# Full text of "Arithmetic of electricity; a practical treatise on electrical calculations of all kinds reduced to a series of rules, all of the simplest forms, and involving only ordinary arithmetic .."

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at jhttp : //books . qooqle . com/ £A«t \ i v ,4*,' \ An Electrical Library* By PROF. T. O'CONOR SLOANE. How to become a Successful Electrician. PRICE tl.OO. Electricity Simplified. PRICE, $1.00. Electric Toy Making, Dynamo Building, etc. PRICE, tl.OO. Arithmetic of Electricity. PRICE, tl.OO. Standard Electrical Dictionary. price, ta.oo. NORMAN W. HENLEY & CO., Publishers, 132 Naeeau Street, New Tork. Arithmetic of .electricity A PRACTICAL TREATISE ON ELECTRICAL CALCULATIONS OF AU, KINDS REDUCED TO A SERIES OF RULES, ALL OF THE SIMPLEST FORMS, AND INVOLVING ONLY ORDINARY ARITH- METIC, EACH RULE 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. 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. 4v.\v. .-«m.^r X 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- 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 Scranton, 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 INTBODUCTOBY. 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. BESISTANCE 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 Yi CONTENTS. CHAPTER IV. POTEOTIAL DIFFERENCE. Drop of Potential in Leads, and Size of Same for Mul- tiple Arc Connections. — Diminishing Size of Leads Progressively CHAPTER V. CIBCT7LAB MILS. The Mil.— The Circular Mil as a Unit of Area.— Cir- cular Mil Rules for Resistance and Size of Leads.- CHAPTER VL 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 VH WOBK AND ENBBGY. 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 60 CONTENTS. Vil CHAPTER VIII. BATTERIES. 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 Mux. — 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 Till 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. \Q 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 yolt, equal to one hundred mil- lions (100,000,000) C. G-. S. units of E. M. F. The last numeral is expressed more briefly as the eighth ZNTBODUCTOBY. 11 power of 10 or 10 8 . Thus the volt is defined as equal to 10 8 0. 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 loooiooos . The exponent indicates the number of ciphers to be placed after 1. When electromotive foroe does work a current is produced. The practical unit of current is the ampebe, equal to & 0. G. S. unit, or 10" 1 C. G. S. unit, ib being expressed by 10~*. 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 Fabad equal to t oooo^oooo or 10"* 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* 0. Gr. S. units. Sometimes, where larger units are wanted, the pre- fix deka, ten times, heka* one hundred times, kilo, 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, cenii, 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. P. of 1.07 volt. A current which in each second deposits .00033 grams 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° O. The capacity of the earth is 1088800 farad. A Leyden jar with a total coated surface of one square meter and glass one mm. thick has a capacity of A microfarad. The last is the more generally used unit of capacity. These practical units are derived from the C. 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~ u ) part of a gram. CHAPTER H. 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 is 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. r = — 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 with the elee* tromottve foree and Inversely with the current. Rale 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 Circuits. 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 B. M. P. 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 1 #- 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 OEM'S LAW. 15 to distinguish it from the resistance of the outer circuit, called external resistance. Resistance in general is denoted by R, electromotive force by E, and current by C. A battery of B 2 ohms; sends a current of .035 ampere through a wire of E 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. P. 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 Eule 3 the resistance is equal to ffc= .157 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 | n opposi- tions add together the E. M. F of all generators of like polarity, thus obtaining two opposed E. M. F.s. Sub- tract the smaller E. M. F. from the larger E. IS. F. to 16 ARITHMETIC OF ELECTRICITY. obtain the effective E. M. F. Then apply Ohm's law 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 effective lattery constants? ■-AAAAAAA/WW Solution: Voltage = (2 + 1) — (1.75 + 1) = .25 volt. Eesistance = J^ + 2+l + 4 = 7^ ohms, or .25 volt, 7j4 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 h- (1}4 + 5) = .02 amperes. A battery of 51 volts E. M. F. and 20 ohms resist- OHM'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 B. M. P. is 51 — 26 = 25 volts. By Eule 3 we have 25 ■*- 1 = 200 ohms, as the total resistance. But the resistance of the batteries (internal resistance) is 20 4- 25 = 45 ohms. The re- sistance of leads, etc. (external resistance), is there- fore 200 — 45 = 155 ohms. POBTIONS OF OlECUITS. All portions of a circuit receive the same current, but the E. M. P., 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 ARITHMETIC OF ELECTRICITY. 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 a , and E 1 , we have the continued proportion, 175 : 100 : 50 : 25 :: 10 volts : E 1 : E a : E 1 . because by the conditions of the problem the total E. M. P. «= 10. Solving the proportion by the regu- lar rule, we find that E 1 — 5.7, E* — 2.8 and E» — 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. P. (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 j4 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 -Aftr 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 0HW8 LAW. 19 total of 205 ohms. To maintain a current of tffo amperes through 205 ohms (Rule 2), an E. M. F. is required equal to i% X 205 volts or 61^ volts. Divided, Branched or Shukt 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, eacn brancn 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, is : tV. 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 A : B : C :: A : is : iV. 20 ARITHMETIC OF ELECTRICITY. Rule 9. To determine the amount of a given cur- rent that will pass through parallel circuits of dUTer- ent resistances, proceed as follows s 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 < 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 1 ohm. They unite at one point. What proportion of a unitary current will pass each branch? Solution: The proportion of currents passed are as irriw : A : g or 3. Reducing to a common de- nominator, these become irSfinr : s tfoWo : WoWo . 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, iri&Tv; No. 2, ri»; No. 3, ttHTO. 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 A : is : A : ih, 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 jh : *fo or reduced to a common denominator and to their lowest terms tHt : irJi*. 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 H, we get H$$ 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- 23 ABITHMETIC 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 hy the reciprocal of the sum or the reciprocals or the resistances. Example. Two leads of a 50 volt circuit (leads differing in potential by 60 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 ft t ts t ioi)o mhos = i8oo t looo t TTnnr ass ioA© mhos. The reciprocal of conductance is resistance. The combined resistance is therefore m* ohms = 14.09 ohms. The current is ^ or 3.5 amperes. Rule 11. The combined resistance or two parallel circuits is round by multiplying the resistance* to- gether, and dividing the product by the sum or the re- sistances. Where there are several circuits, any two can be treated thus, and the result combined in <he same way with another circuit, and so on to get the final resistance. B _ rXrI Example. ^ Four conductors in parallel have resistances of 100 _ 50 — 27 — 19 ohms. What is their combined resistance? OHM' 8 LAW. 23 Solution: Combining the first and second, we have lro*M = 33* ohms. Combining this with the resistance of the third wire, we have |£qp| =■ i*.9 ohms. Combining this with the resistance of the fourth wire, we have ji£pfS = 8.3 ohms. The result is, of course, identical by whatever rule obtained. Rule 12. When all the parallel circuit* 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. o — — R ~" 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 ELECTRICITY. \JfctvK 7 .Zeads. A Ster]* VSAtfWwJ . Jtksistetnee . Solution: The resistance of the motor (Ohm's Law, Rule 3), is found by dividing the E. M. F. by the resistance — 30 -*- 3 = 10 ohms. By Eule 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 flHfths 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 A or as ^. The conductance of the shunt coil must therefore be Hi — t%t = ^fo = A 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 n+aQ = 4 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. P. 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 Biidge principle. CHAPTER IIL besistakoe akd 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 or circu- lar section Inversely as the squares of their diameters, and directly as their lengths. Example. 1. A wire a, 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* : 28* or as 900 : 784. Were they of equal diameter the direct proportion would hold for their lengths: R b : R* :: 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: E b : E* :: 30 a X 315 :28 a 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 glTC the length of the second. Id* — Fd* Examples. 1. There are three wires, a is 2 mils, 5 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 at Solution: Take a and c first and apply the rule, 10 X 4* +2* = 40 feet; then take a and J 10 X 3* -*- 2 a = 22 j£ feet. To prove it compare a and c directly by the same rule 22j£ X 4 a -*- 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 a X 1264 •*- 79 = 7,110,000 •*- 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 1 5. Multiply the square of the diameter of eaeh 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 «*t unity* The term Including the length of a given wire is the one expressing the relative resistance of such wire* Examples. A wir9 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* X 9 = 14,400 = relative resistance of the 3 mile wire. 50 a 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« BE8I8TANCE AND CONDUCTANCE. 29 ameter of a wire of the same material of which the resistance of 1000 feet is 10.09 ohms? (E. E. Day, M. A.). Solution: 10.09 -*• 1.15 = 8.77 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. Rule 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- sectional 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. n _ Sp. R x I Examples. An electro-plater has a bath of sulphate of copper, sp. 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. .\ Eesistance = 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 y 2 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. .-. Eesistance = 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. 31 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. Role 17. The resistance In microhms of a wire of given diameter In centimeters Is equal to the product of the specific resistance by 1.2737 by the length In centi- meters dlylded by the square of the diameter In cen- timeters. ^ n m __, . _ Sp. Re* x «.2737 x L * — -* 5» BZAMPLES. 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 (1 y 2 meter) is 150 centimeters. Its resistance therefore is 1.652 X 1.2737 X 150 «• .01 = 31,561 microhms or .031,561 ohms. TTNTVBB8AL 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 82 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 b 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 tbe above expression tbe values of any factors given. Substitute for factors not given or required tbe figure 1 or unity. Sucb a value deter- mined by division must be given to tbe required factor and substituted In Its place as will make tbe product of tbe left-band factors equal to tbat of tbe rlgbt-band factors. Only one factor can be determined* and all factors not given are assumed to be respectively equal for botb 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* As the specific resistances are identical they are not given. 2 X 70 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 14 Specific Resistances Unknown 7 Lengths: 432 1728 Cross-Sectional areas: 12 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 b, 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 Area, 4 sq. mils. L. Unknown Sp. Res. 7 Resist. 85 ohms 1,000 yards 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. Besistance of Wires Beferred 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 Eules 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 fh x 25 = 1.50 oz. The weights of equal lengths of A and B respectively are as 1.50 : 1.00 which is also the inverse ratio of the resistances of equal lengths. By compound proportion Eule we have E. of " A " : E. 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 Eule for resistance (Eule 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 Resistance of B = 1.50 X 20 -*- 334 = .0899 oi about it ae 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* Examples. A wire has a resistance of ^ftr ohms, what is its conductance? Solution: 126 -*- 18 = 7 mhos. Reduce a conductance of lit to ohms. Solution: 1H = H mhos which gives tt 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)— Eule 2 as " The electromotive force is equal to the current divided by the conduc- tance " (E = — )— Eule 3 as "The conductance is K equal to the current divided by the electromotive 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 = .-£- = 2 mhos. The current = 50 X 2 = 100 amperes. In a circuit a current of 20 amperes is main- tained through 2t ohms. Determine the E. M. F. using conductance. Solution: The conductance = 4 = H mhos. E. M. F = 20 -*- 1* = 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, Dbop op Potential in Leads and Size op Same for Multiple Abo 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. 89 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, (Hi = i ampere). The reciprocal of the current is f ; mul- tiplied by the drop (i 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 ffo = A 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 E. M. F. of the main leads or terminals the factors of the lamps or other appliances, their 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 leads 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 + 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, (tX5 = i = 1?3 ohms) gives the required resistance = lji 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: Rale 22. The resistance of tbe 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 A = 9%. The resistance of the leads = 16g6 x 9 = U9M = \yi ohms 100 — 9 91 ' about. JVote.— 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 DIFFERENCE. 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. P. 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. S&Tmmfm dttiZ+myu. f S&J*unj*. -x+Ka,j, T - -/*K&j>~r- -tsnusDrvp- Starting at group 3 we have 50 lamps in parallel each of 220 ohms resistance, giving a combined re- sistance (Eule 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 *V 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 0—1 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: b . _ j££C4ta._» • -^S- -'SH^Vs* ^Jf±&& &*£!?!$■ <#»*fi^ tev* Irv 4~e* v J CHAPTER V. CIBCULAB MILS. A mil is TtAnr 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 *— .. -.--'jjgtgf. • 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 Eule 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 Us resistances 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 21.5 8, by the dif- ference between 100 and the drop, the whole divided by the resistance of a single lamp multiplied by the drop. __ In x 81.58 X (100— e) 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 VL special systems. Three 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 B. 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 hare 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. Altebnating Cubrent 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 ■+■ 60 = 20. Rule 27. The current in the primary is equal 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 $# — 1 ampere, Ohm's law) = 1000 amperes. 1000 -*-i 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 ABITHMETIC 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 = 321 amperes, drop of primary = 1031 X 3* = 30.9 volts, resistance of pri- mary (Rule 20) 1X30.9 = .9516 ohm. Rale 28. For obtaining the size of the 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 (100—3) -h (50 X 3)] +■ 20 a = 81,637 circular mils. The two last examples may be made to prove each other, thus: The total length of leads is 3600 X2«= 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 8T8TEM8. 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. P. 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 tbe primary are equal In number to tbe product of tbe convolutions of tbe sec- ondary multiplied by tbe ratio of conversion, and vice versa* Examples. The.ratio of conversion of a coil is 20; there are 1000 convolutions of tl\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 VBL work and energy. Ekbbgy 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. P. 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 heat developed in 'circuits to In proportion to the square of the current multiplied by the resistance. Examples. An electric lamp has a resistance of 50 ohms; it is connected to a street main by leads of 24 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 A 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£ : j^ = 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 X 1 : (£)* 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 Jt 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) : (tt) or i: *$ftr = 2025 : 400. The diminution of current there- fore cuts down the heat to I the proper amount. Rule 3 1 • The energy or beat developed In a circuit is In proportion to the square of the electromotive force divided by the resistance. E* H = - K Examples. There are two Grove batteries, each developing 1.98 volts E. M. P. One has an internal resistance of A ohm; the other of i 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 «~£ : «1 *** ^i : 2A as the ratio of heat produced. A battery of one ohm resistance and two volts E. WORK AND ENERGY. 53 M. P. 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 if developed in each case? Solution: if 8 : 1|I or 4:1. Rale 32. The energy or heat developed In a circuit %m In proportion to the K. m. F. multiplied by the current* H = BP Examples. Take data of last problem and solve. Solution: For first battery, by Ohm's law, current ■■ f ampere; for the second, current = I ampere. The heat developed, is by the present rule, in the propor- tion as 1 X 2 : i X 1 or 4 : 1. Compare the heat developed in a 100 volt 200 ohm Jamp and in a 35 volt 35 ohm lamp and in a 50 volt 50 ohm lamp. Solution :The currents (Ohm's law) are : Hfcli and U in amperes — i, 1, and 1 amperes. The heats devel- oped are, by the rule, in the ratio 100 X i : 35 X 1 and 50 X lor50:35 :50. The same problem can be done directly by Rule 31, thus: The three lamps develop heat in the ratio m* : tt* : U* « 50 : 35 : 50. This is the direct and preferable method of calculation. Note, — For " heat," " rate of energy," or " rate of work" can be read in these rules. 54 ARITHMETIC OF ELECTRICITY. The Joule ob 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 C. or w. g. d. 0. or for shortness g. d. 0., 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. C. 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 0. 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 €. G. 8. units (ergs) by multiply- ing by 10 7 and divide by the value of a joule In ergs (4.16 X 10 7 ). The quotient Is joules or water-gram de- grees C. per second* Q _ E X C X 1Q T H " 4.16 XlO* 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 0. 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 -*- 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 ^-j^ or by .24. Q = 0.24 OE. or 0.24 ^ Examples. A difference of potential of 5.5 volts is main- tained at the terminals of a wire of A ohm resist- ance. How many joules per second are developed? Solution: By Ohm's law, current = 5.5 -*- tV = 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 137J4 ohms resistance. What heat does it develop per second? Answer: The resistance = 137J6 X 50. By rules 30 and 34 we have, rate of heat produced = .8* X 137# X 50 = 4400 watts. 4400 X 0.24 = 1056 joules per second. Rules 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 = 137J4 ••- 50 = 2.75 ohms. By rules 31 and 34 rate of heat produced = HO 8 + 2.75 « 4400 watts and 4400 X 0.24 == 1056 joules per second as before. Eules 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«m of rate, not of absolute quantity. Hence watts "per sec- ond " is not stated, as it would be meaningless o* WORK AND ENERGY. 57 redundant, while the joule, denoting an absolute quantity, has to be stated "per second " to indicate the 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* 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 conTectlon, 4000 Joole 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 10" 4 = .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 •+" 4TJW = 76.92, the number of degrees C. above the air to which the conductor could be heated by such a current. WORK AND ENERGY. 69 Besults of this character are only approximate, as the coefficient, dta, is not at all accurate. Rule 37. The cube of the diameter In centimeters of a wire of any given inateriai that will attain a given temperature centigrade under a given current Is equal to the product of the square of the current In amperes x Sp a .Resistance In microhms of the substance of the wire, multiplied by .000391, and divided by the tem- perature in degrees centigrade. d . 3> Q»xSp.Bes. x. 000391 Examples. A lead safety catch is to be made for a current of 7.2 amperes. Its melting point is 335° 0., and its specific resistance 19.85 microhms per cubic centi- meter. What should its diameter be? (Day.) Solution: By the rule, the cube of the diameter = 7.2 2 X 19.85 X .00039 + 335 = .001198. The cube root of this gives the diameter in centimeters. It is .10582 or .106 cm. A copper wire is to act as safety catch for 500 am- peres: melting point 1050° C— Sp. Eesistance 1.652 microhms. What should its diameter be? (Day.) ~ , . ^ , . j« i. 600»x 1.652 x. 000391 Solution: Cube of diameter = j^ ■■ 1B0 » 418 = .1523. The cube root of this is .5341 centi- 1050 * meter, the thickness of the wire sought for. It will be observed that in this rule no attention is paid to the length of the wire, as the effect of ^ a current in melting a wire has no reference to its 60 ARITHMETIC OF ELECTRICITY. length. The time of fusion will vary with the spe- cific heat, but will, of course, be only momentary. WOEK OF A CUBBENT. Role 38* The work of a current In a giren circuit Is proportional to the volt amperes. w — EC Example. A current A of 3.5 amperes with difference of potential in the circuit of 20 volts is to be com- pared to B, a 3 ampere current with a difference of potential of 1000 volts in the circuit; what is the ratio of work produced in a unit of time? Solution: Work of A : work of B :: 3.5 X 20 : 3 X 1000 or as 70 : 3000 or as 1 : 42^- Rule 39. The work or a current In a given ctrcnlt Is equal to the volt -coulombs divided by the acceleration of gravitation (9.81 meters). This gives the result In kilogram-meters. (7.23 foot pounds.) _ Tr _E^C 1 t w -~9^r Example. A current of 20 amperes is maintained in a circuit by an E. M. F. of 20 volts. What work does it do in one minute and a half (90 sec.)? Solution : Work — 20 x 20 x 90 sec -4- 9.81 — 3670 kgmts. and 3670 X 7.23 — 26,534 foot pounds. Note. — This is easily reduced to horse-power. 26,534 foot pounds in 90 sec. = 17,688 foot pounds in 1 min. 1 H. P. = 33,000 foot pounds per min. .\ £$$$$ = .536 H. P. of above current and circuit. The same result can be ob- tained by Rule 41 thus: ^ = .536 H. P. WORK AND ENERGY. 61 Rule 40. To determine work done by a current In a given clrcnit apply Rules 30, 31 or 32 as the case re- quires. These give directly the watts. Multiply by sec- onds and divide by 9.81. The result Is kilogram-me- ters. Examples. 10 amperes are maintained for 55 sec. through 15 ohms. What is the work done? Solution : By rule 32, watts — 10 a X 15 — 1500. Work = 1500 X 55 + 9.81 = 8409 kgmts. 1000 volts are maintained between terminals of a lead of 20 ohms resistance. Calculate the work done per hour. Solution: By Eule 32 watts = 1000* -«- 20 = 50,000. One hour = 3600 sec. Work = 50,000 X 3600 -*- 9.81 = 18,348,623 kgmts. These rules give the basis for determining the ex- pense of maintaining a current. The expense of maintaining a horse-power or other unit of power or work being known the cost of electric power is at once calculable. Electrical Horse-Power. Power is the rate of doing work or of expending energy. In an electric circuit one horse-power is equal to such a product of the current in amperes, by the E. M. F. in volts as will be equal to 746. Rule 41. The electric horse power Is found by multi- plying the total amperes of current by the volts or K. M. F, of a circuit or given part of one and dividing: by 746. MP -&■ 63 ARITHMETIC OF ELECTBICITT. Example. 260 incandescent lamps are in parallel or on mul- tiple arc circuit. Each one is rated at 110 volts and 220 ohms. What electric H. P. is expended on their lighting? Solution: The resistance of all the lamps in par- allel is equal to Hi ohm. The current is equal to 110 + H» = 125 amperes. H. P. = 125 X 110 -*- 746 =» 18A H. P. or 13& lamps to the electrical H. P. As it is a matter of indifference as regards absorp- tion of energy how the lamps are arranged, a simpler rule is the following, where horse-power required for a number of lamps or other identical appliances is required. Rule 42. multiply together the voltage and amperage of a single lamp or appliance ; multiply the prodnct by the nnmber of lamps or appliances and divide by 746* Example. Take data of last problem and solve it. Solution: Current of a single lamp = ffl = % am- pere. H. P. = 110 X y 2 X 250 -*- 746 = 18A H. P. When the voltage and amperage are not given directly, the missing one can always be calculated by Ohm's law and the above rules can then be ap- plied. The same can be done by applying following: Rale 49. To determine the electrical Horse-power apply Rules 30, 3 1, or 32; these give directly the watts; multiplying the result by -i^ or dividing by 746 gives the horse-power. • 4S WORK AND ENERGY. 63 Examples. A current of 10 amperes is maintained through 50 ohms resistance. What is the electrical horse- power? Solution: By rules 30 and 43 we have watts = 10* X 50 = 5000 and electrical horse-power = 5000 -f- 746 or 6.7 H. P. An electromotive force of 1500 volts is maintained in a circuit of 200 ohms resistance. What is the electrical horse-power? Solution: By Bules 31 and 43 we have watts » 1500* -*- 200 = 11,250. Electrical horse-power = 11,250 «•• 746 or 15 H. P. (nearly). Thus the volt-amperes or watts are units of rate of heat energy or of rate of mechanical energy. The ratio of joules per second to a horse-power is 746: 4.16 or 179.3 joules per second = 1 H. P. Other ratios of power and heat units will be found in the tables. Duty and Efficiency of Electric Generators. Rule 44. The duty of an electric generator Is the quo- tient obtained by dividing the total electric energy by the mechanical energy expended In turning the arma- ture. j. e. H. P. (total) v " m.H.F. Examples. A dynamo is driven by the expenditure of 58 H. P. Its internal resistance is 10.7 ohm. Th* 64 ABITHMETIC OF ELECTRICITY. resistance of the outer circuit is 150 ohms and it maintains a current of 16 amperes. What is its duty? Solution: The total electrical H. P. is found by Rules 30 and 43 to be 16* X 160.7 + 746 = 55.1 H. P. Duty = 55.1 * 58.0 = 95*. The result must always be less than unity; if it exceeded unity it would prove that there had been an error in some of the determinations. Rule 45. The commercial efficiency of a generator is the quotient obtained by dividing the electric energy in the outer circuit by the mechanical energy expended In turning the armature. O THff- - e * H * P * (Qjjter circuit) m. H. P. Examples. What is the commercial efficiency of the dynamo just cited? Solution: The electrical H. P. of the outer circuit is found by the same rules to be 16 2 X 150 -«-746 = 61.5 commercial efficiency = 51.5 ■+■ 58.0 = 88.8*. Rule 46. The resistance of the outer circuit Is to the total resistance, as the commercial efficiency Is to the duty. Examples. Take the case of the generator last given and from its duty calculate the commercial efficiency. Solution: 150: 160.7: : x: 95.0. \ x = 88.8 or 88.8*. OHAPTEE VIII. batteries. General Calculations of Current. A battery is rated by the resistance and electro- motive force of a single cell, which factors are termed the cell constants. In the case of storage batteries, whose susceptibility to polarization is very slight, the resistance is often assumed to be neglible. It is not so, and in practice is always knowingly or otherwise allowed for. From the cell constants its energy-constant may be calculated by Eule 31, as equal to the square of its electro-motive force divided by its resistance. This expresses its energy in watts through a circuit of no resistance. There are two resistances ordinarily to be consid- ered, the resistance of the battery which is desig- nated by E or by n E if the number of cells is to be implied and the resistance of the external circuit which is designated by r. Rule 47. The current given by a battery Is equal to Its electro-motive force divided by tbe sum of the exter- nal and Internal resistances. o =_ u B + r 66 ARITHMETIC OF ELECTRICITY. |H$) ($)($)<$)($) Six cells in parallel. Six cells in series. Biz cells— two in parallel, three in series. Six cells— three in parallel, two in series. « Abbangkment of Battery Cells. BATTERIES. 67 Example. A battery of 50 cells arranged to give 75 volts E. M. P. with an internal resistance of 100 ohms sends a current through a conductor of 122 ohms resistance. What is the strength of the current? Solution: Current = 75 -*- (100 + 122) = .338 ampere. This rule has already been alluded to under Ohm's law (page 14). Abbangement of Cells in Battery. In practice the cells of a battery are arranged in one of three ways, a: All may be in series; b: all may be in parallel; c: some may be in series and some in parallel, so as to represent a rectangle, s cells in series by p cells in parallel, the total number of cells being equal to the product of s and p. Other arrangements are possible. Thus the cells may represent a triangle, beginning with one cell, followed by two in parallel and these by three in parallel and so on. This and similar types of arrange- ment are very unusual and little or nothing is to be gained by them. Rale 48. The electromotive force of a battery Is equal to the E. M. F. of a single cell multiplied by the number of cell* In series. Rule 49. The resistance of a battery is equal to the number of its cells in series, multiplied by the resist* ance of a single cell and divided by the number of lta cells in parallel. •o battery _ s_B 68 ARITHMETIC OF ELECTRICITY. Examples. A battery of 50 gravity cells 1 volt, 3 ohms each is arranged 10 in parallel and 5 in series. What is its resistance and electromotive force? Solution: Eesistance = 5 X 3 + 10 = 1.5 ohms. E.M.F.=5X1 = 5 volts. The same battery is arranged all in parallel; what is its resistance and E. M. F. ? Solution: This gives one cell in series. Eesistance = 1 X 3 -*- 50 = .06 ohms. E. M. F. =1X1 = 1 volt. The same battery is arranged all in series; what is its resistance? Solution: This gives one cell in parallel. Eesistance = ^x 8 = 150 ohms. E. M. F. = 50 1 X 1 = 50 volts. The current given by a battery is obtained from these rules and from Ohm's law. Example. 150 cells of a battery (cell constants 1.9 volts, i ohm) are arranged 10 in series and 15 in parallel. They are connected to a circuit of 1.7 ohms resist- ance. What is the current? Solution: The resistance of the battery = ^2Li = .333 ohms. The E. M. F. = 10 X 1.9 = 19 volts. Current = 19 ••- (.333 + 1.7) = 9.34 amperes. BATTERIES. 69 Cells Kequired foe a Given Current. To calculate the cells required to produce a given current through a given resistance and the arrange- ment of the cells proceed as follows. Rule 50. Calculate the cell current through zero exter- nal resistance. Case A. If It Is twice as great or more than twice as great as the current required apply .Rule 51, Case B. IT less than twice as great and more than equal or less than equal and more than one half as great as the current required and so on apply Rule 52. Case C. If the cell current is equal to or Is a unit- ary fraction <* * £, etc.) of the current required apply Rule 53. TVS 9 Rule 51. Case A. Divide the required difference of po- tential of the outer circuit by the voltage of a single cell diminished by the product of the required current mul- tiplied by the resistance of a single cell. Arrange the cells In series. Examples. Five lamps in parallel, each of 100 volts 200 ohms, are to be supplied by a battery whose cell constants are 2 volts i ohm. How many cells and what arrangement are required? Solution: Cell current = | = 10 amperes. The resistance of the five lamps in parallel (Rule 12) = 2p_o _ 4q Q h ms# Th e required current therefore =» W = 2} amperes. As 10 exceeds 2i X 2 (Eule 50) it falls under case A. By Eule 51 the number of cells is 2 -(2ix*) = *f = 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 j = 13.4 ohms. The resistance of the lamps in par- allel is 40 ohms. Hence by Ohm's law the cur- rent **" 40+13.4 = &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 = ^^| or | am- pere. As 10 exceeds i X 2 (Rule 50) it falls again under case A. By Eule 51 cells required = fe_^° xi) = 263. Proof: Current «= foe+iooo = i ampere the current required. 526 is the number of cells multiplied by the Toltage 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. Rule 52. Case B. 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 BATTERLB8. 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 2i X 3 and more than M X 1; therefore it falls under Case B Solution: A group of two cells in parallel gives tH = 8 amperes. 8 exceeds 2i X 2 . •. applying Rule 49 we have number of groups = 100+ [1 - (2j X .125)] = 146 groups in series. Total number of cells = 2 in parallel, 146 in series = 292 cells. Proof: Current = 146 + (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 2J amperes. The cell current is } ampere. The cell current therefore equals (i -*- 2i) A of the required current. This falls under Case C. and is solved by Rule 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 -+• 40 = 20 cells must be placed in parallel. Total cell* -*• 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 slm* ply solved for the total number of cells by dividing the external energy by the cell energy and multiplying by 4. This gives the number of cells. Example. Take as cell constants .75 volt 3 J ohm giving i am- pere. Assume 20 lamps, each 50 volts, 50 ohms and 1 ampere. As J -*• 1 is a unitary fraction (J) Case C. applies. Solution: Cell energy = J X .75 = .375 watts. Ex- ternal energy = 50 x 1 X 20 = 1000 watts. (1000 -*- .375) X 4 =-10,666 cells. Solution by Eule 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 = ( syi^f^) + iooo = 1 am P er ©- 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 = B , ■ 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) *P = 40 ohms. Therefore the efficiency of the battery is 40.0 + (40 + 13.4) = .749 or 74. W. Rale 5 6. To calculate the number of battery cells and tbeir arrangement for a given efficiency : Express. tb© efficiency as a decimal, multiply tbe resistance of tb© external circuit by tbe complement of tbe efficiency (1— efficiency) and divide tbe product by tbe efficiency; this fives tbe resistance of tbe battery* Add tbe two resistances and multiply tbeir sum by tbe current to be maintained for tbe E. M. F. of tbe battery. Arrange tbe cells accordingly as near as possible to tbese 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 = X W = 13* ohms, the resistance of the battery. By Ohm's law the E. M. P. of the battery = (40 + 13*) X 2.5 = 74 ARITHMETIC OF ELECTRICITY. 133i volts. These constants, 13i ohms and 133i volts, require 67 cells in series and 2 in parallel. Proof: a. Of efficiency, by Kule 55, -^^ = .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. Rale 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 3i 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, I ohm. Calculate the cells and their arrangement for 80 per cent. efficiency. Solution: By Rule 56 efficiency = .80 and qq ' = 74 ohms, which is the required resist- ance of the battery; 74+ 30 = 374 ohms are the total resistance of the circuit. By Ohm's law, 37J X 34 = 125 volts, the required E. M. P. of the battery. This requires 63 cells in series, with a resistance for one series of 63 X 4 = 104 ohms. To reduce this to 7j 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 g rou P s * n series. As there are 2 cells in parallel the total cells are 116, of resistance, 58 X A = 4.83 ohms, and of E. M. P., 58 X 2 = 116 volts. Proof: Of efficiency by Rule 55, g^j^ = 86.1*. Of number and arrangement of cells 30 V^gg = 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. Rale 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=.043xH. 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 = 1.14 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: Role 60. The effective rate of work or the rate of work In the external circuit of a battery, Is equal to the 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 = Wk = 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 lbs.) of work are found by multiplying the combining equivalent of the chemical by the number of equivalents In the reaction by the constant .000101867 and dividing by the pro- duct of the E. M. F. by the valency of the element Id question* — ^ Equiv. x n x .000101867 B 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- fore = « x ^» = mQ9 g^ 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 = m4 ^ x x f )10t867 = .0076 grams. Add 56.46# for water of crystallization. All these quantities are for one kilogram-meter of work (7.23 foot lbs.) which may be more or less Effective according to circumstances as developed in Rules 44, 45, and 60. Decomposition of Compounds by the Batteet. 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 Eule 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 calculate the metal or other element lib- erated by a given cnrrent per given time proceed as fol- lows* Calculate the resistance. Determine the counter- electromotive force of the solution by Rule 58 and sub- tract It from the K. M. 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 ^| = 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 + 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 = .24+028 = 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. BATTERIES. 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 flax, 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 &rea of one square centimeter, situated in the center of a circle 10 centimeters in diameter, ELECTBO-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 0. 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 takten 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 Fobce and the Magnetic Cib- 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 fobce 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 substancer. In a bar magnet the return branch of the circuit is through air. ELECTBO-MAGNETS, DYNAMOS AND MOTOBS. 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: Rule 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 be 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* netlzation by magnetizing force* Permeability has to be 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. Role 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 tne area of contact and divided by 72,134,000. In centimeter measurement tne traction in pounds Is equal to tne square of tne 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 (8 » x 081); for dynes of traction the divisor Is 25.132 <8») 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* X 3 = 499,230,000. This di- vided by 72,134,000 gives 6.914 lbs. 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—300. From them a number Is taken (40,000 lines per square Inch for cast Iron or ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 87 100,000 lines per square Inch for wrought Iron are good general averages) and is multiplied by the cross-sec- tional area of the magnet core. Rale 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 0.31 32 divided by the product of its cross-sectional area and permeability. b. 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. = ~^jp = 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 Cibcuit 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 op Lines op Fobce. 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 saps pins 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 11 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 Rule 68 the reluctance of the armature core is ^ £ im X .3132 = .000008. The reluctance of a single air gap is i 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 0058 1 x2 = 100 (about). The coefficient of leakage = 100 f ^° = 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. ^ = i or .5 nearly. Prom 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 ^^ = 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 TO. 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 ■*■ 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 fob Magnetic Cibcuits. 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. P.). 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. Prom 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 \ 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 1| inches of the diameter of the armature core embedded or included within or embraced by them. One-half of this ELECTRO-MAGNETS, DYNAMOS AND MOTORS. 9? 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 ^ x 8 ^ 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 =5 air gaps reluctance (2). The ratio jj of the table for determining the leak* age between cylindrical magnet legs is 4 x 5 8 14 - = .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) + 64.06 = 2.08 (5). To find the relative reluctance of the magnet core whose yoke may be taken as of mean length inohes 94 ARITHMETIC OF ELECTRICITY. and of area equal to that of the core (3.14 X 2 2 = 12.56) we have to first determine the permeability. g %ffi = 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 * + J£*£* M - .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, bat 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 (£) ■■ j^g — .56 giv- ing from the proper table 6.000 (about), ^ 10 = 30 relative permeance of air space between legs. For air gaps reluctance £ -*- 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 ~ .007148 and permeance = -^^ = 139. Coefficient of leakage — ^±*> = 1.21. If the depth of the air gaps was reduced to J 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 Abmatures. 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. Bale 7 1 • One -volt E. RE* 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. P. 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 -8 = 98 X 10" 2 = Ak volts E. M. F. generated, or 1W0W0V0 = Tfifr 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 8 , and 10 8 X 10- 8 — 1 volt Rule T2. The capacity of an armature for current Is determined by tbe cross-section of its conductors. This should be such as to allow 520 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 th 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 21' X 3.14 = 1387 square mils. By the rule the allowable current in amperes for a single lead of such wire is -W = 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. Bale 78* a. The B. RE. F. varies directly with the square of the else 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 E. 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 of 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 Rule 73, sections a and b, 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 ^;^ =2.6 turns. For 100 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, DYliAmkJ^MOTbS& 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 }4 the ampere capacity; this reduction (by Rule 73 b) 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 A of that required. We must therefore divide the number of turns in the trial armature by .89018 2 , giving for the number of turns ^ = 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 = , ^^xi64xi5 = 99>?8 VOlt8. 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 ~jjp==.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 a = .02371 102 • £a£fxkMT$cr or electricity. and .01190 + .02371 = i (nearly), showing that the new armature has one half the ampere capac- ity of the old, or 40 X J = 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 Rule 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 2j£. (2J^)$ = 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 s : 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 .244* -*- .154 2 inches = .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 14j£ 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. Rale 74* The cross-sectional area of the field-magnet cores Is equal to the lines of force In the field divided by the magnetic flax (column B) for the material selected and corresponding: to the chosen permeability (m) 9 mul- tiplied bjr the leakage coenlcient. 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 core3 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 8 ' 45 ^ff ) 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) = m^ (taking r = 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 178x90 oo- 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 A inch for clearance or windage giving (.224 X 2) + .1 = about .6 inch for its depth. Its 106 ARITHMETIC OF ELECTRICITY. reluctance is 6 *^ 182 *=* .00108. As there are two air 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 f 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 H¥* = 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 **5P for the length of the armature wire, because it is in parallel, and therefore is i the length and i 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* X 5104 : a 2 X *¥*, and solving we find a? = 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 louamperes = * °^ m * P ro P er ty 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 = l 2 X 1, The armature resistance is .0419 ohms. There- fore the field resistance is -~- 9 = 24 ohms. The cur- rent through this is equal to W = 4 amperes (nearly). Therefore ^4** = 2910 turns of wire are needed. The length of such wire will be * ^g 2910 = 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. 6. 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 Rules 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 75. The cross-section of the feeder in cir- cular mils is equal to the product of 10.79 times the current in amperes times the length of the con- ductor 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. Referring 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: Rule 76. The drop in volts on any conductor ia found by multiplying- together 10.79, the current in ampere* and it« length in feet, then divide this product by it* 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 Cars. 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 cor on a level track in average condition, multiply the weight of the ear in tons by 80. Example. Find the force required to drag a car weighing 7 tons on a level track. Solution: 7 X 30 = 210 lbs. 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 hone 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 33,000 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 9X30 = 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 And 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 weight 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 33,000 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 -*- 23,100 = 45.021 H. P. Ans. CHAPTER XL ALTERNATING CURRENTS. By far the greater part of calculations in the do- main of alternating currents lie in the relalm 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 : Rale 80. A. Having: g-lven the cycles per second, to find the alternations per minute multiply the cycles per second by 120. B. Having: flriven the al- ternations per minute, to find the cycles per second divide the alternations per minute by 120* 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: 15,000-1-120 = 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, diTide the revolutions per minute by 60 and multiply the quotient by the number of pairs of poles in the Held. Example. Find the frequency of a 16-pole alternator run- ning at 937.5 revolutions per minute. ALTERNATING CURRENTS. 117 Solution: 937.5 -f- 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. Rale 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. Note that these llgrures 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. Rale 88. The E. M. P. of self-induction of a tfiven coil Is found by multiplying: together 12.5664, the total number of tarns in the coil, the number of tarns per centimeter lensrth of the coil, the see- 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 1,000,000,000, multiplied by the time taken by the current to reach its maximum value* Example. Find the volts of counter E. M. P. 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 9 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 la found by multiplying together 12.56649 the total number of turns in the coil, the number of turns per centimeter length, 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. P. 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. P., 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. Rale 86. The resistance dne to ■elf-induction equals 6.2832 times the product of the frequency and the Inductance. Example. Find the inductive resistance of a circuit whos9 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 coll or circuity 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 -r- 20 = .181 second. Ans. CHAPTER 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: Rale 88. The quantity stored equal* the product of the E. M. P. applied and the capacity of the condenser* Q, = ESC* Rule 89. The capacity of a condenser equals the quantity stored divided by the applied S3. M. P. C = -. B Rule 90. The E. M. F. applied to a condenser equals the quantity stored divided by its capacity* 11 = —. C The quantity stored in a condenser is measured in coulombs (i. e., ampere-seconds) ; the E. M. P. 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 Y2 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 ~- 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 -5- .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 be 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 01. The alternating current in a circuit hav- ing capacity equals the product of 6.2832, 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 H. 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 by 6.2832 times the product of 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. Rale 03. 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 -f- .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 he 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. Tfte impedance is the apparent resistance of an alternating current circuit CONDENSERS. 125 Rale 94. To And the impedance of a circuit whose ?hmic resistance can be neglected and which has an inductance and a capacity in series, calculate both the Inductive resistance and the capacity re* sistance; their difference will 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 -r- 6.2832 X 60 X .000100 = 1 -r- .0377 = 26.52. Impedance = 26.52 — 7.42 = 19.1 ohms. Current, by Ohm's Law, = 40 -r- 19.1 = 2.08 am- peres. Ans. CHAPTER XIII. DBMOKSTBATIOlSr OF BULES. 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 1 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- Raie 8, page 1 9. 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 nnity. 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. Rale io f 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 ii f page 22. This is a form of Rule 10. Call the two resistances x and y. The sum of their reciprocals is l+\ which is the conductance of the two parallel circuits or parts of circuits. Reducing them to a common denominator we have: iL -f «_ which equals ^^, whose reciprocal is ^-. Rale it, 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 ^j^^r* As the resistance of a conductor DEMONSTRATION OF RULES. 129 raries also with its length and specific resistance we have as the expression for resistance: Sp. Bea. x 1.2737 X 1 d 9 Bole 18, page 32. Assume two wires whose lengths are I and Z 1 , their cross sectional areas a and #!, their specific resistances 8 and s l9 and their resis- tances r and r v From preceding rules we have for each wire: r = s i (1) and r x = 8 t J 1 (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: 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. Gall the drop e, the combined resistance of the lamps R, and the resistance of the leads x. Then as the whole resistance is expressed as 300 (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 : re or eR Rule 25, page 44. From Rule 22 we have: 130 ARITHMETIC OF ELECTRICITY. Call the resistance of a single lamp r, then we have by Eule 12: R =£ (2) Substituting this value of B in equation (1) we have: _ er ~ n X (100— €) (3) Prom Rule 24 we have, calling the cross-section a: „ l X 10.79 Substituting for z its value from equation (3) we have: „ l X 10.79 X n X (100-e) ° " (5) er 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: „ l X 21.58 X n X (100-g) a = ■ 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 a E = (V E t , and E = %?.B lf 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 BULE8. 131 Rule 87, 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 0. G. S. units, multiplied by the resistance in 0. G. S. units and divided by 4.16 X 10 7 , the latter division efEecting 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. Eesist. X 10 8 X ^ 1 ampere = 10" 1 0. G. S. unit. If we express the current in amperes we must multiply it by 10" x , in other words take one-tenth of it. Our expression then becomes for heat developed in one second , Sp. Resist. X 10* X 4 <*)• X n d* X 4.16 X W The area of one centimeter of the wire is nd square centimeters. The heat developed per square centimeter is found by dividing the above expression by ic d giving: ay- 1 Sp. Resist. X 10* X 4 w* d« X 4.16 X 10* The heat developed is opposed by the heat lost which we take as equal to dW per square centimeter per degree Cent, of excess above surrounding me- dium. Therefore taking t° as the given tempera* 132 ARITHMETIC OF ELECTRICITY. ture cent, we may equate the loss with the gain thus: t° _ /c \« x Sp Resist. X 10* X 4 4000 "~ U0>? «■ d« X 4.16 X 10* c» X Sp. Resist. X 10* X 4 X 40000 - »« X 4.16 X 10 7 Xt° * c* X Sp. Resist. X .00089 t° m Rule 511, 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: n nE c= ^r+f (i) which reduces to: Cr w ~ e-cr (2) but C r = e. •'* n " E-CiJ (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 Jg = (1 ~^ncy >Xr ' The rest of the rule is deduced from Ohm's law. DEMONSTRATION OF RULES. 133 Rule 57, 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 A = 3 ohms, and adding to them 27 cells in series with a resistance of 27 x£ = 4£ ohms, a total of 7* 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. volte X^coulomb. or f()r one coulomb = «tfta Thig ex . presses the work done by one coulomb. Let the volts = E, and equate these two expressions: i = •000010384 X H X 424, which reduces to E - H X .043. Auieei, page 78. For the work (in kilogram- meters) done by a current (volt-coulombs) we have the general expression: _ Yolte x coulombs A11 EQ Wa3 m — or ra (i) Making W ■■ 1 (i. e. one kilogram-meter) and transforming, we have, as the coulombs correspond- ing to 1 kilogram-meter: «- X (2) 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 ~r or, indicating weight by G, Qsx .000010884 X equiv. x number equlv. y 0.81 valency E (3) but .000010384 X 9.81 = .000101867. Q _ equlT. X n X .000101867 •"• Ex valency (4) Bale 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 jin 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 &. 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 i -«- a? = ^. = diameter of X X 9 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 (|y) f — |;. 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 POWEK8 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 a 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"* indicates t4tf; 10 -8 indicates nftnr 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. S-UXlO^^^j^orj^S^. Re- gard must be paid to the decimal point as is done here. NOTATION IN P0WEB8 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*) + (54 X 108 )= 105 X W. (9.1 X 10-*) + (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*) - (51 X 10*) = 3 X 10«. 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 10») - (5 X 10 7 ) = (5 X 100 X IOt) - (5 X 10*)= 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* and 5 X 10 8 . 5 X 10* = 5 X 10* X 10"* : (5 X 10» X 10-^ + (5 X 10-t) = 500005 X 10-*. This may be reduced to a fraction ^ = 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*) X (9 x 10 8 ) = 225 X 10 1 *. (29 X 10-*) X (11 X 10 7 ) = 319 X 10*. (9 X 10 8 ) X (98 X 10-«) = 882 X 10 8 . 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-* (28 X 10" 8 ) •*• (5 X 10 8 ) = 5.6 X 10-*. if i TABLES 1 .8 S I f I i t ¥ t § § I 140 IL— EQUIVALENTS OF UNITS OF ABEA. Square Millimeter Square Gentlmet'r Circular MIL Square MIL Square Square Foot Square Millimeter 1 0.01 1978.6 1560.1 .00155 .0000108 Square Centimeter 100 1 197,861 155,007 .155007 .001076 QrcularMfl. .00060T .0000051 1 .78540 8X10-* Square MIL .000645 .0000065 1.2788 1 .000001 Square Inch 645.182 6.451 1,278,288 1,000,000 1 .006944 Square Foot «*,898.9 928.989 144 1 HL- EQUIVALENTS OF UNITS OF VOLUME. Cublo Inch Fluid Ounce Gallon Cubio Foot Cubic Yard Cu. Cen- timeter liter Cubio Meter Cubic Inch 1 .554112 .004329 .000578 16.8862 .016886 Fluid Oz. 1.80469 1 .007812 .001044 29.5720 .029572 Gallon 281 128 1 .188681 .00495 8785.21 8.78521 .008785 Cubic Ft 1728 957.506 7.48052 1 .087037 28315.8 28.8158 .028815 Cubic Yd. 46,656 25,852.6 201.974 27 1 764,505 764.505 .764505 Cu. Cent! .061027 .088816 .000264 .000085 1 .001 .000001 liter 61.027 88.8160 .264189 .085817 1000 1 .001 Cu. Meter 61027 88816 264.189 85.8169 1.8080 1000 1 141 IV.— EQUIVALENTS OF UNITS OP WEIGHT. Grain. Troy Ounce. Pound Ays. Ton. Milli- gram. Gram. Kilo, gram. Metric Ton. Grain 1 .020888 .000148 64.799 .064799 .000065 TroyOunce 480 1 .068641 81,108.5 81.1085 .081104 PoundAvs. 7,000 14.6888 1 .000447 458.598 .458598 .000454 Ton 82,666.6 2240 1 .001016 1.01605 Milligram .015482 .000082 .000002 1 .001 .000001 Gram 15.4828 .082151 .002205 1000 1 .001 Kilogram 15,482.8 82.1507 2.20462 .000984 1,000,000 1000 1 .001 Hetrio Ton 82,150.7 2204.62 .98421 1,000,000 1000 1 142 V.— EQUIVALENTS OF UNITS Erg. Meg- Grain-de- gree C. Kilogram- degree O. Pound- degree O. Pound- degree F. Erf. 1 .000001 Meg.-erg. 1,000,000 1 .024068 .000024 .000068 .000096 Gram-degree 0. 41.6487 1 .001 .002206 .008968 Ktiogram-degreeC. 41,648.7 1000 1 2.2046 8.9688 Pound-degree C. 18,846.6 468.69 .40869 1 1.8 Pound-degree P. 10,470.1 261.996 .261996 .666666 *\ Watt-Seeond. 10' 10 .24063 .000241 .000681 .000966 Gram-oentlmeter. 981 .000981 .0000286 Kilogram-meter. 98.1X10* 98.1 2.86108 .002861 .005206 .009870 Foot-Pound. 18.6626 .826426 .000626 .000720 .001296 Hone-Power-Sec. English. 7469.48 179.486 .179486 .8967 .71248 Hone-Power-8ee. Metric 7867.6 177.076 177.076 .890876 .70276 143 OF ENERGY AND WORK. Watt- Second. Gram- Centim'tr. Kilogram- meter. Foot- Ponnd. Horse- power- second English. Horse- power- second Metric. 10-' .001019 Erg. .1 1019.87 .010194 .078784 .000184 .000186 Meg-erg. 4.16487 42,858.6 .428686 8.06866 .00667 .006647 Gram-degree O. 4154.87 428.686 8068.66 5.67 5.64708 Kflogram-degTeeC. 1884.66 192.114 1889.6 2.52668 2.66149 Pound-degree O. 1047.08 106.780 772 1.40864 1.42806 pound-degree F. 1 10,198.7 .101987 .787887 .0018406 .0018602 Watt-Second. .000098 1 .00001 .000072 Gram-Centimeter. 9.81 100,000 1 7.28828 .018152 .018884 Kilogram-meter. 1.85626 18,825.8 .188268 1 .0018182 .001848 Foot-Pound. 746.948 76.0692 660 1 1.01888 Horse-Power-Sec JEngHafr, 786.76 75 642.496 .986856 1 Horse-Power-Sec. Metric. 144 VL- TABLE OF SPECIFIC RESISTANCES IK MICBOHMS AND OF COEFFICIENTS OF SPECIFIC RESISTANCES OF METAIA. Spedflo Besist- anoe.Mi- crohms. Ooeffl- dents of Sp. Bes. Spedflo Resist- ance, Mi- crohms. Coeffi- cients of Sp. Bee. Annealed Silver. .., Hard Silver.. Annealed Copper. . , Hard Copper , Annealed Gold Hard Gold Annealed Aluminum Compressed 23no... Annealed Platinum " Iron...., 1.621 1.4KB 1.616 1.662 2.081 2.118 2.945 6.689 9.168 9.826 .9412 1.0228 1.0000 1.0228 1.2877 1.8107 1.8224 8.6204 6.6671 6.0796 Annealed Nickel. Oompres'dTtn... * Lead. " Antimony " Bismuth liquid Mercury 2 Silver, 1 Platinum. German Silver .. ., 2 Gold, 1 Silver.. 12.60 18.86 19.86 86.90 182.70 99.74 24.66 21.17 10.99 7.7970 8.2678 12.2884 22.2158 82.1170 61.7206 16.2699 18.1002 6.8008 SPECIFIC RESISTANCE OF SOLUTIONS AND LIQUIDS. XAnHOBSSBf AKD OTHD8. Names of Solutions. Temper- ature Centi- grade. Temper- ature Fahren- heit Spedfle Resistance. Ohms. Copper Sulphate, concentrated 9« 48.2° 29.82 " with an equal volume of water «« «♦ 46.64 •• with three volumes of water <« «i 77.68 Common Salt, concentrated 18° 66.40 6.98 '* with an equal volume of water . . a it 6.00 " with two volumes of water. . . . t« it 9.24 " with three volumes of water. . tt <• 11.89 Zinc Sulphate, concentrated 14" 67.2° 28.00 44 with an equal volume of water . . t« «< 22.76 •* with two volumes of water <« (« 29.75 Sulphuric Add, concentrated 14.8° 14.6* 67.8° 68.1° 6.82 50.5*, Spedfle Gravity 1.898. . . 1.086 44 29.6*, Spedfle Gravity 1.216. . . 12.8» 64.6° .88 " 12* Spedflo Gravity 1.080... 12.8° 66.0° 1.868 Nitric Add, Spedflo Gravity 1.86 (Blavier) .... 14° 67.2° 1.46 U M «4 M 84° 76.2° L2S Distilled Water, (Temp'tura unknown) (Poufllet) 988. 14i> m— RELATIVE RESISTANCE AND CONDUCTANCE OF PX7RE COPPER AT DIFFERENT TEMPERATURES. II ji Relative Resistance. Relative Conductance 11 n fi Relative Reslstanoe. Relative Conductance 0* 82* 1. t 16° 60.8* 1.06168 .9419 1 88.8 1.00881 .99620 17 62.6 1.06568 .98841 2 86.6 1.00766 .9925 18 64.4 1.06969 .98494 8 87.4 1.01186 .98878 19 66.2 1.07866 .98148 4 89.2 1.01515 .98608 20 68. 1.07754 .92804 5 41 1.01896 .98189 21 69.8 1.08169 .92462 6 42.8 1.0228 .97771 22 71.6 1.08568 .92120 7 44.6 1.02668 .97406 28 78.4 1.08964 .91782 8 46.4 1.08048 .97042 24 75.2 1.09866 .91445 9 48.2 1.08485 .96679 25 77. 1.09769 .9111 10 60 1.08822 .96819 26 78.8 1.10162 .90776 11 61.8 1.04210 .95960 27 80.6 1.10667 .90448 12 68.6 1.04699 .95608 28 82.4 1.10972 .90118 18 65.4 1.0499 .96247 29 84.2 1.11882 .89784' 14 57.2 1.06881 .94898 88 86. 1.11785 .89457 15 69 1.06774 .94541 146 VIIL-AMKEIOAN WIRE GATJGB TABLE. Properties of Copper Wire : Spedflo Grayity, 8.878 ; Specific CondacttTftj, 1.766 at T5°. F. i SlDL WeiOUT A.SD Ltssni, B*MM4irC*\. M 1 & B B o Plnm- «ter In Mill, Square of Diameter M 'v Qrtifirt Foot. Po'nfo pet 10O0 Fu#t Fcot per Pound. Ohm-' per 1000 Feet par Ofcua Ohm a p*r Pound. 1*1- u 4i 0000 460.000 211000.0 447T.2 689 60 1.561 .001 19920,7 /» T85 430 (KM} 4i^.f4<i 167W4 9 8550.5 607. -22 1.971 .068 15304,0 000125 , 262 00 364.800 138079.0 2>l.'i - 402.25 £.486 .080 12084.2 ,000198 1 208 «> 85*4.900 liiWlrt r, 2286 2 819,17 3.183 ,101 9945,8 ,000815 165 1 289.800 Ni^4.4P 1770.9 252 98 8.052 .127 7882,9 .000501 lft< *J 2.17 aso MAT* X 1404.4 200,68 4,904 .100 6251.4 .0O0T99 108 B ^>.4^» 02638.68 lUM 150.09 6.2S6 .202 4907.8 .001268 31 4 204.810 41742,67 683.3 126.17 7.025 .254 8931.6 .00*016 60 D 181.940 88102.16 700.4 100.06 8.005 .321 8117J .008206 52 6 tag .aw 26250 48 000.4 70.84 12,604 ,404 2472,4 ,005095 41 T m.'^t 20H16.72 44W.4 62.02 15.898 .AO0 1060.6 ,008106 82 e 138,490 16300.64 848.8 49.90 20,040 .643 1660.0 .01269 26 • 114.430 13094.22 8T7.1 82.58 25,265 .811 1238,8 ,02043 20 10 101.880 10881.07 218,7 81.88 81.867 1.023 977.9 moe 10 13 90.742 8234 11 174.3 24.80 44u;ij 1.2*9 775,5 .061B1 13 12 BO SOS fa-lit w i*M 19.74 00.650 l r 62P 015.02 .03287 10,9 Vi n.96i 0178,89 108.6 1 :>«;:> 68.898 2 r 048 4*VJ5 .18087 8,1 14 64.034 4106.70 86, ST 12-41 80.080 3.585 8St,flQ .20830 6.4 Lfi 07,088 8256 76 68.83 8.84 101,626 8,177 806.74 .88133 0.1 IS 00,820 2082.67 54. OT 7.81 128.041 4.683 WM JMtt 4,0 IT 45.267 2048.19 ■l;\ B8 6.18 161 + 501 0.183 192.91 ,88744 M ]8 40,308 1624 88 84.87 4.01 908.868 li.Wrt 15299 1.8818 8,0 19 85.880 1202,45 26.00 3.780 204 136 3.477 117,96 2.2393 1.96 90 fti.wi 1021.51 21.00 8.086 824.045 10.394 90,21 8 8488 1.60 21 28.462 B10, 08 17,14 2.448 40r497 IK. 106 70,80 5.3580 1.2B n 25.84T 642.47 18.08 1.942 014.033 u; ftifl 60,51 s.&ow 1.08 a\ bjri 0O9 40 10,77 1.080 640.778 20 842 41 ■>■ 13 384 .80 JS 20 100 404 ol 8.05 1,221 819.001 26.284 88.05 21.024 .03 ar» n.800 820 41 6,77 .067 fi^.iifi 88.185 80.18 84 298 .00 so 16,940 254 0* 5.38 .768 1302,083 41,739 an tr;( 54.410 .40 ST 14.105 201 49 4 26 .608 1544.73T 0-2, 6*7 1>M 86.057 ,81 2B 12.641 159.79 8 39 ,484 2066 116 60.440 15 05 187,288 .25 29 u aw 126.72 2.69 ->-"r 260* J 67 83.752 11.04, 213. J04 .30 Hi' 10.020 100.50 2.11 .802 8811.208 100.641 9.466 840.805 .16 HI 8.92b 79. Tl X.«7 /m 4W LiMi 133.191 7. DOS 507,286 .IB 82 T.90D 03,20 1.83 .100 5268.158 f.v- nn .v;^> 8S4.267 .098 ftg 7 ii- 1 BO, IS 1.00 .151 6623.51 T iu«an 4.721 1403,78 .078 14 6,304 88.14 .847 .121 h2r}4.4&-t 267,160 8.748 224)7 .03 ,00$ 85 0,614 81.62 .638 .094 10688.80 886,61 2,809 3088,12 ,049 B6 5.000 25,00 .025 .075 18883.88 424.60 2.800 5661.71 .000 m 4,458 19. ea .ISO .060 16666.66 535,83 1,ftfl8 3922.20 .081 »s 8. B65 15,72 .315 ,045 22222,22 675,23 1.481 16000.0 ,046 a© 8,08! 13.47 .200 .m% 26*16,79 Bl,1M 1.174 22410.5 .0341 40 8.144 0.88 .210 .030 38333 83 1074.11 .981 80303,8 .015 147 EL— CHEMICAL AND THEBMO-CHEMICAL EQUIVALENTS. FORMATION Or OlLDBS. 1 Name of Compound. Formula. Valency. Chemical Equiv- alents. Combin- ing Weights. Thermo- Chemical Equiv- alents. Water Iron Protoxide HSO FeO Fe»03 ZnO CuO HgO II II III II II II 18 72 160 81 79.4 216 9 86 68.8 40.6 89.7 108 84.6 84.6 Iron Sesquioxide 81.9x8 PfncPxM* 48.2 Copper Oxide Mercury Oxide 19.2 16.6 Formation or Salts. Name of Base. Va- lency. Nitrates Sul- phates Chlo- rides Cya- nides. Iron n Formula. Chemical Equivalents Combining Weights Thermo-Chemical Equlv'lts Fe sr 90 18.9 FeSO* 186 68 12.6 FeClt 127 68.6 00 66 8.2 Zinc n Formula Chemical Equivalents Combining Weights Thermo-Chemical Equlv'lts ZnfN03)» 94.6 9.8 ZnSO* 161 80.6 11.7 ZnCls 186 68 66.4 ZnCyt 117 68.6 7.8 Copper n Formula. Chemical Equivalents .... Combining weights Thermo-Chemical Equlv'lts Cu(NOS)8 187.4 98.7 7.6 CuSO* 169.4 79.7 9.2 CuClt 184.4 67.2 81.8 CuCy* 126.4 62.7 7.8 Mercury II Formula. Chemical Equivalents Combining weights Thermo-Chemical Equlv'lts 163 7.6 HgSO* 280 140 9.2 186.6 9.46 V 126 16.6 148 X.— CHEMICAL AND KLECTEO-CHEMICAL EQUIVALENTS. Name Hydrogen Gold SflYar... Copper (Caprie) , Mercury (Mercuric). " (Mercuroui) Iron (ferric) , ** (ferrous) Nickel , Zinc Lead , Oxygen .... , Chlorine Symbols H An Ag On.. Hg„ Hg. Fe... 7e„ Ni Zn Pb O CI Valen- I III I II n i m ii u h ii n i Chemical Equivalent* 1 196.6 108 68 200 200 66 66 60 66 207 16 86.6 Combining Weights 1 65.6 108 81.5 100 200 18.7 28 29.6 82.6 108.5 8 86.5 Electro- Chemical Equivalent* .0106 .6877 1.184 .8807 1.06 2.10 .1964 .294 .8098 .8418 1.0868 .064 .8728 XL— MAGNETIZATION AND MAGNETIC TRACTION. B B. Dynes Grammes Kilogrs. Founds Lines per lines per per per per per eq. cm. sq. in. sq. oentlm. sq. oentlm. sq. oentlm. sq. Inch* 1,000 6,450 89,790 40.66 .0466 .677 2,000 12,900 159,200 162.8 .1628 2.808 8,000 19,860 868,100 866.1 .8661 5.190 4,000 25,800 686,600 648.9 .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 46,150 1,950,000 1,987 1.987 28.26 8,000 51,600 2,547,000 2,696 2.696 86.95 9,000 58,060 8,228,000 8,286 8.286 46.72 10,000 64,500 8,979,000 4,056 4.066 57.68 11,000 70,950 4,816,000 4,907 4.907 69.77 12,000 77,400 5,780,000 5,841 5.841 88.07 18,000 88,850 6,725,000 6,855 6.856 97.47 14,000 90,800 7,800,000 7,550 7.650 118.1 15,000 96,750 8,958,000 9,124 9.124 129.7 16,000 108,200 10,170,000 10,890 10.89 147.7 17,000 109,660 11,500,000 11,720 11.72 166.6 18,000 116,100 12,890,000 18,140 18.14 186.8 19,000 122,550 14,680,000 14,680 14.68 208.1 20,000 129,000 15,920,000 16,280 16.28 280* 149 2 JL— PERMEABILITY OP WROUGHT AND CAST IRON. 8QUAR1 CKCTIMBTEB WtAflURMCENT. Annealed Wrought Iron. Gray Cast Iron. B M H B M H 5,000 9,000 10,000 11,000 12,000 18,000 14,000 15,000 16,000 8,000 2,250 2,000 1,692 1,412 1,088 828 526 820 161 90 54 80 1.66 4 5 6.5 8.5 12 17 28.5 60 105 200 850 666 4,000 5,000 6,000 7,000 8,000 9,000 10,000 11,000 800 500 279 188 100 71 68 87 5 10 21.5 42 80 127 18S 292 17,000 18,000 19,000 20,000 BQUAKB INCH MmABTTBXMXST. Annealed Wrought Iron. Gray Cast Iron. B. /<• H. B, /<• H, 80,000 40,000 50,000 £0,000 70,000 80,000 90,000 4,660 8,877 8,081 2,159 1,921 1,409 907 408 166 76 85 27 6.5 10.8 16.5 27.8 86.4 56.3 99.2 245 664 1,661 8,714 5,165 25,000 80,000 40,000 60,000 60,000 70,000 768 756 258 114 74 40 82.7 89.7 155 489 807 1,480 100,000 110,000 120,000 180,000 140,000 160 PEBMJUBUJTY 07 SOFT OHABOOAL WBOUGHT IROH (tHKiVOSD BXDWBLL.) SQUAB! < B M H 7,890 1899.1 8.9 11,660 1121.4 10.8 16,460 888.4 40 17,880 160.7 116 18,470 88.8 208 19,880 46.8 427 19,820 88.9 666 0QVARS IXOH MMAltr B. M- H. 47,414 1897 26.0 ft 104 1122 M.l 99,191 888 268 111,189 160 788 118,604 88.8 1886 124,021 46.8 2740 127,106 88.9 8768 (-Both in lines of foroe. B — Magnetic Flux. ) H — Magnetlxing Fort*, f fl — g the Permeability or multiplying power of tbo oora. 151 XIIL-MAGNETIO BELUCTANCE OF ATE BETWEEN TWO PARALLEL OYLINDEBS OF IBON. b p _ Ihoh TJhitb. Batio of least distance apart to circumference. 0.1 .1964 5.1055 0.0771 12.968 0.2 .270T 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 0.1598 6.278 0.6 .4861 2.2988 0.1717 5.825 0.8 .4887 2.0465 0.1924 5.198 1.0 .5816 1.8807 0.2098 4.777 1.8 .5684 1.7996 0.2288 4.571 1.4 .6007 1.6645 0.2865 4.228 1.6 .6289 1.5902 0.2476 4.089 1.8 .6541 1.5287 0.2575 8.888 2.0 .6774 1.4764 0.2667 8.750 4.0 .8857 1.1968 0.8290 8.040 6.0 .9819 1.0782 0.8669 2.726 8.0 1.0047 .9958 0.8955 2.628 10.0 1.0544 .9484 0.4151 2.409 In this table in columns 2 and 8 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 .69855 1 .95820 1* 1.0177 If 1.0978 .70717 I .96850 H 1 0192 1* 1.1019 .72806 1 .97006 n 1 0226 1* 1.1068 .74185 ♦ .97468 u 1 0260 If 1.1087 .76478 f .97798 n 1 0608 If 1.1107 .79870 I .98055 H 1 0879 If 1.1119 .88268 ft .98258 n 1 0491 1ft 1.1129 .89090 n 1 0199 2 1.1287 .98462 n 1 0888 152 XV.-8TANDARD AND BIRMINGHAM WIBB GAUGES. Btavdabd. BnUfXNGHAM. Number of Gauge. Diameter in Mils. Square of Diameter or CtroTrMils. Number of Gauge. Diameter in Mils. Square of Diameter or CtroTr Mils. 0000000 500 250000 0000 454 806116 000000 464 215296 000 425 180625 00000 489 186824 00 880 144400 0000 400 160000 840 115600 000 879 188884 1 800 90000 00 848 121104 2 294 80656 824 104976 8 269 67081 1 800 90000 4 288 56644 2 276 76176 5 220 48400 t 258 68504 6 808 41209 4 289 58824 7 180 82400 5 818 44944 8 165 27225 e 198 86864 9 148 21904 T 176 80976 10 184 17956 8 160 25600 11 120 14400 9 144 20786 19 109 11881 10 128 16884 18 095 9025 11 116 18456 14 088 6889 12 104 10816 15 072 5184 18 099 8464 16 065 4226 14 080 6400 17 068 8864 15 079 5184 18 049 2401 16 064 4096 19 049 1764 17 056 8186 20 085 1225 18 048 2804 21 089 1024 19 040 1600 22 028 784 SO 086 1296 28 025 625 81 069 1024 84 022 484 22 028 784 25 020 400 98 024 576 26 018 824 24 022 484 95 020 400 96 018 824 I ■ © 8 a S° ^ g o g S3 8 S a is I III •"10 »5 me E w 153 e i b a § s a 8 ||S§S3§sgss g g 8 § S S sasssessss § § I I 1 g © »-l « CO •* & IIIS 2 I H I i i 5 I I I I I I § i I I § IlllllHiS! gigfifgggs' oo k- t- S co ss 'S I 1 154 XVH.-WATTS AND HORSE POWER TABLES FOB 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 496 amperes at 101 volts. Solution: The h. p. for 400 amp. is 66.3, for 80 amp. it is 4.82 and for 6 amp., .845. Adding these quantities gives 61.865 h. p., which we will call #1.4 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., 56 volts, take 7 amp. at 560 volte«5.16 h. p.; multiply by 1-10, which gives .516 h. p. To find h. p. at 9 amp., 1,300 volts, take 9 amp. at 120 volts=1.45 ; multi- ply by 10=14 6 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 ":i6 Watts. 105 h.p. .fil Watts. no h.p. .147 2 200 .268 210 .282 220 , .295 8 300 .402 315 .422 ' 330 ' .442 4 400 .686 420 .563 440 .690 6 600 .670 525 .704 560 .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 2,2u0 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 6.90 60 5000 6.70 5,250 7.04 5,600 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 i 9,450 12.7 9,900 13.3 100 10,000 13.4 10,500 14.1 11,000 14.7 . 200 20.000 28.8 21,000 31,500 28.2 22,000 29.5 300 30,000 40.2 42.2 33 000 44.2 400 40,000 63.6 42,000 56.3 44,000 59.0 600 50,000 67.0 52,500 70.4 55,000 73.7 600 60,000 70,000 80.4 63,000 84.5 66,000 88.5 700 93.8 73,500 98.5 77,000 108 800 80,000 90,000 107 84,000 113 88,000 118 900 121 94,500 127 99,000 133 1,000 100,000 134 105,000 141 110,000 147 155 HORSE POWER AT VARIOUS PRESSURES AND CURBENTS. (Continued.) 115 volts. 120 volts. 125 volts. Amp. ms h :& Watts. 120 h -.?61 Watts. 135 h.p. .168 2 HO .808 240 .322 m .335 3 345 .462 36^ .483 375 .505 4 AW .617 480 .644 610 .670 5 575 .770 60 .805 G25 .840 6 M0 .925 720 .966 750 1.01 7 905 1.08 840 1.13 875 1.18 8 wo 1.23 Ofti 1.29 UNO 1.34 9 1,035 139 l,<£fl 145 U35 1.51 10 1,150 1.54 uuo 161 ym 168 20 a>aoo 3.08 2,*w 8.22 £.500 3.35 80 3,450 4.62 Moo 483 3,750 5.15 40 4,600 6.17 4,#>0 6.44 5,HD 6.70 60 &.T50 7.70 (1,1 mo 8.04 ti.250 8.40 60 6,900 9.25 7,300 966 7. , ii 101 70 B.O50 10.8 M<JU 11.3 8,750 11.8 80 9,200 12.3 0,000 12.9 lOiitf 13.4 90 IO&jQ 13.9 iu y m 14 5 ]],£5G 15.1 100 11,500 15 4 ]2a 0& 16.1 13,500 16.8 800 12U0O 30.8 S4,< 00 82.2 85,000 335 800 H4.5O0 462 ieutto 483 :;t,:^) 505 400 4*s,nno 617 ft&OQQ 64.4 fAi.Hi) 67.0 500 57,500 77.0 enw* 80.4 UL*,-VHJ 840 600 OftJ-00 92.5 72, mQ 96.6 75,000' 1)1 700 mi r*fl 108 &1.0G 113 K.b 118 800 88,000 123 8fi.ro * 129 KXMOO 134 900 i*i;i, on 139 lOh.H^I 145 Ufc.fiOn 351 1,000 115,000 154 120,000 161 l£5,u0 168 156 HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. (Continued.) 200 volts. 210 volts. 220 volts. Amp. Watts. 200 .$» Watts. 210 .£2 Watts. 220 h .& 2 400 .586 420 .663 440 .690 8 800 .804 630 .845 660 .885 4 800 1.07 840 1.13 88') 1.18 6 1,000 1.34 7, aw 1.41 1,100 1.47 A 1,200 1.61 l*S80 1.69 1,320 1.77 7 1,400 1.88 3,470 1.97 1,640 2.06 8 1,600 2.14 i.i>M) 2.25 1.760 2.36 9 1,800 £41 KW0 263 1,960 2.65 10 2,000 £.iW 2,lo0 2.82 2,200 2.95 20 4,000 &.*i 4^1(0 5.68 4,400 5.90 80 6,000 BjCM MOD 845 6,600 8.85 40 8,000 io.t 8.4<i0 11.3 8,800 11.8 60 lo,000 l'j.4 10,600 14. 1 11,000 14.7 60 12,000 16 + l 12.600 16.9 13,200 17 7 70 14 MO IBJ J 4, 7(0 19.7 15,400 20.6 80 hi.iJOO 2L4 1(3 ^W 22.5 17,600 23.6 90 18.U0O £4 1 18,9(0 25.3 19,800 26.5 100 KkOOO 26.3 0,000 28.2 22,000 295 200 ^OJKIO 63.6 42,000 66.3 44,000 59.0 800 "MHO 80.4 (S^OtiO 84.6 66,000 88.5 400 8O.C0O 107 84.0(10 113 88,000 118 600 lnyjco 134 ] 05,0(0 141 110,(K0 147 800 lflMoo 161 1W *0 169 132,000 177 700 u\m 188 l47,rlJ0 197 154,000 206 800 mjm 214 168,000 225 176,000 236 900 I'vtOO 241 189,000 253 198,000 265 1,000 fen J, 000 268 2lU,WJ0 282 220,000 295 157 HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. (Continued.) 230 volts. 240 volts. 260 volts. Amp. Watts. 230 h.p. .808 Watts. MO .fc Watts. 300 h :& 2 |80 .617 480 .644 m .670 8 090 .926 720 .966 750 1.01 4 920 1.28 MO 1.29 1,000 1.34 5 1,150 1,54 1.200 1.61 i, aw 1.68 6 1,330 LSfi U40 1.98 LBOO 2.01 7 1,010 %M L680 2.26 h7*0 2&* 8 1.840 8,47 1,930 2.68 3,1)00 2.68 9 2,'*7G 2.77 2, IftO 2.90 3,2*0 8.02 10 2,300 3.08 2J00 3.22 : T ',:Vi0 3a^ 20 4,000 6 17 4.800 6.44 5.000 6.70 80 CkflOO M6 7,2*0 9.66 7..M-) 10.1 .40 0,200 12.3 ^000 12.9 10,000 13.4 60 11,600 15.4 la^oo 16.1 IB, W0 16 8 60 13 + Jto0 18 5 H iOl 19.3 14,000 20.1 70 lti,100 si .a J'. SQ0 22.6 17,600 23.6 80 38,400 24,7 M.i.:,'(J0 268 solooo 26.8 90 20,7111 27.7 2^*10 29.0 23,500 80.2 100 2&(HJU 30. B 84,000 82.2 *sr».UXJ 33.8 XX) uloqq nt.7 4HOU0 64.4 60,000 67.0 300 O'.i ' ■ xi 92.5 72,1100 96.6 76.000 101 400 02,000 123 fl(t,( IX) 129 100,000 134 600 115,000 164 120,000 161 1-5.000 168 600 138,000 18A 144,000 193 160.000 201 700 161,000 216 IfW.lOO 226 175,0 236 800 184,000 247 l!--;.^K) 258 2T0,000 268 900 207,000 277 819,000 290 226.000 802 1,000 2W*X) 808 240,000 332 260,000 336 153 HORSE POWER AT VARIOUS PRESSURES AND CURRENTS. (Continued.) aoo volte. 550\ olta. GOOv Gltfi. Watts Wflita Watt* Amp. and k. w. BOO h -£ and k. w. 5 II fcutt. ,74 and Li . w. eoo b jSi 2 j,IH«0 l£i 1,110 1,47 1,300 l.ttl 8 I..VII 2 01 1,660 a.st l.MO S.41 4 3 HI in 2. m 2,3i] 2.95 2,400 &gg fi BJ500 3.35 a 710 ::."'!■ :mr. 402 e liJUNI 4 + <J2 3,40 4 43 3.6U0 4M 7 3,500 4. OS 3,860 6,1* 4 300 503 8 4JXN» 5 t ;w 4jm» 590 4600 643 ft 4,500 6.44 4,ftri0 £.04 5,400 7 24 li) ftk.w. 0.7 fc.5 k,w. 7-4 6 k. w. 8,04 an 10 13 4 IUD 14.7 12 16,1 30 15 201 I&fl 231 10 n.i 40 30 sas 330 30.3 *4 SS.& 50 3ft IW .ft 27,* 30 .ft ;<ii 40 t 2 60 30 40,2 33.0 44.3 30 4S + B 70 95 4flB 38.5 51.0 42 5<n so 40 63 fi 44 fift.O 43 04.3 ftO 45 txu 43.5 flft.4 M 72.4 10Q fiO A7 55 74.0 00 80.4 200 100 134 no 147 120 101 300 Lflfl 201 Iflft 221 iao HI 4UJ sou 3(58 390 3»5 2411 333 600 200 aat 375 3«S 300 402 600 »«0 *03 330 441 300 483 700 ■{5*1 4*9 MB5 MO 430 ftfl3 suo 403 53fl 440 590 480 *W3 qui 4^ fclitt 405 6H4 540 724 i,ouo 50LI O 550 740 600 804 INDEX. Alternating current sys- tem 47-49 Alternating currents .... 115 Amperage of armature ... 99 Ampere 11-12 Ampere-second 57 Ampere-turns 87 Ampere-turns, rules for calculating 87 Armature amperage of, on short circuit 99 Armature calculations. .99-103 Armature, capacity of . . . . 99 Armature, general fea- tures of 96 Armature, resistance of . . 98 Armatures, rules for cal- culating 98-99 Armature, voltage of.... 99 Armature winding iron or copper 96 Bbidge, Wheatstone, prin- ciple of 25 Batteries, illustrations of arrangements 66 Batteries or generators in opposition 15 Batteries, storage, resist- ance of. 65 Battery arrangement and size for given efficiency. 73 Battery, arrangement of cells in 67 Battery calculations, dis- crepancies in 72 Battery, chemicals con- sumed in 78 Battery constants 65 Battery, current of 68 BatterV, effective rate of work of .,,,.., 7,7-78 Battery, efficiency, to cal- culate 72 Battery, electromotive force of 67 Battery, how rated 65 Battery, resistance of . . . . 67 Battery, to calculate cur- rent and arrangement. 69-72 Battery, to calculate its voltage 76-77 Battery t work of. 77-78 Calobib, gram and kilo- gram, defined 54 Calorie, relation to watts and ergs 54-55 Capacity of armature winding 97 Cars, power to move .... Ill Cell constants 65 Chemicals consumed in a battery 78 Circuits, divided, branched or shunt 19-25 Circuits, portions of 17-19 Circuits, single conductor, closed 14-15 Circular mils calcula- tions 44-45 Circular mils 43-45 Condensers 121 Conductance 36-37 Conductors of same mate- rial 26 Conductors, resistance of different 26 Contact, area of in elec- tro-magnets and arma- tures 86 Convection and radiation, loss of heat by 58 Conversion, ratio of . , . . t 47 100 INDEX. Converters, rale for wind- ing 49 Cores of field-magnet, to calculate 104 Counter-electromotive force of plating-bath 80 Current, battery required for a given 60-72 Current, distribution In parallel circuits 20-21 Current, its heating effect and energy 50 Currents passed by branches of a circuit. • 10 Current, work of 00 Current yielded by a bat- tery 65 Definitions, general 9 Demonstrations of rules. 127 Derived units 10 Dimensions, units of 10 Drop of potential In leads 88-41 Drum type closed circuit armatures 98-99 Duty and commercial ef- ficiency 63-64 Duty of generators 63 Efpictivd B. M. F 16 Efficiency, its relation to resistances 64 Efficiency of generators, commercial 64 Efficiency of generators, electrical 63 Efficiency, to calculate battery for a given.. 73-74 Electrical railways 109 Electro-magnets and dy- namos 82 Electro-magnets, general rules for 85 Electro-magnets, to mag- netize 86-87 Electro-magnets, traction of 85-86 Electromotive force of battery 67 Electro - plating calcula- tions 79 Energy defined 9 Energy in circuit, rules for determining 50-53 Energy of current 50-53 B. 11 F., its meaning... 10 E. M. F., primary and sec- ondary 47 Erg 54 Fobcb defined 9 Force, magnetic 84 Fundamental units 10 Field-magnet cores, to cal- culate 104 Field-magnet for dynamo or motor 104-107 Field, unit Intensity of magnetic 82-83 Genbbatom, efficiency of, 63-64 Generators or batteries In opposition 15 Hbat, absolute quantity in circuit 54 Heating effect of current, 50-53 Heat, specific 57 Horse-power, electrical . . 61 Horse - power, reduction from kilogram-meters . . 60 Horse-power, to calculate for lamps 62 Joulb, his law of heating, effect of currents 50 Joule or gram-calorie.... 54 Kapp, Gisbert, his line of force 107-108 Kilogram-meters 60 Lsads, tapering in size. . 41-42 Leakage between cylindri- cal magnet leg, to cal- culate 91 Leakage between flat mag- net surfaces, to calcu- late 91 Leakage of lines In a mag- netic circuit 95 Leakage of lines of force. 89 Line of force, Kapp's. . 107-108 Lines of force 82 Lines of force cut per sec- ond for one volt 96 Lines of force, leakage of. 89 Lines of force, to diminish leakage of 94-95 INDEX. 161 Magnetic circuit 84 Magnetic circuit calcula- tions 88-89 Magnetic circuit, the law of 85 Magnetic circuits, four parts of 88 Magnetic circuits, general calculations for 92 Magnetic field 82-83 Magnetic flux 82-83 Magnetic force .-. 84 Magnetic potential, aver- age difference of 90 Magnetism, no insulator of 84 Magnetizing force 87 Magnet legs, long and short 94 Mass defined 9 Metal 8, deposition of, by battery 79 Mho, unit of conductance 36 Mil, circular, calculations based on 44-45 Mils, circular 43 Mils, circular, applied to alternating current ... 48 Multiple arc connections, to calculate 38-42 Nbutbal wire in three wire system 46 Notation in powers of ten 136 Ohm 11-12 Ohm's law 13-25 Ohm's law, its universal application 14 Ohnrs law, six expres- sions of 13-14 Parallel connections of equal resistance 23 Parallel leads, resistance of 22 Permeance 83-84 Permeability 85 Permeability, average range of 104 Potential, diagram for cal- culating fall of 41 Potential difference .... 38-42 Potential difference, drop, or fall of 17 Potential, drop of in leads, S8-42 Power to move cars Ill Powers of ten, notation in 136 Primary E. M. F 47 Proving armature calcula- tions 103 Radiation and convection, loss of heat by 58 Railways, electric 109 Rate of heat-energy, units of 63 Ratio of conversion 47 Reluctance 83-84 Reluctance, calculation of. 87 Resistance 26-35 Resistance and efficiency, how related 64 Resistance defined 9 Resistance of battery 67 Resistance of circuit, note relative thereto 14 Resistance of parallel leads 22-23 Resistance referred to weight of conductor . . . 35-36 Resistance specific 29 Resistances, two in bat- tery circuits 65 Resistance, universal rule for 31-32 Rules, demonstrations of. 127 Safety-catches for fuses, how calculated 59 Secondary E. M. F 47 Self- Induction 117 Series winding for dyna- mos 106 Shunt circuits 19 Shunt circuits, resistance of 22 Shunt winding for dyna- mos 107 Sizes of feeders 109 Space defined 9 Specific heat 57 Specific resistance 30-31 System, alternating cur- rent "47-49 Systems, special 46-49 System, three wire 46-47 Tables — American wire gauge table 146 INDEX. Tables — Continued. Chemical and electro- chemical equivalents . . 148 Chemical and thermo- chemlcal equivalents . . 147 Current capacity of bare or insulated overhead wires 153 Equivalents of units of area 140 Equivalents of units of energy and work . . . 142-143 Equivalents of units of length 139 Equivalents of units of volume 140 Equivalents of units of weight 141 Magnetic reluctance of air between two parallel cylinders of iron 151 Magnetization and mag- netic traction 148 Permeability of soft char- coal wrought Iron .... 150 Permeability of wrought and cast iron 149 Relative resistance and conductance of pure copper at different tem- peratures 145 Specific resistance of solu- tions and liquids 144 Standard and Birmingham wire gauges 152 Table of specific resist- ances in microhms and of coefficients of specific resistances of metals. . 144 Table of the sixth roots. . 151 Three wire system 48 Three wire system, saving in size of wires 46 Three wire system, the neutral wire 46 Time defined 9 Units, concrete statement of 12 Units, fundamental 10 Units, original and de- rived 10 Volt 10-12 Voltage of battery, to cal- culate 76-77 Volt amperes.' 56-60 Watt 56-60 Watts and horse power tables for various pres- sures and currents 154 Weight defined 9 Weight of conductor, re- sistance referred to . . . 35-36 Wheats tone bridge, prin- ciple of 25 Winding, series and shunt 106-107 Wire, rule for heating of, by a current 59 Work defined 9 Work of current 60 Scientific and Practical Books Published by The Norman W. Henley Publishing Co. 132 Nassau St., New York, U. S. A. flQP'Any of these books will be sent prepaid on receipt of price to any address in the world. HQp'We will aend/rtt to any address our catalog of Scientific and Practical Books. ASKZK80N. Perfumes and Their Preparation, Containing complete directions for making handker* chief perfumes, smelling salts, sachets, fumigating pastils; preparations for the care of the skin, the mouth, the hair; cosmetics, hair dyes, and other toilet articles. 300 pages. 82 illustrations. $3.00. BABB. Catechism on the Cumbustion of Coal and .the Prevention of Smoke. A practical treatise for all interested in fuel econ- omy and the suppression of smoke from stationary steam-boiler furnaces and from locomotives. 85 illus- trations. 349 pages. $1.50. BARROWS. Practical Pattern Making. A fully illustrated book which teaches you just what you should know about pattern making. It contains a detailed description of the materials used by pattern makers, also the tools, both those for hand use, and the more interesting machine tools; having complete chapters on the Band Saw, the Buzz Saw and the Lathe. Individual patterns of many different kinds are fully illustrated and described, and the mounting of metal patterns on plates for molding machines is included. Price, $2.00. Battleship Chart. An engraving which shows the details of a battle- ship of the latest type, as if the sides were of glass ana you could see all the interior. The finest piece of work that has ever been done. So accurate that it is used at Annapolis for instruction purposes. Shows all details and gives correct name of every part. .28 x '2 inches— plate paper. 60 cents. ** GOOD, PRACTICAL BOOKS BAUBB. Marine Engine* and Boiltrt: Their De- sign and Conitrueuon. A large practical work of 722 pages, 650 illustra- tions, and 17 folding plates, for the use of students, engineers and naval constructors. $9.00 net. BAXTSB, JB. Switchboards. The only book dealing with this important part of electrical engineering. Takes up all sixes ana kinds from the single dynamo in the engine room to the largest power plant work. Includes divert and alter- nating currents; oil switches for high tension; arc and incandescent lighting; railway work, and all the rest, except telephone work. $1.50. BAXTBB, JB, Commutator Construction. The business end of a dynamo or motor is the com- mutator, and this is what is apt to give trouble. This shows how they are made, why they get out of whack and what to do to put 'em right again. Price, 26 cents. BBHJAimr. Modern Mechanism. A practical treatise on machines, motors, and the transmission of power, being a complete work and a supplementary volume to Appleton s Cyclopedia ox Applied Mechanics. Bound in half Morocco. Price, $5.00. BLACSAXXt Air-Brake Catechism. This book is a complete study of the air-brake equipment, including the £. T. Equipment, as well as all the latest devices. All parts of the air brake, their troubles and peculiarities, and a practical way to find and remedy them, are explained. This book contains over 1,500 questions with their answers. 812 pages. $2.00. BLACSAXX. Hew York Air-Brake Catechism. This is a complete treatise on the New York Air- Brake and Air-Signalling Apparatus, giving a detailed description ot all the parts, their operation, troubles, and the methods of locating and remedying the same. 200 pages, fully illustrated. $1.00. BOOTH AND KEBSHAW. Smoke Prevention and Fuel Economy. As the title indicates, this book of 107 pages and 75 illustrations deals with the problem of complete combustion, which it treats from the chemical and mechanical standpoints, besides pointing out the eco- nomical and humanitarian aspects of the question. S2.60. GOOD, PRACTICAL BOOKS BOOTH. Steam Pipes: Their Design and Construe- A treatise on the principles of steam conveyance and means and materials employed in practice, to secure economy, efficiency, and safety. A book which should be in the possession of every engineer and contractor. $2.00. BUCHETTL Engine Tests and Boiler Efficiencies. This work fully describes and illustrates the meth- od of testing the power of steam engines, turbine and explosive motors. The properties of steam and the evaporative power of fuels. Combustion of fuel and chimney draft; with formulas explained or practically computed. 255 pages. 179 illustrations. $3.00. BYRON. Physios and Chemistry of Mining. For the use of all preparing for examinations in Mining or qualifying for Colliery Managers' Certifi- cates. $2.00. Car Charts. Shows and names all the parts of three types of cars. Passenger — Box — Gondola. Printed on heavy plate paper and mailed in a tube. 25 cents each. Set of 8 for 50 cents. COCKIN. Practical Coal Mining. An important work, containing 428 pages and 213 illustrations! complete with practical details, which will intuitively impart to the reader, not only a general knowledge of the principles of coal mining, but also considerable insight into allied subjects. The treatise is positively up to date in every instance, and should be in the hands of every colliery engineer, geologist, mine operator, superintendent, foreman, and all others who are interested in or connected with the industry. $2.50. COLYDf. American Compound Locomotives. The latest and most complete book on compounds. Shows all types, including the balanced compound which is now being used. Makes everything clear by many illustrations, and shows valve setting, break- downs and repairs. $1.50. v COLVItf AND CHENEY. Engineer's Arithmetic A companion to Machine Shop Arithmetic, arranged for the stationary engineer. Shows how to work the problems of the engine room and shows "why." Has steam tables and a lot of other useful information that makes it popular with practical men, 50 cents. GOOD, PRACTICAL BOOKS OOLVXY. Link Xotioae and Yalta Setting . A handy little book for the engineer or machinist that clears up the mysteries of valve setting. Shows the different valve gears in use, how they work and why. Piston and slide valves of different types are illustrated and explained. A book that every rail- road man in the motive power department ought to have. 60 cents. COLVIK AHD CHEVEY, Xachine 8hop Arithmetic Most popular book for shop men. Shows how all shop problems are worked out and "why." Includes change gears for cutting any threads; drills, taps, shink and force fits; metric system of measurements and threads. Used by all classes of mechanics and for instruction by Y. M. C. A. and other schools. 60 cents. COLVIK. The Railroad Pockotbook. Different from any book you ever saw. Gives clear and concise information on just the points you are interested in. It's really a pocket dictionary, fully illustrated, and so arranged that yon can find just what you want in a second without an index. Whether you are interested in Axles or Acetylene; Compounds or Counter Balancing; Rails or Reducing Valves; Tires or Turntables, you'll find them in this little book. It's very complete. Flexible cloth cover. 200 pages. $1.00. Interleaved with ruled pages for notes. $1.50. 00LYDT-8TABEL. Thread! and Thread Cutting. This clears up many of the mysteries of thread- cutting, such as double and triple threads, internal threads, catching threads, use of hobs, etc Contains a lot of useful hints and several tables. Price, 25 cents. COLVIK. Turning and Boring Tapers, A plainly written explanation of a subject that puz- zles many a mechanic This explains the different ways of designating tapers, gives tables, shows how to use the compound rest and gives the tapers mostly used. Price, 25 cents. CBAVE. American Stationary Engineering. A new book by a well-known author. Begins at the boiler room and takes in the whole power plant Con- tains the result of years of practical experience in all sorts of engine rooms and gives exact information that cannot be round elsewhere It's plain enough for prac- tical men and yet of value to those high in the pro- fession, Baa a oomplrtfl *^"«^ for a license GOOD, PRACTICAL BOOKS DALBY. Train Rules and Dispatching. Contains the standard code for both single and double track and explains how trains are handled un- der all conditions. Gives all signals in colors, is illus- trated wherever necessary, and the most complete book in print on this important subject. Bound in fine sea' 1 flexible leather. 221 pages. $1.50. ENGSTROM. Bevel Gear Tables. No one who has to do with bevel gears in any wfcy should be without this book. The designer and drafts- man will find it a great convenience, while to the machinist who turns up the blanks or cuts the teeth, it is invaluable, as all needed dimensions are given and no fancy figuring need be done. $1.00. FOWLER. Boiler Boom Chart. ( An educational chart showing in isometric perspec- tive the mechanisms belonging in a modern boiler- room. Each part is given a reference number, and these, with the corresponding name, are given in a glossary printed at the sides. The chart, therefore, serves as a dictionary of the boiler-room, the names of more than two hundred parts being given on the list. 25 cents. FOWLER. Locomotive Breakdowns and Their Remediei. ■ This work treats in full all kinds of accidents that are likely to happen to locomotive engines while on the road. The various parts of the locomotives are discussed, and every accident that can possibly happen, with the remedy to be applied, is given. 250 pages. GODDARD. Eminent Engineers. An intensely interesting account of the achieve- ments of thirty of the world's best known engineers. Free from tiresome details and giving just the facts you want to know in an entertaining manner. Por- traits are given and the book is an inspiration for both old and young. $1.50. GRIMSHAW. Saw Filing and Management of Saws. A practical handbook on filing, gumming, swaging, hammering, and the brazing of band saws, the speed, work, and power to run circular saws, etc., etc. $1.00. GRIMSHAW. "Shop Kinks." This shows special methods of doing work of vari- ous kinds, and reducing cost of production. Has hints and kinks from some of the largest shops in this country and Europe. You are almotf sure to find some that apply to your work, and in such a way to save time and trouble. 400 pages. Price, $1.50. , # GOOD, PRACTICAL BOOKS GBIH8HAW. Engine Bonner's Catechism. Tells how to erect, adjust, and run the principal steam engines in use in the united States. 886 pases, $8.00. ^ GBIHSHAW. Steam Engine Catechism. A series of direct practical answers to direct prac- tical questions, mainly intended for young engineers and for examination questions. Nearly 1,000 questions with their answers. 418 pages. $8.00. GBXX8HAW. Locomotive Catechism. This is a veritable encyclopedia of the locomotive, is entirely free from mathematics, and thoroughly up to date. It contains 1,600 questions with their an- swers. 450 pages, over 800 illustrations. $8.00. HABBISOtf. Electrio Wiring, Diagrams and Switchboards. A thorough treatise covering the subject in all its ' branches. Practical every-day problems in wiring are presented and the method of obtaining intelligent results clearly shown. 870 pages. 105 illustrations. $1.60. Honley's Twentieth Century Book of Beoeipts, formulas and Processes. Edited by G. D. Hiscox. The most valuable Tech- no-Chemical Receipt Book published. Contains over 10,000 selected scientific chemical, technological and practical receipts and processes, including hundreds of so-called trade secrets for every business. 900 pages. Price, $8.00. HISCOX. Gas, Gasoline, and Oil Engines. Every user of a gas engine needs this book. Sim- ple, instructive, and right up to date. The only com- plete work on the subject. Tells all about the run- ning and management of gas engines. Includes chap- ters on horseless vehicles, electric lighting, marine propulsion, etc 450 pages. Illustrated with 851 en- gravings. $2.50. Henley's Encyclopedia of Practical Engineering and Allied Trades. Edited by Joseph G. Hoenzb, Complete in five volumes. Each volume contains 600 pages and 600 illustrations. Bound in half morocco. Price, $6.00 per volume, or $86.00 for the complete set of five volumes. GOOD. PRACTICAL BOOKS HISCOX. Compressed Air in All Its Applications. This is the most complete book on the subject of Air that has ever been issued, and its thirty-five chap- ters include about every phace of the subject one can think of. It may be called an encyclopedia of com- pressed air. It is written by an expert, who, in its 825 pages, has dealt with the subject in a comprehen- sive manner, no phase of it being omitted. 545 illus- trations, 820 pages. Price, $5.00. HISCOX. Mechanical Movements, Powers, and Devices. Almost every mechanic is interested in mechanical movements, and the designer finds new problems every day. A book of this kind used as a reference is in- valuable in solving some of the many problems in mechanics, especially in designing new machinery. 400 pages. Price, $3.00. HISCOX. Mechanical Appliances, Mechanical Move- ments and Novelties of Construction. This is a supplementary volume to the one upon mechanical movements. Unlike the first volume, which is more elementary in character, this volume contains illustrations and descriptions of many combinations of motions and of mechanical devices and appliances found in different lines of Machinery. Each device being shown by a line drawing with a description showing its working parts and the method of operation. 896 pages, 1,000 specially made illustrations. Price, $8.00. 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 and 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. HOBABT. Brazing and Soldering. A complete course of instruction in all kinds of hard and soft soldering. Shows just what tools to use, how to make them and how to use them. Price, 25 cents. HOBNEB. Modern Milling Machines: Their Design, Construction and Operation. This work of 804 pages is fully illustrated and de- scribes and illustrates the Milling Machine in every detail. $4.00. GOOD, PRACTICAL BOOKS HOBNER. Practical Metal Turning. A work covering the modern practice of machining metal parts in the lathe. Fully illustrated. $3.60. HOBNER. Tools for Machinists and Wood Work- en, Including Instruments of Measurement. A practical work of 340 pages fully illustrated, Jiving a general description and classification of tools or machinists and woodworkers. $3.60. Horse Power Chart. Shows the horse power of any stationary engine without calculation. No matter what the cylinder diameter or stroke; the steam pressure or cut-off; the revolutions, or whether condensing or non-condensing, it's all there. Easy to use, accurate and saves time and calculations. Especially useful to engineers and designers. 60 cents. Inventor's Manual; How to Make a Patent Pay. This is a book designed as a guide to inventors in perfecting their inventions, taking out their patents and disposing of them. 119 pages. Cloth, $1.00. HLEDTHAHS. Boiler Construction. The only book showing how locomotive boilers are built in modern shops. Shows all types of boilers used; gives details of construction; practical facts, such as life of riveting punches and dies, work done per day, allowance for bending and flanging sheets and other data that means dollars to any railroad man. 421 pages. 834 illustrations. Six folding plates. $3.00. KBATT88. Linear Perspective Self -Taught. The underlying principle by which objects may be correctly represented in perspective is clearly set forth in this book; everything relating to the subject is shown in suitable diagrams, accompanied by full ex- planations in the text. Price. $2.60. LE VAN. Safety Valves; Their History, Invention, and Calculation. Illustrated by 69 engravings. 151 pages. $1.50. MARKHAM. American Steel Worker. The standard work on hardening, tempering and annealing steel of all kinds. A practical book for the machinist, tool maker or superintendent. Shows just how to secure best results in any case that comes along. How to make and use furnaces and case harden; how to handle high-speed steel and how to temper for all classes of work. $?.50. GOOD, PRACTICAL BOOKS MATHOT. Modern Gas Engines and Produoer Gat Plants. A practical treatise of 320 pages, fully illustrated by 175 detailed illustrations, setting forth the princi- ples of gas engines and producer design, the selection and installation of an engine, conditions of perfect operation, producer-gas engines and their possibilities, the care of gas engines and producer-gas plants, with a chapter on volatile hydrocarbon and oil engines. $2.50. MEINHA&DT. Practical Lettering and Spacing. Shows a rapid and accurate method of becoming a good letterer with a little practice. 60 cents. PAB8ELL & WEED, Gas Engine Construction. A practical treatise describing the theory and prin- ciples of the action of gas engines of various types, and the design and construction of a half-horse-power gas engine, with illustrations of the work in actual progress, together with dimensioned working drawings giving clearly the sizes of the various details. 300 pages. $2.50. FEBBIGO. Change Gear Devices. A book for every designer, draftsman and mechanic who is interested in feed changes for any kind of machines. This shows what has been done and how. Gives plans, patents and all information that you need. Saves hunting through patent records and rein- venting old ideas. A standard work of reference. $1.00. PERRIGO. Modern American Lathe Practice. A new book describing and illustrating the very lat- est practice in lathe and boring mill operations, as well as the construction of and latest developments in the manufacture of these important classes of machine tools. 800 pages, fully illustrated. $2.50. PERRIGO. Modern Machine Shop Construction, Equipment and Management. The only work published that describes the Modern Machine Shop or Manufacturing Plant from the time the grass is growing on the site intended for it until the finished product is shipped. Just the book needed by those contemplating the erection of modern shop buildings, the rebuilding and reorganization of old ones, or the introduction of Modern Shop Methods, Time and Cost Systems. It is a book written and illus- trated by a practical shop man for practical shop men who are too busy to read theories and want facts. It GOOD, PRACTICAL BOOKS U the most complete all-around book of its kind erer published. 400 large quarto pages, 226 original and specially-made illustrations. $5.00. PBATT. Wiring a Home. Shows every step in the wiring of a modern house and explains everything so as to be readily under- stood. Directions apply equally to a shop. Price, 25 cents. BEAGAH, JR. Electrical Engineers' and Students' Chart and Hand-Book of the Brush Arc Light System. Bound in cloth, with celluloid chart in pocket. 60 cents. BICHABD8 AHD COLVI2J. 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 tune in the drawing room and mistakes in the shops. Con- tains practical examples of various classes of work. 60 cents. BOTJUXION. 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. BOTJUXION. 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. Shows exactly what is needed for all grades of the work from the Kindergarten to the High and Normal School. Gives itemized lists of everything needed and tells just what it ought to cost. Also shows where to buy sup- plies. $2.00. SAUNIEB. Watchmaker's Hand-Book. Just issued, 7th edition. Contains 498 pages and is a workshop companion for those engaged in watch- making ana allied mechanical arts. 250 engravings and 14 plates. $3.00. 8L0ANZ. Electricity Simplified. The object of "Electricity Simplified" is to make the subject as plain as possible and to show what the modern conception of cectricity is. 168 pages. $1.00. GOOD, PRACTICAL BOOKS SLOANS* How U Become a Snooeiafnl Sleotrlelan. It is the ambition of thousands of young and old to become electrical engineers. Not every one is pre- pared to spend sereral thousand dollars upon a col- lege course, even if the three or four years requisite are at their disposal. It is possible to become an elec- trical engineer without this sacrifice, and this work is designed to tell "How to Become a Successful Elec- trician" without the outlay usually spent in acquiring the profession. 189 pages. $1.00. SLOANS. Arithmetic of Electricity. A practical treatise on electrical calculations of all kinds reduced to a series of rules, all of the simplest forms, and involving only ordinary arithmetic; each rule illustrated by one or more practical problems, with detailed solution of each one. 138 pages. $1.00. SLOANS. Electrician's Handy Book. An up-to-date work covering the subject of practical electricity in all its branches, being intended for the every-day working electrician. The latest and best authority on all branches of applied electricity. Pocket- book size. Handsomely bound in leather, with title and edges in gold. 800 pages. 600 illustrations. Price, $3.60. SLOANS. Electric Toy Making, Dynamo Building, and Electric Motor Construction, This work treats of the making at home of electrical toys, electrical apparatus, motors, dynamos, and in- struments in general, and is designed to bring within the reach of young and old the manufacture of genu- ine and useful electrical appliances. 140 pages. $1.00. SLOANS. Bubber Hand Stampa and the Manipula- tion of India Bubber. A practical treatise on the manufacture of all kinds of rubber articles. 146 pages. $1.00. SLOANS. Liquid Air and the Liquefaction of Oases. Containing the full theory of the subject and giving the entire history of liquefaction of gases from the earliest times to the present. 865 pages, with many illustrations. $2.00. SLOAXE. Standard Electrical Dictionary. A practical handbook of reference, containing defi- nitions of about 6,000 distinct words, terms, and phrases. 682 pages. 398 illustrations. $8.00. GOOD, PRACTICAL BOOKS •TARBUCK. Modern Plumbing Illustrated. 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. 800 pages. 66 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. 60 cents. Tractive Power Chart A chart whereby you can find the tractive power or drawbar pull of any locomotive, without making a figure. Shows what cylinders are equal, how driving wheels and steam pressure affect the power. What sized engine you need to exert a given drawbar pull or anything you desire in this line. Printed on tough jute paper to stand rolling or folding. 60 cents. T/sheb. The Modern Machinist. A practical treatise embracing the most approved methods of modern machine-shop practice, and the ap- plications of recent improved appliances, tools, and devices for facilitating, duplicating, and expediting the construction of machines and th*ir parts. 857 engravings. 822 pages. $2.60 VAH DEBVOORT. Modern Machine Shop Tools; Their Construction, Operation, and Manipula- tion. An entirely new and fully illustrated work of 656 pages and 678 illustrations, describing in every de- tail the construction, operation, and manipulation of both Hand and Machine Tools. Includes chapters on filing, fitting, and scraping surfaces; on drills, ream- ers, taps, and dies; the lathe and its tools; planers, shapers, and their tools; milling machines and cutters: Sear cutters and gear cutting; drilling machines and nil work; grinding machines and their work; hard- ening and tempering; gearing, belting, and transmis- sion machinery; useful data and tables. $4,00. GOOD, PRACTICAL BOOKS WALLIS-TAYLOR. Pocket Book of Refrigeration and Ice-Making. This is one of the latest and most comprehensive reference books published on the subject of refriger- ation and cold storage. $1.50. "WOOD. Walschaert Locomotive Valve Gear. The only work issued treating of this subject of valve motion. 150 pages. $1.50. WOODWOETH. American Tool Making and Inter- changeable Manufacturing. 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. WOODWOETH. 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 wor* can secure many valuable suggestions. Thoroughly modern. 384 pages. 505 illustrations. $3.00. WOODWOETH. 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 and 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. Cushing, 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 BOOS 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 SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. DEC 16 1932 SEP 15 1937 AUG * ** o*r> " * 'AP™' LD 21-50m-8,-32 7B 09539 3S3491 UNIVERSITY OF CALIFORNIA LIBRARY