IEEETRANSACTIONSON INDUSTRY APPLICATIONS, VOL. IA-10, NO.2,MARCH/APRIL 1974 261 ABC Short Circuit Learn in an Hour, It Use Anywhere, Memorize Formula It No MOON H. YUEN, SENIOR MEMBER, IEEE Abstract-Short circuit ABC-learn it in a_n-hour, use it anywhere, INFINITEBUS memorize no formula. The MVA method for solving industrial Isc = V/Z power system short circuits appropriatelyfitsthisdescription. Indeed, (V7t}IfSC= {15V5Vvz solving short circuit problems with the MVA method is as easy as Z=0.01OHMS V3 EiSC = E2/Z @13.8KV VASC = E2/Z learningtheABC's. KVASC = 1000(KV) 2/Z MVASC = KV2/Z INTRODUCTION FAULT S HORT CIRCUIT studies are necessary for any power E = rV distribution system to determine switchgear rating for SC=SHORTCIRCUITCURRENTIN AMPERES protective relaying, and to determine the voltage drop during E =LINETOLINEVOLTAGEINVOLTS V =LINETONEUTRAL VOLTAGEINVOLTS starting oflarge motors. One line diagrams are not complete Z =LINETONEUTRALIMPEDANCEIN OHMS unless the short circuit values are solved at various strategic MVASC=SHORTCIRCUITMVA Fig. 1. Onelinediagram. points. No substation equipment, motor control centers, breaker panels, etc., can be purchased without knowledge of of the complete short circuit information of the entire power distribution system. } SYSTEM Knowing how to calculate short circuit problems is a must 500MVA 500 12 for every electrical engineer. To learn itmaybe easy forsome, difficult for others. However, to do the problems anywhere TRANSFORMER in or out of the office where the references are not available 5X0=MVQ.A1 J} 50/0.1= 2 may not be an easy task because the conventional methodsof F F 13.8KV solvingshort circuitsinvolve too many formulas. To memorize MOTOR } them at all times isimpractical forthe majority. 50 MVA 50/0.2 = 250 Xd = 0.2 WHAT REALLY IS THE MVA METHOD? MVA1,2 = MVA1 x MVA2/MVAl + MVA2 Basically, the MVA method is a modification ofthe Ohmic = 500x500/50+ 500 = 250 method in which the impedance of a circuit is the sum ofthe MVA1+3= 250+ 250 = 500 impedances of the various components of the circuit. Since, 1SC= 500x1000/V X13.8 = 20900AMPS Fig.2. Impedancediagram. Fig. 3. MVAdiagram. by definition, admittance is the reciprocal of impedance, it follows that the reciprocal ofthe system admittance isthe sum ofthe reciprocals ofthe admittancesofthe components. Also, y admittance ofa circuit by definition, the admittance ofa circuit or component isthe Zohms impedance inohms maximum current or KVA at unit voltage which would flow Zpu impedance in perunit through the circuit or component to a short circuit or fault KV line to linevoltage when supplied from a source of infinite capacity. Refer to KVAsc short circuit KVA Fig. 1. MVAsc short circuit MVA y1 ~~~~~~~M~V(A1s)c (13.8)2/0.01 = 19 000(for Fig. 1). Zohms Practically, the MVA method is used by separating the KVAsC = 1000X (KV)2 X Y (2) circuit into components, calculating each component with its own infinite bus as shown in Figs. 2 and 3. Fig. 2 isatypical MVAsc =(KV)2 X Y (3) impedance diagram of a one line diagram. Fig. 3 is an MVA diagram. The conversion from a one line diagram to an MVA MVAsC -* (4) diagram issimple arithmetic. Zpu Component 1, the system, is normally given a short circuit MVA rating. So, one merely writes down 500, which is its PaperTOD-73-132,approvedbythePetroleumandChemicalIndustry Committee of the IEEE Industry Applications Society forpresentation system short circuit MVA. Sometimes, if the system MVA is at the 1973 Petroleum and Chemical Industry Conference, Houston, not available, but its voltage and impedance are given, the Tex., September 17-19. Manuscript released for publication October 31, 1973. short circuit MVA can be calculated with the application of Theauthoriswiththe BechtelCorporation,SanFrancisco,Calif. (3). 262 IEEETRANSACTIONSON INDUSTRY APPLICATIONS, MARCH/APRIL 1974 Next, for component 2, use (4). The short circuit MVA of A.CONVERTTOMVA'S the transformer is equal to its own MVA base divided by its ff~~~ 1500 own per unit impedance. (Use reactance X with the MVA I ssnn & method.) Next, for component 3, again use (4). The short circuit (69)2 3.87 MVA contribution of the motor is equal to its own MVA base divided by its own per unit impedance. (Use reactance 15 Xwith the MVA method.) 0.076 Now, let us examine the MVA diagram, Fig. 3. If a short circuit is taken at point F, there will be a series flow ofMVA1 15 and MVA2, and their combination will be in parallel with 0.2 MVA3. The question now is: how do you combine the \L.. Xd = 0.2 MVAF = 228 MVA values in series and in parallel? The answer is again (a) (b) simple arithmetic Fig.4. TheABCofthe MVA. (a)Onelinediagram. (b)MVAdiagram. (MVAI) X (MVA2) series MVA, 2 --((MVA1) + (MVA2) (5) B. COMBINETHEMVA'S parallel MVAI+2 =MVA, +MVA2. (6) 1. SERIES: MVA1,2 =MVA1 xMVA2/MVA1+MVA2 2. PARALLEL:MVA1+2 = MVA1 + MVA2 From (5) and (6), it can easily be recognized that seriesMVA 3. DELTATOWYE combinations are exactly as resistances computed in parallel. ParallelMVAcombinations are exactly as resistances computed in series. Yl = SIDl 2= S/132 The MVASc at point F of Fig. 2 then can be calculated as F~~3TlFn~~~.Li Y3 = S/D3 follows: S = (D1 x D2 + (D12 x D3) + (D3 x Dj) Y = WYE 1 500X 500 D = DELTA MVAl,2 = 500 + 500 250 (a) (b) Fig.5. The ABC of the MVA. (a) Delta connection. (b) Wye con- MVASC =MVAr + MVA3 = 250 + 250=500. nection. The term with the asterisk is the new MVA1 value which is the result of combining MVA1 and MVA2. After the opera- tion, the new MVA, which is 250 MVA, replaces the old MVA1 and MVA2. This scheme of replacing old quantities with new quantities relates to computer datamemory storage system. At this point the short circuit MVA is solved. To find the current value, only the voltage value isrequired. Forexample, B ifthe voltage is 13.8 kV,the currentIC is A+B MVA X 1000 S00 X 1000 _ / T=IAAAx+BBB-= A-A+BB= AxK -- - =2on90nA. / WHERE B>A Nf-X KV V--X 13.8 II I I THE ABC OF THE MVA I B/'A 2 3 4 5 6 7 8 910 15 20 25 30 40 50 60 80 100 Up to now, the reader has spent about 15 min in slow read- Fig.6. Seriescombinationratio. ing. He has found that there has been nothing new, and the formulas are no more than good old Ohm's Law arithmetics. Incoming line short circuit duty in MVA is normally given Now, he can forget the formulas and start the ABC. by power companies. Therefore,use the value as given and no conversion is required. However,ifimpedance or reactance at A. ConverttoMVA's the terminal is given, fimd its short circuit MVAby dividingits Convert all one line components to short circuit MVA's. (KV)2 by itsohms. Equipment such as generators, motors, transformers, etc., are As conversion is being made, an MVA diagram is being normally given their own MVA and impedance or reactance developed. One line diagram 4(a) is replaced with MVA ratings. The short circuit MVA of each is equal to its MVA diagram 4(b). rating divided by its own per unit impedance or reactance. For a feeder where voltage is given and its impedance or re- B. CombineMVA's actance is known, its short circuit MVA is equal to (KV)2 1) SeriesMVA's are combined asresistances in parallel. divided byitsimpedance orreactance in ohms. 2) Parallel MVA's are added arithmetically. Refer to Fig. YUEN: SHORTCIRCUIT ABC 263 13.8KV 20MVA G 300MVA Xd"=0.1 F1 _ 13.8KV I 13.81t X=0.019OHMS X=0.019OHMS X=0.019OHMS I.JA 20MVA 20MVA ,.A.L_ rvyyr% X=0.1 X=0.1 F2 4.16KV | MVA 2MVA 6MVA 1MVA X=0.067 Xd"=0.2 Xd"=0.2 Xd"=0.25 F3 ALLMOTORS 50-200HP TOTAL=1MVA Xd"=0.25 Fig. 7. Onelinediagram. 4(b) for the following: C.REDUCETHEMVADIAGRAM 1500 X 1230 MVA1,2= =675 2 1500 + 1230 (this is the new MVA1 ) 675X 198= MVA, 3 = 153 ,3 675 + 198 MVA1+4 =MVA, + MVA4 = 153 + 75 = 228 228 X 1000 I12 Vf/_Xx1212 =1I1I000A. For MVA1,2, add MVA, and MVA2 in series. For MVA1+4, add MVA1 and MVA4 in parallel. I12 is the short circuit current at 12 kV. Fig.8. MVAdiagram. 3) Delta to wye conversions are rarely used in industrial power distribution systems, but they are again simple arith- curve. Read the horizontal value 0.8, which is the result of metic. Refer to Figs. 5(a) and (b). B/A +B; then 4) The only point that needs more attention is the series combination ifa slide rule is not available. The attempt here is T=A(B/A +B) to be able to solve most short circuit problemswith reasonable = 10 X 0.8 =8. accuracy without the use ofa slide rule. With the aid of the curve in Fig. 6, let us analyze the series It is also noted that when combining two quantities in series, combination the result is always smaller than the smallest of the two. The (MVA1) X (MVA2) example shows the result to be 8 when combining 10 and 40. MVAMI 2 = (MVA1)+ (MVA2) C. ReduceMVA Diagram LetA =MVA1,B= MVA2, and T= total MVA, so that Reducing an MVA diagram takes the same reduction process A X B A(B)_ required for the per unit impedance diagram, except that MVA quantities are used instead of per unit impedances or A+B A+B'AB reactances. B/A +B is plotted as a constant on a log-log scale base from Fig. 7 is a typical distribution one line diagram including a 1 to 100 which is the ratio of B/A. (Refer to Fig. 6.) For delta connected feeder system. Reactances only are used for example, let A = 10,B = 40, and B/A = 4. Read 4, at hori- practical purposes. Fig. 8 is an MVA diagram that shows all zontal scale. From 4 project upward until it intersects the elements in the one line in MVA quantities. Fig. 9(a) shows 264 IEEETRANSACTIONS ON INDUSTRY APPLICATIONS, MARCH/APRIL 1974 1 300 1 3 EF1 13.8KV F1 | 13.8KV 6 20196 2i . 198 6[ 3 396 3 396 - (a) 11 3.2 8 39 8142 10000 419 69 62 4 1 300 1 1 300 F1 8KV Fl 13.8KV Fl 13.8KV 1 0 3 396 6 20196 v 2 198 6[ 2 233 3yDS3 3.91600x01006 =-3961 l 3 3 F1=533MVA Y4=4SD4= 3.961199x66106 =20196 (a) S 3.96 x 106 =66=- 1996 =20196 1 300 S = (Y3) X(Y4)+(Y3)X(Y6)+ (Y4) X (Y6) = (10000) x(196)+(10000) x(196)+ (196) x(196) 1 296 _" 2 198 = 3.96 x106 (b) 3 396 FF2 3 396 _4.16KV F2 1,- 4.16KV 11 33 8 39 11 3.2 8 39 1 4 3[39 1 220 F2 4.16KV F2 J 4.16KV 8 8 42 (c) F2=262MVA Fig.9. MVAreduction. (b) 1 the filrst step reduction sequence. Note that there are three 1 faults tobe calculated,F1,F2, andF3. ~iI2E 11 [1 8 The first step reduction combines the series and parallel II components so that the simplest diagram canbe accomplished, 4.16KV F3 480V and that any fault can be solved in random fashion. Fig. 9(a) W-- ~~~X-111 8 12 shows that items 4 and 5 have been combined to make anew 4; items 6 and 7 have been combined to make anew 6;items 8-10 have been combined to make a new 8. Fig. 9(b) con- verted a delta to a wye configuration. Fig. 9(c) is further 1 reduced to Fig. 10(a), indicating items 2 and 4 have been F3.4-.J 480V combined to make a new 2. Figure 10(a) thenis the simplest 480V 12 diagram that would allow the solving of any of the three F3=18MVA faults in random selection. Figs. 10(a)-(c) show the reduction processin solving faultsF1,F2,andF3,respectively. (C) Fig. 10. MVAreduction. WHY THE MVAMETHOD? There are many reasons why the MVA method is recom- mended forindustrial power short circuit calculations. 1) It does not require a common MVA base as required by the per unitmethod. YUEN: SHORTCIRCUITABC 265 500MVA SYSTEM OHMIC METHOD X100K0VxA(bKV)2 1000x(13.8)2 SYSTEM 500000 *^ ~ 13.8KV 1. = 0.38 OHMS X=0.151ohms X - 0.151 OHMS 13.8KV FEEDER 5000KVA 2. f-.w-.y-Y-' X=0.055 1000x(X P.U.) x(KV)2 TRANSFORMER KVAb 100x0.055x(2.4)2 F 2.4KV 3. 5000 0.063 OHMS 1000x(XP.U.)x(KV)2 MOTOR KVAb 1000x0.16x(2.4)2 2500 O 2500KVA 4. =0.369OHMS {J Xd=0.16 d (a) Fig. 11. Onelinediagram forcomparisonofmethods. PERUNITMETHOD (500000KVABASE) 2) It is not necessary to convert impedances from one X B=ASKEVKAVAS.xC.(1) voltage to another asrequired by the Ohmicmethod. SYSTEM _ 500000x1 1. 500000 = 1.000 3) The conversion formulas as used for both the Ohmic and the per unit methods are complex and not easy to x U (OHMS)x(BASEKVA) 13.8KV FEEDER 1000x (KV)2 (0.151)x(500000( memorize. 2. (13.8)2x1000 = 0.396 4) Both the Ohmic and the per unit methods usually end XpU (X P.U.) x(BASEKVA) up with small decimals resulting from converting impedances TRANSFORMER KVAT (0.055) x(500000) from one voltage to another or from converting impedances 3. 5000 5.500 to the same common base. Therefore, one canmakemistakes 3. Xp(U(XP.U)x(BASEKVA) in the decimals,withresultingwronganswers. MOTOR (0.16)x(500000) 5) The MVA method utilizes large whole numbers denoting 4. 2500 =32.000 MVA quantities. With a little practice, one can estimate the (b) result by lookingat the combination. For example, 10 and 10 in series become 5; 10 and 100 in series become 9.1; and 10 MVAMETHOD and 10 000 in series give 10. A small number combined with MVA1 = 500 too large a number, 100 times larger or more,willhave no ef- SYSTEM fect onthe smallnumber. In order to further prove the preceding points, it isnecessary MVA2_ (KV(2 (13.8)2 to give the following comparison ofmethods that are utilized 13.8KVFEEDER XOHMS 0.151 2. =1260 in solving industrial powersystem short circuits. MVAT 5 MVA3 0.05 COMPARISON OF METHODS TRANSFORMER Xp 0.055 =91 A one line diagram, Fig. 11,isshown. Solve the three-phase MVAm 2.5 fault at pointFwith andwithout motor contribution. MOTOR 4 XP.U. 0.16 Note that reactances only are considered in the three cases =15.6 4. being compared. It is felt that using impedances would give (c) the same result, but would complicate the calculations. It is Fig. 12. (a) Ohmic conversion. (b) Per unit conversion. (c) MVA also widely recognized and acceptable by industries to use conversion. reactances only in calculating industrial power system short circuits, in that it would result in a higher short circuit value, perhaps by 0-3 percent in most cases. Reference [11 tion of August, 1965. As noted, the three-phase fault has been solved to be 228 MVA at the 12-kV bus. Since the exemplifies the use ofreactancesrather thanimpedances. positive sequence fault is equal to the negative sequence fault, Figs. 12(a)-(c) tabulate the conversion calculations for the therefore, three methods. Figs. 13-15 show the three methods utilized for comparison. Fig. 16 tabulates the results of the three MVAX1 =MVAX2 = 228. methods. The zero sequence fault MVA, however, must be calculated, CAN PHASE-GROUND FAULT BE SOLVED? and its MVA value then is combined with the positive and The answer, of course, is yes. Solving phase-ground fault is negative MVAvalues. aseasy assolvingthree-phase fault. Refer to Fig. 4(a) again. During a fault on the 12-kV bus, Refer to Figs. 4(a) and (b). This problem istaken from the only the transformer and the motor contribute to zero California State Professional Engineer Registration Examina- sequence MVA's. The deltaprimary ofthe transformer blocks 266 IEEETRANSACTIONS ON INDUSTRY APPLICATIONS, MARCH/APRIL 1974 [email protected] 3-PHASEFAULTMVA=228 [email protected] SYSTEM X1,2 [email protected] 198 MVAXOT THEREFORE: 1 <> X1,2 (0.531)x(0.03)@2.4KV -J MVAX1 MVAXX22 22_ [email protected] 12KV MVAXo=MVAXOT+MVAXOM CABLE [email protected] =198 +150 [email protected] 150 MVAXOM 348 2 MVAF=(KV)2= (2.4)2=72.8 XF 0.079 15 150= o MOTORZEROSEQUENCEREACTANCE TRANSF X4 [email protected] Fig. 17. Zero sequencefaultpower. 3 XF+M=(0.079)x(0.369)=0.0656OHMS F 2.4KV (0.079)+(0.369) MOT~O~R~~~~~~M~VA~F+~MF=0(KV.)20=6(2.54)62==88 4 OHMS2.4 OHMS13.8x(2.4/13.8)2 228 1 228 228 SPOESQIUTEINVCEE POWER =OHMS138x(0.03) 13.8 NEGATIVE 228 2 228 228 SEQUENCE Fig. 13. Ohmic method reactance diagram. POWER ZERO 38 3 348 348 SEQUENCE POWER PERUNITMETHOD BASEMVA=500 F0 SYSTEM xi=1.000 MVA1,2=228/2=114 X2=0.396 MVA13 =114X348/114+348=86 MVAFO =3X86=258 IFO=258X1000/(3x12=12400A CABLE X3=5.500 Fig. 18. Phase-ground faultofMVA circuit. XF 6.896 WITHOUT MVAF=500/6.896 =72.6 MOTOR CONTRIBUTION TRANSF any zero sequence power flowing from the system and across X4=32.000 the transformer. Therefore, Fig. 17 shows the zero sequence F2.4KV (6.896)x(32.000) power circuit F+M (6.896)+(32.000) MOTOR MVAXOT =MVAX1 =MVAX2 = 198 500 WITHMOTOR MVAF+M 5.67=88- CONTRIBUTION (the transformer zero sequence reactance is equal to its posi- Fig. 14. Reactance diagram. tive and negative reactances) MVAXOM =- = 150MVA MVAMETHOD MVA METHOD ~~~~~~~~~~~~~0.1 (since the zero sequence reactance ofthe motor is about 2 of its positive sequence reactance). The total zero sequence fault power then is equal to the sum,whichis l 1 358 J MVAXOT+MVAXOM = 198 + 150= 348. 3 91 1 72.6 The phase-ground fault power is obtained with the use of Fig. 18. Since these are three branchesinparallel,the simplest KV F Jl 2.4KV F J 2.4KV approach is to take one branch out ofthe circuit and solve its MVA value, then multiply the value by 3, which gives the 4 15.6 4 15.6 fimal answer MVA1+4 =88-2 Fig. 15. MVAdiagram. MVA1,2 = 228/2 = 114 114 X 348 MVAI,3 =114 = 86 348 + METHODS OHMIC PERUNIT MVA MVAFO =3X 86=258 FAULCOTNDITION METHOD METHOD METHOD [email protected] 72.8MVA 72.6MVA 72.6MVA 258X 1000 WITHOUTMOTORCONTRIBUTION IFO 12400A. = [email protected] 88MVA 88MVA 88.2MVA WITHMOTORCONTRIBUTION The problem, as shown in Fig.4(a), is also solvedwiththe per Fig. 16. Resultofcomparisonofmethods. unit method as Appendix I. This gives further comparisonof YUEN: SHORTCIRCUIT ABC 267 SYSTEMSYMETRICAL SMVA SHORTCIRCUIT-500MVA Vd= SMVA+MVASSC- -(26) X=0OHM SMVA=STARTINGMVA 7.5MVATRANSF MVASC=SHORTCIRCUITMVA X=0.0833 W Vd=INSTANTANEOUS VOLTAGEDROP 69/13.8KV >v_fy^ Vd = 21/21 +71 =21/92= 0.228 I 13.8KV BUS F1 VT13.8=1-0.228=0.772 X1=0.655 (I)V-(86 MVA0=3X (86+480 )=221 MVA (76MVA) VT13.2=0.772x 132.23) X2=0.655 348 1=221x100 =10600 X=0.19OHM =0.805OR80.5% (71 MVA) F2 (SHRTCIRCUIT) VT13.8=MOTORTERMINAL x0 =0.428 [email protected] (L.\ (KV)2 (12)2 VT13.2=MOTOR TERMINAL Xf=f0.100 40=-~.=3-.3 =480 4000HPINDUCTIONMOTOR [email protected] 3.5MVA STARTINGMVA=21 AT 13.8KV (19.3MVAAT13.2KV) al=Ia2I ao x1+X2+EX0+3Xf 0.655+0.6515.0+0.428+0.3 -2._0=380.491 Fig. 23. Motorstartingvoltagedrop calculation. Ifpu=31 3x0.492=1.472A If=1.472x7230=10600A the amount of calculation involved between the MVA and the Fig. 19. Phasetoground faultwith addedfaultreactanceXf. per unit methods. Phase-ground fault with an added fault neutral reactance also can be calculated with the MVA method. Fig. 19 il- a V lustrates the preceding problem with an added fault neutral 3 cb m|V reactance Xf. Note that usingboth the MVA and the per unit methods obtain the same result except that the MVAmethod If V shows much less calculation. OPERATORS 10 a= 0 =ea=.0.500+J0.866 ZERO l a2 240 ee2a= 0.500 J0.866 TWO-PHASE TO GROUND FAULT a3 1.0 a4-a Can two-phase to ground fault be solved with the MVA method? The answer is again,yes. Fig. 20 shows a two-phase POS. CONNECTIONSBETWEEN to ground fault diagram and connections between sequence SEQUENCENETWORKS 12 networks. As indicated,If is the fault current between phases NEG. C,B,and ground. In order to develop an MVA equation for two-phase to Fig. 20. Twolinestoground fault. ground faultcalculations, the classical symmetrical component equations are utilized as basis. Relationships between phase and sequence quantities are expressed by the following: Ea=1.0 ( 7 7 Ii = xl-+E(XX22)+(XXO0) = 00..1665555+x00..442288 = 1.09 Va = VO + VI + V2 (7) ($ 0.655+ X2 X0 0.655 xi 0.428 0 _1__X=0_+IXx22 = 1.09x 0.6~5~50.+6505.428 = 0.659 Vb = Vo +a2V1 +aV2 (8) V, = VO +aVI +a2V2 (9) Ea=1.0 IF=310=3x0.66=1.978 0.655 1 =1.978xIpu Ia =IO+I+I2 (10) =1.978x7230=14,300A Ib =Io +a2Il +aI2 (11) Fig. 21. Two-phase toground faultbyperunitmethod. I, =Io +aI +a2I2* (12) MVA 228x576 163.5 From the preceding equations, the following relationships 1 228+576 are obtained: MVAxx0 = MVA11x MVAMV+AMVA2 VO = 3 (Va + Vb + VC) (13) 163.5x 348+228 V1 = 3 (Va +aVb +a2VC) (14) MVAx1 MVAX1 MVA =3=MV9A8x.8 =3x98.8=296-.04 V2 = 3 (Va +a2Vb +aVc) (15) IF=14260A IO =I (Ia +Ib +Ic) (16) Fig. 22. Two-phase togroundfaultbyMVAmethod. =13 I, (Ia +aIb +a2IC) (17) I2 = (Ia +a2Ib +aIC). (18) 268 IEEE TRANSACTIONS ON INDUSTRYRPPLICATIONS,MARCH/APRIL 1974 us a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 10 20 30 40 50 60 70 80 (a) y MVASC SOLVEFORXWITH "C" VS=y5M~VASC+SMVA (01.04) (SMVA,1.04) x=y(SMVA) EQUATION FOR LINE"C", 104 SUBSTITUTEXINTO"A" y=MX= IMV-A, X y 1.04 y 1.04 MVAw-y(SMVA) MVAsc whileM=-x = SMVA (SMVA,0) (MVAC,0) y(MVA.)=1.04[MVA.-y(SMVA)] EQUATION FORLINE"A", y=1.04MVASW/MVASC+SMVA y-Yl Y2-Yl x-X1 X2-x1 SMVA=STARTINGMVA y-0 1.04-0 X-MVAJ - -MVAsc MVA,C= SHORTCIRCUITMVA Vs = Y=STARTINGVOLTAGE y 1.04 OR MVA, -X= MVASC 1.044x71 71+21 =0.805 =80.5%APPROX. (b) Fig. 24. (a)Voltagedropbygraphicsolution. (b)Graphicsolution provenbyanalyticgeometry. Connections between sequence networks can be found from From the preceding analysis, (22)-(25) can be derived for (13)-(18): the application ofboththe per unitmethod Vb=O Vc=< Va? Ia0= E I1 = (22) Io +I +I2 = Q (19) I, + (X2)(XO) (X2)+ (XO) VO = V1 = V2. (20) Equations (19) and (20) are satisfied if the sequence net- Io =I, x X2 (23) works are connected as follows: XO +X2 IF =Ib +IC and the MVAmethod and by addition of(11) and (12) MVA1 +(MVA2 +MVAO) Ib +Ic =-2Io + (a2 +a)(II +I2)_ (24) MVA1 X (MVA2 +MVAo) Since a2 +a =-1, therefore MVAo i cIc =If= 2Io (II MVAX0 = (MVAx1) MVAo + MVA2 (25) Since Fig. 21 shows the use of (22) and (23) for solving the two- IO+Il +I2 =0 phase to ground fault of the same problem, the per unit 1I +I2 =-IO method. Fig. 22 indicates the utilization of(24) and (25) for therefore solving the two-phase to ground fault of the same problem, the MVA method. It is obvious as shown, thatthe use ofthe IF=2Io-(-Io)=3IO. (21) MVAmethod is simpler. YUEN: SHORT CIRCUIT ABC 269 120 115 110 7 104% 100 -+ 90 EXAMPLE MOTOR STARTrING VOLTAGE DROP 80 - 1. Ss_Aeoet-1iO.tpAse;rh1-aa,to-rlrliz"ncA-e"tfaodnIs.71MoVnIn-AfoUJ)Df4%shoonrtthecir3/cuviotltasgcealescaalned.swing it 2. Set Operator "B'' on 21 starrting MVA on short circuit scale. 3. Swing Operator "C" so its hhaairline falls on the intersection of Operator "B" and 104% voDitage. 0-i 60 -1 4. Read 80.5% voltage on crosssing point made by Operations "A" and "C". 5. If it is desired to find whhat starting voltage can be obtained O,A) with increasing bus voltagees to 107% and 109% (corresponding O V \ to 22% and 5% tap settings above normal transformer voltage). 40 -t A. oSfwinOgperOapteorrato"rB""Ca"ndso1077i%tsvholatiarglei.ne falls on intersection aW-, Read 82.5% on crossing point made by Operators "A" and 'C". 0 B. Swing Operator "C" so its hairline falls on intersection of Operator "B" and 1091%/.voltage. Read 85% on crossing poDint made by Operators "A" and "C". 20 -t 0 I 4., I I I I I 1 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 SHORT CIRCUIT MVA, Fig. 25. Sliderulemodel. MVAMETHOD FOR INSTANTANEOUS VOLTAGE ESTIMATE Fig. 24(a) shows a graphic solution of the problem. Figure Large motors are frequently connected to power systems 24(b) illustrates the validity ofthe graphic solution by analytic consisting of complicated networks of lines and cables for geometry. Fig. 25 shows a slide rule made for the sole which a calculation of the voltage drop would be difficult. purpose of solving instantaneous voltage drop in startinglarge Yet, it may be critical to know approximately what the motors. The instruction for the use ofthe slide rule as shown voltage at certain bus must be. This is because the voltage is self-explanatory. affects the motor torque in a square function; i.e., motor Fig. 26 is a compilation of standard industrial nominal torque varies as the square of the voltage for a 10-percent voltages and motor terminal voltages. Note that the unique voltage drop relationship between the nominal and terminal voltages is 4 percent different. This unique relationship aids the slide torque a(E)2 rule operation in solving instantaneous voltage drop during T= (0.9)2 =0.81 or81 percent. motor starting. The torque loss is 19 percent. The voltage drop may be estimated with reasonable ac- THE SIMPLE COMPUTER TIME SHARE PROGRAM curacy, however, if the short circuit MVA is known at the Why is it necessary to develop acomputer program for such point ofpower delivery. When motor starting MVA is drawn an easy method? The answer is simply economics. True, the from a system, the voltage drop in per unit of the initial MVA method enables the engineer to quickly calculate the voltage is approximately equal to the motor starting MVA faults on a power system, but how about documentation? divided by the sum ofthis MVAand the short circuit MVA After the engineer finishes his calculation, the result needs MVAs26 typing, proofreading, etc. From manual calculation to print- VUM=VAs +MVAs(26) ing, the estimated cost for aproblem as shown in Appendix II, having 12 components and three faults, is approximately $100 Fig. 23 shows an example applying the MVA method in including the engineer's pay. But the time share program estimating the voltage at the 13.8-kV bus when a large motor solution from input to print-out costs approximately $24 is started. ($12 for engineering time and $12 for computer and terminal 270 IEEETRANSACTIONSON INDUSTRY APPLICATIONS, MARCH/APRIL 1974 A=TRANSFORMERSECONDARYVOLTAGE CONCLUSION B=MOTORTERMINALVOLTAGE The paper described a unique easy to learn and easy to A/B=480/460=2400/2300=4160/4000=12000/11500 =13800/13200=22900/22000=104% remember method for solving industrial power distribution WHENONE21/2TRANSFORMERTAPISUSED: system short circuit problems. The examples given proved its A/B=490/460=2460/2300=4260/4000=12300/11500 effectiveness in terms of speed, accuracy, and-economy over =14100/13200=23500/22000=107% other conventional Ohmic and per unit methods. The writer 2-21/2TAPSABOVEWILLGIVE-A/B=110% has been using it for the past twenty yearsformany projects, Fig.26. Percent motor terminal voltage related to transformer sec- ondary voltage. small and large, and found it most effective because it seldom required one to memorize formulas aswith other methods. The MVA method also has been taught in various evening time). As the. problem involves more components and more schools and corporation sponsored seminars, including the faults, the cost differential increases noticeably. University of California Extension, ITC College, Bechtel Appendix II is a time share computer solution of Fig. 7. Corporation, Pacific Gas and Electric Co., etc., in the San The program itself is a conversational type. The user of the Francisco Bay area for the past sevenyears. program can input the data by a prepared tape or by typing the data as the computer is ready to receive the data. REFERENCES The input data are separated into two sections. Section one 1] ElectricalPowerDistribution forIndustrialPlants,IEEEStandard is to use lines 200 through 399 for MVAitems as shownwith 141, 1969,ch.IV. the use of the MVA method. For example, item 1 is 300 [2] IEEERedBook,IEEEStandard 141, 1969,sect.IV. MVA, item 2 is 200 MVA, etc. Section two is to use lines APPENDIX I 400 through 999 for command sequence. For example, line The following problem is taken from the Califomia State 400 data 1, 4, 5 instructs the computer to combine items 4 P.E. Registration Examination of August, 1965. The problem and 5 in series; the first number, 1, is for a series operation. willbe solved withthe per unit method. In line 400 data, there is a 2, 8, and 9, which instructs the computer to combine items 8 and 9 in parallel; the first UTILITY SYSTEM number, 2, is for a parallel operation. A 3, 3, 4, 6 command OTOR N instructs the computer to convert a delta towye operation of 6x9=K3V.87 Ohms / I-I- items 3,4,and 6. 1500MVA <,.. Fault Refer to Appendix II. The fault 1 result is 533.4 MVA, 15MVA I5M\A which is close to the manual solution result of 533 MVA. 69/12KV 12KV x=0.075 Xd = 0.200 Note the computer asked for a kilovolt input. The user A. Solve for 3-phase fault Xdo" = 0.100 entered the voltage. The computer then asked whether the B. Solve for single line to ground fault A. 3-PHASE FAULT user prefers an interrupting duty or amomentary duty compu- KVAbase ==150MVA tation. As shown, fault 1 requires an interrupting calculation KVbase 69 and 12 respectively aAnNdSIthSetacnodmaprudtelrategsatvereaqusierreiemsenotfoofutmpuulttipsleileercst.ionsThteo mceoemt- X baeg= (KKVVA)2basxe1000 = (691)52000x01000 = 31.8 Ohm puter solution sequence is exactly as shown on Fig. 10(a) for Xbase12 = = (1211)5520000x000100000 = 0.96 Ohms fault 1,manual solution. base69 = 3KKVWAx9bVaase = 150000 = 1250 A For fault 2, line 410 data are replaced with new commands basel2 --= i/7= xI5000102 = 7230 A as shown. Notice that the sequence follows the MVAdiagram, Fig. 10(b). The result of fault 2 is 261.9 MVA which was utiuittiylppiat.1y. 0° x KKVVAAbSa.s.e1=501050000000 = O.OOp.u. manually solved to be 262 MVA, Fig. 10(b). The computer Xtransf. p.u. = Brated p.u. KKVVKAAV1b5ra0a0st0eed = O.075x19050000 =0.750 p.u. again asked for akilovolt inputand a4.16 wasgiven. The next question again was for interrupting duty or Uin p U. rabtaedseOhms = 3.87 = 0.121 p.u. momentary duty. The answer was momentary so the com- motor dl" = 0.200 x 15105000000 = 2.00 p.u. puter gavetwo answers that are in accordance with [2]. For fault 3, line 410 data are replaced with new commands E = 1.0 p.u. thatfollowthesequenceasshown on Fig. 10(c). The computer (aj)j (b) Xa = Xut. + Xline + Xtransf asked for a kilovolt input, and 0.48 was given. Because 0.48 <X =u0.100 + 0.121 + 0.750 u1t . ~~~0~.=971 kV is a low-voltage system, the computer automatically Xb = 0.200 printed out five answers to suit the user's choice. The multi- pliers are all in accordance with the IEEE Red Book #141, 0Xtransf. <X1fdaidi Xeegg = 0.29.79171x 2 = 0.655 Section IV. ifault = E = 1.0 = 1.525 A Xlines 8eg After the MVA method is mastered in about an hour's time, it will take no more than fifteenminutes tolearn touse the computer program. Appendix III is a pre-made graph for fault 3 ph Ca 12KV fault p.u. x base 12 KV quickestimateofinstantaneousvoltageinstartinglarge motors. = 1.525 x 7230 = I,_LOOO A