model.As shown in figure 5.6, the solution is numerically stable in the beginning, but numericaloscillations eventually start building up. It was hoped that the numerical oscillations could beavoided if the currents in subroutine "source" from the preceding time step were replaced bypredicted values, with linear extrapolation. However, numerical oscillations still appeared withthis approach.The final cure for the numerical oscillations was an "occasional" switch to the criticaldamping adjustment scheme [16], which replaces the trapezoidal rule of integration with step sizeAt by two integration steps of step size At/2 with the backward Euler method. This switch wasdone after every 100 time steps, and produces the stable solutions shown in figures 5.4 and 5.5.O^0.02^0.04^0.06^0.08^0.1^0.12Time (s)Figure 5.6: Numerical OscillationsChapter 6CONCLUSIONS6.1 Results in ThesisSix-phase synchronous machines are increasingly being used as part of 12-pulse converter-fed high-power drives. This kind of arrangement, used in power plant auxiliary systems, is verybeneficial for the reduction of harmonic losses and torque pulsations. The aim of this thesis was toderive a mathematical model for investigating the transient performance of a six-phasesynchronous machine and to develop a model for the simulation of such machines with the EMTP.The first part of this thesis was devoted to the derivation of the equivalent circuit of a six-phase synchronous machine with 30° displacements between the two sets of three-phasewindings. This equivalent circuit includes the effects of stator winding mutual leakage couplingbetween the two three-phase winding sets. Since the machine inductances are all functions of therotor position, Park's coordinate transformation was used to project the rotating fluxes onto thefield axis, where they appear as stationary during steady-state operation. The six-phase machinereactances and resistances were then derived by relating their values to known three-phasemachine values. These parameters were derived assuming the six-phase machine was constructedfrom a three-phase machine by splitting the stator windings into two equal three-phase sets whileretaining the same number of turns per pole.The transient performance of the six-phase synchronous machine was simulated by solvingthe differential equations of the machine, for the case of a six-phase short-circuit at the terminalsof the machine. To solve the ordinary differential equations, the fourth-order Runge-Kuttamethod was used. The obtained curves for both, the short-circuit armature currents and the field,current were consistent with the behaviour of short-circuit currents for three-phase machines[17] .4748The EMTP model for a six-phase synchronous machine presented in this thesis was basedon the idea of representing the six-phase machine as two three-phase machines in parallel. Themagnetic coupling effect between the two three-phase sets, which was created by currents comingfrom both machines, was represented by injecting to the terminals of each machine the currentcoming from the other machine. This method of current injection was achieved by adding aFORTRAN subroutine inside the EMTP source code. This subroutine uses the values of thecurrents at the terminals of the machines, and after a phase rotation, generates six user-definedcurrent sources to be injected back into the two machines at each time step. The test validation ofthe proposed EMTP model was carried out by simulating a six-phase sudden short circuit at theterminals of the machine. The results from this simulation were then compared against thesolution obtained from solving the differential equations with a fourth-order Runge-Kutta method.The EMTP model gave, practically, the same results as those obtained from solving thedifferential equations with the Runge-Kutta method. The results of this comparison confirm theaccuracy of the proposed EMTP model. The problem of numerical oscillations was solved byusing the critical damping adjustment scheme.6.2 Suggestions for Future WorkIn a paper presented at the European EMTP User Group Meeting in October 1992 [18], H.Knudsen describes an ongoing Ph.D. project, involving the Danish utility company ELKRAFTand the Technical University of Denmark, on modelling of six-phase synchronous machines with 2three-phase windings spatially displaced by 30° . In this work the six-phase machine equationswere solved using the "MODELS" input option for control system simulations. With thisapproach the results appear in the network modelled by the EMTP in the next time step, ascontrolled current sources. Knudsen felt that CDA is needed in the synchronous machineequations of MODELS to suppress numerical oscillations at time of discontinuities (switchclosing and opening). However, the program version ATP which he uses does not have CDA49implementation. There is also the problem of introducing a time delay into the interaction betweenthe synchronous machine equations in MODELS and the network equations in the EMTP, whichmay cause numerical instabilities.In order to overcome the above limitations, the six-phase synchronous machine equationsshould be solved simultaneously with the network. This is possible in UBC's EMTP versionMicroTran with subroutine CONNEC [12], which represents the network as a Theveninequivalent circuit, seen from the machine terminals, to the machine equations inside thesubroutine. MicroTran also uses the CDA scheme for the suppression of numerical oscillations,which is not available in the EMTP version ATP. Implementation of the six-phase machineequations with the trapezoidal rule of integration, and with the backward Euler method for theCDA scheme, is a major effort, which was beyond the scope of this thesis.50REFERENCES[1]. Barton T.F., "The Double Winding Generator," General Electric Rev., June 1929, pp.302-308.[2]. Liyu, C., L. Fahai, Y. Bingshou and G. Jingde, "A Mathematical Model for Converter-FedSynchronous Machines With Dual Three-Phase Windings," International Conference onElectrical Machines, Zurich, Switzerland, August 29, 1991, pp. 347-351.[3]. Kataoka, Teruo and E.H. Watanabe, "Steady-State Characteristic of a Current-SourceInverter/Double-Wound Synchronous Machine System for Ac Power Supply," IEEETransactions on Industry Applications, Vol. 16, No. 2, 1980, pp. 262-270.[4]. Park, R.H., "Two-Reaction Theory of Synchronous Machines: Generalized Method ofAnalysis-Part 1," Transactions MEE, Vol. 29, No. 33, 1929, pp. 716-730.[5]. Schiferl, R.F., "Six Phase Synchronous Machine With AC and DC Stator Connection,"IEEE Transactions on Power Apparatus and Systems, Vol. PAS-102, No.8, August 1983,pp. 2685-2700.[6]. Bunzel, E. and G. Muller, "General Analysis of a 6-Phase Synchronous Machine,"International Conference on Electrical Machines, Zurich, Switzerland, August 29, 1991,pp. 333-340.[7]. Fuchs, E.F. and L.T. Rosenberg, "Analysis of Alternator With Two Displaced StatorWindings," IEEE Transactions on Power Apparatus and System, Vol. 93, No. 6, 1974,pp. 1776-1786.[8]. Moustafa, E., J.P. Chassande and M. Poloujadoff, "A d-q Modelling of A Six-PhaseArmature Synchronous Machine," Paper B6.3, 16th University Power EngineeringConference, Sheffield, England, April 1981.[9]. Alger, P.L., Induction Machines, 2nd ed., Gordon and Breach Publishers, New York,1970.[10]. Press, W.H., et. al., Numerical Recipes, Cambridge University Press, New York, 1986.[11]. Dommel, H.W., "Digital Computer Solution of Electromagnetic Transients in Single andMultiphase Networks," IEEE Trans., Vol. PAS-88, April 1969, pp. 388-399.51[12]. Dommel, H.W., Electromagnetic Transients Program Reference Manual (EMTP TheoryBook), Bonneville Power Administration, Portland, U.S.A., August 1986.[13]. Bompa, L., "EMTP Model for Multiple Coil Synchronous Machines," 14th EMTPEuropean User Group Meeting, Trondheim, Norway, May 1988.[14]. Lima, J.A., "Numerical Instability Due to EMTP_TACS Inter-relation," EMTPNewsletter, Vol. 5, No. 1, Jan. 1985, pp. 21-33.[15]. Araujo, A.E.A., H.W. Dommel and J.R. Marti, "Numerical Instabilities in Power SystemTransient Simulations,"^Proceedings of the TASTED International Conference,Vancouver, Canada, August 7, 1992, pp. 176-180.[16]. Marti, J.R. and Lin-Jiming, "Suppression of Numerical Oscillations in the EMTP," IEEETrans. Power Systems, Vol. 4, May 1989, pp. 739-747.[17]. Adkins, B. and Harley, R.G., The General Theory of Alternating Current Machines, JohnWiley & Sons, Inc., New York, 1975.[18]. Knudsen, H., "Implementation of machine model of a 6-phase synchronous machine with2 three-phase windings spatially displaced by 30° using EMTP MODELS," EMTPAutumn Meeting, Heverlee, Belgium, November 9, 1992.APPENDIX AFourth-order Runge-Kutta ProcedureThis Appendix shows the FORTRAN listing of the simulation program and itsassociated subroutines for the solution of the six-phase synchronous machine currentequations for the case of a six-phase short circuit at the terminals of the machine, aspresented in chapter 4. The FORTRAN program "Solve.for" solves the short-circuitcurrent equations with the fourth-order Runge-Kutta method, and gives the results as id1 ,iql and i01 or id2 i92 and i02 . The program "Phasel.for" transforms the d, q and 0quantities of windings a,b,c into phase quantities. The program "phase2.for" transformsid2 5 i92 and ice into the phase quantities x, y and z.PROGRAM SOLVE.FORC This program solves the short-circuit current of the six-phase machine andC gives the result in d, q and 0 quantities for winding a,b,c and x,y,z.C The method used is fourth-order Runge-Kutta.CPARAMETER(NVAR=9)DIMENSION ISTART(NVAR)EXTERNAL DERIVSREAL*8 XBEGIN,XEND,ISTART,PI,DEITATOPEN(10,FILE='solve 1. OUT',STATUS='UNKNOWN')XBEGIN=0.0D0DELTAT=5.0D-5NSTEP=1600XEND=NSTEP*DELTATC ISTART(1) is the initial value of I(1)--x(0)C ISTART(2) is the initial value of I(2)--xdot(0)ISTART(1)=0.0ISTART(2)=0.0ISTART(3)=0.052ISTART(4)=0.0ISTART(5)=456.35ISTART(6)=0.0ISTART(7)=0.0ISTART(8)=0.0ISTART(9)=0.0CALL RKDUMB(ISTART,NVAR,XBEGIN,XEND,NSTEP,DERIVS)ENDSUBROUTINE DERIVS(T,I,DIDT)DIMENSION I(*),DIDT(*)REAL*8 I,DIDT,TDIDT(1)=-118.6*1(1)+1795.9*1(2)+66.2*1(3)+1415.99(4)+2.49(5)+* 3.2*1(6)+1015.4*I(7)-1101.3DIDT(2)=-2110.8*1(1)-106.69(2)-1736.8*1(3)+76.99(4)-1251.4*1(5)* -1251.4*1(6)+3.5*I(7)+0.0DIDT(3)=66.2*I(1)+1415.9*1(2)-118.6*1(3)+1795.9*1(4)+2.4*1(5)+* 3.2*1(6)+1015.4*1(7)-1101.3DIDT(4)=-1736.8*1(1)+76.99(2)-2110.8*1(3)-106.69(4)-1251.4*1(5)* -1251.4*1(6)+3.5*I(7)+0.0DIDT(5)=40.3*1(1)-2471.5*1(2)+40.39(3)-2471.5*I(4)-10.09(5)+* 3.8*I(6)-1562.8*1(7)+4552.1DIDT(6)=35.8*1(1)-2191.7*1(2)+35.89(3)-2191.7*1(4)+2.6*1(5)-* 13.5*1(6)-1385.8*I(7)-1166.1DIDT(7)=4825.2*1(1)+37.3*I(2)+4825.2*1(3)+37.3*1(4)+3138.8*1(5)+* 3138.8*I(6)-11.0*1(7)+0.0DIDT(8)=-8055.55*I(8)DIDT(9)=-8055.55*I(9)RETURNENDSUBROUTINE RKDUMB(ISTART,NVAR,X1,X2,NSTEP,DERIVS)PARAMETER (NMAX=10)INTEGER CONTREAL*8 ISTART,V,DV,X1,X2,H,XDIMENSION ISTART(NVAR),V(NMAX),DV(NMAX)DO 11 I=1,NVARV(I)=ISTART(I)11 CONTINUEX=X1H=(X2-X1)/NSTEPCONT=0WRITE(10,100) X1,V(1),V(2),V(8)C WRITE(10,100) X1,V(3),V(4),V(9)53DO 13 K=1,NSTEPCALL DERIVS(X,V,DV)CALL RK4(V,DV,NVAR,X,H,V,DERIVS)WRITE(10,100) X,V(1),V(2),V(8)C WRITE(10,100)X, V(3), V(4), V(9)IF(X+H.EQ.X)PAUSE 'Stepsize not significant in RKDUMB.'X=X+H13 CONTINUE100 FORMAT(F16.6,2X,F10.4,2X,F10.4,2X,F10.4)RETURNENDSUBROUTINE RK4(Y,DYDX,N,X,H,YOUT,DERIVS)PARAMETER (NMAX=10)REAL*8 Y,DYDX,YOUT,YT,DYT,DYM,X,H,H6,BH,XHDIMENSIONY(N),DYDX(N),YOUT(N),YT(NMAX),DYT(NMAX),DYM(NMAX)HH=H*0.5H6=H/6.XH=X+HHDO 11 I=1,NYT(I)=Y(I)+HH*DYDX(I)11 CONTINUECALL DERIVS(XH,YT,DYT)DO 12 I=1,NYT(I)=Y(I)+HH*DYT(I)12 CONTINUECALL DERIVS(XH,YT,DYM)DO .13 I=1,NYT(I)=Y(I)+H*DYM(I)DYM(I)=DYT(I)+DYM(I)13 CONTINUECALL DERIVS(X+H,YT,DYT)DO 14 I=1,NYOUT(I)=Y(I)+H6*(DYDX(I)+DYT(I)+2. *DYM(I))14 CONTINUERETURNENDPROGRAM PHASEl.FORC This program transforms the d, q and 0 (1) quantities intoC phase quantities a, b and c.DIMENSION TINV(3,3),X(3),Y(3)54REAL*8 BETA,LAMDA,TIME,TINV,OMEGA,X,PI,YOPEN(12,FILE='PHASE1.OUT',STATUS='UNKNOWN')OPEN(10,FILE=1 SOLVE1.OUT',STATUS='OLD')PI=DACOS(-1.0D0)WRITE(*, *) 'OMEGA,LAMDA'READ(*,*) OMEGA,LAMDA300 READ(10,*,END=200) TIME,X(1),X(2),X(3)BETA=OMEGA*TIME+LAMDATINV(1,1)=DSQRT(2.D0)/DSQRT(3.D0)*DCOS(BETA)TINV(1,2)=DSQRT(2.D0)/DSQRT(3.D0)*DSIN(BETA)TINV(1,3)=1/DSQRT(3.D0)TINV(2, 1)=D S QRT(2 . DO)/D S QRT(3 .D0)*DCO S(BETA-2. DO*PI/3 .D0)TINV(2,2)=DSQRT(2.D0)/DSQRT(3.D0)*DSIN(BETA-2.D0*PI/3.D0)TINV(2,3)=1/DSQRT(3.D0)TINV(3, 1)=D S QRT(2 . DO)/D SQRT(3 . DO)*DCO S(BETA+2 DO*PI/3 .D0)TINV(3,2)=DSQRT(2.D0)/DSQRT(3.D0)*DSIN(BETA+2.D0*PI/3.D0)TINV(3,3)=1/DSQRT(3.D0)CALL MULTVECTOR(TINV,X,Y)WRITE(12,100) TIME,Y(1),Y(2),Y(3)100 FORMAT(F16.6,2X,F16.4,2X,F16.4,2X,F16.4)GO TO 300200 CONTINUEENDSUBROUTINE MULTVECTOR(A,X,Y)DIMENSION A(3,3),X(3),Y(3)REAL*8 A,X,SUM,YDO 1 I=1,3SUM=0.D0DO 2 K=1,3SUM=SUM+A(I,K)*X(K)2 CONTINUEY(I)=SUM1 CONTINUERETURNENDPROGRAM PHASE2.FORC This program transforms the d, q, and 0 (2) quantities intoC the phase quantities x, y and z.DIMENSION TINV(3,3),X(3),Y(3)REAL*8 BETA,LAMDA,TIME,TINV,OMEGA,X,PLYOPEN(12,FILE—'phase2.0UT',STATUS='UNKNOWN')55OPEN(10,FILE-2solve2. out', S TATUS='OLD')PI=DACOS(-1.0D0)WRITE(*, *) 'OMEGA,LAMDA'READ (*, *) OMEGA,LAMDA300 READ(10, *,END=200) TIME, X(1), X(2), X(3)BETA=OMEGA* TIME+LAMDATINV(1, 1)=D S QRT(2. DO)/D SQRT(3 .D0)*DCO S(BETA-PI/6.D0)TINV(1,2)=DSQRT(2.D0)/DSQRT(3.D0)*DSIN(BETA-PI/6.D0)TINV(1,3)=1/D S QRT(3 .D0)TINV(2, 1)=D S QRT(2. DO)/D SQRT(3 .D0)*DCOS(BETA-5.DO*PI/6.D0)TINV(2,2)=DSQRT(2.D0)/DSQRT(3.D0)*DSIN(BETA-5.D0*PI/6.D0)TINV(2,3)=1/D S QRT(3 .D0)TINV(3, 1)=D SQRT(2 . DO)/D SQRT(3 .D0)*DCO S(BETA+5 DO*PU6. DO)TINV(3,2)=DSQRT(2.D0)/DSQRT(3.D0)*DSIN(BETA+5.D0*PI/6.D0)TINV(3 ,3)=1/D S QRT(3 .D0)CALL MULTVECTOR(TINV,X,Y)WRITE(12,100) TIME, Y(1), Y(2), Y(3)100 FORMAT(F16.6,2X,F16.4,2X,F16.4,2X,F16.4)GO TO 300200 CONTINUEENDSUBROUTINE MULTVEC TOR(A, X, Y)DIMENSION A(3 ,3), X(3), Y(3 )REAL*8 A, X, SUM, YDO 1 I=1,3SUM=0.D0DO 2 K=1,3SUM= SUM+A(I,K) * X(K)2 CONTINUEY(I)=SUM1 CONTINUERETURNEND56APPENDIX BSubroutine SourceThis Appendix shows the FORTRAN listing of the subroutine Source. Thissubroutine uses as input the values of the currents at the terminals of the two machines ateach time step. The current values are generated by EMTP and stored in x(1) to x(6). Thesubroutine then uses these values, along with the constant parameters x(7) to x(9) neededfor the definition of [D, and [D2 ], to generate six user-defined current sources to beinjected back into the network at each time step. This subroutine also uses a linearextrapolation, to reduced the severity of the time delay between the EMTP and thesubroutine Source.PROGRAM SOURCE.FORSUBROUTINE SOURCE(X,LX,SOURC,T,DELTAT,IHALF,KREAD)DIMENSION X(*), SOURC(*),AUX1 (1 0)IMPLICIT REAL*8 (A-H2 O-Z),INTEGER*4 (I-N)SAVE AUX1KREAD=OSOURC(1)=X(1)*X(7)+X(2)*X(8)+X(3)*X(9)SOURC (2)=X( 1 )*X(9)+X(2)*X(7)+X(3 )*X(8)S OURC (3 )=X( 1 )*X(8)+X(2)*X(9)+X(3 )*X(7)SOURC(4)=X(4)*X(7)+X(5)*X(9)+X(6)*X(8)SOURC(5)=X(4)*X(8)+X(5)*X(7)+X(6)*X(9)SOURC(6)=X(4) *X(9)+X(5)*X(8)+X(6)*X(7)C^^Linear Extrapolation for improving the time delay.IF(X(10).EQ.0.0D0) RETURNIF(T.EQ.0.0D0) THENDO 15 I=1,6AUX 1 (I)=S OURC(I)15 CONTINUEEND IF57DO 20 I=1,6AUX2=2.DO*SOURC(I)-AUX1(I)AUX1(I)=S OURC (I)SOURC(I)=AUX220 CONTINUEEND58APPENDIX CEMTP Input FileThis Appendix shows the EMTP input listing for the simulation of a six-phase shortcircuit at the terminals of a six-phase synchronous machine. "Cards" 4, 5 and 6 define themachine self and mutual impedances. The six switches at the terminals of the machinerepresent the six-phase short circuit.EMTP INPUT FILE^ Case identification cardSIX-PHASE SYNCHRONOUS MACHINE MODEL, SHORT CIRCUIT TESTTO THE TERMINAL OF THE MACHINES^ Time card50.E-6 0.12** ^ Lumped RLC branch1GM1-aNODEla .03320 .5200 1GM1-bNODE1b .03320 .5200 1GM1-cNODElc .03320 .5200 1GM2-aNODE2a .03320 .5200 1GM2-bNODE2b .03320 .5200 1GM2-cNODE2c .03320 .5200 1$ = = End of level* 1: Linear and nonlinear elements Time-controlled switchNODEla .000000 100.0NODElb .000000 100.0NODElc .000000 100.0NODE2a .000000 100.0NODE2b .000000 100.0NODE2c .000000 100.0*$ = = = End of level 2: Switches and piecewise linear elements59S^M Design parameters (Card 7)S M^Node names for armature windings (Card 1)60.0. 391.920^30.0S M^Node names for armature windings (Card 2)GM2-b^.1250^.480 1**50 GM2-a**** S^M Node names for armature windings (Card 1)50 GM1-a^60.0 391.920^0.0*S M Node names for armature windings (Card 2)GM1-b^.1250 .480^1*S M Node names for armature windings (Card 3)GM1-c*S M Impedances and time constants (Card 4)3.9265 3.9265 3.9265 3.9265 4.0896 4.1107 0.00558 0.00826*S M Impedances and time constants (Card 5)1.8142 0.0^1.8142 0.0^1.0^2.0496 0.0^0.00872*0.0^0.000^0.16515 0.001010100000*60*^S M^Node names for armature windings (Card 3)GM2-c*^S M^Impedances and time constants (Card 4)3.9265^3.9265 3.9265 3.9265 4.0896 4.1107 0.00558 0.00826*^S M^Impedances and time constants (Card 5)1.8142 0.0^1.8142^0.0^1.0^2.0496^0.0^0.00872^0.0^0.000^0.16515^0.001010100000*^S M^Design parameters (Card 7)1 GM2-a-12 GM2-b-13 GM2-c-14 GM1-a-15 GM1-b-16 GM1-c-1$ = = = End of level 3: Sources ^GM1-a$ = = = End of level 4: User-defined voltage output.USERI GM1-aNODElaI GM1-bNODElbI GM1-cNODElcI GM2-aNODE2aI GM2-bNODE2bI GM2-cNODE2cP -0.9106836P 0.2440169P -0.3333333P 0.00000.END61

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