LIBRARY UNIVERSITY OF CALIFORNIA. Deceived Ut^fa/i ,189% Accession No. 7 <3 3 Cla *s No. Library of the aid li EXAMPLES IN ELECTRICAL ENGINEERING PHYSICAL AND ELECTRICAL ENGINEERING LABORATORY MANUALS. Vol. I. ELEMENTARY PHYSICS. By JOHN HENDERSON, B.Sc. (Edin.), A.I.E.E., Lecturer in Physics, Manchester Municipal Technical School. Crown 8vo, 2s. 6d. LONGMANS, GREEN, & CO. LONDON, NEW YORK, AND BOMBAY. EXAMPLES IN ELECTRICAL ENGINEERING SAMUEL JOYCE, A.I.E.E. 11 LECTURER TO THE SENIOR CLASSES IN ELECTRICAL ENGINEERING MUNICIPAL TECHNICAL SCHOOL, MANCHESTER LONGMANS, GREEN, & CO. LONDON, NEW YORK, AND BOMBAY 1896 All rights reserved Engineering Library PREFACE MANY of the examples in the following pages have been col- lected during the past few years to illustrate the author's lectures to Advanced and Honours Students in Electrical Engineering, though the majority are here published for the first time. Originally intended as a collection of exercises, the explana- tory matter forming the bulk of the text was, however, found necessary to make the book more complete in itself, though it is not intended to act as a full treatise on the subject. These explanations, together with the tables at the end of the book, will, it is hoped, be found very useful by draughtsmen and 'others engaged in electrical machine design. The author's best thanks are due to such writers as have been made use of, too numerous to mention by name ; and also to two of his third-year students, Messrs. A. B. Mallinson and W. K. Meldrum, for many carefully executed diagrams. Lastly, and not the least, the author's thanks are due to his friend Mr. E. S. Shoults, for considerable assistance in checking examples. S. JOYCE. LATCHFORD HOUSE, GREENHEYS, MANCHESTER, July, 1896. NOTE. The author will be much obliged if readers will kindly notify any errors that may have escaped observation. CONTENTS CHAPTER PAGE I. SIMPLE CIRCUITS i II. DISTRIBUTION. MULTIPLE WIRE CIRCUITS 15 III. CIRCUITS CONTAINING MORE THAN ONE E.M.F. ... 27 IV. DYNAMOS AND MOTORS 35 V. ARMATURE WINDING 55 VI. THE MAGNETIC CIRCUIT 65 VII. MAGNET COIL WINDINGS 82 VHI. ARMATURE REACTIONS 86 IX. EFFICIENCY 94 X. VOLTMETERS AND THEIR RESISTANCE COILS 102 XL ELECTRIC TRACTION, RAILWAYS, ETC 106 XII. ALTERNATING CURRENT CIRCUIT 113 XIII. IMPEDANCE COILS AND TRANSFORMERS 145 XIV. EFFECTS OF CAPACITY . . 160 XV. REGULATING RESISTANCES 174 XVI. PRIME MOVER COSTS . 181 EXAMPLES , . . 186 ANSWERS 205 TABLES 209 INDEX 236 EXAMPLES IN ELECTRICAL ENGINEERING. CHAPTER I. SIMPLE CIRCUITS. The Simple Electric Circuit consists of a conducting path, insulated so as to confine the electric flux as much as possible to that path. In it the relationship known as Ohm's Law holds good, viz. If e be stated in volts, R in ohms, then C will be in amperes. The value of e in the above expression must be considered as being the algebraic sum of all the electro-motive forces (E.M.F.) acting in the circuit, and will be called the effective or active E.M.F. The resistance of a simple circuit may be composed of a number of separate resistances in series ; in which case R will be the sum of all these other resistances, thus R = ri + r^ 4- ; 3 + r, and therefore C = r-L 4- 1\ 4- r s + r, or e = Cfi + O a + Cr 3 + O 4 and each of these quantities may be called a potential differ- ence, or P.D., for short. The E.M.F. e is thus the sum of all the P.D.'s, or is equal to 20 writing the P.D. Oj = v lt we have that e = ^ + z/ 2 + v a 4- 4 The Rate of doing Work. The amount of work done 2 Examples in Electrical Engineering. in a circuit is equal to the quantity, Q, of electricity passed multiplied by the E.M.F. And since Q is the product of the current and the time, then the work done, W, in the time t will be W = eCt from which the rate of doing work, or power, or activity, is '-?-?/:* and as e = Oj + O 2 -f O 3 + O 4 , etc., we have P = eC = CV a + CV 2 + CV 8 -f CV 4 , etc. which gives the distribution of energy expenditure round the circuit, and states that the whole rate of doing work in a circuit is equal to the sum of all the activities in the separate parts of the circuit. The activity or power P is measured in watts, one watt being the activity when the product eC is unity. Owing to the values chosen for the volt and the ampere, the watt is equivalent to the y^ part of the mechanical horse-power, or H.P. = Calculation of Resistance. The value of the resistance of a circuit is determined by the nature of the material of which that circuit is composed, by its temperature, and by the dimensions of the circuit; the relationship between these quantities being that where L and a are respectively the length and sectional area (average if not uniform) measured in terms of the same unit, and p represents the resistivity at a certain temperature of the material; that is to say, j> is the actual resistance in ohms of the unit cube of the material. The term resistance has come more into general use than conductance, and represents the opposite idea, the relationship between them being Simple Circuits. 3 where m stands for the conductivity, that is the actual con- ductance, in mhos of the unit cube. Thus, the value of the resistance of a circuit can be expressed in two ways, viz. R = If = L = ohms a am and similarly the conductance may also be written am a K = -r- = T = mhos L Lp Thus, if the dimensions of a circuit, both electrical and mechanical, be known, its resistance or conductance can at once be found. And in cases where the whole resistance is made up of a number of items, we can find the value of each item r by inserting in the equation r J* a the proper values for /, a, and p, the latter being possibly different for each item of the circuit, either on account of temperature or on account of difference of material. The tables on page 209 will give all necessary data for these values. Different Kinds of Circuits. There are two main ways in which lamps are connected to form a circuit : the series and the parallel. The series circuit is arranged thus FIG. i. 4 Examples in Electrical Engineering. and the lamps are generally arcs, each taking such and such a current at a P.D., usually from 45 to 50 volts in the case of continuous currents. The current in the circuit is that taken by any lamp, and is constant ; that is, fixed in value for that particular circuit. The E.M.F. is that required by each lamp multiplied by the number of lamps and added to the P.D. required for the dynamo and leads. The parallel circuit is arranged thus ^ . c >; ) C ) o o o f_ c ) c ) C ) FIG. 2. and the lamps are generally incandescents, arranged as shown in Fig. 2, at x. These lamps may generally be described as being so many 16 candle-power (c.p.), 60 watt, 100 volt lamps, for example. They thus all take the same current, viz. , . watts required by it current in each lamp = - volts and the whole current is the sum of all the currents, or is watts n X volts Simple Circuits. 5 In some cases there are arc lamps arranged two in series as shown at y, Fig. 2. These lamps will be 10 ampere lamps, for example, and the current in that case will be 10 x half the number of lamps. Calculation of Resistance of Parallel Circuit. In considering the value as to resistance or conductance of a number of items in parallel, it is generally simplest to take the sum of all the values of current in amperes through each item of the parallel circuit due to a common P.D. For example, What is the resistance of four lamps in parallel having resistances respectively 100 ohms, 50 ohms, 25 ohms, and 10 ohms? Consider the common P.D. to be, say, 100 volts, whence the currents respectively will be i ampere, 2 amperes, 4 amperes, and 10 amperes, making a total of 17 amperes, which current is due to a P.D. of TOO volts acting upon the united resistances in parallel, which must conse- quently have the value R = - = - = 5-88 ohms. The direct calculation of conductance is simpler, since the total conductance is obviously the sum of all the conduct- ances. Thus, the respective conductances are 0*01, 0-02, 0*04, and 0*1 mho, or the total conductance is o'lj mho. Now, conductance is the reciprocal of resistance, and resistance is the reciprocal of conductance, or R But the total conductance is the sum of all the separate con- ductances, or is K = ki + &J 4- 3 4- 4 , etc., which can be replaced byK = -+--f- + I , etc. And as the resist- r \ r -2 ?s 7*4 ance is the reciprocal of conductance, we have K 6 Examples in Electrical Engineering. which may be expressed in words : " The resistance of a number of items in parallel is equal to the reciprocal of the sum of the reciprocals of their separate resistances." In the parallel or multiple-parallel system of distribution the circuit is always composed of a number of items in series, such as the resistances respectively of the dynamo, the con- ducting wires, and the lamps; and as these latter generally have to work at a fairly constant P.D., notwithstanding varia- tions in the current, it becomes a question of great importance to know what size the leads must be to prevent more than a certain small loss of pressure in them. As a good com- pound dynamo gives a fairly constant P.D. at its terminals for all values of the current, its own resistance may be disre- garded, and the terminal P.D. taken as the whole E.M.F. of the outside circuit. A few typical cases of this kind will render this clear. i Example I. What must be the sectional area in square inches of the copper leads to convey the current required by 200 lamps in parallel, all situated at the far end of the leads, each lamp taking 60 watts at 100 volts? The distance from the dynamo to the lamps is 100 yards, and the loss along the leads is not to exceed 2 volts. Solution. Since the loss in the leads may be 2 volts, and is equal to the current multiplied by the resistance of the leads, i.e. the P.D. required to send the current through the leads, or O,, it is necessary first to find the total current. Each lamp is a 60 watt, 100 volt, and therefore takes -~ =0-6 ampere ; and the whole current is consequently 0-6 X 200, or = 120 amperes. Therefore the P.D. required by the leads is 1 2 or,, and must not exceed 2 volts ; or r l -~^ = 0-0166 ohm which is the resistance of a certain length of wire, viz. 2 x 100 yards. We can write this resistance also in terms of its length, sectional area, and specific resistance, or 2 X TOO x 36 x 0-00000067 1 ~ a Simple Circuits. 7 where a is the sectional area required in square inches, and 2 x ioo X 36 x 0-00000067 a = . ' = 0-2895 sq. inch > 0-0166 which would be slightly larger than a y stranded cable. Example II. If in the above question the resistance of the dynamo be 0-013 ohm, what will be the total E.M.F. in the whole circuit, the P.D. at the dynamo terminals, and the P.D. at the lamps at full load and at half load, when the lamps are supposed to vary one volt up or down, from ioo volts as an average ? Solution, Neglecting the slight difference made in the total current when the P.D. at the lamp terminals varies slightly, and the shunt current of the dynamo, we have that at full load C = i2oa, loss in leads = 2 volts, P.D. at lamps is 99 volts, and E.M.F. total must be equal to the sum of these plus the P.D. required to send the current through the dynamo. And this last is 120 X 0-013, or 1*56 volt. Whence E = 99 -j- 2 -j- 1-56 = 102*56 volts. Again, at half load only ioo lamps are in circuit, and C = 60 amperes; whence the loss in leads is only 60 x 0*01666 = i volt, and the P.D. required for the dynamo resistance is 60 x o'oi3 = 0-78 volt. The whole E.M.F. is then the sum of these, viz. 1*78, plus the P.D. at the lamp terminals, which is now ioo volts, and total E.M.F. = 10178 volts. Also the P.D. at the dynamo terminals is in each case the sum of the P.D.'s required by the leads and the lamps. At full load this is 2 + 99 = 101 volts, and at half load i + ioo, or, again, 101 volts. Thus the P.D. at the dynamo terminals is to be kept constant, at 101 volts, for all loads. Example III. What will be the diameter of the copper rods to convey 5000 amperes to an aluminium furnace at a distance of 7 yards from the dynamo terminals, if the loss along the leads is to be only \ volt ? Solution. The P.D. required by the leads is 0-125 volt, and is equal to 5000^, whence 8 Examples in Electrical Engineering. 0-125 r, = 0*00002 c; ohm 5000 and the leads are 2 x 7 X 36 = 504 inches long, or have a resistance of 0*000025 = 0*0000000496 ohm per one inch. Therefore their sectional area will be 0*00000067 - = 1-3-e sq. inches 0*0000000496 and if circular bars, their diameter will be d = 2 x \j - = 2 x A/4*3 7T = 2 x 2-073 = 4 >J 46 inches Example IY. What will be the loss of pressure at half load and full load in the leads in a parallel circuit running 40 arc lamps arranged two in series, and taking 1 2 amperes each, when the dynamo is kept at a terminal pressure of 102 volts, and is 50 yards from the lamps, the sectional area of the leads being 0*3 sq. inch ? Solution. At full load the current will be 40 C = 12 x = 240 amperes and at half load C = 120 amperes. The leads are 50 X 2 X 36 = 3600 inches long, and therefore have a resistance Lp __ 3600 x 0*00000067 1 ~ a 0*3 = 0*00804 ohm Therefore the loss of pressure at full load will be 0*008 x 240 = 1*92 volt, and at half load 0*008 X 120 = 0-96 volt nearly. In simple series circuits there is no variation in the loss of pressure in the leads, etc., with variation of load since the current is constant. It is, then, only necessary to keep the loss of energy in the leads within a reasonable percentage of the whole activity. Thus in the following examples : Simple Circuits. 9 Example Y. A series circuit consists of 60 ten-ampere arc lamps, spaced 100 yards apart, and taking an average of 49 volts each. What will be the diameter of the single copper leads required in order that the energy waste in them will be within 2 J per cent, of the energy required by the lamps ? Solution. The expenditure of energy in the lamps is at the rate of 60 X 49 X 10 = 29,400 watts, and 2 \ per cent, of this is - 3 x 29,400 =735 watts, which is a loss of 73-5 volts, as the current is 10 amperes, and consequently their resistance is 7-35 ohms. The length of the leads is 60 x 100 X 36 = 216,000 inches long, and thus the resistance per inch is 7*35 -, -- = 0*000034 ohm 216,000 ri-U 1 0'00000067 . Ihus the sectional area is - = 0-0197 sq. inch, 0-000034 whence the diameter will be diameter = 2 x \/ - = 2 x Vo"oo626 7T = 0*158 inch or about No. 8 B.W.G. Example YI. What will be the brake H.P. of engine required to run the above lamps if the loss in the dynamo itself is 150 volts, and only ^ of the power received is con- verted into electricity ? Solution. The activity in the lamps is 60 x 49 X 10 = 29400 watts, and the loss in the leads is 735 watts. The loss in the dynamo is 10 x 150 = 1500 watts; whence the whole activity is 31,635, or the electrical H.P. is = 42-41 But this is only T ^ of the H.P. at the engine pulley, whence the brake H.P. of engine must be 42-41 X ^ = 47'i3 H.P. IO Examples in Electrical Engineering. Example YII. What will be the E.M.F. required to send a current of 10 amperes through a circuit consisting of a dynamo whose resistance is 7 ohms, of i^ mile of leads of No. 6 B.W.G. = 0*203 i ncn diameter, and 80 lamps each of 2-5 ohms resistance? What will be the H.P. of engine re- quired if the dynamo turns -~ of the power received into electricity? And what will be the cost of i Board of Trade unit in the lamps if i mechanical H.P. cost twopence per hour? Solution. Total resistance of circuit is 7 + 80 x 2-5 + leads and the resistance of the leads is i '5 X 1760 X 36 x 0-00000067 = 4x7 0-203 X 0-203 X 22 = 1*96 ohm or whole resistance is 208-96 ohms ; whence the E.M.F. is 10 x 208-96 = 2089-6 volts and the total activity is 2089-6 x 10 = 20,896 watts and therefore the engine power must be 20,896 X^X^ = 31-12 H.P. Thus the cost per hour's run will be 31- 12 H.P. at 2^., or 62-24 ^ r 2 -> oooo / B D F FIG. 4. Let AB represent the terminals of a dynamo supplying current to a number of lamps, the nearest being situated at a distance AC from the dynamo, and any lamp being at a distance CG from its neighbour. Let the resistance of the leads from A to the first lamp be t\ ohms lead and return, and the resistance of the lead and return between any two lamps be r^ ohms. Now, all the lamps obviously cannot be at the same P.D., 1 6 Examples in Electrical Engineering. since there will be a loss of pressure in between each lamp and the next nearer the dynamo, of an amount equal to r<> times the current flowing to all the lamps beyond the point under consideration. Thus, if the current taken by each lamp be C amperes, neglecting the slight variation of current due to the fact that all lamps are not at the same P.D., the loss of volts between CG = r. 2 x 3Q and similarly the loss between GH and HJ, will be r X 2C, and' r 2 X C. There will also be a loss of r^ x 4C volts between the terminals of the dynamo and the nearest lamp. If e represents the P.D. at the dynamo terminals, the P.D. at the terminals of the four lamps, C, G, H, and J, will be respectively e - ?\ X 4C e - /-! X 4C - r. 2 X $C e r^ x 4-C 7*2 X $0 r. 2 x 2C, and e - ?\ x 4C - ?' 2 x sC - n X 2C - r 2 x C In order that the lamps may not differ from one another by more than quite a small amount in brightness, it is necessary that the difference of P.D. between the nearest lamp C and the farthest E may not be more than 2 per cent, or say 2 volts on 100. Furthermore, it is obvious that this comparative uni- formity between the nearest and farthest lamps will be quite unaffected by the loss along AC. Thus, it is general to fix the P.D. between the terminals of one lamp, viz. that nearest the dynamo, and cause the others to date from that. Adding up all the losses between the lamps, it will be seen that the total loss is 7-3(3 C + 2C + C), or 6/- 2 C volts, which is only half the loss which would have taken place had the whole current of 4C amperes been flowing all along the leads from C to E. From this follows a rule for ascertaining the loss along leads having a uniformly distributed load. The loss of pressure in volts along leads carrying current to lamps uniformly distributed along their length is equivalent to the product of the whole resistance of the leads and the average current. The above is the loss along that part of the leads bounded by the nearest and farthest lamp, and is independent of the Distribution, 1 7 loss along the part AC (Fig. 4), which is equal to ri times the whole current. Thus, calling R x and R 3 the resistances re- spectively of the parts of the leads in between the dynamo terminals and the nearest lamp and in between the nearest and furthest lamps, and C the total current, then the loss from the dynamo to the nearest lamp is and the loss along the leads between the lamps is Uniformity of brightness in the lamps dictates that the latter should not be more than, say, 2 volts on 100; but this condition will not be affected by the loss RiC, provided the middle or some other convenient lamp be maintained at a certain agreed upon P.D., irrespective of the number of lamps in circuit. Economy of cost is the only consideration which will determine the loss along Rj, and this will be considered later on. It is usual to distinguish between these two parts of a circuit by calling that part of a system of leads which carries the same current throughout its whole length and is not tapped anywhere, a feeder ; whilst that part of a system which is tapped at intervals along its length, and consequently carries a current which is not the same in all parts, is called a distributor. A circuit may consist of a number of distributors connected together, and is then called a distributing network. Each portion of such network will have to be supplied by its own feeder to maintain the required P.D. Fig. 5 will render this clear, where the double line of conductors bounding the figure represents the distributing network, which is connected to the station, or electricity works, ; S, by the feeders Sa, S, etc. Each point a, &, c, etc. is called a feeding centre, and is kept at a fixed P.D. by manipulation of the feeders in the station i8 Examples in Electrical Engineering. the lamps situated in between a and b will be partly supplied by the feeder Sa and partly by S. The total current taken off the distributors between a and b being called C amperes, and being the same as that taken off the distributors between any two other feeding centres, it follows that each feeder will have to carry C amperes ; and if R, be the resistance of a feeder, lead and return, the loss of pressure in the feeder will f FIG. 5. be CRy, an amount which is not necessarily the same for all feeders, as their lengths may be different, or they may not be .at all times equally loaded. As the load between b and c (Fig. 5) is equally divided, the Distribution. 1 9 C current in the distributors will be amperes at the feeding centres, but gradually falling till some point is reached about the middle of the distance be, say at x, where the current will be zero. In order, then, to state the maximum difference of P.D. between any two lamps, it is necessary to allow that the dis- Q tributor in between a and b has* a mean current of - amperes, 4 and a length equal to half the distance be, say bx. Calling R cZ the resistance of distributor ab, lead and return, the maximum difference of P.D. between a lamp at b and a lamp at x will be C* R CR Loss of volts along distributors = - x = ^- d volts 4-2 o The determination of the size of the distributors depends upon the difference of P.D. allowable between any two lamps, and has already been considered. But in determining the size of the feeders other considerations are involved, such as economy of cost of transmission. If the feeders be of very small sectional area, they will cost little for material, but will waste much in heating, and the whole yearly cost made up of interest on cost of -cables and laying, and value of energy wasted may be very large. On the other hand, if the feeder be of large sectional area it will cost much, whilst the energy wasted may be very small, but the total yearly cost may again be great. It is therefore necessary to take some intermediate size of leads which shall have the yearly cost as small as possible. This was first pointed out by Sir William Thomson (now Lord Kelvin), whose results may be expressed in the following simple rule The size of the feeders must be such that the interest for one year on money lying idle in their cost and laying together with depreciation, must be equal to the value of the energy wasted in them during one year. Professor Baily has pointed out a simple way of ascertaining that size of cable which gives the desired equality. Construct 2O Examples in Electrical Engineering. two curves of cost (per mile, say) of cable for one year, thus ct Size of Gable FIG. 6. The curve b represents the cost of waste energy in the cable, whilst a represents the interest, etc., on the cost of the cable. The point y, where these two curves cut, shows where the two items are equal and also the size of cable. In calculating out the loss in the feeders it is necessary to allow that no central station is on full load for all the twenty- four hours. It thus comes about that the current taken in the above calculation of waste in the cables must not be the full current ; it should, in fact, be more nearly the square value when this can be ascertained. A little consideration will at once show that strict adherence to Kelvin's law is not always desirable, seeing that where power is very cheap and materials and cost of laying very great, the most economical size of conductor might be such Distribution. 2 1 that the energy waste was excessive, and though of no con- sequence as energy, yet producing a dangerous rise of tem- perature on the conductors. A good many safety regulations limit the rise in temperature of the conductors to 150 Fahr. as a maximum, thus permitting a possible difference between the air and the conductor of about 75 Fahr., and that for a current twice as large as the normal maximum. Since the heating is, other things being equal, proportional to the square of the current, it follows that the rise in temperature due to the maximum working current must not be more than one-fourth part of the 75 Fahr., or about 1 8 Fahr. From a large number of experiments on conductors of all kinds, Mr. Kennelley has discovered a rule connecting the diameter of the conductor and the current to be carried consistent with the fact that the rise in temperature shall be within this limit, viz. C = 56o^or*=|^ thus indicating that the current density should decrease as the diameter increases. As, however, most Fire Office rules fix the current density at 1000 amperes per square inch, irre- spective of the size of the conductor, it follows that though for small sizes there is a very high degree of safety, neverthe- less for conductors exceeding \ sq. inch sectional area the allowable current would produce a greater rise in temperature than that allowed by Kennelley's rule. Thus, when the sectional area is - sq. inch, Kennelley's rule would allow about 17 per cent, more current for safety than the Phoenix Fire Office ; when the area is \ sq. inch, the Fire Office rule would be equivalent to the safe temperature rise; whilst at i sq. inch area, the Fire Office rules allow 50 per cent, more current than is consistent with the maximum allowable rise in temperature. Example XII. The sectional area of the copper con- ductors forming a system of distributors is \ sq. inch, and they are loaded with one 5o-watt ico-volt lamp per yard 22 Examples in Electrical Engineering. of street. What must be the distance between the feeding centres, which are maintained at a P.D. of 100 volts, in order that the maximum difference between any two lamps shall not exceed 2 volts ? Solution. Neglecting the small difference in the total current made by the fact that ail lamps are not run at exactly the same P.D., viz. 100 volts, each lamp takes 0*5 ampere, and consequently the total current supplied by any feeder will be amperes, where L is the distance in yards between the two feeding centres. The resistance of distributor, lead and return, in between the feeding centres will be 2L x resistance per yard, or 2L X 36 X 4 X 0*00000067 = 0-00019296!; ohm and the loss of volts along half of L will be L X = 2 8 whence L 2 = - .. = 165,803 0-00009648 and L = 407 yards Example XIII. What will be the maximum difference in volts between two lamps if the distance between two feeding centres is a quarter of a mile, the size of distributors is J sq. inch sectional area, and current is taken off at the rate of \ ampere per foot of street ? Solution. The maximum current in distributors is \ X i X 13,200 =110 amperes and therefore the average value is 55 amperes. The resistance of i mile (lead and return) of distributor of sq. inch sec- tional area is 4 x 1320 x 12 x 0*00000067 = 0*04244 ohm whence the loss of volts is 0-04244 X 55 = 2-334 volts, which is consequently the maximum difference between two lamps. Distribution. 23 Multiple Wire Systems. In some cases the distributors are more than two in number, three or even five conductors being used. This method was first proposed by Dr. John Hopkinson. The connections are made as shown in Fig. 7, taking the five-wire system as an example T A Q| R r. 6 \ ft I H \ rY > < > N t ^ f P u nT > c ! u Q| R FIG. 7. The five wires constituting the distributors are shown at AG, BH, CK, DL, EM, the outermost conductors AC and EM being considerably larger in sectional area than the three others. The object of the system is to gain the advantages of transmission of energy from the central station at a high pressure, say 400 volts, and nevertheless provide for the use of loo-volt lamps. A three-wire system can do the same, using 2oo-volt lamps, which are now coming in. To transmit the same amount of energy at an E.M.F. of 400 volts, will only require one- sixteenth the weight of copper for the same percentage loss as on a loo-volt circuit. Lamps may be connected in various ways, as at N, loo-volt lamps in parallel between the con- ductors; as at O, two arc lamps being connected in series between each pair of distributors ; and as at P, where 200- volt lamps are connected between the middle and outside wires. 24 Examples in Electrical Engineering. The objects of the middle wires BH, CK, and DL, are (1) To enable the equality of P.D. between the wires to be maintained. This is done by introducing between each of them in the station a dynamo or other device, maintained at P.D., of 100 volts each. (2) To prevent any want of equality in the numbers of the lamps connected between the parallels causing fluctuations in the P.D. between the parallels, by giving the excess current another path back to the station than through the lamps on the lightly loaded side. When the whole load has been judiciously distributed between the parallels this excess of current is not large, and consequently the middle wires need not be so large as the outermost. Indeed, if all the loads be equally balanced, then the middle wires may be of so small a section as to practically vanish ; but in practice this cannot be ensured, and the middle wires are made of such a size as to carry an agreed-upon percentage of the full load. Like the simple parallel system, the distributors are fed by feeders connected to large dynamos running at a P.D. of 400 volts, plus the economical loss of volts in the feeders, con- veying the current to the distributors at feeding centres as shown at QQ and RR. Example XI Y. Compare two systems. No. i. A two- wire system to supply current to a circle of distributors 4700 yards round, at the rate of one 5o-watt loo-volt lamp per foot of street, and having ten feeding centres at equal distances round the circle, the station being at an average distance of 800 yards from the distributors. There is not to be more than 2 volts difference between any two lamps, and the pres- sure at the station end of the feeders may be 10 per cent, more than at the distributors. No. 2. The same town to be supplied on the five-wire system with the same conditions ; provision, of course, to be made for the connection of the distributors to the potential equalizers in the station. Compare the weights of copper used in the two cases. Distribution. 2 5 Solution. No. i. The distance between the feeding centres is 470 yards, whence the average current in the distributors is -^- X i '5 X = 176 amperes, say and thus the resistance of the distributor must be 2x2 ^ = 0*02273 ohm for a 2-volt drop. And as the length of distributor from a feeding centre to feeding centre is (lead and return) 2 x 47 X 36 = 33,840 inches the resistance per one inch is 0*02270 = 0*0000006712 ohm 33840 whence the sectional area is i sq. inch. The feeders have to carry a current of 4 x 176 = 704, say 705 amperes, and are 800 yards long, with 10 volts loss. Thus the resistance is - = 0*01418 ohm 70S or 0*01418 _ 7 = 0*000000240 ohm per i inch 800 X 2 x 36 and the sectional area is - f 2*72850. inches 246 Taking i cubic inch of copper as 0*32 lb., the total weight of distributors will be 4700 x 36 X 0*32 X 2 _ ~2~2lo~ = 48-34* and the feeders IOX 8oox 3 6.JLLX-7'8xoj = (ons 2240 Or the whole weight of copper in No. i = 27274 tons. 26 Examples in Electrical Engineering. In No. 2, as the outside wires will be at 400 volts P.D., the current will only have an average value of \ as much as in No. i, or 44 amperes. Also the fall of P.D. may now be 8 volts, being only 2 per cent, as before. If the feeding centres are at the same distance apart, the sectional area of the distributors will thus be only y 1 ^ that of No. i, or y 1 ^ sq. inch. But their length will be increased by 2 x 800 yards to connect them to the potential equalizers in the station ; or the whole weight of outside wires will be ( 47 00 + 800) X 2 X 3 6 X Q'32 X yV = 2240 and allowing \ the section for the three middle wires each, their combined weights will be half as much as the outsides, or 176 ton. The feeders also carry \ the current, and with four times the loss of volts they will likewise be yg the area of those in No. i, or weight will be }- = 14*04 tons, or the total weight will be 19*33 tons, as against 27274 tons for No. i. CHAPTER III. CIRCUITS CONTAINING MORE THAN ONE E.M.F, IN some cases there is more than one E.M.F. in the circuit, and the consideration of the magnitude of the phenomena set up will depend upon whether all the E.M.F.'s are in one direction, or whether some are opposed to the others. The cases where all the E.M.F.'s are in one direction have already been dealt with, and we will now 'deal with some cases where the effective or active E.M.F. is the difference between an applied E.M.F. and another E.M.F., due to the nature of the circuit, acting against the applied E.M.F., and consequently called Back E.M.F. Let E = an E.M.F. applied to a circuit to (i) overcome the B. E.M.F., and (2) leave, sufficient over to send the necessary current through the resistance of the circuit. e = the B.E.M.F., such as that possessed by chemical cells, running motors, arc lamps, etc. e = the active E.M.F., which is equivalent to the differ- ence between E and e, or = E e, and which is also equal to the product of the resistance of the circuit and the current passing. Thus E = e + e, or e E e, and consequently the current, which will be produced by E in a circuit of resistance R and containing the B. E.M.F. e, will be c-^" e - < R -R The resistance R may be composed of various items as before. 28 Examples in Electrical Engineering. Secondary Batteries. In the case of charging a Secondary Battery, the circuit must consist of a dynamo, the Secondary Battery, and the necessary connecting cables. Each of these items will have their own resistances, that of the Secondary Battery being composed of a number of items in series, each being the resistance of a single cell. In addition to this, each cell will have an E.M.F., which must be overcome in charging, and this E.M.F. is a quantity which changes with the state of charge, being small when the charge is low, and rising to a higher value as the charge increases. This rise of E.M.F. is partly due to actual chemical change, and partly due to a variable temporary increase of local resistance at the surface of the plates. It is, however, customary to refer to it on the whole as an increase of the B. E.M.F. The curve in Fig. 8 shows the values of this E.M.F., corresponding with different charges. H U R 54 FIG. 8. From this it will be observed that, though a cell may require 2-3 volts or so to overcome its B.E.M.F. when charging, it Circuits containing more than one E.M.F. 29 cannot give more than a fraction more than 2 volts for even a short time when discharging. We may thus be required to state what E.M.F. a dynamo must produce in order to charge so many secondary cells with such and such a current ; or we may have to find what will be the value of the current through the cells when charged throughout with a constant E.M.F. Example XY. What must be the E.M.F. to charge with 100 amperes a battery of 54 cells at an average B. E.M.F. of 2*3 volts each, and each having an internal resistance of 0*0004 ohm, when the resistance of the dynamo is 0*02 ohm and that of the leads is 0*03 ohm ? Solution. The total B. E.M.F. is e = 54 x 2 -3 = 124-2 volts and the active E.M.F. required for the resistance of the circuit will be the product of current irito the sum of all resistances, or e = 1 00(0*02 + 0-03 + 54 X 0-0004) = 100 (0*0716) = 7*16 volts whence the applied E.M.F. must be E = 124-2 + 7*16 = 131*36 volts Example XYI. What will be the current through the circuit at the beginning and end of charging in the case of a battery of 53 cells, each of 0*0002 ohm internal resistance, charged by a dynamo at an E.M.F. of 135 volts constant, when the dynamo resistance is 0-015 onm > an d that of the leads is 0*025 ohm, if the charging is continued for 7 hours, the value of e per cell rising from 2*1 volts to 2-35 volts? Solution. The B.E.M.F. at starting to charge will be 2'i x 53 = ni'3 volts, and at the end of 7 hours will be 2 '35 X 53 = 124-55 volts. The total resistance is 53 X 0-0002 + 0*015 ~T" 0*025 = 0*0506 ohm the active E.M.F. at the start of charging will thus be e = E - e = 135 - 111*3 = 23*7 volts 3O Examples in Electrical Engineering. and at the end of charging e = 135 - 124-55 = 10-45 volts whence the current at starting will be c - ^- , = 468 amperes 0*0506 and at end of run is 10*45 c = -~ = 206 amperes 0*0506 Example XYIL In the last example (XVI.), if the dynamo has to be run so as to produce 135 volts always, and it is desired that the current shall not exceed 200 amperes, what will be the value of the necessary resistance to be inserted (i) at starting, (2) at end of charging? Solution. The active E.M.F. at starting e - 237 volts and at end of run is e = 10-45 whence the total resistance at starting must be 2 37 = 0*1185 ohm 200 J and at end of run 10*45 = 0*05225 ohm 200 But the resistance of the circuit is already 0*0506 ohm, whence the added resistance must be 0*1185 0*0506 = 0*0679 onm at start and 0*05225 0*0506 = 0*00165 ohm at end of charging. Example XYIII. A battery of 55 cells has a discharging E.M.F. of 2*05 volts per cell at start, and gradually falling to 1*9 volts each cell at end of 7 hours' run. The internal resist- ance of each cell is 0*0004 ohm. The resistance of the leads from the cells to the lamps is 0-005 onm - What number of Circuits containing more than one E.M.F. 31 cells must be used to supply 400 5o-watt loo-volt lamps if the P.D. at the lamp terminals is to be kept at as nearly 100 volts as possible all the time ? Solution. The lamps are to take 100 volts, and will also require X 400 = 200 amperes, neglecting the slight differ- ence made by the P.D. at their terminals not always being exactly 100 volts. The loss of volts along the leads is 0*005 x 200 = i- volt and the loss in the cells is 0-0004 X 200 = 0-08 volt per cell or the available E.M.F. per cell is 2-05 0-08 = 1:97 volt at start, or 1-90 0-08 = 1*82 volt at end of run The E.M.F. required for leads and lamps is to be 100 + i = ioi volts whence the number of cells required will be IOI r8 7 and IOI = 51-2, say 51 at start 1-82 = 55-4, or 55 cells at end Motors. In running motors we are also dealing with cir- cuits containing a B. E.M.F. A motor is a dynamo caused to run by sending through its armature a current at sufficient E.M.F. to overcome the B. E.M.F., and leave sufficient to maintain that current through the resistance of the circuit. We have, then, as before, that the current in the armature will be c- E "- R R being the resistance of the whole circuit, and e the B.E.M.F. 32 Examples in Electrical Engineering. produced by the rotating armature. The part of the energy which is supplied to the motor that is converted into mechani- cal work is Ce, and of this a part is expended in friction of various kinds in the motor itself; this part will be dealt with later on, and at present will simply be expressed as a waste in watts or horse-power. The difference between Ce and this waste is the available power of the motor, and is usually expressed as so many brake horse-power (B.H.P.), Example XIX. A motor having a resistance of 0*025 ohm, and producing a B.E.M.F. of 90 volts at a given speed, is connected by leads of 0-005 onm resistance to a dynamo of resistance = 0*02 ohm. It is required to produce a B.H.P, of 7 H.P. What must be the E.M.F. of the dynamo if the waste in internal friction, etc., in the motor is 300 watts ? Solution. The total mechanical power is Ce = C X 90 watts of which 300 watts are wasted in the motor, leaving 7 x 746 watts available at the brake. Therefore C x 90 - 300 = 7 x 746 9 ^ = 6 1 -35 amperes And this current is produced through the resistance of the whole circuit by an active E.M.F e = C (resistance of circuit) or e = 6l '35(' 2 5 4- 0-005 + ' 02 ) = 3' 6 75 volt s whence the total E.M.F. must be 90 -f 3*0675 = 93*0675 volts Example XX. In the above question, what will be the' brake horse-power of the motor if the total E.M.F. of the Circuits containing more than one E.M.F.- 33 dynamo is 95 volts, assuming the same speed and loss due to friction, etc. ? Solution. Here the active E.M.F. will be e = 95 ~ 9 = 5 volts and consequently the current will be C = = 100 amperes '5 therefore the total mechanical power is 100 x 90 and of this 300 watts is still the waste, leaving for brake power 9000 300 = 8700 watts or 8 7^ = 1 1 -6 H.P. at brake 746 Arc Lamps. In the case of arc lamps, the energy required for the vaporization of the carbons is generally considered as made up of the product of the current and an E.M.F. at the crater, called a back E.M.F., the magnitude of which is about 39 volts for a continuous current arc ; but which appears to be as low as 30 volts (virtual) for some alternating current arcs. Thus, in an arc lamp, part of the E.M.F. supplied is con- cerned in sending the current through the resistance of the circuit, and the remainder is to overcome the back E.M.F. Example XXI. An arc lamp has a B. E.M.F. of 39 volts, and is connected to leads at P.D. of 50 volts. The resistance of the lamp leads is o'n ohm, the resistance of the lamp coil is 0-09 ohm, and the carbons have a resistance of 0*08 and 0*12 ohm respectively, whilst the arc itself has a resistance of 0*1 ohm. What must be the resistance included in the circuit so that the lamp may take a current of 10 amperes? Solution. The available or active E.M.F. is e= 50 39 = ii volts D 34 Examples in Electrical Engineering. and since the current is to be 10 amperes, we have the total resistance must be e n -^ = i' i ohm C 10 But the resistance of the lamp itself is o-ii -f 0-09 + o'o8 + 0*12 -f- o'i = 0-5 ohm whence the additional resistance required must be i' i 0*5 = 0*6 ohm CHAPTER IV. DYNAMOS AND MOTORS. IN questions relating to the proportions of parts of dynamos and motors we have to consider the relationship between the armature and magnet windings, etc. The electro-motive force generated in a revolving armature can be arrived at from the data of its parts. Thus, the whole generated E.M.F. is io c for a Bipolar machine, whether drum or gramme wound, where N = whole number of c.g.s. magnetic lines which traverse the armature core, n = the revolutions per second of the armature, and w the whole number of wires counting all round the circumference of the armature, io 8 = the number of c.g.s. units of E.M.F. , equivalent to the practical unit, the volt. The factor N stands for the rate of cutting lines in c.g.s. lines per second. The number of lines N is made up of the product of the sectional area of the iron armature core and the number of c.g.s. lines per unit of sectional area. It is very usual to call the number of c.g.s. lines per square centimetre the value of J3, or /? = c.g.s. lines per sq. cm. of armature iron For different types of armature there are different values of ft most suitable. Thus, for gramme or cylinder armatures, ft 36 Examples in Electrical Engineering. may range from 14,000 to 18,000 sq. cms., or even higher; whilst, owing to the necessity of passing the lines through the same size of air gap in the drum as in the cylinder type, we must keep the value of ft for drums at about 9000 to 12,000 per sq. cm., though in some types of drum higher values can be used with advantage. The speed n is limited by the strength of the structure, on the one hand, to not more than 3000 to 4000 feet per minute (roughly, 1500 to 2000 cms. per second) circumferential speed ; or, on the other hand, to far smaller values where slow speeds are desirable for direct coupling, to slow-speed engines, for example. The number of wires all round, w, is determined by the E.M.F. desired, and their size by the current to be carried. As the commutator connects the armature to the outside circuit in two parallels in Bipolar machines, the size of wire on the armature is that to carry half the current from the machine. No fixed rule can be stated for the relationship between the current and the size of wire, as this will depend upon the size of machine, the efficiency desired, and the rise in temperature allowable. This part of the subject is considered in a later chapter. The whole E.M.F. E is the total volts generated by the armature, and, of course, a part of this is required to circulate the current in armature itself. Thus, it r a = resistance of armature winding measured from brush to brush, and C = total current in amperes in the armature, then Cr n volts will be lost in the armature, and the differ- ence E Cr a is the terminal P.D. of the dynamo, in the case of a magneto machine or a shunt dynamo. The resistance r a is obviously the resistance of J- the whole length of wire upon it (since the two halves, and in parallel, and each half is half the resistance of the whole), plus any contact resistance at the brushes and in the leads to the brushes. , I (UNIVERSITY \ Dynamos and Motors. 37 Leaving for the present the exact statement of the relation- ship between the armature and magnet windings, we will take a few simple examples. Example XXII. What will be the sectional area of iron required in the armatures respectively of two dynamos, one ring wound and the other drum wound, each having 240 wires all round, running at 1200 revolutions per minute, and giving a total E.M.F. each of 120 volts? Solution. In each case we have that the number of lines N required will be nw and since n = ^ - =20 revolutions per second XT 120 x io 8 N = = 2x00.000 lines total 240 X 20 Now, the ring-wound armature may have! ft = say 15,000, whilst the drum may have ft only 10,000, say. But N = fta whence in the ring type 2,500,000 a a = - - = i66f sq. cms. 15,000 and in the drum 2.SOO.OOO a = ~ = 250 sq. cms. 10.000 * * area of iron core. Example XXIIL What is the total E.M.F. produced by a dynamo, having an armature ring or gramme wound with 204 wires all round, when ft = 16,000, area of cross- section of iron core is 125 sq. cms., and speed is 25 revo- lutions per second ? Solution. Na = 16^000 X 125 X 25 x 204 = IQ2 yolts IO 8 IO 8 38 Examples in Electrical Engineering. Example XXIY. : At what speed must the above armature be run so as to give 102 volts terminal pressure when the resistance of the armature is 0*02 ohm, the current is 100 amperes, and the value of ft remains at 16,000? Solution. The loss of volts in armature resistance will be O' ft = 100 x 0*02 = 2 volts whence the total generated E.M.F. must be E = 102 + 2 = 104 volts thus the speed Eio 8 104 x io 8 ~- = 25 ' 49 revo!utlons P er sec " or 1529*4 per minute. Example XXY. What is the value of ft in an armature core, drum wound, 25 cms. diameter, with 7 cm. hole in the centre, and composed of 500 discs half a millimetre thick, if wound with 180 wires all round, and run at 1200 revolutions per minute, producing an E.M.F. of 175 volts at the terminals, when the resistance of the winding is 0*032 ohm, and the current is 100 amperes ? Solution. The total generated E.M.F. is E = 175 + = 175 -f 100 X 0*032 = 178-2 and therefore ._ 178*2 x io 8 N = - s = 4, an< 3 triat f i ts shunt coil is 12*5 ohms. At what speed will it run as a motor, and what current will it take at a P.D. of 50 volts to develop 12 brake H.P., when H.P. is lost in friction in the gearing, and 0*352 H.P. in the machine? Solution. The total rate of doing work by the motor must be 12 + o'5 4- 0-352 = 12-852 H.P., or 12*852 x 746 = 9588 watts, which must be the product of the current in the arma- ture and the B.E.M.F. produced. The shunt coil takes ^ = 4 amperes and thus the total current will be C = C +4. Dynamos and Motors. 53 As the resistance of the armature is 0*0147 onm > the l ss f volts therein will be C rt x 0-0147 volts, and thus = 50 - and C rt must be = 9588, or (50 - C n X o-oi47)C rt = 9588 or - 0-OI47C, 2 + 5oC ft - 9588 = o whence C rt = 204 amperes, at which current the loss in the armature is 204 x 0-0147 = 3 vo lt s leaving 50 3 = 47 volts for e The total rate of doing work mechanically is 204 x 47 9588 watts and the speed is 4.7 X 720 = 628-4 per minute 53 Example XXXI Y. A lo-ampere series machine which gives 1000 volts total E.M.F. at a speed of 1000 revolutions per minute, is supplied with a current of 10 amperes to run as a motor, and has to lift a weight of 3 tons at the rate of 50 feet per minute. If there is a loss due to friction in the gearing of ^ H.P., and 0-32 H.P. lost in the friction, etc., in the motor, what will be the speed and P.D. of supply mains required if the resistance of the armatureTis 3-5 ohms, and that of the magnet coils 2-5 ohms? Solution. To lift 3 tons 50 feet per minute is doing work at the rate of 3 X.2240XJO = iQ . i8 H p 33,000 and, adding the losses, the total H.P. will be 10-18 -f 0-5 -f- 0-32 = ii H.P. or in watts = 8206, which is done by a current of 10 amperes against a B.E.M.F. of liM = 820-6 volts 54 Examples in Electrical Engineering. eh the speed will be X 1000 = 820*6 revolutions per minute to produce which the speed will be 820-6 1000 To send 10 amperes through the resistance of the machine will require a P.U. of io(r a -f- r m ) = 10 X 6 = 60 volts whence the P.D. at the motor terminals will have to be 820*6 + 60 = 88 1 volts, say Example XXXY. The length of one wire on the surface of an armature is 20 cms., the mean diameter is 21 cms., the speed is 1200 revolutions per minute, ancl the density of lines in the air gap is 2291 per sq. cm. The angle of polar embrace is 135, and there are 360 wires all round. If this machine be used to give a total current of 50 amperes, what will be the H.P. required to drive it if H.P. is lost in various ways? Solution. Applying the formula on p. 50, we have 1 200 21 3917 ' 7 X ~6o~ X 7o^7 X 50 X 20 X 360 X 135 X 2291 H P = oo 30 47 io 15 = 6*01 H.P., to which must be added \ H.P. as above for friction, etc., or total H.P. required = 6-51 H.P. Example XXXYI. What will be the drag on one wire of a dynamo armature which gives a current of 800 amperes, and has an average magnetic field of 3500 c.g.s. lines per sq. cm. of air gap ? The length of the polar bore is 80 cms., and the machine is bipolar. Solution.- Using the formula given on p. 49, we have pull is P = CL/30 X 0*0000002244 pounds and as the current here is the current in each wire, it is consequently 400 amperes, and P = 400 X 80 x 3500 x 0*0000002244 = 25*1328 pounds CHAPTER V. ARMATURE WINDING. IN considering the size of wire on a given armature to carry a given current, we have to allow a certain temperature rise, and consequently have to take into account all the sources of heating. These are mainly threefold, viz. (a) The C/R a loss due to the effect of the current in the armature on the resistance of the armature ; (b) The H loss due to molecular friction in the iron core, brought about by the changing of the magnetic flux ; (c) The F loss due to Foucault or eddy currents in the core, and other moving parts of the armature. The energy developed in the armature due to these effects has to be got rid of at some reasonable temperature, and it is customary to fix a limiting difference of temperature between the surface of the armature and the outside air. In calculating this temperature difference it is of course necessary to bear in mind that some part of the heating is got rid of in other ways than by convection and radiation from the outside surface of the winding. Thus, part is lost by con- duction through the spider to the spindle, the winding to the commutator. Experience shows that this part so conducted away may be about \ of the total loss. Mr. Esson has shown that the surface of an armature re- volving at a circumferential speed of 3000 feet per minute, can dissipate energy at the rate of 0*00444 watts per sq. cm. of its surface covered by winding for every i C. temperature difference between that surface and the air. The correspond- ing figure for a stationary surface, such as a field-magnet coil, is 0*00281 watts per sq. cm. per i C. temperature difference. 56 Examples in Electrical Engineering. A combination of these figures will give the rate of loss of heat for any armature in terms of the circumferential speed. Thus the increase in rate of cooling due to the speed of movement through the air of 3000 feet per minute (1500 cms. per second) is 0*00444 0*00281 = 0*00163 watts per sq. cm. per i C., or 0*00163 = 0*000000543 watts per sq. cm. extra per 1 C. difference of temperature per i foot per minute of speed through the air or the loss for any speed may be put down as 0*00281 -f- 0*0000005 43/ watts per sq. cm. per i C, where f is the circumferential speed in feet per minute. For a speed of/= 3000 feet per minute, we have that the total difference in temperature between the air and the armature surface will be W T = x - degrees Centigrade A X 0*0044 where W = that part (say, J) of the whole expenditure of energy in the armature, which is considered to be radiated from its surface, expressed in watts ; and A = the surface area in sq. cms. of the armature. Similarly, for a magnet coil we have T = JL A X 0*00281 where A is the outside surface of the magnet winding, not including the cheeks of the bobbin, expressed in sq. cms. The loss W is approximately f (C ft 2 R a + H + F), and of these items the first is, of course, quite easily found, allowance being made that, the temperature being high, we have to employ a high value for p. The hysteresis loss H is due to the fact that molecular friction in the iron causes heating when the iron core as a Armature Winding. 57 mass is rotated by the rotation of the armature, whilst the molecules of iron are themselves stationary. The effect pro- duced is the same as if the molecules were rotated whilst the armature core remained stationary. In a bipolar field the armature core has all the magnetized molecules thus rotated once per revolution, and the value of the energy so expended has been found by Professor Ewing for all conditions of magnetization. The table on p. 214 gives these values ex- pressed in ergs per cc. per cycle of change, one cycle corre- sponding to one revolution of an armature in a bipolar field. Now, as one watt-second is io 7 ergs, we have that the total ergs expended in the armature per revolution multiplied by the speed in revolutions per second and divided by io 7 will give the rate of loss in hysteresis in watts. Calling the value of the ergs per cc. per cycle ^, we have that the total ergs per revolution will be ^V, when V is the bulk of the armature iron in cc. If, then, n be the revolutions per second, we have the hysteresis loss is H = _- watts io 7 The other loss, viz. that due to eddy currents, F, is almost impossible of calculation in an armature. Only a small part of it is in the laminated iron core, the major part being in other conducting masses, such as the winding, end plates, spider and spindle, etc. Due precautions being taken to minimize it as much as possible, we shall assume in these pages that it is equivalent to that part of the total loss which is got rid of in other ways than by radiation directly from the surface of the winding. Therefore we may now put the value of W as W = C ft 2 R ft + H and write the temperature rise as being T = C " 2R " + H d Centigrade A X 0-00444 for a bipolar armature revolving at 3000 feet per minute cir- cumferential speed. OF THB UNIVERSITY 58 Examples in Electrical Engineering. In finding the area of the surface of the armature, we must only take that part of the surface which is exposed to the air. In the case of a drum armature this is easy, but in a ring or gramme armature, some care is required in calculating the true size of that part of the surface which is inside the cylinder. In all cases the surface should be considered as smooth, no allowance being made for corrugations due to the curved surface of the wires ; for the coefficients taken are stated higher to allow for such irregularity. Example XXXYII. What will be the hysteresis loss in an armature ring wound, 25 cms. long, 25 cms. diameter, with a T 5-cm. hole (all dimensions being those of solid iron) ? The speed is 1226 revolutions per minute, in a field of 4 million c.g.s. lines the value of the ergs per cc. per cycle is 10,750 per cc. for this particular iron. Solution. The sectional area of the armature is (25 15) X 25 = 250 sq. cms. whence the density of lines is A \f T rfi (5 = = 16,000 per sq. cm. corresponding to which the ergs per cc. per cycle are given as 10,750. The number of cycles per second will be the revolutions per second, or 1226 " = -SQ- = 20 ' 43 And the number of cc. of iron in the core will be V = 5X25X2oX7r = 7855 cc. whence we have that 7855 x .075 X 20-43 = I0 7 say, 173 watts. Example XXXYIII. If the resistance of the armature in the last example is 0*0212 ohm when hot, and it is used to Armature Winding. 59 light 200 5o-watt loo-volt lamps at a terminal pressure of 102 volts, the resistance of the shunt coil being 51 ohms, what will be the rise in temperature due to hysteresis and CV a losses, if the eddy current loss is got rid of by conduc- tion, etc. ? The surface of the armature is 3675 sq. cms., and the circumferential speed is to be taken as 3000 feet per minute. Solution. The current taken by the lamps is 200 x 50 . = 100 amperes 100 and that taken by the shunt coil is 2 amperes, whence the total current in the armature is C,,, = 1 02 amperes and thus Ci,V a = (io2) 2 0*0212 = 22o'5, say 221 watts and the loss due to H has been found to be 173 watts, whence the total watts W = 221 + i?3 = 394 watts and thus the temperature rise will be . 0-00444 X 3675 Example XXXIX. Suppose the armature mentioned in Example XXXVII. were replaced by a drum having as before 25 cms. length of iron in it, but having only a 7-cm. hole, and run at a density of only 11,000 c.g.s. lines per sq. cm., corre- sponding to which the value of ergs per cc. per cycle is 5900 (1) What will be the temperature rise with the same winding as to size of wire and the same current at the same E.M.F., and what the speed ? and (2) What would be the temperature rise and output if the current and the speed are to be the same as when a gramme armature ? (3) Also, what would be the value of the current to make the temperature rise the same and the speed the same as in the case of the gramme ? 60 Examples in Electrical Engineering. The surface area of the drum may be taken as 4000 sq. cms., and the variations in the speed are not supposed to affect the rate of cooling. Solution. (i) As a drum there will be a sectional area of the armature of (25 7)25 = 450 sq. cms. whence at /? = 11,000 the total number of lines will be N = fia = n,ooo x 450 = 4,950,000 The total generated E.M.F. was before equal to 102 volts for the terminal P.D. plus the loss in the armature of 102 x 0*0212 = 2*16 volts, or 104*16 total to produce which E.M.F. the drum will have to run at a speed inversely as the number of lines, or speed will be ;/ = 20-43 X - - = 1 6*5 revolutions per second the volume of the drum core will be 4 4 or V = 2 5 X 25 x 22 x_g5 _ 7 X 7 X 22 xj5 = 4X7 4X7 Thus the hysteresis loss is H = 5900 . *JI'3J_4 *2^S = 1 10 watts I0 7 And since in question (i) the current is to be the same as before, we have that the C a V n loss is 221 watts as before, whence the rise in temperature will be T = ^- - = 1 8-6 C. 4000 X 0*00444 (2) If the speed is to be the same as before, there will be a greater E.M.F. total in the ratio of Armature Winding. 61 whence with the same current the output will be 100 x 12674 = I2 >6?4 watts as against only 10,200 for the ring winding. But at this speed of 20*43 revolutions per second the hysteresis loss will be H = 5 or the total watts expended will be 136 4- 221 = 357 watts and the consequent temperature rise will be T = - 257 - = 20 C. 4000 X o '00444 (3) If, however, the temperature rise is to be the same as for the ring armature at the same speed, the current may be larger, such that the new C tt V tt loss added to the H loss will make the heating such as to cause a temperature rise of 24-1 C. Thus the total watts will be W = T X A X 0-00444 = 24*1 X 4000 x 0^00444 = 428 watts of which 136 watts are for hysteresis, leaving 292 watts for C V a . If the armature resistance be taken as 0*0212 ohm as before, then O'O2I2 / and C a = V -^- = 117 amperes Example XL. An armature working at a density of 8000 lines per sq. cm. and giving a certain E.M.F. is run at half the speed with a density of 16,000 per sq. cm. : what will be the ratio of the watts lost in hysteresis in the two cases ? Solution. At the higher speed and ft = 8000, h is 2860 (p. 215) ergs per cc. per cycle; and at the lower speed and ft 16,000, h is 10,400 ergs. But as the cycles per second are 62 Examples in Electrical Engineering. only half as many in the second case as in the first, the two losses will be 2860 in the first case to 10 ^- = 5200 in the second case, or as 2860 5^60 ='5499, say 0-55 or the loss in H in the high speed low density machine is only 0*55 as great as in the slow speed high density. Multipolar Machines. When the dynamo is not bipolar, it may be so wound that either the current or E.M.F. is pro- portionally increased. Thus, with what is called the parallel winding, the coils are connected up as before to a commu- tator, which then has as many pairs of brushes as there are pairs of poles, unless the commutator is cross-connected, when all the positive brushes can be replaced by one brush, and similarly all the negative brushes by another brush. Using the same symbols as before, and N still standing for the number of lines entering the armature from or leaving the armature towards any one pole piece, the E.M.F. will be _ E = - g- as before but the armature is now in zp parallels instead of 2 parallels, where/ is the number of pairs of poles, and can consequently carry / times as much current if the wire be the same size, or where Q is the current from the machine with / pairs of poles, and C the current from a bipolar machine with the same value for N, etc. On the other hand, with the series winding, which is to be preferred to the parallel for various reasons, the E,M.F is increased in proportion to the number of pairs of poles, whilst the current-carrying capacity of the armature is the same as before, being only in 2 parallels whatever the number of poles. Thus the E.M.F. will be volts Armature Winding. 63 the current being simply the same as in a bipolar machine having the same size of wire. The effect on the hysteresis loss in the core is the same in both cases and may be found by taking the product of the volume of the armature into h, n, and^ and dividing by io 7 hVnp or H = c- watts 10* for both series or parallel windings. Virtually the effect of the multiplicity of poles is the same as having p equal dynamos connected on the one hand in parallel and on the other in series. Example XLI. A six-pole dynamo has an armature core composed of 600 discs of iron 0*25 mm. thick, and having a radial depth of 7 cms. It is wound with wire which will carry 50 amperes, and has 540 wires all round. If the density of magnetic flux through the iron is 15,000 per sq. cm., and the speed is 6 revolutions per second, what will be the E.M.F. and current if connected up in the parallel method ? Solution. The total value of lines is N = 2 X 600 X 0-25 x o'i X 7 X 15,000 = 3,150,000 lines and the whole E.M.F. will be ' E = 3,150,000x^^540 = io2yolts As each wire will carry 50 amperes, the current, as an ordinary bipolar machine, would be 100 amperes; but there are now 2/ = 6 parallels in the armature, and thus the current is C = 6 x 50 = 300 amperes Example XLIL What will be the hysteresis loss in the above armature if the outside diameter of the core is 70 cms. ? Solution. Wnp H ~ 64 Examples in Electrical Engineering. and h from the table on p. 216, is 9300. The volume of the core is V = 600 x 0-025 x 7 X - 7 - x 63 = 20,790 cc. and thus CHAPTER VI. THE MAGNETIC CIRCUIT. THIS can best be illustrated and explained by a comparison with the electric circuit. The quantities corresponding to conductivity and resistivity, E.M.F., and current, are called, in the magnetic circuit, permeability and reluctivity, magneto- motive force (M.M.F.), and magnetic flux respectively. Per- meability has this difference from conductivity, that it has no fixed value in the so-called magnetic materials (like the corre- sponding value of electric conductivity, which has a fixed value for a given temperature quite irrespective of the magnitude of the current or electric flux), but varies with the variation of the magnetic flux even when the temperature is kept con- stant. In the non-magnetic materials, however, the value of the permeability is independent of the value of the magnetic flux, even when this is confined to the same path, and so varies in density. The use of the idea of current-density is not so usual as to have caused the adoption of a symbol for its indication ; but for purposes of comparison we shall make use of D, as standing for the density in amperes per sq. cm. The corre- sponding symbol very frequently used in magnetic measure is /3, and stands for the number of c.g.s. lines of magnetic flux per sq. cm. of cross-section. It is very usual to confine ft to the magnetic density in iron or the other magnetic metals, but for sake of uniformity we shall not so restrict its use, but let it stand for the density of magnetic lines in any material. It has been shown before f 66 Examples in Electrical Engineering. that we may write the relationship between the three quantities , i Examples in Electrical Engineering. Thus in a composite magnetic circuit tfo/Xo when the same total flux N circulates round the whole circuit. The following comparison of the Electric and Magnetic Circuits will serve to sum up the previous considerations : ELECTRIC CIRCUIT. C = Current or whole electric flux in amperes. D = = amperes per sq. cm. of section, or current density. m = conductivity in mhos of the material of the circuit ; it is the actual conductance in mhos of the unit cube. It is constant for a constant temperature irrespective of D. u = = the resistivity of the ma- m terial or actual resistance in ohms of the unit cube. It is constant subject to the constancy of m. R = = = resistance in ohms a am of a circuit or part of a circuit of length / and area a made of material having resistivity p, MAGNETIC CIRCUIT. N = Magnetic flux in c.g.s. lines total. N ft = magnetic density or c.g.s. a lines per sq. cm. ju = permeability of the material of the circuit, and is the actual permeance of the unit cube. It is constant only for air and the non-magnetic materials having a value of unity (l). It is variable for the magnetic materials, and dependent upon the value of . ; = - = the reluctivity of the M material or actual reluct- ance of the unit cube being constant, and of value unity (i) for air and the non - magnetic materials ; but varies with /* in the magnetic materials, ac- cording to the value of j8. 3& = = = reluctance of a cir- a ap cuit or part of a circuit of length / and area a made of material having a per- meability ft, The Magnetic Circuit. 73 ELECTRIC CIRCUIT. i/p = the D.P. in volts required to send a current of C ampei - es through a circuit or part of a circuit. can readily be calculated from any of the above re- lations. E = the whole E.M.F. in the cir- cuit, being the algebraic sum of all the P.D.s round the circuit, or the sum of all such quantities as Though these P.D.s may be required distributed round the circuit E may be gene- rated at one part, viz. the armature of a dynamo, for example. MAGNETIC CIRCUIT. = The magnetic P.D. required to send a magnetic flux through a circuit or part of a circuit. It is equal N/ to o-SNE = 0-8. ap (o can only be calculated for the non-magnetic ma- terials where /j. = I . For the magnetic materials refer- ence has to be made to the tables for the value of /*, etc. By reference to the /() column of the tables, the value of o> in ampere- turns per cm. length can readily be found, and cal- culation thereby much sim- plified. = the whole M.M.F., being the algebraic sum of all the magnetic P.D.'s round the circuit, or being equal to Though these items may be required distributed round the whole circuit, the whole of fft may be produced at on e part, viz. the magnet coils of a dynamo, for example. In the electric circuit it is quite easy to confine the electric flux to a very definite path, because some materials (air, for example) are practically of infinite resistivity. In the magnetic circuit, however, air and the other so-called non-magnetic materials have a permeability which is, at the very best, only gfoVo as great as that of iron, the very best magnetic con- ductor we have. There is thus no possibility of magnetically insulating any circuit to anything like the perfect extent attain- able in the electric circuit ; and we have consequently to allow for some leakage of lines. The permeability of iron varies 74 Examples in Electrical Engineering. from about 3000 to 30 or so at such values of ft as are con- venient for general work, and thus the leakage of lines may often be as great even as if we had no better insulator than manganin, or such like, for the protection of the copper con- ducting path in the electric circuit. This leakage of lines is particularly the case in dynamos and such apparatus where the magnetic circuit is composite in character, and when the mag- netic P.D. between point and point may be very great. Thus, if the permeance of the armature and air gaps in series be represented by u, and that of the surrounding air space be called w, these two permeances are in parallel, and are subjected to nearly the same magnetic P.D. Consequently, if N be the total magnetic lines through the armature core and air gaps, the magnetic P.D. between the polar faces will be , and this u magnetic P.D. will produce a flux of - lines through the surrounding air space. Now, these two fluxes have to flow in the magnet limbs, which will consequently carry, not N lines, but N 4-N^ = N In their paper on dynamos in 1886 the Drs. Hopkinson chose to represent this fact in the very simple way of stating the lines N m in the magnet limbs as being v times the lines in the armature core ; the value of v depending upon the relation- ship of w to u. Thus, when w = u, v = 2, a value met with in such multipolar machines as the old pattern of Ferranti alternator. The value of w is, however, generally much less than u in a well-proportioned dynamo, some measured values of v being given in the following table : Type of machine. Value of v. Edison, Hopkinson, single magnet, drum '32 Siemens ,, ,, ,, ... ... '30 Phoenix over type ,, ,, gramme "32 Phoenix double magnet ,, '40 Manchester,, -49 Victoria Mordey, 4 pole, ring or gramme '40 Ferranti, 16 pole, disc ... 2 - oo Manchester with polar supports 1*5 to 1*60 The Magnetic Circuit. 75 Professor Forbes, many years ago, showed how a sufficiently exact approximation to the value of v for practical purposes could be calculated from the drawing of the machine. Thus, the same magnetic P.D. which is required between the two polar faces to produce N lines in the armature and air gaps, is also producing a magnetic flux through the very imperfectly magnetically insulating air which surrounds the dynamo, and which additional magnetic flux has to be ac- commodated in the magnet limbs, thereby causing the value of (3 in them to be greater, and making the value of the magnetizing force for that part of the circuit greater than would otherwise be the case. We have, then, to consider the magnetic circuit of the dynamo as composed of a number of items, each of which may be different, either on account of material or the fact that a greater flux of lines has to be accommodated. This composite magnetic circuit consists of (i) The armature, which is almost invariably made of a good quality of wrought iron. The sectional area is easily found, as has been indicated previously. The average length of the path of magnetic lines can be found with sufficient exactness by taking the mean semi-circumference in the case of a bipolar machine, whether drum or gramme ; or for a multipolar machine it is mean circumference where / is the number of pairs of poles. (2) The iron to iron space, called the air gaps, though mostly occupied by the winding, etc., on the armature. The sectional area of this part is equivalent to the area of the curved polar surface, increased by a fringe of 0*8 x the iron to iron space. This increase is to take into consideration the spreading of the lines round the edges, and was first pointed out by the Drs. Hopkinson. The length of this space is twice the iron to iron space. (3) The magnet limbs, which may be wrought iron, cast iron, or cast steel. Their sectional area is easily found from the 76 Examples in Electrical Engineering. drawing, and may vary in different parts, as, for instance, when the pole-pieces are of cast iron. The length is the same as that of a line drawn slightly nearer the centre of the machine than the centre of the limb, and bounded by the centre of the polar face on the one hand, and by the yoke on the other, taken twice over. (4) The yoke, which is generally counted separately, either on account of difference of material or difference in sectional area. The area of section is easily found, whilst the length is that of a curved line joining the extremities of the lines representing the path through the magnet limbs. Considerable experience and discretion is required to accu- rately fix these lengths. Using the following symbols to represent these various measurements, viz. 4, A a , and /* a for the armature, through which a total flux of N c.g.s. lines is passing, and in which, consequently, there N is a magnetic density of /3 a = -r- ** / g , A^ for the iron to iron space or air gaps. In these the value of /A is i, and consequently need not appear; and the magnetic flux is also N, as through the armature. The N density of lines in the gap is thus p ff = A ff l m , A m , p m for the magnet limbs, through which a greater number of lines than N, viz. j/N, will flow, and conse- Nv quently the magnetic density is /? ?w = -r /, Ay, and ^ for the yoke, in which again the total flux is Nv, Nv and the density p y = A - A and the ampere-turns required per cm. length will be 0*8 x 14,127 = 11,301. But as the gap is only 0*1 cm. long the additional ampere-turns will be only 1130-1 ampere-turns for air space. Thus the difference will be an addition of 1130-1 2-28 = 1127*82 ampere-turns* * Strictly speaking, owing to leakage lines caused by the increased magnetic P.D. between the ends of the gap and points near to the gap, there will be leakage of lines outside the area of gap taken, and these lines will have to be accommodated in the iron part of the ring, thus making the ampere-turns for the iron greater than before. The Magnetic Circuit. 79 Example XLYI. What will be the ampere-turns required to magnetize a dynamo which has a ring armature 25 cms. diameter, with a i5-cm. hole, and built up of 1000 discs of iron 0-25 mm. thick, if the total magnetic flux through it is 4 million lines? The diameter of the bore is 27 cms., its length is 27 cms., and the angle of embrace is 135 degrees. The length of the two magnet legs together is 124 cms., with a sectional area of 415 sq. cms., that of the yoke being 54 cms. and 729 sq. cms. respectively. Both magnet legs and yoke are wrought iron, and the leakage coefficient v 1-4. Solution. Using symbols previously mentioned, we have A rt = (25 i5)(iooo X 0*25 X o'i) = 250 sq. cms. and also 7T 25 + 15 4 = - X -- = IOTT - 31 5 cms. say thus 4 X IO H fc-isS- = l6 ' 000 g whence the ampere-turns required for the armature will be " = *J(P) = 4/(i6,ooo) = 3i'5 X 39-84 - 1254-9 say 1255 ampere-turns for armature. For the air gap we have l g = 27 25 = 2 cms. and as the angle of embrace is 135, the curved length of the pole-piece is ,. X TT x 27 = 31 '8 cms. The length parallel with the shaft is 27 cms., and thus the area of the polar surface is 31-8 X 27 = 858-6 sq. cms. to which must be added the fringe all round of o'8 times the iron to iron space, or in this case of 0*8 cm. all round, or 0*8(2 x 27 + 2 x 31*8) = 94*08 sq. cms. 80 Examples in Electrical Engineering. making A <, = 858-6 -j- 94 = 9527, say, 953 sq. cms. o 4 X io 6 p a - = 4200 say 953 and thus whence the ampere-turns for gap = (Og - o'8 x 2 x 4200 = 6720 Again, for the magnet legs we have the total number of line Nv = i '4 X 4 X io 6 = 5-6 x io fi and as A w = 415, we have 5-6 x io 6 corresponding to which /(/?) =10-9 nearly enough, and the ampere-turns required will be m m = mf(P) ==i24Xio'9 = i352 ampfere-turns say Lastly, for the yoke there will be the same total flux, 5 '6 x io 6 , whence for which m = 2-12 and therefore ^ = 54x2-12 = 115 ampere-turns say. Thus the whole magnetizing force will be in ampere-turns Armature ... ... ... 1255 Air gaps ... ... ... 6720 Magnet legs ... ... ... 1352 Yoke ... ... ... 115 9442 ampere-turns total Example XLYII. The length of magnetic circuit of a transformer iron core is 50 cms. Its sectional area is 300 sq. cms., and the maximum value of the current allowed for magnetization is 0*15 ampere. How many turns of wire must The Magnetic Circuit. 81 the primary coil have if the total magnetic flux is 1,200,000 lines ? Solution. 12 x io 5 Density, j3 = = 4000 for which /(/?) is i '24, and therefore the ampere-turns re- quired lf( = 5 X 1-24 - 62 whence, the current being 0-15, the total turns must be 62 w = = 413 say 0-15 CHAPTER VII. MAGNET COIL WINDINGS. WE have seen in the previous chapter how to ascertain the magnetizing force required in ampere-turns, and will now pro- ceed to show how the size of wire and number of turns or convolutions may be found. There are two main cases (1) That in which the current to be used is a fixture. (2) That in which the P.D. available is fixed. The first is simple, and can be easily disposed of. Let ffi = the ampere-turns to be provided, and C the fixed current, then the number of turns is, of course The size of wire is then fixed, partly by the magnitude of the current to be carried, and partly by the fall of potential allow- able. Generally this last is so limited that the expenditure of energy is so small as to make the heating negligible ; but in any case the dimensions of the bobbin, or former, can be made such as to satisfy Mr. Esson's rule, given on p. 56. The coils which mostly come under this, the " series," class, are those of ammeters, cut-outs, series-wound dynamos, series coils on compound dynamos, and independent coils generally. In case (2), which may be called the " shunt " class, we have different conditions to take. A certain E.M.F., say E volts, is to set up such a current that this current, multiplied by the turns of wire, shall be equal to JW, the ampere-turns required ; or Cw = m Magnet Coil Windings. 83 Since the whole resistance R will be R = wr where r is the resistance of the mean turn, we have and wr r or r= E whence a simple rule : Shunt Coil. The resistance of the mean turn is equal to the P.D. at the terminals divided by the ampere-turns required ; or is the ratio of the E.M.F. to M.M.F. It is easy in most cases to fix or find the length of the mean turn, say /, whence the sectional area of the wire will be when p is the resistivity at the particular temperature to which the coil is to be raised. The diameter of the wire, if circular in section, is then Or, putting in the previously found value for the resistance of the mean turn, we have when, of course, / and p must be in cms. or inches, according as d is wanted in cms. or inches. Also p should have the value corresponding to the temperature to which the coil is allowed to rise. To find the number of turns, w, we must again have recourse 84 Examples in Electrical Engineering. to the formula of Mr. Esson, and make the total expenditure of energy E E TT or - R wr such as to make no greater rise of temperature than may have been previously agreed upon. A little investigation will at once show that the number of turns can be varied considerably without altering the value of ffi when once the right size of wire has been found, provided the length of the mean turn be not thereby altered. Thus with only one turn the energy waste will be E 2 r and may be hopelessly large ; but with a suitable number of turns, /, this waste will be reduced to E 2 wr or to times its first value. Since, however, the ampere- w turns are Cw and C = wr we have in both cases that E E $&, = - x w = rw r whatever value w may have. The number of turns it is possible to get on a coil is found when the size of the coil is known, and the depth to which it is possible to wind it. Thus let d^ stand for the diameter of the covered wire (easily found for double cotton covering by adding 0-02" to the diameter of the bare wire in inches). Let L stand for the length of the bobbin or former which may be Magnet Coil Windings. 85 occupied by wire, and D- the depth to which the winding may be laid. Then there will be D each of 1J i -r layers #1 - turns d and LD W = ^ all quantities being, of course, expressed in the same units. The value of ffll for dynamos, etc., is calculated as shown in a previous chapter ; whilst fflt for ammeters, voltmeters, and other small instruments, is usually found in the first instance by experiment. Many ammeters and voltmeters of the magnetic type are alike in size of parts whether ammeter or voltmeter, being only different in the winding. Now, it is evident that, with all the different ranges of reading required, it is not pos- sible to provide exactly any particular number of ampere-turns with any particular current or P.D. Thus all ammeters and voltmeters of this class have some power of adjustment that enables the nearest value of $$l to be made use of. In ammeters it is not always possible to make the product cw = the value of $t desired, and owing to the trade range in sizes of wires for voltmeters, it is not always possible to find that size in stock that will make the right value of resistance of mean turn. Thus both ammeters and voltmeters are calcu- lated for on the assumption that the value of J& is variable within certain limits. In some instruments of the permanent magnet class ^l ranges round 300 ampere-turns, whilst in some of the electro-magnet type it may range round 600 ampere- turns for ammeters, and about 400 ampere-turns for volt- meters. CHAPTER VIII. ARMATURE REACTIONS. So far we have only considered the simplest case of dynamo- winding, and will now proceed to discuss the calculations for different types of machines. In order to appreciate what it is we have to allow for in a dynamo at full load, we shall have to consider the effect of the current in the armature and the magnetic circuit. In Fig. 15 is shown a section through a FIG. 15. dynamo, with the wire on the field magnets and armature indicated by lines. The direction in which the current flows in the field-magnet winding is indicated by arrows. By applying Dr. Fleming's rule for the direction of the E.M.F. in any wire on the armature, we are able to mark these wires to show the direction in which the current would flow if the brushes were con- nected. Thus, on connecting the brushes, a number of things will happen Armature Reactions. 87 (1) Due to the resistance of the armature winding, we shall get a P.D. at the brushes which will be less than the whole generated E.M.F. by an amount which may be considerable at full load, though hardly noticeable at light load. Thus, calling the whole E.M.F. = E, the current in the armature C a , and its resistance r a , we have that the value of the P.D. at the brushes e= E - C a r lt (2) In addition to this loss of volts in the armature resist- ance, we have to allow for the effect of the current in the armature winding as a M.M.F. acting in the magnetic circuit. Thus, with the brushes on the diameter of commutation, corre- sponding with no load, as shown in Fig. 15, we see that the armature winding and current produce a M.M.F., causing a magnetic flux at right angles to the flux set up by the magnets, and acting in a direction up the page, whilst the flux due to the magnet winding acts from right to left, the resultant flux through the armature will, of course, be a flux having a direction somewhere in between these two directions, or inclined to the horizontal line in the figure by an amount which will depend upon the relative magnitudes of the two M.M.F.'s. Thus, if the armature current and number of windings be very large, we shall have the resultant magnetic flux taking a direction chiefly down the length of the page ; the dynamo would then be a very bad one, and spark considerably. On the other hand, if the M.M.F. due to the magnet-field winding greatly predominates, we should have the resultant flux nearly horizontal as before taking any current from the armature. This state of affairs is that to be desired in all good dynamos and motors where sparkless collection among other advantages is wanted. In the best of machines, however, there is some distortion, and it is this we have next to consider. When a current is flowing in the armature the direction of the resultant flux will be as shown in Fig. 16, where a dotted line carried round the magnetic circuit indicates the path of the lines of force. The brushes have been adjusted so as to bear upon 88 Examples in Electrical Engineering. a diameter at right angles to the direction of flux through the core ; that is upon the new diameter of commutation. Two things will now be noticed FIG. 16. (1) The cross magnetic effect of the armature will now be somewhat decreased, being now only that due to the wires on the armature contained by the line ad or be; and (2) The remainder of the winding on the armature, viz. that contained by the line ab or de, is now a M.M.F. causing a magnetic flux in a direction parallel to the flux caused by the magnet winding, but in an opposite sense, or tending to directly decrease its magnitude. Since the value of the whole generated E.M.F. depends upon the value of the magnetic flux through the armature, this E.M.F. will no longer be as large as before a current was taken from the machine. We must thus allow that the generated E.M.F. is now less than E by an amount, say E,, and can write the terminal P.D. e=E- C a r a - E x In a great many cases we are required to produce a machine which will give a constant value of terminal P.D. e l at all loads, and in order to do this it will be necessary to make allowance for the reductions indicated in the above equation. In some cases, however, it is desirable to make a correction also for the loss along leads, at the distant end of which lamps are to be kept at a constant P.D., and we may then add to the above losses the loss along the leads, or Cr ( , making now e = E - C a r a - Cr, - E, Correction for these losses is commonly made by adding to the M.M.F. of the magnet-winding by adding a series coil to the shunt, thus compounding the machine. Now, such series winding has the effect, if connected as "long shunt," of in- creasing the effective resistance of the armature, and our equation will now stand e = E - C a r a - C a r m - Cr, - E, where r m is the series-coil resistance. When, however, the connections are made "short shunt," the series coil acts as an increase in the resistance of the leads, and the resultant reduction of terminal P.D. is slightly different, viz. e = E - C a r a - Cr m - Cr t - E, There is, however, a more important difference between the two cases : whereas the P.D. at the terminals of the shunt coil is practically constant in the case of "long shunt," it is gradually increased from no load to full load with the " short shunt" connections, by an amount equal to the loss in the series coil, and thus the amount of series winding necessary in the case of " short shunt " is not quite so great as in " long shunt," when other things are equal. We have, then, that the generated E.M.F. E = e+ C a r a + C m r m + Cr, + E x putting C m for the current in the series coil, which may be either C or C a . Of these items we can have generally the most accurate knowledge, save in the case of Ej. The exact effect of the band of winding contained by the lines ab or dc in Fig. 1 6, called the "demagnetizing band," is somewhat difficult to predetermine. Its magnitude depends upon the position of the brushes, and this depends upon the exact dis- tortion produced by the cross magnetizing band. This distor- tion can be found by plotting out on the drawing of the 9O Examples in Electrical Engineering. machine the path of the lines for every value of the current in the armature, and thus finding the diameter of commutation for every value of the current. Considerable experience and care are required for this, and it will be out of place to further consider it here. An example of the ascertainment of the value of the neces- sary series winding, when all the above losses have been found, will serve to sum up the matter. A dynamo is required to give a terminal P.D. of e volts at all values of current, from zero to C. amperes. In the first place, at very small output the whole generated E.M.F. need not be appreciably larger than e, and consequently a calcula- tion must be made of the M.M.F. necessary to produce such a value of N that the generated E.M.F. = ^, or N = el0 * nw which will require a M.M.F. of $1 ampere-turns. The winding to produce this must be the shunt coil. In the second place, we have to allow that the losses at full load make the necessary generated E.M.F. to be E volts, and thus necessitate a greater value of magnetic flux, or N, = - TIW to produce which the M.M.F. will be larger, and may be called #fc. Thus we have the additional M.M.F. required at full load is /&i J& ampere-turns which must be supplied by the series coil in addition to that required to balance the demagnetization. The series M.M.F. is a product of the series turns into a current, which may be that in the main circuit (short shunt), or may be that in the armature (long shunt). In the latter event it will be noticed that when compounding has been accurately carried out for all the usual items, the P.D. at the terminals of the shunt coil is a constant at all loads (except when compounding for loss in Armature Reactions. 91 the leads). But in the short shunt combination the P.D. at the shunt terminals will only be e volts at very small loads, and will increase to e 4- Cr m volts at full load, thus increasing the M.M.F. and making consequently less series winding neces- sary. This alteration of M.M.F., due to the shunt coil, is of very great importance in all cases where there is imperfect compounding as any defect. Both over- or under-compound- ing is much exaggerated. A little consideration will show that it is hardly possible to produce a perfectly constant P.D. at the terminals by compound winding. For if we calculate the additional ampere-turns to be supplied by the series coil in, say, two portions, viz. one for half-load, and the other as a further addition for the other half- load to full load, we shall find that the latter exceeds the former. Thus, let the extra M.M.F. required for half-load be J& ; this is due to an increase in the total flux necessary to increase the generated E.M.F. by an amount equal to the sum of a number of items, viz. C a r a + C m r m + Cr t which will be all nearly enough half the amounts for full load, whilst the other item, E lf will be less than half, since it will be due to a demagnetizing band of less turns than at full load, and carrying only half the current. The items for the further increase up to full load will all be higher than for the first half-load, since the resistances will all be slightly higher, and also the demagnetizing band will have increased in number of turns as well as having been doubled in current ; or ^& 2 , the increase of magneto-motive force for the second half of full load, will be considerably greater than J&i. But both are produced by the increasing current flowing in coils of fixed number of turns, which will cause the additional ampere-turns to be simply proportional to the current. And also the relationship between the magnetic flux produced and the ampere-turns producing it is such as to make the additional effect of J& 2 actually less than ^ instead of greater. It thus follows that if a dynamo is to give, say, e volts P.D. at its terminals at a certain speed at no lead and also at full 92 Examples in Electrical Engineering. load, it must give a P.D. which is greater than e at the inter- mediate loads. Consequently it is customary to fix upon two points, say quarter-load and three-quarter-load, and so wind as to produce the same P.D. at the terminals at these points, letting the P.D. be slightly less than this at no load and full load, but slightly higher at half-load. In the above considerations it is sometimes necessary to ask ourselves what effect in a composite magnetic circuit a given increase or decrease of M.M.F. will have. It is very easy to find what increase or decrease of M.M.F. a given increase or decrease of magnetic flux will require ; but the converse is a much more difficult problem, since the relationship between all the items of the circuit is at once altered as soon as the flux is altered. ARMATURE REACTIONS IN A MOTOR. FIG. 17. If in the above diagram DD represent brushes bearing on the commutator of a machine used as a dynamo, we shall have the current flowing as shown by the arrow when the direction of rotation is clock- wise, as indicated by the slope of the brushes. In order to cause rotation in the clock-wise direction, a pull will have to be supplied to the belt to overcome the drag between the current in the wires under the pole-pieces and the lines of force passing into and out of the armature. When, Armature Reactions. 93 however, the diagram is taken to represent a motor supplied with current to flow through the circuit in the direction as shown in the figure, this drag will now be the moving force, and the armature will revolve in a counter clock-wise direction, to admit of which the brushes MM will have to be used sloping in the opposite sense to the dynamo brushes, but bearing upon the same point on the commutator, since the nature of the magnetic forces acting in the magnetic circuit is still unchanged. But as the armature is now revolving in the opposite sense to what it was before the case, the E.M.F. generated will also be opposite to that previously generated, and will, consequently, be against the E.M.F. of the source supplying the motor with power, which will thus have to exceed that generated by the amount required for the resist- ance of the armature coils. It thus follows that, as in a dynamo so in a motor, the current in the armature produces a demagnetizing effect upon the mag- netic circuit. But whereas this demagnetizing effect is dis- advantageous in a dynamo, it actually tends to aid constancy of speed in a motor by reducing the generated E.M.F. for a given speed somewhat in the proportion required by a constant E.M.F. of supply. The above diagram is due to Mr. Swinburne. CHAPTER IX. EFFICIENCY. THE ratio of the useful expenditure of energy in a system to the whole energy is called the " efficiency ; " but we have, in the many and various cases to be considered, to sometimes separate the parts of a system and consider them inde- pendently, as will be seen in the examples taken. As a simple case, we may take a single cell used to light a single lamp. Let the total E.M.F. of the cell be E volts, its internal resistance be r ohms, whilst that of the lamp is R ohms. Then the current flowing through the circuit will be and the rate of doing work in the lamp is E 2 C2R = ( RT7)* R whilst the whole activity is C 2 (R + r) -= ^Trf X (R + r) the efficiency is then the ratio of that part of the energy con- cerned in lighting the lamp to the whole activity, or is (R + _ (R + r) a Or we may split our circuit up into items, and consider the ratio of the useful energy in the lamps to that in the lamps Efficiency. 95 plus that in the leads, thus getting the efficiency of distribution. Or we may consider the ratio of the energy given out by the cell to the whole energy produced by it, in which case we have the electrical efficiency of the cell or other machine. Or, again, we may include the energy required to run the generator a dynamo, for example against the various sources of friction, and thus state the efficiency as the ratio of the energy delivered from the terminals to the energy put in at the pulley, thus getting the mechanical efficiency, or commercial efficiency. In the case of motors, we can take the ratio of the power at the pulley to the power supplied at the terminals, which is again the commercial efficiency. On the other hand, we may regard the motor as a machine for the conversion of electrical energy into mechanical energy, whether useful or not, and so obtain the electrical efficiency. With secondary cells we may, on the one hand, regard the cell as a store of quantity of electricity, and state the quantity efficiency as the ratio of the coulombs got out to the coulombs put in ; or we may consider the cell as a store of energy, and state the energy efficiency as the ratio of the watt-hours got out to the watt-hours put in. Also, as a dynamo and engine will be required for charging, we may have a combined efficiency of the whole apparatus, called the plant efficiency, and being the ratio of the watts got out to the power required to run the dynamo stated in watts. I. Efficiency of Distribution and Transmission. (a) In a simple parallel system, with all the lamps at the ends of the leads, if r t = resistance of the leads, C the current through the lamps, and e the P.D. at the lamps, we have energy used in the lamps = eQ, ,, wasted in leads = CV ? and the efficiency will be eC e cC + CV, - e 4- Cr t which is the same thing as the ratio of the useful P.D. to the total P.D. g6 Examples in Electrical Engineering. (b} Simple parallel system, with the lamps uniformly dis- tributed along the length of the leads. Here the loss of volts in the leads will be C C 2 making the energy waste in the leads = ;- z . But the energy expended in the lamps will be C x ( C by the amount taken by the shunt coil, viz. , the waste in 'i the armature being consequently e 1 Whilst the shunt-coil waste is , and the r o electrical efficiency = And Wj having the same meaning as previously, the- eC mechanical efficiency = W, (d) Compound dynamo, short shunt. Here the output is eC, as before. The current in r m is C amperes, but that in r (l is C + shunt current, or whence the various losses are C 2 r w in the series coil, in the shunt coil, and in the armature, making the electrical efficiency eC the mechanical efficiency being eC c/g-/^ H c Efficiency. 99 (e) Compound dynamo, long shunt. The current in the shunt is , and flows through both r m and ;, in addition to the main-circuit current C ; whence the f-* electrical efficiency = and the mechanical efficiency = 3 + (r.+rj+w, ' S ' (f) A series motor. The electrical efficiency is the ratio of that part of the energy supplied to the machine, which is actually converted into mechanical work, whether useful or waste, to the whole energy supplied. Calling E the P.D. of supply, with C the current, and r a and r m having the values above, we have that the back E.M.F. of rotation is c = E - C(r B + r m ) and the whole rate of doing work mechanically is G ; whence the electrical efficiency = ~ = ~ The mechanical efficiency is EC - Cfo + rj - W, EC or EC Since of the total rate of doing work mechanically the part W 1 being internal friction, etc,, is wasted ; M is direct mechanical friction, H is molecular friction, but F is a load on the arma- ture of exactly the same nature as a closed secondary on a dynamotor armature. ioo Examples in Electrical Engineering. (g) Shunt motor. Calling E the P.D. at which a current of C amperes is supplied, we have that of C amperes, the p 1 shunt coil takes -, leaving the E current in the armature = C r f due to which the loss in r a is f C - J / volts leaving the B.E.M.F = e = E-( C- E }r a \ rj whence the electrical efficiency = EC and the mechanical efficiency EC (ti) Compound motor, short shunt. Calling EC the supply as before, we have the shunt current is r g Whence the armature current is r. and the B.E.M.F. _ e = & r s and therefore the electrical efficiency is EC whilst the mechanical efficiency EC Efficiency. IOI (k) Compound motor, long shunt. Here the shunt current is and the B.E.M.F. e=E-(c--J r m - (c - - J r a and the electrical efficiency EC whilst the mechanical efficiency EC (/) Secondary battery. Calling the charging current C amperes and the time of charging / hours, the corresponding quantities for discharging being c and f lt we have the ampere hours put in = ct and output = ^A, whence the /, quantity efficiency - and may be as much as 100 per cent. Considering, however, the fact that the B.E.M.F. of the cell is higher on charging than the available E.M.F for discharging, we have the energy put in is Cfc and that got out is C^ when e and e l are respectively the E.M.F. to charge against a certain B.E.M.F. and resistance, and the P.D. available for use at the terminals on discharging. The energy efficiency is = p V (m) Plant efficiency. This must include all sources of loss and give the ratio of The useful work done in a given time The energy supplied to the prime mover for the same time CHAPTER X. VOLTMETERS AND THEIR RESISTANCE COILS. IN the following examples notice will be taken of the effect of temperature on the readings of voltmeters of the magnetic class. As was mentioned in the chapter dealing with the winding of coils, there is no difference between an ammeter and a volt- meter in point of shape, etc. There is, however, this im- portant difference between them that, whilst the former has its dial graduated in terms of the current passing through its coil, the dial of the latter is marked in terms of the product of the current through the coil into the resistance of that coil, such resistance being taken at the temperature at which the instrument was calibrated. Thus the readings of an ammeter are independent of the value of its resistance, but those of a voltmeter depend upon the resistance of its circuit, and will only be correct when that resistance has the value it had when the dial was marked. It is, therefore, the object of the instru- ment-maker to so contrive matters that the resistance of the voltmeter circuit shall vary as little as possible. Alteration in temperature may be produced in two ways, viz. by external means such as atmospheric variation, or variation due to locality ; or by internal means such as the expenditure of energy in the coil due to the current through it. We will call the first of these the " temperature " error, and the second the "heating" error. The employment of a material such as manganin or constantan, will make both these errors quite small, or even negligible as far as mere temperature variation of resistance (T.V.R.) is concerned, though the actual heating may be very large, and even dangerous. For the energy waste Voltmeters and their Resistance Coils. 103 in a coil of a given size producing a given value of ampere- turns will be directly proportional to the resistivity of the material of the wire. When the resistivity is very high, it is not easy to make an instrument coil of reasonable dimensions without excessive heating and waste of energy, and owing to the want of a material of low specific resistance and small or negative T.V.R., we are obliged to make a compromise between the two extremes, and wind that part of the instrument which is to produce the magnetic effect with wire of small resistivity in order to get large ampere-turns with small heating ; and the rest with wire of small T.V.R., and having with advantage a high resistivity. The two things to be aimed at are, firstly, to get the requisite M.M.F. with small energy waste, and then to get the necessary freedom from variation of resistance by addition to the instrument resistance of a coil having a constant, or fairly constant, value of resistance. This is effected by making the working coil of the instrument of copper wire to get the maximum M.M.F. with the least energy waste, and by adding to the instrument so wound an extra coil or resistance of, say platinoid or manganin, materials having very small coefficients of T.V.R., and thus causing the effective total resistance to remain nearer a constant then if copper alone had been employed. The effect of so adding to the resistance of the working coil is twofold ; firstly, the current through the instrument due to a certain P.D. will be decreased, and consequently the deflection for a given number of volts will be less than before, or the total reading of the instrument will be increased ; and, secondly, there is the ten- dency to greater constancy in total resistance. The increase of total reading power is obviously proportional to the increase of resistance ; thus if the extra coil have a resistance equal to that of the instrument's working coil, the instrument will read twice as much ; or, more generally, if the extra coil be n times the resistance of the working coil, the total reading power will be (n -h i) times the original. The T.V.R. of the whole com- bination is, on the other hand, inversely as the number of times the total resistance is increased if the material of the extra coil has no T.V.R. If the T.V.R. of the extra coil be IO4 Examples in Electrical Engineering. important, having a value c say, whilst the T.V.R. of the working coil is b, and the extra coil has a resistance n times that of the working coil then the combined T.V.R. will be b+nc a = ~~ per cent, per i C. when b and c are stated in values per cent, per i C. The plus sign is mostly the case, but when carbon or manganin is employed for the extra resistance, then c is negative, and by adjusting n so that nc = I) we shall obtain a system of perfectly constant resistance. It thus follows that the value of an indication on a volt- meter in volts is found by multiplying the reading as marked on the dial by the ratio of the present resistance total to the total resistance when calibrated, or present resistance volts = deflection x -r-=- resistance when calibrated The deflection being of course marked by a number which is equal to the product of the resistance of the instrument when calibrated into the current which a given P.D. produces through the instrument. The value of the present resistance depends upon the temperature and the T.V.R.. values for which are given on A, p. 209. Thus let R T be the resistance at which the instrument was calibrated (temperature, say T C.), and let R, be the resistance at some other temperature / C., which may be either higher or lower than T C. Let a be the T.V.R. per cent, per i C. Then, nearly enough for practical purposes when the difference t r may be zero, positive when t is greater than T, or negative when t is less than r. Similarly, the value of any indication on the voltmeter will be always proportional to its resistance at the time of taking Voltmeters and their Resistance Coils. 105 a reading, and if we call the value per one division k volts, when the instrument is at the correct temperature we shall then have that the value of one division at any other tempera- ture will be TOO CHAPTER XL ELECTRIC TRACTION, RAILWAYS, ETC. THE application of the electric motor to tramways and other tractive work renders it necessary to ascertain the horse-power required for any particular case, and the data necessary for this purpose may now be discussed. In order to draw a car along a line, the mechanical pull required will depend upon the following things (1) The weight of the car. (2) The condition of the line. (3) The inclination of the line. By the condition of the line is meant its smoothness and freedom from dirt or other obstruction. In the best possible cases of railway work the pull required to keep a ton weight of car in motion on a level line may be less than 10 Ibs., but in some cases, even with grooved rails, the pull may be 15 to 20 Ibs. in good cases, and even 40 Ibs. or more in cases where dust or road grit has collected in the rails. Curves also cause a local increase of resistance to traction. In order to allow for the inclination of the line we have to take into consideration the effect of gravity, which effect may be against us in going up an incline, but which is for us in going down an incline. It is seldom that the inclination is so much as to render it necessary to distinguish between the length of the actual distance gone and the true horizontal distance, and consequently we may write the pull due to gravity as being 2240W Electric Traction, Railways, etc. 107 where n is the number of feet along the line for a rise or fall of i foot, this pull being 4- or according as the inclination is up or down. We shall thus have the pull required to move a car of W tons weight along a line of i in n inclination with a resistance to traction of x Ibs. per ton as being in pounds taking the -f sign when going up hill and the when coming down hill. Again, the rate of doing work is proportional to the speed of running, and we have the horse-power required will be the pull multiplied by the distance gone through in feet per minute and divided by 33,000. Thus, if S = the speed in miles per hour, we have the horse-power is H.P. = 88PS 33OOO and this is the useful mechanical rate of doing work, and to it must be added that required to run and excite the motor and that required to transmit the power of the motor to the axle of the car. In general the conditions are so disadvantageous to the high- speed motor, that in practice it is found that for every horse- power put into the motor-terminals we may get out from 60 to 80 per cent, as useful work to run the car ; the lower figure is frequently exceeded, but the higher can only be attained under the best conditions, and with large motors. Calling this factor , we have that the electrical power put into the motors must be or, in full 100 / 88SP y H.P. required = V 33 ) which is converted into watts by multiplying by 746. Now, this power is a product of volts and amperes, and it is simply io8 Examples in Electrical Engineering. a matter of convenience as to what the E.M.F. may be. The motor is in any case arranged with its gearing so that a given current through its armature means a given pull for traction, and the value of this pull per ampere can usually be stated or fixed. From this and the actual pull required we can readily find the current required to move the car, and can add that to overcome internal friction, etc. From the current and the speed of running we can next find the useful rate of doing work, and thence the B. E.M.F. of the motor. In the case of shunt motors this B. E.M.F. will be practically constant, being in any case proportional to the speed, and the cars will run at a nearly constant maximum speed both up and down hill, and on the level if supplied with a constant E.M.F. A knowledge of the electrical data will then enable us to ascer- tain the necessary increase of volts on account of internal losses, and to state the exact value of the P.D. of the mains, In some cases the tractive pull per ampere is not a constant, since with series motors the drag on the wires is increased by an increase of current both on account of that increase itself and also on account of the consequent increase of magnetic flux through the armature. When series motors are run off mains at a constant P.D., we shall have their speeds varying by an amount which is greater than in the case of shunt motors, having a practically constant excitation. Thus, at places on the line taking a heavy load, the extra current in the magnet coils will cause the speed to go down, since other- wise the B. E.M.F. of the motor would soon exceed that of the supply mains, were that possible. The tractive force being thus increased in a proportion more nearly as the square of the current, makes the series motor particularly suitable for overcoming steep gradients and other heavy loads with less heating and waste of energy than would be the case with a shunt motor, for the series motor will automatically reduce its speed up hill, and by increasing its field strength take less increase of current to produce the necessary tractive pull. When getting up speed the car has to store an amount of energy, equivalent in foot pounds to half its mass, multiplied by the square of its velocity in feet per second. The extra Electric Traction, Railways, etc. 109 pull required to do this will depend upon the time taken to attain the speed, and we can write the work to be stored in foot pounds, as W x 2240 / S_x 5280 \' 2 = K 2g \ 3600 / When g is taken as 32, we can simplify this to K = WS 2 X 75 '3 foot pounds where W = tons weight of car and S the speed to be got up in miles per hour. Now, if this store of energy is to be acquired in ;/z, minutes, we must remember that the car in gradually getting up the speed of S miles per hour will go through a distance of Sx 5280 since its average speed during the time will be - if the acceleration be uniform. And through this distance a pull over and above that required to overcome the tractive resist- ance will be required of K -j pounds or the extra pull during the time of m, minutes, to get up speed of S miles per hour will be WS 2 X 75-3 WS - ' D = x 1711 pounds 448^ m Or if speed is to be attained in / seconds, the extra pull is 1027 WS ~ -- pounds If also / be the pull per ampere, the current required will then be n . amperes 1 10 Examples in Electrical Engineering. to which must be added the extra current to get up speed, or r 102 7 WS 1 = -- lt> am P eres The value of/ the pull per ampere may be nearly a constant for all values of current in the case of a shunt motor ; but in the case of a series motor it has more nearly the value where n speed per minute and a and b are constants, de- pending upon the size of the machine. In questions of electric traction involving a knowledge of the value of the pull produced by a given current in the armature, or of finding the value of the current required to give a certain pull, we must be careful to distinguish between the shunt, compound, and series types of motors. N FIG. 18. Shunt Motor. When the E.M.F of .supply is constant, we may regard the magnetizing force, acting upon the magnetic Electric Traction , Raihvays, etc. in circuit as practically constant, subject to the fact that distance from the generating station will mean a lowering of the P.D. at the terminals of the shunt coil by the ohmic loss in the leads. The pull or torque is always proportional to the product of current and lines total ; and these last should be constant in a shunt motor with constant P.D. at the terminals were it not for the fact that armature reactions produce a demagnetizing effect, and the curve of relationship between N and C, the current through the machine will be of the nature of the curve shown in Fig. 18, and consequently the pull per ampere will be decreasing slightly as the current increases. Thus at times of very large current, as at starting, the actual pull may not be very large, and the maximum pull or torque may never be great. Compound Motor. Taking next the case of a motor com- pounded for ordinary work, that is for constant speed with varying load, we find that the effect of the compounding series coil is to still further increase the demagnetizing effect of the current, with the result that the curve of relationship between N and C will be still more drooping at high values of C than that shown in Fig. 18, with the effect of a still further reduced pull under similar conditions. But when the series-magnet winding is connected so as to magnetize the circuit, we may have this effect made, either to give a constant value of N for all values of C (in which case there will always be a constant pull per ampere, whatever value the current may have), or to actually produce an increase of lines with an increase of current, making the curve of N C rise for an increase of C, and thus giving a pull per ampere increasing with the current. It will not be necessary to inves- tigate this action in combination with the shunt coil, but to go on to consider the effect of a series coil alone. Series Motor, Here the value of N is constantly increasing with the current, and would have a curve of relationship similar to the /2H curve for a composite magnetic circuit, were it not for the fact of demagnetization causing a tendency to drop at high values of C. Taking the shape of NC, as shown in Fig. 19, we may also plot a curve, showing the relationship Xu\BR A Vy*X. OF \ ( UNIVERSITY J 112 Examples in Electrical Engineering. between C and the product CN, as shown in the upper curve. The pull exerted is directly proportional to the ordinate of the second curve, and can be found from the nature of the winding C FIG. 19. and gearing employed. Calling the total pull for tractive effect,/ = 0CN x io~ 6 , where a is a constant depending upon the type of machine, we have from a knowledge of the value of p in pounds that the product and a reference to the curve will give the value of C and N corresponding with this product. It is also clear that the pull will reach a maximum value, and the performance of the motor is thereby limited. CHAPTER XII. ALTERNATING CURRENT CIRCUIT. WHEN the current in a circuit changes periodically in direction, passing through a set of cyclic changes of value, it is termed an alternating current, and, owing to the disturbing influence of the medium which surrounds the circuit, the usual relationship known as Ohm's Law requires modification before it can completely represent the true state of affairs. The manner of change of value of the current from time to time is different in different types of machinery, but is very frequently of a nature simply expressed as a sine function of the time. Such a current can be represented in all possible values by the curve in Fig. 20, where ordinates above and below the line Of represent the instantaneous values of the current in amperes corresponding with any instant in time marked along Of. Such instantaneous values can be represented by the ex- pression Q = C m sin *?t where Q is the value at the instant in time /, C m the maximum value to which the current ever attains, and T is the time in seconds taken to go through one complete cycle, or change from zero rising to a positive maximum, declining to zero, reversing and rising to a negative maximum, again declining to zero. The curve can be drawn from a table of natural sines by dividing the horizontal line into 360 parts corresponding to the 360 degrees in a complete revolution, and marking points at a distance above or below the horizontal equal to C m times the natural sine of the angle. The circular measure of 360 i 1 1 4 Examples in Electrical Engineering. degrees is 2tr, and it thus follows that when t has such values as to make - x 2?r equivalent to convenient angles, the curve may be easily drawn in. When / = r the angle = 2?r = 360. FIG. 20. The time T is called the time of one complete cycle or period, and ranges with different makers from -^ second to T ~ second, though smaller values were used in past times. It has a very general value in English practice of either -~ or ~Q second, the value ^^ being American. It is thus customary to describe the circuit as one having, for example, 100 periods per second j or to say it has a frequency or periodicity of so and so per second. Calling the frequency p periods per second, we may put instead of the above expression C t - C sn where / stands for - Alternating Current Circuit. 115 Now, it will be quite evident that in some cases the rapidity with which the current changes in value will materially affect the conditions of working ; and it will be necessary to investi- gate as fully as possible the effect of any such disturbance for the different conditions possible. There are four ways in which an electric current can become manifest to us (1) Chemical. (2) Thermal. (3) Magnetic. (4) Electrostatic. And we will examine the effect produced for each of these considered separately. I. The Chemical Effect. With continuous or uni-directed currents the chemical change produced is always proportional to the strength of the current and the time during which it flows. It is also dependent upon the direction of flow, being of the nature of a solution where the current enters, and a deposition where the current leaves the circuit. Thus, if a current of one ampere be passed for the space of time of / seconds through a solution of copper sulphate, it will dissolve from the plate of copper at which it enters the solution, and deposit upon the plate at which it leaves the solution, an amount of copper which may be expressed as / x 0*00032959 gramme Thus, during the time from o to A in Fig. 20, we have the current flowing all in one direction; and if such current be passed through a solution of copper sulphate, it will cause a deposit of copper on one plate and a solution of copper from the other plate, which will be in amount equal to the product of the average strength of the current into the time, viz. - seconds, and into the numeric used above (the electro- chemical equivalent of copper). But if the circuit be not broken at the instant when the current becomes zero, the contrary effect will be produced, and there will now be a re-solution 'of copper from the plate which before received the deposit, and a deposition on the plate from Ii6 Examples in Electrical Engineering. which the copper was dissolved ; and this reverse effect will be of exactly the same magnitude as the previous effect if the current be maintained for the next or second half-period. It thus becomes apparent that an alternating current cannot produce any chemical effect ; for what is done during any half- period is exactly undone during the next. When, however, the current is " rectified " by a commutating device, timed so as to exactly reverse the connections at the instant the current changes in direction, we shall get a total chemical effect which will be proportional to the product of the time and the average value of the current, or is C,,/o'ooo32959 grammes of copper Now, it can be shown, by carefully measuring the area of the half-wave from o to A, that its average height is 0*6366 times C m the maximum height, and we thus have that the amount of copper deposited by an alternating current which has been rectified is 0-6366 x C m X 0-00032959 x /grammes and knowledge of the electro-chemical equivalent of any other material will similarly give the sum-total of the effect pro- duced, as, for instance, in charging a secondary battery. II. Thermal Effect. This is, at any instant, proportional to the rate of expenditure of energy, and is independent of the direction of the current ; thus, it is proportional to In order to find an expression for the average effect so pro- duced, it is necessary to draw a curve which represents the quantity Q 2 R, and to find the average height of such a curve. Thus, if C m = 10 amperes, and R be i ohm, this curve of energy expenditure will have zero values coinciding with the zero values of current, and will rise to a maximum of 100 at the time when the current is maximum. If now the length corresponding to half a wave be divided up into 18 parts, and the area of each such slice be taken, all such areas being added, it will be found to amount to 899*6, when the width of each slice is called unity, and its height taken as that of its middle Alternating Current Circuit. 117 value as marked on the scale of the diagram. The average height is consequently 899-6 IT = 49 ' 95 which would have come out = 50 exactly, if we had been able to estimate values more exactly. Now, this 50 is the average rate of expenditure of energy in the circuit, and consequently is of the form but we know R to be unity, whence it follows that the C 2 must be = 50, or the effect produced is the same as would be the case had a current of magnitude 4/50 = 7*071 amperes been maintained constant for the corresponding time. But this current is 0*7071 x the maximum and is also the >v/(mean square) It is called the " virtual " value, and is really the value about which we should know. As far, then, as working in the circuit in heating it, the alternating current which follows a sine law has an effect which is the same as would be produced by a continuous current having a magnitude of 0-7071 x C M and lasting for the same length of time. From this value the maximum can be found by multiplying the virtual by U. T ^ ?T , or 1-414, or max. C = C virtual x i'4 T 4 Before going on to consider the magnetic effect in detail, it will be as well to point out that an electro-dynamometer will be affected by the value of the square of the current at any instant as with continuous currents, and will thus read a quantity which is this Vrnean square, or virtual value. 1 1 8 Examples in Electrical Engineering. It follows from this that the rate of expenditure of energy in heating a circuit will be proportional to the product of the resistance in ohms into the square of the electro-dynamometer reading. And also, when the current has been rectified, the electro- dynamometer will still read the virtual value, or 0*7071 x maximum But it has been shown that the chemical effect produced by a rectified current is only proportional to 0-63 6 6 X maximum whence it follows that the dynamometer will indicate a value which is , or I'll times too much 0-6366' III. The Magnetic Effect. This may take place under circumstances which may be summed up in the following order, beginning with the condition where no magnetic effect can be produced : (a] No magnetic effect possible ; circuit non-inductive. (F) Magnetic effect set up in a medium which has a uni- formly constant permeability say air where /A = i and which is also non-conducting. (c) Medium having constant /x = i, but being conducting. (d) Medium having variable value of /x, as in a closed iron circuit. Non-conducting, i.e. laminated. ( / S i / 4 / - // \ \ i '/ / f \ 1 \ \ / , ^ }\ 8 t- ^ 7 / \ t \ ^ ^ ^ \ \ /^ / I \ / f * X S "^^ ^ / j / / 2o ^ ' i E FIG. 22. get the value of E, the E.M.F. which must be supplied to or impressed upon the circuit. This has been done after the manner indicated by Dr. Barfield in I886, 1 and the curve E so produced is marked in full line. It will now be noticed that this E curve does not coincide in phase with the C curve which it produces ; in fact, the C curve lags after E by an amount AB, which we will call the angle X. A 1 "Industries," vol. i. p. 17. Alternating Current Circuit. 123 little consideration will show that this lag X will depend upon the relative values of RC and e, since E partakes most in character and phase with that one which is of greatest value. Thus, when e is zero, as in a non-inductive circuit, E = RC, and there is no lag. But when RC is very small and e very large, then E is equal and opposite to e, and there is a lag of practically X = 90. Now, if we regard RC as a sine curve, it is clear that the opposite of e must be a cosine curve ; and as E is the sum of these two, we have that E = RC, (i sin / + llw cos / when RC, (l is the maximum value of RC and e 1)W the maximum value of ej. Now, this can be shown to be equivalent to E- V(R 2 C 2 TO + f lm ) sin (7* + where X is an angle whose tangent is Calling RC,,, = 'e the active E.M.F., we may write shortly that the impressed E.M.F. E = *Jr + 2 in magnitude and has a phase-position which is A degrees earlier than the current, X being such an angle that e B.E.M.F. tan X = - = : ^-TT-ET e active E.M.F. Thus, when the circuit is of such a nature that a very powerful magnetic field can be set up by a small current, and the resist- ance is also small, there is a considerable lag of the current after the E.M.F., and such E.M.F. is very much greater than that necessary to merely cause the current to flow in the ohmic resistance of the circuit. But again, should the resistance of the circuit be very large, so that the active E.M.F. RC required is large compared to the B.E.M.F. set up by the 124 Examples in Electrical Engineering. changing magnetism, then the current does not lag much, and E is only slightly larger than that required for RC. This can also be graphically illustrated by remembering that the E.M.F. to be supplied, E, is the resultant of the other two E.M.F.'s which are present, viz. e and e, which two are also at right angles. Thus we can construct a figure which shall show this resultant E.M.F., as is shown in Fig. 23, which shows the FIG. 23. magnitude of the sum of e and e, and also the nature of its position. Considering the phase-position of e to be repre- sented by the vertical line, and remembering that the current C is in phase with e, it is clear from the three triangles in Fig. 23 that the difference in phase, as indicated by the departure from the vertical of line E = \f e 1 4- 2 , is least of all when is small compared with e, is very marked when e is comparable to e in value, whilst, however large may become, the sum E = J(e 2 + e 2 ) can never be more inclined to e than the horizontal position, or can never differ in phase from e by more than 90. Further investigation will also show that the rate of expen- diture of energy in the circuit will depend upon the lag ; being greatest when there is no lag, and least when the lag is large. Alternating Current Circuit. 125 If the instantaneous values of the product of E and C be taken for the three possible cases, viz. no lag, small lag, and large lag, we shall get the waves of watts, as represented in Figs. 24, 25, 26, where the areas enclosed by the dotted lines E 1 w c ^ 06 12 ,, - f S" ^ 64 & V \ /v \ / \N 48 5 / / A '/ \ \ . 3~ / \ N 16 2 6 1 o o a i i6 e 24 a M 4 40 5 48 6 66 7 6* 8 72 9 80 |0 8B II 96 12 104 , / . v \ , \ . -. J - - N V - - J I . "7 _ - - - ^ \ ~ Z . (2. \ v /' \ 2 \ /, N \ / / \ \ y \' // V / x s /, X ^ ^ ^> FIG. 24. W will represent the work done. It will be noticed, in the case of no lag, that all the work is positive that is done in the circuit, and it can be shown that the average height of the watts area is then E V C V = watts the voltmeter reading multiplied by the ammeter reading. But in the case of small lag there is part of the area of work marked on the lower part of the zero line and called negative ; this must be understood as representing the case when the circuit is not receiving energy, but is giving it back to the engine, the resultant work being the difference between the + 126 Examples in Electrical Engineering. and areas, an amount which can be shown to be equal as an average value to E t ,Q, x cos X = watts Lastly, in the case of maximum lag = 90 the -f and W RC E ^ \ GO 12 \ , '. ICO 10 / \ 90 9 ^ \ > t - - ~ / 70 7 1 / : s / >^ 50 5 / / \ \ / v \ I \ / 50 3 . 20 2 / 2 f \ \ . ! / i Q \ \ / v \ j / y 20 2 . j s / \ V / / 4 J 50 5 \j j / 60 6 / N / 70 7 t / / 00 8 SO 9 N S K30 10 < \ v^ ^ ^^ II \ / 12 13 \ / 14 --, / .C.RC FIG. 25. areas are equal, and consequently the work done nil. can also be represented as watts = E B C y X cos X This where X = 90 and cos X = o. Such a condition is, of course, not absolutely attainable, though in some cases the lag is very nearly 90. The coefficient cos X has been called by Dr. Fleming the power-factor, and though it cannot be directly measured, it is of great importance. As shown above, its value can be found Alternating Current Circuit. 127 from a knowledge of the ratio of the back E.M.F. to the active E.M.F., or of e e tan A =-, whence cos A = > The active E.M.F. is easily found for any circuit whose resist- W Ec 45090 42585 40080 37575 35070 325 65 30060 27555 2505010 225 45 .3 20040 8 175357 150306 125255 100 20 4 75 "5 3 50 10 2 25 5 1 O 25 5 ' 50 w 2 75 15 3 100 204 125 25 5 15030 6 175557 200408 225 45 9 2505010 275 55 30060 32565 35070 37575 40080 42585 450 90 " ? \ /' 7 / / \ / \ / \ / ,-" v, t ' \ 1 / A / ^ { / \ ^ ^ / \ \ , / / 's. \ I \ i /] \ \ / / 3 ^ / / A \ \ 1 j t * \ ^ A \ ~7 . / / * / \ \ ^ / \ \ 1 / ,' 1 \ / ' \ \ S : ; ^ ^ } / I \ / "t . / \ s / FIG. 26. ance and current are known ; and we will now proceed to ascer- tain . The magnitude of the B.E.M.F. set tip in a circuit of constant //, = i. Consider an infinite solenoid having no magnetic core, and being d cms. = mean diameter, i.e. d is the diameter to the middle of the depth of winding. w = number of convolutions per cm. of length of coil. 128 Examples in Electrical Engineering. The value of the magnetic density set up inside this solenoid will be IO when C is the instantaneous value of the current in amperes. The total number of lines in the plane section of the core at right angles to the axis will be __ 47T N = Cw x - 10 4 and this will alternate in value and direction with the current, having a magnitude at any instant which will be 47T d^TT . 27T N ( . = C m w x -- sin / 10 4 r Putting values to the quantities of w = i turn per cm. length d = 2 cms., and C m = i ampere then the maximum value of N will be 47T 2 -^ X i X i X i = 3*948 lines and a curve can be drawn coinciding in phase with C, and having a maximum value of 3*948, or N, = 3*948 sin / Now, the E.M.F. which a changing magnetic field can set up is, in volts, equal to the number of lines gained or lost per second multiplied by the number of turns or convolutions so gaining or losing lines, divided by io 8 . As before shown, the lines are changing at the greatest rate at the instant when they are zero that is to say, the slope of the lines' curve is then a maximum. Alternating Current Circuit. 129 Let the periodic time T = j second, then the number of lines corresponding to the instant in time 0-000027 second after the zero value, will be +N N (2 X VI 1- Too" = 3*948 sin 0-017452 X 0*000027 130 Examples in Electrical Engineering. which angle is in circular measure, and must be multiplied by iSc- to get degrees, or N, = 3-948 sin ( 0-017452 x - 1 ^ \ 7T = 3-948 sin i But sin i is 0*01745, whence the value of N f is N; = 3-948 X 0*01745 = 0-0689 nne In Fig. 27 is given a magnified part of the curve of lines showing these values, which are taken for a very small part of the curve, in order to approximate the more nearly to the absolute slope at the instant. This slope is called, in the language of the differential calculus -TT- when 1300 / / 1600 f / 1400 i 1200 t / 1000 1 / pnrv / 600 / / / i / o - HH I / i / 1 1 ICOO I 1 1400 i / I / / / / f 1 ^ / / / ? i s^ ' a/V) / ^ 3AOO _. ^^ ' 4000 4200 ^ =: , s-\ /3 / FIG. in friction in ergs per cubic centimetre per cycle, called h as before, for dynamo calculations. Now, the nature of the motion of the iron particles will depend upon the change from moment to moment of the Alternating Current Circidt. 139 E.M.F. acting in the circuit, and when this is of the simple sine curve nature, it is found that the value of j3 also changes after the manner of a sine curve. This is shown by the fact that the shape of the secondary curve of E.M.F. is practically a sine curve. Working backwards it is then easy, as was first pointed out by Mr. Evershed, to plot the corresponding values of current required when the curve of lines has been drawn. The cycle of magnetism in Fig. 30 is supposed to be that found when the iron has settled down to a steady state, and it will thus be seen that the manner of change is such as would be met with in going round the figure in a counter-clockwise direction. It thus becomes apparent that, owing to hysteresis, when (3 is zero, but becoming of positive magnitude, then C has to be of quite a considerable value, and does not increase nearly in accordance with the rise of /3 from o to maximum. Also that when the maximum /? has been reached (which always coincides with the maximum value of the current), the current declines in value much more quickly than ft having even to become of opposite sign before /? has again become zero. The reverse process takes place for the negative half wave of ft and a complete wave is shown in Fig. 31, where the current curve C is drawn to a larger scale than ft for con- venience. It will be noticed (1) That the value of C is very small, owing to the high value of fji, which is thousands of times what it would be in air. (2) That the C curve does not coincide with ft though, of course, e will have its customary position or lag 90 after ft Consequently, when the B.E.M.F. is the major item in deter- mining E, E will still come in nearly opposite phase to e, and consequently C will not lag so much after E as would have been the case with a similar air core. This decreased lag is of great importance, as, together with the altered shape of the current curve, it determines the supply of energy which is wasted in hysteresis, in addition to that necessary for C 2 R. Now, we can easily draw in such a curve as that shown, and know its maximum value. But as indicated on an electro- dynamometer we shall of course get its virtual value, and this 140 Examples in Electrical Engineering. will no longer be 07071 x maximum value. Thus, in such cases as impedance coils made with a closed iron core, and where the current is made fairly large by making the value of FIG. 31. w small, this must be remembered. Owing mainly to the irregular shape of the C curve in such a circuit it is general to make impedance coils with an open iron circuit, the ad- vantage of which will be indicated in the section devoted to the consideration of the open iron circuit. (e) Consideration of the closed iron circuit when conducting, or having eddy current in the iron core, or a secondary load. The nature of the disturbance set up by Foucault currents, or a closed secondary will be of precisely the same character as in the other case considered ; it is only necessary to point out that, as in general the magnitude of C m , as we will call the current required to produce the flux in the iron, is very small, and lags little after E, whilst the current to balance the Foucault losses has a magnitude which is also small, but which is practically in phase with the impressed E.M.F., their sum is consequently larger than either, and lags less than C m , whilst when there is a secondary load to balance there is practically Alternating Current Circuit. 141 no lag of the current after E, for even a small value of the necessary balancing current. (f) Medium composite in character, as, for instance, in the case of an open iron circuit impedance coil. Here, as in the composite circuit of a dynamo, calculation has to be made for the several items. Thus for the straight iron core of length / and sectional area a, to be magnetized to a total of N lines, we have with a given winding of coil surrounding it, that the M.M.F. required will be Co// = a reference to the tables being necessary to find the values of C w corresponding to various values of /?. In addition to this there will be required a current to send the same lines through the air space. This may be found by considering the reluctance of the air space, as given in the formula M I>2 73^ d(2l - d) where / and d now must be considered to have the values for the iron core. To force the lines through this space additional ampere-turns will be required of magnitude or Now, that part of the current C, which is required for the iron core, will be, as indicated, a saw-tooth curve like that in Fig. 31, will lag somewhat after E, but not nearly to 90, and will also be small in magnitude compared to Q, which will be a sine curve, will be large, and will lag practically 90 after E. The total magnetizing current will consequently lag much, will partake mostly of the sine curve shape of Q and will, on the whole, be very large compared to the case of a closed iron circuit. An example will render this clear. Let the iron core 142 Examples in Electrical Engineering. (laminated perfectly) be 30 cms. long and 5 cms. in diameter. Let the value of (3 be 5000 total, as an average of maximum value throughout the whole length. Let there be 4 turns per cm. length of core. Then C#// = {f (13) = C X 4 X 30 = 30 x 1-41 = 0-35 ampere maximum value of the saw-tooth shaped curve for the iron core. The total number of lines produced will be a maximum of ^ X ^ X 2 2 N = 5000 x - - - = 98,200 And the reluctance of the air space will be 1-273x30 1-273 X3Q_ Whence the ampere-turns required will be dwt = 0-8 x 0-1388 X 98,200 0-8 x 0-1388 x 98,200 and Cj = - - = 90-86 amperes 4 X 3 maximum value. Now, the 0-35 amperes required for the iron core will lag after E considerably less than 90; but the 90*86 amperes, which is a sine curve, will lag 90, and as its magnitude is so large compared to that for the iron, we shall get the real magnetizing current required, being practically a sine curve and lagging all but 90 after E, whilst its maximum value will only be slightly less than 91 amperes. Indeed, we may say that the iron core has nullified the reluctance of the core, and all we have to consider is the outer air space. Had there been no iron core the reluctance of the air core would have been and the current required would have been nearly 1000 amperes, It is for this reason, amongst others, that open iron circuit impedance coils are generally used instead of those with a closed iron core. The advantages are Alternating Current Circuit. 143 (1) Current curve is practically a sine curve. (2) Easy shape to wind and adjust. (3) Small quantity of iron required. The Hedgehog transformer of Mr. Swinburne is an example designed to obtain a large all-day efficiency. This will be more fully dealt with in the section on transformers. 1 9 8 7 6 5 *s V .4 f: i / ^ I ^ II /t ?* / , - ~~- ~~~~ I ^- ^ / ^ ^ ^\ econ,(ta,ry L octet, FIG. 32. (g) When the medium of an open-iron circuit coil is conduct- ing, as when imperfectly laminated or when containing a closed secondary circuit, we have the same balancing currents required. Only here such currents are no longer the only ones of any magnitude, and their effects in determining the total current are considerably reduced by the already large mag- netizing current. It thus follows that even at very large 144 Examples in Electrical Engineering. values of the secondary load there is still a very considerable lag of C after impressed E.M.F. In some curves, showing the relationship between the power factor and various loads for closed-iron and open-iron circuit transformers, published by Dr. Fleming in 1892, this great difference is very apparent. For convenience of reference they are here reproduced in Fig. 32, where the relationship is shown for two closed-iron transformers, viz. the Westinghouse and Kapp ; and for one open-iron transformer, the Hedgehog. CHAPTER XIII. IMPEDANCE COILS AND TRANSFORMERS. Impedance Coils, or choking coils, as they are sometimes called, are pieces of apparatus designed to make use of the property of self-induced E.M.F., by introducing such an E.M.F. into a circuit acting so as to decrease the active E.M.F. with- out at the same time absorbing any power except that inci- dental to the properties of the materials employed. Thus, we may have a pair of mains at an alternating P.D. of 100 (virtual) volts, and it is desired to connect to these a lamp taking a current of 10 amperes at a P.D. of 40 volts. With a continuous current circuit one way is to introduce a resistance in series with the lamp of 6 ohms, but then the energy wasted in this resistance will be at the rate of 600 watts. Or a number of secondary cells may be put in circuit, having together an E.M.F. acting against that of the mains of about 60 volts. In this case there will be a storage of the energy not required by the lamp, only a small part being wasted in heating, and other losses peculiar to secondary batteries. With the alterna- ting current it is equally possible to introduce a dead resist- ance of 6 ohms, and waste the same 600 watts ; but we can introduce a similar B. E.M.F., as in the case of the secondary battery, the problem being to know how much ? Remembering, then, that this B. E.M.F. is to be produced by the passage of the current through an inductive circuit, we have that or e = V(E 2 -"?), or d = V(E 2 -?) 146 Examples in Electrical Engineering. and inserting values in this last, we have e 1 = >v/(ioo X 100 40 X 40) = ^(8400) = 91-65 This value is, however, slightly too large, since the very coil which is to produce it will have a resistance, and so absorb a small E.M.F. in becoming heated. In fact the impedance coil is also a resistance coil, and the effect of its own resistance should be allowed for as a resistance in series with the external circuit This resistance of the coil is usually very small, com- pared to the resistance of the rest of the circuit, in order that the power wasted may be small. Calling *i the active E.M.F. required by the coil itself, we then have e 1= VJE'-'^T^?} and all these values may be in either virtual or maximum values, as we please, provided they are all either the one or the other. Transformers. An unloaded transformer, that is with open secondary circuit, is simply a case of an impedance coil designed, as a rule, to take the minimum current and produce a B. E.M.F. differing from E the impressed volts by as small an amount as possible. In as far as, then, as it is unloaded, it can be dealt with in precisely the same manner as the impedance coils discussed above. In allowing for the effects of eddy currents it has been pointed out that the disturbing effect produced by them had to be balanced (automatically) by the increase of the main current by an amount such as would produce ampere-turns magnetizing force sufficient for the purpose of overcoming the demagnetizing effect due to the eddy currents. A little con- sideration will show that the current taken from the secondary coil of a transformer will be of an exactly similar nature and will similarly have to be balanced by a corresponding increase of current in the primary. The consequence of this is that the current in the primary coil is not only variable in magnitude but is variable in phase. Impedance Coils and Transformers. 147 As with impedance coils, there are two cases : the closed iron circuit and the open iron circuit. Consider a simple type of closed iron circuit transformer wound so as to have the same number of turns on the primary, as on the secondary. The current required on open secondary circuit will be of a kind represented in Fig. 31, including the two items (1) True magnetizing current ; and, (2) Balance for eddy currents. The sum of these two will be a curve partaking most largely of the nature of the true magnetizing current, since the current to balance eddies is small and also lags due to self-induction in the core. The phase position of this no-load current, which we will call C , will be after the primary E.M.F. by an amount less than 90 on account of the hysteresis, but will be very small in magnitude. Of the two items of C the magnetic effect of the balance for eddy currents is nullified by the eddy currents, thus leaving only C m , the true magnetizing current to be looked upon as causing so many ampere-turns to act upon the magnetic cir- cuit, setting up an alternating magnetic field in the core which produces the E.M.F. in the secondary as an E.M.F. coinciding with the B.E.M.F. in the primary, and as we have already seen this will have a phase position approximately 180 after the primary E.M.F. So far as the secondary is on open circuit, we get no effect due to its presence except in so far as that presence has pro- bably increased the eddy current loss somewhat. But when the secondary is connected, as is most usual, to anon-inductive circuit, we get a current flowing in it which is in phase with its own E.M.F., but in contrary phase to the primary E.M.F. This current will, of course, mean so many ampere-turns acting upon the magnetic circuit, and these would produce a very serious disturbance were it not for their effect upon the momen- tary value of the B.E.M.F., causing the primary current to increase in magnitude and lag less. This increase is of exactly such a nature as would be produced by an additional 148 Examples in Electrical Engineering. current in the primary acting in contrary phase to the secon- dary current, and of such a magnitude as to produce ampere- turns exactly balancing the magnetic disturbance set up by the secondary current. The real new primary current will, of course, be the sum of this balancing current C 6 , and the original no-load current C , which, it must be remembered, is itself the sum of two items previously stated. In the case of a transformer without iron core and with no eddy current waste, there will then be only two currents to consider, viz., the current C w to set up the necessary magnetic flux, a current lagging almost 90 after the primary E.M.F.; and the balance for secondary load, a current which is in phase with the primary E.M.F., and thus 90 apart from C OT . Calling the latter Q, we shall have that the current at any time in the primary coil will be C X and its phase position will be after the primary E.M.F. by an angle whose tangent is approximately W 5 , which shows that ^6 when C 6 is large compared to C Hl the lag of C^ after the primary E.M.F. may be almost nothing. In the case of the closed iron circuit type this is most marked ; for even at as small a value of C 5 as corresponds to only one-tenth of the full secondary load, we find the ratio of C m to C 6 so small as to make no appreciable lag of C p after primary E.M.F. But in the case of the no iron core or open iron circuit transformer, cm. is so large as to prevent the ratio of C m to C 4 ever becoming very small, even with a value of C 6 corre- sponding to full secondary load. In the case of the open iron core we have C^ being composed of the sum of several items, viz.: (*) C^f to magnetize the iron core : this is small in magni- tude, is saw-toothed in shape, and lags less than 90 after the primary E.M.F. ( 2 ) C wa to magnetize the air space : this is large, is prac- tically sinusoidal, and lags almost 90 after the primary E.M.F. (3) C& to balance the effects of secondary load and any Impedance Coils and Transformers. 149 eddy current waste there may be in the whole apparatus : this is practically sinusoidal, and lags hardly at all after the primary E.M.F. The sum of all these is mainly a sine curve, even when C& is quite small ; and its phase position is mainly determined by the relationship of C MMt to Q. But as C 6 never attains such a size compared to C mtt as to make Q "H 22 = < o'3 <> r so ^6 it follows that there is always considerable lag of C p after the primary E.M.F. This is strikingly shown in the curves reproduced from Dr. Fleming's paper on Transformers, on p. 143, Fig. 32. In transformer design care must be taken that the loss of volts in resistance is very small \ that the loss of power in hysteresis and eddy currents is small; and that, above all, these last are small, inasmuch as they are a constant load and so very materially affect the efficiency when taken over a period of time during which the load varies considerably. The various losses may be thus stated (i) Drop of volts in the primary coil, causing the B. E.M.F. e, which is a measure of the magnetic flux set up, to differ from E the impressed E.M.F. This loss is such as to cause the E.M.F. producing power of the transformer to be smaller than E, by the amount CpRp taken in its proper phase, and making the ratio of ~ slightly less than unity. It is, in the closed iron circuit type, very small at no load, since, then, C p is very small ; and it only reaches a value of importance when the secondary is on full load. But in the open iron circuit type there is not nearly so large a variation in the value of C p for various secondary loads, and so not as great an alteration in the primary loss of volts. (2) The drop of volts due to ohmic resistance of the secondary, viz. QR,, is always proportional to the secondary load in all types, and can be prearranged to have any value desired. 150 Examples in Electrical Engineering. (3) A drop in the secondary due to the slightly inductive character of the winding. This will increase with the load being such as to make the effective E.M.F. in the secondary less than where w s and w p stand for the number of secondary and primary turns, in as far as the secondary coil acts in the same manner as an impedance coil in its own circuit. For this reason the number of windings w, should be kept small. (4) There is also a drop in E.M.F. on full load, due to leakage of magnetic lines. It has already been shown that a fully loaded transformer has two oppositely acting magnetizing forces in the circuit, and if these be not evenly distributed round the magnetic circuit the consequence is evident polarity, followed by inequality of the magnetic flux through the coils. In a good closed iron type with the primary and secondary coils well intermingled along the magnetic circuit this drop can be kept to as low as ~ per cent. (5) The loss due to eddy currents in the core ; and, in the case of open iron circuit types, due to eddy currents in the winding. Mr. Alexander Siemens has pointed out how this loss may be arrived at in the case of a core composed of fine iron wires. The following formula is adapted from his paper on the subject, and is the value of the watts lost in eddy currents per cubic centimetre of iron 6-6//3V 2 F = - - - watts per cc. of iron io 12 where d stands for diameter of wire in cms. p periods c/> per second. ft ,, c.g.s. lines per sq. cm. of iron section. Mr. Evershed has also shown how to express the loss when the iron core is composed of thin iron plates, viz. : F = -^-rr watts per cc. of iron 11 io where / is the thickness of plate in cms. Impedance Coils and Transformers. 151 (6) The hysteresis loss due to molecular friction in the iron. This may be very accurately found in the way shown for the armatures of dynamos by means of the table on p. 2 14. Thus H = -y = hysteresis loss in watts total Calling the total induction in the iron of a transformer N c.g.s. lines, and using the symbols already mentioned, we have, neglecting the slight difference at no load between and E,, E,, X io 8 4 '443 X p and N = fta, ft being quite small, ranging from as low as 3000 to 6000, and seldom exceeding the latter, whence E,, x io 8 7 ^ = 4^rxTx^ which gives an important product of the two unknown quan- tities in terms of known, or easily fixed values. Only ex- perience can fix the relationship between the primary turns, and the sectional area of the core. On the one hand, if w p be large we shall have, with a given weight of copper, a high resistance and large copper loss ; but the iron losses H and F will be small. If, however, a be made large, the copper losses may be very small, but the iron losses will be large. The relationship between these two losses, copper and iron, will depend greatly upon the average output or load-factor. When the transformer is to be on full load always, then the highest efficiency will be attained when the copper losses are equal to the iron losses. But in general the conditions of supply render it impossible that the transformer can be on full load for more than a very small part of the 24 hours, whilst for a very large part it may be practically unloaded. The all-day efficiency can then be made higher by making the constant iron-losses smaller at the expense of large copper losses, which will only be of importance in themselves for the short times of high and full load. In this, of course, due regard must be had for a reasonable constancy in available secondary P.D. 152 Examples in Electrical Engineering. Example. A transformer for 40 amperes and 100 volts secondary output. In order that the efficiency at full load may be, say 95 per cent, it follows that the input must exceed the output by about 200 watts, which is to cover both the copper losses and the iron losses. If these be about equal, we must allow for something like 79 watts in hysteresis, from which, if the value of /? be fixed, we can find the volume of iron. Thus from we have HX_1 7 h f and if ft be 6000, h 1820. Putting these values and 79 for H, we have, if/ = 80 en per second. 7 V = If the primary E.M.F. E^ = 2000 volts virtual, we have from io where N is the maximum induction, that 2000 x io 8 = 562 ' 7 ' OC and as ft = 6000, we can put w p a = 93,800 where a is the sectional area of the iron core in sq. cms. Taking this to be 155 sq. cms., we have from the previously found volume that the average length of the iron core must be V 5425 - = ^ - = 35 cms. i55 If the core is composed of stampings of the shape and size Impedance Coils and Transformers. 153 shown below, we shall require a pile of them 22*2 cms. of iron, not counting lamination. 21 MA }** - ^< 7 X- * FIG. 33. The total volume of iron in the above plates is (21 x 14- 7 X 7) 22-2 = 5439 cc. but as the magnetic induction cannot be taken as filling the iron into the corners, the amount under magnetic influence will be more nearly that previously found, viz. 5425 cc. The sectional area is 7 X 22-2 = 155-4, say 155 sq. cms. as was desired. Of the spaces for winding we will allow that owing to superior insulation the space occupied by the primary coils will be i4'5 sq. cms. out of the total of 24*5 sq. cms., and that the secondary takes 8 sq. cms., leaving 2 sq. cms. for an insu- lating partition between the two coils. From w p a = 93,800 putting a =-. 155, we have w p 606 say 154 Examples in Electrical Engineering. which number of wires has to go into the space of 14*5 sq. cms., or each wire will take up a room of = ' 2 392 sq. cm. area whence the diameter of the covered wire, if round, will be d = ^0-02392 = 0*154 cm. covered wire and if the covering increases the bar diameter by 0-07 cm., the diameter of the bare wire will be 0*154 0*07 = 0*084 cm. which is practically equivalent to a No. 21 B.W.G., which is 0*032" diameter, and has a resistance of about u ohms per 1000 feet warm. Allowing 2 -8 cms. for lamination, the average length of one turn on the primary coil will be 72 cms. if the primary be wound inside the secondary. 72 cms. = 28-5 inches say whence the total length of the primary wire will be 606 x 28*5 inches = 17271 or 1440 feet. Whence the resistance RP = 1*44 X n = 15*4 ohms say Again, the mean length of the secondary turn will be 76 cms., or 30 inches, and i 606 *=- 2Q v>* = 20 3i say The space to be filled by these 31 turns is 8 sq. cms., whence each wire will occupy a room of o = 0*258 sq. cms. and if the wire be square the size will be ^0*258 - 0*507 cm. square covered, which will admit of a bare wire of 0*437 cm. side of square, about equivalent to B.W.G. 6J, and having a sectional area of 0*1909 sq. cm., and a resistance of about 0*2 8 W per 1000 feet warm. Impedance Coils and Transformers. 155 The total length of secondary wire will be 30X31 = 78 feet say 12 whence R, = 0*28 x o'oyS = 0*022 ohm. If, then, the maximum secondary current be 40 amperes, there will be a full load ohmic drop of 40 x o - o22 = o*S8 volts and a full load secondary copper loss of (4o) 2 x 0*022 = 35-2 watts, say 36 Also, at full load the current in the primary coil will be C s C, = ^ + C M where C Hl is to include the magnetizing current giving the loss in hysteresis and also the current required to balance the eddy current loss in the core. In as far as these are losses they can easily be found, as shown below ; but as they are very small compared to the secondary balancing current, they hardly at all affect the value % of the primary copper loss. Thus, for example, corresponding to ft = 6000 every centimetre length of the iron core will require 1*624 ampere turns, or the maximum magnetizing ampere turns on the primary will be 35 x 1-624 = 56*84 which must be produced by a current of , = 0*0937 amperes (maximum value) But this current is not a sine curve, and its virtual value will be greater than 0*707 times the maximum, or the virtual value will be, say 0*075 amperes. Furthermore it is not in phase with the secondary balance, which has a virtual value at full load of 4 = 2 amperes Thus, including the hysteresis and eddy current losses, the I $6 Examples in Electrical Engineering. full load primary current cannot be greater than 2*1 amperes virtual, if so much, and therefore the primary full load copper loss is (2*1)- x 14 = 62 watts say Using the formula on p. 150 for the eddy current loss, and taking the thickness of the stampings as 0*03 cms., we have the loss per cc. is 80 x 80 x 6000 x 6000 x 0*03 x 0-03 r, - = 0'0020 I0 11 whence the total eddy current waste will be 5425 x 0*002 = 10*85 watts say ii watts. Thus, the total output will be 40 x 100 watts, and the input will be 40 x ioo -f 79 -f 62 4- 36 + n = 4188, which is within the prescribed limit. If kept on full load the efficiency will be 4000 100 X 4188 = 95 ' 5 P ercent But when it is remembered that during a day of 24 hours a transformer on a lighting circuit is very variably loaded and may be practically on open secondary circuit for several hours, during which time the iron losses are constantly to be supplied, it will be apparent that the all-day efficiency or ratio of Units paid for during 24 hours Units put in during 24 hours will be a very different value to the full-load efficiency. Consider, then, the case of the above transformer as on a circuit where it is On No load. A load. i load. load. } load. Full 1 ad. For 8hrs. 4 hrs. 4 hrs. 4 hrs. 3 hrs. I hr. We shall have the following table of outputs and inputs for Impedance Coils and Transformers. 157 24 hours, the total full load copper loss being 98 watts, a loss which will vary practically inversely as the square of the load ; and the iron loss equal to 90 watts, whatever the load. Load. , Hours. Output watt-hours. Input = watt-hours. Output + iron loss + copper loss. Total. 8 O o + 8 x 90 + o = 720 T'. 4 4 X 400 = 1600 1600 + 4 X 90 + 4 x 0-98 = 1964 1 4 4 X 1000 = 4000 4000 + 4 x 90 + 4 x 6'i =4385 i 4 4 X 1334 = 5336 533 6 + 4 X 90 + 4 X 10-8 = $739 i 3 3 X 2000 = 6000 6000 + 3x90 + 3x25 = 6345 I i i x 4000 = 4000 4000 + i x 90 + i x 98 = 4188 Output 20936 Input 23341 _ . ioo X 20936 And all-day efficiency = - 23341 = 89-6 per cent.. which is a very different thing to the 95*5 per cent, found above for the full-load efficiency, being even less than the efficiency if the transformer were kept on quarter load always, which would be over 91 per cent. Example XLYIII. A circuit has a B.E.M.F. of 1600 volts and a resistance of 20 ohms. What E.M.F. alternating will be required to send through it a current of 10 amperes? Solution. The relationship between the E.M.F. required E the B.E.M.F. e and active E.M.F. e - RC, is that whence since E = V(* 2 + 2 ) e - RC = 20 x 10 = 200 E = V(i6oo) 2 4- (200)" = 1612 -45 volts Example XLIX. In the above circuit what will be the current if the alternating E.M.F. supplied is 2000 volts, and what will be the rate of doing work ? Solution e = 158 Examples in Electrical Engineering. or e - V(20oo)- (1600)- = 1200 volts but - 20 and the rate of doing work is C 2 R = 72,000 watts or 96-48 HP. Example L. The P.D. at the terminals of a house con- nected to an alternating system of distribution is 100 volts. It is required to run 2 arc lamps, taking 34 volts each, in series off these terminals. If the lamps can be regarded as simple resistances of 3^4 ohms each, what will have to be the value of the B.E.M.F. of the choking coil to be used in series with them? Solution. Since each lamp takes 34 volts, and is equivalent to a resistance of 3*4 ohms, they are taking a current of 10 amperes, and when in series will require 68 volts, which is the value of R.C., the active E.M.F. required. Thus, since E = 100 we have . ? = ^/((ioo) 2 - (68) 2 ) = v^ioooo - 4624) = A/(5376) = 73'3 2 volts B.E.M.F. Example LI. In the last question, what is the saving effected, by using the choking coil instead of a non-inductive resistance, if the choking coil consumes energy at the rate of 30 watts ? Solution. The total expenditure with a non-inductive resistance would be CE = 1000 watts, and with the choking coil it is only C 2 R + 30 or 100 x 6-8 -f- 30 = 710 watts whence the saving is 1000 710 = 290 watts Impedance Coils and Transformers. 159 Example LIL What will be the angle of lag of the current in Example L. ? Solution, the lag is such that tan A = - tan A = 7 -|| 2 = 1-078 68 whence A = 47 about. CHAPTER XIV. EFFECTS OF CAPACITY. THE properties of a condenser may be easily studied by refer- ence to the water analogy suggested by Dr. Fleming, as a mechanical representation of the electrical conditions. D M, M 2 P FIG. 34- In Fig. 34 is represented a system of tubes containing water, divided at D by a flexible diaphragm of indiarubber, and having in each leg a water-meter, Mj and M 2 . A tap and pump are also placed in one leg ; the tap to represent a switch Effects of Capacity. 161 in an electric circuit, and the pump corresponding to the electro-motive force. First of all suppose the diaphragm removed. On working the pump a current of water will flow round the circuit in one direction or another, according to the rotation of the pump. This current will be indicated on the water-meters, and will be at the rate of so many gallons per minute corresponding to an electrical effect of so many coulombs per second. This flow of water is constant in direction and in quantity, so long as the pump be rotated in one direction at the same rate. If, however, the pump be replaced by a simple piston re- ciprocating in one of the tubes, we shall have the current of water constantly varying in direction, according as the piston is moving up or down ; this reciprocating motion of the piston corresponds to an alternating E.M.F., and the alternating flow produced by it will not make any permanent record on the meters, since as much flows through in one sense as flows back in the other sense. The force urging the piston on its course will vary from point to point of the stroke, according as the mass of water is moving with or against it, and according to the resistance to the flow due to friction against the sides of the tube. This friction corresponds to resistance and the momentum of the water to the B.E.M.F. of self-induction. But though the meters record nothing, they may be replaced by suitable mechanism indicating the amplitude of the recipro- cating motion, which is a measure of the work-doing property of the moving mass of water. When, however, the diaphragm is replaced we have a very different state of affairs. Formerly there was no obstruction to the motion of the water right round the circuit ; but now, on rotating the pump, water will, let us say, be forced up the right side, causing the elastic diaphragm to swell out, as shown by the dotted line in the figure. Meter No. 2 will record the quantity of water required to do this as a quantity of water flowing into the diaphragm ; but meter No. i will also record it as a quantity of water flowing out of the diaphragm. For however long the pump be worked the diaphragm will only become distended to an extent equal to the force exerted by M 1 62 Examples in Electrical Engineering. the pump, supposing, of course, that its strength is such as not to admit of rupture by any force the pump can exert. Thus, there will be a flux of water on the one hand into the dia- phragm, and on the other hand out of the diaphragm for a short time only; and the rate of flow will have been larger at the start w r hen the elastic force exerted by the diaphragm was small, than at the close when the elastic force due to stretching is equal, and opposite to the force exerted by the pump, thus leaving no available force for urging water along the circuit. Whilst still keeping the pump in rotation if the tap be closed, the stretched condition of the diaphragm may be preserved indefinitely, although the pump be afterwards stopped. A careful examination will disclose the fact that the only difference now existing in the apparatus is this stretched con- dition of the diaphragm : not a store of water, for as much obviously flowed in as flowed out, but a power to do work by relaxing a store of energy. The above conditions correspond to that of the continuous- current circuit, and the flux of water which lasts for a short time only after the starting of the pump is called the " dis- placement current." When the pump is now replaced by a reciprocating piston, we have a representation of the alternating-current circuit. Now, the diaphragm will be stretched first in one direction and then in the other, and the water-meters will, as before, be affected by a quantity of water surging to and fro in the tubes. The rate at which this quantity of water will be flowing in any one direction will depend upon the force of the piston and upon the elastic force of the diaphragm. When unstretched, the water will flow into the stretching diaphragm at a great rate, even though the force due to the piston may be small ; but as the diaphragm becomes distended, the rate of flow into it will be small, and ultimately reach a zero value as the elastic force of the diaphragm equals that of the piston at the end of its stroke. As the pressure on the piston which caused it to move forwards is relaxed, allowing it to return to its starting position, the elastic force of the diaphragm causes water to flow Effects of Capacity. 163 back, slowly at first, but reaching a maximum rate as the piston reaches its starting-point (the middle of its total stroke), and then again declining to zero as the piston moves on its opposite stroke, the diaphragm now becoming distended in the opposite way. This can better be seen by reference to Fig- 35 j where P is the reciprocating piston, D the diaphragm, B B, r - TJ- ,.-' / ( / ( i \ \ P /\. D ic, > \ ( / n Xx --. \ FIG. 35- and ABC represent the course of the piston. As it starts from B the diaphragm is unstretched as at B 15 and the rate at which water will flow into it is greatest ; as the piston reaches C the diaphragm is distended to Q, and the rate of flow is a minimum, having declined from a maximum, but being always in the same direction as the force applied to the piston. As the piston, displaced thus far by a force, moves back to B by a removal of that force, the water now flows out or back towards the piston, not away from it as heretofore, till the piston reaches B, when the diaphragm is now as at B x the rate of flow, however, being now a maximum, but declining to zero again, though still towards the piston, as the piston moves towards A, due to a reverse force to that which urged it from B to C. The diaphragm is now as at A x exerting a force away from the piston, and as the latter returns, due to the gradual reduction of the outside force acting on it, the water now flows away from the piston at a slow rate at first, but reaching the maximum again as the piston returns to B, when the diaphragm is once more unstretched. Calling the outside force which urges the piston along as equivalent to the alternating E.M.F. E, and the current of water which flows into the diaphragm as the current C, there 164 Examples in Electrical Engineering. being no resistance to the flow and no magnetic effect set up, we thus have the following order of effects E.M.F. - E. Current - C. Zero but rising to a + maximum then declining though still + to zero reversing and rising to a maximum and declining to zero again. A + maximum but declining to zero then reversing and rising again to a maximum declining whilst still to zero reversing and rising to a + maximum again. and so on, which can best be indicated by plotting the curves of E.M.F. and current on the assumption that the variations of strength of E.M.F. are sinusoidal, as is done in Fig. 36, where E is the E.M.F. supplied and C the current, the circuit FIG. 36. being supposed to be unable to set up any magnetic effect and to offer no resistance to the passage of the current, a condition of affairs which would correspond, on the water analogy, to a massless fluid moving in frictionless tubes. Effects of Capacity. 165 Now, the degree to which the diaphragm can be distended by any force will depend upon its size and upon the material of which it is made ; or its capacity depends upon its con- struction. Similarly the electric condenser has a capacity which is proportional to its area, inversely proportional to its dielectric thickness and also proportional to the property of the material of which it is made. Thus the quantity of water required to distend the diaphragm so that its elastic force is equal and opposite to the unit force acting on the piston will be a measure of the capacity of the system. Also, in the electric circuit we may say that the quantity of electricity in coulombs required to strain the dielectric to an extent equal and opposite to the E.M.F. of i volt is a measure of the capacity of the condenser; and when that quantity is unity (i coulomb) the condenser is con- sidered to have unit capacity, or a capacity of i farad. Thus i farad is that capacity which requires i coulomb to pro- duce a P.D. of i volt between its two coatings. But this quantity Q coulombs must, in an alternating-current circuit, be passed into, or out of, the condenser each quarter period, and therefore the average current flowing into, or out of, the condenser will be C = T= r 4 4 where K is the capacity in farads and E M is the maximum value of the E.M.F. supplied to the condenser. Or, putting / = -, we have the average current and thus the virtual current and maximum current will be C. - 4'44^KE, (l and C m = 27r/KE, w But if E be stated as the virtual E.M.F., then C, = 27T/KE, 1 66 Examples in Electrical Engineering. And we have seen that the E.M.F. supplied to the condenser reaches its maximum value, 90 degrees, later than the maximum value of the current due to it, the condenser itself exerting an E.M.F. in the circuit, always equal and opposite to the supplied E.M.F. It is necessary to make a very careful dis- tinction between these two E.M.F.s; that supplied to charge the condenser is of course that of the dynamo, or other gene- rator, and is called, as before, the impressed E.M.F. ; that exerted by the condenser is of the nature of a back E.M.F., and may be called the B. E.M.F. of the condenser, K. Thus, if a condenser be in a circuit having an E.M.F. of E ( , volts, it will take a current C, = 27T/KE, amperes and will exert a B.E.M.F. of With a perfect condenser having plates of some material offer- ing no resistance to the flow of the charge into them, and with a dielectric of perfect insulating properties, this B. E.M.F. K V will be exactly equal and opposite to E w the impressed E.M.F. But since some E.M.F. is required to maintain the momentary charging current, and there is a small actual leakage from one set of plates to the other (in both cases causing heating), it follows that the impressed E.M.F. E, must slightly exceed the B. E.M.F. K P , the difference of phase being also slightly less than 90 degrees. Now, since the supplied E.M.F. lags 90 after the current in the condenser, and the condenser B. E.M.F. is equal and oppo- site, it follows that the current in a condenser lags 90 after its own B. E.M.F. Therefore the effect of capacity in a circuit is to oppose the effect of self-induction, as may be easily seen by reference to the following diagram (Fig. 37), where the current through a circuit is marked as C amperes, the circuit being able to set up a magnetic field of such magnitude as to cause a B. E.M.F. of self-induction of c volts, and also possessing a capacity such as will give K volts, which E.M.F. may now be Effects of Capacity. 167 called a forward E.M.F., in contradistinction to the B.E.M.F. of self-induction e. To maintain the current C amperes through \ \ \ \ FIG. 37. the circuit, the impressed E.M.F. E will thus have to be suffi- cient at any moment to (a) send C amperes through R ohms, (b) to nullify e, and (K 4 44/lv whence it follows, that when the capacity is such as to make the circuit non-inductive in spite of a magnetic effect ~~70 8 ~ ~ = 27T/K whence C,,io 8 i\. = 1 68 Examples in Electrical Engineering. or if N and C be maximum values, then C m io* i = N w X ~^~ 2 or K = ?r x ^ Now, the capacity of a condenser in farads is A K = i 1-131 x io 13 x d where A = the area of one set of coatings in sq. cms. d = the distance between one set of coatings and the other in cms. and / = that property of the dielectric placed in between the two coatings, which is called its specific in- ductive capacity. The values of / for different materials are : Air i Mica 5 Flint Glass 6*51013 Glass 3 Ebonite ... 2*5 to 3-1 Sulphur 2-8 to 3-8 Paraffin wax ... ... ... ... ... about 2 Paper baked in paraffin oil about 2' 7 to 3 In constructing a condenser, the conductors are composed of thin sheets of metal, usually tinfoil, piled upon each other with some dielectric in between. In order to obtain the large surface required for even a small capacity, the sheets are made of a reasonable size, and interleaved with each other, as shown in section in Fig. 38, where the lines show the metal sheets, and the spaces the dielectric. It will be noticed that five alternate plates are connected together by the terminal block T 1} and are interleaved with the other six, these last being also all connected together by the terminal block T 2 . In such an arrangement use is made of each side of each plate, and thus the surface of such a condenser is found by the products of half the number plates into the surface of Effects of Capacity. 169 one plate counting both sides. If, consequently, each plate measures 24 cms. by 48 cms., the area per plate will be 2304 T - T FIG. 38. sq. cms. If now the dielectric be of oiled paper, 0-015 cm. thick, each piece of paper and plate of tinfoil will be, say 100 0*017 thick, or there will be = 5882 plates, or 2940 say connected to one terminal per metre thick, and taking the specific inductive capacity as 3, the capacity per metre thick will be 2940 X 2304 K = 3 x - 13 = 0-000119 farad 1*131 X io 13 X 0*015 or to make a condenser of i farad capacity would take a thick- ness of about 8400 metres, which gives some idea of the great size of a condenser to have so large a capacity. Also, if used on a circuit working at 2000 virtual volts at 100 c/) per second, the current taken will be C, = 27T/KE, = 628 X i X 2000 = 1,256,000 amperes to which should be added such current as may leak from one set of plates to the other, a current which will probably be almost in phase with the E.M.F. supplied, and consequently lagging 90 after the condenser current. Use of Condensers for modifying the effect of self-indue- 170 Examples in Electrical Engineering. tion. We have already seen that the nature of the B.E.M.F., set up by a condenser, enables it to oppose the E.M.F. of self- induction when the condenser is placed in series with the circuit. We may thus say that capacity in series with an in- ductive circuit renders that circuit non-inductive by nullifying the E.M.F. of magnetic action, and thus leaving the circuit in a condition which can be described fully by saying its resist- ance is R ohms, it requires simply RC volts, but can, never- theless, exert a magnetic effect. In practice, one of the most interesting cases is that of the shunt circuit of a watt-meter for measurement of alternating power; this circuit must be able to produce a magnetic field, and must also be perfectly non- inductive with its current consequently in phase with the P.D. at its terminals. Condensers are, however, sometimes used in parallel with a circuit the properties of which they are desired to modify. The circuit to which a condenser is a shunt is, however, un- altered in any way, the main effect is in the generator and leads connecting it to the circuit and its shunting condenser. Thus, if we have a circuit such as the primary of an un- loaded open-iron circuit transformer, we have seen that the current taken by it will be large, but will lag almost 90 after the E.M.F. supplying it. This current will not necessarily mean any great waste of energy in the transformer, but the loss in the connecting leads and dynamo armature will be a considerable item. If now a condenser be connected across the primary terminals, the dynamo will have to supply through the leads a current such as will be determined by the size of the condenser, but coming 90 earlier than the E.M.F., and consequently opposed to the transformer current. The current, therefore, which flows in the leads and armature will be the sum of these two, and may be almost zero when the condenser current is equal to the transformer current, being only so large as will, in combination with the E.M.F., repre- sent the waste in the transformer and condenser. We have seen that the value of the current flowing into a condenser is C tt = 27T/KE, Effects of Capacity. 171 whence in this case we must make the capacity such as will take this current C tf equal to that required by the unloaded transformer at a P.D. of E fl volts, or K- C "x 1 -E, X 2,r> where C K stands for the so-called " magnetizing " current of the open-iron circuit transformer. The use of condensers for the above purpose was proposed by Mr. Swinburne, in conjunction with his " hedgehog " trans- former, but very few are in use. It is clear from the above that a condenser in parallel with a circuit tends to make the whole combination non-inductive, rather by rendering the apparent resistance larger, than by opposing any B.E.M.F. in the circuit. In systems using concentric cables, the capacity of the cable considered as two conducting masses intimately associated but separated by the insulating covering on the inner conductor, is a shunt to the circuit supplied by those leads, and so may modify the effect of self-induction in the circuit. Messrs. Cowan and Still have published the values of the capacity of a few cables, viz. For yf concentric cables the capacity per mile is British Insulated Wire Co. (paper insulation) ... 0-31 microfarad Glover and Co. (vulcanized rubber) ... ... 0*615 (diatrine) ... ... ... 0-315 ,, This capacity will take a current coming 90 earlier than the impressed E.M.F., which current must be added to the current supplied through the cable to the load at the far end. When the load current is lagging much, due to self-induction, the total current will be, as shown above, the difference between the two, and therefore may be zero. Example LIII. A circuit supplied from an alternator takes a current of 10 amperes at a pressure of 100 volts, its resistance being, however, only 5 ohms. Find the lag of the current and the capacity in microfarads of the condenser which must be used in series with the circuit in order that the current may be 2O I K = ^>-^ x -= 0-000367 farad 86'6 27Tioo 172 Examples in Electrical Engineering. 20 amperes, and not lag at all. There are 100 periods per second. Solution. The active E.M.F. will be 50 volts, whence the B.E.M.F. e is /y/ioooo 2500 = 86'6 volts, whence there will be a lag of current after the E.M.F. such that tan A = - 5 or X = 60. In order to reduce this lag to zero, and thus make the current 20 amperes, the capacity will have to be 20 X 27TIOO or 367 microfarads, a capacity which could only be made in a size of about 25 cubic feet. Example LIY. An unloaded open-iron circuit transformer is shunted by a condenser of 30 microfarads capacity, and appears to take a negligibly small current. It is supplied at 2000 volts off a circuit, having 80 periods per second. What is the true magnetizing current ? Solution. It is clear that the condenser current is equal to the magnetizing current as, being in opposite phase, it has just nullified it. Now, the current taken by such a condenser will be ^4 2?r^KE = x 80 x 0*00003 x 2000 - 30*17 amperes which is, therefore, the true magnetizing current. Example LY. A concentric main six miles long, having a capacity of 07 microfarad per mile, is supplying energy to an inductive load, taking 10 amperes and lagging almost 90. If the P.D. is 1000 volts, what is the indicated current where there are 100 c/5 per second? Solution. The current taken by the main on account of capacity will be C = 27T/KE = X ioo X 4'2 X io- x 1000 = 2-64 amperes which current will be almost in opposite phase to that in the load, whence the apparent current will be the difference or 10 2*64 = 7*36 amperes Effects of Capacity. 173 Example LYI. A concentric cable is 12 miles long, and is connected to an alternating P.D. of 10,000 volts. If the capacity is 1*5 microfarad per mile, what will be the current supplied, the circuit working at 80 CO per second ? Solution. Total capacity is 12x1*5=18 microfarads, whence the current taken is C = 27rK/E = x o'ooooiS x 80 x 10,000 = 90-5 amperes Example LYII. Two condensers of \ microfarad capacity each are connected to an alternating P.D. of 50 volts at 225 CO per second. What will be the current taken (a) when the two condensers are in parallel ? (b) when in series ? Solution. In parallel the total capacity will be f micro- farad, since virtually the surface of the condenser is now double that of either. Thus the current will be 44 - X 0-000000667 X 50 x 225 = 0-047 ampere But when in series the effect is the same as making the thick- ness of the dielectric twice as much while the surface remains the same. The capacity is then only -| microfarad, and thus the current will be a quarter of that in parallel, or 0*0117 ampere. CHAPTER XV. REGULATING RESISTANCES. IN order to produce desired variations in the manner of working of various types of electrical machinery, it is sometimes neces- sary to employ resistance coils connected in circuit with the machine to be regulated. For this and other purposes it may be necessary to provide resistances calculated to effect the following (1) To replace some device cut out of circuit; (2) To regulate and vary the speed of a motor. (i) The value of a resistance to replace a bank of lamps, an arc lamp, a motor, or other device which requires when working a P.D. of E volts, and takes a current of C amperes, is simply R = P ohms This resistance must, of course, be made of such shape, size, and material, that it will be able to carry the current for the length of time it is to be in circuit without overheating. Prac- tice varies considerably in this respect, the rise in temperature allowed ranging from quite a few degrees above the surrounding air to a temperature quite sufficient to discolour paper. No rule can be given for the size of wire or current density, as a great deal depends upon the spacing of the wires, or the degree to which the convolutions of the spirals are separated from one another. With a fair amount of spacing, however, a current ranging from \ to of that required to melt the wire employed according to the table of fusing currents determined by Mr. Preece, may be taken as reasonable. The materials usually Regulating Resistances. 175 employed are German silver, platinoid, iron, and manganin, the latter having the advantage of constancy of resistance in spite of variation of temperature. (2) The resistance employed to vary and regulate the speed of a motor is usually combined with a set of resistances and switch contacts to enable the motor to be gradually introduced into the circuit, so that excessive current through the motor on the one hand, and sudden demand on the generator on the other hand may be avoided. In all motors the tendency to rotate is due to a pull on the armature conductors by the poles of the magnets, the pull depending directly upon the strength of the current and the strength of the magnetic field. Thus, whatever be the magni- tude of the current in the armature, there will be little or no pull on its conductors if the strength of the magnetic field be small. It thus follows that in starting a motor the first thing to see to is production of a strong magnetic field. In the case of a shunt motor, the switch for introducing it into circuit is, therefore, arranged so as to, first of all, make circuit with the shunt coil ; then the armature is introduced into the circuit in series with a considerable resistance, in order that the current may be only a small fraction of that at full load. In this way the two essentials to rotation are produced without any sudden strain either on the motor or generator. In the event of there being no load on the pulley the motor now starts, and will soon reach a speed near the maximum possible ; but if there be a load on the pulley, the torque due to the small armature current may not be sufficient to produce rotation. In this case the switch has to be moved over several contacts, cutting out of circuit a resistance at each step till the current through the armature is sufficient to produce rotation. Regulation of speed is effected in either of two ways, or sometimes by both com- bined, viz. : (a) Reduction of the E.M.F. available for the armature by resistance. (b) Reduction of the strength of the magnetic field by intro- duction of resistance into the shunt circuit. (c) A combination of the two previous methods. 176 Examples in Electrical Engineering. Fig. 39 will illustrate this. The dynamo D has one terminal connected directly to the one brush of the motor M, to which is also connected one end of its shunt coil as usual. The other terminal of the dynamo is connected to a contact arm A, which may be brought into contact with either of the contact blocks i, 2, 3, 4, 5, 6, or 7, or may be off any of them, when circuit will then be broken. The other brush of the motor is disconnected from the other end of the shunt coil, and is connected to one end of the series of resistance coils placed between the contact blocks ; the shunt coil is connected to some part of the resistance coils, as shown at B. If now the contact arm be upon block marked a, there will be no circuit, SJvunt FIG. 39. but when contact is made with block i, the armature and shunt coil will then be in circuit, the armature current, however, being quite small as the coils between B and block i have a resis- tance which is large relatively to the armature, but not so large as to make much reduction in the shunt current. Thus there will be produced a considerable field strength, and also some armature current. The resistance now in circuit is such as to prevent the motor running at anything like full speed since the loss of pressure in the resistance coils will very considerably reduce the E.M.F. available for overcoming the full B.E.M.F. of rotation ; but on moving the arm A over to block 3 or 4, the resistance in series with the armature is considerably Regulating Resistances. 177 reduced, and the shunt is now directly connected to the dynamo, this position of the switch being that intended for the slowest speed of running, since the shunt is now strongest, and the armature has still considerable external resistance. It is from this point onwards that the double regulation comes in ; for as the arm A is moved onwards towards block 7, the shunt is being made gradually weaker, necessitating an increase of speed in the armature to give the required B.E.M.F., whilst at the same time there is less and less drop of pressure in external resistance till the arm A is on block 7, when the armature then has the full P.D. of the dynamo, and will, consequently, run fastest. In calculating the magnitude of the resistance coils to be inserted in a switch of the above type it is as well to bear in mind that the current taken by the motor depends mainly upon the demand made upon it at the pulley, and may reach the maximum value even at a slow speed when probably the demand for work may be large. Thus all the coils between blocks 4 and 7 may have to carry the full current, and must consequently be of such a size as to prevent overheating. The coils between contacts i and 4 must also be made of such size as will make them quite safe at any current they may have, the largest value of course being such as the full P.D. can produce through the whole external resistance with the arma- ture of the motor stationary ; these coils, though not intended generally to be left in circuit for any length of time, are frequently so left, and should therefore not be so slender as to risk fusion. As to the value in ohms, a good deal depends upon the range of regulation required, bearing in mind that in any case such regulation is very wasteful of power since any not used by the motor is dissipated in heating the coils. Let the value of the resistance between blocks i and 4 be called R ohms and that between 4 and 7 = r ohms, the resist- ance of the armature being R a and that of the shunt coil R 4 . For the present we will leave out of consideration the effect of the resistance in the shunt coil, and deal only with that in the armature. Calling the dynamo P.D. E volts and the maximum value N 178 Examples in Electrical Engineering. of the motor armature current C tt amperes, it is first of all necessary that R -f- r shall be so large that C = ^ . = the starting current may be less than C fl , or certainly not greater than it. Next we will suppose that the switch has reached the contact block 4, and that the motor is so loaded that it is taking a current of C amperes. Under these circumstances there is a loss of pressure in the external resistance and in the armature, neglect- ing the shunt current, of C(R tt + r) volts leaving E - C a (R rt -t* *) * i the speed being nearly proportional to the B.E.M.F. e. Thus, by making C a (R a -f r) any desired proportion of E, the speed can be thereby reduced in the same proportion. Next consider that the arm A has reached the block 7, then the value of e will be simply E - C a R tt and the speed will be normal. But with the switch connections as shown in Fig. 39, p. 176, the shunt coil has now a resistance inserted in it, the effect of which is to reduce the magnetizing force, and therefore the magnetic field, but not in the same proportion, since the magnetic circuit is composite, and conse- quently the various items of it are altering in relative impor- tance. Thus, whereas the air gap was, at full excitation, a certain large proportion of the whole circuit, it is now relatively much more important, since, due to reduction of magnetic density, the permeability of all the iron items has increased. The exact effect of such reduction of magnetizing force is not capable of direct calculation, since the law connecting j3 and /x is not easily stated in symbols; and, moreover, the greater effect of armature reactions at reduced excitation renders the problem a very difficult one. We can, however, form an approximate estimate of such effect by remembering that, initially, the reduction of lines will be always less than the Regulating Resistances. 179 reduction of magnetizing force. But, counting armature re- actions as more important at low inductions, we shall be very near the truth in assuming that the field falls off directly as the magnetizing force. Thus in the above case the magnetizing force is proportional to the shunt current, which is as a maximum and as a minimum r JL- w STF? If the speed be n at the point when the switch arm is on block 4, when the shunt gets the full P.D., and the armature has the resistance r in series, then this will be the slowest speed if the motor be so loaded as to take the full armature current C a . But when all the resistance is out of the armature circuit and inserted in the shunt, the speed will have increased from its former value to E - C a R R, ;/1 = E and whence a = < E - X F r p j. *', E-C.R. + -1 from which it will be seen that in any case the shunt effect is small. But in the simpler case, where the shunt coil is always at the full P.D. of the circuit, then E - CR n ~ E - C a (R rt + r) and r= 1 So Examples in Electrical Engineering. In all cases ;/ x stands for the maximum speed, and n for the slowest speed. The difference is not considerable between the two cases, the resistance r required to vary the speed from i to 2 being about 8 per cent, less when the shunt is connected as shown in Fig. 39, than when the shunt always has the full P.D. at its terminals. Example LYIII. A shunt motor having R, = 50 ohms, E = 100 volts, C a = 100 amperes, and R a = 0*02 ohm, is re- quired to have a speed ranging from 600 to 1200. Find the value of the regulating resistances required. (a) When the shunt always has the full P.D. 100 volts; (b) When the shunt has the P.D. at its terminals varied by the same resistance as in the armature circuit. Solution. =. 2, whence in case (a) the resistance to be inserted will be 2 TOO -f 2OO -j-4 I 06 ; = o's-? ohm 200 200 and in case (b) sooo 100 4000 r = - = - 5 = 0*485 ohm 100 2 4- 10000 10098 CHAPTER XVI. PRIME MOVER COSTS. IN most forms of heat engine the consumption of fuel may be divided up into two items, one to run the engine idle, and the other to supply the output. The latter is generally represented by a regular constant amount for every actual brake horse- power hour, whilst the former is also generally nearly a con- stant, bearing some relationship to the difference between full load consumption and idle consumption. The actual con- sumption for any output is of course the sum of these two. Calling B = the maximum brake H.P. of the engine; b consumption per actual B.H.P. hour; a = a factor such that a(bty gives the idle or no load con- sumption, we have the actual consumption as being b(aR + x) where x = the load on the brake in horse-power. The values of these constants will depend greatly upon the size of the engine, and may be taken to have the following values for steam and gas engines of the usual sizes employed in electric lighting. FOR STEAM ENGINES. FOR GAS ENGINES. a = 0*3 to 0*4. a 0*16 to 0*28. l> 17 to 3 Ibs. of coal per hour. b = 12 to 18 cubic feet of gas per hour, 1 82 Examples in Electrical Engineering. The smaller the engine in each case the higher the value to be taken. If, now, we call z pence the price per unit, either pound of coal or cubic foot of gas, of fuel, we can write the cost of one hour as being zb(a% -f x) And the cost of i B.T.U. will be found by putting for x the actual B.H.P. required, with the particular gearing and dynamo to give an output of i B.T.U. ; the cost of i B.T.U. thus varying from a somewhat large value at small load to lower values at full load. Example LIX. What will be the cost of fuel for u B.T.U. from a dynamo of 90 per cent, commercial efficiency, when run at full load of ioo a at 100", by a gas engine in which a = 0*2 and b = 15, the price of gas being 2S. $d. per 1000 cubic feet ? The gas engine has a maximum power of 15 B.H.P. Solution. The H.B.P. required will be 1000 100 746 XIOX ~ 9 o ='4 74 = * whence the cost of fuel per hour will be zb(afc -f x) where z is 27 1000 or penny = 7ooo X T5(o ' 2 X X 5 + I474 x 266-1 = 7 -185 pence 1000 >which is the cost of 10 B.T.U., whence the fuel cost per B.T.U. = 07185 pence. Example LX. If with the above engine the average load is only 3 B.T.U. when the commercial efficiency of the dynamo is only 80 per cent., find the fuel cost of i B.T.U. Prime Mover Costs. 183 Solution. Here the B.H.P. required is 1000 100 74 6*3X 8o - = 5-*3 = * and the cost of fuel per hour is 1000 X '5 (' 2X '5 +5 ''3) = 3-33 Pence which is the cost of 3 units or rn pence per B.T.U. Example LXI. Find the cost of running the above dynamo by a steam engine having the following particulars. a 0-4 and b = 2-5, B = 15 when the cost of coal is 0-043 pence per Ib. Solution. (i) on full load of 10 B.T.U. the B.H.P as before will be 14*74 whence cost of fuel will be 0-043 X 2-5 (0-4 x 15 + 1474) = 2-23 pence or only 0-223 pence per B.T.U. (2) But on load of 3 B.T.U, the B.H.P. required = 5-23 as before and the cost of i hour will be 0-043 X 2-5 (0-4 x 15 + 5' 2 3) = I<2 7 pence or only 0-402 pence per B.T.U. Example LXII. A large gas engine having a = o'i6, b - 12, and of 200 B.H.P., is used to run a dynamo which has a commercial efficiency of 50 per cent, at an output of 7 B/T.U. per hour, and 95 per cent, at 140 B.T.U. per hour. The price of gas is 2S. 6d. per 1000; find the cost of gas per B.T.U. at an output of 7 and 140 B.T.U. per hour. Solution. (i) At 7 B.T.U. per hour the horse-power re- quired is 7 X looo ioo _ ~746^ C 50 " whence the cost of gas per hour will be X 12(0-16 x 200 + 18-7) = 18-28 pence IOOO 184 Examples in Electrical Engineering. or 18-28 = 2-61 pence per i B.T.U. (2) At 140 B.T.U. horse-power required is 140 x 1000 100 746 " C "95" = and cost per hour is then 30 IOOO or X I2(o'i6 x 200 -f 198) = 82-8 pence o .o = o'59 pence per i B.T.U. 140 It will be seen from the foregoing examples that the cost of fuel will depend very much upon the extent to which the machinery can be loaded. Large engines and dynamos run to supply only a small fraction of their full power will be very wasteful, and so run up the cost of generation. To the cost of the fuel must also be added the cost of attendance, lubrication, etc., all of which items become still more costly as the load goes down. Thus, though at times of full load the cost of fuel may be but a fraction of a penny per B.T.U., the average cost is much increased by the, to a large extent, unavoidable waste at times of light load. The following example will illustrate this. Example LXIII. The average pressure generated in a central station is 430 volts, and the system is a five-wire net- work maintained at 400 volts between the outers. At full load there are 10,000 30- watt lamps in use. The dynamos have an average efficiency of conversion of 92 per cent. The station is on full load for 2 hours, on half load for 8 hours, on ^ load for 6 hours, and runs idle for the rest of the 24 hours. One mechanical brake H.P. costs 1*75 pence per hour for wages, coal, oil, etc., at full load; costs 2*5 pence at half load, and 6 pence at -^ load, whilst at no load the cost of running is Prime Mover Costs. 185 i$s. per hour. What should be the price charged per B.T.U., in order that there may be a profit of 10 per cent. ? Time hrs. Output Watts watts. generated. Brake H.P. required. Total cost for time in pence. B.T.U. paid for. 2 300,000 $$ x 300,000 = 3 22 >5oo W x w = 469*95 @ 175 pence per hour = 1644-8 6OO 8 150,000 ^ x 150,000 = 161,250 W x ifl^ = 2 34'97 @2'5^. perhr. = 4699'4 1200 6 30,000 3 2 5 2 5 46-995 @ 6d. per hr. = 1691-82 1 80 8 o @ 15*. hour = H40 O T<~ito1 ^^cf Tntnl unite 9475 pence j sold 1980 whence the cost per unit is ^P = 478 pence, and to pay 10 per cent, the price must be 5*258 pence. EXAMPLES. CHAPTER I. 1. The suspension of a moving-coil galvanometer is made of platinoid. It is 3| inches long, 0*0005 mcn thick, and 0*004 inch wide. If platinoid will be 20 times the resistance of copper, what is the resistance of this suspension ? 2. What relationship will the resistance of the above suspension have to the resistance of the coil, which is I inch wide and 2\ inches long, being wound with 400 turns of copper wire 0*002 inch diameter ? 3. What will be the E.M.F. required to send a current of 10 amperes through a circuit consisting of a dynamo of resistance of 7 ohms, i| mile of leads of No. 6 B.W.G., and 80 lamps each of 2^ ohms resistance? State the loss in the dynamo and in the leads, and give the horse-power of engine required, if the dynamo converts T 9 S of the power received into elec- tricity. Also if the lamps are paid for at so much per B.T.U. in the lamps, what will be the cost of I B.T.U. if a mechanical horse-power costs twopence per hour ? 4. What size of leads will be required to keep 400 6o-watt loo-volt lamps at a P.D. of 100 volts, if the dynamo is 170 feet from the lamps, has a resistance of o'Oig ohm, and produces a total E.M.F. of 106 volts? 5. How many secondary cells would be required to run the lamps in the above question when each cell has a resistance of 0*00007 ohm, and an E.M.F. of 2*05 volts at the start, but falling to 1-92 volt at the end of discharge ? The lamps are connected to the cells by the same leads as above. 6. In the above question the numbers of cells do not come out exactly, but as whole numbers have to be used, find what compensating resistances would be necessary to make the current exactly right. 7. A series dynamo has a resistance of 10 ohms, and sends a current of 12 amperes through 40 arc lamps taking 48 volts each, and equal to 1000 candle-power, all connected in series on a circuit of a total length of mile Examples. 1 87 of No. 9 B. W.G. What will be the horse-power of engine required if the dynamo has a frictional loss of f horse-power ? 8. What will be the horse-power of a dynamo to run 40 incandescent lamps in series each of 1000 candle-power, and taking 65 amperes each at 48 volts, if they are on a circuit \ mile long of No. 3 B.W.G., and the resistance of the dynamo is 1*25 ohm, with a friction loss of 3! horse- power? 9. A dynamo has an armature resistance of 0*042 ohm, and is required to run 300 i6-candle-power 5o-watt lamps at the end of leads which are ' OI 333 onm resistance. The total current is 150 amperes. What will be the reading of a voltmeter connected to the terminals of the dynamo, and what percentage is this P.D. of the whole generated E.M.F. ? 10. What will be the respective resistances of two bars, one of copper and one of aluminium, but both the same weight and the same length ? 11. A pair of conductors, I square inch sectional area, conveys 400 amperes to a distance of f of a mile from the dynamo. If the load at the far end requires a P.D. 410 volts, find the P.D. required at the dynamo. 12. In a secondary battery the lead lugs connecting cell to cell are \ x I inch in section, and 10 inches long each. If there are 54 cells joined in series, find the resistance of these connections. 13. If the cells in the above question have 15 plates each cell, each plate being 10 inches square and a quarter of an inch distant from the next, find the resistance of the electrolyte in each cell if the resistance of one cubic inch of the solution is 3 ohms. 14. Neglecting the E.M.F. of the cells in the above questions, find the E.M.F. required to send a current of 100 amperes through the mere resist- ances of the connections and electrolyte if all the cells are in series. 15. The connections on a switch-board are made of rod brass, and have an aggregate length of 27 feet. They are to carry 150 amperes, and the drop of pressure over all is not to exceed half a volt. Find the sectional area of the brass. 16. A standard manganin resistance of o'ooi ohm is 20 inches long. It is made of sheet metal T ' s inch thick. How wide will it be ? 17. An arc-light circuit consists of 60 lamps in series, each taking 10 amperes. Each lamp has the positive carbon 18 inches long and \ inch diameter, and the negative carbon 12 inches long and / 6 inch diameter. Neglecting the effect of^heating, find the loss of energy in the carbons alone. *\ 1 8. If the arc lamps in the above question be evenly spaced along a circuit at a distance of 30 yards apart, and the cable employed is T 7 g S. W.G., what will be the total E.M.F. required if each lamp takes 45 volts in addition to the drop in the carbons ? V 1 88 Examples in Electrical Engineering. 19. What will be the nearest cable of strands of S.W.G. copper to make leads to convey 600 amperes a distance of 30 yards so that there shall not be more than i '26 volt lost in the lead and return ? 20. A motor taking current of 100 amperes at a P.D. of 2000 volts is run from a dynamo 8 miles away. If the loss in the leads is not to be more than 10 per cent., what will be the nearest cable of S.W.G. wire, and what will have to be the P.D. at the dynamo? 21. If the power in the last question be transmitted at 20,000 volts, what will then be the size of the bare solid conductor employed with only 5 per cent, loss of pressure ? 22. Find the horse-power of engine required to run 1000 i6-candle- power lamps taking 100 volts and 0*6 ampere each. There is a loss in the leads of i volt, and the dynamo gives out T 9 5 of the power received by it. 23. If all the above lamps are replaced by arcs in series, taking 10 amperes and 50 volts each of 900 candle-power, so as to give twice the total light, and if the loss in the leads be 4 volts, and the dynamo gives out 85 per cent, of the power received by it, find the horse-power of engine required, and the number of lamps. 24. In the calibration of an ammeter to read 2000 amperes, what will be the smallest power required, if a number of secondary cells be used in parallel, each cell giving 2 volts ? 25. A street 4 miles long is to be supplied with 50 arc lamps per mile, each taking 10 amperes and 45 volts. The current is supplied by 4 dynamos, one to each mile of street, the leads being respectively i, 2, 3, and 4 miles, lead and return, of -^ square inch sectional area. If the mechanical efficiency of each dynamo be 90 per cent., and its resistance 10 ohms, find the E.M.F. of each, and the power required to drive each. 26. What size of leads will be required to send current to a number of 98-volt lamps from a dynamo giving 200 amperes at 100 volts ? 27. A current is required to be sent a distance of 50 yards to 400 lamps, taking o'6 ampere at 100 volts each, from a dynamo at P.D. of 101 volts. What will be the size of leads required ? 28. How many incandescent, and how many arc lamps, will be required to give together 20,000 candle-power for 20,000 watts? Each arc gives 2000 candle-power for 500 watts, and each incandescent 16 candle-power for 60 watts. CHAPTER II. 29. The distance between the feeding centres of a two-wire system of distributors is 400 yards, and current is required at the rate of I ampere per yard. What will be the size of the distributors if no two lamps are to Examples. 1 89 differ by more than 2 volts? And what will be the size of the feeders if the loss along them is not to exceed 10 volts, and the station is an average of 500 yards distant from any feeding centre ? 30. Two parallel mains running round a district are of \ square inch sectional area, and are connected to the station by feeders J square inch sectional area at points 400 yards apart. Each feeder has to convey 400 amperes to the mains, and there are 5 pairs, of lengths respectively of 100, 200, 300, 400, and 500 yards. If the P.D. at each feeding centre is main- tained at 204 volts, what will be the P.D. in the station at the ends of each pair of feeders ? 31. In the above circuit, what is the maximum difference in volts between any two lamps at full load ? 32. The feeding centres in a two-wire system are 500 yards aparU What must be the sectional area of the distributors, so that there may not be more than 2 volts difference between any two lamps if the full load is at the rate of I ampere per yard of street ? 33. The distributors in a street 1200 feet long are connected to feeders midway. The load is equal to one 48-watt loo-volt lamp per yard of street, and the feeding centre is maintained at 101 volts. Find the size of the distributors when the furthest lamps are at a P.D. of 99*5 volts at full load. 34. A system of distributors is \ square inch sectional area, and the load is one 5o-watt loo-volt lamp per yard of street. If the feeding centres are maintained at 101 volts, find the distance between them so that the drop in the distributors shall not be more than i| volt. 35. A street 503 yards long is supplied with current at the rate of I ampere per yard from a 200- volt circuit. What will be the size of leads if the dynamo is connected at the middle of the street, and the difference between any two lamps is not to be more than 3*5 volts ? Also, if the loss in the feeders is 7 volts, what will be the horse-power required at the dynamo terminals ? 36. In a five- wire system the outers are I square inch sectional area. What will be the maximum distance allowable between the feeding centres if no two lamps are to differ by more than \\ volt? There are an average of four o'6-ampere lamps in series per yard of street. 37. If the distance between the feeding centres be 400 yards on a three- 1 wire circuit running two 2OO-volt lamps in series per foot of street, each lamp taking 0*3 ampere, what must be the size of the outers if no two lamps are to differ by more than 2 volts ? 38. In a three-wire system running 200- volt lamps, two in series between the outers, the feeding centres are 500 yards apart and supply a maximum current of 500 amperes each. If the outers are square inch sectional area, Examples in Electrical Engineering. what is the P.D. between the terminals of the poorest lamp when the feeding centres are kept at 410 volts ? 39. A district is to be supplied with current at the rate of I ampere per yard of street, the system is two-wire with loo-volt lamps, and the feeding centres 600 yards apart. What will be the current supplied by each feeder, and what size must the distributors be in order that no two lamps may differ by more than I volt ? 40. If the feeders in the last question are 700 yards long on an average, and the loss along them is 10 volts, find the weight of copper per mile of street. 41. A district takes \ ampere per yard of street, using 2OO-volt lamps on a two-wire system. If the feeding centres are 600 yards apart, what current will they supply, and what will be the size of the distributors in order that no two lamps may differ by more than 2 volts ? 42. If the feeders in the above case be 700 yards long, find the weight of copper per mile of street with 20 volts lost in feeders. 43. A district is supplied with two loo-volt lamps in series on a three- wire system at the rate of \ ampere per yard of street. The middle wire is \ the section of the outers, and the feeders are 600 yards apart. If no two lamps are to differ more than I volt, find the size of the outers. Also find the weight of copper per mile of street, if there be a loss of 20 volts in the feeders, which have an average length of 700 yards. 44. If \ ampere per yard of street be supplied on a three- wire system with two 2OO-volt lamps in series, with feeders 600 yards apart, find the size of the distributors so that no two lamps shall differ more than 2 volts. And if 40 volts be lost in the feeders which are an average of 700 yards long, and the middle wire be \ the area of the outers, find the weight of copper per mile of street. 45. If the system be three-wire, with the outers i square inch sectional area, and the middle wire \ square inch area, the feeders being 600 yards apart, what may be the maximum load in order that no two lamps may differ by more than 2 volts? Each lamp is for 200 volts, and takes 0*3 ampere. 46. A current is required along a street 2000 feet long at the rate of I ampere per lo feet. Find the sectional area of conductors, so that there shall be only 2 volts difference between any two lamps. 1000 feet of copper i square inch section is O'OoS ohm . (a) When current is supplied at one end of the street. (b) When current is supplied midway. (City Guilds Examination.} Examples. 191 CHAPTER III. 47. A secondary battery of 54 cells has a resistance per cell of 0*0002 ohm and an E.M.F. of from 2 volts at start to 1-85 volt at the end of a run. What will be the current sent through a bank of lamps of 2 ohms resistance 70 yards away, connected by leads of 0-02 ohm resistance ? 48. Find the speed at start and end of charging for a dynamo to send 150 amperes through the above cells. The dynamo produces T \j volt per revolution per minute, is separately excited, and has armature resistance cro2i ohm. In charging each cell has B. E.M.F. ranging from 2*2 to 2*4 volts, and the battery is 20 yards away, and connected by leads of 0*0004 ohm per yard. 49. What resistance should be introduced into the circuit on starting to charge in order to keep the current 150 amperes, though the dynamo runs at the higher speed ? 50. Find the E.M.F. required to charge 108 cells in two parallel with 200 amperes each cell, if the B. E.M.F. is 2*3 volts each. Each cell has resistance 0*001 ohm, that of the dynamo being 0*005 ohm > and leads ox> I ohm. 51. Find the E.M.F. required to charge 57 cells in series, each having B. E.M.F. of 2*3 volts and 0*0005 onm resistance. The dynamo = 0*03 ohm, and leads 0^04 ohm, and current 100 amperes. 52. What will be the electrical efficiency of the above cells when dis- charging at 100 amperes through lamps on the above leads, each cell giving an average E.M.F. of 1*9 volt? Also state the P.D. at the lamp terminals. 53. If a dynamo gives 100 volts at 1000 revolutions, at what speed must it run to charge 54 cells at 50 amperes, each cell having B. E.M.F. 2*2 volts and 0*005 ohm internal resistance when the leads = 0*01 ohm ? 54. If the above cells are discharged at 50 amperes, the E.M.F. of each being 2 volts at start, falling to 1*8 volts at the end of run, what per- centage of the power put in is utilized ? CHAPTERS IV. TO IX, 55. What will be the brake horse-power of a motor series wound having armature resistance) 0^025 ohm, series coil 0*025 ohm, when running at 1 200 revolutions per minute with a current of 50 amperes supplied at an E.M.F. of loo volts? The electrical loss in the machine is \ of the friction losses. 56. In the above motor what will be the value of the total flux through 1 92 Examples in Electrical Engineering. the armature if there be 360 wires all round ? Also find the density in the air-gap, if the polar angle is 135 degrees, the bore is 22 cm. diameter, and length 20 cm. 57. In the above motor, what will be the pull in pounds on any one wire ? 58. A motor is supplied with 150 amperes at 100 volts. The magnets are shunt wound to 40 ohms, and the armature resistance O'OI ohm. If I horse-power is lost in various frictions, what will be the brake horse- power, and what the mechanical efficiency ? 59. If the above motor armature has 120 wires all round, find the total lines, and the density in the air-gap', if the polar angle is 140 degrees, the diameter and length of bore respectively 30 cm. and 25 cm., and the speed 1 200 per minute. 60. A machine separately excited gives 100 volts P.D. at 15 revolutions per second when run as a dynamo. The armature resistance is 0*015 ohm, the watts lost in eddy currents = 120, in hysteresis = 180, and in mechanical friction = 150. At what E.M.F. and current will it run, at the same speed as an unloaded motor ? 61. A magneto machine gives 50 volts as a dynamo at no load. The armature resistance is I ohm, and 125 watts are lost in various frictions. Find the E.M.F. and current required to run it as an unloaded motor at the same speed. 62. A shunt dynamo having armature resistance o'O2 ohm, shunt resist- ance 40 ohms, when run on open circuit at 20 revolutions per second, gives P.D. of 100 volts. If it be separately excited at 100 volts, and run as a motor at the same speed, to work a crane lifting 3 tons, at the rate of 30 feet per minute, what will be the current taken, and E.M.F. required, if the crane has a friction loss at the rate of 200 pounds per ton lifted, and the mechanical friction of the motor is equal to 150 watts, the hysteresis loss 130 watts, and eddy current loss 170 watts? Assume that there is no armature reaction. 63. What is the commercial efficiency of the above combined crane and motor ? 64. The E.M.F. produced by a separately excited machine when run as a dynamo is proportional to the speed, and is 100 volts at 1000 revolu- tions per minute. The armature resistance is 0*04 ohm. The loss due to hysteresis is proportional to the speed, and is 120 watts at 1000 per minute. The eddy current loss, however, is 80 watts at 1000 revolutions per minute, and is proportional to the square of the speed. The mechani- cal-friction loss varies directly as the speed, and is 1 10 watts at 1000 revo- lutions. If the above machine be supplied with current at 100 volts, at what speed will it run, and what current will it take (a) unloaded, (l>) loaded to 2 brake horse-power? Examples. 193 65. A machine has armature resistance of 0*005 ohm, and shunt resist- ance 25 ohms, and is separately excited at 100 volts. If run at 600 per minute as a dynamo, it gives 100 volts on open circuit, the eddy current loss being 250 watts, the mechanical friction 300 watts, and hysteresis loss 400 watts. When supplied at 100 volts, what will be the speed and current required to run as an unloaded motor ? 66. A series dynamo gives a current of 12 amperes, and is required to run a series motor of I brake horse-power. The dynamo has 360 wires all round, and a resistance of 1*3 ohm, with one million lines through the armature. The resistance of the motor is also 1*3 ohm, and \ horse-power is lost in various internal frictions. At what speed must the dynamo run, and what will be the P.D. at the motor terminals, the electrical efficiency of the dynamo, and the electrical and mechanical efficiency of the motor ? 67. A series motor produces a B.E.M.F. of 94 volts, and is supplied with loo amperes at a P.D. of 100 volts. If horse-power is lost in various frictions, what will be the brake horse-power, and the electrical and mechanical efficiency ? 68. A motor has an armature resistance of 3 ohms, with 3*6 million lines through it. The shunt coil is 750 ohms, and J a horse-power is lost in various frictions. If the armature has 720 wires all round, and the machine is supplied with 15 amperes at 550 volts, find the speed, the brake horse-power, and the electrical and mechanical efficiency ? 69. A series motor has 240 wires all round the armature, with a resist- ance of O'o8 ohm, and 22 million lines through it. The magnet winding is 0*12 ohm, and the machine is to give ipo brake horse-power when taking a current of 150 amperes. If n be the revolutions per second, and the mechanical friction loss is ioo#, the hysteresis loss Son, and the eddy current loss is 3 2 , find the E.M.F., the speed, and electrical and mechanical efficiency. 70. A series machine has 140 wires all round the armature, which is 222 '2 square centimetres sectional area, with a density of 18,000, and a speed of 1200 per minute. If the P.D. at the terminals for a current of loo amperes is 106 volts, what would be the electrical efficiency as a dynamo, and also as a motor at the same speed and current ? Also, if the friction loss is horse-power, in both cases, state the brake horse-power as a motor and mechanical efficiency, both as dynamo and motor. 71. A series dynamo, when giving a current of 10 amperes, has a P.D. at its terminals of 1000 volts at a speed of 1040 per minute. The armature resistance is 1*8 ohm and magnet resistance 2*2 ohms. At what speed will it run as a motor if supplied with 10 amperes at a P.D. of iooo volts ? The internal friction losses are horse-power. Also give electrical efficiency as a dynamo, the brake horse-power and electrical and me- chanical efficiency as a motor, OF THR CTNIVERSITY- 194 Examples in Electrical Engineering. 72. What would be the E.M.F. required to run the above motor at 1040 revolutions per minute ? and what would then be its brake horse- power and electrical and mechanical efficiency, the internal friction being 400 watts ? 73. An armature running at 20 revolutions per second has 96 wires all round, and 4 million lines through it; its resistance is I '05 ohm, and the shunt magnet coil 20 ohms. If 200 amperes be passed through the armature, what E.M.F. will be required to run it as a motor? Also if the hysteresis and eddy current losses are together equal to 300 watts, and a horse-power is lost in mechanical friction, what will be the brake horse- power, and the electrical and mechanical efficiency ? 74. A crane is to lift I ton at the rate of 500 feet per minute, and has a friction loss of 5 horse-power. It is run by a motor having 6 million lines through the armature, which has 400 wires all round, and a resistance of 0*05 ohm. There is a loss of f of a horse-power in mechanical friction, and 500 watts in iron losses. The magnets are series wound to o - o8 ohm. If it is supplied at 600 volts, what will be the current taken, the speed, the brake horse-power, the electrical and mechanical efficiency of the motor, and the plant efficiency ? 75. A motor with ring armature i6J inches mean diameter and 120 wires all round runs in a field of 3000 lines per square centimetre at 800 revolutions per minute. The length of the bore is 16^ inches, its diameter being 17 inches, and the curved length of the polar face 22 inches. The resistance of the armature is 0*024 ohm, the shunt coil being 48 ohms. Half a horse-power is lost in mechanical friction, and 400 watts in iron losses. If the armature be supplied with a current of 200 amperes, what will be the pull on one side of the belt on a pulley I foot mean diameter ? Also give the brake horse-power and the electrical and mechanical efficiency. 76. A mechanical contrivance requires 100 horse-power, and wastes 15 horse-power in friction. If run by a series motor having armature resist- ance 0*06 ohm, magnet coil 0*08 ohm, 360 wires all round the armature and 18 million lines through it, ij horse-power lost in friction, 600 watts in iron, and supplied with a P.D. of 700 volts, find the speed, the current taken, the electrical and mechanical efficiency of the motor, and the plant efficiency. 77. A compound dynamo to give P.D. of 105 volts and a current of 90 amperes is to have 92 per cent, electrical efficiency. If N be about 3 million lines, the speed noo per minute, and half the loss is to be in the armature, the rest in the magnets divided as f in the shunt and in the series coil, find the wires all round, and the resistances of the armature, shunt, and series coils. 78. A shunt motor has the shunt resistance 750 ohms, armature resist- ance 2 ohms, and N 4*2 million. If supplied with 15 amperes at 500 Examples. 195 volts, find the brake horse-power, the speed, and the electrical and me- chanical efficiency, if there be 720 wires all round the armature, and the friction losses are equivalent to 300 watts. 79. The magnetic circuit of a dynamo has reluctance = 0^0025, and an average of 5 million C.G.S. lines through it. Find the size of wire for a shunt winding, if the average length of one turn is 30 inches, and the P.D. at the terminals is 400 volts. 80. In the above question, what number of turns will be required on the shunt coil so that the energy waste shall not exceed 600 watts ? 81. A dynamo takes 8000 ampere turns, and produces a terminal P.D. of 100 volts. If the average length of one turn on the shunt coil is 2 feet, find the size of wire required. 82. An armature running at 18 revolutions per second, has 288 wires all round, and a resistance of 0*025 ohm. Find N so as to give 70 amperes at 100 volts. 83. The electrical efficiency of a dynamo is 98 per cent., and mechanical efficiency 93 per cent. If the output be 200 units, what power is wasted in resistance, and what power in mechanical and other frictions ? 84. A drum armature is 3 feet in diameter, with a lo-inch hole and 3*8 feet length of iron parallel to the shaft. It runs at 400 revolutions per minute, and is worked at such a density that the hysteresis loss is 5000 ergs per cubic centimetre per cycle. Find the hysteresis loss in watts. 85. If the resistance of the above armature be 0*005 ohm, the current 600 amperes at P.D. of 440 volts, shunt resistance no ohms, eddy current loss 2 horse-power, and mechanical friction 2j horse-power, find the mechanical efficiency. 86. A shunt dynamo has electrical efficiency 95 per cent., and gives an output of 20 units. If there is an equal loss in the shunt and in the armature, and the P.D. is 100 volts, find the resistances of the armature and shunt coils. 87. If the mechanical efficiency of the above dynamo is T 9 , what power must be wasted in other ways than in resistance ? 88. A shunt dynamo has magnet coil 33 ohms resistance, and armature o'oo) ohm, and gives current to 500 6o-watt loo-volt lamps, with a terminal P.D. of 101 volts. Find the electrical efficiency; and if the mechanical efficiency is 81 per cent., what horse-power of engine will be required, and what power is lost otherwise than electrically? 89. Two series machines, each having armature resistance 0*025 ohm, and magnet resistance 0*025, also N is 5 million lines, and the current 100 amperes. They are connected together by leads of 0*1 ohm resist- ance. Each machine has a frictional loss of 600 watts* which may be considered to be independent of the speed. If one machine be run as a 196 Examples in Electrical Engineering. dynamo by an engine of 15 brake horse-power, at what speed will they run, and what will be the brake horse-power of the motor when the current is 100 amperes ? Also state the ratio of the power given out by the motor to the power received by the dynamo. 90. In the last question, what brake horse-power engine would be required to run the dynamo if the motor be used so as to give out 15 brake horse-power, the current and losses being the same as before ? Also state the speeds. 91. A series motor is to give 50 brake horse -power at a speed of 600 revolutions per minute. Its armature has 720 wires all round, is f ohm resistance, and has 12 million lines through it. If the field coils have a resistance of 0*833 ohm, find the E.M.F. required to supply the above power through leads 2\ ohms resistance, the friction losses being 2 horse- power. 92. How many ampere-turns will be required on each limb on a Manchester-type machine having N = 4/2 X io 6 , A = 280, / 35, A g 1620, /,, 2-1, A; 587, l p 35, A m 294, /, 26, v - 1-5 ? The pole-pieces are of cast iron, and magnet cores wrought iron. 93. A shunt dynamo to give 310 volts has the length of the average turn on the magnet coils I metre, and is of the single-magnet type. It also has the following dimensions : l a 70, A a 1400, N 21*6 millions, l g 2*4, A,, 4000, /, 240, A m 2700. If v = I '4, and the magnets are wrought iron wound with 2000 turns of wire, what will be the shunt resistance at a temperature of 35 C. ? 94. A single-magnet bipolar dynamo has the pole-pieces of cast iron and yoke of wrought iron, on which is placed the magnet winding. The armature resistance is 0-024 olim hot, with 144 wires all round, and N = 7 million lines at no load. It is to give 200 amperes at a P.D. of 120 volts, and has the following dimensions: l a 40, A a 450, l g 2*5, A g 2300, IP 80, A^ 1300, /, 76, A m 700, with v I '43. Its shunt coil has 4000 turns, each of 45 inches average length, and the series coil is to be o'Oi ohm resistance hot. If the brushes are lead forward 6 wires on the armature at full load, find the windings of the series coil, the size of the shunt wire, and the electrical efficiency at full load. The shunt coil is connected "long shunt." 95. An armature, gramme wound, for a bipolar-field is required to carry 100 amperes at a P.D. of 100 volts. It has the following dimensions : N = 4 millions, n 20*43, w I2 *> eac ^ being 80 cms. average length and 0*24 sq. cms. sectional area. The iron stampings are 25 cms. diameter with 15-cm. hole, and make altogether 25 cms. length of iron parallel to the spindle. If the over-all surface is 3675 sq. cms., what will be the rise in temperature above the air on account of copper and hysteresis losses, neglecting the heating due to eddy currents ? Take i cc, of copper to be o '000002 ohm resistance. Examples. 197 96. A drum armature for bipolar-field is 25 cms. diameter with 7-cm. hole, and has 25 cms. length of iron parallel to the shaft. It runs at 20-43 revolutions per second in a field of 4-95 X io 6 lines, and is wound with 102 wires all round, each of 0*3 sq. cm. sectional area. Each convolution has an average length of 135 cms. If p = 0-000002, and there be the same rise in temperature as in the gramme armature above, find the maximum current allowable, neglecting the eddy current loss. 97. The magnetic system of the gramme armature in question 95 above has the following dimensions : 4 31, A a 250, l g 2, A^ 950, l m 170, A m 415, all the iron parts being charcoal iron. The leakage factor is I "4. Find the size of shunt wire required if the average turn is 3 feet long, and the P.D. at the terminals is 100 volts, the maximum temperature of the shunt coil being 35 C. 98. Find the size of shunt wire for a dynamo having the following dimensions : Armature ring wound, w = 252, l a 37, A a 137, diameter of core iof ins., bore n| ins., and length of 9 ins. Magnet limbs are of cast iron, 9 ins. X 7 ins. area and 6^ ins. apart, with no polar projections, l m 100. Speed 1000 per minute to produce 97 volts on open circuit. One turn on the shunt coil is 40 ins. long, and the coils rise to a temperature of4OC. Leakage factor I ^. 99. If the armature resistance in the above machine is 0*07 ohm, and the shunt coil 42 ohms, find the number of series turns to keep the P.D. of 97 volts at 70 amperes. Allow the series coil to be 0*04 ohm resistance, and the lead of the brushes 20 degrees. The connections are to be " long shunt." 100. A dynamo is to generate 118 volts, and has 17,080 lines per square centimetre in the armature, which has an area of 480 sq. cms., and runs at 720 revolutions. The average length of I coil on the (ring wound) armature is 46 ins., and the size of conductor 0*072 sq. in. If the machine be bipolar, and gives a current of 200 amperes, find the P.D. at the terminals at 35 C. 101. A dynamo has 200 wires all round the armature, which has a resistance of 0*03 ohm, an area of 250 sq. cms. at a density of 17,000, and is to run 200 half- ampere i6-candle-power lamps at the end of leads of i sq. in. area and 20 yards long. At what speed must she run to keep the P.D. at the lamps at 100 volts? 102. A shunt dynamo gives a terminal P.D. of 216 volts at a current of 50 amperes. Its armature has 7 million lines through it, and 160 wires all round, each coil being 4 feet long, and bipolar ring wound. It runs at 1 200 revolutions, and the shunt resistance is 108 ohms. Find the size of wire on the armature at a temperature of 35 C. 198 Examples in Electrical Engineering. CHAPTER X. 103. A voltmeter coil has a core inch diameter, and can be wound to a depth of \ inch. It is i'2 inch long, and requires about 300 ampere- turns to produce a deflection to the full extent of the scale. What size of wire will be required to make the instrument read 120 volts total, and what will be the energy wasted in the coil if wound with copper wire covered with silk to a radial depth of crooi inch ? 104. A certain type of instrument wound as an ammeter to read 20 amperes total has 40 turns of wire on its coil. The length of the mean turn is 6 inches, and the diameter of the wire 0*2 inch. What is the waste of energy at full reading ? 105. If the space for winding in the above instrument is 2 inches by \\ inch, find the size of wire and total waste of energy to make it read as a voltmeter to 20 volts with the coil wound with copper, and having an extra coil to increase the reading up to 120 volts wound with manganin. 106. How much will it cost a central station to keep an indication of 105 volts on each of two voltmeters for a year of 8000 hours, if the cost of a B.T.U. is 2f pence, and interest on the first cost of instrument is to be at the rate of 5 per cent, per annum ? (1) A magnetic instrument costing 10, and having a resistance of 2100 ohms. (2) A hot-wire instrument costing 7, and having a resistance when in use of 470 ohms. 107. An instrument is wound with copper wire, and has a resistance of 18557 ohms at o C. Its dial is divided into 120 divisions, and 37^5 milli-amperes gives a deflection of 75 divisions. What will be the constant of this instrument as a voltmeter at 20 C., and what resistance should the extra coil of manganin have to increase the range to 2400 volts ? Also state the T.V.R. with the extra coil in circuit if manganin is of constant resistance. 108. A voltmeter is wound with copper, and has a resistance of 2000 ohms at the temperature at which it was calibrated to read 120 volts total. It is first connected to the terminals of a dynamo in a hot engine-room, where its resistance rises to 2200 ohms. It is then taken outside, where the temperature is so low as to make its resistance only 1800 volts, and connected to the end of the circuit, but on the dynamo side of the load. The leads have a resistance of 0*02 ohm, and the lamps take loo amperes. If the true P.D. at the dynamo is 102 volts, what will the instrument read in the two cases ? 109. A voltmeter of 1500 ohms resistance is connected to the terminals Examples. 1 99 of a lamp. An ammeter connected in the main circuit reads O'6 ampere. If the voltmeter indicates 150 volts, what is the resistance of the lamp? no. An instrument reads 10 milli-amperes for 100 divisions deflection at 15 C, being wound with copper to a resistance of 100 ohms. What will it read as a voltmeter at 25 C., when in circuit with a manganin extra coil of 2900 ohms resistance ? in. A voltmeter to read 100 volts requires 1000 ampere-turns. If the average turn of wire be 3 inches long, find the size of wire required. 112. Find the diameter of copper wire for a voltmeter requiring from 300 to 400 ampere-turns, reading 60 volts total, if the mean diameter of the winding is f inch. 113. A voltmeter requires 750 ampere-turns, and reads 120 volts total. If the mean turn is 6 inches long, find the diameter of copper wire required. 1 14. If only one-fourth part of the resistance of the above instrument is to be upon the working coil, the rest being of manganin on an extra coil, find the diameter of the copper wire required for the instrument and the total resistance, if the waste is 8 watts. CHAPTER XL 115. A train weighs 120 tons, and is to go at the rate of 25 miles per hour. The resistance to motion on the level is 25 Ibs. per ton; the gearing is such that I ampere gives a pull of 10 Ibs. The line is divided into four stages, viz. level, up an incline of I in 480, up incline of I in 348, and down incline of I in 960. Find the current required for each stage, going and returning, and also state the horse-power. 116. The resistance to motion on the level is 30 Ibs. per ton, and the train weighs 100 tons. I ampere gives a tractive force of 20 Ibs. Find the current, the E.M.F., and horse-power required to run both ways on the line in the previous question. 117. If the E.M.F. of supply is 500 volts, find the current required to move a train of 200 tons weight at a speed of 50 miles per hour both on the level and up and down an incline of I in 200. The resistance to motion on the level is 15 Ibs. per ton. 1 1 8. In the last question find the total current to get up full speed in five minutes. 119. Find the current required to run a car of 12 tons weight at 10 miles per hour on the level and up and down inclines of I in 200 and I in 80, if the resistance to motion on the level is 25 Ibs. per ton, and the E.M.F. of supply is IQO volts, 2OO Examples in Electrical Engineering. 1 20. For the above car find the extra current required to get up full speed in half a minute. 121. A car weighing 10 tons runs up an incline of I in 100. The motor is geared to give a pull of 10 Ibs. per ampere. If the resistance to motion on the level be 30 Ibs. per ton, find the current required. Also give the horse-power for a speed of 4 miles per hour. 122. A tram-car, weighing 6 tons empty, starts empty, on a level line, picking up 20 persons halfway along. On reaching an incline up I in 50, 10 more persons get on. Fifteen persons get off when the car is going down I in 100. What will be the current required on the level both empty and loaded ; on incline up I in 50 loaded ; and down I in 100 loaded as above? The tractive force required is 25 Ibs. per ton on the level, and I ampere gives a pull of 8 Ibs. Also state the horse-power required for the various conditions for a uniform speed of 6 miles per hour. Take 20 persons to one ton. 123. A series dynamo has the following relationship between N in the armature and the current C : 10 12 13 14 16 18 N 0-90 1-5 1-85 2-i . 2-3 2-44 2-45 2-4}. 2-26 r8 where N is in millions of lines. The armature has 500 wires all round, and a resistance of 077 ohm. The pull exerted as a motor is 4XC X NX io~ 6 Ibs. If supplied with an E.M.F. of 300 volts, find the current required as a motor to haul a car 3 tons in weight on level, and up and down incline of I in 150. Also give the speed of the car and the horse- power required. The resistance to traction on the level is 25 Ibs. per ton. 124. A series dynamo of 0^15 ohm resistance is supplied with current at 440 volts, and has the following relationship between C and N : C 10 26 50 75 ioo 120 140 150 160 180 I 200 I 220 240 250 N I 4 | 8 | 12 | 14-8 | 167 | 177 | 18-1 | 18-2 | 18-1 | 17-5 | i6'6 | 15-3 | 13-4 | 12 the induction being in millions of lines. If the tractive force exerted by it is O'5CN X 10 ~ 6 Ibs., find the C required to run a train of 50 tons weight on level line, taking 20 Ibs. per ton ; the maximum speed in miles per hour, and the revolutions per second of the motor. 125. With the above motor and train, give the C required, the maximum speed of train and of motor to go up an incline of I in 200. 126. For the same motor and train, find the maximum speed of train, the C required, and the speed of the motor to go down an incline of f in 200. Examples. 201 127. Again, with the same train, find the shortest time in which the maximum speed can be attained on the level line. 128. A shunt motor, working at 100 volts, has 200 wires all round the armature and the following relationship between the current in the arma- ture C and the magnetic flux N : C N N being given in millions of lines. The armature resistance is O'I2 ohm, and that of the shunt 50 ohms. The pull exerted is O-8CN X io~ 6 Ibs. Find the motor speed, the current required, and the speed of the car to run on the level if the car weighs 4 tons and resistance to motion is 25 Ibs. per ton. 129. With the same motor and car, find the current required, speed of car and motor to go up incline of I in 200. 130. For what current will the speed of the above motor be slowest ? 6 10 20 30 4 So 60 70 80 86 2-485 2-4825 2*475 2*46 2*4375 2-4023 2-3525 2-2823 2-17 2-0075 CHAPTERS XII. TO XIV. 131. The maximum value of the alternating E.M.F. in a circuit is 2820 volts, and the resistance is 100 ohms, being non-inductive. What will be the average value of the current ; and if the terminals of the circuit be con- nected to an electrostatic voltmeter, what will it indicate ? 132. A transformer has a primary resistance 12 ohms, a secondary resistance 0*02 ohm. It is supplied with a virtual P.D. of 2010 volts at the primary. What will be the P.D. at the secondary terminals when running 50 lamps taking O'6 ampere each, the ratio of transformation being 20 to I, and the magnetizing current negligible ? 133. A circuit has 10 ohms resistance and a B. E.M.F. of 20 volts : what will be the current through it if connected to an alternating P.D. of 29 volts as indicated on a hot-wire voltmeter ? 134. What will be the horse-power supplied to a circuit having a B.E.M.F. of loo volts virtual, and a resistance of I ohm, when connected to a dynamo producing a maximum E.M.F. of 200 volts alternating? 135. A circuit has a resistance of 0-015 ohm > and takes a current of 50 amperes when connected to an alternating P.D. of 60 volts. Find the B.E.M.F. V 136. An alternating P.D. of 100 volts virtual is supplying a current of 12 amperes to two 40- volt lamps connected by a switch with two choking coils, in such a manner that when the switch is on contact No. I, the two lamps in series are in series with coil x ; but when the switch is on contact 2O2 Examples in Electrical Engineering. No. 2, only one lamp is in circuit, and is in series, with the two coils x and y in series. Find the B.E.M.F. which the coils x and y must have. 137. What must be the maximum magnetic flux in the cores of the coils x and y in the previous question, if they have each 200 convolutions of wire, and negligible resistance, there being 80 complete cycles per second ? 138. A transformer is to supply current to one hundred 6o-watt loo-volt lamps, and the secondary resistance is 0*107 ohm. The primary has 20 times as many convolutions as the secondary, and its resistance is 6 ohms. The magnetizing current is 0*8 ampere, and lags practically 90 after the primary E.M.F. What must be the P.D. at the primary terminals to give the lamps 100 volts? 139. A transformer is working off mains kept at a P.D. of 1015 volts. The resistance of its primary coil is 13 ohms, and that of the secondary o*li ohm. If the ratio of transformation is 10 to I, what will be the P.D. at the secondary terminals, when the secondary current is 20 amperes, and the primary magnetizing current is 0-5 ampere lagging 90 ? 140. With the above transformer, what should the primary terminal P.D. be to make the secondary P.D. exactly 100 volts ? 141. A transformer with 9-6 to I ratio of transformation is to give loo amperes at a secondary P.D. of 205 volts. The secondary resistance is 0*035 ohm, and that of the primary 3 ohms. It is connected to the station by leads having a total resistance of 4 ohms. Find the P.D. at the station end, at zero load, half-load, and full load, when the magne- tizing current may be neglected. 142. A transformer has an output of 10 units at full load, with a loss of o'l unit in the iron, and 0*1 unit in the copper. What will be the all-day efficiency if run on a load factor of? Hours 10 4 6 Load 143. A transformer has to convert from 1000 volts to 200 volts, giving a secondary current of 60 amperes. The primary resistance is 075 ohm, the secondary being o - O2i ohm. What will be the efficiency at g, , J, and full load, if there be an iron loss of 120 watts? 144. If the above transformer be used on a load factor thus Hours 16 * Load j o i what will be its all-day efficiency ? Examples. 203 145. The primary coil of a closed iron circuit transformer is to work at P.D. of 2050 volts. The secondary is to give a maximum of 50 amperes, at a P.D. of 101 volts. The secondary resistance is O'oiS ohm, and the primary 8 ohms. If the iron loss is 140 watts, what will be the efficiency of conversion at full load, 5, and ^ load ? And if employed on a load diagram as under, what will be the all-day efficiency ? Hours I i I I I i I 2 I 4 I 2 I 14 Load i 0-9 075 0-5 0-4 0-2 o 146. A transformer has a secondary current of 400 amperes, the resistance of the secondary 0*0015 ohm, and the primary 0*8 ohm, with 20 times as many turns as the secondary. If the iron loss is two-thirds of a horse-power, and the magnetizing current is 5 amperes lagging nearly 90 after the primary E.M.F., what will be the efficiency at full, , ^ and 5 ' load? 147. A number of open iron circuit transformers are supplied by a concentric cable 5 miles long, and having a capacity of 2 microfarads per mile. The total magnetizing current is 20 amperes lagging practically 90- The cable has a resistance of 4 ohms, and the P.D. at the transformers is to be 2020 volts. Find the P.D. in the station and the capacity of the shunting condenser required to reduce the current in the cable to practically zero. The dynamo is 140 CO 'per second. CHAPTER XV. 148. A series motor which requires 50 amperes at 400 volts has a resistance of O'8 ohm. It is to have a starting and regulating resistance such that the starting current cannot exceed 30 amperes, and the speed may range from 200 to 800. Find the values of the resistances required, and state the maximum power for each speed with a current of 50 amperes. 149. A shunt motor having R s 35, R 0*05, and running at 800 when supplied at 120 volts, is required to range in speed from 100 to 800. The full-load current is 100 amperes. What will be the value of the regulating resistance required, arranged so that the shunt is always at a P.D. of 120 volts ? Also, if there be a friction loss of 800 watts proportional to the speed, what will be the brake horse-power at slow and high speeds with full current in each case? 150. A shunt motor is to have a regulating resistance arranged so as to insert resistance in the shunt circuit as it is removed from the armature circuit. R s = 25, R a = O'O2, and the full-load current is 250 amperes at loo volts. The speed is to range from 200 to 600. Find the value of the resistance. 204 Examples in Electrical Engineering. CHAPTER XVI. 151. A gas-engine consumes when on full load three times as much gas as when running idle. At full load it gives 5 horse-power at the pulley, and runs a dynamo having R* 50 ohms and R rt 0*3 ohm, giving 100 volts. Find the average cost of I B.T.U. for a run of 12 hours, if on no load 2 hours, \ load 3 hours, % load 6 hours, and full load of 40 sixty-watt lamps for I hour. The dynamo has a friction loss of 300 watts, and each brake horse-power on the engine takes 18 cubic feet of gas costing 3 shillings per 1000. 152. A steam-engine has a no-load loss equivalent to T 3 5 its maximum brake power, which is 200 horse, for each of which the consumption of coal is i '8 Ib. per hour, costing 15 shillings per ton. The friction load attached to it is 20 horse-power, and useful load 25 amperes at 400 volts for 5 hours. Find the average cost of I B T.U. during the run. ANSWERS. 1. 23-45 ohms. 2. ^gR of coil. 3. 2089-6 volts, 7 volts in dynamo, 19-6 volts in leads, 31-12 H.P., 3-112 pence. 4- 0-4557 sq. in. 5- 53 or 54. 6. 0-000916 ohm at start, and 0-00555 ohm at finish. 7- 33-69 H.P. 8. 179 H.P. 9. 102 volts ; 94 per cent. 10. 2 to- I. JI - 435 '472 volts. 12. 0-018 ohm. 13. 0-000535 ohm. 14. 4-7 volts. 15. 0*272 sq. in. 16. 5-76 ins. 17. 1746 to 3492 watts. 1 8. 2894 or 37 volts. 19. f} S.W.G. 20. fi S.W.G. ; 2185-6 volts. 21. About No. 13 S.W.G. 22. 90-26 H.P. 23. 28-5 H.P., 36 lamps. 24. 5-3 H.P. 25. E.M.F. 2364, 2378, 2392, and 2406 volts; power 35-2, 35-4, 35-6, 35'8 H.P. 26. 0-804 sq. in. 27. 0-578 sq. in. 28. 8 arcs, 272 incandescents. 29. 0-4824 sq. in., 0*9648 sq. in. 30. 207-86, 211-72, 215-58, 219-44, 223-29 volts. 31. 3-86 volts. 32. 0-7545 sq. in. 206 Examples in Electrical Engineering. 33. 0-3087 sq. in. 34. 498 yds. 35. 0-432 sq. in., 138-7 H.P. 36. 1 290 yds. 37. 0-22 sq. in. 38. 202-73 volts. 39. 600 amps., 2*17 sq. ins. 40. 83-3 tons. 41. 300 amps., 0*5427 sq. in. 42. 20-7 tons. 43. 0-5427 sq. in., 21-93 tons. 44. 0*144 S< 1 i n -> 5'69 tons. 45. I -82 amps, per yard of street. 46. 1*6 sq. in., 0*4 sq. in. 47. 53'i amps, to 49-1 amps, at end of run. 48. 1259-7 and 1367-7 per min. at end. 49. 0-072 ohm. 50. 141 volts. 51. 140-95 volts. 52. 93*6 per cent., 101 '45 volts. 53. 1328 per min. 54. 77-2 per cent. 55. 6-03 H.P. 56. 2291 per sq. cm. 57. 0-2566 Ibs. 58. 18-47 H. P., 91-8 per cent. 59. 4,105,208; 4032. 60. 100-0675 volts ; 4-5 amps. 6 1. 56-25 volts and 2-5 amps. 62. 101*08 volts, 56*62 amps. 63. 79 per cent. 64. 998-76 per min. and 3-1 amps. ; 992*752 per min. and 18-119 amps. 65. 599715 and 9*5 amps. 66. 30*05 per sec. ; 90*18 volts, 83*3 per cent., 82*7 per cent., and 68 -9 per cent. 67. 12*1 H.P., 94 per cent., and 90*27 per cent. 68. 19-568 per sec., 9*18 H.P., 87*7 per cent., and 83*06 per cent. 69. 540*576 volts, 9*67 per sec., 94*4 per cent., and 92-0 per cent. 70. 94*64 per cent, dynamo, 94*91 per cent, motor; 14*3 H. P., 90-6 per cent, as dynamo and 90*6 per cent, as motor. 71. 960 per min., 96*15 per cent, as dynamo, 12*35 H.P., 96 per cent., and 92*2 per cent. 72. 1080 volts, 13*4 H.P., 96*29 per cent., and 92*59 per cent. 73. 79*8 volts, 19-68 H.P., 94*3 per cent., and 90*2 per cent. 74. 53-84 amps., 24*7 per sec., 38*93 H.P., 98*83 per cent., 89*9 per cent., and 78*3 per cent. Answers. 207 75- 367'35 Iks., 28-5 H.P., 94-6 per cent., and 91-3 per cent. 76. 10-56 per sec., 111-07 amps., 97-7 percent., 95-94 per cent., and 81-549 per cent. 77. 200 wires ; R, 39-8 ohms ; R rt 0-0467, R )M 0*0156 ohm. 78. 8-65 H.P., 15*6 per sec., 90*0 per cent., and 86'o per cent. 79. 0-00054 sq. in. area. 80. 6650. 81. 0-0419 in. diameter. 82. 1,962,770. 83. 5-49 H.P. in res., and 14-7 H.P. in frictions. 84. 2337. 85. 96-6 per cent. 86. 0*0125 ohm, and 18-9 ohms. 87. 1-56 H.P. 88. 96-3 per cent., 50*1 H.P., 7-88 H.P. 89. Dynamo 21*18 per sec., motor 17*18 per sec., 10*71 H.P., 71-4 per cent. 90. 19*29 H.P., dynamo 27*58 per sec., motor 23*58 per sec. 91. 1043*5 vol ts. 92. 7500. 93. 46 ohms. 94. Shunt wire 0*066 in. diameter, series 30 convolutions, 92*8 per cent. 95. 23*08 C. 96. 148*9 amps. 97. 0*057 in. diameter. 98. No. 15 S.W.G. 99. 96 convolutions. 100. 115*24 volts. 101. 730*2 per min. 102. 0*0081 sq. in. 103. 0*00224 in. diameter ; 2*14 watts. 104. 2*2 watts. 105. 0*0143 in. diameter, 11*5 watts. 106. (a) i nearly, (b) 2 los. od. 107. 38.000 ohms ; 0*0199 per cent. 108. 92*72 in station ; 1 1 I'll volts outside. 109. 300 ohms. no. 30*0388 volts total. in. 0*005 in. diameter. 112. 0*0031 in. diameter. 113. 0*0056 in. diameter. 114. 0*01 12 in. diameter ; 1800 ohms. 115. Level, 300 amps., 200 H.P. up I in 480, 356 amps., 237*3 H.P. ; up i in 384, 370 amps., 246*6 H.P. ; down 1 10960,272 amps., 181*3 H.P. ; up i in 960, 328 amps., 218*6 H.P. ; down I in 384, 230 amps., 153*3 H.P., down I in 480, 244 amps., 162*6 H.P. 208 Examples in Electrical Engineering. 116. Level 150 amps., 80 H.P. : up i in 480, 173-3 amps., 92-4 H.P. ; up I in 384, I79'i amps., 95*52 H.P. ; down I in 960, 138*3 amps., 7376 H.P. ; up i in 960, i6i'6 amps., 86'i6 H.P. ; down I in 384, 120*8 amps., 64-4 H.P. ; down I in 480, 126*6 amps. ; 67-52 H.P. The E.M.F. in all cases being 397*8 volts. 117. 596*8 amps., 1042 amps., 151-1 amps. 118. 1732 amps. 119. Level 59*68 amps., up i in 200, 86*53 amps., down i in 200, 32 8 amps., up I in 80, 126*5 amps., down i in 80, put on brake. 1 20. 82 amps. 121. 52-4 amps., 5-589 H.P. 122. Level, empty 1875 amps., loaded 21*875 amps., 2-4 H.P. and 2'8 H.P., up i in 50, 65-43 amps., and 8*376 H.P. ; down i in 100, 2*193 amps., and 0-28 H.P. 123. Level 8*7 amps., 1504 ft. per min., 3*41 H.P. ; up i in 150, 12*3 amps., 1317 ft. per min., 478 H.P. ; down i in 150, 2*7 amps., 1426 ft. per min., and 1*077 H.P. 124. 115 amps. ; 24-4 miles per hour, 6*71 revolutions per sec. 125. 177 amps., 23-56 miles per hour, 6*5 revolutions per sec. 126. 64 amps., 31*45 miles per hour, 8*69 revolutions per sec. 127. 3*05 mins. 128. 54*2 amps., 2165 ft. per min., 19-6 revolutions per sec. 129. 77 amps., 2273 ft - P er min., 20-53 revolutions per sec. 130. About 42 amps. 131. 17*76 amps., 1993 volts. 132. 99 volts. 133. 2-1 amps. 134. 13-4 H.P. 135. 60 volts. 136. x 60 volts, y 31*65 volts. 137. * 84,375, .727,630. 138. 2020*5 VOltS. 139. 99-24 volts. 140. IO22*2 VOltS. 141. 1968, 2021-55 and 2075*1 volts. 142. 95*51 per cent. 143. 92-4, 95-7, 97'3 and 97'5 P er cent - 144. 93*6 per cent. H5- 95'5 93'9> 7 8>0 and all-day 86 -7 per cent. H 6 - 97'3 8 > 94'7 88-41, and 79-36 per cent. 147. 2O2O volts, 1*25 microfarad. 148. I3!j ohms starting, 5-4 ohms regulating, 6 H.P. slow, and 24*1 H.P. full speed. 149. 1-9 H.P. and 14*3 H.P. 150. 0*252 ohm. 151. 2*21 pence. 152. i -35 pence. ( 209 ) SPECIFIC RESISTANCE OF VARIOUS MATERIALS AT o C., WITH TEMPERATURE VARIATIONS. Temperature B 5,c Material. Ohms per cubic inch. Ohms per cubic centimetre. variation of resistance per cent, per degree S|.s SPli.S U 3,0 Centigrade. > 3 "* O.O Copper 0*00000063 0-0000016 0*388 increase 0.320 Iron 0-00000405 O-OOOOIO4 Q'453 0-280 Aluminium O-OOOOOII4 0-0000029 0*390 0-092 Mercury 0-00003666 0-0000943 0*072 0*489 Lead 0-00000780 O'OOOO2OO 0*387 0*410 Platinum ... 0-00000353 0-0000089 0*247 0-828 Platinoid 0-000014 0-000034 0*021 German silver O-OOOOO82 O'OOOO2I 0*044 0*30 Manganin O-OOOOlS 0-000044 0*002 decrease Brass O-OOOOO28 O-OOOOO72 0*290 Platinum-silver ... 0*0000095 0-000024 0*031 increase Carbon (arc lamp) j 0-0017 to 0*0034 0*0044 to 0-0086 >o*o52 decrease Sulphuric acid solution i* i sp. g. 3-33 8'45 2-2 I'2S P .g. 2-62 6*66 2-2 VALUES OF THE SPECIFIC RESISTANCE OF COMMERCIAL COPPER AT VARIOUS TEMPERATURES. Temperature degrees. Resistance of one cubic Resistance of one cubic Cent. Fahr. centimetre in ohms. inch in ohms. 32 0*00000l6o 0-00000063 5 41 0-OOOOOI63 0-00000064 10 5 0-OOOOOI67 0*00000065 15 O*COOOOl69 0-00000067 20 68 o 00000172 o 00000068 25 77 0-OOOOOI76 0*00000069 30 86 O'OOOOOI79 0*0000007I 35 95 0*00000l82 O*OOOOOO72 40 104 0-00000186 0*00000073 45 H3 0-OOOOOI89 0-00000074 5o 122 0-00000193 O*OOOOOO76 55 131 o 00000196 0*00000077 60 140 0*00000200 0-00000078 65 H9 O*OOOOO2O3 O-OOOOO079 70 158 0-00000207 O'OOOOOOSl 75 I6 7 O*OOOOO2IO O-OOOOOO82 80 176 0*00000214 0-00000084 85 185 0*00000217 0*00000085 90 I 94 O*OOOOO22I 0*00000087 95 203 0*00000224 o*oooooo88 100 212 0-00000228 0*00000090 2io Examples in Electrical Engineering. VALUES OF H CORRESPONDING TO VARIOUS VALUES OF ft FOR CYCLIC CHANGES IN ft AS A SINE-FUNCTION OF THE TIME, AND HAVING MAXIMA OF 1000, 2000, etc., to 7000. The values are given for each hundred ft for a complete half-period. Values for the other half-period can be easily put in by reversing the signs. The hysteresis loss h in ergs per cc. per cycle corresponding with these values of the cyclic j8H curve are given in the table of values of j8 and H on pp. 214-218. H ft H ft H ft H 0'49 600 072 900 0-61 400 0'20 100 0'55 700 075 800 0-42 300 -0-28 2OO 300 0'59 0-62 800 900 0-80 700 600 0-23 0*07 200 100 0^36 0-42 400 0-66 IOOO 0-83 500 0*07 -0'49 500 0-69 o 0*62 IIOO 0-88 1900 0-90 ! 900 - 0-28 100 0-65 1200 0-90 1800 072 800 -0'34 200 0*67 1300 0*92 1700 0-55 700 -0-38 300 0-69 1400 0*95 1600 0-40 600 -0-42 400 072 1500 0-97 1500 O'26 500 0-46 500 074 1600 I '00 1400 0-16 400 O*5O 6OO 076 1700 1-03 1300 0*05 300 - 0'53 700 078 1800 1-05 1200 0-04 200 - 0-56 800 0-80 I9OO I -08 IIOO 0-14 IOO - 0'59 900 0-83 2000 ri2 IOOO -0-21 0'62 1000 0-85 0-83 IOOD 09 29OO I -06 1400 -0-45 100 0-85 1700 ii 2800 0-80 I3OO -0-49 2OO 0-87 1800 13 2700 o'6i 1200 -o*S3 300 0-88 1900 '14 26CO o - 46 IIOO -0-56 400 0-90 2OOO 16 2500 0'33 IOOO o'6o 500 0-91 0-92 2100 2200 18 20 2400 2300 0-21 O'll 900 800 -0-63 -0-66 700 0-94 23OO 22 2200 O'O 700 -0-68 800 0-96 2400 24 2IOO -0-08 600 -071 900 1000 o'97 0-99 25OO 2600 26 27 2000 I9OO -0-15 0*22 500 400 -074 076 IIOO I -01 2700 2 9 I800 -0-27 300 -079 1200 I '02 2800 'SI 1700 -0-32 200 -0-81 1300 1-04 2900 33 1600 -0-37 IOO -0-82 1400 I -06 3000 '35 1500 0-41 -0-83 1500 1-07 Cyclic fiH Values. 21 I ft H ft H ft o'gi 2100 20 3900 100 0-92 2200 22 3800 200 0-94 2300 2 3 3700 3 00 0-95 2400 ' 2 5 3600 4OO 0*96 25OO 26 35oo 500 0-98 2600 28 3400 600 0-99 2700 30 3300 700 I '00 2800 31 3200 800 I -01 2900 '33 3100 9OO I'O2 3OOO '35 3000 1000 1-03 3100 '37 2900 IIOO 1-05 3200 38 2800 I2OO 1-07 330 40 2700 1300 1-09 3400 42 2600 1400 ITO 350 '44 2500 1500 I'll 3600 46 2400 1600 I'M 3700 48 2300 I7OO 1-14 3800 "So 2 2OO I800 1-15 3900 '53 2100 1900 1-17 4000 56 2000 2OOO 1-18 H ft H I'25 1900 -0-55 I -00 1800 - 0-58 0-80 1700 0*60 o - 65 1600 0-62 0-52 1500 0-65 1400 0*67 0-27 1300 070 0-17 1200 - 072 0-08 IIOO -074 O'OO 1000 076 - 0-07 900 - 078 - 0-14 800 - 079 - 0'20 700 - 0-81 - 0-25 600 -0-83 - 0-31 500 - 0-84 - 0-36 400 - 0-85 - 0*40 300 -0-86 - O'44 200 -0-88 - 0-47 100 0-90 - 0-51 -0-91 0-98 2600 1-27 4900 i'34 2400 -0-68 0'99 2700 1-29 4800 i -08 2300 070 '00 2800 I' 3 I 4700 0-90 2200 072 oi 2900 I'33 4600 074 2IOO -074 'O2 3OOO I'35 4500 0-60 2000 -075 '02 I 3IOO I'37 4400 0*46 1900 - 077 03 | 3200 I- 3 8 4300 o'34 I800 -078 04 3300 I-40 4200 O'22 1700 -0-80 'OS 3400 I- 4 2 4100 OT2 I6OO -0-81 06 ! 3500 I'44 4000 O'OO 1500 -0-82 07 3600 1*46 3900 -0-08 1400 -0-83 08 3700 I- 4 8 3800 -0-17 1300 - 0-84 TO 3800 r50 3700 -0-24 1200 -0-85 Tl 3900 i'S3 3600 0-30 IIOO -0-86 T2 4000 i-*55 35oo -0-36 1000 -0-88 *3 4IOO i'57 3400 - 0-40 9OO -0-89 'H 4200 i'59 3300 -0-44 800 - 0-90 "IS 4300 r6i 3200 -0-47 700 0-91 T7 4400 1-63 3100 o'5o 600 0-92 18 4500 1-65 3000 -0-54 500 -0-93 '20 4600 1-67 2900 -0'57 400 -0-94 21 23 4700 4800 170 172 2800 2700 -0-59 O'62 300 200 -0-95 0-96 2 4 % 4900 5000 '"74 176 2600 2500 0*64 -0-66 100 -0-97 - 0-98 212 Examples in Electrical Engineering. CYCLIC )8H VALUES. ft H ft H ft H H I -06 3100 1-42 59oo I-72 2900 -0-75 100 1-07 3200 ; i'43 5800 1-40 2800 076 200 08 3300 i'45 5700 1-16 2700 -077 300 09 3400 1-46 5600 0-94 26OO -079 400 10 3500 1-48 55oo 0-78 2500 -0'80 500 | II 3600 1-50 5400 j 0*62 2400 -0-81 600 12 3700 ! 1-52 5300 ! 0-47 2300 -0-82 700 '13 3800 i'54 5200 0-37 2200 -0-84 800 14 3900 1 1-56 5100 0-27 2IOO -0-85 coo 15 4000 1-58 5000 0-17 2OOO -0-86 1000 1-16 4100 i '60 4900 0-08 1900 -0-87 IIOO 1-17 4200 1-62 4800 O'OO 1800 -0-88 1200 1-18 4300 1-64 4700 - O'lO 1700 -0-89 1300 1-20 4400 1-66 4600 -0-17 I6OO 0*90 14OO 1*21 4500 1-68 4500 0-24 I5OO 0-91 1500 I '22 4600 170 4400 -0-28 1400 - 0-92 1600 1-23 4700 1-72 4300 -0-33 1300 -o'93 I7OO I'24 4800 1-74 4200 -0-37 1200 -0-94 I800 I-2 5 4900 1-76 4100 0-42 IIOO -o'95 1900 1-26 5000 1-78 4000 0-46 IOOO 0-96 2000 I*27 5100 i -80 3900 -0-49 900 -0-97 2100 1-28 5200 1*3 3800 -0-52 800 -0-98 22OO 1-29 5300 1-85 3700 0-56 700 -0-99 2300 r30 5400 1-88 3600 -0-59 600 O'lOO 2400 I-3I 5500 1-91 1 3500 ; 0-62 500 O'lOI 2500 I-32 5600 2-00 34 oo 0-64 400 0-102 2600 i'34 5700 2'10 3300 0-67 303 O-IO3 2700 i '35 5800 2-20 3200 0-69 2OO O-IO4 2800 1-36 5900 2-3C 3100 -0-72 100 - 0-105 2900 1-38 6000 2-40 3000 -0-74 o o*io6 3000 i '40 i I"I2O 1800 I'296 3600 536 5400 1-890 100 I-I28 1900 I-308 ; 3700 552 5500 1-910 2OO 1-136 2OOO I-320 i 3800 568 5600 1-940 3 00 I-I44 2100 I'332 3900 584 5700 1-960 4 00 I-I52 22OO i '344 4000 600 5800 1-980 500 600 1*160 1-170 2300 1 2400 & 4100 4200 620 640 5900 6000 2-010 2*050 700 1-180 ! 2500 1-380 4300 660 6100 2-080 800 1-190 2600 i '394 4400 680 6200 2'IIO 900 1-200 ! 27OO 1-408 4500 700 6300 2^140 IOOO I'2IO 2800 1-422 4600 718 6400 2- 1 80 IIOO 1-220 2900 1-436 4700 736 6500 2"2IO I2OO I-230 3000 i '450 4800 754 6600 2-250 1300 I-24O 3100 1-464 4900 772 6700 2*290 1400 I-250 32OO 1-478 5000 790 6800 2*340 1500 I-26O 3300 1-492 5100 810 6900 2-380 1600 I-272 3400 1-506 5200 830 7000 2-440 1700 1-284 3500 1-520 5300 860 6900 1-960 Cyclic ftH Values. 213 ft H ft H ft H ft H 6800 i-6co 5000 0-460 33 - 0-830 1600 - 0-972 6700 1-380 4900 ~ 0-49 2 3200 0-840 1500 0-980 6600 1-190 4800 -0-524 ; 3100 0-850 1400 - 0-988 6500 I'OOO 4700 - 0-556 1 3000 - 0-860 1300 - 0-996 6400 0-840 4600 -0-588 2900 - 0-870 1200 "004 6300 0-720 4500 0-620 2800 - 0-880 I IOO -012 6200 0-590 4400 - 0-644 2700 - 0-890 1000 -O2O 6100 0-440 4300 - 0-668 ' 2600 0-900 900 - -030 6000 0-320 4200 0-692 2500 0-910 800 -040 5900 O'2OO 4100 - 0-716 2400 0-918 700 '050 5800 O'lIO 3000 0-740 2300 0-926 600 '060 5700 o-ooo 3900 -0752 2 2OO - 0'934 500 'O7O 5600 O'lOO 3800 -0-764 2100 0-942 400 - -080 55 0-200 3700 - 0-776 2 COO - 0-950 300 1*090 5400 0-252 3600 -0-780 I9OO - 0-956 200 I 'IOC 53 - 0-304 3500 1 0-810 I800 0-962 I ioo I'lIO 5200 - 0-356 3400 0-820 1700 -0-968 ! o I'I20 5100 - 0-403 1 2t4 Examples in Electrical Engineering. ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES. ft . Density in c.g.s. lines per sq. cm. area. H in c.g.s. per cm. length. ft Ampere-turns per cm. length. M Permea- bility. Hysteresis loss in ergs per cc. per cycle. &P Product of density and ampere-turns per cm. length. 100 0-17 0-I40 5 88 5 14 200 0-32 0-260 625 10 52 3 00 0-44 0-356 685 17 107 400 0*54 0-432 740 25 172 500 600 0-60 0-65 0-476 0-520 $ 37 5 238 3 I2 700 800 0-69 0-76 O*552 0-584 1014 1052 65 80 386 467 900 0-78 0-624 H54 97 5 6l 1000 0-83 0-664 1204 115 664 1 100 0*87 0-696 1264 134 765 I2OO 0-91 0-728 1318 155 873 1300 0-94 0-752 1382 177 978 1400 0-97 0776 1443 200 1086 1500 0-99 0-796 I 5 I 5 222 1194 1600 02 0-816 1568 245 1305 1700 04 0-836 1634 270 1421 I800 07 0-856 1682 295 1540 1900 09 0-872 1743 32O 1656 2000 II 0-888 1802 345 1776 2IOO ' 1 3 0-904 1858 372 1898 2200 15 0-922 1913 400 2024 2300 17 0-940 1966 430 2l62 2400 19 0-958 2117 460 2304 2500 '22 0-976 2049 490 2440 2600 2 4 0-992 2096 520 2579 2700 26 I '01 2142 550 2727 2800 2 9 1-03 2170 580 2889 2900 '3 1 1-05 2214 610 3045 3000 '34 1-07 2238 640 3216 3100 36 i '09 2279 670 3379 32OO 38 10 2318 700 3520 3300 40 '12 2357 735 3696 3400 42 14 2394 770 3862 3500 '44 16 2430 805 4065 3600 '47 18 2448 840 4233 3700 '49 20 2486 877 4440 3800 'Si 21 2516 915 4590 3900 '53 '22 2548 952 4758 Annealed Charcoal Iron. 215 ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES. /? Density in c.g.s. lines per sq. cm. area. H in c.g.s. per cm. length. /ft Ampere-turns per cm. length. M Permea- bility. h Hysteresis loss in ergs per cc. per cycle. Product of density and ampere-turns per cm. length. 4000 i'SS 24 2580 990 4960 4100 i'57 26 26ll IO22 5166 42OO i'59 27 2641 1055 5346 4300 1-61 29 2670 1092 55 6 7 4400 1-63 3 2699 1130 578i 4500 1-65 32 2726 1170 5940 4600 4700 1-67 1-69 4 2780 1210 1252 6H5 6392 4800 172 3 8 2790 1295 6604 4900 174 '39 28l6 1337 6811 5000 176 '41 2841 1380 7040 5100 178 42 2865 1420 7242 5200 i -80 '44 2889 1460 7488 5300 1-82 46 2912 1505 7738 5400 1-85 48 2918 1550 7992 5500 T RR 50 2 9 2S 1590 8250 5600 1-91 '53 2931 1630 8556 5700 1-94 '55 2 93 8 1677 8835 5800 i '97 58 2944 1725 9140 5900 2 '00 60 2950 1772 9445 6000 2-03 62 2955 1820 9744 6lOO 2'06 65 2961 1867 10065 6200 2'10 68 2955 1915 10416 6300 2'13 71 2952 1962 10773 6400 2-17 74 2949 2OIO 1 1 1 10 6500 2*21 77 2941 2060 ii55 66OO 2-25 80 2938 2110 11880 6700 6800 2-28 2'32 11 2933 2930 2l6o 2210 12261 12662 6900 2- 3 6 89 2923 226O 13041 7000 2-4I i '93 2310 13496 7100 2-46 1-97 2886 2362 13987 7200 2-5I 2-01 2868 2415 14457 7300 2- 5 6 2-05 2851 2470 H965 7400 2-62 2'10 2824 2S 2 5 ^S 10 7500 2-6 7 2-14 2808 2587 16050 7600 273 2-18 2783 2650 16598 7700 278 2-23 2769 2705 17171 7800 2-84 2-27 2746 2760 17721 7900 2-90 232 2723 2810 18328 Sooo 2- 9 6 2'37 2703 2860 18944 8100 3'01 2-4I 2691 2939 19521 8200 3'06 2'45 2679 301.8 20090 8300 3'12 2-50 2660 3097 20750 8400 3'i8 2'5S 2641 3176 21420 216 Examples in Electrical Engineering. ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES. Density in c.g.s. lines per sq. cm. area. H in c.g s. per cm. length. Ampere-turns per cm. length. M Permea- bility. h Hysteresis loss in ergs per cc. per cycle. Product of density and ampere-turns per cm. length. 8500 3' 2 5 2-60 2615 3255 22100 8600 3*3* 2*65 2598 3334 22790 8700 3*37 270 2581 3413 23490 8800 3-43 275 2565 3492 24200 8900 3-50 2'80 2542 357i 24720 9OCO 3-56 2-85 2528 25650 9100 3-62 2-90 2513 3730 26390 9200 3-68 2'95 2500 3810 27140 9300 376 3'01 2473 3890 27993 9400 3-85 3'08 2441 3970 28952 9500 3-92 3-14 2423 4050 29830 9600 4-00 3 -20 2400 4130 30730 9700 4-08 2377 4210 9800 4-16 3'35 2355 4290 32830 9900 4-27 3 '42 2318 4370 33858 IOOCO 4'37 3-50 2288 445 35000 IOIOO 4-48 3'59 2254 4535 36259 10200 4*60 3-68 2217 4620 37536 10300 471 377 2187 4705 38831 IO4OO 4-82 3-86 2157 4790 40147 10500 4-96 3-96 2117 4875 41580 10600 5'7 4-06 2090 4960 43036 IO7OO 5'2O 4-16 2057 5045 445 i 2 I0800 5'33 4-27 2026 46116 10900 4'37 1995 5215 47633 IIOOO 5-60 4-48 1963 5300 49280 1 1 100 573 4'59 1937 5390 50949 II2OO 5-87 470 1908 5480 52640 II3OO 6-03 4-83 1873 5570 54579 II400 6'2O 4-96 1838 5650 56544 JI500 6 '40 5-12 1797 5750 58880 II600 6-60 5-28 1757 5840 61248 II700 6-85 1707 5930 64116 IISOO 7-10 5-68 1662 6020 67024 II900 7-36 5'88 1617 6110 69972 I2OOO 7'6o 6-08 1579 6200 72960 I2IOO 7-90 6-32 1531 6300 76472 12200 8-20 6-56 I 4 8 7 6400 80032 12300 8-50 6-80 1437 6500 83640 12400 8-90 7-12 1393 6600 88288 12500 9'2O I35 1 6700 92000 12600 12700 9-60 I0'0 8-00 1312 1270 6800 6900 96768 101600 12800 10-4 8-32 1231 7000 106496 12900 10-8 8-64 1192 7100 111456 Annealed Charcoal Iron. 217 ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES. a Density in c.g.s. lines per sq. cm. area. H in c.g.s. per cm. length. Ampere-turns per c.m. length. Permea- bility. h Hysteresis loss in ergs per cc. per cycle. ft/ft Product of density and ampere-turns per cm. length. I3OOO 11-3 9-04 II 5 I 72OO II7520 I3IOO 11-7 9-36 III5 7300 I226l6 I32OO I2'2 976 1082 7400 128820 13300 I2'7 lO'I 1047 7500 134330 13400 I3-2 ID'S 1015 7600 140700 I35 137 I0'9 986 7700 I47I50 13600 I4-2 1 1 -3 958 7800 153680 13700 14-8 11-8 926 7900 I6I600 13800 15-4 12-3 8 9 6 8000 169740 13900 16-1 12-8 86 3 8lOO 177920 14000 16-9 13-5 839 8200 189000 I4IOO I7'8 14-2 793 8310 2OO22O I42OO 187 14-9 759 8420 211580 14300 197 157 725 8530 224510 14400 20-7 16-5 695 8640 237600 14500 14600 21-8 23-0 18-4 665 636 8750 8860 252300 278640 14700 24-2 19*3 608 8970 283710 14800 25-5 20-4 580 9080 301920 14900 26-8 21-4 555 9190 318860 15000 28-6 22-8 524 9300 340000 I5IOO 30-0 24-0 503 9410 362400 15200 31-5 25-2 482 9520 383040 15300 33' i 26-4 461 9630 403920 15400 35'o 28-0 440 9740 431200 15500 37'i 29-6 419 9850 458800 15600 39*2 3 J '3 398 9960 488280 15700 41*5 33*2 379 10070 521240 15800 43 '8 361 I0l8o 553000 15900 46-8 37'4 34 1 IO29O 594660 16000 48-8 39-0 321 10400 624000 I6IOO 53*2 42-5 303 I05IO 684250 16200 567 45'3 285 10620 733860 16300 60-7 48-5 270 10730 790550 16400 65-0 52-0 256 10840 852800 16500 69-9 55'9 238 10950 922340 16600 60- 1 221 Ilo6o 997660 16700 16800 87-0 64*5 69-6 207 193 III70 II280 I077I50 1169280 16900 93-0 74*4 II390 1257360 I7OOO lOO'O 80-0 170 II500 1360000 I7IOO 107-0 85-5 159 Il62O 1462050 17200 115-0 92-0 149 II740 1582400 17300 123-0 98-4 141 II860 1702320 17400 131-0 \ 104-0 133 11980 1809600 2i8 Examples in Electrical Engineering. ANNEALED CHARCOAL IRON. VALUES OF MAGNETIC PROPERTIES. ft Density in c.g.s. lines per sq. cm. area. H in c.g.s. per cm. length. Ampere-turns per cm. length. M Permea- bility. * Hysteresis loss in ergo per cc. per cycle. ft/ft Product of density and ampere-turns per cm. length. 17500 140 112 125 I2IOO I96OOOO 17600 149 119 117 I 22 2O 2094400 17700 158 126 III 12340 2230200 17800 167 133 106 12460 2367400 17900 176 140 ioi 12580 2506000 18000 186 148 96 I27OO 2664000 18100 196 156 90 12820 2823600 l8200 211 1 68 86 12940 3057600 18300 226 180 80 13060 3294000 18400 244 195 71 13180 3588000 18500 260 208 69 13300 3848000 18600 276 220 67 13420 4O92OOO 18700 293 234 65 13540 4375800 18800 3 11 248 64 13660 4662400 18900 330 264 60 13780 4989600 19000 350 280 54 13900 5320000 I9IOO 372 297 I4O2O 5672700 I92OO 396 316 48 I4I40 6067200 19300 422 337 45 14260 6504100 19400 448 358 43 14380 6945200 19500 476 380 14500 7410000 19600 505 404 39 14620 7918400 19700 537 429 36 14740 8451300 19800 575 460 34 14860 9I08000 19900 609 487 32 14980 9691300 20000 650 520 3i I5IOO IO4OOOOO ^HY THB Magnetic Properties of Soft Grey Cast Iron. CVERSITY MAGNETIC PROPERTIES OF SOFT GREY CAST IRON. . ft Density in c.g.s. lines per sq. cm. H c.g.s. per cm. length. fft Ampere-turns per cm. length. M Permeability. 3000 i-5 I'2 2000 3IOO 1-9 1*5 1695 3200 2-3 1*8 1391 3300 27 2-2 I2 4 I 3400 3 - i 2 '5 1096 3500 3'5 2'8 1009 3600 3*9 3'i 923 3700 4*3 3'4 860 3800 47 3-8 808 3900 4' i 764 4OOO 5*5 4'4 727 4100 5'9 47 695 42OO 6-3 5' 666 4300 67 5 '4 641 4400 7-1 57 619 4500 7-6 6-1 592 4600 8'! 6-5 568 4700 87 7-0 540 4800 9'3 7'4 516 4900 I0'0 8-0 490 5OOO 107 8-6 467 5100 11-6 9*3 439 5200 12-5 I0'0 416 5300 i3'5 io'8 392 5400 H'5 11-6 372 5500 12-6 348 5600 17-1 137 327 5700 18-5 V 14-8 37 5800 2O'O i6'o 290 5900 21-6 17-3 274 6OOO 23'2 18-5 258 6100 2 4 -8 19-8 246 62OO 26-5 21'2 234 6300 28-2 22'S 223 6400 30-0 24-0 213 6500 3 2-0 25'6 203 6600 6700 36;4 29-I 6800 307 178 6900 40-8 3 2-6 174 7000 43'2 162 220 Examples in Electrical Engineering. MAGNETIC PROPERTIES OF SOFT GREY CAST IRON. . ft H fft Density in c.g.s. lines per sq. cm. c.g.s. per cm. length. Ampere-turns per cm. length. M Permeability. 7100 45 '9 367 155 7200 48-6 38-9 I 4 8 7300 Si-7 41-4 141 7400 54'9 43 '9 135 7500 58-6 46-8 129 7600 62-3 49'8 123 7700 67-2 537 117 7800 70*2 S6-i III 7900 743 59'4 1 06 8000 78-5 62-8 102 8100 827 66-1 9 8 8200 86-9 69-5 94 8300 91 'i 72-8 9i 8400 95 '4 76'3 89 8500 99'8 79'8 86 8600 104-2 83-3 83 8700 108-9 87-1 80 8800 113-6 90-8 77 8900 118-6 94-8 74 9000 123-7 98-9 72 9100 129-5 103-6 70 92OO 135-6 108-4 68 9300 141-9 113-5 66 9400 148-3 118-6 64 9500 154-8 123-8 61 9600 161-4 129*1 59 9700 168-0 134-4 57 9800 1747 1397 55 9900 181-5 H5'2 54 IOOOO 188-5 1 5'% 53 IOIOO 195-8 156-6 5i 10200 203-2 162-5 49 10300 210-8 168-6 48 10400 219-0 i75'2 47 10500 228-0 182-4 45 IO6OO 238-0 190-4 44 10700 248-0 198-4 43 IO8OO 259-0 207-2 42 10900 272-0 217-6 40 IIOOO 288-0 230-4 39 Details of Conductors. 221 DETAILS OF CONDUCTORS. S.W.G. Diameter of each wire. Diameter of the strand. Area. Resistance at 60 F. Weight. Per looo yards. Inch. Inch. Square inches. Per looo yards. Ohms. Ibs. 7/0 0'500 ... 0-196349 0*1246 2270 6/0 0*464 0-169093 0*1446 1955 S/o OH32 ... 0-146574 0*1669 1694 40 0-400 0*125663 0*1946 1454 3/0 0-372 ... 0-108686 0*225I 1256 2/0 0-348 0-095114 0-2572 1099 I/O 0-324 ... 0*082447 0*2967 953 0-300 ... 0*070685 0*3461 817 2 0-276 ... 0*05982 0-4089 691 3 0-252 ... 0*04987 0-4905 576 4 0-232 ... 0*04227 0-5787 488 5 0'2I2 ... 0*03529 0-6931 405 6 O-I92 ... 0*0289 0*8450 334 7 0-I76 ... 0*0243 1*0056 281 8 o'i6o ... O*O2OI 1*2168 232 9 0-144 ... 0*0163 1-5022 188 10 0-128 ... O-OI28 1*9012 148 ii O'ii6 ... 0-0105 2'3I50 122 12 0-104 0-0085 2*8800 9 8 13 0-092 ... 0*0066 3-6803 7 6 H 0*080 0-0050 4-8673 58 15 0-072 ... 0-0040 6-0089 47 16 0-064 ... 0-0032 7-6049 37 17 0-056 ... 0-0024 9'933 2 28 18 0-048 ... 0*0018 i3"5!98 21 19 0*040 ... 0*0012 19*4697 I4-5 20 0-036 ... 0*0010 24'0354 11*7 21 0-032 ... 0-0008 30-422 9'3 22 0-028 ... 0-0006 39-729 7*1 3/25 0-020 0-042 0-00096 25'955 ii 3/24 0*022 O'O22 0-00116 2i*454 I3-5 3/23 0-024 0-024 0-00138 18-026 16 3/22 0-028 O-O28 0-00188 13*243 22 3/21 0-032 0-032 0*00246 10-144 28 3/20 0-036 0-036 0*00311 8-01 18 36 3/19 0-040 O-O4O 0*00384 6-4899 44 3/iS 0-048 0-048 0*0055 4-5066 64 222 Examples in Electrical Engineering. DETAILS OF CONDUCTORS. S.W.G. Diameter of each wire. Diameter of the strand. | A Resistance at Area - 60 F. Weight. Inch. Inch. Square inches. Per looo yards. Per 1000 yards. Ohms. Ibs. 7/25 0*020 0'060 O'OO22 II*I24 25-5 7/24 O*O22 O'O66 0*0027 9*I952 3^5 7/23 0-024 0-072 0*0032 7-7256 37 7/22 0-028 0-084 0*0043 5-6757 5i 7/21* 0-030 0*090 0*0050 4"9445 58 7/21 7/20 0-032 0-033 0*096 0-099 0*0057 0*0064 4:346o 4*0864 66 75 7/20 0*036 O'loS 0*0072 3-4336 84 7/19 0-040 0-120 0*0089 2*7813 104 7/18 0-048 0-144 0*0129 I '93 I 4 149 7/17 0-056 0-168 0*0175 1*4190 203 7/16 0*064 0-192 0*0229 1*0864 266 7/15 0*072 0-216 0*0290 0-8584 336 7/H 0-080 0-240 0-0358 0*6953 415 7/13 0*092 0-276 0-0474 0-5257 549 7/12 0-104 0-312 0-0606 0*4114 701 7/i i 0-116 0-348 0-0754 0*3307 872 7/10 0-128 0-384 0-0918 0*2716 1062 7/9 0-144 0-432 0-1162 0*2146 1343 7/8 0-160 0-480 0>I 435 0*1752 1660 7/6 0-192 0-576 0-2067 0*1207 2390 19/24 0*022 O'HO 0-0073 3-3877 85 19/23 O'O24 0-120 0-0087 2-8463 101*5 19/22 0-028 0-I40 0-0119 2-0910 138 19/21 0-032 0-160 0-0156 1-6011 180 19/20 0'036 0-180 0-0197 1-2650 228 19/19 0-040 O'2OO 0-0244 1-0247 282 19/18 6*048 0*240 0-0351 0*7115 406 19/17 0*056 0-280 0-0478 0-5228 553 19/16 0-064 0-320 0-0624 0-4002 722 i9/*5 0-072 0-360 0-0790 0-3162 914 19/14 0-080 0-400 0-0976 0*2561 1128 I9A3 0*092 0-460 0-1290 0*1937 1491 19/12 O'IO4 0-520 0-1649 Q'^S 1906 19/11 0-116 0-580 0-2052 0*1218 2372 19/10 0*128 0-640 0-2498 0*1000 2888 19/9 0-144 O'72O 0-3162 0*07906 3655 19/8 0-160 0-800 0-3904 0*06406 45i3 19/7 0-176 0-880 0-4724 0*05292 546i Details of Conductors. DETAILS OF CONDUCTORS. 223 S.W.G. Diameter of each wire. Diameter of the strand. Area. Resistance at 60 F. Weight. Inch. Inch. Square inches. Per 1000 yards. Per 1000 yards. Ohms. Ibs. 37/24 0-022 0*154 0*0143 I-7396 I6 5 37/23 O-O24 O'l68 0-OI7I 1*4647 I 9 8 37/22 0-028 0*196 0-0233 1-0737 270 37/21 0-032 0-224 0-0304 0*8222 352 37/20 0-036 0*252 0*0386 0-6496 446 37/19 0-040 0-280 0*0476 0-5262 550 37/i8 0-048 0-336 0-0686 0-3654 793 37/17 0-056 0-392 0-0934 0*2684 1080 37/i6 0-064 0-448 0-1220 0*2055 1410 37/15 0-072 0-504 0-1544 0*1624 1785 37/14 0-080 0-560 0-1906 0-I3I5 2203 37/13 0-092 0-644 0*2521 0-09947 2914 37/12 0*104 0-728 0*3221 0-07783 3723 37/n 0-116 0*812 0*4008 0*06265 4633 37/iQ 0-128 0-896 0*4880 0-05138 5641 37/9 0-144 I -008 0*6176 0*04060 7140 37/8 0-160 I'I20 0-7625 0*03288 8815 61/24 0-022 0-198 0*02378 l '$S l 275 61/23 0-024 0*216 0*02831 0*8865 327 61/22 0-028 0*252 0*03854 0-6583 446 61/21 0-032 0-288 0*0503 0-4987 572 61/20 0-036 0-324 0*0637 0-3940 736 61/19 0-040 0-360 0*0786 0*3191 909 61/18 0-048 0-432 0*1132 O*22l6 1309 61/17 61/16 0-056 0-064 0-504 0-576 0-1541 0*2013 0*l628 0*1246 1781 2327 6i/i5 0-072 0-648 0*2548 0*09850 2945 61/14 0*080 0*720 0-3I45 0-07979 3636 61/13 0-092 0-828 0*4160 0*06033 4809 61/12 0-104 0*936 0-5316 O*O472I 6i45 61/11 0-116 1*044 0*6614 0*03795 7646 61/10 0-128 I*I52 0-8053 0*03Il6 9309 91/18 0-048 0-528 0-1692 0*14857 1956 91/17 0-056 0*616 0*2302 9*10915 2661 91/16 0*064 0*704 0*3007 0-08357 3476 9I/I5 0-072 0*792 0-3806 0*06603 4400 91/14 0-080 0*880 0-4699 0*05348 5432 9i/i3 0-092 I*OI2 0-6215 0*04044 7185 91/12 0*104 1*144 0*7942 0*03164 9181 91/11 0-116 1*276 0*9881 0-02543 11422 224 Examples in Electrical Engineering. SQUARES. | 1 2 \ 3 4 5 6 7 8 9 1 23 4 5 6 7 8 i i 1 ro 1*000 *020 I*040 1-061 1-082 103 124 i-i45 1-166 1-188 2 4 6 8 10 3 15 17 19 *i 1*210 232 I -254 1-277] 1-300 "323 346 1-369 1-392 1*416 2 5 7 9 ii 4 16 18 21 2 1*440 464 1*488 i -513! i -538 563 588 1*613 1-638 1-664 2 5 7 10 12 5 i7 20 22 '3 1-690 * 7 l6 17421769 1796 823 850 1*877 1*904 1*932 ! C 8 ii 13 6 18 22 24 '4 I-960 * 9 88 2-0162-0452-074 103 132 2*161 2-190 2-220 3 6 9 2 14 7 20 23 26 '5 2*250 280 2-310 2-341 2*372 "403 "434 2-465 2-496 2-528 3 6 9 2 15 19 22 2528 6 2*560 592 2*624 2-657 2*690 723 756 2789 2-822 2-856 ! 7 10 3 16 20 23 26 30 7 2-890 924 2*958 2-9933*028 063 098 3-I33J3-I68 3^04 I 7 10 4 17 21 24!28 31 8 (3-240 *2 7 6 3-312 3 '349; 3 "386 "423 460 3-4973-534 3-572 \- 7 ii 5 18 22 26 30 33 9 13-610 2'0 I4-000 648 "040 3-686 4*080 3-725J3764 4*121 4*162 803 202 8423-881 "2444-285 3*9203*960 4 -32614 -368 i 8 i- 8 12 16 12 16 19 20 23 27 25 29 3i 35 33 37 2-1 2*2 4-410 4-840 4-452 J-88 4 4-494 4*928 4'537 4-973 4-580 5*018 623 063 666 108 47094752 5*1535-198 3796 5-244 1- 9 I 9 13 17 13 18 21 22 26 30 27 31 34 39 3640 2*3 5-200 5'336 5'382 5 "429; 5 "476 523 570 5*6175*6645712 5 9 14 19 23 28 33 38 42 5*760 5-8o8 5^56 5 '90S 5 '954 003 052 6*101 6*1506*200 5 io 15 20 24 29 34 39 44 2'5 6*250 6-300 6-350 6*401 6-452 503 '554 6-605 6-656 6*708 5 1 15 20 25 31 36 41 46 2-6 6*760 6-812 6*864 6-917 6*970 023 076 7*129 7*182 7-236 5 ii 16 21 26 32 37 42 48 27 7*290 7-4537-508 56: 7-6187-673 7728 7*784 5 ii 16 22 27 33 38 44 49 2-8 7-840 7-896 7-952 8 -009: 8 -066 8-1808-237 8*294 8-352 6 ii 17 23 28 34 4 46 51 2-9 8-4IO 8-468 8*526 8-5858-644 8703 87628*821 8*880 8*940 6 12 18 24 29 35 4147 53 3-0 g-OOO 9-060 9-120 9'i8ij9*242 9-303 9*3649*425 9*486 9-548 6 12 18 24 3 37 43149 55 3-1 9"6io 9-672 9734 97979-860 9-923 9-986 10*05 10*11 10*18 6 13 i 19 25 2 3 31 3 38 44 4 5 50 57 5 6 3 '2 10-24 10-30 10-37 10-43 1'5 10-56 10*63 10-69 1076 10*82 i 2 < 3 4 5 5 6 3 '3 10-89 10*96 II'02 11-0911-16 11*22 11-29 11-36 11-42 11-49 i 2 ; 3 4 5| 5 6 3'4 11-56 11-63 1170 117611*83 11*90 11*97 12-0412-11 12-18 i 2 < 3 4566 3'5 12*25 12-32 12*39 12*46 12*53 1 2 -60 12-67 1274 12-82 12*89 i 2 ' 4 4566 3-6 12-96 13-03 13*10 13-18 i3'25 IS'S 2 13-40 13-47 13-54 13-62 i 2 3 4 4567 37 13*69 1376 13-84 14*06 14*14 14-21 14-29 14-36 i 2 *: 4 5 5 6 7 14*44 I4'52 H-59 ! 4 *82 14-90114-98 15*05 I5-I3 i 23 4 5 5| 6 7 3'9 IS' 2 15-29 15-37 15-4415-52 I5-60 i5-68!i S - 7 6 15-84 15-92 I 2 2 3 4 5667 4-0 16*00 16*08 16*16 16-24 16*32 I6* 4 16*48 16-56 16-65 16-73 I 2 2 *: 4 5667 16-8 16*89 16-97 17-06 17-14 17*2 17-31 17-39 17-47 I7-56 I 2 2 3 4 5677 4-2 17*64 1772 17-81 17-8917*98 1 8*06 i8*is|i8*2 3 18-32 18*40 I 2 3 3 4 5678 4'3 18*4 18-58 18-66 1875 18*84 1 8*9 19*01 19*10 19-18 19*27 I 2 3 3 4 5678 4'4 19*36 i9'45 I 9'5^ 19*62 ,197 19*8 19*89 19-98 20-07 20*16 I 2 3 4 5 5678 4-5 20*2 20*34 20*42 20*52 20 '6 20*7 2079 20-88 20-98 21-07 I 2 3 4 5 5678 4'6 21*1 21 -2 q 21 '34 21-44 21 '5 21*6 21*72 21 *8l 21*90 22-00 I 2 3 4 5 6 7\ 7 8 47 22*0 22-18 22-28 22*3722*4 22-5 22-66 2275 22*85 22*94 I 2 3 4 5 6 7 8 9 4*8 23*0 23-14 23-23 23-3323-4 23 "5 23-62 2372 23-81 23*91 I 2 3 4 5 6 7 8 9 4'9 24*0 24*11 24*21 24*3024*4 24 '5 24 *6c 24*70 24-80 24*90 I 2 3 4 5 6 7 8 9 25*00 25*10 25-20 25 "3C >25'4 25-5 25*60 2570 25-81 25-91 I 2 3 4 5 6 7 8 9 5'i 5' 2 26*0 27-0 26-11 27*14 26.21 27^ 26*3226*4 27 -3527 '4 26-5 27'5 26*6312673 27-672777 26 '8c 27-88 26*94 27-98 I 2 I 2 3 4 3 4 5 5 6 7 6 7 8 9 8 9 5'3 28-0 28*20 28-30 28*4128-5 28-6 287328*84 28-94 29*05 I 2 3 4 5 6 7 9 10 5'4 2 9 -I 29-27 29-38 29-48 29-5 297 29*81 29*92 30*03 30-I4 1234 6 7 8 9 10 Squares. SQUARES. 225 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 5*5 30-25 30-36 30*47 30-58 30-69 o-8c 0-91 31-02 3i*i4 31-25 I 2 3 4 6 7 8 9 10 c*o 31-36 ?i"47 31-58317031-81 1-92 2-04 32-15132-2632-38 I 2 3 5 6 7 8 9 10 57 5'8 I' 9 6-0 32-49132-60 33 '64i3376 34'8i|34'93! 36-0036-12! 3272 32 -83132-95 33' 8 733*99|34* 35 '5i35*i 6 35*28 36-2436-3636-48 3-06 6'6o 3-1833-20,133-41 4 > 3434 < 46|34*57; 5-52135-643576; 67236-84^6-97 33-52 34 '69 35-88 37*09 I 2 I 2 I 2 I 2 3 5 4 5 4 5 4 5 6 6 6 6 7 8 7 8 7 8 7 9 9 10 9 JI IO II 10 II 6-1 37*2i 37 *33|37 '45; 37*58! 3770 7-82 7-95 38-07 38-19 38-32 I 2 4 5 6 7 9 10 II 6-2 38 "44138 "56^38 "69138 '81538 "94 9-06 9 -I9i39-3ii39 "44 39*5ili 3 4 5 6 6 9 IO II 6-4 39-6939-82 40-9641-09 39-9440-0740-20 41-2241-3441-47 0-32 i -60 o -45:40-58 4070.40-83 i 3 i 73'4i -86 41 -99:42-121 1 3 4 5 4 5 6 6 8 9 8 9 10 II 10 12 6-5 42-2542-38 42-5142-644277 2-90 3 -03 43 -1643 -30 43-43, i 3 4 5 7 8 910 12 6-6 43'5643'69 43-82 43-9644-09 44-22 44 -36 44-49 44-62 4476! i 3 4 5 7 8 9 ii 12 67 6-8 44 '8945-02 46-2446-38 45-1645-29145-43 46 -51 46 '65146 79 6^2 57045-8345-97146-1011 3 4 5 4 5 7 7 8 9!n 12 8 lOJII 12 6-9 47-61 14775 47 -89J48 -02)48 -16 8-30 8 -44 48 -58 48 72J48-86 i 3 4 6 7 8 10 ii 13 49-00 49-14 49-28149-4249-56 970 9-8449-9850-13 50-27 i 3 4 6 7 8 xolii 13 7*1 50-4150-55 50-69 50-84 50-98 I-I2 51-27151-41 5^55 5i7o i 3 4 6 7 9 ioUi 13 7*2 7*3 51-8451-98 53-29153*44 52-I3 52-27J52'42 53-58 5373 S3** 2*56 4'02 527152-85 54*17154*32 53-0053-14 54-4654-61 t 3 i 3 4 6 4 6 7 7 9 io!i2 13 9 10 12 13 7*4 7'5 54*76 56'25 54*9i 56-40 55-0655-20 56-5556*70 56*85 5'50 7-00 55-65|55-8o 57-1557-30 55*9556-io 57-46157-61 i 3 2 3 4 6 5 6 7 8 9 10 12 13 9 II.I2 14 7-6 77 5776 59*29 57*9 r 59 '44 58-06 59-60 58-22 5975 58-37 59-9I 58*52 58-68 a "22 58-83 60-37 58-98 60-53 59* I 4 60-68 2 3 2 3 S 6 5 6 8 8 9 ii 12 14 9 ii 12 14 7-8 60-84 7-9 162-41 6i"oo 62-57 61-15161-31 6273:62-88 2 6304 6I-62 63-20 617861-94 63-3663-52 62-09 63-68 62-25 63*84 2 3 2 3 5 6 5 6 8 8 9 ii 13 14 ID 11)13 14 8-0 64-00:64-16 64-3264-48 64-64 6 4 -8o 64-96 65-12 65-29 2 3 5 6 8 10 II I 3 J 4 8-2 65-616577 67-24,67-40 68-8969-06 65'93!66-io 67-576773 69-2269-39 66-26 67-90 69-56 66-42 68-06 6972 66-59 68-23 69-89 6675 68-39 70-06 66-91 68-56 70-22 67-08 6872 70-39 2 3 2 3 2 3 5 7 5 7 5 7 8 8 8 10 II 10 12 10 12 J 3 *5 i3 IS 13 i5 8-4 70-567073 70-90 71-06171-23 71-46 71*57 7174 71-91 72-08 2 3 5 7 8 10 12 H 15 72 -25 172 -42 72-59 727672-93 73 "1C 73*27 73'44 7362 73*79 2 3 5 7 9 10 12 H 15 8-6 73-9674*i3 74*30 74*48 74-65 74-8 75 "oo 75*i7 75*34 75*52 2 3 5 7 9 10 12 14 16 87 75-697S-86 76-04 76*21 76-39 76-5 7674 76-91 77-09 77-26 2 4 5 7 9 I 12 14 16 8-8 77-4477-62 7779 77*97 7 8-i 78-3 78-50 78-68 78-85 79*03 2 4 5 7 9 I 12 14 16 8-9 79-21179-39 79*57 7974179-92 Jo*i 80-28 80-46 80-64 80-82 9 4 5 7 9 i 1314 16 9-0 Si'oo 8ri8 81-36 81-548172 81-90 82-08 82-26 82*45 82-63 2 4 5 7 9 I 13 14 16 9-1 82-81 82-99 83*I7 83*3683-5; 837 8 3 ' 91 84-09 84-27 84-46 2 4 5 7 9 I 13 15 16 9*2 9'3 84-64 86-49 84-82 86-68 85*01 86-86 87-0587-24 87-4 37-61 g* 86-12 87-98 86-30 88-17 2 4 2 4 6 7 6 7 9 9 I 13 I 13 15 17 *5 17 9'4 88-36 88-55 8874 88-92 8 9 -i 89-3 89-49 89-68 89-87 90-06 2 4 6 8 9 I 13 i5 17 9'5 90-25 90-44 90-63 90*82 91-0 91-2 91*39 91*58 9178 91-97 2 4 6 8 10 I 13 15 17 9*6 97 9-8 92-16 94-05 96-04 92*3 q 94-28 96-24 92-54 94-48 96-42 9274 94 "6/ 96-6' 92-9 94*8 96-8 93*i 95 -of. 97*o 93*3293*5i 95-26)95-45 97-2297-42 9370 95 "65 97-61 93*90 95*84 97-81 2 4 2 4 2 4 6 8 6 8 6 8 10 10 10 12 14 12 14 12 14 IS 17 16 18 16 18 9'9 98-01 98-21 98-41 98-60 >98'8 99 -oc 99-2099-40 99-60 99-80 i 2 4 6 8 10 12 I^ 16 18 I II 226 Examples in Electrical Engineering. RECIPROCALS OF NUMBERS FROM 1000 TO 9999. 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 10 O'OOIOOOO 9901 9804 9709 9615 9524 9434 9346 9259 9174 918 27 36 45 55 64 7382 II 12 o '0009091 0-0008333 9009 8264 8929 8850 8197 8130 8772 8065 8696 8000 8621 7937 8547 8475 8403 7874 7813 7752 8 15 6 *2 23 30 19 26 38 32 45 5361 68 38 45^1 58 13 0-0007692 7634 7576 75i9 7463 7407 7353 7299 7246 7194 5 " l6 22 27 33 38 44 49 14 0-0007143 7092 7042 6993 6944 6897 6849 680367576711 5 10 14 19 24 29 33 38 43 o '0006667 6623 6579 6536 6494 6452 6410 6369 6329 6289 4 8 13 17 21 2 5 29 33 38 i6 17 0*0006250 0*0005882 6211 5848 61736135 5814 5780 6098 5747 6061 6024 5682 5988 5952 56505618 5917 5587 4 7 3 6 II 15 10 13 18 16 22 26 20 23 29 33 26 29 18 0-0005556 5525 5495 5464 5435 545 5376 5348 5319 5291 3 6 9 12 15 17 20 23 26 19 0-0005263 5236 5208 5181 5128 5102 5076,5051 5025 3 5 8 ii 13 16 18 21 24 20 0-0005000 4975 49504926,4902 4878 4854 4831 4808 4785 2 5 7 10 12 14 17 I 9 21 21 22 23 0-0004762 0-0004545 0-0004348 4739 4525 4329 4717 4695 4673 450544844464 431042924274 4651 4444 4255 4630 4425 4237 46084587 44054386 42194202 4566 4367 4184 2 4 2 4 2 4 I 1 5 7 II 10 13 15 12 14 II 13 17 20 16 18 14 16 24 25 0-0004167 0-0004000 4149 3984 4132 3968 45 3953 4098 3937 4082 3922 4065 3906 40494032 38913876 4016 3861 2 3 2 3 H 8 10 12 9 ii 13 IS 12 14 26 0-0003846 3831 38173802 3788 3774 3759 3745373 1 3717 I 3 4 6 7 8 10 II 13 27 28 0-0003704 0-0003571 3690 3559 3676 3663 35463534 365 352i 3636 359 3623 3497 36io|3597 348413472 3584 3460 i 3 X 2 4 5 4 5 7 6 8 9 7 9 II 12 10 II 29 0-0003448 3436 3425 3413 34oi 339 3367133^6 3344 X 2 3 5 6 7 8 9 10 30 0*0003333 3322 3311 3300 3289 3279 3268 3257J3247 3236 I 2 3 4 5 6 7 9 10 31 32 0-0003226 0-0003125 3215 35 3205 3195 3106 3096 3185 3086 3175 3077 3 l6 5 3067 3I553I45 30583049 3135 3040 I 2 I 2 3 4 3 4 5 5 a'? 8 9 8 9 33 34 0-0003030 0-0^02941 3021 2933 3012 3003 2924 2915 2994 2907 2985 2899 2976 2890 29672959 2882 2874 2950 2865 I 2 I 2 3 4 3 3 4 4 5 6 5 6 7 8 7 8 35 o '0002857 2849 2841 2833 2825 2817 2809 2801 2793 2786 X 2 2 3 4 5 6 6 7 36 0-0002778 2770 2762 2755 2747 2740 2732 2725 2717 2710 X 2 2 4 5 5 6 7 37 0-0002703 2695 26882681 2674 2667 2660 2653 2646 2639 2 4 4 5 6 6 38 39 40 0-0002632 0-0002564 0-0002500 2558 2494 26182611 24882481 2604 2538 2475 2597 2532 2469 2591 2525 2463 2584 25 J 9 2457 2577 2513 245 1 2571 2506 2445 X I X 2 2 2 3 3 3 4 5 4 4 4 4 5 f 5 6 5 5 41 42 43 0-0002439 0-0002381 0-0002326 2433 2375 2320 24272421 23702364 2315 2309 2415 2358 2304 2410 2353 2299 2404 2398 2392 2387 23471234223362331 2294 2288 228312278 X X X 2 2 2 3 3 3 3 4 3 4 3 4 5 5 4 5 4 5 44 0-0002273 2268 22622257 2252 2247 2242 2237 22322227 X 2 3 3 4 4 5 45 0-0002222 2217 22122208 2203 2198 2193 2188 2183 2179 o X 2 3 3 4 4 46 O-O002I74 2169 2165 2160 2155 2151 2146 2141 2137 2132 o I 2 3 3 4 4 47 0-0002128 2123 2119 2114 2110 2105 2101 2096 2092 2088 o I 2 3 3 4 4 48 0-0002083 207- 2075 2070 2066 2062 2058 2053 2049 2045 o I 2 3 3 3 4 49 0*0002041 2037 2033 2028 2O2/ 2O2O 20l6 2012 2008 2004 I 2 2 2 3 4 50 0-0002000 1,96 1992 1988 198^ I 9 80 1976 1972 1969 1965 I 2 2 2 3 4 5* 0-OOOI96l 1957 1953 1949 1946 1942 1938 1934 i93i 1927 I I 2 2 2 3 3 3 52 53 o "000192;: 0*000l887 ^ 1916 1880 1912 1876 I908 1873 1905 l86 9 1901 1866 1898 1862 1894 1859 1855 I I I I I I 2 2 2 3 2 2 3 3 3 3 54 o '0001852 1848 1845 1842 1838 1835 1832 1828 1825 1821 X X I 2 2 2 3 3 N.B. Three zeros follow the decimal point in the reciprocal of any four figure whole number except the number 1000. NOTE. Numbers in difference columns to be subtracted, not added. Reciprocals of Numbers from 1000 to 9999. RECIPROCALS OF NUMBERS FROM 1000 TO 9999. 227 55 56 % g 61 62 63 64 65 66 67 63 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 5 86 87 88 89 90 9i 92 93 94 95 96 97 98 99 1 3 3 1808 1776 1745 1715 1686 1658 1631 1605 1580 4 1805 1773 1742 1712 1684 1656 1629 1603 1577 1553 *5 2 9 1506 1484 1462 1441 1420 1401 1381 1362 *344 1326 1309 1292 1276 1259 1244 1229 5 6 7 1795 1764 1733 1704 1675 1647 1621 1595 1570 8 9 1 2 3 4 5 6 7 8 9 0-0001818 o '0001786 0*0001754 0*0001724 0-0001695 0-0001667 0-0001639 0-0001613 0-0001587 0-0001563 0-0001538 0-0001515 0-0001493 0-0001471 0-0001449 0*0001429 0-0001408 0-0001389 0-0001370 0-0001351 0-0001333 0*0001316 0-0001299 0-0001282 0-0001266 0-0001250 0-0001235 0*0001220 0-OOOI205 0-OOOII90 0-OOOII76 o '0001163 0*0001 i4c 0*0001136 0*0001124 0*0001111 0*0001099 0*0001087 0*0001075 0*0001064 0-000105; 0*0001042 0-0001031 0*0001020 O'OOOIOIO 1815 1783 1751 1721 1692 1664 1637 1610 1585 1560 1536 1513 1490 1468 1447 1427 1406 1387 1368 1350 1332 1314 1297 1230 1264 124^ '233 1218 1812 1779 1748 1718 1689 1661 1634 1608 1582 1558 1534 rS 11 1488 1466 1445 1425 1404 'SfS 1366 1348 1330 13" 1295 1279 1263 1247 1232 1217 1202 1188 1174 1160 1147 H34 II2I 1109 I0 9 6 1085 802 770 739 709 1681 1653 1626 1600 1575 155 I5 2 7 1504 1481 1460 H39 1418 1399 1379 1361 1342 1325 1307 1290 1274 1258 1242 1227 1799 I7 67 1736 1706 1678 1650 1623 1597 1572 1548 1524 1502 1479 1458 1437 1416 1397 1377 1359 1340 1323 1305 1289 1272 1256 1241 1225 I2II II 9 6 Il82 1168 "55 1142 1129 1116 1104 1092 1792 1761 1730 1701 1672 1645 1618 1592 1567 1543 1520 1497 1475 1453 1433 1412 1393 1374 1355 1337 1319 1302 1285 1269 1253 1238 1222 1208 "93 1179 1166 1152 "39 1126 1114 IIOI 1089 1078 1066 J 55 1044 1033 1022 1012 1002 1789 1757 1727 1698 1669 1642 1616 1590 1565 i54i 1517 1495 1473 I45 1 I43 1 1410 i39i 1372 1353 1335 1318 1300 1284 1267 1252 1236 1221 1206 1192 II 7 8 1164 " S o H38 1125 III2 1 100 1088 1076 1065 I0 5^ 1043 1032 IO2I IOII 1001 O I o o o I I 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 I 2 3 3 3 3 2 3 2 3 2 3 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 I 2 X 2 X 2 I I J-OO-D 1531 1508 I 4 86 1474 1443 I 4 22 1403 1383 I3 6 4 1346 1328 I3II 1294 1277 I26l 1245 1230 1215 1200 1186 1172 H59 H45 H33 1 120 1107 IO95 I08 3 1072 1000 1049 1038 IO28 1017 1007 1 54 1522 1499 1477 1456 1435 1414 1395 1376 1357 1339 1321 I 34 1287 1271 1255 1239 122^ 1209 "95 1181 1167 "53 1140 1127 1115 no; 1091 1079 1067 1056 1045 1034 1024 1013 1003 o o O O o o O o o O O o c I I I I I I I O I O I I I 2 I I I I I I I I I I I I O O I I I I "99 1198 1183 1170 1156 H43 1130^ 1117 1105 1093 1081 1070 105^ 1047 1036 1026 1015 1005 1189 1175 1161 n 4 8 "35 1 122 IIIO 1098 1086 1074 1063 1052 IO4I IO3O 1019 1009 1171 "57 1144 1131 1119 1106 1094 1082 1071 I0 59 1048 1037 1027 1016 1006 O o o O o o O o o I I I 1 o o 1068 1057 1046 1035 1025 1014 1004 1U ^ 1062 1050 1040 IO29 ioi 1008 I o I I I I I I I X N.B. Three zeros follow the decimal point in the reciprocal ol any four-figure whole number except the number 1000. NOTE. Numbers in difference columns to be subtracted, not added. 228 Examples in Electrical Engineering. NATURAL TANGENTS. 1 2 .30 4 5 6 7 .go .90 oooo 0017 0035 0052 0070 0087 0105 0122 0140 oi57 I 0175 0192 0209 0227 0244 0262 o>79 0297 0314 0332 2 0349 0367 0384 0402 0419 0437 0454 0472 0489 0507 3 0524 0542 0559 0577 0594 0612 0629 0647 0664 0682 4 0699 0717 0734 0752 0769 0787 0805 0822 0840 0857 5 0875 0892 0910 0928 0945 0963 0981 0998 1016 1033 6 1051 1069 1086 1104 1122 "39 H57 "75 1192 1210 7 1228 1246 1263 1281 1299 I3U 1334 i35 2 1370 1388 8 1405 1423 1441 H59 1477 H95 1512 1530 1548 1566 9 1584 1602 1620 1638 1655 1673 1691 1709 1727 1745 10 1763 1781 1799 1817 1835 1853 1871 1890 1908 1926 ii 1944 1962 1980 1998 20l6 2035 2053 2071 2089 2107 12 2126 2144 2162 2180 2199 2217 2235 2254 2272 2290 13 2309 2327 2345 2364 2382 2401 2419 2435 2456 2475 14 2493 2512 253 2549 2568 2586 2605 2623 2642 2661 15 2679 2698 2717 2736 2754 2773 2792 2811 2830 2849 16 2867 2886 2905 2924 2943 2962 2981 3000 3019 3038 17 3057 3076 3096 3H5 3134 3153 3172 3191 3211 3230 18 3249 3269 3288 3307 3327 3346 3365 3385 3404 3424 19 3443 3463 3482 3502 3522 3541 35 6 i 358i 3600 3620 20 3640 3659 3679 3699 3719 3739 3759 3779 3799 3819 21 3839 3859 3879 3899 3919 3939 3959 3979 4000 4020 22 4040 4061 4081 4101 4122 4142 4163 4183 4204 4224 23 4245 4265 4286 4307 4327 4348 4369 4390 44" 443 i 24 25 '4452 4663 4473 4684 4494 4706 4515 4727 4536 4748 4557 4770 4578 479i 4599 4813 4621 4834 4642 4856 26 4877 4899 4921 4942 4964 4986 5008 5029 5051 573 27 5095 5"7 5139 5161 5^4 5206 5228 5250 5272 5295 28 5317 5340 5362 5384 5407 5430 5452 5475 5498 5520 29 5543 5566 5589 5612 5635 5658 5681 5704 5727 5750 3 5774 5797 5820 5844 5^7 5890 59H 5938 596i 5985 31 6009 6032 6056 6080 6lO4 6128 6152 6176 6200 6224 32 6249 6273 6297 6322 6346 6371 6395 6420 6445 6469 33 6494 6519 6544 6569 6594 6619 6644 6669 6694 6720 34 6745 6771 6796 6822 68 47 6873 6899 6924 6950 6976 35 7002 7028 7054 7080 7107 7133 7159 7186 7212 7239 36 7265 7292 7319 7346 7373 7400 7427 7454 748i 75o8 37 7536 7563 7590 7618 7646 7673 7701 7729 7757 7785 3* 7813 7841 7869 7898 7926 7954 7983 8012 8040 8069 39 8098 8127 8156 8185 8214 8243 8273 8302 8332 8361 40 8391 8421 8451 8481 8511 8541 857i 8601 8632 8662 4i 8693 8724 8754 8785 8816 8847 8878 8910 8941 8972 42 9004 9036 9067 9099 9i3i 9163 9195 9228 9260 9293 43 9325 93 S8 9391 9424 9457 949 9523 9556 9590 9623 44 9657 9691 9725 9759 9793 9827 9861 9896 993 9965 Natural Tangents. NATURAL TANGENTS. 229 1 .go . 3 o 4 5 6 7 8 . 9 o 45 I'OOOO 0035 0070 0105 0141 0176 0212 0247 0283 0319 46 i '0355 0392 0428 0464 0501 0538 0575 0612 0649 0686 47 i -0724 0761 0799 0837 0875 0913 0951 0990 1028 1067 *r/ 4.8 i'iio6 1145 1184 1224 1263 1303 1343 1383 1423 1463 *T*J 49 1-1504 1544 1585 1626 1667 1708 1750 1792 1833 1875 50 1-1918 1960 2002 2045 2088 2131 2174 2218 2261 2305 CI 2349 2393 2437 2482 2527 2572 26l7 2662 2708 2753 *3 C2 '2799 2846 2892 2938 2985 3032 3079 3127 3175 3222 J 53 3270 33 J 9 3367 34i6 3465 35H 3564 3613 3663 3713 54 3764 38i4 3865 3916 3968 4019 4071 4124 4176 4229 JT^ 55 4281 4335 4388 4442 4496 4550 4605 4659 4715 4770 56 4826 4882 4938 4994 5051 5108 5166 5224 5282 5340 J 57 '5399 5458 5517 5577 5637 5697 5757 5818 5880 594i 58 6003 6066 6128 6191 6255 6319 6383 6447 6512 6577 3 CO 6643 6709 6775 6842 6909 6977 7045 7H3 7182 7251 60 7321 739i 7461 7532 7603 7675 7747 7820 7893 7966 61 8040 8115 8190 8265 8341 8418 8495 8572 8650 8728 62 8807 8887 8907 9047 9128 9210 9292 9375 9458 9542 63 1-9626 97" 9797 9883 9970 0057 0145 0233 0323 0413 64 2-0503 0594 0686 0778 0872 0965 1060 "55 1251 1348 65 2-1445 1543 1642 1742 1842 1943 2045 2148 2251 2355 66 2*2460 2566 2673 2781 2889 2998 3109 3220 3332 3445 67 2-3559 3 6 73 3789 3906 4023 4142 4262 4383 454 4627 68 2-4751 4876 5002 5129 5257 5386 5517 5 6 49 5782 59i6 69 2-6051 6187 6325 6464 6746 6889 7034 7179 7326 70 2-7475 7625 7776 7929 8083 8239 397 8556 8716 8878 7i 2-9042 9208 9375 9544 97H 9887 0061 0237 0415 0595 72 3-0777 0961 1146 1334 1524 1716 1910 2106 2305 2506 73 3-2709 2914 3122 3332 3544 3759 3977 4197 4420 4646 74 3-4874 5105 5339 5576 5816 6059 6305 6554 6806 7062 75 37321 7533 7848 8118 8391 8667 8947 9232 9520 9812 76 4-0108 0408 0713 1022 1335 l6 53 1976 2303 2635 2972 77 4'33i5 3662 4015 4374 4737 5107 5483 5864 6252 6646 78 47046 7453 7867 8288 8716 9152 9594 0045 0504 0970 79 5-I446 1929 2422 2924 3435 3955 4486 5026 5573 6140 80 5-67I3 7297 7894 8502 9124 9758 0405 1066 1742 2432 Si 6-3138 3859 4596 5350 6122 6912 7920 8548 9395 0264 82 7'"54 2066 3002 3962 4947 5958 6996 8062 9158 0285 83 8-1443 2636 3863 5126 6427 7769 9152 0579 2052 3572 84 9-5I44 9-677 9-845 10-02 10-20 10-39 10-58 10-78 10-99 11-20 85 H'43 n-66 11-91 I2"l6 12-43 12-71 13-00 13-30 13-62 13-95 86 14-30 14-67 15*06 15H6 15-89 16-35 16-83 I7-34 17-89 18-46 7 19-08 19-74 20-45 21'20 22-02 22-90 23-86 24-90 26-03 27-27 88 28-64 30-14 31-82 33^9 35-80 38-19 40-92 44-07 47-74 52-08 89 57-29 63-66 71-62 81-85 95-49 114-6 143-2 191*0 286-5 573-0 230 Examples in Electrical Engineering. NATURAL SINES. 1 2 .30 4 5 6 r .go .90 0000 0017 0035 0052 0070 0087 OIO5 OI22 0140 0157 I 0175 0192 0209 0227 0244 0262 0279 0297 0314 0332 2 0349 0366 0384 0401 0419 0436 0454 9471 0488 0506 3 0523 0541 0558 0576 0593 0610 0628 0645 0663 0680 4 0698 0715 0732 0750 0767 0785 0802 0819 0837 0854 5 0872 0889 0906 0924 0941 0958 0976 0993 IOII 1028 6 1045 1063 1080 1097 IIJ 5 1132 1149 1167 1184 1201 7 1219 1236 1253 1271 1288 1305 1323 1340 1357 1374 8 1392 1409 1426 1444 1461 1478 1495 *5*3 1530 1547 9 1564 1582 1599 1616 1633 1650 1668 1685 I7O2 1719 10 1736 1754 1771 1788 1805 1822 1840 1857 1874 I8 9 X ii 1908 1925 1942 1959 1977 1994 201 1 2028 2045 2062 12 2079 2096 2113 2130 2147 2164 2181 2158 2215 2232 13 2250 2267 2284 2300 2317 2334 2351 2368 2385 2402 14 2419 2436 2453 2470 2487 2504 2521 2538 2554 2571 15 2588 2605 2622 2639 2656 2672 2689 2706 2723 2740 16 2756 2773 2790 2807 2823 2840 2857 2874 2890 2907 17 2924 2940 2957 2974 2990 3007 3024 3040 3057 3074 18 3090 3107 3123 3140 3156 3173 3190 3206 3223 3239 19 3256 3272 3289 3305 3322 3338 3355 337i 3387 3404 20 3420 3437 3453 3469 3486 3502 3535 3551 3567 21 3584 3600 3616 3633 3649 3665 3681 3697 37H 3730 22 3746 3762 3778 3795 3811 3827 3843 3859 3875 3891 23 3907 3923 3939 3955 397i 3987 4003 4019 4035 4051 24 4067 4083 4099 4H5 4131 4H7 4163 4179 4195 4210 25 4226 4242 4258 4274 4289 4305 4321 4337 4352 4368 26 4384 4399 4415 4431 4446 4462 4478 4493 4509 4524 27 4540 4555 4571 4586 4602 4617 4633 4648 4664 4679 28 4695 4710 4726 4741 4756 4772 4787 4802 4818 4*j3 29 4848 4863 4879 4894 4909 4924 4939 4955 4970 3 5000 5015 5030 5045 5060 5075 5090 5105 5120 5135 31 5150 5165 5180 5195 5210 5225 5240 5255 5270 5284 32 5299 53H 5329 5344 5358 5373 5388 5402 5417 5432 33 34 5446 5592 5606 5476 5621 5490 5635 5505 5650 5519 5664 5534 5678 5548 5693 5563 5707 5577 35 5736 5750 5764 5779 5793 5807 5821 5835 5850 5864 36 37 5878 6018 5892 6032 5906 6046 5920 6060 5934 6074 5948 6088 5962 6101 5976 6115 5990 6l29 6004 6i43 38 6157 6170 6184 6198 6211 6225 6239 6252 6266 6280 39 6293 6307 6320 6 334 6347 6361 6374 6388 6401 6414 40 6428 6441 6455 6468 6481 6494 6508 6521 6534 6547 4* 6561 6574 6587 6600 6613 6626 6639 6652 6665 6678 42 6691 6704 6717 6730 6743 6756 6769 6782 6794 6807 43 6820 6833 6845 6858 6871 6884 6896 6909 6921 6934 44 6947 6959 6972 6984 6997 7009 7022 7034 7046 7059 Natural Sines. NATURAL SINES. 231 1 2 . 3 o .40 6 6 7 .QO .90 45 7071 7083 7096 7 108 7120 7133 7H5 7157 7169 7l8l 46 7193 7206 7218 7230 7242 7254 7266 7278 7290 7302 47 73H 7325 7337 7349 7361 7373 7385 7396 7408 7420 48 7431 7443 7455 7466 7478 7490 7501 7513 7524 7536 49 7547 7556 7570 758i 7593 7604 7627 7638 7649 50 7660 7672 7683 7694 7705 7716 7727 7738 7749 7760 5i 7771 7782 7793 7804 7815 7826 7837 7848 7859 7869 5 2 7880 7891 7902 7912 7923 7934 7944 7955 7965 7976 53 7986 7997 8007 8018 8028 8039 8049 8059 8070 8080 54 8090 8100 8111 8121 8131 8141 8151 8161 8171 8181 55 8192 8202 8211 8221 8231 8241 8251 8261 8271 8281 56 8290 8300 8310 8320 8329 8339 8348 8358 8368 8377 57 8387 8396 8406 8415 8425 8434 8443 8453 8462 8471 58 8480 8490 8499 8508 8517 8526 8536 8545 8554 8563 59 8572 8581 8590 8599 8607 8616 8625 8634 8643 8652 60 8660 8669 8678 8686 8695 8704 8712 8721 8729 8738 61 8746 8755 8763 8771 8780 8788 8796 8805 8813 8821 62 8829 8838 8846 8854 8862 8870 8878 8886 8894 8902 63 8910 8918 8926 8934 8942 8949 8957 8965 8973 8980 64 8988 8996 9003 9011 9018 9026 9033 9041 9048 9056 65 9063 9070 9078 9085 9092 9100 9107 9114 9121 9128 66 9135 9H3 9150 9157 9164 8171 9178 9184 9191 9198 67 9205 9212 9219 9225 9232 9239 9245 9252 9259 9265 68 9272 9278 928-; 9291 9298 9304 93" 9317 9323 9330 69 9336 9342 9348 9354 9361 9367 9373 9379 9385 9391 70 9397 9403 9409 9415 9421 9426 9432 9438 9444 9449 71 9455 9461 9466 9472 9478 9483 9489 9494 9500 9505 72 95" 95 * 6 9521 9527 9532 9537 9542 9548 9553 9558 73 95 6 3 9568 9573 9578 9583 9588 9593 9598 9603 9608 74 9613 9617 9622 9627 9632 9636 9641 9646 9650 9655 75 9659 9664 9668 9673 9677 9681 9686 9690 9694 9699 76 9703 9707 97" 9715 9720 9724 9728 9732 9736 9740 77 9744 1 9745 9751 9755 9759 9763 9767 9770 9774 9778 78 978i 97*5 9789 9792 9796 9799 9803 9806 9810 9816 9820 9823 9826 9829 9833 9836 9839 9842 9845 80 9848 9851 9854 9857 9860 9863 9866 9869 9871 9874 Si 9877 9880 9882 9885 9888 9890 9893 9895 9898 9900 82 9903 9905 9907 9910 9912 9914 9917 9919 9921 9923 83 9925 9928 9930 9932 9934 9936 9938 9940 9942 9943 84 9945 9947 9949 9951 9952 9954 9956 995? 9959 9960 85 9962 9963 9965 9966 9968 9969 9971 9972 9973 9974 86 9976 9977 9978 9979 9980 9981 9982 9983 9984 998; 87 9986 9987 9988 9989 9990 9990 9991 9992 9.93 9993 88 9994 9995 9995 9996 9996 9997 9997 9997 9998 9998 89 9998 9999 9999 9999 9999 rooo I OOO rooo rooo roco nearly. nearly. nearly. nearly- nearly. 232 Examples in Electrical Engineering. LOGARITHMS. 1 Q 8 4 5 e 7 8 1 13 3 4 5 e 7 8 e 10 OOOO 0043 0086 0128 017: 0212 0253 0294 03340374 48 12 17 21 25 29 33 37 | II 0414 0453 0492 0531 0569 060 7 0645 0682 07190755 48 II 15 19 23 26 3 34 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 31 13 "39 H73 1206 1239,1271 1303 1335^367 1399' 1430 10 13 16 19 23 26 29 14 1461 1492 1523 155311584 I6l 4 164411673 17031732 3 6 9 12 15 18 21 24 27 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 36 8 ii 14 17 2O 22 25 16 17 2041 2304 2068 2330 2095 2122 2355 2380 2148 240- 2175 2430 2201 2227 2455(2480 22532279 2504 2529 3 5 2 5 8 ii 7 10 13 12 16 18 15 17 21 24 20 22 18 2553 2577 2601 2625 2648 2695 2718 27422765 2 5 7 9 12 14 16 19 21 19 2788 2810 2833 2856 2878 2900 2923,2945 2967 2989 2 4 7 9 II 13 16 18 20 20 3010 3032 3054 3075 3096 3 II8 3139 3160 3181,3201 2 4 6 8 II 13 15 17 19 21 22 23 24 3222 3f4 3617 3802 3243 3444 3636 3820 3263 3284 3464 3483 3655 3674 38383856 3304 3502 3692 3874 3324 3522 37" 3892 3345 3365 35413560 3729 3747 3909 3927 3385 3404 3579i3598 3766(3784 39453962 2 4 2 4 2 4 4 6 8 6 8 6 7 5 7 10 10 9 9 12 14 12 14 II 13 II 12 16 18 IS 17 IS 17 14 16 25 3979 3997 4014 4031 4048 4065 4082 4099 41164133 3 5 7 9 10 12 14 15 26 4150 4166 4183 42OO 4216 4232 4249 4265 4281 4298 3 5 7 8 10 ii 13 15 27 43 J 4 4330 43464362 4378 4393 4409 4425 4440 4456 3 5 6 8 9 J 3 14 28 4472 4487 45024518 4533 4548 4564 4579 4594 4609 3 5 6 8 9 ii 12 14 29 4624 4639 46544669 4683 4698 4713 4728 4742 4757 3 4 6 7 9 1012 13 30 477i 4786 4800 4814 4829 4843 4857 4871 4886 4900 3 4 6 7 9 10 ii 13 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 3 4 6 7 8 10 II 12 32 5051 5065 50795092 5105 5 JI 9 5132 5145 5159 5172 3 4 5 7 8 9 ii 12 33 5185 5198 5211 5224 5237 5250 5263 5276 52^9 5302 3 4 6 8 9'io 12 34 5315 532853405353 5366 5378 5391 5403 54i6 5428 3 4 6 8 9|io ii 35 544i 5453 6465 5478 5490 5502 55M 5527 5539 555i 2 4 6 7 9 ! io ii 36 5563 5575 5599 5611 5623 5635 5647 5658 5670 2 4 6 7 8 10 II 37 5682569457055717 5729 5740 5752 5763 5775 5786 2 3 6 7 8 9 10 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 2 3 6 7 8 9 10 39 5911 5922 5933 ! 5944 5955 5966 5977 5988 5999 6010 2 3 5 7 8 9 10 40 6021 6031 604216053 6064 6075 6085 6096 6107 6117 2 3 4 5 6 8 9 10 4i 6128613861496160 6170 6180 6191 6201 6212 6222 2 3 4 5 6 7 8 9 42 43 44 45 6232 6243 63356345 64356444 65326542 6253 6355 6454 655 1 6263 6 A 5 6464 6561 6274 6375 6474 657i 6284 6385 6484 6580 6294 6395 6493 6590 6304 6314 6325 6405 6415,6425 6503 6513:6522 6599 660916618 2 2 2 2 3 4 3 4 3 4 3 4 5 5 5 5 6 7 67* 6 7 8 9 8 9 8 9 8 9 46 6628J6637 6646 6656 6665 6675 6684 6693 6702 6712 2 3 4 5 6 7 7 8 47 6721 6730 6739 6749)6758 6767 6776 6785:6794:6803 2 3 4 5 5678 48 49 6812 6902 6821 683016839:6848 6911 6920 6928 6937 6857 6946 68666875 6955^964 688416893 697256981 2 2 3 4 3 4 4 4 5 6 5 6 7 8 7 8 50 6 99 c 699817007 7Oi6 i 7O2.i 7033 7042 7050 705917067 2 3 3 4 5 6j 7 8 51 7076 70847093 7101 7110 7118 7126 7135 7143 7152 2 3 3 4 5678 52 7160716817177 71857193 7202 72107218:722617235 2 2 3 4 5 6{ 7 7 53 724317251:7259 7267 7275 7284 7292173001730817316 2 2 3 4 5 6! 6 7 54 73247332:7340 7348 7356 1 7364 73727380 7388 7396 2 2 ^ 4 5 6J6 7 Logarithms. LOGARITHMS. 233 1 2 3 4 5 6 7 8 9 12 3 4 5 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 I 2 2 3 4 5 5 6 7 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 755i 2 2 3 4 5 5 6 7 % 7559 7634 7566 7642 7574 7649 7582 7657 7589 7664 ^ 7679 7612 7686 7619 7694 7627 7701 2 2 3 2 ; 4 4 5 5 4 5 6 7 6 7 59 7709 7716 7723 773i 7738 7745 7752 7760 7767 7774 2 ' 4 4 5 6 7 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 2 < 4 4 5 6 6 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 2 4 4 5 6 6 62 7924 793 1 7938 7945 7952 7959 7966 7973 7980 7987 2 < 3 4 6 6 63 7993 8000 8007 8014 8j2I 8028 8035 8041 8048 8055 2 < 3 4 5 6 64 8062 8069 8075 8082 8089 8096 5 102 8109 8116 8122 2 ^ 3 4 5 6 65 8129 8136 8142 8149 8156 1162 8169 8176 8182 8189 2 3 4 5 6 66 8i95 8202 8209 8215 8222 8228 8235 82 4 ! 8248 8254 2 3 4 5 6 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 83*9 2 < 3 4 5 6 68 83 2 5 833i 8338 8344 835i 8357 8363 8370 8376 8382 2 ; 3 4 4 5 6 69 8388 8395 8401 8407 8414 8420 5426 8432 8439 8445 2 3 4 4 5 6 70 845i 8457 8463 8470 8476 8482 8488 8494 8500 8506 2 3 4 4 5 6 7 1 72 8513 8573 8519 8579 8525 8585 8531 859i 8537 8597 US! ss si 55 8615 8561 8621 8567 8627 2 2 3 3 4 4 4 4 5 5 5 S 73 8633 8630 8645 8651 8657 8663 8669 8675 8681 8686 2 3 4 4 5 5 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 2 3 4 4 5 5 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 2 3 3 4 5 5 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 2 3 3 4 5 5 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 2 3 3 4 4 5 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 2 3 3 4 4 5 79 8976 8982 8987 8993 8998 9004 9009 9 OI 5 9020 9025 2 3 3 4 4 5 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 2 3 3 4 4 5 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 2 2 3 3 4 4 5 82 9*3 8 9143 9149 9154 9159 9165 9170 9175 9180 9186 2 2 3 3 4 4 5 83 9 I 9 I 9196 9201 9206 9212 9217 9222 9227 9232 9238 2 3 3 4 4 5 84 9213 9248 9253 9258 9263 9269 9274 9279 9284 9289 3 3 4 4 5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 934 3 3 4 4 5 86 9345 9350 9355 936o 9365 9370 9375 9380 9385 9390 I 3 3 4 4 5 87 9395 9400 945 9410 9415 9420 9425 9430 9435 9440 2 3 3 4 4 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 2 3 3 4 4 89 9404 9499 9504 9509 9513 95i8 9523 9528 9533 9538 o 2 3 3 4 4 90 9542 9547 9552 9557 9562 9566 957i 9576 958i 9586 2 3 3 4 4 9 1 959 9595 9600 9605 9609 9614 9619 9624 9628 9633 2 3 3 4 4 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 2 3 3 4 4 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 2 3 3 4 4 94 973 1 9736 974i 9745 975 9754 9759 9768 9773 2 3 3 4 4 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 98U o 2 3 3 4 4 96 9823 9827 9832 9836 Q8 4 i 9845 9850 9854 9859 9863 2 3 3 4 4 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 2 3 3 4 4 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 2 3 3 4 4 99 9956 9961 9965 9969 9974 9978 9983 99-7 9991 9996 2 3 3 3 4 234 Examples in Electrical Engineering. ANTILOGARITHMS. 1 2 3 4 5 6 7 8 9 1 2 34 5 7 ! 8 9> oo 1000 1002 1005 1007 1009 1012 1014 1016 1019 102 1 O I I I 1222- oi 1023 IO26 1028 1030 r 33 1035 1038 1040 1042 1045 o o I 2: 2 2 "02 1047 1050 1052 1054 1057 1059 1062 1064 106711069 I 2 2 2 *3 1072 1074 1076 1079 1081 1084 10861089 1091 1094 o I 2 2 2 04 1096 1099 1 102 1104 1107 1109 III2 III4 1117 III9 o 2 2 2 3 'OS 1 122 1125 1127 1130 1132 "35 "38II40 U43 n 4 6 2 2 2 2 06 II 4 8 5i "53 1156 1150 1161 1164^167 1169 1172 2 2 2 2 07 "75 1178 1180 1183 1186 1189 1191 iig/ 1197 1199 o 2 2 2 2 08 1202 1205 1208 I2II 1213 1216 1219 1222 1225 1227 2 2 2 3 09 1230 1233 1236 1239 1242 1245 1247 1250 1253 1256 2 2 2 3 10 1259 1262 1265 1268 I2 7 ! 1274 1276 1279 1282 1285 2 2 2 3 II 1288 1291 129^ 1297 I 3 00 1303 1306 1309 1312 1315 o 2 2 2 2 3 *I2 1318 1321 1324 1327 1330 1334 1337 1340 1343 1346 o 2 2 2 2 3 ;i3 1352 1355 1358 1361 1365 1368 1371 1374 1377 o 2 2 2 3 3 1380 1384 1387 1390 1393 1396 1400 1403 1406 1409 2 2 2 3 3 15 1413 1416 1419 1422 1426 1429 I43 2 J 435 1439 1442 2 2 2 3 3 16 1445 1449 1452 X 4S5 1459 1462 1466 1469 1472 1476 o 2 2 2 3 3 17 J 479 1483 1486 1489 1493 1496 1500 1507 1510 2 2 2 3 3 18 1514 1517 1521 1524 1528 I53 1 1535 S 1542 1545 2 2 2 3 3 19 1549 1552 1556 1560 1563 1567 1570 1574 1578 1581 o 2 2 2 3 3 20 1589 1592 IS96 1000 1603 1607 1611 1614 1618 2 2 2 3 3 *2X 1622 1626 1629 1633 l6 37 1641 1644 1648 1652 1656 2 2 2 2 3 3 22 1660 1663 1667 1671 1675 1679 1683 1687 1690 1694 o 2 2 2 2 3 3 23 1698 1702 1706 1710 1714 1718 1722 1726 1730 1734 o 2 2 2 2 3 4 24 1738 1742 1746 1750 I7S4 1758 1762 1766 1770 1774 o 2 2 2 2 3 4 25 1778 I7 82 1786 1791 1795 1799 1803 1807 1811 1816 2 2 2 2 3 4 26 1820 1824 1828 1832 1837 1841 1845 1849 1854 1858 o 2 2 3 3 3 4 2 7 1862 1866 1871 1875 1879 1884 1888 1892 1897 1901 o 2 2 3 3 3 4 28 1905 1910 1914 1919 1923 1928 1932 1936 1941 1945 2 2 3 3 4 4 2 9 1950 1954 1959 1963 I 9 68 1972 1977 1982 1986 1991 2 2 3 3 4 4 3 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 o 2 2 3 3 4 4 *3* 2042 2046 2051 2056 2061 2065 2070 2075 2080 2084 o 2 2 3 3 4 4 32 2089 2094 2099 2104 2109 2113 2118 2123 2128 2133 2 2 3 3 4 4 "33 2138 2143 2148 2153 2I 5 8 2163 2168 2173 2178 2183 2 2 3 3 4 4 '34 2188 2193 2198 2203 2208 22I-: 2218 2223 2228 2234 2 2 3 3 4 4 5 *35 2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 2 2 3 3 4 4 S 36 22911 22 9 6 2301 2307 2312 2317 2323 2328 2333 2339 2 2 3 3 4 4 S '37 2344 2350 2355 2360 2 3 66 2371 2377 2382 2388 2393 2 2 3 3 4 4 5 38 2399 2404 2410 2415 2 4 2I 2427 2432 2438 2443 2449 2 2 3 3 4 4 5 "39 2455 2460 2466 2472 2477 2483 2489 2495 2500 2506 2 2 3 3 4 5 S 40 2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 2 2 3 4 4 5 5 42 2570 2630 2 Ff 2636 2582 2642 2588 2649 2594 2655 2600 266l 2606 2667 2612 2673 2618 2679 2624 2685 2 2 2 2 3 3 4 4 4 4 If 43 '44 '45 2692 2754 2818 2698 2761 2825 2704 2767 2831 2710 2773 2838 2716 2 7 80 2844 2723 2 7 86 28 5 I 2729 2793 2858 2735 2799 2864 2742 2805 2871 2748 2812 2877 2 3 2 3 2 3 3 3 3 4 4 4 4 4 5 1 t 46 2884 2891 2897 2904 2 9 II 917 2924 2931 2938 2944 2 3 3 4 S 5 6 *47 2951 2958 2965 2972 2979 985 2992 2999 300 6 3 OI 3 2 3 3 4 5 5 6 48 '49 3020 3090 3027 3097 3034 3041 3I05J3H2 3048 055 3126 3062 3133 30693076 31413148 3083 3155 2 3 2 3 4 4 4 5 4 S 6 6 6 6 A ntilogarithms. ANTILOGARITHMS. 235 1 Q 3 4 5 6 7 8 9 12 34 5 6 7 8 9 '50 3162 3170 3177 3184 3192 3i99 3206 3 2 i4 3221 3228 i i 2 3 4 4 5 6 7 '51 3236 3243 3251 3 2 58 3266 3273 3281 3289 3296 334 I 2 2 3 4 5 5 6 7 '52 33ii 3319 3327 3334 3342 335 3357 3365 3373 338i I 2 2 3 4 5 5 6 7 "53 3388 3396 3404 3412 3420 3428 3436 3443 3451 3459 I 2 2 3 4 5 6 6 7 '54 3467 3475 3483 349i 3499 35* 35i6 3524 3532, 3540 I 2 2 3 4 5 6 6 7 "55 3548 3556 3565 3573 358i 3589 3597 3606 3614 3622 I 2 2 3 4 5 6 7 7 56 3631 3639 3648 3656 3664 3673 3681 3690 3698 3707 I 2 3 3 4 5 6 7 8 3 3715 3802 3724 3733 3811)3819 374i 3828 3750 3837 3758 3846 3767 3855 3776 3864 3784 3873 3882 I 2 I 2 3 3 3 4 4 4 n 7 8 7 8 '59 3890 3899 3908 391713926 3936 3945 3954 3963 3972 I 2 3 4 5 5 6 7 8- 6o 398i 3990 3999 4009 4018 4027 4036 4046 4055 4064 I 2 3 4 5 6 6 7 8 61 4074 4083 4093 4102 4111 4121 413 4140 4i5o 4159 I 2 3 4 5 6 7 8 9, 62 63 ! 4178 4276 4188 4285 4198 4295 4305 4217 4315 4227 4325 4236 4335 4246 4345 4256 4355 I 2 I 2 3 4 3 4 5 5 6 7 6 7 8 9 8 9 64 4365 4375 4385 4395 4406 4416 4426 4436 4446 4457 I 2 3 4 5 6 7 8 9 65 4467 4477 4487 4498 4508 4519 4529 4539 4550 456o I 2 3 4 5 6 7 8 9, 66 457i 458i 4592 4603 4613 4624 4634 4645 4656 4667 I 2 3 4 5 6 7 9 io 67 68 69 70 4677 4786 4898 5012 4688 4797 4909 5023 4699 4808 4920 5035 4710 4819 4932 5047 4721 4831 4943 5058 4732 4842 4955 5070 4742 4853 4966 5082 4864 4977 5093 4764 4875 4989 5105 4775 4887 5000 5"7 I 2 I 2 I 2 I 2 3 4 3 4 3 5 4 5 6 6 7 8 7 8 7 8 7 8 Q IO J 9 io 9 io- 9 ir 71 5129 5MO 5152 5164 5176 5188 5200 5212 5224 5236 I 2 4 5 6 7 8 IO It 72 5248 5260 5272 5284 5297 5309 532i 5333 5346 5358 I 2 4 5 6 7 9 IO II 73 5370 5383 5395 5408 5420 5433 5445 5458 5470 5483 I 3 4 5 6 8 9 10 II 74 5495 5508 5534 5546 5572 5585 5598 5610 I 3 4 5 6 8 9 10 12 75 5623 5636 5&49 5662 5675 5689 5702 5715 5728 574i i 3 4 5 7 8 9 10 12" 76 5754 5768 578i 5794 5808 ^821 5834 5848 5861 5875 i 3 4 5 7 8 9 II I2-- 5888 5902 59i6 5929 5957 5970 59845998 6012 i 3 4 5 7 8 io II 12 78 6026 6039 6053 6067 6081 6095 6109 61246138 6152 i 3 4 6 7 8 io II 13; 79 6166 6180 6i94 6209 6223 6237 6252 62666281 6295 i 3 46 7 9 io II 13 80 6310 6324 6339 6353 6368 6383 6397 64126427 6442 1 3 46 7 9 io 12 13- 81 82 83 6457 6607 6761 6471 6622 6776 6486 6501 6637)6653 6792(6808 6516 6668 6823 6531 6683 6839 6546 6699 6855 65616577 67146730 68716887 6592 6745 6902 2 3 2 3 2 3 II 8 8 8 9 ii 9 ii 9 ii 12 14 12 14, 13 14 84 6918 6934 6950 6966 6982 6998 7015 70317047 7063 2 3 56 8 10 II 13 IS 85 7079 7096 7112 7129 7145 7161 7178 71947211 7228 2 3 5 7 8 10 12 !3 IS 86 87 7244 7261 7430 7278 7447 7295 7464 73" 7482 7328 7499 7345 73627379 75i6 7534:7551 7396 7568 2 3 2 3 57 5 7 8 9 10 12 10 12 13 is 14 16 88 7586 7603 7621 7638 7656 7674 7691 7709^7727 7745 2 4 5 7 9 II 12 14 16- '89 7762 7780 7798 7816 7834 7852 7870788917907 7925 2 4 5 7 9 II 12 14 16 90 7943 7962 7980 7998 8017 8035 8054 8072 8091 8 1 io 2 4 6 7 9 .113 IS 17 gj 8128 8i47 8166 8185 8204 8222 $241 826018279 8299 2 4 6 8 9 II I3il5 17 92 8318 8337 8356 8375 8395 8414 8433 84531847218492 24 68 10 12 1415 I 7 '93 8511 8531 8551 8570 8590 8610 8630 8650*86708690 2 4 6 8 IO 12 1416 18 *94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 2 4 6 8 IO 12 14 16 18- '95 8913 8933 8954 8974 8995 9016 90369057 9078 9099 2 4 6 8 IO 12 15 17 19- 9 6 98 9120 9333 9550 9i4i 9354 9572 9162 9376 9594 9183 9397 9616 9204 9419 9638 9226 9661 924719268 9290 946219484)9506 9683 9705J9727 93H 9528 9750 2 4 2 4 2 4 6 8 7 9 7 9 ii ii ii 13 15 17 19- 13 15:17 20 13 16)18 20- '99 |977 2 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 5 7 9 ii 14 16 1 8 20* INDEX Activity, or rate of doing work (P), I measurement of, by ammeter and voltmeter, 12 Alternating-current circuit, 113 shape of wave, 1 14 value of B.E.M.F. in finite solenoid, 131 - chemical effect, 115 - thermal effect, 116 virtual value of, 117 magnetic effect of, 118-144 B.E.M.F. of self-induction, 120, 127 - power in circuit, 125 - lag of, 123 lag of, with small B.E.M.F., 125 medium B.E.M.F., 126 large B.E.M.F., 127 Ammeter, winding of coil of, 85 Ampere-turns magnetizing force, 70 required for armature, 77 a j r gapSf 77 magnet limbs, 77 yoke, 77 compounding, 90 by transformer iron, 155 voltmeters and ammeters, 85 Amperes required by tramcar, 109 Arc lamps, 33 Area of armature iron, 75 air gaps, 75 magnet limb, 75 Armature, reactions in, 86 wire, pull on, 48 total pull on, 49 power to drive, 50 winding of, 55 losses in, 55 temperature rise of, 56 hysteresis loss, calculation of, 57 eddy-current loss, calculation of, 57 cross magnetizing wires on, 88 demagnetizing wires on, 88 Index. 237 0, the magnetic density, 65, 68 B.E.M.F. of secondary cell, 28 self-induction, 120 , magnitude of, 127 Capacity in alternating-current circuit, 160 in parallel with circuit, 170 in series with circuit, 169 , current due to, 165 , E.M.F. *due to, 167 of concentric cables, 171 size of I farad condenser, 169 , specific inductive, values of, 168 Circuit, simple, I , series, 3 , parallel, 4 , multiple wire, 15 magnetic, compared with electric, 72 Coil winding, series class, 82 , shunt class, 83 Condenser, use of, in alternating-current circuit, 169 , size of i farad, 169 Conductivity (m), 2 Conductance, 2 Cross magnetizing wires on armature, 88 Demagnetizing wires on armature, 88 Distribution, 15 Distributor, definition of, 17 Drop of volts in distributor, 16 in feeder, 16 in armature, series, 36 , shunt, 43 , compound, long shunt, 44 short shunt, 45 in transformer primary, 149 secondary, 149 due to B.E.M.F. of secondary, 150- due to magnetic leakage, 150 Dynamo, formula for E.M.F. of, 35 -, P.D. at terminals, 36 , separately excited, 40 , series, 41 , shunt, 42 , compound, long shunt, 43 , short shunt, 44 , calculation of amount of series winding, 86-92-. Economical size of feeders, 19 Eddy-current loss in armature, 57 in transformer, 150 , current to balance, 136 Efficiency of distribution and transmission, 95 magneto machine, 97 Index. Efficiency of series dynamo, 97 shunt dynamo, 97 compound dynamo, short shunt, 98. , long shunt, 99 series motor, 99 shunt motor, 100 compound motor, short shunt, 100 , long shunt, 101 secondary battery, 101 combined plant, 101 transformer, all-day, 156 Electro-motive force, E.M.F. defined, i E.M.F. active (e), 27 back (c), 27 total (E), 27 Feeder, definition of, 17 -, economical size of, 19 , working current in, 19 Field-magnet winding, series, 39-82 , shunt, 83 , compound, 90 Five-wire system, 23 Gap, air or iron to iron space, area of, 75 Gas-engine, cost of gas consumed by, 181 H, strength of magnetic field, or magnetizing force, 67 Heating of armature, 56 voltmeter, 102 Hysteresis loss in transformer, 151 in armature, 57 , effect of on shape of current curve in alternating-current circuit, 137 in cyclic change of magnetization, 138 Impedance coils, 145 , why open iron circuit type, 140 Insulation, electric resistance of, 10 , magnetic, 73 Impressed E.M.F. alternating, how found, 122 Kelvin's law of economy, 19 Kennelley's safety rule, 21 Leakage of magnetic flux, 73 coefficient (j/), 74 drop in transformer, 150 Length, effective, of solenoid, 132 of magnetic path through armature, 75 air gaps, 75 magnet limbs, 75 yoke, 75 Index. 239 Magnet coil, series, size of wire for, 82 , shunt, size of wire for, 83 Magnetic properties of iron, 69 circuit compared with electric, 72 of dynamo, analysis of, 75 of finite solenoid, 131 Magnetizing force (H) physical measure, 68 (ffil) ampere-turns, 70 Motors, 31 , armature reactions in, 92 , compound winding of, 93 , B.E.M.F. of, 51 Ohm's law for continuous currents, I Ohmmeter, connections of, 14 Open iron circuit transformer, 143 Periodicity or frequency, 1 14 Permeability (/*), 66 Potential difference (P.D.) definition of, I Power to drive armature, 50 Power or activity, (P), 2 Power factor (<) in alternating-current circuit, 126 Powermeter, connections of, 14 Powermeter for alternating current circuit, 1 70 Prime mover costs, 181 Pull on armature wire, 48 Quantity- efficiency of secondary battery, 101 Railway, electric, 106 tractive force required, 106 power required, 107 pull to get up speed, 109 E.M.F. required, 107 current, 109 shunt motor on, 1 10 compound motor on, 1 1 1 series motor on, 112 Rate of doing work, or activity, I Regulation of speed of motor, 1 74 Reluctance magnetic (2&), 72 of air space surrounding coil, 131 Resistance to start a motor, 174 in series with motor, 1 76 in shunt coil of motor, 177 calculation of series circuit, 2 parallel circuit, 5 of insulation, 10 measurement of by ammeter and voltmeter, 12 Resistivity (p) definition of, 2 Safety rules, 21 Secondary battery, E.M.F. curves, 28 240 Index. Solenoid, infinite, magnetic properties of, 67 , finite, effective length of, 131 Specific inductive capacity, values of, 168 Steam engine, cost of fuel in, 181 Temperature rise, Esson's rules, 55 Transformer, balancing of secondary output in closed iron circuit type, 14 1 in open iron type, 143 , relationship between power factor and output, 143 , design of, 148 , eddy-current loss in, 150 , hysteresis loss in, 151 , all-day efficiency of, 156 Voltmeter, size of wire for coil of, 85 , resistance of, 102 , extra coil for, 104 , effects of temperature on, 102 , heating error in, 102 , temperature error in, 102 , constant of, at any temperature, 105 Work, rate of doing, I Watt, defined, 2 THE END. PRINTED BY WILLIAM CLOWES AND SONS, LIMITED, LONDON AND BECCLES. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. ENGINEERING LIBRARY OCT 8 1948 JUL 5 1950 f< LD 21-100m-9,'47(A5702sl6)476 VB 24105