UC-NRLF SB 270 fiOl OF An Electrical Library. By PROF. T. O'CONOR SLOANE. How to become a Successful Electrician. PRICE $1.00. Electricity Simplified. PRICE, $1.00. Electric Toy Making, Dynamo Building, etc. PRICE, $1.00. Arithmetic of Electricity. PRICE, $1.00. Standard Electrical Dictionary. PRICE, $3.00. NORMAN W. HENLEY & CO., Publishers, 132 Naau Street, New York. Arithmetic of .Electricity A PRACTICAL TREATISE ON ELECTRICAL CALCULATIONS OF ALL KINDS REDUCED TO A SERIES OF RUIZES, ALI, OF THE SIMPLEST FORMS, AND INVOLVING ONLY ORDINARY ARITH- METIC, EACH RULB ILLUSTRATED BY ONE OR MORE PRAC- TICAL PROBLEMS, WITH DETAILED SOLUTION OF EACH, FOLLOWED BY AN EXTENSIVE SERIES OF TABLES. BY T. O'CONOR SLOANE, A.M., E.M., Ph.D. s AUTHOR OF Standard Electrical Dictionary, Electricity Simplified, Electric Toy Making, etc. TWENTIETH EDITION NEW YORK : THE NORMAN W. HENLEY PUBLISHING COMPANY 132 NASSAU STREET 1909 COPYRIGHTED, 1891, BY NORMAN W. HENLEY & CO. COPYRIGHTED, 1903, BY NORMAN W. HENLEY & CO. COPYRIGHTED, 1909, BY THE NORMAN W. HENLEY PUBLISHING CO. \ PREFACE. The solution of a problem by arithmetic, although in some cases more laborious than the algebraic method, gives the better comprehension of the subject. Arith- metic is analysis and bears the same relation to algebra that plane geometry does to analytical geometry. Its power is comparatively limited, but it is exceedingly instructive in its treatment of questions to which it applies. In the following work the problems of electrical en- gineering and practical operations are investigated on an arithmetical basis. It is believed that such treatment gives the work actual value in the analytical sense, as it necessitates an explanation of each problem, while the adaptability of arithmetic to readers who do not care to use algebra will make this volume more widely available. In electricity there is much debatable ground, which has been as far as possible avoided. Some points seem quite outside of the scope of this book, such as the intro- duction of the time-constant in battery calculations. Again the variation in constants as determined by dif- ferent authorities made a selection embarrassing. It is believed that some success has been attained in over- coming or compromising difficulties such as those sug- gested. 383491 iv PREFACE. Enough tables have ^een introduced to fill the limits of the subject as here treated. The full development of electrical laws involves the higher mathematics. One who would keep up with the progress of the day in theory has a severe course of study before him. In practical work it is believed that such a volume as the Arithmetic of Electricity will always have a place. We hope that it will be favorably received by our readers and that their indulgence will give it a more extended field of usefulness than it can pretend to deserve. PREFACE TO TWENTIETH EDITION. The steady progress of electrical science in conjunc- tion with a continued demand for this work have made advisable a revision and extension of this book. The author feels that in the matter which has been added much more could have been said on the subjects treated of, but, since a full exposition of each theme would alone fill a volume, it is hoped that the practical value of the rules, etc., will atone for the brevity of the text. In the preparation of this edition the author would express his indebtedness to A. A. Atkinson's excellent work on Electrical and Magnetic Calculations and also to the instruction papers of the Electrical Engineering Course of the International Correspondence School of S'cranton, Pa. He would also express his thanks to Henry V. A. Parsell, for his valued advice and assistance in the preparation of the manuscript. THE AUTHOB. CONTENTS. CHAPTER L INTRODUCTORY. Space, Time, Force, Resistance, Work, Energy, Mass and Weight. The Fundamental Units of Dimension, and Derived Units, Geometrical, Mechanical, and Elec- trical. C. G. S. and Practical Electrical Units. Nomen- clature. Examples from Actual Practice 9 CHAPTER IL OHM'S LAW. General Statement. Six Rules Derived by Transpos- ition from the Law. Single Conductor Closed Circuits. Batteries in Opposition. Portions of Circuits. Di- vided Circuits, with Calculation of Currents Passed by Each Branch, and of their Combined Resistance 13 CHAPTER III. RESISTANCE AND CONDUCTANCE. Resistance of Different Conductors of the same Mater- ial. Relations of Wires of Equal Resistance. Ratio of Resistance of Two Conductors. Specific Resistance. Universal Rule for Resistances. Resistance of Wires Referred to Weight. Conductance. Ohm's Law ex- pressed in Conductance .26 vi CONTENTS. CHAPTER IV. 4 POTENTIAL DIFFERENCE. Drop of Potential in Leads, and Size of Same for Mul- tiple Arc Connections. Diminishing Size of Leads Progressively 38 CHAPTER V. CIRCULAR MILS. The Mil. The Circular Mil as a Unit of Area. Cir- cular Mil Rules for Resistance and Size of Leads 43 CHAPTER VI. SPECIAL SYSTEMS. Three Wire System. Rules for Calculating Leads in Same. Alternating Current System. Ratio of Conver- sion. Size of Primary Wire. Converter Winding 46 CHAPTER VII. WORK AND ENERGY. Energy and Heating Effect of the Current. Differ- ent Rules Based on Joule's Law. The Joule or Gram- Calorie. Quantity of Heat Developed in an Active Cir- cuit in a Unit of Time. Watts and Amperes in Rela- tion to Time. Specific Heat. Heating of Wire by a Current. Safety Fuses. Work of a Current. Elec- trical Horse-Power. Duty and Efficiency of Electrical Generators . 50 CONTENTS. Mil CHAPTER VIII. BATTEBIES. Arrangement of Battery Cells. General Calcula- tions of Current. Rules for Arrangement of Cells in a Battery. Battery Calculations for Specified Electromotive Force and Current. Efficiency of Batteries. Chemistry of Batteries. Calculation of Voltage. Work of Batteries. Efficiency of Bat- teries, to Calculate. Chemicals Consumed in a Battery. Decomposition of Compounds by a Bat- tery. Electroplating 66 CHAPTER IX. ELECTRO-MAGNETS, DYNAMOS AND MOTORS. The Magnetic Field and Lines of Force. Per- meance and Reluctance. Magnetizing Force and the Magnetic Circuit. General Rules for Electro- Magnets and Ampere-Turns for Given Magnetic Flux. Magnetic Circuit Calculations. Leakage of Lines of Force. Example of Calculation of a Mag- netic Circuit. Dynamo Armatures. Voltage and Capacity of Armatures. Drum Type Closed Cir- cuit Armatures. Field Magnets of Dynamos. The Kapp Line 82 CHAPTER X. ELECTRIC RAILWAYS. Sizes of Feeders. Power to Move Cars 109 viii CONTENTS. CHAPTER XI. ALTERNATING CURRENTS. Self-Induction 115 CHAPTER XII. CONDENSERS. Page 121 CHAPTER XIII. DEMONSTRATION OF RULES. Some of the Principal Rules in the Work Demon- strated 127 CHAPTER XIV. NOTATION IN POWERS OF TEN. The Four Fundamental Operations of Addition, Subtraction, Multiplication, and Division in Powers of Ten 136 TABLES. A Collection of Tables Needed for the Operations and Problems Given in the Work . . 139 ARITHMETIC OF ELECTRICITY. CHAPTER L INTRODUCTORY. SPACE is the lineal distance from one point to another. Time is the measure of duration. Force is any cause of change of motion of matter. It is expressed practically by grams, volts, pounds or other unit. Resistance is a counter-force or whatever opposes the action of a force. Work is force exercised in traversing a space against a resistance or counter-force. Force multi- plied by space denotes work as foot-pounds. Energy is the capacity for doing work and is measurable by the work units. Mass is quantity of matter. Weight is the force apparent when gravity acts upon mass. When the latter is prevented from moving under the stress of gravity its weight can be appreciated. 10 ARITHMETIC OF ELECTRICITY. Physical and Mechanical calculation, are based on three fundamental units of dimension, as follows: the unit of time the second, T; the unit of length the centimeter, L; the unit of mass the gram, M. Concerning the latter it is to be distinguished from weight. The gram is equal to one cubic centi- meter of water under standard conditions and is invariable; the weight of a gram varies slightly with the latitude and with other conditions. Upon these three fundamental units are based the derived units, geometrical, mechanical and electrical. The derived units are named from the initials of their units of dimension, the C. G. S. units, indi- cating centimeter-gram-second units. In practical electric calculations we deal with certain quantities selected as of 'convenient size and as bearing an easily defined relation to the fundamental units. They are called practical units. The cause of a manifestation of energy is force; if of electromotive energy, that is to say of electric energy in the current form, it is called electromotive force, E. M. F. or simply E. or difference of poten- tial D. P. What this condition of excitation may be is a profound mystery, like gravitation and much else in the physical world. The practical unit of E. M. F. is the VOLT, equal to one hundred mil- lions (100,000,000) 0. G. S. units of E. M. F. The last numeral is expressed more briefly as the eighth INTRODUCTORY. 11 power of 10 or 10 8 . Thus the volt is defined as equal to 10 8 C. G. S. units of E. M. F. This notation in powers of 10 is used throughout C. G. S. calculations. Division by a power of 10 is expressed by using a negative exponent, thus 10~* means ioooagoog Tlie exj)onent indicates the number of ciphers to le placed after 1. When electromotive force does work a current is produced. The practical unit of current is the AMPERE, equal to & C. G. S. unit, or 10' 1 C. G. S. unit, & being expressed by 10" 1 . A current of one ampere passing for one second gives a quantity of electricity. It is called the COULOMB and is equal to 10' 1 C. G. S. units. A coulomb of electricty if stored in a recipient tends to escape with a definite E. M. F. If the recipient is of such character that this definite E. M. F. is one volt, it has a capacity of one FARAD equal to TOOOO^OOOO or 10~ 9 C. G. S. unit. A current of electricity passes through some substances more easily than through others. The relative ease of passage is termed conductance. In calculations its reciprocal, which is resistance, is almost universally used. A current of one ampere is maintained by one volt through a resistance of one practical unit. This unit is called the OHM and is equal to 10 9 C. Gr. S. units. Sometimes, where larger units are wanted, the pre- fix deka, ten times, lieka* one hundred times, Tcilo, one 12 ARITHMETIC OF ELECTRICITY. thousand times, or mega, one million times are used, as dekalitre, ten liters, kilowatt, one thousand watts, megohm, one million ohms. Sometimes, where smaller units are wanted, the prefixes, deci, one tenth, centi, one hundredth, milli, one thousandth, micro, one millionth, are used. A microfarad is one millionth of a farad. For the concrete conception of the principal units the following data are submitted. A DanielFs battery maintains an E. M. F. of 1.07 volt. A current which in each second deposits .00033 grains copper (by electro-plating) is of one ampere intensity and from what has been said the copper deposited by that current in one second cor- responds to one coulomb. A column of mercury one millimeter square and 106.24 centimeters long has a resistance of one ohm at 0. The capacity of the earth is itUooo farad. A Leyden jar with a total coated surface of one square meter and glass one mm. thick has a capacity of -h microfarad. The last is the more generally used unit of capacity. These practical units are derived from the 0. G-. S. units by substituting for the centimeter (C.) one thousand million (10 9 ) centimeters and for the gram, the one hundred thousand millionth (10~ n ) part of a gram. CHAPTER II. OHM'S LAW. THIS law expresses the relation in an active electric circuit (circuit through which a current of electricity is forced) of current, electromotive force, and resistance. These three factors are always pres- ent in such a circuit. Its general statement is as follows : In an active electric circuit the current is equal to the electromotive force divided by the resistance. This law can be expressed in various ways as it is transposed. It may be given as a group of rules, to be referred to under the general title of OHM'S LAW. Rule 1. The current is equal to the electromotive force divided by the resistance. C = R Rule 2. The electromotive force ts equal to the cur- rent multiplied by the resistance. E = C R Rule 3. The resistance is equal to the electromotive force divided by the current. ~ C Rule 4. The current varies directly with the electro* motive force and inversely with the resistance. 14 ARITHMETIC OF ELECTRICITY. Rule 5. The resistance varies directly \vitli tlic elec* troiiiotivc force and inversely with the current. Rule 6. The electromotive force varies directly with the current and with the resistance. This law is the fundamental principle in most electric calculations. If thoroughly understood it will apply in some shape to almost all engineering problems. The forms 1, 2, and 3 are applicable to integral or single conductor circuits; when two or more circuits are to be compared the 4th, 5th and 6th are useful. The law will be illustrated by examples. SINGLE CONDUCTOR CLOSED CIKCUITS. These are circuits embracing a continuous con', ducting path with a source of electromotive force included in it and hence with a current continually circulating through them. EXAMPLES. A battery of resistance 3 ohms and E. M. F. 1.07 volts sends a current through a line of wire of 55 ohms resistance ; what is the current? Solution : The resistance is 3 + 55 = 58 ohms. By rule 1 we have for the current J ^/- giving .01845 Ampere. Note. A point to be noticed here is that whatever is included in a circuit forms a portion of it and its resistance must be included therein. Hence the resistance of the battery has to be taken into account. The resistance of a battery or generator is sometimes called internal resistance OHM'S LAW. 15 to distinguish it from the resistance of the outer circuit, called external resistance. Kesistance in general is denoted by R, electromotive force by E, and current by C. A battery of R 2 ohms; sends a current of .035 ampere through a wire of R 48 ohms; what is the E. M. F. of the battery? Solution: The resistance is 48 + 2 = 50 ohms. By Rule 2 we have as the E. M. F. 50 X .035 = 1.75 volts. A maximum difference of potential E. M. F. of 30 volts is maintained in a circuit and a current of 191 amperes is the result; what is the resistance of the circuit? Solution: By Rule 3 the resistance is equal to ohms. In the same circuit several generators or gal- vanic couples may be included, some opposing the others, i. e. connected in opposition. All such can be conceived of as arranged in two sets, distrib- uted according to the direction of current produced by the constituent elements, in other words, so as to put together all the generators of like polarity. The voltages of each set are to be added together to get the total E. M. F. of each set. Rule 7. Where batteries or generators are In opposi- tion, add together the E. M. F of all generators of like polarity, thus obtaining two opposed E. OT. F.s. Sub- tract the smaller E. JJI. F. from the larger E. OT. F. to 16 ARITHMETIC OF ELECTRICITY. obtain the effective i:. Itt. F. Then apply Ohm's la\v on this basis of E. M. F. It will be understood that the resistances of all batteries or generators in series are added to give the internal resistance. EXAMPLES. There are four batteries in a circuit: Battery No. 1 of 2 volts, y 2 ohm; Battery No. 2 of 1.75 volts, 2 ohms; Battery No. 3 of 1 volt, 1 ohm; Battery No. 4 of 1 volt, 4 ohms constants; Batteries 1 and 4 are in opposition to 2 and 3. What are the eifective tattery constants? Solution: Voltage = (2 + 1) (1.75 + 1) = .25 volt. Kesistance = ^ + 2+1 + 4 = 7^ ohms, or .25 volt, 7^ ohms constants. What current will such a combination produce in a circuit of 5 ohms resistance? Solution: By Ohm's law, Rule 1, the current =* .25 * (7^ + 5) = .02 amperes. A battery of 51 volts E. M. F. and 20 ohms resist- S LAW. 17 ance has opposed to it in the same circuit a battery of 26 volts E. M. F. and 25 ohms resistance. A current of ^ ampere is maintained in the circuit. What is the resistance of the wire leads and con- nections ? Solution: The effective E. M. P. is 51 26 = 25 volts. By Eule 3 we have 25 + i = 200 ohms, as the total resistance. But the resistance of the batteries (internal resistance) is 20 + 25 = 45 ohms. The re- sistance of leads, etc. (external resistance), is there- fore 200 45 = 155 ohms. PORTIONS OF CIRCUITS. All portions of a circuit receive the same current, but the E. M. F., in this case termed preferably dif- ference of potential, or drop or fall of potential, and the resistance may vary to any extent in differ- ent sections or fractions of the circuit. Ohm's Law applies to these cases also. EXAMPLES, An electric generator of unknown resistance main- tains a difference of potential of 10 volts between its terminals connected as described. The terminals are connected to and the circuit is closed through a series of three coils, one of 100 ohms, one of 50 ohms, and one of 25 ohms resistance. The connec- tions between these parts are of negligibly low re- 18 ABITHMETIC OF ELECTKICITY. sistance. What difference of potential exists be- tween the two terminals of each coil respectively? Solution: The solution is most clearly reached by a statement of the proportion expressed in Rule 6, viz. : The electromotive force varies directly with the resistance. The resistance of the three coils is 175 ohms; calling them 1, 2, and 3, and their differences of potential E 1 , E 2 , and E 8 , we have the continued proportion, 175 : 100 : 50 : 25 :: 10 volts : E 1 : E 2 : E 8 . because by the conditions of the problem the total E. M. F. = 10. Solving the proportion by the regu- lar rule, we find that E 1 = 5.7, E 8 = 2.8 and E 3 = 1.4 volts. The same external circuit is connected to a battery of 30 ohms resistance. The difference of potential of the 100 ohm coil is found to be 30 volts. What is the difference of potential between the ter- minals of the battery, and what is the E. M. F. of the battery on open circuit, known as its voltage or E. M. F. (one of the battery constants)? Solution: The total external resistance is 100 + 50 + 25 = 175 ohms. By Rule 6, we have 100 : 175 :: 30 volts : x = 52^ volts, difference of potential between the terminals of the battery. The current is found by dividing (Rule 1), the difference of potential of the 100 ohm coil by its resistance. This E. M. F. is 30. The current therefore is $ amperes. The to- tal resistance of the circuit is that of the three coils or 175 ohms plus that of the battery or 30 ohms, a OHM'S LAW. 19 total of 205 ohms. To maintain a current of ffo amperes through 205 ohms (Rule 2), an E. M. F. is required equal to i 3 & X 205 volts or 61^ volts. DIVIDED, BRANCHED OR SHU.NT CIRCUITS. A single conductor, from one terminal of a gener- ator may be divided into one or more branches which may reunite before reaching the other ter- minal. Such branches may vary widely in resist- ance. Rule 8. In divided circuits, eacli branch passes a portion of a current inversely proportional to its re* sistance. EXAMPLES. A portion of a circuit consists of two conductors, A and B, in parallel of A = 50, and B = 75 ohms, respectively; what will be the ratio of the currents passing through the circuit, which will go through each conductor? Solution: The ratio will be current through A : current through B :: 75 : 50, which may be ex- pressed fractionally, & : T^. Where more than two resistances are in parallel, the fractional method is most easily applied. Three conductors of A = 25, B = 50, and = 75 ohms are in parallel. What will be the ratio of cur- rents passing through each one? Solution: Fractionally AzBrC::^:^:^. 20 ARITHMETIC OF ELECTRICITY. It tile 9. To determine the amount of a given cur- rent that will pass through parallel circuits of differ* ent resistances, proceed as follows : Take the resist- ance of each branch for a denominator of a fraction having 1 for its numerator. In other words, for each branch write down the reciprocal of its resistance. Then reduce the fractions to a common denominator, and add together the numerators. Taking this sum of the numerators for a new common denominator, and the original single numerators as numerators, the new fractions will express the proportional cur- rents as fractions of one. If the total amperage is given, it is to be multiplied by the fractions to give the amperes passed by each branch. The solution can also be done in decimals. EXAMPLES. A lead of wire divides into three branches; No. 1 has a resistance of 10,000 ohms, No. 2 of 39 ohms, and No. 3 of i ohm. They unite at one point. What proportion of a unitary current will pass each branch ? Solution: The proportion of currents passed are as rotary : *V : % or 3. Reducing to a common de- nominator, these become T^Hhny * AWir : V&iW. The proportions of the numerators is the one sought for; taking the sum of the numerators as a common de- nominator, we have in common fractions the follow- ing proportions of any current passed by the three branches. No. 1, n^mr; No. 2, Ttt***r; No. 3, 1170000 118 OT5T9* Four parallel members of a circuit have resistances respectively of 25, 85, 90, and 175 ohms; express OHM'S LAW. 21 decimally the ratio of a unitary current that will pass through them. Solution: The ratio is as & : s : ir& : rf?, or reduc- ing to decimals (best by logarithms), .04 : .011765 : .011111 : .0057. Adding these together, we have .068576, which must be multiplied by 14.582 to pro- duce unity. Multiplying each decimal by 14.58 (best by logarithms), we get the unitary ratio as .5832 : .17153 : .1620 : .08310, whose sum is 1.0000. Unless logarithms are used, it is far better to work by vulgar fractions. A current of .71 amperes passes through two branches of a circuit. One is a lamp with its con- nections of 115 ohms resistance; another is a resist- ance coil of 275 ohms resistance. What current passes through each branch? Solution: The proportions of the current are as TIT : T&S or reduced to a common denominator and to their lowest terms ^ftr : ^ff*. Proceeding as before, and taking the sum of the numerators (55 + 23 78), as a common denominator, we find that the lamp passes H, and the resistance coil H of the whoie current. Multiplying the whole current, .71 by f$, we get nn amperes, or i ampere for the lamp, leaving .21 or a little over I ampere for the resistance coil. Another problem in connection with parallel branches of a circuit is the combined resistance of parallel circuits. This is not a case of summa- 22 ARITHMETIC OF ELECTRICITY. tion, for it is evident that the more parallel paths there are provided for the current, the less will be the resistance. Rule 1 0. In shunt circuits, the resistance of the com- bined shunts is expressed by the reciprocal of the sum of the reciprocals of the resistances. EXAMPLE. Two leads of a 50 volt circuit (leads differing in potential by 50 volts), are connected by a 20 ohm motor. A 50 ohm lamp and 1000 ohm resistance coil are connected in parallel or shunt circuit there- with, what is the combined resistance? and the total current? Solution: The reciprocal of resistance is conduc- tance, sometimes expressed as mhos. (Rule 19.) The conductance of the three shunts is equal to A + ifo + raW mhos = riHhr + -Ms + irinr = yHv mhos. The reciprocal of conductance is resistance. The combined resistance is therefore -W ohms = 14.09 ohms. The current is Jj!L or 3.5 amperes. Rule 11. The combined resistance of two parallel circuits is found by multiplying the resistance* to- gether, and dividing the product by the sum of the re- sistances. Where there are several circuits, any t\vo can be treated thus, and the result combined in *he same way with another circuit, and so on to getthe final resistance. B _ R ^ r x rl EXAMPLE. Four conductors in parallel have resistances of 100 50 27 19 ohms. What is their combined resistance? OHM'S LAW. 23 Solution: Combining the first and second, we h ave 1^ = 33i ohms. Combining this with the resistance of the third wire, we have ^|^^ = 14.9 ohms. Combining this with the resistance of the fourth wire, we have ^$S = 8.3 ohms. The result is, of course, identical by whatever rule obtained. Rale 12. When all the parallel circuits are of uni- form resistance, as in multiple arc incandescent light- ing, the resistance of the combined circuits is found by dividing the resistance of one circuit by the number of circuits. _ n EXAMPLES. There are fifty lamps of 100 ohms resistance each in multiple arc connection. What is their com- bined resistance? Solution: W = 2 ohms. A motor can take 3 amperes of currents at 30 volts safely without burning out or heating injuriously. A 110 volt incandescent circuit is at hand. The motor is to be connected across the leads so as to receive the above amperage. A shunt or branch of some resistance is carried around it, and a resistance coil intervenes between the united branches and one of the main leads. The resistance of the coil is 20 ohms. What should the resistance of the shunt be? 24 ARITHMETIC OF ELECTEICITY. A. -VWWW: UAA/WSAA) rfe.si.sta.njce Solution: The resistance of the motor (Ohm's Law, Rule 3), is found by dividing the E. M. F. by fthe resistance 30 * 3 = 10 ohms. By Rule 5 the resistance of the coil in series (20 ohms) must be to the combined (not added) resistance of the motor and shunt coil, as 110 CO (total voltage minus voltage for motor) : 30 (voltage for motor) or 20 : x :: 80 : 30 . . x = 7.5 combined resistance of parallel or shunt coil and motor. The reciprocal of 7.5 (conductance, Rule 19), may be expressed as HSths of the combined (in this case added) conductances of shunt coil and motor. The conductance of the motor is equal to the reciprocal of 10 which may be expressed as & or as ^. The conductance of the shunt coil must therefore be H# ~~ ^ = TO = ih> mho. The reciprocal of this gives the resistance of the shunt coil which is 30 ohms. The total current going through the system by Ohm's law is amperes. The resistance of the shunt coil 30 ohms is to that of the motor in parallel with it 10 ohms as the current received by the motor is to OHM'S LAW. 25 that received by the coil, a ratio of 30 : 10 or 3:1 giving 3 amperes for the motor and 1 ampere for the coil. This is a proof of the correctness of operations. Two conductors through which a current is passing are in parallel circuit with each other. One has a resistance of 600 ohms. The other has a resistance of 3 ohms. A wire is carried across from an intermediate point of one to a corresponding point of the other. It is attached at such a point of the first wire that there are 400 ohms resistance be- fore it and 200 after it. Where must it be connected to the other in order that no current may pass? Solution: The E. M. F. up to the point of con- nection of the bridge or cross wire is to the total E. M. F. in the 600 ohm wire as 400: 600 or as 2: 3. The other wire which by the conditions has the same drop of potential in its full length must be divided therefore in this ratio. The bridge wire must therefore connect at 2 ohms from its begin- ning, leaving 1 ohm to follow. The principle here illustrated can be proved generally and is the Wheat- stone Bi jdge principle. CHAPTER III. KESISTANCE AND CONDUCTANCE. RESISTANCE OF DIFFERENT CONDUCTORS OF THE SAME MATERIAL. Conductors are generally circular in section. Hence they vary in section with the square of their diameters. The rule for the resistance of conduc- tors is as follows: Rule 13. The resistance of conductors of identical material varies inversely as their section, or if of circu- lar section inversely as the squares of their diameters, and directly as their lengths. EXAMPLE. 1. A wire #, is 30 mils in diameter and 320 feet long; another b, is 28 mils in diameter and 315 feet long. What are their relative resistances? Solution: Calling the resistances R* : R b we would have the inverse proportion if they were of equal lengths R b : R* :: 30 2 : 28 2 or as 900 : 784. Were they of equal diameter the direct proportion would hold for their lengths: R b : R a :: 315 : 320. Com- bining the two by multiplication we have the com- pound proportion R b : R a :: 900 X 315 : 784 X 320 or as 283,500 : 250,880, or as 28 : 25 nearly. The RESISTANCE AND CONDUCTANCE. 27 combined proportions could have been originally expressed as a compound proportion thus: K b : R a :: 30* X 315 : 28 2 X 320. For wires of equal resistance the following is given. Rule 14. The length of one wire multiplied by the square of the diameter of the other wire must equal the square of its own diameter multiplied by the length of the other If their resistances are equal. Or multiply the length of the first wire by the square of the diameter of the second. This divided by the length of the second will give the square of the diameter of the first wire; or divided by the square of the diameter of the first will give the length of the second. Id" I'd 8 EXAMPLES. 1. There are three wires, a is 2 mils, I is 3 mils, and c is 4 mils in diameter; what length must b and c have to be equal in resistance to ten feet of af Solution: Take a and c first and apply the rule, ]0 X 4 9 -2 s = 40 feet; then take a and I 10 X 3 a * 2 2 = 22^ feet. To prove it compare a and c directly by the same rule 22*4 X 4 2 -*- 3 2 = 40. As this gives the same result as the first operation, we may regard it as proved. A conductor is 75 mils in diameter and 79 feet long; how thick must a wire 1264 feet long be to equal it in resistance? Solution: 75 2 X 1264 * 79 = 7,110,000 -5- 79 = 90,000. The square root of this amount is 300 which is the required diameter. 28 ARITHMETIC OF ELECTRICITY. For problems involving the comparison of wires of unequal resistance the rule may be thus stated: Rule 15. Multiply the square of the diameter of each 'wire by the length of the other. Of the two products divide the one by the other to get the ratio of resist- ance of the dividend to that of the divisor taken at unity. The term including the length of a given wire is the one expressing the relative resistance of such wire. EXAMPLES. A wire is 40 mils in diameter, 3 miles long and 40 ohms resistance. A second wire is 50 mils in diameter and 9 miles long. What is its resistance? Solution: 9 X 40 2 = 14,400 relative resistance of the first wire. 3 X 50 2 = 7,500 relative resistance of second wire. 14,400 -*- 7,500 = 1.92 ratio of resistance of second wire to that of first taken at unity. But the latter resistance really is 40 ohms. Therefore the resistance of the second wire is 40 X 1.92 = 76.80 ohms. The result may also be worked out thus: 40 2 X 9 = 14,400 = relative resistance of the 3 mile wire. 50 2 x 3 = 7500 = relative resistance of the 9 mile wire. 14,400 * 7500 = 1.92 = ratio of 9 mile (dividend) to 3 mile (divisor) wire. .-. 40 ohms X 1.92 = 76.8 ohms. A length of a thousand feet of wire 95 mils in diameter has 1.15 ohms resistance; what is the di- RESISTANCE AND CONDUCTANCE. 29 ameter of a wire of the same material of which the resistance of 1000 feet is 10.09 ohms? (R. E. Day, M. A.). Solution: 10.09 *- 1.15 = 8. 7? ratio of resistances. If we divide 1000 by 8. 77 we obtain a length of the first wire which reduces the question to one of iden- tical resistances. 1000 + 877 = 114 feet. Then applying Rule 14, 114 X 95 * 1000 = 1037.88. This is the square of the diameter of the other wire. Its square root gives the answer: 32.2 mils. SPECIFIC RESISTANCE. Specific resistance is .the resistance of a cube of one centimeter diameter of the substance in ques- tion between opposite sides. It is expressed in ohms for solutions and in microhms for metals. From it may be determined the resistance of all volumes, generally prisms or cylinders, of substance. Very full tables of Specific Resistance are given in their place. Rale 16. The resistance of any prism or cylinder of a substance is equal to its specific resistance multiplied by its length in centimeters and divided by its cross- sectioiial area in square centimeters. If the dimensions are given in inches or other units of measurements they must be reduced to centimeters by the table. Sp. R x I * EXAMPLES. An electro-plater has a bath of sulphate of copper, gp. resistance 40 ohms. His electrodes are each 1 80 ARITHMETIC OF ELECTRICITY. foot square and 1 foot apart. What is the resist* ance of such a bath ? Solution : By the table 1 square foot = 929 sq. cent, and 1 foot = 30.4797 cent. .-. Resistance = 40 X 30.4797 + 929 = 1.31 ohms. Where the electrodes in a solution are of uneven size take their average size per area. The facing areas are usually the only ones calculated, as owing to polarization the rear faces are of slight efficiency, and where the electrodes are nearly as wide as the bath or cell the active prism is practically of cross- sectional area equal to the area of one side of a plate. In a Bunsen battery the specific resistances of the solutions in inner and outer cells were made alike, each equalling 9 ohms. The central element was a YZ inch cylinder of electric light carbon. The outer element was a plate of zinc 6 inches long bent into a circle. When there were 2 inches of solution in the cell what was the resistance? Solution : Area of carbon = f X 2 = 3.14 square inches. Area of zinc = 2 X 6 = 12 square inches. This gives an average facing area of (12 + 3.14) -*- 2 = 7.57 square inches = 48.38 sq. cent. The distance apart = ^ inches (nearly) = 1.9049 cent. .-. Resistance = 9 X 1.9049 -*- 48.38 = .354 ohms. For wires, the specific resistance of metals being given in microhms, the calculation may be made in microhms, or in ohms directly. As wire is cylindri- RESISTANCE AND CONDUCTANCE. 3] cal a special calculation may be made in its case to reduce area of cross section to diameter. This may readily be taken from the table of wire factors, thus avoiding all calculation. Rule 17. The resistance In microhms of a wire of given diameter In centimeters Is equal to tlie product of the specific resistance by 1.2737 by the length In centi- meters divided by the square of the diameter In cen- timeters. _ Sp. Res, x '.2737 x L d EXAMPLES. The Sp. Res. of copper being taken at 1.652 mi- crohms what is the resistance of a meter and a half of copper wire 1 millimeter thick? Solution: The diameter of the wire (1 millimeter) is .1 centimeter. The square of .1 is .01. The length of the wire (iy 2 meter) is 150 centimeters. Its resistance therefore is 1.652 X 1.2737 X 150 -H .01 = 31,561 microhms or .031,561 ohms. UNIVERSAL RULE FOR RESISTANCES. Into the problem of resistances of one or two wires eight factors can enter, these are the lengths, sec- tional areas, specific resistances and absolute resist- ances of two wires. Their relation may be ex- pressed by an algebraic equation, which by transposi- tion may be made to fit any case. The rule is arithmetically expressed by adopting the method of cancellation, drawing a vertical line and placing on 32 ARITHMETIC OF ELECTRICITY. the left side, factors to be multiplied together for a divisor, and on the right side factors to be multi- plied together for a dividend. In the expression of the rule as below the quotient is 1, in other words the product of all the factors on the left hand of the line is equal to that of all the factors on the right hand. Calling one wire a and the other b we have the following expression: Resistance of 1) Specific Resistance of a Length of a Cross-sectional area of b Resistance of a Specific Resistance of b Length of b Cross-sectional area of d Rule 18. Substitute in the above expression the values of any factors given. Substitute for factors not given or required the figure 1 or unity. Such a value deter- mined by division must be given to the required factor and substituted in its place as will make the product of the left-hand factors equal to that of the right-hand factors. Only one factor can be determined, and all factors not given are assumed to be respectively equal for both conductors. EXAMPLES. If the resistance of 500 feet of a certain wire is 09 ohms what is the resistance of 1050 feet of the same wire? Solution: The cross sectional areas and specific resistance not being given are taken as equal. (This of course follows from the identical wire being re- ferred to.) The vertical line is drawn and the values substituted : RESISTANCE AND CONDUCTANCE. 33 Resistances : Eesistance of .09 required wire Lengths : 500 1050 (Other factors omitted as unnecessary.) 1050 X .09 * 500 = .189 ohms. What is the diameter of a wire 2 miles long of 23 ohms resistance, if a mile of wire of similar ma- terial of seventy mils diameter has a resistance of 10. 82 ohms? Solution. We use for simplicity the square of the diameter in place of the cross sectional area of the known wire, thus: Resistances : 23 Lengths : 1 Areas : Unknown 10.82 2 70 2 As the specific resistances are identical they are not given. 2 X ?0 2 X 10.82 * 23 X 1 = 4610 square of diameter required : 4610^ = 68 mils. What must be the length of an iron wire of cross- sectional area 4 square millimeters to have the same resistance as a wire of pure copper 1000 yards long, of cross-sectional area 1 square millimeter, taking the conductance of iron as } that of copper? (Day). Solution: 34 ARITHMETIC OF ELECTRICITY. Specific Resistances : 1 Lengths : 1000 Cross-sectional areas : 4 7 (i.e. the reciprocal of conductance) Unknown 1 As the resistances are identical they are not given. Solving we have 1000 X 4 * 7 = 571f yards. There are two conductors, one of 35 ohms resist- ance, 1728 feet long and 12 square millimetres cross- sectional area and specific resistance 7: the other of 14 ohms resistance, 432 feet long and 8 square milli- metres cross-sectional area. What is its specific resistance? Resistances : 35 Specific Resistances Unknown Lengths: 432 Cross- Sectional areas : 12 14 7 1728 8 By cancellation this reduces to 14X8-*-5X3 7.4 Specific Resistance. In these cases it is well to call one wire a and the other #, and to arrange the given factors in two columns headed by these designations. Then the formula can be applied with less chance of error. Thus for the last two problems the columns should be thus arranged. RESISTANCE AND CONDUCTANCE. 35 a b a 6 Area. 4 sq. mils. L. Unknown Sp. Res. 7 1 1,000 yards 1 Resist. 35 olims L. 1,728 feet Area, 12 sq. mils. Sp. Res 7 14 ohms 432 feet 8 sq. mils, required From such statements of known data the formula can be conveniently filled up. RESISTANCE OF WIRES REFERRED TO WEIGHT. The weight of equal lengths of wire is in propor- tion to their sections. The problems involving weight therefore can be reduced to the Rules al- ready given. Problem. A wire, A, is 334 feet long and weighs 25 oz.; another, B, is 20 feet long and weighs 1 oz. what are the relative resistances? Solution: 20 feet of the wire " A" weigh ^ x 25 = 1.50 oz. The weights of equal lengths of A and B respectively areas 1.50 : 1.00 which is also the inverse ratio of the resistances of equal lengths. By compound proportion Rule we have R. of " A " : R. of "B" :: 1 X 334 :: 1.50 X 20; reducing to 16.7 : 1.5 or 11.1 : 1.0 or the wire " A " has about eleven times the resistance of the wire " B." Solution: By general Rule for resistance (Rule 18). Taking 1.50 : 1.00 as the ratio of cross-sec- tional areas and taking the resistance of the long wire A as 1 we have : 36 ARITHMETIC OF ELECTRICITY. Resistances : 1 Lengths : 20 Cross-sectional area : 1.50 Unknown 334 1 Eesistance of B = 1.50 X 20 * 334 = .0899 01 about fr as before. CONDUCTANCE. Conductance is the reciprocal of resistance and is sometimes expressed in units called MHOS, which is derived from the word ohm written backwards. Rule 19. To reduce resistance in ohms to conductance in mhos express its reciprocal and the reverse. K =R EXAMPLES. A wire has a resistance of t 1 ^ ohms, what is its conductance? Solution: 126 -*- 18 = 7 mhos. Eeduce a conductance of lit to ohms. Solution: lit = If mhos which gives if ohms. It is evident that the data for problems or that constants could be given in mhos instead of ohms. In some ways it is to be regretted that the positive quality of conductance was not adopted at the out- set instead of the negative quality of resistance. One or two illustrations may be given in the form of examples involving conductance. Express Ohm's law in its three first forms in conductance. RESISTANCE AND CONDUCTANCE. 37 Solution: This is done by replacing the factor "resistance" by its reciprocal. Thus, Rule 1 reads for conductance: "The current is equal to the electromotive force multiplied by the conduc- tance" (C = EK) Rule 2 as "The electromotive force is equal to the current divided by the conduc- C tance " (E = ) Rule 3 as " The conductance is K equal to the current divided by the electromotive C force." (K = ) E A circuit has a resistance of .5 ohm and an E. M. F. of 50 volts; determine the current, using con- ductance method. Solution: The conductance = .-J- = 2 mhos. The current = 50 X 2 = 100 amperes. In a circuit a current of 20 amperes is main- tained through 2f ohms. Determine the E. M. F. using conductance. Solution: The conductance = 4 = tt mhos. E. M. F = 20 * H = 52 volts. Assume a current of 30 amperes and an E. M. F. of 50 volts, what is the conductance and resis- tance? Solution: Conductance = 30 * 50 = .6 mho. Resistance = 1 -* .6 = 1.667. CHAPTER IV. POTENTIAL DIFFERENCE. DROP OF POTENTIAL IN LEADS AND SIZE OF SAME FOR MULTIPLE ARC CONNECTIONS. SUBSIDIARY leads are leads taken from large sized mains of constant E. M. F. or from terminals of constant E. M. F. to supply one or more lamps, motors, or other appliances. A constant voltage is maintained in the mains or terminals. There is a drop of potential in the leads so that the appliances always have to work at a diminished E. M. F. The E. M. F. of the leads is known, the requisite E. M. F. and resistance of the appliance is known, a rule is required to calculate the size of the wire to secure the proper results. It is based on the princi- ple that the drop or fall in potential in portions of integral circuits varies with the resistance. (See Ohm's law). A rule is required for a single appli- ance or for several connected in parallel. Rule 20. The resistance of the leads supplying any ap- pliance or appliances for a desired drop in potential within the leads is equal to the reciprocal of the cur- rent of the appliances multiplied by the desired drop in volts. POTENTIAL DIFFERENCE. 39 EXAMPLE. A lamp, 100 volts X 200 ohms, is placed 100 feet from the mains, in which mains a constant E. M. F. of 110 volts is maintained. What must be the resist- ance of the line per foot of its length; and what size copper wire must be used? Solution: The lamp current is obtained (Ohm's law) by dividing its voltage by its resistance, (&$ = \ ampere). The reciprocal of the current is f; mul- tiplied by the drop (f X 10 = 20) it gives the resist- ance of the line as 20 ohms. As the lamp is 100 feet from the mains there are 200 feet of the wire. Its resistance per foot is therefore ^ = -fa ohm or it is No. 30 A. W. G. (about). For several appliances in parallel on two leads a similar rule may be applied. There is inevitably a variation in E. M. F. supplied to the different appli- ances unless resistances are intercalated between the appliances and the leads. Rule 21. The K. in. F. of the main leads or terminals the factors of the lamps or other appliances, tbeir num. ber and the distance of their point of connection are given. The combined resistance is found by Rules 8 to 12. Then by Rule 20 the resistance of the lead* is cal- culated. EXAMPLE. A pair of house leads includes 260 feet of wire, or 130 in each lead. Six 50 volt 100 ohm lamps are connected thereto at the ends. The drop is to be 5 40 ARITHMETIC OF ELECTRICITY. volts, giving 55 volts in the main leads. Eequired the total resistance of and size of wire for the house leads. Solution: The resistance of six 100 ohm lamps in parallel is 100 -s- 6 = 16.66 ohms. The current re- quired is by Ohm's law 50 *- 16.66 or 3 amperes. Its reciprocal multiplied by the drop, (1X5=1 = l^i ohms) gives the required resistance = IJ/s ohms. This, divided by 260 feet gives the resistance per foot as .0064 ohm, corresponding by the table to No. 18 A. W. G. A rule for the above cases is sometimes expressed otherwise, being based on the proportion: Eesist- ance of appliances is to resistance of leads as 100 minus the drop expressed as a percentage is to the drop expressed as a percentage. This gives the fol- lowing: Rule 22. The resistance of the leads is equal to the combined resistance of the appliances multiplied by the percentage of drop and divided by 100 minus the percentage of drop. Problem. Take the data of last problem and solve. Solution: The percentage of drop is & = 9$. The resistance of the leads = 16 - 66 x 9 = 149.94 = \y z O hms 100 9 91 about. Note. To obtain accurate results the figures of percen- tage, etc., must be carried out to two or more decimal places. Rules 20 and 21 are to be preferred to any percen- tage rule. Also see Rule 23. POTENTIAL DIFFEEENCE. 41 Where groups of lamps are to be connected along a pair of leads but at considerable intervals, the succeeding sections of leads have to be of diminish- ing size. The same problem arises in calculating the sizes of street leads. The identical rule is ap- plied, care being taken to express correctly the ex- act current going through each section of the lead. The calculation is begun at the outer end of the leads. A diagram is very convenient; it may be conventional as shown below. EXAMPLES. At three points on a pair of mains three groups of fifty 220 ohm lamps in parallel are connected; a total drop of 5 volts is to be divided among the three groups, thus: 1.6 volts 1.6 volts 1.8 volts-. The initial E. M. F. is 115 volts; what must be the resistances of the three sections of wire? Solution: The following diagram gives the data as detailed above: 1. 2. 3. Starting at group 3 we have 50 lamps in parallel each of 220 ohms resistance, giving a combined re- sistance (Rule 12) of 4.4 ohms and a total current (Ohm's law) of 110 -*- 4.4 = 25 amperes. The re- sistance of section 2 3 is by the present rule $5 X 42 ARITHMETIC OF ELECTRICITY. 1.8 = .072 ohms. Taking group 2 the current through this group of lamps is 111.8 * 4.4 = 25.41 amperes. The section 1 2 has to pass also the cur- rent 25 amperes for group 3 giving a total current of 25 + 25.41 = 50.41 amperes. The resistance of section 1 2 is therefore^ X 1.6 = .0317 ohm. Taking group 1 the current for its lamps is 113.4 *- 4.4 = 25.7 amperes. The total current through section 01 is therefore 25 + 25.4 + 25.7= 76.1 amperes. The resistance of the section is ^ X 1.6 = .021 ohms. Arranging all these data upon a diagram we have the full presentation of the calcu- lation in brief as below: r CHAPTER V. CIRCULAR MILS. A MIL is ntar of an inch. The area of a circle, one mil in diameter, is termed a circular mil. The area of the cross-section of wires is often expressed in circular mils. Thus a wire, 3 mils in diameter, has an area of 9 circular mils, as shown in the cut. A circular mil is .7854 square mil. Rules for the sizes of wires for given resistances are often based on cir- cular mils, and include a constant for the conduc- tivity of copper. By the table of specific resistances, the values found can be reduced to wires of iron or other metals. 44 ARITHMETIC OF ELECTRICITY. A commercial copper wire, one foot long, and one circular mil in section, has a resistance of 10.79 ohms at 75 F. This is, of course, largely an as- sumption, but is taken as representing a good aver- age. No two samples of wire are exactly alike, and many vary largely. From Rule 13, and from the above constant, we derive the following rules: Rule 23. The resistance of a commercial copper wire is equal to its length divided by the cross-section in cir- cular mils, and multiplied by 10.79. EXAMPLE. A wire is 1050 feet long, and has a cross-section of 8234 circular mils. What is its resistance? Solution: 1050 X 10.79 *- 8234 = 1.37' ohms. Rule 24. The cross-section of a wire in circular mils is equal to its length divided by its resistance, and mul- tiplied by 10.79. EXAMPLE. A wire is 1050 feet long, and has a resistance of .68795 ohms. What is its cross-section in circular mils? Solution: 1050 X 10.79 -*- .68795 = 16,468 circu- lar mils. Rule 25. The cross-section of the wire* of a pair of leads in circular mils for a given drop expressed in per- centage is equal to the product of the length of leads by the number of lamps (in parallel), by 21.58, by the dif- ference between 1OO and the drop, the whole divided by the resistance of a single lamp multiplied by the drop. __ Inx 21.58 X (100- e) A ~~ er CIRCULAR MILS. 45 EXAMPLE. Fifty lamps are to be placed at the end of a double lead 150 feet long (= 300 feet of wire). The resist- ance of a lamp is 220 ohms. What size must the wire be for 5% drop? Solution: 150 X 50 X 21.58 X (100 5) -*- (220 X 5) = 13,977.9 circular mils. In these calculations and in the calculations given on page 48 it is important to bear in mind that the percentage is based upon the difference of potential at the beginning of the leads or portion thereof under consideration; in other words upon the high- est difference of potential within the system or the portion of the system treated in the calculation. CHAPTER VI. SPECIAL SYSTEMS. THKEE WIRE SYSTEM. As there are three wires in the three wire system, where there are two in the ordinary system, and as each of the three wires is one quarter the size of each of the two ordinary system wires, the copper used in the three wire system is three-eighths of that used in the ordinary system. In the three wire system the lamps are arranged in sets of two in series. Hence but one-half the current is required. The outer wires, it will be no- ticed, have double the potential of the lamps. Hence to carry one-half the current with double the E. M. P., a wire of one quarter the size used in the ordinary system suffices. Rule 26. The calculations for plain multiple arc work apply to the three wire system, as regards size of each of the three leads, If divided by 4. While the central or neutral wire will haye noth- ing to do when an even number of lamps are burning on each side, and may never be worked to its full capacity, there is always a chance of its having to carry a full current to supply half the lamps (all on SPECIAL SYSTEMS. 47 one side). Hence it is made equal in size to the others. ALTERNATING CURRENT SYSTEM. The rules already given apply in practice to this system also, although theoretically Ohm's law and those deduced from it are not correct for this current. A calculation has to be made to allow for the conversion from primary to secondary current in the converter as below. Note. The ratio of primary E. M. F. to secondary is ex- pressed by dividing the primary by the secondary, and is termed ratio of conversion. Thus in a 1000 volt system with 50 volt lamps in parallel the ratio of conversion is 1000 -*- 50 = 20. Rule 27. The current in tbe primary is eqnal to the current in the secondary divided by the ratio of conver- sion. EXAMPLES. On an alternating current circuit whose ratio of conversion = 20, there are 1000 lamps, each 50 volt; 50 ohms. When all are lighted what is the primary current ? Solution : By Ohm's law the secondary current is 1000 x 1 (each lamp using -f$ = 1 ampere, Ohm's law) = 1000 amperes. 1000 -- 20 = 50 amperes is the primary current. The current being thus determined the ordinary rules all apply exactly as given for direct current work. 48 ARITHMETIC OF ELECTRICITY. Given 650 lamps, 50 volt 50 ohms each, 3600 feet from the station. The primary circuit pressure is 1031 volts. A drop of 3$ is to be allowed for in the primary leads. What is the resistance of the primary wire? Solution: Current of a single lamp = 50 -*- 50 = 1 ampere; current of 650 lamps = 650 amperes, cur- rent of primary 650 -*- 20 = 32 amperes, drop of primary = 1031 X 3# = 30.9 volts, resistance of pri- mary (Rule 20) X 30.9 = .9516 ohm. Rule 28. For obtaining tlie size of tlie primary wire In circular mils, calculate by Rule 25, and divide the result by the square of the ratio of conversion. EXAMPLE. Take data of last problem and solve. Solution: [3600 X 650 X 21.58 X (1003) *- (50 X 3)] -* 20 2 = 81,637 circular mils. The two last examples may be made to prove each other, thus: The total length of leads is 3600 X 2 == 7200 feet. If of 1 mil thickness its resistance would be 7200 X 10.79 = 77,688 ohms. As resistance varies inversely as the cross sectional area we have the proportion .9516 : 77,688 :: 1 : x which gives x = 81,639 cir- cular mils corresponding within limits to the result obtained by Rule 28. In all cases of this sort where percentage is ex- pressed the statement in the last paragraph on page SPECIAL SYSTEMS. 49 45 should be kept in mind. The ratio of conversion must be based on the E. M. F. at the coil (in this case on 1031 31 = 1000 volts) not on the E. M. F. at the beginning of the leads or portion thereof con- sidered in the calculation. The percentage of drop must be subtracted before the ratio of conversion is calculated. For winding the converters, the following is the rule : Rule 29. The convolutions of the primary are equal in number to the product of the convolutions of the sec- ondary multiplied by the ratio of conversion, and vice versa. EXAMPLES. The ratio of conversion of a coil is 20; there are 1000 convolutions of th,e secondary. How many should there be of the primary? Solution: 1000 X 20 = 20,000 convolutions. There are in a coil 5000 convolutions of the pri- mary; its ratio of conversion is 50. How many con- volutions should the secondary have? Solution: 5000 *- 50 = 100 convolutions. CHAPTER VIL WORK AND ENERGY. ENERGY AND HEATING EFFECT OF THE CURRENT. IT has been shown experimentally by Joule that the total quantity of heat developed in a circuit is equal to the square of the current multiplied by the resistance. This is equal, by Ohm's law, to the square of the E. M. F. divided by the resistance, which again reduces to the E. M. F. multiplied by the current. Each of these expressions has its own application, and they may be given as three rules. Rule 30. The energy or neat developed in "circuits is in proportion to the square of the current multiplied bjr the resistance. EXAMPLES. An electric lamp has a resistance of 50 ohms; it is connected to a street main by leads of 21 ohms re- sistance. What proportion of heat is wasted in the house circuit? Solution: The current being the same for all parts of the circuit, the heat developed is in proportion to the resistance, or as 2i : 50 equal to 1 : 20. The WORK AND ENERGY. 51 heat developed in the wire is wasted, therefore the waste is > of the total heat developed. The internal resistance of a battery is equal to that of 3 meters of a particular wire. Compare the quantities of heat produced both inside and outside the battery when its poles are connected with one meter of this wire with the quantities produced in the same time when they are connected by 37 meters of the same wire. (Day.) Solution: The relative currents produced in the two cases are found (Ohm's law) by dividing the E. M. F. of the battery (a constant quantity = E) by the relative resistance. As the battery counts for the resistance of 3 meters of wire, the relative resistances are as 4 : 40. "Were the same current de- veloped in both cases these figures would give the desired ratio. But as the current varies it has to be taken into account. To determine the relative bat- tery heat only, we may neglect resistance of the bat- tery, as it is a constant for both cases, the battery remaining identical. By Ohm's law the currents are in the ratio of f : ^ and their squares are in pro- portion to f| : T |J 7 = 100 : 1. By the rule this is the proportion of the heats developed in the battery alone, with the short wire 100, with the long wire 1. "For the heating effects on the outside circuit, as resistance and current both vary, the full for- mula of the rule must be applied. The ratio of the heat in the short wire connection to that in the long 52 ARITHMETIC OF ELECTRICITY. wire connection is as (f) 2 X 1 : (^) 2 X 37 = 100 X 1 : 37 X 1. The ratio therefore is as 100 is to 37 for the total heat produced in the circuit which includes battery and connections. Owing to irregular working of a dynamo, an in- candescent lamp receives sometimes the full amount or i ampere of current; at other times as little as i$ ampere. What proportion of heat is developed in it in both cases, assuming its resistance to remain sensibly the same? Solution: By the rule the ratio is (i) a : ($f) or J: ^AT = 2025 : 400. The diminution of current there- fore cuts down the heat to $ the proper amount. Rule 31. The energy or heat developed in a circuit is in proportion to the square of the electromotive force divided by the resistance. E" ~B EXAMPLES. There are two Grove batteries, each developing 1.98 volts E. M. F. One has an internal resistance of -A- ohm; the other of * ohm. They are placed in succession on a circuit of 2 ohms resistance. What is the ratio of heats developed by the batteries in each case? Solution: As the E. M. F. is constant it may be taken as unity. Then for the two cases we have 2A> * 24 ssa ^i %ds as the ratio of heat produced. A battery of one ohm resistance and two volts E. WORK AND ENERGY. 53 M. F. is put in circuit with 4 ohms resistance. Another battery of 4 ohms and 1 volt is connected through 1 ohm resistance. What ratio of heat is developed in each case? Solution: i|_ 2 : 1|1 or 4 : 1. Kule 32. The energy or heat developed In a circuit I in proportion to the E. 31. F. multiplied by the current. H = Ef EXAMPLES. Take data of last problem and solve. Solution: For first battery, by Ohm's law, current = ampere; for the second, current = $ ampere. The heat developed, is by the present rule, in the propor- tion as | X 2 : X 1 or 4 : 1. Compare the heat developed in a 100 volt 200 ohm lamp and in a 35 volt 35 ohm lamp and in a 50 volt 50 ohm lamp. Solution :'The currents (Ohm's law) are : M&M and ! in amperes = I, 1, and 1 amperes. The heats devel- oped are, by the rule, in the ratio 100 X | : 35 X 1 and 50 X 1 or 50 : 35 : 50. The same problem can be done directly by Rule 31, thus: The three lamps develop heat in the ratio m* : & : 3$' = 50 : 35 : 50. This is the direct and preferable method of calculation. , For " heat," "rate of energy," or "rate of work" can be read in these rules. 54 ARITHMETIC OF ELECTRICITY. THE JOULE OR GRAM-CALORIE. The last rules and problems only touch upon the relative proportions of heat; they do not give any actual quantity. If we can express in units of the same class a standard quantity of heat, then by deter- mining the relation of the standard to any other quantity, we arrive at a real quantity. Such a stan- dard is the joule, sometimes called the " calorie" or "gram-calorie." A joule is the quantity of heat required to raise the temperature of 1 gram of water 1 degree centigrade. It is often expressed as a water-gram-degree 0. or w. g. d. 0. or for shortness g. d. C., from the initials of the factors. It is unfortunate that it is called the calorie as the name is common to the water-kilogram- degree C. or kg.d. 0. The joule is equal to 4.16 X 10 7 or 41,600,000 ergs. It will be remembered that practical electric units are based on multiples of the C. G. S. units of which the erg is one. The joule comes in the C. G. S. order. Therefore to determine quantities of heat the following is a general rule when the practical electric units are used. Rule 33. Obtain the expression of rate of heat devel- oped, or of rate of energy, or of rate of work In volt amperes. Reduce to C. G. S. units (ergs) by multiply- ing by 1O T and divide by the value of a joule in ergs (4.1 6 X 1O 7 ). The quotient is joules or water-gram de- grees C. per second. Q _ E x C X 10 7 * ~ 4.16 X 10 7 WORK AND ENERGY. 55 EXAMPLE. A current of 20 amperes is flowing through a wire. What heat is developed in a section of the wire whose ends differ in potential by 110 volts? Solution: The rate of energy in watts or volt- amperes = 110 X 20 = 2200. In the C. G-. S. units this is expressed by (110 X 10 8 ) X (20 X 10" 1 ) = 2200 X 10 7 ergs, per second; . . quantity of heat = 2200 X 10 7 -s- 4.16 X 10 7 = 528.8 joules or gram-degrees- centigrade per second. As 10 7 *- 10 7 = 1 the rule can be more simply stated thus: Rule 34. The quantity of heat produced per second in a circuit by a current is equal to the product of the watts by ^^g or by .24. Q, = 0.24 CE. or 0.24 1? EXAMPLES. A difference of potential of 5.5 volts is main- tained at the terminals of a wire of > ohm resist- ance. How many joules per second are developed? Solution: By Ohm's law, current = 5.5 *- & = 55 amperes. By the rule 55 X 5.5 X 0.24 = 72.6 joules per second. Note. The energy of a current is given by Rules 30, 31 and 32 in watts, so that all cases are provided for by a com- bination of one or the other of these rules with Rule 34 An example will be given for each case. 56 ARITHMETIC OF ELECTRICITY. A current of .8 ampere is sent through 50 lamps in series, each of 137^ ohms resistance. What heat does it develop per second? Answer: The resistance = 137^ X 50. By rules 30 and 34 we have, rate of heat produced = ,8 2 X 137^ X 50 = 4400 watts. 4400 X 0.24 = 1056 joules per second. Kules 31 and 34. Fifty incandescent lamps, 110 volt, 137^ ohms, each are placed in parallel. What heat per second do they develop? Solution: By Ohm's law total resistance t- 50 = 2.75 ohms. By rules 31 and 34 rate of heat produced = HO 2 -*- 2.75 = 4400 watts and 4400 X 0.24 = 1056 joules per second as before. Rules 32 and 34. Through 50 incandeseent 110 volt lamps a current of .8 ampere is passed, the lamps being in series. What heat per second do they develop? Solution: By rules 32 and 34 rate of heat -= 110 X 50 X .8 = 4400 watts and 1056 joules per second as before. These three examples are purposely made to refer to the same set of lamps, to show that rules 30, 31, and 32 are identical. Each fits one of the three forms of statement of data. They also are designed to bring out the fact that the unit "watts," being based partly on amperes, includes the id<*a of rate, not of absolute quantity. Hence watts "per sec- ond " is not stated, as it would be meaningless ox WOBK AND ENERGY. 57 redundant, while the joule, denoting an absolute quantity, has to be stated "per second" to indicate tlie rate. There is such a unit as an "ampere- second," viz., the "coulomb," but there is no such thing as an "ampere per second," or if used it means the same as an " ampere per hour," "ampere per day" or "ampere." The same applies to watts and to power units such as "horse-power." SPECIFIC HEAT. The specific heat of a substance is the ratio of its capacity for heat to that of an equal quantity of water. It almost invariably is referred to equal weights. Here it will be treated only in that con- nection. The coefficient of specific heat of any substance is the factor by which the specific heat of water (= 1 or unity) being multiplied the specific heat of the sub- stance is produced. Rule 35. The weight of any substance corresponding to any number of joules multiplied by its specific heat gives the corresponding weight of water, and vice versa. EXAMPLE. A current of .75 amperes is sent for 5 minutes through a column of mercury whose resistance was 0.47 ohm. The quantity of mercury was 20.25 grams, and its specific heat 0.0332. Find the rise of temperature, assuming that no heat escapes by radiation. (Day.) 58 ARITHMETIC OF ELECTRICITY. Solution: By Eules 30 and 34, we find rate of heat = .75 2 X .47 = .264375 watts; .264375 X .24 = .06345 joules per second. The current is to last for 300 seconds .'. total joules = .06345 X 300 = 19.035 joules. This must be divided by the specific heat of mercury to get the corresponding weight of mer- cury; 19.035 -* .0332 = 573 gram degrees of mercury. Dividing this by the weight of mercury, 20.25 grams, we have 573 * 20.25 = 28 0. Rule 36. By radiation and convection, 4000 joule about is lost by any unpolished substance in the air for each square centimeter of surface, and for each degree that it is heated above the atmosphere. EXAMPLE. A conductor of resistance, 8 ohms, has a current of 10 amperes passing through it. It is 1 centi- meter in circumference, and 100 meters long. How hot will it get in the air? Solution: By Eule 30, etc., the heat developed per second in joules is 10 2 X 8 -*- 4.16 = 192.3 joules. The surface of the conductor in centimeters is 10,000 X 1 = 10,000 sq. cent. It therefore develops heat at the rate of 192.3 X 1Q-* = .01923 joules per second per square centimeter of surface. When the loss by radiation and convection is equal to this, it will re- main constant in temperature. Therefore .01923 * 4 The required current therefore = W = 21 amperes. As 10 exceeds 21 X 2 (Rule 50) it falls under case A. By Rule 51 the number of cells is 8 _*ft Xt) = ^ = 66.6 or 67 cells, as a cell cannot be divided. The cells must be in series. 70 ARITHMETIC OF ELECTRICITY. Proof: The E. M. F. of the 67 cells in series = 67 X 2 = 134 volts; their resistance = 67 X = 13.4 ohms. The resistance of the lamps in par- allel is 40 ohms. Hence by Ohm's law the cur- rent = ^ + 13 4 - = 2.51 amperes, the current required. The same lamps are placed in series. Calculate the cells of the same battery required. Cell current = 10 amperes. Current required = ^ X g or | am- pere. As 10 exceeds * X 2 (Rule 50) it falls again under case A. By Eule 51 cells required = a_^ xi) = 263. Proof: Current = - 52 ^ 1(m = I ampere the current required. 526 is the number of cells multiplied by the voltage of one cell; 52.6 is the number of cells multiplied by the resistance in ohms of a single cell; 1000 is the resistance of a single lamp, 200 ohms., multiplied by the number, 5, of lamps in series. "Whenever the arrangement and number of cells of a battery has been calculated the calculation should be proved as above. Rnle 52. Case 15. Group two or more cells In parallel so as to obtain by calculation from them through no ex- ternal resistance a current twice as great or more than twice as great as the required current. Then treating the group as if it was a single cell apply Rule 51 to de- termine the number of groups in series. EXAMPLE. Assume the same lamps in parallel, requiring the BATTEEIES. 71 current already calculated of 2i amperes. Assume a battery of constants 1 volt .25 ohm, giving a cell current of 4 amperes. This is less than 2$ X 2 and more than 2i X l; therefore it falls under Case B. Solution: A group of two cells in parallel gives .T^S = 8 amperes. 8 exceeds 2i X 2 . '. applying .Rule 49 we have number of groups = 100 -*-[! (2i X .125)] = 146 groups in series. Total number of cells = 2 in parallel, 146 in series = 292 cells. Proof: Current = 146 -H (40 + 18.25) =2.5 am- peres. Rule 53. Case C. Place as many cells in series as will give twice the required voltage. Place as many cells in parallel as will give a resistance equal to that of the external circuit. EXAMPLE. Assume the same lamps in parallel. Assume a battery of cell constants, 1 volt, 4 ohms. The lamp current is 21 amperes. The cell current is * ampere. The cell current therefore equals ( -* 2|) & of the required current. This falls under Case C. and is solved by Eule 53. Solution: Voltage required 100. By the rule cells in series = 100 X 2 == 200. These have a re- sistance of 800 ohms. To reduce this to the resist- ance of the outer circuit, viz., 40 ohms, 800 -f- 40 = 20 cells must be placed in parallel. Total cells =- 20 x 200 = 4000 cells. 72 ARITHMETIC OF ELECTRICITY. Proof: Current = 200 *- (40 + 40) = 2.5. Rule 54. All cases coming under Case C. may be sim- ply solved lor tlie total number of cells by dividing tli external energy by tlie cell energy and multiplying by 4. 'I'll is gives tiic number of cells. EXAMPLE. Take as cell constants .75 volt M ohm giving i am- pere. Assume 20 lamps, each 50 volts, 50 ohms and 1 ampere. As i -*- 1 is a unitary fraction () Case C. applies. Solution: Cell energy = i X .75 = .375 watts. Ex- ternal energy = 50 x 1 X 20 = 1000 watts. (1000 *- .375)X 4-10,666 cells. Solution by Rule 53: Voltage required taking lamps in series 20 x 50 = 1000. To give twice this voltage requires 2000 *- .75 = 2667 cells in series whose resistance is 2667 X 1.5 = 4000 ohms. To reduce this to 1000 ohms we need 4 such series of cells in parallel giving 10,668 cells. Proof: Current = ^ + 1000 = 1 ampere. Slight discrepancies will be noticed in the current strength given by different calculations. This is unavoidable as a cell cannot be fractioned or di- vided. EFFICIENCY OF BATTERIES. Rule 55. The efficiency of a battery is expressed by di- viding the resistance of the external circuit by the total resistance of the circuit. Efficiency = ^- . xv -t- r BATTERIES. 73 EXAMPLE. A battery consists of 67 cells in series of constants 2 volts i ohm. It supplies 5 lamps in parallel, each 100 volts 200 ohms constants. What is its effi- ciency? Solution: The resistance of the battery is 67 X | = 13.4 ohms. The resistance of the lamps is (Rule 12) *$* = 40 ohms. Therefore the efficiency of the battery is 40.0 -*- (40 + 13.4) = .749 or 74.9*. Rule 5 6. To calculate the number of battery cells and their arrangement for a given efficiency : Express the efficiency as a decimal, multiply the resistance of the external circuit by the complement of the efficiency (1 efficiency) and divide the product by the efficiency; this gives the resistance of the battery. Add the two resistances and multiply their sum by the current to be maintained for the K. M. F. of the battery. Arrange the cells accordingly as near as possible to these require- ments. EXAMPLES. Five lamps, each 100 volts 200 ohms in parallel are to be supplied by a battery of cell constants 2 volts .4 ohm. The efficiency of the battery is to be as nearly as possible 75. Calculate the number of cells and their arrangement. Solution: The constants of the external circuit are 40 ohms (Rule 12) and 100 volts. Applying the rule we have [40 X (1 .75)] *- .75 = ^ = 13* ohms, the resistance of the battery. By Ohm's law the E. M. F. of the battery = (40 + 13*) X 2.5 = 74 ARITHMETIC OF ELECTRICITY. 133* volts. These constants, 13* ohms and 133* volts, require 67 cells in series and 2 in parallel. Proof: a. Of efficiency, by Rule 55, 4Q ^ 13i = .75 or 75#. b. Of number of cells and of their arrange- ment 67 X 2 = 134 volts; (67 X .4) -*- 2 = 13.4 ohms; 134 *- (13.4 + 40) = 2.5 amperes. Rule 57. Where a fractional or mixed number of cells in parallel are called for to produce a given efficiency, take a group of the next highest integral number of cells in parallel and proceed as in Rule 51. EXAMPLE. Assume a current of 3* amperes to be supplied through a resistance of 30 ohms, absorbing 100 volts E. M. F. Let the cell constants of a battery to supply this circuit be 2 volts, * ohm. Calculate the cells and their arrangement for 80 per cent, efficiency. Solution: By Rule 56 efficiency = .80 and 80 X so"' 80 * = ^ ohms, which is the required resist- ance of the battery; 7*+ 30 = 37* ohms are the total resistance of the circuit. By Ohm's law, 37* X 3* = 125 volts, the required E. M. F. of the battery. This requires 63 cells in series, with a resistance for one series of 63 X * = 10* ohms. To reduce this to 7^ ohms ^ = 1.4 cells in parallel are required. As this is a mixed number we take the next highest in- tegral number and place 2 cells in parallel. The constants of this group of 2 cells are 2 volts, & BATTERIES. 75 ohm. Applying Rule 51 we have for the number of such groups in series; 2 _ ( x A) = 58 groups in series. As there are 2 cells in parallefthe total cells are 116, of resistance, 58 X A = 4.83 ohms, and of E. M. F., 58 X 2 = 116 volts. Proof: Of efficiency by Rule 55, ^TtM = 86.1*. Of number and arrangement of cells M"^ = 3.33 amperes. It is to be observed that the efficiency thus obtained is far from what is required. In most cases accuracy can only be attained by arranging the battery irregularly, which is unusual in prac- tice. An example will be found in a later chapter. CHEMISTRY OF BATTERIES. One coulomb of electricity will set free .010384 milligrams of hydrogen. The corresponding weights of other elements or compounds are found by multi- plying this factor by the chemical equivalent, and dividing by the valency of the element or metal of the base of the compound in question. An element or other substance in entering into any chemical combination develops more or less heat, always the same for the same weight and com- bination. The atomic weight of an element or the molecular weight of a compound divided by the val- ency of the element or metal of its base gives the original chemical equivalent. 76 ARITHMETIC OF ELECTRICITY. The quantities of heat evolved by the combination of quantities of substances expressed by their original chemical equivalents multiplied by one gram are termed the thermo-electric equivalents of the ele- ments or substances in question. In the tables it is expressed in kilogram degrees 0. of water (kilogram- calories). From the thermo-electric equivalent of a combi- nation we find the volts evolved by it or absorbed by the reciprocal action of decomposition. Rule 58. The volts evolved by any chemical com- bination or required for any chemical decomposition are equal to the thermo-electric equivalent in kilo* gram-calories multiplied by .043. E = .043 X H. EXAMPLES. What number of volts is required to decompose water ? Solution: From the table we find that the com- bination of one gram of hydrogen with eight grams of oxygen liberates 34.5 calories. Then 34.5 X .043 = 1.48 volts. Rule 59. To determine the voltage of a galvanic couple subtract the kilogram calories corresponding to decompositions in the cell from those corresponding to combinations in the cell for effective energy and multi- ply by .043 for volts. EXAMPLES. Calculate the voltage of the Smee couple. Solution: In this battery zinc combines with BATTERIES. 77 oxygen, giving out 43.2 calories and combines with sulphuric acid, giving 11.7 more calories; a total of 54.9 calories. An equivalent amount of water is at the same time decomposed acting as counter-energy of 34.5 calories. The effective energy is 54.9 34.5 = 20.4 calories. The voltage = 20.4 X .043 = .877 volts. Calculate the voltage of the sulphate of copper battery. Solution: Here we have combination of zinc with sulphuric acid as above 54.9 calories; decomposition of copper sulphate 19.2 + 9.2 = 28.4 .-. 54.9 28.4 = 26.5 calories effective energy 26.5 X .034 = 1.1^ volts. It will be noticed that these results are approxi- mate. Some combinations are omitted in them as either of unknown energy, or of little importance. WORK OF BATTERIES. The rate of work of a battery is proportional to the current multiplied by the electro-motive force. The work is distributed between the battery and the external circuit in the ratio of their resistances as by Eule 55. The horse-power, and heating power are calculated by Rules 30-43, care being taken to dis- tribute the energy acording to the resistance by the following rule: Rule 6O. The effective rate of work or the rate of work in the external circnit of a battery, is equal to t lie 78 ARITHMETIC OF ELECTRICITY. total rate multiplied by the efficiency of the battery ex- pressed decimally. EXAMPLE. 25 cells of 2 volt 1 ohm battery are arranged in series on an external circuit of 250 ohms resistance. What work do they do in that circuit? Solution: The current (Ohm's law) = ~^| = 1 1.818 amperes. Total rate of work = 1.818 X 50 volts = 90.9 watts. Efficiency of battery = m = 90 per cent, (nearly). Effective rate of work = 1.818 X 50 X .90 = 81.81 watts. CHEMICALS CONSUMED IN A BATTERY. Rule 61. The chemicals consumed in grams by a bat- tery for one kilogram-meter (7.23 foot Ibs.) of work are found by multiplying the combining equivalent of the chemical by the number of equivalents in the reaction by the constant .OOO1O1867 and dividing by the pro- duct of the 10. M. F. by the valency of the element In question. - _ Equiv. X n X .000101867 E x valency EXAMPLES. What is the consumption of zinc and sulphate of copper per kilogram-meter of work in a Daniel's battery? Solution: Take the E. M. F. as 1.07 volt. The equivalent of zinc, a dyad, is 65 and one atom en- ters into the reaction. The zinc consumed there- _ 65 XIX. = BATTERIES. 79 The equivalent of copper sulphate, is 159.4. One equivalent enters into the reaction carrying with it one atom of the dyad metal copper. The weight consumed therefore = m " *^ x x f 1M867 = .0076 grams. Add 56.46$ for water of crystallization. . All these quantities are for one kilogram-meter of work (7.23 foot Ibs.) which may be more or less effective according to circumstances as developed in Rules 44, 45, and 60. DECOMPOSITION OF COMPOUNDS BY THE BATTERY. In cases where a compound has to be decomposed by a battery two resistances may be opposed to the work. One is the ohmic resistance of the solution, which is calculated by Rule 16. The other is the electromotive force required to decompose the solu- tion. This is best treated as a counter-electromotive force. Then from the known data the current rate is calculated, and from the electro-chemical equiva- lents the quantity of any element deposited by a given number of coulombs is determined. Rule 62. To calcnlate the metal or other element lib- erated by a given current per given time proceed as fol- lows: Calcnlate the resistance. Determine the counter- electromotive force of the solution by Rule 58 and sub- tract it from the E. Iff. F. of the battery or generator. Apply Ohm's law to the effective voltage thus deter- mined and to the calculated resistances to find the cur- rent, multiply the electro-chemical equivalent of the element by the coulombs or ampere-seconds. 80 ARITHMETIC OF ELECTRICITY. EXAMPLE. A bath of sulphate of copper is of specific resis- tance 4 ohms. The electrodes are supposed to be 10,000 sq. centimeters in area and 5 centimeters apart. Two large Bunsen elements in series of 1.9 volts .12 ohms each are used. What weight in milligrams of copper will be deposited per hour? Solution: By Rule 16 the resistance of the solu- tion is ^j| = 0.023. The electro-chemical equiva- lent of copper is .00033 grams. The thermo-electric equivalent for copper from sulphate of copper is 19.2 -f 9.2 = 28.4 calories. The E. M. F. corresponding thereto = 28.4 X .043 = 1.22 volts counter E. M. F. The E. M. F. of the bat- tery = 1.9 X 2 = 3.8 volts, giving an effective E. M. F. of 3.8 1.22 = 2.58 volts. The resistance of the battery = .12 X 2 = .24 ohms. The current = = 9.8 amperes. This gives per hour 9.8 X 3,600 = 35,280 coulombs, and for copper deposited .00033 X 35,280 = 11.64 grams. In many cases one electrode is made of the mater- ial to be deposited and being connected to the car- bon end of the battery or generator is dissolved as fast as the metal is deposited. In such case there is no counter electro-motive force to be allowed for. BATTEEIES. 81 EXAMPLE. Take the last case and assume one electrode (the anode) to be of copper and to dissolve. Calculate the deposit. Solution: Current = 3.8 -*- (.24 + .023) = 14.4 amperes = 51,840 coulombs per hour = .00033 X 51,840 = 17.10 grams of copper. CHAPTER IX. ELECTRO-MAGNETS, DYNAMOS AND MOTORS. THE MAGNETIC FIELD AND LINES OF FORCE. A CURRENT of electricity radiates electro-magnetic wave systems, and establishes what is known as a field of force. The field is more or less active or in- tense according to the force establishing it. The intensity of a field is for convenience expressed in LINES OF FORCE. These are the units of magnetic intensity, often called units of magnetic flux, and the line as a unit is comparable to the ampere X 10, which is the C. G-. S. unit of current. A line of force is that quantity of magnetic flux which passes through every square centimeter of normal cross- section of a magnetic field of unit intensity. The line is at right angles to the plane of normal cross- section of such field. Such intensity of field exists at the center of curvature of an arc of a circle of ra- dius 1 centimeter, and whose length is 1 centimeter, when a current of 10 amperes passes through this arc. Practically it is the amount which passes through an area of one square centimeter, situated in the center of a circle 10 centimeters in diameter, ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 83 surrounded by a wire through which a current of 7.9578 amperes is passing. The plane of the circle is a cross-sectional plane of the field; a line perpen- dicular to such plane gives the direction of the lines of force, or of the magnetic flux. This cross-sectional area is often spoken of as the field of force. As a field exists wherever there are lines of force, there are in each magnetic circuit either an infinite number of fields of force, or a field of force is a volume and not an area. The number of lines of force or of magnetic flux per unit cross-sectional area of the magnetic cir- cuit, i. e. per unit area of magnetic field, expresses the intensity of the field. In soft iron, it may run as high as 20,000 or more lines per square centimeter of cross-section of the iron which is magnetized. Just as we might speak of a bar of copper acting as conductor for 20,000 C. G. S. units of current, or 2000 amperes, so we may speak of the iron core of a magnet carrying 20,000 lines of force. PERMEANCE AND RELUCTANCE. This action of centralizing in its own material lines of force is analogous to "conductance." It is termed PERMEANCE. Its reciprocal is termed RELUCTANCE, which is precisely analogous to " resistance. " Iron, nickel, and cobalt possess high permeance; the permeance of air is taken as unity. At a low degree of magnetization, soft iron pos- 84 ARITHMETIC OF ELECTRICITY. sesses 10,000 times the permeance of air. At high degrees of magnetization, it possesses much less in comparison with air, whose permeance is unchanged under all conditions. There is no substance of infinitely high reluctance, which is the same as saying that there is no insula- tor of magnetism. MAGNETIZING FORCE AND THE MAGNETIC CIR- CUIT. The producing cause of the magnetic flux or mag- netization just described is in practice always a cur- rent circulating around an iron core. The name of MAGNETIZING FORCE is often given to it. It is the analogue of electro-motive force, and is measured by the lines of force it establishes in a field of air of standard area. A high value for the magnetic force is 585 lines per square centimeter. It is proportional to the amperes of current and to the number of turns the conductor makes around it. Its intensity is often given in ampere-turns. Magnetization always implies a circuit. As far as known, magnetic lines of force cannot exist without a return circuit, exactly like electric currents. But owing to the imperfect reluctance of all materials, the lines of force can complete their circuit through any substance*. In a bar magnet the return branch of the circuit is through air. ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 85 In the same magnetic circuit, the planes of nor- mal cross-section lie at various angles with each other. The law of a magnetic circuit is exactly compar- able to Ohm's law. It is as follows: Rale 63. The magnetization expressed in lines of force is equal to the magnetizing force divided by the reluctance or multiplied by the permeance of the entire circuit. This rule would he of very simple application, ex- cept for the fact that reluctance increases, or per- meance decreases, with the magnetization, and the rate of variation is different for different kinds of iron. Rule 64. Permeability is the ratio of magnetization to magnetizing force, and is obtained by dividing mag- netization by magnetizing force. Permeability has to he determined experimentally for each kind of iron. It is simply the expression of a ratio of two systems of lines of force. It always exceeds unity for iron, nickel, and cobalt. The specific susceptibility of any particular iron to magnetization is its permeability. The susceptibil- ity of a portion, or of the whole of a magnetic cir- cuit is its permeance. GENERAL RULES FOR ELECTRO-MAGNETS. The traction of a magnet is the weight it can sustain when attached to its armature. It is pro- 86 ARITHMETIC OF ELECTRICITY. portional to the square of the number of lines of force passing through the area of contact. Rule 65. The traction of a magnet In pounds is equal to the square of the number of lines of force per square inch, multiplied by the area of contact and divided by 72,134,000. In centimeter measurement the traction in pounds is equal to the square of the number of lines of force per square centimeter multiplied by the area of contact and divided by 11,183,000. The traction in grams is equal to the latter dividend divided by 24,655 .s JT A 981); for dynes of traction the divisor is 25.132 EXAMPLES. A bar of iron is magnetized to 12,900 lines per square inch; its cross-section is 3 square inches. What weight can it sustain, assuming the armature not to change the intensity of magnetization? Solution: 12,900 2 X 3 = 499,230,000. This di- vided by 72,134,000 gives 6.914 Ibs. traction. A table calculated by this rule is given. A diminished area of contact sometimes increases trac- tion, and a non-uniform distribution of lines may occasion departures from it. The above rule and the table alluded to are practically only accurate for uniform conditions. The reciprocal of the rule is applied in determining the lines of force of a magnet experimentally. Rule 66. The lines of force which can pass through a magnet core with economy are determined by the tables, keeping in mind that it is not advisable to let the per- meability fall below 200 30O. From them a number is taken (40,000 lines per square inch for cast iron or ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 87 1OO,000 lines per square incli for wrought iron are good general averages) and is multiplied by the cross-sec- tional area of the magnet core. Rule 67. To calculate the magnetizing force in am- pere turns required to force a given number of magnetic lines through a given permeance, multiply the desired lines of force by the reluctance determined as below. Rule 68. a. The reluctance of a core or of any portion thereof for inch measurements is equal to the product of the length of the core or of the portion thereof by O.3132 divided by the product of its cross-sectional area and permeability. 6. The reluctance for centimeter measurements is equal to the length of the core divided by the product of 1.2566, by the cross-sectional area and the permea- bility. EXAMPLES. 440,000 lines are to be forced through a bar of wrought iron 10 inches long and 4 square inches in area; calculate its reluctance and the magnetizing force in ampere turns required to effect this mag- netization. Solution: The reluctance (a) = 10 X .3132 *- (4 X permeability). 440,000 lines through 4 square inches area is equal to 110,000 lines through 1 square inch; for this intensity and for wrought iron the permeability = 166. 166 X 4 = 664. The reluc- tance therefore = 3.132 -* 664 = .0047. The mag- netizing force in ampere turns = 440,000 X .0047 = 2068. The same number of lines are to be forced through a bar 25.80 square centimeters area and 88 ARITHMETIC OF ELECTRICITY. 25.40 centimeters long. Calculate the ampere turns. Solution: 440,000 lines through 25.80 sq. cent. = 17,054 through 1 sq. cent., for which the permeability = 161. The reluctance therefore, (b) = 25.40 - (1.2566 X 25.80 X 161) =.0048. The ampere turns = 440,000 X .0048 = 2112. MAGNETIC CIRCUIT CALCULATIONS. Practically useful calculations include always the attributes of a full magnetic circuit, because mag- netization can no more exist without a circuit than can an electric current. In practice an electro- magnetic circuit consists of four parts: 1, The magnet cores; 2 and 3, the gaps between armature and magnet ends; 4, the armature core. To cal- culate the relations of magnetizing force to magne- tization the sum of the reluctances of these four parts has to be found. A further complication is introduced by leakage. The permeability of well magnetized iron being so low, not exceeding 150 to 300 times that of air, a quantity of lines leak across through the air from magnet limb to magnet limb. Leakage is included in the sum of the reluctances by multiplying the reluctance of the magnet core by the coefficient of leakage, which is calculated for each case by more or less complicated methods. For parallel cylindrical limb magnets the calculation is ELECTRO-MAGNETS, BYNAMOS AND MOTORS. 89 exceedingly simple. The calculation in all cases is simplified by the fact already stated, that in air permeability is always equal to unity, whatever the degree of magnetization. For copper and other non-magnetizable metals the variation from unity is so slight that it may, for practical calculations, be treated as unity. LEAKAGE OF LIKES OF FORCE. Leakage is the magnetic flux through air from surfaces at unequal magnetic potential, such as north and south poles of magnets. It is measured by lines of force and is proportional to the relative permeance of its path. The coefficient of leakage of a magnetic circuit is the quotient obtained by dividing the total magnetic flux by the flux through the armature. The total magnetic flux is the maximum flux through the magnet core. Rule 69. To obtain the coefficient of leakage divide the permeance of the armature core and of the two gaps plus one-half the permeance of air between magnet limbs by the permeance of the armature core and of the two gaps. EXAMPLE. The total flux through an armature core is found to be at the rate of 70,000 lines per square inch, and the armature core is 3 inches diameter and 10 inches long. The average length of travel of the magnetic 90 ARITHMETIC OF ELECTRICITY. lines through it is H inches. The air gaps are 10 X 3 inches area and i inch thick. The permeance be- tween the limbs of the magnet is 500. Calculate the coefficient of leakage. Solution: 70,000 lines per square inch gives a permeability of 1,921. By Kule 68 the reluctance of the armature core is ^ J* im X .3132 = .000008. The reluctance of a single air gap is ~ X .3132 = .0052. Thus the armature reluctance is so small that it may be neglected. The permeance of the two air gaps is given by 005a 1 xg = 100 (about). The coefficient of leakage = 100 + ) 250 = 3.5. As the coefficient of leakage is the factor used in these calculations, the permeance of the leakage paths is the desired factor for its determination. In the case of cylindrical magnet cores parallel to each other, they are obtained from Table XIII. given in its place later. It is thus calculated and used. The least distance separating the cores (b) is di- vided by the circumference of a core (p) giving the ratio (~) of least distance apart to perimeter of a core. The number corresponding in columns 3 or 5 is multiplied by the length of a core. The product is the permeance. Columns 2 and 4 give the re- luctance. To reduce to average difference of mag' netic potential divide by 2. ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 91 EXAMPLE. Calculate the permeance between the legs of a magnet, 3 inches in diameter and 12 inches high and 5 inches apart. Solution: The perimeter = 3 X 3.14 = 9.42. ^ = 1 or .5 nearly. From the table of permeability we find 6.278. Multiplying this by 12 we have 6.278 X 12 = 75.336, the permeance. Dividing by 2 we have ^f^ = 37.668, the permeance for use in the calculation of leakage coefficient. It will be observed that this calculation is based entirely on the ratio stated, and that absolute di- mensions have no effect on it. For flat surfaces, parallel and facing each other, the following method precisely comparable to the rule for specific resistance is used: Rule 70. The permeance of the air space between flat parallel surfaces is equal to their average area multi- plied by 3.193 and divided by their distance apart, all in inch measurements. EXAMPLE. Determine the permeance between the two facing sides of a square cored magnet 15 inches long, 3 inches wide and 8 inches apart. Solution: 3 X 15 = 45 (the average area); 45 X 3.193 -s- 8 = 17.96. For use in calculations it should be divided by 2 giving 8.98. This division by 2 is to reduce it to the average difference of mag- netic potential between the two magnet legs. 92 ARITHMETIC OF ELECTRICITY. CALCULATIONS FOR MAGNETIC CIRCUITS. A magnetic circuit is treated like an electric one. The permeance (analogue of conductance) or reluc- tance (analogue of resistance) is calculated for its four parts, magnet core, two air gaps, and armature core. The leakage coefficient is determined and ap- plied. The requisite magnetizing force is calcu- lated in the form of ampere turns (the analogue of volts of E. M. F.). The preceding leakage rules cover the case of parallel leg magnets. For others a slight change is requisite in the leakage calcula- tions, but in practice an average can generally be estimated. EXAMPLE. Assume the magnet and armature of a dynamo. The magnet is of cast iron, each leg is cylindrical in shape, 4 inches in diameter and 20 inches high. From center to center of leg the distance is 9 inches. The armature core of soft wrought iron is 4 inches in diameter and 8 inches long, the pole pieces curv- ing around it are 4 inches, measured on the curve inside, by 8 inches long. The air gap is i inch thick. Calculate the reluctance of the circuit and the ampere turns for 500,000 lines of force. Solution: The pole pieces approach within 2$ inches of each other. This leaves l inches of the diameter of the armature core embedded or included within or embraced by them. One-half of this -ELECTRO-MAGNETS, DYNAMOS AND MOTOES. 92 amount may be taken and added to 2i giving 3J as the average depth of core for an area 4 X 8 = 32 square inches. The lines per square inch of arma- ture core are ^^ = 15,625 lines per square inch. By the table of permeability, 4650 is given for per- meability for 30,000 lines in soft iron. For 15,625 lines per square inch 9,000 can safely be taken for permeability. Its relative reluctance is therefore 32x^000 = .000011 relative armature core reluctance. (i) The relative reluctance of one air gap (permeabil- ity = 1) is i -*- 32 = .0078 and .0078 X 2 = .0156 =s air gaps reluctance (2). The ratio - of the table for determining the leak- age between cylindrical magnet legs is 4 X 5 3 ^ = .4. 5 is the distance between the legs. Permeance cor- responding thereto is 6.897, which multiplied by 20, the length of the legs, and divided by 2 for average magnetic potential difference gives 68.97 for rela- tive effective permeance (3). The relative reluctance of the air gaps and arma- ture core is .015611; the reciprocal or permeance is 64.06 (4). For coefficient of leakage we have (64.06 -f- 68. 97) s- 64.06 = 2.08 (5). To find the relative reluctance of the magnet core whose yoke may be taken as of mean length 9 inches 94 ARITHMETIC OF ELECTRICITY. and of area equal to that of the core (3.14 X & 12.56) we have to first determine the permeability. 12 ' 56 = 40,000 lines per square inch, corresponding to a permeability of 258. For the effective reluc- tance of the magnet core introducing the factor of leakage (2.08) we have the expression (20+ 12 2 5 + X 9) ^ 8 2 - 08 = .0314. To get ampere turns, we add the reluctances of circuit, multiply by .3132 and by the required lines, (.000011 + .0156 + .0314) X .3132 X 500,000 = 7362 ampere turns required. In the above calculations, the multiplication by .3132 was omitted to save trouble, relative reluc- tances only being calculated, until the end when one multiplication by .3132 brought out the ampere turns. The leakage appears excessive partly be- cause of the high reluctance of the two air gaps. These should be increased in area and reduced in depth if possible. The leakage is also high on account of the legs of the magnet being close to* gether. Were these separated, a larger armature core might be used, justifying a lower speed or rota- tion of armature, reducing reluctance of air gaps by increasing their area, and reducing leakage between magnet legs by increasing their distance. The magnet legs might also be made shorter, thus reduc- ing leakage. ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 95 Thus assume the magnet core of the same cross- sectional area, but only 10 inches long and with a distance apart of legs of 7 inches, giving a 7 X 10 inch armature core and pole piece areas (air gap areas) of 7 X 10 = 70 sq. inches. For leakage ratio we have () = ^ = .56 giv- ing from the proper table 6.000 (about), ***** 10 = 30 relative permeance of air space between legs. For air gaps reluctance -5- 70 = .00357 which for the two gaps gives .00714 relative reluctance. Treating the armature core as a prism 7 X 10 = 70 sq. inches area and 5 inches altitude, we have for lines per sq. inch 500,000 * 70 = 7000 giving it about 9000 and reluctance as 5 *- (70 X 9,000) = .000008 reluctance. Air gaps and armature core reluctance .007143 and permeance = -^^ 139. Coefficient of leakage = 1 -^^ ) = 1.21. If the depth of the air gaps was reduced to $ inch the coefficient of leakage would then be about 1.11. Every surface in a magnet leaks to other surfaces and the leakage from leg to leg is sometimes but one third of the total leakage. In practice the total leakage often runs as high as 50#, giving a coefficient of 2.00 and in other cases as low as 25#, giving a co- efficient of 1.33. 96 ARITHMETIC OF ELECTRICITY. DYNAMO AKMATUKES. An armature of a dynamo generally comprises two parts the core and the winding. The core is of soft iron. Its object is to direct and concentrate the lines of force, so that as many as possible of them shall be cut by the revolving turns or convolu- tions of wire. The winding is usually of wire. It is sometimes, however, made of ribbon or bars of copper. Iron winding has also been tried, but has never obtained in practice. The object of the wind- ing is to cut the lines of force, thereby generating electro motive force. The number of the lines of force thus cut in each revolution of the armature is determined from the intensity of the field per unit area, and from the position, area and shape of the armature, coils and pole pieces. The number thus determined, multiplied by the number of times a wire cuts them in a second, and by the effective number of such wires, gives the basis for determin- ing the voltage of the armature. Rule 71. One volt E. OT. F. Is generated by the cutting of 10 8 (100,000,000) lines of force in one second. EXAMPLES. A single convolution of wire is bent into the form of a rectangle 7 X 14 inches. It revolves 25 times a second in a field of 20,000 lines per square inch. What E. M. F. will it develop at its terminals? Solution: The area of the rectangle is 7 X 14 = 98 ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 97 square inches. Multiplying this by the lines of force in a square inch, we have 98 X 20,000 = 1,960,000. Each side of the rectangle cuts these lines twice in a revolution, and makes 25 revolutions in a second. This gives 25 X 2 X 1,960,000 = 98,000,000 lines cut per second, corresponding to 98 X 10* X 10"* = 98 x 10~ 2 = t& volts E. M. F. generated, or 4Afttfftft - & volts. The field of the earth in the line of the magnetic dip .5 line per square centimeter. Calculate a size, number of layers, and speed of rotation for a one volt earth coil. Solution : We deduce from the rule the following : Area of coil X revolutions per sec. X convolutions of wire X .5 X 10~ 8 = .5. We may start with revolu- tions per second, taking them at 20. Next we may take 50,000 convolutions. 20 x 50,000 X .5 = 500,000. This must be multiplied by 200 to give 10 8 ; in other words, the average area within the wire coils must be 100 square centimeters, or 10 X 10 centimeters. 2 x 100 X 2000 X 500 X .5 = 10", and 10" X 10' 8 = 1 volt. Rule 72. The capacity f an armature for current is determined by tli e cross-section of its conductors. This should be such as to allow 52O square mils per ampere = 1923 amperes per square inch area. EXAMPLE. A drum armature coil is of 4 inches diameter, and is wound with wire -$fa of the periphery of the 98 ARITHMETIC OF ELECTRICITY. drum in diameter; the wire is 100 feet long. Its E. M. F. is 90 volts. What is the lowest admissible external resistance? Solution: The circumference of the drum is 3.14 X 4 = 12.56 inches. The diameter of the wire is ^ = .0418 in. or 42 mils. The area of the wire is 2 1 2 X 3.14 = 1387 square mils. By the rule the allowable current in amperes for a single lead of such wire is *ffi = 2.66 amperes. But on a drum armature the wire lies with two leads in parallel. Hence it has double the above capacity or 2.66 X 2 = 5.32 amperes. The resistance of such wire may be taken at .137 ohm. By Ohm's law the total re- sistance for the current named must be ^ or 17 ohms. The external resistance is given by 17 ^- .137 = 16.863 ohms. These two rules enable us to calculate the ca- pacity of any given armature. Certain constants depending on the type of armature have to be intro- duced in many cases. DRUM TYPE CLOSED CIRCUIT ARMATURES. For these armatures the following rules of varia- tion hold, when they do not differ too much in size, and are of identical proportions. Rule 73. a. The E. M. F. varies directly with the qu arc of the size of core and with the number of turns of wire. ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 99 b. The current capacity varies with the sixth root of the size of core for identical !:. M. F. c. The resistance varies directly with the cube of the number of turns and inversely with the size of core. d. The amperage on short circuit varies directly with the cube of the size and inversely with the square ot the number of turns. In these rules the proportions of the drum are supposed to remain unchanged. Size may be re- ferred to any fixed factor such as diameter, as lineal size is referred to. These rules enable us to calculate an armature for any capacity and voltage. As a starting point a given intensity of field, speed of rotation, and num- ber of turns of wire and size of wire has to be taken. The wire is selected to completely fill the periphery of the drum. Then a trial armature is calcu- lated of the required voltage and its amperage is calculated. With this as a basis, by applying Eule 73, sections a and #, the size of an armature for the desired current capacity is calculated, the E. M. F. being kept identical. As a standard for medium sized machines 20 of the turns of wire may be con- sidered inactive. EXAMPLE. Calculate a 100 volt, 20 ampere armature, whose length shall be twice its diameter, to work at a speed of 15 revolutions per second. Solution: Take as intensity* of field 20,000 lines per square inch. Allow 80 of active turns of 100 ARITHMETIC OF ELECTRICITY. wire. Start with a core 8 X 16 = 128 square inches, including 128 X 20,000 = 256 X 10* lines of force. The given speed is 15 rotations per second. For the number of active turns of wire per volt we have to divide 10 8 or 100,000,000 by one half the lines of force cut by one wire per second. This number is 256 X 10 4 X 15, or 38,400,000; and ^JoSw = 2 ' 6 turns - For 10 volts > therefore, 260 active turns are needed. If one half the lines were not taken the result would be one half as great, be- cause each line cuts each line of force twice in a revolution, and in the computation a single cutting per revolution only is allowed for. The reason for thus taking one half the lines cut by a single wire as a base is because in the drum arma- ture the wires work in two parallel series, giving one half the possible voltage. The actual turns are 260 *- .80 = 325, say 324 turns. Assume it to be laid in two layers giving 162 turns to the layer. The space occupied by a wire is equal to the peri- meter divided by the number of wires or T? = .154 in. Allowing 25# for thickness of insulation, lost space, etc., we have .115 in. or 115 mils as the diameter of the wire. In the drum armature as just stated the wire is parallel, so that the area of one lead of wire has to be doubled, giving 10,573 X 2 square mils as the area of the two parallel leads. This is enough for 40 amperes or double the amper- age required. This capacity is reached by taking ELECTRO-MAGNETS, D?-NA#f)A.A&J)WOrOSc,. 101 520 square mils per ampere as the proper cross-sec- tional area of the wire. (Rule 72.) We must therefore reduce the size to give ^ the ampere capacity; this reduction (by Rule 73 Z>) is in the ratio 1 : j* ~ 1 : .89018 ; the size therefore is 8 X .89018 diameter by 16 X .89018 length = 7.12 X 14. 24 inches. Applying a for voltage we have for the same num- ber of turns on the new armature a voltage in the ratio of 1 : .89018 2 or about & of that required. We must therefore divide the number of turns in the trial armature by .89018 2 , giving for the number of turns H^ = 409, say 410 turns. To prove the operation we first determine the voltage of the new armature. Its area is 7.12 X 14.24 = 101.4 square inches including 2,028,000 lines of force. The active wires are 410 X .8 = 328. We have for the voltage = .^,000^1^x15 = 99>78 volts. The relative capacity of the wire is deduced from the square of its diameter. The circumference of the new armature is 7.12 X 3.14 = 22.3568. There are 205 turns in a layer giving as diameter of wire ^^ = .1091 mils. This must be squared, giving .0119, and compared with the square of the corre- sponding number for the original armature. This number was 25 -* .162 = .154 inch. .154 2 = .02371 102 A?1T^'IFT-:C OF ELECTRICITY. and .01190 *- .02371 = i (nearly), showing that the new armature has one half the ampere capac- ity of the old, or 40 X | = 20 amperes as re- quired. The gauge of the wire is reached by making the same allowance for insulation and lost space, viz., 25#. .1091 X .75 = .0818 in. or 81.8 mils diameter, for size of wire. Of course there is nothing absolute about 25# as a loss coefficient; it will vary with style of insulation and even to some extent with the gauge of wire. But as Eule 73 is based upon the assumption that this loss is a constant proportion of the diameter of the wire, too great a variation of sizes should not be allowed in its application. In other words the trial armature should be as near as possible in size to the final one. Suppose on the other hand that an armature for 100 amperes was required. This is for 2^ times 40 amperes (the capacity of the first calculated or trial armature). Applying b we extract the 6th root of 2^. = 1.1653 (by logarithms or by a table of 6th roots). The size of the new armature is therefore 8 X 1.1653 by 16 X 1.1653 or 9.3224 X 18.6448 inches. Applying a for voltage we have for the same num- ber of turns of the new armature a voltage in the ratio of 1.1653 2 : 1 or 1.358 times too great. We ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 103 must therefore multiply the original turns by the reciprocal of 1.358, giving ^ = 239 turns. To prove the voltage, we multiply 239 by .8 for the active turns of wire, giving 191.2 turns. The area of the armature is 9.32 X 18.64 = 172.7 square inches. For voltage this gives 172.7 X 191.2 X 20,- 000 X 15 X 10- 8 = 98.6 volts (about). To prove the capacity we must divide the circum- ference of the new armature, 9.32 X 3.14 = 29.26 inches, by the turns of wire in one layer, 4* = 120 turns (about). This gives a diameter of 244 mils (nearly). The ratio of capacities of the original and this wire is .2442 -5- .1542 i nc h es = .059536 -*- .02371 =2.51 corresponding to 40 X 2.51 = 100 amperes. These results, owing to omissions of decimals, do not come out exactly right and it is quite unneces- sary that they should. The accuracy is ample for all practical purposes. For armature dimensions it would be quite unnecessary to work out to the sec- ond decimal place. It would answer to take as ar- mature sizes in the two cases given 7 X 14^ inches and 9^ X 18# inches. It is also to be noted that a very low rate of rota- tion was taken. 25 to 30 turns per second would not have been too much. The latter would give double -the voltage and the same amperage. FIELD MAGNETS OF DYNAMOS. The calculation for a magnetic circuit given on 104 ARITHMETIC OF ELECTRICITY. pages 92 et seq., is intended to supply an example of the calculation of the circuit formed by a field mag- net and its armature, such as required for dynamos. The leakage of lines of force is and can only be so incompletely calculated that it is probably the best and most practical plan to assume a fair leakage ratio and to make the magnet cores larger than re- quired by the lines of force of the armature in this ratio. ( A low multiplier to adopt is 1.25, which is lower than obtains in most cases; 1.50 is probably a good average. Rule 74. The cross-sectional area of the field-magnet cores is equal to the lines of force in the field divided by the magnetic flux (column B) for the material selected and corresponding to the chosen permeability GU.), mul- tiplied by the leakage coefficient. A good range for permeability is from 200 to 400 giving for wrought iron from 100,000 to 110,000 lines of force per square inch and for cast iron from 35,000 to 45,000 lines per square inch; for the field from 15,000 to 20,000 lines per square inch may be taken. The permeability table gives data for different qualities of iron. EXAMPLE. Taking the 100 volt 100 ampere armature last cal- culated, determine the size of field-magnet cores to go with it, and the ampere turns and other data. Solution: Assume 20, 000 lines of force per square ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 105 inch in the field, 45,000 in cast iron and 110,000 in wrought iron core and a leakage coefficient of 1.25. We have for total lines of force passing through armature 172.7 X 20,000 = 3,454,000; cross-sectional area for cast iron core ^^ X 1.25 = 96 square inches; cross-sectional area for wrought iron core X 1.25 = 39 square inches. As length of cores we may take 20 inches with a distance between them of 10 inches. Assume wrought iron to be selected. If cylindrical they would be 7 inches in diameter to give the required cross-sectional area. The yoke connecting them would average in length 10 + 7 = 17 inches, giving for magnet cores and yoke a length of 17 + 20 + 20 = 57 inches. The reluctance of cores and yoke (Rule 68) - Vx'aro (taking A = 200) which reduces to .00132 (1). The armature area is 172.7 inches. As average length of the path of lines of force through it 5 inches may be taken. As it passes only 20,000 lines of force per square inch of field its permeability is high, say 9000. Its reluctance is given by ^^'Soo* This is so low that it may be neglected. The area of each air-gap may be taken as 173 square inches, and of depth of two windings plus about -rV inch for clearance or windage giving (.224 X 2) + .1 = about .6 inch for its depth. Its 106 ARITHMETIC OF ELECTRICITY. reluctance is ^-~^=* .00108. As there are two air 17o gaps we may at once add their reluctances giving .00216(2). " By Rule 67 the ampere turns are equal to the product of the reluctances (1) and (2), by the lines of force giving (.00121 + .00216) X (172.7 + 20,000) = 11640 ampere turns. The proper size of wire for series winding may be determined by Sir William Thompson's rule that in series wound dynamos the resistance of the field mag- net windings should be & that of the armature. The length of the wire in the armature is equal approxi- mately, to the circumference 9.32 X 3.14 = 29.26 multiplied by the number of turns (240) giving 29.26 X 240 = 7022 inches. The wire turns on the field magnets are found by dividing the ampere turns by the amperes giving ^W 4 = 232 turns. The circumference of the mag- net leg is 7.0 X 3.14 = 22 inches. The total length of wire is therefore, approximately, 232 X 22 = 5104 inches. To compare the resistances we must use *^P for the length of the armature wire, because it is in parallel, and therefore is \ the length and j the re- sistance of the full wire in one length. Dividing by 4 introduces this factor. As the resistances of the wires are to be in the ratio of 2 : 3, we have by Rule 13 (calling the thickness of armature wire 244 X .75 = 183 mils to allow for ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 107 insulation, etc.), 2:3 :: 183 2 X 5104 : z 2 X ***-, and solving we find x> = 73026 . . x = 270 mils. For shunt winding Sir William Thompson's rule is that the product of armature and field resistance should equal the square of the external resistance. The latter may be taken (Ohm's law) as equal to = l ohm - Fr P erl y the armature resist- ance should be allowed for, but it is so small that it need not be included. We have therefore, arma- ture resistance X field resistance = I 2 X 1. The armature resistance is .0419 ohms. There- fore the field resistance is -^gj = 24 ohms. The cur- rent through this is equal to *$ = 4 amperes (nearly). Therefore n | 39 = 2910 turns of wire are needed. The length of such wire will be ^ ^^ = 5335 feet. The resistance is about 4.4 ohms per 1000 feet corresponding to about .48 mils diameter. THE KAPP LINE. Mr. Gisbert Kapp, C. E. who has given much in- vestigation to the problems of the magnetic circuit and especially to dynamo construction, is the orig- inator of this unit. He considered the regular C. G. S. line of force to be inconveniently small. He adopted as a line of force the equivalent of 6000 C. G. S. lines and as the unit of area one square inch. Therefore to reduce Kapp lines to regular lines of force they must be multiplied by 6000, and ordi- 108 ARITHMETIC OF ELECTRICITY. nary lines of force must be divided by 6000 to obtain Kapp lines. These lines are often used by English engineers. The regular system is preferable and by notation by powers of ten can be easily used in all cases. CHAPTER X. ELECTRIC RAILWAYS. SIZES OF FEEDERS. To calculate the sizes of feeders for a trolley line Eules 23, 24, and 25 in Chapter V. will be found use- ful in conjunction with the following ones: A. For load at end of feeders: Rule 7o. The cross-section of the feeder in cir- cular mils is equal to the product of 1O.79 times the current in amperes times the length of the con- dnctor in feet, divided by the allowable drop in volts. . EXAMPLE. What should be the cross section of a feeder 3,000 feet long carrying 90 amperes with 35 volts drop? Solution: 3,000 X 90 X 10.79 = 2,913,300. Di- viding this by 35 gives 83,237 circular mils. This would correspond to a No. 1 wire, which has 83,694 cir. mils area. All computations of this kind should be checked by table XVI. of current capacity on page 153 in order to be sure that the wire will not become heated above the allowable limit. 110 ARITHMETIC OF ELECTRICITY. Beferring the above example to this table, it is seen that a No. 1 wire will carry 80 amperes with a rise in temperature of 18 F., and 110 amperes with a rise of 36 F. Hence the current of 90 amperes will cause a rise of 24 F. This is calculated by simple proportion; subtracting 80 from 90 and from 110 gives 10 and 30 as the respective differences and shows 90 to lie at just one-third the distance from 80 to 110. Hence the resulting temperature rise will be at one-third the distance between 18 and 36. The difference between these last two figures is 18, one-third of which is 6. Add this 6 to 18 gives us 24 F. as the answer. It is often desirable to compute the drop on a feeder carrying a given current; this is done by the following : Rale 76. The drop in volts on. any conductor is found by multiplying together 1O.79, the current in amperes and its length in feet, then divide this product by its area in circular mils. EXAMPLE. What is the drop on a feeder 2,800 feet long, of 105,592 cir. mils area and carrying a current of 125 amperes ? Solution: 10.79 X 125 X 2,800 = 3,776,500. Dividing this product by 105,592 gives 35.76 volts drop. B. For a uniformly distributed load: ELECTRIC RAILWAYS. Ill The effect of a uniform distribution of load along a main or trolley wire is the same as that of half the total current passing the full length of the wire; hence we require but half the cross-section needed to deliver the current at the extremity of the wire. This is readily done by substituting the constant 5.4 in place of 10.79 in the foregoing rules. In designing electric railway circuits where the track forms the path for the return current the rails should be of ample area and well bonded, with an extra bare wire connected to the bonds and materially reducing the drop in the track circuit POWER TO MOVE CABS. At ordinary speeds on a level track in average con- dition it is safe to assume that the force necessary to move a car is 30 pounds per ton of weight of car. Rale 77. To find the force required to pull or push a car on a level track in average condition, multiply the weight of the car in tons by 3O. EXAMPLE. Find the force required to drag a car weighing 7 tons on a level track. Solution: 7 X 30 = 210 Ibs. Ans. Should it be required to find the force needed to start a car on a level, or to propel it when round- ing a curve, substitute the constant 70 in place of 30 in the foregoing rule. 112 ARITHMETIC OF ELECTRICITY. EXAMPLE. What force is needed to start an 8 ton car on a level track? Solution: 8 X 70 = 560 pounds. Ans. As the above does not take into account the speed of the car we shall have to add this factor in order to find the horse power needed to move it; we will also allow for the efficiency of the motors. Rule 78. To find the horse power required to move a. car along? a level track multiply together the dis- tance in feet traveled per minute and the force in pounds necessary to move the car (as found by Rule 77), and divide the result by 33,OOO times the ef- ficiency of the motors* EXAMPLE. What horse power is needed to propel a loaded car weighing 9 tons along a level track at the rate of 800 feet per minute, with motors of 70 per cent, efficiency ? Solution : Force to move car is 9 X 30 = 270 pounds. The product of 800 X 270 = 216,000 foot pounds per minute. Dividing this by 33,000 gives 6.54 H. P. required to propel the car. Dividing by the efficiency .70 gives 9.34 H. P. to be delivered to the motors. It will be noted in this solution that the quantity 216,000 should, according to the rule, have been di- vided by the product of 33,000 times .70; it was, ELECTRIC RAILWAYS. 113 however, divided by these two factors successively in order to show the difference between the power actu- ally moving the car and that supplied to the motors. In computing the power taken by a car ascending a grade the equivalent perpendicular rise of the car together with its weight in pounds have to be consid- ered in addition to the factors involved in the rule just preceding. Rule 79. To find the horse power required to pro- pel a car up a grade, take the product of the perpendicular distance in feet ascended by the car in one minute multiplied by its freight in pounds; to this add the product of the horizontal distance in feet traveled in one minute multiplied by the force in pounds required to propel the car; divide this sum by 33,OOO times the efficiency of the mo- tors. Note. The grade of a road or track is generally stated as being so many per cent. This means that for any given horizontal travel of a car its change of altitude when referred to a fixed horizontal plane is expressed as a certain percentage of the horizontal travel. For illustration; if a car while traveling horizontally 100 feet has a total vertical rise (or fall) of 7 feet, the incline on which it moves is termed a 7 per cent grade. EXAMPLE. Find the electrical horse power taken by the motors of an 8 ton car to propel it up a 5 per cent grade at 114 ARITHMETIC OF ELECTRICITY. a speed of 1,000 feet per minute, the motors having 70 per cent efficiency. Solution: Perpendicular rise of car is 1,000 feet X -05 = 50 feet. Weight of car in pounds is 8 X 2,000 = 16,000 pounds. Product of lift and weight is 50 X 16,000 = 800,000 foot pounds. Force re- quired to propel car is 30 X 8 = 240 pounds. Pro- duct of force and distance is 240 X 1,000 =240,000 foot pounds. Sum of the two products is 800,000 + 240,000 = 1,040,000 total foot pounds. Product of 33,000 by efficiency is 33,000 X .70 = 23,100. Electrical H. P. is the quotient of 1,040,000 -f- 23,100 = 45.021 H. P. Ans. CHAPTEE XL ALTERNATING CURRENTS. By far the greater part of calculations in the do- main of alternating currents lie in the realm of trigonometry and the intricacies of the calculus. On this account it is hoped that the following pres- entation of some of the simpler formulae may prove welcome to the craft. A current flowing alternately in opposite directions may be considered as increasing from zero to a certain amount flowing in, say, the positive direc- tion, then diminishing to zero and increasing to an equal amount flowing in the negative direction and again decreasing to a zero value. This action is repeated indefinitely. The sequence of a positive and negative current as just described is called a cycle. The frequency of an alternating current is the number of cycles passed through in one second. An alternation is half a cycle. That is to say, an alter- nation may be taken as either the positive or the negative wave of the current. 116 ARITHMETIC OF ELECTRICITY. The frequency may be expressed not only in cycles per second but in alternations per minute. Since one cycle equals two alternations we can interchange these expressions as follows: Rule 8O. A. Having given the cycles per second, to find the alternations per minute multiply the cycles per second by 12O. B. Having given the al- ternations per minute, to find the cycles per second divide the alternations per minute by 12O. EXAMPLES. If a current has 60 cycles per second, how many alternations are there per minute? Solution: 60X120 7,200 alternations. A current has 15,000 alternations per minute; how many cycles per second are there ? Solution: 15,000-^120 = 125 cycles per second. A bipolar dynamo having an armature with but a single coil wound upon it (like an ordinary mag- neto generator) gives one complete cycle of current for every revolution of the armature. That is to say, its frequency equals the number of revolutions per second. A four-pole generator will have a fre- quency equal to twice the revolutions per second, etc. Rule 81. To find the frequency of any alternator, divide the revolutions per minute by 6O and multiply the quotient by the number of pairs of poles in the field. EXAMPLE. Find the frequency of a 16-pole alternator run- ning at 937.5 revolutions per minute. ALTERNATING CURRENTS. 117 Solution: 937.5 -r- 60 = 15.625 rev. per second. 15.625 X 8 = frequency of 125 cycles per second. Electrical measuring instruments used on alternat- ing currents do not indicate the maximum volts or amperes of such circuits, but the effective values are what they show. These effective values are the same as those of a continuous current performing the same work. Rule 82. The maximum volts or amperes of an alternating: current may be found by ' multiplying: the average volts or amperes by 1.11. Reciprocally, the average values can be found by taking; .707 times the maximum values. Xote that these figrures are strictly true only for an exactly sinusoidal current. EXAMPLE. Find the maximum pressure of an alternating current of 55 volts. Solution: 55 X 1.11 = 61.05 volts. Ans. SELF-INDUCTION. In an alternating current circuit the flow of a current under a given voltage is determined not only by the resistance of the conductor in ohms but also by the self-induction of the circuit. Suppose a current to start at zero and increase to 10 amperes in a coil of 1,000 turns of wire. This magnetizing force, growing from zero to 10,000 ampere-turns, surrounds the coil with lines of force whose action 118 ARITHMETIC OF ELECTRICITY. upon the current in the coil is such as to resist its increase. Conversely, when the current is decreas- ing from 10 amperes to zero, the lines of force change their direction and tend to prolong the flow of current. This opposing effect which acts on a varying or an alternating current is caller the counter E. M. F. or E. M. F. of self-induction and is meas- ured in volts. Rule 83. The K. M. F. of self-induction of a given coil is found by multiplying tog-ether 12.5664, the total number of turns in the coil, the number of turns per centimeter length of the coil, the sec- tional area of the core of the coil in square centi- meters, the permeability of the magnetic circuit and the current; divide the resulting product by 1,OOO,OOO,OOO, multiplied by the time taken by the current to reach its maximum value. EXAMPLE. Find the volts of counter E. M. F. in a coil of 300 turns wound uniformly on a ring made of soft iron wire, the ring having a mean circumference of 60 centimeters and an effective sectional area of 25 square centimeters; its permeability to be taken as 200, and a current of 5 amperes in the coil requires .02 second to reach its maximum value. Solution: Product of 12.5664 X 300 X 5 X 25 X 200 X 5 is 471,240,000. Product of 10 X -02 is 20,000,000. Dividing the former product by the latter gives 23.562 volts, Answer. The coefficient of self-induction or, as it is more ALTERNATING CURRENTS. 119 frequently termed, the inductance of a coil, is meas- ured by the number of volts of counter E. M. F. when the current changes at the rate of one ampere per second. (Atkinson.) The unit of inductance is the henry. Rule 84. The inductance of a coil is found by multiplying? tog-ether 12.5664, the total number of turns in the coil, the number of turns per centimeter lens th, the sectional area of the core of the coil in square centimeters, and the permeability of the magnetic circuit; divide the resulting; product by 1,000,000,000. EXAMPLE. Find the inductance of the coil specified in the preceding example. Solution: The factors are the same as before, omitting the current and time. Product of 12.5664 X 300 X 5 X 25 X 200 is 94,248,000. Dividing this by 1,000,000,000 gives the inductance .094248 henrys. The E. M. F. of self-induction may be computed when the inductance, the current and the time taken for the current to reach its maximum are known. Rule 85. To find the E. M. F. of self-induction divide the product of the inductance and current by the time of current rise. EXAMPLE. Using again the data of the foregoing examples, find the counter E. M. F., the inductance being 120 ARITHMETIC OF ELECTRICITY. .094248 henrys, the current 5 amperes, and the time .02 second. Solution: 5 X .094248 = .47124; dividing this by .02 gives 23.562 volts, as before. Rule SO. The resistance due to self-induction equals 6.2S32 times the product of the frequency and the inductance. EXAMPLE. Find the inductive resistance of a circuit whose frequency is 60 cycles per second and the inductance is .05 henry. Solution: 6.2832 X 60 X -05 = 18.8496 ohms. Ans. The time constant of an inductive circuit is a measure of the growth or increase of the current. It is the time required by the current to rise from zero to its average value. The average value of an alternating current is .634 times its maximum value. It must not be confused with its effective value, which is .707 times the maximum. The average value may be obtained by multiplying the effective value, as shown by instruments, by .897. Rule 87. To find the time constant of a coil or circuit, divide its inductance by its resistance. EXAMPLE. What is the time constant of a coil whose induct ance is 3.62 henrys and resistance is .20 ohms. Solution: 3.62 -~ 20 = .181 second. Ans. CHAPTEE XII. CONDENSERS. A condenser, though it will allow no current to pass through it, yet it will accumulate or store up a quantity of electricity depending on various factors which the following rules will show: Rule SS. The quantity stored equals the product of the K. M. F. applied and the capacity of the condenser. Rule 89. The capacity of a condenser equals the quantity stored divided by the applied E. M. F. E Rule 9O. The E. M. F. applied to a condenser equals the quantity stored divided by its capacity. The quantity stored in a condenser is measured in coulombs (i. e., ampere-seconds) ; the E. M. F. in volts, and the capacity in farads. Condensers in practical use have, however, so small a capacity that 122 ARITHMETIC OF ELECTRICITY. it is usually stated in microfarads and the quantity in microcoulombs. EXAMPLES. A battery of 30 volts E. M. F. is connected to a condenser whose capacity is one half microfarad. What quantity of electricity will be stored? Solution: 30 volts X .0000005 farads = .000015 coulombs. This solution could also be given direct- ly in micro-quantities, thus: 30 volts X % micro- farad = 15 microcoulombs. A condenser is charged with 7.5 microcoulombs under an E. M. F. of 15 volts. What is its capacity ? Solution: 7.5 microcoulombs -r- 15 volts = .5 microfarad. Ans. What E. M. F. is required to charge a condenser whose capacity is .1 microfarad with 21 microcoul- ombs of electricity? Solution: 21 microcoulombs -f- .1 microfarad = 210 volts. Ans. By connecting condensers in parallel the resulting capacity is the sum of their individual capacities. When they are connected in series the resulting capa- city equals 1 divided by the sum of the reciprocals of their individual capacities. It will be noticed that these laws of condenser connections are the inverse of those for the parallel and series connection of re- sistances. CONDENSERS. 123 When applying a direct current to a condenser, as in the above examples, it flows until the increasing charge opposes an E. M. F. equal to that of the charging current. With an alternating current a charge would he surging in and out of the condenser, so that a real current will be flowing on the charging wires in spite of the fact that the actual resistance of a con- denser, in ohms, is practically infinite. Rule 91. The alternating: current in a circuit hav- ing capacity equals the product of 6.2S32, the fre- quency, the capacity, and the applied voltage. EXAMPLE. Find the current produced by an E. M. F. of 50 volts and a frequency of 60 cycles per second in a circuit whose capacity is 125 microfarads. Solution: The capacity 125 microfarads equals .000125 farads. 6.2832 X 60 X .000125 X 50 = 2.3562 amperes. Ans. Rule 92. The alternating: E. M. F. required to he impressed upon a circuit of a given capacity in order to produce a certain current is equal to the current divided hy 6.2S32 times the product ol the capacity and the frequency. EXAMPLE. Find the E. M. F. necessary to produce an alter- nating current of 50 amperes at 50 cycles per sec- ond in a circuit of 80 microfarads capacity. Solution: 6.2832 X .000080 X 50 = .0251328 124 ARITHMETIC OF ELECTRICITY. Dividing the current 50 amp. by .0251328 gives 2,000 volts, nearly. Since, in a condenser circuit, a real current flows under a given E. M. F., the circuit may be treated as though it was of a resistance such as would allow the given current to flow. Rnle 93. The resistance due to capacity equals 1 divided by the product of 6.2832, the frequency and the capacity. EXAMPLE. Find the capacity resistance of a circuit having a frequency of 60 cycles per second and a capacity of 50 microfarads. Solution: 6.2832 X 60 X .000050 = .01885. 1 -T- .01885 = 53 ohms, very nearly. By comparing this Eule 93 for capacity resistance with Eule 86 on page 120, which is for inductive re- sistance, it will be seen that they are mutually recip- rocal and hence the effect of capacity is directly opposite to that of self-induction and vice versa. It follows from this that it is possible, by the proper proportioning of the inductance and the capacity, to have their effects neutralized, and when this adjustment is effected the current will be con- trolled by the volts and ohmic resistance the same as if it were a direct current circuit. The impedance is the apparent resistance of an alternating current circuit. CONDENSERS. 125 Rule 94. To flnd the impedance of a circuit vrliose r>liiiiic resistance can be neglected and ivhicrli has an inductance and a capacity in series, calculate both the inductive resistance and the capacity re- sistance; their difference Trill be the impedance. EXAMPLE. Find the current produced by an alternating E. M. F. of 40 volts on a circuit of slight ohmic resist- ance whose capacity is 100 microfarads, the frequency being 60 cycles per second, and having in series an inductance of .02 henry. Solution: Inductive resistance is 6.2832 X 60 X .02 = 7.42 ; capacity resistance is 1 -=- 6.2832 X 60 X .000100 = 1 -^ .0377 = 26.52. Impedance = 26.52 7.42 = 19.1 ohms. Current, by Ohm's Law, = 40 -f- 19.1 = 2.08 am- peres. Ans. CHAPTER XIII. DEMONSTRATION OF RULES. In the following chapter we give the demonstra- tion of some of the rules. As this is not within the more practical portion of the work, algebra is used in some of the calculations. It is believed that rules not included in this chapter, if not based on experi- ment, are such as to require no demonstration here. Rule i to 6, pages 13 and 14. Ohm's law was determined experimentally, and all the six forms given are derived by algebraic transposition from the first form which is the one most generally ex- pressed. Rule 8, page 19. This is simply the expression of Ohm's law as given in Rule 1, because in the case of divided circuits branching from and uniting again at common points, it is obvious that the difference of potential is the same for all. Hence the ratio as stated must hold. Rule 9, page 20. This rule is deduced from Rule 8. It first expresses by fractions the relations of the current. Next these fractions are reduced to a common denominator, so as to stand to each other in 128 ARITHMETIC OF ELECTRICITY. the ratio of their numerators. By applying the new common denominator made up of the sum of the numerators the ratio of the numerators is unchanged, and the ratio of the new fractions is the same as that of their numerators, while by this operation the sum of the new fractions is made equal to unity. Thus by multiplying the total cur- rent by the respective fractions it is divided in the ratio of their numerators, which are in the inverse ratio of the resistances of the branches of the circuit and as the sum of the fractions is unity, the sum of the fractions of the current thus deduced is equal to the original current. Rule 10, page 22. Resistance is the reciprocal of conductance. By expressing the sum of the recip- rocals of the resistances of parallel circuits we ex- press the conductance of all together. The recipro- cal of this conductance gives the united resistance. Rule 11, page 22. This is a form of Rule 10. Call the two resistances x and y. The sum of their reciprocals is * + * which is the conductance of the two parallel circuits or parts of circuits. Reducing them to a common denominator we have: -. + .*. which equals ~~, whose reciprocal is Rule 17, page 31. Taking the diameter of a wire as d, its cross sectional area is -. The resistance is inversely proportional to this or varies directly with =^. As the resistance of a conductor DEMONS TEA TION OF E ULES. 129 Taries also with its length and specific resistance we have as the expression for resistance: Sp. Bes. X 1.2737 X 1 d a Rule 18, page 32. Assume two wires whose lengths are I and Z 15 their cross sectional areas a and #!, their specific resistances s and 5 1? and their resis- tances r and r^. From preceding rules we have for each wire: r = s ~ (1) and n = s, ^ (2). Dividing (1) by (2) we have: If we take the reciprocal of either member of this equation and multiply the other member thereby it will reduce it to unity, or: -*v v v^-1 r x *x * i t x a . - For convenience this is put into a shape adapted for cancellation. Rule 20, page 38. This is merely the expression of Ohm's Law, Eule 3. Rule 22, page 40. Call the drop e, the combined resistance of the lamps E, and the resistance of the leads x. Then as the whole resistance is expressed as 1 00 (because the work is by percentage) the differ- ence of potential for the lamps is 100 e. By Ohm's law we have the proportion: 100-0 : e :: R : x or eR x ~ lOO^e Rule 25, page 44. From Eule 22 we have: 130 ARITHMETIC OF ELECTRICITY. Call the resistance of a single lamp r, then we have by Rule 12: E =S (2) Substituting this value of R in equation (1) we have: er ~ wx(lOO e) (3) From Rule 24 we have, calling the cross-section a: i x 10.79 ~~F~ (4) Substituting for x its value from equation (3) we have: I X 10.79 X n X (100 e) a = J /K\ (5) But as I expresses the length of a pair of leads, not the total length of lead but only one-half the total, the area should be twice as great. This is effected by using the constant 10.79 X 2 = 21.58 in the equation giving: I X 21.58 X n X (100-e) er Rule 28, page 48. Assuming the converter to work with 100$ efficiency (which is never the case), the watts in the primary and secondary must be equal to each other or: C 2 R = (V R x , and R = ^ R x , or the resistances of primary and secondary are in the ratio of the squares of the currents. The direct ratio is expressed by the ratio of conversion, when squared it gives the ratio of the squares as required. DEMONSTRATION OF RULES. 131 Rule 37, page 59. Let d = diameter of the wire in centimeters. The resistance of one centimeter of such a wire in ohms = Sp. Resist. X 10* 6 X ^. The specific resistance is here assumed to be taken in microhms. The quantity of heat in joules devel- oped in such a wire in one second is equal to the square of the current in C. G. S. units, multiplied by the resistance in C. G. S. units and divided by 4.16 X 10 7 , the latter division effecting the reduc- tion to joules. 1 ohm = 10 9 C. G. S. units of re- sistance. Multiplying the expression for ohmic resistance by 10 9 we have: Sp. Resist. X 10* X ~_. 1 ampere = lOi 0. G. S. unit. If we express the current in amperes we must multiply it by 10" 1 , in other words take one-tenth of it. Our expression then becomes for heat developed in one second /JL.Y x- Sp- Resist - x iQ 3 x 4 V 10 / " IT d a X 4.16 X 10* The area of one centimeter of the wire is *d square centimeters. The heat developed per square centimeter is found by dividing the above expression by ird giving: sjz_ \ * y Sp. Resist. X 10 3 X 4 \10) ir' d 3 X 4.16 X 10 7 The heat developed is opposed by the heat lost which we take as equal to njW per square centimeter per degree Cent, of excess above surrounding me- dium. Therefore taking t as the given tempera* 132 ARITHMETIC OF ELECTEICITY. ture cent, we may equate the loss with the gain thus: t _ /_ \ a 4000 ~ \10j x Sp Resist. X 10 3 X 4 ' IT* d 3 X 4.16 X 10 7 _ c 8 X Sp. Resist. X 10 3 X 4 x 40000 _ ** x 4.16 X 10' X t c a X Sp. Resist, x .00039 t Rule 52, page 69. Call the external resistance r; number of cells n; resistance of one cell R; E. M. F. of one cell E; E. M. F. of outer circuit e. Then from Ohm's law we have: nE ~ nR + r (1) which reduces to: n =!E=QR (2) but C r = e. '; n = W=cJi (3) Rule 54, page 72. This rule is deduced from the following considerations. The current being con- stant the work expended in the battery and external circuit respectively will be in proportion to their differences of potential or E. M. F's. But these are proportional to the resistances. Therefore the resistance of the external circuit r should be to the resistance of the battery R as efficiency: 1 effi- ciency or r : R :: efficiency : 1 efficiency or r - The rest f th from Ohm's law. DEMONSTRATION OF EULES. 133 Rule 5 7, page 74. This rule gives the nearest ap- proximation attainable without irregular arrange- ment of cells. By placing some cells in single series and others two or more in parallel, an almost exact arrangement for any desired efficiency can be ob- tained. Such arrangement are so unusual that it is not worth while to deduce any special rule for them. Thus taking the example given on page 74 the impossible arrangement of 1.4 cells in parallel and 63 in series would give the desired current and efficiency. The same result can be obtained by taking 72 cells in 36 pairs with a re- sistance of 36 x & = 3 ohms, and adding to them 27 cells in series with a resistance of 27 X = 4^ ohms, a total of 7l ohms. The E. M. P. is equal to (36 + 27) X 2 = 126 volts. The total cells are 72 + 27 = 99. Rule 58, page 76. Ono coulomb of electricity lib- erates from an electrolyte .000010384 gram of hydrogen. This has been determined experimen- tally. Let H be the heat liberated by the chemical combining weight of any body combining with another. H is taken in kilogram calories. Hence it follows that for a quantity of the substance equal to .000010384 gram X chemical combining weight, the heat liberated will be equal to H X .000010384, which corresponds to a number of kilogram meters of work expressed by .000010384 X H X 424. The work done by a current in kilogram-meters = 134 ARITHMETIC OF ELECTRICITY. volts X^oulombs OJ . f()r presses the work done by one coulomb. Let the volts = E, and equate these two expressions: ofi = .000010384 X H X 424, which reduces to E = H X .043. Rule 6i, page 78. For the work (in kilogram- meters) done by a current (volt-coulombs) we have the general expression: _ volts x coulombs _ E Q ~W~ - r 9^ (1) Making W = 1 (i. e. one kilogram-meter) and transforming, we have, as the coulombs correspond- ing to 1 kilogram-meter: _ 9.81 -r (3) One coulomb of electricity liberates a weight (in grams) of an element equal to the product of the following: .000010384 X equivalent of element in question X number of equivalents -*- valency of the element. Therefore, the coulombs corresponding to one kilogram-meter, liberates this weight multi- plied by ~ or, indicating weight by G-, G -Q 000103 ^ X equiv. X number equiv. x 9.81 valency E (3) but .000010384 X 9.81 = .000101867. _ _ equiv. X n X .000101867 * E X valency (4) Rule 73, page 98-99. The voltage of an armature of DEMONSTRATION OF RULES. 135 a definite number of turns of wire and a fixed speed, varies with the lines included within its longitu- dinal area, as such lines are cut in every revolution. These lines vary with its area, and the latter varies with the square of its linear dimensions. To maintain a constant voltage if the size is changed, the number of turns must be varied in- versely as the square of the linear dimensions. This ensures the cutting of the same number of lines of force per revolution. If, therefore, its size is reduced from x to ^ the turns of wire must be changed from x to oft. The relative diameters of the two sizes of wire is found by dividing a similar linear dimension by the rela- tive size of the wire. But | * x* = ^ = diameter of the wire for maintenance of a constant voltage with change of size. The capacity of a wire varies with the square of its diameter and (-^) 2 = ^. Therefore the amperage, if a constant voltage is maintained, will vary inversely as the sixth power of the linear dimensions of an armature. CHAPTER XIV. NOTATION IN" POWEKS OF TEN. THIS adjunct to calculations has become almost indispensable in working with units of the C. G. S. system. It consists in using some power of 10 as a multiplier which may be called the factor. The number multiplied may be called the characteristic. The following are the general principles. The power of 10 is shown by an exponent which indicates the number of ciphers in the multiplier. Thus 10 2 indicates 100; 10 8 indicates 1000 and so on. The exponent, if positive, denotes an integral number, as shown in the preceding paragraph. The exponent, if negative, denotes the reciprocal of the indicated power of 10. Thus 10~ 2 indicates TST>', 10' 3 indicates rrsW and so on. The compound numbers based on these are re- duced by multiplication or division to simple expres- sions. Thus: 3.14 X 10 7 = 3.14 X 10,000,000 = 31,400,000. 3.14X10-^ 1 ^^or 1 -^^. Re- gard must be paid to the decimal point as is done here. NOTATION IN POWERS OF TEN. 137 To add two or more expressions in this notation if the exponents of the factors are alike in all re- spects, add the characteristics and preserve the same factor. Thus: (51 X 10 6 ) + (54 X 106 )= 105 X 10. (9.1 X 10- 9 ) + (8.7 X 10- 9 ) = 17.8 X 10-'. To subtract one such expression from another, subtract the characteristics and preserve the same factor. Thus: (54 X 10 6 ) - (51 X 10 6 ) = 3 X 10 6 . If the factors have different exponents of the same sign the factor or factors of larger exponent must be reduced to the smaller exponent, by factor- ing. The characteristic of the expression thus treated is multiplied by the odd factor. This gives a new expression whose characteristic is added to the other, and the factor of smaller exponent is preserved for both. Thus: (5 X 10 7 ) + (5 X 10 9 ) = (5 X 10 7 ) + (5 X 100 X 10 7 ) = 505 X 10 7 . The same applies to subtraction. Thus: (5 X 109) - (5 x 10 7 ) = (5 X 100 X 107) - (5 X 10 7 ) = 495 X 10 7 . If the factors differ in sign, it is generally best to leave the addition or subtraction to be simply ex- 138 ARITHMETIC OF ELECTRICITY. pressed. However by following the above rule it can be done. Thus: Add 5 X 10' 2 and 5 X 10 3 . 5 X 10 s = 5 X 10 5 X 10- 2 : (5 X 10 5 X lO' 2 ) + (5 X lO' 2 ) = 500005 X lO' 2 . This may be reduced to a fraction ^f = 5000.05. To multiply add the exponents of the factors, for the new factor, and multiply the characteristics for a new characteristic. The exponents must be added algebraically: that is, if of different signs the numer- ically smaller one is subtracted from the other one, its sign is given the new exponent. Thus: (25 X 10 6 ) X (9 X 10 8 ) = 225 X 10 14 . (29 X 10- 8 ) X (11 X 10 7 ) = 319 X 10- 1 . (9 X 10 8 ) X (98 X 10- 2 ) - 882 X 10 6 . To divide, subtract (algebraically) the exponent of the divisor from that of the dividend for the ex- ponent of the new factor, and divide the character- istics one by the other for the new characteristic. Algebraic subtraction is effected by changing the sign of the subtrahend, subtracting the numer- ically smaller number from the larger, and giving the result the sign of the larger number. (Thus to subtract 7 from 5 proceed thus: 5 7 = 2.) Thus: (25 X 10 6 ) *- (5 X 10 8 ) = 5 X 10- 2 (28 X 10- 8 ) * (5 X 10 8 ) = 5.6 X 10' u . TABLES. | ! 140 II. EQUIVALENTS OF UNITS OF AREA. Square Millimeter Square Centimet'r Circular Mil. Square Mil. Square Inch. Square Foot. .0000108 Square Millimeter 1 0.01 1973.6 1550.1 .00155 Square Centimeter 100 1 197,361 155,007 .155007 .C01076 Circular Mil. .000507 .0000051 1 .78540 8X10-' Square Mil. .000645 .0000065 1.2733 1 .000001 Square Inch 645.132 6.451 1,273,238 1,000,000 1 .006944 Square Foot ^,898.9 928.989 144 1 HI. EQUIVALENTS OF UNITS OF VOLUME. Cubic Inch Fluid Ounce Gallon Cubic Foot Cubic Yard Cu. Cen- timeter Liter Cubic Meter Cubic Inch 1 .554112 .004329 .000578 16.3862 .016386 Fluid Oz. 1.80469 1 .007812 .001044 29.5720 .029572 Gallon 231 128 1 .133681 .00495 8785.21 3.78521 .003785 Cubic Ft. 1728 957.506 7.48052 1 .037037 28315.3 28.3153 .028315 Cubic Yd. 46,656 25,852.6 201.974 27 1 764,505 764.505 .764505 Cu. Centi. .061027 .033816 .000264 .000035 1 .001 .000001 Liter 61.027 33.8160 .264189 .035317 1000 1 .001 Cu. Meter 61027 33816 264.189 35.3169 1.3080 1000 1 141 IV. EQUIVALENTS OF UNITS OF WEIGHT. Grain Grain. Troy Ounce. Pound Avs. Ton. Milli- gram. Gram. Kilo- gram. Metric Ton. 1 .020833 .000143 64.799 .064799 .000065 TroyOunce 480 1 .068641 31,103.5 31.1035 .031104 PoundAvs. 7,000 14.5833 1 .000447 453.593 .453593 .000454 Ton 32,666.6 2240 1 .001016 1.01605 Milligram .015432 .000032 .000002 1 .001 .000001 Gram 15.4323 .032151 .002205 1000 1 .001 Kilogram 15,432.3 32.1507 2.20462 .000984 1,000,000 1000 1 .001 Metric Ton 32,150.7 2204.62 .98421 1,000,000 1000 1 142 V.-EQUIVALENT8 OP UNITS Erg. Meg- erg. Gram-de- gree C. Kilogram- degree C. Pound- degree C. Pound- degree F. Er f . 1 .000001 Meg.-erg. 1,000,000 1 .024068 .000024 .000053 .000095 Gram-degree C. 41.5487 1 .001 .002205 .003968 Kilogram-degreeC. 41,548.T 1000 1 2.2046 8.9683 Pound-degree C. 18,846.5 453.59 .45359 1 1.8 Pound-degree F. 10,470.1 251.995 .251995 .555556 v Watt-Second. 10 7 10 .24068 .000241 .000531 .000955 Gram-centimeter. 981 .000981 .0000235 Kilogram-meter. 98.1X10 8 98.1 2.86108 .002861 .005205 .009370 Foot-Pound. 18.5626 .326425 .000826 .000720 .001295 Horse-Power-Sec. English. 7459.43 179.486 .179486 .8957 .71243 Horse-Power-Sec. Metric. 7357.5 177.075 177.075 .890375 .70275 143 OF ENERGY AND WORK. Watt- Second. Gram- Centim'tr. Kilogram- meter. Foot- Pound. Horse- power- second English. Horse- power- second Metric. 10- T .001019 Erg. .1 1019.87 .010194 .078784 .000184 .000186 Meg-erg. 4.15487 42,858.5 .428585 3.06355 .00557 .005647 Gram-degree C. 4154.87 428.535 8068.55 5.57 5.64703 Kilogram-degree C. 1834.65 192.114 1889.6 2.52653 2.56149 Pound-degree C. 1047.08 106.730 772 1.40364 1.42805 Pound-degree F. 1 10,193.7 .101937 .737387 .0018406 .0018592 Watt-Second. .000098 1 .00001 .000072 Gram-Centimeter. 9.81 100,000 1 7.23828 .018152 .018334 Kilogram-meter. 1.35626 13,825.8 .188258 1 .0018182 .001843 Foot-Pound. 745.943 76.0392 550 1 1.01888 Horse-Power-Sec. English. 735.75 75 542.496 .986356 1 Horse-Power-Sec. Metric. 144 VI. TABLE OF SPECIFIC RESISTANCES IN MICROHMS AND OF COEFFICIENTS OF SPECIFIC RESISTANCES OF METALS. Specific Coeffi- Specific Coeffi- Resist- ance. Mi- cients of Sp. Res. Resist- ance. Mi- cients of Bp. Res. crohms. crohms. Annealed Silver 1.521 .9412 Annealed Nickel .... 12.60 7.7970 Hard Silver 1.652 1.616 .0223 .0000 Compres'd Tin Lead .... 18.36 19.85 8.2673 12.2834 Annealed Copper.... Hard Copper 1.652 .0223 " Antimony 85.90 22.2153 Annealed Gold 2.081 .2877 " Bismuth . 182.70 82.1170 Hard Gold 2.118 .8107 Liquid Mercury 99.74 61.7203 Annealed Aluminum.. 2.945 1.8224 2 Silver, 1 Platinum. 24.66 15.2599 Compressed Zinc .... 5 689 8.5204 German Silver .... 21 17 18 1002 Annealed Platinum . . 9.158 5.6671 2 Gold, 1 Silver .... 10.99 6.8008 " Iron 9.825 6.0798 SPECIFIC RESISTANCE OF SOLUTIONS AND LIQUIDS. MATTHIESSEN AND OTHERS. Names of Solutions. Temper- ature Centi- grade. Temper- ature Fahren- heit. Specific Resistance. Ohms. Copper Sulphate, concentrated 9 48.2 29.82 " with an equal volume of water it it 46.54 it with three volumes of water it it 77.68 Common Salt, concentrated 13 554 5.98 it with an equal volume of water . . 1C '< 6.00 " with two volumes of water.... II it 9.24 " with three volumes of water. . II 11.89 Zinc Sulphate, concentrated 14 57.2 28.00 it with an equal volume of water. . '< 22.75 " with two volumes of water it it 29.75 Sulphuric Acid , concentrated 14.3 57.8 5.32 50.5#, Specific Gravity 1.393. . . 14.5 58.1 1.086 " 29.6*. Specific Gravity 1.215. . . 12.3 54.5 .83 tt 12* Specific Gravity 1.080... 12.8 55.0 1.368 Nitric Acid, Specific Gravity 1.36 (Blavier) .... 14 57.2 1.45 H (i ( ti 24 75.2 1.22 Distilled Water , (Temp'ture unknown) (Pouillet) 988. VIL RELATIVE RESISTANCE AND CONDUCTANCE OF PURE COPPEE AT DIFFERENT TEMPERATURES. KATTHTESSEX. $4 go 3* !l Relative Resistance. Relative Conductance II 1 Relative Resistance. Relative Conductance 0' 82' 1. 1. 16 60.8 1.06168 .9419 i 83.8 1.00881 .99620 17 62.6 1.06568 .93841 2 85.8 1.00756 .9925 15 64.4 1.06959 .93494 8 87.4 1.01185 .98878 19 66.2 1.07356 .93148 4 89.2 1.01515 .98508 20 68. 1.07754 .92804 5 41 1.01896 .98189 21 69.8 1.08152 .92462 6 42.8 1.0228 .97771 22 71.6 1.08558 .92120 7 44.6 1.02663 .97406 23 78.4 1.08954 .91782 8 46.4 1.03048 .97042 24 75.2 1.09356 .91445 9 48.2 1.08485 .96679 25 77. 1.09759 .9111 10 50 1.08822 .96319 26 78.8 1.10162 .90776 11 61.8 1.04210 .95960 27 80.6 1.10567 .90443 12 58.6 1.04599 .95603 28 82.4 1.10972 .90113 18 55.4 1.0499 .95247 29 84.2 1.11882 .89784 14 57.2 1.05381 .94893 36 86. 1.11785 .89457 15 59 1.05774 .94541 146 VIII. AMERICAN WIBE GAUGE TABLE. Properties of Copper Wire : Specific Gravity, 8.8T8 ; Specific Conductivity, 1.T66 at 75, F. I SIZE. WEIGHT AND LENGTH. RESISTANCE. i'i| I Gauge Num III! lll Diam- eter in Mils. Square of Diameter or circular Mils. Grains per Foot. Po'nds per 1000 Feet. Feet per Pound. Ohms per 1000 Feet. Feet Ohm Ohms per Pound. 0000 460.000 211600.0 4477.2 639.60 1.564 .051 19929.7 .0000785 480 000 409.640 167804.9 3550.5 607.22 1.971 .063 15804.9 .000125 262 00 864.800 138079.0 2815.8 402.25 2.486 .080 12534.2 .000198 208 824.950 105592.5 2236.2 819.17 8.183 .101 9945.8 .000815 165 1 289.300 83694.49 1770.9 252.98 8.952 .127 7882.8 .000501 180 2 257.630 66873.22 1404.4 200.63 4.994 .160 6251.4 .000799 108 8 229.420 52688.58 1118.6 159.09 6.285 .202 4957.8 .001268 81 4 204.810 41742.57 888.2 126.17 7.925 .254 8931.6 .002016 65 5 181.940 88102.16 700.4 100.05 9.995 .821 8117.8 .008206 62 6 162.020 26250.48 555.4 79.84 12.604 .404 2472.4 .005098 41 7 144.280 20816.72 440.4 62.92 15.898 .609 1960.6 .008106 82 8 128.490 16509.68 849.8 49.90 20.040 .648 1555.0 .01289 26 9 114.480 13094.22 277.1 89.58 25.265 .811 1288.8 .02048 20 10 101.890 10381.57 219.7 81.88 81.867 1.028 977.8 .03259 16 11 90.742 8234.11 174.2 24.89 40.176 1.289 775.5 .05181 18 12 80.808 6529.93 188.2 19.74 50.659 1.626 615.02 .08287 10.2 18 71.961 6178.89 109.6 15.65 68.898 2.048 488.25 .18087 8.1 14 64.084 4106.75 86.87 12.41 80.580 2.585 886.80 .20880 6.4 15 57.068 8256.76 68.88 9.84 101.626 8.177 806.74 .83183 5.1 16 50.820 2582.67 64.67 7.81 128.041 4.582 248.25 .52638 4.0 17 45.257 2048.19 48.88 6.19 161.551 5.188 192.91 .83744 8.2 18 40.308 1624.38 84.87 4.91 208.666 6.536 152.99 1.8312 2.5 19 85.390 1252.45 26.50 8.786 264.186 8.477 117.96 2 2392 1.96 20 81.961 1021.51 21.60 8.086 824.045 10.394 96.21 8.3488 1.60 21 28.462 810.09 17.14 2.448 408.497 18.106 76.80 5.8539 1.28 22 25.347 642.47 13.59 1.942 514.938 16.525 60.51 8.5099 1.08 23 22.571 509.45 10.77 1.589 649.773 20.842 47.98 18.884 .80 24 20.100 404.01 8.55 1.221 819.001 26.284 88.05 21.524 .63 25 17.900 820.41 6.77 .967 1084.126 88.185 80.18 84.298 .60 26 15.940 254.06 5.88 .768 1302.088 41.789 28.93 54.410 .40 27 14.195 201.49 4.26 .608 1644.737 52.687 18.98 86.657 .81 28 12.641 159.79 8.89 .484 2066.116 66.445 15.05 137.283 .25 29 11.257 126.72 2.69 .884 2604.167 83.752 11.94 , 218.104 .20 80 10.025 100.50 2.11 .802 8311.258 105.641 9.466 849.805 .16 81 8.928 79.71 1.67 .289 4184.100 188.191 7.508 557.286 .13 82 7.950 63.20 1.83 .190 5268.158 168.011 5.952 884.267 .098 83 7.080 50.18 1.06 .151 6622.517 211.820 4.721 1402.78 .078 84 6.804 89.74 .847 .121 8264.468 267.165 8.743 2207.98 .062 85 5.614 81.52 .658 .094 10688.30 386.81 2.969 3583.12 .049 86 6.000 25.00 .525 .075 18333.33 424.65 2.355 5661.71 .039 87 4.453 19.88 .420 .060 16666.66 535.88 1.868 8922.20 .081 88 8.965 15.72 .815 .045 22222.22 675.22 1.481 15000.5 .025 ' 89 8.531 12.47 .266 .088 26315.79 851.789 1.174 22415.5 .020 40 8.144 9.88 .210 .080 8883388 1074.11 .981 85803.8 .015 147 IX. CHEMICAL AND THEEMO-CHEMICAL EQUIVALENTS. FOBMATIOIT or OXIDES. Name of Compound. Formula. Valency. Chemical Equiv- alents. Combin- ing Weights. Thermo- Chemical Equiv- alents. Water HO II 18 9 84.5 Iron Protoxide FeO II 72 86 84.5 Iron Sesquioxide . . FeO3 III 160 53 3 81.9x8 Zinc Oxide Zn O 11 81 40 5 48 2 Copper Oxide CuO II 79.4 89.7 19.2 Mercury Oxide HgO II 218 108 15.5 FOBMATION or SALTS. Name of Va- Nitrates Sul- Chlo- Cya- Base. lency. phates rides nides. FOBMTJLA. Fe (N03)2 FeSO* FeCia FeCy Chemical Equivalents 180 186 127 112 Combining Weights Thermo-Chemical Equiv'lts 90 13.9 68 12.5 68.5 00 66 8.2 FOBMTTLA Zn(NO3)2 ZnSO ZnC12 ZnCy Chemical Equivalents 189 161 136 117 Combining Weights Thermo-Chemical Equiv'lts 94.5 9.8 80.5 11.7 68 66.4 58.5 7.8 FOBMTTLA. Cu(NO3)2 CuSO* CuCl* CuCy* Copper II Chemical Equivalents Combining Weights 187.4 93.7 159.4 79.7 134.4 67.2 125.4 62.7 Thermo-Chemical Equiv'lts 7.5 9.2 81.8 7.8 FOBMTTLA. Hg(N03)2 HgSO* HgC12 HgCya Mercury II Chemical Equivalents Combining Weights Thermo-Chemical Equiv'lts 324 162 7.5 280 140 9.2 271 135.5 9.45 252 126 15.5 148 X. CHEMICAL AND ELECTEO-CHEMICAL EQUIVALENTS. Name Symbols Valen- cies Chemical Equivalents Combining Weights Electro- Chemical Equivalents Hydrogen H I 1 1 .0105 Gold Au III 196.6 65.5 .6877 Silver Ag I 108 108 1.134 Copper (Cupric) Cu,, II 63 81.5 .8307 Mercury (Mercuric) . . . " (Mercurous).. Iron(ferrlc) " (ferrous) Hg,, Hg. Fe,,, Fe, n i in ii 200 200 56 56 100 200 18.7 28 1.05 2.10 .1964 .294 Nickel Ni ii 59 29.5 .8098 Zinc Zn ii 65 82 5 .3418 Lead Pb ii 207 108.5 1.0868 Oxygen ii 16 8 084 Chlorine Cl i 85.5 35.5 .8728 XI. MAGNETIZATION AND MAGNETIC TRACTION. B Lines per eq. cm. B,, Lines per sq. in. Dynes per sq. centim. Grammes per sq. centim. Kilogrs. per eq. centim. Pounds per sq. inch. 1,000 6,450 89,790 40.56 .0456 .577 2,000 12,900 159,200 162.3 .1628 2.808 8,000 19,850 858,100 865.1 .8651 5.190 4,000 25,800 636,600 648.9 .6489 9.228 5,000 82,250 994,700 1,014 1.014 14.89 6,000 88,700 1,482,000 1,460 1.460 20.75 7,000 45,150 1,950,000 1,987 1.987 28.26 8,000 51,600 2,547,000 2,596 2.596 86.95 9,000 58,050 8,223,000 8,286 8.2S6 46.72 10,000 64,500 8,979,000 4,056 4.056 57.68 11,000 12,000 18,000 70,950 77,400 88,850 4,815,000 5,730,000 6,725,000 4,907 5,841 6,855 4.907 5.841 6.855 69.77 88.07 97.47 14,000 90,300 7,800,000 7,550 7.550 113.1 15,000 96,750 8,953,000 9,124 9.124 129.7 16,000 103,200 10,170,000 10,890 10.89 147.7 17,000 109,650 11,500,000 11,720 11.72 166.6 18,000 116,100 12,890,000 18,140 18.14 186.8 18,000 122,550 14,630,000 14,680 14.68 208.1 20,000 129,000 15,920,000 16,280 16.28 280* 14S 2.IL PERMEABILITY OF WROUGHT AND CAST IRON. SQUARE CENTIMETER MEASUREMENT. Annealed Wrought Iron. Gray Cast Iron. B u H B J* H 5,000 8,000 1.66 4,000 800 5 9,000 2,250 4 5,000 500 10 10,000 2,000 5 6,000 279 21 5 11,000 12,000 13,000 14,000 15,000 16,000 1,692 1,412 1,083 823 526 820 6.5 8.5 12 17 28.5 50 7,000 8,000 9,000 10,000 11,000 133 100 71 63 87 42 80 127 183 292 17,000 161 105 18,000 90 200 19,000 54 850 20,000 80 666 SQUARE INCH MEASUREMENT. Annealed "Wrought Iron. Gray Cast Iron. B. /c, H. B. /*- H. 80,000 40,000 50,000 CO.OCO 70,000 80.000 90,000 100,000 110,000 120,000 180,000 140,000 4,660 8,877 8,081 2,159 1,921 1,409 907 408 166 76 85 27 6.5 10.3 16.5 27.3 86.4 56.8 99.2 245 664 1,581 8,714 5,185 25,000 80,000 40,000 60,000 60,000 70,000 763 756 258 114 74 40 32.T 89.7 155 439 807 1,480 150 PERMEABILITY OP SOFT CHARCOAL WROUGHT IRON (8II1LFOBD BITJWELL.) BQUABE CENTIMETER MEASUBE. B t* H 7,890 11,560 15,460 17,880 18,470 19,880 19,820 1899.1 1121.4 886.4 150.7 88.8 45.8 88.9 8.9 10.8 40 115 208 427 585 BQUAKE INCH MIA8UBEM1NT. B. //, H. 47,414 74,104 99,191 111,189 118,504 124,021 127,165 1897 1122 888 150 88.8 45.8 88.9 25.0 60.1 256 788 1885 2740 8758 B Magnetic Flux. ) _ VBoth in lines of force. H Magnetizing Force. J H the Permeability or multiplying power of the cor. 151 XIII.-MAGNETIC BELUCTANCE OF AIE BETWEEN TWO PAEALLEL CYLINDEES OF IBON. b p Batio of least CBNTOTETKB UNITS. INCH UNITS. distance apart to circumference. 0.1 .1954 5.1055 0.0771 12.968 0.2 .2707 8.6917 0.1066 9.877 0.8 .8251 8.0768 0.1280 7.815 0.4 .8688 2.7158 0.1450 6.897 0.5 .4046 2.4716 0.1593 6.278 0.6 .4861 2.2983 0.1717 5.825 0.8 .4887 2.0465 0.1924 5.198 1.0 .5816 1.8807 0.2093 4.777 1.2 .5684 1 .7996 0.2238 4.571 1.4 .6007 1.6645 0.2365 4.228 1.6 .6289 1.5902 0.2476 4.089 1.8 .6541 1.5287 0.2575 8.883 2.0 .6774 1.4764 0.2667 8.750 4.0 .8857 1.1968 0.3290 8.040 6.0 .9319 1.0782 0.3669 2.726 8.0 1.0047 .9958 0.3955 2.528 10.0 1.0544 .9484 0.4151 2.409 In this table in columns 2 and 3 the Unit length of a cylinder is taken as 1 centi- meter ; in columns 4 and 5 as 1 inch, p circumference of cylinder b = shortest distance apart. XIV.-TABLE OF 6TH BOOTS. Num- ber Sixth Boot Number Sixth Boot Num- ber Sixth Boot Number Sixth Boot 1 .69355 i .95320 1* 1.0177 ii 1.0978 I .70717 1 .96350 H 1.0192 If 1.1019 f .72306 1 .97006 1* 1.0226 if 1.1063 I .74185 f .97463 u 1.0260 i? 1.1087 i .76473 i .97798 i| 1.0308 1$ 1.1107 i .79370 1 .98055 i* 1.0879 if 1.1119 i .88263 ft .98258 n 1.0491 ift 1.1129 i .89090 ii 1.0^99 2 1.1237 1 .93462 if 1.0888 152 XV.-STANDARD AND BIRMINGHAM WIRE GAUGES. STANDARD. BIRMINGHAM. Number of Gauge. Diameter in Mils. Square of Diameter or Circ'l'r Mils. Number of Gauge. Diameter in Mils. Square of Diameter or Circ'l'r Mils. 0000000 500 250000 0000 454 206116 000000 464 215296 000 425 180625 00000 432 186824 00 880 144400 0000 400 160000 840 115600 000 372 138384 1 800 90000 00 348 121104 2 284 80656 324 104976 3 259 67081 1 800 90000 4 238 56644 2 276 76176 5 220 48400 8 252 63504 6 203 41209 4 282 53824 7 180 82400 5 212 44944 8 165 27225 6 192 36864 9 148 21904 T 176 80976 10 184 17956 8 160 25600 11 120 14400 9 144 20736 12 109 11881 10 128 16384 13 095 9025 11 116 13456 14 083 6889 12 104 10816 15 072 5184 18 092 8464 16 065 4225 14 080 6400 17 058 3864 15 072 5184 18 049 2401 16 064 4096 19 042 1764 IT 056 8136 20 035 1225 18 048 2304 21 032 1024 19 040 1600 22 028 784 20 086 1296 23 025 625 21 032 1024 24 022 484 22 028 784 25 020 400 28 024 576 26 018 324 24 022 484 25 020 400 26 018 824 153 3 S 8 3 8 i 1 1 s s ill CO ^ t-. 154 XVII.-WATTS AND HORSE POWER TABLES FOR VARIOUS PRESSURES AND CURRENTS. These tables will be found very convenient for quickly finding the watts and electrical horse power on lighting and power circuits. To find the watts or h. p. for any current up to 1,000 amperes at a standard voltage add the watts or h. p. corresponding to the units, tens and hundreds digits of the current. Example : Find the electrical h. p. of 436 amperes at 10$ volts. Solution: The h. p. for 400 amp. is 56.3, for 30 amp. it is 4.22 and for 6 amp., .845. Adding these quantities gives 61.365 h. p., which we will call 61.4 h. p., as the tabular values are computed to three figures only, which are sufficient for engineering purposes. To find values for voltages higher or lower than in the tables, select a voltage 1-10 or ten times that required, and multiply the re- sult by 10 or 1-10. Thus : to find h. p. at 7 amp., fi5 volts, take 7 amp. at 550 volts*5.16 h. p.; multiply by 1-10, which gives .516 h. p. To find h. p. at 9 amp., i,200 volts, take 9 amp. at 120 volts=1.45 ; multi- ply by 10=14 5 h. p. To read in kilowatts place a decimal point before the watts when less than 1,000 in value, or substitute it for the comma in the larger values, HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. 100 volts. 105 volts. 110 volts. Amperes. Watts. 100 Mi Watts. 105 h.p. .141 Watts. 110 h.p. .147 2 200 .268 210 .282 220 .295 3 300 .402 315 .422 330 .442 4 400 .636 420 .563 440 .590 5 500 .670 525 .704 550 .737 6 600 .804 630 .845 660 .885 7 700 .938 735 .985 770 1.03 8 800 1.07 840 1.13 880 1.18 9 900 1.21 945 1.27 990 1.33 10 1,000 1.34 1,050 1.41 1,100 1.47 20 2,000 2.68 2,100 2.82 2,2uO 2.95 30 3,000 4.02 3,150 4.22 3,300 4.42 40 4,000 5.36 4,200 563 4,400 5.90 50 5000 6.70 5,250 7.04 5,500 7.37 60 6,000 8.04 6,300 8.45 6,600 8.85 70 7,000 9.38 7,350 9.85 7,700 10.3 80 8,000 10.7 8,400 11.3 8,800 11.8 90 9,000 12.1 9,450 12.7 9,9(10 13.3 100 10,(K)0 13.4 10,500 14.1 11,000 14.7 . 200 20.000 28.8 21,000 28.2 22,000 29.5 300 30,000 40.2 31,500 42.2 33000 44.2 400 40,000 53.6 42,000 56.3 44,000 59.0 500 50,000 67.0 52,500 70.4 55,000 73.7 600 60,000 80.4 63,000 84.5 66,000 88.5 700 70,000 93.8 73,500 98.5 77,000 103 800 80,000 107 84,000 113 88,000 118 900 90,000 121 94,500 127 99,000 133 1,000 100,000 134 105,000 141 110,000 147 155 HORSE POWER AT VARIOUS PRESSURES AND CUBBENTS. (Continued.) 115 volts. 120 volts. 125 volts. Amp. Watts. 115 Mi Watts. 120 h. p. .161 Watts. 125 h.p. .168 2 230 .308 240 .322 25') .335 3 345 .462 36> .483 375 .505 4 460 .617 480 .644 .670 5 575 .770 6 .805 625 .840 6 690 .925 730 .fc66 750 1.01 7 805 1.08 840 1.13 875 1.18 8 920 1.23 96 1.29 1,000 1.34 9 1,035 1.39 1,<>80 145 1,125 1.51 10 1,150 1.54 1.200 161 1,250 168 20 2.300 3.08 2,400 3.22 2.500 3.35 30 3,450 4.62 3,600 483 3,750 5.C5 40 4,600 6.17 4,#iO 6.44 S.rOO 6.70 50 5,750 7.70 6,(KIU 8.04 6.250 8.40 60 6,900 9.25 7,200 966 7.5 10.1 70 8.050 10.8 8,400 11.3 8,750 11.8 80 9.200 12.3 9,600 12.9 10 POO 13.4 90 10350 13.9 1K800 145 11,250 15.1 100 11,500 154 12,(00 16.1 12,500 16.8 200 23, (00 30.8 24.' 00 32.2 25,000 335 300 34,500 462 36,000 483 37,500 505 400 46/00 617 48,000 64.4 50.(00 67.0 500 57.500 77.0 60(K<0 80.4 62,500 840 600 69.(00 92.5 72,i M) 96.6 75,000 1)1 700 80500 lf'8 84,0 113 87,5<0 118 800 92,000 123 96, > 129 100.000 134 900 103, 00 139 118,000 145 112.50) 151 1,000 115,000 154 120,000 161 125,ttO 168 156 HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. (Continued.) 200 volts. 210 volts. 220 volts. Amp. Watts. 200 h.p. .268 Watts. 210 h.p. .282 Watts. 220 h.p. .295 2 400 .536 420 .563 440 .690 3 600 .8U4 630 .845 660 .885 4 800 1.07 840 1.13 88' 1 1.18 5 1,000 1.34 1,050 1.41 1,100 1.47 6 1,200 1.61 1,260 1.69 1,320 1.77 7 1,400 1.88 1,470 1.97 1,540 2.06 8 1,600 2.14 1.680 2.25 1.760 2.36 9 1,800 241 1,890 253 1,980 2.65 10 2,000 2.68 2,100 2.82 2,200 2.95 20 4,000 5.36 4,200 5.63 4,400 5.90 30 6.UOO 8.04 6,300 845 6,600 8.85 40 8,000 10.7 8,400 11.3 8,800 11.8 50 li,000 13.4 10,500 14.1 11,000 14.7 60 12,CM) 16.1 12,600 16.9 13,200 177 70 14,0(0 18.8 14,700 19.7 15,400 20.6 80 16,000 21.4 16.800 22.5 17,600 23.6 90 18,000 241 18,900 26.3 19,800 26.5 100 20,000 26.8 21,000 28.2 22,000 295 200 40,000 53.6 42,000 56.3 44,000 59.0 300 60,000 80.4 63,000 84.5 66,000 88.5 400 80,000 107 84,000 113 88,000 118 500 lOO.OCO 134 105,000 141 110,0( 147 600 120,000 161 126,000 169 132,000 177 700 140,00) 188 147,000 197 154,000 206 800 160,000 214 168,000 225 176,000 236 900 180,000 241 189,000 253 198,000 881 1,OUO 200,000 288 210,000 282 220,000 295 157 HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. (Continued.) 230 volte. 240 volts. 250 volts. Amp. Watts. 230 ":& Watts. 240 h-p. .322 Watts. 250 h.p. .&5 2 460 .617 480 .644 500 .670 3 690 .925 720 .966 750 1.01 4 920 1.23 960 1.29 1,000 1.34 5 1,150 1.54 1,200 1.61 1,250 1.68 6 1,380 1.85 1,440 1.93 1,500 2.01 7 1,610 2.16 1,680 2.25 1,760 235 8 1,840 2.47 1,920 2.58 2,000 2.68 9 2,070 2.77 2,160 2.90 2,2AO 3.02 10 2,300 3.08 2,400 3.22 2,500 335 20 4,600 6.17 4,800 6.44 5.000 6.70 30 6,900 9.*5 7,200 9.66 7,500 10.1 40 9,200 12.3 9,600 12.9 10.000 13.4 50 11,500 15.4 12,000 16.1 12,500 168 60 13,800 185 1440) 19.3 15,000 20.1 70 16,100 21.6 16,800 22.5 17,500 23.6 80 18,400 24.7 19,200 258 20,000 26.8 90 20,700 27.7 21,600 29.0 22,500 30.2 100 23.000 30.8 24,000 32.2 20.000 33.6 no 46,000 61.7 48000 64.4 50,000 67.0 300 69000 92.5 72,000 96.6 75.000 101 400 98.000 123 96.1 111 129 100,000 134 500 115,000 154 120,000 161 1 5000 168 600 138,000 185 144,000 193 150.000 201 700 161.COO 216 168.000 225 175,01 236 800 184.000 247 192,000 258 2v a line drawing with a description showing its working parts and the method of operation. 396 pages, 1,000 specially made illustrations. Price, $3.00. HISCOX. Modern Steam Engineering in Theory and Practice. This book has been specially prepared for the use of the modern steam engineer, the technical students, and all who desire the latest snd most reliable infor- mation on steam and steam boilers, the machinery of power, the steam turbine, electric power and lighting plants, etc. 450 pages, 400 detailed engravings. $3.00. HOBART. Brazing and Soldering. A complete course of instruction in a!l kinds of hard and soft soldering. Shows just what tools to use, how to make them and how to use them. Price, 25 cents. HORNER. Modern Milling Machines: Their Design, Construction and Operation. This work of 304 pages is fully illustrated and de- scribes and illustrates the Milling Machine in every detail. $4.00. GOOD, PRACTICAL BOOKS HORNER, Practical Metal Turning. A work covering the modern practice of machining metal parts in the lathe. Fully illustrated. $3.50. HORNER. 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Electrical Engineers' and Students' Chart and Hand-Book of the Brush Arc Light System. Bound in cloth, with celluloid chart in pocket. 50 cents. RICHARDS AND COLYIN. Practical Perspective. Shows just how to make all kinds of mechanical drawings in the only practical perspective isometric. Makes everything plain so that any mechanic can un- derstand a sketch or drawing in this way. Saves time in the drawing room and mistakes in the shops. Con- tains practical examples of various classes of work. 60 cents. ROUILLION. Drafting of Cams. The laying out of cams is a serious problem unless you know how to go at it right. This puts you on the right road for practically any kind of cam you are likely to run up against. Price, 25 cents. ROUILLION. Economics of Manual Training. The only book that gives just the information need- ed by all interested in manual training, regarding buildings, equipment and supplies. 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A comprehensive and up-to-date work illustrating and describing the Drainage and Ventilation of dwell- ings, apartments, and public buildings, etc. Adopted by the United States Government in its sanitary work in Cuba, Porto Rico, and the Philippines, and by the principal boards of health of the United States and Canada. The standard book for master plumbers, architects, builders, plumbing inspectors, boards of health, boards of plumbing examiners, and for the property owner, as well as for the workman and his apprentice. 300 pages. 55 full-page illustrations. $4.00. Tonnage Chart, Built on the same lines as the Tractive Power Chart; it shows the tonnage any tractive power will haul under varying conditions of road. No calcula- tions are required. Knowing the drawbar pull and grades and curves you find tonnage that can be hauled. 50 cents. Tractive Power Chart. A chart whereby you can find the tractive power or drawbar pull of any locomotive, without making a figure. 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A practical treatise of 560 pages, containing 600 illustrations on the designing, constructing, use, and installation of tools, jigs, fixtures, devices, special appliances, sheet-metal working processes, automatic mechanisms, and labor-saving contrivances; together with their use in the lathe, milling machine, turret lathe, screw machine, boring mill, power press, drill, subpress, drop hammer, etc., for the working of met- als, the production of interchangeable machine parts, and the manufacture of repetition articles of metal. $4.00. WOODWORTH. Dies, Their Construction and Use for the Modern Working of Sheet Metals. A new book by a practical man, for those who wish to know the latest practice in the working of sheet metals. It shows how dies are designed, made and used, and those who are engaged in this line of woris can secure many valuable suggestions. Thoroughly modern. 384 pages. 505 illustrations. $3.00. WOODWORTH. Hardening, Tempering, Annealing, and Forging of Steel. A new book containing special directions for the successful hardening and tempering of all steel tools. Milling cutters, taps, thread dies, reamers, both solid and shell, hollow mills, punches and dies, and all kinds of sheet-metal working tools, shear blades, saws, fine cutlery and metal-cutting tools of all descriptions, as well as for all implements of steel, both large and small, the simplest and most satisfactory hardening and tempering processes are presented. The uses to which the leading brands of steel may be adapted are concisely presented, and their treatment for working under different conditions explained, as are also the special methods for the hardening and tempering of special brands, 320 pages. 250 illustrations. $2.50. JUST PUBLISHED HYDRAULIC ENGINEERING By Gardner D. Hiscox, a practical work on hydraulics and hydrostatics in principle and prac- tice, including chapters on : The measurement of water flow for power arid other uses; siphons, their use and capacity; hydrau- lic rams; dams and barrages; reservoirs and their construction; city, town and domestic water sup- ply; wells and subterranean water flow; artesian wells and the principles of their flow; geological conditions ; irrigation water supply, resources and distribution in arid districts; the great projects for irrigation. Water power in theory and practice, water wheels and turbines; pumps and pumping devices, cen- trifugal, rotary and reciprocating; air-lift and air- pressure devices for water supply hydraulic power and high-pressure transmission ; hydraulic mining; marine hydraulics, buoyancy displacement, tonnage, resistance of vessels and skin friction. Relative velocity of waves and boats ; tidal and wave pow- er, with over 300 illustrations and 36 tables of hydraulic effect. A most valuable work for study and reference. About 400 octavo pages. Price, $4.00. THE TELEPHONE HAND-BOOK By H. C. Gushing, Jr., E.E., and W. H. Radcliffe, E.E. A practical reference book and guide for tele- phone wiremen and contractors. Every phase of telephone wiring and installation commonly used to-day is treated in a practical, graphic and concise manner. Fully illustrated by half-tones and line- cut diagrams, showing the latest methods of install- ing and maintaining telephone systems from the simple two-instrument line, intercommunication systems for factories, party lines, etc. 175 pages, 100 illustrations, cloth, pocket size. $1.00. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 5O CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. 6 1932 LD 21- 7B 0958' UNIVERSITY OF CALIFORNIA LIBRARY