i 3. GENERATION OF E.M.F. 15. A closed conductor, convolution or turn, revolving in a magnetic field, passes during each revolution through two positions of maximum inclosure of lines of magnetic force A in Fig. 5, and two positions of zero inclosure of lines of mag- netic force B in Fig. 5. 1 cm.3 refers to a cube whose side is 1 cm., and should not be confused with cu. cm. 12 ELEMENTS OF ELECTRICAL ENGINEERING Thus it cuts during each revolution four times the lines of force inclosed in the position of maximum inclosure. If 3> = the maximum number of lines of force inclosed by the conductor, / = the frequency in revolutions per second or cycles, and n = number of convolutions or turns of the con- ductor, the lines of force cut per second by the conductor, and thus the average generated e.m.f. is, E = 4 fn$ absolute units, = 4fn3> ID"8 volts. FIG. 5. — Generation of e.m.f. If / is given in hundreds of cycles, <£ in megalines, E = 4n$ volts. If a coil revolves with uniform velocity through a uniform magnetic field, the magnetism inclosed by the coil at any instant is, $ COS T where $ = the maximum magnetism inclosed by the coil arid T = angle between coil and its position of maximum inclosure of magnetism. The e.m.f. generated in the coil, which varies with the rate of cutting or change of $ cos T, is thus, e = EQ sin T, where EQ is the maximum value of e.m.f., which takes place for T = 90°, or at the position of zero inclosure of magnetic flux since in this position the rate of cutting is greatest. 2 Since avg. (sin T) = -, the average generated e.m.f. is, GENERATION OF E.M.F. 13 Since, however, we found above that E = 4 fn& is the average generated e.m.f., it follows that EQ = 2 irfn& is the maximurft, and e = 2 7r/n$ sin r the instantaneous generated e.m.f. The interval between like poles forms 360 electrical-space de- grees, and in the two-pole model these are identical with the mechanical-space degrees. With uniform rotation, Fig. 6, the space angle, r, is proportional to time. Time angles are designated by 6, and with uniform rotation & = r, T being measured in elec- trical-space degrees. FIG. 6. — Generation of e.m.f. by rotation. The period of a complete cycle is 360 time degrees, or 2 TT or -- seconds. In the two-pole model the period of a cycle is that of one complete revolution, and in a 2 np-pole machine, — of that np of one revolution. Thus, 6 = 2 wft e = 2>jrfn3> sin 2 irft. If the time is not counted from the moment of maximum Lnclosure of magnetic flux, but ti = the time at this moment, we have e = 2 irfn$ sin 2 IT} (t - ti) or, e = 2-jrfn& sin (6 — 0i), where 0i = 2 irfli is the angle at which the position of maxi- mum inclosure of magnetic flux takes place, and is called its phase. These e.m.fs. are alternating. If at the moment of reversal of the e.m.f. the connections between the coil and the external circuit are reversed, the e.m.f. in the external circuit is pulsating between zero and EQ, but has the same average value E. If a number of coils connected in series follow each other H ELEMENTS OF ELECTRICAL ENGINEERING successively in their rotation through the magnetic field, as the armature coils of a direct-current machine, and the connections of each coil with the external circuit are reversed at the moment of reversal of its e.m.f., their pulsating e.m.fs. superimposed in the external circuit make a more or less steady or continuous external e.m.f. The average value of this e.m.f. is the sum of the average values of the e.m.fs. of the individual coils. Thus in a direct-current machine, if $ = maximum flux in- closed per turn, n = total number of turns in series from com- mutator brush to brush, and / = frequency of rotation through the magnetic field. E = 4/n$> = generated e.m.f. ($ in megalines, / in hundreds of cycles per second). This is the formula of the direct-current generator. EXAMPLES 17. (1) A circular wire coil of 200 turns and 40 cm. mean diameter is revolved around a vertical axis. What is the horizontal intensity of the magnetic field of the earth, if at a speed of 900 rev. per min. the average e.m.f generated in the coil is 0.028 volt? 4027T The mean area of the coil is — j— = 1255 sq. cm., thus the terrestrial flux inclosed is 1255 H, and at 900 rev. per min. or 15 rev. per sec., this flux is cut 4 X 15 = 60 times per second by each turn, or 200 X 60 = 12,000 times by the coil. Thus the total number of lines of magnetic force cut by the conductor per second is 12,000 X 1255 H = 0.151 X 108 H, and the average generated e.m.f. is 0.151 H volts. Since this is = 0.028 volt, H = 0.186. 18. (2) In a 550-volt direct-current machine of 8 poles and drum armature, running at 500 rev. per min., the average vol- tage per commutator segment shall not exceed 11, each armature coil shall contain one turn only, and the number of commutator segments per pole shall be divisible by 3, so as to use the machine as three-phase converter. What is the magnetic flux per field pole? 550 volts at 11 volts per commutator segment gives 50, or as next integer divisible by 3, n = 51 segments or turns per pole. POWER AND EFFECTIVE VALUES 15 8 poles give 4 cycles per revolution, 500 rev. per min. gives 50%Q = 8.33 rev. per sec. Thus the frequency is/ = 4 X 8.33 = 33.3 cycles per second. The generated e.m.f. is E = 550 volts, thus by the formula of direct-current generator, E = 4/n, or, 550 = 4 X 0.333 X 51 , = 8.1 megalines per pole. 19. (3) What is the e.m.f. generated in a single turn of a 20-pole alternator running at 200 rev. per min., through a magnetic field of 6.4 megalines per pole? 20 y 200 The frequency is / = V£~£ ~ = 33-3 cycles- ^ X. OU e = EQ sin r, EQ = 2 irfn3>, $ = 6.4, n = l, f = 0.333. Thus, EQ = 2 TT X 0.333 X 6.4 = 13.4 volts maximum, or e = 13.4 sin 0.