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TRANSMISSION LINE FORMULAS 
 
 FOR 
 
 ELECTRICAL ENGINEERS 
 
 AND 
 
 ENGINEERING STUDENTS 
 
 BY 
 
 HERBERT BRISTOL DWIGHT, B. Sc. 
 
 ASSOCIATE OF THE AMERICAN INSTITUTE OF 
 ELECTRICAL ENGINEERS 
 
 NEW YORK 
 
 D. VAN NOSTRAND COMPANY 
 
 25 PARK PLACE 
 
 1913 
 
Engineering 
 Library 
 
 V 
 
 COPYRIGHT, 1913, BY 
 D. VAN NOSTRAND COMPANY 
 
 Stanbopc jpress 
 
 F. H. GILSON COMPANY 
 BOSTON, U.S.A. 
 
PREFACE. 
 
 THE object of this book is to compile a set of instructions 
 for engineers, which will enable them to make electrical 
 calculations for transmission lines with the least possible 
 amount of work. 
 
 The chart and working formulas have for the most part 
 been developed independently by the author. Where the 
 same or similar methods have been previously published, 
 the fact is generally stated in the footnotes, but it has not 
 been found possible to make these references absolutely 
 complete. 
 
 The second part of the book is for reference and contains 
 the derivation of the principal formulas used in connec- 
 tion with transmission lines. As many recent articles on 
 transmission lines make use of formulas which are only 
 roughly approximate, or are even incorrect, a reliable col- 
 lection of formulas, with the method of obtaining them, 
 should be found valuable. 
 
 It should not be presumed, because the second part of 
 the book requires the use of the integral calculus, that 
 the working formulas will require a knowledge of higher 
 mathematics. The first five or six chapters are complete 
 in themselves, and are planned for the use of those who 
 have an ordinary acquaintance with alternating-current 
 calculations. 
 
 H. B. DWIGHT. 
 
 HAMILTON, CANADA, 
 August, i pi 2. 
 
 in 
 
 266994 
 
CONTENTS. 
 
 PART I. WORKING FORMULAS. 
 
 CHAPTER PAGE 
 Frontispiece REGULATION CHART 
 
 PREFACE iii 
 
 I. INTRODUCTION i 
 
 II. ELEMENTS or A TRANSMISSION LINE 4 
 
 III. REGULATION CHART 8 
 
 IV. FORMULAS FOR SHORT LINES 16 
 
 V. K FORMULAS 25 
 
 VI. CONVERGENT SERIES 41 
 
 PART II. THEORY. 
 
 VII. CONDUCTORS 53 
 
 VIII. TRANSMISSION LINE PROBLEMS 57 
 
 IX. REACTANCE OF WIRE, SINGLE-PHASE 62 
 
 X. SKIN EFFECT 67 
 
 XI. REACTANCE OF CABLE, SINGLE-PHASE 76 
 
 XII. REACTANCE OF TWO-PHASE AND THREE-PHASE LINES 79 
 
 XIII. CAPACITY OF SINGLE-PHASE LINE 84 
 
 XIV. CAPACITY OF TWO-PHASE AND THREE-PHASE LINES 93 
 
 XV. THEORY OF CONVERGENT SERIES 98 
 
 PART III. TABLES. 
 TABLE 
 
 REGULATION CHART 
 
 I. FORMULAS FOR SHORT LINES 104 
 
 Conditions given at Receiver End. 
 
 II. FORMULAS FOR SHORT LINES 106 
 
 Conditions given at Supply End. 
 
 III. K FORMULAS 108 
 
 Conditions given at Receiver End. 
 
 IV. K FORMULAS 112 
 
 Conditions given at Supply End. 
 V 
 
VI CONTENTS 
 
 TABLE PAGE 
 
 V. CONVERGENT SERIES 1 16 
 
 Conditions given at Receiver End. 
 
 VI. CONVERGENT SERIES 117 
 
 Conditions given at Supply End. 
 
 VII. RESISTANCE OF COPPER WIRE AND CABLE 118 
 
 VIII. RESISTANCE OF ALUMINUM CABLE 120 
 
 IX. REACTANCE OF WIRE, 25 CYCLES 121 
 
 X. REACTANCE OF CABLE, 25 CYCLES 122 
 
 XI. REACTANCE OF WIRE, 60 CYCLES 124 
 
 XII. REACTANCE OF CABLE, 60 CYCLES 125 
 
 XIII. CAPACITY SUSCEPTANCE OF WIRE, 25 CYCLES 127 
 
 XIV. CAPACITY SUSCEPTANCE OF CABLE, 25 CYCLES 128 
 
 XV. CAPACITY SUSCEPTANCE OF WIRE, 60 CYCLES 130 
 
 XVI. CAPACITY SUSCEPTANCE OF CABLE, 60 CYCLES 131 
 
 XVII. POWER FACTOR TABLE 133 
 
 ALPHABETICAL INDEX 135 
 
Transmission Line Formulas 
 
 PART I. 
 WORKING FORMULAS. 
 
 CHAPTER I. 
 
 INTRODUCTION. 
 
 THE determination of the electrical characteristics of 
 transmission lines is a problem of considerable practical 
 importance to engineers. It occurs frequently in electri- 
 cal engineering work, and various methods have been pro- 
 posed for carrying out the calculations with greater or less 
 degrees of accuracy. Unfortunately, most of the methods 
 so far presented have been such as to require special mathe- 
 matical skill in using unfamiliar forms such as hyperbolic 
 sines, etc. Even the approximate methods, whose results 
 are not intended to be reliable for lines of considerable 
 length, are often too cumbersome to be used by an engi- 
 neer who has not time to make himself thoroughly familiar 
 with them. The result has been that engineers have often 
 been satisfied with calculations for practical cases which 
 were not nearly as correct as they might easily have been. 
 
 Working methods have been developed, and are presented 
 in the following chapters, for solving problems in connec- 
 tion with actual transmission lines. As these working 
 methods involve comparatively simple operations in algebra 
 
2 , TRANSMISSION .LINE FORMULAS 
 
 .>\ .,,- -i {..*;/.\ 
 
 and arithmetic only, they should be found useful by all 
 electrical engineering students and engineers. Groups of 
 problems with answers are added to provide practice in the 
 use of the formulas. The working formulas can be imme- 
 diately used by electrical engineers without the delay 
 caused by working out the true meaning and correct oper- 
 ation of long and intricate systems of calculation. They 
 are also arranged to require the minimum amount of labor 
 for routine work where many lines are to be calculated. 
 
 The first method, which is described in Chapter III, is in 
 the form of a chart which gives the regulation or voltage 
 drop of a line, and which also shows directly the required 
 size of conductor for given conditions. 
 
 In Chapter IV are given formulas for distribution lines 
 and transmission lines only a few miles long. These are 
 extended in Chapter V by means of the constant K to 
 apply to transmission lines of any length in ordinary 
 practice. 
 
 For purposes of checking different formulas, and for 
 the calculation of extremely long and unusual lines, the 
 fundamental formulas of transmission lines are expressed 
 by rapidly converging series in Chapter VI. While these 
 series require much more arithmetical work than the K 
 formulas, they will give the exact results to any degree of 
 accuracy desired. The method of convergent series in- 
 volves the use of complex numbers, that is, numbers in 
 which "j" terms appear. They are easier to handle, 
 however, than logarithms, sines and cosines of angles, or 
 hyperbolic functions, and therefore the use of these other 
 mathematical functions has been avoided. 
 
 Each of the above groups of working formulas is printed 
 in a table, ready for practical use. The tables will be 
 
INTRODUCTION 3 
 
 found in the collection at the back of the book, as well as 
 in the separate chapters describing them. 
 
 When any formula is given which uses approximations, 
 the limits of its accuracy should be clearly stated so that 
 one can tell at a glance whether the method is sufficiently 
 accurate for the purpose in hand, or whether a longer 
 method giving greater accuracy is desirable. This is espe- 
 cially necessary in the calculation of transmission lines, 
 because approximate formulas are quite permissible for 
 lines only a few miles long, but become very untrust- 
 worthy when the length is increased to one hundred miles 
 or more. For this reason, each table of formulas has its 
 percentage and range of accuracy printed in a prominent 
 position, so that the most suitable method for any case 
 may be instantly chosen. 
 
CHAPTER II. 
 ELEMENTS OF A TRANSMISSION LINE. 
 
 THE essential elements of a transmission line have been 
 described many times, but a short discussion of them, 
 with an explanation of some of the terms used in connec- 
 tion with the subject, may be useful before proceeding 
 with the actual calculations. 
 
 A transmission line consists of two or more conductors 
 insulated from each other so that they can carry energy by 
 electric currents to some more or less distant point. 
 
 The conductors may be solid copper wires, copper cables, 
 or aluminum cables. The diameters and resistances of 
 various standard conductors are given in Tables VII and 
 VIII, pages 118 to 120. It will be noted that the exact 
 value of the resistance of a conductor differs slightly when 
 a direct current, and an alternating current of 25 or 60 
 cycles, is flowing. This is due to the "skin effect," by 
 which an alternating current tends to flow near the surface 
 of a conductor, as explained in Chapter X. The drop in 
 voltage due to resistance is proportional to the current 
 and is in phase with it when the current is alternating. 
 
 Only overhead lines, carrying alternating currents, will 
 be considered in this book. Such lines are supported by 
 poles or steel towers at a considerable height above the 
 ground. The conductors are separated from each other by 
 a distance which may be several inches or several feet. 
 This distance is called the " spacing" of the conductors 
 
 4 
 
ELEMENTS OF A TRANSMISSION LINE 5 
 
 and it has an important bearing on the electrical charac- 
 teristics of the line. 
 
 An alternating magnetic field is formed around, and in- 
 side of, conductors carrying alternating currents. This 
 field generates a voltage along the conductors which is 
 proportional to the current, like the voltage drop due to 
 resistance, but which is 90 out of phase with the current. 
 This voltage is called the reactance drop. Tables of re- 
 actance of transmission lines will be found in Part III. 
 
 Since the voltage drop in a transmission line is due to 
 resistance and reactance, a simple line may be considered 
 to be made up of the elements shown in Fig. i. If R is 
 
 Resistance 
 
 Receiv- 
 
 Suppty Eg E er0 r 
 
 Resistance ^^Sr i Load. 
 
 Fig. i. 
 
 the total resistance, the voltage drop in phase with the 
 current / will be IR, and if X is the total reactance, the 
 voltage drop in quad- 
 rature with the current 
 will be IX. 
 
 The vector diagram of 
 the above quantities will 
 be as in Fig. 2. The cur- 
 rent is in general not in 
 phase with the voltage Fig - 2- 
 
 E, but lags behind it by an angle 6, according to the 
 power factor, cos 6, of the load. The resistance drop IR 
 will therefore not be added directly to E, but must be 
 added vectorially, along with the reactance drop IX, as in 
 
6 TRANSMISSION LINE FORMULAS 
 
 Fig. 2. It is evident that the voltages E and E 8 , and the 
 power factors cos 6 and cos 0, at the two ends of the line, 
 are not the same in value. 
 
 A long transmission line acts as a condenser and this 
 fact also must be taken into account. A condenser con- 
 sists of two electrical conductors placed close together but 
 insulated from each other so that a direct current cannot 
 pass between them. However, if an alternating voltage 
 be applied between them, a charge of electricity propor- 
 tional to the electrostatic capacity of the condenser will 
 flow into and out of the conductors. The result is that an 
 alternating current will appear to flow between them, pro- 
 portional to the t capacity susceptance of the condenser. 
 This current, called the charging current, will be 90 
 out of phase with the voltage, and, unlike most currents 
 in ordinary practice, it will lead the voltage in phase, in- 
 stead of lagging behind it. The amount of the charging 
 current may be determined by means of the tables of 
 capacity susceptance of transmission lines, in Part III. 
 
 A current in phase with the voltage will flow between 
 the conductors, but it is only noticeable at very high 
 voltages. Part of it is a leakage current flowing over the 
 insulators, and part is a discharge through the air, and 
 produces the glow called corona, on high-voltage con- 
 ductors. 
 
 The elements of a transmission line accounting for the 
 leakage current and charging current are shown in Fig. 3, 
 in which resistances and condensers are shunted across the 
 line all along its length. 
 
 Considering for the present that the voltage of the line 
 is the same at all parts and is equal to , the current in 
 phase with E flowing across from one conductor to tjie 
 
ELEMENTS OF A TRANSMISSION LINE 
 
 other will be EG, where G is the total conductance between 
 the wires. So also, if B is the capacity susceptance of the 
 
 Supply 
 
 P Receiv- 
 E er. 
 
 Fig. 3. 
 
 line considered as a condenser, EB will be the value of the 
 shunted current in quadrature 
 with E. 
 
 The vector diagram for the 
 line indicated in Fig. 3 (neg- 
 lecting the voltage drop in the 
 conductors) is shown in Fig. 4. 
 It is seen that the current I 8 at 
 the supply end is different in 
 magnitude and phase from the current / at the receiver. 
 
 In order to calculate the combined effect of the above 
 phenomena, formulas must be used which will take into 
 account the fact that the resistance, capacity, etc., are 
 uniformly distributed along the line, and that the line cur- 
 rent and voltage are different at all parts of the line. 
 
 Fig. 4. 
 
CHAPTER III. 
 REGULATION CHART. 
 
 THE characteristic of a transmission line which limits 
 the load it may carry is its regulation, or the variation in 
 voltage which occurs when the load is thrown on and off. 
 This is especially true when the load has a low power 
 factor, which is the case in most instances at the present 
 time. 
 
 For estimating the regulation of a line, or the size of 
 conductor required, the regulation chart which forms the 
 frontispiece of the book may be used, and it will give the 
 required result much quicker than any method of calcula- 
 tion. An extra copy of the chart is inserted at the back 
 of the book; it may be found useful for cutting out and 
 mounting on cardboard. 
 
 The chart is accurate within approximately \ of i% of 
 full line voltage, when the regulation is less than 10% and 
 the line is not more than 100 miles long. 
 
 In using the chart,* one places a straightedge across it 
 from the point on the left corresponding to the spacing of 
 the transmission line, to the point on the right, correspond- 
 ing to the resistance of the conductor per mile. The reg- 
 ulation voltage, V, per total ampere per mile of line, is 
 then read directly from the chart for the power factor of 
 load considered. The regulation is taken as the change in 
 
 * The process of using the chart is similar to that used with the trans- 
 former regulation and efficiency charts published by J. F. Peters, Electric 
 Journal, December, 1911. 
 
 8 
 
REGULATION CHART 9 
 
 load voltage when the load is thrown on or off, assuming 
 constant supply voltage. 
 
 The total regulation is quickly figured on the slide rule 
 from the following formula for two-phase (four wire) or 
 three-phase lines: 
 
 i . w u looo K.V.A. X IV 
 Regulation Volts = - 
 
 where K.V.A. = Kilovolt-amperes of load, at the receiver 
 
 end. 
 
 E = Line voltage at the load, or receiver end. 
 / = Length of line in miles. 
 
 For single-phase lines use 2V instead of F, making the 
 formula as follows: 
 
 _> . . _, u looo K.V.A. X I X 2 F 
 Regulation Volts = - 
 
 The regulation volts may be expressed as a percentage 
 of E to give the per cent regulation, and a formula is given 
 on the chart for obtaining this result directly. 
 
 The line drop, or difference in voltage between the supply 
 end and the receiver end of the line, is the same as the 
 regulation for lines less than about 20 miles long, but for 
 longer lines the effect of the charging current must be taken 
 into account by the formula 
 
 Line Drop = Regulation Volts 
 
 aooo/ 
 
 where K = 2.16 for 60 cycles, 
 
 and K = .375 for 25 cycles. 
 
 It is seen that the voltage due to the charging current 
 is proportional to the line voltage E, and to the square 
 of the number of miles, but is independent of the size or 
 spacing of the conductors, within the assigned limit of ac- 
 curacy. The constant K does not need to be used in the 
 
10 TRANSMISSION LINE FORMULAS 
 
 v 
 
 formula for regulation, since the charging current is present 
 at both no load and full load. 
 
 In selecting the spacing point on the chart, one notes 
 whether the frequency is 25 or 60 cycles, and whether the 
 conductor is of copper or aluminum. The spacing points 
 are the same for both solid wire and cable. When the 
 wires of a three-phase line are not spaced at the corners of 
 an equilateral triangle, but are at irregular distances a, b, 
 and c from each other, as in Figs. 5 and 6, the equivalent 
 spacing _ 
 
 5 = v abc 
 should be used. 
 
 ........... c ......... ~ 
 
 Fig. 6. Irregular Flat Spacing. 
 
 K -- a ..... -><- ...... a ----->j 
 
 I ............ 2a ............ j 
 
 Fig. 5~ Irregular Triangular Spacing. Fig. 7. Regular Flat Spacing. 
 
 With regular flat spacing, as in Figs. 7 and 8, the equa- 
 tion for the equivalent spacing becomes simply 
 
 s = i. 26 a. 
 
 It makes no difference whether the plane of the wires 
 with flat spacing is horizontal, vertical or inclined. 
 
 The spacing of a two-phase line is the average distance 
 between wires of the same phase. The distance between 
 wires of different phases is not considered. 
 
 The points marked on the resistance scale at the right 
 of the chart are for cables at 20 C., assuming hard-drawn 
 copper of a conductivity equal to 97.3% of the Annealed 
 
REGULATION CHART II 
 
 Copper Standard, and hard-drawn aluminum of 60.86% 
 conductivity, and allowing an increase of i% 
 in resistance for the effect of spiralling of the 
 wires in the cable. However, these resistance 
 points are placed on the chart for convenience 
 only, and are not essential. If other assump- 
 tions are made, or if other sizes of conductor 
 are used, all that is needed is to find the re- 
 sistance of the conductor per mile, and use the 
 corresponding resistance point on the chart Fi s- 8 - Regular 
 to find "V." FtatSpactag - 
 
 One of the most common problems in estimating new 
 projects is to determine the size of wire needed for any 
 given value of regulation, and the chart will be found 
 especially applicable to this work. "V" is first found from 
 the equation, 
 
 v = % Reg'n X E 2 
 ~ 100,000 K.V.A. X I ' 
 
 Then lay a straightedge through "V" and the point for 
 the spacing to be used, and the nearest size of conductor 
 can be seen, at a glance, on the resistance scale at the right. 
 The chart is quite as useful for finding the voltage drop, 
 or required size of conductor, for distribution lines a few 
 hundred feet long as it is for transmission lines many miles 
 
 long. 
 
 PROBLEM A. 
 
 Find, by means of the chart, the regulation and line drop for the 
 following set of conditions: 
 
 Length of line 100 miles. 
 
 Spacing 8 feet. 
 
 Conductor No. 3 copper cable. 
 
 Load (measured at receiver end) , 3000 K. V. A., 
 66,000 volts, 90% P.F., three phase, 60 cycles. 
 
12 TRANSMISSION LINE FORMULAS 
 
 Lay a straightedge from the 8-foot spacing point (60 cycles, copper 
 conductor) to the point on the resistance scale for No. 3 copper 
 cable. It is found to cross the 90% P.F. line at the reading 1.344. 
 Then, by the formula on the chart, 
 
 ^ T. , ,. 100,000 X 3000 x 100 x 1.344 
 
 Per cent Regulation = z 223 
 
 66,000 X 66,000 
 
 = 9.26%, or approximately 9.3%. 
 
 The calculated value of the regulation of this line is 9.40% (Chap. 
 VI, Prob. 2), so that the error involved in using the chart is less than 
 I of i% of line voltage. 
 
 The per cent line drop, according to the chart, is 
 9.26 100 x 2.16 x T<T = 7-io%. 
 
 As the calculated value is 7.08% (Chap. VI, Prob. 2), the error 
 from the chart is less than rV of i% of the line voltage. 
 
 PROBLEM B. 
 
 1 To find the size of copper required to give approximately 10% 
 voltage drop in the following case: 
 
 Length of line 3 miles. 
 
 Flat spacing as in Fig. 7. Wires 2 feet apart. 
 Load (measured at receiver end), 250 K.V.A., 
 2200 volts, 85% P.F., three phase, 60 cycles. 
 First, find V from the formula on the chart. 
 
 V = 10 X 2200 X 2200 = 
 
 100,000 X 250 X 3 
 
 The equivalent spacing is 1.26 X2, or 2.52 feet. The proper 
 spacing point will therefore be just below the spacing point for 2^ 
 feet, copper conductor, 60 cycles. Lay a straightedge from this 
 point to the reading 0.64 on the line for 85% P.F. and it cuts the 
 resistance scale at 0.36 ohm per mile. The nearest size of copper is 
 
 seen to be No. ooo. 
 
 PROBLEM C. 
 
 Find the voltage drop of the following two-phase line: 
 
 Length of line 80 miles. 
 
 Spacing 10 feet. 
 
 Conductor No. oo aluminum cable. 
 
 Load (measured at receiver end), 15,000 K.V.A., 
 100,000 volts, 95% P.F., two phase, 25 cycles. 
 
REGULATION CHART 13 
 
 Laying a straightedge across the chart from the lo-foot spacing 
 point for 25 cycles and aluminum conductor, to the resistance point 
 for No. oo aluminum, the value of V for 95% P.F. is found to be 
 0.750. Then the line drop, in volts, is equal to 
 
 1000x15.000x80x0.750 _ o o o8 o o8 
 
 100,000 
 
 = 9000 240 
 = 8760 volts. 
 
 The calculated value is 8810 volts (Chap. VI, Prob. i). The 
 error from the chart is 0.05% of line voltage. . 
 
 PROBLEM D. 
 
 Find the regulation of the following single-phase line: 
 
 Length of line 15 miles. 
 
 Spacing 3 feet. 
 
 Conductor No. o copper wire. 
 
 Load (at receiver end), 300 K.V.A., 50% P.F., 
 
 11,000 volts, single phase, 60 cycles. 
 From the chart, V = 0.851. 
 Therefore Regulation = io.oo Xoo X i S X a Xo.8 S i _ 6 % 
 
 11,000 X 11,000 
 
 [Calculated value, 6.40% (Chap. IV, Prob. A). Error from chart, 
 0.07% of line voltage.] 
 
 PROBLEM E. 
 
 Find the K.V.A. which can be delivered at the end of the following 
 line, with 8% regulation: 
 
 Length of line 75 miles. 
 
 Spacing 8 feet, regular flat spacing. 
 
 Conductor No. oo aluminum cable. 
 
 Character of load (at receiver end), 
 
 88,000 volts, 85% P.F., three phase, 25 cycles. 
 Equivalent spacing 5 = 8 X 1.26 = 10.08 feet. 
 
 V = 0.755. 
 
 KVA = 8 X 88,000 X 88,000 
 100,000 X 0.755 X 75 
 = 10,900. 
 
14 TRANSMISSION LINE FORMULAS 
 
 PROBLEMS, CHAPTER III. 
 
 (REGULATION CHART.) 
 
 1. Find the size of copper cable which is needed to deliver 200 
 K.V.A. at a distance of 3 miles with 10% drop or less. 
 
 Spacing of line 2 feet. 
 
 Character of load (at receiver end), 2200 volts, 
 80% P.F., three phase, 60 cycles. 
 
 [Ans. No. o.] 
 
 2. Assuming No. o copper cable for the previous problem, find 
 the volts drop in the line. 
 
 [Ans. 219 volts. Calculated, 221 volts (Chap. IV., Prob. i). 
 Error 0.1% of line voltage.] 
 
 3. Find the size of copper required for a drop of 6% or less in the 
 following case: 
 
 Length of line 5000 feet. 
 
 Spacing 18 inches. 
 
 Load (at receiver end), 75 Kw. (79 K.V.A.), 
 95% P.F., 2000 volts, single phase, 60 cycles. 
 
 [Ans. No. 4 copper.] 
 
 4. Assuming No. 4 copper wire of 1.312 ohms per mile, find the 
 per cent line drop and the supply voltage. (See Elec. Journal, Apr., 
 1907, p. 231.) 
 
 [Ans. 543%, 2110 volts, calc. 5-43%, 2109 volts (Chap. IV., 
 Prob. 3)-] 
 
 5. Find the required size of copper for 9% regulation in the fol- 
 lowing case : 
 
 Length of line 25 miles. 
 
 Spacing 3 feet. 
 
 Load (at receiver end), 2500 K.V.A., 
 
 20,000 volts, 60% P.F., three phase, 25 cycles. 
 
 [Ans. No. o copper.] 
 
 6. Assuming No. o copper wire of 0.520 ohm per mile, find the 
 per cent regulation. (See Elec. Journal, Apr., 1907, p. 231.) 
 
 [Ans. 8.42%, calc. 8.51% (Chap. IV, Prob. 5). Error 0.09% 
 of line voltage.] 
 
REGULATION CHART 15 
 
 7. Find the per cent regulation and the voltage drop of the fol- 
 lowing line: 
 
 Length of line 75 miles. 
 
 Spacing, 8 feet, regular flat spacing. 
 
 Conductor No. oo aluminum cable. 
 
 Load (at receiver end), 10,000 K.V.A., 
 
 88,000 volts, 85% P.F., three phase, 25 cycles. 
 [Ans. 7-36% Reg'n., 7.15% line drop, calc. 7-35% Reg'n., 7.13% 
 
 line drop (Chap. V, Prob. 3).] 
 
 8. Find the K.V.A. which can be delivered at the end of the fol- 
 lowing line, with 10% regulation, at 75% and at 90% P.F.: 
 
 Length of line 100 miles. 
 
 Spacing 10 feet. 
 
 Conductor No. oooo aluminum cable. 
 
 Receiver voltage 110,000 volts, three phase, 
 
 60 cycles. 
 [Ans. 14,300 K.V.A., 16,300 K.V.A.] 
 
CHAPTER IV. 
 FORMULAS FOR SHORT LINES. 
 
 THE effect of capacity is inappreciable with short lines 
 as it amounts to only y 1 ^ of i% for a line about 20 miles 
 long. Thus distribution lines and many short transmission 
 lines can be quite accurately calculated without considering ' 
 the line capacity at all. The formulas in Tables I and II, 
 pp. 1 8 and 20, enable one to solve many problems con- 
 nected with such lines. 
 
 The formulas are divided into two groups, those in 
 Table I being used when all the particulars describing the 
 load, such as K.V.A., voltage and power factor, are speci- 
 fied at the receiver end. Table II is used when these par- 
 ticulars are specified at the supply end. 
 
 One first finds the quantities P and <2'or P s and Q s . These 
 are the values of in-phase current, and reactive or quadra- 
 ture current, at the point where the conditions are speci- 
 fied. It is to be noted that only values of current 
 expressed in total amperes are to be used in connection 
 with the formulas in this book. The number of amperes 
 per wire is never used in the calculations (except with 
 single phase lines), but if it is desired to be known, it 
 may be determined from the formulas at the bottom of 
 the tables. 
 
 The next step is to find the quantities A and B, or F and 
 G. One is then ready to find the value of any of the ten 
 quantities, whose formulas are given in the tables. It 
 
 16 
 
FORMULAS FOR SHORT LINES 17 
 
 should be remembered that each of these ten quantities 
 may be determined independently of all the others. Thus 
 it is not necessary to work out the first six equations in 
 order to obtain the value of the seventh, since the seventh, 
 like any of the others, may be calculated directly. 
 
18 TRANSMISSION LINE FORMULAS 
 
 TABLE I. FORMULAS FOR SHORT LINES. 
 CONDITIONS GIVEN AT RECEIVER END. 
 
 These formulas are exact when the line is short. When the line is 20 
 miles long, they are correct within approximately T ^ of i% of line voltage. 
 Conditions given: 
 K.V.A. = K.V.A. at receiver end. 
 
 E = Full load voltage at receiver end. 
 cos 6 = Power factor at receiver end. 
 K.W. = K.V.A. cos 6. 
 
 r = Resistance of conductor per mile. (From Tables VII-VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 I = Length of line in miles. 
 
 1000 K.V.A. cos 6 
 Then P= = = In-phase current at receiver end (m 
 
 total amps.). 
 
 _. looo K.V.A. sin 6 _ f .. 
 
 Q = - = Reactive current at receiver end (in 
 
 total amps.) when current is lagging 
 
 1000 K.V.A. sin 6 , . 
 
 = -- = - when current is leading. 
 hi 
 
 Find the following quantities: 
 
 Three phase or two phase. Single phase. 
 
 A=E + Prl + Qxl. A = E + 2 Prl + 2 Qxl. 
 
 B = Pxl-Qrl. . B = 2 Pxl-2Qrl. 
 
 Formulas (capacity neglected) : 
 
 (1) Voltage at supply end = A -\ -- -. 
 
 2 A. 
 
 B z 
 
 (2) Regulation of line = A H -- j E. (Same as line drop.) 
 
 2 A. 
 
 (3) Per cent regulation of line = - - - - '- per cent. 
 
 (Same as per cent line drop.) 
 
 (4) K.V.A. at supply end = rr^- X K.V.A. 
 
 L 
 
 (5) K.W. at supply end = (AP - BQ). 
 
 IOOO 
 
 (6) Power factor at supply end = ( Ap ~ ^Q) E 
 
 (in decimals). 
 
FORMULAS FOR SHORT LINES 
 AP-B 
 
 (7) In-phase current at supply end = 
 
 (8) Reactive or quadrature current at supply end 
 
 in total amperes.* 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 
 (9) K.W. loss in line = - (AP - BQ - EP). 
 
 (10) Per cent efficiency of line = . _ ^ per cent. 
 
 Total amps. 
 * Amperes per wire, three phase = -- -j= - 
 
 V 3 
 
 Total amps. 
 Amperes per wire, two phase = - 
 
20 TRANSMISSION LINE FORMULAS 
 
 TABLE II. FORMULAS FOR SHORT LINES. 
 CONDITIONS GIVEN AT SUPPLY END. 
 
 These formulas are exact when the line is short. When the line is 20 
 miles long, they are correct within approximately tV f z % f nne voltage. 
 
 Conditions given: 
 
 K.V.A. = K.V.A. at supply end. 
 
 E s = Full load voltage at supply end. 
 cos 6 = Power factor at supply end. 
 K.W. = K.V.A. cos 6. 
 
 r = Resistance of conductor per mile. (From Tables VII-VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 I = Length of line in miles. 
 
 loco K.V.A. cos 9 
 
 Then P a = - ^ = In-phase current at supply end 
 
 s 
 
 (in total amps.). 
 
 _. 1000 K.V.A. sin 9 _ 
 
 Q 8 = - = - = Reactive current at supply .end 
 c-s 
 
 (in total amperes) when current is 
 lagging. 
 
 1000 K.V.A. sin 9 , . . , ,. 
 
 = -- - when current is leading. 
 E 8 
 
 Find the following quantities: 
 
 Three Phase or Two Phase. Single Phase. 
 
 F = E 8 - P 8 rl - Q 8 xl F = E 8 -2 P 8 rl - 2 &*' 
 
 G = Q 8 rl - P 8 xl G = 2 Q 8 rl - 2 P 8 xl. 
 
 Formulas (capacity neglected): 
 
 G 2 
 
 (1) Voltage at receiver end = F + 
 
 (2) Regulation of line = E 8 F p. (Same as line drop.) 
 
 2 
 
 (3) Per cent regulation of line = - per cent. 
 
 (Same as per cent line drop.) 
 F+- 
 
 (4) K.V.A. at receiver end = = X K.V.A. 
 
 L 8 
 
FORMULAS FOR SHORT LINES 21 
 
 (5) K.W. at receiver end = ^^ (FP 8 GQ 8 ). 
 
 i (FP 8 -GQ g )Es 
 
 (6) Power factor at receiver end = y ^ \ x K.V.A. 
 
 (in decimals). 
 (7) In-phase current at receiver end = FPs " Qa in total amperes.* 
 
 , GPs + FQ S 
 
 (8) Reactive or quadrature current at receiver end = ^r 
 
 in total amperes.* 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 
 (9) K.W. loss in line = ^ (E 8 P S - FP a + GQs). 
 
 (10) Per cent efficiency of line = ^ p ^ P er cent - 
 
 Total amps. 
 * Amperes per wire, three phase, = ^= 
 
 _ Total amps. 
 Amperes per wire, two phase, - 
 
22 TRANSMISSION LINE FORMULAS 
 
 PROBLEM A. 
 
 
 
 Find the regulation of the following single-phase line: 
 Length of line ....................... 15 miles. 
 
 Spacing ............................. 3 feet. 
 
 Conductor .......................... No. o copper wire. 
 
 Load (at receiver end), 300 K.V.A., 11,000 volts, 50% P.F., single 
 
 phase, 60 cycles. (Same as Prob. D, Chap. III.) 
 From the tables in Part III, 
 
 r = 0.5331. 
 * = 0.686. 
 
 P = looo X 300 X 0.50 = 6 
 
 11,000 
 
 = looo X 300X0.866 = 
 
 11,000 
 
 A = 11,000 + 2 X 13-64 X 0.5331 X 15 
 
 + 2 X 23.62 X 0.686 Xi5 
 = 11,704. 
 B = 2 X 13.64 X 0.686 X 15 - 2 X 23.62 X 0.5331 X 15 
 
 = " 97 ' 
 
 Per cent regulation = I0 ( 11,704 + 97 X 97 -- 11,000 ) 
 11,000 \ 2 X 11,704 / 
 
 = 6.40%. 
 
 
 PROBLEM B. 
 
 Calculate the volts drop for the following case, where all the 
 conditions are specified at the supply, or generator, end of the line: 
 Length of line ........................ 10 miles. 
 
 Spacing .............................. 3 feet. 
 
 Conductor ........................... No. 2 copper wire. 
 
 Quantities measured at supply end: 600 K.V.A., 6600 volts, 
 80% P.F., three phase, 60 cycles. 
 
 From the tables hi Part III, 
 
 r = 0.8469. 
 * = 0.714. 
 
FORMULAS FOR SHORT LINES 23 
 
 _ looo X 600 X 0.80 _ 
 
 Pe ~ 6600 ~ 72>73 ' 
 
 n _ looo X 600 X 0.60 _ 
 V* ~ ~ ~ 54-55- 
 
 6600 
 
 F = 6600 - 72.73 X 8.47 - 54-55 X 7-14 
 
 = 5594-5- 
 G = 54-55 X 8.47 - 72.73 X 7-i4 
 
 = - 57- 
 Voltage at receiver end 
 
 = 5594-5 + 0-3 
 = 5595 approximately. 
 Drop in volts = 6600 5595 
 
 = 1-005 volts. 
 
 Per cent drop = I0 X I0 * 
 
 5595 
 = 17.96% of receiver voltage. 
 
 PROBLEMS, CHAP. IV. 
 
 (FORMULAS FOR SHORT LINES.) 
 
 1. Find the voltage drop in the following case: 
 Length of line ......................... 3 miles. 
 
 Spacing ............................... 2 feet. 
 
 Conductor ............................ No. o copper cable. 
 
 Load (at receiver end), 200 K.V.A., 2200 volts, 80% P.F., three 
 
 phase, 60 cycles. (Prob. 2, Chap. III.) 
 
 [Ans. 221 volts.] 
 
 2. Find (a) the P.F. at the supply end, 
 
 (b) the per cent efficiency of the line, for the case in Prob. i. 
 [Ans. (a} 78.7% P-F. (b) 92.3% efficiency.] 
 
 3. Find the supply voltage in the following case: 
 
 Length of line ........................ ...... 5000 feet. 
 
 Spacing .................................... 18 inches. 
 
 Conductor, No. 4 copper wire of 1.312 ohms per mile. 
 Load (at receiver end), 75 Kw., 2000 volts, 95% P.F., single phase, 
 60 cycles. (Prob. 4, Chap. III.) 
 
 [Ans. 2109 volts.] 
 
24 TRANSMISSION LINE FORMULAS 
 
 4. Find the volts drop and the watts loss, in the following line: 
 Length of line . . 20 miles. 
 
 Spacing 5 feet. 
 
 Conductor No. i aluminum cable. 
 
 Two phase, 25 cycles. 
 
 K.V.A. at supply end 10,000. 
 
 Volts at supply end 35,000. 
 
 P.F. at supply end 80%. 
 
 [Ans. 5950 volts, 1770 Kw.j 
 
 5. Find the per cent regulation of the following line: 
 
 Length of line 25 miles. 
 
 Spacing 3 feet. 
 
 Conductor, No. o copper wire of 0.520 ohm per mile. 
 
 Load (at receiver end), 2500 K.V.A., 20,000 volts, 60% P.F., 
 three phase, 25 cycles. (Prob. 6, Chap III.) 
 
 {Am. 8.51%.] 
 
CHAPTER V. 
 K FORMULAS. 
 
 WHEN a transmission line is more than 20 miles long, 
 the formulas for short lines given in Chapter IV are no 
 longer accurate, and other formulas must be used, which 
 will take into account the capacity of the line. Such 
 formulas, called K formulas, will be found in Tables III 
 and IV, pp. 28 to 35. The same tables will be found in 
 Part III at the end of the book. 
 
 The K formulas will be found very similar to those of 
 the last chapter, and while they require more arithmet- 
 ical work, they should not be found any more difficult to 
 understand. No more values of line constants need to be 
 looked up for the K formulas than for the " Short Line" 
 formulas. The capacity of the line does not enter into the 
 calculations, since its effect is allowed for by means of the 
 constant K which is the same, at any one frequency, for 
 all values of line capacity. 
 
 The formulas of this chapter assume that the leakage 
 current is zero; that is, that no power is lost from leakage 
 over the insulators or from corona. This is a correct 
 assumption to make for all voltages except the very highest 
 in use. If it is desired to make allowance for corona loss, 
 the formulas of Chapter VI should be used. 
 
 The accuracy of the K formulas is given as approxi- 
 mately YO of i% of line voltage for lines up to 100 miles 
 long and with regulation up to 20%, and as \ of i% for 
 lines up to 200 miles long, and with the same regulation. 
 
 25 
 
26 TRANSMISSION LINE FORMULAS 
 
 These limits are close enough for commercial work, so that 
 the K formulas can be recommended for all ordinary 
 engineering calculations of the performance of long power 
 transmission lines under steady conditions, where the 
 corona loss is small. The accuracy of the electrical calcu- 
 lations will be better than the accuracy with which the 
 resistance and the physical dimensions of the line are 
 generally known. 
 
 The K formulas are well adapted to the solution of 
 long transmission lines which have substations at inter- 
 mediate points between the ends. In such cases each 
 section of the line between substations must be calculated 
 separately, beginning with the end where conditions are 
 known. The first step is to find the voltage, in-phase 
 current and quadrature current at the first substation. 
 The load taken by the substation, expressed as in-phase 
 current and quadrature current, must be added to, or sub- 
 tracted from, the above values of current. When condi- 
 tions are given at the receiver end and one is proceeding 
 toward the supply end, the substation load must be added 
 to the line load. When conditions are given at the supply 
 end, the substation load must be subtracted from the line 
 load, since one is proceeding away from the supply. Hav- 
 ing thus found complete conditions at one end of the 
 second section of the line, the calculation of this section 
 may be taken up in the same way as for the first section. 
 In this manner the entire line may be calculated and the 
 voltage and current at the unknown end may be deter- 
 mined. 
 
 Examples are worked out, which will give a clear idea of 
 the manner in which the K formulas are used. Many 
 other such examples have been calculated and carefully 
 
K FORMULAS 27 
 
 compared with the fundamental formulas. As these ex- 
 amples have covered the range of practicable transmission 
 lines, a sound basis is afforded for the estimate of the 
 accuracy of the K formulas and for the statement that 
 they are sufficiently reliable for all ordinary engineering 
 purposes in the calculation of electric power transmission 
 lines. 
 
28 TRANSMISSION LINE FORMULAS 
 
 TABLE III. ^ FORMULAS FOR TRANSMISSION LINES. 
 CONDITIONS GIVEN AT RECEIVER END. 
 
 Accurate within approximately T V of i% of line voltage up to 100 miles, 
 and ^ of i% up to 200 miles, for lines with regulation up to 20%. 
 
 ^ = 6 (cycles) 2 # =2>l6 for 60 cycles. K = 0.375 for 25 cycles. 
 10,000 
 
 Conditions given: 
 
 K.V.A. = K.V.A. at receiver end. 
 
 E = Full load voltage at receiver end. 
 cos 6 = Power factor at receiver end. 
 K.W. =K.V.A. cos0. 
 
 r = Resistance of conductor per mile. (From Tables VII-VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 I = Length of transmission line in miles. 
 Then P = ^- . . . cos _ ^ j^gg curren t a t receiver end 
 
 (in total amps.) . 
 Q = 1000 K.V.A. sing = Reactiye current afc rece i ve r end 
 
 (in total amps.), when current is 
 lagging. 
 
 1000 K.V.A. sin & . L . , ,. 
 
 = -- = - , when current is leading. 
 
K FORMULAS 
 
 29 
 
 Find the following quantities: 
 
 Full Load. 
 
 . 3 
 
 Load. 
 
 i-*(J-Vi. 
 
 \ioooj 
 
 . The above are for two- and three-phase lines. For single-phase 
 lines use 2 r and 2 x in place of r and #. 
 
30 TRANSMISSION LINE FORMULAS 
 
 TABLE III. (Continued.} 
 
 CONDITIONS GIVEN AT RECEIVER END. 
 Formulas: 
 
 Full Load. No Load. 
 
 Voltage at receiver end. 
 
 :* (,)&--_ 
 
 (for constant supply voltage). 
 Regulation at receiver end in volts, for constant supply voltage. 
 
 N.B. The regulation at receiver end may be expressed as a percentage 
 of E. 
 
 Voltage at supply end. 
 
 (for constant receiver voltage) . 
 Regulation at supply end in volts, for constant receiver voltage. 
 
 f*\ L. B * B * 
 
 (6) A H -- j AQ -- j- 
 
 2 A 2 Ao 
 
 N.B. The regulation at supply end may be expressed as a percentage 
 
K FORMULAS 31 
 
 Current at supply end in total amperes* 
 
 (7) VC 2 + D 2 . (8) Co 2 + A 2 . 
 
 K. V.A . at supply end. 
 
 ( 9 ) - 
 
 1000 2A 
 
 .JF". a/ supply end. 
 (n) (4C + 3D). (12) ^ 
 
 . (10) -^-(^o+^j- )VCo 2 +A> 2 . 
 
 Power factor at supply end, in decimals. 
 
 AC + BD A Q C +B D 
 
 (13) 7 - 
 A + 
 
 In- phase current at supply end in total amperes.* 
 
 AC+BD , ,, A C +B D 
 
 Reactive current at supply end in total amperes* 
 , . BC-AD , Q x ^o 
 
 (I7) "' 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 K.W. loss in line. 
 (19) (AC + BD -EP). (20) (A C + B D ) 
 
 IOOO IOOO 
 
 [same as No. 12] 
 Per cent efficiency of line. 
 , . 100 EP 
 
 AC+BD 
 
 Total amps. 
 * Amperes per wire, three-phase, = -p 
 
 Total amps. 
 Amperes per wire, two phase, = 
 
32 TRANSMISSION LINE FORMULAS 
 
 'TABLE IV. K FORMULAS FOR TRANSMISSION LINES. 
 CONDITIONS GIVEN AT SUPPLY END. 
 
 Accurate within approximately T V of i% of line voltage up to 100 miles 
 and \ of i% up to 200 miles, for lines with regulation up to 20%. 
 
 T ^ 6 (cycles) 2 , , , .. 
 
 K = . K 2.16 for 60 cycles. K 0.375 for 25 cycles. 
 
 10,000 
 
 Conditions given: 
 
 K.V.A. = K.V.A. at supply end. 
 
 E 8 = Full load voltage at supply end. 
 cos 9 = Power factor at supply end. 
 K.W. = K.V.A. cos0. 
 
 r = Resistance of conductor per mile. (From Tables VII-VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 / = Length of transmission line in miles. 
 
 Then P 8 = -^~~ = In-phase current at supply end (in 
 
 c-s 
 
 total amps.). 
 
 Q 8 = ' ' = Reactive current at supply end (in 
 
 c-s 
 
 total amps.), when current is lagging 
 
 1000 K.V.A. sin 9 . . , ,. 
 
 = = , when current is leading. 
 
K FORMULAS 33 
 
 Find the following quantities: 
 
 Full Load. 
 
 / 
 
 Vh-i, _LY. 
 
 3 \ioooy 
 
 = ES ' - I - I * 
 X \IOOOj 
 
 . The above are for two- and three-phase lines. For single- 
 phase lines use 2 r and 2 x in place of r and #. 
 
34 TRANSMISSION LINE FORMULAS 
 
 TABLE IV. (Continued.) 
 
 CONDITIONS GIVEN AT SUPPLY END. 
 Formulas: 
 
 Full Load. No Load. 
 
 Voltage at receiver end. 
 
 (for constant supply voltage). 
 Regulation at receiver end in volts, for constant supply voltage. 
 
 N.B. The regulation at receiver end may be expressed as a percentage 
 of E. 
 
 Voltage at supply end. 
 
 F+ 
 
 (4) E.. (5) E 08 = 2 ^E a 
 
 F +^L 
 
 2F 
 
 (for constant receiver voltage). 
 Regulation at supply end, in volts, for constant receiver voltage. 
 
 N.B. The regulation at supply end may be expressed as a percentage 
 ofE a . 
 
' K FORMULAS 35 
 
 Current in total amperes.* _ 
 
 ( 7 ) VM* + N* at receiver end. (8) Vjjf 2 + N Q 2 at supply end. 
 K.V.A. 
 
 ( ) __L_ {p+**.} VM* + N* at receiver end. 
 w ' 1000 \ 2 Fj 
 
 (10) - E 8 Vlf 2 + No 2 at supply end. 
 
 tf.TF. 
 
 (ll ) _J_(FM + Gtf) at receiver end. (12) -^- E 8 M at supply end. 
 
 7 1000 
 Power factor, in decimals. 
 
 FM + GN - at receiver end. 
 
 * *' / fT 2 \ 
 
 f F + ^ ) 
 
 2V 2 
 
 In-phase current in total amperes* 
 
 ( IS ) FM + ^ N at receiver end. (16) Jfo at supply end. 
 
 ^ + 5 
 
 Reactive current in total amperes* 
 ( I7 ) GM ~ - V at receiver end. (18) ]V at supply end. 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 K.W. loss in line. 
 ( IQ ) _I_ ( Es p 8 - FM - ON). (20) 
 
 Per cent efficiency of line. 
 
 ( 2I ) f J?p P er cent - 
 
 Total amps. 
 * Amperes per wire, three phase, = 
 
 Total amps. 
 Amperes per wire, two phase, = - - 
 
36 TRANSMISSION LINE FORMULAS 
 
 PROBLEM A. 
 
 Find by the K formulas, the line drop in the following case: 
 
 Length of line 100 miles. 
 
 Spacing 8 feet. 
 
 Conductor No. 3 copper cable. 
 
 Load (at receiver end), 3000 K.V.A., 66,000 volts, 90% P.F., 
 
 three phase, 60 cycles. (Prob. A, Chap. III.) 
 From the tables, r = 1.078, x = 0.840. 
 
 1000 X 3000 X 0.90 
 
 P = -f *= = 40.91. 
 
 66,000 
 
 O = looo X 3000 X Q.4359 = o 
 66,000 
 
 2 l6 
 
 A = 66,000 66,000 X : 
 
 100 
 
 +40.91 X 107.8(1 --X ) 
 \ 3 loo/ 
 
 + 19.81 X8 4 (i-^X 
 \ 6 100 
 
 = 70,580 volts. 
 
 . 1.078 .. 2.16 
 B = 66,000 X e- X 
 0.840 100 
 
 + 40.91 X 84(1 -\ X^-J 19-81 X 107.8 ( i- - X ) 
 \ 6 100 / V 3 I00 / 
 
 = 3150 volts. 
 
 Supply voltage = 70,580 + = 7o,6 5 o. 
 
 Line drop = 4650 volts = 7.05%. 
 
 By the fundamental formulas, using the same line constants, the 
 line drop is 7.08% (Prob. 2, Chap. VI). The discrepancy is 0.03% 
 
 of line voltage. 
 
 PROBLEM B. 
 
 Find by the K formulas the voltage at the supply end of the 
 following line: 
 
 Total length of line ......................... 300 miles. 
 
 Spacing ................................... 12 feet. 
 
K FORMULAS 37 
 
 Conductor, 266,800 c.m. aluminum cable. 
 
 Load at receiver end of line, 9000 K.V.A., 80% P.F. (lagging), 
 
 100,000 volts, three phase, 60 cycles. 
 
 Load taken by a substation at the middle of the line, 150 miles 
 from either end, 2000 K.V.A. at the line voltage and at 70% 
 P.F. (lagging). 
 Solution of first section of line. 
 
 r = 0.3410 x = 0.791 / = 150, 
 P = 72 Q = 54. 
 
 From the K formulas, 
 
 A = 100,000 4860 + 3560 + 6360 
 
 = 105,060 volts. 
 B = 2100 -f 8470 2670 
 
 = 7900 volts. 
 C= 72 -3-50+ 1.13 -0.57 
 
 = 69.06 amperes. 
 D = 1.51 - 54 + 2.63 -f 80.59 
 = 30.73 amperes. 
 
 A H = Ei = 105,360 volts. 
 
 AC + BD 
 
 In-phase current = - 
 
 4 I -^ 
 
 Reactive current 
 
 = 71.15 amps 
 BC - AD 
 
 ' 2A 
 
 = 25.46 amps 
 Solution of second section of line. 
 Conditions at middle of line : 
 
 Ei = 105,360. 
 In-phase current of substation load 
 
 looo X 2000 X 0.70 
 
 = = 13.29 amps. 
 
 105,360 
 
 Reactive current of substation load 
 
 looo X 2000 X 0.7 141 
 
 L - a - = 13.56 amps. 
 
 105,360 
 
38 TRANSMISSION LINE FORMULAS 
 
 s 
 
 PI = 71.15 -|- 13.29 = 84.44 amps. 
 Qi = 25.46 + 13.56 = 11.90 amps. 
 Then by the K formulas, 
 
 Ai = 105,360 5120 + 4180 1410 
 = 103,010 volts. 
 
 BI = 2200 + 9940 + 590 
 
 = 12,730 volts. 
 
 T? 2 
 
 AI -\ j- = 103,800 volts 
 
 2 AI 
 
 = voltage at the supply end of the line. 
 
 [By the fundamental formulas, supply voltage = 103,900 volts 
 (Prob. B, Chap. VI).] 
 
 PROBLEMS, CHAP. V. 
 
 (K FORMULAS.) 
 
 1. Find, by means of the K formulas, the voltage drop from 
 the supply end to the receiver end (the line drop) of the following 
 line: 
 
 Length of line 200 miles. 
 
 Spacing 9 feet. 
 
 Conductor No. ooo aluminum cable. 
 
 Load (at receiver end), 4500 K.V.A., 66,000 volts, 80% P.F., 
 three phase, 60 cycles. 
 
 Ans. 6650 volts. 
 [By the fundamental formulas, 6700 volts. (Chap. VI, Prob. A).] 
 
 2. Find the regulation of the line in Prob. A, Chap. V. [See 
 Prob. A, Chap. III.] 
 
 Ans. 9.37%. 
 
 [By the fundamental formulas 9.40%. (Chap. VI, Prob. 2.) 
 Error 0.03% of line voltage.] 
 
 3. Find the per cent regulation and voltage drop of the following 
 line: 
 
 Length of line 75 miles. 
 
 Spacing, 8 feet, regular flat spacing. 
 
 Conductor No. oo aluminum cable. 
 
 Load (at receiver end), 10,000 K.V.A., 88,000 volts, 85% P.F., 
 three phase, 25 cycles. (Prob. 7, Chap. III.) 
 
 Ans. 7-35% reg'n, 7.13% drop. 
 
'K FORMULAS 39 
 
 4. Find, by the K formulas, the per cent voltage drop, the per 
 cent loss, and the power factor at the supply end of the following 
 line: 
 
 Length of line 100 miles. 
 
 Spacing 6 feet. 
 
 Conductor No. oooo copper wire. 
 
 Take r = 0.267, * -7 2 7> & = 6.03 X lo" 6 . 
 Load (at receiver end), 100 amperes per wire, 60,000 volts, 
 95% P.P., three phase, 60 cycles. [Problem of Pender and 
 Thomson, Proc. A. /..., July, 1911.] 
 
 Ans. 13-09% drop, 7.61% loss, 96.58% P.F. 
 [Calc. by series, 13.03% drop, 7.60% loss, 96.66% P.F. (Prob. 4, 
 Chap. VI).] 
 
 5. Find, by the K formulas, the K.V.A. and voltage at the supply 
 end, and the efficiency of the following line: 
 
 Length of line 250 Km. = 155.34 miles. 
 
 Spacing 6 feet. 
 
 Conductor No. ooo copper wire. 
 
 Total resistance of one conductor .... 51.5 ohms. 
 Total reactance of one conductor .... 48 . o ohms. 
 Load (at receiver end), 15,000 K.V.A., 86,600 volts, 80% P.F., 
 
 three phase, 25 cycles. 
 
 [See page 91, "Application of Hyperbolic Functions," by A. E. 
 Kennelly, University of London Press, 1912.] 
 
 Ans. 15,130 K.V.A., 97,920 volts, 89.68%. 
 [By series, 15,153 K.V.A., 97,934 volts, 89.71%.] 
 
 6. Find (a) star voltage at supply end at full load, 
 
 (6) star voltage at supply end at no load, 
 
 (c) regulation volts (star) at the supply end, 
 
 (d) amperes per wire at supply end at full load, 
 
 (e) power factor at supply end at full load, 
 (/) loss in line at full load, 
 
 (g) efficiency of the transmission line, 
 
 (h) amperes per wire at supply end at no load (i.e., the 
 
 "charging current"), 
 (i) power factor at supply end at no load, 
 
 (7) loss in line at no load, 
 
40 TRANSMISSION LINE FORMULAS 
 
 for the following line: 
 
 Length of line 300 miles. 
 
 Spacing 10 feet. 
 
 Conductor, No. ooo copper cable of 0.330 ohms per mile. 
 Load (at receiver end), 18,000 K.V.A., 104,000 line volts, 
 
 90% P.F., three phase, 60 cycles. 
 [Prob. A, page 2, G. E. Review Supplement, May, 1910.] 
 
 Answers: 
 
 By K Formulas. 
 
 (a) 69,820 
 
 (b) 48,610 
 
 (c) 21,210 
 
 (d) 97.0 
 
 (e) 92.2 
 (/) 2530 
 (2) 86.5 
 ( 91-3 
 
 W 7.1 
 
 0') 940 
 
 7. Find, by the K formulas, the voltage at the supply end of 
 the following line: 
 
 Total length of line 400 miles. 
 
 Spacing 15 feet. 
 
 Conductor No. oooo copper cable. 
 
 Load at receiver end of line, 5000 K.V.A., 85% P.F. (lagging) 
 
 110,000 volts, three phase, 60 cycles. 
 
 Load taken by a substation at the middle of the line, 200 miles 
 from either end, 2500 K.V.A., at the line voltage and at 90% P.F. 
 
 (lagging). 
 
 Ans. 89,720 volts. 
 
 [By the convergent series, 90,190 volts (Prob. 6, Chap. VI). Error 
 0.5%.] 
 
 A method of operation of transmission lines is coming 
 into prominence, by which the voltage of the line is kept 
 constant at all points, and the inconveniences due to poor 
 
 By Fundamental 
 Formulas. 
 
 Error in per cent 
 of full voltage 
 or current. 
 
 69,670 volts 
 
 0-3% 
 
 48,950 volts 
 
 0.6% 
 
 20,720 volts 
 
 o.9% 
 
 96 . 59 amps. 
 
 o.5% 
 
 92.35% 
 
 0-2% 
 
 2440 Kw. 
 
 0.6% 
 
 86.90% 
 
 o.5% 
 
 90.97 amps. 
 
 o-5% 
 
 6-47% 
 
 o.7% 
 
 860 Kw. 
 
 0.5% 
 
K FORMULAS 4OA 
 
 regulation are obviated. Synchronous machinery, con- 
 sisting of either synchronous motors or generators, is 
 installed in the stations throughout the transmission 
 system in sufficient quantity to hold the voltage at a 
 constant value by controlling the amount of leading or 
 lagging current supplied to the line. This method will 
 probably come into considerable favor, for there seems to 
 be practically no limit to the extent of a transmission 
 system operated at constant voltage. 
 
 The following problem outlines the method of calcula- 
 tion for such cases. 
 
 A line 400 miles long has a substation at the middle, 
 200 miles from either end. A load of 10,000 Kw. is 
 taken at the receiver end of the line, and 8000 Kw. at 
 the substation. Find the power factor which is required 
 for these loads, in order that the voltage at the generator, 
 substation, and receiver end may be 110,000 volts, the 
 following data being given: 
 
 Conductor, 250,000 c.m. copper cable, 1 4-foot spacing, 
 3-phase, 60 cycles, 
 
 r = 0.2284, x = 0.813. 
 
 ist section of line, / = 200 miles. 
 By the K formulas, 
 
 n 1000 X 10.000 
 
 P = - -=90.91; 
 
 Q is unknown; 
 
 A = 110,000 9500 + 3910 + 160.3 Q 
 = 104,410 + 160.3 Qj 
 
40B TRANSMISSION LINE FORMULAS 
 
 B = 2670 + 14,570 - 43-o Q 
 = 17.240 - 43-o Q- 
 
 Now the voltage at the substation end of the ist section 
 of the line is 110,000; that is, 
 
 . 
 
 +-^ = no,ooo; 
 
 squaring both sides, A 2 + B 2 = 121 X io 8 . 
 This gives a quadratic equation in Q, 
 
 2.754 Q 2 + 3198 Q - 90,150 = o, 
 
 from which ()= 27.53 total amps.; 
 
 Power factor = = 95.7%, 
 
 94.99 
 
 and, as Q is positive, this is a lagging power factor. 
 The power factor obtained by the hyperbolic formulas 
 
 is 95-9%- 
 
 Using the above value of Q, we obtain 
 
 C = +82.78, 
 D = + 90.59. 
 
 In-phase current at substation end of ist section 
 
 = + 95- 11 '- 
 Reactive current at substation end of ist section 
 
 = ~ 77-53- 
 2nd section of line, / = 200 miles. 
 
K FORMULAS 400 
 
 In-phase current in ist section = +95.11, 
 In-phase current of substation load 
 
 1000 X 8000 
 
 - = + 72.73 
 110,000 
 
 In-phase current at substation end of 2nd section 
 
 = 167.84. 
 Reactive current at substation end of 2nd section 
 
 = Q*- 77-53> 
 
 where Q x is the unknown reactive current of the sub- 
 station load. 
 
 By the K formulas, as before, 
 
 A I = 95>3 + J 6o.3 Q x , 
 B l = 32,910 - 43.0 Q x , 
 
 and voltage at generator end of line 
 
 AI-\ l = 110,000. 
 
 24l 
 
 From the above we obtain as before a quadratic equa- 
 tion in Q x , which gives 
 
 &=+ 65.59. 
 
40D TRANSMISSION LINE FORMULAS 
 
 Power factor of substation load = 74.3%, lagging. 
 The power factor obtained by the hyperbolic formulas 
 is 74.6%, 
 
CHAPTER VI. 
 CONVERGENT SERIES. 
 
 THE mathematical expression for finding the operating 
 characteristics of a transmission line, in which exact ac- 
 count is taken of all the electrical properties of the line, 
 has been published many times. It involves the use of 
 hyperbolic sines and cosines, as well as of complex quan- 
 tities,* and, without some special arrangement, cannot be 
 directly applied to the calculation of a particular case. 
 For this reason, most of the systems so far published for 
 calculating transmission lines have used approximate for- 
 mulas which have been based on the hyperbolic formulas. 
 In a few cases, an attempt has been made to devise a 
 system pi working which would give the exact results of the 
 fundamental hyperbolic formulas, but generally the labor 
 required in using the systems is so great as almost to pro- 
 hibit obtaining the exact result, or else the accuracy of 
 the work is seriously impaired by the necessity of inter- 
 polating values, from tables of hyperbolic functions which 
 have been recently prepared for this purpose and are not 
 as large and complete as they should be for good working. 
 
 The original hyperbolic formulas can be expressed in 
 the form of convergent series.f In this form they do not 
 
 * The hyperbolic formulas are given in Chap. XV. 
 
 f Prof. T. R. Rosebrugh, Applied Science Magazine, University of To- 
 ronto, March, 1909; Prof. T. R. Rosebrugh, Proc. A. I.E. E., Nov., 1909, 
 p. 1460; J. F. H. Douglas, Electrical World, April 28, 1910; Dr. C. P. 
 Steinmetz, Electrical World, June 23, 1910; Dr. C. P. Steinmetz, "Engi- 
 neering Mathematics," Chap. V, 1911. 
 
 41 
 
42 TRANSMISSION LINE FORMULAS 
 
 involve hyperbolic or trigonometrical functions, and so do 
 not require any mathematical tables, the only operations 
 being multiplication and addition. The series can be car- 
 ried to any accuracy desired by merely using enough of the 
 terms, which diminish very rapidly when commercial fre- 
 quencies are involved. 
 
 The fundamental formulas as expressed by convergent 
 series have been rearranged, and some new convergent 
 series have been added, to make the formulas as tabulated 
 in this chapter directly applicable to the exact solution of 
 all the problems treated by the K formulas. Exactly 
 the same final formulas in A, B, C, D, etc., are used with 
 the convergent series as with the K formulas. 
 
 Unlike the K formulas, which are expressed in .the 
 simplest algebraical form, the convergent series involve 
 the use of complex numbers, that is, numbers containing 
 the well-known "j" terms. No difficulty should be expe- 
 rienced on this account, however, as the rules for using 
 complex quantities are quite straightforward, and even 
 one who has never worked with them should be able to 
 make use of the formulas described in this chapter by 
 closely following the instructions. 
 
 Each of the complex quantities, (A +JB), (P JQ), 
 Z = (r + jx) I* Y = (g +jb) lj etc., is composed of two 
 parts, the first, a so-called "real" term, and the second, a 
 j term. In adding complex numbers, the j terms must 
 be kept separate from the others. Thus 
 
 4 +j 5 added to 7 +73 = n +78. 
 
 In multiplying two complex quantities, the simple rules 
 
 * The notation Z = (r + jx) I, etc., is used in accordance with the 
 resolution adopted by the International Electrotechnical Commission in 
 Turin, Sept., 191 1. 
 
CONVERGENT SERIES 43 
 
 of ordinary algebra are followed, and it must be remem- 
 bered that 
 
 j-xj-'f 
 
 = -i, 
 and, therefore, 
 
 -jXJ=+i 
 
 j* =+J, etc. 
 
 Thus (4 +j 5) X (7 +y 3) is worked out as follows: 
 
 4+75 
 7+73 
 + 28+jf35 
 
 - 15 +y 12 
 + 13+747- 
 
 In using the convergent series, E, P, and Q are the same 
 as used with the K formulas, E being expressed as a real 
 number without any j term. Z is equal to (r -\-jx) I, 
 where r and x are taken from Tables VII to XII, Part III, 
 for resistance and reactance per mile. F is equal to 
 (g -\-jb) I. The leakage conductance, g, per mile, should 
 be estimated from the most suitable test data available, 
 giving insulator leakage and corona loss under conditions 
 similar to those of the line considered. The capacity sus- 
 ceptance, 6, per mile, will be found in Tables XIII to 
 XVI, Part III. 
 
 After F and Z have been written down in the form of 
 complex numbers, the product FZ should be found, as 
 described above for the multiplication of complex quan- 
 tities. From this is obtained 
 
 FZ FZ FZ 
 
 - , - , and , 
 24 6 
 
44 TRANSMISSION LINE FORMULAS 
 
 each expressed as a complex number of a single real term 
 and a single j term. Multiplying the last two together 
 
 F 2 Z 2 F 2 Z 2 
 
 gives -, from which may be written 
 
 2-3-4 2.3.4.5 
 
 down. In most cases no more terms need to be calcu- 
 lated, even for very accurate work, but this is to be de- 
 termined while doing the work, as one usually figures out 
 the terms of these series until they become too small to be 
 
 FZ 
 
 considered when added to 
 
 2 
 
 By addition of terms obtained above, the values of 
 
 FZ , F 2 Z 2 FZ F 2 Z 2 
 
 1 h etc. and 1 + etc. 
 
 2-3-4 2-3 2-3.4.5 
 
 are obtained, each as a complex number of two terms. 
 
 /FZ \ 
 
 Multiply E by the value found for ( + etc.) and add 
 
 it to E. Multiply (P - jQ) by Z, or (r +jx) /, and by the 
 
 /Y7 \ 
 
 value of I + etc.) and add it to (P - jQ) Z. The 
 
 above quantities are added together to give A -\-jB, the 
 sum of all the real parts being equal to A , and the sum of all 
 the j terms being equal to B. 
 Similarly, C +JD is found by adding 
 
 (P -y, (P -JQ) (^ + etc.), EY, and EY (~ + etc.). 
 
 These values of A, B, etc., are inserted in equations i to 
 21 given with the K formulas, in exactly the same way 
 as the values of A, B, etc., found according to the second 
 page of Table III. Each step of the above procedure is 
 shown in the examples in this chapter. 
 
 The use of Table VI is the same as that of Table V de- 
 scribed above. 
 
CONVERGENT SERIES 45 
 
 TABLE V. CONVERGENT SERIES FOR TRANSMISSION LINES. 
 CONDITIONS GIVEN AT RECEIVER END. 
 
 The convergent series give the results of the fundamental formulas as 
 accurately as desired, if a sufficient number of terms is used. 
 
 When conditions are given at the receiver end, the same as with the K 
 formulas, find the quantities: 
 
 Full Load. 
 
 
 1 ( P if)} 7\^ 
 
 2-3-4 2-3.4.5.6 } 
 YZ Y 2 Z 2 Y 3 Z 3 ^ 
 
 
 ~T \f JV) ^ 1 I 
 
 i n ( P iD"> ( T 
 
 ' 2 3 ' 2-3.4.5 ' 2.3.4-5-6-7' 
 
 
 -r JMJ \Jr JV) \ i - 
 
 I 1 1 ..- t~ CtC. 1 
 
 
 \ 
 
 II 1 1 etc 
 
 1 
 
 1 YZ 
 
 2-3 2-3.4-5 2.3.4-5-6.7 
 No Load. 
 
 V%72 V373 \ 
 1 1 r+r 1 
 
 r 
 
 u I ]DV> -c-l I | 1 
 
 / y 
 
 ,_l_ in n F.V T 4- 
 
 2.3.4 2.3.4.5-6 y' 
 
 Z Y^Z 2 Y Z Z 3 \ 
 
 f . 
 
 V 2-3 2-3.4.5 2. 3. 4'5'6. T I* 
 
 where Z = (r +y) /. 
 
 r = resistance of conductor per mile. 
 
 x = reactance of conductor per mile. 
 
 I = length of transmission line in miles. 
 
 Y=(g+jb)l. 
 
 g = leakage conductance of conductor per mile. 
 
 b = capacity susceptance of conductor per mile. 
 
 Use A, B, C, D, etc., with the equations in the third and fourth pages 
 of Table HI to solve transmission line problems. 
 
 j2 
 
 Note i. In the formulas, A -\ r is used instead of VA 2 + B 2 . 
 
 2 A 
 
 This approximation may be used for very accurate work, as it is correct 
 within approximately T ^ of i% when the regulation is not more than 20%. 
 Note 2. The above are for two- and three-phase lines. For single- 
 phase lines use 2 r and 2 * in place of r and x, and use | g and | b in place 
 of g and b. 
 
46 TRANSMISSION LINE FORMULAS 
 
 TABLE VI. CONVERGENT SERIES FOR TRANSMISSION LINES. 
 
 CONDITIONS GIVEN AT THE SUPPLY END. 
 
 The convergent series give the results of the fundamental formulas as 
 accurately as desired, if a sufficient number of terms is used. 
 
 When conditions are given at the supply end, the same as with the 
 K formulas, find the quantities: 
 
 Full Load. 
 
 ( 
 
 2-3 2-3.4.5 2.3-4-S 
 No Load. 
 
 + JG, = E. (i - h YZ + Y>Z* - g Y'Z* + ^ YW - etc.), 
 
 , +jlf, = E,Y(I -$YZ + Y*2?-^- YW + -,- Y<Z> - etc.), 
 \ -"-5 o-'S 2035 / 
 
 where Z = (r +jx) I. 
 
 r = resistance of conductor per mile. 
 
 x = reactance of conductor per mile. 
 
 I = length of transmission line in miles. 
 
 Y=(g+jb)l. 
 
 g = leakage conductance of conductor per mrle. 
 
 b = capacity susceptance of conductor per mile. 
 
 Use F, G, M, N, etc., with the equations in the third and fourth pages 
 of Table IV to solve transmission line problems. 
 
 Note i. In the formulas, F H -- - is used instead of Vp* + G 2 . This 
 
 2 F 
 
 approximation may be used for very accurate work, as it is correct within 
 approximately yf ^ of i% when the regulation is not more than 20%. 
 
 Note 2. The above are for two- and three-phase lines. For single- 
 phase lines use 2 r and 2 x in place of r and x, and use \ g and \ b in place of 
 g and b. 
 
CONVERGENT SERIES 47 
 
 PROBLEM A. 
 
 Find the line drop, by means of the convergent series, for the 
 following line: 
 
 Length of line 200 miles. 
 
 Spacing 9 f ee t- 
 
 Conductor No. ooo aluminum cable. 
 
 Load (at receiver end), 4500 K.V.A., 66,000 volts, 80% P.F., 
 
 three phase, 60 cycles. 
 From the tables, 
 
 r = 0.5412, * = 0.784, 6=549Xio- 6 . 
 Then p = looo X 4500 X 0.8 ^^^ 
 
 06,000 
 looo X 45QO X 0.6 = 
 
 66,000 
 Z = 108.24+.; 156.8. 
 
 Y = -j-j 0.001098. 
 
 YZ = -0.17216 +j 0.11885. 
 
 YZ 
 
 = 0.08608 +j 0.05942. 
 
 YZ 
 
 = - 0.04304 +y 0.02971. 
 
 4 
 
 YZ 
 
 = 0.02869 +.70.01981. 
 
 6 
 
 0.00059 .70.00085. 
 + 0.00124 j 0.00085. 
 F 2 Z 2 
 
 = + 0.00065 j 0.00170. 
 2-3.4 
 
 YZ 
 
 = 0.08608 +j 0.05942. 
 
 IY7 \ 
 
 h etc. = - 0.08543 +^'0.05772. 
 
 \ 2 / 
 
 F 2 Z 2 
 = + 0.00013 j 0.00034. 
 
 2.3.4-5 
 YZ 
 
 = 0.02869 +y 0.01981. 
 
 2 '3 
 
 ( +etc.)= 0.02856+70.01947. 
 
 \2 3 / 
 
48 TRANSMISSION LINE FORMULAS 
 
 Now E = 66,000. 
 
 (YZ \ 
 
 -- h etc. J = 0.08543 +.;' 0.05772. 
 
 ~ + etc.) = -5640 +.7-3810. 
 
 P-JQ= 54-55 -3 40.91- 
 
 708.24 +y 156.8. 
 
 + 5900 -j 4430. 
 + 6420+^8550, 
 
 (P - JQ) Z= + 12320 + j 4120. 
 
 YZ \ 
 
 etc. ] = 0.02856 + 7*0.01947. 
 * 
 
 \23 
 
 80+^*240. 
 -350 -./i 20. 
 
 (V7 \ 
 
 + etc. ) = - 430 +j 120. 
 2*3 / 
 
 E = 66,000. 
 
 E( h etc.J = 5640+^3810. 
 
 ( p -jQ)Z= 12320+^4120. 
 (P - JQ) Z>(^- + etc.) = - 430 +j 120. 
 
 A+jB= 78,320 - 6070+^*8050 
 = 72,250+^8050. 
 
 B z 
 
 A H = 72,700 volts. 
 
 2 ^1 
 
 Line drop = 6700 volts. 
 
 PROBLEM B. 
 
 Find, by the convergent series, the voltage at the supply end of 
 the following line : 
 
 Total length of line 300 miles. 
 
 Spacing 12 feet. 
 
 Conductor, 266,800 c.m. aluminum cable. 
 
 Load at receiver end of line, 9000 K.V.A., 80% P.F. (lag- 
 ging), 100,000 volts, three phase, 60 cycles. 
 
CONVERGENT SERIES 49 
 
 Load taken by a substation at the middle of the line, 1 50 miles 
 from either end, 2000 K.V.A., at the line voltage and at 70% 
 P.F. (lagging). (Prob. B, Chap. V.) 
 Solution of first section of line: 
 
 r = 0.3410, x = 0.791, b = 5.44 X io~ 6 , / = 150. 
 Z= 51.15-!-.; 118.65. 
 y = +j 0.000816. 
 YZ = 0.09682 +,70.04174. 
 
 ( + -^- + etc. ) = - 0.04809 +j 0.02053. 
 
 \ 2 2 3 '4 I 
 
 F__, -- F 2 Z 2 -I- etc.) = - 0.01607 +j 0.00689. 
 2-3 2-3.4-5 / 
 
 E = + 100,000. 
 
 E ~ + etc = - 4810 +j 2050. 
 
 (P-jQ)Z= 10,090 +./5780. 
 
 (P - jQ) Z ( + etc. 
 
 - 200 - 20. 
 
 A+jB= 105,080 +77810. 
 
 B z 
 
 A + = 1 = 105,370 volts. 
 
 In a similar manner, it is found that 
 
 C+jD = 69.08 + j 30.36 amps. 
 
 In-phase current, - = 71.15 amps. 
 
 A + JL ; 
 
 2 A 
 
 Reactive current, - - = 25.15 amps. 
 
 A+- 
 2 A 
 
 Solution of second section of line: 
 Conditions at middle of line, 
 
 Ei = 105,370 volts. 
 In-phase current of substation load 
 
 1000 X 2000 X 0.70 
 = - - L - = 13.29 amps. 
 
 105,370 
 
 Reactive current of substation load 
 
 1000 X 2000 X 0.7141 
 _ - i * . = I3>55 amps< 
 
 105,370 
 
50 TRANSMISSION LINE FORMULAS 
 
 Current of substation load = 13.29 j 13. 55. 
 Current of first section = 71.15 -\-j 25.15. 
 
 Pi JQi = 84.44 +j 1 1. 60. 
 1 = 4- 105,370. 
 
 (YZ \ 
 
 4- etc. J - 5070 4- J 2, 160. 
 
 (Pi ~~ JQi) Z = 2 >94 4- J 1 0,6 10. 
 
 (Pi JQi) Z i- '- 4- etc.] = - 120 -y 150. 
 
 * * 3 / 
 
 ^i -}-jBi = 103,120 4~J 12,620 volts, 
 
 yli -!> j- = 103,900 volts 
 
 = voltage at the supply end of the line. 
 
 PROBLEMS, CHAP. VI. 
 
 (CONVERGENT SERIES.) 
 
 1. Find, by the convergent series, the voltage drop of the follow- 
 ing line: 
 
 Length of line 80 miles. 
 
 Spacing 10 feet. 
 
 Conductor No. oo aluminum cable. 
 
 Load (at receiver end), 15,000 K.V.A., 100,000 volts, 95% 
 P.F., two phase, 25 cycles. (Prob. C, Chap. III.) 
 
 Ans. 8810 volts. 
 
 2. Find, by the convergent series, the per cent line drop and the 
 per cent regulation of the following line: 
 
 Length of line 100 miles. 
 
 Spacing 8 feet. 
 
 Conductor No. 3 copper cable. 
 
 Load (at receiver end), 3000 K.V.A., 66,000 volts, 90% P.F., 
 three phase, 60 cycles. (See'Prob. A., Chap. Ill, and Prob. A, 
 Chap. V.) 
 
 Ans. 7.08% drop, 9.40% reg'n. 
 
 3. Find, by the convergent series, the K.V.A. and voltage at the 
 supply end, and the efficiency of the following line: 
 
CONVERGENT SERIES 51 
 
 Length of line 250 Km. = 155.34 miles. 
 
 Spacing 6 feet. 
 
 Conductor No. ooo copper wire. 
 
 Total resistance of one conductor, 51.5 ohms. 
 
 Total reactance of one conductor, 48.0 ohms. 
 
 Total susceptance of one conductor, 3.724 X icf 4 mhos. 
 
 Load (at receiver end), 15,000 K.V.A., 86,600 volts, 80% 
 
 P.F., three phase, 25 cycles. (Prob. 5, Chap. V.) 
 Ans. 15,153 K.V.A., 97,934 volts, line; 56,542 volts, star; 
 
 89.71%- 
 
 4. Find, by the convergent series, the per cent voltage drop, the 
 per cent loss, and the power factor at the supply end of the following 
 line: 
 
 Length of line 100 miles. 
 
 Spacing 6 feet. 
 
 Conductor No. oooo copper wire. 
 
 Take r 0.267, x -7 2 7> ^ = 6.03 X io~ 6 . 
 Load (at receiver end), 100 amperes per wire, 60,000 volts, 
 95% P.F., three phase, 60 cycles. [Prob. 4, Chap. V.] 
 
 Ans. 13.03% drop, 7.60% loss, 96.66% P.F. 
 
 5. Find, by the convergent series, 
 
 (a) star voltage at supply end at full load, 
 
 (b) star voltage at supply end at no load, 
 
 (c~) regulation volts (star), at the supply end, . 
 
 (d) amperes per wire at supply end at full load, 
 
 (e) power factor at supply end at full load, 
 (/) loss in line at full load, 
 
 (g) efficiency of the transmission line, 
 
 (ti) amperes per wire at supply end at no load (i.e., the 
 "charging current"), 
 
 ({) power factor at supply end at no load, 
 
 (/) loss in line at no load, for the following line: 
 
 Length of line 300 miles. 
 
 Spacing 10 feet. 
 
 Conductor, No. ooo copper cable of 0.330 ohm per mile. 
 Load (at receiver end), 18,000 K.V.A., 104,000 volts, 90% P.F., 
 three phase, 60 cycles. (Prob. 6, Chap. V.) 
 
52 TRANSMISSION LINE FORMULAS 
 
 Ans. (a) 69,670 volts, (b) 48,950 volts, (c) 20,720 volts, (d) 
 96.59 amps., (g) 92.35%, (/) 2440 Kw., (g) 86.90%, (A) 
 90.97 amps., (i) 6.47%, (;) 860 Kw. 
 
 6. Find, by the convergent series, the voltage at the supply end 
 of the following line: 
 
 Total length of line 400 miles. 
 
 Spacing 15 feet. 
 
 Conductor No. oooo copper cable. 
 
 Load at receiver end of line, 5000 K.V.A., 85% P.P. (lagging), 
 
 110,000 volts, three phase, 60 cycles. 
 
 Load taken by a substation at the middle of the line, 200 miles 
 from either end, 2500 K.V.A. at the line voltage and at 90% P.F. 
 (lagging). (Prob. 7, Chap. V.) 
 
 Ans. 90,190 volts. 
 
PART II. 
 THEORY. 
 
 CHAPTER VII. 
 
 CONDUCTORS. 
 
 THREE main classes of conductors are used for overhead 
 lines for the transmission of electric power; namely, copper 
 wires, copper cables and aluminum cables. The cables used 
 are generally strands of seven wires; that is, they consist 
 of a central straight wire with six 
 wires wound spirally around it, as 
 indicated by the cross section in 
 Fig. 9. 
 
 From this figure it is seen that .. 
 the maximum diameter of a y-wire 
 strand is equal to 3 times the di- 
 ameter of one of the wires. 
 
 The outside wires do not follow 
 a straight path parallel to the 
 central wire and the axis of the cable, but lie in a spiral 
 around it, as mentioned above. As there is always a 
 slight insulating film of oxide on any wire, the current 
 flowing in the cable tends to stay in the individual wires, 
 and so follows the longer path. Thus, the resistance of a 
 cable is greater than that of a solid wire of the same area 
 of cross section. The amount of the difference depends on 
 the number of wires in the cable and the pitch of the 
 
 53 
 
 2s 
 
 Fig. 9. 7-Wire Strand. 
 
54 TRANSMISSION LINE FORMULAS 
 
 spiralling, but an average value of i% is assumed in mak- 
 ing up the tables in Part III. The cross section of the 
 cable is assumed to be equal to the sum of the cross sections 
 of the individual wires. The weight per unit length of the 
 cable calculated from this cross section must be increased 
 by the same percentage as the above increase in resistance, 
 due to the extra length of the outside wires. Since the 
 cross section in Fig. 9 does not cut the outside wires ex- 
 actly at right angles, their sections as shown in the figure 
 
 are really ellipses, and the di- 
 ameter of the cable is slightly 
 greater than 6 pi. However, 
 this difference is small and has 
 been neglected in the figures 
 for diameter of cable tabulated 
 in Part III. 
 
 The number of wires in a 
 strand varies in practice ac- 
 
 j<_ 2$ J cording to the degree of flexi- 
 bility and mechanical strength 
 
 Fig. 10. 19- Wire Strand. J 
 
 desired by the user. The num- 
 ber of wires per strand in the tables represents average 
 practice for overhead lines. The larger cables often have 
 19 or even 37 wires. 
 
 The section of a i9-wire strand is shown in Fig. 10, and 
 it is seen that the maximum diameter is 5 times the diam- 
 eter of one of the individual wires. The same increase of 
 i% in resistance is allowed as with a 7-wire strand. 
 
 There is only a very slight difference in the reactance 
 and capacity of a 7-wire and a i9-wire strand of the same 
 sectional area, so that values listed for 7 wires may be 
 used for 19 wires, and vice versa, without very much error. 
 
CONDUCTORS 55 
 
 The resistances for direct current tabulated in Part III 
 have been calculated in accordance with the recommenda- 
 tions of the Bureau of Standards for the preparation of wire 
 tables.* The Standardization Rules of the* American In- 
 stitute of Electrical Engineers are in agreement with these 
 recommendations. According to the Bureau of Standards 
 and the A. I. E. E., the "Annealed Copper Standard," 
 which is of 100% conductivity, is represented by a resis- 
 tivity of 0.153022 ohm per meter-gram at 20 C. This is 
 equivalent to 1.72128 micro-ohms per centimeter cube at 
 20 C., assuming a density of 8.89. This is the same as 
 the resistivity of Matthiessen's Standard at 20 C., for- 
 merly used by the A. I. E. E. The conductivity of hard 
 drawn copper recommended for wire tables by the Bureau 
 of Standards is 97.3%, this value representing an average 
 for good commercial copper. The average conductivity 
 given by the Bureau of Standards for hard drawn alu- 
 minum on the centimeter cube basis, assuming a density of 
 2.699, is 60.86%. The above values have been used in 
 preparing the tables in Part III, i% being added to the 
 resistance for the effect of spiralling, as already noted. 
 
 If it is desired to calculate the resistance of copper con- 
 ductors for other temperatures than 20 C., the tem- 
 perature coefficient, a 2 o, for hard drawn copper of 97.3% 
 conductivity should be used in connection with the formula 
 
 R t = RW { i + 2o (t - 20) } 
 
 where / is the temperature in degrees Centigrade lor which 
 the resistance R t is desired and where 
 
 2o = 0.00383. 
 
 * Bulletin of the Bureau of Standards, Vol. VII, pp. 71-126, Washington, 
 1911; Proc. A. I. E. ., Dec., 1910. 
 
56 TRANSMISSION LINE FORMULAS 
 
 For other initial temperatures and other conductivities, 
 temperature coefficients should be used as given in the 
 table of temperature coefficients in Part III, which is 
 taken from Appendix E of the Standardization Rules of the 
 A. I. E. E. 
 
 For the temperature coefficient of hard drawn aluminum, 
 a value of 
 
 2o = 0.0039, 
 
 which is recommended by the Bureau of Standards, may 
 be used. 
 
CHAPTER VIII. 
 
 TRANSMISSION LINE PROBLEMS. 
 
 WHEN conditions are given at the receiver, or load, end of 
 a transmission line, the convergent series of Table V give 
 at once the voltage, A +JB, and the current, C +JD, at 
 the other end of the line. By putting the load current 
 equal to zero, we obtain the following expression for the 
 no-load voltage at the supply end: 
 
 CV7 17272 \ 
 
 i+ + JL ^ L - + etc.) 
 2 2-3.4 / 
 
 Thus the ratio of the voltages at the two ends of the line at 
 no load is 
 
 E \ 22-3-4 
 
 which is independent of the voltage E, and depends only 
 on the constants of the line. 
 
 The absolute value of a complex quantity like the volt- 
 age AQ -\-jBo, is its total numerical value independent of its 
 phase relation. This is the same, in the case of the voltage 
 A Q +JB , as its measured value, and is equal to 
 
 or + - 
 
 2 AQ 
 
 to a very close approximation when BQ is smaller than A . 
 Since the two complex quantities making up equation (i) 
 are equal in all respects, their absolute values are eaual, 
 and hence 
 
 CV7 V%7 2 \ 
 
 i + + J - jL - + etc.) (2) 
 2 2-3-4 / 
 
 57 
 
58 TRANSMISSION LINE FORMULAS 
 
 When the line is carrying full load, the measured value 
 of the receiver voltage is E, and of the supply voltage, 
 
 B 2 
 
 A H -- j- If the load be thrown off and the supply volt- 
 
 2 A. 
 
 B 2 
 
 age be kept constant at A -\ -- - > then the receiver voltage 
 
 2 A. 
 
 will rise to a value E Q . The line is now at no load, and the 
 ratio of the voltages at the two ends is, by equation (2), 
 
 I YZ Y 2 Z 2 \ 
 
 = absolute value off i + H ---- (- etc. I 
 
 V 2 2 3 4 / 
 
 A + 
 Thus Eo = 
 
 (equation 2, Table III). 
 
 We are now in a position to obtain the regulation of the 
 line, since by the definition in the A. I. E. E. Standardiza- 
 tion Rules, 
 
 En E 
 
 Per cent regulation = *-= X 100. 
 
 rL 
 
 Thus, the regulation volts at the receiver end which are 
 to be expressed as a percentage of E, are 
 
 as in equation 3, Table III. 
 
TRANSMISSION LINE PROBLEMS 
 
 59 
 
 It is often desirable to find the regulation of a line at the 
 supply end, that is, the per cent change in supply voltage 
 from full-load conditions to no-load conditions, when the 
 receiver voltage is kept constant. If the receiver voltage 
 is E, we have seen in the preceding paragraph that the full- 
 load supply voltage is equal to 
 
 2 A 
 
 and the no-load supply voltage is 
 E 0a =A + - 
 
 The per cent regulation at the supply end is 
 
 E s E 0a 
 
 E 8 
 
 Xioo, 
 
 and the regulation volts at the supply end are, thus 
 
 77 
 
 E a 
 
 2 A 
 
 A 
 
 AQ 
 
 2A 
 
 as in equation 6, Table III. 
 
 In the expression C +JD for current at the supply end, 
 the quantity C denotes the 
 component of current which is 
 in phase with the voltage E at 
 the other end of the line. We 
 can, however, find the com- 
 ponent of supply current which 
 is in phase with the supply 
 voltage, by first finding the k 
 watts at the supply end. 
 
 Let the supply voltage be 
 
 E 9 = A +JB 
 
 Fig. n. 
 
60 TRANSMISSION LINE FORMULAS 
 
 and let its phase be denoted by the angle 0, Fig. n, where 
 
 7? 
 
 tan = > 
 
 r> 
 
 and, therefore, sin = 
 
 ^4 
 
 and cos = 
 
 Similarly, let the current at the supply end be C + JD, 
 at a phase angle 0, where 
 
 and, therefore 
 and 
 
 VC 2 + D 2 
 
 The watts at the supply end are equal to the current, 
 multiplied by the voltage, multiplied by the power factor; 
 that is, 
 
 Watts = absolute value of I a X absolute value of E s 
 
 X cos (0 - 0) 
 = \/C 2 + D 2 X VA 2 + B 2 X (cos cos 0+sin sin 0) 
 
 = VC 2 + > 2 X 
 
 BD 
 
 as in equation n, Table III. 
 
 The quadrature volt-amperes, or reactive power, are 
 given by the following equation : 
 
TRANSMISSION LINE PROBLEMS 6l 
 
 Reactive power 
 
 = absolute value of I 8 X absolute value of 8 Xsin (0 0). 
 
 = VC 2 + D 2 X VA 2 + B 2 (sin 6 cos - cos 6 sin 0) 
 
 j x c__ ^ x z> . ) 
 
 = C - ylZ). 
 
 When the expression BC AD has a positive value, the 
 current at the supply end is lagging behind the supply 
 voltage, and when the expression has a negative value, the 
 current leads the voltage in phase. 
 
 We can now obtain the in-phase component of current, 
 which is equal to watts divided by voltage (equations 15 
 and 1 6), and in the same way the quadrature component 
 of current, which is equal to reactive power divided by 
 voltage (equations 17 and 18). The power factor at the 
 supply end is equal to watts divided by volt-amperes 
 (equations 13 and 14). Since the power supplied is known, 
 being AC + BD, and the power delivered at the receiver is 
 also known, being equal to EP, their difference represents 
 the loss of power in the line due to resistance of the con- 
 ductors, leakage over the insulators and corona loss. 
 
 The equations in F, G, M and N are quite similar to the 
 above equations in their derivation, and they give the solu- 
 tions of similar problems when conditions are given at the 
 supply end of the line. 
 
CHAPTER IX. 
 REACTANCE OF WIRE, SINGLE-PHASE. 
 
 Effect of Flux in Air. Let there be an alternating 
 current, /, in the transmission line wire, A, indicated in 
 
 Fig. 12. 
 
 The magnetic field set up 
 by the current at P, a dis- 
 tance x away from the wire, 
 will be at right angles to 
 the wire. The intensity of 
 the field will be equal to the 
 force on a unit magnetic pole 
 at P due to the current in the wire. The force due to the 
 current in a short length, dl, of the wire will be 
 
 cos 6 = - cos 6 d8, 
 r 2 x 
 
 since dl cos = r dd 
 
 , x 
 
 and 
 
 cos0 
 
 The total force at P due to the current in the wire A is 
 equal to 
 
 1C 
 
 C~ 2 I cosOdd , , . .v 
 
 (where re is a constant) 
 
 J <* ^ 
 
 TT 
 
 I sin 0"P 
 
 2/ 
 
 X 
 
 62 
 
REACTANCE OF WIRE, SINGLE-PHASE 63 
 
 where / is measured in absolute electromagnetic units. 
 When I is in amperes, the field at distance x is 
 
 - lines per sq. cm. (i) 
 
 10 x 
 
 r s-AV ---! 
 
 _A 41 $5. 
 
 _ r-r -J~. ^^ 
 
 Fig. 13. 
 
 In Fig. 13 is shown the cross section of a single-phase 
 transmission line. The lines of force in the path of thick- 
 ness dx surrounding the wire A are 
 
 21 j 
 dx 
 
 IOX 
 
 per centimeter of the transmission line. These lines cut 
 the wire A and produce an alternating voltage in it which 
 is 90 out of phase with the current and is equal to 
 
 2/ 
 
 ju dx X icr 9 volts, 
 
 x 
 
 where co = 2 TT X number of cycles per second, and where 
 / is in amperes. 
 
 The voltage drop between the wires A and B, due to 
 flux in the air produced by the current in A, is obtained by 
 integrating the above expression from x = p to x = s. 
 The integration is not carried beyond x = s, since flux 
 which cuts both A and the return wire B does not produce 
 any voltage between them. The voltage drop is equal to 
 
 * 2 7 s 
 
 jw dx X io~ 9 = y 2 / loge - X io~ 9 . 
 
TRANSMISSION LINE FORMULAS 
 
 There will be an equal drop due to the flux produced by 
 the current in the wire B, so that the total drop due to 
 flux in air is 
 
 jco4/log e - X lo- 9 
 p 
 
 volts per centimeter of line, 
 
 2jw X 741-1 logio- X 
 p 
 
 (2) 
 
 volts per ampere per mile of single-phase line. 
 Effect of Flux in the Conductor. Let i be the current 
 
 per unit area of section at any point in the wire shown in 
 
 Fig. 14. (For the present 
 assume that i is the same at 
 all points of the section.) 
 
 The total area of section of 
 the wire is irp 2 and therefore 
 the total current in the wire 
 is 
 
 / = Trip 2 . 
 
 Fig. 14. Section of Wire. 
 
 The total current inside the 
 circle of radius x is 
 
 /l = irix 2 . 
 
 This is the only current forcing flux around the circular 
 path of width dx t since currents flowing nearer the surface 
 of the wire do not tend to produce magnetic lines in a 
 path which does not surround them. Thus the flux density 
 at the radius x is 
 
 n I _ /> /w/f/v*'" 
 
 10 X 
 
 IOX 
 
 2irix 
 
 IO 
 
REACTANCE OF WIRE, SINGLE-PHASE 65 
 
 
 The total flux in the outer ring of the section is 
 
 dx _ TTJ (p 2 x 2 ) 
 10 10 
 
 This cuts the element dx of the wire and produces a voltage 
 along it equal to 
 
 juTri (p 2 x 2 ) io~ 9 volts per cm. (3) 
 
 This voltage leads the current by 90 in phase at all sec- 
 tions. It is greatest at the center and zero at the surface 
 and so is an unbalanced voltage; it therefore causes a local 
 quadrature current to flow along the center of the wire and 
 return near the surface. 
 
 Let the local current at the element dx be i( X ) per unit 
 area of section. Then the average voltage drop along the 
 wire due to the flux inside it and the resulting local bal- 
 ancing current, is equal to 
 
 juILi = juiri (p 2 - x 2 ) io~ 9 + i (x ) r, 
 
 where L\ is the self -inductance of the wire due to the 
 above-mentioned flux inside it, and where r is the specific 
 resistance of the metal in centimeter units, that is, the re- 
 sistance of a centimeter cube of the metal. The current 
 i (X ) adjusts itself so that the drop is the same at all parts 
 of the section. From the last equation, we have 
 
 jwrip^Li juiri (p 2 - X 2 ) 9 , , 
 
 *(*) = j j ~ X 10 y , (4) 
 
 since / = Trip 2 . 
 
 As i( X ) is a local current in the wire, and does not in- 
 crease or decrease the main current /, its sum when added 
 up all over the section must be zero, and thus 
 
 r. _ 
 2 1^x^( X ) ax = o, 
 . 
 
66 TRANSMISSION LINE FORMULAS 
 
 that is, j ^^ fVLi* - P 2 x icr 9 + x 3 icT 9 ) dx = o 
 r Jo 
 
 Now LI is a constant, independent of x, and so 
 P 4 Z,! - p 4 io~ 9 + - io~ 9 = o. 
 
 2 
 
 Therefore Li = \ X icr 9 (5) 
 
 The voltage drop between the wires A and B due to the 
 flux inside both wires is 
 
 2JuILi = 2/w/ X | X lo" 9 volts per cm. 
 
 = 270) X 80.47 X icT 6 
 
 volts per ampere per mile. The total reactive drop be- 
 tween the wires is thus 
 
 2./ (80.47 + 741.1 logio-) X io~ 6 (6) 
 
 volts per ampere per mile of single phase line. 
 
 This may be written in the following form which is more 
 convenient for computation by means of logarithm tables: 
 
 Reactance drop = 2Ju X 741.1 Iogi X io~ 5 (7) 
 
 0.779 p 
 
 volts per ampere per mile of single-phase line. 
 
 The above is the usual formula for reactance of a single- 
 phase line. The proof is longer than that generally given, 
 but it has the advantage of giving a correct idea of 
 the distribution of current and magnetic flux inside the 
 wire. As the irregular distribution of current produces the 
 "skin effect" described in the next chapter, and necessi- 
 tates slight corrections in the above formula for reactance 
 and in the resistance, the importance of calculating the 
 correct current distribution is evident. The above formula 
 is sufficiently accurate, however, for calculating the tables 
 of reactance of wire in Part III. 
 
CHAPTER X. 
 SKIN EFFECT. 
 
 IN the last chapter a local quadrature current i( X ) was 
 assumed, whose resistance drop balances up the unequal 
 voltages produced at the center and near the surface by 
 the flux inside the wire. This local current, i (X ), when 
 added up over all parts of the section of the' wire, amounts 
 to zero, and so cannot produce magnetic lines in the air 
 outside the wire. But it can produce lines inside the wire, 
 and the effect of these will now be calculated. 
 
 The reactive drop in one wire due to the flux inside it 
 produced by the main current i is 
 
 juirip 2 Li = j(mif^ X \ X io~ 9 volts, 
 where i is in amperes. 
 
 Then at a distance x from the center we have, from equa- 
 tion (4), Chap. IX, 
 
 X X TO-' - - p _ * x 
 2 r 
 
 10" 
 
 This is a lagging current at the center and a leading 
 current at the surface, and it equals zero when integrated 
 over the entire section. 
 
 The current i( X ) , integrated over the circle of radius x } is 
 
68 TRANSMISSION LINE FORMULAS 
 
 The flux density at the element dx, due to the above cur- 
 rent, is 
 
 2 Ii x \ jwirH _ Q / 9 ~x 
 
 - ^ = * - X 10 9 (- p 2 X + X s ). 
 
 iox 10 r 
 
 The flux in the ring outside of the circle of radius x, due 
 to I (x) , is 
 
 ( x) - ^ x 10-9 r (-^4. *s) ^ 
 
 ior J x 
 
 jW^io- 9 / P 4 . p 2 * 2 . p 4 * 4 \ 
 = - r- H --- 1 ---- J 
 
 ior \ 2 2 4 47 
 
 /COTT 2 ^' IO~ 9 / 9 9 4\ 
 
 = < - (- p 4 + 2 p 2 X 2 - ^ 4 ). 
 
 40 r 
 This flux produces a voltage at the element dx, equal to 
 
 U>Vi IO~ 18 / 4 , 9 o 4 \ / \ 
 
 - (- p 4 + 2 p 2 * 2 - ^ 4 ). (i) 
 
 4 P 
 
 A local current, f (2x ), will flow in order to keep the volt- 
 age drop uniform over the section. Let the average drop 
 due to 4>( X ) be 
 
 then 
 
 " Vilo ~ 18 ( P 4 - 2 p 
 
 Integrate ^ (2 x) over the entire surface and as it is a local 
 current 
 
 2irxi(2x)dx = o 
 
 r Jo 
 
 2 
 
SKIN EFFECT 69 
 
 18 / 6 6 " 
 
 Therefore, /Z^P 4 = - ( - - - + 
 
 2 T \2 2 
 
 , . I COTTp 2 I0~ 18 f N 
 
 and 2 = j (3) 
 
 Thus i ( 2x) = - C 7r * I (2 p 4 - 6 p 2 * 2 + 3 x 4 ) (4) 
 
 This current is in phase with the main current and, as it 
 is negative at the center and positive at the surface, it 
 produces a stronger resultant current near the surface of 
 the wire. This is the well-known "skin effect." The 
 effect of the quadrature current i^ is to increase the re- 
 sultant current both at the center and near the surface, 
 but its effect is not as large as that of the in-phase current 
 i (2x ) and so the net result is a crowding of current toward 
 the surface. 
 
 The above process may be continued indefinitely, each 
 step adding a smaller correction than the one before to 
 the current at radius x and to the average drop in the wire. 
 
 Thus the expression for i($ x) , equation (4) , may be inte- 
 grated over the circle of radius #, and will give the value 
 of 1(2 x )' This current produces a flux density at the radius 
 x, and by integrating this over the outer ring of the section, 
 the value of 0@ X ) is obtained. The flux <j> ( 2 X ) produces an 
 unbalanced voltage which must be corrected by a local 
 current i(3 X ), so as to give a uniform drop over the section, 
 due to the inductance Z, 3 . Equating the total local current 
 to zero, as before, gives 
 
 i coVV io~ 27 
 LZ = - 
 
 In the same way it is found that 
 
 . i coVp 6 io~ 36 
 
70 TRANSMISSION LINE FORMULAS 
 
 1 C0 4 7T 48 IO- 45 
 
 and 
 
 8640 r 4 
 
 Let the resistance of the wire per centimeter be R, where 
 R = - ohms per cm., 
 
 COTTp 2 IO~ 9 
 
 and let m = - - - 
 
 r 
 
 - ^ IO ~ 9 
 R 
 
 Then the total drop in the wire is 
 IR 
 
 ( 2 Iog 6 - H --- j 
 
 \ p 2 J 12 
 
 = IR + ;W io~ 9 2 Io 6 - H --- m 
 
 - * + - i2 - ^ 4 ---- ^ (5) 
 48 180 8640 / 
 
 volts per centimeter. 
 The total drop in phase with the current is 
 
 -Y (6) 
 
 / 
 
 12 180 
 
 The total copper loss due to all the currents in the wire is 
 therefore equal to 
 
 PR(I+ -Lm 2 --Lm 4 + ...V 
 
 V 12 180 / 
 
 This can be checked by integrating the losses due to the 
 total in-phase and quadrature currents in all parts of the 
 section of the wire, the above result being obtained by 
 this method also. Thus, in every respect, both as to volt- 
 age drop and watts loss, the resistance of the wire to the 
 alternating current is 
 
 12 
 
SKIN EFFECT 71 
 
 Values of R' for both 25 and 60 cycles are tabulated in 
 Part III. When taking the resistance of a conductor from 
 the tables, R' should always be used for alternating current, 
 and R should be used only when the conductor carries 
 direct current. 
 
 The total drop in quadrature with the current is 
 
 = /co/io" 9 ) 2log - + -(i -- m 2 -\ -- 3-m* - ) (7) 
 
 ( p 2\ 24 4320 I) 
 
 The series i - m 2 + -^-w 4 - 
 
 24 4320 
 
 is thus a correction factor for the term J or 80.47 in the 
 ordinary formula for reactance. Its effect is too small, 
 however, to make any appreciable change in the tabulated 
 values of reactance. 
 
 Proof by Infinite Series. The above formulas for the 
 resistance and inductance of a wire carrying alternating 
 current are sufficiently accurate for transmission line calcu- 
 lations with ordinary frequencies. They may .also be ex- 
 tended to include more terms without undue labor. How- 
 ever, as skin effect formulas are generally obtained and 
 expressed by means of infinite series which can be carried 
 out to any degree of accuracy for high-frequency work, a 
 short outline of the derivation of the infinite series will be 
 given. It will prove a check upon the correctness of the 
 formulas given above, but it will probably not give as 
 clear an idea as they do of the actual distribution of 
 current in the wire. 
 
 Let an alternating current, 7, of sine wave form and of 
 steady value, flow in a round wire of radius p. (See Fig. 14, 
 
72 TRANSMISSION LINE FORMULAS 
 
 Chap. IX.) Let it take up such a distribution that the 
 drop at all parts of the section of the wire, due to re- 
 sistance and to magnetic flux, is the same. Then if i' be 
 the current density at radius x, we may assume 
 
 *' = flo + <*i* 2 + 2* 4 + + a n x 2n +:.-- (8) 
 where OQ, a\, . . . a nj etc., are constants, independent of x. 
 (As the same value of i f would be obtained for both +x 
 and x, only even powers of x need be assumed for the 
 series.) 
 
 The total current in the part of the section inside a 
 circle of radius x will be 
 
 / 
 
 Jo 
 
 = / 2irxi f dx 
 
 i . n ..i ,. 
 
 = 2 TT [ -- 1 --- r H ---- r r (9; 
 \ 2 4 2n / 
 
 The flux density at the radius x is 
 
 iox io \ 2 n 
 
 and the total flux in the outer ring of the section, outside 
 the circle of radius x, is 
 
 2 I f 
 
 r 
 = I 
 
 J x 
 
 , 
 dx 
 
 iox 
 
 io 2 3 n 
 
 The drop at radius x due to the flux 0' is 
 
 jot' X io~ 8 
 and the resistance drop due to the current at the same 
 
SKIN EFFECT 73 
 
 part is i'r. Thus the total drop per centimeter of wire, 
 which is the same at all parts of the section, is 
 V = jwtf icr 8 + i'r 
 
 2' 3 * / 
 
 + r (oo + aix 2 + a*x* + - - + a n # 2n +)" (n) 
 
 The above expression for V is the same for all values of x, 
 and we may therefore equate each coefficient of x 2 , # 4 , etc., 
 to zero. Thus, putting 
 
 corrp 2 ICT 9 
 
 = m, 
 
 r 
 
 we have a\ = 
 
 2p 
 
 np 
 
 ... ... etc. 
 and 
 
 V = fl^+WTo 
 
 Substituting the values of ai, 02, etc., we obtain 
 V OQT -\-jwu lo" 9 
 
 (jm) 2 a p 2 (jm} n ~ l a^ . 
 
 + ' 
 
 (12) 
 
74 
 
 TRANSMISSION LINE FORMULAS 
 
 Now by putting x = p in the expression for /', equation 
 (9), we obtain the value of the total current in the wire, 
 
 2 3 
 fgi + ^ 
 
 n 
 
 Therefore, 
 ^ 
 
 n(jm} n ~ l 
 
 Substituting this value of OQ in equation (12), and putting 
 
 the resistance per centimeter of the wire, we obtain 
 
 (J m Y i , (J m Y i 
 
 + ... + _ + ... 
 
 r j. j.v 
 
 . 
 
 " 
 
 i 6 \J'"'l i 
 
 (W 
 
 + 
 
 _ 
 
 V*Oy 
 
 + 
 
 This expression can evidently be carried to any accuracy 
 desired. It will give the same results as were previously 
 obtained in equation (5), by expanding the denominator as 
 a binomial of the form (i + x)~ l and multiplying by the 
 numerator. This gives 
 
 V = 
 
 or, 
 
 
 
 io 
 
 ~ 9 - 
 
 This is the voltage drop, omitting the effect of the flux 
 
SKIN EFFECT 75 
 
 outside the wire, and is the same as the value previously 
 obtained. (See equations 6 and 7.) 
 
 REFERENCES. 
 
 Maxwell, Elec. and Magn., Vol. II, Para. 689-690. 
 Rayleigh, Phil. Mag., 1886, Vol. 21, page 381. 
 Kelvin, Math. Papers, 1889, Vol. 3, page 491- * 
 Rosa and Grover, Bulletin of Bureau of Standards, Washington, 1911, 
 Vol. 8, No. i, pages 173-181. 
 
CHAPTER XL 
 
 REACTANCE OF CABLE, SINGLE-PHASE. 
 
 As stranded cables are very commonly used for trans- 
 mission lines, it is desirable to have a special formula for 
 the reactance of cables. An outline will be given of the 
 method of obtaining the formula for a seven-wire cable. 
 This formula was used in preparing the reactance tables 
 in Part III. 
 
 A seven-wire strand consists of a central straight wire, 
 with six wires of the same size laid around it in a spiral. 
 The spiralling of the wires increases the resistance of the 
 cable by an amount which is taken as i%. The spiralling 
 also increases the outside diameter of the cable by about 
 T V of i%. (See Chapter VII.) 
 
 In calculating the reactance of the cable, the first step is 
 to plot the flux density at various distances from the center 
 of the cable. (See Fig. 15.) For points entirely outside 
 the cable, the flux density obeys the law 
 
 27 
 
 where x is the distance from the center. The total voltage 
 due to these lines which cut the entire cable is 
 
 2log e - X io~ 9 
 p 
 
 volts per ampere per centimeter. When x is less than the 
 radius of the cable, the flux at the distance x is 
 
 ' 
 
 76 
 
REACTANCE OF CABLE, SINGLE-PHASE 
 
 77 
 
 I (X ) is proportional to the area of conductor inside the 
 circle of radius x, and this must be measured from the 
 diagram of the section of the conductor (Fig. 15)., 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 /' 
 
 ~ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 x 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^^ 
 
 ^ 
 
 
 
 
 
 
 "-* 
 
 -, 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s' 
 
 '^ 
 
 * 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^^ 
 
 -^ 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 /- 
 
 x ^ 
 
 X 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 x~ 
 
 ' -* 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^> 
 
 -^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^ 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^x 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 I.I 
 X 
 
 p 
 
 Fig. 15. 
 
 Plot a curve of flux density, / (x) , for various values of #. 
 Let (x) be the area of the curve of / (x) between the values 
 # and p, where p is the outside radius of the cable. Also, 
 let C( X ) be the length of that part of the circle 2 TX which 
 lies in the section of the wires of the cable. 
 
 As shown in Chapter IX, page 65, the voltage drop 
 along different parts of the cable is not uniform, and must 
 be balanced by a local quadrature current. Thus we have 
 the conditions 
 
 io 
 
 ~ 8 
 
 at any section, 
 
78 TRANSMISSION LINE FORMULAS 
 
 and 2X'z)(z) dx = o. 
 
 o 
 
 Substituting the value of i( X ) from the first equation, we have 
 
 dx ->zCz dx io~ 8 = o. 
 <r 
 
 Now I LI is a constant, and 
 
 2.c (X )dx = A, 
 (T 
 
 the area of the section of the cable. 
 
 By dividing p into a number of equal parts and calling 
 each part dx, the value of 4>( X )C (X ) dx for each successive 
 value of x is found. Adding these together, a close estimate 
 of the value of 
 
 dx 
 
 is obtained. This, divided by I A gives the value 
 
 Li = 0.633 X icr 9 . 
 
 If the conductor were a solid wire of radius p, with a 
 straight line curve of / (x) , 1 would be J X lo" 9 , which 
 would become 80.5 X io~ 6 when reduced to the ordinary 
 formula for reactance per mile. Thus the reactance per 
 mile of cable is 
 
 x = (80.5 X ^3 + 74I .! lo glo -) io- X 2 irf 
 \ 0.500 p/ 
 
 = ( 102 + 74I.I logio-j I0~ 6 X 2 7T/ 
 
 ohms per mile. 
 
 The same process applied to a ig-wire strand gives the 
 formula 
 
 x = ($9 + 741.1 logio-j icr 6 X2irJ 
 ohms per mile. 
 
CHAPTER XII. 
 REACTANCE OF TWO-PHASE AND THREE-PHASE LINES. 
 
 Reactance, Two-phase. The reactive drop in a single- 
 phase line, in which round wire is used, is 
 
 2JI X 2 7r/f 80.5 -f 741.1 log-) io~ 9 
 
 volts per mile of line. In a two-phase four-wire line, the 
 drop will be the same as the above, when / is the current 
 in one phase. 
 
 AT T K.V.A. 
 
 Now 2 / = - = the total amperes. 
 
 hi 
 
 Therefore the reactive drop per mile of line, in absolute 
 value, is 
 
 ^|^ X 2 T/(8o.s + 741.1 log-) io-' 
 \ p/ 
 
 K.V.A., 
 
 ~E~~ Xx > 
 
 wnere x is the tabulated value of reactance. 
 
 Reactance, Three-phase, Irregular Spacing. When the 
 conductors of a three-phase line are spaced so that they are 
 not exactly equidistant, the voltage drop due to reactance 
 is not the same in the different phases. It is the practice 
 with such lines tp interchange, or transpose, the conductors 
 at intervals along the line, so that the different reactive 
 voltages are applied to an equal extent to all three con- 
 ductors. Such a line, when carrying a balanced load (equal 
 currents in each conductor, at 120 in phase from each 
 
 79 
 
8o 
 
 TRANSMISSION LINE FORMULAS 
 
 other), will have the voltages of the three phases equal at 
 the end of the line. 
 
 The average reactance of an irregularly spaced three- 
 phase line, in which the conductors are transposed at regu- 
 lar intervals, may be calculated as follows: 
 
 Fig. i 6. 
 
 17. 
 
 Let Fig. 1 6 represent the spacing of the conductors A, B, 
 and C of a transmission line. Let the currents in the con- 
 ductors be represented by the vectors OP, OQ, and OR 
 (Fig. 17). If the power factor is 100%, these vectors may 
 also represent the star voltages, and the line voltages will 
 be represented by the vectors PQ, QR, and RP. 
 
 Let the current in conductor A be 
 
 and let 
 and 
 
 I A = OP = i .00 / amperes, 
 
 IB = OQ = ( 0.50 + o.S66j) I amperes, 
 
 In = OR = ( 0.50 0.866 j) /amperes. 
 
 Let also 
 
 Voltage from neutral to A = i.oo V. 
 
 Voltage from neutral to B = (- 0.50 + 0.866.7) V. 
 
 And voltage from neutral to C = ( 0.50 0.866.7") V. 
 
 Then the measured, or absolute, value of voltage between 
 B and C is 1.732 V, where V is the star voltage of the line. 
 
REACTANCE OF TWO-PHASE AND THREE-PHASE LINES 8 1 
 The reactive voltage on conductor A per mile is 
 
 1,00/80.5 + 74I.I log \Ij 2 7T/ X IQ- 6 
 
 (due to flux from 7^) 
 + (-0.50 +0.8667) (741-1 logyJ//27r/X io~ 6 
 
 (due to flux from IB) 
 + (- 0.50 - 0.866 j) (741.1 logy JIj 2 TT/ X io~ 6 
 
 (due to flux from I c )j 
 
 where p is the radius of the wire. 
 The reactive voltage on conductor B per mile is 
 
 1.00(741.1 log J7; 2 irf X io~ 6 
 (due to flux from I A) 
 
 (due to flux from IB) 
 + ( 0.50 0.866 j) (741.1 log J Ij 2 irf X I 
 
 (due to flux from /<?). 
 
 The reactive voltage between A and B is therefore 
 (1.50 - o. 
 
 + (0.50 + 0.866 j) 741.1 (log- - log-)// 27T/ X iQ- 6 . 
 \ p pi 
 
 Suppose at the end of one mile the line is transposed 
 so that the above two conductors occupy the positions B 
 
82 TRANSMISSION LINE FORMULAS 
 
 and C. Then the reactive voltage between these conductors 
 for the next mile is 
 
 (1.50 - 0.8667) (80.5 + 74i.i log-) Ij 2 TT/ X iQ- 6 
 
 + (0.50 + 0.8667) 74LI (log- - log-) Ij 2 7T/ X I0~ 6 . 
 
 \ p p/ 
 
 Let the line be transposed again. Then for the third 
 mile the reactance voltage between these same two con- 
 ductors is 
 
 (1.50 - 0.8667) (80.5 + 741.1 log -} Ij 2 TT/ X lo- 6 
 
 + (0.50 + 0.8667) 74I.I (log- - log-) Ij 27T/X IO- 6 . 
 
 \ p P/ 
 
 The total reactive voltage for the three miles is 
 
 (1.50-0.8667) j 80.5X3 + 741.1 (log- + log- + log-) I 
 ( \ P p P/ ) 
 
 X Ij 2 7T/ X XT*. 
 
 /Thus the average reactive voltage per mile, which is the 
 same for all phases, is, in absolute value, 
 
 1.732(80.5 + 741.1- log- -J/27T/X IO- 6 . 
 
 \ P / 
 
 Now 1.732 7 = the total amperes. 
 
 _ K.V.A. 
 
 E 
 Therefore the drop in line voltage, in absolute value, is 
 
 (8o.5 + 7 4 i.ilog^27T/Xio- 6 
 
 in volts per total ampere per mile of transmission line, 
 where 
 
 s = ^ dbc. 
 
REACTANCE OF TWO-PHASE AND THREE-PHASE LINES 83 
 
 Thus the total reactive drop is equal to 
 
 K ' VA - (80.5 + 741.1 log 2 TT/ X io- 6 
 
 E 
 
 K.V.A. 
 
 Xx, 
 
 E 
 
 where x is the tabulated value of reactance. 
 
 Reactance, Three-phase, Regular Spacing. When the 
 conductors are spaced at the corners of an equilateral tri- 
 angle of side 5, then the expression for reactance is the 
 same as the usual formula: 
 
 x = (80.5 + 741.1 logio-J 27T/ 
 
 X 
 
 Reactance, Three-phase, Flat Spacing. When the three 
 conductors lie in one plane (either 
 
 horizontal or vertical), the center j - ? - j 
 one being equidistant from the k-- a -4- ..... a -! 
 other two, as in Fig. 18, the react- Fig. 18. 
 
 ance per mile is 
 
 i a "^i X i X 2 
 80.5 + 741.1 log - - 
 P 
 
 , 1.260 
 
 = 80.5 + 741. i log -- . 
 P 
 
 This is approximately 4% higher in an ordinary case 
 than the reactance for spacing on an equilateral triangle 
 of side a. 
 
 The formulas of this chapter will apply to cable as well 
 as to wire, if the term 80.5 is changed as per Chapter XI. 
 
CHAPTER XIII. 
 
 CAPACITY OF SINGLE-PHASE LINE. 
 
 Capacity of Two Round Wires. The conductors of a 
 transmission line form a condenser, the electrostatic capac- 
 ity of which can be calculated from the dimensions of the 
 line. The simplest line to calculate is a single-phase line 
 consisting of two round wires, and this case will be in- 
 vestigated first. 
 
 Fig. 19. 
 
 Fig. 20. 
 
 Suppose that A and B (Fig. 19) are two long parallel 
 wires of infinitesimal section and that they are spaced a 
 distance t centimeters apart. Let A carry a charge of + q 
 electrostatic units of electricity per centimeter, and let B 
 carry q units per centimeter. 
 
 First, find the force exerted on a unit charge near the 
 wire A . 
 
 From the symmetry of the arrangement it is evident that 
 the resultant force on a unit charge at P (Fig. 20) will be a 
 repulsion away from the wire at right angles to it, since the 
 total effect of the right half of the wire must be equal to 
 the total effect of the left half. The force at right angles 
 
 84 
 
CAPACITY OF SINGLE-PHASE LINE 85 
 
 to the wire exerted by the charge on the element dl will be 
 
 
 = -^ cos dd, 
 
 r\ 
 
 since dl cos = r dB 
 
 V, 
 
 and r = 
 
 cos 
 The total force exerted by the wire will be 
 
 The potential of the point (Fig. 19) midway between 
 the wires, will be zero, since the effect of the positive charge 
 on A will be equal to the effect of the negative charge on B. 
 The potential difference between P and O is the work done 
 in moving a unit charge from one point to the other. The 
 
 2 Q 
 
 force due to the wire A on a unit charge at any point is * , 
 
 acting directly away from A. Therefore the work done 
 in moving a distance dr toward A is 
 
 and thus the total work in moving from P to against the 
 force due to A is 
 
 i_ 
 
 r~2Q, t 
 -* dr = 20 loge 
 r 2ri 
 
 Similarly, the work against the force due to B is equal to 
 
 2r 2 
 
86 TRANSMISSION LINE FORMULAS 
 
 Therefore the potential difference between P and is 
 equal to the total work, and is 
 
 2 q log e - 
 
 Fig. 21. 
 
 At P (Fig. 21) make the angle APD equal to the angle 
 PBD. Then the triangle PBD is similar to the triangle 
 APD, and therefore 
 
 PB ^r* 
 
 AP " n 
 
 p 
 
 If we draw a circle of radius p about the fixed point D, 
 then at any point on this circle similar triangles are formed 
 by p, r\j and r^ as in Fig. 21, and therefore 
 
 r 2 DB 
 
 = = constant. 
 ri p 
 
 where r\ and r 2 are the distances of the point on the circle 
 from A and B respectively. Therefore, the potential 
 
 2#log - 
 fi 
 
 will be the same at all points on this circle. 
 
 Now let a solid cylindrical conductor fill all the space 
 inside the circle of radius p. All points on its surface will 
 be at the same potential. The distribution of potential 
 outside of the cylinder will not be altered from the previous 
 condition when all points on the circle of radius p were 
 
CAPACITY OF SINGLE-PHASE LINE 87 
 
 also at the same potential. The potential of the cylinder 
 will be 
 
 , W , DB 
 
 2qlog e - = 2qlog e 
 
 r\ P 
 
 In the same way, let the wire B be replaced by a solid 
 cylinder of radius p and center E, as in Fig. 22. 
 
 s 
 
 ... t .. 
 
 Fig. 22. 
 
 The potential of this cylinder, which carries q units 
 per centimeter, will be 
 
 AE , DB 
 
 P P 
 
 since the second cylinder is symmetrical with the first. 
 Thus the potential difference between the two cylinders is 
 
 DB a 
 
 4?k)ge = *, 
 
 P v 
 
 where C is the capacity per centimeter of line. 
 
 ThuS If \ C= ^DB- 
 
 4log 
 
 P 
 
 DB FB 
 
 Now - = , 
 
 p FA 
 
 since F is a point on the circle. 
 
 DB DB- p 
 
 rpr, ( 
 
 Therefore 
 
 p DB + p - s 
 
 where 5 is the interaxial distance between the two solid 
 cylinders, and is equal to DE, Fig. 22. 
 
88 TRANSMISSION LINE FORMULAS 
 
 Therefore, 
 
 DB 2 + DB -p - DB -s = DB -p - p 2 
 or DB 2 - DB s + p 2 = o. 
 
 Solving this quadratic equation in DB, we have 
 
 2 
 
 [The negative value of the radical must not be used, 
 since it would give the value of DA instead of DB.] 
 Therefore, 
 
 i 
 
 $4V-' 
 
 \ 2 p * 4 p 
 which may be expressed as 
 
 4 cosh" 1 ( ) 
 
 \2p/ 
 
 or it may be expanded by the Binomial Theorem to give 
 the approximate value 
 
 T 
 
 per centimeter, 
 
 or, very nearly, 
 
 4log e - 
 P 
 
 * It is evident that the expression 
 
 i 
 
 C = 
 
 P 
 
 which is sometimes published, is less accurate than the simpler expression 
 
 4 log,- 
 
CAPACITY OF SINGLE-PHASE LINE 89 
 
 Transferring to other units, 
 
 r = T x * 4343 X 2 ' 54 X I2 X 528 
 
 " 4X9XXO" 
 
 x 
 
 2 
 
 loglo (_) 
 V s/ 
 
 38.83 x io- 9 
 
 , (S p\ 
 
 logio ---) 
 Vp s/ 
 
 farads per mile of single-phase line. 
 The capacity susceptance is 
 
 38.83 X 
 
 fr f v N/ 
 
 2 7T/ C = 2 7T/ X - X 
 
 mhos per mile of single-phase line. The charging current 
 in this line will be 
 
 amperes, 
 
 where b is the tabulated value of capacity susceptance. 
 
 REFERENCE. "A Treatise on the Theory of Alternating Currents," by 
 Alexander Russell, 1904, Vol. I, page 99. 
 
 Capacity of Cable. The formula for capacity of a line 
 using stranded cables will be the same as the above formula 
 for solid wires, p being taken as the maximum radius of 
 the cable. All the electrostatic charge on the cable does 
 not lie at the maximum radius from the center, but as 
 actual cables are generally slightly larger than the calcu- 
 lated diameters in the tables, it will be sufficiently close to 
 take p from the tables and use it in the regular formula for 
 capacity. 
 
90 TRANSMISSION LINE FORMULAS 
 
 Effect of the Earth on Capacity of Line. The effect of 
 bringing a conducting plane, such as the earth, near to two 
 charged wires is to change their electrostatic field and in- 
 crease their capacity. 
 
 Consider two long parallel wires, A and A l (Fig. 23), of 
 infinitesimal section and carrying + q and q units of 
 
 electricity per centimeter 
 respectively. As in Fig. 
 19, the point O midway 
 
 M __ / o N between the two wires 
 
 will be at zero potential. 
 All points having the same 
 
 potential must have equal 
 
 to a constant. It is evi- 
 dent that all points at the same potential as lie in the 
 plane MON, perpendicular to AA 1} since for all such 
 points 
 
 Therefore, the wire A may be replaced by a solid con- 
 ducting plane MN, which will be at zero potential. Thus, 
 when the conducting plane is the earth, its effect is the same 
 as that of a charged wire at a depth below the surface 
 equal to the height of the original wire.* The assumed 
 wire is called an image wire, since it occupies the same 
 position as the image of the real wire, considering the 
 surface of the ground as a mirror. 
 
 In the case of a single-phase transmission line, image 
 wires A' and B' must be assumed for both wires A and B 
 
 * See " Elements of Electricity and Magnetism," by J. J. Thomson, page 
 138. 
 
CAPACITY OF SINGLE-PHASE LINE 
 
 A' 
 
 ----> 
 
 -, ( 
 
 (Fig. 24) and the capacity of the entire system of four wires 
 is then calculated as follows: 
 
 Let h be the distance of the 
 wires from the ground and 5 
 their distance apart from center 
 to center. Let A carry a charge 
 of + q units per centimeter; A', 7 
 q units; B, q units; and 
 B' , + q units. Let a unit charge 
 be carried from the surface of 
 A to that of B. Assuming that Fig 24 
 
 the charges are concentrated at 
 the centers of the wires, the total work done is equal to 
 
 / 
 
 J p 
 
 
 -I 
 
 2 q (s - x) 
 
 dx. 
 
 The sum of the first two integrals has been shown to be 
 approximately 
 
 The sum of the last two integrals is approximately 
 4/r>-f s 2 
 
 2 * log iT7- 
 
 Therefore, the total work is equal to 
 , , zh 
 
 P 
 
92 TRANSMISSION LINE FORMULAS 
 
 Therefore, C is approximately 
 
 ! S . 2 k 
 
 4log e - + 
 
 P 
 
 Taking as an average case, 
 
 h = 360 inches (30 feet), 
 s = 120 inches (10 feet), 
 p = 0.25 inch, 
 
 we have - = 480, 
 
 p 
 
 2 h . 
 
 and = 6. 
 
 s 
 
 Therefore, - = 
 \/4 h 2 + s 2 
 
 Now, Iogio48o = 2.681, 
 
 while logio r^ = 0.006. 
 
 6 
 
 Thus the capacity is changed by the nearness of the 
 ground by less than } of i%, even with the comparatively 
 wide spacing of 10 feet. 
 
 Tests have shown that the effect of the ground in increas- 
 ing the capacity is even less than the above amount, due 
 partly to the fact that the ground is a poor conductor. 
 As the effect of the ground is so slight, it has been neglected 
 entirely in the calculations in this book. 
 
 REFERENCE. For an alternative proof, see "A Treatise on the Theory 
 of Alternating Currents," by Alexander Russell, 1904, Vol. I. 
 
 See also " The Calculation of Capacity Coefficients for Parallel Suspended 
 Wires," by Frank F. Fowle, Elec. World, Aug. 19, 1911. 
 
CHAPTER XIV. 
 CAPACITY OF TWO-PHASE AND THREE-PHASE LINES. 
 
 Capacity, Two-phase. The charging current of a single- 
 phase line was shown in Chapter XIII to be 
 
 *HH) 
 
 amperes per mile of line 
 
 -i* 
 
 where b is the tabulated value of capacity susceptance per 
 mile. 
 
 In a two-phase, four-wire line, each phase is quite similar 
 to a single-phase line, and so the charging current per wire 
 is 
 
 \Eb. 
 
 As this amount of charging current flows in each phase, 
 the total amperes of charging current are 
 
 Eb, 
 
 for a two-phase, four-wire line. 
 
 Capacity, Three-phase, Irregular Spacing. When the 
 wires of a three-phase transmission line are not spaced at 
 the corners of an equilateral triangle, but the transposi- 
 tion of the conductors is carried out at regular intervals, 
 the charging current in the wires will be a balanced, three- 
 phase current, since each wire will have passed through the 
 same average conditions. This is shown in an approxi- 
 mate manner as follows: 
 
 93 
 
94 
 
 TRANSMISSION LINE FORMULAS 
 
 As when calculating the self-induction of an irregularly 
 
 spaced line, consider a line three 
 miles long which is transposed 
 at the end of each mile. 
 
 The work in carrying a unit 
 charge from C to B (Fig. 25) 
 assuming the charges concen- 
 trated at the centers of the 
 wires, is approximately 
 
 Fig . 25 . 
 
 E a = q B 2 loge- - q c 2 loge- + q A 2 loge-- 
 
 p p 
 
 Now q B is a periodic quantity, which alternates in value 
 at the same frequency as the voltage or current. 
 We have q B = CiE 
 
 = _,-v 
 
 27T/ 
 
 where IB' is the charging current flowing into the capacity 
 Ci of the wire B. 
 Thus 
 
 / 2 loge - - /</ 2 loge - + // 2 loge \ 
 p p tv 
 
 2 7T/ 
 
 In the second mile the conductors are transposed into 
 new positions. Let the currents in them remain the same. 
 Therefore, 
 
 E a = =l(l B ' 2 loge - - I C ' 2 log, - + // 2 log -), 
 
 27T\ p CI 
 
 T/ 
 
 and in the third mile 
 
 E a 
 
 ~ - I C ' 2 
 
 p 
 
 " + 
 
 2 loge -) 
 
 al 
 
CAPACITY OF TWO-PHASE AND THREE-PHASE LINES 95 
 
 Let 
 and 
 and 
 
 Then 
 
 Adding together and dividing by 3, we obtain the ap- 
 proximate average value per mile, 
 
 that is, 
 
 where 
 Similarly, 
 
 and 
 
 E a =^(V-v) 2 logA 
 
 5= ^abc. 
 E b = 4( V - I A) 2 log - 
 
 2 7T/ \ / p 
 
 Fig. 26. 
 
 E a = i.ooEj as in Fig. 26, 
 ^b (-5o o.866j) , 
 E c = (o.5o+o.866j)E. 
 
 j 2 irfE (0.50 + 0.8667) 
 5 
 
 P 
 
 (l) 
 
 (0.50 - 0.8667), (3) 
 
96 TRANSMISSION LINE FORMULAS 
 
 also I A + IB + Ic = o, (4) 
 
 since they are currents flowing in a three-phase line. 
 From equations (i) and (3) 
 
 2log e - 
 
 P 
 Adding (4) and (5), we have 
 
 7 '_ _L_ y 2 irfE X 1 .00 
 
 3 2loge~ 
 
 P 
 
 -r, ,,. T , i . . 2 TT/E ( 0.50 -f 0.866 /) 
 From this, I B = j=. X - = 
 
 3 2 log - 
 
 P 
 
 and I ' * x **fE(--SQ-<>.*66fl. 
 
 ^ 2 loge ~ 
 
 P 
 
 The vectors for IA, IB and /c' may now be plotted as 
 in Fig. 27, and it is seen that the vectors are the same 
 , length and are at 120 to each other. 
 Thus the charging current is a bal- 
 . anced three-phase current. 
 
 The power factor of the charging 
 
 current is zero, since the current in 
 any wire I A (Fig. 27) is at right angles 
 to the direction OP (Fig. 26) of the 
 Fig. 27. corresponding star voltage or in-phase 
 
 current. 
 The total amperes of charging current are 
 
 T / 27T/E 
 
 I A - per centimeter 
 
 2 loge- 
 P 
 
CAPACITY OF TWO-PHASE AND THREE-PHASE LINES 97 
 
 , 5 
 logio - 
 P 
 = Eb, 
 
 where b is the tabulated value of capacity susceptance per 
 mile. 
 
 Capacity, Three-phase, Regular Spacing. When the 
 conductors are placed an equal distance, s, from each other, 
 the formula for b is 
 
 , 38.83 X 27T/ Vy ., 
 
 _ o o j_ x 10 a mhos per mile. 
 
 logio ~ 
 P 
 
 Capacity, Three-phase, Flat Spacing. When the wires 
 lie in one plane (either horizontal or vertical), the certer 
 one being at a distance a from the other two (see Fig. 18), 
 and the wires being transposed at regular intervals, the 
 formula for susceptance is 
 
 b= 38.83 X 2 rf _ yjn _. 
 
 , . 
 
 logio 
 P 
 
 mhos per mile of transmission line. 
 
CHAPTER XV. 
 THEORY OF CONVERGENT SERIES. 
 
 THE well-known fundamental formulas for a transmis- 
 sion line without branches, in which the load is delivered 
 only at the end of the line, are as follows: 
 
 E 8 = E cosh VYZ + 7/b| / sinh VYZ, 
 and i a = / cosh VYZ + ~^ sinh VYZ~ 
 
 where E 8 and I 8 are the voltage and current at the 
 
 supply end, 
 E and I are the voltage and current at the re- 
 
 ceiver end, 
 
 Y is the line admittance 
 and Z is the line impedance. 
 
 The above .equations have been published at various 
 times. They are obtained as follows: 
 
 Let r = resistance of conductor per unit length, 
 
 and x = reactance of conductor per unit length. 
 
 Then z = r +jx 
 
 = impedance of conductor per unit length. 
 Let g = leakage conductance from conductor per 
 
 unit length, 
 and b = capacity susceptance of conductor per unit 
 
 length. 
 Then y = g +jb 
 
 = admittance of conductor per unit length. 
 98 
 
THEORY OF CONVERGENT SERIES 99 
 
 Let EI = voltage of line at a distance / from the re- 
 
 ceiver end, 
 and let I l = P l -jQ l 
 
 = current in the line at a distance / from the 
 
 receiver end. 
 
 (Since Ii is usually not in phase with the voltage, it 
 must be expressed as a complex quantity.) 
 
 Now in an element of length, dl, of the line, the voltage 
 consumed by impedance is 
 
 dE l = zl t dl. 
 
 The current consumed by admittance is 
 dl t = yE t dl. 
 
 Thus 1f' = Z/ " (l) 
 
 and ' ^ = yEl : (2) 
 
 Differentiating (i) 
 
 d 2 Et _ dlj 
 dl 2 Z dl ' 
 Substituting (2) in this gives 
 
 This is a differential equation of the second order and 
 may be expressed in the form 
 
 (D* - yz) E, = o, 
 and we have 
 
 and from (i), 
 
 II = z~dT 
 
 (5) 
 
100 TRANSMISSION LINE FORMULAS 
 
 Now at the supply end, 
 yl = Y, 
 
 and zl = Z 
 
 Therefore, E a = A l v + A 2 e ~ ^ (6) 
 
 (7) 
 
 At the receiver end, 
 
 |o 
 
 and E = A 1 +A 2 (8) 
 
 and /=Vf <Xi-^ 2 ). (9) 
 
 From (6) 
 
 (10) 
 
 which, by the definition of cosh 6 and sinh 6, is 
 
 (A 1 + A 2 ) cosh VYZ + (Ai- At) sinh 
 Therefore, from (8) and (9), 
 
 
 E 3 = E cosh VYZ + = 7Z sinh VFZ: (i i) 
 Similarly, from (7), 
 
 = 7 cosh VFZ + = EY sinh VFZ. (13) 
 
 Equations (n) and (13) are the fundamental formulas of 
 transmission lines, as generally written. 
 
THEORY OF CONVERGENT SERIES 
 
 101 
 
 Now 
 
 I . YZ . F 2 Z 2 \ 
 
 = 21 H 1 hetc.l. 
 
 \ 2 2-3 -4 / 
 
 Similarly c VT1 - 
 
 /=7^ / FZ F 2 Z 2 \ 
 
 = 2 v FZ i + H h etc. 1. 
 
 V 2-3 2-3.4.5 / 
 
 Substituting these results in (10) and (12), we can ex- 
 press the fundamental equations as follows: 
 
 and 
 
 /. = 
 
 T 
 
 2-3-4 
 
 r 2-3-4-5 '6 
 
 r cue. i 
 
 FZ 
 
 F 2 Z 2 
 
 F 3 Z 3 
 
 1 r*tr 
 
 
 r 
 
 
 -f-etc. 
 
 2-3 
 
 2-3-4- 
 
 5 2-3.4.5-6 
 
 7 
 
 FZ , 
 
 F 2 Z 2 
 
 I 
 
 etc.) 
 
 T 
 
 H 
 2-3-4 
 
 2.3.4.5-6 
 
 FZ 
 
 F 2 Z 2 
 
 F 3 Z 3 
 
 _ O-Atr 1 
 
 .) (14) 
 / 
 
 .)*(i5) 
 
 2-3 2.3-4-5 2.3.4.5.6-7 
 
 Equations (14) and (15) are the same as those tabulated 
 in Chapter VI for obtaining E a or A + jB, and I 8 or C + JD. 
 For no-load values, all that is necessary is to put the load 
 current, /, equal to zero. 
 
 When conditions are given at the supply end, the same 
 equations for full-load conditions are obtained, except that 
 
 * See references to T. R. Rosebrugh, J. F. H. Douglas, and C. P. Stein- 
 metz, page 41. 
 
102 TRANSMISSION LINE FORMULAS 
 
 the second half of the expressions for voltage and current 
 is negative, since power is now flowing away from the point 
 where the voltage is specified, instead of toward it. Thus 
 we have the series fox F +jG and M +JN. 
 
 At no load, the conditions are really not all specified at 
 the supply end, but the current is specified to be zero at 
 the receiver end, and this necessitates the use of special 
 series. From Table V we have the ratio of the voltage at 
 the two ends of the line, 
 
 E Qs _ A Q +JB<> 
 E ~ E 
 
 , YZ , F 2 Z 2 
 
 H 1 hetc 
 
 2 2.3.4 
 
 This ratio is independent of the voltage E, and depends 
 only on the constants of the line. Thus, if E s , the voltage 
 at the supply end at no load, is given, we can obtain the 
 no-load voltage at the receiver end from the equation, 
 
 etc 
 
 2 2.3-4 
 
 V7 V 2 7 2 X- 1 
 
 or 
 
 2 2-3-4 
 
 which, when expanded by the binomial theorem, gives 
 E Q = 
 
 
 - etc] 
 
 / 
 
 2 24 720 8064 
 
 (16.) 
 as in Table VI. 
 
 The no-load current at the supply end is 
 
 CV7 1727-2 \ 
 
 i+ +- -+etcl 
 
 2-3 2-3.4.5 / 
 
 as in the equation for C Q +JD . 
 
 r ' ' . 
 
THEORY OF CONVERGENT SERIES 103 
 
 Substituting the value of E from equation (16), we have 
 -YZ+^-Y 2 Z 2 - Y*Z 3 +-m-Y*Z*-e 
 
 2 24 720 8064 
 
 2.3 2-3.4.5 2.3.4.5.6.7 
 
 F 4 7 4 \ 
 
 + - ~ - 5 + etc. ). 
 
 2.3.4.5.6.7.8.9 / 
 
 Multiplying the two series together by the ordinary 
 algebraical method, we obtain 
 
 F 4 Z 4 - 
 
 3 i5 3i5 2835 
 
 as given in Table VI of convergent series. 
 
PART III 
 TABLES. 
 
 TABLE I. FORMULAS FOR SHORT LINES. 
 CONDITIONS GIVEN AT RECEIVER END. 
 
 These formulas are exact when the line is short. When the line is 20 
 miles long, they are correct within approximately ^ of i% of line voltage. 
 Conditions given: 
 K.V.A. = K.V.A. at receiver end. 
 
 E = Full load voltage at receiver end. 
 cos 6 = Power factor at receiver end. 
 K.W. = K.V.A. cos 0. 
 
 r = Resistance of conductor per mile. (From Tables VII- VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 / = Length of line in miles. 
 
 1000 K.V.A. cos 8 , ,. 
 
 Then P= - =; - = In-phase current at receiver end (in 
 c, 
 
 total amps.). 
 
 _ 1000 K.V.A. sin _, . , ,. 
 
 Q = - - = Reactive current at receiver end (in 
 
 total amps.) when current is lagging. 
 
 1000 K.V.A. sin , 
 
 = -- - when current is leading. 
 hi 
 
 Find the following quantities: 
 
 Three phase or two phase. Single phase. 
 
 A = E + Prl+QxL A = E + 2 Prl + 2 Qxl. 
 
 B = Pxl- Qrl. B = 2 Pxl- 2Qrl. 
 
 Formulas (capacity neglected) : 
 
 B* * 
 
 (1) Voltage at supply end = A -\ -\ 
 
 R2 
 
 (2) Regulation of line = A + - E. (Same as line drop.) 
 
 (3) Per cent regulation of line = - - - -^ - '- per cent. 
 
 (Same as per cent line drop.) 
 ,. r 104 
 
 : ' ' r"" r ' r ' "' 
 
rrrrffr 
 
 ,,,M,|,,,,,,,,",M, ,,,,,,,,,,,,,,,,,, 
 
 in > i* _ 
 
 T"T" '"J""!"* 1 ! 1 
 4k M M 
 
 T-JL,^ M1 
 
 i/ 
 
 P 
 
 ]i' , , 
 
 . i . i . 1 1 1 1 1 1 
 ? Uj 
 
 ,,,.,, ,,, ,..., j, . .. i 
 
 ro 
 
 ., 1 ,,.. | ... .j ... i ,,., .j .,-,.,,. 
 
 cd 
 
 Ohms Resistano 
 
r ro j- 
 T L L 
 
 I I 
 
 .ro 
 
 les. 
 
 FeetSpacing 
 
 Copper 
 Aluminum 
 
 jo.ro ro 
 
 wNoaxnot^oj K 5J FeetSpacing 
 - FeetSpacing Ml I MM I I i r^^r 
 
 T^rnTTTTTT^ Alumlnum 
 Swo 03 ^ 01 -^ 04 ^oi ~" FeetSpacing 
 25 Cycles. 
 
 Q -<Q 
 
 " JT^ 
 
 ct 1 | 
 
 mn 
 
 fo- 
 
 o 
 
 Z 
 
 i Jo i i ii i i 
 
 I"".!""'"".!" 
 
 CD 00 
 
 T" 1 '".!""''"'! 1 ' J J 
 
 i0 00 "*l . 0* W| 
 
 = Regulation Factor. 
 
 pmr 
 
 |* || 
 
 fe Ii 
 st ^ 
 
 1 
 
 8 
 
 "\ 
 
 
 ^ ^*5 
 
 f^ 
 
 ^ 
 
 
 / 
 
 ; / .^ ? 
 
 
 ^ 
 
 
 \ 
 
 y ii' 
 
 I 
 
 
 . , . . . IJ . . . . 1 
 
 1 1 1 J 1 
 
 rrrf njrrrrrfn 
 
 '1 
 
 1 1 ' 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
 
 O) 
 
 In 
 
 !^ 
 
 m 
 
 fo o 
 
 Conductor per Mile. 
 
TABLES 105 
 
 TABLE I. (Continued.) 
 B 2 
 
 (4) K.V.A. at supply end = -j^- X K.V.A. 
 
 (5) K.W. at supply end = ^ (AP - BQ). 
 
 (6) Power factor at supply end "" 
 
 1000. . . 
 
 (in decimals). 
 (7) In-phase current at supply end = ~ in total amperes.* 
 
 (8) Reactive or quadrature current at supply end = 
 
 in total amperes.* 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 
 (9) K.W. loss in line = -- (AP - BQ - EP). 
 
 (10) Per cent efficiency of line = -^= - ^r per cent. 
 
 AJc n\) 
 
 . Total amps. 
 
 * Amperes per wire, three phase, = - _ 
 
 V 3 
 
 . . . Total amps. 
 
 Amperes per wire, two phase, = -- * 
 
106 TRANSMISSION LINE FORMULAS 
 
 TABLE II. FORMULAS FOR SHORT LINES. 
 CONDITIONS GIVEN AT SUPPLY END. 
 
 These formulas are exact when the line is short. When the line is 20 
 miles long, they are correct within approximately T V of i% of line voltage. 
 
 Conditions given: 
 
 K.V.A. = K.V.A. at supply end. 
 
 E 8 = Full load voltage at supply end. 
 cos 6 = Power factor at supply end. 
 K.W. = K.V.A. cos 6. 
 
 r = Resistance of conductor per mile. (From Tables VII- VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 / = Length of line in miles. 
 
 _. 1000 K.V.A. cos T . 
 
 Then JP = - ~ - = In-phase current at supply end 
 
 /i 
 
 (hi total amps.). 
 
 _ 1000 K.V.A. sin 
 
 Qt = - 5 - = Reactive current at supply end 
 > 
 
 (in total amperes) when current is 
 lagging. 
 
 1000 K.V.A. sin 9 
 = -- = - when current is leading. 
 
 -c-a 
 
 Find the following quantities: 
 
 Three Phase or Two Phase. Single Phase. 
 
 F = E 8 - P 8 rl - Q 8 xl F = E 8 -2 P 8 rl - 2 Q a xl 
 
 G = Q a rl - P 8 xl G = 2 Q a rl - 2 P a xl. 
 
 Formulas (capacity neglected): 
 
 (1) Voltage at receiver end = F -\ ^ 
 
 (2) Regulation of line = E 8 F -- . (Same as line drop.) 
 
 (3) Per cent regulation of line = - -^ - - per cent. 
 
 (Same as per cent line drop.) 
 
 (4) K.V.A. at receiver end = ^ X K.V.A. 
 
TABLES 107 
 
 TABLE II. (Continued.) 
 
 (5) K.W. at receiver end ^^ (FP S GQ a ). 
 
 i (FP 8 - GQ 8 ) E a _ 
 
 (6) Power factor at receiver end : 
 
 (in decimals). 
 (7) In-phase current at receiver end = - -^- - in total amperes.* 
 
 GPs + FQ 
 
 (8) Reactive or quadrature current at receiver end = ^ 
 
 F +TF 
 
 in total amperes.* 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 
 (9) K.W. loss in line = (E S P S - FP S + GQ 8 ). 
 
 (10) Per cent efficiency of line = - ^ per cent. 
 
 Total amps. 
 * Amperes per wire, three phase, = ?=- 
 
 Total amps. 
 Amperes per wire, two phase, = 
 
108 TRANSMISSION LINE FORMULAS 
 
 TABLE in. K FORMULAS FOR TRANSMISSION LINES. 
 CONDITIONS GIVEN AT RECEIVER END. 
 
 Accurate within approximately T V of i% of line voltage up to 100 miles, 
 and \ of i% up to 200 miles, for lines with regulation up to 20%. 
 
 (cycles) 2 
 K = 6 '- . K= 2.16 for 60 cycles. K = 0.375 for 25 cycles. 
 
 Conditions given: 
 
 K.V.A. = K.V.A. at receiver end. 
 
 E = Full load voltage at receiver end. 
 cos 6 = Power factor at receiver end. 
 K.W. = K.V.A. cos 8. 
 
 r = Resistance of conductor per mile. (From Tables VII-VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 / = Length of transmission line in miles. 
 
 ,, 1000 K.V.A. cos 6 
 
 I hen P = - = In-phase current at receiver end 
 
 Hi 
 
 (in total amps.). 
 
 -. 1000 K.V.A. sin ^ 
 
 Q = p; = Reactive current at receiver end 
 
 rL 
 
 (in total amps.), when current is 
 lagging. 
 
 1000 K.V.A. sin . 
 = -= when current is leading. 
 
TABLES 109 
 
 TABLE III. (Continued.} 
 
 Find the following quantities: 
 
 Full Load. 
 
 _ The above are for two- and three-phase lines. For single-phase 
 lines use 2 r and 2 x in place of r and #. 
 
110 TRANSMISSION LINE FORMULAS 
 
 TABLE in. (Continued.) 
 
 CONDITIONS GIVEN AT RECEIVER END. 
 Formulas: 
 
 Full Load. No Load. 
 
 Voltage at receiver end. 
 
 d) 
 
 (for constant supply voltage) 
 Regulation at receiver end in volts, for constant supply voltage. 
 
 A +TA 
 
 (3) - ng.X-A 
 
 N.B. The regulation at receiver end may be expressed as a percentage 
 ofE. 
 
 Voltage at supply end. 
 
 UE.-A + Z.- (5)*.-*+*! 
 
 (for constant receiver voltage). 
 Regulation at supply end in volts, for constant receiver voltage. 
 
 ,,. * , & A B<? 
 (6) A H -- j AQ -- j- 
 
 2 A 2 AQ 
 
 N.B. The regulation at supply end may be expressed as a percentage 
 
TABLES HI 
 
 TABLE III. (Continued.') 
 Current at supply end in total amperes.* 
 
 ( 7 ) VcnTo 2 . (8) 
 
 K.V.A. at supply end. 
 
 fiT.PF. at supply end. 
 
 (n) (4C + 5Z>). (12) 
 
 Power factor at supply end, in decimals. 
 
 AC + BD , , A C +B D 
 
 (13) 
 
 In-phase current at supply end in total amperes* 
 AC+BD , , 
 
 Reactive current at supply end in total amperes* 
 , . BC-AD , O s B 
 
 (I7) "' ( 
 
 2 
 
 When this quantity is positive, the current is lagging. 
 When this quantity is negative, the current is leading. 
 K.W. loss in line. 
 (19) (AC + BD -EP). (20) 
 
 IOOO IOOO 
 
 [same as No. 12]. 
 Per cent efficiency of line. 
 . 100 EP 
 
 percent ' 
 
 Total amps. 
 * Amperes per wire, three-phase, = - -= - 
 
 ^3 
 
 Total amps. 
 Amperes per wire, two phase, = - p 
 
112 TRANSMISSION LINE FORMULAS 
 
 TABLE IV. K FORMULAS FOR TRANSMISSION LINES. 
 CONDITIONS GIVEN AT SUPPLY END. 
 
 Accurate within approximately fa of i% of line voltage up to 100 miles 
 and | of i% up to 200 miles, for lines with regulation up to 20%. 
 
 K = - . K = 2.16 for 60 cycles. K = 0.375 for 25 cycles. 
 10,000 
 
 Conditions given: 
 
 K.V.A. = K.V.A. at supply end. 
 
 E s = Full load voltage at supply end. 
 cos 6 = Power factor at supply end. 
 K.W. = K.V.A. cos0. 
 
 r = Resistance of conductor per mile. (From Tables VII-VIII.) 
 x = Reactance of conductor per mile. (From Tables IX-XII.) 
 I = Length of transmission line in miles. 
 
 Then P 8 = I0 K 'Y' A ' C S = In-phase current at supply end (in 
 
 &8 
 
 total amps.). 
 
 t supply end (in 
 
 total amps.), when current is lagging. 
 
 looo K.V.A. sin 9 , , . . ,. 
 
 = - when current is leading. 
 
TABLES 
 
 TABLE IV. (Continued.) 
 Find the following quantities: 
 
 Full Load. 
 
 iooo 
 
 Kf I 
 
 No Load. 
 
 ( Ti 
 
 Viooo/ ) 
 
 Note. The above are for two- and three-phase lines. For single- 
 phase lines use 2 r and 2 * in place of r and . 
 
114 TRANSMISSION LINE FORMULAS 
 
 TABLE IV. (Continued.} 
 
 CONDITIONS GIVEN AT SUPPLY END. 
 Formulas: 
 
 Full Load. No Load. 
 
 Voltage at receiver end. 
 
 (for constant supply voltage). 
 
 Regulation at receiver end in volts, for constant supply voltage. 
 Go 2 G 2 
 
 N.B. The regulation at receiver end may be expressed as a percentage 
 of E. 
 
 Voltage at supply end. 
 
 (for constant receiver voltage). 
 Regulation at supply end, in volts, for constant receiver voltage. 
 
 F + 1 21 
 (6) E s - - - if- E.. 
 
 N.B. The regulation at supply end may be expressed as a percentage 
 of E,. 
 
TABLES 
 
 TABLE IV. (Continued.) 
 Current in total amperes* 
 
 ( 7 ) VM 2 + N* at receiver end. (8) ^M<? + N<? at supply end.- 
 
 K.V.A. 
 ( Q \ _*_ IP+GL] VM 2 + # 2 at receiver end. 
 
 7 1000 \ 2V I 
 
 ( I0 ) _JL_ a VM 2 + No 2 at supply end. 
 
 ' 
 
 1000 
 
 () 
 
 ^-(^M + G^) at receiver end. (12) 8 M at supply end. 
 
 IOOO 
 
 (13) / rm ^: GN - at receiver end - 
 
 Power factor, in decimals. 
 
 at receiver 
 
 (14) at supply end. 
 
 VM<? + No 2 
 
 In-phase current in total amperes* 
 
 (l ,\ FM + GN at recdver end> ( l6 ) MQ at supply end. 
 
 Reactive current in total amperes* 
 
 ( I7 ) GM-FN at recdver end ( Ig ) jv at supply end. 
 
 When this quantity is positive, the current is lagging. 
 
 When this quantity is negative, the current is leading. 
 
 K.W. loss in line. 
 
 ( IQ ) --(E 8 P 8 -FM-GN). (20) ^ s M (sameasNo.i2). 
 
 Per cent efficiency of line. 
 
 (21) =r-p per cent. 
 
 Total amps. 
 * Amperes per wire, three phase, = - 
 
 O 
 
 Total amps. 
 Amperes per wire, two phase, = - 
 
Il6 TRANSMISSION LINE FORMULAS 
 
 TABLE V. CONVERGENT SERIES FOR TRANSMISSION LINES. 
 CONDITIONS GIVEN AT RECEIVER END. 
 
 The convergent series give the results of the fundamental formulas as 
 accurately as desired, if a sufficient number of terms is used. 
 
 When conditions are given at the receiver end, the same as with the K 
 formulas, find the quantities: 
 
 Full Load. 
 
 2-3-4-S 2-3.4.5.6.7 
 
 2.3 2.3.4.5 2.3.4.5.. 
 
 where Z = (r +./*)* 
 
 r = resistance of conductor per mile. 
 x = reactance of conductor per mile. 
 I = length of transmission line in miles. 
 
 + etc. 
 
 g = leakage conductance of conductor per mile. 
 b = capacity susceptance of conductor per mile. 
 
 Use A, B, C, D, etc., with the equations in the third and fourth pages 
 of Table III to solve transmission line problems. 
 
 2 _ 
 
 Note i. In the formulas, A + 7- is used instead of V A 2 -f- .B 2 . 
 
 zA 
 
 This approximation may be used for very accurate work, as it is correct 
 within approximately T ^ of i% when the regulation is not more than 20%. 
 Note 2. The above are for two- and three-phase lines. For single- 
 phase lines use 2 r and 2 x in place of r and x, and use \ g and % b in place 
 of g and b. 
 
TABLES 117 
 
 TABLE VI. CONVERGENT SERIES FOR TRANSMISSION LINES. 
 
 CONDITIONS GIVEN AT SUPPLY END. 
 
 The convergent series give the results of the fundamental formulas as 
 accurately as desired, if a sufficient number of terms is used. 
 
 When conditions are given at the supply end, the same as with the 
 K formulas, find the quantities: 
 
 Full Load. 
 
 Z3 etc 
 
 2-3-4 2-3- 4-5- 
 
 + F2Z2 + - F3Z3 A + etc 
 2- 3^2- 3- 4. 5 2.3.4.5.6-7 
 
 { , FZ , F 2 Z 2 PZ 3 u \ 
 
 E 3 Y i 4 --- --- -- - -- r- etc. I . 
 
 V - 2-3 2-3.4.5 2-3-4-S-6-7 / 
 
 No Load. 
 iYZ+ ^-Y*Z*- F 3 Z 3 + ^ F 4 Z 4 - etc.Y 
 
 24 720 _ 8004 y 
 
 - I FZ -h F2Z 2 - -- F 3 Z 3 + 
 
 15 3 j s 2835 
 
 where Z = (r +jx) I. 
 
 r = resistance of conductor per mile. 
 x = reactance of conductor per mile. 
 I = length of transmission line in miles. 
 Y=(g+jb)l. 
 
 g = leakage conductance of conductor per mile. 
 b = capacity susceptance of conductor per mile. 
 
 Use F, G, M, N, etc., with the equations in the third and fourth pages 
 of Table IV to solve transmission line problems. 
 
 G 2 / - 
 
 Note i. In the formulas, F -\ -- - is used instead of V F 2 + G 3 . This 
 
 2r 
 
 approximation may be used for very accurate work, as it is correct within 
 approximately T f^ of *% when the regulation is not more than 20%. 
 
 Note 2. The above are for two- and three-phase lines. For single- 
 phase lines use 2 r and 2 x in place of r and x, and use | g and 6 in place of 
 g and b. 
 
n8 
 
 TRANSMISSION LINE FORMULAS 
 
 TABLE VII. RESISTANCE OF COPPER WIRE AND CABLE. 
 
 Data assumed: 
 
 Temperature, 20 C. (68 F.). 
 
 Conductivity of hard drawn copper, 97.3% of the annealed copper 
 
 standard. 
 Increase of resistance of cables due to spiralling, i%. 
 
 Copper Wire. 
 
 
 
 Diam- 
 
 
 Resistance, ohms per mile. 
 
 B. & S. 
 
 Circular 
 
 ctcr 
 
 
 
 gauge. 
 
 mils. 
 
 (2P) 
 
 inches. 
 
 
 Direct 
 current. 
 
 25 cycles. 
 
 In- 
 crease. 
 
 60 
 
 cycles. 
 
 In- 
 crease. 
 
 
 
 
 
 
 
 Per cent 
 
 
 Per cent 
 
 0000 
 
 2ll,6oo 
 
 .4600 
 
 . . . 
 
 .2655 
 
 .2657 
 
 .08 
 
 .2667 
 
 44 
 
 000 
 
 167,800 
 
 .4096 
 
 
 3348 
 
 3350 
 
 05 
 
 .3358 
 
 .27 
 
 00 
 
 133,100 
 
 .3648 
 
 
 .4221 
 
 .4223 
 
 03 
 
 .4229 
 
 I? 
 
 o 
 
 105,500 
 
 3249 
 
 
 5326 
 
 .5327 
 
 .02 
 
 5331 
 
 . II 
 
 I 
 
 83,690 
 
 .2893 
 
 ... 
 
 .6714 
 
 .6714 
 
 .OI 
 
 .6718 
 
 .07 
 
 2 
 
 66,370 
 
 .2576 
 
 
 .8466 
 
 .8466 
 
 .OI 
 
 .8469 
 
 .04 
 
 3 
 
 52,630 
 
 .2294 
 
 
 1.068 
 
 1.068 
 
 
 1. 068 
 
 03 
 
 4 
 
 41,740 
 
 .2043 
 
 
 1.346 
 
 1.346 
 
 
 I.346 
 
 .02 
 
 5 
 
 33,ioo 
 
 .1819 
 
 
 1.697 
 
 1.697 
 
 
 1.698 
 
 .01 
 
 6 
 
 26,250 
 
 .1620 
 
 
 2. I4O 
 
 2. I4O 
 
 
 2.140 
 
 .OI 
 
 7 
 
 20,820 
 
 1443 
 
 
 2.699 
 
 2.699 
 
 
 2.699 
 
 
 8 
 
 16,510 
 
 .1285 
 
 
 3-403 
 
 3-403 
 
 
 3.403 
 
 
TABLES 
 
 119 
 
 TABLE VII. (Continued.) 
 Copper Cable. 
 
 
 
 Diam- 
 
 No. of 
 
 Resistance, ohms per mile. 
 
 B. & S. 
 
 gauge. 
 
 Circular 
 mils. 
 
 eter 
 
 (2p) 
 
 inches. 
 
 wires 
 as- 
 sumed. 
 
 Direct 
 current. 
 
 25 cycles. 
 
 In- 
 crease. 
 
 60 
 
 cycles. 
 
 In- 
 crease. 
 
 
 500,000 
 
 .8lII 
 
 19 
 
 1135 
 
 .1140 
 
 Per cent 
 41 
 
 .1162 
 
 Per cent 
 2-34 
 
 .... 
 
 450,000 
 
 .7695 
 
 19 
 
 .1261 
 
 .1265 
 
 33 
 
 .1285 
 
 1.90 
 
 .... 
 
 400,000 
 
 .7255 
 
 19 
 
 .1419 
 
 .1422 
 
 .26 
 
 .1440 
 
 1.50 
 
 
 350,000 
 
 .6786 
 
 J 9 
 
 .1621 
 
 .1625 
 
 .20 
 
 .1640 
 
 1.16 
 
 
 300,000 
 
 .6211 
 
 7 
 
 .1892 
 
 .1894 
 
 15 
 
 .1908 
 
 85 
 
 
 250,000 
 
 .5669 
 
 7 
 
 .2270 
 
 .2272 
 
 .10 
 
 .2284 
 
 .60 
 
 oooo 
 
 211,600 
 
 .5216 
 
 7 
 
 .2682 
 
 .2684 
 
 .07 
 
 .2693 
 
 43 
 
 ooo 
 
 167,800 
 
 4645 
 
 7 
 
 .3382 
 
 .3384 
 
 05 
 
 3391 
 
 27 
 
 oo 
 
 133,100 
 
 4137 
 
 7 
 
 .4264 
 
 4265 
 
 03 
 
 .4271 
 
 17 
 
 
 
 105,500 
 
 3683 
 
 7 
 
 5379 
 
 .5380 
 
 .02 
 
 5385 
 
 .11 
 
 I 
 
 83,690 
 
 .3280 
 
 7 
 
 .6781 
 
 .6782 
 
 .01 
 
 .6785 
 
 .07 
 
 2 
 
 66,370 
 
 .2921 
 
 7 
 
 8550 
 
 8551 
 
 .01 
 
 .8554 
 
 .04 
 
 3 
 
 52,630 
 
 .2601 
 
 7 
 
 1.078 
 
 1.078 
 
 
 1.078 
 
 -.03 
 
 4 
 
 41,740 
 
 .2317 
 
 7 
 
 1.360 
 
 1.360 
 
 
 1.360 . 
 
 .02 
 
 TABLE VII. (Continued.} TEMPERATURE COEFFICIENTS OF 
 
 COPPER. 
 For different initial temperatures and different conductivities. 
 
 Ohms per 
 meter-gram 
 at 20 deg. 
 cent. 
 
 Per cent 
 conduc- 
 tivity. 
 
 
 
 15 
 
 "20 
 
 25 
 
 "30 
 
 50 
 
 .16108 
 
 95 
 
 .00405 
 
 .00381 
 
 .00374 
 
 .00367 
 
 .00361 
 
 .00336 
 
 15940 
 
 96 
 
 . 00409 
 
 .00386 
 
 .00378 
 
 .00371 
 
 .00364 
 
 .00340 
 
 15776 
 
 97 
 
 .00414 
 
 .00390 
 
 .00382 
 
 00375 
 
 00368 
 
 00343 
 
 .15727 
 
 97-3 
 
 .00415 
 
 .00391 
 
 .00383 
 
 .00376 
 
 .00369 
 
 .00344 
 
 .15614 
 
 98 
 
 .00418 
 
 .00394 
 
 .00386 
 
 .00379 
 
 .00372 
 
 .00346 
 
 15457 
 
 99 
 
 .00423 
 
 .00398 
 
 .00390 
 
 .00383 
 
 00375 
 
 .00349 
 
 .153022 
 
 TOO 
 
 .00428 
 
 .OO4O2 
 
 00394 
 
 .00386 
 
 00379 
 
 00352 
 
 I5I5I 
 
 IOI 
 
 .00432 
 
 .OO4O6 
 
 .00398 
 
 .00390 
 
 .00383 
 
 00355 
 
 where Rt is the resistance at any temperature t deg. cent. 
 
 and Rti is the resistance at any "initial temperature" ti deg. cent. 
 
 From Appendix E, Standardization Rules of the A.I. E. E., June 27, 1911. 
 
L20 
 
 TRANSMISSION LINE FORMULAS 
 
 
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 121 
 
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 TRANSMISSION LINE FORMULAS 
 
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 TRANSMISSION LINE FORMULAS 
 
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INDEX 
 
 PAGE 
 
 Absolute value of a complex quantity 57 
 
 Admittance 98 
 
 Aluminum cables, Table VIII 120 
 
 Annealed copper standard 55 
 
 Capacity 6 1 
 
 Capacity, derivation of formulas: 
 
 Cables 89 
 
 Effect of the earth 90 
 
 Two round wires 84 
 
 Two-phase 93 
 
 Three-phase, irregular spacing 93 
 
 Three-phase, regular flat spacing 97 
 
 Three-phase, regular spacing 97 
 
 Capacity susceptance 6 
 
 of cable, 25 cycles, Table XIV 128 
 
 of cable, 60 cycles, Table XVI 131 
 
 of wire, 25 cycles, Table XIII 127 
 
 of wire, 60 cycles, Table XV 130 
 
 Charging current 6 
 
 Chart, regulation, description of 8 
 
 Complex numbers 42 
 
 Condenser effect 6 
 
 Conductance 43 
 
 Conductivity of copper 55 
 
 of aluminum 55 
 
 Conductors 53 
 
 Convergent series 45, 1 16 
 
 derivation of 98 
 
 description of 41 
 
 Corona 6 
 
 Diameters of wires and cables 118 
 
 Elements of a transmission line 4 
 
136 INDEX 
 
 PAGE 
 
 Fundamental transmission line formulas 41, 98 
 
 Hyperbolic transmission line formulas 41, 98 
 
 Image wire 90 
 
 Impedance 98 
 
 " J" terms 42 
 
 "K" formulas 28, 108 
 
 "K " formulas, description of 25 
 
 Leakage conductance 43 
 
 Leakage current 6 
 
 Line drop 9 
 
 Loss in line, derivation of formula 61 
 
 Magnetic field around wire 5, 62 
 
 Power factor table 133 
 
 Problem, 5000 feet 14, 23 
 
 3 miles, 80% P.F 14, 23 
 
 3 miles, 85% P.F 12 
 
 10 miles 22 
 
 15 miles 13, 22 
 
 20 miles 24 
 
 25 miles 14, 24 
 
 75 miles 13, 15, 3 
 
 80 miles 12, 50 
 
 100 miles, 60,000 volts 39, 51 
 
 100 miles, 66,000 volts n, 36, 38, 50 
 
 100 miles, 1 10,000 volts 15 
 
 155.34 miles 39> 5 
 
 200 miles 38, 47 
 
 300 miles 39*5! 
 
 300 miles, with substation 36, 48 
 
 400 miles, with substation 4. 5 2 
 
 Quadrature volt-amperes, derivation of formula 60 
 
INDEX 137 
 
 .PAGE 
 
 Reactance of cable, 25 cycles, Table X 122 
 
 of cable, 60 cycles, Table XII 12$ 
 
 of wire, 25 cycles, Table IX 121 
 
 of wire, 60 cycles, Table XI 124 
 
 Reactance, derivation of formulas: 
 
 single-phase, wire, effect of flux in air 62 
 
 single-phase, wire, effect of flux in the conductor 64 
 
 single-phase, cable 76 
 
 two-phase 79 
 
 three-phase, irregular spacing 79 
 
 three-phase, regular flat spacing 83 
 
 three-phase, regular spacing 83 
 
 Reactance drop $ 
 
 Reactive power, derivation of formula 60 
 
 Regulation -. 8, 58 
 
 Regulation chart, description of 8 
 
 Resistance of aluminum cable, Table VIII 120 
 
 of copper wire and cable, Table VII 118 
 
 Resistance drop 4 
 
 Series, convergent 45, 116 
 
 derivation of 98 
 
 description of 41 
 
 Short line formulas 18, 104 
 
 description of 16 
 
 Skin effect . 4 
 
 Skin effect, derivation of formula 67 
 
 Skin effect formula, proof by infinite series 71 
 
 Spacing 4, 10 
 
 equivalent 10 
 
 flat 10 
 
 two-phase 10 
 
 Substation loads 26 
 
 Susceptance (see also " Capacity ") 6 
 
 Temperature coefficients 119, 120 
 
 description of 55 
 
 Watts, derivation of formula 59 
 
 Wire and cable tables n 
 
LIST OF WORKS 
 
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