FORMULAE AND TABLES FOR THE CALCULATION OF ALTERNATING CURRENT PROBLEMS McGraw-Hill Book Company Purffis/iers qfBoo/br Electrical World The Engineering and Mining Journal Engineering Record Engineering News Railway Age Gazette American Machinist Signal EnginoGr American Engineer Electric Railway Journal Coal Age Metallurgical and Chemical Engineering Power FORMULAE AND TABLES FOR THE CALCULATION OF ALTERNATING CURRENT PROBLEMS BY LOUIS COHEN .' I McGRAW-HILL BOOK COMPANY, INC. 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. 1913 COPYRIGHT, 1913, BY THE McGRAW-HILL BOOK COMPANY, INC. Stanbopc P. H.G1LSOK COMPANY BOSTON, U.S.A. PREFACE. THE importance of mathematics in the study of engineering problems is being generally recognized now and fully appreciated by engineers. The rapid development in alternating current engineering is largely due to the skillful utilization of mathe- matics in the study of the many complex phenomena which arise in connection with this subject. As a consequence the literature on alternating currents is distinctly mathematical. To be able to utilize fully all the acquired knowledge on this subject, one must have a sufficient mathematical training to enable him to follow the derivations of the many formulae, and be able to inter- pret them correctly. But even the engineer who has the mathe- matical equipment for this work is often obliged to consult a good many books and journals before he finds all of the formulae needed for his particular problem. An attempt has been made here to bring together in a small volume all the important formulae which are necessary and help- ful in the solution of alternating current problems. The aim has been throughout to put the formulae in such form that they can be immediately applied for numerical calculations; and in those cases where it seemed desirable, the formulae are illustrated by numerical examples. In a work of this kind, which is in the nature of a compilation, the author has necessarily drawn freely from all sources of infor- mation, but in most cases the formulae were gone over carefully so as to make certain that they are free from errors. Frequently the formulae were derived independently, and in all cases an effort was made to put them in the simplest possible form. A few of the formulae appear here for the first time. The author is fully aware that accuracy and completeness are of first importance in a bpok of this character. To this end not only were formulae checked with care, but references to original authorities are generally given. The author will appreci- ate a notice of errors discovered in the work, and suggestions 271056 VI PREFACE for additions of other material which might increase the useful- ness of the book. Both the scope and arrangement of the present work seem logical to the author, although a' great deal of material could have been added, but not without the risk of making the book "too cumbersome and expensive for general use. Many thanks are due to Prof. H. H. Norris of Cornell Univer- sity for his careful reading of the manuscript and the many valu- able suggestions he has made. I am also indebted to my friend Dr. P. G. Agnew of the Bureau of Standards for going over care- fully parts of the manuscript and checking some of the numerical examples. The diagrams were prepared by Mr. B. Holland and Mr. A. F. Van Dyck to whom I wish to express my thanks. I wish also to express my appreciation for the valuable assist- ance rendered by the publishers and their technical staff in the editing of this book. LOUIS COHEN. WASHINGTON, D. C,, September, 1913. CONTENTS. CHAPTER I. PAGE RESISTANCE AND EDDY CURRENT LOSSES IN METALLIC CONDUCTORS. 1 Alternating current resistance of straight cylindrical conductors; series formula; high frequency formula; general formula in terms of her and bei functions Current penetration in cylindrical conductors Resistance of looped conductor when the to and fro conductors are close to each other Resistance of concentric conductors; formulae for low-frequency currents, high-frequency currents, and very high-frequency currents Resistance of con- centric conductors with hollow inner conductor Flat conductors; alternating current distribution; alternating current resistance; depth of current penetration Resistance of single layer coils for alternating currents; general formula; approximate formula suitable for high frequencies; approximate formula suitable for low frequencies Alternating current resistance of large slot wound conductors Distribution of induction and eddy currents in a long iron pipe, enclosing an alternating current Effect of grounded lead covering on the resistance and reactance of single conductor cable Effect of an ungrounded lead covering on the resistance and reactance of a single conductor cable Leakage conductance; between a cylindrical conductor and an infinite plane; between two parallel conductors; three core cable Magnetic flux distribution in iron plates; flux density; total flux; equivalent depth of pene- tration; energy loss Magnetic flux distribution in iron cylinders; flux density; total flux; eddy currents; energy loss; equivalent resistance and inductance Magnetic flux distribution in iron core of toroidal coil; total flux; effective resistance Table to facilitate the computation of the resistance of cylindrical conductors, and the magnetic flux distribution in cylindrical conductors Table of specific resistance of metallic wires. CHAPTER II. INDUCTANCE 48 Mutual inductance of two coaxial circles Attraction between circular currents Mutual inductance of two coaxial coils Mutual inductance of concentric coaxial solenoids Self inductance vii Viil CONTENTS PAGE of a single layer coil or solenoid Correction factor for thickness of insulation on wires The self inductance of a circular coil of rectangular section Self inductance of circular ring Empirical formula for the self inductance of coils Self inductance of a rect- angle Mutual inductance of two equal parallel rectangles Self inductance of toroidal coil Inductance of linear conductors Effect of frequency on the inductance of linear conductors Inductance of three-phase transmission lines Mutual inductance of two metallic circuits suspended on the same pole Inductance of grounded conductors Inductance of split conductors; con- ductor split into two parts; conductor split into three parts Self inductance of two parallel hollow cylindrical conductors Self inductance of concentric conductors; triple concentric conductors Self inductance of a circuit formed by three conductors whose axes lie along the edges of an equilateral prism Compound conductors Series or parallel arrangement of inductance Inductance of wires in parallel Table for calculating attractive force between two circular currents Table of values of the end correction in the calculation of the self inductance of solenoids Table of shape factors for certain coil proportions, to be used in the empirical for- mula Table of inductances of complete metallic circuits. CHAPTER III. CAPACITY 86 Ratio of the electrostatic to the electromagnetic unit of capacity Practical unit of capacity Parallel or series arrangement of condensers Capacity of parallel plates; circular plates; correction for edge effects Capacity multiple plate condenser Capacity of concentric spheres; centers of spheres separated by a small dis- tance Capacity of disk insulated in free space Capacity coeffi- cients of two spheres The laws of attraction and repulsion between spheres when the potentials are given Capacity of linear con- ductors; one overhead wire, the ground as the return conductors; two overhead wires; three overhead wires; four overhead wires Capacity of condenser formed by two long parallel cylinders of equal radii; unequal radii; influence of the earth as a nearby con- ducting plane Capacity two metallic circuits suspended on the same pole Capacity three-phase transmission lines Capacity of horizontal antennae Capacity concentric cylinders; axes of cylinders displaced with respect to each other Capacities of cables; concentric cable; triple concentric cable Capacity of two core cable; effective capacity under different conditions Capa- city of three core cables; effective capacity under various conditions List of capacities that can be obtained from a three core cable Table of specific inductive capacities of solids Table of specific inductive capacities of liquids. CONTENTS ix CHAPTER IV. PAGE ALTERNATING CURRENT CIRCUITS 126 Complex waves Characteristic features of different forms of alternating current and pressure curves Effective value of com- plex wave Form factor Curve factor Current and phase angle in circuit of inductance and resistance Current and phase . angle in circuit containing capacity and resistance; voltage across condenser Circuits containing inductance and capacity; res- onance condition; voltage across condenser Impedances in series Parallel circuits each having inductances and resistance A system of branched circuits in parallel each having inductance and resistance Inductance and resistance shunted by condenser; resonance condition Inductive load shunted by condenser Two branch circuits in parallel, one having inductance and resistance, and the other inductance capacity and resistance The branched circuits as well as the main circuits having inductance capacity and resistance Mutual inductance between the branched circuits Power factor; simple wave; complex wave Power trans- mission; maximum power; efficiency; ratio of e. m. f. at receiver and generator Power transmission inductive load; maximum power; efficiency; ratio of e. m. f. at receiver and generator Power transmission calculations; assuming capacity of the line shunted across the middle of the line; half of the line capacity shunted across each end of the line Air core transformer Resonance transformer; condition for maximum current in the secondary circuit Transformer having a condenser only in the secondary circuit Transformer having condenser in the primary circuit Iron core transformer; general design formula; efficiency; ratio of terminal voltages for inductive and non-inductive load; regulation; phase angles Current transformer; ratio of transfor- mation; phase angle. CHAPTER V. TRANSIENT PHENOMENA 171 Rise and decay of currents in an inductive circuit on closing and opening the circuit Rise and decay of current on closing and opening on alternating inductive circuit Current and volt- age on condenser on closing a direct current circuit containing resistance and capacity, no inductance Circuit containing re- sistance inductance and capacity; current and voltage on charging and discharging for logarithmic case; critical case; trigonometric case Frequency of oscillations; damping; logarithmic decrement Effective current Electrostatic energy Alternating current circuits containing resistance, inductance and capacity; transient term on closing and opening circuit; logarithmic case; critical case; trigonometric case Divided circuits; transient terms for change X CONTENTS PAGE in resistance in either branch Transient currents in transmission lines; load short circuited; generator short circuited Mutual in- ductance; values of transient terms which arise on sudden change in resistance in primary or secondary circuit; current rise in circuits when an e. m. f . is introduced in one circuit Oscillation trans- former having inductance and capacity in series; currents and voltages in the circuits; frequencies; ratio of voltages Direct coupling; syntonized circuits; frequencies. CHAPTER VI. DISTRIBUTED INDUCTANCE AND CAPACITY 200 The general equations for current propagation along wires Propagation constant of the line; attenuation constant; velocity constant Approximate values of the propagation constants; the inductance is large; the inductance negligibly small Prop- agation constants expressed as a function of the ratio of inter- axial distance and radius of wire General solution for current and voltage on line expressed in complex quantities and hyperbolic functions Infinite line; current and voltage Line of finite length and open at receiving end Short lines Line grounded at receiving end Line short circuited at receiving end Power transmission; voltage and current given at the receiving end; voltage and current given at the generator Hyperbolic formulae Approximate formulae for short lines Generator voltage and impedance at receiving end known Transient phenomena in circuits having distributed resistance and capacity; cable open at one end and a constant continuous e.m.f. suddenly applied at the other end A charged cable having a constant e.m.f. applied between one end of the cable and ground is suddenly grounded at the other end Circuits containing distributed resistance, inductance and capacity; line oscillations Line is grounded at one end and open at the other end Line grounded at both ends Line open at both ends Line closed upon itself Table of hyper- bolic functions of complex quantities. CHAPTER VII. MATHEMATICAL FORMULAE 264 Exponential and logarithmic formulae Trigonometric formulae Hyperbolic formulae Complex quantities Miscellaneous for- mulae Miscellaneous functions Interpolation formula Table of Naperian logarithms Table of exponential and hyperbolic functions Table of trigonometric functions Table of binomial coefficients for interpolation by differences. INDEX.. 277 ALTERNATING CURRENT PROBLEMS CHAPTER I RESISTANCE AND EDDY-CURRENT LOSSES IN METALLIC CONDUCTORS THE resistance of metallic conductors is higher for alternating currents than for continuous currents due to the so-called skin effect. In the case of alternating currents the distribution of the current is not uniform throughout the entire cross section of the conductor, but tends to concentrate more towards the surface of the conductor, and as a consequence there is a corresponding in- crease in resistance. The increase in resistance for alternating currents depends on the conductivity and permeability of the conductor and on the frequency. The higher the conductivity, permeability and frequency the greater the increase in resistance. Besides the PR loss, there is another source of energy loss in alternating current circuits which does not occur in continuous current circuits. This is the eddy-current loss in outside con- ductors. When an alternating current passes through a con- ductor it creates an alternating magnetic field which will induce currents in conductors placed in that magnetic field and thus causes loss of energy. This is particularly marked in the case of iron bodies placed in a strong alternating magnetic field as in the case, for instance, of a cylindrical iron core inserted within a cylindrical coil. This loss of energy due to eddy currents in out- side conducting bodies may be looked upon as an equivalent in- crease in resistance of the circuit. In fact the outside conductor acts like a closed secondary of a transformer which absorbs energy from the primary. In this chapter on resistance formulae we shall consider not only the resistance of conductors proper, but also the increase in effective resistance caused by eddy-current losses in metallic bodies which are placed in magnetic fields. The general problem of determining the distribution of alter- nating currents in conductors of any shape offers considerable 1 FORMULAE AND TABLES FOR THE mathematical difficulties, and has only been worked out for a limited number of special cases. We shall endeavor to give here all the important available formulae with numerical examples to illustrate their application. Straight Cylindrical Conductors. The problem of determin- ing the effective resistance of wires for alternating currents has engaged the attention of many able physicists and several for- mulae have been deduced for this case. Owing to the importance of this problem in engineering practice, we shall give here all the formulae available, with a brief discussion of their limitation and application. In what follows let R = resistance for alternating currents. RO= resistance for continuous currents, co = 2 TT X frequency = 2 irn. I = length of wire in cm. ju = permeability. p = specific resistance in C.G.S. units. d = diameter of wire in cm. The permeability JJL is assumed to be constant. The form in which the formula for the resistance of cylindrical conductors for alternating currents was first given is as follows:* For non-magnetic materials 1 a, 4 * 4 (1) (2) . ****<* V I . . . (3) ~ "" Formula (1) may also be written in the following form: Formulae (1) to (3) are applicable only for low frequencies. When the diameter of the wire is not very small and the fre- quency fairly high, the series in the above formulae is very slowly convergent, and it will require more terms in the series for an accurate determination of the resistance. * See J. A. Fleming, "The Principles of Electric Wave Telegraphy and Telephony," second edition, p. 117. CALCULATION OF ALTERNATING CURRENT PROBLEMS 3 Example. Copper wire No. B. & S. d = 0.8252 cm., p = 1700, n = 500. R = Ro }1 + 0.081 - 0.0055 + -. . . | . 1.076 R Q . This is a rapidly convergent series. The increase in resistance for this case is about 7.6 per cent. If, however, in the same example we put n = 1500 we get R = Ro\l+ 0.729 - 0.445 +} The series is very slowly convergent; it will require therefore many more terms to determine the value of R accurately. For still higher frequencies or larger diameter of wire, the series in formulae (1) to (3) will be still more slowly convergent, and there- fore the formulae will not be suitable. In the case of magnetic material, as that of iron wire, the formulae is only applicable for very low frequencies. When the frequency is very high, we have the following con- venient formulas,* R = R = 8 p For copper wire, /* = 1, p = 1700 at ordinary temperature, and formula (4) reduces to Formulae (4) and (5) are applicable only when - has a value P much greater than unity, of the order 10 or more. It is obvious that the above formulae (1) to (5) have only a limited application. Formulae (1) to (3) can be used only when is small, and formulae (4) and (5) when -- - is large. Physi- cally it means that the first set of formulae are applicable to cases in which current traverses almost the entire cross section of the conductor, while formulae (4) and (5) are applicable to cases in which the current is confined mostly to a thin surface layer of the conductor. * Lord Rayleigh, "The Self-induction and Resistance of Straight Con- ductors." Philosophical Magazine, Ser. V, May, 1886, Vol. 21, p. 381. 4 FORMULAE AND TABLES FOR THE Recently Professor Pedersen* extended the work on this prob- lem so as to cover all cases of current distribution; he also worked out a valuable set of tables which are reproduced here, see Table II, p. 43. The formula is as follows: R = R r (ma), (6) . ^ _ ma ber (ma) bei' (ma) bei (ma) ber' (ma) , . '^~ ber" (ma) + bei' 2 (ma) where p and a = radius of wire. The value of r (ma) for ma from to 6 is given in Table II and for values of ma larger than 6, the expression for r (ma) as given in (7) reduces to ft r (ma) = 0.35355 ma + 0.25 + - (8) ma In Table I, p. 42, are given the values of m for copper wire for different frequencies. By the aid of Tables I and II and formulae (6) and (8) we can compute quite readily the alternating current resistance of wires of any diameter and all frequencies. To further facilitate the work of computing the resistance of copper wires there are given in Figs. 1 and 2 curves expressing the TT> value 77- as a function of Van for values of Van from 1 to 1000. /to Example. Copper wire No. B. & S., d = 0.8252 cm. p = 1700, n = 500, m = 4.824, ma = 1.99. From Table II we get r (ma) = 1.077, hence R = 1.077 R , which agrees with result obtained for the same example by formula (3). If we put in the above example n = 1500 we have m = 8.346, ma = 3.443. * P. O. Pedersen, Jahrbuch der DrahOosen Telegraphic und Telephonie, Vol. 4, pp. 501-515. CALCULATION OF ALTERNATING CURRENT PROBLEMS 5 4.0 3.5 3.0 2.5 < 2.0 1.5 1.0 y x x X 4 X x X X X / % X X Curve showing the change in resistance of copper wire for alternating currents R = Resistance at frequency (n) n = Frequency ( cycles per sec. ) RO= Resistance for direct current a => Radius of wire in centimeters / Curve showing the change in resistance of copper wire for alternating currents R= KR R = Resistance at frequency ( n ) n = Frequency ( cycles per sec. ) RQ= Resistance for direct current Q> = Radius of wire in centimeters x ' x * X X / 100 200 400 500 600 a FIG. 2. 700 900 1000 6 FORMULAE AND TABLES FOR THE From Table II, r (ma) = 1.470, hence R = 1A7 R Q . For a frequency 100,000, m = 68.22, ma = 28.15. Computing r (ma) by formula (8) we get r (ma) = 10.2, hence R = 10.2 R . Current Penetration in Cylindrical Conductors. It was pointed out above that the increase in resistance for alternating currents is due to the fact that the current is not uniformly dis- tributed over the entire area of the conductor. The intensity of the current diminishes from the surface to the center of conductor. The amplitude of the current at any point distance r from the axis is given by the formula * where IQ = current density on surface of conductor. a = radius of conductor in cm. e = base of (Naperian) logarithms. There is also a change in phase in the current from point to point along the radial axis. The intensity of the current dimin- ishes in a geometric progression as the distance from the sur- face increases in arithmetical progression. For a distance from the surface a r = V the density of the current will be reduced to - = s-== its surface value. V is taken as the 2.77 V 2 7T/ICO measure of the thickness of current penetration, that is, we may consider the current concentrated in a cylindrical shell of thick- ness \~^ ) and the sectional area of the shell as the effective area for the passage of the current. Denoting by 5 the depth of current penetration, that is, the distance from the surface at * See J. J. Thomson, " Recent Researches in Electricity and Magnetism," p. 281. CALCULATION OF ALTERNATING CURRENT PROBLEMS 7 which the current density is - part of the density at the surface,* We have for copper wire (p = 1700) for aluminum wire (p = 2800), = M . \ and for iron wire (p = 10,000, n = 1000) 5= S" Example. Let n = 500. For copper wire, 5=-^ = 0.29 Vsoo that is the current penetrates to 0.29 times the radius below the surface of the conductor. Assuming the radius of the conductor to be 1 cm., the total area of the conductor is TT sq. cms. The area through which the alternating current is passing is TT (I 2 0.7 1 2 ) = 0.496 TT sq. cm. This is about one-half the actual area and there- fore the effective resistance is about double. By formula (6), we have for this case, m = 4.824, ma = 4.824, and r (ma) = 1.98. The alternating current resistance is 1.98 times the continuous current resistance, which agrees closely with the value obtained above. Return Conductor. When conductors are placed close to each other, as in the case of cables, the current distribution is effected by the presence of the return conductor. The greatest current density occurs on the sides of the conductors which face each other. For two conductor cables and low frequencies up to * See " Theorie der Wechselstrome " von J. L. LaCour und O. S. Bragstad, Zweite Auflage, p. 568. 8 FORMULAE AND TABLES FOR THE about 2000 cycles per second we have the following formula for the alternating current resistance.* where d = diameter of wire and h = interaxial distance between wires. For copper cables formula (11) may be put in the following more convenient form For aluminum cables For magnetic materials, the distance between the wires has very little influence on the current distribution, and we may therefore use the same formula as that for a single conductor. Example. Two-conductor copper cable of No. 1 B. & S. wires. Let n = 300, d = 0.73 cm., h = 1 cm. R=R \l + [0.70 + 0.40] 0.0256 - [0.4 + 4.256] 0.00065} = #o \ 1 + 0.028 - 0.0030| = 1.025 R Q , an increase in resistance of about 2.5 per cent. For a single conductor of the same diameter and the same frequency we have R = R (1 + 0.018 - ) = 1.018 Bo, an increase of only 1.8 per cent, hence the presence of the other conductor introduced another increase in resistance of 0.7 per cent. For very high frequencies the resistance of a two-conductor cable is given by the expression: 2 - Formula (12) is applicable only when - iV- is small com- 7T d H pared with unity. * G. Mie, Wied. Ann., 1900. See also "Theorie der Wechselstrome," von J. L. LaCour und O. S. Bragstad, Zweite Auflage, p. 569. CALCULATION OF ALTERNATING CURRENT PROBLEMS 9 Example. Let d = 0.5 cm., h = 0.75 cm., p = 1700 (copper), n = 10 6 . t = 0.05, Trd V n which is small compared with unity, and hence we may use formula (12) for this case. R = Ro (51.1 + 0.125 - 0.335) = 50.9 R . The resistance is about fifty-one times the resistance for con- tinuous currents. For a single conductor of the same diameter and the same fre- quency, the resistance is only about twenty times the continuous current resistance. Resistance of Concentric Conductors. This problem has been investigated in a very able manner by Dr. Alexander Russell,* who derived a general expression for the resistance and inductance of concentric conductors and from it he deduced several simple formulae suitable for low frequencies, high frequencies and very high frequencies. The general expression is somewhat unwieldy and not convenient for numerical calculations, and we shall there- fore give here only the simplified formulae for particular cases. Inner Conductor Solid Metal Cylinder. Let a = radius of inner conductor in cm. b = inner radius of concentric outer cylinder in cm. c = outer radius of concentric outer cylinder in cm. n = frequency. p = resistivity in C.G.S. units. m = 2ir P For direct currents we have The first term in the right hand part of equation (13) is the resistance of inner conductor and the second term the resistance of outer conductor. * Alexander Russell, " The Effective Resistance and Inductance of a Con- centric Main, and Methods of computing the Ber and Bei and Allied Functions," Phil. Mag., April, 1909. 10 FORMULAE AND TABLES FOR THE For low-frequency currents (ordinary commercial frequencies 25 to 60 cycles) where Cm*a*\ 1 + 1927 + Ro11 + * (pl + p * + P3?)}, (14) (7c 2 -6 2 )(c 2 -6 2 ) 6 2 c 4 Pl = ~192~ p2 = 8(c 2 -& 2 )' 4(0* - For high-frequency currents, when ma is greater than 5, that is >5 or p 540 and for copper, n > - the following formula is applicable: R== p_ (1.1. 8\/2ra 2 a 2 ) pm sinh m (c - 6) A/2 + sin m (c - b) V2 2 V2irb coshm (c - 6) A/2 - cosm (c - b) \/2 The first term gives the resistance of the inner conductor and the second term the resistance of the outer conductor. When the frequency is very high and c is not nearly equal to 6, we have the following expression for the resistance of a concentric conductor : (16) 2 27T&2 \a Example. Let the radius of inner conductor, a, be 1 cm. and the external and internal radii of outer conductor 2 and 1.73 cm. respectively. This will give equal sectional areas for the two conductors which is usually the case in practice. If we assume a frequency of 25, we get m =27r\/ = 2 ?r V/ T^ p " 1 / UU = 1.08, and for this value of m with the radii as given in this example, formula (14) is applicable. Introducing these constants in (14), CALCULATION OF ALTERNATING CURRENT PROBLEMS 11 m 4 = 1.36, pi = 0.1302, P2 = 6, P3 = - 48, { = log e ^ = 0.1438 and R = Ri T) L36 x a0004) = # (1.007) + #o (1.00054). The resistance of inner conductor is increased by 0.7 per cent while the resistance of outer conductor is increased only by 0.05 per cent. The increase in resistance of inner conductor is about fourteen times that of outer conductor. If in the above example we make n = 1000, m = 2r V = 7.4, p ma is greater than 5 and we may therefore use formula (15) sinh m (c - b) A/2 = 8.2, cosh m (c - b) \/2 = 8.3, sin m (c - 6) V2 = 0.332, cos m (c - b) \/2 = - 0.943, 8.2 + 0.332 R = 1987 (0.707 + 0.068 + 0.005) + 818 X = 10 9 (1550 + 753). 8.3 + 0.943 For direct current the resistance is R = 10 9 (541 + 541). The resistance of inner conductor was increased to about three times the direct current resistance, while the resistance of outer conductor was increased only to one and five-tenths tunes the direct current resistance. Concentric Conductors with Hollow Inner Conductor. If we denote by a\ and a 2 the internal and external radii of inner conductor, and as before c and 6 denote the outer and inner radii of outer conductor, we have for low-frequency currents correspond- ing to formula (14), 7r(a 2 : + w 4 02 192 a 2 7T (C 2 ~ 6 2 ) 6V 8 (c 2 - 6 2 ) 6V (17) 12 FORMULAE AND TABLES FOR THE x-*- For high frequencies corresponding to formula (16), we have pm sinh m (a? a\) V2 + sin m (a* ai) V2 2 Trai A/2 cosh m(a2 ai) \/2 cos m(az a\) A/2 \ (18) pm sinh m (c b) v 2 + sin m (c b) v 2 2 7r6 V2 cosh m(c - 6) V2 - cos m(c-b) V2 In the above two formulae, (17) and (18), Ri is the resistance of inner conductor and RO the resistance of' outer conductor. Flat Conductors. Resistance and distribution of current density in flat conductors with alternating currents.* Let Fig. 3 represent a section of the conductor, the current flowing in the direction perpendicular to the plane of the paper. We assume that the conductor is very thin compared with its width, as in the case of flat ribbon conductors. v Let 2 a = thickness of conductor in cm. p specific resistance in C.G.S. units. n = frequency. -^ p = permeability. FIG ' 3 - m=2*\J^. The current I x at any point distant x from the center is given by the expression E f cosh 2 mx + cos 2 mx j>? ,^. x ~ p { cosh 2 ma + cos 2 ma ) Formula (19) gives only the amplitude of the current at any point in the conductor, but there is also a continuous change in phase of the current from point to point from the surface to the center of the conductor. The current density at the center of the conductor, x = 0, is (20) E p (cosh 2 ma + cos 2 ma) 2 " * See Ch. P. Steinmetz, "Theory and Calculation of Transient Electric Phenomena and Oscillations," Chapter VII. CALCULATION OF ALTERNATING CURRENT PROBLEMS 13 The mean value of the current for the entire section of the conductor is E \ cosh 2 ma cos 2 ma )? ,~i \ m pma V2 I cos h 2 raa + cos 2 ma ) If we denote by R the direct current resistance and by R the effective resistance for alternating currents, we have R /p. ( cosh 2 ma + cos 2 ma ^ /00 , = ma V2 < - 7-jr - ! - jr - > (22) .Ro ( cosh 2 ma cos 2 ma ) Special Cases. If 2 ma is sufficiently large so that we can neglect cos 2 ma as compared with cosh 2 ma, formula (22) reduces to = 27T 2\a. (23) o > p The increase in resistance is directly proportional to the thickness of plate, the square root of frequency and permeability, and inversely as the square root of the resistivity. If in formula (19) 2 ma is sufficiently large so that cos 2 ma and e -2ma can b e neglected compared with e 2ma , then for points not far below the surface of the conductor, formula (19) assumes the form E E I x =*-*<-*> = --", (24) P P where s = a x is the distance below the surface of conductor. The density of the current diminishes in geometric progression as the distance from the surface increases in arithmetic progression. For the value of ms = 1, the current density is reduced to about one-third its maximum value, and from that point on the current density diminishes with great rapidity; hence we can take the value, . (25) as the measure of the thickness of current penetration, tha*t is, we may consider the current flowing through only two surface layers of the conductor of thickness s given by formula (25). We have obtained an identical expression for the case of cylindri- cal conductors; see formula (10). It is obvious, of course, that in the design of conductors for alternating currents, the mere increase 14 FORMULAE AND TABLES FOR THE in thickness beyond a certain point will not appreciably decrease the resistance, since the current flow will always be practically limited to two surface layers of thickness s = m Example. Copper conductor, p = 1700. For n = 100, s = 0.67 cm. n = 1000, s = 0.21 cm. n = 50,000, s = 0.03 cm. For frequencies, therefore, of 100, 1000 and 50,000 it would be a waste of material to increase the thickness of conductors beyond 1.34 cm., 0.42 cm. and 0.06 cm., respectively. An additional in- crease in thickness of conductor would diminish the direct current resistance, but it would increase in the same proportion the ratio 7-> -5 so that the high-frequency resistance would remain sensibly constant. In the case of iron conductors, even for comparatively low fre- quencies the current flow is limited to very thin surface layers of the conductors. Example. Iron plate, p = 10 4 , /* = 10 3 . For n = 50, s = 0.07 cm. n = 100, s = 0.05 cm. n = 500, s = 0.022 cm. n = 1000, s = 0.016 cm. Resistance of Coils with Alternating Currents. The effect of frequency on the increase of resistance of coils is more marked than that of straight conductors. The problem has been treated mathematically by several investigators and formulae have been derived by which the alternating current resistance of coils can be determined quite accurately. In formulae (26) to (35) given below, the following notation is used. a = radius of coil in cm. r = radius of wire in cm. s = number of turns per cm. p specific resistance in C.G.S. units. o> = 2 TT X frequency = 2 wn. R Q = continuous current resistance. R = alternating current resistance. CALCULATION OF ALTERNATING CURRENT PROBLEMS 15 A very simple formula, but of very limited application, has been given by Prof. Wien:* 0.272 V p / This formula is applicable only when the frequency is low or the radius of the conductor is small, that is, Besides this the coil must be long compared with its diameter. The length of the coil should be greater than seven times its diameter. A general formula derived by the author, f which is applicable for all frequencies and any size of wire, provided only the wind- ings are close together and the coil is fairly long compared with its diameter, is the following: p = P i _ x P x Ti* 2 (a* 2 + & 2 ) cosh4or-cos4&r ; where 16 7T 5 m = ~ and x = 1, 3, 5, 7, . . . . When r is small (1 or 2 mm.) and the frequency is fairly high then, to a very high degree of approximation, 256 s 2 o> 2 7rra ( 28 ) When the frequency is high and the radius of the wire is not very small, so that we may neglect m 4 compared with ^-, formula * M. Wien, Ann. d. Physik, 14, 1, 1904. t Louis Cohen, Bulletin Bureau of Standards, 4, 162, 1907. 16 FORMULAE AND TABLES FOR THE (27) reduces to - (29) P ) When the frequency is so low that -- ^ may be neglected com- pared with w 4 , formula (27) reduces to , 107rW> , qm 1 + -- 2 -- > (30) Prof. Sommerfeld * derived the following two formulae corre- sponding to (29) and (30). For very high frequencies, when ~ V - - > 6, (2-). (3D The value of (2 TITS) has not been determined, but from the experimental results of Black f the following values of < were obtained : 2 rs = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 1.00 = 1, 1, 1.1, 1.27, 1.45, 1.67, 1.92, 2.2, 2.6, 3.1, 3.7. For low frequencies, (32) R = Example. r = 0.05 cm., w = 27rX5X10 5 , s = 8, p = 1700. By formula (29) we get /o _ 2 \/ K \/ i r>5 = 6.43. By formula (31) we have for 2 rs = 0.8, < = 2.6 p rj ric /t U.Uc) . . ^ 1700 =7.02. * Sommerfeld, Ann. d. PAi/s., 24, 609, 1907. See also A. Essau, Jahrbuch d. Drahtlosen Teleg., Vol. 4, 490, 1911. t Black, Ann. d. Physik, 19, 157, 1906. CALCULATION OF ALTERNATING CURRENT PROBLEMS 17 If in the above example we put co = 2 TT X 5 X 10 3 , we may apply formulae (30) and (32). By (30) R 107 (125) 2 X IP" 12 X 64 X 4 7T 2 X 25 X 10 6 _ 1 m7 R Q ~ (1700) 2 By (32) ?L - _u J7T 2 X 47r 2 X 25 X 10 6 X 625 X IP" 8 o " (1700) 2 \ + (2.51) 2 + (2.51) 4 + . . . = 1.036. It is seen that in the above examples the agreement between the two sets of formulae is fairly good. It may be of interest to compare the difference in the increase in resistance for a coil and straight wire. Using the same data as in the above example, r = 0.05 cm. and n = 5 X 10 5 , we have by formula (6) for the case of straight wire, jr = r (ma), (6') m = 2ir r (ma) = 0.35355 ma + 0.25 + = 2.91, hence = 2.91. KQ For the same size of wire and the same frequency the wire wound 73 into a coil, jr- 6.43: hence winding the wire into a coil has more tio than doubled its high-frequency resistance. When the frequency is neither very high nor very low, and the windings are close together, the general formula (27) should be used. Dr. Nicholson * has derived an expression for alternating * J. W. Nicholson, Phil. Mag., 19, 77, 1910. 18 FORMULAE AND TABLES FOR THE current resistance of coils which is only applicable when the windings are at some distance from each other, as follows: (33) ?L = j_ * r 4 7T 2 co 2 _ J_ /r 4 7T 2 co 2 \ 2 flo + 12 P 2 180 \ p 2 I ' where a is the angular pitch of the winding. Formulae (26) to (33) are applicable only for single layer coils. For coils of more than one layer we have the following formula given by Wien: R _ 2 ( - = L+ - " where TI and r 2 are the inner and outer radii, I the length of the coil and ra is the number of layers. This formula holds only for coils of length at least eight times the diameter and when the frequency is not very high, f3ry- -<3y- For flat coils, length of coil small compared with diameter, we have* R 1 . rT* S i i 3 ri 2 ) . . = ' 1 + - Alternating Current Resistance of Large Slot-wound Con- ductors. In the case of alternators, induction motors, etc., the conductors are put into iron slots, that is, the conductor is surrounded by iron on three sides. The eddy current losses in such conductors carrying alternating currents are considerable, in some cases many times the PR loss. That is, the effec- tive resistance for alternating currents, because of the eddy currents, may be several times the resistance as calculated from the dimensions and specific resistance of the conductor. This problem was discussed in an interesting paper by Mr. Field, f who obtained a formula giving approximately the effective resistance of the conductor. He also derived expressions for the * A. Essau, Jahrbuch der Drahtlosen Telegraphic und Telephonic, Vol. 4, p. 490, 1911. , t A. B. Field, Proc. A.I.E.E., 24, 659, 1905. CALCULATION OF ALTERNATING CURRENT PROBLEMS 19 current distribution, density and phase, in the entire sections of the conductors. If we denote by R Q the direct current resistance and by R the alternating current resistance, we have for the m th layer RQ A (m 2 m) (cosh af cos af) (sinh af since/) -f- (sinh 2 af sin 2 af) cosh 2 af cos 2 af (36) FIG. 4. FIG. 5. where / = depth of conductor in cm. (if a laminated conductor take the gross depth). a 0.145 V. n = frequency. r 2 = practically unity for solid conductors, and for laminated conductors is generally equal to the ratio of half the mean length of turn to gross length of core. r\ = for solid conductors, the ratio of net copper measured across the slot, to the slot width, and for laminated con- ductors generally the same multiplied by the ratio of net to gross conductor depth. 20 FORMULAE AND TABLES FOR THE m denotes the m th layer, counting the number of layers from the bottom of the slot upwards. RQ To facilitate the computation of -5- as given by equation (36) K Mr. Field worked out a set of curves, which are reproduced in n Fig. 6, giving the values of -^- for different values of af and m. As an illustration consider the problem of a two-layer winding having the following dimensions: Slot width = 0.65 in. Conductors, 2-1 (0.4 X 0.75 in.). 7-2 = 1, n = ~ = 0.615, / = 0.75 X 2.54 = 1.905. U.oo For 25 cycles, a = 0.145 \/25 X 0.615 = 0.568, af = 1.08. For 60 cycles, a = 0.145 A/60 X 0.615 = 0.88, af = 1.68. From the curves given in Fig. 6 we obtain the following values t R for jc-: 25 cycles. 7- Bottom conductor, m 1, ^- = 1.12. /to Top conductor, m = 2, TT = 2.00. lQ 60 cycles. r> Bottom conductor, m = 1, "5" = 1-55. R Top conductor, m = 2, -5- = 5.6. /to Mr. Field discusses at some length the relative advantages of different form of windings and the original paper should be con- sulted by those who are particularly interested in this subject. Distribution of Induction and Eddy Currents in a Long Iron Pipe, Enclosing a Conductor Carrying an Alternating Current. It is now generally appreciated that single core cables designed to carry alternating currents cannot be laid in separate iron pipes, on account of the large eddy current and hysteresis losses which will be produced by the magnetic induction in the iron pipes. It is CALCULATION OF ALTERNATING CURRENT PROBLEMS 21 Values of af 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.6 1.4 1.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 t2 1.3 Values of af FIG. 6. 1.4 22 FORMULAE AND TABLES FOR THE interesting, however, to know what the energy loss is likely to be in any particular case, and we give here the formulae as developed by Mr. Field* for the flux distribution and energy loss in iron pipes enclosing single core cables which carry alternating current. Let I = current amplitude in -cable, h = thickness of pipe wall in cm., d = internal pipe diameter in cm., L = mean circumferential length of pipe in cm., JLI = mean permeability, 4717,1 ' uT. - m= 0.0002V/, p w = 2 TT X frequency = 2 7m, p = specific resistance (electrical) iron, p _ Im . /cosh mh sin mh L V cosh mh + cos mh ' B x = induction in lines per sq. cm. at point distant x from surface, I 9 = current density in amperes per sq. cm. of the longitudinally flowing current at point distant x from surface, W = eddy current loss per cm. length of pipe expressed in watts. Two cases are considered: (1) Where the pipe is laid in dry concrete or dry earth, so that no eddy currents leak out or leave the pipe. (2) Where the pipe is laid in wet soil, or where the ends are for any other reason at the same potential. CASE 1 I x = Ple m *sm(a>t + mx + $)-e m ( h -*)s[ut+m(h-x) + ]\. (37) The amplitude of I x may be put in the following form: 7 _ -^m . /2 f cosh m (h 2 x) cos m (h 2 x) j x ~ L V cosh mh + cos mh * M. B. Field, U A theoretical consideration of the currents induced in cable sheaths and the losses occasioned thereby," Journal Inst. of Elect. Eng., Vol. 33, pp. 936-963, Apr., 1904. _ If] * / x ~~ L V ... , CALCULATION OF ALTERNATING CURRENT PROBLEMS 23 For the flux distribution we have Pm X 10 8 p c . , , B x = \ e mx (sin cos) (cot + mx -f- ) CO -f e m (*-*) (sin - cos) [coZ + m (h - x) + 0] } , (39) and the amplitude of B x is cosh m (A 2 ap + cos m (h 2 a?) _ ,., cosh mh + cos raft The energy loss is w _ Pmp sinh mh sin mh L ' cosh mh + cos raft CASE 2 We have .^ _ Pmp sinh 2 raft sin 2 raft 2L cosh 2 raft + cos 2 mh J x and #3; are of the same form as (38) and (40) except that we write 2 h in place of h. Example. d=2cm., h = 0.1 cm., 7 =25 amp., p = 27rX60, ^=1000, TJ = ~2 = 1256, p = 10~ 5 ohm, L = TT (d + h) = 6.6 cm. m = 15.4, raft = 1.54. sin 2 raft = 0.061, cos 2 raft = - 0.998, sinh 2 raft = 10.86, cosh 2 raft = 10.9, w 625 X 15.4 X10~ 5 vx 10.86-0.061 -^T - X 10.91-0.998 = 159 X 10 WattS Per Cm * If the pipe is 200 feet long, the energy loss is W = 1590 X 10~ 5 X 200 X 30.5 = 97 watts. For 100 amperes under the same conditions the loss is W = 16 X 97 = 1552 watts. Effect of Grounded Lead Covering of Single Conductor Cable. The lead covering acts as a short-circuited secondary to the e.m.f . generated by the inductive reactance of the line. It reduces therefore the inductance of the line, but increases the energy loss due to the currents in the lead covering. 24 FORMULAE AND TABLES FOR THE Let R = line resistance, Ri = resistance of lead covering (for the same cross-section, four- teen times that of copper), x = line inductive reactance if no lead covering were used and a conductor of the same outside diameter as that of the lead covering were used in the calculations. The formula obtained by Berg* for the effective resistance and effective reactance are as follows: Effective reactance = As a rule XQ is small compared with Ri and we can write, (43) /Y 2 Effective resistance = R + - 5 - Effective reactance = z . To calculate, therefore, the effective resistance and effective reactance of a lead covered cable, we determine the resistances of conductor and resistance lead covering R and Ri, and x = 2irnL. The inductance L is taken as that between conductors of the same diameter as the outside diameter of the lead covering and the dis- tance, the interaxial distance between the conductor and its return. Example. Two No. 0000 B. & S. conductors each enclosed in a lead sheath of external diameter =1" and internal diameter = 0.75". Interaxial distance between the conductors, D = 4.5". Length of conductors = 1000 feet. To determine the effective resistance and reactance at 25 and 60 cycles. We have R = 0.049 ohm, area of lead = J TT X f = 0.343 sq. in., and Ri = 0.332 ohm. At 25 cycles, we have for the inductive reactance, x Q = 2 TT X 25 (2 log e j + O.s) ^^y *'" X ^g = 0.0234 ohm. * Ernst J. Berg, Constants of Cables and Magnetic Conductors, Proc. A.I.E.E., April, 1907. CALCULA T10N OF ALTERNATING CURRENT PROBLEMS 25 Therefore, R ett = 0.049 + (0.0234)' 0.332 = 0.0506 ohm. For a frequency of 60 cycles we have X Q = 0.0561 ohm and R^ = 0.0574 ohm. The resistance is increased by 3 per cent for the 25 cycle fre- quency and by 16 per cent for the 60 cycle frequency. The table given below applies to a single-conductor cable No. 0000 B. & S. having a lead covering T Vin. thick with an outside diameter of 1 in. FREQUENCY = 25. Interaxial distance in inches. 2.13 4 12 24 R 264 0.264 0.264 0.264 2*0 0.0497 0.916 0.1544 0.1904 a-o 2 0.0025 0.0084 0.0238 0.0362 R ett 0.2651 0.2676 0.2743 0.2796 Increase . 0.42% 1.37% 3.9% 5.9% FREQUENCY = 60. Interaxial distance in inches. 2.13 4 12 24 R 0.264 0.264 0.264 0.264 XQ 0.1193 0.220 0.3706 0.4570 Z 2 0.0142 0.048 0.137 0.208 7? efl 0.2701 0.2848 0.3233 0.3536 Increase 2.3% 7.9% 22.5% 33.9% Effect of Ungrounded Lead Covering in a Single-conductor Cable. Professor Berg * has also derived expressions for the eddy current and energy losses in an ungrounded lead covered cable, as follows: Eddy currents in lead = j-^- Energy loss in lead 4ft (44) Effective resistance, R eft = R + * 1. c., p. 24. co 2 L 2 26 FORMULAE AND TABLES FOR THE where co = 2 IT X frequency, .Ri = lead resistance in ohms, R = resistance of conductor proper in ohms, inductance per mile in henrys, 322 L = --j loge T D = interaxial distance between the conductor and its return, DI = external diameter of lead sheath. As an illustration we may use the same data as given in the example for the case of grounded lead covered cable. We have then D = 4.5 in., Z>i = 1 in., 475 hence, R = 0.049 ohm, R! = 0.332 ohm, w = 27rX60, = 0.049 = a()49 CALCULATION OF ALTERNATING CURRENT PROBLEMS 27 It is seen from this example that the increase in resistance is negligible if the lead covering is not grounded. Leakage Conductance.* All insulating materials are more or less conducting to some extent; there is no absolutely perfect insulator. Hence, when a difference of potential exists between two conductors which are embedded in an insulating medium, as in the case of the cores in a cable, there is a current flowing from one conductor to the other through the insulating material separating the two conductors. In consequence of the incomplete insulation and also because of dielectric hysteresis and electro- static radiation, we have an energy component of the current, I=Eg, which is in phase with the e.m.f. We give here some formulae for the calculation of the leakage conductance of linear conductors. The leakage conductance between a cylindrical conductor and an infinite conducting plane with uniform distance d between the plane and the axis of the cylinder is (45) where a = radius of cylinder in cm., d = distance between plane and axis of cylinder in cm., I = length of cylinder in cm., p = specific resistance of the medium in C.G.S. units. The resistance between a cylinder and an infinite plane is the reciprocal of the conductance given by equation (45). , 2d plog e ' * See " Theorie der Wechselstrome," von J. L. La Cour und O. S. Bragstad, Zweite Auflage, pp. 588-596. 28 FORMULAE AND TABLES FOR THE Example. a = 3 cm., d = 20 cm., P = 5 X 10 10 , I = 100 cm., 2d 40 log,-- = log, -o-* 2.59, a o 5X10 l X2.59 , ^ = o x/ inn = 2 X 10 10 absohms, ATT /\ 1UU 10~ 10 g = absmhos. z . cm If we express p in megohms (10 6 ohms) per - - 2 and I in miles, cm. formula (45) reduces to Prof. Kennelly* derived formulae for the leakage conductance of linear conductors in terms of antihyperbolic functions. For the formula corresponding to (45), he gives 2-irl 1.01 /yl -x g = - - = - -t mho per mile. (47) p cosh' 1 - p cosh" 1 - % Example. OTY1 p = 5 X 10 8 megohms per - 2 , approximately the value for gutta- cm. percha, log, = log 10 = 2.302, cosh- 1 ^ = 2.292. By formula (46) we have By formula (47), we get the approximately same result, - 530^29 = 88X10- mho per mile. * Proceedings Am. Phil. Socy., Vol. 48, pp. 142-165, 1909. CALCULATION OF ALTERNATING CURRENT PROBLEMS 29 For two parallel conductors separated by a distance d, and of equal radii a, we have 1.01 1 d+ 2a I 1.01 1 ,, m -7 mho, approx., (49) 2 P log e - 1.01 1 , KA . or g = ^ (50) 2 p cosh" 1 TT ,4 d When - is small, formula (49) is considerably in error. As an example suppose we take - = 2.5. Then log.- = 0.916, log,- - -=0.693, cosh- 1 - = 0.689. a From these values it is obvious that (48) and (50) agree closely while (49) is in error about 35 per cent. For concentric cylinders, we have for the leakage conductance 1.01 Z g= - -^mho, (51) where I is the length of cylinders expressed in miles, p is specific resistance expressed in megohms, D is the inner diameter of outer cylinder, and d is the outer diameter of inner cylinder. In the above conductance formulae it was assumed that the in- sulation is homogeneous, having the same resistivity for the entire depth of the insulation. If this is not the case the computations are very complex. We may obtain approximate results by as- suming that the insulation consists of several layers of different resistivities, and in that case the formula for concentric cylinders is as follows: 30 FORMULAE AND TABLES FOR THE 1.01 1 mho, (52) where d x is the diameter of the x ih insulation layer. For a three-core cable we have Q = 2.02 I mho, (53) FIG. 8. where D = diameter of sheath, a = radius of each core, d = interaxial distance between cores. Magnetic Flux Distribution and Eddy Current Losses in Iron. It is well known that in subjecting iron to an alternating magnetizing force, eddy cur- rents are generated in the iron in planes perpendicular to the direction of the magnetic flux. These eddy currents produce magnetic fields of their own which prevent the mag- netic flux from penetrating completely into the body of the iron. As a consequence we have an uneven distribution of the magnetic flux in the iron and also energy losses. Both of these effects depend on the permeability and conductivity of the iron and on the frequency. The problem of determining the magnetic flux distribution and eddy current losses offers considerable math- ematical difficulties, and it has only been worked out for a few special cases which we shall consider here. Infinite Plates.* Consider an iron plate, Fig. 9, whose sur- face in the direction perpendicular to the plane of the paper is infinite in extent, its thickness being 2 a. Assume m.m.f. acting on the plate producing a magnetic field in the direction perpendicular to the plane of the paper, and let B a cos at denote the magnetic flux density on the surface of the plate, which is produced by the impressed m.m.f. The magnetic flux density at any point in the * See " Theory and Calculation of Transient Electric Phenomena and Oscil- lations " (Chapter VI) by C. P. Steinmetz, and " The Theory of Alternat- ing Currents," Vol. I, Chap. XVI, by Alexander Russell. CALCULATION OF ALTERNATING CURRENT PROBLEMS 31 direction perpendicular to the surfaces, and at distance x from the center is given by B x = Bf, cosh 2 mx -\- cos 2 mx cos (at - cosh 2 ma + cos 2 ma sinh m(a x) sin m (a+ff) -j-sinhm (q-j-a) sinm (a x) cosh m (a z)cosm(a+a;)-|-coshm (a+z)cosm (a x) tan 7s = where m = 2 At = permeability, n = frequency, p = specific resistance in C.G.S. units. jf. (54) , (55) From equations (54) and (55) it is obvious that the amplitude of B diminishes from the surfaces to the center of the plate, and the phase of the magnetic flux changes from point to point along the axis perpendicular to the surfaces. The minimum value of B is at the center of the plate; B n V2 FIG. 9. tan 7 C Vcosh 2 ma + cos 2 ma sinh ma sin ma cosh ma cos ma (56) (57) The mean or average value of the magnetic flux density for the entire section of the plate is m B cosh 2 ma cos 2 ma a V2 am I cos h 2 ma + cos 2 ma ) When 2 ma is large, equation (58) reduces to (58) (59) 32 FORMULAE AND TABLES FOR THE The total magnetic flux in the plate is B a V2 ( cosh 2 ma cos 2 ma ) 2 B t m ) cosh 2 ma + cos 2 ma (60) If the flux density had been uniform throughout the entire section of the plate and of an intensity equal to B a , then the thickness of plate 2 d required to get the same total magnetic flux as that given by equation (60) would have been 2d = 2 ( cosh 2 ma cos 2 ma ^ m V2 t cos h 2 ma + cos 2 ma ) (61) The value of d as given by equation (61) may be considered as the equivalent depth of uniform magnetic flux density. When 2 ma is large, equation (61) reduces to d = 1 m V2 27r V 2fj,n (62) Example. a = 0.025 cm., /* = 2000, p = 10 4 , n = 60. 4 /2 X 10 3 X 60 m = 27r V TnH~ = 21.74. X, cosh 2 mx. cos 2 mx. B* B tl ' 0.000 1.000 1.000 0.972 0.005 1.023 0.976 0.973 0.020 1.403 0.644 0.985 0.025 1.651 0.465 1.000 The equivalent depth, d = 0.0244 cm. For such fine lamination and low frequency, the penetration of magnetic flux density is practically complete. Suppose, however, that in the above example we assume a frequency of 500; then, m = 62.83. CALCULATION OF ALTERNATING CURRENT PROBLEMS 33 X. cosh 2 mx. cos 2 raz. B x B a 0.000 1.000 1.000 0.43 0.005 1.205 0.809 0.44 0.010 1.896 0.309 0.46 0.015 3.368 -0.309 0.54 0.020 6.230 -0.809 0.72 0.025 11.590 -1.000 1.00 The equivalent depth d is 0.012 cm. which is about one-half the actual thickness of plate. The energy loss due to the eddy currents is given by the ex- pression sinh 2 ma sin 2 ma ) ? watts W X10~ 7 cosh 2 ma + cos 2 ma \ cu. cm. (63) When ma is greater than TT, that is 2 y a greater than unity, the factor in the brackets of equation (63) is practically unity, and rap#o 2 X 10- 7 W = cu. cm. HQ being the magnetizing force. When ma is small the expression for W as given by (63) is approximately as follows: K2 ma) 3 X wV#o 2 X 10- 7 watts 24 7T 2 cu. cm. approximately. (65) For small values of ma the magnetic flux is practically con- stant through the entire section of the iron plate, and hence we may assume constant permeability. Hence putting # max = nH 0) equation (65) becomes watts . 1.64(2^ 10-' cu. cm. In the above equation the thickness of the plate 2 a must be ex- pressed in cm. p is the resistivity in C.G.S. units (about 10,000), n is the frequency and B is the maximum value of the flux density. 34 FORMULAE AND TABLES FOR THE The values of equivalent depths d (of uniform magnetization) in cm. for various materials and frequencies are given below:* Frequency. 25 60 1000 10,000 10 Soft iron, M = 10 3 , P = 10 4 Cast iron, /* = 200, p = 10 5 0.0714 0.504 0.922 7.14 0.0460 0.325 0.595 4.60 0.0113 0.080 0.144 1.13 0.0036 0.0252 0.0461 0.357 0.00036 0.0025 0.0046 0.036 i rv4 Copper, u = l, p = Resistance alloys, n = 1, p = 10 5 . . . Iron Cylinders. The problem of magnetic flux distribution and eddy current losses in iron cylinders has been discussed by several writers, and the results have been summarized in a very APPROXIMATE FORMULA Function . x< 1. x > 6. ber (x) bei(x) X(x} Y(x) Zf \ (X) W(x) (z) d (z) w(x) r 4 1 * I n 01 ^9^ -r-4 r>ra /? /v O 7O71 -r 0.046 64 T 2/ T 4 \ T 2 11 I (i n nni7^ft T**^ V27TZ 0.0884 0.046 fa cin ft' R (\ 7O71 r 4 \ 576/ 4 1 + ^ = 1 + 0.03125 z 4 T 2/ T 4 \ T 2 ill ' I ^1 4- OO'i ! T 4 "\ V2wx 0.0884 0.0625 x x 2 ^(, 0.7071 , 0.25 , z 3 .265\ 4 V 192/ 4 .V 1 x z 2 z 3 j x ( 1 -i- ^ \ ^ n -u n m 04 r 4< i Y ^07071 1 a 88 1 al25 2V 1 + 96j 2 (1 !_ J|Z 4 = 1-0.01302 z 4 ~2/ T 4 \ 3-2 fl 1 /I O OO7S1 i-4N 2 0.7071 0.25 z z 2 z 3 0.7071 0.25 si 1 128J 8 (1 zV 11 \ z 2 2.8284 2 0.3536 z 4 z z 2 z 3 1.4142 0.1768 48-1 z z 3 * This table is taken from Steinmetz " Trans. Electrical Phenom.," p. 365. CALCULATION OF ALTERNATING CURRENT PROBLEMS 35 interesting paper by Prof. P. O. Pedersen, which appeared in the Jahrbuch der Drahtlosen Telegraphic und Telephonic, Vol. 4, pp. 501-515. He also worked out very valuable tables to facilitate numerical computations. The formulae and tables given in this section are taken from that paper. In the formulae given in this section, certain functions are employed which are extensions of the ber and bei functions first introduced by Dr. O. Heaviside and Lord Kelvin. The values of all these functions for arguments from to 6 are given in Table II, p. 43. For arguments greater than 6 or less than unity the values of the functions are given by the foregoing approximate expres- sions, the largest error in these approximate formulae being less than 1 per cent. Magnetic Flux Distribution in Iron Cylinders. Assume an iron cylinder subjected to an uniform alternating m.m.f. in the direction parallel to the axis of the cylinder. This can be realized in practice by placing the iron cylinder in a long coil through which alternating current is flowing. Let a = radius of cylinder in cm., A = sectional area of cylinder in sq. cm., H = permeability, p = specific resistance in C.G.S. units, n = frequency, co = 2irn, P /o cos a)t = current in coil, N = number of turns per cm., ir H a = IQ (magnetizing force), the intensity of magnetic field at the surface of cylinder. The intensity of the magnetic field at any point distant r from the axis of the cylinder is given by tf r = tf r cosM-7r), (67) _ ber (mr)bei (ma) bei (mr) ber (ma) a 7r ~ ber (mr) ber (ma) + bei (mr) bei (ma) 11 ::: 3 n I, El II and tan 5 = 5 (ma). ] The density of the eddy currents at any point distant r |; axis of the cylinder is given by J r = I T COS (ut - ]8 r ), 10 where and jY(mr) tanfr X(ma) ber' (mr) bei (ma) bei' (mr) ber' (mr) ber (ma) + bei' (mr) bei (ma) At the axis of the cylinder /o = and tan /3 = ber (ma) bei (ma) The energy loss due to the eddy currents in one centime of cylinder is fJLH AH a 2 w (ma) watts. 8X10 9 The increase in resistance of the coil due to the dissi energy in the iron cylinder or core in one centimeter cylinder is Ru = r^r AN 2 w (ma) ohms. The inductance per centimeter length of coil is 4?r L u = nAN 2 S (ma) henry. (76) The increase in inductance of coil per centimeter length due to presence of iron cylinder within the coil is given by AL (77) Equations (67) and (77) give all the necessary formula for a com- plete study of magnetic flux distribution, energy losses, etc., in CALCULATION OF ALTERNATING CURRENT PROBLEMS 37 an iron cylinder placed in a variable magnetic field whose direc- tion is parallel to the axis of the cylinder. We shall now consider a few examples in illustration of the above formulae. Example. An iron cylinder is subjected to an alternating mag- netizing force of intensity H a in the direction parallel to its axis. What is the intensity of magnetizing force and density of eddy currents at any point along its radius? Let a = 0.5 cm., v = 10 3 (assumed constant for entire section of cylinder), p = 10 4 C.G.S. units, n = 25 cycles per second. We have m = 14, ma = 7, VX (ma) = 22.4. r. Vx (mr). H r H a ' I, H a ' Vy(mr). 0.50 22.4 21.28 1 10.58 0.45 13.86 13.10 0.62 6.51 0.40 8.98 8.44 0.40 4.19 0.35 5.90 5.50 0.26 2.73 0.30 3.87 3.57 0.17 1.77 0.20 1.75 1.61 0.08 0.70 0.10 1.06 0.71 0.047 0.35 0.00 1.00 0.00 0.044 0.00 From an inspection of the fourth column in the above table it is seen that for an iron cylinder of 0.5 cm. diameter even for such low frequency as 25 cycles, the magnetization diminishes very rapidly as we recede from the surface of the cylinder. This shows that even for low frequencies lamination is required to obtain complete penetration and not make the eddy currents excessive. By formula (70) the total magnetic flux in the cylinder is $o = TT (0.25) 2 X 10 8 T a (ma), (ma) = 0.272, $o = 53.3 H a . For zero frequency (ma) = 1 ; hence the effect of the frequency which prevents complete penetration is to reduce the flux to about 27 per cent of its value for steady magnetization. Referring to formula (70), it is evident that if there were no 38 FORMULAE AND TABLES FOR THE iron present, the total flux would be $ = AH a ; hence we may consider the factor /*< (ma) as the equivalent permeability of the iron for the variable magnetizing force. Now it may happen that when the frequency is high and the radius of the cylinder is not very small the quantity /x (ma) would be less than unity, which would give the effect of permeability less than unity. This would mean that the penetration is limited to such thin sur- face layers of the cylinder that the total flux is less than what it would be if the iron were absent and the flux were passing through the air space occupied by the cylinder. For instance in an iron cylinder of 1 cm. radius subjected to alternating magnetizing force, assuming ju = 10 3 , p = 10 4 , we have for n= 100 10 3 10 4 5.10 4 10 5 10 6 ji0 (ma) = 1000 70 22 7 3.2 2.2 0.7 In this particular example at the frequency of one million cycles per second the total flux is less than if the iron were entirely removed. Let us now consider a more practical case that of a sole- noidal coil having a core made up of a bundle of iron wires. We shall find the total flux in the core, the energy loss and the in- crease in resistance of coil due to energy dissipation in the iron core. We shall assume a coil 10 cm. in diameter and 15 cm. long, total number of turns 60, or 4 turns per centimeter, the frequency 60 cycles per second and the current in the coil 2 amperes. The core is made up of 1000 iron wires, insulated from each other, the radius of each wire being 0.1 cm. We shall also assume a constant permeability of 1000 for the entire section of each wire. The magnetizing force /2 X 60 X 10 3 I-2*!/- -- = 21.74. The radius of each wire being 0.1 cm., ma = 2.174. From Table II, page 43, we find (ma) = 0.8062, w (ma) = 0.728. CALCULATION OF ALTERNATING CURRENT PROBLEMS 39 The flux for one wire of the core is $o = TT (O.I) 2 X 10 3 X 10.05 = 254.5 The total flux through the entire core is $ = 254.5 X 1000 = 254,500. The energy loss due to eddy currents in one wire is, by formula (74), ins y fin " TT (10.05) 2 X 0.728 X 15 = 0.026 watt. o X lir The loss of energy in the entire core is W = 0.026 X 10 3 = 26 watts. The increase in resistance of coil due to eddy current losses in one wire is, by formula (75), ft. = 2TX1 327rX60 . w(x). r(x). I (x). li (x). z. 3.00 6153 4990 7214 7199 1 318 951 4226 3.00 3.05 6053 4890 7298 7137 1 334 974 4188 3.05 3.10 5956 4793 7378 7072 1 351 997 4149 3.10 3.15 6862 4700 7452 7005 1 368 1 019 4109 3.15 3.20 5770 4611 7522 6937 1 385 1 042 4070 3.20 3.25 5680 4525 7588 6867 1 402 1 064 4030 3.25 3.30 5593 4443 7649 6797 1 420 1 086 3990 3.30 3.35 5409 4364 7707 6726 1 438 1 108 3949 3.35 3.40 5427 4288 7760 6654 1 456 1 130 3909 3.40 3.45 5348 4215 7810 6583 1 474 1 151 3868 3.45 3.50 5270 4144 7856 6512 1 492 1 172 3828 3.50 3.55 5195 4077 7900 6441 1 510 1 193 3787 3.55 3.60 5123 4012 7940 6371 1 529 1 214 3746 3.60 3.65 5052 3949 7978 6301 1 547 1 234 3707 3.65 3.70 4984 3889 8013 6233 1 566 1 255 3666 3.70 3.75 4917 3831 8045 6165 1 584 1 275 3626 3.75 3.80 4853 3775 8076 6098 1 603 1 295 3586 3.80 3.85 4790 3721 8105 6032 1 622 1 314 3547 3.85 3.90 4729 3669 8132 5968 1 641 1 334 3508 3.90 3.95 4670 3619 8157 5904 1 659 1 353 3470 3.95 4.00 4613 3570 8181 5842 1 678 1 373 3432 4.00 4.05 4557 3523 8203 5781 1 697 1 392 3394 4.05 4.10 4503 3478 8225 5721 1 715 1 411 3357 4.10 4.15 4450 3434 8245 5662 1 734 1 430 3310 4.15 4.20 4399 3391 8264 5605 1 752 1 448 3284 4.20 4.25 4349 3349 8283 5548 1 771 1 467 3248 4.25 4.30 4300 3309 8301 5493 1 789 1 485 3213 4.30 4.35 4253 3270 8318 5439 1 808 1 504 3179 4.35 4.40 4207 3231 8334 5386 1 826 1 522 3145 4.40 4.45 4162 3194 8350 5335 1 845 1 540 3111 4.45 4.50 4118 3158 8366 5284 1 863 1 558 3078 4.50 46 FORMULAE AND TABLES FOR THE TABLE II (Concluded) X. *(*). S (z). (). (*). r (z). I to. h (*). z. 4.50 4118 { 3158 8366 5284 | 863 1 558 3078 4.50 4.55 4075 ( 3123 8381 5235 ] 881 1 576 3046 4.55 4.60 4033 3089 8396 5186 ] 899 1 594 3014 4.60 4.65 3992 3055 8410 5139 ] 917 1 612 2983 4.65 4.70 3952 3022 8424 5092 ] 935 1 630 2952 4.70 4.75 3913 2990 8438 5047 1 953 1 648 2922 4.75 4.80 3874 2959 8452 5002 1 971 1 666 2893 4.80 4.85 3837 2928 8465 4958 1 989 1 684 2864 4.85 4.90 3800 2899 8479 4915 5 007 1 702 2835 4.90 4.95 3764 2869 8492 4873 025 1 720 2807 4.95 5.00 3729 2841 8505 d 4832 043 1 737 2780 5.00 5.05 3695 2813 8518 4792 060 1 755 2753 5.05 5.10 3661 2785 8531 4752 078 1 773 2727 5.10 5.15 3628 2758 f 8544 4713 096 1 791 2701 5.15 5.20 3595 2731 c 8557 4675 114 1 808 2675 5.20 5.25 3563 2706 8569 4637 131 I 826 2650 5.25 5.30 3532 2680 8582 4600 149 1 844 2626 5.30 5.35 3501 2655 8594 4564 166 1 862 2602 5.35' 5.40 3471 2631 8607 4528 184 1 880 2578 5.40 5.45 3441 2606 8619 4493 201 1 897 2555 5.45 5.50 3412 2583 8631 4458 219 1 915 2532 5.50 5.55 3383 2559 8643 4424 236 1 933 2510 5.55 5.60 3355 2537 8655 4391 2 254 1 951 2488 5.60 5.65 3327 2514 8667 4358 2 271 1 969 2467 5.65 5.70 3300 2492 8678 4325 2 289 1 986 2446 5.70 5.75 3273 2470 8690 4294 2 306 2 004 2425 5.75 5.80 3246 2449 8701 4262 2 324 2 022 2404 5.80 5.85 3220 2428 8713 4231 2 341 2 040 2384 5.85 5.90 3195 2407 8724 4200 2 359 2 058 2364 5.90 5.95 3170 2387 8735 4170 2 376 2 076 2345 5.95 6.00 3145 2367 8746 4141 2 394 2 093 2326 6.00 CALCULATION OF ALTERNATING CURRENT PROBLEMS 47 'O oOS V s 'dura; JO 9ST30IOHI '3 [ aoj 9otre:}sis9j p t^* OO o o CO OO 00 *O I> CO '^ CO ooooo ooo I Suiq3i9AY '3uoj !}OOJ 'l 9JIAV B JO O: , t-. l>. l>- 10 OOCOrH-^l'^(CO rHCOl OOOOOOOOOCJrHrHrHi 00 t^ i ( CO (N (M or qout CO T QI '3uo[ $OOJ I ' 9JIA1. B JO 'Q. 10 1~>. 10 1^ "* t^ co co b- -o -- 000 HI -mo 'bs i '3uo[ earn B jo Q mimmmm iu 00 OO rH OO (M O rH CO OS CO GO * C1 !>. CO OS Tf< IClOCOOi-HOSCOOCO^rHrH-rrOSO COOOOO rH rH rH rH (N (N C<| IQ OS OS 03 CO OS IO O "t 1 "^ O O rH rH rH CO CO OS 4 For spaced windings, formula (19) is not recommended, espe- cially when the space between the turns is great as compared with the diameter of the bare con- ductor. For long coils, we may for a first approximation consider F' and F" separately equal to unity, and formula (19) reduces to its first term (20) FIG. 16. + c + Formula (20) should be employed only when the length 6 is at least twice the outside diameter 2 R. The error involved scarcely exceeds 4 per cent with a multiple layer winding, and is within 2 per cent for a single layer solenoid, becoming more accurate as the length of the coil increases. To check the accuracy of formula (19) we will work out by this CALCULATION OF ALTERNATING CURRENT PROBLEMS 61 formula the same examples as those given in illustration of for- mula (13), (16) and (17). Example. a = 10 cm., 6=5, c = diameter of wire = 0.10, N = 50, R = 10.05, 50 + 1.2 + 20.1 F =50 + 1 + 14.07 - L 96 ' F" = 0.5 lo glo (lOO + ^)= L0278, T 4 7r 2 X 100X2500 1.096X1.0278 L= -ISIS"" "TOS" 0.734 mh. By formula (13) the value of the inductance for this coil is 0.7207 mh., hence the error in formula (19) for this case is about 2 per cent. Example. a = 10, 6 = 1, c = 1, N = 100, R = 10.5, 10 + 12 + 21 = 10 + 10 + 14.7 = F" = 0.51o glo (lOO + 14X 5 10 ' 5 )= 1.0566, L = 47r2 >< 1Q OX 104 X 1.239 X 1.0566 = 4.13 mh. iz.o By formula (16) the value of the inductance for this coil is L = 4.008 mh., and the error in formula (19) for this case is 3 per cent. Example. o = 5, 6 = 50, c = 0.4, N = 2000, R = 5.2, F' = 1.007, F" = 1, r 25X4X10 1.007 _ 50 + 0.4 + 5:2- X ^T = 71.14 mh. By formula (17) the value of the inductance for this coil is L = 70.56 mh. which agrees with the above within 0.8 per cent. As an extreme case we may consider that of a single turn. Let a = 10 cm., p = 0.5 cm. log e = log e 160 = 5.075, and by formula (18) we get L = 47r 10 J5.075 - 1.751 = 417.83 cm. 62 FORMULAE AND TABLES FOR THE Working out the same example by formula (19) we find L = 413 cm., the error being only 1 per cent. From these examples it is evident that formula (19) can be relied upon to give results with an accuracy of from 1 to 3 per cent. Table IX, p. 84, is reproduced from Prof. Brooks' and Turner's paper giving the values of F f and F" for various coil proportions. Self Inductance of Rectangle. Let a = length of rectangle, b = breadth of rectangle, p = radius of conductor, d = V a 2 + b\ The formula for the self inductance is L = 4 5 (a + 6) log e a log e (a + d) (21) For a rectangle made up of a conductor of rectangular section L = 4 (a + 6) lo& - a log, ( + 2 d + 0.447 (a + 0) For a = 6, a square, L = 8 ' 2235 0.726). (22) (23) Mutual Inductance of Two Equal Parallel Rectangles. For two equal parallel rectangles of sides a and 6 and distance apart d the mutual inductance, which is the sum of the several mutual inductances of parallel sides, is + 8 [ a 2 + 6 2 + CALCULATION OF ALTERNATING CURRENT PROBLEMS 63 For a square, where a = b, we have M = + 8 a + V2a* + d* a 2 + d 2 - 2 Va 2 + d 2 + d]. (25) FIG. 17. Self Inductance of Toroidal Coil. When the section of the coil is rectangular, and it is wound uniformly with a single layer of n turns, the inductance is given by the following expression : TZ >i* (26) -R ~e FIG. 18. r 2 and r\ are the outer and inner radii and h is the axial depth of the coil. For a coil of circular cross section we have for the in- ductance,* L = 4 7m 2 (R - VR - a 2 ), (27) * "A Treatise on Alternating Current Theory, " by Alex. Russell, Vol. I, p. 52. 64 FORMULAE AND TABLES FOR THE where R is the mean radius of the coil and a is the radius of its cross section. Inductance of Linear Conductors. In problems dealing with the transmission of electrical energy by alternating currents over long distances, we must know accurately the electrical con- stants, inductance, capacity and resistance of the lines. We shall therefore try as far as possible to give here a complete list of all available formulae for the inductance of linear conductors of various arrangement of circuits. The simplest case is that of a single straight cylindrical wire, whose inductance is * 071 2 I < loge 2 ( > approximately. OL 4 (28) I = length of wire in cm. and a = radius of wire in cm. When the permeability of the wire is /* and that of the medium outside the wire is unity, formula (28) assumes the following form: (29) The mutual inductance between two parallel wires separated by a distance d is I + M = = 2 1 < log -r + -j 1 [ approximately. (30) If in the case of two parallel wires one is the return circuit of the other, then the effective inductance of both wires is LI +L 2 2 M and the effective inductance of each wire separately is - : Zi When the two wires are of the same diameter, the effective in- ductance of each wire is L M . Using the values of L and M from (28) and (30) we get (31) * E. B. Rosa and Louis Cohen, Bull Bureau of Stand., Vol. 5, pp. 1-132. CALCULATION OF ALTERNATING CURRENT PROBLEMS 65 Usually -j is very small, hence (32) If the wires are of magnetic permeability /*, (33) The total inductance of the circuit including the to and fro con- ductors is a '4' Equations (28) to (34) give the inductances in centimeters. To express the inductance in mh. per km., we must multiply each of the formulae by the numerical factor 0.1. To express the inductances in mh. per mile, we must multiply each of the formulae by the numerical factor 0.161. If we replace the Naperian logarithms by the common loga- rithms, we may put (34) in the following form: L = 1.483 logio- + 0.161 mh. per mile d (35) = 0.921 logio- + 0.10 mh. per km., d and a being expressed in the same units, inches or centimeters. Formulae (28) to (35) were derived on the assumption of uni- form current distribution over the entire sectional areas of con- ductors, which is only strictly true for continuous currents. For alternating currents formula (34) assumes the form a - W where a is a constant the value of which depends on the frequency. For low frequencies we can take a equal to J and for very high frequencies a is zero. An accurate expression for a has been given by Professor Pedersen,* which is as follows: a = li (ma), (37) 2X10 9 * P. O. Pedersen, Jahrbuch der Drahttosen Tekg., Vol. 4, pp. 501-515. 66 FORMULAE AND TABLES FOR THE where n = frequency in cycles per second, p V = permeability, p = resistivity in C.G.S. units, a = radius of wire in cm. The values of l\ (ma) for values of ma from to 6 are given in Table II, p. 43, Chap. I. For values of ma greater than 6 we have the following approximate expression: 7 , , 1.4142 0.5303 0.75 li (ma) = ---- 7 vr ~ 7 \r (38) ma (ma) 3 (ma) 4 Example. What is the value of a for a copper conductor 0.1 cm. radius and frequency 20,000 ? 30.52, ma = 3.05. From Table II we find h (ma) = 0.4188, hence a = 0.2094. In Table X, p. 85, we give the values of L calculated by formula (35) for various sizes of B. & S. wire, and different distances apart. Inductance of Three-phase Transmission Lines. The in- ductance of one wire of a three-phase circuit arranged on the corners of an equilateral triangle, is the same as in the case of a two-wire circuit with the wires the same distance apart. (39) where d = distance between any two wires and a = radius of wire. To get the inductive drop between two wires, the inductances are combined geometrically, the inductance of circuits (1.2), for example, being (40) If the wires forming the circuit are arranged in a straight line we have for the inductance of the middle wire 4 ' CALCULATION OF ALTERNATING CURRENT PROBLEMS 67 while the inductance of the outer conductors, assuming equal load distribution on all three phases, is as follows: L 2 = L 3 = 21 j log e ~ + 0.625 j - (41) To equalize the inductances of all three phases the wires are transposed, each wire taking the center pin for one-third the dis- tance. Arranged in this way the inductance of each wire is (42) In equations (39) to (42) the length of the lines and the inductances are expressed in centimeters. Expressing the inductances in mh., km. or mile, and at the same time convert the Naperian loga- rithms into common logarithms, equations (40) and (42) assume the following form: Lig = 0.797 logio- + 0.0866 mh. per km. d (40/) = 1.283 logio- + 0.14 mh. per mile. Li2 = 0.797 logio- + 0.166 mh. per km. = 1.283 logio- + 0.27 mh. per mile. (42') Mutual Inductance.* Two metallic circuits A, B and C, D suspended on the same pole and arranged in the form shown in Fig. (19). The mutual inductance between A and CD is the mutual inductance between B and CD is Since the current in A is equal to that in B but of opposite sign, we have for the mutual inductance between the two circuits -^l- (43) tt ) * " Theorie der Wechselstrome," von J. L. La Cour and O. S. Bragstad, pp. 547-555. 68 FORMULAE AND TABLES FOR THE If the circuit CD consists of only one conductor C, the earth being the return conductor, then we may put to a high degree of approxi- mation di = d 3 and M AB _ c 2 1 log e (44) In case it is desired to avoid inductive effects the conductors must be so arranged as to make did = d^d^ so that Such an arrangement of a pair of circuits is shown in Fig. 20. FIG. 19. FIG. 20. Grounded Conductors.* The self inductance of a linear con- ductor suspended at height h above ground, and the ground being the return conductor, is ***i* d (45) = 0.74 logio - + 0.08 mh. per mile; a a is the radius and I is the length of conductor. The mutual inductance between two grounded parallel wires is + 0i + + fa - -i . per mile, (46) * O. Heaviside, Collected Papers, Vol. 1, p. 101. CALCULATION OF ALTERNATING CURRENT PROBLEMS 69 where hi and h z are the heights above ground of wires 1 and 2 respectively and d is the distance between them. When the wires are at the same height above ground, equation (45) reduces to M = 0.74 log y rj~~ m ^* per m ^ e> ( 4 ^) Inductance of Split Conductors.* The use of split con- ductors has been proposed for long distance transmission lines as a means of reducing the inductance and increasing the capacity of the line. Consider first the case of a looped conductor split into .two parts, the two parts of each conductor being placed symmetrically with respect to each other, and the distance between the centers of gravity, D, as shown in Fig. 21. Let a denote the radius of Q FIG. 21. an equivalent conductor whose sectional area is equal to the sum of the sectional areas of the two parts of the split conductor. Then we have for the total inductance of the circuit including the to and fro conductor L = 2 Z j log, ^ + loge ^ + 0.597 j cm. (48) L = 0.322 j log - + loge ^ + 0.597 I mh. per mile. (49) ( a a ) d = 5 a, or Example. Let a = 0.1825 in., D = 48 in., ^ = 5.571, log e ^ = 3.962, L = 3.26 mh. per mile. * Louis Cohen, "Inductance of Split Aerial Conductors," Electrical World, Nov., 1912. 70 FORMULAE AND TABLES FOR THE For a solid conductor of the same radius and the same distance between outgoing and return conductor we find, by formula (34), L = 3.84 mh. per mile. Hence, splitting the conductor into two parts and separating these parts by a distance equal to five times the equivalent radius of solid conductor reduces the inductance by about 15 per cent. Conductor Split into Three Parts.* Let the three parts of each conductor be arranged symmetrically at the vertices of an \ / f- -f ,/ \ FIG. 22. equilateral triangle as shown in Fig. 22. The inductance of the circuit including the to and fro conductor is cm. 0.322 \ | log e - + 1 log e ^ + 0.533 I mh. per mile. / o a o d ) (50) Considering the same example we have worked out for the case of a conductor split into two parts, we have D = 48 in., a = 0.1825 in., d = 5 a and L = 3.07 mh. per mile. By splitting the conductor into three parts and separating the parts by a distance equal to five times the radius, the inductance of the circuit is reduced by about 20 per cent. Self Inductance of Two Parallel Hollow Cylindrical Con- ductors.f Let 02, cti, and 62, &i denote the outer and inner radii of the two cylinders respectively; d is the interaxial dis- tance between the two conductors, and I the length of each con- ductor in centimeters. The total inductance of the circuit is 02 , Ia2 2 -3ai 2 of -of * 1. c. f Alex. Russell, "Theory of Alternating Currents," Vol. I, pp. 55-59. CALCULATION OF ALTERNATING CURRENT PROBLEMS 71 If we put h 2 = 02 2 - ai 2 , fc 2 = 6 2 2 - &! 2 , equation (51) can be written in the following form : , KO , To reduce (50) and (51) so as to give the inductance in mh. per mile it is necessary to multiply the right-hand side of either equation by the numerical factor 0.161. Self Inductance of Concentric Conductors. The following formulae apply to two hollow cylinders, one inclosed in the other, and insulated from it, the current entering one cylinder and returning by the other. Let the outer and inner radii of the outer and inner cylinders be denoted by 6 2 , &i, (h, i and let I be the length in cms. The inductance is 02 1 a 2 2 -3a? cm - 1 (6 2 2 -6i 2 ) 2 &6 6i 2 & 2 2 - In practice the cross-sectional areas of the two cylinders are always equal, that is, and formula (53) becomes 2 . (54) If the inner cylinder were solid, ai would be zero and 6 2 2 would be equal to 6i 2 + i 2 > and formula (54) would reduce to _ The least possible value of L is when 61 = a 2 and in this case L = I (4 log e 2 - 2) = 0.7726 Z. 72 FORMULAE AND TABLES FOR THE For very high frequencies where the current is concentrated along the surface of the conductor, the magnetic force is confined within the space between the conductors, and in that case we have for the inductance L = 2Zlog ^ (56) Equations (55) and (56) may be combined into one by writing them in the following form: where a is a quantity that depends on the frequency. For low frequencies it may be taken equal to unity, and for very high frequencies it is zero. Example. 61 = 0.4 in., a z = 0.2 in., log,- 1 = 0.693, -0578 For low frequencies a = 1 and % L = 0.316 mh. per mile. For very high frequencies a = and L = 0.223 mh. per mile. Triple Concentric Conductors.* The inductance of a triple concentric conductor depends on the current distribution in the three conductors. Consider three hollow cylinders of negligible thickness and radii 7*1, r 2 , r 3 , beginning with the inner one. Let the current 7i flow through the inner conductor and return cur- rents 1 2 and 7s through the middle and outer conductors. Then The inductance of the system is L = 0.322 \ log,- + f^Y log,- > mh. per mile. (58) * "Theory of Alternating Currents," by Alex. Russell, Vol. I, Chap. 2. See also "The Magnetic Circuit," by V. Karapetoff, Chap. XI. CALCULATION OF ALTERNATING CURRENT PROBLEMS 73 If a current 1 2 flows through the middle conductor and /i, 7 3 are the return currents through the inner and outer conductors, L = 0.322 j (j\* loge^ 4/rY loge ^ | mh. per mile. (59) If a current 7 3 flows through the outer conductor and the return currents 7i, 7 2 through the inner and middle conductors, L = 0.322 \ (Y^ loge - + log e - I mh. per mile. ( V 3/ Pi 7*2 ) (60) The Self Inductance of a Circuit Formed by Three Conductors Whose Axes Lie Along the Edges of an Equilateral Prism. We will assume that the three conductors have equal radii a, the interaxial distance between any two conductors being d. Let a current 7i flow through conductor No. 1 and the return currents 7 2 and 7 3 through conductors No. 2 and 3 respectively. The inductance of the system is L = = 0.161 | 4 loge -+ 1 I jl -y^ 3 >mh. per mile. (61) The inductance has a maximum value when either 7 2 or 7 3 is zero and it has a minimum value when 7 2 = 7 3 = J I\. Compound Conductors. A compound conductor consisting of a steel core and a concentric coating of copper is now being used in many cases for transmission purposes. The inductance of a linear conductor con- sists of two parts, one of which is due to the magnetic field outside of the conductor and the other is due to the magnetic field within the conductor. We may call these the external and internal FIG. 23. 74 FORMULAE AND TABLES FOR THE inductances, respectively. In the expression for the inductance of a linear conductor The first term is the external inductance, and the second term the internal inductance. In compounding the conductor only the internal inductance is affected and the term is replaced by the following formula : * where (62) i = radius of core, 02 = radius of shell, 71 = current in core, 7 2 = current in shell, jLti and ^2 are the permeabilities of core and shell, re- spectively. The above formula is somewhat complicated for numerical computation, and Mr. Fowle has supplied a short table in his paper, which we reproduce here, giving the values of L for the standard sizes now in commercial use. TABLE (A) INTERNAL INDUCTANCE OF COPPER-CLAD STEEL WIRES WHEN THE CON- DUCTANCE RATIO OF STEEL TO COPPER is 12 PER CENT Values of n. Internal inductance in C. G. S. units. Per cent 20 30 40 50 L t - = 0.149 MI Li = 0.0506 MI L{ = 0.0208 MI L f = 0.00925 MI + 0.0646 M2 + 0.102M2 + 0.147M2 + 0. 193 M2 * The above formula was first given by Prof. A. Vaschy in his book, " Traite d'electricite et de Magnetisme, " Vol. 1, p. 367 and was independently derived by Mr. F. F. Fowle, Electrical World, Dec. 29, 1910. CALCULATION OF ALTERNATING CURRENT PROBLEMS 75 TABLE (B) INTERNAL INDUCTANCE OF ALUMINUM-STEEL WIRES, WHEN THE CONDUCT- ANCE RATIO OF STEEL TO ALUMINUM is 19.35 PER OINT Values of n. Internal inductance in C. G. S. units. Per cent 20 30 40 L { = 0.127 MI Li = 0.0328 MI Li = 0.00871 MI + 0.0990 M2 + 0.1861M2 + 0.2624 M2 In the above two tables n is the conductance ratio of the com- pound wire to a solid copper wire of equal size. Series or Parallel Arrangement of Inductance. If several inductances are connected in series the total inductance is the sum of the several inductances. L t = Li + L 2 + L 3 + - + L n . (63) For several inductances in parallel, we have for the total in- ductance, -TT7 r^ (64 > -Lil/ 2 L 4 L n -\- - If the mutual inductance between the various inductance coils is taken into account, then for two coils in series we have for the effective inductance, T T I T I r> -\r f&A\ Jut Jui -r~ L/2 -\- Z M } (yQ) or L t = L, + L 2 - 2 M. The first of the above two formulae applies when the magnetic fields of the two coils are aiding, and the second formula applies when the magnetic fields are opposing. For two coils in parallel we have r T 71 T9. (65) For more than two inductances in parallel the formula for the total inductance is generally very complicated if the mutual in- ductance between the various branches is to be included. For the case, however, of cylindrical conductors symmetrically ar- ranged, formulae have been derived for the total inductance of several wires including the mutual inductance between the wires.* * "Inductance and Capacity of Linear Conductors," The Electrician, Feb. 14 and 21, 1913. 76 FORMULAE AND TABLES FOR THE Such an arrangement of circuits is being used for instance in the case of horizontal antennae which generally consist of a number of wires hi parallel suspended at some height above ground. Inductance of Wires in Parallel. The following formulae (66) to (72) were derived on the assumption that all the wires are suspended at the same height above ground, the earth being the return conductor; also that all the wires are of the same size and equally spaced. For two wires in parallel ' (66) where is the total inductance, L is the inductance of each individual wire and M is the mutual inductance between the wires. The values of L and M can be obtained by formulae (45) and (48). ^ Example. a, radius of wire = 0.08 in., d, distance between wires = 24 in., h, height above ground = 80 ft. By (45) and (48) we have A 2h . 1\ OJ /160X12 . _ \ L = 2(\o&-+^= 2 log. (-^g- + 0.25) = 19.15 cm. per cm. ' . 4/i 2 . 4X6400 M= loge-3r = 1& -- o -- = cm * per Cm ' d 2 In the expression for M we neglected T^ as being very small com- pared with unity. We have by (66) P 19.15 + 8.76 = - = - = 13.15 cm. per cm. For three wires in parallel C = orl" J A n, 1 ' (67) Mi is the mutual inductance between wires 1, 2 and 2, 3; M 2 is the mutual inductance between 1 and 3. CALCULATION OF ALTERNATING CURRENT PROBLEMS 77 Since the wires are equally spaced, we may put M 2 = MI log,, 4 = Mi 1.386 and formula (67) becomes _ L(L + Mi-1.386)-2M 1 3L-3Mi- 1.386 Using the same values for L and M as in the previous example we find the value of the inductance for three wires. r - 19-15 (19.15 +8.76 -1.386) -2 (8.76)* _ . 3 X 19.15 - 3 X 8.76 - 1.386 For four wires in parallel r _ (L + Mi) (L + M 3 ) - (M! + M 2 ) 2 , , 4L-4M 2 -2M! + 2 where MI = M& = M 23 = M 34 , M 2 = Mis = M 24 , HH-H-H 6 6 6 cb 234 FIG. 24. Formula (69) may be put in the form r _ (L + MQ (L + M! - 2.2) - (2 MJ - 1.386) 2 4L-4M 1 + 1.15 / Taking again the values of L = 19.15 and MI = 8.76, we get for the inductance of four wires in parallel, L = 10.93 cm. per cm. For five wires in parallel, = -2M iM, - -M 2) +Afi (2M 3 (71) where MI = M i2 = M 23 = M 34 = M 46 , M 2 = Mi 3 = M 24 = M 35 , M 3 = Mi 3 = M 25 , M 4 = M 15 . 78 FORMULAE AND TABLES FOR THE Expressing the inductance in terms of L and MI only, as in the previous cases, formula (71) reduces to JC = - 1.86) + 5 M? - 1.87^-4.18 (72) For the values L = 19.15 and M l = 8.76 r _ 7022.7 + 4899.4 - 8662.3 + 3038.8 - 949 " 1833.6 - 1642 + 383.7 - 16.38 - 4.18 5349.5 ft . = ,_ . Q =9.64 cm. per cm. OO4.o If we have more than five wires in parallel, which is generally the case for antennae of large stations, we can obtain an approximate value of the inductance in the following manner: Suppose we have an antenna consisting of ten wires, determine the inductance of five wires in parallel and denote it by I/, then assume that the antenna consists of two wires separated by a distance 5 d (where d is the distance between adjacent wires) and the inductance of each wire is Z/. By formula (66), the effective inductance of the ten wires in parallel is L' + M' = -2 ' where M' is the mutual inductance between two wires separated by a distance 5 d. Taking the data from the previous examples, we have for the inductance of five wires in parallel L' = 9.64 cm. By (48) we have P 9.64 + 5.54 hence = - = 7.6 cm. per cm. z If the antenna were 200 feet long, we should have = 7.6 X 200 X 30.5 = 0.046 mh. If the antenna consisted of twelve wires, for instance, we could break it up into three groups of four wires each and proceed in the same manner as above using formulae (68) and (70). The above formulae for the inductance of parallel wires are very useful in the determination of the capacity of antennae, as will be shown in Chapter III. CALCULATION OF ALTERNATING CURRENT PROBLEMS 79 TABLE IV (For use in Formula 4.) r-VAadJfl A f-T/AadM-]. A Ios 4 , - d ,J logl l * aTJ 0.020 0.022 0.024 0.026 0.028 1.03365 1 . 13786 1.23314 1.32091 1.40231 10421 9528 8775 8140 0.106 0.107 0.108 0.109 0.110 0.93984 0.95193 0.96395 0.97590 0.98778 1218 1209 1202 1195 1188 0.030 0.032 0.034 0.036 0.038 1.47823 1.54938 1.61635 1.67963 1.73962 7592 7115 6697 6328 5999 0.111 0.112 0.113 0.114 0.115 0.99958 1.01132 1.02298 1.03459 1.04612 1180 1174 1166 1161 1153 0.040 0.042 0.044 0.046 0.048 1.79668 1.85108 1.90309 1.95292 0.00077 5706 5440 5201 4983 4785 0.116 0.117 0.118 0.119 0.120 1.05760 1.06901 .08036 .09166 . 10289 1148 1141 1135 1130 1123 0.050 0.052 0.054 0.056 0.058 0.04681 0.09118 0.13401 0.17542 0.21551 4604 4437 4283 4141 4009 0.121 0.122 0.123 0.124 0.125 .11407 . 12520 1 . 13727 1.14729 1.15826 1118 1113 1107 1102 1097 0.060 0.062 0.064 0.066 0.068 0.25439 0.29213 0.32881 0.36450 0.39927 3888 3774 3668 3569 3477 0.126 0.127 0.128 0.129 0.130 1.16917 1 . 18004 1 . 19086 1.20164 1.21237 1091 1087 1082 1078 1073 0.070 0.072 0.074 0.076 0.078 0.43317 0.46625 0.49857 0.53017 0.56108 3390 3308 3232 3160 3091 0.131 0.132 0.133 0.134 0.135 1.22305 1.23369 1.24429 1.25484 1.26536 1068 1064 1060 1055 1052 0.080 0.082 0.084 0.086 0.088 0.59136 0.62103 0.65013 0.67868 0.70672 3028 2967 2910 2855 2804 0.136 0.137 0.138 0.139 0.140 1.27584 1.28627 1.29667 1.30703 1.31736 1048 1043 1040 1036 1033 0.090 0.092 0.094 0.096 0.098 0.73428 0.76137 0.78802 0.81425 0.84009 2754 2709 2665 2623 2584 0.141 0.142 0.143 0.144 0.145 1.32765 1.33790 .34812 .35831 .36847 1029 1025 1022 1019 1016 0.100 0.101 0.102 0.103 0.104 0.86555 0.87814 0.89065 0.90307 0.91540 2546 1259 1251 1242 1233 1 99fi 0.146 0.147 0.148 0.149 0.150 .37859 .38869 .39875 1.40879 1.41879 1012 1010 1006 1004 1000 0.105 0.92766 80 FORMULAE AND TABLES FOR THE TABLE V NAGAOKA'S TABLE OF VALUES OF THE END CORRECTION K AS FUNCTION T-, DIAMETER OF THE RATIO -T LENGTH (For use in Formula 13.) Diameter K. AI. A,. Diameter K. AI. A,. Length 0.00 1.000000 -4231 +24 0.45 0.833723 -3160 +21 0.01 1.995769 -4207 26 0.46 0.830563 -3139 22 0.02 1.991562 -4181 24 0.47 0.827424 -3117 21 0.03 1.987381 -4157 25 0.48 0.824307 ouyo 21 0.04 1.983224 -4132 25 0.49 0.821211 -3075 21 0.05 0.979092 -4107 +25 0.50 0.818136 -3054 +21 0.06 0.974985 -4082 26 0.51 0.815082 -3033 21 0.07 0.970903 -4056 24 0.52 0.812049 -3012 21 0.08 0.966847 -4032 24 0.53 0.809037 -2991 20 0.09 0.962815 -4008 26 0.54 0.806046 -2971 21 0.10 0.958807 -3982 +25 0.55 0.803075 -2950 +20 0.11 0.954825 -3957 24 0.56 0.800125 -2930 20 0.12 0.950868 -3933 23 0.57 0.797195 -2910 20 0.13 0.946935 -3910 26 0.58 0.794285 -2890 20 0.14 0.943025 -3384 27 0.59 0.791395 -2870 20 0.15 0.939141 -3857 +23 0.60 0.788525 -2850 +19 0.16 0.935284 -3834 23 0.61 0.785675 -2831 19 0.17 0.931450 -3811 26 0.62 0.782844 -2812 20 0.18 0.927639 -3783 24 0.63 0.780032 -2792 19 0.19 0.923854 -5761 24 0.64 0.777240 -2773 19 0.20 0.920093 -3737 +24 0.65 0.774467 -2754 +19 0.21 0.916356 -3713 24 0.66 0.771713 -2735 19 0.22 0.912643 -3689 25 0.67 0.768978 -2716 19 0.23 0.908954 -3664 23 0.68 0.766262 -2697 18 0.24 0.905290 -3641 25 0.69 0.763565 -2679 18 0.25 0.901649 -3616 +23 0.70 0.760886 -2661 +18 0.26 0.898033 -3593 24 0.71 0.758225 -2643 19 0.27 0.894440 -3569 23 0.72 0.755582 -2624 17 0.28 0.890871 -3546 24 0.73 0.752958 -2607 18 0.29 0.887325 -3522 24 0.74 0.750351 -2589 18 0.30 0.883803 -3498 +22 0.75 0.747762 -2571 + 17 0.31 0.880305 -3476 24 0.76 0.745191 -2554 17 0.32 0.876829 -3452 23 0.77 0.742637 -2537 18 0.33 0.873377 -3429 23 0.78 0.740100 -2519 17 0.34 0.869948 -3406 22 0.79 0.737581 -2502 16 0.35 0.866542 -3384 +24 0.80 0.735079 -2486 +19 0.36 0.863158 -3360 22 0.81 0.732593 -2467 16 0.37 0.859799 -3338 23 0.82 0.730126 -2451 16 0.38 0.856461 -3315 22 0.83 0.727675 -2435 16 0.39 0.853146 -3293 23 0.84 0.725240 -2419 17 0.40 0.849853 -3270 +22 0.85 0.722821 -2402 + 16 0.41 0.846583 -3248 23 0.86 0.720419 -2386 16 0.42 0.843335 -3225 21 0.87 0.718033 -2370 15 0.43 0.840110 -3204 21 0.88 0.715663 -2355 16 0.44 0.836906 -3183 23 0.89 0.713308 -2339 17 CALCULATION OF ALTERNATING CURRENT PROBLEMS 81 TABLE V (Continued) Diameter K. A* A* Diameter K. AJ. A* A* Length Length ' 0.90 0.710969 - 2322 + 14 2.50 0.471865 - 9292 + 405 0.91 0.708647 - 2308 16 2.60 0.462573 - 8887 378 0.92 0.706339 - 2292 15 2.70 0.453686 - 8509 355 0.93 0.704047 - 2277 16 2.80 0.445177 - 8154 330 0.94 0.701770 - 2261 14 2.90 0.437023 - 7824 312 0.95 0.699509 - 2247 + 15 3.00 0.429199 - 7512 + 293 0.96 0.697262 - 2232 15 3.10 0.421687 - 7219 275 0.97 0.695030 - 2217 15 3.20 0.414468 - 6944 260 0.98 0.692813 - 2202 14 3.30 0.407524 - 6684 245 0.99 0.690611 - 2188 14 3.40 0.400840 - 6439 230 1.00 0.688423 -10726 +344 3.50 0.394401 - 6209 + 220 1.05 0.677697 -10382 330 3.60 0.388192 - 5989 207 1.10 0.667315 -10052 316 3.70 0.382203 - 5782 195 1.15 0.657263 - 9736 303 3.80 0.376421 - 5587 186 1.20 0.647527 - 9433 290 3.90 0.370834 - 5401 174 1.25 0.638094 - 9143 +278 4.00 0.365433 - 5227 + 168 1.30 0.628951 - 8865 266 4.10 0.360206 - 5059 161 1.35 0.620086 - 8599 255 4.20 0.355147 - 4898 152 1.40 0.611487 - 8343 244 4.30 0.350249 - 4746 141 1.45 0.603144 - 8099 236 4.40 0.345503 - 4605 138 1.50 0.595045 - 7863 +224 4.50 0.340898 - 4467 + 134 1.55 0.587182 - 7639 215 4.60 0.336431 - 4333 125 1.60 0.579543 - 7424 208 4.70 0.332098 - 4208 118 1.65 0.572119 - 7216 198 4.80 0.327890 - 4090 115 1.70 0.564903 - 7018 190 4.90 0.323800 - 3975 102 1.75 0.557885 - 6828 +184 5.00 0.319825 -18321 +2227 -397 1.80 0.551057 - 6644 176 5.50 0.301504 -16094 1830 -306 1.85 0.544413 - 6468 170 6.00 0.285410 -14264 1524 -241 1.90 0.537945 - 6298 161 6.50 0.271146 -12740 1283 -193 1.95 0.531647 - 6137 154 7.00 0.258406 -11457 1090 -153 2.00 0.525510 -11809 +580 7.50 0.246949 -10367 + 937 -127 2.10 0.513701 -11229 539 8.00 0.236582 - 9430 810 -104 2.20 0.502472 -10690 499 8.50 0.227152 - 8620 706 - 86 2.30 0.491782 -10191 465 9.00 0.218532 - 7914 620 2.40 0.481591 - 9726 434 9.50 0.210618 - 7294 10.00 0.203324 82 FORMULAE AND TABLES FOR THE TABLE VI VALUES OF CORRECTION TERM A, DEPENDING ON THE RATIO ^ OF THE DIAMETERS OF BARE AND COVERED WIRE ON THE SINGLE LAYER COIL (For use in Formula 15.) d D' A. AI. d D' A. AI. 1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.85 0.84 0.83 ' 0.82 0.81 0.80 0.5568 0.5468 0.5367 0.5264 0.5160 0.5055 0.4949 0.4842 0.4734 0.4625 0.4515 0.4403 0.4290 0.4176 0.4060 0.3943 0.3825 0.3705 0.3584 0.3461 0.3337 100 101 103 104 105 106 107 108 109 110 112 113 114 116 117 118 120 121 123 124 19ft 0.70 0.69 0.68 0.67 0.66 0.65 0.64 0.63 0.62 0.61 0.60 0.59 0.58 0.57 0.56 0.55 0.54 0.53 0.52 0.51 0.50 0.2001 0.1857 0.1711 0.1563 0.1413 0.1261 0.1106 0.0949 0.0789 0.0626 0.0460 0.0292 0.0121 -0.0053 -0.0230 -0.0410 -0.0594 -0.0781 -0.0971 -0.1165 -0.1363 144 146 148 150 152 155 157 160 163 166 168 171 174 177 180 184 187 190 194 198 7Q 3211 0.78 0.77 0.76 0.75 0.74 0.73 0.72 0.71 0.70 0.3084 0.2955 0.2824 0.2691 0.2557 0.2421 0.2283 0.2143 0.2001 127 129 131 133 134 136 138 140 142 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 -0.1363 -0.2416 -0.3594 -0.4928 -0.6471 -0.8294 -1.0526 -1.3403 -1.7457 1053 1178 1335 1542 1823 2232 2877 4054 CALC ULA TION OF ALTERNA TING C URRENT PROBLEMS 83 TABLE VII VALUES OF THE CORRECTION TERM B, DEPENDING ON THE NUMBER OF TURNS OF WIRE ON THE SINGLE LAYER COIL (For use in Formula 15.) Number of turns. B. Number of turns. B. 1 0.0000 50 0.3186 2 0.1137 60 0.3216 3 0.1663 70 0.3239 4 0.1973 80 0.3257 5 0.2180 90 0.3270 6 0.2329 100 0.3280 7 0.2443 125 0.3298 8 0.2532 150 0.3311 9 0.2604 175 0.3321 10 0.2664 200 0.3328 15 0.2857 300 0.3343 20 0.2974 400 0.3351 25 0.3042 500 0.3356 30 0.3083 600 0.3359 35 0.3119 700 0.3361 40 0.3148 800 0.3363 45 0.3169 900 0.3364 50 0.3186 1000 0.3365 TABLE VIII TABLE OF CONSTANTS FOR FORMULA (16) b c - or r . c b Vi- ft. b c - or r . c b Vi> 1/2- 0.00 0.50000 0.1250 0.55 0.80815 0.3437 0.05 0.54899 0.1269 0.60 0.81823 0.3839 0.10 0.59243 0.1325 0.65 0.82648 0.4274 0.15 0.63102 0.1418 0.70 0.83311 0.4739 0.20 0.66520 0.1548 0.75 0.83831 0.5234 0.25 0.69532 0.1714 0.80 0.84225 0.5760 0.30 0.72172 0.1916 0.85 0.84509 0.6317 0.35 0.74469 0.2152 0.90 0.84697 0.6902 0.40 0.76454 0.2423 0.95 0.84801 0.7518 0.45 0.78154 0.2728 1.00 0.84834 0.8162 0.50 0.79600 0.3066 84 FORMULAE AND TABLES FOR THE TABLE IX SHAPE-FACTORS, F f AND F", FOR CERTAIN COIL PROPORTIONS (For use in Formula 19.) a. 6. c. R. F 1 . r F' F". 19 200 2 20 1.008 1.001 1.009 15 200 10 20 1 .015 1.001 1.016 19 100 2 20 1.015 1.003 1.018 15 100 10 20 1 .028 1.002 1.030 19 50 2 20 1.029 1.006 1.035 15 50 10 20 1 .051 1.004 1.055 19 20 2 20 1.064 1.013 1.078 15 20 10 20 1 .097 1.008 1.106 19 12 2 20 1.095 1.019 1.116 15 12 10 20 1 .129 1.011 1.141 19 8 2 20 1.125 1.026 1.154 15 8 10 20 1 .154 1.013 1.169 19 5 2 20 1.163 1.035 1.204 15 5 10 20 1 .182 1.015 1.200 19 4 2 20 1.182 1.040 1.229 15 4 10 20 1 .190 1.016 1.209 19 3 2 20 1.205 1.045 1.259 15 3 10 20 1 .203 1.017 1.223 19 2 2 20 1.235 1.054 1.301 15 2 10 20 1 .216 1.017 1.237 19 1 2 20 1.277 1.065 1.360 15 1 10 20 1 .232 1.018 1.254 . 19 0.5 2 20 1.302 1.073 1.397 15 0.5 10 20 1 .241 1.018 1.263 19 0.2 2 20 1.320 1.079 1.424 15 0.2 10 20 1 .246 1.019 1.270 19 0.1 2 20 1.326 1.081 1.433 15 0.1 10 20 1 .249 1.019 1.273 SQUARE WINDING-SECTION COILS 17.5 5 5 20 .172 1.023 1.199 19 2 2 20 .235 1.054 1.301 19.5 1 1 20 .292 1.097 1.418 19.75 0.5 0.5 20 .342 1.163 1.561 19.9 0.2 0.2 20 .387 1.29 1.79 19.95 0.1 0.1 20 .407 1.42 2.00 CALCULATION OF ALTERNATING CURRENT PROBLEMS 85 i TABLE X INDUCTANCE PER MILE OF COMPLETE METALLIC CIRCUIT, FORMULA 35 Size, B. &S. Diameter in inches. Distance in inches. Inductance in mh. Size, B. &S. Diameter in inches. Distance in inches. Inductance in mh. 0000 0.46 12 2.707 4 0.204 12 3.231 18 2.969 18 3.492 24 3.154 24 3.678 48 3.600 48 4.124 000 0.41 12 2.782 5 0.182 12 3.305 18 3.043 18 3.566 24 3.228 24 3.751 48 3.675 48 4.198 00 0.365 12 2.857 6 0.162 12 3.380 18 3.118 18 3.641 24 3.303 24 3.826 48 3.750 48 4.273 0.325 12 2.930 7 0.144 12 3.456 18 3.193 18 3.717 24 3.378 24 4.349 48 3.824 48 4.349 1 0.289 12 3.007 8 0.128 12 3.531 18 3.268 18 3.792 24 '3.453 24 3.978 48 3.900 48 4.424 2 0.258 12 3.080 9 0.114 12 3.606 18 3.341 18 3.867 24 3.526 24 4.052 48 3.973 48 4.500 3 0.229 12 3.157 10 0.102 12 3.678 18 3.418 18 3.940 24 3.603 24 4.124 48 4.050 48 4.571 86 FORMULAE AND TABLES FOR THE CHAPTER III CAPACITY THE capacity of a conductor is defined as the ratio of its charge to its potential, when all other conductors are at zero potential. C= (1) or, according to Maxwell, the capacity of a conductor is its charge when its own potential is unity and that of all the other conductors is zero. The capacity between two conductors is the ratio of the charge on either conductor to the difference of potential between .the conductors, assuming also that all other conductors are at zero potential, otherwise the capacity is considerably influenced to an extent depending on the proximity of the other conductors. In general the capacity of a conductor depends on the geometri- cal configuration of its own surface, its distance from other con- ductors and the material medium between them. The problem of determining the capacity of a conductor, or a system of conductors, offers considerable mathematical difficul- ties, and there are not therefore as many formulae available for the calculation of capacities as in the case of inductances. For linear conductors, however, which are the most important in practice, a sufficient number of formulae have been derived to enable us to compute the capacity of any arrangement of circuits. Owing to the extreme importance of the electrical constants of linear conductors in problems of transmission of electrical energy, we shall give here an extensive list of such formulae. The unit of capacity is the capacity of a condenser in which unit charge produces a unit difference of potential between its conductors, which may be either in electrostatic or electromagnetic units. The ratio of the electrostatic to the electromagnetic unit of capacity is equal to F 2 where V = 3 X 10 10 cm. per sec. (velocity of light); hence C=-C B , ( 2 ) CALCULATION OF ALTERNATING CURRENT PROBLEMS 87 The practical unit of capacity is the farad. It is the capacity of a condenser which has a potential difference of one volt when charged with one coulomb. 1 farad == 10~ 9 C.G.S. units. 1 microfarad = KH 5 C.G.S. units. In practice the microfarad is generally used. If C is the capacity in microfarads and C e the capacity in electro- static units, we have 10 15 1 c ' C e . Parallel or Series Arrangement. In electrical circuits we frequently use a number of condensers which may be either con- nected in parallel or in series. For the parallel arrangement the total capacity is the sum of the capacities of the several condensers, that is, C = d + C 2 + <7 3 + - - - + C n . (3) FIG. 25. In the series arrangement the resultant capacity is given by the following expression: or C~i /~i /~ fi | 20304 ... On -r The capacity of three condensers in series for instance is C=7T7T (4) (5) The total capacity of several condensers connected in series is always less than the capacity of the smallest condenser in the series. 88 FORMULAE AND TABLES FOR THE Example. Three condensers of capacities Ci, Cz, C$ connected in series. Let Ci = 0.01 mf., C 2 = 0.05 mf., C 3 = 0.1 mf. 0.01X0.05X0.1 _ 0.00005 f 0.005 + 0.001+0.0005 0.0065 " which is less than the value of Ci, that of the smallest condenser. Capacity of Parallel Plates. If the area of each plate is denoted by A sq. cm. and the distance between them by h cm., the capacity of the condenser is A ~TT 4wh 9 X 10 5 For circular plates of radius r, we have (6) If the space between the plates is filled with a uniform dielectric of specific inductive capacity K, then AK 1 Formulae (6) to (8) are only approximate; they were derived on the assumption that the distribution of the electric force between the plates is uniform throughout the entire area. This, how- ever, is not the case; the lines of force curve around the edges, and the correction due to this effect cannot be neglected in accurate calculations. This correction factor has been determined only for circular plates, and the corrected formula for this case is as follows:* (9) where t is thickness of the plates and h is the distance between them. For very thin plates ~ 1 , /1A , mf - * G. Kirchoff, Gesammelte Abhandlungen, p. 112, "Zur Theorie des Con- densators." See also J. A. Fleming, "The Principles of Electric Wave Telegraphy," p. 176. CALCULATION OF ALTERNATING CURRENT PROBLEMS 89 Example. Capacity of two-plate circular condenser of the fol- lowing dimensions. r = 10 cm., h = 0.2 cm., t = 0.05 cm. air dielectric, K = 1. By formula (7), C = 13.9XlO- 5 mf. By formula (9), C = 14.51 XlO- 5 mf. By formula (10), C = 14.44 X 10~ 5 mf . It is seen that in this particular case the capacity as computed by formula (7) is in error by about 4.5 per cent. The smaller the area of the plates and the greater the distance between them, the larger is the correction term. For a multiple-plate condenser consisting of n plates, the capacity is (n 1) times the expression for capacity given in formulae (6) to (8). When the dielectric of a plate condenser consists of several layers of different specific inductive capacities, we have, on neglecting the correction factor, where hi and Ki denote the thickness and specific inductive capac- ity of layer i, and n is the number of layers, or I^^^^M^^^ FIG. 26. Example. Same constants as in the preceding example. Two-plate circular condenser, r = 10 cm., A = TT X 100, hi = h z = h% = /i 4 = 0.05 cm., h (distance between plates) = 0.2 cm., #1 = 1.5, Ki = 2, K 3 = 2.5, # 4 = 3. 90 FORMULAE AND TABLES FOR THE C== ^XlOO l_ (0.05 0.05 ,0.05 .0.5} 9 X 10 5 "I 1.5 " 2 " 2.5 "" 3 J Capacity of Concentric Spheres. C = ^r^r 9xiQ5 mf . (13) r 2 is the inner radius of outside sphere and r\ is the radius of inner sphere. This formula is exact, no correction term is re- quired. When r 2 is infinitely large which is the capacity of an isolated sphere in infinite space. From formula (14) it is obvious that a sphere will have to be of radius 9 X 10 5 cm., or 9 km., to have a capacity of one microfarad. When the centers of the spheres are separated by a small dis- tance d, the capacity is given by the following expression : rif2 f f ~ (ri - r 2 ) (n* - r 2 3 ) If the dielectric between the spheres is made up of several spherical layers of different specific inductive capacities, the ex- pression for the capacity is as follows: c=-- - - -; (16) T\ is the radius of the sphere which separates the layer i from the layer i + 1. Comparing formulae (13) and (15) it is seen that displacing the centers of the spheres reduces the capacity. Example. TI = 30 cm., r 2 = 28 cm. By formula (13), 30X28 1 _ 1 f 30-289X10 5 9X10 5 CALCULATION OF ALTERNATING CURRENT PROBLEMS 91 If the centers of the spheres are separated by 0.5 cm., that is, in formula (16) d = 0.5, we get 30 X 28 { _ 210 j 30 - 28 1 10,096 ) Displacing the centers of the spheres by 0.5 cm. reduces the capacity by two per cent approximately. Capacity of Disk insulated in Free Space. <> = *> (I?) d = diameter of disk in centimeters. Capacity Coefficients of Two Spheres.* If we apply a differ- ence of potential to two spherical conductors which are separated from each other, the relation between the charges and potentials are given by the following equation: qi = CiVi + CuV 2 and q 2 = C 2 V 2 + C^Vi, (18) where Ci, C 2 and CM are the capacity coefficients of the spheres. Maxwell f derived general formulae for the capacity coefficients Ci, C 2 and C& covering the cases of equal and unequal radii of the two spheres with any distance separation between them. The general formulae, however, as given by Maxwell are somewhat complicated, so we shall give here only the simplified formulae obtained by Russell, which are applicable only for spheres of equal radii. Let the radius of each sphere be a and d the distance between their centers; then, r =\'? 1 f< sinh (2s + 1) a .'I! (19) -Ci2 = x T; 1 > s ^J sinh 2 sa where sinha = - and 4X 2 = d 2 -4a 2 . (20) * Alexander Russell, Journal of the Institution of Electrical Engineers, Feb., 1912. f Maxwell, Electricity and Magnetism, Vol. 1, p. 270, 173. 92 FORMULAE AND TABLES FOR THE Denoting - by y, formulae (19) may be put in the following form: a C 1 = 1 +I+2 + 5 15 49 166 o y 2 y i/ 6 2/ 8 y w y 12 and also = l+-o + OUi A '~dd~ A + 1 dd (21) (22) Ci is the capacity of the sphere when all neighboring conductors are at zero potential. When the charges on the two spheres are + q and q respectively, and V is the difference of potential between them, then -^ is the capacity C between the spheres and this is the capacity generally considered in engineering problems. In this case, if V and V are the potentials of the two spheres, (23) When both spheres and all neighboring conductors are at the same potential V, then q = C,V + CnV. =C l + Cu = C'. (24) V When y is not less than 3 the values of C and C 1 are given by the following formulas: 2C = 3/0/-1) 2 jg + 1 l y C^ = y (t/ + I) 2 _ y- 1 _ 1 a 2/(2/ + l) 2 + l 2/ 2 2/ 8 ' (25) CALCULATION OF ALTERNATING CURRENT PROBLEMS 93 For values of y greater than 8 we have the following formulae: (26) 2C , 1 " 1 " ___. a y y 2 y* y* C C f In Table XI, p. 122, are given the calculated values of - and for values of - from 2 to 1000. a The Laws of Attraction and Repulsion Between the Spheres When the Potentials are Given. If we denote by W the electrostatic energy of the system, we have W = \ C, (TV + TV) + C^FiFj (27) and the force acting between the spheres is F = W = ~ \ A (Vl * + V ** } + BVlVz ' (28) A and B are given by equation (22). In practice three cases may arise: (1) When the potentials are equal and opposite. (2) When they are equal. (3) When one of the spheres is at zero potential. In the first case the force F is attractive and is given by In the second case the force F r is repulsive. 1 (1 ~ y)2 I i 2 ^ / 4 " V /oi\ In the third case the force F" is attractive. * 94 FORMULAE AND TABLES FOR THE When y is less than 2.001 we have the following very approxi- mate formulae: aV*_ (32) F = 2 (d - 2 a) F' = 0.07386 7 2 . For values of - greater than 10, the following formulas give a four- figure accuracy. F= ""i_^ (33) Capacity of Linear Conductors. For the transmission of electrical energy, linear conductors are employed in the form of overhead wires, concentric mains or cables. In some of these cases the derivation of capacity formulae offers considerable analytical difficulties. A sufficient number of formulae have how- ever been derived to cover nearly all cases which are apt to arise in practice. Overhead Wires.* We shall first consider the case of over- head lines where the ground acts as the return conductor, which is generally the case for telegraph lines. If we have a single wire of radius a suspended at height h above ground its capacity is 2 log. I is the length of the line in centimeters; h and a can be expressed in any units so long as they are both expressed in the same units. We may write formula (34) in the following form : 0.0894 /0 _, (7 = - ^-r mf . per mile. (35) * See O. Heaviside, Collected Papers, Vol. I, pp. 42-46. F. F. Fowle, "Calculation of Capacity Coefficients of Wires," Electrical World, Vol. 58, Aug., 1911. CALCULATION OF ALTERNATING CURRENT PROBLEMS 95 Formulae (34) and (35) give only an approximation, the degree of approximation being better the greater the ratio - For large values of - formula (34) gives fairly accurate results. Prof. Kennelly * has obtained an exact formula for this case which is as follows: I 1 C = mf., 2 cosh" 1 - a or we may put it, Example. By (35), By (37), 0.0894 , M C = T mf . per mile. cosh" 1 - a h , - = 5. a (36) (37) C = = 0.0384 mf. per mile. = 0.0390 mf . per mile. The difference in results by the two formulae is about 1.5 per cent. For greater values of - the results by the two f ormulae will be more nearly equal. Formula (35) was derived on the assumption that the con- ductor is suspended in free space and that there are no other con- ductors near by to influence the value of its capacity. This is a condition which is never realized in practice. Generally there are a number of wires suspended on the same pole and the capacity of each conductor is affected by the presence of the other con- ductors. In the following we shall give some formulas for the case of two or more grounded wires suspended on the same pole. A. E. Kennelly, Proc. Am. Phil Soc., Vol. 48, pp. 142-165, 1909. 96 FORMULAE AND TABLES FOR THE Two Grounded Wires. Two wires 1 and 2 are suspended at heights hi and h 2 above ground and at distance d& from each other (see Fig. 27). 2 . r -*r- K \ \ j I , v \ A FIG. 27. The capacities of wires 1 and 2 and the mutual capacity be- tween them are 0.0894 log, 2 2_ D 0.0894 log e ^ mf . per mile, D 0.0894 log e ^- 2 12 mf . per mile, mf . per mile, n A 2hi\ A 2h 2 \ A VV D=(loge- ) (log,- J-(loge-T-). \ i / \ az / \ i2/ (38) where ai and 02 are the radii of wires 1 and 2 respectively. When the wires are of the same radii, di = 0% = a, and at the same heights above ground, hi = h 2 = h, formulae (38) reduce to CALCULATION OF ALTERNATING CURRENT PROBLEMS 97 0.0894 log, Ci = C-2 = , r mf . per mile, (39) 0.0894 log, mf. per mile. Example. Single wire, No. 8 B. & S., suspended 25 feet above ground. a = 0.0642 in., log, = log, = 9.142. 0&Q4 C = ^j^ = 0.00978 mf . per mile. Example. Two grounded wires, No. 8 B. & S., equal heights above ground 25 feet and distance apart 1 foot. As in the preceding example we have = 9.142, log- 3.912. 0.0894X9.142 C 2 = 2 _ (3.912)2 = - 0120 mf - P er 0.0894X3.912 Cl2 = (9.142) 2 - (3.912) 2 = An additional wire on the same pole increases the individual capacity of each wire. In the case considered in above examples, the increase in capacity is about 23 per cent. As we increase the number of wires on the same pole the in- dividual capacities of the several wires are continually increased. We shall give here formulae for the capacities of three and four wires. Though the formulae are not very simple, and are unwieldy for numerical computation, yet the numerical examples worked out here will give some idea of the magnitude of the increase in capacities as the number of wires is being increased. Three Grounded Wires. The wires are at equal heights h above ground, the same distances d apart and of equal radii a.- Wires 1 and 3 are symmetrically situated with respect to wire 98 FORMULAE AND TABLES FOR THE 2, hence their capacities are equal and for the same reason the mutual capacities Ci2 and C^ are equal. 0.0894 j (log.**)'- (log. Vag (\ a I \ d D ~ _ 0.0894 D . ) mf . per "mile, mf. per mile, 2 ) (log. V ^p - loge 2 l\ d & a C 13 i loge D . vxl - log e - X loge J a to D f mf . per mile, , mf. per mile, (40) where Example. Use same constants as in preceding examples. a = 0.0642 in., h = 25 feet, d = 1 ft. = 9.142, log, 3.912, log, 3.219. D = 764.5 + 98.54 - 279.84 - 94.80 = 488.4. 0.0894J (9.142) 2 - (3.912) 2 C 3 = 0.0125 mf. per mile. 0.0894K9.142) 2 - (3.22) 2 J C 2 = - 4884 - - = 0.0134 mf. per mile. C. = C- - 5 ' 92S = - 0.0043 mf. per mile. 0.0894J (3.912)^9.142 X3.912J CALCULATION OF ALTERNATING CURRENT PROBLEMS 99 Four Grounded Wires. Suppose we have four grounded wires placed at the corner of a rectangle as shown in Fig. 28, and we shall assume that the wires are of equal radii r. Let log. = d, The significance of the subscripts to h and d are obvious from the figure. 1 2 -o- FIG. 28. The capacities of the separate wires and the mutual capacities are as follows : - J 2 ) + d (cf - de) -c(df- ce}\ +D,] C Z = C* = \e (a 2 - 6 2 ) + d (be - ad) + c (bd -ac) { +D, - Cu= \d (df -ce)+c (cf -de)+b (-28. 08Q4 By formula (46), C = , .0 ^non = 0.01643 mf. per mile. By formula (47), C = , .Q'OQ = O-O 164 ^ mf - P^ mile. o.4o It is seen from the above example that even when the conductors are brought within 5 feet from the earth the capacity is increased only by about 0.7 per cent, so that in all practical cases we can neglect the earth's influence and use the simple formula (44) or (45). When the radii of the wires are not equal, as has been assumed in the above formulae, we have 0.0894 C = - -, - r-mf. per mile single wire, (49) CALCULATION OF ALTERNATING CURRENT PROBLEMS 103 tn n n d d\ G&2 where a = di and 02 are the radii of the two wires. The capacities of looped conductors can be also expressed very conveniently in terms of anti-hyperbolic functions, as has been done by Prof. Kennelly.* When the radii of the wires are equal we have 0.0894 mf . per mile of single wire. (50) cosh" 1 This formula is equivalent to (45). For unequal radii we have (51) mf. per mile of single wire, which is equivalent to (49). Formulae (45) and (50) are equivalent to each other since Riifl fnsh~l and both formulae are exact. In Table XII, p. 122, the values of log - and cosh" 1 ^- are d 2i O/ given for values of jr from 1.01 to 25. d Formula (44) gives fairly accurate results when ^ is greater Z d than 5, increasing in degree of accuracy as the value of ^ in- 2 d creases. When ^ is small, that is, when the wires are close z d together, the results by formula (44) are considerably in error. * A. E. Kennelly, Proc. Am. Phil Socy., Vol. 48, pp. 142-165, 1909. 104 FORMULAE AND TABLES FOR THE Example. For s- = 10. 2a By formula (44), C = ^Qg!^ = 0-0298 mf. per mile. By formula (45) or (50), C = O'QQQO = 0.0299 mf. per mile. For ^ = 1.5. By formula (44), C = ^T- = 0.0814 mf. per mile. By formula (45) or (50), C = n ' gR7 ri = 0.1031 mf. per mile. The error in formula (44) for this case is about 27 per cent. Example. For wires of different diameters, a\ = 0.5 cm., a z = 1 cm., d = 10 cm. By formula (49), 100 - 1 - 0.25 - = 98.75, log, \a + V^=lj = 5.275, C= 1 ;. = 0.034 mf. per mile single wire. 2 (o.ZtO) Working out the same example by formula (50) we get - 1 ^ = cosh- 1 ^ = 2.9932, l.U cosh- 1 ^ = 2.924, 2.924) The results by formulae (49) and (50) agree exactly. CALCULATION OF ALTERNATING CURRENT PROBLEMS 105 Two Metallic Circuits Suspended on the Same Pole.* If we have two circuits suspended on the same pole, the capacity of either circuit will be sensibly influenced by the proximity of the other circuit and the capacity will be larger than that given by formula (44). Suppose we have two circuits (1, 2) and (3, 4) on the same pole, as shown in Fig. 29. The capacity of circuit (1, 2) may be expressed by the formula 6 X 0.179 d 5 , d Q A d s d 2 log e I log e -p-p a 3 a \ didj mf . per mile of loop conductor. (52) FIG. 29. If circuit (3, 4) were removed to an infinite distance from circuit (1, 2) we should have-i^ = 1 and log -^j = 0. Equation (52) would reduce to 0.179 C = 7- mf . per mile of loop conductor, which is formula (44). If the circuits are separated by a distance of 5 feet or more, the capacity is modified but slightly on account of the proximity of the other circuit. If, however, the circuits are brought close together, the capacity may be affected materially. * Louis Cohen, London Electrician, Feb. 14, 1913. 106 FORMULAE AND TABLES FOR THE Example. a = 0.1 in. d = 2 ft. d & = 2 ft. di = I ft. d 2 = 3 ft. d 3 =.2.24 ft. d 4 = 3.6 ft. log e | 6 = log ^ = log, 240 = 5.480, 5 = loge^- = Iog e 240 = 5.480. l oge = log 1.866 = 0.624. ttitt4 C = 4X5.48X5^48 -(0.624)' = - 0824 mf ' per mile ' lo P wire ' If circuit (3, 4) were removed, the capacity of circuit (1, 2) would be C = ^ ' , ^ = 0.00812 mf. per mile of loop wire. The presence of circuit (3, 4) has increased the capacity of (1, 2) by 1.5 per cent. Three-phase Transmission Lines.* If the spacings are symmetrical, as in the case when the wires are placed at the vertices of an equilateral triangle, and the voltages on the lines are balanced and equal, the capacity for each wire is the same as for a single-phase transmission line, n 0.0894 , C = - -T mf . per mile. The value of C as given above is the capacity between each wire and neutral, and d distance between wires, a = radius of wire. Charging current per wire is EaC j = For unsymmetrical spacing, the calculations of the capacities or charging current in a three-phase line is very much involved, and is not of any great practical importance. It is therefore omitted here. * F. A. C. Perrine and F. G. Baum, Trans. Am. Inst. E. E., May, 1900. CALCULATION OF ALTERNATING CURRENT PROBLEMS 107 We may get an idea of the error introduced by using formula (44) for unsymmetrical spacings by determining the limiting values of the capacities as given by that formula. As an illustration let us consider the case of the three wires arranged in a horizontal plane. The distance between wires is 12 feet, and the conductors being No. B. & S., the radius a = 0.163 inches, taking d = 12 feet we have log,- =6.78, C = 0.0132 mf. per mile. For d = 24ft., log.- = 7.37, C = = 0.0121 mf. per mile. \ FIG. 30. Prof. Karapetoff* states that Mr. J. G. Pertsch, Jr., found that, with certain simplifying assumptions, and when the wires are transposed, the equivalent spacing for the inductance and capac- ity is equal to the geometric mean of the three actual spacings, or d eq = V di2 d<& C?13. * For a detailed discussion of the charging currents in a three-phase line with unsymmetrical spacings, see "The Electric Circuit" by V. Karapetoff, p. 199; and an article by G. S. Humphrey, Electrical World, Vol. 58, p. 1300, Nov. 25, 1911. See also "Inductance and Capacity of Three-phase Transmis- sion Lines," by Louis Cohen, Electrician, July 18, 1913. 108 FORMULAE AND TABLES FOR THE In the above example for instance, d eq = ^12X12X24 = 15.12, log e = loge-' = 7.012, = 0.0128 mf. per mile. Capacity of Horizontal Antennae.* A horizontal antenna is generally made up of a number of wires in parallel suspended at a considerable height above ground. To obtain an expression for the capacity of such a network of conductors is rather a difficult matter. We may, however, obtain an approximate value of the capacity of antennae by an indirect method. For linear conductors we have the following relation between the inductance and capacity: = ' LC' LV 2 ' where L is the inductance in cms. per cm., C is the capacity in cms. per cm. expressed in electromagnetic units and V is the veloc- ity of light (V = 3 X 10 10 cms. per cm.). The relation given by equation (54) is only strictly true when there is no energy loss in the conductor, that is with perfect conductivity. The expression for C as given by equation (54) gives fairly accurate results if in the evaluation of L we neglect that part of the inductance which is due to the magnetic field within the conductor, and consider only the inductance which is due to the external magnetic field of the conductor. In Chapter II we have given a set of formulas, (65) to (71), for the calculations of the inductances of two, three, four and five wires in parallel, and we have also indicated a method for the approximate determination of the inductance of any number of wires in parallel; hence, knowing the inductance we can use for- mula (54) to determine the capacity. As an illustration let us consider an antenna of the following constants : a (radius of wire) ^= 0.08 in., d distance between adjacent wires = 2 feet, h height above ground = 80 feet, 10 wires in parallel and 150 feet long. * Louis Cohen, Electrician, Feb. 21, 1913. CALCULATION OF ALTERNATING CURRENT PROBLEMS 109 The self inductance of a single wire, neglecting the internal mag- netic field of the conductor, is 2h 160 X 12 =18.65. The mutual inductance between two adjacent wires is 4X6400 ,, , M = to- = 8.76. By formula (70), Chapter II, we find the inductance of five wires in parallel, ' = 9.51 cms. per cm. The mutual inductance between two wires placed at a distance of 5 d is M' 5.54 cms., and the total inductance of the 10 wires in parallel is r L' + M' 9.51 + 5.54 = ~ - = - jr - = 7.52 cms. per cm. The capacity of the antenna is 10 15 1 10 15 X 150 X 30.5 C = 9 X 10 20 X 7.52 9 X 10 20 X 7.52 = 0.00068 mf. Concentric Cylinders. The capacity of a condenser formed by two concentric cylinders is Kl 210*2 0.0894 K mf . per mile, (55) where r 2 = inside radius of outer cylinder, 7*1 = radius of inner cylinder, K = specific inductive capacity. If the dielectric between the cylinders is not homogeneous, but 110 FORMULAE AND TABLES FOR THE consists of several cylindrical layers of different specific inductive capacities, formula (55) assumes the form 0.0894 (56) 0.0894 Ti . 1 , r 2 . 1 - + ^-log e - + r A 2 TI A Example. ri, radius of inner cylinder = 1 cm., r 2 , inner radius of outer cylinder = 2 cm. The dielectric consists of two layers each one-half cm. in radial thickness, the specific inductive capacities being 5 and 3, respec- tively. Introducing these values in formula (56) we get 0.0894 0.0894 C = 1.5 1 - 2 = 0.0811+0.0959 = a505 mf ' per mile ' A uniform dielectric of specific inductive capacity 3 or 5 would give, in the above example, a capacity of 0.39 mf. per mile or 0.65 mf. per mile, respectively. If the axes of the cylinders are displaced with respect to each other by a distance d, the capacity is given by = - . lo ge (/3 + V>_l) r 2 2 + n 2 - d 2 where e=--2^r If d 2 is small compared with r 2 2 n 2 , then approximately, 0.0894 K ,- Q , mf . per mile. (58) Another approximate formula which has been derived for this case is as follows: C = i + - - m f. per mile. (59) CALCULATION OF ALTERNATING CURRENT PROBLEMS 111 By comparing formulae (58) and (59) with (55) it is obvious from inspection that the capacity is a minimum when the axes of the cylinders coincide. As an illustration of the relative accuracy of formulae (58) and (59) the computed values of the capacities by formulae (57), (58) and (59) for - = 4 and different values of d are given in the following table: (K = 1.) r z - TI. d. Formula 57. Formula 58. Formula 59. 4 1 0.0645 0.0645 0.0645 4 1 1 0.0676 0.0676 0.6750 4 1 2 0.0837 0.0829 0.0773 4 1 2.5 0.1111 0.1054 0.0837 The results by formulae (57) and (58) agree very closely for all values of d, while formula (59) is applicable only when d is small. For comparatively large values of d, the error in formula (59) is considerable. Capacities of Cables.* In the discussion of capacities of cables it will be convenient to introduce the term "effective capacity," which we shall designate throughout by C e . By the term effective capacity we mean that factor which, multiplied by the time rate of variation of electromotive force, will give the value of the charging current, that is, 7 = C e 2 irnE. In the case of a simple condenser, the effective capacity is merely the capacity of the condenser, but in the case of cables the effective capacity varies with any alteration in the other parts of the circuits as for instance when the lead sheath is grounded or insulated, or any of the other cores are grounded, etc. In engineering problems we are mostly interested in knowing the values of the charging currents for a certain e.m.f. and a certain frequency, hence it will be very useful to obtain formulae for the effective capacity of any cable system. * Alex. Russell, "Theory of Alternating Currents," Vol. I, Chapters IV andV. L. Lichtenstein, "Capacity of Cables," Elektrotech. Zs., Vol. 25, pp. 106- 110 and 124-126, Feb., 1904. L. Lichtenstein, Beitrage zur Theorie der Kabel; Dissertation Koniglichen Technischen Hochschule in Berlin, 1908. 112 FORMULAE AND TABLES FOR THE Concentric Cable. We shall consider first the case of a simple cable having a lead sheath and an iron sheath and insulating dielectrics between them. Let TI = the radius of conductor, r 2 = the inner radius of lead sheath, r% = the outer radius of lead sheath, r 4 = the inner radius of iron sheath. FIG. 32. KI is the specific inductive capacity of insulating material between conductor and lead sheath and K z is the ^specific inductive capacity of insulating material between lead and iron sheath. If the lead sheath is grounded the capacity of the cable is simply that of a cylindrical condenser formed by conductor and lead sheath, that is, by formula (55) ! 0.0894 mf . per mile. (60) If, however, the lead sheath is insulated, charges are induced on the outer surface of the lead sheath and the inner surface of the CALCULATION OF ALTERNATING CURRENT PROBLEMS 113 iron sheath and thus another condenser is formed between lead and iron sheaths whose capacity is f. per mile. (61) The effective capacity of the cable is that of the two condensers in series, that is, If KI = K 2 we have #0.0894 C e #0.0894 mf . per mile. , 7- 2 r 4 lOge The capacity of the cable is reduced when the lead sheath is in- sulated. Example. ri = 1 cm., 7*1 = 1.5 cm., r 3 = 1.75 cm., r 4 = 2.5 cm., loge - = 0.405, loge - = 0.356, log, = 0.761. C = 0.221 K mf. per mile, C' = 0.251 K mf. per mile, C e = 0.118# mf. per mile. In this case, by insulating the lead sheath, the capacity of the cable is reduced to about 50 per cent of its value when the lead sheath is grounded. Triple Concentric Cable. Let ri = the radius of inner conductor, r 2 = the inner radius of second conductor, r 3 = the outer radius of second conductor, n = the inner radius of third conductor, r 5 = the outer radius of third conductor, r 6 = the inner radius of lead sheath. 114 FORMULAE AND TABLES FOR THE We have in this case three cylindrical condensers formed be- tween the several conductors, whose capacities are, n Ki 0.0894 Ci = mf . per mile, K 2 0.0894 . 62 = mf . per mile, K 3 0.0894 . ., (7 3 = m f. per mile. FIG. 33. We shall assume that the three conductors are supplied from a three-p'hase generator delivering an electromotive force of pure sine form, the stator being star-connected. If the neutral point is grounded, the charging currents have the values CALCULATION OF ALTERNATING CURRENT PROBLEMS 115 7 2 = / ST\) 7 3 = JEfx C 2 V3 sin co C 3 sin ( coZ + ) > , ( \ 6 /) in ( ut + -^- s sn (64) The charging currents consist of two components differing in phase from each other and we cannot, therefore, speak of effective capacity in the case of a triple concentric cable. The effective values of the charging currents may be obtained graphically. If the stator winding is insulated, neutral point not being grounded, we have for the charging currents i V3 cos J 2 = (65) J 3 = -co In this case also the charging currents can be best obtained graphically. Capacity of Two-core Cable. The effective capacity of a two-core cable can be expressed for all possible arrangements in terms of Ci and Ci2, where Ci is the capacity of No. 1 conductor, when No. 2 and the sheath are grounded and Ci 2 is the mutual capacity between the two conductors. The values of Ci and d 2 in terms of the dimensions of the cable are, log. 0.0894 K /. R*-d*y L d* + R*\ ( log <-flH-( log <-2d/r) -Ca- - per mile, mf . per mile, (66) where R is internal radius of sheath, r is the radius of either conductor, d is the distance between centers of sheath and either conductor. 116 FORMULAE AND TABLES FOR THE It is assumed that both conductors are of the same radius and are at the same distance from center of sheath. If the sheath is grounded and the two conductors are con- nected to a generator supplying an alternating e.m.f., we shall have for the charging current and C e (effective capacity) = J (Ci - C 12 ). (67) FIG. 34. If we introduce the values of Ci and Cm from (66) we shall have mf . per mile. (68) (69) Let us suppose now that conductor 2 and the sheath are both grounded. In this case the capacity of No. 1 conductor is simply Ci and that of No. 2 conductor, C i2 ; that is the charging currents on the two conductors and sheath are 71 = CiwE cos co, 7 2 = CizuE cos ut, I 8 = _ (d + Ci2) uE cos at. If one terminal of generator is connected to the two conductors in parallel and the other terminal to the sheath which is grounded, the effective capacity of either conductor is C e = Ci + Cl2 and the total charging current on the two conductors is 7 = 2 (Ci + Ci 2 ) uE cos at. (70) Evidently the charging current on sheath is 7. = - 7 = - 2 (Ci + Ci 2 ) wE cosorf. CALCULATION OF ALTERNATING CURRENT PROBLEMS 117 In all the above cases we assumed that the lead sheath was grounded. If the lead sheath is insulated, and one terminal of alternator is connected to the two conductors in parallel and the other to ground, we have for the effective capacity where Co = 2 (Ci + CM) and C is capacity of cylindrical condenser formed between lead and iron sheaths. Example. r = 0.5 cm., d 1.0 cm., R = 2.0 cm. (4 -1)X 0.0894 X i = _ ) - t er m _ Cl2 - - - 0.016 K mf. per mile. For the case given by formula (68) <7 e = 0.051 K mf. per mile. For the case given by formula (70) C e = 0. 134 Kmt. per mile. For the case given by formula (71) we must also determine the value of C. We shall assume that the outer radius of lead sheath is 2.3 cm. and inner radius of iron sheath is 2.6 cm.; then, C = ' 89 9 4 5 = 0.732 K mf . per mile, 2 (Ci + Ci 2 ) = 0.134 K mf. per mile C'C 1 and C e = ^fr = 0.114 K mf . per mile. o -f- OQ Capacity of Three-core Cables. As in the case of a two-core cable we can express the capacities of the various possible ar- rangements in terms of Ci and C&. It is assumed that the conductors are symmetrically placed with respect to the axis of the sheath. 118 FORMULAE AND TABLES FOR THE The formulae for C\ and Ci2 are an 2 ai2 2 an* 2 ai2 3 (72) FIG. 35. where an = 2 log* Rr R = radius of sheath, r = radius of conductor, d = distance between axes of conductor and axis of sheath. An equivalent formula to (72) is the following: 1 1 61og e 3 RWr 1 61og e -d 1 6 log. R*-d 2 (73) Formulae (72) and (73) were both derived by the aid of the prin- ciples of images and give identical results. The values of Ci and Ci2 in formulae (72) and (73) are given in electrostatic units. To convert them to microfarads per mile we must multiply by the numerical factor 0.179 and, since the con- ductors are embedded in a dielectric, we must multiply by K the CALCULATION OF ALTERNATING CURRENT PROBLEMS 119 specific inductive capacity of dielectric. As an illustration we shall consider a cable of the following dimensions: r 0.5 cm., R = 2.5 cm., d = 1.3 cm. ai = 2 log* 3.65 = 2.6, ecu = log* 1.66 = 0.51, = j(2.6) 2 - (Q.51) 2 jO.: 1 (2.6) 3 - 3 X 2.6 X (0.51 1 (0.51) 2 - 2.6 X 0.51 1 K X 0.179 C 12 - ( 2 . 6) 3_ 3X2>6X(0>51)2 + 2(0 . 51)2 = -0.0119mf.permile. By formula (73) we find for the same example, 7?2 _ ^72 = 0.63, R*d* d V3 and Ci = 0.074 K mf . per mile, C M = - 0.0119 K mf. per mile. From this example it is seen that both formulae give identical results. We shall assume now that the three cores of the cable are con- nected to a three-phase star-connected generator supplying an e.m.f. of pure sine form, the neutral point and the lead sheath being grounded. In this case the effective capacity of each con- ductor is C. = Ci- Ci 2 . (74) The charging currents on the three conductors are equal to each other in amplitude and differ, of course, in phase by 120 degrees; that is, /i = wC e E m cos coZ, 7 2 = wC e E m cos U* - -j, . 7 3 = uCeEm COS f Cot ~ If we ground one conductor and connect the other two con- 120 FORMULAE AND TABLES FOR THE ductors to a single phase generator, the effective capacity of each conductor is C e = Cl ~ C *. (76) The charging currents on the two conductors are equal and oppo- site in sign; /I = 7 2 = CeO)E m COS w. A three-core cable is supplied with a single phase alternating cur- rent, two conductors connected in parallel for the transmission of the outgoing current and the third conductor being the return, the lead sheath grounded. Such an arrangement offers the somewhat peculiar condition that for the same e.m.f. the charging currents on the conductors differ when the neutral point of the winding is grounded or insulated. That is to say, the effective capacities of the conductor depend on the electrical condition of the stator winding, grounded or insulated. In the first case when the neutral point of stator winding is grounded the effective capacities of the three conductors, 1, 2 and 3, are | Ci, i Ci, J (Ci 2 C^), respectively. The charging currents are cos orf, /S = - i (Ci - 2 Cu) wE m COS CO*. The charging current on the inner surface of the sheath is Is = - i (Ci + 2 Ci 2 ) wE m cos co, and /I + /2 + /8 + /. = 0. It is evident from (77) that the currents in the outgoing and return conductors are not equal. When the stator winding is insulated, the effective capacities of the three conductors, 1, 2 and 3, are i (C[ - C 12 ), J (Ci - C B ) and f (C 12 - d), respectively. The charging currents are II = /2 = | (Ci - C i2 ) uE m COS 0)t,\ /8 = i(C u -Ci)tf w cos*. ) Evidently, as required, CALCULATION OF ALTERNATING CURRENT PROBLEMS 121 The above example shows that the effective capacity depends not only on the arrangement of the conductors but also on the electrical condition of the stator winding with respect to the cable, that is grounded or insulated. A list of the capacities that can be obtained from a three-core cable is given in Russell's " Alternating Current Theory "as follows: (1) Capacity between 1 and 2 (3 grounded) = J (Ci 12). (2) Capacity between 1 and 2, 3 = f (Ci - Cy. (3) Capacity between 1 and S (2 and 3 insulated) = (Ci -C 12 ) (Ci + 2C 12 ) Ci + C 12 (4) Capacity between 1 and S, 2 (3 insulated) _ (Ci - C M ) (Ci + Cn) (5) Capacity between 1 and S, 2, 3 = Ci. (6) Capacity between S and 1, 2 (3 insulated) _2(C 1 -C 12 )(C 1 +2C 12 ) Ci (7) Capacity between 1, S and 2, 3 = 2 (d + C 12 ). (8) Capacity between S and 1, 2, 3 = 3 (Ci + 2 Cu). Example. Taking the values of Ci and C& as given on page 119, Ci = 0.074 mf. per mile, Ci2 = - 0.0119 mf. per mile, we get (1) Capacity between 1 and 2 = 0.043 mf. per mile. (2) Capacity between 1 and 2, 3 = 0.0573 mf. per mile. (3) Capacity between 1 and S (2 and 3 insulated) = 0.0695 mf. per mile. (4) Capacity between 1 and S, 2 (3 insulated) = 0.072 mf. per mile. (5) Capacity between 1 and S, 2, 3 = 0.074 mf. per mile. (6) Capacity between S and 1, 2 (3 insulated) = 0.1165 mf. per mile. (7) Capacity between 1, S and 2, 3 = 0.1242 mf. per mile. (8) Capacity between S and 1, 2, 3 = 0.1863 mf. per mile. 122 FORMULAE AND TABLES FOR THE TABLE XI VALUES OF C AND C" For use in Formulae (25) and (26) _d C a C' a i V = a' C a C' a 2.0 6 1 4.6647 0.6931 2.6 0.8552 0.7340 2.0 S 1 4.0891 0.6931 2.7 0.8275 0.7401 2.01 3.5134 0.6931 2.8 0.8044 0.7460 2.0 3 1 2.9378 0.6931 2.9 0.7847 0.7517 2 .Oil 2.3626 0.6932 3.0 0.7677 0.7572 2.01 1.7896 0.6942 3.1 0.7528 0.7625 2.02 1.6191 0.6946 3.2 0.7396 0.7677 2.03 1.5202 0.6954 3.3 0.7278 0.7726 2.04 .4506 0.6961 3.4 0.7172 0.7774 2.05 .3970 0.6968 3.5 0.7076 0.7820 2.06 .3536 0.6975 3.6 0.6989 0.7864 2.07 .3171 0.6983 3.7 0.6909 0.7907 2.08 .2857 0.6990 3.8 0.6836 0.7948 2.09 .2582 0.6997 3.9 0.6768 0.7988 2.10 .2337 0.7004 4.0 0.6705 0.8026 2.15 1.1413 0.7040 5.0 0.6263 0.8345 2.20 1.0775 0.7075 6.0 0.6006 0.8577 2.25 1.0291 0.7117 7.0 0.5836 0.8753 2.30 0.9911 0.7144 8.0 0.5712 0.8891 2.35 0.9594 0.7178 9.0 0.5626 0.9001 2.40 0.9326 0.7211 10.0 0.5556 0.9092 2.45 0.9095 0.7245 - 100 0.5051 0.9901 2.50 0.8892 0.7277 1000 0.5005 0.9990 TABLE XII d 2~a' -* cosh. A. d 2a' *.'-. cosh-'^. 1.01 0.7031 0.1413 3.5 1.9459 1.9248 1.05 0.7419 0.3149 4.0 2.0794 2.0634 1.1 0.7885 0.4435 4.5 2.1972 2.1846 1.2 0.8755 0.6224 5.0 2.3025 2.2924 1.3 0.9555 0.7564 5.5 2.3979 2.3896 1.4 .0296 0.8670 6.0 2.4849 2.4779 1.5 .0986 0.9622 7.0 2.6390 2.6339 1.6 .1631 .0470 8.0 2.7726 2.7687 1.7 .2238 .1232 9.0 2.8903 2.8873 1.8 .2809 .1929 10.0 2.9956 2.9932 1.9 .3350 .2569 12 3.1780 3.1763 2.0 .3863 .3170 14 3.3322 3.3309 2.2 .4816 .4255 16 3.4657 3.4648 2.4 1.5686 .5216 18 3.5835 3.5827 2.6 1.6487 1.6096 20 3.6888 3.6882 2.8 1.7228 1.6886 25 3.9119 3.9116 3.0 1.7918 1.7627 CALCULATION OF ALTERNATING CURRENT PROBLEMS 123 TABLE XIII SPECIFIC INDUCTIVE CAPACITIES OF SOLIDS (AIR = UNITY) Substance. Specific induc- tive capacity. Authority. Calcspar parallel to axis 7.5 Calcspar perpendicular to axis .... 7.7 Caoutchouc 2.12-2.34 Caoutchouc, vulcanized 2.69-2.94 Celluvert, hard gray 1 . 19 Celluvert, hard red 1 .44 Celluvert, hard black 1 .89 Celluvert, soft red 2. 66 Ebonite 2.08 Ebonite 3.15-3.48 Ebonite 2.21-2.76 Ebonite 2.72 Ebonite 2.56 Ebonite 2.86 Ebonite 1.9 Fluor-spar 6.7 Fluor-spar 6.8 Glass,* density 2.5 to 4.5 5-10 Double extra dense flint, density 4.5 9 . 90 Dense flint, density 3.66 7.38 Light flint, density 3.20 6.70 Very light flint, density 2.87 6.61 Hard carbon, density 2. 485 6.96 Plate 8.45 Mirror 5.8-6.34 Mirror 6.46-7.57 Mirror 6.88 Mirror.., 6.44-7.46 Romich and Nowak. Romich and Nowak. Schiller. Schiller. Elsas. Elsas. Elsas. Elsas. Rossetti. Boltzmann. Schiller. Winkelmann. Wiillner. Elsas. Thomson (from Hertz's vibrations). Romich and Nowak. Curie. Various. Hopkinson. Hopkinson. Hopkinson. Hopkinson. Hopkinson. Hopkinson. Schiller. Winkelmann. Doule. Elsas. * The values here quoted apply when the duration of charge lies between 0.25 and 0.00005 of a second. J. J. Thomson has obtained the value 2.7 when the duration of the charge is about of a second; and this ia confirmed by Bloudlot, who obtained for a similar duration 2.8. 124 FORMULAE AND TABLES FOR THE TABLE XIII (Continued) SPECIFIC INDUCTIVE CAPACITIES OF SOLIDS (AIR = UNITY) Substance. Specific induc- tive capacities. Authority. Guttapercha 3.3-4.9 Submarine cable data. Gypsum . 6 33 Curie. Mica 6 64 Klemencic. Mica . . 8 00 Curie. Mica 7.98 Bouty. Mica 5 66-597 Elsas Mica 4 6 Romich and Nowak Paraffin 2 32 Boltzmann. Paraffin 1.98 Gibson and Barclay. Paraffin 2 29 Hopkinson. Paraffin, quickly cooled, translu- cent 1.68-1.92 Schiller.* Paraffin, slowly cooled, white Paraffin fluid, pasty.. 1.85-2.47 1 98-2 08 Schiller. Arons and Rubens. Paraffin, solid... 1.95 Arons and Rubens. Porcelain 4.38 Curie. Quartz along the optic axis Quartz, transverse 4.55 4 49 Curie. Curie. Resin 2 48-2.57 Boltzmann. Rock salt 18 Hopkinson. Rock salt 5.85 Curie. Selenium 10 2 Romich and Nowak. Shellac 3.10 Winkelmann. Shellac 3 67 Doule. Shellac . 2.95-3.73 Wiillner. Spermaceti 2.18 Rosetti. Spermaceti 2 25 Felici. Sulphur 3 84-3 90 Boltzmann. Sulphur. 2 24 J. J. Thomson. Sulphur 2.94 Bloudlot. * The lower values were obtained by electric oscillations of duration of charge about 0.0006 second. The larger values were obtained when duration of charge was about 0.02 second. CALCULATION OF ALTERNATING CURRENT PROBLEMS 125 TABLE XIV SPECIFIC INDUCTIVE CAPACITIES OF LIQUIDS Substance. Specific induc- tive capacity. Authority. Alcohols: Amyl 15-15.9 Cohen and Arons. Ethyl 24-27 Various. Methyl 32.65 Tereschin. Propyl 22.8 Tereschin. Anilin 7.5 Tereschin. Benzene 1.93-2.45 Various. Hexane, between 11 and 13 C.. . . Octane, between 13.5 and 14C.. . Decane, between 13.5 and 14.2 C. Amylene, between 15 and 16. 2 C. Octylene,betweenll.5andl3.6C. Decylene, between 16.7 C Oils: Arachid 1.859 1.934 1.966 2.201 2.175 2.236 3.17 Landolt and Jahn. Landolt and Jahn. Landolt and Jahn. Landolt and Jahn. Landolt and Jahn. Landolt and Jahn. Hopkinson. Castor 4 6-4 8 Various Colza 3.07-3 14 Hopkinson Lemon 2 25 Tomaszewski Neatsf oot 3.07 Hopkinson. Olive 3.08-3.16 Arons and Rubens ] Hop- Petroleum Petroleum ether Rape seed 2.02-2.19 1.92 2 2-3 kinson. Various. Hopkinson. Various Seasame 3 17 Hopkinson Sperm 3 02-3 09 Hopkinson Turpentine 2.15-2.28 Various. Vaseline 2.17 Fuchs. Ozokerite 2.13 Hopkinson. Toluene 2.2-2.4 Various. Xylene 2 3-2 6 Various 126 FORMULAE AND TABLES FOR THE CHAPTER IV ALTERNATING CURRENT CIRCUITS IN alternating current circuits the distribution of the currents and potentials in any arrangement of circuits is governed by the frequency of the impressed electromotive force, the resistances, self and mutual inductances and capacities of the various branches of the circuit. A variation in any of the electrical constants may result in a change in the currents, in magnitude as well as in phase, in all the branches of the circuit. Every alternating current problem must therefore be considered separately and careful account must be taken of all the electrical constants of the circuit. Since there is a wide range of possible arrangements of circuits, we shall require a considerable number of formulae to cover all the more important cases which may arise in practice. It is also to be noted that the permanent value of the current, which in the case of alternating currents is a constant periodic function of the time, is reached only after an appreciable time interval when the circuit is closed. At the moment of closing or opening, or on any change in the electrical condition of circuits, a transient phenomena occurs which is usually active only for a very short interval. The formulae for transient phenomena will be considered in Chapter V, while in this chapter we shall give a collection of formulae for the permanent values of current and potential distribution in various arrangements of circuits. In this and the succeeding chapters the following notation will be used : Notation. r = resistance. L = inductance. C = capacity. N = frequency, co = 2 vN. x m = wL = inductive reactance. CALCULATION OF ALTERNATING CURRENT PROBLEMS 127 x c = -7 ^ = capacity reactance. coC x = x m x c reactance of circuit having inductance and ca- pacity. Z = r + j% impedance. z = Vr 2 + x 2 = absolute value of impedance. Y = -^ = admittance. Li y = Vg 2 -\- b 2 = absolute value of admittance. g = conductance. b = susceptance. 1 rjx -' g = -g r- 2 = -g = effective conductance in mhos, energy 7* ~T~ X component of current. /v /v 6 = -5 ; 5 = -s = effective susceptance in mhos, wattless r 2 + a; 2 z 2 component of current. r = 2 *. , 2 = 2 = effective resistance in ohms, energy component of e.m.f . s = o , 79 = 9 = effective reactance in ohms, wattless g 2 + b 2 y 2 component of e.m.f. Any alternating current wave can be resolved by the aid of Fourier's theorem into a series of waves, each one being a sine function in its variation with respect to time; hence in the dis- cussion of alternating current problems it is nearly always justi- fiable to assume that the impressed electromotive force is of sine form. If it is a complex wave it can be resolved into its several harmonic components, and each one treated as a separate source of e.m.f. If E is a periodic function with respect to time we can put it in the following form : ^ 3 )H ---- (1) The alternating electromotive force and current curves which occur in practice have only odd harmonics. A complete dis- 128 FORMULAE AND TABLES FOR THE cussion of the theory of harmonic analysis is beyond the scope of this book, and to give only an outline of the subject would not be of any practical value. The reader will find a good discussion on the general subject of harmonic analysis in Byerly's Fourier's Theory and Spherical Harmonics, and its application to the analysis of alternating current curves is quite fully discussed in the " Theorie der Wechselstrome," by J. L. LaCour and 0. S. Bragstad. In alternating currents we have to distinguish between the maximum value, the effective value, which is the square root of the mean square, and the mean or average value. In the case of a A simple harmonic function the effective value is 7= where A is the V2 maximum value, and the mean or average for a half period is - 7T The arithmetic mean for a whole period is zero. In Table XV we give the relation between effective, mean and maximum values for various shapes of curves. TABLE XV CHARACTERISTIC FEATURES OF DIFFERENT FORMS OF ALTERNATING CURRENT AND PRESSURE CURVES Area of Ratio of Ratio of Ratio of Ratio of squared Name of curve. average to maximum effective to maximum maximum to effective effective to average responding value. value. value. value. maximum value. Sinusoid 0.637 707 1.414 1.112 0.500 Semicircle 0.785 0.835 1.198 1.063 0.697 Parabolic curve 0.666 0.730 1.369 1.096 0.533 Triangle 0.500 0.577 1.732 1.155 0.333 Approximate rectangle. 0.856 0.889 1.124 1.038 0.791 Rectangle 1.000 1.000 1.000 1.000 1.000 In the case of complex waves consisting of several harmonies the effective value is** E ea = V(#i 2 +#3 2 + #5 2 + . - ), (2) where E\, E z , E$, etc., are the maximum values of the several harmonic components. * See "Theorie der Wechselstrome," von J. L. LaCour und O. S. Bragstad, Ch. XII. CALCULATION OF ALTERNATING CURRENT PROBLEMS 129 Form factor. The ratio of the effective value of a periodic curve to its mean value is called the form factor, because it varies with the form of the curve, the more peaked the curve the greater the value of the form factor. For a pressure curve the form factor is given by the expression (3) Edt For a sine curve the form factor is 1 2 TT V2 ' * 2\/2 = 1.112. Curve factor. The ratio of the effective value to the maxi- mum value of the fundamental harmonic is called the curve factor. E m As an illustration consider the case of an e.m.f. curve having three upper odd harmonics. The maximum values of the fundamental and the harmonics are E! = 100, # 3 = 40, E, = 20, E 7 = 10. The effective value is # efl = V (100) 2 + (40) 2 + (20) 2 + (10) 2 = 110. The curve factor is e _ IT lop Alternating Current Circuits. When an electromotive force of sinusoidal form, E cos ut, is impressed on a circuit containing inductance and resistance, it generates a current in the circuit. C S "* - the current lagging behind the electromotive force by the angle 130 FORMULAE AND TABLES FOR THE Example. r = 5ohms, L = 0.01 henry, w = 2^X60, E = 500 volts. x m = coL = 3.73. 500 V(3.77) 2 + (5) 2 = 80 amp. tan ^ = 0.754, = 37 3'. The maximum value of the current is 80 amperes and it lags 37 3' behind the e.m.f. If the circuit contains capacity ^reactance and resistance, the current is Tjl cos (ut + ^), (6) the current leading the electromotive force by the angle \f/. The voltage on the condenser is x c E Example. r = 5 ohms, C = 0.002 farads, co = 2 IT X 60, E = 500 volts. x c = 4 = 1-324, E 500 V(1.324) 2 = ^P = 0.265, I = 14' o = 96.7 amp., 51'. The current leads the e.m.f. by the angle 14 51 r . The voltage across the condenser is 1.324 X 500 . e , = 128 volts. V (1.324) 2 + (5) 2 When the circuit contains both capacity reactance and inductive reactance. we have E * vA^ 1 -* // x V (x m - x c ) tan 7 = CALCULATION OF ALTERNATING CURRENT PROBLEMS 131 When x m = x c , equation (8) reduces to 7 I = cos wt. (9) The circuit acts as if there is no reactance, and it is said to be in resonance with the frequency of the impressed electromotive force. The condition for resonance then is or or J^ C' 1 1, VLC (10) 30 28 26 24 22 20 18 4-> I" 12 10 $ / Curve I, R = 10 Ohms . II, R= 5 HI, JJ=3.5 - IV, #=2.0 n ss \\ 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 86 38 40 i FIG. 36. Resonance curves for different values of resistance in the circuit. In any circuit containing inductance and capacity it is always possible to adjust those constants so as to produce resonance condition for a given frequency. For a given electromotive force, the current rises as the resonance condition is approached, which is more pronounced the smaller the resistance in the circuit. In Fig. 36 are plotted a series of resonance curves for different resistances. 132 FORMULAE AND TABLES FOR THE The voltage across the condenser is x c Esm(at-y) ~ For resonance condition and small resistance, the voltage across the condenser may exceed many times the voltage of the generator. Example. w =2irX60 C = 20mf. 1325 = 66.2 # sin o>*. The voltage across condenser is about 6600 volts, 66 times the electromotive force of the generator. If the electromotive force is a complex wave, that is, of the form E = EI cos (co + i) +#3 cos (3 ut + as) + E$ cos (5 coi + #5) the current generated in the circuit is , __ EI cos (co< + cti 7i) , E 8 cos (3 at + 3 73) V(z w -z c ) 2 + r 2 ^5 COS (5 (0^ + <^5~ 7s) i ^m "o" 5 x m ^~ -^ - tan 73 = tan 75 = (12) It may happen that the inductance and capacity of the circuit are of such magnitudes as to bring the circuit into resonance for / 7* some harmonic of the wave, for instance we may have 3 x m - o = 0, that is, the circuit is in resonance for the third harmonic. In that case the current produced by the third harmonic in the elec- tromotive force wave may be larger than that produced by the fundamental, though the amplitude of the e.m.f . in the fundamental may be much larger than that of the third harmonic. CALCULATION OF ALTERNATING CURRENT PROBLEMS 133 Example. C = 7.77 mf. L = 0.1 henry. co = 2 TT X 60. Ei = 100, Ez = 10, r = 10 ohms. = 0.33 amp. 10 = = lamp. The current due to the first harmonic of the e.m.f. curve is three times as large as the current due to the fundamental, though the maximum value of the fundamental in the e.m.f. curve is ten times as large as the first harmonic. If there are several impedances Zi, 2 2 , 23 connected in series in the circuit, the resultant is the algebraic sum of the separate impedances. Denoting by z~ t the total impedance, we have 2* = '3i + 3z + 23 + - = (TI + r 2 + r 3 + ) i + x 2 + x 3 + . . . ), (13) and the absolute value of the current, / = - JL (14) ) 2 + (ri + r 2 + r 3 + - - - ) 2 The e.m.f. E n across any impedance coil, say the nth coil, is -v/r 2 4- r 2 E n = ^ ^^ ' (15) x 2 + z 3 + ) 2 + (ri + r 2 + r 3 + ) 2 Parallel Circuits.* Two circuits in parallel, each having resistance and inductance. Denoting the reactances and im- pedances of the two branches by Xi, x 2 and 2i, Z 2 respectively, the * Alex. Russell, "Theory of Alternating Currents," Vol. I, Ch. VII. Andrew Gray, "Absolute Measurements in Electricity and Magnetism," Vol. II, Pt. I, Ch. IV. 134 FORMULAE AND TABLES FOR THE currents in the two branched circuits and the main circuit are E 1 1 cos (co t an0 n r 2 or we may write 2 21 22 E cos M - ) 7 '.. . o = / o . = r cos W ~ w V r 2 + a; 2 2 where jTi i J5 2 2 '02 r = r - rr ; - r-ris the equivalent resistance of the circuit and 2 + 2 is the ec l uivalent react - ance of the circuit. If the resistances and reactances of the two branched circuits are equal, the above values reduce to r = \ r, x = i x. Example. Xi = 5 ohms, ri = 3 ohms, x 2 = 3 ohms, r 2 = 5 ohms. 2l = z 2 = 5.83, E = 500 volts. E Ii = = 85.8 amp., 2l 7 2 = = 85.8 amp., 22 = tan- 1 = 59 3', _^ __ 3>~~ 02 = tan-i = 31, CALCULATION OF ALTERNATING CURRENT PROBLEMS 135 XQ = 5 + 3 34 17 17 + (A) 2 8 500 = 166.4 amp., = - 1 - = = tan = tan- 1 1 = 45. The current in the main circuit is less than the sum of the currents in the two branches, and this is because the phase angles in the two branches are different. -AAAAAT FIG. 37. Two Branch Circuits, One having Resistance Only and the Other Inductance Only. One branched circuit has a resistance r 2 and the other an inductance LI of negligible resistance, the main circuit inductance and resistance being r and L (see Fig. 37). The total impedance of circuit A B is given by the following expression : This is a maximum when co 2 Li (L 2 + 2 LC 2 r c . (18) For the value of r 2 given by equation (18), the current in the main circuit is a minimum. Hence shunting an inductance coil with a resistance sometimes increases the apparent resistance of the circuit, a result which has been noticed in practical work. 136 FORMULAE AND TABLES FOR THE Example. Lo = L 2 =0.1 henry,' r = 50 ohms, r 2 = 129 ohms, co = 2w X 60. By (17) we have Z 2 = (68.1) 2 + (61.75) 2 = 8450.67, Z = 91.9 ohms. If the inductance L% were not shunted by any resistance, the impedance of the circuit would be Z 2 = co 2 (Li + L ) 2 + r 2 = (75.4) 2 + (50) 2 = 8185.16, Z = 90.5 ohms. Shunting the inductance by a resistance increases the impedance of the circuit instead of decreasing as we would ordinarily expect. A System of Branched Circuits in Parallel Each having Resis- tance and Inductance. Let z\, %, 0,3 ... denote the impedance of the respective branches and ZQ the total impedance of the cir- cuit. The currents in the respective branches and the main circuit are as follows: TjJ /! = --cos(coZ - ) I 2 = Ex c cos co, (21) _ o/c o/ m T i ** /oo^^ tan ^ = - = -- (22) r 2 I r 2 I ^^ U/jji When - = 2 *_7 2 (23) *^c ^m I ' the circuit is in resonance, that is, the main current is in phase with the electromotive force and Er lo = -^2 2 cos ut. (24) T \ X m From (23) it is obvious that for this arrangement the resonance condition depends not only on the inductance and capacity, but also on the resistance. For every increase in r, x c must be increased and the capacity decreased to maintain the resonance condition. By varying x c we can maintain the current in the main circuit in phase with the e.m.f. for any variation in the resistance of the receiving circuit. As an illustration we assume a value of x m = 15, and calculate the values of x c required to maintain the resonance condition in the circuit for values of r from 1 to 10 ohms. The results are plotted in the curve given in Fig. 39. The e.m.f. E Q across the resistance r is Er .cos(co-0). (25) ,' + r' It is also to be noted that in such an arrangement of circuits, at resonance condition, the currents in the branches are larger than the currents in main circuit. Example. x m = 25, r-10, f - TowSlTnyS = - 0345 ' # = 100 volts, x c \ "<->/ -f- v^-w 100 3.7 amp., CALCULATION OF ALTERNATING CURRENT PROBLEMS 139 7 2 = 100 X 0.0345 = 3.45 amp., 100 X 10 625OO The currents in the branch circuits are considerably larger than the current in the main circuit. 21 20 19 -1 17 16 15 / / Curve giving Variation in Xc with r so as to Maintain Main Circuit Current in Phase with e.m.f. / ' ^ / ^ / x ^ x 7 ^x ^ / i x X x ? _' ^ ^ , . ,~- 1 2 3 4 5 r 6 7 8 8 1C FIG. 39. FIG. 40. Inductive Load Shunted by Condenser. Assume an elec- tromotive force of sine form impressed on the circuit and a react- ance x in series with alternator (see Fig. 40). Denote by /i the current in resistance branch, 7 2 the current in condenser, 7 the main current, E the voltage of alternator and E Q the voltage on the receiving circuit. 140 FORMULAE AND TABLES FOR THE The distribution of the currents and voltage are as follows: Ex c (x e - tan x c (x (x c - / 2 =- E Vrt 2 + V j z c '(z + Zi) - xixol 2 + ri 2 (x c - cos (coi i), 0i 4- i/O- , x l tan y = > T cos (otf tan 7 = Ex c n 2 + (26) If we make x c = Xi XQ = x the denominators in the above equation reduce to x and the equations simplify to 1 1 = sinotf, x (27) From these equations we can draw the interesting conclusion that for a constant value of E the current in the receiving circuit /i is constant, and independent of any variation in the load r\. In other words this arrangement of circuits acts like a transformer converting a constant potential into a constant current. Circuits in Parallel. Two branch circuits in parallel, one having inductance and resistance and the other inductance capacity and resistance (see Fig. 41). Denote the current in branch (1) by 7i, that in branch (2) by 7 2 and the current in main circuit by /o; then CALCULATION OF ALTERNATING CURRENT PROBLEMS 141 E Sl 2 tanT= - (28) FIG. 41. The circuit is in resonance when n I o T = o. (29) Putting for brevity + = k, equation (29) becomes + fcr and - liVT 11 2/b (30) The resonance condition depends on the resistance of the branch circuits as well as their inductances and capacities. If the con- stants of the circuits are of such magnitudes as to make 4 & 2 r 2 2 greater than unity, the value of x 2 as given by (30) is an imaginary quantity, that is to say it is not possible to produce resonance in the circuit. If 4 & 2 r 2 2 is less than unity, x% has two distinct 142 FORMULAE AND TABLES FOR THE values. The system is double periodic; for the same constants we may have resonance at two different frequencies. As an illustration, let k = 0.1, r 2 5. - 1 Vl - 0.2 OK ** = 2x0.1 - 95; that is, if the frequencies are such as to make x% = 5 or 95, the circuit is in resonance and the current in the main circuit is /o = r 2 2 \ i?+^j cos co. (31) The Branch Circuits as Well as the Main Circuit having In- ductance, Capacity and Resistance. The currents in the branch circuits are denoted by I\ and 7 2 respectively, and the current in the main circuit by 7o- Vc 2 + d 2 E + cos (ut + Vc 2 + d 2 cos(o>Z + (32) X O =LO u- 7 Q U where tan0=-, and c = r d = n FIG. 42. , tan \f/z TZ CALCULATION OF ALTERNATING CURRENT PROBLEMS 143 When the inductances and capacities are so adjusted that each branch is separately in resonance, we get Xi = X Z = XQ = 0, c = TO (ri+ r 2 ) + rir 2 , d = 0, and V, /2 = Er 2 TO (r. + r 2 ) + ri En r<> (n + r 2 ) + n E (n + r 2 ) p.ns t,\t. (33) TO (n + r 2 ) + Mutual Inductance Between the Branched Circuits.* We shall now consider the case of two branched circuits, in which t AT FIG. 43. there is also mutual inductance between the two branches. De- noting the currents in the branched circuits by /i and 7 2 respec- tively and the current in the main circuit by I we have the following formulae for the currents in the circuits : Vfco 2 (M 2 -L 1 L 2 )+r 1 r 2 p+a, 2 (L 1 r 2 +L 2 r 1 ) 2 cos coc EV(Li- VJco 2 (M 2 -L 1 L 2 )+r 1 r 2 j 2 +co 2 (L 1 r 2 +L 2 r 1 > 2 E - 2 r 2 ) Vjco 2 (M 2 -L 1 L 2 )+7- 1 r 2 j 2 +co 2 (L 1 r 2 +L 2 r 1 ) 2 CO (LiT-2 + cos(coi (34) tan \I/2 = tan r 2 . * J. J. Thomson, "Recent Researches in Electricity and Magnetism," Ch. VI; Lord Rayleigh, Phil. Mag., May, 1886. 144 FORMULAE AND TABLES FOR THE It is interesting to note that in such an arrangement of circuits, the currents in the branched circuits may, under some circum- stances, be considerably larger than the current in the main cir- cuit. Considering only the amplitudes of the currents, we have V(L 2 - M) 2 o> L 2 - 2 (n + r 2 ) s h /o \/(Li - M) 2 o; 2 L 2 - 2 M) 2 co 2 + (n + r 2 ) (35) (36) If we neglect the resistances of the two branches compared with the reactances we have 7i = L 2 -M and -^ = (37) As an extreme case we may take the coefficient of coupling equal to unity, that is, M = VLiL 2 , and Li+L 2 -2 -VL, Li + L 2 - 2 i - VL 2 (38) /- If LI and L 2 are not much different from each other the denomi- nators in equations (38) are very small and consequently the ratios -^ and ~ are large; that is, the currents in the branch cir- -/o -to cuits are considerably larger than the current in the main circuit. As an illustration let us take the following constants: Li = 0.1 henry, L 2 = 0.01 henry, M = 0.03 henry. /o 0.02 _ 7 2 0.07 0.05 ' / 0.05 1.4. The current in branch 2 is 1.4 times the current in the main cir- cuit. Power Factor.* If an alternating e.m.f., e = E V2 cos at, * See C. P. Steinmetz "Theory and Calculation of Alternating Current Phenomena," 3d edition, Ch. XXVII; also "Theorie der Wechselstrome," von J. L. LaCour und O. S. Bragstad, Ch. XII. CALCULATION OF ALTERNATING CURRENT PROBLEMS 145 produces in a circuit a current, i = lV2 cos (at =t 0), where is the phase angle lagging or leading, the power p = ei = EIlcos + cos (2 = 7 Ir COS0 = -g' The factor cos is called the power factor. Substituting (40) in (39) we get 40) p=Pl+- -^ (41) [ COS0 In an ordinary alternating current circuit the power fluctuates between the values and PJl I- (42) [ COS0J When the phase angle is zero, cos = 1, the fluctuation of the power is between 2 P and 0. If the phase angle is 90 degrees, p = P (1 + cos 2 o>Z), and the average or effective power P = 0. If the pressure curve is not a simple harmonic curve but con- tains also harmonics, the average power supplied to the circuit is given by expression, P = EJi COS 0i + #3/3 COS 03 + EJ&OS 05 + - - - = 7cos0, (43) where E and I are the effective values of the e.m.f . and the current, and ., (44) COS 2 0! + COS 2 03 + COS 2 5 + f) 146 FORMULAE AND TABLES FOR THE where = 1 j^> (52) The efficiency for maximum power supplied to receiving circuit is Pmax = *_ . For comparison we will give here a set formulae for a non-induc- tive line corresponding to equations (46) to (53), which are as follows: Power supplied to receiving circuit is (56) 148 FORMULAE AND TABLES FOR THE and it is a maximum when r = TO- Total power supplied is (57) -r + r.-27' The efficiency for maximum power supply to receiving circuit is P 1 * max -* 120 110 100 90 80 70 60 50 40 30 20 10 100 90 80 ' 70 60 | 50 l 40 30 20 10 - *^- _ ^ X ^ ^ / / Sx ^ ^ s., ^ / *$> / X \ ^ / / \ 7 / / ^ ^ ^~ ^~- . 1 / ^ ^^ "^ I / f s ^ 1 *. *^ ~-^^ 1 / f / Inductive Line, r=4Ohms Curve, I Efficiency II Current III Voltage E r u IV Power f^ , / 7 // iff f 700 650 'geoo ^550 1 500 450 2 400 350 ^300' >250 1200 150 100 50 0123456789 10 11 12 13 14 15 16 17 18 19 20 r FIG. 45. Power transmission characteristic curves. As an illustration we give here a set of curves showing the variation with the load of /, E r , TQ and P r for an inductive line of the following constants: E = 1000 volts, r = 4 ohms, x = 8. Inductive Load. Let Xi, r\ represent the reactance and resist- ance of transmission line, x z , r 2 the reactance and resistance of receiving circuit, E g = e.m.f. of generator, E r = e.m.f. at receiving circuit. 7TT Tjl T &a &r + r 2 ) (59) r 2 CALCULATION OF ALTERNATING CURRENT PROBLEMS 149 The phase angle fa between E g and I is given by Xi + %2 tan fa = -r and the phase angle fa between E r and I is given by *v2 tan fa = We may put equation (59) in the following form : I = E \/ ^ 2 + ^ 2 (60) V (1 + rift + i6 2 ) 2 + (sift - ribz) 2 and =^= = * a _ (61) where ft = o , 2 5 = effective conductance of receiving circuit, r 2 2 + # 2 2 r = effective susceptance of receiving circuit. 2 _ The power supplied to receiving circuit, P = E r *g 2 = E g *g**, (62) and it has its maximum value when ft = Vft 2 + (bi + 6 2 ) 2 ; (63) and for this value of ft we have max = x 1 =- (64) V 2 ft (ft^i 2 + rj The maximum power that can be supplied to the receiving cir- cuit is E 2 Pmax = o / / . N (65) 2 (ft2i 2 + n) For constant susceptance 6 2 the efficiency has its maximum value when 02 = &2 and (66) If we do not neglect the capacity of the line, the problem be- comes somewhat more complex, but it may be simplified very materially by the following considerations : 150 FORMULAE AND TABLES FOR THE In engineering practice, power transmission problems frequently present themselves in the following way. Given the values of the voltage and current and their phase relation at the receiving circuit, what will be the corresponding values of voltage and current at the generator end, the power transmitted, efficiency, etc.? And vice versa, if the e.m.f. and current are given at the generator end what will be the corresponding values at the re- ceiving end? In other words given the constants of the line and electrical conditions at one end of the line to determine all the other factors of the problem. As stated before, the exact solution of the problem necessitates taking into account the effect of the distributed inductance and capacity of the line which are worked out in Chapter VI. For short lines, however, we can get fairly accurate results by assuming localized inductance and capacity. Different formulae may be obtained depending on the assump- tions made regarding the capacity distribution, and we may have the following cases: (1) Capacity of line neglected entirely. (2) Entire capacity of line shunted across the middle of the line. (3) Half of line capacity shunted across at each end of the line. (4) Four-sixths of line capacity shunted across the middle of the line and one-sixth of line capacity at each end of the line. We shall give here formulae corresponding to the first three cases, and work out some examples to compare the closeness of approximation of the above three cases.* The fourth case was not considered for the reason that the formulae are somewhat complex and the results obtained by this method are not more accurate than those obtained by the method considered in case (3). It is not considered advisable to multiply the number of formulae unnecessarily. * Ch. P. Steinmetz, "Theory and Calculation of Alternating Current Phenomena," 3d edition, Chap. XIII. F. A. C. Perrine and F. G. Baum, The Use of Aluminium Line Wire and Some Constants for Transmission Lines, Trans. Am. Ins. E. E., May, 1900. P. H. Thomas, Output and Regulation in Long Distance Lines, Trans. Am. Ins. E. E., June, 1909. P. H. Thomas, Calculation of High Tension Lines, Trans. Am. Inst. E. E. t June, 1909. CALCULATION OF ALTERNATING CURRENT PROBLEMS 151 We shall use the following notation: Eg = Eg' jE " = e.m.f . at generator, I = lg jl g " = current at generator end, E r = Er j# r " = e.m.f . at receiving end of line, I r = Ir j7 r " = current at receiving end of line. The double sign in the above notation is to indicate whether the phase angle is leading or lagging; the plus sign meaning phase angle leading, and the minus sign phase angle lagging. Suppose the voltage and current are given at the receiving end of the line and the phase angle of the current with respect to the voltage is leading, then, E r = E r ', If on substituting these values in the formulae we obtain for the e.m.f. and current at generator end, Eg=Eg'-jE g ", it would mean that the e.m.f. at generator end lags behind the pi n e.m.f. at receiving end by an angle tan" 1 -^-, , while the current I " at generator leads the e.m.f. at receiving end by an angle tan" 1 ~r L Q Case I. Line capacity neglected entirely. Suppose that the admittance of receiving circuit g jb and E r are given, we have I r =E r (gjb)=I r 'jI,". Eg and I g have the following values: /. = Ir, E g = \E r Leo//' + rl/l +j \LuI r ' rl r ' Where a double sign appears in the above equation, the upper sign is to be used when the current at the receiver end is leading and the lower sign when the receiving current is lagging. If the voltage and current are given at the generator end, that is Eg and // jl g " given, we have = (GQ} TJ< f T7 1 i T T If T f t c T T / i T//I! \"'sJ E r = \EgL(*I 9 -rI I -j 152 FORMULAE AND TABLES FOR THE The double sign in the terms of equation (69) has the same sig- nificance as in (68). Example. Power to be transmitted over a three-phase line 50 miles long using hard-drawn stranded copper wire No. 000, tri- angularly spaced 10 feet spacing between wires. Frequency 60 cycles. At the receiving end we have the following conditions: Line voltage 30 kv. between wire and neutral, load current at re- ceiving end 150 amperes at 90 per cent power factor, lagging cur- rent. Phase angle lagging 25 50'. Calculating the inductance and capacity of the line by formulae given in Chapters II and III we find, r = 0.33 ohm per mile, L = 2.13 mh. per mile, C = 0.014 mf . per mile. The total resistance, inductance and capacity of the line are 16.5 ohms, 0.1065 henry and 0.7 mf. respectively. The electrical conditions at the receiving end are E r = 30 kv., I r = I/ - jl r " = 135 - j 65.4 amp. In formula (68) we neglect the capacity of the line entirely, hence introducing the values of r, L and w we get I g =I r = 135 - j 65.4 amp., E g = 530 + 2.63 + 2.23J + j {5.42 - 1.08J = 34.86 +.7 4.34. The absolute value of e.m.f. at generator is E g = vX/2 + E g "* = 35.1 kv. The voltage at generator leads the voltage at receiver by an angle hence the current at generator end lags behind the generator e.m.f. by the angle 25 50' + 7 5' = 32 55'. The power factor at the generator end is cos (32 55') = 0.84. The efficiency of transmission CALCULATION OF ALTERNATING CURRENT PROBLEMS 153 Example. The same conditions as in above example except that the line is 150 miles long. We have as in the previous example, I g =L = 135 - j (65.4) amp., E g - 44.58 +j 13.02 kv. The absolute value of e.m.f. at generator is E g = V(44.58) 2 + (13.02) 2 = 46.4 kv. The generator e.m.f. leads the voltage at the receiving end by the angle The current at generator end lags behind E g by the angle 25 50' + 16 16' = 42 6'. The power factor at generator end = cos (42 6') = 0.742. The efficiency of transmission is 30 X 150 X 0.9 46.4X150X0.742 = 7SA per Cent ' L r J ~2~'T =^X , 2 ' 2 T * 1 J T* ^7 I C Vr.I, 7 \ \ . . FIG. 46. Case II. Capacity of the line shunted across the middle of the line. If the admittance of receiving circuit g jb and the voltage E r are given, we have I r = E r (gjb) = I r ' db jl r ". The voltage and current at the generator end are given by the formulae I g = \KIr'^\CurI r "\ +JlCurI r 'KIr"+E r Cul =l g 'jl g ", (70) E g = E r + ;jr (// + //) - JLu (/," //Of + j \ i r (//' I r ff ) + i Leo (// + //) \ = E g ' jE g ", (71) where K = 1 - J 154 FORMULAE AND TABLES FOR THE The double signs in the terms of equations (70) and (71) have the same significance as in equations (68) and (69). The upper sign is to be used when I r " is positive, that is, the receiving current leading voltage at receiving end, and the lower sign is to be used when I r " is negative. If the voltage and current are given at the generator end, that is, Eg and I g = IJ jl g " are given, the current and voltage at the receiving end are given by the formulae: I r = I KI=F J rCuI g "\ + j\ KI g " = lr'jlr", (72) E T = E g - \\r (7/ + //)- JTxo (/," db 7/0} - JlMIr" /,") + i Txo (/'+//) J =E g 'jE,, (73) where as before K = 1 \ CT/co 2 . As an illustration and for comparison we shall work out ex- amples given above by formulae (70) and (71). r = 16.5 ohms, L = 0.1065 henry, C = 0.7 mf. K = 0.995, Ceo = 264 X 1Q- 6 . E r = 30 kv., I r = 135 - j 65.4 amp. The phase angle at the receiving end is 25 51' lagging. 7, H 134.3 + 0.14J + j ;o.29 -65.07 +7.92 j = 134.44 -j 56.86 amp.; absolute value of I g = V (134.44) 2 + (56.80) 2 = 146 amp. E g = (30 + 2.22 + 2.45) + j (-1.01 + 5.41) = 34.67 + j 4.4kv; absolute value, E g = V (34.67) 2 + (4.4) 2 = 34.95 kv. r> or> The phase angle between I g and E r is tan -1 ' . = 32 16 r , lagging. 1O4.4 4 4 The phase angle between E g and E r is tan -1 = 7 15', leading. o4:.b7 I g lags behind E g by the angle 23 16 r + 7 15 r = 30 31'. Power factor at generating end = cos (30 31') = 0.862. Efficiency of transmission, 150 X 30 X 0.9 146X34.95X0.862 = 92.1 percent. CALCULATION OF ALTERNATING CURRENT PROBLEMS 155 Example. The same conditions as in the example above except that the length of the line is 150 miles long. E r = 30 kv., I r = 135 - j 65.4 amp., K = 0.952. I g =(128.52+1.26) + j(2.61-62.26+23.76) = 129.8 - j 35.9 amp. Absolute value of current at receiving end, I g = V(129.8) 2 + (35.9) 2 = 135 amp. Eg =(30 + 6.55+6.1) +j(- 2.51 + 15.95) =42.65 +j 13.44 kv. Absolute value of voltage, Eg = V (42.65) 2 + (13.44)2 = 44.7 kv. 35 9 The phase angle between I g and E r is tan" 1 = 15 30' lagging. The phase angle between E g and E r is 10 A A ^^^iWr leading. I g lags behind E g by the angle 15 30' + 17 30' = 33. Power factor at generating end, cos (33) = 0.84. Efficiency of transmission 150 X 30 X 0.9 11 = 135 X 44.7 X 0.84 = = 8 per Lr E, h FIG. 47. Case III. Half of line capacity shunted across at each end of the line. As in previous cases / and II we shall assume that E r and /,. are given in magnitude as well as phase relation, that is, / r =I/+j(/ r "). (74) 156 FORMULAE AND TABLES FOR THE The values of the voltage and current at the generator end are given by the formulae E g = \E r K + rl r ' - Lul r "l +j li CW E r + rl r " + Leo//} = #,'+j(tf,") (75) I g = 5//-pV'C+7 r "J =l g 'jl a ". (76) The values of I r " and E g " are to be taken positive or negative depending on the sign of these quantities in the parentheses in (74) and (75). If Eg and I g are given in magnitude as well as in phase relation, I g =I g '+j(I a "}. (77) We have the following expressions for E r and I r : E r = \E g K - rl g f + Lco//'J +j\$E a Cur - Leo// - rl a "\ = E r '+j(E r "), (78) Ir=\I '+$Er"C<*]+j\I g "-$E a C<*-$E r 'C<*} = //+j(zb// ; ), (79) where K = l-\LCu\ In the following two examples we shall use the same data as in examples considered under Case II. Example. Fifty-mile transmission line. r = 16.5 ohms., L = 0.1005, C = 0.7 mf., K = 0.995, E r = 30 kv., I r = 135 - j 65.4 amp. E g = { 29.85 + 2.23 + 2.63 { + j {0.065 - 1.08 + 5.42 j = 34.7 +j 4.4 kv. Absolute value of voltage is E g = V(34.7) 2 + (4.4) 2 = 35 kv. I g = \ 135-0.58 l+j [3.96+4.58- 65.4 j = 134.4 -j 56.86 amp. Absolute value of current is I g = V(134.4) 2 + (56.S6) 2 = 145.9 amp. E g leads E r by the angle tan- 1 7 = 7 14', I g leads E r by the angle tan- 1 = 22 57', I g lags behind E g by the angle, 22 57' + 7 14' = 30 11'. CALCULATION OF ALTERNATING CURRENT PROBLEMS 157 Power factor at generating end, cos (30 110 = 0.864. Efficiency of transmission, 150 X 30 X 0.9 12 = 145.9 X 35 X 0.864 = 9L8 per Cent ' Example. One hundred fifty-mile three-phase transmission line. R = 49.5 ohms, L = 0.3195, C = 0.21 mf., K = 0.952. E r = 30 kv., I r = 135 - j 65.4 amp. . E g = S28.5 + 6.7 + 7.9J + j {0.6 - 3.25 + 16.25J = 43.1 + 13.6 kv. The absolute value of the generator voltage, E g = V (43.1) 2 + (13.6) 2 = 45.2 kv. I g = J135 - 5.4} +j {11.9 + 17.05 - 65.4J = 129.6 - j 36.5 amp. Absolute value of current, I g = V(129.6) 2 + (36.5) 2 = 134.7 amp. E g leads E r by the angle tan- 1 ^ = 17 30', I g lags behind E r by the angle tan ~ 1 ^- a = 15 45', izy.o Ig lags behind E g by the angle 17 30' + 15 45' = 33 15'. The power factor at generator end, cos (33 15') = 0.836. The efficiency of transmission, 150 X 30 X 0.9 134.7 X 45.2 X 0.836 = 79 ' 5 per Cent ' 158 FORMULAE AND TABLES FOR THE s - city end I* F 3 oo o "tf co "" " IO Th rH O O if T-H 05 8*5 O CO T-I ; ^H Oi o S CX a 03- 9 ^-| o3O0 O5 00 ^_| ^^ ^^ O5 C S ^T^ ^ Irt S $ri "^ O ^ O G Y- w *-* CALCULATION OF ALTERNATING CURRENT PROBLEMS 159 TRANSFORMERS Any two electrical circuits which are magnetically interlinked so that when a current flows in one circuit it will induce an e.m.f. in the other circuit constitutes a transformer. For all commer- cial work the iron-core transformer is used. In high-frequency work, as in the case of wireless telegraphy for instance, air-core transformers are used exclusively. We shall first give the equation governing the action of an air- core transformer, and then give a brief discussion of the theory of the iron-core transformer. Air-core Transformer.* Denote by LI, Ri and L 2 , R 2 the total inductance and resistance of the primary and secondary cir- cuit respectively and by M the mutual inductance between the two circuits. We shall also assume an e.m.f. of sine wave applied to the primary circuit. The equations expressing the reactions in the transformer circuits are dt dt (80) from which we obtain the values of the currents in the two cir- cuits as follows : (81) * Alex. Russell, "Theory of Alternating Currents," Vol. 1, Chap. X. Clerk Maxwell, " A Dynamical Theory of the Electromagnetic Field," Phil. Trans., 1865, p. 475. 160 FORMULAE AND TABLES FOR THE Ri + * ~ 7) (82) 2 /* 2 7 2 2 _1_ 7? 2 = jg * 2 2 == -L/ 2 CO -j- /t/2 . It2 The value of the current in the primary circuit is the same as MVR 2 that in a simple circuit whose resistance is Ri -\ ^ and in- Tl yC0 ft T ductance is LI ^ -, that is, the secondary circuit influences the primary circuit in the same way as if its resistance were increased and the inductance decreased. In other words the effective resistance in R\ + ^ instead of Ri and effective inductance is LI ^-^ instead of LI. If we neglect the re- 2 sistance of the secondary circuit, the effective inductance of the primary circuit is 7l/f2/,27" / M 2 \ (83) The quantity cr is called the leakage factor. Resonance Transformer.* If we introduce capacities in the primary and secondary circuits, the expressions for the cur- rents have exactly the same form as that given by equations (81) and (82) except that LICO and L 2 co are replaced by the terms LICO -^ and L 2 co ^ where Ci and C 2 are the capacities in C ico C 2 co primary and secondary circuits respectively. Now suppose the primary and secondary circuits are separately tuned so as to be in * G. W. Pierce, "Theory of Electrical Oscillations in Coupled Circuits," Proc. Am. Acad., Vol. 46, pp. 293-322, Jan., 1911. J. S. Stone, "The Maximum Current in the Secondary of a Transformer," Phys. Rev., Vol. 32, pp. 398-405, April, 1911. G. Benischke, "Der Resonanztransformotor," E. T. Z., 1907, p. 25. J. Bethenod, "tiber den Resonanztransformotor," Jahrbuch der Drahtlosen Telegraphic, Vol. I, pp. 534-570. CALCULATION OF ALTERNATING CURRENT PROBLEMS 161 resonance for the frequency of the impressed e.m.f. Then -ei = o, z = " (84) (85) and equations (81) and (82) reduce to , ER% cos a)t A -f M 2 co EM co sin <*>t 1 2= Current in Secondary t-t to Co * en cr< -3 ooto c / ^ 3 - ^ ^ / \ ^ fe / \ ^ 1 '^ \ ^>^_ "^ -^. ^~. 1 Current variation in the Secondary of a resonance Transformer R i = 20 Ohms R 2 =12.5 \ "^^ / / / 1 Mu 012 4 56 7 8 9 10 11 12 13 14 15 16 17 18 19 20 FIG. 49. Current variation in the secondary of a resonance transformer for different degrees of coupling. The current in the secondary circuit is a maximum when MV = RiR 2 and E sin at , . If we have only a condenser in the secondary as shown in Fig. 50, we have the following expressions for the currents in the primary and secondary circuits and the voltage on the condenser. E V I C 1 2J CO j x (87) 162 FORMULAE AND TABLES FOR THE E Mu sin (ut ) FIG. 50. The voltage on condenser is given by n 'M V/{(M'- (88) (89) The condition for maximum currents in the circuits, that is, the resonance condition, is or l- = 0. CL 2 (1 - , where .K 2 = M (90) If the constants of the circuits are fixed so as to satisfy equation (90), and if we neglect RiR 2 which is generally small, we have the following expressions for the currents and voltage in the circuits: EMu cos cot sin ( 2 V J M 2 co 2 + #ifl 2 - \ ^1W/J [ \ UiCO/ (94) (95) If we adjust the inductance and capacity of the primary circuit so as to be in resonance for the frequency of the e.m.f. impressed on the circuit, that is, if we make L\u 7^ = 0, equations (94) and (95) reduce to E VL 2 = EM* 2 + R^R^l + The coupling or the mutual inductance which gives the maximum current in the secondary circuit is determined by the equation + L 2 2 o> 2 = BA. (97) For the value of Mco, given by equation (97), we have for the currents in the primary and secondary circuits EZ * l V2 (Z 2 2 + R 2 Z 2 ) ' R 1 As an illustration we may use the following constants: #1 =' 15 ohms, R 2 = 200 ohms, LI = 1 mh., L 2 = 2 mh., C = 0.001 mf., co = 10 6 . By formula (97) the value of M which makes 7 2 a maximum is M = 0.1736 mh. CALCULATION OF ALTERNATING CURRENT PROBLEMS 165 Introducing these constants in equations (96) we get the fol- lowing results: M. fc /2- mh. 0.1 0.15 0.1736 0.25 0.50 1.00 6.1 XlO- 2 # 5.1 XlO- 2 # 4.48X10~ 2 # 2.80XlO- 2 # 0.78xlO- 2 # 0.20XlO- 2 # 3.2 X10~ 3 # 3.84xlO- 3 # 3.89XlO- 3 # 3.47xlO- 3 # 1.96XlO- 3 # i.ooxio- 3 # The Iron-core Transformer.* In the iron-core transformer we are chiefly concerned with efficiency ratio, regulation and exciting current. It is preferable to consider the transformer from the point of view of the leakage inductance, as this method leads to simpler formulae than does the treatment from the mutual inductance point of view, and at the same time brings out more clearly the relations between the various factors in the transformer. The general design formula is E = irVZnQN X 10~ 8 , (99) where E is the effective voltage induced in a winding, n the fre- quency, 3> the maximum total flux linking with the winding and N the number of turns. Example. In a 60-cycle transformer whose core is 10 cm. square and which is worked at 10,000 lines per centimeter, the volts per turn are E = 7rV2 X 60 X 10 2 X 10,000 X 10~ 8 = 2.7. Let Ri, R 2 = primary and secondary resistance of windings, Xij X 2 = primary and secondary leakage reactances of windings, Zi, Z 2 = primary and secondary internal impedances, /o = exciting current, I = load current measured on primary side, * William Cramp and C. F. Smith, "Vector and Vector Diagrams Applied to the Alternating Current Circuit," Ch. V. C. P. Steinmetz, "Theory and Calculation of Alternating Current Phe- nomena," 3d edition, C. XIV and XV. V. Karapetoff, "The Electric Circuit." 166 . FORMULAE AND TABLES FOR THE EI, E 2 = primary and secondary e.m.f.'s, E 2 ,o = secondary e.m.f . at no load, 6 = phase angle of connected load, being positive when the secondary voltage leads the secondary cur- rent, cos 6 being the power factor of the load, 7 = angle by which EI leads the exciting current, a = angle between EI and E 2 reversed, K = ratio of turns, primary to secondary. Mathematically it is convenient to refer the impedances to the primary. This may be done since an impedance Z 2 in the second- ary is equivalent to an impedance K 2 Z 2 in the primary. Hence we may write R = R! + K Z R 2 , K*X (100) Z = Z l In computing the efficiency of ordinary power transformers it is usual to neglect the copper loss due to the exciting current and, unless the contrary is distinctly stated, the efficiency refers to unity power factor load. Hence the formula is simply, output KE 2 I . . _ . . per cent efficiency = -* = ^^ T , T9P , - -, -- (101) input KE 2 I + PR + core loss The core loss is practically equivalent to the power consumption on open secondary. In an ideally perfect transformer EI would be equal to KE 2 , but in the actual transformer these differ by the impedance drops in the windings due to the load current and to the ex- citing current, or & - KE 2 = (7 + 7) Zx + K Z IZ 2 = 7 Zi + 7Z, (102) where the dotted capitals indicate that the quantities are vector quantities. This is the general equation for every form of trans- former. The leakage reactance of the windings is ordinarily determined by the ' 'short-circuit" test, in which one winding is short-circuited and rated current passed through the windings by applying a voltage to the other winding. From the value of this voltage, the power required and the current, the resistance R and the reactance X, equivalent to a simple impedance, may be readily computed. In the derivation of the formulae for regulation, ratio and CALCULATION OF ALTERNATING CURRENT PROBLEMS 167 phase angle, slight approximations are necessary to put the formulae in usable form, but in all practical cases the formulae are more accurate than the measurements of the quantities in- volved can be made experimentally.* The ratio of terminal volt- ages at any load is Ei _ K i IRcosO + IXsmO (IR sin 6 - IX cos 0) # 2 ~ KH E 2 . IQ (Ri cos 7 + Xi sin 7) t T? ~~ > J&2 for non-inductive load, E,_^ , IR ~ { " E* and for no load, EI _ ZQ QRi cos y -j- Xi sin 7) The regulation at any power factor expressed in per cent is per cent regulation = 100( 2> r, - ) \ A^ 2 / "sin (IRsmd - IXcosd) 2 for non-inductive load, f IR I 2 X* 1 per cent regulation = 100 i ^^- -f- ^ ^ 2 ^ 2 X . ^ /VIZ/2 ^ IV H/2 J The phase angle of the transformer is given by sin a =^ \Xcosd Rsmdl + ^- jXiCOS7 i2isin7j ; (108) and for the non-inductive load, sina =-^-+-^(^Licos7 ^isin7), (109) and for no load, sin a = j=r (X cos 7 Ri sin 7). (110) * A derivation of these formulae showing the magnitude of the approxima- tion involved will appear in Vol. 10 of the Bulletin of the Bureau of Standards, by P. G. Agnew and F. B. Silsbee. 168 FORMULAE AND TABLES FOR THE The reversed secondary voltage leads the primary voltage for in- ductive loads and ordinarily at no load, but lags behind it for non-inductive loads. Example. Consider a 2 kva., Wir-volt, 60-cycle transformer. Let #1 = 6, #2 = 0.065, X = 15, 7 = 0.12, cos 7 = 0.4, K = 10, 7 = 2, R = R! + K*R 2 = 12.5. If we assume the reactance drop to be divided equally between primary and secondary, Xi = 7.5. The computations are correct to 0.01 per cent, although the measurements are not. Then by (105) the no-load ratio is & f 1+ 0.12 (6X0.4 + 7.5X0.92)1 # 2 ,o V 100 J The regulation at 0.6 power factor, lagging current, is by (106) 25X0.6+30X0.8 , (25X0.8-30X0.6) 2 ft 100 [ t 2x100x10,000 For leading current the terms in (106) which contain sin be- come negative so that at the same power factor leading current the regulation is 100 (- 0.0090 + 0.0007) = - 0.83 per cent. The minus sign shows that under this condition the secondary voltage is increased by the leading load current. At non-inductive load the regulation is by (107) ( 25 900 ) 100 1000 ~ = *' 55 per cent ' The phase angle at 0.6 power factor, lagging current is by (108) 30 X 0.6 - 25 X 0.8 , 0.12(7.5X0.4-6X0.92) -Tooo- ~looo~ ~ a 023 ' a =-8', the minus sign meaning, according to the convention adopted, that the reversed secondary voltage leads the primary voltage. From formulae (109) and (110) we have for non-inductive load, 3 a . 0.12 (7.5 X 0.4 - 6 X 0.92) , 1000 + - ~ - = a 297 ' a = l 42 ' CALCULATION OF ALTERNATING CURRENT PROBLEMS 169 the reversed secondary voltage lagging; and for no load 0. 12 (7.5 X 0.4 - 6 X 0.92 , sm a = - 10QQ - = - 0.0003, a = 1 , the reversed secondary voltage leading. The Current Transformer.* While the foregoing relations hold also for the current transformer, the conditions of use and the relations desired are so different that a different method of treatment is far more convenient. The resistance and reactance connected to the secondary are kept as low as possible, and the iron is worked at low flux densities, the density being practically proportional to the current. The ratio and phase angle of the current transformer are independent of the resistance and react- ance of the primary. Let 1 1, 1 2 = primary and secondary currents, K = ratio of turns, secondary to primary, M = wattless component of exciting current, F = core loss component of exciting current, i total secondary reactance = tan- 1 j T-T -j total secondary resistance 6 = angle between /i and 7 2 . The ratio of transformation is J t _ M sin + F cos and the phase angle is given by (112) If the load connected to the secondary circuit is non-inductive and if, as is usually the case in high-grade transformers, the leakage reactance of the secondary winding may be neglected, equations (111) and (112) become f x =X + f, (113) Iz *2 tan 0=|^. (114) A./2 Equations (111) to (114) neglect second-order terms, but are suffi- ciently accurate for all practical work. * P. G. Agnew, Bull. Bureau of Standards, 7, 431, 1911. Reprint No. 164. 170 FORMULAE AND TABLES FOR THE Example. A 1000 to 5 ampere current transformer has 198 secondary turns and one primary turn and, when the total re- sistance and the total reactance of the secondary circuit are 0.4 and 0.5 ohm respectively, the core loss is 0.02 watt, and the ex- citing current is 8 amperes for a secondary current of 2.5 amperes. Then K = 198, sin = = = 0.78, cos < = 0.62. VO.16 + 0.25 The voltage induced in the primary is T J 2.5 VO.16+0.25 = 0.0081 . Hence F, the power component of the exciting current, is 0.02 -f- 0.0081 = 2.47 amp. and M, the wattless component, is \/8 2 (2.47) 2 = 7.6 amp. Sub- stituting these values of F } M and < in equations (111) and (112), we have for the ratio and phase angle, /! . 7.6X0.78 + 2.5X0.62 7- = iy o ~r o c 12 ^.O and the reversed secondary current leading the primary current. CALCULATION OF ALTERNATING CURRENT PROBLEMS 171 CHAPTER V TRANSIENT PHENOMENA IF the electrical conditions of a circuit are disturbed in any way, as for instance by a change in the electrical constants of the cir- cuit, or a change in the electromotive force acting on the circuit, a readjustment of the current and potential in the circuit will necessarily follow. The permanent state, however, is not reached instantaneously; it requires an appreciable time interval before electrical equilibrium is again established. The electric phe- nomena which occur in the time interval before the permanent state is reached again have been properly designated Transient Electric Phenomena. These phenomena are of frequent occur- rence and of considerable importance in electrical engineering, particularly whenever high frequencies or high potentials are employed. It is not within the scope of this book to enter into any discus- sion of the physical interpretation of these phenomena but, as in the previous chapters, we shall state only the problem, that is, the electrical conditions governing the production of these phenomena, and give the mathematical formulae expressing the distribution of the current and potential in the circuit. Wherever desirable we shall illustrate the formulae by numerical examples and curves. In this chapter we shall limit ourselves to a consideration of circuits of localized inductance and capacity. The transient phenomena which occur in a circuit of distributed inductance and capacity will be discussed in Chapter VI. INDUCTANCE AND RESISTANCE IN DIRECT CURRENT CIRCUITS* If we close suddenly a direct current circuit containing resist- ance and inductance the current does not assume its permanent value instantaneously. It requires a certain interval of time to * Ch. P. Steinmetz, "Theory and Calculation of Transient Electric Phe- nomena and Oscillations," Ch. III. J. L. LaCour and O. S. Bragstad, "Theorie der Wechselstrome," Ch. XXIV. 172 FORMULAE AND TABLES FOR THE build up the magnetic field in the inductance coil, and during that interval there is a back e.m.f. produced by the inductance which prevents the current from rising immediately to its permanent value. The value of the current at any instant of time after closing the circuit is given by I = 7.(l -*'), (1) Tp where I, = = is the value of the current in the steady state, it and the time is counted from the instant of closing the circuit. At t = 0, 7 = 0, t oo , I = 1 8 , permanent value -5 is called the time constant of the circuit, and its reciprocal K T) value a = -j- is called the damping factor of the circuit. The Li time required for the current to rise to n per cent of its steady value is 100 The smaller the inductance and the larger the resistance the quicker will the current attain its permanent value. In a per- fectly non-inductive circuit the current would assume its perma- nent value instantaneously. Example. When L = 0. 1 henry, R = 5 ohms, what is the time required for the current to rise to 90 per cent of its permanent value? By equation (2), T = ^ log, * = 0.02 X 2.3 = 0.046 sec. o i u.y If in the above example we make L = 1 henry, then the time re- quired for the current to rise to 90 per cent of its permanent value is 0.46 second, ten times as large as in the former case. In 0.046 second the current would in this case rise only to 21 per cent of its permanent value. On removing the e.m.f. from a direct current circuit, the cur- rent does not vanish immediately. The magnetic field of the CALCULATION OF ALTERNATING CURRENT PROBLEMS 173 inductance produces an e.m.f. which causes a current I to flow in the circuit the value of which is given by the expression _R I = I.e l , (3) 1.0 0.9 0.8 0.7 0.6 0.4 0.3 0.2 0.1 _ - _ - 1 ^ ^ - ^ ^ , / / / Rise of Current in Inductive Circuit / / #=10 Ohms, L= 0.5 Henry / / / 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0,14 0.15 0.16 0.17 0.18 0.19 .20 Time in Seconds FIG. 53. where I 8 is the steady value of the current before the e.m.f. is removed. The e.m.f. generated by the magnetic field of the in- ductance coil is a R F -r E ~ L di (4) 1.00 \ Decay of Current in Inductive Circuits B=10 Ohms L=0.5 Henry 0.01 O.OS 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19.20 Time in Seconds FIG. 54. In Figs. 53 and 54 two curves are given showing the rise and decay of the currents in an inductive circuit on closing and open- ing the circuit. 174 FORMULAE AND TABLES FOR THE CLOSING AN ALTERNATING CURRENT INDUCTIVE CIRCUIT Let us assume that at the instant of closing the circuit the e.m.f. of the generator is e = E sin (ut + ^) = E sin ^ =0 . (5) The current in the permanent state is /' = / sin (at + $ - 0) = / sin (f - 0) f =0 , (6) , E wL where / = , = > tan = ~ W 2 L 2 The current, however, will not rise instantly to its permanent value as given by (6) owing to the back e.m.f. generated by the magnetic field of the inductance coil. For a short interval of time after closing the circuit the value of the current will be given by the expression: _R t i = I sin (at - $ + 0) - Ie L sin (^ - 0). (7) In short-circuiting an alternating current inductive circuit, in which there is no external e.m.f. acting on it, the current grad- ually dies out in accordance with the expression: _R t i=Ie L sin(^-0). (8) Direct Current Circuits Containing Resistance and Capacity Only, no Inductance. On closing the circuit the current and potential difference at condenser terminals are given by the ex- pressions /-{', o) i ) (10) where E is the impressed e.m.f. and R and C are the resistance and capacity of the circuit. Equation (9) does not apply for very E small values of t, because at t = 0, equation (9) gives 7 = ^, that is, at the moment of closing the current jumped instantly from zero to finite value and this is not possible. To get the value of the current for the very small values of time the inductance of CALCULATION OF ALTERNATING CURRENT PROBLEMS 175 the circuit must be taken into account though it may be ex- tremely small, and in that case we get t 77F R -57; j ? * Example. I = R R E= 1000 volts, R = 100 ohms, C = 10 mf., L = 1 mh. (ID t -RC R t L e / 1 1 io- 4 0.990 0.368 10 6.22 1(H 0.904 96 9.04 2X10- 4 0.818 .... 282 8.18 5X10- 4 0.606 394 6.06 8X10- 4 0.449 551 4.49 10- 3 0.368 632 3.68 1.5X19- 3 0.223 777 2.23 2X10- 3 0.135 865 1.35 3X10- 3 0.050 950 0.50 4X10- 3 0.018 982 0.18 5X10-3 0.007 993 0.07 It is to be noticed that the factor e L in equation (11) is only operative for an extremely small value time interval, and after that it is entirely negligible and equation (9) may be used. CIRCUITS CONTAINING RESISTANCE, INDUCTANCE AND CAPACITY IN SERIES* The problem of determining the current and potential in a circuit containing resistance R, inductance L and capacity C, either on charging the condenser when the circuit is closed, or on * W. Thomson, Transient Electric Currents, Phil Mag., Ser. 4, Vol. 5, p. 393, 1853. J. A'. Fleming, "The Principles of Electric Wave Telegraphy," Ch. I. C. P. Steinmetz, "Theory and Calculation of Transient Electric Phenom- ena," Sect. I, Ch. V. H. Armagnat, "The Theory, Design and Construction of Induction Coils." Translated and edited by O. A. Kenyon. 176 FORMULAE AND TABLES FOR THE discharging the condenser when the circuit is short-circuited, resolves itself into the study of the differential equation 2 y ^ (logarithmic case). (II) R=2\/^ (critical case). (13) (III) R < 2 y -~ (trigonometric case). We shall give below the formulae for the current and potential for all three cases either on charging or discharging. CASE I. R > 2 y ~ , LOGARITHMIC CASE On charging, the current in the circuit and the voltage across condenser at any instant of time are as follows: R-s R+S R-S 2L - (R - S) e * L ' (15) where S When the condenser is discharging we have for the current and condenser voltage the following equations: R ~ 8 * (17) CALCULATION OF ALTERNATING CURRENT PROBLEMS 177 where E Q is the condenser voltage before the condenser is begin- ning to discharge. The expressions for the currents on charge and discharge of condenser are identical in form but of opposite signs, that is, the currents are exactly the same but in opposite directions. The voltage across condenser increases in one case in exactly the same way as it decreases in the other. 1.001 9 1 x ^w 0.8 1 0.7 I f \ s ^^^^^ ^-^ ,^- ^> ^" * * ^spj* x^ ^ Current and Potential on Charging Circuit containing luctance Capacity and Resistance ,R> 2 1^~( Logarithmic Oas) 5 1 / t >< ^ In. ^ r x ^ fi 1 / / / "^ ^ ^ 2J / / ^^ * x If X 2 4 6 8 10 12 14 16 18 " t X FIG. 55. 20 22 24 26 28 30 32 34 36 38 40 Seconds An examination of the expressions for the current (14) and (16) shows that the current has zero values for the time i = and t = GO ; hence, at some instant of time the current has a maximum value. That is, the current is unidirectional, gradually increasing from zero to its maximum value and then decreasing to zero again. The time at which the current is at its maximum value is given by the expression L ' ||. (18) Example. Suppose we have a circuit of the following constants: L = 2 mh., C = 2 mf ., R = 100 ohms, E = 100 volts. 8 8 x 10 " 3 c ^=5750, 2x10 = 44,250. = 77. 178 FORMULAE AND TABLES FOR THE The time in which the current will rise to its maximum value is L ' R + S 2 X 10~ 3 177 t = -olog, ^ ~ = == loge-^- = 5.3 X 10~ 5 sec. o it o / / Z6 CASE H. H 2 = , CRITICAL CASE The current and voltage across condenser on charging are . R (20 ) The current and condenser voltage on discharging are CASE HI. R < 2 y ~i TRIGONOMETRIC CASE The current and voltage across condenser terminals on charging are /=-e- n = 1 - 7 =- (27) 2irVLC 7-> The quantity ^rj- denoted by a is called the damping factor. Z h If the resistance is small - is small compared with unity and equa- tions (24) and (25) reduce to / =-# C/3e- 12 13 1* 180 FORMULAE AND TABLES FOR THE rp where T is the duration of a complete period. The term -~- = - = -r is denoted by 5 and is called the logarithmic decre- ment of the oscillation per half period. Hence, 7l _/2 _/3_ _ . T T j etc. e , J2 -13 i\ and 8= log, = 10^, etc. (32) ^2 ^3 Since n = - == , approximately, the expression for the loga- 2 7T v Li\j rithmic decrement may be put in the form (33) As an illustration we may take the following constants: L = 1 mh., C = 0.001 mf., R = 100 ohms, E = 1000 volts. 1 is negligible compared with -=- . The ratio of two successive half waves is 5 = loge 1.164 = 0.152. The small difference in the values of 5 arises on account of carry- ing only to two decimal places the values of e~ at in calculating the values of 7i, 7 2 , etc. If we introduce a hot-wire ammeter in the oscillating circuit, the current indicated is the effective value of the square root of the mean square of the discharge current. Generally the time interval between two successive discharges is sufficiently long to CALCULATION OF ALTERNATING CURRENT PROBLEMS 181 allow complete discharge of the condenser. That is, the current is practically zero before another discharge takes place. To get the effective value of the current we may therefore integrate equation (24) from zero to infinity and divide by the time interval between two successive discharges. The square root of the value thus obtained will be the effective value of the discharge current. If we denote by N the number of discharges per second, we have (34) If the condenser is charged N times a second to a potential E Q the power supplied is . (35) We can also get an expression for the effective value of the current in terms of the first maximum value of the oscillation, as follows: INI& i ~ ' eff " / ~8ri X l /\ ' (36) where 8 is the logarithmic decrement, n is the frequency of oscil- lations and N is the number of discharges. For oscillatory currents of high frequency such as are generally employed in wireless telegraphy - is generally a small quantity not greater than 0.1, 7T and often much smaller than that, hence we may neglect (-) in \7T/ comparison with unity, and equation (36) reduces to rt Example. Let us take the same data as in the illustration given on page 180. E Q = 100 volts, C = 10~ 3 mf., L = 1 mh., R = 100 ohms, I I = 0.92, n =^, d = 0.157. Z 7T Assuming N = 1000 we have by (34) = 0.071 amp. 182 FORMULAE AND TABLES FOR THE By (36) we have e 5 = 1.17. , 8 _ 10 3 (0.92) 2 X 1.17 1 5.04 X 10~ 3 = = 5.04 X 10~ 3 , approximately. Jeff = 0.071 amp., which is in agreement with the value obtained by formulae (34). It is also to be noted that the term (-) is very small compared with unity, and formula (37) would have given the same results. Alternating Current Circuits Containing Resistance, Induc- tance and Capacity in Series. When an alternating e.m.f. is introduced into a circuit having a resistance R, inductance L and capacity C the current and condenser voltage do not attain their permanent values instantly. It requires a certain time interval, which is generally very small in all practical cases, before the current and voltage across condenser have built up to their maxi- mum value. The current and condenser voltage may be expressed as the difference of two quantities, , where the subscript s denotes the steady or permanent values and the subscript v denotes only the variable values of the current and voltage which are present during the transition period. I v and E v diminish in accordance with an exponential law with re- spect to time and consequently disappear very rapidly. Formulae for I 8 and E 8 have been given in Chapter IV, hence we shall give here only formulae for I v and E v . As in the previous section we have to distinguish here also three distinct cases depending on the relative magnitudes of the electrical constants L, C and R. Let us consider first the trigonometric or oscillatory case <2W^ CALCULATION OF ALTERNATING CURRENT PROBLEMS 183 We shall assume that at the instant of closing the circuit the steady values of the voltage and current are E a = #max sin (coZ+ ^) = # max sin ^ =0 , where sin (wt + ^ 0) = 7 max sin (^ For the limiting conditions given by (39) we have the following formulae for / and E v : /L._. ) (40) Ds^-^sin^+^+^^y-^sin^ j, PC ) (41) where Any change in the electrical constants of the circuit will also introduce a transient term. The current and potential cannot pass over instantly from one permanent value to another, it re- quires a certain time interval for the readjustment. During the transition period the current and potential can also be expressed by formulae (38), but in this case the values of I v and E v are as follows : E sin ft + A7 y g sin (/^ - 7) > , (42) ^ =-'? T^l^ sin tf - T ) + ^ sin /3^ , (43) where A# = (E 2i , A/ = (7 2>8 - Ji,.)(=o). Equations (44) give the variation in the permanent values of the current and condenser potential in the circuit at the time t = 0. 184 FORMULAE AND TABLES FOR THE For the critical case R = 2 y -^ the value of I v is given by the formula I v = - cos *-*- (45) Closing on Alternating Inductive Circuit FIG. 56. When the resistance is still further increased, so that R > 2 y -^, we have the logarithmic case, and the value of I v is CALCULATION OF ALTERNATING CURRENT PROBLEMS 185 [X )-H TT +^j Umax Sin M < (46) As in the trigonometric case we have here also for any change in load condition a transition period during which we have, be- sides the term representing the permanent value of the current, another term which is effective for a short period only, the transient term. The value of the transient term which we designated throughout by / is as follows: (47) ^(L \* / J L \* / J ) where - \ 1 J (48) represent the differences in 7 and # between the steady values before and after the change. The values of a and /? in equations (46) and (47) are given by Divided Circuits.* We shall first consider the simplest case, the circuits having only inductance and resistance. Suppose we have a continuous current flowing through the circuits and either the main circuit is short-circuited, or a varia- tion in the resistance of any of the branches is made. The problem is to determine the values of the transient terms of the currents in the branches. Let E = e.m.f. LO, Li, L 2 = inductances of main and branch circuits respectively. RO, Riy R 2 = resistances of main and branch circuits respectively. The permanent values of the currents in branches 1 and 2 are T , r t N ~W 2 = ~W' ^ ) where R* = RoR 2 + R^ * See Ch. P. Steinmetz, "Theory and Calculation of Transient Electric Phenomena and Oscillations," Section I, Ch. IX. 186 FORMULAE AND TABLES FOR THE Now let us suppose that there is a change in resistance in branch 1 or 2. The currents in the two branches are as follows: (50) T I" X. ... (51) where // and 7 2 ' are the permanent values of the currents in the two branches before the change in the resistance, and //' and / 2 " are the permanent values of the currents after the change. Rv 00000 FIG. 57. = L L 2 4" RoRi H~ 2L 2 (L, + Lo), 2L 2 As an illustration let us consider the following example: LO = 5 henrys, LI = 2.5 henrys, L 2 = 10 henrys, RO = 10 ohms, RI = 2 ohms, Rz = 25 ohms, and we shall assume that the value of Rz is suddenly changed to 15 ohms. CALCULATION OF ALTERNATING CURRENT PROBLEMS 187 The permanent values of the currents before and after the change in the resistance are // = 7.81 amp., h' = 0.625 amp., /i" = 7.5 amp., /," = 1.00 amp. We also have L 2 = 87.5, jR 2 = 200, /S 2 = 267.5. Xi=- 1.42, X 2 =-1.63, Ki = 0.47, K 2 = - 0.12, Introducing these values in equations (50) and (51) we get Ji = 7.5 + 0.70 e- 1 - 42 ' - 0.39 e- 1 - 63 ', 7 2 = 1.00 - 0.33 e- 1 - 42 ' - 0.045 e- 1 - 63 '. For t = which are the permanent values of the current before the change in resistance. For t = oo 71 = 7.5 amp. "1 which are the permanent values of the current 7 2 = 1.00 amp. J after the change in resistance. If the circuit is short-circuited we must put in equations (50) and (51) /i" and 7 2 " separately equal to zero, and the equations reduce to , 1^2/1' + Iz f x , Kill + ^2 XJ fi = ~ - Y~ e -- - ~ c /! = 7.81 1 7 2 = 0.625 J 7 - _ lzl l% * t , l^z H" 1^2/2^ x j *2 JV- 7^ ^ T T7- T7- ^ ' Xvi /V2 -tVi /V2 Using the same constants as in the above example, we get /i = 0.53 c- 1 - 42 ^ 7.28 C- 1 - 63 ', 7 2 = - 0.25 e- 1 - 42 ' + 0.87 e- 1 - 63 '. Transient Currents in Transmission Lines.* The capacity of the transmission line is represented by a condenser shunted across the middle of the line. We shall give here formulae for the transient currents and potentials which may arise either when the load is short-circuited or the generator is short-circuited. * See J. L. LaCour and O. S. Bragstad, "Theorie der Wechselstrome," Ch. XXIV. 188 FORMULAE AND TABLES FOR THE Consider first the case of the transient phenomena which occur in transmission line when the load is short-circuited, as shown diagrammatically in Fig. 58. FIG. 58. Let 7 1>8)0 , 7 a ,., , 7 C , S>0 , E Ct8>0 represent the permanent values of the currents in the different branches and the condenser potential at the instant that the line is short-circuited at the receiving end. Then we have the following formulae for the transient values of the currents and potential. W (54) (55) = (56) where a = Ei Rz 2Li"2L 2 8= ,i = - 2 =2o: and = tan- 1 - Li Lz The values of a and /3 given above have been obtained on the . . , , , RI Rz assumption that -j = j LI L 2 CALCULATION OF ALTERNATING CURRENT PROBLEMS 189 An examination of equation (56) makes it appear evident that no great rise in potential can occur on the line. As an illustration consider the case of a single-phase line 100 miles long x made up of No. 000 B. & S. with wires 96 inches interaxial distance. The resistance per mile R = 0.33 ohm. inductance per mile L = 2.06 mh. Capacity per mile C = 0.0145 mf . The total resistance, inductance and capacity of the line are 33 ohms, 206 mh. and 1.45 mf. respectively. We have therefore Lj = L 2 = 103 mh. a = -- = 80.1, ' V/103 X 10-3 X 1.45 X 1** ~ (8 ' 1)2 " 36 ' 6 X 10 ' /3C = 0.005, = l.004 = 1, approx. EC., = - c- 80 | sin ft + E C ,,, Q sin (ft + = - c- 80 1 200 7 C , 8>0 sin ft + E c , 8>0 sin (pt + 4>) I . If 7 c>8 ,o is numerically equal to one hundredth of E c>8i0 , that is, ET 7 Ci8>0 = -jg^j the rise in potential is only about three times the normal value. For the case where the generator is short-circuited we may neglect L 2 compared with R 2 , and in that case the transient cur- rent in the branches and the potential is given by the following expressions: *)|j (58) 1 sin $ + e ,.,o V 1 + I* sin (|M + 0) K (59) 190 where FORMULAE AND TABLES FOR THE a ~h tan _ ,v2 FIG. 59. If a direct e.m.f . is acting on the line, then E Cta>Q = E and 7 c>8i0 = 0, where E is the e.m.f. impressed on the line. Equation (59) reduces to = 6 y 1 + sin E The increase in potential is proportional to (60) This ratio is seldom much greater than unity and increases as a increases. Mutual Inductance.* If we have two circuits connected inductively and impress continuous e.m.f. J s E\ and E 2 on circuits I and II respectively the currents in the circuits are and (61) * See Ch. P. Steinmetz, "Theory and Calculation of Transient Electric Phenomena," Section I, Ch. X. Einfiihrung in die Maxwellsch Theorie der Elektrizitat von Dr. A. Foppl herausgegeben von Dr. M. Abraham, Part III, Ch. II. CALCULATION OF ALTERNATING CURRENT PROBLEMS 191 If there is a sudden change in load conditions, say R% is changed to Rz', the permanent values of the currents in the circuits after the change are (62) Ii" = g and The change, however, from one current value to another due to a change in circuit conditions does'not occur instantly. The cur- rents rise gradually to the changed permanent values; the interval of time required before the perma- nent conditions are established again depends upon the electric constants of the circuits. The expression for the currents in the two circuits at any time t are as follows: (63) FIG. 60. where JLi J-i -f - e" 1 * ra<3 mi miMz , t , m2 mi m 2 A7 2 x 2 1-2 l - W2 mi T -- T f T ff i l2 1% J-2 , m 2 mi ' m\i mX2 ' ^ - (1 + 0! 2 ) + V(a 2 - <*i) 2 -f 4oW 2 1- X 2 2 -o:i) 2 H-4Q: 1 a 2 X 2 1- 7? J? -ftl .fi/ 2 i = 2j- , a 2 = 2^-? IT. - M2 . X 2 (64) (65) If there is a change in the resistance of circuit I from RI to Ri the permanent values of the currents in the circuits after the change are and (66) 192 FORMULAE AND TABLES FOR THE The currents in the circuits before the permanent values are established are given by the expressions. _ _ _ eXl( + __ M (67) mi m? mi mz = A7.rn.m3 fXil _ ^ mi m 2 mi mi where A/i = Ii Ii" and mi, m2, Xi, \2 have the same signifi- cance as given in equation (65). If one of the circuits is suddenly opened, we shall merely have to put either 7 2 " = in equations (63) and (64) or 7/' = in equation (67) and (68) depending upon whether we open circuit II or circuit I. Example. Li = 0.1 henry, Ri = 10 ohms, L 2 = 0.2 henry, # 2 = 25 ohms, M = 0.1 henry, EI = 100 volts, E 2 = 200 volts. Now suppose we suddenly change R 2 to 40 ohms. We have 100 r , 200 7 200 K h f = -jQ- = 10 amp., U = -^ =8 amp., 7 2 " = -^ =5 amp., 10 25 - (50 + 62.5) + V(62.5 - 50) 2 + 4X 50X 62.5 X 0.5 _ = ~ 1-0.5 " _ (50 + 62.5) - V (62.5 -50) 2 +4X5QX62.5X0.5^ 1 -0.5 _ - 0.1 X 65 + 10 _ - 0.1 X 385 + 10 = - 0.1 X 65 - 0.1 X 385 A7 2 = 8-5 = 3; hence, 7i = 10 + ^ e~* 5 ' j^g ~ 385 ' > 7 2 = 5 + 1-26 e- 65 ' + 1.74 - 385 '. For t = 71 = 10 amp. \ permanent values before change in 72 = 8 amp. J resistance of circuit II. For t = oo 7i = 10 amp. \ permanent values after change in I z = 5 amp. J resistance of circuit II. CALCULATION OF ALTERNATING CURRENT PROBLEMS 193 If in the above example we had changed R\ from 10 to 20 ohms and kept R 2 constant, we should have had 7,'= = 10 amp., / 1 " = = mi, w 2 , Xi, X 2 have the same values as above. By (67) and (68) we get 7 2 = 8 + 1.56e-* 5 <- 1.56 e- 385 '. For t = 7/ = 10. amp. 1 permanent values before the change in 1 2 = 8 amp. J resistance of circuit I. For t = oo 7i" = 5 amp. j permanent values after the change in 7 2 " = 8 amp. J resistance of circuit I. Let us now consider the case where one circuit has no e.m.f. acting on it and an e.m.f. EI is introduced in the other circuit. Suppose E% = 0. There is no current in either circuit before the e.m.f. is introduced in cifcuit I; that is, we have 7/ =0 , 7 2 '=0, Ii"=^> 7 2 " = 0. Jtti Equations (67) and (68) reduce for this case to the following: " (69) The values of mi, m^ and Xi, X 2 are given by (65). The current in the secondary circuit first commences to rise, reaches a maxi- mum value and then diminishes to zero again. The time in which it reaches its maximum value is given by = L j-Io6^- (71) X 2 Xi X 2 As an illustration we may use the same constants as in the pre- vious example, p mi =-0.54, m 2 = 0.74, Xi = -65, X 2 =-385, 7i =|10 - 5.78-* 5 <- 4.22oo COOOCOCO*O^fCOC^i 1O ? i i j 3, 3, xxxxxxxxxxx C^ t^t^^HC5 i- ^- c> G> XXXXXXXXXXX xxxxxxxxxxx i-IOOOOOOOOOO ooooooooooo xxxxxxxxxxx TtftlTTTTTT S22SSSSS2S3 d o d o o d d o d d d T-HC3COIOOOOO 206 FORMULAE AND TABLES FOR THE a a > S S 3 S B e 13 IS 2| wa 8-S xxxxxxxxxxx xxxxxxxxxxx xxxxxxxxxxx IIIIIIIIISI xxxxxxxxxxx T-l i-l T-l rH ^H T-I d O O O O x 3, xxxxxxxxxxx O O O O xxxxxxxxxxx oooo xxxxxxxxxxx o o o o o o o o o o o' xx cqio xxxxxxxxx cot>-ooOi>-OiO5cooo kl XX Illll X X X X X illl X X X X o o o o o o o o o o xx IIIIS X X X X X llll XX XX ^Hr-t.-iOOOO CS O> OS O> d d d o II , X X X X X X X lll Illl X X X X o o o o o o o o o aa88 CALCULATION OF ALTERNATING CURRENT PROBLEMS 207 TABLE XIX* VALUES OP a AND /? PER MILE FOR TRIANGULAR SPACING OF THREE-PHASE TRANSMISSION LINES AT 25 AND 60 CYCLES Size of wire, B. &S. Resist- ance in ohms per mile Spacing between wires in inches. 60 Cycles. 25 Cycles. a a No. 250,000 0.222 ( 72 j 96 1 120 1.144 0.322X10- 3 0.306X10- 3 0.294X10- 3 0.287X10- 3 2.11X10- 3 2.10X10- 3 2.09X10- 3 2.09X10- 3 0.306X10- 3 0.306X10- 3 0.283X10- 3 0.275X10- 3 0.918X10-3 0.912X10- 3 0.907X10- 3 0.906X10-3 0000 0.263 ( 72 ) 96 1 120 1144 0.374X10- 3 0.357X10- 3 0.344X10- 3 0.334X10- 3 2.11X10- 3 2.10X10- 3 2.10X10- 3 2.09X10- 3 0.352X10- 3 0.337X10- 3 0.326X10- 3 0.316X10-3 0.932X10-3 0.927X10- 3 0.922X10- 3 0.915X10-3 000 0.33 ( 72 J 96 1 120 L144 0.457X10- 3 0.437X10- 3 0.421X10- 3 0.408X10- 3 2.12X10-3 2.12X10- 3 2.11X10- 3 2.11X10- 3 0.420X10- 3 0.403X10- 3 0.390X10- 3 0.381X10- 3 0.960X10- 3 0.953X10-3 0.946X10-3 0.941X10-3 00 0.416 { 72 J 96 1 120 U44 0.556X10- 3 0.536X10- 3 0.517X10- 3 0.504X10- 3 2.15X10-3 2.14X10-3 2.13X10-3 2.13X10-3 0.497X10- 3 0.482X10- 3 0.468X10-3 0.456X10- 3 0.995X10-3 0.989X10-3 0.980X10-3 0.972X10-3 0.525 f 72 j 96 1 120 1144 0.678X10- 3 0.653X10- 3 0.633X10- 3 0.615X10- 3 2.18X10- 3 2.17X10-3 2.16X10- 3 2.15X10-3 0.591X10- 3 0.569X10- 3 0.554X10- 3 0.541X10- 3 1.05X10- 3 1.03X10- 3 1.02X10- 3 1.01x10-3 1 0.665 ( 72 J 96 1 120 tl44 0.825X10- 3 0.791X10- 3 0.766X10- 3 0.749X10- 3 2.23X10- 3 2.21X10- 3 2.20X10- 3 2.19X10-8 0.691X10-3 0.670X10-3 0.655X10- 3 0.640X10-3 1. 10 X lO- 3 1.09X10-3 1.08X10- 3 1.07X10- 3 2 0.835 ( 72 J 96 1 120 1 144 0.989X10- 3 0.945X10- 3 0.920X10- 3 0.905X10- 3 2.29X10- 3 2.27X10- 3 2.26X10- 3 2.25X10- 3 0.800X10-3 0.779X10-3 0.756X10-3 0.744X10- 3 1.17X10- 3 1.16X10- 3 1.14X10- 3 1.14X10- 3 * This table can also be used for single-phase lines. 208 FORMULAE AND TABLES FOR THE CALCULATION OF ALTERNATING CURRENT PROBLEMS 209 210 FORMULAE AND TABLES FOR THE CALCULATION OF ALTERNATING CURRENT PROBLEMS 211 _ CO *O CO* CO* CO* CO* 03 CO* 3 M* 0* N * N* S S 3 3 3 S OT X 212 FORMULAE AND TABLES FOR THE The general solution of equations (2) giving the current and voltage at any point on the line is as follows: i--, or I = A^+A 2 e^ ) E=(a,- joj (- A lt rv* + A#v*) t J where co + g Cja + g' (14) 02 = The quantity 01 jaz is frequently expressed in the following form: fli y aa _ ^ __V(Ljco + ^)(Q-co + 6r) = VL^ + ^ (15) and is sometimes denoted by ZQ. The quantity Z is called the initial sending-end impedance. Equations (12) are to be used when the distance is considered from generator end, in the direction of decreasing power, and equations (13) are to be used when the distance is considered from the receiving end, in the direction of increasing power. The solution of equations (2) may also be expressed in terms of hyperbolic functions. 7 = AI cosh Vs + A 2 sinh Fs, ) E=-( = 0, which is approximately the case in cables in which the dielectric losses and the inductance are small, we have (20) = tan" 1 - = tan" 1 1 = 45. * References are given at the end of this chapter. 214 FORMULAE AND TABLES FOR THE Expressed in hyperbolic functions the formulae for the current and potential in an infinitely long line are -p I = 2-r (sinh Vs cosh Vs), ai - jE g p\, E = } {KJE g - K 2 (aj a f aJ ") - K 3 (a 2 // T aj g K 2 (02// =F ai//0 - K 3 (a,I g f db 222 FORMULAE AND TABLES FOR THE HYPERBOLIC FORMULA Another set of formulae corresponding to those of (48) to (52) has been worked out in terms of the hyperbolic functions. If the electrical conditions are given at the receiving end, considering the distance s from receiving end we have 7 = I r cosh Vs + E r (p + jq) sinh Vs, ) /^x E = E r cosh Vs + I r (d t - ja) sinh 7s. ) If Eg and I g are given, that is, the current and voltage at generator end, considering the distance s from generator end we have I = I g cosh Vs Eg (p + jq) sinh Vs, ) ^A\ E = E cosh 7s - I g (ai - joj) sinh 7s. ) Formula (53) corresponds to formula (48) and formula (51) corre- sponds to formula (52), the various terms having the same signifi- cance as in the above equations. The terms sinh 7s and cosh 7s are complex quantities. They can be put in the form of two terms, a real and an imaginary, thus, sinh 7s = sinh (a + ,//3) s = sinh as cos /3s + j cosh as sin /3s, 1 cosh 7s = cosh (a + j]8) s = cosh as cos 0s -j- j sinh as sin /3s. ) The values of sinh as and cosh as can be obtained from a table of exponential functions. A short table, Table XX, is also appended at the end of this chapter giving directly the values of cosh 7s and sinh 7s. The range of values of 7s covered by this table is sufficient for all practical problems in power transmission. Formulae (53) and (54) appear simpler and more compact than the corresponding formulae (48) and (52), but the labor involved in numerical computation by the hyperbolic formulae is not any less than by the other formulae, and there is, therefore, little choice in regard to the use of the two different sets of formulae for numeri- cal problems. It is desirable, however, to be able to check the results by two different methods, so that any possibility of numeri- cal errors may be eliminated. APPROXIMATE FORMULAE FOR SHORT LINES If the electrical conditions are given at the receiving end, = Ir \ 1 +* >( *' 2 y) +jtt/fe*| + E r \ (p + jq) (a (56) CALCULATION OF ALTERNATING CURRENT PROBLEMS 223 If the electrical conditions are given at the generator end, o2 (^2 _ OZ\ ) ^ ' -JV f 1 1 + ^ % ^ ^ + j/3s 2 ( -/, { (ai - jaO (a +#){. ( ^ ) I The above two formulae were given by W. E. Miller.* He states that they can be used with an accuracy of 1 per cent for lines using No. 2 wire up to 120 miles at 60 cycles and 150 miles at 25 cycles. Greater accuracy is obtained if larger wires than No. 2 are used, though the difference is immaterial. \ ILLUSTRATIONS Example. The following example is taken from Mr. Miller's paper. We shall work it out by the formula (48) and compare the results with those obtained by Mr. Miller by the use of the hyperbolic method. Three-phase line, 300 miles long using hard-drawn stranded copper wire, No. 000 B. & S. triangularly spaced, with wires 10 ft. apart. Frequency, 60 cycles. At the receiving end we have the following conditions. Line voltage 104 kv. or 60 kv. between wire and neutral; load current 100 amperes at receiving end at 90 per cent power factor lagging. Then E r = 60 kv., IT = IT - jl r " = 90 - j 43.5 amperes, R = 0.33 ohm per mile, L = 2.13 mh. per mile, C = 0.014 mf. per mile, g = (neglected), a = 0.421 X 10~ 3 , = 2.11 X 10~ 3 , al = 0.126, 01 = 0.633 = 36 16' 5", e < + e -ai = 2.016, cos 0Z = 0.8062, 2*1 - e -i = 0.253, sin 01 = 0.5915, Ki = 1.625, K 2 = 0.204, K 3 = 1.193, KI = 0.149, ai = 392, a 2 = 78, p = 2.44 X 10~ 3 , q = 0.485 X 10~ 3 . Introducing these values in equation (48) we get for current and voltage at generator end I g = 73.94 +j 61.65 amp., Eg = 66.4 +.7 21. 04 kv. * W. E. Miller, Formulae, Constants and Hyperbolic Functions for Trans- mission Line Problems, General Electric Review Supplement, May, 1910. 224 FORMULAE AND TABLES FOR THE The absolute values of current and voltage are / = V (73.94) 2 + (61.65) 2 = 96.3 amp. E a = V(66.4) 2 + (21.04) 2 = 69.6 kv. The generator current leads the voltage at the receiving end by an angle = tan- 1 0.834 = 39 48'. The generator voltage leads the voltage at the receiving end by an angle tan- 1 = tan- 1 0.317 = 17 36'. DO.OO The current at the generator end leads the generator voltage by the angle (39 48' -17 36') = 22 12'. The power factor at the generator end is cos (22 120 = 0.926. Transmission efficiency, 60 X 100 X 0.9 13 69.6 X 96.3 X 0.926 To get the value of the charging current put I r = 0, no load current. Then I = - 2 A3 + j 90.3 = 90.3 amperes per wire. The values obtained by Mr. Miller for this example using the hyperbolic functions are I g = 74.6 +J61.9 amp., E g = 66.2 + j21.0kv. The values differ very slightly from the values obtained by formula (48) and even this slight difference may be accounted for by the fact that in using the hyperbolic tables, interpolations are required, and apparently Mr. Miller only used first differences. It is interesting to obtain the generator current and voltage and the .different phase angles for different load currents. Since we have given in detail the numerical work for the case of 100 am- peres load current, it will be sufficient to simply tabulate the results for other load currents. CALC ULA TION OF ALTERNA TING C URRENT PROBLEMS 225 /,. 25 50 100 150 Ig'+jlg" 16 7+7 83 1 35 .9+j 74 .9 73. 9+/ 61. 65 112. l+j 47. 3 E g '+jE g " 53.2+.; 8. 6 57.9+j 12.6 66. 4+; 21. 04 75. 2+; 29. 3 V7,'2 + 7/" 84.8 amp. 83 amp. 96.3 amp. 121 . 7 amp. VE a ' 2 + E a " 2 53 8kv 59 2 kv 69 6kv 80 7kv Phase angle, I g E r j Phase angle, E g E r j Phase angle, I E g < rj = efficiency of ( transmission . . . . ( 78 40' leading 9 12' leading 69 28' leading 84% 64 18' leading 12 18' leading 52 leading 89.2% 39 48' leading 17 36' leading 22 12' leading 88% 22 54 leading 21 16' leading 138' leading 82.5% As an illustration of the application of formulae (52) and (54) we shall take the following example: A three-phase line made up of No. 0000 B.& S. copper-strand wire triangularly spaced with 10 feet between wires; frequency, 60 cycles; length of line 200 miles. Voltage at generator between wires, 104 kv., between wire and neutral, 60 kv. Ig = 120 amperes at .90 per cent power factor leading. I = 108 +j 52.3, Eg = 60. To determine the voltage and current at receiving end and the various phase angles. R = 0.26 ohm per mile. L = 2.09 mh. per mile. C = 0.0143 mf. per mile. a = 0.344, ft = 2.10 X 10~ 3 . 01 = 0.420 = 24 3' 50", cosjSZ = 0.913, sin/3Z = 0.408, (?i + -i = 2.005, ? 1 - -<*' = 0.1376, K! = 1.831, K 2 = 0.126, K 3 = 0.818, K^ = 0.0561, ai = 396, 02 = 62.5, p = 2.47 X 10~ 3 , q = 0.39 X 1Q- 3 . Introducing the above values in equations (52) we get I r = 97.64- .7 11. 18 amp., E r = 57.74 -j 18.02 kv. 226 FORMULAE AND TABLES FOR THE Using formulae (54) for this example, we find from Table XX by interpolation cosh VI = cosh (0.0688 + ,7 0.42) = 0.915 +j 0.028, sihh VI = sinh (0.0688 +j 0.42) = 0.063 +j 0.409, and using the values of p, q, a x and a 2 as obtained above we get I r = 97.6- j 11.19. E r = 57.7 - j 17.98. The agreement in the values of I r , E r as calculated by the two different formulae is very close indeed. The absolute values of the current and voltage are I r = V(97.6) 2 + (11.2) 2 = 98.2 amp., E r = V (57.7) 2 + (18) 2 = 60.4 kv. The current at receiving end lags behind voltage at generator end by angle tan- 1 ^| = tan- 1 0.101 = 5 46'. y i .o The voltage at receiving end lags behind voltage at generator i o ro end by angle tan' 1 ^^ = tan" 1 0.312 = 17 20'. o/ . / Therefore, the current at receiving end leads the voltage at receiving end by the angle 17 20' - 5 46' = 11 34'. ,_ ffi . 98.2 X 60.4 X 0.98 n on Transmission efficiency = 12 Q x 60 X 90 = per ' GENERATOR VOLTAGE AND IMPEDANCE AT RECEIVING END KNOWN The problem in long distance transmission may also present itself in the following form: Given the potential at the generator end and the impedance at the receiving end, to determine the current at the generator end and the potential and current at the receiving end. Let Eg denote the voltage at the generator end, Z denote impedance at the receiving end. Distance is considered from receiving end. CALCULATION OF ALTERNATING CURRENT PROBLEMS 227 The values of I g , E r , and 7 r are given by the following formulae: Ir = (q-jh)(c + l+jd) / = E,(K 1 +jK t )(a 1 +ja a ) (q - jh) (58) where "'-oT a2 = c^' e+ *-J^S- <> q = ct al cos 01 d(? 1 sin 01 ~ al cos 01, h = Cf? 1 sin 01 + rfe aZ cos 01 + ~ ai sin /?/, KI = ce al cos j8/ de al sin j8Z + ~ aZ cos /3Z, -^2 = ce al sin /3/ + c? aZ cos 01 e~ al sin 01. In terms of hyperbolic functions we obtain the following set of formulae for this problem: E g jcoshFs + si / Z E _ Z sinh VI + Z cosh VI _E g \Z Q sinh Vs + Z cosh Z sinh VI + Z cosh (60) where Z has been put for brevity in place of ai ^2. For / and E at the receiving end, we put s = in formulae (60) and we get lr Z sinh VI + Z cosh VI E = E Z ZosmhVl + ZcoshVl' At the generator end, s = I, we have >i+ z + Z~o (61) Z sinh VI + Z cosh FZ (62) 228 FORMULAE AND TABLES FOR THE Example. Let Z = 540 + j 261, V(540) 2 + (261) 2 = 600, and let us take for the line constants the same as those given in the example on page 223. R = 0.33 ohm per mile, L = 2.13 mh. per mile, C = 0.014 mf. per mile, = 0, 1 = 300 miles, co = 2;rX60, E g = 60 kv., al = 0.126, # = 0.633, & l = 1.134, ~ al = 0.882, cos/3Z = 0.806, sin# = 0.5915, 01 = 392, 02 = 78, 392-J78 + 540+J261 1;i = =-1.4 $ = -3.46, h = 1.47, #1 = -2.04, K 2 = 0.43, 5 2 + h 2 = 14.13, (ai 2 + 02 2 ) ( S 2 + /i 2 ) = 2,257,240. Introducing these values in formulae (58) we get E g (-3.46 -j 1.47) (-2.46 + J2.ll) ^r = - 14 13 ~ = u g (0.82 - j 0.26) = 49.2 - j 15.6 kv. The absolute value of E r is V(49.2) 2 + (15.6) 2 = 51.6 kv. The voltage at receiving end lags behind the generator voltage by the angle tan- 1 ^ = tan- 1 0.317 = 17 36'. , = E (-3.46 -j 1.47) (-0.46 + J2.ll) (392 + j 78) 2,257,240 = gg (1.043^0.987) = 626 absolute value = V (62.6) 2 -f (59.5) 2 = 86.4 amp. The current at receiving end lags behind the generator voltage by the angle tan- 1 ^| = tan- 1 0.934 = 43 3'. = E g (-2.04 + .7 0.43) (392 + j 78) (-3.46 -j 1.47) 2,257,240 absolute value = V(76.8) 2 + (31.7) 2 = 83.1 amp. CALCULATION OF ALTERNATING CURRENT PROBLEMS 229 The current at the generator end leads the generator voltage by the angle tan- 1 ?^ = tan- 1 0.413 = 22 27'. 7o.o The current at the receiving end lags behind the voltage at the receiving end by the angle 43 3' - 17 36' = 24 27'. At the generator end the power factor is cos (22 27') = 0.924 and at the receiving end the power factor is cos (24 27') = 0.910. Efficiency of transmission, 51.6 X 86.4 X 0.910 77 = 60X83.1X0.924 ^percent. Working out the same problem by the hyperbolic formulae we get cosh VI = cosh (al +jpl) = cosh (0.126 +j 0.633) sinh VI = sinh (al +J0J) = sinh (0.126 +j 0.633). From Table XX we find cosh (0. 12 +j 0.64) = 0.808 +j 0.072, cosh (0.12 +j 0.62) = 0.820 +j 0.070, cosh (0.14 + .7 0.64) = 0.810 + j 0.083, cosh (0.14+ j 0.62) = 0.822 + j 0.081 . Therefore, cosh (0.12 + j 0.633) = 0.812 +j 0.072, cosh (0.14 +j 0.633) = 0.814 + j 0.082, and cosh (0.126+ j 0.633) = 0.812 + j 0.075. In a similar way we obtain from the same table by interpola- tion sinh (0.120 +j 0.633) = 0.102 + j 0.597, Z Q = ai - joj = 392 - j 78, Z = 540 + j 261, #, = 60 kv. Introducing the above values in equations (61) and (62) we get ". 60,000 r ~~ (392 -j 78) (0.102 + j 0.597) + (540 + j 261) (0.812 \+j 0.075) 60,000 505.4 +j 478.5 _ 60,000 (505.4 - j 478.5) 484,391 bAe J 59.3 amp. 230 FORMULAE AND TABLES FOR THE 0.812 +j 0.075 + ^"jj 7 '. 2 ^ 1 (0.102 +j 0.597) | Ia = tf a { 0.812 +j 0.075 +(1.198 + j 0.904 ) (0.1 02 +j 0.597)1 (392 - j 78) (0.102 + j 0.597) + (540 + j 261) (0.812 + j 0.075) _ 60,000 (0.394 +j 0.882) 7fi7 505.4 +j 478.5 " = 76 ' 7 505.4 + 7 478.5 The results obtained by the two different sets of formulae are 'in exact agreement. TRANSIENT PHENOMENA IN CIRCUITS HAVING DISTRIBUTED INDUCTANCE AND CAPACITY In Chapter V we gave a brief discussion of transient electrical phenomena. We have shown that if the circuit conditions are disturbed in any way the current and potential do not pass over instantly from one permanent value to another; it always re- quires a certain time interval before the current and potential reach their steady values again. During the transition period the current and potential may be considered as consisting of two components, one of which corresponds to the permanent value, and the other the transient term which disappears more or less rapidly. In the formulae given in Chapter V we have considered only circuits of localized inductance and capacity, so that the space element does not enter into the formulae. We were concerned only with the time element, the time rate at which the transient term disappears. In problems, however, connected with long-distance transmis- sion lines, the distributed character of the inductance, capacity and resistance must be taken into consideration. If, for instance, an e.m.f. is impressed on a long line, every element of the line is not charged to the same potential at the same instant. It requires a certain length of time before the entire line is com- pletely charged, and during that interval the charge and poten- tial differ for different points on the line. In the discussion, therefore, of transient phenomena on lines we must deal with two variables, distance and time. CALCULATION OF ALTERNATING CURRENT PROBLEMS 231 It is to be regretted that the formulae for transient phenomena on transmission lines cannot be presented in the same manner as those given in other chapters. The formulae are generally very long and involved, and if the bare formulae are given leaving out the derivation and the complete discussion of the physical significance of the various factors, their usefulnesss is greatly reduced. The formulae for transient phenomena occurring on a transmission line have their greatest value in enabling us to get a proper appreciation of the phenomena occurring on the line irre- spective of magnitude. It was thought advisable, therefore, not to attempt to cover thoroughly this branch of the subject, but to give only a few formulae for the cases of most common occurence. Those inter- ested in the subject can refer to Professor Steinmetz's book " Transient Electric Phenomena and Oscillations" where the whole subject is discussed in a masterly manner. TRANSIENT PHENOMENA IN CIRCUITS HAVING DISTRIBUTED RESISTANCE AND CAPACITY Assume the inductance of the line negligible as in the case of transatlantic cables. Suppose the cable is open at receiving end, and a constant con- tinuous e.m.f. E is suddenly applied at the sending end. A charging current commences to flow in the cable and propagates itself from point to point on the cable until the whole cable is charged to the impressed potential E. During the charging period the current and potential vary at different points on the cable, so that the current and potential are functions of the time and the distance. Let r = resistance per unit length, C = capacity per unit length, I = length of cable, E = impressed e.m.f, s = distance from sending end to any point on the cable. The potential at any point on the cable distant s from sending end may be put in the form, e=E + e v , (63) 232 where FORMULAE AND TABLES FOR THE ** 25 t (64) The current 2E( -?A COS - -7 H- 25 1 4r(7 COS (5^s\ \2 l) (65) If we put Ir = TI total resistance, 1C = Ci total capacity, then, neglecting upper harmonics, formulae (64) and (65) become approximately, (66) At the end of the line (s = I), the potential is x per cent less than the maximum at the time T x given by X 100 x (67) As an illustration we may take the cable between Ireland and Newfoundland which is 2640 km. long. The total resistance n = 6000 ohms, Ci = 40 mf . The problem is to determine the time required before e reaches the value 0.95 E. We have e v = 0.05 E, x = 5, 4riCi, 400 4X6000X40X10- 6 , 80 96 X 10 -2 X 3.27 = 0.31 second. CALCULATION OF ALTERNATING CURRENT PROBLEMS 233 The time required for the potential to attain 99 per cent of its normal value is T x = 4 X 6000 X 40 X 10~ 6 , 400 A . ftn = 0.499 second. As another illustration let us determine the value of the potential and current for points 100 km., 1000 km. and 2000 km. distant at the time t = 0.1 sec. Neglecting the upper harmonics and using formula (66) we have T 2 9.87 4 X 6000 X 40 X 10~ 6 -i.o2 8= Q.358. 10.28. Rise in Voltage m a Transatlantic Cable 0.1 0.2 0.3 0.4 0.5 j 0.6 0.7 0.8 FIG. 68. Rise in voltage in a transatlantic cable. 0.9 1.0 For s = 100 km., sin (^ ) = sin (0.06) = 0.06, cos (0.06) = 0.998, l>j s = 1000km., sinf | yl = sin (0.60) = 0.56, cos (0.60) = 0.825, s = 2000km., sin/! |) = sin (1.2) = 0.93, cos (1.2) = 0.362. Introducing these values in formula (66) we get, 8 e j 100km. 1000 km. 2000km. 0.973# 0.745 E 0.544# 0.7147 0.5917 0.2597 234 FORMULAE AND TABLES FOR THE where / = In Fig. 68 a set of curves is given showing the rise of poten- tial over the entire length of the cable. In Fig. 69 curves are given for the current as a function of the time for the points on the cable s = 0, s = 0.25 I, s = 0.5 Z, s = 0.75 Z. Current Distribution on a Cable on Sudden Charging 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 t in Seconds FIG. 69. Current distribution on a cable on sudden charging. A CHARGED CABLE HAVING A CONSTANT E.M.F. APPLIED BE- TWEEN ONE END OF THE CABLE AND GROUND AND SUD- DENLY GROUNDED AT THE OTHER END The permanent values of the potential and current at any point on the line distant s from the point where the e.m.f. is applied are --I (68) The values of the transient terms of potential and current at any point on the line are given by the following formulae: CALCULATION OF ALTERNATING CURRENT PROBLEMS 235 2E( 1 ~ - ' "/f r ^ s v * ' w ^ (69) * =~ ? J^' COS (T)- "^'fr) 9 IT 2 in \ ) i r"r>~ * / " "TT^M r + riCl COSf y ) > For the current at the end of the cable we may combine the second equations of (68) and (69) and put them in the following form : / = f (l-2 m - r^'cosrmr), (70) V m=l which gives the arrival curve of the current at the receiving end. The series given by equation (70) is very slowly convergent for small values of t, and makes the calculations very laborious. Equation (70) may be put in another form which gives a rapidly convergent series for small values of t. The equivalent of formula (70) is 41 As an illustration we take the same data as given in the exam- ple given on page 232. n = 6000 ohms, Ci = 40 mf., I = 2640 km. For t = 0.001 second, we have ~t = 0.041. By (70), 7 = ^ Jl - 2e-- 041 cos7r - 2e-- 164 cos27T - 2 e-- 369 cos 3 TT - { = - }1 + 1.92 - 1.7 + 1.38 --!. It requires a very large number of terms to obtain the value of/. 236 FORMULAE AND TABLES FOR THE By formula (71) we have _ 40 X IP" 6 X 6000 4 X 10- 3 ^ - * X 2.9 X e- which shows that for the time t 0.001 sec. the current at the receiving end is negligible. If the cable is grounded all the time at one end and a constant e.m.f. is applied at the other end the permanent values of the potential and current as before will be (72) TI The transient terms for the potential and current are given by the formulae ^-i J -^-t >2*8 (73) i C COS I ~T~ I 1 * COS I ~ ^ J i" * i * ri /27TS COS( CIRCUITS CONTAINING DISTRIBUTED RESISTANCE, INDUCTANCE AND CAPACITY; LINE OSCILLATIONS We have shown in Chapter V that in a circuit containing inductance and capacity, the transient possesses an oscillatory character, unless the resistance exceeds a certain critical value depending on the inductance and capacity of the circuit. In circuits of distributed inductance and capacity, the transient is, in general, oscillatory. The discharge of an accumulated charge of atmospheric electricity on a line, a sudden change in load, an arcing ground, or a break in the circuit will set up electric oscil- lations on the line. Owing to the inductance and capacity of the system a flow of current stores up magnetic and dielectric energy in the system. A sudden change in circuit condition, grounding the line or a break in circuit, for instance, releases the energy stored in the system and it oscillates back and forth until the entire energy is dissipated by the resistance and leakage of the system. CALCULATION OF ALTERNATING CURRENT PROBLEMS 237 Since the maximum magnetic energy must equal the maximum dielectric energy, we have - Vi*. (74) The quantity Z is called the natural or surge impedance, and yo is the natural or surge admittance. In the case of circuits of large inductance and small capacity even moderate currents may cause a dangerous rise in voltage by a sudden change in circuit condition. The general expressions for the oscillating current and voltage are as follows: The values of the constants ?, n , A n , B n , P n and Q n depend on the terminal conditions, and We shall now consider a few specific cases. LINE GROUNDED AT ONE END AND OPEN AT THE OTHER END The terminal conditions are E = when s = 0, and 7 = when s = I I is the length of the line. To satisfy these conditions we must have ^ 7 (2 n 1) TT Q n = and y n l = - - J A 238 FORMULAE AND TABLES FOR THE and equations (76) reduce to A n cos p n t + B n sin (3 n t \ , (78) cos (2 n ** I A n cos ^ + B n sin n = 1 _(2n-l) ri^~ at 2) n- l) 2 7r 2 If we neglect a 2 under the radical, as it is generally small com- 7T 2 pared with the term y^77 > an( i also put ZL = L and ZC = C , so that Lo and C represent the total inductance and capacity of the line, we get j8 is the frequency constant of the line. The fundamental frequency of the oscillations on the line is given by f = A = _ 4\/LoCo' We may therefore write /3 n =27r(2n- !)/!. (82) Equations (78) may now be written in the following form: n = oo COS (83) The current and potential on the line are complex waves con- sisting of fundamental waves and odd harmonics. For the funda- mental wave the current is zero at the open end of the line (s = I) and gradually increases to its maximum value at the grounded end of the line. The potential is zero at the grounded end of the line (s = 0) and gradually increases to its maximum value at the open end of the line (s = I). The wave length of this, the fundamental oscillation, is four times the length of the line. CALCULATION OF ALTERNATING CURRENT PROBLEMS 239 The values of the constants D n and 4> n depend upon the initial conditions, and have to be evaluated independently for each par- ticular problem. In general the constants are determined in the usual manner of evaluating a Fourier Series. We may best indi- cate the process by the following illustration: Let the line be .charged to a uniform potential EQ but with no current in the line before the discharge. The line is then grounded at one end s = while open at the other end s = 1. We have then for t = 0, 1 = for all values of s and for t = 0, E = E for all values of s. From equation (78) it is evident that to satisfy the first condition we must have A n = 0, and to satisfy the second condition we must have > (2n I)TTS .., The values of B n can now be determined in the usual way of evaluating the constants of a Fourier Series, that is, by multiply- ing both sides of the equation by sin 7 and integrating 2. I between the limits and 2 I.* We shall then get for the value of the constant Introducing the value of B n in equation (78) we finally get , 4#o 4 /C . ( ITS . ,1 STTS . / = V T e ) cos ?n sm uit + o cos TTT sm 3 wit 7T T _/v T Zi I, A Zi I, L 1 57TS . - cos -^y sin 5 coif + - 4^/0 \ 7TS .1 . ^ .1 u -~ nt ' sin JT-J cos wit + - sm -^-r cos 2Z 3 2/ i (85) where i = 2 TT/I. LINE GROUNDED AT BOTH ENDS Equations (76), being perfectly general, also apply in this case, but the constants y n and /3 n are different from those discussed in the preceding section. * See Byerley, Fourier Series. 240 FORMULAE AND TABLES FOR THE We have in this case the potential zero at both ends of the line, that is, E = for s = 0, and E = for s = I. To satisfy these conditions we must have Q n = and y n l will have to be a multiple of TT, that is 7 n Z = mr, y n =^> (86) and n = ^. = . - ? neglecting a 2 compared with IVLC The fundamental frequency of the oscillations is given by /i = I- (87) Equations (76) may now be written for this case in the following form: / n = oo E = V 77 -< V D n sin ^sin (Zirnfit - ^ n * C/ v J-Sn COS ^ ' COS ( ^ 7T71/16 Yn) n= 1 The fundamental oscillation of the current has its maximum values at either end of the line but in opposite directions; and it has zero value at the middle of the line. The fundamental oscil- lation of the voltage has its maximum value at the middle of the line and is zero at both ends of the line. The fundamental wave length is half the length of the line. The oscillations contain all the harmonics even and odd. LINE OPEN AT BOTH ENDS In this case we also make use of equations (76) as the general solution, but the terminal conditions for this problem are 1 = when s = and s = I. These conditions are satisfied by equations (76) if we put tVJT P n = and y n l = mr or y n = (89) CALCULATION OF ALTERNATING CURRENT PROBLEMS 241 The value of /3 n is the same as in the previous case, rnr mr i VLC VL O C O V, ^ The fundamental frequency of the oscillation is Introducing the above values, equations (76) assumes the form : Dn gin C os (27m/!* - n ), E= - cos (90) For the fundamental frequency the current is zero at both ends of the line and has its maximum value at the middle of the line. The potential has its maximum value at either end of the line but of opposite signs, and its value is zero at the middle of the line. The fundamental wave length is equal to half the length of the line. The oscillations on the line is of a complex form con- taining a fundamental wave and all of its harmonics, even as well as the odd ones. CIRCUIT CLOSED UPON ITSELF The terminal conditions for this case are 7 = Ii and E Q EI, that is, the current and potential have the same values for the points s = and s = L Both these conditions are satisfied if we put in equations (76), which is the general solution, and hence, y n l = 2tt7r and y n = j > 2mr i VLC VL Q C O (91) The fundamental frequency of the oscillations is /. /5i 1 (92) 242 FORMULAE AND TABLES FOR THE Equations (76), in their application to this case, take the forms n = oo 2 wns , ~ , 2 TTHS ) --- 4 = 1 \ A n cos 2 7ra/i + B n sin x^ ( n - 27ms . 2 ]^nSin-y- n = 1 ' (93) | A n sin 2 irnfit B n cos 2 irnfit \ ; or we may write the above equations in the following form : V D cos (2 7m/!* - F n sin - cos 0n) n= 1 2irns . /0 -. cos j sm (2 7m/i (94) For the fundamental frequency of the oscillations equations (94) reduce to cos - cos - y gT^j A sin-^sin^- (95) CALCULATION OF ALTERNATING CURRENT PROBLEMS 243 O- <-l(M O-*CO cococo o p p o p p d o" o o o iiSJS ooooo ddddd O5O5O5O5O5 OOOOOOOOOO p p ppp ppppp o'oo'od odddd ooooo ooooo i-i i-i H d O O O O d O OOOOO O O O O O O O O O O ooooo ooooo ooooo oco^-c^ oc5cotoo5 T-icotocor^. 88888 d d o' o' o' o o o o Oir.HCOioK O^COlCt^ OOOM-^io i-l C-q I * | ^J< * o' o' o o' o' o' o' odd o' d o' d d 88888 88888 88888 o' d o" o' d d o' o' o' o' d d o' d d 88888 d o' d d d 88888 ddddd I^HT-HOO OOOOO 88888 88888 88888 ddddd dddo'd dddo'd o' d o d o' o o' o o" d o' o' o o' o' ooooo ooooo ooooo ooooo ooooo 244 FORMULAE AND TABLES FOR THE I ^ I + 9 3 oo oo oo'd ddddd do' odd -! iCJioiococp 00000 ooooo ooooo COi-HCO'-HiC OCOCOOCO COOSi-ccMCO * *f CO t-- OS O i i CO -^ CO t^- CO C5 i I CM CO ^ eocococot- t>.t^t^t~-t^ i>.i>.opooop op ooooo o'dddd ddddd d odd do' o'dddd dddo'd d ddddd U5COOOO5O sssss ddddd ddddd 88 g ddddd o'ddo'd d ooooo ooooo ooooo ooooo ooooo o 000 JO CM 0> o' o' d d d T-* CO co rf< oo oo ooooo *OiOiOOiO ^^^^^tl doddo doddo ooooo ooooo ooooo o 333SS o'dddo ooooo ooooo o d d d o d 00000 ooooo t^COCOlM O5 d d d o' o" d d d d d ooooo ooooo o O5-*OOCOOO COt^T-lOOO i-l ooooo ooooo ooooo o < i^ r^ co c ooooo ooooo ooooo ooooo ooooo o ooooo COCOIM (M ^H 8888,8 d d d d d ooooo p p pop pop p p p o' d o' d o' d o' d d d o Ill do o t^ CO *O ^f CO C- o' d d d o' d d o' o' d ooooo ooooo o Coio>oio odood O .t--t>.t--t^ do odd oo CALCULATION OF ALTERNATING CURRENT PROBLEMS 245 XI s + 8 COCO< JSS! 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O 03O5O5 OO o o O COCO t^ 00 OO OO OO OO OO d o o' o' o' ooooo ooooo ooooo ooooo ooooo o 00 e. t>- 1>- CO CO < o' o' d o' o' o o' o' o' d o' d o" d o" o COt^OrJ- CDCD CDiOlO Tj< O IO IO IO d d d d d ooooo ooooo ooooo o ooooo ooooo ooooo ooooo ooooo o 5O CO CD O CO CD O CO iO l^ O ooooo ooooo OS OO 1C CO O3 ooooo t^ COOS CO .-H ooooo ooooo o t>. t>- CO CO CO CD CD CO IO kO IO ooooo ooooo ooooo ooooo ooooo o CM o>cort< ^H O3 O CM *}< CO ^tlOlOlOlO ddddd IOOCD COCO d d o o' d tti-i OOiOCM OO OO t^- t^ t~- o' o d o' d ooooo ooooo o ooooo ooooo ooooo ooooo ooooo ooooo ooooo ooooo s; . o' d d d o d d d d d ooooo ooooo ooooo o CO CD CO 10 10 10 d o' o' o' o' d OO O CM T^ IO *< iO iO iO iO o' o d d d COCO O51OCM iO iO CO CO CO d d d o' d ooooo ooooo IO l>- OO O T-I CO CO CO t> . t-. d d o o' o' ^ CM OS CO CO lOJO-*^-^ d o d d d o o o o O t^-^t-100 Tt< CO CO CO (M OOOOO d o o' o' d d 1OC.OOOCO oSSS^n rt^rt^^ ooooo ooooo ooooo ooooo ooooo o 8SS5SS o' o' d d o' r* co o 10 * < c i COIO^ICM ~H ( i-- 1-* t^ t-- t^- < **< OOcxi co CO v I O OO C CO CO CD 10 ooooo ooooo ooooo ooooo o OOOOO OOOOO OOOOO OOOOO OOOOO *-i CALCULATION OF ALTERNATING CURRENT PROBLEMS 249 li \ , X + X e ooooo ooooo ooooo ooooo ooooo ooooo ooooo ooooo i-o-03 i-5cO"5P- -cx)CiOO rHc^coco-^ do' odd ddddd lilis IslSI 10 OOOOO 8SSSS d d d d d COCO CO CO CO o' o o d d CO COCO COCO d d d d d ddddd CO 10 o O 10 irj Tt< TJH <$< -*tt COCOCOCOCO COCOCOCOCO 00000 O5OCOO51C COCOCOC^C^ cococococo do odd iiiil d d d d d ISSS? d d d d < 3-gggg SssS! SSSSS d d d o' d d d d d d 3SSI d d d d d ddddd d o' d d d o' o' d d d co r-ooooos O i-l CO O COOOOOl 1000 OOOOO 88898 d d d d d cococococo o' o' o" o o' ooooo ooooo ooooo CO CO CO CO CO o' d d o' o' COCO CO CO CO o' d d d d IS533 d d d o o ooooo ooooo ooooo (M ^Hi-H OOO ooooo OTH^H 00 i Tj< -^ -^ CO < O OO O < 8S5 ddddd ooooo ooooo T-l CO -< CO i-l CO CJ d o' o d d 1C Tt< 1 ( OO Tt< GO t-- CO -^ CO OO OOOCOOOO odddd OJ CN CSIM CM d d d d d o ddddd OOCM OO^ *f CO i t O GO d d d d o o OM< oeo CO CO CO CO CO o' d d o d ^H^ t^OO CD CO O O o o' d d d OOOO OrtH O O 1O 1O 1O d d d d d CN O5 lO^H 1-- co co co co co o o d o d !-< O2 COCO O O T-I CO 10 ?i CD COO COCO O^OOCM ooooo ooooo ooooo ooooo o d d d d d o' d d d d CM CM CN CM CN d o d d d C7> CO O ^ 1C O o' o' o o o' o' OCM O idi-l ^OOO O5 CN CN -t^t2 ^St^OO O O O O Jo o'o'do'd odd do' odd do' o'o'do'o' o'do'o'd d SGN -3< O lOiOiO ooooo d d d d d I Ci < >0< o o d o' o d o' d d o' d d d d o C33 <*< t-- 01 b- Ct^ OOO rH t^l>. t^OOOO ooooo ooooo o CNO.-I OO ^ CO CO CM CM CM CM CM CM CN o' d d d d CM CN CM S S 22 d d o' d o' o' O CN OO Tt< O o' d d d o' CN I O O OO O5 O3O50OOO d o' d d d CM CM CM CM CM d d d d o' lOrH r^ CM OO O5 OO^H ^H i-l - 1^ ooooo ooooo ooooo 1O 1O CD CD l~ t^- CN CM CN CM CM CN d o' d o' d o' ^ N O ^ O OO 1O CO CN O GO O o o' o' o' o' o' ooooo h-COOt^-* CN CM CN CM CN d d d o' d ooooo I-H OO-* l> 'CC d o' o' d o' CM l^- CM OO T^ d d d d o' 3 OOO O ooooo o CO i 1 CD T-H CD 10 10 CD CO ooooo ooooo o o >o o *< t-l CM CM CM CM ddddd CM CN CM CN CN d d o' d o' d d d d o' d d o' d d o' d d d o' o' d o' d d ) CN 10 O5CN O J d o o d o > CO CD O O 1O J o' o' o' d o SCNlTjl COOO O5>O2 d d d d d ooooo ooooo ooooo 8 J d o' d d -i CALCULATION OF ALTERNATING CURRENT PROBLEMS 251 I! 1 S * + si PQ 7 s < + H H oooo oo o o o o o 8S o' o o d d (M-* CO *-H CO 1C l I-H C^l CO M* cococoii o o o o oo oo o' o' o o' o o o o o o o ddddd 00 oi oo t^ r^ co co ic ^HCNCO T^lCOt^OO o o t^ r^ to ic co ooooo 2g OOOOO COOOOO JiC-^cO >000 d' d o' o' o" OSOOO^H o'dddd o'ddo'd o'dddd o'o'do'o' do' odd ooooo ooooo ooooo ooooo ooooo S8S ddddd s .SSo-8 o' o' d d d OO OO I s * CO 1C d ( OO 1C * 1 CO I CO T < OOOOO 00a>0?05 OOOOfSr- ooooo ooooo CO^H I 00< o o ooooo ooooo ooooo ooooo liii 3jii o' o' d o' o' ooooo ooooo ooooo ooooo ooooo o'dddd do odd o'ddo'd do'ddd o'o' odd Ol^kCt s OSOCM^CO t^-< C< -^ CO OO O CO 1C t Oi i I ( OOOO 1 I 1 I T * 1-H T-H Cl( H -+, ,-H _*, _^- , )O 5CO poop o' d d d d ^^OCOt^ ppppp d o' d d d -^ ^HOSOOlCCN OOO5.t^t^t^ O O O O O O O O O O O CM Tj< CO OO OS O5 O> O5 OS O O O O O to iO iO O CO ddddd SiSiS d d d o' d ^HCOi-HCD^H IO 1111 o' o' o o o' ooooo CMCOO5CMIO oor^cocoio COCOCOCOCO ooooo ooooo 333^22 CO CO CO CO CO d d d d d O CM -^ CO OO CO CO CM CM CM o o o o' d . co -^ o co r- (M CN O) C^ (M o o o' o o o o o o o MTO COCO ScSw ooooo C^ O CO C^ C o o o' d o OOOOO OOOOO d d o' d d odd d d OCOor--( ) CO COCOCDCO' ooooo > IO O T^ O5 Tt< C 5 oo oo r^- co co 5COCO COCO COt OOCOOOCMCO d d d d d 5CO COCOCOCM < oo op oo oo os o" d o' o' d ss^fegs? CM CM CM (M CM OOOOO - T i I CM CO Tt4 Tt< CM CM CM CM CM d d d d d o o o d d o' d iii ooooo ooooo ooooo ooooo ooooo d d d d d d d d d d do'do'd o'do'o'd do'o'dd do odd do'o'o'd ooooo f^CM h-(M t^. 10 10 -^ -^ co CO CO CO CO CO d o' d d d T-I IOCS COt-. CO CM ^H i I O CO CO CO CO CO d o' o' o' o' ^cScScM CsSS^C^ d d d d d d d d d d O T * CM CO CO CMCMCMCMCM CM CM CM CM CM d d d d d d d d d d O -^ CO C OS 05 05 COCOCOCOCO COCOCOCOCO do odd ddddd do'do'd o'ddo'd o'do'dd o'dodd ooooo ooooo CO CO CO CO CO o' d o' d d CM OO -^f O ^H CM Ttl CO d o' d d d do odd do odd ooooo ooooo ooooo CM CM CM d d d d d O O CM CM CM CO CO CO CO o'dddd ddddd I-HO OSOOCO^S 1 *^ CM l OS GO CO 1OCOCMOOO r-lOCOCM< ooooo ooooo ooooo o o OCM^COOO S.t-.t^b-t^ d d d d d OCM-rfCOOO OCM^t*COOO 0000*0000 05 OS 05 05 05 d d d d d ooooo CALCULATION OF ALTERNATING CURRENT PROBLEMS 253 * o , ^ t H A 1 1 3 % 8SSSS O (M-* DCX3 d d d d d COCOCOCOCO l ! 3! 2? 2 do odd ddodd do'do'd 8SSS8 ddodd COCOCNI i-H^H d d o o' d H CO C5S (M l l C<1 C^l CO C ooooo ooooo ooooo ooooo odd do odd do' do'o'o'd do'do'd >r^T' d 5 M * T}< d d o' o' d d d d o' d odd o' o' ooooo ooooo ooooo ooooo ooooo d o o o o ) t~- O5r-( ( d d d d d CO O i I C T^t -^ ^ Tj< TP lO ^O ^ COi - d o' d d o' CO to t^- Oi r-t-t^t- ooooo coco to T? S o' o' o' d o' d d d d d ooooo COCOOOOCM ooooo o d d o d O * i OJ Tt< to CO d d d d o o' d d d d d ooooo ooooo ooooo d d d d d i-HOOOO OOOOO d d d d d Qf5o>c5 o 5_T?T?: o' d d d o' M< COOOO5O J^ CO CO S S. d d d d d to to to Tf< CO Tt< CO 00 O CM d d d d o o o ooooo ooooo o' d d d o' o' d d o' o' d o d d o y i to OO i < OO ^H CO 0005050 ooooo o CO CO CO CO CO CO CO CO CO CO o d d d d odd o' d o o llslS i|lii iss&l d d d o' d d d d d d ooooo ooooo ooooo ooooo ooooo COCOOCMIO t-O-HCOT o' d d d d d d d d d do' odd do' odd ddddd T* *}< T< T>< T}< CO CO CO CO CO ddddd do' odd d o' d o d o o o o d o' d o' o' o' odd o o' i^gii n co co co co d o' d d d d o' d o' d S'o o^IcN ) T* Tfl T(H Tt< CO COCO ( o' d d o' d o' d o' d d ooooo ooooo ooooo , |S SSScSg d d d o d o o o d d OCXI * COOO ddddd do odd OOOOO OOOOO 1-1 CALCULATION OF ALTERNATING CURRENT PROBLEMS 255 I + t"^ J^x 88388 SS3SS2 dodo'd dddo'd do' odd do'dod ddddd do' odd ddddd ddddd do odd do odd *t< Tt< CO OQ i I Oi !> IO CO i t OS O CO OS 1C O IO O O O 1C O CO !> O COCOCOCOCO CSC^C 10 o kc o 10 os co r>- )-^^fCO CO O4 I-H i i O > CD CD CD CD CD CD CD CD OOOOO OOOOO OOOOO OOOOO OOOOO ooooo O COt^CDO'H do'ddd do odd Oi OO t-- OOO ^^So5 OOOO o'do'dd o o o o S^coos^ coSoooco rH i-H ^H r-t CJCOCO ddddd o'do'dd d d o' o o' d o' o d d oo oodo'd o' o' d d d d d d d d ddddd o ooooo ooooo' C- Oi i ' **< CD OO i ( CO O OO O C>l 0000 rt^^H^-lC Ca(M !>. S >. ?i o d d d d ooooo o i- < 1C CO t"- OO Oi O> < > * I CO IO t- Oi * I C (CO COCO CD cot^l ooooo ooooo COCOO5-IOO Ot-OOO5O ooooo ooooo ooooo IsSss o o o d d d d d d d ooooo ooooo o ooooo ooooo ooooo ooooo 1000 ooooo ooo'o'd ooooo o zJ2t2ttco o d d o' d d d ooooo ooooo ooooo d d d d d d d i d i-i i-i i-i CD CO CO to *O to tO iO iO ooooo ooooo ooooo ooooo ooooo o o o' d o' d o' d o' d d ddddd O i-H C O i ' CO *O d d o d d mt~*.a*T* co -+ ~-~ oi 01 ^-H >C 1C i 10 10 1C OOOOO OOOOO OOOOO -H > Ci OO t->- SCO CO CO ooooo ooooo ooooo ooooo ooooo o >-i(M CO- cocococo do' odd O IO O IO -^ C) CO Tt< 10 CD d d d d d I>OOOO I * Tt I -^ I i 10 d d odd d d d d d o ,_, ^H ,-H ,-< ,-1 OOOOO OOOOO OOOOO OOOOO O d d o' d d d d d d d CD OO Ci i < CO t^. t^. t^- oo co b-OJrH (MCO ooooo ooooo o ooooo ooooo ooooo ^ tO CD CD CD d d d d o' c^cogco 1 3, ddddd d o cor^ooo CO CO CO CO CO d d d d d do odd do odd o'o'o'o'd d - O CO t>- t^ ^I^H^^O ooooo ooooo ooooo T I O> OO CO ^ _b-coeoco to d o' d d d d od o o o oo o o 8 o'o'ddd -! CALCULATION OF ALTERNATING CURRENT PROBLEMS 257 + CQ + 8SSSS dddo'd o o oooo OOU50 d o' d d o o o'ddo'd o'd odd d o' o' o d odd do IC003 cOCOOOO d o o' d o' o d d o' d odd do o'o'o'dd. o ooioor^co CM i -* I-H i-H CO CO CO CO CO co co i"* t*- "^ oo o c usii^co ic t^ oo o I-H co ic co oo CM CM CM co co eo co co co < o o oooo o'oo'o < CM eo ** 10 us - CO 1C **< CO CM i ' N C^ IM (N M W I d o' d d d o COIOKMIM COCX3COOQO OOOOO O T*< O 1^5 T-^ s^^Sci ddddd oooooooooo d o' d o o . dddo'd ooooo o' o' d d o' ooooo O i t CO lO O OOOOO ooddd coor~coo OOOi-HCO-^ O t i I < i < ddddd CO GO O T 'CO i-H i ( i I C^l C^l ddo'od ) CO t^ O CO IT-.COCDIO ICMIMIMO* 1^- OO Oi O < CO . t-- ^HlCO: CM ooddd 338co- oo'dd ddo'o'o o o d OOrt Tjt^ i i ^ CO OO ^< US 1C 1C us d o' o' d d o' o' d d o' iliSi ooooo' t^- t^- t^- CO CO CM CM CM CM CM OOTj< O ICO t-^. O5 i < CM -^ O p ^H T-H ^H oo'ddd COIM t^CJt- lCt>-OOO^H T-H ,-H rt iCO ICO (MiCt^O OOOi^ ddddd O >COCDO Tt< O CI>-OCM i^ 1^1 li CN! C^ O5 -* OOCO Tt< t^ Oi cq dOiCNCO o ^H 1COOCN 1C t^OSt-H^fCD COCO 1 ^^^ dddo'd O5CM US OOO OO i * CO 1C OO d o' d o' d ooooo ooooo ooooo ooooo ooooo CO ^H CO ^H CO r-o>o OO O O T-H i-H T 4 i I TjtecciiM i-iooioo ooooo ^CO(Mi I OOOt^-COiO COC-1OOOO O OOOO OiOiOiOlO 10 IO "* $< Tt< ooooo doodd o'odod d ddddd GO Oi * ' Ol -^ ooo oo dodod -00000 o'odod d i-iOOOO OOOOO OOOOO O SS'oS^ ddddd i-- r~ t^ oo oo d d d o" d OOOOO 0000.-1 o'ddo'd o'ddo'd o'o'do'd o'ddo'd odd do d OOOOO OCOO CO(M OOOOO t^o IM ^o ddddd e-OOOO OOOO * < CO CD OO O COI>-I I <*< 0( o' o' d o o 00 t- < SSSS8 o o' d d o' ooooo OSOOOt^-CO S i o S 5 d o d d d co -< oo 10 C O> o d o o' o IO ifl i 1O IO >O o o' o' o d giro's j o -^ ^ *< Tt< *# dodod o' Jo^2S^ Sooooio caco^ioco oo i 4 C^ CO CO SO-H cj 000 O O 00 O OOOOO OOOOO OOOOO OOOOO O sssss d d d d d d d o' d d d d o' d d d d d d d o' sgsgg sssss s ddddd d I-H' -< -i i-i -! ooooo ooooo ooooo ^t 4 co c^ o> 1O 1O O IO >* d o' d o" d f->oe-> CO >0 1-H(M- O CO lO OO w co co co co d d d d d COO5OSiO Tt< r O i I **< o 10 10 o o o o o o o CS OS 0> C5 03 odddd o >o-*nr-< IIM-HOO ooooo ooooo 000000 I s * CD o *< co O O O O O ^ r-l 12 T-H o o o o co *-H t ( o o i-H ? (M S CO Ot -^ OO-^OOC^t^ t^OCM^l^ OfMOt^O SllfttSSS (M (N CO CO CO -* IT* | * I ^ I >O O "O O CO do'do'd do' odd do'do'd OCOO>OO "*OCOt^O COCOOOOIM o' o d d d d o' o o o' o' d d o' d O 0000 o' d d d o ooooo ^o -<*t ooooo CO CO CO CO CO CM CM CO CO CO COCOCO-*Tt< o o o' d d o' d o' d d 3!sS ooooo do' odd > t^ CO O l^- -^ O >- t"- F d d d d d CO CO 1O 1C TfH d d o' d d o> c^ -*t co oo i i co *o t^ Ci o* o o o o o o o o o O COr-tCO-H< ooooo ooooo ooooo CO CD CO O I s - ^ Tj< fcO *O iO o o" o o o >COOCO o d o' d d o' o o o o o oooooooooo d o d d o OOOCOIOS. 05^-i-^lCDOO t>.00(0000 00 05 0> 05 05 o d o o o d o' d d o o o oo . SScoiot-: COC^TJHIO K OOOO C^'"<~<'-^' M i I-H t-, o <*< co ^H COO CO O O o' d d d d ooooo o 33J338 o' o' d d o o d d o d d d d d d IO CO * < O OO CO -ft< CO * 'OS ^-l^i-l^-iO OOOOOS ,_,' ^H' ^H' ,_' ^H' ^ ^-i _(' i-! O OOOOO OOOOO O COCOCOIMO t^-iOCO^ - t>- OO OO OO d d o d o" iOO>O ^ - o' d d o' d r-( OO Tt< O kO O eo t- g m c^ ^ o So i- S t- co CO CO CO d d d d d d o' d d d 10 S C^l O O d o d d d d SSSSS55 o'dddd OOOOO o 'C^: coco id ) CO ;g OOOOO 100 o o o o CO-* O> t^US CO Tti CN OS t^ 10 CO o' d o' d d d o 0>0-i- CO kC *- co ddddd co to ^ co co co coco d d o' d o d d d o o O O d d o d d d i t co i 1 10 o gss?ss d d d d d S' Oi CC I s * O OO (^ >O g l>- OO O5 - ' Ol 1C O IO CO CO d d d d d ooooo 1 d d d o" o d d d ooooo o OO OO l>- !> l^- t*- ddddd o ooooo ooooo ooooo ooooo o M ** co oo o OSO 03 05 OS O d d d o' d ^ CALCULATION OF ALTERNATING CURRENT PROBLEMS 261 ^3 * e OQ o < T5 H 8SSS 2S3SJSS 8S3SS85 o' d o d d d o' o' o" d d d o d o o d o o o o o p p p p TH TH TH C^ **< CO "5 o o o o o o o o o o O O O O O CO C^l o o oo o o'o'ddd o'dddd oo oo oooo CO CO CO CO * >* * * * >O odo'dd o'o'do'd I-H i I O OO CO CO O I s - ^^ cococococo o' d d d o' o' o' d o' d ddddd ooooo ooooo TH' TH TH TH TH O O O O O ^ CO (M C- ^ Tt< -* 1C U5 US U5 CO CO CO d o' o' d o' d d o' o' d n o' o o' do' o' o' o' ooooo ooooo ooooo 3n$S38S SSSSg S22 ooooo ooooo O O rH rH rt O SI U5 U5 CO < ooooo ooooo ooooo ooooo ooooo O O O O O O O O O O ssssa mm dddod ooooo o'dooo' o o o o S(M-*COOO OIMr-r^l>. Op 00 GO OO OO O5O5O5O5O5 O ooooo ooooo o' d o' d o' d d d o d o o o' o' d H S8 88gg SSSSS 322 do' odd do odd d i-i i-l ,-i .-! ,-i ,-i ,-i ,-i "5-^(NOOO O CM O5 IO T-H t-~ CM t-*- CM t*>- t-H ^ I ) O t^ GO O CM CO O CO GO O5 t OJ ( ooooo CO CO CO CM CN ooooo ooooo 5oooooo< o' d o' o' d OOOCOCOO T-| ,-H' ^i d O o COCNl t*- . t^. C do'o'do ooooo 001 - O5 C^l **< CO O5 i-H CO 1C OO O CM Tt* O OO O CM CO O t^- O5 O 01 t^tl>.t^. OOOOOOOOOJ O5OSO5OO OOOrH^-l 1-1 r-n-H ^-i (M CM OOOOO OOOOO OOOt ^ C5 t^ -fi ^ r 10 ^ co c^ OO 00 GOOD GO s GO CO O OO5CO COO3(MCOO O ?* CO CO CO CO CO CO 1C 1C o' o' d o' o' o' d d d o' d o' o' o' o' o' d o' do' o o o' o' d d III I ooooo CO CM r~ CM CO ooooo ooooo ooooo ooooo o sssss >-! d d d d ooooo o o'do'dd ddo'dd do'o'o'i-* 0300000000 COOOOOOoS ?2t^t2^?2 o" d d o' d d d o o' d o' o' o o d SoOO 5^10 !S82 o' o' o o o d o d d d d 000*00 ooooo oo'o'oo ooooo CQ^O}'*&O* CO O O5 C^J ^ CDOOOiCiOS OO t 2 /v4 /v6 sin x = \j (e xj e~ X1 '). cosx = e A cos x 5 sin a; where tan ^ = -r cos A$mxBcosx = VA 2 + B 2 sin (x 0), T> where tan = -j ^1 sin 2 z - sin 2 ?/ = sin (x + y) sin (x y). cos 2 x cos 2 y = sin (x + z/) sin (x y). cos 2 a: sin 2 ?/ = cos (x + T/) cos (x y). cos 2 a; sin 2 x = cos 2 z = 1 2 sin 2 x = 2 cos 2 x 1, sin 2 z = 2 sin x cos x. (cos z db j sin z) n = cos nx d= j sin wz. 266 FORMULAE AND TABLES FOR THE i. 90 z. 180 x. 270 z. 360 z. sin . . sin x -f-cosx T~sin x cos x isin x cos . . ... -j-cosx -Fsinx cosx -f-sin X -(-(jOS X tan tanx -Fctnx rttanx -T-ctna; rttanx ctn ctnx -Ftanx -l-ctn x -Ftanx -l-ctn x sec -{-sec x -Fcscx sec x -4-P.SP, X -j-sec x cose esc x -j-sec x -Fcsc x sec x irsft x HYPERBOLIC FORMULA sinh (x + y) sinh x cosh y + cosh x sinh y. sinh (x y) = sinh # cosh ?/ cosh x sinh y. cosh (a? + y) = cosh z cosh y + sinh z sinh y. cosh (x y) = cosh x cosh ?/ sinh # sinh y. tanh # + tanh y farcMs + a-^tanhstonhy- , , N tanh a: tanh y tanh ($ #) = - r r^- 1 tanh x tanh y cosh 2 a; sinh 2 a; = 1. 1 tanh 2 x = sech 2 x. 1 ctnh 2 x = csch 2 a;. A sinh x B cosh x = VA 2 B 2 sinh (a; ) n where tanh = -j A cosh x d= B sinh x = VA 2 - B 2 cosh (x 0; r> where tanh = -j sinh a; = J (e* e" 1 ) = sinh ( x) = j sin cosh x = % (e x + e"*) = cosh ( x) = cos (jz). tanh x = 6 a . ^ = tanh ( x). sinh x + sinh t/ = 2 sinh % (x + y) cosh | (a; 2/)- sinh x sinh 2/ = 2 cosh % (x + y) sinh J (a; y). cosh x + cosh j/ = 2 cosh J (a: + y) cosh |(a; y) cosh a; cosh y = 2 sinh J (a; + y) suih J (a; y). cosh x + sinh x = e x . cosh x sinh a; = e~ x . x 2 x* cosh a? = 1 + jrt 4- 7 H" * CALCULATION OF ALTERNATING CVRRENT PROBLEMS 267 cosh (x + jy) = cosh x cos y + j sinh x sin y. sinh (x + jy) = sinh z cos y + j cosh z sin y. cosh (x dh jy) = cos (?/ =F jx). sinh (x d= j?/) = =b j sin (?/ =F jx). sin (a; db jy) = sinh (y =F jx). cos x it COMPLEX QUANTITIES Any function of a complex quantity can be put in the fol lowing form: f(a+jb) =A+jB. Addition : (a+jb) + (c+jd) =A+jB, A = a + c, B = b+d. Subtraction : (a+jb)-(c+jd)=A+jB, A = a c, B = b d. Multiplication: (a+jb)(c+jd) =A+jB, A = ac - bd } B = b* + ad. Division : . ac -j- bd j-. be ad Square root: Va+jb = A+jB, B = Logarithm : log(a+jb) =A+jB, A = J log (a 2 + 6 2 ), B = tan a 268 FORMULAE AND TABLES FOR THE Exponential : jd = A + JB, A = ae c cos d, B = at sin d. A = ae c cos d, B = ae c sin d, = +JB, A = pq a+c cos (6 + d), B = pqe a+c sin (6 + d). E 4 - cos (6 - d), B = . _ aec bfc + afd + e&6? R _ acf + efrc ae d + c 2 + d 2 c 2 + rf 2 ^4. = a + j'6 = r (cos /3 + j sin /3) where r = Va 2 + 6 2 , tan /? = - - (a+jb) n = r n Scos/3-f.7sin/3) n j: i IP \/a-\-jb = r n e n . Roots of unity: , ^i = +1, -i, +j, -j , =+i, -i, +,, -A cyr. w/'T ^ 7T/i/ . ^ TT/v / 7 rv r -i \ VI = cos -- hjsm - =e n (A; = 0,1,2, . . . n 1), CALCULATION OF ALTERNATING CURRENT PROBLEMS 269 MISCELLANEOUS FORMULAE (a + b) n = a + nar-*b + H (n ~ 1} a"~ 2 6 2 + n\a n ~ k b k r , 2 . 21 - [x* < I]. [x* < 1]. 1 1-1 , 1-1-3 , 1-1-3-5 . ---*--- a ,)-, = i T [x 2 < 1]. 1 _ 1^2 2 1-2-5 3 _ 1-2-5-8 4 [x 2 < 1]. [x 2 < 1}. ~ MISCELLANEOUS FUNCTIONS j.4 . /v.8 2 2 4 2 2 2 - 4 2 6 2 8 2 2 2 4 2 6 2 8 2 10 2 X 10 ber'(x) 2! 2 2 -4 2 -6 2 2 2 -4 2 -6 2 -8 2 -10 2 d ber (x) _ x 3 x 7 = 2~ 2 2 -4 2 -6 + 2 2 -4 2 -6 2 -8 2 -10 270 FORMULAE AND TABLES FOR THE For x > 6 we have the following approximate expressions. , f x a cos j8 ber (x) = , > V2irx , . / v 6 a sin j8 bei (x) = , ) where a = 0.707105 x + 0.0884 x~ l - 0.046 x~ s , j8 = 0.707105 a; - 0.39270 - 0.0884 x- 1 - 0.0625 or 2 - 0.046a;- 3 . INTERPOLATION FORMULA where CALCULATION OF ALTERNATING CURRENT PROBLEMS 271 TABLE XXI TABLE OF NAPIERIAN LOGARITHMS TO NINE DECIMAL PLACES FOR NUMBERS FROM 1 TO 100 1 0.000000000 36 3.583518938 71 4.262679877 2 0.693147181 37 3.610917913 72 4.276666119 3 1.098612289 38 3.637586160 73 4.290459441 4 1.386294361 39 3.663561646 74 4.304065093 5 1.609437912 40 3.688879454 75 4.317488114 6 1.791759469 41 3.713572067 76 4.330733340 7 1.945910149 42 3.737669618 77 4.343805422 8 2.079441542 43 3.761200116 78 4.356708827 9 2.197224577 44 3.784189634 79 4.369447852 10 2.302585093 45 3.806662490 80 4.382026635 11 2.397895273 46 3.828641396 81 4.394339155 12 2.484906650 47 3.850147602 82 4.406719247 13 2.564949357 48 3.871201011 83 4.418840608 14 2.639057330 49 3.891820298 84 4.430816799 15 2.708050201 50 3.912023005 85 4.442651256 16 2.772588722 51 3.931825633 86 4.454347296 17 2.833213344 52 3.951243719 87 4.465908119 18 2.890371758 53 3.970291914 88 4.477336814 19 2.944438979 54 3.988984047 89 4.488636370 20 2.995732274 55 4.007333185 90 4.499809670 21 3.044522438 56 4.025351691 91 4.510859507 22 3.091042453 57 4.043051268 92 4.521788577 23 3.135494216 58 4.060443011 93 4.532599493 24 3.178053830 59 4.077537444 94 4.543294782 25 3.218875825 60 4.094344562 95 4.553876892 26 3.258096538 61 4.110873864 96 4.564348191 27 3.295836866 62 4.127134385 97 4.574710979 28 3.332204510 63 4.143134726 98 4.584967479 29 3.367295830 64 4.158883083 99 4.595119850 30 3.401197382 65 4.174387270 100 4.605170186 31 3.433987204 66 4.189654742 32 3.465735903 67 4.204692619 33 3.496507561 68 4.219507705 34 3.526360525 69 4.234106505 35 3.555348061 70 4.248495242 272 FORMULAE AND TABLES FOR THE TABLE XXII EXPONENTIAL AND HYPERBOLIC FUNCTIONS X e* -* Coshz Sinhx X e* IT" Coshx Sinhz 0.00 1.00000 1.00000 1.00000 0.00000 0.50 1.64872 0.60653 1.12763 0.52110 0.01 1.01005 0.99005 1.00005 0.01000 0.51 1.66529 0.60050 1.13289 0.53240 0.02 1.02021 0.98020 1.00020 0.02000 0.52 1.68203 0.59452 1.13827 0.54375 0.03 1.03045 0.97045 1.00045 0.03000 0.53 1.69893 0.58860 1.14377 0.55516 0.04 1.04081 0.96080 1.00080 0.04001 0.54 1.71601 0.58275 1.14938 0.56663 0.05 1.05127 0.95123 1.00125 0.05002 0.55 1.73325 0.57695 1.15510 0.57815 0.06 1.06184 0.94177 1.00180 0.06004 0.56 1.75067 0.57121 1.16094 0.58973 0.07 1.07251 0.93239 1.00245 0.07006 0.57 1.76827 0.56553 1.16690 0.60137 0.08 1.08329 0.92312 1.00320 0.08009 0.58 1.78604 0.55990 1.17297 0.61307 0.09 1.09417 0.91393 1.00405 0.09012 0.59 1.80399 0.55433 1.17916 0.62483 0.10 1 . 10517 0.90484 1.00500 0.10017 0.60 1.82212 0.54881 1.18547 0.63665 0.11 1.11628 0.89583 1.00606 0.11022 0.61 1.84043 0.54335 1.19189 0.64854 0.12 1 . 12750 0.88692 1.00721 0.12029 0.62 1.85893 0.53794 1.19844 0.66049 0.13 1.13883 0.87810 1.00846 0.13037 0.63 1.87761 0.53259 1.20510 0.67251 0.14 1.15027 0.86936 1.00982 0.14046 0.64 1.89648 0.52729 1.21189 0.68459 0.15 1.16183 0.86071 1.01127 0.15056 0.65 1.91554 0.52205 1.21879 0.69675 0.16 1.17351 0.85214 1.01283 0.16068 0.66 1.93479 0.51685 1.22582 0.70897 0.17 1.18530 0.84366 1.01448 0.17082 0.67 1.95424 0.51171 1.23297 0.72126 0.18 1.19722 0.83527 1.01624 0.18097 0.68 1.97388 0.50662 1.24025 0.73363 0.19 1.20925 0.82696 1.01810 0.19115 0.69 1.99372 0.50158 1.24765 0.74607 0.20 1.22140 0.81873 1.02007 0.20134 0.70 2.01375 0.49659 1.25517 0.75858 0.21 1.23368 0.81058 1.02213 0.21555 0.71 2.03399 0.49164 1.26282 0.77117 0.22 1.24608 0.80252 1.02430 0.22178 0.72 2.05443 0.48675 1.27059 0.78384 0.23 1.25860 0.79453 1.02657 0.23203 0.73 2.07508 0.48191 1.27850 0.79659 0.24 1.27125 0.78663 1.02894 0.24231 0.74 2.09594 0.47711 1.28652 0.80941 0.25 1.28403 0.77880 1.03141 0.25261 0.75 2.11700 0.47237 1.29468 0.82232 0.26 1.29693 0.77105 1.03399 0.26294 0.76 2.13828 0.46767 1.30297 0.83530 0.27 1.30996 0.76338 1.03667 0.27329 0.77 2.15977 0.46301 1.31139 0.84838 0.28 1.32313 0.75578 1.03946 0.28367 0.78 2.18147 0.45841 1.31994 0.86153 0.29 1.33643 0.74826 1.04235 0.29408 0.79 2.20340 0.45384 1.32862 0.87478 0.30 1.34986 0.74082 1.04534 0.30452 0.80 2.22554 0.44933 1.33743 0.88811 0.31 1.36343 0.73345 1.04844 0.31499 . 0.81 2.24791 0.44486 1.34638 0.90152 0.32 1.37713 0.72615 1.05164 0.32549 0.82 2.27050 0.44043 1.35547 0.91503 0.33 1.39097 0.71892 1.05495 0.33602 0.83 2.29332 0.43605 1.36468 0.92863 0.34 1.40495 0.71177 1.05836 0.34659 0.84 2.31637 0.43171 1.37404 0.94233 0.35 1.41907 0.70469 1.06188 0.35719 0.85 2.33965 0.42741 1.38353 0.95612 0.36 1.43333 0.69768 1.06550 0.36783 0.86 2.36316 0.42316 1.39316 0.97000 0.37 1.44773 0.69073 1.06923 0.37850 0.87 2.38691 0.41895 1.40293 0.98398 0.38 1.46228 0.68386 1.07307 0.38921 0.88 2.41090 0.41478 1.41284 0.99806 0.39 1.47698 0.67706 1.07702 0.39996 0.89 2.43513 0.41066 1.42289 1.01224 0.40 1.49182 0.67032 1.08107 0.41075 0.90 2.45960 0.40657 1.43309 1.02652 0.41 1.50682 0.66365 1.08523 0.42158 0.91 2.48432 0.40252 1.44342 1.0409C 0.42 1.52196 0.65705 1.08950 0.43246 0.92 2.50929 0.39852 1.45390 1.05539 0.43 1.53726 0.65051 1.09388 0.44337 0.93 2.53451 0.39455 1.46453 1.06998 0.44 1.55271 0.64404 1.09837 0.45434 0.94 2.55998 0.39063 1.47530 1.08468 0.45 1.56831 0.63763 1 . 10297 0.46534 0.95 2.58571 0.38674 1.48623 1.09948 0.46 1.58407 0.63128 1 . 10768 0.47640 0.96 2.61170 0.38289 1.49729 1.11440 0.47 1.59999 0.62500 1.11250 0.48750 0.97 2.63794 0.37908 1.50851 1.12943 0.48 1.61607 0.61878 1.11743 0.49865 0.98 2.66446 0.37531 1.51988 1.14457 0.49 1.63232 0.61263 1.12247 0.50985 0.99 2.69123 0.37158 1.53141 1 . 15983 0.50 1.64872 0.60653 1 . 12763 0.52120 1.00 2.71828 0.36788 1.54308 1 . 17520 CALCULATION OF ALTERNATING CURRENT PROBLEMS 273 TABLE XXII (Continued} EXPONENTIAL AND HYPERBOLIC FUNCTIONS 2 e x e~ x Coshz Sinh x X e x c~ x Cosh* Sinhx 1.00 2.71828 0.36788 1.54308 1.17520 1.50 4.48169 0.22313 2.35241 2.12928 .01 2.74560 0.36422 1.55491 1.19069 1.51 4.52673 0.22091 2.37382 2.15291 .02 2.77319 0.36059 1.56689 1.20630 1.52 4.57223 0.21871 2.39547 2.17676 .03 2.80107 0.35701 1.57904 1.22203 1.53 4.61818 0.21654 2.41736 2:20082 .04 2.82922 0.35345 1.59134 1.23788 1.54 4.66459 0.21438 2.43949 2.22510 .05 2.85765 0.34994 1.60379 1.25386 1.55 4.71147 0.21225 2.46186 2.24961 .06 2.88637 0.34646 1.61641 1.26996 1.56 4.75882 0.21014 2.48448 2.27434 .07 2.91538 0.34301 1.62919 1.28619 1.57 4.80655 0.20805 2.50735 2.29930 .08 2.94468 0.33960 1.64214 1.30254 1.58 4.85496 0.20598 2.53047 2.32449 1.09 2.97427 0.33622 1.65525 1.31903 1.59 4.90375 0.20393 2.55384 2.34991 1.10 3.00417 0.33287 1.66852 1.33565 1.60 4.95303 0.20190 2.57746 2.37557 1.11 3.03436 0.32956 1.68196 1.35240 1.61 5.00281 0.19989 2.60135 2.40146 1.12 3.06485 0.32628 1.69557 1.36929 1.62 5.05309 0.19790 2.62549 2.42760 1.13 3.09566 0.32303 1.70934 1.38631 1.63 5.10387 0.19593 2.64990 2.45397 1.14 3.12677 0.31982 1.72329 1.40347 1.64 5.15517 0.19398 2.67457 2.48059 1.15 3.15819 0.31664 1.73741 1.42078 1.65 5.20698 0.19205 2.69951 2.50746 1.16 3.18993 0.31349 1.75171 1.43822 1.66 5.25931 0.19014 2.72472 2.53459 1.17 3.22199 0.31037 1.76618 1.45581 1.67 5.31217 0.18825 2.75021 2.56196 1.18 3.25437 0.30728 1.78083 1.47355 1.68 5.36556 0.18637 2.77596 2.58959 1.19 3.28708 0.30422 1.79565 1.49143 1.69 5.41948 0.18452 2.80200 2.61748 1.20 3.32012 0.30119 1.81066 1.50946 1.70 5.47395 0.18268 2.82832 2.64563 1.21 3.35348 0.29820 1.82584 1.52764 1.71 5.52896 0.18087 2.85491 2.67405 1.22 3.38719 0.29523 1.84121 1.54598 1.72 5.58453 0.17907 2.88180 2.70273 1.23 3.42123 0.29229 1.85676 1.56447 1.73 5.64065 0.17728 2.90897 2.73168 1.24 3.45561 0.28938 1.87250 1.58311 1.74 5.69734 0.17552 2.93643 2.76091 1.25 3.49034 0.28650 1.88842 1.60192 1.75 5.75460 0.17377 2.96419 2.79041 1.26 3.52542 0.28365 1.90454 1.62088 1.76 5.81244 0.17204 2.99224 2.82020 .27 3.56085 0.28083 1.92084 1.64001 1.77 5.87085 0.17033 3.02059 2.85026 .28 3.59664 0.27804 1.93734 1.65930 .78 5.92986 0.16864 3.04925 2.88061 .29 3.63279 0.27527 1.95403 1.67876 .79 5.98945 0.16696 3.07821 2.91125 .30 3.66930 0.27253 1.97091 1.69838 .80 6.04965 0.16530 3.10747 2.94217 .31 3.70617 0.26982 1.98800 1.71818 .81 6.11045 0.16365 3.13705 2.97340 1.32 3.74342 0.26714 2.00528 1.73814 .82 6.17186 0.16203 3.16694 3.00492 1.33 3.78104 0.26448 2.022Z6 1.75828 .83 6.23389 0.16041 3.19715 3.03674 1.34 3.81904 0.26185 2.04044 1.77860 .84 6.29654 0.15882 3.22768 3.06886 1.35 3.85743 0.25924 2.05833 1.79909 1.85 6.35982 0.15724 3.25853 3.10129 1.36 3.89619 0.25666 2.07643 1.81977 1.86 6.42374 0.15567 3.28970 3.13403 1.37 3.93535 0.25411 2.09473 1.84062 1.87 6.48830 0.15412 3.32121 3.16709 1.38 3.97490 0.25158 2.11324 1.86166 1.88 6.55350 0.15259 3.35305 3.20046 1.39 4.01485 0.24908 2.13196 1.88289 1.89 6.61937 0.15107 3.38522 3.23415 1.40 4.05520 0.24660 2.15090 1.90430 1.90 6.68589 0.14957 3.41773 3.26816 1.41 4.09596 0.24414 2.17005 1.92591 1.91 6.75309 0.14808 3.45058 3.30250 .42 4.13712 0.24171 2.18942 1.94770 1.92 6.82096 0.14661 3.48378 3.33718 .43 4.17870 0.23931 2.20900 1.96970 .93 6.88951 0.14515 3.51733 3.37218 .44 4.22070 0.23693 2.22881 1.99188 1.94 6.95875 0.14370 3.55123 3.40752 .45 4.26311 0.23457 2.24884 2.01428 .95 7.02869 0.14227 3.58548 3.44321 .46 4.30596 0.23224 2.26910 2.03686 .96 7.099330.14086 3.62009 3.47923 .47 4.34924 0.22993 2.28958 2.05965 .97 7.170680.13946 3.65507 3.51561 .48 4.39295 0.22764 2.31029 2.08265 1.98 7.24274:0.13807 3.69041 3.55234 .49 4.43710 0.22537 2.33123 2.10586 1.99 7.315330.13670 3.72611 3.58942 1.50 4.48169 0.22313 2.35241 2.12928 2.00 7.389060.13534 3.76220 3.62686 274 FORMULAE AND TABLES FOR THE TABLE XXII (Continued) EXPONENTIAL AND HYPERBOLIC FUNCTIONS X