Electric Circuits.. Theory & Applications - Vol.1... Short-Circuit Calculations and Steady-State Theory - O.G.C. Dahl - ( McGraw-Hill Book CO. - 1928 - pp.pdf

A lr^ / 939 ELECTRIC CIRCUITS THEORY AND APPLICATIONS VOLUME I SHORT-CIRCUIT CALCULATIONS AND STEADY-STATE THEORY ELECTRIC CIRCUITS THEORY AND APPLICATIONS VOLUME I SHORT-CIRCUIT CALCULATIONS AND STEADY-STATE THEORY BY 0. G. C. DAHL, S.B., S.M. Associate Professor of Electric Power Transmission, Massachusetts Institute of Technology, Member of the American Institute of Electrical Engineers First Edition Impression Thiiii> McGRAW-HILL BOOK COMPANY, NEW YORK AND LONDON 1928 Inc. Copyright, 1928, by the McGraw-Hill Book Company, Inc. PRINTED IN THE UNITE!) STATES OF AMERICA THE MAPLE PRESS COMPANY, YORK, PA. PREFACE The purpose of this treatise is to present the engineering aspects of circuit theory. Although theoretical, the language and viewIt gives the methods point of the book are those of the engineer. and tools necessary for the analysis of modern power-circuit problems. The book has grown out of the author's experience as a teacher and an electrical engineer. It is primarily intended as a textbook for the course "Electric Circuits" included in the postgraduate curriculum of the Electrical Engineering Department of the Massachusetts Institute of Technology. There has been a long-felt need for such a text. It is hoped, however, that the book also may prove of use to students of similar courses at other institutions, to electrical-engineering students in general, and to many practising electrical engineers. Numerous illustrative examples are worked out in the text. These examples are, wherever practicable, based on actual engineering data and are representative of the type of problems with which the electrical engineer to-day deals. The author wishes to express his appreciation and thanks to Mr. G. H. Arapakis, Instructor in Electrical Engineering at the Massachusetts Institute of Technology, who read the manuscript, offered many valuable suggestions, and who worked out and checked several of the numerical problems. Thanks are also due to Dr. E. A. Guillemin, Instructor in Electrical Engineering, for reading and criticizing parts of the manuscript, to Messrs. E. Bramhall, L. A. Bingham, C. V. Bullen, O. W. Walter, R. B. Wright, and D. S. Young, formerly graduate students of electrical engineering, for calculating the numerical data used in the example in the chapter on synchronous-machine charts, and to Mr, H. F. Goodwin of the engineering staff of Jackson and Mor eland, Engineers, Boston, Mass., for preparing some of the more important drawings. Last, but not least, the writer is indebted to the authors of those technical papers and manufacturers' publications which have served as sources of material O. G. C. Dahl. during the preparation of the manuscript. Cambridge Masb., , July, 1928. v — CONTENTS Page Preface v Chat. I. Calculation of Shout-circuit Currents in Networks ... Simplification of Networks— p. 5; Example 1 'Solution of a Simple Network Problem by the Method of Simplification —p. 7; Short-circuit Current Delivered by a Synchronous Machine p. 12; Decrement Factors p. 18; Application of Decrement Factors p. 22; Solutions by Calculating Table Calculation of the Current which the p. 27; Example 2 — — 1 — — — Circuit Breaker in a Feeder has to Interrupt during Short Circuit p. 29. — II. Transformer Impedance and Equivalent Circuits 34 — General Theory of Multicircuit Transformers p. 34; Twocircuit Transformers p. 38; Three-circuit Transformers p. — — —Calculation of Voltages and Voltage Regulation of a Three-circuit Transformer — 46; Four-circuit Transformers — 49; Determination of Separate Leakage Reactances — 51; Single-phase Tests — 53; Three-phase Tests — 59; Example 2—Calculation of Separate Leakage Reactances from Test Data — 53; Example 3 — Calculation of Separate Leakage Reactances from Test Data — 42; Example 1 p. p. p. p. p. p. p. 66. 111. Unbalanced Circuits 68 Solution by Simultaneous Equations Based on — Ohm's and Kirchhoffs Laws— p. 69; Example 1 Unbalanced Loads on V-connected Transformers— p. 74; Method of Symmetrical Phase Coordinates. General p. 77; Analytical Determination of the Components in a Three-phase System p. 81; Example 2 Calculation of Symmetrical Phase Components — — — — a Three-phase System p. 83; Graphical Determination of Components in a Three-phase System p. 84; Analytical Determination of the Components in Two-phase and Fourphase Systems p. 86; Graphical Determination of the Components in Two-phase and Four-phase Systems p. 87; Power in Unbalanced Circuits p. 89; Copper Losses in Unbalanced Circuits p. 90; Component Voltages in a Threephase System in Terms of Impedance Drops. Equivalent Positive-, Negative-, and Zero-sequence Impedance of an Unbalanced Load p. 92; Degree of Unbalance—p. 94. in — the — — — — vii — CONTENTS vjji _ Chap. IV. Page Tnl «« Transformers with Unbalanced Loading Two-circuit Transformers. ' V. Y-Y Connection without Primary Neutral—p. 101; Y-A Connection—p. 107; Three-circuit Transformers—p. Ill; Y-A-Y Connection— p. 112; Example 1— Calculation of Currents for Various Types of Short Circuit on the Secondary Side of a Y-A-Y-connected Bank of Transformers—p. 125; A-A-Y Connection—p. 127. Induction Motors on Unbalanced Voltages. Single-i-hase 138 Operation. General Discussion— p. 138; Three-phase Induction Motor on Unbalanced Voltages. Analytical Treatment—p. 142; . : Balancing Effect of a Three-phase Induction Motor— p. 144; Example 1— Calculation of Balancing Effect—p. 147; Singlephase Operation of a Three-phase Induction Motor—p. 149; Example 2— Comparison of Three-phase and Single-phase Performance—p. 154; Example 3— Effect of Opening One Line when a Three-phase Induction Motor is Supplying a Constant-torque Load—p. 157; The Single-phase Induction Motor— p. 162; Example 4— Calculation of Performance of a Single-phase Induction Motor—p. 164; Approximate Deter- mination of Performance from Circle Diagram— p. 167. 169 in the Steady State Introduction—p. 169; Capacitance Lumped at Center (Nominal T)— p. 170; Capacitance Lumped at Ends (Nominal H)— p. 171; Capacitance Lumped at Center and at Ends— p. of a 172; Example 1— Comparison of Calculated Performances the on Based arc Solutions the when Line Transmission Short Exact and on the Various Approximate Methods— p. 173. VI. The Short Transmission Line VII. The Long Transmission Line in the Steady State General Solution— p. 179; Line Loaded— p. 181 Line Grounded —p. 183; Line Free— p. 184; Load Impedance Equal to Surge Impedance. The Infinite Line—p. 185; Wave Length— p. 187; A Few Remarks on the Hyperbolic Functions. Some ConHyperbolic, version Formulas—p. 187; Numerical Values of j Functions— p. 191; Example 1— Distribution of Voltage, Current, and Power on an Electrically-long Open Line—p. Calculation of Load Position Angle— p. 198; 194; Example Direct-current Lines— p. 199; Example 3— Distribution of 2— Voltage and Current on a Direct-current Line, v > r«— p. 201; Ground-return Telegraph Line with Terminal Example Relays—p. 202; Composite and Bifurcated Lines—p. 205; 4— Equivalent T and n Circuits—p. 208; Example 5— DeterminaRepresentation of tion of an Equivalent II—p. 212; Equivalent Composite Lines—p. 213; Tests for the Determination of Line Constants — p. 215. 178 315. Transformer Harmonics — 225. The Evans and Sels Chart—p. Example 1 Calculation of Data for and Construction of Salient-pole Generator Charts p. Transformers Represented by Cantilever T-circuits Circuits p. 354. 243. General Circuit Constants Conception of General Circuit Constants 246 — Determinap. Simplified Chart p.— . 219. 273. 261 Example 2 Derivation of Formal Expressions for the General Circuit Constants of a Two-circuit Sectionalized Transmission Line with Sending and Receiving Transformers p. 256. 266. Example Connections of Single-phase Transformers— 2 — Calculation of the Third-harmonic Current Entering a 240. 291 The Modified Evans and Sels Chart 293. Example 2 Calculations of an —p. VIII. lation of the — — — IX. 246. 299. — p. Example 1— Third Harmonic In a Feeder Circuit with Metallic Neutral — Generator Harmonics — 222. Loss Circles—p. 296. Solution of Steady-state Transmission Problems 351 Determination of Synchronous-condenser Capacity p. Transmission-line Charts 265 — Diagram Directly Based on Vector Diagram p. Some Additional Remarks. Example 4 of Third-harmonic Currents Circulating in the Windings of a A-A-A-connected Transformer Bank p. — XII. ix Pao« . p. p. Negative Power p. Example 1 —Calculation of the General Circuit Constants of a Two-circuit High-tension Transmission line with Receiving Transformers p. p. 279. Voltage Circle — Circles — p. Voltage Circles—p. Networks in Series and Multiple p. 216. Efficiency Circles — — — p. — X. 322. Transformers Represented by Equivalent p. — — — — — . 273. Example 1 Graphical Solution of a Transmission Problem Involving the Determination of Synchronous-condenser Capacity at Various Loads p. p. Single-phase Transformers — 225. Loss Circles — p. Three-phase 232. — and a Modified Evans and Sels Chart — p. Salient-pole Generators p. 251 Combination of Networks Usually Met with in Power Transmission—p. 259. Synchronous-machine Charts — — 306 Non-salient-polc Generators p. 286. Efficiency Circles p. CONTENTS Chap. 298. 307. 256. — — p. 351. Example 1 Graphical Solution of a Simple Transmission Problem— 271. Synchronous Motors and — — — Condensers p. 241. 325. Evans and XL Sels . Example 3 — CalcuThree-phase Transmission line — p. Third -harmonic Current in the Tertiary Delta of a Y-A-Y-connected Transformer Bank Supplying a HighCalculation tension Transmission Line p. 247. Two Important tion of General Circuit Constants Relations p. Angular Relations 282. 329. 216 Harmonics The Behavior of Harmonics on a Line p. 294. 263. p. Determination of Regula- — — — . 250. 363. Miscellaneous Examples— p. tion by means of Modified Evans and Sels Charts Example 3— Determination of Load and of Voltage Regulation — with the Tirrill Regulators Controlling the Fields of the — Synchronous Condensers out of Service p. • 379 . 371. Load Centers p. 377. of a Transmission Problem by Charts Two Transmission Lines in Two Synchronous-condenser One Generating and Two Involving Scries. 363. Example 2— Determination of Voltage Regulap.CONTENTS x „ Paok lion— p. 5—Solution Example — 365. Two Generating Stations. Example 6—Discussion of the Solution by Charts of a Transmission Problem Involving Two Trans- — Lines in Series. Determinap. Station. 371. Example 7— Numerical Solution of the Transmission mission Problem Discussed Indkx in Example 6 —p. and Two Load Centers —p. Installations. 363. tion of Load Division between Two Parallel Lines Example 4— Discussion of Three Methods by which the Division of Load between Two Parallel Lines may be Determined in a Specific Case— p. 365. 373. Two Synchronous-condenser Installations. . two problems in network solution with which the electrical engineer has to deal. For a more thorough treatment. 3. and C. following. This is particularly so when is fed by several generating stations and supplies numerous loads. (6) solution by trial and In the error. W. 1 . Feldmann. R. "Analytical Solutions." by Fortescue. "Development of Analytical Solutions." by Woodward. is divided into three parts as follows "Testing with Miniature Networks. Jour. Inc. The methods presented in the last two parts are applicable also under normal operating conditions. (c) solution by simplification of the network. D. R. the first two methods will be only briefly indicated. namely. and power for normal operating conditions and (2) the determination of voltages and currents during short circuits.. 1921. 1 J. 2. Woodruff. the reader is referred to other texts 1 and to the numerous papers 2 on the subject in the technical works a network press. netze in Theorie 2 See. L. for instance: Woodward." Julius Springer. Analytical solutions of complicated alternating-current net- may be extremely laborious. L. John Wiley and Sons. Three methods of attack are available: (a) solution by simultaneous equations. Chap. 344. New York. The method Herzog. "Principles of Electric Power Transmission and Distribution." Elec.. "Calculating This article Short-circuit Currents in Networks. 1925. currents. "Die Berechnung Elektrischer Leitungsund Praxis." by Evans. XIV. in general. Evans. Fortesctte.: SHORT-CIRCUIT CALCULATIONS AND STEADY-STATE THEORY CHAPTER OF CALCULATION I SHORT-CIRCUIT CURRENTS IN NETWORKS There are. (1) the determination of voltages. p. 1. 1919. of solution by simplification will be discussed and C. F. Berlin. Jour. H. If. 694. T..." General Elec. Rev. Richter. K. 486. A. 1405. as it is almost universally applied in practical short-circuit calculations. Upon having established the equations.." pp. calculations are carried out between points at which certain electrical conditions are definitely known. it is necessary to assume values of current or power in one or more branches of the network." Elec. 1924." Elektrotek. 149 Elec. 1927. R.. D. however.... Jour. Rev. . p. If. Z.." Elektrotek. Evans.. If the electrical quantities calculated at the points at which conditions are actually known check with the latter. "Zur zeichnerischen Behandlung beliebiger Leitungs- netze. F. and 207. the known electrical conditions are not checked. 1923. on the other hand. The first method involves the setting up of a system of vector equations by applying Ohm's and Kirchhoff s laws. Z. discrepancies between calculated quantities and known quantities still are present. "The Calculation of Direct-current and Alternatingcurrent Networks. this process is an extremely cumbersome one. 320. cies Thomalen. "Beitrag zur Berechnung elektrischer Leitungsnetze. 391. "Alternating-current Secondary Networks. although the process of may be shortened by the use of determinants. a second adjustment of the assumed values with a subsequent recomputation is necessary. "Evolution of Alternating-current Secondary Networks. p. be in conformance with the following general scheme: Making use of the assumed values. In order to use the second method by which the correct solution is approached by steps. 1921. This sometimes (particularly in simple loop circuits) involves coming back to the starting point.. the unknowns are determined by simultaneous solutions. p. D. a correct solution is obtained. 1923. the original assumptions are obviously in error. Chapman." Elec. This shows. (London). that the original assumptions in regard to values of current or power (or both) are correct. The calculations.ELECTRIC CIRCUITS— THEORY AND APPLICATION 2 more in detail. upon recomputation. It is impossible definitely to outline the procedure to be followed without reference to some specific problem. The discrepanelimination between calculated and known conditions may this time be used as a guide in correcting the initially assumed values of current or power. "The Analytical Solution of Networks. p. W. would as a rule. p. then. If the number of equations is large. Blake. Chrustschoff. 1925. and the calculations repeated.. W. know the short-circuit currents at various points in the network Durso that the proper size of circuit breakers may be selected. It is itself. Julius Springer. W. involving simplification of the network loads. is somewhat limited in its application. in order to enable the engineer to select and currents.Elec." by J. This can always be done even if the network is complicated. the solution for short-circuit conditions is a good deal simpler than for normal conditions. "Calculation of Short-circuit Ground Currents on Threephase Power Networks. The determination of short-circuit currents is a frequently It is necessary to recurring problem in electrical engineering. be obtained after one or two recalculations. Rev. Rev.. The currents which the breakers have to interrupt are usually a good deal smaller than the short-circuit currents which they carry however. The operation of many of the protective relays used in networks today also depends upon short-circuit Hence. ing times of short circuit." Cen. these currents may be extremely high and easily equal to many times the current which the breaker normally handles. 479. 1925. 1926. Even initially.SHORT-CIRCUIT CURRENTS By 3 proceeding in this manner. 1925. So -far. When it is desired to determine voltages and currents during short circuits. a correct solution With some reached. the discussion has been devoted to the solution of networks for normal operating conditions. practice in the is eventually trial-and-error method. generally applicable only when the network is fed by a single generating station and supplies a single load.. independent of whether the network is fed by one or more generating stations and supplies one or more The third method. Lewis. however. however. x Usually. as a rule. The solution by this method is obtained by reducing the network between the generating station and the load to a single equivalent impedance. the circuit breakers involved are called upon to stand the maximum short-circuit current which will flow and to interrupt the current after the elapse of a certain time. S. " Single-phase Short-circuit Calculations. The following papers discuss special short-circuit problems: Bekku. Biebmanns.. Elec. p. 472. Berlin. 1 An so. solutions sufficiently accurate for engineering purposes will. . p. excellent treatise on the general matter of short-circuit currents in networks is "tJberstrome in Hochspannungsanlagen.." Gen. the same general methods may be applied. The two methods just described are general and applicable to any type of network. that the in generator voltages are equal. When the assumption is made. If it were attempted to obtain methods would become exceedand require an undue amount of time. These methods have become standardized. and this is always assumed in the calculations. as already mentioned. leakance. and generators should all be included. the network will. transformers. the methods should be as direct and simple as possible. the short circuit is confined to a single point only at a time..y the least. Since. however. knowledge of the values of the short-circuit currents is necessary. the ingly complicated even to complicated networks. station only.4 ELECTRIC CIRCUITS— THEORY AND APPLICATION install the proper relays and to determine their setting. The network may then be reduced to a single equivalent impedance between the hypothetical bus and the point of short circuit. the impedance of lines. obvious that the method of In order to extend the applicability of this method also to the general case where several stations supply the network. The first assumption which is made. The fact really is that it would be impossible in most cases to obtain rigorous solutions. When a short circuit occurs. and the solution hence obtained by the method of simplification. It has become customary to neglect resistance. It is not essential to obtain the short-circuit currents with extreme accuracy. short-circuit calculations have to be performed frequently. In reducing the network. but the error introduced by this assumption should not be a serious one. is that the load currents may be ignored in comparison with the shortcircuit currents. it is assumed that the induced voltages at the various generating stations are equal both simplification it is would be immediately applicable. Of course. magnitude and phase. They are simple and readily applicable entirely rigorous solutions. this will not be quite true. Experience has shown that the approximate methods which are in common use are sufficiently accurate for engineering purposes.e. It is further assumed that the impedance of the short a short circuit is always considered to be a "dead short circuit" in the true sense of the word. by one generating i. be supplying power to several loads. to su. and capacitance of lines and . If the network on which the short circuit occurs were supplied circuit itself is zero. as a rule. As a rule. in order to facilitate the short-circuit calculations. all generators supplying the system may be assumed connected to a hypothetical bus at which the common voltage is maintained. and Fig. Simplification of Networks. A-impedances.SHORT-CIRCUIT CURRENTS feeders and to consider their reactance only. the and generators is neglected. absorb the same amount of active and reactive jjower. 413.. 2. A. A-F Transformation.— Equivalent Y. . The three methods — of transformation are given below. —Whenever or a three-cornered mesh. Similarly. the conversion formulas are as follows Zi Z31Z12 = Z\2 ZsiZi2 + Z%z + Zzi Z12Z23 Z12Z2Z Z\% z3 = 1 in Kennellt.and A-admittances. a Y it is three impedances form a A possible to convert this circuit into The two or a three-cornered star. Fig. the handling of complex quantities is avoided during the process of simplification. 1 circuits (Fig. In terms of impedances. Usually it is necessary to simplify the network by steps through a repeated application of transformations. -f- E. a fact which evidently reduces the amount of labor in no small degree. 1. assigned to the Since the entire network contains reactances only. 1899. —Equivalent Y. generators will be discussed in more detail below. I) will be equivalent as far as conditions at the terminals are concerned. There are three transformations available which may be used during the process of reducing a network to a single equivalent impedance between two points. XXXIV. World and Eng. Vol. This means that the two circuits will offer identical impedances between any pair of terminals and will. The transformers are represented by their equivalent reactance. p. for the same applied voltage. " The Conducting Networks. their resistance of transformers The reactance exciting currents being ignored.." + 2/23 + _ 2/31 (2) SZ Z23 H~ Zz\ ZjzZ zx Z\2 (1) 2Z Z23Z31 (3) 2Z Equivalence of Triangles and Three-pointed Stars Elec. ELECTRIC CIRCUITS— THEORY AND APPLICATION 6 In terms of admittances (see Fig. the conversion formulas are „ = ^12 ZiZ* ZiZ 2 „ Z/23 + Z7Zz — + Z Z = «-iV + Z Z + Z. The converse theorem does not hold. . " 2/23 y 2/122/23 s- + 2/23 2/232/31 + _ 2/312/12 ^12^23 " + AT = 2/31 ^ ' . 62. possible to convert ajgeneral mesh Star-mesh Transformation. v 2/12 2/12 — Y F-A Transformation. it is not^jn general. He derives the necessary formulas and also shows that the mesh. {V) y" in terms of admittances "2/1 . + "2/1 + -2/i + 2/2 + 2/3 2/22/3 2/2 + 2/3 _ _ 2/32/1 2/2 + become 2/12/2 nm (10) 2/22/3 nn (U) ^ _ - 2/3 . Vol. X -y ^31 . p. into its equivalent mesh circuit circuit into its equivalent star.E. in general. ..Z = N z Z 2 8 „ = Z\Zi The conversion formulas -y~ „ 2/31 = \tS) N ._. cir- In 1 See the paper "A New Network Theorem. 1).e. (5j 2/31 2/232/31 + iV 2/312/ 12 .. l Figure 3 shows a star with n rays and cuit. (London). Jour. In terms of impedances (refer to Fig. become 2/12^23 + 2/232/31 + N 2/312/12 . Mr..-.23 #3^1 + ^2^3 « + 2/i2/2 w 2/12 . If the given circuit is a or threecornered star. 916." by A.E. (7J X 2JT ^ / 2/32/1 x 19 (12) —Any star circuit may be converted independent of the number of rays in the originally given star._. The number of its equivalent mesh impedances in the mesh is ~ (n A — 1). Rosen. . In this paper. i. it may be changed into an equivalent A or threecornered mesh. 2. Rosen proves the general conversion from a star circuit to a mesh circuit. the conversion formulas 2). I. cannot be converted to a star. In terms of admittances. if applied to for the Y-A transformation. much and substations. inclusive. inclusive. EXAMPLE 1 This example illustrates the solution of a simple network problem by the method of simplification. N (IfO J K Since these equations are general. —Equivalent mesh impedances corresponding to a given star circuit. The assumptions. the (7) to (9). the corresponding equations become yn ~ Vmn ~ yjya a/13/3 Vl + Vi + y. they will obviously also hold It is easily shown that. simplified. 3. Statement of Problem Figure 4 shows a single-wire diagram (simplified) of a railway electrifica- tion with trolleys. + 2/2 + • • • • • + v* ~ _ VrnVn • + yn sy VmVn ?y n7 U7. and the general admittance equations to equations (10) to (12).SHORT-CIRCUIT CURRENTS 7 terms of impedances. equations reduce to equations impedance general this case. the general conversion formulas from star to mesh are (13) (14) zmn = zm z n {^ +y+ • • • x + £) = ZmZn ^ (15) z34 Fig. feeders. are as follows: . rail and ground return. 000 copper conductor. and power at the train when the bus voltage at the power station is strictly 44. Compute voltage. -k-'6- I 13' TPosition of Train Trolley-Track System Fig... 00 copper): Resistance 1 = 0.M -J^--.. 7 per cent reactance. current. 4 per cent resistance.404 ohm/loop-mile 1 Values taken from tables in "Handbook for Electrical Engineers. 00 copper conductors spaced ft.. John Wiley and Sons. at 80 per cent power factor when trolley voltage is 11. 1917. Consider train in the position shown and with controller set so that impedances are those corresponding to maximum input. Solution Circuit Constants: High-tension Feeders (No. Rail return. New York. Trains.. f 8m 4/w. consider return to be equal diameter conductor spaced 40 ft.630 ohm/loop-mile Trolley-track System: Resistance 1 = Reactance = 2^741 0." by H. Pender. 4.000 volts. while substations 1 and 2 are fed by two identical feeders.000 volts.-a. Inc. -'- STATION • '-Feeder System ~5ulF 5UB- SUB m.-a.000-volt transmission lines 6 two No. trolley at 1 1. approximately.. approximately. 16' 12'- POWER <.000 volts..5)l0-« (40-^^ V 741 logio = 12 + \ 80.000 kv.822 ohm /loop-mile Reactance 1 = 0. maximum input 6. 25 cycles. These feeders may be considered to be on separate pole lines.ELECTRIC CIRCUITS— THEORY AND APPLICATION 8 All circuits single phase.2 STA. Neglect resistance of In computing trolley inductance. . All 44.326 ohm /loop-mile logi„ j + 80.000 volts. Neglect excitation. —Simplified layout ><--j / !- of a single-phase railway electrification. Trolley No.. Transformers at each substation rated at 8.No. Note that substation 3 is fed by a single transmission feeder.5 j 10- 0. Distribution at 44.000 kv.^. 576 + j'7.8 X X 0.1 0.5 = 24. lent impedance between the high-tension bus of the generating station and the Referring to train.U/37-./10.864 Z ae = Zbe = 3. 9 16.5 8.040 = 19.30/37°. .5 12. I 0.30/37-.04 = 9.__„ = 16.288 + J2..000 T /rated= Impedance = Resistance = = Reactance X 11.822 + j0./2.76 ohms = (0.822 +j0.560 = + .000 11 .5 1.76 + .88/37°.520 ohms = (0./5.70.68 + J16.88 /37°.560 ohms = (0./0. Conversion of abc = 9. the network in Fig.000 1Q1 18L8c am P^QQQ 44.000 KAa = 546am P- 1P1 . 181.080 ohms = Zbe = Z fg = 9. to .288 Zab SZ = 6. 4 and to Fig.630)4 = 3.6 = = ohms 20.11 12. equivaSimplification of the Network.630)16 = 13.64.576 + .8 0. impedances of the various sections become series The two by the dotted impedances wherever possible.0.630)8 = 6.07 .14 20.44/37°. Resistance X 1.5 4./7.842 . Fig.000 ao ohms —io X 0.404)4 2 X 10 = 52.040 ohms = (0./1.522 — 1. 56 is A 's abc and beh will now be converted to Y's.992 ohms = (0.5 4. all impedances will be referred to the high-tension side.09 ohms ohms — In reducing the system to a single./2.630)12 = 9.68 = — _ Referred = 8.152 + .5 = 12.326 + j0.32 +j38.60 ohms Zab Z ae Zbe Zbf Zcd Z de Zeh Z gh By adding obtained.520 = + J5. /rated -o .326 + j'0.5 Za % X 12.1*/3r. — 44.5 + .5 _ 24. high-ten- sionside .822 +. 120 4.14/37°.88/37°.000 - X .326 + .096 +.14 = /cos.44/3r.728 +J15.8 „ Reactance = . 14/36°.92 ohms 1./51.30/37°.644 8. g ?? 20.— : SHORT-CIRCUIT CURRENTS 9 Substation Transformers .404)4 2 X 8 = 41. 5a.88/37°.864 +.5 = 3..68 ohms = (0.822 + j'0.5 _ 24.000 } - . .404)4 2 X 6 = 31.44/37°.20 + . the = (0. Train: 6. as indicated lines.5 Zbi X „ ' = Zc _ i X 8. .261 - 24.288 + . 152+^10. impedance between the power station and the .32+j38.040 b f 13.92 OS di 41.08 9.76+JSI.68 +J 16.76-~^ZG5tj4.*s (f)£ »£ "S J ^ ^ A > •». 5. 4 to a single in the reduction of the circuit in train. ^ «6 5?' j ^ •^ & > *M <5S <\i I h ' Fig.32+J 38.76 (a) Train-- 3l. — Diagrams showing successive steps Fig.46 (b) .68 3/.ELECTRIC CIRCUITS— THEORY AND APPLICATION 10 rSMion Bv i aS76+/S. 7 2 X 16.34 187.60 6.2/51°.6 /50°.60 = + j38./36.7 .2/51^8 118.882 =» + j2.14/36^9 9.3 X 92.03 +^147.000/0 _ = V.7 16. = 2.00 /38^6 = 477-7\38°.11 X 10" 3 = 3.03 + . 5e).7 49.6 X 20.11 +.12 = = 92.11 + .77 /51°.6 amp. The equivalent impedance between the high-tension bus of the generating station and the train is.ll Substituting this impedance for the two parallel branches gives the circuit (Fig. - "l0"8. Z = 30.90 9.19 H-jll. ./24.2 49.28 + J10.7 *w 19.02 = 108. Y 23. By addition of the series impedances.00/38^6 ohms Train Current: Tt ~ 11.992 16.2 64.55/50^0 = 20.27/53°.SHORT-CIRCUIT CURRENTS 11 Conversion of beh Zbe = Zbh ~ Zrt - 31. hence.882 at Train. -^ 13 80/ ' = 8.2 118.6 /50°.7/38°.l 19. 2/53°.32 SZ = 116.2 + .68 75. and +/36 56 = ' Power 1.76 = X 19.77/51°^ = _ 31. 16.620/fV7 volts Train Power: Pt - 477.2 /53°./16.! = 187. the final circuit / is obtained.56 ohms which referred to the low-tension side becomes a0 11 Z - - Current.17 +J73.7/53°j ODrt/MO/. Voltage.7 /53°.680 kw.284 + + j'14.2.8 X 49.92 = + J91.50/60°./87. Train Voltage: V = t ItZ t = = 477.80 +.27/53°.19 „ .77 /51°.28 = ft.46 Using these equivalent Y circuits gives the network immediately reduces to the series-parallel circuit shown in Z i(ct)i = ZiWi = 2Z = 55.09 = 23. 5c) which d.284 ohms —The total impedance Z referred to the low-tension side becomes Zo = Z +Z = t 1.374 17./12. 9.50 /60°.63 +J4.50 /60°.63 +J13.2 Z« (Fig. + .6/50^7 " 12-35/59J irT^ra X 118. there is a which the current changes gradually transition period during a final value (the sustained or steady-state current). ^WW The during the s^. Short-circuit Current Delivered The currents which flow when a — short circuit occurs at a point in depend largely upon the general character of the short-circuit currents in the synchronous machines. from an initial to In general.constant coefficients. 6) of fixed and inductance at one end of which a constant sinusoidal voltage is maintained. not be established immediately. and the latter may represents the complementary + (20) is the particular integral. quantities. 6. When a constant sinusoidal voltage is impressed on a constant impedance.— A short circuit is suddenly applied to the lumped impedance by closing the switch S. the sustained short-circuit current will. x~>oor\ o^ transition quite Q) e=Em sin(wt+9) current first flowing instants of the may period easily be considerably greater than the sustained current. it is helpful first to . When a a network will short circuit applied. this is + Ri = Em sin (at + sum of first order with be expressed as Referring to equation (20). is as a rule. review the case where a short- suddenly applied to a lumped impedance (Fig. 2 (i„) is 1 See any standard treatise on differential equations.12 ELECTRIC CIRCUITS— THEORY AND APPLICATION by a Synchronous Machine. i function and is = it it (i t ) is the current. The differential equation applying to this circuit circuit is resistance is up as readily set follows Lj As seen. (19) a linear differential equation of the . 2 The treatise. In order to get an understanding of the character of the short-circuit Fiq. the solution of which the 9) general problem of transients is treated in detail in Vol. 1 the solution of equation (19). This is a condition which would obtain in a synchronous machine. II of this . the former In terms of electrical equal to the transient component of equal to the steady-state current. which is two terms. It will be assumed that the reactance part of this impedance is predominant and that the resistance is quite small. by synchronous currents delivered mac hines. it is easy to obtain a rigorous mathematical solution for the current. the total current consists of a sinusoidally varying component superimposed on a component which decreases exponentially. The amplitude of the steady-state current depends only on the value of the applied voltage and the steady-state impedance of the circuit. i. be readily established rent steady-state theory. it is seen that the . in addition to depending on voltage and impedance. + *.SHORT-CIRCUIT CURRENTS It is well the first known that. hence.. the determinantal determinantal equation obtained derivatives to zero. or steady-state current. this constant becomes y—^-r-— A = and the i . tan. by equating the polynomial of In this particular case.ta"'i) determined by reference to the Assuming zero -initial initial conditions. (23) current.. Assuming that the resistance is small. by applying the ordinary alternating-curThe complete solution may. the exponential. it depends on the angle 6. so that the angle of lag of the steady-state current behind the voltage is nearly 90 deg. The amplitude of the transient. equation becomes Lp where p represents the +R = (21) operator.e. written i = ^ The constant vTO + of integration A 8in is (". hence. on the other hand. differential The exponent. also depends on the point on the voltage wave at which the switch is closed.1 ^ VflM^wL) (d - tan-' °£) (24) given by final solution is ~Em mi(d= sin ^ '- ' e"~^'+ 2 Em V# 2 +(coL) 2 sin (<d+d\ tan" 1 ~\ R (25) As seen. becomes V The is = -\ (22) solution for the particular integral. when the linear 13 differential equation is of complementary function consists of a single The exponent is found by solving the so-called order. . former. The curve shows the steady-state current which will be set up immediately in the circuit in Fig. "Reactance of Synchronous Machines and its Applications.e. the character of the transients produced by a short (polyphase or single-phase) applied at the machine Doherty..). Shirley. R. p. sinusoidal current. on the other hand.. or 180 deg. circuit 1 . 6 when the switch is closed as the applied voltage passes through one of its maxima. 1209.14 ELECTRIC CIRCUITS—THEORY AND APPLICATION transient will entirely disappear when the switch is closed (i. 6 Mathe- matically this curve consists of a sinusoid superimposed on an exponential.E. If the reactance of the synchronous machine 1 were a constant quantity. — Fig. the maximum possible total current will be reached after the elapse of a time corresponding to half a cycle. 7. Figures 7 and 8 illustrate the currents in the two cases. 8. in Fig. the steady-state current is set up immediately. E. In the In the the wave of total current is a completely "offset" wave obtained by adding the exponential transient and the steady-state latter. wave passes through one of its short circuit occurs) as the voltage maximum the switch values (0 = 90 or 270 deg.E. 1918. In this case.I. the initial amplitude of the transient is equal and opposite to the maximum amplitude of the steady-state current. and O. A." Trans. — Fig. If the short circuit occurs at values of voltage intermediate between zero and maximum. The curve shows the current which will flow in the circuit when the switch is closed as the applied voltage passes through zero. E. therefore.). a "partly offset" curve of total current evidently results. is closed on zero voltage (0 = If. It is furthermore seen that the amplitudes decrease as time elapses. A. E. Single- phase Short Circuits. in machines without damper wind- equal to the sum of the leakage reactance of the armature and the leakage reactance of the field. . is not known. As seen. — Symmetrical short-circuit current delivered by circuit applied as the voltage wave passes through a 9. A. this will not give rise to appreciable errors 1 Probably the most up-to-date and rigorous treatment of short-circuit currents in synchronous machines is given in Part IV of a recent paper by R. therefore. Since the flux in the magnetic circuit of the machine cannot change abruptly when the short circuit is applied. 1928. armature leakage reactance may be used in place of the transient reactance. This decrease is caused by the fact that the reactance increases from a low initial value to a much higher value Fig.E. with the exception that. eddy currents set up in the field structure will also affect the value of the tranings. If damper windings are present. an alternator. the sient reactance. If the exact value of the transient reactance.. On account of the change in reactance. This is due to the inductive coupling between the field and armature circuits. for the steady state. Furthermore. mathematically. however. their effect must be taken into account and the value of the transient reactance modified. 1 When a short circuit is applied at such an instant that the transient disappears. As a rule.SHORT-CIRCUIT CURRENTS 15 terminals would be as discussed above. 9. maximum Short value. the total current will have the shape illustrated in Fig. matters become more complicated. the transients would consist of two exponentials instead of merely one. In general. Doherty and C. This total leakage reactance is usually called the transient reactance and is. and a rigorous mathematical solution is not readily obtained.E. this curve is not offset from the zero axis and may be looked upon as being symmetrical with respect to the latter. the initial current will be limited by the total leakage reactance of the machine. IV.I." Jour. Nickle: "Synchronous Machines. the armature leakage reactance will constitute the major part of the transient reactance. Fig. and the net The effect of this is flux in the magnetic circuit decreases. direction. 11. The reactance finally becomes equal to the synchronous reactand steady-state conditions are obtained. be dissymmetrical. the discrepancies. will be in a conservative which is slightly too low is used. This curve is completely offset and is obtained when the short circuit is applied as the voltage wave passes through zero. as shown in . partially offset wave will result. If the short circuit is applied in such a manner that the tran- ance. hence. 11. This curve of a is is symmetrical wave of the type shown in Fig. As time elapses.ELECTRIC CIRCUITS— THEORY AND APPLICATION 16 except. The total wave may be looked upon as consisting sient or exponential terms wave will Fig. a Fig. do not disappear. not completely offset and is obtained when the short circuit applied when the voltage has a value between zero and maximum. 10. perhaps. as illustrated in Fig. — Dissymmetrical short-circuit current delivered by an alternator. in salient-pole Furthermore. any. the total current be offset from the horizontal axis and. since a reactance equivalent to a gradual increase in the reactance of the machine. short circuit occurs when the voltage has a value between zero and maximum. if machines without dampers.— Dissymmetrical short-circuit current delivered by an alternator. a comIf the pletely offset wave is obtained. 10. 9 superimposed on the transient components. the armature reaction builds up. If the short circuit is applied as the voltage of the phase involved passes through zero. These curves are alternating-current component. which decreases with time and eventually disappears. in Fig. By halving the vertical distance between the curves of positive and negative crests. — This diagram shows how the direct-current and the alternating-current component in a dissymmetrical wave of short-circuit current are determined. When the total wave of current. looked upon as being R. or Effective Value Curve ofAlternating-Current Component ofPositive Crests Curve of Negative Crests' Fig. 13. or Effective Value of Wave of To fa/ Current Direct. 12. is known.S. negative crests. M. as indicated in Fig.Current. termed curves of 'positive crests and 12. and also shown separately in Fig. S. 13. as shown. the direct-current component for any value of time may be determined by drawing smooth curves through the maximum positive and negative amplitudes. for instance. respectively. The total dissymmetrical wave of current is made up of a direct-current component. 12. 12.. and an the amplitude of which also decreases as time elapses but which finally reaches a steady state. — Direct-current component in the current wave in Fig. M. the amplitude of the direct-current component is obtained. Each loop of the total current when considered with respect to the direct-current component is then assumed to Obviously this represent one half-cycle of a sinusoidal current. Direct Component Time Fig.SHORT-CIRCUIT CURRENTS Although the complete story of a short-circuit current only when the 17 is actual instantaneous values of current are at various times. it given known has become standardized practice in practical short-circuit calculations to use effective or root-mean-square values exclusively. .Component Wave of Total Current R. 12. 14. the effective value of the total current is equal to the square root of the sum of the squares of the effective values of the compo- The combination is. where a curve of the effective value also has been drawn in. 9.ELECTRIC CIRCUITS— THEORY AND APPLICATION 18 is The effective value of each of these halfby dividing the maximum amplitudes with the direct-current component by \/2. (28) character of the short-circuit cur- by a synchronous machine when a terminals has been briefly described The short-circuit currents which will flow in a network which one or more generators feed will be of the same general On account of the impedance which the network itself type. carried out in accordance with the following equation nents. the direct-current given by h. for instance. Both instantaneous and component with the effective value of the alternat- ing-current component for the various values of time. 12.„ = AB (26) and the effective value of the alternating-current component by /.+ Decrement Factors. In order to obtain data which might be applicable to general short-circuit calculations. It is likewise shown in Fig. ™ = (27) The alternating-current component plotted alone will give a curve of the type shown in Fig. 14. direct-current ofAlternating Current in the wave of short-circuit current in effective values are shown. S. —Alternating-current component Fig. —The rent which will be delivered short circuit is applied at its H*. offers. hence... cycles is calculated respect to the first maximum component is in Fig. The effective value of the total current is obtained by combining the Alternating Component R. Uu = VIL. these short-circuit currents will obviously be lower than if the short circuit were applied at the terminals of the machines. into . or Effective Value Fig. Obviously. M. Considering an approximation. extensive tests have been performed by above. 38 3.98 3.14 3.00 3.17 1.85 7.08 0.28 1.85 4. 0. and.48 3.94 4.06 5.00 5.21 2.00 0.83 3.28 0.90 1.00 1.79 3.01 6.90 2.— SHORT-CIRCUIT CURRENTS the manufacturers.48 1.981 4.94 2.877 3.99 3.27 2.30 0.42 2.44 1. 3.60 3.18 18.43 2.15 16.05 0.17 1.04 1.64 1.50 8.e.28 1.01 1.52 2.60 11.24 2.70 1. for higher values.02 1.66 1. It has been previously Table Time from I. System Short-circuit Current Factors Applicable Three-phase Short Circuits on Three-phase Systems Root-mean-square total current expressed in number of times full-load current for various per cent reactance instant of short 5 per seconds cent circuit.32 5. i.54 2.05 0. The external reactance was varied so that a range of total reactance.53 2.15 4.52 2.820 .54 1.15 0.43 2.75 2.50 5.11 0.74 7. Line-to-neutral short circuit.15 1.90 10.40 8.843 2.67 2.98 2.35 2.32 1.19 1.10 9.00 8 per cent 10 per cent 12 per cent 22.00 15 20 30 40 50 60 75 per per per per per per per cent cent cent cent cent cent cent 100 per cent 12.64 2.38 1.81 7.10 0.11 1.55 2.03 0.20 4.63 1.00 14.65 10.11 6. Symmetrical three-phase short circuit.00 17.89 2.00 21.22 4. Reactances up to 15 per cent were inside the machines.13 2.56 8.10 2. The short circuits were applied directly at the terminals of the machines and also with external reactance between the machines and the short circuit..51 1.00 9.22 0.10 6.13 2.75 11. Single-phase line-to-line short circuit.61 1.943 4.25 1.888 3.904 3.70 8.34 1. 2.82 3.50 2.35 3. namely.862 2.04 2.86 7.42 1.72 1.56 6.41 2.19 3.39 1.40 0. These tests comprised the determination of short-circuit currents delivered by machines of various ratings and designs for three general types of short circuit.23 1.23 6.28 2.919 4.08 0.35 1.20 0.45 3.36 1.66 5.25 0.30 5. external reactance plus machine reactance.80 5.52 6.16 0.86 1. 1.81 1.27 1.55 2.96 4.72 1.83 1.40 1.48 2.84 3.89 3.00 4.56 2.56 1.48 11.15 1.54 9.00 35.71 1.15 7. 15 per cent were inside the machines and the remainder external to the machines.00 1.00 0.00 13.02 0.18 1.52 3.836 2.50 13.87 2.33 7.45 1.55 1.03 1.828 2.42 1. between a very low value and 100 per cent.50 0.13 0. The excitation used in all tests was that corresponding to full load at 80 per cent power factor (lagging).72 5.90 9.70 1.74 5.91 2.47 1.30 4.70 2.88 1.72 2.31 1.55 4.34 2.03 1.39 1. particularly by the General Electric 19 Company and the Westinghouse Electric and Manufacturing Company.50 2.74 2.49 1.79 1.36 6. was covered.18 0.78 1.94 1.61 1. 83 1.70 1.15 0.69 5.00 9.00 2.45 8.16 4.24 3.84 3. were compared by the manufacturers responsible for the tests.00 6.40 0.87 1.19 1.95 1.99 2.44 1.18 7.00 18.84 2.86 2.50 0.48 4.— ELECTRIC CIRCUITS— THEORY AND APPLICATION 20 stated that the magnitude of the short-circuit current depends largely on the value of voltage at the instant the short circuit In order to be on the conservative side.00 3.00 0.70 2. to figure with the maximum possible values.73 5.43 2. The tests results.88 9.67 2.07 3.57 3.26 0.35 2.55 6.80 7.79 4.54 3.52 3.56 1.49 1.01 5.80 14. and a set of values selected which was considered representative of the short-circuit currents which machines of average design would deliver.00 5 per cent 8 per cent 35.45 5.79 2.82 3. therefore.16 2.27 3.75 6.73 2.47 1.36 1.09 6.21 0.62 3. the short circuit was applied in such a manner that maximum possible Oscillograms of this current was obtained in at least one phase.66 10.86 2.30 8.29 1.43 7.12 2.94 1.23 1.15 1.60 1.98 1.08 0.74 1.71 4.36 11. however.02 16.67 1.89 6.66 1.20 0.01 1.66 3.25 0.35 0.15 2.51 1.25 2.68 1.40 3.30 3.81 3.46 4.30 0.00 22.41 1. current were taken and analyzed for their root-mean-square In this values of current in the manner previously described. it was possible to plot a series of curves values short-circuit of current versus time different for values of reactance.48 3.49 2.87 5.42 4.53 12.23 3. averaged up.18 0.90 1.32 12.51 2. seconds 0.40 1.18 7.00 14.95 2.01 6.20 0.35 3.08 3.15 8.84 1.40 2.29 0.04 4.08 7.67 1.45 1.15 5.13 0.16 0.30 4.17 3.48 2.54 1.line Short-circuits on Three-phase Systems Table Time from Root-mean-square total current expressed in number of times full-load current for various per cent reactance instant of short circuit.69 8.02 21.10 0.31 2.00 12.89 1. These data in the System Short-circuit Current Factors Applicable to II.50 2.90 2.95 7.76 5.61 3.82 11.05 0.31 5.92 1. Single-phase Line-to.05 1.15 5.69 3.25 1.00 4.77 2.00 10 per 12 per cent cent 40 50 60 75 100 20 30 15 per per per per per per per per cent cent cent cent cent cent cent cent 17.51 1.50 3.93 11.64 2.46 9.08 1.92 3.70 2.19 5. Different types somewhat and sizes of machines would obviously give different values of short-circuit current.89 3.73 2.62 5.00 3.36 1.00 1.10 13.15 0.73 3.14 0.80 1.24 0.95 2.82 2.84 2.90 9.56 5. whether symmetrical three-phase or single-phase.85 6.96 6. it is advisable occurs.00 7.92 9. In all these tests.63 4.00 4.47 1.07 4.41 3.10 1. root-mean-square of way.38 3.11 1.12 5.86 1.90 1.06 1.20 4.50 4.82 1.98 1.81 .01 6.21 3. 87 17.38 4.76 3.27 6.30 14.60 9.28 1.91 2.48 15.95 3.83 1.40 1.28 2.23 1.85 8. the data are immediately made applicable to all sizes of machines.50 2.09 6. It should be carefully noted that Table III of decrement factors for single-phase line-to-neutral short circuits is of limited appli- Table III.85 4.22 5.81 2.60 8.07 2.nt on p.90 10. 1926.60 9.29 1.00 3.06 10.49 4. and the paper "The Application of Decrement Factors in Short-circuit Studies.50 0.99 3. 1 Instead of giving actual currents in amperes and reactances in ohms.47 1." published by the National Electric Light Association.84 1.60 2. the manufacturers' publications contain this information.53 5.64 1. The factors which indicate the number of times rated current which will flow are called decrement factors.87 2.25 0.45 1. the former are given in terms of rated current of the machine involved and the latter in per cent based on the rating.85 3.18 3.50 6.85 1.00 5.34 1.00 icabilit.60 7.00 21.98 1.40 33.10 8.87 1.71 7.80 12.17 6.90 21.01 4.58 2. / of the factors in this 1 able.67 5. Tables I.46 6.95 2.38 1.32 10. Elec.50 18.20 2.80 2. System Shobt-circtjit Current Factors Applicable to Single-phase Line-to-neutral Short-circuits on Three-phase Systems 1 Time from Root-mean-square total current expressed in number of times .00 3.35 7. New York. and III give the decrement factors for symmetrical three-phase short circuits.85 1.94 2.10 1.00 1.41 1.20 7.10 10. p.10 9.89 6.54 1.91 1.01 4.10 22. 0.52 1.25 4. II.24 1.05 5.70 14.49 1.90 15.10 11.10 5.80 11.52 2.05 12.73 2.19 8.70 16.96 2.08 0.85 1.42 6. See also "Relay Handbook. 213. 1 1924.91 2.98 5.00 4.41 1.46 3.89 1.85 3.28 6. single-phase line-to-line short circuits.10 19.instant full-load current for various per cent reactance of short 5 per seconds cent circuit.51 3.20 14.— SHORT-CIRCUIT CURRENTS 21 form of tables and curves are now available for the engineer who has to perform calculations of short-circuit currents in networks.10 0.26 1.86 1. Woodward.00 Note 1 8 per cent 52.90 33.25 28.29 3.70 1.80 12.30 1.46 1.56 2.23 1.19 6.23 7.25 1.44 1.80 11.10 the limited Some app 10 12 15 per cent per cent per cent 20 30 40 50 60 75 100 per per per per per per per cent cent cent cent cent cent cent 27.78 2.40 1.39 3.77 5. 22.45 1.50 9.35 5.58 8.35 8.83 8.15 0.67 5.83 1.30 0.10 7.80 7.07 6.11 5.20 0.80 2. and single-phase line-to-neutral short circuits.61 5.42 2.06 12.10 12.40 18.65 4.66 2.75 5.50 8.95 10.30 10.05 0.02 5. R.96 7.14 7.42 1.10 17. respectively.01 2.43 5.89 12.03 5.40 25.80 7.55 2. In this manner.02 4.63 5.68 2.52 9. of .05 3.30 12.65 3.84 2.84 1.86 2.93 2.10 3.10 13.59 3.20 1.67 3. Jour.22 1.40 0.70 1.00 0.88 6. See s tatemc.05 5." by W. 7. All reactance values up to and including 15 per cent are inside the generators and. as already stated. The excitation of the generators corresponds to full load 80 per cent power factor 5. 2." loc. and the decrement factor selected. preferably on a base corresponding to the kilovolt-ampere rating of the machine. Woodward. in Short- . The short circuit wave giving maximum 6.ELECTRIC CIRCUITS— THEORY AND APPLICATION 22 cation in practice. they are both expressed in per cent. making use of the experimentally determined decrement factors involve.: cit. R. the prothe after flows which current If an external reactance and 1 cedure will obviously be as follows: The total reactance to the point of short circuit is first determined. is at (lagging). may appropriately be summarized. There are no voltage regulators. — Practical short-circuit of Decrement Factors. There are but few systems in operation which would fully conform with these specifications as to connections and grounding. The impedance at the point of short circuit is zero. a single synchronous machine supplies a short circuit through it is desired to determine the amount of elapse of a certain time (t Q )." Iqc. This reactance equals the sum of the external reactance and the transient reactance of the machine. cit. 3. "The Application of Decrement Factors See also "Relay Handbook. established at the point of the voltage possible instantaneous current. 15 per cent inside the generators and the remainder external to the generators. In adding these reactances. these 1. and capacitance is neglected. The proper table of decrement factors is entered with the percentage total reactance and the time at which the short-circuit The rated current is desired. 4. several assumptions. as follows: Transient characteristics of generators of normal design. leakance. for higher values. circuit Studies. W. The effect of resistance. equal then is factor current of the machine times the decrement which current to the root-mean-square value of the short-circuit will flow at 1 the elapse of the specified time. It can be used only with Y-connected when (a) the neutral points of all generators are solidly grounded and distribution takes place at generator voltage or (b) all transformer connections are Y-Y generators and then only with secondary neutral points solidly grounded and with primary and generator neutral points interconnected. At Application calculations this point. After a while these machines will slow down and eventually come to rest. supply current to the short circuit as if they were operating as generators. the ratio of the current which actually flows and the rated current must obviously equal the inverse ratio of the actual reactance and the Substation Bus 100 per cent reactance. Owing to their inertia. la where 7 represents the and X = /rated initial X 100 (29) Xn symmetrical short-circuit current the total reactance to the point of short circuit. they will continue to run at synchronous speed for a short time and will. during this period. Rointof When a short circuit occurs on Short Circuit a network. will the synchronous motors or condensers contribute to the sustained short-circuit this happens. therefore. . In order to illustrate these methods more fully. in one or more of the stations or at other points of the system. This layout involves two generating 1 If. there are large synchronous motors or condensers operating. the layout shown in Fig. the equation is correct may be appreQ That ciated from the following consideration: If the reactance were 100 per cent. factors in such cases. Either a single decrement factor may be applied to the synchronous machines lumped. or else a separate decrement factor may be applied to each station. Much before however. as a rule. — Simple two-station generating stations. this can be obtained by dividing the rated current by the percentage reactance and multiplying by 100. their own circuit breakers may have opened and disconnected them from the system. Hence. In no event. currents.SHORT-CIRCUIT CURRENTS If it is desired to determine the initial 23 symmetrical value of the short-circuit current. l There are two system used in discussing applica- methods of applying the decrement tion of decrement factors. assuming that the internal voltage of the machine has its rated value. the short-circuit current will. When the reactance is different from 100 per cent and the internal voltage normal. be supplied from several Fig. evidently rated current would flow. these machines may be treated as generators for the first few seconds after the occurrence of the short circuit. 15 will be considered. 15. Figure 16 shows a diagram of the reactances involved. supplies power to the substation over a high-tension transmission network. decrement factors used. since exactly the same methods apply when any number of stations supply the In Fig. and the reactance df the feeder between the substation bus and the point of short circuit. resent the three banks of transformers. or be selected arbitrarily. cd. 15. The network reactances are next combined suitable size so that the circuit is reduced to a single reactance between the hypothetical bus and the point of short circuit. and ed repstations. AB Fig. as shown in Fig. These reactances should kilovolt-ampere base. — Reactance diagram of the in Fig. 17. layout involves several generating stations. Decrement Factors to the System be assumed that the reactances have been cal- Application of It will . supplies power to the latter over a bank of line step-up transformers. 15. with step-up and step-down transformers. — 1. 15. 17. 16. Both stations are connected to a common hypothetical bus at which the is assumed to be The reactances Aa and Be represent the transient reactances of the two The reactances ab.ELECTRIC CIRCUITS— THEORY AND APPLICATION 24 stations only but serves to illustrate the principles. as the combined When the rating of the two stations. a short circuit occurs on a feeder a certain Generating station A distance from the bus of a substation. the reactance be the transmission line. This process of reduction must always be gone through independent of the method of application of Method Lumped. an arbitrary base of is most commonly used. Station B. Fig. system all The be expressed in per cent on a common selected kilovolt-ampere base may be taken as the rating of either generating station. normal voltage of the short-circuited feeder maintained. — Equivalent reactance of the system in Fig. located near the substation. amount which each station supplies next determined by properly apportioning the total current (7 ) between the stations. Method now is and the decrement factor Application of ^ Aerated (31) Decrement Factors to Each Generat- ing Station Separately. All stations contribute their share to this short-circuit current. If portional to the kilovolt-ampere base. . it is necessary first to calculate the initial symmetrical current in the short circuit by equation (29). are given by 7T0(A) — _/T is X ____ Bd m XAd Making use of the initial symmetrical short-circuit currents. The X . the per cent total reactance must be converted to a base equal to the combined capacity of the two stations. in this case. the per cent reactance is directly pro. Let the total reactance on this X base be a single decrement factor is to be applied to the two stations combined.an equivalent reactance between each station and the point of short circuit is calculated as follows: Z7ratedU) o(4) = y= X 100 X X7rated(B) o(B) = 100 0(A<\ y i0(B) (34) (35) . When the voltage is unchanged. (k) entered with the reactance corresponding to the desired short-circuit current is which will flow at then given by la. respectively. let the total reactance calculated on some arbitrary base be In order to determine the proper decrement factor which should be used for each station. Hence.SHORT-CIRCUIT CURRENTS 25 culated on an arbitrary kilovolt-ampere base different from the rating of either station and also different from the combined ratings of the stations.c. the per cent reactance based on the total rating of the machines is given by yr _ y Rating of mach ines Base kilovolt-ampere Y A +B The proper X table or curve time (to) selected.—Also. The the elapse of the time to 2. the initial symmetrical shortcircuit currents supplied by the stations A and B. In the particular case under discussion. Neither method is rigorous even if the general assumptions on which these short-circuit calculations rest are disregarded.o. in general.o. -Zs. is A suitable scheme suggested in Table IV.(B) ~ = #A-*raled(X) V^°/ ^B-tratedCB) («•) The total current flowing in the short circuit is obviously given by the sum of the currents supplied by each machine. the first one in which the decrements The second are applied to the system lumped is the easier one. Short-circuit Calculations Amperes at normal voltage At time = equivalent Station Rated Initial symmetrical to seconds Per cent reactance Decrement Shortcircuit factor amperes A B Evidently. The question then arises which one is preferable in a practical case.o. particularly when the system involves several generating stations.— 26 ELECTRIC CIRCUITS— THEORY AND APPLICATION Since these reactances are computed by means of the rated currents of the machines.o.(B) In carrying out calculations of this type. be entered with these reactances and a separate decrement factor selected for each The short-circuit current which each station at the desired time. one involves a good deal more labor.(A) la. it is convenient. to arrange the calculations in tabular form. = ia. of this time is then obtained by elapse the after station delivers of multiplying the rated current of that particular station by its decrement factor. . of the two methods described above for the application of decrement factors. therefore.c. the short-circuit currents at the time are given t by is. have widely Hence. Table IV. they are obtained on the kilovoltampere base of the respective machines. different decrements is not properly taken into account.(A) + (38) -^B. Letting kA and hB represent the decrement factors of the two stations. Hence. When the single decrement factor is applied to all stations combined. the fact that large and small stations. The tables or curves decrement factors may. p. the first method is the easier and shorter one. "A New Short-circuit Calculating Table." Trans. the voltage drops in the various branches of the circuit may not balance up at all values of time. World ' Schurig. Since.. 797. p. Corbett. D. two methods give results that are Ordinarily. R." Gen. " Calculation of Short-circuit Currents in Alternatingcurrent Systems. for instance. the very close together. p. 1919.. — Solutions by Calculating Table. R. the first method will give slightly higher values of currents than the second. 1922. 1920. cit. In order to facilitate determination of short-circuit currents." Gen.. Elec. W. as already stated. Inaccuracies may consequently be anticipated also with this method.SHORT-CIRCUIT CURRENTS inaccuracies are obviously introduced If. World. W. In spite of the simplifying assumptions. "A Short-circuit Calculating Table. which is the case whenever new extensions and load additions to a system or installations of new circuit breakers and additional equipment are planned. From this standpoint.. 1923. it is also the one which is most frequently used in practice. W. on the other hand.. as a rule.E. This is obviously a great disadvantage where a frequent check-up on short-circuit currents is necessary. W. is It is difficult to predict. particularly at small values of time. Dillard. W. p. "Calculating Part I of this article (byWoodward) discusses testing with miniature networks. O. applied to each station. which method will give the be said that. Rev. in general. Elec. This may be seen. Although this will be so in most cases. calculating tables 1 have been used quite extensively. "A Short-circuit Calculating Table.. Fortescue. 140.I. in Electric . by comparing the results in Example 2 obtained by computations based on both methods. loc.E. Rev. 669. R. It may tive results are obtained. 1923.. a separate factor 27 by lumping the stations. Woodward." and C. L. J. 985. however.. Lewis. E.. 10. currents calculated by the second method may actually be larger than those calculated by the first method." Elec. not be universally true. p. Evans. These 1 Lewis. calculations of short-circuit currents in complicated networks require considerable time and labor. "Experimental Determination of Short-circuit Currents Power Networks. W." Elec. the former is preferable in that conservabetter results. In cases where a large station of low reactance is located in the immediate neighborhood of the it may point of short circuit. A. Short-circuit Currents in Networks. L. as in the ordinary analytical solution. circuit calculations reactances only are considered. reactances. At points where generating stations feed into the actual network. the If values initial The equivalent reactance however. By a convenient plug board or dial arrangement. (or reactances). the generator resistances are adjusted ured. as a rule. Sometimes the calculating table is permanently set up to This is the simplest and involves represent a specific system. Ordinarily. it is obviously possible to obtain values of current for any number of simultaneous short circuits. useful in cases where determination of short-circuit Some of the currents is not confined to any specific system. use calculating tables for short-circuit determination have tables As their systems are extended. with which the decrement tables are entered can be determined by means of the values of current read on the calculating table. interval. sustained short-circuit It is thus seen that the calculating currents will be obtained. a'short circuit occurs at one point only at a time. used. decrement factors must be applied. Such calculating tables are flexible and. the proper direct-current voltages are impressed on the miniature system. When a calculating table of this type is used. short-circuit currents or else sustained the generator to represent used resistance upon the values of corresponding to transient reactances are symmetrical short-circuit currents will be measIf. to correspond to synchronous reactances. necessary fixed resistances on the calculating table in order to keep the miniature system up to date. hence. a The currents single short circuit is all that has to be considered. on the other hand. however.28 ELECTRIC CIRCUITS— THEORY AND APPLICATION consist of a combination of resistances connected so as to repre- Since in practical shortsent the network under consideration. if this is desired. . The currents are then simply recorded by direct-current ammeters. table does not give the solution at the elapse of a certain time If this is required. it is possible to represent these on the calculating table by pure resistances and to use direct current. they simply add the of this type. Many operating companies who resistances of fixed values only. It is obviously also possible to construct calculating tables with a number of variable resistance units. these resistances may be combined and connected so as to represent any arbitrary system or network within the range of the table. short-circuit symmetrical initial measured will represent either It all depends currents. so that. This can be done by using phase shifters 2 to represent the synchronous machines. be simulated merely by impressing an alternating voltage of the correct magnitude. Recently it has been proposed to extend the use of calculating tables also to the solution of networks under normal operating conditions. "A Miniature Alternating-current Transmission System for Network and Transmission-system Problems. Elec. 611.-a. cit. p. 1 In order to make this possible. it is desired to determine the amount of current that the switch S will have to interrupt after a time of 0.C. since it makes it pos- short-circuit problems. "The Solution of Electric Power Transmission Problems Laboratory by Miniature Circuits. cannot..E. A. 1925. 831 1 in the . as well as the reactances of the interconnecting lines." Trans..E. current calculating table will obviously also give the solution of It should actually be superior to the direct-current table for the latter purpose. the alternating- currents. Fig. B.. and power under normal conditions. are indicated on the diagram.SHORT-CIRCUIT CURRENTS manufacturers and consulting-engineering offices 29 have calculating tables of this type. the miniature sys- tem must be built up of impedance units instead of merely resistances. By these phase shifters. its use eliminates the necessity of the assumption that all generator voltages are in phase. The loads in such a system are represented by impedances. ' 1923. in this case. and H. Hazen. he. In addition to being useful for the determination of voltages." Trans.E. Rev. base. and D feed into a transmission network. Furthermore. take resistance into account.. Schurig. "Artificial Representation of Power Systems." Gen. The nominal voltages of the various parts of the system. 1923. All reactances given are on a 20. p. R. sible also to EXAMPLE 2 Statement of Problem Four generating stations A. feeders in power station A. L. Assuming a short circuit on one of the 22-kv. 18. such as motors and condensers. The generating stations and other synchronous machines. . H.2 sec. p.E. voltages of the correct magnitude as well as phase displacement are obtained. 72. and alternating currents must be used instead of direct currents.000-kv. 2 Described in the paper by Spencer and Hazen.I.I. H. O. as shown in the sketch. It must be ascertained that the voltages also have the correct phase displacement relative to each other. A. ' Spencer. Base l 3.. four generating stations and a substation.-a. [7. two have 19.and 4-kv. Equipment in Station C: Total capacity of synchronous machinery 47.9 per cent react- .73 *- m 7.1 Wk 7. synchronous generator with 5. generator with 25.7 13. and one has 12. transformers have 58.-a.500-kv.- WWWW 1.3 per cent reactance.-a.000 kv-a.j Figures indicate Per Cent Reactance 66Vfv.4 per cent reactance. The transformers in this station have 5.3 I Ml/.7 per cent reactance between the 66.500-kv.050 kv. 7 per cent reactance between the 22.5 WW VUMV 65AV.1 / 7 nJo ^! 1 .-a. 6.4 per cent reactance.8 per cent reactance each. 7.-a.3 per cent reactance between the 66.' £_ SUBSTATION E STA TION D 13. and each of the three-circuit transformers has 5.0 1.4 per cent reactance. generators with 18 per cent reactance. two-circuit transformer in this station has 6. STATION C O Synchronous Generorhr © Synchronous Condenser ® Group ofSynchronous 2. windings.71 per cent. on 20. Equipment in Station B: One 37.0 n j.13 per cent. www jkfCv.ELECTRIC CIRCUITS— THEORY AND APPLICATION 30 Equipment in Station A: Two One One The 5. 7.\ ! Machines ~TL J L _£J j STATIONS STATION A — Single-line diagram showing layout of power system containing Fig. Four of the seven 6 %i-kv. with reactance of 4. -66Kv.6 per cent reactance.and 4-kv. ^Feeder 2^6js_/Jsl4 _ AWMA 22Kv. 5. The short-circuit calculations in Example 2 are based on this layout.3 per cent reactance. windings.-a. with combined reactance of 2.6 per cent reactance each. The transformer in this station has 4.5 5. Equipment in Station D: Total capacity of synchronous machinery 134. and 5.812-kv.500 kv.S 66 Kv. 18.000-kv.6 66 XV.and 22-kv. windings. synchronous condenser with 85. Equipment in Substation E: The two 6 % 2 -kv.000 „ = 525 -pA amp. the circuit diagram (Fig. — V^X22 symmetrical short-circuit current corresponding to 4.3 iy — Total reactance of the seven 6 £f i-kv.55 per cent +5._ r Initial h = —^gg— = 11.7-5. .15 per cent 6. each having 24 per cent reactance. the reactances to be assigned to the branches of the equivalent Y become „ Zi — = 5.63 thetical 22-kv. (72)).7 = n „ . transformers in this station have 13 per cent reactance each. equations (70).6^18^85.9 58. 3.4 Each of the three-circuit transformers in this station may be replaced by an equivalent Y-connected network (see Chap.volt winding as No. 19a) results. By going through the steps indicated in Fig. and the 4. __ 2. The seven 2 31 £fi-kv. transformers in station D 6.6 Total reactance of the group of four 2 „ 3. (71). winding as No.63 per cent reactance 525 X 100 „ _. the circuit is reduced to a single reactance between a hypoThis reactance is 4. The various loads connected to the system have not been indicated.0 2 % x -kv. 2. 19a to g. transformers are identical. in other words. _„_ 20.8 T 12. transformers in station D per cent Using the values calculated above for machine and transformer combinations in conjunction with the rest of the data given on the layout (Fig. 1. Solution Total reactance of synchronous machines in station A 1 = 6.7 = ! „ Z2 = 5.SHORT-CIRCUIT CURRENTS ance. 18).3 + 5. Base current = . and Designating the 66-kv.7 - ' = . the 22-kv.05 per cent 2 %].7+6. transformers in station 1 Jl+JL+JL T 19. all generators may be assumed con- all nected to one common bus. II.18 percent 25.0 per cent Total reactance of the group of three iyz =8. bus and the point of short circuit. The magnitude and phase of the voltages at the generating stations will be considered the same. winding as No.3-5.7 Zz = 6.000.340 amp.15 per cent = $i = D 4. 3. as load currents are to be neglected as compared to the short-circuit currents. per cent.-kv. 2 sec. .ELECTRIC CIRCUITS— THEORY AND APPLICATION 32 2. Decrement factors for symmetrical three-phase short circuits and single-phase line-to-line short circuits plotted versus reactance for a definite time to = 0. Time =6.36 5. 20.7S ISff 6.To-L'ine Short Cir •u'rr Sym metrica ' Three-)°haseS> ort Ore•urf a) b £4 10 20 30 60 50 40 Cen+ Reac+ance 10 80 90 100 Per — Fig.78 — Fig. Data for these curves have been obtained from Tables I and II. 19. Diagrams showing successive steps in reducing the system in Fig. 18 to a single equivalent reactance between the hypothetical bus and the point of short circuit.2 second I'° •2 8 % 16 Line .S0 14. 415 amp. B.82 3. —The symmetrical short-circuit current divides among the four generating stations A. . —Total Application of Decrements to the Generating Stations Lumped. lated 4. 56. *^j^ == From B From C 2 310 am P- 1. =3.236 ' ii + 3.51 _ X """ = ^ D ^Q &mp 7 3.6 78.SHORT-CIRCUIT CURRENTS 1. the short-circuit currents supplied by each station at the end of 0.570 3.170 1. and D as follows: initial From A and D X ". hence.900 4. circuit breaker in the feeder will interrupt after 0. Three-phase short circuit 6. C.120 amp.2 sec.415 Line-to-line short circuit 6. B.236 3.310 1.2 3.30 42. 10. short circuit short circuit Per cent equivalent Stations Rated A 664 985 1.2 sec.63 = = 4.TO From A From ^ 4. sec.^ From B and C 7.570 amp.940 amp.460 ' .830 amp. ' Obtaining the decrement factors from the curves (Fig. sec.2 at 0.530 ^In — = = 6.2 sec. 1-55 1 69 Three-phase short circuit Line-to-line short circuit The .410 Total The 52 have to interrupt after circuit breaker in the 22-kv.2 sec. C.55 1.340 2. sec. Application of Decrements to 33 Bach Generating Station Separately.2 at 0. 0.880 amp.7 113 2 1. rated current capacity of the four stations A.000 amp. 20) are obtained the following decrement factors at t = 0. Amperes at Three-phase Line-to-line 22 kv.050 9.120 Amperes Factor Amperes at 0.900 2. are calculated in the table below.69 = = 9.970 1.530 B C D Calcu- Factor reactance at 0.340 amp.15 0. feeder will.2 1.400 amp.120 1. Three-phase short circuit 9.530 2. /rated = 664 + 985 Equivalent reactance From + = 1.82 10. and D combined 2.7 15.420 2. Line-to-line short circuit 10.450 2.415 X X 1. 20) plotted from data in tables I and II.24 0. 4.2 sec.6 per cent the decrement curves (Fig. the n-circuit transformer considered is an air-core transformer. being functions of the saturation and. the self. the instantaneous terminal voltage of each winding can readily be expressed in terms of these constants and the instantaneous currents in the various windings.. the self. which is inherent with iron-core transformers.§ f*+ + Ma H ++ * Si + .and mutual inductances are strictly constant. R and L with appropriate subscripts represent the resistance and self-inductance of the various circuits and the mutual inductance between the two windings designated by the double subscript attached. Employing the classical equations for coupled circuits.§+ dt "-" dt ^'"" dt ^ „ = Srf.and mutual inductances are variables.§ + *. hence.— Consider a transformer having its coils connected in such a manner that there n independent circuits. may. It should be noted that the order M of these subscripts is insignificant. viz. of the instantaneous currents. The difficulty caused by the non-constancy of these fundamental parameters. and mutual exist circuit has inductance with respect to every one of the other circuits. M ln = M nl . The following equations result „ = BA +!-. be removed by the following device If 34 . as is always the case in commercial practice.* + *. in a great many problems.. etc. * +J M" Si + £. • • • • • • • • +M jhiJ4 T " m +Jf +M » " B* + L-W + M-W + ^-§ + *-§ + * *"Tt • • ' (1) (2) (3) In these equations. If the transformer has an iron core. self-inductance. each a definite resistance. In this n-circuit transformer.CHAPTER II TRANSFORMER IMPEDANCE AND EQUIVALENT CIRCUITS General Theory of Multicircuit Transformers.. since . three or n. - c) c M ^ + (M -M )^ + (M 21 c) c + (M u -M. It will now be assumed that the flux confined to the iron core depends only upon the value of the magnetomotive force producing it and is entirely unaffected by the position of this magnetomotive force with respect to the core. assume M c mutual have of the for simplicity that all windings number of turns or that all constants involved are reduced to the same base (referred to the same winding) in the well-known manner by multiplying resistances and self-inductances by the squares of the proper ratios of turns and mutual inductances by the direct ratios of turns. M + c) *i+ c • • -M • •+**„) (4) -M) ^+ 2S ^ + (M + M ~ (tx + t. etc. the inductances corresponding to the differences (Li in air. While this is not precisely true. and (3) may then be rewritten as follows: the same = t>i • • vn • • + + (M = R*i v2 - . — M c ). Equations (1). ffiti Xn + M ^ + (M .)^ + M.M )^ + c • + in) (6) In these equations. + it + 13 • n2 • c) c • + *») ^+ (5) .TRANSFORMER IMPEDANCE 35 The magnetic fluxes which give rise to the flux linkages corresponding to the self.. the error should be very small indeed. but the latter and smaller part determines almost entirely the operating characteristics of the transformer.and mutual inductances may be divided into two components. (M 12 — M c ). are sensibly constant they are due to fluxes which fully or partly exist The last term in each equation represents the voltage quantities. (2). two. the iron flux is assumed to be the same whether produced by a given number of ampere turns in circuit one. The former is by far the greater portion of the total flux. Also.^(i + 1 = RJ n (Mn3 - + (L» -M )|+ (M c M ^+ c) • ' • nl - t. Let the flux in the iron contribute the part inductances.M ^ + (M -M )^ + M ± & + * + *+ (Lx - 12 c) c (L. one which is confined to the iron core and another which wholly or partly exists in air. In other words. equations (4) to (6) inclusive may be rewritten in vector form. These reactances are due to fluxes which wholly or partly exist in air. In is the self leakage reactance of circuit 1.M )I n + jco(M .M )U + + In) (8) + j«M (/i + U + U + = Vi fli/i 12 C 13 • 3 C ln 23 • e ' • c • e M )I c n - +ju>(M nl -M )h + ja(Mn2 + c * c c V n = RJ n +jw(L n - • • c 2 2 C • • c jt*M e (I x M )h c ' • + + h h + + M )I M M M M jo>(M nS S + + /„) C • • • • • • (9) Introducing. in general. and X i2 is the mutual leakage reactance between circuits 1 and 2. X nn X nl X n2 = = = u>(L n X X ln 2n - M c) = <o(M nl = u(M n2 - e) c) = «(M ln = co(M 2n - (10) e) (11) c) (12) In) (13) and E = jwM c c (h +h+h+ • • • + the equations for the terminal voltages reduce to V = l (#! + il u )/i + jXnh + jX 13 +•••+ iX /3 ln 7 n +E c (14) F = 2 (R 2 + jX 22 )/ 2 + jXaJx + jX 23 + /3 • • • + jX 2n7 n + E c (15) 7n = (#» + jX nn )I n + jX nl /i + jX n2I + jZ n3/ + 2 3 • • • +# (16) Here.ilf )Ji + j<»(M 2n .ELECTRIC CIRCUITS—THEORY AND APPLICATION 36 induced in each winding by the flux exclusively confined to the iron core. + MLi . Considering voltages and currents of a single frequency only.M )h +MM21 . Ec is eliminated and a new set obtained giving the difference .M )U + MM .M )I + + In) (7) + jaM (h + h +1* + V = RJ2+ML2 .M )I n + jco(iW .M )I + jco(M . By subtracting each of the above equations from the preceding one. The significance of the other reactances is at once apparent. Hence they are very nearly constant and independent of saturation. namely. of necessity. be In an w-winding defined with respect to some other winding. as follows V -V = [Rv+j(Xn . in general. (In — 2i ) Similarly. for instance.z 23 )]/ + j(X 2l . Hence. = R2 ]X n (m) = Rn + j(X n .X 2n )I n (17) . due. there is a possibility that two or more of these values may coincide. the true leakage reactance of winding 2 with respect to winding X other.V = [R + j(X 22 .: TRANSFORMER IMPEDANCE 37 between the terminal voltages or the effective impedance drop of pairs of windings.X 13 )h + • • • (19) The difference between the self-reactance of a winding and the mutual reactance between this and one of the other windings is the true leakage reactance of the first winding with respect to the is the true leakage reactance of Thus. the first subscript refers to the winding itself. (X nn — Xmn ) is the true leakage reactance of winding n with respect to winding m.i) + j(X . therefore. + j(X xl j(X n2 . and.[R.X 22 ~\~J\X nn (20) (21) 12 ) X mn ) \^^J leakage impedance of winding n ). in general. Zn m ( . winding 1. for instance.J 12 )]/ + . have n — 1 values. represents the with respect to winding m. transformer.X.X zl )h +••'•+ j(X 2n .X 23 )h + + j(X ln . a distinct value for each of the other windings with respect to which the leakage reactance is determined. (X22 — -^12) is 1 with respect to winding 2.X 12 )I 2 +j(X nS .X ln )}I n . the true leakage reactance of one of the windings may.X 32 )] h . It is convenient to introduce symbols for the true leakage impedances and reactances as follows: = Ri Z 2W = R £i ( 2) 2 Z n (m) = R n + jX + iXj 1{2) "f" (1 ) = R. The relative aspect of the leakage reactances should be carefully noted. The leakage reactance of a winding is not a quantity which is dependent upon and characteristic of that winding alone.[R + j(X 3s .X 3n )I n (18) 1 2 j(X ls V\ Vx • 2 • 2 3 7n - • = [Rn 3 3 X nl )]h + + j(X nn .Xn)]h - [R 2 +j(X 22 . Of course. to symmetrical arrangement of the windings. and the second one in parenthesis indicates the winding with respect to which the leakage reactance is considered. it must. the general equations (14) to (16) inclusive reduce to V = V = x (R x 2 (#2 + jX^h + jX + jX22)h + jX 12 I2 21 Ii +E +E c (28) c (29) The difference between the terminal voltages or the impedance drop becomes 7 X V2 Here.ELECTRIC CIRCUITS— THEORY AND APPLICATION 38 Substituting abbreviations. of the constants involved. the subscripts in parentheses may be omitted. thus simpliWith multicircuit transformers. on the other fying the notation. however. Two-circuit Transformers.X 12 ))l2 — Z\(2)I\ — ^2(1)^2 (30> are the true leakage impedances of Zi( 2 and Z 2 (i) ) 1 and 2.. Application of the General Equations. in a two-winding transformer.— In. the case of a two-circuit transformer. + /2 + /3+ •••+/„ = is On (27) The voltage equations previously given (equations (14) to (16) or (23) to (25) inclusive) used in conjunction with one of the current equations (equations (26) or (27)) suffice for the solution any single-phase or polyphase transformer problem. and will of terminal be specifically applied four-circuit transformers. with respect to the other winding.[R2+KX22 . the sum /1 small compared with the load currents.X )h + (19) (23) + ' (24) 2 of the currents in the several windings is • • • (25) equal to the Hence. these equations j(X ln V% — — Vz — Z2(3)l2 #3(2)^3 + ^(-^21 — X3i)Zi = Z n (l)In — V\ Zl( n )Ii X - + j(X 2n Vn — ' The sum + j(X n — X^Ii + j(X n3 . i. respectively. + It+ • • • + J.-7. . the exciting current ignored. three-circuit. the general theory to two-circuit. number conditions. must be known. as well as a sufficient In the following. Ii many 2n )I n ' . of the currents is zero.[Bi+i(In . there can never be any doubt about the proper relative aspect of these impedances. Of course. (26) practical problems.X Sn )I n 13 In to (17) become inclusive + I. since it is this basis. the following relation holds: exciting current. Since.e. windings effective . however.X 2 i)]Ii . is (31) the logical equivalent network of a two-circuit transformer. . hence. while the impedance of the pillar current is make up Z e The the arms of the carrying the exciting given by Z = D Vl ~ IlZl (36) J-e In general. 21. it is assumed that the equivalent impedance splits equally between the two windings. equation (33) reduces to neglected. exerts but a small influence on the actual values of Ignoring excitation. hence. since the exciting current seldom exceeds 5 per cent of the full-load current and. as shown in Fig.V = Vx 2 (Z x (32) (32) gives + Z )h . separate leakage impedances Zx and Z 2 T. the double subscripts must be retained.Z 2I 2 2 1 If excitation is l (30) (31) l taken into consideration. equal to the sum of the separate leakage impedances. the exciting current is frequently This is permissible in very many problems.TRANSFORMER IMPEDANCE 39 hand. represents the equivalent impedance of the transformer. Equations and (32) indicate that a T-circuit.ZJe + Z )I + ZJ = -{Z 2 2 1 2 (33) e In practical calculations. The composition of the equivalent impedordinarily determined taken into ance is at once apparent from the following equation Z12 = Re + jX = R + R + j(X + X e 2 l x (35) 2) — Equivalent Network of Two-circuit Transformers. Vi-V 2 = {Z + Z 2 )h = -(Zx = Z 12 I = -Z12/2 x + Z )I 2 2 (34) X Z i2 Here. may. Very often. This assumption is freings are not equal .V = Z I . however. the sum of the primary and secondary currents equals the exciting current h+U=L Combining equations and (31) . quantity test and is is This by the standard short-circuit the only constant required when the excitation is not account. be written Equation V . power. and efficiency. the separate leakage impedances of the two windand the equivalent T-circuit representing the transformer will consequently be dissymmetrical. current. losses. At other values of saturation (voltage). equation (36) may also be written Zc which again = E « + /*»*' = may be modified to Zc = ^^ + MM * t 12 - |s M + jXlt c) (37) = ja>M 12 (38) e %l W\A-f5tfd Z3 1 VW-^ 4 Fig. depending upon and the position of the curve the shape of the magnetization operating point on the latter. basing the constants of the equivaon the general theory. lent circuit transformers. the representa- approximate. since enough data for a correct determination impedances are seldom at hand. what amounts to the same thing. Furthermore. Hence. the equivalent circuit is exact only at the particular value of saturation (voltage) at which the impedance Z c is determined. The reason is. the representaFor iron-core tion will be exact for air-core transformers. Making use of equation (28). only when the mutual inductance constant quantity or. where the magnetization curve is a straight line. . Although it might be desirable from the standpoint of accuracy for to use several values for Z c when the equivalent circuit is used tion is. more or less the determination of performance at widely different voltages. — Equivalent network of a two-circuit transformer. It is customary. with the exception of such as specifiof the separate cally deal with the distribution of the harmonic components of the exciting currents between primary and secondary circuits. 21. of course. therefore. of necessity. the pillar impedance Z c contains reactance only. Equation (38) shows that. It is a will be constant.40 ELECTRIC CIRCUITS— THEORY AND APPLICATION quently necessary. that the effect this is usually not done. the value of this reactance corresponds to the mutual inductance between the two windings. the symmetrical T-circuit thus obtained is sufficiently accurate for most practical problems. of the exciting current is fundamentally small. Fortunately. The equivalent derived directly from this theory. equation (38) shows the impedance Z c to contain nothing but reactance. exact correspondence is also here obtained only at some particular value of voltage. the actual value of the impedance R\ excitation impedance becomes 1.e. the pillar impedance.. carries no current. giving Z = R c c ^ (40) +jX -v\ (41) + jX = c and Yc ~ Rc c .995 or 2. Obviously. it is immaterial whether this impedance is considered to be 1. since the relation between core loss and saturation is not exactly quadratic. in the iron. Hence. i. The sum of the excitation impedance and the primary leakage impedance from equation (39) corresponding to 5 per cent exciting current is 2. or excitation impedance (admittance). This resistance is given such a value that the fictitious copper loss developed in it by the exciting current is equal to the core loss. To substantiate this. As already mentioned.000 per cent. the difference between the two values being but one-quarter of 1 per cent. hence. consider a transformer having 10 per cent equivalent impedance and 5 per cent exciting current. therefore. equation (36) becomes loss may Z = ¥l c Zi = (#! + jcoLO - Zi (39) *-e The leakage impedance Z\ is negligible compared to the + juLi corresponding to the self-inductance. This be accounted for by assigning also a resistance to the pillar impedance. to determine the pillar of hysteresis and eddy currents circuit.995 per cent. is determined by an open-circuit test. Obviously. one winding is excited while the other is left open and.TRANSFORMER IMPEDANCE 41 impedance only at a saturation corresponding to normal voltage. In practice. This is due to the fact that the classical theory of coupled circuits fails to consider the effect therefore. For this condition. Assuming that the primary leakage impedance equals one-half the equivalent impedance. the excitation impedance is found by simply dividing the normal impressed voltage by the exciting current. as it is commonly termed. makes no provision for the proper consideration of the core loss.000 per cent. 1924. In the case of a three-circuit transformer. V.E. however. an entirely erroneous picture may be had of the conditions at the transformer terminals. and.E. 1925.. Since a T-circuit circuit. 813. may be just the currents that should be considered. impedance drops in circuits 1 and and circuits 3 and 1. . the consideration of excitation is omitted. A. 1925. as indicated by equation (34).. pp. A. This is true when the analysis is concerned with the conditions Where inductive interference is in the power circuit alone.I." Trans. The former case is rather unimportant.E. There are but few problems involving transformers which actually require that the exciting current be taken into account.E. this type exist. Three-circuit Transformers. F. since. is exceedingly important.42 ELECTRIC CIRCUITS— THEORY AND APPLICATION always convertible into an equivalent IImay evidently also be represented by the This is seldom done. and particularly their higher harmonic components. Peters. "Theory of Three-circuit Transformers. Jour. and 71. Lyon Elec." by W. respectively. the When T representation is more convenient. p.. circuits 2 the and 3 Boyajian. A. since very few Y-Y connections of The latter. As examples of power problems where the effect of the excita- tion should be included may be mentioned unbalanced operation of the Y-Y-connected transformer bank without primary neutral and long-distance transmission where transformer banks are connected to long lines of considerable capacitance.I. 22 2 ss s 31 ls Is 23 I s S2 I 2 +E +E +E c (42) c (43) c (44) By taking differences between pairs of the above equations. 12 in Trans. J... the exciting currents. involved. however. latter type of network. is the transformer in general. the general equations (14) to (16) inclusive reduce to — Vi V* Vz = = = + jXu)Ii + jXiJi + jX + jX )/ + jX»Ji + jX (Rz + jX )I + jX Ii + jX (fii (J2. p. 508. the trans- former circuit reduces to a single impedance. become effective 1 2. Discussion "Three-winding Transformers. The lagging exciting current of the transformers tends to reduce the effect of the leading charging current of the line. * Application of the General Equations. if neglected. This impedance is equal to the equivalent impedance Z i2 of the transformer. TRANSFORMER IMPEDANCE V .X + i(X . +i(Zu - *2i)]/i + j(X .X )]Ix + j(X [Rx + j(Xxx .Vi = [A.Xx )]/ [R + j(X . for instance.X 23 13 ) (52) 22 32 13 12 ) (53) 33 13 12 23 ) (54) It is interesting to note the composition of the reactance part of X Consider Zx.X 22 32 )]/ 2 + j(x„ .72 - [R x x F . the mutual reactances would not enter into the effective reactance of the first circuit at The problem then immediately reduces to a two-circuit all. + j(X F .X + X .X 23 )I 12 . respectively.F3 = [12. may be written these 13 . the effective or composite reactance associated with circuit 1 is the leakage reactance of circuit 1 with respect to circuit 2 plus the differential If the third circuit were effect of circuit 3 upon circuits 1 and 2.X + X . and (47).)]7i . (46). the other two circuits.X 13 )]/ 3 - [JR.X 22 12 )]/ 2 -X )/ . be mutual effect upon each would 23 would equal circuit would be its leakage 1 reactance or 13 and the effective third circuit if the alone. Or.V = 2 [Rx 23 22 2 3 22 23 33 3 2s lt 2 12 3 x2 23 The symbols Z x Z 2 and Z 3 . [fix 43 + j(X .X + X . circuit 2 to respect reactance with 23 and 13 did not carry any current.X )]/ -X )I -X -X )h j(X 13 + j(X 33 j(X 12 + j(I u j(X 23 23 23 ls 3 1 3 x)]/i 12 + (45) 3 + (46) + (47) of the currents in the three circuits equals the exciting viz.Xx2 + X .X 12 )J + j(Xu .X« + X . 2 y _ y = 3 [ij 3 x The sum current. X X X X . Ix +h+ = h I3 (48) Introducing this relation in equations (45).Xu)]I* . X transformer problem. .X + X . 2 12 33 3 13 32 will e (49) 2 e now be introduced (50) (51) to represent the relatively complex impedances in the brackets above. X . Here Xu — x2 is the leakage reactance of winding 1 with respect to winding 2. these impedances.)]I.X ))h + j(X [Rt + j(X y -V = [R + j(X .Xtx + X Vi .Xi. + j(X . as follows Zx = Ri Z2 = R2 Z 3 = R3 + j(Xxx .X»)I.X + j(X . its respect to symmetrically located with the same. - [«.Xi.X + Xia .Xx + X . i3 are the mutual reactances between windings 2 and 3 23 and and between windings 1 and 3. Hence. 23 . Z3/3 Vz. When this is the case.V z = Z 2 I 2 . connection with the exciting current in equations (49) to (51) may be written as follows: inclusive jyXis J (^12 J\Xz3 — — ~ -^13) = = X12) = X23) Z2 Zs Z\ — — — ^2(1) •^1(2) Zz( 2 ) Z Z\(Z) ^3(1) 2 (Z) — Z\ —Z — Zz 2 (55) (56) (57) Hence.ZJ Vz = £2/2 . however. and equations (58) to (60) inclusive reduce to . The difference between the mutual reactances appearing in tive reactance. As a rule.Z 2I 2 + V .Vx = Zzlz . the effective reactance of Zi may become negative and have the effect of a capacitive reactance.V = ZJi . If the former are sufficiently predominant in magnitude.Zzlz + V -V = Zzh . Although the exciting current varies slightly with the load. The exciting current. 2(1) ( e (58) (59) (60) These equations properly applied will solve any three-circuit transformer problem where a sufficient number of terminal conditions is known.Z )J . (50). They provide for taking the exciting current into account. be taken equal to the curis impressed on one of the windings.Z IW )I. and (51) may now be written in the following simplified form: V -V 2 = ZJ . is very seldom included in the rent flowing It when rated voltage calculations.Zz 2))Ie . Equations (49). the currents in the three circuits add to zero. it is neglected. it would usually be considered constant and be given the value which corresponds to normal saturation. if this refinement is desired. would. hence.44 ELECTRIC CIRCUITS— THEORY AND APPLICATION Another interesting feature is the possibility of negative effecAgain referring to the impedance Z h it is conceivable that the circuits might be so arranged that the mutual reactances between circuits 1 and 2 and between circuits 1 and 3 are relatively large in comparison with the mutual reactance between circuits 2 and 3. the mutual reactance differences in question can always be determined as the difference between the effective "threecircuit" impedance of one of the windings and the leakage impedance of the same winding with respect to one of the others.ZJx Vx V 2 2 2 (61) (62) (63) . the other two being open.Z I + 1 l 2 s 1 1 l (Z 2 {Zz (Z 1 . Assuming that power is supplied to circuit 1 with circuit 2 short-circuited. obtained at practically zero saturation. 22. (62). the currents in the Since the exciting current for this two windings under test are equal and opposite. the threecircuit transformer may be represented by an equivalent Yconnected network.TRANSFORMER IMPEDANCE Determination Impedances. equations (61). Equations (61) to (63) inclusive show that. ticable to represent the three-circuit transformer It is imprac- by an exact equivalent network when the exciting current is taken into account. (53). and (63) give is = F2 = 73 = 7i (Zi (Z 2 (Z 3 + Z )/ = + Z )/ = + ZJh = 2 x 3 2 Z 12h Z 23 I 2 Z 31 h (64) (65) (66) The equivalent impedances Z 12 Z 23 and Z 31 are thus obtained. to circuit 3 with circuit 1 short-circuited. They are. then to circuit 2 with circuit 3 short-circuited. (72) 3 — Equivalent Network of Three-circuit Transformers. finally. and (54)) to be assigned to the three windings are readily determined by three short-circuit tests. The equivalent of pairs of windings are measured exactly as for the impedance two-circuit transformer. when the effect of excitation is omitted. Inspection of equations (58) to (60) inclusive will show the futility of attempting such representation. The effective impedances of the three circuits are then calculated . In three-phase connections. the individual transformers may also be represented by Y-connected circuits of the type shown in . from Z Z2 Z x s +Z =Z +Z =Z + Z! = Z 2 12 (67) 3 23 (68) 31 (69) Simultaneous solution of these equations gives ^ = Z X% + Z3! . Not so.Z 23 (7Q) Z Z2 = » + Z£ - Z» Z Zy = +Z — Zi 2 3i 23 (71) . . of —The 45 impedances effective (equations (52). hence. and. in this case. condition entirely negligible. as indicated in Fig. This representation is simple and very convenient in many instances. however. 22. 23. 300. It complete three-phase arrangement is indicated in should be noted. is convertible to a A-connected circuit. that this representation is not practical except when conditions are perfectly balanced so that the three-phase problem reduces to a single-phase problem. in general. EXAMPLE 1 Statement of Problem A 2. it is evident that the latter may also be used to represent a three-circuit transformer.100 kv. three-circuit transformers.volt generating station supplies power to two short transmission lines through a bank of three 2. of — • transformer. of Since a Y-connected circuit. z2 j 1 4 vw^- : -X-X YlG.46 ELECTRIC CIRCUITS— THEORY AND APPLICATION Fig. 22. however. — Three-phase Phase c IT arrangement transformers. however. As a rule. — Equivalent network a three-circuit neglected. the equivalent Y-connected network is more convenient from the standpoint of calculation. The . When unbalance is involved. 23. the solution must be based on the equations themselves rather than on an equivalent network. Phase WW-£ffl5>1 Excitation a VW\A-CTtfv>-» 13 Zz Zi T Phase b Z3 13 Z? —WWW- 1 Zt 4 z3 Fig. A Fig.-a. equivalent networks of three-circuit Excitation neglected. and their voltage ratings are 110 The transformers are connected A-A-A.10 6. 24.15 per cent between the 22.TRANSFORMER IMPEDANCE 47 transmission lines are both three-phase.and the 110-kv. 3. windings 6.10 per cent between the 2. Solution Ratios of Transformation: Winding 2 22^ y0D 2.. The voltage on the 22-kv.3- If the 110-kv. windings 6.8 2. Consider the nominal voltages and the base currents calculated from the full rating at the nominal voltages to be the 100 per cent values of voltages and currents. and the high-tension voltage strictly 110 kv.32 per cent 1.15 6.59 6.and the 110-kv. and and the equivalent Y-connected.-a.3.27 per cent 2. The voltage regulation when the 110-kv.15 6.000 volts —Three-circuit transformer layout for Example 1.59 + + + 6.10 6.15 - 6. The reactances of the latter are (equations in Fig. (71).3 Winding 1 Winding 3 _ 110 = 47. load is disconnected. = 6. 24. short-circuit reactances are and the 22-kv. windings 6. load at the transformer terminals is 3.150 kv.10 = = = 3. The voltage of the generator. single-phase network in Fig. the 22-kv.83 per cent 3. load at the transformer terminals 3.150 kv.000 volts 110. circuit.3 Winding 1 Winding 110 22 3 Winding 2 The circuit layout is shown (70).-a. respectively. 2. (72)) Xx = X 2 = X. calculate 1. kv. at 85 per cent power factor (lagging). respectively.. The equivalent 2.300 volts 22. .59 6. 25. at 90 per cent power factor (lagging). The nominal voltages of each single-phase transformer are as follows: and 22 Winding 1 Winding 2 Winding 3 Fig.59 per cent between the 2. 5(0.436 as standard phase. between the voltage at the junction is very small. ^ 3-& kv-a. This voltage.025 ± -s/25. 24 with loads attached.02 X fiP 100.jI X 3 3 = = 13 cos fa V Using 3 at the junction point of the equivalent (a) 3 50 per cent of base current 0. 25.3. being less than 0.j0.002 seen.j'0. 2: point As ± circuit 22 X 5. also given is by V m = V . Fig. — Equivalent network of one of the three-circuit transformers in Fig.4 -y48. to consider the phase angle of the currents with respect to the junction-point voltage as being the same as with respect to the terminal voltages. f.008 V = V5.2 l2 61 -^r' + 127.025 The voltage on the 22-kv.04 kv.527) = 86.32 + j'1.85 .f.jO.88 A/VW-£ffi5Wd- at 30%p.10.41 = 100.527 as standard phase.j0. = 10« - 0. = ^' 9 peT cent 1 = ^Q02 = -(/i + I )' = 49.83 (0-9 t/ _l 611 F 2 _l 2 127 2 - F 2 Squaring gives 100.85 .9 .2 2 2 which again reduces to V\ .021 is.2 per cent therefore. = 50 X 100 per i 2 cos V 2 <f> 2 cent of base current t? 2 Using (b) 2 = 0.9/a = V2 + "r / » .9 2 + = V\ + 122.527).9(0.8 deg. and the terminal voltage V 3 Vm and V2 The displacement between is still smaller.23 X 2 = \/~5.9 per cent 100 = 100.jhX .85 = sin fa 0. .90 sin = <£ 2 0. 100.1 — 2 /1 h t (c) . 1. Vm = + 0.atSS%p. particularly since the resistances have been neglected.436) + 50(0.436)j2. therefore. the phase displacement Vm = is = 22.050 V\ The solution of this equation + 20. It is sufficiently accurate.ELECTRIC CIRCUITS— THEORY AND APPLICATION 48 The voltage which would appear by circuit is given Vm = V .. however. —In Application of the General Equathe case of a four-circuit transformer.10 = 72 + *|? 2 v 2 +i^ v 2 Hence.864 2.4 per cent .45 +J4.9 - 100 = is.23 voltage at generator is. 1925...9 per cent of the 110-kv. Again neglecting the small displacement between the voltages Vi and F 2 as far as the phase of the current is concerned and considering V] as standard phase. 1 See discussion by /1 +h+h+ W.025 = 2.45 = 101.88 +J1A1 = The 102.494(0. = F2 Fi Using V 2 + UZ n = V . windings V = 101. equation (/) gives 7 = 102.5 = 101.27 102. . Lyon Ii = in Trans.115 ± V2 6U9 X 10< 2 = \/57ll5"± The 5. p. hence.300 X 1.9 per cent Four-circuit Transformers.5 per cent (0.E. + jX u ) + jXuh + JX2J2 + jX s J s + E Vi 3 c (73) (74) (75) (76) Also.9 . (/) 50 X 100 ~Trfpj — = 49. + Ec V 3 = h(R 3 + jX s) + jX l3Ii + jX 2SI 2 + jXsJ* + El F 4 = h(R.108 regulation of the 22-kv.J 4 + Ec V 2 = h(R* + jX 22 ) + jXuh + jX i3 h + jX 2 J. (77) .I./6.230 V\ = r^0^9~X~10«~ V = V5. = V\ 102. 7i = 100. therefore. the general equations (14) to (16) reduce to 1 = Ix(Ri + jX n ) + jXiJ» + jX 13 I 3 + jX. it will be assumed that the size and power factor of the 22-kv.436)i3.j0. circuit 101.52 + 266 + ^ + ^^- which reduces to - V$ The solution of this + 93. circuit —— - The = is = 3 JT 1 = T it = is 0.481)j3.436).hZ 2 (e) 12 as standard phase — 102.1 per cent is.j0. hence. With no load on the 110-kv. V. 816.E. neglecting excitation.9 per cent by given I& - V.8 3 The regulation . 1.5/0 = V 2 +^(0. + = (d) . A.j'1. load remains the same in spite of the change in voltage.TRANSFORMER IMPEDANCE The generator voltage 49 given by is V^Vn+jhXi Using V3 as standard phase.360 volts In calculating the regulation.100 10.27 .0. = ' 1Q02 voltage of the 110-kv.9 v -j0. circuit the following equation holds: 3. tions. . first difference. second. The order of the second and third subscripts is unimportant.(Z. will eliminate 7 4 from the ence. I x from the second This gives 12 12 23 ls V . the following method of elimination used First eliminate Ec by taking successive differences. Thus.X„ + X X + Xu .X u )Ii (80) + - + These three equations together with equation (77) are sufficient any case.Z 214 . In the first equation. and that the coefficient of 7\ in X X m X .Xu . and fourth windings were considered as a three-circuit transformer. there is no difference between Z124 and Z U2 It will also be noticed that the coefficient of I 3 in equation (78) is Z 2X3 — Z 2U that the coefficient of 7 4 in equation (79) is Z 3i2 — Z 3 n.F4 = 24 )] X . fundamental equations maybe handled in a variety is some advantage in having them in Since there symmetrical form. in X ances.i.X j(X .V2 = V2 . An examination of these equations shows some interesting facts in regard to the impedto determine the currents.X + X -X . . This will be represented by Z214. the equations of voltage be written as follows: is may V . and I 2 from the third V .X +X — X + X — Xi j(X .V2 = x difference. . This impedance will be represented by Z The impedance R 2 +j(X 22 — i2 + X14 — 24 ) is that which would be assigned to the second winding if the first. equation (80) differences Z 4 i 3 — Z423.hZ 2U + 7. and fourth were considered as a three-circuit transformer.I Z 32 + / 4 (Z 42 .V = V — Vi = x 3 3 IxZxu I 2 Z 231 I 3 Zu 2 .X + X u +j(X tl h[R* + j(X 22 - IJLRi 2i 23 34 24)] 23 24 14 - X U )I be Then : differ- 3 (78) X1O/4 (79) . The second and third subscripts indicate which of the other windings are grouped with the first to form a three-circuit transformer.Z + X . that is. Ri + j(Xu — X i2 — i4 + 24 ) is the impedance that would be assigned to the first winding if the first. second.Z 3l2 ) — I\Zi 32 + I\(Z\\ 3 — £423) ) 3 x 3 (81) (82) (83) .X + X — X + X — X24)] i(X .ELECTRIC CIRCUITS— THEORY AND APPLICATION 50 These five of ways. The first subscript shows to which winding the impedance is attached.V = 2 z I 2 [R 2 I 3 [R 3 + j(X 22 + j(X 3 3 + i(X /4IR4 + j(^44 J 3 [i2 3 33 - 24 )] + 24 23 12 13 )] 23 i2 3 )] 34 13 V3 . The standard formulas found in textbooks on principles and design of transformers are all based on broad assumptions in regard to the distribution of the leakage flux and may easily give results which are in error to a considerable extent. A. fa . "Resolution of Transformer Reactance into Primary and Secondary Reactances. The separeactances of transformer windings cannot. therefore.. experimental methods must be resorted to. there are several tests for this purpose.. 1925. — Determination of Separate Leakage Reactances. Having obtained by test the equivalent pairs of windings. Since the impedances involved in the solution of four-circuit transformer problems are the same as in the three-circuit case. Theoretically. 805. A." Trans. (93) Very little. 92 ) no . is gained by attempting to represent the four-circuit transformer by an equivalent network. be calculated with accuracy. in general. Z124 Z134 Z42I — — — Zl23 Z132 Z423 = —Z2IA + = —Zsu + = — Z241 + ^213 (84) Z312 (85) ^243 (86) — Determination of Impedances. the desired effective impedances of impedances are calculated by Z 124 = Z 214 = Zl2 + Z ~ Zu 41 2 Z24 " 24 — + Zn T *£ - ^ — Z\2 Z 12 ~ Z%\ Zsl Z\\ (88) 2 •Z23 Z 213 = Zn + + Z31 Zsl + Z S21 = (87) (89 ) 2 -f- — Z23 Z 23 ~ Z\2 Zl2 (90) 2 Z34 + Z23 — Z24 (91) 2 Z24 Z432 7 Z413 -f- . ances.E.I.E. if anything. some single-phase 1 and some 1 Boyajian. p. For precise determination of these reactances.Z34 ^ 2 — + Z34 ~r ^41 Z\\ ^34 -g — ^23 ( ~Z — Z3I n . Furthermore. it seems to be doubtful whether more rigorous and reliable formulas are capable of being rate leakage developed. they can be determined in a similar manner.TRANSFORMER IMPEDANCE 51 There are some other interesting relations between these impedFor example. . In particular. for that matter). these various tests may not give identically the same results. The latter corresponds closely to the conditions under which the equivalent impedance is obtained by a short-circuit test. at the other value. When the saturation is decreased from that corresponding to normal operation to approximately zero. 2 Substantiation of this statement will be found in the experiments described in the paper "Separate Leakage Reactance of Transformer Wind- ings. as a rule. while some determine the leakage reactances at a three-phase. since.e. G. cit. l very low (practically zero) saturation. a change in saturation affects the leakage reactance of the winding nearest to the core to a larger extent than the leakage reactance of the winding farther away from the core. A. This is exactly what might be expected. therefore.E. the effect of saturation is absent. 1925. where the two operating points in question lie on each side of the bend of the magnetization curve. It should also be noted. Dahl.E." by O. Since there undoubtedly is some change in the leakage reactances when there is a material change in saturation. i. for instance. will also increase but not necessarily in the same proportion. It should be noted that leakage-reactance changes of the order of magnitude mentioned above manifest themselves only when rather large changes occur in the flux densities. in other words.> above the knee of the magnetization curve." Trans.ELECTRIC CIRCUITS— THEORY AND APPLICATION 52 Some of these tests give the leakage reactances at operating flux density (or any desired value of flux density. 2 1 Dahl. C. some change in the leakage reactances may be expected when one value of the flux density corresponds to a condition where the iron is more or less saturated. however. a large drop in saturation will be accompanied by a slight redistribution of leakage reactance between the windings. "Separate Leakage Reactance of Transformer Wind- ings. O. The separate leakage reactances. It is believed that the increase in equivalent reactance will not exceed 10 per cent and usually will be less than this figure. 785. the iron paths of the leakage fluxes. while. C. In a core-type transformer with cylindrical coils. the leakage reactance remains sensibly constant. the value of the equiva- due to the reduction in reluctance of lent reactance will increase. loc. G. p.. since the leakage flux of the former has a relatively longer path in iron than the leakage flux of the latter.I. that even for quite considerable changes of flux density in the operating region. in others. all assumed that the transformers considered have unity ratio If the ratio is different from unity. The single-phase tests are applicable to both two-circuit and multicircuit transformers and may be used to give the leakage impedance of one winding with respect to any other winding. which determine them at a high reactances at as nearly operating density as possible. It is of transformation. in outlining the tests below. from this standpoint. however. twocircuit transformers will be assumed. will give the maximum accuracy. of the leakage therefore. when the sum The latter. it is 53 desirable to ascertain the values of the leakage Hence. will suffice impedances or the equivalent impedance is known. oscillograms must be taken and the components of the desired frequency singled out by analysis. potential transformers must be used in the test involving measurement of voltage drop due to exciting current and in the parallel-conjunction test. Making use of the component which is the largest percentage of the composite wave. the tests All of the tests. even in a multi-circuit transformer and. are not likely density are preferable. In some tests. it is theoretically immaterial which one of the harmonics is used. same inherent precision. the best tests are either those which involve metering of pure waves of either fundamental or higher-harmonic frequency or those where any distortion likely to occur will be third harmonic. component of the wave is capable of by oscillogram analysis. therefore. it merely means repetition of the same tests with appropriate changes of connections.TRANSFORMER IMPEDANCE Evidently. The determination of the separate leakage impedances is fundamentally a two-circuit problem. In calculating the leakage impedances from test data. so that each winding is considered with respect to every one of the others. voltages and currents of a single frequency must be used. When the waves are distorted. A current transformer must be used in the seriesopposition test. There are two single-phase tests which will give the actual values of the separate leakage impedances and so small that the principal exact determination one which only — will give their ratio. the largest component may be the fundamental. the to give the From the standpoint of precision of measurement. In some of the tests. Single-phase Tests. If it is desired to determine the possible separate leakage impedances involved in a multicircuit transformer. however. . These auxiliary transformers should have the same ratio of transformation as the main transformer under test. Both the measured current and the voltage drop will contain harmonics. = V = For 7 2 = 0. Excitation of One Winding Only. oscillograms must be taken and the desired harmonic components separated out by analysis.54 ELECTRIC CIRCUITS— THEORY AND APPLICATION may introduce errors due to incorrect and also due to the phase displacement between their primary and secondary voltages and currents. Figure 27 . as shown in Fig. one of the windings is excited. by meas- ages of the two windings can be measured as indicated with a voltmeter drawing negligible current. (97) impedance of winding 1 with respect to winding 2 is immediately obtained. In this test.V = 2 [#! + j(Xn - Xti)]Ii = Zi/i (96) Hence. a. the leakage determined. -Single-phase determination of separate leakage impedances uring voltage drop due to exciting current. One terminal The auxiliary transformers ratio — is connected to the excited winding in such a manner that the difference between the terminal volt- of the other winding 7b Oscillograph tip- Fig. Drop Due to Exciting Current. Since the equations are applicable to quantities of a single frequency only. 26. Since the secondary winding carries no current. Vi (Ri 2 (Rt + jX u )Ii + jX I + Ec + jX )h + jXnl. 26.i + E lt t 22 c (94) (95) subtraction of these equations gives Vi . By repeating the test with winding 2 as primary and winding 1 as secondary. the leakage impedance of winding 2 with respect to winding 1 is also readily As seen. and the exciting current measured. the voltage difference thus measured is directly equal to the primary leakage impedance drop due to the The general equations are exciting current. Vx V* = = (R. perhaps. at very high saturations where the third-harmonic component may be the greatest. as shown in Fig.Phase Trans-former Approx. no flux will exclusively exist in the core. This is due. 27. (R 2 may + + jX iX„)Ji 22 )I 2 be written . Series-opposition Test. however. As a rule.f$ ma)t=l2.juM c (I 1 e (Ii - 70 h) (98) (99) . in series opposition windings in opposition. with the b. This is readily seen from the general equations which. — Oscillogram from leakage-impedance shown test using the circuit connections in Fig.jX l2Ix . maximum precision should be obtained by using the fundamental compo- Sfngle. Oscillographic measurements are not necessary in this case. 28. 26. also. Only when the component which is wanted is predominant can exact determination be expected. Even though the fundamental components be used. to the inherent difficulty of analyzing a complex wave with great accuracy.300gausses Fig. 28. therefore. the precision of this test leaves something to be desired. of course. Usually. nents of voltage drop and current in calculating the leakage impedance. except. The fundamental component of the exciting current will always be the largest. as both voltages and currents are sinusoidal. Single-phase determination of separate leakage impedances by the series-opposition test. the fundamental component of the voltage drop will be larger than any of the harmonics. — If the two windings are connected and an alternating-current voltage impressed. -MSiSlSlSiSi^ — Fig.jX l2h +jo>M .TRANSFORMER IMPEDANCE 55 shows an oscillogram taken during a test of this type. V = 7i +F 2 = (Zi + Z 2 )I = Z n I (103) • Vo/f&gre across One Winding Single -Phase Transformer Approx.. and oscillograms can be omitted. which means that These there is no net magnetization of the core as a whole.. = = I2 i. — With parallel conjunction as indicated in Fig. the 1 than those corresponding to . since ^yi joj-fe precision. reduce to = (J2i+jXi)J = Z I = V 2 (R 2 + jXi)I = ZJ Vi (101) t (102) which show that the terminal voltage of each winding is directlyequal to its separate leakage-impedance drop. c. this test is very — Single-phase deter- leakage impedances at zero saturation. ELECTRIC CIRCUITS— THEORY AND APPLICATION 56 Since the two windings carry the same current. . It will be noted that both voltage and current waves are pure sinusoids. . Test. . Thus. .B=0 Fig. . as from leakage-impedance Note the sinusoidal shape connections shown in Fig. I. From the standpoint of sinusoidal. larger parallel-conjunction test. the windings excited in 30 the current will divide . since I (100) the last term in equations (98) and (99) is zero. — Oscillogram of the voltage by the series-opposition and current waves. therefore. therefore.e. suppressed. equations. Circuit is sinusoidal.i and the values obtained may be slightly v=Vj=Vg satisfactory. r t_. the current flowing. . Parallel-conjunction . 30. Fig. . method. The total voltage impressed on the two windings in series evidently equals the equivalent leakage-impedance drop. 28. will also be Exact the flux in the core is current involved are and the voltages measurements of the two readily obtained. however. Figure 29 shows an oscillogram taken during a series opposition test of a small laboratory transformer. mination of the ratio of separate leakage impedances by the It gives. . 29. normal density.. When the applied voltage well as the terminal voltages of the test two windings. however.) Addition of these equations gives Vt + V. The ratio of the currents in such a case would evidently be unity. gives = (i^ + jXx)/! - ZJi - + jX )7 = (# 2 2 2 Hence.e. This test may ~ Z2I2 (106) ^ % be performed at any desired value of saturation by varying the impressed voltage. r. oscillographic The records are necessary. Only when the two windings are identical in every respect and perfectly symmetrically arranged with respect to the core will the currents divide inversely as the leakage impedances. applicable to a single frequency only.TRANSFORMER IMPEDANCE 57 between the two windings in the inverse ratio of their leakage impedances. the current division is given by h h = R* + J( X ™ + X ") + +X 2jwMe {6) Rt +j(X n 21 ) +2juMa which is entirely different from the inverse leakage-impedance ratio. when the saturation is 1 It might be thought that. and.jX 12 I +ja M 2 +jX n )I -jXtJi t e (Ii. oscillograms Figs. and the windings would have the same impedance. As seen. promi- nent harmonics are present even comparatively low (see Fig. since the currents in this test will always be more or less distorted. in general. which were taken when tests of this type were applied to a small laboratory transformer.. By equating equations (a) and (6). serve to illustrate this point. with this connection. Equation (107) is. = 2V = Ztlt + ZJ. ^ Hence. (c) or y^Z_Jt + Z_J. 1 The equations for this case are 7i V* = = + jXu)Ii + jXnh + E + 3X22)12 + jXnli + # (Ri (R 2 c (104) c (105) which. with the windings in parallel opposition. as the following analysis will show. 31). and the impressed voltage equal to the impedance drop of either winding. . the current would divide inversely as the leakage impedances and also that. by subtraction.h) (a) - (6) +jaMc(Ii I. the core as a whole completely demagnetized). 31 and 32. the impressed voltage is equal to one-half the sum of the separate leakage-impedance drops of the two windings. the impressed voltage would equal each of the leakage impedance drops. This is erroneous. The equations for this case are Vt V* = = (J?! (R 2 +jXu)h . The last term in equations (a) and (&) would be zero (i. distortion is larger than shown in Fig. the ratios are as follows: T 1\ n AV2 QV 0A2 ~ x{ " 3zi " -y" f -A. A equal the inverse ratio of the leakage reactances. In this particular to the reactance) can oscillograms be omitted.l Single-Phase Trans-former Approx. 2/ ^o/tage across ^Winding No. Assuming fundamental and third-harmonic components only. necessitating a separate exposure for each winding. The voltage across winding 1 is reversed with respect Note that the to the current and represents a voltage rise instead of a drop. between the recorded quantities for winding 1 and winding 2 have no real significance. 1 Single -Phase Transformer Appro*. 2 Winding No.Curreniin Winding No. The voltage across winding 1 is reversed with respect to the current and represents a voltage rise instead of a drop. 30. J . graph was used. The apparent phase displacements. 32.58 ELECTRIC CIRCUITS— THEORY AND APPLICATION Only when the leakage impedances contain no resistance (or. 31 on account of increase in saturation. A Current in Winding No.2 Voltage across Winding No. The voltages should be very nearly in phase. therefore.200gausses — Oscillogram from leakage-impedance test by the parallel-conjuncFig. when the resistance is entirely negligible as compared in practice. graph was used. therefore. The voltages should be very nearly in phase. two-element oscilloCircuit connections shown in Fig.2 Voltage across Current in (Winding No. case. 30. J f Winding No. two-element oscilloCircuit connections shown in Fig. tion method. the ratio of the effective values of the distorted current will Current in Voltage across Winding No. tion method. B moix= I0. between the recorded quantities for winding 1 and winding 2 have no real significance. The apparent phase displacements. necessitating a separate exposure for each winding. 2 xf ~ T" ±x 1 iff' (108) . 31. B ma)l =4100 gausses — Oscillogram from leakage-impedance test by the parallel-conjuncFig. principle of the method is as follows If sinusoidal voltages are impressed on a Y-A-connected : bank of transformers. the ninth and fifteenth harmonics. The three-phase method by which the separate leakage reactance of the windings of a transformer may be determined makes use of the third-harmonic component which inherently exists in the magnetizing current of a transformer when a sinusoidal voltage is impressed. etc. If the transformers are perfectly balanced and there is no external load on the secondary.TRANSFORMER IMPEDANCE 59 Hence. seventh harmonics. no current other than the third harmonic and its multiples can exist in the delta. The parallel-conjunction test gives the ratio only of the leakage impedances and not their actual values. and the separate leakage impedances determined in this manner. the ratio of the fundamental and the third-harmonic currents in the two windings is the same. will give the desired triple-frequency leakage impedance. equation n = x* 2 (110) vKS'+^W"* which indicates that the effective currents divide inversely as the leakage reactances. are negligible and need not be considered. Usually. The problem is then to measure with precision the proper third-harmonic electromotive force and current. which. where . When the equivalent impedance is known from other tests. however. etc. The method is applicable only when a three-phase bank of three identical trans- — formers The is available. it can be apportioned between the two windings. by simple division. The third-harmonic electromotive force induced per phase of the delta is just balanced by the triple-frequency impedance drop due to the circulatory third-harmonic current. The ratio of the effective currents may be written + + )2 2 ) Substituting for I'i and !'{' (109) reduces to h 72 = by means w+ / ( j (/r) 2 (/n 2 (109) of equation (108). This would be so also for the fifth harmonics. the third- harmonic component of the magnetizing current will be confined it appears as a circulating current. to the delta.. Three-phase Tests. All doubt in regard to the calibration of the instrument transformers is thus eliminated. — To Fig. due to unbalgive good accuracy of measurement. and. it may be impracticable to obtain entirely pure waves. ance and other causes. Two-winding Transformers. The bank is connected Y-A. the quantities of triple frequency which it is desired to measure will be entirely predominant. if used. the bank of resistors may be omitted and the voltage measured between generator and transformer neutrals.60 ELECTRIC CIRCUITS— THEORY AND APPLICATION The three-phase third-harmonic determine the leakage They should Of course. The third-harmonic current in the delta and the third-harmonic electromotive force per phase on the primary side are recorded.winding transformers. instruoscillogram analysis may be necessary. correct determination is highly facilitated even though Furthermore. of connections for three-phase practically obtained by connecting a Y-connected resistor bank between the lines and measuring the voltage between the resistor and transformer neutrals. If the generator is Y-connected and its phase voltage is free from a third harmonic (and multiples). is entirely without a third-harmonic component in its voltage to neutral. — Diagram Oscillograph leakage-impedance test on two. The resistance of the voltmeter and the bank of resistors should be sufficiently high so that the primary third-harmonic current is negligible compared to the current circulating in the delta. No commercial Y-connected generator. are not likely to affect the results seriously. and balanced sinusoidal voltages impressed. 33. While the current separate out the third harmonics . since their secondaries are connected directly to indicating meters. it will be necessary to take oscillographic records and by analysis. ment transformers. however. a. in any event. but. The latter is most tests reactances at normal (or any desired) saturation. As a rule. so this method is scarcely of practical interest. hence. if even the slightest unbalance in the impressed voltages. One to one ratio Fig. respectively.TRANSFORMER IMPEDANCE in the delta is sensibly third 61 harmonic. — Vector diagram of third-harmonic quantities involved in the "two- winding method. The diagram of connections and the third-harmonic vector diagram are given in Figs. the neutrals. 34. at voltage is the sum of two compo- least. r the resistance and F"n' the third-harmonic voltage between of the voltmeter. Figure 35 shows an oscillogram taken during a test of a bank of experimental transformers. Circuit connections shown in Fig.800gciusses m|)lx Winding*! Primary Winding*3 Seconda ry Fig. R is the resistance of the resistors per phase. negligibly small). If the transformers have another the quantities in the various equations given below should all be referred to the same side. this .B =l2. Line Voltage. the resistors. a fundamental and also other harmonics are unavoidable between the two neutrals. Voltage Primary Neutral to Current in Secondary Delta Neutral of Resistor Bank Y-A Connection Approx. then vr = n'„'(i + 1) (in) Since the third-harmonic current in the primary is zero (or. 33 and 34." of transformation is assumed. If -Oscillogram from three-phase leakage-impedance test on two-winding transformers. 33. or the transformers themselves is present. 35. ratio. V = . of connections for three-phase the true triple-frequency leakage reactance.(R2 + jX 2")I'2" = -Z'"l' The (112) triple-frequency leakage impedance and reactance of the secondary windings are now found by To Oscillograph To Oscillograph Fia. winding Fig. Dividing the triple- —Vector diagram of third-harmonic quantities involved in the "three- winding method. The second component fX"!!'" is induced by the part of the third-harmonic flux in the air which produces linkages with the primary winding. a more convenient method may be used . the leakage reactance of the other be found in a similar manner. — Diagram This X'l' is leakage-impedance test on three-winding transformers. 6. Multi-winding Transformers. Applying equations (28) and (29) to third-harmonic quantities. the fundamental reactances are found. remembering that V" and 2" are both zero. may and vice versa. gives nents only. — When the transformers have more than two windings.ELECTRIC CIRCUITS— THEORY AND APPLICATION 62 One component Ec" is induced in each primary winding by the triple-frequency flux <\/c " in the core." frequency reactances by three. 36. By repeating the measurements with the original primary winding as secondary. 37. In order to do this. It was desired to determine the separate leakage reactance of winding 3 with respect to winding 2. 37. As before. of the voltage appearing across the open corner of . for balanced conditions. as both the voltage and the current will be sensibly third harmonic. and the circulating third-harmonic current in it recorded.800gciusses — Oscillogram from three-phase leakage-impedance transformers.TRANSFORMER IMPEDANCE 63 which eliminates the necessity of establishing the artificial neutral on the primary side. An oscillogram from a test at this type is shown in Fig. max JN =l3. the other individual reactances are obtained. The connections are shown in Fig. Power was supplied to the primary Y-connected winding and oscillograms taken of the primary line current. By repeating the measurements with changed connections. One delta is while the other two windings are A-connected. oscillographic records are unnecessary when the transformers are well balanced. ance of the closed delta winding with respect to the open delta winding. closed.B Fig. 2. 38. be equal to three times the third-harmonic electromotive force per phase. and the third-harmonic vector diagram in Fig. 36. EXAMPLE 2 Statement of Problem Three small three-circuit transformers having unity ratio of transformation are used in a research laboratory. the primaries are Y-connected. The windings are designated 1. The other delta is not closed. 36. the third-harmonic voltage appear- ing across the open corner of this delta may. Solution of the vector diagram exactly as in the preceding case gives the leakage reactance of winding 3 with respect to winding 2. therefore. Recording these two quantities makes it possible to determine the leakage reactPrimary Line Current Voltage across Open J^ Current in Closed Approx. 38. 39. In many cases. the transformers were connected as in Fig. be meas- ured and will. Circuit connections shown test on three-winding in Fig. and 3. 54 1. John Wiley . An 3. Inc.16 Bn Bi B7 B> -0.89 0.63 0.08 Resultant Coefficients Expressed in Arbitrary Units Curve analyzed Co 31.02 0.10 0.88 Per Cent of Equivalent Sine Wave Co 100 100 100 C\ 99.99 0.21 0.70 39.4 11.38 -0.10 1.89 -0. VII.05 -0.08 -0..56 -0.48 0.63 -0.15 -1.20 0.17 Co C7 3.19 Cosine Coefficients Expressed in Arbitrary Units Curve analyzed Ii EA2 Bi Bi 0.87 As A7 A. Joseph. & Sons.23 0.83 99.60 37. in Ci 31.30 0. New York.60 0.01 -0.11 0.8 C3 C5 Ci 1.87 0.10 2.53 0. E&2 Harmonics Curve analyzed II JA .42 0. "Graphical and Mechanical Computation.7 96.09 2.71 29.54 iLiPKA.20 -0.20 Ii JA .72 1.41 1. 1918.29 C3 0.4 8.64 ELECTRIC CIRCUITS— THEORY AND APPLICATION winding 2.53 37.30 32.28 0. of the oscillograms results: WindingZ Winding 2 Winding t AVjU/ne) h To Fig.66 0. 39.21 0.28 0. Oscillograph — Diagram To Oscillograph To Oscillograph connections for three-phase leakage-impedance tests considered in Example 2.28 2. and Harmonic analysis by the 11-coordinate schedule method 1 gave the following of the current circulating in winding 3.57 32.29 #A2 0.41 -0." Chap.61 0.85 9.55 32. of Sine Coefficients Expressed in Arbitrary Units Curve analyzed 31.77 Ix j Ta3 As Ai .89 0. -9. 445 The discrepancy. 0.964 = 8. + ®^Y = of 1. Hence. The resistance of winding 3 was ohm per phase.466 -^- 1.785 X 0.466 From Solution v Third-harmonic voltage across corner of open delta = E"A 8. including the resistance of the thermocouple ammeter. when the impressed voltages and the transformers themselves are well balanced.785 amp.336 ohms winding 3 with respect to winding 2 therefore.50 . = 3»(2) "^o - = 0. = 234 volts = 8. Base the calculations on the third-harmonic components and also on meter readings alone and determine the discrepancy between the two values so obtained. is not large. fair accuracy may be obtained from meter readings alone without resorting to oscillogram analysis. The total metering resistance in the corner of winding 3. the triple-frequency reactance becomes x"[t) = ^ 1.466 The fundamental leakage reactance is. I = \j _ 1.781 Third-harmonic current in closed delta iZ = E"A = wT. 0. was 0. 1. Fi(iine) Em Im The 65 resistance of the voltmeter in the corner of winding 2 was sufficiently high so that negligible current would flow in this winding.5542 - z 3 (2) = X (o.66 ohm.445 ohm Using meter readings alone.32 3 37 A3 in Xa(2) .997 8. _ ft1 = 1501 ohms .32 X 0.02 sxim amp. calculate the separate leakage reactance of winding 3 with respect to winding 2. the following values are obtained E A2 8. the above data. as seen.TRANSFORMER IMPEDANCE The actually measured values were .32 volts = 1. The resistance part of this impedance obviously includes the resistance per phase of winding 3 as well as one-third of the metering resistance in the oorner of this winding. .554 +^Y ohms = 1395 ohms — = 0.501* - (0.445 inn 100 — = 1.785 = _ n per cent+ 4.02 volts = 1. This shows that. .465 ohm Discrepancy between the two values 465 — -—-0. 0 3.984 0.fe 111 $ ^ •« „ »°? in •SMS A3 J'" Third-harmon amperes ^.0 234. gausses 11.800 234.02 1.554 1.502 1. The data obtained are given in Table V.308 0.0 8.552 1.88 1. an aver- age value.300 13.100 222.£ 6.689 0. Table VI Oscillogram number 1 2 3 4 Average Line volts impressed 192.501 ^3 .553 1. Compare using third-harmonic quantities as well as meter readings alone.554 1.501 1. o si * S c . 11 Winding 1. As seen.987 4.0 6.66 3 13. open A. V Table Connections GO closed A.100 13.10 1. 11M S •a a c 3 $ „ £ So GO 0) go .32 1.500 1.500 1.0 Approximate Ohms Ohms impedance impedance Tpl" flux density.66 ELECTRIC CIRCUITS— THEORY AND APPLICATION EXAMPLE 3 Statement of Problem In order to determine the possible variation with saturation of the leakage reactance of winding 3 with respect to winding 2 of the laboratory transformers referred to in the preceding example and to obtain.— .0 222.66 4 13.0 4.22 0.553 1.300 210.800 3/ A .tj o <u i SS delta •« o H s 1 11.10 0. average and individual values.3 *£ » ft"S >•* i o 3 §. primary Y.43 0.310 5.2 O.000 gausses approximately. Winding 3. if necessary. additional tests were carried out (see Fig.781 0.300 12. 31'" 1.785 8.691 3.±? . Winding 2.000 to 14.60 0. this range covers what might be termed the operating region of the transformers.66 Calculate the average leakage reactance of winding 3 with respect to 2. These were performed at four values of impressed voltage so as to give flux densities of from 11. 39). Solution Table VI gives the results calculated by the process indicated in Example 2.300 192. OQ 6 11 fl > h » 8" a S c "o « •S < o .66 2 12.0 210. . therefore. impedance This holds for the figures obtained by using third-harmonic quantities as well as for those obtained from meter readings alone.TRANSFORMER IMPEDANCE It appears from the table that the four values 67 of leakage (including metering resistance) practically coincide. Inspection of the figures shows that the average leakage reactance would be the same as calculated from a single test in Example 2. substantiate the statement previously made in this chapter. These results. Hence. the reactance values obtained by eliminating the resistance part from the impedances would also coincide. that the change in leakage reactance is negligibly small for a reasonable variation of saturation in the operating range. This may particularly be the case where heavy single-phase loads. if the circuit has an even number of phases (n is an even integer). The angular displacement between the vectors belonging to consecutive phases will be k(2r/n) in the fcth of the n — 1 symmetrical systems. is that they really sinusoidal unbalanced voltages replaced by n — 1 On the other are balanced systems and may be treated as such. frequently quite laborious. such problems can be solved by the application of Ohm's and Kirchhoff's laws. be symmetrical n-phase systems of voltages and currents and one single-phase system. are connected. leaving only the symmetrical polyphase systems to be considered. and currents difficulties when is rotating machines are involved. which is due to C.E. A much more convenient scheme is the method of symmetricalThis method. p. the n — 1 symmetrical systems will not all be balanced n-phase systems. there are systems where quite severe unbalance may exist. is often inconvenient and may even lead to considerable voltages. the circuit connections are such that the single-phase system disappears. L. however.CHAPTER III UNBALANCED CIRCUITS In the operation of polyphase systems. The solution of problems involving unbalanced impedances. In spite of this fact. In many cases. 1027. to keep metrical short circuits. not be the same. L.. such as electric Unbalance is also caused by disymfurnaces. A. if the circuit has an odd number of phases (n is an odd integer). "Method of Symmetrical Coordinates Applied to the Solution of Polyphase Networks. in general.. 1918. of course. Hence. Fortescue.E. 1 is coordinates. it is usually attempted all loads and. powerful an extremely that in any n-phase system the actual the fact upon depends It and currents may. This method. The essential point. however. When sufficient data are at hand. C. hence. 68 . The phase sequence of these balanced systems will. the n — 1 symmetrical systems will all be balanced n-phase systems. one for handling problems of unbalance. hand. 1 Fortescue." Trans.I. the voltages and currents as nearly balanced as possible. they must be given the value actually corresponding to the particular conditions existing in the Often these impedances cannot be readily ascertained. to a few typical three-phase cases. Unbalanced Three-phase Y-connected Load on Unbalanced Consider the circuit shown in Fig.UNBALANCED CIRCUITS 69 The general problem of unbalance in an n-phase system will not be treated here. circuit. so that no particular difficulty should be encountered in extending the circuits. and solutions for such systems are readily worked out on the same — basis. for instance. nates will be confined to three-phase. at the sending end of the feeder are known. and it is desired to — solve for the currents. may be used in connection with other polyphase systems. Solution by Simultaneous Equations Based on Ohm's and Kirchhoff's Laws. in such cases. Unbalanced Trnpedance — Unbalanced three-phase Y-connected impedance load on unbalanced feeder. No neutral connection. however. This method will be illustrated by application Exactly the same principles. it is exceedingly difficult. the amount of current imbalance. Such problems should be analyzed by the method of symmetrical coordinates. 40. to handle problems concerning unbalanced operation of rotating machines by this method. For this reason. if not impossible. This illustrate the theory to other polyphase systems. . 40.n Fig. All impedances involved are assumed constant and independent If some of the impedances vary with of the current unbalance. and. The line voltages Feeder. and four-phase will cover most practical cases and will also fundamental principles sufficiently well. the method of solution by simultaneous equations for the several phases fails. -romj>-=^! %Za gr-. since such systems are not encountered in The discussion of the method of symmetrical coordipractice. two-phase. Za )(Z 2 + Z„) w.) . as seen. V an = hZa Vbn = hZb V cn = IcZ (10) (11) (12) c and = Van .I (Z + Z Vu = UZt F 23 = h(Z = 7.Ic(Z + Z + Z .7 (Z + Za) + Z + Z + Za)(Z + Z.ELECTRIC CIRCUITS— THEORY AND APPLICATION 70 By taking voltage drops around the circuits. (8). upon elimination of I c + 7 C (Z 2 h) 2 by making use Zb ) (5) of equation (3). the . the following equations may be written: + Z. applicable to any The formulas case. h + h + Ic = Combining equations = 7l2 and (4) (4) (1) gives + Z + Za + Z + Ia{Zy which. or unbalanced.7 = (Z 2 r ic _ ~ (Z + Z + Z + Z + Z )(Zi + Za) + (Z + Z )(Z + Z 7 i(Z + Z . + 6 3 Z + e) (Zi + . Thus. give the currents in terms of the line voltage at the sending end of the feeder and the line and load constants. the currents must add to Hence.) .) 3 b x c Equations 3 c) 6 2 23 b) 2 3 (Z 3 12 c and (9). ( which.) + (Z. + Z )(Z! + Z.Vbn = IaZa ~ IbZb = Vbc Vbn ~ V cn = hZb . the solutions for the currents in phases b and ib c become V u (Zi + Za) .I Z V ca = V cn ~ Van = IcZo ~ IaZa The impressed line voltages F i2 F 23 and F 3i may be (13) Vab c . 3 C w . (8) e) X 2 (7). Knowing the currents. becomes = L Z 7i2 l + Z* + Za +Z + (Z x + b Z a )(Z + Z z3 + z 2 b) + yVn Z + Zh z +z c 2 (6) c 3 or 7 la 7 12 (Z + 3 ~ (Zi +Z + Z 2 tt Z. the voltages to neutral and the line voltages at the load are readily computed.h(Z + Z + Z . (14) e (15) balanced so far established are general When the line and voltages are balanced.F 31 (Z 2 + Z ) b + Z )(Z. are symmetrical.i Since there zero. is b) (1) z e) (2) 1 a) (3) t 2 b) Ie{Z z a c) no neutral connection. In a similar manner. . equations (10) to (15) inclusive are used. In other words. + Z% +Z lm (19) c determined. while evidently the reciprocal is not true. obtained for the voltage + Za) + /o(Zl V n0 points: v When r *&+%&** + %&& V3 the voltage from equations 1 V n0 Z° Zb Zi + Z. h - ^/9qo _ yn0 L = Having obtained the Zz +Z (21) (22) c solutions for the currents. 2 Zb ) Zc) Fn0 + V« + 7n0 (16) (17) (18) and (18). the solution of the problem just discussed may be obtained in a slightly different way. there is obviously no displacement of the load neutral with respect to the neutral point corresponding to the . Then Fl ° = ^20 = 730 = Vl ^g X150 vl Combining equations solution is ^ — = 5 = h(Z = + + 7e(Zs (16). (20) .UNBALANCED CIRCUITS 71 voltages to neutral may be either balanced or unbalanced. the load voltages are calculated as previously indicated. >* + + Zi Zi + and - (18) as follows: V3 + z. It should be noted in this connection that a balanced system of voltages to neutral will always give rise to balanced line voltages. (17). This involves the determination of the displacement of the neutral point of the load from the neutral point corresponding to the applied voltages. the currents are obtained is (16). (4). the following between the two neutral (17). If the impressed voltages to neutral are known. Assume that the latter are actually balanced and that the magnitude of the line voltages is V. V ^150° -y. When the feeder impedances and the load impedances are symmetrical.0 Z2 + Zi z. The currents in the three phases no this circuit. This in accordance with equation (19). i. to know the impressed voltages to neutral or else to have some quantity given besides the line voltIn working out the solution below. 41. The voltage between the two Vn0 neutrals Vn o is now given by = -I n Z n (24) Evidently. namely. is where the impedance combination in the numerator equals zero for this condition.ELECTRIC CIRCUITS— THEORY AND APPLICATION 72 impressed voltages. 41. and the phase currents obtained by equations (20) to (22) inclusive. This is the same assumption that was made in the second solution of the circuit without a neutral connection. consider the circuit shown in Fig. Neutral connection. since an additional unknown. (17). is introduced. Thus. in this case.. longer add to zero but are equal to the neutral current reversed. Upon determining . In order to solve for the currents in knowledge of the impressed line voltages alone does not suffice. however. assumed that the impressed voltages to neutral are strictly balanced. The load voltages are then determined as when the neutrals are isolated. the current in the neutral. it will be ages.e. It is necessary. /„ + +L+ Ib In = (23) a Pa Unbalanced n Impedance z%K Load b Fig. and (18) still hold. and by combining these with equations (23) and (24) the following expression is obtained for the neutral current: In = 1 v V3 Z\ \3 ° + o Za + Z\ + Za\-TE7\o ^T^\ + Zn 150 ° Z\ Z<l + + Z\ ~\- Z a /90 + z7t^ i Za \Z\ -\- Zb Z% -f- Za Z c (25) c the current in the neutral. — Unbalanced three-phase Y-connected impedance load on unbalanced feeder. equations (16). the voltage V n0 is calculated by equation (24). Assume next that there is a neutral connection. these V 12 = la Zl [ -\~ ~y . Z* ) (38) . 42. y .IaZl b "+" Ica. . Three-phase on may V u = hZ ^23 = IbZ2 Vti = LZt 1 is + UZ^ . 42.Z C a Ic = lab. 7 (Zjbc" 7 y "3 "T" _j_ Z/a6 -f- ) />a& — "t" \ ca y J Z&c -r "caj i_ — Zc IafZr 4.: UNBALANCED CIRCUITS Unbalanced Feeder. as in Fig. and reduce to (28). By simultaneous solution of equations (29) to (32) inclusive. rr J — "ab T" ^6c -T ^ca/ \ + ^ Tst^y "be "T ^ea/ (36) /\ Z «+7-x¥t7-) "ab -T ^6c ^ca/ (37) *M ^2 \ ^23 — ^6 #2 4" ( \ ~r/ _i_ r/ r/ "ab ~r "be -T ^ea/J .'* ^ 2 Fig. expressions for the load currents in terms of the line currents are obtained* j lab i-a^ca J-b^bc lj ~T~~7 V £>ab ~T &bc "T "ca — i-b^ab j c ~ Z k *-*ab ca \y J~c"ca ~ 7 "ab ) (1A\ 4Zk 4. (27). Ia « Z ^\g ab ct *-v. Unbalanced A-connected.lea = Ibc — lab = La ~ he I £& lb (26) 2 (27) (28) ca ca IbcZbc ~t~ la = (29) (30) (31) (32). immediately be written down the following equations IabZab Load A-connected — When the unbalanced load 73 Unbalanced Zbc A — Unbalanced three-phase A-connected impedance load on unbalanced feeder."ca Z r "be ^ ' \ J-c"bc j /QQ\ A I *-a"ab /QKN W^v j_ y T" ^6c "T ^ca ZlT~7 Substituting for the load currents in equations (26).I Z + IbcZbc — IcZz + I Z . and (38) are just the equivalent Y-impedances of the delta. V* 1 ) <7 -J. After having determined the line currents.000/460-volt. and (38) might have been written down immediately if the for the unbalanced unbalanced A were replaced by its equivalent Y-circuit. V _l &ab "T 6bc . As a matter of fact. The generator is of sufficient capacity so that its terminal voltages may be assumed to remain constant and balanced at 11. Za = „ Z + Zab Zab Z Zbc . (37). The transformers are fed from a Y-connected generator over feeders having a resistance and a reactance per conductor of 1. 43). — Diagram of V-connected transformers for Example 1. the impedance combinations in (36).74 ELECTRIC CIRCUITS— THEORY AND APPLICATION Comparison of these equations with equations (1). and (3) Y-connected load without neutral shows complete similarity in form. . that the is (40) Zc Uca equivalent line currents are given by equations (7) to (9) inclusive. (34). therefore. (2). Obviously.0 ohms.0 ohm and 2.ilea I most convenient method of attack Y and solve the problem on It seems. 11. equations (36). the load currents (A-currents) are calculated by equations by (33).-a. Finally. the this basis. . (37).1 Generator a 1 Transformers Watts 1.. each transformer: following short-circuit data apply to Volts Amperes 265 9. ~\- „ Z (39) c ZbcZCa + + tJab Zdb ~T~ tJbc Zbc ~T~ ZcaZgb _ — 7 "c rr to replace the A by its Hence. the load voltages are obtained EXAMPLE = = IatZ* (42) hcZbc (43) V ca == J-ca^ca \"^J Vab Vbc 1 Statement of Problem Two identical 100-kv.004 a Loads Z Fig. 60-cycle transformers are oper- The ated in V-connection (Fig. respectively..200 volts between line3. 43. viz. and (35). 9 (14.9) 3.6 sin - 6.6 cos 6)] (d) gives + 2.I A Z Z = 1.0 sin 36°. however.12 26. T Ib 7L7 = g3g = On 3. the generator voltages may be written = Vea + (hi + I*e)Z = Vea + (21 A ~ IB )Z = Va+ Ia(Z.18)(1.IB Z Vi* = VB + Ib(Z.1 Ia = 119. Va will be taken as reference vector. Then (a) Vab = VB + IbZ.8 sin B - 36°.18)(14.5 "X'q = 23. It should.1 Z < X ' ohms 12 -! = V29.1 B = ¥^r = 9. 5.1 - 0. at 90 per cent power factor (lagging) and 71.9 3. be in the neighborhood of 120 deg.91) + j(VB sin + 47.7) - 36°.47.UNBALANCED CIRCUITS 75 The transformers supply power to two single-phase loads drawing currents amp. all referred to the primary side.8 3.5) - 108.18 + j sin (0 . Va = Va/0 = Va +J0 Vb = Vb/B = Ia Ib = = = = VB = 5.0[0.P - ohms = 12. respectively.7 cos + 88.0 /0 = (7a 7i 2 = Vb - + 2. + 2Z) . Determine the voltages at the Vu leads Vav Assume that Solution Ratio of Transformation: 000 -239 MJi 460 " Transformer constants referred to primary a = I1 Z = |^ = e 29.50 . Vca = VA + LiZ e (6) Taking the current directions as shown in the diagram..9 X 2.5 ohms Let Va and Vb be the secondary voltages and I a and Ib the secondary currents./2.7 amp.4 +j0 + (4.J2.0[cos (0 - (e) + j sin 6) (f) .82) (g) + J2. + 2Z) .0 (c) (d) amp.2.5 -i2. The exact phase displacement between Va and Vb is not known.0 = 2.0 amp.6 cos .9)] + 0.0 cos 6 -f- 3.8 cos 8 Making these substitutions in equations (c) and V n = 7. Thus.0 \25°.5 cent power factor (lagging).0 (Vb cos + 88.99 sin - (cos +j sin 0) 6. at 80 per of 119.0 +.0 /0 - (cos d 4.7 sin .0) (&) .1 + J3Q.36°.8.236/63^4 - j(6.236/63^4 ohms 7.0 /0 + 130.99 cos + J30.2) (4.5) - 3. it will be introduced as 8 deg.50 36°. loads and the power delivered to each.6 sin + j(0. At present. 5 = 11.7 volts Using this value of Vb. however.200/0 = + (-0.5 Vn = 11.063. Hence. As a second trial.063 " = 462.5 cos gives (g) = 11. are correct.2 volts sin 11. the vector expression for 7i2 becomes 7i2 = -5.5 volts which is voltages exactly correct. = Vn = 0../104.„_ u = 465 volts 23 9 The amounts of power absorbed by the two loads are.86347b + 45.199.7 1^00 X 0.'(0.200/a = (VA + 136.063. is now Hence.6 = 120 deg.200/0^6 volts Similar substitution in equation (h) gives 7 i2 = 11. the vector expression for 7si becomes Vn = 11.2 = 11.2002 trial-and-error process.200/a VA = Vn = 11. therefore.5 2 from which VA = With 11.82 - 94.5 = 11.669.47b +jl04.i 11. • 120.6 volts expression for 7i2 now becomes 7. assume are balanced.9 +^9.3 - = 0. as found by the second trial.200/0 = (-0.112. it is safe to assume that the values of the Vb.199 5 +. This angle.6) + 45.8 0. =26. = 120. Va and .3 deg.5 p P » = 465 X 119.200/0^6 and ll^OO2 = V\ = -0.50457b + 174. must be modified.866 sin Substituting these values in equation F.5 2 + 46.3) = (V A + 136.5) +.02 from which VB = 11.7 deg.ELECTRIC CIRCUITS— THEORY AND APPLICATION 76 These equations at first that is now be will by a Then solved exactly 120 deg. since the generator voltages The original assumption in regard to the phase angle between — Va and Vb.6 7 i2 and 7 is 3i 119.650.(VA + 94.6 .9 = 498 kw „ .5VB - + + = V% 174.5 + J104.3) 2 + 104.7kw.0.5045 cos .27b 94. Assume = -0.8) volts + .0) 46. should be exactly 120 deg.6 . - .2 volts this value of Va.2002 11.5 X mmx) X 71.6* from which VB = The vector The 11.a = 120.3) +il04.866Fb + 94. hence._ .112.6 = 11.112./(0. Referring these values to the low-tension (secondary) side gives for the load voltages 2 T7 VA = 11.200/120^6 volts calculated phase displacement between 7i 2 and 7 3 i p . VB = Jr p = Pa 462.2 = -5701.200/120^3 volts The calculated phase displacement between - a.5 136.1 +J9640. 239 11.8634 11. = 0. Hence. are considered to be complex quantities which.. Evidently. voltages. General. separately satisfied. b. however. etc. since. at the present. Mr. the unbalanced voltages may each be written in terms of three components as follows: Va = V = V = + + + + + x z (45) y cz ax by (46) dx ey ffz (47) c It will be noted that the components of Vb and V e have been written as a factor times the corresponding components of Va This is perfectly permissible and does not reduce the generality of the resolution if the coefficients a. unless the component voltage systems can be handled more It readily than the original unbalanced voltage system itself. This is the most appropriate resolution and gives a maximum of convenience of handling. Fortescue. may have any arbitrary magnitude and angle consistent with each equation's being b . each one of these voltages may be resolved into any number of arbitrary components. method in his exposition of the of symmetrical coordinates applied to w-phase systems. unbalanced general in the Hence. or balanced is the system zero whether case. provided only the sum of the components is equal to the particular voltage which they repreNothing at all is gained. by such resolution. ConAssume that sider the voltages to neutral in a three-phase system. value different from zero. add not and do these voltages are unbalanced is always system three-phase in a line voltages the sum of the the case this is not not. Thus. will now be shown that the unbalanced voltages whose sum is different from zero may be replaced by two balanced three-phase systems of opposite phase sequence and one single-phase system. Although to zero.UNBALANCED CIRCUITS 77 — Symmetrical Phase Coordinates. calls these coefficients sequence operators and determines their values for resolution into symmetrical polyphase and single-phase systems. + b + e)y + (1 V + V b c) 3Fo (48) (49) . as a rule. phase with the the sum of the voltages to neutral. most circuits are susceptible to easy solution where nothing but balanced three-phase sent. or single-phase quantities are involved. Va + Vb + V c has a definite Method of . Assume that the sum Va + Vb + V = c (1 of the three voltages + a + d)x + is 3F (l+c+/)* = or z V = H(V a -F . This will always hold true when these components each form balanced systems. it is evident that the superposition of the two will give any of the original vectors may appear uncertain when the zero-sequence component is considerable. quantity V is subtracted from each of the three voltages. If this it is Thus. the system whose phase voltages are y z. their resultant {i. for these systems become + b c The ^-system is = = l\i20° 1 /120° = and / = and e + + 1/120° (54) 1\120° (55) called the positive-sequence (or direct-phase) system and has the same phase order as the original unbalanced This assumes that the phase order of the latter is a-b-c in a counterclockwise direction.) (51) and since the single-phase components in the three phases are + considered equal.7 + ) (7.e. Obviously. . l adhered to throughout this discussion... however. viz. Let x be this component. the subtraction of the zero-sequence voltage from the original unbalanced voltages does not necessarily result in the voltages so obtained forming a balanced system.ELECTRIC CIRCUITS— THEORY AND "APPLICATION 78 . it will remain unbalanced. (Va - + Vo) (Vb . are given the same phase sequence. evident that the remaining vectors will form a closed triangle. the sum of the y and z components in the three phases must always be zero. In such cases. - Vo) = (50) This is equivalent to considering a single-phase (uniphase or zerosequence) component Vo to be present in each of the original voltages. and ey fz) will always be a balanced system. Consequently. If these sytems. the correct phase order is the same as that of the vectors found by subtracting the zero1 The phase order sequence component from each of the original vectors. the zero-sequence operators are a From = 1 and equations (48) and (50). therefore. The z-system is termed the negative-sequence (or reverse-phase) system and has a phase order opposite to that of the original vectors. In order to take care of this. the phase sequence of the balanced y and z systems must be taken opposite. (52) now follows + c +f)z = it (1 all 1 that (53) In other words. x = V = H(Va Vb + V. When the phase sequence of the two balanced component systems is opposite and the relative size and phase of their vectors is properly adjusted. The sequence operators. + + b (1 e)y + d = unity. by cz. an assumption which will be vectors. In general. negative-.and negative-sequence components. in general. however. (46). and zero-sequence components) as indicated in the following equations /. the resolution of three-phase voltages and currents only has been considered. Equations (45). The symbol Vq has already been introduced for the x-system or zero-sequence components. . the unbalanced voltages and currents be resolved either into one single-phase and one balanced two-phase system of voltages and currents or into two balanced two-phase systems of opposite phase order. The symbols V+ and V~ with appropriate subscripts representing phase will be used for the yand 2-systems. general equations for the symmetrical phase components of such systems may also be established. be split into positive-. In the two-phase system. respectively. to two-phase and four-phase circuits. therefore. be written in a more convenient nomenclature. Applying a similar reasoning. The problem is next to determine these components when the unbalanced voltages are known. an unbalanced current system may.'+ V Vb = + V> + V Vc = Vt + V.+ V (56) n (57) (58) or Va = Vi/0 + V-/0 vb ^ vt\m° + F = 7+/120 + c +V y-/i2o° F7\l20"° (59) +v + Fo (60) (61) A system of unbalanced currents may be treated exactly as a system of unbalanced voltages.=/+ + I' + h = n + /r + I = 7+ + /J + /o (62) Jo (63) 7 (64) or = = h 7 = 7a C + 77/0 + 7 7+\l20"° + 77/120 + 7 7+/120 + 77\120 + 7 7+/0 = H(h (66) c The zero-sequence component 7 (65) c of current is given +h+7 C) (67) by (68) So far. as follows: Va =V+ + V. and (47) may.UNBALANCED CIRCUITS desired system of unbalanced voltages whose 79 sum is zero. Hence. or positive. The latter resolution may is preferable. voltages Va and 76 may then be The unbalanced two-phase resolved as follows: The Va =Vt+V- Vb = Vt = 7+/0 + y-«ll V b = VfW5 + V-/90° + current components are given la h The most neutral. in (69) (70) by = 7+ + la = 7+/0 + = It + K = ifW 'a/0 (71) + (72) 7 q/90° general four-phase (or quarter-phase) system has a case. thus. It not determined by the same general rule which applies to three-phase and other polyphase systems. the statement is often made that there can be no zero-sequence components in a two-phase system independent therefore. of is whether a three-conductor or a four-conductor circuit is used.ELECTRIC CIRCUITS— THEORY AND APPLICATION 80 should be noted that. be written Va Vb Ve = V++V~+ VI + 7 = Vt + V> + V*b + V = V++V7+ V: + V vd = n+ y~* + V* + ^o (73) (74) (75) < 76 ) or Va Vb V 7d c The = 7+/0 + 7-/0 + Vl/0 + 7„ = 7+\90 5 + 7-/180° + 7:/90^ + V = 7+/180 + V-/0 + 7:/180° + 7 = 7+/90° + 7^ /180° + 7:\90 5 + 7 current equations h /6 7C /<* = = = = n+ + + + it It (78) (79) (80) become /- it 1+ (77) /7 '7 + + io + n + Jo + It + Jo + Ia + Jo si) /: < 82 ) (83) (84) . of the unbalanced vectors. the single-phase component is not found by taking one-half the sum This zero-sequence component. Since the resolution into two balanced two-phase systems is generally used. an additional symbol must be added to the previous notation of plus and minus to indicate the third symmetrical system. An asterisk will be used for this purpose. The general equations for the voltages to neutral may. zero-sequence components may be present which in the voltages to neutral as well as in the currents. if the former resolution is used. Since in a four-phase system there will be three symmetrical component systems. /180° (86) (87) • (88) Components in a Three-phase is given by equation (49). and (61). the positive. Thus.may now be determined by substituting equation (95) in equation (89). Solving for b = F+/120° c equation (89).\m° C -V~a (96) giving V. equations (59).and negative-sequence componentsjnay be expressed in terms of two of the . hv + hv \W .Hv.HV* . (60).y v V3F„\30° K7 \l20° = 3\/3F+\30° (93) (94) gives V+a = H(Va + F6 /120° + F \120°) (95) c y. upon contraction.V+JWV be written + V3IV90 - \/SVc/Z0° F+ + : may which.HV. = F+\120° + 7-/120° -MV.= H(Va + 76X120° + F /120°) c (97) Equations (95) and (97) give the positive. obtained HvSW ./180° + 7:\90° + 7 7+/0 + /-/0 a (85) 7.HV.HVa . ~ HVb + %V = 7+/120° + F-\120° 2 AV (89) a (90) h (91) C Operating on equation (90) by \120° gives -nV \m° + a %n\120° _ By subtracting this from following equation HVa + is 3 V~ is V- (92) and the eliminated. these Analytical Determination of the System. — The zero-sequence voltage Substituting this value in become Hv* .UNBALANCED CIRCUITS 81 or = h = 7 = 7d = /„ C + r /0 + h 7+\90° + + I*J9(F + 7o 7+/180_° + 7-/0 + 7:/180° + 7 7+/90° + 7. = vt + v-g -HVa + %V .HV.Hv*nm° - HV .and negativesequence voltages in terms of the three unbalanced voltages. HVa . If the latter add to zero or if the zero-sequence component be subtracted from the original unbalanced voltages so that the remaining vectors form a closed triangle.xm 5 - c c = V+a . World. First Method: Voltage components: V = H(Va + V + 7 y+a = H(Va + V /120^ + 7 \l20°") V. W./120° primes. upon contraction.and negative-sequence components by means of equations (104) and (106) is particularly convenient when no zero-sequence components are present. : zero by may be written = Va -Vo = V+J_0 + V-/0 = Vb -V = 7+\120° + 7. "Unbalanced Three-phase C) (107) c (108) C (109) Circuits. In this case.. the following equation V^ V'b v = e . . since obviously Va = The components rents may V'a and 7& = V'b of a three-phase . V~a = ^3 (7 -' + 7*W)W° ( 106 ) Determination of the positive. which add to voltages only.7 '\120° = V'm - 7+/0 6 7+/120 (102) or V'a 7+ Solving for + 7 '/60° 6 VSV+W° = (103) gives Ft = 4= F + V3 Substituting this value of ~ ( 7+ v a = Ya ~ a n/60°)/30° (104) in equation (98) gives — ~ ^k — ' f ^h f (105) Hence.: AND APPLICATION ELECTRIC CIRCUITS— THEORY 82 Indicating the unbalanced voltages. p. V. system of unbalanced cur- be determined exactly as discussed for the voltages. for voltages as well as currents will be collected The formulas below for rapid reference.= HiVa + 7>\120 5 + 7 /120°) b 6 1 Lyon. 1304. 1920." Elec. the original voltages Va and 7& are used in the formulas.F = To 7+/i20°_ 5 + v-xm (98) (99) (100) Operating on equation (99) by \120° gives + ^\I20° = V+/J2F 7-/0 (101) Subtracting equation (101) from equation (98) eliminates V~ and leaves . Solution Figure 44 1. In order to check up on the conditions.2 deg. one of the permanently connected line voltmeters reads low. three-phase. 87.6 = 119.: UNBALANCED CIRCUITS 83 Current components: h = H(h + h + Ic) = HV* + ^/i2o° + iSJW I~a = M(h + /6\120"° + 7 /120_°) it (110) (in) ) (112) C Second Method: Voltage components: Vo ^ V'b = H(Ya + Vb = F a . the three-line voltages and two of the voltages to neutral are accurately recorded as follows Vab = V = be Voa = 200 VoltS 230 volts 230 volts Van = Vbn = 1. Determine the symmetrical phase components of the line voltages.Vl33 2 2.Fo =V b - +F (113) c) (114) Vo (115) n = ^= (^ + nV60°)/30 7" = (7 ° Vl + o (116) ^W)W (117) Current components: h = HVa + h + Ic) h- I'b It = -~{I'a + 76760° ) /30° /a ^ = -L EXAMPLE (118) = = /£ /o (119) 7o (120) (/ + / jfteo 5 )^ (121) 5 (122) 2 Statement of Problem In a 230-volt.5 volts . 133 Volts 133 volts Vcn ? of the voltages to neutral. - voltages. cos a = 1( >%33 = 0.752 1002 and phase = 207. What 2.1 a = 41. Y-connected circuit. 3. The is a vector diagram of the line third voltage to neutral is Ven = V230T^T0(P . Determine the symmetrical phase components is the third voltage to neutral. thus indicating the presence of unbalance. 8 +j"130.0 -J87.6 = 11.6 .9 .2 volts V.84 ELECTRIC CIRCUITS— THEORY AND APPLICATION Using Vab as standard phase.25.5\150°) = H(100 -J87.3 \T50^2 \3~0~5 = 19.8 -J63.j'59.j'87.0 .6 = 18.5 -J190.8 .6\90° volts Vt = }i(Van + Fbn /120^ + F „\120°) = H(133\4T\2 + 133\18°.6 .6 .6 volts = -100.8 .103.j"5.j'87.2 + 119.4 = 126.119. c ) ) Vca "-Vab ZOOvolis Fig. Check Vo + Vt + V.5 volts of these voltages become = H(Va „ + Vbn + Vcn) = ^(_jl75.2 133\138^8 119.#9.0 volts y-a = H(V«» + vbn \m° + 1W120 = ^(133\4l^2 + 13 3/101°.6 volts = J 119.2) = 109.6 \180^2 volts Check V+b + Vol = 219.7\30°.6 + 125.07 = 200 + J0. the symmetrical phase components of an unbalanced system of three-phase vectors may be found graphically.0 .07 ^ 200 = Vat Graphical Determination of the Components in a Three-phase System.6 = Van 9.j'42.5/90^ The symmetrical components = 100.8) = -9.7 /30j\= 219. — .2 + .= -J18.8) = ^ (329. The graphical methods are based directly on the previously developed formulas. When the desired degree of precision is not too great.j5.8 + 119.8 .6/0 volts W° V~b = V3 F^W = V3 X 11. 44.6 The positive3.J16. the vector expressions for the voltages to neutral are Van = Vbn = Vcn = 133\41°.5X30 5 = H(100 .6 .J63.6 - 19. and negative-sequence components of the line voltage Vab are y+ = v/ 3F„+„/30° = V3 X 126.4 = 100.6 + 109.8 + 103.J87.6 + J0.4 . The line voltages contain no zero-sequence components.5) = -J18. — Vector diagrams of line voltages and voltages to neutral (Example 2).3\150°.8) = ^(_29. negative-. Graphical determination of positive-. S vj+vtfsd .— Graphical determination of positive. Va '+V6 '\m [30}va Va'+Vb 'l60?-\ Fig. and zero-sequence voltage components in a three-phase system. in a negative direction and the vector V through c 120 deg. in a negative direction and adding them to the vector V a. nents in magnitude and phase. in .. Addition. in a positive direction and V c through 120 deg. is accordance with equations (107). c. as in b. the negative-sequence component of phase a is obtained. It will be noted that the procedure followed . (108). by rotating the vector Vb through 120 deg. 46. as the positive-sequence component of phase a is Finally. By rotating the vector Vh through 120 deg. 45. immediately gives the zero-sequence compo- VC#£0°^V6 \720* Fig. In Fig. the vectors a represent the unbalanced voltages. and (109).e.and negative-sequence volta K e in a three-phase system when the zero-sequence component is either components absent or subtracted from the original unbalanced voltages. in a positive direction and again adding them to the vector V a as shown in d. scale. 45.UNBALANCED CIRCUITS 85 the operations indicated by the equations are simply carried out graphically by drawing the vectors involved to a suitable i. indicated in determined. The former calls for rotation of the vector 7 & through 60 deg. in a positive direction. It is based on equations (116) and (117). in a positive direction. triangles equilateral two particularly suitable These lines divided by V% give the magnitudes of the positiveand negative-sequence components. and rotated on V'b is + 7 \60°. as shown in a. Evidently. /90° gives by equation (70) on (123) 7 b /90° = 7+/0 + Fq/180° Adding this to equation (69) eliminates 7". can always be resolved into The two voltages. It is when only the magnitudes of the components can be obtained by merely constructing the These required. the construction is carried out in accordance with equation (117).ELECTRIC CIRCUITS— THEORY AND APPLICATION 86 Figure 46 illustrates a graphical solution applicable to unbalanced vectors whose sum is zero. and negative-sequence and (70). in a negative direction. Construction of an isosceles triangle with 1 to 4 as the base line by laying off the two 30-deg. are and scaling off the lines 1-4 and 1-5. magnitude and phase. Analytical Determination of the Components in Two-phase and Four-phase Systems. In order to obtain the negative-sequence component. angles immediately accomplishes this end and locates Vt in magnitude and phase. and by constructing an by laying off two 30-deg. This operation is readily carried out by constructing an equilateral triangle on this vector. the line 1 to 4 is then equal to V'a + V3 This quantity is now 76/60°. through 30 deg. equal to V'a b isosceles triangle with this line as base angles 7« As is obtained in seen. this graphical scheme is extremely simple. as indicated. and the following solution for Vt is positive obtained Vt = H(Va +F 6 /90_°) (124) Substituting this value in equation (69) gives for the negative- sequence components V~a = my a + IV\W) (125) . equilateral triangle The vector 7& b. for instance. Operating equations (69) by indicated as components. to be divided by The The line 1 to 5 is is now constructed as indicated in thus rotated 60 deg. respectively. As already stated. there is no necessity of considering any zero-sequence components of voltage and cur- — rent in a two-phase system. + %v d = (129) b By n/90° + 7 q/180° + 7*\90° (130) adding equations (127) and (128) and equations (128) and (129). The zero-sequence voltage is given by V = H(V a + Vb + 7 + V d C (126) ) Substituting this in equations (77) to (80) inclusive gives %V . In a four-phase (quarter-phase) system. as Finally.HV + %V .HV = 7+ /0 + Va/0 + V*/0 (127) -H v + y v -y v -yv = vt\9o° + y-/i8o°+ 7301(128) .V27d /45_ = 4V2nW° from which Vt - M(7 a + n/90° + 7 /180° + C Fd\go*) (133) (134) Addition of equations (127) and (129) eliminates both Vt and 72.}iV . zero-sequence components may be present in the voltages to neutral as well as in the Similar expressions current. graphical solutions may be obtained for the sym- .Av + y v + y v . Vt is obtained.V2 7 a*\45° (132) nv + y v .}iV . these components must be considered. the following equations are obtained: — (131) .y v = V2 v+ W° + V2VZ/45 .HV . Hence.y v = a x 2 b a 2 2 2 c b 2 d 2 c - d \/2 7+/45° Operating on equation (132) by \90° and subtracting from equation (131) eliminates VZ and gives \/2 ya \45"° + V2 7 6 /45° .y v . 7: = ) (136) in either equation (131) H(7o + 76\90°+7 /180^+7d /W) c (137) Graphical Determination of the Components in Two-phase and Four-phase Systems.UNBALANCED CIRCUITS 87 may be written for the current components. —By basing the construction on the estab- lished formulas. in order to cover the general case. 77 becomes Fa = K(7« + 76/180° + 7 + 7d /180° C by substituting equation (134) or equation (132).Hv -KV. giving HVa - yv + y v -y2 v b 2 c d = 27^ /o (135) Hence.a/2F c \45^ .HV a .HVa = a b a 4c a d c b d c 4c b C + V^/0 + V*a /J80^ 7+/180 . in a negative direction and added to V a giving the negativesequence voltage Fl. le) — Graphical determination of positive-. .and negative-sequence voltages of a two-phase system.and negative-sequence voltage components in a two-phase system. Va . as indicated in phase. Vdim td) Fig. used — Graphical is determination of positive. 47. The origidiagram a. The vector a positive direction and added to in accordance with equations (124) nal unbalanced vectors are V b is rotated shown through 90 deg. The process Fiq.. and (125).ELECTRIC CIRCUITS— THEORY AND APPLICATION 88 metrical components of unbalanced two-phase and four-phase systems. obtained in magnitude and rotated through 90 deg. Thus. Vt the vector is V h is . in in Vcim" Vd\90° vb Mo' 4Va Vd/M° -it. negative-. 48. In diagram b. c. and zero-sequence voltage components in a four-phase system. Figure 47 shows such a graphical solution for the positive. In general. and (137). In is vectors after four times V+. to b. P =Pa + P +Pc b (144) . are scalar quantities and not When vectors. by adding the having rotated Vb and Vd through 180 deg. Power Unbalanced in unbalanced — The power per phase of an depends on the .. in a negative direction. in a positive direction. as well as in the subsequent equations for power. and Vd through 90 deg. circuit evidently and power-factor angle of that phase. in a positive direction. negative-. . be written COS d^ + Vala cos *Y? 'a V+Ia cos 6^1 1a Vah cos d Va T + la V+Io cos d + VI 7 /+ Valt cos lo + J-0 V VoI+ cos e Tl la + cos d V ° T i0 d V l T ij Vola cos oil la + + (143) Similar expressions may also be written for phases b and c. this manner The sum of the vectors when laid off in Diagram d gives V~. however. phase a Pa may = Vtli and currents are resolved into and zero-sequence components. and c Pa Pb Pc The more unbalanced three-phase the are given specifi- circuit. V c through 180 deg. Vb is rotated through 90 deg. This construction determines F*. the actual power per phase may be calculated.. These statements are perfectly general and hold for any number The of phases. the power in the. Diagram b gives the sum of the vectors and thus determines the zero-sequence voltage V In c. The by = VJa COS da = VhIb cos 6b = VJc cos 6 (140) (141) (142) C voltages and currents in these equations. diagram e. will be cally confined power in phases a. Diagram a shows the original unbalanced four-phase vectors. voltages positive-. Circuits. the power developed or utilized in the various phases will not be the same when unbalance is present. discussion below. the vector Vb is rotated through 90 deg. in a negative direction. V c through 180 deg. current. The total power in a three-phase system equals the sum of the phase powers. (134). and Vd through 90 deg. (136). based on equations (126).UNBALANCED CIRCUITS 89 Figure 48 shows the construction for a four-phase system. From these equations. Thus.voltage. . "Single-phase Power Production.E. and G. lhus. and three-phase circuit is produced by positive. Hill. negative-sequence.E. 1916. ." Trans.and negative-sequence components only. "Supply of Single-phase (149) Loads from Central Stations. "Single-phase Power Service from Central Stations.. ^+ copper loss in any phase of an unbalThe Fig *49. A.-Vector diagram of positive-..ELECTRIC CIRCUITS— THEORY AND APPLICATION 90 component and power-factor angles by adding equation (143) and the corresponding equations for the two other phases.. . . Philip. Gilman. Alexanderson. The effect of the negative-sequence components is thus to transfer power from one phase to power is This ability of the negativesequence components to transfer power is 1 utilized in the phase balancer.I. and zero-sequence components. and anced circuit equals the product of the another.R. three-phase system. negative-.I. 1329. it will be found that the sums of the terms involving unlike components are zero. p.E.E. E. F. zero-sequence currents in a D^ three-phase system. the total power in an unbalanced equal to the sum of the powers due to the positive-sequence. = 3(P++P-+Po) (145) Hence. A. ... Q f the ac t ual current Carried by that _. Paw = Hra = Phcu. and C. 1315. tor a phase times its resistance.E. L. p. W. 1916.1293. Since the magnitude of like components is the same in all three phases. H.E.cos 0" + V h cos O ) The total power is readily evaluated in terms of the voltages.I. 1916. p. the power in at least one phase will be greater and in at least one phase less than the average power per phase.. P C(CU If there is loss in this is c + + hYu dt+ IT + hYn (It + I' + Io)*r. Copper Losses in Unbalanced Circuits." Trans. (It I~a (146) (147) (148) a neutral connection carrying current. In performing this addition. Fortescue. the subscripts referring to phase may be dropped and the expression for the total power written as p = 3(7+7+ cos 0+ + V-I. A. currents. In such cases." Trans. E. Quite often the zero-sequence components are absent. the copper given by Pn^= 1 ) = Hn = = I\r = IlTn = (h + h + /)*« = 9PQT n Torchio. 0) sin (120° 0)]}r a) sin (240° (153) + 7~7 The sum zero. 49 is the vector diagram of component curMaking use of this diagram. This is true for lines and transformers but is not. and rents. true for rotating machines.0) + {77 sin (120° = 2 - 2 a) e 2 2 c 777 sin (120° The - - a) sin (240° total copper loss in the three phases may - 0)]re (152) be obtained by adding equations (150).0)] sin (240° .a) cos (240° . negative-. Above. . equations (146).0) + 777 cos (120° .0)} ]r [(7t) + (77) + 7 ]r + 2[(7+77 cos (120° . (147). and sum zero- sequence currents considered separately. (151). may (148) for the copper losses per phase Pa(cu) + [(It)* + = = [(It 2[ltla be written + + (7a (la)* + n\Ta + cos a + Itlo cos + 777 la cos 7 cos 0) 2 a sin cos Zo/o sin Pko«) = [{It + /? cos (240° a) + 7„ cos sin 0) 2 ]ra a cos + a sin 0]ra (120° - jS)} (150) + 2 + 7„ sin (120° .a) cos (120° .UNBALANCED CIRCUITS 91 Assume that Fig.a) cos (240° . the total copper loss (exclusive of possible loss in the becomes neutral) *V> = + 3[(7+) 2 (7-) 2 + 7 ]r + + cos(120° 2{7+7"[cos a 7+7 7"7 + [cos [cos 2 cos (120° + a cos - - 0) cos (240° a) + cos(240° - + cos (240° - a) cos (120° + + 0)] .a) sin (120° . Picuy The + sin = 3[(7+) 2 + (7-) 2 + total copper loss in the three phases of the copper losses due to the is (7 2 ) (154) ]r thus equal to the positive-. it has tacitly been assumed that the resistance was the same for the several components of current. Dropping subscripts referring to phase and assuming that the phase resistances are identical.a) + 7+7 cos (240° .0) + {7r sin (240° - - +I a 2 a) 6 2 2 b 7^Zo sin (240° P c «7«) = [{7+ + 77 cos (120° - a) a) sin (120° +7 cos (240° - (151) 0)]ri 2 /3)} + + 7„ sin (240° .0) + 7^7 cos (240° . and (152).0) + a)] cos (120° .a) + 7 +7 cos (120° . in general. [sin a - - + of the trigonometric functions in each of the brackets is Hence.0)} ]r = [C#) + dTY + / oh + 2[ItI* cos (240° . h hZo c )] (162) where 5 Z+ = 3 a Z_ y (Z + Z„/120° + ZM20 = y (Z + Z \l20°" + Z /120°) 3 a Zo = H(Za ) 6 + Zb c + Z e) (163) (164) (165) . and zero-sequence impedance of each phase are identical.— y-c onnected across the three impedances. the total copper loss is given by P (Ctt) - + 3[(7+)V+ Of course. the positive-. edanCea They may be written NttTrftur . Fig. respectively. equation (155) also (7-)V- + Pr (155) ] assumes that the component resistances of the three phases are identical.ELECTRIC CIRCUITS— THEORY AND APPLICATION 92 When these resistances are different. 50. and V c are the voltage drops 50. c (160) c (161) and (158) in equation (159) gives the zero-sequence voltage IcZc) Vo = MVaZa b + ZJ + la + h)Z + = Hidt + = H[It(Z a a + Z6 \120 5 __ + 77 + U)Z + (7+ /120° + 7a/W + U)Z + Z /120°) + l7(Z a + Z>/120° + Z \l20°) + Io(Z a + Z + Z (7+\120° /120° + 7«Z+ + C\ c b f = HZ. as shown in Fig. and Zerosequence Impedance of an Unbalanced Load. (159) e) + F \120°) + y /120°) (157). the component voltages in terms of the actual voltages are given by From V = y (V + V + V Vt = H(Va + 76 /120° Q Va = a 3 b + 7 \120° VsiVa 6 Substituting equations (156). Consider a circuit. Negative-. negative-. Component Voltages in a Three-phase System in Terms of Impedance Drops. ' V a = IaZ a V b = hZ h Vc = (156) (157) IcZ c (158) previous derivations. In other words. V a Vb. consisting of three unbalanced Y-connected impedances whose values are the same — component systems for the of current. Equivalent Positive-. It is seen that. (160). The impedance Z+. In such a circuit. (164). If the circuit is symmetrical. As the above equations indicate and this should be carefully noted and appreciated these equivalent impedances are not positive-. and zero-sequence imped- — — ances in the ordinary sense. are both zero. the positive. the zero-sequence current is suppressed. They are named equivalent positive-. and Za = Zb = Z c . With the A-connection. positive-. fore. 51) may be treated exactly as the Y-connected circuit. all three current components contribute to each compo- nent drop. When the Y-connected circuit has no neutral connection. and the zero-sequence drop by a zero-sequence current only. negative-. If. can be equated to zero. (166). (157). in the sense that each should be used exclusively with the current of the corresponding sequence. and zero-sequence impedances. equation (162). the negative-sequence drop by a negative-sequence current only. and (165)) by a process exactly similar to the one used in determining the symmetrical components of a system of unbalanced voltages or currents (equations (159). giving the zero-sequence voltage in terms of component impedance drops. independent of whether or not the line voltages are balanced. the circuit is symmetrical. Hence. of the unbalanced three-phase circuit.e. voltage . Z-. in general.— A-connected unbalanced impedances. however. negative-. the impedances while Z = Z+ and Z_ Z. negative-. there- drop is *%"» caused by a positive-sequence current Fig. the zero-sequence voltage must always be zero. the positive-sequence An unsymmetrical A-connected circuit (Fig. and Z may be termed the equivalent positive-. and (161)). respectively. and negative-sequence voltage drops in terms of the component currents and a combination of the impedances. the zero-sequence is always zero. and (167) express the zero-. only.. and (158) in equations and (161).and negative-sequence voltage drops in phase a may be written Vt = ItZ + IaZ_ Va = ItZ+ + IaZ + hZ+ + 7 Z_ (166) (167) Equations (162).UNBALANCED CIRCUITS By (160) 93 substituting equations (156).51. and the impedances combined in the same manner. "at the same time. and zerosequence impedances merely because they are evaluated from the unbalanced impedances (equations (163). i.Z. 313. (Degree of unbalance) (Degree of unbalance) 1 See. for instance. the "degree of unbalance" which may be conventional expression for degree said to exist in the circuit. equations in (166) and (167). "Induction cuits. Slepian. to the positive-sequence component as an indication of the degree of unbalance. it is to express.and negaSubstituting." In general. J. the degree of voltage unbalance is given by the degree of negative-sequence unbalance and the of the negative-sequence 1 degree of zero-sequence unbalance.: ELECTRIC CIRCUITS— THEORY AND APPLICATION 94 and the following expression obtained U = for zero-sequence the -—. A of unbalance. - Vi = tt(z V~a = Ii(z+ ^) + -0 -^) I*(z_ -^) + Ia(z (169) (170) symmetrical. therefore. the zero-sequence current must of whether or not the impressed independent always be zero. - = y+ (171) = y^ (172) Motors on Unbalanced Cir- . p.sequence voltages. of Unbalance. the positiveequation from (168) and negative-sequence voltages may be written in terms of impedance drops due to positive. should be fixed by definition. it seems logical in an unbalanced circuit to use the numerical ratio and zero-sequence components. readily seen from equation (168). is This balanced. World. however. 1920. Hence.and negative-sequence currents only. numerically. the degrees of voltage and current unbalance will be different and should be calculated separately.+ current 7 ° Z+ (168) Equations (166) and (167) still hold for the positive. These ratios designating the degree of unbalance in a circuit may also be termed "unbalance factors. Thus. the value of I tive. respectively. When the conditions in a circuit are completely balanced." Elec. —When the voltages and currents in a sometimes convenient circuit are unbalanced. thus. voltages are Z+ and Z_ become zero for a both in that mind is borne when it If the A-circuit symmetrical Degree polyphase is circuit. only positive-sequence voltages and currents are present. At any point in a circuit. It is possible in tance * n <? capacitance vconnected. simultaneously.UNBALANCED CIRCUITS Similarly. The degree of unbalance may have any numerical value from zero to infinity. 52. is considered positive. be encountered only under abnormal Fig. therefore.— Pure inducoperating conditions. however. There is. It is probably safe to say that 100 per cent unbalance is the maximum ^ 3 which is likely to occur in power circuits. which obviously represents the most extreme case. abnor=* mally high degrees of unbalance are seldom I met with. the magnitude of the currents flowing in the two branches of the open delta will be the same. * and even this degree of unbalance will. The values of . 53. as Z a rule. one for the negative- sequence and one for the zero-sequence degree of unbalance. as a matter of fact.inductance and capacitance are such that the two tive sequence unbalance and 100 per ^ * branches have identical cent zero-sequence unbalance to occur numerical reactance. . Evidently. — . no way of combining these into one readily interpretable value. If the inductance and capacitance be given such values that they offer the same numerical reactance. This may seem rather astonishing at first thought. In order to illustrate the possibility of infinite unbalance. The vector diagram of voltages and currents in this circuit is given in Fig. and. the degree of current unbalance at circuit is given 95 any point in a by (Degree of unbalance) (Degree of unbalance) - = = j^ (173) y^ (174) In order to specify the degree of voltage or current unbalance completely. since any inductance would have some resistance and any condenser some leakance. this circuit is only approximately realizable. apparently. Inspection of the vectors representing line currents reveals the fact that they form a balanced system whose phase order is opposite to that of the voltages. for 100 per cent nega. Only when either the negative-sequence or the zero-sequence system is absent will one figure suffice. If the phase order of the voltages. consider the circuit shown in Fig. the currents are all . 52. Balanced three-phase voltages are impressed on a pure inductance and capacitance connected in open delta. such cases. two values must be given. 54. are infinite. Hence. Fig. Fig. circuit is the relative phase relation of the sidering the voltage V a as standard phase. therefore.ELECTRIC CIRCUITS— THEORY AND APPLICATION 96 They contain no positive-sequence or zero-sequence components. tance are connected line to neutral. (111). Diagram of a circuit in which a pure inductance and a pure capaciFig. Fig. neutral are strictly reactance are numerically equal. 53. The voltages to three phase circuit. the currents contain no positive-sequence component. — — The third phase is open. . 53. + /\30°) = ^\60 Hd \9o° + / /<xn Z\150°) =^7|\ 12° and (112) = c o C As seen. (Degree of unbalance) - = = 7+ 7T = °° Figure 54 shows a circuit in which a pure inductance and capacitance are connected line to neutral in two of the phases of a Vca a ^Vab Fig. the component currents are readily deter- \ mined by equations and become U = — 55. so capacitive reactance and the that equal currents flow. 52. \ gram /« = it ^W + 1 ^(I\90 = C (110). Both the negative-sequence and the zero-sequence degree of current unbalance. The degree of (negative-sequence) current unbalance will consequently be negative-sequence currents. 55. The vector diagram for this Vc showing then given in Fig. 54. and it is assumed that the inductive balanced. Vector diaof voltages and currents in the circuit in Fig. Con- phase voltages and line currents. The values of inductance and capacitance are such that the two branches have the same numerical reactance. Vector diagram of voltages and currents in tho circuit in Fig. 54. 57. when phase a voltage unbalance Taking the is determine the degree of short-circuited. 56. I- = ~^ 1° = jc 97 • ~ = °° °° Next. Since the line is balanced. 57. the degree of (zero-sequence) voltage unbalance (Degree of unbalance) = ^ 100 = yp~^ 10 ° = c is 100 per cent Consider next the three-phase Y-connected circuit without Phase c is opened. Vab = 7/0. . standard phase. — Unbalanced three- phase Y-connected circuit without The phase c is idle. a few examples will be given to illustrate cases where 100 Consider first the per cent voltage or current unbalance exists.. 56.e. Single-phase line-toneutral short circuit on a three- phase Y-connected Fig. Y-connected three-phase circuit shown in Fig. circuit. the magnitude of the positivesequence phase voltages is V/y/S and the negative-sequence components are zero. in Fig. Vo The zero-sequence component becomes = H(Ya +7 +7 6 C) 1 = = ~(7+7\60°) 7 V3 \30 Hence. 7/180° V = 7/120 c c a f ! Short """* Circuit In $ kb v : * — Fig. thus reducing the circuit neutral. The line It is desired to voltages are balanced.UNBALANCED CIRCUITS (Degree of unbalance) - = (Degree of unbalance) I- = j+ 7° j^. neutral. be written line voltage Vab as the voltages to neutral may Va = Vb = i. no voltage of zero-sequence can possibly appear. It is desired to determine the degree of current and voltage unbalance. Since there is no neutral connection. 100 = 100 per cent 7/V3 The degree of voltage unbalance. and These give Vo Va + IaZ+ =Hl (Z . the imped- ances are balanced. a zero-sequence voltage will appear in addition to the negative-sequence voltage. the voltage components may . If. This will merely affect the relative magnitudes of the positive. or not. (Degree of unbalance)- = — 100 7+ = ' T . having the values Z a Zb) and Z c respectively. If the impedances are unbalanced.98 ELECTRIC CIRCUITS— THEORY AND APPLICATION to single phase. in spite of the fact that no zero-sequence current flows. The phase currents are h = = Ic = 7a 7/0 7 /180° The positive.and negative-sequence components become (see equations (111) and (112)) 7+ = 1. (166). easily be . no zero-sequence current can flow. on the other hand. found by applying equations (162).and negative-sequence voltages.and negative-sequence impedances per phase are the same. It makes no difference in this connection if the positive. in this case. 6 la The degree — + 7\60°) ^(7/0 = ld/0 in phase a = -^=\30° V3 + 7/60° = -4=/30 c ) of (negative-sequence) current unbalance is.Z + i«z_ = y3 i (z + z \60°) ItZ+ + IaZ = }iIa(Z + Z /60°) = 7+Z_ vt = = itZo a a b) a a 6 a b (167). . Assuming that each branch offers the same impedance to currents of either sequence but that they are unequal. conse- quently. depends on whether or not the three load impedances are balanced and on whether or not each branch has identical values of impedance for current of positive and negative sequences. Z a + Z6\60° It will be seen that the impedance Z c of the idle phase. When the impedances are balanced (Za = Zb = Zc = Z). 1^0 V+ + Z /60° Z a + Z \60 Za 6 c 6 Z a — Zj. it is possible to have 100 per cent voltage and current unbalance occurring at the same time.. the components become of voltage to neutral Vi = ^L\30° N V3 T7 The degree of voltage unbalance is (Degree of unbalance) t - = ytz 100 v+ \ = now — . Referring to the three-phase circuit shown in Fig. UNBALANCED CIRCUITS The degrees of voltage unbalance thus (Degree of unbalance) - T7 V = (Degree of unbalance) become V+ = 99 . What is the degree of voltage unbalance in this case? Using the line voltage Yah = V/0 as standard phase. as might be expected.. 58. therefore.IZ/\/S 100 = 100 per cent m In the given circuit. — Simultaneous line-to-neutral and line-to-line short circuit phase Y-connected on a three- circuit. 58. -fl>"0'0'0'0> Fig. the voltages to neutral are Va = V = Vbc = -V/0 V = h c . does not affect the results. it is assumed that a line-to-neutral short circuit on phase a and a short circuit between lines a and c occur simultaneously. degrees of current unbalance of 100 per cent would obviously be obtained as follows: (Degree of unbalance) (Degree of unbalance) - = = y^ 100 = 100 per cent = ^ 100 = y^| 100 = 100 per cent jjo 100 . have a similar condition o£ current unbalance. 59). It is possible Short C/'rcvf'-f Fig. Consider. (Degree of unbalance) - (Degree of unbalance) As ^ /60° V~ = ^+ 100 = V/3 yrjo = -^ 100 = seen. from which It /o = g/O With these components. both the negative-sequence 100 = 100 per cent y/jr 100 = 100 per cent and zero-sequence degrees of voltage unbalance are 100 per cent in this case. hence. for instance. —Single-phase also to line-to-neutral short circuit on the secondary side of the A-Y-connected bank of transformers. 59.100 ELECTRIC CIRCUITS—THEORY AND APPLICATION The symmetrical components by of these voltages are then given Fo=-|/0 n = ^\6o° VZ = The degrees of voltage unbalance are. a single-phase line-to-neutral short circuit secondary side of a A-Y-connected bank of transformers The secondary currents may be written I a = 1/0 h = Ie = la = on the (Fig. The method of symmetrical coordinates. including dissymmetrical short circuits. four-phase. both two-phase. Furthermore. on account of its decidedly unsatisfactory operating features. which can be just as easily solved without resorting to this device. is due to the fact that the performance when loaded unsymmetrically. three-phase. of to a considerable extent depends on the no-load characteristics of the transformers. — unbalanced. This this connection. the transformers in the various phases are assumed to be strictly identical as far as electrical characteristics are concerned. Obviously. This connection particularly when the unbalance is pronounced. where it highly contributes to the ease and elegance with which solutions may be obtained. Both the two-circuit and three-circuit banks treated in the following are assumed to consist of single-phase transformers. and consequently a limited number of cases only will be treated. however. Symmetrical coordinates are not used in the simpler problems. in order to cover all possibilities which may be encountered in present-day practice. Y-Y Connection without Primary The Y-Y connection without neutral on the primary side is little used in practice. Two-circuit Transformers. the space required for such a comprehensive discussion would be excessive. Also. which then may be applied to the analysis of other cases. is utilized in the more complicated cases. however. and six-phase systems with all possible transformer connections should be considered. from a theoretical standpoint since it where represents the only case it is necessary to take into account therefore. even when the loading is only slightly Neutral. would require the consideration and analysis of a very great number of cases. trate the principles involved These. Thus. will serve to illusand the mode of attack. with each connection.CHAPTER IV TRANSFORMERS WITH UNBALANCED LOADING A complete treatment of the general problem of transformers with unbalanced loads. solutions should be worked out for all conceivable unsymmetrical loads and short circuits. as need may arise. is interesting JL01 . Z I 2a = Znli* + Eb Vu = ZiJic + E Vu = Vzb = jxiJib + Eb F 2c = jx 12 I lc + E Ea = — Z 22 ha .Iu) +Eb -E = Vu . line to neutral. they have been written down without imposing any simplifying assumption in regard to the magnetization characteristics of the transformers. Since no secondary current is drawn from phases b and c. Single-phase Short-circuit. holding for any arbitrary system of primary line voltages. Line to Neutral. —A single-phase short circuit. evidently represents the extreme case of unbalanced loading.: 102 ELECTRIC CIRCUITS— THEORY AND APPLICATION the exciting current in order to arrive at anything like a solution at all.Vu = Zuilu.jx 12I la Ila + lib + hc = 7io5 + Vxbc + V lca = Via (1) 2 Vxh (2) (3) c (4) (5) (6) c (7) (8) (9) These equations are perfectly general. as follows: Vlab V V Ua lbc ~ Via — Vlb = Z\I\a — Z\\I\b ~ Z 2I 2a ~ Eb = V lb . as indicated in Fig. however. the primary current in these phases is exciting current only. The primary current in phase a. Furthermore.+ E c c (10) (11) (12) . 60. 60. 1. . The may be written down immediately transformer equations in Chap. consists of an exciting (2) (J) Fig. II) following equations (see the general = ZJ la .Via = Z XXIu ~ Z J la + Z 2 / 2o . current plus a load component corresponding to the secondary short-circuit current of this phase. The primary line voltages may be expressed in terms of the phase voltages. —Single-phase line-to-neutral Y-Y-connected bank short circuit on the secondary side of a of transformers without primary neutral. . express the induced voltages as a function of the exciting currents. et seq. ™ • and the instantaneous exciting manner for positive values of the argument. Inc. 1917. "Theory and Calculation of Electric McGraw-Hill Book Company. inserted in equation (13). The following equation embodies these assumptions: E = jkl (16) e Then.V lM = (2Z X . Eb + E = jk(I + lb c I u) = -jkl la (17) which.. for instance.(Eb +E c) (13) necessary to introduce an empirical mathematical expression for the magnetization curve of the transformers. equation (18) may be Splitting the current 7i a into its current Iiae written V lab . The instantaneous induced voltage is obtained by differentiation and is given by relates the instantaneous flux current in a quite satisfactory . New York. C. A cut-and-try process or else a simpler expression for the magnetization curve must be used. lish it is not possible from equation (15) to estaba simple vector relation between a voltage and current of the same frequency. the exciting and the load component — ha. in other words. the use of Frohlich's equation in this particular problem complicates the equations to such an extent that an explicit analytical solution for the currents in terms of the impressed voltages cannot be obtained. Assume a straight-line magnetization curve and a true quadrature relation between induced voltage and exciting current. See." p.TRANSFORMERS WITH UNBALANCED LOADING 103 Subtracting equation (12) from equation (10) and making use of the current relation expressed by equation (8) gives Vm .V lca = (2Zi + Z n + jk)I la 2Z 2/ 2a (18) two components. Circuits.<» x • io -vfip! < i5 > Since. 42. - 2Z 2J 2o .V lca = (2Zi + Zn + jk)I lae - (2Z 12 + Zn + jk)I 2a 1 Steinmetz. gives V lab . however.+ Zu)I la In order to proceed further. The well-known Frohlich's equation 1 it is -m. (19) . the latter being equivalent to ignoring the core-loss current. P. ELECTRIC CIRCUITS— THEORY AND APPLICATION 104 expressing the induced voltage in transformer a (equation a relation between the (7)) in terms of its exciting current, in this phase is obtained. component load the exciting current and — (20) = jXnha jXlthae —Zwlta jMiae By + = or j(k + X^hae = —7i%I% a -Z Combining equations [ tr (2^1 2 Ita (19) + — jXnllae ( and Zl1 + 21 ) (21) gives MZ * 4-97 4-7 4- *1 7o (22) From this equation, the current in the short circuit I 2a is the primary line voltages are balanced, the voltage difference in equation (22) is equal to obtained. If Vi* where V - V lca = V37\30° is the value of the line voltage. (23) Vy* has been taken as standard phase, the sequence being a — b — c. Having determined the current ha, the exciting current I u , The total primary current of phase is found from equation (21). a is then calculated by Jla = hae , ~ ha x (24) currents in the other phases, as well as the primary voltages to neutral and the secondary voltages appearing across phases of the b and c, may now easily be computed by making use The appropriate formulas. The heaviest current will flow in phase a, at least when the line voltages are not unbalanced to an excessive degree, which very seldom will be the case in practice. The primary voltages to neutral will be decidedly unbalanced, the voltage across the shortcircuited transformer (phase a) being very small compared to the the flux others, due to the effect of the short circuit which causes In other words, in this transformer to collapse to almost zero. posisymmetrical its from travel will point neutral the primary tion toward line a, almost coinciding with the latter. The accuracy of the results evidently depends on the correctIf ness of the assumption in regard to the magnetization curve. knee, the transformers normally operate considerably below the the straight-line representation may be proper and lead to correct transformers, however, normally operate at a knee of the fairly high saturation, i.e., at or slightly above the of the displacement large magnetization curve. In view of the results. Modern TRANSFORMERS WITN UNBALANCED LOADING 105 primary neutral point, causing considerably more than normal voltage to be impressed on the transformers in the non-shortcircuited phases, large errors may result from the assumption of the magnetization curve's being a straight line. 1 An approximate method applicable to all cases suggests itself, however. This method gives values of current which are conservatively high. If it is assumed that the neutral point is completely displaced, i.e., lies on line a, then the line voltages Viba and Vica are impressed directly on the transformers in phases b and c, respectively. Vv> Vic = Viba = —Viab (approximately) = Vua (approximately) (25) (26) Assuming that the induced voltages do not differ appreciably from the impressed voltages, the exciting currents of transformers b and c (lib and Ii c ) corresponding to the voltages given by equations (25) and (26) are read from a magnetization curve. The current in phase a is then given by I la = ~ha = ~(/l6 + he) (27) the line voltages are balanced and Viab = V/0 taken as standard phase, equations (25) and (26) may be written If Vv> Vic = 7 /180° = 7 /120° (approximately) (28) (approximately) (29) These voltages are equal in magnitude and 60 deg. out of phase. Hence, the corresponding exciting currents will also be equal and 1 In the solution above, the same straight-line magnetization curve, i.e., the same value of k, has been used for all three transformers. Since the transformer in phase a operates at a very much lower flux density than the transformers in phases b and c, it would be appropriate and strictly more accurate to use a different and higher value of k for the former. This value (fci) may be taken from the straight portion of the actual magnetization The coefficient (fc) for the two other transformers may be determined by means of a straight line drawn between the origin and a selected point well up on the actual magnetization curve. This point should correspond, as nearly as possible, to the voltages across phases 6 and c and assumes that these voltages are equal. This, however, will not be the case when the curve. impressed line voltages are unbalanced. When the primary line is balanced, the voltages across the non-short-circuited transformers have strictly the same magnitude and are very nearly equal to the line voltage. When different values of k are used, equations (21) and (22) become j(k! and + Xi 2 )/la e = —Z 2I ia ^ V lab - V lca = -\ 2-^4^¥~ J + 23" + Z n + j*l J- (21a) (22a) ELECTRIC CIRCUITS— THEORY AND APPLICATION 106 The numerical value phase-displaced 60 deg. the short-circuited phase is, Iu = and is, of the current in therefore, V3/ = VSI 16 as seen, very readily calculated. (30) l0 — 2. Single-phase Load, Line to Neutral. A single-phase load of impedance Z, connected line to neutral, as indicated in Fig. 61, will now be considered. (2) S Load of ^Impedance y\ v Fig. 61. Z I —Single-phase line-to-neutral impedance load on a bank of Y-Y-connected transformers without primary neutral. Equations (2), (3), (5), (6), (8), and (9), previously given, may be used also in this case. In addition may be written the general equations Via V 2a = Z U I la = Z 22/ 2o + jXuha + Ea + jXnha + Ea (31) (32) which, in connection with Via = -IzaZ Via = (33) give Zil la - (Z 2 + Z)l2a (34) and Ea = - (Z 22 + Z)Ua ~ jX 12I la (35) By a procedure similar to the one used in the previous case, an expression for the difference between the line voltages Viab and The result is Fica may be formed. V lab - V lca = (2Z'i + Z u )Iia - 2(Z 2 + Z)I 2a - (Eb +E c) (36) Again, in order to obtain an explicit analytical solution, straightline magnetization curves will be assumed. Equation (36) may then be written V lab - Vua = (2Z X = (2Z l + Z n + jk)I - 2(Z + Z)ha + Z +jk)I (2Z + 2Z + Z n +jk)h la 11 2 lae 12 a (37) Introducing the linear relation between induced voltage and exciting current, in equation (35), this reduces to j(k x n )I lae = -(Z 2 Z)7 2a (38) + + + + TRANSFORMERS WITH UNBALANCED LOADING 107 which, combined with equation (37), gives Vidb — Vua — (2Zl + K^t This equation + z) + 2Z - + 2Z+ z » +jk may h (39) be solved for the load current I% a The is then found from equation (38), and the two combined to give the primary current in phase a as given by equation (24). The other quantities of interest may readily be found, as discussed for the short-circuit case. If the load impedance Z is large, so that the current drawn is small, and, as a consequence, the shift of the primary neutral point is not large, the assumption of a straight-line magnetization curve may not give rise to appreciable errors in the results. If, on the other hand, the load impedance is low, the solutions are subject to the same limitations as in the case of a dead short circuit, although to a somewhat lesser degree. 1 If greater accuracy is desired, cut-and-try methods must be resorted to. Y-A Connection. The Y-A connection with unbalanced load on the secondary can be handled without taking the exciting current into account with the same degree of precision as in any single-phase or symmetrical transformer problem where exciting . exciting current of transformer a — currents are neglected. If desired, it is entirely practicable to introduce the effect of this current for most engineering purposes, however, this refinement is unnecessary and will be neglected in the following analysis. ; Considering quantities of fundamental frequency only, the is fixed independent of the load of the secondary. This will be apparent from the following reasoning The displacement of the neutral, if any, is equal to the uniphase voltage. If a uniphase voltage were present, a circulatposition of the primary neutral point : 1 Different values of k may be used for the transformer in phase a and the transformers in phases b and c, also when a single-phase, line-to-neutral load is supplied. The procedure is the same as that discussed in the footnote on p. 105 for the line-to-neutral short circuit. Since the voltages across phases b and c depend upon the load impedance, the straight line representing the magnetization characteristic of the transformers in these phases cannot be located as definitely as in the short-circuit case. With different values of k, equations (38) j(ki y- - v- - T 2Z + xu)l la, ( '^*.+'*i r' and — (Z 2 (39) become + Z)I* 2z " (38a) 2z + z " +*j'(39a) 108 ELECTRIC CIRCUITS— THEORY AND APPLICATION ing current would, of necessity, exist in the delta. Since the Y-connection, however, eliminates the possibility of any uniphase currents' flowing in the primary windings, no uniphase or circula- tory current can flow in the secondary delta, because, in general, ignoring exciting currents, a current in a secondary winding, in order to exist, must have an equivalent component in the corresponding primary winding. Hence, since there are no uniphase currents, there can be no uniphase voltages and, consequently, no displacement of the neutral point. It should.be noted, however, that there is a triple frequency displacement of the neutral point, due to the third-harmonic component of the exciting current which circulates in the secondary delta. This third-harmonic voltage, however, is quite small, being equal to the secondary third-harmonic leakage-impedance drop. Short Circuit Fig. 62.- -Single-phase line-to-line short circuit on the secondary side of a Y-A-connected transformers without primary neutral. bank of — Short Circuit. Consider a Y-A-connected 1. Single-phase bank, as shown in Fig. 62, with a dead single-phase short circuit on the secondary side. Neglecting magnetizing current, the general equations for the primary and secondary terminal voltages may be written V 2b v may be written Fi6 2c In addition to these ZJ + Ea ZJ + E Zxlu + E -Z Zia + Ea -Z /i6 + E -Z I + E = = = = = = Via 2 2 2 Vxa* lc + la + Iu + + lie + Vac + Vua -^26 (41) b c Ila 1 2a (40) la lb = = = V ia + F 2& + 7 2e = < (42) (43) b (44) c (45) (46) (47) (48) (49) TRANSFORMERS WITH UNBALANCED LOADING 109 The equations as written above are not restricted to the case of a single-phase short circuit but hold for any type of load. In the case at hand, the secondary terminal voltage of the short-circuited phase equals zero, i.e., V 2a = 0. Hence, Ea = ZJ la (50) V la = Z l2 I la (51) and The primary line voltages may now be expressed as follows: V lab = Vu - Vu = Z 12I la - ZJu, - Eb = V - V = Z (I - I + Eb - E V lca = V lc - Vi a = ZJu - Z i2 I la + E Fi6 C lb lc lb 1 le ) (52) (53) c (54) c Taking the difference between equations and - (Eb (52) V lab - V lca = 2Z 12 I la - Zi (I + I ) = (2Z 12 + Z )I la - (E + E lb lc b 1 From = V 2b (54) gives +E c) (55) e) equations (49), (44), and (45) is obtained V 2c = Ec = Iu) 2 (/ 16 b -Z + +E + + Z 2 I la +E +E b c (56) which gives E + E = -Z b c 2 Ii a (57) Substituting this in equation (55) gives Vi^ - Vua = (2Z 12 Z 2 I la Zi)I la The primary current of phase a is thus + /» lO — + V lab = 3Z 12 7 la Ua (58) /rr\\ {0\f) W7y OZ/i 2 Evidently, lib = he (60) lib = Iu (61) hence, also, ~ which, in connection with equation (46), gives h The b = Iu = (62) total current in the secondary short circuit — j __ i j J- 2a — T lib — oI 2a —n~ = Vlab i — Vua fTy s , an . (OoJ If the primary line voltages are balanced of magnitude V equations (59) and (63) may be written, using Viab = V/0 as standard phase and assuming the phase sequence to be a — b — c, } 7 .._vg^! (66) 110. ELECTRIC CIRCUITS— THEORY AND APPLICATION The equations show that the primary current short-circuited phase splits equally between the carried two by the others; in other words, the current in the two latter are of one-half the of phase with the current in the magnitude and 180 deg. out former. On the secondary side, the currents add up to a total short-circuit current of three-halves the value of current in the most heavily loaded phase. The calculations are, as seen, extremely simple. The only constant required is the equivalent impedance of the transformers. If the Y-A-connected 2. Single-phase Load on Secondary. bank is loaded with a single-phase load of impedance Z, as shown (short-circuit) — Load of Impedance Zi Y-A-connected transformers on a bank — Single-phase impedance load primary neutral. of Fig. 63. without in Fig. 63, equations (40) to (49) inclusive addition, the following equations may may still be used. In be written: F 2o = - (I 2a ~ I»)Z = HhaZ (66) which, substituted in equation (43), gives Ea = From this, in (Z, + %Z)I connection with equation (67) la (40), is V la = (Z« + %Z)Ii* obtained (68) This equation may be compared with equation (51), which is the corresponding one when there is a single-phase short circuit on the secondary. Since the only difference is the addition of three-halves of the load impedance to the equivalent transformer impedance when the bank is loaded, it may be immediately inferred that, upon following the method of solution used in the previous case, final equations of the same form would be obtained, should be used instead of with the exception that Z t2 Hence, making use of equations (59), (63), (64), merely Zi 2 + %Z . TRANSFORMERS WITH UNBALANCED LOADING and 111 most heavily loaded become (65), the solutions for the current in the transformer, and for the load current, — ~ T lla ~ lab r T / < cn \ 69 ) mm K lco lob 2(Z 12 for the general case, and, lca + %Z) 3(Z 12 (70) + %Z) when the primary line voltages are balanced, Iu = F^° _ / V3(£x2 = - / (71) + ^Z) V3y\3o° 2(Z« , r72 ; + %Z) ^ The relation between the currents in the various phases is evidently as in the short-circuited case and may be computed by equations previously given. Three-circuit Transformers. —The general equations for the which were derived in Chap. II three-circuit transformer will be repeated here for reference. = = Tic = V 2a = V21, = ^2c = V 3o = F 86 = F 3c = Via Vib -f jXi I + jXisIsa + E + jxnly, + jxulsb + E Zulie + JXiJ + JXuIzc + E Z I + jXi I + jX I + E Z I + i^i /i6 + jx Isb + E Z + JX I U + ,7X23/30 + E Z / + JXiJ + yX /2a + Ea Z33/36 + jX U I + .7X23/26 + E Z Z + /i + .7x23/2* + E Znlia 2 2c 22 2a 2 22 2h 2 2 2i"2 C 12 33 33 Via 3a - V 2a = -V Fie Vu F 3c - = F 36 = Vib = V 2c = Vzc = Vu = Since the transformer bank V» Vsb 2b 2Z a (73) b (74) c (75) a (76) b (77) Sa 2S e 23 lb b c e ;'xi 3 3c 3 Vib la la Zilia- Z 2I 2a — Zzlza / 3o ZJi a Zilib - Z2/2* Z 7 26 - Z3/36 Z hb - Zihb Zilu - Z 2/ 2c Z / 2c - Z I Zzlzo - Zili V 2a V 3a = Z F a - 7l. = Z circuit, 2a Z nIib 2 I 2a 3 (78) (79) (80) (81) (82) (83) (84) (85) 2 (86) 3 (87) 2 itself s (88) 3c (89) c (90) represents a symmetrical these equations hold separately for the positive-, negative-, and zero-sequence (uniphase) components as well as for the actual voltages and currents. Also. 64. so that excessive distortion of the voltages is avoided. for winding 1. To the equations which already have been given may be added the following holding for the particular connection being treated: V lab + V + Vua = = Vza + F + V (91) lbc 3& (92) 3c ^ should be remembered that. — Consider first a secondary single-phase short circuit. + Vu) = ZiJio + JZ12/20 + jxij/jo + Eo (94) . heavy circulating currents of fundamental frequency may be set up in the tertiary winding. without reference to phase. = H(Vi a + Vv. The principal purpose of this winding. It transformers connected in a well-known fact that a bank of primary and secondary sides may be operated quite satisfactorily without primary neutral connection if supplied with a tertiary A-connected winding. neglecting exciting currents. Single-phase Short Circuit. is to provide a path of low impedance for the third-harmonic component of the exciting current. short circuits. is Y on the Special relay protection for the tertiary (D is. Hence. by definition. Line to Neutral. of a and it becomes of considerable importance to predetermine with precision the performance of such a bank when it supplies an unbalanced load or is subjected to dissymmetrical required. as shown in Fig. it /i +U+h = (93) The uniphase voltage is. 64. Vv. hence. the sum of the current in the three windings of any one phase must be zero. equal to one-third of the vector sum of the phase voltages and is. in this particular connection. line to neutral. however. therefore. 1. For certain types of unbalanced load and -\ dissymmetrical short circuits. usually (V (3) * 1 Short 1 Circuit" 1 1 — —/ i t Fig. — Single-phase bank of line-to-neutral short circuit on the secondary side Y-A-Y-connected transformers without primary neutral.112 ELECTRIC CIRCUITS— THEORY AND APPLICATION — Y-A-Y Connection. These currents may often exceed by a wide margin the maximum current which can be carried without injury to the winding. JX 13 )I 30 (98) F 20 = (Z 22 — JX 23 )I20 j(Xl2 X 13 )Iw — (Z33 — JX2 3 )1 30 (99) The sum of the uniphase components in the three windings — — + + must be zero. the displacement of the neutral points computed as soon as 7 2o has been determined. be located symmetrically in the equilateral triangle representing the line voltages. windings. Introducing this in Since winding 3 is a closed delta. times 7 2 o. for the other windings. F = Znlzo + jxiJw + JZ23/30 + E V = Z 33 I + jx 13I w + JX2J20 + -So (95) 2o F10 and Vw (96) 30 30 represent the displacement of the primary and secondary neutral points. an expression for E the uniphase voltage induced per phase due to the flux exclusively existing in the iron core. at no load. negative-. /20 = -/30 This relation inserted in equations (98) and (99) gives F10 1^20 From may = = + j(xi2 — x — £13)1/20 = Z3/20 — JX23 + Z22 — JX2 ]ho = Z23/20 [Z 33 23 [Z 33 3 (101) (102) (103) these equations. the neutral point will. is . in each phase of winding 2 given is + + / t + / 7 + 7 = ht 12 2 2 2o (109) . The currents in phase a of winding 2 will equal sum and zero-sequence components by (107) ha + ho = 3/ 20 (108) I* +ho = of the positive-. 7 2o must. respectively. hence. E = — Z33/30 Q jxulw — JX23I20 (97) Substituting equation (97) in equations (94) and (95) gives j(xi2 V10 = (Zn — jxu)I w £23)120 — (Z 33 . Therefore. equation (96). from the no-load position corresponding to the impressed voltages. The sum of the (105) (106) therefore. obtained. be equal to three positive- and negative-sequence two times /20. F30 = 0.TRANSFORMERS WITH UNBALANCED LOADING 113 Similarly. Since there no is /10 + ho + ho = no neutral connection on the primary uniphase can currents flow in the (100) side. readily be may be written for the sum of the phase currents in the three windings: J tt (104) he = ha The following equations ha ha lib + + + /» /» + + + he = 3/20 ho = 37 3 and he are both zero. primary however. If the latter are balanced. 66. III). gives Vit + ZiJ w + Via + £12/20 + Z 2 3/ 20 = (117) . negative-. If the primary line voltages are balanced. F + = V& . and the negative-sequence voltages are The secondary voltages may be computed from the primary voltages by making use of equations (82). (85).114 ELECTRIC CIRCUITS— THEORY AND APPLICATION These equations are evidently satisfied by the systems shown The primary positive. 65. 7 W is zero. and (114).-r * 2a ~*20 —Vector diagram of positive-. Equations (110). by \/3. in connection with equations (103).and negative-sequence phase voltages can readily be evaluated in terms of the line voltages either by analytical or graphical methods. previously described (Chap.r // lga' Ji>a Fig. for phase of the component systems. the primary and secondary currents of these sequences are equal and Furthermore. the magnitude of the positive-sequence voltages is equal to line in Fig. and (88) voltage divided zero.Zx/io + Z / 2 2 2 2 2 2 20 (110) (111) (112) Since there can be no positive. on account of the delta connection. Zero-Sequence Currents *° r.ZJd + Z h+ F « = Via . Positive-Sequence Currents Negative-Sequence Currents r+. (107). and (112) then reduce to F 2+ = 7i+ + Z 12 I 2+ (H3) opposite. a.and negative-sequence current in the tertiary winding. (113). and zero-sequence-current components in the secondary of the transformers in Fig. (111).Zl/l« + Z / a F 20 = Fio . 64. and applying them separately to each Thus. 77 = Via = V10 2 V20 + Z l£ + £2/20 (114) 12 (115) Since the secondary voltage of the short-circuited phase may be written: is zero. the following relation V& + V2I: +V 20 = (116) which. Line to Neutral. of magnitude V = V/0. The other = — -12a = 3720 I la = — 2/on — 1 3a — 1 30 = — + Vu) + Z23 2(7i+ + Vu) 2Z12 + Z23 3(Fx+ (119) 2Z12 -t20 = 2Z12 + (120) Z23 currents as well as the various voltages readily be determined may now by making use of the appropriate equations. If and the primary line voltages are balanced.TRANSFORMERS WITH UNBALANCED LOADING 115 Hence. in this case. ee. the analysis can be carried through exactly as for the single-phase line-to-neutral short circuit Since. Single-phase Load.-^-Single-phase line-to-neutral impedance load on a Y-A-Y-connected bank of transformers without primary neutral. The is . equations (118). the secondary currents are zero in phases b and c. Viab (119). the phase order being a — b — c. (2) a Load of Impedance Z Fig. .When the secondary loaded with a single-phase load of impedance Z. and (121) reduce to c 1 20 — 1 Za — — Ila F\30 \/3(2Z 12 + Z 23 ) \/S F\30° 2^12 = + 1 3a — (123) Z23 2F\30 V3(2Z 12 (122) C +Z (124) 23 ) F\30 c (125) V3(2Z 12 + Z 23 ) — 2. (120). the system of currents illustrated in Fig. 66. also. as shown in Fig. 65 is valid. I20 — — 2Zi2 + (118) Z23 Also. connected line to neutral. and (114). no current of fundamental frequency will flow. Line to Line. indicated in Fig. combined with equations (103). the two neutral points will remain fixed in spite of the short circuit. (107).. Single-phase Short Circuit. fl33> (133) 2F\30° * V3(2Z 12 (134) +^3 + 3Z) n3o° V3(2Z 12 + Z 2 3 + 3Z) __. circuit transformer : .2Z + Z + 3Z 12 23 2 ("^l« + V£) -977 2i2 .Il °° 2Z 12 + Z 23 + 3Z = /3a /30 = /20 — KI7 ^ /lOf^ (1Z9j fion\ (W0) ny (131) ^23 ~f" ^^ When the line voltages are balanced. as well as equations (91) to (115) inclusive. + V& + gives Fit Hence.07 . (113). therefore. 67.116 ELECTRIC CIRCUITS— THEORY AND APPLICATION general equations (73) to (90). + Z12/20 + Vz + t * /2 ° The £12/20 ~ ~ 2Z 12 + la £23/20 + 3Z/ 20 = (127) *_la ~l~ +Z + 23 OCA (128) /""I 3Z currents in the three windings of the loaded phase a are then given by 3(Klo + 'la) t _ 3i 2 o _ I* . winding 2. be reduced. n (W5) - — Consider next a 3. no uniphase current can. the result being i2 ° /2° ila t i3a "J ZZ/12 T" " V3(2Z 12 4^23 + V3F\30° = " 2Z 12 + Z 23 + 3Z = " - j (132) 3Z) . exist in the tertiary circuits. may be expressed in terms of the impedance drop in the load. there can obviously be no uniphase currents in any of these Hence. delta in which then. since it is unloaded. as This problem actually reduces to a twoproblem for the following reason Since there is no neutral^ connection on either the primary or the secondary side. these equations may. single-phase line-to-line short circuit on the secondary side. equation (116) may be modified for the present case as follows: V& ^2o = -ZIta = -3Z/ 20 (126) which. still hold. as in the previous case. In this case. Since the terminal voltage of phase a. All other. In* Fig. The solution. Fig. (Z) — a. = (139) equations (82) and (85) follow V 2a = V la + Z 12/ 2a F 6 = V + Z 12I 2b 2 (140) (141) lb which. = 1 2b = he = 1 2a These equations are may be written + 1 2a = —lib I t + Ifr = —1 2a I t + h7 = (136) Iia~ 2 (137) 2 (138) satisfied by the system shown in Fig. will be worked out below without resorting to this device. however. Vub = -2Z 12 J 2il (143) . inserted in equation (139). —Single-phase line-to-line short circuit on the secondary side of aY-A-Yconnected bank of transformers without primary neutral.TRANSFORMERS WITH UNBALANCED LOADING The 117 currents will contain positive. 67. Since the voltage across the short circuit V2a From - Vv. 68. solution of this case. 68.and negative-sequence components For the secondary currents only. is zero. 67. gives V la . therefore.V lb = -Z 12 (h a - 7 26 ) (142) or. Short Circuit %.and negative-sequence current components in the secondary of the transformers in Fig. —Vector diagram of positive. is so simple and straightforward that nothing is actually gained by considering the component systems of currents and voltages. from each Their magnitude is equal to the current in the short circuit divided The by ^/%. components are equal and displaced 60 deg. although. The analysis of this case will be given as an interesting and instructive application of symmetrical coordinates. in connection with equations (140) and (141). The only constant entering the solution is the equivalent impedance of windings 1 and 2 as determined by a simple short-circuit It is seen that the solution. Equation (139) may now be written is just as readily obtained. . Single-phase Load. Via ~ F = 26 -U Z (147) a which. I*a = I la = — lib lie — — V lab —Iw> (144) 2Z\2 Vlab (145) 2Zi2 = -ho = (146) depends on the value of one of the impressed line voltages and is independent of whether the primary line voltage triangle is symmetrical or not. (137). as shown is in Fig. load of Impedance Z Fig. gives v lb = — Z12 (Iia — lib) — Zlta Via (148) or. 69. 69. actually. test. (140). Line connected line to line to Line. and (141) hold unaltered. in this case.118 ELECTRIC CIRCUITS— THEORY AND APPLICATION Hence. Line to Neutral. v lah = -(2Z + 12 Z)I%a (149) Hence. (138). is bank similar to the solution of the short-circuit case and Equations (136). also in this case. — Single-phase the solution line-to-line impedance load on a Y-A-Y-connected of transformers without primary neutral. Two Single-phase Short Circuits. 4. — on the secondary If a single-phase load side. Iza = — 1 2b Ila = -lib he = -ho = Via* 2Z 12 )+ Z (150) v lab 2^12 ~\~ Z (151) (152) — 5. 70. The secondary currents may be expressed by circuit is ha = ht + ha + ho hb = hf\i20 o + f 2 q/120° he = (153) +I 5 + haVM + ho = 7 2£/120° from which the following relation / 2t\60° ' + (154) 20 (155) results ha/6V (156) -*20 (2) — a. V 2t + V a + V& + Vj + F20 = 2 of the short-circuited phases Hence.or negativesequence currents. The secondary terminal voltages are zero. Fig. by addition.+ : TRANSFORMERS WITH UNBALANCED LOADING 119 single-phase short circuits from line to neutral are not likely to occur in practice on two of the phases simultaneously. give o = v++ +W + / vra + v& + vrb + £12 (/2 2 q/60°) + 2Z 23 /2o (164) . The shown in Fig. the secondary terminal voltages of these sequences of phases a and b may be written Vd = Vd + Zuht = Via = V& V£ V 2J = Vlb Via The zero-sequence (157) +Z I + Z n I +\120° + Z ha/120 12 (158) 2a (159) c (160) 12 voltages on the secondary side are all equal to V = Z 237 20 (161) 20 as previously given by equation (103). V 20 = = Vtt Via Z X2ht ZiJ 2 + = V+ + + + Vi + -g- +Z J 2 20 (162) Z 12 7 2+\l2r £12/2* /120° +Z 23 ho (163) which. Since the tertiary delta carries no positive. 70. from equations (157) to (161). Short Circuit K/«6 ^ ^ — Two single-phase line-to-neutral short circuits on the secondary side of a Y-A-Y-connected bank of transformers without primary neutral. the solutions for the secondary current components become If I20 — — W v lab VZ(Z 12 + 2Z 23 ) (Z12 + ht= - Z 23 )V lab\Z0° \/SZi2(Zl2 I"La — (170) + (171) 2Z23) z it v Xab \Jm° v3Zi2(Zi2 "I" (172) 2Z23) — When the Single-phase Loads. 6. 7 20 = -IA±I^±IA±I» + 2^23 Vi+W* + ^la/60° 2*12 + 2Z23 Z\% Simultaneous solution of equations (162) and (163) gives. I& = —V. for the positive.' C (167) single-phase line-to-neutral impedance loads on a Y-A-Y-con- nected bank of transformers without primary neutral. The short-circuit currents are and the current primary are given by and (154).120 ELECTRIC CIRCUITS— THEORY AND APPLICATION Substituting equation in (156) sequence current in the secondary equation (164).+ + Z 23 /2o/60° (166) Vu + Z 23 /20\60 Fig. Line to Neutral.and negative-sequence currents. now found by equations in the tertiary /30 = —1 20 — I la = — (lia Fit\60° + Z12 and phase a + 7^/60° + 2Z2 (153) of the (168) (169) Iza) the impressed line voltages are balanced. the zerois found in terms of the impressed voltages and the transformer constants. — Two 71. Two transformer bank supplies two single-phase loads connected line to neutral. 71. as shown in Fig. equations (153) to (161) inclusive . Vit + Via + Zulia' + Znlia + Z 23 7 2 o = — Z a Iia ~ V£ + Vrh + Z 12 +£ 72+\l20* Eliminating 7 20 be written 12 /2-/120^.*.Kh~a Yit fc 2 2 (177) 2 (178) Simultaneous solution of equations (177) and (178) gives j.= TRANSFORMERS WITH UNBALANCED LOADING 121 are still valid./ ^ y$ + V* = . — &1&4 + fc. _ _ — — (h V3(fc 2 /c 3 7.Z6 7 20 (174) by means of equation (156).Kht . (156). (184) kiki) Three-phase Load. w (183) — +—— V60°)7. Case of Unbalanced fcifc 4 ) — .= _ (fci + fc 1 + 7 /60°)F 1+ &2&3 Having found the values by equation (*« + A^W^x.*W' — \/3(k2k 3 T -i2o .+ . Line to —Let the secondary of the three-circuit transformer be loaded with three unequal impedances Z a Zb and Z . + Zalta V ~ Zalw (173) = 20 -Z &7 2 +\120° . j. r- j-j + (*. equations (181) and (182) reduce to of rent is found m= <*' + Mgnn. may which also + i2a r _ - fj) (179) kiki - (180) AJfc 4 be written (*4 + *»/60°)7.Z b 7 2i /120° . 1<M .uvt + fc 2 fc 3 — i2a *. c. General Neutral. j _ h(Vii + vd . The secondary voltages of phases a and b may be equated to the impedance drops in the two loads. + 7ia = -fcj + . . connected . these equations may Fit + = -[Z 12 Vig + + Za + Z )W)ht - (Z 23 a + (Z + Z )/W]/ ^1 + V£ = ~[(Z + Z )\l20° + (Z + Z^WW [(Z + Z )/120° + (Z + Z )/60^]7 „ [Z 12 12 +Z„ 23 6 2 (175) 2 (176) 23 12 23 6 6 or. . Thus. 1Q1 .W^u (181) (182) 7 2t and 7 2 „ the zero-sequence curThe other currents may be found as outlined for the short-circuited case. When the impressed line voltages are balanced. in Chap.Zolza. 72. — 1 2a ~T 1 2a + /| = I%b ht\l20° + 72a /120° + 5 he = J £/120° + I2AI2O + (188) 1 2a 2 72 . as this case. negative-. Without resorting to this scheme.= H(Z a + Zb\120° + Z /120°) Zo =H(Z a . as is usually the The secondary currents are. the components of the secondary terminal voltages of phase a are given by v£ = -Zohi ..Z_7 - 2 Vw = -ZMt - 20 (191) 20 (192) Z+ha . c (185) e (186) (187) c '(1) h t i Vlab J V Fig. (166).122 ELECTRIC CIRCUITS— THEORY AND APPLICATION line to neutral. The of Y-A-Y-con- neutrals of the transformer Since the load impedances are supposed to be known. then. case in simpler problems. and (167). since there is no easy and obvious way by which the components can be immediately determined. 72. Ill. is of somewhat more complicated than of the three-circuit transformer problems so far treated. U (189) (190) Making use of equations (162). l — Unbalanced three-phase impedance load on a bank nected transformers without primary neutral. it can be handled with comparative ease by the method of symmetrical coordinates. be expected. as any may shown Although the solution in Fig.z-i 2a .Z I iQ (193) . secondaries and the load are interconnected. and zero-sequence make use impedances of the load previously worked out (Chap. It is convenient in this case of a general unbalanced load to of the equivalent positive-. Z + = H(Za + Zb /120° + Z \120°) Z. the value impedances can readily be evaluated. and zero-sequence components have to be used in this problem. negative-. the solution would be very laborious indeed. The general of these expressions for the secondary currents in terms of their positive-.+ Zb + Z ) III)-.z +i V^ = -Z+I + . in this case..~ Z + ^0 T" ^23 (201) (202) (203) Eliminating 7 2£ and ha. (157).ZJ[ 2 a . By combining equations (103) and (193).(Z + Z Xi )ha ~ Z_J 20 V& = (194) V (195) Xa The displacements primary and secondary neutral points by equations (102) and (103).Z_V - + (z7+l£ " ~ Z 12 2 2 T" -"23 z+)i*t Zo Zl2 7aJ (198) z may now conveniently Vd = kj£ + 2/ 2« Vu . respectively. equations (204) and (205) can readily be solved.Z+I w = -Z+ItZ .TRANSFORMERS WITH UNBALANCED LOADING 123 Similar equations may be written for the other phases but are not necessary for the solution.k*r T _ = kiVia — k Vit ha T + hi = kiVit ror\A\ (204) ki iFwr 3 /r»rvc\ (205) Vit and Via are determined from the impressed line voltages. and the actual secondary currents by equations (156). of the are also.sequence currents are obtained: — kiVia . The uniphase current 7 20 is then computed by equation (196).and negative-sequence currents is obtained. Substituting equation (191) in equation (113) and equation (192) in equation (114) gives -(Z„ + Z 12 )/ 2 + . "T Z+ 7" /m«\ (196) 2o y ^23 Inserting equation (196) in equations (194) and (195) gives Vit v 1 a = = ~Zo{J~^r \Aq T" ^23 Z ( Zo + Z23 - Z + + ( V y }7 \4o / . and (158). the following explicit solutions for the secondary positiveand negative. between equations (199) and (200). an expression for 7 2o in terms of the positive.-ZAq "T ^23 = t-4^7. . Hence.kj i + kxha These equations (197) / ) be written fc 2 (199) (200) where ki k2 kz = j£rh---Z*-Z a ^0 T ^23 = y-^y. given T Ii0 Z_/ 2a = y ^0 + . This also gives a check on the correctness of the general formulas. by means = 0. Za = Zb = Z = Z (212) c Furthermore. It is inconvenient. desired. equations (210). Vi% while Fia (209) if = —V/=\30 c Equations (204) and (205) then reduce to V JJ -& (210) V •^3-^X30° V<5 (211) /C2/C3 /Cj The solutions just worked out for the general unbalanced load evidently also hold for special cases such as a single-phase load or a short circuit from line to neutral and also for a balanced threephase load. Z+ = Z_ = and ZQ = Z (213) Hence. fci = . as illustrated below. When the load is balanced. . the various voltages.124 ELECTRIC CIRCUITS— THEORY AND APPLICATION The primary currents are given by ha = ht + ha = ~ (I«J + ha) I» = + hb = -(7. when one or more of the load impedances are zero or infinite. they lead to indeterminate expressions which have to be especially evaluated. the impressed line voltages are balanced. of the appropriate formulas. since. They are readily applicable to a balanced load. however.(Z + Assuming balanced primary and (196) give Z«) and k 2 = _ I* h = (214) line voltages. =0 vW° In = (215) (216) which evidently is correct and might have been written down immediately by inspection.t\120° + /n = Ixl + hi = -(/it/120^ + U (206) W120 ) (207) / 2a\120°) (208) and the tertiary current by ha = /» = 7 8 c = Having solved are easily determined If 7»0 = -1 20 for the current. (211). to apply them to the former. ..55 per cent = 6.3 x Q 1 . = 5.7 per cent 6. g + Ziz — Z\% ZZiz = g a .000 volts 22.15 per cent 5. Symmetrical short 2. windings 5.000 volts equivalent short-circuit reactances are between the 66-kv. and the 4-kv. line to circuit. The 20.7 X\ +6. windings 5. Z\3 w W .7 + = 2. point is 1. The nominal voltages of each single-phase transformer are as follows: tion in Winding 1 Winding 2 Winding 3 The 66. 3. Z\z g reactances only) are computed — Z23 ~ + Z12 — Z23 _ j w . line to neutral. windings between the 66-kv. and the 22-kv..7 ^ - 6.7 = ^ x2 = 5. 60-cycle transmission line terminates at a substa- which the voltage is stepped down by a bank of transformers to approximately 38 kv. per cent 3. calculate and compare the currents and the displacements of the neutral points for the following conditions of secondary short circuit: The 4. 115-kv. between lines in order to be transmitted over an underground system to a load center. line.-a.7-5. Single-phase short circuit. Solution Ratios of Transformation: Winding 1 Winding~2 _ " 66 22 66 4 22 _ „ Winding 1 _ = Winding 3 Winding 2 _ Winding 3 ~ 4 The composite impedances from the following formulas: V Zi = zV 2 = Zu + (in this case. Hence.TRANSFORMERS WITH UNBALANCED LOADING EXAMPLE 125 1 Statement of Problem A three-phase. Single-phase short circuit. .3 Xa +5.7 per cent between the 22-kv. and the 4-kv.3-5.7 = = t 3.. _ __ _ _ .000 volts 4.3 per cent The secondary neutral grounded. Assuming the high-tension voltages to be strictly balanced and equal to 115 kv.000-volt windings form the tertiary delta.000 kv. transformer bank consists of single-phase three-circuit transformers connected Y-A-Y.15 . . X ' 174 amp - 587 amp.000 volts between Base current referred to high-tension side .126 ELECTRIC CIRCUITS— THEORY AND APPLICATION Balanced impressed voltage V = 115. = I 1.680 amp. Line 1 10 F 20 =0 to Neutral.r. Displacements of neutral points 3. Tertiary currents = Iia — I&> = lie Displacements of neutral points 2.15X587 v 1Q .. Secondary currents |/„| = la = (1.210 volts Single-phase Short Circuit.761 = 3 5.. X X 100 \hc\ = %i 2 = 5T 2X5. |/i6| - — \/3 X 2 100. Line to Line.5 = 9. = Tertiary currents |/ia| = |/»| = |/i«| = X 587 16. 3. \Iia\ = he = lr .174 /2c + = 587)3 5. ' P ' . 2^30"° - +Z V3(2Z 12 Flo F20 = = (e) 23 ) Z3/20 (/) £23/20 (ff) Primary currents = |/i*l = 100.000 —-— V3 X .i-l°°: |/id| = |iie| it- loo = \ i amp.4 = lines amp.4 X 5. Fio = Single-phase Short Circuit. 115 Symmetrical Short Circuit.283 amp. F10 = 12. 1ly . I\a F 20 = 7. inn 100. Ila ~ (d) V^7 2 Primary currents \r i |/io| \r = 1004r?X =.4 F20 = -^joo"4 — = 33-3 per cent Hence.283 amp.i-ir. 1UU.761 5.380 volts ^ = 2Z~ Primary currents |r .7 Secondary currents = |/ 2 »| = |/»| = \hc\ X 1. 20. 1.7 + = . X 7 100 = KO - .4 1/101 Ir . 9. 5.210 volts 7. Single-phase Short-circuit.380 volts amp. 4.line to neutral.525 X 3 = 4. Consider a seccircuits of this connection — ondary single-phase short circuit. 1. . amp. as shown in Fig. in general. Short on a Bank Transformers Circuit (Example op Y-A-Y-connected 1) Single- Quantities calculated Symmetrical Single-phase three-phase line. amp. b 525 amp. 73. 9.761 amp. Oamp. amp.to-neutral short short circuit on circuit phase a phase line- to-line short circuit between phases a and Primary currents: Phase a Phase b Phase c Secondary currents: Phase a Phase b Phase c 1 761 amp. 12. 525 amp. 5. superior to the Y-A-Y connection in that it imposes a less severe duty on the tertiary delta. When supplying unbalanced loads and dissymmetrical short such a nature that zero-sequence currents are produced. 1. Tertiary currents: Phase a Phase b Phase c Displacement of neutral points Primary Secondary amp. amp.5. he = Tertiary currents I »a — lib = Ize = Displacements of neutral points V l0 =0 F 20 = The results are gathered in Table VII so that they may be readily compared. two low-impedance paths are provided for the third-harmonic current. 1. . 1 . . 587 amp.680 amp. volts volts — A-A-Y Connection. Line to Neutral.TRANSFORMERS WITH UNBALANCED LOADING 127 Secondary currents \ha\ = \hb\ = 1.283 amp.680 amp. 4. 587 amp. When two of the windings of a three-circuit transformer are A-connected.174 amp. 1.575 amp. volts volts 9.761 amp. is.680 amp. is very satisfactory from the standpoint of wave shape.283 amp. amp. therefore.283 amp. This connection. Oamp.575 amp. amp. Table VII.283 amp.575 amp. 1 . 5. and the following equations may be written: V - V20 = ZJ W — ^30 = Zzlm — ^20 7 . inserted in equation (217). which. in this case. equal to three times the and negative-sequence zero-sequence primary and secondary windings. (221) I: (2) "(I Jl Fig. no zero-sequence voltage can appear across the terminals of the A-connected windings. \short \Circuif —Single-phase line. Hence. (223). 73. for Z 12/ 2+ (223) Zth+ = V xi VA = V£ current in the short-circuited phase component. gives -(* + £) (222) The relative value of current components in the secondary (winding 2) are the same as for the Y-A-Y connection with a single-phase line-to-neutral short Hence. and (224) gives total - Fit + V iT + (s2Z12 -+- Z2 Z Z S\ 1 -+- (225) ZnJ from which •120 = — 7l+ + + Z2 2Z12 Via Z1Z3 + Vi 2Z12 ~r Z2 Z1Z3 ~r (226) . These components the secondary circuit.128 ELECTRIC CIRCUITS— THEORY AND APPLICATION Obviously.V = ZI 10 • 10 3o /10 + S S0 Z2/20 = -F Z$Iw == V 20 Z1/10 = (217) 20 (218) (219) ho + In = (220) Eliminating 7 3 o by substituting equation (220) in equation (219) gives Zz Zi+ — — nT 1 20 Z. the in only currents exist phase a. F 2« = Vxa - is Positive- + ZJ£ + ZiZjJ + Z ha = Vu + Z 12l£ 2 (224) secondary voltage of the short-circuited phase being The zero. 65. are illustrated in Fig. also. addition of equations (222).to-neutral short circuit on the secondary side of a A-A-Y-connected bank of transformers. Equating the secondary terminal voltage of phase a to the drop across the load impedance gives V*t + V^ + 720 = V. — 2. Equations (217) to (224) inclusive hold also in this case. 74.TRANSFORMERS WITH UNBALANCED LOADING The short-circuit current 1 2a — — — SI 20 129 becomes 3F 1o (227) " + 2Zi2 2/2 + Z\Zs Z\z Using equation (221). Single-phase Load. Consider the secondary loaded by a single-phase load of impedance Z. connected line to neutral. — Single-phase A line-to-neutral bank impedance load on a A-A-Y-connected of transformers. Line to Neutral. depend only on the line voltage impressed on phase a and are independent of the values of the other line voltages. as shown in Fig. the zero-sequence current in the primary is obtained as 7io Z V la = 3 (228) Zu[ 2Zi2 < -\- Zi -\- ZiJ Also. (^10) becomes directly equal to the balanced line voltage (F). 74. Iza = Iao = (1 10 + I20) = Zx —fj—IiQ &1Z Z{Vu (229) Zwi 2Z12 + + z J (2Z + Z )F = — <\Zz Z\A 2Z12 + Z + ZuJ < Z% 13 Jla = — 2/20 + 13 /l0 3 la (230) \ 2 When the primary line voltages are balanced. Since the currents. however. the solutions remain the same. the negativesequence voltage (Via) is zero and the positive-sequence voltage ® Fig.t + Via + Uz n + Z2 + ^V» -3Z/20 = (231) . + Z12I26 (238) .t — 2Zi2 + Z2 + 7ta + M? + 3Z + 2Z12 The load i2a — 3i current 20 Z2 consequently. the tertiary delta can carry no current of fundamental frequency. since neither the primary nor the secondary currents can contain zero-sequence components. as Short Circuit € Fig. This is obviously true. and the solution will be worked out without resorting to this device. and the total current in phase a of the primary are given by I io z T — — -n—3 -/20 z v la — 3 (234) A 2Zi2 Iza = / 30 + Z2 -2/20 (235) 2Zi2 < 3. shown in Fig. the tertiary current. + + z + ^Z13 + Z )V 3 2 (2Z 13 + ZlO = Zi 3 ( 2Z 12 s 3Z la (236) + + ^.+ 3Z 3 by 3F la = — (233) + 2Z 12 Z2 + %^. 75. phase > -*20 si = 3Z Zl .ELECTRtC CIRCUITS— THEORY AND APPLICATION 13(> Hence. Very little is gained in this case by resolving the voltages and currents into symmetrical components.+ 3 The primary zero-sequence 3Z current. —When a 3 Z2 Single-phase Short Circuit.+ 3Z Line Line. The secondary voltages of phases a and b may be written 7 2 a = 7i« + Zi 2 / 2 a (237) F 26 = Vv. given is. 75. X20 V.If— 7 — ^13 /la + +^^13 3 to line-to-line short circuit occurs on the secondary single- side. b -Single-phase line-to-line short circuit on the secondary side of a A-A-Yconnected bank of transformers. + (232) ^. —lib = Via ~ 7l6 2Zi2 (241) he = ~hc = (242) When the primary line voltages are balanced.TRANSFORMERS WITH UNBALANCED LOADING Equating the voltage across the short F 2a circuit to zero gives = V u ~ V* + F 26 = 131 2I 2a Z 12 (239) from which Iia = — lib = I la = v la . connected (244) 2Zu — If a single-phase load line to line. V^ . line-to-line is can be bank of worked out exactly as in the preceding case. 76. + 2Z 12 J 2o (246) Hence. the currents are given by -U = Ila 4. (245) 2Zi2 Single-phase Load. + Z (248) (249) . the solution — Single-phase impedance load on a A-A-Y-connected transformers.v lb (240) 2Zi2 Also.V a = V37 lo /30° (243) In this case. 76. Equation (239) may now be written V^a . using the voltage of phase a as reference vector. the voltage difference in the above equations may be written. lla = —lib = Ila = —lib = lie = -z 2c = ~ Via 2Z 12 Via ~ 2Z 12 Vlb + Z (247) VU. Line to Line. Equations (237) and (238) still hold. as indicated in Fig.V» = -Zha = Via - Vu. V37ia/30° VSVla/S0° = Fig. voltage and current in the secondary windings.ELECTRIC CIRCUITS— THEORY AND APPLICATION 132 When the impressed line voltages are balanced. 77. 77. next single-phase short circuits from line to neutral on two phases. — Two single-phase line-to-neutral short of a circuits on the secondary side A-A-Y-connected bank of transformers. Equating the secondary voltages of the short-circuited phases to zero gives = =* + Fia V{b + Fi+ F 16+ + +Z 12 J 2 + + Zi 2 / 2+\120° Z 12I 2 a + ( ( = V + l + (252) + 7i» Vit 2 Z 12 7 2 q /120 o z° Adding equation ^ Z + + (* xr))ho (252) + Z1Z3' + §ir) (253) and equation (253) gives + + Vit Zx^+XoO 5 o + Z / q/60 + + ^A/ (254) 2 12 2 (z 2 20 Eliminating the positive.and negative-sequence current (using equation (156)). the currents become 1 2a — ~I%i — — I la ~ ~I\b — -n/3Fi„/30 2Zi2 V3> o (250) +Z lo /30° + 2Z X2 (251) Z — Consider 5. shown as in Fig. Equations (153) to (160) inclusive are valid in this case. Line to Neutral. this equation reduces to = 7i+ + 7uT + V£ + Vit + [z 12 + 2(1Z 2 +^Aj/ 20 (255) from which X20 — — v tW° + vweo° Z 12 + 2 U Z % + -^\ +v + 2^Z + Vla 1 Z 12 lb 2 ^ (256) . Equation (222) gives the relation between the zero-sequence (Z)_ (3) (1) Short Circuit Fig. Two Single-phase Short Circuits. (262) + 7! (263) -(Irf 2 Iu = -(7 2+/120° + 7te\120°) (261) . as positive- T + — — 7+ + V3 (z* (Vu + ^)/2o/W +F 16 /60°)/30° + (* + **)/ C 2 n/60 (257) Zl2 *2a — r* -^(Fia +7 o lft \60 )\30° + (z 2 + ^V 2o\60° (258) currents in the primary and tertiary wind- The zero-sequence by -ings are given r •t Z _ — _ raT— ° 10 r J- 20 _ — _ ZziVSW + L 1 Zr Z 12 ^13 ^ia/ 6 0°)— + 2(* + — = ©]+ F Z 3 (F la 7an — -*3a — Zx 1 20 Z (F t\605 1 — Zi 1 zu + + 2(z. the and negative-sequence currents are obtained.TRANSFORMERS WITH UNBALANCED LOADING By 133 simultaneous solution of equations (252) and (253). lb ) (259) 7-/60°) + *g)] (260^) The secondary and given by (153) (154).+' (7 +\120° + 7^/120°) + 7. short-circuit currents are The primary Ua= found by equations currents in the three phases are + I»r)+/l0 -<7. as shown in Fig.phase line.134 ELECTRIC CIRCUITS— THEORY AND APPLICATION When the primary line voltages are balanced.to-neutral short circuits are The secondary terminal may now V{-a a A-A-Y-con- + voltages of the loaded be equated to the drops in the load impedances. connected line to neutral. phases giving v t 1 + V£ + 7i6 referred to in the treatment single. 78. former bank supply two single-phase loads of impedances Z a and Zb. z 12 ht +Z 12 + z 12 i^ 7 2 +\l20° +Z -Z + ^V» = -Za (J£ + ha + 7 «/120° + fz + + 12 (z % 2 2 6 (J 2 +\i20"° + 7 2 a/120° /*>) ^\ + 7 20 ) (267) 7 20 = (268) . 78. respectively. Let the trans6. Line to Neutral. (I) I * ^ Vja * Fig. the equations for the secondary current components reduce to F la \60 — 1 20 c (264) (z 12 ht = - +Z + 2 ^ [z tt+a (z. + Vu (265) ^)] /20= (266) Z 12 [Z 12+ 2(Z 2 + ^)] — 7W Single-phase Loads. — Two single-phase line-to-neutral impedance loads on nected bank of transformers. The same equations which were above of the two applicable. Line to Consider. the given by equasecondary zero- . Equating of the secondary is this to equation (193). finally. the solutions are given by equations (179) and (180) or by equations (181) and (182). these equations Vit + vi 1 Fit + 7i6 which I" + Za + = -F(Z« (Zi.TRANSFORMERS WITH UNBALANCED LOADING Substituting for 7 2o in terms of I change to (156). by means of equation ^ +^ = -\z l2 + za + (z 2 [Z« + (Z 2 + + Z )/120° + U + 2 6 + + z«) W] hi - + Za\ /Wl ha Z6 )\120 5 135 (z 2 + ^r? + Z*) X60 ^+ Z6) /60^1 7^ (269) 5 ]/. the transformer bank loaded with Neutral. and zerosequence impedance of the unbalanced load. this problem can be handled by the same general method used in connection with the unbalanced three-phase load on the Y-A-Y-connected transformer bank. the zero-sequence current in the secondary is found by equation (156). The zero-sequence voltage tion (222).and negative-sequence currents become — *£ 20 ~ Yi i k k —~r kiki _ (k» + fei/60°)7i.and negative-sequence currents are deter- mined. 79) indicates.+ (270) be written V la = V + 7» = V& x + 7* = -Jbi7. ktki . may £ and 1 2„. three unequal impedances Za . and Z c connected line to 7. positive- When the positive.t . The currents in the other windings are determined as outlined for the preceding case.kj^ (271) (272) Simultaneous solution of these equations gives the secondary and negative-sequence currents. as the circuit diagram (Fig. Since they are identical in form to equations (177) and (178). equations (185) to (195) inclusive apply also in this case. Evidently. When the primary line voltages are balanced. — . negative-.kxkt 2 _ Iaa ( 273 ) z (274) General Case of Unbalanced Three-phase Load. Zb.kJto + 7u = -A^J . Making use of the equivalent positive-. the secondary positive. neutral. (282) Z13 Z1Z3 + Z + Zn 2 — Z\z Z 1 = Zn Z1Z3 z\ Z k3 + . which may be written V£ = kj+ + kjfa (278) (279) Where hx Z+Z_ = z fc 2 z 2 + = +z + 2 Zq Z\Z% Z\ (280) - z. as follows: 1 20 + Z+Iu + Z + Z1Z3 Z-Ijfc — Z„ (275) 2 Inserting this in equations (194) and (195) gives Z + Z_ (276) (277) Fig. — Unbalanced nected transformers. (281) - z. 79.ELECTRIC CIRCUITS—THEORY AND APPLICATION 136 sequence current is obtained in terms of the positive.and negative-sequence currents. three-phase impedance load on a bank of A-A-Y-conneutrals of the transformer secondaries and the load The are interconnected. The primary zero-sequence current is given by equation (221) therefore. sequence currents then reduce to If Fi7 = *2+a 7 = .-(/£/120° + 7 -\120°) + 7 7 30 becomes = -|^7 20 = Zu (285) (286) 10 ^^+ + + Z+ zJzo z% l^ x ^\ (287) the impressed line voltages are balanced.+ 7 10 J» = -(J£\120° 2 lie 2 tertiary current U=J 3c = (284) + 7 -/120°) + Zio .and negative-sequence currents. T ilO The total and the I 3a = _ — — ^3 j ~y-l20 Zj{Z-J^ _ Zl3 7 ZufZo + + Z+7 2 o) T rr K Z2 + \ /OQQ\ {1&6) *™ primary currents are Iu = . The secondary zero-sequence currents are determined by equation (275).(/+ +_I2~) . are given and is. V& = V and The solutions for the secondary positive. hence. (200). by equations (204) and (205).TRANSFORMERS WITH UNBALANCED LOADING 137 Equations (278) and (279) are similar to equations (199) and The secondary positive.and negative0.= k\ - k 2k 3 *F& (288) (289) . respectively. 1 Each of the balanced current systems may be looked upon as giving rise to a formance 1 Lyon. "Induction Motors on Unbalanced Circuits. When the motor draws unbalanced currents equivalent to two balanced current systems of opposite phase sequence. p. 30. 1920. The slip of the motor will automatically adjust itself. p. M. "Unbalanced Three-phase Circuits. therefore. W. This field. "Effect of Unbalanced Voltages on the Operation of Induction Motors. p. which is always the case in practice. C. Assuming that the motor is either A-connected or Y-connected without neutral. revolves with synchronous speed with respect to the stator windings. it will "break down" and the and rotor currents. When unbalanced voltages are impressed on a three-phase induction motor. A. World." Elec... World. Dudley. form a closed triangle and may be decomposed into two balanced systems of positive and negative phase sequence.. which. Jour. its per- may be determined by considering the effect of each system separately and superimposing the results. "Induction Motors on Unbalanced Voltages. when the currents in a polyphase induction motor are balanced. Schoenfeld. p. a revolving field is set up in the air gap. until the internal torque. V. J.. and torque is produced in the direction of the revolving field. Vector Method of Analysis of Unsymmetrical Systems. If the motor is incapable of producing an internal torque of the necessary magnitude. The unbalanced current vectors. 1304. 1924. which is proportional to the product of the stator equal to the sum of the load torque and the torque due to the losses and thus sufficient to keep the motor operating. has a sinusoidal space distribution. 1925. It is a well-known fact that." Elec.. 138 ." Elec. Jour.CHAPTER V INDUCTION MOTORS ON UNBALANCED VOLTAGES SINGLE-PHASE OPERATION — General Discussion. rotor will come is to rest. 339. O. It induces currents in the rotor windings. in the ideal case. 313. Slepian. no Uniphase currents of impressed frequency can possibly flow. 1920. the currents which it takes will also be unbalanced." Elec. e. the positivesequence slip. produce oppositely acting torques.. The range between 100 per cent slip (standstill) and 200 per cent slip is obtained by driving the rotor in the Assuming that the motor reverse direction by external means. these fields. 100 Per Cent Slip I Fig. Obviously. must equal the sum of the negative-sequence torque and the torques due to the load and the losses. 80. in conjunction with the currents they induce in the rotor windings. operates on unbalanced voltages and that slip-torque curves. viz. as in Fig. to 200 per cent slip for operation on constant balanced voltages. which is the actual slip of the motor. Figure 80 shows the speed-torque or slip-torque curve of a polyphase induction motor V Negative Sequence Fig. motor operating on bal- When the rotor turns over at synchronous speed. and resultant slip-torque curves of a polyphase induction in the region from motor operating on unbalanced voltages. namely. the negative-sequences lip is evidently 2 — s. in order to make operation possible. the positive-sequence slip) is s. are available for values of balanced voltages equal to . If the slip (i. — Positive-sequence. therefore. is zero.INDUCTION MOTORS ON UNBALANCED VOLTAGES 139 revolving field in the air gap.and the negative-sequence fields. This torque. hence. while the negative-sequence slip has its maximum value. will rotate in opposite directions and will.. negative-sequence. -Slip-torque curve of a polyphase induction anced voltages. 81. 2 or 200 per cent. the positive. The positive-sequence torque acts in the direction of rotation of the rotor. 80. remains sensibly the same. Figure 82 shows the slip-current and slip-power-factor relations torque. the power due to positive. however. Positive.140 ELECTRIC CIRCUITS— THEORY AND APPLICATION the positive.and negative-sequence currents must be combined Assuming that the vectorially in the proper phase relation. the phase position of the currents is immediately ascertained by means of the power- factor angles. and the net power determined for any slip.and negative-sequence currents The may evidently be carried through either analytically or graphically. the complete slip-torque characteristic for the unbalanced operation may be obtained by superposition. One phase (and frequently two. in order to deliver a specified pulley torque. the amounts of heat developed in the three phases of the stator will be unequal. 82. depending on the amount of unbalance) will be heated considerably more than . This shows that unbalance reduces the net torque for all values of slip. The starting torque. as well as the pull-out lower than for balanced operation. 81. the copper losses and. The slip at pull-out.and negativesequence components may readily be calculated. the slip must increase when the impressed voltages become unbalanced. is known. the positive. From the values of — Fig. combination of the positive. Hence. is for positive- and negative-sequence systems. 83). hence. as well as their magnitudes. In order to determine the unbalanced currents. relative phase position of the component voltages. as indicated in the vector diagrams (Fig. current and power factor. as shown in Fig. With unbalanced currents.and negative-sequence current and power-factor curves plotted versus slip of a polyphase induction motor operating on unbalanced voltages.and negative-sequence components of the unbalanced system actually impressed. INDUCTION MOTORS ON UNBALANCED VOLTAGES 141 delivers the same output on balanced voltages. respectively. which. Fig. A part of the may also be due to field distortion caused by the unbalance. in copper by loss. will be identical even when the motor operThis is due to the fact that each ates on unbalanced voltages. under balanced operation. increase in total core loss . of course. the copper loss in each stator phase can readily be calculated. V+ / + c c vb'J~. It is seen that both the total core loss and the total copper loss The increase when the impressed voltages are unbalanced. although higher than for balanced operation.and negative-sequence rotor currents. 3[(/+)V+ + (I-yr~] (1) the equivalent resistances per phase for the and negative-sequence systems. rotor phase carries the different frequency. The core loss will usually increase somewhat when the motor is operated on unbalanced voltages. be taken that the correct rotor resistance is used for each frequency. —Vector diagrams of voltages and currents in a polyphase induction motor operating on unbalanced voltages. The total copper loss is larger for unbalanced than for balanced operation. it is given by positive- Pen where r\ and [r~ are = be computed directly from the Neglecting exciting currents. when the motor When the same effective values of two currents of magnitude equal to the positive. The copper losses in the rotor phases. therefore.and negaThe tive-sequence load components in the stator windings. The aggregate copper loss may and negative-sequence current. assuming the same power output in the two cases. separately. The increase is principally due positive- to the introduction of a double-frequency core loss in the rotor. the positive. is obviously obtained losses due to and adding the copper Care should. is practically without core loss on account of the low frequency of the rotor flux. in each rotor phase calculating. unbalanced currents are determined. 83. . + jY + (xt "Principles of Alternating-current McGraw-Hill Book Company. Note that magnitude as all s. . X2 = stator and rotor reactance per phase impressed at frequency Zi = Z = e ri + jxi = stator impedance equivalent impedance corresponding to slip In subsequent formulas.I e Z. Three-phase Induction Motor on Unbalanced Voltages." York. quantities are reduced to stator.. R.ELECTRIC CIRCUITS— THEORY AND APPLICATION 142 and windage remain sensibly unaltered. R. performance may be calculated by the following equations: 1 Vi = Vi . I2 phase e per phase impressed at frequency Ti r'i = = rotor resistance per phase (effective or ohmic) rotor effective resistance per phase at double impressed frequency Xi. When the induction motor operates on balanced voltages. New + x*Y Machinery. Analytical Treatment. 1916 and 1920. V[ = P = 2 I 2Z e 1 Ilr 2'2" = - U ri + j+jCsi + 2 - r2 ) 2 +s (F{) (1 s (sn + 2 a*)]" its (2) (3) s)sr 2 (zi +xy 2 2 (FI) (1 - «)(4) Vi 1 Lawrenck. Inc. since the is not very large. well as phase of the rotor currents (7 2 ) are referred to stator. The motor will consequently suffer a decrease in efficiency when operated on unbalanced voltages. The following notation will be used in the friction loss will reduction in speed for the same output — discussion below: = impressed voltage per phase = stator and rotor current per I = exciting (no-load) current T-t = internal torque per phase P2 — internal power per phase p = number of poles /1 = impressed frequency s = slip rx = stator effective resistance Vi I\. the performance of the motor may be calculated by means of these equations. is ohmic rotor different for the resistance. Strictly. assumes operation at a small slip. The rotor resistance. the effec- — s. due to a small decrease in inductance effect. Since the exciting current always lags the impressed voltage by a large angle. Equation (3) relates the voltage V[. however. the for the negative. and two systems. For the positive. The exciting currents may be taken into account for both the caused by skin positive.INDUCTION MOTORS ON UNBALANCED VOLTAGES P m 12 ~ TZT. respectively. in general. the slip equals the actual slip while the slip for the negative-sequence system is 2 s. will be neglected. Equations (4) and (5) give the internal power and torque. of course. (2) to (5) inclusive) hold separately for positive. The stator resistance is the same for both systems. and the constants of the motor. Since holding. Hence. and the equivalent impedance of the motor. For the positive-sequence system. 1 7-2 Wi r2 _ ?. the negative-sequence rotor reactance may be slightly less than its positive-sequence reactance. Since the positive-sequence exciting current is quite appreciable . for the positivesequence system alone.and the negative-sequence system. in terms of the rotor current or the voltage V[. namely. « P(Vi) 2 Mi 143 sr 2 («"i + r2 ) 2 + a'fo + x2) 2 ^2 p(7i) 2 s (5) Equation (2) gives the voltage which is effective in producing the load component of the current (rotor current) by subtracting the stator impedance drop due to the exciting current from the impressed voltage. as determined by a test with the rotor blocked. the slip. or they may be completely ignored. the effective rotor resistance at double impressed frequency should be used. the equations above (equations when the motor is operating on an unbalanced circuit and either the component voltages or currents are known.and negative-sequence systems. tive resistance at impressed frequency. This. however. it is usually sufficiently accurate to perform this subtraction algebraically instead of vectorially. This. for balanced voltages and currents. the rotor current. The equivalent + reactance (x x x t ) is usually considered the same for the two systems and taken equal to the reactance at impressed frequency. Assuming that the motor operates on an unbalanced system of voltages whose positive. This.and negative-sequence components are Vf and Vt. since This is due to the fact that the negativeit is always very small. the negative-sequence flux is small. and the (20 to 30 per cent of the full-load current) this into consideration. however. in connection with the fact that the impressed negativesequence voltage is usually much smaller than the positive-sequence voltage. causes the induced voltage of negative sequence to be quite low.)*^ + r&] (17) Motor. negligible. It is excitation required to maintain it. and neglecting negative-sequence excitation. n + -7 + j(xi + x (6) t) p-^ = Pt = be carried through as follows: (7) dtYrJ^ (8) i^-CW'4^ The (9) power and torque are now given by 3(P+ + P.) T = 3(T+ + 7Y) total internal P = The unbalanced (13) stator currents are ha = ht = ht ho = hi 7l5 and the (12) total copper loss Wt + Ila + hi + Iu = let + ht + ha = ht + I£ + hb = hi + hi + Iu (14) (15) (is) becomes + (1. to take the negative-sequence exciting current into account. Consider Induction Three-phase Balancing Effect of a circuit to an feeder over a power supplying a generating station Pcu = + 2 It) ri + 2 (/£) r 2 — . respectively. Consequently.144 ELECTRIC CIRCUITS— THEORY AND APPLICATION it is advisable to take hardly worth while. the calculation n= II may n-itz. sequence equivalent impedance for values of slip in the stable operating range is very small and seldom exceeds 15 per cent of the equivalent impedance for the positive-sequence system. balance at the load is effected. as well as the volt- ages at the load. -IjIZ vt . capacity of the generating station is large compared the If to the load. loc. the positive. 1 If a three-phase induction motor. and an improvement in the voltage resultant negative-sequence current carried reduced. the negativedrops decrease. The Z — Single-line diagram of a three-phase feeder circuit supplying an Fig. Lyon.I+Z eiU (19) is. In order to obtain this balancing effect. unbalanced load. the unbalance in the internal impedance drops of the generators will be inappreciable. in spite of the unbalanced load.INDUCTION MOTORS ON UNBALANCED VOLTAGES 145 unbalanced three-phase load. Figure 84 shows one phase of the system.and negativesequence voltages at the load are given by V£ = V% - IL tZ (18) VLT = -IL JZ The degree of voltage unbalance at the load V7 vL t T 1 See paper by W. A polyphase induction motor is floated on the system in parallel with the load. There is no neutral. induction motor will draw negative-sequence currents which will add to the negative-sequence currents taken by the load in such a manner that the the line is sequence line by As a consequence of this. The currents. Before the induction motor is added. it will tend to restore the balance of the system. is connected to the system at the unbalanced load. and the station voltages remain very nearly balanced. loaded or unloaded. and it is assumed that the impedances of the three conductors are equal and independent of the division of current among them. (20) . it is necessary that there be impedance between the motor and the point at which balanced voltages are maintained. V. 84. will then. in general. be unbalanced. thus. A synchronous motor for instance. Z 1m l£i Z (QfW + Zm This equation shows that if Zm were zero. the negative-sequence currents taken by the load and the induction motor would be Equaexactly equal and opposite and.__. cancel on the line. as previously brought out. exert a pronounced balancing effect. a tendency to balance . has a negative-sequence impedance which is considerably less than its positive-sequence impedance and will. respectively. and this may appropriately be emphasized at this point.and negative-sequence impedances of the induction motor.and the negative-sequence voltage at the load when the induction motor is added. hence.(hi + Im)Z = ImZm (21) (22) where Z% and Z M represent the positive. The machine does not necessarily have to be an induction motor. The new degree of unbalance is most conveniently obtained by combining equations (24). hence. the voltages at the load now become that the currents taken when the motor is V Lt = 7+ .146 ELECTRIC CIRCUITS— THEORY AND APPLICATION The motor is now connected to the circuit. Vm _ VLT Z Z u 4~ Z MZ M vLt vLt'zz M + z M z~M This equation shows that there . The ratio of the positive. If it is assumed by the unbalanced load do not change added. may be used. (23) and Thus. the degree of voltage unbalance tion (25) shows that for Zm = .and the negative-sequence voltage at the load after and before the induction motor is added is now very simply given by V ^ + £1 V Li Vl t 7+ Z+ ZM Z + ZM " Z {Z6) + (24) Equations (23) and (24) show that there is a reduction in both the positive. From equation (22) is obtained will of voltage unbalance and.{ILt + IZ)Z = I M Z M ViZ = . hence. It is evident. ^ 0) be a reduction in the degree up the circuit whenever Z M < Z M The induction motor. that any piece of apparatus whose impedance to negative-sequence currents is smaller than to positive-sequence currents may be used for the purpose of balancing up a circuit. 222 54.000 = Phase voltages at the load VtX = V% . Generator voltages to neutral referred to low-tension side I .5 .000-hp.j41. and 90 amp. 2. 50.INDUCTION MOTORS ON UNBALANCED VOLTAGES 147 also zero.Q) 1. "^ 3 (positive sequence) V3X 7.15 ohm at 120 cycles.08 ohm at cycles.itZ = 1. high-tension feeder to an unbalanced load.30 ohm.3(0. referred to the low-tension side. On open circuit.400 volts.7 .785 shl * ' . Hence the smaller the negative-sequence impedance of the motor is the more effective it is as a balancing agent. What are the degrees of voltage unbalance at the load before and after the induction motor is added? and that the What 2. the station voltmeter indicates 7.200 volts. 1. respectively. \ I Vq~ = ^_t VG — J? Load currents from diagram j/£ = 71. Only the magnitude of the components is required.200 volts. 85./0.386 - = 1. The no-load line current is 75 amp. The station instruments are connected in the circuit at the terminals of the generator. 2.1 power factor and rated voltage. 1.400 The positive. is EXAMPLE 1 Statement of Problem Power is supplied from a 60-cycle station over a three-phase. Power factor at generator C ° 8 dG = [Ti 698. The induction motor has the following constants per phase..200X71. 2. 80.5 and 0.10 ohm at 60 cycles.222 volts = -69. line currents is feeder before the ratio of the negative-sequence currents carried after the induction motor is added? by the and Solution 7 200 ~-tj~ = 3. °. r 2 = 0. Assume that the station voltages are balanced taken by the load do not change when a 1..3 214.200-volt induction motor is floated across the feeder at the load. while that on the low-tension side of the load transformers shows.and negative-sequence components of the unbalanced currents are determined graphically in the diagram.6 ohm.6 = = -IiTiZ .386 volts X X 3 3 = = 69.j34. r 2 = 0.1 volts .4 = 23. the station voltmeter shows 7. amp.8 = 1. With the unbalanced load connected to the feeder.214. Fig. Assume that the slip is zero.2(0.2 amp. Xi = x 2 = 0. per conductor.4 V£ « 00 °' 618 +J0. Balance would be completely restored.5 +J0. Overall ratio of transformation = Unbalanced Load on Line: a.Q) = -34.785 -. The resistance and reactance of the line and transformers are 0. and the line ammeters. assuming Y-connection: r x = 0. the polyphase watt- meter indicates 698 kw. at 0.618)(0.1 referred to low-tension side. It is assumed that the currents taken by the unbalanced same as before.6) + 48.0 2 2 .5 + j'33.148 ELECTRIC CIRCUITS— THEORY AND APPLICATION Degree of voltage unbalance at the load — Unbalanced Load and Three-phase Induction Motor on Line. load remain the HZ may be written 1. is X/ \l~\. 1.995)(0. 86. — Graphical + \=7/.4amperes determination of positive. \l — Vector diagram 1). (a) VlX .222 [a = VL+ + = VlX 75(0./0.VlX + sin 6m = 0. equation (a) motor is added. 1). of positive-sequence voltages and currents (Example to the power-factor angle at the Note that the angle marked 0j corresponds generator before the induction motor is added.1 Using VlX as reference axis and inserting numerical values. 85.1 - j0.0 Hence.and negative-sequence currents (Example Fig.5 +.222* = {VlX + 48.23.995 cos dtt =0. Figure 86 a vector diagram of the positive-sequence quantities after the induction 6.5) + 33.1 amperes — y" Fig. 6 volts Degree of voltage unbalance at the load 7x.2~ _ 24. . since they are based on the assumption that the load currents remained the same after the induction motor was added. which. however. and The ratio of the negative-sequence currents carried after the induction motor is $/-The ratio in per cent is. solved for Vm. The — Single-phase Operation of a Three-phase Induction Motor induction motor operating on a three-phase circuit will con- An tinue to operate on single-phase supply if one of its phases is This statement is based on the assumption that the torque of the load which the motor has to carry does not exceed the breakdown torque for single-phase operation.173 volts = VlT Z ViT.10 percent. 2.625 ohm gives 0-175 ! W +Z- Negative-sequence impedance of motor at zero Z» = 149 °^ = +_$*__. jz 100 = 45. 1 added - is given by the feeder before by Z z + z-M W r y therefore. of necessity. Hence.r = = 54 (6) 54.6 = 0.43 to 2.2-10 Hence. the floating of the induction motor on the circuit in parallel with the unbalanced load reduces the degree of voltage unbalance from 4.175 + j0.43 to 2. calculated results.10 per cent.173 91 _ percent -. If greater accuracy is desired. must be caused by the connection of the induction motor. = 54 - 0. material calculations circuit in parallel improvement in the conditions is realized.6 24. The opening opened.1 + 0.175 slip °2- + J' - 6 +J0. should be taken into account.4 per cent L Resume The show that the floating of the induction motor on the with the unbalanced load more than halves the negativesequence currents carried by the feeder and thus reduces the degree of voltage unbalance at the load from 4. should not be considered exact. Z ri + 2^1 + J'** + X2) 1 = 01 = Substituting in equation F . the change in the load currents. gives = VlX 1.6 X 100 F4 100 - 1.INDUCTION MOTORS ON UNBALANCED VOLTAGES which. If the applied voltages during three-phase operation are badly unbalanced. ance will be obtained by opening one phase of the motor. 87. the positive. Slepian. line cur- This last statement assumes that the three-phase voltages are balanced or. the output is reduced to Hence. these components are all the time equal in magnitude and in a fixed phase position relative to each other. The equations for the stator currents are Iu = hi + hi = h+\120° + JW120° he = hi/120° + Iia\120 ha + h b + ho = he = I lb c (27) (28) (29) (30) (31) See paper by J. Power is supplied to the terminals a and b only. if the degree of unbalance (unbalance factor) of the three-phase line exceeds 70 per cent of the -percentage reactance of the motor. on the other hand. will losses in slip. at least. It is stated in this paper that. nearly balanced. the continuous capacity of the motor may become so severely reduced that it actually may be advantageous to open one phase and allow the motor to run single phase. 1 During single-phase operation. — Circuit diagram of a three-phase Y-connected induction motor operat- ing single-phase. slip. he. 1 . Consider the three-phase induction motor shown in Fig. cit. and be accompanied by an increase and a decrease in efficiency. As a matter of fact. In the single-phase case. 87.and negative-sequence currents.and negativesequence voltages across the stator terminals change with the Rotor Sfaior (2) (I) Fig. better continuous performless than 70 per cent of normal output. there is a definite relation between the positive. The continuous capacity of a polyphase induction motor operating single phase is somewhat less than 70 per cent of its rated capacity.150 of ELECTRIC CIRCUITS— THEORY AND APPLICATION one phase rent. They are not fixed by the applied line voltage as in the unbalanced three-phase case. = I la (Z+ + Z~) (35) . 88..7 X +)Z+ (I. The equa- tions so far established for the stator currents are valid. When the exciting currents are neglected. 88. |/rl - is equal to viz. The -30' *60' *Ja Fig. The relation between the impressed single-phase voltage and the current taken by the motor is obtained by expressing the terminal voltages across the two active phases in terms of their positive.V lb = (7 + . The values of the components as determined by equation (32) may consequently be used in the general three-phase equations for power and torque. the current necessary for negative-sequence excitation may safely be ignored. As in the unbalanced threephase case. Neglecting Exciting Currents. 88. whether the exciting currents are considered or omitted. therefore. be looked upon as representing also the rotor currents referred to stator. the numerical value of the components the actual current taken divided \it\ = by \/3. 87. Fig.. —Vector diagram U JC of stator currents of the induction motor in Fig.I»)Z- + = I laZ+ + haZ. V u = V£ + Vu = hiZ+ + l£Zy* = V£ + V b = I+Z+ + IfZr x (33) (34) Subtracting equation (34) from equation (33) gives V* = V la . vector diagram. Obviously.INDUCTION MOTORS ON UNBALANCED VOLTAGES These equations are satisfied by the system 151 shown of currents in Fig. may. the positive-sequence exciting current may or may not be taken into account.and negative-sequence components.. Analysis of Performance. Thus. It will always be very small. the total positive-sequence — as well as the total negative-sequence stator currents will have equivalent components flowing in the rotor windings. ^ (32) In deriving the formulas required for the calculation of singlephase performance. quite obviously.) + +r + 2 3[(/l) (ri 7!(2n 2 2 (7T) (r! + rj)] (40) r't ) a Y-connected induction motor operates single phase. as given by equations (33) and (34). Vah .) Making use of equations (8) to (13) inclusive and dropping subscripts referring to phase. a voltage will also appear across the idle phase. as shown in Fig. the total internal power and torque become Po = 3 (If)Vl L_!_( /r )v4— -m-sfi-^\ = = Since » —— is T 3P|" (/+). during single-phase operation (assuming the same slip as for threeslip. phase operation) is principally due to the decrease of the positivesequence power and torque. is less than the current taken during three-phase operation on balanced voltages of the same magnitude as the single-phase voltage. voltages will appear from line to neutral on the two active phases.152 ELECTRIC CIRCUITS— THEORY AND APPLICATION VA Vab z+ + zri + — + j(xi+x t) +ri + g-Z-^ + J(x + i &«) F„6 2ri -2 + o + #(*! + + «-^~ ^ o (36) x. This is due to the fact that the positive-sequence current. In The general. 87. This is readily seen._ < + -^- (39) Z~) V3Z+ The total copper loss during single-phase operation when exciting currents are ignored becomes V3 Pcu = = (Z+ + r. value of these voltages depends on the slip. when operated single phase. since.!! ( (37) iT )i i^[?-^i] very much (38) " smaller than - for ordinary values of the effect of the negative-sequence power and torque is The reduction in the total power and torque suffered small. Neglecting exciting When . while the voltage across the idle phase is zero. the = 00). This is due to the fact that. will never be obtained in actual practice.zV— V3 Z+ V3 174- /7i (42) (43) 1 the motor runs at synchronous speed. (42). (46) \/3 Fully balanced voltages. exactly one-half of the applied single-phase voltage appears across each of the active phases. . the phase voltages become positive- ya = ^f Vi = ^f V = c As (47) (48) (49) seen. At synchronous speed. hence. and the and negative-sequence impedances are equal. and (43) reduce to • y = 1^0° rt Vb = F V = c (44) V3 ^ 150 ° (45) V3 Ko6 ~- . and positive-sequence impedance becomes infinite (Z + equations (41). as indicated above. even at synchronous speed. the slip is unity. introduce a negative-sequence component in the voltages to neutral. the three phase voltages form a balanced system.INDUCTION MOTORS ON UNBALANCED VOLTAGES 153 currents. When the motor is at standstill. they are readily expressed in terms of the voltage Vab impressed on the active phases. even if the motor would operate at exactly synchronous speed. For this condition. the motor must draw a certain single-phase current. Va Vb Yc When (41) + Z7^150° Z+ + Z-\60° Z + + Z~ V3 vjw° z+ . part of which will obviously be of negative sequence and. Thus. ( / r M ' 2n+ j + 2~^~s +M*i + Xi) Torque. three-phase Torque. Neglect the no-load current. h -+ = z++z.5 per cent in each case. single-phase (d) Inserting numerical values gives j la-<t> — — 2.200/^/3 V(»- _lz27p_ — o ocw — — . The slip is 2. the magnitude of the voltages becomes equal to the single-phase voltage impressed. 1 +o^)"+ oib amp.154 ELECTRIC CIRCUITS— THEORY AND APPLICATION The voltages existing between the terminals of the active phases and the terminal of the idle phase are given by + Z-/120° _—~~ +z Z+/120 + Z-\l20° V -V = Vj ~^ ^ + Vbc = Vh = Yea -V Z+\120° = c Vab z+ a C (50) (51) At synchronous speed. -' >/(<"+££+£&)* +**"* . three-phase —^ h-* = n + 2 (o) +j(xi s + fc 2 ) Line current. Solution Line current.200-volt circuit. single-phase V . these voltages =V Vbc EXAMPLE (53) become equal and are given by =-^ ca (54) 2 Statement of Problem Compare the line currents and the internal torque developed by the three-phase induction motor described in Example 1 when operated on a balanced three-phase.200-volt circuit and on a single-phase 2.= V . Equations (50) and (51) reduce to V = V ab\120 o V ca = TW120° (52) bc At standstill. 2. as shown iri the vector diagram. and 72 + the stator compo- 1+ nents of the positive-sequence rotor currents. 7 X + represent the positive-sequence stator currents. it is convenient for the analysis to split also each current in this system into two components.975 J Three-phase current Single-phase current Three-phase torque Single-phase torque _ ~ X 378* k X 9. 87. It . 22 _ ' Analysis of Performance.6 598* X 3. I£.15 ] 1. When the total positive-sequence stator currents contain components required for excitation. *te If Components Fig. the current 7ia is decomposed into 7 2« equal in magnitude to 7 2J and displaced 60 deg. Although the total stator currents of negative sequence are reproduced in the rotor. 7 e+.632 598 2 378 X 9. positive-sequence exciting current being taken into account. Fig. in phase a.INDUCTION MOTORS ON UNBALANCED VOLTAGES 155 TV* = k = k Ti-# X 3 X 598 X 2 2 378 X 0. Taking Positive-sequence Exciting Current into Account. Thus.124 = ~ . 89.6 598* X 378 . the latter — must be subtracted from the total values in order to give the system reproduced in the rotor.025 08 _ f^025 X 0. 7 XJ. and ha equal to the positive-sequence exciting current and displaced 60 deg: from The negative-sequence currents in the other phases are let- split up in a similar manner. /£ Components —Vector diagram of stator and rotor currents of the induction motor in Fig. and 72+. in which 7X+. This is indicated in the vector' diagram. 89.0. and positive-sequence fee. from the latter. 72^.08 = k 0.124 3. and the exciting currents. The latter is given by the sum of 7 2a and the negative-sequence component 7 3a i-e. The following equations may be written for the voltages across the active stator phases: Vu = Vxt + Via = LiZ + 7 +Z+ + 7 = V t + 7* = IetZx + I£Z+ + 2 1 tt x (7*T (72r + ha)Z~ + Ig)Z~ (55) (56) Subtraction gives V* = Vla . V3 or.i pz+ +_ 2 ~ Iib)Z ~\(7 2a 7 2a (Z+ Z~) Z~) + — (7 3a Isb)Z + (57) this follows T = ha Vab ~ + + Z- Iea(Zl Z+ Z~) . From equation (58). ~ IeaiZi V3(Z+ + + Z~\ X Z-) omitting reference to phase.U)Z IJiZx + +_ t (72+ .AND APPLICATION ELECTRIC CIRCUITS— THEORY 156 should be borne in mind that none of the negative-sequence components thus obtained are exciting currents. the positive-sequence load component is obtained as Va. They all exist in both stator and rotor. I ea is the current which the motor would actually draw at no load (assuming zero slip).Vlb = = From (/ + ._ C v Q 58 ) Note that the current 7 2a .- 2 '=7i W (61) (62) . the total rotor current equals 7 2o + 73^" (or . as determined by equation (58). This no-load current is made up of a positive-sequence exciting current and a negative-sequence load component of equal magnitude but phase displaced 60 deg. when operated single phase. from the former. does not represent the total rotor current for single-phase operation. The value of the single-phase no-load current is equal to the positivesequence exciting current times v3.mzi + z+ + negative-sequence z-) zcurrent = in Kl z+ + (60) { J phase za is now determined by = %/30° = **L+Z-/30° = (7 + + 7+)/60° Vfr V3 Omitting reference to phase. /+2 The = Jl^ = total -t| Yl V3 .. 7 2o + 7^/60°). this may be written / = 7t + /t 7 = /. 200-volt.000-hp. the (66) ) r'2 t»|>8 p "I 2 r2 Comparison of these equations with equations (37) and (38) shows conclusively that the omission of exciting current from the analysis results in power and torque values which are too optimistic. 2. and the line current taken and compare with the corresponding quantities for three-phase operation.K7-^ 4-J (65) 2 Making use of the approximate relation in equation power and torque formulas may also be written a _.075 ohm per phase Rotor ohmic resistance referred to stator Rotor effective resistance at 25 cycles referred to stator 0. tion motor operates on a balanced circuit and delivers rated output to a constant-torque load. the total copper loss becomes Pcu = 3[(/+)V 1 + (7+)V. Determine to engineering accuracy the slip. the total P.INDUCTION MOTORS ON UNBALANCED VOLTAGES 157 Since the angle between 1% and 7+ is large. (63). 12-pole. the power output. When exciting currents are taken into account.. the efficiency. Stator effective resistance at 25 cycles 0. they may.__4_]_ iS L=L! ft . 25-cycle. This simplifies the calculation of the negative-sequence currents and gives Ii = V(It) 2 + Having determined the currents. Y-connected induc- ri) 3 Statement of Problem A three-phase. (It) 2 (approximately) (63) and negative-sequence load internal power and torque are given by positive- = 3[(/t)V. be added as if they were in quadrature. the total copper loss may be written Pcu EXAMPLE = H(2n + r2 + - 7|r2 (69) 1. One of the lines feeding the motor is suddenly opened and the motor continues to operate single-phase. r i^il (64) -f.i-=^ - (/7)V. p. + 2 (/T) (r x + rQ] (68) Again introducing the approximate relation expressed by equation (63).095 ohm per .083 ohm per phase 0. without appreciable error. rp _ V i. 1 deg.ELECTRIC CIRCUITS— THEORY AND APPLICATION 158 Rotor 50 cycles referred to effective resistance at ohm per phase ohm per phase 0.314 = 1.075 Vl .675 = _ 4Q 228.2745 ± 0.075 X 0.270 75.128.0.246 2 X 0.(K)V 2 2 [r? + On +* 2) P*[r\ + (X! +* 2) P B ] + (T^i ] + 2 ? C } (e) (V^r.007565 0. ^° = yj[ .(F^s + P*\ = (6) slip gives • . is given a = tan- 2 ^ = tan"* °° 14* $£*** = tan.SEEMI ~ \(1 \ X (/) 2 r 12 The phase x s)r 2 0.B (c) where A = 2P 2rir 2 .330 X 0.083 209.2 Solution Three-phase Operation: The internal power per phase given is by - (Fl)-(l •)? Pa = f (ri which may be written + P t (xx + xtf + [P*r\ Solving for the r + ^Y + (Fi)V 2 ] S2 + (xi (a) +xy [2P 2rir 2 2 .330 X 0.08* 228.626 2 ) 251. 2 251.800 = _ 125.075* 3. of this current.2602 P^I^r = ™ _ rT-^~ = .054 = 3.083 0.9857 0.0143 or 1.43 per cent = 0.083 1. 75.246 volts 251. by .125 .800 2X A 8 + + + 251.314 0.140 stator + Xt) Equivalent reactance at 25 cycles (xi No-load line current at rated voltage No-load power Friction and windage loss 0. kw.0.1 X 0. 8.600 = V + ^ 0.2745 ± 0.1 15.626 amp.--2 W(4)'.600 12p00 99.330(0.33 kw. with respect to the induced voltage.5 P ° amp.1. Assuming the leakage reactance to divide equally between stator and rotor gives Zi = = x 2 = 0.0 kw.246 X 0. Vi Xi V3 p2 = _ ~ D m s - = = ohm i )2 70 volts = IeZi = 746_+l8 - 1.313 = 0.0754 .083 .313 ohm jO. 0212 +./0.0523) It + + 1. 2rj Equation r* .59 = 75.1 (cos + sin n + 75.075 +j0. Since. 8.' <f>' ) ri -f I\rt) loss Friction 159 ™ 746 746 = . From this curve the exact slip may be obtained. 100 36 5Q = = = = = 22. however.6 per cent./75.INDUCTION MOTORS ON UNBALANCED VOLTAGES The power factor of the exciting current.9986 jO. 36. the effect of the terms involving the reverse-phase slip s' =2 — s is comparatively small. 0. {1 ' } V " r» +—+2 - ( -s) 2 At .2 3 x 1246 1 59 = ~i + loss.1455 +.313ohm 0.O 209 3(/ It 2 </>» X 3(227.V(/+)» + is ^« .» + r + j)- Crt)*.5(0. Assume./0.93 kw.50 kw.9998) + . 2- (A) approximately correct: (7+) 2 IT tf) cannot be solved for s explicitly. __ 95. Internal torque per phase Single-phase Operation: v p (Z++Z~)* Vs /o.5* + 209.)>}+ (Fj>£|. further. *" .4 per cent. 930 = T /core X 3 __ 1.57 kw.P X X 0.984 = constant. - 8.5 X 2 0. A solution may be obtained by computing for assumed values of s and plotting the values versus slip.1 amp.626 = 0.') Solving for the slip gives « = -f±^/(I) ^T v 2 — s = 1. that the direct-phase exciting current is the same as for three-phase operation. .8 deg. Equation (ft) may be written Equation # (ft) K [JJC + [(* + (It)* j] (It)* {(*i + f) J)2r 2 (2n 2 +4 ( Xl +*. 5. amp. "*" s) v s "'2 - 22-sV X2 '^A» assumes that the following relation (ft) . = 209. Hence.6 + J86. + (K + (/!)* J)r| - (.0 5.0 = 227.626 ohm .0 = 210.075 + 5^|? +j0. depends Core = loss cos#.0212 = n <t>' 88.00 kw. with respect to the induced voltage.. corresponding to a slip of 1.083) 10~ 3 and windage Total losses .5 (cos a 209. = h = = = = = Copper Core loss on the core - 15.075 10~ 3 = 5.075 0. .].93 kw. Assume s' = it Zi = = Z~ = 75. 0.5 amp.1(0. Efficiency X . this slip may be assumed constant for small changes in s without introducing appreciable error.6 + jll. + j sin a) + 75. 095 " 2 2 /une = V3 X V235 + 75.626 + jO.198 volts °-°705 + 0.1982 X 0.0705 = 0.1 .!(i S )T + Z~ ' (approximately) 2 V (1 .100 X D = s s Z+ Z+ = = = ' 0.0114 .2205 ^ - + t {Z + Z~) * +^ = 0.941 + jl. _ = s)K (1 p 2 = M.964 = 75.5 r V.8.100 119.10685 ± 0.700 = 250.700 + 75.100 266 1 088_» = Q 00343 513.3 kw.392 (y[)*~.500 412.166 X 0.1 1. some error in the core loss is not serious unless the efficiency is desired with a great deal of precision._ xr_~ Tr ^ = = 2 101.76 per cent = n r + j + j(xi + x t) = 0.655 9345 .095 ohms V[ .100[0. P = 2 Power output = 995 hp.075 = 4.1982 X 0.00343 0.0176)254. The exact single-phase core loss is not known.100 109.100 X 0.000 513.083 O0T4T K+ (i+)» ^ = 254." /2 Z+ + Z.10685 + Vo. .0176 or s = 1. and there is no data from which it can be accurately computed.083 0.0.0 = 742.964 1.°^0 _ ? " 2 .0705 = + -J* = 0. Z~ = 0.626 = (xi + x 0.500 WBX0 _ n oi*7 : 0. that the core loss increases by 50 per cent when the motor operates single-phase.119.100 - 119.2205 + 4 X C ~ 255.160 ELECTRIC CIRCUITS—THEORY AND APPLICATION + Zy = y.5 per cent of the total losses. 0.1 = 427 amp.795 ^^ + + jO.984 ~ 2ri j'O. 1* X 25 5. = 1.~ 5. It will be noted below that.100 + 0.04865 _ ~ _ 254 700 255.0705 (Fi')V« = 1.252 = 5.04865 = 0.15 = 939 = x e r -i«_ .08925 0.100 watts = 3 X 250.s ~ 1.626 = 4. however.400 + 101.2205 \2n 2 2 2) 209. since the core loss in this case makes up only about 1 1.M?8 _ Z65 am „ 235 amp I.392] 101. It will be assumed.270 - ohm X 0.0. 30 kw. Power output Line current in per cent of . and Slip.8 .000 hp. Friction and windage Total losses = 86. 1 See paper by J. + xj* the slip at which r2 m Vr\ + + (* x (71) *2 ) 2 and the corresponding maximum torque becomes 1 T = 4tt/i As formerly 2[ ri + Vr\ stated. operation is usually insignificant.40 kw.5 187. Efficiency Hiinciency [427*(2 = 742. 1 pull-out torque and slip will be.7 123 93. its pull-out torque and slip are It is easy to examine what the approximate seriously affected. . 1 43 per cent 95 4 per cent Slip . 1 76 per cent 89 5 per cent 1.14) 0.93 X 1.— When a three-phase induction motor is operated single phase.083]10" = 69.90 kw.1* X 0.3-X 1 °° 742 3 g6 3 + r> = 89.INDUCTION MOTORS ON UNBALANCED VOLTAGES Copper loss = = I\(2ri +r + . = 8. .5 per f cent Comparison op Results Single-phase Single-phase Three-phase Items calculated operation operation 227.5 = 8. . the torque + (x x +x (72) t )*] produced by the negative sequence system during unbalanced three-phase or single-phase It will be neglected here.083 0. the internal torque per operation.Hr * r'%) 2 - + + X 161 0. Efficiency three-phase 995 hp.00 kw. For balanced three-phase Pull-out Torque phase is given by = T* p(7i) Wi 2 rt/s (70) 2\ ri+^Y+ / ri+ By differentiation of T 2 with respect maximum torque occurs is found to be to (xi s. .075 8 3 X 75. Slepi an.5 amp. cit. loc. C!ore loss = 5.- 99. 427 amp. the single-phase induction motor problem reduces to the very same problem which its . the induction motor in — question was assumed to be a three-phase machine and it was shown how the analysis of its performance could be carried through with great ease by resorting to symmetrical coordinates. It should be carefully noted. that the equations previously derived are directly applicable only when the single-phase induction motor has a symmetrical polyphase rotor winding. performance can also be calculated by the same general scheme. The Single-phase Induction Motor.and negative-sequence currents flow and thus produce the requisite fields. Sm The maximum The = Wr\ + Zi PiV'i) 4xfx (74) +W> internal torque per phase T = result is is given by 2 1 4[n + Vr\+ {xx + (75) xtf] Hence.ELECTRIC CIRCUITS— THEORY AND APPLICATION 162 The internal torque for single-phase operation 2 T 2 . This pulsating field may be broken up into two oppositely revolving fields of the very same type that would be produced by a three-phase (or any polyphase) winding. In such a motor. which usually is the case. Throughout the preceding discussion of single-phase operation. the single-phase stator winding will set up a pulsating field when it carries current. the slip at single-phase breakdown may be determined by differentiation of this equation. which does not introduce any appreciable error when the slip is not large. however. a three-phase induction pulls out at approximately half the slip and develops approximately half the pull-out torque corresponding to balanced motor three-phase operation. In this manner. when operated single-phase. If the induction motor is actually a single-phase machine.P(F *> w (z+ is then given by ^ ~ + z-y r 8 ' (an Neglecting the term +?+ r ^) i + 4fe + ^« k~z~ *n *ke denominator of equation (73). the single-phase stator winding of the motor may be assumed replaced by an equivalent three-phase winding in which positive. Hence. however. INDUCTION MOTORS ON UNBALANCED VOLTAGES 163 has just been treated in detail. per phase of the equivalent three-phase windings corresponding to impressed frequency.Phase Actual SinglePhase Mo-tor Fig. v (79) . that of the three-phase induction motor operated single-phase. Neglecting exciting currents. 90. 2I la Z e + j2(x + x *. the performance of the single-phase motor may be completely determined by methods and formulas previously given. and power supplied to the single-phase motor. Sfirfor (1) t Rotor Sfator Rotor (1) (2) (2) aa coffitiZpn |tfffi> ?Gflfr %b Three. respectively. The constants per phase of the equivalent windings can be determined by a blocked -test just as readily as if the motor actually were a three-phase machine. When the rotor is blocked the slip is unity for both the positiveand negative-sequence systems. When the constants of the equivalent three-phase windings are known.- Equivalent -Schematic diagram of an actual single-phase induction motor and its three-phase equivalent. the relation between the impressed voltage and the current taken is given by Vai = Iu(Z+ + Z~) = = 2Z lc (r + jx ) = e e 7 l0 [2n +-2r. they are given by ™ V Z = e re 27 — (78) 272 = VZi - . the equivalent constants can be calculated. r*. Omitting subscripts referring to windings and phase. line current. By measuring the impressed voltage. namely.)] (76) Here re and x e are the equivalent resistance and reactance. Figure 90 is a schematic diagram of the actual circuit of a single-phase induction motor and its three-phase equivalent. 2.143 + 694.468 5.8 ohms 2 per cent + z. EXAMPLE 4 Statement The Problem of following data hold for a single-phase induction motor: 30 hp.200-volt line at 2 per cent slip.842 r2 2 At a slip s = z* X 1.468 ohms ri (effective) = 2 = = 2. 60 cycles Frequency Number 6 of poles [ No load | 7. Power 1.200 volts .300 watts Power 900 watts and windage Ohmic resistance between stator terminals Friction Ratio of effective to ohmic resistance. . Rating Voltage.1 ohms ' effi- . Stator reactance rotor reactance referred to Solution Equivalent three-phase constants (from blocked data) 620 = 4 35 ohms = Zt = 21 2 X 21.6 amp.+^+^h)' + *^ + Xi) = Vl0. the output.842 ohms = 3.212 ohms 3.8 same side.316 1.875 watts [ f Blocked 2. .68 ohms • 2 X 21.68 = 13.18 ohms 6307 X L1 = 3.200 volts Voltage Current I I 620 volts Voltage Current 21. Determine the pulley torque. the power factor.307 Rotor Rotor 1.36 amp. giving the effective resistances at impressed frequency r x and r 2 The ohmic rotor resistance r 2 as well as its effective resistance at double impressed frequency r2 are now determined or estimated. f I [ = 60 c Stat° r cles f \ 120 cycles when ohms L1 6.5.= VO^. - 5. . and the ciency when this motor operates on a 2.2 1.35 .6 P = 5S r„ = n H? 09 = 5.164 ELECTRIC CIRCUITS— THEORY AND APPLICATION The equivalent resistance r e is split between stator and rotor.680 r (effective) 2 2 2 o (ohmic) = 912 = —^ r' (effective) = 1.8 = 104.62 2 X P I ' Xe „ V^TTi> = Vu. 98 J 746 mi _ L674] = 236 5 ft _ lb _ .7 51. Taking into Account No-load Current: n- .842 + 3. Neglecting No-load Current: Line current 2 200 V T 01 IO .98 X 746 = 236. Zx+Z.4 = 231. . Tf+w = Pulley torque ._ .84 2 3.468 ^ + .— (1 P P No-load copper - = 2 Core 231.316) loss and windage Friction .7 .1 1 Since. Power factor = . being 43.12>(2 2 2) X 3. t.7 per cent 220Q x 21 12 More Exact Solution. for this particular motor. Power output Wi.97. 5. -.12* X 550 4ir X 60 X 746 Friction and windage torque 1.93 X 746 X 100 .875 Operating at 2 per cent slip loss X 0.-lb.468 = 375 watts .0..122 [~ 1.7.468 + 1.(Z + Z-) vectorially .= + —^— S + j(2xi + 2n V[ = _.895 watts = 9. Pf+w 4*fZ -s) 6 X 900 X 550 4ir X 60 X 0. '* = - 2 = = 8.6 per cent of the reactance.200 .02 6 X 21.I.INDUCTION MOTORS ON UNBALANCED VOLTAGES 165 1 Approximate Solution. X 100 = 849 per Cent 6 0.054.362 = = +r +r 7 2 (2ri 21.1 ft.93 60.7.900 .5.23 hp..4 ft.77 = = 2.316 1550 _ X 60 1.. . > Total internal torque Wil = X 6 4x 2 s - s] 21.61 Vi X xt) * £1 + M£ + j3 X 3.395 watts 600 900 = 6.5 volts = 1L39 amp ' 104. K1 5L7 hp X 3.1 * 7.5 . Efficiency = 100 51. it might perhaps be doubted that algebraic subtraction of the no-load reactance + . 2.77 .375 = 600 watts X Pc = 1.-lb.36 l X V\ V3(Z++Z-) 19. the resistance part of the impedance Z\ Z~ is exceptionally high.5 V3X 1 ' 2.054.(2*1 + xt) algebraically ££ Vi 2.98 X X 550 60 6 loss Copper X loss 2l\n Core 4x xm s)T Total losses X = ^-^^ = = = 5./19. 39 X 1.048.n 2. that the reverse torque is less than 2 per cent of the direct torque.1 = 2.34 per cent _ .4 .36 2.4 This discrepancy is also negligibly small.200 .200 - 7.°.4 volts i V[ = = = = = will give sufficient accuracy.842] 5.316 1 J 1.TP 2. .048.15 amp.4 1ft _ -i V8.4 negUgible. the efficiency and the power factor.041 volts 100 = n 0.98 X - 1] 12. ' .4 X 3.152 (2 + + rj) + X 3.°" 33 = 2.9933)(8.0 - 0. however.316) + loss Friction and windage Total losses drop' from the terminal voltage X 1.36V / It) 2 +.77) = 2.1158 = 2. 2 048 4 2 054 100 =0-30 Per cent onl a . 2.468 (Jr) 2 = 44 8 h r.7.lb - 200.842 _ 0.02 X 60 X 746|_ 2 3 ~ 4tt I X Solc^ - 1 ' 948 = Pulley torque " 247 206.3 +J46. = Vi ~VT = \A\/3 x.152 = 5. from the approximate solution. of little importance as far as torque and output are concerned.1158 .048. It affects.048.257 watts 600 watts 900 watts 6. The no-load power factor is COS •« 11. .048.(/e ) 2 — V147.-lb.05 hp. n (approximately) V§ /7.166 ELECTRIC CIRCUITS— THEORY AND APPLICATION It appears. which is Subtracting the impedance drop algebraically gives - ' V[ The = - 2.200 .6 6 X 550 + 3[(J 2 )V 2 = /f2rx = = 2 3[(/r) (2r! 3[12. therefore. hence. In order to the no-load 07c 8in dn .392 The discrepancy in Vi is. show that the error introduced by this procedure impedance drop is subtracted vectorially below.61 2 X 21.61 +^19. It seems.36 discrepancy in this case is 2.i0.6 ft.36 . is insignificant. Great accuracy in the negativesequence current is.200X7.98 206 ° ft .55 „ + = 19.8 = 12.2 'l (It)^] + 3.2. that it makes very little difference which method is used for the calculation of 7j. therefore. (approximately) Total internal torque r -&!>« ? -«^] X 6 X 550 f ll. • .7. 1 1 = 7i U + 1.041. .36(0.757 watts 9. Power output p = Copper Core loss 4tt X 60 + X 200. and power factor may be read. If the voltage another. Having located is 2 — s. It is assumed that the component voltages impressed are known so that the readings may be properly corrected.05 100 44.8 hp.: INDUCTION MOTORS ON UNBALANCED VOLTAGES 167 Efficiency = 44. Machinery.. usually the rated voltage of the motor.8 X 44.85 X 746 X 100 = 86.2 per cent 53. exciting current . viz. the calculated results are gathered in the following table Power factor = V3 Solution Items calculated Line current Pulley torque Power output Efficiency considered amp 231.1 ft.15 It is interesting to compare the results obtained by the two solutions. Since the negative-sequence for each system 1 See.-lb. say.-lb. R. power. torque.9 per cent 2. for instance. Lawrence. 44. The negative-sequence operating point is located where the slip This would be at point j. when excitation is neglected or taken into account. for instance. Approximate Determination of Performance from Circle Diagram.85 83.and negative-sequence systems. 84 9 per cent 97 7 per cent . 200. the operating points the current. Power factor 2. 1. It is assumed that the reader is familiar with its construction and use. . 21." by .7 hp. Figure 91 shows the well-known circle diagram 1 of the three-phase induction motor. "Principles of Alternating-current R.05 amp. neglected . and the power and torque readings by the square of this ratio. Solution exciting current 21 12 i.8 + 9. The diagram is intended to give the performance for operation on a balanced circuit and is constructed for some specific value of impressed — voltage.8 = X 100 = 53. loc. cit. The is located by means of point e in the diagram. In order to facilitate comparison. The performance of the motor on unbalanced voltages may be approximately determined from the diagram by reading off separately the power and torques due to the positive. 51. The current readings should be multiplied by the ratio of the actual is voltage to the voltage on which the diagram is based.6 ft. positive-sequence operating point the actual slip of the machine at.200 X X 12.2 per cent 86 9 per cent . the scales will no longer be correct but can readily be adjusted so as to correspond to the new value of voltage. 83. rapid. this method of predicting performance on an unbalanced circuit is very simple. the separate systems. When the circle diagram is available. 91. etc. the resultant power and torque and the efficiency. may ~eg ~oe Power Factor 'Cos 9 Slip ~fyeg Efficiency =&/e£ — Circle diagram of a polyphase induction motor. give a precision which is sufficient in many instances. It is constructed on the basis of a blocked test. It should be borne in mind. . for the origin of the diagram. Power Oufput~ef Torque Current Fig. the rotor resistance is very nearly ohmic for the positive-sequence system and should correspond very nearly to double impressed frequency for the negative-sequence system. that the circle diagram in general does not take into account the variation in rotor resistance with slip. Hence. however. in the operating region. and •S//p(t-s)--jf ( Axis ofReactive Current and Power For Operating Point e.. during which the rotor resistance corresponds to impressed frequency. excitation is negligible. power.168 ELECTRIC CIRCUITS— THEORY AND APPLICATION the current for this sequence should be taken as proportional to the line aj rather than to the line oj from Knowing current. the diagram fails to consider the fact that. they may be combined to give the actual unbalanced currents. Or one-half of the total capacitance may be lumped at each end. Jr. As the line becomes longer. In each case. F.CHAPTER VI THE SHORT TRANSMISSION LINE IN THE STEADY STATE — Introduction. New York. is equivalent to lumped impedances. in which case the nominal T-line results. high-voltage come into effect in very critical voltage. all uniformly distributed along the line. for instance. 1 disruptive the when the operating voltage exceeds configuration of the line.. the idea being to approach the actual smooth line more closely. The former is quite The latter will only insignificant and need seldom be considered. inductance. leakance. the only on The capacitance depends The effect of the capacitance on the performance of a very short extremely small." McGraw-Hill Book Company. ^ee. at the center of the line. Inc. considered they are words. A smooth line possesses four distinct electrical parameters or constants: resistance. and inductance line (say. 169 . contained resistance if they as treated are Such lines. The capacitance may also be lumped at several other points along the line. the sum of the separate lumped capacitances is taken equal to the This gives the nominal U-line. The leakance of aerial lines for negligibly small. (about 40 to 100 miles in length) are often considered as having the capacitance lumped at certain points instead of uniformly The total capacitance may be considered lumped distributed. the effect of the capacitance becomes more and more noticeable and should be accounted for. other In only. W. 1920. neither length of such lines.. below 40 miles in length). or feeder. is The power transmission is usually the capacitance very large per unit leakance depends on the power lost over the insulators and the corona loss. Peek. and A rigorous capacitance. Such lines of moderate length at least in an approximate way. therefore. and even there only lines. "Dielectric Phenomena in Highvoltage Engineering. analysis of performance should obviously take all four parameters into account. . is hardly worth while. For lengths between 2 miles and 5 miles (approximately).or n-circuits. and for the case where two-thirds of the capacitance is lumped at the center and one-sixth at each end of particularly the latter. an aerial Hence. are considerably higher than for line. etc. It may be said. equal to one-half the impedance of the line. may chapter. —The >Load — Nominal T representation of a short transmission line. the leakance and capacitance per unit length. below 2 miles in length) can capacitance be neglected. two-thirds of the capacitance be lumped at the center and one-sixth at each end. 92. efficiency. = YVr + (l + ^jlr r (1) (2) Figure 93 gives the vector diagram for this circuit. The admittance Y of the leak is equal to the total admittance of the line. . Lumped Capacitance shown in Fig. If greater accuracy is required it is both easier and better to apply the exact long-line solutions presented in the next total capacitance. however. as a rule. This admittance would. For cables longer than specified above. the voltage and current at the sending end are given by V = (l+ ^jVr + Z(l + Zpjl s J. In the following are presented analytical solutions for the nominal T. exact methods of solution should be employed. at Center (Nominal T). can be solved.ELECTRIC CIRCUITS— THEORY AND APPLICATION 170 For instance. The impedance Z/2 % S VWVAr-6~Q~CRfe- M of each arm circuit is of the T is Z/2 /p 'S ! r Fig. In a cable. 92. By means of equations (1) and (2) any problem involving power transmitted. regulation. that subdivision beyond this and maybe even beyond two lumped capacitances. the calculations of performance may be based on the nominal T. only with very short cables (say. the nominal n. the line. as in the II-line. contain capacitance only. In terms of the voltage and current at the load. + cent voltage regulation is. = Regulation rises to ZY —Vector diagram of voltages and currents in the T-circuit The per 171 line. the voltage at the receiver end Vs V' = l Fig. given in Fig. The nominal Here the architrave impedance Z impedance. while each of the end leaks F/2 at in Fig. (6) .-(i+¥)* + ZI J. equal to the total equal to one-half of the total admittance of the line which the II-circuit replaces. 95. is r 100 (4) y X*=F?rXLoacl — Nominal * representation of a short transmission Capacitance II-circuit is Lumped shown in Fig. 93. — Ends (Nominal II). 9 *—R ->VWV\A-ir"0W0> fc=r=j is (3) hence. —Vector diagram of voltages and currents in the II-circuit in Fig. 94. The sending voltage and current may be written N ZF >. Fig. = y(i The vector diagram + ?f)v + r for the II-circuit (5) r (l is + ?f)l. V z ls S Fig. 95. 92. 94.THE SHORT TRANSMISSION LINE When the load is removed. 94. removed. /.+ ELECTRIC CIRCUITS— THEORY AND APPLICATION 172 The receiver voltage at no load and the voltage regulation are given by the same expressions as for the T-circuit (Equations (3) and (4)). The trates a circuit in is z/2 Is S/J VtlrR —AVWV^75>->^ J/ -£ J. impedance tance of the architraves on each side of the midpoint admitone-half the impedance of the line (Z/2).J2 4 Yz± 3 6 Ir y> fy\Loact 6 ^L Fig. + -*( + Fig. — Improved representation of a short transmission line by lumping capacitance at center and at ends. regulation (9) (4). (8) — Vector diagram of voltages and currents in the circuit in Fig. 96.+ z( a. — Lumped Capacitance at Center and at Ends. 97. 5ZY 36 + lr (7) ) ZY Z Y* 1+ "2+ -36~ 2 2 . + ¥+ ?S-V. Figure 96 illuswhich two-thirds of the line admittance (%Y) lumped at the midpoint and one-sixth (F/6) at each end. -( . 216 . The vector diagram When Z Y 2 6 the load is is shown in Fig. . 96. the . V.—9->VWWV -61^ » . voltage-current equations for this circuit become is The V. the receiver voltage VI rises to v- = 1 — ZY Z Y + ^6 ^ 36 2 2 Substituting this value of receiver voltage in equation is obtained. 97. . 6) = (i (per wire) +^f)vR + Vsz(i + ^)/« 10-*(26 = + jZ9) 0. has the following constants per wire mile: between conductors of No. spacing wire.52 10"* 6 mhos The power received is 4.78 ohm ohm j5.J23. The sending voltage.Q35 + V3(50.635) (31.97855 +^0. d.8 V3 X The Admittance Lumped Sending Voltage: vs = 1 7Y + ^. using the b. One-half of the admittance lumped at each end.600 volts (between lines) .23.97855 (a) + J0.000 volts receiver voltage.j0. 100 miles long with 7-ft. 2.THE SHORT TRANSMISSION LINE EXAMPLE 173 1 Statement of Problem A three-phase. The sending power. by smooth- line theory. Two-thirds of the admittance lumped at the center and one-sixth at each end.886 X 10" 4 (52 +. factor (lagging) The sending current. at the Center. The efficiency of transmission.5 X 0.000 volts sending voltage. at 80 per cent power and 66. The no-load charging current at 66. + j5. 4. g. e.49 -. /.0143 0.= = '( 1 . Receiver current 7r 1.49 66 +il.78 4 \ / +j77. Solution Total impedance and admittance per wire Z = 52 j78 ohms Y = j5.600 kw.000 VS (40. following circuit arrangements: 1.886 Vs = (0.8 = 3. exact solution i.97865 \ 1+ ZY T") M L -tq^i _i_i =(52+j78)(l-K 5 5 r 50. = IB = /*' ouu (0.000 volts between lines.-a. = = = r x y 0. 60-cycle transmission line. Admittance uniformly distributed.500 0. The no-load charging current at 66.e.. 3. c. a. Calculate a.0143)66. The admittance lumped at the center.62 amp.784.5 X = = 31.500 kv. The sending power factor.6) = - + J77. h. The per cent regulation.722 .62) 70.S X 10~ 4 mhos + Total receiver power Pr = X 4. 0143)66.705)10"3 = 3. • - 3./0.5 kw..49 -J23.97855 .768 X 31.003 -j'1.722 X 31. V3V . ^^ + V3 (0.6 X 1.796.152 .96 7=r 7= Is = #.000 Volts Receiver Voltage: «_ ft _ Q YVr = J5.20 amp.0393 +^5. cent (lagging) cos 0s = 7= Is X 70. ' Each End.5 X 100 Ps = 99.7 X 100 = ftn 99.96 ^7865 One-half of the Admittance Lumped Sending Voltage: 2.Vr „ Regulation = —~= .44 . Sending Power Factor: 3.1.781.679 volts (between lines) Vs = Vs at = 01 2L4An° amp (l Sending Current: /. 768 + j 1.06 amp.003 .000 — 66 000 = ^ per c No-load Charging Current at 66./0. + ?P) VR + VSZIR = (0.20 s e. 70.62) 31.4410)10~* X = c.0393 +j5.000 + a/3(52 + j78) (31.705 = 31.7 kw. d.000 Volts Sending Voltage: h.706.0143) (31.5 X ^ 10-<(l + (i + + J5-5X ^> 10-^(52 (c) (.930 = 31.45 per cent (lagging) VsVsIs X 70.772. „ = . —V3 - V3 No-load Charging Current at 66.4410)10-* Is = (-0.66. Efficiency of Transmission OQ Per Cent = PR = 3.3.J23.— : : ELECTRIC CIRCUITS— THEORY 174 AND APPLICATION Sending Current: b.600 0^97865 = 72. Js = j5. .) +i78)^ = (-0.600 X 100 = 94 82 ' c.52 per.784.9) = 70.9 X 1.62) = 31. — . Sending Power Factor: Ps .958 T OA amp..5 X X10~« 66.930)10"3 = 3.679 X 31.-r(i + Y(l+%) = . b. Vb .5 X 10~ 4 X 66.0143) (31. 72 1/ln 14 ° V° u ltS ' 2 — .= .49 .j'23. Total Sending Power: Pa = 3(40. V3 " Ps /.772. d.97855 + .62) = a/3 (40.1.600 X 31. T 8 YVs = r(.5. = on 20.^\ = 20.7 Per cent Voltage Regulation 17' * = F« 7T?Z 1 + = «. Total Sending Power: ps = 3(40. a.152 -jl.06 + V§ .000 = 720.140 .796.781.000 V3 + (0 97g55 +y . Sending Voltage: Vs = ZF + w) Fs + + X 10.02183 + j5A672)10~ 66. Efficiency of Transmission = PR Pi « /. _ on = n952 ° No-load Charging Current at 66.3.078 X = .388)10.0393 = -0.Y (\1 . i .43 per cent + j5. ^°° + X (-0.560 .02183 + .5 Z2 F = 2 + 4 ^0 + ^) = Vs = = b.814 - -J3.245 -y245. +^YZ* = y(i = (-0.1498 +J20. a.8 =.jO.787.4 kw.814 c. X 3.0142) (31.174) 3.00000079 = (-0.49 70. 175 Two-thirds of the Admittance 20. 3.70. Per Cent 2 = Vr T .5 Per Cent Voltage Regulation: + g.781.9787 0.840 -^6. No-load Charging Current at 66.62) Sending Current: Tv Is F = (e) (-429 +j"286)10-* 1+^+^=1 + (-214.5 +il43)10-" + z(\ Ir Total Sending Power: Ps = (70.„ . + ^\ 1 ( + jO./77.600 t.0000019) i V3 (0.000 Volts Receiver Voltage I8 h.j23.j23.0142 = 0.69 + (0.(52 J78) = (102.97858 .4672)10-* .74 amp.74 = /f^^Y\ 0^7865 Lumped " 2L19 21iq amp - at the Center and One-sixth at Each End.078 = 70.97858 (52 + j78)(l - + + (2.MYZ*Y*\Vr + ( 1 + ~2+ -w + "2i6 .000 V3(51.256 +J77.¥ : : : THE SHORT TRANSMISSION LINE c.200 — -___66.62) 31.97858 +J0.000 in — 100 = 66. VR = Vr 72. 100 3.5. Regulation .49 = -^(53.256 +.816)10-8 0.4410)10-« X ^~ 20.12 amp.\ 0.69) (31.174)10-* = 3..560 X 53. 733 = _ .000 Volts Sending Voltage r(i+g)r« = S 3.00715 + j0..000 _ ' ! „. + ~W) Ir v vs + f^) = j5 5 X 10 ~ 4(1 ~ 000596 + J0-00397 + .627 volts (between lines) + llW ZY Z*Y*\ .0142)66.'00477) = 51. . W - = Is . 9./5. per ^ gr.9765 jl.771 c._ = 701 72 165 Volts ' .627 100 X = (lagging) ss e.216 = 0.2254/73^45 + + + (0.97865 VS X 412.4 = V3VsIs X a/3 70. Sending Voltage: Vs = cosh = Zo = 6 VZY _ VR + Z = V(52 7fl sinh ( ff ) X +.2155 1.787.45 (31.45 /5= 0.2173 hyp._ 95 -°o per cent Per Cent Voltage Regulation _ v' Vr ~ - Regulation Vs Hl^Ir 70.562 Sending Current: T Is 0.600 = p- q = __ .6 = " -J3.83 = ^+^) 0^787 = 2L28amP- Solution by Exact Formulas.0658 X 0. 72.0141) 66.97865 jO. A18T86 ohms Calculating the hyperbolic functions by equations (87) and (88) in Chap.10 amp.562 X 53.629 volts (between lines) -/= + cosh .16^-86.3 kw.22) 10~» = 3.000^ _ Q 3 .082.12 ^ Efficiency of Transmission n /.83 amp.6 X 3. 15 0. . VII gives = = = sinh = 7s = cosh + j'0.227 /73°.082.49 Vr sinh d = —^ X 66.0022 X 0.000 . No-load Charging Current at 66. .0658 + j'0. : = Pr — X 100 7§7 4 — 3. X 3.225 4/73°.: Sending Power Factor: Ps = cos 0s e.9765 0.0658 X 0.000 + 70.2155 0.J23. V " 1+ 36 V3(l + 20.0022 X 0.0141 ><* = -*=(53.771 - = 1 -« -»«« 31. „ n T IR (h) ValT^T^ls + (a97865 + *. a. .A16°. No-load Charging Current at 66.000 Volts Receiver Voltage: /i.0643 jO.85 = b.97875 0. F 31.784.2254/73°.0141 = 0. ELECTRIC CIRCUITS— THEORY AND APPLICATION 176 d. + jO. X + j3./20.833 = r 76 4.0832 + .22) Total Sending Power: Ps = (70. + T* 216 20.62) 70. 3.5 = = Vf V^xHH 412 - = = 10"« 0.627 = 7~ZT~Z^ 09787 .000 Volts Sending Voltage: = -0.78)j5. almost exact When the nominal T-circuit or n-circuit is used. 20 83 amp. Two-thirds of admittance lumped at Exact center. .796.784. one- sixth at each end 70.784. 13 per cent 9. the results are obtained.000 volts h. b.97875 \/3Z o cosh0 21.27 amp.: .81 sinh Vs =- Is 0.787.2254/73^.000 Volts Sending Voltage: 20. . . voltage current power power of factor. c.7 kw. .000 Volts Receiver Voltage: 9- = Is sinh 9 Vr To V3 0. THE SHORT TRANSMISSION LINE 177 Sending Power Factor: d. In the case where the capacitance is lumped at three points. Efficiency o.600 volts 70. . .7\16° .629 X 31.10 Efficiency of Transmission: = P» 3. 21.627 volts 70.3 X 100 = VS X 99.3 Per cent Voltage Regulation /. 3. Table VIII Circuit arrangement Admittance Item calculated lumped at center (nominal T) d.28 amp. Table VIII shows the results which have been calculated above using four Comparison of the figures for the approxiwith the corresponding ones for the correct circuit with distributed constants shows that the agreement is very good.10 amp. 99 47 per cent 94. 3. 20. 21.13 per cent VsVsIs " PS 3.81 V3 h. rent at 66. . One-half of admittance lumped at each end (nominal n) . 3.37 per cent 20.^ P« cent No-load Charging Current at 66.20 amp.81 amp. 99 52 per cent 31.85 /90°.27 amp.34 per cent 95.20 per cent 9 43 per cent 95 05 per cent 9. Sending Sending Sending Sending e. No-load charging cur- •/. Ration *Ljl: - = *%£"» «» .74 amp.12 amp. No-load charging current at 66. .06 amp.784.600 X 100 = 95.47 per cent (lagging) 70.000 412. 31.629 volte 31. 21 40 amp. . mate lumped-constant for circuits most engineering purposes. Per cent regulation.3 = 20.781 5 kw.000 volts .4 kw. amp. 99 47 per cent 3. 99 45 per cent 31. 20.047 X 66. discrepancies are not large and the accuracy would undoubtedly be sufficient different circuit arrangements.19 amp. 21. cos ds Ps = 3.679 volts 70. trans- g. .3 kw. No-load Charging Current at 66.96 amp.45 V3 X 36.82 per cent 9 30 per cent 95. Kennelly. E. A. "The Application trical-engineering Problems. which is by far the predominant part of the dielectric admittance. electrically long. resistance. A.CHAPTER VII THE LONG TRANSMISSION LINE IN THE STEADY STATE A rigorous analysis of transmission-line performance should. 1918. recognize and take into account all four parameters of the capacitance. the formulas obtained are then directly applicable to direct currents by substituting zero for the frequency (or angular velocity) in the terms where it appears. more correctly. 1 The following books treat the theory of long transmission lines in the steady state: Kennelly. "Electrical Phenomena in Parallel Conductors. is directly proportional to the frequency. and short." McGraw-Hill Book Com- York. Evidently the directcurrent steady-state performance depends upon resistance and leakance alone. while the physical length and the they may electrical length are identical for direct-current lines... the effect of leakance and may be neglected. pany. understood when it is remembered that the susceptance. viz. it is sufficient to work out the solutions for alternating currents.. Hence. differ greatly for alternating-current lines. A single set of formulas is thus capable of covering the alternating-current as well as the direct-current case. inductance and capacitance have no effect. the electrical length of an alternating-current line depends on the frequency as well as on its actual physical length. as previously pointed out. the line these two parameters should be considered. Hence. 1 should be noted that. 178 . When the symbolic notation is employed a direct-current voltage (or current) may be considered equivalent to an alternatingcurrent voltage (or current) of zero frequency. 1912 and 1925. It line. alternating-current line may Indeed." John Wiley & Sons... New York. New E. E." of Hyperbolic Functions to Elec- McGraw-Hill Book Company. capacitance If is inductance. "Artificial Electric Lines.. leakance. if the line is long or. New York. F. Pebnot. 1917. Inc. Inc. a very short This is readily be electrically-long. Inc.. 98. 98. and C represent the equivalent three-phase constants. be written ^=°(r+j«L)J. however. steady-state case. will line. The four differential equations just given were established for two parallel wires. current. be obtained for a three-phase line if the constants r.: THE LONG TRANSMISSION LINE 179 — General Solution. Equations (1) and (2) may.e. lience. respectively. They are written in terms of instantaneous voltage and current and are applicable to the transient. the resistance per wire mile and the inductance. —Section of a two-wire transmission line. (3) ^= (4) (g + jcoC)Ex = yEx In these equations. in other words. having the following constants per loop mile shown = r ohms = L henries = g mhos = C farads Resistance Inductance Leakance Capacitance At a point x miles from the receiver end R. as a matter of fact. Consider two parallel wires S — R as in Fig. the formulas which are derived below are applicable to either case (and. both voltage and current undergo sinusoidal variations with time. g. The voltage and current are then to neutral and per Hence. however. L. the following differmay be established by considering the change in voltage and current in a differential element dx ential equations = ri Tx + L dt (1) These equations are perfectly general. as well as to the steady state.. &* ) k '+£**{* In the alternating- * 1 R !k dx Fig. for a single-phase Exactly the same equations. also to other wire. E and I may represent either maximum or root-mean-square values of voltage and current. and capacitance per wire mile to neutral.«*7. leakance. . i. (3) and +Ba -f- D (12) a cosh ax (13) (4) in equations (12) and (13) when x = Ba = zl r Da = yE 1 cosh ax (14) r See any standard treatise on differential equations.180 ELECTRIC CIRCUITS— THEORY AND APPLICATION polyphase lines) as long as the proper values are assigned to the constants. 1 tions and may The latter is preferable in line calcula- be written Ex = A Ix The constants = C cosh ax cosh ax + B sinh ax + D sinh ax of integration are evaluated (10) (11) by making use of the terminal conditions. By is and differentiation of equations (3) with respect to x (4) obtained d 2E x ~dtf d 2I x W x =Z dl ^x . (5) dE x . E x = E r and I x = I r and C are immediately obtained as A = E r and C = I r = In order to determine B and D. (15) .-. Their solution may be obtained either in terms of exponential or hyperbolic functions.„>. (7) zyI x =a*I x (8) where = a -\/zy = V (r + juL)(g + jwC) ( 9) The quantity a is termed the attentuation constant of the line. When x Hence A 0. =V ~dx (6) Substituting equation (4) in equation (5) and equation equation (6) gives ^ = zyE x = ^= d (3) in a*E. equations (10) and (11) are differentiated with respect to x -r^ = Aa sinh ax -r-^ = Ca sinh ax ax ax Substituting equations gives. Equations (7) and (8) are recognized as simple linear differential equations of the second order with constant coefficients. steady-state conditions and are perfectly general hold whether the grounded at the receiver end. Although these equations as they stand are convenient for many purposes. These shorter forms. The receiver-end current is then the receiver-end voltage E r divided tions. are — . (16) ~\v~\\Ig + ju)C Zo has the dimensions of an impedance and is called the surge impedance of the line. Line Loaded. which are still more suitable for computawill line is loaded. — Single-phase representation of a long transmission line loaded impedance end and sending end respectively. Inserting the constants in equations (10) and (11) gives E x = E r cosh ax + ZoI r sinh ax (17) Ix = I r cosh ax + sinh ax yf Zo (18) If the conditions at the sending end had been used in determining the constants of integration. with an a. 99. They for sinusoidal. the following equations would have been obtained: Ex = E Ix = s cosh a(l 18 cosh a(l — x) — Z Is sinh a(l — x) — x) — -£ sinh a — x) (I (19) (20) Zo Equations (17) to (20) give the voltage and current distribution along the line with reference to voltage and current at the receiver *-— P^ TT- fs Fig. they may be reduced to more compact forms. 99. or somewhat different for the various conditions that may obtain at the receiver terminals.THE LONG TRANSMISSION LINE 181 from which Jj — —I — a r Zq1 t Z a where -±££ r . free. Let the load at the receiver end be represented by its equivalent impedance a as indicated in Fig. . F sinh (al + _ Er ~ sinh0' _ ~ T lr cosh (al cosh 00 + 0') 0' _ w sinh 8.] I r [cosh aa. cosh 8 X _ cosh (ax + 0' W . (24) + tanh 0' sinh aa. and by making the proper substitution in equations (17) and (18) these may be written Ex = Er cosh ax = I r cosh ax Ix Introducing a quantity load defined sinh ax -\ sinh ax + -=Zo (21) I (22) I the so-called hyperbolic angle of the 0'. . by 6' and in equations (21) Ex = E = E r = = = _ ~ = tanh" (23) (22) gives ax + coth 0' sinh ax] cosh ax sinh 0' + sinh ax cosh [cosh r ^ ~ 1 sinh sinh (ax - + Sffi"? sinh 0') ~ 0' 0' „ ^ slnhT' 8X . ~ ^imhT' cosh 8 _ ir T " cosh 0' a f (26) . cosh ax cosh 0' + sinh ax sinh /r cosh 0' 0') _ . ^n By substituting the values of receiving-end voltage and current from equations (26) and (27) in equations (24) and (25). Hence. the voltage and current at any point are obtained in terms of sending voltage and current. . being equal to the sum of the hyperbolic angle 0' of the load and the hyperbolic angle ax of the portion of the line between the Ir point and the load. The relation between the voltages and currents at the sending and receiving ends is evidently obtained from equations (24) and (25) by placing x — Ea T Is I. _* 9 ~ ^cosh^7 cosh"? The 5's are termed position angles and represent the total hyperbolic angle at the point considered (distance x from receiver end).182 ELECTRIC CIRCUITS— THEORY AND APPLICATION by this impedance. t smn ~ Pr>x = EJ ErI sinh X r _ ~ As of either equations (24) sinh 28 ^7 'sTn¥W COSh 8X 0' cosh 0' = EJ x power at a point &x sinh 28 x (30) °-stih28 8 proportional to the hyper- is The active power P and power Q at any point are most readily computed by multiplying the product of the numerical values of voltage and current at the point by the cosine and sine of the angle between them. It is evident that the following general rule will hold: The ratio of the voltages at two 'points of a line is equal to the ratio of the hyperbolic sines of the position angles at the two points and the ratio of currents is equal to the ratio of the hyperbolic cosines of the same position The angles. sinh _ T lx = T is C0S k cosh 88 ^ — " 8. the sending-end impedance becomes Z = Z tanh s line 5S (32) —A grounded line is equivalent to a loaded where the impedance of the load is zero. These quantities may also be obtained by multiplying the vector current I x by the conjugate of the vector voltage E x The impedance at any point offered by the line (including the load) is given by bolic sine of twice the position angle. similarly. of the load. the voltage and current at any point maybe written in terms of the voltage and current at any arbitrary point. = Z tanh 5* (31) 8X cosh 0' and. The position angle Line Grounded. variation in vector power (volt-amperes) along the line by making use readily obtained or equations (28) T7- and A Vector seen.1 Hence. therefore. T is COSn cosh A* v ' W /OA\ where 0* = ax and = al (35) . 0' is = % tanh. from equations (28) and sinh — ~ (29).THE LONG TRANSMISSION LINE 183 In a similar manner. reactive . the vector and is (25) (29) rir = t. mhfc _ E sinhS. sinh 8X Zx = = ^ ^ ^= cosh I sin x Ir a coth 0' tanh 5. -..184 ELECTRIC CIRCUITS— THEORY AND APPLICATION The sending-end impedance The is receiver voltage is grounded for the is Zs = Z line tanh (36) The evidently zero. A line open at the distant end is equivalent to a loaded line where the impedance of the load is infinite. be quite in operation than unity. . For lines of the lengths and frequencies used for commercial power transmission today. at no load the receiver This difference may. sinhffl+^J cosh Z = Z U = E~ ™bh cosh . for the longest lines appreciable.„. ('•+4) \ zl — ~ U + j£\ E ~ From (39) a . ' this equation .e.A for the zero. larger than the sending voltage.. 8 ~ "sinh 3 cosh 8X . cosh is voltage always is less Hence. cosh The sending-end impedance s The by tanh receiver-end current is is may /. receiver current given by lr K } Zosinhfl The current at any point in terms of the receiver current cosh becomes Ix = I r cosh (38) X — Line Free. the voltage and current at some point on the line are obtained in terms of the voltage and current at the sending end sinhf Em = E »*bJ± = E ~".tt\ . The position angle of such a load is 0' = tanh -1 co = j- Using equations (28) and (29).. . _ T 'wiTs " U _ + j|] X \ open = Z The t ia smn ^ (A(\\ {W) sinh line coth (41) receiver voltage is given Ai be computed the Ferranti (42) effect of the the numerical ratio of receiving to sending voltage when the line is free. (about 300 miles). i. the numerical value of line. . /a . to the impedance of the load of the line (both in The Infinite equal to the surge is magnitude and phase) the position angle of the load becomes infinite. — When impedance r cosh (43) 0* Surge Impedance. in the general case. ^ . attenuation 1 when the load impedance and Such a line is said to and the exponential t~ du is have normal normal called the Evidently. + + + + .1 ~ = tanh" = 1 oo The voltage and current distribution may be obtained from equations (28) and (29).e ° 7 a e-«» (48) (49) here the hyperbolic angle of the line from the sending end to The equations show that the voltage and the point in question. 0' = tanh.e-<«-'»> = E = a e-<*«-* Z. current distribution are exponential surge impedance are equal. . Kennelly's "Artificial Electric Lines. attenuation factor. dt." loc. l . . { ' use of the following relation. 1 See Dr. however: 0' 0' 0' cosh sinh -cosh Limit (sinh cosh sinh B x cosh — may = 0') Limit (cosh 0' 00 —» 0') 00 be simplified by canceling out sinh Hence.-= IX 0„ is = K = l£ = #*-<•-«•> I.„.THE LONG TRANSMISSION LINE The voltage any point at 185 terms of the receiver voltage in becomes Ex = E Load Impedance Equal Line. s (46) . sinh X cosh 0* Ex = Es cosh sinh sinh cosh 6X 0s j Tx — ~ * cosh sinh 0' and numerator and denominator. which may be written * j x sinh (6X _ ~ w sinh (0 * _ ~ j cosh (6X cosh Making (0 + 6') + 0') ~ + 0') _ + 0') ~ * these equations in 0' 8 6X 'cosh cosh 0' cosh 0' + cosh 6X sinh 0' + cosh 6 sinh + sinh d sinh 0' + sinh sinh 0' 0' x . But + sinh cosh = € e _ e -e e e _|_ e -9 |- s 2 — *9 2 ' Inserting this in equations (46) and (47) gives Ex = Eu = E. . the attenuation factor is given by the ratio of two hyperbolic functions. ..e-«0-*> = =E s e. " and by virtue of this fact behave as such a line. . be On the latter the line angle at any point 8X must obviously independent of the position of the point. consequently. Considering infinite. * . the load-position angle 0' will be finite and may be ignored in equations (44) and (45) as compared to 0j.'-mr . It is interesting to note that the line loaded by an impedance equal to the surge impedance behaves.*£-^ cosh + e~ 7 WT0 = 7V+7^ 6X . however (including zero and infinity corresponding to the line being grounded and open respectively). Zx = Z tanh (0 X + oo ) =Z tanh oo = Z (54) Hence. for the infinite line."\d^ Ia Here the (51) a (52) s magnitude while the angle 0„ 2 directly indicates the swing in phase of the voltage and current with respect to the corresponding quantities at real exponential e~ 0ul gives the attenuation in the sending end. Hence.e-f-M = E es I~ = I s e-«>-^ = I s e- » » (57) (58) . then the position angle of this load would always be finite except when the load impedance is equal to the surge impedance. so. For any other load impedance.186 ELECTRIC CIRCUITS— THEORY AND APPLICATION As a rule 6U will be a complex quantity which = eu eul may be written + jdu2 (50) The exponential in equations (48) and (49) may therefore be broken up into two parts and the equations written Eu = Ese-^e-i "* = E e-^\d^ = 7. the impedance offered at any arbitrary point on the line is always equal to the surge impedance. electrically. The sending-end impedance becomes Za = Z tanh (0 + oo ) = Z tanh = Z oo (53) Also. Eu = E~ = E. the hypothetical possibility of a load on the infinite line. the infinite line ance equal to the surge impedance. as an infinite line.and 0.6-^e-^ = I e. Ix ** T e s5 > < e* * . In the latter would itself be a "line loaded by an imped- case. (56) Here the exponentials with negative exponents are negligible compared to the ones having positive exponents. The space in this treatise. which are all beyond the length used in power transmission today. etc. For instance. Z = ZQ s tanh + (fi t 6') = (« Z„ tanh -f $') = Z tanh Z — Wave qo = (59) Length.000167 + jO. have their peculiar and interesting characteristics. however. are not realizable in practice for the fundamental frequency. and of length X/4 a quarter-wavelength line. (62) (63) . is equal to the surge impedance. Att2 = 6U 2 X = ^ = 2t (60) A line of length X is a one-wavelength line.002075 The length necessary to make this line a quarter-wave line is consequently = 757 mileS (61) 2 X 0. including the line. they may easily obtain for some of the harmonics.. These lines.. i. however. identical. It should be noted. Evidently the impedance at any point of the infinite home end.00 2075 A Few Remarks on the Hyperbolic Functions.THE LONG TRANSMISSION LINE 187 These equations for the infinite line and equations (48) and (49) by an impedance equal to the surge impedance for the line loaded are. sinh x = cosh x = *-—ir~'* € ~t * viz. of length X/2 a halfwavelength line. The conception of wave length is based on the line having normal attenuation. The wave length X is then the length in which the phase swing (or phase attenuation. transmission line has an attenuation constant at 60 cycles per second of a = ai + ja 2 = 0.. as seen.e. that. or 2t radians. The hyperbolic functions are most conven** = = i 2^2 = — iently defined by their exponential equivalents. as it is sometimes termed) amounts to 360 deg. on the infinite line or the line whose load impedance equals the surge impedance. Some Conversion Formulas. although quarterwavelength effects. does not permit a detailed discussion of these features. An idea of the length necessary to produce quarter-wave effects may be had from the following example: A certain 220-kv. half-wavelength effects. series the trigonometric of functions are /j»3 /v»5 re = x —^+^- cosz = 1 - sin |j +-|j - • • • • • • (68) (69) from which. (65) the series representing the hyperbolic functions may easily be formed. — sm x = = cos x - (72) — e ix _|_ € j> 2 (73) It may be of interest to compare these important equations with equations (62) and (63). . sinh 3 = 3 + 31 + 5-1+ cosh x = 1 + 2 |j ' ' * • • ( 66 ) 4 + The corresponding well-known + • (67) I-.= l_ a |j + + |j • - • (64) + |?_|!+.: 188 : ELECTRIC CIRCUITS— THEORY AND APPLICATION By making use of the power series of the exponential functions = €- 1 +x+ ^. . (71) may be derived the relation between the trigonometric and the exponential functions.. namely. From equations (72) and (73).. . in conjunction with the following series of the exponentials with imaginary exponents ^=l+jx-^-jf + • • • (70) l *--= 1-jx-^+j^ + . the following useful relation From equations may be obtained ±jx __ cos ± j sin x (62) and (63) may be derived cosh 2 x — sinh x = 1 € x 2 which may (74) (75) be compared with the well-known sin 2 x + cos 2 x = 1 (76) . + jV) [a. + jy) = ±j sinh (x ± jy) ± j(y ± ^J = (87) (92) x sinh cosh x sinh 2/ y (91) a. A few of these are listed below. ± y) tanh (77) — (95) (96) (97) (98) (99) (100) 1 (101) a. (89) a: = ±j (83) (85) (a: sinh x (81) = [a. sinh + (78) cosh x tanh jy fa. ±j cosh tanh! 2 sm sin (90) H)H j # a. (a..THE LONG TRANSMISSION LINE 189 By making use of equations (62) and (63) in conjunction with equation (74) a series of very important and most useful formulas may be obtained. tanh x Vl ~ (102) tanh 2 ... ± jy) ± j{y ± 2tt)] = cosh (x ± jy) ± j(y ± ir) = —cosh (a. x ± ± ± j(y ± 2w)] = sinh (x ± jy) ± j(y ± x)] = -sinh + jy) ± [x ?/ tanh x ± tanh i/ ^ ± tanh x tanh y = +J (*±4) a: cosh x cosh y cosh x cosh sinh x cosh = cosh x = — cosh [a. cosh j|j — = cosh (x ± j2ir) cosh (a. = = = sinh 2x cosh 2x sinh (x cosh (x 2 sinh x cosh x sinh 2 x + cosh 2 x 2 1 = 2 cosh x 2 sinh 2 x + ± ± — 1 (79) 2 cosh 2 k = cosh + 1 (80) y) y) (x ± jV) ± jy) sinh (x ± j2r) sinh (x ± jt) sinh (z cosh (x sinh sinh cosh cosh x sinh y sinh x sinh # (82) = = = j sin y cos y (84) j tan y (86) = smn # cos = cosh x cos = sinh = — sinh ± j cosh ± j sinh 2/ y x cosh (88) (93) (94) a. 1 = sinh jy cosh jy sinh - 2 sinh 2 ~ . + jy) (z | ± — iv) — (x±jy) = e - H[e z (cos y ± j e ^ e x _ e sin y) -x . cos y g = For instance. All of these relations equation (87) sinh (104) 2 *3 ± j cosh y x sin y In determining the position angle of the load.190 ELECTRIC CIRCUITS— THEORY AND APPLICATION cosh x = (103) . follows. First. can easily be proved.e . the angle whose hyperbolic tangent may be real or complex) may be written tanh" 1 u This relation is = hog ° 2 \^ — u = tanh -1 u then x u = tanh x ^ — = e-rq e~ x = e c 2x — 1 ^T1 from which ix — ~ 1 + 1 — tanh x tanh x _ 1 + 1 — u u Taking logarithms gives 2x = log °l-« -z from which equation (106) immediately is (105) u (where u (106) 1 readily proved as follows: If x *. Vl — tanh 2 x x 2 tanh ~ sinh x — = 1 + tanh (a. The logarithm of a complex quantity is given by venient to log (a ±jb) = Va + ft***" "log Va + b ± jtan-^ = 8 2 log 2 1 2 In general. it is necessary to derive a few auxiliary relations which then will be used in obtaining the desired formula. x sinh x cos y - e-*(cos y +2 e~x Sm + j sin y)\ . . however. it is often conmake use of a formula which will be developed below. Making use of equation (106) gives tanh-i (a ± jb) = = V log { * ° 2 3^[log (1 The complex logarithms in by equation (105). however. evaluation of the hyperbolic functions in this manner is laborious. the hyperbolic function can immediProbably the most complete and ately be selected from tables. In general.1 tan -1 be modified. Of course. Numerical Values of Hyperbolic Functions. the one should be chosen which. . If the argument is real. equally well whether the hyperbolic angle is real or complex.THE LONG TRANSMISSION LINE 191 Now. (m (1 ° 7) 2 formula t + V(l tan -1 2 2 1 1 log 1. most accurate table is contained in the Smithsonian Mathemati- . — It is evident that the numerical values of hyperbolic functions may always be determined by resorting to the equivalent exponential formulas These methods will work or by using the hyperbolic series. tanh -1 1 log 2 (a + jb) vTl + a) 2 + = a) 2 l0g 2 Vd ~ a) 2 (l jb) - - a + jb)] can be taken care of log (1 = + j j% ± a this expression b2 ± jtan- The second term +b +b in this ± a a) 2 — 1+a -z i — +T 2 ^ tan ± jtan- 1- ' 1 1 -a ~. giving r tan -1 . Hence. this ratio is complex and of the form a + jb. making it necessary to evaluate tanh -1 (a ± jb). + * may . will give the arc-tangents in the numerator of the second term with the greatest precision. obtained. —1 ^ a — b 2 (108) In applying one of these formulas (equations (107) or (108)). are available. for the particular numerical values of a and 6. —+h a . the position angle of a load impedance is the angle whose hyperbolic tangent is the ratio of the load impedance to the surge impedance (see equation (23)). although any desired degree of accuracy can be Several other methods which are more practical. D. — The curves show the hyperbolic sine. The value always positive. . Reprinted 1920. 1909. Washington. Figure 100 shows graphically the values of the hyperbolic sine. It should be noted that the value of the cosine approaches the value of the sine for large positive arguments. for instance in Pender's "Handbook for Electrical Engineers" and in Hudson's "The Engineer's Manual. The value of the tangent. finally reaching the value of infinity. goes through zero and reaches a maximum value of plus and minus unity corresponding to infinite arguments. 1 nificant The values in this table are carried out to five sigThe Smithsonian Mathematical Tables also places. 100. Shorter tables will be found in most handbooks. From this table. contain a very complete table of the exponential functions carried out to seven places. when the arguments become infinite. C. cosine. which always is given by the ratio of the sine to the cosine. It is unity when the argument is zero and increases in a positive direction for increasing positive as well as negative arguments. It will be noted that the value of the sine passes through zero and extends to plus and cosine.192 ELECTRIC CIRCUITS— THEORY AND APPLICATION cal Tables. the hyperbolic functions may obviously be calculated if this degree of precision is desired." in which the hyperbolic functions are given to four significant places. minus infinity of the cosine is 1 Published by the Smithsonian Institution. Fig. and tangent of real arguments. and tangent of real arguments. Mass. on the other hand. usually necessary to interpolate in requires quite a little as ordinarily used care and time. Kennelly. Kennelly's charts are constructed from the values in his tables and give the same information in an extremely convenient form for engineering use." 1921. . E." Harvard University Press. the hyperbolic angle corresponding to a given function is just as readily determined. polar as well as in rectangular form. lead to error. There are tables with the values in complex hyperbolic sines. are added to or subtracted from the imaginary part of the argument. may. In converting from radians to quadrants. therefore.. E.. One quadrant is equivalent to x/2 radians. In using the tables it is two dimensions. It should be noted that Dr. the imaginary part is multiplied by 2/x. Cambridge. It has been previously demonstrated that the hyperbolic functions repeat themselves or change according lation based when t/2. 1914. Kennelly. be obtained to three significant figures and are. Conversely. or 2x. The values. A. It is to certain rules. the imaginary part is given in terms of radians. cosines..or II-circuit. however. Kennelly's tables and charts. it is necessary to 1 Kennelly. introduces the use of quadrants instead of radians for the imaginary part of the hyperbolic angles. Cambridge. in most of his tables and charts. To do this is much easier when the imaginary part is kept in terms of quadrants. from the charts. a process which Straight-line interpolation Dr. since the changes occur at integral multiples of one quadrant. accurate enough for most engineering purposes. Reprinted tions. t. can. respectively. "Tables of Complex Hyperbolic and Circular FuncHarvard University Press. outlines methods for more accurate interpoin certain cases. while in converting from quadrants to radians the imaginary part is multiplied by ir/2. Mass. A. The value of any hyperbolic function can be read immediately by spotting the argument in the proper chart. If. "Chart Atlas of Complex Hyperbolic and Circular Functions. hence. important to keep this fact in mind and make use of it in obtaining numerical values. Reprinted 1921 and 1924. as a rule..THE LONG TRANSMISSION LINE 193 Complex hyperbolic functions may be determined by Dr. Dr. and also the correction factors necessary in converting a smooth line to its equivalent T. Kennelly. 1914. on Taylor's theorem which may be used when great accuracy is required. 1 His tables give the values of the and tangents. 6 /89°.2\0°. can be obtained to a large EXAMPLE of significant figures. 10 3 V0.55 hyp 0.92 = = . = = = = 10. « = 2ir60 X 29 z = r +juL = y = g +j*>C . If a very high degree of precision is required. Kennelly's charts. The magnitude of the tooth-harmonic voltages is 0. current.. and when the distant end of the line of tooth-harmonic frequency is free. which.6 /89°l)2 0.ELECTRIC CIRCUITS— THEORY AND APPLICATION 194 keep track of integral multiples of t.3 V3./sec. and power distribution on an alternating-current line.8 0.18 = 23.2 23.0142) X 155. 60-cycle transmission line is fed from alternators whose voltage waves contain tooth harmonics of twenty-nine times the fundamental frequency.0142 X 10-« farad} J plot the distribution of voltage. vector power.92 mho/mile 10-«/89°.060 = a = 0. Solution The values of the hyperbolic functions used below are all obtained from Dr.00212 henry 0.12 ohms .0382 hyp /mile + J9.68 155. considerably less convenient. 220-kv.18/89^68 10.00212 0. current.47 (1.930 X +jl5. when an idea It also serves to give electrically of the peculiarities of such a line very long.00021 23. The latter (equaand (88)) involve hyperbolic and trigonometric funcreal arguments only.1416 radians.00021 = X 155.1492\0°.2 X lQ.0525 X 0.930 rad.710. The constants of the line are = = g = L = C = Length r Compute and active power 250 miles 0.12 386.2 4 X 10~ 8 X lQ. Statement of Problem A three-phase.18/89^68 ohms/mile (1. which. or 3.060/89^8 hyp/mile + J0.597 /89°. 1 This example shows the application of the hyperbolic theory to the computation of voltage. 0.47 a = y/^j = = = = + .47 ohm X 10 -8 mho ] per wire mile 1 0. 18/89°.127 1. the complex hyperbolic functions may be calculated from the series (equations quite obviously.5 per cent of the fundamental.127 +. and (66) is by equations (87) and (88).710.127 +J23. when the proper tables are (67)) or tions (87) tions of number used.520)10-8 V 250 Ji X = 102 = + jO.930 X 0. t^ 1^ i-H "* hns 00 ^ 8(NtOi-HOl-l©00O©t^.(NTH«D00tOr-lr-HO5T-l (OOffiOHHiOO^iMHONM^OO^e ©©©©©C^©©©C0O©©Tt<©00©©©t^. to to to <* CO CO i— 001 (N i— T-H i-H 00 I© t>- r>- © © © ^ 00 i— ^^ CO TjliHT(l«OTf(iO»0>^OJNHl01>MM5>0(N COtO Q N«Ni-iO)MMNtOWacOOMO H ** 00 CO i^ 00 CO 00 "5 i—i ©© O © 00 © o 00 00 o o .CO t^ 1^ t^ CO t* 00 CO NT-H CN |(M <N CO rH TtH •* to CO |© CO r^ a> 00 00 00 00 r^ CO CO CO to <tf IM .l> 1> CO h- 3 * ^^ ^ <NC<l©0(N(N©©(N<N©©i-i(Ni-i©^HlNr-< Ph © 8 *» t^ © ^\^ OS 00 00 > T-l 1-H i-H r^. CO N.r^ t>.o '© 1© 1^- to ^i CO © *>• <N o -* 00 © 00 © © © © o O © o© o n© o© n© 1^ to © © to CO * to to Oi CO CO CO i-H o Tt< 00 n © © © to © © t^ CO 00 CO M* CO i-H <N (N «o t>. t» o l> n CO r^ 00 r^ CO h.OS ©OHH00O00 T* rt i-l * Tt< Ttt tO|rJ4 <N * © "* •* * CO * ©* ©* o O o O o o o o o o b.5 a ji a s HJU5HOMOW001U5U30 ONM115W3NCCOONM10 ^HrHrHrHT-I^Hr-HININfNNCN .+ + + + + + + + + + + + + + + + + + to © CNtotOOOOtoOOCOtotOOOO tO © © CO IO tO tOtO©i-HtO©r. 1^ W K It-1 r-f i—i r-i •*• to to 1 g" 2 fc 3 ^ ©© © i-H to © O© tH O co to ©©© 1© to © © © © © 00 © © o <* o © © © o© o© to © 00 © 1^ t^ © 2 S © 00 g © <N 51 <N CO <* * * o i-H O to ©' ©' T-i O ©' ©' © © o 00 o <N o to © © CO o on o CO * 00 (N CO CO <N o •* to © © © t^ 00 00 00 tH CO tO CO OS <N 00 00 * (N CO _ CO ^ O l> © © ©© to © i-Hi-HO©©l-H©0©rH©©Oi-H©©©i-HO IO <*< 5 S Tt< 00 o o CO to i-H 1-H i-H •a i-H S o < > 'o o So © (—> a" CO 00 r^ t^ CO N. b» t>.00 1^ CO (N CO CO CO -n to to >o r. <N CO -* CO CO CO "3 ICO CO CO 00 CO »o Jco o GO < CO tO »-H •4 3 CO 00 O© {5 « S W £j < 2 o X o 10 i-H H O O z o o M h5 H J So hhNNMWt(I^i5iO(D(0 '«^ •»» + H.N » N H M 80rtHHHNNNNM»M CO00 <N <* t^ OS <Nto O ©I© ^> -< •«! &^ o a < "«"» *'-» «*» "«^ •<"» *<*» 't-> •»» •<-» •»>» T-| ffl •<*< ©©©©©©OOOOO© OOOOO © ©©©©©©©oooooooooooo 1 '"tl X tJ< tJ< bO » rt Q ^3 o h .<N IN 00 00 o CO l>- fc5 ft i-H t>.THE LONG TRANSMISSION LINE 195 o <1 o z Oh B © cy B O S eg O « S O O OS o o 00 CO CO © IN H HWO * o o o ^ [CO o to CO to tO l> "* CO NH i—t 00 CO to 00 CO i-i t- to ©MN O H <* © © © © b. 761/897°. —"Voltage and current distribution on an electrically long transmission whose distant end is open. The voltage and current distribution are given &X = t lx Vector power where <j> is referred to ~ at by cosh Ox -"8 1 ^ cosh e sinh 9X t la sinh 9 Px = E x /0 X The voltage Example 1).405\83°.4 386.5/83°. Active power P> „ = EX I Z cos <j> watts (d) . see line. the sending end is used as standard phase (for calculations._ Vs = 635 volts 452. . x (a) (6 (c) Ex /0. plotted in polar coordinates.2\0°. I x /4> volt-amp.650/813°.l 0.8 (of ohms tooth-harmonic frequency) 635/0 — s Fig.5 = Z„ coth — = '- '- s m E ' . 7= Vs = = 1. 101.8 amp.196 Za = ELECTRIC CIRCUITS— THEORY AND APPLICATION E = _0. the distributions which are characteristic of a line which is electrically very long and whose distant end is free. 102.8 miles 0. whose distant end is open..060 = 2. It will . see Example 1). The wave length (as previously defined) corresponding to tooth-harmonic frequency on this line would be 2t = ~ Hencej the line 250 is actually jf^j. This diagram shows strikingly are given in Table IX. 27 = 104. in this case. as seen. a certain number of points so selected that the hyperbolic angle subtended at these points have imaginary parts representing an integral number of quadrants. It is advisable in calculations of this sort to make use of these "quadrantal points" in order to benefit by the greater ease with which the hyperbolic functions may be determined at these points. however. Voltage and current distribution along an electrically long transmission line. will crossing occur of lines belonging to the same spiral.e. 197 and power distribution be noted that points spaced regularly along the line 25 miles apart have been used and. The spirals which. . 101 in polar coordinates. plotted versus distance measured along the line (for calculations. in addition. broaden out when load is applied at the receiver end of the line. The voltage and current distribution have been plotted in Fig. This diagram gives a good picture of the very great variation in voltage and current which It may be noted occurs from point to point on an electrically long line.385 wave lengths long. that both voltage and current go through five maximum and five minimum This will be the case whenever the line subtends a hyperbolic angle values. 175 100 125 150 Miles from Receiving End — Fig. i. are very narrow. versus angular displacement with reference to the sending voltage (of tooth-harmonic frequency). whose imaginary part lies between nine and ten quadrants.THE LONG TRANSMISSION LINE The numerical calculations of the voltage. current. In no case. Figure 102 shows the voltage and current distribution plotted in rectangular coordinate versus distance measured along the line. 198 " ELECTRIC CIRCUITS— THEORY AND APPLICATION power and active power are plotted versus The vector power shows a smooth variation from its maximum value at the sending end to zero at the receiving end. The It should be active power on the other hand is of a "pulsating" nature. noted, however, that the active power continually decreases along the line. Obviously this must always be the case. The distributions of vector distance in Fig. 103. 250 225 200 175 150 50 15 100 125 25 Miles -from Receiving End — Distribution of vector power and active power along an electrically Fid. 103. long transmission line, whose distant end is open, plotted versus distance measured along the line (for calculations, see Example 1). EXAMPLE 2 Statement of Problem Example 1 is delivering 60,000 kw. at 80 per cent Determine The receiver voltage is strictly 220 kv power factor (lagging) the position angle of this load by application of equation (108). The line described in . . If <f> Solution > designates the power-factor angle, the load impedance 2202 X 220, 000/ ^3 0.8 V3 60,000 X 220 X 0.8 /cos- 1 0.8 = is given /cos" 1 0.8 645.3 /36°.86 60 The linear impedance and admittance of the line at ohms 60 cycles per second become z y = = r g = = +jo,C = +jo>L = The 0.127 +JS77 X 0.127 + jO.799 = 0.00212 0.809/80^96 ohm 0.0142 X 10*) 10" 8 10-s = 5.35 X lQ- g /89°.85 (1.47 +./535) (1.47 +JS77 X 60-cycle surge impedance of the line is, mhos hence, 0.809/80°.96 -a/* "Y 5.35 X by 10-V89^85 = 388.5\4°.45 ohms THE LONG TRANSMISSION LINE a |- = Zo = 1.660/41°.31 = + j'1.096 1.246 388.5\4°.45 The angle 0' 645.3/36°.86 199 of the load = tanh" 1 ^ given 0' is - by tan" 1 (1.246 246* .246* El — + + , 1-096* 1.096* In 2.224 = log ooo, jr- tan" 1-/ - .ir +j , x tan _, 1 2.050 - 2.246 _, -0.246 „„„ — xtan 1 , nn „ 1.096 1.096 2 * , +3 tan" 1 ( -0.2243) o 262 = + J1.096) ~+ 1.1175 0.2204 = _ .__ 1100 0.400 +jl.l22 . . , hyp Above, the angle corresponding to a negative tangent was considered negaand less than x/2 radians. This should always be done if the angle corresponding to the positive tangent is taken in the first quadrant, which presumably usually will be the case. It might be noted, however, that the two angles may be taken in the third and second quadrants, respectively, both being considered positive. The results would then be tive = 6' 0.400 - j'2.019 hyp imaginary parts of the two solutions are displaced by t Hence, the second result is obviously also correct, since the values radians. of the hyperbolic tangents repeat themselves when a multiple of ir is added to or subtracted from the imaginary part of the hyperbolic angle. It is immaterial which one of the load position angles above is used in computing the voltage, current, and power distribution on the line. Evidently, the imaginary part of the position angle at any point on the line will This does not matter in the least, howdiffer by v radians in the two cases. In the first place, such a change (or a change by a multiple of ir) ever. does not change the hyperbolic tangents and, hence, the values of impedance (including sending-end impedance) which all depend on these tangents. Secondly, although hyperbolic sines and cosines change sign when the imaginary part of the angles is increased or decreased by x (or a multiple of It is seen that the a-) radians, the computed voltage, current, and power distributions will be correct since they depend on ratios of hyperbolic sines and cosines. Consequently, any effect of the reversal in signs is eliminated. still Direct-current Lines. may —As previously stated, the direct-cur- be looked upon as a special case of the alternating-current problem. Hence, the formulas which have been derived are all applicable to direct current when the frequency is considered zero. In general, therefore, the direct-current problem is simpler, since, in the steady state, the effect of inductance and capacitance is eliminated, resistance and leakance only having to be considered. Furthermore, all quantities involved, including the hyperbolic functions, become real rather than rent problem complex. \ 200 ELECTRIC CIRCUITS— THEORY AND APPLICATION Thus, for a direct-current the attenuation constant line, is real and may be written = a The y/rg now surge impedance, which the surge resistance, becomes (109) may appropriately be termed = J- #o (HO) always possible to solve any problem on distribution of and power along a direct-current line by using real hyperbolic functions only. In this connection, however, one particular case deserves special mention. Assume the line to be loaded with a resistance larger than the surge resistance. The hyperbolic angle of the load then becomes It is voltage, current, 0' = tanh -1 ^Ho = tanh -1 number > (real 1) (111) No such angle exists in the real region, the maximum possible value of the hyperbolic tangent of a real angle being 1. Equation (111), however, may be written = 0' This is ?-° tanh-i + f- = 0" + j| (112) easily verified, since, in general, tanh = 0' a = -=r Rn coth (»'±>!) Hence, e' ±j% = £ coth- 1 = tanh- 1 —° from which equation (112) immediately follows. Evidently 0" = tanh -1 R /<r is always obtainable from a table of real when a > R The voltage and current distribution hyperbolic functions (28) . is now given by equations and (29) + sinhffl, —?°*. + jl) 0" *.-*.—> • i sinh / n ( + , „,/ 0" cosl/k-f- 0" j " T V J coshf a + , an 0" , ± • j 7T t :*+£? + 0") cosh (113 > (0 ^J ±j^\ 2/ ^ ± co -t\ 3 ^ ) - ~ r8 sinh(0 a; sinh (0 + 0") + 0") rm . (1M) THE LONG TRANSMISSION LINE 201 The sending-end impedance becomes (from equation Z = Z a tanh (d + 8'>"±;1) = Z coth (0 + (32)) B") (115) Since 6 X 8, and 0" are real quantities, equations (113), (114), and (115) can all be evaluated by means of real hyperbolic , functions. EXAMPLE 3 Statement of Problem A ground-return telegraph line, 320 km. long, has the following constants: r = 8.5 ohms/km. g = 2.4 X 10~ 6 mho /km. Calculate and plot the voltage and current distribution on this line when 1 is impressed at the sending end and the load resistance is volt direct current a = 6,430 ohms. Solution The attenuation constant and surge resistance are a = Vrg = V8~.5 X 2.4 X 10"« = 0.00452 hyp Am. J R = Jr = >o 8.5 = 1,881 ohms \2.4X10-« 200 100 Kilometers from Sending End — Fig. 104.- Voltage and current distribution along a ground-return telegraph One volt direct current is impressed at the sending end, the line being loaded with a resistance of 6,430 ohms at the receiving end (for calculations, see line. Example 3). It is seen that, in this case, the load resistance is larger than the surge Hence, equations (113), (114), and (115) should be used. resistance. 0" = tanh" 1 — a ° = tanh" 1 = ^4S 6,430 tanh" 1 0.293 = 0.303 hyp J* The entering current becomes /. = Table E. tanh (0 Ro + 0") tanh 1.7494 0.9413 1,881 1,881 = 0.500 X 10 -3 amp. X gives the calculations of the voltage and current distribution, the distance x being measured from the receiver end. Figure 104 shows the shapes of the voltage and current distributions on the line. — ELECTRIC CIRCUITS— THEORY AND, APPLICATION 202 Calculation of Dibect-current Voltage and Current Distribution on a Telegraph Line, the Load Resistance Being Larger than the Surge Resistance (Example 3) Table X. * ax = 8" 8x 80 160 240 320 EXAMPLE 8" hyper- hyper- hyper- bolic bolic bolic radians radians radians 0.3616 0.7232 1.0848 1.4464 0.303 0.303 0.303 0.303 0.303 0.3030 0.6646 1 0262 1.3878 1.7494 kilo- meters + 0x cosh (0* + 8') sinh + 8") 0.3077 0.7147 1.2160 1.8782 2.7886 1.0462 1.2291 1.5744 2.1279 2.9625 . (& Ex Ix, , volts 0.353 0.415 0.532 0.718 1.000 . milli- amperea 0.055 0.128 0.218 0.337 0.500 4 Statement of Problem A ground-return telegraph line (Fig. 105) I miles in total length has in with it at each end a battery and a relay. The voltages of the bat- series es, A Af r'WWV V Fig. 105. P\* .JU —Ground-return telegraph — ** 2L —fwwv i with batteries and relays at each end line (Example 4). Ei and E% volts, respectively, and the resistances of the relays are and az ohms. The line has a resistance of r ohms per mile and a conductance to ground of g mhos per mile. teries are v\ 1. Establish a general expression for the current through the relays, the current at the middle of the line, and the voltage above ground at the middle Treat both the case where the batteries act in conjunction and of the line. the case where the batteries act in opposition. 2. Introducing the following numerical values: I — 90 miles 100 volts 2 <T2 = 400 ohms 20 ohms per mile 5 X 10 -s mhos per mile E =E = x <t\ r g = = — and the voltage at the midpoint for the case where the batteries act in conjunction. compute the relay currents, the current at the midpoint, Solution Derivation of Formulas: The voltage and current at the point 1. Vx = Vb cosh ax Ib cosh ax P are + IbRq sinh ax VB K ax •+- -=- sinh (a) (6) ~ THE LONG TRANSMISSION LINE 203 where R " o The voltages at A "Sj-, B and a ™ Vrg are also given by Va=Ei- <tJa Vb = (±E (c) + «n/a a) where the plus sign applies when E* acts (d) in opposition to Ei, the when it acts in conjunction with E\. Va and I a can be found from equations (a) and by (6) minus substituting x sign = I. Hence, Va = Ei — aj a — (±Et — <t2/b) ( ±Ei ~ = Ia Solving equations Ib cosh cd and (e) (/) + sinh (e) oil sinh al (/) /to and for I a #i( cosh + IbRo cosh al g2/B) cd -Tb gives + ^ sinh od\ - ±# ( 2) (9) + (cti <r 2 ) - El 7b cosh aZ sinh aZ ( (±-S 2 )( cosh aZ * . +<rn) cosh aZ (ffi + ^p + Ro ) + + 2 Cjf ^ Ro sinh ad J } + -Ro (h) sinh al 1 The current at the middle point can be obtained from equation stituting x = Im = Multiplying equation /at cosh (6) by sub- 1/2 giving (i) Ib cosh ^r ^ + Ib cosh2 ^ + ~ sinh -^ cosh = -jf (cosh aZ ™ & + 2\ Ib jf y (i) through by cosh cd gives = ~ sinh + 1) + jy^ ZJXq cosh "* -^r sinh cd + IT sinh "0 = "^"2 — Hence, / = 7* +Ia 2 cosh * ^ Substituting equations CJ) (g) and (A) in equation (J) gives for the current at the midpoint {Ei - -±E,) + ~cd 2 COSh -~- As ( before, the —) (<ri + {Ei <r 2 ) upper sign (+) for conjunction. - ±E») cosh od + ~ cosh aZ is + Ta~o 2 ( -jt- sinhoZ(^i - ±E,) \ + #o ) for opposition of ^ ^ sinh aZ Ei and E 2, the lower sign ELECTRIC CIRCUITS— THEORY AND APPLICATION 204 By a procedure similar to the one used above in deriving IM , the following expression for the voltage at the midpoint is obtained: xl - z cosh 2 cosn and also, — Vm = 2 cosh C#i<r 2 (E! + ±E )+ ^[(E 1 +±E 2) 2. 2 = ] * j ^ ± Exn) - (Eun ± E^) 2 coshoZ - (E 1 ± + o" 2 ) cosh ad + (~ + R 2) Jh? The value 0-2) ( °-Vr^?£= 90\/20 X l\/rg sinh al = = = cosh |? = 1.1030 al 5 2 000 ' X 10~ 6 = and (h) 0.90 hyp 1.0265 + tep + R Q sinh al\ 800 lOO^cosh al X becomes = 1.4331+ 2,080 + 2^) sinh Ia X 1.0265 263.82 o qq 2 = + 2 = 3,280.0 should be 100 0.0804 amp. (h) Ib = 0.0804 amp. (f) lT From equation al\ E 3,280 = From equation \ ohma Since the batteries act in conjunction, the minus sign for used. Hence, from equation (g), equation m 1.4331 of the denominators in equations (g) COSh al al _ \ sinh al Numerical Computations: cosh al From | E ^P sinh . g + +a IB - <nI A 2 cosh (<ri (<ri (I) f M = + Ia Ib at „ , 2 cosh g = °- 0804 , = al o.0730 amp. * cosh r— (ra) VM =0 — Note. In the particular case of equal battery voltages and equal resistances of the relays, i.e., where complete symmetry exists with respect to the midpoint, the problem can be readily solved without resorting to the general expressions derived above. It is evident by inspection that, when such THE LONG TRANSMISSION LINE 205 symmetry is present and the batteries act in conjunction, the voltage at the midpoint must of necessity be zero, while with the batteries in opposition the current at the midpoint must be zero. Recognizing this fact, solutions may be obtained by considering the line split at the midpoint, and treating each half separately, or, still shorter, by treating just one-half of the line, since the numerical values of voltage and current will be the same in the two halves, although their directions (or signs) In the former case (batteries acting in conjunction) the be assumed grounded at the distant end (corresponding to the midpoint), while in the latter case (batteries acting in The solutions can then opposition) the distant end should be assumed free. be written down immediately by making use of the formulas previously may be different. half of the line considered should worked out With the for grounded and open lines. batteries in conjunction, the following equations will apply: IA ^ =IB = <r VA = E - *IA aI ^~~^\ =E/l E( 1 = a im =~I jjr + Ro tanh -^ -^ = cosh -y With the (n) ^ + Rq tanh -~ <r cosh -K- + Ro sinh (0) J (p) al -~ batteries in opposition, the following equations will apply U = -Ib = a + Ro coth -~5 =— a\ 1 a VM = — ^ cosh ~n + Ro COth -^ ^^- = <* sinh -„- (9) ^ + Ro cosh — — M J GO Composite and Bifurcated Lines. When there is an abrupt change in electrical constants at one or more points along a line, Thus, an aerial line and a cable would, it is said to be composite. The steady-state analysis together, constitute a composite line. of the composite line can be carried through without difficulty; the complete numerical solution of a problem involving composite lines, however, is, as might be expected, considerably more laborious than the solution of a simple line problem. Consider the two-section composite line ABCD shown in Fig. This line is loaded at D with an impedance a. The surge 106. impedances are Z i and Z02 and the total hyperbolic angles 0i and 62 of the two sections, respectively. current. By equations (122) and (123). but there can be and power.ELECTRIC CIRCUITS— THEORY AND APPLICATION 206 The position angle of the load 6' At C the position angle given by is = tanh- -£- 1 (116) given by is = 5C 2 + 9' (117) Sw Fig. The impedance obviously given by there no discontinuity at line Zoi tanh 8 A (121) and current on the first section of given by where 8u = du + 8B = aiu + 8B (124) the distances u being figured from the junction point BC. At C —Two-section composite C is is in voltage.1 = 8A total sending-end distribution of voltage the line is 0i + (119) A is obviously (120) 8b impedance ZA = The p- position angle at the sending end and the (118) and the position angle at 8B The with load impedance at the end. 106. Zc This impedance section of the first = may now line Zo 2 tanh5 c be considered to act as a load on the B determined by = tanh. a discontinuity in constants. the voltage and current at the junction point are determined and used as reference voltage and . -^r (128) tanh. 8 (125) 8c 8X (126) 5C where 8X 207 = 0' a 2x + 0' Evidently. is obtained by subtracting the current taken by the load <j\ from the current at B. the procedure that the impedance at C and the impedance A BC &j 2 is as before.: THE LONG TRANSMISSION LINE current in computing the distribution on the second Here the following equations are used sinh Ex - E%mh T T x c cosh cosh = 6X + section. as indicated.1 p- (129) ZB = and 8B = Z 01 In computing the distribution on the second line. 107. Thus. (127) it is and current at D as reference in computing the distribution on section CD and the voltage and It all depends on current at B in connection with section AB. 108. it should be noted that the current at C. —Two-section composite line with load impedance at one end and at the junction point. except <n of the load at the D Zoj Fig. 107. possible to use the voltage If the two-section composite line is loaded also at the junction point. the impedances at C and E These two are first obtained by working back from D and F. by simple transformations of the equations. junction are combined in parallel and the combination treated as a single load on the first section. . and loaded by three loads. which may be used as reference current for this section. what values initially are known. as shown in Fig. Ic = Ib~— (130) If the composite line is bifurcated. as indicated in Fig. to the constants of the nominal T and n. rather than the complete distributions on the lines themselves. 109. the arrangement of sections and any number of branches. Y. _ Equivalent T and smooth short transmission T — The nominal T and n representapreviously discussed in connection with the II Circuits. — Fig. 109. will not be discussed here. line. become unsatisfactory when the line is By applying correction factors. however. Smooth Line z Z/> Y e Equivalent »—ww-t'OWWi) T Z/g — vwwfliWITO i Y' T J3C.• 208 ELECTRIC CIRCUITS— THEORY AND APPLICATION impedances and the load impedance a x are then "combined parallel. The idea of general circuit constants. and 6 represent the total impedand hyperbolic angle of the smooth line. however. . tion of a 2 line. but nevertheless it is always capable of solution by straight-forward methods. They are treated in detail in Chap. latter. the so-called equivalent Tand II-circuit may be obtained. however. let Z. the equivalence. Referring to Fig. Let Z'/2 and Y' represent the impedance of the arms and the admitance. perhaps the most powerful and generally applicable method is making use of the and lines general circuit constants of the various networks involved. By making i- = - Zjb <J\ + i- &c + in (i3i) i&e use of the principles used for the three circuits treatment can be extended to any briefly discussed above. with three impedance loads. Equivalent n Z" \ T* —-Bifurcated composite line _ Fig. If solutions for the conditions at the terminals and at the junctions of a composite network only are desired. 108. giving the impedance at B. indeed. is limited to a single frequency as the correction factors depend upon the electrically long. Equivalent T at x representation of a long transmission line. IX. admittance. The problem may become cumbersome. These represent correctly the smooth line at the terminals. of the equivalent II . all T line only.circuit value for for the conditions of loading. y 2 (132) ~w = Y sinh d (133) 6 For n-circuit.sinh d (136) (137) the corresponding relations become + ?P) + E. respectively. by its operation as far as conditions at the It is sufficient to show that the correct general relation exists between voltage and current at the sending end and the corresponding quantities at the receiver end. The general relations between voltages and currents at the two line are given by ends of the smooth E =E a I. and. = Ir(l+^P)+ErY' hZ' (l + Zf) (138) d39) . Z" = Z?™±° (134) tanh = Y» = Y-^ (135) correctness of these formulas can readily be proved demonstrating the actual equivalence of the smooth line and The equivalent circuits during terminals are concerned. or simply by transformation of the T(Y). equivalent circuits are then obtained as follows: For T-circuit.THE LONG TRANSMISSION LINE 209 tance of the staff leak. = r cosh 6 I r cosh 6 + + Z Ir sinh 6 r f. of the equivalent T and Z" and Y'/2 the architrave impedance and the admittance of the The constants of the pillars. in addition to this. the proof for the II line will may to a II(A)-circuit. tanh 7 z =z7 . that the sending-end impedance has the proper The proof be given here be worked in a similar manner. respectively. = Er (l I. For the T line. grounded. 1 a . . The two = sinh 8 _ sinh — cosh 1\ L 2 )^ ~ = Z ZoSmh . Line Open: ~ **• o~ ~r 2 ' vi Y' ~ f" a ' Y sinh -ap^^nr')-*-*' which is correct for the open line. therefore. 1 + v/ tLJL ZY tanh = i = 1 *'(^) = = f + — sinh = ^-v—sinh -\ cosh N ZF tanh 2Z tanh | / | sinh 20 2 . respectively. identical. end. -g 2 = which 1 is X Z y. 2+rTT^ + ' + 2 tan ^ X ^- Q Z7coth-0— Y' „ /cosh Zo ( sinh0 correct for the cosh 1 + - 1 \ Zo tanh smhTc-osh^j = grounded * line. a 9 sinh sets of equations for terminal voltages and currents are.e. Line Grounded: z Z = ._ — cosh 1 /t ^^Inh-F-^ 1 + v _ y . . and loaded by an impedance a at the distant is free. See equations (83) and (101) used as conversion formulas in tnte proof. immaterial of loading. Next it impedance will is be demonstrated that the correct sending-end i. when the line obtained.ELECTRIC CIRCUITS— THEORY AND APPLICATION 210 It will now be shown that equations (138) and (139) reduce to equations (136) and (137).. = 2 sinh-i - Vr cosh = 2 sinh-* (140) (141) For Wrdrcuit: . (103)._Z> .x * } = „ Zo tanh 5' which is correct for the loaded line. ^ + »'+ COS sinh — 1 — „ = Zo + cosh + sinh + tanh sinh cosh tanh l + Z + coth Z \ ° /sinh 9 -f <r + ^0 sinh -=- 1 + + cosh -Zo ( "2 v? eosh Z Zo tanh ._* Zo tanh = ^ + _ tanh tanh 0' tanh0tan hT' = -~- sinh 0' 0' „.. . So far.. it has been shown how an equivalent T or n corresponding to a given smooth line can be obtained. Conversely. and (104) used as conversion formulas in the proofs. For T-circuit: Zn y^!. Zo tanh (* + „.or n-circuit. _ VZ"F" (142) 2 i ft" cosh (14a) 1 Proof for T-line: 1 4 tanh 2 Z'Y' = 2 tanh | sinh = 1 1 - tanh 2 4 sinh2 2 See equations (102).„.THE LONG TRANSMISSION LINE 211 Line Loaded by an Impedance a: .(I + ?)f _ Z. it is also possible to determine the constants of the conjugate smooth line which corresponds to a given T. The equations below give the necessary relations. = "2 + = Z' .. 799 X X X / 2..00212 henry 0.6 /90 o mho V a81 4778°.8 ohms X l<r /9(P = 1.8 /78°.158 +j'0.3 /84°.814/78^8 5. Statement A of Problem three-phase.08 = 162. Z\ — 1 = tanh 2 2 Z\ wu cosh " $ = ° 2 gives equation (143).3 /90° mho X 10. 5 This example serves to demonstrate the numerical determination of the equivalent 7r-circuit of a long transmission line.315 2.315 X 10"° = 5. Lfne: Z"F" = 2 tanh = sinh T (as in case) Hence.8 X 5.4 = 0.0141 0.315/90°"= Constants of entire Z = Y = = X 0.08 X lQ. Z" Zl sinh 6 y>i q ~ 2 tanh Z which. 200-mile. equation (142) follows. tanh 2 2/ °V cogh2 * 2 gives equation (141).3 /84°. 60-cycle transmission line has the fol- lowing constants per wire mile r = g =0 L = C = ohm 0. solved for Proof for i II .4 hyp line 0.3 200 200 200 X 0.00212 0. 72 A\ = Z§(l • follows.158 +J377 9 = + JaC = V^y = J377 10.0141 X lO" 8 farad Determine the constants of the equivalent n-circuit of this line.416/84^4 hyp 8 .: 212 ELECTRIC CIRCUTTS— THEORY AND APPLICATION from which equation (140) immediately tanh r -nhtf ZQ which.814/78^8 = X 10"« ohm = J5.158 0. solved for EXAMPLE .063 X lQ. Solution Constants per mile z V = = a = r +j<*L = 0. 150-kv.315 X 10. however.1ohms tanh 2 ' r-T^ = 0. be dissymline That is to say.0/79^1 +jl55.2 1.or Il-circuit Tvhich will correctly represent it as far as conditions at the terminals are concerned and at one definite frequency.9 - 213 n 0.537)10-'jnho representation of a composite show the various steps necessary to obtain the equivalent cations of Y-A and A-Y~ transformations. The diagrams by repeated appli- line. metrical.7 162.404/84°. the equivalent T. T .8/78^^^ = 158. These limitations are the same as were previously imposed on the equivalent T and II representing a simple line. II — Equivalent Representation of Composite Lines. in general.3 /89°.and II-circuits of composite lines will. Z— sinh 9 29.THE LONG TRANSMISSION LINE Constants of equivalent Z" = = Y" -2 Fig.00752 +y0.2 II lO-/^ ^ 2in/84°. the impedances of the arms of the and the admittances of the pillars of the II will be unequal.063 X = (0. A composite can always be replaced by a single equivalent T. While a simple line always gave rise to symmetrical T's and n's.537 X 110— Equivalent - 10. "Artificial Electrie Lines.»# r. It consists of three simple line sections and an impedance load at the junction BC. intermediate any discussed in detail. Let it suffice to illustrate it by quoting the necessary formulas for the conversion of a three-section composite line consists of line [Fig. /iaa\ (144) c\a. dt. (b) Fig. Ill] to The an equivalent architrave impedance II. is convenient when a series of simple sections without the composite method. DE The network shown The in (c) results. however. giving the circuit shown in Converting the T-circuit (e). admittances and the pillar tances at II. The impedances AG and GI. GH may DE BC A central II-circuit (a). 111. Evidently the effect of casual impedance loads can also readily be taken into account by this method. • i u by « *" cosh 8C • cosjrTs cosh 5D cosh 5E 8E ^^^ cosh cosh 5B 8C E. will not be This loads." loc.— Three-section composite line. by a Y-A transformation gives (/). An alternative equivalent T and method 1 also available for obtaining the is This II of a composite line. II II obtained. and the load a at BC. 110. Let it be desired to determine the equivalent The reduction may be carried out as follows: Replace each Combine the pillar admitsection by an equivalent II as in (&).. given is cosh «. and IH and HF are added. Each section of the composite line may be replaced by an equivalent T or II and the composite circuit so made up reduced to a single T or n by repeated application of Y-A and A-Y transformations. Z" = &. terminal leaks. smh tA 7 — ~~ 03 * - KKennelly. consider the network shown in Fig.214 ELECTRIC CIRCUITS— THEORY AND APPLICATION One method of obtaining either the dissymmetrical T or II of a composite line suggests itself immediately. the resultant equivalent AIF into a by paralleling the as shown in (g) is Finally. now be changed into a T [GIH in (d)] by a A-Y transformation. A. For instance.k\ . . and capacitance per unit length. the impedance and admittance per unit length can be determined. 215 W ZoitanhS. The by and and (146) Similar at A formulas in equations (145) can be and established (147). the surge impedance Z and hyperbolic angle can readily be calculated. A. for reduction to the equivalent T. " Z^Ttftohi. 1 Kennelly.. If the sendingend impedance of a line is measured with the distant end of the line both grounded and free.|? (150) 1 (151) Knowing the surge impedance and the hyperbolic angle. "Artificial Electric Lines. It is thus possible to obtain the fundamental constants by two simple tests. the impedance and admittance can be found Z = Y = lz = ly = l(g+ ju Q = l( r + faL) = Z d A total (152) (153) Evidently. cit. if the length of the line is also known. for instance by an impedance bridge or by use of the three-voltmeter scheme. etc. sections. and from these the resistance. ~Z (146) (W7) 77 represent the surge impedances of the various 8's represent position angles at the point denoted the subscripts with the line grounded at F in equations (144) Z02. are given by Zg = Z Zf = Z which. The actual measurements may be made by various methods. E. leakance. inductance. grounded and free. solved for Z and 0.THE LONG TRANSMISSION LINE The leak admittances are obtained from Y" = Y" = Zoi. tanh (148) coth 8 (149) give Z = VZ^Zf = tanh. — Tests for the Determination of Line Constants." loc. 1 The sending-end impedances. such as pulsating torque. . B. The harmonics which may be present are usually too small to be of importance as far as the power circuit is concerned." Trans. however. as a rule. The The are: generators. A. p. 216 Artificial Line. 1922. p. "An its fundamental and its higher harmonic Oscillographic Study of Corona. 2. "Wave Form and Amplification of Corona Discharge. F. Jr. It is not at all difficult to obtain the steadyThe impressed distorted voltage state solution of this problem. "Corona Investigation on an A. A. realized under normal operating conditions when well-designed generating equipment and proper transformer connections are used. current. 1921. E.. Peek. 138. J." Trans. F.E. 1 The Behavior of Harmonics on a Line.. Although small. Inouye.E. they may have considerable effect as regards inductive interference with neighboring communication lines. due to resonance.CHAPTER VIII HARMONICS It is desirable to operate alternating-current power circuits with voltages and currents of pure sinusoidal wave shapes. Furthermore the pres- ence of harmonics in rotating machines and transformers may result in other undesirable features.I. 1787.. 1913.E. is — simply resolved into 1 Bennett.E. 1925. The main sources of harmonics 1. Gardner. "Voltage and Current Harmonics Caused by Corona. 897. p. M. transformers. E.. and N. In present-day commercial practice.. vibrations.I. Assume that a nonsinusoidal voltage is impressed on a transmission line and that it is desired to determine the distribution of voltage.. Whitehead.I. W..E. Distorted waves cause increased losses in lines and machines and hence a reduction in the efficiency of transmission and the efficiency of the connected machinery.E.E. and overvoltages. 3.. 1155.I. p.. A. and power on the line. Corona." Trans." Trans. an almost ideal wave shape is. HARMONICS 217 Each harmonic is then in itself a sinusoidal voltage and is treated as such. In polyphase systems this also involves the possibility of different circuits existing for different groups of harmonics. If there is a ground return. even when the ground is dry and its resistivity high. that the ground return circuit acts as a conductor of an immense cross-section. therefore. the constants for the harmonics of the second group are. This is due to the fact that the ground current distributes itself over so large an area. These currents. these are combined according to the wellcomponents.e.e.. the triple-frequency group (i. as a rule. The constants for the triple-frequency group. current. and power distribution of each harmonic frequency is computed separately by making use of the methods and equations developed in Chap. That is to say. on the other hand. cannot flow unless there is a metallic neutral or a circuit for their return. The position of the latter depends on the condition of the soil. seventh..) follow different paths. In carrying through such calculations. in a three-phase system. particularly on its moisture content. The resistance of a ground return is very small and is usually neglected. Evidently. etc. . The harmonics of the other group. The equivalent ground plane may be located above the ground when this is covered with snow and as far down as 400 to 500 ft. eleventh. the voltage. cannot always be obtained with precision. Thus. below the surface in very dry regions. the position of the equivalent ground plane greatly affects the inductance and capacitance of the return circuit. the fifth. known rules for the combination of quantities of different frequencies. readily calculated (or selected from tables). exact calculations can usually be carried through. and may hence vary along the line. If a metallic neutral exists. VI or VII for sinusoidal voltages. being 120 deg. out of phase in the three phases. the proper constants must be used in connection with the various harmonics. however. assumptions as to the position of the ground circuit or "equivalent ground plane" have to be made. ground When the circuit configuration is known. being in phase in the three phases. The harmonics in the former group appear as residuals. Having obtained the separate solutions. follow the same circuit as the fundamental. the third harmonic and its multiples) and the group consisting of the other odd harmonics (i. k -d--- >j ~\P <^F~ ' i —Two metallic hemispheres located at the surface Fig.= (1) 12X2. below the conductor. and the distance between them d. conducting medium extending to infinity in a homogeneous of all directions.54 resistance of 100 miles of No. In order to show that the resistance of the ground as a rule is small compared with the resistance of the outgoing metallic circuit and particularly compared with the reactance of the loop formed by the metallic conductor (or conductors) and the ground Assume a be well to consider a ground miles long with 100 single No. In applying this formula to a practical case. as shown in Fig. the resistance of the ground circuit 500 -2r The 76. The ground . and the plates is 40 sq. plane is 30 ground located equivalent the that further assumed return. 0000 copper conductor. The approximate order ground may circuit is p. the radius of the two hemispheres n and r 2 respectively. it may specific case. the ground plates by equivalent hemispheres of the same surface area as the plates-and an estimated average value of soil are considered replaced resistivity is used.280 X 12 X = 2. ft. 112. 0000 copper conductor of 100 per cent conductivity is 25.8 cm.54^ = is then 76. 112. be 500 assumed to is ground resistivity of the ft.07 ohms 2._ 218 ELECTRIC CIRCUITS— THEORY ANt) APPLICATION of magnitude of the resistance of a be estimated by making use of the formula for resistance between two hemispheres located.9 ohms at 20°C. The radius of the equivalent hemisphere /A and using equation becomes S . The area of It is ohm-cm.8 100 X 5. at the surface of a homogeneous conducting medium extendIf the resistivity of the medium ing to infinity in all directions. ground the return. the resistance of the circuit through which a current would flow from one hemisphere to the other is given by . Since the actual value of the latter is very uncertain.HARMONICS 219 resistance. The complete fundamental. VII. (a) Fig. follows the same circuit as the it is usually. EXAMPLE 1 This example shows how a harmonic of the triple-frequency group may be taken care of when present in a feeder circuit with metallic neutral. ignored. current. hence. The calculations are carried through for a tooth-harmonic of twenty-nine times the fundamental frequency.5^ 10- = (741 log* ^^ + = 2. 113.5^ 10-« 10. the impedance of the loop is affected only to a negligible extent by consideration of the ground resistance. the total 60-cycle reactance X = 100 X 377 X 2. as seen.673 X 10~ 6 = 100. other. and appropriately. Chap. Obviously. The only resistance and inductance need be Calculation of the proper constants is included. only about 2 per cent of the reactance of the loop. 60-cycle feeder circuit with neutral shown is arranged as in Fig. and power distribution on a long line due to a high-frequency harmonic is illustrated in Example 1. (n) circuit with neutral. The inductance of the loop formed by the conductor and the ground return is L = (741 log M y + 80. b. therefore. (W (c) — Configuration of three-phase feeder The conductors a. This harmonic does not belong to the triple-frequency group and. is in this case about 8 per cent of the resistance of the metallic conductor. while the neutral to each . 113. hence. Statement of Problem A short three-phase. and c are completely transposed with respect n maintains its position on the right. circuit in this case is so short that taken into account.8 ohms The ground resistance is. voltage.673 X 6 80.6 henry per mile and. 5m) 10-" X30 12 = three conductors — Dab X j*" X D . Hence.518 ohm per conductor Average fundamental reactance per conductor X' = 2o/741 ^~ 377(741 logio 2 = 1. the neutral will carry no current of fundamental frequency.200 volts and balanced voltages to neutral of 7.259 ohm per mile). the system is balanced both as regards funda- mental and third-harmonic quantities. Solution Since the load is balanced and the conductors a.50M )l0-« logic = 2a/741 X logic q + 80..460 in.. the load is Y-connected and balanced.259 = 0.5)l0- 6 = 4. The voltmeters at the Y-connected generator show balanced line voltages The line of 13. b.ELECTRIC CIRCUITS— THEORY AND APPLICATION 220 The spacing D is 2.300 The average number X"' = 2 log + 80. and (c) any one of the per centimeter becomes / f[21og^+!+!(21ogA +21og _£ + 21og _l + A +21ogJL +2 logi) . The third-harmonic current will divide equally between a.G. resistance = 0.21ogi- 2 log ± - 2 log Average third-harmonic reactance per conductor X'" = 6o/741 logio^y^ = 6 X 377(741 4 + 80. b. and c and return on the neutral n.6 kw.5)l0-« 3 7 logic q ^ 3° + 80. (b). The length of the feeder is 2 miles. and c are completely transposed. The polyphase wattmeter (current coils in line a b and potential coils between a and c and between b and c) shows 6. the value of the phase and line voltages and the total power at the load. (diameter = 0.802 kw. 0000 A. fi=2X 0. and only harmonic present Compute is the third. and the copper conductors are No.700 volts.696 ohms £] . and the ammeter in the neutral indicates 60 amp.5 ft.5)l0-« ohms of third-harmonic flux linkages about (a).. while the single-phase wattmeter (current coil in the neutral n and It is known that the potential coil between c and n) indicates 39. ammeters show balanced currents of 350 amp.W.+ 80. 533 volts ohms .jO./1.236 = 12.6 .622 .518 = 7.I'(R+jX') = 7.349.4 Third-harmonic power factor at the generator ™°-=iM^o- <><***> Fundamental phase voltage at the load V'load = V£ .jl73 = 818.HARMONICS The average number (n) per centimeter of third-harmonic flux linkages about the neutral becomes £+ U{2 log I + § .5^10-« 181 7 logic 3° 23 + 80.163 Fundamental phase voltage at the generator = 13.8515 349.70W - 2 ^|^ = 1.236 volts +.100 .8515 .J292 = 7.2 X 0.j0.4 amp.100 volts line currents /' = V3502 - = 202 349.072 + J17.4(0.20(0.5)l0-« = 4.5 volts = V'J' Line voltages at the load Fime = V3 X 7.300) Third-harmonic phase voltage at the load VULt ~ ^r[4# +j(X'" + 3Zi")] = 1.5243) (0.8) (2.200 = V3 Third-harmonic phase voltage at the generator V'„" Fundamental = yj7.185) = 800 .230 . Fundamental power factor at the generator cos 6' = -= V3 X 6 802 = ' 13.1(2 log K" = - 221 P /H2 log ^ ~ X X D- fl -» 2 log jj- + 2 log j*-) +M Average third-harmonic reactance of the neutral XS' = 6«(741 = 6 X logio 377(741 ~^ + 80. 1918. if present. Third-harmonic power per phase at the load PioU = (800 X 0.6 kw. A W.) possibility of ridding the open-circuit line voltages of all the important harmonics.5 + 12. 1122. and having space distributions in the air gap corresponding to frequencies in this group. etc. the armature reaction will not distort the flux and the voltages will remain sinusoidal. however. that by properly distributing the latter.5243)349. the distribution of air-gap It may be flux due to the field is made very nearly sinusoidal.52 7.650. ninth. Total power at the load P = 3(2. on the other hand.8)20 X lO" 3 = 12. p. the values of these harmonic voltages may be made By making use of this property in conjunction extremely small." Trans.E. may as a rule be considered strictly sinusoidal.222 ELECTRIC CIRCUITS— THEORY AND APPLICATION Phase voltages at the load = V7.8515 + 292 X 0.282 volts Fundamental power per phase at the load Pioad = = (7. when generated by fluxes.4 X lO" 8 2.2362 Fi oad + = 818.5 kw. except at Some of the 1 The harmonics in the air-gap flux and in the induced voltage of a synchronous generator can be predicted with good accuracy from design data.6 + 173 X 0. A. harmonics introduced in the flux will be reflected in extent. will induce harmonic voltages in the armature windings. V. there is obviously excellent third. —The harmonics which appear in the a well-designed generator are usually The line voltages of such machines.230 X 0.368 kw. open-circuit line voltage of windings of a non-salient-pole machine. 1 a non-salient-pole machine carries a balanced load. with the fact that no harmonic of the triple-frequency group (i.E. fifteenth. the armature reaction will distort the air-gap flux to a certain When power factors in the neighborhood of zero. These harmonic fluxes. It is well known. therefore. general method applicable to non-salient-pole machines is given by Lyon in "Application of Harmonic Analysis to the Theory of Synchronous Machines.204.e. By properly designing the shape of the pole shoes of a salientpole machine and by properly arranging and distributing the field Generator Harmonics. . impracticable to get a perfect flux distribution. can appear between lines of a three-phase system.I.204. In a salient-pole machine.37) = 6. so small as to be negligible. stationary with respect to the poles.. but the harmonic components can always be made very small.. it is convenient to consider the unbalanced armature currents equivalent to a positive. This will be the case whether the generator is a salient. second-harmonic eddy currents in the iron. speed with respect to the armature and is stationary with respect to the negative-sequence armature reaction. in many instances. reach pole machine. harmonics in the phase and line voltages are unavoidable. It is assumed that the generator has no neutral connection so that no zero-sequence currents can flow. fore. the latter induces third-harmonic These third-harmonic voltvoltages in the armature windings. should always be rather insignificant.HARMONICS 223 the armature voltages unless suppressed by the connection or There is a chance. therefore. When a synchronous machine carries an unbalanced load. Hence. the currents induced in them are. they will be 120 deg.and a negativesequence balanced system of currents. phase-displaced in the three phases and will appear in the terminal voltages. and also secondharmonic currents in the damper windings. one of these fields rotates at synchronous poles. ages are caused by fluxes of fundamental space distribution. strictly single-phase. if such windings are The field windings represent single-phase windings and present.or a non-salientThese harmonics may. The armature reaction due to the negative-sequence currents sweeps by the poles at twice synchronous speed and induces by transformer action second-harmonic currents in the field windings. machine. currents induced in the iron and the currents in the dampers are of a three-phase or semi-three-phase nature. . while the other rotates with three times synchronous speed with respect to the armature. These hai^nonics. The former tends to compensate for a part of the negative-sequence reaction. In order to get an idea of how these harmonics originate. however. The armature reaction due to the positivesequence currents is stationary with respect to the field poles and produces no flux distortion and hence no harmonic voltages except as discussed above for the balanced load on the salient-pole appreciable magnitudes. theredistribution of the armature windings. that some harmonics may appear in the line voltages of a salient-pole machine under balanced load. The pulsating second-harmonic magnetomotive force of the The eddy field is equivalent to two fields revolving in the air gap in opposite directions with twice synchronous speed with respect to the field Obviously. The second-harmonic currents in the field windings and the second-harmonic currents of single-phase nature in the iron and in the dampers will react back on the armature at fundamental and triple frequency. no currents of this frequency can flow. The magnitude of these harmonic voltages. rapidly decreases at the order goes up. the latter fourth-harmonic currents in the field windings. a seventhharmonic voltage may appear.224 ELECTRIC CIRCUITS— THEORY AND APPLICATION The second-harmonic eddy currents and damper currents of true three-phase nature produce only revolving magnetomotive forces stationary with respect to the negative-sequence reaction and wijl contribute to the compensation of the latter. The fourth-harmonic field currents and the other fourth-harmonic single-phase currents will react back on the armature at third- harmonic voltage is and fifth-harmonic frequency. the iron. If. a harmonic voltage of the next higher order will be introduced in the voltage. If the eddy currents and damper currents are also to some extent of a single-phase nature. The positive- readily seen into a positive- sequence reaction revolves at twice synchronous speed. A fifth-harmonic thus introduced. contribute to the compensation of the negativesequence armature reaction and also induce third-harmonic voltages in the armature windings. These third-harmonic voltages will appear between lines as did the third-harmonic voltage caused by the second-harmonic currents in the field windings. however. In such cases a triple-frequency voltage is the only harmonic voltage introduced by the unbalance. This is by resolving the third-harmonic currents and a negative-sequence system. there is an opportunity for unbalanced third-harmonic currents to flow. if fifth-harmonic unbalanced currents flow. Similarly. therefore. and the dampers. In general. The former induces secondharmonic currents. on the other hand. . therefore. when an unbalanced harmonic current flows in the armature circuit. the negative-sequence reaction with four times synchronous speed with respect to the field poles. similar to that of the field windings proper. produce oppositely revolving fields in the air gap. the single-phase parts will have an effect Hence they will They will. a fifth-harmonic component may be introduced in the phase and line voltages. If the unbalanced load offers infinite impedance to third- harmonic currents. . the trend in transformer design has been a distinct and continuous increase in operating densities with a corresponding increase in voltage or current harmonics. A.I. They could be taken care of and eliminated by proper design.E. The harmonics due to transformers. p. Higher odd in general. p.. p. economical design which it would entail. 351.. F. 351.. 1922. 557. These higher harmonics are." Elec.. therefore. p. 949.E. Wave Shapes Peters. of minor The third harmonic is generally while usually an appreciable by far the fifth also is present. on account of the less saturation. A. "Harmonics in Transformer Magnetizing Currents.E. were negligibly small under ordinary operating conditions. on the other hand.I. C. p. W." Trans.. importance as far as the operation of the transformer is concerned. "Observation of Harmonics in Current and in Voltage of Transformers. 1 Single-phase Transformers. It is fortunate. It is a well-known fact that. most prominent. Quite to the contrary. V . Faccioli.E." Trans." Obtaining Approximate Values of Harmonics.E. their magnitude easily traced. however.I. The harmonics can be decreased only by operating at reduced This is not done. G.E. A. cannot be eliminated at the source. "Transformer Harmonics and Their Distribution. Lyon. "Triple Harmonics in Transformers. G." Trans. J. 1915. The number and magnitude of the harmonics which the exciting current contains depend upon the characteristics of the iron and the — maximum density at which it is operated.225 HARMONICS — Transformer Harmonics. the 1 Frank. J. At normal saturation. The transformer harmonics. Dahl. It was brought out in the preceding discussion that the harmonics in the generator voltages. are caused by definite and inherent properties of iron and steel which cannot be changed by human control. is. World.E.E. the exciting current that it takes will be non-sinusoidal. but also harmonics are very small. as a rule. J.I. . therefore. 809." Jour. however. 1910. the fundamental is about 90 per cent. The distribution of harmonics in single-phase and three-phase transformer banks is discussed below in detail. 1917. A. when a sinusoidal voltage is impressed upon a single-phase transformer. that in a three-phase sysdistribution of the harmonics between primary and secondary circuits can be partially controlled by the connections tem the used.. 1925. O. The curves are rent. as found from analysis of a series of oscillograms. Figure 114 shows an oscillogram of single-phase exciting curThe magnitudes of the fundamental. 115. this is not forced saturations. At such abnormal harmonics may be expected to to percentage decrease. third-harmonic.226 ELECTRIC CIRCUITS— THEORY AND APPLICATION harmonic about 45 per cent. 100 5 |0 15 20 Flux Density in Kilogausses — Fig. the up abnormal values. Harmonic components in the exciting current of a transformer with sinusoidal voltage impressed. 115. These curves show the variation in the harmonics with saturation. are given by the curves of Fig. . The percentage harmonics increase with the flux density as long as third Fig. 114. and fifth-harmonic components. and the fifth harmonic about 15 per cent of the equivalent sinusoidal exciting current. — Oscillogram of the exciting current of a transformer with sinusoidal voltage impressed. Gen. 116. The phase relation between each harmonic current. in single-phase connections all and what ZrSLoad is harmonics follow the same laws said in regard to the third harmonic will hold for any harmonic. Vy V mdiamen M mc 1 f ' 1 20 40 60 80 100 120 140 160 180 Degrees fromZero of ResuHant Wave-Fundamental 200 Scale — Fig. The relative magnitudes as well as phase relationships of the harmonics are correctly reproduced. 117. the results of the analysis of the exciting- current curve in the oscillogram Fig. 114 are plotted in Fig. will flow in the secondary as well as in the primary circuit. and the corresponding third-harmonic electromotive force depends on the triple-frequency impedance of the circuits where it flows as established by equations (6) and (7). Components of exciting current from analysis of the oscillogram in Fig.QZc Fig.160 c § /« 'Su/7brrr 14-0 JlZO § 100 <u i 80 § 60 u: 3rd Ham vnic *S 40 3 20 Sth. for ^. 116. that established for this harmonic. The relative magnitudes and phase relations are correctly repro- duced. the third harmonic alone is mentioned and the equations have been It should be noted. Consider the third-harmonic components in the transformer In general. the third. 117. As a typical example.HARMONICS 227 plotted against fundamental flux density and it will be noted that the percentage harmonics increases with the density. however. In the discussion of distribution of harmonics which follows. — Two-circuit transformer with impedance load on the secondary.ttamo £ v. third-harmonic currents circuit shown in Fig. . instance. 114. the triple-frequency electromotive forces are generated by a triple-frequency flux in the iron core. the opposite significance was attached to these symbols. . third-harmonic impedances third-harmonic leakage impedances third-harmonic mutual reactances mary Zl self leads leads It is apparent that it is impossible to get entirely rid of a small third-harmonic component in the terminal voltages of a transformer. h = third-harmonic electromotive forces induced in the windings by the flux in the core third-harmonic terminal voltages third-harmonic currents Zn Z 2 = Z = Zi X12 X21 = Zq = third-harmonic impedance of generator and pri- = third-harmonic impedance of load and secondary 2 . Assuming unity ratio of transformation. or referring all quantities to the same side. . they must be produced by triple-frequency electromotive forces. the generator voltage cannot directly give rise to thirdHence. the third-harmonic electromotive forces induced in the two windings by this flux are equal in magnitude and phase. The following equations 1 interrelate the third-harmonic harmonic currents. Being sinu- sinusoidal.: ELECTRIC CIRCUITS— THEORY AND APPLICATION 228 The induced voltage of the generator is assumed to be strictly Since third-harmonic currents exist. h . 2 . The complete elimination of the third-harmonic voltage components would require zero triple-frequency impedance 1 Note that in this discussion of transformer sidered convenient to let the symbols of the circuits (or harmonics it has been con- V of terminal voltage represent voltage drops and the symbol Ec of induced voltage represent a voltage rise. since the harmonic voltages are generated in the transformer itself. In the general transformer equations previously developed in Chap. quantities V" V'" _ = — Z'A =R. II. even if the generator voltage is a pure sine wave.+ j(X[" + T/'" V\ i'" Lg 1\ t'"7'" = — — w'" ft\c f" i\ 7" — Zn r t'"?'" 7?'" ay'"t'" jXvili AY'"T'" Xu) (5) 2 in SQ\ (4) Z£ = R + j(Xi" + Xn) The symbols used ro\ (2) these equations have the following meanings: = E2c = E\ c Vi = Yi . soidal. " „/// . however. but also the phase of the impedance which the third-harmonic current must overcome. where no appreciable impedance is offered to the flow of the third-harmonic current. reduces the triple-frequency flux in the core. the actual third-harmonic current in the primary may. while a leading current tends to amplify it. Transposing terms. Not only the magnitude. then.*'£ -jx'A' + z'l' zT + z'g" 7.HARMONICS 229 third. It may be said. be the smaller in the case of a step-down transformer of large ratio. will be sensibly one of the where the circuits) sinusoidal. the triple-frequency voltages will usually be so small as to be entirely negligible in comparison with the fundamental component. therefore.harmonic current flows. The resultant voltages. a never can be fulfilled."' z"i " _i_ v. Hence. the third-harmonic current in the primary will be many times greater than in the secondary when both currents If they are not referred to the same are referred to the same side. may be said that a lagging third-harmonic current. that the distribution of third-harmonic current between the two windings of a transformer operating . Usually. The distribution is inversely proportional to the ratio of the total impedances of the two circuits.v'" . of course. has an important effect on the magnitude of the third-harmonic voltages and Without going into the question in further detail." (8) shows that the distribution of third-harmonic current between the two windings depends upon the leakage impedance of the windings and upon the external impedances. in general. condition which.'. the load impedance will be very much greater than the other impedances involved (the two leakage impedances and the generator impedance) when referred to the secondary side. it currents. side. In single-phase or polyphase connections. equations 1 By i*"(z£ and (3) may be written )+jf "x'£' + zT) + ji'i"xZ EZ = f "(zZ + &' = (2) . of course. in general. z'J t (6) (7) equating these expressions the distribution of the third- harmonic current between the primary and secondary is obtained. ^L rjl'l _ - . The ratio of the currents I? ^22 "v r/'H J'" Equation (8) is „ttt '. xT + xT the secondary is tuned to x'o" = o (io) series resonance for the third-harmonic same time. cit. been settled to at certain points everybody's satisfaction. power is input to the core at fundamental this voltage. then none of the harmonic currents can produce power in conjunction with Hence. Operation in this region is also accompanied by peculiar instability phenomena. 1 There has been considerable discussion of the causes for the harmonics in the magnetizing current and the voltage of a transformer. Let large to the z2" + If. The writer's conception of this question is outlined in the following paragraphs. The minimum fundamental impedance of the load is determined by the rating of the transformer. as far as the writer is aware. may easily be made single-phase largely regulated is of the triple-frequency impedance fundamental and still extremely small to the third harmonic by a suitable combination of inductance and capacitance in series. If the voltage of the generator is strictly sinusoidal.. given If. Frank. frequency only. Much larger currents as well as voltages may be obtained at other capacitive loads and at a second resonance condition. May. at the by (11) 2 A\ ~\- ZlQ It should not be inferred from this statement that the secondary "leakage-resonance" condition gives rise to the maximum amount of third-harmonic current. loc. 2 The relation of the harmonics to the power losses in the core. the load resistance is zero. Pittsfield. presented at the Regional Meeting of the A. distribution current is the being a third-harmonic current. 1927. most engineers now agree they are caused both by the varying permeability of the iron and by hysteresis. While opinions on this question have differed a good deal in earlier years. King E. however. . The load.230 ELECTRIC CIRCUITS— THEORY AND APPLICATION by the magnitude and character of the load. Z'T = R>+ jX'>" + R L + j(X? - X'o") (9) then.E." paper. J. has not. "Instability in Transformer Banks. 2 Excellently discussed by J. however. the third-harmonic current to the primary ratio of the secondary maximum..E. 1 Gould. Mass.I.. the only effect of the sudden introduction of harmonics in the current would be to increase the excitation conductance of slight It would now draw a fundamental and the apparent increase in core loss is the transformer. core is . part of which is absorbed by the core (and this part may be very small if the non-fundamental fluxes are small). or to the fundamental input to the core. a part of this power is expended as copper loss by harmonic currents. is then evidently conveyed to the core as power of fundamental frequency. Fig.HARMONICS 231 The harmonic currents. will necessarily give rise to a copper loss in the circuits where they exist. a part of which is given out to the circuits where the currents of the higher frequencies flow. In the core it is converted to power of other frequencies. . Figure 118 may now be used as a vector diagram of fundamental quantities only. —Vector diagram showing increase in apparent core loss. Neglecting the in the change in the fundamental leakage-impedance drop of the primary. The power corresponding to this copper loss. 118. (12) This power is. magnetizing current. plus the losses in the core caused by the non-fundamental fluxes. Exile -V c ) watts current* Ie . however. 118. Conceive a hypothetical transformer having core loss but requiring no harmonics in the magnetizing current for impressed sinu- "*/ ^ f\r* jc ic Fig. as already pointed out. in reality. The transformer core is in this respect nothing but a frequency and the harmonics add to what may be called the apparent core loss. while the rest is dissipated as copper loss in the circuits carrying the harmonic currents. converter. The vector diagram on open circuit is given in and the exciting current taken is I'e The hypothetical now exchanged for a regular iron core requiring harmonics soidal voltage. while. not utilized as power of fundamental frequency but is converted to power of higher frequencies. In a A-connection it will circulate in the closed delta but cannot escape from this and enter The third-harmonic lines while it the may lines. or both. but will exist as — . In the following discussion of the distribution of the harmonics. cannot be controlled by transformer connections independent of the external circuits. in other words they appear as residuals. the third harmonic and its multiples and those which are not multiples of three.232 ELECTRIC CIRCUITS— THEORY AND APPLICATION — Single-phase Transformers. The third-harmonic current can appear on the lines only in a Y-connection with neutral. this is equivalent to voltage to neutral and line current. and hence are phase-displaced 120 deg. strictly sinusoidal impressed voltages and balanced transformers The ratio of transformation of the transformers is are assumed. generator and the balanced load are both assumed Y-connected for all transformer connections. respectively. voltage. however. therefore. equations for the fifth The as well as for the third harmonics have been established. i. in the three phases. No load. in all cases. It should be noted that voltages and currents of triple frequency are given per transformer. as otherwise the equations are easily open to incorrect interpretation. to neutral and per line. means that the secondary lines are opened at the transformer terminals. Fifth-harmonic are voltages and currents are any connection used. With this connection (Fig.. Since the same laws do not govern the distribution of the two classes of harmonics. When single-phase transformers are connected for three-phase operation. are easily modified to hold when The equations given. cannot appear between be found as a component of the Voltage to neutral. In a Y-connection. either generator or load. for This significance of the symbols used in the equations below should be kept in mind.e. delta-connected. the method of connection constitutes a means by which the distribution of the third harmonics and multiples may be partly This depends controlled independent of the external circuits. The other harmonics which are not multiples of three. assumed to be unity. Three-phase Connections upon the well-known of fact that the triple-frequency voltages and currents in a balanced system are in phase in the three phases. harmonic currents cannot appear on the lines. 119) the third(a) A-A Connection. 233 Conditions are the same whether the transformer bank is loaded or not. and the division secondary is third-harmonic current between primary and of given by t'" y'" i Zj2 1 (14) Z[< t'" The fifth-harmonic currents will appear on the lines. The induced third-harmonic electromotive forces are given by < . (6) A-Y The Eic = Eic V% = E%c following relations hold: = Ii [ ~o — I" %g f+^ 'X21 Connection.k: = cc +<x' = cc +K'%" d3) Fig. (17) I Isolated Neutrals. the transformers are loaded the following equations holpl: TP y — — W%2c = r When f0 + Z^+3ll~ v + zl\+jil?f = fi(^ II Zl II zl + + (15) 3ZI (16) szl When the load is disconnected (at the secondary terminals of the bank). the fifth-harmonic current ceases to flow in the secondary windings. is loaded and when it is on . both when the bank open circuit.HARMONICS circulating currents in the two deltas. while a fifth-harmonic voltage V\ still appears on the secondary side. — With (18) this connec- tion (Fig. the third-harmonic current will be confined to the primary delta. 119. 120). t 21 (19) = T ft.— When the bank is loaded (Fig. the following equations hold for the third harmonic 1/" _ v'" = Zn jh A 12 h = ir r "i 12 When the load is + + ZT + 3Z-) +jI "X' " z + zT + 3Z T ."&£' 1 2 [' (25) 2 2 rj'" ^ ' -^1 removed. j (21) n V3(Zt Zr + + Zj) V 3Z a \30 c (22) Disconnecting the load gives E%c El = -^\30° = Il(*& yj = El + Z^ . 121). 120. rfj? g +i^^r ) 2 \3o c rX. 121. . !'. -V A E% — Jll f. Interconnected Neutrals. rr"' il Zl (20) Fig. The equations for the fifth harmonic when the bank is loaded aie &1C *p — .3-^XlJZV = VSI V + (23) ^W (24) Fig.234 ELECTRIC CIRCUITS— THEORY AND APPLICATION The following equations will hold tp'" Ej\c 7 2 " — = T?'" Ei<Lc = i\ Zin •T n. equations (19) and (20) express the triple-frequency relations. (23). With this connection (Fig. With this connection (c) Y-A Connection. 123). The following equations hold When E\e = Vi = Eu — jh the bank is E%c ss 1 1 Z%2 V3 Il{^f + + II Zl II V3(Zr + 72 Z (28) 2 Zl\ + Zl) + j^= X T 2 /30 c + jllX Yn /30_° (29) /30° (30) Zl) gives y/3Elc/30° Vl = Elo Jl(Zli 3ZI Removing the load Eu = Xvt loaded the fifth-harmonic relations are E yu = VSEl/30° = = (27) = v = + Zl) I (Zu V v -3~i= X nW° = ^r(Z + (31) Zl) \30* (32) — Interconnected Neutrals.: HARMONICS 235 harmonic will follow the same laws as in the case with isolated neutral and is hence determined by equations (21). to the secondary delta both for load and no-load conditions. . 122). the third-harmonic current will be exclusively confined Fig. 123. secondary delta both at no load and when the bank The following equations hold: is loaded. and (24). (22). 122. the third-harmonic current will flow in the primary lines and the Fig. The fifth — (Fig. Isolated Neutrals. :'= k' (35) loaded.JI2X12 = ll{Z\. side = vt = vi" When the bank is #. (31). the equations for the fifth harmonic are Eu = Ei = Ii(Zn -f.236 ELECTRIC CIRCUITS— THEORY AND APPLICATION J2»'" _ T?'" _ j>"( — i2 >^22 zT + The (d) v i v'" + 3Zln tn )+# in 2 v Al2 iti (33) ryl'l /. no third-harmonic current can flow on either Fig. + Zl) + jl\Xl. and (32).' they 7 >" in jlll rjlll z£' (34) + szS! neutral connection does not affect the fifth harmonics. With this con- — Y-Y nection (Fig. Connection. . still follow equations (29). under any condition. however. c (36) + zl zl + zl zl II (37) Disconnecting the load gives Eu = E%c = 1 (Z11 + Zq) V\ = El-jllXl =H(Zl + (38) 1 Zl)' Primary and Generator Neutrals Interconnected. A third-harmonic voltage. All Neutrals Isolated. primary side. 125). 124. the third-harmonic current will flow on the Fig. 125. both at no load and when load for the third harmonic are The equations is put on the bank. (30). which may be very large.Zq) -\. (39) — With this connection (Fig. 124). exists on both sides. and (39). = &' = I["(Z'A' = l't"(Z% + ZT = z"' z[" + z'L' + + z7 + + + Z'o" + 3Z£) 3Z'/:) + jfi'x'U + j/i"x« (44) sz^: (45) 3Zx':' ." (41) The conditions as far as the fifth harmonics are concerned are identical with those existing when all neutrals are isolated.HARMONICS EZ = E\i: = iTiz'll 237 + z7 + (40) ZZ'Z) = Eu . — With (Fig. 127. (37). Fig.jli"X'" = I["(Z[" + z£" + 3Z£) 7. (38). as expressed by equations (36). 126). Neutrals Interconnected on Both Sides." this connection on both sides when loaded.jl'"X% = + Z'l' + 3Z£) 1 (42) 2 JW (43) At no load there is no circuit in which the third-harmonic Equation (35) holds for the voltages. 126. (37). The fifth harmonics behave as in the other Y-Y connections (equations (36). Fig. current can flow. 127). is E'£ Vi" will flow on the = E'2c" = i'i\z% + zT + 3z£) = Eu . and (39)). — When the bank Secondary and Load Neutrals Interconnected. a third-harmonic current will flow the bank is E['J iT_ /. the third-harmonic current secondary side. (38). loaded (Fig. other than the third can flow in the tertiary winding. Designating the tertiary winding as winding 3 and assuming unity ratio of transformation also for this winding. the (/) A-A-A Connection. Neutrals Isolated. the following equations may be written the bank and its multiples. the third-harmonic current will always be confined to the tertiary delta. Z 3 a) and Z 3 (2> represent the third-harmonic leakage impedance of the tertiary winding with respect to windings 1 and 2. No harmonic. respectively. loaded or not. (37). The fifth harmonics behave as in the other Y-Y connections (equations (36).ELECTRIC CIRCUITS— THEORY AND APPLICATION 238 When the load dently hold. 128. third-harmonic current will circulate in the three deltas under all conditions. and (38). (37). (38). — With this connection (Fig. 129). equations (40) and (41) will eviharmonics follow equations (36). Y-Y Connection with Tertiary Delta. is */" Vl T ./// V2 w'" = = Mile yp/ft tLlc v'" r"V" — Jl3 A13 = . (e) With is The removed. Fig. and (39)). fifth (39). . 129.ylll^T-llt — Jl3 A23 = lz (46) Zi3(\) (47) jtll —III l3 ^3(2) (48) In equations (47) and (48). — this connection (Fig. 128). independent of whether Fig. Subtracting equation (50) from equation (49) and equation (51) from equation (50) gives [Ri [Rz + J(Xu ~ X + j(X .X n .X .Z. respectively. Hence.X + X n U . 3 2 where 7"' is * ne total current required for third-harmonic excitaequations which follow. rtr (49) (so) (51) /con (52) /// .X )]/ + j(X u ~ X )7 = + j(Xtt .X 33 (57) 21 13 23 ) 12 13 23 ) (58) 23 2i 31 ) (59) Furthermore.X + X . = Ri Z2 = R 2 Zz = R 3 Zi + j(Xu . the following equations are obtained: [Bj + j(Xn .Xu) (60) .[R + j(X . j(Xi 3 .Z 2(3) = ( i) where Z 2W and Z 2(3 ) i(Xi.X 22 - 21 )]Ii 32 )]7 2 ~ [# 2 [Rz + j(X ~ X 22 12 )]7 2 + j(X 13 .X )]h + j(X« . + JXsM" + jxS'/i" + i^I'/J" (B. third-harmonic currents and reactances will be omitted.X + (56) X .X + X )]7 . Z 2 . II). = Z 2 .X + j(X .X + X .X + .X )7„ = 23 13 [« 2 21 32 12 23 2 33 )] 23 31 3l 21 in the brackets are such as The impedances 22 2 X 23 would be assigned to the windings of a three-circuit transformer (or to the branches network) when exciting current is neglected of its equivalent Y (see Chap.X )h = (54) 3S 3 23 )]/ 3 j(X n 31 Eliminating 7 3 from equation (53) and 7i from equation (54) by means of equation (52).X + X + j(X 22 .HARMONICS The following equations f = tf» = lC = E[ : may 239 be written + jX'AW + JX'M" + jXW («. + iXi'O/i" + JX'A'I[" + jXg%" T T T + 7fin = h 7i + 7 ( Rl /// .X + (55) Zi3 .X 23 )7 = (53) + j(X .X 23 ) . Introducing this abbreviated notation and eliminating 7 3 from equation (56) by using equation (52) gives ZnIi + Z nIi+(Z iW -Z n )h = (62) .X . the triple primes indicating In the tion.[Rz + j(X . (61) represent the leakage impedances of wind- ing 2 with respect to winding 1 and winding 3. to a 150-kv.z + Z Z +r-^r-^ Similarly the currents in winding 2 and 3 — and (55) Z3Z 3(1) Zi 2 Zi(3) — ZlZi (2 ) (66) which gives the distribution of third-harmonic current between the three deltas. therefore. The primary neutral points are connected to the neutral points of the generators. 130) has two 12.. and (65). The secondary (high-tension) neutrals are grounded. The fifthharmonic distribution. .Y-connected transformers. Y-connected. . the transformer operates as if only the primary and secondary circuits were present since no fifth-harmonic current can flow in delta 3 under any conditions. 60-cycle turbo alternators.J. H.— 240 ELECTRIC CIRCUITS— THEORY AND APPLICATION between equations Finally. eliminating 7 2 II ^23^2(1) = ZiZ 23 — Z2Z12(3) +zz 2 (Z 2 4" Zz)Zi -r -J-0 iU z7¥~yi z2 3 x h / = ZsiZs(2) Z3Z 3(1) Z2Z3I + Z3Z] — ^12^K 3) ~ Z ZH i z&n + From /o ZxZ* 2) = j l0 Z\Zi (Zi Z2Z20) 2 3 Z$\Zz(2) -f- — Z3Z3 (i) Z3Z1 ZiZi (2 ZZ + 2 3 + Z )Zi(3) — + Z2Z3 + 2 ) z.z 2 — T i it (63) x become z7z7~ (64) r ° (65) follows that 78 /» — Z 2^2(3) 2Z (i) (62) gives i (Z3 ~r Z\)Zz(2) equations (63).T. Ji Z 2 3Z 2 — i) ( 1^~„ z. is given by equations (15).600-volt. transmission line through two banks of Y.-a. (16). Of course. nal load circuit (17). this statement assumes that no exterconnected to the tertiary delta. ^WM\^I jEWkAAA^— <£5Z Substation Generator Station Fig.000 kv. operating in parallel and supplying energy fLihe. Bus —Circuit diagram showing generating station and substation connected by completely transposed transmission line (Example 2). and is (18) EXAMPLE 2 Statement of Problem A generator station (Fig.Complefe(y Transposed B 1 . As far as the fifth harmonic is concerned.Bus H. (64). 130. 6. HARMONICS 241 The wave shape of the generated voltage.95 = 22.000/86.43 + j'260. l zZ.+ Z"L = |(8.04 jl.7] ohms triple-frequency impedance of the high-tension circuit 1" Z£.43 = 16.i The = ° _i_ T" 7"' l i = ±[(0.4) + 3(4 .65 +j'68. . = 7 1 ^.848 amp - Third-harmonic current entering each conductor of the transmission r tn 2 = r n'Zi. The separate triple-frequency leakage impedances of the transformer windings are + = = Low-tension High-tension ohms ] per phase ohms side 6.65 2 + J370.2 is + 260.7) 3 X X0.30)22.72 °. (95 per cent assumed Compute the third-harmonic current to be of triple frequency entering each conductor of the three- phase line at A.800/22.„ hT = (30. - 3 Statement of Problem A power station contains four generating units.1 + 30. both voltage to neutral and voltage between lines.200-kv. the low-tension triple-frequency impedance per phase referred to the high-tension side becomes 7"' Zi. The transformers have three windings (3.j'1. With the system in operation. 30. Solution Ratio of Transformation a 150 = = 2272 6.848 X 371 "i^o— cir- = noi °. the ammeters in the primary neutrals show In i t i„ 2 = = amp..7 amp. is sinusoidal.7 8.500 volts) connected Y-A-Y and three-phase.490 = 1.490 ohms Average primary third-harmonic current per phase of the paralleled cuits referred to the high-tension side . transformers. The triple-frequency impedance per phase of the generator is 0. 60-cycle turbo alternator (Y-connected) and a bank of three 4.6 + jl.72 + 6.30 ohms.1 on t 30.J540) .4 referred to high-tension J With the lines disconnected from the high-tension buses at A and B.4 = 371 +-J68.211 amp line . EXAMPLE = 0.-a.000 kv.04 = 13. t. the impedance at A of the three conductors in parallel to ground has been measured by an impedance bridge and found to be 4 — ^540 ohms at 180 cycles.600-volt.-a. 6. each of which consists of a 12.6 Combining the two parallel circuits. The substation transformers are A-A connected. 40 amp. while the low-tension neutral points. In every case.3 0. as follows: . per cent 71. With all transformers and both lines in operation. the bank was energized from the low-tension side (winding 1) at rated voltage (sinusoidal) and rated frequency.800 22.000-volt winding for this condition. the triple-frequency separate leakage impedances of winding 2 with respect to 3 and of winding 3 with respect to 2 are. generators. per cent Current in closed A 3 The three-phase data given represent the average of several tests.000-volt line. Calculate the third-harmonic current entering the 22.43 Three-phase Tests Voltage across open A Transformer connections Volts Y-A-open. transmission line. the ammeter in each high-tension neutral registers 0. the sending-end impedance of the three conductors in parallel to ground was measured by an impedance bridge and found to be 20 — j240 ohms at 180 cycles.57 8.5 -^2 = „ Q 3 Q 93 - Referred to the high-tension side.A ELECTRIC CIRCUITS— THEORY AND APPLICATION 242 supply power to a 150-kv. are isolated.A 12 Amperes Third harmonic. and a 22-kv. Calculate the third-harmonic current in each 22.65 5. The 150-kv. as well as the neutral points of the is tion with A-A-connected transformers. 2. line.000 86.500 2 3 Ohms resistance referred to high-tension winding 6.6 91.69 99.000. 1. Solution Ratio of transformation between the 22.and the 86.5 33. The neutral points of the transformers are grounded on the high-tension side. The line voltages of the generators are strictly sinusoidal. which completely transposed. terminates at a substaWith the line disconnected from the high-tension buses at power station and substation.18 93 3 Y-open A- 12 Third harmonic.500-volt windings «23 = 86. Transformer Data Winding Voltage 1 3.6 62 0. 4 amp.4 I3 =famp.62 3.22 = = _ = 3 X 0. to be used in calculating the division of the third-harmonic current (Example 3).300 volts.6 X 0.57 +J84.43 +j'240 = 248.43 2 = 240.880 8. 4 Statement of Problem Three 2.7*240)3 X 4 = 240 .000-volt EXAMPLE line. .652 ohms + The division of the third-harmonic current is given nr /// by in — /// Hence . three-circuit transformers are connected A-A-A. 131. 22. third-harmonic current can enter the 22.0 ohms 5._ _ .93 = 16.0.652 4175am .HARMONICS „».4 No 4.72.4 ohms .18 X 093- * 243 _ . Actual third-harmonic current circulating in the 22.j2. X 2. = .43+j 240 ohms 5.-a.„ 7T * 1.j'2.000 volts.6 2 V240.880 ohms . The .43 + .4 ohms 8.913 X 3.93 a/84.5.6 = „m Z = ° X7 = X7 = Zi" Z3" X X 0.= 0. 000-volt windings 1? = 2..\ Equivalent Third— Harmonic Generator —Equivalent circuit per transformer per phase.6..640 = 2.57 2 = 84.„ P3X84.0 ohms Equivalent line-to-ground impedance at 180 cycles per transformer per phase = (20 . Z 71.100-kv. 24° 2 ° hmS .93 3 X 0. nominal voltages of the windings are as follows: Winding 1 Winding 2 Winding 3 2.87+j 84. The equivalent circuit to be used per transformer per phase in determining the division of the third-harmonic current is as shown in Fig.4 ohms -4"0000'00"6 < 1 > ffiWClfo (T „ ^FUO-JMMohm* . 8. 110. 131. Total triple-frequency impedance in the high-tension circuit per transformer per phase = 240 -. Fio.4 .8.175 X 3./240..000 volts. 84 6 ° hmS " - „ .69 X 799533. 3-kv.27 per cent „ = 6. windings. current.00 per cent with respect to winding 3 = 3. 6.100 ^l$.15 6.10 referred to the primary = 2.59 per cent between the 2.10 per cent between the 2.39 - 2.ELECTRIC CIRCUITS— THEORY AND APPLICATION 244 The equivalent short-circuit reactances are 6. 6.15 3.39 = . is = = = 9.32 per cent 3.00 3.10 6.300 volts are impressed on winding 1 and the two other windings carry balanced loads. since changing to triple-frequency reactances merely involves The distribution of third-harmonic currents a factor of 3 which would cancel out in equation The given reactances are Z12 Z23 Z 3i Zi(2) £2(3) Zi( 3 ) The other reactances required = = = = = = (66).27 X X X 3.= _ n1 913 amp.59 = QQ per cent 2.15 per cent between the 22-kv. 3.10 per cent 3. it is found by oscillographic analysis that the primary delta current contains a third-harmonic component of 7.83 „ Zi Zi — 6. Solution is given by equation (66). 6.10 = = = 3.10 per cent winding 3 =3.59 3 Z 2 (i> = Z (2) = Z 3( = 3 i) 6.Zi(i) ZiZi(i) = = = 6.15 6.20 3. reactances at fundamental frequency may be used.00 3.00 3. The reactance The reactance The reactance winding 2 =3.10 3.10 per cent 6.15 per cent 6. / X X X 3. windings. and the 22-kv.10 The 100 per cent current .20 per cent Evaluating the denominators in equation (66) gives ^23^2(1) ^31^3(2) ZijZki) - ^2^2(3) Z.39 per cent in the solution may be calculated from the ones given above.00 per cent 3. Three of the separate leakage reactances are known.96 per cent 10.59 - 6. windings. . and the 110-kv.15 per cent 3. namely.13 per cent 10.32 3. = 6.48 amp.00 per cent 3.55 per cent . giving the latter in per cent of full-load current.15 ~ 6.83 3.10 ! „ Z = = — +6.59 per cent 3.10 + 6.59 6.15 = 0>_ Q 3.59 s 6. Calculate the third-harmonic currents flowing in the secondary and terAlso determine the total third-harmonic magnetizing tiary windings. Since resistances will be neglected.3-kv.30 per cent of winding 1 with respect to of winding 2 with respect to of winding 1 When 2. ___ . and the 110-kv.15 + - 6. wind- ings are >» T /.867 = is.3 iooxM 913 X 0. >„ T /.82 + 0. — X 10. - - 1ft 166am P- .867 X 100X110 2.833 + 0.833 1QA5 = 0. 913 X 0.82 ».96 total third-harmonic magnetizing current t" = . hence.48 m= 0. T 1 ^T 2 /'/' 3 The X 100 _ = QO oi~q = _ QOO per cent 0. the third-harmonic currents become = 7.833 X 2. 2.867 per cent 9.13 " QQfi = 082 X P er ce 0. °' 795 A Q .HARMONICS 245 In percentage of the full-load current.52 per cent The actual third-harmonic currents flowing in the 22- and 110-kv.3 = • - „_ amp . however. . and Do are Q all general case. Co. = C E + D I r a r The written r (1) r (2) Solving these equations 1 with respect to the receiver voltage and current. r. Note use and (4) from equations by equation (26). equations interrelating these quantities may then be E = AoE + BoI 7. AV *9 Network V Ir Fig. p. be expressed in terms of the voltage and current at the other by very simple relations. for instance. and depend (3) s r r a (4) s constants. It has long been known that the simple relations stated above existed in a network for the conditions specified.. Consider the network shown in Fig. 356.. that. Jour. and Sels. H." Elec. I s and I r denote the voltages and . 1921. "Transmission-line Constants and Resonance.CHAPTER IX GENERAL CIRCUIT CONSTANTS — Conception of General Circuit Constants. It is believed. July. the latter are given by E = DoE — BqIs I = -C E + Ao/ A Bo. in the steady state. that Evans and Sels 2 were the first ones to call attention to the great advantage of using general circuit constants in the 1 (2). respectively. 306. In a network conany combination of constant impedances and in which power is supplied at one point and received at another point. 132.. the sisting of voltage and current at each point can." Elec. p. and let E E s. in establishing equations (3) is made of the relation given (1) and 2 Evans. R.. Jour. 246 . complex quantities in the upon the network impedances. 1921. K. D. They are called general circuit constants. August. 132. and "Transmission Lines and Transformers. currents at the sending and receiving ends. The general circuit receiving voltage original constants are then written In many down by inspection. The general constants can always be determined by working out expressions for the sending voltage and current in terms of the circuit and current (or vice versa). may be worked out with comparative ease. however. Lumped Impedance. —Considering the lumped admittance (Fig. 0. making use of the impedances and admittances which make up the network. then the appropriate formula may in many cases be selected directly from the table. are worked out. for this r = (5) r Ir (6) circuit. even the formulas for the particular configuration are not included in the table. * —Lumped impedance. Furthermore. Admittance. 133). the following equations evidently hold E = E + ZI 8 /. configurations if — 1. A = 1. 133. B = C = Z. — Determination of General Circuit Constants. which quite readily can be done as far as transmission circuits are concerned. If once for all. by making proper use of the possibility of series and multiple combination of networks by pre-established formulas. and tabulated for the circuits most frequently met with in practice. the establishment of the expressions for the is both complicated and laborious. Lumped V ?r ks 2.: GENERAL CIRCUIT CONSTANTS solution of when many problems. D= 1 (7) Ir Is A ' <K Es *r X_JL_i_ —Lumped admittance. They 247 are particularly convenient comes to combination of networks and also in the calculation of circle diagrams for graphical solution of transmission it problems. 134). the following equations may E = E Is = YE + s (8) r T be written: Ir (9) . Fig. Hence. The terms in these expressions may then be collected so that equations of the form (1) to (4) result. 134. cases. numerous network general circuit constants formulas. *_ _t Fig. For the lumped impedance (Fig. the general circuit constants for a symmetrical II become If the II-circuit is are equal A = 1 ^ + B = Z note that the constant B is always equal to the architrave impedance of the II-circuit whether the latter is symmetrical or not.ELECTRIC CIRCUITS— THEORY AND APPLICATION 248 Hence. Z Y)E r 1 + (1 + + T-circuit. T-circuit. the general relations — Dissymmetrical may + (Zi Z 2 Y)I be written Z2 + ZiZ 2 F)/ r (15) (16) r Hence. For this circuit. the admittances of the two leaks and may be put equal to F/2. 4. for the dissymmetrical A = (1 r + ZY x )l r (12) t. —In order to cover the general cal II-circuit. as shown (10) 1 case. It is interesting to -^OOOOO z ? i__2 Fig. 135. 136). the dissymmetrical T-circuit first (Fig. Jl-circuit. will first be considered. + ZY )E + ZI + F + ZY Y )E + (1 2 (Fx r 2 1 Hence. — — tL Fig. + ZF + F + ZFiF 1 £ = Z 2 C = Fi (11) r 2 2 Z> 2 = 1 + ZY (13) X symmetrical. D = F. 135. a dissymmetri- in Fig. for the dissymmetrical T. A = 3. Dissymmetrical II-circuit. — Considering E = s Ia (1 + = YE. I. 136. A = 1 C = Y + ZiY B = Z + x D= 1 Z* +ZY 2 + ZxZiY (17) . C = 0. B = 1. the following relations hold: = = E. Thus. As previously demonstrated. + R + icoL. it is of interest to note that the constant C is always equal to the admittance of the T. impedances of the two leakage separate to the branches are equal 5. therefore. the impedances of the two arms The and may be put equal to Z/2. as indicated in Fig. when referred to the same side. as the case may be. therefore. windings respectively. the equivalent circuit of a two-winding transformer is in general a dissymThe impedances of the metrical T. considerable detail. Z = R Z2 = #2 1 1 and should be determined by +jX + JX2 (19) 1 tests (20) which correctly give the Often no such test data are separate leakage impedances. Most frequently. Hence the admittance used is given by impedance v r ~ R = I + R + j(Xx + X (21) v . viz. independent of whether or not the latter is symmetrical. Since Ri and Xi are small compared to R and X this admittance 1 c c) R l c is c c. . constants of the symmetrical T are given D = C = Y general circuit by l + 7Y ~ (18) In connection with the T-circuit. If the transformer is approximately represented by a II-circuit. they are assumed equal available. since upon the voltage. the admittance of the leak is usually taken as the apparent open-circuit admittance of the transformer at rated impressed voltage. each being one-half the equivalent As also previously discussed in of the transformer. — The Transformer. which evi1 See Chap. hence. 1 The general circuit constants of the transformer are. 39 to 42 inclusive. very nearly equal to the correct admittance v = c It is = - R +jX e R +ju>Mn (22) y ) c unnecessary to determine this admittance with very great it varies with the saturation and. pp.GENERAL CIRCUIT CONSTANTS If the T-circuit are equal is 249 symmetrical. 136. II. depends accuracy.. given by equations (17) or (18). the general circuit constants of the transmission A = cosh B C-**» Two Important Relations.e. any symmetrical for this particular circuit. the total admittance is connected at one. Furthermore. the are always equal.250 AND APPLICATION ELECTRIC CIRCUITS— THEORY dently is possible. circuit is reducible to a symmetrical equivalent n. of the terminals and the general circuit constants should then be determined for this As a matter of fact.BC = is the following: 1 (26) Resorting to the dissymmetrical II. the general circuit constants A and Do of the whole system will be unequal in spite of the fact that each of the three parts comprising the system may in itself be a symof the formulas for the II-circuit. A second relation which is generally true AD . i. although the former little is circuit. the relation just stated must hold in the general symmetrical case. It is seen to hold for the transmission line. currents at the terminals of a long transmission line is given by — E = E cosh 6 + I Z sinh I = E ?^-^ + I cosh0 8 r s r r 6 (23) (24) r Hence. On the other hand. equation (13) or (14) should be used for the Sometimes the transformer general circuit constants. where A = D = cosh 0.. for instance. if the sending and receiving transformers are different. metrical circuit. however. which representation difference in practice it is makes very used unless unusual refinements are required. 2 2 (27) (28) . a transmission line with identical sending and receiving transformers will also have equal A and D constants. Forming the products involved by using equation (13) gives AD = (1 BC = Z(Fi + + ZYJ = 1 + ZYt + ZY + Z*YxY + Y + ZYtY = ZYx + ZY 2 -f Z YtY ZY*){\ 2 2 2 2) Subtraction immediately gives equation (26). The relation between the voltages and 6. Transmission Line. repre- is sented by a cantilever circuit. Inspection shows it to be true Since. general circuit constants A B = Z D= D (25) coshe —When a and line are sinh circuit is symmetrical. equation (26) may be proved for the general case. T representation of the trans- the logical aDd fundamentally correct one. . in Fig. Below. Bz. C<l. + DJ> )/. Two Networks —Two networks in in Series. however. C\. —Let the separate networks shown Bh circuit constants A\. provided power is supplied or received at these points only.)I + D C )E + (CiJ5. a few cases of series and multiple networks will be It should be carefully noted that in determining the general circuit constants of a series network the order of the series treated. D x and following equations hold: E = AiEm + Bil„ (29) — C\Em + D\I„ E m = AlE + B^Ir Im = C<zE 4" Dil (31) s Is (30) r r (32) r Substituting equations (31) and (32) in equations (29) and (30) and collecting terms gives E = s Is = (AiA. not be changed. There are but few cases. 251 —Any number of networks connected in series or multiple may be represented by a single set of general circuit constants between any two terminal points. unless the effect of power supplied or received at the intermediate points can be taken into account by the insertion of a constant impedance at these points. where this can be done. representation as stated is not possible.GENERAL CIRCUIT CONSTANTS Networks in Series and Multiple. 1. the general + BxC^Er + (AA + B D.Z>iZ) 2 2 (35) . Neiwork Is A. 137. If power is supplied or received at other points. r 1 1 2 2 r two networks circuit constants of (33) (34) in series are A = A A* + B C Bo = A B % + 5iZ>2 X X 2 X Co Do = Cx4 2 + £iC = Ci# 2 4. since the equivalent impedances representing generating stations or loads in general would depend on the voltage. (CiA t Hence. sections must. 137 have the general A2 Z> 2 .*BfCt > Neiwork 2 t Im Dj Fig. Er series. in general. The Ir A2 >B2 C2 >D2 > Em E. in Parallel. 138.£ 2 Z) + ZMCgBs + D 2 Z) 3 ) A = 3 Bo s Co Do X 2 2 3 Aj'Bj* Cj> Dj hz > Network 2 AZ >B2 .B {C 2A 3 4.(A*Az +B 2 + CS) = Is 2 3) 4- +D B 1 2 2 3 + D D )]I )]E + B + D D )]I 2 C 3 ))E r (C 2 B 3 4- + D!(C A + D C + B D 4.D 2 D ) = Cx(A 2 A 4. the general BiiCtAz BD [A 1 (A 2 B 3 4- 2 3 2J 3) 2 3 r (44) 2 3 r (45) r 3 circuit constants of three networks in series are A±(A 2 A 4.B 2 C 3 ) [C x (AzB 3 Hence. — Considering the two networks in Fig.D 2 C = Ai(A 2 B + B 2 D 3 ) 4.Di(C [Cx(A 2 A 3 4.D2 shown Jn Ir Fig. 139. the following equations down immediately. Fig. 139. 2. n AzE + B I C E 4~ Dzl 2 (38) 2 (39) 2 3 r 3 r r (40) r (41) Network 3 Network 2 J Im Aj> Bj >Cj> Dj In — A2 >B2 C2 >D2 A3 B3 * => > P The Er.- Substituting equations (40) and (41) in equations (38) and (39) gives Em = = Im +B (A 2 A 3 (C 2A 3 4- 2 r 4- (A 2 B 3 )E r + (C 2 B 3 C )E 3 DC 2 3 + B D )I + D D )I 2 2 3 3 (42) r (43) r Substituting equations (42) and (43) in equations (36) and (37) and collecting terms gives E = s [A. immediately be written down Three Networks in Series.5i(C 5s 4. may be . following equations may E = A\Em + a = Em = Im = En — In = C\Em Is Network 4~ B\I m Dil m (36) (37) A En + B I n C E n 4~ L><J.ZMtVU + Z> 2 C = d(A 2 £ 3 4. as 3) 2 y hz —Two networks in parallel. Two Networks f fy £ written (46) 3) 3) &s 3. Cg .B C 4. t C^» Ir D3 Er — Three networks in series. 3 3) Network in parallel.— : ELECTRIC CIRCUITS— THEORY AND APPLICATION 252 — Figure 138 shows this case.B C 3 ) 4. GENERAL CIRCUIT CONSTANTS 253 E = A E + BJri E = A E + BJ r2 s x r (47) s 2 r (48) = C.DO E + + 52 B D 2 + Dx£ B +£ )(D 2 r t 2 (57) 2 x From g -"2 -D 2 equations (54) and (57). - from equation (A l -A £i 2 i.(!) '" Dl) (56) r (54) and contracting gives . + £>2 r ~ DlE. upon contraction. combined with equations (52) and (48) gives 7.i + 7 s2 = I rl + 7 r2 /. (50).i 7*2 78 7r r (49) r (50) (51) (52) Substituting equation (52) in equation (47) gives E = A. the general circuit constants A Ao — = A m C = 2C — = ^ g 2 Do = D Bo become . the general circuit constants for in parallel are obtained as two networks A A = iB + BiA* 2 fii B = +B 2 B\B 2 B +B x 2 (Ax + c2 + 5xD 2 Do = + Dx5 2 -+- #2 Co = C\ Bi It is useful to -A Bx 2 +B (58) . From *+ ^b'+IT equations (49).DO ){D 2 2 note that. when the two networks operated in parallel are identical.E + Z)x/ rl = C 2 E + D 27 r2 = /. = -[ Ci + C2 r + DJ + Z£. . and (51) + = (d ls S- 4' = *• C 2 )E r is (54) obtained + DJn + DJ (55) r2 which. (58a) . = (d + C 2 )£ Substituting for 7.E + Bdr ~ BJ r2 (53) r a Eliminating I r2 between equations (48) and (53) gives. = (d + C2 + Cz)E r BiBs 2 Z is obtained + Dllrl + DJri + 2>3/r3 (69) (70) . (63). = AxEr The ht + BJr - ^E + e -D2 — " -Mir T Bi solution of this equation with respect to _ AiB 2 B 3 + BiA^Bs + BrBzAs ~ BiB 2 + B 2 B + B. Network Is.C3 >D3 — Three networks in parallel. (64).B. Three Networks in Parallel. The general circuit constants of tion can be obtained from the following relations: this combina- E = A E + BJ rl E. Bj> C. the constants A and D are equal to.ELECTRIC CIRCUITS— THEORY AND APPLICATION 254 Hence.Dj Er Network 3 ^ Ir3 A 3 >B3 ..- 4. Fig. and Co twice the corresponding constants of each network separately. Dj > Network 2 A Es hz lr Irz Az'ByCj. 140. the constant B onehalf. —Figure 140 shows three net- works in parallel. Aj I lr. = A E + Bdri E. = A E + BzIrZ / s 1 r (59) 2 r (60) 3 r (61) = CxE r + DJn = CiJE + T)<J-r1 — C%E + D^I z = Isl ~f~ IsZ + Is3 = Irl -\~ Ir2 I rZ Itl Is2 (62) (63) r ISZ Is (64) r r (65) (66) ~\~ lr Substituting equation (66) in equation (59) gives E = AiE + BJ r r s Bilri ~ B X I rZ (67) Eliminating 7 r2 and I r s by making use of equations (60) and (61) gives E. T + IHr&s Bz E s B3 -E r (68) is„ ~*~ 3 B\BiB% BtB* From +BB + equations (62). and (65) J. -A.BiB% -\. Z)i Bz Bi g< (66).BiBz l l -D )B 9 2 D\B%Bz + D^BiBz -f. + DsBiB-. an equally simple relation holds as in the case of two It may now be inferred that.)(D.B1B3 B1B2B3 D _ B\B% -\. B1B2 + B2B3 -\.^ (D. becomes = /.B\Bz -\- \Er + (72) Inspection of equations (69) and (72) gives the general circuit constants of three parallel networks.B^Bz + BiBz the three networks are identical.)Bi + (A. if n parallel identical networks. = [Ci + C + Cz + -D )B. (A 1 -A i )(D i D = If -D 1 (73) )B 3 + (A i -Ai)(Di -D 2 )Bi + (A t -A )(D B\B2 -\.)B 2 (A 1 -A. (60). the general circuit constants would be A =A Bo -(736) Co = nC Do = D .B2B3 -f. the general circuit constants are given by A =A Co Bo = g (73o) = 3C Do = D Hence. + D 2 BiB.-D. as • AxBiP* + BtAzBs + BxBtA. It B1B2 -\. identical networks are operated in parallel.B1B3 Co = Ci + Ci + Ci + A = .DzB\B% B1B2 -\.DQA. D2 ~ r - P.)(D t l B1B2 + BiB% - t B\Bz DiBtB. + (A.GENERAL CIRCUIT CONSTANTS when combined with equations which. ^ - (Pi - _ 255 and PQ^ (61) (71) Bz Substituting for E.B2B3 -f. + (Ci C2 + Cz)E r + DJ + . the following expression results: /. from equation (69) and reducing.~Ai)(Di-D. . Line J*r A>B>C>D Fig. subscripts r receiving transformers. and line with both sending and circuit The general circuit constants for these be given below. 141 shows this circuit.5 per cent. Figure — Transformer .256 ELECTRIC CIRCUITS— THEORY Combination of Networks Usually mission. 1 and By Transformers repre- using the formulas previously given for the combination of two networks in the general circuit constants for the combination become Evans and Sels in their paper "Transmission Lines and Transformers. 1. The transformers will be replaced by their symmetrical equivalent T-circuits. line with receiving transformers. however. C. Formulas worked out on this basis are also given. 141." give a very complete table of formal expressions for the general circuit constants of various networks and network combinations.. cit.. . B. T-circuit. sented hv by ermivftlent. The error introduced by the use of the simplified formulas is rather insignificant 2 and considerable time and labor are saved in the computations. two or more equal lines in parallel. are greatly simplified when the transformers are represented by their approximate "cantilever" circuits. The formal expressions for the general circuit constants. Transmission Line with Sending Transformers (Exact). The combinations which most frequently present themselves for solution are line with sending transformers. loc. 2 Evans and Sels in their paper. cit.. with this circuit is about 0. 1 —The transmission AND APPLICATION Met With in Power Trans- proper consists of lines and transformers. equivalent T-p. A. will used. two or more unequal lines in parallel or a line itself. Transformers Represented by Equivalent T-circuits.irr>iiit. Subscripts s attached to the transformer impedances and admittances signify sending transformers. and D will represent the general circuit constants of the whether the latter consists of a single line. the leakage impedance of the two windings being considered equal. sectionalized line with one or more sections out of operation. Exact values of the general circuit constants of the combinations are obtained when this transformer circuit is receiving transformers. state that the approximate error loc. — Transmission line with sending transformers. for the transformer series. This circuit is shown in Fig. Transformers repre- — 2.GENERAL CIRCUIT CONSTANTS 257 ^\ + CZjfl + ^Y±\ = B(l + ?£=S + Wl + ^\ A = A(l + B (74) Co = c(l + ^~\ + AK. Trans- formulas for three networks in series the general circuit constants oi this case become . the general circuit constants are obtained as A = A^l + Bo = B^l ^j + BY r + ZYr^ + AZ. —Transmission line with sending and receiving transformers. Er Zr Zr r —Transmission 2 line with receiving transformers. formers represented by equivalent T-circuits. + Transformer Line A'£'C>D 2 Fig. By a process as in the preceding case. 142. Co = c(l + Do = I)(l ^) + DY r + ^pj + CZ {\ + ^pj T Transmission Line with Sending and Receiving Transformers Figure 143 shows this circuit. "Do = D(l ^) + BY. By inserting in the (Exact).(l + ^i) (75) 3. 142. Transmission Line with Receiving Transformers (Exact). — Transformer Line Transformer A>B>CD Fig. 143. sented by equivalent T-circuit. i.A.258 ELECTRIC CIRCUITS— THEORY AND APPLICATION ^)K i+ ^) + H + Ao= i+ ( z { i + ^rXK ^(l + ^r)[cZr(l (l D = F [AZ 8 r + 1 ^) + DYr Mr)] +4f) + D(l + + ^)[c(l + + ^r)] + + ^f) + + ^)[^(i + ^) + K J8(l (76) ^) + DYr (l (1 ] 1 + ^) above combinations. in the rical.. Thus. the constants A and D are equal and the formulas given may be somewhat simplified. consists of (76) may be written A = A[(l + ?01 + ^r) + F BY (l + ^f) + r CZ. for a symmetrical line. the transmission line is symmeta single line only or identical lines in parallel.(l + ?£)] + + ^)(l + ^r) [(77) (?„ = c(i + ^)(i + ^) + .jr.e. Z {l r )] +57x + ^)] + .(l r Z. equations If. (1 y'( i+ Do = a[(i + ^)(l + ^) + + ^f) + ¥ Y. Y. Jr Line A>B>C>D Fig. Transformers represented by cantilever circuit (approximate). 259 in addition to the lines being symmetrical. Transmission Line with Sending Transformers (Approximate). 145. 144. Do = D(l + Z. 145. Using the formulas for two networks in series the general circuit constants become A = A + CZ. 1. —Transmission line with sending transformers. — general circuit constants are . r . equations (77) reduce to !+ Cz4l + ^pj \ (78) 2 Co = c(l + ^) + 2AY(l + ~^j + 5F 2 Transformers Represented by Cantilever Circuits. 144 shows the approximate circuit. —Transmission line with receiving transformers. Bo = B + DZ. Figure — Transformer . The mate). Transmission Line with Receiving Transformers (ApproxiThe approximate circuit is shown in Fig. = C(l + Z. 2. Co 6 (79) t Transformer Z/>?e A'B'C>D Fig.Y ) + AY.GENERAL CIRCUIT CONSTANTS ' If. the sending and receiving transformers are identical. Transformers represented by cantilever circuit (approximate).) + BY. F. the transmission line D = A[(l + ZrYr) + Z8 Yr + BYr + CZ.) + Zr Y + 5F. formers represented by cantilever circuits (approximate). the above formulas in this case. . + CZ (X + Z.F. .) + D(l + Z. the general A = A(l + ZrY + BY + CZ. .)(1 + ^F + r r T) Trans- r a r) ) t [ (81) DF (l + Z. A2Z 2 2 2 (83).) Ao A ] Bo (82) 8] If.: ELECTRIC CIRCUITS— THEORY AND APPLICATION 260 A = A(l+ Z = B + AZr r Yr) + BY r Bo = C(l + Z Y Do = D + CZr Co r r) (80) + BY r Transmission Line with Sending and Receiving Transformers (Approximate).F. the sending and transformers are identical. Figure 146 shows the approximate circuit. + Zr ) +B + CZs Zr Co = A[Ye (1 + ZrYr) + Fr (l + Z 8 YS )] + BYs Yr + C(1 + Z. — Transformer r J-S E.Y. Using the formulas for three networks in circuit constants are obtained as series. Transformer .1 Fig. as follows receiving = Bo = Co = D = Ao + 2ZF) + BY + CZ(l + ZY) + B + CZ A2Y(1 + ZY) + BY + C(l + ZF) A(l + 2ZF) + BY + CZ(1 + ZF) AQ.ocxto<-\ .)(1 + Zr Fr ) Do = A[(l + Z8 F.Y r t) 8) is symmetrical so that reduce to are equal. the expressions for the general circuit constants may be simplified still more. in addition to the line's being symmetrical. and t + BF. ncrcior\ Line ? A'B'CjD . TJ — Transmission line with sending and receiving transformers.(1 + Z Y ) + DZ.il + Zr Yr ) =A(Z. 3. Co =AY (l + Zr Fr + BY„Yr + C(l + Z.Yr BQ = AZr + B + CZ Zr + DZ. + CZr (l + Z. 146.Y r = AZ r Y Do If. 000-kv. A.R.01352) 10" 8 = (0. (per circuit) 0.1)10.3 /84°. 90 Miles E.102 X 412\4°.1 +j377 X 0.1 X 10-8/88^88 10.982/0^21 B = Z sinh 6 = 77. 1). three-phase.1892/84^32 hyp = 0. bank of receiv- cuit. 147. 147).1 X 10-« mho/mile 0.1879/84^39 cosh 6 = 0.01352 X 10-« farad/mile 0. the constants will be calculated for one circuit in connection with one-half the transformer capacity. Constants of line: 90 miles.S. are given below.154 +.154 ohm/mile 0.55 5.Viy = Z = Y? = 2.853 ohm/mile 0. Fig..-a. 220 k v. mils./0.GENERAL CIRCUIT CONSTANTS EXAMPLE 261 1 Statement of Problem Calculate the general circuit constants of a 90-mile.-a. 60-cycle w L/'nes . two-cirtransmission line with a 150.C. — diagram Single-line two of parallel three-phase transmission lines with receiving transformers (Example The constants ing transformers (see Fig. %. / 60 cycles T Er OOOkv-cu of Transformers ISO.000 kv.1 j5.853 = 0. of line and transformers Solution Since the two transmission circuits are symmetrical. 5/79^84 c = fE^e = 000456/88 o 94 6 sinh 6 Receiving transformers: Rating = 75.5 per cent Magnetizing current = 5 per cent R = X mho /mile .e + = a . 0.32 hyp/mile ohms = cd =90 X a = 0.5 per cent = 12 per cent Iron loss =0. 605. 220-kv.866/79^77 ohm/mile (0.982/0^21 A = D = cosh 6 = 0.000 R = X = G = C = 2 = y = cir.6 per cent Copper loss = 0. 84 X 7.8 /4°.985 0.000118 = 1. ohms 196.9 0.22 0.2 2 /220.000456/88^94 X 77.5/87^61 ohms = 7<MXX)>^(rox 0.00302/0^01 + 0.61 ZrYr = = 1 + ^p 1 + 302.982/0°.0030 2 /0°.71 0. 77.00604 0.2/87^82 + J76.00151/0° + 77.X 10.75 (0.00302 /0° .0353 + .837 + 7.30 2.000235 = + jO.0.77. 220 220.94j97 + i0.80 X 10-\83°15 = = = Do = 45.2 X X 7.982/0^21 X 77.000 = 10A0 196.15 .777 +i38.0.223 + j*77.000350 + + X 0.000235 1.95 5 (1.15 mho General circuit constants: Equations (75) will be used.00151/0° = = = + + jO.00151/0° = = = Co = + 76.985 /0°.5 + 2.00591 0.4 .00151 A = 0.00302/0° .4 __ = = .5 /79°.985 ^0.00605\3°.005 S= V3 X ' X e = Zr = Gr X 220.000456/88°.65 X 10—\82°.61 X 1.31 = = Bo = = 10-\83°.0354/176^55 0.5 /88 o .000 X 0.80 X 10.5/87^61 X 1.16)10" 0.B \83°.01 + 0.895 + J76.223 = .6 77.75 X 10.000/V3 9.mho 1Q _6 mho - j7.75/79^85 13.75)10-« = X 7.01 1. 77.006 ^^..985/0^22 + 0.982/0^21 X 1.982/0^21 X 7.80 10.9 amp. 5/79°.000 j- Q OOQ 3.ELECTRIC CIRCUITS— THEORY AND APPLICATION 262 /T _ — 75.5 /87°.12 V3 X 196.00378 0.00302 +J0.605 + J152.1 16.00302/0°.15 = 1 + 0.94 -.00378 0.000V 3 V Vs ^ x X 10.46 1.99104 .Q1 + 0.940 +i45.5 /87°.00302 +^0.21 X 1.9 3./0.93 X^ 196.80 10-\83°. ohms 77.00213 .84 X 1.94 X 1.= 6 = 196^X005 = V3 X Yr = - (0.59)10+0.00343 + 77. V3 X „ = R.9 1Q _6 = 2 2 220. >.01 - 0. Finally. The general circuit constants for each of the five series sections of the network are first written down. ers are identical and have Z ohms impedance of Z ohms and short-circuit Each excitation admittance per phase. -Next. the general circuit constants of the transmission line alone are calculated by means of equations (46) for three networks in series. the formal expressions for the general circuit constants of the entire system have been obtained by applying equations (76) for a transmission line with sending and receiving transformers. 148. H sectionalizing stations at points tance from the sending end. one-third and two-thirds of the transmission distance from the sending end (Example 2).GENERAL CIRCUIT CONSTANTS EXAMPLE 263 2 Statement of Problem The transmission 148 consists of two three-phase circuits with and of the transmission disB and C. tension bus of the receiving transformers one circuit of section when the line is operating with CD disconnected. —Single-line diagram I i =HH of a two-circuit three-phase sectionalized trans- mission line with sending and receiving transformers. Solution The necessary combination of circuit constants is carried through in Table XI. % impedance and Y mhos section of the line has a surge of hyperbolic angle of 6 radians per circuit per phase. The sectionalizing stations are located at points B and C. The banks of sending and receiving transform- line in Fig. . Work out formal expressions for the general circuit constants for the system between the low-tension bus of the sending transformers and the low- <s>Hfr Fig. i?o + 2 1 + '(-¥) + 0)]+ z(l + 2 cosh 2 Y Bo cosh 0(cosh 2 (cosh 2 cosh + + cosh 0(cosh 2 2 cosh 2 2 0) 2 0) 1 + 1 2 sinh 2 0) + YZo + 5 C ° 8h2 6) + 1 +~\\z(l +^f) cosh0(cosh + 2 sinh 0) + (l +-~^\Zo sinh (sinV 0+2 cosh 0)1 + z(\ + ^) [z(l + ~\ —£-.ELECTRIC CIRCUITS— THEORY AND APPLICATION 264 Table XI Sending General Section Section 2 1 T^-oT constants 1 ZY + Za sinh . cosh + Zo sinh 0(sinh 2 sinh i + cosh 0(cosh 2 2 zr> 2 Z_F> '(•+¥) 4 8 ~~Zo~ cosh 9 2T ZY + sinh 6 Z„ cosh 1 Za sinh 2 sinh Zo zr cosh 9 sinh 9 2 2 sinh 9 i «—oA/UVJyf&WV-o IH~ cosh cosh 2 '(+¥) 4 Ao Receiving transformers Section 3 transformers circuit y[" 0) z(l + 2 cosh 2 (l 2 + -^\ cosh 0(cosh 2 cosh 0(cosh 2 (l + 2 sinh e + ^P) ^r^ 2 0) (sinh 2 + 5 sinh 2 0) + YZo sinh + 5 cosh 2 1 0(sinh 2 0) + I" + 2 cosh 2 0) cosh 0(cosh 2 "1 + + I ^) cosh 0(cosh 0)] (l+^f) + + 2 2 sinh 2 0) + (l + ~}z (l+^)[z(l+^) -~-^ (sinh cosh 0(cosh 8 2 + 5 sinh 2 0) 1 2 sinh 0(sinh 2 + 5 cosh 2 0) + + .(sinh + (l 2 5 cosh 2 0) + Y[l+ -^\ (l + ^~\ [ 5 sinh 2 Do : ZY + 2 2 Co ZF sinh 0(sinh 2 + ? (8inh2 ' 0) + <+?) 0) ^)[ (l + ¥)~iV 5 sinh 2 i 2 sinh' 9) + 5 cosh + 5 sinh (sinh 2 Z ZY + ^r)f (l + ^P) (l 0. It is called but reasonable to expect that graphical methods. the analytical methods of no longer suffice. Without exaggeration it may be said that graphical methods constitute one of the electrical engineer's most important and powerful tools for handling many of the problems which he is upon to solve today. If the diagrams and charts involved are constructed and used with care. developed to meet the new demands. fundamentally. however. they are all modifications of the well-known vector diagram which readily can be drawn for the electrical quantities at the sending and receiving end and have. The graphical solutions may be made as accurate as desired.increase as the art progresses. In the following will be discussed: 265 .CHAPTER X TRANSMISSION-LINE CHARTS In modern transmission engineering. its performance may also be determined graphically by a circle diagram. been known for a long time. It is interesting to note that. In such cases. the importance of graphical methods will . Whenever the relations between the electrical quantities at the terminals of a network or a system are expressible in terms of general circuit constants. There are several such diagrams available. or in analyzing the behavior of precision of results fully sufficient for engineering purposes may easily be obtained*. at least in principle. will play an even greater part in solving the still more advanced problems which are bound to confront the profession in the near future. a istics of design. so many results are usually required that analytical solutions become prohibitive on account of the excessive amount of time and labor involved. therefore. it becomes necessary to resort to graphical methods. In studying the electrical character- solution a projected transmission line preliminary to the actual an interconnected transmission system. many of the problems encountered today are so complex that analytical methods often fail because they prove fundamentally inadequate and hence incapable of handling the situation. Furthermore. No doubt. and of current. 475. + A I constants may conveniently J. Splitting the receiving-end current Ir into an in-phase and a quadrature component. A diagram of this type for the transmission line alone is discussed in the paper "A Graphic Method for the Exact Solution of Transmission lines.I. Vol. 515. 1 Let the relations between sending and receiving voltages and currents of the network for which a circle diagram is desired be given by Circle E = AoEr + B t Ir (l) = CoEr + D<>Ir Er = DoE. and power factor at both ends. diagrammes et regulation des lignes de transport d'energie a longue distance. The last-mentioned one is probably the most convenient chart 1. using the receiver voltage Er as standard phase. Thielemans in a series of articles "Calculs. it also gives the effi1 circle ciency of transmission. H. pp. Irp and Irq respectively. 599. IX. however. VII. A much more complete circle diagram of this type is described by L. which is described in this chapter. 3.E. A. as follows: A = B = \A \/a = Do = \Co\h Co \B a \/§ \D (5) \/8 Figure 149 shows the vector diagram corresponding to equaand (2). The modified Evans and Sels chart. 403. 2. — Diagram Based Directly on Vector Diagram. is more uniand somewhat superior to the Thielemans chart for most Sels chart. p.B QI L = -CoE. Suffice it to say that it is an excellent chart. Thielemans' chart assumes constant voltage at one end of the 929.. suitable for a variety of transmission problems. for use in transmission problems. 1921." by C. 878." Revue Generate de I'Electricite. Vol. .E. Holladay. 451. In addition. that the modified Evans and purposes. (2) (3) S (4) t The general circuit be expressed in polar form. 1922. 675.ELECTRIC CIRCUITS— THEORY AND APPLICATION 266 The circle diagram directly based on the vector diagram. The author believes. The no-load sending voltage AJEr is displaced by an angle a from the receiver voltage. pp. 435. Space prevents a detailed discussion of the Thielemans chart in this treatise. Trans. . versally applicable . 785. system. power. a being equal to the angle of the general tions (1) circuit constant A . 1920. It contains loci of the voltage at the other end. The Evans and Sels chart. power scale ds obvious since power is voltage times current. . 2. from the unity power-factor <f>r. laid off the drops. respectively. the sending voltage vector is located as indicated in the figure. due to the total receiver current. in kilowatts The possibility of establishing also the the terminus of the sending voltage or operating point corre- . the power scales are the important ones and the others are usually omitted. the sending active and reactive power corresponding to any load condition at the receiver end. Receiver circle diagram based directly on vector diagram.: TRANSMISSION-LINE CHARTS the component drops BoI rp and 267 BJrq may be laid off as shown. Now assume that the receiver voltage E r is constant. point 0. sending current. The drop B It. in volts. sending power factor. As a matter of fact. Along the lines OL and MN three scales may be constructed 1. Scale of in-phase and quadrature current. This diagram is valid at one definite value of receiver voltage and may be used to determine the sending voltage. and. 3. Using the power scales. This immediately fixes the size and direction of A E r and hence the of Constant Sending Voftage Circles «/ of Constant Sending Current Circles — Fig. and the direction of the drop BoIrq is 90 deg. in amperes. 149. displaced from the former or from the unity power-factor line. is displaced an angle the receiver power-factor angle. Having line. Scale of voltage drop due to in-phase and quadrature cur- rent. at the and kilovolt-amperes. BJrv due to the in-phase Evidently. the direction of the drop current is determined by the angle /? of the general circuit con- stant Bo. hence. respectively. Scale of active and reactive power. receiver. respectively. and the receiver voltage is constant. also. the no-load sending current. one scale can be fixed arbitrarily. Thus. A sufficient number of these circles may be drawn so that the values of the sending voltage can be easily read. The sending current can also be determined graphically. a scale properly reduced but still reading kilowatts and reactive kilovolt-amperes received may be laid off along the lines GH and JK. giving the direcof the general circuit constant Cocurrent Irp due to the in-phase sending of the part tion of the CoE r is D component . scale.268 ELECTRIC CIRCUITS— THEORY AND APPLICATION sponding to any value of active and reactive power received (or to active power and power factor) may immediately be located. by the angle 8. is displaced the receiver power-factor angle 4> r from this same line. the sending voltage for any operating condition at the The loci of constant sending receiving end can be determined. and may be drawn in. and power are interdependent. voltage are circles with the point F as center. where <f>a is the diagram. the others. the angle The sending component of the load current. Ir . respectively. from . The loci of constant sending current are circles about the point F. The terminus of the sending-current vector can then immediately be The active located when the receiver conditions are known. <t> In constructing the diagram. must be calculated on the basis of the chosen Usually it is most convenient to select the power scale. The scales of voltage. is located D GH and the sending current D of the general circuit constant D Irq is displaced 90 deg. This vector is laid off from the point F (or if desired in a separate diagram) its size being fixed by the receiver voltage and its direction by the angle y. The following equaparticularly if coordinated paper is used. however. <f> s $ 8 (6) (7) the sending power-factor angle which is scaled from The accuracy of the sending power determined in mainly on the precision with which the depends manner this angle 8 can be read. due to the total load current. scales tions give the relation between the . the angle The line GH. current. In order rapidly to determine the sending current corresponding to given operating conditions at the receiver end as specified by active and reactive power received. by calculated and reactive power at the sending end may now be P = EJs cos = EJS sin a Q. so that the active and reactive power at the receiver end may be calculated by The and circle (2).TRANSMISSION-LINE CHARTS Power scale Voltage scale Current scale = p watts per inch = ^r Mir = 269 9 volts per inch (8) £. presumably. per inch Mr MN will be located If coordinated paper is used for the diagram chosen.amp. This diagram will have the sending voltage Es as reference vector and will give the receiver voltage corresponding to any value of active and It may also give the receiver reactive power at the sending end. current and power-factor angle. receiving power factor. a circle diagram may be constructed on the basis of equations (3) and (4). the receiving active and reactive power corresponding to any condition of power and reactive power at the sending end. also. diagram is Figure 150 shows this diagram which. does not need a detailed description. It is along one of the coordinate lines and the point convenient then to locate the point F by determining the direction by means of the angle of the vector A E r relative to the line Xr which is given by MN \r = + 90 a - (9) diagram so far discussed was based on equations (1) In identically the same way. valid at one definite value of sending voltage and may be used to determine the receiving voltage. Pr = E Ir Qr = ErI r r cos Sin (10) <j> r (11) <j>r I Circles of Constant -p-Receiver Voltage •Circles of Co Receiver Current —Sending This circle diagram based directly on vector diagram. Fig. 150. and. hence. . receiving current. since the principles involved are exactly the same as in the preceding one. It is possible. the operating point in the sending chart can be determined as follows: First. Since now the receiver voltage is known. i. Hence.ELECTRIC CIRCUITS— THEORY AND APPLICATION 270 The scales are determined scale = p watts per inch Voltage scale = —^ volts per inch Current scale = 4^- Power The angle X„ which may structing the diagram. Sels chart. There are problems. this operating point must lie on the circle representing the proper receiver voltage. evidently also be used in transferring a point from the sending to the receiving chart. and particularly not where corresponding values of sending and receiver power have to be determined at a variety of terminal voltages. subject to the same limitation as in the former. In their present shape. applicable without the strictly imposed condition that one of the voltages must be fixed.e. amp.. provided the latter is constructed for exactly the sending voltage necessary to produce the given receiver conditions. however. If this is the case. by introducing variable scales to make these diagrams universal. . . If by (13) in. this angle may be read in the receiving chart and used in the sending chart so that the direction of the receiver voltage is determined. but its application in this case is. It all depends upon what the conditions are and what results are desired. the diagrams so far discussed are not altogether suitable where power conditions at both ends of the system are required. however. the angular displacement between the sending and receiving voltage must be the same in the two diagrams. for which these diagrams are entirely satisfactory. the then the corresponding sending operating point may be found in the sending diagram (Fig. Second. This will not be discussed here but will be taken up in connection with the Evans and (Fig. 150). the sending operating point is immediately located and the The same scheme may active and reactive power determined. of course. both in magnitude and direction. for instance. per inch conveniently be = 90 made use of in con- +5- an operating point has been located receiver diagram i given by is X. 149). 151 from which the following values are obtained: 1 These tables will be found in "Electrical Characteristics of Transmisby the Westinghouse Electric and Manufacturing sion Circuits.38 X 10"< = 5.00923 Scales for the graphical solution 1 in. Vr (l given + ~) + IbZ = A V + Bo/r (a) r — = l+ l+ YZ Ao = is 5.0 deg.5 ohms Y = J5.38 X 10.0 /66°. with 9-ft.4 /90^ mhos The constants The by of the nominal between the receiving-end and sending-end voltage relation = V.) J at the load is to be maintained = 66. 2.342 ohm = 0." published Company. Determine the per cent regulation for a load of 4. 100 miles long.5 X =0 . 80. mph = 2478in 1. has the following constants per wire mile: r x y = 0.. cent power factor.-a. Statement of Problem A three-phase.9788 +J0. February.38 H = _= 10~V90° == X 86. 60-cycle transmission line.979 /0°.000 volts between lines. (total) 2 = X 10* watts per inch . f Per phase p = i I V 2. spacing between solid conductors of No.90 o^x^ooo V3 X + 0. Solution 1.788 ohm = J5. = 2X10«X . __ K 1 ' fi . . n are Z = 34. Total power and line voltage are indicated on the graphical solution of Fig.- 86.000 — V~r / V 1 + = 10 3 ^ = PZ = I X yz) 2. .2 + j/8. Let the total load vary from to 10. 1922.000 kw.500 kv.505 66.TRANSMISSION-LINE CHARTS EXAMPLE 271 1 This example illustrates the use of the receiver diagram just described It will be noted that all in the solution of a simple transmission problem.8 = 86. Obtaining data from a simple graphical solution based on the nominal n line.5 = 24.5 0. plot sending-end voltage versus received load for 100. and 70 per cent power factor (lagging).38 X 10 -6 mho } \ From Westinghouse Tables 1 (at 25°C.5 - . ? 5 Volt8 per .0/66°.0 X V3 ~~TX66. at 80 per The voltage 1. 90. as Reprint 82.000 Vr= 2 / Angle X r . lines in the diagram superfluous for this particular problem have been omitted. 000 wire.000 kw. 151.000. Receiver circle diagram for the 100-mile transmission line considered in Example 1. 84.000 \<$ ^ 74. The solution is based on the nominal II.000 Syr iX As « "^000 J" 76. 66.000 % 72. of power The curves .000 80.000 volts between lines.000 UnjiHij i> & 64. 2000 4000 6000 - 8000 10000 Received Power in K.W. 152. —'100-mile transmission line (Example Receiver voltage equal to 2).000 66. Receiver voltage equal to 66.000 1 70.000 82. Fig.000 w 68.272 ELECTRIC CIRCUITS— THEORY AND APPLICATION Received Power in kw. — Fig.000 volts. give sending voltage versus power received for values factors ranging from 100 per cent to 70 per cent lagging. New York. K.650 67.250 79. Inc. Regulation for a load of 4. at 80 per cent power-factor.-a. Sels. B. In this article the calculations of the data necessary for the construction of the charts were systematized an excellent manner.000 66 000 rises to 71. Dwight's " Constant-voltage Transmission." John Wiley Sons. 1 t. 530.— This chart was presented by Evans and Sels in 1921. it may appear to differ materially from those previously described. l Although.650 66. R. general circuit constants.450 70. kilowatts 100 per cent 90 per cent 80 per cent 70 per cent power factor power factor power factor power factor 64. Jour..000 10.900 76.100 71. D.400 Figure 152 shows the curves of sending voltage versus received power for the desired values of power-factor.650 65. 1915 .66.350 volts .250 64.. It embodies. December. .850 volts (obtained from the 80 per cent powerfactor curve at 3.000 8. in All chart constants were given in terms of the A voltage-power circle diagram of this type is also described in H. however. E = a A<>Er +BI r = CqEt D Ir Er = DoEa — Bolg Ir = —CoEs + AqI8 Ia -\- (14) (15) (16) (17) 1 Evans.) When the load is removed..979 71.. certain added features such as loci of constant line loss and constant efficiency of transmission. —Let the general equations for the system be Voltage Circles.150 71.450 70. and H.850 2.000 64. .000 4.500 kv. "Circle Diagrams for Transmission Systems.600 kw. p. 1921.350 . Regulation = + YZ = 69. at first glance.300 76.850 0.750 82.850 69.650 68.100 69. Sending-end voltage = 69.500 73.700 66.400 73.200 78. it is fundamentally a vector diagram.850 68.TRANSMISSION-LINE CHARTS 273 Sending-end voltage (volts between lines) Total load.300 64. the receiver voltage = V'r .000 6. 2." Elec.__ " = per cent The Evans and Sels Chart.600 75. ft| : — aib 2 )Er + 62 &f ] - #1 ^« may left-hand side + <$£? + &i (a? 6? The squares on the (23) be completed by the addition of / OX&X v + «2&2\ ^2 2 &?+&i y i / «2&X - «l&2\ 2^2 v + &2 &? r 2 (axbi = / + a 2& 2) 2 (6? The J-rv + (a &x — +& 2 «x&2) 2 2 2 E 2 (24) ) result is \ 6? + 6| *' + J 9 L &1 + = Er &I \ &TF&I (25) . This is aiEr + bj rp . upon squaring.Ira I" &2 a 2 b 2 )Er T ^p _|. (ax + jaJEr + + jb (fei + jl 2 )(I rp (20) rq ) Multiplying out and collecting in-phase and quadrature terms gives = E.bj rq + j(a^E + bjrp + r bj^) (21) a vector equation which. turns into the following algebraic equation: EPS = (a{ + al)E* + 2(a x &i + a 2& 2 )#r/rp (6 may which ''P + 2 + + 2(a 2 &x + &f)A> (6? .ELECTRIC CIRCUITS— THEORY AND APPLICATION 274 Also.)/„ (6x (19) Using the receiver voltage isV as standard phase and introducing the in-phase and quadrature components of the receiver current I r equation (19) becomes .aibt)E Ir 9 + r + bl)Prq (22) be arranged as follows 2(aibx &2 — ^_ r2 j_ 2(a 2 bx i. = E. let the general circuit constants be given in polar and rectangular form as follows A = \A \/a = Co = Do = \B Bo Equation may (14) \/_p \C \/y |Do|/5 = = = = + ja + jb + jc +# ax bx z 2 ci 2 dx (18) 2 then be written E = + ja (a x t 2 )#r + + #. Equation (25) may be converted into a power equation by introducing /rj> The Pr =^and/ -|r (29) ra result is + b\ + b\ m + J Qr [ + 6? + 61 ^J ~ (30) 6f+6i In a polyphase system Pr and Qr are the power (in watts) and power (in volt-amperes). and the reactive voltages are is to neutral.TRANSMISSION-LINE CHARTS If 275 Irp and I rq are taken as variables. Thus. given by its Horizontal displacement = Vertical displacement = and the radius of the circle is \ — * * _i_ jl E r (26) * 2 ?i j_ m %** (27) given by B ~ fiM = vwm (28) A circle diagram based on equation (25) would be a voltagecurrent diagram and might be useful for certain purposes. per phase. however. As a rule. the formulas will be especially to this system and modified to read in terms of total and kilovolt-amperes in connection with line Capital subscripts will designate total power in kilowatts or kilovolt-amperes and line voltage in volts. the three-phase system paramount adapted kilowatts in importance. however. Pr= i» andCr = i» (31) . a voltage-power diagram is required and the equations for this will now be established. equation (25) obviously The position of the center is displacements from the coordinate axes. as follows: represents the equation of a circle. Further comment on the current diagram at this point seems unnecessary since the general discussion of the power diagram which will follow is directly applicable also to the current diagram. respectively. reactive voltage in volts. Since. B B will consequently come out positive and the center will be disIn other placed. and radius C R the radius are given by The last circle . vertically in a positive direction (upward). the sending voltage is constant and the receiver voltage variable. the center will lie in the second quadrant. represent one definite value of sending voltage. will immediately give the sending voltage corresponding to any value of active and reactive power at the receiver end. may be constructed from equation (34). words. the position of the center A family of concentric circles may then be drawn about fixed. In a system with ordinary constants the product «i& 2 will be much greater than a 2 bi. the displacement of the center and tive. is this center. the receiving chart for constant sending In this the voltage. the receiving chart for constant receiver voltage. If .000y 6f (33) - which may be written in simpler terms. £«] 2 = C% (35) equation is the well-known standard expression for a with center in the first quadrant assuming A B and B B posiHence.000 * Q* ^ + T^n 1.A RY + - [Q. the center will be displaced horizontally in a negative direction (to the left).000 2 = (l^o) J' ^^ ( 34 > or. If the receiver voltage E B is constant. as 2 Pb + ER + 1. another chart. [P* . each of which will. Ar = "i^ooo^ = 5 " = ~ pOO^* ~ Cb = n l^oo^* - ~im^WTWE% (36) ^B ^' . 1^00 b\ + 61 1 E SEB I^oo V6T+T (38) 2 Since A R is negative. the receiving voltage being fixed. Figure 153 shows such a diagram. This chart.ELECTRIC CIRCUITS— THEORY AND APPLICATION 276 Inserting equations (31) Fb + 1^00 ~W+W and (32) in equation (30) gives R \ l > 000 L b* + 6* * E%E\ / 1 \*_E] ^l. When the active and reactive power received are known. the receiving voltage can be determined. therefore. Evans and Sels receiver chart for constant receiver voltage. all lie on a straight line passing through the origin. Evans and Sels receiver chart for constant sending voltage. — Fig. — Fig. however. The depend centers will. therefore. If the receiver power and reactive power.TRANSMISSION-LINE CHARTS 277 position of the center evidently will not be fbjed. the sending voltage can be determined. Each of the concentric circles represents constant sending voltage. A family of eccentric circles each representing a definite value of the . Each of the eccentric circles represents constant receiver voltage. but will on the chosen values of receiving voltage. are known. 153. 154. By basing the derivation of the circle equation on equation (16). . The result is /. the receiver voltage sending voltage being fixed. . converted (43) + into b\ a power equation by introducing 7ap It = gandJag = (44) |-] becomes Fl b\ + b\ *• M~ rf 2 bi — rfi&2 «"* i>f -+- b| (45) .(40) The position of the center is determined as follows Horizontal displacement = — = — Vertical displacement and the radius of the circle is given Radius = (41) is 2 . the only difference being the minus sign. (41) 2 (42) 2 by 2^1 Vb\ Equation . . It gives the receiver voltage corresponding to any value of active and reactive power at the receiving end. (16) may be written E = r (rfi + jd2)E - which the sending voltage in + MiLp + jL (6i a (39) q) used as reference vector. diagrams having active and reactive power at the sending end as may coordinates Thus equation also be constructed. all the intermediate steps necessary in the preceding derivation may now be omitted and the final voltagecurrent equation written down immediately by help of equation (25). The charts so far discussed involve receiver power only. 2 ..ELECTRIC CIRCUITS— THEORY AND APPLICATION 278 may be drawn about the centers. This chart is shown in Fig. is Since this equation is entirely similar to equation (20). 154. the receiver voltage being fixed. as shown in Fig.TRANSMISSION-LINE CHARTS 279 Again adapting this to a three-phase system using total power in power in kilovolt-amperes. still simpler. the displacement of the center and the radius are given by A3 ^ " V 1. and reactive power at the sending end. p Fa the result 1 ~ 1^00 is dM + dtbt ^Y + s b* + b\ \ . The angular displacement (phase angle) between the sending and receiving voltages may also be obtained from the preceding charts. . Since A s is positive. and draw the line FO connecting — . 155. Hence the center will be displaced is di& 2 In other words. Angular Relations. a sending chart for constant receiver voltage may be drawn This chart gives the sending voltage as indicated in Fig. the center will vertically in a negative direction. for instance. Take the receiver chart for constant receiver voltage. corresponding to any value of active. This chart immediately gives the receiver voltage corresponding to any value of active and reactive power at the sending end. *S Vbf+1 (AQ) y . and line voltage kilowatts. It lie in the fourth quadrant..000 1 m' _ = D 5s= T^00^ I^00 Cs = poo ^ A= dlhl b\ + dih + b\ * d 2 6i — di6 2 6? + 61 l^oo F2 ** _.A aY + [Qs . \n djbj 1 6? (— which may dib t Ft +6 * Y -+ 1 ' 000 L - M 2 62 V _ J (46) be written or. Km (50) (51) should be noted that n' = n and Cs = C*.BSY = (48) CI Hence. total reactive in volts. a sending chart for conbe constructed. E8 is constant. [Pa . 5 S will always be negative in ordinary systems since greater than d 2 6i. the center will be displaced horizontally in a positive direction. the If the sending voltage stant sending voltage may sending voltage being fixed. 156. If the receiver voltage is constant and the sending voltage variable.000 - l v* ** ~ 1. power and reactive power.280 ELECTRIC CIRCUITS— THEORY AND APPLICATION the center with the origin. Each If the sending of the concentric circles represents constant receiver voltage. the receiver voltage can be determined. are known. from FO in the direction shown. of the eccentric circles represents constant sending voltage. 156. Each If the sending. are known. indicates the phase position of the receiver . active. 155. therefore. This line repreAlso draw a line FR displaced Es -Qonshnf — Fig. — Fig. and reactive power. Evans and Sels sending chart for constant sending voltage. the sending voltage can be determined. as shown in Fig. the The line FR then angle of A . therefore. by an angle a. Evans and Sels sending chart for constant receiver voltage. sents the no-load sending voltage. 157. sending chart for constant sending voltage as shown in Fig. coincide. This figure shows how the displacement between the sending and receiving voltage can be determined in the Evans and Sels receiver chart for constant receiving voltage. the circles must have the same center. 158. it immediately follows that the angular displacements between voltages may be found in the diagram (Fig. will coincide. straight lines radiating from the may be drawn in so that the phase angle between voltages may be read at a glance.e. the angle of the general circuit constant Ao. ' the former on point in the latter. 157. Without resorting to an analytical proof the correctness of this statement may be demonstrated by comparing the diagram (Fig. i. 149 by placing the origin of . 149) and showing that the two are fundamentally the same. Assuming identical power and reactive OL and MN.TRANSMISSION-LINE CHARTS voltage the E r with respect to the rest of the diagram. The diagram (Fig. A family of angle lines. This being the case. 157) may be superimposed on Fig. the lines FO in the two diagrams. as well as the direction of the line FR in Fig. as pointed out. by a comparison with The Fig. The reference line (the direction of the receiver voltage) is located as indicated by means of the angle a. 157 and the direction of the receiver voltage vector E r in Fig. so that the power axes of Fig. consequently. and.. 150. their sending-voltage circles must. 149. 149. . 157) with the original vector diagram (Fig. center to a sending-voltage circle such as 281 Any line from FS represents correctly a sending voltage in magnitude and phase and the phase displacement between the voltages at the two ends is directly given by the angle 6. F Eft= Constant — Fig. 157). in two diagrams. In this case the reference line (the sending-voltage vector) is displaced the angle 5 from correctness of this may be OF verified in the direction shown. of necessity. Evidently the same may be done in the center. 157 coincide with Fig. power scales in the respectively. Hence. The reference line (the direction of the sending voltage) is located as indicated by the angle 5. an operating point from one chart to another. Hence. for constant receiver volt- — In the receiver chart Loss Circles. Loss = P 8 — P r watts per phase difference (52) . they may be of assistance in transthe other. due to the eccentricity of the voltage circles. circles and can The power easily be drawn in. the angular relations may also be obtained from the receiver chart for constant sending voltage (Fig. 154) and from the sending chart for constant receiver voltage (Fig. however. Since. 158.. the use of the angular relations is rather inconvenient in the charts having eccentric voltage circles. age and in the sending chart for constant sending voltage. it is These are concentric possible to locate loci of constant loss. own family of angle lines emanating S — This figure shows how the displacement 6 between the sending and Fig. each one would require its from its own center. it may be more convenient merely to draw the reference lines (one for each center) and use a protractor to read the angles. if it is desired to obtain phase angles from these charts. charts. If a complete set of angle lines were to be drawn in these 156).282 ELECTRIC CIRCUITS— THEORY AND APPLICATION As a matter of fact.e. however. i. the angle of the general circuit constant Do. they would become rather crowded. it is usually more ferring practical to make use of power loss for this purpose. The phase displacements between the sending and receiving voltages when used in connection with their magnitudes make it possible to determine the conditions as regards active and reactive power at one end corresponding to the conditions existing at In other words. lost in transmission is given by the between the sending and receiving power. since. receiving voltage can be determined in the Evans and Sels sending chart for constant sending voltage. jd (/?. in either case.jl + (6i + jb )(di .l)E Irp + aid . take the product of the current and the conjugate of the voltage. and.a di)E I + (d^ + d b )(Prp + 1%) = um + tP .. or vice versa. = Loss (EJs . respectively. it is preferable to tive power.?m (56) and adding 1 It may be noted that the imaginary part of these products represent reacWhen the latter quantity is desired. it makes no difference which product is used. 1 Designating the conjugate of a complex quantity by a dash above the symbol. + IK) . as this procedure will give the reactive power with its correct sign. By means of equation (53).wQ + v***® (55) rLi (aid + a&^E* (6*ci - 6 xc 2 lCl 2 2 2 r 2 rg 2 r 2 r 2 r Rearranging as follows: I F\ + *-EfPr + <S . the loss may be written of the general circuit constants. reactive active The real part will be the same.E (Ir P . the receiver power as the real part of the product of the receiver voltage and the conjugate of the receiver current.jI )U (54) [(AoE r [AoCoErEr r r r (<*i r jEV/rJreal 2 X 2 r rq ) r rq ) t 2) r rq Multiplying out and collecting the real terms gives Lr = + (& + b& + a^ + a d .3d*)E (Irp . . the power loss is obtained in terms of active and reactive power at the receiver end or sending end.jc )E (Irp +jl + + ja*)(di . negative. Use the symbols L r and L 8 for the loss per phase in the two cases.TRANSMISSION-LINE CHARTS 283 In order to derive a general expression for the power loss in terms it is convenient to express the sending power as the real part of the product of the sending voltage and the conjugate of the sending current. in connection with equations (14) and (15).L r may be written Lr = + B Ir)(CoEr + Ddr) ~ #//r]real = = + B C E I + A D E I + BoDolJr [(ai + ja.E*Q = v r ^m . however.EI r (53) r ) teaX By substituting in this equation the sending voltage and current in terms of the receiver voltage and current. As far as the power is concerned. similarly. It is assumed that leading power is considered positive and lagging.)(ci -jct)El + (6 + jb )(Cl . fixes 1 the corresponding operating point in the sending chart. the the third quadrant.ELECTRIC CIRCUITS— THEORY AND APPLICATION 284 in order to complete the squares.000 V v + 2 +W 2 (61) 4WV™ ER (62) constants calculated from the general circuit auxiliary constants are t U = = = = w = V Since t will &1C1 + 2(6iCi dibi + + &2C1 — d\C\ &2C2 + + a. immediately gives the loss corresponding to any condition of active and reactive power at the receiver end.000/ Here LR + wT~a— 2 luv represents the total loss in watts. each representing constant loss. gives P + £. This diagram. results in 12 P* t + 2.* + * = Qr-TfoE} m £ + £±£^](m> Adapting this equation specifically to a three-phase system. the radii of a family of concentric circles. as was previously done for the voltage-power equations. 159. may be computed and plotted center usually lies in as indicated in Fig. a circle t A RL = The Brl = Crl = F2 2.\d\ + a^2 a^d^) 1 63) (64) 02^2 ^2^2 &1C2 be positive and (65) + aidz — aidi (66) w negative for ordinary systems. ™ &R (59) The equation repre- 4v 2 v whose center is fixed (assuming constant receiver voltage) and whose radius depends upon the value of the loss. (60) R —w— t 1. This reduction depends on the relation A D — BC = 1. The displacement of the center and the radius are given by sents.00(to E\ El LR t 2 ~I \1.000v E. as seen. By assuming appropriate values of the loss. Adding the receiver power and the loss determines the sending Sending power and voltage both being known definitely power.000v B 2. . then. the receiver voltage being fixed. Evans and ER . From this diagram the power lost in transmission can be determined when the active and reactive power at the receiving end are known.+ «. + + ^1.E\ 2.000. employed may equation (67) (54)./ ( Es f 2 {w'Y - 4u \_v' The displacement AsL — of the center and the radius V p2 "| (69) are given by t ~.(D E8 . For constant loss increment the spacing between the circles decreases as the loss increases.TRANSMISSION-LINE CHARTS 285 loss in terms of sending active and reactive power be determined from equation (53) by substitution of equaThis gives tions (16) and (17).BJ )(-CoEs s a process similar to the one previously with equation following form: -H . + Z /*)] (67) r eai in connection be brought into the + £*! [ L. may be w - ^"T+[ Qs + 2. . These lpci are concentric circles. which for a three-phase system t 2. - Constant —This figure shows how the loci of constant loss can be located in the Sels receiver chart for constant receiver voltage.000v I written E ^s 2 V fL. v' + f + (i»')» . 159.000 v (70) .4uV 4(d') ! E} - (68) Center of Voltage Ore fes Center of--loss Circles j Fig. The power may U By = [Ej.000*. If the system A = D is symmetrical u = u'. t = = &2C2 + a\ and the auxiliary constants become + a\ — = a^% v = v' = ai&i + w = w' = —bid u u' 1 = 2a 2d 2 = (approximately) (76) (77) a 2 &2 (78) (79) . and w = w' since f This would mission line with identical terminal transformers or a transmission If. the leakance is negligible. in the latter case. For constant loss increment the spacing between the circles From this diagram the power lost in transmission decreases as the loss increases. the center is located in the This will usually be the case. Evans and known.v = v' . as a rule.000 ~ (sSL E* \L 8 1.000\ ]?2 (71) s v' ^ v> (72) 4(y') 2 additional auxiliary constants needed in this case are The = = = u' v' •u/ aibi + + 6 2Ci — diCi d&i (73) a 26 2 (74) + 61C2 — a 2di ai^2 (75) Figure 160 shows a family of loss circles in the sending chart for constant sending voltage.286 ELECTRIC CIRCUITS— THEORY AND APPLICATION B W s _ 2. is 4a- Center of'— -. negative. w).^ Loss Circles 's -Consfard Cenhr of trVoliage Circles — This figure shows how loci of constant loss can be located in the These loci are conSels sending chart for constant sending voltage. As seen. centric * circles. is hold. practically zero. for instance. since w' (as well as first quadrant. in the case of a trans- . can be determined when the active and reactive power at the sending end are Fig. 160. d line alone. drawn. 2. loci of located. The per cent efficiency y is consequently expressed by ' - F. upon rearrangement. The efficiency is the ratio of power received to power supplied. The horizontal displacement of the also a function of the efficiency while the vertical dis- . 4 TO )] -l is the equation of a circle whose radius depends upon the efficiency. + uE + tP r (80) equation (80) may be written — = Pr Pr + U =P t r .PVtLr 100 equation (55). . L by Substituting for r 100 Pr .wQ + V r (81) Which./ TRANSMISSION-LINE CHARTS 287 — In the charts where loss circles may be constant efficiency of transmission may also be These loci will also be circles. equation .. becomes - V + P? '-EWr V + Q? - = --E -E?Qr V r (82) V Adding -(?-)T. this the value of center is 1 ^ +.000y . Efficiency Circles.E + [£]»2v r H^)J + 4e> 2 w 2 -£* (83) in order to complete the squares gives Pr ) ^ + + [*-5-]' L El 2v [f]"(Kf -0] For a three-phase system.+ —H may (84) (84) be modified as follows t Pb+ + E% 2^000t> w [q. [A]'([«-pr- As seen. coordinates of the center and the radius are given by t Ar v — B Rl = . It may be noted that the vertical displacement is the same as for the center of the loss circles. These loci are circles which are eccentric but whose centers all lie on a straight line parallel to the active-power axis.288 ELECTRIC CIRCUITS— THEORY AND APPLICATION placement is fixed (assuming constant receiver voltage). Ordinarily. By assuming values of the efficiency a family of efficiency circles may be drawn as shown in Fig. The circles are eccentric. A Rv will be positive while B Bn will be negative. Evidently.000v w F2 E« (87) * 2. From this diagram the efficiency of transmission can be determined when the active and reactive power received are known. 161. all lie on a straight horizontal line which also passes through the center of the loss circles. This figure shows how loci of constant efficiency of transmission can be located in the receiver chart for constant receiver voltage. Their centers. Hence. however.<onsiarrt — Fig. B . the center (or centers) will lie in the fourth quadrant. maximum obtained at the point where the radius of the becomes zero. This means that efficiency is efficiency circles . It is seen that the greatest efficiency for received is realized receiver end are lagging ment any value of power reactive kilovolt-amperes at the equal to Bv the vertical displace- when the and of the efficiency centers. 161. C - - — (? .000v The l\f-f 100 A =Co20^ + w* 4uv (88) /Er.) (86) 2. 1 P° ~ + I 1 ~~ 100 2. this equation brought into the following form: < + ( 1K '•- ^ ioo) v r J + [ft which. power reactive maximum at efficiency is obviously given by equation (87). for a three-phase system. of it maximum The occurs.) TRANSMISSION-LINE CHARTS [t _ _ /i2°_ AV + W 2 - ±uv 289 = (89) from which 100 r?max / (90) ± V±uv.' E '\ + 0s [ + be written » w' £ V " »J [AKt'+^-^i'+^'-H The displacement of the center (95) 2 + - (93) and the radius are given by B " = ~5ooo7'£ * c* " »<>/[< be w' „V + s^J - may r readily i5o)] + (w ' )2 " *" V (96) .w 2 -f 1 may then be computed from equation Inserting the value efficiency in equation (86) gives the power at which The maximum efficiency (90) using the plus sign in front of the radical. The equation for these circles may be derived by expressing the efficiency as follows: = P ~L * v * 100 (91) may Omitting intermediate steps. Efficiency circles may also be drawn in the sending chart for constant sending voltage.000. w By = 10 . The maximum by putting may efficiency also be calculated from equation (96) 2 + (!. first Figure 162 shows a family of efficiency circles located in the sending chart for constant sending voltage. it 2 t. . u. maximum inserting the value of the (94). This figure shows how loci of constant efficiency of transmission can be located in the sending chart for constant sending voltage. 162. - efficiency in equation sending power may be Evidently. value of (w') 2 ] (98) be used before the radical. the efficiency calculated from equations (90) and That this actually will be the case may (98) should be the same.{w'Y = 1 (4wy —w -w = 2 and that l) 2 - 4m 2 2 ) (100) (101) . w.w 100[t (t 2 resorting to the original expressions for w'. From this diagram the efficiency of transmission can be determined when the active and reactive power at the sending end are known. (99) y ) u'.w)] + [' the solution of which *7m ax Here By = ° < 97 > is 100[* + ± 1 y/lu'v' minus sign should always the 4mV = (w/)2 "" the corresponding determined. v' t ' and can be demonstrated that 4wy (t + V .290 ELECTRIC CIRCUITS— THEORY AND APPLICATION The center is usually located in the quadrant. The loci are circles which are eccentric but whose centers all lie on a straight line parallel to the active-power axis.w . v.V*»E*1 + l) . Centers ofEfficiency Circles Center of Loss Circles—Z i^j^ft-^ff? ^Constant Center of Voltage Circles y — Fig. = 2 + X .(4wy . be shown as follows: From equation (90). t £— + 1 + V4wy . s § § S" S" s" I s s §" |- §• Active Power sf in kw. Evans and Sels sending and receiving chart for constant receiver This chart is for a single circuit of a 240-mile. 220-kv. Parts (a) and (b). of transformers at each end.000 kv. equation (99) reduces.-a. . Negative Power. In constructing the Evans and Sels charts. The calculations for this chart will be found in Example 2. — Some Additional Remarks. — Fig. voltage. thus giving the same maximum efficiency. 163. three-circuit transmission line with 360.TRANSMISSION-LINE CHARTS 291 Hence.. to equation (98) when the minus sign is used in the latter. it is possible and sometimes convenient to draw sending and receiver charts for either constant sending voltage or constant receiving voltage on the same sheet of paper. three-phase. The calculations for this chart will be found in Example 2. Evans and Sels sending and receiving chart for constant sending voltage. three-phase. 156 and also the charts shown in Fig. 155 may be combined. charts of the type shown in Fig.-a. Parts (a) and (&) Figures 163 and 164 show two complete Evans and Sels charts in which the sending and receiving charts have been drawn on the same sheet. three-circuit transmission line with 360. This chart is for a single circuit of a 240-mile. of transformers at each end.. — Fig. 220-kv. It should be noted that the loss and efficiency must be read at the receiver operating point in the chart for constant receiver . 154 and Fig. 153 and in Fig.000 kv.. 2pqooo & &' &- s m f 1~ §f g & £ $ 8 5 Active Power in kw. 164. 292 ELECTRIC CIRCUITS—THEORY AND APPLICATION In other words. E. A. A.I. These efficiency efficiency circles in the negative-power region. complicate the chart in a prohibitive manner. D. p. of at . since a complete family of these circles would be required for each value of the vari- able voltage. p. Whether or not the charts should be drawn to include also negative power depends on the particular problem whose solution Very often a mere inspection of the problem will is desired. Fortunately. it is possible to avoid these difficulties modification of the original Evans and Sels chart. in such cases involved under all conditions. Trans. therefore. E. Nickle (Trans. if to predict definitely the direction of in all lines Obviously. Furthermore they would circles for the other part of the chart. Complete computations found in Example for these charts will be 2. p.. that the subsequent handling and use of the many charts would be rather inconvenient.E. 80.TRANSMISSION-LINE CHARTS voltage. the performance charts just described were end all of the system. may easily be extended to considered. The modified charts are also briefly described in the paper " Power System Transients" by V. 1924) attention was called to the possibility "modifying" the usual circle diagrams so as to make each chart applicable any value of sending and receiving voltage. 1 The by a result is 1 In discussions by F.I. This merely involves completing the voltage and loss circles and drawing a set of In the charts so far.I. Negative power.E. Bush and R. 84. positive for circles are exact images of those values of power simply indicate that power flowing in the is opposite direction. embrace also the negative-power region.. The Modified Evans and not impossible. and 293 at the sending operating point in the chart for constant It is unnecessary to have loss and efficiency sending voltage. siderable amount of tion of the charts. 85. in the first place. Booth.E. A. —The it is safer Evans and Sels based on constant voltage at one In order to study. to make the charts complete. the system at several values of sending. only positive values of power have been The charts. A.and receivingend voltage. however. an operation which is frequently required. In other cases it may be quite difficult. it would be necessary to construct a series of charts so as to cover the of This means that. and. secondly. negative values of power are likely to be encounwhether reveal tered. a conwork would have to be spent in the prepara- desired voltage range.E. 1925. power flow Sels Chart.E. Terman (Trans. 1924) and C.. to have the chart directreading at the nominal value (V B ) of the receiver voltage. —Dividing equation [* + JL T+ U| 1. Voltage Circles. as and the radius is not given by absolute voltage values but depends on the ratio of the sending and the receiving voltage. Pb - + t^L (» The result QR V. Only two and one receiving chart. "*" iei by (34) 2 = + ^ -^l 1. respectively. At other voltages the chart reading must be multiplied by the actual receiver voltage seen. as well as the receiving end of the system. ratios would be concentric and evenly spaced. Obviously this can be accomplished only by introducing a variable power scale depending on the voltage. namely. when < is should be multiplied by the square of the ratio of the actual receiver voltage to the nominal receiver voltage.ELECTRIC CIRCUITS— THEORY AND APPLICATION 294 a type of chart which is applicable to any value of voltage at the sending. are required.ooo/ YfoY \eJ nran uuzj This is the equation of a circle in a coordinate system where active and reactive power received. It is more convenient. when the now will be direct-reading 103 > EB = equal to the nominal For other values of receiver voltage. The equations for the modified chart are easily derived from those previously established for the Evans and Sels chart. The scale would be correct and hence the chart direct-reading at unity receiver voltage. This is readily accomplished by multiplying equation (102) by Vr. A circle diagram based on equation The circles of voltage (102) would be applicable at all voltages. equation (34) charts. rather than at 1 volt. i.000 ' ' (» lmo n J(wJ It is seen that the chart VR . " is _L ^ m V 1.. making interpolation easy.000 J EB gives n l \i. the scales actual receiver voltage receiver voltage. divided by the receiver volt- The age squared are the variables. ment of the center and the radius are given by The displace- .000 J ^ ffe.e. however. entirely independent of the voltages squared. one sending chart and (47). position of the center is. have also been drawn in.000 &? + &§ fi^«- 1.E TRANSMISSION-LINE CHARTS A *R = d B L_y* = * 1. they are actually superfluous since. The circles represent ratios of receiver voltage to sending voltage. radial angle lines Sels charts.000 F28 (» [psHW <w Here the chart is direct-reading when the actual sending voltage equal to the nominal sending voltage.000 1. however.000^^ * 1. Obviously. In the modified charts.000 r* - n' Ks —?— ^i-t-^. the loss can always be obtained as the difference between sending and receiving power.000 + a2&2 V% + 6 ^« 2 62 1. The displacement of the center and the radius are given by is A 3s = -J—v%s = 1.000 & 2 + 6| 1. the modified charts are "angle true. if desired.000 T79 Ks 295 (©" ""' Yine\ UU5.F 2 1." just as well as the original Evans and Sels charts.000 _ »* 1. with these exceedingly convenient to make use of the angular relations. A point may be transferred from the sending chart to the receiving chart and vice versa by means of voltage ratio and angle. only one family of angle lines is required since all voltage circles are concentric. t^ T104Y Uw. ( im} Figures 165 and 166 show two complete modified Evans and In addition to the voltage circles. and the It is efficiency as the ratio of the receiving always well to avoid unnecessary lines and sending power.000 FJ 2 s nron UU8.000 * 1.000 u* 1. R obtained bythe nominal The following equation is obtained: m ®a I' * C10b) vw+bt The equation for the modified sending chart multiplying equation (47) by V%/E% where Vs voltage at the sending end. a process which is both direct and precise. charts it is Although loss and efficiency circles easily can be drawn in.000 — aibl " 1 a 2 &i — ai&2 &2 + &i *#B ~ 1. and to keep the charts . Ks C109) ^ VfeT+T2 ^ /. is is ' 1. Hence. of transformers at each end.000 kv. — Modified Evans and Sels sending chart. the as simple as possible.296 ELECTRIC CIRCUITS— THEORY AND APPLICATION As a matter of completeness.000/ R J +w — 2 4v 2 4ui>. circuit of a 240-mile. necessary formulas for loss and efficiency circles are given below. The calculations for this chart will be found in Example 2.„ a « in kw.000i. «\ ^ [(E B /v B y LR ^1. however.. By multiplying equation (59) by V%/E% the equation for the loss circles in the modified Evans and Sels receiver chart is obtained as (#*/F*) 2 + v 2. — Loss Circles.. three-phase. This chart is for a single Fig. Parts (a) and (c). t 2 2. 165.000/ <» + . VI (HI) . three-circuit transmission line with 360. 220-kv.-a. g § g g 8' s" Active Power s § S ar . three-circuit transmission line with 360. of transformers at each end. variables and Qb The values of the loss L R indi- eated by the loss circles depend on the actual receiver voltage..-a.000 kv. The loss circles are direct-reading when the receiver voltage equals . 166.TRANSMISSION-LINE CHARTS 297 This equation represents a circle whose center is fixed independent of the actual receiver voltage in a coordinate system with Circles give ratio 'of'Sending'Voltage to Receiving Voltage Angle linpsg/ve displacement bettveen Sending IMtOOO Voltage and Receiving Voltage _J BftOOO £f S? I I s' L_ sr §' i" Adive Power * - . Modified Evans and Sels receiving chart. 220-kv.* in kw (%f — Fig. Parts (a) and (c).. The calculations for this chart will be found in Example 2. three-phase. This chart is for a single circuit of a 240-mile. R ~. w2 — The 4lUV (119) position of the center and the radius of the circles are independent of the actual receiver voltage.000y F* 2 2 -\- As seen.000v W is (112) T/2 2. t 2. this equation represents a circle. the values of loss as obtained the nominal voltage.298 ELECTRIC CIRCUITS— THEORY AND APPLICATION In general. The horizontal displacement of the center and .000*/ Bbl - ^_ F 2.vi] + Q& w (j. 2 + equation (85) by V%/E%.000v Crl n (113) * — (114) The equation for the loss circles in the modified sending chart obtained by multiplying equation (69) by V\/E% and becomes Ps {Es/VsY t 2.000v -. should be multiplied by (E a /V a ) 2 displacement of the center and the radius are given by from the The circles .000*/ 2 (117) 8 t CsL Efficiency Circles. + (w'Y - 4u'v' vi WY (118) — The equation for the modified Evans and Sels receiver chart Pn l(E R /v B y 4uV W) The displacement of the loss-circle center nd the radius are given by A 3l = 2 2./ + 2 [ (E s /V 8 ) + t 2 w F (w'Y 'm + t vi (115) in the sending chart -Vs2 (116) 2.000y It is efficiency circles in the obtained by multiplying becomes U^r-tvi W [ (En/Vn) 2 2. 000. The frequency is end is cent) voltage at each and step-down trans60 cycles per second.71 pooT^ ' < m>p vi * a t -. hence. coordinates of the center and the radius are given by -(?-). t 4* + 4* .000*/ L -Io o)] 2+M2 4MV ) F 'J < 123 > This displacement of the center and the radius are. .C.R. c «. the equation for the efficiency circles in the modified sending chart becomes 2 + t 100/ \ (Es /Vs) 2 2.: TRANSMISSION-LINE CHARTS also the radius. each completely transposed. EXAMPLE — 0-A). SMa/v a v> y + 2.000^' F f [2^yi[ + ( < 1 + Lt^t.By moi^f- (™ - x ) +w ~ * iuv (122) multiplying equation (93) by V%/E%. A. three-circuit transmission line with step-up formers. mils.yj 2. Length = 240 miles. however.000 cir. Conductors: 795. Line data Three circuits.St. Statement of Problem A large hydroelectric generating station supplies power over a three-phase.S.. The depend on the value 299 of the efficiency. The nominal (or 100 per 220 kv. 220-kv.000y + (l- 124 > < 125 > j%q)J+ WY ~ 4i*V (126) 2 This example illustrates the calculation and construction of the Evans and Sels and the modified Evans and Sels charts for a typical high. Equivalent equilateral spacing = 29 ft. referred to the high-tension side.voltage transmission circuit. 8781 + jO.00209/85^45 = 0.6 farad per mile Transformer data and receiving transformers.25 = 0.45 X 0. / = „ = R X„ = circuit —120._ 2.45 = (0.0012/89°.00— V 3 X 220 .9 mho per mile a = Vzy = V0..6 per cent Copper loss =0. general circuit constants for a single circuit of the line with terminal transformers. ohms .5 per cent — 12 per cent Equivalent Core loss =0.839 X 5. Solution Part (a).1 C = 0.99)10"4 = 0.5 per cent Exciting current = 5 per cent 360. Prepare sending and receiving charts of the Evans and Sels type for constant (100 per cent) sending voltage and constant (100 per cent) receiving voltage.4814/85°.21 X lQ.1/82^39 = 25. — Constants of a Single Line Circuit without Transformers: = 0.29 Transformer Constants: Capacity per .^ X 0.40 . — Note.1 +J377 X 0.01913 B = Z Q sinh = 401\3°.01382)10"« = 5.45 = 0.117 + jO.000 '—— V3 X 0. ohms 315 220.005 y= X = amp.5 J= 0.000 kv.5015/85^45 hyp z 6 Z = -^ = 103^^\3T45 = 401\3T45 ohms A = D = cosh e = 0. Each chart should be drawn for a single circuit of the line with terminal transformers.000 7= V3 X 120. = __._ 48. type.-a.21 X lQ. Identical sending = Total installation at each end Equivalent R = 0.64 + j'191.84 sinh 9 C = 5^L? = zo 401\3°.01382 X lO-o X mho per mile 10.017 ./88°.1491 +ill.839/82^0 ohm per mile y = (0.831 = 0.4814/85^84 = 193.12 = 315 = AQ .8782/l°. X Compute the a.000 kv.: ELECTRIC CIRCUITS— THEORY AND APPLICATION 300 R = X ohm 0.6 /85°. Prepare a sending and a receiving chart of the modified Evans and Sels c. _ Q1 315 220.117 per mile =0.831 ohm per mile G = 0.-a.00209/85^45 hyp per mile 8 = al = 240 X 0. b. 02301 + bi + fa = B + A2Z + CZ* + 2(0.40) + J273.31) 2 10.662 275.99) 10.8781 + JO._ .1491 4 2 2 2(0.00599 + J0.64 +J191.10 = Do = = + +j"48. = 1488 X 10 mh ° .9940)10.5 (Q.(1.017 25.7 75.01913) (1.01197 - (1.488 . oa v .8542 + CZ(1 + ZY) = A„ + j'0.05X120.31 X 10" s mho Z = R + jX = 2.488 -yi2. . A = = ax + ja* = + A(l 2ZY) + BY + CZ{\ + ZY) + .3 a\ +jd 2 = A(l + 2ZY) + BY 0.il2.006 X 120. + b\ —Constants = 27.8082><10 K' = VTTl = .mho ZY = (2.31)10.662 = aibl + 1' = I = bl Co = 1 + JO for Voltage Circles: + 75.7 = 2 Vb\ +bl = V75.31)10.000 220T000 X 1000 2 301 . single circuit of the line The approximate formulas the transformers at each end.5) (1.0.5) +jll.000472) 0.662 27 °6 b'.06 2 + 273.70.64 +.00599 +J0. IX.000X1000 = .8542 j'0.662 bl = 7l|i =0 3887xl0 3 m-m'= °'6 ~ aih = 002301 X - ' ' b\ = — + ^i°-. (see equations (83)) are used in this computation.488 12.06 + 0-t)2301 X 273.31)10~* = (598.000472) +- (0._ Y 124° X 220.02301 Check on constants AD -B Part b\ (b).00599 (25.01913)(1..06 ci +jl 1. mh0 B = V12.402 . —These include a and Chap. X 273.40 ohms Y = G -jB = (1..40) 2 + jO.04539 +y0./48.99)10.017 + j'48.488 -il2.017 +.1491 Co = = = 27.00599 jO.31)10" 5 + (0. = C(l (0.1491 + = B = = (1. 10 ._ v = 0.5 (1. 1913) (2.4 (2.j'12.000472) + +jc.6 +i47.4 (2.18)10-s = 1.99)10.TRANSMISSION-LINE CHARTS n = G 0.40) (1.488 .017 +. .4882 X 10"* = 12.7 + ZY)* + A2Y(1 -f ZY) + BF + jll.000472 1 + ZY 1 + 2ZY = 1.000* 1n _.64 +jl91.8781 +j'0.7 -' ^7 - 3-635 X 10- The positions of the centers and the radii for both the sending and receiving charts are given in Table XII.07 a2&2 = 0-8542 X 27.8781 +y0.017 + ^48./191.000944) + (25.1.01197 +^0. .40) (0./48.8542 75.000944 6 General Circuit Constants. c e b 7*& 63 160 150 140 130 120 110 100 90 80 70 60 50 1 2 3 4 5 6 7 8 9 10 11 12 *** 1 V 123.790 27.180 06 *-i C I 1 © -3 o £ 5 281.8542 10" 3 + 0.340 228.4uV .940 246.976 23.470 73. aj ft i S as -A II £ 3 149 180 149. Sending and Receiving Chart.04539 X 10" 3 .516 X 10~ 3 u' — a.740 211.820 -18. Receiver Voltage Constant = 220 Kv.043 X 10" . Calculation of Voltage-power Circles in Evans and Sels Chart (Example 2). 1 £3 3 3 48.400 48.06 X 0.180 .820 -18.700 . 1 * o 48.412 X 10" 8 3 % 3. the figures indicating center positions will change sign.820 -18.7 X 0.770 123.400 48.980 is constant = 220 kv.04539 3. The tabulated sendingend figures should be used for the receiver end and vice versa.110 -214.400 39 204 30.100 53. 3 (w') 2 = The is XIII Since the system is symmetrical.06 0.650 -292.530 263.— ELECTRIC CIRCUITS— THEORY AND APPLICATION 302 Table XII.400 48.700 37.412 w' = biCi — biCz aid-i — aidi 273. 180 -120.4122 position of the centers 3 and the radii of the loss and 1U efficiency circles in the receiver chart for constant receiver voltage are given in Tables and XIV.220 6.890 -335. o o IN IN <N "o "8 »i OQ « .04539 X X 10-« 0. ?u a 8 ..471 X 10" 29.400 48.864 81.900 94.820 -18.400 48.170 105. pi-3 +* OS II * S -18.960 158.Aiw _ P + {w') .7 = 29.4 X 61. .424 12.040 9.400 48.8542 X 27.820 -18.65 X -7.796 69.27.47P X 10-« .180 149.02301 X 0.696 68.043 X l O^ = _ ZA)6Z X 4 X 29.330 36.7.65 v' = X dibi 10" 3 + 0. Table XII may still be symmetrical (A = D ).400 $ » -381. Table XIII may be used also for the loss circles in the sending chart when the sending voltage is constant .716 17.150 193.290 6q 05 69 S -«1§! s g.149 .820 % .170 42.400 48.820 -18.9940 X 10" 3 = — 4wV 14. 149.400 48.02301 X X 0. Constants for Loss and Efficiency Circles: When the sending voltage used.14.*..360 140.570 87.904 108.770 18.500 .\C\ + 61.02301 X 273.550 175.400 48.830 - 95.820 -18.820 -18.9940 10" 3 + d2&2 + + 0.06 X 0.820 -18.820 -18. Evidently.2 Ii-h"S II II 69 100 100 100 100 100 100 100 100 100 100 100 100 °63 3 3 a > _0 r«8 kTs - w0a 'o > o >- ft «A ft M o hi JL. .820 -18.180 149 180 149.516' X 10.090 22.02301) a 2C2 X 0.100 .400 48.400 48.880 31.810 -180.770 4.180 149.820 15 240 12.180 149 180 149 180 149 180 149. .180 149.564 48.390 -252. since the system = = = u = = = v = = w = = — 4uv = = = t v? P + vP- + 02^2) 2(6iCi 2(27. 906 -11.690 58.050 78.0 83.906 13.893 -2.906 37.000t> amperes kilovolt-amperes -2.893 -2.893 -2.906 -11.893 -2.300 55.730 98.050 142. provided the signs of the horizontal and vertical displacement of gives the necessary figures for efficiency the center are reversed.906 -11.400 70.906 -11.906 -11.210 91.893 -2.893 2.906 -11. Table circles in the sending chart.893 -2.96 91.906 -11.906 -11. 4wn_.000 5.906 -11.210 111.—— TRANSMISSION-LINE CHARTS 303 = 220 kv.000 7.000 1 2 RL 2.300 153.892. Receiver Voltage Constant = 220 Kv.430 137. Circle ^R — A RL Lb.530 98.000 11.550 IW['-(v-0]'+ .000 \ v t1 + w* — 4»* Chl..890 71.0 87. 2 o [-(?.906 -11.906 -11.j R Table XIV. 1 kilovolt2.893 -2.800 109.893 -2.906 -11.0 85.500 . Calculation of Efficiency Circles in Evans and Sels Chart (Example 2).0 t "*' wEB 2^007 Crv.660 84. kilowatts kilowatts 3 4 5 6 7 8 9 10 CRL —+ Lr 1.01* Bm ~ V.310 120.0 89.000 6.480 88. Receiver Chart.0 86. 1 2.050 131. Receiver Voltage Constant = 220 Kv. Calculation of Loss Circles in Evans and Sels Chart (Example 2).893 -2. Receiver Chart.510 82.300 42.0 84.000i>' -11. XV Table XIII.830 165.680 100 250 108. per cent 1 2 3 4 5 6 7 8 9 10 »c £ 91.150 115.906 -11.5 3.000» kilowatts kilovolt- kilovolt- amperes amperes 69.906 -11.000 4.000 8.906 -11.000 10.0 90.380 124.630 -11.906 -11.460 150.000 9.0 88.906 -11.906 -11. A SV — per cent Circle Bsv fww kilowatts 91. Table XVI — Table XVI.820 -18.440 95.180 149.820 Br = T7. Constants for Voltage Circles.940 246 340 228. Circle per cent 1 2 3 4 5 6 7 8 9 10 11 12 160 150 140 130 120 110 100 90 80 70 60 50 Vr.050 76.180 149 180 149. Sending Chart.560 85.400 48.850 118.906 11.080 126.0 87.180 149.400 48.400 48.0 89.906 11.940 49.170 93.540 142. The finished Evans and Sels chart for constant (100 per cent) sending voltage and constant (100 per cent) receiving voltage are shown in They are constructed from the data calcuFigs. .850 62.820 -18.0 85.820 -18. 163 and 164.890 74.820 -18.820 -18.180 149.740 211.000 yi Es BEr' kilovolt- amperes amperes 149.400 48. Calculation of Efficiency Circles in Evans and Sels Chart (Example 2).820 -18. sending charts and receiving charts have been drawn on the same sheet.000 kilowatts -18.2 mVR 1.820 -18.940 85.000' kilovolt- Cr = n 1.180 149.—— ELECTRIC CIRCUITS— THEORY AND APPLICATION 304 Table XV.906 11.180 149.906 11. respectively.0 83.760 115.180 149 180 149. m.0 90.000 kilovolt-amperes o 2.530 263.906 + ( 1 "i~)] + (w ' ) '" 4 " v Csv. kilo- volts 220 220 220 220 220 220 220 220 220 220 220 220 VR . Part (c). lated in Tables XII to XV inclusive.400 48.170 105.150 193. As seen.0 86. m'.400 48.400 48.0 84.0 88.820 -18.000t> 11.980 . 2. V.906 11. and n' will be the same as calculated in Part (6).000s? V[' w'El - / 2. .470 124. 1 kilovolt- amperes 33.630 109. Constant Sending Voltage = 220 Kv. .570 87.0 1 2 3 4 5 6 7 8 9 10 1 = C„„ *' 69.960 - — Charts. The values of the constants I. Receiver Chart Ar = Es .360 140 770 123. n.550 175.960 158.180 281.96 91.400 48.400 48.820 -18. V. (kilo- volts) 2 48.310 134. .820 -18.400 101.906 11.280 E .906 11. Calculation of Voltage-power Circles in Modified Evans and Sels Chart (Example 2).180 149.400 IV* 1.400 48.790 105.906 11.180 149.400 48.906 11.820 -18. Only voltage circles and angular-displacement lines have been drawn in. however. 165 and 166. are shown in Figs. Charts. Er/Es. the figures in Table XVI may be used gives the position of the center The signs of the displacements of the center. loss and efficiency circles are superfluous and only complicate the chart without being of any real advantage in its use. while loss and efficiency circles are omitted. should be reversed and the voltage ratio of the circles taken as also for the sending chart. Since the system is symmetrical and the nominal voltage at the sending end is the same as at the receiver end. in .TRANSMISSION-LINE CHARTS 305 and the radii in the receiver chart for a nominal receiver voltage (Vr) of 220 kv. As stated in the discussion of the modified Evans and Sels chart. constructed from the data Table XVI. — The modified Evans and Sels charts. the task would be very laborious and would require an undue amount to obtain so When charts are available.. and loads. synchronous condensers. transformers. Bush and R.e. From a complete generator chart (for instance. 201) may be obtained the value of induced voltage (air-gap voltage) and field current for any reactive assumed operating condition and also the angular displacement between the terminal and the induced voltage and between the terminal and the excitation voltage. however. 306 . Trans." by V." i. p. II of this treatise. D. A. the solution of even a problem of considerable complication may be obtained with comparative ease. in general. such as lines. necessary to construct one chart for each value of terminal voltage which it is desired to consider. Booth.E. Unfortunately. 2 Briefly described in the paper. to say the least.CHAPTER XI SYNCHRONOUS-MACHINE CHARTS Very often it is convenient to determine the performance of synchronous machines by graphical methods. Fig. The synchronous machine charts 2 are essentially like the Evans and Sels charts for the transmission line discussed in the preceding chapter. It is particularly important to have performance charts of the machines when it is desired to examine the operating characteristics of a system of which one or several generating stations form an integral part. a single chart of the "modified type" would suffice for all terminal voltages. On account of the non-linearity of the magnetization curve resulting in a variable synchronous reactance. conditions which simultaneously satisfy all links in the system. 1925. 80. If a study were to be made by analytical methods alone. Especially in problems involving stability analyses 1 it is necessary many "operating points. it is. The coordinates are power (kilowatts) and power (reactive kilo volt-amperes).I. generators..E. 1 Discussed in detail in Vol. of time. If it were possible to use a constant synchronous reactance. that graphical solutions throughout are absolutely essential. "Power System Transients. the same zero-power-factor This. will the assumption of a constant synchronous reactance rule. In any generator. of the air gap is constant. of characteristic. and reactive power are given and the constants of the machine known. an induc- In a non-salient-pole machine the reluctance when the small effect of the slots is neglected. as a than can be tolerated. without completing the chart in the negative-power region. power. Non-salient-pole Generators. be assumed. synchronous generator is a small quantity and. lead to inaccuracies larger not introduce appreciable error. power factors is for course. Hence. Neglecting the resistance reduces the labor slightly in the computation of the data for the charts. 167. and the same short-circuit characteristic. the distribution of the flux due to the armature reaction is also sinusoidal and may be considered as a vector. .SYNCHRONOUS-MACHINE CHARTS 307 will. A second advantage of neglecting the resistance is that a chart constructed for a generator may also be used to represent the same generator when operated as a synchronous motor. When terminal voltage. the characteristic of the salient-pole machine differs materially from that of the non-salient-pole machine. due to the fact that. even though these machines may have the same magnetization curve. the resistance of the machines may Usually the resistance of a or may not be taken into account. if neglected. the calculation of induced voltage and excitation voltage is a simple problem which requires nothing but the solution of a simple vector diagram. than other zero. a sinusoidal distribution of the magnetomotive force of the armature reaction — Fig. and hence it is more or less a matter of judgment whether or not the resistance should be taken into account. it is also necessary to have separate charts for non-salientand salient-pole machines. Furthermore. may —Vector diagram of a non-salient-pole generator supplying tive load. In constructing the charts. ELECTRIC CIRCUITS— THEORY AND APPLICATION 308 When computations have to be made. putting the calculations into tabular form and using Figure graphical methods during the process whenever possible. and short-circuit saturation curves. Figure 168 shows the characteristic curves of a non-salient-pole If necessary. The relation between terminal voltage and the voltage induced by the air-gap flux is then given by Ea = V + jI aXa t (1) . During computations. a lot of time can be saved 167 gives the well-known vector diagram of a non-salient-pole machine supplying an inductive load. but it permits Sufficient data for the may the use of the same chart for generators of different ratings. factor. the open-circuit. however. the leakage reactance is determined by generator. for a series of by properly systematizing the work. The following notation will be adopted V = = Ea = Ia = P = Q = P' = Q' = If = If = t Ea A = terminal voltage voltage induced by air-gap flux excitation voltage armature current power at the terminals reactive power with respect to terminal voltage active power in the air gap reactive power with respect to induced voltage (E a ) active actual field current field current from magnetization curve corresponding to the induced voltage (E a ) kl a = armature reaction in terms of equivalent field current xa xs = = armature leakage reactance synchronous reactance computations necessary for the charts be obtained from the rating of the machine and the three characteristic curves. provided they have the same fundamental characteristics.and third-mentioned curves if the leakage reactance is given. namely. the two first-mentioned curves will usually suffice. conditions. zero-powerAs a matter of fact. Evidently. and sufficient data may also be obtained from the first. this is not necessary. it is usually most convenient to express all quantities on a percentage basis. the Potier-triangle method. 60-cycle. X).-a.voltage circles in the Sels receiver chart (as given in vertical displacement of the center and the radii are.5 per By cent. in this case. turbo generator.— Characteristic curves of a 37.000-volt. 14.. 1. » 1. referring to the general equations for the position of the Evans and Chap. given by A = > aib * + q#y? = +6 2 2 b\ B' C' = = a^bi — d\b<i b\ + b\ F = Yl 2 EaV = EaV X« Vfcf+T2 t o xa (3) (4) t (5) .800r.000 400 300 Field Fig.0 Short-circuit ratio Armature leakage reactance =18.m. the general circuit constants of which are A = = C = Do = Bo + ja = 1 + j&2 = jXa = Ci + JC di + jda = 1 ai 2 (2) &1 2 22.p. the horizontal and center and radii of the sending.SYNCHRONOUS-MACHINE CHARTS 309 The chart corresponding to this equation will be the ordinary Evans and Sels chart for a lumped reactance. 500 Amperes 168.500-kv. displacement between (^/) may the be immediately read. and the induced voltage voltage . A . the active and reactive power are given in per cent of the rating of the machine. as far as the E'a circles are concerned. Figure 169 shows a chart based on equation (6). . iamily of radial angle terminal voltage and induced VOltage JtL a voltage. although. If. stated. phase . it is desired to put field-current curves on the same chart. per cent values of the voltages and the leakage reactance are used. The modified chart just referred to would be based on the following equation obtained by multiplying equation (6) by V* /V* p r T+ . . on the other hand. Usually. 169. ^c . [{v /vQ y\ t q _ l(v /v y r t Q in _ /nv/^v x \ ~\T a )\y ) 2 a t W . As already respectively. a modified chart might be constructed so that the same chart could be used for all values of terminal voltage. this cannot. . Since. in general. however. one chart has to be prepared for each value of terminal voltage. generator is terminal neglected. . since the field currents are obtained from a non-linear magnetization curve. As seen. the center is displaced upward and lies on the vertical axis. . since the resistance is drawn for constant terminal Voltage and contains a famneglected. it is usually convenient to work on the percentage basis.310 ELECTRIC CIRCUITS— THEORY AND APPLICATION The equation representing each of the constant obviously becomes Ea circles (6) If the voltages are taken as volts to neutral and the leakage is given in ohms per phase. each representing a showing circles of constant induced voltage and angle lines of constant specified value of air-gap constant displacement between -y . the active power P and reactance reactive power Q represent watts and volt-amperes per phase. It is Fig. The resistance of the li nes j s also drawn in SO that tho % ..-Generator chart of circles. . be done. be considered equal to the 100 per As cent value. 167 repre- and armature currents. as described below. respectively. it is necessary to calculate a sufficient number of points on each. field reaction are rotated clockwise through 90 deg. I^2tI3h JSSS with field-current and arma- SSSSJS^SdST" Also. corresponding to the induced Voltage. therefore. the field current i}. and armature may. current corresponding to induced voltage. the chart being directreading when the terminal voltage has its nominal value. 170 results. The power and reactive-power scales are not constant and depend on the actual value of the terminal voltage. In this diagram. this reduces to ©' +(1H)' -i (10) .. These points are then laid off in the charts and smooth curves drawn in. the actual field current If the excitation voltage field ot Ea The is in phase with relation between field current. In order to obtain the lines of constant field current and angular displacement between the terminal voltage and the excitation voltage. be expressed by reaction // This equation is = //+ jkla = If + jk(Iap + jlaj (8) easily reduced to J-ap i J-aq It 1} k k2 (9) In these equations.SYNCHRONOUS-MACHINE CHARTS V is 311 the nominal terminal voltage of the machine and should. the circles in this chart represent ratios between the induced voltage and the terminal voltage. and the vector klv representing armature reaction. seen. as a matter of convenience. is in phase with the latter. Iap anc Iaq represent components of the current in phase and in quadrature. the diagram shown in Fig. is in quadrature with the armature current. and internal reactive power as expressed by P' = E'jap and Q' = E'a I aq in equation (9). If the senting vectors in Fig. The calculations are comparatively laborious but can be shortened by systematizing and by making use of an auxiliary chart. with the induced voltage Substituting the relations between internal power a E . open. The chart based on equation (12) will be an auxiliary chart. it is current 7/ corresponding to induced voltage. the actual assumed value obtained.ELECTRIC CIRCUITS— THEORY AND APPLICATION 312 Dividing through by (I/) 2 gives seen that this equation represents a circle and is of Q' P' If -7-7 and -f-. which is convenient during the process of gathering data for the actual chart. For any value ** Zs (14) rent If of is P and obtained by entering the magnetization curve. factor . may either be obtained from ar<l Fio.. and instead of assigning specific values of the field current to the circles." It is readily Ealf Ealf evident that a diagram based on this equation may be used for any value of If. magnetization curve If.. power are if resistance identical.and short-circuit characterisFigure 171 gives a short-circuit vector diagram. is obtained (see Fig.— Vector diagram of the open-circuit and the zero-power1 asynchronous generator on short circuit. From the tics. from characteristics the or . The field current I f corre- sponding to the same value of armature current is read either from the short-circuit curve or from the zero-power-factor curve at its point of intersection with the horizontal axis (at zero terminal . which If-A^IaXa is representative of the armature / 'v ~"y -^ /a jcla reaction. The value of the constant k. is neglected. field current corresponding to any power and reactive power can be the internal power P' and the internal of internal The values of reactive power Q' are given P' by =P+ ^^r v (13) a t and Q'^Q-^^Xa Evidently. they are lettered in terms of the ratio between the actual field current and the field variable parameters. 168). E a is read from the circle diagram of induced voltThe corresponding field curages. corresponding to the leakage-reactance drop IaXa. are considered as the the "modified type. From the auxiliary chart. the internal and external Q. 171. preferable and is used in the calculations in In the auxiliary chart. method is believed Example 1. process of computing data for the main generator chart. The armature reaction in terms of equivalent field current is evidently If — If. and draw draw in the family of radial lines indicating In order to get points angular displacement between Ea and V Compute the location of the center for the in these circles. the former of the normal open- circuit voltage. Also. calculated from equation (15) either rents and the armature current them both in terms of rated by using the field curby expressing in amperes. in general. t. and the desired constant k becomes * Ij -?^ = (15) In order to introduce percentages in equation (12) the power (P f ) and reactive power (Q') should be taken in per cent of the rating. the latter either of the rated armature current or of the field current corresponding to normal voltage on open circuit. between Ea and Ea . corresponding to a specific terminal voltage. This chart also contains a family of radial lines which evidently represent angular displacement between If and // or. The Ea circles. k must be calculated from equation (15) with the field currents expressed in per cent of normal field current and the armature The last-mentioned current in per cent of its rated value. may be summed up as follows: Determine xa and k from the characteristic curves. k is If taken as a fraction of the rated armature current. The air-gap voltage (E a ) and the field currents (// and //) should be taken as fractions. the horizonand vertical displacement of the center and the radii of the tal circles are given by A" = (16) ™ B" = C<< - (17) U8) f(ff) Figure 172 shows the auxiliary chart. 313 most convenient to use the rated (100 per cent) value of the armature current.SYNCHRONOUS-MACHINE CHARTS voltage). then. or armature current. . on a percentage basis. what amounts to the same thing. It is. If the field currents are introduced as fractions of normal field current. 172. proceed as follows: Choose values of P and Q.-Cenier . &^T~ph per cent £*a lf Short-Circuit Ratio -1 xs -!00% jc a -?0% k--80% Center-Horizontal Displacement-O Vertical Displacement^ /2S% ' — When this chart is Fig. In obtaining the necessary data. Compute P' and Q' and read // and the angle between E a and E a from the By selecting a number of values for P and Q.314 ELECTRIC CIRCUITS— THEORY AND APPLICATION on the field-current curves. as suggested in Table XVII. . displacement curves may be determined so that these curves may also be drawn in. and one main chart conIt is evident. independent of terminal voltage. it is usually convenient to arrange the figures in tabular form. a auxiliary chart. that structed for each terminal voltage. should be for convenient integral values of field current as well as angular displacement. the same auxiliary chart may be used for all the computations. however. The curves in the main chart. lines give displacement between field current and field current corresponding to induced voltage or between excitation voltage and voltage induced by the air-gap flux. The values of field current and angular displacement obtained in the tables will usually be non-integral. the circles give directly the ratio of actual The angle field current to field current corresponding to induced voltage. in which. however. some of the columns may be omitted if deemed superfluous. used for a non-salient-pole machine. sufficient number of points on the field-current curves are determined so that these curves may be properly located in the main Similarly. The computations have to be repeated for each terminal voltage which it is desired to consider. incidentally. P' . a sufficient number of points on the angularchart. In order to obtain . read the corresponding Ea and Z E' " v from the main chart. t and select the correct 1/ from the magnetization curve. 'Auxiliary chart for a synchronous generator. 173. Salient-pole Generators. the result is then a distortion of the main flux without appreciable change in total strength. the axis of the magnetomotive force of the armature lies midway between poles. the field axis and the axis of the armature magnetomotive force coincide. Such auxiliary plots are indicated in Fig. When the power factor with respect to the excitation voltage is unity. therefore. 173. Obviously. the reluctance of the paths in the interpolar spaces is considerably higher than the reluctance of the air gap under the pole shoes. v t Q 315 ' Calculation of a Generator Chart E> Q' '-Vt P' Eal' Q' E'l' / E° -E' Vt such integral values.— SYNCHRONOUS-MACHINE CHARTS Table XVII. Auxiliary curves of field current and angular displacement between excitation voltage and terminal voltage plotted versus active power for constant These curves make it possible to select convenient integral reactive power. When this power factor is zero. values of field current and angular displacement for location in the main generator charts. —The reluctance which the armature reaction has to overcome in a salient-pole machine is not constant. the result being a weakening of the main flux without distortion. From these curves. the non-integral values may be plotted versus power for constant reactive kilo volt-amperes or versus reactive kilovolt-ampere for constant power. Evidently. . since the latter fixes the space relation of the magnetomotive force of the armature reaction with respect to the field axis. — Fig. which will be set up in the two regions depend on the reluctances and on the power factor of the armature current with respect to the excitation voltage. The relative amounts of reaction flux. the desired integral values may easily be read. 1927. This will not introduce a very great error. "An Extension of Blondel's Two-reaction Theory. "Principles of Alter- netizing coefficients nating-current Machinery. A.I. p.." Jour. only the fundamental component will be considered. p. 927.I.. 1926.." Trans. As seen. A. Fourier series.I. 912. since. Inc. 1 For is between excitation voltage and armature current. 1339. R. multiplying the total reaction by cos It is evident that the axis of the former will lie midway between any angle </> total <f> poles.. Lawrence.I.E. a very comprehensive discussion of synchronous-machine theory has been given in the paper. all concentrated in the magnetomotive force in ampere-turns per pole per phase." by R.E. it convenient to resort to BlondeFs two-reaction method.ELECTRIC CIRCUITS— THEORY AND APPLICATION 316 In order to take the effect of the two reactions into account. "Torque-angle Characteristics under Transient Conditions. The paper is divided into four parts: I. Nickle.E. 1926. IV. p. III. This magnetomotive force is uniform between the pair of slots conThe sidered. Dohekty and C. New York." Trans. A. December. "Steady-state Power-angle Characteristics. Since this chapter was written.E. It calculations. especially in the case of three-phase machines..E. is given sin x by + ^ sin Zx + = sin 5x + • • • (20) To simplify calculations. 1928. E." McGraw-Hill Book Company." Jour.E. slots. 1 The derivation of the formulas for the demagnetizing and cross-magis taken from R. A. respectively. the sum of the third harmonics is zero. Part of the paper presents an extension of Blondel's two-reaction probable that the methods which Doherty and Nickle suggest will improve still further the precision of salient-pole-machine method. A. "Synchronous Machines. is I . while the axis of the latter will coincide with the axis of the poles.| (19) where Ia is the effective current per phase in amperes. II. "Single-phase Short-circuits. in this case. and its shape is rectangular. is V2I. If there are a single pair of Z inductors per pole per phase. due to the maximum phase current. 1920. which represents the space distribution of this rectangular wave.E. as shown in Fig. 174.E. the armature reaction may be resolved into two components by and sin <£. SYNCHRONOUS-MACHINE CHARTS The maximum ordinate of the fundamental = -V2/«f A 317 is 0. and add up to zero.obviously be nSlPT TSZT •* Fig. the reaction per phase may be looked upon as consisting of two rotating magnetomotive forces of sinusoidal space distribution. 174. 174. Rectangular distribution of magnetomotive force produced by a winding concentrated in a single pair of slots. 175. They both armature. coincide. stationary with respect to the field poles. while the other will rotate with twice synchronous speed with respect to the of each field poles. equation (21) should be multiplied by a reduction factor equal to the product of the pitch factor k v and the breadth factor kb. which are deg.9 J„Z (21) 7T In the case where the winding is distributed instead of being concentrated. this reduces to 0A5kpkJaZ (24) . stationary. the reactions which revolve at double synchronous speed with respect to the field are displaced by 240 The reactions. 175. the armature reaction of a three-phase braically. Considering the armature reaction of any one phase as an oscillating vector and resolving it into two oppositely revolving vectors. Its maximum coincides with the pole axis. machine is given by OAbkpkJaZS If Z is considered to be the total number (23) of inductors per pole in all phases. Diagram showing the demagnetizing component of the armature reaction. The maximum amplitude becomes OAbkpkblaZ (22) In a three-phase machine. The shaded portion of this magnetomotive force is effective in producing reaction. each of one-half the magnitude of the oscillating vector. however. and the effect of the three phases is obtained by adding the three reactions algeHence. Fig. revolve with synchronous speed with respect to the One of these component reactions will . b ->l Fig. — — Fig. In reality. — — The coefficient K d is and component called the demagnetizing coefficient equal to 0.45 times the factor by which the direct is of the reaction must be multiplied in order to take into account the ratio of pole arc to pole pitch.318 ELECTRIC CIRCUITS— THEORY AND APPLICATION The direct or demagnetizing and the distorting or crossmagnetizing components are. over the pole shoe in order to find its effect in value average modifying the field. ofFlux Fig. it is necessary to determine its sine law. pole arc by a and the pole pitch by b. The portion of the magnetomotive force which is effective is shaded in the sketch. Fig. to the fundamental component which is shown. the mean value of the effective magnetomotive force is flir = A'd OAbkpkblaZ sin ar/b t <t> cos xdx air 62 air sin = 0A5kJc bI a Z sin <$> &2 aw/b2 K d (k p k b I a Z sin 0) (27) where air sin Kd = 0. The magnetomotive force is not constant over the pole but varies according to the Since it is not constant. 176. 176.45 62 aw/b2 (28) M.F. Fundamenial Comp. Approximate flux distribution produced by the cross component This flux curve contains a big third harmonic in addition of armature reaction. it will be omitted. ture reaction. Diagram showing the cross-magnetizing component of the armaThe shaded Its maximum is located midway between poles. A d = 0A5kpkJ aZ sin A = 0A5k pkbIaZ cos (25) <f> and c The (26) <f> curve representing the direct component of the reaction with respect to the pole is shown in Fig. Fig. 177.M. portion of this magnetomotive force is effective in producing reaction. 177. but since this Designating the fringing is not very definite. respectively. a small amount of fringing should be taken into account. Fig. . 175. T K The factor K = ' c is 0. air sin -r- (30) The complete expression for the cross component of the armature reaction is Ac = 0A5k p kbIaZ cos 4> = KdkpkbZIa cos ax . is shown in Fig. 177. (32) K c is by which the cross component of the armature magnetomotive force must be multiplied in order that it may be used on the open-circuit characteristic. and back The circuit is then comacross the other half of the air gap. a-rrl -v[b-. 176. then through the pole shoe. How much it will be affected by saturation 0. pleted through the armature. obtained by the following integration: M 2 — ~ fF(x)$inxdx \ = -A * sin 2 I c\ =4f-.9 fa 1 . hence. and M mental neglected. whose maximum value is given by equation (26). cross-component is shown in Fig. The mean value the maximum + xdx LJo sin of this sin 2 I xdx J*-% 1 (29) °i] fundamental component is 2/w times value and.45 times the factor . equal to 2 - M = -2 A.SYNCHRONOUS-MACHINE CHARTS 319 The cross-magnetizing component. fo . The cross-component tends to produce a local or component flux which passes across one-half of the air gap. The component flux due to the cross component of the reaction will probably not be as highly influenced by saturation as the flux due to the demagnetizing component. sm T\ called the cross-magnetizing coefficient. a A HJf TT 1 j- T IT . The cross-magnetomotive force will be all terms above the fundaThe amplitude of the fundamental is effective part of the resolved into a Fourier series.r 7T| O sin 7T y <t>) (31) where . The shaded portion of this figure shows the part of this magnetomotive force which is The actual shape of the flux produced by the really effective. . will show that the field-current vector obviously divides the armature-reaction vector (kl a ) into two parts. and the cross-magnetizing component of armature . the by saturation. saturation. the component flux due to the crossmagnetizing reaction is only slightly affected by the degree of pole faces. a somewhat smaller extent than the main flux. therefore. saturation primarily takes place in the stator teeth and in the two components may be about equally influenced If.320 ELECTRIC CIRCUITS— THEORY AND APPLICATION depends upon what parts of the magnetic If circuit saturate first. parts of all desired should be used. Figure 178 shows the vector diagram for the salient-pole machine. to determine the voltage produced of Finally. This method of treating the armature reaction assumes that the effect of the demagnetizing and the cross-magnetizing reaction is equally Examination of the vector diagram affected by saturation. to it. reaction. saturate about circuit 1 A1 —Vector diagram affected still is A Fig. the direction . The reactions are expressed in terms of equivalent field current. equally. The total armature reaction is kl a and the vector representing it is drawn opposite in direction to the armature current from the end of the vector //. 178. the field current corresponding to the induced voltage Ea The total field current is //. the yokes and the poles themselves saturate first. the cross flux obviously. magnetic the it is the lower and less by saturation although. on the other hand. the demagnetizing component of armature reaction is drawn parallel. If in this last-mentioned case. perpendicular to this field current. the magnetization -curve saturated part if by 1 IS 1 of a salient-pole generator supplying an inductive load. the ratio of js which is as -^ c —j Ad — AFc - Making use of this relation. is the ratio of pole arc to pole pitch. Figure 179 shows the same vector diagram with all field-curIt rent vectors rotated through 90 deg. origin indicated the as at On a strip of K . representing the total armature reaction is obtained in the usual manner from an open-circuit and a zero-power-factor curve or else from an open-circuit and a shortIn both cases. the power factor of the current with circuit curve. respect to the excitation voltage will be practically zero. but when it manner described comes to determin- ing the magnitude and direction of the field current. this part of the chart will be the same for a non-salient. for the non-salient-pole machines The location of the center of the constant air-gap voltage E a and their radii are determined by the same formulas (equations (3).pole machine having the same leakage reactance.SYNCHRONOUS-MACHINE CHARTS of the total field-current vector is 321 immediately given when the total armature-reaction vector is drawn. in a clockwise direction. . and. A perpendicular from the armature-reaction vector onto the field-current line immediately fixes the magnitude of the latter. A point corresponding to in 180. is evident that in obtaining data for the salient-pole charts. in general. The units are immaterial. cients c cannot very well be obtained from test data but d and The only information are computed by the given formulas. the same procedure as outlined can. The value of the coefficient A. paper or celluloid two scales may be constructed. but the ratio of the smaller to the This strip may be pinned down larger scale should be as c :K d Fig. circles of auxiliary chart can also be constructed in the for the non-salient-pole machine. be followed. a special device will have to be resorted to. (4). K K necessary for this Fig. in a clockwise direction. there will be no cross-magnetizing action. Hence. and (5)).and a The salient. 179. The coeffihence. — Vector diagram of a salient-pole generator with all field-current vectors rotated through 90 deg. " Trans. Lawton: "An Investigation of Transmission-system Power Limits. the magnitude of the field current. The actual computations found in Example 1. paper is now swung strip of until its edge touches the located point.I.E.. Another and strip of paper. 1926. p. 180. A. At this point. A. us well as the angular displacement between the excitation voltage Pomtatwhich Field Current and Angle . Simplified Chart. L.pole machine. Auxiliary chart for a synchronous generator. Nickle and F. the figure on the larger scale is read. By means of a triangle. This operation is evidently in accordance with the vector diagram and constitutes a very expedient method of determining field current of a salient-pole machine. shown A set of charts for a salient-pole generator is 204 inclusive.ELECTRIC CIRCUITS— THEORY AND APPLICATION 322 internal power and reactive power is spotted in the auxiliary- The chart exactly as for a non-salient. and the induced voltage. This figure illustrates the method for obtaining values of field current and angular displacement in a salient-pole machine. M. are read ^-. . The data main charts are tabulated exactly as for the and the charts constructed in a for the non-salient-pole generators. 199 to for these will be — 1 This chart as developed by J. is read. 1.Point located by P' and Q' — Fig..E. Bryant is described in Appendix I of a paper by C. For non-salient-pole machines it is possible to construct a simplified chart 1 which is applicable at all terminal in Figs. is its edge coincides with the figure on the smaller scale corresponding to the figure just read on the larger scale. a perpendicular is dropped from the located point onto the edge of the strip rotating about the field- which is then rotated until current center. similar manner. pinned down at the center of the field-current circles. = Field Current corresponding to Terminal Voltage Equivalent Field Current corresponding to Leakage Reactance Drop Equivalent Field Current (-Jkla) corresponding to Armature Reaction —Approximate vector diagram of a non-salient-pole generator supplying an inductive load. however. approximate. This latter will simultaneously take into account the effect of the armature reaction and the leakage-reactance drop. Fig. The field current in a non-salient-pole machine may be obtained by vector addition of the field current as obtained from the magnetization curve corresponding to the terminal voltage and a field current which is proportional to the armature current. is 323 is concerned. all quantities will be taken on a percentage basis. The following notation is used P = Q = Vt = l it = I fg = per cent active power per cent reactive power per cent terminal voltage per cent field current to give the voltage as given by the magnetization curve per cent field V t on open circuit current to give normal armature current on short circuit 7/ Ia = = per cent total field current per cent armature current Rotating the field-current vectors through 90 deg. 181. 182. as shown in under load is given by Fig. In setting up the equations for this chart. ' ^ Iff.. the resistance drop being neglected.: SYNCHRONOUS-MACHINE CHARTS voltages also as far as determination of field currents This chart. the field current If = lft+jJfJa = In 100 + JI {r +j v) fg t (33) . Figure 181 shows a vector diagram utilizing this method of determining the field current. = 1 } (35) &) with variable parameters The circles represent P constant ratios sP. Obviously. the machine operates at a high saturation where the magnetization curve is will give very good results. inaccuracies may be anticipated. it on the other hand. As already stated.iQi -Vt <7a Fig. The horizontal and vertical displacement of the center and the radius are given by A = B = 1 C = y^- (36) I (37) (38) ft Figure 183 shows this chart. circle respectively. If. the chart has to be used in conjunction with the magnetization curve. decidedly non-linear. the corresponding value of this chart should. 182. in general. If the machine is field current I ft be looked upon operating on the straight (or almost straight) part of the magnetization curve. As seen. angle lines have also been drawn in giving displacements between excitation voltage and terminal voltage. since for any value of terminal voltage V t must be known.ELECTRIC CIRCUITS— THEORY AND APPLICATION 324 which may be modified to (0+( "-W =/ 7 ™ ' Dividing through by I ft gives [vjWij] + [r&rT. ~ This equation represents a and T. particularly . as being approximate. T /T V Uft/Ifs . —Approximate vector diagram of a non-salient-pole generator with field-current vectors rotated clockwise through 90 deg. of actual field current and current corresponding to the field terminal voltage. i. The chart is applicable at any value of terminal voltage but must be used in conjunction. This value is too low for high saturations. should appropriately have the value n. 183. and core loss can easily be taken into account. however. the and windage.e. at a very low saturation. effect of friction .SYNCHRONOUS-MACHINE CHARTS at low power factors. Fig. The angle lines indicate displacement between excitation voltage and terminal voltage. 184.with the magnetization curve. involve no principles which have not already been covered in the discussion of the generator charts. The reason 325 for the discrepancies in this that the magneto-motive force (or equivalent field current) representing the effect of the leakage-reactance drop is region is always taken equal to its value at short circuit. as illustrated in Fig. the charts for motor action are sending charts. when friction and windage loss and core loss are neglected. while. — The circles in this Simplified chart for non-salient-pole generator. The performance of synchronous motors may be determined by charts of the same general type as those used for the generators. the equivalent field power factor current of the leakage-reactance drop. at zero (lagging). m The chart utilizes the equivalent field current at all saturations. While the charts representing generator action fundamentally are receiving charts. chart represent ratios of actual field current to field current corresponding to terminal voltage as read from the magnetization curve. 1 1 Assuming that the core loss is a function of the terminal voltage. The calculation and construction of the motor charts.. at the higher saturation indicated. — Synchronous Motors and Condensers. The resistance is neglected. as was previously done for the generators. be used to same machine when operating as a give the performance of the motor. to consider the core loss as depending upon the air-gap voltage. when the charts are drawn. designed to act as condensers alone and are not intended to supply mechanical power to external loads. the motor resistance is neglected. then. Leading reactive power on the chart should. . If a parabolic relation is assumed. therefore. S^S^ThSK^hSi power axis. Fig.326 A ELECTRIC CIRCUITS— THEORY AND APPLICATION chart constructed for a generator may. its performance may be determined by selecting operating points along the reactivesion. It is unnecessary. to prepare complete charts A in order to determine the performance of such machines. for voltage control in transmission systems are. generator. If the synchronous motor where it appropriately should operates as a condenser and all losses have the value n. is constant for each terminal voltage and maybe added on to the input power. when operating overexcited. Operating points for motor action are then located to the left of the vertical or reactive- power axis. The value of the shunt resistance may be based on the opencircuit core loss at normal voltage. The synchronous condensers used family of V curves suffices. It may be more nearly correct. in other words. however. the generator charts may directly be used If to represent motor action without exten- provided the sign of the reactive power is reversed. as from this the reactive current and reactive kilovolt-amperes may be obtained for any value of field current.-Magnetization recalled that a synchronous { it h u J non-salient-pole ""^ curve of The simplified motor.rent is lagging when the machine operates »S^iS. be considered as lagging and vice The correctness of this is obvious versa. provided the induced-voltage circles and the field-current curves are extended into the negative-power region. 184. underexcited. a constant shunt resistance at the air-gap terminal of the leakage impedance. while the curage-reactance drop at all sat. in general. The sum of these losses. the core loss may be taken care of by attaching. are ne gi ec te( i. as a rule. draws a leading current. in the circuit representing the motor. Figure 185 shows a vector diagram of a synchronous condenser operating overexcited and drawing a leading current.. therefore. therefore. In order to calculate these curves. it may be impracticable to obtain the curves by test. it is necessary to have the open-circuit characteristic of the machine and to know the leakage reactance and the coefficient k representing the effect of the armature reaction.-kIa Ek Vt --* -JJaXa kla. it is most . The coefficient k may be obtained either from an open-circuit and a short-circuit characteristic or teristic in from an open-circuit and a zero-power-factor characthe manner previously described in connection with the generator charts. they are almost always determined by calculation. because. the bottom point obviously being obtained directly from the magnetization curve. as a rule. In carrying through the calculations for a family of V curves. Usually. 4 . the V curves. for all practical purposes. the V curves are plotted in terms of reactive current They may also be plotted in terms of reacversus field current. 185. —Vector diagram of a synchronous condenser operating overexcited. It is sufficient. ~ kh (40) which 2/ represents the field current corresponding to the air-gap voltage (E a ). The leakage reactance may be obtained in the ordinary manner by means of the Potier triangle. when plotted in this manner. in tive power versus field current. to use reactive current rather than reactive power.SYNCHRONOUS-MACHINE CHARTS The V curves may be obtained by test or by calculation. become straight lines. 327 With large machines. and for such machines. to calculate but one point on each slope of the V. I* Fig. The airgap voltage (E'a) is obtained from E'a = V -jlaxa and the field current is given ls (39) by = f. It is advantageous. when an open-circuit characteristic and a zeropower-factor curve are available. as obtained from the magnetization curve. The latter covers a voltage range 200 20 Fiq. December. using data from the other characteristic curves. Figure 186 shows the characteristics of a large synchronous condenser of modern design. L. reactive current. A. 186. 1927." Jour. In this paper. Points on any V cu rve may easily be checked by calculation.E. although be used. "Synchronous Condensers. p. P. attention is called to the fact that a more economical design results when the lagging kilovolt-ampere capacity is less .328 ELECTRIC CIRCUITS— THEORY AND APPLICATION practical calculate to points corresponding to 100 per cent any arbitrary value may. With smaller condensers. Large synchronous condensers of modern design usually have same kilo volt-ampere rating overexcited and underexcited. 40 60 100 80 120 Per Cerrr Field Amperes 160 — Characteristic curves of a large synchronous condenser.. 1 This is due to the fact that these machines have the 1 Alger. of course. 1330. from 60 to 180 per cent of normal. on the other hand..E. zero-powerfactor and short-circuit saturation curves are given and also a complete family of V curves. it frequently happens that the rating underexcited is somewhat less than the rating overexcited.I. The open-circuit. 4 per cent. EXAMPLE 1 This example illustrates in detail the calculations of the necessary data for the construction of charts for a salient-pole generator. 85. 187. field current. = 1. In addition.8 power-factor generator. — Typical 140 160 180 200 220 : Norma l-Mage Field Currenf saturation and synchronous-impedance curves of a large salient-pole. Fig. Transient reactance = 30 per cent. It is suggested that the use of static reactors in parallel with the synchronous condenser be considered in rases where a large value of lagging reactive power is require^. 80. 60-cycle.0 Short-circuit ratio = 0. 40 60 80 100 120 Field Current in Per Cent of No-Load Fig.SYNCHRONOUS-MACHINE CHARTS 329 such constants that the field current. 90. 0. than the leading. Statement of Problem The characteristic curves of a large salient-pole generator are given in The ratio of pole arc to pole pitch is %. . and 70 per cent of normal. Armature resistance Armature leakage reactance = 23 per cent. would have to be decreased to a prohibitive degree. the exciters and the control equipment (Tirrill regulators) are not sufficiently flexible to vary the field current through such an extreme range. 95. Prepare charts giving the performance of this generator at terminal voltages of 100. It might even be necessary to reverse the It is always undesirable to operate a synchronous machine at a very low excitation. 187. in order to give the same rating underexcited as overexcited. Radii of circles in per cent placement of per cent E'a = 200 per cent E'a = 100 per cent E'a = 50 per cent 435 392.372 = .372 2tt/6 Cross-magnetizing coefficient (see equation (32)) _ Ke = 0.5 213 870 826 782 739 696 608. Calculation of Generator Chart Circles of Induced Voltage Vertical dis- v t. 174 152. obtained by construction when the charts are being drawn.5 195. vL&-x sm yJ )V2 2wl {3-. since these circles are concentric and The radii of intermediate circles are more conveniently equidistant. Horizontal displacement of center Vertical displacement of center Radius of circles = Ea V = = — Xa t Table XVIII gives the necessary values calculated from these formulas.5 314. 80 per cent coefficient (see equation (28)) Kd = b2 0. per cent 100 95 90 85 80 70 (Example 1).5 278. Leakage reactance xa = 23 per cent Leakage-reactance drop at rated armature current xa Ia = 23 per cent From the magnetization curve.112 Kd 0. hence.8 435 413 391 369. = _ °.45' = ax/62 0. however.301 Circles of induced voltage.2 be noted that the radii have been computed for only three values of This is fully sufficient.4 217. .ELECTRIC CIRCUITS— THEORY AND APPLICATION 330 Solution Necessary Generator Constants. the field current corresponding to this drop is The coefficient k // = 20 per cent k representing the effect of the armature reaction = //-// _ /„ Demagnetizing 100 .5 184.8 center.45*" *" 6 = 0.1lo 112 0.5 348 304. It will E'a . Table XVIII.9ra 1 axl .5 352.5 206.20 100 100 = is. Sin ¥j 1 Ke 0. SYNCHRONOUS-MACHINE CHARTS Auxiliary Chart. —Auxiliary chart for salient-pole generator.8 //A (jrj = 125 //A per {yj cent in Fig. 187 and the calculations in Curves of Constant Field Current. 188. 188 rj§ jk : 40 w ^ 20 ^ -sL' 20 vO C "S 40 -g 60 £ 80 v/ § VX\ 3^ c 9?" <=fr b 120 •5 140 1 £ 160 4 ><* KV* > "g p s\ 180 20C 24C — — 26C — 22C 28() 3 —4- '¥ 1: s — 30U o I ff*=- ^- ^ ^ j- »*7r 60 80 00 20 140 60 80 WO J20 i 40 \/£0 1 60 - J4)0 p' of Rating Active Internal Power jsfj?T in Per Cent Fia. Horizontal displacement of center Vertical displacement of center Radii of circles The = X ^j = — 100 auxiliary chart 100 r 80 - 60 = is //A shown 100 331 = 100 -r- 125 per cent 0. By using the method previously described for the salient-pole machine and the arrangement suggested in Table XVII. are given in Fig. The characteristic curves Example 1. the necessary numerical data are . 1 206.0 - 64.0 44.0 62.0 18.8 39.3 5.5 - 131.3 91.1 91.0 21.8 70. per cent per cent z< Vt p.8 3.0 94. - 25 75 27.3 80.0 100 - 02.6 139. ) [ 23.4 9.7 18.0 6.0 25 50 3.1 252.23.0 62.0 54. 62.5 1 08.5 180.5 24. per degrees cent per cen per cen per cent 77.4 - 21.0 50 129.0 26.0 29.0 125.7 8.2 112.0 25 25 3.0 14.0 11.0 85. P'.0 75 75 9.3 50 - 100 I 40.) 91 5 () 100.0 26.0 100 100 75 - 100 100 - 25 100 100 -\- 75 17.0 112.3 73. 44.0 55.0 89.0 6.0 1 192.4 89.: //.— Calculation of Generator Chart (Example current Curves Vt.0 84.0 5.0 9.5 95.0 75 75 13.0 132.5 100 - 12. ()100 2.0 10.2 31.0 166.0 11.0 117.2 18.0 50 8.2 - 0. c 53.8 133. S 202 100 100 ) 100 25 100 25 100 25 100 25 100 25 25 100 100 100 100 100 100 + 75 + 50 + 25 25 70.7 74..6 3.0 75 10.5 .0 1 18. J 76 53 88.0 75 50 9.1 -211.3 25 -162.0 50 112.1 114.0 24.0 63.0 75 -100 -125 -150 8.6 264.3 13.1 Z.5 18.0 100 100 13.0 59.0 1 47. C - 90. -201.9 12.3 17.0 9.9 111.0 44.5 10.1 100.0 54.9 137.0 100 100 -\- 50 16.0 46.4 30.0 85.0 51.0 4..2 1 40.5 141 100 -10( 123. 3 t> 82.2 1 26.3 57.6 102.5 13.0 51.8 1 13. 100 +101 100 + + + 100 100 - 100 100 //.2 27.5 1 48.1 85.5 34.5 85.3 174.0 100 75 100 75 100 75 100 75 100 75 100 75 100 75 100 75 100 75 + + + 79.0 17.4 80.0 21. > 83 2 94. Kl'f 25 100 100 - 134.8 68.1 209.4 55.0 10.3 162.f 100. 42.' degrees 98.0 56.5 102.0 74.0 15.0 148.0 50 106.0 +140. E'j'f 25 100 100 77.0 90.0 15.9 49.0 4. 25 88.0 4.6 12.3 Q' .1 95.2 118. C 166 100 -12J -15C 25 +10C 128.1 145.3 2.0 23.0 38.8 4.7 246.7 -100.0 f f 95. 1 09.5 142.3 -- 61.0 60.2 206. *Sill 5() lll.0 72 5 39.4 7.5 78.8 148.7 129.0 75 flOO 18.f 124 7*> 117.0 8.0 25 94.0 67.0 102.0 1 35.5 4.0 3.3 83. per cent 21.2 100 50 100 50 + + + 9.8 150.0 74.0 75 25 11.7 148.0 96.9 1.Ea V.6 101. Ea per cent per cent degrees o ) 26.0 27.8 79.7 22.2 42.2 52.5 136.7 - 1 14.0 27.6 94. -160.0 12.0 75 50 12.3 39.6 0.7 132.0 50 8.0 2.0 11.0 12.7 7.9 43.7 + + + 6.8 24.2 107.4 149.3 171.3 -93.21.0 75 8.78.1 6 8 142.1 304.0 180 60.4 -136.4 62.5 146.0 68.6 15.4 51.5 1. C 164. + + + + 25 4.0 -124.0 100 100 100 -I- 25 15.5 100.0 81.8 1.0 61.3 20.8 24.0 5S14. per cenl Q.0 - 22.0 57. 71.1 63.6 103.0 5.0 25 - 27.4 12.5 25 -203.3 150.0 -128.0 J 5.3 2.0 144.4 105.7 31.8 124.7 -173.0 132.1 20.1 29.8 50 200.0 100 100 100 25 100 100-50 97.3 6.0 25 2.6 -.0 49.3 119.6 34. - + + + 75 50 42.6 -100 -125 25 -150 50 +100 50 16.5 1 31.0 3.0 4.2 82.5 50 + + + + 66.0 34.4 15.7 127.0 39.0 12.2 147.2 180 2 60.0 26.3 84.5 56.0 7.7 .0 4.5 109.2 28.3 72.0 45.8 50 -207.2 111.0 113.3 78.0 91.8 -125 5.8 77.6 2.0 16.0 13.5 18. 22.0 34.3 130.0 .5 75.7 83.0 50 11.9 21.2 75.6 - + + + + 69 75 100 100 P' «» Field- 1).9 46.0 21.5 218.49.> 105.8 27.8 32.5 13.0 10.0 18.3 98.9 123. q.5 50 -150 75 +100 14. 88.0 145.6 04..5 8 5 10.0 6.0 60.0 36.1 118.0 245. 7 124.0 92.0 50 50 123.3 112.0 87.2 50 72.5 77.7 112.0 .0 - 62.0 30.1 89.0 323.3 168.5 100 50 100 100 -100 5.0 1 15.2 7.5 +129.0 19.5 41.4 If.1 -166.0 - 5.0 .5 61.9 25 50 100 142.2 38.0 25 100.7 52.5 + + + + 75 100 100 49.2 117.0 - -- 53.0 - 48.8 93.6 1 51.0 16.0 25.0 165.7 5.0 58.8 1.6 // per cenl +145 + + + -123.2 1 06.0 71.3 309.4 87.5 78.5 3.0 > 117.3 92.0 75 101.3 57.0 1 70.4 4.6 84.0 3.6 84.5 8.C 332 C S J( C ( < ELECTRIC CIRCUITS—THEORY AND APPLICATION Table XIX.71 30.0 23.0 76.39.5 58.0 32.0 70.6 100.2 10.6 25.4 f f 300.6 95.8 1 52.0 108.0 50 - 75 6.0 204.5 50 50 25 7.2 119.0 41. 0 214.0 100 82.5 26.6 108.0 25.4 144.2 164.0 41.6 193.7 98.0 34.1 161.0 200 75 23.6 339.0 176.3 14.3 110.0 152.0 68.8 129.8 8.0 93.5 60.6 164.0 22.5 110.7 222.0 47.0 19.0 56.3 175 -272.0 100 100 Q'.0 122.0 200 27.0 108.0 125 50 16.0 45.0 150 50 19.9 -150 +100 26.8 198.8 130.0 46.1 134.0 10.4 131.0 89.8 22.7 99.0 54.0 35.0 125 -197.0 58.7 104.4 190.0 150 -212.0 26.6 182.7 127.0 17.9 12.7 151.0 12.5 175 92.0 28.4 88.0 177.0 6.9 - - 9.2 176.0 40.5 202.5 215.9 181.1 123.0 37.1 87.9 19.9 75 If.9 112.7 181.2 101.0 47.0 175 -231.2 155.0 88.0 115.0 56.1 154.7 102.0 37.2 100.2 75 100 100 P'.0 118.8 14.0 69.0 73.0 92.0 25 75 21. Z Vt' degrees 155.0 9.6 8.3 173.9 148.5 150 175 100 229.5 81.SYNCHRONOUS-MACHINE CHARTS 333 Table XIX.0 360.2 245.0 5. per cent - per degrees cent 100 125.7 133.0 -118. 62.0 34.6 66.6 18.7 175 100 cent 84.0 8.0 154. per cent -100 11.5 154.0 63.7 20.0 69.0 -146.5 62.2 183.9 180.0 115.3 100 - Eal'f 25.0 36.0 232.6 -125 11.5 31.0 54.1 134.5 140.0 44.0 27.0 62.0 76.5 153.1 362.6 264.5 97.8 199.0 59.0 175.3 100 per cent Field- 1).0 79.0 + 51.0 124.2 91.0 -126. per cent 100 100 100 100 100 100 100 100 100 125 100 125 100 125 100 125 100 125 100 125 100 125 100 125 100 125 100 125 100 125 100 150 100 150 100 150 100 150 100 150 100 150 100 150 100 150 100 150 Q.0 100 + + + cent 89.0 167.0 111.9 150 175 81.5 91.4 .3 200.0 200 200 200 64.4 per cent 19.8 125 -237.8 168.0 47.0 23.1 132.4 12.0 29.5 -107.7 23.3 382.0 125 -100 14.0 10.9 31.0 203.0 -294.0 185.5 5.9 - 15.0 265.6 9.4 172.5 173.0 44.2 164.0 20.4 136.0 312.1 94.0 28.8 167.0 42.0 49.5 119.5 64.0 51.7 104.5 200 25 28.0 -31.3 175 + + 152.1 31.0 191.0 97.7 175 -150 200 +100 34.2 150 100 100 - - //.0 60.0 161.0 56. (Continued) — Q' P' Vt.0 80 69.2 209.0 41.5 100 90 4.0 64.5 57.5 150.5 173.0 150 -174.6 60.7 97.1 149.7 211.0 125 17.3 170.0 31.3 175 100 175 100 100 -100 20.0 75 175 + + + f Eal'f per -111.4 159.0 30.0 + - 6. -Calculation of Generator Chart (Example current Curves.0 36.0 200 50 24.2 120.0 8.5 125 50 20.0 125 75 15.8 115.0 50 23.0 66.1 119.5 175 -125 19.2 129.5 + + degrees 7.2 180 +100.0 108.0 70.3 161.1 102.0 79.0 175 - + + + per Ea L E'l 66.5 130.0 21.0 32.5 -125 13.8 129.3 222.0 100 -184. per cent p.0 49.3 289.3 199.0 30.0 52.0 27.7 159.0 107.0 152.0 175 23.1 183.5 13.0 34.0 150 92.0 34.0 70.6 62.0 116.0 71.0 25.7 5.0 +194.2 189.0 78.7 127.9 91.9 161.8 210.2 89.5 250.6 177.5 150 -150 175 +100 31.0 28.5 -159.5 27.2 100 200 100 200 100 200 100 200 100 200 100 200 100 200 100 200 90 15.1 75.0 39.0 158.0 47.7 26.1 91.3 130.2 11.6 177.8 7.5 -180.9 17.4 274.0 67.0 77.0 33.4 99.0 200 50 30.0 51.0 26.2 228.4 96.0 46.0 21.0 62.1 -150 +100 22.2 156.4 25.0 175 -158.0 200 -100 21.0 67.0 13.0 125 75 21.3 151.5 87.1 151.0 120.0 10.6 245.3 150 -125 16.9 171.2 291.0 125 25 16.0 12.0 105.0 175 75 - - 7.0 171.0 125 + + + - + + + - 338.2 23.4 28.0 175 25 26.2 100 90 274.5 100 -225.0 71. per cent 12.8 109.5 246.9 221.0 70.2 135.0 200 32.8 188.0 65.5 -139.7 146.0 32.0 59.6 125.5 81.9 60 74.0 180 20.5 95.7 84.0 200 25 26.0 175 -193.7 178.9 20.0 174.0 + + + + + 45.0 238.8 24.0 56.2 132.2 100.0 119.3 8.0 78.5 118.6 150 -253.1 8.0 150 76.7 120.2 10.0 125 25 18.0 65.6 150.0 175 50 27.2 200 -125 200 -150 +100 130.2 16.7 81.7 15.0 72.9 12.9 108.0 37.5 -124.2 310.0 -215.0 46.5 25 /'/ 153.6 75 18.5 180 17.3 228.0 116.0 150 - per cent 5.6 -147.3 124.0 150 25 20.0 38.5 138.2 84.4 94.6 -10.6 -100 17.0 175 25.5 154.0 66.2 152.6 113.0 -253.0 175 50 22.5 180 +146.0 30.3 122.7 30.8 154.0 19.0 248.5 139.5 172.8 27.5 10.7 16.6 126.3 +13.0 100.0 45.0 41.5 203. 71. 0 60 40 10.4 14.2 26.6 40 6.6 2.6 8.0 85.0 13.5 82.0 42.4 81.9 122.0 60 20 9.0 195.0 16.8 105.9 5. C.0 60 20 10.4 36.6 -232.1 92.2 86.0 12.0 +141.5 12.3 15.2 19.0 5.7 86.0 4.7 19.0 56.7 40 80 5.5 21.0 60 60 8.0 73.7 79.7 45.8 67.2 -138.2 169.1 -197.7 109.6 170.7 6.9 63.0 51.5 28.8 11.5 4.0 40.5 85.2 69.8 75.3 170.0 157.0 6.3 40 60 7.3 11.1 153.0 .7 142.0 115.2 90.3 67.3 24.0 57.4 71.3 100.6 155.6 21.0 5.5 170.3 90.0 122.0 90.6 17.3 102.0 25.6 162.0 20 90 -100 -120 20 -140 20 -160 40 +100 2.0 64.2 10.3 10.3 125.5 18.5 -162.8 40 100.0 60 9.6 25.0 20 90 20 80 2.E« Vt' degrees //.0 48.0 25.1 226.0 267.6 40 4.3 166.2 8.9 312.1 15.2 93.6 6.7 1.8 4.9 - 21.0 18.5 86.0 68.5 40 95.0 76.7 40 40 7.0 +160.0 156.2 97.3 34.2 158.2 7. per cent per cent per cent 40 79.0 110.7 171.0 3.0 60.3 17.5 +192.5 122.7 20 4.9 141.7 110.334 ELECTRIC CIRCUITS— THEORY AND APPLICATION Table XIX.8 20 2.0 37.0 7.9 83.8 -102.3 1.2 237.7 194.0 20 60 4.0 60 40 8.2 28.3 -237.6 169.1 32.8 2.7 101.5 149.0 60 90 90 + + + + + + + + 80 13.0 11.8 -195.0 40 -100 40 -120 40 -140 •40 -160 60 +100 5.5 373.0 262.0 20 40 3.0 7.9 72.1 184.3 62.0 73.5 -171.0 35.5 11.9 100.2 1.4 46.2 270.0 61.1 44.0 3.2 58. — 34.8 22.0 55.4 79.0 86.9 264.0 103.1 45.2 228.8 69.3 106.0 130.1 92.8 110.2 149.3 70.3 100.5 20 2.1 63.5 135.0 79.5 14.8 58.6 136.5 22.3 138.3 60 105.1 11.9 40 4.0 + + + h Ea -3.3 17.1 7.1 51.0 165.5 20 90 40 90 40 90 40 90 90 90 90 40 90 40 90 40 90 40 90 40 90 90 90 57.0 20 2.3 80.5 24.8 31.0 87.6 76.6 309.0 85.8 30.3 1.2 373.6 37.0 80 110.0 71.1 112.0 60 12.0 82.5 131.7 6.5 12.5 135.0 20 95.6 35.0 10.3 110.9 102.4 44.8 121.8 45.6 4.0 192.2 40 4.2 76.5 70.5 26.0 9.0 -.8 -132.5 86.0 74.0 40 80 8.9 136.5 3.6 20 20 6.1 -129.0 85.2 54.o 40 5.0 92.1 180.0 75.5 14.2 78.0 109.4 73.0 13.0 164.7 30.0 3.7 40.3 111.2 96.0 90 90 .2 20 3.6 25.0 -128.0 78.0 32.0 159.5 9.4 53.1 168. 20 40 100.8 8.7 loi. {Continued) Q.5 68.6 100.9 111.7 21.7 41.3 4.5 82.0 6.8 221.0 164.3 123.2 5.2 2.2 93.0 26.0 8.6 116.0 37.5 139.5 222.2 + + + + + - - - t If degrees per cent L.2 121.0 9.1 74.3 101.5 4.2 204.8 181.0 199.0 80 20 P' E'a If + + + + + - 70.9 79.4 158.2 130.0 7.2 100.4 4.9 126.7 45.5 -200.0 30.0 60 -100 60 -120 7.2 92.9 222.2 2.0 17.1 109.0 174.2 86.8 40 90 - 9.0 2.6 87.4 120.0 15.0 86.4 121.9 3.6 134.0 60 60 11.6 14.5 67.7 135. percent P.3 34. Vt.0 95.4 -160.4 67.6 5.5 42.0 60 80 7.9 73.3 66.3 110.3 88.9 42.4 2.3 59.4 137.0 ( Field- 1).4 1.0 45.0 3.2 88.4 52.8 25.4 148.2 39.3 134.8 20 84.2 119.7 12.0 168.0 80.1 149.0 47.6 +124.8 166.7 91.3 22.1 80.5 19.5 120.1 68.5 56.9 85.2 +130.0 17.5 85.6 116.5 4.0 20 3.6 4.6 20 60 90 20 2.5 18.8 -245. E'a per cent per + + go 90 degrees cent - 90 90 l'„ P'.2 20 40 2.4 99.0 1.0 119.0 48.6 17.5 176.0 164.0 40 40 90 60 90 60 90 60 90 60 90 60 90 60 90 60 90 60 90 60 90 60 90 65.9 -165.0 117.0 85.0 60 7.3 73.0 169.9 — - 125. per cent Calculation of Generator Chart (Example current Curves.0 61.8 27.0 6.8 49.5 169.9 63.7 126.0 80.0 61.0 60 - _QL E'jf per cent per cent + + 35.0 10.4 4.2 40 92.9 125.0 8.7 155.0 304.0 20 90 20 115.0 81.5 45.5 105.8 62.0 84.3" + + + - - 198. per cent 61.0 16.2 98.5 67.1 115.2 -108.4 4.74.0 +181.0 5.0 40 6.2 20 3.1 2.0 25.6 64.4 + + + - 28.4 7.4 28.1 25.3 + + + + + 13.2 49.0 90 -100 -120 -140 -160 20 +100 90 20 90 20 90 20 90 20 90 20 90 90 90 90 90 90 20 90 20 + + + + - + + 72.5 29.9 81.5 90.5 60.7 20.1 T6.7 59.5 51.3 225.5 155. 0 20.8 75.5 94.7 109.1 -139.2 198.6 76.4 16.0 44.0 120 60 16.0 140 101.0 11.8 83.0 32.0 87.0 88.3 40.8 123.8 75.0 66.0 17.3 173.6 119.9 8.8 150.6 178.4 + + - 42.1 58.2 105.8 120 40 17.3 11.1 27.0 93.0 16.2 104.2 145.3 103.4 179.3 381.4 167.5 114.0 29.0 100 14.9 81.6 105.7 88.0 100 -261.4 72.6 131.0 296.6 32.0 100 40 17.3 18.1 226 91.0 203.2 87.0 67.6 230.2 75.5 15.0 20.0 157.3 107.5 -116.0 92.7 13.0 45.0 80 90 -100 10.3 80 -160 8. per cent per cent per cent 90 60 6.0 34.3 15.5 16.4 5.2 21.0 49.0 24.2 240.0 140 94.4 35.6 40.5 79.5 88.2 164.9 100 +100 21.0 100 -73.0 83.5 79.0 25.3 163.6 21.0 4.5 170.0 80 60 15.1 -120 9.6 20 degrees 2.0 120 119.5 22.0 +134.8 116.2 38.5 17.2 52.0 44. If degrees 80.0 12.5 130.6 38.0 60 17.0 100 90 100 74.5 37. Q'.5 71.4 206.0 227.6 88.2 9.0 100 90 100 90 90 90 90 90 100 90 100 - + + + + 80 20.0 80.3 4.8 100 //.7 45.9 206.2 60 18.0 100 98.4 240.5 11.7 179.0 3.0 157.0 113.5 65.6 134. per Q.6 8.2 10.0 9.5 17.7 131.0 104.8 67.9 158.9 111.8 132.9 162.8 176.5 49.6 28.6 102.2 18.4 25.1 72.0 6.0 62.8 120 -120 14.2 51.7 246.4 120.6 174.2 255.0 87.0 14.3 17.0 37.7 123.9 105.2 100.0 62.5 80.9 21.7 182. per cent 90 -140 -160 80 +100 90 80 90 80 90 degrees cent //.0 55.3 180.2 25.8 -243.0 92.6 28.0 80 80 80 122.5 .9 137.7 271.0 120 134.0 94.3 328.0 4.0 294.4 342.7 7.0 25.5 73.0 80 90 20 12.5 - 49.6 136.6 60 13.7 85. (Continued) — P' Ea.6 59.9 33.0 80 90 80 40 11.3 120 -140 13.0 80 13.5 174.2 22.0 39.2 191.5 189.4 15.6 88..7 241.3 79.0 165.1 4.7 80 -140 9.0 80 90 80 60 11.9 78.0 100 - 98.4 28.0 100 12.0 5.7 270.3 14.0 5.6 192.3 58.3 79.4 -146.1 - per cent 3.9 66.7 283.6 48.0 28.1 247.6 -156. Ea B'j't - 90 335 92.1 66.2 130.2 125.0 38.7 10.0 73.1 56.5 179.0 140 60 21.0 14.2 81.0 80 80 + + + 40 14.— SYNCHRONOUS-MACHINE CHARTS Table XIX.0 56.0 163.3 25.6 + + + - 9.0 120 80 23.1 198.0 29.5 94.5 -126.3 174.2 46.7 - 20 20.5 29.9 106.8 95.3 320.5 38.4 258.9 136.5 25.8 154.0 +102.2 30.0 90 -100 100 -120 100 -140 100 -160 120 +100 90 120 90 120 90 120 90 120 90 120 90 120 90 120 90 120 90 120 90 120 90 100 90 100 90 100 90 100 90 90 90 + + + + - 40 11.7 56.6 67.7 117.3 70.0 50.0 120 129.0 6.0 140 20 22.0 -168.0 53.4 173.3 213.0 12.0 132.0 21.5 5.9 92.0 80 80 16.4 97.2 23.9 1.9 90.2 97.5 88.6 6.5 168.0 88.0 120 60 22.0 85.0 —111.9 165.0 80 90 80 + 20 13.0 33.2 87.2 178.5 22.9 424.4 80.0 156.0 120 18.1 95.0 6.9 180.0 100 -224.2 10.1 61.0 85.3 142.7 137.0 57.1 255.0 17.2 22.0 312.5 158.0 10.1 85.5 112.9 68.9 43.1 167.6 15.5 21.5 133.2 116. per cent Calculation of Generator Chart (Example current Curves.6 173.9 173.5 106.4 82.0 21.5 24.0 30. P'.6 169.0 169.5 79.0 140 90 90 90 90 140 90 140 90 140 90 140 90 140 + + 80 27.0 33.5 144.0 174.2 74.0 12.5 23.1 81.6 114.0 80 69.5 6.0 9.0 70.0 19.0 1.1 359.1 154.0 9.0 62.8 135.0 -251.0 120 40 21.0 80 127.0 35.7 166.0 172.6 40.7 32.1 71.0 100 Ealf per cent per cent -206.6 108.0 120 124.3 109.1 39.9 87.8 92.0 100 89.2 63.2 121. V.0 15.0 36.0 48.1 113.1 96.8 144.0 59.0 176.3 - + + + + + - 171.5 17.0 15.9 20 17.8 125.3 -201.3 182.9 31.0 236.7 120 -160 13.0 133.0 69.3 64.6 187.1 394.6 148.8 71.0 44.5 -179.2 140 +100 29.2 78.0 13.9 27.5 19.4 8.6 119.0 + + + 68.6 77.0 80 90 80 80 12.9 61.5 153.6 96.0 120 80 15.9 126.0 + + + + 94.8 138.0 120 20 19. per cent P.6 82.9 88.6 6.5 53.0 60 90 60 6.0 41.0 22.6 155.7 -236.1 28.0 76.0 60.8 63.0 91.0 92.5 39.1 176.9 6.9 409.4 31.3 8.5 38.5 118.0 113.4 128.0 -214.0 25.0 148.0 53.0 8.2 181.0 43.3 100 5 87.5 20 per cent 165.0 - + + + - + + - 13.5 67.0 20.0 140 + + 40 24.5 103.0 + + + + 1).9 99.5 82.0 120 90 -100 14.9 5.8 170.2 154.9 11.0 100 59.0 107.5 202.8 75.0 7.0 27.9 152.0 140 90 140 78.0 213.8 100 90 Field- 87.2 46.7 19.5 89.7 34.0 121.4 79.2 128.5 30.0 80 90 80 80 10.7 179.3 207.0 138.2 200.4 150.0 100 -188.4 160.0 133.9 178.9 183.0 30.3 187.3 77.0 120 99.0 36.0 80 132.5 87.0 17.5 -270. 2 -120 16.2 274.0 13.5 159.0 9.8 52.5 122.6 167.0 75.0 87.0 204.0 33.5 163.7 87.1 35.0 35.0 160 80 30.6 215.0 112.0 283.2 77.5 173.9 -160 15.0 87.9 -184.9 131.9 116.2 212.5 84.8 -158.0 180 -288.0 140 130.6 247.0 87.2 167.0 -102.6 229.0 160 127.0 .0 98.2 191.5 209.3 200 75 30.0 43. per cent 18.5 98.336 ELECTRIC CIRCUITS— THEORY AND APPLICATION Calculation op Generator Chart (Example current Curves.0 180 60 32.0 80 180 + + + + - 103.0 237.0 54. Vt.0 119.0 7.0 200.0 200 -100 200 -120 200 -140 200 -160 +100 23.1 205.6 201.0 96.0 312.0 46.9 200 29.5 53.6 22.5 17.0 -109.0 180 20 28.2 116.0 79.0 -136.0 68.0 -184.7 94.5 -102.5 10.8 152.0 -288.0 8.3 128.3 215.5 49.5 - 101.7 152.5 39.0 40 24.0 78.0 92.0 27.0 330.6 94.0 200 36.0 21.8 128.8 155.3 28.0 180 27.0 64.6 429.0 140 .3 209.0 -102.4 92.4 85.0 92.4 222.4 44.0 28.0 -309.5 -251.6 220.3 140.0 36.0 68.5 96.0 -106.0 31.1 258.0 160 28.0 -212.0 128.2 182.0 -108.3 103.5 202.0 96.0 13.0 -117.7 36.5 200 32.0 - 94.8 140 90 P' P'.5 86.0 + + 119.0 33.5 28.5 32.3 190.4 90. per cent P.7 185.5 111.8 80 17.0 140 90 160 90 160 90 160 90 160 90 160 90 160 90 160 90 160 90 160 90 160 90 160 90 160 90 180 90 180 90 180 90 180 90 180 90 180 90 180 90 180 90 180 90 180 90 90 -100 -120 180 -140 180 -160 200 +100 90 200 90 200 90 200 90 200 90 200 90 200 90 200 90 200 90 200 90 200 80 per cent 140 160 80 per cent 140 90 90 Eal'f' 111.5 96.1 80.9 236.0 70.2 14.4 6.3 -143.6 269.0 73.1 19.4 82.0 160 24.0 160 60 21.0 23.0 - 89.0 183.0 -153.5 347.4 8. {Continued) Table XIX.0 140 135.4 34.0 -171.2 198.2 189.4 201.3 106.0 294.6 22.0 -201.6 91.0 180.0 182.8 186.8 -140 18.1 199.0 200 34.0 160 132.0 15.0 12.0 93.0 354.6 25.3 34.9 249.6 80 22. per cent 19.6 214.0 119.0 160 20 25.0 124.5 180 - 75.0 160 + + + + - 11.0 61.0 105.1 121.0 69.2 19.0 178.5 86.0 22.2 408.8 193.5 62.1 158.9 80 60 20.0 24.1 123.0 120.5 25.0 322.E« //' degrees 9.6 207.0 -233.8 85.1 197.0 52.0 1.5 + + + + Eat per 40 140 90 £ Vt' 120.0 144.3 +100 32.4 33.0 200 40 27.125.0 57.5 140 76.0 20.0 129.5 5.8 98.2 184.0 -160 17.0 -110.2 224.5 42.8 Ea Field- 96.0 -113.3 162.0 202.3 86.5 50.6 173.5 227.8 191.5 7.4 66.0 47.0 79.1 11.0 -216.7 186.0 81.0 70.6 90.0 200 20 31.2 107.0 180 -253.0 -242.9 101.3 218.0 30.3 270.2 187.5 8.0 241.5 58.6 180.2 191.6 200 21.5 45.8 31.0 75.0 56.5 -140 15.7 20.6 103.0 95.9 245.0 11.4 320.0 -103.0 62.7 177.5 114.0 236.9 36.4 +100 35.0 9.1 100.4 94.0 160 -100 19.5 64.3 225.0 30.0 234.8 -104.0 25.6 42.7 187.0 36.0 200 22.0 63.5 74.3 103. per cent 90 140 90 140 90 140 90 140 90 140 — Q.4 194.5 96.7 295.6 194.0 180 83.0 138.3 75.0 68.3 50.5 33.0 180 -162.0 180 136.8 133.7 110.0 37.9 254.2 111.0 42.0 269.0 33.3 -305.7 307.9 117.5 +280.0 80.0 160 137.0 200 22.0 72.8 99.5 225.0 67.5 204.3 287.9 376.0 .0 90.4 217.0 90.0 -110.4 183.0 6.5 23.7 329.0 42.7 1.0 98.0 137.5 31.0 - 206.6 225.3 182.5 113.5 381.5 80.0 133.6 211.7 183.0 160 80 20.5 55.0 40 + 47.6 142.2 -106.0 -346.0 160 90 Eaff 106.1 99.0 65.4 117.3 31.5 -134.5 220.0 222.4 63.2 35.0 71.0 17.2 161.2 190.5 -120 18.0 +186.8 52.8 207.5 per cent degrees per cent 85.4 205.6 250.5 26.8 121.0 83.0 272.3 211.1 172. per cent - -100.5 39.2 172.5 150.9 228.0 180 40 30.4 180.3 259.8 105.0 28.9 141.0 180 -190.0 93.1 135.8 115.0 205.0 180 25.0 51.2 200 80 24.0 72.2 46.1 200 60 25.0 261.0 -113.9 288.2 -126.3 112.0 160 40 27.0 253.5 20 60 23.0 200 20 28.1 185. - 182.0 97.7 17.0 328.0 180 155.5 82.0 15.6 -268.7 90 - per cent 60 90 90 degrees cent -100 17.0 180 -222.0 57.1 68.0 58.0 90 90 Q' Q'.0 40.0 - 1).0 -274.7 247.0 13.2 49.2 18.0 160 40 22.0 197.5 14.2 108.0 92.0 26.7 199.1 89.0 160 60 20 23.0 191.5 28.0 87.6 21.5 25.0 11.0 53.1 236.9 55.8 126.0 180 -325.2 85.0 10.0 19.0 -103. 0 23.5 + + 49.0 80 -100 7.5 76.0 + 100 + 75 + 50 + 25 + + - 115.0 78.2 90.2 78.7 5.0 22.3 188.0 1.3 121.0 217.0 189.5 50 25 9.0 19.0 3.9 104.4 8.0 28.2 20.0 52.2 19.6 60.6 49.0 339.8 50 45. Ea P'.5 53.2 50 -145.5 15 3 283.0 50 50 8.4 2.5 14.0 116.0 100 80 100 80 80 80 + + + 80 100 80 100 80 100 80 100 80 100 80 100 - 80 100 75 26.0 -113.5 94.5 178.7 -171.9 17.8 46.8 87.2 8.2 73.0 150.9 75 80 per cent per cent 50 80 .5 99.5 62.7 40.0 +101.5 75 75 75 117.0 91.0 123.0 206.5 12.5 14.3 10.2 99.1 137.8 117.1 326.2 + + + 16.7 28.0 78.0 75 25 16.3 25 -133.8 114.6 -108.0 11.0 195.3 50 25 - 36.0 100 +100 29.0 ' 35.0 68.2 75 +100 23.6 103.5 -120.0 6.1 2.0 47.0 68.0 25 7.8 57.8 143.0 100 25 68.5 6.3 83.3 9.4 182.0 60.8 193.1 50 -239.5 +176.8 45.2 61.6 85.8 172.2 9.8 192.5 181.0 166.E« Eal'f' per -138.3 4.4 127.8 + 48.8 52.5 90.2 180.6 - 32.3 - 39.0 15.8 46.0 141.0 75 80 -100 11.0 74.5 93.0 3.0 50 -190.0 58.0 49.8 -114.4 81.0 140.0 115.5 75 75 20.0 64.7 108.2 315.8 101.2 124.0 7.0 51.2 14.0 50 55.0 107.5 119.7 -115.5 +145.5 65.6 36.5 +265.0 11.0 80.1 97.5 -107.0 100 75 16. percent P.0 80 75 80 75 80 75 80 80 50 80 75 80 75 80 75 80 75 + + + 116.5 25 50 4.7 175.1 81.5 44.3 -114.0 59.0 45.0 67.6 15.0 158.0 80 -100 25 -125 25 -150 50 +100 80 50 80 50 80 50 80 50 80 50 80 50 80 50 80 - + + + + - degrees 41.2 12.0 12.5 62.5 25 6.0 -119.5 100 -100 14.5 75 50 18.5 1.0 87.5 44.8 87.0 71.3 164.6 61.5 129.5 116.7 117.1 71.7 25 -183.5 95.1 -114.1 61.6 83.5 37.0 126.5 191.0 98.0 239.3 264.1 71.1 256.5 75 101.5 198.3 34.8 165.7 120.5 75 75 12.3 58.6 -104.8 278.9 38.0 9.4 5.0 69.7 -131.5 116.0 218.8 89.3 26.0 25 87.2 76.2 -104.8 11.2 81.0 43.5 105.5 4.8 13.1 -230. per cent per degrees cent per cent per cent + + 80 50 65.9 86.0 8.0 + + + - .5 7.0 25 25 4.4 18.8 106.0 - 95.5 75 -150 10.5 106.9 27.4 -251.3 104.— SYNCHRONOUS-MACHINE CHARTS Table XIX.2 + + + + - 49.0 +180.0 100 50 24.5 66.0 63.0 // per cent 90.0 + + - 76.5 192.5 25 6.0 -181.0 171.6 192.0 75 15.8 97.5 117.0 + + 43. 25 + + + + 261.8 186.0 100 19.5 79.9 20.5 -156.2 85.2 73.8 186.0 109.5 17.1 28.9 70.5 104.0 8.0 2.4 59.2 - .6 132.8 2.0 34.0 25 75 4.4 25 A E'a Field- E'alf -136.8 137.5 -115.0 73.8 103.8 178.4 52.3 155.5 191.7 45.2 3.0 123.0 63.5 164.0 89.0 6.6 227.5 109.2 30.5 25.8 7.0 36.6 318.5 50 75 8.4 74.0 103.4 20.0 57.0 50 13.1 271.0 5.0 9.7 327.8 - 29.5 100 18.0 3.0 101.4 97.0 110.8 65.0 26.0 32.8 107.8 67.1 82.3 201.5 - 23.0 75 14.0 91.5 168.0 165.5 221.5 53.2 45.8 122.0 220.7 50 -150 6. per cent cent 80 337 55 2 44.0 50 94.4 11.6 50 10.5 100 50 17.1 220.5 145.4 23.0 -125 10.5 191.8 94.5 91.5 92.0 25 6.6 63.8 5.0 5.4 159.9 75.7 29.5 100 25 22.8 112.0 -111.2 42.5 75.6 145.4 + + + 143.8 63.0 80 - 80 80 80 80 80 80 80 25 80 25 80 25 80 25 80 25 80 25 80 25 80 25 80 80.0 95.0 51.0 -69.6 53.1 -112.7 66.0 22.9 6. Vt.0 43.5 30.0 140.5 32.0 45.0 55. (Continued) — P' Q.5 94.8 22.5 39.8 178.4 180.1 50 -125 6.2 29.5 12. E'a.2 88.8 197.3 74.0 25 3.0 26.5 21.1 -114.6 -116.8 158.0 35.8 +226.4 194.0 15.8 2.0 14.0 192.0 102.0 13.2 28.5 11.2 34.8 114.0 +163.2 27.5 75 124.0 146.2 12.3 188.3 200.0 105.7 180.5 19.5 75 50 13.8 63.0 37.5 110.0 82.6 55.8 152.0 80.8 85.2 120.8 22.4 23.5 17.8 122.0 7.9 -115.0 96.6 51.0 3.5 26.5 75 58.0 25 - 24.5 18.2 81.0 1).7 64.0 70.0 158.7 21.5 36.0 4.8 264. per cent Calculation of Generator Chart (Example current Curves.2 -201.2 4.5 19.0 -100 -125 -150 108.1 degrees 173.0 56.5 25 72.3 54.7 9.0 13.5 204.8 176.0 71.6 135.5 17.4 15.5 92. 0 125 -100 18.0 75 23.2 136.3 132.5 -103.0 56.1 -151.1 227.3 215.2 -111.7 195.5 55. Auxiliary Curves.3 39.5 28.0 5.5 88.3 -192. 201 to 206 inclusive.5 20.6 81.7 62.0 150 -216.5 107.0 150 97.2 149.6 19. .0 24.5 79. per cent 80 100 80 100 80 125 80 125 80 125 80 125 80 125 80 125 80 125 80 125 80 125 80 125 80 125 80 150 80 150 80 150 80 150 80 150 80 150 80 150 80 150 80 150 80 150 80 150 Q.0 150 25 31.3 207.5 -117.0 86.8 163.0 7.0 43.0 3.0 91.0 125 + + 32.2 262.6 17. The radial.5 184.1 175.1 249. 0'.8 150 -262.8 115.3 -123. Main Generator Charts.8 48.8 67.0 135.3 59. The circles of constant induced voltage are drawn by using the figures in Table XVIII.5 150 50 33.0 150 75 36.5 78.8 - 23.0 24.0 -100 21.5 125 75 19.5 212.2 -287.0 43.0 8.0 24. — P' Vt.2 301.8 302. Points on curves of constant field current and constant displacement beween excitation voltage and terminal voltage are read from the auxiliary curve sheets and laid off in the main charts.5 -125 20.0 358.0 42.3 -150 19.4 -267.5 -133.5 72.0 117.0 33.4 168.0 101.0 15.6 206.0 138.5 71.5 125 50 28.5 71.5 18.0 14.1 346.0 82.3 244.5 -126.0 34.5 38.3 173. per cent per cent E.2 120.1 74.0 87.0 97.1 13.0 8. The data for these curves are taken from Table XIX.0 -108.0 150 28.0 55.5 -127.8 265.5 -115.5 207. 90.0 56.0 - - a If per cent Field- L.0 -150 16. .4 34.0 per cent P'.4 218.5 223. and the curves of constant angular displacement between excitation voltage and terminal voltage.0 -237.9 195.0 100 -217.1 -139.0 83.0 5.0 4.0 125 25 26. (Continued) Table XIX. per degrees cent -125 14 -150 13.5 60.5 125 + + + - + + + - 75 121.3 240.0 114.2 30.3 171.0 364.3 322.5 103.7 64.0 130.2 +100 35.2 179.0 -104.2 222. In order to obtain a maximum number of such values in any region.5 192.0 198.8 150 -176. The figures for the other values of terminal voltage are not included.5 -122.0 55.0 21.5 147.5 50 24.8 241.5 94.6 78.0 £ E' degrees cent 119.0 68.8 186.— 338 ELECTRIC CIRCUITS— THEORY AND APPLICATION Calculation of Generator Chart (Example current Curves.0 223.8 200. Figures 189 to 200 inclusive show the auxiliary curves of field current and angular displacement from which the desired integral values may be obtained.2 -126.8 146.1 50.0 320.0 1).8 191.0 224.8 269.0 7.5 125 24.8 192. per cent P.5 210 10.8 150 150 122.0 150 -311.0 57.1 29.3 203.5 113. and 80 per cent.4 158.7 264.0 25 26.0 276.0 11.3 187.1 -132.8 378. B'jf E'alf per cent per 100 126.0 30.0 140.2 78. The complete charts are shown in Figs.0 125 50 21. Table XIX contains the data for terminal voltages of 100.0 110.0 101.2 24. Smooth curves are then drawn through the points.5 125 127.2 201.0 206. equiangular lines indicating phase displacement between the induced voltage and the terminal voltage are located by means of a protractor.0 16.2 95.5 125 25 22.0 125 67.0 72.7 220.0 76.2 -115.0 - 1.2 230.3 145.7 169.5 55.1 36.0 80.0 39. per cent Ea.0 84.0 27.8 209.3 301.9 62.0 51.0 26.0 34.2 +100 40.2 208.2 -125 17.0 58.0 9.3 30.Ea Vt' degrees per cent 6.0 162.5 65.7 236.0 16.0 68.4 -112.3 91.6 106.0 101.0 20.9 380.0 28.5 -123. - obtained for the curves of constant field current.0 74.0 50.8 405. the curves have been plotted both versus reactive power for constant active power and versus active power for constant reactive power.5 70.0 38.0 42.0 91.5 123.5 83.0 41.0 202. . 190. 201). Auxiliary field-current curves for salient-pole generator at 100 per cent terminal voltage. 20 40 60 80 100 120 140 160 Field Current in Per — 180 200 220 Cent of Normal 240 260 280 300 320 340 Fig. These curves are plotted versus reactive power for constant active power from data in Table XIX. From these curves integral values of field current have been selected for location in the main generator chart at 100 per cent terminal voltage (Fig. From these curves integral values of field current have been selected for location in the main generator chart at 100 per cent terminal voltage (Fig.SYNCHRONOUS-MACHINE CHARTS 100 120 140 160 180 Field Current in Per Cent of — ZOO 339 220 Normal Fig. These curves are plotted versus active power for constant reactive power from data in Table XIX. 201). Auxiliary field-current curves for salient-pole generator at 100 per cent terminal voltage. 189. Auxiliary angular-displacement curves for salient-pole generator These curves are plotted versus active power at 100 per cent terminal voltage.o 1 I |j ^S i0£%^ . 30 ^mZ^ <£?£ I § ^ 1 ^^^ %<&&^- ! « tm& 20 10 *0 40 20 60 80 120 100 160 140 200 180 Active Power in Per Cent of Rati ng — Fig. 201). From these curves for constant reactive power from data in Table XIX. integral values of angular displacement between excitation voltage and terminal voltage have been selected for location in the main generator chart at 100 per cent terminal voltage (Fig. 201). Auxiliary angular-displacement curves for salient-pole generator These curves are plotted versus reactive power at 100 per cent terminal voltage. .I 20 %_% &'& 10 120 Aciiv 100 80 La — g g 60 i 40 20 Reactive Power n g rJ^S perce 20 in it 40 60 80 100 Per Cent of Rating Lead ng i Fig. I. 192.340 ELECTRIC CIRCUITS— THEORY AND APPLICATION 70 XT' £ a 60 Terminal Volfagi V( =100per cent -§§ E 50 .flM>i /. From these curves integral for constant active power from data in Table XIX.50 1^40 J^ •sflO 1* Ji. 70 ' Terminal Voltage J£ =100per cent J 1 60 ^ if? £ 1 •S 1 . values of angular displacement between excitation voltage and terminal voltage have been selected for location in the main generator chart at 100 per cent terminal voltage (Fig. 191. These curves are plotted versus active power for constant reactive power from data in Table XIX. . These curves are plotted versus reactive power for constant active power from data in Table XIX.SYNCHRONOUS-MACHINE CHARTS 200 Field Current in Per Cent of 220 240 260 341 280 300 320 340 Normal — Fig. 203). From these curves integral values of field current have been selected for location in the main generator chart at 90 per cent terminal voltage (Fig. Auxiliary field-current curves for salient-pole generator at 90 per cent terminal voltage. 194. 193. 120 — 140 160 180 200 Field Current in Per Cent 220 240 of Normal Fig. 203). From these curves integral values of field current have been selected for location in the main generator chart at 90 per cent terminal voltage (Fig. Auxiliary field-current curves for salient-pole generator at 90 per cent terminal voltage. ^^ ^3^ ^jp-" 40-J ent^. From these curves integral values of angular displacement between excitation voltage and terminal voltage have been selected for location in the main generator chart at 90 per cent terminal voltage (Fig. 203). 196. constant reactive power from data in Table XIX. Auxiliary angular-displacement curves for salient-pole generator These curves are plotted versus active power for at 90 per cent terminal voltage. values of angular displacement between excitation voltage and terminal voltage have been selected for location in the main generator chart at 90 per cent terminal voltage (Fig. na/Vo 'fage r t*9t perc <nt # • <# JfcZ Is. These curves are plotted versus reactive power at 90 per cent terminal voltage. From these curves integral for constant active power from data in Table XIX. V S" Term.342 ELECTRIC CIRCUITS— THEORY AND APPLICATION 80 60 40 20 100 160 140 120 ZOO 180 Active Power in Per Cent of Rafmg — Fig. TO :l» V ^ /. 195. 'O. 203). AchjIVPOM "120 100 80 40 60 Lagging — 20 20 Reactive Power in 40 60 80 100 Per Cent of Rating Leading Auxiliary angular-displacement curves for salient-pole generator Fig. . These curves are plotted versus reactive power for constant active power from data in Table XIX.. SYNCHRONOUS-MACHINE CHARTS 80 20 100 120 140 160 180 200 Field Current in Per Cent of — 220 240 260 343 280 500 320 34d Normal Fig. These curves are plotted versus active power for constant reactive power from data in Table XIX. 198. Auxiliary field-current curves for salient-pole generator at 80 per cent terminal voltage. . From these curves integral values of field current have been selected for location in the main generator chart at 80 per cent terminal voltage (Fig. 205) . 197. Auxiliary field-current curves for salient-pole generator at 80 per cent terminal voltage. 205) 20 40 60 80 100 120 140 160 180 200 Field Current in Per Cent of — 220 240 260 280 300 320 340 Normal Fig. From these curves integral value's of field current have been selected for location in the main generator chart at 80 per cent terminal voltage (Fig. . 200. / -fe* /$ / 20 '.3 10 ' <$ ^^ 4& \ffi m^Terminal Volfage V^ = /) im ^ "0 20 Wpercent i>^ 40 60 80 100 120 140 160 180 200 220 Active Power in Per Cent of Rating — Auxiliary angular-displacement curves for salient-pole generator Fig. constant reactive power from data in Table XIX. From these curves integral values of angular displacement between excitation voltage and terminal voltage have been selected for location in the main generator chart at 80 per cent terminal voltage (Fig. 120 100 80 60 40 20 Reactive Power Lagging — 20 in 40 60 80 100 Per Cent of Rating Leading Fig.344 ELECTRIC CIRCUITS— THEORY AND APPLICATION TO c f60 I I A v*- 50 / :40 i : J . Auxiliary angular-displacement curves for salient-pole generator at 80 per cent terminal voltage. 199. These curves are plotted versus reactive power for constant active power from data in Table XIX. From these curves integral values of angular displacement between excitation voltage and terminal voltage have been selected for location in the main generator chart at 80 per cent terminal voltage (Fig. 205). 205). These curves are plotted versus active power for at 80 per cent terminal voltage. -s . Curved angle lines represent displacement in electrical degrees between excitation voltage and terminal voltage. Straight angle lines represent displacement in electrical degrees between voltage induced by air-gap flux and terminal voltage.Powerin Per Cent of Rating — Fig. Characteristic curves of this generator are given in Fig. 187 and calculations in Example 1.— SYNCHRONOUS-MACHINE CHARTS (5»i 345 c rfl^S • V*\ ~ loU $k 140 = - 01 V .= 120 & — ^s r~ ^Vl S-A "" un ^ ^E •'£ 9 . Chart for a salient-pole generator at 100 per cen terminal voltage.A> yva p£ * ?nrT a- ^ ^v 20. = r> W- 3jSt ©» o* ° * $r __. . Circles represent voltage induced by air-gap flux in per cent of normal terminal voltage. 201. Tables XVIII and XIX. -A?? U 'iU -Hi> 160 ~W 4' 180 20 40 60 80 100 120 Terminal Vo Ifaqe 140 160 180 200 70JW ce/7/- r ' 220 240 260 260 300 Active. Curves represent field current in per cent of field current at normal open-circuit voltage.= l?0 ia) c ^ — b> ^^ £> v^ -t- o S J* °- H n - > 40 " IP — -- >EjL r^ W5 12^ 60 80 i%3. Circles represent voltage induced by air-gap flux in per cent of normal terminal voltage. Curved angle lines represent displacement in electrical degrees between excitation voltage and terminal voltage. Curves represent field current in per cent of field current at normal Straight angle lines represent displacement in electrical open-circuit voltage. 202. .ELECTRIC CIRCUITS— THEORY AND APPLICATION 346 200 — a 4> 160 — 140 -h „ s\\ b° -U-U-H 120 100 -- - "W * 80 = = 60 iV\ -- [$$ 40 _= 20 -= SRtck \q\ oS?i gM$ =: -H^^n)? 20 — 40 — W — 60 xfk AV JQA. Characteristic curves of this generator are given in Fig. Fiq. degrees between voltage induced by air-gap flux and terminal voltage. 187. >X\ $M £ rS^r — ^I^nrV' ou ™ Tt — — 1 T3- \ lE — I |_ ] 140 — i— -Fr 160 — ?f ~H& i -T 1 5° ' : ***= -55>< Ter 1 200 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 Active Power in Per Cent of Rating — Chart for a salient-pole generator at 95 per cent terminal voltage. Characteristic curves of this generator are given in Fig. Straight angle lines represent displacement in electrical degrees between voltage induced by air-gap flux and terminal voltage. Tables XVIII and XIX. Curves represent field current in per cent of field current at normal open-circuit voltage. 203. cent \o "0 ~4r A fflO 20 40 60 80 100 CO 140 Active Power — 180 ZOO 220 Per Cent of Rating 160 in 240 260 280 300 Fig. ?n -- S )5aP ^ .SYNCHRONOUS-MACHINE CHARTS — 4W 347 :vyv\ • . Circles represent voltage induced by air-gap flux in per cent of normal terminal voltage. i V = ^w^ ioofh Vr> 80-- \>TN - ~~ 60 y$ A$^o <s 15 40-^ 1 ° = ?ft-« 'v^v5 = o> T$rr WMi <u a> ^ \xjk\ lW$ . eVt = 90pe. Chart for a salient-pole generator at 90 per cent terminal voltage. c 0-~M. # SXs! <T\V\ (60 "T T> <~ $kv 120= . . 187. 40 fin 80 =: Sm — DjUq = = : $*6 4 ^w -^ v * M0 160 -\% $> HpT k£ — r» o V? — "\ L £L ' >a St tI ^° ffi 1 — ^3B • / /A/A. and calculations in Example 1. Curved angle lines represent displacement in electrical degrees between excitation voltage and terminal voltage. Curved angle lines represent displacement in electrical degrees between excitation voltage and terminal voltage.348 ELECTRIC CIRCUITS— THEORY AND APPLICATION 20 40 60 80 100 120 Active — 160 140 Power in 180 200 220 240 260 280 300 Per Cent of Rating Fig. degrees between voltage induced by air-gap flux and terminal voltage. . Chart for a salient-pole generator at 85 per cent terminal voltage. 204. Characteristic curves of this generator are given in Fig. Straight angle lines represent displacement in electrical open-circuit voltage. Circles represent voltage induced by air-gap flux in per cent of normal terminal Curves represent field current in per cent of field current at normal voltage. 187. Chart for a salient-pole generator at 80 per cent terminal voltage. 205. and calculations in Example 1. Curves represent field current in per cent of field current at normal open-circuit voltage. . Tables XVIII and XIX. Straight angle lines represent displacement in electrical degrees between voltage induced by air-gap flux and terminal voltage.SYNCHRONOUS-MACHINE CHARTS 80 100 120 Active Power — 140 in Per 160 180 ZOO 220 240 349 260 260 300 Cent of Ra+ing Fig. Curved angle lines represent displacement in electrical degrees between excitation voltage and terminal voltage. 187. Characteristic curves of this generator are given in Fig. Circles represent voltage induced by air-gap flux in per cent of normal terminal voltage. 350 ELECTRIC CIRCUITS— THEORY AND APPLICATION 100 120 140 Active Power — in 160 180 ZOO Per Cent of Rating Chart for a salient-pole generator at 70 per cent terminal voltage. Fig. 206. Circles represent voltage induced by air-gap flux in per cent of normal terminal Curves represent field current in per cent of field current at normal voltage. Straight angle lines represent displacement in electrical open-circuit voltage. Curved degrees between voltage induced by air-gap flux and terminal voltage. angle lines represent displacement in electrical degrees between excitation Characteristic curves of this generator are given voltage and terminal voltage. in Fig. 187. CHAPTER XII SOLUTION OF STEADY-STATE TRANSMISSION PROBLEMS In order to illustrate the solution of problems of electric power transmission in the steady state, particularly by graphical methods, a series of examples is presented below. In most instances, the modified Evans and Sels chart is used to give the performance of the transmission line and transformers. necessary, the generators are represented described in the preceding chapter. by When charts of the type Determination of Synchronous-condenser Capacity. —In the usually required to maintain operation of a transmission line, it is constant voltage at the receiver end at all loads. This can be done in three ways: 1. By varying the sending voltage. 2. By keeping the sending voltage constant and using a synchronous condenser at the receiving end. 3. By a combination of the two methods just mentioned. The first method is not practical in long-distance transmission. Constant receiving voltage would, as a rule, necessitate variation in the voltage at the sending end over a very wide range, a feature which is not desirable. At light loads, the sending voltage would have to be quite low in order that the Ferranti effect At heavy lagging loads, of the line be properly compensated for. on the other hand, the sending voltage would have to be very high in order to offset the line drop. Since the factor of safety of the insulators of a high-voltage system cannot be too liberal, for economic reasons, the high voltage might subject the insulators near the sending end of the line to excessive stresses. Further- more, an undue amount of corona formation might also result on Variation in sending voltage with load can this part of the line. be obtained through control of the generator fields by means of compounded ^or 1 Tirrill regulators. trical "Handbook for by H. Pender and "Standard Handbook for Elec- description of Tirrill regulators see, for instance, Electrical Engineers," Engineers." 351 ELECTRIC CIRCUITS— THEORY AND APPLICATION 352 In most transmission systems operating today, the voltage is kept constant at the sending end as well as at the receiving end, giving what The sending is usually termed constant-voltage voltage is kept at the desired value transmission. 1 (in most cases, equal to the receiving voltage) by means of shunt-wound Tirrill regulators in the field circuits of the generators. The receiving voltage is controlled by synchronous condensers. By field adjustment, as governed by Tirrill regulators, the synchronous condensers may be made to draw a certain amount of leading or lagging A leading current taken by the synchronous condensers current. will produce a drop in the line which counteracts the drop due to a lagging load. If, on the other hand, the synchronous-condenser current is lagging, the drop in the line will tend to compen- sate for the Ferranti effect and for the drop due to a leading load. method of keeping the receiving voltage constant involves the use of compounded regulators at the sending end and synchronous condensers at the receiving end. The variation of sending voltage is, in this case, limited, so that no undesirable stress is imposed on the insulators. The compounded sending The third voltage contributes to the maintenance of constant receiving The additional regulation required at certain loads is taken care of by the synchronous condensers whose fields, as in voltage. the previous case, are controlled by shunt-wound Tirrill regulators. 2 It is obvious that the size of the synchronous condenser installation necessary to insure constancy of the receiver voltage when the sending voltage is compounded to a certain extent than when this voltage is constant. A problem which often arises is, therefore, to determine the proper size of the synchronous-condenser installation. If the is less capacity of the synchronous condensers depended only on their adequacy to give constant receiver voltage at full load and at no load, it could be definitely determined by means of a receiver In deciding, however, on the synchronous-condenser installations in connection wi,th a long transmission line or an chart. For analysis of operation with compound regulators see L. F. Woodruff, "Regulator Settings for Long Lines," Elec. World, Vol. 84, pp. 441 and 460 1924. Dwight, H. B., "Transmission Systems with Over-compounded Voltages," Trans. A.I.E.E., p. 95, 1926. 1 Dwight, H. B, "Constant-voltage Transmission," John Wiley Inc., 2 New York, 1915. See handbooks for electrical engineers, loc. cit. & Sons, STEADY-STATE TRANSMISSION PROBLEMS interconnected transmission system, stability must be taken into account. the 353 question of system When this factor, which is of extreme importance in large systems, is properly considered, entirely possible that synchronous condensers which would have ample capacity for the required steady-state voltage regulation would be wholly inadequate from the standpoint of Hence, only with comparatively short lines and in stability. 1 small systems which inherently possess a considerable stability margin is it safe to base the capacity of the synchronous condens- it is ers on considerations Fig. 207. of voltage regulation alone. synchronous-condenser capacity by means — Determinationmodified Evans and of of the Sels chart. Evidently, in such cases, the installation should be equal to the capacity required at full load or at no load, depending on which is the larger. As mentioned in the preceding chapter, synchronous condensers sometimes have a slightly smaller rating when operating underexcited than when operating overexcited. If necesLarge present-day sary, this must be properly allowed for. condensers, however, usually have the same rating over- and underexcited. Figure 207 shows how the synchronous-condenser capacity can be determined by means of the modified Evans and Sels chart. Assume that the system is to operate with 100 per cent voltage at each end at all loads. Point a is the full-load operating point, 1 The question of this treatise. of power-system stability is discussed in detail in Vol. II ELECTRIC CIRCUITS— THEORY AND APPLICATION 354 by means of active and reactive power of the load. Points b and d are the line operating points at full load and no load, respectively. Neglecting the losses in the synchronous condenser, the vertical line ab gives the required capacity (leadlocated and the line cd the required capacity (lagging) Obviously, the installation should, in this case, be equal to the kilo volt-amperes given by the full-load requirements. Ignoring the synchronous condenser losses does not introduce any appreciable error. ing) at full load, at no load. EXAMPLE 1 Statement of Problem Line: Three-phase. Length = 200 miles. Conductors: Three aluminum-steel cables spaced 210 in. ter 0.954 in. 61 (steel - 7; in a horizontal plane. Diame- The total number of strands in the concentric-lay cables is aluminum - 54; all strands same diameter). The conduc- tors are completely transposed. Load: Full load = 40,000 kw. at 75 per cent power factor. Transformers: Two banks of transformers in parallel at each end of line. Each bank has a capacity of 20,000 kw. at 75 per cent power factor and possesses 0.8 per cent resistance and 12 per cent reactance. The kilovolt-amperes required for excitation are 1,000 kv.-a. per bank, and the core loss 0.5 per cent of the rated capacity (at 75 per cent power factor). The receiving banks step the down to 15 kv. Transformer Connections: voltage A — Y; neutral grounded. Conditions of Operation: Frequency 60 cycles. 150 kv is held on the high-tension side of the sending transformers by adjustment of the generator fields; 15 kv. is held on the low-tension side of the receiving transformers by adjustment of the synchronous-condenser fields. Desired Results: 1. Necessary capacity of synchronous condensers to insure constant voltage operation. 2. Plot of reactive kilovolt-amperes of the synchronous condensers versus load (at 75 per cent power factor). Use the chart, which is based directly on the vector diagram. In calculating the resistance of the conductors, neglect the iron core. The skineffect ratio may, to a sufficient degree of approximation, be considered the same as for a solid aluminum wire of the same diameter. Assume that the iron core increases the internal inductance by 10 per cent. The leakance should be based on the corona loss under average weather conditions. STEADY-STATE TRANSMISSION PROBLEMS 355 Solution Resistance: The overall diameter (d) of a concentric-lay cable =d d i(2n is given by 1 + b) * (a) where = di diameter of each strand of layers over core n = number = b 1, 2, 2.155, and 2.414 for 1, 2, 3, and 4 wires in the the central steel wire, in this case, Hence, strand cable has four layers. If 0.954 0.954 U1 2X4 + core, respectively. considered as the "core," the 61- is = 0.106 in. 9 1 Area of aluminum A = 54 X,X 0.106' 4 =0476h, 2 A solid aluminum conductor of 61 per cent conductivity and 0.1662 square (No. 0000 A.W.G.) has a resistance of 0.424 ohm per mile in. cross-section at 20°C. 2 The resistance of the given conductor = R' 0.424 ^^ = X 0.1482 is, therefore, ohm /mile — Correction for Spiraling. The resistance R' should be corrected for the Obviously, the actual length I of a strand (see Fig. 208) is larger than the length measured axially along the conductor. effect of the spiraling. Fig. 208. — Sketch indicating relation between pitch and length of a strand in a concentric-lay cable. The size of the correction which should be applied to the resistance computed above depends on the average pitch factor of the cable. The pitch factor of any layer is given by l ' -P - V 1 + 0£)' (i,) where P = dp I 1 Pender, H., = = the pitch of the layer the pitch diameter of the layer the actual length of the strands "Handbook for Electrical Engineers," following. 2 See "Handbook for Electrical Engineers," loc. cit. p. 1873 and ... ELECTRIC CIRCUITS— THEORY AND APPLICATION 356 The minimum pitch is 10.1 times aluminum part of the given cable, the pitch diameter. 1 dp, inches Hence, for the P, inches Outer layer Second layer Third layer 8.56 6.42 4.28 . . Average pitch factor A U . X (* .846Y L , , /tXO. 636 \ 2 42 j + CTT^X 0.424V 1 V + {—&&-) J 1 = 1.048 Since, however, there undoubtedly is some conduction axially from strand to strand in each layer, a correction corresponding to the average pitch The correction may more factor 1.048 would probably be somewhat high. appropriately be taken proportional to l+k(pwhere k is a factor less than unity, which, assumed equal to %. Hence, in this case, arbitrarily will p'«-l+?2<£0« = The corrected d-c. resistance R = p'R' Skin-effect Ratio. = is, (c) 1) be 1.032 therefore, 1.032 X 0.1482 = 0.153 ohm/mile — The "skin-effect" impedance ratio is given by 1 or Jo(ar) 2 Ji(ca) id) where « n p <o = — = = 4tT(XO} v \45° (*) the permeability of the conductor the resistivity of the conductor 2ir/ = the angular velocity The resistivity of aluminum ohm-cm. at 0°C. At 20°C, of 61 per cent conductivity is 2.607 X 10 -6 See "Handbook for Electrical Engineers," loc. cit. Kennelly, A. E., F. A. Laws, and H. P. Pierce, "Experimental Researches on Skin Effect in Conductors," Trans. A.I.E.E., p. 1953, 1915. 1 2 STEADY-STATE TRANSMISSION PROBLEMS p = Po (i + a 4) = 2.607 X ~. - 2.83 - 1.279 X X = 1.0876/33°. 18 = o^sos^rM = X 20) 2,830 abohm-cm. 2.54r ^45* = 1.55\45° 1 55\45° X 2 fl o Effective resistance of R = X 0.954 Jo(ar) 1 0.00423 ohm-cm. Ji(ar) ~Z = + 10-*(1 10"* 357 X ^= 1.0876 /33°. 18 = 1.071/16°.0 = 1.03 0.7868\27°.86 +J0.295 aluminum = = 1.03« 1.03 X 0.1576 0.153 0.158 ohm/mile = Internal reactance X» = 0.295i2 = = 0.295 X 0.153 ^ = ~^? = 0.0001198 henry /mile 0.0452 ohm/mile Internal inductance = Li This internal inductance is slightly larger than the internal inductance obtained on the assumption of uniform current distribution (the latter being 0.0000805 henry /mile). This shows that the formula used is not well applicable to stranded conductors, so far as inductance is concerned. Internal inductance computed from uniform current distribution will, therefore, be used. Inductance: Equivalent spacing D = 210-^2 = 264.8 L = ^7411og 10 ^ = (741 log 10 ^| + 80.5)l0-« = (2,033 + +80.5)l0-« 80.5)10"« This inductance would be correct core. As previously internal inductance in. stated, it will by 10 per = 0.002114 henry /mile if the aluminum conductor had no iron be assumed that the core increases the cent. On this basis, the total inductance becomes L = (2,033 + 80.5 X 1.10)10^ = = 0.002122 0.00212 henry /mile 1 Convenient tables of Bessel functions suitable for skin-effect calculations be found in the paper "Experimental Researches on Skin Effect in Conductors," by Kennelly, Laws, and Pierce, loc. cit. A very complete will " Funktionentafeln mit Formeln und Kurven," by E. Jahnke and F. Emde, published by B. G. Teubner, Leipzig, 1923. See, also, standard treatises on Bessel functions. set of tables is also included in 6 kv.7 assuming 1 in. upon location depending r J The voltage will.03882X10" = 0-08882X10^ = c = U . (/) by 25 )^(logio P." loc. symmetrical case. must be calculated separately for each conductor. Inc. the spacing is dissymetrical. the gradients at the three conductors have In the identical magnitudes. per inch disruptive critical voltage eo = is given g inoSrln~ by = 123m 5r logio — (h) Corona depends fundamentally on the voltage gradient at the conductors. however. F. using equation (g). 2 See Woodruff. New York. "Dielectric Phenomena in High-voltage Engineering. as an approximation. the difference between the two total inductances is only about one- half of 1 per cent. the corona loss may be calculated either by When (/) in conjunction with equation (h) or by equation (g).. cit.. Hg. 2 equation When 1 Peek.m = density factor 459 + t m = irregularity factor of the wires / = frequency in cycles per second r = conductor radius in inches & . therefore. L. Jr. 1920.o logl0 Q mm x 1Q _6 farad/mi]e 0^77 Leakance: Leakage over insulators will be assumed negligible. VIII." McGraw-Hill Book Company.. by computation of the actual voltage gradients at the conductors. Chap. Capacitance: 0. "Principles of Electric Power Transmission and Distribution.y^g _ g omo 8)n0-'kw. the corona loss per mile of conductor is given by or + 25) >J^ (e - p = p = M70 ^ + ?90(/ e 6 )n0-%w. the voltage gradient is not the same at each conductor. D = e e g Po The conductor spacing in inches = system voltage in kilovolts to neutral — disruptive critical voltage in kilovolts to neutral = gradient at the conductors in kilovolts per inch = 53. Peek's empirical formulas will be used.. The corona loss. The corona" loss may be computed by Temperature (t) = 40°F. and each conductor has the same corona loss..ELECTRIC CIRCUITS— THEORY AND APPLICATION 358 As seen. (g) Where = . . over the entire line. the spacing is symmetrical. Barometric pressure (o) = 29. therefore. W. be assumed to be constant equal to 150 kv. 1 For a three-phase line with symmetrical spacing (conductors located at the corners of an equilateral triangle). 1 logio- ^o4.. F. 0141 X 10~« farad J 0.4 = 0.799 = 0. = 162.08 X 10. there is no corona The loss.477 g mod.814/78 /°- X X = j'5. Designating the conductors a.814/78.°8 ohm j377 X 0. per inch Q < Since gb ln 17. and c and letting b represent the middle conductor.315790° = 2. leakance.5 kv..6ohms .416/844° hyp 3 . In order to check up if corona occurs at all. therefore.08 X lQ.063 X 10-a /90° mho . io» °.6 X 1.065 X 0.s /84°4 hyp 391\5°.158 +^0. it is sufficient to calculate this gradient.158 ohm ] L I a z = r y = g - C = +jo>L = = +juC = y/Ty = J* = per wire mile 0.8 81 0.315 X 10~« = 5.gdn2^ D (t) + gjn2^ (j) + 9b + g = ga (k) c Simultaneous solution of these equations gives for the voltage gradient at the middle conductor ^ —F\150° = 9b .85 = 48.814 /78°.8 : STEADY-STATE TRANSMISSION PROBLEMS 359 In the case of a horizontal arrangement of wires. maximum gradient occurs at the middle conductor. b.85 (as an average for stranded conductors) 53. as in the present case..0141 10.7 = 1ftfi _ 1065 459+40 _ = S 0. (J) fr- V3rln-^ -S/2r Inserting numerical values gives 150\150 = V3 X g Y2^^WW7 = 310 \ 150 ° kv Per - inch 0./84°.s \/a81 4778o 8 .315/90° Constants of the Entire Line: Z =200 X 0.315 X 10^/90° = 1. 8/78°.00212 0.158 +JS77 X 0.9 m — mS = assumed X X 29.8 ohms Y = 200 X 5. will be zero. the following equations may be written the Vab = V/0 = rUg* - g b )ln^ Vu = F\120° = rUgb - gc )ln . Constants to Be Used: r = 0.200a = 200 X 2.00212 henry 0.315 X 10-«/90^ mho 10"6 5. 000 kv. 0.4 = per transformer per phase = .38 No-load power factor cos ft.2 =» 2 e = / 20. On these assumptions. Will further consider the losses in the synchronous condenser to be constant.008 _ . 210.995) — j'3. per phase Synchronous Condenser Losses. Cond. 209. — Fig.85(0.77 — J7.404/84°. — The circuit which presents itself for Solution of the Equivalent Circuit.063 X 10. 1 ' 000X S + J50. is shown in Fig. per bank per phase in parallel I e = 0.537)10~ 3 0.75 ohms 1A1 o 101.12 V3 X (0. synchronous condenser.__ = mho (at full load) .9 +j'155.8 /89°. The circuit includes transmission line.8/78^8 X 7^T4 0>416 = 158.000 t= 102. per phase 150 Line Rec.6 ohms per phase X 20. the constant current required to supply the losses — becomes T _ 30.75 150. which.0/79?! ohms 29. will be modified to that shown in Fig. Equivalent circuit of system to be solved in Example 1. Trans7? ww\MT5T5> Z" 2 -Load Syn.ELECTRIC CIRCUITS— THEORY AND APPLICATION 360 Constants of the Equivalent II: ^p = 0.1 9 XI Y tanh o —-^ = 0.000 X -7= VZ X X amp.000 = 0.1 = 0.416/84°.66 amp.100 Exciting current /. 6.537 Transformer Constants: X 10.000 X V3 X 0.005 -^-poo - ninft 0.210/84°.2 X 1. solution .6 Since two banks in parallel Zt = 3.04 /0 = 4.00752 +i0. 102. receiving transformers. for the sake of simplicity. Two banks - 01 !° :L V3 X 150 = 3.7 sinh Z" = Z = 162. Xt = = - V3 X 150 150.6 0.2 ohms 102.-a.62/0 amp.000 = D = Rt -.6 ——X 0. and assume that its power factor is 4 per cent. and load.83 amp.385 jO. Will assume size of synchronous condenser = 30. 209.3 /90° 0. 000 25.3 +.200 7.-a. Figure 212 gives the desired relation between reactive kilovolt-amperes and load at 75 per cent power factor. Combined equivalent line and transformer impedance +Z T = 33.700 +49. I 6 in load impedance. 210.65 + J46.537 volts The receiver diagram. I if and 7g are independent of the load and to a constant current /'. 209..77 . The currents at the receiving end are as follows: Current Current Current Current Current I\ in equivalent line leak. 2.04 +i38. Fig. the receiver-end line leak and the transformer excitation admittance have been moved to the point where load and synchronous condenser are attached. Z yHjT TV: z — As seen. Approximate circuit representing system in Fig.000 40.620 15.66 + = 6. I» in synchronous condenser conductance.480 +20.7 ohms Currents I\. 211.5 + 0.62 Voltage drop in line and transformer impedance due to this current I'(Z" +Z T ) = (6. of the synchronous condensers . Z" I' = 11 +I +I 2 3 may be combined = 0./205. The Synchronous Condenser -10.800 +58.000 30.'2.240 +27.9 amp.7) = -7.180 20. 1\ in transformer excitation admittance.000 - 4.300 5.9)(33.000 50.100 +34. the accuracy of the solution is not seriously reduced.000 _.3 + J205.000 45. Fig.000 10. Reactive Kilovolt-amperes of capacity of the synchronous condenser installation should be at least 41. 7 4 in synchronous condenser susceptance.04 +j38.000 35. The necessary synchronous condenser data are obtained from the diagram and collected in the following table Load Kilowatts at 75 Per Cent Power Factor 1. per phase 4.j7. is based on the calculations above.400 +41.000 + + 1.700 kv.: STEADY-STATE TRANSMISSION PROBLEMS 361 Although the placing of the line leak and the transformer excitation admittance in parallel with the k>ad itself involves an approximation.000 +13.800 +. 000 40. 211. 10. X.362 ELECTRIC CIRCUITS— THEORY AND APPLICATION Load In Kilowatts 50.000 I j 1 o 40.000 "S 5 i % 10.000 50. . -Capacity of synchronous condensers necessary to maintain constant receiver voltage for various sizes of load at 75 per cent power factor (lagging). receiver chart based directly on the vector diagram is used.000 % Power Factor Fig.. The trt 50. Graphical solution of the transmission problem in Example 1.000 ' Load in 20. 212. This curve is for the system treated in Example 1.000 | 30.000 Kilowatts at 75 30.000 OS n -• — »ooon. This type of chart is fully discussed in Chap.000 1 20.j'(z'+**) Ter*^s f fcceivir>9 tna En d — Fig.000 . 000 kw. V ^r = 130 per cent or Es = 130 per cent tlR When the load is it removed = 85-5 per cent Sir Hence.e. The low-tension voltage at the sending end is maintained by automatic field control of the generators while the voltage of the low-tension bus at the receiving end is regulated by three 30. 165 and 166. i. Hence. therefore. 220-kv. the regulation is which is thrown EXAMPLE is automatically 105. .. ER = The voltage Q-g^ = 152 per cent regulation.-a. a. synchronous condensers. transmission line is normally operated at terminal voltages of 220 kv. the sending voltage adjusted to 90 per cent.-a. Regulation = 152 is — 100 = 52 per cent Upon disconnecting the load. 165 and The characteristics of the synchronous condensers are given in Fig.) can be located immediately. the latter being kept constant Solution Since the receiver voltage is normal. When there is no voltage regulator at the sending end.2 - 100 = 5. Calculate the per cent regulation when a load of 100. (lagging) is removed. When the sending voltage is controlled by a compound regulator so adjusted that the sending voltage at no load is 90 per cent.000 reactive kv.000-kv. the modified Evans and Sels charts. b. 2 Statement of Problem A 240-mile.STEADY-STATE TRANSMISSION PROBLEMS Determination of Regulation. and 45. 6. 166) .000 lagging kv. illustrate the solution Sels charts of EXAMPLE 363 —The two examples given below by means of the modified Evans and problems involving voltage regulation.2 per cent 105. Figs. ER = 00 = —^ Regulation = In this case. and 45. operated at normal (100 per cent) by adjustment of the sending The performance of this line including transformers is given by voltage. The performance of the line including transformers is given by the modified Evans and Sels charts (Figs. 60-cycle transmission line is receiver voltage. 240-mile. and shows that a. 60-cycle. 1 86. the receiver chart is direct-reading and the operating point (100. 3 Statement of Problem A three-phase.-a.000 kw. Er = Vr = 100 per cent.2 per cent obviously independent of the amount of load off. 500 kw.850 .0 68.-a.700 kv. when b. and line data at no load.5 Table XXI. it is seen that at 100 per cent voltage and 85 per cent field current. This gives a. Assuming the Thrill regulators adjusting the fields of the synchronous condensers to be out of service (so that the field current remains constant). amperes -39.910 -35.000 -43. In order to locate definitely the new operating point. Pr = 95. per cent 3) 1 (Er/VrY.000 -35. ? a.500 29.8 5.1 100 91 26. as obtained b.700 kv. V t. (100 per cent is 90. and XXI give synchronous-condenser data as obtained from Tables the V-curves at 85 per cent field current.800 61. 85.000 53.9 Es Er Qe per cent 76.200 9. it is best to prepare an auxiliary plot of fine kilovolt-amperes and condenser kilovolt-amperes versus receiving voltage and obtain the intersection between the two curves. calculate the voltage regulation of the line when the 90 per cent power factor load determined in (a) is thrown off. at 90 per cent power factor When the load is removed and the field currents of the synchronous condensers are kept constant.000 kv.300 -26. the load operating point is located where the vertical distance between the powerfactor line and the 100 per cent voltage circle equals 61.5 68.2 90 100 110 120 130 140 Es.-a.900 (Example Qr Er.6 50.0 27. kilovolt| amperes amperes 143 125 111. the leading kilovolt-amperes taken are 68.500 .500 4. the receiving voltage will change until the reactive kilovolt-amperes supplied by the line equal the reactive kilovoltamperes taken by the synchronous condensers at that particular voltage.46 -28. Qc 3) Qr.-a.3 4.6 per cent or 61.8. Table XX.0 32.—— ELECTRIC CIRCUITS— THEORY AND APPLICATION 364 What load at 90 per cent power factor (lagging) can the line supplythe synchronous condensers operate at 85 per cent field current and the low-tension sending and receiving voltage are both the equivalent of 220 kv.9.).700 49.500 14. per cent Synchronous-condenser Data (Example I a per cent . per cent 100 100 100 100 100 Line Data at kilovolt- 68.880 -26. XX from the receiver chart.6 55. kilovolt- per cent 70 80 90 100 110 No Load . If a 90 per cent power-factor line is drawn in the receiver chart. Solution From the V-curves of the synchronous condensers.800 . As seen.9 per cent 150 140 "I 130 £ 120 Synchronous Line Receiver ikv-a Condenser kv-<r 66°/ ^s Uf~ pi-Zero Powe Es =IO 0% 1 100 b 90 1 (X Norn >a/000%. 80 70 60 o o o c> * . A load is very often supplied over two parallel lines from the same generating station or substation. The example below discusses some methods which may be used to determine the division of load between the parallel lines in EXAMPLE a specific case. curves give the solution of Example 3. If the parallel lines have identical electrical constants. 213. the power and power factor (lagging) of the load. they will not carry an equal share of the load. The voltages on the high-tension side of the sending transformers and on the low-tension side of the receiving transformers have their normal (100 per cent) values.STEADY-STATE TRANSMISSION PROBLEMS 365 Figure 213 shows the plotted curves. the lines have widely different In constants. Hence. on the other hand. The constants of the two lines are widely different. they intersect at 125. a considerable distance apart. outline the procedure to be followed in order to determine the load division between the two lines. If.9 per cent of normal receiving voltage.9 - 100 = 25. Knowing the constants of the lines and transformers. 4 Statement of Problem A generating station and a substation.)voli ige 220 kv. are connected by two three-phase transmission lines on separate pole lines on separate rights of way. The load division can be determined by calculation or by the use of charts. such cases. ^ «* Reacf.ve kv-a Lagging o o N o o * o c>^ «° Leading — Fig. the problem of determining the load division between the two lines arises. Part (6). Regulation = 125. they will obviously divide the load equally. . — Determination of Load Division between Two Parallel Lines. Determination of voltage regulation by obtaining intersection between locus of reactive power at the receiver end of the line and locus of These reactive power of the synchronous condensers at constant field current. Both lines are tied into the high-tension buses at each end. First it is necessary. Bi. of the 2 —System considered system between points S and B in Example (see Fig.D 2 Bx ~A (Ai j- 2 )(D 2 n TTg -f- + DiB +B . 214.DJ (c) ' 2 (d) 2 Letting Z r and Yr represent the equivalent impedance and excitation admittance. power between them. is given.%JLq + bj c) « (M .366 ELECTRIC CIRCUITS—THEORY AND APPLICATION Solution /. IX. respectively. By calculation using general circuit constants. however. of the receiving transformers. equations (75)) A = A{l+^)+B'.. to determine the current taken by the synchronous condenser in order that the total current at B may be obtained.bJc) + j(N + bj c ) "(i) . the general circuit constants of the two lines in series with the receiving transformers. .r D'o (l + (e) r voltage of the lines (at S) (g) r ^) + CU?r(l + (/) ?£-') (h) In order to obtain the (at R) must be calculated. 4.Y B = i?. become (Chap. equations . distribution of E. C ]f Du General circuit constants of line 2: A 2 B 2 C 2 D 2 General circuit constants of the two lines in parallel [Chap.(ai +ja 2 )Eb + (6x + jb 2 )(ILp .(i+^)+z/. the receiving voltage = AoEb + BJb = AoEb + B (Il + Ic) . .jILq + jl c) = (aiEb + bJLp + b 2ILq .bj c) +j(a 2Eb + b 2ILp . it will be assumed that the synchronous condenser current (Ic) is leading. L /he 1 Line Fig. 214). . IX. (58)] A' = B' n' Co Z>'„ = A^B + BtA ""bTTbT" 2 2 (a) BiB 2 B +B ±n T Cj — r 1^1 = (b) 2 t B. General circuit constants of line 1: Ai.(i c„ = Do = The sending + ^) + AX (l + ££-') c. i. Since the power factor of the load is lagging.e. STEADY-STATE TRANSMISSION PROBLEMS 367 Since the absolute value of E, is known, this equation is readily solved by squaring and gives for the synchronous condenser current Nbi 01 - Mb + /[" 2 , 2 hl+bl V|_ b2 - Mb iV&i Y _ M + n* * bl J + - e; 0') b\ The sum of this current and the load current is the total current at B. The voltage and current on the high-tension side of the receiving transformers are given by ^)E + Z (l + ^)h + +¥-> E = (l+ b r 7r = YrEb (k) r (I) the load voltage Eb has been taken as reference vector. In the it is convenient to use E r for this purpose, and I r as found from equation (I), should, therefore, be inserted in the equation below in its proper phase relationship with respect to this voltage. So far, following, , E, + BJn = A E + B = AiE r 2 r 2 I r2 (m) which, in conjunction with In hi = + (n) Ir gives (A 2 In = Ir2 = (At - At)Er + Bi + B 2 -A Bx 2 )E T +B B 2 Ir + BJr (p) 2 Since 7 r i and I r2 are obtained with reference to E r the phase displacements and 4>r2 at the receiving ends of the two lines are readily determined. The total amounts of power supplied to the substation transformers by the , 4> r i lines are, therefore, = 3EJn PR2 = 3EJ r2 Pri By cos 4>n cos 0r 2 (q) (r) on equivalent Il-circuits. each line has been replaced by its equivalent II. In order to determine the synchronous-condenser current, the two n-circuits 2. calculations based As shown in Fig. 215, I g yz&x I me J Ij < ^ J J L/'ne? -£• J —> Zz — Equivalent circuit of the system in Fig. 214. The two parallel transmission lines have been replaced by their equivalent Il-circuits. Fig. 215. are combined into a single n, as indicated in Fig. 216, transformers replaced by their equivalent^T-circuit. and the receiving 368 ELECTRIC CIRCUITS— THEORY AND APPLICATION The voltage and current on the high-tension side of the receiving transformers are given by equations (fc) and {I) as before. The voltage on the high-tension side of the sending transformers is y = B. (l+Zg' )Er + ZJr (s) &&£ — Fig. 216. -Circuit representing the system in Fig. 215 obtained by combining the two equivalent II's in parallel and representing the receiver transformers by their equivalent T-circuit. which, when combined with equations e - [(' + -¥)( i + - -¥) + z " Es = = («i (ai may + ja )E + + ja )E + (I), changes into + [ z -( j + ^~°) r '] E> ^) +R 1+ ( This equation, when reduced, and (fc) 1+ ( ^)] / . (t) be written 2 b (bx 2 b (bx Comparison shows that equations + jb )h + jb )(I p - jh + jlc) 2 2 (u) b and q are identical, (i) (tt) The syn- chronous-condenser current, therefore, is determined by equation (j). Knowing the total current (II + Ic) on the load side of the receiving transformers, the voltage and current on their high-tension side are calcuThe ratio of the currents carried by the lated by equations (k) and (I). architraves of the two equivalent n's (Fig. 215) equals the inverse ratio of Thus, the architrave impedances. /i = U The sum Zj (v) Zi of the architrave currents is given /i +h = (Y x + Y,)E r Solving for Ii and I 2 from equations = II (v) and 2 1 /r2 = = (x) 2 at the receiver end of each line IrX (w) r (w) gives Z 2 (Y + Y )Er + Z 2I r z +z Zx(Yx + Y 2 )E r + Zxlr Zx + Z 2 t The current by +I is, (y) hence, Z +Y „1„ +z + Yx^Er + Zx + z Zx(Yx + Y + y ]e, + Zx Zx Zx + Z +z J Z (Yx Zx 2) 2 . . 2 2 (2) 2 2) 2 2 (aa) 2 and the power supplied to the substation transformers by the by equations (q) and (r). lines is given STEADY-STATE TRANSMISSION PROBLEMS S. By 369 the use of charts. In order to solve this problem by means of charts, it is necessary to have at least three charts available, namely, a receiver chart for the system between the high-tension bus of the sending transformers and the low-tension — Fig. 217. Modified Evans and Sels receiver chart for the system in Fig. 214 between the high-tension bus at the sending end (S) and the low-tension bus at the receiving end (B). bus of the receiving transformers (i.e., between points S and B) and one receiver chart for each transmission line proper (i.e., between S and R). Assume that these charts Evans and are of the modified *o 'Pb'Qb a Sels type. r* >QR 1 i _1 m ^s Os en a nPW **N^ "P^ -J rV ¥r? (b) la) Fig. 218. P* ^S —•Modified Evans and Sels charts for the performance of the receivingend transformers. By spotting active and reactive power taken by the load in the chart for the system between S and B (Fig. 217), the synchronous-condenser kilovoltamperes and the active and reactive power (Pb and Qb) at the low-tension side of the receiving transformers are determined. The current at this point ELECTRIC CIRCUITS— THEORY AND APPLICATION 370 may now be calculated, and the voltage and current on the high-tension side by means of equations (k) and (I). Having the voltage and current, the active and reactive power are also readily This power and kilovolt-amperes, however, can be determined calculated. of the transformer obtained — 'Modified Evans and Sels receiver chart for one of the transmission between high-tension buses. Operating points must lie on the dotted circle representing the correct ratio between sending and receiving voltage. Fig. 219. lines if performance charts for the receiving transformers alone between points R and B) are available. Figure 218 shows a receiving and sending chart for these transformers. The operating point (Pb and Qb) can be directly spotted in the receiving chart a and transferred to the sending chart 6 by means of voltage ratio and still more readily (i.e., Ms zEn Fig. 220. — Graphical determination of load parallel lines. distribution between the two This gives the voltage and active and reactive power (Er, Pr, and Qs) on the high-tension side of the receiving transformers. Figure 219 shows a receiving chart for one of the transmission lines. Since Er is known, the ratio Es/Er is given and, hence, the circle on which the operating point must he. This circle is shown dotted. The load will divide between the two lines in such a manner that the following two condiangle. tions are satisfied: STEADY-STATE TRANSMISSION PROBLEMS 1. The sum of the power supplied by each line must equal the 371 total power at R. The displacement between the sending and receiving voltages must be same for the two lines. When the power is determined in accordance with these conditions, the values obtained for the reactive power will be such that the condition that the reactive power supplied by the two lines must equal the total reactive power at R will automatically be satisfied. The correct power division can be obtained most conveniently by an auxiliary plot (Fig. 220). Values of power corresponding to points along the proper voltage-ratio circle in the two receiving charts are plotted versus angular displacement. By addition, a curve of the total power versus angle is obtained. On this curve, the correct Pr is spotted, and the amounts of power Pm and Pri supplied by each line determined, as indicated in the 2. the figure. — Miscellaneous Examples. In order to illustrate still further the use of charts for the solution of steady-state transmission problems, a few examples are given below. These involve the determination of voltages, loads, power, and power factor of generators, condensers, field taken kilovolt-amperes reactive currents of generators, by synchronous and synchronous condensers, etc. EXAMPLE 6 Statement of Problem A generating station long transmission lines Fig. 221. G supplies power to two loads L\ and The sections AM and MB (Fig. 221). L 2 over the AM and MB — Transmission system considered in Example 5. and identical in every respect. Their steady-state pergiven by the modified Evans and Sels charts (Figs. 165 and 166). The voltage at A is regulated by automatic field control of the generators and is normally 110 per cent. The voltages at and B are regulated by the are of equal length formance is M synchronous condensers Ci and C 2 and are normally 100 per cent. The nominal (100 per cent) voltage is 220 kv. referred to the high-tension sides. The load L\ is 100,000 kw. at 85 per cent power factor (lagging) and the load L 2 40,000 kw. at unity power factor. ; , a. For operation as specified above, what are the reactive kilovolt-amperes taken by the synchronous condensers? What are the power and the power factor at the generators? ELECTRIC CIRCUITS— THEORY AND APPLICATION 372 b. voltage, power, and power (The loads L Y the synchronous condenser Ci is shut down? and B are 100 per cent of normal.) are as before, and the voltages at What would be the effect on the steady-state factor at and L 2 A if M Solution a. Entering the receiver chart with 40,000 kw. at unity power factor, the reactive kilo volt-amperes taken by the synchronous condenser C 2 is obtained as the vertical distance between the load operating point and the 100 per cent voltage circle. The angle between the sending and receiving voltage is also read. This gives Line Pr = Pl2 = 40,000 kw. Qr = Qa = -16,300 kv.-a. MB S = zf EjR 13.2 By entering the sending chart with the angular displacement 13.2 deg., is found on the 100 per cent the sending operating point of the line voltage circle. Thus, ,/ D /Ps = 44,000 kw. t~_ „„„ , Line \ _ MB MB Qs = +25,000 . I kv.-a. M takes 100,000 kw. at 85 per cent power factor. The load at The reactive power taken by this load is consequently —62,000 kv.-a. Entering the receiver chart is 144,000 kw. power received over the line with 144,000 kw., the reactive kilovolt-amperes supplied by this line are The AM It is given by the vertical distance between the and the 110 per cent voltage circle. The angular displace- found to be +45,000 kv.-a. horizontal axis ment is also read. » Hence, PR = AM Line 144,000 kw. Q R = +45,000 kv.-a. E = 50.3 deg. ZJ &R The given reactive kilovolt-amperes of the synchronous condenser C\ are now by Qci = Qr ~ Qs - Ql = 45,000 - 25,000 + 62,000 by means = +82,000 kv.-a. AM The sending operating is located in the sending chart point of the line of angular displacement 50.3 deg. and voltage ratio which now is £ = ra = 9 °- 9 P ercent be remembered that the sending voltage in this case is different from the nominal voltage; the chart readings must be multiplied by the It should square of the ratio of these voltages. Ps = 132,000 X Hence, = 160,000 kw. Qs = -37,000 X 1.10 = -44,800 kv.-a. = 96.3 per cent (lagging) 4>s = 1.102 2 cos — , = nM^7 STEADY-STATE TRANSMISSION PROBLEMS The 373 desired results are = -16,300 kv.-a. = +82,000 kv.-a. Pa = 160,000 kw. at A = 96.3 per cent (lagging) Qcz Qci Power factor When the synchronous condenser C x is shut down, the power received over the line remains 144,000 kw. The reactive kilovolt-amperes, however, are now —37,000 kv.-a., namely, the sum of the reactive kilovoltamperes taken by the load Li and the reactive kilovolt-amperes supplied to the line MB. Locating the receiver operating point as given by the above given values of power and reactive kilovolt-amperes in the receiver chart, and the angle between them are read. the ratio of the voltages at A and b. AM M Hence, PR = 144,000 kw. Qa = -37,000 Line AM -^r- kv.-a. = 141.5 per cent = 34.5 deg. VjR A S £jr The sending and the chart is now entered by the angular displacement 34.5 deg. ratio ER Es 100 70.6 per cent 1.415 This gives Ps = 79,000 X 1.4152 = 158,000 kw. Qs = -40,500 X 1.4152 = -81,000 kv.-a. cos <j>s = — - . i The rrrt^ =89.0 per ceht (lagging) ( 81 \ desired results are Power factor at i Ea = Pa = at A = 141.5 per cent 158,000 kw. 89.0 per cent (lagging) Hence, the shutting down of the synchronous condenser affects the power A but very little. The voltage and power factor, however, are affected The voltage has to be materially increased to transmit the desired power, while the power factor is somewhat decreased. to a considerable extent. EXAMPLE 6 Statement of Problem A hydro generating station of known kilovolt-ampere capacity at A supplies power over the transmission line AB to the load L& and over the line to the load LN (Fig. 222). A steam generating station is feeding into the system at B. Synchronous condensers Cb and Cn of ample capacity are installed at the loads. The voltages at the points A, B(M), and A^ have their normal (100 per cent) values. MN 374 ELECTRIC CIRCUITS— THEORY AND APPLICATION The power taken by the load Lb and its power factor (lagging) are known. Assuming that modified Evans and Sels charts for the two transmission lines, charts at 100 per cent terminal voltage for the generators, and V-curves for the synchronous condensers are available, describe the procedure which should be followed in order to determine Fig. 222. a. —Transmission system considered The maximum unity power-factor steady state at load LN in Example which can be carried The in the N without overloading the generators. b. The power and reactive kilovolt-amperes and the hydro generators. c. 6. reactive kilovolt-amperes and field field current of the current of the synchronous condensers. Solution The load on any generator geuttruiiur will tor, limited nimteu by its Kiiovort-ampere kilovolt-ampere rating, uy us rating. It it supply maximum power, therefore, when operating at unity power facprovided such operation is possible. is Kv-a. of Hydra Generators &. "Sending Operating ^Point — Fig. 223. Determination of active and reactive power at point A in the sysin Fig. 222. The operating point is obtained at the intersection between the proper voltage-ratio circle in a modified Evans and Sels sending chart and an arc representing rated kilovolt-amperes of the hydrogenerators. tem The hydro generators may not be factor, since fixing the voltages at A able to supply power at unity power and B also fixes the sending reactive kilovolt-amperes of the line AB corresponding to any particular value of power. Obviously, the sending reactive kilovolt-amperes of the line and the Modified Evans and Sels sending chart for the line (see Fig. 225. The active and reactive power at 222). as Fig. the steam generators can be operated at unity power factor and. — M-N Fig. 222) From this chart the active and reactive power at B are obtained.STEADY-STATE TRANSMISSION PROBLEMS 375 The sending reactive kilovolt-amperes of the generators must be the same. shown in Fig. operating point is located by striking an arc about the origin of the sending chart with a radius equal to the kilovolt-ampere capacity of the hydro generators. 224. are located in this chart and the displacement between the terminal voltages read so as to make transfer to the receiver chart possible. hence. . supply to the system an amount of power equal to their kilovolt-ampere rating. 223. AB The sending operating point of the line is transferred to the receiver chart for this line Fig. 224 by means of voltage ratio (Ea/Eb = 100 per cent) and angular displacement between Ea and Eb. —Modified Evans and Sels receiver chart for the line A-B (see Fig. M Since there is a synchronous condenser at B. 376 ELECTRIC CIRCUITS— THEORY AND APPLICATION The power supplied to the line MN is Pm =Pb + given by Pgb - Plb (a) MN The sending operating point of the line is located as shown in Fig. maximum unity-power- N without overloading the generators. are obtained. 222) in the generator chart for the proper terminal voltage (solution of Example 6. . Part (6)). 227. 222). — Modified Evans and Sels receiver chart From The value this chart the active of the power Pn for the line and reactive power at so determined factor load which can be connected at is the N M-N (see Fig. be determined by locating the active and reactive power at A (see Fig. Hence. provided. 226.Generators vt =100% — Diagram showing how the field current of the hydrogenerators may Fig. 226). Fig. 225 by means of power Pm and voltage ratio {En /Em = 100 per cent) and then transferred to the receiver chart in the ordinary manner (Fig. L N = Pn W Operating Point of Hydro . of course. the synchronous condensers at this point are sufficiently large to compensate for the reactive kilovolt-amperes. It should be noted that the plied to the load at maximum amount of power which can be sup- N is actually independent of the power factor of this load. where Qm is the sending reactive power of the line Qb and Qn are the receiving reactive power of lines ' AB and MN. their field currents are determined from the V-curves. 201). using the following data: Hydro Generating Station: Installation: 4 — 32. tively.Qm (c) and Qcn = Qn (d) MN as found in Fig. the condenser operating point is located on the 100 per cent voltage curve. 225. 166). Assuming that the kilovolt-amperes taken are leading. 228. Performance given by the chart (Fig.-a. as determined in Figs. generators. 224 and 226. 227). Part (c)). Transmission Lines and Transformers: Performance given by the modified Evans and Sels charts.— Diagram showing how the field current of the synchronous condensers are determined by means of a family of F-curves (solution of Example 6. B is immediately determined.000 ky .STEADY-STATE TRANSMISSION PROBLEMS 377 6. Steam Generating Station: Installation: 2 — 30. the field current of the hydro generators The and N c. The power and reactive kilovolt-amperes of the hydro generators are found at the sending operating point of the line AB (Fig. 223). 228. respec- Operating Point ofSynchronous Condensers Field Current in Per Cent Fig.-a. . By locating a point corresponding to this power and reactive power in the generator chart' for 100 per cent terminal voltage (Fig. 165 and Fig. and the field current can immediately be read EXAMPLE 7 Statement of Problem Work out the numerical solution of Example- 6. (Fig. as shown in Fig. Having determined the reactive kilovolt-amperes of the synchronous condensers. generators.000 kv. reactive kilovolt-amperes taken are found by the synchronous condensers at by Qcb =Qb - Qlb . and Qlb is the reactive power taken by the load at B. synchronous condenser. In per cent of the total generator installation. Z/ En P N = 82. .-a.500 Q™ = Their field current = 146 per cent of normal Synchronous condenser at Qcb = 38. - ._2 ftft 128.booand the field = 1ft 19 .-a. at 90 per cent power factor (lagging).9_ -°_ perCent = +68.-a.000 + 60. synchronous condensers.500 kv.500 127.ga) c.000 kw.-a.000 In per cent of the installation this — 68. Q M = +13.378 ELECTRIC CIRCUITS— THEORY AND APPLICATION Synchronous Condensers: Installation at B: 3 Installation at N: 1 — 30.600 Xi00 907000 and the field .-000- 11.5 deg.-a.500 kv. Solution Following the procedure outlined in Example 6 gives a. Q N = +5. 186).000 kw. = 90. these figures become PoA = X 127. Plb Load o. 43.000 = 87. at N L N = 82. 6percent current becomes If(CN) =61.-a.500 X 100 ^-2s.000 kw.000 kw. = kv.500 kv.600 - 13.000 .2 per cent of normal. P M = 117. kv.000 100 = ".000 kv.600 becomes ~ 7bd per cent current I/(cb) = 88. B: 90.000 Per Cent and X 100 "128.000 + B takes 43.000 kw. or in per cent 5. Eur = 28.8 per cent of normal Synchronous condenser at N takes Q CN = +5.000 kw.-a. Pb = 117. Performance given by the V-curves Load at (Fig. Reactive power Qga = —11.5 deg. — 28.90. = .000 kv. QB = +38.600 kv.-a.000 kv. Pa = A" Qa = -11.-a.000 kw. is I/(.-a. Qlb = -43.000 kw.000 kv. Hydro generators supply Power Pga = 127. determination of the motor on unbalanced of rents in networks. 42. 293. of load. P.. D. W. 51 Bryant. line-to-neutral. J. 256 Chapman. J. 279 constant receiver voltage.. 270. 273.. 338 motor and consynchronous denser. position. 291-292. 306-350 transmission line. 289. 180 205-208 line-to-line short circuits on three-phase Evans and Blake. M. 151-154 taking positive-sequence excit- Calculations system ages. 2 Chart. 293 main generator. 155 Analytical. 306 C Calculation of a generator chart. E. short-circuit currents. 81-83 in two-phase and four-phase volt- cur- short-circuit short-circuit of factors ing current into account. 86-87 treatment. for constant sending voltage.. Table I. neglecting exciting currents. 90 Alger. 3 Bifurcated and composite lines. three- on three-phase systems. W. V. auxiliary. R. 144-147 Bekku. 306 Boyajian. 328 Analysis of induction-motor performance. 250. 325 synchronous machine. 293 receiving. 304. 322 Bush. 290 Admittance. 182 short circuits Booth. 276 for constant sending voltage. 142-144 lines. 2 Circle diagram. 266. B K. 3 Bennett. 247 Alexanderson. 353 modified. 182 Balancing effect of a three-phase induction motor. 2 Blondel's two-reaction method. 266 Charts and negative power.. E. 1-34 276 sending. to systems. lumped. three-phase induction Angle. Attenuation. 316 II.. 265-305 Chrustschoff. Table XVIII. Table III.. 21 29-33 Cantilever circuits. Table Angles. F. S. for constant receiver voltage. F. 280 329-350 single-phase. 330 simplified. 322-325 Thielemans'. 265 for 379 . 216 Biermanns. current to to single-phase. constant. 330 Sels. A. 279 generator. calculation of. 293...INDEX maximum Calculation of efficiency of transmission. 305. L. D.. 20 systems. 19 components in a three-phase system. applicable phase short circuits on threephase systems. T.. hyperbolic. 273-291. 250. neutrals isolated. analytical determination of 86-87 lines. 167-168 based directly on vector diagram. three- negative-sequence of unbalanced load. Table XIII. power 84- of. 170 equivalent. 211. 246 determination of. two. 64 Crests. 92-94 positive. general circuit. 27 transformer. curves of negative. effected by saturation. . 112-124 Y-Y. 64 sine. determination of. 92-94 equivalent of. symmetrical. sending voltage. 294 voltage-power. 302. XV. 282-291. graphical determination Copper losses. transformers repre- sented by.. 233 Y-A. equivalent. Tables XIV. 303. isolated neutrals. 232 A-A. 351-354 Condensers. 298 calculation of. XVI. 259-263 equivalent. calculation of. 248 equation of. line grounded. 238 Connections. 107 isolated neutrals. 17. 286. 304 loss. 87- 205- equivalent representation of. cross-magnetizing. in unbalanced circuits. 90-92 Corbett. 235 Y-A-Y. long transmission line. A-A. 17 Cross-magnetizing coefficient. 127-137 A-Y. cosine. 267 Circles. and synchronous motors. 210 loaded by an impedance 211 open. 64 resultant. 90-92 n. 304 Components. attenuation. 213-215 Circuits. of. analytical determination in graphical determination 89 Composite and bifurcated 208 cantilever. 17 positive-sequence of unbalanced equivalent zero-sequence impedance. T. 325-329 Connection.and four-phase system?. 296-299 calculations of. 246-263 conception of. 268 efficiency. 256-259 T. Tables XII. synchronous. 318 Coefficients. three-phase. 256 T and n. 92-94 equivalent impedance. load. impedance. L. 83 in a three-phase system. 211 unbalanced. transformers represented by. J. 236 with tertiary delta. 248 Coefficient. 81-83 86 Circuit. 34-67 208-212 T. 231 Cosine coefficients.A. 92-94 lines. 179 in phase system in terms of impedance drops. 180 -voltage transmission. er. capacity. and transformer impe- dance. 89-90 319 demagnetizing.and n. 210 unbalanced. 247 per loop per mile. cantilever. copper losses in. Condenser. Core loss. approximate deter- mination of induction-motor performance from. 232-245 Constant. 68-100 in. 319 reaction. 320 Current. 352 Constants. 238 A-A-Y. 303 voltage. all neutrals isolated. 64 Component voltages. dissy metrical. of singlephase transformers.380 ELECTRIC CIRCUITS— THEORY AND APPLICATION Circle diagram. G. 212 lines. 42-46 to two-circuit transformers.. copper loss due load. 308 stator. open-circuit-saturation. 129 short circuit. 308 of constant field current. 151 short-circuit. neutral. 357 "Engineers Manual. application to generating stations. 308 Curves. in separate leakage 381 A-A-Y connection. 318 Determinantal equation. 13 Equations.. line to neutral. 15. determinantal. copper loss. isolated neutrals. 135 F.. 22-27 Dahl. E. W. 132 A-Y connection. damper. in networks. 217 representation of composite 213-215 T and n circuits. harmonic. 185 exciting. 273.. 12-18 two single-phase eddy. M. 233 Demagnetizing coefficient. Table to short circuits. 227 Equivalent ground plane. 54 harmonic components in oscillogram of. 1-34 150 Curve. 225 Decrement.INDEX Current. O. 13 Diagram. 18-22 application of. 316 Dudley. 94-100 A-A connection. Effect. 130 line. 208-212 proof for T-line. 226-227 number of harmonics in. 127 reactances. directly on. 225 curves of constant. 338 non-salient-pole characteristic. circle. The. 138 Dwight. 231 rotor. Ferranti. 192 Equation. 266-273 Dillard. line three-phase connections of singlephase transformers. line to line to neutral. application to four-circuit transformers. E. long transmission line.. differential. table of. calcu- lation of. 352 machine. 131 of. 238 Delta. 233 Currents. 224 exciting. 155 to. 184 Efficiency circles. 52. 33 factors. A. 134 field. 331 short-circuit. B. 224 short-circuit-saturation. equations of. 331 field-current. 27 Doherty. 157 drop due to. 308 zero-power-factor. R. calculations generator chart. 238 A-A-Y connection. . 127-137 general case of unbalanced threephase load. loads. to each generating station. 49-51 to three-circuit transformers. delivered by a synchronous machine. 211 n line. H. Y-Y connection with. single-phase line to neutral. tertiary. D C. 179-180 general transformer. positive-sequence. 232 A-A-A connection." by Hudson. second-harmonic. 25 system lumped. 14. auxiliary. 266-273 diagram based circle vector. line to line. 38-42 of third-harmonic. E in XIX. line to neutral. 24 Degree of unbalance. 286 equations 298 332-338 Emde. 265 based directly on vector diagram. equations of distribution. neutrals isolated. 273 Examples. conversion. 292. 338 Generators. 353 modified. 86-87 Frank. D. 77. 20 line short circuits line-to-neutral short-circuits symmetrical of phase coordinates. F. 25 Feldmann. 222 Generator harmonics. R. determination of. conception of. 230 Graphical determination of the components in a three-phase system. equivalent. 222. tests cable to single-phase line-to- on threephase systems. 307315 salient-pole. 273. 24 system short-circuit current appli- on three-phase systems. 187-191 Faccioli. 226-227 currents.. L. 184 Fluxes.and four-phase systems. 249 transmission line. 266. 22-27 method Company. 231 transformer. 225. 225 of. 19 Feeder. non-salient-pole. 293 Evans. 223 main. 222-224 groups. 250 General Factors.. C. Table II. 90 Gould. 216-245 behavior of on a line. 291. M. complex. 304. 248 the transformer. 217 H "Handbook neers. 2. 315-322 Gilman. King E. unbalanced. 192 Harmonic. 73 Y-connected load on. 256 chart. 1.382 ELECTRIC CIRCUITS— THEORY AND APPLICATION Evans and Sels. J. 191-194 straight-line interpolation. air-gap. hyperbolic. 187-191 Fortescue. Table I. 230 Functions.. 27. 193 exponential equivalents. 247 lumped impedance.. decrement. C. 216-222 generator.. 371-378 Exponential equivalents of hyperbolic functions. 246. of. 223 in salient-pole machine. 187-191 numerical values of. 77 to system lumped. 222 Harmonics. 21 three-phase short circuits on three-phase systems. 18-22 application General circuit constants. in oscillogram of single-phase exciting current. 251-256 T-circuit. 87-89 methods. E." for Electrical Engi- by Pender. 246 determination of. R. 248 n-circuit. 247 load 1 Ferranti effect. 27. 69-73 Electric by.. 225 voltages. copper loss. unbalanced three-phase A-connected on. Table III. 90 Four-phase and two-phase systems. 216 General circuit constants. network in series and multiple. 84-86 in two. 68. charts. miscellaneous. of steadystate transmission problems. air-gap fluxes. 265 Ground plane. analytical determination of components in. 305. 250 two important relations. 224 action of in Formulas. 247 lumped admittance. 193 G Gardner. 223 in synchronous machine. F. action of. 1. 246-263 non-salient-pole machine. J. 270. 217 . harmonic. 19 to generating stations. 222 components. 142-144 unbalanced load and three-phase induction motor on line. line loaded by an. 187-191 383 Induction motors on unbalanced and voltages. 225 Hazen. under balanced load. 217 transformer. 216 in exciting current. solution of. solution of. 68 68 2. G.. 159-161 three-phase operation. 233 triple-frequency group. analytical treatment. determination 51-67 Inouye." 192 Hyperbolic. 225 in polyphase systems. L. 215. 357 K straight-line interpolation. load equal to surge impedance. 364 equivalent T and n circuits. determination of. J. 193 exponential equivalents. 182 functions. tests for the determination of. 357 Kirchhoff's law. 216 of. 148 Impedance. 29 Herzog.INDEX Harmonics. 181 transformer. 217 in single-phase transformers. Table XXI. H. E. single-phase operation. pull-out torque slip. Law.. in per cent of equivalent sine wave. 64 in salient-pole machine. 167. 225232 in three-phase connections of sin- gle-phase transformers. 149154 three-phase induction motor on unbalanced voltages. W. 191-194 remarks on. single-phase operation. 138-142 2. 45. 193. C. 138-168 analysis of single-phase perform- ance. angle of load. H. 215 data at no-load. R. Ohm's. 210 infinite. 5.. 211 surge. 193 Kennelly. 27 Line. 2. "Engineer's Manual. 90 Holladay.. 185-187 lumped.. 316 Laws. 356. 155 balancing effect of a three-phase induction motor. N. 223 main sources of. F. 357 Lawton. H. 151 taking positive-sequence exciting current into account. 232 A-Y connection. 216222 constants... 3. 247 <r. complex. L. 185. W. 161 single-phase operation of a three- phase induction motor. 185-187 . 178. 232- 245 A-A connection. 187191 numerical values of.. 226 Hudson. A. Kirchhoff's. 322 Lewis. 142. neglecting exciting currents. Induction motor. 234 Jahnke. R. 1 Hill. 356. The.. 34-67 Impedances. three-phase connections of single-phase transformers. behavior of harmonics on. Interconnected neutrals. and equivalent circuits.. 68 Lawrence. infinite line. E. A. 158 Induction motors on unbalanced voltages.. 214. F. 144-147 general discussion. on unbalanced feeder. 260 Lines. 290 Method. 185-187 line to neutral. transformers with. W. 169 T-. 308 simplified chart for. division. 183-184 in the steady state. 178-215 equivalent T and n circuits. 285 Lumped admittance. parallel lines. 63 Load. Blondel's two-reaction. 169 T-. 169 n-. 42. graphical. 322-325 salient-pole. 184 line grounded. 247 Lyon. 256. Table X. 187-191 wave length. 208212 general solution. 144-147 effect of . 195 direct-current. A-connected. be- 365— 371 general case of unbalanced threephase. 211 long transmission. 191-194 remarks on hyperbolic functions. 148 three-phase. 179-180 direct-current lines. Joseph. 225 M hyperbolic angle of. unbalanced. 106. balancing a three-phase. current line to calculation and power distribution Table IX. free. 183 load impedance equal to surge impedance. 178-215 loaded. 129 three-phase. 182 impedance equal to surge impedsingle-phase. 306-350 short-circuit current. 284. 121 two Long transmission ance. 169-177 smooth. proof for. 199-201 two of parallel. 282-286 equations. 118.. line to line. 131 single-phase. 134 Machine. on unbalanced feeder. 135 induction three-phase and motor on line. 101-137 line. 49. 183-184 line loaded 181-183 ratio of voltages. 211 transmission. gen- eral case of unbalanced. 259 with sending transformers. 315 synchronous. 247 impedance. non-salient-pole. unbalanced threephase. line to neutral. 205-208 equivalent representation of. 184-185 general solution. 202 composite and bifurcated lines. 187 Loss circles. characteristic. 179-181 grounded. 265 Motor induction. 213-215 direct-current. 365- 371 Lipka. Loads. 317 Maximum efficiency of transmission. 250 transformers. 138. 316 Methods. 73 Y-connected. conversion formulas. 205-208 constants per loop per mile.ELECTRIC CIRCUITS— THEORY AND APPLICATION 384 Line. determination tween two of. reactions. 179 differential equations. 222. of voltage. 12-18 three-phase. proof for. 82. 145. V. neutral. 212 short transmission. 181-183 nominal II-. 259 with sending and receiving transformers. 307 curves. determination load division between. receiving with 257. 115. 257. charts. 69-73 Loading. 120. line to neutral. 185-187 numerical values of hyperbolic functions. 179 line free. 199-201 in the steady state. loaded by impedance a. composite and bifurcated. parameters of. 289. 169. 135 single-phase load. 5-11 star-mesh transformation. 293 sequence.. single- phase operation and unbalanced 149-154 of. F. Motors. charts for. 178 method of. usually met with power transmission. 385 Networks. 127 two single-phase law... 169. 316. negative. H. induction. 171 Non-salient-pole generators. 78 Power. 68 Parameters. 106. 132 Network. 77 for. harmonics in. 291-293 transmission. on unbalanced voltages. 118.and T-circuits. 170 simplification of. three-phase. of three-circuit 42 reduction 2. for Electrical Engi- neers. equivalent. F. 252 two networks in series. 68 for. 217 Position angles. 322 Nominal n. synchronous. 351. 121. 252 two in series. 254 series. separate leakage reactances. on unbal- anced analytical voltages. 169 Peek. Jr. sequence. 254 in series. 356. single-phase opera- 138-168 tion. A. 251 Y-A transformation. 62-63 T-line. in Pernot. 256 transformation. three in parallel. E. W. of smooth line. 9 Networks. 322- N O Negative power and charts. 151. 77 n. proof combination in series Phase. F. 142-144 170 line. 78 Ohm's Neutral. 251-256 three networks in parallel. 224 system. 252 simplification of. 169 Pierce. 293. 252 two in parallel. 307-315 machine. 170 capacitance lumped at center. 15. 355 "Handbook short circuits. remarks. 42. H. on line. 148 single-phase induction. load.. 170. 8. 45-49 4 of short- 1-34 A-Y of. 357 Polyphase systems. 171-172 treatment. J. 216." 192 Peters. equivalent. 256 circuits. 325- 329 Multi-winding transformers. 182 Positive-sequence. circuit. nominal. 112. and condensers. general case of unbalanced.INDEX Motor induction. 358 Pender. . capacitance lumped at ends. 151. 211. combination of networks usually met with in. 308 Operators. Open-circuit-saturation curve. 6 Nickle. 5 and multiple. P. 120. 224 system. loads. line to. 162-167 three-phase induction. 134 transformers.. calculation circuit currents in.. 39of. 115. three-phase load. 129 short circuit. symmetrical coordinates. C. simplified chart 325 n-phase systems. 6 252 248 208 212 line. 225 of two-circuit transformers. in parallel. 364 of synchronous condenser data. 224 exciting current. 293 currents. 112. 291.ELECTRIC CIRCUITS— THEORY AND APPLICATION 386 R Short-circuit-saturation curve. 37 Reaction. 51-67 true leakage. 227 J.. 169 approximate. H. 64 wave. zero. R. 26 226- induction motor. leakage. 165 by simultaneous equations. 138 270. 165 load. Richter. 14 Short-circuit calculations. to line table of line data at no-load.. 311. 363- chart. armature. per cent components approximate 165 solution. 225-232 three-phase connections 292. 162-167 ing affected by saturation. 305. O. Table IV. 351 Resistance. 118. 6 tion. 115. single-phase opera- torque pull-out S rotor current. modified. surge. determination of. 315-322 266. har- monic transient. Solution. neutral. 320 three-phase machine. 36 cy. 149-154 analysis of induction-motor performance neglecting ex- 364 Regulators. Rosen. of network. 164 more exact solution. 2 Rotor current. 108 line Schurig. 353 negative. 116. 131 365 Tirrill. slip. 69-77 long transmission line. to 127 H. 138. 256 Sels. Simplified chart.. line to line. 308 Simplification. exciting current. 64 citing currents. 92-94 Shirley. 15 harmonics in equivalent. 322-325 Sine coefficients. 129 of a three-phase operation induction motor. neutral. transformers. 102.. Sequence. 273 Sels and Evans. based on Ohm's and Kirchhoffs Smooth line. 9 Reactances. 106. equivalent resistance. 161 Smithsonian Mathematical Tables. 351-378 general. K. 130 C. and 161-162 Salient-pole generators. 155 impedance of unbalanced 92-94 load. taking positive-sequence exciting current into account. 317 Regulation. 151-154 155-157 156 Schoenfeld. A. 200 Resultant coefficients. cross-magnetiz- of. 179-181 of steady-state transmission problems. 246. . determination leakage. 27. 191-192 parameters of. 92-94 zero. triple-frequen- 62 mutual leakage. of. 64 Single-phase. E. 155 impedance of unbalanced load. laws. separate of. 150 . in.. 232-245 Slepian. O. operator. of singlephase induction motor. 304. Slip. line to line. 156 Saturation. 77 positive. 29 273. and pull-out torque.. 94. O. impedance of unbalanced load. 36 self leakage. 56 short circuit. and 33 by. 329-350 non-salient-pole generators. 2 Three-circuit. 266 Third-harmonic. 238 A-Y connection. transformers. 223 for parallel-conjunction. equations of. E. 353 Systems. 322-325 synchronous motors and condensers. determination 363-365 determination of synchronouscondenser capacity. 228 voltages. connections of single-phase 307-315 salient-pole 211 of. 256-259 Terman. isolated neu- trals. 233 interconnected neutrals. 228 coordinates.. for the determination of line constants. C. method of. 387 generators. 232-245 322 data. 27- Spencer. 34-38 Thielemans' chart. distribution of harmonics. 315- transformers. all neutrals isolated. 215 Steinmetz. 232 A-A-A connection. 293 Terminal voltages. 236 .INDEX Solution of steady-state transmission problems. 87-89 A-A connection. F. 181 nation the for single-phase. n-circuits. H. 29 Stability. A. leakage separate of reactances. calculating table. 306-350 calculation of generator chart. 103 Surge impedance. Symmetrical phase. 235 interconnected neutrals. 56-59 series-opposition. 325 System stability. P. Table XX. 364 reactive kilovolt-amperes. H. table. 6 Station. 53 Symbols. 234 Y-A connection. 237 primary and generator neutrals interconnected. calculating.. third-harmonic. 55-59 application of decrement factors 33 Tests. to.. Test. 248 equation Star-mesh transformation. 77 Synchronous-condenser data. 227 meaning of symbols. 59-60 Theory. for the determi- nation generating. graphical determination of components in. 27-33 solutions by. harmonics in polyphase. 361 machine charts. the determination of separate leakage reactances. general of multicircuit transformers. three-phase. for the determination • separate of leakage react- ances. 217 two. 351354 miscellaneous examples. 235 Y-Y connection. 325-329 motors and condensers. T-circuits. 236 neutrals interconnected on both Table. 200 of determi- separate leakage reactances. isolated neutrals. 111 -phase. 25.. meaning of. 378 by Solutions T. sides. 223 Thomalen. 170 -circuit. 242 simplified chart.and four-phase. 371of regulation. transformers represented 353 resistance. regulators. tests for determination of separate leakage reactances. 231 harmonics. for deter- mination of separate leakage reactances. 6 Y-A. 54 equivalent network of threecircuit transformers. exact. transmission line with. 6 Transformer. 62-63 receiving and sending. 51-67 drop due to exciting current. connections of singlephase transformers. 53-59 three-circuit transformers. 51 separate leakage reactances. 3942 four-circuit transformers. scheme. and slip. Transformer impedance. equivalent. 259 with sending. proof for. and equivalent circuits. the. 249 core loss. 60-62 Transformers. trans- mission line with. pull-out. 59-63 two-circuit transformers. secondary and load interconneutrals nected. application of general equations. 81- mination of separate leakage reactance. for deter- in. 73 Y-connected load. 42-44 determination of three-phase tests. neutrals isolated. for determina- components 83 graphical ing parallel-conjunction test. 351 in. determination tests for separate of leakage reactances. 161 Transformation. Philip. four-circuit. 260 transmission line with. on unbalanced feeder. 49-51 multicircuit. 55 single-phase tests. 34- 38 multi-winding. determination of separate leakage reactances. 211 Torchio. 257 with sending. 49-51 harmonics in per cent of equivalent sine wave. 256 single-phase. multi-wind- components 84-86 unbalanced. 90 Torque. 38-42 two-winding transformers. 34-67 determination of impedances. 225 impedance. general theory of. approximate. 56 series-opposition test. 217 load. approximate. 5 star-mesh. 259-263 transmission line with receiving. for de- with tertiary delta. 257 with transmission line. Y-Y connection. A-connected load. 259 equivalent T-circuits. 225 action of. harmonics. 256 sending and receiving. reactions. 238 Three-phase. approximate. 64 tion of separate leakage reactances.388 ELECTRIC CIRCUITS— THEORY AND APPLICATION Three-phase. determination of separate leakage reactances. analytical determination of transformers. 135 machine. on unbalanced feeder. 317 system. general case of unbalanced. 69-73 -voltmeter. 215 Tirrill. 257 represented by cantilever circuits. 225-232 three-phase connections of 232~ . 62-63 termination of separate leak- age reactances. 45-49 two-circuit transformers. 256-259 transmission line with sending and receiving. A-Y. 237 line to neutral. 245 . triplefrequency group. 45. T-line. degree calculation of efficiency circles. 118. calculation of voltage-power circles. determination of the in three-phase systems. line to neutral. Table VII. Triple-frequency. circuits. group. 303 Y-Y con- single-phase loads. 111 two single-phase of voltage. three-circuit. 256 single-phase load. 15 Transmission. 112 ponents loads. 101-107 determination two-winding. line to neutral. 299 negative power. 257 with sending and receiving transformers. equivalent network 389 of separate leakage reactances. transformers. 257 exact. 256 problems. line to neutral. 111 application of general equations. 132 -winding transformers. 266-273 the Evans and Sels chart. 129 equivalent T-circuits. 134 short circuits. combination of networks usually met with in. Unbalanced 94-100 68-100 of. 291-293 voltage circles. 130 Y-A-Y 351-378 monics. 316 250 calculation in. 296. 86-87 -reaction method. Tables XIV. 112-124 transformers. 217 127-137 short-circuit. 39—42 two-circuit. 302 Table XVI. 127 line. 101 Y-A-Y connected. 282-291. Table XII. line to neutral. 131 by represented 303. 118. Two- line to line. har- and four-phase systems. 42-49. line to neutral. 273-29 i. 293 loss circles.INDEX Transformers. 53-62 U Unbalance. line to short circuits. analytical determination of components in. 101-137 A-A-Y connection. 101-107 -phase and four-phase systems. 304 circles. line to neutral. constant-voltage. solution of steady- on a bank nected of Y-A-Y con- transformers. graphical determination of com- -circuit three-circuit transformers. BlondeFs. 134 charts. 127-137 general case of unbalanced three-phase load. 224 Table VII. 38-42 single-phase short circuit. 87-89 without primary neutral. determination of separate leakage re- actances. 132 connection. nection neutral. loss Table XIII. Transmission-line charts. two-circuit state. Y-Y connection without primary neutral. 304 diagram based directly on vector diagram. 352 265-305 of XV. short circuit on a bank of. 106. reactance. 294 with receiving transformers. line to neutral. 135 power. 53-62 with unbalanced loading. 260 circle of. 107-111 Y-Y without connection mary pri- neutral. 257 approximate. 42-46 equivalent network 45 of. line to line. 81 analytical components . 102 Y-A connection. 273 modified. 120. 127 Transient. 142-144 Zero-power-factor curve. terminal. 1. short-circuit. R. 22. 120 Table IX. loads. 77 power in... 187 symmetrical phase coordinates. single-phase. 224 connection. 120 short circuits. 217 primary opera- two101-107 neutral. two circuits. two short- to line. 27 21. tests by. to neutral. 216 Woodruff. 118-121 loads. line to neutral. 138-168 operation of three-phase induction motor on. 107 single-phase short circuit. 83 degree of unbalance. 352 calculation of line to neutral. 1. W. single-phase. line to neutral. 352. 90-92 current components. 89-90 solution by simultaneous equations based on Ohm's and Kirchhoff's laws. 294 two transmission. 291. single-phase line circuit. 195 direct-current.. 112-115 distribution short circuits. 223 unbalanced. 19 Westinghouse facturing Whitehead. B. 86-87 and four-phase systems. with tertiary delta. 294 equations. 112-115 Voltage circles. 73 Y-connected load on unbalanced feeder. 308 saturation. 69-77 unbalanced three-phase A-connected load on unbalanced feeder. calculation of dis- unbalanced three-phase load. line to line. 238 tion of induction motors on. 112-115 load. line 121 single-phase transformers. single-phase single-phase. of of. 223 third-harmonic. Table X.390 ELECTRIC CIRCUITS— THEORY AND APPLICATION Unbalanced copper losses in. F. circuit transformers. 273. 118 line to line. short 118 two single-phase line to neutral. analytical de- termination of the components in two-phase and four-phase system. 112-124 general case of unbalanced three- phase load. 222 without sinusoidal. Wave Electric and Manu- Company. 82 111 of transformation. 121 line to line. 202 Y-Y triple-frequency. line to neutral. Y-A connection. 358 Woodward. general case tribution of. 84-86 in two- W circuits. harmonic. J. 69-73 voltage components. L. 108- 87-89 method length. 94-100 graphical determination of the components in a three-phase system. of. 236 Voltages. 6 Y-A-Y connection. neutrals lated. 56 iso- . Documents Similar To Electric Circuits.. Theory & Applications - Vol.1... Short-Circuit Calculations and Steady-State Theory - O.G.C. 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