ELECTRICAL
ENGINEERING
THE THEORY AND CHARACTERISTICS
OF ELECTRICAL CIRCUITS
AND MACHINERY
BY
CLARENCE V. CHRISTIE, M. A., B. Sc.
ASSOCIATE PROFESSOR OP ELECTRICAL ENGINEERING, MCGILL
UNIVERSITY, MONTREAL, CANADA
SECOND EDITION
REVISED AND ENLARGED
SECOND IMPRESSION
McGRAW-HILL BOOK COMPANY, INC.
239 WEST 39TH STREET. NEW YORK
LONDON: HILL PUBLISHING CO., LTD.
6 & 8 BOUVERIE ST., E. C.
1917
'?/:/
Engineering
Library
COPYRIGHT, 1913, 1917, BY THE
McGRAw-HiLL BOOK COMPANY, INC.
n j C
THE MAPLE PRESS YORK PA
PREFACE TO THE SECOND EDITION
THE second edition contains all the material in the original text
but much of it has been rewritten and a great deal of new material
added.
The more important additions include sections on complex
alternating waves and wave analysis, on polyphase alternating
current circuits, on the construction of the characteristic curves
of direct-current generators and motors, on the design of direct-
and alternating-current machinery, on the Blondel diagram for
the synchronous motor, on the symbolic method of analysis of
the induction motor, on alternating-current commutator motors,
and finally a chapter on electrical measuring instruments.
The chapter on direct-current machinery has been entirely
rewritten and much enlarged, and to make it complete a short
chapter outlining the design of a direct-current generator has
been added.
This chapter and the other sections dealing with design are not
intended to cover the work required in a course on design but only
to give the student some idea of the formulae and constants
involved.
The chapter on measuring instruments has been placed at the
end of the book because some of the principles involved cannot
be well understood by the student until he has mastered the
theory of the more important electrical machines.
It was originally intended to introduce problems at the end of
each chapter to be worked out by the student and also to give a
list of references to the most important articles covering the sub-
ject matter of the chapter, but there was danger of making the
volume too cumbersome. A number of good books of problems
have been published recently and the student will find good lists
of references at the ends of the various sections in the Standard
Handbook.
My thanks are due to those teachers who have offered valuable
criticisms of the arrangement and method of treatment in the
first edition. , *
CLARENCE V. CHRISTIE.
McGiLL UNIVERSITY, MONTREAL,
August, 1917.
PREFACE TO THE FIRST EDITION
THIS book has been compiled as a foundation for lecture courses
for junior and senior students in Electrical Engineering.
The theory and characteristics of electrical machines are
developed from the fundamental principles of electrostatics and
electromagnetics. Only the more standard types have been dis-
cussed since familiarity with the principles of .their operations will
guide the student to a complete understanding of other machines
which differ only in minor respects. This general groundwork
may be extended to suit the requirements of particular classes.
C. V. C.
McGiLL UNIVERSITY, MONTREAL,
October 14, 1913.
Vll
CONTENTS
PA.OE
PREFACE v
CHAPTER I
ELECTROSTATICS
1. Electrification 1
2. Electrical conductors and insulators 1
3. Electrostatics and electromagnetics 2
4. Laws of electrostatics 2
5. Coulomb 2
6. Electrostatic field 3
7. Field surrounding a point charge 3
8. Dielectric flux from a unit charge 4
9. Field between two point charges 4
10. Field between parallel plates 5
11. Potential 5
12. Volt 6
13. Induced charges of electricity 6
14. Equivalent charges 7
15. Distribution of potential in the space surrounding a point charge . 7
16. Potential at a point due to a number of charges 8
17. Potential at a point due to a charged surface 8
18. Equipotential surfaces 9
19. Potential gradient 9
20. Potential and potential gradient near a charged sphere 10
21. Potential and potential gradient between two charged spheres . . 11
22. Capacity 12
23. Farad 13
24. Dielectric permeance 13
25. Condenser 14
26. Parallel plate condenser 14
27. Capacity of concentric cylinders 15
28. Potential and potential gradient between concentric cylinders . . 16
29. Capacity of parallel conductors 17
30. Potential and potential gradient between parallel conductors . . 19
31. Dielectric field and equipotential surfaces between parallel con-
ductors 21
32. Capacity of a single wire to earth s 22
33. Capacity of a sphere to earth 23
34. Condensers in multiple 24
35. Condensers in series 24
36. Energy stored in a condenser 25
ix
x CONTENTS
i
PAGE
37. Stresses in an electrostatic field 27
38. Force exerted on a dielectric by an electrostatic field 28
39. Effects of introducing dielectrics of various specific inductive
capacities into a uniform field 30
40. Effect of introducing conductors into electrostatic fields .... 34
41. Graded insulation for cables 35
42. Air films in generator slot insulation . . . 39
43. Condenser bushing 40
44. Dielectric strength 41
45. Breakdown 43
46. Dielectric losses 44
47. Surface leakage 45
48. Corona 45
CHAPTER II
MAGNETISM AND ELECTROMAGNETICS
49. Magnetization 46
50. Laws of magnetism 47
51. Magnetic field 47
52. Magnetic flux 47
53. Flux from unit pole 48
54. Magnetic potential 48
55. Magnetomotive force 49
56. Permeability 50
57. Magnetic reluctance 50
58. Permeance 51
59. Electromagnetics 51
60. Laws of induction 52
61. Unit of electromotive force 52
62. Force exerted by a magnetic field on an electric circuit 53
63. Unit current 54
64. Transformation of mechanical energy to electrical energy .... 54
65. Electric power and energy 55
66. Intensity of magnetic fields produced by electric currents .... 56
67. Magnetomotive force of a solenoid 61
68. Examples 61
69. Energy stored in the magnetic field ., . > . '. . 66
70. Stress in the magnetic field -., 67
71. Force between parallel wires carrying current 68
72. Magnetic characteristics 69
73. Hysteresis . ... . 71
74. Magnetic materials 73
75. Effect of chemical composition and physical treatment on hystere-
sis loss 74
76. Theories of magnetism 75
77. Lifting magnets 76
CONTENTS xi
CHAPTER III
ELECTRIC CIRCUITS
PAGE
78. Ohm's Law 79
79. Joule's law . 79
80. Heat units 80
81. Examples 80
82. Resistance 80
83. Conductance 81
84. Effect of temperature on resistance 81
85. Properties of conductors 83
86. Resistance of conductors 84
87. Drop of voltage and loss of power in a distributing circuit ... 86
88. Current-carrying capacity of wires 87
89. Examples 87
90. Kirchoff's laws 88
91. Examples 89
92. Resistances in series 90
93. Resistances in parallel 90
94. Potentiometer 91
95. Inductance 92
96. Examples 93
97. Inductance of circuit^ containing iron 94
98. Mutual inductance and self -inductance 95
99. Self-inductance of continuous-current circuits 97
100. Example 101
101. Inductance of parallel conductors 102
CHAPTER IV
ELECTRIC CIRCUITS (CONTINUED)
102. The sine wave of electromotive force and current 104
103. The average value of a sine wave 106
104. The effective value of a sine wave 106
105. Inductance in alternating-current circuits . 107
106. Resistance and reactance in series 109
107. Capacity in alternating-current circuits Ill
10&. Resistance and condensive reactance in series 112
109. Resistance, inductance and capacity in series 113
110. Vector representation of harmonic quantities 115
111. Power and power factor 116
112. Examples 122
113. Numerical examples 126
114. Circuit constants ....: 131
115. Rectangular coordinates 134
116. Examples in rectangular coordinates . 138
117. Kirchoff's laws applied to alternating-current circuits 141
xii CONTENTS
CHAPTER V
COMPLEX ALTERNATING-CURRENT WAVES
PAGE
118. Complex alternating waves 142
119. Examples 143
120. Analysis of alternating waves 146
121. Example of analysis 149
CHAPTER VI
POLYPHASE ALTERNATING-CURRENT CIRCUITS
122. Polyphase Alternating-current circuits 152
123. Three-phase circuits 154
124. Electromotive forces, currents and power in three-phase circuits 156
125. Measurement of power in polyphase circuits 159
126. Examples 164
CHAPTER VII
DIRECT-CURRENT MACHINERY
127. The direct-current dynamo : 165
128. Yoke 165
129. Pole pieces . . . : 165
130. Armature core , 167
131. Armature winding 167
132. Ring windings 167
133. Drum winding 168
134. Multiple-drum windings 171
135. Equalizer rings 171
136. Series-drum windings 172
137. Double windings 174
138. Commutator 176
139. Brushes and brush holders 178
140. Field windings 178
141. Direction of rotation of generators and motors 179
142. Generation of electromotive force 180
143. Effect of moving the brushes 181
144. Building up of electromotive force in a self -excited generator . . 182
145. Armature reaction and distribution of magnetic flux 182
146. No-load saturation curve 186
147. Load saturation curves 187
148. Voltage characteristic or regulation curve . . . . > . . . . . 188
149. Field characteristic or compounding curve 190
150. Voltage characteristic of a shunt generator . 191
151. Effect of change of speed on the voltage of a shunt generator . .192
152. Effectof saturation on the voltage characteristic of a shunt generator. 193
CONTENTS xiii
PAGE
153. Compound generator 194
154. Voltage characteristic of a compound generator 195
155. Short-shunt and long-shunt connection 196
156. Regulation ....... 196
157. Series generator . .... . ... . . . 196
158. Electric motors . . . . . 198
159. Types of motors . 198
160. Speed equation of a motor 199
161. Methods of varying speed \ ...... 199
162. Speed characteristics of motors 201
163. Torque equation . . v . . ... . . 203
164. Torque characteristics of motors . 204
165. Construction of the speed characteristics 204
166. Construction of the speed characteristic for a series motor . . . 205
167. Variation of speed of a shunt motor with line voltage 206
168. Variation of speed with temperature of the field coils 207
169. Construction of the torque characteristics 207
170. Starting torque 208
171. Motor starter with no-voltage release 209
172. Adjustable speed operation 210
173. Multiple-wire systems of speed control 211
174. Ward Leonard system of speed control 211
175. Speed control of series motor 212
176. Interpole motors .*....... 213
177. Applications of motors . 213
178. Losses in direct-current machinery 215
179. Shunt-field loss ! . .... f 215
180. Series-field loss 215
181. Armature copper loss 216
182. Loss at the brush contacts *i . . - 216
183. Hysteresis loss ...... . . . . . , . . . . . . . , . . 217
184. Eddy current loss 218
185. Pole-face loss . . , . ..... . . . . , . . .... ... 219
186. Brush-friction loss v ......... . 220
187.. Bearing-friction and windage losses . ; ;.. . 220
188. Eddy-current losses in the armature copper . . . ... . . . . 221
189. Constant losses and variable losses 222
190. Core loss. . . . . ... . . . . , . .'. 222
191. Efficiency. . . . . . , . . . < . .-;:. 222
192. Commutation 224
193. Commutating electromotive force . . . 231
194. Conditions essential to sparkless commutation 233
195. Calculation of the reactance voltage for a full-pitch multiple
winding . ...._. . . . V , -, . . . 233
196. Current density at the brush contacts . . . . . . . . .-. ,.. . 236
197. Burning of the brush and commutator 236
198. Poor commutation resulting from too many coil sides per slot . . 237
199. Interpoles 237
xiv CONTENTS
PAGE
200. Compensating windings . . .. ... 4 ; ". . ,. . . 239
201. Flashing . .... > v- . .>. ; . . .";*....:.'. 240
202. Parallel, operation . . v V- : . i _. -v.'v. ;..'.. 242
203. Parallel operation of compound generators ..... .. .;; - . . . . 243
204. Storage batteries ... .^ ..:;.-..... 245
205. Applications :.-; - . . . . . . 246
206. Boosters ... ..:.... 247
207. Balancers . V . . 249
208. Rosenberg generator for train lighting ...-.. 250
209. Homopolar generators 251
210. Limits of output of electric machines 252
211. Temperature limits . . . . ... . . . 253
212. Temperature of commutators 255
213. Temperature of cores . . <' . . . .. . , ; . . 255
214. Temperature of other parts ... t ... . . . 255
215. Ventilation 256
216. Semi-enclosed and totally enclosed machines . . . ...... . 256
CHAPTER VIII
DESIGN OF A DIRECT-CURRENT GENERATOR
217. Symbols 258
218. Design of direct-current machinery 259
219. Magnetic leakage ;, ... 263
220. Design of a direct-current generator ..;... 264
221. Flux in the air spaces between the teeth . . 267
222. Effect of tooth taper , . . . 268
223. Ampere-turns per inch for an air path . -.,;. . 268
224. Air gap ^ . .. : X,,. . 269
225. Length of the air gap V 269
226. Design of poles and yoke 270
227. Determination of the no-load saturation curve . . : . , . . 270
228. Field winding -. - 273
229. Current density in field windings . . . ., . /. -.274
230. Size of wire for the field winding -. > ; ; : .;. ^,... . . 274
231. Resistance of the armature winding . . . 275
232. Determination of the magnetomotive force of the series winding 275
233. Design of the series-field winding 277
234. Design of the commutator . . 278
235. Losses , , .... .V>- .'V . 279
CHAPTER IX
SYNCHRONOUS MACHINERY
236. Alternator 280
237. Types of alternators 280
CONTENTS xv
PAGE
238. Electromotive force equation 284
239. Form factor 286
240. Polyphase alternating-current generators . 286
241. Alternator windings " .'.... 289
242. Distribution factors 290
243. Multiple-circuit windings 294
244. Short-pitch windings 294
245. Effect of distributing the winding 297
246. Harmonics due to the teeth 297
247. Effect of third harmonics in three-phase alternators 299
248. General electromotive force equation 300
249. Rating of alternators. . . . S * . 300
250. Comparative ratings of an alternator wound single, two- and
three-phase ' 301
251. Armature reaction 302
252. Armature reactance 304
253. Leakage fluxes 304
254. Polyphase armature reaction 305
255. Single-phase armature reaction 309
256. Electromotive forces in the alternator 311
257. Armature resistance 312
258. Vector diagram of electromotive forces and magnetomotive
forces 313
259. Voltage characteristics 315
260. Compounding curves 317
261. Tests for the determination of the regulation of alternating-
current generators 317
262. Regulation 321
263. Short-circuit of alternators 323
264. Synchronous motor 327
265. Vector diagrams for a synchronous motor 328
266. Characteristic curves 330
267. Compounding curves 330
268. Load characteristics 331
269. Phase characteristics 333
270. Torque 334
271. Blondel diagram for a synchronous motor 334
272. Synchronizing power 338
273. Construction of the characteristic curves of a synchronous motor
from the Blondel diagram 339
274. Load characteristics 340
275. Phase characteristics or "V" curves and compounding curves. . 341
276. Starting synchronous motors 342
277. Self -starting motors 343
278. Synchronous phase modifier 345
279. Parallel operation of alternators .,,',.. 346
280. Effect of inequality of terminal voltage 346
281. Effect of inequality of frequency 347
xvi CONTENTS
PAGE
282. Effect of difference of wave form 350
283. Conclusions 351
284. Hunting . .." 351
285 . Frequency of hunting . . . 352
286. Design of alternating-current generators and motors 354
287. Electromotive force equation 354
288. Output equation 355
289. Flux densities 356
290. Current densities 356
291. Insulation for high-voltage alternators 356
292. Extra insulation required under special conditions 357
293. Armature windings 357
294. Slots per pole 357
295. Regulation 357
296. Excitation regulation 358
297. Excitation .358
298. Losses 359
299. Ventilation 359
300. Cylindrical rotors 361
CHAPTER X
TRANSFORMERS
301. The constant-potential transformer 363
302. Vector diagrams for the transformer 366
303. Exciting current 368
304. Leakage reactances 368
305. The transformer as a circuit 370
306. Examples ; . . 371
307. Measurement of the constants of a transformer 374
308. Regulation , , 376
309. Voltage characteristics 378
310. Losses in transformers 378
311. Hysteresis loss , v., ... 378
312. Eddy-current loss in transformer 379
313. Extra losses 380
314. Efficiency 381
315. Types of transformer 383
316. Methods of cooling ,.'......,. 384
317. Transformer connections . . . 386
318. Single-phase transformers on polyphase circuits 387
319. Star-star connection . . 388
320. Delta-star and star-delta connection . . 390
321. Delta-delta connection . . 391
322. Open-delta connection ' 391
323. "T" connection . . 392
324. Transformation from two-phase to three-phase 393
CONTENTS xvii
PAGE
325. Single-phase power from three-phase circuits 395
326. Multiple operation of transformers 395
327. Booster transformers 39(5
328. Auto-transformers. .'. 396
329. Instrument transformers 397
330. Current transformers. '. ... 399
331. Example ............. ^ 401
332. The constant-current transformer 401
333. Regulation 402
334. Power factor . ... . ... . . 403
335. Induction regulator 403
336. Design of a transformer 405
337. Reactance 408
338. Design of a shell-type, water-cooled transformer, single-phase, 60
cycles, 1,000 kva., 22,000 to 2,200 volts for power transmission . 4.10
CHAPTER XI
CONVERTERS
339. Types of converters 414
340. Synchronous converter 414
341. Ratios of e.m.fs. and currents 415
342. Two-phase or quarter-phase converter 417
343. Three-phase converter 418
344. n-phase converter . . . . . . . . . . . . 419
345. Wave forms of currents in the armature coils . .'.'.'. . . . . 421
346. Heating due to armature copper loss . . ' 422
347. Armature reaction . . . 426
348. Control of the direct-current voltage '. . , . . . . 429
349. Methods of varying the alternating voltage ...:..... 430
350. Compounding by reactance ; . . 430
351. Synchronous booster converter /i . . . . .431
352. Direct-current booster converter . .x 431
353. Split-pole converter . . .' '. . 432
354. Frequencies and voltages . . .... . . ... ...... . . 432
355. Outputs and efficiencies -.<'. : . . . . 433
356. Overload capacity 433
357. Dampers , . .-.^. . . . . . . . 433
358. Starting . . ..... ..... . 433
359. Alternating-current self-starting 433
360. Direct-current self-starting 434
361. Starting by an auxiliary motor 434
362. Brush lifting device 434
363. Bucking and flashing 435
364. Parallel operation 435
365. Inverted converter 435
366. Double-current generator. ...'.. 436
xviii CONTENTS
J*AGE
367. Three-wire generator 437
368. Parallel operation of three-wire generators 439
369. Frequency converters i. . 439
370. Mercury vapor rectifier 440
371. Currents and voltages ; v ...... 441
372. Losses and efficiency : 442
373. Three-phase rectifier 442
374. Hot-cathode argon-filled rectifier 442
CHAPTER XII
INDUCTION MOTOR
375. Induction motor 444
376. The stator 444
377. Revolving magnetomotive force and flux of the stator 447
378. The rotor 450
379. Slip 450
380. Magnetomotive force of the rotor 451
381. Electromotive force and flux diagram for the induction motor . . 451
382. Proof that the locus is a circle 454
383. Magnetomotive force diagram = .-.-... . 454
384. Stator current diagram 454
385. Rotor electromotive force and current ' . . . . 456
386. Rotor input 456
387. Rotor copper loss and slip 457
388. Rotor output 457
389. Torque : . . . ... 457
390. Rotor efficiency 458
391. Modification of diagram . . . < 458
392. Interpretation of diagram 460
393. Construction of diagram from test for a three-phase motor . . .461
494. Test of an induction motor 462
395. Analysis by rectangular coordinates 465
396. Characteristics of an induction motor by the symbolic method . 470
397. Methods of starting 473
398. Applications . ......... 475
399. Speed control of induction motors . . . . . .-;.-. 476
400. Voltage control . . ". :. 476
401. Rotor resistance control 476
402. Pole-changing ',."*- 476
403. Cascade control or concatenation . . . .,;,.. 476
404. Exciting current 478
405. Leakage reactances , . '. j. . 479
406. Stator and rotor resistances 480
407. Effect of change of voltage and frequency . . . . . . V -. . . 481
408. Single-phase induction motor -.. 482
409. Horizontal field at slips >,......,. 484
CONTENTS xix
PAGE
410. Starting single-phase induction motors 484
411. Comparison of single-phase and polyphase motors 485
412. Induction generator . . . 486
413. Asynchronous phase modifier 488
414. Phase converter 491
415. Induction frequency converter .'.... ... . . 493
CHAPTER XIII
ALTERNATING-CURRENT COMMUTATOR MOTORS
416. Motor characteristics 494
417. Alternating-current series motor ;. 494
418. Design for minimum reactance 497
419. Compensating windings 497
420. Commutation ..... . . ... . ... . ... . . . 499
421. Speed control 500
422. Polyphase commutator motor 501
423. Repulsion motor ... , . ........... 501
424. Commutation . .'. ... . . . . . . . . . 503
425. Compensated repulsion motor '. . ..... . , 504
426. Alternating-current commutator motors with shunt characteristics 505
427. Single-phase induction motor with repulsion starting 506
CHAPTER XIV
TRANSMISSION SYSTEMS
428. Transmission line 507
429. Relative amounts of conducting material for single-, two- and
three-phase transmission lines 507
430. Reactance 508
431. Capacity ......... 509
432. Voltage and frequency . .... . 510
433. Spacing of conductors . . 510
434. Single-phase transmission line . . . . . .'. 510
435. Three-phase transmisson line . 517
436. Application of a synchronous phase modifier to a transmission
system .......... 521
437. High-voltage direct-current system .... . . . 524
438. Advantages and disadvantages of the Thury or series system . . 525
CHAPTER XV
ELECTRICAL INSTRUMENTS
439. Electrical instruments 527
440. Direct-current voltmeters and ammeters 527
441. Voltmeter multipliers 527
xx CONTENTS
PAGE
442. Ammeter shunts . . . . . . * v v * , 528
443. Thomson inclined-coil ammeter 528
444. Weston soft-iron-type ammeters and voltmeters 528
445. Electrodynamometer-type voltmeter 529
446. Hot-wire ammeters and voltmeters 529
447. Dynamometer-type wattmeter 530
448. Induction-type wattmeter 531
449. Power-factor meters 532
450. Frequency meters 533
451. The Weston frequency meter 534
452. Synchroscope 534
Tirrill regulator 535
454. Automatic voltage regulator for alternating-current generators . 537
INDEX . 541
ELECTRICAL ENGINEERING
CHAPTER I
ELECTROSTATICS
1. Electrification. Bodies which are charged with electricity
are said to be electrified. Charges are of two kinds called posi-
tive and negative. Bodies which have a positive charge are
acted upon by forces tending to make them give up their charge;
bodies which have a negative charge are acted upon by forces
tending to convey a positive charge to them equal to their nega-
tive charge. These forces are exerted through the medium sepa-
rating the charges and the medium is in a state of stress.
The body with the positive charge is at a higher potential than
the body with the negative charge and the difference of potential
between the two is a measure of the tendency for electricity to
pass from one to the other.
2. Electrical Conductors and Insulators. If two metallic
bodies charged to different potentials are joined by a metal wire,
electricity will flow from one to the other until the potential of
both is the same and the transfer of electricity will take place
almost instantaneously. The metal wire is therefore a good con-
ductor of electricity; or, it offers a low resistance to the passage
of electricity through it.
If the two charged bodies had been joined by a glass rod, there
would have been no transfer of electricity between them, or, it
would have taken place so slowly that it could only be detected
by the most delicate instruments. Glass is therefore a very bad
conductor; or, it offers a very high resistance to the passage of
electricity. It is called a non-conductor or insulator.
As all materials conduct to a certain extent, it is not possible
to divide them absolutely into conductors and insulators, but,
since the resistance of a good insulator is many million times
that of a good conductor, they may be so divided for practical
purposes.
I
2 ELECTRICAL ENGINEERING
In the first class are silver, which is the best conductor, copper
and other metals, graphite, impure water and solutions of salts.
In the second class are air, which when dry is an almost perfect
insulator, glass, paraffin, ebonite, porcelain, rubber, shellac, oils and
the numerous insulating compounds used in electrical engineering.
3. Electrostatics and Electromagnetics. Electrostatics com-
prises phenomena related to electric charges at rest and to the
stresses produced in the fields surrounding them. These phe-
nomena become of great importance where very large differences
of potential must be provided for, as for example in the design
and operation of all high voltage apparatus and systems.
Electromagnetics comprises phenomena related to electricity
in motion, that is, to currents of electricity and the magnetic
fields produced by them. Almost all the problems to be solved
by the electrical engineer come under this head.
4. Laws of Electrostatics. First Law. Like charges of elec-
tricity repel one another; unlike charges attract one another.
Second Law. The force exerted between two charges of elec-
tricity is proportional to the product of their strengths and is
inversely proportional to the square of the distance between
them; it also depends on the nature of the medium separating
them.
This law can be expressed by the formula,
where q and #1 are the charges of electricity, r is the distance
between them in centimeters, ft is a constant depending on the
medium separating the charges and is called its specific inductive
capacity or dielectric constant. The unit of quantity is so chosen
that the dielectric constant for air is unity; for all other sub-
stances it is greater than unity. The table on page 43 gives
the dielectric constants for some of the most common dielectrics.
/ is the force in dynes exerted between the two charges; if the
charges are of the same kind the force between them is a repulsion
and / is positive.
5. Coulomb. One electrostatic unit of quantity is that quan-
tity which, when placed at a distance of 1 cm. in air from a
similar quantity, repels it with a force of one dyne.
The practical unit of quantity is the coulomb; one coulomb is
3 X 10 9 electrostatic units; it is the quantity of electricity which
ELECTROSTATICS 3
passes a point in a conductor when one ampere flows for one
second (see Art. 63).
6. Electrostatic Field. Any space in which electrostatic forces
act is called an electrostatic field. The direction of the force at
any point in the field is the direction in which a unit positive
charge placed at the point tends to move and its intensity is the
force in dynes exerted on the unit charge.
The electrostatic field is conveniently represented by lines of
electrostatic induction or dielectric flux drawn in the direction of
the force. In air the number of lines per square centimeter is
equal to the force in dynes at the point and in a medium of di-
electric constant k the number of lines per square centimeter is
equal to k times the force. This may be stated in another way:
Unit electrostatic force produces one line of dielectric flux per
square centimeter in air and k lines per square centimeter in a
medium of dielectric constant k.
The electrostatic force at a point is expressed in dynes and is
represented by $; the dielectric flux density at a point is expressed
in lines per square centimeter and is represented by 3D.
7. Field Surrounding a Point Charge. At a distance r cm. in
air from an isolated charge q, a unit charge is acted on by a force
ff = | dynes; (2)
and the dielectric flux density produced at the point is
3) = $ \ lines per square centimeter. (3)
This density is produced over the surface of a sphere of radius r
and therefore the total dielectric flux from the charge q is
i mes .
Consider the same charge surrounded by a medium of dielectric
constant k.
The force exerted on a unit charge at a distance r cm. from q is
=
the dielectric flux density produced is
> = $k = ~ lines per square centimeter,
and therefore the dielectric flux from charge q is, as before,
lines. (4)
4 ELECTRICAL ENGINEERING
Fig. 1 represents the electrostatic field about a positive point
charge. The lines of flux extend out radially in all directions.
8. Dielectric Flux from a Unit Charge. The dielectric flux
from a unit charge is 4?r lines by equation 4. Thus if a dielectric
flux \f/ starts from any surface the positive charge on that surface
^
is q = -^ units, and if a flux ^ ends on any surface the negative
charge on that surface is q = -r- units.
-Q
FIG. 1. Field surround- FIG. 2. Field between two Fio. 3. Field between
ing a point charge. point charges. two parallel plates.
A unit positive charge is always associated with each 4rr lines
leaving a surface and a unit negative charge with each 4?r lines
entering a surface.
9. Field between Two Point Charges. The field between two
point charges q and q consists of curved lines extending from
the positive, to the negative charge.
The direction and intensity of the force at any point such as
P may be obtained by combining the forces exerted at the point
by the two charges as shown in Fig. 2. If a unit positive charge
is placed at the point P it is repelled by q with a force /i = ~ 2
ELECTROSTATICS 5
and is attracted by q with a force / 2 = ^y The resultant
force at the point is 9> and it is the vector sum of f\ and / 2 .
If D cm. is the distance between the charges, the force at a
point on the line joining the charges distant r cm. from the charge
q is
^ = ^ + (^~p dynes.
Since the medium is air the flux density at any point is numeric-
ally equal to the force ^ at the point.
10. Field between Parallel Plates. The field between two
parallel plates A and B, Fig. 3, charged respectively with +g
and q units per square centimeter consists of straight lines
normal to the plates except near the edges where they become
slightly curved.
The dielectric flux density is uniform throughout the volume
between the plates, and its value is
3D = 4:irq lines per square centimeter. (5)
If the medium between the plates is air, the electrostatic force
at any point is
ff = 3D = 4:irq dynes. (6)
If the medium has a dielectric constant k, the force is
> 4:irq , ,.
* == fc = "T ynes * ( }
11. Potential. The difference of potential between two points
is measured by the work done in carrying a unit charge from one
to the other against the electrostatic force in the field; it is there-
fore the line integral of the force between the points.
The difference of potential between the two plates in Fig. 3 is
f
Jo
(8)
where t cm. is the distance between the plates.
Difference of potential tends to cause electricity to flow from
one point to the other and is therefore called electromotive force.
Unit difference of potential (electrostatic) exists between two
points when one erg of work is done in conveying unit charge
from one to the other.
The earth is usually taken as the arbitrary zero of potential,
and the potentials of other bodies are given with reference to it.
6
ELECTRICAL ENGINEERING
If a difference of potential is produced between two points on
a conductor, electricity will flow from the point of high potential
until the potentials of all parts of the conductor are the same.
When, however, the difference of the potential is maintained
by an electric generator a current of electricity flows continuously
from one point to the other.
When a difference of potential is produced between two con-
ductors separated by an insulating material, the material is sub-
jected to a stress and lines of dielectric flux pass through it.
12. Volt. The practical unit difference of potential or electro-
motive force is called the volt. One electrostatic unit is equal
to 300 volts.
13. Induced Charges of Electricity. When a positively
charged body A, Fig. 4, is brought near to an insulated conductor
BC, a negative charge is induced on the nearer side B and an
equal positive charge on the farther side C. The explanation is
FIG. 4. Induced charges.
FIG. 5. Induced charges.
that the potential of the end B due to the charge on A is higher
than that of C, but, since BC is a conductor, as soon as there is
a difference of potential between two points on it, electricity
flows from the region of high potential at B to the region of low
potential at C; thus a positive charge appears at C and an equal
negative charge at B and these charges so distribute themselves
over the surface of BC that the potential at every point on it is
the same, being the sum of the potentials due to the charge on A
and the two charges on BC.
The same result is illustrated in Fig. 5, which shows the dielec-
tric flux produced.
The flux from A extends out radially in all directions. In the
region surrounding BC the lines are deflected and a large number
pass through the conductor J5C, since its dielectric constant is
infinite and therefore it offers an easy path. For every 4r lines
ELECTROSTATICS 1
entering at B a unit negative charge appears on the surface,
and for every 4?r lines leaving at C a unit positive charge appears.
14. Equivalent Charges. A uniformly distributed charge on
the surface of a sphere acts as though it were concentrated at the
center. If a sphere of radius R cm. has a charge Q uniformly
distributed over its surface, the density of the charge is
and since 4?r lines emanate from unit charge, the flux density at
the surface of the sphere is j~~ = lines per square centimeter.
If the charge Q is concentrated at the center of the sphere, then,
by formula (3), the flux density at a distance R cm. from the
charge is - and therefore the uniformly distributed charge on
the surface of the sphere may be represented by an equal charge
concentrated at the center.
Similarly a uniformly distributed charge on the surface of a
cylinder may be represented by an equal charge uniformly dis-
tributed along 'the axis of the cylinder.
15. Distribution of Potential in the Space Surrounding a Point
Charge. In Fig. 6 a positive charge q is placed at the point
i !
FIG. 6. Distribution of potential about a point charge.
at an infinite distance from all other charges. The potential at
a point PI, distant 7*1 cm. from 0, is the work done in moving
unit charge from a point of zero potential to the point PI against
the forces in the field. The intensity of the force at a distance
r cm. from is by equation 2
$ = ^ dynes;
the work done in moving a unit charge against this force through
a distance dr is
dr
8 ELECTRICAL ENGINEERING
and the work done in moving the charge from a point of zero
potential to PI is
rj r ^
ei = 5 dr = \ q.
Jri J ri * r 2
-H]; '
= ^ ergs; (9)
therefore, the potential at a point due to a charge of electricity
is equal to the charge in electrostatic units divided by its distance
in centimeters from the point.
16. Potential at a Point Due to a Number of Charges. If
there are several charges q\, q%, q$, etc., in the field, the potential
at any point is the sum of the potentials due to the separate
charges and is
where ri, r 2 , r 3 , etc., are the distances from the various charges to
the point (Fig. 7).
B
A
FIG. 7. Potential due to a num- FIG. 8. Potential due to a charged
ber of charges. surface.
17. Potential at a Point Due to a Charged Surface. In Fig. 8
AB is a surface with a non-uniform distribution of charge over it;
if dq is a small element of charge at a distance r cm. from the
point P, the potential at P due to the charge dq is
dq
de = i
r
and the potential due to the total charge on the surface is
e =
(ID
ELECTROSTATICS
9
18. Equipotential Surfaces. Surfaces of which all points are
at the same potential are called equipotential surfaces.
In Fig. 9, A is an isolated sphere of radius R, charged with
Q units of electricity. Any spherical surface drawn about the
center of A is an equipotential surface. The potential of surface
(1) is * that of (2) is and that of the sphere A is -
7*1 7*2 it
The difference of potential between surfaces (1) and (2) is
and is the work that must be done in taking a unit charge
from any point on (2) to any point on (1). It makes no differ-
ence what path the charge follows, because its path can always
be resolved into two displacements, one
along the equipotential surface and the
other normal to it; no work is done in
moving along the equipotential surface,
since there is no opposing force and there-
fore all the work is done in the displace-
ment along the normal.
The electrostatic force on an equipoten-
tial surface is normal to the surface, there-
fore lines of induction pass normally
through equipotential surfaces.
Electric conductors not carrying current are equipotential sur-
faces, since if differences of potential did exist electricity would
flow from the points of high potential until the potential became
uniform throughout the conductor.
19. Potential Gradient. The potential gradient at a point is
the space rate of change of potential at the point measured in
the direction of the electrostatic force.
The difference of potential between two points is
FIG. 9. Equipotential
surfaces.
f
Jo
(12)
where ^ is the electrostatic force at any point and D is the dis-
tance between the points. The potential gradient at any point is
de , 1Q v
= dr =
and is equal to the electrostatic force at the point.
In practical units the potential gradient is expressed in volts
per centimeter or kilovolts per centimeter.
10
ELECTRICAL ENGINEERING
20. Potential and Potential Gradient Near a Charged Sphere.
Determine the potential and potential gradient in the field
surrounding an isolated sphere of radius R charged with Q units
of electricity (Fig. 10).
The charge Q is uniformly distributed over the surface of the
sphere and may be considered as an equal charge Q concentrated
at the center (see Art. 14).
The potential at a point P, distant r cm. from the center of
the sphere is
Q
e =
The potential gradient at P is
de
Q
(14)
and is equal to the electrostatic force at the point.
FIG. 10. Potential and potential gradient near a charged sphere.
Curve 1, Fig. 10, shows the variation of potential with distance
measured from the center of the sphere. The equation of this
curve is
-7- -
Across the sphere the potential is constant and is equal to 75;
but near the surface it falls off very rapidly.
The potential gradient at any point is represented by the slope
of the tangent to the potential curve at that point; its values are
plotted in curve 2. The equation of this curve is
de Q
ELECTROSTATICS
11
21. Potential and Potential Gradient between Two Charged
Spheres. Determine the potential and the potential gradient at
any point between two spheres A and B (Fig. 11) of radius R cm.
if the distance between their centers is D cm. and they have
charges of Q and Q respectively. This last condition means
that all the lines of dielectric flux which leave the sphere A fall on
the sphere B.
FIG. 11. Potential and potential gradient between two charged spheres.
At a point P on the line joining the centers of the two spheres
and at a distance of r cm. from the center of A the potential due
to the charge Q on A is , and that due to the charge Q on B
s
-Q
; the actual potential at P is, therefore,
.Q ' Q
~ r D -r
(15)
Midway between the spheres the potential is
Q Q
D/2
At the surface of A it is
D -D/2
Q
= 0.
(16)
R D - R
The potentials at all points between A and B are plotted in curve
1 (Fig. 11).
12 ELECTRICAL ENGINEERING
The potential gradient at P is
_. _9_ / , jQ_\
r 2 (Z)-r) 2 \r 2 h (Z> - r) 2 /
Midway between the spheres the potential gradient is
/ I \ 8Q,
V(D/2) 2 t (D/2) 2 / " D 2
which is its minimum value and at the surface of either sphere it
(19)
R 2 (D - E) 2
its maximum value.
The potential gradient is plotted in curve 2, Fig. 11.
It has been assumed that the charges on the spheres are uni-
formly distributed, this is not correct as may be seen by an inspec-
tion of the field between two point charges in Fig. 2. The
flux and the charges are concentrated on the sides of the spheres
which are facing one another and the centers of the charges are
drawn together through a very short distance. Where the dis-
tance between the spheres is large compared to the radius, the
error is small.
22. Capacity. Conductors of different sizes and shapes have
different capacities for storing electricity. If two conductors of
different capacities are charged with equal quantities of electric-
ity, they will be raised to different potentials, the one of small
capacity will be raised to a high potential and the one of large
capacity to a low potential.
The potential to which a body is raised varies directly as the
quantity of electricity stored in it and inversely as its capacity
and may be expressed by the formula
E = |> (20)
where Q is the quantity of electricity, or charge,
C is the capacity of the body,
E is the potential.
The capacity therefore is
C = | (21)
and is equal to the charge divided by the potential, or it is equal
to the charge per unit potential.
ELECTROSTATICS 13
A body has unit capacity (electrostatic) when one unit of elec-
tricity is required to raise its potential by unity.
A sphere with a radius of 1 cm. has unit capacity, because
if a charge of one unit be given to it it will act as though it
were concentrated at the center and will produce at the surface
unit potential. The capacity of a sphere of radius R cm. is R
electrostatic units.
23. Farad. The practical unit of capacity is the farad. A
conductor has a capacity of one farad when one coulomb of elec-
tricity is required to raise its potential by one volt.
1 coulomb 3 X 10 9
1 farad
1 volt } is proportional to the difference of
potential E and to the permeance of the path (P.
The permeance (P is directly proportional to the cross-sectional
area of the path and to its dielectric constant and is inversely
proportional to the length of the path. It is usually difficult to
calculate the permeance directly, since the section of the path
varies throughout its length. In such cases the dielectric flux
t
\I/ or the electric charge Q = j- is assumed and the potential dif-
ference E calculated as in the preceding examples.
14
ELECTRICAL ENGINEERING
The permeance is then
""I-T- 4 -* 7 '
(23)
and is equal to the capacity of the system consisting of the two
conductors and the dielectric between them multiplied by 4n-.
25. Condenser. An electrical condenser is an arrangement of
conductors which is capable of storing a large quantity of elec-
tricity, or which has a large capacity. It is one in which a large
flux is produced when a given potential difference is applied to
its terminals.
The capacity of a condenser is measured by the amount of
charge necessary to raise its potential by unity, and is therefore
equal to the positive charge on it at any time divided by the
potential difference between its terminals; this relation is ex-
pressed by the equation
26. Parallel Plate Condenser. Determine the capacity of a
condenser formed of two parallel plates each having an area of
FIG. 13. Parallel plate condensers.
A sq. cm. separated by t cm. of material of dielectric constant k
Fig. 13 (a).
If a difference of potential E is applied between the terminals
a dielectric flux ^ is produced passing from the plate of high
^
potential to the plate of low potential; a positive charge Q = j-
theref ore appears on one plate and a negative charge Q =
47T
Q
on the other; the density of the charge on the plates is q = -r
ELECTROSTATICS 15
units per square centimeter, and the dielectric flux density be-
4irQ \b
tween the plates is 3D = faq = -j- = -^ lines per square centi-
meter. The thickness of the dielectric is assumed to be so small
that the lines of flux travel directly across from one plate to the
other.
The intensity of the electrostatic force is
(24)
~ K " Ak
The difference of potential between the plates is the work
done in carrying a unit charge from one plate to the other against
the force ff; it is therefore
Ak
since if is constant; and the capacity of the condenser is
C- Q
E
Q Ak
- electrostatic units. (25)
lirQt
Ak
The capacity therefore varies directly with the area of the
plates and with the dielectric constant of the material separating
them, and inversely as the distance between them.
When the plates are separated by air the capacity is
'
In order to increase the capacity of such a condenser a large
number of plates are used joined in multiple and separated by
very thin sheets of dielectric as shown in Fig. 136.
27. Capacity of Concentric Cylinders. Determine the capac-
ity of a condenser formed of two concentric cylinders, Fig. 14,
of radii Ri and R 2 cm.
If a charge of q units per centimeter length is given to the
inner cylinder, lines of dielectric flux pass out radially and pro-
duce, at a distance r cm. from the axis of the cylinder, a flux
density
2<7 ,.
3D = -j = lines per square centimeter.
16 ELECTRICAL ENGINEERING
If the dielectric constant of the material is k, the electrostatic
force at the point is
D
and the difference of potential between the two cylinders is
The capacity per centimeter length of the condenser is
C = ^ = ^ - -=- = - ^-=- electrostatic units. (28)
This is the case of a single-conductor cable with a lead sheath,
and the capacity is usually expressed in farads per mile; it is
r _k __ 2.54 X 12 X 5,280
2io g | 9X10U : v
k 2.54 X 12 X 5,280
/\
2 X 2.303 log, %
Rl
= 3.82 X ^-5- X 10- 6 farads per mile. (29)
, ri2
logic -
KI
28. Potential and Potential Gradient between Concentric
Cylinders. Assume that the potential of the inner cylinder in
Fig. 14 is E and that of the outer one is zero.
The charge per centimeter length is
q = CE = ^-=- by equation (28).
2 log f 2
Hi
The potential at radius r is
E dr
(*
J<
, 2
log "7
[logr] - -- (30)
ELECTROSTATICS
17
The potential gradient at radius r is
= ?? = ?2 = _A
dr AT
(31)
it has its maximum value at the surface of the inner cylinder.
E
{/max, D
(32)
Curve 1, Fig. 14, shows the variation of the gradient from one
conductor to the other.
The area under the curve represents the difference of potential,
E, between the conductors.
4. Curve 1,0
Area = Difference of Potential between Cylinder!
^- Potential =0
-Potential = E
k Dielectric Constant
FIG. 14. Concentric cylinders.
The ends of the outer cylinder are shown turned out at a
radius not less than the distance between the conductors. This
prevents concentration of flux at the ends which might cause a
breakdown if the difference of potential were very great.
29. Capacity of Parallel Conductors. Determine the capacity
of two parallel cylindrical conductors A and B, Fig. 15(6) of
radius R cm. suspended in air at a distance of D cm. between
centers.
When a difference of potential E is applied between them a
dielectric flux \j/ passes across from A to B and a positive charge
t
Q = - - appears on A and an equal negative charge Q =
~r7T
If q is the charge per centimeter length on A, the electrostatic
force at any point Q, on the line joining the centers, distant r cm.
from A and D r cm. from B is the resultant of the forces exerted
by the charges on A and B acting independently.
18
ELECTRICAL ENGINEERING
From each centimeter length of A, 4^-q lines of dielectric flux
extend out normally and produce at the point Q a flux density
47T# 20 .-.
>A = 2^: =: -^ lmes P er square centimeter.
The electrostatic force at the point Q is
and acts from A to B.
= ~ dynes
FIG. 15. Electrostatic field between two parallel cylindrical conductors.
The charge q on B produces at Q a force
in the same direction as & A .
The resultant force at Q is
2?
dynes.
(33)
The work done in moving a unit positive charge from B to A is
equal to the difference of potential between the two conductors;
it is
- R
"
r,D-jR
2 9 [log -^- -
,
= 4q log
(34)
ELECTROSTATICS 19
The capacity of the two conductors per centimeter length is
C = ^ = ^ n = ~ =5 electrostatic units. (35)
Mi . , Lf K , , D ft
q log ^ 4 log
The capacity per mile of a line consisting of two conductors is
c = 2.54 X 12 X 5,280
4 X 2.303 log D p R X 9 X10 11
1
19.4 X 10~ 9 . , , 0fl ,
=r = farads. (36)
i Iy it
log* -a- ^
It is sometimes useful to separate the capacity of the line into
the capacities of the two conductors forming it. The potential
of the point midway between the conductors is zero and the
potential of conductor A is
.
and, therefore, the capacity of A per centimeter length between
the conductor and the neutral or point of zero potential is*
C A = ~ = - j? - 5 = =r -- 5 electrostatic units (37)
Hi A n i Lf K n . L) K
2q log ^ 2 log --
and the capacity of each conductor in farads per mile is
_ 38.8 X 10-
D-R
log M -^
30. Potential and Potential Gradient between Parallel Con-
ductors. The potential at the point Q, Fig. 15(6), is the work
done in moving a unit positive charge from the neutral point to
the point Q; it is
e =
20 ELECTRICAL ENGINEERING
, , TjJ
q = CE = - yr --- - from equation (35),
and, therefore,
1 r\cr _
- 2 * ^ ~ *" '- (39)
This equation is plotted in curve 1, Fig. 15 (a).
TjJ 77f
The potential of conductor A is and that of B is
2 2
The potential gradient, at any point Q, which is equal to the
intensity of the electrostatic field at the point, is given by the
equation
or expressed in terms of the potential of the conductor A it is
*- *'
, D-R(r(D-
log -?r
For small values of r and 72 this may be written
E A
9 = ~ J^'
If E and ^ A are expressed in volts the potential gradient is
expressed in volts per centimeter.
Equation 40 is plotted in curve 2, Fig. 15 (a).
If E is taken as 200,000 volts, D as 200 cm. and R as 1 cm.,
the maximum gradient, which occurs at the surface of the con-
ductor is
' ' (43)
= 19 ' 000 volts per centimeter and the
ELECTROSTATICS 21
minimum gradient, which occurs at the point midway between the
conductors, is
D
D - R
D D
2 X 2
Dlog
- R
R
v , 1 _ onn 1A
X -^ = 3,800 volts per centimeter.
log 199 5U
If the two conductors had been suspended in a medium of
dielectric constant k and the same difference of potential applied
between them the capacity and the charge per centimeter would
have been increased in the ratio k but the potential and potential
gradient would remain as before.
31. Dielectric Field and Equipotential Surfaces between Par-
allel Conductors. The field between two parallel conductors is
shown in Fig. 15(d). It is obtained by superposing the fields due
to the conductors A and B separately. At the point P in Fig.
15(6) the two forces acting are /i = due to the charge on A
and / 2 = due to the charge on B. These two forces may be
combined vectorially to give the resultant force 3>, which is
tangent to the line of force passing through P. The lines of
force are excentric circles cutting the conductors normally and
passing through points slightly displaced from their centers.
A number of equipotential surfaces are also shown in Fig.
15 (d)', they are excentric cylinders surrounding the conductors
with their centers on the line joining the conductor centers.
The equations of these surfaces may be found as follows:
In Fig. 15(c) the point P has coordinates r*i and r 2 referred to
the centers of the two conductors, and coordinates x and y re-
ferred to the rectangular axes. It is required to determine the
equation of th equipotential surface passing through P.
The potential at P is
e =
ri r 2
2CE log (45)
22 ELECTRICAL ENGINEERING
and , . r 2 e
or
= 6 2CE fc _ a constant for the surface.
?*i
Expressing r\ and r 2 in terms of a: and y,
r 2 = ATI or r 2 2 = /b 2 ri 2
(D - xY + y 2 = k 2 (x 2 + y 2 )
or, simplifying,
(A; 2 - 1) x 2 + 2Dx + (k 2 - 1) y 2 = D 2
2 2D . , 2 = D 2
X + fc 2 - 1 x + y " fc 2 - 1
-! ' (fc 2 -l) 2 ' k z - 1 ' (fc 2 -!) 2 (fc 2 -!) 2
This is the equation of a circle with its center at the point
and of radius 2 _ ' (47)
Thus the equipotential surfaces are cylinders (the traces in the
xy plane are circles) and may be determined by assuming any
e
potential e and substituting the value k = 2CE in the equations
above.
32. Capacity of a Single Wire to Earth. Determine the capac-
ity of a single wire of radius R cm., suspended at a height H cm.
above the earth. If the wire is raised to a potential E above
the earth potential, lines of dielectric flux will pass from the wire
to earth as shown in Fig. 16. If ^ is the flux from each centi-
&
meter length of A, then q = j- is the charge on each centimeter
of A, and a corresponding negative charge q appears on the
earth, but it is not evenly distributed being most concentrated
directly beneath the wire.
The flux passing from the wire A to the earth would be un-
changed if the charge on the earth were collected on a second
wire B placed at a distance H cm. below the earth. Its potential
would be E.
ELECTROSTATICS
23
The difference of potential between A and B is 2E and from
equation (34)
2E = 4q log
2H - R
thus the capacity of the single wire with respect to the earth per
centimeter length is
a 1
r - q
C ~
n . 2H - R " . 2H - R
2glog ~^- 2 log
e.s. units, (48)
R R
and the capacity in farads per mile is
38.8 X 10~ 9
rrteth-n ' '. 'isi'iii"^.') 'I/ 1
v
FIG. 16. Capacity of a wire
to earth.
FIG. 17. Capacity of a sphere
to earth.
33. Capacity of a Sphere to Earth. In Fig. 17, A is a sphere
of radius R cm. at a height H cm. above the earth. If A is
raised to a potential E above earth potential, a dielectric flux
\I/ is produced passing from the sphere to the earth as shown
\L
and a positive charge Q = j- appears on A and an equal negative
4?r
charge on the earth.
Without changing the distribution of flux in any way the nega-
tive charge on the earth may be assumed to be collected on a
sphere B similar to A but placed at a distance H cm. below the
surface. The distribution of potential between two such spheres
was worked out in Art. 21. The potential midway between
them is zero, the potential at the surface of A is
E = ""
- R by e( l uation
24 ELECTRICAL ENGINEERING
and therefore the capacity of the sphere is
_ Q _ Q 1
E Q
R
Q
1
electrostatic units. (50)
2H - R R 2H - R
When H is very large compared to R the capacity is
as in Art. 22.
34. Condensers in Multiple. If a number of condensers of
capacities Ci, C 2 and Cz are connected in multiple, as shown in
Fig. 18, and a difference of potential E is applied to the terminals,
each condenser receives a charge proportional to its capacity,
Qi = CiE,
Qz = C-2.E,
Qs =
and the total charge on the system is
Q = Qi + Q* + Q 3 = E (Ci + C 2 + C 3 ).
Foffe
Fio. 18. Capacities in multiple. FIG. 19. Capacities in series.
The capacity of the system is
(51)
and therefore the capacity of a number of condensers connected
in multiple is equal to the sum of their separate capacities.
35. Condensers in Series. When a number of condensers of
capacities Ci, Cz and Cs are connected in series, as in Fig. 19, and
a difference of potential E is applied to the terminals of the sys-
tem, a charge Q appears on each condenser and the potential E
ELECTROSTATIC^ 25
divides up among the condensers in inverse proportion to their
capacities.
The drop of potential across condenser (1) is
Q
that across (2) is
Et ,o_,
and that across (3) is
Es = Q_.
but
771 T7T I T7T I T7T
& = til -\- &2 -\- J&s
1
and, therefore, the capacity of the system consisting of three con-
densers in series is
r _ Q __ L .
J - + - + -
Ci L>2 ^8
When two equal condensers of capacity C are connected in
series, their combined capacity is
(53)
i
C + C
and is equal to one-half of the capacity of either condenser alone.
38. Energy Stored in a Condenser. When a condenser is
being charged, work is done in rais-
ing the charge through the difference
of potential between the terminals,
and this amount of energy is stored
in the electrostatic field of the con-
Fio. 20.
denser.
In Fig. 20, PN is a condenser of capacity C formed of two
parallel plates separated by t cm. of a dielectric of constant k.
When a potential difference e is produced between the plates by
the generator G, electricity flows from N to P until the charge
on P is
26 ELECTRICAL ENGINEERING
if the potential e is increased by de the charge q is increased by
dq = C de and the work done in raising the charge dq through
the difference of potential e is
dw = e dq = Ce de.
The total work done in charging the condenser with a quantity
of electricity Q, or to a difference of potential E, is
W = I dw = f Ce de
J Jo
~*v - c [lt
= C ~ ergs. (54)
Thus the work done in charging a condenser, or the energy
stored in the electrostatic field of the condenser, is equal to one-
half of the capacity multiplied by the square of the difference of
potential between the terminals.
Equation (54) may be expressed in two other forms by substi-
tuting for E its value -^',
W = QE, (55)
W =i| 2 - . . ' (56)
If the area of the plate P is A sq. cm., then the dielectric flux
density between the plates is
3D = -p lines per square centimeter,
and the electrostatic force is
^ = T dynes.
The potential difference between the plates is
and the capacity of the condenser by equation (25) is
Ak
L "47?
ELECTROSTATICS 27
Substituting these values for E and C in equation (54) gives
a fourth expression for the energy stored in the field, namely
w = \ x 55 x (r
* rrvrt \ Iv .
- AI.X s=t ergs. (57)
Since the volume of the field is At c.c. and the flux density is uni-
form, the energy stored per cubic centimeter of the field is
3D 2
W =
or
$ z k
w = -g^- ergs. (59)
Thus, the energy stored per cubic centimeter in an electro-
static field is equal to the square of the dielectric flux density
multiplied by ^-y, or is equal to the square of the intensity of
k
the electrostatic force multiplied by TT-*
O7T
From equation (54) a very useful definition of capacity may be
obtained,
C = ^ (60)
or the capacity of a condenser is equal to twice the energy stored
in its field divided by the square of the difference of potential
across its terminals, or the capacity is equal to twice the energy
stored when the difference of potential is unity.
37. Stresses in an Electrostatic Field. The energy stored in
an electrostatic field is
E 2
W = C~ ergs;
and the energy stored per cubic centimeter is
> 2
*,= ergs.
These two equations represent the potential energy of the field.
Stresses exist throughout the field tending to reduce the potential
energy to a minimum; first, there is a tension along the lines of
induction tending to shorten them and to draw the bounding
28 ELECTRICAL ENGINEERING
surfaces of the field together and so reduce the volume to zero;
second, there is a pressure at right angles to the lines tending
^.^ mim mm ^ to spread them apart and so reduce the
"" density in the field. Since the system is in
equilibrium these two stresses are of equal
magnitude.
To obtain an expression for the stress per
FIG 21 square centimeter on the bounding surfaces,
consider the parallel plate condenser in
Fig. 21. The energy stored in the field is by equation (57)
T2
W = At X A
i:
If a force of P dynes is applied to one of the plates and the dis-
tance between the plates is increased by amount dt, the work
done is P dt ergs. The charges on the plates are assumed to re-
main constant and therefore the flux density remains constant,
but the volume of the field is increased by the amount A dt c.c.,
3D 2
and the energy stored in it is increased by A dt X TT-T, but the
increase in the stored energy is equal to the work done by the
force P and, therefore,
T)2
Pdt = Adt X^
and
P = A X J^ dynes. (61)
This is the pull exerted by the field on each plate of the condenser
tending to draw them together.
The pull per square centimeter is
(62)
thus, the pull per square centimeter on any charged surface is
equal to the square of the induction density at the point divided
by Sirk.
This is the value of the tension along the lines of force tending
to shorten them, and also the value of the pressure at right angles
to the lines tending to spread them apart.
38. Force Exerted on a Dielectric by an Electrostatic Field.
Fig. 22 shows two metallic plates of area A sq. cm. separated
ELECTROSTATICS 29
by a dielectric of constant k and thickness t cm. Determine
the pull required to remove the dielectric from the condenser
against the opposing force due to the field.
(a) Assume, first, that the condenser has a charge Q and is not
connected to any source of potential.
The energy stored in the condenser is
4 v/ 1 ^ xi /L/ J\. K
when the dielectric is removed the stored energy is
Wz = 2 Cl = 2 ~A :
where C\ is the capacity before and C 2 is the capacity after the
dielectric is removed.
Dielectric I
Constant k !"'*''*
-Q
3
Dielectric
Constant &
I
h
1
1
, J
d 1
(a) FIG. 22. (6)
The change of stored energy is
W W - 2r ^ 2 /i _ IN \
A \ ~ k)'
and this is equal to the average pull P multiplied by th* distance
through which it is exerted = Pd.
Therefore, the average pull is
Z> * ''I *Jiti/\g /^ JL v _. -jj,. f)O/ \ ^AQ\
but Q = Ci^Ji = . . 1 , where Ei is the initial difference of
potential between the plates; and, therefore,
9-7T/ A 2 fr 2 T^ 2 / 1\ ATc (1? 1^7^ 2
DiT\(i xj. Iv JCfl / 4 X\ ^1/v v./v J.>ZI/i . /r>^\
^"l 1 ~ jfc) = ~ STT^ dyneS< (6 4)
Ad
The final potential difference between the plates is
C
30 ELECTRICAL ENGINEERING
(by Assume, second that the difference of potential EI is
maintained constant by a generator, Fig. 22(6), and find the
pull required to remove the dielectric.
Initial stored energy is
ir s *Ci*. *>!;
the final stored energy is
C2 ~2T = 4^ ~2~"
The loss of stored energy is /
This amount of energy is given back to the generator supply-
ing the charge and as before the average pull required is
_ _ .
d '
The ratio of the pulls required in the two cases (a) and (6) is
Ak (k - 1) ffi 2
P_ Sirtd
P 1 == A (k - 1) ES ''
Sirtd
39. Effects of Introducing Dielectrics of Various Specific In-
ductive Capacities into a Uniform Field. 1. Fig. 23 shows a
parallel plate condenser with air as dielectric.
t = distance between plates in centimeters.
A = area of each plate in square centimeters.
E difference of potential between plates.
^ = dielectric flux.
^
3) = -7 = dielectric flux density.
fL
$ = 3D = electrostatic force in the field.
Since the force is constant throughout the field, therefore,
E = $t
and the electrostatic force or the stress in the air is
If E is expressed in volts, the potential gradient or the stress
may be expressed as
E
g = volts per centimeter.
ELECTROSTATICS
31
Assuming that E 25,000 volts and t = 1 cm. the gradient or
E 25 000
stress is g = y = ^ = 25,000 volts per centimeter. Air has
L -L
a dielectric strength of 31,000 volts per centimeter and it will
therefore not break down in this case.
2. In Fig. 24 a sheet of glass of thickness 0.3 and dielectric
constant k = 6 is introduced into the field as shown and the
difference of potential is the same as before.
K=6,
ft
1|LJ|
Or 21
FIG. 23.
FIG. 24.
FIG. 25.
The dielectric flux density 3D is constant throughout the field;
the electrostatic force in the air is
the drop of potential across the air portion of the field is
E l = Q.7t X 5 A = 0.7* X 3D;
the electrostatic force in the glass is
- 5 j?
= k == 6'
The drop of potential across the glass is
= 0.3* X
X if = 0.05* X
o
the difference of potential between the plates is
E = EI ~\~ EZ
= 0.7J X 3D + 0.05* X >
and the dielectric flux density is
0.75Z
thus the stress in the air is
and is greater than it was in the first case.
Again, assuming E = 25,000 volts and t = 1 cm., the gradient
in the air is g = | - = ^ X 25,000 = 33,000 volts and the air
616 , t
will break down.
32 ELECTRICAL ENGINEERING
In this case by introducing a dielectric of high specific inductive
capacity and much greater dielectric strength than the air, the
air is made to break down. It now becomes a conductor and
the full stress of 25,000 volts comes on the glass. The potential
25 000
gradient in the glass becomes g ' Q = 83,000 volts per
U.o
centimeter.
3. Fig. 25 shows the same pair of plates with three sheets of
dielectric introduced between them, of thickness ti, fa and t s and
dielectric constants fci, & 2 and & 3 respectively.
3D = dielectric flux, which is constant throughout the field.
3D
^i = =- = stress in layer (1),
/C]
F 2 = -- = stress in layer (2),
K%
3D
$3 = Y~ = stress in layer (3),
#3
3D
EI = $iti j- ti drop of potential across (1),
KI
3D
E z = g^ 2 = fa = drop of potential across (2),
KZ
3D
E 3 = 3^3 = =- 1 3 = drop of potential across (3);
&3
the difference of potential between the plates is
E = Ei + #2 + #3
' If ' 1?
/x/2 fv
and the dielectric density is
E '
h + b + !i
and the total flux is
? -A=E<5>
where
l , 2 ,
fci fa fc 3
= $ D k = k = 5 A k.
FIG. 26.
t
FIG. 27.
I
*-
--*
a
d
f
FIG. 28.
The permeance of the path is increased and an increased flux
is produced passing between the plates, but since the increased
flux is all confined to the dielectric there is no increase in the stress
in the air.
5. In Fig. 27, the same block of dielectric is shown with a
groove cut in it. The air in the groove is very highly stressed
34
ELECTRICAL ENGINEERING
and as the difference of potential between the plates is increased,
this air will be the first to break down.
6. Fig. 28 is another example of the same^ thing. The stress on
the air in the pockets a and b is high. As the difference of poten-
tial is increased this air will break down and become conducting
and there will be a discharge over the surface acb at a potential
difference much lower than that required to break over the shorter
surface df, where the stress is uniform.
7. In Fig. 29, A is a small piece of material with a high dielectric
constant such as a drop of oil. It offers a local path of high
FIG. 29.
FIG. 30.
FIG. 31.
FIG. 32.
permeability, and a greater flux density is produced in it than in
other parts of the field, and therefore the stress in the air at its
surface will be greater than the average stress throughout the
field.
8. In Fig. 30 the drop of oil is replaced by a knob on the con-
ductor forming one boundary of the field. Since the dielectric
constant of the conducting knob is infinity the stress in the air at
its surface will be greater than in the case of the oil.
9. Fig. 31 shows the field at the edge of the condenser in case
(1). In this region the dielectric flux density is not uniform, and
just at the edges of the plates it is greater than in the main body
of the field. The electrostatic stress is also greater than the
average value.
This condition may be corrected by turning out the edges of
the plates or electrodes at a radius not less than the distance
between them as shown in Fig. 32.
40. Effect of Introducing Conductors into Electrostatic Fields.
An insulated conductor may be introduced into an electrostatic
field without changing the flux distribution if it is placed entirely
on an equipotential surface. Fig. 33(6) shows such a case.
The conductor takes the potential of the surface on which it lies.
In Fig. 33 (c) the conductor does not lie along one of the original
equipotential surfaces and the distribution of flux is changed as
ELECTROSTATICS
35
shown. The conductor is an equipotential surface and the lines
of force enter it normally. Its potential is fixed by the relative
capacities between it and the two plates forming the boundaries
of the field. If it is placed symmetrically between the plates
its potential is the mean of the potentials of the plates.
In Fig. 15 which represents the field between two parallel
cylindrical conductors an insulated conducting cylinder may be
Volts
(a)
Volts
placed on any one of the equipotential surfaces without changing
the field; and if this cylinder is then connected to a source of
potential of the same value, the original conductor may be re-
moved without changing the field external to the cylinder.
41. Graded Insulation for Cables. Fig. 34 (a) shows a single-
conductor cable insulated with a dielectric of constant ki = 6;
Maximum Gradient
FHH
O) (6)
FIG. 34. Graded insulation for cables.
the radius of the conductor is R\ = 0.4 cm. and the inside radius
of the sheath is R* = 2.4 cm. If the sheath is grounded and the
conductor is raised to a potential E, the potential at any point
in the dielectric at a distance r cm. from the center of the con-
ductor is
e = E
, R*
log-
by equation (30)
36 ELECTRICAL ENGINEERING
and the potential gradient or stress at the point is 7
g = -T- = - J ~- volts per centimeter by equation (31).
dr , rt4
The gradient is plotted in the figure and the hatched area under
the curve represents the difference of potential E.
The maximum gradient or stress occurs at the surface of the
conductor where r = R\\ it is
i log
(68)
If the maximum allowable stress is assumed to be 100,000 volts
per centimeter, then the maximum difference of potential at
which the cable can be operated safely is
= 100,000 X 0.4 log ~ = 72,500 volts. (69)
In this case the stress in the outer layers of the dielectric is far
below the maximum allowable stress and the material is not
used to the best advantage.
The capacity of this cable per centimeter length, equation (28)
is
C = - ^-5- = - ^-7 = 1.585 electrostatic units.
21og| 21og||
Fig. 34(6) shows the same cable insulated with three layers of
materials of dielectric constants ki = 6, A; 2 = 4 and k z = 2;
the outside radii of the three layers are R 2 = 0.6, Rs = 1.2 and
fl 4 = 2.4 cm.
If q is the charge per centimeter length, the dielectric flux den-
sity at radius r is
3D = lines per square centimeter
and the electrostatic force or stress in the medium is
% = Y dynes,
where k has different values in the three dielectrics.
ELECTROSTATICS 37
In order to make the stresses in the various parts of the insula-
tion equal it would be necessary to place next to the conductor
a material of high dielectric constant ki and to decrease this con-
stant gradually in succeeding layers in inverse proportion to the
distance from the center of the conductor. This would be a
very expensive process and it is not necessary since good results
can be obtained by using three or four layers of dielectric.
The stress in the outer layer is
*-%> (70)
and the drop of potential across it is
f*R*
= JJ
2g #4
the stress in the second layer is
and the drop of potential across it is
2q 2q , R s , n .
r ctr = T~ log -p j \i&)
the stress in the inner layer is
and the drop of potential across it is
Si = B 'g dr = | log |. (75)
The difference of potential between the conductor and the
sheath is
^(7 i xL2 ^w(7 i -it^S ^^/ -i jTL4 / \
The capacity of the cable per centimeter length is
0.85 electro-
2 / log + - log + - loe ^ static units.
\6 g 0.4 T 4 g 0.6 T 2 S 1.27
38 ELECTRICAL ENGINEERING
The maximum stress or gradient in (1) is
2q 2CE
S^nax. = 7 p = T p electrostatic units,
or max . = -j-g- volts per centimeter; (78)
the maximum gradient in (2) is
O/^T?
fiTmax "-jjTfl-; (79)
and the maximum gradient in (3) is
0max. = -j-g-. (80)
If the three materials have the same dielectric strength, the
thickness should be so chosen that the three maximum stresses
are equal. The required condition is that JdRi = k 2 R z =
ksRs. In the present case the three materials are assumed to
have the same maximum allowable stress of 100,000 volts per
centimeter.
If the allowable maximum stress in layer (2) had been only
50,000 volts per centimeter or one-half of that in (1), J? 2 should
have been chosen so that k 2 R 2 = 2kiRi and would have been
1.2 instead of 0.6.
The maximum difference of potential which can be applied to
this cable can be found by substituting numerical values in any
one of the equations for maximum gradient.
From equation (78)
= 2CE
and thus
E = m ax. X -^r 1 = 100,000 X s^r/rb = 141,000 volts.
^lU i /\
The gradient from conductor to sheath is shown by the dis-
continuous curve in Fig. 34(6), and the area under the curve
represents the safe operating voltage of the cable.
In this case by grading the insulation on the cable without
changing the outside diameter the safe operating voltage has
been increased by 95 per cent, and at the same time the capacity
has been decreased about 45 per cent.
ELECTROSTATICS
39
42. Air Films in Generator Slot Insulation. Fig. 35 shows a
section of one slot of a three-phase, 11, 000- volt generator with
a chain winding. The insulation on the coils in addition to the
insulation between turns consists of ti = 0.2 cm. of molded mica
with a dielectric constant ki = 4, t% = 0.4 cm. of varnished cloth
of dielectric constant k 2 = 5, and an outer wrapping of fiber of
thickness 3 = 0.1 cm. and dielectric constant k 3 = 2. A film
of air of thickness 4 = 0.05 cm. is assumed to be included in the
slot. Determine the maximum stress in the air.
'Copper
Double Cotton Cover
(Neglected)
Molded Mica
0.2 Cm., 7ci = 4
Varnished Cloth
0.4 Cm.,/,; 2= 5
Fibre 0.1 Cin.,& 3 > 2
Core
Film 0.05 Cm. ,fc 4-1
FIG. 35. Slot insulation for an 11,000 volt generator.
The maximum voltage to neutral, i.e., between any coil and
the core, is found at the terminal coil of any one of the phases
and its value is ~- X \/2 = 8,960 volts (see Art. 123). The
side of the coil and the side of the slot may be considered as form-
ing the plates of a condenser with four layers of insulation and
the stress in the air may be found from equation (67),
121
8,960
8,960
0.23
= 39,000 volts per centimeter.
This is well above the critical stress in air (31,000 volts per
centimeter) and the air will therefore break down and ozone and
nitric acid will form, which will attack the insulation. If the air
is continually renewed by the expanding and contracting of the
coil due to alternate heating and cooling, the corrosive action
may go on until the insulation is broken down and the coil be-
comes grounded. The corners of the conductors should be
rounded off to prevent local concentration of flux and increased
stress as shown in Fig. 35.
40
ELECTRICAL ENGINEERING
43. Condenser Bushing. The leads from transformers, in
some cases must be insulated for very high voltages and it is diffi-
cult to make the terminals of reasonable size. This can be ac-
complished only by making each part of the insulating material
take care of its proper proportion of the total stress. One
method of obtaining this result is illustrated in Fig. 36, which is a
cross-section of a condenser bushing.
Insulating
Compound
Metal Sheets
IHHHHHh*-
Equivalent Circuit
Transformer
(Grounded)
nductor
FIG. 36. Condenser bushing.
The conductor is a hollow tube of sufficient radius to reduce the
stress at its surface to a safe value and it is placed as far away
from the other terminal as possible. The tube is wrapped with
layers of insulating material separated by their sheets of metal.
Adjacent metal sheets with the dielectric between them form
condensers which should all have approximately the same capac-
ity, but since, due to the increase in diameter, the outer plates
ELECTROSTATICS 41
tend to have greater areas than the inner ones they must be made
shorter. The lengths of succeeding plates are reduced by equal
amounts in order to get as long a surface leakage path as possible
and if necessary the stresses can be equalized by varying the
thickness of the dielectric. The whole bushing is covered with a
cylinder rilled with an insulating compound.
44. Dielectric Strength. The dielectric strength of an insulat-
ing material is measured by the difference of potential required to
puncture it, but this is a quantity which is dependent to a large
extent on the conditions under which the stress is applied. The.
test is usually made by applying an alternating voltage to the
specimen, which is placed between two electrodes. Frequencies
of from 40 to 60 cycles per second are generally used and the
shape of the voltage wave is approximately sinusoidal. In
this case the ratio of the maximum voltage, which is responsible
for the puncture, to the effective voltage, which is indicated by
the voltmeter, is \/2 to 1.
The dielectric strength is variously expressed in volts per mil,
volts per inch, volts per millimeter, or volts per centimeter.
The most important conditions affecting the test are: (1)
thickness of the specimen; (2) shape of the electrodes; (3) medium
in which the test is made; (4) temperature; (5) pressure; (6) time
of application of the voltage; (7) frequency of the impressed
voltage; (8) wave shape of the impressed voltage.
1. Fig. 37 shows the variation of apparent dielectric strength
with thickness.
When an alternating voltage is impressed, the changes in the
direction of the stress cause a loss in the dielectric resulting in
a rise of temperature, which decreases the dielectric strength. A
thin specimen can more easily radiate this heat than a thicker one
and so its temperature does not rise to the same extent. Further,
it is more difficult to make thick sheets of dielectric as homoge-
neous and free from flaws as thin ones. Both of these con-
ditions tend to make the thin specimen show greater strength
under test.
When a considerable thickness of dielectric is required it is
advisable to build it up of a number of thin sheets held by an
insulating varnish as this makes the product more flexible
and it is not likely that a flaw will extend through more than
one sheet.
2. If the electrodes are not properly shaped, excessive stresses
42
ELECTRICAL ENGINEERING
may occur at certain points and cause a breakdown at a lower
voltage than normal. For example, transformer oil which will
stand 225 to 250 volts per mil when tested between half-inch
discs will only stand about 200 volts per mil when tested between
half-inch spheres.
3. If a sheet of material is tested between spherical electrodes
under oil it will show a lower dielectric strength than it would if
tested in air, since the air will break down first and so give the
effect of flat electrodes.
4. In general an increase of temperature results in a decrease
in dielectric strength unless the increase of temperature reduces
the amount of moisture in the dielectric. It is a complex
phenomenon.
20 40 60 80 100 120
Thickness in Mils.
\
\
S^
20 40 60 80 100 120
Time in Seconds
FIG. 37. Typical curve of disruptive
strength vs. thickness.
FIG. 38. Typical curve of puncture
voltage vs. time.
5. The dielectric strength of air and oil increase directly with
the pressure.
6. Fig. 38 shows the variation of dielectric strength with
time of application of the voltage.
A given material will withstand a much higher voltage for a
short time than it will for an extended period. For this reason, in
making high-voltage tests on machines the stress must be applied
for a specified time.
7. Increase of frequency results in an increase of loss and a
rise in temperature, which decreases the dielectric strength.
8. The breakdown is the result of the maximum value of the
impressed voltage. If the wave is peaked the ratio of maximum
to effective value will be greater than with a sine wave and the
material will break down with an effective voltage lower than
normal.
In the following table are given approximate values of the
dielectric constants and dielectric strengths for materials in com-
mon use in electrical engineering.
ELECTROSTATICS 43
TABLE OF DIELECTRIC CONSTANTS AND DIELECTRIC STRENGTHS
Material
Dielectric
constant
ft
Dielectric
strength,
volts per mil
Remarks
Air
1
79
31 000 volts per cm
Asphalt
2 7
30
60 mils thick
Cellulose acetate
300-1 800
Cloth oiled
750-1 000
10 mils thick
Cloth, varnished
3 5-5 5
500-1 300
5-16 mils thick
Cotton impregnated
86
7 mils thick
Cotton not impregnated
21
7 mils thick
Ebonite
1.9-3.5
1,700-3,750
Up to 20 mils
Fiber
2.0
150-250
Fullerboard
7 5 at 100C.
Boiled in transformer
Fullerboard varnished . . .
Glass
2.9 at 25C.
5.5-10.0
300
150-300
oil.
Gutta percha
3.0-5.0
Mica
2 5-6.0
1,000-2,500
Molded mica or micanite.
Paraffin wax . .
2.5-6.0
1 9-2.3
900
300
Porcelain
4 0-6.0
J300
100 mils thick
Rubber .
2 0-3 O)
\ 220-240
500 mils thick
Rubber compounds
Shellac
3.0-4.0J
2.75
300-500
Slate
6 0-7
5-10
Transformer oil
Vacuum
2.5
9994
225-250
200
Between 0.5-in. discs
0.2 in. apart.
Between 0.5-in.
spheres 0.15-in.
apart.
45. Breakdown. When a discharge takes place between two
electrodes in a gaseous or liquid dielectric, the natural circulation
of the material heals the break and the insulating qualities are
not impaired. If the discharge in oil is very heavy some of the
oil becomes carbonized and the carbon particles tend to line up
in the field of greatest stress and reduce the dielectric strength.
Particles of moisture in oil due to their high dielectric constant
seek the most intense parts of the field and cause an increase of
stress and tend to cause breakdown.
One-tenth of 1 per cent, of water in transformer oil will reduce
its dielectric strength to 20 per cent, of its original value.
44 ELECTRICAL ENGINEERING
With porcelain or glass when breakdown occurs the material
is ruptured and its insulating value destroyed.
In the case of cotton or silk fabrics the overstress causes local
breakdown accompanied by carbonization and a weak spot de-
velops which gradually extends until complete rupture occurs.
46. Dielectric Losses. If a constant voltage is impressed on
the terminals of a condenser (Fig. 13) a dielectric field is set up
and the dielectric material is in a state of stress. So long as the
voltage is maintained constant there is not consumption of energy
due to this cause. When, however, an alternating voltage is
impressed the stresses in the dielectric alternate in direction. If
the dielectric were perfectly homogeneous all the energy stored
in the field during the increase of voltage would be returned as it
decreased again and no energy would be consumed. This occurs
in the case of gaseous and of some liquid dielectrics. In solid
dielectrics some energy is consumed in reversing the stress in the
material and appears as heat causing a rise in the temperature of
the dielectric. The loss of energy is proportional to the square
of the impressed voltage, that is, to the square of the dielectric
stress and is called the dielectric hysteresis loss; it, however,
differs from magnetic hysteresis in that it causes a slight lag of
flux in time behind the voltage. This is noticeable only at high
frequencies.
A further energy loss occurs in dielectrics due to the leakage
of current between the electrodes. This current is directly pro-
portional to the impressed voltage and is inversely proportional
to the resistance of the dielectric. The resulting loss is propor-
tional to the square of the current and therefore to the square of
the voltage. It occurs with both direct and alternating voltages.
The insulation resistance of dielectrics or insulating materials
is a very complex quantity and varies through a wide range with
temperature and other conditions. Fig. 68 shows the variation
of the resistance of slot insulation with temperature. The pres-
ence of minute quantities of moisture in the dielectric reduces
the insulation resistance to very low values.
There is no direct relation between the insulation resistance of a
material and its dielectric strength. For instance, dry air has a
very high insulation resistance but very low dielectric strength.
Ordinarily the dielectric losses are small, but in the case of
high-voltage cables where the stresses are very high they may be
large enough to cause a dangerous rise in temperature which in
ELECTROSTATICS 45
turn will decrease the insulation resistance and further increase
the loss.
47. Surface Leakage. When dielectric surfaces are clean and
dry, very little surface leakage takes place but if dust or mois-
ture is present the leakage of current may become serious.
The flashing-over of clean dielectric surfaces is due to the
breakdown of one dielectric, usually air, at the surface of a
stronger dielectric. In Fig. 28 leakage occurs over the long
surface acb due to the breaking down of the air in pockets at a
and 6; this air becomes conducting. The shorter surface df does
not exhibit the phenomenon to nearly so marked an extent.
In the ordinary pin-type insulator, discharges over the surface
take place due to the breakdown of the air in the pockets between
the petticoats combined with the presence of dust and moisture.
48. Corona. When the electrostatic stress or potential gradi-
ent at any point in air exceeds about 30,000 volts per centimeter
a brush discharge takes place and the air becomes conducting
and luminous. The discharge does not necessarily extend from
the positive to the negative electrode but exists only in the region
where the dielectric strength of the air has been exceeded.
In the case of two parallel wires suspended in air the discharge
or corona first appears at any rough points on the wires and fin-
ally forms a luminous envelope about them which increases in
diameter as the voltage is raised.
If the increase in the effective diameter of the wire due to the
presence of the corona results in a decrease of the stress a flash-
over between the wires will not result; but if the increase in di-
ameter so decreases the distance between the conductors that the
stress at the surface of the conducting envelope is not reduced
below the breakdown stress of the air, the diameter of the envel-
ope increases until a flash takes place from one wire to the other.
Corona is accompanied by a loss of energy proportional to the
square of the voltage rise above the critical voltage at which the
discharge begins.
CHAPTER II
MAGNETISM AND ELECTROMAGNETICS
49. Magnetization. When bodies are magnetized magnetic
forces act at every point throughout their volume and lines of
magnetic induction pass through them. There are two kinds of
magnetic poles just as there are two kinds of electric charges;
as a positive electric charge appears where a dielectric flux leaves
a surface, so a positive magnetic pole appears where a magnetic
flux leaves a surface. The positive magnetic pole is called a
north pole. Similarly a negative magnetic pole or south pole
appears where a magnetic flux enters a surface. -
FIG. 39. Magnet.
FIG. 40. Magnet with armature on.
Thus a body which is magnetized has a north pole at one part
of its surface and an equal south pole at another part unless the
magnetic path forms a closed circuit as in Fig. 40.
Fig. 39 represents a horseshoe magnet. The lines of magnetic
induction pass through it in the direction shown, leaving the sur-
face at N and entering it again at S. Thus TV is a positive mag-
netic pole or a north pole and S is a negative magnetic pole or a
south pole.
Fig. 40 represents the same magnet with its armature on. The
lines of magnetic induction pass in the same direction as before,
but the circuit is closed and the poles do nor appear until a gap
is made by removing the armature,
46
MAGNETISM AND ELECTROMAGNETICS 47
60. Laws of Magnetism. First Law. Like magnetic poles
repel one another; unlike magnetic poles attract one another.
Second Law. The force exerted between two magnetic poles
is proportional to the product of their strengths and is inversely
proportional to the square of the distance between them. This
law can be expressed by the formula
where m and mi are the pole strengths, r is the distance between
them in centimeters, and / is the force exerted between them in
dynes. If m and mi are like poles the force is a repulsion and
/ is positive.
The unit of pole strength is denned as follows: A magnetic
pole has unit strength if, when placed at a distance of 1 cm.
from a similar pole, it repels it with a force of one dyne.
The force exerted on a unit pole at a distance of r cm. from a
pole of strength m is
/ = 2 dynes. (82)
51. Magnetic Field. The space surrounding a magnetic pole
or a current of electricity in which magnetic forces act is called
a magnetic field. The direction of the force at any point in the
field is the direction in which a unit north pole placed at the point
would tend to move and its intensity is the force in dynes exerted
on the unit pole.
The magnetic field is represented by lines of magnetic induc-
tion or magnetic flux drawn in the direction of the force.
Unit magnetic force or unit magnetizing force produces one
line of magnetic flux per square centimeter in air and p lines per
square centimeter in a magnetic material of permeability /z.
The magnetic force at a point is expressed in dynes and is
represented by 3C; the magnetic flux density at a point is ex-
pressed in lines per square centimeter and is represented by (B.
Fig. 41 shows the magnetic fields produced in certain cases.
The lines of induction are all closed lines and extend from a
north to a south pole in air and from a south to a north pole in-
side the magnetic material or the generator of m.m.f.
52. Magnetic Flux. The total number of lines of magnetic
induction passing through a given section is called the magnetic
flux through the section and is represented by <.
48
ELECTRICAL ENGINEERING
The unit of magnetic flux, which is one line, is called the max-
well.
53. Flux from Unit Pole. At every point on a sphere of 1 cm.
radius, surrounding a unit pole as center, a similar, unit pole is
repelled with a force of one dyne. There must therefore be one
line of induction per square centimeter passing through the
surface, and since the surface is 4?r sq. cm. the total flux from the
unit pole is
= 4rr lines.
The flux from a pole of strength m is
<1> = 4rrm lines.
(a) Field of a Bar Magnet
( b ) Field of a Solenoid
(c) Field near arilsolated North Pole
( d) Field of a Dynamo
FIG. 41. Magnetic fields.
Thus a unit north pole is associated with each 4?r lines leaving
a surface and a unit south pole with each 4?r lines entering a
surface.
64. Magnetic Potential. The magnetic potential of an iso-
lated magnetic pole is the work done in carrying a unit north
pole from an infinite distance to the point against the forces in
the magnetic field.
Fig. 42 shows a north magnetic pole of, strength m. Its field
MAGNETISM AND ELECTROMAGNETICS 49
extends out radially in all directions and the magnetic force at a
distance of r cm. from m is
m ,
5C = -2 dynes.
The magnetic potential of the point P at a distance of ri cm.
from m is the work done in carrying a unit north pole from an
infinite distance to the point against the force of repulsion of m.
The work done is
Jjr f , f" dr m
W = OC dr = m- 2 = -
Jri Jri i
Therefore the magnetic potential of a point at a distance of ri cm.
from an isolated magnetic pole of strength m is
(83)
FIG. 42. Magnetic potential.
The difference of magnetic potential between the points P and
Q in Fig. 42 is
MT C r \n j f r ' dr m m
M = I 5C dr = I m - T =
Jr, Jr, ^ ri r 2
Thus the difference of magnetic potential between two points is
the line integral of the magnetic force between them.
55. Magnetomotive -Force. The difference of magnetic poten-
tial or the line integral of the magnetic force between two points
is called the magnetomotive force (m.m.f.) between the points.
It causes magnetic flux to pass from one to the other.
Unit m.m.f. will produce one line of magnetic flux through a
(cm.) 3 of air and /* lines through a (cm.) 3 of a magnetic material
of permeability /*. It is called the gilbert.
Since m.m.f. is the line integral of the magnetic force or
M = I 5C dr,
50 ELECTRICAL ENGINEERING
therefore the magnetic force is the space rate of change of m.m.f.
or
30 = w' (84)
and thus magnetic force is the m.m.f. per centimeter.
56. Permeability. Permeability is the ratio of the magnetic
conductivity of a substance to the magnetic conductivity of air
and is represented by /*.
Lines of magnetic flux pass through air or any other substance
except iron, nickel or cobalt, as easily as they do through a
vacuum. The permeability of such substances is for all practical
purposes the same and is taken as unity. Iron and its compounds
and to a lesser degree nickel and cobalt are found to allow mag-
netic lines to pass through them much more easily than empty
space; that is, a giveHm.m.f. will produce a much larger flux
through a volume of iron than it will through an equal volume of
air. The permeability of the iron is therefore greater than that
of the air and is expressed by some number greater than unity.
The permeability of magnetic materials is not constant but
varies with the induction density as shown in Art. 72.
67. Magnetic Reluctance. The reluctance of a magnetic
circuit may be defined as the resistance offered by the circuit to
the passage of magnetic flux through it and is represented by (R.
If a m.m.f. M is applied to a path of length I cm., of uniform
section A sq. cm. and of permeability /*, the m.m.f. per centimeter
will be -y and this will produce through each square centimeter a
M
flux or flux density
the reluctance is equal to the m.m.f. divided by the flux.
Assuming the flux to be unity in the last equation the reluc-
tance of the path may be defined as the m.m.f. required to pro-
duce unit flux through it.
58. Permeance. The permeance of a magnetic path is the
reciprocal of its reluctance and is represented by (P; thus
and assuming that the m.m.f. acting is unity, the permeance
may be defined as the flux produced through the path by unit
m.m.f.
The permeance of a path of uniform section is
and is directly proportional to the sectional area and to the per-
meability, and is inversely proportional to the length of the path.
59. Electromagnetics. The region surrounding a conductor
carrying a current of electricity is a magnetic field. A current
of electricity therefore represents a m.m.f. If the conductor
is isolated from other magnetic forces the lines of force will form
circles around it.
Maxwell's Corkscrew Rule. The direction of the current and
that of the resulting magnetic force are related to one another
as the forward travel and the twist of an ordinary corkscrew.
This rule is illustrated in Fig. 43.
The symbol represents a current flowing down and O a
current flowing up.
52 ELECTRICAL ENGINEERING
Faraday discovered that a current is induced in a closed coil
of wire when a magnet is brought near it. The same effect is
noticeable if a coil of wire carrying current is moved to or from
the closed coil, or if the second coil is fixed in position and the
current in it is varied. The induced current only exists while
the magnet or inducing coil is moving with respect to the fixed
coil or while the current in the inducing coil is varying.
FIG. 43. Magnetic flux produced by an electric current.
The induced current is due to the fact that a difference of
potential or electromotive force is produced in the circuit by
changing the number of lines of magnetic flux threading through
it or by causing lines of magnetic flux to cut across it.
60. Laws of Induction. First Law. A change in the number
of lines which pass through a closed circuit induces a current
around the circuit in such a direction as to oppose the change in
the flux threading the circuit.
Second Law. The electromotive force induced around a closed
circuit is equal to the rate of change of the flux which passes
through the circuit; or the electromotive force induced in a con-
ductor is equal to the rate at which it cuts across lines of magnetic
flux.
61. Unit of Electromotive Force. The absolute unit of elec-
tromotive force (e.m.f.) is the e.m.f. induced in a coil of one turn
when the flux threading the coil is changing at the rate of one
line per second; or it is the e.m.f. induced in a conductor when
it is cutting one line per second. It is called the abvolt.
The practical unit is the e.m.f. produced by cutting 10 8 lines
per second and is called the volt. E.m.f. is commonly called
voltage.
To change from absolute units of e.m.f. to volts divide by 10 8 .
MAGNETISM AND ELECTROMAGNETICS 53
If a coil of wire has n turns and the flux through it is changing
at the rate "-? lines per second, the e.m.f. induced in the coil is
e = n - absolute units. (SI)
at
The negative sign is used because when the flux is decreasing
the induced e.m.f. is in the positive direction, that is, it tends to
prevent the decrease of the flux.
62. Force Exerted by a Magnetic Field on an Electric Cir-
cuit. Every part of an electric circuit situated in a magnetic
field is acted upon by a force at right angles both to the direction
of the current and to the lines of force, and the circuit as a
whole is acted upon by forces tending to move it into the position
where it will include the greatest possible flux.
Force Force=fflil Force- JI Cos
(a) (b) (c) (d) (e)
FIG. 44. Force on an electric conductor in a magnetic field.
In Fig. 44, (a) represents a uniform magnetic field between two
unlike poles and (6) represents the field surrounding a conductor
carrying current. If the conductor is placed in the uniform field
the resultant distribution will be as shown in (c). The intensity
of the field above the conductor will be greater than that below
and a force / will act on the conductor at right angles to it and to
the lines of flux, tending to push it out of the field. This force is
directly prop6rtional to the intensity of the field or the flux den-
sity, to the length of the conductor in the field and to the strength
of the current, or
/ = (&II dynes, (92)
where , therefore
* dw = / d, (94)
and the work done in moving a current across a magnetic field is
equal to the product of the current and the flux cut.
In moving completely across the pole face the work done is
W = f / d = /*,
where is the flux from the pole.
MAGNETISM AND ELECTROMAGNETICS 55
If I is expressed in absolute units and $ in maxwells, W is in
ergs.
If the motion through the distance dx takes place in time dt sec.,
the work done is
dw = I d = I dfifc = el dt, (95)
where e = -jr is the e.m.f. generated in the conductor and I dt
= dq is the quantity of electricity raised through the difference
of potential e. Therefore, the mechanical work supplied to
move the conductor through the distance dx against the force /
is used up in doing the electrical work of raising a quantity of
FIQ. 46. Transformation of mechanical energy to electrical energy.
electricity dq through a difference of potential e, or in driving
a current / against an e.m.f. e for a time dt. Thus mechanical
energy is transformed into electrical energy. This is what takes
place in an electric generator.
If electric power is supplied to drive the current / against the
e.m.f. e -j-, for a time dt energy is supplied
dw = eldt = ftldt = Id-
the conductor exerts a force / = (BZ I dynes through a distance dx
and does mechanical work,
/ dx = &II dx = / d.
This is the action of an electric motor.
65. Electric Power and Energy. When an electric circuit
carrying a current / absamp. incloses a flux , which is not pro-
56 ELECTRICAL ENGINEERING
duced by the current, the potential electrical energy of the system
is
w = I ergs. (96)
This amount of energy must have been expended in bringing the
electric circuit into its position.
Power is the rate of flow or the rate of transformation of
energy. The mechanical power required to move the con-
ductor AB across the field and thus to vary the flux < is
n dw , d . r dx , ,
P = ~Ji = I -T7 = (&II -r = fv dyne-cm, per second,
where v = ~r is the velocity of the conductor normal to the field.
The electric power generated during the motion is
D dw r d j
P = ~j~ = I -=- el ergs per second,
where e = -r is the e.m.f . generated in the conductor by cutting
the flux.
The electric power in a circuit is the product of the current and
the e.m.f. in the circuit.
The practical unit of power is the watt. It is the power in a
circuit carrying one ampere when the e.m.f. across it is one volt.
P Watts = ^Volts X J Amperes. (97)
The kilowatt, which is 1,000 watts, is more commonly used
where the amounts of power are large. One horsepower is
equivalent to 746 watts.
The electric energy transformed in a circuit is the product of
the power and the time. The practical units of electric energy
are the watt-second or joule, the watt-hour and the kilowatt-hour.
The absolute unit of electric energy is the erg.
1 watt-sec. = 1 volt X 1 amp. X 1 sec.
= 1 abvolt X 10 8 X 1 absamp. X 10" 1 X 1 sec.
- 10 7 ergs.
66. Intensity of Magnetic Fields Produced by Electric Currents.
The following cases are of special importance: (A) At the center of a
circular loop of wire carrying a current I absolute units (Fig. 47).
MAGNETISM AND ELECTROMAGNETICS 57
If 3C is the field intensity at 0, the center of the loop, a magnetic pole
of strength m placed at this point will be acted upon by a force
/ = w3C dynes
in a direction perpendicular to the plane of the coil.
The pole will produce at the wire a field of intensity
^i = dynes,
and a flux density
where r cm. is the radius of the loop.
m ,.
= -^ lines per square centimeter,
Flux
FIG. 47. Intensity of the magnetic field at the center of a circular coil of wire.
This field will act on the wire with a force
/i = (BZ7 dynes
in a direction perpendicular to the plane of the coil, where I = 2-nr is
the length of the wire in centimeters.
Substituting the values of (B and I gives
/i = X 2irrl = m -- dynes,
but the forces / and f\ are equal and therefore
and
27T/
m3C = m. >
2irl
3C = dynes. (98)
The flux density at the center of the loop is
27T/
(B = 3C = lines per square centimeter.
If / is expressed in amperes the field intensity at the center of the
loop is
dynes . (99)
58
ELECTRICAL ENGINEERING
(B) At a distance of r cm. from a long straight wire carrying a current
of 7 absolute units (Fig. 48).
If 3C is the intensity of the field, the work done in moving a unit mag-
netic pole around the wire at a distance r cm. from it against the force
JC is
w = 2irr3 ergs.
The work done is equal to the product of the current and the flux cut
and, therefore,
2xr3C = 4x7,
and
21
3C = dynes;
(100)
the intensity of the field varies directly as the strength of the current
and inversely as the distance from the wire.
The flux density at distance r is
27
(B = 3C = lines per square centimeter.
If 7 is in amperes the field intensity is
27
-y- dynes,
3C
and the flux density is
(B =
0.27
lines per square centimeter.
(101)
(102)
(103)
Unit
Pole
Motion
Unit current would produce a flux density
of 2 lines per square centimeter at a distance
of 1 cm. from a straight wire and it would
produce this flux through a distance of 2ir
cm., or it would produce a flux density of 4x
lines per square centimeter through a dis-
tance of 1 cm. in air. Thus one absolute
unit of current represents a m.m.f. of 4*-
gilberts and one ampere represents a m.m.f.
of 0.47T gilberts.
(C) Between two parallel wires A and B,
Fig. 49, at a distance of D cm. apart and
carrying equal currents 7 but in opposite
directions.
The field intensity or magnetic force at
P int P . distot * Cm " frOm A and D .~ ' Cm '
from B is the resultant of the magnetic forces
due to the currents in A and B. Since these
forces act in the same direction at all points between the wires they
can be added directly; the field intensity is by equation 100
straight wire.
3C =
27
-f
27
D -r
dynes,
(104)
MAGNETISM AND ELECTROMAGNETICS 59
and the flux density at P is
27 27
(B = 3C = h > _ lines per square centimeter. (105)
\ (D) At any point on the axis of a short coil of radius r cm. (Fig. 50).
n = number of turns in the coil,
7 = current in the coil in absolute units.
FIG. 49. Magnetic field between two parallel wires.
Take any point P on the axis at a distance x cm. from the plane of the
coil and let 3C be the field intensity there. If a pole of strength m is
placed at P it will be acted on by a force of w3C dynes perpendicular to
the plane of the coil. The force exerted on the coil by the pole is equal
and opposite to the force exerted on the pole by the coil.
//cos e
i00000to000000B
D
FIG. 50. Intensity of the magnetic FIG. 51. Magnetic field in a solenoid,
field on the axis of a short coil.
The field intensity at the wire due to the pole m at P is -^ dynes and
the flux density is - lines per square centimeter, where d cm. is the dis-
tance from the point to the wire; the length of wire is 2irrn cm. and there-
fore the force exerted on it is
/ = ^ X 2 irrnl dynes.
This force acts at right angles to the lines of flux and may be resolved
into two components, / cos e in the plane of the coil and / sin 6 perpen-
60 ELECTRICAL ENGINEERING
dicular to the plane of the coil. The component / cos 6 taken around
the loop is zero and therefore the component / sin 6 is equal in magnitude
to the force w3C.
The field intensity or magnetic force at P is, therefore,
/sinfl 2irrnl .
3C = - = ^ sin e
m d z
r 2irr 2 nl
snce
V 2 + x 2
(E) On the axis of a long solenoid. In Fig. 51 AB is a solenoid of
length I cm. and radius r cm. If n is the number of turns in the sole-
noid, the number of turns in the section CD of width dx is j dx.
The field intensity at P due to the section CD is, by equation (106),
where / is the current in the solenoid.
The field intensity due to the complete solenoid is
J.1
2 27iT 2 / n
l(r . + ir l)KX j d*.
o
'2
To integrate this let angle CPT = 6, then
x = r cot 0,
(r 2 + z 2 ) % = r 3 (l + cot 2 o)* = r 3 cosec 3 6 = ~
rfx = - r cosec 2 d de = -
if the solenoid is assumed to be very long the limits may be taken
as and TT, and, therefore,
27rn7 f .
= -- sm
2irn/ . 10
= - - [COS e\l
dynes. (107)
If the current is expressed in amperes
0.47Ttt7 , /ino\
OC = j dynes, (108)
MAGNETISM AND ELECTROMAGNETICS 61
and the field intensity on the axip of a long solenoid is proportional to
the product of amperes and turns or ampere-turns and is inversely pro-
portional to the length of the solenoid.
The field intensity throughout the volume enclosed by the solenoid
is practically uniform except near the ends and can be expressed by
equation 108.
67. Magnetomotive Force of a
Solenoid. The m.m.f. of a solenoid
is the line integral of the magnetic /
forces along any closed path through \
it and is measured by the work done
in carrying a unit magnetic pole
around the closed path (Fig. 52). FlG - 52. Magnetomotive
. , force of a solenoid.
The work done is equal to the
product of the current and the flux cut, and thus
m.m.f. = 4irnl, where / is in absolute units,
or
m.m.f. = OAirnl, where / is in amperes.
The m.m.f. is proportional to the ampere-turns of the coil. It
does not make any difference whether it is a small current in a
large number of turns or a large current in a small number of
turns.
The m.m.f. of one ampere in one turn or one ampere-turn is
0.4T gilberts.
In studying the characteristics of electrical machinery it is
more convenient to use the ampere-turn as the unit of m.m.f.
Thus the m.m.f. of a field coil of n turns carrying a current /
amp. is specified as nl ampere-turns instead of QAirnl gilberts.
68. Examples. 1. The solenoid in Fig. 53 has nl ampere-turns and is
wound on a ring of non-magnetic material. The m.m.f. of the solenoid is
M = QAirnl gilberts.
If I cm. is the mean length of the path and A sq. cm. is the sectional area
of the path, the reluctance is
~~ A'
and the flux inside the ring is
_ M _ 0.47m/,.
The flux density in the ring is
(B -j lines per square centimeter,
62
and the magnetizing force is
ELECTRICAL ENGINEERING
4> 0.4rm/
and is the m.m.f. per centimeter.
If the solenoid is wound on an iron ring of permeability M, the reluctance
is reduced and becomes
(Ri = --,
the flux is increased to
and the flux denaty or induction density is increased to
CB I= -;
the magnetizing force remains the same as before,
_ (Bi _ $1 _ 0.47m/
\JL AH I
FIG. 53. Ring
solenoid.
FIG. 54. Solenoid
FIG. 55. Solenoid wound
on an iron bar.
2. Fig.^54, represents a solenoid of n turns carrying a current of / amp.;
the m.m.f. is M = 0.47m/ gilberts. If I cm. is the length of the solenoid
and A sq. cm. is its sectional area, the reluctance of the path through it is
The m.m.f. required to drive the flux < through this reluctance is
MI = -r gilberts.
A.
But the flux going out at one end has to pass, around through the air and
in at the other end as shown. The reluctance of this return path is difficult
to calculate and is not of great practical importance; its length is greater
than I but its sectional area is very much greater than A and its reluctance
is small compared to (Ri. Thus in the case of long solenoids of small section
the reluctance of the return path may be neglected and the assumption may
be made that the whole m.m.f. M is utilized in driving the flux through the
reluctance (Ri. Thus
*
MAGNETISM AND ELECTROMAGNETICS 63
the induction density in the solenoid is
0.47m/
and the magnetizing force at any point inside the solenoid is
3C = (B = ' , ft* dynes, as in equation (108).
If an iron bar of length I and permeability /* is placed in the solenoid the
reluctance of the path through the solenoid is reduced in the ratio - and the
reluctance of the return path is no longer negligible in comparison with it and
its value must be calculated (Fig. 55).
A.SQ.OI
FIG. 56. Ring with an
air gap.
FIG. 57. Magnetic circuit
of a dynamo.
3. In Fig. 56 a solenoid of nl ampere-turns is wound on an iron ring of
permeability IJL and a section of length 1 2 is cut from the iron. If l\ is the
mean length of the path through the iron and its sectional area is A sq.
cm. the reluctance of the path through the iron is
h
^4?
the reluctance of the path through the air is
and the total reluctance of the path is
(R = (Ri +
M 2 = 2 = j l t .
4. Fig. 53 represents the magnetic circuit of a bipolar dynamo. It con-
sists of a number of parts of different materials as follows :
One yoke y of section A y , length l v , and permeability MJ/,
Two pole cores c,c of section A c , length l c , and permeability MC,
Two pole pieces p,p of section A p , length l p , and permeability MP,
Two air gaps g,g of section A g , length 1 , and permeability n a = 1,
One armature a of section A a , length l a , and permeability MO-
The reluctance of these parts are, respectively,
ZO7 O7 o; 7
,., y , (r, _ * L C , "lp m Al>g m l>a
(Jiy = . > Ot c ~ A ' Olp = ~7 ' W.g = ~r> Cn a = 1 ;
AyfJLy A C fJL C ApfJLp Ag A a fJLa
and the reluctance of the whole circuit is
tK ^ (jiy "i (j\c i~ Orp \~ (Jig \~ CHa
The m.m.f. M is provided by field coils placed on the pole cores as shown,
and the flux through the circuit is
= M = M
01 Ol tf + (R c + (R p + (R ff + (R
and is equal to the m.m.f. divided by the total reluctance; and the m.m.f.
is
= My + M c + Mp + Mg + Ma,
where M v is the part of the total field m.m.f. required to drive the flux
through the yoke, etc.
The m.m.f. M g required to drive the flux across the air gaps is sometimes
as much as 80 per cent, of the total m.m.f.
5. Determine the reluctance of the ring in Fig. 58 made up of three parts
of lengths li, h and h cm. respectively and sectional areas Ai, At and A 3
sq. cm. and permeabilities MI> M2 and MS. The m.m.f. of the solenoid is M.
The reluctance of section (1) is (Ri = -^ ;
lz
the reluctance of section (2) is (R 2 = A ;
the reluctance of section (3) is (R 3 = ^ ',
the flux through section (1) is $1 = /o~J
M
the flux through section (2) is * 2 = ^-;
the flux through section (3) is 3 = ^r f
MAGNETISM AND ELECTROMAGNETICS
the total flux through the ring is
= M M_ M_
(Hi (R 2 (Ra
65
j_ j_\
i (R 2 (R 3 /
(110)
But the flux is equal to the m.m.f. divided by the reluctance of the ring, or
M
and therefore
M
(HI)
FIG. 58. Parallel magnetic paths. FIG. 59. Series-parallel magnetic circuit.
6. Fig. 59 shows a ring of iron with a piece set in made up of three
parts of different permeabilities. The lengths, sections and permeabilities
are indicated and the reluctances of the various parts are (Ri = -r^
AlMl
and (R 4 = -7 Determine the reluctance of the
circuit.
Let M the m.m.f. applied to the ring,
$ = flux through the ring,
MI = m.m.f. consumed in section (1),
M 2 = m.m.f. consumed in section ab,
then MI = 3>(Ri, and M 2 = >(R b.
The reluctance of the section ab consisting of three paths in multiple must
be found. The flux $ passing through it divides into three parts which are
inversely proportional to the reluctances of the paths,
therefore,
and
(Rs' 4 (R 4 '
(Rab =
(R 2 (R 3 (R4
66 ELECTRICAL ENGINEERING
The reluctance of the whole circuit is
M - Af i + M
= j tfoV.and the work done is equal
to the product of the current and the increase of the flux,
, . 7 Aiirn^Au, . 7 .
aw = ni a = = 1 di ergs.
The work done while the current is building up to its full value
/is
i
,-^ON
-g ergs (112)
(113)
,
where = j is a constant, if // is constant, and is called the
inductance of the circuit.
MAGNETISM AND ELECTROMAGNETICS 67
This amount of energy is stored in the magnetic field of the
solenoid and may be expressed in other forms.
The energy stored is
TF = ^p r J er S s '
but the magnetic force in the solenoid is
5C --j-'
therefore,
or since the induction density is
(B = jitfC,
the energy may be expressed as
W - 2 A]
- Q Yit.
% 07TJU
The product Al represents the volume of the magnetic field
and therefore the energy stored in the field per unit volume is
W ,,TP2
_ /Zuv
or
or
5m
STT
ergs. (116)
When the magnetic field is produced in air the field intensity
3C is equal to the flux density (B, and the energy stored per cubic
centimeter is
3C 2
w = -g^ergs, (117)
or
(B 2
w = ergs. (118)
70. Stress in the Magnetic Field. The force between two
magnetic poles is not exerted at a distance but is transmitted
through the medium separating them and the medium is stressed.
68
ELECTRICAL ENGINEERING
There is a tension along the lines of force or induction tending
to shorten them and draw the boundary surfaces of the field
together and so decrease the magnetic energy stored in the field;
there is also a pressure at right angles to the lines tending to
spread them apart, and so decrease the flux density in the field
and, therefore, also the energy stored. Since the medium is in
equilibrium these two forces are equal in magnitude.
The pull on the bounding surfaces is usually expressed in dynes
per square centimeter and may be found as follows:
The energy stored in a magnetic field in air was found to be
(& 2
ergs in Art. 69
07T
Fig. 61 represents a horseshoe magnet with its armature re-
moved a distance x. If (B is the flux
density in the field between the magnet
poles and the armature and A sq. cm. is
the area of the two pole faces, the volume
of the field is Ax c.c. and the energy
stored in it is
W = $- Ax ergs.
O7T
If a pull P is exerted on the armature
and it is moved through a distance dx, the
work done is P dx and this is equal to the
increase in the energy stored in the field;
therefore,
FIG. 61. Pull of a
magnet.
Pdx = -
and
A
dynes;
this is the pull of the magnet on the armature.
The pull per square centimeter is
P 6
P = -A = Q
(119)
(120)
and this is the tension along the lines of induction in the field and
is also equal in magnitude to the pressure at right angles to the
lines.
71. Force between Parallel Wires Carrying Current. Parallel
wires carrying currents /i and /2 absamp. will either be attracted
MAGNETISM AND ELECTROMAGNETICS
69
or repelled depending on whether the currents are in the same or
opposite directions. Fig. 62 shows the fields in the two cases.
When the currents are in the same direction the lines of force
combine to form lines surrounding the two wires and these lines
tend to shorten, or the circuit tends to change so that the reluc-
tance is a minimum and the flux a maximum. When the currents
are in opposite directions the lines of force pass between the wires
and they tend to spread out resulting in a repulsion between the
wires. In this case again the forces tend to decrease the reluc-
tance of the circuit by increasing its section.
p - d-
FIG. 62. Force between two parallel wires carrying current.
To obtain an expression for the force between the two wires
refer to Fig. 62. A and B are the two wires carrying currents /i
and / 2 respectively and d cm. is the distance between their centers.
The field intensity and flux density at wire B produced by the
current /i in A is
27
(Bi = ~~T~ lines per square centimeter.
This field acts on the wire B carrying current 1% witha force
f = (81/2 =
O7" T
dynes per centimeter length. (121)
This is the force of repulsion between the two wires.
72. Magnetic Characteristics. If an iron ring, Fig. 60, which
has been completely demagnetized, is gradually magnetized, by
increasing the current in the exciting coil from zero, the induction
density in the ring increases with the magnetizing force as shown
in curve 1, Fig. 63. With feeble magnetizing forces the gradient
of the curve is small and the permeability is low; as the force
increases the curve becomes very steep and nearly straight and
the permeability increases rapidly and becomes very large; as
the force is further increased the curve rounds off and the gradi-
ent becomes small again and a large increase in magnetizing
force is required to produce any considerable increase in density,
the permeability decreases to a low value and the material is
70
ELECTRICAL ENGINEERING
said to be saturated. Curve 1 is called a "magnetization curve"
or " saturation curve," or (B-3C curve of the material.
Curve 2, Fig. 63, shows the relation between the permeability
/T>
M = and the induction density (B; when the density is low
Magnetizing Force = 3C
FIG. 63. Magnetic characteristics.
(Bi
Magnetic Induction = (B
the permeability is low; as the density is increased the per-
meability increases until it reaches a maximum at the point m
where the tangent from the origin touches curve 1; above this
point the permeability decreases again.
In Figs. 64 and 65 are shown magnetization curves for materials
used in electrical machine design. These curves are plotted with
100 200
Permeability - ft.
400 500 600 700 800 900 1000 1100 1200 1300
in
100 120 140 160 180 200 220 240 2GO
Ampere Turns per Inch.
FIG. 64. Magnetic characteristics of cast iron.
induction density B expressed in lines per square inch on a base
of ampere-turns per inch instead of magnetizing force JC. This
results in a change of soales only since B lines per square inch
= (2.54) 2 (B, and T ampere-turns per inch corresponds to a
magnetizing force 5C = ' ..
MAGNETISM AND ELECTROMAGNETICS
71
Permeability curves for cast iron and cast steel are also shown.
For cast iron the maximum value of n occurs at a density of about
4,000 lines per square centimeter or 25,000 lines per square inch
and for steel at about 6,500 lines per square centimeter or 40,000
lines per square inch. Densities below these points are not of
very great importance.
Permeability *=
CO 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300
H 140
400
Ampere Turns per Inch
FIG. 65. Magnetic characteristics of cast steel.
600 650
73. Hysteresis. If the exciting current i or magnetizing force
5C, of the solenoid in Art. 69, is increased the flux density (B in-
creases, following the curve OA in Fig. 66 until it reaches its
maximum value (B TO at A. If 3C is now gradually reduced to
zero, (B will not come back on AO but will follow the curve AR
and when 3C is zero (B will have the value (B r represented by OR.
This is called the residual magnetism; it is not a fixed quantity
but depends on the degree of magnetization of the material.
To remove the residual magnetism 5C is reversed and increased
to the point C. (B is now zero. The magnetizing force repre-
sented by OC is called the coercive force. As 5C is still increased
(B increases to its maximum value (B m again at the point A' with
the same magnetizing force as at A. Again reducing 3C, (B follows
the curve AiRiCi to A and closes the loop. This closed curve
is 'called a hysteresis loop.
The area of the hysteresis loop represents the amount of
energy consumed in carrying the material through the cycle of
magnetism.
72
ELECTRICAL ENGINEERING
From Art. 64 the work done in increasing the flux threading
the solenoid by an amount d<$> is
, ni ,
dw = d<}> ergs,
Where n is the number of turns on the coil and i is the current
in amperes when the flux threading the coil is 0. If the ring
has a constant section of A sq. cm. and a length of I cm., then,
4'7r??'Z' "*^*
since 3C = -^r and d$ X Ad(&, the work done is
10Z
Al
dw = -
FIG. 66. Hysteresis loop.
and the work done during a complete cycle is
W =
JAl C m
dw = ^p I 3Cd(B ergs.
4?T Jffi m
AZis the volume of the ring in cubic centimeters and therefore
the loss in ergs per cycle per cubic centimeter of iron is
i
W h = -r-
(122)
Referring to Fig. 66 and starting from the point Ci, the energy
per cubic centimeter supplied to increase the induction density
MAGNETISM AND ELECTROMAGNETICS 73
from to (B TO is -r (area OCiAM) ergs. While (B decreases from
(B TO to (B r , energy is given back = -r (area AMR). The energy
4r7T
required to remove the residual magnetism is ^ (area ROC).
Similarly below the axis the energy consumed is
j (area OCAiMi area AiMiRi + area ORiCi) ergs.
Thus the hysteresis loss per cubic centimeter per cycle is j (area
of loop) ergs.
The area of the loop increases faster than the maximum den-
sity. Steinmetz gives the following equation for the loss per
cubic centimeter per cycle in terms of the maximum density:
COA = liffim^ergs, (123)
where rj is called 'the hysteretic constant of the material.
The hysteresis loss in a volume of V c.c. of iron at a frequency
of / cycles per second is
W h = i?(B n 1 - 8 /V r 10- 7 watts. (124)
The following table gives ordinary values for the hysteretic
constant for commercial materials.
Good dynamo sheet steel . 002
Fair 0.003
Silicon steel 0.00076-0.004
Cast iron 0.011-0.016
Cast iron 0.011-0.016
Cast steel 0.003-0.012
Good commercial iron after assembly . 0027
Iron can be carried through a cycle of magnetism or hysteresis
cycle (1) by applying an alternating m.m.f. and producing an
alternating flux as in the transformer; (2) by producing a revolv-
ing flux which cuts stationary iron as in the stator of an induction
motor; and (3) by revolving iron in a stationary magnetic field
as in direct-current generators and motors.
The energy expended in hysteresis heats the iron and causes
a loss in efficiency.
74. Magnetic Materials. The most important magnetic mate-
rials used in electrical design are cast iron, cast steel, ordinary
sheet steel and silicon sheet steel.
74 ELECTRICAL ENGINEERING
Cast iron has low permeability and large hysteresis loss. It
is used for parts of the magnetic circuit where the induction den-
sity is low and is in a constant direction. It is cheap and can be
made into castings of complex form.
Cast steel has a much higher permeability and smaller hys-
teresis loss. It replaces cast iron where greater strength or
greater permeability is required or when the appearance would
be improved by reducing the section.
Sheet steel is used for parts carrying alternating fluxes of all
densities. It is necessary to use thin sheets in such places to
reduce the loss due to eddy currents set up in the iron as it cuts
across the flux (see Art. 184). To further reduce the eddy-cur-
rent loss the iron should have as high an electrical resistance as
possible.
Silicon steel contains a small percentage of silicon which has
the effect of reducing the hysteresis loss and increasing the elec-
trical resistance and so reducing the eddy-current loss. Much
higher flux densities can thus be used and the weight and cost of
machines reduced. Silicon steel is used principally for trans-
formers, where very high efficiency is required. It is quite
expensive.
75. Effect of Chemical Composition and Physical Treatment
on Hysteresis Loss. Hysteresis loss is dependent on:
1. Chemical composition of the iron.
2. Heat treatment.
3. Mechanical treatment.
1. The general effect of impurities is to decrease the permea-
bility and increase the iron loss but there are exceptions.
Carbon decreases the permeability and lowers the saturation
point; it increases the residual magnetism and the coercive force.
These effects are greater in hardened steel than in soft iron.
Such materials are valuable for permanent magnets but are not
used in rotating machinery.
Silicon in certain percentages has a very beneficial effect; 2.5
to 4 per cent, of silicon alloyed with the iron increases the per-
meability and decreases the hysteresis loss to a marked degree.
It also increases the electrical resistance.
T. D. Yensenhas obtained samples of iron, alloyed with certain
percentages of silicon and melted in vacuo, which exhibit mag-
netic qualities far superior to any commercial materials. These
MAGNETISM AND ELECTROMAGNETICS
75
materials are only in the experimental stage. The table below
shows a comparison between various steels.
Aluminum has an effect similar to that of silicon but the im-
provement is not so great.
2. Annealing the material after manufacture has a very im-
portant effect. In certain cases it has reduced the loss in sheet
iron to half.
It is found that the hysteresis loss in some steels increases
with continued heating during use. This is known as aging and
may have very serious results.
Silicon steel is non-aging.
3. Punching, hammering and bending increase the hysteresis
loss in sheet steel.
The following table shows a comparison between Yensen's
experimental steels and some commercial materials.
Material
Maximum
permeability
Density for
maximum
permeability
in lines
per square
centimeter
Hysteresis loss in
ergs per cubic
centimeter per
cycle
Electrical
resistance
in
microhms
per
centimeter
cube
/o
^max. "
10,000
max.=
15,000
Experimental steel with 0.15 per
cent, silicon
66,500
63,300
3,850
3,160
6,500
6,500
7,000
4,000
286
280
3,320
2,260
916
1,025
5,910
3,030
11.8
48.5
11.0
51.0
Experimental steel with 3.40 per
cent, silicon
Standard transformer steel
4 per cent, silicon steel
76. Theories of Magnetism. The magnetization of iron is
accompanied by molecular changes and energy is consumed.
Ewing was the first to formulate a theory to explain the resulting
phenomena. He imagined iron to be made up of magnetic
molecules. In unmagnetized material these molecules are ar-
ranged in groups, in which the elementary magnetic poles neu-
tralize one another. When a weak magnetizing force or m.m.f . is
applied, the less stable groups are broken up and their component
molecules turned with their magnetic axes in the direction of the
impressed m.m.f., making the material as a whole slightly mag-
netic. This stage is represented by the lowest section, A, of the
magnetization curve, Fig. 63. If the magnetizing force is re-
moved, the molecules return to their original groupings and no
magnetism remains. When, however, an increasingly powerful
76 ELECTRICAL ENGINEERING
m.m.f. is applied, the more stable groups break up and a large
proportion of the molecules turn in the direction of the external
field and form new stable groupings ; the (B-3C curve rises sharply
and the permeability is large. This is section B of the curve.
If at this point the m.m.f. is reduced to zero, only a part of the
molecules resume their old groupings, the majority remaining in
their new positions. In this way the phenomenon of residual
magnetism is accounted for.
With an ever increasing external m.m.f. more and more of the
molecules line up and the material approaches saturation.
Beyond this point any increase in m.m.f. only produces an in-
crease in flux density comparable with that produced in a non-
magnetic material (section C).
The later electron theory of magnetism replaces Ewing's
magnetized molecules by electromagnets. It is assumed that
each molecule of iron is the seat of an electric current or^whirl of
electrons and therefore exerts a small m.m.f. An external m.m.f.
tends to draw these into line with itself as discussed above and
thus increases the resultant m.m.f. of the system and co'rise=
quently the flux and induction density. With this assumption-
the conception of permeability is unnecessary as increasing
permeability is replaced by increasing m.m.f. which has the same
effect.
77. Lifting Magnets. Design a direct-current electromagnet
capable of lifting a block of cast iron weighing 2,000 Ib.
The pull of a magnet is given, by equation (119),
(B 2 A
P = ^ dynes
where (B is the flux density in the air gap in lines per square
centimeter and A is the area of the pole surfaces in square
centimeters.
1 Ib. = 444,000 dynes
and therefore
p = ^ = 444,000 X 2,000 = 888 X 10 6 dynes.
07T
Assuming an air-gap density of 6,000 lines per 'square centi-
meters (39,000 lines per square inch) the required pole area is
888 X 10 6 X Sir
-- (6 000) 2 = 6 6 sq< Cm * = Sq> m '
If the magnet is made in the shape shown in Fig. 67 (a) the
MAGNETISM AND ELECTROMAGNETICS
77
area of each leg is
9S.5
4$ sq. in. The remaining dimensions
may be chosen to make the magnet of a suitable shape but space
must be left for the winding.
The number of ampere-turns required for the exciting winding
may be found as follows:
Assume that the contact between the magnet and the mass to
be raised is not perfect and that an air gap of J^-in. is left. The
ampere-turns required for the gap = 0.3132 X 39,500 X0.25
X 2 = 6,170. (Art. 223). Length of the path through the steel
magnet = 36 in.; the ampere-turns per inch required for a
density of 39,500 lines is 17 (Fig. 65) and the total ampere-
turns = 17 X 32 = 544. Length of the path in the cast iron
Sliding
Non-Magnetic Plates'
FIG. 67. Lifting magnets.
may be assumed to be 24 in. and the section 60 sq. in.; the
flux density in the cast iron is
39,500_X 48
60
= 31,600; the
ampere-turns per inch = 32 (Fig. 64) and the total ampere-
turns for the cast iron = 32 X 24 = 768.
The exciting ampere-turns for the magnet = 6,170 + 544
+ 768 = 7,482.
In order that the coil may not become overheated it is neces-
sary to limit the power loss to a value which can be radiated from
the external cylindrical surface of the coil without too great a
temperature rise. For a magnet in continuous service a loss of
about 0.7 watts can be taken care of by each square inch of the
radiating surface. The length of the coil is 12 in. and the per-
iphery is 44 in. and thus the radiating surface is 44 X 12 = 528
sq. in. The allowable loss is therefore 528 X 0.7 = 370
watts. If the voltage is 220 volts, as is usually the case, the cur-
370
rent is / = ^^ = 1.7 amp.
220
and the resistance of the coil is
= -^=130 ohms.
78 ELECTRICAL ENGINEERING
, , e , m ampere-turns 7,480
The number of turns = T = - Current - = -^- = 4,400,
the length of the mean turn is 36 in. = 3 ft. and the length of
wire is 4,400 X 3 = 13,200 ft.
The resistance of a wire is give by equation (131).
length in feet
D
JK = p
section in circ. mils
The specific resistance p may be taken as 12 for a temperature
of 60C. and substituting in the equation above the section of
the conductor in circular mils may be obtained :
130 = 12 X 4
circ. mils
and
., 12 X 13,200 1 oon
circ. mils = - ' - = 1,220.
Referring to the wire table on page 85, it is seen that No. 19
wire has a section of 1,288 circ. mils and it would therefore be
used.
Lifting magnets are usually made ring-shaped as shown in
Fig. 67(6) and a large factor of safety is required on account of the
varied nature of the material to be handled.
CHAPTER III
ELECTRIC CIRCUITS
78. Ohm's Law. When an e.m.f. is applied to the terminals
of a conductor, a current is produced which is directly propor-
tional to the e.m.f. and is inversely proportional to the resistance
of the conductor;
/ = -5 amp. ( 125)
XL
where / is the current in amperes,
E is the e.m.f. in volts,
R is the resistance in ohms.
This is Ohm's Law.
A conductor has a resistance of one ohm, when an e.m.f. of
one volt is required to drive a current of one ampere through it.
When, therefore a current / flows through a resistance R,
e.m.f. is consumed; ^
E = IR volts. (126)
79. Joule's Law. Whenever a current flows through a re-
sistance, electric energy is transformed into heat energy. The
power or the rate at which energy is transformed in the circuit
is equal to the product of the current and the e.m.f. consumed in
driving the current through the resistance of the circuit.
P = El watts,
but E = IR and, therefore,
P = PR watts; (127)
thus, the power consumed in the circuit is equal to the square of
the current multiplied by the resistance. This is Joule's Law.
The power consumed in the resistance of circuits represents a
loss of power except in such cases as the incandescent lamp, where
it is utilized in producing light, or the electric heater, where the
heat developed is applied to a useful purpose.
79
80 ELECTRICAL ENGINEERING
80. Heat Units. When a current of / amp. flows through a
resistance of R ohms for a time t sec. heat is developed
H = PRt watt-sec, of joules. (128)
The practical heat units are the British thermal unit and the
calorie.
The British thermal unit, B.t.u., is the heat energy required
to raise 1 Ib. of water 1F.
1 B.t.u. = 1,055 watt-sec. = 1.055 kw.-sec. (129)
The calorie is the heat energy required to raise 1 gram of water
1 calorie = 4.2 watt-sec. = 0.0042 kw.-sec. (130)
81. Examples. 1. If electric energy costs lf cts. per kilowatt-hour,
what will it cost to raise the temperature of 100 gal. of water 60F.?
One gallon of water weighs 8.4 Ib.
The energy required to raise 100 gal. = 840 Ib. of water 60F. is
840 X 1,055 X 60 watt-sec. = - ' * kw.-hr.
and the cost is
840X1,055 X60 _ _
1,000 X 3,600
2. If 2 tons (2,000 Ib.) of coal are required per month to heat a house, what
would it cost to supply the same amount of heat electrically at 1^ cts. per
kilowatt-hour?
Assume 1 Ib. of coal to give 10,000 B.t.u.
Heat energy in 2 tons of coal = 2 X 2,000 X 10,000 B.t.u.
2 X 2,000 X 10,000 X 1.055
The equivalent kilowatt-hours = -
3,600
2 X 2,000 X 10,000 X 1.055
and the cost = - - X 0.015 = $175.83.
82. Resistance. The resistance of a conductor varies directly
as its length and inversely as its sectional area; it also depends on
the material of which the conductor is made;
# = P j-> (131)
where
I is the length of the conductor,
A is the sectional area,
p is the specific resistance or resistivity of the material.
ELECTRIC CIRCUITS 81
The specific resistance or resistivity may be expressed in ohms
per centimeter cube or per inch cube, which is the volume resis-
tivity; or in ohms per meter-gram, i.e., the resistance of a uniform
round wire 1 meter long weighing 1 gram, this is called the mass
resistivity.
In engineering problems wires are usually specified by gage
numbers, their lengths are given in feet and their sectional areas
in circular mils. A circular mil is the area of a circle one mil or
one-thousandth of an inch in diameter. The specific resistance
is then the resistance of a wire one foot long and one circular
mil in section.
83. Conductance. The reciprocal of the resistance of a con-
ductor is called its conductance and is represented by the letter
G, where
o- l -
-R
The reciprocal of the specific resistance or resistivity of a material
is called its conductivity and is represented by the Greek letter
7, where
1
7 = - '
P
Conductivity should be expressed in per cent, of the conduc-
tivity of the International Annealed Copper Standard taken as
100 per cent. This standard is expressed in terms of mass re-
sistivity as 0.15328 ohms (meter-gram), at the standard tempera-
ture 20C. This is equivalent to 10.371 ohms per circ. mil foot
at 20C.
The conductivity of metals is usually decreased by the presence
of impurities. Drawing or other cold working makes metals
harder, stronger and slightly more dense and decreases their con-
ductivity. The conductivity of copper wires ranges from 96 to 98
per cent. Aluminum wire has a conductivity of 60 or 61 per cent.
84. Effect of Temperature on Resistance. The resistance of
all pure metals increases with increase of temperature. The
variation can be expressed by the formula
R t = Ro(l + at), (132)
where R is the resistance at a chosen standard temperature, R t
is the resistance at a temperature t higher and a is the tem-
perature coefficient of resistance. It is the increase in resist-
82 ELECTRICAL ENGINEERING
ance per degree rise in temperature expressed as a fraction of the
resistance at the standard temperature. If the Centigrade scale
is used and RQ is the resistance at 0C., the value of a for copper
of 100 per cent, conductivity is 0.00427. This value may be con-
sidered as constant over the range of temperature from to
100C. With the Fahrenheit scale and R Q taken as the resist-
ance at 32F., the value of a is -^-^ = 0.00237.
l.o
If the temperature tC. is taken as standard the formula may
be written
R t = R tj {l +a(t- O}, (133)
where a is not the same as before but is smaller because the in-
crease of resistance is expressed as a fraction of the resistance at
tiC. which is greater than the resistance at 0C.
When ti is taken as the standard temperature 20C. the value
of a is
Ro
R (1 + a X 20) 1_ + 2Q
OiQ
234<+20
Similarly, starting from this standard temperature coefficient
at 20C., the value of a at any temperature t is found as
RZQ OizQ R%Q 1
a t = 0:20
5 + t - 20
0.00393
This result is correct if the conductivity is 100 per cent. When
it is less than 100 per cent, the value of a must be varied propor-
tionally. For copper of 95 per cent, conductivity
in.- 1 -~ (135)
'
0.95 X 0.00393
If the resistance at tiC. is known, but the corresponding value
of a tl is not known, the resistance at any other temperature t
may be found from
1 + 0.00427* \ D / 234.5 + t\ ,
0.00427 J = fl (mri)' for 10 per
cent, conductivity (136)
and
ELECTRIC CIRCUITS 83
1 + 0.00427 7 \ 234.5 +
234.5
where 7 is the conductivity expressed as a decimal fraction.
If the resistance R tl at a known temperature ti is given and also
the resistance R t at an unknown temperature t, this tempera-
ture t may be found as follows:
Rt^ R (l+ 0.00427* ) _ 234.5 + t
Ru "
or
and
R (I + 0.00427*i) 234.5 +
R t Rti t ti
R t
234.5
t-t, = ^^ (234.5 + i).
K t
If the conductivity is 7,
R t - R ti /234.5
(138)
(139)
85. Properties of Conductors. The properties of the most im-
portant electrical conductors are given in the table below.
Material
Temperature coefficient
Density, grams
per (cm.) 3
Resistance, ohms per
(cm.) 3 at 0C.
Aluminum
Cooper .
0.00390
0.00393
0.00387
0.00381
0.00072
0.00367
0.00377
0.00500
2.70
8.89
11.36
1.69-1.75
13.55
21.20-21.70
10.40-10.60
19.30
2.64
1.59
18.40-19.60
4.10-5.00
94.07
9.00-15.50
1.50-1.70
5.50
> X 10-"
Lead
Magnesium . . .
Mercury
Platinum
Silver ....
Tungsten
The temperature coefficients of all pure metals are positive and
are quite large. Certain alloys such as manganin have very
small temperature coefficients and are used in the design of elec-
trical instruments where the resistance must remain constant
over the ordinary range of temperature.
Carbon unlike the metals has a negative temperature, coeffi-
cient and it is not constant.
The resistance of insulating materials decreases very rapidly
with increase of temperature, but the variation is not regular and
84
ELECTRICAL ENGINEERING
cannot be expressed by a simple equation. Fig. 68 shows the
variation of the resistance of slot insulation with temperature.
The resistance is expressed in megohms or millions of ohms.
86. Resistance of Conductors. The resistance of a circular
mil foot of copper of 100 per cent.
conductivity at 20C. is 10.371 ohms;
therefore, the resistance of a copper
wire at 20C. is, by equation (131),
length in feet
200
180
B160
140
loo
I 80
S 60
R = 10.371
= 10.371
section in circular mils.
10 20 30 40 50 60 70 80 90 100
Temperature Degrees Centigrade
FIG. 68. Variation of resist-
ance of slot insulation with
temperature.
circ. mils.
In the case of rectangular conductors
such as busbars, the section is ex-
pressed in square mils and the value
of p is then the resistance of a square
wire, one mil on each side and one foot long.
The following table gives ;the specific resistance of copper at
various temperatures and the corresponding values of the tem-
perature coefficient.
Temperature
in degrees C.
Resistance of one
circular mil foot in
ohms
Resistance of one
square mil foot in
ohms
Temperature
coefficient a
9.560
7.52
0.00427
15
10.160
7.98
0.00401
20
10.371
8.13
0.00393
25
10.550
8.30
0.00385
50
11.590
9.11
0.00351
75
12 . 620
9.85
0.00323
In this table and in wire tables generally the conductivity is
assumed as 100 per cent. If the conductivity of a given wire is
less than this the specific resistance must be changed accordingly;
thus, the resistance of one circular mil foot of copper of 98 per
cent, conductivity at 25C. is ~ '' = 10.76 ohms; the resistance
of one circular mil foot of aluminium of 60 per cent, conductivity
at 25C. is = 17.58 ohms.
ELECTRIC CIRCUITS
85
TABLE OF COPPER WIRE OF 100 PER CENT. CONDUCTIVITY
Ameri-
can Wire
gage
(B. & S.)
No.
Diameter
in mils
at 20C.
Section at 20C.
Resistance,
ohms per
1,000 ft. at
20C.
Current capacity,
amperes
Circular
mils
Square
inches
Rubber-
covered
Other
insulation
0000
460.00
211,600.0
0.1662
0.04901
225
325
000
409.60
167,800.0
0.1318
0.06180
175
275
00
364.80
133,100.0
0.1045
0.07793
150
225
324.90
105,500.0
0.08289
,0.09827
125
200
1
289.30
83,690.0
0.06573
0.1239
100
150
2
257.60
66,370.0
0.05213
0.1563
90
125
3
229.40
52,640.0
0.04134
0.1970
80
100
4
204.30
41,740.0
0.03278
0.2485
70
90
5
181.90
33,100.0
0.02600
0.3133
55
80
6
162.00
28,250.0
0.02062
0.3951
50
70
7
144.30
20,820.0
0.01635
0.4982
8
128.50
16,510.0
0.01297 -
0.6282
35
50
9
114.40
13,090.0
0.01028
0.7921
10
101.90
10,380.0
0.008155
0.9989
25
30
11
90.74
8,234.0
0.006467
1.260
12
80.81
6,530.0
0.005129
1.588
20
25
13
71.96
5,178.0
0.004067
2.003
14
64.08
4,107.0
0.003225
2.525
15
20
15
57.07
3,257.0
0.002558
3.184
16
50.82
2,583.0
0.002028
4.016
6
10
17
45.26
2,048.0
0.001609
5.064
18
40.30
1,624.0
0.001276
6.385
3
5
19
35.89
1,288.0
0.001012
8.051
20
31.96
1,022.0
0.0008023
10.15
21
28.46
810.1
0.0006363
12.80
22
25.35
642.4
0.0005046
16.14
23
22.57
509.5
0.0004002
20.36
24
20.10
404.0
0.0003173
25.67
25
17.90
320.4
0.0002517
32.37
26
15.94
254.1
0.0001996
40.81
27
14.20
201.5
0.0001583
51.47
28
12.64
159.8
0.0001255
64.90
29
11.26
126.7
0.0000995
81.83
30
10.03
100.5
0.0000789
103.2
86
ELECTRICAL ENGINEERING
87. Drop of Voltage and Loss of Power in a Distributing Circuit
The distributing circuit in Fig. 69 delivers 20 kw. at 220 volts to a re-
ceiver circuit 1,000 ft. distant; if the size of the conductors is No. 1
B. & S., determine the voltage required at the generating end of the
line and the power lost in the line.
Eg is the generator voltage,
E is the receiver voltage = 220 volts,
I is the load current,
r is the resistance of each conductor.
FIG. 69. Distributing circuit.
The power delivered is
the current is therefore
El = 20,000 watts;
20,000
= 220 = 90<9
the resistance of each conductor at 25C. is
r = 0.126 ohms;
the voltage drop in each conductor is
e l = Ir = 90.9 X 0.126 = 11.48 volts;
the generator voltage is therefore
E = E + 2Ir = 220 + 22.96 = 242.96 volts;
the drop of voltage in the circuit is
2ei = 27r = 22.96 volts
per cent< = 9-4 ^ per cent '
22 96
The loss of power in the circuit is
2/ 2 r = 2 X 90.9 2 X 0.1261 = 2,100 watts;
the power delivered by the generator is
E g l = 242.96 X 90.9 = 22,100 watts;
therefore the power loss* is
2 100
22TOO * 10 ^ per Cent * = 9 ' 45 per cen ^'
and the efficiency of the transmission is 90.55 per cent.
ELECTRIC CIRCUITS 87
If the drop of voltage had been limited to 10 volts what size of wire
would have been required?
The drop in voltage is
1 000
2Ir = 2 X 90.9 X 10.55 ^ -~r = 10 volts:
circ. mils
therefore the required section in circular mils is
1 000
A = 2 X 90.9 X 10.55 -^ = 192,000 circ. mils.
88. Current-carrying Capacity of Wires. The energy con-
sumed in the resistance of a wire raises the temperature of the
wire until the point is reached where the heat radiated and con-
ducted from the wire is equal to the heat generated in it. When
the wire is bare the heat will escape easily into the air, but when
it is covered with insulating material the heat cannot escape so
easily and for a given current density the temperature rise will
be greater. This increase in temperature decreases the resist-
ance of the insulating material and so decreases its insulating
properties; in extreme cases the insulation may be charred and
rendered useless. The last two columns of the table in Art. 86
give the values of current which can be carried safely by differ-
ent sizes of wire. With rubber insulation the allowable current
is about 25 per cent, less than with weatherproof insulation be-
cause the rubber is more easily affected by heat. When the in-
sulated wires are inclosed in conduits the current-carrying capac-
ity is less than that given in the table.
89. Examples. 1. If the resistance of a copper conductor at 25C. is
10 ohms, determine its resistance at 65C.
fles = #2 5 {l + 25 (65 - 25)} = 10{l + 0.00385(65 - 25)} =
11.55 ohms.
2. Determine the resistance of a copper wire of 97 per cent, conductivity,
100,000 circ. mils in section and 50 ft. in length at 50.C.
, The resistance is = 0.00597 ohms.
3. Determine the resistance of an aluminum bar 0.75 in. by 0.375 in. by
100 ft. at 25C., if the conductivity is 60 per cent.
The resistance is
Q O 100
4. If the resistance of the shunt-field winding of a generator is 30 ohms
at a temperature of 25C. and after running under load is found to be 31.5
ohms, determine the average temperature of the winding.
88 ELECTRICAL ENGINEERING
If / is the average temperature of the winding when hot, 'equation (133)
gives
Rt = #25(1 + 0.00385(f - 25)},
or substituting
31.5 = 30{l + 0.00385(< - 25)},
the rise in temperature is
31.5 _
0.00385
and the average temperature of the winding is
t = 25 + 13 = 38C.
This method is used in measuring the temperature rise in the field and
armature windings of electrical machines.
90. Kirchoff's Laws. Two laws enunciated by Kirchoff are of
great value in solving problems dealing with continuous-current
circuits.
First Law. The algebraic sum of all currents flowing toward
or away from any junction is zero.
Second Law. The algebraic sum of the e.m.fs. acting in a
closed circuit is equal to the algebraic sum of the products of
the current and resistance in the several parts of the circuit.
V V H
() (b)
FIG. 70. Kirchoff's laws.
The first law is illustrated by the circuit ABCD in Fig. 70 (o)
The current i divides at B into two parts ii and i 2 ., iis numerically
equal to the sum of i\ and i z and therefore the algebraic sum of all
currents at the junction B is
i i\ i<2. = 0.
The second law is illustrated by the circuit FGHK in Fig.
70(6). Between the points F and G an e.m.f. E is applied which
drives a current i around the circuit in a clockwise direction
against the e.m.f. e of the battery between H and K. r\ and r 2
are the resistances of the circuit.
ELECTRIC CIRCUITS
89
The e.m.f. acting in the direction of the current is E e and
is consumed in driving the current i through the resistances
ri and r z and by Ohm's law
E e =
+ ir 2 .
91. Examples. (A) Fig. 71 shows a three-wire system supplied by two
similar 110-volt generators. The resistances in the various parts of the
circuit are indicated. Find the currents in the three lines, assuming the
direction of flow to be as indicated by the arrows.
From circuit abgf, 110 = ii(n + r 3 + RJ + i 2 r 4 or 110 = 8.3ii + 0.4i 2 .(\)
From circuit fgcd, 110 = i 3 (r 2 + n + R 2 ) - i 2 r 4 or 110 = 10.3i 3 - 0.4i 2 .(2)
From circuit abed, 220 = ii(ri + r 3 + Ri) + is(Rz + r& + r 2 )
or 220 = 8.3ii + 10.3i 3 . (3)
At the point g, i\ = i* + is (4)
Substituting (4) in (1) 110 = 8.7i 2 + 8.3i 3 (5)
Solving (2) and (5), i 2 = 2.4 amp., i 3 = 10.9 amp. and ii = i 2 + is =
13.3 amp.
(B) Find the currents in the various branches of the circuit in Fig. 72.
.** r 8 =o.i
, C .
FIG. 71.
FIG. 72.
Equating the voltages around the various circuits to zero.
(ABCDG) 2i + 5ii + Mi = 100.
(ABFDG) 2i + 4i 2 + 5i 5 = 100.
(BCF) 5ii + 3i 3 - 4i 2 = 0.
(FCD) 3i 4 - 5i 5 - 3i 3 = 0.
(BCDF) 5ii + 3i 4 5i 5 4i 2 = 0.
Equating the currents at the junction points to zero.
(-B) i ii i 2 = 0, or i = i\ -f~ i 2 .
f/~1\ n
\ff/ *i i 3 ^4 = u, or i\ = ii i 3 .
(i)) i i 4 i 5 = 0, or i = i 4 + i 5 .
/P\ ... . ... . ...
\" / ^2 T *3 *6 == ", or 15 = i 2 -f~ i 3 .
(1) *
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Substituting (7) and (9) in (4),
3 (ii -is) - 5 (i 2 + i s ) - 3i 3 = 0, or 3ii - 5i 2 - Ili 3 = 0. (10)
Substituting (7) and (9) in (5),
5ii + 3 (ii - is) - 5 (i a + is) - 4i 2 = 0, or 8ii - 9i 2 - 8i 3 = 0. (11)
90 ELECTRICAL ENGINEERING
Solving (3), (10) and (11) simultaneously,
ii. - 6.77 amp., i 2 = 8.66 amp. and i a = 1.76 amp.
From (6), i = 15.43; from (7), u = 8.53; from (9), i* = 6.90.
92. Resistances in Series. If a voltage E is applied across
a circuit consisting of a number of resistances Ri, R 2 and R s con-
nected in series (Fig. 73) the total resistance of the circuit will
be equal to the sum of the resistances of the different parts.
The drop of voltage across the resistance Ri is
where I is the current in the circuit.
Similarly ^ = JR ^
EZ = IRz,
E = EI + Ez +
and the resistance of the whole circuit is
R*. (141)
FIG. 73. Resistances in series. FIG. 74. Resistances in parallel.
93. Resistances in Parallel. Determine the resistance of
t a circuit consisting of a number of resistances Ri, R% and R s
(Fig. 74) connected in parallel.
If an e.m.f E is applied to the circuit a current / will flow
which wijl divide up into branch currents /i, / 2 and 7 3 . By
Ohm's law
I = i
I >
'-!
ZI3
and ^r
where E is the resistance of the whole circuit,
ELECTRIC CIRCUITS
91
but
therefore,
/ = /I + /2 + /3
\RI RZ R&
R =
E
\RI RZ RZ>
-
RI
(142)
If the conductances G, GI, G 2 and G 3 are used instead of the terms
ID> ^D~> D~> an d ^ET the result can be written
K ll 12 ^3
R = -
+ 2 + G s
or
(143)
that is, the conductance of a circuit consisting of a number of
parallel branches is equal to the sum of the conductances of the
branches.
94. Potentiometer. The potentiometer
shown diagrammatically in Fig. 75 may
be used to obtain a variable voltage from
a constant voltage supply in a continuous-
current system. By changing the position
of the point B any voltage E AB from zero
to the value of the constant supply voltage
E may be applied to {he load resistance RI.
The resistance of the potentiometer from
A to B is R 2 and from B to C is R 5 . The
various currents are indicated in the figure.
FIG. 75. Potentiometer.
The resistance of the circuit between A and B is
1
the resistance between A and C is
R
AC
RAB + RBC =
i*i
K\ K
.
92 ELECTRICAL ENGINEERING
and the current from the mains is
T E E
1 = ~P~ = P p ~ J
JtiAc K\KZ
the drop of potenti?! from B to C is
j7i TD &K$
&BC ***J r T-.
K\K'z
^ i r> ~T -3
and the drop across the load circuit is
77T _ ~T7I _ TjJ _ Tjl
, ,-,
There is a continuous waste of power in the resistances R 2 and # 3 and
the efficiency of the arrangement is always less than the ratio of the
variable voltage E AB to the constant voltage E.
95. Inductance. When a current of electricity flows in a cir-
cuit, lines of magnetic induction interlink with the circuit; if the
current remains constant the flux threading the circuit is con-
stant, but when the current varies the flux varies proportionally
and in doing so generates in the circuit an e.m.f. proportional to
the rate of change of the interlinkages of flux and turns or to the
product of the turns and the rate of change of the flux. This
e.m.f. has a value
6 = n -57 c.g.s. units, (144)
where n is the number of turns in the circuit and -57 is the rate
of change of the flux interlinking with the turns. The negative
sign is used because the e.m.f. produced opposes the change in
the current and, therefore, the change in the flux.
The inductance of a circuit is defined as the rate of change of
the flux interlinkages with the current in the circuit and is repre-
sented by <, thus,
d
= n-f-. c.g.s. units.
ai
This quantity is not a constant except in circuits which do not
contain any magnetic materials and in which therefore the flux
varies directly with the current.
In such circuits the inductance may be defined as the number
ELECTRIC CIRCUITS 93
of interlinkages of flux and turns for unit current in the circuit,
or,
= ^ c.g.s. units, (145)
where i = current in c.g.s. units,
= flux produced by current i,
n interlinkages for current i.
Since = -n -jr = n-jr / -j-> the e.m.f. generated in a circuit due
ai at / at
to its inductance is
e= n-jT = -r c.g.s. units, (146)
It is equal to the product of the inductance of the circuit and the
rate of change of the current. When e is expressed in volts and
i in amperes the inductance is in henrys and is represented by L
to distinguish it from the inductance < expressed in c.g.s. units.
Equation (146) may then be written
e = - L j t volts. (147)
The henry is defined as the inductance of a circuit in which a
counter e.m.f. of one volt is generated when the current is chang-
ing at the rate of one ampere per second.
The relation between the two units of inductance is found as
follows :
_ di
e = -T c.g.s. units,
= < ^.10~ 8 volts, and i in c.g.s. units,
di
= ~r 10~ 9 volts, in c.g.s. units, i in amperes,
and, therefore,
1 henry = 10 9 c.g.s. units of inductance. (148)
96. Example. (a) Find the inductance of an endless solenoid in the form
of a ring, Fig. 53.
n = number of turns in solenoid = 1,000.
r = mean radius of the ring = 10 cm.
Z = 27iT = mean length of the flux path through the solenoid.
,4 = sectional area of the solenoid = 5 sq. cm.
i = current in solenoid in c.g.s. units.
94 ELECTRICAL ENGINEERING
The flux through the solenoid is
m.m.f. 4r7rru ,.
^ = T7T lmes >
reluctance I/A
the flux per unit current is
the number of interlinkages of flux and turns per unit current is
Z
and is equal to the inductance of the circuit in c.g.s. units; thus, the induct-
ance of a circuit is proportional to the square of the number of turns.
Substituting the values given above the inductance of the solenoid is
4 X 3.14 X (1,000) 2 X 5 =
2 X 3.14 X 10
and the inductance in practical units is.
L = 10 6 X ID- 9 = 0.001 henry.
(6) Find the energy stored in the magnetic field of the solenoid when the
current has reached a value / = 10 c.g.s. units.
The energy stored when any current i is flowing is equal to the work done
in building up the current i against the counter e.m.f. of inductance e =
-j-' it is therefore
dt
W = \ eidt = I ^ i dt = \ i di
J J dt Jo
When the current is i = / = lOc.g.s. units, the energy stored is
/2 1Q2
W = -2 = 10 6 X -y = 5 X 10 7 ergs
= 5 X 10 7 X 10~ 7 = 5 watt-sec.
97. Inductance of Circuits Containing Iron. As stated in Art.
95, the inductance of circuits containing magnetic materials is
not constant, since the flux does not vary directly with the cur-
rent. The relation between flux and current may be represented
by a magnetization curve or saturation curve such as those shown
in Art. 72, but these curves cannot be represented by mathemat-
ical equations and for engineering calculations it is necessary to
choose some constant average value to represent the inductance
01 a circuit, that is, to assume that the material has a constant
permeability M and then to apply a correction to the result.
If a solenoid of n turns is wound on an iron ring of section A
ELECTRIC CIRCUITS 95
sq. cm. and mean length I cm. and permeability /*, the flux pro-
duced by a current i c.g.s. units is
,.
lines,
Ait
the flux per unit current is
.,
c.g.s. units,
&
and the inductance is
= ni = - = -- c.g.s. units.
It is proportional to the square of the number of turns, as seen
in the last example, but it also depends on the permeability of the
material and is therefore not a constant quantity.
When the current in the solenoid is small, the iron is unsatur-
ated, the permeability is high, the flux per unit current is large
and therefore the inductance is large. When the current is large,
the iron is saturated, the permeability is low, the flux per unit
current is small and therefore the inductance is small.
Thus the inductance of circuits containing iron or other mag-
netic materials is not constant but, above a certain point,
decreases as the current increases and the induction density
increases.
The permeability curves in Figs. 64 and 65
show the relation between permeability and
induction density for cast iron and cast steel.
98. Mutual Inductance and Self-induc-
tance. If two electric circuits as A and B in
Fig. 76 are interlinked with a magnetic circuit
of constant permeability /*, or constant mag-
netic reluctance (R, the mutual inductance of
one circuit with respect to the second circuit
may be defined as the number of interlinkages of the second
circuit with the flux produced by unit current in the first circuit.
It is equal to the mutual inductance of the second circuit with
respect to the first and is represented by 3TC.
The flux produced by unit current in the n\ turns of the circuit
A is
96 ELECTRICAL ENGINEERING
and the inductance of A is
1 = ni4>i = - ^- c.g.s. units; (149)
ot
if all the flux produced by A passes through B, then the mutual
inductance of A with respect to B is
3TlAs = n 2 (j)i = - 1 ^ c.g.s. units. (150)
ot
Similarly, the flux produced by unit current in B is
02 = ^
and the inductance of B is
(R
again if all the flux produced by B passes through A, the mutual
inductance of B with respect to A is
_
ot
and is equal to the mutual inductance of A with respect to B,
3K BA = m AB = 2HX.
The leakage flux, that is, the flux which passes out between the
two circuits and links with only one of them, has been assumed to
be very small. In this case
If the leakage flux is not negligible, the flux $, produced by
unit current in A can be divided into two parts Ml and 8l) Fig.
77, of which Ml interlinks with B and Sl forms a local leakage
circuit about A and does not interlink with B; Sl is the self-
inductive flux produced by unit current in A.
The self-inductance of a circuit is the number of interlinkages
with the circuit of the flux, produced by unit current in it, which
does not interlink with any other circuit; it is the number of
interlinkages of the circuit with the stray flux or the leakage flux
produced by unit current in itself.
& v and therefore
Turns
O
Ocoil
O B
^v
QTurns
FIG. 77. Self and mutual inductance.
When any magnetic materials are present the reluctance of the
magnetic circuit is not constant but varies with the flux density,
that is, with the current and therefore the inductances, the mutual
inductance and the self-inductances are all variable quantities.
For any value of current i\ in A producing a flux (f> M linking
B the rate of change of flux generates in B an e.m.f. which may
be expressed as
e =n d$M = ^ dii , 152 )
dt dt
and the mutual inductance of A with respect to B is
3H = n^- (153)
-p- can, however, only be evaluated on the assumption that the
flux is proportional to the current or that the reluctance of the
magnetic circuit is constant. In practical applications allowance
must be made for the errors resulting from this assumption.
99. Self -inductance of Continuous -current Circuits. The
self-inductance of continuous-current circuits is apparent only
when the current is increasing or decreasing. The two most im-
98 ELECTRICAL ENGINEERING
portant cases are when the current is starting and when it is
stopping.
Starting of Current. When a constant e.m.f. E is impressed
on a circuit of resistance R and inductance L the current does not
immediately reach a steady value on account of the opposing
e.m.f. due to inductance. If at time t after E is impressed the
current is changing at the rate -T ,; the e.m.f. of inductance is
T di
b= ~ L dt
By Lenz's law it opposes the impressed e.m.f. and is therefore
negative.
The e.m.f. acting on the circuit is
and the current is
therefore
. _ E _ L di
~R~ R dt
or, transposing,
di _ Rdt.
E~ L'
the integral of this is
Rt
i /. M Kt . . n
loge (i - ft) = - - + loge C,
where loge C is the constant of integration. This reduces to
Rt
"
#_i_^~ T >
or ^ = p + Ge
R
when t = 0, i = 0, and C = ^
Substituting this value the current is
\t\
(154)
ELECTRIC CIRCUITS 99
W
When t = a and i = ^> and the current has reached its steady
rt
value, which .is independent of the inductance of the circuit;
calling this value I and substituting
(155)
The expression is called the time constant of the circuit.
Stopping of Current. If the impressed ' e.m.f. E is removed
from a circuit of resistance R and inductance L, when it is carry-
E
ing a current 7 = j? an d the circuit is closed through a resistance
Ri, the current will not at once fall to zero.
The e.m.f. of inductance at time t after the circuit is closed is
. . ,. di
et= - L dt
and it opposes 1 the decrease of current.
The currejit in the circuit at this instant is
T di
L dt
1T+TS'
or, transposing,
A (R + fli) .
i " L ***
the integral of this is
logei= -
T * I * v '6 v - / )
where log C is the constant of integration. This reduces to ,
i = Ce~ L '
when t = 0, i = 7, and C = 7.
Substituting this value, the current is
i=Ie f j (156)
or
< ; -i'~^'- <167)
The e.m.f. generated in the, coil, at the time t, is
" / T> I 7") \ 7vT -^ k I * ** 1 "" , t / -t P* O \
66 = i (ti + /ti) = E, ^ e L (158)
100 ELECTRICAL ENGINEERING
When t = 0, the generated e.m.f. is
e b = # 7 ^^-; (159)
therefore the e.m.f. generated in the circuit is greater than the
DID
impressed e.m.f. in the ratio ^ -
If, when the impressed e.m.f. is removed, the circuit is short-
circuited, RI = and the generated e.m.f. is
e b = E,
that is, the e.m.f. does not rise.
But if the circuit is broken Ri = & and e b = <* , and danger-
ous e.m.f s. are generated in the circuit. A large value of RI
causes the current to decrease rapidly, but it also causes a high
e.m.f. to be generated.
The energy supplied to the circuit while the current is starting
is
W = f Ei dt }
Jo
but
therefore,
/ = (8 dx lines.
The flux between the wires per centimeter length produced by
the current in one wire is
D
- J
x '..~.~* ~ R
27 log
R
and the flux of self -inductance of the circuit per centimeter length
is
JT1 D
^ = 2
ELECTRIC CIRCUITS
103
Xbut only half of this Q.UX surrounds each wire and therefore the
inductance of each wire per centimeter length is
4>
(160)
2 j[)
I/i = -j 2 log -j5 c.g.s. units.
The flux inside the conductor has been
neglected, but its inductive effect can be
calculated very easily if it is assumed that
the current is distributed uniformly over
the section of the wire. In Fig. 79 the
section of wire A is shown enlarged.
The flux density at distance x cm. from
the center of A is
(g = ?L' FIG. 79. Flux inside a
X ' conductor.
where I' is the current inside the radius x cm. and its value is
/' = / -~2 c.g.s. units.
The flux in the ring of radius x, width dx and length 1 cm. is
. , 21' 2Ix , ..
d = dx = -pg- dx lines;
this flux surrounds only the current F and is equivalent to a
smaller flux surrounding the current I, of value
, x 2 2lx*
and the flux equivalent to the flux inside the whole section is
f" f'2I ..
02 == d
and the inductance per centimeter due to the flux inside the con-
ductor is 02 i
2 = -7- = o c -g- s - units. (161)
/ z
The inductance of each wire per centimeter is therefore
= 1 + 2 = 2 log^ + ^c.g.s. units,
(162) V
and the inductance per mile of wire is
L = 2.54 X12X 5,280 (2 X 2.303 lo glo ^ +
= ^0.74 Iog 10 ^ + 0.0805) 10~ 3 henrys.
10' 9 henrys
(163)
CHAPTER IV
ELECTRIC CIRCUITS (Continued)
ALTERNATING-CURRENT CIRCUITS
102. The Sine Wave of Electromotive Force and Current. If
the coil in Fig. 80 rotates with a constant angular velocity in
a uniform magnetic field between the poles N and S, an alter-
nating e.m.f. will be generated between the terminals t and t'.
Referring to the figure it can be seen that in position (1) the e.m.f.
generated in the coil is zero, since the conductors forming it are
moving parallel to the lines of magnetic flux and therefore are
not cutting them; in position (2) it is positive and increasing; in
Rotation
FIG. 80. Generation of an alternating e.m.f.
(3) it is positive and has reached its maximum value since the
conductors are cutting perpendicularly across the flux; in (4) it is
positive and decreasing; in (5) it is zero; in (6) it is negative and
increasing; in (7) negative and maximum; in (8) negative and de-
creasing and in (1) is zero again, having gone through one com-
plete cycle. This cycle is represented in Fig. 80.
If = the maximum flux inclosed by the coil, n = the number
of turns on the coil and co = the angular velocity in radians per
second, then at time t sec. after the position of maximum in-
closure the coil has moved through angle 6 = ut radians, and the
flux inclosed is
= cos cot;
104
ELECTRIC CIRCUITS 105
the e.m.f. generated in the coil at this instant is
e = n j-. (< cos co)
at
= con$ sin ut absolute units
= ow 10~ 8 sin ut volts. (164)
When = ~, the e.m.f. has its maximum value
2
E m = con$ 10- 8 volts,
and therefore ^ . , /ic\
e = E m sin ut', (165)
and the e.m.f. generated in the coil varies as a sine wave.
The number of cycles through which the e.m.f. passes in one
second is called its frequency and since one cycle represents 360
/electrical degrees, the frequency may be expressed as
/ = -^ cycles, (166)
and therefore .
CO = 2-7T/.
Substituting this value of co in equation (164) gives
e = 2irfn3> 10~ 8 sin 2irft
= E m sin 6,
where = Zirft is the angle turned through in time t sec. after the
position of zero e.m.f.; the maximum value of the e.m.f. is
E m = 2irfn3> 10~ 8 volts. (167)
In a two-pole alternating-current generator one cycle corre-
sponds to one revolution of the coil and
/ = rev. per sec.,
and one electrical space degree is equal to one mechanical space
degree. With a p-pole alternator the e.m.f. passes through one
cycle for each pair of poles and the frequency is
/ ? rev. per sec.,
A
and one electrical space degree is less than one mechanical space
2
degree in the ratio -
If a non-inductive resistance of R ohms is connected across the
terminals of the coil in Fig. 81 an alternating current will flow in
the coil of instantaneous value
E m sin 6
I m sin
106
ELECTRICAL ENGINEERING
E
where I m = -^ is the maximum value of the current. The cur-
K
rent and voltage waves pass through their zero and maximum
values together and are therefore in phase. They are represented
by the two curves in Fig. 81.
FIG. 81. Resistance in alternating-current circuits.
103. The Average Value of a Sine Wave. The average value
of the ordinate of a sine wave i = I m sin 6 can be found by in-
tegrating over one-half wave; it is
1C* .
Jo
sin d6
= -f[ - cos $H
= [1 + 1] = - I m = 0.6377,
TT 7T
(169)
that is, the average value of the ordinate of a sine curve is
2
- times the maximum ordinate.
7T
104. The Effective Value of a Sine Wave. The effective value
of an alternating current is the value of continuous current which
would have the same heating effect.
When an alternating current i = I m sin 6 flows through a
resistance R } energy is transformed into heat at the instanta-
neous rate i 2 R watts ; the average rate of transformation of energy
is
1 C* 1 C*
- I i 2 # d0 = - I J
VO ^JO
MR
sin 2 B d6
- cos 26) d6
ELECTRIC CIRCUITS 107
27T
r 2P
H
27T
'*R=PR: (170)
thus the alternating current i = I m sin has the same average
heating effect or consumes the same average power as a con-
tinuous current j= ; the value I = -~= is therefore called the
V2 V2
effective value of the alternating current i = I m sin 6
The effective value of the alternating e.m.f. e = E m sin 8 is
(171)
The effective value of any alternating quantity is the square
root of the mean of the squared instantaneous values taken over
one complete cycle; it is equal to the maximum value divided
by \/2 only in the case of sine waves.
Alternating-current voltmeters and ammeters indicate the
effective values of voltage and current regardless of the wave
form.
105. Inductance in Alternating-current Circuits. When a cur-
rent flows in a conductor a magnetic field is produced in the space
surrounding it ; as long as the current remains constant this field
does not react on the electric circuit, but when the current varies
the flux linking with the circuit also varies and induces in the
conductor an e.m.f. opposing the change in the current and con-
sequently the change in the flux. This action is due to the inertia
of the niagnetic field and is analogous to the action of the fly-
wheel in mechanics. The inertia of the flywheel opposes any
change in speed just as the inertia of the magnetic field opposes
any change in current. Energy is stored in the flywheel as the
ppeed increases and is given back as it decreases and the only loss
of energy is that due to friction. Similarly energy is stored in
the magnetic field as the current increases and is returned to the
electric circuit as the current decreases, and the only loss of energy
is that due to hysteresis and eddy currents in the iron parts of
the magnetic circuit.
108
ELECTRICAL ENGINEERING
The energy stored in the flywheel is
w - 4
(172)
where / is its moment of inertia and co is its angular velocity.
The energy stored in the magnetic field is
W ==- watt-sec.,
where L is the inductance of the circuit in henrys and i is the
current in amperes.
The inductance of the coil opposes the change in current by
generating a back e.m.f.
di
T
= " L
dt
If an alternating current i = I m sin 6 = I m sin 2-n-ft is flowing
through a circuit of inductance L henrys and negligible resist-
ance an e.m.f. of inductance will be set up
di
= L -r (I m sin 2wft)
- 2irfLI m cos 2irft.
2irfLI m sin (2irft. - 90) ;
(173)
FIG. 82. Inductance in alternating-current circuits.
it is a sine wave of e.m.f. lagging behind the current wave by
90 degrees (Fig. 82).
In order to drive the current through the circuit an e.m.f.
must be applied to the terminals, which at every instant will be
equal and opposite to the back e.m.f. b', its instantaneous value
e = - e b = 2wfLI m sin (2irft + 90), (174)
a sine wave of e.m.f. leading the current wave by 90 degrees.
ELECTRIC CIRCUITS 109
This e.m.f. is consumed by the inductance of the circuit. Its
maximum value is
E m = 2irfLI m , (175)
where the term 2irfL is called the inductive reactance of the cir-
cuit and is denoted by X. It is of the nature of resistance and
is expressed in ohms.
Thus, in a circuit of reactance X and negligible resistance the
impressed e.m.f. is 90 degrees ahead of the current, or the cur-
rent lags 90 degrees behind the impressed e.m.f.
106. Resistance and Reactance in Series. If a circuit con-
tains a resistance R and a reactance X in series and carries an
alternating current i = I m sin 2irflj determine the value of the
impressed e.m.f. and its phase relation with the current (Fig. 83).
Curve , 6
Curve (4),
Curve
Curve
FIG. 83. Resistance and reactance in series.
To drive a current i = I m sin 2irft (curve 1) through a resist-
ance R an e.m.f. is required equal to
e R = iR = ImR sin 2irft (curve 2),
a sine wave in phase with the current with a maximum value I m R'
The inductance of the circuit sets up a back e.m.f.,
e b = - Lj t = 27r/L7 w sin (2ir/* - 90) (curve 3),
a sine wave lagging 90 degrees behind the current with a maxi-
mum value 2irfLI m = I m X.
110 ELECTRICAL ENGINEERING
To overcome this back e.m.f. due to inductance an e.m.f. must
be impressed on the circuit equal and opposite to 65.
T di
ex = - eb = L dl
= 2irfLI m sin (27r# + 90)
= I m X sin (2ttft + 90) (curve 4).
This is a sine wave 90 degrees ahead of the current wave, with a
maximum value ImX and is the e.m.f. consumed by the reactance
in the circuit.
The instantaneous value of the impressed e.m.f. (curve 5) is
the sum of the instantaneous values of e R and e x , thus
e = e R + e x = Ri + L^ (176)
= I m R sin 2irft + I m X sin (2nft + 90)
= I m R sin 2wfl + I m X cos 2irft
= I m (R sin B + X cos 6)
= I m Z sin (0 + 0), (177)
where Z = \/ R 2 -f- X 2 is called the impedance of the circuit and
is expressed in ohms and $ is the angle of lead of the impressed
e.m.f. relative to the current.-
X R
sm = ^=== and cos =
The impressed e.m.f. is therefore a sine wave leading the cur-
rent by an angle <. Its maximum value is
E m = I m V R 2 + X 2 = I m Z, (178)
and its effective value is
J. _ I^Z
? V2 V2 IZ '
where I is the effective value of the current.
The component of E in phase with the current is
T?
E l = IR = - R = E cos 0;
&
and the component of E 90 degrees ahead or in quadrature ahead
of the current is
E 2 = IX = X = #sin<.
ELECTRIC CIRCUITS 111
107. Capacity in Alternating-current Circuits. The charge
on a condenser or the quantity of electricity stored in it is pro-
portional to the difference of potential between its terminals;
thus,
q = Ce,
where q is the charge,
e is the difference of potential between the terminals,
and C is the capacity of the condenser.
A condenser has a capacity of one farad when one coulomb
of electricity stored in it produces a difference of potential of
one volt between its terminals.
If an alternating e.m.f. e = E m sin 2irft is impressed on the
terminals of a condenser of capacity C farads, a current i flows.
At any time t the charge on the condenser is
q = Ce,
but the charge is the amount of electricity which has flowed into
the condenser and its value is
= ft A
Jo
and therefore
fi dt = Ce.
Differentiating with respect to t gives
and substituting the value of e,
i = Cj (E m sm2irft)
= 2irfCE m cos 2irft
= 2irfCE m sin (27T/Y + 90), (179)
and thus the current flowing into the condenser is a sine wave
leading the impressed e.m.f. by 90 degrees. Its maximum value
is
(180)
-A-c
the condensive reactance of the
and is expressed in ohms.
where X c = ~ ^ is called the condensive reactance of the circuit
112
ELECTRICAL ENGINEERING
In Fig. 84 curve 1 represents the impressed e.m.f. e E m
sin 2irft and curve 2 the current i = I m sin (2irft + 90). Be-
tween the points a and b the current is positive and the e.m.f. is
increasing; from 6 to c the current is negative and the e.m.f. is
decreasing.
FIG. 84. Capacity in alternating-current circuits.
108. Resistance and Condensive Reactance in Series. If an
alternating current i = I m sin 2wft flows in a circuit consisting
of a resistance R in series with a condenser of capacity C farads
and reactance of X c = , , ~, ohms, determine the magnitude of
the impressed e.m.f. and its phase relation with the current
(Fig. 85).
Curve (190)
/ f (X - X c ) 2
and from equation (162)
T7T 13
" Wl ^^
where
z =
is the impedance of the circuit.
ELECTRIC CIRCUITS
115
The impressed e.m.f. is therefore a sine wave of maximum
value E m = I m Z = I m VR* + (X - X c ) 2 , and leads the current
X X
wave by an angle < = tan" 1 ^ ~*
If X c > X, < is negative and the impressed e.m.f. lags behind
the current.
If X c = X, = and the e.m.f. is in phase with the current,
and the impedance of the circuit is Z = R.
110. Vector Representation of Harmonic Quantities. A sine
wave may be obtained from a circular locus as shown in Fig. 86.
s
)=wt
\/
\_
sz
FIG. 86. Sine wave.
The radius of the circle is the maximum value of the sine func-
tion and is called its amplitude. If the radius vector is rotated
at uniform angular velocity o> in the counter-clockwise direction,
its vertical projection at any time t is
y = R sin co = R sin 6,
where time is measured from the horizontal. Plotting the values
of y on a base of angle 6 for a complete revolution of the radius
vector gives the cycle shown.
To represent the sine function
e = E m sin 0,
a circle of radius E m is taken and the vertical projections of the
revolving vector are plotted on base of angle (Fig. 87 (a)).
To represent a sine function
i = I m sin (B 4>),
a circle of radius I m is taken, but since i does not pass through
zero until angle 6 = , the sine wave is displaced to the right by
angle as shown and the e.m.f. wave ^= E m sin leads the cur-
rent wave i = I m sin (6 0) by angle . Instead of using com-
plete circles to represent sine waves, their radii E m and I m can
116
ELECTRICAL ENGINEERING
be used as in Fig. 87(6) or since alternating quantities are repre-
sented by their effective values instead of their maximum values
the two vectors OE and 01, Fig. 87 (c), are used to represent the
W
two sine waves in Fig. 87 (a). OE = E = j= is the effective
V2
value of the e.m.f . e = E m sin 6, and 01 = I = ^ is the effect-
V2
ive value of the current i = I m sin (0 <). The vector 01 is
behind the vector OE by angle < and so indicates the relative
phase of the two quantities. The counter-clockwise direction is
FIG. 87. Vector representation of harmonic functions.
taken as the direction of advance in phase since it was adopted
at the International Electrotechnical Congress at Turin and is
now standard.
111. Power and Power Factor. In continuous-current cir-
cuits the power consumed is the product of the impressed e.m.f.
and the current, and is-
P = El watts.
(192)
In alternating-current circuits the power varies with time; its
instantaneous value is
ei watts,
(193)
where e and i are the instantaneous values of the e.m.f. and cur-
rent in the circuit.
Two cases will be considered, first, when the current and e.m.f.
are in phase and, second, when the current lags behind the e.m.f.
Case I. If an e.m.f. e = E m sin 6 is impressed on the terminals
of a circuit of resistance R a current i = I m sin 6 will flow in
phase with the e.m.f.
ELECTRIC CIRCUITS 117
The instantaneous power in the circuit is
p = ei = E m l m sin 2 = ^^ (1 - cos 26). (194)
The values of e, i and p for a complete cycle are plotted in Fig. 88.
The power varies with twice the frequency of the current from
to E m l m , but never becomes negative. The area beneath the
power curve represents the energy consumed in the circuit during
one cycle.
if
= -I P
" f/ U
FIG. 88. Power in a non-inductive circuit.
The average power is
P
= l ("T^ (i - c ^ de
= ErnL.nl- ^ sin 26
E m L
27T
EJ[,
M
E m I m
(195)
2 V2 V2
therefore, the average power in an alternating-current circuit is
equal to the product of the effective values of the e.m.f. and the
current, if they are in phase.
Since
P = El
and
E = IR, .'
therefore,
P = PR, (196)
118
ELECTRICAL ENGINEERING
and the power is equal to the square of the effective value of the
current multiplied by the resistance.
Case II. If an e.m.f. e = E m sin 6 is impressed on a circuit
containing a resistance R, and an inductive reactance X, the cur-
rent will lag behind the e.m.f. by an angle 4> = tan- 1 TT and its
instantaneous value will be
i = I m sin (0 0).
The instantaneous power in the circuit is
p = ei = E m l m sin B sin (B 0)
E m l m
cos cos (2 e
(197)
The values of e, i and p for one cycle are plotted in Fig. 89.
The maximum value of power is m (1 + cos 0), which is less
than in Case I, and the power curve falls below the base twice in
each cycle. The energy consumed per cycle is the difference
between the positive and negative areas intercepted by the power
curve.
FIG. 89. Power in an inductive circuit.
The average power is
P =- Fp dd
* Jo
(*
cos0 cos (20 0)[d0
sin (20 -_0)T
, sin (-
E m l,
2* J
E m l m r
2x
27T
cos
7T COS
E m l m r sin(-
cos
2
EI cos 0;
(198)
ELECTRIC CIRCUITS
119
therefore, the average power is the product of the effective values
of the e.m.f. and current multiplied by the cosine of the angle of
phase difference between them.
From the vector diagram, in Fig. 89, it is seen that
E cos = Ei = IR,
and therefore the power is
p = EJ = PR,
and is equal to the active or in-phase component of the e.m.f.
multiplied by the current or is equal to the square of the current
multiplied by the resistance of the circuit as found in equation
FIG. 90. Power in an inductive reactance.
FIG. 91. Power in a condensive reactance.
196; thus, all the power consumed in the circuit is consumed by
the resistance. Reactance or self-inductance does not consume
power since the energy stored while the current is increasing is
given back while it is decreasing and the e.m.f. consumed by self-
inductance is a wattless e.m.f. Similarly condensive reactance
does not consume power since the energy stored while the e.m.f.
is increasing is given back to the circuit while the e.m.f. is decreas-
ing and the e.m.f. consumed by condensive reactance is wattless.
These results are illustrated in Fig. 90 and Fig. 91.
120 ELECTRICAL ENGINEERING
In Fig. 90 are plotted the values of e, i and p for an inductive
circuit without resistance in which the current lags 90 degrees
behind the impressed e.m.f. The power curve cuts off equal
areas above and below the base line and therefore the average
power is zero. The area below the line represents the energy
given back by the magnetic field while the current is decreasing
from its maximum value I m to zero and the area above the line
represents the energy stored in the magnetic field while the cur-
rent is increasing again to its maximum value. The amount of
energy in each case is L -~- watt-sec.
In Fig. 91 are plotted the values of e, i and p for a circuit con-
taining a condensive reactance but without resistance. The cur-
rent leads the impressed e.m.f. by 90 degrees and the average
power is again zero, so that no energy is consumed in the circuit.
The positive area cut off by the power curve represents the energy
stored in the electrostatic field of the condenser while the e.m.f.
is increasing and the negative area represents the energy re-
turned to the circuit while the e.m.f. is decreasing. The maxi-
mum amount of energy in each case is C ~- watt-sec, where
E m is the maximum e.m.f.
R
\M
E ' -
E E.1X ^E ^ E!=IX A^>*
=P Kinfh \^^ \\
S
=ESin0 \
I - * >L^i
Ei=IR = E Cos
(a) (6) (c)
FIG. 92. Power in alternating-current circuits.
Fig. 92 illustrates various methods of representing the power
in a circuit; in (a) it is the product of the impressed e.m.f., the
current and the cosine of the angle of phase difference between
them,
P = E X I X cos 0; (199)
in (6) it is the product of the current and the in-phase component
of the e.m.f.,
P = /X#cos0 = /X#i = PR', (200)
in (c) it is the product of the e.m.f. and the in-phase component
of the current,
P = E X / cos (201)
ELECTRIC CIRCUITS
121
The apparent power in a circuit is the product of the impressed
e.m.f . and the current and is expressed in volt-amperes or kilovolt-
amperes.
The power factor of a circuit is the ratio of the true power to
the apparent power and is
P El cos
EI = ~ET~ =COS
(202)
15
10 20 30 40 50 60 70 80 90
Per Cent Power Factor - In-Phase Component - Cosine
FIG. 93.
therefore, the power factor is the cosine of the angle of phase
difference between the current and the impressed e.m.f.; it is
usually expressed in per cent, and may be either leading or
lagging.
The sine of the angle of phase difference between the current
and the impressed e.m.f. is called the reactive factor of the
circuit.
When the current is in phase with the e.m.f., the power factor
is unity or 100 per cent, and the reactive factor is zero.
122
ELECTRICAL ENGINEERING
When the current leads the e.m.f . by 30 degrees the power fac-
tor is cos 30 = 0.866 = 86.6 per cent, leading and the reactive
factor is sin 30 = 0.500 = 50 per cent.
When the current lags 60 degrees behind the e.m.f. the power
factor is cos 60 = 0.500 = 50 per cent, lagging and the reactive
factor is sin 60 = 0.866 = 86.6 per cent.
When the power factor is 90 per cent, the reactive factor is
*Vl - O9 2 = 0.436 = 43.6 per cent.
When the power factor is 99 per cent, the reactive factor is
Vi - Q.99 2 = 0.141 = 14.1 per cent.
From this last example it may be seen that very great care
must be exercised in determining reactive factors from power-
factor readings when the power factor is near unity; an error of 1
.per cent, in reading the current, e.m.f. or power might cause a
reactive component of 14.1 per cent, to be missed entirely.
These and similar problems may be solved quickly by using
the chart in Fig. 93.
112. Examples. 1. If an alternating e.m.f. of effective value E is im-
pressed on a non-inductive circuit of resistance R, a current 7 will flow in
phase with e.m.f., where
-I
The vector diagram is shown in Fig. 94.
I I
R E
FIG. 94.
FIG. 95. FIG. 96. FIG. 97.
FIG. 98.
2. If an alternating e.m.f. E is impressed on a circuit of reactance X and
V
negligible resistance, a current 7 = will flow lagging 90 degrees behind the
X
e.m.f. (see Fig. 95).
3. If an alternating e.m.f. E is impressed on a circuit of resistance R and
reactance X, a current 7 will flow lagging behind the e.m.f. by angle <, where
tan* = -
ELECTRIC CIRCUITS
123
The vector diagram for the circuit is shown in Fig. 96. The e.m.f. con-
sumed in the resistance is EI = IR volts in phase with the current and is
represented by the vector OEi.
The e.m.f. consumed by the reactance is E 2 = IX volts leading the cur-
rent by 90 degrees represented by OE Z . The impressed e.m.f. is OE = E
and is the vector sum of EI and E z ', therefore
E
= iz,
where
Z = \/R 2 + X 2 is the impedance of the circuit.
4. Fig. 97 shows the vector diagram of a circuit containing a condenser of
reactance X c , when an alternating e.m.f . E is impressed on its terminals and
ET
a current I = -~- flows through it leading the e.m.f. by 90 degrees.
A c
6. Fig. 98 shows the diagram for the same circuit with a resistance R
added in series.
The e.m.f. consumed by the resistance is OEi = EI = IR, in phase with
01 = /.
The e.m.f. consumed by the reactance is OE 2 = E 2 = IX C , lagging 90
degrees behind OI.
The impressed e.m.f. is
OE = E = \/Ei 2 + E 2 2 = I\/R 2 + X e 2 = IZ,
lagging behind OI by angle , where
. E 2 IX. X e
T
t:
FIG. 99.
6. If an alternating e.m.f . E is impressed on a circuit containing a resist-
ance R) an inductive reactance X = 2irfL and a condensive reactance X e =
connected in series, determine the magnitude and phase relation of the
current and draw the vector diagram for the circuit (Fig. 99).
01 = I is the current taken as horizontal.
OEi = EI = IR is the e.m.f. consumed by the resistance and is in phase
with 7.
OE Z = E 2 = IX is the e.m.f. consumed by the inductive reactance and
leads 01 by 90 degrees.
OE 3 = EZ = IX e is the e.m.f. consumed by the condensive reactance and
lags behind 01 by 90 degrees.
124
ELECTRICAL ENGINEERING
OE = E = IZ is the e.m.f. impressed on the circuit, or the e.m.f. con-
sumed by the impedance Z. It is the vector sum of the three components
Ei f E 2 and E 3 and leads the current by angle 0; therefore,
and
z =
+ (X - X c )
(X - x c
where L is the inductance of the circuit in henrys, C is the capacity in farads
and /is the frequency of the impressed e.m.f.
The angle of phase difference between the e.m.f. and current is , where
tan =
E l
IR
X - X c
TT^
and
If X > X c the current lags behind the e.m.f.;
if X = X c the current is in phase with the e.m.f.;
if X < X c the current leads the e.m.f.
1 R
E III
J X
^70
J tf
^^ I
FIG. 100.
7. An alternating e.m.f. E is impressed on the terminals of the circuit in
Fig. 100 consisting of a resistance R and an inductive reactance X in parallel.
The main current I has two components,
and
ET
/i = -~, in phase with E,
Tjl
/2 = ^, 90 degrees behind E.
From the vector diagram
7 1 * + / 2 2 = #\/-^- 2 +^- 2
E
VR* + x*
RX
and lags behind E by an angle , where
tan _ h = E / X - %
The impedance of the circuit is
E RX
1 VR* + x 2
8. In Fig. 101 a third branch is connected in parallel with the two in
example (7) containing a condensive reactance X c > X.
ELECTRIC CIRCUITS
The main current 7 has three components,
1 1 = , in phase with E,
pi
/ 2 = , 90 degrees behind E,
1 3 = > 90 degrees ahead of E.
I
I
125
and
FIG. 101.
From the vector diagram
and lags behind E by an angle , where
7 2 - 7 3 _ 1/X - 1/X.
7/72
tan =
The impedance of the circuit is
FIG. 102.
9. Find the magnitudes of the currents in the various parts of the circuit
in Fig. 102 and their phase relations with the impressed e.m.f.
E
I\ = y and lags behind the impressed e.m.f. by an angle i = TT
7, =
E
Xt
and leads the e.m.f. by an angle < 2 , where tan
126 ELECTRICAL ENGINEERING
I = \//i 2 + I* 2 + 2/i/ 2 cos (0i + 02) (see vector diagram) and lags
behind the e.m.f. by an angle 0, where
/i sin 0! - / 2 sin 2
tan = j ; y '
/i cos 0i + / 2 cos 2
The power consumed in the circuit is
> = El cos
= /: 2 #i + / 2 2 #2 watts.
113. Numerical Examples. 1. If an alternating e.m.f. of 200 volts at a
frequency of 60 cycles per second is impressed on a circuit consisting of a
resistance of 10 ohms in series with an inductance of 0.1 henry and a capacity
of 100 microfarads, (a) determine the current in the circuit and its phase
relation with the impressed e.m.f., (6) the e.m.f. consumed in each part of the
circuit, (c) If the impressed e.m.f. is maintained constant and the frequency
is varied determine the maximum value of the current, (d) Plot the current
and the various e.m.fs. on a frequency base.
(a) The inductive reactance of the circuit is
X = 27T/L = 2 X 3.14 X 60 X 0.1 = 37.6 ohms;
the condensive reactance is
1 10 6
Xc = = 2 X 3.14 X 60 X 100 =
the impedance of the circuit is
Z = VR 2 + (X - X c ) 2 = VlO 2 + (37.6"- 2^4)2 = 15 ohms;
and therefore the current is
T E 200
J-3-15 ==13.3 amp.
The current lags behind the e.m.f. by an angle 0, where
X -X c 37.6 - 26.4
tan0= -3- ~-
and
= 48 18'.
(6) The e.m.f. consumed in the resistance is
Ei = IR = 13.3 X 10 = 133 volts;
the e.m.f. consumed in the inductive reactance is
E z = IX = 13.3 X 37.6 = 500 volts;
and the e.m.f. consumed by the condensive reactance is
E 3 = IX C = 13.3 X 26.4 = 350 volts.
The impressed e.m.f. is
E = VES+ (E t - 'E 3 ) 2 = A/133 2 + (500 - 350) 2 = 200 volts.
The vector diagram is shown in Fig. 103.
(c) The current in the circuit at any frequency is
7=1=
If E is maintained constant at 200 volts and / is varied, / varies.
When / = 0, ^ = cc and / = 0;
ELECTRIC CIRCUITS
127
when/ = oc, 2irfL = oc and / = 0;
when 2irfL =
27T/C
or/ =
2 X 3.14 I^-3Ampe
/ Ei=IR=-133 Volts
= IX C =350 VoltB
FIG. 103.
The e.m.f. consumed in the resistance is
Ei = IR = 20 X 10 = 200 volts;
the e.m.f. consumed in the inductive reactance is
E, = IX = 20 X 2 X 3.14 X 50 X 0.1 - 628 volts;
and the e.m.f. consumed in the condensive reactance is
1
E a = IX C = 20 X
E 2 "IX- 628 Volts
20 Amperes
2 X 3.14 X 50 X
E- E!= IR-200 Volts
VE 8 -IX C - 628 VoltB
FIG. 104. Resonant circuit.
= 628 volts.
M.F.Cojisunied in the
Inductive Eeactance = e 2
Effective Value
E 2 = 028 Volts
urrent i, Effective Value
I20 Amperes
Impressed E.M.F. -
E.M.F. Consumed in
the Kesistnuce- e,
Effective Value
E - 200 Volts
E.M.F. Consumed in
the Condensive
Reactance e 3
Effective Value '
Es-628 Volts
FIG. 105. E.m.fs. and currents in a
resonant circuit.
The vector diagram for the circuit is shown in Fig. 104, and the current and
e.m.f. waves are shown in Fig. 105. A series circuit in which the inductive
128
ELECTRICAL ENGINEERING
reactance and the condensive reactance are equal at a certain frequency is
said to be in a state of resonance for that frequency. In commercial circuits
the capacity is usually so small that resonance cannot occur at ordinary
frequencies, but when any high frequency e.m.fs. are produced in the circuit
resonance may occur and very large e.m.fs. may appear and break down the
insulation.
(d) In the table below are given the values of the circuit constants for
frequencies from to 200 cycles per second and the corresponding values of
/, Ei, E 2 , E 3 and cos .
Current Lagging
Cycles per Second
FIG. 106.
/
X
Xc
x-x e
(X-XJ* Z'
,|,
Si
E 2
Ez
Cos
0.0
cc
a
oc
oc
oc
0.0
0.0
0.0
200.0
0.000
10
6.3
159.0
-152.7
23,400
23,500
153.0
1.3
13.0
8.2
208.0
0.065
40
25.1
39.8
-14.7
216
316
17.8
11.2
112.0
281.0
446.0
0.560
50
31.4
31.4
0.0
100
10.0
20.0
200.0
628.0
628.0
1.000
60
37.6
26.4
11.2
126
226
15.0
13.3
133.0
502.0
353.0
0.670
100
62.8
15.9
56.9
3,240
3,340
57.7
3.5
35.0
218.0
55.2
0.170
200
125.6
7.9
117.7
13,800
13,900
118.0
1.7
17.0
213.0
13.4
0.085
ec
oc
0.0
oc
oc
oc
K
0.0
0.0
200.0
0.0
0.000
These quantities are plotted on a frequency base in Fig. 106. J 2 reaches
its max. when/ = 51.4 and E 3 reaches its max. when/ = 49.2 cycles per
second.
2. If an alternating e.m.f. E = 200 volts at a frequency/ = 60 cycles per
second is impressed on the circuit in Fig. 107, determine the value and phase
relation of the main current and the currents in the three branches.
The first branch is a resistance R = 40 ohms; the second branch is an
inductance L = 0.1 henry and has a reactance X = 2irfL = 37.6 ohms at
60 cycles; the third branch is a capacity C 100 microfarads or 10~ 4 farads
and has a reactance X c = fn = 26.4 ohms.
ELECTRIC CIRCUITS
129
The current in the resistance is
E 200
/1== =~ =
in phase with the impressed e.m.f.; the current in the inductive reactance is
E 200
/2 = X = 37^6 = 5 ' 3 amp "
90 degrees behind the impressed e.m.f.; the current in the condensive react-
ance is
E 200
90 degrees ahead of the impressed e.m.f.
The main current is
/ = V/i 2 + (/a - /s) 2 = V5 2 + (5.3 - 7.6) 2 = 5.5 amp.,
leading the impressed e.m.f. by an angle <, where
tan * = Ap[? = 2.3 o.46,
J 1 O
and therefore
= 24 42'.
If the e.m.f. impressed on the circuit is maintained constant and the fre-
quency is varied find the magnitude of the main current when it is in phase
with the e.m.f.
The current at any frequency is
= V/l 2 +(/2- / 3 )
I s =7.6 Amperes
C =10'* Farads
=
X c = 26.4 Ohms
1= j.o Amperes
E=200 Volts
1=5 Amperes
=> 2442'
IjsaS.3 Amperes
FIG. 107.
the angle of lag of the current behind the e.m.f. is
When the current is in phase with e.m.f.
= and X = X c or 2irfL
the frequency is therefore
1
2*VLC ~ 2 X 3.14 V 0.1 X HT*
~ 5
130
ELECTRICAL ENGINEERING
The main current at a frequency of 50 cycles is
7 = 7=? = 200
111 R 40
the current in the inductive reactance is
E 200
X 2X3.14X50X0.1
the current in the condensive reactance is
_ E._ 200
*3 -y " <
AC 1
- = 6.36 amp.
= 6.36 amp.;
2 X 3.14 X 50 X 10- 4
and the current in the lead between the first and second branches is zero.
3. If an alternating e.m.f. E = 100 volts is impressed on a circuit, Fig.
108, made up of two parallel branches 1 and 2, the first with resistance
Ri = 4 ohms and inductive reactance Xi = 3 ohms and the second with
resistance R z = 6 ohms and inductive reactance X z = 8 ohms; determine the
magnitude of the main current / and of the two branch currents I\ and 7 2
and their phase relations with the impressed e.m.f. E. The vector diagrain
is shown in Fig. 109.
I (1)
(2)
II
I2 <
Ri=4
E = 100 Volts
1
R 2 = 6
1 =10
FIG. 108.
1-29.9
FIG. 109.
The current in branch 1 is
E
1 1
= 100
~ \/4 2 + 3 2
and it lags behind the impressed e.m.f. by angle
= 20 amp.,
The current in branch 2 is
E
100
/2 =
and it lags behind the impressed e.m.f. by angle
02 = tan"" 1 ^ = tan" 1 % = tan" 1 % = sin" 1 % = cos" 1 2.
Kz
The main current / is the vector sum of I\ and /2 and may be found as
/ =\/(Ii sin 0i + /2 sin 02) 2 + (/i cos 0i -f- I z cos
and it lags behind the impressed e.m.f. E by angle
tan
cos i + h cos
=, tan- = tan- 0.909 = 42 18'.
ELECTRIC CIRCUITS 131
114. Circuit Constants. A continuous-current circuit has two
constants :
impressed e.m.f.
resistance R, r = - >
current
~ current
conductance G, g = .-
impressed e.m.f.
and the conductance is the reciprocal of the resistance or
Continuous-current circuits also have inductance and electro-
static capacity, but these do not affect the flow of current except
at the instant of opening or closing the circuit.
An alternating-current circuit has six so-called constants:
1. resistance R, r,
2. reactance X, x,
3. impedance Z, z,
4. admittance Y, y,
5. conductance G, g,
6. susceptance B, b.
1. The resistance of a circuit consumes a component of e.m.f.
in phase with the current and so consumes power. In circuits
which are partially inclosed in iron an alternating magnetic flux
is produced in the iron and a loss of power occurs due to hysteresis
and eddy currents. These iron losses are sometimes included
with the copper loss and charged against the resistance. This
gives a value of resistance greater than the true ohmic resistance
and is called the effective resistance of the circuit. Since the
hysteresis and eddy-current losses vary both with the frequency
and the induction density in the iron, the effective resistance is
not a constant quantity.
The active component of the impressed e.m.f. or the component
in phase with the current is
E l = IR }
and the resistance is
EI active component of impressed e.m.f.
K = -=.- = -
I current
= m -phase component of impressed e.m.f. (2Q3]
current
2. The reactance of a circuit consumes a component of e.m.f.
in quadrature with the current, leading in the case of circuits of
132 ELECTRICAL ENGINEERING
large inductance and lagging in circuits of large electrostatic
capacity, but it does not consume any power.
The inductive reactance of a circuit is
X = 2-n-fL ohms,
where / is the frequency of the impressed e.m.f. and L is the in-
ductance in henrys. Commercial circuits are operated at a fixed
frequency and so / is constant.
The inductance of a circuit in air or any non-magnetic material
is constant but in an iron-clad circuit it varies with the current,
decreasing as the current increases since the permeability of the
iron decreases as the flux density in it increases.
Since inductive reactance consumes a component of e.m.f. in
quadrature ahead of the current it is taken as a positive reactance.
The condensive reactance of a circuit is
Xc =
where C is the capacity of the circuit in farads. The capacity of
a circuit does not vary with the current or e.m.f. and thus
the condensive reactance is constant so long as the frequency is
constant.
Condensive reactance is taken as a negative reactance since it
consumes a component of e.m.f. in quadrature behind the cur-
rent. Thus when inductive reactance and condensive reactance
are connected in series they oppose and the reactance of the
circuit is
X = X L - X c = 27T/L - ~~ ohms.
In series-parallel circuits the reactance is a complex function of
the resistances and reactances of the various branches.
The reactance of any circuit is
reactive component of impressed e.m.f.
X =
current
quadrature component of impressed e.m.f.
current
3. The impedance of a circuit includes both the resistance and
the reactance; it is
ELECTRIC CIRCUITS 133
4. The admittance of a circuit is
Y = : .; (206)
impressed e.m.f.
it is the reciprocal of the impedance and thus
Y Z = VW^' . . (207)
The admittance has two components, conductance and sus-
ceptance.
Fig. 110 shows a circuit of impedance Z = \/R 2 + X 2 in
which the current lags behind the impressed e.m.f. by an angle
FIG. 110.
. The current I can be resolved into two components in phase
and in quadrature with the impressed e.m.f. E. The in-phase or
active component of current is
7i = 7 cos = | cos = E~ = EG, (208)
where
is the conductance of the circuit.
The quadrature of reactive component of current is
7 2 = I sin = | sin = E^ 2 = EB, (210)
where
X X
B = & = w*^
is the susceptance of the circuit.
The total current
but, by equation (206), I = EY, and therefore
Y = G 2 + B*. (212)
134r ELECTRICAL ENGINEERING
5. The conductance of a circuit is, from equation (208),
~ _ active component of current
impressed e.m.f.
6. The susceptance of a circuit is, from equation (210),
/? reac tive component of current ( I ""I
(a) (&1 (c)
FIG. 111.
In Fig. 111(6) which represents the e.m.f. and current in a cir-
cuit of impedance Z \/r 2 + x 2 the axis is chosen in the direction
of the current and the e.m.f. is resolved into two components, e\ in
phase with the current and e 2 in quadrature ahead of the current.
The absolute value of the e.m.f. is
E = Vei 2 + 6 2 2 ,
and it leads the current, which was chosen as axis by an angle 0,
where tan =
ei
Thus when the e.m.f. is expressed as the sum of two components
at right angles both its magnitude and its phase are known.
To distinguish between horizontal and vertical components
the prefix j = V 1 is added to all vertical components and
the expression for the e.m.f. above is
E = e t + je 2 . (215)
The dot is placed under the E to show that it is expressed in rec-
tangular coordinates and serves to distinguish it from its absolute
value.
ELECTRIC CIRCUITS 135
Since e\ = E cos < = Ir and e 2 = # sin < = 7z,
$ = E cos + jE sin
= /r + j/z
(216)
and therefore the impedance in rectangular coordinates is
Z = r + j*. (217)
In Fig. lll(c) the e.m.f. is chosen as axis and the current is
behind it in phase by an angle and has two components i\ in
phase with the e.m.f. and iz in quadrature behind it.
The current may be written
/ = ii ~ #2, (218)
and this equation indicates that the current has a value
/ = ii* + iV,
and that it is behind the chosen axis (in this case the e.m.f.) in
phase by an angle 4>, where
tan < = -?
*i
Since i\ = I cos = Eg and iz = I sin = Eb, where Y =
Vg 2 + 6 2 is the admittance of the circuit, equation (218) may
be written
1 = 1 cos 4> jl sin
= Eg - jEb
= E (g - jb), (219)
and therefore the admittance in rectangular coordinates is
Y = g - j&.
Admittance and impedance are not alternating quantities and
their components are independent of the axis of reference, but
they can be represented in rectangular coordinates.
A current multiplied by an impedance gives an e.m.f. displaced
from it in phase by an angle whose tangent is the ratio of the re-
actance to the resistance.
A current divided by an admittance gives an e.m.f. displaced
in phase by an angle whose tangent is the ratio of the susceptance
to the conductance.
Similarly an e.m.f. divided by an impedance or multiplied by
an admittance gives a current.
136
ELECTRICAL ENGINEERING
By definition a vector multiplied by j is turned through 90
degrees in the counter-clockwise direction; when multiplied by
j z it is turned through 180 degrees and its sign is reversed.
Therefore
ji = - 1
and
j = V^n. (220)
Taking this value for j alternating quantities expressed in rec-
tangular coordinates referred to a given axis can be added, sub-
tracted, multiplied and divided and the results obtained are
expressed in rectangular coordinates referred to the same axis.
I=-irfjs
FIG. 112.
It is not necessary to choose either the current in a circuit or
the e.m.f. as axis and any other line may be taken as shown in
Fig. 112, but the e.m.f. must now be expressed as the sum of two
components along and perpendicular to the new axis, thus
E = e l + je*,
and similarly the current is
T- 1_
The e.m.f. has an absolute value
77? f o" I o
j = y/ Ci ~\~ 2
and is ahead of the axis by an angle
i 2 ,* i
0i = tan" 1 - = cos" 1 -
The current has an absolute value
and is ahead of the axis by an angle
02 = tan" 1 ^ =
ELECTRIC CIRCUITS 137
The e.m.f. leads the current by an angle
= i 02.
The impedance of the circuit in rectangular coordinates is
$ ei + je 2 ei
= r + jx,
where the resistance of the circuit is
_ eiii + eziz _ ejii + ^2^
ti 2 + ta 2 I 2
and the reactance of the circuit is
_ ezii eiiz _ ei ei
+ i 2 2 I 2
The admittance of the circuit is
Y - i = ^ + J^ 2 _ ^i + JJ2 ei jet
~ E ~ ei+ je 2 ~ ei + je z ei - je 2
_ e\i\ + 62^2 _ -e^ii
= 2 J
= g - jb;
the conductance is
eiii + ^2^2 _ eiii + 62^
6i 2 + 2 2 : .E' 2
and the susceptance is
7 _
:
The power factor of the circuit is
COS = COS $1 2
= cos 0i cos 02 + sin 0i sin
The power consumed in the circuit is
P = EI cos
= eiii + erf* (221)
and is the sum of the products of the components of the e.m.f.
and current which are in phase. The products of the components
of the e.m.f. and current which are in quadrature, namely,
and 62*1, do not represent power consumed.
138 ELECTRICAL ENGINEERING
The power may also be represented as
75 __ T _|
or
p = E 2 g = eiii + e 2 i 2 . (222)
116. Examples in Rectangular Coordinates. 1. Find the cur-
rent in the circuit in Fig. 113 in terms of the impressed e.m.f. and
the constants of the circuit.
E = E, + # 2 = / {(n + r 2 ) + ./(si + x 2 )(.
t ^AA/ V JiflflQQft>
r 2 X 2
FIG. 113.
The impedance of the circuit is
Z = (f! + r a ) + ./(zi -f a;,)
and its absolute value is
+ (xi
and the absolute value of the current is
T7T 777
T j it/
Z =
the power factor of the circuit is
The vector diagram is drawn taking the current as the axis.
2. Solve the circuit in Fig. 114.
where
7*1
Qi = ., ; -- 1> and DI =
2 2
ELECTRIC CIRCUITS
139
the main current is
and its absolute value is
= E V(gi
(6
* I.
FIG. 114.
The admittance of the circuit is
Y (0i ~i~ 2) j
and its absolute value is
Y = V(gi + g*) 2 + (bi + 6 2 ) 2 .
The power factor of the circuit is
cos =
The vector diagram is drawn with the e.m.f. as axis.
I A/wA-^m?N-, i s
FIG. 115.
3. Solve the circuit in Fig. 115.
/2 / . = E
r . $* _ u
63)!,
and therefore
+ jh) - j(bt + 6s)
140 ELECTRICAL ENGINEERING
where
__ 2 + gs , Y _ fr 2 + 63
ai
_
" (92 + R s
! X=100hm. T
g at 60 Cycles 1 X
X = 600hms
*p at 250 Cycles
at 250 Cycles
.. (6)
:
8 I 4-
'
Fundam<
and
10
> E
fifth
Curr
J
()
Harmonic
juts and
...M.i'.
10 I -I
10
(d) i
:ntal Currents
E.M..F.
FIG. 120.
3. If the same e.m.f. wave is impressed on the parallel circuit AB in Fig.
120, find the current in the three branches at a frequency of 50 cycles per
second.
The resistance of branch (1) is 10 ohms; the reactance of (2) is 10 ohms at
10
146 ELECTRICAL ENGINEERING
%.
50 cycles and 50 ohms at 250 cycles; the condensive reactance of (3) is 10
ohms at 50 cycles and 2 ohms at 250 cycles. ^~
The current /i in (1) consists of a fundamental of -777 = 10 amp. and a
10
fifth harmonic of y^ = 1 amp. in phase with their respective e.m.fs. The
equation of the current is
ii = 10\/2 sin + \/2 sin 50.
The wave shape is the same as that of the impressed e.m.f. Resistance,
therefore, does not affect the wave shape. 10 ~
The current 7 2 in (2) consists of a fundamental of -^ r- = 10 amp. and a
10
fifth harmonic of ^ = 0.2 amp. in quadrature behind the e.m.fs. pro-
ducing them. The equation of the current is
it = 10\/2 sin (8 - 90) + 0.2 X \/2 sin (56 - 90)
The fifth harmonic is not nearly so prominent as in the resistance circuit.
Inductive reactance thus tends to eliminate the harmonics in irregular waves
and makes them more sinusoidal in form. in ~
The current 7 3 in (3) consists of a fundamental of ^TTT = 10 amp. and a
10
fifth harmonic -^ = 5 amp. in quadrature ahead of the e.m.fs. producing
them. The equation of the current is
i z = lO-s/2 sin (6 + 90) + 5\/2 sin (50 -f 90).
The fifth harmonic is much more prominent in the capacity circuit than in
either of the others. Capacity, therefore, tends to exaggerate the har-
monics in a peaked wave.
The main current I consists of a fundamental and a fifth harmonic. The
fundamental is the resultant of the fundamental currents in the three
branches and from Fig. 120 (d) is found to be 10 amp. in phase with the
fundamental e.m.f. The fifth harmonic is the resultant of the fifth harmonic
currents in the three branches and from Fig. 120 (e) it is found to be 4.8 amp.,
leading the fifth harmonic e.m.f. by approximately a quarter of a cycle. The
equation of the resultant current is
i = 10\/2 sin 6 + 4.8 X \/2 sin (50 -f- 90).
This current is shown in Fig. 120 (c).
20. Analysis of Alternating Waves. The majority of alter-
nating-current problems are solved on the assumption that the
waves of e.m.f. and current are sine waves or such that they
can be represented by sine waves with sufficient accuracy. In
certain cases, however, it is necessary to know the values of the
more important harmonics in order to understand the phenomena
completely.
The most general expression for a univalent periodic function
may be given in the form of an infinite trigonometric series or
Fourier series,
y = do + i cos -f- 02 cos 26 -\ \- a n cos nB -f- 61
sin B + 6 2 sin 26 -f + b n sin nO (224)
COMPLEX ALTERNATING-CURRENT WAVES 147
where ao, a\, a 2 , 61, 62, etc., are constants; or combining the sine
and cosine terms it may be expressed as
y = a -f Ci sin (0 + ) (225)
where
, 2 - = On t = 6n
tan n = ^- (226)
If the ordinates of such a function for one complete cycle of
the fundamental, are given in form of a curve as obtained from
an oscillograph record or tabulated as on page 149, the constants
a n and b n for any harmonic can be found as explained below.
To find the coefficient ao integrate equation (224), from to 2ir.
r/2,r
ydd = I (ao + a\ cos + a 2 cos 20 -f + + a n cos
Jo
nd + &i sin + & 2 sin 20 + + + b n sin n0)d0 = a [0]o ir = 27ra ,
since all the other terms vanish between these limits.
Thus a = o~ = average value of the ordinate between and 2ir.
The coefficient a n is found by multiplying all the terms in equa-
tion (224) by cos n8 and integrating from to 2ir. This gives
A n = I y cos nddd = I (ao cos nQ + ai cos nd- cos H | h
Jo Jo
a n cos 2 n0 -f &i cos n& sin H h + b n cos nd sin n0)d0
I a cos nd + -jjcbft'Oi - 1)0 + cos (n + 1)0} + -h +
/>
^(1 + cos 2n0) + Trfsin (n + 1)0 - sin (n - 1)0} + + +
I
and oM\ o / a\ 27r
an = \2^/ = avg> ^ cos n ^o
b n is found by multiplying equation (224) by sin n0 and integrat-
ing as before.
2* 2*
B n = I y sin n0d0 = I (a sin nB + 0,1 sin n0 cos + + +
Jo Jo
a n sin n0 cos nd + 61 sin n0-'sin + + + b n sin 2 n6)dO
148 ELECTRICAL ENGINEERING
= I [a sin nS + ^{sin (n -f 1)0 + sin (n - 1)0} + + +
^(sin2n0) + ~ {cos (n - 1)0 - cos (n + 1)0} + + +
^(1 cos 2n0)Jd0
and
6n = 2 \2W = 2 avg * ^ sin n ^o*
Applying these results,
a = avg. (y)l'
ai = 2 avg. (y cos 0)5 T 61 = 2 avg. (y sin 0)^
a 2 = 2 avg. (y cos 26)1' 6 2 = 2 avg. (y sin 20)o x
a n = 2 avg. (y cos n0)o 7r 6 n = 2 avg. (# sin 710)^
Practically all of the waves met with in electrical engineering
are symmetrical waves having the two half waves of the same
shape but opposite in sign. The values of the ordinates from
180 to 360 degrees are the same as those from to 180 degrees
but their signs are different.
In symmetrical waves only those harmonics can exist which
reverse in sign when is increased by 180 degrees. Therefore,
all the even harmonics are absent and there is no constant term.
The general expression for a symmetrical alternating wave
may be put in the form
y = a\ cos + ^3 cos 30 + a 5 cos 50 + + +
61 sin + 6 3 sin 30 ++ (227)
or combining the sine and cosine terms
y = ci sin (0 + i) + c 3 sin (30 + #3) + c 5 sin
(30 + B ) (228)
where the values ci, c 3 , i, $3, etc., are found as indicated in
equation 226.
In order to analyze such a wave into its component harmonics
it is necessary to know the values of the ordinates for one-half
wave only and the constants are determined from the following
equations,
ai = 2 avg. (y cos 6)' bi = 2 avg. (y sin 6)'
a 3 = 2 avg. (y cos 30)^ 6 3 = 2 avg. (y sin 30)*
a 5 = 2 avg (y cos 50)* 65 = 2 avg. (y sin 50)*
COMPLEX ALTERNATING-CURRENT WAVES 149
121. Example of Analysis. One-half of an alternating-current
wave is shown as curve 1, Fig. 121, and it may be represented by
the expression,
y = ai cos + a s cos 30 + a 5 cos 50 + + +
61 sin 6 + 6 3 sin 30 + 6 8 sin 50 + +
or by
y = Ci sin (0 -f
*
fl
3
CO
CO
1C
5
S
a
8
53
a
8
.2
a
o
c
%
A
a.
6
a
Si
s>
5
a
a
Q
0.0
0.0
,000
0.000
0.00
0.00
1.000
0.000
0.00
0.00 1.000
O.OCO! 0.00
0.00
10
5.0
25.0
0.985
0.174
4.85 0.871 0.866
0.500
4.33
2.50 0.643
0.766
3.22
3.84
20
8.2
67.2
0.940
0.342
7.70 2.80 0.500
0.866
4.10
7.10-0.174
0.985
- 1.43
8.07
30
40
9.7
11.0
94.0
121.0
0.8660.500
0.7660.643
8.40
8.42
4.85J 0.000
7.07i-0.500
1.000
0.866
0.00
- 5.50
9.70-0.866 0.500
9.53-0.940-0.342
- 8.40
-10.34
4.85
- 3.76
50
12.0
144.0
0.6430.766
7.71 9.19-0.866
0.500
-10.40
6.00-0.342'-0.940
- 4.10-11.28
60
13.7
187.7
0.5000.866
6.85 11.86-1.000
0.000
-13.70
O.OOi 0.500 ! -0.866
6.85-11.85
70
15.1
228.0
0.3420.940
5.16
14.15-0.866
-0.500
-13.08
- 7.55 0.985-0.174
14.87
- 2.62
80
17.7
313.3
0.1740.985
3.70
17.40-0.500
-0.866
- 8.85
-15.32 0.766
0.643
13.56
11.37
90
25.7
660.5
0.000
1.000
0.00
25.70
0.000
-1.000
0.00
-25.70 0.000
1.000 0.00
25.70
100
30.5
930.2
-0.174'0.985
- 5.28
30.00
0.500
-0.866
15.25
-26.40-0.766
0.643-23.40
19.60
110
38.5
1482.2
-0.3420.940- 13.30
36.00
0.866
-O f 500
33.35
-19.25-0.985
-0.174-37.90
- 6.70
120
44.0
1936.0
-0.5000.866- 22.00
38.10
1.000
0.000
44.00
0.00-0.500-0.866
-22.00-38.20
130
46.0
2116.0
-0.6430.766- 28.60
35.30
0.866
0.500
39.85
23.00 0.342; -0.940
15.70-43.20
140
40.8
1664.6
-0.7660.643- 31.30
26.20
0.500
0.866
20.40
35.30 0.940-0.342
38.35
-13.83
150
30.0
900.0
-0.866
0.500- 26.00
15.00
0.000
1.000
0.00
30.00 0.866
0.500
25.98
15.00
165
19.0
361.0
-0.9400.342- 17.87
6.50
-0.500
0.866
- 9.50
16.46 0.174
0.985
3.30
18.70
170
10.0
100.0
-0.985
0.174- 9.85
1.74
-0.866
0.500
- 8.66
5.00-0.643
0.766
- 6.43
7.66
Total 376. 9 11330.7
-101.41282.73
91.59
50.37
1
7.83
-16.65
ai = 2 avg. (y cos 0)'
03 = 2 avg. (y cos 30 J i a& = 2 avg. (y cos 50 j *
/
101.41v
11.27
-= 2
/91.5
) - 10.17
= 2
/7.83>
\ 18 ;
- 0.87
" "V 18 ) ~
V 18
61 = 2 avg. (y sin J
6s = 2 avg. (y sin 30)'
65 - 2 avg. (y sin 50)'
_ /282.73s
= 31.41
= 2
/50.3S
) = 5.596
= 2
/-16.65 N
-1.85
"A 18 /
V 18
^ 18
)
ci = V a\
2 + 612 = 33.3
cs - Vaz 2 + W = 11.6
C5 = >
/as 2 + 6s 2 - 2.05
sin 0i
ai - 11.27
1.338
sin03=^ =
10.17
0.877
sin 05
as
0.87
= 0.434
ci 33.3
11.6
es
2.05
cos = b
31.41
CU4
63 5.596
65
-1.85
a 33.3 ~
S ca 11.6
COS 05
C5
2.05
= - 0.902
tan0!--
i -11.27
Q
.359
as
10.17
= 1.82
tan 06= r- S =
06
0.87
i 31.41
5.596
-1.85
--0.47
0! - -
- 19 45'
03 - 61 10'
08 - 154 49'
150
ELECTRICAL ENGINEERING
The fundamental is yi = Ci sin (0 ~f 0i) = 33.3 sin (0 19 45'),
Curve 2, Fig. 121; the third harmonic is 3/3 = Cs sin (30 +#3)
= 11.6 sin (30 + 61 10'), Curve 3, Fig. 121; the fifth harmonic
is 2/5 = c 5 sin (50 + 6 ) = 2.05 sin (50 + 154 49'), Curve 4, Fig.
25 50 75 100 125 150 175 200
FIG. 121.
121; and the complete expression for the alternating-current
wave is y = 33.3 sin (0 - 19 45') + 11.6 sin (30 + 61 10')
+ 2.05 sin (50 + 154 49') + higher harmonics which are
negligible.
no
100
40 50
FlG. 122.
The maximum ordinate taken from the plotted curve is
2/ma*. = 46;
The average ordinate is found from column (1)
376.9
2/avg.
18
= 20.9;
COMPLEX ALTERNATING-CURRENT WAVES 151
the effective value is obtained from column (3)
_ 11,330.7
2/eff. -
and the form factor is
The effective value may also be found by plotting the values of
y on polar coordinate paper as in Fig. 122 and measuring the area
of the curve A p by means of a planimeter.
The area of a polar curve is
A p =
and therefore the effective value is
o
The wave analyzed above was taken from an oscillograph
record of the exciting current of a transformer.
CHAPTER VI
POLYPHASE ALTERNATING-CURRENT CIRCUITS
122. Polyphase Alternating-current Circuits. In Art. 102
the generation of an alternating e.m.f. was discussed. Fig. 123
represents a simple single-phase alternator. The field poles are
excited by direct current and an alternating e.m.f. is produced in
the winding which is carried on the armature. It does not make
Sl Armature Winding f l
FIG. 123. Single-phase alternator, revolving field type.
any difference whether the poles are fixed and the armature wind-
ing moves or the winding is fixed and the poles move.
The winding starts at Si and ends at f\ and the equation of the
e.m.f. between s\ and /i is
-pi
ei = E m sin 6, effective value = % = E
A BC D
FIG. 124. Two-phase winding.
The positive direction of e\ is from Si to /i. It is here assumed
that the flux in the air gap is so distributed that the e.m.f. is a
sine wave.
If a second winding is placed on the same armature, Fig. 124,
but displaced 90 electrical degrees from the first, it will have an
152
POLYPHASE ALTERNATING-CURRENT CIRCUITS 153
e.m.f. generated in it of the same value and wave form as the
first but displaced from it in phase by 90 degrees (Figs. 125 and
126).
Phase 2 starts at s 2 and ends at/2 and the e.m.f. generated in it
is
tm
02 = E m sin (6 90), effective value
The machine is called a two-phase alternator.
\/2
E =
ElectricalNDegrees
FIG. 125. E.m.f. waves of a two-
phase alternator.
FIG. 126.
The two windings are usually entirely separate and their ends
are brought out to four terminals ABCD.
If any two terminals as B and C are joined, the e.m.f. between
A and D is
e A D = ei e 2 = E m sin E m sin (6 90)
= E m (sin + cos 0) = V2#m sin(0 + 45);
it leads e\ in phase by 45 degrees, its maximum value is \2-6Jm and
its effective value is E AD =
E
V2E. This value
A B C D F
can also be obtained by subtract-
ing the two vectors as shown in
Fig. 126.
The middle points mi and ra 2
of the two windings are some-
times connected together and a
fifth terminal F used as shown
in Fig. 127. The common term-
inal F is called the neutral point
of the winding and may be connected to earth. The e.m.f. from
each of the other terminals to F is the same,
_ E
/I
mi,m% f
(((OTTOCOTIT) 2
0(footfo(ftr v -
8 l
g
FIG. 127.
154
ELECTRICAL ENGINEERING
The four e.m.fs. E AC , E C B, E B D and E DA are equal, since each is
7?
the vector difference of two e.m.fs. of effective value -~ at right
angles to one another, and these four e.m.fs. are also at right
angles to one another and form a four-phase or quarter-phase
system.
The effective value of each of the four e.m.fs. is
E
123. Three-phase Circuits. If three similar windings are
placed on the same alternator armature displaced 120 electrical
degrees from one another, Fig. 128, and the ends of the windings
FIG. 128. Three-phase winding.
are brought out to terminals, the machine is a three-phase alter-
nator. The e.m.fs. generated in the three windings are dis-
placed 120 degrees.
The e.m.f. in phase 1 is e\ = E m sin 0, effective value E = ~~^=;
the e.m.f. in phase. 2 is e 2 = E m sin (0 120), effective value
E
E = p; and the e.m.f. in phase 3 is e 3 = E m sin (0 240),
effective value E =
ABC
FIG. 129. Delta connection.
The windings may be interconnected in two ways. (1) Join
/i to $2, /2 to s s and / 3 to Si, and connect the three junctions to
the terminals A, B and C (Fig. 129) This is called the "delta"
connection or ring connection and is represented by A.
POLYPHASE ALTERNATING-CURRENT CIRCUITS 155
The resultant e.m.f. around the closed circuit at any instant is
+ e 2 + .
The current in phase 1 is
i\ = Im sin (B ), effective value 7i =
the current in phase 2 is
iz = I m sin (B $ 120), effective value 1 2 =
the current in phase 3 is
? 3 = I m sin (B < 240), effective value J 3 = ='
POLYPHASE ALTERNATING-CURRENT CIRCUITS 157
The effective value of the currents in the three phases is the same
and is 7 =
The current in line A is
i A = u - * 2 = 7_sin (0 - ) - I m sin (0 - 4> - 120)
= V37 m sin (0 - + 30), (233)
and its effective value is
IA = V3 ^%= \/3 7.
The current in line B is
i B = i 2 - i s = I m sin (0 - 4> - 120) - I m sin (B - - 240)
= A/37 sin (0-0-90), (234)
and its effective value is
I B = V3I-
The current in line C is
ic = is ii = Im sin (6 - < - 240) - I m sin (0 - <)
= V3Jm sin ((9 - - 210), (235)
and its effective value is
Ic = V3/.
Thus the currents in the three lines are equal in magnitude and
are 120 degrees out of phase with one another. If the effective
value of the current in each of the lines is represented by Ii then
Ii = I A = IB = I c = V3I.
The effective values of all these quantities are shown in the vector
diagram in Fig. 132.
158
ELECTRICAL ENGINEERING
The power supplied by the alternator is
P = Eili cos < + E z lz cos + E 3 Is cos <
= 3EI cos $ (236)
= \/ZE t Ii cos (237)
and is equal to \/3 times the product of the terminal e.m.f., the
line current and the power factor.
If the system is not balanced, that is, if either the currents or
the power factors in the three phases differ from one another the
line currents will not be equal and they will not be displaced in
phase by 120 degrees.
If the current in phase 1 is
i\ = I mi sin (0 fa)
and the current in phase 2 is
*2 = / 2 sin (0 - 02 - 120)
the current in line A is
= I mi sin (0 - fa) - I m2 sin (0 - fa - 120). (238)
Fig. 133 shows a star- or F-connected three-phase system
FIG. 133.
Using the same notation as before and taking the results ob-
tained in Art. 123,
the e.m.f. between lines A and B is
e A B = ei-e 2 = \/liE m sin (0+30), effective value E AB =
the e.m.f. between lines B and C is
e B c = e 2 -e3 = \/3E m sin (0-90), effective value E BC =
the e.m.f. between lines C and A is
Sin (0-210), effective value E CA =
POLYPHASE ALTERNATING-CURRENT CIRCUITS 159
the current in line A is
i A = i\ = I-m sin (B 0), effective value I A = 7= = //;
the current in line B is
Im
IB = ii = Im sin (B 120), effective value I B = ~~7^ = Ii',
the current in line C is
ic = is = Im sin (6 240), effective value I c = ~^
Fig. 134 is a vector diagram for this circuit.
E A8
FIG. 134.
125. Measurement of Power in Polyphase Circuits. The
power in two-phase circuits (Fig. 135) can be measured by-
connecting wattmeters in the two phases and taking the sum of
Phase 1
Phase 1
Load
FIG. 135. Measurement of power in two-phase circuits.
the readings. If the two phases are similarly loaded a single
meter will suffice if its reading is doubled.
In the three-phase circuits (Fig. 136) the total power is given
by the sum of the two wattmeter readings, if they are connected,
160
ELECTRICAL ENGINEERING -
as shown at a, with their current coils in two of the lines and their
voltage coils connected across from these lines to the third line.
Terminal Voltage
Line Current
FIG. 136. Power in balanced three-phase circuits.
Consider first the case where the loads on the three phases
are balanced. Referring to the vector diagram, Fig. 137, for
the Y connected, circuit,
Wi = E AB I A cos (30 + 0) - \/3EI cos (30 + 0).
W 2 = E CB I C cos (30 - 0) = \/3EI cos (30 - <#>).
FIG. 137.
Adding,
Wi + TF 2 =
{cos (30 + 0) + cos (30 - 0)}
{(cos 30 cos sin 30 sin <) + (cos 30
cos + sin 30 sin $) }
= \fSEI (2 cos 30 cos 0) =3#/cos< = \/3#Jj cos <
and this is the total power in the circuit.
The power factor of the circuit can also be found from the two
wattmeter readings.
Subtracting,
W 2 - Wi = \/3EI (2 sin 30 sin 0) = \/3EI sin 0.
and
W 2 -
sn
_
: : tan
POLYPHASE ALTERNATING-CURRENT CIRCUITS 161
Gl-
and
tan = V3
W 2 -
COS =
(239)
(240)
When the power factor is 50 per cent., $ = 60 degrees and
the reading of wattmeter Wi becomes zero,
Wi = E AB I A cos (30 + 60) = 0.
For lower power factors the reading of W\ would be negative
and the total power in the circuit is given by the difference of
the readings.
In some cases it may be difficult to tell whether the readings
should be added or subtracted. The curve in Fig. 138 makes it
possible to determine the sign of Wi. It shows the relation
between power factor and the ratio of the readings of the two
meters, cos cj> vs.
w.
-1.0
-0.5 ...
Ratio n *
W 2
FIG. 138. Watt-ratio curve.
0.5
1.
The equation of the curve may be obtained from equation (240) .
Cos
1
V (W 2 -I- Wtf + 3(TF 2 - W i) 2
n
162
ELECTRICAL ENGINEERING
Above 50 per cent, power factor W\ is positive and v^- is positive.
Wi
At 50 per cent, power factor Wi and wr become zero. Below
W i
50 per cent, power factor W\ and r- are negative. If in any
W 2
case it is doubtful whether W\ is positive or negative, try it
both ways and calculate the corresponding value of cos $ from
equation (241) and then referring to Fig. 138 see which assump-
tion agrees with the watt-ratio curve.
If the current and voltage readings are also given the deter-
mination of cos is simplified.
True power Wi W\
Power factor = cos * = Apparent - power -
The following readings taken on a three-phase induction motor
running light show the value of the watt ratio curve.
E
I
Wi
W 2
W\
WI
Cos =
W z Wi
Cos from
Fig. 138
Correct
value
of TFi
VZEI
20
1.01
+4
19
+0.21
0.656 0.66
+4
-4
-0.21
0.428
0.35
40
1.24
+5
40
+0.125
0.528
0.60
o
-0.125
0.410
0.41
-5
In the first case the wattmeter readings must be added while in
the second they must be subtracted.
It is possible to determine the total power in a balanced
three-phase circuit with one wattmeter without opening the
circuit to change the current coil from one line to the other.
This may be done in two ways. (1) Referring to Fig. 136(6),
first connect the current coil in line A with the potential coil
between A and B, the reading is
= E AB I A cos (30 +
V3EI cos (30 + 0);
then with the current coil still in line A connect the potential
coil from B to C; the reading is
W
E AC I A cos (30 -
and the power is the sum of the tw readings.
POLYPHASE ALTERNATING-CURRENT CIRCUITS 163
2. The second method requires an ammeter and voltmeter
in addition to the wattmeter. Connect the wattmeter with its
current coil in line A and its potential coil from B to C; the
reading is
W = E BC I A cos (90 - 0), Figs. 136(c) and 137,
= EJi sin .
Sin
W
is therefore known and cos < can be calculated.
The power in the circuit is ^3E t Ii cos .
This latter method does away with the difficulty encountered
in the case of low power factors.
The two wattmeters connected as in Figs. 136 (a) and 139,
give the total power in the circuit, whether the loads on the
B i,
FIG. 139.
phases are balanced or unbalanced. Take the case of the delta-
connected circuit, Fig. 139, and consider the instantaneous values
of voltages and currents and power.
e 2A =
e 22 - e 2 i
and
but
w z = e 2 i 2 + e 3 ^3
+ e z + 6 3 = 0, and
(e z + e 3 ),
therefore w\ + ^2 = v\i>\ + e 2 i 2 + e s i s = the sum of the in-
stantaneous values of power in the three phases.
When measuring power in a three-phase, four-wire system, the
two-wattmeter method will give accurate results only when the
loads are balanced. For unbalanced loads three wattmeters
must be used connected in the three lines and with their potential
coils connected between the lines and neutral.
164 ELECTRICAL ENGINEERING
126. Example. A 220-volt, three-phase alternator supplies the following
loads :
1. Twenty-five kilowatts at 100 per cent, power factor for lighting.
2. A 100-hp. (output) induction motor of 92 per cent, efficiency and 90
per cent, power factor.
3. A number of small induction motors with a combined output of 50 hp.,
average efficiency = 80 per cent, and average power factor = 75 per cent.
If the loads on the three phases are balanced, find:
(a) The output of the alternator in kilowatts,
(fe) The line current,
(c) The power factor of the combined load.
(a) The output of the alternator = 25.0 + Q^ x i QQQ + Q~80 X FOOO
= 25.0 + 81.2 + 46.7 = 152.9 kw.
25 X 1 000
(6) The in-phase current per line for (1) = = ' = 66 amp.
\/o X 2i2i(j
The quadrature current per line for (1) =0
SI 2 X 1 000
The in-phase current per line for (2) = ' - = 213 amp.
\/o X 2\j
213
The total current per line for (2) = -pr^r = 237 amp.
u.y
The quadrature current per line for (2) = V(237) 2 - (213) 2 = 103 amp.
46 7X1 000
The in-phase current per line for (3) = ^= j = 123 amp.
*\/3 X 220
123
The total current per line for (3) = Q-^. = 164 amp.
The quadrature current per line for (3) = V(164) 2 - (123) 2 - 108 amp.
The total in-phase current per line = 66 + 213 + 123 = 402 amp.
The total quadrature current per line = + 103 + 108 =211 amp.
The current per line = V(402) 2 + (211) 2 = 455 amp.
402
(c) The power factor of the load is X 100 = 88.3 per cent.
CHAPTER VII
DIRECT-CURRENT MACHINERY
127. The Direct-current Dynamo. A direct-current dynamo
consists of an electric circuit, connected to a commutator and
tapped by brushes, revolving in a magnetic field which is produced
by stationary electric circuits.
Such a machine is illustrated in Fig. 140 and comprises the
following parts:
1. Yoke 1
2. Pole pieces > Magnetic circuit.
3. Armature core J
4. Armature winding ) ^. , . , , . .,
_ ~ > Revolving electric circuit.
5. Commutator J
6. Brushes and brush holders. Collecting apparatus.
7. Field winding. Stationary exciting circuit.
128. Yoke. The yoke serves mechanically as the frame of the
machine and magnetically to carry the flux from pole to pole; in
small machines it is usually made of cast iron but in machines
where great weight is objectionable it is made of cast steel which
has greater strength and permeability.
129. Pole Pieces. The pole pieces or pole cores are usually
made of cast steel or sheet steel and are bolted to the yoke. For
small machines the yoke and poles are sometimes cast in one
piece. All solid poles must have laminated pole faces bolted to
them in order to reduce the eddy-current loss due to local varia-
tions of the magnetic density in the pole faces as the armature
teeth move across them.
The pole cores carry the field windings of the machine. Solid
poles are made circular and so have the greatest section for a
given perimeter and require the smallest length of field copper.
Laminated poles must, however, be made rectangular.
The section of the pole face is made much greater than that of
the pole core in order to reduce the flux density in the air gap.
165
166
ELECTRICAL ENGINEERING
DIRECT-CURRENT MACHINERY
167
130. Armature Core. The armature core carries the rotating
electric circuit in slots punched out on its periphery. It is built
up of sheets of steel about 0.014 in. in thickness. Alternate
sheets are coated with an insulating varnish to increase the re-
sistance in the path of the induced eddy currents. Open spaces
are left in the core, called vent ducts (V.V., Fig. 140), which
allow air to circulate through the armature and carry off the heat
generated due to the iron and copper losses. The number of
vent ducts required depends on the length of the armature.
The armature punchings are carried on a spider S and are kept
in place by heavy end plates which have projections on their outer
edges to support the end connections of the armature coils.
131. Armature Winding. The armature winding is the seat
of the generated e.m.f. It must be tapped at certain points by
brushes, in order that the machine may supply power to an
external receiver circuit. The winding consists of a number of
coils of one or more turns, connected together to form a con-
tinuous winding; leads are run from their junctions to the com-
mutator bars from which the current is collected by the brushes.
The coils forming the winding must be so connected together
that the e.m.fs. generated in coils between brushes of opposite
polarity will all act in the same direction.
FIG. 141. Bipolar ring winding.
The earliest type of armature winding was the ring winding,
Figs. 141 and 142, but this has been replaced by the various forms
of drum windings, a few of which are illustrated in Figs. 144 to 151.
132. Ring Windings. In the bipolar ring winding, Fig. 141,
all the conductors on each half of the armature are connected in
series between the brushes. When the brushes are placed on the
neutral line, that is, in such a position that the coil being corn-
mutated is not generating any e.m.f., the e.m.fs. generated in all
conductors under one pole will act in the same direction and will
combine to give the terminal e.m.f. of the generator. The e.m.fs.
generated under the other pole will be equal in magnitude but
168
ELECTRICAL ENGINEERING
will act in the opposite direction. Thus, there is no e.m.f . tending
to cause current to circulate through the winding at no load and
there are two paths in multiple for the current flowing through
the armature.
The connection. from one conductor to the next is run through
inside the armature, where it cannot cut magnetic flux, and con-
sequently one-half of the winding is not effective in generating
e.m.f. This extra wire increases the resistance of the armature
and adds to the weight and cost of the machine. The ring wind-
ing has the further disadvantage, that it is very difficult to replace
injured coils. On the other hand, the voltage between adjacent
coils is so low that very little insulation is required between them.
This type of winding is now obsolete.
FIG. 142. Six-pole ring winding.
Fig. 142 shows a six-pole ring winding with 36 coils connected
to a commutator with 36 bars. This winding must be tapped at
six equidistant points by brushes; there are six paths in parallel
through the armature from positive to negative terminals and the
voltage of the machine is that generated in one-sixth of the
winding.
FIG. 143.-
O)
-Armature coils.
133. Drum Winding. In drum-wound machines the whole of
the armature winding is carried in slots on the outside of the
armature core and both sides of any coil are effective in generat-
DIRECT-CURRENT MACHINERY
169
ing e.m.f. The single coils are of the shapes shown in Fig. 143
and may consist of one or more turns.
The conductors forming the two sides of a coil must be situated
in fields of opposite polarity in order that the e.m.fs. generated
Generator
I -^ ^^ 1
U-^-J
Equalizer Rings
I , ,., t .- 1
N
S
N
S
N
S
FIG. 144 AND 145. Six-pole multiple-drum winding.
in them may act in the same direction. One side of a coil is
placed in the top of a slot and the other side in the bottom of a
slot in a similar position under the next pole.
170
ELECTRICAL ENGINEERING
According to the way in which the end connections are brought
out to the commutator bars and the coils are connected together,
drum windings are divided into two classes, multiple or lap wind-
Generator
-
LJ-
FIG. 146. Six-pole multiple-drum winding. Fractional pitch.
ings, as illustrated by coils a and b in Fig, 143 and the windings in
Figs. 144 to 146 and series or two-circuit windings, as illustrated
by the coils c and d in Fig. 143 and the windings in Figs. 149 and
150.
DIRECT-CURRENT MACHINERY 171
134. Multiple -drum Windings. In the multiple winding the
two terminals of a coil are connected to adjacent commutator
bars. Fig. 144 represents a multiple winding for a six-pole
machine with 72 conductors and 36 slots. The sides of a coil are
placed in slots 1 and 7 and the terminals are connected to bars
1 and 2. The same winding is shown in Fig. 145 and the direc-
tions of the currents are shown by arrowheads. The brushes are
placed on the no-load neutral points and therefore directly under
the centers of the poles and as many sets of brushes are required
as there are poles.
Tracing through the winding from a positive to a negative
brush only one-sixth of the conductors are taken and there are
therefore six paths in multiple through the armature winding
from the positive to the negative terminal of the machine and
each conductor carries only one-sixth of the current passing
through the armature. The number of paths is equal to the
number of poles as in the ring winding.
Generally in multiple windings the coil pitch is almost equal to
the pole pitch and the windings are called full-pitch windings.
When the coil pitch is less by one or more teeth than the pole pitch
the winding is called a fractional pitch or short-chord winding.
Fig. 146 shows a fractional pitch, multiple-drum winding. Frac-
tional-pitch windings have shorter end connections than full-pitch
windings and have a smaller inductive e.m.f. generated during
commutation since the two coils in one slot are not commutated
at the same time and thus the inductive flux is that due to half
the ampere-turns acting in the case of a full-pitch winding.
135. Equalizer Rings. In multiple-wound machines, if there
is any irregularity in spacing the brushes or if the air gaps under
all the poles are not of the same depth, the e.m.fs. generated in
the different sections of the winding will not be equal and the un-
balanced e.m.f. will tend to cause current to circulate through the
brushes even when the machine is not carrying any load. To
keep these circulating currents out of the brushes similar points
under the different pairs of poles which should normally be at the
same potential are joined together by heavy copper connections
called equalizer rings and these provide a path of very low re-
sistance which the circulating currents follow in preference to the
comparatively high-resistance path through the brushes (Figs.
144 and 145).
Equalizer rings should be provided on all multiple-wound
172
ELECTRICAL ENGINEERING
machines except the smallest sizes. They are connected to every
third or fourth coil. Windings to be used with equalizer rings
must have the number of coils a multiple of the number of pairs
of poles.
Fig. 147 shows the circulating currents in a four-pole machine,
in which the gaps under the two lower poles have been shortened,
due to wear of the bearings. The voltages generated in the two
lower sections of the winding are assumed to be 10 per cent,
higher than those in the upper sections. Fig. 148 shows three
equalizer rings connected to this machine to keep the circulating
currents out of the brushes.
FIG. 148.
136. Series-drum Windings. In the series winding the ter-
minals of a coil are connected to two commutator bars approxi-
mately twice the pole pitch apart. Fig. 149 represents a series or
two-circuit winding for a six-pole machine with 44 conductors
and 22 slots. One side of a coil is placed in the top of slot 1 and
the other side in the bottom of slot 5 and the terminals of the
coil are connected to commutator bars 1 and 8.
Tracing out the winding from the positive brush BI to the nega-
tive brush B z one-half of the armature conductors are taken in.
There are therefore but two circuits in multiple between terminals
independent of the number of poles and the winding is called a
two-circuit or series winding.
Only two sets of brushes are required to collect the current but
when the current is large it is usual to employ other sets of
brushes as shown at B z , # 4 , B$ and J5 6 and as many sets of brushes
as there are poles may be used.
Series windings are used in small high-voltage machines or
where it is desirable to use only two sets of brushes, as in small
railway motors; but in large multipolar machines with many sets
DIRECT-CURRENT MACHINERY
173
of brushes the current does not divide equally between brushes
of the same polarity and commutation is unsatisfactory.
The number of coils in a series winding must be one more or one
less than a multiple of the number of pairs of poles, or
N = C\ 1,
LJT_ g- g+ y- g+ y-
f r
N
s
N
S
N
S
FIG. 149. Six-pole series-drum winding, retrogressive.
where
N = number of armature coils,
p/2 = numbers of pairs of poles,
and C = a constant whole number = avg. winding pitch.
Each coil may have any number of turns.
174
If
ELECTRICAL ENGINEERING
AT=C|-1, (241)
the winding starting from bar 1 goes once around the armature
and is connected to bar 2. It is therefore called a progressive
winding (Fig. 150).
Generator
FIG. 150. Six-pole series-drum winding, progressive.
If w-cjj+i,
(242)
the winding starting from bar 1 and going once around the arma-
ture is connected to the bar before 1 and it is called a retrogres-
sive winding (Fig. 149).
137. Double Windings. If space is left between adjacent
coils of a multiple winding, a second winding may be placed on
DIRECT-CURRENT MACHINERY
175
the same core. The second winding may be entirely separate
from the first, that is, each of the two windings closed upon itself;
or after passing through the first winding the circuit enters the
second and after passing through the second reenters the first.
Generator
r i
FIG. 151. Sixpole multiple-drum winding, duplex, doubly reentrant.
In the first case the winding is duplex doubly reentrant and in the
second case duplex singly reentrant. Duplex multiple windings
have twice as many circuits in multiple between terminals as there
are poles. Such windings are suitable for large low-voltage ma-
176
ELECTRICAL ENGINEERING
chines used in electrolytic work. The brushes must be wide
enough to collect current from both sections of the winding at the
same time. Fig. 151 shows a duplex doubly reentrant winding
for a six-pole machine with 72 conductors and 36 slots.
Similarly the series winding may be made double by placing a
second winding in alternate slots and connecting it to alternate
commutator bars. The second winding is in multiple with the
first and there are four paths in multiple between terminals.
138. Commutator. The commutator is one of the most im-
portant parts of a direct-current machine. It consists of a num-
ber of bars of hard-drawn copper, insulated from one another by
thin sheets of mica or other insulating material, and built up into
the form of a cylinder (Fig. 152). The bars are held together by
Micanite
FIG. 152. Commutator.
a cast-iron spider from which they are insulated by micanite " V"
rings. The terminals of the coils forming the armature winding
are connected to the bars either directly by soldering them into
slots in the bars or by means of vertical connectors called commu-
tator risers. In order that the brushes, which collect the current
from the commutator, may run smoothly without vibrating or
chattering the commutator surface must be perfectly round and
smooth.
The function of the commutator is illustrated in Fig. 153. The
current from the machine is I amp. and the current in each
conductor is I c = ^ am P- During the time taken for the brush
to move across the insulation between bars 1 and 2 the current
DIRECT-CURRENT MACHINERY
177
in coil c must change from I c in one direction to I c in the opposite
direction. This reversal of the current is called commutation and
to be satisfactory it must be effected without sparking. The
brushes are shown placed on the neutral line and the coils short-
circuited are not cutting any flux and therefore have no e.m.f.
generated in them due to rotation. If the short-circuited coil had
no inductance the current would reverse completely due to the
contact resistance between the brush and the commutator.
deb
i i i 3 i 2
U
2l c
Q
FIG. 153. Commutation.
In (1) the current in coil c is 7 C ; in (2) it is zero since current I c
from bar 1 goes through one-half of the brush-contact area and
current I c from bar 2 goes through the other half, and the drop of
voltage on both sides is the same and therefore there is no voltage
available to drive the current through the resistance of the coil.
Between (1) and (2) the resistance from bar 2 to the brush is
greater than the resistance from bar 1 to the brush and so part of
the current from 2 flows through the coil c. Between (2) and (3)
the resistance from bar 2 to the brush is less than from bar 1 and
part of the current from 1 flows through the coil c. In (3) the
current in c is I c but in the opposite direction from that in (1)
and commutation is complete.
The self -inductance L of the coil opposes any change of current
by generating a back e.m.f. L -r volts; when the current is large
and the time of commutation short this back e.m.f. is large and
the current will not be reversed when the brush breaks contact
with bar 1 and sparking will occur.
To counteract the effect of self-inductance the brushes in a gen-
erator are moved ahead of the neutral in the direction of rotation
and back in a motor. The short-circuited coil is then in a field
which generates in it an e.m.f. due to rotation which opposes the
back e.m.f. of self-inductance, or, as it is usually called, "the re-
12
178
ELECTRICAL ENGINEERING
actance voltage of the coil," and assists commutation. The prob-
lem of commutation is discussed fully in Art. 192.
139. Brushes and Brush Holders. The brushes collect the
current from the moving commutator and from them it passes to
the receiver circuit.
Brushes were at first made of copper because it had a low re-
sistance and large current-carrying capacity but commutation of
large currents was not satisfactory. Carbon brushes were then
introduced and commutation was greatly improved due to the
action of the high-resistance contact film between the brush and
commutator. A much better contact surface was also obtained
and the wear on the commutator was reduced. But since carbon
will carry only about 30 to 50 amp. per square inch while copper
will carry 150 to 200 amp. per square inch a much larger brush
area is required and a larger commutator.
In order to maintain a good contact between the brush and
commutator a spring is used exerting on the brush a pressure of
about 1^2 "to 2 Ib. per square inch of contact area.
The brush holders are made of brass and carry part of the cur-
rent but leads are connected directly from the brushes to the main
leads of the machine to prevent any drop of voltage which might
occur due to poor contact between brushes and holders. '
Load
Source of . .
Power 1 ( I Load
*
FIG. 154. Separate excitation.
Load
Load
FIG. 155. Shunt or self-excitation.
140. Field Windings. The field winding is a stationary .elec-
tric circuit consisting of one or more coils of wire placed on
each of the field poles. They are supplied with current and pro-
vide the m.m.f. necessary to drive the magnetic flux through the
DIRECT-CURRENT MACHINERY
179
machine. The methods used in calculating the number of
ampere-turns required to produce the flux in a machine is worked
out in Art. 227.
According to the manner of exciting the fields, direct-current
machines are divided into magneto machines in which the
flux is produced by permanent magnets; separately excited ma-
chines (Fig. 154) in which the flux is produced by a winding sup-
plied with current from some source outside the machine; shunt
machines in which the field winding is connected across the arma-
ture terminals and receives a small current at the full-machine
voltage (Fig. 155) ; series machines in which the field winding is
connected in series with the armature and carries the full arma-
ture current (Fig. 156); and compound machines in which the
field has both a shunt and a series winding (Fig. 157).
FIG. 156. Series excitation.
(a) Long Shunt (6) Short Shunt
FIG. 157. Compound excitation.
According to the number of poles machines are divided into
bipolar and multipolar machines but the bipolar type is adapted
only for small sizes.
141. Direction of Rotation of Generators and Motors. Fig.
158 represents either a generator or motor. The directions of the
currents in the armature are shown by the dots and crosses and
the directions of rotation by arrows.
With currents as shown, the direction of rotation of a generator
is counter-clockwise and of a motor is clockwise.
These results are obtained by examining the fields produced by
the armature currents.
180 ELECTRICAL ENGINEERING
Take for example the conductor A. Its field combines with
the main field and produces a strong field below the conductor
and a weak field above it. There is therefore a force / acting on
the conductor tending to move it up. This is the force that must
be overcome by the engine driving the generator in order to de-
velop electric power and therefore the rotation of a generator is
against this force and is counter-clockwise.
In the case of the motor, the force / on the conductor is the me-
chanical force developed and the rotation is in the direction of this
force and is clockwise.
To reverse the direction of rotation of a motor it is necessary to
reverse either the amature current or the field current but not
both.
Motor
FIG. 158. Direction of rotation.
142. Generation of Electromotive Force. The e.m.f . generated
in the armature of a direct-current generator or motor is
8 = Zn3> 2- 10- 8 volts,
Pi
where Z = number of conductors on the armature,
n speed of armature in r.p.s.,
$ = flux crossing the air gap from one pole,
p = number of poles,
and pi = number of paths in multiple between terminals.
In 1 sec. each conductor cuts n$p lines of force and thus the
average e.m.f. generated in each of the Z conductors is
e = n 10~ 8 volts.
Between the terminals there are conductors connected in series
Pi
and therefore the e.m.f. between terminals is
8 = Zn$> 2- 10- 8 volts. (243)
Pi
DIRECT-CURRENT MACHINERY
181
This is, the e.m.f. equation of a direct-current generator; it may
be written
(244)
where
Q
np
K = Z 10~ 8 is a constant of the machine.
Pi
The e.m.f. is therefore directly proportional to the speed and to
the flux crossing the air gap.
Fig. 159 shows the relation between
e.m.f. and speed, for a constant value
of . The locus is a straight line
passing through the origin.
In a generator the current flows in
the direction of the generated e.m.f.
but in a motor an e.m.f. is impressed
on the terminals of the armature and
drives a current against the e.m.f.
generated in the motor. The e.m.f.
100
^9 76
h
3
^50
s
O
z
/
'/
Z
1 25 50 75 100
Speed n
FIG. 159. Variation of gener-
ated e.m.f. with speed.
generated in a motor armature is therefore called a back e.m.f.
143. Effect of Moving the Brushes. The equation
8 =
- 10
Pi
Maximum Voltage
Generator
Generator
holds only if the brushes are on the no-load neutral points.
When the brushes are moved ahead of the neutral points or
behind them the e.m.f. between terminals is decreased. This
may be seen by reference to Fig. 160. With the brushes on
the neutral points the
e.m.fs. generated in all
the conductors in series
between terminals act in
the same direction and
combine to give the
maximum e.m.f. Wlien
the brushes are moved
i_ i j_v r v
ahead the e.m.f. be-
FIG. 160. Effect of moving the brushes. tween terminals is only
that generated in con-
ductors a-b or d-c since the resultant of the e.m.fs. generated
in conductors a-d and b-c is zero. Thus advancing the brushes
corresponds to a decrease in the number of armature conductors.
182 ELECTRICAL ENGINEERING
144. Building up of Electromotive Force in a Self -excited
Generator. In a self-excited generator at rest there is no flux
crossing the gap except the residual magnetism. When the
armature is rotated only a small e.m.f. is generated in it and a
very small current is produced in the field winding. If the
m.m.f . of this current is in the direction of the residual magnetism,
it will increase the flux and the e.m.f. will increase and gradually
build up to its full value.
If, however, the current opposes the residual magnetism, it will
cause it to decrease and the e.m.f. will not build up until the field
winding is reversed.
If there is no residual magnetism, the e.m.f. cannot build up
until power is supplied to the winding from some outside source to
start the flux.
145. Armature Reaction and Distribution of Magnetic Flux.
Fig. 161 shows approximately the distribution of flux in a two-
pole machine with the fields excited but without current in the
t!urve(2), B
Curve (1), My
)OOOOOOOOOOOOOOOC>QgOOOOOOOOOOOOOOOOji
-i ^^ - \r.
(a) (6)
FIQ. 161. Distribution of flux and m.m.f. at no load.
armature winding. Curve 1 shows the m.m.f. acting at each
point of the armature circumference; under the north pole it is
positive and has a constant value; under the south pole it is
negative but of the same magnitude; beyond the pole tips its
value may be represented by the straight line which passes
through zero midway between the poles. The m.m.f. is expressed
in ampere-turns and is denoted by M/.
The flux density produced by this m.m.f. is at every point di-
rectly proportional to the m.m.f. and is inversely proportional to
the reluctance of the path. It is of constant value over the pole
face if the air gap is uniform but falls off rapidly beyond the pole
tips due to the increased reluctance of the air path and to the
decrease in the m.m.f. acting. Midway between the poles it is
zero. It is represented by B and its values are plotted in curve
2. The total flux entering the armature is represented by the
DIRECT-CURRENT MACHINERY
183
area under curve 2 and this is the value of $ which appears in the
e.m.f, equation.
When, however, the armature is carrying current it exerts a
m.m.f. called armature reaction, which combines with the
m.m.f. of the field winding and changes both the distribution
and the total value of the flux entering the armature.
Curve (1),M
Curve (2), B
FIG. 162. Armature m.m.f. and flux.
Fig. 162 (a) shows the distribution of flux produced by the
armature m.m.f. acting alone, and the values of the m.m.f. of the
armature at all points around the circumference are plotted in
curve 1, Fig. 162(6). The brushes are placed on the no-load
neutral points. The armature m.m.f. M a is a maximum in line
with the brushes and falls off as a linear function to zero under
the center of the poles. The distribution of the flux produced by
the armature m.m.f. is shown in curve 2, Fig. 162(6).
Curve(l), M-M/ + M a
Curve (2) B
FIG. 163. Distribution of flux and m.m.f. under load, with the brushes on the
no-load neutral points.
Fig. 163 (a) represents the conditions when the m.m.f s. of field
and armature are acting together and with the brushes still on the
no-load neutral points. Curve 1, Fig. 163(6), shows the resultant
m.m.f. acting at each point. Its ordinates are represented by M
and they are the sum of the corresponding ordinates M/ and M a .
The m.m.f. across the pole face is no longer constant but is
184
ELECTRICAL ENGINEERING
decreased over one-half and increased over the other half by the
same amount. The flux density B (curve 2) at each point is still
proportional to the m.m.f. and inversely proportional to the
reluctance of the path, but, since part of the path is made up of
a magnetic material, due to the effect of saturation, the increase
of flux under one-half of the pole is less than the decrease under
the other half and consequently the total flux is decreased.
If it were not for the effect of saturation the ordinates of curve
2, Fig. 163(6), could be obtained by adding the corresponding
ordinates of Fig. 161(6) and Fig. 162(6).
The neutral points are no longer midway between the poles but
have been shifted in the counter-clockwise direction in Fig. 163 (a)
and to the right in Fig. 163(6). To prevent sparking the brushes
must be moved up to or a little v beyond the load neutral points.
In a generator the brushes must be moved forward in the direc-
tion of rotation and in a motor must be moved backward against
the direction of rotation as indicated in Fig. 164.
With the brushes midway between the poles the direction of
the armature m.m.f. is at right angles to the field m.m.f. and it
therefore does not directly oppose it but causes a distortion of the
flux and a decrease due to saturation. In this case the m.m.f. of
the armature is cross-magnetizing.
Generator
Curved ).M=M/+M a
Curve(2), B
FIG. 164. Distribution of flux and m.m.f. under load, with the brushes
advanced to the pole tips.
When, however, the brushes are moved into the fringe of lines
at the pole tips, as in Fig. 164(a), the two m.m.fs. are no longer at
right angles and as seen in Fig. 164(6) the part of the armature
m.m.f. which is subtracted from the field m.m.f. is much greater
than the part which is added to it and therefore the resultant
m.m.f. is reduced and the flux is both distorted and decreased.
Referring to Fig. 164 (a) the armature conductors may be
separated into two groups ; namely those between a and d included
DIRECT-CURRENT MACHINERY 185
in the double angle of advance a with their return conductors from
c to b and those under the pole between b and a with their return
conductors between d and c. Th,e first group acts in direct
opposition to the field m.m.f. and decreases the flux crossing the
air gap. They are therefore called the demagnetizing ampere-
turns of the armature. This demagnetizing m.m.f. increases as
the shift of the brushes is increased and it also increases directly
with the armature current.
The second group exerts a m.m.f. at right angles to the field
m.m.f. and distorts the flux as in Fig. 163 (a) and causes a de-
crease due to saturation. They are called the cross-magnetizing
ampere-turns of the armature.
For sparkless commutation without the use of interpoles the
brushes must be moved ahead of the neutral points in order that
the coils short-circuited by them may be cutting the fringe of flux
at the pole tips. E.m.fs. are thus generated in the coils opposing
the back e.m.fs. due to inductance, and they aid in reversing the
current. As the armature current is increased a point is finally
reached where the armature m.m.f. is so strong that it over-
FIG. 165. Distribution of flux and m.m.f. with the brushes under the centers
of the poles.
balances the field m.m.f. at the pole tips and therefore no revers-
ing field is left and commutation is not possible without interpoles.
It is of no use to move the brushes further ahead because that only
increases the demagnetizing component of armature m.m.f. and
decreases the flux more. In direct-current machines without
interpoles the armature ampere-turns per pole at full load
should not exceed 70 per cent, of the field ampere-turns per pole.
Fig. 165 shows the effect of moving the brushes to the center
of the poles. The whole armature m.m.f. is demagnetizing and
the flux is reduced to a small value. There is no difference of
potential between the brushes since the sum of the e.m.fs.
generated in one-half of the conductors in series between the
186
ELECTRICAL ENGINEERING
brushes is exactly equal and opposite to that generated in the
other half.
146. No-load Saturation Curve. The no-load saturation
curve of a generator shows the relation between the voltage
generated at no load and the field current. It is sometimes
plotted with the values of $ g , the useful flux as ordinates and the
field m.m.f. or field ampere-turns per pole as abscissae.
To obtain the no-load saturation curve, the generator is run at
constant speed and a variable resistance is connected in series
with the field winding. As the current is raised from a very
low value the flux and the generated voltage increase directly
as the current while the iron parts of the magnetic circuit are
unsaturated, but as the flux density increases the magnetic cir-
cuit becomes saturated and a greater increase in current is re-
quired to produce a given increase in voltage than on the lower
part of the curve (Fig. 166).
o a
Field Ampere Turns
o 0i 02 abed
Field Ampere Turns
FIG. 166. No-load saturation curve. FIG. 167. Full-load saturation curve.
Shunt-excited machines are operated at a point, slightly above
the knee of the saturation curve to insure stability, that is, in
order that slight changes of speed may not cause large changes in
voltage.
Machines with compound windings must be operated at a
lower point on the saturation curve in order that the required in-
crease of voltage may be obtained without supplying too large
a m.m.f. in the series field windings.
If the field current is gradually decreased from its maximum
the flux and the generated voltage fall off, but they come down on
a curve slightly above the original curve, which cuts the vertical
axis at a point above the origin. The value of flux represented
by this intercept is the residual magnetism in the machine.
DIRECT-CURRENT MACHINERY 187
147. Load Saturation Curves. The curve showing the rela-
tion between the terminal voltage and the field current, when the
generator is delivering a given load current is called a load satura-
tion curve (Fig. 167). The load saturation curves may be ob-
tained from the no-load saturation curve if the values of the
demagnetizing and cross-magnetizing armature ampere-turns
are known and also the resistance of the armature.
If the brushes are moved ahead of the no-load neutral position
by an angle a electrical degrees the demagnetizing ampere-turns
per pole = jorj X armature ampere-turns per pole (Fig. 164).
If a = 18 degrees or 10 per cent, of the pole pitch, the demagnet-
izing ampere-turns per pole = 0.2 X armature ampere-turns
per pole.
To overcome the effect of the demagnetizing ampere-turns an
equal number of ampere-turns must be added to the field winding.
This number should be increased 15 or 20 per cent, to make allow-
ance for the leakage of flux between the pole and armature (Art.
219).
The remaining armature ampere-turns per pole are _cross-
180 - 2a
magnetizing. They are ir^r X armature ampere-turns
loU
per pole, or when a = 18 degrees, the cross-magnetizing turns =
0.8 X armature ampere-turns per pole.
The cross-magnetizing ampere-turns produce a very high
saturation in the teeth under one-half of the pole tips and so in-
crease the reluctance of the magnetic circuit as a whole and de-
crease the flux. This effect is very difficult to determine accu-
rately but in normally designed machines in which the field
ampere-turns per pole = approximately 150 per cent, of the
armature ampere-turns per pole, the cross-magnetizing ampere-
turns may be replaced by 40 per cent, of their number of demag-
netizing ampere-turns.
This value can be applied only for points on the no-load satura-
tion curve just above the knee. Where the saturation is very
high the effect of the cross-magnetizing turns is relatively larger
and a larger percentage than 40 must be used. On the other
hand, for points below the knee of the saturation curve much
smaller percentages must be used.
In the case of the generator designed in Chapter VIII, the arma-
ture ampere-turns per pole = 4,860 and the angle of brush lead
188 ELECTRICAL ENGINEERING
a = 18 electrical degrees; the demagnetizing ampere-turns per
pole, = 0.2 X 4,860 = 972, and the cross-magnetizing turns
= 0.8 X 4,860 = 3,888. Here the full voltage point is just at
the knee of the saturation curve and 30 per cent, of 3,888 = 1,166
demagnetizing ampere-turns are taken to represent the cross-
magnetizing effect.
The demagnetizing effect varies directly with the current but
the cross-magnetizing effect increases at a more rapid rate. If
the two effects are equal at full load, then on overloads the cross-
magnetizing effect will be greater than the demagnetizing effect
while at light loads it will be less.
In Fig. 167, os is the no-load saturation curve plotted on a base
of field ampere-turns. At no load the field m.m.f. required for
rated voltage is oa. Let ab be the demagnetizing ampere-turns per
pole at full load and be be the demagnetizing ampere-turns equiva-
lent to the cross-magnetizing turns. In order to generate the same
voltage under load as at no load the field m.m.f. per pole must be
increased by the amount ac = mp and p is a point on the curve
OiSi showing the relation between the voltage generated under
load and the field m.m.f. Curve OiSi would be parallel to os if
the cross-magnetizing effect were constant but at the lower points
on the curve it becomes very small and the intercept 001 may be
taken as = ab.
The terminal voltage E will be less than the generated voltage
8 by the resistance drop in the armature and series field, Ir =
pq. The curve 0282 is drawn parallel to the curve o^i at a dis-
tance pq = Ir below it. This is the full-load saturation curve for
the generator.
A saturation curve for half load could be obtained by joining
the center points of all the lines such as mq, m^i, etc.
148. Voltage Characteristic or Regulation Curve. The voltage
characteristic of a direct-current generator is the curve showing
the relation between the terminal voltage and the current output
or load.
The voltage generated in the armature is
8 = Zn10- 8 = Kn$ volts. (Art. 142)
Pi
Take first the case of a separately excited generator in which the
speed n and the field current // are both kept constant.
DIRECT-CURRENT MACHINERY 189
At no load the flux crossing the air gap under each pole is 3>
and the voltage generated is
8 = Kn$Q]
this is also the terminal voltage at no load.
As the generator is loaded the terminal vpltage decreases due
to two causes: (a) armature reaction, and (6) armature resist-
ance.
(a) When current flows in the armature, the armature m.m.f.,
which is partly demagnetizing and partly cross-magnetizing,
reduces the flux crossing the gap and thus reduces the generated
voltage. The loss of flux increases at a more rapid rate than the
current. The consequent loss of voltage is called the drop due
to armature reaction.
(6) It is necessary to distinguish between the voltage, ,
generated in the armature and the terminal voltage, E. At no
load they are the same but when current flows in the armature
part of the generated voltage is consumed in driving the armature
current / through the resistance of the armature winding and
brushes r. This resistance drop is Ir and increases directly with
the current.
The terminal voltage is E = 8 Ir. (245)
In Fig. 168 are shown the no-load and load saturation curves
plotted on the same base. The constant exciting ampere-turns
= oa. A vertical Kne through a cuts the various saturation
curves at ra, n, r and t which represent the terminal voltages corre-
sponding to the respective loads. These points are plotted on a
load base as curve m\r and this is the voltage characteristic of
the generator. It is sometimes called the external characteristic
to distinguish it from the corresponding curve of generated
voltages niik.
The straight line ol represents the armature resistance drops
at all loads = Ir.
The voltage generated in the armature at any load is 8 = E +
Ir and by adding the ordinates of ol to those of m\r the curve mik
is obtained showing the relation between the generated voltage
and the load. This is called the internal voltage characteristic.
At full load the total drop of voltage below the no-load value
is mr and it is made up of two parts, mk the drop due to armature
reaction and kr the armature resistance drop.
190
ELECTRICAL ENGINEERING
149. Field Characteristic or Compounding Curve. The field
characteristic of a generator is the curve showing the increase of
field current required under load to maintain a constant terminal
voltage at constant speed.
In a separately excited or self-excited generator the terminal
voltage can be maintained constant as the load current increases
if the field current is increased to such a value that the increase
in field m.m.f. will not only overcome the effect of armature
m.m.f. but will produce an increase in flux to provide the extra
voltage to supply the armature resistance drop.
M Load
l^load
Fia. 168. Voltage characteristics FIG. 169. Field characteristics,
separately excited.
Referring to Fig. 167 the terminal voltage at no load = am
and the field m.m.f. = oa; to obtain an equal terminal voltage
at full load the field m.m.f." must be increased by an amount ad;
of this ac is required to overcome armature reaction and cd is
required to supply the armature resistance drop.
In Fig. 169 the horizontal line through the no-load voltage
point m cuts the various load saturation curves at e, f and g.
The corresponding values of field m.m.f. or field current plotted
on a load base give the curve af which is the field characteristic
of the generator.
DIRECT-CURRENT MACHINERY
191
The increase of field current under load is obtained by gradually
cutting out resistance from the field rheostat in series with the
field winding (Fig. 170). This regulation of the voltage must
be done by hand and cannot take care of sudden changes of load.
150. Voltage Characteristic of a Shunt Generator. In the
shunt generator, armature reaction and armature resistance
cause a decrease in terminal voltage
under load, but a third condition
must also be taken into account.
The field circuit is connected across
the terminals of the armature and the
current in it is proportional to the
terminal voltage; thus, when the
terminal voltage decreases due to
armature reaction and armature resistance, the field current also
decreases and causes a further decrease in the flux; and therefore
the terminal voltage of a shunt-excited generator is less than it
would be if the machine were separately excited (Fig. 171).
The voltage characteristic of a shunt generator may be obtained
from the load saturation curves, Fig. 171, by drawing a straight
line om from the origin through the no-load voltage point m.
The slope of this line represents the resistance of the shunt field
FIG. 170. Shunt generator.
FieM Current O S ^ Load Full Load 1% Load 2 Load
FIG. 171. Voltage characteristics of a shunt generator.
winding and its intersections with the various saturation curves
give the terminal voltages corresponding to the respective loads.
There are two values of voltage corresponding to each value of
load. The method of obtaining the voltage characteristic is
shown.
The load current can be increased by decreasing the resistance
in the load circuit until the point M is reached. If the load
resistance is further decreased the armature current increases
for an instant and then decreases as its m.m.f. wipes out part of
192
ELECTRICAL ENGINEERING
the flux and causes the generated voltage to decrease. Finally,
when the load resistance is zero and the generator is short-cir-
cuited the terminal voltage and the field current become zero
and the residual flux produces a small voltage which is consumed
in driving the current os through the armature resistance.
The maximum current output of a shunt generator is usually
many times full-load current and can be reached only with small
machines of poor regulation.
The internal voltage characteristic can be obtained by adding
the armature resistance drops to the terminal voltages; that is,
by adding the ordinates of curves 1 and 2, Fig. 171, curve 3 is
obtained.
151. Effect of Change of Speed on the Voltage of a Shunt
Generator. In Fig. 172 ('a) oba is the no-load saturation curve
of a shunt generator at normal speed. The point a is the normal
10 20 30 40 SO 60 70 80 90100110120130140150
Percent Field Current
40 50 60 70 80 20 1Q(|
Percent Speed
FIG. 172. Variation of voltage of a shunt generator with speed.
voltage point and the point 6 is a point below the knee of the
curve. The slope of the line oa represents the resistance in the
shunt field for normal operation and the slope of ob represents
the resistance when operating with the magnetic circuit un-
saturated. obiai is a no-load saturation curve for 90 per cent,
of normal speed, each of its ordinates is 90 per cent, of the corre-
sponding ordinate of oba.
From these curves it is seen that a drop of speed of 10 per
cent, causes a drop of voltage pq = 14 volts = 14 per cent, when
16 X 100
the machine is saturated and a drop p\qi = 16 volts = ^
= 19 per cent, when it is unsaturated. Therefore, a saturated
shunt generator is more stable than an unsaturated generator,
that is, its voltage changes less due to slight changes in speed.
DIRECT-CURRENT MACHINERY
193
At normal saturation to counteract the 10 per cent, drop in
speed the field current must be increased 43 per cent.
The saturation curves for speeds from 80 to 40 per cent, of
normal are also shown. They are intersected by the line oa
at a 2 , a 3 , etc., and these points show the no-load voltages cor-
responding to the various speeds.
The voltage speed curve at no load is shown in Fig. 172(6).
The voltage becomes zero in this case at 40 per cent, of normal
speed, since the resistance line oa is tangent to the saturation
curve at this srjeed and so does not cut it. The effect of residual
magnetism has been neglected; it tends to change the curve as
indicated by the dotted lines. The line oa in Fig. 172(6), shows
the voltage speed curve at no load for a separately excited
generator.
FIG. 173. Effect of change of speed
on the voltage characteristic of a shunt
generator.
FIG. 174. Effect of saturation on
the voltage characteristic of a shunt
generator.
In Fig. 173 are shown the no-load saturation curves for speeds
of 100 per cent., 90 per cent, and 80 per cent, of normal and the
corresponding voltage characteristics, which may be obtained
as in Fig. 171. The curves for .the low speeds fall off very quickly
since the field m.m.f. is decreased more than the speed; the
maximum current outputs are also much reduced. The voltages
ori, or 2 and or 3 produced by the residual magnetism are di-
rectly proportional to the speed as are also the currents osi,
os 2 and os 3 produced by them.
15"2. Effect of Saturation on the Voltage Characteristic of. a
Shunt Generator. In Fig. 174 is shown the no-load saturation
curve ra of a shunt generator intersected by three resistance lines
oa corresponding to normal resistance, 06 to 125 per cent, of
normal and oc to 160 per cent. The characteristic curves for
13
194
ELECTRICAL ENGINEERING
the lower saturations fall off very quickly since the field m.m.f.
is weak but they all pass through a single point s since the voltage
produced by the residual flux is the same for all. An unsaturated
shunt generator does not tend to maintain a constant terminal
voltage under load, that is, its voltage regulation is poor.
153. Compound Generator. Increase of field m.m.f. under
load can be obtained automatically by placing a series winding
on the field poles in addition to the shunt winding. The series
winding carries the load current.
If the series winding is designed so that the terminal voltage
is the same at full load as at no load the generator is flat-com-
pounded, if the terminal voltage at full load is higher than at no
load the generator is over-compounded.
In Fig. 179 are shown the voltage characteristics or regulation
curves of a generator: (1) self -excited; (2) separately excited;
(3) flat-compounded; and (4) over-compounded. If a generator
is flat-compounded so that it gives the same voltage at full-load
as at no load it will be slightly over-compounded at half load.
Referring to Figs. 167 and 169 for a flat-compound machine
the series winding at full load must provide a m.m.f. equal to
ad. The number of turns per pole on the series winding may be
-4-
FIG. 175. Compound generator
(short shunt).
FIG. 176. Compound generator
(long shunt).
found by dividing the required ampere-turns ad by the load
cuj;ren\j\ In designing a series field winding it is advisable to
put on some extra ampere-turns to make up for any decrease in
voltage due to decrease in the speed of the prime mover under
load. After the machine is completed the series excitation may
be corrected by connecting a shunt across the terminals of the
series winding (Figs. 175 and 176). By varying the resistance
of the shunt any required portion of the current may be taken
from the series winding. In the case of a generator supplying
a rapidly fluctuating load the shunt to the series winding must
be designed with its inductance in the same ratio to the induce
DIRECT-CURRENT MACHINERY
195
tance of the series winding as its resistance bears to the resist-
ance of the series winding in order that the variable current may
divide up in the correct proportions to give the required
compounding.
154. Voltage Characteristic of a Compound Generator.
The voltage characteristic for a flat-compound generator may be
O Field M.M.F. a di d d 2 O JLoad Full Load 1^ Load
s a \ ffy Load Current
FIG. 177. Voltage characteristic of a flat compound generator.
obtained from the saturation curves in Fig. 177 (a). At full
load the series excitation is ad and the terminal voltage is dm 2 =
the no-load voltage am; at half load the series excitation is adi =
77 and the voltage is dimi slightly greater than am: at one and
A
one-half load the series excitation is ad% = %ad and the terminal
voltage d 2 m 3 is less than am. The voltage characteristic is shown
in Fig. 177(6).
m 3
E&
.
FIG. 178. Voltage characteristic of an over-compound generator.
The characteristic for a 10 per cent, over-compound generator
may be obtained in a similar way from Fig. 178 (a). At full
load the series excitation is af and the terminal voltage is /ra 2 =
110 per cent, of the no-load voltage am; at half load the series
excitation is a/i = -^ and the voltage is f\m\] m\ lies slightly
196
ELECTRICAL ENGINEERING
The voltage charac-
Compound
ompound
^Separately Excited
^Shunt Excited
above the straight line joining m and w 2 .
teristic is shown in Fig. 178(6).
The increase of excitation for full load of includes a small
component which is supplied by the shunt winding, since as the
voltage across it has increased 10 per cent., its current and m.m.f.
have increased in the same proportion. Allowance should be
made for this in designing series windings.
155. Short-shunt and Long-shunt Connection. In compound-
wound generators the shunt winding may be connected across the
armature terminals, Fig. 175, called short-shunt, or outside of the
series winding, Fig. 176, called long-shunt. The characteristic
curves are not affected to a great extent by the difference in con-
nection since with the short-shunt the voltage across the shunt
winding is higher than with the long-shunt by the resistance drop
in the series winding and the
current in the series winding
is less by the amount supplied
to the shunt winding.
156. Regulation. The
regulation of a separately
excited or shunt-excited gen-
erator is denned as the rise
in voltage when full load is
taken off expressed as a per
cent, of full-load voltage.
Fig. 179 shows the characteristic curves of a generator: (1)
separately excited, (2) shunt-excited, (3) flat-compounded and
(4) over-compounded. The per cent, voltage regulation of the
be
separately excited generator is ;. X 100 per cent. ; the regulation
C J
of the shunt generator is -^ X 100 per cent. The regulation of
the flat-compounded generator is the per cent, variation of volt-
age from the normal no-load or full-load value; it is - X 100
per cent. The regulation of the over-compounded generator is
rr X 100 per cent., where gh is the maximum intercept between
the voltage characteristic and the straight line va.
57. Series Generator. The series generator, Fig. 180, is
excited by a series winding carrying the load current and has no
25 50 75
Per Cent Load
100 125
FIG. 179. Voltage characteristics of
generators.
DIRECT-CURRENT MACHINERY
197
shunt winding. Its characteristic curves are shown in Fig. 181.
The effect of residual magnetism is neglected. Curve 1 is a no-
load saturation curve obtained by exciting the generator from a
separate source. Curve 2 is the internal voltage characteristic
showing the relation between generated voltage and load when
the generator is self-excited,
and the horizontal distance
between curve 1 and curve 2
represents the effect of arma-
ture reaction. The armature
reaction due to full-load cur-
rent is represented by mp, and
assuming that the armature
reaction varies directly as the
load current the other points on curve 2 may be found easily.
Curve 3, the external characteristic, is obtained by subtracting
from the ordinates of curve 2 the corresponding resistance
drops in the armature and series field shown in curve 4.
The terminal voltage of a series generator can be varied by con-
necting a variable resistance in shunt to the series winding. The
FIG. 180. Series generator.
100 150
Percent Load Current
FIG. 181. Voltage characteristics of a series generator.
slope of the line oq represents the resistance in the load circuit
at full load; as this resistance is increased the load current and
the terminal voltage decrease. When the resistance line oC
is tangent to the voltage characteristic the machine is unstable
and is liable to lose its voltage. The load resistance at which this
condition occurs is called the critical resistance for the generator.
198 ELECTRICAL ENGINEERING
It is an indefinite quantity on account of the effect of residual
magnetism.
158. Electric Motors. In generators mechanical power is
supplied and electrical power is generated. The speed is fixed by
the prime mover and is constant. The terminal voltage is ap-
proximately constant in the shunt generator and flat-compound
generator and increases with load in the over-compound generator
and the series generator. The generated voltage is always greater
than the terminal voltage by the drop in the armature resistance ;
it is
8 = E + Ir.
In motors electrical power is supplied and mechanical power is
generated. The impressed voltage is fixed by the supply circuit
and is constant. The speed is either approximately constant as
in the shunt motor or decreases with load as in the compound
motor and series motor. The voltage generated in the armature
has the same equation as the voltage in a generator, but it is a
back voltage and opposes the current; the impressed voltage E is
greater than the back-generated voltage by the armature resist-
ance drop; thus,
, E = & + Ir, (246)
or Bt*^ 8 = E - Ir. (247)
Field Rheostat VH, Rheostat **. Field Rheostat
FIG. 182. Shunt FIG. 183. Series FIG. 184. Corn-
motor, motor. pound motor.
159. Types of Motors. There are three types of direct-cur-
rent motors corresponding to the three types of generators, shunt,
compound and series. The shunt motor has its field circuit con-
nected across the line in shunt with the armature and therefore
has a, constant excitation (Fig. 182). The compound motor
has a series winding carrying the load current in addition to its
shunt winding. The excitation therefore increases with load
(Fig. 184).
The series motor has its field circuit in series with the armature
and has no shunt winding. Its excitation is zero at no load and
increases directly with the load current (Fig. 183).
DIRECT-CURRENT MACHINERY
199
160. Speed Equation of a Motor. The impressed voltage is
E = 8 + /r,
and
where,
therefore
and
8 = ZnQ 10~ 8 = Kn$> volts,
Pi
K = Z- P - 10~ 8 is a constant;
E = Kn& + Ir
E - Ir
n =
(248)
this is the speed equation.
The term Ir is small in comparison to E and may be neglected
except at heavy loads.
The speed equation may therefore be written
n =
E
(249)
thus, the speed of a motor is directly proportional to the impressed
voltage and is inversely proportional to the flux crossing the air
gap.
161. Methods of Varying Speed. The speed can be varied in
three ways: (1) by varying the impressed voltage E, (2) by vary-
ing the flux 3> by reducing the field current, (3) by shifting the
brushes.
if
[I R
FIG. 185.
Impressed Voltagre E
FIG. 186. Variation of speed
with impressed voltage.
1. The voltage impressed on the armature can be varied by
introducing a resistance R, Fig. 185, in series with the armature.
If E s is the supply voltage, the voltage impressed on the motor is
E = E 8 - IR.
200
ELECTRICAL ENGINEERING
By increasing R, E can be reduced to any required value and
any speed from standstill to normal speed can be obtained but
no increase of speed above normal.
Fig. 186 shows n as a function of E; the locus is a straight line
passing through the origin.
This method of varying speed is uneconomical as a large
amount of power is lost in the control resistance; it is the product
of the current input and the voltage consumed in the resistance
and is
= I X IR = PR watts.
The resistance R must not be connected in series with the field
winding as it would then decrease the field current and therefore
the flux and tend to cause an increase in speed.
If the speed is reduced to 50 per cent, of normal in this way the
efficiency of the motor is less than 50 per cent.
With a variable voltage supply a wide range of speed can be
obtained efficiently (see Arts. 173 and 174).
^-
s*
^"
/
/
j
/
/
/
1
Field Current \f
FIG. 187.
Field Currant "
FIG. 189.
2. When varying the speed by field control the full line voltage
is impressed on the armature and a resistance R is connected in
series with the field winding, Fig. 182. As this resistance is
increased the field current // decreases according to equation
E
and the flux decreases with the field current as shown by the
saturation curve of the machine in Fig. 187. Since the speed
varies inversely as 3>, Fig. 188, the variation of speed with field
current will be represented by a curve of the shape shown in
Fig. 189.
DIRECT-CURRENT MACHINERY 201
By this method the speed can be increased to any required
value, and it tends to approach infinity when the field current is
zero but is limited by the residual magnetism.
In machines of ordinary design the speed can be increased
satisfactorily only about 70 per cent, above normal speed by
field weakening. Beyond this point it is not possible to get
sparkless commutation of full-load current, since the armature
m.m.f. is strong enough to overcome the weak field m.m.f. and
wipe out the commutating field and at the same time the time of
commutation is decreased.
By using interpoles which neutralize the effect of armature
m.m.f. and provide a commutating field, the speed can be in-
creased to four or five times normal by field weakening without
injurious sparking.
The loss in power in the resistance controlling the field current
is very small since the power used for field excitation is only 1 or 2
per cent, of the rated output.
3. At no load with the brushes on the neutral line, all the con-
ductors on each half of the armature are effective in generating
the back e.m.f. and this is therefore the position of minimum
speed Fig. 160(a).
When the brushes are moved back against the direction of
rotation, Fig. 160(6), only the belts of conductors under the poles
are effective in generating the back e.m.f. and the result is the
same as though the flux had been decreased. The speed is there-
fore increased.
Under load, when current is flowing in the armature, the con-
ductors between the poles exert a demagnetizing m.m.f. and cause
a decrease in the flux; the speed therefore rises more than at no
load.
This method of speed variation is not used to any extent since
it interferes with commutation and causes injurious sparking.
162. Speed Characteristics of Motors. The speed charac-
teristic is the curve showing the variation of speed with armature
current. The speed equation was found to be
n = E ~ r r.p.s. (Art. 160)
Shunt Motor. As the motor is loaded and / increases the speed
is affected in two ways: (1) the resistance drop Ir increases and
tends to cause a proportionate decrease in speed; (2) the flux <
202
ELECTRICAL ENGINEERING
is decreased by the armature demagnetizing and cross-magnetiz-
ing m.m.f.; therefore the armature reaction tends to increase
the speed. The voltage drop at the brush contacts (Art. 182)
increases with load but not in direct proportion. It acts with
the armature resistance to decrease the speed. If the motor is
operated at a point just above the knee of the saturation curve
the drop in speed due to the armature resistance will be greater
than the rise due to armature reaction and the speed character-
istic will fall as shown in Fig. 190 curve 1.
The speed regulation of a motor is the rise in speed when full
load is thrown off expressed as a per cent, of full-load speed.
The speed regulation of shunt motors of large size is from 2 to
5 per cent.
\
1) Shunt
2) Compound
3) Differential Compound
4) Series
(1) Shunt
(2) Series
(3) Compound
(4) Differential Compound
n
* % H Full
Load Load Load Load
Armature Current I
'FiG. 190. Motor speed charac-
teristics.
H * % Full I*
Load Load Load Load Load
Armature Current I
FIG. 191. Motor torque charac-
teristics.
Compound Motor. In the compound-wound motor the arma-
ture resistance causes a drop in speed and the m.m.f. of the series
winding overcomes the effect of the armature m.m.f. and increases
the flux and thus decreases the speed more than in the shunt
motor. A typical speed characteristic of a compound motor is
shown in curve 2, Fig. 190.
If the series winding is reversed, its m.m.f. opposes the field
m.m.f. and thus decreases the flux and causes the speed to in-
crease with load as shown in curve 3, Fig. 190.
The motor is then called a "differential compound" motor and
may be designed to give constant speed under all loads. If the
series-field winding is strong the motor is unstable and tends to
run at excessive speed.
DIRECT-CURRENT MACHINERY 203
Series Motor. At no load the series motor tends to run at a
very high speed limited only by the residual magnetism or the
torque required to overcome the losses. As load is applied the
current and flux increase and the speed falls rapidly until the mag-
netic circuit of the machine becomes saturated; the speed charac-
teristic then becomes almost horizontal (Fig. 190, curve 4).
163. Torque Equation. The torque of a motor is proportional
to the product of the flux crossing the air gaps and the current
in the armature. Its equation is derived as follows:
The voltage impressed on the armature is
E = 8 + Ir,
where
8 =
Pi
is the back voltage generated in the armature and Ir is the voltage
consumed by the armature resistance.
The power input to the armature is
El = 87 + I 2 r watts. (250)
The power lost in the armature winding is I 2 r watts, and thus
the electric power transformed into mechanical power is
7 = ZnQl^lQ- 8 ;
Pi
this is the power output in watts neglecting the friction losses.
The motor output in horsepower is
746 746
If the torque developed is T lb.-ft., then the output in horse-
power is
Zn$/^-10~ 8
2irnT Pi
550 746
and the torque is
550 ""~ / - 10 8
T = ^ TZfi- - = 0.1177Z7^- 10- 8 lb.-ft. (251)
2irn 74o pi
This is the torque equation of a direct-current motor.
T = 0.1177 X 10- 8 X (Z ) X(p$) lb.-ft., (252)
204 ELECTRICAL ENGINEERING
where Z is the sum of all the currents in the armature conduc-
tors and p$ is the sum of the fluxes crossing the air gaps under all
the poles.
The torque equation may be written
T = K3>I lb.-ft., (253)
where K = 0.1177 Z^- 10~ 8 is a constant.
Pi
The torque of a motor is therefore directly proportional to the
armature current and to the flux in the air gap.
This is the torque developed at the shaft of the motor. The
torque output is less due to the iron and friction losses.
164. Torque Characteristics of Motors. The torque charac-
teristic is a curve showing the relation between the torque and
the armature current.
The torque equation is
T = K&I. (Art. 163.)
Shunt Motor. In the shunt motor is almost constant under
load, decreasing only by a small percentage due to armature re-
action. The torque therefore varies almost directly with the
current (Curve 1, Fig. 191).
Series Motor. In the series motor the flux increases almost in
direct proportion to the current, while the magnetic circuit is un-
saturated, and therefore the torque is proportional to the square
of the current; at heavy load, when the magnetic circuit becomes
saturated, the flux becomes almost constant and the torque then
increases in direct proportion to the current. Curve 2, Fig. 191,
is a typical torque-current curve for a series motor.
Compound Motor. In the compound motor the flux increases
with load but not in direct proportion to the current ; thus, the
torque-current curve (curve 3) lies between those of the shunt
and the series motor.
If the series winding is reversed, as in the differential compound
motor, the flux decreases with load and the torque-current curve
falls below that of the shunt motor (Curve 4).
165. Construction of the Speed Characteristics. The speed
characteristic of any motor can be constructed if the no-load
saturation curve is given and the magnitudes of the armature
reaction and armature resistance. In Fig. 192 omn is the no-
DIRECT-CURRENT MACHINERY
205
load saturation curve of a motor, flux <1> vs. field current //. The
field current is maintained constant and the flux at no-load is
$0 = oa and the speed is UQ.
The effect of armature reaction at full load is represented by
mp and it reduces the flux to the full-load value oai. At no load
the impressed voltage and the back voltage are equal and may
be represented by oa. The required back voltage at full load
is less than at no load by the armature resistance drop Ir and
is represented by oa^ The speed at full load is therefore reduced
. ,, .. 00,2 , , oa 2
in the ratio to a value n = n Q
'Compound
Load Full Load
w
Load
Current
FIG. 192. Construction of speed characteristics.
If the resistance drop is very small it may happen that the
point a 2 comes above ai, in this case the speed rises above the no-
load value. Such a condition often exists at fractional loads.
With overloads the demagnetizing effect increases directly with
the current and reduces the flux density considerably but on
this account the cross-magnetizing effect is relatively smaller
and the increased resistance drop always results in a decrease
of speed. This is the case of the shunt machine and the resulting
speed characteristic is as shown in Fig. 192(c).
In the compound motor the effect of armature reaction will be
opposed by the m.m.f. of the series winding and the flux will be
reduced less than in the shunt motor or it may be increased. If
a series winding is added to the shunt motor above exerting at
full load a m.m.f. rag, Fig. 192(6), greater than the armature
reaction mp, the flux at full load will be increased to the value oai,
and the speed will be reduced to n = n 2 - Other points on
the speed characteristic can be found in the same way, Fig. 192(c).
166. Construction of the Speed Characteristic for a Series
Motor. In Fig. 193 (a) om is the no-load saturation curve of a
206
ELECTRICAL ENGINEERING
series motor obtained by separately exciting the fields and driving
as a generator at the full-load speed of the motor. If mp repre-
sents the effect of armature reaction at full load, oai is the back
voltage at full load at a speed n f and it may be taken to represent
also the flux at full load. The impressed voltage is greater than
oai by the resistance drop Ir and may be represented by oa.
This value is constant.
O K Load Full Load
Field Current or Lead Current
(a)
Load Current
FIG. 193. Construction of speed characteristics or a series motor.
At one-half load the armature reaction may be taken as
mp
and the corresponding flux is oa 3 ; the resistance drop is only
aa\
half of its former value = 0,0,2 = ~? and so the back voltage
must be oa 2 . The speed is therefore increased to n = n f X
The complete characteristic is shown in Fig. 193(6).
100
90
80
70
r
ll
^~
==>-
90
80
M
|50
k
20
10
~7
/
r
~V
x
/,
-
'
/
?'
y
/
-^
/
--,/_
1
^4
1
f
10 20 30 40 50 60 70 80 90 100
Percent Line Voltage
10 20 30 40 60 60 70 80 90 100
Percent Field Current or Line Voltage
() (6)
FIG. 194. Variation of speed with line voltage (shunt motor without load).
167. Variation of Speed of a Shunt Motor with Line Voltage.
If the voltage impressed on the terminals of a shunt motor
decreases, the speed decreases but not in direct proportion.
Referring to Fig. 194, (a) shows a no-load saturation curve
DIRECT-CURRENT MACHINERY 207
plotted with flux on a base of field current or line voltage and (b)
shows motor speed on a voltage base.
When the line voltage is 100 per cent., the field current, the
flux and the speed are all 100 per cent. If the line voltage drops
to 90 per cent., the field current drops to 90 per cent, and the flux
to 97 per cent;, due to the decrease in voltage impressed on the
armature the speed tends to drop to 90 per cent, but this is partly
counterbalanced by the drop in flux to 97 per cent. The speed is,
therefore, 100 X 0.90 X JT|= = 92.9 per cent. When the volt-
age falls to 50 per cent, the flux is 75.5 per cent, and the speed is
100 X 0.5 X - = 62.2 per "cent.
If the saturation curve passes through the origin, the speed
voltage curve in (b) tends to cut the speed axis about 55.5 per
cent, speed. This is a theoretical point for a motor without
friction and without residual magnetism.
If a residual flux of 10 per cent, is assumed as indicated by the
dotted part of the saturation curve the speed voltage curve drops
off to the origin along the dotted part of the curve in (b). The
motor becomes unstable on the lower voltages.
168. Variation of Speed with Temperature of the Field Coils.
If the speed of a motor is measured immediately after being
started and again after running under load for some time it will
be found to have increased. If the rise in temperature of the field
is 40C., the increase in resistance will be approximately 40 X
0.4 = 16 per cent, and the field current will decrease 16 per cent.;
this results in a decrease of 7 or 8 per' cent, in the flux and an
increase of 7 or 8 per cent, in speed. The amount of the speed
change depends on the saturation of the machine.
169. Construction of the Torque Characteristic. In Art. 165
the method of obtaining the flux in the air gap at any given value
of load current is indicated. If curves showing the variation of
flux with load current are plotted as shown in Fig. 195, the torque
characteristics of the motors may be found by multiplying the
values of flux by the corresponding values of current. This
follows from the torque equation
T = K$I.
The torque at some particular load must be known in order to fix
the scale of the curve.
208
ELECTRICAL ENGINEERING
The values of torque obtained in this way represent the torque
developed at the shaft of the motor. The external torque or
available torque is less by the torque required to supply the iron
and friction losses.
Flux=0
% Load Full Lend
Compound Motor
U)
Load Full Load
ries Motor
FIG. 195. Construction of torque characteristics.
170. Starting Torque. In Fig. 196 are shown typical speed
and torque characteristics for shunt, series and compound motors
with the same full-load speed and torque.
When starting, a motor must often exert a torque considerably
in excess of the full-load value in order to overcome friction and
the inertia of the load. At the same time the starting current
must be kept within reasonable limits in order to prevent fluc-
tuations of the voltage of the system. If the required starting
torque is just equal to full-
load torque, each type of
motor would draw full-load
current. If the starting cur-
rent is limited to 150 per cent.
of full-load current, the start-
ing torque developed by the
X X H
Load Current
X H
Load Current
FIQ. 196. Speed and torque
characteristics.
, . 111
three motors would be ap-
. , , , . . _
. proximately, shunt 140 per
cent, of full-load torque, com-
pound 160 per cent., depending on the strength of the series
winding and series 200 per cent. Thus for loads requiring
starting torques in excess of full-load torque the series and com-
pound motors have the advantage over the shunt motor that
they can develop the required torque with smaller currents and
thus with smaller demands on the supply system.
When starting shunt or compound motors the field rheostat
DIRECT-CURRENT MACHINERY
209
Field Rheostat;;
FIG.
Starting Resistance
197. Starting a shunt
motor.
should be entirely cut out in order to give as strong a field as
possible (Fig. 197).
171. Motor Starter with No-voltage Release. Fig. 198 shows
the connections of a motor starter with a no-voltage release.
When the main switch is closed, current cannot flow to the motor
until the starting handle is moved to contact C, which allows
current to flow through the field A
winding and no- voltage release B and
through the total starting resistance
R s to the motor armature. The
auxiliary contact A is connected to
contact C but is placed in such a
position that, when the motor is to be
disconnected the starting handle
opens the circuit at A instead of at C and the spark produced
by the field discharge burns only the small contact A which is
easily renewed. In starting up the handle should be held on C
for a short time to allow the field of the motor to build up. As
the motor speeds up and develops its back voltage the handle is
gradually moved over to the final position in which all the start-
ing resistance is cut out and the
armature is connected directly
to the supply. The starting re-
sistance is cut into the field cir-
cuit but its effect in reducing the
field current is usually negligible.
An iron armature is carried by
the starting handle and in the
final position it completes the
magnetic circuit of the no-
voltage release and the handle
is held in the running position.
If the power is interrupted for
any reason, or if the field circuit
opens, the no-voltage release becomes demagnetized and a
spring pulls the handle back to the starting position.
The full starting resistance R s must have such a value that the
starting current will not be greater than one and one-fourth to
one and one-half times full-load current,
The starting resistance is designed only for starting duty and
must not be used to operate the motor at low speeds as it would
14
FIG.
198. Direct-current motor
starter.
210 ELECTRICAL ENGINEERING
become overheated and burn out. It is mounted in a ventilated
box on the back of the starter.
Example. A 100-hp., 230-volt, shunt motor has an efficiency of 90 per
cent, and its armature resistance is r = 0.03 ohm. Find (a) the starting
resistance to limit the starting current to one and one-fourth times full-load
current and (6) the armature current if the motor were connected directly
to the supply while at standstill.
(a) The full-load current is 030 v Q on = ^ am P-J tne allowable start-
ing current is % X 360 = 450 amp.
The starting resistance must therefore be
OOA
* R > = 7^ == 0.51 ohm.
4oU
This value includes the small armature resistance.
(6) If full voltage were applied to the armature at rest the current would be
limited by the armature resistance only and would tend to reach a value
230
7 = Q^y5 = 7,600 amp. This would burn out the armature winding unless
the circuit was protected by circuit-breakers or fuses.
172. Adjustable Speed Operation. In Art. 161 three methods
of varying motor speeds were discussed: (a) field control for in-
creasing the speed; (b) armature control for decreasing the speed;
and (c) shifting of the brushes which is possible only in special
cases.
For lathes where the cutting speed must be constant although
the diameter of the material changes field control must be used.
b Armature con,trol is not satisfactory as may
/^~ ~~^\ be seen from the following example. Refer-
ring to Fig. 199, abc is a piece of metal to be
turned. Assume that the motor driving the
.x , lathe is operating on a 115-volt system and
that while cutting through the projection be-
. 199 * tween a and b the armature current is 100
amp., the voltage across the armature terminals is 75 volts and
the speed is 300 r.p.m. This would require a resistance in series
with the armature R = r~7 ~ 0.4 ohms to take up the extra
\a
k I Tool
40 volts.
After passing the point 6 the cut is light and the current is
reduced to 10 amp., the drop across the series resistance is
10 X 0.4 = 4 volts and the voltage impressed on the armature is
111; the speed therefore rises to 300 X - = 444 r.p.m. and the
DIRECT-CURRENT MACHINERY
211
point of the tool is liable to be broken when it strikes the point
a again.
Field control alone cannot, however, give a wide enough
range for all cases since the speed can be increased only about
70 per cent, above normal without trouble due to commutation.
Where speed ranges of 3 to 1 or 4 to 1 are required it is necessary
to use interpole motors or to have a multiple-wire system of
supply.
173. Multiple-wire Systems of Speed Control. In Fig. 200
G is the main generator giving 250 volts between terminals.
FIG. 200. Multiple-wire system.
FIG. 201. Three- wire generator.
BiBzBz is a balancer set with three armatures on the same
shaft giving voltages of 60, 80 and 110. With this system a com-
plete range of voltages from 60 to 250 is available for the motor
armature and the intermediate speeds can be obtained by varying
the field of the motor which is permanently connected across the
outer wires at 250 volts. The possible speed range is about 6
to 1, but the system is very complicated and expensive.
Three-wire systems are often used instead of four- wire systems.
That shown in Fig. 201 is very common. A three- wire generator
(Art. 367) giving 230 volts ..
Supply]
between outers and 115 volts
to neutral is used. A speed
range of 4 to 1 is possible but
the motors must be specially
designed to give a 2 to 1 range
by field control.
174. Ward Leonard System
of Speed Control. This system is illustrated in Fig. 202. MI GI
is a high-speed motor-generator set. The field of the generator
is arranged so that it can be varied through a wide range and
reversed. Its terminal voltage is impressed on the armature of
Motor Generator Set Adjustible Speed Motor
FIG. 202. Ward Leonard system.
212 ELECTRICAL ENGINEERING
the motor M in which speed control is required. A uniform
variation of speed from a maximum in one direction through
zero to a maximum in the opposite direction may be obtained.
The field of the motor M is permanently connected across the
supply lines.
175. Speed Control of Series Motor. The speed of a series
motor can be varied by connecting a resistance in series with the
armature but in cases where a number of motors are connected
to the same load, as in electric-railway operation, series-parallel
control may be used resulting in a more efficient speed variation.
@ VW W^M^+W^ Start
Speed
Speed
. Full Speed
W
R_
FIG. 203. Series-parallel control.
In Fig. 203 M i and M% are two similar motors with their equal
starting resistances Ri and R*. In position (1) the whole of the
line voltage (500 volts) is taken up by the starting resistances
Ri and R* and the motors are at rest. In position (2) the starting
resistances are short-circuited, the current is the same as before
and each motor is running at half speed and develops a back
voltage of 250 volts. In (3) the motors are in parallel and the
starting resistances are in series again. The current to each
motor is the same as before but the current from the line is
doubled. The speed is the same as in (2). In (4) the resist-
ances are short-circuited again and the motors are running at full
speed with 500 volts impressed on their terminals. The loss of
energy in the control resistances is just half what it would have
been if the motors had been connected in parallel at the start.
DIRECT-CURRENT MACHINERY 213
The current per motor is kept constant so that the rate of accel-
eration of the car may be constant.
If speeds higher than full speed are required they may be
obtained by connecting resistances in shunt to the motor fields.
When it is necessary to start a series motor very slowly the
armature is shunted by a resistance which reduces the starting
torque.
When it is necessary to bring a series motor to rest quickly it
may be disconnected from the line and a resistance connected
across the armature terminals. The motor acts as a generator
and slows up very quickly. The power developed is wasted in
heating the resistance. This is one form of dynamic braking.
176. Interpole Motors. Interpole motors are fitted with
small poles, midway between the main poles, excited by the load
current (Fig. 224) . The interpole m.m.f . increases directly with
the load ; it is stronger than the armature m.m.f. and neutralizes
it in the space directly beneath the interpole and produces a flux
in the proper direction to assist commutation. The brushes are
placed on the no-load neutral and the motor may be operated in
either direction. Since commutation is taken care of by the
interpole field speed ranges of 4 to 1 or even 6 to 1 can be ob-
tained by field control without serious sparking.
To prevent saturation of the interpoles at heavy loads they
must be designed with very low flux densities at normal load.
If shunts to the interpole windings are used the ratio of their
inductance to the inductance of the interpole windings must be
the same as the ratio of their resistances, in order that sudden
variations of load may be taken care of.
177. Applications of Motors. When deciding on the type
of motor to be employed in a given case three points are of special
importance: (1) speed whether constant, variable or adjustable;
(2) starting torque whether greater or less than full-load
torque; (3) starting current whether it is liable to disturb the
voltage of the system.
The types of motor available are: (1) the shunt motor which
runs at approximately constant speed and develops a torque
proportional to the current; the speed may be adjusted to any
suitable value within limits; (2) the series motor with variable
speed and torque proportional to the square of the current below
saturation; if twice full-load torque is required at start it draws
about one and one-half times full-load current; (3) the compound
214 ELECTRICAL ENGINEERING
motor with variable speed and a torque characteristic between
the shunt and the series. The compound motor has the advan-
tage over the series motor that it approaches a limiting speed at
light load set by its shunt excitation.
Shunt motors are used for lathes, boring mills, and all constant
speed machine tools, for driving line shafting when the starting
load is not too heavy, for fans, centrifugal pumps, etc.
Series motors are used in electric-railway operation and for
cranes, hoists, etc., where very large starting torque is necessary
and where constant speed is not required.
When a load comes on a series motor it responds by decreasing
its speed and supplying the increased torque with a small in-
crease of current, thus preventing a sudden shock on the system.
A shunt motor under the same conditions would hold its speed
nearly constant and would supply the required torque with a
large increase of current and would thus make a heavy demand on
the system.
Series motors must not be used for belt drives or in any case
where the load may be removed suddenly since they run at excess-
ive speed at light loads. %
Compound-wound motors are used in classes of work where
constant speed is not necessary and where a fairly large starting
torque is required, except in those cases in which series motors
must be employed on account of their very large starting torque.
For express elevators a compound-wound motor is employed,
the series winding is required to give the large torque at start and
is cut out after a certain speed is reached. The motor then runs
at constant speed as an ordinary shunt machine.
For operating shears, punches, etc., where a high maximum
load must be carried for a short period a compound motor, with
a flywheel attached, is used. The drooping speed characteristic
is necessary to enable the flywheel to give up part of its stored
energy to supply the peak of the load and so relieve the supply
system. A shunt motor would not drop in speed and the fly-
wheel would not be of any value. A series motor would not be
suitable as it would run at excessive speeds before and after the
cut.
In rolling mills where the load fluctuates very rapidly a similar
compound motor with a heavy flywheel is used.
Compound motors are often installed in positions where the
constant speed characteristic of the shunt motor would be more
DIRECT-CURRENT MACHINERY 215
suitable but where the series winding is required to supply the
torque to overcome the inertia of the heavy moving parts, as in
the case of heavy planers, etc.
178. Losses in Direct-current Machinery. The main power
losses in direct-current generators and motors may be divided
into copper losses, iron losses and friction losses and these may be
subdivided as follows:
Copper losses :
Shunt-field copper loss.
Series-field copper loss. *
Armature copper loss.
Iron losses:
Hysteresis loss.
Eddy-current loss.
Friction losses:
Brush-friction loss.
Bearing-friction loss.
Windage loss.
179. Shunt-field .Loss. The shunt-field copper loss is 7/ 2 r/
watts, where // is the field current and r/ is the resistance of the
winding at the operating temperature of the machine. This loss
can be expressed as E/I/, where E/ = J/r/ is the voltage im-
pressed on the field winding. All the energy supplied to the field
winding is transformed into heat and is wasted, since no energy
is required to maintain the magnetic flux after it is once
established.
The shunt-field loss is constant under all conditions of load in
the shunt motor and the flat-compound generator; it decreases
slightly with load in the shunt generator and increases with load
in the over-compound generator.
The loss ranges from about 1 per cent, of full-load output in
the case of large high-speed machines to 5 per cent', in small low-
speed machines.
If there is a rheostat connected in series with the shunt-field
winding, the power wasted in it should be included in the shunt-
field copper loss.
180. Series-field Loss. The series-field copper loss is I, 2 r t
watts, where I s is the current in the series winding and r a is its
resistance. This loss increases as the square of the load current
of the machine. In interpole machines the resistance r, will in-
216
ELECTRICAL ENGINEERING
elude the resistance of the interpole winding. The loss in the
shunts to the series winding or in the series-field rheostat must
be included in the series-field loss.
181. Armature Copper Loss. The armature copper loss pro-
per is I a 2 r a watts, where I a is the armature current and r a is the
resistance of the winding not including the brush contacts. It
increases as the square of the load current.
In addition to the loss due to the load current there are extra
losses produced by local currents in the armature coils short-
circuited by the brushes. If the e.m.fs. induced in the various
parts of the armature winding connected in multiple between the
terminals are not all equal circulating currents will flow both at
no load and under load and cause loss.
182. Loss at the Brush Contacts. Closely associated with
the armature copper loss and usually included with it is the loss
at the brush contacts. Between the carbon brush and the copper
commutator there is a drop of
voltage which is more of the
nature of a back voltage than a
voltage consumed by resistance.
This drop varies with the current
density and is also affected by
the direction of the current flow.
Fig. 204 shows the relation be-
0.3
0.6
Q
(1) Carbon to Copper -Negatire Brush
(2) Copper to Carbon -Positire Brush
20 30 40
Amperes per Sq. In.
FIG. 204. Brush contact drop.
tween contact-drop and current
density for an ordinary hard
carbon brush. The drop is
higher at the negative brush,
that is, from carbon to copper, than at the positive brush. At
low densities it increases with the current but above 35 or 40
amp. per square inch it becomes almost constant.
The loss of power at the brush contacts is el a , where e is the
drop of voltage at the positive plus the negative brush and may
be taken as approximately 2 volts except for very soft carbon or
graphite brushes or for the combined metal and carbon brushes,
where it is very much lower.
With a current density of 35 amp. per square inch and a con-
tact drop of 1 volt per brush the power loss is 35 watts per square
inch of brush contact. This is a reasonable value. Where the
contact drops are smaller higher current densities may be used.
In low-voltage machines this loss of power is an appreciable
DIRECT-CURRENT MACHINERY
217
per cent, of the output and may be very serious, but for high-
voltage machines it is less important.
With copper brushes the contact drop is very small and densi-
ties up to 150 amp. per square inch are used.
183. Hysteresis Loss. The hysteresis loss is due to the rever-
sal of the magnetic flux in the armature iron as it moves across a
pair of poles.
The loss per cubic centimeter of iron per cycle is
WH = TjCB 1 - 6 ergs (equation 123),
where (B is the maximum induction density in lines per square
centimeter and TJ is the hysferetic constant for the iron. The
value of tj for ordinary armature iron after assembly may be
taken as 0.0027.
The number of cycles of magnetism per second or the frequency
is
7}
f = r.p.s. X number of pairs of pole s = n -
(254)
If the induction density B is expressed in lines per square inch,
the hysteresis loss per cubic inch of iron per second is
2 ) (2.54) 3 / ergs per second
.o4 /
/ 10~ 7 watts. (255)
W h =
FIG. 205. Distribution of flux in the armature core.
In transformers where the induction density is very nearly uni-
form throughout the volume of the iron, this -value multiplied
by the volume of the iron would give the hysteresis loss very
closely; but in the case of dynamos the induction density is not
218 ELECTRICAL ENGINEERING
uniform but varies from a maximum at the roots of the teeth
to a very low value at the bottom of the armature core, as indi-
cated in Fig. 205. The loss can only be determined very approxi-
mately by making separate calculations for the teeth and the core.
The hysteresis loss varies directly with the frequency, that is,
with the speed of the machine; it increases as the 1.6th power of
the flux density and thus of the voltage and it increases to some
extent under load due to the distortion of the flux. The loss is
increased in sections where the density is increased more than it
is decreased where the density is decreased.
The hysteresis loss may be very seriously increased by careless
handling of the materials during assembly.
In some cases the hysteresis loss has been found to increase
after the machine has been in operation for a short time. This is
due to aging of the iron and will increase the temperature rise of
the machine. Silicon steel in additio-n to having a low hysteresis
loss and high electrical resistance is non-aging.
184. Eddy-current Loss. The eddy-current loss is due to
electric currents set up in the
armature iron by the e.m.fs. gener-
ated in it as it cuts across the flux.
In Fig. 206 abed represents a sec-
tion of an armature punching of
thickness t in. If the flux density
in the gap is B lines per square
inch and the edge ab is moving with
FIG. 206. Eddy-current loss in a ve locity of v in. per second across
the armature core.
the flux, then, the e.m.f. generated
in the length ab is
e = BtvlQ' 8 volts.
This e.m.f. will cause a current to circulate through the iron as
indicated by the arrows and its value will be
i = kye = kyBtvlO' 8 amp.
where 7 is the electric conductivity of the iron and k is a constant
depending on the dimensions of the section.
The loss in the section will be
p = i* r = J = 10 -i 6 = fc lT W watts (256)
KJ Icy
where ki = k X 10~ 16 is a constant.
DIRECT-CURRENT MACHINERY
219
The eddy-current loss, therefore, varies as the square of the
induction density, that is, the square of the terminal voltage,
the square of the thickness of the plates, and the square of the
speed; it also depends on the conductivity of the iron used. It
cannot be calculated accurately in the case of an armature where
the induction density varies throughout the volume.
The loss increases under load due to the distortion of the field
but it tends to decrease as the temperature rises and decreases
the conductivity.
The eddy-current loss is reduced by building up the armature of
thin punchings insulated from one another by varnish which in-
creases the resistance in the path of the eddy currents. If the
slots are filed out after assembly to remove rough edges, adjacent
plates may be short-circuited and the value of the lamination
of the core partly lost. If the punchings are assembled under too
great pressure the resistance between them will be reduced and
the loss increased.
I I
i 1
LJ
a
in
CA;
FIG. 207. Eddy-current loss in the pole face.
Eddy-current losses also occur in the armature end plates,
spider and bolts, due to leakage fluxes, and they are comparatively
large since these parts are not laminated.
185. Pole-face Loss. Eddy currents are also produced in
the pole faces due to local variations of the induction density as
the armature teeth move across them (Fig. 207). The pulsations
220 ELECTRICAL ENGINEERING
of flux do not extend far below the surface of the pole face since
they are opposed by the eddy currents. The frequency is very
high, being proportional to the product of the revolutions per
second and the number of teeth. Formulae have been derived
for the calculation of the pole face losses but they are very com-
plicated and must be applied with great care.
To reduce the loss the pole faces of all direct-current machines
should be laminated. Machines with long air gaps have small
pole-face losses.
186. Brush -friction Loss. The brush-friction loss in foot-
pounds per second is equal to the product of the total brush
pressure in pounds, the peripheral speed of the commutator in
feet per second and the coefficient of friction between the brush
and the commutator.
With carbon brushes the pressure should be from 1.5 to 2
Ib. per square inch; this value multiplied by the area of all the
brushes gives the total brush pressure in pounds. In railway
motors, where there is a great deal of vibration, pressures up to
5 Ib. per square inch are used in order to insure good contact.
The coefficient of friction between an ordinary hard carbon
brush and the commutator may be taken as 0.3, for graphite
brushes it is about 0.25 and for copper brushes 0.2.
The brush friction loss varies directly with the speed but is
independent of the load.
187. Bearing-friction and Windage Losses. These two losses
are difficult to separate and only their combined value can be
obtained by test.
The bearing-friction loss is proportional to the (speed) '* for
bearing velocities up to 2,000 ft. per minute and is approximately
proportional to the speed for higher values. It can be calculated
approximately.
The windage loss varies as the (speed) 3 but is usually small up
to peripheral speeds of 6,000 ft. per minute. It depends on the
construction of the machine and cannot be calculated with any
degree of accuracy.
The combined friction and windage losses vary from about H
per cent, of the rated output in large slow-speed machines to 2
or 3 per cent, in small high-speed machines.
In very high-speed machines the windage loss may become the
larger part of the combined loss.
The friction and windage losses are independent of the load.
DIRECT-CURRENT MACHINERY
221
188. Eddy-current Losses in the Armature Copper. In
addition to the losses discussed above under certain conditions
large eddy-current losses may occur in the armature copper both
at no load and under load. There are two causes of this loss.
1. Some of the flux from the field poles passes down into the
slots and cuts the armature conductors inducing eddy currents
in them in the directions indicated in Fig. 208(1). The propor-
tion of the flux entering the slots depends on the length of the air
gap, the width of the slots and the saturation of the teeth.
In direct-current machines of moderate size the resultant loss is
small. In turbo alternators with very wide slots and long air
(i) (2)
FIG. 208. Eddy-current losses in armature conductors.
gaps the loss tends to be large and the conductors in the upper
part of the slot should be laminated vertically. Under load the
loss tends to increase, due to the increased saturation of some of
the teeth.
2. A second and more important cause of eddy currents in
the copper is due to the unequal distribution of flux in the arma-
ture teeth near the pole tips Fig. 208(2). The flux density in
tooth a is much lower than that in 6; the magnetic potential of
the roots of the two is the same, but since a larger m.m.f. is
required for b than for a the magnetic potential at the top of 6
must be greater than that at the top of a and flux will pass across
the slot as indicated. The flux density and the e.m.f. induced in
the conductor at the top of the slot will be greater than at the
bottom and eddy currents will flow in the direction of the e.m.f.
in the upper part of the conductor returning in the lower part.
The loss due to these currents is especially large in machines with
deep conductors. It is a function of the tooth saturation and
therefore increases very rapidly with increase of voltage; it
increases with load due to field distortion and also increases with
speed. In cases where it is liable to be large conductors should
be laminated in a horizontal plane.
222 ELECTRICAL ENGINEERING
Three laminated conductors are shown in -Fig. 208.
These eddy-current losses in the copper cannot be calculated
but may be kept small by proper design.
189. Constant Losses and Variable Losses. The losses are
sometimes divided into two groups, the constant losses and the
variable losses.
The constant losses are those which do not vary to any great
extent under load and include the shunt-field copper loss, the
iron losses and the friction and windage losses.
The variable losses are those which increase with load, namely,
the armature copper loss and the series-field copper loss.
190. Core Loss. The core loss includes all the losses located
in the armature core except the armature copper loss. It may
be obtained by running the machine without load, measuring the
input and subtracting the shunt-field copper loss and the fric-
tion and windage losses. It therefore includes the no-load iron
losses and in addition the eddy-current losses in the pole faces
and in the armature conductors and any losses which may be
produced by currents circulating in the armature windings.
The core loss increases to some extent under load due to field
distortion.
In designing machines an approximate value for the core losses
may be obtained from curves such as those in Fig. 257, compiled
from tests on standard machines.
191. Efficiency. The efficiency of a machine may be va-
riously expressed, as,
output _,
n = - ~~ 100 per cent,
input
output
= - . , , - 100 per cent,
output + losses
input losses .,__ /OCT\
= - - 100 per cent. (257)
input
The efficiency varies with the output; at light loads it is low on
account of the constant losses; between three-fourths load and
full load it is maximum and the constant losses and variable losses
are nearly equal; above full load it decreases again due to the
rapid increase of the variable losses. The limit of the efficiency
which can be reached commercially depends on the output, the
voltage and the speed. A higher efficiency can be obtained with
large machines than with small machines. A higher efficiency
DIRECT-CURRENT MACHINERY
223
can be obtained with high-voltage or high-speed machines than
with low-voltage or low-speed machines.
For 220-volt direct-current motors the full-load efficiency
ranges from about 85 per cent, for small sizes to 93 per cent, for
large sizes.
15000
14000
40 80 120 100 200 240 280 320 360 400
Amperes Output
FIG. 209. Characteristic curves of a
200 kw. compound-wound generator.
20 40 60 80 120 160
Amperes
200
FIG. 210. Characteristic curves
of a 75-h.p. railway motor.
/
/
/
/
\
/
/
90
\
E
ffic
en
cy
7
^
^
.
g 800 >>80
5 H 70
\
t
/
f
fr
^
7
,lc I
\ 3 1 2 J 1 | |
**' ]S
V
d
TP
C
nh
b
TP
. Ic'
,t i'
Ic -
i Ic
r L
Ic ,
!L^bi
I 3 Pl^sJUj
2I C
FIG. 214.
2I C
FIG. 215.
8*0
FIG. 216.
The resistance of the brush contact plays a very important part
in commutation; it tends to reduce the current in the short-cir-
cuited coil to zero and then to build it up in the opposite direction.
It would produce complete commutation if it were not opposed
by the effects of the resistance and inductance of the coil. Its
function is illustrated in Fig. 215. If the resistance of the total
brush contact is r c , then in Fig. 214 the drop of potential between
the brush BI and bar 1 is 2l c r c . As soon as the brush touches bar
2 commutation begins and the brush-contact resistance must be
separated into two parts, r\ the resistance from the brush to bar 1
and r 2 the resistance from the brush to bar 2. If at the instant
represented in Fig. 215 the current in the coil is i, then the current
flowing from bar 1 to the brush is I c + i and the drop of potential
is (I e -[- i)ri m j the current from bar 2 to the brush is I c i and
the drop of potential is (I e i)r%. Since the resistance ri is in-
creasing while r 2 is decreasing, the current from bar 2 will increase
while that from bar 1 will decrease and the current in the coil will
decrease. Neglecting the resistance and inductance of the coil
15
226
ELECTRICAL ENGINEERING
the current flowing in the coil will be zero when ri and r 2 are equal
and when therefore half of the time of commutation has passed.
Any further increase in the resistance ri will cause part of the
current from coil 6 to flow through coil c in order to reach the
brush BI by the path of least resistance. As the resistance 7*1
still increases, more and more current flows through c until ri
becomes infinite as the brush breaks contact with bar 1 and the
total current I c from b flows through c. Commutation is then
complete.
In Fig. 217, curve (1), the current in coil c is plotted on a time
base for half of one revolution; it is reversed in the time T, rep-
[Time of Commutation
(1) Commutation by Contact
Kesistauce only
(2) Effect of the Besistance
of the Coil
(3) Effect of the Inductance
Of the Coil
(4) Effect of a Negative
Commntating E.M.F.
(5) Over-Commutatioa
Time
resented by OT, during which the coil moves across the brush B\,
and it must vary according to a straight line law. This can be
proved as follows:
If Fig. 215 represents the condition t sec. after the beginning
of commutation, neglecting the resistance and inductance of
the coil, the drop of potential from the commutator to the brush
must be the same at both sides, or
= r c
but
therefore, /j , ^ r
Solving for i this gives
T - t
T
and
T - 2t
T
t
T
which is the equation of a straight line.
T
When t = -,i = 0, and when t = T, i = - I c .
4
(258)
(259)
DIRECT-CURRENT MACHINERY
227
If the resistance of the coil is taken into account, the drop of
potential across TZ (Fig. 216) must be greater than the drop
across ri by the amount required to maintain the current i through
the resistance r; therefore,
(I c +
and
r (Tt - t 2 ) + r c T 2
(260)
T
When t = , i = 0, and when t = T, i = - I c .
The current therefore passes through zero at the same instant
as before and is completely reversed in the same time, but the
variation does not follow a straight line law but a curve as shown
in curve (2), Fig. 217. The effect of the coil resistance is very
small and may be neglected.
The effect of the inductance of the armature coil must next
be considered. Armature coils are partially surrounded by iron
and therefore have a large inductance which is proportional
to the square of the number of turns in the coil. With full-pitch
(ft)
Full Pitch Winding
(a)
FIG. 218. Inductive flux in a
full-pitch winding.
I ' 02
Short Pitch Winding
(a)
FIG. 219. Inductive flux in a
short-pitch winding.
drum windings, Fig. 218, both the coils in one slot will be short-
circuited at one time and the inductive flux linking the slot part
of each of them will be almost twice as large as in the case of
fractional pitch windings (Fig. 219). The flux around the end
connections is approximately the same in the two cases. The
inductance of the coil is, therefore, partly self-inductance and
partly mutual inductance between adjacent coils but the resultant
effect may be considered as due to an inductance L,
228 ELECTRICAL ENGINEERING
When the current in a coil of inductance L henrys is changing
at the rate -r amp. per second, an e.m.f. L -j volts is generated
in a direction opposing the change of current.
In Fig. 216 the drop of potential from bar 2 to the brush is the
same by the two paths, one through the resistance r 2 and the
other through the coil in series with the resistance n; and thus
(I c + i) ri + ri + L ^ = (I c - i) r 2 ,
or, substituting the values of ri and r 2 found above,
T rt'i T
(I. + i) r c - + ri + L = (/. - i) r c - (261)
The complete solution of this equation is difficult but a partial
solution, for the instant when commutation is complete, may be
obtained easily.
Equation (261) may be written,
T di il c ~ i\ rn /Ic + i\
L dt = - n + rcT rid " rcT (f^i)
when t = T and i = I c
L d ~ t = rl c + 27 c r c - r c T X ; (262)
this last term is indefinite but its value may be found by differ-
entiating the numerator and denominator with respect to the
independent variable t.
di
r/ c -f a"! di di
r -
and substituting this in equation (262),
T ^ = rI c + 2I c r c + r c T^
dt at
or
di = Ic(r + 2r c )
dt L - r c T '
If L = r c T, ~r = a , the reactance voltage L ^ = oc , and spark-
ing will occur.
di
If L>r c T, -r is positive and the current tends to increase in-
stead of reversing and sparking results.
DIRECT-CURRENT MACHINERY
229
If L c ;
27 27
~f is the average rate of change of current and -jjfr L is called the
average reactance voltage, while 2l c r c is the voltage drop at one
brush contact.
Therefore, the condition for sparkless commutation is that the
average reactance voltage must be less than the voltage drop at
one brush contact. This, however, applies only to the case in
which no flux is cut by the short-circuited coil and where there-
fore no e.m.f. due to rotation is generated in the coil either assist-
ing or opposing commutation. The brush-contact resistance is
the only factor operating to cause the current to reverse.
Due to the effect of inductance, the current does not decrease
so quickly as in curve 1, Fig. 217, but follows curve 3, and the
rate of change of current becomes very rapid as the brush breaks
contact with bar 1 and sparking is liable to occur. The current
density in the brush tip also becomes very high, tending to burn
both the brush and the commutator.
No-Load Neutral
Load Neutral
Load Neutral
FIG. 220. Commutating field.
When the inductance and the reactance voltage are large it is
necessary to have an e.m.f. generated in the coil to assist commu-
tation. The brushes of a generator are, therefore, moved ahead
in the direction of rotation (back in the case of a motor) , so that
the coil when short-circuited is cutting the fringe of lines from
the pole tip (Fig. 220).
If at the time represented in Fig. 216 there is an e.m.f. e gen-
erated in the coil assisting commutation, equation (261) can be
written
230 LECTRICAL ENGINEERING
This equation cannot be solved in general but it is possible to de-
termine the value of e required at any instant to cause the current
to vary as a linear function of time from I c to I c in the time T.
On this assumption
. T T - 2t
* = I --T-
and di ', _ 2I<
dt~ T'
Substituting these values in equation gives
or
e = 7 c j^-r(l -2^) }, (264)
which gives at the beginning of commutation t = 0,
e = I c (^ - r) , (265)
and at the end of commutation t = T,
e T = I c (^ + r) . (266)
This e.m.f. is proportional to the current I c but is independent
of the brush resistance r c . The average value of this e.m.f. is
called the commutating e.m.f. in the coil and is represented by
E c .
If the commutating e.m.f. is less than that required to reverse
the current completely in time T, commutation is imperfect
and there is a tendency to spark, and if the e.m.f. is so large that
the current is more than reversed there is a tendency to spark due
to over-commutation.
The contact resistance helps to prevent sparking when the
e.m.f. generated in the coil by rotation is either too great or too
small to produce perfect commutation.
When commutation is produced by the high-resistance brush
contact without the aid of any e.m.f. generated in the coil, it is
called "natural" or "resistance" commutation; when it is as-
sisted by an e.m.f. generated in the coil, it is called "forced" or
' ' voltage ' ' commutation.
Resistance commutation can never be perfect unless the induct-
ance of the coil is negligible, but at light loads it will reverse
DIRECT-CURRENT MACHINERY 231
the current without injurious sparking. Assume that the brushes
of a generator delivering half load are set on the corresponding
neutral line and that commutation is satisfactory. If the load
is increased the increased m.m.f. of the armature causes the neu-
tral line to move ahead so that the coil short-circuited is cutting a
field of such a direction as to tend to maintain the current or even
to increase it. The reversal of the current is therefore retarded
and there is a greater tendency to spark than before. If the
load is reduced the neutral line falls behind the brushes and a
voltage assisting commutation is generated in the coil.
Voltage commutation is also limited in its application and as
the current in the armature is increased a point is reached (usu-
ally about 25 per cent, overload) beyond which sparkless com-
mutation is impossible, since when the current is increased a
stronger field is required to reverse it, but the stronger current in
the armature increases the m.m.f. of the armature and moves the
neutral line ahead of the brushes and at the same time decreases
the flux. The brushes have to be advanced further and the de-
magnetizing effect is increased. When the armature m.m.f. is
large enough to overbalance the field m.m.f. the flux at the pole
tip is wiped out and voltage commutation is impossible. Moving
the brushes further ahead only decreases the flux.
To take full advantage of voltage commutation it would be
necessary to vary the position of the brushes with varying load,
but this is not practicable, and therefore the brushes must be set
to give good commutation at some intermediate load and the
resistance of the brush contact must be relied on to prevent spark-
ing above and below this point. Modern machines are designed
to give good commutation at all loads from no load to 25 per cent,
overload with fixed brush position.
193. Commutating Electromotive Force. The commutating
e.m.f. is the e.m.f. generated in the short-circuited coil due to
cutting flux and depends on a number of factors.
There are two fluxes to be considered: (1) that produced by the
field m.m.f.; and (2) that produced by the armature m.m.f.
These two fluxes are represented in Figs. 161 to 164 and their
resultant in the interpolar space is shown in Fig. 220. The e.m.f.
generated in the coil due to cutting the field flux tends to assist
commutation, while that generated by cutting the armature flux
opposes commutation. The resultant of the two is called the
commutating e.m.f., E c . This e.m.f., therefore, depends on the
232
ELECTRICAL ENGINEERING
field excitation, the position of the brushes, and the load, that is,
the armature current.
When the brushes are on the " no-load" neutral there is no
e.m.f. assisting commutation; if the machine is carrying load at
the time there will be an armature flux both in the interpolar
space and around the end connections and this flux being sta-
tionary is cut by the short-circuited coil and generates in it an
e.m.f. opposing commutation. E c is therefore negative.
When the brushes are moved ahead to the position where the
e.m.f. due to cutting the field flux is equal to the e.m.f. due to
cutting the armature flux, the commutating e.m.f. is zero. This
position is called the load neutral. In this case commutation is
carried out by the brush-contact resistance and will be satis-
factory only if the reactance voltage is very low.
When the brushes are moved ahead of this load neutral to a
position where the commutating e.m.f. is positive and equal to the
reactance voltage commutation is perfect.
X^C
ER
X M H I ll^LoadO
tttuiit Generator
M * H llHLwuJO M H K l i^LoadO X X K
Flat-Coinpouud Over-Compouud Interpole Generator
FIG. 221. Reactance voltage.
In Fig. 221 the reactance voltage E R and the commutating
voltage E c are shown plotted on a load base for different types of
generators.
The reactance voltage is directly proportional to the current
and is represented by a straight line passing through the origin.
For the shunt generator the commutating e.m.f. varies from a
maximum at no load to zero at about 50 per cent, overload.
The brushes are set to give' perfect commutation at 75 per cent.
load. At 25 per cent, overload the armature m.m.f. pretty well
wipes out the main field near the pole tip and only a very small
commutating e.m.f. is left. Sparking will not be serious so long
as the difference between E R and E c is not greater than the voltage
drop at one brush contact. The reactance voltage at full load
should be limited to this or a lower value.
For a flat-compounded machine the conditions are much better,
the commutating e.m.f. does not decrease to any great extent
DIRECT-CURRENT MACHINERY 233
with load and the difference between E R and E c is very much less;
larger values of reactance voltage may, therefore, be allowed.
For an over-compounded machine conditions are still further
improved; E c increases with load and satisfactory commutation
may be obtained with reactance voltages more than double the
drop at the brush contact.
194. Conditions Essential to Sparkless Commutation. It is
necessary to limit the armature strength at full load to such a
value that it will not interfere too greatly with the flux produced
by the field winding. For shunt machines the field ampere-
turns required to drive the flux through the gap and teeth should
be = 1.2 (armature ampere-turns per pole) + the demagnetizing
ampere-turns per pole. This insures good commutation from
no load to 25 per cent, overload with fixed brush position pro-
vided the reactance voltage does not exceed the limits specified
below. Lower values of armature strength tend to give better
commutating conditions.
For flat-compounded machines the field ampere-turns for the
gap and teeth should be = 1.2 (armature ampere-turns per pole).
The reactance voltage must also be limited approximately to
the following values.
In shunt machines from 1 to 1.25 volts except with high-
resistance brushes.
In 10 per cent, over-compounded generators from 2.5 to 3.0
volts.
In shunt motors where any large variation of speed by field
control is to be obtained values of reactance voltage below 1
volt should be used or a very weak armature.
In interpole machines satisfactory commutation may be
obtained with reactance voltages up to 15 volts.
195. Calculation of the Reactance Voltage for a Full-pitch
Multiple Winding. In Fig. 218 (a) is shown one coil, c, of a full-
pitch double-layer winding, short-circuited by the brush and at
(b) is shown a section through a slot. If the number of turns per
coil is n, the number of conductors per slot is 2n. Here n = 2.
The flux of self and mutual inductance is shown in the figure
and may be divided into two parts: (1) that surrounding the slot
part of the coil; and (2) that around the end connections.
Let e be the
234 ELECTRICAL ENGINEERING
number of lines of force which link 1-in. length of the end connec-
tions for each ampere conductor in the corresponding group of
end connections simultaneously short-circuited.
Then, with a full-pitch winding and a brush covering only one
bar the total inductive flux linking the coil, c, when carrying a
current i amp., is
6, = 2(2m> s L c ) + 2(m> e L e ) = 2m(20 s L c + e L e ),
where L c = length of the imbedded part of one conductor in
inches, and L e = length of one end connection.
The inductance of the coil is
L = ^ 10- 8 = 2n*(2 s L c + e L e )lQ-* henrys
u
and the average reactance voltage is
E R =^L (page 229).
The time of commutation is
width of brush or bar
T =
peripheral velocity of commutator
width of bar
r.p.s. X no. of bars X width of bar
1 1
r.p.s. X no. of bars r.p.s. X no. of coils
armature conductors r.p.m. X Z
~2rT
From tests made on a large number of machines Hobart found
that , has a value of approximately 10 and (f> e a value 2.
Therefore, the average reactance voltage is
7? "*- c " XN x 'fr" A ^' o^s/onr _i_ or ^Mn-8
CIR \ ZiH \Z\jLi c "f~ Zij e ) 1 1U
= 10 C -.^ Le 10~ 8 X / c ^ X r.p.m. X n
lo
= Y^ 10~ 8 X ampere conductors on . the
lo
armature X r.p.m. X turns per coil.
For the ordinary type of coil L e may be taken as = 2L C and
the average reactance voltage may be expressed by
E R = 0.8 X 10~ 8 X L c X ZI C X r.p.m. X n (267)
= 0.8 X 10~ 8 X gross length of core X ampere conduc-
tors on the armature X r.p.m. X turns per coil.
DIRECT-CURRENT MACHINERY
235
An increase in the brush width does not change the reactance
voltage because although it increases the number of coils short-
circuited at any instant it also increases the time of commutation
in the same proportion.
If there are six coil sides per slot, Fig. 222, and the brush covers
three segments then the flux surrounding the slot part of the coil
will be three times as great as before since the m.m.f . is increased
to three times, while any increase in the width of the path across
the slot will be counterbalanced by a
proportional increase in the depth.
Similarly, the flux about the end con-
nections will be about three times as
great; therefore, the inductance is in-
creased three times but the time of com-
mutation is also increased three times
and the reactance voltage remains as
before.
With a short-pitch winding (Fig. 219)
the slot flux is reduced to about half the
value with a full-pitch winding. This is due to the increased
reluctance of the path. The flux about the end connections is
decreased only slightly, due to their shorter length.
The reactance voltage of a short-pitch winding is
^ 10 -s x Z I C X r.p.m. X n
FIG. 222.
lo
or again substituting L e = 2L C
E R = 0.46 X 10- 8 X L c X ZI e X r.p.m. X n. 268;
The ratio -I-T rrjr-was assumed to be approximately 3.5 in
the derivation of the formulae above. If very deep and narrow
slots are used the flux s will be increased and therefore also the
reactance voltage.
In the case of a series or two-circuit winding there are p/2 coils
short-circuited in series and the reactance of this set of coils is
p/2 times the reactance of one coil but the commutating e.m.f . is
also increased in the proportion p/2. Since both these voltages
are greater than before it is difficult to keep their difference with-
in the value which can be taken care of by the brush-contact
resistance. On the other hand, if the commutating field is too
strong, the reactance of p/2 coils in series tends to prevent the
236 ELECTRICAL ENGINEERING
current from growing to a large value. Values of reactance
voltage calculated as for full-pitch windings, equation (267),
should be kept about 50 per cent, lower for series windings.
196. Current Density at the Brush Contacts. There are two
currents in the coil short-circuited by the brush: (1) the working
current or load current; and (2) the short-circuit current pro-
duced by an active e.m.f. in the coil.
When the e.m.f. generated 'in the coil is equal and opposite to
the reactance voltage there is no circulating current produced in
the coil and the load current reverses at a uniform rate due to the
action of the contact resistance (page 226) ; the current density
is then uniform over the brush face and the loss of energy is a
minimum.
When the. two e.m.fs. in the coil do not balance their result-
ant tends to cause a current to circulate through the coil and
brush, and the current density is no longer uniform and the loss
at the contacts is increased above normal.
This circulating current crosses the brush contact twice and
the contact resistance tends to keep the current down to a reason-
able value. When the unbalanced e.m.f. is not greater than 2
to 2.5 volts the current will not be large.
Reducing the width of the brush may in some cases so cut down
the short-circuit current that the maximum current density is
decreased although the average is increased.
The average current density in machines without interpoles
is from 35 to 40 amp. per square inch but higher values are some-
times used with low-contact resistance brushes.
Unequal distribution of current between brushes on the same
arm or between arms of the same polarity causes the density to
vary from the average. This is particularly true of machines
with series windings and gives rise to selective commutation.
In the alternating-current series commutator motor in which
the circulating currents tend to become very large the contact
resistance is reinforced by the addition of high-resistance leads
between the coils and the commutator bars.
197. Burning of the Brush and Commutator. An apparent
electrolytic action takes place under the brushes in the direction
of the current flow, carbon is deposited on the commutator under
the negative brush and the brush is burned, and copper is de-
posited on the positive brush and the commutator is burned.
The burning is dependent on the energy consumed at the par-
-f
DIRECT-CURRENT MACHINERY 237
ticular point on the brush contact and this is proportional to the
product of the current density at the point and the voltage drop
at the contact.
Severe burning of the commutator results in high mica which
spoils the contact between the brush and the commutator and
increases the liability to spark. To pre-
vent the commutator burning in grooves
the brushes should be staggered in sets as
shown in Fig. 223.
Even a very slight deposit of copper on
the brush face reduces the brush-contact
resistance very materially. On the other
hand, the darkening or polish on the commu-
tator surface increases the contact resistance.
198. Poor Commutation Resulting from Too Many Coil
Sides per Slot. When a number of coil sides are placed in one
slot, they are not all short-circuited at once and therefore they
will be commutated in fields of different strengths and the com-
mutating e.m.fs. will be different since the intensity of the field
varies very rapidly as the pole tip is approached. The first
coil to be commutated will be in the weakest field and may not
have its current completely reversed, while the last may be over-
commutated. If there are six coil sides per slot every third
commutator bar is liable to be burned due to improper
commutation.
It is therefore not advisable to reduce the number of slots
too far. Small machines should have at least 12 slots per pole
and large machines at least 14 slots per pole.
199. Interpoles. Interpoles are small poles placed midway
between the main poles of either motors or generators. They are
magnetized by a winding connected in series with the armature
and carrying the load current. Fig. 224 shows an interpole
motor or generator. The brushes are fixed on the no-load
neutral points. The interpoles have the same effect as moving
the brushes since they move the poles magnetically.
The m.m.f. of the interpole winding must oppose the m.m.f.
of the armature and must be strong enough to overbalance it
and produce a field under the interpole of the proper intensity to
reverse the current in the short-circuited coil. Since the inter-
pole winding is in series with the armature the commutating field
increases with load and satisfactory commutation up to and
238 ELECTRICAL ENGINEERING
beyond the overload limits of output set by armature heating
can be obtained.
The interpole provides a much more definite means of con-
trolling the commutating e.m.f. than is the case where reliance
is placed on shifting the brushes. As the output, the speed or the
voltage of direct-cunent machines is increased the reactance
voltage is increased and a limit is reached beyond which satis-
factory commutation can no longer be obtained economically by
shifting the brushes and interpoles must be supplied. They are
required for all large reversible motors, where the brushes must
be fixed on the no-load neutral and where therefore no commutat-
ing e.m.f. can be obtained without them. They are also required
FIG. 224. Interpole generator or motor.
for adjustable-speed shunt motors where a large speed range is
to be obtained by field control. With interpoles a speed range of
3 to 1 or even 4 or 5 to 1 is possible.
Interpoles are not so necessary on generators as on motors but
they are useful where very wide variations of load, are met with
and especially if the variations are rapid. The interpole m.m.f.
follows exactly the fluctuations of the armature m.m.f. but if the
changes of load are very rapid, it is difficult for the interpole flux
to build up rapidly enough to give perfect commutation, since
this flux has to pass through the solid yoke and eddy currents
are induced which tend to retard it.
If the interpole is provided with a shunt, it is necessary to
have the resistance of the shunt so adjusted that the interpole
winding will receive its proper proportion of the current, but in
addition the shunt must be designed with an inductance propor-
tional to the inductance of the interpole winding to insure the
proper division of the current when the load is fluctuating
rapidly.
DIRECT-CURRENT MACHINERY 239
Interpoles must be designed with very low flux densities at
normal load, so that, under heavy overloads, they will not be-
come saturated and will be able to carry the flux required to
produce the increased commutating e.m.f. Interpoles should
be provided on all machines where the reactance voltage is over
3.5 or 4 volts and with them satisfactory commutation can be
obtained with reactance voltages up to 15 volts. The interpole
shunt must, however, be very carefully adjusted or the difference
between the commutating voltage and the reactance voltage
may be so large that serious circulating currents will be
produced.
When interpoles are provided the ratio of the field ampere-
turns per pole for the gap and teeth to the armature ampere-
turns per pole may be as low as 0.8 and it is no longer necessary
to limit the armature strength to such a low value as that given
for non-interpole machines on page 261.
The interpole m.m.f. wipes out the armature flux in the com-
mutating zone but does not prevent the distortion of the flux
under the poles (Fig. 224). This distortion is more serious be-
cause of the increased armature strength and is liable to produce
flashing due to the generation of excessive voltages between
adjacent commutator bars (see Art. 200).
200. Flashing. When an arc passes over the surface of the
commutator from a positive to a negative brush, the machine is
said to "flashover." A severe flash is equivalent to a dead
short-circuit and may have serious effects.
Flashing is usually due to a combination of a high voltage
between adjacent commutator bars with a partial short-circuit
between bars due to the collection of carbon or graphite dust on
the mica insulation.
The causes of flashing are distinct from the causes producing
sparking at the brushes but a flash may sometimes be started
by a spark originating under the brush due to poor commutation.
Fig. 225 shows the flux distribution in the air gap at no load
and at full load. At no load the flux density under the pole
face is uniform and the voltage generated between bars is com-
paratively low; under load the armature m.m.f. causes the flux
density to increase at one pole tip ti and to decrease at the other
tip t. The voltage between bars at t\ will be very much increased
and, if the commutator is dirty, it may cause a flash locally be-
tween bars and this flash may be carried around in the direction
240 ELECTRICAL ENGINEERING
of rotation and result in a complete short-circuit between brushes
either momentary or permanent depending on conditions.
When the load comes on a motor very suddenly the armature
m.m.f. causes a part of the flux to swing quickly across from
t to ti and, as this movement is opposite to the direction of rota-
tion, the flux is cut more rapidly than under ordinary conditions
and the tendency to flash is increased. In the case of a generator
the swing of flux is in the direction of rotation and the voltages
are not so high.
If the load is suddenly removed from a generator the swing of
flux is opposite to the direction of rotation and the voltage be-
tween bars may be very high.
Full Load Field Form
No-Load Field Form
FIG. 225.
When the magnetic circuit is highly saturated and the machine
has what is called a stiff field, the flux peak at ti is not so high and
there is less danger of flashing.
A motor operating at high speed due to field weakening is
peculiarly liable- to flash; the reduced saturation allows the arma-
ture m.m.f. to produce a high flux peak while the increased speed
counterbalances the decreased main flux; and at the same time
commutating conditions are bad.
Interpole machines are more likely to flash than those without
interpoles because of their increased armature strength. In
non-interpole machines the field m.m.f. per pole is from 30 to
40 per cent, greater than the armature m.m.f. per pole while in
interpole machines the field m.m.f. per pole is approximately
equal to the armature m.m.f. and may even be less. Commuta-
tion is taken care of by the interpole flux but the armature m.m.f.
which is all cross-magnetizing, since the brushes are on the no-
load neutral, produces a large distortion of the flux under the
poles due to the relatively great armature strength.
DIRECT-CURRENT MACHINERY 241
If load is suddenly applied to an interpole motor the flux swings
rapidly across the pole and piles up at ti and induces a high volt-
age between bars ; at the same time sparking at the brushes may
occur due to the fact that the interpole commutating flux cannot
build up as fast as the armature current since the interpole field
winding has a large inductance and the solid yoke opposes the
required change of flux by the production of eddy currents. A
flash may therefore result. Further, if the interpole shunt is not
properly designed it may happen that the rapidly changing cur-
rent will follow the path of low inductance through the shunt
rather than the proper path through the winding. Interpoles
should be designed with inductive shunts to take care of such
currents.
In the case of an interpole generator flashing is more severe
when load is removed suddenly as the swing of flux is then in the
direction to produce the greatest voltage between bars; at the
same time the interpole commutating field cannot disappear
immediately and sparking occurs at the brushes due to over-
commutation of the current.
Fall Load Field Form
No-Load Field Form
FIG. 226. Compensating winding.
Where extreme variations of load are liable to occur with great
rapidity it is advisable to supply the machines with compensating
windings placed in slots in the pole faces, Fig. 226, to counteract
the armature m.m.f. over the whole armature surface and so
prevent any distortion of the flux.
The maximum value of volts per bar under normal operating
conditions should not exceed 30 volts for moderate capacities
or 28 volts for large capacities. The average volts per bar will
be about half of the maximum value and should not exceed about
15 volts. For small machines operating under steady load higher
values may be used but the smaller values are preferable if they
16
242
ELECTRICAL ENGINEERING
can be obtained without too great a sacrifice, especially when the
machine may experience sudden changes of load.
201. Compensating Windings. In cases where the average
voltage between bars is already so high that it is necessary to
limit the field distortion to prevent flashing, compensating wind-
ings should be provided. They are windings placed in slots in
the pole faces, which exert a m.m.f. equal and opposite to the
armature m.m.f. and so prevent any distortion of the field. The
compensating winding carries the load current and may there-
fore be designed with a smaller number of turns than the arma-
ture. Interpoles are required in addition to provide the corn-
mutating e.m.f . ; but since the armature m.m.f. is already neutral-
ized by the compensating winding, the m.m.f. of the interpole
winding is only that required to produce the flux necessary for
commutation, through the magnetic circuit of the interpole.
Fig. 226 shows a machine provided with interpoles and a com-
pensating winding.
Generators with compensating windings require only , weak
series fields to maintain constant terminal voltage since the
armature reaction is neutralized. The resistance drop is in-
creased to a slight extent.
202. Parallel Operation. In power houses in which the load
varies at different hours of the day a number of generators are
usually installed. When the
load is light one generator is
operated and supplies the
demand and when the load
increases a second machine is
started up and connected in
parallel with the first and its
excitation is adjusted until it
takes its proper share of the
load. Fig. 227 shows two
shunt generators (1) supplying
power and (2) ready to be connected in parallel with it. Before
closing the switches Si and Sz which connect the second machine
to the load it is necessary that its polarity be correct and that its
terminal voltage be the same or a little higher than that of (1).
If the field rheostat of (2) is so adjusted that the voltage of (2)
is the same as the voltage of (1) and switch Si is closed, then, if
there is no voltage across $2, it may be closed. But if the voltage
8.
IjmjOQOO J UlflJLfiJLfiJL
.FiG. 227. Parallel operation of shunt
generators.
DIRECT-CURRENT MACHINERY
243
across S% is found to be about double the terminal voltage, the
polarity of (2) must be reversed before closing switch $2- After
closing 2 the field rheostat of (2) must be adjusted until (2) takes
its proper share of the load.
If the voltage of (2) is the same as the voltage of (1) when
the switch is closed, (2) will not take any load but will run idle.
If the voltage of (2) is less than the voltage of (1), machine (2)
will run as a motor driving its prime mover and will draw power
from (1). If, however, the terminal voltage of (2) is higher than
that of (1), machine (2) will supply part of the load and will
relieve (1) until the voltages of the two become the same. Fig.
228 represents the voltage characteristics of the two machines
plotted on the same base. If the terminal voltage is E, (1) sup-
plies a current Ji and (2) a current 7 2 and the total current
d
^
==:
" -.
(1)
^
' !
(2)
C
-^.^
(3)
--. ^^
To-
~"-<
^"
^^
T
j j
i+T
Amperes
(1) Generator (1)
(2) Generator (2')
(3) Station
FIG. 228.
Amperes
(1) Generator (1)
(2) Generator (2)
(3) Station
Fia. 229.
supplied by the station is / = 7i + / 2 . The machine with the
flatter characteristic will supply the greater amount of power.
If the two machines are rated at the same current output, (2)
can be made to take its share of the load by cutting out resist-
ance from its field rheostat and so raising its voltage character-
istic and inserting resistance in the field circuit of (1) and lowering
its characteristic as shown in Fig. 229.
Shunt generators will operate in parallel and divide up the
load in proportion to their capacities if their voltage character-
istics are similar, that is, if their terminal voltage falls from no
load to full load by the same amount and in the same manner.
If the characteristics are different a proper division of load can be
obtained by regulating the field rheostats.
203. Parallel Operation of Compound Generators. Fig. 230
shows two compound-wound generator's connected in parallel and
244
ELECTRICAL ENGINEERING
supplying power to a load circuit. Their voltage characteristics
are shown in Fig. 231. Assume that the prime mover of (1) runs
for an instant at a slightly increased speed; the voltage of (1)
rises and it takes more than its share of the load; the voltage of
(2) falls because its load is decreased and its series excitation is
decreased. Machine (1) therefore takes more of the total load
and its voltage rises higher until it supplies all the load and in
FIG. 230.
addition drives (2) as a motor. Since the current in (2) is re-
versed the m.m.f . of its series winding is also reversed and it runs
as a differential motor driving its prime mover at a high speed
until the load on (1) becomes so great that the protective appa-
ratus opens the circuit and shuts down the system.
To get over this difficulty the equalizer connection ee, Fig. 232,
is used. It is a conductor of low resistance connecting the series
windings of the two machines in multiple. Now if the prime mover
of (1) runs above normal speed
the voltage of (1) rises and it
takes an increased load. The
increase of current does not
all go through the series wind-
ing of (1) but divides between
the windings of (1) and (2) in
inverse proportion to their re-
sistances and so prevents any
decrease of the voltage of (2) .
Thus with an equalizer con-
nection (2) will still hold its
load. The resistances of the series windings must be adjusted so
that the load current will divide between them in such propor-
tion that each machine will supply its proper share of the load.
Before connecting machine (2) in parallel with (1) which is
delivering power, first close switches >S 2 and /S 3 , Fig. 232, and
Si
JULfiJL&JUL
e
m
^3
Equalizer Connection
. .
>
j.ooojm.
w
FIG. 232. Parallel operation of com-
pound generators.
DIRECT-CURRENT MACHINERY 245
adjust the shunt field of (2) until its terminal voltage is the same
as that of (1). The excitation of (2) is now provided partly by
its shunt field and partly by its series field carrying part of the
load current. After checking the polarity to see that it is correct
close switch Si and adjust the shunt field of (2) until the machines
divide the load in proportion to their capacities.
From the above discussion it is seen that two compound-
wound generators connected in parallel form an unstable system
unless an equalizer connection is placed between their series
windings.
204. Storage Batteries. A storage battery is an apparatus
in which electrical energy can be stored to be used at some later
time.
Batteries are made up of a number of cells connected in series
multiple according to the voltage and current required.
Each cell is composed of two plates or electrodes of suitable
materials immersed in an electrolyte. The most commonly used
storage battery has a positive plate of lead peroxide Pb0 2 and a
negative plate of sponge lead Pb immersed in dilute sulphuric acid
H 2 SO 4 .
When the battery is discharging the electrolyte combines with
the active materials of the electrodes and when it is being charged
the electrodes are reduced to their original condition and the
materials taken from the electrolyte are returned to it.
The main chemical changes taking place are represented by the
following formula:
charge
PbO 2 + Pb + 2H 2 SO 4 = 2PbS0 4 + 2H 2 O. (269)
discharge
Capacity.- The unit of capacity of a storage cell is the ampere-
hour and it is generally based on the 8-hr, discharge rate. An
800-amp.-hr. battery will give a continuous discharge of 100
amp. for 8 hr. If, however, the rate of discharge is increased the
ampere-hour capacity of the battery decreases. At a 6-hr, dis-
charge rate the capacity is only about 95 per cent., at a 4-hr,
discharge rate it is^about 80 per cent, and at a 1-hr, rate it is only
50 per cent, of its 8-hr, rating. Thus the battery mentioned above
would give a continuous discharge of 400 amp. for only 1 hr.
246
ELECTRICAL ENGINEERING
The capacity of a cell is proportional to the area of the plates
exposed to the electrolyte and for an 8-hr, discharge rate a
current density of from 40 to 60 amp. per square foot of positive
plate is common practice.
Voltage. The voltage of a cell depends on the character of the
electrodes, the density of the electrolyte and the condition of the
cell but is independent of the size. The variation of the terminal
voltage of a cell during charge and discharge is shown by the
curves in Fig. 233. On charge the voltage begins about 2 volts
and rises to 2.5 volts or a little above. When the charging cir-
cuit is opened the voltage falls to 2.1 volts and during discharge
falls off gradually to about 1.9 volts. Beyond this point the fall
of voltage is very rapid and discharge should not be continued
after the voltage has fallen to 1.7 volts.
-==- Battery
EM: Variable Load
FIG. 233. Voltage characteristics of a FIG. 234. Battery with a shunt-
storage cell. generator.
The required battery voltage is obtained by connecting a num-
ber of cells in series and the required current is obtained by con-
necting a number of plates or cells in multiple.
205. Applications. Batteries are installed in direct-current
power stations to store energy during periods of light load and to
deliver energy in parallel with the generators during periods of
heavy load. When the load is light the generators charge the
battery and when the load is heavy the charge is given up and so
the load on the generators is maintained nearly constant and they
can be operated at maximum efficiency. The result is that the
voltage regulation of the system is improved.
Batteries are also installed in electric-railway substations to
prevent large variations of the load on the feeders supplying
them and so regulate the substation voltage.
A third very important application of storage batteries is in la-
DIRECT-CURRENT MACHINERY
247
ternating-current power stations where they provide an auxiliary
supply of direct current in case of a breakdown of the exciters and
may thus prevent a shutdown of the whole system.
Fig. 234 shows a battery connected across the terminals of a
shunt generator. At normal load the battery voltage and the
generator voltage are equal and the battery floats on the line
neither giving nor receiving power. If, however, the load in-
creases the generator voltage falls and the battery discharges and
supplies part of the extra load and so relieves the generator and
prevents any large drop in its terminal voltage. The battery in
this way takes care both of sudden overloads and continuous
overloads. During periods of light load the generator voltage
rises above the battery voltage and the battery charges.
206. Boosters. Boosters are direct-current generators con-
nected in series with the line to raise or lower the voltage. They
are low-voltage machines of large current capacity and are
usually driven at constant speed by shunt motors.
-=-Battery
Load
>
*
Booster
FIG. 235. Shunt booster.
Load,
FIG. 236. Compound booster.
Boosters are used to compensate for line drop in distributing
systems by adding an equal voltage to the circuit and they are
also used very extensively to regulate the charge and discharge
of storage batteries in parallel with the generators in constant-
voltage systems. They may be either shunt, series, or compound-
wound.
The shunt booster, Fig. 235, is an ordinary generator with its
field connected across the station busbars and its armature in
series with the generator armature. Its function is to raise the
voltage impressed on the battery in order to send current into
it to charge it. The booster voltage is controlled by a field
rheostat.
Compound boosters are automatic in their action and are
divided into two classes, non-reversible and reversible, depending
on the relative strengths of their shunt and series windings.
248
ELECTRICAL ENGINEERING
Fig. 236 shows a non-reversible automatic booster. The
shunt field / is connected across the station busbars; the series
field s carries the load current of the generator and it opposes the
shunt field but has at all times a smaller m.m.f. and thus the
booster voltage is always in the direction of the generator
voltage.
When the load current increases, the increase of current
through the series field decreases the booster voltage and allows
the battery to discharge; when the load decreases, the booster
voltage rises and causes the. battery to charge. The current
from the generator remains practically constant regardless of the
fluctuations of the load on the system.
The reversible booster is similar in construction to the non-
reversible booster but has a stronger series field. At normal load
the shunt and series fields are of equal strength and the booster
voltage is zero. The battery then floats on the line and neither
charges nor discharges. An increase of load above normal
increases the strength of s and overpowers / and so discharges
the battery. When the load decreases s becomes weaker than
/ and the booster causes the battery to charge.
Booster
FIG. 237. Series booster.
Insulated Negative Feeder
Negative Booster
FIG. 238. Negative booster.
Thus by installing an automatic booster the battery is made
more sensitive to variations of load and a better regulation of the
generator load and voltage is obtained.
If the load is a combined lighting and power load the lighting
circuit can be supplied at constant voltage from the generator
terminals and the power circuit connected outside the battery.
Sudden variations of load will be cared for by the battery and the
voltage on the lighting circuit will not be affected.
Series boosters are used in electric-railway engineering and in
general power distribution to raise the voltage on certain sections
of the line as in the case of a long feeder supplying power to an
outlying section, as shown in Fig. 237. The booster field is
DIRECT-CURRENT MACHINERY 249
connected in series with the line and its voltage increases with
the load and so neutralizes the line drop.
In cases where insulated negative feeders are connected to the
track return at various points to carry the current back to the
substations and so reduce the danger of electrolysis, a booster
connected in series with the negative feeder can be made to
reduce its effective resistance to a very low value and so increase
its capacity. The use of such negative boosters, Fig. 238, is
very common and results in a large saving in conductor material
in the return feeders.
207. Balancers. A direct-current compensator or balancer
comprises two or more similar direct-current machines directly
coupled to each other and connected in series across the outer
conductors of a multiple-wire system of distribution for the pur-
pose of maintaining the potentials of the intermediate wires.
They may be either shunt-wound or compound-wound.
FIG. 239. Balancers.
The generator G, in Fig. 239 (a), develops a voltage of 230
volts between the outer wires and the function of the balancer
set BiBz is to maintain the potential of the neutral wire midway
between the outers. When the loads on the two sides are bal-
anced, no current flows in the neutral wire and the two machines
run light as motors; when the load on one side is heavier than on
the other, as shown, the voltage across BI is lower than that
across B 2 and the neutral point is shifted; the machine B 2 runs
as a motor driving B\ as a generator and partially restores the
balance.
The connection shown in Fig. 239(6) gives better results.
The field of the motor B 2 is excited from the heavily loaded side
and it therefore tends to run at an increased speed, while the field
of BI is excited from the lightly loaded side and its voltage there-
fore rises due to increase in speed and increase in excitation.
The balance cannot, however, be perfect since a slight inequality
of voltages is necessary to make the balancer act.
250
ELECTRICAL ENGINEERING
Where perfect balance is required the machines BI and B 2
must be compound-wound and connected as shown in Fig. 239(c).
The current in the neutral wire flows through the series windings
of the two machines, strengthening the field of BI and raising the
voltage and reducing the field of B 2 and increasing the speed of
the set. The automatic balancing action in this case depends
on an unbalance of current in the two outers and not on an
unbalance of voltage between the outers and neutral as in (a)
and (6).
208. Rosenberg Generator for Train Lighting. For train
lighting it is necessary to have a generator which will give
approximately constant current or voltage independent of the
speed. The Rosenberg generator shown diagrammatically in
Fig. 240 gives a constant current at all speeds above a certain
minimum.
A A is a battery which supplies the lighting system when the
train is at rest or running at slow speeds. The field winding //
is excited by the battery and
its m.m.f. Mf remains con-
stant. The main brushes
which supply the load are
placed under the centers of
the poles and are connected
to the battery terminals and
to the load, brush B% directly
and brush BI through the
aluminum cell C, which has
the property of allowing cur-
rent to flow from the genera-
tor to the battery or load but
offers a high resistance to the
flow of current in the opposite
direction. The auxiliary brushes bb are placed on the commuta-
tor at right angles to the main brushes and are short-circuited.
As the armature rotates its conductors cut the flux produced
by the field m.m.f. M f and a voltage is generated between the
brushes bb but none between the main brushes BiB 2 . A short-
circuit current flows through the armature winding and the
brushes bb and exerts a m.m.f. M b at right angles to the field
m.m.f. This m.m.f. produces a comparatively large flux.
06 through the armature and pole faces. This cross-flux is
FIG. 240. Rosenberg train-lighting
generator.
DIRECT-CURRENT MACHINERY
251
cut by the armature conductors and a voltage is generated be-
tween the brushes BiB 2 and a current / flows to the load. Cur-
rent / flowing in the armature conductors exerts a m.m.f. M B
opposing the field m.m.f. M/ and reducing it to a comparatively
small value.
Above a certain speed, which depends on the field excitation,
the current remains approximately constant independent of the
speed.
The load current is limited by the fact that the m.m.f. M B
must always be less than M f in order that the m.m.f. M b and the
cross-flux 06 may exist.
If the direction of rotation changes, the direction of the e.m.f.
between bb changes, the directions of Mb and <& change and
therefore the direction of the voltage from B% to BI remains as
before.
Notches are cut in the pole faces above the brushes to prevent
large voltages being generated in the coils which are being
commutated.
The current output may be varied by varying the excitation.
.Field Winding
( Ct) Kadial Type (&) Axial Type
FIG. 241. Homopolar generators.
209. Homopolar Generators. Homopolar or acyclic genera-
tors are designed with the armature conductors revolving in a
unidirectional field and thus the e.m.fs. generated in them do not
alternate during the revolution. Two types have been de-
veloped, the radial and the axial, so-called from the direction of
flow of the armature currents.
In the radial type, Fig. 241 (a), the armature consists of one
or more discs rotating in the magnetic field and an e.m.f. is
generated between the center and the edge of the disc. Current
is collected by brushes rubbing on the external periphery of the
discs.
252 ELECTRICAL ENGINEERING
The e.m.fs. generated are very low and a number of discs, each
supplied with brushes, must be connected in series making a very
complex and expensive construction.
In the axial type, Fig. 241(6), the discs are replaced by one or
more copper cylinders with brushes at each end. The direction
of the armature current is along the axis of the machine. Here
again the e.m.f. is low and it is difficult to arrange enough cylin-
ders in series to give the required e.m.f.
Such machines can be used where very large currents at low
voltage are required as in electrolytic processes.
210. Limits of Output of Electric Machines. -The factors
which limit the output of electric machines are:
1. Regulation.
2. Efficiency.
3. Heating.
4. Commutation.
1. In motors the regulation is a speed regulation. With
increase of load the speed falls off and the increased torque is
obtained at a decreased speed. In constant-speed work the
shunt motor is used but if it is overloaded its speed falls to a
value too low for satisfactory operation.
In generators the regulation is a voltage regulation. As the
load is increased the voltage falls off and a point is finally
reached where the voltage is so low that the power supplied is
unsatisfactory.
2. The efficiency of a machine increases with increasing load
to the point where the variable copper losses are equal to the con-
stant losses. Above this point the efficiency decreases due to the
rapid increase of the variable losses.
With properly designed machines the output is limited by either
heating or commutation beiore the regulation or efficiency becomes
too poor.
3. All the losses of power in a machine are converted into heat
and raise the temperature of the various parts until the point is
reached where the rate at which heat is being radiated or carried
off by the ventilating apparatus is equal to the rate at which heat
is being generated. The temperature will then remain constant.
When a machine is overloaded its losses increase and consequently
its temperature rises above normal.
DIRECT-CURRENT MACHINERY 253
If a machine operates at a high temperature for any
length of time permanent injury to the insulating materials
will result.
4. Sparking will occur in a machine when the field cut by the
coil which is being commutated is not strong enough to reverse
the current in the time of commutation. Sparking will therefore
occur in generators or motors at heavy load when the armature
m.m.f. is so great that it wipes out the field under the pole tip or
weakens it to such an extent that it cannot produce the required
commutating e.m.f. Motors will also spark at high speed since
the time of commutation is reduced, especially when the high
speed is produced by field weakening.
Take for example a shunt motor rated at normal speed as
10 hp., 110 volts, 80 amp., 400 r.p.m. and suppose the temperature
rise to be 50C. If the motor is operated at half speed of 200
r.p.m. by reducing the impressed voltage to half, the rating may
be taken as 5 hp., 55 volts, 80 amp., but the temperature rise
will be greater than before because the armature copper loss is the
same, the field copper loss is the same and the iron and friction
losses are less due to the low speed, but the ventilation is only
about half as good as before.
When operated at twice full speed, produced by field weaken-
ing, the rating may be taken as 10 hp., 110 volts, 80 amp.,
but the temperature rise will be less than before, because the
armature copper loss is the same, the field copper loss is reduced
to about one-quarter of its normal value and the iron and friction
losses are increased, but the ventilation is very much improved.
The rated output of the machine for normal temperature rise
might be increased but due to the higher speed and consequent
reduced time of commutation sparking would occur.
211. Temperature Limits (Standardization Rules A. I. E. E.
1914). The capacity of a machine is limited by the maximum
temperature at which the materials in the machine, especially
those employed for insulation, may be operated for long periods
without deterioration. When the safe limits are exceeded,
deterioration is rapid. The insulating materials become per-
manently damaged by excessive temperature, the damage
increasing with the length of time that the excessive temperature
is maintained and with the amount of excess temperature until
finally the insulation breaks down.
The actual temperature attained in the different parts of a
254 ELECTRICAL ENGINEERING
machine and not the rises in temperature affect the life of the
insulation of the machine. The safe operating temperatures of
the various parts of a machine are often expressed in terms of the
allowable temperature rise above the temperature of the sur-
rounding air. The temperature of this air is called the ambient
temperature and 40C. is taken as its standard value. The
allowable temperature rise plus 40C. gives the maximum allow-
able temperature.
A machine may be tested at any convenient ambient tem-
perature but the permissible rises of temperature must not exceed
those given in column 2 of the table on page 255.
As it is usually impossible to determine the maximum tem-
perature attained in insulated windings, it is convenient to apply
a correction to the measured temperature to cover the errors due
to the fallibility in the location of the measuring devices as well
as inherent inaccuracies in measurement and method.
The two most usual methods of measuring temperature
rises are: (1) by thermometer, and (2) by the increase of
resistance.
1. Thermometers are applied to the hottest accessible parts
of the machine. In this case the hottest spot temperature for
windings should be estimated by adding 15C. to the highest
temperature observed.
2. The measurement of the increase of temperature by increase
of resistance is only applicable to windings of comparatively
high resistance. The resistance is measured before operation
at the ambient temperature and again after operation. The
rise of temperature can then be calculated as shown in Art. 84.
A hot-spot correction of 10C. should be added.
The following table gives the limits for the hottest spot tem-
peratures of insulations. The permissible limits are indicated
in column 1 of the table. The limits of temperature rise per-
mitted under rated load conditions are given in column 2 and are
found by subtracting 40C. from the figures in column 1. What-
ever be the ambient temperature at the time of the test, the rise
of temperature observed must never exceed the limits in column
2 and the highest temperature must never exceed the limits given
in column 1.
Table of hottest spot temperatures and corresponding per-
missible temperature rises.
DIRECT-CURRENT MACHINERY
255
Class
Description of insulation
Column 1,
highest
permissible
temperature
for the hot
spot,
degrees C.
Column 2,
highest
permissible
temperature
rise for the
hottest spot
above 40C.
A,
A,
B
C
Cotton, silk, paper and other fibrous materials
not so treated as to increase the thermal
limit.
95
105
125
No limit i
55
65
85
s specified
Similar to AI but treated or impregnated or
permanently immersed in oil and including
enameled wire. . . . ...
Mica, asbestos or other material capable of re-
sisting high temperatures in which any class
A material or binder, if used, is for structural
purposes only and may be destroyed without
impairing the insulating or mechanical
qualities. .
Fireproof and refractory materials as mica,
porcelain etc
212. Temperature of Commutators. The observable tempera-
ture shall in no case be permitted to exceed the values given in
the table above for the insulation employed, either in the com-
mutator or in any insulation whose temperature would be affected
by the heat of the commutator. For commutators so constructed
that no difficulties from expansion can occur, the following tem-
perature limits have been suggested.
Current per brush arm
200 amp. or less
200 to 900 amp.
900 amp. and over
Maximum permissible temperature
130C.
130C. less 5 for each 100 amp. in-
crease above 200
95C.
213. Temperature of Cores. The temperature of those parts
of the iron core in contact with insulating materials must not
exceed the limits of temperature and temperature rise permitted
for those materials.
214. Temperature of Other Parts. Other parts (such as
brush holders, brushes, bearings, pole tips, cores, etc.), whose
temperature does not affect the temperature of the insulating
material, may be operated at such temperatures as shall not be
injurious in any respect. But no part of continuous duty machin-
256
ELECTRICAL ENGINEERING
ery subject to handling in operation, such as brush rigging, shall
have a temperature in excess of 100C.
215. Ventilation. The increase in output per pound of active
material in modern machines has been largely due to improved
methods of ventilation.
Inlets of sufficient size and proper location are provided; the
cool air is drawn in either by natural suction or by fans placed on
the rotating member; and the heated air is expelled through
outlets in such a direction that it will be thrown completely
away from the machine.
The air ducts through the active material are of two kinds,
radial ducts, Fig. 242, and axial ducts, Fig. 243. They must be
of sufficient section to carry the required amount of air and must
have sufficient surface to allow the heat to pass from the copper
and iron to the air.
J _ I.
FIG. 242 Axial ducts.
FIG 243. Radial ducts.
About 100 cu. ft. of air per minute per kilowatt lost should be
provided.
Care must be taken that the end bells or overhanging frame
do not deflect the hot air back into the machine.
A generator when driven by a belt may run cool due to windage
from the belt, whereas, if it is direct-connected to a motor or
steam engine it may run hot due to poor ventilation or due to
the transfer of hot air from the driving machine.
Machines of short length and large diameter are easy to cool.
216. Semi-enclosed and Totally Enclosed Machines. When
a machine is partially enclosed it is more difficult to get rid of the
heat due to the losses and unless a more efficient system of ven-
tilation is provided the output must be decreased 15 or 20 per
cent, to keep within the allowable limits of temperature rise.
DIRECT-CURRENT MACHINERY 257
When totally enclosed the rating must be still further reduced.
For the same temperature rise a totally enclosed machine can give
about 70 per cent, of the output of an open machine if the speed
is increased about 20 per cent. The reduction of output reduces
the copper losses and the increase of speed decreases the flux per
pole and so decreases the iron losses but not in direct proportion.
17
CHAPTER VIII
DESIGN OF A DIRECT-CURRENT GENERATOR
217. Symbols. The subscripts used are c for core, g for air gap, t for
teeth, p for pole and y for yoke.
A = sectional area in square inches.
AT = ampere-turns.
AT g = ampere-turns per pole for the gap.
B = flux density in lines per square inch.
B a g = actual gap density.
Bat = actual tooth density.
B g = apparent gap density.
B t = apparent tooth density.
D a = external armature diameter.
D e = commutator diameter.
I = load current.
I c = current per conductor.
L c = axial length of core.
L g = gross iron in the frame length.
L n = net iron in the frame length.
S = number of commutator segments.
Tf = field turns per pole.
Z = total conductors on the armature.
d = slot depth.
p = number of poles.
pi = number of paths through armature.
q = ampere conductors per inch.
s = slot width.
t = tooth width at top.
5 = air-gap clearance.
17 = efficiency.
X = slot pitch.
v = leakage factor.
T = pole pitch.
= magnetic flux.
a = useful flux per pole.
^ = per cent, pole enclosure.
The meaning of other symbols used will be explained in the context.
Figs. 244, 245, 246, 247, 252, 257 are reproduced from the "Standard
Handbook," Section 8, by A. M. GRAY.
258
DESIGN OF A DIRECT-CURRENT GENERATOR 259
218. Design of Direct-current Machinery. The design of
electrical machinery is based on a number of formulae or equa-
tions which can be very easily derived and on a number of em-
pirical relations obtained from experience, which are usually
expressed in the form of curves or limiting values.
Following are the most important of these fumiamental rela-
tionships and limits. They are indicated by Roman numerals
for purposes of reference.
/. The e.m.f. equation:
generated e.m.f. = 8 = Zn^ g - 10~ 8 volts (Art. 142),
or
c
useful flux = ff = - li nes< (270)
Zn " 10~ 8
II. The output equation:
output in watts = 8/ = ( )(op)n X 10~ 8
10~ 8
60
r.p.m.) (271)
= a constant X electric loading X magnetic loading X r.p.m.
ZI C = total ampere conductors on the armature is called the
electric loading.
4> g p = total flux crossing the gaps under the poles is called the
magnetic loading.
A large electric loading requires a large amount of copper in
the machine, while a large magnetic loading requires a large
amount of iron.
The output equation can be written in a more easily applicable
form:
10~ 8
watts = --(TrD a q)(B g ^TL c p) (r.p.m.)
'QQ-vDoQ B a t~L e p X r.p.m.
60X10 8
or
r.p.m.
D a 2 L c qB g \l/ X r.p.m.
V 1H7
(272)
260
ELECTRICAL ENGINEERING
where,
D a 2 L c = volume of the active material in the armature.
B g = apparent density in the air gap.
\f/ = per cent, pole enclosure.
ZI
q = fp= ampere conductors per inch of the armature
irJJa
periphery.
///. Flux Densities. The Qux density in the air gap is limited
by the maximum allowable flux density in the roots of the teeth,
which is dependent on the frequency of the reversals of the
magnetism.
The approximate limits of the maximum tooth density are
150,000 lines per square inch for 30 cycles per second and 125,000
lines for 60 cycles. The frequency is given as
/ = ~ cycles per second.
Zi
Having fixed the density at the roots of the teeth the air-gap den-
sity will depend on the diameter of the armature.
Gap Density In
&OX10*
20 40 60 80 100
Armature Diameter In Inches = D
Ampere Conductors per Inch = Q
o i i
N
1' 10
s
^~~
^=
,---'
- '
,
.
.>-
-H
^f
X*
-Scale
B
s
j
/
8
calJ B
2000
4000
;
UO
JO 1
i
ot
(1
4C
1
)
HOj
-Ttn
I
H.
M
^
s^
.
i u
^
X
/*
'Be
ale
A
1
"/
/
s
/
/
j
/
i
a
1
tt_
J
) L-
i
/
a.i
jf
__
1
^
^
'
ti
(
Scale
A
> 20 40 60 80 10
Kilowatts
FIG. 245.
> 20 40 CO 30 10
Watts
R.P.M.
FIG. 246
FIG. 244.
Fig. 244 shows the relation between B g and D a for a frequency
of 30 cycles per second. Lower values of B g should be used for
higher frequencies.
The ordinary flux densities in other parts of the magnetic
circuit are:
Flux density, lines per square
inch, depending on the frequency
70,000-100,000
Armature core
Pole (cast steel)
Yoke (cast iron)
Yoke (cast steel)
90,000- 95,000
35,000- 40,000
70,000- 80,000
DESIGN OF A DIRECT-CURRENT GENERATOR 261
IV. Ampere Conductors per Inch. The value of q the
ampere conductors per inch of armature periphery is limited
partly by heating and partly by commutation and it depends
on the output of the machine. Fig. 245 shows the relation
between q and kilowatt output for machines in satisfactory
operation.
V. Electric Loading. For the most economical construction
there is a more or less fixed relation between the electric and the
magnetic loading.
Therefore, referring to equation (271), it is seen that the
electric loading, or the total ampere conductors on the armature,
. watts
depends on the ratio -
r.p.m.
The relation between ampere conductors, ZI C and watts per
revolution is shown in Fig. 246.
VI. Current Density. The current density used in the arma-
tures of direct-current generators varies from 2,000 to 3,000 amp.
per square inch, that is, 600 to 400 circ. mils per ampere. It is
limited by the allowable temperature rise. The value to be used
on a given machine depends on the specific electric loading q
and on the peripheral speed. Fig. 247 shows the relation
, , . ampere conductors per inch .
between the ratio - -, - ^ - and the peripheral
circular mils per ampere
speed in feet per minute for
a temperature rise of 40C.
VII. Choice of Number of
Poles. The number of field
ampere-turns per pole is usu-
ally about 50 per cent.
greater than the armature
ampere-turns per pole. If
the number of poles is too
small the required ampere-
turns per pole will be large
and the poles must be made
long to carry the windings.
The number of poles may be fixed approximately by the two
relations :
Z*4
/
9 n
&
Y
ko
I
,
7
5
B 1 JJ
(V <
/
^
/
o 10
O
\
/
\>
/^
P V
.
/
/
V
Sv
/
/
/
x
^
/
f
/
a
/
/
^
fS
ftt
$
^
'/
c
,--
2
I
^~~-
7{ o
(
1
\
'
4
1
l
a 8
Peripheral Velocity in Ft. per Mio.
FIG. 247.
(a) Ratio
Ues between L1 and L7 usuall y-
ength
(6) Armature ampere-turns per pole should not exceed 7,500.
262 ELECTRICAL ENGINEERING
VIII. Commutation. To provide a commutating field, the
. field ampere-turns per pole for the gap and teeth
armature ampere-turns per pole ~ should not
be less than 1.2. If the brushes are advanced so that part of the
armature m.m.f. is demagnetizing the field ampere-turns must
be increased by the amount of the armature demagnetizing am-
pere-turns.
IX. Reactance Voltage. The formulae for the reactance
voltage were derived in Art. 195.
E R = K X L c X ZI C X r.p.m. X n X 10~ 8 volts
where K is a constant.
For a full-pitch multiple winding K = 0.8, and for a short-
pitch multiple winding K = 0.46.
With the armature strength limited as indicated in VIII above
and the brushes advanced into a suitable commutating field,
the reactance voltage with a full-pitch winding should not ex-
ceed 1.5 to 2 vol^s for shunt machines. In compound-wound
machines values up to 3.5 volts may be taken care of in certain
cases. However, when the reactance voltage is much above 2.5
volts it may be advisable to use interpoles.
With short-pitch windings the two sides of the short-circuited
coil are not in commutating fields of equal strength and slightly
lower limits should be used.
With series windings since a number of coils are short-circuited
in series the reactance voltage calculated by the formula for the
full-pitch winding should be kept about 50 per cent, lower than
for the multiple winding.
X. Slots. The ratio -= - rVrr varies from 2.5 to 3.5.
slot width
slot width
rheratl maximum tooth width == l Approximately.
Slots per pole should be greater than 12 for small machines
and 14 for large machines.
XI. Commutator. The commutator diameter is from 60 to
75 per cent, of the armature diameter. If possible the peripheral
speed of the commutator should not exceed 3,500 ft. per minute
but values up to 5,000 may be used in special cases.
XII. Brushes. The brush arc should not cover more than
three segments and should not subtend more than one-twelfth
of the pole pitch.
DESIGN OF A DIRECT-CURRENT GENERATOR 263
The current density in the brushes depends on the kind of
brush used. For ordinary brushes from 35 to 40 amp. per square
inch may be used and the contact resistance drop may be taken
as 1 volt per brush. This gives a loss of 35 to 40 watts per square
inch of brush contact, which is satisfactory. Where brushes with
lower contact resistances are used the current density may be
higher.
The brush pressure is from IJ^j to 2J^ Ib. per square inch de-
pending on the service.
The coefficient of friction between the brush and the commuta-
tor may be taken as 0.3 for ordinary hard carbon brushes.
219. Magnetic Leakage. Since there is no material through
which magnetic flux cannot pass, it is not possible to confine all
the flux produced in a generator to the magnetic circuit. In
practice the main circuit is made of such low reluctance that only
a small portion of the flux leaves it.
FIG. 248. Leakage flux about a
bipolar dynamo.
FIG. 249. Leakage flux about a
multipolar dynamo.
Figs. 248 and 249 show the leakage fluxes about the magnetic
circuits of a bipolar and multipolar generator.
The principal part of the leakage occurs between the pole tips
because the m.m.f. consumed between these points is from 60 to
80 per cent, of the total m.m.f. ; it includes the m.m.f. required to
drive the flux across the two gaps and through the teeth and
armature core. As a result the flux passing through the field
poles and yokes is greater than the flux crossing the gap into the
armature by an amount depending both on the mechanical con-
struction of the machine and on the load.
The dispersion coefficient or leakage factor is the ratio of the
flux through the field poles to the flux crossing the gap into the
armature, that is, the ratio of the total flux to the useful flux;
thus the leakage factor is
< P le
264 ';;.; ELECTRICAL ENGINEERING
In preliminary calculations the following values of leakage
factor may be used :
Four-pole machines up to 10 in. armature diameter ........ 1 .25
Multipolar machines from 10 to 30 in. armature diameter. . . 1.2
Multipolar machines from 30 to 60 in. armature diameter. . . 1 . 18
Multipolar machines over 60 in. armature diameter ........ 1 . 15
Under load the armature exerts a m.m.f. which in part is
demagnetizing and opposes the field m.m.f. and in part is cross-
magnetizing and increases the reluctance of the magnetic- circuit
(Art. 145). Thus under load a greater proportion of the field
m.m.f. is required for the air gaps, teeth and armature than at no
load and the leakage flux is increased in proportion. The leakage
factor is therefore greater under load than at no load.
In the case of the flat-compound generator, designed in this
chapter armature reaction consumes a component of the full-
load field m.m.f. = 2,117 ampere-turns per pole. At no load
the m.m.f. consumed in the gap, the teeth and the armature core
is 5,928 ampere-turns per pole and the leakage factor is taken as
1.18. At full load the leakage factor is increased to
220. Design of a Direct-current Generator. Determine the
dimensions, and characteristics of a generator of the following
rating: 250 kw., 250 volts, 400 r.p.m. (Fig. 250).
Armature design.
Watts 250,000
_
Rrn ~ 400
Ampere conductors = ZI C = 78,000, Fig. 246.
Ampere conductors per inch = q = 690, Fig. 245.
ZI C 78,000
Armature circumference = irDa = = Ann = llo m.
5
I -I O
Armature diameter = D a = - = 36 in.
Apparent gap density = B g = 55,000, Fig. 244.
Pole enclosure = \f/ = 0.7, assumed.
watts 60.8 X 10 7
D * L * = x equation (272) -
250,000 60.8 X 10 7 _ v _1 __ 1t .
Frame length = L c = 400 X 55j0 00 X 0.7 X 690 X (36) 2 ~
Number of poles = p = 8, assumed.
7 T 78 000
Armature ampere-turns per pole = -g- 5 = 2x8 = 4,860.
DESIGN OF A DIRECT-CURRENT GENERATOR 265
TrD a IT X36
Pole pitch = r = = g
14.13 in.
Pole pitch
14.13
11
1.3.
Frame length L c
Area of the air gaps = A g = ^rL c = 0.7 X 14.13 X 11 = 109 sq. in.
Useful flux per pole = < = B g A g = 55,000 X 109 = 6 X 10 6 lines.
Winding assumed to be multiple, short pitch, with one turn per coil.
Reactance voltage = E R = 0.46 X 10" 8 X L c X ZI e X r.p.m. X n,
equation (268) = 0.46 X 10~ 8 X 11 X 78,000 X 400 X 1 = 1.58. With a
full-pitch winding it would have been 2.7 volts.
FIG. 250. Magnetic circuit.
Number of path in multiple between terminals = pi =8.
Total conductors on the armature =
7 S_ 250 X 10 8
Z = = 77^ o = 624.
Conductors per slot = 4, assumed.
Number of slots = -7- = 156.
156
Slots per pole = - = 19.5.
o
Peripheral velocity of the armature in feet per minute =
36
TT X X 400 = 3,800.
Ampere conductors per inch
Circular mils per ampere for 4 C ' rlse = L4 ' Fl ' 247 '
Circular mils per ampere = ^-7 = y-j = 494.
266
ELECTRICAL ENGINEERING
Total load current
watts
volts
61,750 circ. mils = 0.0485 sq. in.
~ r / 1,000
Current per conductor = l c = = ^ = 125 amp.
Section of conductor = 125 X 494
Slot P itch - * - numbefo'f slots ' ^ " ' 725 in '
Slot width = 0.5 X slot pitch, assumed = 0.5 X 0.725 = 0.3625.
Conductor dimensions = width X depth = x X y = 0.0485 sq. in.
Thickness of insulation in width = 0.112 in. (Fig. 251.)
Allowance for clearance = 0.040 in.
}.375->|
nductor O.l * 0.5
Half Lapped Tape 0.012
Half Lapped Tape 0.012
Empire Cloth 0.010
Papw 0.010
FIG. 251. Details of slot and insulation.
Width of two conductors = 2z = 0.3625 - 0.152 = 0.2105.
Width of conductor = x = 0.105 =0.1 in.
Slot width =s = 2X0. 1+0. 112+ 0.040 = 0.352 = 0.35 in.
Depth of conductor = y = ' ^ = 0.485 = 0.5 in.
Circular mils per ampere = 494 X - = 510.
0.0485
4 10 6
Amperes per sq. inch = - X^rx = 2,500.
TT OlU
Thickness of insulation in slot depth = 0.216 in.
Thickness of wedge = 0.2.
Depth of two conductors = 1.0.
Allowance for clearance = 0.040.
Depth of slot = 1.456 = 1.45 in.
Maximum tooth width = t = \ s = 0.725 0.35 = 0.375 in.
Diameter at bottom of slots = (36 - 2 X 1.45) = 33.10 in.
vx 00 -I Q
Minimum tooth width = *' 0.35 = 0.67 - 0.35 = 0.32 in.
Tooth taper = * = ^
x.
faun.
156
0.375
0.32
1.17.
Number of vent ducts in the core = 3 - = 6 X 10 6 lines.
6 X 10 6
Apparent minimum flux density in the teeth = ^ = 131,000 lines
per square inch, Art. 221.
6 X 10 6
Apparent maximum flux density in the teeth = = 152,000 lines
per square inch.
, p X n 8 X 400
Frequency of the reversals of magnetism = / = ^ " = 2 X 60 =
27 cycles per second.
Flux density in the core = 85,000 lines per square inch, assumed.
Flux in the core = c = | ff = 6 \ ^ = 3 X 10 6 .
3 X 10 6
Core area = - oe Ann = 35.4 sq. in.
OO,UUU
core area 35.4
Depth of iron below slots = y -- = Q-Q^ = 3.92 in.
Internal diameter of armature = J36 - 2(1.45 + 3.92)} = 25.16 = 25 in.
QC _ 25
Depth of iron below slots = - ^ --- 1.45 = 4.05 in.
Core area = A c = 4.05 X 9.05 = 36.7 sq. in.
6 X 10 6
Flux density in the core = B c = ^ Q A n = 82 >000 line s per square inch.
L /\ oO. I
221. Flux in the Air Spaces between the Teeth. If the flux
density in the teeth is above 100,000 lines per square inch it is
necessary to take account of the fact that the path through the
teeth is paralleled by an air path consisting of the slots, the vent
ducts and the insulation between punehings. This path has
usually a much larger section than the path through the teeth
and consequently at high densities where the permeability of the
iron is low it will carry a considerable part of the flux.
In the case of the generator designed:
Maximum tooth area per pole = 46 sq. in.
Gap area = 109 sq. in.
Area of path through slots and vent ducts = 63 sq. in.
Minimum tooth density = 131,000 lines per square inch.
Ampere-turns per inch for this density = 800.
Flux density in the parallel air path = 800 X 3.2.
Flux in the parallel path = 800 X 3.2 X 63 = 160,000.
The average ampere-turns per inch for the teeth will be greater than 800
due to the tooth taper (Art. 222) and it will be better to assume that the
268
ELECTRICAL ENGINEERING
flux carried by the air path is somewhat greater than 160,000, assume
200,000, then flux in the teeth = 6 X 10 6 - 200,000 = 5.8 X 10 6 ..
Maximum flux density in the teeth = 147,000 lines per square inch.
222. Effect of Tooth Taper. When the diameter of the arma-
ture is small or the slots are deep, the flux density at the roots
of the teeth is much greater than that at the tops. If the ampere-
turns for the teeth are calculated by using the maximum density,
the result is too large, while if the minimum density is used the
result is much too small. The curves in Fig. 252 have been
calculated for various tooth tapers, k = ', and give the average
ampere-turns per inch. Corresponding to the actual tooth den-
sities at the root.
icono*
100 120 140 160 180 200
Turns per In. *
FIG. 252. Magnetization curves for tapered teeth.
With a flux density of 147,000 lines per square inch at the root
and a tooth taper k = 1.17 the ampere-turns per inch required =
1,065.
223. Ampere -turns per Inch for an Air Path. When a flux
density B lines per square inch is to be produced in an air path,
the ampere-turns per inch required may be found as follows :
Taking a path 1 in. long and 1 sq. in. in section < = J5; I = 1
DESIGN OF A DIRECT-CURRENT GENERATOR 269
in. = 2.54 cm., A = 1 sq. in. = (2.54) 2 sq. cm., ju = 1; and
substituting
0.4 X 3.14 (nl)
B =
2.54
(2.54) 2 X 1
and the ampere-turns per inch.
2 54
(nl) per inch = B X
0.4 X 3.14 X (2.54) 2 ~ 3.2* (273)
224. Air Gap. The air gap is the most important part of the
magnetic circuit as it requires the largest proportion of the field
m.m.f.
The section of the air gap is A g = \f/rL c and the apparent flux
density is B g = -/- = ~rj~' This would be the correct density
if the flux were uniformly distributed, but due to the presence
of the slots and vent ducts the whole space is not utilized and the
actual flux density is B ag = CB g , where C is a constant called
the Carter coefficient and depends on the tooth width, slot width
and gap length. Referring to Fig. 253,
C =
t+fs
(274)
and the values of / which depend principally on the ratio
slot width s , x . . ,
gap len~th = 5 mav obtained from the curve.
FIG. 253. Carter's coefficient.
The ampere-turns per inch required for the air gap = 0.3132 X
Z? /nr o
T" Jj Q,Q vyX)/7
B ag -zr^ -5-77 (Art. 223) ; and the ampere-turns per pole for
O*j O.w
the gap
AT = ~^ 6 ' (275)
226. Length of the Air Gap.
Armature ampere-turns per pole = 4,860.
Ampere-turns per pole (gap + teeth) = AT 0+t = 1.2 X 4,860 = 5,832.
270 ELECTRICAL ENGINEERING
Ampere-turns per pole for the teeth = AT t = 1,550.
Ampere-turns per pole for the gap = AT g 4,282.
gX 5000 X
Gap length = 5 = 0.22.
s _ 0.35 _
I ~ 022 ~ L59 '
/ = 0.76, Fig. 253.
t + s 0.375 + 0.35
* +/s 0.375 +0.76 X0.35
'Ampere-turns for the gap
_ CB d _ 1.13 X 55,000 X 0.22
~" A7(r ~~!2~ 3.2
226. Design of Poles and Yoke.
Density in the pole core = 95,000 lines per square inch (assumed).
Leakage coefficient = v = 1.18, Art. 219.
Flux in the pole = p = 6 X 10 6 X 1.18 = 7.08 X 10 6 lines.
Section of the pole core = Q 5 QQQ = 74.6 sq. in.
Assume a round pole 10 in. in diameter.
Section of pole = A p = -^- X 10 2 = 78.5 sq. in.
Flux density in the pole core = B p = ^^ = 90,000 lines per
square inch.
The field ampere-turns per pole are approximately 50 per cent, greater
than the armature ampere-turns per pole = 1.5 X 4,860 = 7,290; about
1,000 ampere-turns can be placed on 1 in. of winding space and, there-
7 290
fore, the space for the shunt coil is -TQQQ =7.3 in. ; adding 30 per cent.
to this to take care of the series field the total winding space is 10 in. and
allowing 1 in. for the pole face, the length of the pole is l p = 11 in.
Yoke material = cast iron.
Yoke density = 40,000 lines per square inch (assumed).
Flux in yoke = 7 ' 08 2 X 10 * = 3.54 X 10 6 .
3 54 X 10 6
Section of the yoke is 4Q QQQ = 88 - 5 S< 1- in -> take A v = 95 sc l- in -
S 54 X 10 6
Density in the yoke = B v = - g ~ - = 37,300.
227. Determination of the No-load Saturation Curve. Fig.
250 shows the dimensions of the magnetic circuit of the eight-
pole, 250-kw., 250-volt, 400-r.p.m. generator. It is required to
obtain the no-load saturation, curve for this machine. The
armature has a multiple winding with 624 conductors.
DESIGN OF A DIRECT-CURRENT GENERATOR 271
The voltage generated in the armature is
8 = Zn g - 10~ 8 = 624 X X , X f 10~ 8 volts,
PI uU o
and therefore
4> g = 24,0008.
To produce the rated voltage, 250 volts, a flux is required in
the gap, g = 24,000 X 250 = 6 X 10 6 lines.
The leakage factor may be assumed to be 1.18, Art. 219, so
that the flux per pole is
p = 1.1800 = 1.18 X 6 X 10 6 = 7.08 X 10 6 .
The sections and lengths of the various parts of the magnetic
circuit may be obtained from the design sheet or from the sketch.
10 20 30 40 50 60 70 80 90 100
Ampere Turns per Inch
25 50 75 100 125 150 175 300 225 250
Ampere Turus per lucb
FIG. 254. Magnetization curves.
Yoke.
Material = cast iron.
Section = A v = 95 sq. in.
Length of magnetic path = l y = 12 in., Fig. 250.
Flux in yoke = j> y = - = - ^ = 3.54 X 10 6 lines.
3 54 X 10 6
Flux density = B y = ^ = 37,300 lines per square inch.
Ampere-turns per inch = 65, Fig. 254.
Ampere-turns per pole for the yoke = AT y = 65 X 12 = 780.
Pole.
Material = cast steel.
Section = ,A P = 78.5 sq. in.
Length = 11 in.
Flux = P = 7.08 X 10 6 lines.
Flux density = B p = -^^TK = 90,000 lines per square inch.
272 ELECTRICAL ENGINEERING
Ampere-turns per inch = 48, Fig. 254.
Ampere-turns per pole for the pole core = 48 X 11 = 528.
Air gap.
Section = A g = 109 sq. in.
Length = 6 = 0.22 in.
Flux = fa = 6 X 10 6 lines.
i A \S 1 A6
Apparent flux density = B g = -~ = ^ = 55,000 lines per sq. inch.
Carter coefficient C = 1.13.
Actual flux density = B ag = CB a = 1.13 X 55,000 = 62,150 lines per
square inch.
Ampere-turns per inch = -^ = ' = 19,400.
O.6 O.
Ampere-turns per pole for the gap = 19,400 X 0.22 = 4,270.
Teeth.
Material = sheet steel.
Maximum section = 46 sq. in.
Minimum section = At = 39.5 sq. in.
Length of magnetic path = It 1.45 in.
Flux = fa = 6 X 10 6 lines.
6 X 10 6
Apparent maximum flux density = on g = 152,000 lines per square
inch.
Apparent minimum flux density = -r^ = 131,000 lines per square
inch.
Actual maximum flux density = 147,000 lines per square inch, Art. 221.
Tooth taper = k = 1.17.
Ampere-turns per inch = 1,065, Fig. 252.
Ampere-turns per pole for the teeth = AT t = 1,065 X 1.45 = 1,550.
Armature core.
Material = sheet steel.
Section = A c = 36.7 sq. in.
Length = l c = 6 in., Fig. 250
Flux = c = f = ^i? 8 = 3 X 10 6 lines.
3 X 10 6
Flux density = B c = oa - = 82,000 lines per square inch,
oo./
Ampere-turns per inch = 18.
Ampere-turns per pole for the core = AT e = 18 X 6 = 108.
The ampere-turns per pole required for a voltage of 250 volts at no load is
AT V + AT P + AT X ATt + AT C = 780 + 528 + 4,270 + 1,550 + 108
= 7,236.
These results are tabulated below and also the results of similar
calculations made for 225 volts and 275 volts.
From these results the saturation curve for the machine is
plotted in Fig. 256. Volts vs. field ampere-turns per pole.
DESIGN OF A DIRECT-CURRENT GENERATOR 273
No-load voltage = 8
225
250
275
Useful flux per pole = 4>o
5.4 X 10
6 X 10
6.6 X 106
Leakage factor
1.18
1.18
1.18
Length
Area
Density
AT
Density
AT
Density
AT
Yoke
12.00
95.0
33,600
600
37,300
780
41,000
1,000
Pole
11.00
78.5
81,000
280
90,000
528
99,000
1,000
Air gap
0.22
109.0
1.13X49,500
3,840
1.13X55,000
4,270
1.13X60,500
4,700
Tooth (min.) . .
1.45
39.5
136,800
152,000
167,000
133,000
830
147,000
1,550
160,000
2,000
Armature core.
6.00
36.7
73,800
80
82,000
108
90,200
200
Field-ampere turns per
pole '
5,630
... j 7,236
8,900
I
^
cxio 6
h a/
f
01!
/
A
*i
ffl
-1
S
/
' 3
/
<^>-f >
c =3400
b = P840
c
250
225
200
175
160
l
> 100
75
50
25
TO!
Field Ampere Turns per Pole foi
Teeth and Armature
> Gap
Field Ampere Turns per Pole
FIG. 256. Saturation curves.
FIG. 255. Cross-magnetizing effect.
228. Field Winding. The field coils must be designed with
the required numbers of ampere-turns and with sufficient surface
to radiate the heat due to the field copper loss.
If the coils are made thick it will be difficult to find space for
them between the poles and they will be liable to get hot in the
center layers. A good average thickness is 2 in.; assuming a
current density in the field copper of about 800 amp. per square
inch or 1,500 cir. mils per ampere and a space factor of 0.6 foi
the winding, it is possible to place 800 X 0.6 X 2 = approximate 1 ^
1,000 ampere-turns on each inch of the coil space.
The radiating surface of the field coil is taken as the external
cylindrical surface; this neglects the surface of the ends of the
coil and the internal surface from which heat is transferred to the
pole. The efficiency of the external radiating surface depends on
the speed of the armature and its consequent fanning action.
About 0.6 watts per square inch can be radiated from an ordinary
field coil for a temperature rise of 40C. For very short thick
18
274 ELECTRICAL ENGINEERING
coils a higher value may be used since the neglected surface of
the ends is so large.
229. Current Density in Field Windings. The current den-
sities employed in shunt-field windings vary from 500 amp. per
square inch for small slow-speed machines to 1,000 amp. per
square inch for large high-speed machines with ventilated field
coils.
230. Size of Wire for the Field Winding. Let E f = voltage
across one coil, // = current, T/ = number of turns, l m = mean
turn in inches, A = section of wire in circular mils and Rf
resistance of the coil.
_ v T
19 ' 7 T
At 60C. R f = 12 X --J-- = m j L
and
Ef = IfRf = IfTf -j^-
or, the section of the wire is
A = i& X l m ; (276)
tif
therefore the size of the wire to be used is fixed when the ampere-
turns and the voltage per coil are known.
Design of the field winding.
Voltage taken up in the field rheostat = 20 per cent, of 250 = 50 volts.
250 5Q
Voltage across one field coil = Ef = - = 25 volts
o
Ampere-turns per coil I/Tf = 7,236.
Thickness of coil = 2 in. (assumed).
Mean turn = 38.3 in.
7 236
Section of the wire = A = -~- X 38.3 = 11,200 circ. mils
This is between No. 9 and No. 10. B. & S. wire (Art. 86).
Either use an intermediate size or wind half of*each coil with No. 9 and
half with No. 10.
Diameter of No. 9 d.c.c. = 0.126.
Diameter of No. 10 d.c.c. = 0.114.
Average = 0.12.
Winding space = 7.2 X 2 sq. in.
72
Turns per layer = Q-^ = 60.
2
Layers = ^^ = 16-
Turns per coil = T f = 60 X 16 = 960
Current = // = -- = 7.5 amp.
DESIGN OF A DIRECT-CURRENT GENERATOR 275
Power lost = 25 X 7.5 = 188 watts.
Radiating surface = 7.2 X 3.14 X 14.2 = 320 sq.'in.
1 oo
Watts per square inch = ^Q = ^-59.
Temperature rise = 40C. (Art. 228).
231. Resistance of the Armature Winding. The length of a
single armature conductor is given approximately by the empiri-
cal formula,
I = 1.35r + L e + 3 in. (277)
If the section of the conductor is A circ. mils, its resistance at
60C. is approximately ..;
length in feet . rt 12 / ,
r c = P - -^-. - : --- T- = 12 - r = -,- ohms.
section in circ. mils A A
g
The number of conductors in series between brushes is and the
Pi
number of paths in multiple is p\ and therefore, the armature
resistance is
Z
r c
r - _. _ r
~ Pi " Pi' '
(278)
624 1.35 X 14.13 + 11+3
Armature resistance = r a = -55- X - a , n CAn - = 0.0052 ohms.
8^ ol7,500
Armature resistance drop at full load Ir a = 1,000 X 0.0052 = 5.2 volts.
Armature copper loss = IJr a = 1,000 2 X 0.0052 = 5,200 watts.
232. Determination of the Magnetomotive Force of the Series
Winding. In the case of a flat-compound generator the series
excitation must be great enough to counteract the demagnetizing
and cross-magnetizing m.m.f. of the armature and the armature
resistance drop.
Demagnetizing Armature Magnetomotive Force. If the brushes
are advanced 10 per cent, of the pole pitch the demagnetizing
71 78 000
ampere-turns per pole = 0.2 -~ = 0.2 in the gap at no load may be taken
as proportional to chfe; due to cross-magnetizing a flux propor-
tional to area hgj is lost and a flux proportional to area glf is
gained; the result is a loss of flux and voltage proportional to
hgj glf. The flux in the gap is now ofa where fa is found by
making the area hjiffi = area hgj area glf. The length kg\
represents the effect of the cross-magnetizing armature m.m.f.
as a number of demagnetizing ampere-turns = 600 ampere-
turns in this case. The series field m.m.f. required to overcome
cross-magnetizing = 600 X 1.25 = 750 ampere-turns per pole.
The loss of flux due to oross-magnetizing does not increase
directly as the current but at a faster rate; it also depends very
materially on the saturation of the magnetic circuit especially
the teeth; with low flux densities the flux added on one side is
approximately equal to the flux subtracted on the other side and
the cross-magnetizing effect is very small.
To obtain accurate results it is necessary to take account of
the increase of the leakage factor under load due to the presence
of the armature m.m.f. (Art. 219). In this case the increase is
from the assumed no-load value 1.18 to 1.25. The flux densities
in the pole and yoke are increased in the same proportion and an
increase of 170 ampere-turns per pole is required on the field on
this account.
DESIGN OF A DIRECT-CURRENT GENERATOR 277
In constructing the full-load saturation curve of a genera-
tor as in Art. 147, it is convenient to treat the demagnetizing
ampere-turns as a constant and to assume that they can be over-
come by an equal number of series-field ampere-turns; the extra
turns required on account of the leakage factor = 0.25 X 972 =
243 may be included under cross-magnetizing as also the extra
turns for the pole and yoke. With this understanding the series-
field turns required to overcome armature demagnetizing =
972 and to overcome armature cross-magnetizing = 750 + 243
+ 170 = 1,163. The cross-magnetizing armature ampere-turns
per pole = 3,888 and the series-field turns required to overcome
them = 1,163 = approximately 30 per cent, of the cross-magnet-
izing ampere-turns per pole. For points higher up on the satura-
tion curve a larger percentage than 30 would be required. The
leakage factor decreases with a decrease of field m.m.f. and so the
cross-magnetizing effect may still be assumed to be negligible
at the lower points on the full-load saturation curve.
Fig. 256 shows the no-load and full-load saturation curves for
the 250-kw., 250- volt generator; oa = 7,236 = the field ampere-
turns per pole required at no load to produce the rated voltage
250 volts; mp. = the field m.m.f. required at full load to over-
come armature reaction = 972 + 1,163 = 2,135; pq = the vol-
tage consumed in the resistance of the armature and series field
= 8 volts; q is therefore a point on the full-load saturation curve.
The other points may be found as explained in Art. 147. The
series-field ampere-turns required to overcome the resistance
drop = pn = 865 and, therefore, the ampere-turns per pole re-
quired in the series-field winding for flat-compounding = mn =
ad = 2,135 + 865 = 3,000.
233. Design of the Series-Field Winding.
The series-field ampere-turns per pole = 3,000.
The load current = 1,000 amp.
Turns per series coil = j 2 ^ = 3 '
Use 3^ turns per coil and shunt the part of the current not required,
through a diverter.
The current in the winding = ' 1 = 860 amperes.
Current density in the series winding = 1,000 amp. per square inch
(assumed).
Size of conductor = ., nnn = 0.86 sq. in.
1,UUU
278 ELECTRICAL ENGINEERING
Use three strips 2.5 X 0.125 in. connected in parallel.
Actual section of conductor = 3 X 2.5 X 0.125 = 0.94 sq. in.
Length of the mean turn = 36 in. =3 ft.
Resistance of one coil at 60C. = 12 X -- ^~r = 1-03 X 10~ 4 ohms.
0.94 X -10 6
7T
Voltage drop in one coil = 860 X 1.03 X 10~ 4 = 0.088 volts.
Power lost = 0.088 X 860 = 75 watts.
The cylindrical radiating surface of the coil = 2.5 X 40 = 100 sq. in.
Watts per square inch of surface = y^ = 0.75, which is quite satisfactory.
The voltage drop in the series-field winding = 8 X 0.088 = 0.7 volts.
Power lost in the series-field winding = 8 X 75 = 600 watts.
234. Design of the Commutator.
Winding is one turn multiple, short pitch.
7 AQ A.
Numbers of commutator segments = 2 = ~W =312.
Diameter = 0.75D a = 0.75 X 36 = 27 in.
o 1 A vy 1 07
Width of segment -f mica = ~^~ " = 0.271 in.
Width of mica = ^2 in. = 0.031 in.
Width of segment = 0.24 in.
Brush arc should not cover more than three segments = 3 X 0.271 =
0.813 in.
Take brush arc = 0.75 in. = % in.
Sets of positive brushes = 4.
1 000
Current per set = ' = 250 amp.
Current density in the brush = 35 amp. per square inch.
250
Section of brushes per set = -^- = 7.5 sq. in.
oo
Length of brushes per set = r= = 10 in.
Brushes per set = 5 (% in. X 2 in.).
Length of commutator = 12 in.
27
Peripheral velocity = 3.14 X ^ X 400 = 2,825 ft. per second.
Drop of voltage at each brush contact = 1 volt.
Loss of power at the brush contacts = 1,000 X 1 X 2 = 2,000 watts.
Brush pressure = 2.5 Ib. per square inch.
Total brush pressure = 2.5 X 7.5 X 8 = 150 pounds.
Coef . of friction = 0.3.
Brush-friction loss = 150 X 0.3 X 2,825 = 127,000 ft.-lb. per minute.
Radiating surface of commutator = cylindrical surface = 3.14 X 27 X
12 = 1,020 sq. in.
Total watts lost = 2,000 + 2,900 = 4,900 watt.
4 900
Watts per square inch = ' ^Q 4.8.
Temperature rise will be under 40C. (Fig. 247).
DESIGN OF A DIRECT-CURRENT GENERATOR 279
235. Losses and Efficiency. The core losses cannot be calculated ac-
curately by any formula but fairly satisfactory values may be obtained
by the use of curves such as those in Fig. 257 compiled from the results
of tests on completed machines. The loss in the teeth must be cal-
culated separately from that in the core since the densities are different.
Cycles per Second
15 20 25 30 40 50 00 80 100
5 10 15 20 25 30 35 40 45 50
"Watts per Lb.
FIG. 257. Iron-loss curves for direct- current machines.
Weight of iron in the teeth
- 156 X 0.35 X 1.45 X 9.05 X 0.28 = 200 Ib.
= | (
where 0.28 is the weight of a cubic inch of iron.
400 8
The frequency is -^ X 7, = 27 cycles per second.
Watts lost at 140,000 lines per square inch = 10 watts per pound.
Watts lost in the teeth = 200 X 10 = 2,000 watts.
Ar X 33A 2 TT X 25 2 \
Weight of iron in the core below the teeth = ( - -. - -- r ) X
9.05 X 0.28 = 940 Ib.
Watts lost per pound at density of 80,600 lines per square inch = 3.3.
Loss = 940 X 3.3 = 3,100 watts.
Total core loss = 2,000 + 3,100 = 5,100 watts.
Shunt-field copper loss = EI f = 250 X 7.5 = 1,875 watts.
Series-field copper loss = 600 watts.
Armature copper loss = 5,200 watts.
Brush-contact loss = 2,000 watts.
Brush-friction loss = 2,900 watts.
Windage and journal friction loss = 1 per cent. = 2,500 watts (assumed).
Total losses = 20,175 watts.
Efficiency at full load = outpu t Bosses 10 P er Cent ' =
250 000
250,000 + 20,175 X 10 per Cent ' = 92 ' 6 per Cent '
CHAPTER IX
SYNCHRONOUS MACHINERY
236. Alternator. An alternating-current generator or alter-
nator in its simplest form consists of an open coil of wire revolving
at uniform speed in the magnetic field between a pair of unlike
poles (Fig. 258).
The fields are excited by direct current from a separate machine
called an exciter at 125 or 250 volts.
FIG. 258. Single-phase alternator, revolving-armature type.
Between the slip rings a and b an alternating e.m.f . is generated
of instantaneous value
= n ~ 10~ 8 volts
at
(279)
where n is the number of turns in the coil and -77 is the rate of
change of the flux interlinking with the coil or the rate at which the
coil is cutting the flux. The result is the same if the armature is
stationary and the field revolves.
237. Types of Alternators. There are three principal types
of alternators:
(a) Revolving armature.
(b) Revolving field.
(c) Inductor.
280
SYNCHRONOUS MACHINERY
281
Type (a) is illustrated in Fig. 258. The field poles are sta-
tionary and the armature revolves between them. The ends of
the winding are brought out to two slip rings in single-phase
machines and to three or more slip rings in polyphase machines
and the current is collected by copper or carbon brushes.
The armature is necessarily of small size since the peripheral
speed is limited and there is very little space for insulation. The
armature conductors are also acted upon by centrifugal forces
which tend to throw them out of the slots. The revolving arma-
ture is therefore only suitable for machines of small size and low
voltage. It is, however, necessary in the case of rotary converters
where the same armature winding carries both alternating and
direct currents.
(6) The revolving field type illustrated in Fig. 259 and Fig.
260, is almost universal for all sizes and voltages. The armature
is the stationary part and the field poles revolve. This type has
many advantages over the revolving armature type. (1) It
FIG. 259. Single-phase alternator, revolving-field type.
requires only two slip rings even for polyphase machines and these
slip rings carry only the direct current supplied to the field wind-
ing, while the load current is taken off from stationary terminals.
(2) There is much more space for the armature windings and
they are relieved from all centrifugal strains. They can, there-
fore, be much better insulated and ventilated. The field wind-
ings are made of copper strap and the revolving member is very
rugged and is not affected by strains due to rotation; thus, much
higher peripheral speeds may be used than with type (a), with
consequent increase in economy of material.
The revolving-field members are made in two forms, the
salient pole rotor and the cylindrical rotor. The rotor with
salient poles, Figs. 259 and 260, is used almost universally for
machines of all but the largest outputs and highest speeds.
282
ELECTRICAL ENGINEERING
Cylindrical rotors, Fig. 335, are employed for turbo-alternators
of very large output which must run at extremely high speeds in
order to obtain the maximum output per pound of material.
The peripheral speeds run up to 25,000 ft. per minute and very
large stresses are developed due to centrifugal force. The rotors
are made of solid steel castings or of a number of steel discs
bolted together and have slots milled out to carry the field wind-
ings. The shaft is bolted or dovetailed to the ends of the rotor.
Windln,
Blip Rings for Excitation
iDBUla
Field.8plder
FIG. 260. Revolving-field alternator.
The cylindrical rotor gives an air gap of practically uniform
length and reluctance over the whole periphery and the field
m.m.f. is distributed. It lends itself more readily to analytical
treatment than the salient pole rotor with its varying gap re-
luctance and concentrated field m.m.f. The vector diagrams in
Figs. 303 to 306 give very satisfactory results for machines with
cylindrical rotors but one more approximate for those with salient
poles.
(c) The inductor alternator is almost obsolete. In these
machines the field and armature windings are both stationary
and a part of the iron of the magnetic circuit revolves, producing
a periodic pulsation of the reluctance of the magnetic circuit and
consequently a variation of the flux linking with the armature
winding. Fig. 261 represents one type of inductor alternator:
SYNCHRONOUS MACHINERY
283
ff is the stationary field winding which produces the magnetic
flux <,<, in the direction indicated. a,b, a,6, are the armature
coils which may be connected either in series or in multiple. /
is the revolving part of the magnetic circuit and is called the in-
ductor. The polar projections on it are all north poles but the
amount of flux issuing from 7 and linking with the armature
coils a,a, 6,6, depends on the relative position of the inductor
projections and the projections carrying the armature coils.
In Fig. 262, curve 1 represents the variation of the flux a
interlinking with one armature coil a starting from the position
when this flux is maximum, Fig. 261. Curve 2 represents the
e.m.f. generated in coil a by the varying flux. As 0^ decreases
FIG. 261. Inductor alternator.
FIG. 262. Fluxes and e.m.fs. in an
inductor alternator.
an e.m.f. is generated in the positive direction and as it increases
again an e.m.f. is generated in the negative direction. Thus al-
though the flux does not reverse its direction and never reaches
zero, an alternating e.m.f. is generated in the armature coil.
If the pole pieces are properly shaped a sine wave of e.m.f. will
be produced. Curves 3 and 4 represent the variation of flux
and e.m.f. in coil b. The e.m.f. in coil 6 is displaced 180 degrees
from that in a and before the two are connected in series the
terminals of one must be reversed.
Inductor alternators were very heavy and expensive and have
been superseded by the other types.
Any one of these three types may be wound as single-phase or
polyphase machines.
Alternators are divided into classes depending on the type of
prime mover employed: (a) Engine type, (6) waterwheel type,
(c) steam turbine-driven type.
284 ELECTRICAL ENGINEERING
(a) Engine-type alternators are direct-connected to recipro-
cating steam engines and they require a large flywheel effect to
ensure uniform angular velocity.
(6) Waterwheel-type alternators range in speed from 50
to 400 r.p.m. depending largely on the available head of water.
The angular velocity of water turbines is uniform and flywheels
are not required. Water wheel alternators may be either hori-
zontal or vertical.
(c) Steam turbine-driven alternators or turbo-alternators are
extra high-speed generators usually 750 or 1,500 r.p.m. for 25
cycles and 1,800 or 3,600 r.p.m. for 60 cycles, corresponding to
four-pole and two-pole rotors respectively. Cylindrical rotors
are ordinarily employed on account of their greater mechanical
strength and smaller windage loss. On account of the large
output per pound of material forced ventilation is required to
keep the temperature within the required limits and therefore
such machines must be entirely enclosed.
238. Electromotive Force Equation. Fig. 258 represents a
two-pole, single-phase alternator. The armature winding is a
single coil of n turns revolving at a constant speed. The magnetic
field is assumed to be uniform.
The e.m.f. generated in the winding goes through one complete
cycle during each revolution, and thus the frequency in cycles
per second is equal to the speed in revolutions per second or / =
rev. per sec.
The angular velocity of the coil is co radians per second, and
therefore
w = 27r X rev. per sec. = 2irf (280)
If is the maximum flux inclosed by the coil, that is, the flux
inclosed when the coil is vertical, as shown, and time is measured
from this instant, then at time t, after the coil has turned through
an angle 6, the flux inclosed is = $ cos 0, and the e.m.f. gener-
ated in the coil is
e= -rAo- 8
at
but d = ut = 2irft, and thus
e = - n d j(3>c
at
= 2vfn3> 10~ 8 sin 2irft volts (281)
= E m sin 6.
SYNCHRONOUS MACHINERY 285
This is a sine wave of maximum value
E m = 27r/7i$ 10~ 8 volts (282)
and effective value
V
E = 5| = 4.44/71$ 10- 8 volts (283)
This is the e.m.f. equation for an alternator which produces a
sine wave of e.m.f. and has a concentrated winding, that is, all
the turns wound in a single coil.
This result may also be obtained as follows. The flux cut per
second by each turn of the coil is 4/$ lines, and therefore the
average e.m.f. generated in the coil is
E avg = 4/n$ 10- 8 volts, (284)
but for a sine wave the ratio of the maximum to the average
ordinate is ~' and therefore the maximum e.m.f. is
E m = I E aug = 27r/7i$ 10~ 8 ,
and the effective value is as before
Tjl
E = ^ = 4.44/7i$ 10- 8 .
The e.m.f. generated in an alternator is directly proportional
to the frequency /, to the number of turns in series n and to the
flux under each pole $.
In a two-pole machine one revolution or 360 mechanical de-
grees corresponds to one cycle or 360 electrical degrees and the
frequency is equal to the number of revolutions per second ; in a
p-pole machine the e.m.f. goes through a complete cycle when
fY\
the coil moves across a pair of poles and thus through ~ cycles
in one revolution. In this case 360 mechanical degrees = | X
IY\
360 electrical degrees or one mechanical degree = electrical
2i
degrees.
The frequency in a p-pole alternator in cycles per second is
/ = X rev. per sec. (285)
286
ELECTRICAL ENGINEERING
239. Form Factor. If the flux in the air gap is not so distrib-
uted as to give a sine wave of e.m.f., the average value of the
generated e.m.f. is still given by equation (284),
E avg
10-
but the ratio of maximum to average value is not ~ and the ratio
of effective to maximum value is not ~~7s'
The effective value of the general alternating wave can be
expressed as
E = 4yfn$ 10~ 8 volts, (286)
where 7 is called the form factor of the wave and is defined as
the ratio of the effective value to the average value of the ordi-
nate of the wave.
The form factor of a rectangular wave is 1.00 and for all other
waves is greater than 1.00. For a sine wave it is
E m
7 =
1.11.
240. Polyphase Alternating-current Generators. If the arma-
ture of an alternator carries two similar windings displaced 90
electrical degrees from one another, Fig. 263, the winding is
A BCD
. * 2 A A o
FIG. 263. Two-phase alternator.
a two-phase winding and the alternator a two-phase alternator.
Two e.m.fs. are produced equal in value but displaced 90 degrees
in phase.
If the armature carries three similar windings displaced from
one another by 120 electrical degrees (Fig. 264), the winding is a
three-phase winding and the alternator a three-phase alternator.
Three e.m.fs. are produced equal in value but displaced 120
degrees in phase.
SYNCHRONOUS MACHINERY
287
If the windings of the three phases start at i, s% and 83 and end
at /i, /2 and /s, the phases may be interconnected in two ways :
(1) join /i to s 2 ,/ 2 to s 3 and/s to Si, this is the " delta" connection,
Fig. 265; (2) join si, s 2 and s 3 together and /i, / 2 and / 3 to the
three terminals, this is the "star" or Y connection (Fig. 266).
For a given number of turns per phase the Y connection gives
a higher terminal voltage than the delta connection and a corre-
spondingly smaller current output.
Rotation
A >/i "/.
FIG. 264. Three-phase alternator.
'/.
FIG. 265. Delta connection. FIG. 266. Star or "Y" connection.
,
A
/I
/,
Si
Jfo
1
i
Four Pole, Single Phase
Chain Winding
*2
ur Pole, Two Phase,
Chaiu Winding
i
our 1
3 l/i 1
'ole, Three Phase,
Chain Winding
FIG. 267. Concentrated chain windings.
The e.m.fs. currents and power in polyphase circuits are dis-
cussed in Art. 124.
The electrical power developed at any instant in a single phase
alternator is the product of the instantaneous values of the e.m.f .
and current (Fig. 89). It pulsates between a maximum positive
TJI T rr T
value "X m (1 + cos <) and a negative value " (1 cos 0)
and its average value is El cos .
The power developed in a three phase alternator is the sum of
the instantaneous powers developed in the three phases ; its value
288
ELECTRICAL ENGINEERING
at any instant is z\i\ + eziz + etfz = 3EI cos $ and remains
constant.
The armature m.m.f. of the single-phase alternator pulsates
likewise between nl m and zero where n is the number of turns on
the armature, but it remains fixed in direction relative to the
armature and revolves at synchronous speed relative to the poles.
s f
Four Pole, Single Phase, Double Layer Winding
Four Pole, Two Phase, Double Layer Winding
Four Pole. Three Phase. Double Layer Winding
FIG. 268. Concentrated double-layer windings.
The armature m.m.fs. of the various phases of a polyphase
alternator combine to produce a constant m.m.f. of armature
reaction fixed in direction relative to the poles and revolving at
synchronous speed relative to the armature.
The conditions of operation of a polyphase alternator or syn-
chronous motor are thus much more satisfactory than those of
SYNCHRONOUS MACHINERY
289
the single-phase machines. Single-phase machines are built
only in the smaller sizes except in special cases as for instance to
supply power to a single-phase electric-railway system.
241. Alternator Windings. There are a great many special
alternator windings but the majority of them come under the two
classes of chain windings and double-layer windings.
\ (
\ C
s if
FIG. 269. Four-pole, single-phase, chain winding distributed in six slots
per pole.
Fig. 267 shows four-pole single-, two- and three-phase chain
windings for armatures with one slot per phase per pole. These
windings are all concentrated windings.
Fig. 268 shows the corresponding double-layer windings.
Fig. 269 shows a four-pole single-phase chain winding distrib-
uted in six slots per pole. Fig. 270 shows a winding for the same
machine using only four of the six slots per pole as explained in
Art. 242.
FIG. 270. Four-pole, single-phase chain winding using only four of the six
slots per pole.
Fig. 271 shows a two-phase chain winding for the armature in
Fig. 289. The windings are distributed in three slots per phase
per pole.
Fig. 272 shows a three-phase chain winding for the same arma-
ture, distributed in two slots per phase per pole.
19
290
ELECTRICAL ENGINEERING
Figs. 273 tp 276 show the double-layer windings corresponding
to Figs. 269 to 272.
242. Distribution Factors. The windings in Figs. 267 and
268 are all concentrated windings, that is, they are placed in one
slot per phase per pole.
. w
FIG. 271. Four-pole, two-phase, chain winding distributed in three slots per
phase per pole.
When a winding is made up of a number of coils placed in sepa-
rate slots the e.m.fs. generated in the various coils are displaced
in phase and the terminal e.m.f. is less than if the winding had
been concentrated. The factor by which the e.m.f. of a concen-
trated winding must be multiplied to. give the e.m.f. of a distrib-
J/ 2 1/3
FIG. 272. Four-pole, three-phase, chain winding distributed in two slots per
phase per pole.
uted winding of the same number of turns is called the distribu-
tion factor for the winding and it is always less than unity.
When a single-phase winding is distributed in two slots per
pole spaced at 90 degrees the e.m.fs. in the two coils are 90 de-
grees out of phase. If the effective value of the e.m.f. generated
SYNCHRONOUS MACHINERY
291
FIG. 273. Four-pole, single-phase, double-layer winding distributed in six slots
per pole.
FIG. 274. Four-pole, single-phase, double-layer winding using only four of the
six slots per pole.
FIG. 275. Four-pole, two-phase, double-layer winding distributed in three slots
per phase per pole.
FIG. 276. Four pole, three-phase, double-layer winding distributed in two slots
per phase per pole.
292
ELECTRICAL ENGINEERING
in each coil is e, then the terminal e.m.f. is e t -\/2e, Fig. 277,
and the distribution factor is
8 = L =
2e
2e
0.70/7
When a single-phase winding is distributed in three slots per pole
spaced at 60 degrees the terminal e.m.f. is the sum of three e.m.fs.
e at 60 degrees to one another. It is e t = 2e } Fig. 278, and the
distribution factor is
= 0.666.
0.705
FIG. 278.
When a single-phase winding is distributed in six or more slots
per pole the distribution factor may be taken as
8 = = o.64.
7T
In Fig. 279 the semi-circumference represents the e.m.f. of the
concentrated winding and the diameter represents the e.m.f. of
the distributed winding.
The e.m.fs. in the coils from b-c add very little to the terminal
e.m.f. and this part of the winding is usually omitted and the
terminal e.m.f. is decreased in the ratio = cos 30 = 0.866, or
oc
is decreased 13.4 per cent, while the resistance and reactance of
the winding are decreased 33 J^ per cent.
The distribution factor for this case is
g _ 06 _ 3|le
4e 4:6
Figs. 270 and 274 show four-pole single-phase windings with
only four of the six slots per pole used.
SYNCHRONOUS MACHINERY
293
The terminal e.m.f. of a two-phase winding distributed in two
slots per phase per pole is made up of two e.m.fs. of value e
displaced 45 degrees from one another. It is e t = 1.848 e, Fig.
280, and the distribution factor is
e t 1.848e
2e
0.924.
FIG. 279.
FIG. 280.
FIG. 281.
For a two-phase winding with three slots per phase per pole,
Figs. 271 and 275, the terminal e.m.f. is made up of three e.m.fs.
displaced 30 degrees from one another. It is e t = 2.733e, Fig.
281, and the distribution factor is
2 " 733e = 0.911.
FIG. 282.
FIG. 283.
The terminal e.m.f. of a three-phase winding distributed in
two slots per phase per pole, Figs. 272 and 276, is made up of two
294
ELECTRICAL ENGINEERING
e.m.fs. of value e displaced 30 degrees from one another. It is
e t = 1.9316, Fig. 282, and the distribution factor is
* 1.93'le
2e
= 0.966.
With three slots per phase per pole the factor is
6 = 0.96, Fig. 283.
The following table gives the distribution factors for single-,
two- and three-phase windings.
Slots per phase
per pole
Distribution factor
Single-phpse
Two-phase
Three-phase
1
1.000
1.000
1.000
2
0.705
0.924
0.966
3
0.666
0.911
0.960
4
0.653
0.906
0.958
6
0.64
0.903
0.956
6
0.83, if only two-thirds of slots are used.
243. Multiple -circuit Windings. The windings already dis-
cussed are all single circuit, that is, all the turns of one phase are
connected in series. In low- voltage machines with a large current
output it is necessary to connect the coils forming each phase in
multiple circuit. When connected two-circuit the terminal e.m.f.
is reduced to one-half and the current output is doubled; the
power output, therefore, remains the same.
The e.m.fs. generated in the sections of the windings which are
connected in multiple must be of the same value and must be in
phase or circulating currents will flow. It is also necessary that
the resistances and reactances of the sections be of the same
value or one part of the winding will supply more current than
the other.
Fig. 284 shows a four-pole three-phase double-layer winding
with one slot per phase per pole connected Y and A single
circuit and two circuit. A winding may be connected with as
many circuits in multiple as there are pairs of poles.
244. Short-pitch Windings. The pitch of a winding is the
distance between the two sides of one of the coils forming the
winding. When the coil pitch is equal to the pole pitch or the
SYNCHRONOUS MACHINERY
295
distance between the centers of adjacent poles, the winding is full
pitch. When the coil pitch is less than the pole pitch the winding
is fractional pitch or short pitch.
In Fig. 286 abcdf shows the distribution of flux under two
adjacent poles of an alternator. The area under the section of
Star or Y
Delta
1
M '
k I I < f f
??
Si
S 2 *S 3 Two Circuit
J
f
1
'f. 1
Two Circu:
Star or Y
T? I
Two Circuit
Delta
FIG. 284. Four-pole, three-phase, multiple-circuit windings.
the curve abc or cdf multiplied by the length of the pole parallel
to the shaft gives the flux 3> crossing the air gap under each pole.
As the side g of the coil cuts across the flux in the gap an e.m.f.
is generated in it of the same wave shape as the flux distribution.
If gh is a full-pitch coil the side h will occupy a position under the
adjacent pole similar to that of g and the e.m.fs. generated in the
290
ELECTRICAL ENGINEERING
two sides will be of the same value and wave shape but displaced
180 degrees in phase; they therefore act in the same direction
around the coil and add directly to give the terminal e.m.f. If e
is the effective value of the e.m.f. generated in one side of the
coil the terminal e.m.f. is E = 2e. With a full-pitch concentrated
winding the wave form of the generated e.m.f. is the same as the
wave of flux distribution under the poles.
E.M.F. between Terminals
E.M.F. in g or h
FIG. 285. Full-pitch coil.
If the coil pitch is less than the pole pitch by an angle a, the
e.m.f. wave generated in the side h leads the e.m.f. in g by an
angle a, Fig. 285, and the terminal e.m.f. is the vector sum of two
e.m.fs. of effective value e displaced in phase by an angle a. It is
E = 2e cos -
(287)
and is less than the e.m.f. generated in the full-pitch winding in
Ot
the ratio cos ~ * 1.
E.M.F. between
Terminals
E=2e Cos .5
FIG. 286. Short-pitch coil.
Fractional-pitch windings are sometimes used in order to elimi-
nate certain harmonics from the e.m.f. wave of the generator.
Take the case of a machine with the wave of flux distribution,
shown in Fig. 287, consisting of a fundamental and a fifth har-
monic. With a full-pitch winding the e.m.f. wave would consist
SYNCHRONOUS MACHINERY 297
of a fundamental and the prominent fifth harmonic. If, how-
ever, the coil pitch is made only 80 per cent, of the pole pitch the
e.m.f. in one side of the coil will lead that in the other by 36
degrees and the fifth harmonics in the two sides will be in direct
opposition and will disappear (Fig. 288). The terminal e.m.f.
will consist only of the fundamental and it will be decreased in
ratio cos 18 degrees:!. To eliminate an nth harmonic the coil
pitch must be either lengthened or shortened by - th of the pole
pitch. Thus the wave form of the e.m.f. generated in a short-
pitch winding is not the same as the wave of the flux distribution
in the air gap.
E.M.F.between-
Terminals
.Curve of Flux Distribution
FIG. 287. Flux wave with fifth FIG. 288. Elimination of fifth harmonic,
harmonic.
Similarly the wave form of any distributed winding differs
from the wave of flux distribution, since the terminal e.m.f. is
the sum of a number of waves displaced from one another. A
fully distributed winding gives an e.m.f. wave of approximately
sine form at no load regardless of the flux distribution.
245. Effect of Distributing the Winding. (1) The core is used
to better advantage since a number of small slots evenly spaced
are used instead of a few large ones. (2) The copper is evenly
distributed over the armature surface and thus the copper loss is
also distributed and the heat developed by it can more easily be
dissipated. A higher current density in the copper can, therefore,
be used. (3) The self-inductive reactance is very largely de-
creased by distributing the winding in a large number of slots,
since the coefficient of inductance of a coil is proportional to the
square of the number of turns. (4) The terminal e.m.f. is de-
creased as shown in Art. 242 but the wave form is made more
nearly sinusoidal.
246. Harmonics Due to the Teeth. Fig. 289 shows two posi-
tions of the armature of an alternator relative to the field poles.
In A there are four teeth under the pole and three slots, while in
298 ELECTRICAL ENGINEERING
B there are three teeth and four slots. The reluctance of the
magnetic circuit of the generator is a minimum in A and the
flux in the circuit is a maximum, while in B the reluctance is a
maximum and the flux is a minimum. If there are a teeth or
slots per pole the flux per pole pulsates once in the distance of a
slot pitch or 2a times in two pole pitches. The frequency of the
pulsations is 2af where / is the normal frequency of the machine.
The flux per pole consists of a constant value $ and superim-
posed on it an alternating flux of amplitude $1 and frequency 2af.
UITUTJTLT
m
&
FIG. 289. Harmonics due to the teeth.
Measuring time from the instant represented in A when the
flux is maximum, the flux at any time t or angle is
P = $ + $1 cos 2a0,
and the flux enclosed by the coil at angle is
p cos 8 .= $ cos 6 + $1 cos 2a0 cos B
= $ cos B + ^{cos (2a + 1)0 + cos (2o - 1)0} ;
Zi
the e.m.f. generated in a coil of N turns concentrated in a single
slot is
e = - N^-( P cos 0) = - N-j-A $ cos + ^{cos (2o + 1)0
at at/ L z
-f- cos (2a
= - ^[- $sin + ^{ - (2a + 1) sin (2o + 1)0
L ^
- (2a - 1) sin (2a -
but
dd _
and therefore,
e = 2T/ATJ$ sin + 5 ^>i sin (2a + 1)0
+ ^ lS in (2a-l)0|
SYNCHRONOUS MACHINERY 299
a fundamental and two harmonics the (2a + l)th and the (2a
l)th. With three slots per pole the tooth harmonics are the
seventh and the fifth; and with six slots per pole the harmonics
are the thirteenth and the eleventh.
The pulsations of the flux are opposed by the large inductance
of the field winding and they are therefore very small and in
addition an nth harmonic current is opposed by a reactance in
the alternator armature and load circuit which is n times as
great as the reactance for the fundamental current.
If the width of the pole face plus a small allowance for fringing
is a multiple of the slot pitch the variation of the gap reluctance
is negligible and the pulsation of flux does not occur.
247. Effect of Third Harmonics in Three-phase Alternators.
If the e.m.f . per phase in a three-phase delta-connected alternator
contains a third harmonic, triple-frequency currents will circulate
through the closed delta at all times. Referring to Fig. 129,
the resultant e.m.f. around the closed circuit at any instant is
ei + ^2 + 3 and this was shown to be equal to zero in the case
of three similar sine waves displaced at 120 degrees, Art. 123.
The third harmonics are, however, not combined at 120 degrees
but at 3 X 120 = 360 degrees and are in phase and the resultant
e.m.f. is three times the magnitude of the third harmonic of one
phase and this will cause a third harmonic of current to circulate
through the closed winding. This current may be of the order
of full-load current in the case of alternators of low reactance.
The reactance of the alternator winding to the triple-frequency
current is three times that opposed to the current of fundamental
frequency but only the true reactance and not the synchronous
reactance is effective in limiting the circulating current. The
third harmonic of e.m.f. does not appear at the terminals since
it is consumed in producing the circulating current.
If the alternator is connected F, the third harmonic e.m.f. will
not appear in the e.m.f. between terminals, since this e.m.f. is
the difference of two e.m.f s. at 120 degrees to one another or the
sum of two e.m.fs. at 60 degrees, the third harmonics will be
combined at 3 X 60 = 180 degrees and will therefore neutralize
one another. If, however, the neutral is connected to ground at
both the generator and receiver ends. A third harmonic of
current may flow in the neutral supplied from the three phases.
Alternators should, whenever possible, be connected Y instead
of A to reduce the danger of circulating currents.
300 ELECTRICAL ENGINEERING
248. General Electromotive Force Equation. The e.m.f.
equation, derived in Art. 239,
E = 4yfn3> 10~ 8 volts
which applies only to concentrated windings may be extended to
include all windings by introducing the distribution factor 5.
Thus the general equation for the effective value of the e.m.f.
between terminals of an alternator is
E = 45yfn3> 10~ 8 volts, (288)
where
/ = frequency in cycles per second,
n = number of turns in series between terminals,
3> = flux from one pole,
7 = form factor of the e.m.f. wave,
8 = distribution factor of the winding.
This equation holds both for the single-phase alternator and for
any phase of a polyphase alternator with n turns in series per
phase.
If the winding is short pitch the e.m.f. is reduced in the ratio
cos o 1 where the coil pitch is 180 a electrical degrees.
4
249. Rating of Alternators. Alternators are designed to give
a certain terminal voltage and to supply any current up to a
certain maximum or full-load current.
The output is
P = nEI cos 6 watts,
where
E is the voltage per phase,
/ is the full-load current per phase,
cos 6 is the power factor of the load, and
n is the number of phases.
The power output, therefore, depends on the voltage which is a
fixed quantity, the current which is variable and is limited by the
allowable temperature rise caused by the copper losses and other
losses in the machine, and the power factor of the load over which
the designer has no control.
. Alternators should, therefore, be rated not in watts or kilowatts
which depend on the power factor but in volt-amperes or kilovolt-
amperes.
SYNCHRONOUS MACHINERY 301
A machine rated at 1,000 kva. can supply 1,000 kw. to a non-
inductive load at unity power factor or it can supply 1,000 X
0.80 = 800 kw. to an inductive load of 80 per cent, power factor.
250. Comparative Ratings of an Alternator Wound Single-,
Two- and Three-Phase. Take the case of a machine with six
slots per pole. Let e be the effective value of the e.m.f. generated
in each coil of the winding and I be the current per conductor.
The current will be the same in the three cases for the same tem-
perature rise.
When wound single-phase using all the slots the distribution
factor is 0.64 and the terminal e.m.f. is
E = 6e X 0.64
and the output is
P'i = El cos = 3.84e7 cos <,
where cos $ is the power factor of the load.
When wound single-phase 'using only four slots per pole the
terminal e.m.f. is
E = 4e X 0.83
and the output is
Pi = El cos = 3.3207 cos <.
When wound two-phase with three slots per phase per pole the
distribution factor is 0.91, the e.m.f. per phase is
E = 3e X 0.91
and the output is
P 2 = 2J5J7 cos = 5.46e/ cos .
When wound three-phase with two slots per phase per pole the
distribution factor is 0.96, the e.m.f. per phase is
E = 2e X 0.96
and the output is
P 3 = 3EI cos = 5.76e/ cos 0.
Taking the three-phase rating as 100 the comparative ratings
are as given below.
Number of phases Rating
Three-phase 100.0
Two-phase 95 .
Single-phase using all the slots 67 .
Single-phase using only four slots per pole 57 . 7
302
ELECTRICAL ENGINEERING
In practice an alternator is given the same rating two- and three-
phase and 65 per cent, of that rating single-phase.
251. Armature Reaction. The flux distribution in the air gap
of an alternator at no load is symmetrical about the center line of
the pole and usually follows approximately a sine wave. The
e.m.f. generated in the armature is also a sine wave (see Fig.
285).
When current flows in the armature winding, the m.m.f. of the
armature combines with the m.m.f. of the field and changes both
CdH-^s
77 <,T7
Alte-uator
Synchronous
Motor
S I Alternator
FIG. 290. Armature reaction and reactance.
the magnitude and distribution of the flux crossing the air gap
and cut by the armature conductors. It thus changes both the
magnitude and the wave form of the e.m.f. generated. These
results are termed the armature reaction.
Armature reaction depends not only on the intensity of the
current in the armature but also on its phase relation with the
generated e.m.f. Fig. 290 illustrates armature reaction in a
machine with a single-phase concentrated winding.
In (a) the armature coil is shown in the position of zero e.m.f;
if the current is in phase it is also zero.
SYNCHRONOUS MACHINERY 303
In (b) the e.m.f. is maximum and the current is maximum.
The m.m.f. of the armature is cross-magnetizing, that is, it de-
creases the flux over one-half of the pole and increases it over the
other half. The useful flux is only decreased by the small amount
lost due to the higher saturation, and, therefore, decreased permea-
bility over the half of the pole where the density is increased.
The flux distribution no longer follows a sine wave and the e.m.f.
will not be a sine wave.
In (c) the current is maximum but lags 90 degrees behind the
generated e.m.f. The m.m.f. of the armature acts directly
against the m.m.f. of the field. It is, therefore, demagnetizing
and decreases the flux but does not distort it.
In (d) the current is maximum and leads the e.m.f. by 90
degrees. The m.m.f. of the armature acts directly with the field
and magnetizes it. The useful flux is increased and is not
distorted.
The following results have been obtained :
1. A current in phase with the generated e.m.f. is cross-
magnetizing and only decreases the flux to a very slight extent.
2. A current lagging 90 degrees behind the generated e.m.f.
demagnetizes the field and decreases the flux and decreases the
generated e.m.f.
3. A current leading the generated e.m.f. by 90 degrees
magnetizes the field, increases the flux and increases the generated
e.m.f.
If the current lags behind the e.m.f. by angle , it may be
resolved into two components, / cos in phase with the e.m.f.
and, therefore, cross-magnetizing and I sin lagging 90 degrees
behind the e.m.f. and demagnetizing.
The effect of armature reaction increases almost directly with
the current until the magnetic circuit becomes saturated after
which it increases much faster than the current. Due to satura-
tion the cross-magnetizing effect increases faster than the current ;
the demagnetizing m.m.f. increases directly with the current and
the decrease of flux caused by it would be proportional to the
current, if it were not for the change in the leakage factor of the
machine. The increase of the leakage factor under load is
largely due to the presence of the demagnetizing armature m.m.f.
and the resulting decrease of flux must be charged against it.
In salient pole machines the cross-magnetizing m.m.f. acts on
a path of much larger reluctance than the demagnetizing m.m.f.
304 ELECTRICAL ENGINEERING
and its effect is correspondingly smaller. In machines with cylin-
drical rotors the reluctance of the air gap is uniform over the
whole periphery and the two m.m.fs. therefore act on similar
paths.
252. Armature Reactance. The flux produced by the cur-
rent in the armature coil in Fig. 290 may be separated into two
parts as shown.
Part (1) is the flux of armature reaction which crosses the gap
and interferes with the flux threading the field circuit. Its effect
is either cross-magnetizing, demagnetizing or magnetizirig.
Part (2) is the flux which only interlinks with the coil itself and
does not interfere with the flux produced by the field m.m.f. It
is the self-inductive flux of the coil and generates in the coil an
e.m.f. of self-inductance, which consumes a component of the
e.m.f. generated by rotation. This e.m.f. called the armature
reactance drop is equal to the product of the armature current I
and the armature reactance x and leads the current by 90 degrees.
The reactance is x = 2irfL, where L is the inductance of the
armature. L and x both decrease as the armature current in-
creases due to the increased saturation and, therefore, decreased
permeability of the leakage path surrounding the armature con-
ductors. They also vary as the armature is rotated, when the
conductor is under the pole the reluctance of its local leakage
/Tooth Tip Leakage Flux
Tooth Tip Leakage Flux
One Slot per Phase per Pole Three Slots per Phase per Pole
FIG. 291. Slot leakage flux and tooth- tip leakage flux.
path is minimum and L and x are large, when between the poles
the reluctance is maximum and L and x are reduced. An aver-
age value of x is chosen to represent the armature reactance. At
light loads when the current is small the reactance drop will be
greater than the value corresponding to the average reactance
and when the current is large it will be smaller.
253. Leakage Fluxes. The leakage fluxes of armature react-
ance may be separated into three parts: (a) the slot leakage flux,
(b) the tooth-tip leakage flux, Fig. 291, and (c) the end connection
leakage flux, Fig. 292.
SYNCHRONOUS MACHINERY
305
(a) The slot leakage flux is directly proportional to the number
of ampere conductors in the slot, that is, to the total current in the
slot; it increases directly with the slot depth and is inversely
proportional to the slot width. If the slots are partly closed, the
component of the flux across the narrow opening is very large.
The slot leakage flux is independent of the position of the slot
relative to the pole; due to saturation of the teeth it does not
increase directly with the current and therefore the component
of the armature reactance due to it decreases with increase of
current.
(6) The tooth tip leakage
follows paths such as those
shown in Fig. 291. It cannot
be calculated so easily or so
accurately as the slot leakage
and it depends to a limited
extent on the position of the
FIG. 292. End-connection leakage flux.
slot relative to the pole, espe-
cially if the air gap is short.
This does not apply to ma-
chines with cylindrical rotors.
(c) The end-connection
leakage flux follows a path entirely in air. It is proportional to
the number of ampere conductors in the phase belt but when
the winding is distributed in a number of slots the length of the
leakage path is increased. The great length of the end connec-
tions makes this component of the leakage flux comparatively
large. Since the path never becomes saturated this flux is
directly proportional to the current. Distributing the winding
in a number of slots per phase per pole decreases all of these
leakage fluxes by increasing the length of the leakage paths.
254. Polyphase Armature Reaction. If n is the number of
turns per phase per pair of poles on a two-phase alternator and
i\ = Im cos 6 is the current in phase 1 and iz = I m cos (0 90)
= I m sin 6 is the current in phase 2, the m.m.fs. of the two phases
are mi = ra'i = nl m cos 6 and m 2 = nl m sin 0. The two m.m.fs.
are in quadrature in time and space but combine to give a con-
stant m.m.f. nl m fixed in position relative to the field m.m.f.
and revolving synchronously backward relative to the armature.
This can be seen by reference to Fig. 293. AB is the winding
of phase 1 and the current is assumed to lag behind the e.m.f. by
20
306
ELECTRICAL ENGINEERING
angle <. At the instant represented the current is maximum and
the m.m.f. of the coil is maximum and acts in direction OF. The
current in phase 2 is now zero. The m.m.fs. acting are shown in
Fig. 294.
FIG. 294.
In Fig. 295 the coil A B has moved through angle 6 and its cur-
rent has decreased to I m cos 6 and its m.m.f. to nl m cos 6. The
current in coil CD has a value I m sin 8 and its m.m.f. is nl m sin 6.
FIG. 295.
FIG. 296.
The component of the m.m.f. of phase 1 in direction OY is
nl m cos B - cos 6 = nl m cos 2 6 and the component in direction O^Tis
nl m cos 6- sin 0.
The component of the m.m.f. of phase 2 in direction OF is
nl m sin B - sip- 3 = nl m sin 2 6 and the component in direction OX is
SYNCHRONOUS MACHINERY
307
-f- nl m sin 0-cos 6. The resultant m.m.f. of the two phases in di-
rection OF is nl m cos 2 + nl m sin 2 = nl m and in the direction
OX is nl m cos 6 - sin nl m cos sin = 0.
Thus the resultant armature m.m.f. is nl m in fixed direction
relative to the field m.m.f. and, therefore, revolving synchron-
ously relative to the armature (Fig. 296).
The direction of the resultant armature m.m.f. relative to the
field m.m.f. is determined by the angle of phase difference between
the current and the e.m.f. generated at no load.
If the current is in phase with the e.m.f., the armature m.m.f.
acts at right angles to the field m.m.f. and is, therefore, cross-
magnetizing only; if the current lags by angle $, the armature
m.m.f. can be separated into two components nl m sin which is
demagnetizing and nl m cos which is cross-magnetizing (Fig.
294).
Y
Cos(0-240)
X
FIG. 297.
FIG. 298.
If the two-phase winding, Fig. 295, is replaced by a three-phase
winding, Fig. 297, with the first phase AB in the same position
as before and the other phases CD and EF displaced 120 degrees
and 240 degrees from it, the m.m.fs. of the three phases will be
respectively nl m cos 0, nl m cos (0 120) and nl n cos (0 240)
and will act in the directions represented. As before is meas-
ured from the instant of maximum current and the currents in
the three phases are assumed to lag behind the e.m.fs. by angle .
The sum of the components of m.m.f. in direction OF is
nl m cos 2 + nl m cos 2 (0 - 120) + nl m cos 2 (0 - 240) = %nl m .
and the sum of the components in the direction OX is
nl m cos sin -f nl m cos (0 - 120) sin (0 - 120)
+ nl m cos (0 - 240) sin (0 - 240) = 0.
308
ELECTRICAL ENGINEERING
Thus, the resultant m.m.f. of the armature of a three-phase alter-
nator is
M a =%nl m , (289)
where n is the number of turns in series per phase and I m is the
maximum value of the armature current.
The armature m.m.f. is fixed in direction relative to the fields
and revolves synchronously relative to the armature (Fig. 298).
@@(B
w ^rn^r
flTTp
44
ll-0.86l m C
r... 8 , I TT Tj
Time, = ._[_ _Q
HUH
S.O
Q
Armature M.M.F.with the Current in Phase with js
the Generated Voltages per Phase. Cross-MagnetlzingOH
*i = . QQOOOOO
2.2 ~ vOOlW * M * * ^ "^ * Kl '
Time, ^*0
HUH
HI IH 5
Armature M.M.F.with the Current in Quadrature
Behind the Generated Voltages per Phase. Demagnetizing]
Voltages per Phase
urrents per Phase
(in Phase)
JOC
Currents per Phase
(Lagging)
HUH
a Currents per Phase
(Leading)
Armature M.M.F.with the Current in Quadrature
Ahead of the Generated Voltages per Phase. Magnetizing
FIG. 299. Armature m.m.f. of a three-phase alternator.
Fig. 299 shows the same facts in another way in the case of a
three-phase alternator with its winding distributed in six slots
per pole.
(1) is a section through the conductors indicating the directions
and intensities of the e.m.fs. for the time t = 0.
SYNCHRONOUS MACHINERY
309
(2) shows the e.m.f. waves generated in the three phases.
(3) shows the armature winding.
(4), (5) and (6) represent the armature m.m.fs. corresponding
to the three instants, t = 0, t = ti and t = t 2 or 6 = 0, = 30
degrees and 6 = 60 degrees, with the currents (7) in phase with
the generated e.m.f. The armature m.m.f. is cross-magnetizing;
its value represented by the area under the m.m.f. lines is approx-
imately constant and it is fixed in position relative to the poles
and therefore revolves relative to the armature.
(8) and (9) show the armature m.m.f. and currents at time t =
0, with the currents in quadrature behind the e.m.fs. and there-
fore demagnetizing.
(10) and (11) show the armature currents leading and the
armature m.m.f. magnetizing,
255. Single -phase Armature Reaction. In a single-phase
alternator with n armature turns per pair of poles carrying a
current i = I m sin 0, the armature m.m.f. varies from a maximum
value nl m to zero; it is fixed in direction relative to the armature
and revolves relative to the poles. It produces a double-fre-
quency pulsation of the field and a third harmonic of e.m.f.
FIG. 300. Single-phase armature reaction.
Fig. 300 shows four instants during the revolution of a single-
phase alternator in which the current is assumed to be in quadra-
ture behind the generated e.m.f. In (1) the e.m.f. is zero and
the current is maximum, the armature m.m.f. is demagnetizing
and reduces the flux. In (2) the current is zero and the flux has
its maximum value. In (3) the current is maximum again and
the flux is reduced as in (1). In (4) the current is zero again.
The values of the flux crossing the gap for one revolution or
one cycle are shown in Fig. 301. The flux may be separated into
two parts, the constant value $ and the alternating flux of
amplitude $1, which goes through two complete cycles during one
cycle of the current or e.m.f. Single-phase armature reaction
310
ELECTRICAL ENGINEERING
thus produces a double-frequency pulsation of the flux crossing
the gap.
If time t and angular displacement of the rotating field are
measured from position (1), the flux inclosed by the coil at this
instant may be represented by
x
&
0,
T
0min
Cycle
1 Cur
Cycle
D @
Resultant
E.M.F. Wave
SO 90 180 270 360
h Angular Displacement =
FIG. 301. FIG. 302.
At time t and .angle after position (1) the flux inclosed is
= 4>o cos 6 = $ cos 0 d
6 " ~ U dt = ~ n dt
cos - y (cos 30 + cos 0) 1
= n \ 3> sin ~ (3 sin 30 + sin 0) I -^
2 } at
but = 2wft and = 2irf.
and therefore
e = 2-n-fn j ( ~j sin % i sin 30
I \ 2 /
i in
= E sin d - E sin 30.
m m
The generated e.m.f. consists of an e.m.f. of fundamental
frequency having a maximum value E m = 2irfn ( < ~) and a
third harmonic of maximum value E = Sirfn 0ij the twoe.m.fs.
m
pass through zero together but in opposite directions and the
resultant wave is symmetrical with a peak at the center (Fig. 302) .
Thus single-phase armature reaction produces a double-fre-
quency pulsation of the field and a third harmonic of e.m.f.
SYNCHRONOUS MACHINERY 311
Since the magnetic circuit is surrounded by a field winding,
consisting of a large number of turns, the pulsations will be less
than the above results indicate. The variation of the flux linking
with the field winding induces in it e.m.fs. and currents which
oppose the variation and limit it to a small value.
In a machine with a large number of field turns the pulsation
produced by armature reaction up to full-load current is very
small and the armature reaction may be considered as constant in
value with reference to the fields.
In the case of a short-circuit, however, where from three to five
times full-load current flows in the armature, a large pulsation of
flux is produced in the magnetic circuit and very large e.m.fs.
and currents may be induced in the field windings.
If a short-circuit occurs on one phase only of a three-phase
alternator, the armature reaction can be separated into the ordi-
nary three-phase armature reaction with equal currents and a
single-phase armature reaction due to the excess of the short-
circuit current over normal current acting in the turns of one
phase. This single-phase armature reaction produces a double-
frequency pulsation of the field and a third harmonic of e.m.f.
in all the phases.
The effects of single-phase armature reaction are relatively
greater in machines with a small number of turns on the fields, as
turbo alternators.
256. Electromotive Forces in the Alternator. In studying
the performance of an alternator it is necessary to determine the
relation between the terminal e.m.f. E, the e.m.f. E\ generated by
rotation and e.m.f. E Q generated at no load.
EI is the e.m.f. generated in the armature by the rotation of the
flux produced in the air gap by the resultant of the m.m.fs. of
the field and armature. It is the vector sum of the terminal
e.m.f. E and the e.m.f. consumed by the impedance of the arma-
ture. The armature impedance is Z = \/r 2 + x 2 , or expressed
in rectangular coordinates Z = r + jx, where r is the resistance
of the armature and consumes a component of e.m.f. Ir in
phase with the current /, and x is the true self-inductive reactance
of the armature and consumes a component of e.m.f. Ix in quad-
rature ahead of the current.
The generated e.m.f. thus is
& = + IZ
= $ + I(r+jx), (290)
312 ELECTRICAL ENGINEERING
and the terminal e.m.f . is the vector difference between the e.m.f.
generated in the armature by rotation and the impedance drop
E = Et- I (r+jx). (291)
Eo is the e.m.f. generated at no load due to cutting the flux pro-
duced by the field m.m.f. M f acting alone. Under load current
flows in the armature and exerts a m.m.f. M a , which is either
cross-magnetizing, demagnetizing or magnetizing depending on
the phase relation of the current and the terminal e.m.f. This
armature m.m.f. combines with the field m.m.f. and changes both
the intensity and the distribution of the flux in the gap, so that
under load the e.m.f. generated in the armature is not the same as
at no load. .
The difference between the two is the e.m.f. consumed or the
e.m.f. not generated due to the presence of the armature reaction.
This e.m.f. is proportional to the current and can be expressed as
the product of the current / and a component of reactance x' '. It
is Ix' and is in quadrature ahead of the current. Thus
E Q = E l + jlx'
= $ + I(r+jx) + jlx f
= E + I{r+j(x + x')}
= E + I(r+jx 8 ). (292)
The total reactance of the armature x s is called the synchronous
reactance and consists of two components, x which represents the
effect of the armature leakage flux and which has been called the
armature reactance and x' which represents the effect of armature
reaction.
The quantity Z 8 = r + jx 8 is the synchronous impedance of the
armature and consumes a component of the no-load e.m.f. IZ 8 =
I (r + fas) which is the synchronous impedance drop in the
armature.
Thus the e.m.f. generated at no load is the vector sum of the
terminal e.m.f. and the synchronous impedance drop
E Q = E + I(r+jx s ).
257. Armature Resistance. The resistance of the armature r,
used in calculations for regulation and efficiency, is not the true
ohmic resistance as measured by passing direct current through
the winding but is from 20 to 100 per cent, greater than this
value. This increased resistance is sometimes called the effective
resistance of the armature.
SYNCHRONOUS MACHINERY
313
The loss of power which is charged against the armature copper
is the increase in the total losses in the machine due to the presence
of the armature winding carrying current. This loss includes (1)
the true 7 2 r loss due to the passage of the load current, (2) any
increase in the iron losses due to the distortion of the flux by the
armature currents and (3) eddy-current losses in the armature
copper, such as those discussed in Art. 188.
In ordinary cases it is advisable to add about 50 per cent, to
the resistance measured by direct current; a slight error will not
affect the efficiency of the machine to any great extent since the
copper loss is usually considerably less than the iron losses.
258. Vector Diagram of Electromotive Forces and Magneto-
motive Forces. In Fig. 303 (a) is shown the vector diagram of an
alternator supplying an inductive load.
FIG. 303.
E = terminal e.m.f.
7 = armature current lagging behind E by angle .
IT = e.m.f. consumed by armature resistance, in phase
with 7.
Ix = e.m.f. consumed by armature reactance, in quadra-
ture ahead of the current.
Eb = Ix = counter e.m.f. of armature reactance due
to the armature leakage flux + jE sin <;
the synchronous impedance drop is
/Z. = Ir+jlx 8 ;
and the no-load e.m.f. is
E = E + !? 8 = (E cos + Ir) + j (E sin + la?.);
or taking the absolute value
E Q = (E cos (^ + 7r) 2 + (E sin + Ix s ) 2 . (293)
259. Voltage Characteristics. The relation between the ter-
minal e.m.f. and armature current of an alternator, with a fixed
value of field current and a given load power factor, is called
the " regulation curve" or "voltage characteristic" for the given
power factor.
When the power factor is unity and the current is in phase with
the terminal e.m.f. E } it lags by an angle < (Fig. 304) behind the
316
ELECTRICAL ENGINEERING
no-load e.m.f. E and there is thus a small demagnetizing effect
proportional to / sin < , which decreases the flux and a large cross-
magnetizing effect proportional to / cos fa which changes the
distribution of the flux but only decreases it slightly due to satura-
tion. Thus even with non-inductive load the armature reaction
causes a decrease in the flux crossing the air gap, and the e.m.f. EI
generated by rotation is less than the no-load e.m.f. E . In
addition, the armature reactance x and the resistance r both con-
sume components of e.m.f. proportional to the current.
120
10. 20 30 40 50 60 70 80 90 100
Percent Load Current
10 20 30 40 50 60 70 80 90 100
'Percent Load Current
FIG. 308. Voltage characteristics of FIG. 309. Compounding curves of an
an alternator. alternator.
Therefore, at non-inductive load the terminal e.m.f. falls with
increasing current as shown in curve (1), Fig. 308, which is the
voltage characteristic for unity power factor.
With inductive load the demagnetizing effect is increased and
the terminal e.m.f. falls off more, curve (2). With a capacity
load in which the current leads the terminal e.m.f. the armature
m.m.f. is magnetizing and so raises the terminal e.m.f., curve (3).
These voltage characteristics are calculated from equation
(293) on page 315. The value of field current // is chosen and the
corresponding value of E Q obtained from Fig. 310. Any required
power factor cos is taken, the current I is varied and the values
of E obtained and plotted as ordinates.
SYNCHRONOUS MACHINERY
317
260. Compounding Curves. The "compounding curves" or
" field characteristics" show the relation between the field current
and armature current for a constant terminal e.m.f. at any re-
quired power factor.
Fig. 309 shows the compounding curves for unity power factor,
curve (1), 80 per cent, power factor lagging, curve (2), and 80 per
cent, power factor leading, curve (3).
At non-inductive load an increase of field current is required as
the load current increases to maintain a constant terminal e.m.f.
With inductive load a much larger increase of field current is
required to counteract the effect of the lagging current.
With capacity load the field current must be decreased in order
to maintain a constant terminal e.m.f.
The same results are shown in the diagrams, Figs. 304, 305, and
306.
These compounding curves can also be calculated from equation
Eo = \/(E cos + Ir) 2 + (E sin + Ix s ). z
a b c
Field Kxcitatiou
field Ejcitation \ f My
FIG. 310. Saturation curves.
E remains constant, a certain value of cos is chosen, / is
varied and the value of Eo is calculated. The corresponding value
of // is obtained from the saturation curve, Fig. 310, and is
plotted on the" armature-current base.
261. Tests for the Determination of the Regulation of Alter-
nating-current Generators. The standards committee of the
American Institute of Electrical Engineers suggests the three
following methods of determining the regulation of alternating-
current generators. They are given in the order of preference.
318
ELECTRICAL ENGINEERING
Method (a). The regulation may be measured directly by
loading the generator with the specified load and power factor
and then reducing the load to zero. The difference between the
voltage readings at no load and full load expressed as a per cent,
of the full-load voltage is the regulation under the specified
conditions. The two voltage readings must be taken under the
same conditions of speed and excitation.
This method cannot be applied generally for shop tests on
large generators and either method (6) or (c) must be used.
Method (6). The regulation is determined from data obtained
from the no-load saturation curve and the full-load saturation
curve for zero power factor (Fig. 310). The latter curve may be
obtained by loading the generator with under-excited synchron-
ous motors, which can be made to give a very low power factor.
From these two curves points on the load saturation curves
for any other power factor can be obtained as follows: With
FIG. 311.
excitation oa the open-circuit voltage is am = EQ and the termi-
nal voltage with full-load current at zero power factor is aA: 3 .
The drop of voltage is wfc 3 and this is the voltage consumed by
the synchronous reactance of the armature, since under the
condition of zero power factor it is directly subtracted from the
open-circuit voltage if the resistance drop is neglected.
Fig. 311 shows the vector diagram for this case.
EQ = am = no-load or open-circuit voltage.
/ = full-load current in quadrature behind EQ.
Ix 8 = voltage consumed by the synchronous reactance in
quadrature ahead of /.
E = ak z = terminal voltage; it is obtained by subtracting
= Ix 8 from EQ. *
SYNCHRONOUS MACHINERY
319
To obtain the corresponding point on the load saturation curve
for any other power factor the vector k 3 m Ix a must be sub-
tracted from EQ in its proper phase relation.
The diagram in Fig. 3 12 (a) may be applied for any power
factor. The horizontal line through a is taken as the direction
of the constant current vector / and a circle of radius am = E Q
is described about a. The terminal voltage line for any power
factor cos 4> makes an angle < with the current line; the voltage
consumed by synchronous reactance as = Ix s is se.t up in quadra-
ture ahead of I and through its extremity s a line is drawn parallel
to the terminal voltage to cut the circle of radius EQ aim. From
m the line mk s is drawn parallel to as to cut the terminal voltage
line at & 3 . The vector ak$ then represents the terminal voltage
E under the given conditions.
This diagram may be replaced by the simplified diagram in
Fig. 312(6) which gives the same results. The circle of radius
Eo is drawn about m as center instead of a.
If it is desired to take account of the resistance drop the dia-
grams in Fig. 312 may be replaced by those in Fig. 313.
(a)
FIG. 313.
In Fig. 310 load saturation curves for unity power factor
(0 = 0) and 80 per cent, power factor ( = 36.8 degrees) are
shown.
The values of regulation corresponding to the various power
factors and with a fixed field excitation oa are as follows :
Regulation at unity power factor = ~
Regulation at 80 per cent, power factor =
Regulation at zero power factor
100 per cent. = lOO per cent. = 8.7 per cent.
100 per cent. = |^100 per cent. = 26.5 per cent.
* 100 per cent. = ^100 per cent. = 43.0 per cent.
320
ELECTRICAL ENGINEERING
When expressing the regulation of a generator for various
power factors it is usual to make the comparison on a basis of
equal terminal voltages = rated voltage and not under the condi-
tion of fixed excitation.
To obtain these values of regulation the various saturation
curves must be extended to cut the rated voltage line.
The regulation at unity power factor
The regulation at "80 per cent, power factor
The regulation at zero power factor
= - 100 per cent. = 7 per cent.
100 per cent. =16 per cent.
100 per cent. = 25 per cent.
These values could not be obtained from a single vector dia-
gram since the synchronous reactance drop is not constant but
decreases as the saturation increases.
FlG. 314.
Method (c). If it is not possible to obtain the full-load satura-
tion curve at zero power factor from test, it may be constructed
from the no-load saturation curve and the short-circuit curve
(Fig. 314). The short-circuit curve is obtained by short-circuit-
ing the generator through an ammeter. A very low value of
field excitation must be taken at start and gradually increased
until the armature current is about twice full-load current. The
SYNCHRONOUS MACHINERY 321
curve showing the relation between the armature current and
the field current is a straight line passing through the origin and
it may be extended to any desired length.
Since the armature is short-circuited and there is therefore no
terminal voltage the no-load voltages EQ (curve (1)) must be
consumed as synchronous impedance drops by the armature
currents (curve (2)). The ordinates of curve (1) divided by the
corresponding ordinates of curve (2) give the values of the syn-
chronous impedance corresponding to the various values of
exciting current (curve (3)).
If these values of Z s or x s are multiplied by full-load current
and subtracted from the ordinates of curve (1), the result is
curve (5) which may approximate very closely to the full-load,
zero-power-factor, saturation curve. Using this curve the regu-
lation can be obtained as in method (&).
In machines of low saturation, low reactance and small mag-
netic leakage, a curve such as (4) obtained by moving curve (1)
to the right through a constant distance OB will be fairly close
to the required zero-power-factor saturation curve, but in ma-
chines of high saturation large reactance and large magnetic
leakage the curve obtained in this way is much too high. This
latter is the method suggested by the standards committee.
In Fig. 310 the broken line shows the upper part of a no-load
saturation curve calculated using the full-load leakage factor.
When this correction is made the horizontal distance between the
no-load and the full-load zero-power-factor saturation curves is
more nearly constant but is still considerably increased on the
upper part of the curve.
The triangle pqr, Fig. 310, has as its horizontal side the line pq
which represents the effect of armature reaction expressed in
ampere-turns and as its vertical side the line qr representing
a voltage drop, which is the true reactance drop Ix. The distance
between the two curves measured in the direction pr should be
very nearly constant.
262. Regulation. The voltage regulation of an alternator is
expressed as the per cent, rise of the terminal voltage when full
load is removed without changing the speed or field excitation.
The regulation is usually expressed for 100 per cent, power-factor
and 80 per cent, power-factor loads. Usual values are 6 to 9 per
cent, at 100 per cent, power factor and 15 to 25 per cent, for 80
per cent, power factor. Where automatic voltage regulators are
21
322 ELECTRICAL ENGINEERING
used to maintain constant terminal voltage close regulation is not
necessary. In modern machines of large output values of
regulation as low as 14 or 15 per cent, at 100 per cent, power
factor are considered satisfactory. Such poor regulation in-
dicates a large synchronous reactance which limits the short-
circuit current to one and one-half or two times the full-load
current and the instantaneous short-circuit current to six or eight
times full-load current. This minimizes the stresses on the end
turns which are proportional to the square of the current. In
many cases reactance coils are connected in series with alternators
for the same purpose.
The voltage regulation depends mainly on two factors: (a)
the ratio of field strength to armature strength, and (6) the degree
of saturation of the iron parts of the magnetic circuit.
The excitation regulation of an alternator is the per cent.
increase of field excitation required to maintain constant terminal
voltage as the load increases from zero to full load. In Fig. 310
the excitation regulation at 100 per cent, power factjr is -
100 per cent. = 14 per cent, and at 80 per cent, power factor it is
o>c
- X 100 per cent. = 43 per cent.
oo>
The extra excitation has three components:
1. That required to counteract the effect of armature reaction
especially its demagnetizing component.
2. That required to make up for the voltage drop due to the
reactance and resistance of the armature winding.
3. That required to provide for the increase in the magnetic
leakage.
The excitation regulation of an alternator under different con-
ditions of load and power factor may be obtained from the com-
pounding curves, Fig. 309, or from to no-load and full-load satura-
tion curves, Fig. 310.
(a) Fig. 315 (a) shows four no-load saturation curves drawn for
a machine with its magnetic circuit unsaturated but with the
length of the air gap increased by equal amounts from 100 per
cent, in (1) to 400 per cent, in (4), the armature strength remain-
ing constant. The field strength increases directly as the air-gap
length and the regulation as represented by ^^> -~A - and
604
is very much improved. In alternators of ordinary design
SYNCHRONOUS MACHINERY
323
constants the field strength is two to three times the armature
strength and the air gaps are much longer than in direct-current
generators.
Fig. 315(6) shows four saturation curves for a machine with a
fixed gap but with increasing saturation of the iron parts of the
magnetic circuit. The voltage regulation is very much improved
by saturation.
Such machines require a minimum of iron in the magnetic
circuit and are light in weight but it is difficult to obtain any
considerable increase of voltage without overheating the fields.
1 2 3
Field Excitation
Field Excitation u Field Excitation
Effect of Saturation on Regulation
Effect of Gap Length on Regulation
(a) (b) to
FIG. 315. Regulation of alternators.
In Fig. 315(c), Oci and Ocs are saturation curves for two
machines with low densities in the iron; the gap in the second one
is twice the length of that in the first. Oc 2 and Oc4 are curves
for similar machines with saturated magnetic circuits. The
improvement in the regulation due to saturation is not so great
in the machine with the long gap as in the other.
263. Short-circuit of Alternators. If the impedance of the
load circuit of an alternator, with normal excitation, is gradually
reduced to zero, the short-circuit current is limited only by the
synchronous impedance of the armature; it is
E
Z s
(294)
and ranges from one and one-half to three or four times full-load
current, the larger values occurring in machines of low synchron-
ous reactance and good regulation.
In the case of sudden short-circuits the momentary current
ET
will be much larger than the permanent value -~r because the
&8
component x' of the synchronous reactance, which represents
324
ELECTRICAL ENGINEERING
the effect of armature reaction, does not act instantaneously to
limit the current. It represents a change in the flux which inter-
links with the field circuit of the machine and on account of the
inductance of the field winding with its large number of turns,
this change of flux cannot take place instantaneously but may
take several seconds to become complete. The initial wave of
current is limited only by the true impedance of the armature
and is
E * (295)
The amplitude gradually decreases as the armature reaction
reduces the flux in the field and the generated e.m.f. E Q . With
very large armature currents the path of the armature reactance
flux becomes saturated and x is reduced below its normal value.
Armature Current in Phase 1
"Instantaneous Maximum
, Short Circuit Current Approximately 2^/2 1 sc
1 Transient Field Current
(a)
Phase 2
Armature Current in Phase 2
. Pulsating Field Current 6 )
Normal Field
Current
Normal Field Current
Field Current
C*)
FIG. 316. Short-circuit currents in a two-phase alternator.
J '
I'sc = 7 = will represent the short-circuit current
\/r 2 -j- x 2 x
only when the short-circuit occurs at the maximum point of the
generated e.m.f. wave as illustrated in Fig. 316(a). Neglecting
the armature resistance the generated e.m.f. is consumed by
the back e.m.f. of armature inductance = L ~v' The cross-
hatched area under the e.m.f. wave is I edt and is = I L-j-dt
JO JV2I'.< di
= \/2I' to L and it causes the current to increase from to
SYNCHRONOUS MACHINERY 325
V/2/'c- The succeeding quarter wave of e.m.f. is negative and
reduces the current to zero again; it then reverses and grows in
the opposite direction to a value slightly less than before since the
armature reaction will have caused a slight reduction in the flux
and the generated e.m.f. The current is practically symmetrical
about the zero line and its amplitude decreases until the reduction
of flux by armature reaction is complete after which it remains
constant. The transition from the initial to the permanent value
will take from a few hundred to thousands of cycles depending on
the inductance of the field winding. In Fig. 316 it is represented
as covering only a few cycles.
Fig. 316(6) shows the short-circuit current in the second phase
of the machine. The short-circuit occurs at the zero point of
the generated e.m.f. wave and the positive area under the wave
is twice as great as before and builds the current up to double the
value in (a). The current then decreases and it alternates not
about the line of zero current as before but about a line starting
above the zero line by an amount \/21 8C o n the current scale and
sloping down to coincide with the zero line after a few cycles.
The armature current may be considered as being composed
of a transient direct current, shown as a broken line, and, super-
imposed on it, the alternating current l' 8C with its decreasing
amplitude.
The m.m.f. of the transient armature current, being fixed in
position on the armature, revolves relative to the field poles and
produces a single-frequency pulsation of the field flux which dies
out with the passing of the transient current. This pulsation of
the field causes the generated e.m.f. to become unsymmetrical
about the zero line as shown and so after the first cycle the arma-
ture current crosses the zero line and after a few cycles becomes
symmetrical.
The magnitude of the transient armature current depends on
the point of the voltage wave at which the short-circuit takes
place and therefore the short-circuit currents in the various
phases of a polyphase alternator will not be identical.
In machines of high self-inductance and low armature reac-
/' / _1_ yf
tion the ratio y^ = will be small but in large low-fre-
J-sc ^
quency turbo-alternators with low self-inductance and high arma-
ture' reaction the ratio may be very large In extreme cases the
instantaneous short-circuit current may reach twenty to thirty
326 ELECTRICAL ENGINEERING
times full-load current. Machines should be designed so that
the instantaneous short-circuit current will not exceed five to
eight times full-load current or if this is not possible external
reactance must be added to limit the currents.
Circuit breakers are provided with time limits so that they will
not open the circuit until the current has fallen to its steady
value.
The pulsation of the field flux due to the m.m.f. of the tran-
sient armature current induces in the field winding an alternat-
ing current of normal frequency which opposes the pulsation of
the flux; this pulsating field current is superimposed on the nor-
mal field current giving the resultant current shown in Fig. 316(c).
In some cases the maximum field current may be ten to fifteen
times its normal value.
If the field poles are solid or if they are provided with damper
windings, the alternating current instead of appearing in the field
winding may appear as eddy currents in the pole faces and dam-
pers and the field current will scarcely be affected at all.
ted Voltage
7\
u
u
u
* Current
1 A A A A A //
V \
11
V
V
V
o
V
V
V
V
V
Field Current with Double Frequency Pulsation
(6)
FIG. 317. Single-phase short circuit currents.
Fig. 3 17 (a) shows the short-circuit current of a single-phase
alternator in which the circuit was closed at the maximum point
of the voltage wave. There is no transient term and the current
falls off from the large initial value to the permanent value as in
the case of the polyphase machine. The current wave has,
however, a decided peak showing the presence of a third harmonic
due to the double-frequency pulsation of the field current as
shown in Fig. 317(fe). This phenomenon was discussed in Art.
255. If the circuit is closed at any other point on the voltage
wave the current will have a transient term and the field current
will have a single-frequency pulsation in addition to the double
frequency pulsation; the single-frequency pulsation disappears
after a few cycles while the double-frequency pulsation decreases
SYNCHRONOUS MACHINERY
327
from a maximum at the first instant to its permanent value.
Single-phase generators should always be provided with dampers
which reduce the pulsation of the field current and the resulting
third harmonic.
A short-circuit of one phase of a polyphase alternator is similar
in its effects to a short-circuit of a single-phase alternator but in
addition a third harmonic appears in the voltages of the phases
not short-circuited.
264. Synchronous Motor. A synchronous motor is similar to
an alternator in construction and may be either single-phase or
polyphase. The single-phase motor is not self-starting and must
Direction of JJevolving Field and Torque
Current :i;i Phase 1
t in Phi
(c) (d)
FIG. 318. Two-pole, two-phase synchronous motor.
be brought up to synchronous speed before being connected
to the supply. It is, therefore, not used except in special cases.
The polyphase motor when connected to the supply will acceler-
ate and run up to synchronous speed but only a low voltage
should be impressed on it at start or very large lagging currents
will be drawn from the supply lines. In many cases, however,
polyphase synchronous motors are started by auxiliary motors
to prevent voltage fluctuations on the line.
. Fig. 3 18 (a) represents a two-phase, two-pole motor. The
328 ELECTRICAL ENGINEERING
armature is stationary and is supplied with two-phase alternating
currents, Fig. 318(6). The armature m.m.f. is constant in value
as in the alternator and revolves at synchronous speed in the anti-
clockwise direction and produces a revolving field of constant
value. Figs. 3 18 (a), (c) and (d) represent the armature m.m.f. at
the instants (1), (2) and (3).
The speed of the field is directly proportional to the frequency
of the impressed e.m.f. and inversely proportional to the number
of pairs of poles; it is
n = ~ = rev. per sec. (296)
P P
This is the speed at which the motor operates and it is constant
independent of the impressed e.m.f. of the field excitation and of
the load.
265. Vector Diagrams for a Synchronous Motor. If an alter-
nating e.m.f. E is impressed on the terminals of a synchronous
motor of resistance r and synchronous reactance x s and a current I
flows in the armature, the phase relation of the current and the
impressed e.m.f. depends on the field excitation.
M/
Power Factor -100 Parcent Power Factor -50 Percent Lagging
' M / " _ oOLag
FIG. 319. FIG. 320.
In Fig. 319:
E = impressed e.m.f., which remains constant.
7 = armature current, in this case, in phase with E.
Ir = component of the impressed e.m.f. consumed by the
resistance, in phase with the current.
Ix 8 = component of the impressed e.m.f. consumed by
the reactance x sj in quadrature to the current.
IZ a = component of the impressed e.m.f. consumed by
the synchronous impedance, JZ a = /(r + jx t ).
SYNCHRONOUS MACHINERY 329
gf Q = component of the impressed e.m.f. consumed
by the counter e.m.f. of the motor.
E Q = counter e.m.f. of the motor, generated in the
armature by cutting the flux produced by the field
m.m.f.
Mf = field m.m.f. in quadrature ahead of the generated
voltage EQ. This is the value of field excitation
required to make the power factor of the motor
unity under the assumed conditions of load.
Fig. 320 is the vector diagram when the current has the same
value as before but lags behind the impressed e.m.f. by angle
4> = 60 degrees.
Fig. 321 is the vector diagram when the current leads by angle
< = 60 degrees.
Eo
Power Factor=50 Percent Leading
60 Lead
*,
FIG. 321.
Referring to these diagrams it is seen that the field excitation
required in a synchronous motor to produce a leading power factor
or to cause the current to lead the impressed e.m.f. is greater than
that required to produce a lagging power factor or to cause the
current to lag behind the impressed e.m.f.
If, therefore, the field current of a synchronous motor is varied,
there is no change in speed as in the direct-current motor, but the
generated e.m.f. EQ changes both its value and its phase relation
with the impressed e.m.f. E and allows leading or lagging currents
to flow to make up for the change in excitation; when the field
current is decreased a component of current 90 degrees behind the
impressed e.m.f. flows in the armature and magnetizes the field
330 ELECTRICAL ENGINEERING
and when the field current is increased a component of current
90 degrees ahead of the e.m.f. flows and demagnetizes the field.
266. Characteristic Curves. The most important character-
istic curves of the synchronous motor are (1) the compounding
curves, or the relation between field current and armature current
for given values of power factor, (2) the load characteristics
showing the relation between armature current and output and
power factor and output for a given value of field excitation, and
(3) the phase characteristics or V curves showing the relation
between armature current and field current for given values of
motor output.
To predetermine these curves it is necessary to know the re-
sistance r and synchronous reactance x s of the armature and to
have the no-load saturation curve showing the relation be-
tween the generated voltage EQ and the field current // or field
m.m.f.M/.
The impressed e.m.f. E is constant.
267. Compounding Curves. The voltage equation of the
synchronous motor is
E = E' + IZ 8 (297)
or E' = E - IZ a (298)
If the current / is taken as the line of reference, the impressed
e.m.f. can be expressed in rectangular coordinates as
E = E cos + jE sin 0,
where is the angle of lag of the current.
The impedance drop in the armature is
Ei = I?s = Ir + jlx.
The e.m.f. consumed by the generated e.m.f. is, therefore,
E' = (E cos - 7r) + j(E sin - Ix.) (299)
and its absolute value is
E' = V(E cos - Ir) 2 + (E sin - 7z fi ) 2 (300)
This relation can also be obtained by reference to the vector
diagram in Fig. 322.
but , , T-, T
oa = ob ab = E cos I r ,
and ad = bf - cf = E sin - Ix,.
therefore,
cos - Ir) 2 + (E sin - Ix 8 Y
SYNCHRONOUS MACHINERY 331
Fig. 327 represents the compounding curves for unity power
factor, 80 per cent, power factor leading and 80 per cent, lagging.
To predetermine these curves the impressed e.m.f. E is main-
tained constant, a definite value of power factor is chosen for
each curve, the armature current / is
varied and the values of E'Q are calcu-
lated from equation (300). The
values of field current // correspond-
ing to the calculated values of E'o
are obtained from the saturation
curve, Fig. 325, and are plotted as
ordinates.
268. Load Characteristics. The power input to the motor
armature is the product of the current and the in-phase com-
ponent of impressed e.m.f.; it is
Pi = El cos 0. (301)
The electrical power transformed into mechanical power is the
product of the current and the in-phase component of the
e.m.f. E'o, consumed by the generated e.m.f.; it is
P = I(E cos - Ir) = El cos - 7 2 r, (302)
and is less than the power input by the armature copper loss.
The power output is less than the mechanical power developed
by the amount of the constant losses in the motor, namely, the
iron, friction and windage losses; the output, therefore, is
PZ = P constant losses
= PI I 2 r constant losses
= El cos I 2 r constant losses. (303)
Fig. 328 represents the load characteristics for a given value of
field current //. The value of // chosen here is such that the
back e.m.f. of the motor is greater than the impressed e.m.f.
and the motor is over excited at light load. If a lower value of
// had been chosen the power factor would never have reached
100 per cent, and would have been lagging through the whole
range of load. Since // is constant E' Q is constant. / is varied
and the corresponding values of cos are obtained from equation
300. These values are substituted in equation 303 and the
values of / and cos < are plotted on a base of power output.
At light loads the power factor is low and leading and the motor
332
ELECTRICAL ENGINEERING
is over-excited ; as the load is increased the power factor increases
until it reaches 100 per cent, at a value of load depending on the
field excitation; beyond this point the power factor decreases
again and becomes lagging. The current increases continually
but is finally limited by the synchronous impedance of the
armature.
FIG. 323.
These results may be verified by referring to the diagrams in
Fig. 323. In (2) the current / is in phase with the impressed
e.m.f . E, that is, the power factor is 100 per cent, and = 0. In
(3) the power output is increased and the current is assumed to
have increased to double the value in (2) ; the synchronous imped-
ance drop is also doubled ; E'o remains constant in all cases and its
position can be found by drawing arcs of two circles of radii E' Q
and IZ a about the two ends of the vector E. The synchronous
impedance drop can be resolved into its two components Ir and
Ix a by setting off the constant angle a = tan" 1 . The current
X 8
I is then drawn in the direction of Ir and it lags behind E by the
SYNCHRONOUS MACHINERY 333
angle 0. The power factor is cos and is lagging. The power
output, neglecting the constant losses, is equal to the product of
the current and the in-phase component of the e.m.f., E'o, con-
sumed by the generated voltage EQ or it is equal to the product
of the e.m.f., E'o, and the component of the current in phase
with it, and since E'o is constant the power output may be
represented by the in-phase component, of the current OP in,,
each of the diagrams.
In (1) the power is decreased and the current is taken as half
of that in (2); the position of E'ois found as before and the current
leads E by the angle $.
The power output cannot increase indefinitely but reaches its
maximum when the decrease in power factor overcomes the in-
crease in current or when the component of the current in phase
with the generated e.m.f. begins to decrease; the motor then be-
comes unstable and falls out of synchronism and stops. In (4)
the current is increased to three times the value in (2) its compo-
nent OP in phase with E' Q is greater than in (3) and therefore the
power output is still increasing. In (5) the current is four times
its value in (2) but the component OP in phase withE'ois less than
in (4) and the output is therefore decreasing. The position of
maximum output lies between (4) and (5).
The maximum power output of a synchronous motor is far
beyond the limits set by temperature rise.
269. Ph^se Characteristics. If the field excitation of a motor
with constant output is varied, the armature current changes
both its value and its phase relation with the impressed e.m.f.
For each output there is a certain value of field excitation which
makes the current a minimum and brings it in phase with the
e.m.f. ; as the excitation is decreased below this value the current
increases and becomes lagging; as the excitation is increased the
current increases and becomes leading.
In Fig. 329 (a) are shown the phase characteristics for outputs,
PI = or at no load, P 2 = full load and P 2 = twice full load.
For each curve the output Pz = El cos I 2 r constant
losses is kept constant; thus,
PZ + I 2 r + constant losses
cos =
EI
E' Q = V(E cos - 7r) 2 + (E sin - Ix 8 ) 2 .
As / varies the corresponding value of cos < is found from equa-
tion (304) and by substituting /, cos < and sin in equation (300),
334 ELECTRICAL ENGINEERING
the values of E' Q are found. These are replaced by the corre-
sponding values of field current // obtained from the no-load
saturation curve.
The lowest point on each curve represents the smallest current
input for the given output and thus represents the condition of
unity power factor. The curve joining these lowest points is the
compounding curve for unity power factor. If the phase char-
acteristics are very steep a slight change in field excitation pro-
duces a large change in armature current, or a large component
of wattless current is required to correct for a slight variation in
field excitation. This is the case in a motor with small synchro-
nous reactance or small armature reaction and the motor is un-
stable. If the synchronous reactance is large only a slight change
in armature current is produced by a change in field excitation
and the phase characteristics are flat and the motor is stable.
270. Torque. Since the speed of the synchronous motor is
constant, the torque developed at any output is directly propor-
tional to the output.
If T is the torque in pounds at 1-ft. radius developed in the ar-
mature, the electrical power transformed into mechanical power is
X 746 Watts = EI COS * ~
and thus the torque developed is
and the torque available for the load is
m EI cos < Pr constant losses xx 33,000 1U . /onA s
27T X r.p.m. ^W
271. Blondel Diagram for a Synchronous Motor. In Fig. 324
the triangle OAB is a voltage diagram for a synchronous motor.
OA = E is the constant impressed voltage, AB = E is the back
voltage of the motor and OB = E' = IZ 8 is the synchronous
impedance drop due to current 7. The angle OAB = a is called
the coupling angle.
The current I lags behind E by angle 4> and behind E' by angle
= tan
OC and AC are two straight lines making equal angles 6 with
the ends of the constant impressed e.m.f. vector OA.
SYNCHRONOUS MACHINERY
335
The current may be represented by the vector OB if its scale
is Z a times the voltage scale and the phase relation of the current
and the impressed e.m.f. is found as the inclination of OB to OC
which is equal to the phase angle <. The component of the
current in phase with the impressed e.m.f. is Oa, the projection
of OB on OC. '
For constant power input to the motor this in-phase current
Oa must remain constant and the locus of B is therefore a straight
line through a perpendicular to OC.
E ^ A E A
FIG. 324. Blondel diagram.
The current Oa may be expressed in terms of E and Eo by
dropping a perpendicular from A on OC at 6, Fig. 324 (2), remem-
bering that voltage values are changed to current values by
dividing by Z s .
77? 77?
Oa = Ob + ba = ^ cos 0+-^ cos {180 - (0 + a)}
77? 77?
= -^ cos -^ cos (B + a)
and the power input is
77*2
Pi = E X Oa = ~ cos -
cos (6 + a)
(307)
A similar expression for the power output may be obtained by
drawing the line cBd making angle 9 with AB, Fig. 324(3) and
dropping perpendiculars from and A on it.
336 ELECTRICAL ENGINEERING
The component of current in phase with the generated voltage
is
cB = cd Bd = Y cos (0 <*) -~ cos
and the power output is
777 77T Try 9
P 2 = E X cB = -^ cos (0 - a ) - ~- cos 0. (308)
It is now required to find the equation of the locus of B which
will make P 2 constant as the excitation and therefore the gener-
ated voltage EQ varies and the current varies.
Take two rectangular axes, Fig. 324(4), the X-axis along AC
and the F-axis at right angles to it, then
x 2 + 2/ 2 = #o 2 (309)
and
x = E cos (0 - a). (310)
Substituting these in (308)
D ( -> i 9\ cos
Pi = -yT ^- (Z 2 + 2/ 2 ) -0T
63 ^
and
P 2 r
cos cos cos 2
and completing squares
-7- ~ P&
(SID
v ;
..
2 cos 6*7 4 cos 2 cos 2 cos 2
this is the equation of the locus of B and it is a circle of radius
R - - = u - P 2 r (312)
- .
and having as its center the point f ^ - z' Oj which is the point
C.
Thus the lines* of constant power output P% are circles of radius
R = \/^ - P 2 r drawn about the point C, Fig. 324(5). The
r \ 4
circle corresponding to P 2 = passes through and A.
The maximum possible power output would correspond to the
SYNCHRONOUS MACHINERY 337
case where the radius = and is represented by the point C.
This maximum power P 2 is found by equating the equation for
the radius to zero
and therefore
P 2 = f? (313)
This is the maximum power output with maximum excitation
and is far beyond the practical limit of output.
The maximum output with a given excitation I/ corresponding
to a generated voltage E Q may be found by differentiating P 2
with respect to a,
777 77F "Jjt 2
P 2 = =-2 cos (6 a) ^?-cos B
63 * AS
sin (0 a) = 0, for maximum; (314)
da Z 8
the condition of maximum output is
sin (B a) = 0, or a = 6
and substituting this in the equation for P 2 ,
- ^cos (315)
This corresponds to the condition when B falls on AC. For
positions of B to the right of AC the power decreases again. If a
motor is loaded until the coupling angle a. becomes greater than
6 it will fall out of step and stop. The line AC, therefore, marks
the limit of stable operation.
When the point B falls on the line OC the current is in phase
with the impressed voltage and the power factor is unity. This is
the position of minimum current for a given output. For points
to the right of OC the current is lagging and for points to the left
the current is leading.
pi
The value P 2 = T~ may be found by differentiating
EE E 2
^- -fj- cos 6 with respect to Eo,
= -=- = cos B = for maximum;
22
338 ELECTRICAL ENGINEERING
E
therefore, 2Eo cos 6 = E and EQ r - - and the maximum
2i COS u
value of P 2max . is
E E 2 cos e
p I \
2 " Z 8 2 cos " \2 cos B)
Z 8 2 cos \2 cos B Z 8
E 2 E 2 E 2 , ,
= p: -- 7- = -7- as before.
2r 4r 4r
To find the relation between x a and r which makes P2 ma x. a
_ r r
maximum, substitute Z 8 = v r 2 + x a 2 and cos 6 = ~- =
/ 2 .
v T ~r
assume r to be constant and differentiate P 2 max . with respect
to x s .
2max.
dx e
= / 2 i l\% ~t~ ( 2 _L iV2 = f r m aximum,
and assuming EQ = E, which would be approximately correct
for normal operation,
-x.Vr 2 + x, 2 + 2rx s =
and from this
x, = V3r.
This is a much lower value of x 8 than is usually found in such
machines.
272. Synchronizing Power. By synchronizing power is meant
the change in power per degree change of the coupling angle a, it
represents the stiffness of the coupling of the motor and w&s
found in equation (314)
dPz EE$ /n ,.
_ = -_ sin (e. - a) ;
da. Zj 8
it is directly proportional to the excitation or EQ and is inversely
proportional to the synchronous impedance and therefore to the
ja 1 Jf
armature strength. It varies from a maximum value -~ sin
6 y 8 when a = to zero when a = 6. If a becomes greater
"
than 6 the machine cannot pull into step again.
SYNCHRONOUS MACHINERY
339
273. Construction of the Characterictic Curves of a Synchro-
nous Motor from the Blondel Diagram. A three-phase, 150-kw.
(output) synchronous motor has a resistance per phase r = 0.04
ohms and a synchronous reactance x a = 0.6 ohms. The con-
stant losses may be taken as 5 per cent, of the rated output =
7,500 watts.
500
400
300
300
100
X
X
X"
>
/
/
T
/
/
/
A
,
^_
'
5 10 15 20 2?
Impressed Voltage per Phase
E=^f =** Volts
Power Output per Phase at
Full Load
P 2 = 60 Kilowatts
Constant Loss per Phaeo
Pl = 2.6 Kilowatts
Resistance per Phasa
= 0.04 Ohms
cbroiious Keartanc*
per Phase
0?.=0.6 Ohms
FIG. 325.
In constructing the Blondel diagram it is as well to work on
one phase only with a full-load output of 50 kw. Assuming that
the armature is F-connected, the impressed voltage per phase
440
is E = -f= = 254 volts and this is maintained constant. The
\/3
no-load saturation curve is shown in Fig. 325.
Perdent P. F.
U Load I La *
Current
FIG. 326.
In Fig. 326 the horizontal 6 A = E = 254 volts and the two
1*
lines OC and AC are drawn making angles = tan" 1 = tan -1
15 with the ends of OA ; the point of intersection C is not shown.
340 ELECTRICAL ENGINEERING
The length of each of the two sides OC and AC is
_A_ 254
2 cos 0.04 1>yuo -
The circles representing constant power outputs per phase
from to 110 kw. are shown. Their radii may be found by
substituting the given values of output in the equation.
B = Z > '
but here P% is the power developed at the shaft and is greater
than the output by the amount of the constant losses per phase
= 2,500 watts.
For example, the radius of the circle for full-load output is
* - v - ** = o
= 1,776.
and the radius for zero output is
0.60 /
R ~ KoiV
2 - (2,500) (0.04) = 1,899.
4
274. Load Characteristics. To construct the load character-
istics, take any value of motor back voltage EQ and draw a circle
about A with EQ as radius. Here EQ is taken as 278 volts cor-
responding to a field current // = 10 amp. and this gives unity
power factor at full load. At the point B, Fig. 326, AB represents
the constant . motor voltage EQ = 278, OB represents the syn-
chronous impedance drop E 1 = IZ a at full load and it also repre-
129 129
sents the current. OB = 129 volts or = -~- = 7:-^ = 215 amp.;
Z/ a U.O
the power factor = 100 per cent. At the point Bi the output
= 90 kw. per phase, the current = -^ = 7^- = 413 amp. and
& 8 U.O
the power factor cos BOBi = cos 19.5 degrees = 0.943 = 94.3
per cent, lagging.
At #3 the output = 0, the current = ^ = ^ = 40 amp.
/j a U.O
and the power factor cos = cos BOB 3 = cos 75 degrees =
0.259 = 25.9 per cent.
The maximum power output for the given excitation occurs at
#2 where the circle cuts the line AC. Here output = 106.5 kw.
SYNCHRONOUS MACHINERY
341
per phase, the current = 605 amp. and the power factor = 79 per
cent. To the right of AC operation is unstable.
These values of current and power factor are plotted on an
output base in Fig. 328.
40 80 120 100 200 240 280 820 800 400
Amperes Load Current
FIG. 327. Compounding curves of a
synchronous motor.
10 20 80 40 60 70 80 90 100 110
Kilowatts Output per Phase
FIG. 328. Load characteristics of a
synchronous motor.
275. Phase Characteristics or "V" Curves and Compounding
Curves. To obtain, the "V" curve for full-load output take
points as B, B and B*> on the 50-kw. circle. At B the current
1 2Q
-TT = 215 amp., the power factor = cos = 1.00, the
s
motor back-voltage = AB = 278 volts and the corresponding
5 10 15 20 25
Amperes Field Current
"V" curves.
FIG. 329.
5 10 15 20
Amperes Field Current
w
Power-factor curves.
field current is 10 amp., Fig. 325; at B the current is 250 amp.,
the power factor is cos 30 degrees = 0.866 = 86.6 per cent, lead-
ing, the motor voltage is 350 volts and the field current is 13.8
amp.; at B*, the current is 250 amp. the power factor is 86.6 per
cent, lagging and the field current is 7.4 amp. The "V" curve?
342 ELECTRICAL ENGINEERING
for zero output, full load and twice full load, are shown in Fig.
329 (a). The minimum armature current for each output occurs
where the power circle cuts the line OC, that is, where the power
factor is unity. The corresponding values of the power factor
are plotted on a field current base in Fig. 329(6).
Compounding Curves. In Fig. 326(6) circles are drawn about
center O with radii corresponding to currents from one-half
to twice full-load current. Using the points where these cut the
constant power-factor lines the compounding curves in Fig. 327
are drawn for unity power factor, 80 per cent, power-factor lead
and 80 per cent, power-factor lag.
The theoretical maximum possible power output of this motor
E 2 C254D 2
would be P = - r = 4 V Q Q4 = 403,000 watts = 403 kw. per
phase which is more than eight times full load. Such an output
would require a motor back voltage = AC = 1,905 volts.
276. Starting Synchronous Motors. The single-phase syn-
chronous motor is not self-starting and must be brought up to
synchronous speed before being connected to the supply. This
is due to the fact that single-phase armature reaction does not
produce a revolving field. Single-phase motors are, therefore,
not used except in special cases and in small sizes.
Polyphase synchronous motors are inherently self -starting and
when connected to the supply will accelerate and run up to syn-
chronous speed, but only a low voltage should be impressed at
start or very large lagging currents will be drawn from the supply
lines. In order to do away with the large starting currents a
great many polyphase synchronous motors are started by auxil-
iary motors; they must be brought up to synchronous speed and
synchronized just as alternators. If the exciter for the motor is
mounted on the same shaft it may be used as a starting motor
or, if the motor is one unit of a motor-generator set, the genera-
tor may be used as a starting motor, but in both cases a supply
of direct current is necessary.
An induction motor mounted on the same shaft may be used
to start the synchronous motor but it must have a smaller num-
ber of poles and therefore a higher synchronous speed than the
motor to be started. The synchronous motor is raised above
synchronous speed, the induction motor is then disconnected,
and the synchronizing switch is closed as the motor passes through
synchronous speed.
SYNCHRONOUS MACHINERY
343
277. Self-starting Motors. In order to improve the starting
torque of synchronous motors and also to prevent hunting;
short-circuited grids are placed in the pole faces or between the
poles and sometimes complete squirrel-cage windings are carried
by the rotors (Fig. 330). The field winding at start may be
either open or short-circuited. In the following discussion it will
be assumed that the field circuit is open. When a motor at rest
is connected to a polyphase supply torque is developed in two
ways: (1) the revolving field sweeps across the pole faces and
grids and generates e.m.fs. and currents in them; these currents
FIG. 330. Rotor of a synchronous motor with a squirrel-cage winding.
react on the field and produce torque which causes the rotor to
follow the field. The rotor can, however, never be brought up to
synchronous speed by this torque because the e.m.fs. and currents
are induced only below synchronous speed. The motor starts as
a regular induction motor and when it is nearly up to speed the
field locks with the revolving armature in the position of minimum
reluctance.
2. The second method of producing torque for starting depends
on hysteresis; as the armature north pole passes a field pole it
induces a south pole in it and the attraction between the two
tends to make the rotor follow the armature field. Due to
hysteresis this induced field pole persists after the armature pole
has passed; the succeeding armature south pole first 'repels the
field pole then while passing changes it to a north pole and at-
344 ELECTRICAL ENGINEERING
tracts it in the direction of rotation. The resulting torque is
small at low rotor speeds but becomes greater as synchronous
speed is approached and the field pole remains longer under the
influence of the armature pole. When the armature pole is
moving very slowly across the field pole the two lock in the posi-
tion of minimum reluctance.
Since the grids or squirrel-cage windings are carrying current
only while starting up, or when there is relative motion between
the armature reaction flux and the field poles, as when there is
hunting, it is not objectionable if they have a comparatively high
resistance; in fact, the higher the resistance the greater will be the
starting torque, within limits. The bars of the squirrel cage can
therefore be made of any suitable material and steel is often
employed but the end rings are generally made of brass.
When the motor is running at synchronous speed the field
circuit may be closed and the impressed e.m.f. raised to its full
value. On closing the field circuit it may be found that the
polarity is wrong and the rotor will then drop back in phase by
180 degrees and this change will be accompanied by a sudden
rush of current. To avoid this it is better to excite the fields
through a large resistance just before synchronism is reached,
then increase the field to normal and raise the impressed voltage
to full value. When starting in this way the motor draws a very
large lagging current since the impressed voltage is consumed by
the synchronous impedance of the armature, and the power factor
is very low. The impressed voltage at start must be reduced
to about one-third of its full value in order to reduce the starting
current. It is the flux of armature reaction which produces the
starting torque and, therefore, a motor with high armature reac-
tion will give better starting torque than one of low armature
reaction.
When the motor is running below synchronous speed large
voltages sometimes reaching 5,000 volts or more are induced in
the open field winding by the revolving armature flux. Such
voltages are dangerous and may puncture the insulation of the
field or endanger the lives of operators. Both the magnitude and
frequency of these induced voltages becoine zero when the motor
reaches synchronous speed.
The potential stresses may be reduced by breaking up the field
into sections, by means of a suitable switch, during the starting
period.
SYNCHRONOUS MACHINERY 345
The approach of synchronism may be recognized by the sound
or an alternating-current voltmeter connected across a section of
the field winding will indicate synchronism by a zero reading.
If the field winding is short-circuited while coming up to syn-
chronism these large induced voltages will not exist but the start-
ing torque of the motor will in general be reduced. The short-
circuited field winding tends to act in somewhat the same way as
the grids but its inductance is so high that the flux produced
through it by the armature m.m.f . is small and this reduction of
the flux in the field poles more than counterbalances the torque
due to induced currents. In the majority of cases, however, it is
better to start the motor with the field winding short circuited
since the safety of operation is thereby increased. When syn-
chronous speed is reached the full excitation should be applied
and then the impressed voltage raised to full value.
When individual exciters are used the motor field may be
short-circuited by reducing the exciter field to its lowest value.
278. Synchronous Phase Modifier. Since by varying the field
excitation of a synchronous motor the power factor can be made
either leading or lagging, such machines can be used to improve
the power factor of transmission lines or distributing circuits by
drawing wattless leading currents to compensate for the wattless
lagging currents required by the load. The fields must be over-
excited and the synchronous reactance should not be very large.
This is one of the most important characteristics of the synchro-
nous motor and is being applied to an ever-increasing extent.
A synchronous motor used in this way is called a synchronous
phase modifier and is usually operated without load drawing the
required wattless leading current and the small power current
supplying its own losses. In some cases, however, it may be
advantageous to supply some load from it.
In the case of long-distance transmission lines the synchronous
phase modifier is used to obtain a constant terminal voltage
under all conditions of load with a constant impressed voltage at
the generating station. Assume a transmission line with a con-
stant impressed voltage of 110,000 volts. Under normal condi-
tions the terminal voltage at no load rises to 120,000 due to the
compounding effect of the line capacity, while at full load 80 per
cent, power factor the terminal voltage falls to 90,000 volts.
To maintain a constant terminal voltage of 100,000 a synchro-
nous phase modifier must be installed in the terminal station;
346 ELECTRICAL ENGINEERING
during periods of light load it is operated under excited drawing
a large lagging current to pull down the voltage from 120,000 to
100,000; while under load it is over-excited and draws a leading
current to raise the terminal voltage from 90,000 to 100,000.
The constant voltage regulation is made automatic by install-
ing a voltage regulator operating on the field circuit of the phase
modifier and having its alternating-current control magnet con-
nected through a potential transformer across one phase of the
circuit in which the voltage is to be maintained constant.
279. Parallel Operation of Alternators. Before an alternator
is connected in parallel with another machine which is supply-
ing power, the incoming machine must be adjusted to give the
same voltage, must have the same
frequency and must be in phase.
The condition of synchronism may
be indicated by incandescent lamps or
some form of synchroscope (Art. 452).
In Fig. 331 when the incoming
machine B is in synchronism with A
FlQ ' 331 p 7rli 1 i t e e i niat0rsin there is no voltage across the switch
and the lamps are dark. The two
machines are assumed to have been adjusted to give equal
voltages.
If the frequency of B is lower or higher than that of A, there
will be a slow pulsation of the light showing the difference between
the two frequencies. When the pulsations are very slow and the
periods of darkness long the switch may be closed and the two
machines will operate in parallel. Lamps are not very satis-
factory, since they do not show whether the incoming machine
is running too slow or too fast.
280. Effect of Inequality of Terminal Voltage. If two alter-
nators are operating at the same frequency and are in phase but
have not their fields adjusted to give the same terminal voltage,
a wattless current will flow between the two machines leading and
magnetizing in the machine of lower field excitation and lagging
and demagnetizing in the machine of higher field excitation.
If E B , the voltage of B } is lower than E A , the voltage of A, then
E' the difference between the two will act in the local circuit
through the two armatures in series and will produce a current
F lagging nearly 90 degrees behind E l and E A and leading E B
(Fig. 332).
SYNCHRONOUS MACHINERY 347
The circulating current is
7 + 7
4 A r Z B
where Z 4 and Z B are the synchronous impedances of the two
machines.
This current lowers the terminal voltage of A since it is lagging
in A and raises the terminal voltage of B since it is leading in B
and the two are made equal to the load voltage E.
I
FIG. 332.
By adjusting the field rheostats the wattless circulating cur-
rents can be eliminated for any load, but if the two machines
have different voltage characteristics, as the load varies wattless
currents will circulate to correct for the differences of excitation.
With machines of reasonably high armature reaction the
wattless cross currents are small even with large variations of
excitation.
When two similar alternators in parallel are supplying an
inductive load, they should operate at the same power factor.
If one has a lower excitation than the other it will not supply
its proper share of the watless kva. and will operate at a higher
power factor than the second machine.
281. Effect of Inequality of Frequency. Two alternators
operating in parallel must have the same average frequency, but
one may instantaneously drop behind or run ahead of the other.
Alternators driven by water turbines or steam turbines or
electric motors will have a constant angular velocity but when the
prime movers are steam engines or gas engines the angular veloc-
ity will pulsate about its average value during each revolution.
If two machines in parallel are excited to give the same termi-
nal voltage and one falls behind the other, a power cross-current
will circulate through the armatures and transfer energy from the
leading to the lagging machine.
348 ELECTRICAL ENGINEERING
Fig. 333 shows the case of two machines of which A is ahead
of its normal position by angle a and B is behind by angle a.
A is therefore ahead of B by angle 2a and the resultant voltage
causing a current to circulate through the two armatures is
E' = 2E sin a.
The circulating current is
E r 2E sin a E sin a
2Z
where Z = Z A = Z B is the synchronous impedance of each of the
two machines which are assumed to be similar. Under ordinary
circumstances the current I' lags nearly 90 degrees behind E' ; it
. . _. I x ,
'l,
"E B (Referred to the Load Circuit)
FIG. 333.
is approximately in phase with the terminal voltage E A of A and
in phase opposition to E B ] and it therefore transfers power from
the leading to the lagging machine causing the leading machine
to drop back in phase and the lagging machine to come up.
The component of I' in phase with E f represents the power
consumed by the passage of I' through the resistance of the local
circuit through the two armatures in series; the component in
quadrature behind E' represents the transfer of power required
to keep the machines in step.
Take the case of two similar 60-cycle alternators driven at
240 r.p.m. by single-cylinder steam engines of 1 per cent, speed
variation. The driving effort and the speed reach a maximum
twice during the revolution and fall to a minimum twice. For
one-quarter of the revolution the alternator A runs at an average
speed Y of 1 per cent, above normal and it therefore moves
9
ahead of its normal position by T = Zn ^ a mechanical
degree. If at the same time the second machine B is running
slow it will fall behind its normal position by an equal angle and
the maximum phase difference between the two machines will be
SYNCHRONOUS MACHINERY 349
of a mechanical degree. These alternators would have 30
poles and therefore the maximum phase displacement is
20 X \ = 20 X T = - 6 - 75 electrical degrees = 2a.
The circulating current is by equation (317)
T' =-
= Z
sin 3.37 = I 8C sin 3.37
where I 8C is the short-circuit current with full excitation and may
be from two to six times full-load current; assuming I 8C = 47 the
circulating current is
F = 41 X 0.059 = 0.247
and is approximately 25 per cent, of full-load current. With
lower values of synchronous reactance and consequently higher
values of 7 SC , the cross-currents tend to become very large.
Machines with very small synchronous reactance or armature
reaction are not suitable for parallel operation. Too high syn-
chronous reactance reduces the cross-currents and the synchroniz-
ing power too much.
These power cross-currents when of great magnitude tend to
tear the machines out of synchronism and they also cause fluctua-
tions of the voltage.
If the two machines above had been synchronized so that their
angular velocities rose and fell together, there would not have
been any power currents circulating between them but in that
case the frequency of the voltage of the system would pulsate
and power currents would circulate between the alternators and
any synchronous motors operating on the system.
Alternators driven by steam engines or gas engines must be
provided with flywheels large enough to reduce the pulsations of
angular velocity during the revolution to a negligible amount.
If the speed characteristics of the prime movers are not the
same and the speed of one machine tends to fall below the other
as the load on the system is increased, then the machine driven
by the prime mover of closer speed regulation takes more than
its share of the load and so relieves the other machine and keeps
its speed up.
Thus to insure a proper division of the load beween alter-
nators operating in parallel it is necessary that their prime movers
350 ELECTRICAL ENGINEERING
have similar speed characteristics, that is, that their speed shall
fall under load by the same amount and in the same manner. It
is, therefore, preferable that the prime movers have drooping speed
characteristics. The voltage characteristics have no effect on
the division of the load but they do affect the amount of the
reactive cross-currents between the machines.
Let R and X represent the resistance and reactance of the
circuit through two alternators in parallel including the lines con-
necting them which may be very important in the case of alter-
nators in different stations.
The circulating current is
E' E'R-X E'R . E'X
R+jX R* + X 2
the quadrature component of the current which gives the power
transfer is I Q = P2.i_ vV "^' ^^' anc ^ ^ us ^ ma ^ ^ e seen
reactance or armature reaction is necessary in alternators to be
operated in parallel.
For a given value of R the synchronizing current I Q is maxi-
mum when X = R] this result is obtained by differentiating /
with respect to X and equating to zero.
dI Q E'{(R 2 + X z ) - X(2X)}
" - = ' for maximum
and therefore R 2 - X 2 = 0, or X = R.
Such a low value of X would allow excessive currents to flow
between the machines. In the majority of cases X is many times
greater than jR, but when two stations are to be operated in
parallel R and X may approach equality, especially with under-
ground distribution, and R may even be the greater. In such a
case the operation would be improved by inserting extra reactance
to make X = R and so increase the synchronizing current.
282. Effect of Difference of Wave Form. If two machines
in parallel are adjusted to give the same effective value of voltage
but have different wave shapes, then, since, due to the presence
of the higher harmonics, the voltages are not equal at every
instant, reactive cross-currents will flow to correct these in-
equalities in voltage. These currents will usually be very small
since the voltages producing them are small and they are of high
frequency and thus the path through the two machines offers a
SYNCHRONOUS MACHINERY 351
high impedance to them; the impedance is, however, only the true
impedance and not the synchronous impedance.
If two F-connected alternators with neutrals grounded are
connected in parallel and one has a prominent third harmonic in
the voltage wave of each phase while the other gives a true sine
wave, then, although the terminal voltages of the two machines
may have the same effective values and the same wave form, still
third harmonic circulating currents will flow between them
returning through the neutral connections.
283. Conclusions. Three kinds of circulating currents may
exist in parallel operation of alternators: (1) reactive currents
transferring magnetization between the machines due to a differ-
ence in excitation; (2) active currents transferring power between -
the machines due to phase displacements between their voltages ;
and (3) higher-frequency reactive currents due to differences of
wave form.
The reactive cross-currents may be eliminated by regulating
the excitation, that is, by electrical means, but the power cross-
currents are due to differences in speed or phase and can be
corrected only by regulating the speed of the prime movers.
Parallel operation of alternators is therefore not so much an
electrical as it is a mechanical problem.
When, therefore, an alternator is to be connected in parallel
with machines supplying the load, the incoming machine must
be brought up to synchronism and the switch connecting it to
the load circuit closed, then the governor of its prime mover
must be adjusted so that it supplies its proper share of the load
and its field must be adjusted so that it supplies its proper share
of the reactive current required by the load. Ordinarily the
machines will be operated at approximately the same power fac-
tor. Further adjustment may be necessary as the load changes.
284. Hunting. If two alternators are operating in parallel
and one drops behind the other in phase due to a sudden decrease
in the speed of its prime mover, the second machine supplies
power to pull it into synchronism again. The impulse received
causes it to swing past its mean position and it oscillates a few
times before falling into step.
If the action producing the speed pulsation is repeated periodic-
ally and coincides with the natural period of the machine the
oscillations instead of dying out will increase in amplitude until
they are limited by the losses in the pole faces and the dampers or
352 ELECTRICAL ENGINEERING
until the machines fall out of step. When the oscillations tend to
become cumulative the machines are said to be hunting.
Hunting may occur in a similar way in the case of a synchro-
nous motor supplied by an alternator. If the load on the motor
suddenly increases it falls back in phase to receive the extra
power required and oscillates about its final phase position before
running in synchronism again. This oscillation may become con-
tinuous as in the case of alternators in parallel.
The amplitude of the oscillations in hunting is very much re-
duced by the use of dampers in the form of grids or squirrel-cage
windings placed in slots in the pole faces on the rotor. At
synchronous speed the armature reaction flux is stationary rela-
tive to the fields and, therefore, does not produce any current in
the grids but if the machine falls below or runs above synchro-
nous speed, the flux sweeps across the grids and produces e.m.fs. in
them and large currents flow which react on the field and tend to
hold the machine exactly in synchronism.
Dampers are applied to alternators only in cases where hunting
is liable to occur, due to the fact that the prime movers are re-
ciprocating engines or gas engines but they should be used on all
large synchronous motors since these are usually high-speed
machines with small moment of inertia and a high natural fre-
quency of oscillation which is more liable to coincide with some
forced frequency on the system. The dampers are also required
to make the motors self-starting.
285. Frequency of Hunting. Referring to Fig. 333, the power
supplied by A to pull B up in phase again is
P 8 = El' cos ,
but
7/ _ E' _ 2E sin a _ E .
~2Z 8 " ~^zT~ ~^ S] a '
and thus the synchronizing power is
E z . E 2 'E 2 ,
P a = =- sin a cos a = p-^- sin 2a = ^r 2a,
/j s *i* AL S
since for small angles the sine is approximately equal to the angle,
here a is expressed in electrical radians.
Changing a into mechanical radians,
(318)
SYNCHRONOUS MACHINERY 353
The synchronizing torque is
P 8 33,000 = 2Z S _ 33,000 =
1 s ~ 27T X r.p.m. 746 2^ X r.p.m. 746
3.52 E*p
Z 8 X r.p.m.
and it is directly proportional to the angular displacement a.
Since the sum of the moments of the external forces acting on
a rigid body is equal to the moment of inertia of the body about
the axis of rotation multiplied by its angular acceleration, th e
equation for the resulting motion may be expressed as
T j^ a 3.52 E 2 p , qQ v
1 8 = 1 375- = 'rjr Oi. (3lV)
at* L s X r.p.m.
where / is the moment of inertia of the rotating field member
including the flywheel. The negative sine is used because the
torque tends to decrease the angular displacement a.
This equation may be written
/J2 Q KO 77'2P
II CX. O.iJA -C/ i 7 /orrk\
W = " Z I X r m a = ~ ' ^ ^
This is the equation of a simple harmonic motion of periodic
time
sec. (321)
and therefore the natural frequency of hunting is
, 1 1
fk = =
The frequency of hunting is directly proportional to the voltage
E or to the air-gap flux 3> and is inversely proportional to the
square root of the moment of inertia / and the synchronous
impedance of the armature and connecting feeders.
The natural frequency may be increased by increasing E,
that is, increasing the voltage of the system but this is not usu-
ally practicable and it may be decreased by inserting reactance
in the lines between the machines.
The natural frequency of oscillation of an alternator should
not approach within 20 per cent, of any forced frequency which is
liable to occur on the system and the flywheel is usually designed
so that it increases / to such a value that the natural frequency
is at least 20 per cent, below the lowest forced frequency.
The same formula holds in the case of a synchronous motor
23
354 ELECTRICAL ENGINEERING
operating on a large constant-voltage system. E may be taken as
the motor voltage which is dependent on the excitation. When
the motor is under-excited E is low and the synchronizing power
transfer per degree angular displacement is small and the natural
period of oscillation is low. The electromagnetic coupling is soft
and the motor may be thrown out of step by a sudden increase
of load. When the motor is over-excited the synchronizing
power is large and the natural frequency is high. The electro-
magnetic coupling is stiff and the motor is stable in operation.
If, however, any forced frequency occurs approximating the
natural frequency, serious hunting will occur.
Hunting may sometimes be reduced or eliminated by changing
the field excitation and thus changing the natural period of oscil-
lation of the machine.
Machines with high armature reaction are much less liable to
hunt than machines of low armature reaction since the high arm-
ature reaction reduces the circulating currents produced by
changes in phase and lowers the natural frequency of oscillation.
286. Design of Alternating-current Generators and Motors.
The design of alternating-current generators and synchronous
motors is similar to that of direct-current generators and motors
but without the difficulties and limitations due to commutation.
Much larger outputs, higher voltages and greater peripheral
speeds must be provided for in certain cases. The voltage regu-
lation of alternators is not so good as that of direct-current gen-
erators, since a large component of the generated voltage is con-
sumed by the armature reactance. The regulation also depends
on the power factor of the load. To obtain reasonable values of
regulation the air gap must be made long and therefore the ratio
of field ampere-turns to armature ampere-turns must be large.
For machines of very large output as turbo-alternators the periph-
eral speed becomes a limiting factor, the frame must be made
long and forced ventilation is necessary to prevent undue rise of
temperature.
Following are some of the formulae and constants involved.
287. Electromotive Force Equation.
E.m.f. per phase = E = 4dyfnW-* (Art. 248)
or assuming a distribution factor 6 = 0.96 and a form factor
7 = 1.11, the e.m.f. is
E = 4.26/n4>10- 8 volts. (323)
S YNCHRONO US MA CHINER Y
355
288. Output Equation.
Output in volt-amperes = p'EI (where p' = number of phases)
= 4.26p'fnl 10~ 8 = 4.26 X
r.p.m. X p
120
X Bt
X
X 10- 8
4.26
120 X 2
4.26
120 X 2
or
B a =
X 10- 8 X TT X r.p.m. X B a ^q X prX Dale
X 10~ 8 X 7T 2 X r.p.m. X B g ^q X D a 2 L c ,
_ volt-amperes X 5.7 X 10 8 .
r.p.m. X Bg^q
average gap density, varies from 40,000 to
(324)
60,000 lines per square inch, but is usually below 50,000.
^ = pole enclosure = 0.60 to 0.65.
q = ampere conductors per inch, ranges from 400 to 1,000; a
good average value is 500.
L c . frame length
= ratio j ; , ranges from 0.6 to 2.0; a good average
is 1.
p = number of poles = is fixed by the frequency and
the r.p.m.
Approximate rated speed in r.p.m. of prime movers for gener-
ators of 250-, 500- and 1,000-kw. capacity.
Prime mover
Rated speed in r.p.m.
250-kw.
500-kw.
1,000-kw.
Moderate-speed steam engine
High-speed gas engine
160
250
350
1,000
2,500
3,000
120
200
270
1,800-
2,500
80
120
220
1,200
1,900
High-speed steam engine
De Laval steam turbine
Curtis steam turbine
Parsons steam turbine
The speed of hydraulic turbines is controlled by the head of
water available.
Peripheral speed varies over a wide range; for low-speed ma-
chines a good average value is 3,500 ft. per minute but in very
large turbo-alternators it may be as high as 25,000 ft. per minute-
356
ELECTRICAL ENGINEERING
289. Flux Densities. The following table from the "Standard
Handbook" gives the average flux densities used in the various
parts of the magnetic circuit.
Part
Frequency
Flux density in lin
es per square inch
Ordinary iron
Silicon steel
Armature core ....
.
25 cycles per second; 60,000 to 70,000
70,000 to 80,000
60 cycles per second
50,000 to 60,000
60,000 to 70,000
Armature teeth... .
25 cycles per second
100,000 to 120,000
60 cycles per second
90, 000 to 110,000
Pole core
95,000 to 110,000 with wrought iron or steel poles.
VnVp
90,000 to 100,000 for cast steel.
30,000 to 35,000 for cast iron.
290. Current Densities. Current densities in the armature
conductors range from 1,500 to 2,500 amp. per square inch; the
lower values should be used in machines which are difficult to
cool, that is, machines with long cores or in high-voltage machines
in which the thick slot insulation restricts the radiation of the
heat due to the copper loss. Short machines of large diameter
are easy to cool since the end connec-
tions are spread out and have a good
radiating surface.
Current density in field windings
varies from 1,200 to 2,500 amp. per
'square inch depending on the con-
struction, insulation, available space
and on the temperature rise allowed.
291. Insulation for High-voltage
Alternators. The very best materials
must be used because the space is limited. Micanite or Bakel-
ized paper tubes may be used or the insulation may be built
up of a number of layers of empire cloth or similar material.
The curve in Fig. 334 shows approximately the thickness of the
slot insulation for various voltages. The built-up insulations
may require more space than indicated by the curve,
0.3
^^
o
fl
(!>
^^
5* 2
5
/
X
Thickii
j
/
5000 10000 15000 20000
Effective E.M.F. in Volts
FIG. 334. Thickness of slot
insulation for alternators.
SYNCHRONOUS MACHINERY 357
The winding should not be distributed in more slots than
is necessary since so much of the slot width is required for
insulation.
292. Extra Insulation Required under Special Conditions.
If a three-phase, star-connected, 11,000-volt generator is operated
with its neutral grounded through a resistance of 10 ohms, and a
ground occurs on one line, the short-circuit current flowing through
the ground resistance will raise the potential of the neutral point
and will seriously increase the potential stresses from the other
11 000
windings to the core. The voltage per phase is ~=- = 6,350
v 3
volts and taking the resistance in the short-circuit as 12 ohms the
6 Q FAQ
current is ' = 530 amp. This will raise the neutral point
to 530 X 10 = 5,300 volts above ground and may increase the
stress from the ungrounded phases to the core to approximately
6,350 + 5,300 - 11,650 volts.
293. Armature Windings. Armature windings are discussed
in Arts. 241 to 245. They should be distributed rather than
concentrated to dissipate the heat more readily and to improve
the wave form and reduce the reactance. Full-pitch windings
give the highest voltage but fractional-pitch windings as low as
66 or even 50 per cent, of full pitch are used to eliminate
objectionable harmonics or to reduce the length of the end
connections.
294. Slots per Pole. Large slow-speed machines with many
poles have one to three slots per phase per pole, while high-speed
machines have from four to nine slots per phase per pole.
_. . tooth width
Ratio , , . ,,, vanes from 1 to 2 for open slots and the
slot width
slot depth
ratl vanes from 2 to 3 - 5 or 4 -
295. Regulation. The voltage regulation depends on: (1)
armature resistance, (2) armature reactance, (3) armature reac-
tion and (4) load power factor.
1. Armature resistance is determined more from considerations
of current density and temperature rise than of regulation.
2. The armature reactance drop is produced by the various
leakage fluxes and depends on the shape of the slots, the length of
the gap, the pitch and distribution of the winding (see Arts. 252
and 253).
358 ELECTRICAL ENGINEERING
3. The effect of armature reaction is determined by the
field ampere-turns
> armature ampere-turns' the length of the ^ and the
flux densities in the gap, the teeth and the poles. The ratio-
field ampere-turns
- usually lies between 2.0 and 3.0, the
armature ampere-turns
lower value being for high-speed machines.
4. The effect of load power factor or regulation is discussed in
Art. 261.
For the best regulation use a high air-gap density, a long gap
field ampere-turns
and a large ratio -
armature ampere-turns
Air gaps are much longer than in direct-current machines
ranging from 0.5 or lower to 2.5 in. in some turbo-alternators.
Eighty-five or 90 per cent, of the field m.m.f. may be required
for the gap.
Short gaps make the regulation poor and result in distortion
of the gap flux and the wave of generated e.m.f.
Ordinary values of regulation are 5 to 8 per cent, at 100 per cent,
power factor and 15 to 25 per cent, at 80 per cent, power factor.
With automatic voltage regulators, close regulation is not
necessary and may be a distinct disadvantage since the armature
current under conditions of short-circuit is likely to reach dan-
gerous dimensions in large low-reactance alternators.
296. Excitation Regulation. Excitation regulation or the per
cent, increase of field excitation to maintain constant voltage
from no load to full load ranges from 10 to 15 per cent, for non-
inductive load to 25 to 35 per cent, for 80 per cent, power-factor
load.
297. Excitation. Alternators and synchronous motors are
excited by a separate direct-current shunt generator called an
exciter, usually at 125 or 250 volts.
The exciter should preferably be driven by a separate prime
mover to reduce the variation of the alternator voltage with
speed. When the exciter is mounted on the same shaft as the
alternator a 1 per cent, decrease in speed produces more than 1
per cent, decrease in the exciter voltage and more than 2 per cent,
decrease in the alternator voltage.
The rating of the exciter is usually from 1 to 2 per cent, of the
alternator rating unless it is necessary to maintain the voltage
to very low power factors, when it must be larger.
SYNCHRONOUS MACHINERY 359
298. Losses. The losses in alternators are similar to those in
direct-current generators.
The field copper loss is usually greater than in direct-current
machines since for good regulation the field must be stronger.
This loss is about 1 per cent, for non-inductive loads but will be
greater for inductive loads.
Armature copper loss is similar to that in direct-current ma-
chines but for the large outputs conductors of large section must
be used and they must be laminated to prevent large extra losses
due to eddy currents as discussed in Art. 188. Eddy-current
and hysteresis losses in the teeth and other metal parts near the
conductors which are due to the presence of the alternating cur-
rent in the conductors add to the copper loss and increase the
apparent or effective resistance of the armature. In some cases
it may be 50 per cent, greater than the true ohmic resistance.
The no-load core loss may be from one and a half to four times
the full-load copper loss depending on the rated speed and fre-
quency. With silicon -steel this loss may be reduced by 30 or 40
per cent. The increase of core loss under load may be as great
as 25 or 50 per cent, in some cases.
Windage and bearing-friction losses in moderate-speed ma-
chines range from 0.5 to 1 per cent, but in high-speed turbo-alter-
nators with fans for cooling this loss may reach 1.5 per cent.
299. Ventilation. The rate at which heat is dissipated from
a surface is proportional to the difference in temperature between
the surface and the cooling air. A continuous circulation of cool
air is required to prevent dangerous temperature increases in gen-
erators and motors.
There is very little difficulty experienced in cooling slow-speed
machines as the losses per pound of material are small and the
radiating surfaces are large. The rotor with its salient poles acts
as a fan and blows air over the armature core and end connec-
tions. If necessary, fans may be added to direct the flow of air
and increase the cooling effect.
The insulation on the coils is a poor heat conductor and most
of the heat generated by the copper losses passes along the con-
ductors to the end connections where the insulation is thinner
and is there radiated to the air. The end connections must be
very well ventilated.
When the frame length is small and the diameter is large, the
end connections are spread out and the machine is easy to cool.
360
ELECTRICAL ENGINEERING
When the frame length is great and the diameter is small there
is very little space for the end connections and they are difficult
to cool. Radial vent spaces are left in the core by placing special
vent plates between blocks of punchings at suitable intervals.
The vent spaces must be wide enough to allow the air to pass
through them freely. The cooling air carries off the heat from
the iron due to the core losses and also part of that due to the
copper losses. The remainder of the heat due to the core losses
is carried along the laminations to the air gap or to the frame.
This is called radial ventilation (Figs. 242 and 336).
Another method of cooling makes use of axial ducts instead of
or supplementing the radial ducts (Fig. 243). The axial ducts
supply air which passes over the edges of the punchings and is
very effective as a cooling agent, since the iron core conducts heat
from 20 to 100 times better along the laminations than across
them, depending on the method of stacking.
High-speed, steam turbine-driven alternators of large out-
put are very difficult to cool. To limit the peripheral speed
Steel Line.
Two Pole, Cylindrical Eotor
(5)
Four Pole, Cylindrical Eotor
FIG. 335.
Two Pole, Cylindrical Rotor
the diameter must be made small and consequently the frame
must be long. Forced ventilation is necessary and the machines
must be completely enclosed to give control of the cooling air
and to prevent objectionable noise. Openings are provided for
the passage of cooling air. Fans are placed on the rotor and
the air, after it has been cleaned and cooled, is drawn along the
air gap and forced out through the stator radial ducts into pas-
sages which carry it to the outlet. Axial ducts may be provided
in the core if it is considered necessary.
The rotor with its small diameter and restricted space for field
copper is difficult to cool. Most of the heat developed in it is
SYNCHRONOUS MACHINERY
361
carried to the air gap and dissipated there. In some cases axial
ducts are cut at the bottom of the rotor slots and these are very
effective (Fig. 335).
About 100 cu. ft. of cooling air per minute are required per
kilowatt lost to limit the temperature rise at full load to permis-
sible values.
For example, in a 25,000-kva. turbo-alternator of 97 per
cent, efficiency the losses amount to 750 kw. and about 75,000
cu. ft. of air per minute are required. With a velocity of 5,000 ft.
75 000
per minute the section of the air passages must be , ' - = 15
O,UUU
sq. ft. Air may be admitted from both ends but even so the
gap must be made very long to give the required area. The gap
length in some cases must be determined in this way.
In. the circumferential system of ventilation air is admitted at
one side of the machine and passes circumferentially through the
stator ducts to the other side where it is discharged. Air must
at the same time be forced along the air gap to cool the rotor
(Fig. 336).
Stator Ducts
FIG. 336. Circumferential ventilation.
300. Cylindrical Rotors. The rotors of steam-turbine alter-
nators present new features in design. To keep the peripheral
speeds down to safe values, that is, below 25,000 ft. per minute,
the diameter must be made small and the length great. The
field winding is distributed and must be completely imbedded in
slots. The rotor teeth must be designed with sufficient strength
to stand the enormous centrifugal forces acting on them. At
25,000 ft. per minute a single pound of material is acted upon by
a force of over 1 ton. Fig. 335 shows three sections of cylindrical
362 ELECTRICAL ENGINEERING
rotors. The four-pole rotor in (6) is ventilated by means of axial
ducts below the slots. A section of the winding in one slot is
shown. To force the winding solidly into place pressure is ap-
plied to the bronze wedge and then the steel liners are driven into
place.
Since the space for the field winding is limited high cur-
rent densities must be used and the rotors run hot. Only those
materials which can stand high temperatures must be used for
insulation.
CHAPTER X
TRANSFORMERS
301. The Constant-potential Transformer. The constant-
potential transformer consists of one magnetic circuit interlinked
with two electric circuits, the primary circuit which receives
energy and the secondary circuit which delivers energy. Its func-
tion is to transform electric power from low voltage and large
current to high voltage and small current, or the reverse. In
step-up transformers the primary is the low-voltage (L.V.) side
and the secondary is the high-voltage (H.V.) side. In step-
down transformers the primary is the high-voltage side.
In the following discussion letters with the subscript 1 will be
used to represent primary quantities and with the subscript 2 to
represent secondary quantities.
Primary
*
^ Core
s?
**<
Secondary
I
> E 2
l
, <
' <
, <
Ei E'<
jg
/
FIG. 337. Constant potential transformer.
FIG. 338.
Fig. 337 represents a transformer. The core is made up of
thin sheets of iron or steel of high permeability with small hystere-
sis and eddy-current loss.
The primary winding consists of n\ turns in series and has a
resistance of r\ ohms, a self -inductive or leakage reactance of x\
ohms and thus an impedance of Zi = \A"i 2 + #i 2 ohms.
The secondary winding consists of n 2 turns in series. Its re-
sistance is r 2 ohms, its reactance is x 2 ohms and its impedance is
Z 2 = Vr 2 2 + ^2 2 ohms
363
364 ELECTRICAL ENGINEERING
When an alternating e.m.f. EI is impressed on the primary
winding with the secondary open, a current /o flows in the pri-
mary and produces an alternating flux through the core of maxi-
mum value <. The current Jo is called the exciting current of
the transformer and consists of two components (Fig. 338) I M
in phase with the flux <, called the magnetizing current, and I c
in quadrature ahead of the flux and in phase with the impressed
e.m.f., called the core-loss current. The product of EI and I c
is the power wasted in the core loss, that is, in supplying the
hysteresis and eddy current losses of the transformer. The ex-
citing current, therefore, lags by an angle , which is less than
90 degrees, behind the impressed e.m.f. Cos is the no-load
power factor of the transformer.
The excitng current /o is from 3 to 10 per cent, of full-load
current and the no-load power factor is of the order of 30 per
cent.
The alternating flux produced by the magnetizing current
links with the secondary winding and induces in it an e.m.f.,
dd>
e 2 = n z -7r 10~ 8 volts.
If the frequency is / cycles per second and the flux follows a
sine wave of maximum value , the instantaneous e.m.f. induced
in the secondary is
e 2 = n 2 -j ($ sin 2-7r/010~ 8 volts
= - 2irfn z 3> cos 27r/<10" 8 volts
2irfn 2 3> sin (2irft - 90) 10~ 8 (325)
this is a sine wave of e.m.f. in quadrature behind the flux, of
maximum value
#2max. = 2irfn 2 3> 10~ 8
and effective value
10~ 8 = 4.44/n 2 $ 10~ 8 volts. (326)
v 2
The flux also links with the primary winding and induces in it
an e.m.f. of instantaneous value
10- 8 sin (2irft - 90),
a sine wave of maximum value
10~ 8
TRANSFORMERS 365
and effective value
E b = 4.44/n^ 10- 8 volts. (327)
This e.m.f. induced in the primary is almost equal in value and
opposite in phase to the impressed e.m.f., the vector sum of the
two being the small component of impressed e.m.f. required to
drive the exciting current through the impedance of the primary
winding. Thus
This component has been neglected in Fig. 338. The induced
e.m.fs. Eb and E% are directly in phase since they are produced
by the same flux, and their intensities are in the ratio of the turns
on the two windings; therefore,
lO- 8 n x .... ,- OQ .
m = ~~ = ratl of turns. (328)
10~ 8 n 2
If the secondary is connected to a receiver circuit of impedance
Z = R + jXy a current I \ flows in it. The primary current is at
the same time increased by a component /', the primary load
current, -which exerts a m.m.f. equal and opposite to that of the
secondary current.
Thus
nil' = n 2 /2-
and
Y, = = ratio of transformation (329)
J. 11/2
The resultant m.m.f. acting on the magnetic circuit of the trans-
former is still that of the primary exciting current and the flux
threading the two windings remains almost constant.
The primary current under load is I\ and has two components
/o the exciting current, which is proportional to the flux, and /'
the load current which is proportional to the secondary current.
The exciting current /o can be expressed as the product of the
primary induced e.m.f. Eb and the primary exciting admittance
YQ = go - jb ; thus
7o = E b (go - jbo) = E' (go - jb ), (330)
where E f is the component of impressed e.m.f. required to over-
come the induced e.m.f. Eb.
The primary load current is /' = 1 2, and is opposite in phase
to/o.
366 ELECTRICAL ENGINEERING
As the load on the transformer is increased, the primary in-
duced e.m.f. decreases (except when the power factor of the load
is leading) because a larger component of impressed e.m.f. is
consumed in driving the current through the primary impedance,
thus,
E' = - E b = E, - IiZi. (331)
A smaller flux, is, therefore, required and a smaller exciting
current. The decrease in flux from no-load to full-load non-in-
ductive is 1 or 2 per cent, and for an inductive load of 50 per cent,
power factor is only 5 or 6 per cent. With anti-inductive load the
flux increases.
The secondary induced e.m.f., which is proportional to the
primary, decreases with it.
The secondary terminal e.m.f. E is less than the secondary
induced e.m.f. by the e.m.f. consumed by the secondary imped-
ance thus
E = E 2 - I 2 Z 2 . (332)
FIG. 339. Vector diagram of a transformer with inductive load.
302. Vector Diagrams for the Transformer. Fig. 339 is the
vector diagram of e.m.fs. and currents in a transformer with a
load impedance Z = R + jX and a load power factor cos 6 =
E = secondary terminal e.m.f.
/2 = secondary current lagging behind E by angle 0.
2 = e.m.f. consumed by secondary resistance, in phase with
/2.
2^2 = e.m.f. consumed by secondary reactance, in quadra-
ture ahead of 7 2 .
-^2 = e.m.f. consumed by secondary impedance.
TRANSFORMERS
367
EI = e.m.f. induced in the secondary winding by the alter-
nating flux <. #2 = jp + /20"2 + j%z).
E b = e.m.f. induced in the primary winding by the alter-
r>.-,
nating flux $
-
b -
$ = flux linking both primary and secondary windings, in
quadrature ahead of Eb and EZ.
Io = primary exciting current leading the flux < by an angle
a = 90 , where cos do is the primary power factor
at no load.~~
/' = primary load current in phase opposition to 7 2 , /' =
I\ = total primary current.
FIG. 340. Transformer with non- FIG. 341. Transformer with a load
inductive load. power factor of 50 per cent, leading.
E f =
= e
1 =
6 2 =
component of primary impressed e.m.f. required to
overcome the primary induced e.m.f Eb.
s.m.f. consumed by primary resistance, in phase with
r
e.m.f. consumed by primary reactance, in quadrature
ahead of /i.
e.m.f. consumed by primary impedance,
primary impressed e.m.f. EI = E' + li(r\ + jxi).
angle of lag of the primary current behind the primary
impressed e.m.f. ; cos 61 = primary power factor,
angle of lag of the secondary current behind the sec-
ondary induced e.m.f.
368
ELECTRICAL ENGINEERING
Fig. 340 shows the vector diagram of the transformer with a
non-inductive load and Fig. 341 with a capacity load of 50 per
cent, power factor leading.
The secondary current and secondary terminal e.m.f. are the
same in the three cases and it may be seen that the required pri-
mary impressed e.m.f. is greatest in the case of the inductive
load and least in the case of the capacity load.
303. Exciting Current. When a sine wave of e.m.f. is im-
pressed on the primary winding of a transformer, a sine wave of
flux must be produced linking with the primary winding. The
exciting current which produces the flux cannot be a sine wave
on account of the lag of flux due to hysteresis. This is shown in
Fig. 342. Curve (1) abcf is a hysteresis loop for the transformer
(1 ) Hysteresis Loop
.6
(2) Flux in Core
(3) Exciting Current
f 4) bine Wave Equivalent to(8)
FIG. 342. Exciting current.
iron, plotted with values of flux as ordinates on a base of exciting
current. Curve (2) is the sine wave of flux in the core and curve
(3) is the wave of exciting current. The method of obtaining
curve (3) can be seen from the figure. The maximum values of
flux and current must occur together; when the flux is zero the
current has a value oa or oc and when the current is zero the
flux has its residual value od or og.
For purposes of .analysis the current wave (3) is replaced by
the equivalent sine wave (4). The current wave (4) leads the
flux wave (2) by an angle a, which is called the angle of hysteretic
advance. If the eddy-current loss is small enough to be neglected,
a = 90 0o, where cos is the no-load power factor.
304. Leakage Reactances. Figs. 343 and 344 chow the leak-
age paths around the windings of a "shell-type" and "core-type"
transformer. Since the low-voltage windings are placed next to
the iron, the leakage path surrounding the low-voltage winding is
TRANSFORMERS
369
of slightly lower reluctance than that surrounding the high-vol-
tage winding and the reactance is correspondingly larger.
In the shell-type transformer the two windings are divided into
a number of sections and high-voltage and low-voltage coils
placed alternately to reduce the reactances.
The reactance voltage of a transformer with full-load current
is about 10 per cent, of the total voltage. If, therefore, full vol-
FIG. 343. Leakage fluxes in a shell- FIG. 344. Leakage fluxes in a core-
type transformer. type transformer.
tage were impressed on a transformer primary with the secondary
short-circuited about ten times full-load current would flow.
When, however, full voltage is impressed on the primary with
the secondary open, the exciting current which is less than 10
per cent, of full-load current flows. Thus the open-circuit react-
ance of a transformer is of the order of one hundred times the
short-circuit reactance.
The vector diagram in Fig. 345 shows the various fluxes
present in a transformer.
24
370 ELECTRICAL ENGINEERING
= flux linking both primary and secondary windings.
$L 2 = leakage flux surrounding the secondary winding; it in-
duces in the secondary an e.m.f. causing the secondary
reactance drop I 2 x 2 .
$, = total flux linking the secondary winding 8 = $ + &LZ',
it induces in the secondary an e.m.f. E 8 in quadrature
behind 3> 8 . E 8 = $ + / 2 r 2 .
& LI = leakage flux surrounding the primary winding; it induces
in the primary an e.m.f. causing the primary reactance
drop IiXi.
$>P = total flux linking the primary winding < P = < -f $L V
it induces in the primary an e.m.f. E P in quadrature
behind $ P .
E'p component of the impressed e.m.f. required to overcome
E P . ]'p = E P = EI J\TI.
305. The Transformer as a Circuit. The circuit in Fig. 346
represents a transformer supplying a load of resistance R and
7")
reactance X, or of power factor cos 6 = . =
The relations between the various quantities may be expressed
by the following equations :
(333)
(334)
W = -E* = ~h {(r + R) +j(x 2 + X)} (335)
/fc2 *^2
/' = !/ (336)
/o = ^(^0 - J6o) (337)
7i = /o + /' (338)
#1 = 7i?i -r- W (339)
and these equations may be combined as follows :
Vi = Mi + W
= (/o + /'Xn + jxd + ~/2 ( (r s + fl)
7o(r, + j*0 + /'ri + (r 2 + B)
(340)
TRANSFORMERS
371
This equation may be represented by the circuit in Fig. 347;
secondary quantities are represented by equivalent primary
quantities :
E.m.f. EZ in the secondary is equivalent to e.m.f. E' = E z
in the primary.
FIG. 346. Circuit diagram of a trans- FIG. 347. Equivalent circuit diagram
former. of a transformer.
E.m.f. E in the secondary is equivalent to e.m.f. E in the
72-2
primary.
Current 7 2 in the secondary is equivalent to current I' Iz
in the primary.
Resistance r 2 in the secondary is equivalent to resistance
") TZ in the primary.
Reactance x 2 in the secondary is equivalent to reactance
( ) 2 in the primary.
\722/
FIG. 348. Simplified circuit diagram of a transformer.
If the exciting current is neglected, equation 340 becomes
(341)
and the simplified circuit diagram in Fig. 348 may be used. The
resulting error is very small.
306. Examples. 1. A stepdown transformer with a ratio of. turns of
10 : 1 delivers 100 kw. at 2,000 volts to a receiver of power factor 80 per
cent, lagging. Determine the primary impressed voltage, the current and
the power factor.
372 ELECTRICAL ENGINEERING
The primary impedance is
Zi = r l + jx l = 50 + 80; ohms.
The secondary impedance is
?2 = r 2 + jx z = 0.6 + 0.8; ohms.
The primary exciting admittance is
Yo = go - jb = (2 - 6;) 10~ 6 .
I 00 r)f)f)
The current output or secondary current of the transformer is nnn ' n OA
z,UUU X O.oU
= 62.5 and taking this as axis the various quantities may be expressed in
rectangular coordinates as follows:
Secondary current
7 2 = 62.5 + Oj
Secondary terminal e.m.f.
E = 2,000 (cos 6 + j sin 0) = 1,600 + 1,200.;', where cos 6 = 0.8
is the load power factor.
Secondary impedance e.m.f.
I,Z 2 = 62.5 (0.6 + 0.8.7) = 37.5 + 50j.
Secondary induced e.m.f.
$1 = E + h?2 = 1,637.5 + 1,260;.
Primary induced e.m.f.
E b = E' = # 2 = 10# 2 = 16,375 + 12,500j.
7l2
Primary exciting current
/ = ^'F = (16,375 + 12,500y) (2 - 6j) 10~ 6 =0.11 - 0.07;.
Primary load current
Total primary current
1 1 = I' +/ o=6.36 -0.07j.
Primary impedance e.m.f.
I,Z, = (6.36 - 0.07J) (50 + 8Q/) = 324 + 505j.
Primary impressed e.m.f.
#! = E' + /ii = 16,699 + 13,005;.
Taking absolute values,
primary impressed e.m.f.
E, = V(16,699 2 +(13,005) 2 = 21,160 volts,
primary current
/i = \/(6^36) 2 + (O.'OT) 2 = 6.36 amp.,
exciting current
/o = V (0.1 1) 2 +"(007) 2 = 0.13 amp..
primary induced e.m.f.
^' = V(16,375) 2 + (lI3~00) 2 = 20,60p volts.
TRANSFORMERS 373
The primary impressed e.m.f. is inclined to the axis of coordinates at
an angle 0', where
tan 6' = i|^i = 0.7785 and therefore 0' = 3750'.
16,o9y
The primary current is inclined to the axis at an angle 0", where
tan 6" = - ~ = - 0.011 and therefore 0" = - 40'.
6.36
The angle of phase difference between the primary current and the
primary impressed e.m.f. is 0i = 0' - 0" = 38 30', and the primary power
factor is cos 0i = cos 38 30' = 0.782, or 78.2 per cent.
The regulation of the transformer under these conditions of loading is
2,116 - 2,000
- X 100 per cent. = 5.8 per cent.
z,uuu
Primary copper loss is
/iVi = (6.36J 2 X 50 = 2,020 watts.
Secondary copper loss is
/ 2 2 7- 2 = (62.5) 2 X 0.6 = 2,340 watts.
Iron loss is
E'*g = (20, 600) 2 X 2 X 10~ 6 = 850 watts.
The efficiency is therefore
output
77 = ^-r-, - 100 per cent.
output + losses
= roo,6ro?^2io x 10 per cent " = 95 per cent "
2. If the transformer in example 1, with 2,000 volts impressed, is used
as a step-up transformer to charge a cable system of negligible resistance
and supplies 5 amp., determine the secondary terminal e.m.f.
Primary impedance is now Z\ = 0.6 + 0.8.;.
Secondary impedance is Z 2 = 50 + 8Q/.
Primary exciting admittance is Y ' = 10 2 F = (2 6j)10~ 4 .
Let the secondary terminal e.m.f. be E and take it as the real axis, the
other e.m.fs. and currents may then be expressed in rectangular coordinates
Secondary terminal e.m.f.
E = E + Oj.
Secondary current
/ 2 = + 5j.
Secondary impedance e.m.f.
/ 2 2 = 5j (50 + 80j) = - 400 + 250J.
Secondary induced e.m.f.
E 2 = $ + h?2 = E - 400 + 250j.
Primary induced e.m.f.
E' = 1/10 # 2 = O.IE - 40 + 25j.
Primary load current
/' = 10 7 2 = + 50j.
374 ELECTRICAL ENGINEERING
Primary exciting current
/
p-; the primary exciting admittance is Y = VfToM- &o
TFo TFo
the core-loss current is 7 C = = Eig , and thus, <7o = jjn',
TRANSFORMERS 375
the magnetizing current is I M = V/o 2 I c 2 = Eib , or b may be
found as 60 = X/F 2 go 2 ', the no load power factor is cos 0o =
^-j- and the angle of hysteretic advance is a = 90 0o, if the
&llQ
eddy-current loss is neglected.
In the short-circuit test, Fig. 350, the secondary is short-
circuited and a voltage E sc is impressed on the primary of such a
value that full-load current /i flows; the secondary will then carry
full-load current 1 2', readings are taken of E 8C , Ii and the power
W
input W 8C : the power factor is found as cos 6 8C = ^ 8 l Theex-
XWi
citing current is so small that it may be neglected.
Since the terminal voltage is zero, the impressed voltage E tc
is consumed by the impedance of the transformer and is the full-
load impedance drop. When expressed as a per cent, of the
rated primary voltage, it is the per cent, impedance drop. The
per cent, resistance drop can be found by multiplying by cos 6 8C
and the per cent, reactance drop by multiplying by sin 0, c .
E sc IlZi -\ /2^2
-+" )'+('+'*
where r, x and Z are the equivalent resistance, reactance and im-
pedance of the transformer as primary quantities.
The power input W 8C = primary and secondary copper losses
plus a very small core loss, which may be neglected.
W 8C = 7M + / 2 V 2 = IS n +
and thus
W 8C
to determine the values of TI and r 2 , it may be assumed that the
two windings have been designed for the same current density
and that therefore the two copper losses are approximately the
same and
376
ELECTRICAL ENGINEERING
In a similar way x = Xi + ( ) x 2 may be separated into i
XTio/
its
two components x\ = ~ x 2 = ^' This is correct if the leakage
fluxes about the two windings are equal, which may be assumed
to be the case, since the m.m.fs. are equal and the leakage paths
are very similar.
308. Regulation. The regulation of a transformer is the rise
in secondary terminal voltage, when full load is thrown off, ex-
pressed as a per cent, of full-load voltage. The regulation of a
transformer is very much better than that of an alternator be-
cause the resistance and reactance drops are very much smaller.
The regulation at 100 per cent, power factor ranges from 1.2
to 2 per cent, for large 25-cycle transformers and from 0.75 to 1.6
per cent, for large 60-cycle power transformers. Larger values
are found in some small transformers. The regulation of large
power transformers may be as high as 6 or 7 per cent at 80 per
cent, power factor.
If the terminal voltage at full load is E = 100 per cent, and the
impedance drop is expressed in per cent., the vector sum of the
two will give the terminal voltage at no load in per cent. Taking
.
s,
y
2
3 4
o a
1 *
xKso c
T
4
J
o
C
^
^
Lag
Lead
Per Cent Power Factor
FIG. 351. Regulation of a transformer.
the current as axis and a load power factor cos 6, the secondary
terminal voltage is E = 100 cos 8 -fjlOOsinfl; the impedance
drop is IZ = Ir + jlx, all in per cent., and the terminal voltage
at no load is E<> = E + IZ = (100 cos 6 + Ir) + j(lQQ sin +
Ix); its absolute value is
E = A/(100 cos B + Ir) 2 + (100 sin 6 + Ix) 2 (342)
and the per cent, regulation is E 100.
A formula giving the regulation directly may be found by re-
ferring to the vector diagram in Fig. 351 (a). The per cent.
TRANSFORMERS 377
regulation for a load power factor cos 6 is represented by the
length ad =
D = a b + be + cd = Ir cos + /z sin 6 + cd.
From Pig. 351 (6), cd X ck = 7f and cd = =
(Ix cos - Jr sin BY , . . 4 .
^ , , - and the per cent, regulation is
T T , '(!% cos Ir sin 0) 2 , ._.
D = 7r cos + Is sin + (343)
where Ir is the per cent, resistance drop and Ix the per cent, re-
actance drop in the transformer.
The last term is negligible except near unity power factor
and if it is neglected, the per cent, regulation is
D = Ir cos 6 + Ix sin 6 = IZ (^ cos 6 + ~ sin d\
= IZ (sin cos e + cos sin 6) = IZ sin (0 + <) ; (344)
\ r
the angle < = tan -1 - is shown in Fig. 351.
x
The regulation is maximum when sin (0+ <) = 1, or 6 = 90
, that is, at a load power factor cos 6 = cos (90 0) = sin =
-y and is equal to the per cent, impedance drop IZ.
Take for example a single-phase, 60-cycle, 33, 000- volt trans-
former of 100 kva. output with a resistance drop of 1.5 per cent.
and a reactance drop of 4 per cent., and calculate the regulation
at 100 per cent, power factor and 80 per cent, power factor lag
and lead, using equation (343) :
At 100 per cent, power factor, the regulation
= 1.5 + ^ = 1.5 + 0.08 = 1.58 per cent.
At 80 per cent, power factor lag the regulation
= 1.2 + 2.4 + 0.026 = 3.626 per cent.
At 80 per cent, power factor lead, the regulation
= 1.5 X 0.8 - 4 X 0.6 + (IX 2
= 1.2 - 2.4 + 0.08 = - 1.12 per cent.
378
ELECTRICAL ENGINEERING
50 Percent Power Factor Leadi
80 Percent Power Factor
ower Factor Lagging
aggi
Amperes Load Current I
FIG. 352. Voltage characteristics of a
transformer.
The maximum regulation is Z> max . = IZ = vlr 2 + Ix 2
*\/1.5 2 + 4 2 = 4.27 per cent, and corresponds to a load power-
T i 5
factor, cos 6 = -^ = 7-5= = 0.35 = 35 per cent.
The complete regulation curve for this transformer is shown
in Fig. 351(c). For leading power factors the regulation becomes
negative, that is, the voltage
rises with load.
309. Voltage Characteris-
tics. The voltage character-
istics of a transformer show
the relation between the
secondary terminal voltage
and the secondary load cur-
rent for given values of power
factor, with a constant im-
pressed voltage Ei. Typical
voltage characteristics are
shown in Fig. 352 plotted in
per cent. Since the no-load
secondary voltage is constant = EQ = Ei, it has been taken as
100 per cent, and the values of E calculated from the equation,
E Q = 100 = V(E cos e + /r) 2 + (E sin + Ix) 2 .
310. Losses in Transformers. The losses in a transformer
are the copper losses and the iron losses.
The total copper loss is
W c = I&i + 7 2 V 2 (345)
and is almost evenly divided between the primary and the second-
ary since the two windings are designed for approximately the
same current density.
The iron loss consists of the hysteresis and eddy-current losses
and does not vary with load. It can be measured by applying
full voltage to the primary with the secondary open and reading
the watts input, r **\
W o = EJ Q cos = Eil c = Ei*g Q . ^346)
Wo includes a small copper loss /oVi which may be neglected.
311. Hysteresis Loss. The hysteresis loss is the energy con-
sumed in magnetizing and demagnetizing the iron. It is directly
TRANSFORMERS
379
proportional to the frequency/, varies as the 1.6th power of the
maximum induction density m l ' & ergs per second (348)
= rjfV(8> m l - 10- 7 watts.
At constant frequency the hysteresis loss varies approximately
as the 1.6th power of the impressed voltage.
312. Eddy-current Loss in Transformer Iron. The eddy-
current loss is the energy consumed by the currents induced in
the iron of the transformer by the alternating flux cutting it.
In Fig. 353
t = thickness of sheets in centimeters,
($>> = maximum induction density,
/ = frequency,
7 = electric conductivity of iron.
AB is a section of length 1 cm., depth 1
cm. and width dx cm., parallel to the edge
of the plate and distant x cm. from the
center. DC is a similar section on the
other side.
The. maximum 'flux inclosed by ABODA9>2($> m x lines. Of this
(& m x lines cut across the section AB and generate in it an e.m.f. of
effective value,
dE = \/2Trf($> m x c.g.s. units.
The conductance of the section A B is 7 dx, and thus the current
induced in it is
dl = dEy dx = \/2irf($> m yx dx c.g.s. units.
The power consumed by the induced current in section is
dw e = dE dl = 2Tr 2 f 2 (& m 2 yx 2 dx ergs per second
-*-&-
1
H-
A"*
kr*>
dx
t
i
i
1
c
i
B
FIG. 353. Eddy current
loss in sheet iron.
380 ELECTRICAL ENGINEERING
and thus the power consumed per square centimeter of the plate
is
= ir 1 / 2 * 2 7 TT ergs per second. (349)
The power consumed per cubic centimeter of iron is
t D
Thus the eddy-current loss for a volume of V c.c. is
ITT " m f ' i
Tr e = - ^ ergs per second
= ^ 2( V^ 2F 10-7 watts. (351)
D
The eddy-current loss, therefore, varies as the square of the
frequency, the square of the maximum induction density and the
square of the thickness of the plates. It also varies directly with
the electric conductivity of the iron.
To reduce the eddy-current loss the core is usually built up of
sheets of about 0.014 in. thick insulated with varnish. The
electric conductivity of ordinary sheet iron is about 10 5 . Silicon
steel has a conductivity in some cases as low as 2 X 10 4 and it
may be used in sheets 0.020 in. thick.
With a sine wave of impressed voltage the eddy-current loss
varies as the square of the voltage, but if the wave is peaked the ^
loss may be greater, { p**^* & <**&* %*4*>* j {4 ^ Jb^U
Since iron has a positive temperature coefficient for resistance x
the eddy-current loss will decrease slightly as the transformer
heats up.
313. Extra Losses. When the windings of a transformer are
made up of a number of sections connected in multiple, if the
impedances of the sections are not the same, circulating currents
will flow and increase the copper losses. They will exist at no
load as well as under load.
If wide copper strip is used for the winding, eddy currents may
flow in the copper due to the fact that one side is in a stronger
TRANSFORMERS 381
field than the other. This may increase the copper loss as much
as 20 per cent, in extreme cases. It is not advisable to use strip
of width greater than 0.50 in. in 60-cycle transformers or greater
than 0.75 in. in 25-cycle transformers.
314. Efficiency. The efficiency of a transformer is
output
r; = - , : -- ; -- 100 per cent.
output -f copper loss + iron loss
EI 2 cos 9
= - - 100 per cent.
EI 2 cos e + I\r l + 7fr 2 + k
= - EI 2 C osO
EI 2 cos e + l\r + k
where r = ( ) r\ + r 2 is the equivalent secondary resistance
\Yl\J
and k = constant iron loss.
The efficiency is very high and remains high over a wide range
of load; it reaches its maximum value, for a given load power
factor, when the variable copper loss = the constant iron loss.
This may be shown by differentiating rj with respect to Iz and
equating the result to zero.
^L
dI 2
{ (EI 2 cos 8+I 2 2 r+k)E cos 6-EI 2 cos 8 (E cos 6 + 2J 2 r) 1
(#J 2 cos0 + / 2 2 r + /c) 2
for maximum efficiency.
Therefore, E cos (EI 2 cos + IJr + k - EI 2 cos 9 - 2I 2 2 r)
= and I 2 2 r = k, that is, the copper loss = the iron loss. Below
one-fourth load the efficiency falls off rapidly due to the constant
iron losses and above full load it falls due to the large copper
losses.
The efficiency also depends on the power factor of the load,
decreasing with it but not in direct proportion.
In the smaller sizes the efficiency at unity power factor varies
from 93 per cent, at one-fourth load to 97 per cent, at full load
and in the larger sizes from 97 per cent, at one-fourth load to
99 per cent, at full load.
Take for example a transformer rated at 100 kva. with a full-
load copper loss of 2,000 watts and an iron loss of 1,600 watts.
The efficiency at full-load unity power factor is
100,000
100,000 + 2,00 +
100per cent> = 96 ' 5 per cent ' ;
382
ELECTRICAL ENGINEERING
the efficiency at full-load 80 per cent, power factor is
100,000 X 0.80
77 = 100,000 X 0.80 + 2,000 +^600 X< I0per Cent ' = 9
the efficiency at one-fourth load unity power factor is
100,000
000
= 93 ' 6
Cent '
1,600
In the case of distributing transformers which are connected
to the supply lines at all times but are delivering power for only a
few hours during the day, the all-day efficiency is of more impor-
tance than the actual efficiency. It is the ratio of the energy out-
put during the day to the energy input.
50 100
Percent Load
FIG. 354. Efficiency and loss curves of a 550-kw., 10,500-volt, 60-cycle, air-
blast transformer.
If the transformer mentioned above is operated at full-load
unity power factor for 6 hr. during the day and is unloaded the
remainder of the time, the all-day efficiency is
,, , energy output .,__
17 all-day = - -^ - 100 per cent.
energy input
100,000X6
100,000X6+2,000X6+1,600X24
When a good all-day efficiency is required, the iron losses should
be kept small by using the best grades of iron or by reducing the
flux density.
Fig. 354 shows efficiency and loss curves for a large transformer.
TRANSFORMERS
383
315. Types of Transformers. Transformers may be divided
into two general types depending on the arrangement of the core
and the windings: the core type, (Fig. 355); and the shell type
(Fig. 356).
In the core type a single ring of iron is surrounded by two
groups of windings, while in the shell type a single ring of copper
formed of the two windings is surrounded by two or more rings
of iron. The core type has a long magnetic path and a short
nil ill ill | ill! I!,* |
FIG. 355. Core-type transformer. FIG. 357. Three-phase core-type trans-
formers.
Unit of Equivalent Single
Phase Group
FIG. 356. Shell-type transformer.
FIG. 358. Three-phase shell-type trans-
former.
mean turn of winding and the shell type has a short magnetic
circuit and a long electric circuit. The core type is more suitable
for small high-voltage transformers as it is very easy to insulate
the high-voltage from the low-voltage coils by placing a cylinder
of insulating material between them. Large power transformers
are usually made of the shell type but core-type transformers can
be designed with equally good characteristics.
Three-phase transformers are built of both the core and the
shell types (Figs. 357 and 358) . In Fig. 358 a three-phase trans-
384 ELECTRICAL ENGINEERING
former is compared with one of the equivalent group of single-
phase transformers. Since the fluxes in the three phases are
combined at 120 degrees in the common paths a considerable
saving in core material is possible.
In the majority of polyphase systems groups of single-phase
transformers are used instead of three-phase units but the three-
phase units have some advantages which tend to increase their
application.
A three-phase transformer is less expensive than the equivalent
single-phase group requiring a smaller weight of iron, a single
tank, and fewer high-voltage brushings, since the connections
between phases are made inside the tank. The floor space re-
quired is very much less and the bus-bar and switching layouts
are simplified, as is also the cooling water system. A spare three-
phase unit must be installed instead of a spare single-phase unit
but the extra capacity may be very valuable at times of overload.
For very large outputs three-phase units may be too heavy to
handle easily in case repairs are necessary and single-phase units
may be more satisfactory.
If one phase of a three-phase delta-connected transformer is
damaged, operation may be carried on with reduced output
using the other two windings in open-delta but both the primary
and secondary windings of the damaged phase must be discon-
nected and short-circuited.
316. Methods of Cooling. Very small transformers do not
require any special method of cooling but are so designed that the
exposed surface is large enough to radiate the heat generated by
the power losses in the windings and core without a temperature
rise exceeding the limits consistent with the life of the insula-
tion. Transformers must be designed to operate for 24 hr. at
full load with a temperature rise not exceeding 55C. above the
ambient temperature of 40C.
Since the output and losses in a transformer increase in pro-
portion to its volume or as the cube of its linear dimensions
while the radiating surface increases only as the square of the
linear dimensions, as the output is increased special methods of
cooling must be adopted.
Transformers up to 500 kw. are usually immersed in tanks
containing oil of good, insulating qualities. This oil serves the
double purpose of increasing the insulation of the transformer and
conducting away the heat developed by the losses. Such trans-
TRANSFORMERS
385
formers are called oil-insulated self-cooled transformers. The
cases are made with deep corrugations to give a larger radiating
surface exposed to the air.
The oil, as it comes into contact
with the coils and core, becomes
heated, rises to the top of the tank
and flows outward to the sides where
it gives up its heat to the air and falls
again to the bottom of the tank. A
continuous circulation of oil is thus
maintained and care must be taken
that paths of ample section are left
for the passage of the oil through the
transformer (Fig. 359).
Where larger self-cooled units are
required, as in the case of large out-
door substances, special methods of
design are necessary to provide sufficient radiating surface. Two
successful types are shown in Fig. 360, the tubular tank and
the external radiator tank.
The tubular tank has a large number of tubes welded to it,
which draw the hot oil from the top of the tank and feed it back
(
Ml ]
LeVe
1
s
^N
^
f
Y
1
Cc
il
X
i
J
I
1
1
I
t
t
L
/-
t
t
t
tj
|E
I
1
1
j
(
^
1
r
C
/
are
\
<
Y
y
adiator
Tubular tank. External radiator tank.
FIG. 360. Oil-insulated self-cooled transformers.
near the bottom after its temperature has been reduced. The
tubes are very efficient radiators and the effective surface of the
tank may be increased many times by their use. Self-cooled
units up to 2,000 kva. may be built in this way.
25
386 ELECTRICAL ENGINEERING
A second method of increasing the radiating surface of the
tank is shown in Fig. 360. Radiators made of corrugated iron
with very large surfaces are used to replace the tubes in the last
design. They are made so that they can be easily detached
when the transformer is to be shipped. Units of this type with
an output of 5,000 kva. are in successful operation.
These special tanks are very bulky and are not suitable for
installation where space is limited.
In the case of large banks of transformers it may be econom-
ical to provide a forced oil circulation. The hot oil is pumped
from the transformers to a chamber where it is cooled by an air
blast or in some other way.
A second method of getting rid of the heat developed by the
losses is to blow air through the transformers to cool them.
Transformers cooled in this way are termed "air-blast" and are
used on electric locomotives where their light weight is an advan-
tage and in places where oil cannot be used on account of the dan-
ger of fire. They cannot be operated safely above about 30,000
volts as the insulation rapidly deteriorates due to ozone set free
in the air at high voltages. Oil is then a necessary protection.
For air-blast transformers about 150 cu. ft. of air are required
per minute per kilowatt lost to keep the temperature within the
permissible limits.
The majority of large transformers are cooled by placing in
the upper part of the tank cooling coils which carry a continuous
flow of cold water, which conveys the heat away from the oil.
With the incoming water at 25C., about Y gal. per minute pe-
kilowatt lost is required. If too much water is supplied the trans-
former may be cooled below the temperature of .the air and mois-
ture may collect on it and cause a breakdown of the insulation.
The cooling coils should have a surface in contact with the oil of
about 1.0 sq. in. per watt lost.
Impurities in the water are liable to collect in the pipes and
interfere with the flow.
Combined self-cooling and water-cooled transformers are
sometimes built. The radiating surface of the tank is large
enough to take care of the losses up to half load but above this
point water must be circulated through the coils to keep down the
temperature.
317. Transformer Connections. If the primary and sec-
ondary windings are each divided into a number of coils, which
TRANSFORMERS
387
can be connected either in series or parallel, a number of differ-
ent ratios of transformation can be obtained.
Take for example a standard lighting transformer with two coils
for 110 volts on the low-voltage side and two coils for 1,100 volts
Ratio 2200:220=10:1 Ratio 2200:110=20:1 Ratio 1100:220=5:1
FIG. 361. Transformer connections.
tRatio 1100:110=10:1
on the high-voltage side. The four possible connections are
shown in Fig. 361.
If small percentage changes 01 ratio are required for line regula-
tion a number of taps are brought out from the primary or
secondary winding so that the number
of turns in use may be changed by the
required amount (Fig. 362).
The secondary voltage may be in-
creased by cutting out some of the pri-
mary turns or it may be decreased by
cutting out some of the secondary turns.
318. Single -phase Transformers on
Polyphase Circuits. Single-phase trans-
formers are used in groups on polyphase circuits. The principal
transformations are two-phase to two-phase, three-phase to
three-phase, two-phase to three-phase or three-phase to two-
phase, and three-phase to six-phase.
FIG. 362. Variable ratio
transformer.
Star- Star Delta-Star Star- Delta Delta-Delta
FIG. 363. Single-phase transformers on three-phase circuits.
The transformation from two-phase to two-phase needs no
discussion. The two phases are usually entirely separate but
they may be interconnected if required.
Transformation from three-phase to three-phase can be
388 ELECTRICAL ENGINEERING
carried out in a large number of ways, the most important of
which are, star-star, delta-star, star-delta, delta-delta, open-
delta and tee (Figs. 363, 366 and 367).
Transformation from two-phase to three-phase is obtained by
the use of the Scott connection, Fig. 369, and is reversible.
Six-phase power is required for the operation of many rotary
converters. It can be obtained from a three-phase circuit by
the double-delta connection, the diametrical or double-star
connection and the ring connection, Fig. 370. Other methods
of connection are possible such as the double tee. Six-phase
power can be obtained from two-phase circuits by using two
Scott connected banks of transformers.
319. Star-star Connection. The star-star connection may be
used to transform three-phase power to three-phase power at
another voltage but it has some disadvantages which restrict its
use to special cases.
In Art. 303 it was shown that, when a sine wave of voltage is
impressed on a transformer and a sine wave of flux is produced in
the core, the exciting current is not a sine wave but has the shape
shown in Fig. 342. When such a wave is analyzed as in Art.
121 it is found to consist of a fundamental sine wave with a very
prominent third harmonic and higher harmonics of smaller
amplitude*
The fundamental sine wave of exciting current or the equiva-
lent sine wave in Fig. 342 may be separated into two sine waves
as in Fig. 364 the core-loss current, curve 3, in phase with the
impressed e.m.f., curve 1, and the magnetizing current, curve 4,
in quadrature behind the impressed e.m.f. and in phase with the
flux, curve 2. Due to saturation a third harmonic magnetizing
current, curve 5, must be added to the fundamental to produce
the sine wave of flux. The shape of the resultant magnetizing
current wave, curve 6, is therefore symmetrical with a peak at the
center. The wave of exciting current, curve 7, is found as the
sum of 3 and 6, and has the shape shown in Figs. 342 and 364.
The amplitude of the third harmonic increases with the degree of
saturation of the core.
In the three branches of the star connection, the three funda-
mental currents are displaced at 120 degrees while the third
harmonic currents are displaced 120 X 3 = 360 degrees and are
therefore in phase. For the third harmonics the ungrounded
three-phase circuit is like a single-phase circuit without any
TRANSFORMERS
389
return. When the neutral point is not grounded the third har-
monic and its multiples in the exciting current cannot flow and
are therefore suppressed.
The result is that the flux wave is no longer a sine wave but
may be considered to be made up of a sine wave, curve 3, Fig.
365, and a third harmonic, curve 4. The third harmonic flux
is opposite in phase to the missing third harmonic of magnetizing
current and the resultant flux wave, curve 2, is flat on the top
and has very steep sides.
The sine wave of 'flux induces a sine wave of voltage, curve 8,
in the two windings, in quadrature behind itself and the third
(1)= Impressed Voltage
(2) = Flux
(3)= Core Lpss Current
( 4) = Fundamental Wave of Magnetizing Current
( 5 ) = Third Harmonic- Magnetizing Current
(C)= Resultant Magnetizing (Jurrent
(7)=Eiciting Current
FIG. 364.
l)=Impressed Voltage
Flux
( 3)= Fundamental Ware of Flux
(4;=Third Harmonic Flux
(6)=Core Loss Current
(G)=Magnetizing Current
(7)=xclting Crrent
(8)=Fundamental Induced Voltage
(9)=Thlrd Harmonic Induced
Voltage
(10)=Resultant Induced Voltage
FIG. 365.
harmonic flux induces a third harmonic voltage, curve 9, in
quadrature behind itself. The two components of the induced
voltage combine to form the peaked wave, curve 10.
The maximum value of the third harmonic voltage induced in
the secondary is 2?r X 3/n 2 $3 10~ 8 = 67r/n 2 $3 10~ 8 volts, where 3> 3
is the maximum value of the third harmonic flux. Since the
secondaries are star-connected, the third harmonic cannot appear
in the voltage between lines but it will increase the maximum
value of the voltage in each transformer and may cause a break-
down of the insulation. The effective value of the secondary
voltage may be increased only 5 or 10 per cent., while the maxi-
mum value is increased 40 or 50 per cent. This is a very dan-
390 ELECTRICAL ENGINEERING
geroiis condition in high-voltage transformers where the factor
of safety is small.
If the primary neutral is connected to the generator neutral the
third harmonic of exciting current can flow, returning along the
neutral and there will be no distortion of the voltage wave.
If the neutral on the secondary side only at the generating end
of a transmission line is grounded and the neutral on the primary
side at the receiving end, the third harmonics in the induced
secondary voltage waves will cause third harmonic currents to
flow in the three lines and since they are in phase with one another
they may produce inductive disturbances in neighboring tele-
phone lines.
The star-star connection cannot be used to supply a three-
phase four-wire system since the three outers pulsate against the
neutral connection at triple-frequency.
In star-star connected three-phase core-type transformers, on
account of the mutual inductance between phases the third
harmonic in the induced voltage wave is suppressed.
Another objection to the use of this connection is the fact that,
if one transformer is short-circuited, the voltage across the other
two increases by 73 per cent, unless the primary and generator
neutrals are grounded. In any case if one transformer of the
group is damaged the whole group must be shut down.
320. Delta-star and Star-delta Connections. The delta-star
connection is used at the generating end of high-voltage trans-
mission lines and the star-delta connection at the receiving end.
The advantage of the star connection for the high-voltage side
is that very high line voltages can be obtained with moderate
transformer voltages, the ratio of the line voltage to the trans-
former voltage being \/3:l, the cost of insulating the trans-
former is therefore less than if it were designed for the full-line
voltage and the conductors being designed for larger currents are
of greater section and better able to stand mechanical strains.
In stepping up with the delta-star connection the required
third harmonic exciting current can flow in the three transformers
and the induced voltages will be sine waves, if the impressed vol-
tages are, and the secondary neutral can be grounded without
causing any inductive disturbances.
When stepping down with the star-delta connection the re-
quired third harmonic currents cannot flow in the primary but the
third harmonic e.m.fs. induced in the three transformers are in
TRANSFORMERS 391
phase and cause a current to circulate through the delta and this
acts as a magnetizing current and supplies the missing component
of flux. The exciting current is therefore divided between the
primary and the secondary.
321. Delta-delta Connection. This connection is used very
extensively even in the case of very high-voltage transmission
lines, where it is thought advisable to make provision for an in-
crease of the transmission voltage to meet the requirements of
future extensions of the system. In such a case the secondary
may be changed to star and extra insulation added to the line.
There is no trouble from wave distortion as in the star-star
connection and if one transformer burns out the system can still
be operated without any change in connections with the two re-
maining transformers in open delta but the load must be de-
creased to about 60 per cent, for the same temperature rise.
a & A B aa&AU
-/TrcwrosTON? -/WOTOTN 'TOwsw^ -/TOWWBN
LJLJ / ? Uek*J IT IT
assimjMiiLa / / p^Msp^ fww^
C C
FIG. 366. Open-delta connection. FIG. 367. Tee connection.
322. Open-delta Connection. If one transformer of a delta-
delta bank is put out of commission the system can still be
operated with the other two transformers without any change in
connections. This arrangement of two transformers on a three-
phase system, Fig. 366, is called the "open-delta." connection.
It may be used as an emergency connection in case of accident
or it may be employed in a new installation for temporary opera-
tion and a third transformer added to complete the delta-delta
bank when conditions warrant the increase in capacity.
In this system each transformer carries the full-line current
and for a given output the transformer currents in the open-delta
bank will be 73 per cent, greater than in the closed- delta bank.
To supply a load of 300 kva. from a three-phase system, three
100-kva. transformers are required connected delta-delta or
two 173-kva. transformers connected in open-delta and therefore
the open-delta bank must have a capacity of 2 X 173 = 346 kva.,
392
ELECTRICAL ENGINEERING
which is 15 per cent, greater than the closed-delta bank. Thus
when one transformer of a delta-delta bank burns out the capacity
100
of the bank is reduced to -z-^ = 5% P er cent.
Fig. 368 shows a vector diagram for an open-delta system with
a balanced inductive load. E\, E 2 and E s are the e.m.fs. gen-
erated in the three transformers of a closed delta; they are also
the terminal e.m.fs. at no load in the open delta. 7'i, 7' 2 and 7' 3
are the load currents in the closed delta.
FIG. 368. Vector diagram for an open delta.
Transformer No. 3 is disconnected and the two remaining
transformers carry currents /i = I\ 7' 3 and ^ 2 = I'z /Y
The terminal e.m.f. of No. 1 is E AB = EI IiZ, where Z is the
impedance of each of the transformers: the terminal e.m.f. of
No. 2 is EEC = EZ I 2 Z and the e.m.f. between the lines C and
A is EGA = (E A B + E BC ) since the sum of tKe three must
equal zero.
The three terminal e.m.fs. are slightly unbalanced under load
but not enough to have any serious effect on operation. If an
extra impedance = %Z is added to the primary at b the three
secondary line e.m.fs. will be equal. Here Z is the impedance of
one transformer.
The impedance drops in Fig. 368 are very much exaggerated
and the terminal e.m.fs., therefore, appear to be very badly
unbalanced.
323. "T" Connection. The "T" connection is a second
method of obtaining a three-phase to three-phase transformation
using only two transformers. One transformer is designed for
only 87 per cent, of the three-phase e.m.fs. while the second one
TRANSFORMERS
393
is designed for 100 per cent, and has a tap at its center point
to which one terminal of the 87 per cent, winding is connected
(Fig. 367). Between the lines A, B, C three equal e.m.fs. are
obtained at no load and they become only slightly unbalanced
under load. If the short-circuit impedance of the 87 per cent,
transformer is equal to one-half the impedance of the 100 per
cent, transformer the e.m.fs. will be balanced. External imped-
ance may be added to obtain this condition.
The two halves of the winding of the 100 per cent, transformer
carry currents which are out of phase and they should be well
interleaved.
324. Transformation from Two-phase to Three-phase. The
Scott connection, Fig. 369, is an arrangement of two transformers
by means of which three-phase power can be obtained from a
two-phase circuit. This connection is reversible.
E OA I 3 E,
EFB
FIG. 369. Two-phase to three-phase (Scott connection).
Two similar transformers are used with their primaries con-
nected to the two-phase system. The secondary AB of one has
a tap at its center point and the secondary DC of the other has a
tap 86.6 per cent, from the end C. The two taps are connected
at F and three-phase power can be taken off from the terminals
A, B and C. The section DF is not used.
ff the e.m.f. generated in the winding DC is
e D c = E m sin d,
and that in A B is
e AB = E m sin (0 - 90),
then the e.m.fs. generated in the various sections are
ep c = 2~E m $me,
e FA = M E m sin (0 + 90)
f> 1 / Jj 1 ,-ii-r-v fQ (\f\\
"FB /2 E'm siu. \v \y\jj f
394 ELECTRICAL ENGINEERING
100
the neutral point of the three-phase system is at 0, which is
57.7 per cent, from the end C, and thus
Tji
e 0c = -p- sin
"V 3^
e FO = ( ^ --- -^ }E m sin 6 = - -= E m sin 0.
\ 2 -v/3/ 2V3
The three-phase star voltages are eo A , e OB and e oc ',
e OA = e QI , + e PA = - ~~ sin + sin (0 + 90)
-^W - 3 cos - ^ sin = -p sin (0 + 120)
\/3\ 2
e FB = - --^=, sin + -^ sin (0 - 90)
2\/3 2
= |(-
cos - J$ sin - - sin (0 + 240).
Thus the three star voltages COA, COB and eoc are equal in value
and displaced in phase by 120 degrees and form a three-phase
system.
These star voltages may be combined to form the three delta
or line voltages, e AB = E m sin (0 90), e B c = E m sin (0 + 30)
and e CA = E m sin (0 + 150).
If the loads on the three phases are balanced the terminal vol-
tages will be very slightly unbalanced, since the impedance from
to the two terminals A and B is greater than that from to C.
To correct this an external impedance Z c may be added at C and
its value must be such that
Z c + ZQC ZOF ~\~Z FA
and taking Z as the short-circuit impedance of each transformer
Z _/V3 1
vf~\2~ v
z -ff
2 2V3 2
and
Z c = -^ +- -- ^L = - ~ l Z = 21.2 per cent, of Z.
2V3 2 V3 2 V3
Fig. 369 shows a vector diagram for a balanced load of power
factor cos 0.
The power delivered by the secondary DC is
Pi = Ere 1 1 cos = - El cos
3 - Co D -^^-
4
TRANSFORMERS 395
where E and I are the line voltage and line current respectively.
The total power is \/3EI cos 6 and thus each transformer supplies
one-half of the load.
ULj 6
f
3^ S 4
Diametrical ox Double-Star Ring
FIG. 370. Three-phase to six-phase.
325. Single-phase Power from Three-phase Circuits. Single-
phase power can be drawn from any phase of a three-phase circuit
but when the amounts of power required are large this leads to an
unbalance of the terminal voltages and to unequal heating of the
three phases. Fig. 371 shows one method of connecting three
transformers on a three-phase four-wire system to supply a single-
phase load by means of which equal currents are drawn from the
three phases. The power factors of the loads on the three trans-
formers are very different but the temperature rise will be the
same. The single-phase voltage is only double the voltage of one
transformer and therefore the transformer capacity must be 150
per cent, of the load.
Thtee-Pkase, Four Wire
A T
njoTOn njwftn rowoci
no LJ 110 I 1 110 I
1 Bitigle-Ehase *
FIG. 371. Three-phase to single-phase.
326. Multiple Operation of Transformers. In order that two
transformers connected in multiple shall divide the load in pro-
portion to their capacities it is necessary (1) that they have the
same ratio of turns and (2) that the per cent, resistance drops
and the per cent, reactance drops are the same in both.
If the ratios are not the same they may be equalized by using
an auto-transformer and differences in the impedances may be
corrected by adding external impedance to one transformer.
In three-phase systems only those groups of transformers can
396 ELECTRICAL ENGINEERING
be operated in multiple which give the same change in phase be-
tween the primaries and secondaries, a star-star bank can be
operated in multiple with a similar star-star bank or with a
delta-delta bank, but neither of these banks can be operated in
multiple with a delta-star or star-delta bank. One delta-star
bank may be operated in multiple with another delta-star bank,
or it may be operated in multiple with a star-delta bank with
its secondary voltages reversed, if the ratios are such as to make
the voltages equal.
327. Booster Transformers. Booster transformers are con-
nected with their primaries across the line and their secondaries
. in series with the line to raise or
lower the voltage (Fig. 372).
If the primary becomes discon-
nected while the load is -jm, the
booster transformer becomes *a series-
^ n _
FIG. 372. Booster transformer, transformer with open secondary.
The load current acts as an exciting
current and very high voltages may be generated in the high-
voltage winding, resulting in danger to persons and damage to
the insulation. To disconnect a booster transformer under load
the secondary must be short-circuited at the same instant that
the primary is opened.
328. Auto -transformers. An auto-transformer has only one
winding. The primary includes all the turns, while the second-
ary includes only a part of them. The secondary voltage is usu-
ally made variable by bringing out a number of taps.
Auto-transformers may be used in almost every case in which
ordinary transformers are used but are not satisfactory where it
is desirable to have the secondary insulated from the primary.
They are used very extensively to obtain a variable voltage
for starting induction motors, synchronous motors, single-phase-
series motors, etc., and as balance coils on three-wire distributing
circuits. They are necessary when transformers of unequal ratios
are to be operated in multiple. Three-phase auto-transformers
are often used to interlink two systems of unequal voltage.
Fig. 373 shows an auto-transformer for starting and operating
single-phase series motors from a high-voltage trolley in electric-
railway service. The section of the winding between T and M
carries the primary current, which is small, while the section
from M to R carries the difference between the secondary and
TRANSFORMERS
397
primary currents which is very large if the ratio of voltages is
large and, therefore, it must be made of large current capacity.
The copper loss in an auto-transformer is smaller than in an
equivalent two-coil transformer and the efficiency is therefore
higher but this advantage decreases as the ratio of turns increases.
Auto-transformers may be used either to step up or step down
the voltage.
When starting induction motors or synchronous motors, a low
voltage must be impressed at start to prevent objectionable
current surges. From 40 to 70 per cent, of full voltage is usually
impressed at the first step depending on the size of motor.
Starters for medium-sized motors will usually have only two taps
while for large-sized motors three or more taps may be required.
T Trolley 3
R Rail
Auto-Transfoimer for
Single- Phase Electric
Eailway Service
Star- Connected
Auto-Starter
FIG. 373.
FIG. 374.
Auto -Transformer
as Balance Coil for
a Three- Wire System
FIG. 375.
For three-phase motors three auto-transformers connected in
star may be used or two connected in open delta or tee (Fig. 374).
Fig. 375 shows an auto-transformer used as a balance coil on a
220-volt, three-wire system.
329. Instrument Transformers. Instrument transformers are
of two kinds, shunt or potential transformers and series or current
transformers.
Potential transformers are used to reduce the line voltage to a
value suitable for operating, indicating, integrating and record-
ing instruments as voltmeters, wattmeters, power-factor meters,
etc., and relays for voltage regulation or protection. They also
serve to insulate these instruments from the high-voltage circuit.
The ratio of primary impressed voltage to secondary terminal
voltage should be as nearly as possible constant and the phase
angle between these two voltages should be small. Fig. 376
shows a vector diagram for a potential transformer with a 1 to 1
ratio and non-inductive load. This is the usual case since the
voltage coils of instruments are practically non-inductive. The
same symbols are used as in Art. 302.
398
ELECTRICAL ENGINEERING
E' .
The ratio ^r is constant and is equal to the ratio of turns but
Ez
f JT'
the ratio ^ is always greater than -g- on account of the imped-
ance drops in the primary and secondary. Thus to obtain the
primary impressed voltage E\ from the secondary terminal volt-
age E } it is necessary to multiplyEby the nominal ratio and by a
FIG. 376. Instrument transformer.
FIG. 377.
ratio correction factor ki. This factor varies with the impressed
primary voltage or secondary terminal voltage and it also varies
with the secondary load, that is, with the number and type of
instruments connected to the secondary. The ratio correction
factor is usually plotted on a base of secondary terminal voltage
for a specified secondary load resistance. Fig. 378 shows values
of ki for a Weston portable transformer it increases with the
secondary terminal voltage E.
50 75 100
Secondary Volts
FIG. 378. Ratio-correction curves for a Weston portable potential transformer.
The angle of phase shift or the phase angle of the transformer
under given conditions is the angle by which EI and E are out of
line and is shown as 7 in the diagram. The phase angle depends
on the impedance drops in the transformer and increases with the
secondary load current and with the secondary terminal voltage.
The phase angles of standard potential transformers are very
small, unless they are overloaded, and they can be neglected
except in the most accurate measurements or at low power fac-
TRANSFORMERS 399
tors. Phase-angle corrections do not apply to voltmeter read-
ings but only to the indications of wattmeters and power-factor
meters which are affected by change of phase.
Potential transformers are designed with' very low impedances
by subdividing the coils well and placing them as close together
as possible. The very best insulating materials must be used.
They are designed with very low flux densities to prevent distor-
tion of wave form due to the exciting current.
Calibrations must be made with a definite instrument load in
the secondary.
330. Current Transformers. Current transformers are con-
nected with their primaries in series with the circuit and are used
to reduce the line current to a value suitable for current relays
and for indicating and integrating instruments. They also
insulate the instruments from the high-voltage circuits.
The ratio of primary to secondary current should be as nearly
as possible constant and the phase angle between them should
be very small.
Fig. 376 may also be used as a vector diagram for a current
transformer. The meter load is again assumed to be non-induc-
tive and its resistance is given as the ratio of secondary terminal
ET
voltage to secondary current = j-. The exciting current /o
*2
added to the primary load current prevents the ratio of currents
from being constant and equal to the ratio of turns and it also
causes the currents to be out of line by the phase angle /?.
The exciting current can be kept low by using iron of very
high permeability and low loss and designing the secondary with
a low impedance. The core-loss current is nearly in phase with
the primary load current and has a large effect on the ratio but
little effect on the phase angle while the magnetizing current has
a large effect on the phase angle and a small effect on the ratio.
Increasing the secondary load resistance by adding extra instru-
ments increases the voltage generated in the secondary for a given
value of secondary current and so increases the flux required
and the exciting current. This increases both the phase angle
and the ratio correction. Reactance added to the secondary cir-
cuit causes the exciting current to swing more nearly into line
with the load current and therefore increases the ratio correction
but decreases the phase angle.
Fig. 379 shows ratio and phase angle curves for a Westo.n
400
ELECTRICAL ENGINEERING
standard current transformer. To find the primary current,
corresponding to a given secondary current I 2, multiply 7 2 by the
nominal ratio and by the ratio correction factor & 2 .
The ratio correction factors and phase angles are smaller on
60 cycles than on 25 cycles since the flux and exciting currents
for a given load current are less. The phase angles of current
transformers are greater than those of potential transformers
and a correction should always be made for them.
01 2345
Secondary Amperes
FIG. 379. Ratio-correction and phase-angle curves for a Weston portable
current transformer.
The primary winding of a current transformer usually consists
of a single turn in series with the line while the secondary may
have any number of turns depending on the reduction in current
required. The current in the primary is the load current of the
circuit and does not depend in any way on the transformer. It
sets up a magnetic flux in the core which links with both the pri-
mary and secondary windings. If the secondary were open-
circuited all the primary m.m.f . would be magnetizing and would
produce a very large flux in the core and in a transformer having
a large ratio thousands of volts would be induced in the secondary,
resulting in danger to operators and damage to the instrument.
The iron core is designed to carry only the flux required with
closed secondary and it would be very highly saturated under
open-circuit conditions and would become hot due to excessive
iron losses. The transformer would require to be recalibrated
before its readings could be relied on. Current transformers
should always be provided with a short-circuiting link which
must be closed before the secondary is opened.
TRANSFORMERS 401
Potential transformers are built for all voltages up to 100,-
000 volts but for the high voltages they become very expensive.
The secondary is usually designed for 100 volts.
Current transformers are built with a rating of about 25 volt-
amp, and a secondary current of 5 amp. The primary current
ranges from 5 amp. to 3,000 or more. On circuits up to 20,000
volts they are usually air-cooled but above this they are oil-
immersed. For primary currents above 600 amp. no primary
winding is supplied but the transformer is slipped over the
bus-bar which then forms its single turn.
331. Example. In the circuit in Fig. 377(o) it is required to find the
primary voltage E\, current I\, power factor cos 61 and power Wi, knowing
the secondary voltage E, current / and power W. The phase angle of the
potential transformer is 7, Fig. 376, and the ratio correction factor is fcij
the phase angle of the current transformer is /3 and the ratio correction
factor is k%.
The primary voltage is E\ E X nominal ratio X k\.
The primary current is /i = / X nominal ratio X k z .
W W
The secondary power factor is cos 6 = ~^j, or 6 = cos l -prf
Referring to Fig. 377(6) the primary angle of lag is d\ = 9 + + 7 and the
primary power factor is cos (6 + /3 +7).
The secondary voltage always lags behind the primary voltage and the
secondary current always leads the primary current.
Errors due to the phase angles /3 and 7 are much more serious on low power
factors than on high power factors. When using two wattmeters to read
three-phase power, phase-angle corrections should always be made.
332. The Constant-current Transformer. The constant-cur-
rent transformer is shown diagrammatically in Fig. 380. The
primary coil P is fixed in position and receives power at constant
voltage. The secondary coil S is movable and regulates for
constant current in the receiver circuit which it supplies irrespec-
tive of the load. The transformer is used to obtain a constant
current for series arc-light circuits.
When the secondary coil is close to the primary there is very
little leakage and most of the flux produced by the primary links
with the secondary and the secondary voltage is, therefore, a
maximum. Primary and secondary currents are in opposite
directions and the two coils repel one another. The weight W
is so adjusted that the pull due to it together with the force of
repulsion of the coils just balances the weight of S and allows
the coil to remain in such a position that the required current
flows in it. Fig. 381 shows the flux in the core.
26
402
ELECTRICAL ENGINEERING
If the resistance or impedance of the load circuit decreases due
to the cutting out of one or more arc lamps an increase of the
current in both secondary and primary follows and the repulsion
between the coils, which is proportional to the product of their
currents, increases. The secondary, therefore, rises and increases
the leakage reactances of both coils and so less of the primary
magnetism links with the secondary; its voltage is, therefore,
decreased and its current drops to the required value. The
moving arm must be designed to give the required regulation with
a fixed weight W.
FIG. 380. Constant-current trans-
former.
FIG. 381. Flux in the core of a con-
stant-current transformer.
Such an arrangement regulates for constant current between
the limits of secondary voltage set by the two extreme positions
of the moving coil.
Neglecting the primary exciting current the relation between
the terminal voltages may be written :
Eij /2, 7*1 and r 2 are constant and Xi and x z increase as E decreases.
333. Regulation and Power Factor. The secondary current
should be maintained within 0.1 amp. of the rated current from
no load to full load.
The full-load power factor of an alternating arc-lamp system
seldom exceeds 70 per cent, and is much lower at light loads.
In series tungsten-lamp systems the power factor varies from
about 23 per cent, at one-fourth load to 85 per cent, at full load.
TRANSFORMERS
403
334. Cooling. Small constant-current transformers are cooled
by natural air circulation; above 100 kva. they are oil-immersed
and may be water-cooled.
335. Induction Regulator. Induction regulators are special
transformers used to vary the voltage of an alternating-current
distributing circuit or the voltage impressed on a rotary con-
verter.
/Secondary Coll S
^Primary Coil P
FIG. 382. FIG. 383.
Single-phase induction regulator.
FIG. 384.
There are two types of induction regulators, single-phase and
polyphase.
1. The single-phase regulator is illustrated in Figs. 382 to 384.
The primary coil P is carried on a movable core built of laminated
iron and is connected across the line. The secondary coil S is
carried on a stationary core and is connected in series with the
line to raise or lower the voltage (Fig. 385).
Secondary
Line
Load
FIG. 385.
S 2 Si
FIG. 386.
The primary exciting current produces an alternating flux,
which induces a voltage in the secondary. This secondary volt-
age varies with the position of the primary winding, but it is
always in phase with or in opposition to the impressed voltage or
line voltage. In Fig. 382 all the primary flux (neglecting leakage)
passes through the secondary coil and the secondary voltage is a
maximum and is in opposition to the impressed voltage, and so
gives the minimum load voltage.
In Fig. 386 OP is the impressed voltage or line voltage, OSi is
404 ELECTRICAL ENGINEERING
the maximum secondary induced voltage and S\P is the load
voltage.
The load current of the circuit flows in the secondary, and
there must also be a load current in the primary of equal and
opposite m.m.f. as in the ordinary transformer.
When the movable core is turned through an angle 6, Fig. 383,
only part of the primary flux passes through the secondary and
the secondary voltage is reduced approximately in the ratio
1 : cos 0, but is still in opposition to the line voltage. The load
voltage is represented by S^P, Fig. 386.
With the primary coil at right angles to the secondary coil,
Fig. 384, none of the primary flux passes through S and there is
no secondary induced voltage.
When the core is turned through 180 degrees the secondary
voltage is again maximum, but is in phase with the impressed
voltage and so raises the load voltage to S$P. Thus the total
variation of the load voltage is from SiP to S&P and is equal to
twice the maximum secondary voltage.
In Fig. 384 the m.m.f. of the load current in the secondary
cannot be opposed by any primary m.m.f. since the coils are at
right angles. To supply the m.m.f. required to balance the
secondary load m.m.f., and so prevent a large reactance drop in
the winding, the coil T called the "tertiary" coil is placed on the
movable core at right angles to the primary coil. It is short-
circuited and exerts a m.m.f. equal and opposite to the second-
ary m.m.f., and so reduces the secondary reactance to the value
corresponding to the leakage flux.
The only current carried by the primary coil in this position
is the exciting current.
In intermediate positions of the rotor the secondary m.m.f. is
partly balanced by the induced current in the tertiary or com-
pensating coil and partly by a load current in the primary coil.
2. The polyphase induction regulator has a polyphase wind-
ing on the moving core, which is connected to the polyphase
supply. The secondary or stator is wound with the same number
of phases, but the phase windings are kept entirely separate so
that they can be connected in the different lines to raise or lower
the voltage.
When polyphase currents flow in the primary windings a
revolving magnetic field is produced of constant value as in the
alternator or induction motor. This field cuts the secondary
TRANSFORMERS 405
windings and generates e.m.fs. in them of the same frequency as
the primary impressed e.m.fs., but less in the ratio of turns.
As the primary is turned the magnitude of the revolving field
is not changed, and therefore the magnitudes of the secondary
e.m.fs. are not changed but their phase relations with the im-
pressed e.m.fs. are changed and the load voltage is varied as
shown in Fig. 387. By turning the primary through 180 degrees
the phase of the secondary e.m.f . is changed from direct opposition
to the line voltage to direct addition to it and thus the load volt
tage is varied by an amount equal to twice the secondary voltage.
The rotor must be clamped in the required position or it will
tend to rotate at a high speed in the direction opposite to that of
the revolving field.
Si
FIG. 387.
The advantage of the induction regulator over a transformer
with variable voltage taps on the secondary or primary is that
the variation of voltage is uniform over the entire range. Regu-
lators are, however, very expensive and require a large exciting
current and have large leakage reactances.
The regulator is operated either by hand Or by means of a
small motor placed on the top or it can be made automatic if
required.
336. Design of a Transformer. Given the type and method
of cooling, number of phases, frequency, capacity in kilovolt-
amperes, ratio of voltages and the class of service, a transformer
may be designed using the equations and constants listed below.
The e.m.f. equation is
E = 4.44/n$10- 8 volts Art. 301. (353)
Volts per turn = e = g^put in. volt-amperes,
A/
where k is a constant for which the following approximate values
may be used :
For shell-type power transformers, k = 25,
For core-type power transformers, k = 50,
For core-type distributing transformers, k = 80.
406
ELECTRICAL ENGINEERING
Flux densities. The flux densities to be used depend on the
quality of the iron, the frequency, and the service in which the
transformer is to be placed.
With good alloyed iron densities from 65,000 to 90,000 are
used. The lower values should be used for 60-cycle distributing
transformers to keep down the iron losses. When using silicon
steel with its low iron losses higher densities may be employed
but care must be taken that the magnetizing current does not
become too large since the permeability of silicon steel is compara-
tively low.
Ratio of losses. In distributing transformers the core loss is
from 30 to 50 per cent, of the copper loss while in power trans-
formers the two are approximately equal.
Knowing the voltage and the volts per turn the number of
turns may be found and substituting in equation (353) the flux
3> is determined. Dividing 3> by the assumed flux density gives
the section of the magnetic circuit. Fig. 388 shows the magnetic
circuits of shell- and core-type transformers and their relative
dimensions.
Watt* par Pound Wait, pal Pound
0.4 0.8 1.2 1.8 1.0 2.4 2.8 3.2 0.4 0.8 1.2 1.6 2.0 2.1 2.8 3.2 3.8
100000
4 8 12 IB 20 24 23 32 36 40
Volt Amp.ro | or Pound
Core Type
FIG. 388.
FIG. 389. Magnetic characteristics of
transformer sheets.
Before deciding on the size of the opening in the iron the sec-
tions and arrangement of the winding must be determined.
Design of the windings.
(a) Volts per coil should not exceed 5,000.
(b) Volts per layer should not exceed 350.
(c) Current density should lie between 625 and 1,250 amp. per
square inch or 2,000 to 1,000 circ. mils per ampere. In large
water-cooled transformers higher densities may be used up to
2,000 amp. per square inch.
(d) The strip used in the windings should not be wider than
0.5 in. for 60 cycles nor 0.75 in. for 25 cycles to prevent excessive
eddy currents in the copper.
TRANSFORMERS 407
(e) Coils should not be made more than 1 in. in thickness and
must have a surface in contact with the oil of at least 2 to 3 sq.
in. per watt copper loss, the lower value being for water-cooled
transformers.
Copper losses should be calculated at 75C. and if they are
found to be too large either the current density must be decreased
or the radiating surface increased by further subdivision of the
winding. Decrease of current density will improve the efficiency
while an increase in the number of coil groups will decrease the
reactance and therefore improve the regulation but will not affect
the efficiency.
(/) The spacings between coils and between coils and iron de-
pend on the voltage and approximate values may be obtained
from Fig. 393. The fullerboard insulation between the high-
voltage (H. V.) and low-voltage (L. V.) coils may be assumed to
take up from one-third to one-half the available space. Spacing
blocks are used to strengthen the coils and to keep the paths for
oil circulation open. Low-voltage coils are always placed next
the iron and the windings are arranged in groups of high- and
low-voltage coils to reduce the reactance.
Having determined the space required for the windings and the
opening in the iron the size of plates may be found and the total
volume and weight of iron calculated.
Fig. 389 shows characteristic curves for good alloyed iron at
25 and 60 cycles plotted with flux density against watts iron loss
per pound and volt-amperes per pound. Knowing the flux dens-
ity the watts iron loss per pound may be obtained and this value
multiplied by the total weight of iron gives the total iron loss or
core loss in the transformer. If the iron loss is too large it may
be decreased by using a lower flux density or by making the core
of silicon steel with lower losses. If the total flux in the trans-
former is decreased the number of turns must be increased and
this will result in increased copper losses.
The iron must have a radiating surface in contact with the oil
of at least 1 sq. in. per watt lost in the core.
Self-cooled transformers are placed in corrugated tanks and
are immersed in oil. From 5 to 8 sq. in. of tank surface are
required for each watt lost depending on the depth of the
corrugations.
In water-cooled transformers with cooling coils \Y in- m
408
ELECTRICAL ENGINEERING
diameter a coil surface of 1 sq. in. per watt lost is required and
about Y gal. of water per minute per kilowatt lost.
337. Reactance. The reactance of the windings of a trans-
former can be determined very accurately since the leakage paths
are of simple form.
Fig. 390 shows a section through the coils of a shell-type trans-
former and an enlarged section of one group of coils, w = width
of the iron opening, i and a 2 are the widths of high-voltage and
low-voltage coils respectively, and di and d 2 are the thicknesses
of these coils. The distance between the coils s. When ai
and a 2 are only slightly smaller than w the leakage flux may be
assumed to cross the opening from one side to the other but in
high-voltage transformers where a\ is much less than w, a more
approximate expression for the length of the leakage path in air
may be found by inspection.
Core
Section of Winding
High Voltage Ooil
FIG. 390.
Assuming that the two windings are divided into the same
number of coils = g, let t\ be the number of turns in one primary
coil carrying current /i and t 2 be the number of turns in one
secondary coil carrying current /2, then t-Ji = falz. The leakage
flux about any coil may be separated into two parts: (a) that
crossing the coil, and (6) that passing through between the high-
and low-voltage coils. One-half of the flux between the coils is
assumed to surround each of them.
At a distance y from the inside of coil A the m.m.f. acting is
ti Ii -T and this produces a flux, in the path of section l\dy and
i
length w, of a value
d'
w
hdy
TRANSFORMERS 409
Where li is the length of the mean turn of the coil; this flux
surrounds only -4^ turns and is equivalent to a flux d surround-
ing ti turns, where
, . y __ brtj
and the flux equivalent to the total flux crossing the coil is
* = f4 = ^ rVdy = ^ *
J^O ttX*l 2 JO W 3
The flux in the path between the coils is
w
and the part of this surrounding coil A is
2 w 2
The total leakage flux surrounding coil A is
and the self-inductance of the coil is
A = ^^ = - - / - - 1 -f- j absolute units, if all dimensions
are in centimeters.
Using inch units, the inductance of each coil is
A = - - (~ + -)2.54 absolute units,
iv \ o 2i'
or
= 3.2 X 10~ 8 -^-^ (V + ^) henry.
If \ o 2tl
If the g coils are connected in series the inductance of the winding
is
and the reactance is
X, - 27T/L! = 27T/ X 3.2 X lO- 8 ^- 1 + flf
= 20.2 X 10- 8 / - (i 1 + ohms, (355)
g w \ 6
410
ELECTRICAL ENGINEERING
where tti = tig = total primary turns. The reactance is in-
versely proportional to the number of coil groups.
The reactance of the secondary winding is
20.2X10-*/ 71 ? 2 --. (t 2 + 4'
X
IV
ohms.
(356)
The reactance of a core-type transformer may be worked out in a
similar way.
FIG. 394.
FIG. 395.
FIG. 396.
FIG. 397.
338. Design of a Shell-type, Water-cooled Transformer,
Single-phase, 60 Cycles, 1,000 kva., 22,000 to 2,200 Volts for
Power Transmission.
Volts per turn = Voutput in volt-amperes =
k
25
40.
Number of primary turns
22,000
40
550, assume 540. Make the pri-
mary winding of six coils four of 92 turns and two outer coils of 86 turns,
this leaves space for extra insulation on the end turns to take care of strains
due to switching or lightning.
Circular mils per ampere = 1,000.
Current density = 1,270 amp. per square inch.
T>- 1,000,000 ._.
Primary current = = 45.4 amp.
Section of conductor = 45.4 X 1,000 = 45,400 circ. mils = 0.0368 sq. in.
Size of strip = 0.5 X 0.08 bare or 0.515 X 0.095 d.c.c.
Actual current density = 1,170 amp. per square inch.
Actual circular mils per ampere = 1,085.
Insulation between layers = 2 ply 0.007 fullerboard.
Thickness of coil = di = 0.515 in.
Width of coil = ai = 0.095 X 92 + 2 X 0.007 X 91 = 10.014 = 10.5 in.
co allow for warping (Fig. 398).
The end coils have 5 ply 0.007 fullerboard between turns 1 to 11 and 3 ply
between turns 11 to 31 making the coil width 10.5 as before.
TRANSFORMERS
411
Number of secondary turns
540
~r = 54.
Number of secondary coils = 6.
Turns per coil = 18, connect coils in series multiple.
Secondary current = 454 amp.
Current per coil = 227 amp.
Amperes per square inch = 1,270 as in primary.
Section of conductor = 0.2 sq. in. = 0.4 in. X 0.5 in.
Use four strips 0.4 X 0.125 in parallel separated by 1 ply 0.007 fuller-
board and taped with half-lapped tape.
Insulation between layers = 2 ply 0.007 fullerboard.
High Voltage
Coil Section
Low Voltage
Coil Section
FIG. 398.
in. block + 2 (H) in. fullerboard + % in. block
Thickness of low voltage coil = d 2 = 0.425 in.
Width of low voltage coil = a 2 = 18(4 X 0.125 + 3 X 0.007 + 0.025) -f
17 X 2 X 0.007 = 10.066.
Take the width = 10.5 in. to allow for warping (Fig. 398).
Spacing
H.V. to L.V. =
1.0 in.
H.V. to H.V. = % in. block -f-^ in. fullerboard + % in. block = % in.
L.V. to L.V. = % i n . block.
L.V. to iron = Y in. block +H in. fullerboard + % in. block = % in.
H.V. to iron = 1 in. (Fig. 393).
H.V. to iron at top and bottom of cores = Z) = 4 in. (Fig. 393).
Arrangement of coils )
Opening in iron = 12.5 X 20 | Fig. 394.
Size of plates 5 X 25 and 5 X 17.5 j
Flux density = 80,000 lines per square inch, assumed.
Ei 22,OOOX10 8
Flux required =
Section of iron
15.3 X10 6
80,000
4.44X60X540 =
= 191 sq. in. = (5+5)XAX0.9 (Fig. 391).
412 ELECTRICAL ENGINEERING
191
Height of tongue = h = 1Q Q Q = 21.2 in.
Weight of iron = 2 30 X 22.5 -20 X 12.5 - - X4
X21.2X0.9X0.28= 5,538 Ib.
where, 0.28 is the weight of 1 cu. in. of iron in pounds.
Watts lost per pound = 0.70 (Fig. 389).
Total iron loss = 0.70 X 5,538 = 3,877 watts.
3 877
Core-loss current = I c = 22~000 = - 176 am P-
Volt-amperes per pound = 11.5 (Fig. 389).
Volt-amperes at no load = 11.5 X 5,538 = 63,680.
Exciting current = 22 QOQ = ^ = 2 ' 89 amp '
2 89
Per cent, exciting current = TK~! X 100 = 6.36 per cent, of full-load current.
Magnetizing current = I M = Vl* -/J = ^(2.89)2 - (0.176) 2 =2.88 amp.
3 877
No-load power factor cos = ,> 6 Q Q 100 per cent. = 6 per cent.
T fy OQ
Primary exciting admittance = Y = Vg 2 Q -j- &J = ^ = 22QQQ = 1.31 X
io- 4 .
Conductance = g = = 22 OQQ = 0.08 X 10~ 4 .
Susceptance = 6 = j=r = = 1.3 X 10~ 4 .
Radiating surface of the iron = 8,900 sq. in. (Fig. 395).
8 900
Square inches per watt iron loss = ' -- = 2.3.
OjO I /
Length of the mean turn of high-voltage winding = 2K + 2L + irai
= 117 A in. = 9.8 ft. (Figs. 391 and 395).
Resistance of the H.V. winding at 75C. = n = p -. - rr- = 13.2
ClrC. mils*
9.8 X 540 1 ACt ,
lM5~X45A = L42 Ohms '
Primary copper loss = I^n = (45.4) 2 X 1.42 = 2,940 watts.
Length of the mean turn of L.V. winding = 9.8 ft.
Resistance of the L.V. winding = r 2 = T7J2 = 0.0142 ohms.
Secondary copper loss = 7 2 2 r 2 = /i 2 ri = 2,940 watts.
Total copper loss = 5,880 watts.
Surface of each coil = mean turn X width X 2 = 117.4 X 10.5 X 2
= 2,470 sq. in.
Area covered by spacing blocks = 360 sq. in. (Fig. 397).
Radiating surface per coil = 2,470 - 360 = 2,110 sq. in.
Radiating surface of H.V. winding = 2,110 X 6 = 12,660 sq. in.
12 660
Square inches per watt lost = QA = 4-3.
Square inches per watt lost on L.V. winding = 4.3.
Total watts lost = 3,877 + 5,880 = 9,757.
, Cooling coil surface required = 9,757 X 1.0 = 9,757 sq. in.
TRANSFORMERS 413
Diameter of pipe = 1.25 in.
9 757
Length of pipe = ^35 x'3.14 X 12 = 2 7 ft *
Diameter of case = 5 ft.
Length of one turn of cooling coil = 3.14 X 5 = 15.7 ft.
207
Number of turns of cooling coil = y^ = 13 turns.
Efficiency at full-load unity power factor = i Qoo'oob'-u 9 757
X 100 per cent. = 99 per cent.
(540) 2 X 117 4 1
Primary reactance = xi = 20.2 X 10~ 8 X 60 X j^s X 6
(f> + l)=3.7o hm ,
Primary reactance drcp = I&i = 45.4 X 3.7 = 168 volts
Secondary reactance = X* = 20.2 X 10~ 8 X 60 X (18) 2 X X
(f- 5 +') = 0.035 ohms.
Secondary reactance drop = 7 2 z 2 = 454 X 0.035 = 15.9 volts
2^jQ X 100 = 0.72 per cent.
Total reactance drop = 1.48 per cent.
ft A K
Primary resistance drop = Itfi = 45 .4 X 1.42 = 64.5 volts = 22 QO Q
X 100 = 0.294 per cent.
6.45
Secondary resistance drop = 7 2 r 2 = 454 X 0.0142 = 6.45 volts =
X 100 = 0.294 per cent.
Total resistance drop = 0.588 per cent.
(1 48) 2
Regulation at unity power factor = 0.588 + 2QQ = 0.6 per cent.
Regulation at 80 per cent, power factor = 0.588 X 0.8 + 1.48 X 0.6
(1.48 X 0.8 - 0.588 X 0.6) 2
+ - 2QQ ~ = 1-35 per cent.
CHAPTER XI
CONVERTERS
339. Types of Converters. Converters are rotating machines
which change electrical energy from one form to another. They
are of several types :
1. A synchronous converter or rotary converter converts from
an alternating to a direct current or vice versa.
2. A frequency converter converts the power of an alternating-
current system from one frequency to another with or without a
change in the number of phases or in the voltage.
3. A rotary phase converter converts from an alternating-cur-
rent system of one or more phases to an alternating-current sys-
tem of a different number of phases but of the same frequency.
The most usual change is from
single phase to polyphase (see
Art. 414). Changes from one
polyphase system to another are
usually carried out by means of
static transformers (Art. 318).
Any of these conversions can
be made with motor-generator
sets.
340. Synchronous Converter.
The synchronous converter or
rotary converter is a combina-
tion of a synchronous motor and a direct-current generator. It
receives alternating current and converts it into direct current.
The fields are excited by a shunt winding connected between
the direct-current brushes or from a separate exciter. In some
cases a series winding is added for compounding.
The armature winding is an ordinary direct-current winding
and may be either series or multiple. It is connected to a com-
mutator and taps are taken out from it at equidistant points and
connected to slip rings. The alternating current is delivered to
the slip rings either single-phase, two-phase, three-phase or six-
414
FIG. 399. Ring-wound rotary
converter.
CONVERTERS
415
phase and drives the armature as a synchronous motor. The
same armature conductors generate and carry the direct current.
Fig. 399 shows a ring-wound bipolar armature tapped for
single-, two-, or three-phase currents; single-phase 1 to 2 or 3 to 4;
two-phase 1 to 2 and 3 to 4; three-phase 1 to 5, 5 to 6 and 6 to 1.
Fig. 400 shows a six-circuit multiple winding for a six-pole,
three-phase rotary converter and Fig. 401 shows a two-circuit or
series winding for an eight-pole, three-phase converter.
FIG. 400. Multiple-drum winding for a three-phase rotary converter.
In a series-wound armature the total number of coils must be
divisible by the number of phases and in a multiple- wound arma-
ture the number of coils per pair of poles must be divisible by the
number of phases.
Six-phase converters are operated from three-phase circuits.
Practically all converters above 200 or 300 kw. output are now
built with six collector rings and six phases. Three methods of
connecting the supply transformers are shown in Fig. 370.
341. Ratios of E.M.Fs. and Currents. With the brushes
fixed on the no-load neutral line, the direct and alternating e.m.fs.
generated in the converter bear a definite relation to each other
and one cannot be varied without varying the other. At unity
416
ELECTRICAL ENGINEERING
power factor the alternating and direct currents in the armature
also bear a definite relation to each other if the current required
to supply the converter losses is neglected.
Since the alternating and direct currents flow in. the same arma-
ture conductors and in opposite directions, the e.m.f . consumed in
the armature is small and the power loss is small. In the follow-
ing analysis these quantities will be neglected and the alternating
current will be assumed to be in phase with the impressed e.m.f.
r
FIG. 401. Two-circuit, retrogressive winding for a three-phase rotary converter.
This condition can be obtained in practice by properly adjusting
the exciting current.
Take first the case of the single-phase converter, Fig. 402.
Let E = direct voltage of the converter.
/ = direct-current output.
Ei = effective value of the alternating supply voltage,
whether single phase or polyphase.
1 1 = alternating current in the supply lines.
/' = alternating current in the armature winding.
The voltage between the leads li and Z 2 or between the slip
rings Ri and R 2 is alternating and reaches its maximum value
CONVERTERS
417
when li and 1 2 are under the brushes and it is then equal to the
direct voltage of the machine. Therefore, in a single-phase con-
verter the direct voltage is equal to the maximum value of the
alternating voltage and thus
E = \/2#i (357)
or
E (358)
Neglecting losses and phase displacements the output of the con-
verter is equal'to the volt-amperes input or
77T 7? 7"
Jiil -= Hi il\
and the alternating current in the line is
(359)
The alternating current in the winding is
r = - 1 ,
p
where p is the number of circuits in multiple through the winding.
In the bipolar machine in Fig. 402, p = 2 and, therefore,
$.*-**. ^
Fio. 402. Single-phase converter.
FIG. 403. Two-phase or quarter-phase
converter.
342. Two -phase or Quarter-phase Converter. When four
collector rings R i} R 2 , Rs and R^, Fig. 403, are connected to four
equidistant points li, 1 2 , h and Z 4 , the machine is a two-phase or
quarter-phase converter. The two voltages- Ri to R 2 and R s to
R are equal and are in quadrature, forming a two-phase system;
27
418
ELECTRICAL ENGINEERING
the four voltages RI to R 3) R 3 to R 2 , Rz to R and R 4 to Ri are
all equal and form a four-phase or quarter-phase system.
The voltage between lines or the voltage per phase of the two-
phase supply is E
"'
The voltage between adjacent rings or the quarter-phase vol-
tage is Ei E
'
Assuming the volt-amperes input two-phase to be equal to the
direct-current output, that is,
2E 1 I l = EI,
the alternating current per line is
77T T 77T T T
T & &* 1 foao\
7 i = ^r jjr 7 (363)
The alternating current in the winding is the resultant of two
currents = ,- in quadrature and its value is therefore
2 2\/2
F =
(364)
343. Three-phase Converter. With three collector rings Ri,
R 2 and R 3 , Fig. 404, connected to three equidistant points h, 1 2
and 3 the machine is a three-phase converter.
Fro. 404. Three-phase converter. FIG. 405. E.m.fs. and currents
in a three-phase converter.
.The e.m.f. between each of the rings and the neutral point or
the "star" e.m.f. is equal to half of the single-phase voltage. It
is shown as E s in Fig. 405.
CONVERTERS
The e.m.f. between rings or " delta" e.m.f. is
Ei = V3E S = ^! = 0.612E.
2\/2
The power input is
i = 3^Ji = 3EJ'
419
(366)
and is equal to the output EI.
Thus the line current is
and the alternating current in the winding is
I' = L = j = 0.5457.
A/3 3V3
(368)
FIG. 406. n-phase converter.
FIG. 407.
344. n-phase Converter. For an n-phase converter, Fig. 406
the winding must be tapped at n equidistant points. The e.m.f.
between each of the rings and the neutral point or the "star"
e.m.f. is as before
The e.m.f. between rings or the e.m.f. between lines is the vector
difference between two e.m.f s. E s displaced at -- radians (Fig.
407). Thus'.;.
2E a sin - = n
The power input is
nE s Ii =
and is equal to the output EI.
,_
\/2
(369)
420
ELECTRICAL ENGINEERING
Therefore the alternating current in the line is
7 El = 2\/2
nE K E n '
h =
n
and the alternating current in the winding is
El El \/2I
I' =
. 7T
n sin -
n
(370)
(371)
The values obtained above for the e.m.fs. and currents in single-
phase, two-phase and three-phase converters can also be ob-
tained by substituting the proper values of n in equations 369,
370 and 371 ; single-phase n = 2, two-phase or four-phase n = 4,
and three-phase n = 3. The results are tabulated in Fig. 408.
Type
Single-
phase
n = 2
Three-
phase
n = 3
Two-phase
or four-
phase
n = 4
Six-phase
n = 6
n-phase
E.m.f. between collector
E
VZE
E
E
777 V
JT, gin
rings or line e.m.f. E\. . .
V2
2V2
2
2-\/2
n
Current per line I\
V2/
2V2/
7
V2/
V2
2\/2/
3
V2
3
n
V2I
Current in the winding /'
I
2\/2/
I
V2/
, if
V*
3\/3
2
3
n sin -
n
FIG. 408.
These ratios of currents only hold on the assumption that the
power factor is unity and that the efficiency is 100 per cent.
The power factor can be maintained approximately unity by
adjusting the excitation, but the power losses cannot be elimi-
nated and the values of alternating current in the, table must be
increased by the small component required to supply the losses in
the machine. When the power factor is not unity the reactive
or wattless components of current must be added to the power
components.
The ratios of e.m.fs. are* the ratios of the generated e.m.fs.
and can only approximately represent the ratios of terminal
CONVERTERS
421
e.m.fs. since components of e.m.f. are consumed in the resistance
and reactance of the armature. The ratios also depend on the
assumption that the alternating e.m.f. wave is a sine wave. If
the wave is peaked, the ratio of the effective value to the maxi-
mum value is less than 7= and the values of the alternating
e.m.fs. must be reduced. If the wave is flat topped the ratio of
effective to maximum value is greater than 7= and the values of
the e.m.fs. must be increased.
FIG. 409.
(2) Alternating Current
(D-
FIG. 410. Current in coil c, Fig.
409, at unity power-factor.
FIG. 411. Current in coil a, Fig.
409, at unity power-factor.
345. Wave Forms of Currents in the Armature Coils. The
current in any armature coil is the difference between the alter-
nating-current input and the direct-current output.
In Fig. 409, li and 1% are the two leads of one of the n-phases of
a converter, a is the coil next to one lead and c is the coil in the
center of the phase. The alternating e.m.f. and the power com-
ponent of the alternating current in the phase li to 1 2 are maximum
when this section of the winding is midway between the brushes
and they are both zero when the center coil c is under the brush.
Fig. 410 shows the resultant of the alternating and direct
422
ELECTRICAL ENGINEERING
current's in coil c during one revolution. The alternating current
is opposed to the direct current and is zero when the direct current
reverses as the coil passes under the brush. The current in the
center coil is, therefore, less than the current in any other coil
in the phase when the power factor is unity.
Fig. 411 shows the current in coil a next to one of the leads.
The alternating and direct currents are not directly opposing and
the resultant current is greater than in the center coil c.
The coils next to the leads, therefore, carry larger currents than
the coils in the center of the phases and they rise to a higher
temperature.
The worst condition of local heating occurs in the coil next the
lead of a single-phase converter, Fig. 412. The alternating cur-
rent has its maximum value when the direct current reverses.
Alternating Current
77
\\
Q) Direct Current
Resultant Current
FIG. 412. Current in the coil next to
the lead in a single-phase converter.
x NS ^ Alternating Currei
\@ Kesultant Current
\ \ 180
\\
(TXDirect CurrentX
FIG. 413. Current in coil c, Fig. 409,
at 70 per cent, power factor.
When the power factor is not unity the minimum resultant
current will not occur in the center coil of a phase but in a coil
displaced from it by the angle of lag or lead of the current. Fig.
413 shows the current in coil c when the power factor is 70 per
cent, and a component of lagging current equal to the power
current flows in the armature.
346. Heating Due to Armature Copper Loss. Take the case
of a two-pole, n-phase armature, Fig. 409, and use the same
notation as before. In the center coil c of the phase, the direct
current is ^ and the effective value of the alternating current is
nsin -
n
CONVERTERS
423
The instantaneous value of the alternating current is
i = A/2 F sin =
21
n sin -
n
sin 6,
and the instantaneous value of the resultant current is
27
(372)
(373)
In an armature coil d displaced by angle from the center of
the phase the alternating current is
i = A/27' sin (0 - 0) (374)
and the instantaneous value of the resultant current is
sin (0 - 0) - 75
7T
n sin -
n
4 sin (0 - 0)
- 1
sin -
71
(375)
The effective value of the resultant current is
7 j /v
-^-JliW. 5 /-J, nsi
4 sin (0 - 0)
(376)
n
|ip
TTjO
16 sin 2 (0 - 0) _ 8 sin (0 - 0) 1
7T /i
n sin -
71
o o 7T
7i 2 sin 2 -
n
- cos 2(0 - 0)( 8 sin (0 - 0)
+ 1
Jo[
o o If
7i 2 sin 2 -
n
. TT \av
n sin -
n J
" 8
sin 2
(9 - ft) } , 8 cos (.0 - 0) , fl '
n 2 sin 2 -
n
2~ H
7i sm
n
[" 8r
16 cos
~h TT
*
n 2 sin 2 -
71
. 7T
7i sm -
n
16 cos
(377)
n TT sin
n n
424 ELECTRICAL ENGINEERING
Since ^ is the current in the coil when the machine is operating as
a direct-current generator, the ratio of the power lost in the coil
when operating as a converter to that lost when operating as a
direct-current generator with the same output is
16 cos ft (37g)
and this is the ratio of the coil heating under the two conditions.
This ratio is a maximum for the coil next to the alternating
leads h or Z 2 , where = , and it is
IV
, . 7T . 7T
n 2 sin 2 - mr sin -
n n
7T
16 cos -
n + 1. (379)
It is a minimum for the center coil of the phase, where (3 = 0,
and is
h = - -^ : + 1. (380)
n 2 sin 2 mr sin -
n n
The ratio of the total power lost in the armature of the con-
verter to that lost when the machine is operating as a direct-
current generator with the same output is found by integrating
the ratio h$ over one-half phase from (3 = to = and taking
the average. It is
/-- // 8 16 cos/3 A ,.
hftdp = -
/o I n
2 sin 2
7T
7T
ITT sm -
r*
n
7T
n\
8/3
16
sin |8
+ 18
f
w\
n 2 sin 2 -
mr
7T
sm -
\
n
n
Jo
' 8
16 ,
^
S + l, (381)
n 2 sin 2 -
n
and this is the relative armature heating under the two conditions.
To get the same loss in the armature of a converter and the
CONVERTERS
425
same heating as in the direct-current generator, the armature
current and the output may be increased in the ratio ~~/~'
V H
The values of H and / for the various polyphase converters
are tabulated in Fig. 414.
Type
Direct-
current
generator
Single-
phase
n = 2
Three-
phase
n = 3
Two-phase
or four-
phase
n = 4
Six-phase
n = 6
Relative armature heat-
ing H
1 00
1 37
55
37
26
Rating by armature heat-
1
me /
1 . 00
0. 85
1 34
1 64
1 96
e v#
FIG. 414.
For a single-phase converter, n = 2, the value of , is 0.85,
and, therefore, the output of a machine as a single-phase converter
is only 85 per cent, of its output as a direct-current generator for
the same temperature rise.
For a three-phase converter 7= = 1.34 and therefore the out-
put is 34 per cent, greater than as a direct-current generator.
For a six-phase converter the output is increased 96 per cent.
These values only hold if the alternating current is in phase
with the impressed e.m.f. When leading or lagging currents flow
in the armature the heating is very largely increased and the rating
must be decreased.
The reduction in rating due to reactive lagging or leading cur-
rents may be found as follows :
If the alternating current lags behind the voltage by an angle
0, then, since the ratio of the power components of the alternating
and direct currents must be the same as before, the alternating
current input is increased by the amount of the reactive lagging
current and its effective value is
/' =
n sin - cos
n
(382)
426 ELECTRICAL ENGINEERING
and the instantaneous value of the resultant current in the coil
d is
27 /
sin (0 - |8 - ) - -; (383)
\ v *r/ rt
n sin - cos
n
its effective value is
ZZJL _ iBeoBjfjE^. (384)
n 2 sin 2 - cos 2 ?ITT sin -
n n
the ratio of the power lost in the coil of a converter to that of a
direct-current generator is
8 16cos(0+0) + 1(38g)
n 2 sin 2 - cos 2 < WTT sin - cos
2 J n n
and the relative armature heating is
n 2 sin 2 - cos 2
(386)
To find the power factor at which the rating of the converter
is reduced to that of the direct-current generator, equate H to
unity and solve for cos <.
Thus,
~ + 1 = 1 and cos = - . (387)
n 2 sin 2 - cos 2 \/2n sin -
n n
When n = 3, cos = 0.85 and when n = 6, cos < = 0.745;
therefore the rating of a machine as a three-phase converter will
be the same as when operated as a direct-current generator when
the power factor is 85 per -cent, and the rating of a machine as a
six-phase converter will be the same as when operated as a direct-
current generator when the power factor is about 75 per cent.
Synchronous converters should be operated at approximately
unity power factor at full load and overload.
347. Armature Reaction. The armature reaction of a rotary
converter is the resultant of the armature reactions of the machine
as a direct-current generator and as a synchronous motor. The
direct-current brushes are usually placed on the no-load neutral
points and therefore the direct-current exerts a m.m.f. in
CONVERTERS
427
quadrature behind the field m.m.f. (Fig. 415). The power com-
ponent of the armature current in a synchronous motor exerts
a m.m.f. in quadrature ahead of the field m.m.f. and it is there-
fore opposed to the m.m.f. of the direct current.
If Z is the number of conductors on the armature of a bipolar
generator and ^- is the direct current in each conductor, the arma-
z
Z I
ture m.m.f. is the resultant of ^ m.m.fs. of magnitude ~ uni-
formly distributed over the circumference of the armature and
Chord
FIG. 415. M.m.fs. in a direct-current generator.
FIG. 417.
it is therefore less than that of a concentrated winding in the
2
ratio - (Fig. 416). The m.m.f. of the direct current in the arma-
7T
ture of a converter is thus
Z I 2 - ZI
(388)
in quadrature behind the field m.m.f.
If the machine is connected as an n-phase converter, the num-
9
ber of turns per phase is ~- and the effective value of the alternat-
ing current in each is
/' =
V2I
n sin -
n
428
ELECTRICAL ENGINEERING
and the m.m.f. per phase in effective ampere-turns is
ZT_ * ZI
2n ~~
m'
(389)
These ampere-turns are distributed over -th of the circumfer-
ence of the armature and their resultant is, Fig. 417,
r ^
2r sin -
m =
ZI f chord
2n ' arc
ZT
2n
417),
ZI
r* if
2r sin -
. 7T
sin-
n
27TT
n
ZI
The maximum value of the m.m.f. per phase is
/ Z/
m m = V 2m = ampere-turns.
7m
(390)
(391)
To find the resultant m.m.f. of the alternating current in the
armature it is necessary to combine n m.m.fs. of maximum
ZI 2ir
value m m = - - displaced in direction by angle and displaced
TT/t' TL
in phase by -th of a period or by angle
In Fig. 418 phase 1 is shown in the
position of maximum m.m.f. if the
power factor is unity. The direction of
the m.m.f. is OB and it is in quadrature
ahead of the field m.m.f. If time and
angular displacement are measured from
OB > then at time * and angle e the m.m.f.
of phase 1 is m m cos 6 in direction OB\
and its component in direction OB is m m cos 2 6.
At time t the m.m.f. of phase 2 is m m cos 10 -\ -- j making an
angle (0 -\ -- ) with OB and its component in direction OB is
\ 77 /
Pioi8.Synchronou8
motor.
CONVERTERS 429
The resultant m.m.f. of the n phases in the direction OB at
any time t is
M a = m m COB 0+coB e + ~ + + + cos
= w w X n X average (cos) 2 = ra m =
since the average cos 2 is = J^.
The resultant m.m.f. of the n phases in the direction at right
angles to OB or in line with the field m.m.f. is zero.
Therefore the resultant m.m.f. of the alternating current in
the converter armature has a constant value
,, . n ZI n ZI /or^x
^ = m "2 = -2 = 27 (392)
and is in quadrature ahead of the field m.m.f. It is thus equal to
the m.m.f. of the direct current and is opposed to it and the re-
sultant armature reaction of the direct current and of the cor-
responding power component of the alternating current is zero.
The armature reaction due to the power current required to
supply the losses remains but it is very small and produces only
a slight distortion of the flux in the air gap.
The effective armature reaction of a six-phase converter is
from 7 to 20 per cent, of that of the corresponding direct-current
generator.
When the power factor is not unity the wattless currents in
the armature exert m.m.f s., as in the synchronous motor, which
act in line with the field m.m.f. and are either magnetizing or
demagnetizing but are not distorting.
Thus in the rotary converter there is very little field distortion
or very little shifting of the neutral points under load. As a
result the limit of overload set by commutation is much higher
than in the direct-current generator. This is very important in
the case of converters supplying railway loads where the load
factor is usually below 50 per cent. The overload capacities for
short periods must be high. In some cases when using interpoles
momentary overloads of 200 per cent, are permitted.
348. Control of the Direct-current Voltage. Since the ratio of
alternating- to direct-current voltage is fixed, for a given flux
distribution, the direct voltage cannot be controlled as simply
as in the direct-current generator. There are three practicable
methods of control:
430 ELECTRICAL ENGINEERING
(a) Variation of the alternating voltage.
(6) Variation of the direct-current voltage by means of a
direct-current booster.
(c) Variation of the flux distribution as in the split-pole
converter.
349. Methods of Varying the Alternating Voltage. The im-
pressed alternating voltage may be varied: (1) by variable ratio
supply transformers; (2) by induction regulators in the supply
lines; (3) by introducing reactances in the supply lines and draw-
ing lagging or leading currents through them by varying the
field excitation of the converter ; and (4) by a synchronous booster.
The first method has the -disadvantage of a step-by-step
regulation and the contacts are liable to be burned as the circuit
is opened when changing from one tap to another; the second
method requires expensive apparatus but gives very good regula-
tion and may be made automatic; the third is inexpensive and
is made automatic by placing a series winding on the converter
poles. The fourth method requires the use of an additional
synchronous machine. It is very satisfactory and is being em-
ployed very extensively especially in large installations.
350. Compounding by Reactance. When the field current of
a synchronous converter is varied, reactive lagging or leading
currents flow in the armature and magnetize or demagnetize the
field and bring it back to its former strength and the direct-
current voltage remains as before. To vary the direct-current
voltage the impressed alternating voltage must be varied. This
may be accomplished by introducing reactance coils in the supply
lines and placing a series-field winding on the converter poles.
The e.m.f. of inductance lags 90 degrees behind the current;
thus, when the converter is under-excited a component of current
lagging 90 degrees behind the, impressed e.m.f. flows through
the reactance coils, and the e.m.f. of inductance due to it lags
180 degrees behind the impressed e.m.f. and therefore opposes
and lowers it. When the converter is over-excited and a com-
ponent of current 90 degrees ahead of the e.m.f. flows through
the reactance, the e.m.f. of inductance due to it is in phase with
the impressed e.m.f. and raises it.
Therefore, when reactance coils are connected in the supply
lines an increase of the field excitation raises the impressed e.m.f.
and so raises the direct-current voltage, and a decrease of the ex-
citation lowers the impressed e.m.f. and also the direct-current
CONVERTERS
431
voltage. The result is the same as in the direct-current generator
but it is produced in a somewhat different way.
The series winding causes the direct-current voltage to rise
automatically with load. The excitation is so adjusted that at
light load the converter is under-excited and the power factor is
low and lagging while at full load it operates at a power factor of
about 98 per cent, leading. This results in only a slight increase
in the heating.
In some cases the required reactance may be provided in the
step-down transformers but in 25-cycle systems it is difficult to
provide sufficient reactance in this way.
Alternating
Booster
Rheostat
FIG. 419. Synchronous booster converter.
351. Synchronous Booster Converter. The synchronous boos-
ter converter is a combination of an ordinary converter and an
alternating-current generator on the same shaft with the same
number of poles and having its armature winding connected in
series with the converter armature. The excitation of the booster
is so arranged that it can be reversed, and therefore the booster
voltage reversed, giving a uniform voltage variation of double
the booster voltage. The booster may be shunt-excited and have
its voltage controlled by an automatic regulator or it may be
series- wound. Fig. 419 shows the diagram of connections for a
shunt-excited booster.
352. Direct-current Booster Converter. The direct-current
voltage can be varied by inserting a booster in the direct-current
432
ELECTRICAL ENGINEERING
(a) Menu do Voltage
(b) Maxinaun^ j2,c Voltage
_
(c) Mioimum do Voltage
circuit. It may be direct-connected to the converter or driven
by a separate motor. In either case additional floor space is
required and the commutator of the booster must have the same
current capacity as that of the converter.
353. Split-pole Converter. In the split-pole converter the
variation of direct-current voltage is secured, not by a variation
of the impressed alternating voltage but by changing the shape
I . of the magnetic field so that the total flux
/ \ and the direct-current voltage are changed
/ \ while the alternating voltage remains un-
changed. This is possible because in three-
phase windings third harmonics and their
multiples do not appear in the terminal
voltage.
If the pole is made in three parts the exci-
tation of the various parts may be changed
so as to introduce a third harmonic in the
flux wave. With the three sections equally
excited the flux wave is flat as shown at (a)
in Fig. 420 and the direct-current voltage
has its mean value. If the excitation of 1
and 3 is increased while that of 2 is de-
creased a third harmonic is produced in the
flux wave and in the generated voltage but
will not appear in the terminal voltage of the
three-phase converter with its taps at 120 degrees. The total
flux and therefore the direct-current voltage is increased in pro-
portion to the area under one-half wave of the third harmonic
flux (Fig. 420 (6)). If the excitation of 1 and 3 is decreased and
that of 2 is increased the flux and the direct-current voltage are
decreased as shown at (c). A variation of the direct-current
voltage of 20 per cent, in either direction may be obtained.
The space required for the three-part pole is large and the con-
verter must be made of large diameter to give room for it.
Similar results can be obtained by using a two-part pole but the
flux wave is not symmetrical and it is more difficult to maintain
a suitable commutating field, but the machine has a smaller
diameter and is less expensive.
354. Frequencies and Voltages. Converters are built for 25
and 60 cycles and for voltages 250 and 600 volts. Some 25-
cycle converters have been built for 1,200 and 1,500 volts, but
Thrae Part Pole
FIG. 420. Split-pole
converter.
CONVERTERS 433
where such high direct-current voltages are required it is more
usual to connect two converters in series or to use motor-genera-
tor sets.
The design of converters is similar to that of direct-current
generators and no special difficulties are met in 25-cycle conver-
ters up to 600, volts but in 60-cycle converters for 600 volts and
in 25-cycle converters for 1,200 volts, since the number of poles
is fixed by the frequency and the speed, it is difficult to leave
space enough between direct-current brushes for a sufficient
number of commutator bars. The bars must be made narrow
and the number of volts per bar is high and the danger of flash-
over is great. However, by increasing the commutator periph-
eral speed to 5,500 ft. per minute the distance between neutral
points is satisfactory and the number of volts per bar is reduced
to a safe value. Sixty-cycle 600-volt converters are now giving
perfect satisfaction in outputs up to 2,000 kw. but commutating
poles are required on all above 500 kw.
355. Outputs and Efficiencies. Six-hundred-volt, 25-cycle
converters are built in sizes up to 4,000 kw. and range in efficiency
from 95 to 96 per cent. ; 600-volt 60-cycle converters are built up
to 2,000 kw. and with efficiencies from 91 to 94 per cent. Con-
verters for 250 volts are built up to 1,000 kw. with efficiencies of 93
to 95.5 per cent, for 25 cycles and 90.5 to 94 per cent, for 60 cycles.
356. Overload Capacity. Commutating-pole converters will
stand momentary overloads of 100 per cent, and in special cases
of 200 per cent. The armature reaction is much smaller than in
direct-current generators and therefore the interpoles do not re-
quire so much excitation and can be designed to takei ^are of very
heavy overloads without becoming saturated.
357. Dampers. Synchronous converters should always be
provided with damper windings m to prevent hunting. A com-
plete squirrel-cage winding is used with non-interpole converters
but when interpoles are required separate grids are placed in the
main pole faces. A short-circuited winding surrounding the in-
terpole would delay the growth of flux in it and so interfere with
commutation.
358. Starting. Synchronous converters may be started: (1)
from the alternating-current end, (2) from the direct-current end,
and (3) by an auxiliary induction motor.
359. Alternating-current Self-starting. If low alternating
voltage is applied to the slip rings the converter will start as an
28
434 ELECTRICAL ENGINEERING
induction motor. The damper windings in the pole faces act
as the secondary and the rotating armature as the primary.
When it is nearly up to synchronous speed, it falls into step as
explained in Art. 277. The field circuit is then closed and full
voltage impressed. For small converters one low-voltage tap
only is required but for the larger ratings two starting taps are
brought out from the supply transformers.
While coming up to speed alternating voltages are induced in
the field winding. Their magnitude depends on the ratio of field
turns to armature turns and on the slip of the rotor behind the
revolving field, and they disappear when synchronous speed is
reached. To prevent dangerous induced voltages in the field
winding it is usually broken up into a number of sections during
starting by means of a field break up switch, or it may be short-
circuited.
The converter may be brought up to speed very quickly in this
way since it does not require to be synchronized, but there is no
way of predetermining the polarity of the direct-current brushes.
If the polarity is wrong the converter may be forced to slip a pole
by reversing the field switch while the armature is still connected
to the low-voltage taps. The switch must then be returned to its
original position.
When starting from the alternating-current end a large current
at low power-factor is drawn from the lines and it may be objec-
tionable.
360. Direct-current Self -starting. Converters may be started
as shunt motors from the direct-current end if suitable -power
is available in the station but a longer time is required to put
them in operation than with alternating-current starting and
they must be synchronized. The starting current is low.
361. Starting by an Auxiliary Motor. If an induction motor
with a smaller number of poles than the converter, and conse-
quently a higher synchronous speed, is mounted on the same
shaft it may be used as a starting motor. This method requires
synchronizing and thus takes considerable time but the starting
current is small and there is no trouble with the polarity. The
extra starting motor increases the cost of the equipment.
362. Brush Lifting Device. When starting a converter as an
induction motor the revolving armature flux induces voltages in
the coils short-circuited by the direct-current brushes and spark-
ing results but it is not usually serious. In commutating-pole
CONVERTERS 435
converters the sparking tends to be very much worse since the
reluctance of the path through the short-circuited coil is reduced
by the presence of the interpole iron and therefore the flux and the
induced voltage are increased. To prevent serious sparking a
brush-lifting device is employed which raises all the brushes
except two narrow ones which are required to indicate the
polarity.
363. Bucking or Flashing. Bucking or flashing may be the re-
sult of poor commutating conditions causing bad sparking under
the brushes, or of high voltage between adjacent bars; or of low
surface resistance between adjacent brushes. These causes are
usually all present to a greater or lesser degree when flashing
takes place.
Commutating conditions in converters are normally better
than in direct-current generators since the resultant armature
reaction is small ; but very heavy overloads may distort the field
and cause sparking ; or sparking may result from mechanical de-
fects as wrong brush setting, poor contact, high mica, etc.
Comparatively high voltage between bars is the usual condition
in rotary converters since the space between neutral points is
limited especially for the higher frequencies and voltages.
Any increase of impressed voltage due to disturbances on the
lines will increase the volts per bar or a sudden increase of cur-
rent, due to a short-circuit, may distort the field and increase the
voltage between certain bars. This tends to cause local sparking
between these bars. A spark starting under the brush due to
any cause may then be carried round to the next brush especially
if the commutator is dirty. Such a flash-over, if severe, will
short-circuit the converter and cause it to buck, fall out of
synchronism and stop.
364. Parallel Operation. When two or more converters are
connected to the same direct-current bus-bars they should be sup-
plied from separate banks of transformers to prevent large cur-
rents circulating 'between them due to differences of counter e.m.f.
or to variations in the resistance of the direct-current brush con-
tacts. The separate transformer banks permit slight corrections
to be made in the voltages of the converters.
365. Inverted Converter. Where a small alternating-current
load is to be supplied from a direct-current system, a rotary con-
verter may be used as an inverted converter to transform direct
current to alternating current.
436 ELECTRICAL ENGINEERING
The ratios of the voltages are the same as under normal oper-
ating conditions but the ratios of the currents vary since it is not
possible to eliminate or control the wattless components of the
alternating current. These components depend on the character
of the load and are not affected by varying the exciting current.
When changing from alternating current to direct current the
speed of the converter is fixed by the frequency of the system and
remains constant. When changing from direct current to alter-
nating current the speed is not fixed but depends on the excita-
tion and varies as the field strength varies. When the load is
inductive the wattless lagging current demagnetizes the field and
so raises the speed of the converter and the frequency of the alter-
nating current. This may increase the lagging current and so
raise the speed more until it gets beyond safe limits. The in-
verted converter has thus the two disadvantages: (1) that it
tends to run at dangerous speeds, and (2) that it supplies a cur-
rent of varying frequency. It must, therefore, be provided with
some means of cutting off the load when the speed rises above a
certain value or with some means of limiting the speed.
' If the converter is excited by a direct-current generator
mounted on the same shaft any increase in speed raises the exciter
voltage at a higher rate and, therefore, the field of the converter
is strengthened and the speed is limited.
Rotary converters to be used as inverted converters should be
shunt-wound to secure as constant a speed and frequency as
possible.
If two converters are operating in parallel and the alternating-
current circuit breakers of one open, this converter will run in-
verted as a direct-current shunt motor and if it is under-excited
the speed will rise. If it is in a position to supply alternating
current to a reactive circuit the rise in speed may be serious.
This condition may be prevented if the direct- and alternating-
current circuit-breakers are interlocked or if the direct-current
breakers are provided with a trip operated by a reverse-current
relay.
366. Double-current Generator. If mechanical power is sup-
plied to drive a rotary converter it can be used as a double-current
generator to supply direct current from the commutator and
alternating current from the slip rings.
The two currents in this case flow in the same direction in the
armature conductors and the losses are increased and the arma-
CONVERTERS
437
ture reaction is the sum of the reactions due to the two currents.
The voltage regulation is, therefore, poorer than in the converter.
367. Three-wire Generator. The three-wire direct-current
generator is similar in construction to a single-phase or quarter-
phase rotary converter. It is used to supply a three-wire system
with from 220 to 280 volts between outer wires and from 110 to
140 volts between each of the outer wires and the neutral wire.
In order to obtain a point at a potential midway between the
potentials of the direct-current brushes special transformers called
compensators are used. They have a single winding tapped at
the center and are connected by means of slip rings across points
on the armature winding 180 electrical degrees apart (Fig. 421).
The neutral wire of the system is connected to the central point
of the compensator and its potential is maintained almost mid-
way between the potentials of the outer wires. When more than
one compensator is used the center points of all the compensators
are joined together before being connected to the neutral wire
(Fig. 422).
FIG. 421. Three- wire generator with FIG. 422. Three- wire generator with
a single compensator. two compensators.
In some cases the compensators are connected directly to the
armature windings and rotate with it and their neutral points are
connected and brought out to a single slip ring.
The voltage between the termirials of the compensators is
alternating and when the loads on the two sides of the three-wire
system are equal and no current flows in the neutral wire the only
current in the compensator is the very small exciting current.
When the loads are unequal the unbalanced current I N flows in
the neutral wire as shown in Fig. 421. On reaching the com-
pensator it divides into two parts which flow through the winding
and up to the positive brush by the path of least resistance.
Since the current in the neutral wire is a direct-current the react-
438
ELECTRICAL ENGINEERING
ance of the compensator does not oppose it and the only voltage
drop is that due to resistance.
The actual amount of current carried by the various sections
of the armature winding is very difficult to calculate and it varies
from instant to instant due to the change in the relative positions
of the direct-current brushes and the compensator connections.
The average current carried by each half of the compensator
.:... '.'In
winding is .-*
The unbalancing of the currents in the sections of the armature
winding produces an unbalancing of the armature reactions and
results in a slight unbalancing of the voltages between the neutral
point and the brushes. With two or more compensators, Fig.
422, this unbalancing is reduced due to the more even distribu-
tion of the current.
Positive
Neutral
FIG. 423. Parallel operation of three-wire generators.
Machines can be designed to give a regulation of 2 per cent,
or less with an unbalanced load of 25 per cent.
With this system the voltages on the two sides cannot be regu-
lated independently and the flexibility of the three-wire system
supplied by t,wo generators in series is lost, but there is a corre-
sponding gain in space and cost of machines.
The capacity required in the compensators is small. With 25
CONVERTERS 439
per cent, unbalancing of the loads the required compensator
capacity is less than 10 per cent, of the generator capacity.
368. Parallel Operation of Three-wire Generators. Fig. 423
shows the diagram of connections for parallel operation of two
compound-wound three-wire generators, which are provided with
interpoles.
The series windings must be divided into two groups, one on
either side of the system, and two equalizer connections are re-
quired. Half of each interpole winding must be connected on
one side of the system and half on the other. In the case of the
series-field windings it is sufficient to supply the south poles from
one side and the north poles from the other.
To connect No. 2 in parallel with No. 1 bring it up to speed and
close switches Si, Si, thus exciting its series windings, adjust the
voltage by means of the shunt-field rheostat #, close $ 2 , Sz and
finally close the neutral switch $ 3 .
Circuit-breakers B, B with overload and reverse-current trip
coils should be connected on each side to protect the machines.
369. Frequency Converters. Frequency converters are used
where power is transmitted at 25 cycles and is required by the
consumer at 60 cycles or where two systems of different fre-
quencies are to be linked together in a high-voltage network.
The most usual form of frequency converter is a synchronous
motor-generator set, a 25-cycle motor direct-connected to a 60-
cycle alternator. The numbers of poles on the two machines
must be in the ratio of 25 to 60. When a 10-pole motor is used the
alternator must have 24 poles and the speed is fixed at 300 r.p.m.
When frequency converters are to be operated in parallel they
must be synchronized on both the 25-cycle and 60-cycle ends.
If the motor is synchronized first there is only one chance in five
that the alternator is in synchronism, while if the alternator is
synchronized first there is only one chance in twelve that the
motor is in synchronism.
If the motor is synchronized and it is found that the generator
is out of synchronism the circuit must be opened and the motor
allowed to slip back a pair of poles at a time until the correct
position is reached.
When linking up two systems by a frequency converter, the
converter must be connected to one of them and then the second
system brought up to the position of synchronism.
The induction frequency converter is discussed in Art. 415.
440 ELECTRICAL ENGINEERING
370. Mercury Vapor Rectifier. A rectifier acts as an electric
valve allowing current to flow through it in only one direction
and it may therefore be used to rectify an alternating current,
that is, to change it to a uni-directional current.
The mercury vapor rectifier, which is used very largely to con-
vert alternating current to direct current for charging storage
batteries, supplying arc lamps and many other purposes, is shown
in Fig. 424 (a). It consists of an exhausted bulb B which has two
projections on its sides containing the positive terminals or anodes
A and A' which are made of graphite, and two projections on the
bottom containing mercury ; C is the negative terminal or cathode
and S is a third anode used only for starting. The large upper
space in the bulb is the cooling chamber in which the mercury
vapor, which has been heated by the passage of electricity, is
condensed and from which it falls down into the cathode again.
The two anodes are connected to the terminals of the trans-
former TT f and the load circuit is connected between the center
point of the transformer and the cathode C in series with the
sustaining coil F.
The operation of the rectifier depends on the fact that current
can pass through the tube in one direction only, from either of
the anodes to the cathode, and it can only pass in this direction
after an arc has been formed at the cathode in such a direction
as to make the mercury negative. In the opposite direction the
tube is a very good insulator and a difference of potential of
thousands of volts would be required to produce a current. The
starting arc is produced by impressing a voltage between the two
mercury terminals C and S through the starting resistance r s
and tipping the tube until the mercury forms a bridge and closes
the circuit. Current then passes and when the circuit is opened
by raising the tube to the vertical position an arc is formed and
the cathode is said to be excited. If at this instant either of the
anodes is at a higher potential than the cathode, current will
flow from it and will continue to flow so long as the difference of
potential is greater than the counter e.m.f . of the rectifier. When
the terminal T of the supply transformer is positive, current
flows from it to A through the tube to C, through the sustaining
coil F and load circuit to the terminal 0. When the voltage
reverses and T becomes negative, T' becomes positive and cur-
rent flows from it to A ' through the tube to C and through the
coil F to the terminal 0.
CONVERTERS
441
If there were no drop of voltage in the rectifier the current wave
would be the same shape as the voltage wave and in the load cir-
cuit it would vary from zero to a positive maximum. There is,
however, a drop of voltage of from 14 volts to 25 volts in the com-
mercial rectifier, which remains approximately constant inde-
pendent of the load, and if the sustaining coil were left out of
the circuit the current through the bulb would drop to zero as
soon as the potential of T had fallen below the counter e.m.f. of
the rectifier and load circuit and would remain at zero until the
potential of T r rose to a value greater than this counter e.m.f.
In the meantime the cathode would have lost its excitation and
r
n
n
V,
L
/
\
L
\
L
i
\
y
\
f
\
r
\
f
\
^
^
"^
^
"^
^-^
^x
Impreesed
Voltage
Current through
Electrode A
Current through
- Electrode A'
Rectified Current
in the Load Circuit
o
Direction
of Current
(a) F (6)
FIG. 424. Single-phase mercury vapor rectifier.
FIG. 425. Three-phase
mercury vapor rectifier.
the tube would have to be tipped again before any current could
flow. To prevent the current in the tube from falling to zero
and so to insure continuous operation the reactance coil F is
introduced. Its action is as follows: While current is flowing
from T energy is stored in the magnetic field of the coil and when
the potential of T becomes too low to force the current against
the counter e.m.f. of the converter the coil discharges its stored
energy and maintains the current until the potential of T r rises
and current flows from it to the load. The effect of the sustain-
ing coil is therefore to spread out the two halves of the current
wave so that they overlap and produce in the load circuit a direct
current with only a slight pulsation (Fig. 424(6)).
371. Currents and Voltages. The voltage is controlled by a
regulating reactance connected in the alternating-current supply
circuit and in the ordinary rectifiers the direct voltage ranges
from 20 to 52 per cent, of the alternating voltage while the alter-
nating current ranges from 40 to 66 per cent, of tfte direct current.
Rectifiers are designed for direct currents of 10, 20, 30 or 40
442
ELECTRICAL ENGINEERING
amp. and can be built to operate on any required voltage and
any frequency.
372. Losses and Efficiency. Since the counter e.m.f. of the
rectifier is approximately constant independent of the load, the
power losses vary directly as the current instead of as the square
of the current. Neglecting the losses in the sustaining coil and
regulator the efficiency of the rectifier is constant at all loads and
is higher the higher the voltage. Values up to 80 per cent, are
reached with rectifiers supplied from a 220-volt alternating-cur-
rent circuit and delivering 110 volts direct current.
The power factor of the rectifier is high and under ordinary
conditions may be assumed to be about 90 per cent.
373. Three-phase Rectifier. The three-phase rectifier, Fig.
425, does not require any sustaining coil since the voltage waves
of the three phases overlap and there is therefore no tendency for
the load current to fall to zero and allow the cathode to lose its
excitation. It must, however, be started in the same way as the
single-phase rectifier.
FIG. 426. Hot-cathode half-wave rectifier.
FIG. 427. Hot-cathode full-
wave rectifier.
374. Hot-cathode Argon-filled Rectifier. Another type of
rectifier suitable for low-voltage large-current circuits has been
developed recently, which depends on the principle that current
can flow from a cold to a hot electrode but not in the reverse
CONVERTERS 443
direction (Fig. 426). The rectifier consists of a glass bulb filled
with argon gas with a graphite anode mounted on a heavy
tungsten lead and two tungsten cathodes. One cathode A is
used for starting and the other B for operating. Cathode A is
similar to the filament of a tungsten lamp and is excited from a
40-watt transformer which raises it to a very high temperature
and heats the operating cathode B. If the bulb is connected in
series with a battery or other load across an alternating-current
supply only one-half of each alternating wave can pass. If
required, two half-wave rectifiers may be combined in one, Fig.
427, and both halves of the alternating wave passed through to
the battery. Fig. 426(6) and Fig. 427(6) show the methods of
connecting the half-wave and full-wave rectifier in the circuit.
The current may be regulated by a resistance in series as shown
or by means of an auto-transformer in the supply circuit.
In rectifiers for very small currents it is necessary to keep the
exciting cathode in operation at all times but with larger currents
it is only required for starting. When the exciting cathode is
used continuously the loss in it and in the supply transformer is
offset by a low energy loss in the arc between the electrodes. The
arc drop in this case is from 4 to 8 volts while without the exciting
electrode it is 10 to 14 volts.
The rectifier will start on voltages as low as 20 volts, and it
is used principally for charging small storage batteries.
CHAPTER XII
INDUCTION MOTOR
375. Induction Motor. Fig. 428 shows an induction motor
of the squirrel-cage type. It consists of two main parts, the
primary or stator and the secondary or rotor. The stator is
exactly similar to the armature of a synchronous motor. The
rotor which takes the place of the rotating field member of the
synchronous motor is not excited by direct current but has cur-
rents induced in it by transformer action from the stator; thus
FIG. 428. Induction motor.
the transfer of power from the stator to the rotor is similar to
the transfer of power from the primary to the secondary in a
transformer. The rotor is, however, free to move and there is an
air gap in the magnetic circuit and therefore the magnetizing
current is large, the leakage reactances are large and the power
factor is low.
376. The Stator. The primary or stator consists of a winding
carried in slots on the inner face of a laminated iron core. The
444
INDUCTION MOTOR
445
winding is similar to an alternator or synchronous motor winding
and the coils are connected in groups according to the number of
phases and poles, one group per phase per pair of poles.
The stator is supplied with polyphase alternating currents and
a revolving m.m.f. is produced similar to the m.m.f. of armature
reaction in an alternator, which produces a magnetic field re-
volving at a constant speed called the synchronous speed of the
motor.
In Fig. 429, (a) represents the stator winding of a two-pole,
two-phase induction motor, (6) the currents supplied to the two
Phase 1, z,=I m Co80 ,
Current / Phase 2 2=I m sin
(2) (3) (4)
FIG. 429. Revolving m.m.f. and flux in a two-pole, two-phase induction motor.
phases and (c) the fluxes produced by the resultant stator m.m.f.
at the points (1), (2), (3) and (4) of the cycle.
The windings start at Si and s z and finish at /i and / 2 respec-
tively. A positive current is one which enters at Si or s 2 and a
negative current is one which enters at f\ or / 2 .
Referring to Fig. 429 (c) it is seen that the north pole makes one
complete revolution in the anti-clockwise direction while the
current in phase 1 passes through one cycle.
Fig. 430 (a) represents the stator of a two-pole -three-phase
induction motor, 430(6) the currents supplied and 430(c) the
fluxes corresponding to the points (1), (2), (3) and (4) on the
cycle. The north pole as before makes one revolution during
one cycle.
446
ELECTRICAL ENGINEERING
Fig. 431 (a) represents the stator of a four-pole, two-phase
induction motor, (&) the currents supplied and (c) the fluxes
produced.
Phases, Z 3 =I m Cos (0-340)
Cos (0-120)
FIG. 430. Revolving m.m.f. and flux in a two-pole, three-phase induction motor.
(2) (3) (4)
CO
FIG. 431. Revolving m.m.f. and flux in a four-pole, two-phase induction motor.
Fig. 432 (a), (6) and (c) represent a similar set of conditions
for a four-pole, three-phase stator.
In Fig. 431 (c) and Fig. 432(c) it is seen that, while the current
INDUCTION MOTOR
447
goes through one cycle, the revolving field turns through the angle
occupied by one pair of poles.
If the stator winding has p-poles, the revolving field turns
SfiO 2
through T~ degrees, that is, through - of one revolution during
one cycle of the current.
If the frequency of the supply is / cycles per second the revolv-
2f 120f
ing field makes r.p.s. or r.p.m. The synchronous speed of
an induction motor is, therefore,
AT 120 /
N = - - r.p.m.
P
(393)
FIG. 432. Revolving m.m.f. and flux in a four-pole, three-phase induction
motor.
377. Revolving Magnetomotive Force and Flux of the Stator.
In Fig. 433, OX is the direction of the m.m.f. of phase 1 of the
two-phase motor in Fig. 429 and its value at any instant is
mi = nj m sin (0 + 90) = nj m cos 6,
^ where HI is the number of turns per phase and ii = I m sin (0 +
90) is the current in phase 1.
At the same instant the m.m.f. of phase 2 is
m 2 = nil m sin 0, in direction OF,
where
*2 = I m sin is the current in phase 2.
448
ELECTRICAL ENGINEERING
The resultant m.m.f. of the two phases is
m =
+ m 2 2 = nj m Vcos 2 + sin 2 =
and makes an angle 6 with the OX axis.
The resultant m.m.f. is, therefore, constant in value, being
equal to the maximum m.m.f. of one phase, and it revolves at
synchronous speed in the anti-clockwise direction.
FIG. 433.
FIG. 434.
This constant m.mj. acting on a path of constant reluctance
produces a field of constant strength revolving with the m.m.f.
and, therefore, revolving at synchronous speed relative to the
winding of the stator. The flux linking with each phase of the
stator is an alternating flux which reaches its maximum when
the current in the phase is maximum and is therefore in phase
with it.
Figs. 430 (a), (b) and (c) represent respectively the winding of a
three-phase, two-pole stator, the currents supplied and the
m.m.fs. produced. Fig. 434 shows the m.m.fs. of the three
phases as vectors.
The currents are:
^'i = I m cos 6, in phase 1,
i? = I m cos (0 120), in phase 2,
& 3 = I m cos (0 - 240), in phase 3.
INDUCTION MOTOR 449
The m.m.f. of phase 1 is nj. m cos 6 in direction OA.
The m.m.f. of phase 2 is nj. m cos (6 120) in direction OB.
The m.m.f. of phase 3 is nil m cos (0 240) in direction OC.
The resultant m.m.f. in the horizontal direction is
m x = nj m cos B + nj m cos (0 - 120) cos 120 + nj m cos (0 - 240)
cos 240
= nj m \ cos ~(cos cos 120 + sin sin 120)
I a
- (cos cos 240 + sin sin 240)
= ni TO cos = 2 nim cos
The resultant m.m.f. in the vertical direction is
m Y = nj m {cos (0 - 120) sin 120 + cos (0 - 240) sin 240}
= nj m \ ^r-(cos cos 120 + sin sin 120) -^TT ( cos e cos 24
( 4
+ sin sin 240)
Q
sin = ^ w t 7 TO sin 0.
The resultant m.m.f. of the three phases is
m = vW 2 + m y 2 = %wil V cos 2 + sin 2 = %nil m
and makes an angle with the OX axis.
The resultant m.m.f. is, therefore, constant in value being
equal to % times the maximum m.m.f. of one phase and it re-
volves at synchronous speed.
This constant m.m.f. produces a field of constant strength
revolving at synchronous speed. The revolving field links suc-
cessively with the windings and generates e.m.fs. in them. The
flux linking with any phase is maximum when the current in that
phase is maximum and, therefore, the flux and current are in
phase.
If a four-pole stator, Fig. 431, had been chosen instead of the
two-pole stator, the m.m.fs. of the two-phase windings would
have been combined at 45 degrees instead of 90 degrees and the
resultant m.m.f. and flux would not remain constant but would
pulsate four times during each revolution. The flux threading
any phase would, however, still vary according to a sine law and
would be in phase with the current in that phase,
29
450
ELECTRICAL ENGINEERING
To reverse the direction of rotation of a two-phase induction
motor, it is necessary to reverse one phase only.
To reverse a three-phase motor any two leads may be inter-
changed.
378. The Rotor. The secondary or rotor is made in two
forms : (a) the wound rotor, and (6) the squirrel-cage rotor. The
wound rotor consists of a laminated iron core with slots carrying
the winding, which must have the same number of poles as the
stator winding but may have a different number of phases. It is
usually wound for three phases and the ends of the windings are
brought out to slip rings so that resistances may be inserted in the
windings for starting and the terminals short circuited under
running conditions.
OD
irjril I
FIG. 435. Squirrel-cage rotor.
The squirrel-cage rotor winding consists of a number of heavy
copper bars short-circuited at the two ends by two heavy brass
rings, Fig. 435. The construction is very rugged and there is
nothing to get out of order.
When the rotor with its closed windings is placed in the revolv-
ing magnetic field produced by the stator currents, the flux cuts
across the conductors on the rotor and generates e.m.fs. in them.
Currents flow in the rotor equal to the e.m.fs. divided by the rotor
impedances. These currents reacting on the magnetic field
produce torque and the rotor revolves in the direction of the field.
At no load the rotor runs almost as fast as the field and very small
e.m.fs. and currents are induced in its conductors. When the
motor is loaded the rotor lags behind the field in speed and large
currents are induced to give the required torque.
379. Slip. The difference between the synchronous speed or
speed of the stator field and the speed of the rotor is called the
slip and is expressed as a per cent, of synchronous speed.
INDUCTION MOTOR 451
A six-pole, 60-cycle motor has a synchronous speed
If the speed of the rotor at full load is 1,176 r.p.m., the slip is
X 10 P er cent - = 2 per cent.
1,JUU
The slip at full load varies from 2 to 5 per cent, in motors designed
for constant speed.
The rotor speed may be expressed as
S = (I - s) N r.p.m. (394)
380. Magnetomotive Force of the Rotor. The frequency of the
e.m.fs. and currents induced in the rotor windings at a slip s is sf
if / is the frequency of the e.m.fs. impressed on the stator.
The polyphase currents in the rotor windings produce a re-
sultant m.m.f. revolving relative to the rotor at a speed - =
sN r.p.m. But the rotor itself is revolving at a speed S =
(1 s) N r.p.m. and therefore the rotor m.m.f. is revolving
at a speed sN + (1 s) N = N, that is, at the same speed as
the stator m.m.f. The two m.m.fs. are therefore stationary rela-
tive to each other and they are nearly opposite in phase as in the
transformer.
381. Electromotive Force and Flux Diagram for the Induction
Motor.
Let TI = stator resistance per phase.
LI = stator self -inductance per phase.
Xi = 2-jrfLi = stator reactance per phase.
Zi = -\Ai 2 + #i 2 = stator impedance per phase.
r 2 = rotor resistance per phase.
L 2 = rotor self-inductance per phase.
. x 2 = 27r/L 2 = rotor reactance per phase at standstill.
sxz = 27rs/L 2 = rotor reactance per phase at slip s.
Z'z = v^2 2 + #2 2 = rotor impedance per phase at standstill.
Z 2 = \/7*2 2 + 2 #2 2 = rotor impedance per phase at slip s.
In Fig. 436
Ei = the e.m.f. impressed on one phase of the stator.
1 1 = the current in one phase of the stator. /i = I M + I'-
I M = the magnetizing current in one phase of the stator.
I' = the load current in one phase of the stator.
452
ELECTRICAL ENGINEERING
the current in one phase of the rotor; 7 2 = I', where
the turns per phase per pair of poles on the stator,
the turns per phase per pair of poles on the rotor,
the flux per pole linking one phase of the stator which
would be produced by the current Ii acting alone.
the part of $1 which crosses the gap and links with one
phase of the rotor. $i g = v&\, where vi is a constant.
FIG. 436. Currents, fluxes and e.m.fs. in an induction motor.
>i L = the part of $1 which does not cross the gap, but is the
leakage flux of self -inductance of the stator.
2 = the flux per pole linking one phase of the rotor, which
would be produced by the current 1 2 acting alone.
$2 = $20 + 3>2L-
> 2 <7 = the part of $2 which crosses the gap and links with one
phase of the stator. ^> 2ff = ^2^2, where v z is a constant.
> 2 L = the part of $2 which does not cross the gap, but is the
leakage flux of self -inductance of the rotor.
$ a = the actual flux per pole crossing the gap and linking
with one phase of both stator and rotor ; it is the result-
ant of 3>i a and 20.
$, = actual flux per pole linking one phase of the stator it ; is
the resultant of < and $IL-
INDUCTION MOTOR
453
E
E 8
3> r = the actual flux per pole linking one phase of the rotor;
it is the resultant of 3> ff and $ 2L .
Ei g = the back e.m.f. generated in one phase of the stator by
the flux 3> ff .
= the e.m.f. of self-inductance generated in one phase of
the stator by the leakage flux $11,. EIL = Ii%i>
= the back e.m.f. generated in one phase of the stator by
the flux $ 8 . E s is the resultant of Ei g and EIL.
EI = the e.m.f. impressed on one phase of the stator. It
must be exactly equal and opposite to E 3 if the stator e-
sistance drop / \r i is neglected. This drop is of the order
of 2 per cent, at full load.
Since EI is constant, E s and $ s must be constant.
E 2g = the e.m.f. generated in one phase of the rotor by the
flux <1> . E 2a = sEig = sE 2 . where E 2 is the e.m.f.
HI
which would be generated in one phase of the rotor by
the flux at standstill.
FIG. 437.
E 2L = the e.m.f. of self-inductance generated in one phase of
the rotor by the leakage flux $ 2L . E 2L = sI 2 x 2 .
E r = the e.m.f. generated in one phase of the rotor by the
flux r . E r is the difference between E 2g and E 2L and is
equal to 1 2 r 2 . It is in phase with the rotor current I 2
and lags 90 degrees behind the flux 3> r .
As the induction motor is loaded the end d of the vector $1 fol-
lows a circle passing through / and having its center on of pro-
duced (Fig. 437).
454
ELECTRICAL ENGINEERING
382. Proof that the Locus is a Circle. From d draw dk at
right angles to fd to cut of produced in k. Then the semicircle
fdk is the locus of d.
In the triangles aob and fdk
therefore
and
Z oab = Z fdk, being right angles,
< oba = Z dfk,
fk_fd_ fd fd
ob ab ac cb oh cb
ob X
Vi-of X v*- oh
o/i c6 0/1
_ .
1
a constant
since of is constant.
Therefore the locus of d is a circle described on the diameter fk.
383. Magnetomotive Force Diagram. Since the magnetic
circuit of the machine is not saturated by the fluxes $ 8 and r , the
flux diagram, Fig. 437, may be replaced by the m.m.f . diagram,
Fig. 438.
total m.m.f. of the stator per phase.
nil' = m.m.f. of the load component of stator current.
n-J-M = m.m.f. of the magnetizing current.
n 2 J 2 = m.m.f. of the rotor per phase ; it is equal and opposite
to nj f .
FIG. 438. M.m.f. diagram.
FIG. 439. Stator current diagram.
384. Stator Current Diagram. The m.m.f. diagram, Fig.
438, may be replaced by the stator current diagram, Fig. 439.
J x = total current t in one phase of stator. I\ = I M + /'
I M = magnetizing current in one phase of stator.
I' = load component of current in one phase of stator.
7 2 = current in one phase of rotor. 7 2
INDUCTION MOTOR
455
fd = /' = J 2 = load component of stator current per phase
and represents the rotor current per phase.
To obtain an expression for the diameter of the circle, consider
the special case where the motor is at standstill and the rotor and
stator resistances are both zero.
/
FIQ. 440. Induction motor at standstill.
Referring to Fig. 440
E 1 = E 8 = E lg
and
therefore,
rl ^2;
#1
- I'x,
if the magnetizing current I M is neglected in comparison with the
load component of stator current /'.
This value of I' is represented by fk and is the diameter of the
circle. Its value is
- (395)
The diameter of the circle, therefore, varies directly as the
impressed e.m.f. and inversely as the sum of the reactances and
therefore as the frequency of the supply.
Thus as the motor is loaded the end d of the vector representing
the stator current follows a semicircle of diameter
ft
456
ELECTRICAL ENGINEERING
385. Rotor Electromotive Force and Current. The e.m.f.
generated in one phase of the rotor at slip s by the flux 3> ff is
E 2g = sK$ g , where K is a constant,
= sEz, where E z = K$ g is the e.m.f. which would
be generated in the rotor at standstill by the flux ff . Ez does
not remain constant as the motor, is loaded, since the flux < a does
not remain constant but decreases about 30 per cent, from no load
to standstill when the rotor is locked and slip is unity. Up to
full load & g and E z may be considered to remain constant.
The impedance of the rotor at slip s is
Z 2 - Vr 2 2 4- s 2 z 2 2 (369)
The rotor current is
T Oi^ 2
= ~Z~
The rotor power factor is
cos 62 =
E!
(397)
(398)
FIG. 441.
386. Rotor Input. The power transferred from the stator to
the rotor per phase is the product of the back voltage E', gener-
ated in the stator by the flux $ g , the load component of the
stator current I' and the cosine of the angle between them ; it is
p r = ET cos 02 (see Fig. 441) (399)
but E' = E = E-
n 2
and T/ ^2 r .
INDUCTION MOTOR 457
therefore,
p r = ET cos 02 = E 2 h cos 2 = #2/2 cos <9 2 . (400)
n 2 HI
Thus the power input to the rotor per phase is the product of the
e.m.f. which would be generated in the rotor by the flux <$> ff at
standstill and the power component of the rotor current.
The total power input to the n phases of the rotor is
P r = nE 2 I 2 cos 6 2 (401)
387. Rotor Copper Loss and Slip. The power consumed by
the rotor copper loss per phase is
/ 2 2 r 2 = sE 2 I 2 cos 6 2 , (402)
and for the n phases it is
L r = n/ 2 V 2 = nsE 2 I 2 cos 2 . (403)
01 . ,. rotor copper loss L r snE 2 I 2 cosB 2 ,._..
Slip = the ratio - P- - = ~ = pr^ -~ = s. (404)
rotor input P T nE 2 I 2 cos 02
388. Rotor Output. The rotor output per phase is
p = p r I 2 2 r 2 = E 2 I 2 cos 02 sE 2 I 2 cos 2
= (I - s) # 2 /2 cos 2 (405)
and the total rotor output is
p = np = n(l - s) #2/2 cos 2 watts (406)
* (1 " S \ff 2 C S * 2 h.p. (407)
From equations (401) and (406) the rotor output is
P- (l-s)P r =^P r
and it is equal to the rotor input multiplied by the rotor speed in
per cent, of synchronous speed.
389. Torque. If T is the torque in Ib.-ft. the output may be
expressed as
2irST
~ 33,000 P '
Thus the torque is
, _ n (1 - s) E 2 I 2 cos 2 33,000 _ _ ^ nE^Ii cos 2
' '
7.05 = 7.05 lb .. ft . (408)
N sync, speed
458
ELECTRICAL ENGINEERING
The torque of an induction motor is usually expressed not in
pounds-feet but in synchronous watts, that is, in terms of the
power which would be developed at synchronous speed.
The torque in synchronous watts is
sync, watts
= PX
N
S
(409)
and it is equal to the power input to the rotor.
390. Rotor Efficiency. Neglecting all losses except the rotor
copper loss the rotor efficiency is
.,
77 r = =- 100 per cent.
P r
n(l s) E 2 h cos 2
cos
100 per cent.
S
= (1 s) 100 per cent. = - 100 per cent.
(410)
that is, the rotor efficiency is equal to the rotor speed in per cent,
of synchronous speed and, therefore, the efficiency of an induction
motor is always less than the speed in per cent, of synchronous
speed.
391. Modification of Diagram. When an induction motor is
running without load, a current 7 flows in each phase of the
stator which has two components, Fig. 442, I M the magnetizing
current 90 degrees behind the impressed e.m.f. E\, and I P the
power component in phase with EI.
Ip ^,
_ a r
FIG. 442. FIG. 443. Circle diagram of an induction motor.
The product nEJ P (where n is the number of phases) is the
power required to supply the no-load losses. These are the iron
loss and a small copper loss in the stator and the friction and
windage losses of the rotor. The iron loss in the rotor may be
neglected since the rotor frequency is low.
INDUCTION MOTOR 459
The current required to supply the stator losses has no cor-
responding component in the rotor, but the power to overcome
the friction and windage losses must be transferred from the
stator to the rotor and therefore requires a current in
the rotor.
As the motor is loaded and slows down the stator iron loss re-
mains nearly constant, the friction and windage losses decrease
and the rotor iron loss increases. At standstill the friction and
windage losses are absent but the rotor iron loss is large since the
rotor frequency is the same as the stator frequency. The iron
friction and windage losses are therefore considered to remain
constant and the small component of rotor current required to
supply the friction and windage losses is neglected.
The diagram, Fig. 439, must therefore be changed to Fig. 443
by the addition of I P the power component of the stator current
per phase at no load.
The diameter of the circle is raised through the distance af =
IP and of now represents not the magnetizing current I M but the
no-load current /o = V^M 2 + Zp 2 -
If os represents the stator current per phase at standstill and
sw is its power component, then, since there is no output, the
power input is consumed by the losses. Therefore, input =
losses = nEiSw; constant losses = nEJ. P = nEiVW and copper
losses = nEiSV = n(I" i r l +. 7 2 2 r 2 ).
The stator copper loss is taken as nl' 2 ri because the stator cop-
per loss at no load is included in the constant losses. It is there-
fore assumed that /iVi = IO*TI + I /2 n + which is approximately
correct up to full load.
Divide sv at t so that st : tv = 7 2 2 r 2 : 7'Vi, then, stator-load
copper loss = nEitv and rotor copper loss = nEist.
Join ft and from d any point on the circle to the left of s draw
dmpqr perpendicular to the diameter fk. It is to be shown
that mp is the stator current required to supply the rotor copper
loss for a rotor current represented by fd and that pq is the cur-
rent to supply the corresponding stator load copper loss.
The rotor copper loss is
n/ 2 V 2 = n( I'Yri = KI' 2 = Kfd = K(fa + qd)
W2 '
= K X fq(fq + qk) = K X fq X D = fq X a constant,
since K and D are both constants.
460 ELECTRICAL ENGINEERING
Therefore the rotor copper loss is proportional to fq ; but
mp _ st mp st
and since st represents the rotor copper loss for a current fs, mp
represents the rotor copper loss for a current fd.
Similarly pq represents the stator copper loss for stator cur-
rent fd.
392. Interpretation of Diagram. At any value of stator cur-
rent od = /i, Fig. 443.
nEi dr = stator input in watts,
nEi qr = constant losses,
nEipq = stator copper loss,
nEi mp = rotor copper loss,
nEi dm = rotor output in watts = mechanical load,
dr ,
-, = power factor,
dm _ .
-j- = efficiency,
mp _ rotor copper loss _ ,.
dp rotor input
dm _ rotor output _ actual speed _ J? __ -i
dp rotor input synchronous speed N
The torque corresponding to output nEi dm is
-, nEidm 33,000 , ,
T = . X n , - ^ lb. at 1 ft. radius.
746 2ir (r.p.m.)
At synchronous speed this torque would represent an output
N dr)
nEi dm X TT = nEi dm X 3*- = nE\ dp watts = rotor input,
o dm
The torque in synchronous watts is equal to the watts input to
the rotor = nEidp.
At standstill the torque in synchronous watts is nEiSt and
represents the starting torque of the motor.
The maximum value of torque in synchronous watts is nEi X
maximum value of dp.
The maximum output in watts is nE\ X maximum value of dm.
For average 25-cycle motors, starting torque is one and one-half
to two and one-half times full-load torque ; starting current is six
to eight times full-load current ; and maximum running torque is
INDUCTION MOTOR 461
two and one-half to three and one-half times full-load torque.
For 60-cycle motors, starting torque is one to one and one-half
times full-load torque ; starting current is five to six times full-load
current ; and maximum running torque is. two to two and one-half
times full-load torque.
393. Construction of Diagram from Test for a Three-phase
Motor. 1. Run the motor light at rated voltage and rated fre-
quency. Read impressed voltage, current and watts input
Ei, IQ and W Q .
J /o = of on the diagram.
V 3E
= IP = af = wv on the diagram.
2. Lock the rotor and impress reduced voltage and raise it until
twice full-load current flows in the stator. Read impressed vol-
tage, current, watts input and torque, EL, IL, W L and T L . To
get the value of locked current at rated voltage, raise the values
of I L in the ratio EI : E L . To get the values of locked watts and
locked torque at rated voltage, raise the values of W L and T L in
the ratio Ei 2 : E L 2 . To get accurate results for the circle diagram
it is better to reduce the values of watts and torque to terms
of power current per phase. This is done by dividing the values
of W L and T L by \/3E L .
Plot on a base of impressed voltage,
These three loci should be straight lines passing through the
origin and can be produced until they cut the ordinate at the
rated voltage of the motor.
The following results are obtained.
Value of I L at rated voltage = os on the diagram.
W L
Value of ._ at rated voltage = sw on the diagram.
3. Measure the resistance of the stator per phase = r\. Then
the stator copper loss locked at rated voltage is 3 os 2 n watts
and - ,- = vt on the diagram.
The rotor copper loss, locked at rated voltage, is \/3Ei X st
and is known since st = sw wv vt.
462
ELECTRICAL ENGINEERING
The rotor copper loss also represents the starting torque in
synchronous watts and therefore if T\ is the value of T L at rated
voltage
Ti X 2ir X (sync, speed r.p.m.) \/3#i X s
33,000 - Sh uld equal ~ "
The circle diagram can be drawn in from the values obtained
above.
The motor has been assumed to be F-connected and the vol-
tage EI is the line voltage.
394. Test of an Induction Motor. The following readings
were obtained from tests on a 30-hp., 440-volt, 60-cycle, three-
phase star-connected induction motor, having eight poles and a
synchronous speed of 900 r.p.m.
Resistance of stator between terminals at normal operating
temperature = 0.50 ohms.
Resistance of stator per phase = r\ = 0.25 ohms.
RUNNING SATURATION OR NO-LOAD TEST
Impressed
e.m.f. = E
Exciting current =
Jo
Watts input = Wo
550
23.0
2,920
480
15.8
1,750
440
14.0
1,475
405
12.5
1,350
380
10.9
1,250
290
7.8
890
COCKED TEST
Impressed
e.m.f. => EL
Stator
current = IL
Watts input '
= WL
Starting
torque = TL
Ib.-ft.
. WL
TL
VSEL
VZEL
307.5
121.0
32,100
145
60.3
0.271
276.5
108.5
25,400
120
53.2
0.251
257.5
100.0
21,000
99
47.3
0.244
233.5
88.0
16,500
84
40.9
0.207
201.5
74.0
11,550
60
33.2
0.170
164.0
59.0
7,350
35
25.9
0.124
440.0
170.0
61,600
297.5
81.0
0.390
INDUCTION MOTOR
463
These values are plotted in Fig. 444 against impressed e.m.f.
and the values at 440 volts are obtained by producing the curves.
W" T 7
The loci of IL, /OET and .--, are straight lines passing through
the origin.
With line voltage E = 440, the impressed voltage per phase =
: = JU^ = 254.
V3
Amperes Exciting Current = I o
5 10 15 20 25 30 35 40
UJ
400
I
300
200
3
100
X
500 1000 1500
Watts Input at No-load = W
81 100
MO 170 Amperes
600
440
50 100 150 200 250 300 350 390
FIG. 444.
At No Load.
Stator current or exciting current = 7 = 14.0 amp. = of
(Fig. 445).
Watts input = Wo = 1,475 = iron loss -f friction and wind-
age losses and these are assumed to be constant.
Wo 1,475
Power factor = cos
= 0.137
V3EI Q V3 X 440 X 14
= 13.7 per cent.
In-phase current = I P = I cos = 1.92 amp. = fa = vw.
Magnetizing current = I M = V/o 2 Ip 2 = \ 14 2 1.92 2 =
13.8 amp. = oa.
464
ELECTRICAL ENGINEERING
Rotor Locked.
Stator current = I L = 170 amp. = os.
Watts input = W L = 61,600 = constant losses + stator cop-
per loss + rotor copper loss.
W L 61,600
Power factor = cos L
= 0.476
V3EI L V3 X 440 X 170
= 47.6 per cent.
In-phase component of stator current = I L cos 8 L = 81 amp.
= sw.
StanditlU
E,=.
FIG. 445.
Quadrature component of stator current = I L sin B L = 149
amp. = ow.
Stator copper loss = 3/Vi = 3 X 170 2 X 0.25 = 21,700
21 700
watts, represented by the current tv = 7=^ : - = 28.4
V3 X440
amp.
Rotor copper loss = 61,600 - 1,475 - 21,700 = 38,425, rep-
38,425
resented by the current st =
\/3 X440
= 50.5 amp.
The points o, /, a, s, t, v and w are therefore fixed and the circle
may be drawn in, as shown in Fig. 445.
The starting torque at 440 volts = 0.39 X V3 X 440 = 297.5
Ib.-ft. and the corresponding output at synchronous speed =
2 X 3.14 X 900 X 297.5
= 51
' = 51 X
33,000
746 = 38,046 synchronous watts = rotor input = rotor copper
loss and should agree with the value 38,425 above. This serves
as a check on the circle.
INDUCTION MOTOR 465
For full-load output of 30 hp. the vertical intercept between
the circle and the line fs must be the required in-phase current
30 X 746
= 7= - = 29.3 amp. = dm. /The full-load point d is thus
V3 X 440 I
fixed and the corresponding stator current = l\ = od = 40 amp.,
power factor = cos 0i = --, = 0.8625 = 86.25 per cent., effi-
j od j
ciency rj = -? 100 per cent. = 84.3 per cent., speed = S = -7 X
sync, speed = 0.9275 X 900 = 835 r.p.m., and the slip = s =
^ = 0.0725 = 7.25 per cent.
dp
At one-half load, I\ 23 amp., cos 0i = 75 per cent., 17 = 84.0
per cent., S 870 r.p.m., s = 3.3 per cent.; at one and one-half
load, 7i = 65 amp., cos 0i = 85.5 per cent., 77 = 77.5 per cent.,
S = 779 r.p.m., s = 13.4 per cent.
Maximum torque or pull-out torque is represented by the maxi-
mum intercept between the circle and the line/ = 73.6 amp. =
73.6 X \/3 X 440 =~5fi,000 "sync. ; watts~^l407o Ib.-ft. The
slip for maximum torque is 36.4 per cent.
Maximum output or stalling load is represented by the maxi-
mum intercept between the circle and the line/s = 57.6 amp. =
57.6 X V3 X 440 = 44,000 watts = 58.9 hp. and the corre-
sponding slip is 24.4 per cent.
395. Analysis by Rectangular Coordinates. Using the method
of rectangular coordinates called the symbolic method the per-
formance characteristics of an induction motor can be determined
if the constants, of the motor are known.
Let
Y = 9 ~ fo = stator exciting admittance per phase at rated
voltage.
Z\ = 7*1 + jxi = stator impedance per phase.
?z = r 2 + jsx 2 = rotor impedance per phase at slip s.
Assume that the ratio of turns is HI : n 2 = 1 : 1 and take as real
axis of coordinates the e.m.f. generated in the stator by the flux
of mutual inductance. The quantities used refer to one phase
of the stator and the corresponding phase of the rotor.
E' = e.m.f. generated in the stator.
Ez = E f = e.m.f. generated in the rotor at standstill.
sE 2 = e.m.f. generated in the rotor at slip s.
These three e.m.fs. lie along the real axis.
30
466 ELECTRICAL ENGINEERING
The following equations show the relations between the various
e.m.fs. and currents at slip s.
Rotor current:
where
Stator load current :
/' = j 2 = E '(a l -ja 2 ). (413)
Stator exciting current:
7 = E'Y = E' (g - jb.) (414)
Total stator current:
/i = /'+/ = E'{ (a, + g)-j (a, + b) } = E'(b, - jb 2 ) (415)
where 61 = 0,1 + Q and 6 2 = a 2 + & (416)
E.m.f. impressed on the stator:
#i = J' + /!?! = E' + E' (b, - J6 2 )(n + jxi)
= #'{(1 + bin + 6ja?i) + j (6ix, - 6 2 rO = E' ( Cl + jc 2 )
(417)
where Ci = 1 + 6iri + b 2 Xi and c 2 = 61X1 b z ri (418)
The constant absolute value of the impressed e.m.f. is
E l = E' Vc7T~^. (419)
Thus the e.m.f. generated in the stator is
W = pJL=. (420)
Vci 2 + c 2 2
Substituting this value for E' in the equations above the abso-
lute values of the various quantities as slip s can be obtained.
Rotor current:
/ 2 = /' = E' aS + a 2 2 = E, - (421)
Vci 2 + c 2 2
Exciting current:
(422)
+ c 2
Stator current:
-. (423)
INDUCTION MOTOR 467
The torque in synchronous watts, was found in Art. 389 to be
equal to the rotor input in watts, which is
P r = nE 2 h cos 02 = n#'X#'ai = n#' 2 ai = n#i 2 2 ^ 2 (424)
Ci T 2
where n is the number of phases.
The torque in pounds-feet is from Art. 389.
r = 7 .05 = 7 - 05 ^^-, (425)
N N ci 2 + c 2 2
The rotor output is
p = (441)
The slip corresponding to maximum output may be found by
differentiating P with respect to s and equating to zero ; the oper-
ation is rather long and the result only will be given here :
Smax. output = / == == ^' (442)
TY+ V (ri + r 2 ) 2 + (xi + z 2 ) 2
Substituting this value of s in equation (435) the maximum
output or stalling load is found :
-- =-. (443)
r a ) + V(ri + r a )+(si + * 2 )}
These equations are quite accurate except when the exciting
current is very large. When calculating the stator power factor
and the stator current up to full load the exciting current cannot
be neglected.
From equations (437) and (443) it is seen that both the maxi-
mum torque and the maximum output are proportional to the
square of the impressed e.m.f. and therefore any decrease in the
line e.m.f. will seriously affect the motor characteristics.
396. Characteristics of an Induction Motor by the Symbolic
Method. Before applying this method the constants g, b, TI, r z ,
Xi and # 2 must be determined. Taking as an example the motor
of Art. 394, the constants may be found as follows:
440
Impressed e.m.f. per phase = E\ -7= = 254 volts.
v 3
Stator resistance per phase = r x = 0.25 ohms.
Stator current per phase at no load = 7 = 14 amp.
Watts input at no load = W = 1,475.
Power factor at no load = cos = 7=- ~ = -7
V3#/o V 3 X 440 X 14
= 0.137 = 13.7 per cent.
Stator exciting admittance per phase = Y = ^- = oc7 = 55.2
X 10- 3 .
Stator exciting conductance per phase = g = Y cos 6 = 7.6
X 10- 3 .
Stator exciting susceptance per phase = b = V~Y 2 g 2 =
54.0 X 10~ 3 .
INDUCTION MOTOR 471
Stator current per phase locked = I L = 170 amp.
Watts input locked = W L = 61,600.
i i A W L 61,600
Power factor locked - COB fc -^- = V x 44Q x 17Q =
0.476 = 47.6 per cent.
T7
Equivalent impedance per phase locked = Z L = -y- 1 =
1.49 ohms.
Equivalent resistance per phase locked = R L = Z L cos 6 L =
0.71 ohms.
Equivalent reactance per phase locked = X L = \/Z L 2 R L 2 =
1.31 ohms.
Rotor resistance per phase referred to the stator = r 2 = R L
ri = 0.71 - 0.25 = 0.46 ohms.
Assuming that the rotor reactance per phase referred to the
stator is equal to the stator reactance per phase :
Xi = x 2 = -^ - = ~V~ = 0.655 ohms.
Z Jj
The characteristic curves may be obtained by substituting any
values of slip in the equations in Art. 395 using the constants
found above.
It will be sufficient to check the results of this method at one
or two values of slip against the results obtained from the circle
diagram.
When
s = 0.05:
01 = -Tr-a I* = - 095 > * = ~2-n 2 = 0.00764,
r 2 2 + s 2 x z 2 r z 2 + s 2 x 2 2
bi = ai + g = 0.1026, 6 2 = 2 + b = 0.0626,
Cl = 1 + & iri + 6 2 ^i = 1.066, c 2 = biX! - b 2 ri = 0.0514.
ft 2 J_ 7 2
Stator current = /i = EiJ~~ \ = 28.6 amp.
\ Ci i C 2
The check values from the circle diagram will be taken at this
value of stator current.
Power factor = cos B l = c ^ ~ ^2 = Q g25 =
82.5 per cent, (circle gives 81.8 per cent.)
Output = P = nE^ l ~^ a ^ = 15,300 watts = 20.5 hp.
(the circle diagram gives 15,225 watts = 20.4 hp.).
472 ELECTRICAL ENGINEERING
J, - a842 ' 84 ' 2 per
cent, (circle gives 85 per cent.)
Torque . r .Z^*!.-^_. 126 .5 Ib.-ft.
(circle gives 125.5).
Slip at 28.6 amp. from the circle is 4.55 per cent.
When s = 0.10:
ai = 0.212, 2 = 0.0302, bi = 0.22, 6 2 = 0.085, ci = 1.11,
c 2 = 0.123. 7i = 53.4 amp., cos 61 = 88.4 per cent., P =
29,250 watts, 77 = 81.3 per cent., T = 255 Ib.-ft.
The values obtained from the circle diagram for a stator cur-
rent 1 1 = 53.4 are cos 0i = 88.1 per cent., P = 29,000 watts,
17 = 80.9 per cent., T = 252 Ib.-ft., s =0.10.
At standstill when s = 1:
oi = 0.716, 2 = 1.02, 61 = 0.724, 6 2 = 1.075, Cl = 1.885,
c 2 = - 0.2185, /i = 168 amp. and T = 300 Ib.-ft.; the cor-
responding values from the circle diagram are I\ = 170 and
T = 297.5 Ib.-ft.
Slip for maximum torque, equation (431), is
Smax. torque = / = 0.345 = 34.5 per cent.
Vri 2 + (xi + Z 2 ) 2
Maximum torque, equation (437), is
=
max .
N 2[n
From the circle diagram, !F max . = 440 Ib.-ft. and the corre-
sponding slip = 36.4 per cent.
Slip for maximum output, equation (442), is
s max. output = - ,- ===== = 23.6 per cent.
Maximum output, equation (443), is
Pmax. = r J 1 l = =; = 44,000 watts
= 58.9 hp.
From the circle diagram, P max = 44,000 watts = 58.9 hp. and
the corresponding slip is 24.4 per cent.
From the calculations above it is seen that the two methods
check very satisfactorily.
INDUCTION MOTOR
473
397. Methods of Starting. Except in the case of small ma-
chines, induction motors should not be started by connecting
them directly to the mains, since the large starting current at
low power factor disturbs the voltage regulation of the system.
Two methods of reducing the starting current are in use. (1)
The voltage impressed on the stator is reduced by using an in-
duction starter which is simply an auto-transformer with one or
more taps (Fig. 374). (2) Resistance is inserted in series with
the rotor windings.
1. When the impressed voltage is reduced, the starting current
is reduced in proportion to it, but the starting torque is reduced
as the square of the voltage.
FIG. 446. Starting on reduced FIG. 447. Starting torque with various rotor
voltage. resistances.
In Fig. 446 E\.s\t\ represents the starting torque of the motor
at full voltage and osi represents the starting current, neglecting
pi
the exciting current, and -^- s 2 2 represents the starting torque at
&
half voltage and os 2 represents the starting current.
Since os 2 = ~- the starting current is reduced to one-half its
value at full voltage, but the starting torque is reduced to one-
quarter. The power factor is not changed.
Thus starting with reduced voltage gives very small starting
torque and low power factor.
A squirrel-cage rotor may be used.
2. When resistance is inserted in the rotor windings the start-
ing current is reduced and is brought more nearly in phase and
the starting torque is increased.
In Fig. 447 s 2o represents the starting torque wheD the rotor
circuits are closed without any starting resistance.
s it i is the starting torque when resistance Ri is inserted.
474
ELECTRICAL ENGINEERING
s 2 tz is the starting torque when resistance R 2 >Ri is inserted.
s 3 Z 3 is the maximum possible starting torque and is obtained
by inserting a resistance R S >R 2 ; it is the same as the maximum
running torque of the motor. R s -}- r% = A/ri 2 + (#1 + ^2) 2 >
equation (441). os , osi, os 2 and os 3 are the corresponding stator
currents and cos 0o, cos 0i, etc., are the power factors at start.
The curves in Fig. 448 are the "speed-torque" characteristics
for the motor operating with the various resistances in the rotor.
The maximum torque is the same in all cases but it is reached at
different speeds. One current
curve holds in all cases.
If a resistance R 4 > R z is
inserted in the rotor windings
the starting current is further
reduced and the power factor
is improved but the starting
torque is decreased. R 4 may
be made of such value that
the starting torque s 4 4 is
equal to full-load torque and
the starting current oSi is
equal to full-load current.
Curve (4) is the speed-torque
characteristic for this case.
Thus by inserting resistance
in the rotor any starting torque up to the maximum running
torque or "pull-out" torque may be obtained. The starting
current is reduced and the power factor is improved.
In starting a heavy load resistance R s is used and the motor
gives its maximum torque at start. The resistance is then cut
out gradually as the speed increases and the motor operates with
short-circuited rotor with characteristics as shown in curve (0).
If the load to be started is not very great and a large starting
current at low power factor is objectionable, resistance R* is used
and the motor starts with full-load torque and draws full-load
current.
This second method of starting requires a wound rotor with
slip rings and large starting resistances which is much more
expensive than a squirrel-cage rotor.
For the same line current, resistance starting gives about four
times the torque given at reduced voltage.
25 50 75 100 125 150 175 200 225 250
Torque in Percent of Full Load
FIG. 448. Speed-torque character-
istics of an induction motor with various
rotor resistances.
INDUCTION MOTOR
475
398. Applications. The constant-speed or squirrel-cage in-
duction motor takes the place of the direct-current shunt motor
and has very similar characteristics. It is of much more simple
and rugged construction than the shunt motor and the wear
and danger due to sparking are entirely eliminated.
It should be used where fairly constant power is required for
long periods, where good speed regulation is required, where
starting is infrequent and only average starting torque is neces-
sary, where the motor is exposed to dust or to inflammable
materials or is not easily inspected. It is suitable for driving line
shafting, for high- and low-speed centrifugal pumps, blowers, fans,
etc. It must be started on reduced voltage except for the small-
est sizes.
The variable-speed induction motor has a wound rotor with its
terminals connected to slip rings so that resistance may be intro-
duced to vary the speed or to give a large starting torque.
100
90
I 80
s 70
o 60
700
GOO
50 *' 500
3 p-
40 a 400
| 30 1 300
W 20 m 200
10
100
50
100
250
350
150 200
Torque Ft. Lb
FIG. 449. Characteristic curves of a three-phase, 60-cycle, 220-volt, 20 horse-
power induction motor.
It should be used where frequent starts under load are neces-
sary, or where the motor is large enough to have a bad effect on
the regulation of the system due to the large starting current at
low power factor, as for cranes, elevators, hoists, etc.
The squirrel-cage motor with a comparatively high-resistance
rotor can be used where fairly large starting torque is required
and where a wound-rotor motor is not advisable, as in cement
mills, etc.
Fig. 449 shows the characteristic curves of a three-phase, 60-
cycle, 220-volt, 20-hp. induction motor with a squirrel-cage rotor.
476 ELECTRICAL ENGINEERING
399. Speed Control of Induction Motors. The induction
motor is inherently a constant-speed motor like the direct-current
shunt motor but its speed may be varied in four principal ways:
(1) By variation of the impressed voltage, (2) by inserting resist-
ance in the rotor windings, (3) by changing the number of poles
and (4) by cascade control or concatenation.
400. Voltage Control. Since the torque for a given slip is
proportional to the square of the impressed voltage, the speed
may be varied through a small range by variation of the voltage.
This may be accomplished (a) by introducing resistance in series
with the stator windings or (b) by using a variable ratio trans-
former or compensator.
Control by series resistance is very simple but is inefficient
while control by a compensator is more expensive but gives
higher efficiency. Both methods to be effective require a motor
having a high-resistance rotor with its resulting large slip. A
squirrel-cage rotor may be used.
401. Rotor Resistance Control. The speed of an induction
motor may be varied by using a wound rotor with slip rings and
connecting external resistances into the circuit. With suitable
resistances any required speed-torque curve may be obtained
(Art. 397). This method of control corresponds to the use
of resistance in series with the armature of a direct-current
shunt motor. It is very inefficient and the speed changes
with load.
402. Pole-changing. By the use of special windings the num-
ber of poles on an induction motor may be changed and therefore
also its synchronous speed. Two or at most three synchronous
speeds may be obtained by this means. Take for example an
eight-pole motor with a double-layer winding of two-thirds coil
pitch; its synchronous speed may be doubled by reconnecting
for four poles with a coil pitch of one and one-third. A squirrel-
cage rotor is generally used but if intermediate speeds are re-
quired a wound rotor must be used and its poles changed at the
same time as those on the stator. Resistance may then be
inserted in the rotor circuits. Such a motor is very expensive.
403. Cascade Control or Concatenation. When operating in
cascade two similar induction motors with wound rotors are
rigidly connected to the same shaft. The stator of the first
motor is connected to the line ; the stator of the second motor is
connected to the rotor winding of the first motor and receives
INDUCTION MOTOR 477
power from it; the rotor of the second motor is closed through
starting resistances (Fig. 450).
The frequency of the e.m.fs. generated in the rotor of an in-
duction motor is s/, where / is the frequency of the supply and s
is the slip. Thus the frequency impressed on the stator of the
second motor is the frequency of slip of the first motor. The
speed of the two motors is always the same and thus at no load
(1 - s )/ = s f and s = 0.5.
FIG. 450. Cascade control of induction motors.
Therefore, two similar motors connected in cascade tend to
approach a speed of half synchronous speed at no load and fall
below this speed under load. Speeds below half synchronous
speed are obtained by inserting resistance in the rotor windings
of the second motor. For speeds above half synchronous speed
the stator of the second motor must be connected to the line and
the rotor of the first motor closed through resistances.
This method of control is used for some three-phase traction
systems and is very similar to the series parallel control of direct-
current series motors. The induction motor does not, however,
tend to increase its speed indefinitely. If it operates above syn-
chronous speed, it becomes a generator and feeds back power to
the line acting as a brake.
If the two motors have different numbers of poles, the no-load
speed when connected in cascade will not be half synchronous
speed but the speed of the set will be that of a motor having as
many poles as the two motors combined. For example, if on a
60-cycle system a 12-pole motor is connected in cascade with a
4-pole motor connected to run in the same direction the speed
120 X 60
will be that of a motor with 12 + 4 = 16 poles, that is, ^
= 450 r.p.m. When the two motors tend to run in the same
direction they are in direct concatenation. If the 4-pole motor
is connected to run in the opposite direction from the 12-pole
478 ELECTRICAL ENGINEERING
motor, the resulting speed will be that corresponding to 12 4 =
8 poles. This is called differential concatenation.
With this combination of motors, four speeds may be obtained,
120 X 60
450 r.p.m. with direct concatenation, : - = 600 r.p.m.
i&
120 X 60
using only the 12-pole motor, io_4 = 900 r.p.m. with dif-
120 X 60
ferential concatenation and finally : r - = 1,800 r.p-m.,
using only the 4-pole motor. Intermediate speeds can be ob-
tained inserting resistances in the rotors.
When connected in cascade the impedance of the second motor
is added to the rotor of the first and its exciting current is supplied
through the first motor resulting in very low power factor. With
two similar motors in cascade each will operate at about half
voltage and since the output varies as the square of the voltage
the combined output of the set will be about half that of each
motor separately. The losses will be those of two motors and
therefore the efficiency will be very low. The maximum or
breakdown output will be reduced in the same proportion.
404. Exciting Current. The exciting current or no-load cur-
rent of an induction motor is larger than that of a transformer
and is of the order of 25 per cent, of full-load current. The
power component supplies the iron losses of the stator and the
friction and windage losses of the rotor. The quadrature com-
ponent or magnetizing current is large because the magnetic cir-
cuit contains an air gap and the iron on both sides is cut away
to form the slots for the rotor and stator windings. The large
magnetizing current lowers the power factor of the motor par-
ticularly at light loads and is therefore objectionable. To reduce
it the air gap is made as short as possible, being determined
largely by the necessity for mechanical clearance. The slots may
be partly closed to increase the effective gap area; the rotor slots
are often completely closed. Closing the slots, however, adds to
the reactance of the motor.
Low-speed motors have larger magnetizing currents than high-
speed motors and therefore their power factors tend to be lower. \
Unless the teeth of the stator and rotor are properly designed
the reluctance of the magnetic circuit of the motor will vary and
the flux will vary. If the stator and rotor have the same number
of teeth, then in position (6), Fig. 451, the reluctance will be a
INDUCTION MOTOR
479
maximum and the flux a minimum while in (c) the flux will be
maximum. The rotor will tend to lock in the position of maxi-
mum flux (a) and the torque will be reduced. Such pulsations of
flux may be prevented by designing the rotor and stator with
different numbers of teeth and making the width of the rotor
teeth a multiple of the stator slot pitch.
(a) Slot Leakage Flux
( C ) Zig-Zag Leakage Flux
FIG. 451.
405. Leakage Reactances. The diameter of the circle, in the
diagram for the induction motor, is inversely proportional to the
sum of the reactances, and it is therefore necessary for the de-
signer to determine these quantities very accurately and they
should ordinarily be kept as low as possible.
The leakage fields are: (1) that about the end connections, (2)
the slot leakage, (3) the zig-zag or tooth-tip leakage and (4) the
belt leakage.
480 ELECTRICAL ENGINEERING
The end-connection flux may be reduced by using fractional-
pitch windings but this results in increased magnetizing current.
The slot leakage flux, Fig. 451, is directly proportional to the
slot depth and is inversely proportional to the slot width. Deep
narrow slots, therefore, give large slot leakage. When the slots
are partly closed, to decrease the magnetizing current, the slot
leakage is very much increased. The use of magnetic wedges in
the slots gives the same result.
The zig-zag leakage flux, Fig. 451(6), corresponds to the tooth-
tip leakage flux in alternators and is usually quite large since the
gap is short. It also depends on the relative position of the rotor
and stator teeth. In position (6), Fig. 451, the leakage flux is
maximum and in (c) it is minimum. Such a large pulsation of
the flux would have very serious results. The rotor would tend
to lock in the position of maximum flux especially when being
started at reduced voltage with large current. The rapid pulsa-
tion of the flux might also produce an objectionable noise.
If the numbers of teeth on the stator and rotor have no common
divisor, only one tooth will be in the position of maximum flux at
any time and the tendency to lock will be negligible. If less than
20 per cent, of the teeth are in the position of maximum flux at
any time the tendency to lock will not be serious. This condition
can easily be fulfilled in a squirrel-cage rotor but with wound
rotors the number of slots on the rotor and stator must be divis-
ible by the number of poles and phases and conditions are not
so good. However, wound-rotor machines are usually started
with resistance in the rotor circuits and the starting current and
the zig-zag leakage are therefore small.
The belt-leakage flux is due to the fact that the stator phases
are not fixed in position relative to the rotor phases as in the trans-
former and increased leakage results when they are not directly
opposite.
When the sum of the reactances of a motor is small, the diame-
ter of the circle is large and the pull-out torque is large but the
starting current is also high.
406. Stator and Rotor Resistances. The stator resistance n
causes a loss of power and should be kept as small as possible.
The rotor resistance r 2 likewise causes a loss of power but it may
be necessary to make it comparatively large in order to get suit-
able characteristics.
The slip is proportional to the rotor copper loss and therefore
INDUCTION MOTOR 481
for low slip the rotor resistance must be low. However, by
increasing the rotor resistance the starting torque is increased and
the starting current is reduced and where large starting torque
is required a comparatively high rotor resistance is necessary.
With wound rotors external resistance may be introduced at
start and then cut out, but with the squirrel-cage rotor the resist-
ance cannot be varied.
If the starting torque of a squirrel-cage rotor is found to be too
small, it may be increased by slotting the rotor end rings and
so increasing the resistance.
If the bars of a squirrel-cage rotor are made very deep, good
starting torque can be obtained without reducing the efficiency.
At start the frequency of the rotor current is high and the slot
leakage flux is large, the lower parts of the bars have a high
reactance and carry very little current. Due to the unequal
current distribution the effective rotor resistance is increased and
the Pr loss is increased directly increasing the starting torque.
Near synchronous speed the rotor current and frequency both
become low and the current is uniformly distributed and the
effective rotor resistance approaches its true value. The effi-
ciency is therefore not reduced but due to the high slot leakage
the reactance is increased and the maximum output is decreased.
407. Effect of Change of Voltage and Frequency. If the
voltage impressed on an induction motor is decreased 10 per
cent, while the frequency is maintained constant, the flux is
decreased 10 per cent, and the exciting current is decreased more
than 10 per cent. ; the iron loss is decreased nearly 20 per cent. ;
the diameter of the circle is decreased 10 per cent. For a given
output the current is increased 10 per cent, and the copper loss is
increased 20 per cent. ; the slip is increased about 20 per cent. ;
the speed is reduced. The efficiency is decreased slightly. The
power factor at light load is increased due to the decreased excit-
ing current but above full load it is decreased due to the increased
current and decreased circle diameter. The maximum output
and maximum torque are decreased 20 per cent.
If the frequency is increased 10 per cent, while the voltage is
kept constant, the flux is decreased 10 per cent. ; exciting current
is decreased more than 10 per cent. ; the iron loss remains about as
before; the diameter of the circle is decreased 10 per cent. For
the same output the current is as before ; the copper loss and slip
are as before but the speed is higher by 10 per cent. The effi-
31
482
ELECTRICAL ENGINEERING
ciency is as before. The power factor is improved at light loads.
Maximum output and torque are decreased less than 10 per cent.
A 10 per cent, increase in voltage would be pretty well offset
by a 10 per cent, decrease in frequency but the increased losses
and lower speed would cause the motor to heat up.
408. Single -phase Induction Motor. 1 The stator of a single-
phase induction motor has a single winding with any number of
pairs of poles.
The rotor is either of the squirrel-cage type or is wound with
the same number of poles as the stator but with any number of
phases.
Y
FIG. 452. FIG. 453. FIG. 454. FIG. 455. FIG. 456.
FIGS. 452 TO 456. Single-phase induction motor.
Fig. 452 shows the relative directions of the currents in the two
windings at standstill. The stator carries a current /i which
consists of two components, /' the load component and I M the
magnetizing current. The rotor carries a current 7 2 opposite in
phase to /' and equal to it in m.m.f. If the ratio of turns is
assumed to be ft i : n z = 1 : l;then/ 2 = 7'. The motor at stand-
still is a transformer with a short-circuited secondary.
The flux which crosses the air gap and links with both stator
and rotor is produced by the stator exciting current. It is
always directed along the line YOY. There is no component of
flux in the horizontal direction XOX and therefore no torque is
exerted tending to turn the rotor in either direction. Thus the
rotating field which is produced in the polyphase induction motor
does not exist in the single-phase motor at standstill. The single-
phase induction motor, therefore, has no starting torque. If,
however, it is started in either direction it will develop torque
and will accelerate and come up approximately to synchronous
speed at no load.
Fig. 453 represents the motor with the rotor open-circuited and,
therefore, without current in its windings. The stator carries
only the magnetizing current.
Fig. 454 represents conditions at synchronous speed at the
INDUCTION MOTOR 483
instant when the stator magnetizing current is maximum. The
stator flux is then maximum downwards.
The rotor conductors moving at synchronous speed cut the
stator flux and an e.m.f . is generated in them proportional to the
product of flux and speed. Since the flux is alternating the e.m.f.
generated is of double frequency and produces a current of double
frequency in the closed-rotor winding. The current produces a
flux the rate of change of which through the rotor windings gener-
ates in them an e.m.f. equal and opposite to the e.m.f. generated
by rotation. This flux must, therefore, be of the same value as
the stator flux and it is in phase with the rotor current.
The rotor current goes through two complete cycles during one
revolution. In Fig. 454 it is maximum and is opposed to the
stator current, but the e.m.f. impressed on the stator is constant
and the stator flux is constant, and, therefore, a current must flow
in the stator to balance the m.m.f . of the rotor current I' M . Since
the ratio of turns has been taken as 1 : 1 the increase in stator
current is I' M and the total stator current at synchronous speed is
IM + I' M . In the position shown the rotor flux is not produced
because the rotor m.m.f. is opposed by an equal and opposite
m.m.f. on the stator.
Fig. 455 represents conditions after the rotor has turned
through one-half a revolution and the stator current has passed
through 'one-half cycle. The rotor current is in the same direc-
tion as before and has completed one cycle.
Fig. 456 represents conditions midway between Fig. 454 and
Fig. 455. The stator current is zero and the rotor current is
maximum and exerts a m.m.f. in the horizontal direction. There
is no stator m.m.f. opposing it and a flux is produced of the same
value as the stator flux in Fig. 454 or Fig. 455. Since the reluc-
tance of the path for the horizontal flux is the same as that for the
vertical stator flux, the rotor magnetizing current I' M must be
equal to the stator magnetizing current at standstill I M , and,
therefore, at synchronous speed the stator magnetizing current
is 21 M and is double its value at standstill.
Thus at synchronous speed there is a resultant m.m.f. of con-
stant value revolving at synchronous speed and the magnetic
field of the single-phase motor is identical with that of the poly-
phase motor, Fig. 457. The m.m.f. to produce the vertical field is
supplied by the true stator magnetizing current, while the m.m.f.
to produce the horizontal field is provided by an equal stator
484
ELECTRICAL ENGINEERING
magnetizing current, in phase with the true stator magnetizing
current, which induces in the rotor the rotor magnetizing current.
When the rotor runs at synchronous speed its conductors do
not cut this revolving flux and the only current in the rotor is the
double-frequency magnetizing current.
When the rotor runs at a slip s below synchronous speed the
rotor conductors cut the flux and currents are produced in them
and torque is developed just as in the case of the polyphase motor.
FIG. 457. Revolving field of a
single-phase induction motor at
synchronous speed.
FIG. 458. Revolving field of a
single-phase induction motor at
slip s.
409. Horizontal Field at Slip s. When the rotor runs at a
speed S = (I s) X synchronous speed, the e.m.f. generated in
it due to cutting the stator flux is less than at synchronous speed
in the ratio 1 s : 1 and the horizontal flux and the rotor mag-
netizing current are less in the same ratio.
The stator current is I M + (1 s) I M + I' and the rotor cur-
rent is (1 s) I M + 7 2 . The frequency of the rotor magnetizing
current is (2 s)f and the frequency of the rotor load current is
sf, where / is the frequency of the e.m.f. impressed on the stator.
The revolving field at slip s is not constant in value but has the
horizontal axis shorter than the vertical in the ratio 1 s : 1,
Fig. 458. The field follows an elliptical instead of a circular locus.
The torque which is proportional to the product of the rotor
load current and the horizontal field is less than that produced in
the polyphase motor in the ratio 1 s : 1.
410. Starting Single-phase Induction Motors. In order to
obtain the torque required to start a single-phase induction motor
a component of flux in quadrature in time and in space with the
stator flux must be produced at standstill. It has been shown
that when once the motor is started the rotor produces the
required quadrature flux and thus the torque to carry the load.
INDUCTION MOTOR
485
Two principal methods are employed to produce the quadrature
flux at standstill: (1) phase splitting and (2) shading coils.
1. If the two stator windings of a two-phase induction motor
are connected to a single-phase supply, phase 1 directly and phase
2 through a suitable resistance or condensive reactance, the flux
produced by phase 2 will have a component in quadrature in time
with phase 1 and will thus give the required starting torque,
Fig. 459. When the motor has come up to half speed, the starting
winding is cut out and the motor runs as a single-phase motor on
phase 1. This method of starting is called phase splitting. The
second winding need not have as many turns as the first but it
should be placed at 90 electrical degrees to it.
Fig. 459(6) shows a three-phase motor connected to a single-
phase circuit. Two phases are used for normal operation and
the third phase is used for starting only.
Stator Pole
Starting
Wlndi
(a)
Starting
Winding
FIG. 459.
FIG. 460.
2. The shading coils, Fig. 460, are short-circuited coils surround-
ing part of each pole of the stator. Currents are induced in them
and oppose the increase and decrease of the flux in the parts of the
poles which they inclose. Thus the north pole in section A will
reach its maximum value before it is maximum in section B.
When the north pole has decreased to zero in B it will be in-
creasing in section C and thus there is a rotation of the mag-
netic field and torque is produced. When the motor is started
the short-circuited coils may be opened and they will then be idle
and will not cause any power loss.
411. Comparison of Single-phase and Polyphase Motors.
Take the case of a two-phase motor operating on a single-phase
circuit using only one phase of the stator winding.
The slip single-phase is less than two-phase since the whole
rotor corresponds to one phase of the stator and thus the rotor
current and rotor copper loss are decreased.
The efficiency is lower because the output decreases more than
486
ELECTRICAL ENGINEERING
i 60
50
|
40
the losses. For a given impressed e.m.f. and frequency the iron
and friction losses remain practically constant.
The power factor is lower because the magnetizing current is
approximately doubled.
A given motor wound single-phase can be operated at higher
densities than when wound polyphase since the losses are less
and its ventilation is the same, and in this way its output may be
made from 65 to 75 per cent, of its output polyphase.
The torque at any speed can be increased by introducing re-
sistance into the rotor windings,
but this changes the maximum
torque since the torque is propor-
tional to 1 s.
Fig. 461 shows typical speed-
torque curves of a single-phase
induction motor with various
external resistances connected in
the rotor windings.
A single-phase induction mo-
tor is usually either a two-phase
or three-phase motor operated
on a single-phase circuit using
only part of the stator winding.
If one phase of a two-phase
motor is opened at light load, the magnetizing current of the
other phase is doubled and the motor runs as a single-phase
motor. If two phases of a three-phase motor are opened the
motor runs as a single-phase motor with the magnetizing current
in the third phase trebled. In both cases the flux distribution
and flux densities remain approximately the same as before.
412. Induction Generator. If the stator of an induction
motor is connected to the supply lines and its rotor is driven
above synchronous speed, the machine will develop electrical
power and supply it to the system.
The stator flux is not affected by the increase in the speed of
the rotor, but revolves in the same direction as when the machine
operates as a motor. The slip is, however,- reversed and the
e.m.fs. and currents induced in the rotor are reversed. Thus
the direction of torque and power is reversed and the mechanical
power supplied to drive the rotor is transformed into electrical
power and supplied over the lines to the load.
20 40 60 80 100 120 140 100 180
Torque in Percent of Full Load
FIG. 461. Speed-torque curves of a
single-phase induction motor.
INDUCTION MOTOR
487
The power transferred from the rotor to the stator depends on
the slip just as in the induction motor. Using the same notation
as in Art. 386, the power transferred to the stator is
P = nEzh cos 6 2
7- 2 2 + s V
Thus to increase the power delivered by the generator its speed
must be increased. If therefore an induction generator is con-
nected to a prime mover of variable speed, it will supply power
almost in proportion to the increase of its speed above synchro-
nous speed.
The characteristics of an induction generator may be deter-
mined from a circle diagram (Fig. 462). The diagram is con-
structed as for a motor but since the slip is negative the complete
circle is required. Points below the line oa represent negative
slip and therefore power output as a generator.
FIG. 462. Circle diagram for an induction generator.
Assuming the generator to be three-phase star-connected, the
voltage per phase is E and the voltage between terminals is
E = V3# G .
For the point d:
Input = \/3E X dm watts = mechanical power
supplied.
Rotor copper loss = \/3E X pm watts.
488
ELECTRICAL ENGINEERING
Output to stator = \/3E X dp watts
Stator copper loss = \/3E X qp watts.
Stator core loss = \/3E v watts
Output = A/2tf
Efficiency
qp watts.
X rq watts.
X dr watts.
j 100 per cent.
dm
The magnetizing current oa at no load and or under load must
be supplied by a synchronous machine operating in parallel with
the induction generator (Fig. 463). The voltage and the fre-
quency of the system are also
fixed by this synchronous gen-
erator and are not affected by
the induction generator.
The induction generator has
two very serious disadvantages ;
it is not self-exciting, and it
cannot supply reactive currents
Rotor
Driven
'abore
Synchrom
Speed
Induction. Generator
FIG. 463. Induction generator.
to an inductive load. It must,
therefore, be operated in parallel
with an alternator of sufficient
capacity to supply both the
magnetizing current for the induction generator and the reactive
currents required by the load. The exciting kilovolt-amperes at
no load may be as low as 12 per cent, of the full-load rating but
under load it will never be less than 25 per cent. If, however,
the output of the induction generator is small in comparison with
the capacity of the synchronous machines on the system this will
not have serious effects.
The induction generator has some good points: (a) it does not
require to be synchronized ; (6) there is no danger of trouble due
to hunting, since it does not operate at constant speed ; (c) in case
of short-circuits it loses its excitation and dogs not tend to supply
power to the fault.
In construction an induction generator is similar to an induc-
tion motor but the exciting current must be kept as low as pos-
sible by making the air gap short. Since no starting torque is
required, the rotor may be made of very low resistance and is
usually of the squirrel-cage type.
413. Asynchronous Phase Modifier. The asynchronous phase
modifier or phase advancer is a machine for use with induction
motors to improve their power factor ; it fills the place of the ex-
INDUCTION MOTOR
489
citer for a synchronous motor. A number of different types have
been designed but only the simplest one will be discussed here.
It consists of a direct-current drum armature with a commutator
and three sets of brushes per pair of poles displaced at 120 degrees.
The stator is merely a frame with laminations but without slots
or windings except in the larger sizes where a compensating wind-
ing may be required to take care of commutation. For small
sizes the stator may be omitted if the armature windings are
placed in totally closed slots. The iron above the slots serves
to complete the magnetic circuit.
Transformer
Induction
Motor
Horse-power
Starting Switches
Resistances
FIG. 464. Asynchronous-phase modifier.
Referring to Fig. 464, A is the main motor, which must have a
wound rotor and starting resistances R. The switch S is used
to disconnect the phase advancer and close the rotor circuits
while starting up. When full speed is reached, the switch S is
opened, connecting the terminals of the rotor to the three brushes
on the phase advancer B. If the armature is at rest it acts as a
three-phase reactance and produces a revolving field of the fre-
quency of slip. When the armature is driven in the direction
of the revolving field and at the same speed the reactance be-
comes zero. If the speed is then increased, the reactance is
reversed and acts as a capacity reactance causing the rotor
current 7 2 to lead the rotor induced e.m.f. E 2o) and so improves
the stator power factor.
The change in phase is illustrated in Fig. 465. (a) is the dia-
gram without the phase advancer and (b) represents the case
where the phase advancer is driven at such a speed that the
stator current 1 1 is, in phase with the impressed e.m.f. E\. The
490
ELECTRICAL ENGINEERING
rotor current 7 2 has then two components, I't the component ex-
erting a m.m.f. equal and opposite to that of the stator current
/i and I'M, the rotor magnetizing current filling the place of
the ordinary stator magnetizing current. The rotor reactance
drop is shown to be reversed and is approximately sIi(x B x 2 )
where X B is the condensive reactance of the phase advancer wind-
ing at the given speed. By driving the phase advancer at a still
higher speed the primary power factor may be made leading.
FIG. 465.
The capacity of the phase advancer is very small, being equal
to the required change in the stator reactive volt amperes multi-
plied by the slip. This may be understood from the following:
A stator e.m.f. E\ Q corresponds to a rotor e.m.f. E Zg = sE% =
Ei g .
A stator current I' corresponds to a rotor current /
/'. And, therefore, the stator volt-amperes Ei g l' correspond
to the rotor volt-amperes E^glz = sEi g ! f .
The phase advancer is driven by a small high-speed, squirrel-
cage induction motor C connected to the supply lines directly or
through a transformer. It supplies the friction windage and iron
losses of the phase advancer but the copper loss is supplied from
the rotor of the main motor.
INDUCTION MOTOR 491
The capacity of the main induction motor is very largely in-
creased by the elimination of the reactive currents from the
stator. The supply lines are also relieved of a large component
of current.
To raise the power factor of a 1,000-kva. induction motor from
86.6 per cent, to 100 per cent., the reactive kilo volt-amperes re-
quired in the stator = 1,000 Vl -0.866^= 1,000 X 0.50 = 500
kva. Assuming a slip of 3 per cent, the capacity of the phase
advancer is only 0.03 X 500 = 15 kva. A K-hp- driving motor
would supply the losses. The power developed by the induction
motor is increased from 1,000 X 0.866 = 866.0 kw. to 1,000 kw.,
that is, by about 15 per cent.
To reduce the required phase advancer capacity the slip of the
motor should be made as small as possible by decreasing its rotor
resistance.
If the equipment of a shop consists of one large motor and a
number of smaller ones, a phase advancer may be applied to the
large motor to make its power factor leading and so supply the
reactive currents for all the motors.
414. Phase Converter. To change from one polyphase system
to another, as from two-phase to three-phase or three-phase to
six-phase, stationary transformers should be used; but when a
change from single-phase to two-phase or three-phase is required,
transformers are no longer satisfactory since, due to the pulsating
nature of single-phase power, energy storage must take place
and this can best be accomplished by means of the momentum of
a revolving machine.
To change from single-phase to two-phase an ordinary two-
phase induction motor with a squirrel-cage rotor may be used
as a phase converter. The connections are shown in Fig. 466 (a).
One phase A of the stator winding is connected to the single-
phase supply and the converter runs as a single-phase induction
motor and near synchronous speed a revolving field is produced
of approximately constant value (Art. 409). This field cuts the
second stator phase B and generates in it a voltage in quadrature
with the supply voltage and with approximately the same effect-
ive value at no load, Fig. 466(6).
One phase of the two-phase system receives its power directly
from the single-phase supply and the second phase is supplied
from the phase B of the converter. Under load the voltage of
phase 1 remains constant but that of 2 decreases due to increased
492
ELECTRICAL ENGINEERING
slip and to the impedance drop in the converter and it falls back
in phase to E f 2) Fig. 466 (c). This unbalance of the two-phase
system may be corrected for a given load by changing the connec-
tions as indicated by the broken lines in Fig. 466 (a). The voltage
across phase A under load is increased by changing the trans-
former tap, thus raising the voltage generated in B and in addi-
tion a component oa of the supply voltage is connected in series
with B. The voltage of phase 2 is thus changed from E f z back
to E 2 .
Single-Phas
[000000000000
Two-Pbase
(a)
Three-phase Motor
FIG. 466.
FIGS. 466 AND 467. Phase converters.
FIG. 467.
When changing from single-phase to three-phase a similar
phase converter may be used but phase B must be wound for
only 87 per cent, of the voltage of phase A. Fig. 467 (a) shows
the connections required for operating a three-phase induction
motor from a single-phase circuit and the equilateral triangle
abc represents the balanced three-phase voltages at light load.
Under load the voltage E B decreases and falls back in phase and
the point c moves to c'. To balance the system the connection
for phase B must be changed from d to d f and the voltage im-
pressed on A must be increased by changing to tap b f . This,
however, gives a balance only at one particular load.
In both cases the momentum of the rotor stores and returns
energy and makes possible the change from pulsating single-
phase power to constant polyphase power.
By using a phase converter three-phase induction motors can
be operated from a single-phase supply circuit for the propulsion
of electric locomotives. A single converter may be used to
supply a number of induction motors and it may be designed as a
high-speed machine and will therefore be light and cheap; and
INDUCTION MOTOR 493
since it exerts very little torque the shaft may be made small.
To reduce the losses and slip the rotor should be made of very
low resistance. The weight of the phase converter may be only
about 25 or 30 per cent, of the combined weight of the motors
which it supplies.
The function of the phase converter is reversible and when the
motors are running above synchronous speed they become gen-
erators supplying single-phase power back to the system and
give a large and uniform braking effect.
415. Induction Frequency Converter. The induction fre-
quency converter may be used instead of the synchronous-motor
generator set to convert power from one frequency to another
or to link up two systems of different frequencies. It consists
of an induction motor with a wound rotor driven by a synchro-
nous motor connected to the 25-cycle supply lines. The stator
of the induction motor is also connected to the supply and
produces a revolving field. At standstill the frequency of the
e.m.fs. generated in the rotor windings is 25 cycles. When
the rotor is driven backward at synchronous speed the fre-
quency is 50 cycles and when driven at 140 per cent, of synchro-
nous speed it is 60 cycles. If a receiver circuit is connected to
the rotor slip rings 60-cycle power can be supplied to it.
Twenty-five-sixtieths or five-twelfths of the power output is
supplied to the rotor by transformer action from the stator and
the remaining seven-twelfths is supplied by the synchronous
motor as mechanical power.
The principal disadvantage of the induction frequency con-
verter is its poor voltage regulation. Due to the presence of the
air gap in the magnetic circuit the reactances are large and the
e.m.fs. consumed by the reactances are large.
The exciting current required by the induction motor may be
provided by over-exciting the fields of the synchronous motor
and making it draw a leading current. In this way the power
factor of the set may be made unity, but the increased current
in the synchronous-motor windings increases the copper losses
and the heating.
Since the rotor of an induction motor may be wound for any
number of phases irrespective of the number of stator phases the
induction frequency converter may be used to change the number
of phases as well as the frequency. The same result may of
course be obtained with the synchronous-motor generator set.
CHAPTER XIII
ALTERNATING-CURRENT COMMUTATOR MOTORS
416. Motor Characteristics. Direct-current motors are of
three types, shunt, compound and series. The methods of vary-
ing the speed of such motors were discussed in Arts. 161 and 172.
Of the alternating-current motors, the synchronous motor runs
at a constant speed at all loads and this speed cannot be varied.
The induction motor with a low-resistance rotor also runs ap-
proximately at constant speed and this can be decreased only by
using a wound rotor with slip rings and inserting resistance in
the rotor windings to increase the slip. This is analogous to the
introduction of resistance in series with the armature of a direct-
current motor and is inefficient. There is no way of increasing
the speed above synchronous speed corresponding to the field
control in a shunt motor.
The single-phase synchronous motor and induction motor can-
not start under load but the polyphase motors exert large starting
torque. The current required, however, is greater than for a
similar direct-current motor since it is not in phase with the im-
pressed voltage and the torque per ampere is therefore less.
The induction motor with a comparatively high-resistance
rotor has characteristics very similar to those of the direct-current
compound motor. The speed falls off under load and the start-
ing torque is good.
To design an alternating-current motor with characteristics
similar to the direct-current series motor a commutator and
brushes must be added. A large number of motors of this kind
have been developed and the principles of their operation are
discussed below.
By adding a commutator and brushes to the rotor of a single-
phase induction motor the power factor may be compensated and
the speed may be made adjustable from values above synchro-
nous speed down to half synchronous speed.
417. Alternating-current Series Motor. The alternating-cur-
rent series motor is very similar to the direct-current series motor
and can be operated on direct-current with increased efficiency
and output.
494
ALTERNATING-CURRENT COMMUTATOR MOTORS 495
If a direct-current series motor is connected to an alternating-
current supply circuit it will rotate since the currents in the field
and armature reverse together and therefore the torque is always
in one direction, but it will be very inefficient and will spark very
badly.
With alternating current flowing in the field winding an alter-
nating magnetic flux is set up through the magnetic circuit and
causes very large losses due to hysteresis and eddy currents. To
reduce these to a minimum the whole magnetic circuit of an
alternating-current series motor must be laminated. The field
circuit must be very heavily insulated to prevent short-circuits
between turns which would burn out the motor on account of
the large induced currents.
The relation between the e.m.fs. and current in the direct-
current series motor is given by the equation
E = 8 + I(r a + TV), (444)
where E = impressed e.m.f.,
8 = counter e.m.f. generated by rotation,
/ = current,
r a = resistance of the armature,
r/ = resistance of the field.
In the alternating-current series motor the alternating flux sets
up large e.m.fs. of inductance in both the field and armature
windings, which consume components of the impressed e.m.f. in
quadrature ahead of the current. If L/ is the inductance of the
field and L a the inductance of the armature, their reactances are
x f = 27r/L/and x a = 2irfL a) respectively, where / is the frequency
of the impressed e.m.f.
ry a Current I
FIG. 468. Vector diagrams of a single-phase series motor.
Fig. 468 shows the vector diagram for the motor.
ox = I = current in field and armature.
oa = Ir f = e.m.f. consumed by the resistance of the field.
496 ELECTRICAL ENGINEERING
ab = Ixf = e.m.f. consumed by the reactance of the field.
be = Ir a = e.m.f. consumed by the resistance of the armature.
cd = Ix a = e.m.f. consumed by the reactance of the armature.
dk = 8 = e.m.f. generated in the armature due to rotation,
in phase with the field flux and, therefore, in phase with
the current, neglecting the hysteretic lag.
ok = E = impressed e.m.f.
cos kox = cos cj> = load power factor.
cos dox = cos <$> s = power factor at start.
Taking the current as the real axis the relation between the
current and the impressed e.m.f. can be expressed in rectangular
coordinates as
g = g + I(r a + r f ) + jl(x a + x f ) (445)
and taking absolute values
E = {& + I(r a + TV)) 2 + \I( Xa + z,)) 2 (446)
At standstill
E = lV(r a + r f y + (x a + xJY' (447)
and the current is
/ = E (448)
+ r f y + (x a + x f Y
Full voltage can usually be impressed on the motor at stand-
still without causing any injury since the current is limited by
the large impedance.
The power factor under running conditions is
cos =
but 8 = kn$, where is the maximum value of the flux per pole,
n is the motor speed in revolutions per second and k is a constant
depending on the number of turns in the armature winding and
on the shape of the flux wave. The flux $ is almost propor-
tional to the current I and the generated e.m.f. may be expressed
as
8 = k'nl.
Substituting this value for 8 in equation and eliminating I
k'n + r a + r/ ,. _ n ,
cos $ = , J =; (450)
V(k'n + r a + r f y + (x a + */)
the power factor, therefore, increases with increasing speed and
approaches unity. At low speed and at standstill it is low on
ALTERNATING-CURRENT COMMUTATOR MOTORS 497
account of the reactances in the field and armature and for satis-
factory operation it is necessary to make these reactances as low
as possible.
418. Design for Minimum Reactance. The inductance of any
coil is proportional to the square of the number of turns and is
inversely proportional to the reluctance of the magnetic circuit
through it. To reduce the inductance L/ of the field winding it
is designed with a small number of turns but this reduces the
field m.m.f . and in order to obtain the required flux the reluctance
of the magnetic circuit must be made very low. For this purpose
large sections of high permeability are used, the slots are partially
closed and the air gap is made as short as possible.
The reactance of the winding is proportional to the product of
the inductance and the frequency and therefore the frequency
should be low. Motors are usually designed for 25 cycles since
that is the lowest standard frequency, but they will operate on
15 cycles or on direct current with a much improved efficiency
and power factor and a larger output. The frequency of the
supply does not affect the speed of the motor directly, but it
does indirectly since the reactance drop decreases with the fre-
quency and, therefore, the speed for a given current increases.
FIG. 469. Four-pole, single-phase, series motor with compensating winding.
419. Compensating Windings. The armature inductance and
reactance cannot be decreased by reducing the number of turns
on the armature since, for a given impressed voltage, that would
increase the speed of the motor, and, further, since the field is
made comparatively weak the armature must be made corre-
32
498
ELECTRICAL ENGINEERING
spondingly strong in ampere-turns in order to produce the re-
quired torque.
The armature m.m.f. as in direct-current machines is cross-
magnetizing and distorts the main field and so weakens it and
interferes with commutation. The flux produced by it is alter-
nating and induces in the armature a back e.m.f. of armature
inductance. This flux of armature reaction or armature induct-
ance may be reduced as in the direct current generator by the
use of a compensating winding. The compensating winding is
placed in slots in the pole faces as shown in Fig. 469. It is dis-
tributed over the whole periphery of the armature and exerts a
m.m.f. opposing the armature m.m.f. and so limiting the cross
flux to a very small value and reducing the armature inductance
and reactance in the same proportion.
The m.m.f. of the compensating winding can be produced in
two ways illustrated in Fig. 470 and Fig. 471. The first is called
inductive compensation and the second conductive compensation.
Field
Winding,
Variable Ratio Transformer
FIG. 470. Induc-
tively compensated
series motor.
FIG. 471. Con-
ductively compen-
sated series motor.
FIG. 472. Series
motor with variable
voltage supply.
1. In the inductively compensated series motor the compensat-
ing m.m.f. is produced by short-circuiting the compensating coil.
It then acts as the closed secondary of a transformer of which the
armature is the primary. The m.m.f. of the compensating wind-
ing is almost equal to the m.m.f. of the armature but can never be
greater than it and, therefore, over-compensation is not possible.
The combined reactance of the armature and compensating wind-
ing corresponds to the reactance of a transformer on short-circuit.
The mutual flux is almost destroyed but the leakage fluxes remain.
2. In the conductively compensated motor the compensating
coil is connected in series with the field and armature and the
amount of compensation can be varied. When the m.m.f s. of
the two windings are equal there is no mutual flux and the com-
bined reactance is a minimum. When the m.m.f. of the compen-
sating winding is stronger than that of the armature the armature
reaction flux is reversed but the reactance of the compensating
ALTERNATING-CURRENT COMMUTATOR MOTORS 499
winding is increased and so part of the advantage is lost, but the
flux due to over-compensation assists commutation of the load
current in the same way that interpoles do and is thus a great
advantage.
A conductively compensated motor can be operated on direct
current but an inductively compensated motor cannot since the
compensating winding would not be effective and sparking would
occur.
420. Commutation. Satisfactory commutation is very much
more difficult to obtain in the alternating-current series motor
than in the direct-current motor because, as may be seen in Fig.
473, the short-circuited coil is in the position of the short-circuited
secondary of a transformer with the main field as primary and
tends to have as many ampere-turns induced in it as there are on
FIG. 473. FIG. 474.
a pair of field poles. This large short-circuit current interferes
with commutation and must be reduced as far as possible. For
this purpose high-resistance leads, called preventive leads, are
connected between the coils and the commutator bars, as shown
in Fig. 474, and narrow carbon brushes of high contact resistance
are used. The short-circuit current must pass through two resist-
ance leads in series and is thus greatly reduced while the load
current is carried by two or more in multiple. The resistance of
one of the leads must be very much higher than that of an arma-
ture coil in order to reduce the current sufficiently.
There are losses in the leads due to the resultant of the two
currents flowing in them, but by increasing the resistance up to a
certain point the short-circuit current is reduced and the com-
bined loss is reduced.
The resistance leads are not made of large enough capacity to
carry the current continuously but under running conditions any
one lead is in circuit for only a very short time. If the motor
is stalled with power on the leads are likely to be destroyed.
The torque of the motor is very much improved by the use of
500
ELECTRICAL ENGINEERING
resistance leads since without them the large short-circuit current
would weaken the main field and decrease the torque.
Fig. 475 shows the characteristic curves of a 150 hp., single-
phase series motor. The torque and speed curves are very
much the same shape as those of the direct-current series motor.
The power factor approaches unity at light load when the
speed is high as explained above, but at full load it is still very
good, reaching 90 per cent, in some cases. At start and at low
speeds it is low because the reactance of the motor is constant.
E>U 100
46 90
40 80
^
*^\
,
/
/
200
180
160
140 sj
120 <2
I
100 M
80 8
4
60 >
40
20
Jgvt
M
"ri=^
^
^
F is the flux in F, which is in phase with the current.
E F is the component of impressed e.m.f. across the termi-
nals of F.
E s is the e.m.f . generated in the armature by rotation and
transferred to the compensating coil.
ALTERNATING-CURRENT COMMUTATOR MOTORS 503
EC is the component of impressed e.m.f. across the termi-
nals of C ; it is equal to E s if the coil C has the same num-
ber of turns as the armature, and it is in phase with it.
E = ^/E^ + E c 2 is the constant line voltage impressed
on the motor.
The e.m.f. consumed by the impedance of the armature and com-
pensating winding is neglected. As the speed increases the e.m.f.
EC increases and E F decreases, the flux in the main field F de-
creases and the current and torque decrease.
At start, when the drop across C is small, the current is large
and the main field F is very strong. The repulsion motor, there-
fore, gives a good starting torque. The field at start will be
decreased to a certain extent by the current in the short-cir-
cuited coil undergoing commutation, as in the series motor.
424. Commutation. In the single-phase series motor and the
repulsion motor there are two currents to be commutated: (1)
the load current and (2) the short-circuit current produced in
the coil under the brush by the alternating flux of the main field.
1. To reverse the load current a m.m.f. is required opposing
the m.m.f. of armature reaction and strong enough to produce a
flux in the opposite direction to the armature reaction flux. Such
a flux can be produced by interpoles placed between the main
poles and excited by a winding in series with the main field or it
can be produced by a compensating winding. The conductively
compensating winding is the only one which can give perfect
commutation since its m.m.f. can be made stronger than the
armature m.m.f. Commutation is assisted by the use of high-
resistance carbon brushes.
2. To eliminate the short-circuit current in the coil under the
brush an e.m.f. must be generated in the coil equal and opposite
to the e.m.f. producing the short-circuit current. The neutraliz-
ing e.m.f. cannot be generated by the alternation of a magnetic
flux through the coil, since that would require a flux equal and
opposite to the field flux and would destroy the field of the motor.
The required e.m.f. can, however, be generated by the rotation
of the armature through a commutating field of the proper inten-
sity and position, but the field must be in quadrature with the
main field in both time and space. In the repulsion motor under
running conditions such a field is produced in the compensating
coil C. The intensity of the field varies with the speed of the
504 ELECTRICAL ENGINEERING
motor. Near synchronous speed the e.m.f. is entirely neutral-
ized and the current is wiped out. Below synchronous speed the
current is reduced and above synchronous speed another current
is produced as objectionable as before and commutation becomes
bad again. At standstill no neutralizing e.m.f. is produced.
In the single-phase series motor there is no field in quadrature
with the main field in time and so no neutralizing e.m.f. can be
produced, but the short-circuit current is reduced by using high-
resistance preventive leads as explained in Art. 420.
Thus near synchronous speed the commutation of the repulsion
motor is better than that of the series motor.
Since preventive leads are not used in the repulsion motor the
short-circuit current in it at start will be greater than in the series
motor and will weaken the main field and decrease the starting
torque.
While running the short-circuit current is not so great, since it
cannot reach its maximum value on account of the self -inductance
of the coil.
The repulsion motor cannot be operated more than 40 per cent,
above synchronous speed on account of commutation troubles.
425. Compensated Repulsion Motor. The main field wind-
ing, F, or the torque field of the repulsion motor may be omitted
if two extra brushes are placed on the commutator at right angles
to the main brushes as shown in Fig. 481. The armature winding
() (6)
FIG. 481. Compensated repulsion motor.
is thus made to carry the exciting current and the total reactance
of the motor is decreased and the power factor improved. It is
not usually desirable to design the armature with the proper
number of turns for excitation and the motor Fig. 481 (a) may
be replaced by that in (6), in which the exciting brushes are
supplied from taps on the secondary of a transformer with its
primary in series with the compensating coil C. With this
arrangement the speed may be controlled by changing the tap to
ALTERNATING-CURRENT COMMUTATOR MOTORS 505
which the exciting brush is connected. These motors have the
serious objection that four sets of brushes are required per pair
of poles.
A large number of motors, differing in certain details from the
simple series and repulsion motors described herfe, have been de-
signed and have good operating characteristics, but a discussion
of them is beyond the scope of this book.
426. Alternating-current Commutator Motors with Shunt
Characteristics. The single-phase induction motor has a speed
characteristic similar to a shunt motor but the speed cannot be
adjusted and the motor has no starting torque.
If the squirrel-cage rotor is replaced by a direct-current arma-
ture with two sets of short-circuited brushes at right angles as
shown in Fig. 482 the characteristics of the motor are not changed
but it is possible to improve its power factor and to adjust its
speed.
When the motor is at rest there is no e.m.f. between the xx
brushes and no current flows and therefore there is no flux along
the xx axis. There is flux along the yy axis and current flows
1C
(a) (b) (c)
FIG. 482. FIG. 483. Single-phase commutator motor with
power-factor compensation.
through the yy brushes. If the motor is started in either direc-
tion the conductors cut the yy flux and an e.m.f. is generated
between the xx brushes in phase with the yy flux and proportional
to the speed. This e.m.f. causes a current to flow in the xx
brushes which produces a torque field along the xx axis and thus
torque is developed as in the single-phase induction motor
(Art. 408).
There are three ways of improving the power factor of the
motor: (1) By introducing in the circuit of the yy brushes a
stator winding in the xx axis, Fig. 483 (a); (2) by introducing in
the circuit of the xx brushes a stator winding in the yy axis,
506
ELECTRICAL ENGINEERING
Fig. 483(6); and (3) introducing in the circuit of the xx brushes
an e.m.f. in phase with the line voltage, Fig. 483 (c).
There are likewise three methods of controlling or adjusting
the speed: (1) By introducing in the circuit of the xx brushes a
stator winding in the xx axis, Fig. 484 (a) ; (2) by introducing in
the circuit of the xx brushes a variable inductance, Fig. 484(6);
and (3) by introducing in the circuit of the yy brushes an e.m.f.
in phase with the line voltage, Fig. 484 (c).
Variable Inductance
(a) W (0}
FIG. 484. Single-phase commutator motor with speed adjustment.
427. Single -phase Induction Motor with Repulsion Starting.
The repulsion motor in Fig. 479 (a) may be changed into a single-
phase induction motor by short-circuiting the commutator.
When the brushes of such a motor are set at the proper angle to
the stator winding a good starting torque is produced. A centri-
fugal governor may be set to short-circuit all the segments of
the commutator and to raise the brushes when a certain speed
is reached and the motor will then operate as an induction
motor at approximately constant speed.
CHAPTER XIV
TRANSMISSION SYSTEMS
428. Transmission Line. The transmission line carries the
electrical energy from the generating station to the receiving
station or substation, where it is either transformed into mechani-
cal energy or distributed to the customers throughout the district.
The most important characteristics of a transmission line are:
(1) reliability, (2) regulation and (3) efficiency.
1. To insure reliability of service lines should, wherever pos-
sible, be installed in duplicate and all the necessary protective
devices applied.
2. For good regulation the reactance of the line should be as
small as possible and therefore the frequency should be low. The
capacity of a line draws a leading current, which partially coun-
teracts the drop in voltage due to reactance and so improves the
regulation.
3. The power losses in a line are the resistance loss, which varies
as the square of the current, and the comparatively small losses
due to leakage over the insulators and to the formation of corona
around the conductors.
To reduce the power loss the resistance of the line should be
made as low as possible. This can be done by increasing the
cross-section of the conductors, but the increased cost of the
material required soon overcomes the saving due to the increase
in efficiency.
For a given loss and a given voltage between lines power can
be transmitted with a smaller amount of conducting material
three-phase than either single-phase or two-phase.
429. Relative Amounts of Conducting Material for Single-,
Two- and Three-phase Transmission Lines.
Let P = power input to the line in watts.
p = per cent, loss of power in the line resistance due to
full-load current.
/ = full-load current.
507
508 ELECTRICAL ENGINEERING
cos B = power factor.
r = resistance of each conductor.
n number of conductors in the system.
The loss in the line is
and the resistance of each conductor is
100 rc/ 2
For the same voltage E between conductors the current is
P
E cos B
P
, single-phase,
-, two-phase,
2E cos
p
I = 7= , three-phase,
V3E cos
and the resistance per conductor is
pP X E 2 cos 2 6 . nn _ pE 2 cos 2 .
r = ICQ x 2P 2 = ~T ' single-phase,
pP X 4# 2 cos 2 , p# 2 cos 2 e .
100 X 4P 2 ~P ' two-phase,
pP X 3# 2 cos 2 S n m p# 2 cos 2 ., ,
r = 100 X 3 P 2 ~P ' three-phase.
Since the single-phase line has only two conductors while the
two-phase line has four the amount of copper required for both is
the same. The three-phase line consists of three conductors of
the same section as the two-phase conductors and, therefore, the
amount of copper required for a three-phase line is only 75 per
cent, of that required for a two-phase or single-phase line with the
same per cent, power loss and the same maximum voltage between
lines.
430. Reactance. The inductance of a line per mile of con-
ductor is (equation 163)
74 logic p + 0.0805 Y 10~ 3 henrys
where D is the distance between conductors,
and R is the radius of the conductors.
The reactance of the line per mile of conductor is
X = 27T/L ohms.
L = (o
TRANSMISSION SYSTEMS 509
The reactance could be decreased by decreasing the distance
between the conductors or by increasing the radius, but these
quantities are fixed by other considerations than the reactance
and reactance drop.
431. Capacity. The capacity of a line per mile of conductor
between the conductor and neutral is
00 O
C = - 10- 9 farads (equation 38).
, D ri
logio ft
This value applies for each conductor of a single-phase or poly-
phase line. If the conductors of a three-phase line are suspended
in one plane instead of in the form of an equilateral triangle the
capacity of the central conductor is slightly greater than that of
the others, but since all lines are transposed the total capacity of
each of the three is the same and is given with sufficient accuracy
by the formula above if the distance D is taken as the shortest
distance between conductors.
The capacity reactance per mile of conductor is
and the charging current per mile of conductor is
where e is the voltage between conductor and neutral.
For transmission lines up to 50,000 volts the capacity is very
small and its effect on the regulation may be neglected. If, how-
ever, any part of the transmission is carried out through under-
ground cables, the capacity may be very largely increased and
may not be negligible. Above 50,000 volts the capacity of the
line must be considered in calculating the regulation. For lines
up to 100 miles in length and for voltages up to 100,000 volts the
capacity of each conductor may be considered as a condenser
connected at the center of the line between conductor and neutral.
If more accurate results are necessary the fact that both the re-
actance and the capacity of the line are distributed over the whole
length must be taken into account.
Due to the presence of the charging current in a line the current
flowing into the receiving circuit may be very much larger than
the current entering the line at the generating station.
510
ELECTRICAL ENGINEERING
432. Voltage and Frequency. Voltages up to 150,000 volts
are now in use for the transmission of large amounts of power
over long distances. The voltage employed in any given system
is as a general rule approximately one thousand times the length
of the line in miles.
Power is usually generated and transmitted at either 25 cycles
or 60 cycles. With 25 cycles the reactance drop in the line is less
Distance between Lines - Incbe
o 8 S g
^
^
^
^
"^
^
S*
'
'
, "
-^
) 10 20 30 40 50 00 70 80 90 100 11
Kilovolts between Lines
FIG. 485. Spacing of conductors.
than with 60 cycles and therefore the voltage regulation is better.
In the case of very long high-voltage lines the increased charging
current at the higher frequency may counteract the larger react-
ance drop of voltage. Where power is required for lighting 60
cycles is necessary unless frequency chargers are installed.
433. Spacing of Conductors. The distance between the con-
ductors of a transmission line depends both on the voltage and
also on certain points in the mechanical design, such as the mate-
Eo
FIG. 486. Single-phase transmission line.
rial of the conductor, length of span and the amount of sag al-
lowed. The curve in Fig. 485 gives approximately the relation
between the spacing of the conductors and the voltage.
434. Single-phase Transmission Line. 1. A single-phase trans-
mission line, Fig. 486, delivers 5,100 kw. to a receiver circuit at
TRANSMISSION SYSTEMS 511
60,000 volts. If the power factor of the load is 85 per cent.,
find the generator voltage.
r = resistance of the line = 20 ohms.
x = reactance of the line = 50 ohms.
The power delivered to the receiver circuit is
P = El cos = 5,100,000 watts,
where E = 60,000 is the receiver voltage
and cos = 0.85 is the power factor;
the current is therefore
P 5,100,000
" E cos ~~ 60,000 X 0.85 "
The vector diagram is drawn with the current 01 = I as hori-
zontal.
The receiver e.m.f. OE = E leads the current by -an angle
and has two components
OEi = EI = E cos < in phase with / and
OE 2 = E 2 = E sin in quadrature ahead of /.
The voltage consumed in the resistance of the line is Ir in
phase with 7; the voltage consumed in the reactance of the line
is Ix in quadrature ahead of /.
The component of the generator e.m.f. in phase with I is
E l + Ir = E cos + Ir
and the component in quadrature ahead of / is
E 2 + Ix = E sin + Ix,
and therefore the generator e.m.f. is
E G = \/(E cos + Ir) 2 + (E sin + Ix)' 2 ,
or substituting the numerical values
E G = V(60,000 X 0.85 + 100 X 20) 2 +(60,000 X 0.52+100X50) 2
= 64,000 volts.
The e.m.f. consumed in the line is
iVr 2 + x 2 = 100 V20 2 + 50 2 = 5,400 volts.
The loss of power in the line is
7 2 r = 100 2 X 20 2 = 200,000 watts.
- 200 kw.
The power factor at the generator is
cos to * l^+Jr = 53,000 = Q ^ = ^
JQ D4,UUU
512 ELECTRICAL ENGINEERING
Using rectangular coordinates and taking the current as axis,
the e.m.f. at the receiver terminals is
E = E cos + jE sin ,
the e.m.f. consumed in the impedance of the line is
E' = IT + jlx,
and the generator e.m.f. is
E G = E + E' = (E cos + Ir) + j(E sin 4> + Ix),
and its absolute value is
E G = \/(E cos + Ir) 2 + (E sin0 + Ix) 2 .
The capacity of the line has been neglected in this example.
2. A transmission line of impedance Z = r + jx delivers power
to a receiver circuit of admittance F = g jb at a constant
voltage E. If the capacity of the line is assumed to be concen-
trated at the center determine the charging current of the line,
the total current delivered by the generator and the terminal
voltage of the generator.
The condensive reactance of the line is
4 ' j-fir
^2
Js
<
E <
\
vvW\
1
= 0.85;
therefore, the current in the receiver circuit is
P 10,000,000
= E cos ~ 100,000 X 0.85 ~ L1/ amp ' ;
TRANSMISSION SYSTEMS 515
the power component is
I P = /cos = 117 X 0.85 = 100 amp.;
the quadrature component is
I w = I sin = V/ 2 - Ip 2 = A/117 2 - 100 2 = 60 amp.
The admittance of the receiver is
1 H7
~ E~ 100,000 ~
the conductance is
/ cos 100
and the susceptance is
7 sin 60
"- = =
The generator voltage is, by equation (453),
E = E j 1 + (r + jx) (g - jb + j |) + j |(r + jx)*(g - jb) )
and substituting the values obtained above
+ (65.4 + 154.5J) (O.OOI - 0.0006J +
00097^
+ - - j (65.4 + 154.5J) 2 (0.001 - 0.0006J)
and simplifying
E Q = ^(1.137
and the absolute value is
E Q = #(1.137) 2 + (0.114) 2
= 100,000 X 1.142 = 114,200 volts.
The current from the generator is, by equation (542),
lo = E [g-jb+jb. { 1 + r -^p- (g-jb)
= & [o.OOl - 0.0006? + 0.000275.?
= ^(0.000984 - 0. 00029;'),
and its absolute value is
I Q = 100,000 V(0.000984) 2 + (0.00029) 2 = 102.6 amp.
516 ELECTRICAL ENGINEERING
The charging current of the line is, by equation (541),
I c =jb c E [1 + !+&(,_#))
= E (- 0.0000158 + 0.00031J),
and its absolute value is
I c = 100,000 V(0.0000158) 2 + (0.00031) 2 = 31 amp.
= 26.5 per cent, of the receiver current.
At no load E = 100,000, g = 0, b = and the generator vol-
tage is
and its absolute value is
E Q = 1 00,000 V(a9786) 2 -{- (0.0098) 2 = 97,860 volts.
The power factor at the generator may be found by reference
to the diagram in Fig. 488.
EG
E leads E by an angle <', where
. / JL _L .^ 4-\ -* f\
tan ,' ,. - = 0.10;
I G lags behind E by an angle <", where
tan-0" = ~j^ = 0.0296;
and I G lags behind E G by an angle G = cos 7 30' = 0.99
= 99 per cent.
The impedance drop in the line is found very approximately as
E z =
= 117V(65.4) 2 + (154.5) 2 = 19,650 volts
= 19.6 per cent, of the receiver voltage.
TRANSMISSION SYSTEMS
517
The power loss in the line is found approximately as
T T
Pr = Ir 2 ~ + I' 2 -
2 2
= (102.6) 2 34 + (117) 2 34
= 826,000 watts
= 826 kw.
= 4.13 per cent, of the output.
435. Three-phase Transmission Line. A three-phase, trans-
mission line delivers 30,000 kva. at 100,000 volts, 60 cycles to a
receiving circuit of 88 per cent, power factor; determine the
voltage and current and power factor at the generating station,
the charging current and charging kilovolt-amperes of the line and
the efficiency of the transmission. Determine also the rise in
voltage at the terminals of the receiving circuit if full load is
suddenly removed.
Length of line 100 miles.
Conductor = No. 0000 B. & S. copper of 97 per cent, conduc-
tivity.
Diameter of conductor = 2R = 0.46 in.
Distance between conductors = 100 in.
FIG. 489. Three-phase transmission line.
In Fig. 489
E = voltage between lines at the receiving end = 100,000.
I rvr\ 000
e = voltage between lines and neutral = -- W- = 57,600.
\/6
E = voltage between lines at the generating station.
e = voltage between lines and neutral =
ET
.
V3
I = load current in each conductor.
I c = charging current in each conductor.
I G = current per conductor at the generating station.
518 ELECTRICAL ENGINEERING
Y = \/V + b 2 = admittance of each phase of the receiving
circuit.
Z \/r 2 -j- x z = impedance of each line including the step-
up and step-down transformers.
b c = capacity susceptance of each conductor to neutral, which
is assumed to be concentrated at the center of the line.
Resistance of a single conductor of 97 per cent, conductivity
at 20C., 100 miles long = r' = 0.2667 X 100 = 26.67 ohms
(article 86).
30,000,000
= x 100,000 = 173 amp '
Resistance drop in each conductor = Ir' = 173 X 26.67 =
4,620 volts = ' 10Q QOQ = 8 per cent.
Assuming a resistance drop in step-up and step-down trans-
formers of 0.5 per cent, each, the total resistance drop per line is
8 + 2(0.5) = 9 per cent.
g
r = equivalent resistance of each line = 26.67 X 5 = 30
o
ohms.
Inductance of each conductor is
L = (0.74 log + 0.0805)10- 3 X 100 = 0.203 henry.
O.Zo
Inductive reactance of each conductor = x r = 2irfL = 2 X
3.14 X 60 X 0.203 = 77 ohms.
Reactance drop per conductor = lx' = 173 X 77 = 13,321
13,321 X 100 00
V lts= 100000" = 23.1 per cent.
Reactance drop in each transformer = 3.45 percent, (assumed).
Total reactance drop per line = 23.1 + 2(3.45) = 30 per cent.
77 X 30
x = equivalent reactance of each line = ^-^ = 100 ohms.
Zo. 1
Capacity of each conductor to neutral is
00 O
C = - -A-OQ X 10 ~ 9 X 10 = !- 475 X 10 ~ 6 farads -
i 1UU (J.AO
logl ~~
TRANSMISSION SYSTEMS 519
Capacity reactance of each conductor to .neutral = x c =
10 6
= 1,800 ohms.
2 X 3.14 X 60 X 1.475
b c = capacity susceptance per line = = 0.000555.
x c
T 17S
Y = load admittance per phase = - = IQQ QQQ = 0-003.
cos = load power factor = 0.88.
g = load conductance per phase =- - - = Y cos =
6
0.003 X 0.88 = 0.00264.
7 sin 6 ., .
b = load susceptance per phase = - - = Y sin 9 =
6
0.003 X 0.475 = 0.001425.
Voltage between lines and neutral at the generating station at
full load is
ft* = 6 1 + (r + jx)(g - jb +j equation (454).
= e[l + (30+100.?) 1 0.00264 -j (0.001425 -0.0002775) J
= e(1.1942 + 0.23;)
and its absolute value is
e = D2 + 023 - 70,000 volts
V3
230
and it leads the terminal e.m.f. e by angle 6' = tan" 1
tan- 1 0.1925 = 10.9 deg.
E G = voltage between lines = \/3e G = \/3X 70,000= 121,600 volts.
The current per conductor at the generating station is
J =
= e[o.00264-0.001425j + 0.000555J X
j i + SO^IOQ, ^ OQ264 _ . 001425j ) j ]
= e [o.00258 - O.OOOSlj] .
and its absolute value is
IG = - -W- Jo.00258 2 + 0.00081 2 = 161 amp.
520 ELECTRICAL ENGINEERING
00081
It lags behind the terminal e.m.f. e by angle 0" = tan -1 f ^^ KO
O.OOZoo
= tan~! 0.314 = 17.45 degrees.
I lags behind e G by angle G = d' + 0" = 10.9 + 17.45 =
28.35 degrees and the power factor at the generator is cos =
cos 28.35 degrees = 0.88 = 88 per cent.
The charging current per line may be taken as ijie product of
the capacity susceptance b c and the average of the two voltages
e and e, thus,
7c = + bc , 70,000 + 57,600 x Q Q ^
= 63,800 X 0.000555 = 35.4 amp.
and the charging kilo volt-amperes for the system
3 X 63,800 X 35.4 ,_,
= 6 > 800 kva -
The copper losses in the three lines and the step-up and step-
down transformers may be taken as 3(/ G 2 + I 2 ) = 3(16 I 2 +
O
173 2 )15 = 2,520,000 watts = 2,520 kw. = 0>88 X 100
= 9.5 per cent.
Assuming the iron losses in the transformers to total 1 per
cent, the total loss is 10.5 per cent, and the efficiency of the
transmission from the generator busbars to the low-voltage bus-
bars in the receiving station is 89.5 per cent.
If the load is suddenly removed, the rise in the receiver voltage
may be found from the following relations :
g = 0, b = 0, e G is assumed to be maintained constant, and
e = e jl + (r+jx) (j|) } = e{l + (30 + 100J) (0.0002775J) 1
= 6(0.97225 -O.OOSj)
and taking absolute values
70,000 = e = e\/ (0.97225) 2 + (0.008) 2 = e X 0.9723
0^23 - = 72 > 00
and E = \/3e = \/3 X 72,000 = 124,700 volts.
The regulation is > ~J [Q Q > 000 X 100 per cent. = 24.7 per
cent.
TRANSMISSION SYSTEMS 521
436. Application of a Synchronous Phase Modifier to a Trans-
mission System. In the transmission system worked out in the
last article, the regulation is poor and the efficiency is low. These
can both be improved by installing in the receiving station a
synchronous machine to operate as a phase modifier. It should
be over-excited and draw full leading kilovolt-amperes at periods
of full load to keep the receiver voltage up and should be under-
excited and draw lagging kilovolt-amperes at periods of light load
to keep the voltage down. With such an equipment the voltage
at the generating station may be maintained at a suitable con-
stant value and the receiver voltage may likewise be maintained
constant at some lower value by means of an automatic vol-
tage regulator excited from the constant receiver voltage and
operating on a shunt to the field rheostat of the phase modifier.
The equation e G = e j 1+ (r-\- jx) (g jb + j ~) |was derived
on the assumption that the capacity susceptance was connected
at the center of the line, but the same result is obtained if one-half
of this susceptance is connected at the receiving end. When the
synchronous phase modifier is added, Fig. 490 (a), the equation
becomes
e
= e [l + (r + jx) (g - jb + j ^ j&.) } (455)
where b s is the susceptance per phase of the phase modifier;
b s is positive when drawing a leading current and negative when
drawing a lagging current.
The rating of the phase modifier required for any system can
best be found by a graphical construction and equation (455)
used only as a check.
1. If it is desired to operate the system of Art. 435 with a con-
stant voltage E = 100,000 volts and a constant generating sta-
tion voltage E G = 110,000 volts, the rating of the required phase
modifier may be found as follows :
In Fig. 490
Oa = receiver voltage to neutral = e = -- h= = 57,600 volts.
ab = receiver current per phase = I = 173 amp. at full load.
Cos 6 load power factor = 0.88.
ac = resistance drop per line due to the load current = Ir =
173 X 30 = 5,200 volts,
522
ELECTRICAL ENGINEERING
cd = reactance drop per line due to the load current = Ix =
173 X 100 = 17,300 volts. _
ad = impedance drop per line = / \/r 2 + z 2 = 173-\/30 2
= 173 X 104.4 = 18,060 volts leading the current by angle a =
x 100
/ *J\J
Assuming the line capacity to be represented by the susceptance
^ at the receiver, Fig. 490 (a), the charging current flowing over
the line is e X ^ = 57,600 X 0.0002775 = 16 amp. = ah.
Radius =e G =63300
FIG. 491.
ak = impedance drop due to current ah = 16^30 2 + 100 2 =
1,670 volts. If dg is drawn through d making angle a = tan -1
with the charging current line and dl is cut off equal to ak,
then 01 is the voltage to neutral required at the generating sta-
tion, without the phase modifier = 70,000 volts from the dia-
gram. This value was also obtained in Art. 435.
i 1 1 n nnn
About draw a circle of radius e G =
= 63,360
=
to cut dg at /; then the line If represents the impedance drop per
line produced by the leading current flowing to the phase modifier.
TRANSMISSION SYSTEMS 523
Since ad is the impedance drop due to the load current the line
If will represent the kilovolt-ampere rating of the phase modifier
to the same scale that the line ad represents the load kilovolt-
amperes of 30,000. The rating of the phase modifier is therefore
-^ X 30,000 = 12,500 kilovolt-amperes.
ad
At no load the voltage required at the generating station,
without the phase modifier, is represented by the line Ok but
since e G is maintained constant at 63,360 volts, the phase modifier
must be under-excited and draw a lagging current to cause an
impedance drop represented by kf, where /'is the point of inter-
section of the e G circle with the line kaf making angle a with
ag. kf is approximately equal to If and the phase modifier will
be carrying full lagging current.
It is not usual to maintain constant voltage over such a wide
range and the rating of the phase modifier may be reduced con-
siderably especially if a greater difference between e G and e is
allowable.
2. If it is required to maintain a constant receiver voltage, E
= 100,000 from full load of 30,000 kva. at 88 per cent, power
factor to one-fourth load of 7,500 kva. at 75 per cent, power
factor, determine the rating of the phase modifier and the most
suitable voltage at the generating station.
The point li in Fig. 491, corresponds to point I in Fig. 490 and
is found in the same way; and angle d l = Q = cos" 1 0.88.
At one-fourth load of 7,500 kva. at 75 per cent, power factor.
a& 2 = r- = r- = 43.25 amp. = load current.
Cos 6 2 = 0.75 = load power factor.
IT
Angle d 2 ab 2 = angle diabi = a = tan" 1 -
Points d 2 and Z 2 are found in the same way as d and I, Fig. 490.
To enable the phase modifier to be used to its full capacity
leading at full load and full capacity lagging at one-fourth load,
the generator voltage must have the value represented by the
radius of a circle drawn about 0, which will make the two inter-
cepts /]Zi and / 2 ?2 equal. The required value of e G is 65,350 volts
and E = \/3 X 65,350 = 113,000 volts. The capacity of the
phase modifier is ~r X 30,000 = 9,100 kva. and its suscept-
ance per phase at full load is b a = > = 0.00091.
524 ELECTRICAL ENGINEERING
Referring to Fig. 490 (a) the current in each line at full load is
f . b c , , }
IG = e I g jb H-j ^ + A j
= 6(0.00264 - j(0.001425 - 0.0002775 - 0.00091))
= e (0.00264 - 0.00025J)
-2
and I G = 57,600 \0.0264 + 0.00025 = 153 amp.
The full-load copper loss = 3 X (153) 2 X 30 = 2,115,000
2115
watts = 2,115 kw. = 30)00( ^ x Q88 X 100 = 8 per cent.
Taking the transformer iron losses as 1 per cent., the efficiency
of the transmission is 91 per cent. The losses in the phase
modifier should be subtracted.
437. High-voltage Direct-current System. The transmission
of power over long distances by direct current is limited by the
difficulty of obtaining a sufficiently high voltage. In the Thury
system the required line voltage is obtained by connecting a
number of series-wound generators in series and the line current
is automatically maintained constant. Series machines are
built generating from 1,500 to 5,000 volts with a single commuta-
tor. A single generating unit may have two commutators and
two units may be driven by one prime mover giving a direct
voltage up to 20,000 volts per set. If five such machines are
connected in series the voltage between lines will be 100,000 volts.
To maintain constant line current with variable load, the gen-
erator voltage must be varied; this is accomplished by means of
an automatic regulator which moves the brushes and varies the
resistance in a rheostat or diverter shunting part of the field
winding. In some cases the speed of the prime mover is varied.
The regulator may be set for any required current. In the sys-
tems at present in operation currents from 50 to 200 or 300 amp.
are employed.
It is impossible to insulate rotating machines for the high
voltages required for transmission and it is therefore necessary
to mount the generators on insulated platforms and to couple
them to the prime movers by insulating couplings.
The various generating units need not be included in a single
station but may be located at any point along the proposed line.
In this way a number of small generating stations may be linked
up into one large system and since the current is constant they
cannot be overloaded under any condition.
TRANSMISSION SYSTEMS
525
Fig. 492 shows the layout of a 60, 000- volt, 150-amp. system.
At periods of light load any generator may be taken out of
service by disconnecting the regulator and moving the brushes
over to the position of zero voltage and then closing the short-
circuiting switch. To put it in service again it is brought up to
speed, the short-circuiting switch is opened and the regulator set
for the required current.
The substation equipment is very similar to that in the gen-
erating station and like it may be located in one or more stations.
If it is necessary to operate the motors at constant speed a regu-
lator is provided which maintains the speed constant by moving
the brushes and shunting part of the motor field current.
Q \r Short Circuiting
Bwitch
Generator Out of Service
(a)
10000 Volts 8000 Volts 2000 Volts
150 Amperes 150 Amperes 150 Amperes 160 Ampei
Generating Station
150 Amperea
" f
Or Stations
20000 Volte
150 Amperes
V R
L@jaJ
FIG. 492. Thury or series system.
The motors must drive through insulating couplings and are
put in operation by opening a short-circuiting switch and moving
the brushes to the proper position. The regulator is then con-
nected in and takes care of the speed.
In the majority of terminal stations the power which has been
transmitted by direct current is converted into alternating cur-
rent for local distribution, the series motors driving alternat-
ing-current generators.
438. Advantages and Disadvantages of the Thury or Series
System. Advantages. 1. Power factor is always unity as there
is no reactance drop. Cables carrying direct currents can be
laid in iron pipes if necessary.
2. Higher voltages may be used for the same line insulation.
Direct voltages may be double the effective alternating voltages
since there are no dielectric losses in the cables.
3. Two conductors only have to be insulated and the center
point of the system may be grounded to reduce stresses.
526 ELECTRICAL ENGINEERING
4. A number of stations can be operated in series and a new
station may be connected to the line at any point. The indi-
vidual stations are entirely independent of local overloads or lack
of demand since the current is maintained constant and regula-
tion does not enter into the question.
5. Switching arrangements are very simple.
6. Cost of right-of-way will usually be very small since the
line may be placed entirely underground making it possible to
cross country where a right-of-way could not be obtained. In
crossing large bodies of water high-voltage submarine cables may
be used saving the expense of step-down and step-up transformers
required in a similar case in the alternating-current system. To
supply electric-railway substations the line could be placed under-
ground along the right-of-way, a single conductor being tapped
at each substation.
7. A single conductor with earth return may be used either
as the ordinary method of operation or in case of emergency.
All other grounds must then be removed and some of the appar-
atus insulated for the full-line voltage.
Disadvantages. 1. Insulated floors and couplings are required.
2. Units are of moderate size. A single unit delivering 250
amp. at 20,000 volts has a rating of only 5,000 kw.
3. Line loss is constant independent of the load.
4. Special regulating devices are required to keep the motor
speed constant.
5. Motors have no overload capacity, since the current is
constant.
CHAPTER XV
ELECTRICAL INSTRUMENTS
439. Electrical Instruments. A complete study of electrical
instruments is beyond the scope of this book but in the following
pages the principles of a number of the more important types
used in the measurement of electrical quantities are discussed.
440. Direct-current Voltmeters and Ammeters. The majority
of direct-current voltmeters and ammeters are of the movable-
coil permanent-magnet type (Fig. 493). A light rectangular coil
of wire M, wound on an aluminum frame, turns in the field of a
permanent magnet NS. The field is made uniform by placing
an iron cylinder C between the poles and this gives the instru-
ment a uniform scale. The coil is supported on jewelled bearings
Current Terminal
Current Terminal
FIG. 493. Weston perma-
nent-magnet type of direct-
current voltmeter or ammeter.
FIG. 494. Ammeter
shunt.
FIG. 495.
and its motion is controlled by two spiral springs s, s, above and
below, coiled in opposite directions. The springs are used to
lead the current in and out of the coil. The aluminum frame
acts as a damper and makes the instrument dead beat.
A large resistance r must be inserted in series with the coil to
limit the current in it to a suitable value. The resistance re-
quired varies from 50 to 150 ohms per volt and the current
ranges from 7 to 20 milliamp. at full-scale deflection.
441. Voltmeter Multipliers. If it is desired to change the
range of a voltmeter an extra resistance r m , Fig. 495, must be
connected in series with the movable coil. If the resistance r m
527
528 ELECTRICAL ENGINEERING
of the multiplier is exactly equal to the resistance of the coil plus
any resistance r in series with it, then for a given line voltage the
current is reduced to half and the scale reading to half. Such
a multiplier doubles the range of the instrument and all readings
must be multiplied by 2. Multipliers from 2 to 1 up to 20 to 1
are used with direct-current voltmeters.
442. Ammeter Shunts. The ammeter is exactly similar to
the voltmeter in construction (Fig. 493). A shunt of known re-
sistance r a , Fig. 494, is placed in the circuit and carries the cur-
rent to be measured. A voltmeter or millivoltmeter is connected
across the terminals and the pointer indicates the drop of voltage
across the shunt. This drop is Ir s and is proportional to the
current since the resistance r a is constant. By properly arrang-
ing the resistance of the shunt and the resistance r in series with
the movable coil the instrument may be made to read directly
in amperes.
If a shunt has a resistance r 8 = 0.001 ohm and carries a cur-
rent of 100 amp., the voltage drop is 0.1 volts = 100 millivolts.
Instead of calibrating the scale in millivolts it may be calibrated
directly in amperes.
If the ammeter is provided with a number of shunts it may be
used to measure a large range of currents. For the smaller cur-
rent ranges the shunt is usually placed inside the meter case.
The permanent magnet type is standard for direct-current in-
struments but cannot be used for alternating currents.
443. Thomson Inclined-coil Ammeter. Fig. 496 shows a sec-
tion of an inclined-coil ammeter or voltmeter. The current to
be measured flows through the inclined coil and produces a mag-
netic field perpendicular to the plane of the coil. A soft-iron
core is placed in the field and tends to turn until it becomes par-
allel to the direction of the flux. The motion is controlled by a
spiral spring and the counterweight balances the moving parts.
The motion of the pointer is damped by a light aluminum vane.
In the ammeter the stationary coil consists of a few turns of large
wire, while the voltmeter has a large number of turns of small
wire. The scale is not uniform. Inclined-coil meters may be
used for both alternating and direct currents but they are not so
good as the permanent-magnet direct-current meters.
444. Weston Soft-iron-type Ammeters and Voltmeters. This
type of instrument is illustrated in Fig. 497. A is a fixed tri-
angular piece of soft iron bent into the form of a cylinder and B
ELECTRICAL INSTRUMENTS
529
is a movable piece carried on a shaft. A magnetic field is pro-
duced by the coil C which carries the current to be measured or a
current proportional to the voltage to be measured. The two
soft-iron pieces become magnetized in the same direction and
repel one another, producing a deflection of the pointer., Motion
is controlled by a spiral spring and oscillations are damped out
by a vane enclosed in an air chamber. The iron pieces are
shaped to give a uniform scale.
Such instruments may be used to measure direct currents as
well as alternating currents.
FIG. 496. Thomson inclined-
coil ammeter.
FIG. 497. Weston soft- FIG. 498. Elec-
iron type ammeter or volt- tro dynamometer-
meter, type voltmeter.
445. Electrodynamometer-type Voltmeter. The electrodyna-
rnometer-type voltmeter, Fig. 498, depends on the interaction
between two coils carrying a current proportional to the voltage
to be measured. The permanent magnet of Fig. 493 is replaced
by a coil of wire in series with the moving coil. No iron is used
in the instrument. This construction is not suitable for am-
meters since the current must be carried by a moving contact
between the coils. It can be .used for voltmeters but other types
are more satisfactory. Wattmeters are, however, designed on
this principle.
446. Hot-wire Ammeters and Voltmeters. The expansion of
a wire, due to the heat produced by the passage of current through
it, is utilized in this type of meter (Fig. 499). AB is a wire of
platinum-silver alloy which carries the current to be measured.
As it expands the tension on the fine phosphor-bronze wire CD is
reduced and this enables the spring S to pull the silk fiber FE to
the left. The silk fiber passes around a pulley on the shaft of
the moving system and causes the pointer to move over the
scale. The motion of the disc is damped by the aluminum disc
which turns between the poles of the permanent magnet shown
in Fig. 499 (a).
530
ELECTRICAL ENGINEERING
For voltmeters a large resistance R is connected in series with
the hot wire so that the current is proportional to the voltage to
be measured. The zero can be adjusted by the screw at A
n.A
-Pointer
Hot Wire -Platinum Siver
^Phosphor Brnze Wire
Permanent
agnet
Terminal
Series Resistance
W
R
Terminal
FIG. 499. Hot-wire ammeter or voltmeter.
447. Dynamometer-type Wattmeter. The dynamometer prin-
ciple is satisfactory when applied to wattmeters, Fig. 500, because
the two coils are not connected in series but carry different
currents. The series coil of large wire carries the line current
and takes the place of the permanent magnet in the direct-
current voltmeter, Fig. 493. The voltage coil is exactly similar
to the moving coil in the direct-current voltmeter and is supported
and controlled in the same way.
(tt) Uncompensated (&) Compensated
FIG. 500. Dynamometer type wattmeter.
The torque exerted on the moving coil is proportional to the
product of line current and line voltage, that is, to the power in
the circuit.
If necessary a shunt may be used in parallel with the current
coil and a multiplier in series with the voltage coil.
ELECTRICAL INSTRUMENTS
531
A wattmeter connected as in Fig. 500 (a) has a slight zero error,
because when no current is being supplied to the load, the current
for the voltage coil passes through the series coil and the watt-
meter indicates a small
E 2
amount of power = Ei = -5- where
JK
E is the line voltage and R is the resistance of the voltage coil.
To correct this error a third coil C called the compensating coil
Fig. 500(6) is added. It has the same number of turns as the
series coil but is wound in the opposite direction and its m.m.f.
E
neutralizes the m.m.f. of the small current i = -5"
K
For ordinary measurements of power the voltage is applied
between terminals A and B, but when the meter is to be cali-
brated, using separate voltage and current sources, no zero error
is introduced and a third terminal D must be provided instead of
A. In series with it is a resistance r c equal to the resistance of
the compensating coil.
Dynamometer-type wattmeters can be calibrated with direct
current and used on alternating-current circuits.
Voltage Coil Compensating
O E o ^^ Coil
Eesultant
Flux
(a)
Current Coil
FIG. 501. Induction- type wattmeter.
448. Induction-type Wattmeter. An induction wattmeter is
shown in Fig. 501. The voltage coils VV are connected in series
across the line and the current coils are connected in series with
the line. Both sets of coils are wound on the same laminated
iron core.
The voltage coils on the two legs are wound in the same direc-
tion and their m.m.fs. add. The circuit is highly inductive and
532 ELECTRICAL ENGINEERING
the current through it is nearly 90 degrees behind the line volt-
age. The flux 4> v is in phase with the current and is therefore in
quadrature behind the line voltage.
The current coils on the two legs are wound in opposite direc-
tions and both tend to produce a flux 4> c in the gap in mechanical
quadrature with the flux v . The flux < c is in phase with the line
current and with non-inductive load it is 90 degrees ahead of the
flux fl . A revolving field is thus produced in the space between
the poles.
A light aluminum cup or disc D is pivoted in this field and
tends to turn with it but is opposed by spiral springs. The
torque is proportional to the product of the voltage and the line
current, that is, to the power in the circuit.
The current in the circuit VV and the flux v , Fig. 501 (a), induces in
the coil S an e.m.f. and current approximately 90 degrees behind
it and this current produces a component of flux < s in phase with
itself. s combines with v and produces a resultant potential
flux exactly in quadrature behind E. s may be adjusted by
adjusting the resistance r in series with the coil S.
Polyphase wattmeters are made in the same way and consist
of two single-phase elements exerting torque on a single disc.
Induction-type ammeters and ' voltmeters have also been
constructed. All meters of this type are accurate only for the
frequency for which they have been designed.
If the controlling springs are removed from a wattmeter and a
registering mechanism is added the meter becomes a watt-hour
meter and registers energy consumption.
449. Power-factor Meters. Power-factor meters indicate
the power factor of a circuit; they are constructed on the same
principles as wattmeters, of both electrodynamometer and induc-
tion types and may be either single-phase or polyphase.
The circuits of a Weston single-phase power-factor meter are
shown in Fig. 502. The current coil F is similar to that in the
wattmeter, Fig. 500, but the moving system is made up of two
ELECTRICAL INSTRUMENTS
533
similar coils C\ and C 2 . The common terminal of the two coils
is connected to one side of the line and the other terminals are
connected through a resistance R and an inductance L to the
other side of the line. The current in Ci is therefore in phase
with the line voltage and the current in 2 is in quadrature behind
the line voltage.
At non-inductive load the currents in Ci and F are in phase
and C\ will be forced around until its plane is parallel to the plane
of F. If the load current lags 90 degrees behind the voltage the
current in C 2 is in phase with that in F and C 2 is turned into the
plane of F. For intermediate power factors the moving element
takes up intermediate positions.
TS/ R
V r
A A A A A
ww
\
"1
Til
/TTJOOOTN
L
%
E3
F
C
FIG. 502. Weston single-phase power-factor motor.
This instrument can be used as a two-phase power-factor meter
if Ci and C 2 are each connected across one phase in series with a
suitable resistance.
A three-phase meter has three coils on the moving element
connected in star to the three lines.
450. Frequency Meters. The vibrating reed frequency meter
(Fig. 503) contains a number of steel strips of different lengths
fixed at one end and free to vibrate at the other. An electro-
magnet is placed behind the strips and is excited from the circuit
of which the frequency is required. The strip with a natural
period corresponding to that of the magnetic field will be set
in vibration. The ends of the reeds are painted white and when
one is in vibration it shows as a white band. The periods of the
reeds are adjusted during manufacture by attaching minute
weights to the free ends.
534
ELECTRICAL ENGINEERING
451. The Weston Frequency Meter. The elements of a Wes-
tern frequency meter are shown in Fig. 504. Two stationary coils
A and B are fixed at right angles ; a resistance R and inductance
X are connected in series across the line ; the coil A is connected
in series with an inductance X A across the resistance R and coil
B is connected in series with a resistance R B across the induc-
tance X. The series inductance X a is added to damp out
higher harmonics.
Alternating Current
Supply
FIG. 503. Vibrating reed frequency meter.
FIG. 504. Weston fre-
quency meter.
The moving element consists of a soft-iron core C mounted on
a shaft without control of any kind and it takes up a position in
the direction of the resultant of the two fields.
When the frequency increases, the current in A decreases and
that in B increases and the core and pointer move with the
resultant field.
1234 Incomiug Machine
FIG. 505. Synchroscope.
452. Synchroscope. A synchroscope indicates: (1) whether
the incoming machine is running too fast or too slow, and (2)
the exact instant when synchronism is reached.
One form of synchroscope is shown in Fig. 505. It has a lam-
inated bipolar magnetic circuit M excited by field coils FF, which
are connected across the alternating-current busbars at 1 and 2
and produce an alternating field. The moving core C is also
laminated and carries two windings A and B at right angles to
ELECTRICAL INSTRUMENTS 535
one another. Their common terminal is connected to one side
of the incoming machine at 4. The other terminals of A and B
are connected respectively through a reactance x and a resistance
r to the other side of the machine at 3.
The current in F is in quadrature behind the line voltage, the
current in A is in quadrature behind the machine voltage and the
current in B is in phase with the machine, voltage.
When the incoming machine is exactly in synchronism the coil
A takes the position shown in the figure since the current in A
is in phase with the current in F. When the machine is 90 degrees
behind the position of synchronism the current in B is in phase
with the current in F and the armature turns through 90 degrees
and brings the coil B in line with the poles.
For intermediate phase relations the armature takes inter-
mediate positions such that the revolving field produced by the
armature winding is in line with the field poles when the current
in F is maximum. The phase relation is indicated by a pointer
on the dial of the synchroscope.
When the frequency of the incoming machine is lower than
that of the line, the phase of the current in A continually falls
behind that of F and the pointer rotates in the direction marked
"Slow." When the incoming machine is running too fast, the
rotation of the pointer is in the opposite direction marked " Fast."
When the machine is running exactly at synchronous speed
and is exactly in phase, the pointer is vertical and stationary.
The switch can then be closed and the speed and excitation ad-
justed until the machine takes its proper share of the load and
operates at the proper power factor. .
The synchroscope described is for a single-phase circuit. It
can be used for a two-phase machine by connecting the two coils
A and B to the two phases of the machine and the coils FF across
one phase of the line.
For three-phase systems the armature carries a three-phase
winding connected to the three phases of the machine.
453. Tirrill Regulator. The Tirrill regulator is an automatic
voltage regulator designed to maintain a steady voltage at the
terminals of a direct-current generator irrespective of ordinary
load fluctuations or changes in generator speed. It can also be
made to compensate for line drop by increasing the generator
voltage as the load increases.
The regulator controls the voltage by rapidly opening and clos-
536
ELECTRICAL ENGINEERING
ing a shunt circuit across the field rheostat of the generator. The
rheostat is so adjusted that when in circuit it tends to reduce the
voltage considerably below normal and when short-circuited the
voltage tends to rise above normal. The relative lengths of
time during which the short-circuit is closed or opened determines
the average value of the field current and therefore the value of
the terminal voltage.
The method of operation of the regulator is illustrated in Fig.
506. The regulator consists essentially of two magnets control-
ling two sets of contacts. The main control magnet has two in-
dependent windings, one, the potential winding, connected across
Compensating Resistance Shunt
'O
Potential Winding
Compensating Winding
Main Control Magnet
FIG. 506. Automatic voltage regulator.
the generator terminals and the other across a shunt in the load
circuit. The latter is the compensating winding and is used only
when a rise of voltage with load is required. The relay magnet
is differentially wound and controls the circuit shunting the field
rheostat. The operation is as follows: When the short-circuit
across the field rheostat is opened the voltage tends to fall below
normal. The main control magnet is weakened and allows the
spring to pull out the movable core until the main contacts are
closed. This closes the second circuit of the differential relay
and demagnetizes it. The relay spring then lifts the armature
and closes the relay contacts. The field rheostat is short-cir-
cuited and the field current and terminal voltage tend to rise.
The main control magnet is strengthened and opens the main
contacts allowing the differential relay to open the short-circuit
across the field rheostat. The terminal voltage falls again and
ELECTRICAL INSTRUMENTS
537
this cycle of operations is repeated at a very rapid rate main-
taining a steady voltage at the generator terminals. When the
compensating winding is not used the terminal voltage is main-
tained constant.
When it is necessary to compensate for line drop and maintain
a constant voltage at the receiver end of the line, the compen-
sating winding is connected across a shunt in the load circuit.
The resistance of the shunt is adjusted to give the required com-
pounding. The compensating winding opposes the action of the
potential winding on the main control magnet so that as load
increases a higher potential is necessary at the generator ter-
minals in order to close the main contacts and open the shunt
across the field rheostat. Thus, the generator voltage rises with
load. The condenser connected across the relay contacts serves
to reduce the sparking when the circuit is opened.
Main
Contacts
Compensating
Winding^
A.C. Field A.C. Generator
Bheostat
FIG. 507. Automatic voltage regulator for alternating-current generators.
454. Automatic Voltage Regulator for Alternating-current
Generators. The regulator described in Art. 453 may be modi-
fied to regulate the voltage of alternators. The desired voltage
is maintained by opening and closing a short-circuit across the
exciter field rheostat.
The method of operation of the regulator can be understood
from the diagram of connections shown in Fig. 507. The direct-
current control magnet is connected to the exciter busbars and
has a fixed core in the bottom and a movable core in the top
attached to a pivoted lever, at the other end of which is a spring
538 ELECTRICAL ENGINEERING
which closes the main contacts. The alternating-current control
.magnet has a potential winding connected across one phase of the
alternator and it may also have a compensating winding con-
nected through a current transformer to one of the feeders. The
core is movable and is connected to a pivoted lever controlled by
a counterweight. The relay magnet is differentially wound and
is connected as shown.
Operation. The direct- and alternating-current control mag-
nets are adjusted for the required voltage by means of the coun-
terweight. The exciter field rheostat is then set to reduce the
voltage about 65 per cent, below normal. This weakens both of
the control magnets and the spring closes the main contacts and
demagnetizes the relay magnet. The pivoted armature is re-
leased and the relay contacts are closed and thus short-circuit
the exciter field rheostat and immediately raise the exciter voltage
and the alternator voltage. When the alternator voltage reaches
the value for which the regulator is adjusted, the main contacts
open again, the relay magnet is again magnetized and the short-
circuit on the exciter field rheostat is opened. This reduces the
voltage as before and the cycle of operations is repeated at a
very rapid rate and maintains a constant voltage at the terminals
of the alternator if the compensating winding is not connected.
Three-phase Generator
LJ, rr^ri
1 .....
( a ) .S.ingle-phase ( ft } Three . phase
FIQ. 508. Line compensators.
When it is necessary to maintain a constant voltage at the
receiver end of the line the compensating winding is connected as
shown. As the load increases it brings the main contacts closer
together and so increases the time of short-circuit of the field
rheostat and thus increases the terminal voltage of the alternator.
Using a current transformer and a dial switch any line drop up to
15 per cent, can be compensated for, but only at a given power
factor. When the power factor of the load varies through a
wide range a line compensator, Fig. 508 (a), should be used in con-
ELECTRICAL INSTRUMENTS 539
junction with the potential coil and the compensating coil should
be disconnected.
The line compensator forms a local circuit with the same vol-
tage characteristics as the main line. The shunt transformer TI
gives a secondary voltage proportional to the generator voltage.
The current transformer T 2 produces through the circuit rx a cur-
rent proportional to the load current, r is a resistance which
consumes a voltage proportional to and in phase with the resist-
ance drop in the line and this voltage is transferred to the com-
pensator circuit by the potential transformer TV # is a reactance
and consumes a voltage proportional to the reactance drop in
the line. This voltage is transferred to the compensator circuit
by the transformer T* which also forms the reactance. Thus
there are in the compensator circuit three voltages proportional
respectively to the generator voltage and the resistance and
reactance drops. If the same proportions are maintained in each
case, the voltage between the terminals AB will be proportional
to the voltage at the end of the line, and, therefore, if the poten-
tial coil of the regulator is connected across AB it will maintain
a constant voltage at the receiver end of the line. In the case
of transmission lines of very high voltage a correction must be
made for the capacity of the lines. With a three-phase alter-
nator, as in Fig. 507, two current transformers must be used
connected as shown in Fig. 508(6).
One automatic voltage regulator can be applied to control the
voltage of a system operating two or more alternators in parallel.
The regulator may also be applied to the exciter of a syn-
chronous-phase modifier to maintain a constant voltage at the
receiver end of the line.
INDEX
Absampere, definition of, 54
Abvolt, definition of, 52
Adjustable speed operation, 210
Admittance, 131, 132, 135
stator exciting, 465
Air-blast transformers, 386
All-day efficiency, 382
Alternator windings, 289
Alternators, rated speed of, 355
rating of, 300
types of, 280
Aluminum, properties of, 83
Alternating-current commutator
motor with shunt character-
istics, 505
series motor, 494
speed control of, 500
Ammeters, 527
Ampere, definition of, 54
turn, 61
Analysis of complex waves, 146
Argon filled rectifier, 442
Armature copper loss, 216
core, 165
reaction in alternating-current
generators, 302
in direct-current generators,
182
in polyphase generators, 305
in single-phase generators, 309
in synchronous converters,
426
resistance, 275, 312
windings, 167
Asynchronous phase modifier, 488
Automatic voltage regulator, 535
Auto-transformer, 396
B
Balancers, 249
Bearing friction loss, 220
Blondel diagram for synchronous
motor, 334, 339
Boosters, 247
negative, 249
series, 248
Booster transformers, 396
British thermal unit, 80
Brush contact drop, 216
contact loss, 216
effect of moving the, 181
friction loss, 220
lifting device, 434
Brushes, 178, 262
Bucking, 435
Burning of brushes, 236
Calorie, 80
Capacity, 12
in alternating-current circuits,
111
of concentric cylinders, 15
of single conductor cable, 16
of parallel conductors, 17, 19
unit of, 13
Carter's fringing constant, 269
Cascade operation of induction
motors, 476
Charging current of transmission
line, 509
Circle diagram of induction motor,
458, 460, 461
Coercive force, 71
Commutating e.m.f., 230, 231
Commutation, 224, 253
in alternating-current series
motor, 499
in repulsion motor, 503
time of, 225
Commutator, 176
peripheral speed of, 262
541
542
INDEX
Complex alternating waves, 142
analysis of, 146
average value of, 142
effective value of, 142
form factor of, 142
Compensating winding, 242, 498
Compensators for three-wire genera-
tors, 437-8
Compound excitation, 179
Compounding converters, 430
curve of alternator, 317
direct-current generator, 190
synchronous motor, 330, 342
Concatenation of induction motors,
476
differential, 478
Condenser, 14
bushing, 40
energy stored in, 25
Condensers in multiple, 24
series, 24
Conductance, 81, 131, 134
Conductivity, 81
standard of, 81
Conductors, 1
Constant current transformer, 401
potential to constant current,
140
transformer, 363
Converters, frequency, 414, 493
phase, 414, 491
split-pole, 432
synchronous, 414
booster, 431
Cooling of transformers, 384
Copper losses, 215
properties of, 83
Core loss, 222
current, 364
type transformer, 383
Corona, 45
Coulomb, definition of, 2
Cross-magnetizing, 185, 187, 275
Current capacity of wires, 85, 87
densities in alternators, 356
brush contacts, 236
direct-current armatures, 261
field windings, 274
transformer, phase angle of, 400
Current transformer, ratio correc-
tion factor of, 400
unit of, 54
Cylindrical rotors, 282, 361
D
Damping grids, 343,. 436
Demagnetizing m.m.f., 185, 187, 275
Design of alternators, 354
direct-current machines, 264
transformers, 405
Dielectric constant, 2, 43
flux, 4
hysteresis, 44
losses, 44
permeance, 13
strength, 41-43
Dispersion coefficient, 263
Distribution factors, 290
Double current generators, 436
windings, 174
Drum winding, 168-172
E
Eddy-current loss in armature cop-
per, 221
in direct'current machines, 218
in transformers, 379
Effective value of a sine wave, 106
complex wave, 142
Efficiency of direct-current ma-
chines, 222, 252
transformers, 381, 382
Electric energy, 55
loading, 259, 261
power, 55
Electrical measuring instruments,
527
Electrification, 1
Electrodynamometer voltmeter, 529
Electromagnetics, 2, 51
Electromotive force equation of an
alternator, 284, 300
a direct-current generator, 180
unit of, 52
Electrostatic field, 3, 4, 5
stresses in, 27
INDEX
543
Electrostatics, laws of, 2
Equalizer connection, 242
rings, 171
Equipotential surfaces, 9, 21
Equivalent sine wave, 142
Excitation of alternators, 358
regulation, 322
Exciting current of induction motor,
478 : v "
transformer, 368
Farad, definition of, 13
Field characteristic of direct-current
generator, 190
windings, 178
Flashing, 239
Form factor of alternating waves,
142, 286
Fourier series, 146
Frequency converter, 493
meters, 533
of hunting, 352
G
Induced charges, 6
Inductance, 92, 107
of armature coil, 227
of iron-clad circuits, 94
Induction frequency converter, 493
generator, 486
motor, 444
analysis by rectangular co-
ordinates, 465
applications of, 475
circle diagram of, 454
single-phase, 482
speed control of, 476
starting, 473, 481
synchronous speed, 447
regulator, 403
type wattmeters, 531
Inductive compensation, 498
Inductor alternator, 282
Insulation, breakdown of, 43
of alternators, 356
resistance, 44
Insulators, 1
Interpole motors, 213
Interpoles, 237
Graded insulation for cables, 35
H
Harmonics in e.m.f. wave, 296, 297
Heat units, 80
Heating, 253
of converters, 422
Henry, definition of, 93
High voltage direct-current system
of transmission, 524
Homopolar generators, 251
Hot cathode rectifier, 442
wire instruments, 529
Hottest spot temperatures, 255
Hunting, frequency of, 352
of synchronous machines, 351
Hysteresis, 71
loss, 271, 378
Hysteretic constants, 73
Joule, definition of, 56
Joule's law, 79
K
Kirchoff's laws, 88
applied to alternating-current
circuits, 141
Lead, properties of, 83
Leakage fluxes, 304
reactances of induction motors,
479
of transformers, 368
Lifting magnets, 76
Limits of output of direct-current
machines, 252
544
INDEX
Line compensator, 538
Load characteristics of synchronous
motors, 331, 340
Losses in alternators, 359
direct-current machines, 215
transformers, 378
Open-circuit test of transformers,
374
-delta connection, 391
Output equation of alternating-cur-
rent generator, 359
direct-current generator, 259
M
Magnetic characteristics, 69
of cast iron, 70
of cast steel, 70
field, 47, 67
energy stored in, 66
flux, 47
leakage, 263
loading, 259
materials, 73
potential, 48
Magnetism, laws of, 47
theories of, 75
Magnetization, 46
curves for cast iron, 271
,cast steel, 271
sheet steel, 271
Magnetizing current, 364
force, 47
Magnetomotive force, 49
unit of, 61
Manganese, properties of, 83
Maxwell's corkscrew rule, 51
Mercury, properties of, 83
vapor rectifiers, 440
Motor starter, 209
Multiple-circuit windings, 294
wire systems, 211
Mutual inductance, 95
N
N-phase converter, 419
No load neutral, 183
No-voltage release, 209
Ohm, definition of, 79
Ohm's law, 79
Parallel operation of alternators, 346
direct-current generators, 242
synchronous converters, 435
three-wire generators, 439
transformers, 395
Permeability, 50
Permeance, dielectric, 13
magnetic, 51
Phase angle of instrument trans-
formers, 398, 400
characteristics of synchronous
motor, 333, 341
converter, 491
splitting, 485
Platinum, properties of, 83
Pole face loss, 220
pieces, 165
Polyphase alternating-current cir-
cuits, 152
armature reaction, 305
commutator motor, 501
Potential, 5
gradient, 9-11, 16, 19
transformer, 397
Potentiometer, 91
Power, electric, 55
-factor, 116, 121
meters, 532
in alternating-current circuits,
116
in three-phase circuits, 158
measurement of, 159
units of, 56
Progressive winding, 174
R
Kadiator tank, 386
Ratio correction factors for instru-
ment transformers, 398,
400
INDEX
545
Reactance, 50, 131, 132
condensive, 111
inductive, 109
of transformers, 408
voltage, 228, 233
Reactive factor, 121
Rectangular coordinates, 134
Rectifier, hot cathode, argon filled,
442
mercury vapor, 440
Regulation, 252
curves of alternators, 317
direct-current generators, 188,
196
transformers, 376
Repulsion motor, 501
compensated, 504
Resistance, 79, 80
commutation, 230
temperature coefficient of, 81
Resistances in parallel, 90
series, 90
Resonant circuit, 128
Reversing induction motor, 450
Revolving field, 307
Rosenberg generator, 250
Rotor, 450, 475
Saturation curve, no load, 186, 270
under load, 187
Scott connection, 393
Self -inductance, 95
Self-starting synchronous motors,
343
Semi-enclosed motors, 256
Separate excitation, 179
Series excitation, 179
Series field-copper loss, 215
Series-parallel speed control, 212
Shading coils, 485
Shell-type transformers, 383
Short-circuit currents of alternators,
323
test of transformers, 375
pitch windings, 171, 235, 294
Shunt excitation, 179
-field copper loss, 215
35
Silicon steel, 74
Silver, properties of, 83
Sine wave, 104
average value of, 106
effective value of, 106
Single-phase induction motor, 482
Slip, 450
Spacing of conductors, 510
Sparkless commutation, 233
Specific inductive capacity, 2
resistance, 81
Speed characteristics of motors, 201
construction of, .204
equation of direct-current
motor, 199
methods of varying, 199
variation with line voltage, 206
Split-pole converter, 432
Starter for direct-current motor,
209
Starting induction motors, 473, 481
synchronous converters, 433
motors, 342
torque, direct-current motor,
208
induction motor, 469
Stator of induction motor, 444
Storage batteries, 245
Surface leakage, 45
Synchronizing power, 338
Synchronous booster converter, 431
motor, 327
phase modifier, 345
speed, 328, 447
Synchroscope, 534
Tee connection, 392
Temperature limits, 253
Third harmonics in alternators, 299
Thompson inclined coil ammeter,
528
Three-wire generators, 437
Thury system, 524
Time constant of a circuit, 99
Tirrill regulator, 535
Tooth taper, 268
Torque characteristics of motors, 204
546
INDEX
Torque, characteristics of motors,
construction of, 207
starting, 208, 469
Totally enclosed motors, 268
Transformer, 363
constant current, 401
constants of a, 374
Transmission systems, 507
Tubular tank, 385
Tungsten, properties of, 83
Turbo-alternators, 284
"V" curves for synchronous motor,
341
Vector diagrams for alternators, 313
synchronous motors, 328
transformers, 366
Vent ducts, 167
Ventilation, 359
Volt, definition of, 6
Voltage characteristics of alterna-
tors, 315
Voltage characteristics of direct-
current generators, 188-193
transformers, 378
Voltmeters, 527
W
Ward Leonard system, 211
Watt, definition of, 56
-ratio curve, 161
Wattmeter, dynamometer type, 530
induction type, 531
Weston soft iron instruments, 528
Windage loss, 220
Wire table, 85
Wound rotor, 474
"Y" connection, 155
Yoke, 165
Zig-zag leakage flux, 479
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